vector-valued laplace transforms and cauchy problems: second edition

Monographs in Mathematics

Managing Editors: H. Amann Universität Zürich, Switzerland J.-P. Bourguignon IHES, Bures-sur-Yvette, France K. Grove University of Maryland, College Park, USA P.-L. Lions Université de Paris-Dauphine, France Associate Editors: H. Araki, Kyoto University F. Brezzi, Università di Pavia K.C. Chang, Peking University N. Hitchin, University of Warwick H. Hofer, Courant Institute, New York H. Knörrer, ETH Zürich K. Masuda, University of Tokyo D. Zagier, Max-Planck-Institut Bonn

Vol. 96

Vector-valued

Second Edition

Wolfgang Arendt Charles J.K. Batty

Frank NeubranderMatthias Hieber

Laplace Transforms andCauchy Problems

Wolfgang ArendtAngewandte AnalysisUniversität Ulm89069 UlmGermany

Charles J.K. BattySt. John’s CollegeOxford OX1 3JPUK

Matthias HieberFachbereich MathematikTU DarmstadtSchlossgartenstr. 764289 DarmstadtGermany

Frank NeubranderDepartment of MathematicsLouisiana State UniversityBaton Rouge, LA 70803USA

ISBN 978-3-0348-0086-0 e-ISBN 978-3-0348-0087-7 DOI 10.1007/978-3-0348-0087-7

© Springer Basel AG 2011

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Contents

Preface to the First Edition ix

Preface to the Second Edition xii

I Laplace Transforms and Well-Posedness of Cauchy Problems 1

1 The Laplace Integral 5

1.1 The Bochner Integral . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 The Radon-Nikodym Property . . . . . . . . . . . . . . . . . . . . 15

1.3 Convolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.4 Existence of the Laplace Integral . . . . . . . . . . . . . . . . . . . 27

1.5 Analytic Behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1.6 Operational Properties . . . . . . . . . . . . . . . . . . . . . . . . . 36

1.7 Uniqueness, Approximation and Inversion . . . . . . . . . . . . . . 40

1.8 The Fourier Transform and Plancherel’s Theorem . . . . . . . . . . 44

1.9 The Riemann-Stieltjes Integral . . . . . . . . . . . . . . . . . . . . 48

1.10 Laplace-Stieltjes Integrals . . . . . . . . . . . . . . . . . . . . . . . 55

1.11 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2 The Laplace Transform 63

2.1 Riesz-Stieltjes Representation . . . . . . . . . . . . . . . . . . . . . 65

2.2 A Real Representation Theorem . . . . . . . . . . . . . . . . . . . 68

2.3 Real and Complex Inversion . . . . . . . . . . . . . . . . . . . . . . 73

2.4 Transforms of Exponentially Bounded Functions . . . . . . . . . . 77

2.5 Complex Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 80

2.6 Laplace Transforms of Holomorphic Functions . . . . . . . . . . . . 84

2.7 Completely Monotonic Functions . . . . . . . . . . . . . . . . . . . 89

2.8 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

vi CONTENTS

3 Cauchy Problems 107

3.1 C0-semigroups and Cauchy Problems . . . . . . . . . . . . . . . . . 108

3.2 Integrated Semigroups and Cauchy Problems . . . . . . . . . . . . 121

3.3 Real Characterization . . . . . . . . . . . . . . . . . . . . . . . . . 132

3.4 Dissipative Operators . . . . . . . . . . . . . . . . . . . . . . . . . 137

3.5 Hille-Yosida Operators . . . . . . . . . . . . . . . . . . . . . . . . . 141

3.6 Approximation of Semigroups . . . . . . . . . . . . . . . . . . . . . 145

3.7 Holomorphic Semigroups . . . . . . . . . . . . . . . . . . . . . . . . 148

3.8 Fractional Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

3.9 Boundary Values of Holomorphic Semigroups . . . . . . . . . . . . 171

3.10 Intermediate Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 184

3.11 Resolvent Positive Operators . . . . . . . . . . . . . . . . . . . . . 188

3.12 Complex Inversion and UMD-spaces . . . . . . . . . . . . . . . . . 197

3.13 Norm-continuous Semigroups and Hilbert Spaces . . . . . . . . . . 201

3.14 The Second Order Cauchy Problem . . . . . . . . . . . . . . . . . . 202

3.15 Sine Functions and Real Characterization . . . . . . . . . . . . . . 217

3.16 Square Root Reduction for Cosine Functions . . . . . . . . . . . . 222

3.17 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230

II Tauberian Theorems and Cauchy Problems 239

4 Asymptotics of Laplace Transforms 243

4.1 Abelian Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 244

4.2 Real Tauberian Theorems . . . . . . . . . . . . . . . . . . . . . . . 247

4.3 Ergodic Semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . 261

4.4 The Contour Method . . . . . . . . . . . . . . . . . . . . . . . . . . 272

4.5 Almost Periodic Functions . . . . . . . . . . . . . . . . . . . . . . . 288

4.6 Countable Spectrum and Almost Periodicity . . . . . . . . . . . . . 295

4.7 Asymptotically Almost Periodic Functions . . . . . . . . . . . . . . 306

4.8 Carleman Spectrum and Fourier Transform . . . . . . . . . . . . . 318

4.9 Complex Tauberian Theorems: the Fourier Method . . . . . . . . . 325

4.10 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329

5 Asymptotics of Solutions of Cauchy Problems 337

5.1 Growth Bounds and Spectral Bounds . . . . . . . . . . . . . . . . . 338

5.2 Semigroups on Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . 351

5.3 Positive Semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . 352

5.4 Splitting Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 360

5.5 Countable Spectral Conditions . . . . . . . . . . . . . . . . . . . . 371

5.6 Solutions of Inhomogeneous Cauchy Problems . . . . . . . . . . . . 378

5.7 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386

CONTENTS vii

III Applications and Examples 397

6 The Heat Equation 4016.1 The Laplacian with Dirichlet Boundary Conditions . . . . . . . . . 4016.2 Inhomogeneous Boundary Conditions . . . . . . . . . . . . . . . . . 4086.3 Asymptotic Behaviour . . . . . . . . . . . . . . . . . . . . . . . . . 4126.4 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

7 The Wave Equation 4177.1 Perturbation of Selfadjoint Operators . . . . . . . . . . . . . . . . . 4177.2 The Wave Equation in L2(Ω) . . . . . . . . . . . . . . . . . . . . . 4237.3 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427

8 Translation Invariant Operators on Lp(Rn) 4298.1 Translation Invariant Operators and C0-semigroups . . . . . . . . . 4308.2 Fourier Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . 4358.3 Lp-spectra and Integrated Semigroups . . . . . . . . . . . . . . . . 4418.4 Systems of Differential Operators on Lp-spaces . . . . . . . . . . . 4498.5 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 458

A Vector-valued Holomorphic Functions 461

B Closed Operators 467

C Ordered Banach Spaces 477

D Banach Spaces which Contain c0 481

E Distributions and Fourier Multipliers 485

Indexes 493Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533

Prefaces

Preface to the First Edition

Linear evolution equations in Banach spaces have seen important developmentsin the last two decades. This is due to the many different applications in thetheory of partial differential equations, probability theory, mathematical physics,and other areas, and also to the development of new techniques. One importanttechnique is given by the Laplace transform. It played an important role in theearly development of semigroup theory, as can be seen in the pioneering monographby Hille and Phillips [HP57]. But many new results and concepts have come fromLaplace transform techniques in the last 15 years. In contrast to the classicaltheory, one particular feature of this method is that functions with values in aBanach space have to be considered.

The aim of this book is to present the theory of linear evolution equations ina systematic way by using the methods of vector-valued Laplace transforms.

It is simple to describe the basic idea relating these two subjects. Let A be aclosed linear operator on a Banach space X. The Cauchy problem defined by A isthe initial value problem

(CP )

{u′(t) = Au(t) (t ≥ 0),

u(0) = x,

where x ∈ X is a given initial value. If u is an exponentially bounded, continuousfunction, then we may consider the Laplace transform

u(λ) =

∫ ∞

0

e−λtu(t) dt

of u for large real λ. It turns out that u is a (mild) solution of (CP ) if and only if

(λ− A)u(λ) = x (λ large). (1)

Thus, if λ is in the resolvent set of A, then

u(λ) = (λ− A)−1x. (2)

x PREFACES

Now it is a typical feature of concrete evolution equations that no explicit infor-mation on the solution is known and only in exceptional cases can the solution begiven by a formula. On the other hand, in many cases much can be said aboutthe resolvent of the given operator. The fact that the Laplace transform allowsus to reduce the Cauchy problem (CP ) to the characteristic equation (1) explainsits usefulness. The Laplace transform is the link between solutions and resolvents,between Cauchy problems and spectral properties of operators.

There are two important themes in the theory of Laplace transforms. Thefirst concerns representation theorems; i.e., results which give criteria to decidewhether a given function is a Laplace transform. Clearly, in view of (2), suchresults, applied to the resolvent of an operator, give information on the solvabilityof the Cauchy problem.

The other important subject is asymptotic behaviour, where the most chal-lenging and delicate results are Tauberian theorems which allow one to deduceasymptotic properties of a function from properties of its transform. Since in thecase of solutions of (CP ) the transform is given by the resolvent, such results mayallow one to deduce results of asymptotic behaviour from spectral properties of A.

These two themes describe the essence of this book, which is divided intothree parts. In the first, representation theorems for Laplace transforms are given,and corresponding to this, well-posedness of the Cauchy problem is studied. Thesecond is a systematic study of asymptotic behaviour of Laplace transforms firstof arbitrary functions, and then of solutions of (CP ). The last part contains appli-cations and illustrative examples. Each part is preceded by a detailed introductionwhere we describe the interplay between the diverse subjects and explain how thesections are related.

We have assumed that the reader is already familiar with the basic topicsof functional analysis and the theory of bounded linear operators, Lebesgue inte-gration and functions of a complex variable. We require some standard facts fromFourier analysis and slightly more advanced areas of functional analysis for whichwe give references in the text. There are also four appendices (A, B, C and E)which collect together background material on other standard topics for use invarious places in the book, while Appendix D gives a proof of a technical result inthe geometry of Banach spaces which is needed in Section 4.6.

Finally, a few words should be said about the realization of the book. Thecollaboration of the authors is based on two research activities: the common workof W. Arendt, M. Hieber and F. Neubrander on integrated semigroups and thework of W. Arendt and C. Batty on asymptotic behaviour of semigroups overmany years. Laplace transform methods are common to both.

The actual contributions are as follows.Part I: All four authors wrote this part.Part II was written by W. Arendt and C. Batty.Part III was written by W. Arendt (Chapters 6 and 7) and M. Hieber (Chap-

ter 8).

PREFACES xi

C. Batty undertook the coordination needed to make the material into aconsistent book.

The authors are grateful to many colleagues and friends with whom theyhad a fruitful cooperation, frequently over many years, which allowed them todiscuss the material presented in the book. We would especially like to acknowledgeamong them H. Amann, B. Baumer, Ph. Benilan, J. van Casteren, R. Chill, O. El-Mennaoui, J. Goldstein, H. Kellermann, V. Keyantuo, R. deLaubenfels, G. Lumer,R. Nagel, J. van Neerven, J. Pruss, F. Rabiger, A. Rhandi, W. Ruess, Q.P. Vu,and L. Weis

Special thanks go to S. Bu, R. Chill, M. Haase, R. Nagel and R. Schnaubelt,who read parts of the manuscript and gave very useful comments.

The enormous technical work on the computer, in particular typing largeparts of the manuscript and unifying 4 different TEX dialects, was done with highcompetence in a most reliable and efficient way by Mahamadi Warma. To him goour warmest thanks.

The authors are grateful to Professor H. Amann, editor of “Monographs inMathematics”, for his support. The cooperation with Birkhauser Verlag, and withDr. T. Hintermann in particular, was most enjoyable and efficient.

Ulm, Oxford, Darmstadt, Baton RougeAugust, 2000

Wolfgang ArendtCharles Batty

Matthias HieberFrank Neubrander

xii PREFACES

Preface to the Second Edition

Ten years after the publication of the first edition of this monograph, it is clearthat vector-valued Laplace transform methods continue to play an important rolein the analysis of partial differential equations and other disciplines of analysis.Among the most notable new achievements of this period are the characterizationof generators of cosine functions on Hilbert space due to Crouzeix, and quantitativeTauberian theorems for Laplace transforms with applications to energy estimatesfor wave equations.

In this second edition, the new developments have been taken into accountby updating the Notes on each Chapter and the Bibliography. For example, thecharacterization of generators of cosine functions on Hilbert space by a purelygeometric condition on the numerical range is precisely stated in Theorem 3.17.5.The main text has not been substantially changed, except in Section 4.4 wheresome results are now presented in quantitative forms. Their applications in thestudy of damped wave equations are explained in detail in the Notes of the section.

A few minor mathematical gaps and typographical errors have been corrected,and we are grateful to M. Haase, J. van Neerven, R. Schumann and D. Seifert foralerting us to some of them.

September 2010 The Authors

Part I

Laplace Transforms andWell-Posedness of Cauchy

Problems

As a guide-line for Part I, as well as for the entire book, we have in mind theformula

u(λ) = R(λ,A)x (3)

saying that a mild solution of the Cauchy problem, u′(t) = Au(t) with initial valuex, is given by the resolvent of the underlying operator A evaluated at x. Thus,if we want to find solutions, we first have to characterize those functions whichare Laplace transforms; i.e., we study representation theorems. Correspondingly,on the side of evolution equations, we investigate existence and uniqueness ofsolutions of the Cauchy problem. Other subjects treated here include regularityand positivity.

Part I contains three chapters as follows:

1. The Laplace Integral

2. The Laplace Transform

3. Cauchy Problems

We start with an introduction to the vector-valued Lebesgue integral; i.e., theBochner integral. For our purposes it suffices to consider functions defined on thereal line. Then we introduce the Laplace integral and investigate its analytic prop-erties, giving special attention to its diverse abscissas. This will play an importantrole when solutions of the Cauchy problem are considered, as the abscissas giveinformation about the asymptotic behaviour for large time. Operational proper-ties of the Laplace integral are also discussed. Finally, we introduce functions of(semi) bounded variation defined on the half-line and the Laplace-Stieltjes trans-form. They will be needed when we study resolvent positive operators (Section3.11) and Hille-Yosida operators (Section 3.5).

The vector-valued Fourier transform on the line is introduced in Section 1.6and we prove the Paley-Wiener theorem for functions with values in a Hilbertspace. This is the first of several representation theorems for Laplace transformswhich we present in this book.

In Chapter 2, real representation theorems are the central subject. We prove avector-valued version of Widder’s classical theorem which describes those functionswhich are Laplace transforms of bounded measurable functions. The vector-valuedversion (Section 2.2) will lead directly to generation theorems in Chapter 3 forsemigroups and integrated semigroups (Section 3.3) and for cosine functions (Sec-tion 3.15). A particularly simple representation theorem is valid for holomorphicfunctions (Section 2.6). The Laplace transform is an isomorphism between certainclasses of holomorphic functions defined on sectors in the complex plane. This willlead directly to the generation theorem for holomorphic semigroups in Section 3.7.The third representation theorem is a vector-valued version of Bernstein’s theo-rem describing Laplace-Stieljes transforms of monotonic functions (Section 2.7). It

3Part I

4

has its counterpart for Cauchy problems in Section 3.11 where resolvent positiveoperators are considered.

The uniqueness theorem for Laplace transforms is easy to prove (Section1.7), but it has striking consequences. It gives directly an approximation result forsequences of Laplace transforms. In Chapter 3 we find its counterpart for Cauchyproblems in the form of the Trotter-Kato theorem (Section 3.6).

For Cauchy problems, the most satisfying situation is when there correspondsexactly one (mild) solution to each initial value. This notion of well-posedness isequivalent to existence of a C0-semigroup (Section 3.1). We also consider weakerforms of well-posedness which are characterized by the existence of integratedsemigroups. In applications, they allow one to describe precise regularity for cer-tain partial differential equations in Lp(Rn), and Chapter 8 is devoted to this.Here in Part I, there are three situations where integrated semigroups occur in anatural way. Operators satisfying the Hille-Yosida condition generate locally Lip-schitz continuous integrated semigroups. Using convolution properties establishedin Section 1.3, we prove a beautiful existence and uniqueness theorem due to DaPrato and Sinestrari for the inhomogeneous Cauchy problem defined by such op-erators. The second interesting class of examples are resolvent positive operatorswhich always generate twice integrated semigroups. This will be proved in Section3.11. In Chapter 6 a resolvent positive operator will provide an elegant transitionfrom elliptic to parabolic problems. Finally, in Section 3.14 we show that the sec-ond order Cauchy problem is well-posed on a space X if and only if the associatedcanonical system generates an integrated semigroup on the product space X ×X.

In Section 3.10 we show that integrated semigroups and semigroups are equiv-alent, up to the choice of the underlying Banach space. This choice is particularlyinteresting in the context of the second order Cauchy problem. In Section 3.14 wewill show the remarkable result that the space of well-posedness is unique, and wefind the phase space associated to the second order problem. In the applicationsto the wave equation given in Chapter 7 we will see how this space is well adaptedto perturbation theory, allowing us to prove well-posedness of the wave equationdefined by very general second order elliptic operators.

Special attention is given to C0-groups; i.e., to Cauchy problems allowingunique mild solutions on the line. In Section 3.9 we study when a holomorphicsemigroup of angle π/2 has a boundary group. This problem will occur againin Section 3.16 where we investigate which cosine functions allow a square rootreduction. A striking theorem due to Fattorini shows that on UMD-spaces a squareroot reduction is always possible; i.e., each generator A of a cosine function is of theform A = B2 − ω where B generates a C0-group and ω ≥ 0. This beautiful resultconcludes the three sections on the second order Cauchy problem, applications ofwhich will be given in Chapters 7 and 8.

Part I

Chapter 1

The Laplace Integral

The first three sections of this chapter are of a preliminary nature. There, wecollect properties of the Bochner integral of functions of a real variable with valuesin a Banach space X . We then concentrate on the basic properties of the Laplaceintegral

f(λ) :=

∫ ∞

0

e−λtf(t) dt := limτ→∞

∫ τ

0

e−λtf(t) dt

for locally Bochner integrable functions f : R+ → X. In Section 1.4 we describe theset of complex numbers λ for which the Laplace integral converges. It will be shownthat the domain of convergence is non-empty if and only if the antiderivative of f isof exponential growth. In Section 1.5 we discuss the holomorphy of λ �→ f(λ) and

in Section 1.7 we show that f is uniquely determined by the Laplace integrals f(λ)(uniqueness and inversion). In Section 1.6 we prove the operational properties ofthe Laplace integral which are essential in applications to differential and integralequations. In particular, we show that the Laplace integral of the convolutionk ∗ f : t �→ ∫ t

0k(t − s)f(s) ds of a scalar-valued function k with a vector-valued

function f is given by

(k ∗ f)(λ) = k(λ)f(λ)

if f(λ) exists and k(λ) exists as an absolutely convergent integral. In Section 1.8 weconsider vector-valued Fourier transforms and we show that Plancherel’s theoremand the Paley-Wiener theorem extend to functions with values in a Hilbert space.Finally, after introducing the basic properties of the Riemann-Stieltjes integral inSection 1.9, we extend in Section 1.10 the basic properties of Laplace integrals toLaplace-Stieltjes integrals

dF (λ) :=

∫ ∞

0

e−λt dF (t) := limτ→∞

∫ τ

0

e−λt dF (t)

of functions F of bounded semivariation.

W. Arendt et al., Vector-valued Laplace Transforms and Cauchy Problems: Second Edition,Monographs in Mathematics 96, DOI 10.1007/978-3-0348-0087-7_1, © Springer Basel AG 2011

5

6 1. THE LAPLACE INTEGRAL

If f is Bochner integrable, then the normalized antiderivative t �→ F (t) :=∫ t

0f(s) ds is of bounded variation. We will see that f(λ) exists if and only if dF (λ)

exists, and in this case f(λ) = dF (λ). Thus, the Laplace-Stieltjes integral is a nat-ural extension of the Laplace integral. This extension is crucial for our discussion ofthe Laplace transform in Chapter 2 since there are many functions r : (ω,∞)→ Xwhich can be represented as a Laplace-Stieltjes integral, but not as a Laplace inte-gral of a Bochner integrable function. Examples are, among others, Dirichlet series

r(λ) =∑∞

n=1 ane−λn = dF (λ), where F is the step function

∑∞n=1 anχ(n,∞), or

any function r(λ) = dF (λ), where F is of bounded semivariation, but not theantiderivative of a Bochner integrable function.

1.1 The Bochner Integral

This section contains some properties of the Bochner integral of vector-valuedfunctions. We shall consider only those properties which are used in later sections,and we shall assume that the reader is familiar with the basic facts about measureand integration of scalar-valued functions.

Let X be a complex Banach space, and let I be an interval (bounded orunbounded) in R, or a rectangle in R2. A function f : I → X is simple if it isof the form f(t) =

∑nr=1 xrχΩr(t) for some n ∈ N := {1, 2, . . .}, xr ∈ X and

Lebesgue measurable sets Ωr ⊂ I with finite Lebesgue measure m(Ωr); f is a stepfunction when each Ωr can be chosen to be an interval, or a rectangle in R2. HereχΩ denotes the characteristic (indicator) function of Ω. In the representation of asimple function, the sets Ωr may always be arranged to be disjoint, and then

f(t) =

{xr (t ∈ Ωr; r = 1, 2, . . . , n)

0 otherwise.

A function f : I → X is measurable if there is a sequence of simple functionsgn such that f(t) = limn→∞ gn(t) for almost all t ∈ I. Since any χΩ for Ω measur-able is a pointwise almost everywhere (a.e.) limit of a sequence of step functions,it is not difficult to see that the functions gn can be chosen to be step functions.When X = C, this definition agrees with the usual definition of (Lebesgue) mea-surable functions. It is easy to see that if f : I → X, g : I → X and h : I → Care measurable, then f + g and h · f are measurable. Moreover, if k : X → Y iscontinuous (where Y is any Banach space), then k ◦ f is measurable whenever f ismeasurable. In particular, ‖f‖ is measurable. If X is a closed subspace of Y , andf is measurable as a Y -valued function, then f is also measurable as an X-valuedfunction.

To verify measurability of a function we often use the characterization givenby Pettis’s theorem below. We say that f : I → X is countably valued if there isa countable partition {Ωn : n ∈ N} of I into subsets Ωn such that f is constanton each Ωn; it is easy to see that f is measurable if each Ωn is measurable (and

1.1. THE BOCHNER INTEGRAL 7

conversely, {t ∈ I : f(t) = x} is measurable whenever f is measurable andx ∈ X). Also, f : I → X is called almost separably valued if there is a null setΩ0 in I such that f(I \ Ω0) := {f(t) : t ∈ I \ Ω0} is separable (equivalently,f(I \Ω0) is contained in a separable closed subspace of X); f is weakly measurableif x∗ ◦ f : t �→ 〈f(t), x∗〉 is Lebesgue measurable for each x∗ in the dual space X∗

of X .Here and subsequently, 〈·, ·〉 denotes the duality between X and X∗. For a

subset D of X, we shall denote the closure of D in X by D. For x ∈ X and ε > 0,we shall let BX(x, ε) := B(x, ε) := {y ∈ X : ‖y − x‖ < ε} and B(x, ε) := {y ∈ X :‖y − x‖ ≤ ε}. We shall also use this notation when X = Rn or X = C, when itwill be implicit that the norm is the Euclidean norm.

Theorem 1.1.1 (Pettis). A function f : I → X is measurable if and only if it isweakly measurable and almost separably valued.

Proof. If f is measurable, then there exist a null set Ω0 and simple functions gnsuch that gn → f pointwise on I \ Ω0. The simple functions x∗ ◦ gn converge tox∗ ◦ f on I \ Ω0 for all x∗ ∈ X∗. Therefore, f is weakly measurable. The valuestaken by the functions gn form a countable set D and f(I \ Ω0) ⊂ D. Thus, f isalmost separably valued.

To prove the converse statement one may replace X by the smallest closedsubspace which contains f(I \ Ω0) and then choose a countable dense set {xn :n ∈ N}. By the Hahn-Banach theorem, there are unit vectors x∗n ∈ X∗ with|〈xn, x

∗n〉| = ‖xn‖. For any ε > 0 and x ∈ X there exists xk such that ‖x−xk‖ < ε.

Hence,

supn|〈x, x∗n〉| ≤ ‖x‖ ≤ ‖xk‖+ ε = |〈xk, x∗k〉|+ ε

≤ |〈x− xk, x∗k〉|+ |〈x, x∗k〉|+ ε

≤ supn|〈x, x∗n〉|+ 2ε.

So‖x‖ = sup

n|〈x, x∗n〉| for all x ∈ X. (1.1)

This implies that t �→ ‖f(t)−x‖ = supn |〈f(t)−x, x∗n〉| is measurable for all x ∈ X.Let

Δ := {t ∈ I \ Ω0 : ‖f(t)‖ > 0} and Δn,ε := {t ∈ Δ : ‖f(t)− xn‖ < ε}for ε > 0 and n ∈ N. The sets Δn,ε are measurable and

⋃nΔn,ε = Δ. For

fixed ε > 0, the sets Ω1,ε := Δ1,ε and Ωn,ε := Δn,ε \⋃

k<n Δk,ε (n ≥ 2) form ameasurable partition of Δ. Define a measurable, countably valued function gε onI by gε :=

∑∞i=1 xiχΩi,ε

. Let t ∈ I \ Ω0. If t �∈ Δ, then f(t) = 0 = gε(t). If t ∈ Δ,then there exists n ∈ N such that t ∈ Ωn,ε. Hence,

‖f(t)− gε(t)‖ < ε for all t ∈ I \ Ω0.

This shows that f is the uniform limit almost everywhere of a sequence of mea-surable, countably valued functions.

Let (In) be an increasing sequence of bounded subintervals of I with I =⋃n In. For each n ∈ N, define a simple function hn := g2−nχHn

, where Hn := In ∩⋃kn

i=1Ωi,2−n and kn is chosen such that the Lebesgue measure m(In \Hn) < 2−n.If t ∈ ⋂∞

n=k Hn for some k ≥ 1, then

‖f(t)− hn(t)‖ = ‖f(t)− g2−n(t)‖ < 2−n

for all n ≥ k. Thus, limn→∞ hn(t) = f(t) for all t ∈ ⋃∞k=1

⋂∞n=k Hn. For k ≥ j,

m

(Ij \

∞⋂n=k

Hn

)≤

∞∑n=k

m(In \Hn) < 2−k+1.

Hence, Ij \⋃∞

k=1

⋂∞n=k Hn is null, for each j. This shows that limn→∞ hn(t) = f(t)

for almost all t ∈ I.

Corollary 1.1.2. Let f : I → X. Then the following statements hold:

a) The function f is measurable if and only if it is the uniform limit almosteverywhere of a sequence of measurable, countably valued functions.

b) If X is separable, then f is measurable if and only if it is weakly measurable.

c) If f is continuous, then it is measurable.

d) If fn : I → X are measurable functions and fn → f pointwise a.e., then fis measurable.

Proof. The statement b) is an immediate consequence of Pettis’s Theorem 1.1.1.For d), observe first that f is weakly measurable. Define Ω0 := ∪nΩn where Ωn is anull set such that fn(I \Ωn) is separable. Then m(Ω0) = 0 and Δ :=

⋃n fn(I \Ω0)

is separable. Since the least closed subspace containing Δ is separable and containsf(I \ Ω0) it follows that f is almost separably valued. Thus, f is measurable. Iff is continuous, then f is weakly measurable and the countable set {f(t) : t ∈Q} is dense in the range of f . Again by Pettis’s theorem, f is measurable. Oneimplication of a) was established in the proof of Pettis’s theorem and the conversefollows from d).

Pettis’s theorem can be improved considerably in the following way. A subsetW of X∗ is called separating if for all x ∈ X \ {0} there exists x∗ ∈ W such that〈x, x∗〉 �= 0.

Corollary 1.1.3. Let f : I → X be an almost separably valued function. Assumethat x∗ ◦ f is measurable for all x∗ in a separating subset W of X∗. Then f ismeasurable.


Proof. Changing f on a set of measure 0 and replacing X by a subspace, we canassume that X is separable. Let

Y := {x∗ ∈ X∗ : x∗ ◦ f is measurable}.Then Y is a subspace of X∗ which contains W . By the Hahn-Banach theorem, Yis weak* dense in X∗. Let Y1 = Y ∩ BX∗(0, 1). We show that Y1 is weak* closed.Let x∗ be in the weak* closure of Y1. Since X is separable, the weak* topology onBX∗(0, 1) is metrizable (see [Meg98, Theorem 2.6.23], for example). Thus, thereexists a sequence (x∗n)n∈N in Y1 converging to x∗. Hence, x∗n ◦f → x∗ ◦f as n→∞pointwise on I. Thus, x∗ ◦ f is measurable; i.e., x∗ ∈ Y1. This proves the claim. Itfollows from the Krein-Smulyan theorem (Theorem A.6) that Y is weak* closed.Since Y is weak* dense, we have Y = X∗; i.e., f is weakly measurable. Now theresult follows from Theorem 1.1.1.

For a simple function g : I → X, g =∑n

i=1 xiχΩi, we define∫

I

g(t) dt :=

n∑i=1

xim(Ωi)

where m(Ω) is the Lebesgue measure of Ω. It is routine to verify that the definitionis independent of the representation g =

∑ni=1 xiχΩi , and the integral so defined

is linear.A function f : I → X is called Bochner integrable if there exist simple

functions gn such that gn → f pointwise a.e., and limn→∞∫I‖f(t)−gn(t)‖ dt = 0.

If f : I → X is Bochner integrable, then the Bochner integral of f on I is∫I

f(t) dt := limn→∞

∫I

gn(t) dt.

It is easy to see that this limit exists and is independent of the choice of thesequence (gn). If Ω is measurable with finite measure, then χΩ can be approximatedin L1-norm by step functions, and it follows that the functions gn can always bechosen to be step functions. The integral

∫If(t) dt lies in the closed linear span of

{f(t) : t ∈ I}. The set of all Bochner integrable functions from I to X is a linearspace and the Bochner integral is a linear mapping. When X = C, the definitionsof Bochner integrability and integrals agree with those of Lebesgue integrationtheory.

When I is a rectangle, we may denote a Bochner integral by∫If(s, t) d(s, t).

It is one of the great virtues of the Bochner integral that the class of Bochnerintegrable functions is easily characterized.

Theorem 1.1.4 (Bochner). A function f : I → X is Bochner integrable if and onlyif f is measurable and ‖f‖ is integrable. If f is Bochner integrable, then∥∥∥∥∫

I

f(t) dt

∥∥∥∥ ≤ ∫I

‖f(t)‖ dt. (1.2)

Proof. If f is Bochner integrable, then there exists an approximating sequenceof simple functions gn. Thus f and ‖f‖ are measurable. The integrability of ‖f‖follows from ∫

I

‖f(t)‖ dt ≤∫I

‖gn(t)‖ dt+∫I

‖f(t)− gn(t)‖ dt.

Moreover, ∥∥∥∥∫I

f(t) dt

∥∥∥∥ = limn→∞

∥∥∥∥∫I

gn(t) dt

∥∥∥∥ ≤ limn→∞

∫I

‖gn(t)‖ dt

=

∫I

‖f(t)‖ dt.

To prove the converse statement, let (hn) be a sequence of simple functionsapproximating f pointwise on I \ Ω0, where m(Ω0) = 0. Define simple functionsby

gn(t) :=

{hn(t) if ‖hn(t)‖ ≤ ‖f(t)‖(1 + n−1),

0 otherwise.

Then ‖gn(t)‖ ≤ ‖f(t)‖(1 + n−1) and limn→∞ ‖gn(t)− f(t)‖ = 0 for all t ∈ I \Ω0.Because the functions ‖f‖ and ‖gn − f‖ are integrable and ‖gn(t) − f(t)‖ ≤3‖f(t)‖, we can apply the scalar dominated convergence theorem and obtain thatlimn→∞

∫I‖gn(t)− f(t)‖ dt = 0.

Example 1.1.5. a) Let X be the Lebesgue space L∞(0, 1) of all (equivalence classesof) bounded measurable functions from (0, 1) to C. Let f : (0, 1) → L∞(0, 1) begiven by f(t) := χ(0,t). Then f is not almost separably valued since ‖f(t)−f(s)‖ =1 for t �= s. Thus, f is not measurable and therefore not Bochner integrable.

b) Let X be the Banach space c0 of all complex sequences x = (xn)n∈N such thatlimn→∞ xn = 0, with ‖x‖ = supn |xn|. Identify X∗ with the space �1 of all complexsequences a = (an)n∈N such that ‖a‖ :=

∑∞n=1 |an| < ∞. Let f : [0, 1] → c0 be

given by f(t) := (fn(t))n∈N where fn(t) := nχ[0, 1n ](t). Let x

∗ = (an)n∈N ∈ �1. Then

t �→ 〈f(t), x∗〉 = ∑∞n=1 nanχ[0, 1

n ](t) is measurable on [0, 1]. Since c0 is separable,it follows from Pettis’s theorem that f is measurable. Moreover,∫ 1

0

|〈f(t), x∗〉| dt ≤∞∑

n=1

|an| = ‖x∗‖ <∞.

However, ‖f(t)‖ = n for t ∈ ( 1n+1 ,

1n ], so t �→ ‖f(t)‖ is not integrable on [0, 1].

Thus, f is not Bochner integrable on [0, 1].

Now we will consider the behaviour of the Bochner integral under linearoperators. The following result is a straightforward consequence of the definitionof the Bochner integral, and we shall use it frequently without comment, especiallyin the case of a linear functional (Y = C).

Proposition 1.1.6. Let T : X → Y be a bounded linear operator between Banachspaces X and Y , and let f : I → X be Bochner integrable. Then T ◦f : t �→ T (f(t))is Bochner integrable and T

∫If(t) dt =

∫IT (f(t)) dt.

We shall also need a version of Proposition 1.1.6 for a closed operator A onX (see Appendix B for notation and terminology).

Proposition 1.1.7. Let A be a closed linear operator on X. Let f : I → X beBochner integrable. Suppose that f(t) ∈ D(A) for all t ∈ I and A ◦ f : I → X isBochner integrable. Then

∫If(t) dt ∈ D(A) and

A

∫I

f(t) dt =

∫I

A(f(t)) dt.

Proof. Consider X ×X as a Banach space in the norm ‖(x, y)‖ = ‖x‖+ ‖y‖. Thegraph G(A) of A is a closed subspace of X ×X . Define g : I → G(A) ⊂ X ×X byg(t) = (f(t), A(f(t))). It is easy to see that g is measurable and∫

I

‖g(t)‖ dt =∫I

‖f(t)‖ dt+∫I

‖A(f(t))‖ dt <∞.

By Theorem 1.1.4, g is Bochner integrable. Moreover,∫Ig(t) dt ∈ G(A). Applying

Proposition 1.1.6 to the two projection maps of X ×X onto X shows that∫I

g(t) dt =

(∫I

f(t) dt ,

∫I

A(f(t)) dt

).

This gives the result.

Now we give vector-valued versions of two classical theorems of integrationtheory.

Theorem 1.1.8 (Dominated Convergence). Let fn : I → X (n ∈ N) be Bochnerintegrable functions. Assume that f(t) := limn→∞ fn(t) exists a.e. and there existsan integrable function g : I → R such that ‖fn(t)‖ ≤ g(t) a.e. for all n ∈ N.Then f is Bochner integrable and

∫If(t) dt = limn→∞

∫Ifn(t) dt. Furthermore,∫

I‖f(t)− fn(t)‖ dt→ 0 as n→∞.

Proof. The function f is Bochner integrable since it is measurable (by Corollary1.1.2) and since ‖f‖ is integrable (because ‖f(t)‖ ≤ g(t) a.e.). Define hn(t) :=‖f(t) − fn(t)‖ for t ∈ I. Since |hn(t)| ≤ 2g(t) and hn(t) → 0 a.e., the scalardominated convergence theorem implies that

∫I‖f(t) − fn(t)‖ dt → 0 as n → ∞.

By (1.2), ∥∥∥∥∫I

f(t) dt−∫I

fn(t) dt

∥∥∥∥→ 0.

Theorem 1.1.9 (Fubini’s Theorem). Let I = I1 × I2 be a rectangle in R2, letf : I → X be measurable, and suppose that∫

I1

∫I2

‖f(s, t)‖ dt ds <∞.

Then f is Bochner integrable and the repeated integrals∫I1

∫I2

f(s, t) dt ds,

∫I2

∫I1

f(s, t) ds dt

exist and are equal, and they coincide with the double integral∫If(s, t) d(s, t).

Proof. Since any measurable function is almost separably valued, we may assumethat X is separable.

The scalar-valued case of Fubini’s theorem implies that ‖f‖ is integrable onI,

∫I2‖f(s, t)‖ dt exists for almost all s ∈ I1, and for each x∗ ∈ X∗ the repeated

integrals ∫I1

∫I2

〈f(s, t), x∗〉 dt ds,∫I2

∫I1

〈f(s, t), x∗〉 ds dt

exist and are equal. It follows from Theorem 1.1.4 that f : I → X is Bochnerintegrable and

∫I2f(s, t) dt exists for almost all s ∈ I1, and from Theorem 1.1.1

that s �→ ∫I2f(s, t) dt is measurable. Moreover,∫

I1

∥∥∥∥∫I2

f(s, t) dt

∥∥∥∥ ds ≤∫I1

∫I2

‖f(s, t)‖ dt ds <∞,

so Theorem 1.1.4 shows that∫I1

(∫I2f(s, t) dt

)ds exists. Since∫

I2

∫I1

‖f(s, t)‖ ds dt =∫I1

∫I2

‖f(s, t)‖ dt ds,

it follows similarly that∫I2

(∫I1f(s, t) ds

)dt exists. For any x∗ ∈ X∗,⟨∫

I1

(∫I2

f(s, t) dt

)ds, x∗

⟩=

∫I1

∫I2

〈f(s, t), x∗〉 dt ds

=

∫I

〈f(s, t), x∗〉 d(s, t)

=

⟨∫I

f(s, t) d(s, t), x∗⟩

=

∫I2

∫I1

〈f(s, t), x∗〉 ds dt

=

⟨∫I2

(∫I1

f(s, t) ds

)dt, x∗

⟩.

The Hahn-Banach theorem implies that∫I1

(∫I2

f(s, t) dt

)ds =

∫I

f(s, t) d(s, t) =

∫I2

(∫I1

f(s, t) ds

)dt.

Let L1(I,X) denote the space of all Bochner integrable functions f : I → X,and let

‖f‖1 :=

∫I

‖f(t)‖ dt.

In the usual way, we shall identify functions which differ only on sets of measurezero. Then ‖ · ‖1 is a norm on L1(I,X).

Theorem 1.1.10. The space L1(I,X) is a Banach space.

Proof. Let (fn) be a sequence in L1(I,X) with∑ ‖fn‖1 < ∞. By the monotone

convergence theorem for series of positive scalar-valued functions,∑ ‖fn(t)‖ <∞

a.e.,∑∞

n=1 ‖fn(·)‖ is integrable, and∫I

∞∑n=1

‖fn(t)‖ dt =∞∑

n=1

∫I

‖fn(t)‖ dt.

Hence,∑∞

n=1 fn(t) converges a.e. to a sum g(t) in the Banach space X . By Corol-lary 1.1.2, g is measurable. Moreover, ‖g(t)‖ ≤∑∞

n=1 ‖fn(t)‖, so ‖g‖ is integrable.By Theorem 1.1.4, g is integrable. Finally,∥∥∥∥∥g −

k∑n=1

fn

∥∥∥∥∥1

≤∫I

‖g(t)−k∑

n=1

fn(t)‖ dt ≤∫I

∞∑n=k+1

‖fn(t)‖ dt→ 0

as k →∞. Thus, L1(I,X) is a Banach space.

By the definition of Bochner integrability, the simple functions are dense inL1(I,X) and, by the remarks following the definition, the step functions are dense.It follows easily that the infinitely differentiable functions of compact support arealso dense in L1(I,X).

We shall be particularly interested in the case when I = R+ := [0,∞). Iff ∈ L1(R+, X), an application of the Dominated Convergence Theorem showsthat ∫ ∞

0

f(t) dt = limτ→∞

∫ τ

0

f(t) dt. (1.3)

When f ∈ L1loc(R+, X) (i.e., f is Bochner integrable on [0, τ ] for every τ ∈ R+),

the limit in (1.3) may exist without f being Bochner integrable on R+. If thelimit exists, we say that

∫∞0

f(t) dt converges as an improper (or principal value)integral, and we define ∫ ∞

0

f(t) dt := limτ→∞

∫ τ

0

f(t) dt.

When f ∈ L1(R+, X), i.e.∫∞0‖f(t)‖ dt <∞, we say that the integral is absolutely

convergent .For 1 < p < ∞, let Lp(I,X) denote the space of all measurable functions

f : I → X such that

‖f‖p :=

(∫I

‖f(t)‖p dt)1/p

<∞.

Let L∞(I,X) be the space of all measurable functions f : I → X such that

‖f‖∞ := ess supt∈I‖f(t)‖ <∞.

Note that the spaces Lp(I,C) (1 ≤ p ≤ ∞) are the usual Lebesgue spaces whichwe shall denote simply by Lp(I). With the usual identifications, each of the spacesLp(I,X) becomes a Banach space. The proofs of completeness are similar to thescalar-valued cases.

The proof of Theorem 1.1.4 shows that the simple functions are dense inLp(I,X) for 1 < p < ∞ (so Lp(I,X) can also be defined in a similar way to theBochner integrable functions). It follows that the step functions, and the infinitelydifferentiable functions of compact support, are also dense. By considering suchfunctions first and then approximating, one may show as in the scalar-valued case,that if f ∈ Lp(I,X) and

ft(s) :=

{f(s− t) if s− t ∈ I,

0 otherwise,

then t �→ ft is continuous from R into Lp(I,X) for 1 ≤ p <∞.We have presented the theory above in the case when I is an interval in R

(or, for Fubini’s theorem, I is a rectangle in R2). Almost all the integrals whichappear in this book will indeed be over intervals in R (or repeated integrals inR2). However, the entire theory works, with no changes in the proofs, when I is ameasurable set in Rn (or in Rm×Rn, in Fubini’s theorem). Since the step functionsare dense in each of the spaces Lp(I × J,X) for 1 ≤ p <∞, it is easy to see fromFubini’s theorem that there is an isometric isomorphism between Lp(I×J,X) andLp(I, Lp(J,X)) given by f �→ g, where

(g(s))(t) := f(s, t).

This enables many properties of the spaces Lp(I,X) when I is a rectangle in Rn

to be deduced from the case n = 1.Finally in this section, we introduce notation for spaces of continuous and

differentiable functions. Let I be an interval in R. We denote by C(I,X) the vectorspace of all continuous functions f : I → X. For k ∈ N, we denote by Ck(I,X) thespace of all k-times differentiable functions with continuous kth derivative, and we

1.2. THE RADON-NIKODYM PROPERTY 15

put C∞(I,X) :=⋂∞

k=1Ck(I,X). When I is compact, C(I,K) is a closed subspace

of L∞(I,X) and therefore a Banach space with respect to the supremum norm‖ · ‖∞.

When I is not compact, we let Cc(I,X) be the space of all functions inC(I,X) with compact support, and C∞c (I,X) := Cc(I,X) ∩ C∞(I,X). ThusC∞c (I,X) is a dense subspace of Lp(I,X) for 1 ≤ p <∞. When I = R+ or I = R,we shall also consider the space C0(I,X) of all continuous functions f : I → X suchthat lim|t|→∞,t∈I ‖f(t)‖ = 0 and the space BUC(I,X) of all bounded, uniformlycontinuous functions f : I → X. These are both Banach spaces with respect to‖ · ‖∞, and C0(I,X) ⊂ BUC(I,X) ⊂ L∞(I,X).

When X = C, we shall write C(I) in place of C(I,C), etc., and we shallextend this notation to cases when I is replaced by an open subset Ω of Rn. Notethat C∞c (Ω) coincides with the space D(Ω) of test functions on Ω (see AppendixE), and we shall use both notations according to context. Furthermore, when Ω isany locally compact space, we shall let C0(Ω) be the Banach space of all continuouscomplex-valued functions on Ω which vanish at infinity, with the supremum norm.When K is any compact space, we shall let C(K) be the Banach space of allcontinuous complex-valued functions on K, with the supremum norm.

1.2 The Radon-Nikodym Property

In this section we consider properties of functions F obtained as indefinite inte-grals. If f : [a, b] → X is Bochner integrable, we say that F : [a, b] → X is anantiderivative or primitive of f if

F (t) = F (a) +

∫ t

a

f(s) ds (t ∈ [a, b]).

Given a function F : [a, b]→ X and a partition π, a = t0 < t1 < . . . < tn = b,let

V (π, F ) :=n∑

i=1

‖F (ti)− F (ti−1)‖.

Then F is said to be of bounded variation if

V (F ) := V[a,b](F ) := supπ

V (π, F ) <∞,

where the supremum is taken over all partitions π of [a, b].We say that F is absolutely continuous on [a, b] if for every ε > 0 there

exists δ > 0 such that∑

i ‖F (bi)− F (ai)‖ < ε for every finite collection {(ai, bi)}of disjoint intervals in [a, b] with

∑i(bi − ai) < δ. We say that F is Lipschitz

continuous if there exists M such that ‖F (t)−F (s)‖ ≤M |t− s| for all s, t ∈ [a, b].Clearly, any Lipschitz continuous function is absolutely continuous.

Proposition 1.2.1. Let F : [a, b] → X be absolutely continuous. Then F is ofbounded variation. Moreover, if G(t) := V[a,t](F ), then G is absolutely continuouson [a, b].

Proof. Take ε > 0, and let δ be as in the definition of absolute continuity of F . Then∑i V[ai,bi](F ) ≤ ε whenever {(ai, bi)} is a finite collection of disjoint subintervals

of [a, b] with∑

i(bi − ai) < δ. In particular, F is of bounded variation on anysubinterval of length less than δ. Since [a, b] is a finite union of such intervals, Fis of bounded variation on [a, b]. Moreover,∑

i

|G(bi)−G(ai)| =∑i

V[ai,bi](F ) < ε.

Thus, G is absolutely continuous.

A point t ∈ [a, b] is said to be a Lebesgue point of f ∈ L1([a, b], X) if

limh→01h

∫ t+h

t‖f(s) − f(t)‖ ds = 0. It is easy to see that any point of continu-

ity is a Lebesgue point, and the following proposition shows that almost all pointsare Lebesgue points.

Proposition 1.2.2. Let f : [a, b]→ X be Bochner integrable and F (t) :=∫ t

af(s) ds

(t ∈ [a, b]). Then

a) F is differentiable a.e. and F ′ = f a.e.

b) limh→01h

∫ t+h

t‖f(s)− f(t)‖ ds = 0 t-a.e.

c) F is absolutely continuous.

d) V[a,b](F ) =∫ b

a‖f(s)‖ ds.

Proof. To show a) and b) let gn be step functions such that

f(t) = limn→∞ gn(t) a.e. and lim

n→∞

∫ b

a

‖f(t)− gn(t)‖ dt = 0.

For h > 0,∥∥∥∥∥ 1h∫ t+h

t

f(s) ds− f(t)

∥∥∥∥∥ ≤ 1

h

∫ t+h

t

‖f(s)− f(t)‖ ds

≤ 1

h

∫ t+h

t

‖f(s)− gn(s)‖ ds

+1

h

∫ t+h

t

‖gn(s)− gn(t)‖ ds+ ‖gn(t)− f(t)‖.

Since gn is a step function and s �→ ‖fn(s) − gn(s)‖ is Lebesgue integrable, itfollows from Lebesgue’s theorem [Rud87, Theorem 8.17] that

lim suph↓0

∥∥∥∥∥ 1h∫ t+h

t

f(s) ds− f(t)

∥∥∥∥∥ ≤ lim suph↓0

1

h

∫ t+h

t

‖f(s)− f(t)‖ ds

≤ 2‖gn(t)− f(t)‖

for all t ∈ [a, b] \ Ωn and some null set Ωn. Taking the limit as n → ∞ yields theright-differentiability of F and

limh↓0

1

h

∫ t+h

t

‖f(s)− f(t)‖ ds = 0

for all t ∈ [a, b] \⋃n∈N Ωn. The left-hand limits are similar.For c), let ε > 0. There exists δ > 0 such that

∫Ω‖f(s)‖ ds < ε whenever

μ(Ω) < δ. If {(ai, bi)} is a finite collection of disjoint subintervals of [a, b] with∑i(bi − ai) < δ, then taking Ω =

⋃i(ai, bi), we deduce that

∑i

‖F (bi)− F (ai)‖ =∑i

∥∥∥∥∥∫ bi

ai

f(s) ds

∥∥∥∥∥ ≤∫Ω

‖f(s)‖ ds < ε.

To prove the statement d), observe first that, for any partition π of [a, b],

V (π, F ) =∑i

∥∥∥∥∥∫ ti

ti−1

f(s) ds

∥∥∥∥∥ ≤∫ b

a

‖f(s)‖ ds.

Thus, V (F ) ≤ ∫ b

a‖f(s)‖ ds. Conversely, given ε > 0, we may choose a step function

g such that∫ b

a‖f(s) − g(s)‖ ds < ε. There is a partition π of [a, b] such that g is

constant on each interval (ti−1, ti). Then∫ b

a

‖f(s)‖ ds− V (F ) ≤∫ b

a

‖f(s)‖ ds− V (π, F )

≤∫ b

a

‖g(s)‖ ds+ ε−∑i

∥∥∥∥∥∫ ti

ti−1

f(s) ds

∥∥∥∥∥=

∑i

(∥∥∥∥∥∫ ti

ti−1

g(s) ds

∥∥∥∥∥−∥∥∥∥∥∫ ti

ti−1

f(s) ds

∥∥∥∥∥)

+ ε

≤∫ b

a

‖f(s)− g(s)‖ ds+ ε

≤ 2ε.

Since ε > 0 is arbitrary, this completes the proof of d).


In the scalar case, the fundamental theorem of calculus [Rud87, Theorem8.18] states that any absolutely continuous function F : [a, b]→ C is differentiable

a.e., f := F ′ is Lebesgue integrable, and F (t)− F (a) =∫ t

af(s) ds for all t ∈ [a, b].

We will see below (Example 1.2.8) that the fundamental theorem does not holdfor Lipschitz continuous functions with values in arbitrary Banach spaces. Thefollowing weaker statement holds for all Banach spaces.

Proposition 1.2.3. Let F : [a, b] → X be absolutely continuous, and suppose that

f(t) := F ′(t) exists a.e. Then f is Bochner integrable and F (t) = F (a)+∫ t

af(s) ds

for all t ∈ [a, b].

Proof. Since f(t) = limn→∞ n(F (t+ 1/n) − F (t)), it follows from Corollary 1.1.2that f is measurable. Let G(t) := V[a,t](F ), so G : [a, b]→ R is absolutely contin-uous by Proposition 1.2.1. Hence G is differentiable a.e. and G′ ∈ L1[a, b]. Since

‖F (t+ h)− F (t)‖ ≤ V[t,t+h](F ) = G(t+ h)−G(t),

‖f(t)‖ ≤ G′(t) a.e. Hence ‖f‖ ∈ L1[a, b], so f is Bochner integrable by Theorem1.1.4. For x∗ ∈ X∗,

〈F (t), x∗〉 = 〈F (a), x∗〉+∫ t

a

〈f(s), x∗〉 ds

=

⟨F (a) +

∫ t

a

f(s) ds, x∗⟩

by the scalar fundamental theorem of calculus. By the Hahn-Banach theorem,F (t) = F (a) +

∫ t

af(s) ds.

Let I be any interval in R. A function F : I → X is said to be absolutelycontinuous if it is absolutely continuous on each compact interval of I. We nowconsider the property that every absolutely continuous function F : I → X isdifferentiable a.e. It is easy to see that this property is independent of the intervalI, so it is a property of X alone.

Proposition 1.2.4. For any Banach space X the following are equivalent:

(i) Every absolutely continuous function F : R+ → X is differentiable a.e.

(ii) Every Lipschitz continuous function F : R+ → X is differentiable a.e.

Proof. Clearly, (i) implies (ii). Assume that statement (ii) holds and let F : R+ →X be absolutely continuous. By Proposition 1.2.1, F is locally of bounded variationand G is absolutely continuous where G(t) := V[0,t](F ). Let h(t) := G(t)+ t. Thenh is strictly increasing, h(0) = 0, and h(R+) = R+. Moreover,

‖F (t)− F (s)‖ ≤ G(t)−G(s) ≤ h(t)− h(s)


for all t ≥ s ≥ 0. Hence, F ◦h−1 : R+ → X is Lipschitz continuous. By assumption,F ◦h−1 is differentiable a.e. Since |h(t)−h(s)| ≥ |t−s|, h−1 maps null sets to nullsets. Moreover, h is differentiable a.e., since G is absolutely continuous. It followsfrom the chain rule that F is differentiable a.e.

Definition 1.2.5. A Banach space X is said to have the Radon-Nikodym propertyif the equivalent conditions of Proposition 1.2.4 are satisfied.

By the remarks above, the space X has the Radon-Nikodym property if andonly if every Lipschitz continuous function F : [0, 1] → X is differentiable almosteverywhere. It is clear that a closed subspace of a space with the Radon-Nikodymproperty has the Radon-Nikodym property.

Next we exhibit a class of spaces having the Radon-Nikodym property.

Theorem 1.2.6 (Dunford-Pettis). Let X = Y ∗ where Y is a Banach space, andsuppose that X is separable. Then X has the Radon-Nikodym property.

Proof. By Proposition 1.2.4, it suffices to show that every Lipschitz function F :R+ → X is differentiable a.e. We may assume that F (0) = 0 and the Lipschitzconstant is 1, so that ‖F (t) − F (s)‖ ≤ |t − s|. For y ∈ Y , the function 〈y, F (·)〉 :R+ → C is Lipschitz with Lipschitz constant ‖y‖, so there exists gy ∈ L∞(R+)such that ‖gy‖∞ ≤ ‖y‖ and

〈y, F (t)〉 =∫ t

0

gy(s) ds (t ∈ R+).

Moreover, gy is unique up to null sets.Since Y ∗ is separable, Y is also separable (see [Meg98, Theorem 1.12.11]).

Let D be a countable dense subset of Y . Suppose that y =∑n

i=1 αiyi for somen ∈ N, αi ∈ Q+ iQ and yi ∈ D. Then

〈y, F (t)〉 =n∑

i=1

αi〈yi, F (t)〉 =∫ t

0

n∑i=1

αigyi(s) ds,

so gy(s) =∑n

i=1 αigyi(s) a.e. Hence,∣∣∣∣∣

n∑i=1

αigyi(s)

∣∣∣∣∣ ≤∥∥∥∥∥

n∑i=1

αiyi

∥∥∥∥∥ (1.4)

for almost all s ∈ R+. This holds for all possible n ∈ N, αi ∈ Q+ iQ and yi ∈ D,but there are only countably many such possibilities. Hence there is a null subsetΩ of R+ such that (1.4) holds for all s ∈ R+ \ Ω and all n ∈ N, αi ∈ Q + iQ andyi ∈ D. It follows immediately that (1.4) holds for all αi ∈ C. This shows that fors ∈ R+ \ Ω, the map y �→ gy(s) from D to C extends to a unique f(s) ∈ Y ∗ = Xwith ‖f(s)‖ ≤ 1. For y ∈ D, 〈y, f(·)〉 is measurable and bounded, and

〈y, F (t)〉 =∫ t

0

〈y, f(s)〉 ds. (1.5)

By density of D and the dominated convergence theorem, (1.5) is valid for y ∈ Y .Since Y is weak* dense in X∗ = Y ∗∗ and f is separably valued, it follows fromCorollary 1.1.3 that f : R+ → X is measurable. Since f is bounded, f is locallyBochner integrable and it follows from (1.5) that

F (t) =

∫ t

0

f(s) ds.

By Proposition 1.2.2 a), F is differentiable a.e.

Corollary 1.2.7. Every reflexive space has the Radon-Nikodym property.

Proof. Since a continuous function has separable range, it suffices to show thatevery separable reflexive space has the Radon-Nikodym property. This followsfrom Theorem 1.2.6.

Next we give some examples of spaces which do not have the Radon-Nikodymproperty.

Example 1.2.8. a) Let X = C[0, 1] and define F : [0, 1] → C[0, 1] by F (t)(s) :=(t− s)χ[0,t](s). For 0 ≤ t1 ≤ t2 ≤ 1,

F (t2)(s)− F (t1)(s) =

⎧⎪⎨⎪⎩t2 − t1 (0 ≤ s ≤ t1),

t2 − s (t1 < s ≤ t2),

0 (t2 < s ≤ 1).

Thus ‖F (t2) − F (t1)‖∞ = |t2 − t1|, so F is Lipschitz continuous. However,limh→0

1h (F (t + h) − F (t)) does not exist in the norm topology, so C[0, 1] does

not have the Radon-Nikodym property.

b) It follows from a) that L∞(0, 1) does not have the Radon-Nikodym propertyeither. However, when F is as in a) and L∞(0, 1) is identified with L1(0, 1)∗,the weak* derivative F ′w∗(t) := w*- limh→0

1h(F (t + h) − F (t)) = χ[0,t] exists in

L∞(0, 1). Note that F ′w∗ is not measurable and therefore not Bochner integrable(Example 1.1.5 a)). We shall see in Section 1.9 that F ′w∗ is Riemann integrable

and F (t) =∫ t

0F ′w∗(s) ds in the sense of Riemann integrals.

More generally, it follows from the next result that every Banach space con-taining a closed subspace isomorphic to c0 (see Example 1.1.5 b) and AppendixD) does not have the Radon-Nikodym property.

Proposition 1.2.9. The space c0 does not have the Radon-Nikodym property.

Proof. Let F (t) := (Fn(t))n∈N, where Fn(t) :=1n sin(nt) (n ∈ N). Then F : R+ →

c0 is Lipschitz continuous since

|Fn(t)− Fn(s)| =∣∣∣∣∫ t

s

cos(nr) dr

∣∣∣∣ ≤ |t− s| (t, s ≥ 0, n ∈ N).

1.3. CONVOLUTIONS 21

However, F is nowhere differentiable, since F ′n(t) = cos(nt) and (cos(nt))n∈N �∈ c0.Thus, c0 does not have the Radon-Nikodym property.

It follows from Theorem 1.2.6 that the space l1 = c∗0 has the Radon-Nikodymproperty. However, L1(0, 1) does not, even though c0 is not a closed subspace ofL1(0, 1) (see Appendix D).

Proposition 1.2.10. The space L1(0, 1) does not have the Radon-Nikodym property.

Proof. In fact, define F : [0, 1] → L1(0, 1) by F (t) = χ[0,t]. Then F is clearlyLipschitz continuous. Let 0 < t < 1. Then F is not differentiable at t. In fact,∥∥∥∥ 1h (F (t+ h)− F (t))− 1

2h(F (t+ 2h)− F (t))

∥∥∥∥1

= 1

for 0 < h < (1− t)/2.

1.3 Convolutions

For k, h ∈ L1(R), standard arguments with Fubini’s theorem (see [Rud87, Theorem7.14]) show that the convolution

(k ∗ h)(t) :=∫R

k(t− s)h(s) ds

exists a.e. and k∗h ∈ L1(R). Moreover, convolution is commutative and associative.In this section, we consider convolutions involving vector and operator-valued

functions. If k : R→ C and f : R→ X are measurable, we define the convolutionby

(k ∗ f)(t) =∫R

k(t− s)f(s) ds (1.6)

whenever this exists (as a Bochner integral). Since (k ∗ f)(t) = ∫Rk(s)f(t− s) ds,

we may write f ∗ k in place of k ∗ f .Proposition 1.3.1. Let k, h ∈ L1(R) and f ∈ L1(R, X). Then

a) (k ∗ f)(t) exists for almost all t ∈ R and k ∗ f ∈ L1(R, X).

b) h ∗ (k ∗ f) = (h ∗ k) ∗ f a.e.

Proof. These results may be deduced from the vector-valued version of Fubini’stheorem (Theorem 1.1.9) in the same way as in the scalar case. (Alternatively, theymay be deduced from their scalar cases, using Theorem 1.1.4 and the Hahn-Banachtheorem).

Many standard facts about convolutions of scalar-valued functions extend tothe vector-valued case. We summarize some of them here.

Proposition 1.3.2. a) (Young’s inequality) Let 1 ≤ p, q, r ≤ ∞ satisfy 1/p +1/q = 1 + 1/r. If k ∈ Lp(R) and f ∈ Lq(R, X), then k ∗ f ∈ Lr(R, X) and

‖k ∗ f‖r ≤ ‖k‖p‖f‖q.

b) Let 1 < p, p′ < ∞ satisfy 1/p + 1/p′ = 1. If k ∈ Lp(R) and f ∈ Lp′(R, X),then k ∗ f ∈ C0(R, X).

c) If k ∈ L1(R) and f ∈ L∞(R, X), or if k ∈ L∞(R) and f ∈ L1(R, X), thenk ∗ f ∈ BUC(R, X).

d) If k ∈ L1(R) and f ∈ C0(R, X), or if k ∈ C0(R, X) and f ∈ L1(R), thenk ∗ f ∈ C0(R, X).

Proof. Under the assumptions of a), the scalar-valued version of Young’s inequality[RS72, Section IX.4] shows that (|k| ∗ ‖f‖)(t) exists a.e., and |k| ∗ ‖f‖ ∈ Lr(R). If(|k| ∗ ‖f‖)(t) exists, then (k ∗ f)(t) exists and

‖(k ∗ f)(t)‖ ≤ (|k| ∗ ‖f‖)(t).

Moreover, k ∗ f is weakly measurable by the scalar-valued theory, and hence mea-surable by Pettis’s theorem (Theorem 1.1.1). Thus, k ∗ f ∈ Lr(R, X).

The proofs of the remaining parts are similar to the scalar-valued case [Rud62,Theorem 1.1.6]. Suppose that 1 ≤ p <∞ and 1/p+ 1/p′ = 1. For t, h ∈ R,

‖(k ∗ f)(t+ h)− (k ∗ f)(t)‖ ≤∥∥∥∥∫

R

(k(t+ h− s)− k(t− s)) f(s) ds

∥∥∥∥≤

(∫R

|k(s+ h)− k(s)|p ds)1/p

‖f‖p′→ 0

as h → 0. This shows that k ∗ f is uniformly continuous. When k ∈ L∞(R) andf ∈ L1(R, X), uniform continuity is established in a similar way with the rolesof k and f reversed. Boundedness follows from a). This proves c). For b) and d),observe that k ∗ f has compact support if both k and f have compact support,and then the results follow by density arguments.

When k and f are defined on R+, or on [0, τ ] where τ > 0, then k ∗ f maybe defined by (1.6) by taking k(t) and f(t) to be zero for other values of t. Then

(k ∗ f)(t) =∫ t

0

k(t− s)f(s) ds (t ≥ 0).

It is immediate that Propositions 1.3.1 and 1.3.2 remain valid in these contexts.Note that if k ∈ L1

loc(R+) and f ∈ L1loc(R+, X), then k ∗ f ∈ L1

loc(R+, X).

As a tool, we shall need the notion of regularization from harmonic analysis.A mollifier is a sequence (ρn)n∈N in L1(R) of the following form. The function ρ1 ∈L1(R) satisfies

∫Rρ1(t) dt = 1, and ρn ∈ L1(R) is given by ρn(t) = nρ1(nt) (t ∈

R, n ∈ N). The next lemma shows that any mollifier acts as an approximate uniton various function spaces.

Lemma 1.3.3. Let (ρn)n∈N be a mollifier.

a) Let f ∈ BUC(R, X). Then limn→∞ ‖ f ∗ ρn − f‖∞ = 0.

b) Let f ∈ L1(R, X). Then limn→∞ ‖ f ∗ ρn − f‖1 = 0.

Proof. a) Let ε > 0. There exists c > 0 such that 2‖ f ‖∞∫|s|≥c

|ρ1(s)| ds ≤ ε. Then

for t ∈ R, n ∈ N,

‖(f ∗ ρn)(t)− f(t)‖ =

∥∥∥∥ ∫R

(f(t− s)− f(t))ρn(s) ds

∥∥∥∥=

∥∥∥∥ ∫R

(f(t− s

n)− f(t))ρ1(s) ds

∥∥∥∥≤

∫|s|≤c

∥∥∥ f (t− s

n

)− f(t)

∥∥∥ |ρ1(s)| ds+

∫|s|>c

∥∥∥ f (t− s

n

)− f(t)

∥∥∥ |ρ1(s)| ds≤ sup

|h|≤c/n

‖ f (t− h)− f(t) ‖ ‖ρ1‖1 + ε

≤ 2ε

for all sufficiently large n, since f is uniformly continuous.b) Let f ∈ L1(R, X). Then

‖ f ∗ ρn − f ‖1 =

∫R

∥∥∥∥ ∫ (f(t− s)− f(t))ρn(s) ds

∥∥∥∥ dt

=

∫R

∥∥∥∥ ∫R

(f(t− s

n

)− f(t))ρ1(s) ds

∥∥∥∥ dt

≤∫R

∫R

∥∥∥ f (t− s

n

)− f(t)

∥∥∥ dt |ρ1(s)| ds.

As observed in Section 1.1, limn→∞∫R‖ f(t− s

n )− f(t) ‖ dt = 0 for all s ∈ R. Nowthe claim follows from the dominated convergence theorem.

Lemma 1.3.3 is also valid for the spaces Lp(R, X) (1 < p <∞) (see Remark1.3.8 b)). The notion of mollifier can be extended to a family {ρε : 0 < ε ≤ 1}where ρε(t) = ε−1ρ(t/ε), and Lemma 1.3.3 remains valid in that case.

The theory of vector-valued convolutions and mollifiers extends, with almostno changes, to the case of functions on Rn for n ≥ 1.

Now we move on to consider convolutions of vector-valued functions withoperator-valued functions.

The space of all bounded linear operators from a Banach space X into a Ba-nach space Y is denoted by L(X,Y ), or simply by L(X) when Y = X. A functionT : R+ → L(X,Y ) is strongly continuous if t �→ T (t)x is continuous for all x ∈ X.By the uniform boundedness principle, a strongly continuous function T is locallybounded. Note also that ‖T‖ is lower semi-continuous and hence measurable.

We state the convolution results for strongly continuous functions T : R+ →L(X,Y ), but they are also valid for T : (0,∞) → L(X, Y ) if T is strongly con-tinuous on (0,∞) and bounded on (0, 1). There are similar results for compactintervals [0, τ ].

Proposition 1.3.4. Let f ∈ L1loc(R+, X) and let T : R+ → L(X,Y ) be strongly

continuous. Then the convolution

(T ∗ f)(t) :=∫ t

0

T (t− s)f(s) ds

exists (as a Bochner integral) and defines a continuous function T ∗ f : R+ → Y .

Proof. Fix t ≥ 0. First, we show that s �→ T (t − s)f(s) is measurable on [0, t].When f(s) = χΩ(s)x for some measurable Ω ⊂ R+ and x ∈ X, then

T (t− s)f(s) = χΩ(s) · T (t− s)x.

This is measurable, being the product of a measurable scalar-valued function anda continuous vector-valued function. By linearity, T (t− ·)f(·) is measurable whenf is a simple function. For measurable f , there is a sequence of simple functionsfn → f a.e. and then T (t − s)fn(s) → T (t − s)f(s) s-a.e., so T (t − ·)f(·) ismeasurable.

Since‖T (t− s)f(s)‖ ≤ ‖T (t− s)‖ ‖f(s)‖

it follows from Theorem 1.1.4 that (T ∗ f)(t) exists. Continuity of T ∗ f followsfrom the dominated convergence theorem (Theorem 1.1.8).

Now we state the analogue of Proposition 1.3.2 for operator-valued functionson R+.

Proposition 1.3.5. Let f ∈ L1loc(R+, X) and T : R+ → L(X,Y ) be strongly contin-

uous.

a) (Young’s inequality) Let 1 ≤ p, q, r ≤ ∞ satisfy 1/p + 1/q = 1 + 1/r. If∫∞0‖T (t)‖p dt <∞ and f ∈ Lq(R+, X), then T ∗ f ∈ Lr(R+, Y ) and

‖T ∗ f‖r ≤ ‖f‖q(∫ ∞

0

‖T (t)‖p dt)1/p

.

b) Let 1 < p, p′ < ∞ satisfy 1/p + 1/p′ = 1. If∫∞0‖T (t)‖p dt < ∞ and f ∈

Lp′(R+, X), then T ∗ f ∈ C0(R+, Y ).

c) If∫∞0‖T (t)‖ dt < ∞ and f ∈ BUC(R+, X), or if T is bounded and f ∈

L1(R+, X), then T ∗ f ∈ BUC(R+, Y ).

d) If∫∞0‖T (t)‖ dt < ∞ and f ∈ C0(R+, X), or if limt→∞ ‖T (t)‖ = 0 and

f ∈ L1(R+, X), then T ∗ f ∈ C0(R+, Y ).

Proof. The proofs are similar to Proposition 1.3.2, with the exception of the uni-form continuity of T ∗ f when ‖T‖ is integrable and f is bounded and uniformlycontinuous. Then, for 0 ≤ t ≤ t+ h,

‖(T ∗ f)(t+ h)− (T ∗ f)(t)‖

≤∥∥∥∥∥∫ t+h

t

T (s)f(t+ h− s) ds

∥∥∥∥∥+

∥∥∥∥∫ t

0

T (s)(f(t+ h− s)− f(t− s)) ds

∥∥∥∥≤

(∫ t+h

t

‖T (s)‖ ds)‖f‖∞ +

(∫ ∞

0

‖T (s)‖ ds)sups≥0

‖f(s+ h)− f(s)‖

→ 0

uniformly in t as h→ 0.

If f or T is more regular, then T ∗ f is continuously differentiable. We givetwo such results.

Proposition 1.3.6. Let T : R+ → L(X,Y ) be strongly continuous and bounded,

x ∈ X, f ′ ∈ L1loc(R+, X), f(t) = x+

∫ t

0f ′(s) ds (t ≥ 0). Then T ∗ f ∈ C1(R+, Y )

and(T ∗ f)′(t) = (T ∗ f ′)(t) + T (t)x.

Proof. Let u(t) := (T ∗ f ′)(t) + T (t)x. Then by Proposition 1.3.4, u ∈ C(R+, Y ).By Fubini’s theorem we have∫ t

0

u(s) ds =

∫ t

0

∫ s

0

T (r)f ′(s− r) dr ds+

∫ t

0

T (s)x ds

=

∫ t

0

∫ t

r

T (r)f ′(s− r) ds dr +

∫ t

0

T (s)x ds

=

∫ t

0

∫ t−r

0

T (r)f ′(s) ds dr +∫ t

0

T (s)x ds

=

∫ t

0

T (r)(f(t− r)− x) dr +

∫ t

0

T (s)x ds

= (T ∗ f)(t) (t ≥ 0).

By Proposition 1.2.2 a), this proves the claim.

Proposition 1.3.7. Let T : [0, τ ]→ L(X, Y ) be Lipschitz continuous with T (0) = 0,and let f ∈ L1([0, τ ], X). Then T ∗ f ∈ C1([0, τ ], Y ).

Proof. First, suppose that f ∈ C1([0, τ ], X). By Proposition 1.3.6, T ∗ f has aderivative g ∈ C([0, τ ], Y ). For 0 ≤ r ≤ t ≤ τ ,∥∥∥∥∫ t

r

g(s) ds

∥∥∥∥ = ‖(T ∗ f)(t)− (T ∗ f)(r)‖

≤∫ t

r

‖T (t− s)f(s)‖ ds+∫ r

0

‖(T (t− s)− T (r − s))f(s)‖ ds

≤∫ t

r

L (t− s)‖f(s)‖ ds+∫ r

0

L (t− r)‖f(s)‖ ds≤ L (t− r)‖f‖1,

where L is a Lipschitz constant for T , so that ‖T (t2) − T (t1)‖ ≤ L|t2 − t1| and,in particular, ‖T (s)‖ ≤ Ls since T (0) = 0. It follows that ‖g(s)‖ ≤ L‖f‖1 for alls ∈ [0, τ ].

Now, consider f ∈ L1([0, τ ], X). There is a sequence (fn) in C1([0, τ ], X) suchthat ‖fn − f‖1 → 0. By Proposition 1.3.5, ‖(T ∗ (fn − f))(t)‖ → 0. By the firstparagraph,

‖(T ∗ fn)′ − (T ∗ fm)′‖∞ ≤ L ‖fn − fm‖1→ 0,

so ((T ∗ fn)′) converges uniformly to a function g ∈ C([0, τ ], Y ). Since

(T ∗ f)(t) = limn→∞(T ∗ fn)(t) = lim

n→∞

∫ t

0

(T ∗ fn)′(s) ds =∫ t

0

g(s) ds,

it follows that T ∗ f ∈ C1([0, τ ], Y ).

Remark 1.3.8. a) There is an analogous theory of operator-valued convolutions onR. One may define

(T ∗ f)(t) :=∫R

T (t− s)f(s) ds

for almost all t, if T : R→ L(X, Y ) is strongly continuous,∫R‖T (t)‖p dt <∞ and

f ∈ Lq(R, X) where 1/p+ 1/q ≥ 1, and Proposition 1.3.5 is valid in this case.

b) Convolutions of scalar, vector or operator-valued functions can sometimes beconsidered as Bochner integrals with values in a function space. Suppose that0 < τ ≤ ∞, k ∈ L1(0, τ ) and f ∈ Lp((0, τ), X) where 1 ≤ p < ∞. DefineG : (0, τ)→ Lp((0, τ), X) by

G(s)(t) :=

{f(t− s) (0 < s < t < τ)

0 otherwise.

1.4. EXISTENCE OF THE LAPLACE INTEGRAL 27

Then G is continuous and ‖G(s)‖p ≤ ‖f‖p. We can therefore form the Bochnerintegral

∫ τ

0k(s)G(s) ds in L1((0, τ ), X). Then∫ τ

0

k(s)G(s) ds = k ∗ f a.e. in (0, τ).

This can be proved by first considering step functions, or by considering integralsof the form

∫ τ

0〈(k ∗ f)(t), x∗〉g(t) dt for arbitrary x∗ ∈ X∗ and g ∈ Lp′(0, τ). Thus,

‖k ∗ f‖p ≤ ‖k‖1‖f‖p, a special case of Young’s inequality.This approach can also be used for convolutions on R, when the spaces

Lp(R, X) can be replaced by BUC(R, X) or C0(R, X). For example, this leadsto a very short proof of Lemma 1.3.3 for all these spaces.

The same idea can be used in the operator-valued case. Suppose that 0 <τ ≤ ∞, T : (0, τ) → L(X, Y ) is strongly continuous and f ∈ Lp((0, τ), X), where1 ≤ p <∞. Let

H(s)(t) :=

{T (s)f(t− s) (0 < s < t < τ ),

0 otherwise.

ThenH(s) ∈ Lp((0, τ), Y ), ‖H(s)‖p ≤ ‖T (s)‖ ‖f‖p, andH : (0, τ)→ Lp((0, τ), Y )is continuous. If

∫ τ

0‖T (s)‖ ds <∞, one may form the Bochner integral

∫ τ

0H(s) ds

in Lp((0, τ), Y ) and it coincides a.e. with T ∗ f .

1.4 Existence of the Laplace Integral

Let X be a complex Banach space and L1loc(R+, X) := {f : R+ → X : f is Bochner

integrable on [0, τ ] for all τ > 0}. This section is concerned with the existence ofthe Laplace integral

f(λ) :=

∫ ∞

0

e−λtf(t) dt := limτ→∞

∫ τ

0

e−λtf(t) dt

for f ∈ L1loc(R+, X) and λ ∈ C. Note that

∫ τ

0e−λtf(t) dt exists as a Bochner

integral, and if∫∞0

e−λtf(t) dt exists as a Bochner integral then it agrees with thedefinition above, by the dominated convergence theorem.

Of special interest will be the abscissa of convergence of f , given by

abs(f) := inf{Re λ : f(λ) exists}.

It will be shown that the set of those λ ∈ C for which the Laplace integral convergesis either empty or a right half-plane whose left boundary point abs(f) coincideswith the exponential growth bound of the antiderivative

t �→ F (t)− F∞ (t ≥ 0),

where F (t) :=∫ t

0f(s) ds, F∞ := limt→∞ F (t) if the limit exists, and F∞ := 0

otherwise. Therefore, one particular result of this section will be that a locallyBochner integrable function f is Laplace transformable if and only if its normalizedantiderivative F : t→ ∫ t

0f(s) ds is exponentially bounded.

Proposition 1.4.1. Let f ∈ L1loc(R+, X). Then the Laplace integral f(λ) converges

if Re λ > abs(f) and diverges if Reλ < abs(f).

Proof. Clearly, f(λ) does not exist if Reλ < abs(f). For λ0 ∈ C define G0(t) :=∫ t

0e−λ0sf(s) ds (t ≥ 0). Then, for all λ ∈ C and t ≥ 0, integration by parts gives∫ t

0

e−λsf(s) ds =

∫ t

0

e−(λ−λ0)se−λ0sf(s) ds

= e−(λ−λ0)tG0(t) + (λ− λ0)

∫ t

0

e−(λ−λ0)sG0(s) ds. (1.7)

If f(λ0) exists, then G0 is bounded. Moreover, it follows from (1.7) that f(λ) existsif Reλ > Reλ0 and

f(λ) = (λ− λ0)

∫ ∞

0

e−(λ−λ0)sG0(s) ds (Reλ > Reλ0). (1.8)

This shows that f(λ) exists if Reλ > abs(f).

If f(λ) converges for all λ ∈ C, then abs(f) := −∞. If the domain of conver-gence is empty, then abs(f) :=∞. A function f is called Laplace transformable ifabs(f) <∞.

It follows from Proposition 1.4.1 that the interior of the domain of con-vergence of f(λ) is the open right half-plane {Reλ > abs(f)}. As the followingexample shows, the domain of convergence may or may not include points on theboundary abs(f) + iR.

Example 1.4.2. Let f(t) := (1 + t)−1. Then abs(f) = 0 since the Laplace integral

f(λ) converges for λ > 0 but not for λ = 0. If λ = ir (r �= 0), then integration by

parts implies that f(ir) converges and

f(ir) =1

ir− 1

ir

∫ ∞

0

e−irt

(1 + t)2dt.

Thus, the domain of convergence of f(λ) is {Reλ ≥ 0, λ �= 0}. For f(t) := 1,

the domain of convergence of f(λ) is the open half-plane {Re λ > 0}; for f(t) :=(1 + t2)−1 it is the closed half-plane {Reλ ≥ 0}.

From the proof of Proposition 1.4.1 and the uniform boundedness principleone also obtains the remarkable result that the abscissa of convergence of f(λ) is

given by

abs(f) = inf

{λ ∈ R : sup

t>0

∣∣∣∣∫ t

0

e−λs〈f(s), x∗〉 ds∣∣∣∣ <∞ for all x∗ ∈ X∗

}. (1.9)

To see this we denote the right-hand side of (1.9) by absw(f). Clearly, absw(f) ≤abs(f). Assume that absw(f) < abs(f). Then there exists λ0 such that absw(f) <λ0 < abs(f) and supt>0 |〈G0(t), x

∗〉| < ∞ (x∗ ∈ X∗), where G0 is as in theproof of Proposition 1.4.1. It follows from the uniform boundedness principle thatG0 is bounded. Thus, by (1.7), f(λ) exists for all λ > λ0. Since this contradictsλ0 < abs(f) one obtains that absw(f) = abs(f).

Next abs(f) will be described by the exponential growth of f and its an-tiderivatives. For f : R+ → X the exponential growth bound is given by

ω(f) := inf

{ω ∈ R : sup

t≥0‖e−ωtf(t)‖ <∞

}.

It is obvious thatabs(f) ≤ abs(‖f‖) ≤ ω(f). (1.10)

It will be shown in Example 1.4.4 below that there are cases in which one has strictinequalities in (1.10). In fact, it is possible that abs(f) = −∞ and ω(f) = ∞.It will be shown next that abs(f) is determined by the exponential growth of

the antiderivative t �→ F (t) − F∞ (t ≥ 0), where F (t) :=∫ t

0f(s) ds, F∞ :=

limt→∞ F (t) if the limit exists, and F∞ := 0 otherwise.

Theorem 1.4.3. Let f ∈ L1loc(R+, X). Then abs(f) = ω(F − F∞).

Proof. Suppose that abs(f) <∞. For λ0 > abs(f) define G0(t) :=∫ t

0e−λ0sf(s) ds

(t ≥ 0). Then G0 is continuous and convergent as t → ∞, so G0 is bounded. Toprove that abs(f) ≥ ω(F −F∞) one considers the three cases abs(f) > 0, abs(f) =0, and abs(f) < 0. First, let abs(f) > 0. Then F∞ = 0 and, for λ0 > abs(f),integration by parts gives

F (t) =

∫ t

0

f(s) ds =

∫ t

0

eλ0se−λ0sf(s) ds = eλ0tG0(t) − λ0

∫ t

0

eλ0sG0(s) ds.

It follows that ‖F (t)‖ ≤ Ceλ0t + C(eλ0t − 1) ≤ 2Ceλ0t for t ≥ 0, where C :=sups≥0 ‖G0(s)‖. This shows that abs(f) ≥ ω(F − F∞) if abs(f) > 0. Second,let abs(f) = 0. If F∞ = 0, then the same procedure as above yields abs(f) ≥ω(F − F∞). If abs(f) = 0 and limt→∞ F (t) = F∞ exists, then it follows from thecontinuity of F that supt≥0 ‖F (t) − F∞‖ < ∞. Thus, abs(f) = 0 ≥ ω(F − F∞).This shows that abs(f) ≥ ω(F −F∞) if abs(f) = 0. Third, let abs(f) < 0. Chooseabs(f) < λ0 < 0. For r ≥ t ≥ 0 one has

F (r)− F (t) =

∫ r

t

eλ0se−λ0sf(s) ds

= eλ0rG0(r)− eλ0tG0(t) − λ0

∫ r

t

eλ0sG0(s) ds.

Since G0 is bounded and limr→∞ F (r) = F∞, it follows that

‖F∞ − F (t)‖ =∥∥∥∥eλ0tG0(t) + λ0

∫ ∞

t

e−|λ0|sG0(s) ds

∥∥∥∥ ≤ 2Ceλ0t

for t ≥ 0. This proves that abs(f) ≥ ω(F − F∞).To show the reverse inequality, suppose that ω(F − F∞) < ∞ and let ω >

ω(F − F∞). Since F is continuous, there exists M ≥ 0 such that ‖F (t) − F∞‖ ≤Meωt for all t ≥ 0. Let λ > ω > ω(F − F∞). Using the fact that F − F∞ is anantiderivative of f , integration by parts yields∫ t

0

e−λsf(s) ds = e−λt(F (t)− F∞) + F∞ + λ

∫ t

0

e−λs(F (s)− F∞) ds.

Hence, f(λ) exists for λ > ω(F −F∞) and is given by f(λ) = F∞+λ (F − F∞)(λ).

This shows that abs(f) ≤ ω(F − F∞). Since λF∞(λ) = λ∫∞0

e−λtF∞ dt = F∞ ifReλ > 0,

f(λ) = λF (λ) if Reλ > max{abs(f), 0}. (1.11)

If ω ≥ 0, then the triangle inequality implies that ω(F ) ≤ ω if and only ifω(F − F∞) ≤ ω. Thus, a locally Bochner integrable function f is Laplace trans-

formable if and only if its antiderivative F (t) =∫ t

0f(s) ds is exponentially bounded;

and,abs(f) ≤ ω ⇐⇒ ω(F ) ≤ ω (if ω ≥ 0). (1.12)

The following is a first example of a function which is Laplace transformable butnot exponentially bounded; i.e., abs(f) < ω(f) =∞.

Example 1.4.4. For t ≥ 0 let f(t) := eteet

cos(eet

). Then ω(f) =∞. Since

F (t) =

∫ t

0

f(s) ds =

∫ t

0

esees

cos (ees

) ds =

∫ eet

e

cos(u) du = sin eet − sin(e),

it follows that F∞ = 0. Thus, by Theorem 1.4.3, abs(f) = ω(F ) = 0.

Finally, the results of this section will be formulated for strongly continuousoperator-valued functions T : R+ → L(X,Y ). We define the exponential growthbound of T by

ω(T ) := ω(‖T‖) = inf

{ω ∈ R : sup

t≥0‖e−ωtT (t)‖ <∞

}.

By the uniform boundedness principle,

ω(T ) = sup{ω(ux) : x ∈ X},

where ux(t) := T (t)x.

If T : R+ → L(X,Y ) is strongly continuous, and λ ∈ C, then∫ t

0e−λsT (s) ds

denotes the bounded operator x �→ ∫ t

0e−λsT (s)x ds, and we define

abs(T ) := inf

{Re λ :

∫ t

0

e−λsT (s) ds converges strongly as t→∞}

= sup{abs(ux) : x ∈ X}.Here and in what follows, to say S : R+ → L(X,Y ) converges strongly as t → ∞refers to the strong operator topology; i.e., it means that limt→∞ S(t)x exists inY for all x ∈ X .

Proposition 1.4.5. Let T : R+ → L(X,Y ) be strongly continuous, let S(t) =∫ t

0T (s) ds, and S∞ be the strong limit of S(t) as t→∞ if it exists, and S∞ := 0

otherwise. Then

a) limt→∞∫ t

0e−λsT (s) ds exists in operator norm whenever Reλ > abs(T ).

b) abs(T ) = inf

{λ ∈ R : supt>0

∣∣∣∫ t

0e−λs〈T (s)x, y∗〉 ds

∣∣∣ <∞for all x ∈ X and y∗ ∈ Y ∗

}.

c) abs(T ) = ω(S − S∞).

Proof. If∫ t

0e−λ0sT (s) ds converges strongly, then it is uniformly bounded in oper-

ator norm. Thus, a) follows from (1.7), while b) is immediate from (1.9).To prove c), it is possible to repeat the proof of Theorem 1.4.3. Alternatively,

one may deduce c) from Theorem 1.4.3 as follows. Let ux(t) := T (t)x, vx(t) :=S(t)x− S∞x, and

vx(t) :=

{S(t)x− lim

s→∞S(s)x if the limit exists,

S(t)x otherwise.

By Theorem 1.4.3, abs(ux) = ω(vx). Moreover, ω(vx) and ω(vx) coincide if eitherof them is strictly positive, since vx − vx is a constant function.

First suppose that abs(T ) < 0. Then ω(vx) = abs(ux) ≤ abs(T ) < 0, solimt→∞ vx(t) exists for all x ∈ X . Thus, S(t) converges strongly, so vx = vx and

abs(T ) = sup{abs(ux) : x ∈ X}= sup{ω(vx) : x ∈ X}= sup{ω(vx) : x ∈ X}= ω(S − S∞).

Next, suppose that ω(S − S∞) < 0. Then again, S(t) converges strongly and, asabove, abs(T ) = ω(S − S∞).

Next, suppose that abs(T ) > 0. Then

abs(T ) = sup{abs(ux) : x ∈ X, abs(ux) > 0}= sup{ω(vx) : x ∈ X, ω(vx) > 0}= sup{ω(vx) : x ∈ X, ω(vx) > 0}= ω(S − S∞).

Finally, suppose that ω(S − S∞) > 0. Then the same argument as in theprevious paragraph can be applied to show that abs(T ) = ω(S − S∞).

Remark 1.4.6. If S∞ is the norm-limit of S(t) as t → ∞ if this exists and S∞ ∈L(X,Y ) is arbitrary otherwise, and S∞ is as in Proposition 1.4.5, then it is trivial

that ω(S − S∞) = ω(S − S∞).

If T : R+ → L(X,Y ) is strongly continuous and abs(T ) < ∞, we define theLaplace integral of T by

T (λ) :=

∫ ∞

0

e−λsT (s) ds := limt→∞

∫ t

0

e−λsT (s) ds (Reλ > abs(T )),

where the right-hand integral is interpreted as above and the limit exists in oper-ator norm (Proposition 1.4.5).

1.5 Analytic Behaviour

Let f ∈ L1loc(R+, X). In this section it will be shown that λ �→ f(λ) is holomorphic

for Reλ > abs(f). In general, f need not have a singularity on the boundaryabs(f)+ iR of its domain of convergence and may be extended holomorphically toa strictly larger half-plane. However, it will be shown that abs(f) is a singularity

of f if X is an ordered Banach space with a normal positive cone and f(t) ≥ 0a.e. (see Appendix C). We put N0 := N ∪ {0}.Theorem 1.5.1. Let f ∈ L1

loc(R+, X) with abs(f) < ∞. Then λ �→ f(λ) is holo-morphic for Reλ > abs(f) and, for all n ∈ N0 and Re λ > abs(f),

f (n)(λ) =

∫ ∞

0

e−λt(−t)nf(t) dt (1.13)

(as an improper Bochner integral).

Proof. Define qk : C→ X (k ∈ N0) by

qk(λ) :=

∫ k

0

e−λtf(t) dt = limN→∞

N∑n=0

λn

n!

∫ k

0

(−t)nf(t) dt.

1.5. ANALYTIC BEHAVIOUR 33

The limits exist uniformly for λ in bounded subsets of C. By the Weierstrassconvergence theorem (a simple special case of Vitali’s Theorem A.5), the functions

qk are entire and q(j)k (λ) =

∫ k

0e−λt(−t)jf(t) dt for all j ∈ N0.

Let Reλ > λ0 > abs(f) and define G0(t) :=∫ t

0e−λ0sf(s) ds. By Proposition

1.4.1, G0 is bounded and integration by parts gives

f(λ)− qk(λ) =

∫ ∞

k

e−(λ−λ0)se−λ0sf(s) ds

= −e−(λ−λ0)kG0(k) + (λ− λ0)

∫ ∞

k

e−(λ−λ0)sG0(s) ds.

It follows that qk converges to f uniformly on compact subsets of {λ : Reλ >

abs(f)}. Again by the Weierstrass convergence theorem, f is holomorphic and

q(j)k (λ)→ f (j)(λ) as k →∞, for Reλ > abs(f).

If abs(f) <∞, then the abscissa of holomorphy of f is denoted by

hol(f) := inf{ω ∈ R : f extends holomorphically for Reλ > ω}.

By Theorem 1.5.1,

hol(f) ≤ abs(f). (1.14)

In general, equality does not hold in (1.14) (see Example 1.5.2 below). However weshall see that equality does hold for positive functions on ordered spaces (Theorem1.5.3) and for exponentially bounded functions which extend holomorphically intoa sector {| arg(λ)| < α} for some 0 < α < π

2 (see Section 2.6). Furthermore, inChapter 4 (Theorem 4.4.19) we shall see that

abs(f) ≤ hol0(f)

whenever f is exponentially bounded, where

hol0(f) := inf{ω ∈ R : f has a bounded holomorphic extension for Reλ > ω}

is the abscissa of boundedness of f . The following example shows that it mayhappen that hol(f) < abs(f) < ω(f).

Example 1.5.2. Let f(t) = et sin et (t ≥ 0). Obviously, ω(f) = 1. Since F (t) =∫ t

0f(s) ds = cos 1 − cos et, one obtains from Theorem 1.4.3 that abs(f) = 0. It

follows from

∫ t

0

e−λsf(s) ds =

∫ t

0

e−λs d

ds(− cos es) ds

= −e−λt cos et + cos 1−∫ t

0

λe−λs cos es ds

= −e−λt cos et + cos 1− λ

∫ t

0

e−(λ+1)s d

ds(sin es) ds

= −e−λt cos et + cos 1− λe−(λ+1)t sin et

+λ sin 1− λ(λ+ 1)

∫ t

0

e−(λ+2)ses sin es ds

thatf(λ) = cos 1 + λ sin 1− λ(λ+ 1)f(λ+ 2)

if Reλ > 0. Thus, f has a holomorphic extension to Reλ > −2, then to Reλ > −4,etc. This shows that hol(f) = −∞.

Another remarkable property of this function is that abs(f) < abs(|f |).Clearly,

∫∞0

e−λt|f(t)|dt =∫∞1

u−λ| sinu| du converges for λ > 1. Since∫ nπ

1

1

u| sinu| du ≥

√2

2

n−1∑j=1

∫ (4j+3)π/4

(4j+1)π/4

1

udu ≥

√2

2

n−1∑j=1

2π

4j + 3,

it follows that abs(|f |) = 1.

Theorem 1.5.3. Let X be an ordered Banach space with a normal positive cone.Assume that abs(f) <∞ and f(t) ≥ 0 a.e. Then hol(f) = abs(f). If, in addition,

abs(f) > −∞, then f has a singularity at abs(f).

Proof. If abs(f) = −∞, then the statement follows from (1.14). If abs(f) > −∞,then one may assume that abs(f) = 0. Otherwise replace f by t �→ e−ωtf(t), where

ω := abs(f). By (1.14), λ �→ f(λ) is holomorphic for Reλ > 0. Assume that f isholomorphic at λ = 0. Then there exists a full circle with centre at 1 and radius1 + 2λ0 for some λ0 > 0 such that f is holomorphic on it (the extension of f topoints in the complex plane with Reλ ≤ 0 is denoted by the same symbol). Hence,

f(−λ0) =∞∑k=0

(−1)k(1 + λ0)k 1

k!f (k)(1).

Next it will be shown that∫∞0

eλ0t〈f(t), x∗〉 dt converges for all x∗ ∈ X∗, which isa contradiction to (1.9). Let g(t) := 〈f(t), x∗〉. Since X∗+ is generating (PropositionC.2), one can assume that x∗ ≥ 0 and thus g(t) ≥ 0. Then, by Theorem 1.5.1 andthe monotone convergence theorem,

g(−λ0) =∞∑k=0

(1 + λ0)k 1

k!

∫ ∞

0

e−ttkg(t) dt

=

∫ ∞

0

e−te(1+λ0)tg(t) dt =

∫ ∞

0

eλ0tg(t) dt.

1.5. ANALYTIC BEHAVIOUR 35

The following example shows that there are positive functions with hol(f) <ω(f).

Example 1.5.4. Let Ω :=⋃

n∈N[n, n + e−n3

) and f(t) := et2

χΩ(t). Then f(t) ≥ 0

and ω(f) = ∞. However,∫∞0

e−λsf(s) ds =∑∞

n=0

∫ n+e−n3

ne−λses

2

ds converges

for all λ ∈ R. Hence, hol(f) = abs(f) = −∞.

Finally in this section, we consider operator-valued functions. Let T : R+ →L(X,Y ) be strongly continuous, where X and Y are arbitrary Banach spaces, andassume that abs(T ) <∞.

As in Section 1.4, let ux(t) := T (t)x (x ∈ X, t ≥ 0), and

T (λ) := limt→∞

∫ t

0

e−λsT (s) ds (Reλ > abs(T )),

where the integral is interpreted as in Section 1.4 and the limit exists in operatornorm (Proposition 1.4.5).

Since T (λ)x = ux(λ), it follows from Theorem 1.5.1, (1.8) and PropositionA.3 that T : {Reλ > abs(T )} → L(X,Y ) is holomorphic. We define

hol(T ) := inf{ω ∈ R : T extends to a holomorphic function from

{Reλ > ω} into L(X,Y )}.

Proposition 1.5.5. Let T : R+ → L(X,Y ) be strongly continuous with abs(T ) <∞.Then

hol(T ) = sup {hol(ux) : x ∈ X} .Proof. Since T (λ)x = ux(λ) for Reλ > abs(T ), it is clear that hol(T ) ≥ hol(ux)for all x ∈ X. We have to show that if ω < abs(T ) and each ux extends holomor-phically to {Reλ > ω} and T (λ)x = ux(λ) (Reλ > ω), then T is a holomorphicfunction of {Re λ > ω} into L(X,Y ).

Take λ0 with Reλ0 > abs(T ). For |λ− λ0| < abs(T )− ω,

T (λ)x =

∞∑n=0

ux(n)

(λ0)

n!(λ− λ0)

n.

Moreover, for 0 < r < abs(T )− ω,

sup

{∥∥∥∥∥m∑

n=0

ux(n)

(λ0)

n!(λ− λ0)

n

∥∥∥∥∥ : m ∈ N, |λ− λ0| < r

}<∞

for each x ∈ X. By the uniform boundedness principle, T (λ) ∈ L(X,Y ) and T isbounded on {|λ−λ0| < r}. It follows from Proposition A.3 that T : {Reλ > ω} →L(X,Y ) is holomorphic.

1.6 Operational Properties

The importance of Laplace integrals in applications to differential equations lies inthe fact that they transform the analytic operations of differentiation, integrationand convolution into algebraic operations of multiplication. In this section, weestablish these and other basic properties of Laplace transforms.

Proposition 1.6.1. Let f ∈ L1loc(R+, X), μ ∈ C and s ∈ R+. Let

g(t) := e−μtf(t) (t ≥ 0),

fs(t) := f(s+ t) (t ≥ 0),

hs(t) :=

{f(t− s) (t ≥ s),

0 (0 ≤ t < s).

Let λ ∈ C. Then

a) g(λ) exists if and only if f(λ+ μ) exists, and then g(λ) = f(λ+ μ).

b) fs(λ) exists if and only if f(λ) exists, and then

fs(λ) = eλs(f(λ)−

∫ s

0

e−λtf(t) dt

).

c) hs(λ) exists if and only if f(λ) exists, and then hs(λ) = e−λsf(λ).

Proof. These results follow immediately from the formulae:∫ τ

0

e−λtg(t) dt =

∫ τ

0

e−(λ+μ)tf(t) dt,∫ τ

0

e−λtfs(t) dt = eλs(∫ s+τ

0

e−λtf(t) dt−∫ s

0

e−λtf(t) dt

),∫ τ

0

e−λths(t) dt = e−λs

∫ τ−s

0

e−λtf(t) dt (τ > s).

Proposition 1.6.2. Let f ∈ L1loc(R+, X) and T ∈ L(X,Y ), and let (T ◦ f)(t) =

T (f(t)). Then T ◦ f ∈ L1loc(R+, Y ). If f(λ) exists, then T ◦ f(λ) exists and equals

T (f(λ)).

Proof. By Proposition 1.1.6, T ◦ f ∈ L1loc(R+, Y ) and∫ τ

0

e−λt(T ◦ f)(t) dt = T

∫ τ

0

e−λtf(t) dt.

The second statement follows on letting τ →∞.

1.6. OPERATIONAL PROPERTIES 37

Proposition 1.6.3. Let f ∈ L1loc(R+, X) and A be a closed operator on X. Suppose

that f(t) ∈ D(A) a.e. and A ◦ f ∈ L1loc(R+, X). Let λ ∈ C. If f(λ) and A ◦ f(λ)

both exist, then f(λ) ∈ D(A) and A ◦ f(λ) = A(f(λ)).

Proof. By Proposition 1.1.7,∫ τ

0

e−λt(A ◦ f)(t) dt = A

∫ τ

0

e−λtf(t) dt.

Since A is closed, the second statement follows on letting τ →∞.

Now we consider convolutions.

Proposition 1.6.4. Let k ∈ L1loc(R+), f ∈ L1

loc(R+, X), λ ∈ C, and suppose that

Reλ > max(abs(|k|), abs(f)). Then (k ∗ f)(λ) exists and (k ∗ f)(λ) = k(λ)f(λ).

Proof. Replacing k(t) by e−λtk(t) and f(t) by e−λtf(t), and using Proposition1.6.1, we may assume that λ = 0.

First, we give the simple proof in the case when f ∈ L1(R+, X). Then Fubini’stheorem gives that k ∗ f ∈ L1(R+, X) and

(k ∗ f)(0) =

∫ ∞

0

(k ∗ f)(t) dt

=

∫ ∞

0

∫ t

0

k(t− s)f(s) ds dt

=

∫ ∞

0

(∫ ∞

s

k(t− s) dt

)f(s) ds

= k(0)f(0).

Now, assume only that f(0) exists. Replacing f(t) by f(t) − e−tf(0) (and

using the previous case), we may assume that f(0) = 0. Let ε > 0. There exists Ksuch that

∥∥∫ τ

0f(s) ds

∥∥ < ε whenever τ > K. Then∫ τ

0

(k ∗ f)(t) dt = (1 ∗ (k ∗ f))(τ) = (k ∗ (1 ∗ f))(τ).

Hence, ∥∥∥∥∫ τ

0

(k ∗ f)(t) dt∥∥∥∥ ≤

∥∥∥∥∥∫ K

0

k(τ − t)

(∫ t

0

f(s) ds

)dt

∥∥∥∥∥+

∥∥∥∥∫ τ

K

k(τ − t)

(∫ t

0

f(s) ds

)dt

∥∥∥∥≤ M

∫ τ

τ−K

|k(s)| ds+ ε

∫ τ−K

0

|k(s)| ds,

where M := supt≥0

∥∥∥∫ t

0f(s) ds

∥∥∥ <∞. Letting τ →∞,

lim supτ→∞

∥∥∥∥∫ τ

0

(k ∗ f)(t) dt∥∥∥∥ ≤ ε‖k‖1.

Since ε > 0 is arbitrary, it follows that (k ∗ f)(0) = 0, as required.

As a corollary, we recover a simple result which was already observed in(1.11).

Corollary 1.6.5. Let f ∈ L1loc(R+, X) and let F (t) =

∫ t

0f(s) ds. If Reλ > 0 and

f(λ) exists, then F (λ) exists and F (λ) = f(λ)/λ.

Proof. This is immediate from Proposition 1.3.1 with k(t) = 1.

Corollary 1.6.6. Let f : R+ → X be absolutely continuous and differentiable a.e.

If Re λ > 0 and f ′(λ) exists, then f(λ) exists and f ′(λ) = λf(λ)− f(0).

Proof. By Proposition 1.2.3, f ′ ∈ L1loc(R+, X) and f(t) − f(0) =

∫ t

0f ′(s) ds. The

result follows from Corollary 1.6.5.

Now we want to consider the substitution of√λ for λ; we will find a function

h such that h(λ) = f(√λ). For this we first calculate the Laplace integral of a

special function.

Lemma 1.6.7. Let s > 0 and

φs(t) =e−s2/4t

√πt

,

ψs(t) =se−s2/4t

2√πt3/2

(t > 0).

Then

φs(λ) =1√λe−s

√λ,

ψs(λ) = e−s√λ (Reλ > 0).

Proof. First, we show that, for α > 0,∫ ∞

0

e−((α/u)−u)2 du =

∫ ∞

0

α

v2e−(v−α/v)2 dv =

√π

2.

The first equality follows from the change of variable v := α/u. Taking the average

1.6. OPERATIONAL PROPERTIES 39

and making the change of variable w := u− α/u, this gives∫ ∞

0

e−((α/u)−u)2 du =1

2

∫ ∞

0

(1 +

α

u2

)e−(u−α/u)2 du

=1

2

∫ ∞

−∞e−w2

dw

=

√π

2.

Now, for λ > 0,

φs(λ) = e−s√λ

∫ ∞

0

1√πt

e−(√λt−s/(2

√t))2 dt

= e−s√λ 2√

πλ

∫ ∞

0

√λ

2u2e−(s

√λ/(2u)−u)2 du

= e−s√λ 2√

πλ

√π

2

=e−s

√λ

√λ

,

ψs(λ) = e−s√λ

∫ ∞

0

s

2√πt3/2

e−(√λt−s/(2

√t))2 dt

= e−s√λ 2√

π

∫ ∞

0

e−(s√λ/(2u)−u)2 du

= e−s√λ.

For Reλ > 0, the results follow by uniqueness of holomorphic extensions.

Proposition 1.6.8. Let f ∈ L1loc(R+, X) with ω(f) <∞, and let

g(t) =

∫ ∞

0

e−s2/4t

√πt

f(s) ds,

h(t) =

∫ ∞

0

se−s2/4t

2√πt3/2

f(s) ds.

Then g(λ) = f(√λ)/√λ and h(λ) = f(

√λ), whenever Re λ > (max{ω(f), 0})2.

Proof. If Reλ > (max{ω(f), 0})2, then Re√λ > max{ω(f), 0}, and Fubini’s theo-

rem and Lemma 1.6.7 give

g(λ) =

∫ ∞

0

∫ ∞

0

e−s2/4t

√πt

e−λtf(s) dt ds

=

∫ ∞

0

e−s√λ

√λ

f(s) ds

= f(√λ)/√λ.

Similarly, h(λ) = f(√λ).

1.7 Uniqueness, Approximation and Inversion

In this section we shall show that any f ∈ L1loc(R+, X) with abs(f) <∞ is uniquely

determined by its Laplace transform f and we shall give the Post-Widder inversionformula (Theorem 1.7.7). Other inversion theorems appear in Section 2.3. Thefollowing elementary statement will be used in the proofs of many results in thisand the following sections.

Lemma 1.7.1. Let a, b > 0 and define λn := a+nb, e−λn(t) := e−λnt (n ∈ N0, t ≥

0). Then {e−λn: n ∈ N0} is total in L1(R+).

Proof. By the Stone-Weierstrass theorem, the linear span of the set P := {t �→a−1tbn/a : n ∈ N0} is dense in C[0, 1] and thus in L1(0, 1). Now the statement fol-lows from the fact that Φ : L1(0, 1)→ L1(R+) defined by (Φg)(t) := ae−atg(e−at)is an isometric isomorphism which maps P onto the exponential functions {e−λn

:n ∈ N0}.Proposition 1.7.2. Let f ∈ L1

loc(R+, X) with abs(f) < ∞, let a > abs(f), b > 0

and λn := a+ nb. If f(λn) = 0 for all n ∈ N, then f(t) = 0 a.e.

Proof. One can assume that a > max{abs(f), 0}. Define F (t) :=∫ t

0f(s) ds (t ≥ 0).

Then 0 = f(λn) = λnF (λn) for all n ∈ N0 (see Corollary 1.6.5). It follows fromTheorem 1.4.3 that a > ω(F ). Thus G(t) := e−atF (t) is continuous and bounded

on R+, and G(nb) = F (a+nb) = 0 (n ∈ N). For x∗ ∈ X∗ define gx∗(t) := 〈G(t), x∗〉(t ≥ 0). Then gx∗ ∈ L∞(R+) = L1(R+)

∗ and

〈e−nb, gx∗〉 =∫ ∞

0

e−nbtgx∗(t) dt = 〈G(nb), x∗〉 = 0.

Since {e−nb : n ∈ N} is a total subset of L1(R+) by Lemma 1.7.1, it follows thatgx∗(t) = 0 for all t ≥ 0 and x∗ ∈ X∗. This implies that F (t) = eatG(t) = 0 for allt ≥ 0 and thus f(t) = 0 a.e.

Because of its importance we reformulate this result in the following formwhich is used frequently in the book.

Theorem 1.7.3 (Uniqueness Theorem). Let f, g ∈ L1loc(R+, X) with abs(f) < ∞

and abs(g) < ∞, and let λ0 > max(abs(f), abs(g)). Suppose that f(λ) = g(λ)whenever λ > λ0. Then f(t) = g(t) a.e.

Remark 1.7.4. A sequence (λn)n∈N of complex numbers is called a uniqueness se-quence for the Laplace transform if f = 0 a.e. whenever f ∈ L1

loc(R+, X), abs(f) <

Reλn for all n, and f(λn) = 0 for all n. It was shown in Proposition 1.7.2 that

1.7. UNIQUENESS, APPROXIMATION AND INVERSION 41

equidistant sequences λn = a + nb (b > 0) are examples of uniqueness sequences.In particular, this shows that a function of the form λ �→ q(λ) sin(λ) cannot havea representation as a Laplace transform. A finite sequence (λ1, · · · , λn) is not auniqueness sequence since the function λ �→ (λ+μ)−2n(λ−λ1) · · · (λ−λn) (Reλ >−μ) is the Laplace transform of the convolution product of the functions t �→(1 − (λi + μ)t)e−μt (1 ≤ i ≤ n). A characterization of uniqueness sequences willbe given in the Notes to this section.

One can deduce the following fundamental result on approximation from theuniqueness theorem by a quotient argument (see also the proof of Vitali’s theoremin Appendix A).

Theorem 1.7.5 (Approximation). Let fn ∈ C(R+, X) with ‖fn(t)‖ ≤ Meωt forsome M > 0, ω ∈ R and all n ∈ N. Let λ0 ≥ ω. The following are equivalent:

(i) The Laplace transforms fn converge pointwise on (λ0,∞) and the sequence(fn)n∈N is equicontinuous on R+.

(ii) The functions fn converge uniformly on compact subsets of R+.

Moreover, if (ii) holds, then f(λ) = limn→∞ fn(λ) for all λ > λ0, where f(t) :=limn→∞ fn(t).

Proof. The space c(X) := {(xn)n∈N : xn ∈ X and limn→∞ xn exists} is a closedsubspace of l∞(X) := {(xn)n∈N : xn ∈ X and supn∈N ‖xn‖ < ∞}. Define w :R+ → l∞(X) by w(t) = (fn(t))n∈N. Assume that (i) holds. The equicontinuity of

fn implies the continuity of w, and the convergence of fn(λ) implies that w(λ) =

(fn(λ))n∈N ∈ c(X) for all λ > λ0. Consider the quotient mapping q : l∞(X) →l∞(X)/c(X). Then (q ◦ w)(λ) = q(w(λ)) = 0 for all λ > λ0. Since q ◦ w : R+ →l∞(X)/c(X) is continuous, it follows from the uniqueness theorem that q◦w(t) = 0for all t ≥ 0; i.e., w(t) ∈ c(X) for all t ≥ 0. Hence (fn)n∈N converges pointwise.Since (fn)n∈N is equicontinuous, this implies uniform convergence on each compactsubset of R+.

Conversely, assume that (ii) holds. Clearly, uniform convergence implies equi-

continuity. Let f(t) = limn→∞ fn(t). Then limn→∞ fn(λ) = f(λ) for all λ > λ0 bythe dominated convergence theorem.

We point out that one cannot omit the condition that the sequence (fn)n∈Nis equicontinuous. To give an example, let X = C and fn(t) := eint (t ≥ 0, n ∈ N).

Then fn(λ) = 1λ−in

converges to 0 as n → ∞ for all λ > 0. But fn(t) does notconverge as n→∞ if t ∈ R+ \ 2πZ.

Another quotient argument enables us to deduce the converse of Proposition1.6.3 from the uniqueness theorem.

Proposition 1.7.6. Let A be a closed linear operator on X, let f, g ∈ L1loc(R+, X)

such that abs(f) < ∞ and abs(g) < ∞, and let ω > max{abs(f), abs(g)}. Thenthe following assertions are equivalent:

(i) f(t) ∈ D(A) and Af(t) = g(t) a.e. on R+.

(ii) f(λ) ∈ D(A) and Af(λ) = g(λ) for all λ > ω.

Proof. The implication (i) ⇒ (ii) has already been proved in Proposition 1.6.3.(ii) ⇒ (i): Let G(A) be the graph of A, which is a closed subspace of X ×X,

and let q : X × X → (X × X)/G(A) be the quotient map. Define h : R+ →(X × X)/G(A) by h(t) = q(f(t), g(t)). Then h(λ) = q(f(λ), g(λ)) = 0 for allλ > ω, by (ii). By the uniqueness theorem, h(t) = 0 a.e. This proves that (i) istrue.

Recall from Proposition 1.2.2 that t is a Lebesgue point of f ∈ L1loc(R+, X)

if limh→01h

∫ t+h

t‖f(s) − f(t)‖ ds = 0 and that almost all points t are Lebesgue

points of f . We now prove the Post-Widder inversion formula.

Theorem 1.7.7 (Post-Widder). Let f ∈ L1loc(R+, X). Assume that abs(f) < ∞

and that t > 0 is a Lebesgue point of f . Then

f(t) = limk→∞

(−1)k 1

k!

(k

t

)k+1

f (k)

(k

t

).

Proof. To explain the structure of the proof, we first consider the special case whenf is a bounded continuous function. By Theorem 1.5.1,

(−1)k 1

k!

(k

t

)k+1

f (k)

(k

t

)=

∫ ∞

0

ρk(s)f(s) ds,

where ρk(s) := 1k!

(kt

)k+1e−ks/tsk (s > 0). The functions ρk are “approximate

Dirac δ-functions”; i.e., ρk ≥ 0,∫∞0

ρk(s) ds = 1, and for all ε > 0 and all openintervals I ⊂ R+ containing t we have

∫s/∈I ρk(s) ds < ε for all sufficiently large k

(see below). Since f is assumed to be bounded and continuous, it follows from∥∥∥∥∫ ∞

0

ρk(s)f(s) ds− f(t)

∥∥∥∥ =

∥∥∥∥∫ ∞

0

ρk(s)(f(s)− f(t)) ds

∥∥∥∥≤ 2‖f‖∞

∫s/∈I

ρk(s) ds+ sups∈I

‖f(s)− f(t)‖

that∫∞0

ρk(s)f(s) ds → f(t) as k → ∞. This proves the statement for boundedand continuous functions f .

Now let f ∈ L1loc(R+, X) and max(abs(f), 0) < ω <∞. By (1.12), ω > ω(F ),

where F (s) :=∫ s

0f(r) dr. In the following let t > 0 be a fixed Lebesgue point of

f , and let k ∈ N such that k > ωt. Let

G(s) :=

∫ s

t

(f(r)− f(t)) dr = F (s)− F (t)− f(t)(s− t) (s ≥ 0).

1.7. UNIQUENESS, APPROXIMATION AND INVERSION 43

Since ω > ω(F ), there exists M > 0 such that ‖G(s)‖ ≤ Meωs for all s ≥ 0.Since 1

k!λk+1

∫∞0

e−λssk ds = 1 for λ > 0 (by induction and integration by parts),it follows from Theorem 1.5.1 and integration by parts that

Jk := (−1)k 1

k!

(k

t

)k+1

f (k)

(k

t

)− f(t)

=1

k!

(k

t

)k+1 ∫ ∞

0

e−ks/tskf(s) ds− f(t)

=1

k!

(k

t

)k+1 ∫ ∞

0

e−ks/tsk(f(s)− f(t)) ds

=1

k!

(k

t

)k+1

k

∫ ∞

0

e−ks/tsk−1(st− 1

)G(s) ds

=kk+2

k!t

∫ ∞

0

e−kuuk−1(u− 1)G(ut) du.

Let ε > 0. Since t is a Lebesgue point of f , there exists 0 < δ < 1 such that

1

t‖G(ut)‖ =

∥∥∥∥1t∫ tu

t

(f(r)− f(t)) dr

∥∥∥∥ ≤ ε

3|u− 1|

if |u− 1| ≤ δ. Define

J1,k :=kk+2

k!t

∫ 1+δ

1−δ


Then

‖J1,k‖ ≤ ε

3

kk+2

k!

∫ 1+δ

1−δ

e−kuuk−1(u− 1)2 du

≤ ε

3

kk+2

k!

∫ ∞

0

e−ku(uk+1 − 2uk + uk−1

)du

=ε

3

kk+2

k!

((k + 1)!

kk+2− 2k!

kk+1+

(k − 1)!

kk

)=

ε

3

for all k ∈ N with k > ωt. Let

J2,k :=kk+2

k!t

∫ 1−δ

0


If k > 1/δ, the function u �→ e−kuuk−1 is increasing on (0, 1− δ). Thus,

‖J2,k‖ ≤ kk+2

k!te−k(1−δ)(1− δ)k−1

∫ 1−δ

0

(1− u)‖G(ut)‖ du =: bk.

Since bk+1/bk =(1 + 1

k

)k+2

eδ−1(1 − δ) → eδ(1 − δ) < 1 as k → ∞, one obtains

that ‖J2,k‖ < ε/3 for all sufficiently large k. Finally let

J3,k :=1

t

1

k!kk+2

∫ ∞

1+δ


The function u �→ e−muum is decreasing on (1 + δ,∞) for all m ∈ N. Choosek0 > tω and let k > k0. Then

‖J3,k‖ =

∥∥∥∥kk+2

k!t

∫ ∞

1+δ

e−(k−k0)uuk−k0e−k0uuk0−1(u− 1)G(ut) du

∥∥∥∥≤ kk+2

k!te−(k−k0)(1+δ)(1 + δ)k−k0

∫ ∞

1+δ

e−k0uuk0−1(u− 1)Meωut du

=: ck.

Since ck+1/ck =(1 + 1

k

)k+2e−1−δ(1 + δ) → e−δ(1 + δ) < 1, one obtains that

‖J3,k‖ < ε/3 for all sufficiently large k. It follows from Jk = J1,k +J2,k + J3,k thatJk → 0 as k →∞.

1.8 The Fourier Transform and Plancherel’s Theorem

In this section, we give a brief summary of the properties of vector-valued Fouriertransforms, and we extend Plancherel’s theorem and the Paley-Wiener theorem(characterizing the Laplace transforms of L2-functions) to functions with valuesin a Hilbert space.

For f ∈ L1(R, X), the Fourier transform of f is the function Ff : R → Xdefined by

(Ff)(s) :=∫ ∞

−∞e−istf(t) dt.

We also define

(Ff)(s) :=∫ ∞

−∞eistf(t) dt = (Ff)(−s) = (F f)(s),

where f(t) := f(−t).Many properties of the Fourier transform on L1(R, X) can be proved in ex-

actly the same way as for the scalar-valued case (some can also be proved by ap-plying linear functionals and using the scalar-valued results and the Hahn-Banachtheorem), and we quote some of them here. Proofs of the scalar-valued cases maybe found in standard textbooks such as [Rud87], [RS72], [Yos80], [Rud91].

Theorem 1.8.1. Let f ∈ L1(R, X) and g ∈ L1(R).

a) F(g ∗ f)(s) = (Fg)(s)(Ff)(s).

1.8. THE FOURIER TRANSFORM AND PLANCHEREL’S THEOREM 45

b)∫∞−∞ g(t)(Ff)(t) dt =

∫∞−∞(Fg)(t)f(t) dt.

c) (Riemann-Lebesgue Lemma) Ff ∈ C0(R, X).

d) (Inversion Theorem) If Ff ∈ L1(R, X), then f = 12πF(Ff) a.e.

For scalar-valued functions, Plancherel’s theorem [Rud87, Theorem 19.2]shows that Ff ∈ L2(R) and ‖Ff‖2 =

√2π‖f‖2 whenever f ∈ L1(R)∩L2(R), and

hence the restriction of F to L1(R) ∩ L2(R) has a unique extension to a boundedlinear operator on L2(R) (also denoted by F), 1√

2πF is a unitary operator on

L2(R), and Theorem 1.8.1 b) holds for f, g ∈ L2(R). Hence, F−1 = (2π)−1F ,where (Ff)(t) = (Ff)(−t).

Plancherel’s theorem is not true for vector-valued functions, except when thespace X is a Hilbert space (see the Notes). If X is a Hilbert space with innerproduct (·|·)X , then L2(R, X) is also a Hilbert space with inner product

(f |g)L2(R,X) :=

∫ ∞

−∞(f(t)|g(t))X dt.

As observed in Section 1.1, the simple functions are dense in L2(R+, X). Inthis Hilbert space context, this can also be shown by computing the orthogonalcomplement of the simple functions.

Theorem 1.8.2 (Plancherel’s Theorem). Let X be a Hilbert space. Then Ff ∈L2(R, X) and ‖Ff‖L2(R,X) =

√2π‖f‖L2(R,X) for all f ∈ L1(R, X) ∩ L2(R, X).

The restriction of F to L1(R, X) ∩ L2(R, X) extends to a bounded linear operatorF on L2(R, X) and 1√

2πF is a unitary operator on the Hilbert space L2(R, X).

Moreover, ∫ ∞

−∞((Ff)(t)|g(t))X dt =

∫ ∞

−∞(f(t)|(Fg)(−t))X dt (1.15)

for all f, g ∈ L2(R, X).

Proof. Let f ∈ L1(R, X) ∩ L2(R, X). To prove that

‖Ff‖L2(R,X) =√2π‖f‖L2(R,X),

it suffices to assume that X is separable, since f is almost separably valued. Let{en : n ∈ N} be an orthonormal basis of X , and let fn(t) := (f(t)|en)X . Thenfn ∈ L1(R) ∩ L2(R) and ((Ff)(s)|en)X = (Ffn)(s). Now, using the scalar-valuedPlancherel theorem,∫ ∞

−∞‖(Ff)(s)‖2 ds =

∫ ∞

−∞

∞∑n=1

|(Ffn)(s)|2 ds

= 2π

∫ ∞

−∞

∞∑n=1

|fn(t)|2 dt

= 2π

∫ ∞

−∞‖f(t)‖2 dt.

This proves the first part of the result. Since L1(R+, X) ∩ L2(R+, X) is dense inL2(R+, X), F extends uniquely to a bounded linear operator on L2(R, X) suchthat 1√

2πF is an isometry.

One may prove (1.15) in a similar way, using Parseval’s formula and the cor-responding scalar-valued result (Theorem 1.8.1 b)). This implies that the adjointoperator of F in the sense of Hilbert spaces is F , where (Ff)(t) = (Ff)(−t).Hence, FF = 2πI so F is surjective, and it follows that F is surjective.

We remark that Plancherel’s theorem extends to functions of several variableswith values in a Hilbert space X: the normalized Fourier transform (2π)−n/2F isa unitary operator on L2(Rn, X). This can be deduced from the scalar-valued caseas in the proof of Theorem 1.8.2. Alternatively, it may be deduced from Theorem1.8.2 by induction, using the identification L2(Rn+1, X) = L2(R, L2(Rn, X)).

When f ∈ L1(R+, X), we consider f as being a member of L1(R, X) withf(t) = 0 for t < 0, so

(Ff)(s) =

∫ ∞

0

e−istf(t) dt = f(is),

where, as usual, f is the Laplace transform of f . Similarly, if t �→ e−atf(t) belongsto L1(R+, X) (for example, if f ∈ L2(R+, X) and a > 0), then its Fourier trans-

form is s �→ f(a + is). Thus, Plancherel’s theorem can be used to study Laplacetransforms of functions in L2(R+, X) when X is a Hilbert space.

Let C+ := {λ ∈ C : Reλ > 0} andH2(C+, X) be the space of all holomorphicfunctions g : C+ → X such that

‖g‖2H2(C+,X) := supα>0

∫ ∞

−∞‖g(α + is)‖2 ds <∞.

For scalar-valued functions, the Paley-Wiener theorem [Rud87, Theorem 9.13]

shows that g ∈ H2(C+) := H2(C+,C) if and only if g = f |C+for some f ∈ L2(R+).

Then

supα>0

∫ ∞

−∞|g(α+ is)|2 ds = lim

α↓0

∫ ∞

−∞|g(α+ is)|2 ds.

Moreover, g has Ff as a boundary function in the sense that g(α+ is)→ (Ff)(s)s-a.e. and in L2-norm, as α ↓ 0. In addition, g is the Poisson integral of Ff :

g(α+ is) =α

π

∫ ∞

−∞

(Ff)(r)α2 + (s− r)2

dr

[Dur70, Chapter 11], [Koo80, Chapter VI].Again, these results are not true for vector-valued functions in general, but

they are true in the case of Hilbert spaces.

1.8. THE FOURIER TRANSFORM AND PLANCHEREL’S THEOREM 47

Theorem 1.8.3 (Paley-Wiener Theorem). Let X be a Hilbert space. Then the map

f �→ f |C+is an isometric isomorphism of L2(R+, X) onto H2(C+, X). Moreover,

for f ∈ L2(R+, X),

f(α+ is) =α

π

∫ ∞

−∞

(Ff)(r)

α2 + (s− r)2dr. (1.16)

As α ↓ 0, ‖f(α+is)−(Ff)(s)‖ → 0 (s)-a.e. and∫∞−∞ ‖f(α+is)−(Ff)(s)‖2 ds→ 0.

Proof. Let f ∈ L2(R+, X). For α > 0,

f(α+ is) =

∫ ∞

0

e−ist(e−αtf(t)

)dt.

By Plancherel’s Theorem 1.8.2,∫ ∞

−∞

∥∥∥f(α+ is)∥∥∥2 ds = 2π

∫ ∞

0

e−2αt‖f(t)‖2 dt ≤ 2π‖f‖22.

Thus f ∈ H2(C+, X). Moreover, Plancherel’s theorem and the dominated conver-gence theorem give∫ ∞

−∞‖f(α+ is)− (Ff)(s)‖2 ds = 2π

∫ ∞

0

|e−αt − 1|2‖f(t)‖2 dt→ 0

as α ↓ 0. For α > 0 and x ∈ X,

(f(α+ is)|x)X =α

π

∫ ∞

−∞

((Ff)(r)|x)Xα2 + (s− r)2

dr =

(∫ ∞

−∞

(Ff)(r)α2 + (s− r)2

dr

∣∣∣∣x)X

,

which establishes (1.16). The proof that ‖f(α+ is)− (Ff)(s)‖ → 0 a.e. is similarto the scalar-valued case.

Conversely, let g ∈ H2(C+, X). Then g is separably valued, so we may assumethat X is separable. Let {en : n ∈ N} be an orthonormal basis of X, and letgn(λ) = (g(λ)|en)X . Then gn ∈ H2(C+), so the scalar-valued case implies that

there exists fn ∈ L2(R+) such that gn = fn|C+. Moreover,∫ ∞

0

∞∑n=1

|fn(t)|2 dt = limα↓0

∞∑n=1

∫ ∞

0

e−2αt|fn(t)|2 dt

=1

2πlimα↓0

∞∑n=1

∫ ∞

−∞|gn(α + is)|2 ds

=1

2πlimα↓0

∫ ∞

−∞‖g(α+ is)‖2 ds <∞.

Hence∑∞

n=1 |fn(t)|2 converges, and therefore∑∞

n=1 fn(t)en converges to a sumf(t) in X , for almost all t. Now, f ∈ L2(R+, X) and, for λ ∈ C+,

f(λ) =

∞∑n=1

fn(λ)en = g(λ).

1.9 The Riemann-Stieltjes Integral

This section is an introduction to the Riemann-Stieltjes integral∫ b

ag(t) dF (t) of

a vector-valued function F and a scalar-valued function g on [a, b]. Such integralsplay an important role in the approach to Laplace transform theory taken inChapter 2. More precisely, let f : R+ → X be a bounded measurable function.

Then t �→ F (t) :=∫ t

0f(s) ds is Lipschitz continuous and we will show in Sections

1.10 and 2.1 that

f(λ) =

∫ ∞

0

e−λtf(t) dt =

∫ ∞

0

e−λt dF (t) = TF (e−λ)

for all λ > 0, where TF : g �→ ∫∞0

g(t) dF (t) is a bounded linear operator from

L1(R+) to X and e−λ(t) := e−λt. Thus, the Laplace integrals f(λ) are evaluationsof a bounded linear operator TF at the exponential function e−λ. Since the mapΦS : F �→ TF turns out to be an isometric isomorphism between the Lipschitzcontinuous functions F : R+ → X and L(L1(R+), X) (see Section 2.1), manyfunctional analytic arguments can be applied to Laplace transform theory.

A function F : [a, b] → X is of bounded semivariation if there exists M ≥ 0such that ‖∑i(F (ti) − F (si))‖ ≤ M for every choice of a finite number of non-overlapping intervals (si, ti) in [a, b]. Recall from Section 1.2 that F is of boundedvariation if there exists M ≥ 0 such that

∑i ‖F (ti) − F (ti−1)‖ ≤ M for every

finite partition a = t0 < t1 < . . . < tn = b of [a, b]. Further, F is of weakbounded variation if x∗ ◦ F : t �→ 〈F (t), x∗〉 is of bounded variation for everyx∗ ∈ X∗. The set of functions F : [a, b]→ X of bounded semivariation is denotedby BSV([a, b], X). A function F : R+ → X is in BSVloc(R+, X) if it is of boundedsemivariation on every compact subinterval of R+.

As remarked in Section 1.2, any Lipschitz function is of bounded variation,and it is easy to see that any function of bounded variation is of bounded semivari-ation. We show in the following proposition that the function F : [0, 1]→ L∞[0, 1]defined by F (t) = χ[0,t] is of bounded semivariation. Since ‖F (t) − F (s)‖ = 1 forall t �= s, F is not of bounded variation, not separably valued, and not measur-able (see Example 1.1.5). Hence, functions of bounded semivariation may not bemeasurable.

Proposition 1.9.1. Let X be an ordered Banach space with normal cone. Let F :[a, b]→ X be increasing. Then F is of bounded semivariation.

Proof. Let (si, ti) (i = 1, 2, . . . , n) be disjoint intervals in [a, b]. Then

0 ≤n∑

i=1

(F (ti)− F (si)) ≤ F (tn)− F (s1) ≤ F (b)− F (a).

Hence ∥∥∥∥∥n∑

i=1

(F (ti)− F (si))

∥∥∥∥∥ ≤ c‖F (b)− F (a)‖,

1.9. THE RIEMANN-STIELTJES INTEGRAL 49

where c is a constant associated with the normal cone (see Appendix C).

In the context of Proposition 1.9.1, it is easy to see directly that F is of weakbounded variation. In fact, there is the following general result.

Proposition 1.9.2. A function F : [a, b] → X is of bounded semivariation if andonly if it is of weak bounded variation.

Proof. Assume that F is of weak bounded variation. Let SΩ :=∑

i(F (ti)−F (si)),where Ω is the union of finitely many disjoint intervals (si, ti) in [a, b]. For eachx∗ ∈ X∗, there exists Mx∗ := V[a,b](x

∗ ◦ F ) such that |〈SΩ, x∗〉| ≤ Mx∗ for all

such Ω. It follows from the uniform boundedness principle that F is of boundedsemivariation.

Now, let F be of bounded semivariation. Then there exists a constant M ≥ 0such that ‖∑i(F (ti)− F (si))‖ ≤ M for any choice of a finite number of dis-joint intervals (si, ti) in [a, b]. To obtain the weak bounded variation, one writesx∗ = x∗1 + ix∗2, where x∗j is a real-linear functional, and distinguishes betweenthe subintervals on which the numbers 〈F (ti)− F (ti−1), x

∗j 〉 are either positive or

negative.

Let F, g be two functions defined on an interval [a, b], one with values in Xand the other with values in C. If π denotes a finite partition a = t0 < t1 <. . . < tn = b of [a, b] with partitioning points ti and with some intermediate pointssi ∈ [ti−1, ti] (i = 1, . . . , n), we denote by |π| = maxi(ti− ti−1) the norm of π, andby

S(g, F, π) :=n∑

i=1

g(si) (F (ti)− F (ti−1))

the Riemann-Stieltjes sum associated with g, F and π. We say that g is Riemann-Stieltjes integrable with respect to F if∫ b

a

g(t) dF (t) := lim|π|→0

S(g, F, π)

exists in the norm topology of X. Here π runs through all partitions of [a, b]with intermediate points, and the limit must be independent of the choice ofintermediate points.

It is immediate from the definition that the set of all functions g which areRiemann-Stieltjes integrable with respect to a fixed function F is a linear space,and the Riemann-Stieltjes integral is linear in g (and also in F ). If F is of bounded

variation, g is bounded and∫ b

ag(t) dF (t) exists, then∥∥∥∥∥

∫ b

a

g(t) dF (t)

∥∥∥∥∥ ≤ supt∈[a,b]

‖g(t)‖ V[a,b](F ). (1.17)

If F is of bounded semivariation, then it follows from the proof of Proposition1.9.2 that ∥∥∥∥∥

∫ b

a

g(t) dF (t)

∥∥∥∥∥ ≤ 4M supt∈[a,b]

‖g(t)‖, (1.18)

where

M := sup

{∥∥∥∥∥∑i

(F (ti)− F (si))

∥∥∥∥∥ : (si, ti) disjoint subintervals of [a, b]

}.

(1.19)

The Riemann-Stieltjes integral respects closed operators; there are easy ana-logues of Propositions 1.1.6 and 1.1.7, both when g is scalar-valued and F isvector-valued and in the alternative case.

When F (t) = t, we write S(g, π) for S(g, F, π) and call it the Riemann sumassociated with g and π. We say that g is Riemann integrable on [a, b] if g isRiemann-Stieltjes integrable with respect to F (t) = t, and we write∫ b

a

g(t) dt := lim|π|→0

S(g, π).

In the scalar-valued case, g : [a, b]→ C is Riemann integrable if and only if gis bounded and continuous a.e., and the Riemann and Lebesgue integrals are thenequal [Rud76, Theorem 11.33]. By applying linear functionals, it follows that ifg : [a, b]→ X is Riemann integrable, then it is bounded and weakly measurable. IfX is separable, then g is measurable by Pettis’s theorem 1.1.1 and hence Bochnerintegrable by Theorem 1.1.4. Whenever g is both Riemann and Bochner integrable,the two integrals coincide (so our notation should not cause confusion). However,Riemann integrable functions with values in an inseparable space may be nowherecontinuous and not even measurable (see Example 1.9.7 below).

Now we return to Riemann-Stieltjes sums and integrals of two functions Fand g. Let π be a partition of [a, b] with partitioning points a = t0 < t1 < . . . <tn = b and intermediate points si ∈ [ti−1, ti]. If one chooses s0 = a and sn+1 = b,then we obtain a partition π′ with partitioning points a = s0 ≤ s1 ≤ . . . ≤ sn+1 =b, with intermediate points ti ∈ [si, si+1], and |π′| ≤ 2|π|. Moreover,

S(F, g, π) = g(b)F (b)− g(a)F (a)− S(g, F, π′).

It follows that if g is Riemann-Stieltjes integrable with respect to F , then so isF with respect to g (and vice versa, by symmetry) and the following formula ofintegration by parts holds:∫ b

a

g(t) dF (t) = g(b)F (b)− g(a)F (a)−∫ b

a

F (t) dg(t). (1.20)

Example 1.9.3. Let a ≤ c ≤ d ≤ b and let I be an interval with endpoints c andd. Let F : [a, b] → X. If a �∈ I and b �∈ I, then F and χI are Riemann-Stieltjesintegrable with respect to each other if and only if F is continuous at c and d, andthen ∫ b

a

χI(t) dF (t) = −∫ b

a

F (t) dχI(t) = F (d)− F (c).

If a = c ∈ I and b �∈ I, then F and χI are Riemann-Stieltjes integrable withrespect to each other if and only if F is continuous at d, and then∫ b

a

χI(t) dF (t) = F (d)− F (a),

∫ b

a

F (t) dχI(t) = −F (d).

If a �∈ I and b = d ∈ I, then F and χI are Riemann-Stieltjes integrable withrespect to each other if and only if F is continuous at c, and then∫ b

a

χI(t) dF (t) = F (b)− F (c),

∫ b

a

F (t) dχI(t) = F (c).

Proposition 1.9.4. Let F : [a, b] → X and g : [a, b] → C. If one function iscontinuous and the other is of bounded semivariation, then F and g are Riemann-Stieltjes integrable with respect to each other.

Proof. a) Assume that F is of bounded semivariation and that g is continuous. Letε > 0. Then there exists δ > 0 such that |g(s1)−g(s2)| < ε whenever |s1−s2| < δ.

Let πj , (j = 1, 2), be two partitions of [a, b] with |πj | < δ/2. Let a = t0 <t1 < . . . < tn = b be the partitioning points of π1 and π2 together. Then

S(g, F, πj) =

n∑i=1

g(sj,i)(F (ti)− F (ti−1))

where sj,i, ti and ti−1 belong to the same subinterval of πj . In particular, |s1,i −s2,i| < δ. For x∗ ∈ X∗,

|〈S(g, F, π1)− S(g, F, π2), x∗〉| =

∣∣∣∣∣n∑

i=1

(g(s1,i)− g(s2,i))〈F (ti)− F (ti−1), x∗〉∣∣∣∣∣

< εn∑

i=1

|〈F (ti)− F (ti−1), x∗〉|

≤ 4εM‖x∗‖,where M is defined by (1.19). Now, Cauchy’s convergence criterion implies that∫ b

ag(t) dF (t) exists.

b) Assume that F is continuous and g is of bounded semivariation. By Propo-sition 1.9.2, g is of bounded variation. Given ε > 0 there exists δ > 0 such that‖F (s1)− F (s2)‖ < ε whenever |s1 − s2| < δ. Similarly to a), one shows that

‖S(F, g, π1)− S(F, g, π2)‖ < εV[a,b](g)

whenever |π1| < δ/2 and |π2| < δ/2, where V[a,b](g) is the total variation of g.

Hence∫ b

aF (t) dg(t) exists.

Corollary 1.9.5. Let g : [a, b] → C be piecewise continuous, and F : [a, b] → Xbe continuous and of bounded semivariation. Then F and g are Riemann-Stieltjesintegrable with respect to each other.

Proof. Since g is piecewise continuous, g = g1 + g2 where g1 is continuous andg2 is a step function. Now the result follows from Proposition 1.9.4 and Example1.9.3.

Corollary 1.9.6. Let X be an ordered Banach space with normal cone. Let f :[a, b]→ X be increasing. Then f is Riemann integrable.

Proof. This is immediate from Propositions 1.9.1 and 1.9.4.

Example 1.9.7. Let f : [0, 1] → L∞[0, 1] be defined by f(t) := χ[0,t], so f is in-creasing, nowhere continuous and not measurable (Example 1.1.5 a)). By Corollary1.9.6, f is Riemann integrable and∫ 1

0

f(t) dt = limn→∞

1

n

n∑i=1

χ[0,i/n].

This shows that∫ 1

0f(t) dt is the function s �→ 1−s. Similarly, if F (t) :=

∫ t

0f(r) dr,

thenF (t)(s) = (t− s)χ[0,t](s)

(see Example 1.2.8 b)).

Proposition 1.9.8. Let F : [a, b]→ X be of bounded semivariation and g : [a, b]→ Cbe of bounded variation. Then gF is of bounded semivariation.

Proof. There exists M > 0 such that |g(t)| ≤M and ‖F (t)‖ ≤M for all t ∈ [a, b].The function gF is of bounded semivariation since it is of weak bounded variation.This follows from the assumptions and the estimates:∑

i

|〈g(ti)F (ti)− g(ti−1)F (ti−1), x∗〉|

≤∑i

|〈(g(ti)− g(ti−1))F (ti), x∗〉|

+∑i

|〈g(ti−1)(F (ti)− F (ti−1)), x∗〉|

≤ M‖x∗‖∑i

|g(ti)− g(ti−1)|+M∑i

|〈F (ti)− F (ti−1), x∗〉| ,

for all x∗ ∈ X∗.

In the remainder of this section, we give some results which reduce Riemann-Stieltjes integrals to Riemann or Bochner integrals when g or F has a derivativein an appropriate sense.

Proposition 1.9.9. Let F : [a, b]→ X be of bounded semivariation and g ∈ C1[a, b].Then Fg′ is Riemann integrable and

∫ b

a

F (t) dg(t) =

∫ b

a

F (t)g′(t) dt.

Proof. a) We show that hF is Riemann integrable for each h ∈ C[a, b]. If h is astep function, then hF is of bounded semivariation and hence Riemann integrable,by Proposition 1.9.4. Since each continuous function h on [a, b] is a uniform limitof step functions and F is bounded, the claim follows since a uniform limit ofRiemann integrable functions is Riemann integrable.

b) There exists M > 0 such that |g′(t)| ≤M and ‖F (t)‖ ≤M for all t ∈ [a, b].For ε > 0 there exists δ > 0 such that |g′(s) − g′(s′)| < ε whenever |s − s′| < δ.Let π be a partition of [a, b] with |π| < δ and with partitioning points ti andintermediate points si. By the mean value theorem, there exist s′i ∈ (ti−1, ti) suchthat g(ti) − g(ti−1) = g′(s′i)(ti − ti−1). Let π′ be the partition with partitioningpoints ti and intermediate points s′i, so |π′| = |π|. Then

S(F, g, π′) =∑i

F (s′i)g′(s′i)(ti − ti−1)

= S(Fg′, π′).

Letting |π| → 0, it follows that∫ b

aF (t) dg(t) =

∫ b

aF (t)g′(t) dt.

Proposition 1.9.10. Let F : [a, b] → X be of bounded semivariation and g, h ∈C[a, b]. Then G(t) :=

∫ t

ah(s) dF (s) is of bounded semivariation on [a, b] and

∫ b

a

g(t) dG(t) =

∫ b

a

g(t)h(t) dF (t).

Proof. Let M be such that |h(t)| ≤M for all t ∈ [a, b]. Let π be a partition of [a, b]with partitioning points ti and intermediate points si. By (1.17), for x∗ ∈ X∗,

∑i

|〈G(ti)−G(ti−1), x∗〉| =

∑i

∣∣∣∣∣∫ ti

ti−1

h(s) d〈F (s), x∗〉∣∣∣∣∣

≤ MV[a,b](x∗ ◦ F ).

It follows from Proposition 1.9.2 that G is of bounded semivariation. By Proposi-

tion 1.9.4,∫ b

ag(t) dG(t) and

∫ b

ag(t)h(t) dF (t) both exist.

For ε > 0 there exists δ > 0 such that |g(s′)−g(s)| < ε whenever |s′−s| < δ.

If |π| < δ, then ∣∣∣∣∣⟨S(g,G, π)−

∫ b

a

g(t)h(t) dF (t), x∗⟩∣∣∣∣∣

=

∣∣∣∣∣∑i

∫ ti

ti−1

(g(si)− g(t))h(t) d〈F (t), x∗〉∣∣∣∣∣

≤ εMV[a,b](x∗ ◦ F ).

It follows that ⟨∫ b

a

g(t) dG(t)−∫ b

a

g(t)h(t) dF (t), x∗⟩

= 0.

The result follows from the Hahn-Banach theorem.

The following result gives analogues of a special case of Proposition 1.9.10and of Proposition 1.9.9, with Riemann integrals replaced by Bochner integrals.

Proposition 1.9.11. Let g : [a, b]→ C and F : [a, b]→ X. If F is an antiderivative

of a Bochner integrable function f and if g is continuous, then∫ b

ag(s) dF (s) exists

and equals the Bochner integral∫ b

ag(s)f(s) ds. If F is continuous and g is abso-

lutely continuous, then∫ b

aF (s) dg(s) equals the Bochner integral

∫ b

aF (s)g′(s) ds.

Proof. Assume that g is continuous and that there exists a Bochner integrable func-tion f such that F (t) = F (a) +

∫ t

af(s) ds for all t ∈ [a, b]. Then F is of bounded

variation (Proposition 1.2.2) and the Riemann-Stieltjes integral∫ b

ag(s) dF (s) ex-

ists by Proposition 1.9.4. The Bochner integral∫ b

ag(s)f(s) ds exists by Theorem

1.1.4, since g is bounded and measurable. For ε > 0 there exists δ > 0 such that|g(s′)− g(s)| < ε whenever |s′ − s| < δ. For any partition π with |π| < δ,∥∥∥∥∥S(g, F, π)−

∫ b

a

g(s)f(s) ds

∥∥∥∥∥ =

∥∥∥∥∥∑i

∫ ti

ti−1

(g(si)− g(s))f(s) ds

∥∥∥∥∥≤ ε

∫ b

a

‖f(s)‖ ds.

Letting |π| → 0 and ε→ 0, the result follows.The proof of the second statement is analogous and is omitted.

Combining Propositions 1.9.4, 1.9.8 and 1.9.9 with the integration by partsformula (1.20) one obtains the following statement which will be used frequentlyin later sections. Let F : [0, t]→ X be of bounded semivariation. Then∫ t

0

e−λs dF (s) = e−λtF (t)− F (0) + λ

∫ t

0

e−λsF (s) ds (1.21)

1.10. LAPLACE-STIELTJES INTEGRALS 55

for all λ ∈ C. One should notice that the integral∫ t

0e−λsF (s) ds is a Riemann

integral if F is of bounded semivariation, by Propositions 1.9.4 and 1.9.9. If F isalso continuous, then the integral can be taken in the Bochner sense.

1.10 Laplace-Stieltjes Integrals

This section contains the essential properties of the Laplace-Stieltjes integral

dF (λ) :=

∫ ∞

0

e−λt dF (t) := limτ→∞

∫ τ

0

e−λtdF (t),

where F ∈ BSVloc(R+, X); i.e., F : R+ → X is of bounded semivariation on eachcompact subinterval of R+. First we observe that the Laplace-Stieltjes integral isa generalization of the Laplace integral.

Proposition 1.10.1. Let f ∈ L1loc(R+, X) and F (t) :=

∫ t

0f(s) ds. For λ ∈ C, dF (λ)

exists if and only if f(λ) exists, and then dF (λ) = f(λ).


0

e−λt dF (t) =

∫ τ

0

e−λtf(t) dt (τ ≥ 0),

and the result follows by letting τ →∞.

The results obtained in Sections 1.4, 1.5 and 1.6 for Laplace integrals carryover to Laplace-Stieltjes integrals with only minor modifications of the proofs, andwe give these below, starting with elementary properties.

Throughout this section, integrals over R+ are to be understood as improperRiemann-Stieltjes (or Riemann) integrals. Thus∫ ∞

0

g(t) dF (t) := limτ→∞

∫ τ

0

g(t) dF (t),∫ ∞

0

F (t) dt := limτ→∞

∫ τ

0

F (t) dt.

It is easy to see that∫∞0

F (t) dt exists if F ∈ BSVloc(R+, X) and ‖F (t)‖ ≤h(t) (t ≥ 0) for some h ∈ L1(R+).

Recall from Section 1.4 that the exponential growth bound of a function F ∈BSVloc(R+, X) is defined by

ω(F ) := inf

{ω ∈ R : sup

t≥0‖e−ωtF (t)‖ <∞

}.

It follows from (1.21) that abs(dF ) ≤ ω(F ) and

dF (λ) = −F (0) + λ

∫ ∞

0

e−λtF (t) dt (Reλ > ω(F )). (1.22)

Note that (1.22) is a generalization of both (1.11) and Corollary 1.6.5.

Proposition 1.10.2. Let f ∈ BSVloc(R+, X) and F (t) :=∫ t

0f(s) ds. Then F is

locally Lipschitz continuous, and

df(λ) = −f(0) + λdF (λ) = −f(0) + λ2F (λ)

whenever Reλ > ω(f).

Proof. Since f is locally bounded, (1.17) implies that F is locally Lipschitz con-tinuous and ω(F ) ≤ ω(f). In particular, F ∈ L1

loc(R+, X) ∩ BSVloc(R+, X). By(1.21) and Proposition 1.9.10,∫ τ

0

e−λs df(s) = e−λτf(τ)− f(0) + λ

∫ τ

0

e−λs dF (s).

Letting τ →∞ gives

df(λ) = −f(0) + λdF (λ)

whenever Reλ > ω(f). By (1.22), dF (λ) = λF (λ).

Now we give a generalization of Proposition 1.6.1 a).

Proposition 1.10.3. Let F ∈ BSVloc(R+, X), μ ∈ C and let G(t) :=∫ t

0e−μs dF (s)

(t ≥ 0). For λ ∈ C, dG(λ) exists if and only if dF (λ + μ) exists, and then

dG(λ) = dF (λ+ μ).


0

e−λt dG(t) =

∫ τ

0

e−(λ+μ)t dF (t),

and the result follows on letting τ →∞.

For F ∈ BSVloc(R+, X), let

abs(dF ) := inf{Reλ : dF (λ) exists

}.

Proposition 1.10.4. Let F ∈ BSVloc(R+, X). Then dF (λ) converges if Re λ >abs(dF ) and diverges if Re λ < abs(dF ).

Proof. Clearly, dF (λ) does not exist if Reλ < abs(dF ). For λ0 ∈ C define G0(t) :=∫ t

0e−λ0s dF (s) (λ ∈ C, t ≥ 0). Then, by Proposition 1.9.10,∫ t

0

e−λs dF (s) =

∫ t

0

e−(λ−λ0)s dG0(s) (λ ∈ C , t ≥ 0).

Integration by parts (1.20), and Proposition 1.9.9, yield∫ t

0

e−λs dF (s) = e−(λ−λ0)tG0(t) + (λ− λ0)

∫ t

0

e−(λ−λ0)sG0(s) ds. (1.23)

1.10. LAPLACE-STIELTJES INTEGRALS 57

If dF (λ0) exists, then G0 is bounded. Therefore, dF (λ) exists if Reλ > Reλ0 and

dF (λ) = (λ− λ0)

∫ ∞

0

e−(λ−λ0)sG0(s) ds (Reλ > Re λ0). (1.24)

This shows that dF (λ) exists if Reλ > abs(dF ) and, as for the Laplace integral(see (1.9)),

abs(dF ) = inf

{λ ∈ R : sup

t>0

∣∣∣∣∫ t

0

e−λs d〈F (s), x∗〉∣∣∣∣ <∞ for all x∗ ∈ X∗

}. (1.25)

Theorem 1.10.5. Let F ∈ BSVloc(R+, X) and let F∞ := limt→∞ F (t) if the limitexists, F∞ := 0 otherwise. Then abs(dF ) = ω(F − F∞).

Proof. For λ0 > abs(dF ) define G0(t) :=∫ t

0e−λ0sdF (s) (t ≥ 0). Then G0 is

bounded. To prove that abs(dF ) ≥ ω(F − F∞) one considers the two casesabs(dF ) ≥ 0 and abs(dF ) < 0. First, let abs(dF ) ≥ 0 and λ0 > abs(dF ). Itfollows from Proposition 1.9.10, integration by parts (1.20), and Proposition 1.9.9that

F (t) = F (0) +

∫ t

0

eλ0s dG0(s) = F (0) + eλ0tG0(t)− λ0

∫ t

0

eλ0sG0(s) ds

for all t ≥ 0, so supt≥0 ‖e−λ0t(F (t) − F∞)‖ < ∞. Thus ω(F − F∞) ≤ abs(dF ) ifabs(dF ) ≥ 0. Second, let abs(dF ) < 0. Choose abs(dF ) < λ0 < 0. For r ≥ t ≥ 0one has

F (r)− F (t) =

∫ r

t

eλ0s dG0(s) = eλ0rG0(r)− eλ0tG0(t) − λ0

∫ r

t

eλ0sG0(s) ds.

Thus,

limr→∞F (r) = F∞ = F (t)− eλ0tG0(t)− λ0

∫ ∞

t

eλ0sG0(s) ds

exists and supt≥0 ‖e−λ0t(F (t) − F∞)‖ < ∞. Therefore, ω(F − F∞) ≤ abs(dF ) ifabs(dF ) < 0.

To show the reverse inequality let ω > ω(F − F∞). Then there exists M ≥ 0such that ‖F (t)−F∞‖ ≤Meωt for all t ≥ 0. Let λ > ω > ω(F −F∞). Integrationby parts (see (1.21)) yields∫ t

0

e−λs dF (s) = e−λt(F (t)− F∞) + F∞ − F (0) + λ

∫ t

0

e−λs(F (s)− F∞) ds.

Hence, dF (λ) exists for λ > ω(F − F∞) and is given by

dF (λ) = F∞ − F (0) + λ

∫ ∞

0

e−λs(F (s)− F∞) ds. (1.26)

This shows that abs(dF ) ≤ ω(F − F∞).

Note that (1.26) is a generalization of (1.22).

Theorem 1.10.6. Let F ∈ BSVloc(R+, X) and assume that abs(dF ) < ∞. Then

λ �→ dF (λ) is holomorphic for Re λ > abs(dF ), and

dF(n)

(λ) =

∫ ∞

0

e−λt(−t)n dF (t) (Reλ > abs(dF ), n ∈ N0)

(as an improper Riemann-Stieltjes integral).

Proof. Let qk(λ) :=∫ k

0e−λt dF (t). It follows from (1.18) that

qk(λ) = limN→∞

N∑n=0

λn

n!

∫ k

0

(−t)n dF (t).

By the Weierstrass convergence theorem (a special case of Vitali’s Theorem A.5),qk is entire and

q(j)k (λ) =

∫ k

0

e−λt(−t)j dF (t)

for all j ∈ N0. Let Reλ > λ0 > abs(dF ), and define

G0(t) :=

∫ t

0

e−λ0s dF (s).

Then G0 is bounded and it follows from Proposition 1.9.10, (1.20) and Proposition1.9.9 that

dF (λ)− qk(λ) =

∫ ∞

k

e−(λ−λ0)s dG0(s)

= −e−(λ−λ0)kG0(k) + (λ− λ0)

∫ ∞

k

e−(λ−λ0)sG0(s) ds.

Hence qk converges to dF uniformly on compact subsets of {λ : Reλ > abs(dF )}.Again by the Weiestrass convergence theorem, dF is holomorphic and q

(j)k (λ) →

dF(j)

(λ) as k →∞, if Reλ > abs(dF ).

Finally in this section, we consider operator-valued Laplace-Stieltjes integrals.Let S : R+ → L(X,Y ) be a function. By the uniform boundedness principle,S ∈ BSVloc(R+,L(X,Y )) if and only if vx := S(·)x ∈ BSV(R+, Y ) for all x ∈ X.When S ∈ BSVloc(R+,L(X,Y )), we let

ω(S) := inf

{ω ∈ R : sup

t≥0‖e−ωtS(t)‖ <∞

},

abs(dS) := inf

{Reλ :

∫ t

0

e−λs dS(s) converges strongly as t→∞}

= sup{abs(dvx) : x ∈ X}.The following analogue of Proposition 1.4.5 follows from Theorems 1.10.5 and1.10.6.

1.11. NOTES 59

Proposition 1.10.7. Let S ∈ BSVloc(R+,L(X,Y )) and let S∞ be the strong limitof S(t) as t→∞ if it exists, and S∞ := 0 otherwise. Then

a) limt→∞∫ t

0e−λs dS(s) exists in operator norm whenever Reλ > abs(dS),

b) abs(dS) = ω(S − S∞).

Proof. If∫ t

0e−λ0s dS(s) converges strongly as t→∞, then it is uniformly bounded

in operator norm. Thus, a) follows from (1.23). Hence,

abs(dS) = inf

{Re λ :

∫ t

0

e−λs dS(s) converges in norm as t→∞}

= ω(S − S∞),

by Theorem 1.10.6, where S∞ is the norm limit of S(t) as t → ∞, if this exists,

S∞ := 0 otherwise. It is trivial that ω(S − S∞) = ω(S − S∞), so b) is proved.

1.11 Notes

Section 1.1The Bochner integral is an extension of the Lebesgue integral to functions with valuesin Banach spaces. Introduced around 1930 by Bochner, it has become a widely usedintegral in infinite dimensional applications. Much of Section 1.1 follows the treatmentof the Bochner integral in Chapter III of [HP57], where many references to the originalliterature can be found. Comprehensive treatments of the Bochner integral and vector-valued measures, as well as references to the literature are contained in the monograph[DU77] by Diestel and Uhl. Corollary 1.1.3 is taken from [Are01].

One reason for introducing the Riemann integral here is that increasing functionswith values in an ordered Banach space with normal cone are always Riemann integrable(Corollary 1.9.6). However, if the space is not separable, Riemann integrable functions arenot necessarily Bochner integrable and their antiderivative may be nowhere differentiable.One way to circumvent these difficulties is to consider generalizations of the Riemannintegral. This allows a version of the fundamental theorem of calculus where all continuousfunctions f : [0, 1] → X (where X is a Banach space) with f(0) = 0 are differentiablein the mean and coincide with the generalized Riemann integral of their derivatives (see[BLN99]).

Section 1.2The Radon-Nikodym property was identified in the 1970s as an important property inthe theory of vector measures and also in the geometry of Banach spaces. Our treatmentis based on [DU77].

Section 1.3Most of the results are vector-valued versions of standard material. Proposition 1.3.7 iscontained in [KH89].

Section 1.4The Laplace transform has a long history, dating back to Euler’s paper ‘De constructione


aequationum’ from 1737, Lagrange’s ‘Memoire sur l’utilite de la methode de prendre lemilieux entre les resultats de plusieurs observations’ from 1773, and Laplace’s ‘Memoiresur les approximations des formules qui sont finctions de tres grands nombres’ from1785. Since then it has been widely used in mathematics and engineering (in particularin ordinary differential, difference and functional equations, electrical engineering andapplications to signal processing problems). Modern Laplace transform theory began toemerge at the end of the 19th century when Heaviside popularized a user-friendly andpowerful operational calculus within the engineering community in connection with hisresearch in electromagnetism [Hea93]. Since his methods were to a large degree based onpurely formal operations with few mathematical justifications, many mathematicians atthe beginning of the 20th century began to strive for a solid mathematical foundationof Heaviside’s operational calculus by virtue of the Laplace transform. These effortsculminated in Widder’s books ‘The Laplace Transform’ [Wid41] and ‘An Introduction toTransform Theory’ [Wid71] as well as Doetsch’s ‘Theorie und Anwendung der Laplace-Transformation’ [Doe37] and his monumental, three volumed ‘Handbuch der Laplace-Transformation’ [Doe50]. These monographs have been among the best introductionsto the subject and have become classic texts. A first comprehensive look at Laplacetransform theory for functions with values in a Banach space X is contained in Hille’smonograph ‘Functional Analysis and Semi-Groups’ from 1948 [Hil48]. Many historicalnotes on Laplace transform theory can be found in the books of Doetsch and in surveyarticles by Deakin [Dea81], [Dea82], and Martis in Biddau [Bid33].

One weakness of Laplace transform theory—compared to Heaviside’s operationalcalculus—are the restrictions concerning the growth of the functions at infinity. To removethese restrictions, Vignaux introduced in 1939 an asymptotic version of the Laplacetransform [Vig39], [VC44]. For an extension of the asymptotic Laplace transform tovector-valued functions, and references to the literature, see [LN99] and [LN01].

The characterization of the abscissa of convergence by the exponential growth ofthe antiderivative are vector-valued versions of classical results due to Landau (1906) andPincherle (1913) that can be found in [Doe50, Volume I,Theorems 2.2.7 and 2.2.8], or[HP57, Section 1.6.2].

Section 1.5Theorem 1.5.1 is due to Pincherle and Landau (1905); Theorem 1.5.3 is due to Landau(1906). The proofs given here, as well as Example 1.5.2, follow [Doe50, Volume I, Sections3.2–3.4] where further references to the classical literature can be found.

Section 1.6The results of this section are straightforward vector-valued versions of standard resultsin the classical theory of Laplace transforms (see [Doe50, Volume I, Sections 2.14, 2.15],for example).

Section 1.7Theorem 1.7.5 has been proved by Ti-Jun Xiao and Jin Liang [XL00], but a special casewas given by Kurtz [Kur69] and the general result was mentioned by Chernoff [Che74,p.106]. In fact, the proof of [Che68, Proposition] carries over to the case considered inTheorem 1.7.5. The short proof given here appeared in [Bob97b] and [Are01].

The Inversion Theorem 1.7.7 is due to Post (1930) and Widder (1934) (see [Wid41,Section 7.6], and [Doe50, Volume I, Section 8.2]). Numerically more efficient Post-Widdertype inversion formulas for the Laplace transform can be found in [Jar08, Theorem 4.1]

1.11. NOTES 61

and [JNO08]. They are derived from rational approximation methods for operator semi-groups developed by Hersh and Kato [HK79] and Brenner and Thomee [BT79] (see also[Kov07]).

The Uniqueness Theorem 1.7.3 was mentioned first by Pastor [Pas19] in 1919 andis a special case of the following result of Shen [She47] (see also [Doe50, Volume I, Section2.9], [BN94], [Mih09]).

Theorem 1.11.1. Let f ∈ L1loc(R+, X) with abs(f) < ∞. Let (λn) be an infinite sequence

with no accumulation point and Reλn ≥ a > 0 for all n ∈ N and some a > abs(f). If

∞∑n=1

1−∣∣∣∣λn − 1

λn + 1

∣∣∣∣ = ∞,

then (λn)n∈N is a uniqueness sequence; i.e., f(λn) = 0 (n ∈ N) implies that f = 0.Conversely, let (λn) be a sequence with λn �= 0 and | arg(λn)| ≤ θ < π/2 (n ∈ N).

If (λn) has no accumulation point and the sum above is finite, then there exists 0 �= f ∈L1

loc(R+, X) with f(λn) = 0 for all n ∈ N.

Consider the horizontal sequences λn := a + nγb for a, b > 0. If 0 < γ ≤ 1 orγ < 0, then (λn) is a uniqueness sequence. If γ > 1 or a = 0 and γ < −1, then {λn}is the set of zeros of a non-trivial Laplace transform. For example, if f(t) = 1√

tsin( 1

t),

then f(λ) =√

πλe−√

2λ sin(√2λ) which has zeros for λn = 2n2π2 (n ∈ N). The vertical

sequences λn = 1 + inγ are uniqueness sequences if 0 < γ ≤ 12. If γ > 1

2, then {λn} is

the set of zeros of a non-trivial Laplace transform.

Uniqueness sequences are important in the discussion of Cauchy problems which arewell-posed in the regularized sense (see [Bau01] or [LN99] for definitions and references).

Section 1.8For 1 ≤ p ≤ 2, a Banach space X is said to have Fourier type p if the Fourier transformon L1(R, X)∩Lp(R, X) extends to a bounded linear operator of Lp(R, X) into Lp′(R, X),where 1/p+1/p′ = 1. The Hausdorff-Young inequalities show that C has Fourier type forevery p ∈ [1, 2]. Every Banach space has Fourier type 1, and a space with Fourier typep also has Fourier type q whenever 1 ≤ q ≤ p. Theorem 1.8.2 shows that Hilbert spaceshave Fourier type 2, and conversely Kwapien [Kwa72] showed that a space with Fouriertype 2 is isomorphic to a Hilbert space. A space of the form Lp(Ω, μ), where 1 ≤ p < ∞and (Ω, μ) is any measure space, has Fourier type min(p, p′). The spaces with non-trivialFourier type (i.e., Fourier type p for some p > 1) have been characterized by Bourgain[Bou82], [Bou88] (see also [Pis86]). Every superreflexive space (a Banach space with anequivalent uniformly convex norm) has non-trivial Fourier type, but there exist reflexivespaces which do not have non-trivial Fourier type and there exist non-reflexive spaceswhich do have non-trivial Fourier type.

A Banach space X is said to have the analytic Radon-Nikodym property (ARNP)if each function g ∈ H2(R, X) has a boundary function, i.e. limα↓0 g(α+ is) exists s-a.e.This property was first considered by Bukhvalov [Buk81], [BD82] using functions on theunit disc rather than C+, and Hp-spaces for p �= 2, but this formulation is equivalent.Thus, Theorem 1.8.3 shows in particular that Hilbert spaces have the (ARNP). Everyreflexive space has the (ARNP), and more generally, any space with the Radon-Nikodymproperty, and also any space of the form L1(Ω, μ), has the (ARNP). On the other hand,


c0 does not have the (ARNP), and there exist spaces with non-trivial Fourier type whichdo not have the (ARNP) (see [HN99]).

Section 1.9This section contains some of the basic properties of the Riemann-Stieltjes integral; see[HP57] and [Wid41] for further details and references to the original literature.

Section 1.10In the classical Laplace transform literature, many authors preferred Laplace-Stieltjesintegrals

∫∞0

e−λt dF (t) since they include Laplace integrals∫∞0

e−λtf(t) dt (when F is

differentiable a.e.) and Dirichlet series∑∞

i=1 aie−λti (when F is a step function). The

importance of the Laplace-Stieltjes integral for our purposes is that many classical re-sults for Laplace-Stieltjes integrals of complex-valued functions F can be extended tofunctions with values in arbitrary Banach spaces X, whereas vector-valued extensions ofthe corresponding Laplace transform results often require additional assumptions on X(see, for example, [Zai60]). All results of this section are vector-valued versions of classicalresults for Laplace-Stieltjes transforms in [Wid41] or [Wid71].

Chapter 2

The Laplace Transform

In this chapter the emphasis of the discussion shifts from Laplace integrals f(λ) and

dF (λ) to the Laplace transform L : f �→ f and to the Laplace-Stieltjes transform

LS : F �→ dF . The Laplace transform is considered first as an operator acting onL∞(R+, X) and the Laplace-Stieltjes transform as an operator on

Lip0(R+, X) :=

{F : R+ → X : F (0) = 0, ‖F‖Lip0(R+,X) :=

supt,s≥0

‖F (t)− F (s)‖|t− s| <∞

}.

These domains of L and LS are relatively easy to deal with and have immediateand important applications to abstract differential and integral equations.

The following observation is the key to one of the basic structures of Laplacetransform theory. If f ∈ L∞(R+, X), then t �→ F (t) :=

∫ t

0f(s) ds belongs to

Lip0(R+, X) and

L(f)(λ) =∫ ∞

0

e−λtf(t) dt =

∫ ∞

0

e−λt dF (t) = TF (e−λ),

where TF : g �→ ∫∞0

g(s)dF (s) is a bounded linear operator from L1(R+) into

X, and where e−λ denotes the exponential function t �→ e−λt. The operatorTF is fundamental to Laplace transform theory. In Section 2.1 it is shown thatΦS : F �→ TF is an isometric isomorphism between Lip0(R+, X) and L(L1(R+), X)(Riesz-Stieltjes representation theorem). This representation is crucial for the fol-lowing reason. The main purpose of Laplace transform theory is to translate prop-erties of the generating function F into properties of the resulting function λ �→r(λ) =

∫∞0

e−λt dF (t) and vice versa. Since F (t) = TFχ[0,t] =∫∞0

χ[0,t](s) dF (s)

and r(λ) = TF e−λ =∫∞0

e−λs dF (s), the generating function F as well as theresulting function r are evaluations of the same bounded linear operator acting ondifferent total subsets of L1(R+).

W. Arendt et al., Vector-valued Laplace Transforms and Cauchy Problems: Second Edition, 63Monographs in Mathematics 96, DOI 10.1007/978-3-0348-0087-7_2, © Springer Basel AG 2011

64 2. THE LAPLACE TRANSFORM

In Section 2.2, the range of the Laplace-Stieltjes transform acting onLip0(R+, X) is characterized. It is shown that a function r : R+ → X has aLaplace-Stieltjes representation r = LS(F ) for some F ∈ Lip0(R+, X) if and onlyif r is a C∞-function whose Taylor coefficients satisfy the estimate

‖r‖W := supn∈N0

supλ>0

λn+1

n!‖r(n)(λ)‖ <∞. (2.1)

This can be rephrased by saying that the Laplace-Stieltjes transform is an isometricisomorphism between the Banach spaces Lip0(R+, X) and

C∞W ((0,∞), X) := {r ∈ C∞((0,∞), X) : ‖r‖W <∞}.If the Banach space X has the Radon-Nikodym property (see Section 1.2), then(and only then) “Widder’s growth conditions” (2.1) are necessary and sufficient forr to have a Laplace representation r = L(f) for some f ∈ L∞(R+, X); i.e., Banachspaces with the Radon-Nikodym property are precisely those Banach spaces inwhich the Laplace transform is an isometric isomorphism between L∞(R+, X) andC∞W ((0,∞), X). For X = C, this is a classical result usually known as “Widder’sTheorem”.

If r = LS(F ) for some F ∈ Lip0(R+, X), then the inverse Laplace-Stieltjestransform has many different representations. A few of them, such as

F (t) =1

2πi

∫Γ

eλtr(λ)

λdλ = lim

n→∞

∞∑j=1

(−1)j+1etnjr(nj)

= limk→∞

(−1)k 1

k!

(k

t

)k+1dk

dtk

(r(λ)

λ

) ∣∣∣∣λ=k/t

,

will be proved in Section 2.3.In Section 2.4, the results of the previous sections are extended to functions

with exponential growth at infinity; i.e., we investigate the Laplace transformacting on functions f with ess supt≥0 ‖e−ωtf(t)‖ <∞.

In applications it is usually impossible to verify whether or not a givenfunction r satisfies Widder’s growth conditions (2.1). Thus, in Sections 2.5 and2.6 some complex growth conditions are discussed which are necessary (and ina certain sense sufficient) for a holomorphic function r : {Re λ > ω} → X tohave a Laplace representation. In Section 2.5, the growth condition considered issupReλ>ω ‖λ1+br(λ)‖ <∞ for some b > 0.

In Section 2.6, we discuss functions r which are holomorphic in a sectorΣ := {| arg(λ)| < π

2 + ε} and satisfy supλ∈Σ ‖λr(λ)‖ < ∞. We will see thatany such r is the Laplace transform of a function which is holomorphic in thesector {| arg(λ)| < ε}. The final class of functions which we will consider are thecompletely monotonic ones; i.e., C∞-functions r with values in an ordered Banachspace such that (−1)nr(n)(λ) ≥ 0 for all n ∈ N0 and λ > ω. In the scalar case,

2.1. RIESZ-STIELTJES REPRESENTATION 65

Bernstein’s theorem states that a function r is completely monotonic if and onlyif it is the Laplace-Stieltjes transform of an increasing function. In Section 2.7 weinvestigate for which ordered Banach spaces Bernstein’s theorem holds.

2.1 Riesz-Stieltjes Representation

In the following sections the emphasis will be on the properties of the Laplace

transform L : f �→ f and the Laplace-Stieltjes transform LS : F �→ dF . As isthe case with all linear operators, the choice of the domain is crucial. For theLaplace-Stieltjes transform LS the most convenient choice of the domain space is

Lip0(R+, X) :=

{F : R+ → X : F (0) = 0, ‖F‖Lip0(R+,X) :=

supt,s≥0

‖F (t)− F (s)‖|t− s| <∞

}.

If F (t) =∫ t

0f(s) ds for f ∈ L∞(R+, X), then F ∈ Lip0(R+, X) and∫ ∞

0

e−λt dF (t) =

∫ ∞

0

e−λtf(t) dt (λ > 0),

by Proposition 1.10.1. Thus, any result for LS acting on Lip0(R+, X) translatesinto one for L acting on L∞(R+, X). However, since there are Banach spacesin which not every Lipschitz continuous function is the antiderivative of an L∞-function (see Section 1.2), the Laplace-Stieltjes transform is a true generalizationof the Laplace transform. It is the generalization needed to deal effectively withLaplace transforms of vector-valued functions.

In this section we investigate the Riesz-Stieltjes operator ΦS which assignsto F ∈ Lip0(R+, X) a bounded linear operator TF : L1(R+)→ X such that

TF f :=

∫ ∞

0

f(s) dF (s) := limτ→∞

∫ τ

0

f(s) dF (s),

when f ∈ L1(R+) is continuous. It will be shown that ΦS is an isometric isomor-phism between Lip0(R+, X) and L(L1(R+), X), the space of all bounded linearoperators from the Banach space L1(R+) into X (Riesz-Stieltjes representation).This observation is fundamental for the whole chapter. To see why the Riesz-Stieltjes representation is such an important tool, observe that

F (t) = TFχ[0,t] (t ≥ 0) , and dF (λ) = TF e−λ (λ > 0).

Thus, if one knows F , then the operator TF is specified on the set of characteristicfunctions χ[0,t] (t > 0), which is total in L1(R+). Therefore, TF and, in partic-

ular, the Laplace integrals TF e−λ = dF (λ) (λ > 0) are completely determined.

Conversely, the Laplace integrals dF (λ) determine TF on the set of exponentialfunctions e−λ (λ > 0), which is also total in L1(R+) (Lemma 1.7.1). Hence, the

Laplace integrals dF (λ) determine the properties of TF and, in particular, theproperties of F (t) = TFχ[0,t] (t ≥ 0).

Theorem 2.1.1 (Riesz-Stieltjes Representation). There exists a unique isometricisomorphism ΦS : F �→ TF from Lip0(R+, X) onto L(L1(R+), X) such that

TFχ[0,t] = F (t) (2.2)

for all t ≥ 0 and F ∈ Lip0(R+, X). Moreover,

TF g = limt→∞

∫ t

0

g(s) dF (s) :=

∫ ∞

0

g(s) dF (s) (2.3)

for all continuous functions g ∈ L1(R+).

Note that it is part of the claim that the improper integral in (2.3) converges.We shall call the isomorphism ΦS the Riesz-Stieltjes operator.

Proof. Let D := span{χ[0,t) : t > 0}, the space of step functions, which is densein L1(R+). For each f ∈ D there exists a unique representation

f =

n∑i=1

αiχ[ti−1,ti),

where 0 = t0 < t1 < . . . < tn, αi ∈ C (i = 1, . . . , n). Let F ∈ Lip0(R+, X). DefineTF : D → X by

TF (f) = TF

(n∑

i=1

αiχ[ti−1,ti)

):=

n∑i=1

αi(F (ti)− F (ti−1)).

Then,

‖TF (f)‖ ≤ ‖F‖Lip0(R+,X)

n∑i=1

|αi|(ti − ti−1) = ‖F‖Lip0(R+,X)‖f‖1.

Hence, TF has a unique extension TF ∈ L(L1(R+), X). Moreover,

‖TF ‖ ≤ ‖F‖Lip0(R+,X).

Conversely, if T ∈ L(L1(R+), X), let F (t) := Tχ[0,t) for t ≥ 0. Then fort > s ≥ 0,

‖F (t)− F (s)‖ = ‖Tχ[s,t)‖ ≤ ‖T‖ ‖χ[s,t)‖1 = ‖T‖(t− s).

2.1. RIESZ-STIELTJES REPRESENTATION 67

Thus, F ∈ Lip0(R+, X) and ‖F‖Lip0(R+,X) ≤ ‖T‖. It follows from the definitionsthat T = TF and if T = TG then F = G. This shows that F �→ TF is an isometricisomorphism.

Finally, let g ∈ L1(R+) be a continuous function and let F ∈ Lip0(R+, X).Take t > 0, and let π be a partition of [0, t] with partitioning points 0 = t0 < t1 <. . . < tn = t and intermediate points si ∈ [ti−1, ti]. Let

fπ :=n∑

i=1

g(si)χ[ti−1,ti).

Thus, S(g, F, π) = TF (fπ). As |π| → 0, ‖fπ − gχ[0,t)‖1 → 0, so∫ t

0

g(s) dF (s) = TF (gχ[0,t)).

As t→∞, ‖gχ[0,t) − g‖1 → 0, so∫ ∞

0

g(s) dF (s) = TF (g).

We conclude this section by discussing convergence of functions and theirLaplace-Stieltjes transforms. In fact, the Laplace-Stieltjes transform allows us togive a purely operator-theoretic proof of the following approximation theorem.Note, however, that the essential implication (i) ⇒ (iv) can also be obtained withthe help of Theorem 1.7.5 (which may easily be strengthened by merely consideringconvergence on a sequence of equidistant points).

Theorem 2.1.2. Let M > 0, Fn ∈ Lip0(R+, X) with ‖Fn‖Lip0(R+,X) ≤ M for alln ∈ N, and rn = LS(Fn). The following are equivalent:

(i) There exist a, b > 0 such that limn→∞ rn(a+ kb) exists for all k ∈ N0.

(ii) There exists r ∈ C∞((0,∞), X) such that rn → r uniformly on compactsubsets of (0,∞).

(iii) limn→∞ Fn(t) exists for all t ≥ 0.

(iv) There exists F ∈ Lip0(R+, X) such that Fn → F uniformly on compactsubsets of R+.

Moreover, if r and F are as in (ii) and (iv), then r = LS(F ).

Proof. By the Riesz-Stieltjes Representation Theorem 2.1.1, there exist Tn ∈L(L1(R+), X) such that ‖Tn‖ = ‖Fn‖Lip0(R+,X) ≤ M, Tne−λ = rn(λ), andTnχ[0,t] = Fn(t) (n ∈ N, t ≥ 0, λ > 0). Each of the statements imply that theuniformly bounded family of operators Tn converges on a total subset of L1(R+)

(see also Lemma 1.7.1). By equicontinuity (see Proposition B.15), for any uni-formly bounded sequence of operators, the topology of simple convergence on atotal subset equals the topology of simple convergence and the topology of uniformconvergence on compact subsets. Thus there exists T ∈ L(L1(R+), X) such thatTng → Tg as n→∞ for all g ∈ L1(R+) (simple convergence). For all b > 0 the setsKb := {χ[0,t] : 0 ≤ t ≤ b} and Eb := {e−λ : 1

b≤ λ ≤ b} are compact in L1(R+)

(continuous images of compact sets are compact). Hence, Tn → T uniformly onKb and Eb (uniform convergence on compact subsets). Now the statements followfrom the Riesz-Stieltjes representation.

2.2 A Real Representation Theorem

In this section the range of the Laplace-Stieltjes transform LS : F �→ dF acting

on Lip0(R+, X) will be characterized. Since λ �→ dF (λ) = λF (λ) is holomorphicand, by Proposition 1.7.2, functions like λ �→ (sinλ)x (x ∈ X) cannot be in therange of LS , the range must be a proper subset of C∞((0,∞), X). The followingobservations will lead to a complete description of the range.

Let F ∈ Lip0(R+, X) and TF := ΦS(F ), where ΦS is the Riesz-Stieltjesoperator of Section 2.1. Define

r(λ) := dF (λ) =

∫ ∞

0

e−λt dF (t) (λ > 0).

Then, by Theorem 1.10.6, r ∈ C∞((0,∞), X) and

r(n)(λ) =

∫ ∞

0

e−λt(−t)n dF (t) = TFkn,λ,

where kn,λ(t) := e−λt(−t)n (t ≥ 0, λ > 0, n ∈ N0). Since ‖kn,λ‖1 =∫∞0

e−λttn dt= n!/λn+1 and ‖TF ‖ = ‖F‖Lip0(R+,X), it follows that

‖r(n)(λ)‖ ≤ ‖F‖Lip0(R+,X)n!/λn+1

for all n ∈ N0 and λ > 0. Thus, r is a C∞-function whose Taylor coefficients satisfy

‖r‖W := supλ>0,k∈N0

λk+1

k!‖r(k)(λ)‖ ≤ ‖F‖Lip0(R+,X).

This shows that the Laplace-Stieltjes transform LS : F → dF maps Lip0(R+, X)into the space

C∞W ((0,∞), X) := {r ∈ C∞((0,∞), X) : ‖r‖W <∞}.In 1936, Widder showed that the Laplace transform maps L∞(R+,R) ontoC∞W ((0,∞),R). The following result is the vector-valued version of Widder’s clas-sical theorem.

2.2. A REAL REPRESENTATION THEOREM 69

Theorem 2.2.1 (Real Representation Theorem). The Laplace-Stieltjes transformLS is an isometric isomorphism between Lip0(R+, X) and C∞W ((0,∞), X).

Proof. We have already shown that LS maps Lip0(R+, X) into C∞W ((0,∞), X) and

that ‖LS(F )‖W ≤ ‖F‖Lip0(R+,X). If LS(F ) = dF = 0 for some F ∈ Lip0(R+, X),

then TF e−λ =∫∞0

e−λt dF (t) = dF (λ) = 0 for all λ > 0. Since the exponentialfunctions e−λ (λ > 0) are total in L1(R+) (Lemma 1.7.1), it follows that TF = 0.In particular, TFχ[0,t] = F (t) = 0 for all t ≥ 0. Thus, LS is one-to-one.

The hard part of the proof is to show that LS is onto. Let r ∈ C∞W ((0,∞), X).Define Tk ∈ L(L1(R+), X) by

Tkf :=

∫ ∞

0

f(t)(−1)k 1

k!

(k

t

)k+1

r(k)(k

t

)dt (k ∈ N0).

The operators Tk are uniformly bounded by ‖r‖W since ‖Tkf‖ ≤ ‖r‖W ‖f‖1 forall f ∈ L1(R+). We will show below that Tke−λ → r(λ) as k → ∞ for all λ > 0.Since the exponential functions e−λ (λ > 0) are total in L1(R+) it then followsfrom Proposition B.15 that there exists T ∈ L(L1(R+), X) with ‖T‖ ≤ ‖r‖W suchthat Tkf → Tf for all f ∈ L1(R+). In particular,

r(λ) = limk→∞

Tke−λ = Te−λ.

The Riesz-Stieltjes Representation Theorem 2.1.1 then yields the existence of someF ∈ Lip0(R+, X) with ‖F‖Lip0(R+,X) = ‖T‖ ≤ ‖r‖W such that Tg =

∫∞0

g(t) dF (t)for all continuous functions g ∈ L1(R+). Hence, for all λ > 0,

r(λ) = Te−λ =

∫ ∞

0

e−λt dF (t) = dF (λ).

Thus, LS is onto and ‖LS(F )‖W = ‖dF‖W = ‖F‖Lip0(R+,X) for F ∈ Lip0(R+, X).It remains to be shown that Tke−λ → r(λ) as k →∞ for all λ > 0. Observe

that

Tke−λ =

∫ ∞

0

e−λt(−1)k 1

k!

(k

t

)k+1

r(k)(k

t

)dt

= (−1)k 1

(k − 1)!

∫ ∞

0

(e−λk/uuk−1

)r(k)(u) du

= (−1)k 1

(k − 1)!

[k−1∑j=0

(−1)j dj

duj

(e−λk/uuk−1

)r(k−j−1)(u)

∣∣∣∣∣∞

u=0

+(−1)k∫ ∞

0

dk

duk

(e−λk/uuk−1

)r(u) du

].

To discuss the derivatives of u �→ e−λk/uuk−1, define G(x, u) := e−x/u(ux

)k−1.

Then G(sx, su) = G(x, u) for all s > 0. Differentiating both sides of the last


equality with respect to s and then setting s = 1 yields x∂G∂x (x, u)+u∂G

∂u (x, u) = 0

or 1x

∂G∂u (x, u) = − 1

u∂G∂x (x, u). This implies that

∂

∂u

(e−x/uu

k−1

xk

)= − ∂

∂x

(e−x/uu

k−2

xk−1

).

By induction on j, it follows that

∂j

∂uj

(e−x/uu

k−1

xk

)= (−1)j ∂j

∂xj

(e−x/uu

k−j−1

xk−j

)(0 ≤ j ≤ k),

or∂j

∂uj

(e−x/uuk−1

)= (−1)jxkuk−j−1 ∂j

∂xj

(e−x/u

xk−j

). (2.4)

Hence,

h(u) :=k−1∑j=0

(−1)j ∂j

∂uj

(e−x/uuk−1

)r(k−j−1)(u)

=k−1∑j=0

xk ∂j

∂xj

(e−x/u

xk−j

)uk−j−1r(k−j−1)(u).

Since

‖uk−j−1r(k−j−1)(u)‖ ≤ ‖r‖W (k − j − 1)!

u,

one obtains that

‖h(u)‖ ≤k−1∑j=0

‖r‖W (k − j − 1)!

uxk

∣∣∣∣ ∂j

∂xj

(e−x/u

xk−j

)∣∣∣∣ .It follows that limu→∞ h(u) = 0 = limu→0 h(u). Therefore, letting x = λk,

Tke−λ =1

(k − 1)!

∫ ∞

0

dk

duk

(e−λk/uuk−1

)r(u) du.

Since by (2.4),

∂k

∂uk

(e−x/uuk−1

)= (−1)k x

k

u

∂k

∂xk

(e−x/u

)=

xk

uk+1e−x/u,

it follows that

Tke−λ =λkkk

(k − 1)!

∫ ∞

0

e−λk/u 1

uk+1r(u) du

=λkkk+1

k!

∫ ∞

0

e−λkttk−1r

(1

t

)dt.

2.2. A REAL REPRESENTATION THEOREM 71

Define f(t) := 1t r(

1t ) and s := 1

λ . Then

Tke−λ =s

k!

(k

s

)k+1 ∫ ∞

0

e−kt/stkf(t) dt

= s(−1)k 1

k!

(k

s

)k+1

f (k)

(k

s

).

Finally, one concludes from the Post-Widder Inversion Theorem 1.7.7 that

limk→∞

Tke−λ = sf(s) = r

(1

s

)= r(λ)

for all λ > 0.

For later use in Section 2.5, we observe that in the Widder conditions it isnot necessary to consider all values of k.

Proposition 2.2.2. Let r ∈ C∞((0,∞), X), and suppose that limλ→∞ r(λ) = 0 and

there exist M > 0 and infinitely many integers m such that supλ>0 ‖λm+1

m! r(m)(λ)‖≤M . Then r ∈ C∞W ((0,∞), X) and ‖r‖W ≤M .

Proof. It suffices to show that if ‖r(m)(λ)‖ ≤ Mm!/λm+1, for all λ > 0, then‖r(k)(λ)‖ ≤Mk!/λk+1 for all λ > 0 and 0 ≤ k < m. Let

r(λ) :=(−1)m(m− 1)!

∫ ∞

λ

(λ− μ)m−1r(m)(μ) dμ.

Note that the integral is absolutely convergent, r(m)(λ) = r(m)(λ), and the sub-stitution t = λ/μ gives

‖r(λ)‖ ≤Mm

∫ ∞

λ

(μ− λ)m−1

μm+1dμ =

Mm

λ

∫ 1

0

(1− t)m−1 dt =M

λ.

Hence r − r is a polynomial and limλ→∞(r − r)(λ) = 0, so r = r. It follows that

‖r(k)(λ)‖ =

∥∥∥∥ (−1)m(m− k − 1)!

∫ ∞

λ

(λ− μ)m−k−1r(m)(μ) dμ

∥∥∥∥≤ Mm!

(m− k − 1)!

∫ ∞

λ

(μ− λ)m−k−1

μm+1dμ

=Mk!

λk+1

for λ > 0 and 0 ≤ k < m.

Now it will be shown that the Laplace transform is an isometric isomorphismbetween L∞(R+, X) and C∞W ((0,∞), X) if and only if the Banach space X has the


Radon-Nikodym property. Recall from Section 1.2 that X has the Radon-Nikodymproperty if every F ∈ Lip0(R+, X) is differentiable a.e., or equivalently if everyabsolutely continuous function F : R+ → X is differentiable a.e. As shown inTheorem 1.2.6 and Corollary 1.2.7, every separable dual space (for example, l1)and every reflexive Banach space have the Radon-Nikodym property. However,L1(R+) and c0 do not have the property (Propositions 1.2.9 and 1.2.10).

Theorem 2.2.3. Let X be a Banach space. The following are equivalent:

(i) X has the Radon-Nikodym property.

(ii) The Laplace transform L : f �→ f is an isometric isomorphism betweenL∞(R+, X) and C∞W ((0,∞), X).

(iii) The Riesz operator Φ : f �→ Rf , Rfg :=∫∞0

g(t)f(t) dt is an isometricisomorphism between L∞(R+, X) and L(L1(R+), X).

Proof. Define the normalized antiderivative I : L∞(R+, X) → Lip0(R+, X) by

I(f) := F , F (t) :=∫ t

0f(s) ds (t ≥ 0). Then I is one-to-one and ‖I(f)‖Lip0(R+,X) ≤

‖f‖∞ for all f ∈ L∞(R+, X). If I is onto, thenX has the Radon-Nikodym property(see Proposition 1.2.2). Conversely, if X has the Radon-Nikodym property andF ∈ Lip0(R+, X) then f(t) := F ′(t) exists for almost all t ≥ 0. Since f(t) =

limh→0F (t+h)−F (t)

h a.e., one concludes that ‖f‖∞ ≤ ‖F‖Lip0(R+,X). In particular,f ∈ L∞(R+, X) and by Proposition 1.2.3, F = I(f). Thus X has the Radon-Nikodym property if and only if I is an isometric isomorphism.

The Riesz-Stieltjes operator ΦS : F �→ TF , where TF g =∫∞0

g(t) dF (t) forall continuous g ∈ L1(R+), is an isometric isomorphism between Lip0(R+, X) andL(L1(R+), X), and the Laplace-Stieltjes transform

LS : F �→ dF , dF (λ) =

∫ ∞

0

e−λt dF (t),

is an isometric isomorphism between Lip0(R+, X) and C∞W ((0,∞), X). When F =I(f), TF g =

∫∞0

g(t)f(t) dt for all g ∈ L1(R+), by Proposition 1.9.11 and conti-nuity in L1-norm. Now the statements follow from the fact that Φ = ΦS ◦ I andL = LS ◦ I on L∞(R+, X).

Example 2.2.4. a) Consider X = L1(R+). Let F (t) := χ[0,t] (t ≥ 0) and r(λ) :=

e−λ (Reλ > 0), where e−λ(t) = e−λt. Then F ∈ Lip0(R+, L1(R+)) and

r(λ) =

∫ ∞

0

e−λt dF (t) = dF (λ).

Since F is nowhere differentiable (see Proposition 1.2.10), there does not existf ∈ L∞(R+, L

1(R+)) such that

r(λ) =

∫ ∞

0

e−λtf(t) dt.

2.3. REAL AND COMPLEX INVERSION 73

b) Consider C0(R+) as a subspace of L∞(R+). Define F : R+ → C0(R+) byF (t)(s) := (t − s)χ[0,t](s), and f : R+ → L∞(R+) by f(t) := χ[0,t]. Then F ∈Lip0(R+, C0(R+)) and F (t) =

∫ t

0f(s) ds as a Riemann integral in L∞(R+), but F

is nowhere differentiable and f is not measurable (see Examples 1.2.8 and 1.9.7).Moreover,

1

λe−λ =

∫ ∞

0

e−λt dF (t) =

∫ ∞

0

e−λtf(t) dt

as (improper) Riemann-Stieltjes and Riemann integrals, but λ �→ 1λe−λ is not the

Laplace transform of any function in L1(R+, L∞(R+)).

2.3 Real and Complex Inversion

We have shown in Section 2.2 that the Laplace-Stieltjes transform LS is an iso-metric isomorphism between Lip0(R+, X) and C∞W ((0,∞), X). In this section wewill derive several representations of the inverse Laplace-Stieltjes transform L−1

S .

Theorem 2.3.1 (Post-Widder Inversion). Let F ∈ Lip0(R+, X), r = LS(F ), andt > 0. Then

F (t) = limk→∞

(−1)k 1

k!

(k

t

)k+1dk

dλk

(r(λ)

λ

) ∣∣∣∣λ=k/t

.

Proof. Since ω(F ) ≤ 0 and F (0) = 0, it follows from (1.22) that

r(λ)

λ=

∫ ∞

0

e−λtF (t) dt

for all λ > 0, where the integral is an absolutely convergent Bochner integral. Nowthe statement follows from Theorem 1.7.7.

Applying Leibniz’s rule (f · r)(k) = ∑kj=0

(kj

)f (k−j)r(j) to f(λ) := 1

λand

r one can rewrite the Post-Widder inversion of the Laplace-Stieltjes transform as

F (t) = limk→∞

k∑j=0

(−1)j 1j!

(k

t

)j

r(j)(k

t

)(t > 0). (2.5)

Compared to the Post-Widder inversion, it is remarkable that in the followingPhragmen-Doetsch inversion formula only the values r(k) for large k ∈ N areneeded and that the convergence is uniform for all t ≥ 0.

Theorem 2.3.2 (Phragmen-Doetsch Inversion). Let F ∈ Lip0(R+, X) and r =LS(F ). Then ∥∥∥∥∥∥F (t)−

∞∑j=1

(−1)j+1

j!etkjr(kj)

∥∥∥∥∥∥ ≤ c

k‖r‖W

for all t ≥ 0 and k ∈ N, where c ≈ 1.0159..., and ‖r‖W = ‖F‖Lip0(R+,X).

Proof. By the Riesz-Stieltjes Representation Theorem 2.1.1 and the Real Rep-resentation Theorem 2.2.1, there exists T ∈ L(L1(R+), X) such that r(λ) =∫∞0

e−λt dF (t) = Te−λ (λ > 0), Tχ[0,t] = F (t) (t ≥ 0) and ‖T‖ = ‖r‖W =‖F‖Lip0(R+,X). Thus,∥∥∥∥∥∥F (t)−

∞∑j=1

(−1)j+1

j!etkjr(kj)

∥∥∥∥∥∥ ≤ ‖T‖∥∥∥∥∥∥χ[0,t] −

∞∑j=1

(−1)j+1 1

j!etkje−kj

∥∥∥∥∥∥1

.

Define pk,t(s) := 1− e−ek(t−s)

=∑∞

j=1(−1)j+1 1j!etkje−kj(s). Then,

‖χ[0,t] − pk,t‖1 =

∫ t

0

|pk,t(s)− 1| ds+∫ ∞

t

|pk,t(s)| ds

=

∫ t

0

e−ek(t−s)

ds+

∫ ∞

t

(1− e−ek(t−s)

)ds

=1

k

∫ ekt

1

e−u

udu+

1

k

∫ 1

0

1− e−u

udu

≤ 1

k

(∫ ∞

1

e−u

udu+

∫ 1

0

1− e−u

udu

)for all t ≥ 0 and k ∈ N. Now the claim follows from the fact that

∫∞1

1ue−u du +∫ 1

01−e−u

u du = −2Ei(−1) + γ ≈ 1.0159..., where Ei(z) is the exponential integraland γ is Euler’s constant (see [Leb72, Section 3.1]).

The following corollary shows that the Phragmen-Doetsch inversion is invari-ant under exponentially decaying perturbations for small values of t.

Corollary 2.3.3. Let F ∈ Lip0(R+, X), r = LS(F ), and q(λ) = r(λ)+a(λ) (λ > 0),where a : (0,∞) → X is a function such that lim supn→∞

1n log ‖a(n)‖ ≤ −T for

some T > 0. Then

F (t) = limk→∞

∞∑j=1

(−1)j+1

j!etkjq(kj)

for all 0 ≤ t < T .

Proof. Let 0 < T0 < T and choose k0 such that ‖a(k)‖ ≤ e−T0k for all k ≥ k0.Then, ∥∥∥∥F (t)−

∞∑j=1

(−1)j+1

j!etkjq(kj)

∥∥∥∥≤

∥∥∥∥F (t)−∞∑j=1

(−1)j+1

j!etkjr(kj)

∥∥∥∥+

∥∥∥∥ ∞∑j=1

(−1)j+1

j!etkja(kj)

∥∥∥∥≤ 2

k‖r‖W +

∞∑j=1

1

j!etkje−T0kj ≤ 2

k‖r‖W + ee

−(T0−t)k − 1.

2.3. REAL AND COMPLEX INVERSION 75

The Post-Widder inversion and the Phragmen-Doetsch inversion are calledreal inversions of the Laplace-Stieltjes transform since they use only propertiesof r(λ) for large real λ. For the following complex inversion formula we use thefact that if r(λ) =

∫∞0

e−λt dF (t) (λ > 0) for some F ∈ Lip0(R+, X), then radmits a holomorphic extension for Reλ > 0 which we denote by the same sym-bol (see Theorem 1.10.6). We shall give here a proof based on the Riesz-Stieltjesrepresentation, but we shall give another, rather simple, proof in Section 4.2.

Theorem 2.3.4 (Complex Inversion). Let F ∈ Lip0(R+, X) and r = LS(F ). Then

F (t) = limk→∞

1

2πi

∫ c+ik

c−ik

eλtr(λ)

λdλ,

where the limit is uniform for t ∈ [0, a] for any a > 0, and c > 0 is arbitrary.

Proof. By the Riesz-Stieltjes Representation Theorem 2.1.1, there exists T ∈L(L1(R+), X) such that r(λ) = Te−λ (Reλ > 0) and F (t) = Tχ[0,t] (t ≥ 0).Thus,∥∥∥∥∥F (t)− 1

2πi

∫ c+ik

c−ik

eλtr(λ)

λdλ

∥∥∥∥∥ ≤ ‖T‖∥∥∥∥∥χ[0,t] − 1

2πi

∫ c+ik

c−ik

eλte−λ

λdλ

∥∥∥∥∥1

.

Now the statement follows from the next lemma.

Lemma 2.3.5. Let t ≥ 0 and a, c > 0. Then the functions

hk,t :=1

2πi

∫ c+ik

c−ik

eλte−λ

λdλ

converge towards χ[0,t] in L1(R+) as n→∞, uniformly for t ∈ [0, a].

Proof. Let ‖hk,t − χ[0,t]‖1 = Ak + Bk , where Ak :=∫ t

0|hk,t(s) − 1| ds and Bk :=∫∞

t|hk,t(s)| ds. We show first that limk→∞Ak = 0. The residue of the function

λ �→ eλ(t−s)/λ at the point 0 is 1. By Cauchy’s theorem,

hk,t(s)− 1 =1

2πi

(∫Γ+

−∫Γ−−∫Γ0

)eλ(t−s)

λdλ,

where Γ± := {λ : λ = u± ik; 0 ≤ u ≤ c}, Γ0 := {λ : λ = keiu; π/2 ≤ u ≤ 3π/2}.Along Γ+, and similarly along Γ−, it follows from 0 ≤ s ≤ t that∣∣∣∣∣

∫Γ+

eλ(t−s)

λdλ

∣∣∣∣∣ =∣∣∣∣∫ c

0

e(u+ik)(t−s)

u+ ikdu

∣∣∣∣ ≤ cec(t−s)

k.

Along Γ0, for 0 ≤ s < t,∣∣∣∣∫Γ0

eλ(t−s)

λdλ

∣∣∣∣ =∣∣∣∣∣∫ 3π/2

π/2

ek(t−s)eiu du

∣∣∣∣∣ ≤∫ 3π/2

π/2

ek(t−s) cosu du.


Hence,

Ak =

∫ t

0

|hk,t(t− s)− 1| ds

≤∫ t

0

(cecs

πk+

1

2π

∫ 3π/2

π/2

eks cosu du

)ds

→ 0

as k → ∞, uniformly for t ∈ [0, a] for all a > 0, by the monotone convergencetheorem, or by explicit estimation.

In order to estimate Bk, we define Γ± := {λ : λ = u± ik ; c ≤ u ≤ k}, Γ0 :={λ : λ = k

√2eiu ; −π/4 ≤ u ≤ π/4}. By Cauchy’s theorem,

hk,t(s) =1

2πi

(−∫˜Γ+

+

∫˜Γ−

+

∫˜Γ0

)eλ(t−s)

λdλ.

Along Γ+, and similarly along Γ−, it follows from s− t ≥ 0 that∣∣∣∣∣∫˜Γ+

eλ(t−s)

λdλ

∣∣∣∣∣ =

∣∣∣∣∣∫ k

c

e(u+ik)(t−s)

u+ ikdu

∣∣∣∣∣ ≤ 1

k

∫ k

c

e−u(s−t) du

=e−c(s−t) − e−k(s−t)

k(s− t).

Along Γ0,∣∣∣∣∫˜Γ0

eλ(t−s)

λdλ

∣∣∣∣ =

∣∣∣∣∣∫ π/4

−π/4

ek√2(t−s)eiu du

∣∣∣∣∣ ≤∫ π/4

−π/4

ek√2(t−s) cos u du

= 2

∫ π/4

0

ek√2(t−s) cos(u) du ≤ π

2ek√2(t−s) cos(π/4) =

π

2e−k(s−t).

Hence, for all t ≥ 0,∫ ∞

t

|hk,t(s)| ds ≤ 1

π

∫ ∞

t

e−c(s−t) − e−k(s−t)

k(s− t)ds +

1

4

∫ ∞

t

e−k(s−t) ds

=1

π

∫ ∞

0

zk(s) ds+1

4k,

where zk(s) := 1ks(e

−cs − e−ks) ≤ e−cs for k ≥ c by the mean value theoremapplied to e−x over [cs, ks]. By the dominated convergence theorem, or by explicitestimation, Bk → 0 as k →∞, uniformly for t ∈ [0, a] for all a > 0.

2.4. TRANSFORMS OF EXPONENTIALLY BOUNDED FUNCTIONS 77

2.4 Transforms of Exponentially Bounded Functions

So far in this chapter, Laplace transforms have been considered for bounded orglobally Lipschitz continuous functions. We shall now adapt the results of theprevious sections to functions with exponential growth at infinity, by an elemen-tary “shifting” procedure (see Proposition 1.6.1 a) and Proposition 1.10.3). Moreprecisely, for ω ∈ R we consider the Laplace-Stieltjes transform acting on

Lipω(R+, X) :=

{G : R+ → X : G(0) = 0,

‖G‖Lipω(R+,X) := supt>s≥0

‖G(t)−G(s)‖∫ t

seωr dr

<∞}

and the Laplace transform acting on

L∞ω (R+, X) :=

{g ∈ L1

loc(R+, X) : ‖g‖ω,∞ := ess supt≥0

‖e−ωtg(t)‖ <∞}.

It is easy to see that

‖G‖Lipω(R+,X) =

⎧⎪⎪⎨⎪⎪⎩sup0≤s<t

‖G(t)−G(s)‖(t− s)eωt

if ω ≥ 0,

sup0≤s<t

‖G(t)−G(s)‖(t− s)eωs

if ω ≤ 0.

It is clear that the multiplication operator Mω : g �→ e−ω·g(·) is an isometricisomorphism between L∞ω (R+, X) and L∞(R+, X), and we now set up the corre-sponding isomorphism between Lipω(R+, X) and Lip0(R+, X).

For G ∈ Lipω(R+, X) and f ∈ BSVloc(R+), it follows from the definition ofthe Riemann-Stieltjes integral that∥∥∥∥∥

∫ b

a

f(t) dG(t)

∥∥∥∥∥ ≤ ‖G‖Lipω(R+,X)

∫ b

a

|f(t)|eωt dt (0 ≤ a ≤ b). (2.6)

Let

(IωG)(t) :=

∫ t

0

e−ωs dG(s).

Then (2.6) implies that

IωG ∈ Lip0(R+, X) and ‖IωG‖Lip0(R+,X) ≤ ‖G‖Lipω(R+,X).

Similarly if F ∈ Lip0(R+, X) and

(JωF )(t) :=

∫ t

0

eωs dF (s),

then JωF ∈ Lipω(R+, X) and ‖JωF‖Lipω(R+,X) ≤ ‖F‖Lip0(R+,X). Moreover, JωIωG= G and IωJωF = F , by Proposition 1.9.10. Hence, Iω is an isometric isomorphismof Lipω(R+, X) onto Lip0(R+, X).

Note that if G ∈ L∞ω (R+, X) then ω(G) ≤ ω and abs(dG) ≤ ω by Theorem1.10.5. Thus, the Laplace-Stieltjes transform

(LS,ωG)(λ) := dG(λ) =

∫ ∞

0

e−λt dG(t)

exists for λ > ω. By Proposition 1.10.3,

(LS,ωG)(λ) = (LSIωG)(λ− ω). (2.7)

Let

C∞W ((ω,∞), X) :=

{r ∈ C∞((ω,∞), X) :

‖r‖W := supλ>ω,k∈N0

(λ− ω)k+1

k!‖r(k)(λ)‖ <∞

}.

This is a Banach space, and it is clear that the shift Sω : r �→ r(· − ω) is anisometric isomorphism of C∞W ((0,∞), X) onto C∞W ((ω,∞), X). The equation (2.7)may be written as LS,ω = Sω ◦ LS ◦ Iω.

Now we can give the following reformulation of the Real Representation The-orem 2.2.1.

Theorem 2.4.1. Let ω ∈ R. The Laplace-Stieltjes transform is an isometric iso-morphism of Lipω(R+, X) onto C∞W ((ω,∞), X). In particular, for M > 0 andr ∈ C∞W ((ω,∞), X), the following are equivalent:

(i) ‖(λ− ω)k+1 1k!r

(k)(λ)‖ ≤M (λ > ω, k ∈ N0).

(ii) There exists G : R+ → X satisfying G(0) = 0 and ‖G(t + h) − G(t)‖ ≤M

∫ t+h

teωr dr (t, h ≥ 0), such that r(λ) =

∫∞0

e−λt dG(t) for all λ > ω.

Proposition 1.6.1 a) gives

Lω = Sω ◦ L ◦Mω

where L and Lω are the Laplace transforms on L∞(R+, X) and L∞ω (R+, X). HenceTheorem 2.2.3 can be reformulated as follows.

Theorem 2.4.2. Let M > 0, ω ∈ R. If X has the Radon-Nikodym property thenfor any r ∈ C∞W ((ω,∞), X) the following are equivalent:

(i) ‖(λ− ω)k+1 1k!r(k)(λ)‖ ≤M (λ > ω, k ∈ N0).

(ii) There exists g ∈ L1loc(R+, X) with ‖g(t)‖ ≤ Meωt for almost all t ≥ 0 such

that r(λ) =∫∞0

e−λtg(t) dt for all λ > ω.

2.4. TRANSFORMS OF EXPONENTIALLY BOUNDED FUNCTIONS 79

As in Theorem 2.1.1 one shows that there exists an isometric isomorphismΦS,ω between the spaces Lipω(R+, X) and L(L1

ω(R+), X), where

L1ω(R+) :=

{h ∈ L1

loc(R+) : ‖h‖ω,1 :=

∫ ∞

0

eωt|h(t)| dt <∞}.

The isomorphism ΦS,ω assigns to every function G ∈ Lipω(R+, X) an operatorT ∈ L(L1

ω(R+), X) with ‖T‖ = ‖G‖Lipω(R+,X) such that

Th =

∫ ∞

0

h(t) dG(t)

for all continuous functions h ∈ L1ω(R+), Tχ[0,t] = G(t) for all t ≥ 0, and Te−λ =

dG(λ) if Reλ > ω.The inversion theorems in Section 2.3 all remain valid, with almost no changes

in the proofs (the version of Theorem 2.3.4 for Lipω(R+, X) can be deduced directly

from the case ω = 0 by using the isomorphism Iω). Thus, if r = dF for someF ∈ Lipω(R+, X), then

F (t) = limk→∞

(−1)k 1

k!

(k

t

)k+1dk

dλk

(r(λ)

λ

) ∣∣∣∣λ=k/t

. (2.8)

If c > max(ω, 0), then

F (t) = limk→∞

1

2πi

∫ c+ik

c−ik

eλtr(λ)

λdλ, (2.9)

where the limit exists uniformly on compact subsets of R+. Finally,

F (t) = limk→∞

∞∑j=1

(−1)j+1 1

j!etkjr(kj), (2.10)

where the limit exists uniformly on R+.The following is a consequence of the Phragmen-Doetsch inversion (2.10).

Proposition 2.4.3. Let ε > 0 and f ∈ L1loc(R+, X) with abs(f) <∞. The following

are equivalent.

(i) lim supλ→∞1λ log ‖f(λ)‖ ≤ −ε.

(ii) f = 0 a.e. on [0, ε].

Proof. Let F (t) :=∫ t

0f(s) ds and G(t) :=

∫ t

0F (s) ds. Since abs(f) < ∞, ω(F ) <

∞ by Theorem 1.4.3 and hence G ∈ Lipω(R+, X) for some ω ∈ R. By Corollary1.6.5 and Proposition 1.10.1,

f(λ) = λF (λ) = λdG(λ) = λ2G(λ)

for Reλ > ω. Define

r(λ) :=1

λf(λ) = F (λ) = dG(λ)

for λ > ω. If (i) holds, then lim supλ→∞1λlog ‖r(λ)‖ ≤ −ε. Let 0 < ξ < ε. Then

there exist M,λ0 > 0 such that ‖r(λ)‖ ≤ Me−λξ for all λ > λ0. Let t ∈ [0, ξ).Then, for λ0 < k ∈ N,∥∥∥∥∥∥

∞∑j=1

(−1)j+1

j!etkjr(kj)

∥∥∥∥∥∥ ≤M∞∑j=1

1

j!e(t−ξ)kj = M

(ee

(t−ξ)k − 1)→ 0

as k → ∞. Since r = dG, it follows from (2.10) that G = 0 on [0, ξ) for all0 < ξ < ε. Thus, G = 0 on [0, ε] and hence f = 0 a.e. on [0, ε], by Proposition1.2.2. This proves that (i) ⇒ (ii).

Suppose that (ii) holds. Then F = 0 on [0, ε]. Thus

r(λ) =

∫ ∞

0

e−λtF (t) dt =

∫ ∞

ε

e−λtF (t) dt = e−λε

∫ ∞

0

e−λtF (t+ ε) dt.

Since t �→ F (t+ε) is exponentially bounded, it follows that ‖ ∫∞0

e−λtF (t+ε) dt‖ ≤C for some C > 0 and therefore ‖r(λ)‖ ≤ Ce−ελ for all sufficiently large λ. Thisproves that (ii) ⇒ (i).

If f ∈ L1loc(R+, X) with abs(f) <∞, then it follows from Corollary 1.6.5 and

the exponential boundedness of F that there exist M,λ0 > 0 such that ‖f(λ)‖ ≤M for all λ > λ0. Thus, lim supλ→∞

1λ log ‖f(λ)‖ ≤ 0. This and the previous

proposition yield the following corollary.

Corollary 2.4.4. Let f ∈ L1loc(R+, X) with abs(f) < ∞. Then the following are

equivalent:

(i) lim supλ→∞1λ log ‖f(λ)‖ = 0.

(ii) For every ε > 0, the restriction of f to [0, ε] does not vanish a.e.

2.5 Complex Conditions

It was shown in the previous section that a holomorphic function q : {Reλ > ω} →X has a Laplace-Stieltjes or multiplied Laplace representation

q(λ) =

∫ ∞

0

e−λt dF (t) = λ

∫ ∞

0

e−λtF (t) dt

if there exists a constant M > 0 such that the Taylor coefficients 1k!q

(k)(λ) arebounded by M/(λ− ω)k+1 for all λ > ω and k ∈ N0. Since only properties of thefunction q along the real half-line (ω,∞) are involved, Widder’s growth conditions

2.5. COMPLEX CONDITIONS 81

are also referred to as “real conditions”. In many instances, these real conditionsare too difficult to be checked because all derivatives of q have to be considered,whereas the growth of q in a complex half-plane Reλ > ω can be estimated. Inthese cases one can apply the following representation theorem.

Theorem 2.5.1 (Complex Representation). Let ω ≥ 0, let q : {Reλ > ω} → Xbe a holomorphic function with supReλ>ω ‖λq(λ)‖ <∞ and let b > 0. Then there

exists f ∈ C(R+, X) with supt>0 ‖e−ωtt−bf(t)‖ < ∞ such that q(λ) = λbf(λ) forReλ > ω.

Proof. Let α > ω and define

f(t) := limR→∞

1

2πi

∫ α+iR

α−iR

eλtq(λ)

λbdλ =

1

2π

∫ ∞

−∞e(α+ir)t q(α + ir)

(α+ ir)bdr.

Observe that the latter integral is absolutely convergent, by the assumption on q, sothe limit exists uniformly for t in compact subsets of R+. Hence, f is continuous onR+. By applying Cauchy’s theorem over rectangles with vertices α±iR, β±iR, andusing the assumption on q, it is easy to see that the definition of f is independentof α > ω.

For α > ω and R > 0, let Γα,R be the path consisting of the vertical half-line{α + ir : r < −R}, the semicircle {α + Reiθ : −π

2 ≤ θ ≤ π2 }, and the half-line

{α+ ir : r > R}. By Cauchy’s theorem,

f(t) =1

2πi

∫Γα,R

eλtq(λ)

λbdλ

=1

2π

∫ −R

−∞e(α+ir)t q(α+ ir)

(α+ ir)bdr

+1

2π

∫ π/2

−π/2

e(α+Reiθ)t q(α+Reiθ)

(α+Reiθ)bReiθ dθ

+1

2π

∫ ∞

R

e(α+ir)t q(α+ ir)

(α+ ir)bdr.

Hence,

‖f(t)‖ ≤ Meαt

π

∫ ∞

R

dr

rb+1+

M

2π

∫ π/2

−π/2

e(α+R cos θ)t

Rbdθ

=Meαt

πbRb+

Meαt

πRb

∫ π/2

0

eRt cos θ dθ,

whereM := supReλ>ω ‖λq(λ)‖. Choosing R = 1/t, we obtain that ‖f(t)‖ ≤ Ctbeαt

for some C independent of α > ω. Hence, ‖f(t)‖ ≤ Ctbeωt.

Given λ with Reλ > ω, choose ω < α < Re λ. By the dominated convergencetheorem and Fubini’s theorem,∫ ∞

0

e−λtf(t) dt = limR→∞

∫ ∞

0

e−λt 1

2πi

∫ α+iR

α−iR

eztq(z)

zbdz dt

= limR→∞

1

2πi

∫ α+iR

α−iR

q(z)

(λ− z)zbdz.

By Cauchy’s residue theorem around the path consisting of the semicircle {α +Reiθ : −π/2 ≤ θ ≤ π/2} and the line-segment {α+ ir : −R ≤ r ≤ R},

1

2πi

∫ α+iR

α−iR

q(z)

(λ− z)zbdz =

1

2π

∫ π/2

−π/2

q(α+Reiθ)Reiθ

(λ− α−Reiθ)(α +Reiθ)bdθ +

q(λ)

λb

→ q(λ)

λb

as R→∞, using the assumption on q.

We mention that Theorem 2.5.1 does not hold for b = 0. In fact, Desch andPruss [DP93] construct a scalar-valued holomorphic function q on C+ satisfying

supReλ>0

‖q(λ)‖(1 + |λ|) <∞

such that q is not the Laplace transform of a function f ∈ L∞loc(0,∞).On the other hand, if λq(λ) and λ2q′(λ) are bounded on the right half-plane,

then q is the Laplace transform of a bounded continuous function, as we show inthe following corollary.

Corollary 2.5.2 (Pruss). Let q : {Reλ > 0} → X be holomorphic. If there existsM > 0 such that ‖λq(λ)‖ ≤ M and ‖λ2q′(λ)‖ ≤ M for Re λ > 0, then thereexists a bounded function f ∈ C((0,∞), X) such that q(λ) =

∫∞0

e−λtf(t) dt forReλ > 0. In particular, q ∈ C∞W ((0,∞), X).

Proof. It follows from Theorem 2.5.1 that there are functions fi ∈ C(R+, X) (i =0, 1) and C > 0 such that ‖fi(t)‖ ≤ Ct for t > 0,

q(λ) = λ

∫ ∞

0

e−λtf0(t) dt, and λq′(λ) = λ

∫ ∞

0

e−λtf1(t) dt

for Reλ > 0. By Theorem 1.5.1,

q′(λ) =∫ ∞

0

e−λtf0(t) dt− λ

∫ ∞

0

e−λttf0(t) dt =

∫ ∞

0

e−λtf1(t) dt.

Integration by parts (or Corollary 1.6.5) yields

λ

∫ ∞

0

e−λt

(∫ t

0

f0(s) ds− tf0(t)

)dt = λ

∫ ∞

0

e−λt

∫ t

0

f1(s) ds dt.

2.5. COMPLEX CONDITIONS 83

Since the Laplace transform is one-to-one, it follows that tf0(t) =∫ t

0f0(s) ds −∫ t

0f1(s) ds. Thus, f0 ∈ C1((0,∞), X) and tf ′0(t) = −f1(t). Therefore, ‖f ′0(t)‖ ≤ C

for all t > 0 and

q(λ) = λ

∫ ∞

0

e−λtf0(t) dt =

∫ ∞

0

e−λtf ′0(t) dt (Reλ > 0).

Remark 2.5.3. If f ∈ L∞((0,∞), X), then r = f is holomorphic on the righthalf-plane and

‖λr(λ)‖ ≤ |λ|Reλ

‖f‖∞,

‖λ2r′(λ)‖ ≤( |λ|Re λ

)2

‖f‖∞ (Reλ > 0).

In particular, λr(λ) and λ2r′(λ) are bounded on each sector Σα = {reiγ : r >0, |γ| < α} where α ∈ (0, π/2). In Corollary 2.5.2 the estimate is required uni-formly on the right half-plane, which is more. On the other hand, continuity isobtained as an additional result.

We close this section with a characterization of Laplace transforms of func-tions in L1

loc(R+, X) with ‖f(t)‖ ≤ Mtn for some M,n ≥ 0 and almost all t ≥ 0(if X has the Radon-Nikodym property) or the Laplace-Stieltjes transforms of

functions H : R+ → X with H(0) = 0 and ‖H(t) −H(s)‖ ≤ M∫ t

srn dr for some

M > 0 and all 0 ≤ s ≤ t (for general X).

Corollary 2.5.4. Let M > 0, n ∈ N0, and r ∈ C∞((0,∞), X). The following areequivalent:

(i) ‖λk+n+1

(k+n)! r(k)(λ)‖ ≤M (λ > 0, k ∈ N0).

(ii) There exists H : R+ → X satisfying H(0) = 0 and ‖H(t) − H(s)‖ ≤M

∫ t

srn dr (0 ≤ s ≤ t), such that r(λ) =

∫∞0

e−λt dH(t) for all λ > 0.

Proof. By the Real Representation Theorem 2.2.1, the statement holds for n = 0.Therefore, let n ≥ 1. To show that (i) implies (ii), define

m(λ) := (−1)n∫ ∞

λ

1

(n− 1)!(u− λ)n−1r(u) du

for λ > 0. Then, m(k)(λ) = r(k−n)(λ) for all k ≥ n and λ > 0. Since ‖λk+1

k!m(k)(λ)‖

= ‖λk+1

k!r(k−n)(λ)‖ ≤ M for all λ > 0 and k ≥ n, it follows from Proposition

2.2.2 that m ∈ C∞W ((0,∞), X) and ‖m‖W ≤ M . By Theorem 2.2.1, there existsG : R+ → X with G(0) = 0 and ‖G(t)−G(s)‖ ≤M |t− s| for all t, s ≥ 0 such thatm(λ) =

∫∞0

e−λt dG(t) for all λ > 0. By Theorem 1.5.1 and Proposition 1.9.10,

r(λ) = m(n)(λ) =

∫ ∞

0

e−λt(−t)n dG(t) =

∫ ∞

0

e−λt dH(t),

whereH(t) :=∫ t

0(−s)n dG(s). Now the statement (ii) follows from ‖H(t)−H(s)‖ =

‖ ∫ t

s(−r)n dG(r)‖ ≤M

∫ t

srn dr for all 0 ≤ s ≤ t.

To show that (ii) implies (i), let x∗ ∈ X∗. The function x∗ ◦ H is lo-cally Lipschitz continuous, hence absolutely continuous and differentiable a.e. Ifh(t) := d

dt〈H(t), x∗〉, then |h(t)| ≤ Mtn‖x∗‖ and 〈r(λ), x∗〉 = ∫∞

0e−λth(t) dt, by

Proposition 1.9.11. Hence,∣∣∣∣⟨ λk+n+1

(k + n)!r(k)(λ), x∗

⟩∣∣∣∣ =

∣∣∣∣ λk+n+1

(k + n)!

∫ ∞

0

e−λt(−t)kh(t) dt∣∣∣∣

≤ M‖x∗‖.

Now (i) follows from the Hahn-Banach theorem.

2.6 Laplace Transforms of Holomorphic Functions

In this section those functions are characterized which are Laplace transforms ofholomorphic, exponentially bounded functions defined on some open sector Σα :={reiγ : r > 0,−α < γ < α} for some 0 < α ≤ π/2. The closure of Σα is denotedby Σα. We shall use the same notation for 0 < α < π. Note that Σπ

2= C+ :=

{Reλ > 0}.Theorem 2.6.1 (Analytic Representation). Let 0 < α ≤ π

2 , ω ∈ R and q : (ω,∞)→ X. The following are equivalent:

(i) There is a holomorphic function f : Σα → X such that supz∈Σβ‖e−ωzf(z)‖

<∞ for all 0 < β < α and q(λ) = f(λ) for all λ > ω.

(ii) The function q has a holomorphic extension q : ω + Σα+π2→ X such that

supλ∈ω+Σγ+π2

‖(λ− ω)q(λ)‖ <∞ for all 0 < γ < α.

Proof. Assume that (i) holds. Let 0 < β < α. Then there exists M > 0 such that‖f(z)‖ ≤M |eωz | for all z ∈ Σβ \ {0}. Define paths Γ± by Γ± := {se±iβ : 0 ≤ s <∞}. By Cauchy’s theorem,

q(λ) =

∫ ∞

0

e−λtf(t) dt =

∫Γ±

e−λzf(z)dz

= e±iβ

∫ ∞

0

e−λse±iβ

f(se±iβ) ds (2.11)

for all λ > ω. Let 0 < ε < π2− β, and let λ ∈ C with −π

2− β + ε < arg(λ− ω) <

π2 −β−ε. Then −π

2 +ε < arg((λ−ω)eiβ) < π2 −ε, so Re((λ−ω)eiβ) ≥ |λ−ω| sin ε.

It follows that

‖e−λseiβf(seiβ)‖ ≤Me−s|λ−ω| sin ε.

2.6. LAPLACE TRANSFORMS OF HOLOMORPHIC FUNCTIONS 85

Consequently, the integral

q+(λ) := eiβ∫ ∞

0

e−λseiβf(seiβ) ds

is absolutely convergent and defines a holomorphic function in the region −π2−

β + ε < arg(λ− ω) < π2 − β − ε, with ‖(λ− ω)q+(λ)‖ ≤M/ sin ε. Similarly,

q−(λ) := e−iβ

∫ ∞

0

e−λse−iβ

f(se−iβ) ds

defines a holomorphic function in the region −π2 +β+ ε < arg(λ−ω) < π

2 +β− ε,with ‖(λ − ω)q−(λ)‖ ≤ M/ sin ε. By (2.11), both q+ and q− are extensions of q,and together they define a holomorphic extension q to ω + Σ π

2 +β−ε, satisfying‖(λ − ω)q(λ)‖ ≤ M/ sin ε in the sector. Since β < α and 0 < ε < π

2 − β arearbitrary, this proves (ii).

Assume that (ii) holds. Let 0 < γ < α and δ > 0. There exists M > 0 suchthat ‖(λ− ω)q(λ)‖ ≤M for all λ ∈ (ω + Σγ+π

2) \ {ω}. Consider an oriented path

Γ (depending on γ and δ) consisting of

Γ± :={ω + re±i(γ+π/2) : δ ≤ r

}and Γ0 :=

{ω + δeiθ : −γ − π

2 ≤ θ ≤ γ + π2

}.

Let 0 < ε < γ and consider z ∈ Σγ−ε. For λ = ω + re±i(γ+π/2) ∈ Γ±,

Re(λz) = ωRe z + r|z| cos(arg z ± (γ + π/2))

≤ ωRe z − r|z| sin ε.

Hence,

‖eλz q(λ)‖ ≤ eωRe ze−r|z| sin εM

r(λ ∈ Γ±). (2.12)

This shows that

f(z) :=1

2πi

∫Γ

eλz q(λ) dλ (2.13)

is absolutely convergent, uniformly for z in compact subsets of Σγ , and thereforedefines a holomorphic function in Σγ . By Cauchy’s theorem, this function is in-dependent of δ > 0, and also independent of γ < α so long as arg z < γ (here weuse the assumption on q to estimate the integral of eλz q(λ) over arcs {ω + Reiθ :γ1 +

π2 ≤ |θ| ≤ γ2 +

π2 }). Hence (2.13) defines a holomorphic function f ∈ Σα.

To estimate f(z), we choose δ = |z|−1, and choose γ and ε such that γ < αand | arg z| < γ − ε. On Γ0, λ = ω + |z|−1eiθ (−γ − π/2 ≤ θ ≤ γ + π/2), so∥∥∥∥ 1

2πi

∫Γ0

eλz q(λ) dλ

∥∥∥∥ ≤ 1

2π

∫ γ+π/2

−γ−π/2

eωRe zecos(arg z+θ)M dθ

≤ Me1+ωRe z. (2.14)

On Γ±, λ = ω + re±i(γ+π/2), and the estimate (2.12) gives∥∥∥∥∥ 1

2πi

∫Γ±

eλz q(λ) dλ

∥∥∥∥∥ ≤ 1

2π

∫ ∞

|z|−1

eωRe ze−r|z| sin εM

rdr

=MeωRe z

2π

∫ ∞

1

e−r sin ε

rdr

≤ MeωRe z

2π sin ε. (2.15)

Now (2.14) and (2.15) establish that

supz∈Σγ−ε

‖e−ωzf(z)‖ <∞

for any 0 < ε < γ < α.Next we will show that f(λ) = q(λ) whenever λ > ω. Given such λ, choose

0 < δ < λ− ω, and 0 < γ < α. Then λ is to the right of the path Γ, and Fubini’stheorem and Cauchy’s residue theorem imply that

f(λ) =

∫ ∞

0

e−λt 1

2πi

∫Γ

eμtq(μ) dμ dt

=1

2πi

∫Γ

q(μ)

λ− μdμ

= q(λ) + limR→∞

1

2πi

∫˜ΓR

q(μ)

λ− μdμ,

where ΓR := {ω +Reiθ : −γ − π/2 ≤ θ ≤ γ + π/2}. Then∥∥∥∥∫˜ΓR

q(μ)

λ− μdμ

∥∥∥∥ ≤∫ γ+π/2

−γ−π/2

M

|ω +Reiθ − λ| dθ→ 0

as R→∞. This proves that f(λ) = q(λ).

When f is as in Theorem 2.6.1 (i), it is an easy consequence of Cauchy’sintegral formula for derivatives that

supz∈Σβ

∥∥∥zke−ωzf (k)(z)∥∥∥ <∞

for all 0 < β < α.Recall from Sections 1.4 and 1.5 that, for f ∈ L1

loc(R+, X),

ω(f) := inf{ω ∈ R : sup

t≥0‖e−ωtf(t)‖ <∞

},

abs(f) := inf{Re λ : f(λ) exists

},

hol(f) := inf{ω ∈ R : f has a holomorphic extension for Reλ > ω

}.

2.6. LAPLACE TRANSFORMS OF HOLOMORPHIC FUNCTIONS 87

Moreover, hol(f) ≤ abs(f) ≤ ω(f). We will now show that equalities hold when fis holomorphic and exponentially bounded on a sector.

Theorem 2.6.2. Let 0 < α < π/2, let f : Σα → X be holomorphic, and suppose that

there exists ω ∈ R such that supz∈Σα‖e−ωzf(z)‖ < ∞. Then hol(f) = abs(f) =

ω(f).

Proof. By Theorem 2.6.1, there exists γ > 0 such that f has a holomorphic exten-sion (also denoted by f) to ω+Σγ+π/2 and C := supλ∈Σγ+π/2

‖(λ−ω)f(λ)‖ <∞.

By definition of hol(f), f also has a holomorphic extension to hol(f) + Σπ/2 =

{Reλ > hol(f)}.Let ω′ > hol(f). There exists γ′ > 0 such that

ω′ + Σγ′+π/2 ⊆ (ω +Σγ+π/2) ∪ (hol(f) + Σπ/2).

Hence, f is holomorphic on ω′ +Σγ′+π/2 and continuous on the closure. Let

U :={λ ∈ (ω′ + Σγ′+π/2) ∩ (ω +Σγ+π/2) : |λ− ω′| < 2|λ− ω|} .

If λ ∈ U , then ‖(λ − ω′)f(λ)‖ ≤ 2C. Moreover, (ω′ + Σγ′+π/2) \ U is compact.

Hence, supλ∈ω′+Σγ′+π/2‖(λ−ω′)f(λ)‖ <∞. It follows from Theorem 2.6.1, and the

fact that the Laplace transform is one-to-one, that supz∈Σβ‖e−ω′zf(z)‖ < ∞ for

some β > 0, and in particular, ω(f) ≤ ω′. Since this holds whenever ω′ > hol(f),

it follows that ω(f) ≤ hol(f), completing the proof.

In the remainder of this section we will consider asymptotic behaviour of f(t)as t → ∞ and as t → 0. In the case of holomorphic functions defined on a sectorit can be described completely in terms of the Laplace transform. This is not thecase in general, and in Chapter 4 a systematic treatment of this question will begiven. However, here we can use contour arguments directly on the basis of therepresentation formula (2.13).

First we show that asymptotic behaviour along one ray and on the wholesector are equivalent. This is a consequence of Vitali’s theorem (Theorem A.5).

Proposition 2.6.3. Let 0 < α ≤ π and let f : Σα → X be holomorphic such that

supz∈Σβ

‖f(z)‖ <∞

for all 0 < β < α. Let x ∈ X.

a) If limt→∞ f(t) = x, then lim |z|→∞z∈Σβ

f(z) = x for all 0 < β < α.

b) If limt↓0 f(t) = x, then lim |z|→0z∈Σβ

f(z) = x for all 0 < β < α.

Proof. a) Let fk(z) = f(kz). It follows from Vitali’s theorem that limk→∞ fk(z) =x uniformly on compact subsets of Σα. Let 0 < β < α. Let ε > 0. There existsk0 ∈ N such that ‖fk(z) − x‖ ≤ ε whenever z ∈ Σβ , 1 ≤ |z| ≤ 2, k ≥ k0. Letz ∈ Σβ , |z| ≥ k0. Choose k ∈ N such that k ≤ |z| < k + 1. Then

‖f(z)− x‖ = ‖fk(z/k)− x‖ ≤ ε.

This proves a).b) This follows by applying a) to the function z �→ f(z−1).

Now we consider the asymptotic behaviour of f(t) as t→∞ and t ↓ 0.Theorem 2.6.4 (Tauberian Theorem). Consider the situation of Theorem 2.6.1,and let x ∈ X.

a) One has limt↓0 f(t) = x if and only if limλ→∞ λq(λ) = x.

b) Assume that ω = 0. Then limt→∞ f(t) = x if and only if limλ↓0 λq(λ) = x.

Proof. We can assume that ω = 0 for both cases a) and b) by replacing f(z) bye−ωzf(z) otherwise. Replacing f(t) by f(t) − x, we can also assume that x = 0.For simplicity, we shall denote the function q of Theorem 2.6.1 by q.

Assume that limλ→∞ λq(λ) = x. Let 0 < γ < α. By Proposition 2.6.3,lim |λ|→∞

λ∈Σγ+π/2

λq(λ) = x. Let ε > 0. There exists δ0 > 0 such that ‖λq(λ)‖ ≤ ε

whenever |λ| ≥ δ0, λ ∈ Σγ+π2. Let 0 < t ≤ 1/δ0. Now we choose the contour Γ as

in the proof of Theorem 2.6.1, (ii) ⇒ (i), with δ = 1/t. Then∥∥∥∥ 1

2πi

∫Γ0

eλtq(λ) dλ

∥∥∥∥ =

∥∥∥∥∥ 1

2πi

∫ γ+π/2

−γ−π/2

eeiθ

q

(eiθ

t

)ieiθ

tdθ

∥∥∥∥∥≤ ε

2π

∫ γ+π/2

−γ−π/2

ecos θ dθ

≤ ε e,

and ∥∥∥∥∥ 1

2πi

∫Γ±

eλtq(λ) dλ

∥∥∥∥∥=

∥∥∥∥∥ 1

2πi

∫ ∞

1/t

et·re±i(γ+π/2)

q(re±i(γ+π/2))re±i(γ+π/2) dr

r

∥∥∥∥∥=

∥∥∥∥ 1

2πi

∫ ∞

1

ese±i(γ+π/2)

q(s

te±(γ+π/2))

s

te±i(γ+π/2) ds

s

∥∥∥∥→ 0

as t ↓ 0 by the dominated convergence theorem. It follows from the representation(2.13) that lim supt↓0 ‖f(t)‖ ≤ ε e.

2.7. COMPLETELY MONOTONIC FUNCTIONS 89

The converse implication is easy and does not depend on holomorphy. Assumethat limt↓0 ‖f(t)‖ = 0. Let ε > 0. There exists τ > 0 such that ‖f(t)‖ ≤ ε for allt ∈ [0, τ ]. Then

lim supλ→∞

‖λq(λ)‖ ≤ lim supλ→∞

{‖λ

∫ τ

0

e−λtf(t) dt‖+ ‖λ∫ ∞

τ

e−λtf(t) dt‖}

≤ ε+ lim supλ→∞

λ

∫ ∞

τ

e−λtMeωt dt

= ε+ lim supλ→∞

Mλ

λ− ωe−(λ−ω)τ = ε,

where ω > ω(f) and M is suitable. This completes the proof of a).

The assertion b) is proved in the same way as a).

2.7 Completely Monotonic Functions

Throughout this section, X will be an ordered Banach space with normal cone (seeAppendix C). Let f : R+ → X be increasing. Then f is of bounded semivariationon each interval [0, τ ], by Proposition 1.9.1. Assume that ω(f) < ∞. Then theLaplace-Stieltjes transform

df(λ) = limτ→∞

∫ τ

0

e−λt df(t) =:

∫ ∞

0

e−λt df(t) (2.16)

converges on the half-plane {Reλ > abs(df)}, and defines a holomorphic function

df on {Reλ > abs(df)}. Recall from Theorem 1.10.5 that abs(df) <∞ if and onlyif ω(f) <∞.

Theorem 2.7.1. Let f : R+ → X be an increasing function. Assume that −∞ <

abs(df) <∞. Then abs(df) is a singularity of df .

Proof. Replacing f(t) by∫ t

0e− abs(df)s df(s), we can assume that abs(df) = 0.

Assume that df has a holomorphic extension to a neighbourhood of 0. Then thereexists δ > 0 such that

df(−δ) =∞∑

n=0

(−1)n(1 + δ)n(df)(n)(1)

n!.

Let x∗ ∈ X∗+. Then

〈df(−δ), x∗〉 =∞∑

n=0

(1 + δ)n

n!

∫ ∞

0

e−ttn d〈f(t), x∗〉.

Since all expressions are positive we may interchange the sum and the integral andobtain ∫ ∞

0

eδt d〈f(t), x∗〉 =

∫ ∞

0

e−te(1+δ)t d〈f(t), x∗〉

=∞∑

n=0

(1 + δ)n

n!

∫ ∞

0

e−ttn d〈f(t), x∗〉

= 〈df(−δ), x∗〉 <∞.

Since X∗+ spans X∗ (see Proposition C.2), it follows that abs(x∗ ◦ f) ≤ −δ forall x∗ ∈ X∗. It follows from (1.25) that abs(df) ≤ −δ, which contradicts theassumption.

Corollary 2.7.2. Let f ∈ L1loc(R+, X) such that f(t) ≥ 0 a.e. Assume that −∞ <

abs(f) <∞. Then abs(f) is a singularity of f . Hence, hol(f) = abs(f).

Proof. This is immediate from Proposition 1.10.1 and Theorem 2.7.1.

Our aim is to characterize functions of the form df where f : R+ → X isincreasing. Then

(−1)ndf (n)(λ) =

∫ ∞

0

e−λttn df(t) ≥ 0

for all n ∈ N0, λ > ω. Thus df is completely monotonic in the sense of the followingdefinition.

Definition 2.7.3. A function r : (ω,∞) → X is completely monotonic if r isinfinitely differentiable and

(−1)nr(n)(λ) ≥ 0 for all λ > ω, n ∈ N0. (2.17)

In the following, we shall assume that ω = 0 for simplicity (otherwise, we

can replace r(λ) by r(λ+ ω) and f(t) by∫ t

0e−ωs df(s)). Recall that by Theorem

1.10.5 abs(df) ≤ 0 if and only if ω(f) ≤ 0.

Definition 2.7.4. We say that Bernstein’s theorem holds in X if for every com-pletely monotonic function r : (0,∞) → X there exists an increasing function

f : R+ → X such that ω(f) ≤ 0 and r(λ) = df(λ) for all λ > 0.

Bernstein’s theorem does hold in X = R; this is just Bernstein’s classicaltheorem from 1928 [Ber28]. Here we will prove it, as a special case of Theorem2.7.7, with the help of the Real Representation Theorem 2.2.1.

Definition 2.7.5. The space X has the interpolation property if, given two sequences(xn)n∈N, (yn)n∈N in X such that

xn ≤ xn+1 ≤ yn+1 ≤ yn (n ∈ N) (2.18)

there exists z ∈ X such that

xn ≤ z ≤ yn for all n ∈ N. (2.19)

Examples 2.7.6. a) Assume that X = Y ∗ where Y is an ordered Banach space withnormal cone. Then X has the interpolation property.

Proof. Let x∗n ≤ x∗n+1 ≤ y∗n+1 ≤ y∗n (n ∈ N). Replacing x∗n by x∗n − x∗1 and y∗n byy∗n − x∗1 we can assume that x∗n ≥ 0. Define z∗ ∈ X∗ by 〈x, z∗〉 = supn∈N〈x, x∗n〉.Then z∗ is linear and positive, and hence continuous (see Appendix C).

b) If X is reflexive, then X has the interpolation property. This follows from a).

c) Each von Neumann algebra (i.e., a ∗-subalgebra of L(H) which is closed inthe strong operator topology, where H is a Hilbert space) has the interpolationproperty. This follows from a) and [Ped89, Theorem 4.6.17].

d) Every σ-order complete Banach lattice (i.e., a Banach lattice in which eachcountable order-bounded set has a supremum) has the interpolation property.

e) If X has order continuous norm (i.e., each decreasing positive sequence con-verges) then X has the interpolation property.

f) The space C[0, 1] does not have the interpolation property.

See the Notes for further comments on the interpolation property.

Now we can formulate the following characterization, which is the main resultof this section.

Theorem 2.7.7. Bernstein’s theorem holds in X if and only if X has the interpo-lation property.

The proof of Theorem 2.7.7 will be carried out in several steps. On the waywe will prove a characterization of completely monotonic functions which is validwithout restrictions on the space. First, we study convex functions.

Let J ⊂ R be an interval. A function F : J → X is called convex if

F (λs+ (1− λ)t) ≤ λF (s) + (1− λ)F (t)

for all s, t ∈ J, 0 < λ < 1. Many order properties of convex functions carry overfrom the scalar case since for x ∈ X we have

x ≥ 0 if and only if 〈x, x∗〉 ≥ 0 for all x∗ ∈ X∗+.

For example, a twice differentiable function F is convex if and only if F ′′ ≥ 0.

Lemma 2.7.8. Let [a, b] be a closed interval in the interior of J and let F : J → Xbe convex. Then F is Lipschitz continuous on [a, b]. Moreover, if F (J) ⊂ X+ andF (a) = 0, then F is increasing on [a, b].

Proof. Let c < a, d > b such that [c, d] ⊂ J . Then for a ≤ t < s ≤ b,

F (a)− F (c)

a− c≤ F (s)− F (t)

s− t≤ F (d)− F (b)

d− b.

Since the cone is normal this implies that F is Lipschitz continuous on [a, b]. Thesecond assertion is easy to see.

We notice in particular that every convex function defined on an open intervalis continuous.

Let −∞ < a < b ≤ ∞ and let f : [a, b) → X+ be increasing. Then f isRiemann integrable on [a, t] whenever a ≤ t < b (see Corollary 1.9.6). Let

F (t) :=

∫ t

a

f(s) ds (a ≤ t < b). (2.20)

Then F : [a, b)→ X+ is convex.

If X has the interpolation property, then the following converse result holds.

Proposition 2.7.9. Assume that X has the interpolation property. Let F : [a, b)→X+ be convex such that F (a) = 0, where −∞ < a < b ≤ ∞. Then there exists anincreasing function f : [a, b)→ X+ such that (2.20) holds.

Proof. The following two properties follow from convexity:

a) Let a ≤ s < b. Then the difference quotient

1

h(F (s+ h)− F (s))

is positive and increasing for h ∈ (0, b− s).

b) Let a ≤ s < s+ h ≤ t < t+ k < b. Then

1

h(F (s+ h)− F (s)) ≤ 1

k(F (t+ k)− F (t)) . (2.21)

Put f(a) = 0. It follows from the interpolation property, a) and b) that for eacht ∈ (a, b) there exists f(t) ∈ X such that

1

h(F (s+ h)− F (s)) ≤ f(t) ≤ 1

k(F (t+ k)− F (t)) (2.22)

whenever a ≤ s < s + h ≤ t < t + k < b. It follows from (2.21) and (2.22) thatf : [a, b)→ X+ is increasing.

Let G(t) :=∫ t

af(s) ds. We show that F = G. Let a < t < b. Let a ≤ t0 <

t1 < . . . < tn = t be a partition of [a, t]. Setting hi := ti − ti−1, we obtain from


(2.22) that

n∑i=1

f(ti−1)(ti − ti−1) ≤n∑

i=1

1

hi(F (ti−1 + hi)− F (ti−1))hi

=n∑

i=1

(F (ti)− F (ti−1))

= F (t)− F (a) = F (t).

It follows from the definition of the Riemann integral that G(t) ≤ F (t). Also by(2.22),

n∑i=1

f(ti)(ti − ti−1) ≥n∑

i=1

F (ti)− F (ti−1)

ti − ti−1(ti − ti−1)

= F (t).

Hence G(t) ≥ F (t).

Next, we prove a converse version of Proposition 2.7.9.

Proposition 2.7.10. Assume that for every convex function F : R+ → X+ withF (0) = 0 and ω(F ) = 0 there exists an increasing function f : R+ → X+ such

that F (t) =∫ t

0f(s) ds (t ≥ 0). Then X has the interpolation property.

Proof. Let xn ≤ xn+1 ≤ yn+1 ≤ yn (n ∈ N). We can assume that x1 ≥ 0 (replacingxn by xn − x1 and yn by yn − x1 otherwise). Define f : R+ → X by

f(t) :=

⎧⎪⎪⎪⎨⎪⎪⎪⎩xn if t ∈ [n−1

n, nn+1

); n ≥ 1,

yn if t ∈ [n+1n , n

n−1 ); n ≥ 2,

y1 if t ∈ [2,∞),

0 if t = 1.

Then f ∈ L1loc(R+, X). Let F (t) :=

∫ t

0f(s) ds. Then F : R+ → X+ is convex and

F (0) = 0. By assumption, there exists an increasing function g : R+ → X such

that F (t) =∫ t

0g(s) ds (t ≥ 0). Then

F (t− h)− F (t)

−h ≤ g(t) ≤ F (t+ h)− F (t)

h

for all t > 0 and h > 0 small enough. It follows that g(t) = F ′(t) whenever Fis differentiable at t. Consequently, g(t) = xn if t ∈ (n−1

n, nn+1

) and g(t) = yn if

t ∈ (n+1n

, nn−1

). Hence, xn ≤ g(1) ≤ yn. Thus, z := g(1) interpolates between thetwo sequences.

For completeness, we also give the usual representation of convex functionsas a corollary of Proposition 2.7.9.

Corollary 2.7.11. Assume that X has the interpolation property. Let F : (a, b)→ Xbe convex, and let c ∈ (a, b). Then there exist x ∈ X and an increasing functionf : (a, b)→ X such that

F (t) = F (c) + (t− c)x+

∫ t

c

f(s) ds

for all t ∈ (a, b).

Proof. We may assume that c = 0. It follows from convexity that

1

t(F (0)− F (−t)) ≤ 1

s(F (s)− F (0))

whenever 0 < s < b, 0 < t < −a. Moreover, the left-hand difference quotient isdecreasing in t, and the right-hand one is increasing in s. By the interpolationproperty, there exists x ∈ X such that

1

t(F (0)− F (−t)) ≤ x ≤ 1

s(F (s)− F (0))

for all 0 < t < −a, 0 < s < b. In particular, the function

G(t) := F (t)− F (0)− tx (t ∈ (a, b))

is positive, convex and satisfies G(0) = 0.By Proposition 2.7.9, there exist increasing functions f1 : [0, b) → X+ and

f2 : [0,−a)→ X+ such that

G(t) =

∫ t

0

f1(s) ds for t ∈ [0, b) and

G(−t) =

∫ t

0

f2(s) ds for t ∈ [0,−a).

We can assume that f1(0) = f2(0) = 0. Let f(t) := f1(t) for t ∈ [0, b) and

f(t) := −f2(−t) for t ∈ (a, 0). Then f is increasing and G(t) =∫ t

0f(s) ds for all

t ∈ (a, b).

Now we will study completely monotonic functions. We need the followingformulas (2.23) and (2.24) (the latter is merely needed for n = 1 and n = 2). In

the remainder of this section we shall sometimes use loose notation such as r(λ)λ

to

denote the function λ �→ r(λ)λ , and

(r(λ)λ

)′and

(r(λ)λ

)(n)

to denote its derivatives

of orders 1 and n.

Lemma 2.7.12. Let r ∈ C∞((0,∞), X). Then

(−1)nn!

λn+1

(r(λ)

λ

)(n)

=

n∑m=0

(−1)mm!

λmr(m)(λ) (2.23)


and (λk+n

(r(λ)

λn

)(k))(n)

= λkr(k+n)(λ) (2.24)

for all λ > 0, k, n ∈ N0. In particular, if r is completely monotonic, then λ �→r(λ)/λ is also completely monotonic.

Proof. The first formula (2.23) is immediate from Leibniz’s rule. It follows that ifr is completely monotonic, then λ �→ r(λ)/λ is also completely monotonic.

We show by induction over n that (2.24) holds for all k ∈ N0. It is obviousfor n = 0. Moreover,

λkr(k+1)(λ) = λk

(λr(λ)

λ

)(k+1)

= λk

{λ

(r(λ)

λ

)(k+1)

+ (k + 1)

(r(λ)

λ

)(k)}

=

(λk+1

(r(λ)

λ

)(k))′

for λ > 0. This shows that (2.24) holds for n = 1.Now assume that (2.24) holds for a fixed n ∈ N and k ∈ N0. Then, applying

(2.24) to r′ yields

λkr(k+n+1)(λ) =

(λk+n

(r′(λ)λn

)(k))(n)

(2.25)

for λ > 0. Observe that(λk+n+1(r(λ)/λn+1)(k)

)′=

(λn · λk+1(r(λ)/λn+1)(k)

)′= nλn−1

(λk+1(r(λ)/λn+1)(k)

)+ λn

(λk+1(r(λ)/λn+1)(k)

)′= nλn−1

(λk+1(r(λ)/λn+1)(k)

)+ λnλk(r(λ)/λn)(k+1),

by applying (2.24) for n = 1 to the function r(λ)/λn instead of r. Hence,(λk+n+1(r(λ)/λn+1)(k)

)′= nλn+k

(r(λ)/λn+1

)(k)+ λn+k

(r′(λ)/λn − nr(λ)/λn+1

)(k)= λn+k (r′(λ)/λn)

(k)

for λ > 0. It follows from (2.25) that(λk+n+1(r(λ)/λn+1)(k)

)(n+1)

=(λn+k(r′(λ)/λn)(k)

)(n)

= λkr(λ)(k+n+1).

Thus, (2.24) holds when n is replaced by n+ 1.


Proposition 2.7.13. Let F ∈ Lip0(R+, X) and let

r(λ) = λdF (λ) = λ

∫ ∞

0

e−λt dF (t) (λ > 0).

Then r is completely monotonic if and only if F is convex and F (t) ≥ 0 (t ≥ 0).

Proof. Assume that r is completely monotonic. Note that r(λ)λ =

∫∞0

e−λt dF (t).Thus, by the Post-Widder formula (Theorem 2.3.1), for t > 0 we have F (t) =limk→∞ Fk(t), where

Fk(t) := Gk(k/t), Gk(λ) :=(−1)kk!

λk+1(r(λ)/λ2

)(k).

By Lemma 2.7.12, λ �→ r(λ)/λ2 is completely monotonic, and it follows that

Fk(t) ≥ 0. We show that Fk is convex; i.e., that F ′′k (t) = −(kt−2G′k(k/t)

)′ ≥ 0.

Let H(λ) := −λ2kG′k(kλ). Then F ′′k (t) =ddtH(1/t) = −t−2H ′(1/t). Thus it

suffices to show that H ′(λ) ≤ 0 or equivalently 2λkG′k(kλ) + λ2k2G′′k(kλ) ≥ 0 forλ > 0. Letting μ := kλ we have to show that

(μGk(μ))′′= 2G′k(μ) + μG′′k(μ) ≥ 0 (μ > 0).

This is true since (2.24) for n = 2 gives

(μGk(μ))′′ =

(−1)kk!

(μk+2(r(μ)/μ2)(k)

)′′=

(−1)kk!

μkr(k+2)(μ) ≥ 0 (μ > 0).

This proves one implication.Conversely, suppose that F is convex and F (t) ≥ 0 for all t ≥ 0. Let x∗ ∈ X∗+.

Then x∗ ◦ F is convex, positive and Lipschitz continuous. There is an increasing,bounded function g : R+ → R+ such that g(t) = d

dt〈F (t), x∗〉 a.e., and 〈F (t), x∗〉 =∫ t

0g(s) ds for all t ≥ 0 (see Proposition 2.7.9). We may assume that g(0) = 0. By

Proposition 1.10.1 and (1.22),

〈r(λ), x∗〉 = λ〈dF (λ), x∗〉 = λg(λ) = dg(λ) (λ > 0).

Hence, x∗◦r is completely monotonic for all x∗ ∈ X∗+ and therefore r is completelymonotonic.

Next we prove a representation theorem for completely monotonic functionsdefined on R+ (and not merely (0,∞)).

Proposition 2.7.14. Let r ∈ C∞(R+, X) such that (−1)nr(n)(λ) ≥ 0 (λ ≥ 0). Thenthere exists a convex function F ∈ Lip0(R+, X) such that F (t) ≥ 0 (t ≥ 0) and

r(λ) = λdF (λ) (λ > 0). (2.26)


Proof. It follows from (2.23) that for k ∈ N and λ > 0,

pk(λ) :=(−1)kk!

λk+1

(r(λ)

λ

)(k)

=

k∑m=0

(−1)mm!

λmr(m)(λ) ≥ 0.

Moreover, limλ↓0 pk(λ) = r(0). It follows from (2.24) for n = 1 that

p′k(λ) =(−1)kk!

λkr(k+1)(λ) ≤ 0 (λ > 0).

Thus 0 ≤ pk(λ) ≤ r(0) for all λ > 0. Since the cone is normal, this implies that the

function r(λ)λ is in C∞W ((0,∞), X). By Theorem 2.2.1, there exists F ∈ Lip0(R+, X)

such that r(λ)λ = dF (λ) (λ > 0). It follows from Proposition 2.7.13 that F is

positive and convex.

Theorem 2.7.15. A function r : (0,∞) → X is completely monotonic if and onlyif there exists a convex function F : R+ → X+ satisfying F (0) = 0 and ω(F ) ≤ 0such that

r(λ) = λ

∫ ∞

0

e−λt dF (t) (λ > 0). (2.27)

In that case, F is uniquely determined by r.

Proof. a) Assume that r is of the form (2.27). Let x∗ ∈ X∗+. Then there exists anincreasing function f : R+ → R+ such that f(0) = 0 and

〈F (t), x∗〉 =∫ t

0

f(s) ds (t ≥ 0).

Thus

〈r(λ), x∗〉 =∫ ∞

0

e−λt df(t) (λ > 0).

Hence, 〈r(·), x∗〉 is completely monotonic and

〈(−1)nr(n)(λ), x∗〉 = (−1)n(

d

dλ

)n

〈r(λ), x∗〉 ≥ 0.

Since x∗ ∈ X∗+ is arbitrary, it follows that r is completely monotonic.

b) Conversely, let r be completely monotonic. By Proposition 2.7.14, thereexists a convex functionG ∈ Lip0(R+, X) such thatG(t) ≥ 0 (t ≥ 0) and r(λ+1) =λ∫∞0

e−λt dG(t) (λ > 0). Let

F (t) :=

∫ t

0

(1− (t− s))es dG(s).


Then F is positive and convex. In fact, let x∗ ∈ X∗+. Then there exists an increas-

ing function g : R+ → R+ such that 〈G(t), x∗〉 =∫ t

0g(s) ds and g(0) = 0. By

Proposition 1.9.10, Fubini’s Theorem and (1.20),

〈F (t), x∗〉 =

∫ t

0

esg(s) ds−∫ t

0

(t− s)esg(s) ds

=

∫ t

0

(esg(s)−

∫ s

0

erg(r) dr

)ds

=

∫ t

0

∫ s

0

er dg(r) ds (t ≥ 0).

Thus x∗ ◦ F is positive and convex for all x∗ ∈ X∗+, so F is positive and convex.By Proposition 1.10.1 and (1.22),

〈r(λ+ 1), x∗〉 = λg(λ) =

∫ ∞

0

e−λt dg(t)

for λ > 0. By Proposition 1.10.3, for λ > 1,

〈r(λ), x∗〉 =∫ ∞

0

e−λtet dg(t) =

∫ ∞

0

e−λt df(t),

where

f(t) :=

∫ t

0

es dg(s) = etg(t)−∫ t

0

esg(s) ds,

by (1.20). Since 〈F (t), x∗〉 =∫ t

0f(s) ds, it follows that r(λ) = λ

∫∞0

e−λt dF (t)

for λ > 1. By Theorem 2.7.1, abs(dF ) is a singularity of dF . Moreover, applyingProposition 2.7.14 to r(· + δ) shows that r has a holomorphic extension to {λ ∈C : Reλ > δ} for all δ > 0, and hence to {λ ∈ C : Reλ > 0}. It follows from

the uniqueness of holomorphic extensions that abs(dF ) ≤ 0 and dF (λ) = r(λ) forλ > 0. By Theorem 1.10.5, ω(F ) ≤ 0 (actually, ω(F ) = 0 unless r ≡ 0). Finally,uniqueness of F follows from the Post-Widder formula (Theorem 2.3.1).

Theorem 2.7.16. Assume that X has the interpolation property. Let r : (0,∞)→ Xbe completely monotonic. Then there exists an increasing function f : R+ → X+

such that f(0) = 0, ω(f) ≤ 0 and

r(λ) =

∫ ∞

0

e−λt df(t) (λ > 0).

Proof. By Theorem 2.7.15, there exists a convex function F : R+ → X+ satisfyingF (0) = 0 and ω(F ) ≤ 0 such that r(λ) = λ

∫∞0

e−λt dF (t) for all λ > 0. ByProposition 2.7.9, there exists an increasing function f : R+ → X+ such that


F (t) =∫ t

0f(s) ds (t ≥ 0). We can assume that f(0) = 0. Let ω > 0. There exists

M ≥ 0 such that ‖F (t)‖ ≤Meωt. Since f is increasing we have

t

2f(t/2) ≤

∫ t

t/2

f(s) ds ≤ F (t).

It follows that ω(f) ≤ 0 (actually, ω(f) = 0 unless r ≡ 0). By Proposition 1.10.2,∫ ∞

0

e−λt df(t) = λ

∫ ∞

0

e−λt dF (t) = r(λ) (λ > 0).

Now we can prove Theorem 2.7.7.

Proof of Theorem 2.7.7. One direction is given by Theorem 2.7.16. In order to provethe other, assume that Bernstein’s theorem holds inX . We show thatX has the in-terpolation property. Let F : R+ → X+ be convex such that F (0) = 0 and ω(F ) =

0. By Proposition 2.7.10, it suffices to show that F (t) =∫ t

0f(s) ds (t ≥ 0) for some

increasing function f : R+ → X . By Proposition 2.7.13, r(λ) := λ∫∞0

e−λt dF (t)defines a completely monotonic function on (0,∞). By assumption, there existsan increasing function f : R+ → X such that

r(λ) =

∫ ∞

0

e−λt df(t).

We may assume that f(0) = 0. Let H(t) :=∫ t

0f(s) ds. Using Proposition 1.10.2

and (1.22), λ2H(λ) = df(λ) = r(λ) = λ2F (λ) for all λ > 0. It follows from theuniqueness theorem that H(t) = F (t) for all t ≥ 0.

If r : (0,∞) → X is completely monotonic, there may be many increasing

functions f : R+ → X+ such that r = df . However, if X has order continuousnorm, then we may pick out a normalized version of f .

Let f : R+ → X be increasing and assume that X has order continuous norm.For t ≥ 0 we define f(t+) = lims↓t f(s), and for t > 0 we let f(t−) = lims↑t f(s).We say that f has a jump at t > 0 if f(t+) �= f(t−).

Lemma 2.7.17. Assume that X has order continuous norm and that f : R+ → Xis increasing. Then the number of jumps of f is countable.

Proof. Let τ > 0 and J := {t ∈ (0, τ) : f(t+) �= f(t−)}. Let ε > 0 and Jε :={t ∈ J : ‖f(t+) − f(t−)‖ ≥ ε}. We claim that Jε is finite. Otherwise thereexist tn ∈ Jε (n ∈ N), tn �= tm for n �= m. Let xn = f(tn+) − f(tn−). Then∑m

n=1 xn ≤ f(τ) − f(0) for all m ∈ N. Since X has order continuous norm, thesum

∑∞n=1 xn converges. Hence, ‖xn‖ → 0 as n → ∞. This is a contradiction.

Since J =⋃

n∈N J1/n, it follows that J is countable.


We continue to assume that X has order continuous norm. Let f : R+ → Xbe increasing. We define the normalization f∗ : R+ → X of f by

f∗(t) =

{f(0+) if t = 0,12(f(t+) + f(t−)) if t > 0.

The function f is called normalized if f = f∗.It follows from the definition of the Riemann-Stieltjes integral that∫ t

0

g(s) df(s) =

∫ t

0

g(s) df∗(s)

for every t > 0 and every continuous function g : [0, t]→ C. In fact, one may takea sequence of partitions (πn)n∈N with intermediate points which avoid the jumpsof f). Then S(g, f, πn) = S(g, f∗, πn) for all n ∈ N, and so∫ t

0

g(s) df(s) = limn→∞S(g, f, πn) = lim

n→∞S(g, f∗, πn) =

∫ t

0

g(s) df∗(s).

In conclusion, we obtain the following result.

Theorem 2.7.18 (Bernstein’s theorem). Assume that X has order continuousnorm. Let r : (0,∞) → X be completely monotonic. Then there exists a uniquenormalized increasing function f : R+ → X such that f(0) = 0, ω(f) ≤ 0 and

r(λ) =

∫ ∞

0

e−λt df(t) (λ > 0).

Proof. Since X has the interpolation property, existence follows from Theorem2.7.16. For uniqueness, suppose that r(λ) =

∫∞0

e−λt df(t) (λ > 0). By Proposition

1.10.2, r(λ) = λ∫∞0

e−λt dF (t) (λ > 0) where F (t) :=∫ t

0f(s) ds. It follows from

Theorem 2.7.15 that F is uniquely determined by r. Since

F ′(t+) := limh↓0

1

h(F (t+ h)− F (t)) = f(t+)

if t ≥ 0, and

F ′(t−) := limh↓0

1

h(F (t)− F (t− h)) = f(t−)

if t > 0, the normalized function f is also unique.

2.8 Notes

Section 2.1Representation of operators from a space of the form L1(Ω, μ) into a Banach space

2.8. NOTES 101

X by vector measures is a classical subject (see [DU77, Section III.1]). In view of theapplications to Cauchy problems, Stieltjes integrals seem more appropriate than vectormeasures in our context. In the context of Laplace transform theory, the Riesz-StieltjesRepresentation Theorem 2.1.1 appeared in a paper of Hennig and Neubrander [HN93](see also [Neu94] and [BN94]). For a discussion of the representation of bounded linearoperators in L(Lp(R+), X) as functions of bounded p′-variation (1/p+ 1/p′ = 1, p′ > 1),see the work of Weis [Wei93] and Vieten [Vie95].

Section 2.2For real-valued functions, Theorem 2.2.1 was proved by Widder in 1936 [Wid36] (see also[Wid41]). In trying to extend scalar-valued Laplace transform theory to vector-valuedfunctions, Hille [Hil48] remarks on several occasions that Widder’s theorem can be liftedto infinite dimensions if the space is reflexive, but not in general (see [Hil48, p.213] or[Miy56]). In fact, it was shown by Zaidman [Zai60] (see also [Are87b] or Theorem 2.2.3)that Widder’s theorem extends to a Banach space X if and only if X has the Radon-Nikodym property (for example, if X is reflexive). In 1965, Berens and Butzer [BB65]gave necessary and sufficient complex conditions for the Laplace-Stieltjes representabil-ity of functions in reflexive and uniformly convex Banach spaces. However, these resultswere of limited applicability. In general, important classes of Banach spaces that appearin studying evolution equations do not possess the Radon-Nikodym property. As a conse-quence, in the 1960s and 1970s Laplace transform methods were applied mainly to specialvector-valued functions, like resolvents and semigroups, which have nice additional alge-braic properties. In the theory of C0-semigroups the link between the generator A andthe semigroup T is given via the Laplace transform

(λ− A)−1x =

∫ ∞

0

e−λtT (t)x dt (x ∈ X).

The crucial algebraic property which made it possible to extend classical Laplace trans-form results to this abstract setting is the algebraic semigroup law T (t+ s) = T (t)T (s),(t, s ≥ 0). Hille and Phillips comment in the foreword to [HP57] that “.... in keepingwith the spirit of the times the algebraic tools now play a major role....” and that “.... theLaplace-Stieltjes transform methods..... have not been replaced but rather supplemented bythe new tools.” The major disadvantage of the “algebraic approach” to linear evolutionequations becomes obvious if one compares the mathematical theories associated withthem (for example, semigroup theories, cosine families, the theory of integro-differentialequations, etc.). It is striking how similar the results and techniques are. Still, withouta Laplace transform theory for functions with values in arbitrary Banach spaces, everytype of linear evolution equation required its own theory because the algebraic propertiesof the operator families changed from one case to another. In the late 1970s, in searchof a general analytic principle behind all these theories, the study of Laplace transformsof functions with values in arbitrary Banach spaces was revitalized by Sova (see [Sov77]up to [Sov82]). An important result for Laplace transforms in Banach spaces is Theorem2.6.1, proved by Sova in 1979 [Sov79b], [Sov79c]. This analytic representation theorem isbehind every generation result for analytic solution families of linear evolution equations.

The Real Representation Theorem 2.2.1 shows that the statement of Widder’s The-orem extends to arbitrary Banach spaces if the Laplace transform is replaced by theLaplace-Stieltjes transform. It is due to [Are87b] where it was deduced from the scalarresult by Widder [Wid41] by duality arguments. The proof of Theorem 2.2.1 given here is

a modification of Widder’s original proof given in [Wid41]; see [HN93]. Further extensionsof these results are given in [Bob97a], [Bob97b], [Kis00], [Bob01] and [Cho02].

The characterization of the range of the Laplace-Stieltjes transform acting onLip0(R+, X) given in Theorem 2.2.1 is based on the Post-Widder inversion formula inTheorem 1.7.7. Corresponding to other inversion formulas, equivalent descriptions canbe formulated. Employing the complex inversion formula (see [Sov80b], [BN94]), or thePhragmen-Doetsch inversion (see [PC98]), one can prove that the following growth andregularity conditions are equivalent.

Theorem 2.8.1. Let r : (0,∞) → X be continuous. The following are equivalent:

(i) r ∈ C∞((0,∞), X) and

supλ>0k∈N0

∥∥∥∥λk+1

k!r(k)(λ)

∥∥∥∥ < ∞.

(ii) limλ→∞ r(λ) = 0 and r has an extension to a holomorphic function r : {Reλ >0} → X such that, for all γ > 0, supRe λ>γ ‖r(λ)‖ < ∞ and

sups>0k∈N0

∥∥∥∥ 1

2π

∫ ∞

−∞

r(γ + it)

(1− ist)k+2dt

∥∥∥∥ < ∞.

(iii) supλ>0 ‖λr(λ)‖ < ∞ and

supλ>0k∈N

∥∥∥∥∥∞∑j=1

(−1)j−1

(j − 1)!ejkλr(jλ)

∥∥∥∥∥ < ∞.

For a discussion of the Lp-conditions∫ ∞

0

∥∥∥∥∥(k

t

)k+11

k!r(k)

(k

t

)∥∥∥∥∥p

dt ≤ M for all k ≥ 0,

and their connection to the representability of r as the Laplace transform of a function ofbounded p-variation (p > 1), see [Wid41, Chapter VII], [Lev69], [Sov81a], [Wei93], and[Vie95]. It is shown in [KMV03] that a function r ∈ C∞((0,∞), X) is the finite Laplace-Stieltjes transform r(λ) =

∫ τ

0e−λt dF (t) of a Lipschitz continuous function F : [0, τ ] → X

with ‖F (t)− F (s)‖ ≤ M |t− s| for all 0 ≤ t, s ≤ τ if and only if

supk∈N0

supλ>k/τ

∥∥∥∥λk+1

k!r(k)(λ)

∥∥∥∥ ≤ M

andsupk∈N

supλ∈(0,k/τ)

∥∥∥τ−keλτr(k)(λ)∥∥∥ < ∞.

Section 2.3Theorem 2.3.2 goes back to Phragmen’s proof of the Uniqueness Theorem 1.7.3 (see[Phr04]), and to Doetsch [Doe37] who recognized the usefulness of the formula as aninversion procedure (see also [Doe50, Volume I, Section 8.1]). The Phragmen-Doetschinversion formula shows that a Laplace transformable function f is determined by the

2.8. NOTES 103

values of f(λn), where λn = n ≥ n0. An extension of the Phragmen-Doetsch inversionto arbitrary Muntz sequences (λn) ⊂ R+ (i.e., λn+1 − λn ≥ 1 and

∑∞n=1 λ

−1n = ∞), has

been obtained by Baumer [Bau03] (see also [BLN99]). There does not seem to be anyinversion formula that holds for arbitrary uniqueness sequences (see Theorem 1.11.1).Corollary 2.3.3 is taken from [BN96] and is one of the key ingredients in the theory ofasymptotic Laplace transforms (see [LN99], [LN01]). Whereas the complex inversion for-mula in Theorem 2.3.4 (the proof given here is from [HN93]) is in general affected byexponentially decaying perturbations of the Laplace transform, the following modifica-tion, due to Lyubich [Lyu66], gives a complex inversion formula which holds locally evenif the transform undergoes such perturbations.

Theorem 2.8.2. Let τ > 0, ω > 0, F ∈ Lip0(R+, X), and q(λ) =∫∞0

e−λt dF (t) +a(λ) (λ > 0), where a ∈ L1

loc(R+, X) is a function with lim supλ→∞1λlog ‖a(λ)‖ ≤ −τ .

Then

H(μ) :=1

2πi

∫ ∞

ω

eμtq(t)

tdt

is well defined for Reμ < 0, has a holomorphic continuation to the sliced half-plane{μ : Reμ < τ} \ [0, τ), and

F (t) = limε→0

(H(t+ iε)−H(t− iε)) for all t ∈ [0, τ).

Haase [Haa08] has given a different approach to Theorem 2.3.4 and Lemma 2.3.5.

Section 2.4With the exception of Proposition 2.4.3 which is due to Doetsch (see [Doe50, Volume I,Section 14.3]) and Corollary 2.4.4, the results are straightforward reformulations of themain theorems of the sections 2.1–2.3. Using a Phragmen-Doetsch type inversion formulaalong sequences (λn) ⊂ R+ with λn+1 − λn ≥ 1 and

∑∞n=1 λ

−1n = ∞ (Muntz sequences),

one can strengthen the statement of Proposition 2.4.3 as follows (see [Bau03]).

Theorem 2.8.3. Let 0 ≤ τ and let f ∈ L1loc(R+, X) with abs(f) < ∞. Then the following

are equivalent:

(i) f(t) = 0 almost everywhere on [0, τ ] and τ ∈ supp(f).

(ii) Every Muntz sequence (βn) satisfies lim supn→∞1βn

log ‖f(βn)‖ = −τ.

(iii) For every Muntz sequence (βn) there exists a Muntz subsequence (βnk) such that

limk→∞

1

βnk

log ‖f(βnk )‖ = −τ.

(iv) There exists a Muntz sequence (βn) with lim supn→∞1βn

log ‖f(βn)‖ = −τ.

(v) lim supλ→∞1λlog ‖f(λ)‖ = −τ .

As a consequence of these equivalences one obtains the following short proof ofTitchmarsh’s theorem (see [Bau03], [BLN99] or [MB87, Section VI.7]).

Corollary 2.8.4 (Titchmarsh’s Theorem). Let k ∈ L1[0, τ ] with 0 ∈ supp(k) and f ∈L1([0, τ ], X). If k � f = 0 on [0, τ ], then f = 0.

Proof. We extend k and f by zero to R+. Then, by Proposition 2.4.3 and Corollary

2.4.4, lim supλ→∞1λlog |k(λ)| = 0 and lim supλ→∞

1λlog ‖k � f(λ)‖ ≤ −T . By taking

subsequences, it follows from the theorem above that there exists a Muntz sequence (βn)such that limn→∞ 1

βnlog |k(βn)| = 0 and

−τ ≥ limn→∞

1

βnlog ‖k � f(βn)‖ = lim

n→∞1

βnlog |k(βn)|+ lim

n→∞1

βnlog ‖f(βn)‖

= limn→∞

1

βnlog ‖f(βn)‖.

Thus, f = 0 on [0, τ ].

A function k ∈ L1loc(R+) with abs(k) < ∞ is a regularizing function if

lim supλ→∞

1

λlog |k(λ)| = 0,

or, equivalently, if 0 ∈ supp(k) (by Corollary 2.4.4). By the Titchmarsh-Foias theorem(see [BLN99]), the condition 0 ∈ supp(k) is necessary and sufficient for the convolutionoperator K : f → k ∗ f , (k ∗ f)(t) :=

∫ t

0k(t − s)f(s) ds to be an injective operator on

C(R+, X) with dense range in the Frechet space C∗(R+, X) of all continuous functionsg : R+ → X such that g(0) = 0, equipped with the seminorms ‖g‖n := supt∈[0,n] ‖g(t)‖.Moreover, ‖f‖K,n := supt∈[0,n] ‖Kf(t)‖ defines a family of seminorms on C(R+, X) and

K extends to an isomorphism between the Fr echet completion C[k](R+, X) of C(R+, X)with respect to that family of seminorms and the Frechet space C∗(R+, X). Typicalexamples of regularizing functions are

k(t) =tb−1

Γ(b)with k(λ) =

1

λb(b > 0), or

kδ(t) =1

2πi

∫ω+iR

etλ−λδ

dλ with kδ(λ) = e−λδ

(0 < δ < 1).

Note that k1/2(t) =1

2√πt−3/2e−1/4t (see Lemma 1.6.7).

If k is a regularizing function, then the elements of the Frechet space C[k](R+, X)are called k-generalized functions. A k-generalized function u is said to be Laplace trans-formable if the continuous function f := k ∗u ∈ C∗(R+, X) is Laplace transformable andthe Laplace transform of u is defined as

u(λ) :=f(λ)

k(λ).

Let H = {λ : Reλ > ω} and m : H → C be holomorphic. A meromorphic functionq : H → X is said to have an m-multiplied Laplace representation if there exists f ∈C∗(R+, X) with abs(f) ≤ ω such that mq = f on H. If m = k for some regularizingfunction k, then the meromorphic function q has a Laplace representation q = u foru = K−1f ∈ C[k](R+, X) (see [Bau97], [BLN99], and [LN99]).

Section 2.5Theorem 2.5.1 is a standard result of Laplace transform theory. Corollary 2.5.2 is due to

2.8. NOTES 105

Pruss [Pru93], the proof given here is from [BN94]. Corollary 2.5.4 is a special case ofresults in [DVW02] (see also [DHW97]).

Theorem 2.5.1 can be interpreted in terms of k-generalized functions and Laplacetransforms (see the Notes of Section 2.4; we use the same notation here). Let q : H → Xbe holomorphic with supλ∈H ‖λq(λ)‖ < ∞. As shown in Theorem 2.5.1, for all b > 0 there

exists f ∈ C∗(R+, X) such that q(λ) = λbf(λ) on H. Thus, q(λ) = u(λ) = f(λ)

k(λ), where

k(t) = 1Γ(b)

tb−1 and u = K−1f ∈ C[k](R+, X) coincides with the b-th (distributional)derivative of f . More generally, if q is a meromorphic function on some half-plane H withvalues in X for which λ → λk0(λ)q(λ) is holomorphic on H and

supλ∈H

‖λk0(λ)q(λ)‖ < ∞

for some regularizing function k0, then it follows from Theorem 2.5.1 that there existsf ∈ C∗(R+, X) such that 1

λk0(λ)q(λ) = k(λ)q(λ) = f(λ) or q(λ) = u(λ), where k :=

1 ∗ k0 and u ∈ C[k](R+, X) is a generalized function such that f = k ∗u. Notice that if kiare regularizing functions and k1 ∗ k2 = k3, then C[k1](R+, X) is continuously embeddedin C[k3](R+, X). Thus, a faster growing q will have a less regular u such that q = u.

Section 2.6Theorem 2.6.1 is due to Sova [Sov79b] and Theorem 2.6.2 is taken from [Neu89b].

Section 2.7.In 1893, Stieltjes proved in a letter to Hermite that a bounded continuous functionf : R+ → R is positive if and only if f (n)(λ) ≥ 0 for all n ∈ N0 and all λ sufficiently large(see [BB05]). Bernstein proved his theorem in 1928 [Ber28].

The characterization of those ordered Banach spaces in which Bernstein’s theorem(Theorem 2.7.7) holds is due to Arendt [Are94a].

The interpolation property is of particular interest for spaces of the form C(K),where K is a compact space. Then it can be described in terms of K: the space C(K)has the interpolation property if and only if K is an F -space (i.e., if A,B ⊂ K are openand disjoint Fσ-sets, then A ∩ B = ∅). Note that C(K) is σ-order complete if and onlyif K is quasi-stonean (i.e., if A ⊂ K is an open Fσ-set, then A is open). For example,K := βN \N is a F -space which is not quasi-stonean (where βN denotes the Stone-Cechcompactification of N). Whereas every quasi-stonean space K is totally disconnected(i.e. the connected component of each point x is {x}), there exist connected compact F -spaces. One reason why these spaces have been studied is that C(K) has the Grothendieckproperty (see Section 4.3) if K is an F -space. We refer to the article by Seever [See68]for this and further information.

The interpolation property is also equivalent to two other vector-valued versionsof classical theorems; namely, Riesz’s representation theorem for positive functionals onC[0, 1] and Hausdorff’s theorem on the moment problem. More precisely, the followingis proved in [Are94a].

Theorem 2.8.5. Let X be an ordered Banach space with normal cone. The following areequivalent:

(i) X has the interpolation property.

(ii) For every positive T ∈ L(C[0, 1], X) there exists an increasing function f : [0, 1] →X such that Tg =

∫ 1

0g(t) df(t) for all g ∈ C[0, 1].


(iii) For each completely monotonic sequence (xn)n∈N in X there exists an increasingfunction f : [0, 1] → X such that xn =

∫ 1

0tn df(t) (n ∈ N).

Here, a sequence x = (xn)n∈N is called completely monotonic if (−Δ)kx ≥ 0 for all k ∈ Nwhere Δ : XN → XN is given by Δx = (xn+1 − xn)n∈N.

Bernstein’s theorem in ordered Banach spaces with order continuous norm (The-orem 2.7.18) is proved in [Are87a] with the help of the classical scalar theorem. A firstvector-valued version of Bernstein’s theorem is due to Bochner [Boc42]. But Bochnerconsidered convergence in order, whereas for our purposes norm convergence of Riemann-Stieltjes sums and improper integrals is essential in order to make the results applicableto operator theory. Here we deduce Bernstein’s theorem from the Real RepresentationTheorem 2.2.1.

One can obtain Widder’s theorem (the scalar case of Theorem 2.2.1) as an easycorollary of Bernstein’s classical result (see [Wid71, Section 6.8]). However this argumentis restricted to the scalar case. On the other hand, it is possible to deduce the vector-valued version of Theorem 2.2.1 from the scalar case by a duality argument (see [Are87b]and the Notes of Section 2.2).

Chapter 3

Cauchy Problems

In this chapter we study systematically well-posedness of the Cauchy problem.Given a closed operator A on a Banach space X we will see in Section 3.1 thatthe abstract Cauchy problem{

u′(t) = Au(t) (t ≥ 0),

u(0) = x,

is mildly well-posed (i.e., for each x ∈ X there exists a unique mild solution) ifand only if the resolvent of A is a Laplace transform; and this in turn is the sameas saying that A generates a C0-semigroup. Well-posedness in a weaker sense willlead to generators of integrated semigroups (Section 3.2). The real representationtheorem from Section 2.2 will give us directly the characterization of generators ofC0-semigroups in terms of a resolvent estimate; namely, the Hille-Yosida theorem.When the operators are not densely defined, we obtain Hille-Yosida operatorswhich are studied in detail in Section 3.5. Also for results on approximation ofsemigroups in Section 3.6 we can use corresponding results on Laplace transformsfrom Section 1.7. Much attention is given to holomorphic semigroups which areparticularly simple to characterize by means of the results of Section 2.6. Weconsider not only holomorphic semigroups which are strongly continuous at 0,but more general holomorphic semigroups which will be useful in applications tothe heat equation with Dirichlet boundary conditions in Chapter 6. When theholomorphic semigroup exists on the right half-plane, the boundary behaviour isof special interest. If the semigroup is locally bounded, then a boundary C0-groupis obtained on the imaginary axis. This case is particularly important for fractionalpowers (see also the Notes of Section 3.7) and for the second order problem (Section3.16). When the holomorphic semigroup is polynomially bounded we obtain k-times integrated semigroups where the k depends on the degree of the polynomial.A typical example is the Gaussian semigroup. Its boundary is governed by theSchrodinger operator iΔ, which we study in Section 3.9 and in Chapter 8. The last


108 3. CAUCHY PROBLEMS

three sections are devoted to the second order Cauchy problem; i.e., to the theoryof cosine functions. A central result will be to establish a unique phase space onwhich the associated system is well-posed. This is a particularly interesting specialcase of the intermediate spaces which are constructed in Section 3.10 for integratedsemigroups.

In two places we will give results for UMD-spaces which are not valid ingeneral Banach spaces: in Section 3.12 where we establish a particularly simplecomplex inversion formula for semigroups, and in Section 3.16 where we proveFattorini’s remarkable theorem on the square root reduction.

There is no special section on perturbation theory, but we prove perturbationresults for Hille-Yosida operators, integrated semigroups and generators of cosinefunctions in the corresponding sections. For holomorphic semigroups we considernot only “relatively small perturbations” but also “compact perturbations” withrespect to A. This chapter contains some interesting examples of holomorphicsemigroups in Sections 3.7 and 3.9, but for real applications we refer to Part III.

Throughout this chapter we will make extensive use (sometimes without com-ment) of notation, terminology and basic properties of closed operators which maybe found in Appendix B. In some examples we shall use some basic notions of dis-tributions and Sobolev spaces which may be found in Appendix E.

3.1 C0-semigroups and Cauchy Problems

Let A be a closed operator on a Banach space X. We consider the abstract Cauchyproblem

(ACP0)

{u′(t) = Au(t) (t ≥ 0),

u(0) = x,

where x ∈ X. By a classical solution of (ACP0) we understand a function u ∈C1(R+, X) such that u(t) ∈ D(A) for all t ≥ 0 and (ACP0) holds.

If a classical solution exists, then it follows that x = u(0) ∈ D(A). It will beuseful to find a weaker notion of solution where x may be arbitrary. This can bedone by integrating the equation. Assume that u is a classical solution. Since A isclosed, it follows from Proposition 1.1.7 that∫ t

0

u(s) ds ∈ D(A) and A

∫ t

0

u(s) ds = u(t)− x (t ≥ 0). (3.1)

Definition 3.1.1. A function u ∈ C(R+, X) is called a mild solution of (ACP0) if(3.1) holds.

The following assertion shows that mild and classical solutions differ merelyby regularity.

Proposition 3.1.2. A mild solution u of (ACP0) is a classical solution if and onlyif u ∈ C1(R+, X).

3.1. C0-SEMIGROUPS AND CAUCHY PROBLEMS 109

Proof. Assume that u ∈ C1(R+, X). Let t ≥ 0. Then

1

h(u(t+ h)− u(t)) =

1

hA

∫ t+h

t

u(s) ds

for all h �= 0 small enough (h > 0 if t = 0). Since A is closed, it follows that

u(t) = limh→0

1

h

∫ t+h

t

u(s) ds ∈ D(A) and

u′(t) = Au(t).

Next we want to characterize mild solutions with the help of Laplace trans-forms. Let u ∈ C(R+, X). Recall from (1.12) that abs(u) <∞ if and only if∥∥∥∥∫ t

0

u(s) ds

∥∥∥∥ ≤Meωt (t ≥ 0) (3.2)

for some M,ω ≥ 0. As before, we denote by

u(λ) :=

∫ ∞

0

e−λtu(t) dt (λ > ω)

the Laplace transform of u.

Theorem 3.1.3. Let u ∈ C(R+, X) such that (3.2) holds. Then the following as-sertions are equivalent:

(i) u is a mild solution of (ACP0).

(ii) u(λ) ∈ D(A) and λu(λ)− Au(λ) = x for all λ > ω.

Proof. (i) ⇒ (ii): Let u be a mild solution. Let λ > ω. We know from (1.11) that

u(λ) = λ

∫ ∞

0

e−λt

∫ t

0

u(s) ds dt.

Since A is closed, it follows from Proposition 1.6.3 that u(λ) ∈ D(A) and

Au(λ) = λ

∫ ∞

0

e−λtA

∫ t

0

u(s) ds dt

= λ

∫ ∞

0

e−λt(u(t)− x) dt

= λu(λ)− x.

(ii) ⇒ (i): Let v(t) :=∫ t

0u(s) ds. Then by (1.11), v(λ) = u(λ)/λ ∈ D(A) and

Av(λ) = Au(λ)/λ = u(λ)− x/λ = f(λ) (λ > ω),

where f(t) := u(t)− x (t ≥ 0). It follows from Proposition 1.7.6 that v(t) ∈ D(A)and Av(t) = f(t) = u(t)− x for all t ≥ 0; i.e., u is a mild solution of (ACP0).

Let u be a mild solution of (ACP0) satisfying (3.2). Assume that ω < λ ∈ρ(A). Then it follows from Theorem 3.1.3 that u(λ) = R(λ,A)x. Thus the Laplacetransform of a mild solution is the resolvent applied to the initial value. This leadsus to consider operators whose resolvent exists on a half-line and is a Laplacetransform.

For this, let T : R+ → L(X) be strongly continuous. Recall from Proposition1.4.5 that abs(T ) <∞ if and only if∥∥∥∥∫ t

0

T (s)x ds

∥∥∥∥ ≤Meωt‖x‖ (t ≥ 0, x ∈ X) (3.3)

for some ω ≥ 0, M ≥ 0. In that case, abs(T ) ≤ ω and

T (λ)x := limt→∞

∫ t

0

e−λsT (s)x ds

defines a bounded operator T (λ) ∈ L(X) whenever Reλ > abs(T ). Moreover,T : {Reλ > abs(T )} → L(X) is holomorphic (see Section 1.5).

As in Sections 1.4 and 1.5, we denote by ω(T ) the exponential growth boundof T , and by hol(T ) the abscissa of holomorphy of T . Recall that hol(T ) ≤ abs(T ) ≤ω(T ).

We will consider Laplace transforms of operator-valued functions on manyoccasions in this and subsequent chapters. The following definition will be helpful.

Definition 3.1.4. Let λ0 ∈ R and let R : (λ0,∞) → L(X) be a function. Wesay that R is a Laplace transform if there exists a strongly continuous functionT : R+ → L(X) such that abs(T ) ≤ λ0 and

R(λ) = T (λ) (λ > λ0).

The following proposition is a simple consequence of the uniqueness theorem.

Proposition 3.1.5. Let T : R+ → L(X) be strongly continuous such that abs(T ) <∞. Let ω > abs(T ). Then the following hold:

a) If B ∈ L(X) such that BT (λ) = T (λ)B for all λ > ω, then BT (t) = T (t)Bfor all t ≥ 0.

b) In particular, if T (μ)T (λ) = T (λ)T (μ) for all λ, μ > 0, then T (t)T (s) =T (s)T (t) for all t, s ≥ 0.

Proof. a) For x ∈ X and λ > ω, one has∫ ∞

0

e−λtT (t)Bxdt = T (λ)Bx = BT (λ)x =

∫ ∞

0

e−λtBT (t)x dt.

It follows from the uniqueness theorem that T (t)Bx = BT (t)x for all t ≥ 0.b) Let μ > ω. It follows from a) that T (μ)T (t) = T (t)T (μ) for all t ≥ 0.

Fixing t ≥ 0 and applying a) to B := T (t) shows that T (s)T (t) = T (t)T (s) for alls ≥ 0.

Now we introduce C0-semigroups.

Definition 3.1.6. A C0-semigroup is a strongly continuous function T : R+ → L(X)such that

T (t+ s) = T (t)T (s) (t, s ≥ 0),

T (0) = I.

In the next theorem we show that C0-semigroups are exactly those stronglycontinuous operator-valued functions whose Laplace transforms are resolvents. Itis remarkable that C0-semigroups are automatically exponentially bounded.

Theorem 3.1.7. Let T : R+ → L(X) be a strongly continuous function. The fol-lowing assertions are equivalent:

(i) abs(T ) <∞ and there exists an operator A such that (λ0,∞) ⊂ ρ(A) and

T (λ) = R(λ,A) (λ > λ0)

for some λ0 > abs(T ).

(ii) T is a C0-semigroup.

In that case, ω(T ) < ∞, {Reλ > hol(T )} ⊂ ρ(A) and T (λ) = R(λ,A) wheneverReλ > hol(T ).

Proof. a) Let T be a C0-semigroup. We show first that ω(T ) < ∞. Let M :=sup0≤t≤1 ‖T (t)‖. Then M < ∞ by the uniform boundedness principle. Let ω =logM . Let t ∈ R+. Take n ∈ N0 and s ∈ [0, 1) such that t = n + s. Then‖T (t)‖ = ‖T (s)T (1)n‖ ≤MMn = Meωn ≤Meωt.

b) Assume that abs(T ) <∞. Let μ > λ > abs(T ). Then integration by partsyields for x ∈ X,

T (λ)x− T (μ)x

μ− λ=

∫ ∞

0

e(λ−μ)tT (λ)x dt−∫ ∞

0

1

μ− λe(λ−μ)te−λtT (t)x dt

=

∫ ∞

0

e(λ−μ)t

∫ ∞

0

e−λsT (s)x ds dt

−∫ ∞

0

e(λ−μ)t

∫ t

0

e−λsT (s)x ds dt

=

∫ ∞

0

e(λ−μ)t

∫ ∞

t

e−λsT (s)x ds dt


=

∫ ∞

0

e−μt

∫ ∞

t

e−λ(s−t)T (s)x ds dt

=

∫ ∞

0

e−μt

∫ ∞

0

e−λsT (s+ t)x ds dt.

On the other hand,

T (μ)T (λ)x =

∫ ∞

0

e−μt

∫ ∞

0

e−λsT (s)T (t)x ds dt.

So it follows from the uniqueness theorem (Theorem 1.7.3) that (T (λ))λ>abs(T )

is a pseudo-resolvent (see Appendix B) if and only if T satisfies T (s + t) =T (s)T (t) (s, t ≥ 0).

Now assume that T is a pseudo-resolvent. Then T (0) is a projection. More-over, T (0)x = 0 if and only if T (t)x = T (t)T (0)x = 0 for all t ≥ 0. Thus bythe uniqueness theorem, T (0)x = 0 if and only if T (λ)x = 0 (λ > ω(T )). ByProposition B.6, (T (λ))λ>abs(T ) is a resolvent if and only if T (0) = I. This provesthat (i) ⇔ (ii).

c) It follows from (i) and Proposition B.5 that {Reλ > hol(T )} ⊂ ρ(A) andT (λ) = R(λ,A) whenever Reλ > hol(T ).

Definition 3.1.8. Let T be a C0-semigroup. The generator of T is defined as theoperator A on X such that (ω(T ),∞) ⊂ ρ(A) and T (λ) = R(λ,A) for all λ > ω(T ).

Thus, an operator A is the generator of a C0-semigroup if and only if itsresolvent is a Laplace transform in the sense of Definiton 3.1.4.

In the following proposition we collect the diverse relations of a C0-semigroupand its generator. These properties will be used frequently without further refer-ence.

Proposition 3.1.9. Let T be a C0-semigroup on X and let A be its generator. Thenthe following properties hold:

a) limλ→∞ λR(λ,A)x = x for all x ∈ X; in particular, A is densely defined.

b) For all x ∈ X, the function ux(t) := T (t)x is a mild solution of (ACP0).

c) R(λ,A)T (t) = T (t)R(λ,A) for all λ ∈ ρ(A) and t ≥ 0.

d) x ∈ D(A) implies T (t)x ∈ D(A) and AT (t)x = T (t)Ax.

e)∫ t

0T (s)x ds ∈ D(A) and A

∫ t

0T (s)x ds = T (t)x− x for all x ∈ X and t ≥ 0.

f) Let x, y ∈ X. Then x ∈ D(A) and Ax = y if and only if∫ t

0T (s)y ds =

T (t)x− x for all t ≥ 0.

g) Let x ∈ X. Then x ∈ D(A) if and only if y = limt↓0 1t (T (t)x− x) exists. In

that case, Ax = y.


h) T (·)x is a classical solution of (ACP0) if and only if x ∈ D(A).

i) If λ ∈ C then (eλtT (t))t≥0 is a C0-semigroup and A+ λ is its generator.

j) Let x ∈ X and λ ∈ C. Then x ∈ D(A) and Ax = λx if and only if T (t)x =eλtx for all t ≥ 0.

Proof. a) follows from the following Abelian argument. There exist M ≥ 0 andω ∈ R such that

‖T (t)‖ ≤Meωt (t ≥ 0).

Let ε > 0 and x ∈ X . There exists τ > 0 such that ‖T (t)x−x‖ ≤ ε for all t ∈ [0, τ ].Therefore

lim supλ→∞

‖λR(λ,A)x− x‖

≤ lim supλ→∞

∥∥∥∥λ∫ ∞

0

e−λt(T (t)x− x) dt

∥∥∥∥≤ lim sup

λ→∞

{λ

∫ τ

0

e−λtε dt+ λ

∫ ∞

τ

e−λt(Meωt + 1) dt ‖x‖}

= ε.

b) By Theorem 3.1.3, a function u ∈ C(R+, X) is a mild solution if andonly if u(λ) = R(λ,A)x = T (λ)x for all λ > ω(T ). So the claim follows from theuniqueness theorem.

c) follows from Proposition 3.1.5.d) follows from c) (by Proposition B.7).e) follows from b).

f) Let x ∈ D(A). Then by d), T (s)Ax = AT (s)x. Hence by e),∫ t

0T (s)Axds

= A∫ t

0T (s)x ds = T (t)x − x. Conversely, let x, y ∈ X such that

∫ t

0T (s)y ds =

T (t)x− x for all t ≥ 0. Then

R(λ,A)y = λ

∫ ∞

0

e−λt

∫ t

0

T (s)y ds dt

= λ

∫ ∞

0

e−λt(T (t)x− x) dt

= λR(λ,A)x− x.

Thus x ∈ D(A) and y = λx− (λ− A)x = Ax.

g) Let x ∈ D(A). Then by f), 1t(T (t)x−x) = 1

t

∫ t

0T (s)Axds→ Ax as t→ 0.

Conversely, let x, y ∈ X such that y = limt↓0 1t (T (t)x−x) = limt↓0 1

tA∫ t

0T (s)x ds.

Since A is closed, it follows that x ∈ D(A) and Ax = y.

h) Let x ∈ D(A). Then T (t)x = x +∫ t

0T (s)Axds by e). Thus, T (·)x ∈

C1(R+, X) and the claim follows from Proposition 3.1.2. Conversely, if T (·)x is aclassical solution, then x = T (0)x ∈ D(A) by definition.

i) follows from Theorem 3.1.7.j) Replacing A by A − λ we may assume that λ = 0. Now the claim follows

from f).

Property g) is sometimes expressed by saying that A is the infinitesimalgenerator of T .

Since almost all the C0-semigroups which arise naturally from differentialoperators cannot be written down explicitly, we do not give examples in this sec-tion. However the reader who wishes to see explicit examples may look already toExamples 3.3.10, 3.4.8, 3.7.5, 3.7.6 and 3.7.9, and to various examples in Chapter5.

We note here that if T : R+ → L(X) satisfies T (t+ s) = T (t)T (s) (t, s ≥ 0)and limt↓0 ‖T (t)x − x‖ = 0 (x ∈ X), then T is a C0-semigroup. To see this, wehave to show that T is strongly continuous at t > 0. Right-continuity followsimmediately from the estimate ‖T (t + h)x − T (t)x‖ ≤ ‖T (t)‖ ‖T (h)x − x‖. Forleft-continuity, note that the assumptions imply that there exist M > 0 and δ > 0such that ‖T (h)‖ ≤ M whenever 0 < h < δ (otherwise, there exist tn ↓ 0 suchthat ‖T (tn)‖ → ∞ and, by the uniform boundedness theorem, there exists x ∈ Xsuch that (T (tn)x) is unbounded, which is a contradiction). Hence for 0 < h < δ,we have ‖(T (t− h)x− T (t)x‖ ≤ ‖T (δ − h)‖ ‖T (t− δ)‖ ‖x− T (h)x‖ → 0 as h ↓ 0.Since T (0)T (t)x = T (t)x, letting t ↓ 0 shows that T (0) = I.

The following result characterizes C0-semigroups which are norm-continuouson R+. It also describes the situation when the generator A of a C0-semigroup Tis bounded. Since A is closed, this is equivalent to saying that D(A) = X.

Theorem 3.1.10. Let A be the generator of a C0-semigroup T . The following as-sertions are equivalent:

(i) The operator A is bounded; i.e., D(A) = X.

(ii) limt↓0 ‖T (t)− I‖ = 0.

In that case, T (t) = etA :=∑∞

k=0tkAk

k!(t ≥ 0).

Proof. (i) ⇒ (ii): Assume that A is bounded. Then clearly, T (t) :=∑∞

k=0tkAk

k!defines a continuous mapping T : R+ → L(X) such that T (0) = I and ‖T (t)‖ ≤et‖A‖. Let λ > ‖A‖. Then∫ ∞

0

e−λtT (t) dt =∞∑k=0

Ak

k!

∫ ∞

0

e−λttk dt =∞∑k=0

Akλ−(k+1) = R(λ,A).

Thus, T is a C0-semigroup and A is its generator by Definition 3.1.8.(ii) ⇒ (i): It follows from Proposition 4.1.3 or direct computation as in the

proof of Proposition 3.1.9 a) that limλ→∞ ‖λR(λ,A)− I‖ = 0. Thus, there existsλ > ω(T ) such that ‖λR(λ,A)− I‖ < 1/2. This implies that λR(λ,A) is invertiblein L(X). In particular, D(A) = λR(λ,A)X = X .

Now we consider uniqueness of mild solutions of (ACP )0.

Proposition 3.1.11. Let T be a C0-semigroup and A be its generator. Let τ >0, x ∈ X. Let u ∈ C([0, τ ], X) such that

∫ t

0u(s) ds ∈ D(A) and

A

∫ t

0

u(s) ds = u(t)− x

for all t ∈ [0, τ ]. Then u(t) = T (t)x.

Proof. Let v(t) =∫ t

0(u(s) − T (s)x) ds. Then by hypothesis and by Proposition

3.1.9 e), v(t) ∈ D(A) (0 ≤ t ≤ τ). Moreover, v′(t) = Av(t) (0 ≤ t ≤ τ) and

v(0) = 0. We show that v ≡ 0. Let S(t)y :=∫ t

0T (s)y ds. Then S(t)y ∈ D(A) and

AS(t)y = T (t)y − y for all y ∈ X , by Proposition 3.1.9 e). Let 0 < t ≤ τ, w(s) :=S(t− s)v(s), 0 ≤ s ≤ t. Then

w′(s) = −T (t− s)v(s) + S(t− s)v′(s)= −T (t− s)v(s) + S(t− s)Av(s)

= −T (t− s)v(s) + AS(t− s)v(s)

= −v(s).

Since w(t) = w(0) = 0, we conclude that

0 = w(t) =

∫ t

0

w′(s) ds = −∫ t

0

v(s) ds.

Since t ∈ (0, τ ] is arbitrary, it follows that v(s) = 0 for s ∈ [0, τ ].

Proposition 3.1.9 and Proposition 3.1.11 show in particular that the abstractCauchy problem (ACP0) is well-posed (in the sense of mild solutions) wheneverthe operator A generates a C0-semigroup T . Moreover, the orbits are given byT (·)x where x is the initial value.

Now we show the converse assertion. If (ACP0) is mildly well-posed (i.e.,for each x there exists a unique mild solution), then the operator generates aC0-semigroup. More precisely, we have the following result.

Theorem 3.1.12. Let A be a closed operator. The following assertions are equiva-lent:

(i) For all x ∈ X there exists a unique mild solution of (ACP0).

(ii) The operator A generates a C0-semigroup.

(iii) ρ(A) �= ∅ and for all x ∈ D(A) there exists a unique classical solution of(ACP0).

When these assertions hold, the mild solution of (ACP0) is given by u(t) = T (t)x.

Proof. (i) ⇒ (ii): Let ux be the mild solution for the initial value x ∈ X. It followsfrom uniqueness that ux(t) is linear in x. So for each t ≥ 0 there exists a linearmapping T (t) : X → X such that T (t)x = ux(t) for all x ∈ X. We show that T (t)is continuous. Denote by Φ : X → C(R+, X) the mapping Φ(x) = ux. Note thatC(R+, X) is a Frechet space for the topology of uniform convergence on intervalsof the form [0, τ ] where τ > 0. The mapping Φ is linear. We show that Φ has aclosed graph. In fact, let xn → x in X and uxn

→ u in C(R+, X). Let t > 0. Then∫ t

0uxn

(s) ds converges to∫ t

0u(s) ds as n → ∞. Since A

∫ t

0uxn

(s) ds = uxn(t) −

xn and since A is closed, it follows that∫ t

0u(s) ds ∈ D(A) and A

∫ t

0u(s) ds =

limn→∞ uxn(t)− xn = u(t)− x. Thus u(t) = T (t)x; i.e., u = Φ(x). It follows from

the closed graph theorem that Φ is continuous. This implies that T (t) ∈ L(X) forall t ≥ 0.

Let u be a mild solution of (ACP0) with initial value x. Then it is easy tosee that u(·+ s) is a mild solution for the initial value u(s). So uniqueness impliesthat T (t+ s)x = T (t)T (s)x.

We have shown that T is a C0-semigroup. Let B be the generator of T . ThenR(λ,B) = T (λ) (λ > ω(T )). On the other hand, by Theorem 3.1.3, T (λ)x ∈ D(A)and (λ − A)T (λ)x = x for all x ∈ X and λ > ω(T ). Thus, D(B) ⊂ D(A) and(λ − A)R(λ,B)x = x (x ∈ X) if λ > ω(T ). If we show that (λ − A) is injective,then it follows that λ ∈ ρ(A) and R(λ,A) = R(λ,B). Thus, A = B.

Assume that λ > ω(T ) and let x ∈ D(A) such that (λ − A)x = 0. Thenu(t) := eλtx is a mild solution. Thus, T (t)x = eλtx. Since ω(T ) < λ, it followsthat x = 0.

(ii) ⇒ (iii) follows from Proposition 3.1.9 b) and Proposition 3.1.11.(iii) ⇒ (i): Let λ ∈ ρ(A). Let x ∈ X. There exists a classical solution v of

(ACP0) with initial value R(λ,A)x. It is easy to check that u(t) := (λ − A)v(t)defines a mild solution of (ACP0) with initial value x. This shows existence. Inorder to show uniqueness, let u be a mild solution for the initial value x = 0. Thenv(t) :=

∫ t

0u(s) ds defines a classical solution for the initial value 0. Hence v(t) = 0

for all t ≥ 0 by assumption. It follows that u(t) = 0 (t ≥ 0).

As a corollary of Theorem 3.1.12 we show that one can also characterize C0-semigroups and their generator by property e) of Proposition 3.1.9. This will beuseful later.

Corollary 3.1.13. Let A be a closed operator on X and T : R+ → L(X) be strongly

continuous such that∫ t

0T (s)x ds ∈ D(A) and

A

∫ t

0

T (s)x ds = T (t)x− x

for all x ∈ X, t ≥ 0. Assume that T (t)x ∈ D(A) and AT (t)x = T (t)Ax for allx ∈ D(A), t ≥ 0. Then T is a C0-semigroup and A is its generator.

Proof. Let x ∈ X. Then u(t) := T (t)x defines a mild solution of (ACP0). As in the


proof of Proposition 3.1.11, u is the unique mild solution. Now the claim followsfrom Theorem 3.1.12.

We show by an example that the condition that A has non-empty resolventset cannot be omitted in assertion (iii) of Theorem 3.1.12; i.e., it can happenthat the abstract Cauchy problem (ACP0) is well-posed in the sense of classicalsolutions without A being the generator of a C0-semigroup.

Example 3.1.14. Let B be a densely defined closed operator on a Banach space Ysuch that D(B) �= Y . Consider the operator A on X := Y × Y given by

A =

(0 B0 0

)with domain Y ×D(B). Then A is closed and densely defined. Moreover, for all(x, y) ∈ D(A),

u(t) = (x+ tBy, y) (t ≥ 0)

is the unique classical solution of (ACP0). However, there does not exist a mildsolution for an initial value (x, y) if y ∈ Y \D(B). This is easy to see.

Now we consider the inhomogeneous Cauchy problem. In contrast to the ho-mogeneous case, we consider this on a bounded interval [0, τ ] where τ ∈ (0,∞), butresults on R+ can be deduced by letting τ vary. We shall apply Laplace transformtechniques to inhomogeneous Cauchy problems on R+ in Section 5.6.

Let A be a closed operator and let f ∈ L1([0, τ ], X) where τ > 0. We considerthe inhomogeneous Cauchy problem

(ACPf )

{u′(t) = Au(t) + f(t) (t ∈ [0, τ ]),

u(0) = x,

where x ∈ X. A function u ∈ C([0, τ ], X) is called a mild solution of (ACPf ) if∫ t


u(t) = x+A

∫ t

0

u(s) ds+

∫ t

0

f(s) ds (t ∈ [0, τ ]).

Assume that f ∈ C([0, τ ], X). Then we define a classical solution as a functionu ∈ C1([0, τ ], X) such that u(t) ∈ D(A) for all t ∈ [0, τ ] and such that (ACPf )is valid. Note that in that case, since Au(t) = u′(t) − f(t) (t ∈ [0, τ ]), one hasu ∈ C([0, τ ], D(A)), where D(A) is seen as a Banach space with the graph norm.Since A is closed, the proof of Proposition 3.1.2 is also valid in the inhomogeneouscase, so the following holds.

Proposition 3.1.15. Let f ∈ C([0, τ ], X) and u ∈ C([0, τ ], X) be a mild solution of(ACPf ). Then u is a classical solution if and only if u ∈ C1([0, τ ], X).


In the case when A generates a C0-semigroup there always exists a mildsolution.

Proposition 3.1.16. Let A be the generator of a C0-semigroup T on X. Then forevery f ∈ L1([0, τ ], X) the problem (ACPf ) has a unique mild solution u given by

u(t) = T (t)x+

∫ t

0

T (t− s)f(s) ds (t ∈ [0, τ ]). (3.4)

Sometimes, (3.4) is called the variation of constants formula for the solution.

Proof. Uniqueness: Let u1, u2 ∈ C([0, τ ], X) be two mild solutions of (ACPf ). Then

u := u1 − u2 ∈ C([0, τ ], X), u(0) = 0 and A∫ t

0u(s) ds = u(t) for all t ∈ [0, τ ]. It

follows from Proposition 3.1.11 that u ≡ 0.Existence: We have seen that T (·)x is a mild solution of the homogeneous

Cauchy problem. It remains to show that v(t) :=∫ t

0T (t − s)f(s) ds is a mild

solution of (ACPf ) with initial value x = 0. Extending f by 0 to R+, Proposition1.3.4 shows that v ∈ C([0, τ ], X). Using Proposition 3.1.9 e) and Fubini’s theoremwe obtain

A

∫ t

0

v(s) ds = A

∫ t

0

∫ s

0

T (s− r)f(r) dr ds

= A

∫ t

0

∫ t

r

T (s− r)f(r) ds dr

=

∫ t

0

A

∫ t−r

0

T (s)f(r) ds dr

=

∫ t

0

(T (t− r)f(r)− f(r)) dr

= v(t)−∫ t

0

f(r) dr.

This proves the claim.

Corollary 3.1.17. Let A be the generator of a C0-semigroup. Let x ∈ D(A), f0 ∈X, f(t) = f0 +

∫ t

0f ′(s) ds (t ∈ [0, τ ]) for some function f ′ ∈ L1([0, τ ], X). Then

the function u defined by (3.4) is a classical solution of (ACPf ).

This follows from Proposition 3.1.16, Proposition 1.3.6 and Proposition 3.1.15.This result will later be extended to a class of operators which are not densely de-fined (Theorem 3.5.2).

Finally, given a closed operator A on X , we consider the Cauchy problem onthe line

ACP0(R)

{u′(t) = Au(t) (t ∈ R),

u(0) = x,

where x ∈ X. A function u ∈ C(R, X) is called a mild solution of ACP0(R) if∫ t


A

∫ t

0

u(s) ds = u(t)− x for all t ∈ R.

Proposition 3.1.18. Assume that A is an operator such that A generates a C0-semigroup T+ and −A generates a C0-semigroup T−. Define

U(t) =

{T+(t) if t ≥ 0,

T−(−t) if t < 0.(3.5)

Then U : R → L(X) is strongly continuous, U(0) = I and U(t + s) = U(t)U(s)(t, s ∈ R).

Proof. Note first that T+(t)T−(s) = T−(s)T+(t) for s, t ≥ 0, by Proposition 3.1.5.The only assertion which is not obvious is to show that U(t− s) = U(t)U(−s) ift ≥ 0, s ≥ 0. We can assume that 0 ≤ s ≤ t (replacing A by −A for the othercase). Let x ∈ X, t ≥ 0, v(s) := T+(t− s)x for s ∈ [0, t]. Then for 0 ≤ r ≤ t,

−A∫ r

0

v(s) ds = −A∫ r

0

T+(t− s)x ds = −A∫ t

t−r

T+(s)x ds

= T+(t− r)x− T+(t)x = v(r)− T+(t)x.

Thus, v is a mild solution of the problem{v′(s) = −Av(s) (0 < s ≤ t),

v(0) = T+(t)x.

It follows from Proposition 3.1.11 that v(s) = T−(s)T+(t)x. Hence,

U(t− s)x = T+(t− s)x = v(s) = T−(s)T+(t)x = U(−s)U(t)x.

Definition 3.1.19. An operator A on X is said to generate a C0-group if A and−A generate C0-semigroups. In that case, the function U : R → L(X) defined by(3.5) is called the C0-group generated by A.

Proposition 3.1.20. A closed operator A generates a C0-group if and only if forevery x ∈ X there exists a unique mild solution u of ACP0(R). In that case,u(t) = U(t)x (t ∈ R), where U is the C0-group generated by A. If x ∈ D(A), thenU(·)x ∈ C1(R, X), U(t)x ∈ D(A) for all t ∈ R and d

dtU(t)x = AU(t)x (t ∈ R).

Proof. Assume that ACP0(R) is mildly well-posed; i.e., for all x ∈ X there existsa unique mild solution of ACP0(R). Then it is clear that for each x ∈ X thereexists a mild solution of

(CP )±

{u′(t) = ±Au(t) (t ≥ 0),

u(0) = x.

The solutions of (CP )+ and (CP )− are both unique. In fact, let u ∈ C(R+, X)be a mild solution of (CP )+ with initial value u(0) = 0. Then extending u by 0on (−∞, 0) one obtains a mild solution of ACP0(R). Hence u ≡ 0 by assumption.The same argument is valid for (CP )−. Now it follows from Theorem 3.1.12 thatA and −A are both generators of C0-semigroups.

Conversely, if A generates a C0-group, then it is easy to see that U(·)x isa mild solution of ACP0(R). The remaining properties follow directly from thecorresponding results for semigroups.

If A generates a C0-semigroup T , then mild solutions of ACP0(R) can bedescribed differently.

Definition 3.1.21. A function u ∈ C(R, X) is called a complete orbit of T if

u(t+ s) = T (t)u(s) for all t ≥ 0, s ∈ R.

Proposition 3.1.22. Let A be the generator of a C0-semigroup T and let u ∈C(R, X), x = u(0). Then u is a mild solution of ACP0(R) if and only if u isa complete orbit.

Proof. Assume that u is a complete orbit. Let x = u(0). Since T (t)x = u(t) fort ≥ 0, we have

A

∫ t

0

u(s) ds = u(t)− x (t ≥ 0).

For t < 0 we have T (−t)u(t) = u(0) = x and

A

∫ t

0

u(r) dr = A

∫ 0

−t

u(r + t) dr = −A∫ −t

0

u(r + t) dr

= −A∫ −t

0

T (r)u(t) dr = u(t)− T (−t)u(t) = u(t)− x.

Thus u is a mild solution of ACP0(R).Conversely, assume that u is a mild solution of ACP0(R) with x = u(0). Let

s ∈ R. Then for t ≥ 0,

A

∫ t

0

u(r + s) dr = A

∫ s+t

s

u(r) dr = u(t+ s)− u(s).

Thus u(·+ s) is a mild solution of (ACP0) for x = u(s). Hence u(t+ s) = T (t)u(s)for t ≥ 0.

If a C0-semigroup T extends to a C0-group, then T (t) is invertible for allt > 0. The following converse statement is sometimes useful (and will be neededin Proposition 4.7.2, for example).

Proposition 3.1.23. Let A be the generator of a C0-semigroup T . If there existst0 > 0 such that T (t0) is invertible, then A generates a C0-group.

3.2. INTEGRATED SEMIGROUPS AND CAUCHY PROBLEMS 121

Proof. a) Let t > 0. We show that T (t) is invertible. Let T (t)x = 0. Choose n ∈ Nsuch that nt0 > t. Then T (nt0)x = T (nt0 − t)T (t)x = 0. Since T (nt0) = T (t0)

n

is invertible, it follows that x = 0. Thus, T (t) is injective. Let y ∈ X. Let x :=T (nt0 − t)T (nt0)

−1y. Then T (t)x = y. Thus, T (t) is surjective.b) Define U(t) := T (t) for t ≥ 0 and U(t) := T (−t)−1 for t < 0. Then

U : R → L(X) satisfies U(t + s) = U(t)U(s) for all t, s ∈ R. Let x ∈ X, t0 ∈ R.Let t1 > max{−t0, 0}. Then

limt→t0

U(t)x = T (t1)−1 lim

t→t0T (t+ t1)x = U(t0)x.

Thus U is strongly continuous.c) We show that the generator of (U(−t))t≥0 is −A. Let x ∈ D(A). Then

limt↓0

1

t(U(−t)x− x) = lim

t↓0U(−1)

(1

tT (1− t)x− T (1)x

)= T (1)−1(−AT (1)x) = −Ax.

Conversely, if x ∈ X such that y := limt↓0 1t (U(−t)x − x) exists, it follows as

above that −y = limt↓0 1t(T (t)x − x). Thus, x ∈ D(A) and −Ax = y. Now the

claim follows from Proposition 3.1.9.

3.2 Integrated Semigroups and Cauchy Problems

Let T be a C0-semigroup on a Banach space X with generator A. For k ∈ N wedefine S : R+ → L(X) by

S(t)x :=

∫ t

0

(t− s)k−1

(k − 1)!T (s)x ds (t ≥ 0, x ∈ X).

By Theorem 3.1.7, there exist M,ω ≥ 0 such that ‖T (t) ‖ ≤ Meωt for all t ≥ 0.Taking Laplace transforms and integrating by parts yields

R(λ,A) = λk

∫ ∞

0

e−λtS(t) dt (λ > ω). (3.6)

Here, the Laplace integral is understood in the sense of Section 1.4, but, for eachx ∈ X and λ > ω, one has R(λ,A)x = λk

∫∞0

e−λtS(t)x dt as an absolutelyconvergent Bochner integral.

The above formula (3.6) is the basic idea behind the following definition.Consider an arbitrary strongly continuous function S : R+ → L(X). We recall fromProposition 1.4.5 that abs(S) < ∞ if and only if there exist constants M,ω ≥ 0such that ∥∥∥∥∫ t

0

S(s) ds

∥∥∥∥ ≤Meωt (t ≥ 0). (3.7)


In that case, the Laplace integral

S(λ)x :=

∫ ∞

0

e−λtS(t)x dt := limτ→∞

∫ τ

0

e−λtS(t)x dt

exists for all λ ∈ C with Reλ > ω and all x ∈ X and defines a bounded operatorS(λ) on X. Hence, the following definition is meaningful.

Definition 3.2.1. Let A be an operator on a Banach space X and k ∈ N0. We call Athe generator of a k-times integrated semigroup if there exist ω ≥ 0 and a stronglycontinuous function S : R+ → L(X) such that abs(S) ≤ ω, (ω,∞) ⊂ ρ(A) and

R(λ,A) = λk

∫ ∞

0

e−λtS(t) dt (λ > ω). (3.8)

In this case, S is called the k-times integrated semigroup generated by A. If k = 1we also use the notion once integrated semigroup.

By Theorem 3.1.7, a 0-times integrated semigroup is the same as a C0-semigroup. The discussion above shows that if A generates a 0-times integratedsemigroup, then A generates a k-times integrated semigroup for every k ∈ N. Thesame argument shows that if A generates a k-times integrated semigroup, then Agenerates an n-times integrated semigroup for every n > k.

As in the situation of C0-semigroups we collect diverse relations of an inte-grated semigroup and its generator.

Lemma 3.2.2. Let k ∈ N and let S be a k-times integrated semigroup on X withgenerator A. Then the following hold:

a) R(μ,A)S(t) = S(t)R(μ,A) (t ≥ 0, μ ∈ ρ(A)).

b) If x ∈ D(A), then S(t)x ∈ D(A) and AS(t)x = S(t)Ax for all t ≥ 0.

c) Let x ∈ D(A) and t ≥ 0. Then∫ t

0

S(s)Axds = S(t)x− tk

k!x.

In particular, ddt (S(t)x) = S(t)Ax+ tk−1

(k−1)!x.

d) Let x ∈ X and t ≥ 0. Then∫ t

0S(s)x ds ∈ D(A) and

A

∫ t

0

S(s)x ds = S(t)x− tk

k!x.

In particular, S(0) = 0.

e) Let x, y ∈ X such that∫ t

0S(s)y ds = S(t)x− tk

k!x for all t ≥ 0. Then x ∈ D(A)and Ax = y.


Proof. By definition and assertion (3.7), there exist constants M,ω ≥ 0 such that

(ω,∞) ⊂ ρ(A) and || ∫ t

0S(s) ds|| ≤Meωt for t ≥ 0. In the following let λ > ω.

a) follows from Proposition 3.1.5.b) is implied by a) (by Proposition B.7).c) Let x ∈ D(A). Integrating by parts yields

λk+1

∫ ∞

0

e−λt tk

k!x dt = R(λ,A)(λ− A)x

= λk+1

∫ ∞

0

e−λtS(t)x dt− λk

∫ ∞

0

e−λtS(t)Axdt

= λk+1

∫ ∞

0

e−λtS(t)x dt

−λk+1

∫ ∞

0

e−λt

∫ t

0

S(s)Axds dt.

The uniqueness theorem implies the assertion.d) Let μ ∈ ρ(A) and x ∈ X . By a) and c) we have∫ t

0

S(s)x ds = μR(μ,A)

∫ t

0

S(s)x ds−R(μ,A)S(t)x+tk

k!R(μ,A)x.

Hence∫ t

0S(s)x ds ∈ D(A) and

(μ−A)

∫ t

0

S(s)x ds = μ

∫ t

0

S(s)x ds− S(t)x+tk

k!x.

e) Let x, y ∈ X such that∫ t

0S(s)y ds = S(t)x− tk

k!x for all t ≥ 0. Then

R(λ,A)(λx− y) = λk+1

∫ ∞

0

e−λtS(t)x dt− λk

∫ ∞

0

e−λtS(t)y dt

= λk+1

∫ ∞

0

e−λtS(t)x dt− λk+1

∫ ∞

0

e−λt

∫ t

0

S(s)y ds dt

= x.

Hence x ∈ D(A) and λx− y = λx− Ax, which implies that Ax = y.

Remark 3.2.3. Observe that in contrast to the situation of C0-semigroups, gen-erators of k-times integrated semigroups for k ≥ 1 need not be densely defined.However, assertion d) of Lemma 3.2.2 implies that S(t)x ∈ D(A) for t ≥ 0 andx ∈ X .

We saw in Theorem 3.1.7 that C0-semigroups are precisely those operator-valued functions whose Laplace transforms are resolvents R(λ,A) of operatorsA. By definition, k-times integrated semigroups are exactly those operator-valuedfunctions whose Laplace transforms are λ−kR(λ,A) for operators A. In the follow-ing proposition we show that this property corresponds to the functional equation(3.9) for S.


Proposition 3.2.4. Let S : R+ → L(X) be a strongly continuous function satisfying

‖ ∫ t

0S(s) ds ‖ ≤Meωt (t ≥ 0) for some M,ω ≥ 0. Let k ∈ N. For λ > ω set

R(λ) := λk

∫ ∞

0

e−λtS(t) dt.

Then the following assertions are equivalent:

(i) There exists an operator A such that (ω,∞) ⊂ ρ(A) and R(λ) = (λ− A)−1

for λ > ω.

(ii) For s, t ≥ 0,

S(t)S(s) =1

(k − 1)!

[∫ t+s

t

(s+ t− r)k−1S(r) dr

−∫ s

0

(s+ t− r)k−1S(r) dr

], (3.9)

and S(t)x = 0 for all t ≥ 0 implies x = 0.

Proof. We first claim that {R(λ) : λ > ω} is a pseudo-resolvent if and only if (3.9)holds. Since

R(λ)

λk

R(μ)

μk=

∫ ∞

0

e−λt

∫ ∞

0

e−μsS(t)S(s) ds dt (λ, μ > ω),

the claim follows from the uniqueness theorem (Theorem 1.7.3) provided we areable to prove that

1

μ− λ

1

λk

1

μk(R(λ)−R(μ)) (3.10)

equals the term∫ ∞

0

e−λt

∫ ∞

0

e−μs 1

(k − 1)!

∫ s+t

t

(s+ t− r)k−1S(r) dr ds dt

−∫ ∞

0

e−λt

∫ ∞

0

e−μs 1

(k − 1)!

∫ s

0

(s+ t− r)k−1S(r) dr ds dt. (3.11)

Notice that (3.10) equals

1

μk

1

μ− λ

(R(λ)

λk− R(μ)

μk

)+

1

μ− λ

(1

μk− 1

λk

)R(μ)

μk=: I + II.

As in the proof of Theorem 3.1.7 we see that term I equals

1

μk

∫ ∞

0

e−μt

∫ ∞

0

e−λsS(t+ s) ds dt

=

∫ ∞

0

e−λt

∫ ∞

0

e−μs

∫ t+s

t

(t+ s− r)k−1

(k − 1)!S(r) dr ds dt.

Hence, it remains to show that term II is equal to the second term in (3.11). Thisfollows from the following computation:

−II =

k−1∑j=0

λ−(j+1)μ(j−k)S(μ)

=k−1∑j=0

λ−(j+1)

∫ ∞

0

e−μs

∫ s

0

(s− r)k−j−1

(k − j − 1)!S(r) dr ds

=k−1∑j=0

∫ ∞

0

e−λt tj

j!dt

∫ ∞

0

e−μs

∫ s

0

(s− r)k−j−1

(k − j − 1)!S(r) dr ds

=

∫ ∞

0

e−λt

∫ ∞

0

e−μs

∫ s

0

(s+ t− r)k−1

(k − 1)!S(r) dr ds dt.

Finally, recall that {R(λ) : λ > ω} is the resolvent of an operator A in X ifand only if KerR(λ) = {0} (see Proposition B.6). This is equivalent to the factthat S(t)x = 0 for all t ≥ 0 implies x = 0.

In contrast to the situation for semigroups the functional equation (3.9) inProposition 3.2.4 does not imply that abs(S) <∞. This will be shown at the endof this section in Remark 3.2.15.

It follows from Proposition 3.1.5 or from the above Proposition 3.2.4 that fora k-times integrated semigroup S we have

S(t)S(s) = S(s)S(t) (s, t ≥ 0). (3.12)

The above elementary properties of integrated semigroups will be used in thefollowing without further notice.

A particular example of an integrated semigroup is the antiderivative of asemigroup which is not necessarily strongly continuous at 0. In order to make thismore precise, consider a strongly continuous function T : (0,∞)→ L(X) satisfying

a) T (t+ s) = T (t)T (s) (s, t > 0),

b) there exists c > 0 such that ||T (t)|| ≤ c for all t ∈ (0, 1],

c) T (t)x = 0 for all t > 0 implies x = 0.

Then by the proof of Theorem 3.1.7, there exist constants M,ω ≥ 0 such that||T (t)|| ≤Meωt for all t > 0. For t ≥ 0 set

S(t) :=

∫ t

0

T (s) ds.

Then (S(t))t≥0 satisfies condition (ii) of Proposition 3.2.4 with k = 1. Hence, thereexists an operator A such that (ω,∞) ⊂ ρ(A) and

R(λ,A) = λ

∫ ∞

0

e−λtS(t) dt =

∫ ∞

0

e−λtT (t) dt (λ > ω). (3.13)

Definition 3.2.5. Let T : (0,∞) → L(X) be a strongly continuous function satis-fying assumptions a), b) and c) above. Let A be defined as in (3.13). Then T iscalled a semigroup and A is called its generator.

We will see in the following Section 3.3 that a semigroup T on X is a C0-semigroup on X if and only if D(A) = X.

Proposition 3.2.6. Let A be the generator of a k-times integrated semigroup S onX for some k ∈ N and let a ∈ C. Then A − a generates a k-times integratedsemigroup Sa on X which is given by

Sa(t) = e−atS(t) +k∑

j=1

(kj

)aj

∫ t

0

(t− s)j−1

(j − 1)!e−asS(s) ds.

Proof. Taking Laplace transforms of Sa we obtain, for μ sufficiently large∫ ∞

0

e−μtSa(t) dt =

∫ ∞

0

e−(μ+a)tS(t) dt

+

∫ ∞

0

e−(μ+a)tS(t) dt

k∑j=1

(kj

)ajμ−j

=R(μ+ a,A)

(μ+ a)k1

μk

k∑j=0

(kj

)ajμk−j

=R(μ+ a,A)

(μ+ a)k(μ+ a)k

μk=

R(μ,A− a)

μk.

Hence, the assertion follows directly from Definition 3.2.1.

Proposition 3.2.7. Let A be an operator on X and let μ ∈ ρ(A), k ∈ N. Then Agenerates a k-times integrated semigroup S on X if and only if there exists ω ∈ Rsuch that (ω,∞) ⊂ ρ(A) and R(·, A)R(μ,A)k is a Laplace transform T in the senseof Definition 3.1.4. In that case, ω(T ) <∞ if and only if ω(S) <∞.

Proof. By Proposition 3.2.6, the operator A generates a k-times integrated semi-group if and only if A−μ does so. By Definition 3.2.1 and Proposition 1.6.1 a), thisis equivalent to λ �→ (λ − μ)−kR(λ,A) being a Laplace transform. The resolventequation implies that

R(λ,A)R(μ,A)k =R(λ,A)

(μ− λ)k− R(μ,A)

(μ− λ)k− R(μ,A)2

(μ− λ)k−1− . . .− R(μ,A)k

(μ− λ)(3.14)

for λ, μ ∈ ρ(A), λ �= μ. The first assertion follows easily. Moreover, each step inthe passage between S and T preserves exponential boundedness.

In the following we characterize those operators which generate a k-timesintegrated semigroup for some k ∈ N simply by the fact that the resolvent ispolynomially bounded on a half-plane. The real characterization given in the nextsection determines precisely the order of integration.

Theorem 3.2.8. Let A be an operator and let k ∈ N.

a) Assume that there exists ω ≥ 0, M ≥ 0, b > 0 such that λ ∈ ρ(A) and‖R(λ,A)‖ ≤ M |λ|k−1−b whenever Re λ > ω. Then A generates a k-timesintegrated semigroup S satisfying ω(S) ≤ ω.

b) Conversely, if A generates a k-times integrated semigroup S such that ω(S) <∞, then for ω > max{ω(S), 0} there exists M such that λ ∈ ρ(A) and‖R(λ,A)‖ ≤M |λ|k whenever Reλ > ω.

Proof. a) Apply Theorem 2.5.1 to q(λ) := λbR(λ,A)/λk.b) Let max{ω(S), 0} < ω1 < ω. There exists M1 ≥ 0 such that ‖S(t)‖ ≤

M1eω1t (t ≥ 0). Hence, by Proposition B.5, λ ∈ ρ(A) and

‖R(λ,A)‖ =

∥∥∥∥λk

∫ ∞

0

e−λtS(t) dt

∥∥∥∥≤ |λ|kM1(Reλ− ω1)

−1 ≤ |λ|kM1(ω − ω1)−1

whenever Reλ > ω.

We now turn our attention to the inhomogenous Cauchy problem

(ACPf )

{u′(t) = Au(t) + f(t) (t ∈ [0, τ ]),

u(0) = x,(3.15)

where τ > 0, f ∈ L1([0, τ ], X), x ∈ X and A is assumed to be the generator ofa k-times integrated semigroup S on X for some k ∈ N. Recall from Section 3.1that by a mild solution of (ACPf ) we understand a function u ∈ C([0, τ ], X) such

that∫ t

0u(s) ds ∈ D(A) and u(t) = A

∫ t

0u(s) ds + x +

∫ t

0f(s) ds for all t ∈ [0, τ ]

and that by a classical solution of (ACPf ) we mean a function u ∈ C1([0, τ ], X)∩C([0, τ ], D(A)) satisfying (ACPf ) for all t ∈ [0, τ ]. For x ∈ X consider the functionv given by

v(t) := S(t)x+

∫ t

0

S(s)f(t− s) ds (t ∈ [0, τ ]). (3.16)

It follows from Proposition 1.3.4 that v ∈ C([0, τ ], X).

Lemma 3.2.9. a) If there is a mild solution u of (ACPf ), then v ∈ Ck([0, τ ], X)and u = v(k).

b) If there exists a classical solution u of (ACPf ), then v ∈ Ck+1([0, τ ], X) andu = v(k).

Proof. a) For 0 ≤ s ≤ t ≤ τ set w(s) := S(t− s)∫ s

0u(r)dr. Since

∫ s

0u(r)dr ∈ D(A)

for s ∈ [0, τ ], it follows from Lemma 3.2.2 c) that

w′(s) = −S(t− s)A

∫ s

0

u(r) dr − (t− s)k−1

(k − 1)!

∫ s

0

u(r) dr + S(t− s)u(s)

= − (t− s)k−1

(k − 1)!

∫ s

0

u(r) dr + S(t− s)

(x+

∫ s

0

f(r) dr

)(s ∈ [0, t]).


Since 0 = w(0)− w(t) = − ∫ t

0w′(s) ds we have∫ t

0

S(t− s)

(x+

∫ s

0

f(r) dr

)ds =

∫ t

0

(t− s)k−1

(k − 1)!

∫ s

0

u(r) dr ds.

Using this and Proposition 1.3.6, it follows that

u(t) =dk+1

dtk+1

(∫ t

0

(t− s)k−1

(k − 1)!

∫ s

0

u(r) dr ds

)=

dk

dtk

(S(t)x+

∫ t

0

S(s)f(t− s) ds

)= v(k)(t).

b) This follows immediately from Proposition 3.1.15.

Lemma 3.2.10. Let v be defined by (3.16). Assume that v ∈ Ck([0, τ ], X). Thenu := v(k) is a mild solution of (ACPf ). Moreover, if v ∈ Ck+1([0, τ ], X), thenu := v(k) is a classical solution of (ACPf ).

Proof. By Fubini’s theorem,∫ t

0

v(s) ds =

∫ t

0

S(s)x ds+

∫ t

0

∫ t−r

0

S(s)f(r) ds dr

for t ∈ [0, τ ]. By Lemma 3.2.2 d),∫ t

0S(s)x ds ∈ D(A),

∫ t−r

0S(s)f(r) ds ∈ D(A)

and A∫ t−r

0S(s)f(r) ds = S(t−r)f(r)− (t−r)k

k! f(r). By Proposition 1.1.7,∫ t

0v(s) ds

∈ D(A) and

A

∫ t

0

v(s) ds = v(t)− tk

k!x−

∫ t

0

(t− r)k

k!f(r) dr. (3.17)

Since A is closed and v ∈ Ck([0, τ ], X), it follows from (3.17) that v(j−1)(t) ∈ D(A)for t ∈ [0, τ ] and that

Av(j−1)(t) = v(j)(t)− tk−j

(k − j)!x−

∫ t

0

(t− r)k−j

(k − j)!f(r) dr (3.18)

for j = 1, . . . , k. Since v(0) = 0, this implies that v(j)(0) = 0 for j = 1, 2, . . . , k−1.It now follows from (3.18) for j = k that u := v(k) is a mild solution of (ACPf ). Ifv ∈ Ck+1([0, τ ], X), we may differentiate (3.18) once more and see that v(k)(t) ∈D(A) and that

Av(k)(t) = v(k+1)(t)− f(t) (t ∈ [0, τ ]). (3.19)

Hence u := v(k) satisfies u′(t) = Au(t) + f(t) for t ∈ [0, τ ]. Also, by (3.18) forj = k, u(0) = v(k)(0) = x.

Combining the above Lemmas 3.2.9 and 3.2.10 with Lemma 3.2.2 c) andProposition 1.3.6, we obtain the following corollary.


Corollary 3.2.11. Let A be the generator of a k-times integrated semigroup on Xfor some k ∈ N.

a) Assume that x ∈ D(Ak+1). Then there exists a unique classical solution of(ACP0).

b) Assume that f ∈ Ck+1([0, τ ], X) and that there exist xj ∈ D(A) for j =0, 1, . . . , k satisfying x0 = x, xj+1 = Axj + f (j)(0) (j = 0, 1, . . . , k). Thenthere exists a unique classical solution of (ACPf ).

c) Assume that f ∈ Ck([0, τ ], X) and that there exist xj ∈ D(A) for j =0, 1, . . . , k − 1 satisfying x0 = x, xj+1 = Axj + f (j)(0) (j = 0, 1, . . . , k − 1).Then there exists a unique mild solution of (ACPf ).

Remark 3.2.12. We remark that in contrast to the case of a C0-semigroup (seeCorollary 3.1.17) a mere regularity condition on the function f does not sufficeto ensure the existence of a classical solution of (ACPf ). Indeed, let A be thegenerator of a once integrated semigroup S on X such that A does not generatea C0-semigroup on X. Then there exists y ∈ X such that S(·)y /∈ C1([0, τ ], X).Consider the function f defined by f(t) := y for t ∈ [0, τ ]. If x = 0, then v(t) =∫ t

0S(s)y ds, but v /∈ C2([0, τ ], X).

In the following we turn our attention to the converse of Corollary 3.2.11; i.e.,we are aiming to show that A is the generator of an integrated semigroup wheneverthe associated Cauchy problem (ACP0) admits a unique classical solution for allinitial data x belonging to the domain of some power of A. To this end, we restrictourselves to the case of generators of exponentially bounded k-times integratedsemigroups; i.e., we assume that the function S in Definition 3.2.1 satisfies inaddition the property that ‖S(t)‖ ≤Meωt for all t ≥ 0 and some suitable constantsM,ω ≥ 0.

For x ∈ X consider then the “(k + 1)-times integrated version” of (ACP0),which is to find v ∈ C1(R+, X) ∩ C(R+, D(A)) satisfying

(ACPk+1)

⎧⎨⎩v′(t) = Av(t) +tk

k!x (t ≥ 0),

v(0) = 0.(3.20)

Assume that A generates an exponentially bounded k-times integrated semigroupS on X and define v by v(t) :=

∫ t

0S(s)x ds for t ≥ 0. Then by Lemma 3.2.2 d),

v′(t) = S(t)x = A

∫ t

0

S(s)x ds+tk

k!x (t ≥ 0) (3.21)

and v(0) = 0. Hence v is a classical solution of (ACPk+1). It is unique by Lemma3.2.9 and exponentially bounded since S is so. We have therefore proved the impli-cation (i)⇒ (ii) of the following result. Recall that a 0-times integrated semigroupis the same as a C0-semigroup, so the following may be compared with Theorem3.1.10.


Theorem 3.2.13. Let A be a closed operator on X and let k ∈ N0. The followingassertions are equivalent:

(i) A generates an exponentially bounded k-times integrated semigroup on X.

(ii) For all x ∈ X there exists a unique classical solution of (ACPk+1) which isexponentially bounded.

(iii) ρ(A) �= ∅ and for every x ∈ D(Ak+1) there exists a unique classical solutionof (ACP0) which is exponentially bounded.

Proof. The remarks before Theorem 3.2.13 imply the assertion (i)⇒ (ii). Moreover,the implication (i) ⇒ (iii) follows from Corollary 3.2.11. For the proof of theimplication (ii) ⇒ (i) we need the following “uniform exponential boundednessprinciple”.

Lemma 3.2.14. Let X,Y be Banach spaces and let V : R+ → L(X,Y ) be a func-tion. Assume that V (·)x is exponentially bounded for all x ∈ X. Then there existconstants M ≥ 0, ω ∈ R such that

‖V (t) ‖ ≤Meωt (t ≥ 0).

Proof. Observe that, for n ∈ N, the set Xn defined by

Xn :={x ∈ X : ‖V (t)x ‖ ≤ nent for all t ≥ 0

}is a closed subset of X. The hypothesis implies that X =

⋃n∈N Xn. Hence, by

Baire’s theorem, there exists n0 ∈ N such that Xn0has non-empty interior. It

follows that there exist z ∈ X, ε > 0,M ≥ 0 and ω ∈ R such that

‖V (t)x ‖ ≤Meωt (t ≥ 0)

provided ‖x− z ‖ ≤ ε. For ‖ y ‖ ≤ 1 we have

εe−ωt‖V (t)y ‖ ≤ e−ωt‖V (t)(εy + z) ‖+ e−ωt‖V (t)z ‖ ≤ 2M

for t ≥ 0. Thus ‖V (t) ‖ ≤ 2Mεeωt for t ≥ 0.

The above Lemma 3.2.14 enables us now to prove the implication (ii) ⇒ (i)in Theorem 3.2.13.

(ii) ⇒ (i): Denote by V (·)x the solution of (ACPk+1). The mapping V (t) :X → D(A) is linear by uniqueness. We even have that V (t) ∈ L(X,D(A)). Indeed,the space C(R+, D(A)) is a Frechet space for the seminorms

pm(f) := sup0≤t≤m

‖f(t)‖D(A).

Define a mapping Φ : X → C(R+, D(A)) by Φ(x) = V (·)x. Then Φ is closed andthe closed graph theorem implies that Φ is continuous (see the proof of Theorem

3.1.12). In particular, the mapping X → D(A), x �→ V (t)x is continuous for t ≥ 0.The hypothesis together with Lemma 3.2.14 implies that ‖V (t) ‖ ≤Meωt (t ≥ 0)for suitable constants M,ω ≥ 0. Therefore Q(λ)x := λk+1

∫∞0

e−λtV (t)x dt is welldefined for λ > ω. Since∥∥∥∥∫ t

0

AV (s)x ds

∥∥∥∥ =

∥∥∥∥V (t)x− tk+1

(k + 1)!x

∥∥∥∥ ≤ (Meωt +

tk+1

(k + 1)!

)‖x‖,

it follows from Theorem 1.4.3 that abs(AV (·)x) ≤ ω. By Proposition 1.6.3,Q(λ)x ∈ D(A) for all x ∈ X, all λ > ω and

(λ− A)Q(λ)x = λk+2

∫ ∞

0

e−λtV (t)x dt− λk+1

∫ ∞

0

e−λtAV (t)x dt

= λk+2

∫ ∞

0

e−λtV (t)x dt− λk+1

∫ ∞

0

e−λt d

dtV (t)x dt

+ λk+1

∫ ∞

0

tk

k!e−λtx dt

= x

for λ > ω. In order to show that λ−A is injective for λ > ω assume that (λ−A)x =0 for some x ∈ D(A) and λ > ω. Then the solution V (t)x of (ACPk+1) is given

by V (t)x =(∫ t

0(t−s)k

k!eλs ds

)x. Since ‖V (t)x ‖ ≤Meωt‖x‖ for all t ≥ 0, it follows

that x = 0. Hence, R(λ,A) = Q(λ) for λ > ω and V is a (k + 1)-times integrated

semigroup generated by A. By hypothesis S(t)x := ddtV (t)x = AV (t)x + tk

k!xexists for all t ≥ 0 and all x ∈ X and V (0) = 0. Integrating by parts shows thatR(λ,A) = λkS(λ), so S is a k-times integrated semigroup generated by A.

(iii) ⇒ (ii): For the time being, assume that 0 ∈ ρ(A). Let x ∈ X and let ube the solution of (ACP0) with initial value u(0) = A−k−1x. Then v given by

v(t) := u(t)− A−k−1x− tA−kx− . . .− tk−1

(k − 1)!A−2x− tk

k!A−1x

is an exponentially bounded solution of (ACPk+1). Let v be another solution of(ACPk+1). Then u = v− v solves (ACP0) with initial value u(0) = 0. Hence, u ≡ 0and we have proved (ii) provided 0 ∈ ρ(A). In the case where 0 �∈ ρ(A), we havethat 0 ∈ ρ(A− μ) for some μ ∈ C. Hence, the preceding argument shows that (ii)holds for (A − μ). The implication (ii) ⇒ (i) and Proposition 3.2.6 imply that Agenerates a k-times integrated semigroup which, as we have seen, implies (ii).

Remark 3.2.15. a) The assumption that ρ(A) �= ∅ in Theorem 3.2.13 (iii) cannotbe omitted even if k = 0 (see Example 3.1.11).

b) The assumption of exponential boundedness in Theorem 3.2.13 (ii) and (iii)cannot be omitted as the following example shows: Let 1 ≤ p <∞ and X be the

space �p of all complex sequences x = (xn)n∈N such that ‖x‖ := (∑∞

n=1 |xn|p)1/p <∞. Define the closed operator A on X by

D(A) := {x ∈ X : (anxn) ∈ X}, Ax := (anxn)n∈N

where an := n + ien2

for n ∈ N. For t ≥ 0 set S(t)x :=((∫ t

0esan ds)xn

)n.

Then S(t) ∈ L(X) for t ≥ 0 and S(·)x is strongly continuous. For x ∈ X let

v(t) :=∫ t

0S(s)x ds. We verify that v′(t) = Av(t)+ tx for t ≥ 0; i.e., v is the unique

solution of (ACP2). However, v is not exponentially bounded if xn �= 0 for all

n ∈ N. Observe also that S satisfies S(s)S(t) =∫ s+t

sS(r)dr− ∫ t

0S(r)dr. However,

S is not Laplace transformable since v is not exponentially bounded.

3.3 Real Characterization

In Section 3.1 (respectively, 3.2) we proved that the Cauchy problem

u′(t) = Au(t) (t ≥ 0), u(0) = x,

possesses a unique classical solution for all x ∈ D(A) (respectively, x ∈ D(Ak+1))provided the operator A generates a C0-semigroup (respectively, k-times inte-grated semigroup) on X . It is therefore interesting to characterize generators ofC0-semigroups (respectively, integrated semigroups) by properties of the operatorsA or their resolvents.

In the following we characterize generators of C0-semigroups (respectively,exponentially bounded integrated semigroups) in terms of estimates for the resol-vents and all their powers for real λ. Recall that an operator A was defined tobe the generator of a k-times integrated semigroup S on X for some k ∈ N0 if(ω,∞) ⊂ ρ(A) for some ω ≥ 0 and there exists a strongly continuous functionS : R+ → L(X) satisfying abs(S) ≤ ω and

R(λ,A) = λk

∫ ∞

0

e−λtS(t) dt (λ > ω).

By applying the Real Representation Theorem 2.4.1 to the special case of re-solvents, we obtain the following characterization. Here and in what follows, weuse the notation (R(λ,A)/λk)(n) to denote the nth derivative of the functionλ �→ R(λ,A)/λk. Note that the first derivative of R(λ,A) is −R(λ,A)2 (see Corol-lary B.3).

Theorem 3.3.1. Let A be a linear operator on X. Let M ≥ 0, ω ∈ R and k ∈ N0.Then the following assertions are equivalent:

(i) (ω,∞) ⊂ ρ(A) and

supn∈N0

supλ>ω

‖(λ− ω)n+1(R(λ,A)/λk)(n)/n!‖ ≤M.

3.3. REAL CHARACTERIZATION 133

(ii) A generates a (k + 1)-times integrated semigroup Sk+1 on X satisfying

‖Sk+1(t)− Sk+1(s)‖ ≤M

∫ t

s

eωrdr (0 ≤ s ≤ t).

Proof. The implication (i) ⇒ (ii) follows from Theorem 2.4.1 and assertion (1.22).Conversely, assume that (ii) holds. By Definition 3.2.1, there exists ω′ ≥ ω such

that R(λ,A)λk = λ

∫∞0

e−λtSk+1(t) dt for all λ > ω′. By Proposition B.5, (ω,∞) ⊂ρ(A) and (i) follows from Theorem 2.4.1 and (1.22).

When k > 0 and X �= {0}, conditions (i) and (ii) of Theorem 3.3.1 cannotbe satisfied for ω < 0.

Note that, given a linear operator A satisfying condition (i) above, one cannotimprove the order of integration of Sk+1, in general (see Example 3.3.10 below).However, if A is densely defined, we obtain the following characterization.

Theorem 3.3.2. Let A be a densely defined operator on X. Let M ≥ 0, ω ∈ R andk ∈ N0. Then the following assertions are equivalent:

(i) (ω,∞) ⊂ ρ(A) and

supn∈N0

supλ>ω

‖(λ− ω)n+1(R(λ,A)/λk)(n)/n!‖ ≤M.

(ii) A generates a k-times integrated semigroup Sk on X satisfying

‖Sk(t)‖ ≤Meωt (t ≥ 0).

The following lemma will be useful in the proof of Theorem 3.3.2. In additionto the space Lipω(R+, X) defined in Section 2.4, we also set

C1ω(R+, X) :=

{f ∈ C1(R+, X) : f(0) = 0, sup

t≥0||e−ωtf ′(t)|| <∞

}= C1(R+, X) ∩ Lipω(R+, X).

Lemma 3.3.3. For ω∈R, the space C1ω(R+, X) is a closed subspace of Lipω(R+, X).

In particular, if S ∈ Lipω(R+,L(X)), then {x ∈ X : S(·)x ∈ C1(R+, X)} is aclosed subspace of X.

Proof. Let (fn) ⊂ C1ω(R+, X) such that (fn) converges to f in Lipω(R+, X).

Then fn(t) =∫ t

0f ′n(s) ds for t ≥ 0. Since sups∈[0,t] ‖f ′n(s) − f ′m(s)‖ ≤ eωt‖fn −

fm‖Lipω(R+,X) for n,m ∈ N0, it follows that (f ′n) converges uniformly on com-

pact sets to a function g ∈ C(R+, X) and that f(t) =∫ t

0g(s) ds for t ≥ 0. Hence

f ∈ C1ω(R+, X). The final statement follows, since x �→ S(·)x is continuous from

X into Lipω(R+, X).


Proof of Theorem 3.3.2. Assume that (ii) holds. Then A also generates a (k + 1)-times integrated semigroup Sk+1 on X which in addition satisfies assertion (ii) ofTheorem 3.3.1. Hence assertion (i) follows from that theorem.

Conversely, assume that (i) holds. By Theorem 3.3.1, A generates a (k + 1)-times integrated semigroup Sk+1 on X such that Sk+1 ∈ Lipω(R+,L(X)) with‖Sk+1‖Lipω(R+,L(X)) ≤M . By Lemma 3.2.2 c),

Sk(t)x :=d

dtSk+1(t)x (3.22)

exists for all x ∈ D(A) and t �→ Sk(t)x is continuous. By Lemma 3.3.3, thedefinition of Sk(t)x given in (3.22) is also meaningful for x ∈ D(A) and t �→ Sk(t)xis also continuous for x ∈ D(A). By assumption D(A) = X and therefore A is thegenerator of the k-times integrated semigroup Sk on X which clearly satisfies

‖Sk(t)‖ ≤Meωt (t ≥ 0).

Notice that the special case k = 0 in Theorem 3.3.2 is precisely the classi-cal Hille-Yosida theorem (in the general form presented here due to Hille, Yosida,Feller, Miyadera and Phillips), which we state explicitly due to its special impor-tance.

Theorem 3.3.4 (Hille-Yosida). Let A be a densely defined operator on X. Then Agenerates a C0-semigroup on X if and only if there exist constants M ≥ 0, ω ∈ Rsuch that (ω,∞) ⊂ ρ(A) and

‖ (λ− ω)n+1R(λ,A)(n)/n! ‖ ≤M (λ > ω, n ∈ N0). (3.23)

It is immediate from Theorem 3.3.2 and the relation

(−1)nR(λ,A)(n)/n! = R(λ,A)n+1 (λ ∈ ρ(A), n ∈ N0) (3.24)

(see Corollary B.3) that the generator of a C0-semigroup T of contractions maybe characterized as follows.

Corollary 3.3.5. Let A be a densely defined operator on X. Then A generates aC0-semigroup on X satisfying ‖T (t)‖ ≤ 1 for all t ≥ 0 if and only if (0,∞) ⊂ ρ(A)and

‖λR(λ,A) ‖ ≤ 1 (λ > 0). (3.25)

It is possible to express the semigroup in terms of the resolvent via “Euler’sformula” for exponentials, which is well known in the scalar case.

Corollary 3.3.6. Let A be the generator of a C0-semigroup T . Then

T (t)x = limn→∞

(I − t

nA)−n

x (3.26)

for t > 0 and x ∈ X.

3.3. REAL CHARACTERIZATION 135

Proof. By (3.24) we have

(I − t

nA)−n

= λnR(λ,A)n =(−1)n−1

(n− 1)!λnR(λ,A)(n−1)

where λ = nt . Thus assertion (3.26) is precisely the Post-Widder inversion formula

proved in Theorem 1.7.7.

For a densely defined operator A on X denote its adjoint by A∗. ThenR(λ,A)∗ = R(λ,A∗) for all λ ∈ ρ(A) = ρ(A∗) (see Proposition B.11). As a directconsequence of Theorem 3.3.1 and Theorem 3.3.4 we obtain the following result.

Corollary 3.3.7. Let A be the generator of a C0-semigroup on X. Then the adjointA∗ of A generates a once integrated semigroup on X∗.

We remark that if the underlying Banach spaceX is reflexive, then the adjointA∗ of A even generates a C0-semigroup on X∗. This follows from Theorem 3.3.4,since A∗ is densely defined (see Proposition B.10). In fact, the following propositionshows that operators satisfying the Hille-Yosida condition (3.23) acting on reflexivespaces are necessarily densely defined.

Proposition 3.3.8. Let A be a linear operator on a reflexive Banach space X. As-sume that there exist constants M,ω ≥ 0 such that (ω,∞) ⊂ ρ(A) and ‖λR(λ,A)‖≤M (λ > ω). Then A is densely defined.

Proof. Let x ∈ X and for n ∈ N with n > ω set an := R(n,A)x. By assumption,(nan)n∈N,n>ω is a bounded sequence. Let z be a weak limit point of the relativelyweakly compact set {nan : n ∈ N}. Since nan −Aan = x and limn→∞ an = 0, theclosedness of A implies that x = z. But z is in the weak closure of D(A) and hencein the norm closure of D(A).

Corollary 3.3.9. Let X be a reflexive Banach space and assume that A generates aC0-semigroup on X. Then the adjoint A∗ of A generates a C0-semigroup on X∗.

Example 3.3.10. Let 1 ≤ p <∞ and X := Lp(R). Consider the operator Apf := f ′

on Lp(R) with domain D(Ap) := W 1,p(R). For the definition of the Sobolev spaceW 1,p(R), see Appendix E. Then Ap generates the C0-semigroup Tp on Lp(R) givenby

(Tp(t)f)(x) = f(x+ t) (x ∈ R, t ≥ 0).

Identifying Lp(R)∗ with Lp′(R) where 1/p + 1/p′ = 1, Corollary 3.3.9 impliesthat −Ap′ generates a C0-semigroup on Lp′(R), provided p > 1. In fact, this isevident since Tp′ extends to a C0-group. If p = 1, then by Corollary 3.3.7, −A∞generates a once integrated semigroup S on L∞(R), where A∞f := −f ′ withdomain D(A∞) := W 1,∞(R). This is given by (S(t)g)(x) =

∫ t

0g(x− s) ds.

As a further consequence of Theorem 3.3.1 and Theorem 3.3.2 we note thefollowing corollary.

Corollary 3.3.11. Let T be a semigroup on X in the sense of Definition 3.2.5 andlet A be the generator of T . Then T is a C0-semigroup on X if and only if D(A)is dense in X.

Finally, given an operator A in X and a closed subspace Y of X, we definethe part AY of A in Y by

D(AY ) := {y ∈ Y ∩D(A) : Ay ∈ Y }AY y := Ay. (3.27)

If D(A) ⊂ Y , then ρ(A) ⊂ ρ(AY ) and R(λ,AY ) = R(λ,A)|Y for all λ ∈ ρ(A).An important case is Y = D(A) when A is not densely defined. Then it may wellhappen that AY is not densely defined either (a concrete example is the Poissonoperator considered in Section 6.1). Nevertheless the following holds true.

Lemma 3.3.12. Let A be an operator on X such that (ω,∞) ⊂ ρ(A) and

M := supλ>ω

‖λR(λ,A)‖ <∞

for some ω ∈ R. Let Y = D(A). Then

a) limλ→∞ λR(λ,A)x = x for all x ∈ Y .

b) D(AY ) is dense in Y .

c) If A satisfies the Hille-Yosida condition (3.23), then AY generates a C0-semigroup on Y .

Proof. a) The assumption implies that limλ→∞R(λ,A)x = 0 for all x ∈ X. Hence,limλ→∞ λR(λ,A)x = limλ→∞(x + R(λ,A)Ax) = x if x ∈ D(A). Consequently,limλ→∞ λR(λ,A)x = x if x ∈ D(A) = Y .

b) Since R(λ,A)x ∈ D(AY ) if x ∈ Y , this follows from a).c) This follows from the Hille-Yosida theorem.

Although the part of an operator A in D(A) may not be densely defined, weobtain the following result from the proof of Theorem 3.3.2.

Corollary 3.3.13. Let A be an operator satisfying the equivalent conditions of The-orem 3.3.1 for some k ∈ N0 and let Y = D(A). Then the part AY of A in Ygenerates a k-times integrated semigroup on Y .

Proof. By Lemma 3.2.2, Sk+1(·)x ∈ C1(R+, X) provided x ∈ D(A). It follows fromLemma 3.3.3 that Sk+1(·)x ∈ C1

ω(R+, X) for all x ∈ D(A) = Y . For t > 0 andx ∈ Y let Sk(t)x = d

dtSk+1(t)x. Since Sk+1(t)D(A) ⊂ D(A) for t > 0, it follows

that Sk(t)x ∈ Y for all x ∈ Y . Thus, Sk : R+ → L(Y ) is strongly continuous and

R(λ,A)x = λk+1

∫ ∞

0

e−λtSk+1(t)x dt = λk

∫ ∞

0

e−λtSk(t)x dt

for all x ∈ Y and λ sufficiently large.

3.4. DISSIPATIVE OPERATORS 137

Proposition 3.3.14. Let T be a C0-semigroup on X with generator A. Let X :=D(A∗). Then T (t) := T (t)∗|X� defines a C0-semigroup whose generator A isthe part of A∗ in X .

The C0-semigroup T is known as the sun-dual of T .

Proof. It follows from Lemma 3.3.12 that the part of A∗ in X generates a C0-semigroup T on X . For x ∈ X and x∗ ∈ X , Corollary 3.3.6 shows that⟨

x, T (t) x∗⟩

= limn→∞

⟨x, (I − t

nA )−nx∗

⟩= lim

n→∞⟨x, ((I − t

nA)−n)∗x∗⟩

= limn→∞

⟨(I − t

nA)−nx, x∗⟩

= 〈T (t)x, x∗〉= 〈x, T (t)∗x∗〉.

It follows that T (t) = T (t)∗|X� for all t > 0.

3.4 Dissipative Operators

In the previous section we saw that, by the Hille-Yosida characterization, thegenerator of a contraction semigroup may be characterized in terms of a resolventestimate for real λ. It is the aim of this section to give a second characterizationof such semigroups, which turns out to be quite useful in particular when dealingwith differential operators. In order to do so, we define for x ∈ X the subdifferentialdN(x) of the norm N : X → R+, N(x) = ‖x‖ at x by

dN(x) := {x∗ ∈ X∗ : ‖x∗‖ ≤ 1, 〈x, x∗〉 = ‖x ‖} . (3.28)

The Hahn-Banach theorem implies that dN(x) �= ∅ for all x ∈ X .

Definition 3.4.1. An operator A on X is called dissipative if for every x ∈ D(A)there exists x∗ ∈ dN(x) such that

Re〈Ax, x∗〉 ≤ 0. (3.29)

A useful characterization of dissipative operators is the following.

Lemma 3.4.2. An operator A on X is dissipative if and only if

‖(λ− A)x‖ ≥ λ‖x‖ (x ∈ D(A), λ > 0) (3.30)

or equivalently,

‖x− tAx‖ ≥ ‖x‖ (x ∈ D(A), t > 0).

Proof. Assume that A is dissipative. Let x ∈ D(A), t > 0. Let x∗ ∈ dN(x) satisfy(3.29). Then for t > 0,

‖x− tAx‖ ≥ Re〈x− tAx, x∗〉= ‖x‖ − tRe〈Ax, x∗〉≥ ‖x‖.

Conversely, let x ∈ D(A) and assume that ‖x − tAx‖ ≥ ‖x‖ (t > 0). Choosex∗t ∈ dN(x− tAx) and let x∗ be a weak* accumulation point of x∗t as t ↓ 0. Then‖x∗‖ ≤ 1. Since ‖x− tAx‖ = 〈x− tAx, x∗t 〉, letting t ↓ 0 shows that ‖x‖ = 〈x, x∗〉.Thus, x∗ ∈ dN(x). Moreover,

‖x‖ ≤ ‖x− tAx‖ = Re〈x, x∗t 〉 − tRe〈Ax, x∗t 〉≤ ‖x‖ − tRe〈Ax, x∗t 〉.

Thus, Re〈Ax, x∗t 〉 ≤ 0. Letting t ↓ 0 shows that Re〈Ax, x∗〉 ≤ 0.

Example 3.4.3. a) Dissipative operators acting on Hilbert spaces or Lp-spaces maybe characterized as follows:

(i) Let A be an operator on a Hilbert space H. Denote by (·|·) the inner productin H. Then A is dissipative if and only if Re(Ax|x) ≤ 0 for all x ∈ D(A).

(ii) Let Ω ⊂ Rn be open, 1 < p < ∞, set X := Lp(Ω) and identify X∗ withLp′(Ω) where 1/p+ 1/p′ = 1. For f ∈ X \ {0} we define sign f ∈ L∞(Ω) by

(sign f)(x) :=

⎧⎨⎩0 if f(x) = 0,f(x)

|f(x)| if f(x) �= 0.

Then dN(f) = ‖f‖−(p−1)p

{sign f · |f |p−1

}, where f denotes the complex con-

jugate function of f . Therefore an operator A on X is dissipative if and onlyif

Re

∫Ω

Af · sign f · |f |p−1 dx ≤ 0 (3.31)

for all f ∈ D(A).

b) If A is dissipative, then cA is dissipative for all c > 0.

c) If B ∈ L(X), then B − ‖B‖ is dissipative.

Lemma 3.4.4. Let A be a densely defined dissipative operator on X. Then A isclosable and A is dissipative.

Proof. Let (xn) ⊂ D(A) such that xn → 0 and Axn → y for some y ∈ X asn → ∞. We show that y = 0. To this end, let z ∈ D(A). It follows from Lemma

3.4. DISSIPATIVE OPERATORS 139

3.4.2 that ‖(I − tA)x‖ ≥ ‖x‖ for all t > 0 and all x ∈ D(A). Hence ‖ xn +tz ‖ ≤ ‖xn + tz − tA(xn + tz) ‖ for all n ∈ N and all t > 0. This implies that‖ tz ‖ ≤ ‖ tz − ty − t2Az ‖ and hence ‖ z ‖ ≤ ‖ z − y − tAz ‖ for all t > 0. Lettingt→ 0 we obtain ‖ z ‖ ≤ ‖ z − y ‖ for all z ∈ D(A). Since D(A) is dense, it followsthat y = 0 which means that A is closable. Taking limits in (3.30) shows that Ais dissipative.

The following theorem due to Lumer and Phillips characterizes generators Aof C0-semigroups of contractions in terms of dissipativity of A.

Theorem 3.4.5 (Lumer-Phillips). Let A be a densely defined operator on X. ThenA generates a C0-semigroup of contractions on X if and only if

a) A is dissipative, and

b) (λ−A)D(A) = X for some (or all) λ > 0.

Proof. Let A be the generator of a C0-semigroup of contractions. Then assertionb) holds by the Hille-Yosida theorem (Corollary 3.3.5). Moreover, the Hille-Yosidatheorem combined with Lemma 3.4.2 implies assertion a).

In order to prove the converse implication note that by Lemma 3.4.2 we have

‖ (λ−A)x ‖ ≥ λ‖x ‖ (x ∈ D(A), λ > 0). (3.32)

Since (λ0 − A)D(A) = X for some λ0 > 0, it follows from (3.32) that λ0 − A isinvertible and that ‖R(λ0, A) ‖ ≤ λ−1

0 . We show that this property holds for allλ > 0. In fact, let Λ := ρ(A) ∩ (0,∞). Then Λ �= ∅ and therefore A is closed.Furthermore let (λn) ⊂ Λ such that limn→∞ λn = λ > 0. By Corollary B.3,dist(λn, σ(A)) ≥ ‖R(λn, A) ‖−1 ≥ λn for all n ∈ N and it follows that λ ∈ Λ. Thisshows that Λ is closed in (0,∞). Since Λ is obviously open, it follows that Λ =(0,∞) and therefore (0,∞) ⊂ ρ(A). Inequality (3.32) implies that ‖R(λ,A)‖ ≤ λ−1

for all λ > 0 and the Hille-Yosida theorem finally implies the assertion.

By the same proof with Theorem 3.3.1 replacing the Hille-Yosida theoremwe obtain the following characterization in the case when D(A) is not necessarilydense.

Corollary 3.4.6. Let A be an operator on X. The following assertions are equiva-lent:

(i) A is dissipative and (λ−A)D(A) = X for some (or all) λ > 0.

(ii) A generates a once integrated semigroup S satisfying

‖S(t)− S(s)‖ ≤ |t− s| (t, s ≥ 0).

In concrete examples, dissipativity is often relatively easy to verify whereasthe range condition b) in Theorem 3.4.5 is hard to show. However, in the followingexample of the Dirichlet-Laplacian, the range condition is just a consequence ofthe Riesz-Frechet lemma.


Example 3.4.7 (The Laplacian with Dirichlet boundary conditions). Let Ω ⊂ Rn

be an open set. Consider the operator A : D(A)→ L2(Ω) defined by

D(A) := {u ∈ H10 (Ω) : Δu ∈ L2(Ω)},

Au := Δu.

Here, H10 (Ω) is the Sobolev space defined in Appendix E, and Δu is defined to be∑n

j=1D2ju in the sense of distributions (see also Appendix E). Denoting by (·|·)

the inner product in L2(Ω), we see that (Au|u) = ∫Ω(Δu)u = − ∫

Ω|∇u|2 ≤ 0 for

u ∈ D(A). Hence, by Example 3.4.3 a), the operator A is dissipative. In order toprove the range condition b) of Theorem 3.4.5, let f ∈ L2(Ω). Then the mappingΦ : v �→ ∫

Ωfv defines a continuous linear functional on the Hilbert space H1

0 (Ω).By the Riesz-Frechet lemma, there exists a unique u ∈ H1

0 (Ω) such that Φ(v) =(v|u)H1

0 (Ω) for all v ∈ H10 (Ω). Here,

(v|u)H10 (Ω) :=

∫Ω

uv +

n∑j=1

∫Ω

DjuDjv

denotes the inner product in H10 (Ω). Considering in particular v ∈ D(Ω), it follows

that u − Δu = f in D(Ω)′. This implies that u ∈ D(A) and u − Au = f . Obvi-ously, D(A) is dense in L2(Ω) and by the Lumer-Phillips theorem, A generates acontraction semigroup on L2(Ω). We call A the Laplacian with Dirichlet boundaryconditions on L2(Ω), and we denote it by ΔL2(Ω). We remark that in the casewhere Ω is a bounded domain with boundary of class C2, it can be shown (see[Bre83, Theoreme IX.25]) that

D(ΔL2(Ω)) = H2(Ω) ∩H10 (Ω).

Example 3.4.8. Consider the Hilbert space X := L2(0, 1) and the operator

D(A) := {u ∈ H1(0, 1) : u(0) = 0},Au := u′.

Then Re (u|Au) = 12

∫ 1

0(u(x)u′(x)+u′(x)u(x)) dx = 1

2 |u(1)|2 ≥ 0. For f ∈ L2(0, 1)and λ ∈ C, define u by

u(x) :=

∫ x

0

e−λ(x−y)f(y) dy (x ∈ (0, 1)).

Then u ∈ D(A) and λu + u′ = f . Hence, the range condition (λ + A)D(A) = Xis fulfilled for all λ ∈ C and by the Lumer-Phillips theorem, −A generates acontraction semigroup T on X. It is not difficult to see that T is given by

(T (t)f)(x) =

{f(x− t) (t ≤ x),

0 (t > x).(3.33)

3.5. HILLE-YOSIDA OPERATORS 141

In fact, the mapping x �→ ∫∞0

e−λt(T (t)f)(x) dt =∫ x

0e−λ(x−y)f(y) dy belongs to

D(A) and (λ + ∂∂x )

∫∞0

e−λt(T (t)f)(x) dt = f(x) for x ∈ (0, 1). Since (λ + A) is

invertible for all λ ∈ C, we see that (λ+ A)−1 =∫∞0

e−λtT (t) dt.The representation (3.33) implies that T also defines a C0-semigroup of pos-

itive contractions on Lp(0, 1) for 1 ≤ p <∞. Its generator is given by −Ap where

D(Ap) = {u ∈W 1,p(0, 1) : u(0) = 0} and Apu = u′. (3.34)

3.5 Hille-Yosida Operators

In this section we consider operators which satisfy the Hille-Yosida condition (3.23)but are not necessarily densely defined.

Definition 3.5.1. A linear operator A on X is called a Hille-Yosida operator ifthere exist ω ∈ R,M ≥ 0 such that (ω,∞) ⊂ ρ(A) and

‖ (λ− ω)nR(λ,A)n ‖ ≤M (n ∈ N0, λ > ω). (3.35)

We note that by the Hille-Yosida theorem and the identity

(−1)nR(λ,A)(n)/n! = R(λ,A)n+1,

the class of densely defined Hille-Yosida operators coincides with the class of gener-ators of C0-semigroups on X. We also observe that by Theorem 3.3.1 an operatorA on X is a Hille-Yosida operator if and only if A generates a once integratedsemigroup S on X satisfying

‖S(t)− S(s) ‖ ≤M

∫ t

s

eωr dr (0 ≤ s ≤ t) (3.36)

for some ω ∈ R,M ≥ 0. The above estimate (3.36) implies in particular thatS is a locally Lipschitz continuous function on R+. This fact will be of crucialimportance in the proof of the following result on the inhomogeneous Cauchyproblem for operators which are not necessarily densely defined. More precisely,consider the problem

(ACPf )

{u′(t) = Au(t) + f(t) (t ∈ [0, τ ]),

u(0) = x,(3.37)

where f : [0, τ ] → X is a given function and x ∈ X. When D(A) = X, then theinhomogeneous problem can be solved in the classical sense by means of the varia-tion of constants formula provided x ∈ D(A) and f ∈ C1([0, τ ], X) (see Corollary3.2.11 b)). Note that this method cannot be used when D(A) �= X and whenf(t) �∈ D(A). The method which we use in the following to treat (ACPf ) is based


on the fact that a Hille-Yosida operator generates a once integrated semigroupwhich is locally Lipschitz continuous. Employing the results of Section 3.2 in thepresent situation, we see that existence and uniqueness results for (ACPf ) areequivalent to the fact that v given by

v(t) = S(t)x+

∫ t

0

S(t)f(t− s)ds

is sufficiently regular. More precisely, the following holds true.

Theorem 3.5.2 (Da Prato-Sinestrari). Let A be a Hille-Yosida operator on X andlet τ > 0.

a) Let f ∈ L1([0, τ ], X) and x ∈ D(A). Then there exists a unique mild solutionof (ACPf ).

b) Let f(t) = f0 +∫ t

0g(s) ds where f0 ∈ X and g ∈ L1([0, τ ], X). Let x ∈ D(A)

and assume that Ax+f0 ∈ D(A). Then there exists a unique classical solutionof (ACPf ).

Remark 3.5.3. a) Note that x ∈ D(A) is a necessary condition for a mild solution

u to exist, because limt→01t

∫ t

0u(s) ds = x and

∫ t

0u(s) ds ∈ D(A) by definition of a

mild solution.

b) If a classical solution u of (ACPf ) exists, then x ∈ D(A) and

Ax+ f(0) ∈ D(A). (3.38)

In fact,Au(0) + f(0) = u′(0) = lim

t→0

1t(u(t)− u(0)) ∈ D(A).

Proof of Theorem 3.5.2. Let A be a Hille-Yosida operator on X . By Theorem 3.3.1,A generates a once integrated semigroup S on X which satisfies estimate (3.36).In order to prove assertion a), it suffices by Lemma 3.2.9, Proposition 1.3.7 andLemma 3.2.10 to show that t �→ S(t)x belongs to C1([0, τ ], X). This follows fromthe assumption that x ∈ D(A), Lemma 3.2.2 c) and Lemma 3.3.3. Thus, we haveproved assertion a).

In order to prove assertion b), it suffices by Lemma 3.2.9 and Lemma 3.2.10to verify that v ∈ C2([0, τ ], X). Since x ∈ D(A) it follows from Lemma 3.2.2 c)and Proposition 1.3.6 that

v′(t) = x+ S(t)Ax+ S(t)f0 +

∫ t

0

S(s)g(t− s) ds.

Now, by Proposition 1.3.7 the convolution term on the right hand side abovebelongs to C1([0, τ ], X) and t �→ S(t)(Ax + f0) belongs to C1([0, τ ], X) by the

3.5. HILLE-YOSIDA OPERATORS 143

argument given in the proof of assertion a), since Ax+ f0 ∈ D(A) by assumption.

Given a Hille-Yosida operator A, we consider now the problem whether ornot the sum A + B of A and some operator B is again a Hille-Yosida operator.We start with the following renorming lemma.

Lemma 3.5.4. Let A be a Hille-Yosida operator satisfying estimate (3.35) for someM > 0 and ω = 0. Then there exists an equivalent norm | · | on X such that‖x‖ ≤ |x| ≤M‖x‖ for x ∈ X and

|λR(λ,A)x| ≤ |x| (x ∈ X,λ > 0).

Proof. For μ > 0 and x ∈ X set

‖x‖μ := supn≥0

‖μnR(μ,A)nx‖.

Then‖x‖ ≤ ‖x‖μ ≤M‖x‖ and ‖μR(μ,A)‖μ ≤ 1. (3.39)

Let λ ∈ (0, μ] and set y := R(λ,A)x. It follows that y = R(μ,A)(x + (μ − λ)y)and hence that ‖y‖μ ≤ 1

μ‖x‖μ + (1− λμ )‖y‖μ. Therefore, ‖λR(λ,A)‖μ ≤ 1 and it

follows from (3.39) that

‖λnR(λ,A)nx‖ ≤ ‖λnR(λ,A)nx‖μ ≤ ‖x‖μ (0 < λ ≤ μ). (3.40)

Hence ‖x‖λ ≤ ‖x‖μ for 0 < λ ≤ μ. Defining

|x| := limμ→∞ ‖x‖μ,

the assertion follows by taking n = 1 in (3.40) and letting μ→∞.

Theorem 3.5.5. Let A be a Hille-Yosida operator on X and let B ∈ L(D(A), X).Then A+B is a Hille-Yosida operator.

Proof. Replacing A by A − ω we may assume that the estimate (3.35) is satisfiedfor A with ω = 0. Denote by | · | the norm introduced in Lemma 3.5.4. It followsfrom that lemma that

|λR(λ,A)| ≤ 1 (λ > 0).

Note that λ−(A+B) = (I−BR(λ,A))(λ−A) for λ > 0. Since |BR(λ,A)| ≤ |B|/λ,the operator I −BR(λ,A) is invertible for λ > |B| and

|(λ− (A+B))−1| ≤ |(λ−A)−1| |(I −BR(λ))−1| ≤ 1

λ

1

1− |B|λ−1=

1

λ− |B|for those λ. Hence |(λ−|B|)R(λ,A+B)| ≤ 1 for λ > |B|. Returning to the originalnorm we have for x ∈ X ,

‖(λ− |B|)nR(λ,A+B)nx‖ ≤ |(λ− |B|)nR(λ,A+B)nx| ≤ |x| ≤M‖x‖

for λ > |B|. Thus A+B is a Hille-Yosida operator.

Taking into account the Hille-Yosida theorem and the fact that generators ofC0-semigroups are densely defined we obtain the following corollary.

Corollary 3.5.6. Let A be the generator of a C0-semigroup on X and let B ∈ L(X).Then A+B generates a C0-semigroup.

Next we consider perturbation by operators defined on the domain of thegiven operator. If A is a closed operator, we consider D(A) with the graph norm‖x‖D(A) := ‖x‖+ ‖Ax‖ for which it is a Banach space. Two operators A, defined

on X, and A, defined on a second Banach space X , are called similar if there existsan isomorphism U : X → X such that

D(A) = UD(A) and U−1AUx = Ax for all x ∈ D(A).

In that case, A and A have similar properties. For example, if A generates a C0-semigroup T , then A generates the C0-semigroup T on X given by

T (t)y = UT (t)U−1y (y ∈ X, t ≥ 0).

Similarly, if A generates an integrated semigroup S, then A generates the inte-grated semigroup S on X given by

S(t)y = US(t)U−1y (y ∈ X, t ≥ 0).

Theorem 3.5.7. Let A be an operator such that (ω,∞) ⊂ ρ(A) and M :=supλ>ω ‖λR(λ,A)‖ < ∞ for some ω ∈ R, and let B ∈ L(D(A)). Then thereexists a bounded operator C ∈ L(X) such that A + B and A + C are similar.In particular, if A is a Hille-Yosida operator, then A + B is also a Hille-Yosidaoperator.

For the proof we need the following well-known result.

Lemma 3.5.8. Let U, V ∈ L(X). If I − UV is invertible, then so is I − V U .

Proof. One has (I − V U)−1 = I + V (I − UV )−1U .

Proof of Theorem 3.5.7. Choose λ0 > ω and let S := (λ0 − A)BR(λ0, A) ∈ L(X).Choose λ > λ0 such that ‖SR(λ,A)‖ < 1. Then I − (λ0 −A)BR(λ,A)R(λ0, A) =I − SR(λ,A) is invertible. It follows from Lemma 3.5.8 that I −BR(λ,A) is alsoinvertible. Let C := (λ − A)BR(λ,A) ∈ L(X). We show that A + B and A + Care similar. Let U := I − BR(λ,A). Then U is an isomorphism on X such thatUD(A) = D(A). Moreover,

U(A+ C)U−1 = U(A− λ+ C)U−1 + λ

= U [A− λ− (A− λ)BR(λ,A)]U−1 + λ

= U(A− λ)[I −BR(λ,A)]U−1 + λ

= U(A− λ) + λ = A− λ+B + λ = A+B.

3.6. APPROXIMATION OF SEMIGROUPS 145

This proves the claim. Now the second assertion follows from Theorem 3.5.5.

Finally, we collect several examples of Hille-Yosida operators.

Example 3.5.9. a) Let A be the generator of a C0-semigroup on X. Then theadjoint A∗ of A is a Hille-Yosida operator on X∗.

b) As a concrete example, consider X = L∞(R) and define A by Au := −u′ withD(A) := W 1,∞(R). By Example 3.3.10, (0,∞) ⊂ ρ(A) and ‖R(λ,A)‖ ≤ 1/λ forλ > 0.

c) Let A be the generator of a semigroup T in the sense of Definition 3.2.5. Since

S(t) :=∫ t

0T (s) ds fulfills assumption (ii) of Theorem 3.3.1 for k = 0 it follows from

that theorem that A is a Hille-Yosida operator.

d) Let X := C[0, 1] and define an operator A on X by

Au := −u′D(A) :=

{u ∈ C1[0, 1] : u(0) = 0

}.

Then D(A) = {u ∈ C[0, 1] : u(0) = 0} �= X, (0,∞) ⊂ ρ(A) and ||R(λ,A)|| ≤ 1/λfor λ > 0. In fact, for f ∈ X and λ > 0 set

u(x) :=

∫ x

0

e−λyf(x− y) dy (x ∈ [0, 1]).

Then u ∈ D(A), (λ− A)u = f and

supx∈[0,1]

|u(x)| ≤ ‖f‖∫ ∞

0

e−λy dy =1

λ‖f‖,

which implies the assertions above.

3.6 Approximation of Semigroups

In this section we study convergence of semigroups. It is interesting that we obtainthe main result (Theorem 3.6.1) directly as a consequence of the approximationtheorem for Laplace transforms given in Section 1.7, which in turn was proved bya simple functional analytic argument. At the end of the section we give a secondproof of the Hille-Yosida theorem, as a simple corollary of the approximationtheorem.

Let Tn be a C0-semigroup on X with generator An (n ∈ N). We suppose that

‖Tn(t)‖ ≤M (t ≥ 0, n ∈ N). (3.41)

If for each x ∈ Xlim

n→∞Tn(t)x =: T (t)x

converges uniformly on [0, τ ] for each τ > 0, then it is easy to see that T is aC0-semigroup. Denote its generator by A. Then it follows from the dominatedconvergence theorem that limn→∞R(λ,An)x = R(λ,A)x for all x ∈ X, λ > 0.We now show the converse assertion.

Theorem 3.6.1 (Trotter-Kato). Let Tn be a C0-semigroup on X with generatorAn (n ∈ N) and suppose that (3.41) holds. Let A be a densely defined operator onX. Suppose that there exists ω ≥ 0 such that (ω,∞) ⊂ ρ(A) and

limn→∞R(λ,An)x = R(λ,A)x (3.42)

for all x ∈ X, λ > ω. Then A is the generator of a C0-semigroup T and

T (t)x = limn→∞Tn(t)x (3.43)

uniformly on [0, τ ] for all τ > 0 and all x ∈ X.

Proof. a) Let x ∈ X. We show that the sequence (Tn(·)x)n∈N is equicontinuouson R+. Since D(A) is dense in X, we can assume that x ∈ D(A). Let ε > 0. Letμ > ω, y := (μ−A)x. Choose n0 ∈ N such that M‖R(μ,A)y −R(μ,An)y‖ ≤ ε/2for all n ≥ n0. Then (by Proposition 3.1.9 f)) for n ≥ n0,

‖Tn(t)x− Tn(s)x‖ = ‖Tn(t)R(μ,A)y − Tn(s)R(μ,A)y‖≤ ‖Tn(t)R(μ,An)y − Tn(s)R(μ,An)y‖+ ε/2

=

∥∥∥∥∫ t

0

Tn(r)AnR(μ,An)y dr

−∫ s

0

Tn(r)AnR(μ,An)y dr

∥∥∥∥+ε

2

≤ M |t− s| ‖AnR(μ,An)y‖+ ε/2

= M |t− s| ‖μR(μ,An)y − y‖+ ε/2.

Since supn∈N ‖R(μ,An)y‖ <∞, there exists δ > 0 such that ‖Tn(t)x−Tn(s)x‖ ≤ εwhenever |t− s| ≤ δ and n ≥ n0. Since Tn(·)x : R+ → X is continuous for n < n0,this shows that the sequence is equicontinuous.

b) Now it follows from Theorem 1.7.5 that Tn(t)x converges uniformly on[0, τ ] as n→∞ for all x ∈ X and all τ > 0. It is clear that T (t)x := limn→∞ Tn(t)x(x ∈ X) defines a C0-semigroup T on X . By the dominated convergence theoremone has∫ ∞

0

e−λtT (t)x dt = limn→∞

∫ ∞

0

e−λtTn(t)x dt

= limn→∞R(λ,An)x = R(λ,A)x (x ∈ X,λ > ω).

By Definition 3.1.8 this means that A is the generator of T .

In order to apply Theorem 3.6.1 it is useful to give other criteria equivalentto (3.42).

3.6. APPROXIMATION OF SEMIGROUPS 147

Proposition 3.6.2. Let ω ∈ R. Let A and An be operators such that (ω,∞) ⊂ ρ(A)and (ω,∞) ⊂ ρ(An) for all n ∈ N. Assume that supn∈N ‖R(λ,An)‖ < ∞ for allλ > ω. Then the following assertions are equivalent:

(i) limn→∞R(μ,An)x = R(μ,A)x for all x ∈ X and all μ > ω.

(ii) limn→∞R(μ,An)x = R(μ,A)x for all x ∈ X and some μ > ω.

(iii) For all x ∈ D(A) there exist xn ∈ D(An) such that limn→∞ xn = x andlimn→∞Anxn = Ax.

(iv) There exists a core D of A such that for all x ∈ D there exist xn ∈ D(An)such that limn→∞ xn = x and limn→∞Anxn = Ax.

Proof. (i) ⇒ (ii) is trivial.(ii) ⇒ (iii): Let x ∈ D(A). Then xn := R(μ,An)(μ− A)x ∈ D(An), xn → x

by hypothesis, and Anxn = μxn − (μ− A)x→ Ax as n→∞.(iii) ⇒ (iv) is trivial.(iv)⇒ (i): Let μ > ω. Let x ∈ D. By hypothesis, there exist xn ∈ D(An) such

that xn → x and Anxn → Ax. Let yn := (μ− An)xn. Then yn → y := (μ− A)x.Hence,

lim supn→∞

‖R(μ,An)y −R(μ,A)y‖

≤ lim supn→∞

(‖R(μ,An)(y − yn)‖+ ‖R(μ,An)yn −R(μ,A)y‖

)= lim sup

n→∞‖R(μ,An)yn −R(μ,A)y‖

= lim supn→∞

‖xn − x‖ = 0.

Since D is a core and μ − A is surjective, {(μ− A)x : x ∈ D} is dense in X (seeAppendix B), and (i) follows by approximation.

Corollary 3.6.3. Let A be a densely defined operator on X. Let An ∈ L(X) suchthat

‖etAn‖ ≤M (t ≥ 0, n ∈ N),

where M ≥ 0. Assume that (ω,∞) ⊂ ρ(A) and limn→∞Anx = Ax for all x ∈D(A). Then A generates a C0-semigroup T and for all x ∈ X, limn→∞ etAnx =T (t)x uniformly for t ∈ [0, τ ] for all τ > 0.

It is easy to deduce the Hille-Yosida theorem (Corollary 3.3.5) from Corollary3.6.3. In fact, let A be a densely defined operator on X such that (0,∞) ⊂ ρ(A)and ‖λR(λ,A)‖ ≤ 1 for all λ > 0. Denote by

An := n2R(n,A)− nI (n ∈ N)

the Yosida approximation of A. Then

‖etAn‖ = e−nt‖etn2R(n,A)‖ ≤ e−ntetn2‖R(n,A)‖ ≤ 1 (t > 0, n ∈ N).

Then limn→∞Anx = Ax for all x ∈ D(A). In fact, Anx − Ax = nR(n,A)Ax −Ax→ 0 by Lemma 3.3.12. Now it follows from Corollary 3.6.3 that A generates acontractive C0-semigroup.

3.7 Holomorphic Semigroups

This section is devoted to the study of holomorphic semigroups. This class ofsemigroups plays an important role in the theory of evolution equations. Indeed,the modern treatment of linear and nonlinear parabolic problems is based on thetheory of holomorphic semigroups. When compared with arbitrary C0-semigroups,holomorphic C0-semigroups show many special properties. We only mention here

a) characterization results involving only a resolvent estimate (for example,Theorem 3.7.11 and Corollary 3.7.17),

b) regularity properties of solutions of the Cauchy problem (Corollary 3.7.21and applications in Chapters 6 and 7),

c) determination of the asymptotic behaviour of the C0-semigroup by spectralconditions on the generator (Theorem 5.1.12 and Theorem 5.6.5).

Throughout this section, let Σθ := {z ∈ C \ {0} : | arg z| < θ} be the sector in thecomplex plane of angle θ ∈ (0, π]. Recall from Definition 3.2.5 that a semigroupT was defined to be a strongly continuous mapping (0,∞) → L(X) satisfying a)T (t+ s) = T (t)T (s) for all s, t > 0; b) ‖T (t)‖ ≤ c for all t ∈ (0, 1] and some c > 0;and c) T (t)x = 0 for all t > 0 implies x = 0.

Recall also from Corollary 3.3.11 that a semigroup is a C0-semigroup if andonly if its generator A is densely defined.

Definition 3.7.1. Let θ ∈ (0, π2 ]. A semigroup T on X is called holomorphic of

angle θ if it has a holomorphic extension to Σθ which is bounded on Σθ′ ∩{z ∈ C :|z| ≤ 1} for all θ′ ∈ (0, θ).

If no confusion seems likely, we denote the extension of T to Σθ also byT . Also, if we do not want to specify the angle θ in Definition 3.7.1, we call asemigroup T holomorphic if it is holomorphic of angle θ for some θ ∈ (0, π2 ].

Proposition 3.7.2. Let θ ∈ (0, π2 ] and let T be a semigroup on X with generator A.Assume that T is holomorphic of angle θ. Then the following hold:

a) T (z + z′) = T (z)T (z′) (z, z′ ∈ Σθ).

b) For all θ′ ∈ (0, θ) there exist M ≥ 0, ω ≥ 0 such that ‖T (z)‖ ≤MeωRe z forall z ∈ Σθ′ .

3.7. HOLOMORPHIC SEMIGROUPS 149

c) Let α ∈ (−θ, θ). Denote by Tα the semigroup given by Tα(t) := T (eiαt) (t ≥0). Then eiαA is the generator of Tα.

d) If T is a C0-semigroup, then

limz→0,z∈Σθ′

T (z)x = x

for all x ∈ X and all θ′ ∈ (0, θ).

Proof. a) For fixed z′ ∈ (0,∞) consider the holomorphic functions z �→ T (z + z′)and z �→ T (z)T (z′) for z ∈ Σθ. Since the two functions coincide on (0,∞), theidentity theorem for holomorphic functions implies that T (z + z′) = T (z)T (z′)for z ∈ Σθ and z′ ∈ (0,∞). For fixed z ∈ Σθ the two holomorphic functionsz′ �→ T (z + z′) and z′ �→ T (z)T (z′) coincide for z′ ∈ (0,∞) and the assertionfollows from the identity theorem for holomorphic functions.

b) Let θ′ ∈ (0, θ),

M := supz∈Σθ′ ,|z|≤1

‖T (z)‖, ω′ := max{logM, 0}.

Then ‖T (z)‖ ≤ Meω′|z| for all z ∈ Σθ′ . In fact, let z = teiβ where |β| ≤ θ′.

Applying the proof of Theorem 3.1.7 a) to Tβ one obtains

‖T (z)‖ = ‖Tβ(t)‖ ≤Meω|z|.

Since |z| ≤ Re z/ cos θ′ for z ∈ Σθ′ , the claim follows with ω := ω′/ cos θ.c) Let Aα be the generator of Tα. For R > 0, let ΓR be the contour consisting

of the line segments {t : 0 ≤ t ≤ R} and {teiα : 0 ≤ t ≤ R} and the arc {Reiϕ :0 ≤ ϕ ≤ α}. Cauchy’s theorem implies that

∫ΓR

exp(−λe−iαz)T (z)x dz = 0 forλ > 0 and x ∈ X. Letting R→∞, we obtain

eiαR(λ,Aα)x = eiα∫ ∞

0

e−λtTα(t)x dt =

∫ ∞

0

exp(−λe−iαt)T (t)x dt

= R(λe−iα, A)x,

by Theorem 3.1.7. This implies that Aα = eiαA.d) Let θ′ ∈ (0, θ). By b) there exist ω ≥ 0 and M ≥ 0 such that ‖e−ωzT (z)‖ ≤

M for all z ∈ Σθ′ . It follows from Proposition 2.6.3 b) that

limz→0,z∈Σθ′

e−ωzT (z)x = x

for all x ∈ X . This implies the claim.

We note that in the situation of Proposition 3.7.2, Tα is a C0-semigroup foreach α ∈ (−θ, θ) whenever T is a C0-semigroup; i.e., if D(A) is dense.

Next, we define bounded holomorphic semigroups.

Definition 3.7.3. Let θ ∈ (0, π2 ]. A semigroup T is called a bounded holomorphicsemigroup of angle θ if T has a bounded holomorphic extension to Σθ′ for eachθ′ ∈ (0, θ).

We denote the extension of T to Σθ by T again. If we do not want to specifythe angle, we call T a bounded holomorphic semigroup if T is a bounded holomor-phic semigroup of angle θ for some θ ∈ (0, π

2].

Some caution is required concerning this terminology. If T is a boundedsemigroup which is holomorphic, then it is not necessarily a bounded holomorphicsemigroup since it is just bounded on R+ and may not be bounded on a sector.For example, let X = C and T (t) = eit (t ≥ 0).

The following result is an immediate consequence of Proposition 3.7.2 b).

Proposition 3.7.4. An operator A generates a holomorphic semigroup if and onlyif there exists ω ≥ 0 such that A−ω generates a bounded holomorphic semigroup.

Next we give some examples of holomorphic semigroups.

Example 3.7.5 (Selfadjoint operators). Let A be a selfadjoint operator on a Hilbertspace H. Assume that A is bounded above by ω. Then A generates a boundedholomorphic C0-semigroup of angle π/2 satisfying

‖T (z)‖ ≤ eωRe z (Re z > 0).

Proof. By the Spectral Theorem B.13, we can assume that H = L2(Ω, μ) and thatA is given by

D(A) = {f ∈ H : mf ∈ H}, Af = m · f,where m : Ω→ (−∞, ω] is measurable. It is easy to see that

(T (z)f)(x) := ezm(x)f(x) (x ∈ Ω, Re z > 0),

defines a holomorphic C0-semigroup on H, whose generator is A.

Example 3.7.6 (Gaussian semigroup). Let X be one of the spaces Lp(Rn) (1 ≤ p <∞), C0(Rn) or BUC(Rn). Then

(G(t)f)(x) := (4πt)−n/2

∫Rn

f(x− y)e−|y|2/4t dy (t > 0, f ∈ X, x ∈ Rn)

defines a bounded holomorphic C0-semigroup of angle π/2 on X. Its generator isthe Laplacian ΔX on X with maximal domain; i.e.,

D(ΔX) = {f ∈ X : Δf ∈ X},ΔXf = Δf,

where we identify X with a subspace of D(Rn)′, and Δf =∑n

j=1D2j f (see Ap-

pendix E).

Proof. a) Let kt ∈ S(Rn) be given by

kt(x) =1

(4πt)n/2

e−|x|2/4t (t > 0, x ∈ Rn).

Then G(t)f = kt ∗ f ∈ X. Note that Fkt = ht, where ht(x) := e−t|x|2 . Hence,ht+s = hths. Recall that F is an isomorphism from S(Rn)′ onto S(Rn)′ suchthat F(ψ ∗ f) = Fψ · Ff for all f ∈ S(Rn)′, ψ ∈ S(Rn) (see Appendix E).Thus, F(G(t)f) = ht · Ff . It follows that G(t + s) = G(t)G(s) (t, s > 0).Since {kt : t > 0} is an approximate identity (see Lemma 1.3.3), it follows that‖kt ∗f −f‖X → 0 as t ↓ 0 for all f ∈ X. We have shown that G is a C0-semigroup.

b) The function kz is also defined for Re z > 0 and z �→ kz : C+ → L1(Rn) isa holomorphic function satisfying supz∈Σθ

‖kz‖L1(Rn) < ∞ for each 0 < θ < π/2.This shows that G(z)f := kz ∗ f defines a holomorphic extension of G to C+ withvalues in L(X) such that supz∈Σθ

‖G(z)‖ <∞ for each 0 < θ < π/2.c) We identify the generator of G.

First step: Let f ∈ X such that Δf ∈ X. We show that Δ(G(t)f) = G(t)(Δf).Let m(x) = −|x|2. Then

F(Δ(G(t)f)) = mF(G(t)f) = mhtFf = htmFf = htF(Δf) = F(G(t)Δf).

This proves the claim.Second step: Let ψ ∈ S(Rn). Then∫ t

0

G1(s)Δψ ds = G1(t)ψ − ψ,

where G1 is the Gaussian semigroup on L1(Rn). In fact,

F(∫ t

0

G1(s)Δψ ds

)(x) =

∫ t

0

F(G1(s)Δψ)(x) ds

=

∫ t

0

e−s|x|2(−|x|)2(Fψ)(x) ds

= (e−t|x|2 − 1)(Fψ)(x)= (F(G1(t)ψ − ψ))(x).

The claim follows from the uniqueness of Fourier transforms.Third step: Let f ∈ X, t > 0. We show that

Δ

∫ t

0

G(s)f ds = G(t)f − f.

Then it follows from Corollary 3.1.13 (using also the first step) that the generatorofG is the Laplacian with maximal domain. Let ψ ∈ S(Rn). Then Fubini’s theorem

gives 〈Δψ,G(s)f〉 = 〈G1(s)Δψ, f〉 and⟨ψ,Δ

∫ t

0

G(s)f ds

⟩=

⟨Δψ,

∫ t

0

G(s)f ds

⟩=

∫ t

0

〈Δψ,G(s)f〉 ds

=

∫ t

0

〈G1(s)Δψ, f〉 ds

=

⟨∫ t

0

G1(s)Δψ ds, f

⟩= 〈G1(t)ψ − ψ, f〉= 〈ψ,G(t)f − f〉,

by the second step. This proves the claim.

Remark 3.7.7. a) Let X = Lp(Rn) (1 < p < ∞). Then D(ΔX) = W 2,p(Rn). Infact, R(1,ΔX) is given by the Fourier multiplier x �→ (1 + |x|2)−1 (see AppendixE). It is easy to see that the function mjk(x) := −xjxk(1 + |x|2)−1 satisfies thecondition of Mikhlin’s theorem (Theorem E.3), so it is a Fourier multiplier forLp(Rn) for j, k ∈ {1, . . . , n}. One verifies easily that F(DjDkR(1,ΔX)f) = mjkFffor all f ∈ S(Rn). Thus DjDkR(1,ΔX) : S(Rn)→ S(Rn) has a bounded extensionto Lp(Rn). This means that

D(ΔX) = R(1,ΔX)Lp(Rn) ⊂W 2,p(Rn).

b) If X = L1(Rn), C0(Rn) or BUC(Rn), then D(ΔX) is not a classical functionspace. For example, if X = L1(Rn), then D(ΔX) � W 2,1(Rn). Similarly, if X =C0(Rn), then D(ΔX) contains functions which are not in C2(Rn). See [DL90,Chapter II, Section 3, Remark 5].

Modifying the Banach space in Example 3.7.6 we obtain an example of aholomorphic semigroup which is not a C0-semigroup. Another example will begiven in Chapter 6. Let Cb(Rn) be the Banach space of all bounded continuouscomplex-valued functions on Rn with the supremum norm.

Example 3.7.8. Let X = Cb(Rn) or L∞(Rn). Define the Gaussian semigroup G onX as in Example 3.7.6. Then G is a bounded holomorphic semigroup which is nota C0-semigroup. Its generator is the operator ΔX defined as in Example 3.7.6.

Example 3.7.9 (Poisson semigroup). Let X be one of the spaces considered inExample 3.7.6. Let

pt(x) = cnt

(t2 + |x|2)(n+1)/2(x ∈ Rn, t > 0)

where cn := Γ((n+ 1)/2)/π(n+1)/2. Then pt ∈ L1(Rn) and (Fpt)(x) = e−t|x| (t >0). Similarly to Example 3.7.6, one shows that

T (t)f := pt ∗ f (t > 0)

defines a C0-semigroup on X, which is called the Poisson semigroup. It is again abounded holomorphic C0-semigroup of angle π/2 on X. Its holomorphic extensionto the sector Σπ/2 is bounded on Σθ for θ < π/2 and is given by

T (z)f := pz ∗ f,where

pz(x) := cnz

(z2 + |x|2)(n+1)/2

for x ∈ Rn and Re z > 0. Its generator is the operator AX defined by

AXf := F−1(−| · |Ff),D(AX) := {f ∈ X : F−1(| · |Ff) ∈ X},

where F now denotes the Fourier transform in S(Rn)′.A more explicit description of the operator ALp(R) is of particular interest.

Observing that−(−i sign(ξ))(iξ) = −|ξ| (ξ ∈ R),

it follows from (E.19) that

ALp(R) = −H ∂

∂x,

where H denotes the Hilbert transform defined by

(Hf)(x) := limε→0,R→∞

1

π

∫ε≤|y|≤R

f(x− y)

ydy.

Since the Hilbert transform acts as a bounded operator on Lp(R) for 1 < p < ∞by Proposition E.5, it follows that the domain D(ALp(R)) of ALp(R) coincides withW 1,p(R). However, ALp(R) is not a first-order differential operator. We shall returnto the relationship between the Poisson and Gaussian semigroups in Example 3.8.5.

Remark 3.7.10. Let G be the Gaussian semigroup on L1(Rn). It follows from theexplicit formula for kz given in Example 3.7.6a) that for Re z > 0

‖G(z)‖L(L1(Rn)) = ‖kz‖L1(Rn) =

( |z|Re z

)n/2

. (3.44)

Thus, by Proposition 3.7.2c), Gα defined by Gα(t) := G(eiαt) for |α| < π2 is an

example of a bounded C0-semigroup, which due to (3.44), is not a contractionsemigroup.

The following characterization of a bounded holomorphic semigroup in termsof a single resolvent estimate for its generator is of fundamental importance.

Theorem 3.7.11. Let A be an operator on X and θ ∈ (0, π2 ]. The following asser-tions are equivalent:

(i) A generates a bounded holomorphic semigroup of angle θ.

(ii) Σθ+π2⊂ ρ(A) and

supλ∈Σθ+π

2−ε

‖λR(λ,A)‖ <∞ for all ε > 0.

Proof. In order to prove the assertion (i) ⇒ (ii), note that if λ0 ∈ ρ(A) andλ �→ R(λ,A) has a holomorphic extension to some open connected set Ω containingλ0, then by Proposition B.5, Ω ⊂ ρ(A) and the extension is the resolvent. Thus(i) ⇒ (ii) follows immediately from Theorem 2.6.1. In order to prove the converseimplication (ii) ⇒ (i), note that by Theorem 2.6.1 there exists a holomorphicfunction T : Σθ → L(X) which is bounded on Σα for 0 < α < θ such that

R(λ,A) =

∫ ∞

0

e−λtT (t) dt (Reλ > 0). (3.45)

The proof of Theorem 3.1.7 shows that T is a semigroup. This proves (i).

In particular, a densely defined operator A satisfying (ii) of Theorem 3.7.11 isa Hille-Yosida operator, by Example 3.5.9 c). Moreover, the bounded holomorphicsemigroup T generated by A is a C0-semigroup if and only if D(A) is dense in X,by Corollary 3.3.11.

We note from (2.13) that the semigroup T generated by A is given by

T (z) =1

2πi

∫Γ

eλzR(λ,A) dλ (z ∈ Σα), (3.46)

if 0 < α < θ, where the contour Γ consists of

Γ± := {re±γ : δ ≤ r} and Γ0 := {δeiθ′ : |θ′| ≤ γ}

with α+ π2< γ < θ + π

2and δ > 0.

If we do not want to specify the angle of holomorphy in the above theorem,then it suffices to verify condition (ii) above on a right half-plane. In fact, thefollowing holds true.

Corollary 3.7.12. For an operator A on X the following are equivalent:

(i) A generates a bounded holomorphic semigroup on X.

(ii) {z ∈ C : Re z > 0} ⊂ ρ(A) and

M := supReλ>0

‖λR(λ,A)‖ <∞.

Proof. By Theorem 3.7.11, we only have to prove that (ii) implies the secondassertion of Theorem 3.7.11. Set c := 1

2M . For s ∈ R \ {0} and −c|s| < r ≤ 0, letλ := c|s| + r + is. Then |λ − (r + is)| = c|s| ≤ 1

2‖R(λ,A)‖−1. By Corollary B.3,r + is ∈ ρ(A) and

‖ (r + is)R(r + is, A) ‖ ≤ 2M|r + is||s| ≤ 2M(c+ 1).

Thus, (ii) is satisfied with θ = arctan c.

Remark 3.7.13. Suppose that an operator A satisfies the equivalent conditions (i)and (ii) of the above corollary. For Y := D(A) let AY be the part of A definedas in (3.27). It follows from Lemma 3.3.12 and Corollary 3.7.12 that in this caseAY generates a bounded holomorphic C0-semigroup TY on Y . Moreover, TY (t) =T (t)|Y for t ≥ 0, where T is the semigroup generated by A.

Corollary 3.7.14. Let A be an operator on X such that σ(A) ⊂ iR. Assume thatthere exists a constant M > 0 such that

‖R(λ,A)‖ ≤ M

|Reλ| (Reλ �= 0). (3.47)

Then A2 generates a bounded holomorphic semigroup of angle π/2 on X.

Proof. Let θ ∈ (0, π/2] and λ ∈ Σθ+π/2. Then there exist r > 0 and ϕ ∈ (0, π/2)such that λ = r2e2iϕ. Observe that

λ− A2 = (reiϕ +A)(reiϕ − A).

The assumption implies that λ ∈ ρ(A2) and that

R(λ,A2) = −R(reiϕ, A)R(−reiϕ, A).

The resolvent estimate (3.47) implies that

‖R(λ,A2)‖ ≤ M2

(r cosϕ)2=

M2

(cosϕ)21

|λ| (λ ∈ Σθ+π/2).

Hence, the assertion follows from Theorem 3.7.11.

Applying Corollary 3.7.14 to the situation of generators of bounded C0-groupswe immediately obtain the following result.


Corollary 3.7.15. Let A be the generator of a bounded C0-group U on X. Then A2

generates a bounded holomorphic C0-semigroup T of angle π/2 on X. Moreover,for t > 0,

T (t) =

∫R

kt(s)U(s) ds,

where kt(s) = (4πt)−1/2e−|s|2/4t.

Proof. The fact that A2 generates a bounded holomorphic C0-semigroup of angleπ/2 is immediate from Corollary 3.7.14. Define T (0)x = x and

T (t)x =

∫R

kt(s)U(s)x ds =

∫R

k1(s)U(s√t) ds

=

∫ ∞

0

kt(s)(U(s)x+ U(−s)x) ds.

Then T is strongly continuous by the dominated convergence theorem. By Propo-sition 1.6.8,

T (λ) =1

2√λR(√

λ,A)+

1

2√λR(√

λ,−A)= R(λ,A2)

for λ > 0. Thus T is a C0-semigroup generated by A2.

Let X be any of the spaces considered in Example 3.7.6 in the case n = 1,and let U be the shift group: (U(t)f)(x) = f(x − t) (see Example 3.3.10). Thenthe C0-semigroup constructed in Corollary 3.7.15 is the Gaussian semigroup.

The following proposition will be useful in Chapter 6 when we are dealingwith the holomorphic semigroup generated by the Laplacian subject to Dirich-let boundary conditions on spaces of continuous functions. Note that A is notnecessarily densely defined.

Proposition 3.7.16. Let A be a dissipative operator on X and assume that A gen-erates a holomorphic semigroup T on X. Then ‖T (t)‖ ≤ 1 for all t > 0.

Proof. By the remark after Theorem 3.7.11, A is a Hille-Yosida operator. Hence,there exists λ0 > 0 with λ0 ∈ ρ(A). By the proof of the Lumer-Phillips theorem,we have (0,∞) ⊂ ρ(A) and ||λR(λ,A)|| ≤ 1 for all λ > 0. By the proof of Corollary3.3.6, we have T (t)x = limn→∞ ‖(I − t

nA)−nx‖ for t > 0 and x ∈ X. Hence,

‖T (t)‖ ≤ lim supn→∞

∥∥(I − tnA

)−n∥∥ ≤ 1

for t > 0.

Applying Corollary 3.7.12 to the operator A − ω for suitable ω, in view ofProposition 3.7.4, we obtain the following characterization results for holomorphicsemigroups.

Corollary 3.7.17. Let A be an operator on X, and a ∈ R. Then A generatesa holomorphic semigroup if and only if there exists r > 0 such that {λ ∈ C :Reλ > a, |λ| > r} ⊂ ρ(A) and

supReλ>a|λ|>r

‖λR(λ,A)‖ <∞.

Modifying the proof of Corollary 3.7.12 to the situation of holomorphic semi-groups which are not necessarily bounded, we obtain the following result.

Corollary 3.7.18. Let A be the generator of a semigroup T on X. Then T is holo-morphic if and only if there exists r > 0 such that {is : s ∈ R, |s| > r} ⊂ ρ(A)and

sup|s|>r

‖sR(is, A)‖ <∞.

We now characterize bounded holomorphic semigroups in terms of the be-haviour of ‖tAT (t)‖ for positive t.

Theorem 3.7.19. Let T be a bounded semigroup on X with generator A. Then T isa bounded holomorphic semigroup if and only if T (t)x ∈ D(A) for all t > 0, x ∈ X,and

supt>0

‖ tAT (t) ‖ <∞. (3.48)

Proof. Suppose that T is a bounded holomorphic semigroup of angle θ. Then T isnorm-differentiable on (0,∞), so T (t)x ∈ D(A) for t > 0 and x ∈ X . By Cauchy’sintegral formula for the derivative,

AT (t) = T ′(t) =1

2πi

∫|z−t|=t sin θ/2

T (z)

(z − t)2dz.

Hence,‖tAT (t)‖ ≤ (sin θ/2)−1 sup

z∈Σθ/2

‖T (z)‖ <∞.

Conversely, let M := supt>0 {‖T (t) ‖, ‖ tAT (t) ‖}. By assumption, T (t)x ∈D(A) for all t > 0 and x ∈ X. Since AnT (t) = (AT (t/n))n ∈ L(X) for n ∈ N, wehave ∥∥∥∥ AnT (t)

n!

∥∥∥∥ =

∥∥∥∥ (AT (t/n))n

n!

∥∥∥∥ ≤ (nt M)n

n!≤(eM

t

)n

for all n ∈ N.

If |z| < t2eM

, then T (t+z) :=∑∞

n=0 zn AnT (t)

n!converges in norm and ‖T (t+z)‖ ≤

1 + M . Then for s ∈ [0, t2eM

), T (t + s) = T (t + s). In fact, let x ∈ D(A) and

u(s) := T (t + s)x. Then u(s) ∈ D(A) and u′(s) = Au(s) (s ∈ [0, t2eM

)). Sinceu(0) = T (t)x, it follows from Proposition 3.1.11 that u(s) = T (s)T (t)x = T (t+s)x.By uniqueness of analytic extensions, we obtain a bounded, holomorphic extensionT of T to the sector Σθ where θ = arctan 1

2eM .

Remark 3.7.20. A slight modification of the above proof implies the followingassertion:Let A be the generator of a bounded holomorphic semigroup T onX . Then T (t)x ∈D(An) for all x ∈ X , t > 0 and n ∈ N, and we have

supt>0

||tnAnT (t)|| <∞ (n ∈ N).

Theorem 3.7.19 and Remark 3.7.20 have several important consequences forthe regularity of the solution of the associated Cauchy problem. In contrast tothe situation of C0-semigroups where we obtain a classical solution of the Cauchyproblem only if the initial condition x belongs to the domain D(A) of the generatorA, we see that if A generates a holomorphic C0-semigroup, then for all x ∈ X weobtain a solution which is differentiable for t > 0. More precisely, the followingholds.

Corollary 3.7.21. Let x ∈ X and assume that A generates a holomorphic C0-semigroup on X. Then there exists a unique function u ∈ C∞((0,∞), X)∩ C(R+, X) ∩ C((0,∞), D(A)) satisfying{

u′(t) = Au(t) (t > 0),

u(0) = x.

The phenomenon described in Corollary 3.7.21 is frequently called thesmoothing effect of holomorphic C0-semigroups.

We now consider the inhomogeneous Cauchy problem

(ACPf )

{u′(t) = Au(t) + f(t) (t ∈ [0, τ ]),

u(0) = x,

associated with the generator A of a holomorphic semigroup, and we establish thatthe variation of constants formula holds when x ∈ D(A).

Proposition 3.7.22. Let A be the generator of a holomorphic semigroup T on X.Let f ∈ L1((0, τ), X) and x ∈ D(A). Then (ACPf ) has a unique mild solution uwhich is given by

u(t) = T (t)x+

∫ t

0

T (t− s)f(s) ds. (t > 0)

Proof. By the remark following Theorem 3.7.11, A is a Hille-Yosida operator.Hence, the first assertion follows from the Da Prato-Sinestrari Theorem 3.5.2.The formula for u follows from Lemma 3.2.9, since the once integrated semigroupgenerated by A is given by S(t)x =

∫ t

0T (s)x ds and the derivative of S ∗f is easily

seen to be T ∗ f .

The following perturbation result for generators of holomorphic semigroupsis particularly useful when dealing with lower order perturbations of differentialoperators. Consider a closed operator A on X. Then a mapping B : D(A)→ X iscontinuous (with respect to the graph norm on D(A)) if and only if

‖Bx‖ ≤ c‖Ax‖+ b‖x‖ (x ∈ D(A)),

for suitable c, b ≥ 0. One frequently says that B is a relatively bounded perturbationof A in that case.

Theorem 3.7.23. Let A be the generator of a holomorphic semigroup on X. LetB : D(A) → X be an operator such that for every ε > 0 there exists a constantb ≥ 0 such that

‖Bx ‖ ≤ ε‖Ax ‖+ b‖x ‖ (x ∈ D(A)). (3.49)

Then A+B generates a holomorphic semigroup.

Proof. Assume first that A generates a bounded holomorphic semigroup on X .Corollary 3.7.12 implies that there exists θ ∈ (0, π

2] such that

Σθ+π/2 ⊂ ρ(A) and supλ∈Σθ+π/2

‖λR(λ,A)‖ =: M <∞.

It follows from the assumption on B that given ε > 0, there exists b ≥ 0 such thatfor x ∈ X

‖BR(λ,A)x ‖ ≤ ε‖AR(λ,A)x ‖+ b‖R(λ,A)x ‖= ε‖λR(λ,A)x− x‖+ b‖R(λ,A)x‖≤ ε(M + 1)‖ x ‖+ bM

|λ| ‖x ‖ (λ ∈ Σθ+π/2).

Choosing ε := (2(M + 1))−1, it follows that ‖BR(λ,A) ‖ < 3/4 whenever |λ| >4bM and hence that I − BR(λ,A) is invertible, with ‖(I − BR(λ,A))−1‖ < 4.Since

λ− (A+B) = (I −BR(λ,A))(λ− A) (λ ∈ Σθ+π/2),

it follows that λ− (A+B) is invertible for λ ∈ Σθ+π/2 with |λ| > 4bM and that

‖R(λ,A+B) ‖ ≤ 4M

|λ| (λ ∈ Σθ+π/2, |λ| > 4bM).

By Corollary 3.7.17, A+B generates a holomorphic semigroup.If A generates a holomorphic semigroup which is not bounded, choose ω ∈ R

such that A−ω generates a bounded holomorphic semigroup onX (see Proposition3.7.4). The first part of the proof implies that A+B−ω, and therefore also A+B,generates a holomorphic semigroup on X .

Note that in the situation of Theorem 3.7.23, A and A + B have the samedomain. Thus, the semigroup generated by A is a C0-semigroup if and only if theone generated by A+B is a C0-semigroup.


Example 3.7.24 (First order perturbations of the Laplacian). Consider the Gauss-ian semigroup G with generator ΔX on any of the spaces X of Example 3.7.6. Forj = 1, 2, . . . , n, let Uj be the C0-group on X defined by

(Uj(t)f)(x) = f(x1, . . . , xj − t, . . . , xn)

with generator −Dj , and let Tj be the holomorphic C0-semigroup with generatorD2

j (see Corollary 3.7.15). Then T1, . . . , Tn commute and G(t) = T1(t) · · ·Tn(t).Since

(Tj(t)f)(x) =

∫R

(4πt)−1/2e−(xj−s)2/4tf(x1, . . . , s, . . . , xn) ds,

(DjTj(t)f)(x) = −∫R

(xj − s)

4√πt3/2

e−(xj−s)2/4tf(x1, . . . , s, . . . , xn) ds.

Hence by Young’s inequality (see Proposition 1.3.2),

‖DjTj(t)‖L(X) ≤∫R

|s|4√πt3/2

e−s2/4t ds

=2√πt

∫ ∞

0

ue−u2

du

=1√πt

.

Since ‖Tj(t)‖L(X) = 1, it follows that ‖DjG(t)‖L(X) ≤ 1√πt, and hence

‖DjR(λ,ΔX)‖L(X) =

∥∥∥∥∫ ∞

0

e−λtDjG(t) dt

∥∥∥∥L(X)

≤ 1√π

∫ ∞

0

e−λt

√t

dt

=c√λ

for all λ > 0, for some constant c.

Now let B be a first-order differential operator of the form

(Bf)(x) =

n∑j=1

bj(x)(Djf)(x) + b0(x)f(x),

for some bj ∈ L∞(Rn) (j = 0, 1, . . . , n) (and bj continuous if X = C0(Rn); bj


uniformly continuous if X = BUC(Rn)). For f ∈ D(ΔX),

‖Bf‖ = ‖BR(λ,ΔX)(λ−ΔX)f‖

≤n∑

j=1

‖bj‖∞‖DjR(λ,ΔX)‖L(X) ‖(λ−ΔX)f‖+ ‖b0‖∞‖f‖

≤n∑

j=1

‖bj‖∞c√λ

‖ΔXf‖+⎛⎝ n∑

j=1

‖bj‖∞c√λ+ ‖b0‖∞

⎞⎠ ‖f‖.Since λ may be chosen arbitrary large, this establishes (3.49), and Theorem 3.7.23shows that ΔX +B generates a holomorphic C0-semigroup on X.

We shall extend Example 3.7.24 to more general differential operators inSection 7.2.

We now prove a second perturbation theorem for holomorphic semigroupswhere the norm estimate (3.49) is replaced by compactness.

Theorem 3.7.25 (Desch-Schappacher). Let A be the generator of a holomorphicC0-semigroup T . Let B : D(A) → X be a compact linear operator where D(A)carries the graph norm. Then A + B generates a holomorphic C0-semigroup S.Moreover, T (t)− S(t) is compact for each t > 0.

Proof. By Corollary 3.7.17, there exist r > 0, M > 0 such that λ ∈ ρ(A) and‖λR(λ,A)‖L(X) ≤ M whenever |λ| ≥ r, Reλ > 0. Since D(A) is dense in X,it follows that lim |λ|→∞

Reλ>0λR(λ,A)x = x for all x ∈ X (see Lemma 3.3.12). Since

λR(λ,A)x−x = AR(λ,A)x, it follows that lim |λ|→∞Reλ>0

R(λ,A)x = 0 in D(A) for all

x ∈ X. By Proposition B.15, the convergence is uniform on compact subsets of X.Since B : D(A)→ X is compact, it follows that

lim|λ|→∞Reλ>0

‖R(λ,A)B‖L(D(A)) = 0.

Consequently, there exists r1 ≥ r such that ‖R(λ,A)B‖L(D(A)) ≤ 12 whenever

|λ| ≥ r1, Reλ > 0. Denote by ID(A) the identity map on D(A). It follows that(ID(A) −R(λ,A)B)−1 exists in L(D(A)) and

‖(ID(A) −R(λ,A)B)−1‖L(D(A)) ≤ 2 (|λ| ≥ r1, Reλ > 0).

Thus (λ− (A+B)) = (λ−A)(I −R(λ,A)B) is invertible and

R(λ,A+B) = (ID(A) −R(λ,A)B)−1R(λ,A) (Reλ > 0, |λ| ≥ r1).

Moreover, for |λ| ≥ r1, Reλ > 0,

‖R(λ,A+B)‖L(X,D(A)) ≤ 2‖R(λ,A)‖L(X,D(A))

≤ M1 := 2

(M

r1+ 1 +M

),

since

‖R(λ,A)x‖D(A) = ‖R(λ,A)x‖X + ‖AR(λ,A)x‖X≤ M

|λ| ‖x‖X + ‖λR(λ,A)x− x‖X

≤ M

r1‖x‖X + (M + 1)‖x‖X .

Hence for x ∈ X, |λ| ≥ r1, Reλ > 0,

‖λR(λ,A+B)x‖X = ‖(A+B)R(λ,A+B)x− x‖X≤ ‖R(λ,A+B)x‖D(A)

+ ‖B‖L(D(A),X)‖R(λ,A+B)x‖D(A) + ‖x‖X≤ (

M1(1 + ‖B‖L(D(A),X)) + 1) ‖x‖X .

Now it follows from Corollary 3.7.17 that A + B generates a holomorphic C0-semigroup.

It remains to show the last assertion. Denote by K(X) the closed subspaceof L(X) consisting of all compact operators and by q : L(X) → L(X)/K(X) thequotient mapping. For λ > λ0 := max{ω(T ), ω(S)}, we have∫ ∞

0

e−λt(S(t)− T (t)) dt = R(λ,A+B)−R(λ,A)

= R(λ,A+B)[(λ− A)− (λ− A−B)]R(λ,A)

= R(λ,A+B)BR(λ,A) ∈ K(X).

Since S and T are holomorphic, the function U := S − T is norm-continuous

on (0,∞). Since (q ◦ U)(λ) = q(U(λ)) = 0 for all λ > λ0, it follows from theuniqueness theorem that q ◦ U ≡ 0; i.e., U(t) ∈ K(X) for all t > 0.

We should point out that in the situation of Theorem 3.7.25 the norm esti-mate (3.49) is not true in general; see the Notes for further information.

3.8 Fractional Powers

A particularly interesting example of a holomorphic semigroup is the family offractional powers of a sectorial operator. Consider an operator B on X for which(−∞, 0] ⊂ ρ(B) and supλ≤0(1 − λ)‖R(λ,B)‖ < ∞. It follows from Corollary B.3that B is sectorial in the sense that there exist constants M > 0, ϕ ∈ (0, π) suchthat

σ(B) ⊂ Σϕ and ‖R(λ,B)‖ ≤ M

1 + |λ| , λ ∈ C\Σϕ. (3.50)

3.8. FRACTIONAL POWERS 163

Let Γ be the downward path consisting of {se±iϕ : s ≥ r} and {reiθ : −ϕ ≤θ ≤ ϕ}, where r > 0 is chosen so small that σ(B) is to the right of Γ. Then thefractional powers (B−z)Re z>0 of B are defined by

B−z :=1

2πi

∫Γ

λ−zR(λ,B) dλ, (Re z > 0). (3.51)

Here, λ−z = exp(−z(log |λ|+iθ)) if λ = |λ|eiθ, −π < θ < π. Note that the integralis absolutely convergent, uniformly for z in compact subsets of C+, and thereforez �→ B−z is holomorphic from C+ to L(X). By Cauchy’s theorem, the definitionof B−z is independent of the choices of ϕ and r. Moreover, when z = n ∈ N, Γmay be replaced by a closed contour around 0 and then the residue theorem showsthat

B−z = − 1

(n− 1)!

(d

dλ

)n−1

R(λ,B)

∣∣∣∣∣λ=0

= (−1)nR(0, B)n = B−n

in the usual sense.Now, assume for the time being that 0 < Re z < 1. Then

B−z = limϕ↑πr↓0

1

2πi

∫Γ

λ−zR(λ,B) dλ

= −e−iπz

2πi

∫ ∞

0

s−z(s+B)−1 ds+eiπz

2πi

∫ ∞

0

s−z(s+B)−1 ds

=sinπz

π

∫ ∞

0

s−z(s+B)−1 ds. (3.52)

In the particular case when X = C and B = 1, (3.52) gives∫ ∞

0

s−z(s+ 1)−1 ds =π

sinπz(0 < Re z < 1). (3.53)

Hence,

‖B−z‖ ≤ | sinπz|π

∫ ∞

0

s−Re z M

1 + sds = M

| sinπz|sin(πRe z)

(0 < Re z < 1). (3.54)

Theorem 3.8.1. Let B be an operator on X such that (−∞, 0] ⊂ ρ(B) andsupλ≤0(1 − λ)‖R(λ,B)‖ < ∞. Then the family (B−z)Re z>0 defines a holomor-phic semigroup on X of angle π/2. If B is densely defined, then (B−z)Re z>0 is aholomorphic C0-semigroup.

Proof. We have observed above that z �→ B−z is holomorphic for Re z > 0, and itfollows from (3.54) that B−z is uniformly bounded in Σθ ∩ {z ∈ C : |z| < 1} forθ ∈ (0, π/2). To verify the semigroup property, let Γ and Γ′ be two contours as in

the definition of B−z, with Γ to the left of Γ′. For Re z1 > 0 and Re z2 > 0, theresolvent identity and Fubini’s theorem give

B−z1B−z2 =1

(2πi)2

∫Γ

∫Γ′λ−z1μ−z2R(λ,B)R(μ,B) dμ dλ

=1

(2πi)2

∫Γ

∫Γ′λ−z1μ−z2

(R(λ,B)−R(μ,B)

μ− λ

)dμ dλ

=1

2πi

∫Γ

λ−z1

(1

2πi

∫Γ′

μ−z2

μ− λdμ

)R(λ,B) dλ

+1

2πi

∫Γ′μ−z2

(1

2πi

∫Γ

λ−z1

λ− μdλ

)R(μ,B) dμ

=1

2πi

∫Γ

λ−z1λ−z2R(λ,B) dλ

= B−(z1+z2).

Here, we have used Cauchy’s integral formula (after changing to a closed contouraround 0) to see that

1

2πi

∫Γ′

μ−z2

μ− λdμ = λ−z2 and

1

2πi

∫Γ

λ−z1

λ− μdλ = 0.

This proves the first assertion.Now suppose that B is densely defined. We have to show that lim z→0

z∈Σθ

B−zx =

x, for all θ ∈ (0, π/2) and all x ∈ X . Since B−z is uniformly bounded on Σθ ∩{z ∈ C : |z| < 1}, we may assume that x ∈ D(B) (see Proposition B.15). For0 < Re z < 1, it follows from (3.52) and (3.53) that

B−zx− x =sinπz

π

∫ ∞

0

s−z((s+B)−1x− (s+ 1)−1x

)ds

=sinπz

π

∫ ∞

0

s−z

s+ 1(s+B)−1(I −B)x ds.

Hence,

‖B−zx− x‖ ≤M| sinπz|

π

∫ ∞

0

s−Re z

(s+ 1)2ds ‖(I −B)x‖.

It follows that lim z→0z∈Σθ

‖B−zx− x‖ = 0 for all θ ∈ (0, π/2).

Now suppose that B is an operator on X such that (−∞, 0) ⊂ ρ(B) andM := supλ<0 ‖λR(λ,B)‖ <∞. We shall show that B has a special type of squareroot, which has interesting properties for semigroup generators.

For ε > 0, (−∞, 0] ⊂ ρ(ε+B) and

supλ≤0

(1− λ)‖R(λ, ε+B)‖ = supλ≤0

(1− λ)‖R(λ− ε, B)‖ ≤M/ε.

Hence we can define (ε + B)−1/2 ∈ L(X) as above, and then ((ε + B)−1/2)2 =(ε+B)−1. In particular, (ε+B)−1/2 is injective. Let (ε+B)1/2 be the algebraicinverse of (ε+B)−1/2, so

D((ε+B)1/2) = Ran((ε+B)−1/2),

(ε+B)1/2((ε+B)−1/2y) = y (y ∈ X).

Then (ε+B)1/2 is a closed operator on X, ((ε+B)1/2)2 = ε+B, and for x ∈ D(B),

x = (ε+B)−1/2((ε+B)−1/2(ε+B)x),

so

(ε+B)1/2x = (ε+B)−1/2(ε+B)x

=1

π

∫ ∞

0

s−1/2(s+ ε+B)−1(ε+B)x ds. (3.55)

Proposition 3.8.2. Let B be a densely defined operator on X such that (−∞, 0) ⊂ρ(B) and supλ<0 ‖λR(λ,B)‖ < ∞. Then there is a unique closed operator B1/2

such that

a) (B1/2)2 = B, and

b) For x ∈ D(B),

B1/2x = limε↓0

(ε+B)1/2x =1

π

∫ ∞

0

s−1/2(s+B)−1Bxds.

Moreover, D(B) is a core for B1/2, and D(B1/2) = D((ε+B)1/2) for all ε > 0.

Proof. Since

‖s−1/2(s+ ε+B)−1(ε+B)x‖ ≤Ms−3/2 (‖x‖+ ‖Bx‖) (0 < ε < 1)

and

‖s−1/2(s+ ε+B)−1(ε+B)x‖ = ‖s−1/2(x− s(s+ ε+B)−1x)‖≤ (M + 1)‖x‖s−1/2,

one can apply the dominated convergence theorem and take limits in (3.55) asε ↓ 0. Thus, we let

B1/2x := limλ↓0

(ε+B)1/2x =1

π

∫ ∞

0

s−1/2(s+B)−1Bxds


for x ∈ D(B). Moreover, for ε > 0,

B1/2x− (ε+B)1/2x

=1

π

∫ ∞

0

s−1/2((s+B)−1Bx− (s+ ε+B)−1(ε+B)x

)ds

= − ε

π

∫ ∞

0

s1/2(s+ ε+B)−1(s+B)−1x ds.

Let

Sε := − ε

π

∫ ∞

0

s1/2(s+ ε+B)−1(s+B)−1 ds.

Then Sε ∈ L(X),

‖Sε‖ ≤ ε

π

∫ ∞

0

M2

s1/2(s+ ε)ds = M2ε1/2,

by (3.53), and

B1/2x− (ε+B)1/2x = Sεx (x ∈ D(B), ε > 0).

We define

B1/2 := (ε+B)1/2 + Sε with D(B1/2) = D((ε+B)1/2).

Since (ε + B)1/2 is closed and Sε is bounded, B1/2 is closed. Moreover, b) holds.Since (ε + B)1/2 is densely defined and invertible, D(B) = D(((ε + B)1/2)2) isa core for (ε + B)1/2 (see Appendix B) and hence for B1/2. It follows that thedefinition of B1/2 is independent of ε.

Let x ∈ D(B). Then (ε+B)1/2x→ B1/2x and

B1/2(ε+B)1/2x = (ε+B)x+ Sε(ε+B)1/2x→ Bx as ε ↓ 0.

Since B1/2 is closed, it follows that B1/2x ∈ D(B1/2) and (B1/2)2x = Bx.Let y ∈ D((B1/2)2). By Lemma 3.3.12, ε(ε + B)−1y → y and B(ε(ε +

B)−1y) = ε(ε+B)−1(B1/2)2y → (B1/2)2y as ε→∞. Since B is closed, y ∈ D(B).Thus B = (B1/2)2.

Finally, let B be any closed operator such that Bx = B1/2x for all x ∈ D(B)

and B2 = B. Then (B + i)(B − i) = B2 + I = B + I, which is invertible, so B + i

is invertible. Since B is densely defined and ρ(B) is non-empty, D(B2) = D(B)

is a core for B (see Appendix B). Hence, B is the closure of B1/2|D(B), and thisproves uniqueness.

Now suppose that A is the generator of a bounded C0-semigroup T on X . If0 ∈ ρ(A), then the theory above can be applied to B := −A, so (−A)−z is defined

for Re z > 0. Substituting (s−A)−1 =∫∞0

e−stT (t) dt into (3.52), it is not difficultto see that

(−A)−z =1

Γ(z)

∫ ∞

0

tz−1T (t) dt (3.56)

for 0 < Re z < 1, and hence for Re z > 0 by uniqueness of holomorphic extensions.We shall not use this.

Proposition 3.8.2 shows that (−A)1/2 is defined whenever A generates abounded C0-semigroup.

Theorem 3.8.3. Let A be the generator of a bounded C0-semigroup T on X, anddefine

S(t)x =

⎧⎪⎨⎪⎩∫ ∞

0

te−t2/4s

2√πs3/2

T (s)x ds (t > 0),

x (t = 0).

Then S is a bounded holomorphic C0-semigroup of angle π/4, and the generator ofS is −(−A)1/2. Furthermore, for x ∈ D(A), u(t) := S(t)x is the unique boundedclassical solution of the second order Cauchy problem{

u′′(t) = −Au(t) (t ≥ 0),

u(0) = x.(3.57)

Moreover, if T is a bounded holomorphic C0-semigroup of angle θ ∈ (0, π/2], thenS is a bounded holomorphic C0-semigroup of angle ( θ2 + π

4 ).

Proof. For z ∈ Σπ/4, let ψz(s) =ze−z2/4s

2√πs3/2

(s > 0). Then ψz ∈ L1(R+), ‖ψz‖1 =

|z|Re(z2) ,

∫∞0

ψz(s) ds = 1, and lim z∈Σθz→0

∫∞δ|ψz(s)| ds = 0, for δ > 0 and 0 ≤ θ <

π/4.

For t ∈ R+, S(t)x =∫∞0

ψt(s)T (s)x ds (x ∈ X) and ψt(λ) = e−t√λ (λ > 0)

(Lemma 1.6.7). Hence ( ψt1 ∗ ψt2)(λ) = ψt1+t2(λ). It follows from the uniquenesstheorem that ψt1 ∗ ψt2 = ψt1+t2 (t1, t2 ≥ 0). Now,

S(t1)S(t2)x =

∫ ∞

0

∫ ∞

0

ψt1(s)ψt2(r)T (s+ r)x dr ds

=

∫ ∞

0

∫ t

0

ψt1(s)ψt2(t− s) ds T (t)x dt

= S(t1 + t2)x.

For z ∈ Σπ/4, let S(z)x =∫∞0

ψz(s)T (s)x ds. Then S(·) is holomorphic, ‖S(z)‖ ≤|z|

(Re z)2 sups≥0 ‖T (s)‖ and

‖S(z)x− x‖ =

∥∥∥∥∥∫ ∞

0

e−z2/4s

2√πs3/2

(T (s)z − z) ds

∥∥∥∥∥→ 0 (z ∈ Σθ, z → 0).


Thus S is a bounded holomorphic C0-semigroup of angle π/4.Let B be the generator of S and let x ∈ D(A). For t > 0, we can differentiate

through the integral sign and obtain

BS(t)x =d

dt(S(t)x) =

∫ ∞

0

e−t2/4s

2√πs3/2

(1− t2

2s

)T (s)x ds,

0 =

∫ ∞

0

e−t2/4s

2√πs3/2

(1− t2

2s

)x ds.

Hence

BS(t)x =

∫ ∞

0

e−t2/4s

2√πs3/2

(1− t2

2s

)(T (s)x− x) ds.

Since ‖T (s)x − x‖ ≤ s‖Ax‖ supt≥0 ‖T (t)‖, the dominated convergence theoremgives, on letting t ↓ 0, that x ∈ D(B) and

Bx =

∫ ∞

0

T (s)x− x

2√πs3/2

ds

=

∫ ∞

0

1

π

∫ ∞

0

λ1/2e−sλ dλ (T (s)x− x) ds

=1

π

∫ ∞

0

λ1/2(R(λ,A)x− λ−1x) dλ

=1

π

∫ ∞

0

λ−1/2R(λ,A)Axdλ

= −(−A)1/2x.

Since D(A) is a core for −(−A)1/2, it follows that B extends −(−A)1/2. However,B + i is invertible (since B generates a bounded holomorphic semigroup) andi − (−A)1/2 is invertible (since I − A = ((−A)1/2 − i)((−A)1/2 + i)), so B =−(−A)1/2.

Let x ∈ D(A) = D(((−A)1/2)2). Then it is immediate that u(t) := S(t)x isa bounded classical solution of (3.57).

Let u1 be any bounded solution of (3.57). Take μ > 0, and let

v(t) := R(μ,A)(u(t)− u1(t)) (t ≥ 0).

Then v(0) = 0, v is bounded, and

v′′(t) = −Av(t) = (I − μR(μ,A))(u(t)− u1(t)),

which is bounded. Since

v′(t) = v(t+ 1)− v(t)− 1

2

∫ t+1

t

(t+ 1− s)v′′(s) ds,

it follows that v′ is bounded.Let

w(t) :=

{v′(−t) + (−A)1/2v(−t) (t ≤ 0),

S(t)v′(0) (t > 0).

Since (−A)1/2R(μ,A) is bounded (by the closed graph theorem, or by direct cal-culation), w is bounded. For t ≤ 0,

w′(t) = −v′′(−t)− (−A)1/2v′(−t)= Av(t)− (−A)1/2v′(−t)= −(−A)1/2w(t).

Also, w′(t) = −(−A)1/2w(t) for all t ≥ 0, since −(−A)1/2 generates S. It followsfrom Proposition 3.1.11 that

w(t+ s) = S(t)w(s) (t ≥ 0, s ∈ R).

Now we extend w to C by

w(λ+ s) := S(λ)w(s) (λ ∈ Σπ/4, s ∈ R).

This is well defined since S(λ1 + λ2) = S(λ1)S(λ2). Moreover, w is holomorphic,and bounded since λ may be chosen in Σπ/8, where S is bounded. By Liouville’stheorem, w is constant, so S(t)v′(0) = v′(0) for all t ≥ 0. Now,

v′(t) = −(−A)1/2v(t) + w(−t)= −(−A)1/2v(t) + v′(0) (t ≥ 0),

v(0) = 0.

By Proposition 3.1.16,

v(t) =

∫ t

0

S(t− s)v′(0) ds = tv′(0).

But v is bounded, so v′(0) = 0 and hence v(t) = 0. Since R(μ,A) is injective, itfollows that u1(t) = u(t).

Now suppose that T is a bounded holomorphic C0-semigroup of angle θ ∈(0, π/2]. Let α ∈ (−θ, θ). An application of Cauchy’s theorem shows that

S(t) =

∫ ∞

0

te−t2/4reiα

2√πr3/2eiα/2

T (reiα) dr (t > 0).

Now let

S(z) =

∫ ∞

0

ze−z2/4reiα

2√πr3/2eiα/2

T (reiα) dr

for α2 − π

4 < arg z < α2 +

π4 . This defines a holomorphic extension of S to this sector,

and it is bounded in each proper subsector. Varying α provides a holomorphicextension of S to Σ( θ

2+π4 ) which is bounded on Σθ′ for θ

′ < ( θ2 + π4 ).


Example 3.8.4. Let (Ω, μ) be a measure space, X := L2(Ω, μ), m : Ω → R+ bemeasurable, and A be defined by

D(A) := {f ∈ X : mf ∈ X}Af := mf.

Then−A generates the C0-semigroup T (t)f = e−tmf . The operator A1/2 of Propo-sition 3.8.2 is given by

D(A1/2) = {f ∈ X : m1/2f ∈ X} and A1/2f = m1/2f.

The semigroup S generated by −A1/2, as in Theorem 3.8.3, is given by S(t)f =

e−tm1/2

f .

Example 3.8.5. Let X be any of the spaces of Example 3.7.6, and let T be theGaussian semigroup on X. Then T (s)f = ks∗f , where ks(x) = (4πs)−n/2e−|x|

2/4s.Hence, the holomorphic semigroup S of Theorem 3.8.3 is given by S(t)f = ht ∗ f ,where

ht(x) =

∫ ∞

0

te−t2/4s

2√πs3/2

e−|x|2/4s

(4πs)n/2ds.

Putting r = (t2 + |x|2)/4s gives

ht(x) =Γ(n+1

2 )t

π(n+1)/2(t2 + |x|2)(n+1)/2.

Thus, S is the Poisson semigroup considered in Example 3.7.9. Note that althoughthe generator ΔX of T is a second order differential operator, the generator AX =−(−ΔX)1/2 of S is not a first order differential operator.

It should be mentioned that (3.57) is an abstract elliptic equation. For ex-ample, if T is the Gaussian semigroup on C0(Rn) (Example 3.8.5), then lettingu(t, x) := (S(t)f)(x) for f ∈ C0(Rn), x ∈ Rn and t > 0, u is a solution of

utt +Δu :=∂2u

∂t2+

n∑j=1

∂2u

∂x2j

= 0 on (0,∞)× Rn;

i.e., u is a solution of the Laplace equation.

The wave equation

utt = Δu

will be treated in Sections 3.14–3.16 and Chapter 7.

3.9. BOUNDARY VALUES OF HOLOMORPHIC SEMIGROUPS 171

3.9 Boundary Values of Holomorphic Semigroups

Let T be a holomorphic C0-semigroup on X of angle π2. In this section we are

interested in the behaviour of T (is + t) as t tends to 0 and we ask under whatcircumstances the “boundary value” T (is) of T (which will be defined preciselybelow) exists and defines a C0-group. We also address the converse problem: WhichC0-groups are obtained as boundary values of holomorphic C0-semigroups?

As in Section 3.7, we let Σϕ := {z ∈ C \ {0} : |argz| < ϕ} be the sector in thecomplex plane of angle ϕ ∈ (0, π). Furthermore, we set Σ+

ϕ := Σϕ∩{z ∈ C : Im z ≥0}, Σ−ϕ := Σϕ ∩ {z ∈ C : Im z ≤ 0} and define D by D := {z ∈ Σπ/2 : |z| ≤ 1}.The following result gives an answer to our first question above.

Proposition 3.9.1. Let A be the generator of a holomorphic C0-semigroup T on Xof angle ϕ ∈ (0, π/2]. Then the following are equivalent:

(i) eiϕA generates a C0-semigroup T (eiϕ·) on X.

(ii) supz∈Σ+ϕ∩D ||T (z)|| <∞.

In this case, the C0-semigroup T (eiϕ·) is given by

T (eiϕs)x = limt↓0

T (t+ eiϕs)x (x ∈ X, s ≥ 0). (3.58)

The C0-semigroup S(s) := T (eiϕs) defined by (3.58) is called the boundarysemigroup of T . The following lemma will be useful in the proof of Proposition3.9.1.

Lemma 3.9.2. Let A be the generator of a holomorphic C0-semigroup T on X ofangle ϕ ∈ (0, π/2]. Assume that eiϕA generates a C0-semigroup S on X. Then

T (t+ eiϕs) = T (t)S(s) for all s, t ≥ 0.

Proof. Obviously, the resolvents of A and eiϕA commute. By Proposition 3.1.5,S(s)T (t) = T (t)S(s) for all s, t ≥ 0. Fix a, b ≥ 0 and denote by B the generator ofthe C0-semigroup V on X defined by V (t) := S(bt)T (at). For x ∈ D(A) we haveddtV (t)x = (a + beiϕ)AV (t)x. Hence d

dtV (t)x|t=0 = (a + beiϕ)Ax and thereforeB extends (a + beiϕ)A. It follows that V (t) = T ((a + beiϕ)t) for all t ≥ 0 (seeProposition 3.7.2 c)). In particular, we have V (1) = T (a + beiϕ) = S(b)T (a) =T (a)S(b) for all a, b ≥ 0.

Proof of Proposition 3.9.1. Assume that (i) holds. Let z ∈ Σ+ϕ ∩ D. There exist

a, b ∈ [0, 1] such that z = a + beiϕ. It follows from Lemma 3.9.2 that T (z) =T (a+ beiϕ) = T (a)T (beiϕ). Hence, ‖T (z)‖ ≤ ‖T (a)‖ ‖T (beiϕ)‖ ≤M for a suitableM and all z ∈ Σ+

ϕ ∩D. This implies (ii).In order to prove the converse implication, fix R > 0. Then there exists

MR > 0 such that ‖T (z)‖ ≤ MR whenever z ∈ Σ+ϕ and |z| ≤ R. For x ∈ X,

0 < t < t′ ≤ 1 and s ≥ 0 satisfying |t+ eiϕs| ≤ R, we have

‖T (t+ eiϕs)x− T (t′ + eiϕs)x‖ ≤ ∥∥T (t+ eiϕs)(x− T (t′ − t)x

)∥∥≤ MR‖x− T (t′ − t)x‖.

It thus follows that T (eiϕs)x := limt↓0 T (t+ eiϕs)x exists uniformly in s ∈ [0, R].Consequently, the mapping R+ → L(X), s �→ T (eiϕs) is strongly continuous. Itis easy to see that T (eiϕ·) is a C0-semigroup. Denote by B its generator. Letx ∈ D(A) and τ > 0. Then∫ τ

0

T (eiϕs)eiϕAxds = limt↓0

∫ τ

0

eiϕAT (t+ eiϕs)x ds

= limt↓0

∫ τ

0

d

ds

(T (t+ eiϕs)x

)ds

= limt↓0

(T (t+ eiϕτ)x− T (t)x)

= T (eiϕτ)x− x.

It follows from Proposition 3.1.9 f) that x ∈ D(B) and Bx = eiϕAx. We haveshown that B is an extension of eiϕA. Since ρ(B) ∩ ρ(eiϕA) �= ∅, it follows thatboth operators are equal.

Remark 3.9.3. Note that the above result may be easily modified to the case whereeiϕA and e−iϕA generate C0-semigroups. Indeed, assuming that

supz∈Σϕ∩D

||T (z)|| <∞, (3.59)

it follows that e±iϕA generate C0-semigroups on X which are given by T (e±iϕs)x:= limt↓0 T (t+ e±iϕs)x. In particular, if (3.59) is satisfied for ϕ = π

2, then iA and

−iA generate C0-semigroups T (±is) on X and we call S(s) := T (is) (s ∈ R), theboundary group of T .

For 1 ≤ p ≤ ∞, let Δp be the Laplacian on Lp(Rn) with maximal domain:

Δpf := Δf,

D(Δp) := {f ∈ Lp(Rn) : Δf ∈ Lp(Rn)}, (3.60)

where Δf is defined in the distributional sense. We proved in Example 3.7.6 thatfor 1 ≤ p < ∞ the operator Δp is the generator of the Gaussian semigroup Tp

on Lp(Rn). Moreover, Δ∗p = Δp′ , where 1/p + 1/p′ = 1. Although Δ∞ does notgenerate a C0-semigroup on L∞(Rn), it does generate a holomorphic semigroupT∞ given by

T∞(z)f := kz ∗ f (f ∈ L∞(Rn)),

(see Example 3.7.8).

We wish to determine whether iΔp generates a C0-semigroup. By duality, wemay restrict ourselves in the following to the case when 1 ≤ p ≤ 2. Since FT2(z)f =

e−z|·|2Ff and (2π)−n/2F is unitary on L2(Rn), we have ||T2(z)||L(L2(Rn)) = 1 forall z ∈ C with Re z > 0. Hence, by Proposition 3.9.1 the operator iΔ2 generates aC0-semigroup on L2(Rn). By Remark 3.7.10,

‖T1(z)‖L(L1(Rn)) =

( |z|Re z

)n/2

(Re z > 0).

Hence, by the Riesz-Thorin interpolation theorem (see [Hor83, Theorem 7.1.12])

‖Tp(z)‖L(Lp(Rn)) ≤( |z|Re z

)n(1/p−1/2)

(1 ≤ p ≤ 2,Re z > 0). (3.61)

In fact, we will now show that a multiple of the above upper bound will also serveas a lower bound for ||Tp(z)||L(Lp(Rn)). Then we can apply Proposition 3.9.1 andobtain the following result.

Theorem 3.9.4 (Hormander). Let 1 ≤ p <∞. Then the operator iΔp generates aC0-semigroup on Lp(Rn) if and only if p = 2.

Proof. By Proposition 3.9.1, it suffices to show that a multiple of the above upperbound (3.61) will also serve as a lower bound for ||Tp(z)||L(Lp(Rn)). More precisely,we prove in the following that if 1 ≤ p <∞, then

‖Tp(z)‖L(Lp(Rn)) ≥ 2−n/2p

( |z|Re z

)n|1/p−1/2|(Re z > 0). (3.62)

We already observed that it suffices to consider the case where 1 < p ≤ 2.

Fix z ∈ C+ and consider the function f : Rn → C defined by

f(x) := exp

(−|x|2z

).

Taking p′ such that 1p + 1

p′ = 1 we verify that

‖f‖p′ =(π

p′

)n/2p′ ( |z|2Re z

)n/2p′

. (3.63)

Let x, y ∈ Rn and recall that z ∈ C+. Then

−|x− y|2z

− |x|2

z= −2Re

(1

z

) ∣∣∣x− y

2

∣∣∣2 + 2i(x− y

2

)y Im

(1

z

)− |y|

2

2Re

(1

z

).


Moreover,

Tp′(z/4)f(y)

=1

(πz)n/2

∫Rn

exp

(−|x− y|2

z− |x|

2

z

)dx

=1

(πz)n/2

∫Rn

exp

(−2|x|2 Re

(1

z

)+ 2ixy Im

(1

z

))dx

× exp

(−|y|

2

2Re

(1

z

))=

1

(πz)n/2

(1

4Re(z−1)

)n/2 ∫Rn

exp

(−|x|

2

2

)exp

(ixy

Im(z−1)

(Re(z−1))1/2

)dx

× exp

(−|y|

2

2Re

(1

z

))=

1

(πz)n/2

(1

4Re(z−1)

)n/2

(2π)n/2

× exp

(−|y|

2

2

(Im(z−1))2

Re(z−1)

)exp

(−|y|

2

2Re

(1

z

)),

where in the last step we used the fact that the function x �→ exp(−|x|2/2) is aneigenvector of the Fourier transform. Hence,

|Tp′(z/4)f(y)| =( |z|2Re z

)n/2

exp

(− |y|22Re z

)(y ∈ Rn),

and since∫Rn exp(−|x|2/2) dx = (2π)n/2, we obtain

‖Tp′(z/4)f‖p′ =( |z|2Re z

)n/2 (2π

p′

)n/2p′

(Re z)n/2p′.

Combining this equality with (3.63) we see that

‖Tp′(z/4)‖L(Lq(Rn)) ≥ ‖Tp′(z/4)f‖p′‖f‖p′ = 2−n/2p

( |z|Re z

)n(1/p−1/2)

.

Finally, since ‖Tp′(z)‖L(Lp′ (Rn)) = ‖Tp(z)‖L(Lp(Rn)), it follows that

‖Tp(z)‖L(Lp(Rn)) ≥ 2−n/2p

( |z|Re z

)n|1/p−1/2|(Re z > 0).

Interesting examples of boundary values of holomorphic C0-semigroups occuralso in connection with fractional powers of operators and the so-called Riemann-Liouville semigroup. Indeed, consider in Lp(0, 1) the operator

Au := u′ with domain D(A) := {u ∈W 1,p(0, 1) : u(0) = 0}. (3.64)

We showed in Example 3.4.8 that −A generates a C0-semigroup T on Lp(0, 1) (1 ≤p <∞), which may be represented by

T (t)f(x) =

{f(x− t) (t ≤ x),

0 (t > x).(3.65)

Since T (t) = 0 for t ≥ 1, we see that abs(T ) = −∞. Now inserting (3.65) in (3.56)we see that

A−zf(x) =1

Γ(z)

∫ x

0

(x− y)z−1f(y) dy (x ∈ (0, 1),Re z > 0, f ∈ Lp(0, 1)).

By Theorem 3.8.1, (A−z)Re z>0 is a holomorphic C0-semigroup of angle π/2. ThisC0-semigroup is called the Riemann-Liouville semigroup. By Proposition 3.9.1and Remark 3.9.3, the question whether or not the Riemann-Liouville semigrouppossesses a boundary group is equivalent to

supz∈Σπ/2∩D

||A−z|| <∞. (3.66)

When (3.66) holds true, we denote the boundary group by (S(s)) := (Ais)s∈R. Inorder to show the estimate (3.66) we make use of the transference principle due toCoifman and Weiss [CW77]. We will use it in the following form [Ama95, ChapterIII, Example 4.7.3 c)]. For a measure space (Ω, μ) and 1 ≤ p ≤ ∞, we denote byLp(Ω, μ) the usual Banach space of equivalence classes of p-integrable functions(bounded functions when p =∞).

Theorem 3.9.5 (Coifman-Weiss). Let (Ω, μ) be a σ-finite measure space and let1 0 and s ∈ R let A−t+is be defined as in(3.51). Then Aisf := limt↓0 A−t+isf ∈ Lp(Ω, μ) for f ∈ Lp(Ω, μ) and there existsa constant M , depending only on p, such that

‖Ais‖L(Lp(Ω,μ)) ≤M(1 + s2)eπ|s|/2 (s ∈ R).

Thus, for 1 < p < ∞, the Riemann-Liouville semigroup admits a boundarygroup. For p = 1 the situation is different. Indeed,

||A−z||L(L1(0,1)) = supy∈[0,1]

1

|Γ(z)|∫ 1

y

|(x− y)z−1| dx

=1

|Γ(z)| supy∈[0,1]

∫ 1

y

(x− y)Re z−1 dx =1

|Γ(z)|1

Re z,

which by Proposition 3.9.1 implies that we do not have a boundary value forp = 1. In summary, we have proved the following result for the Riemann-Liouvillesemigroup on Lp(0, 1).

Theorem 3.9.6. Let 1 ≤ p < ∞ and denote by G the generator of the Riemann-Liouville semigroup on Lp(0, 1). Then iG is the generator of a C0-group on Lp(0, 1)provided 1 < p < ∞. If p = 1, then iG does not generate a C0-semigroup onL1(0, 1).

We now consider the converse to the situation described in Proposition 3.9.1;namely, we ask for conditions on the boundary group itself which imply that Agenerates a holomorphic C0-semigroup. We begin with the following result.

Theorem 3.9.7. Let A be an operator on X and let ϕ ∈ (0, π/2). Assume that e±iϕAgenerate bounded C0-semigroups on X. Then A generates a bounded holomorphicC0-semigroup of angle ϕ.

Our proof of Theorem 3.9.7 is based on the following version of the Phragmen-Lindelof theorem.

Theorem 3.9.8 (Phragmen-Lindelof ). Let ϕ ∈ (0, π/2] and let h : Σϕ → X becontinuous on Σϕ and holomorphic in Σϕ. Set α := π

2ϕ . Assume that for all ε > 0there exists a constant Cε > 0 such that

‖h(z)‖ ≤ Cεeε|z|α (z ∈ Σϕ).

If ‖h(re±iϕ)‖ ≤M for all r > 0, then ||h(z)|| ≤M for all z ∈ Σϕ.

For a proof of Theorem 3.9.8 we refer to [Con73, Cor.6.4.4].

Proof of Theorem 3.9.7. Denote by T+, T− the C0-semigroups generated by A+ :=eiϕA and A− := e−iϕA, respectively. Let M ≥ 0 such that ||T±(t)|| ≤ M for allt ≥ 0. Then

‖R(λ,A±)‖ =∥∥∥∥∫ ∞

0

e−λtT±(t) dt∥∥∥∥ ≤ M

Reλ(Reλ > 0).

For λ ∈ Σ+ϕ this implies that

‖R(λ,A)‖ = ‖R(λ, eiϕA−)‖ = ‖R(λe−iϕ, A−)‖ ≤ M

Re(λe−iϕ)≤ M

|λ| cosϕ.

Similarly, ‖R(λ,A)‖ ≤ M|λ| cosϕ if λ ∈ Σ−ϕ . Thus we have

‖(I − zA)−1‖ = ‖z−1R(z−1, A)‖ ≤ M

cosϕ(z ∈ Σϕ).

For n ∈ N and z ∈ Σϕ set Tn(z) := (I − znA)−n. Then ‖Tn(z)‖ ≤M for z = re±iϕ

and

‖Tn(z)‖ ≤(

M

cosϕ

)n

for z = re±iα, r ≥ 0, |α| < ϕ.

It follows from the Phragmen-Lindelof Principle 3.9.8 that ‖Tn(z)‖ ≤M for all z ∈Σϕ and all n ∈ N. The Hille-Yosida Theorem 3.3.4 implies now that A generates aC0-semigroup T . By Corollary 3.3.6, we have limn→∞ Tn(t)x = T (t)x for t ≥ 0 andx ∈ X. It thus follows from Vitali’s theorem (see Theorem A.5 and PropositionA.3) that T has a holomorphic extension T to Σϕ satisfying ‖T (z)‖ ≤ M for allz ∈ Σϕ.

Consider now the case where e±iϕA generate C0-semigroups T±ϕ which arenot necessarily bounded. In this case, there exist constants M,ω ≥ 0 such that‖T±ϕ(t)‖ ≤ Meωt for t ≥ 0. It follows that e±iϕ(A − μ) generate bounded C0-semigroups for μ := ω

cosϕ. Theorem 3.9.7 implies now that A − μ generates a

bounded holomorphic C0-semigroup of angle ϕ. We have thus proved the followingresult.

Corollary 3.9.9. Let ϕ ∈ (0, π/2) and assume that e±iϕA generate C0-semigroupsT±ϕ on X. Then A generates a holomorphic C0-semigroup of angle ϕ with bound-ary semigroups T±ϕ.

The following result is a consequence of the above Corollary 3.9.9 and Propo-sition 3.9.1.

Corollary 3.9.10. Assume that A generates a C0-semigroup T and that iA generatesa C0-group U . Then T has a holomorphic extension to Σπ/2 and U is the boundarygroup of T .

Proof. By Corollary 3.9.9, the operator e±iπ/4A generates a holomorphic C0-semi-group of angle π/4. Thus eiθA generates a C0-semigroup for all θ ∈ (−π/2, π/2). Itfollows from Corollary 3.9.9 again that A generates a holomorphic C0-semigroupof angle π/2. By Proposition 3.9.1, U is its boundary group.

Next we consider spectral conditions on A which imply that A generates aholomorphic C0-semigroup on X under the assumption that iA generates a C0-group U on X . An obvious necessary condition for this is that the spectrum of Ais located in a left half-plane. However, this condition is not sufficient, in general.Indeed, consider the generator G of the Riemann-Liouville semigroup on Lp(0, 1)for 1 1, σ(A) is empty by Theorem 3.1.7.Moreover, iA generates a group by Theorem 3.9.6. However, A does not generatea C0-semigroup. Nevertheless, if σ(A) is contained in some left half-plane, then Agenerates a holomorphic C0-semigroup on X provided U satisfies a certain growthcondition.

Theorem 3.9.11. Let A be an operator on X such that iA generates a C0-group Uon X. Assume that there exists a dense subspace Y of X such that for all x ∈ Ythere exist constants C ≥ 0 and k ∈ N (depending on x) such that

‖U(t)x‖ ≤ C(1 + |t|)k (t ∈ R).

If σ(A) ⊂ {λ ∈ C : Reλ ≤ b} for some b ∈ R, then A generates a holomorphicC0-semigroup of angle π/2 on X (whose boundary group is U).

The key of the proof of Theorem 3.9.11 is the following result of Phragmen-Lindelof type.

Proposition 3.9.12. Let r : Σπ/2 → X be continuous. Assume that r is holomorphicin Σπ/2 and that there exist constants C,M ≥ 0, R0 > 0, k ∈ N such that

‖r(λ)‖ ≤ C

| sinϕ|k (Reλ ≥ 0, Im λ �= 0, |λ| ≥ R0, arg λ = ϕ) and

‖r(is)‖ ≤ M (s ∈ R).

Then ||r(λ)|| ≤M for all λ ∈ C with Reλ ≥ 0.

Proof. For R ≥ R0 and k ∈ N consider the holomorphic function

Φ : D0R := {z ∈ Σπ/2 : |z| < R} → X, λ �→

(1− λ2

R2

)k

r(λ).

Let λ := Reiϕ for ϕ ∈ (−π/2, π/2). If ϕ �= 0, we obtain

‖Φ(λ)‖ = |1− ei2ϕ|k‖r(λ)‖ = 2k| sinϕ|k‖r(λ)‖ ≤ 2kC.

Moreover, Φ(R) = 0 and

‖Φ(is)‖ = ∣∣(1 + s2

R2

)k∣∣ ‖r(is)‖ ≤ 2kM (|s| ≤ R).

The maximum principle implies that ‖Φ(λ)‖ ≤ 2k max{C,M} for all λ ∈ D0R.

Letting R→∞, we deduce that ||r(λ)|| ≤ 2k max{C,M} provided Reλ ≥ 0. Nowthe Phragmen-Lindelof Principle 3.9.8 implies that

‖r(λ)‖ ≤M

for all λ ∈ C with Reλ ≥ 0.

Proof of Theorem 3.9.11. Replacing A by A−ω, we may assume that σ(A) ⊂ {λ ∈C : Reλ < −δ} for some δ > 0. By assumption, iA generates a group and wetherefore have sups∈R,|s|≥w ||sR(is, A)|| <∞ for suitable w ≥ 0. It follows that

M := sups∈R

||sR(is, A)|| <∞.

Thus the second assumption of Proposition 3.9.12 is satisfied for the functionλ �→ λR(λ,A)x (x ∈ X). In order to verify the first assumption in this propositionlet x ∈ Y . Then, by hypothesis, there exist constants k ∈ N, C ≥ 0 such that


‖U(t)x‖ ≤ C(1 + |t|)k (t ∈ R). For λ of the form λ = reiϕ, where r ≥ 1 andϕ ∈ (0, π/2] we therefore obtain

‖λR(λ,A)x‖ =

∥∥∥∥λ∫ ∞

0

eiλtU(−t)x dt∥∥∥∥ ≤ |λ|C∫ ∞

0

e−| Imλ|t(1 + tk) dt

≤ C|λ|(

1

| Imλ| +k!

| Imλ|k+1

)≤ C(1 + k!)

| sinϕ|k+1.

Similarly, for λ = reiϕ, with r ≥ 1 and ϕ ∈ [−π/2, 0) we have

‖λR(λ,A)x‖ ≤ C(1 + k!)

| sinϕ|k+1.

Hence both assumptions of Proposition 3.9.12 are satisfied for the function λ �→λR(λ,A)x and it follows from that proposition that

‖λR(λ,A)x‖ ≤M‖x‖ (Reλ ≥ 0, x ∈ Y ).

Since Y is dense in X , we conclude that

‖λR(λ,A)‖ ≤M (Reλ ≥ 0).

It follows from Corollary 3.7.12 that A generates a bounded holomorphic C0-semigroup and from Corollary 3.8.9 that the angle is π/2 and U is the boundarygroup.

We finally turn our attention back to the question raised at the beginningof this section: under which conditions on the holomorphic C0-semigroup T the“boundary value” of T exists and is again a C0-semigroup. In the following, weweaken the sense of “boundary value” of T and also allow integrated semigroupsas “boundary values” for holomorphic C0-semigroups T . We say that an operatorA on X generates a k-times integrated group on X for some k ∈ N0 if A and −Agenerate k-times integrated semigroups on X. A k-times integrated group is calledexponentially bounded if the integrated semigroups generated by A and −A areexponentially bounded.

Theorem 3.9.13. Let γ ≥ 0 and k ∈ N. Assume that A generates a holomorphicC0-semigroup T of angle π/2. Then the following assertions hold:

a) Assume that there exist constants M,ω ≥ 0 such that

‖T (z)‖ ≤ Meω|z|

(Re z)γ(Re z > 0).

Then iA generates an exponentially bounded k-times integrated group pro-vided k > γ.

b) Assume that iA generates an exponentially bounded k-times integrated groupon X. Then there exist constants M,ω ≥ 0 such that

‖T (z)‖ ≤ Meω|z|

(Re z)k(Re z > 0).

Proof. a) Let x ∈ X, k ∈ N and let λ0 > ω. Then

R(λ0, A)kx =

∫ ∞

0

e−λ0uuk−1

(k − 1)!T (u)x du.

Hence,

R(λ0, A)kT (z)x =

∫ ∞

0

e−λ0uuk−1

(k − 1)!T (u+ z)x du (Re z > 0).

Setting z = t+ is the assumption implies that

‖R(λ0, A)kT (z)‖ ≤ M

(k − 1)!

∫ ∞

0

e−λ0uuk−1 eω|u+t+is|

(u+ t)γdu

≤ M

(k − 1)!

∫ ∞

0

e−λ0uuk−1−γeω|u+t+is| du <∞, (3.67)

provided k > γ. Since z �→ R(λ0, A)kT (z) is holomorphic and bounded in a rect-

angle of the form {t + is : 0 < t < 1,−R < s < R} for some R > 0 it follows bydominated convergence that

limt↓0

R(λ0, A)kT (t+ is)x

exists for all s ∈ R and all x ∈ X. For s ∈ R and x ∈ X set

S(s)x :=

⎧⎨⎩i−k limt↓0

R(λ0, A)kT (t+ is)x (s ≥ 0),

(−i)−k limt↓0

R(λ0, A)kT (t+ is)x (s < 0).

In order to show that iA generates a k-times integrated semigroup it suffices byProposition 3.2.7 to verify that R(·, iA)R(iλ0, iA)

k is a Laplace transform. Notefirst that by the estimate (3.67) there exists a constant C ≥ 0 such that ‖S(s)‖ ≤Ceω|s| for s ∈ R+. We claim that

S(λ) = R(λ, iA)R(iλ0, iA)k (λ > ω).

In fact, by Fubini’s theorem, the representation formula (3.46) for holomorphic


C0-semigroups and Cauchy’s theorem, we have

ik∫ ∞

0

e−λsS(s) ds = limt↓0

∫ ∞

0

e−λsR(λ0, A)kT (t+ is) ds

= limt↓0

∫ ∞

0

e−λsR(λ0, A)k

∫Γ

eμ(t+is)R(μ,A) dμ ds

= limt↓0

∫Γ

∫ ∞

0

e−λseμis ds eμtR(λ0, A)kR(μ,A) dμ

= limt↓0

∫Γ

1

λ− iμeμtR(λ0, A)

kR(μ,A) dμ

= limt↓0

R(λ, iA)R(λ0, A)kT (t)

= R(λ, iA)R(λ0, A)k (λ > ω),

where Γ denotes the path defined in (3.46). Thus we have∫∞0

e−λsS(s) ds =

R(λ, iA)R(iλ0, iA)k for λ > ω. The corresponding result for −iA is proved inexactly the same way.

b) By rescaling, we may assume that σ(A) ⊂ {λ ∈ C : Reλ ≤ −1} (seeProposition 3.1.9 i) and Proposition 3.2.6). We subdivide the proof into two steps.

Step 1: By assumption, A generates a holomorphic C0-semigroup T of angle

π/2. Hence, the function S : z �→ S(z) :=∫ z

0(z−ξ)(k−1)!

k−1T (ξ) dξ is holomorphic in

the open right half-plane. We claim that there exist constants M,ω ≥ 0 such that

‖S(z)‖ ≤Meω|z| (Re z > 0). (3.68)

Integrating by parts, using dn

dξnT (ξ)A−n = T (ξ), gives

S(z)x = T (z)A−kx− zk−1

(k − 1)!A−1x− · · · − A−kx (x ∈ X).

Hence, in order to prove (3.68) it suffices to show that

‖T (z)A−k‖ ≤Meω|z| (Re z > 0), (3.69)

for suitable constants M,ω ≥ 0. Obviously (S(t))t≥0 is the k-times integratedsemigroup generated by A. For z ∈ {μ ∈ C : Reμ > 0} set z = t + is for t > 0and s ∈ R. Let (R(s))s∈R be the k-times integrated group generated by iA. Forx ∈ D(Ak) set

T (is)x :=dk

dskR(s)x = R(s)(iA)kx+

k∑n=1

sn

n!(iA)n−1x (s ∈ R).

ThenT (t+ is)x = T (t)T (is)x = T (is)T (t)x (t > 0, s ∈ R), (3.70)

because the function v given by v(s) := T (is)T (t)x−T (t+ is)x is the unique mildsolution of the problem u′(s) = iAu(s), u(0) = 0, by Lemma 3.2.9 and Lemma3.2.10.

We consider first the case where Re z ≥ 1. By (3.70) we have for z = t+ is

‖T (z)A−k‖ ≤ ‖T (t− 1/2)A−k‖ ‖T (is)A−k‖ ‖AkT (1/2)‖.

Since t �→ ‖T (t − 1/2)A−k‖ and s �→ ‖T (is)A−k‖ are exponentially bounded weobtain (3.69) in the case where Re z ≥ 1.

Next, we consider the case where 0 < Re z ≤ 1. For x ∈ X we have by(3.70): T (z)A−2kx = T (t)A−kT (is)A−kx. It then follows that there exist constantsM1, ω1 ≥ 0 such that

‖T (z)A−2k‖ ≤M1eω1|z| and ‖T (is)A−k‖ ≤M1e

ω1|s| (3.71)

for s ∈ R and z ∈ C satisfying Re z > 0. For z ∈ Ω := {μ ∈ C : 0 ≤ Reμ ≤ 1} andx ∈ X set

f(z) := (cos z)−2ω1T (z)A−2kx.

Then f is holomorphic in the interior of Ω and continuous on Ω, by (3.70). For|s| ≥ 1 we have

‖f(z)‖ ≤ 22ω1‖T (z)A−2kx‖ |es − e−s|−2ω1 ,

which by (3.71) implies that z → ‖f(z)‖ is bounded for z ∈ Ω. For z = is andz = 1 + is we have

‖f(is)‖ ≤ 22ω1‖T (is)A−k‖ ‖A−kx‖

|es + e−s|2ω1≤M2‖A−kx‖

‖f(1 + is)‖ ≤ 22ω1‖T (is)A−k‖ ‖T (1)A−kx‖

|es + e−2ie−s|2ω1≤M2‖A−kx‖

for a suitable constant M2 ≥ 0. The three-lines lemma [Con73, Theorem VI.3.7]now implies that there exists a constant M ≥ 0 such that

‖T (z)A−2kx‖ ≤Me2ω1|z|‖A−kx‖ (Re z > 0).

Since D(Ak) is dense in X, the claim follows.Step 2: Set

Sk+1(z) :=

∫ z

0

S(ξ) dξ (Re z > 0).

Cauchy’s integral formula implies that

T (z) =(k + 1)!

2πi

∫γz

Sk+1(ξ)

(ξ − z)k+2dξ (Re z > 0),

where γz denotes the path defined by the circle with centre z and radius r = Re z2 .

It follows that

T (z) =(k + 1)!

2πi

∫γz

Sk+1(ξ)− Sk+1(z)

(ξ − z)k+2dξ

=(k + 1)!

2πrk+1

∫ 2π

0

(Sk+1(z + reiϕ)− Sk+1(z)

)e−iϕ(k+1) dϕ.

It follows from the definition of Sk+1 and (3.68) that

‖Sk+1(z + reiϕ)− Sk+1(z)‖ ≤M3reω3|z| (Re z > 0)

for suitable M3, ω3 ≥ 0. Inserting this in the above representation of T (z), itfollows that

‖T (z)‖ ≤M4eω4|z|

rk= M4

2keω4|z|

(Re z)k(Re z > 0)

for suitable constants M4, ω4 ≥ 0. The proof is complete.

As an application of Theorem 3.9.13 we consider once again “boundary val-ues” of the Gaussian semigroup. More precisely, let 1 ≤ p <∞ and let the operatorΔp be defined as in (3.60). Then the following corollary holds true.

Corollary 3.9.14. Let 1 ≤ p <∞ and let k ∈ N0. Then iΔp generates an exponen-tially bounded k-times integrated group on Lp(Rn) if k > n|1

2− 1

p|. Moreover, the

order of integration is optimal in the sense that iΔp does not generate a k-timesintegrated semigroup on Lp(Rn) if k < n|12 − 1

p |.

Proof. It follows from (3.61) and by duality that the assumption of Theorem 3.9.13a) is satisfied for γ = n|1/2 − 1/p|. Hence, the first assertion above follows fromthis theorem. Conversely, suppose that iΔp generates an exponentially boundedk-times integrated semigroup S on Lp(Rn) for some 1 ≤ p < ∞ and k < n|1/2−1/p|. Let J be the conjugation on the complex space Lp(Rn); Jf = f . Then−iΔp = J(iΔp)J , which generates the k-times integrated semigroup JS(·)J , soiΔp generates an exponentially bounded k-times integrated group. By Theorem3.9.13 b), there exists a constant M > 0 such that ‖Tp(z)‖ ≤ M/(Re z)k forz ∈ Σπ/2 with |z| = 1, where Tp is the Gaussian semigroup as in Example 3.7.6.However, by (3.62) we have

‖Tp(z)‖ ≥ 2−n/2p(Re z)−n|1/2−1/p| (z ∈ Σπ/2, |z| = 1),

which yields a contradiction.

More general results for differential and pseudo-differential operators will begiven in Section 8.3.


3.10 Intermediate Spaces

It turns out that k-times integrated semigroups are the same as C0-semigroups upto the choice of the underlying Banach space. This will be made precise in thissection.

Throughout this section, Z,X and Y are Banach spaces. We write Z ↪→ Xif Z ⊂ X and there is a constant c such that ‖x‖X ≤ c‖x‖Z for all x ∈ Z. If in

addition Z is dense in X we write Zd↪→ X.

The following lemma is a consequence of the closed graph theorem.

Lemma 3.10.1. If Z ↪→ Y and Z ⊂ X ↪→ Y , then Z ↪→ X.

Let A be an operator on X . If Z ↪→ X we denote by AZ the part of A in Z;i.e., D(AZ) := {x ∈ D(A) ∩ Z : Ax ∈ Z}, AZx := Ax. If A is closed, then AZ isclosed. The following is easy to prove (see also Proposition B.8).

Lemma 3.10.2. Let A be an operator on X, Z ↪→ X. Let μ ∈ ρ(A) such thatR(μ,A)Z ⊂ Z. Let B be an operator on Z. Then B = AZ if and only if μ ∈ ρ(B)and R(μ,B) = R(μ,A)|Z .

Let A be a closed operator on X and k ∈ N. Then D(Ak) is a Banach spacefor the norm ‖x‖Ak := ‖x‖ + ‖Ax‖ + . . . + ‖Akx‖. Moreover, D(Ak) ↪→ X . Wedenote by Ak the part of A in D(Ak); i.e., Ak is the operator on the Banach spaceD(Ak) given by Akx = Ax, D(Ak) = D(Ak+1). If ρ(A) �= ∅, then Ak and A aresimilar (see Section 3.5 for the definition). In fact, Ak = U−1AU where U maybe taken as U = (μ− A)k for any μ ∈ ρ(A). In particular, σ(Ak) = σ(A) and Ak

generates an (exponentially bounded) m-times integrated semigroup on D(Ak) ifand only if A generates an (exponentially bounded) m-times integrated semigroupon X .

The following result on the spectrum of intermediate operators is of generalinterest.

Proposition 3.10.3. Let A be an operator on X, Z ↪→ X. Assume that R(μ,A)Z ⊂Z for some μ ∈ ρ(A) and that D(Ak) ⊂ Z for some k ∈ N. Then σ(AZ) = σ(A)and R(λ,AZ) = R(λ,A)|Z for all λ ∈ ρ(A).

Proof. a) Let B = AZ and λ ∈ ρ(A). Iterating the resolvent equation

R(λ,A) = R(μ,A) + (μ− λ)R(μ,A)R(λ,A)

gives

R(λ,A) =

k∑j=1

(μ− λ)j−1R(μ,A)j + (μ− λ)kR(μ,A)kR(λ,A). (3.72)

Since R(μ,A)Z ⊂ Z and R(μ,A)kX = D(Ak) ⊂ Z, it follows that R(λ,A)Z ⊂ Z.Hence by Lemma 3.10.2, λ ∈ ρ(B) and R(λ,B) = R(λ,A)|Z .

3.10. INTERMEDIATE SPACES 185

b) In order to prove the converse, we observe that ρ(A) = ρ(Ak) since A andAk are similar operators. Let Y := D(Ak) with the graph norm. Then Y ↪→ Zby Lemma 3.10.1, D(Bk) ⊂ Y and Ak = BY . Moreover, R(μ,B)Y = R(μ,A)Y =D(Ak+1) ⊂ Y . It follows from a) (applied to B instead of A) that ρ(B) ⊂ ρ(Ak) =ρ(A).

Our aim is to prove the following result.

Theorem 3.10.4 (Sandwich Theorem). Let A be an operator on X and let k ∈ N.The following assertions are equivalent:

(i) The operator A generates a k-times integrated semigroup S such that ω(S) <∞.

(ii) There exists a Banach space Y and the generator B of a C0-semigroup V onY such that

a) D(Bk) ⊂ X ↪→ Y ,

b) R(λ,B)X ⊂ X for some λ ∈ ρ(B), and

c) A = BX .

(iii) There exists a Banach space Z such that

a) D(Ak) ⊂ Z ↪→ X,

b) R(λ,A)Z ⊂ Z for some λ ∈ ρ(A), and

c) AZ generates a C0-semigroup U on Z.

Proof. (i) ⇒ (ii): Assume that A generates a k-times integrated semigroup Ssatisfying ‖S(t)‖ ≤Meωt (t ≥ 0) where M,ω > 0. For x ∈ D(Ak), let

T (t)x := S(t)Akx+tk−1

(k − 1)!Ak−1x+ . . .+ tAx+ x (t ≥ 0). (3.73)

By Lemma 3.2.2 (see also Lemma 3.2.10), v(t) := T (t)x is a mild solution of(ACP0). Hence, s �→ v(t+s) is a mild solution of (ACP0) with initial value T (t)x.By Theorem 3.1.3,

R(λ,A)T (t)x =

∫ ∞

0

e−λsT (t+ s)x ds (t ≥ 0, λ > ω) (3.74)

for all x ∈ D(Ak).Fix μ0 > b > ω, and define a norm ‖ · ‖Y on X by

‖x‖Y := supt≥0

‖e−btT (t)R(μ0, A)kx‖X , (3.75)

and let Y be the completion of (X, ‖ · ‖Y ). We claim that

‖(λ− b)R(λ,A)x‖Y ≤ ‖x‖Y (λ > b, x ∈ X). (3.76)

In fact,

‖e−btT (t)R(μ0, A)kR(λ,A)x‖X

=

∥∥∥∥e−bt

∫ ∞

0

e−λsT (s+ t)R(μ0, A)kx ds

∥∥∥∥X

=

∥∥∥∥∫ ∞

0

e−(λ−b)se−b(t+s)T (s+ t)R(μ0, A)kx ds

∥∥∥∥X

≤ ‖x‖Y∫ ∞

0

e−(λ−b)s ds =‖x‖Yλ− b

.

Hence R(λ,A) has a unique extension R(λ) ∈ L(Y ) and ‖(λ − b)R(λ)‖L(Y ) ≤ 1(λ > b). Then (R(λ))λ>b is a pseudo-resolvent on Y .

Next we show that

limλ→∞

‖λR(λ)y − y‖Y = 0 (3.77)

for all y ∈ Y . Since lim supλ→∞ ‖λR(λ)‖L(Y ) < ∞, it suffices to prove (3.77) for

y ∈ X, X being dense in Y . Let x := R(μ0, A)ky. Let ε > 0. There exists M ′ such

that ‖T (t)x‖X ≤M ′eωt (t ≥ 0). Then∥∥∥∥e−bt

∫ ∞

0

λe−λs(T (t+ s)x− T (t)x) ds

∥∥∥∥X

≤ e−bt

∫ ∞

0

λe−λsM ′(eω(t+s) + eωt

)ds

= M ′e−(b−ω)t

(λ

λ− ω+ 1

).

Hence, there exists t0 such that

supλ>2ω

supt>t0

∥∥∥∥e−bt

∫ ∞

0


∥∥∥∥X

< ε. (3.78)

Since t �→ T (t)x is uniformly continuous on [0, t0 +1], there exists τ > 0 such that‖T (t+ s)x− T (t)x‖X < ε whenever s ∈ [0, τ ], t ∈ [0, t0].

For λ > 2ω and 0 ≤ t ≤ t0,∥∥∥∥e−bt

∫ ∞

0


∥∥∥∥X

≤ e−bt

∫ τ

0

λe−λsε ds+ e−bt

∫ ∞

τ

λe−λsM ′(eω(t+s) + eωt) ds

≤ ε+M ′(

λ

λ− ωe−(λ−ω)τ + e−λτ

). (3.79)

3.10. INTERMEDIATE SPACES 187

By (3.74), (3.75), (3.78) and (3.79),

lim supλ→∞

‖λR(λ,A)y − y‖Y= lim sup

λ→∞supt≥0

∥∥e−bt (λR(λ,A)T (t)x− T (t)x)∥∥X

= lim supλ→∞

supt≥0

∥∥∥∥e−bt

∫ ∞

0

λe−λs (T (t+ s)x− T (t)x) ds

∥∥∥∥X

≤ ε.

Since ε > 0 is arbitrary, the claim is proved.It follows from (3.77) and Proposition B.6 that there exists a densely de-

fined operator B on Y such that (b,∞) ⊂ ρ(B) and R(λ,B) = R(λ) (λ > b).By the Hille-Yosida theorem, B generates a C0-semigroup V on Y satisfying‖V (t)‖L(Y ) ≤ ebt (t ≥ 0). It follows from Lemma 3.10.2 that A = BX . By defini-

tion, ‖R(μ0, A)ky‖X ≤ ‖y‖Y (y ∈ X). Hence, R(μ0, B)kY ⊂ X ; i.e., D(Bk) ⊂ X

and the proof of (ii) is complete.(ii) ⇒ (iii): Let Z :=

(D(Bk), ‖ · ‖Bk

). Then AZ = Bk which is similar to

B, so AZ generates a C0-semigroup U on Z. By Proposition 3.10.3, ρ(A) = ρ(B)and R(λ,A) = R(λ,B)|X for all λ ∈ ρ(B). It follows that R(λ,A)Z ⊂ Z for allλ ∈ ρ(A).

(iii) ⇒ (i): Note that D(Ak) ↪→ Z by Lemma 3.10.1, and R(λ,AZ) =R(λ,A)|Z for λ ∈ ρ(A) = ρ(AZ), by Proposition 3.10.3. Let μ ∈ ρ(A). Thenλ �→ R(λ,A)R(μ,A)k is the Laplace transform of t �→ U(t)R(μ,A)k. Now, (i)follows from Proposition 3.2.7.

Corollary 3.10.5. Let A be the generator of an exponentially bounded k-times in-tegrated semigroup T on X and let B ∈ L(X,D(Ak)). Then A + B generates ak-times integrated semigroup S on X satisfying ω(S) <∞.

Proof. We use the notation of Theorem 3.10.4 (iii). The operator B|Z is bounded.By Corollary 3.5.6, AZ + B|Z generates a C0-semigroup on Z. It is clear that(A+B)Z = AZ +B|Z . So it will follow from Theorem 3.10.4 that A+B generatesan exponentially bounded k-times integrated semigroup on X once we have provedthat A +B satisfies conditions a) and b) of Theorem 3.10.4 (iii). It is easy to seethat D((A+B)k) ⊂ D(Ak), so a) is satisfied.

In order to show b), take μ ∈ ρ(A). Then C := (μ − A)kB ∈ L(X) and by(3.73)

R(λ,A)B = R(λ,A)R(μ,A)kC

=R(λ,A)C

(μ− λ)k−

k∑j=1

(μ− λ)j−k−1R(μ,A)jC.

Since λ �→ λ−kR(λ,A) is the Laplace transform of an exponentially bounded func-tion, lim supλ→∞ ‖λ1−kR(λ,A)‖ <∞. This implies that limλ→∞ ‖R(λ,A)B‖ = 0.

Consequently, (I −R(λ,A)B) is invertible for large λ. Then by Lemma 3.5.8, (I −BR(λ,A)) is also invertible for large λ. Hence, (λ−(A+B)) = (I−BR(λ,A))(λ−A)is invertible for large λ. Hence, there exists λ ∈ ρ(A + B) ∩ ρ(AZ + B|Z) ∩ ρ(A).Let y ∈ Z, x = R(λ,A + B)y. Then (λ − A)x = y + Bx ∈ Z. Hence x =R(λ,A)(y +Bx) ∈ Z by Proposition 3.10.3.

3.11 Resolvent Positive Operators

In this section we assume that X is an ordered Banach space with normal coneX+ (see Appendix C).

Definition 3.11.1. An operator A on X is called resolvent positive if there existsω ∈ R such that (ω,∞) ⊂ ρ(A) and R(λ,A) ≥ 0 for all λ > ω.

If A generates a C0-semigroup T , then A is resolvent positive if and only ifT is positive (i.e. T (t)X+ ⊂ X+ for all t ≥ 0). In fact, if T is positive, then

R(λ,A) =

∫ ∞

0

e−λtT (t) dt ≥ 0

for all λ > ω(T ). The converse follows from Euler’s formula

T (t)x = limn→∞

(I − t

nA)−n

x

(see Corollary 3.3.6). But there are interesting examples of resolvent positive op-erators which do not generate C0-semigroups; see Section 6.1 for an example.

Let A be a resolvent positive operator. Then for λ > ω one has

(−1)nR(λ,A)(n) = n!R(λ,A)n+1 ≥ 0 (3.80)

for all n ∈ N (see Appendix B). Thus, the function R(·, A) is completely monotonic(cf. Section 2.7).

We first use Bernstein’s theorem for real-valued functions to prove some gen-eral properties of resolvent positive operators.

Proposition 3.11.2. Let A be a resolvent positive operator. Denote by

s(A) := sup{Re λ : λ ∈ σ(A)}

the spectral bound of A. Then s(A) <∞ and

R(λ,A) ≥ R(μ,A) ≥ 0

whenever s(A) < λ < μ. Moreover, if λ ∈ R ∩ ρ(A) such that R(λ,A) ≥ 0, thenλ > s(A). Finally, s(A) ∈ σ(A) if s(A) > −∞.

3.11. RESOLVENT POSITIVE OPERATORS 189

Proof. Let

s := inf{ω : (ω,∞) ⊂ ρ(A) and R(λ,A) ≥ 0 for all λ > ω}.By assumption, s <∞. Replacing A by A− ω, we may assume that s ≤ 0.

a) Let s < λ < μ. Then

R(λ,A)−R(μ,A) = (μ− λ)R(λ,A)R(μ,A) ≥ 0.

Thus, R(·, A) is a decreasing function on (s,∞).b) Assume that s > −∞. Then s ∈ σ(A). In fact, if s ∈ ρ(A) then R(s,A) ≥ 0.

Moreover, for μ < s sufficiently close to μ one has

R(μ,A) =

∞∑n=0

(s− μ)nR(s,A)n+1 ≥ 0.

This contradicts the definition of s.c) Let Hs := {λ ∈ C : Reλ > s}. We claim that Hs ⊂ ρ(A); this and b)

establish that s = s(A). Denote by Ω0 the connected component of Hs ∩ ρ(A)containing (s,∞). If Hs �⊂ ρ(A), there exist μn ∈ Ω0 such that μ := limn→∞ μn ∈Hs \ ρ(A). Then by Corollary B.3,

supn∈N

‖R(μn, A)‖ =∞.

By the uniform boundedness principle, there exist x ∈ X, x∗ ∈ X∗ such thatsupn∈N |〈R(μn, A)x, x

∗〉| = ∞. Since X = spanX+ and X∗ = spanX∗+ (Proposi-tion C.2), we can assume that x ∈ X+, x∗ ∈ X∗+. By Bernstein’s Theorem 2.7.7,there exists an increasing function α : R+ → R such that α(0) = 0 and

〈R(λ,A)x, x∗〉 =∫ ∞

0

e−λt dα(t) = dα(λ)

for all λ > s. It follows from uniqueness of holomorphic extensions that

〈R(μn, A)x, x∗〉 = dα(μn)

for all n ∈ N. Consequently,

|〈R(μn, A)x, x∗〉| ≤ dα(Reμn) = 〈R(Reμn, A)x, x

∗〉 ≤ 〈R(λ,A)x, x∗〉,where λ := infn∈N Reμn > s (since Reμ = limn→∞Reμn > s). This is a contra-diction.

d) In order to prove the remaining assertion, assume that there exists λ ∈ρ(A) such that λ < s(A) and R(λ,A) ≥ 0. Let μn ↓ s(A). Since s(A) ∈ σ(A),

‖R(μn, A)‖ → ∞ as n→∞.

As in a), we haveR(λ,A) ≥ R(μn, A) ≥ 0 (n ∈ N).

This is impossible.

From Theorem 3.11.2 and its proof, we note the following.

Corollary 3.11.3. Let A be a resolvent positive operator. Then

|〈R(λ,A)x, x∗〉| ≤ 〈R(ω,A)x, x∗〉 (3.81)

whenever Reλ ≥ ω > s(A), x ∈ X+ and x∗ ∈ X∗+. In particular, for each

ω > s(A),sup

Reλ≥ω‖R(λ,A)‖ <∞.

We need the following identity.

Lemma 3.11.4. Let A be an operator and λ ∈ ρ(A). Then for all m ∈ N

(−1)mλm+1 (R(λ,A)/λ)(m)

/m! =m∑

k=0

λkR(λ,A)k+1. (3.82)

Proof. This is immediate from Leibniz’s rule, since

(−1)kR(λ,A)(k)/k! = R(λ,A)k+1.

Now we can prove the following generation theorem.

Theorem 3.11.5. Let A be a resolvent positive operator. Then A generates a twiceintegrated semigroup which is Lipschitz continuous on bounded intervals. If D(A)is dense, then A generates a once integrated semigroup.

Proof. Considering A− ω instead of A, we can assume that s(A) < 0 (see Propo-sition 3.2.6). Let m ∈ N, λ ≥ 0. Then

m−1∑k=0

λkR(λ,A)k+1 = R(0, A)− λmR(λ,A)mR(0, A). (3.83)

For m = 1, this is just the resolvent equation. Then (3.83) follows by induction.Consequently,

0 ≤m−1∑k=0

λkR(λ,A)k+1 ≤ R(0, A)

for m ∈ N, λ ≥ 0. It follows from Lemma 3.11.4 that

supλ>0,m∈N0

‖λm+1(R(λ,A)/λ)(m)/m!‖ <∞.


Now the claim follows from Theorem 3.3.1 and Theorem 3.3.2.

More generally, if A is resolvent positive, it follows from Corollary 3.3.13that the part of A in Y = D(A) generates a once integrated semigroup. Thefollowing example shows that a resolvent positive operator does not generate aonce integrated semigroup in general.

Example 3.11.6. Let X = C[−1, 0]× C and A be given by

D(A) := C1[−1, 0]× {0}, A(f, 0) := (f ′,−f(0)).Then ρ(A) = C and R(λ,A)(f, c) = (g, 0) with

g(x) := eλx(∫ 0

x

e−λyf(y) dy + c

)for all λ ∈ C. Thus, A is resolvent positive. Let eλ ∈ C[−1, 0] be given by

eλ(x) := eλx (λ > 0, x ∈ [−1, 0]).Then (eλ, 0) = R(λ,A)(0, 1). One has eλ =

∫∞0

λ2e−λtkt dt where kt ∈ C[−1, 0]is given by kt(x) := 0 if x + t ≤ 0 and kt(x) := x + t otherwise. If A were thegenerator of a once integrated semigroup, then λ �→ eλ/λ would be a Laplacetransform. Hence, k : R+ → C[−1, 0] would be differentiable. But d

dtkt(x) does

not exist at x = −t if t ∈ (0, 1).

The following result shows that the situation is different if X has order con-tinuous norm.

Theorem 3.11.7. Let A be a resolvent positive operator and assume that X hasorder continuous norm. Then A generates a once integrated semigroup.

Proof. Replacing A by A−ω, we may assume that s(A) ≤ 0. Let x ∈ X+. Then byTheorem 2.7.18, there exists a unique normalized increasing function Fx : R+ → Xsuch that Fx(0) = 0 and

R(λ,A)x =

∫ ∞

0

e−λt dFx(t) (λ > s(A)).

It follows from uniqueness that Fx(t) is additive and positive homogeneous in xfor all t ≥ 0. Hence, there exists a positive linear operator S1(t) ∈ L(X) such thatFx(t) = S1(t)x for all t ≥ 0, x ∈ X. Since S1(·) is increasing, we can define the

Riemann integral S2(t)x :=∫ t

0S1(s)x ds for all x ∈ X (see Corollary 1.9.6). Then

R(λ,A)x = λ

∫ ∞

0

e−λtS1(t)x dt = λ2

∫ ∞

0

e−λtS2(t)x dt

for all x ∈ X, λ > max{0, s(A)}, where the first integral is understood as animproper Riemann integral (see (1.22) and Proposition 1.10.2). We have to showthat S1 is strongly continuous on R+.

Since S2(·)x is continuous, S2 is the twice integrated semigroup generatedby A. Thus for x ∈ D(A), S2(·)x is continuously differentiable by Lemma 3.2.2.Hence S1(·)x is continuous if x ∈ D(A).

Let x ∈ X+ and μ > s(A). Then R(μ,A)S1(·)x is a normalized increasingfunction on R+ and∫ ∞

0

e−λt d(R(μ,A)S1(t)x) = R(μ,A)

∫ ∞

0

e−λt d(S1(t)x)

= R(μ,A)R(λ,A)x

= R(λ,A)R(μ,A)x

=

∫ ∞

0

e−λt d(S1(t)R(μ,A)x)

for all λ > s(A). It follows from uniqueness of the representation (see Theorem2.7.18) that

S1(t)R(μ,A)x = R(μ,A)S1(t)x.

Since the norm is order continuous, y+ := lims↓t S1(s)x exists. Moreover,

R(μ,A)y+ = lims↓t

R(μ,A)S1(s)x

= lims↓t

S1(s)R(μ,A)x

= S1(t)R(μ,A)x

= R(μ,A)S1(t)x.

Since R(μ,A) is injective, it follows that y+ = S1(t)x. In the same way one showsthat lims↑t S1(s)x = S1(t)x if t > 0. We have shown that S1 is strongly continuouson R+.

The closure of the domain of a resolvent positive operator is of a very specialnature if the underlying space is a Banach lattice with order continuous norm.

Let X be a complex Banach lattice; i.e., the complexification of a real Banachlattice. A subspace J of X is called an ideal if

a) x ∈ J implies Rex ∈ J ; and

b) if x, y ∈ X are real, |y| ≤ |x| and x ∈ J , then it follows that y ∈ J .

In a space X := Lp(Ω, μ), where (Ω, μ) is a σ-finite measure space and 1 ≤p <∞, every closed ideal J is of the form

J = {f ∈ Lp(Ω) : f |Ω0= 0 a.e.} ,

where Ω0 is a measurable subset of Ω (see [Sch74, p.157]).

Theorem 3.11.8. Let X be a Banach lattice with order continuous norm. If A is aresolvent positive operator, then D(A) is an ideal.

Proof. a) Note that by definition X is the complexification of a real Banach latticeXR. Since the resolvent leaves XR invariant we have Rex ∈ D(A) whenever x ∈D(A). Now observe that if J is a closed ideal of XR, then J ⊕ iJ is a closed idealof X. These remarks show that we can assume that X is a real Banach lattice,which we do.

b) We can also assume that s(A) < 0 (replacing A by A− ω otherwise).

c) Let 0 ≤ y ≤ R(0, A)x where x ∈ X+. We claim that y ∈ D(A). For λ > 0we have

0 ≤ λR(λ,A)y ≤ λR(λ,A)R(0, A)x = R(0, A)x−R(λ,A)x ≤ R(0, A)x.

Since the order interval [0, R(0, A)x] is weakly compact [AB85, Theorem 12.9],there exists a weak limit point z of λR(λ,A)y as λ→∞. In particular, z ∈ D(A).Then

R(0, A)y −R(λ,A)y = λR(0, A)R(λ,A)y

has R(0, A)z as weak limit point. By the inequality above,

limλ→∞

R(λ,A)y = 0,

so R(0, A)y = R(0, A)z. Since R(0, A) is injective, it follows that y = z ∈ D(A).

d) Let y ∈ D(A). Then |y| ∈ D(A). In fact, there exists x ∈ X such thaty = R(0, A)x. Hence, |y| ≤ R(0, A)|x| and the claim follows from c).

e) If y ∈ D(A), then |y| ∈ D(A). This follows from d) since the absolute valueis a continuous mapping.

f) Let 0 ≤ y ≤ x ∈ D(A). Let xn ∈ D(A) such that limn→∞ xn = x. Itfollows from e) that |xn| ∈ D(A). We have

y ∧ |xn| ≤ |xn| = |R(0, A)Axn| ≤ R(0, A)|Axn|.

It follows from c) that y ∧ |xn| ∈ D(A). Hence, y = limn→∞(y ∧ |xn|) ∈ D(A).

g) Let |y| ≤ |x|, where x ∈ D(A). Then 0 ≤ y+ ≤ |x|, so y+ ∈ D(A), by e)and f). Similarly, y− ∈ D(A), and therefore y = y+ − y− ∈ D(A).

In some special cases, densely defined resolvent positive operators are auto-matically generators of C0-semigroups.

Theorem 3.11.9. Let X = C(K) where K is a compact space. Let A be a denselydefined resolvent positive operator. Then A generates a positive C0-semigroup.

Proof. Since D(A) is dense, there exists a strictly positive function u ∈ D(A);i.e., u(x) ≥ ε > 0 for all x ∈ K and some ε > 0. We can assume that s(A) < 0(replacing A by A−ω if necessary). There exists v ∈ C(K) such that u = R(0, A)v.


Then for λ > 0, n ∈ N and f ∈ C(K) with ‖f‖∞ ≤ 1,

|(λR(λ,A))nf | ≤ (λR(λ,A))n|f |≤ 1

ε(λR(λ,A))nu

≤ 1

ε(λR(λ,A))nR(0, A)|v|

=1

ε

(R(0, A)|v| −

n−1∑k=0

λkR(λ,A)k+1|v|)

≤ 1

εR(0, A)|v|,

using (3.83). It follows that

‖(λR(λ,A))n‖ ≤ 1

ε

∥∥R(0, A)|v|∥∥∞for all λ > 0 and n ∈ N. Now the claim follows from the Hille-Yosida theorem.

Now we return to the case when X is an arbitrary ordered Banach space withnormal cone. We consider the inhomogeneous Cauchy problem

(ACPf )

{u′(t) = Au(t) + f(t) (t ∈ [0, τ ]),

u(0) = u0,

where A is a closed operator on X, τ > 0 and f ∈ C([0, τ ], X). Recall from Section3.1 that a mild solution of (ACPf ) is a function u ∈ C([0, τ ], X) such that∫ t

0

u(s) ds ∈ D(A) and u(t)− u0 = A

∫ t

0

u(s) ds+

∫ t

0

f(s) ds

for all t ∈ [0, τ ]. The function u is called a classical solution if in addition u ∈C1([0, τ ], X). In that case, since A is closed, it follows that u ∈ C([0, τ ], D(A)) and(ACPf ) is satisfied. The following result is a special case of a sharper version ofCorollary 3.2.11 c) which is valid for generators of Lipschitz continuous integratedsemigroups.

Theorem 3.11.10. Let A be a resolvent positive operator. Let u0 ∈ D(A), f0 ∈ X

such that Au0 + f0 ∈ D(A). Let f(t) = f0 +∫ t

0f ′(s) ds where f ′ ∈ L1((0, τ), X).

Then (ACPf ) has a unique mild solution.

Proof. Denote by S the twice integrated semigroup generated by A. Let

v(t) = S(t)u0 + (S ∗ f)(t).By Lemmas 3.2.9 and 3.2.10, there exists a unique solution if and only if v ∈C2([0, τ ], X). By Proposition 1.3.6, one has S ∗ f ∈ C1([0, τ ], X) and

d

dt(S ∗ f)(t) = (S ∗ f ′)(t) + S(t)f0.

Now it follows from Lemma 3.2.2 c) that v ∈ C1([0, τ ], X) and

v′(t) = S(t)(Au0 + f0) + tu0 + (S ∗ f ′)(t).Since S is Lipschitz continuous on [0, τ ] (by Theorem 3.11.5), it follows from Propo-sition 1.3.7 that S ∗ f ′ ∈ C1([0, τ ], X). Since Au0 + f0 ∈ D(A), it follows fromLemma 3.3.3 that S(·)(Au0 + f0) ∈ C1([0, τ ], X). The proof is complete.

Theorem 3.11.10 will be used in Section 6.2 to solve the heat equation with in-homogeneous boundary conditions. The following result shows that mild solutionsof the inhomogeneous problem are positive if the initial value and the inhomogene-ity are positive. In Section 6.2 this will be used to prove the parabolic maximumprinciple.

Theorem 3.11.11. Let A be a resolvent positive operator, τ > 0, f ∈ C([0, τ ], X+),and u0 ∈ X+. Let u be a mild solution of (ACPf ). Then u(t) ≥ 0 for all t ∈ [0, τ ].

Proof. Denote by S the twice integrated semigroup generated by A. It follows fromTheorem 2.7.15 that S is an increasing convex function. Let

w(t) := S(t)u0 +

∫ t

0

S(t− r)f(r) dr.

It follows from Lemma 3.2.9 that w ∈ C2([0, τ ], X) and u(t) = w′′(t). Thus, itsuffices to show that w is convex. We know this already for the first term of w.Define S(t) := S(t) for t ≥ 0 and S(t) = 0 for t < 0. Since S is increasing and

convex and S(0) = 0, it follows that S(t) : R→ L(X) is also convex. Hence,∫ t

0

S(t− r)f(r) dr =

∫ ∞

0

S(t− r)f(r) dr

is convex in t ≥ 0.

Next we consider a simple perturbation result.

Proposition 3.11.12. Let A be a resolvent positive operator. Let B : D(A) → Xbe linear and positive (i.e., Bx ≥ 0 if x ∈ D(A) ∩ X+). If the spectral radiusr(BR(λ,A)) < 1 for some λ > s(A), then A + B is resolvent positive and s(A +B) < λ.

Notice that BR(λ,A) is a linear, positive mapping on X and so it is auto-matically continuous.

Proof. Let x ∈ D(A). Then

(λ− (A+B))x = (I −BR(λ,A))(λ− A)x.

Let

Sλ := (I −BR(λ,A))−1 =

∞∑n=0

(BR(λ,A))n.

Then Sλ is a bounded, positive operator on X and

R(λ,A)Sλ(λ− (A+B))x = x

for all x ∈ X and

(λ− (A+B))R(λ,A)Sλy = y for all y ∈ X.

Thus, λ ∈ ρ(A + B) and R(λ,A + B) = R(λ,A)Sλ ≥ 0. If μ > λ, then byProposition 3.11.2, μ ∈ ρ(A) and BR(μ,A) ≤ BR(λ,A) and so r(BR(μ,A)) ≤r(BR(λ,A)) < 1. Replacing λ by μ, it follows that μ ∈ ρ(A+B) and R(μ,A+B) ≥0 for all μ ≥ λ.

The following example shows that, in the theorem above, A+B may not begenerator of a C0-semigroup even if A generates a positive C0-semigroup.

Example 3.11.13. Let α ∈ (0, 1). Define the operator A by

Af(x) := −f ′(x) + α

xf(x) (x ∈ (0, 1])

on C0(0, 1] := {f ∈ C[0, 1] : f(0) = 0} with domain D(A) := {f ∈ C1[0, 1] :f ′(0) = f(0) = 0}. Then A is resolvent positive but not the generator of a C0-semigroup. Moreover, s(A) = −∞.

Proof. Let A0f := −f ′ with domain D(A0) = D(A). Then A0 is the generator ofthe C0-semigroup (T (t))t≥0 given by

(T (t)f)(x) =

{f(x− t) (x ≥ t),

0 (x < t).

Moreover, σ(A0) = ∅ and

(R(λ,A0)f)(x) = e−λx

∫ x

0

eλyf(y) dy (λ ∈ C, f ∈ C0(0, 1]).

Let B : D(A0)→ C0(0, 1] be given by

(Bf)(x) :=α

xf(x) (x > 0), (Bf)(0) := 0.

Let g ∈ C0(0, 1], f := R(0, A)g. Then

|(Bf)(x)| =∣∣∣∣αx

∫ x

0

g(y) dy

∣∣∣∣ ≤ α‖g‖∞.

Thus ‖BR(0, A0)‖ ≤ α < 1. Now Proposition 3.11.12 implies that A = A0 +B isresolvent positive.

3.12. COMPLEX INVERSION AND UMD-SPACES 197

It remains to show that A is not the generator of a C0-semigroup. One caneasily check that for all λ ∈ C one has λ ∈ ρ(A) and

(R(λ,A)g)(x) = e−λxxα

∫ x

0

y−αeλyg(y) dy

=

∫ x

0

xα(x− t)−αg(x− t)e−λt dt (g ∈ C0(0, 1]).

Suppose that A generates a C0-semigroup T . Then

(R(λ,A)g)(x) =

∫ ∞

0

e−λt(T (t)g)(x) dt

for sufficiently large λ. It follows from the uniqueness theorem (Theorem 1.7.3)that

(T (t)g)(x) =

{xα(x− t)−αg(x− t) (x ≥ t),

0 (x < t).

This does not define a bounded operator on C0(0, 1].

3.12 Complex Inversion and UMD-spaces

In this section, we apply the complex inversion formula for Laplace transforms(Theorem 2.3.4) to orbits of C0-semigroups; i.e., to solutions of well-posed Cauchyproblems. In general this produces a representation only of classical solutions interms of the resolvent of A, but we shall see in Theorem 3.12.2 that there is a classof Banach spaces where the representation holds for mild solutions.

Proposition 3.12.1. Let T be a C0-semigroup on a Banach space X with generatorA, and let ω > ω(T ) and t ≥ 0. Then

T (t)x = limk→∞

1

2π

∫ k

−k

e(ω+is)tR(ω + is, A)x ds

for all x ∈ D(A).

Proof. Replacing T (t) by e−αtT (t) where ω(T ) < α < ω, we may assume thatω(T ) < 0 < ω. Let x ∈ D(A) and define F (t) = T (t)x−x. Then F is differentiablewith F ′(t) = T (t)Ax. Since F ′ is bounded, F ∈ Lip0(R+, X). The Laplace-Stieltjestransform of F is

dF (λ) = R(λ,A)Ax.

By Theorem 2.3.4,

T (t)x− x = limk→∞

1

2π

∫ k

−k

e(ω+is)tR(ω + is, A)Ax

ω + isds

= limk→∞

1

2π

∫ k

−k

e(ω+is)t

(R(ω + is, A)x− x

ω + is

)ds

= limk→∞

(1

2π

∫ k

−k


)− x,

where we have used a standard contour integral.

For f ∈ L2(R, X) and 0 < ε < R, let

(HεRf)(t) :=1

π

∫ε≤|t−s|≤R

f(s)

t− sds = (ψεR ∗ f)(t) (t ∈ R),

where

ψεR(t) :=

⎧⎨⎩1

πtif ε ≤ |t| ≤ R,

0 otherwise.

Then HεR ∈ L(L2(R, X)), since ψεR ∈ L1(R, X) (see Proposition 1.3.2).

The Banach space X is said to be a UMD-space if

Hf := limε↓0

R→∞HεRf

exists in L2(R, X) for each f ∈ L2(R, X). Then by the Banach-Steinhaus theorem,H is a bounded linear operator, known as the Hilbert transform, on L2(R, X).

When f(t) = χ[a,b](t)x, then Hf exists in L2(R, X). Since the step functionsare dense in L2(R, X), it follows that X is a UMD-space if sup0<ε<R<∞ ‖HεR‖ <∞. It follows easily from Plancherel’s theorem that any Hilbert space is a UMD-space (see also Proposition E.5). Any space of the form Lp(Ω, μ) for 1 < p < ∞is a UMD-space. If X is any UMD-space, then X is reflexive and X∗ is also aUMD-space. See, for example, [Fra86, Section II].

Theorem 3.12.2. Let T be a C0-semigroup on a UMD-space X with generator A,and let ω > ω(T ) and t > 0. Then

T (t)x = limk→∞

1

2π

∫ k

−k


for all x ∈ X.

3.12. COMPLEX INVERSION AND UMD-SPACES 199

Proof. Replacing T (t) by e−ωtT (t), we may assume that ω(T ) < 0 = ω. For k > 0and t ∈ R, define

Tk(t) :=1

2π

∫ k

−k

eistR(is, A) ds,

Sk(t) :=1

2π

∫ k

−k

eistR(is, A)2 ds.

Since Tk(t)x→ T (t)x for all x ∈ D(A) (Proposition 3.12.1) and D(A) is dense, itsuffices to show that

supk‖Tk(t)‖ <∞

for each t > 0. Integration by parts gives

Tk(t) =1

2πit

(eiktR(ik,A)− e−iktR(−ik, A))+ 1

tSk(t).

Since ‖R(±ik,A)‖ ≤ ∫∞0‖T (s)‖ ds <∞, it suffices to show that

supk‖Sk(t)‖ <∞.

Since X is reflexive, T ∗ is strongly continuous (Corollary 3.3.9 and Proposi-tion 3.3.14). Let x ∈ X, x∗ ∈ X∗ and t ∈ R. Using Fubini’s theorem,

〈x, Tk(t)∗x∗〉 =

1

2π

∫ k

−k

∫ ∞

0

eiste−isr〈x, T (r)∗x∗〉 dr ds

=1

2π

∫ ∞

0

eik(t−r) − e−ik(t−r)

i(t− r)〈x, T (r)∗x∗〉 dr

= limε↓0

R→∞

1

2i

(eikt〈x,HεR(fk)(t)〉 − e−ikt〈x,HεR(f−k)(t)〉

),

where

fa(r) :=

{e−iarT (r)∗x∗ (r ≥ 0),

0 (r < 0).

It follows that

Tk(t)∗x∗ =

1

2i

(eiktH(fk)(t)− e−iktH(f−k)(t)

)t-a.e.,

where H is the Hilbert transform on L2(R, X∗). Hence,(∫ ∞

−∞‖Tk(t)

∗x∗‖2 dt)1/2

≤ ‖H‖2

(‖fk‖2 + ‖f−k‖2)≤ M‖H‖ ‖x∗‖,

where M := (∫∞0‖T (s)‖2 ds)1/2 <∞.

For x ∈ X and x∗ ∈ X∗, Fubini’s theorem gives

〈Sk(t)x, x∗〉 =

1

2π

∫ k

−k

∫ ∞

0

eist〈R(is, A)e−isrT (r)x, x∗〉 dr ds

=

∫ ∞

0

⟨ 1

2π

∫ k

−k

eis(t−r)R(is, A)T (r)x ds, x∗⟩dr

=

∫ ∞

0

〈T (r)x, Tk(t− r)∗x∗〉 dr.

Now the Cauchy-Schwarz inequality gives

|〈Sk(t)x, x∗〉| ≤

(∫ ∞

0

‖T (r)x‖2 dr)1/2(∫ ∞

0

‖Tk(t− r)∗x∗||2 dr)1/2

≤ M2‖H‖ ‖x‖ ‖x∗‖.Thus,

‖Sk(t)‖ ≤M2‖H‖.Example 3.12.3. Theorem 3.12.2 is not valid if the assumption that X is a UMD-space is omitted.

Let X := L1(R) and T be the C0-semigroup of invertible isometries on Xdefined by

(T (t)f)(r) := f(t+ r) (t ≥ 0, r ∈ R).

Let

Tk(t) :=1

2π

∫ k

−k

e(1+is)tR(1 + is, A) ds.

A routine calculation shows that

(Tk(t)f) (r) =1

π

∫ ∞

−∞f(r + s+ t)φk,t(s) ds,

where

φk,t(s) :=

⎧⎨⎩e−s sin(ks)

s(s ≥ −t),

0 (s < −t).Hence,

‖Tk(t)‖ = ‖φk,t‖1 =

∫ ∞

−t

e−s

∣∣∣∣ sin(ks)s

∣∣∣∣ ds≥ e−1

∫ 1

0

| sin(ks)|s

ds = e−1

∫ k

0

| sin s|s

ds→∞

as k → ∞. Thus, {Tk(t) : k ≥ 0} is not uniformly bounded and therefore notstrongly convergent on X.

3.13. NORM-CONTINUOUS SEMIGROUPS AND HILBERT SPACES 201

3.13 Norm-continuous Semigroups and Hilbert Spaces

In this section, we shall consider C0-semigroups T = (T (t))t≥0 which are norm-continuous for t > 0 . This class contains all holomorphic C0-semigroups, includingmany examples arising from differential operators. In general, there is no knowncharacterization of such semigroups in terms of the generator and resolvent, butthere is a simple characterization in the case of Hilbert spaces.

Proposition 3.13.1. Let T be a C0-semigroup on a Banach space X with generatorA, and suppose that T is norm-continuous for t > 0. Let ω > ω(T ). Then ‖R(ω+is, A)‖ → 0 as |s| → ∞.

Proof. Since T : (0,∞)→ L(X) is norm-continuous,

R(ω + is, A) =

∫ ∞

0

e−(ω+is)tT (t) dt

as an L(X)-valued Bochner integral. Let H(t) := e−ωtT (t) if t > 0, and H(t) := 0if t < 0. Then H ∈ L1(R,L(X)), and R(ω + is, A) = (FH)(s). The result followsfrom the Riemann-Lebesgue lemma (Theorem 1.8.1).

It follows from Neumann series expansions (Corollary B.3) that if ‖R(ω +is, A)‖ → 0 as |s| → ∞, then for any real a, {λ ∈ σ(A) : Reλ ≥ a} is bounded,and ‖R(α + is, A)‖ → 0, uniformly for α > a, as |s| → ∞.

The converse of Proposition 3.13.1 does not hold in general, but we will nowprove that it is true when X is a Hilbert space.

Theorem 3.13.2. Let T be a C0-semigroup on a Hilbert space X with generatorA. Let ω > ω(T ), and suppose that ‖R(ω + is, A)‖ → 0 as |s| → ∞. Then T isnorm-continuous for t > 0.

Proof. Replacing T (t) by e−ωtT (t), we may assume that ω(T ) < 0 = ω. Letx ∈ D(A) and F (t) := t2T (t)x. Then F is differentiable and F ′(t) = 2tT (t)x +t2T (t)Ax. Hence F ∈ Lip0(R+, X) with Laplace-Stieltjes transform

dF (λ) = −2 d

dλ(R(λ,A)x) +

d2

dλ2(R(λ,A)Ax)

= 2R(λ,A)2x+ 2R(λ,A)3Ax

= 2λR(λ,A)3x.

By Theorem 2.3.4,

t2T (t)x = limk→∞

1

π

∫ k

−k

eistR(is, A)3x ds. (3.84)

Let M :=(∫∞

0‖T (t)‖2 dt)1/2 < ∞. Given ε > 0, there exists N such that

‖R(is, A)‖ < ε/4M2 whenever |s| > N . For k > N , t, t0 ≥ 0 and y ∈ X,∣∣∣∣∣( ∫

N≤|s|≤k

(eist − eist0

)R(is, A)3x ds

∣∣∣∣y)X

∣∣∣∣∣≤ 2

∫N≤|s|≤k

‖R(is, A)2x‖ ‖R(is, A)∗y‖ ds

≤ ε

2M2

(∫ ∞

−∞‖R(is, A)x‖2 ds

)1/2(∫ ∞

−∞‖R(is, A)∗y‖2 ds

)1/2

=πε

M2

(∫ ∞

0

‖T (r)x‖2 dr)1/2(∫ ∞

0

‖T (r)∗y‖2 dr)1/2

≤ πε‖x‖ ‖y‖,

where (·|·)X denotes the inner product onX and we have used the Cauchy-Schwarzinequality and Plancherel’s theorem for Hilbert spaces (Theorem 1.8.2). Hence,∥∥∥∥∥

∫N≤|s|≤k

(eist − eist0

)R(is, A)3x ds

∥∥∥∥∥ ≤ πε‖x‖.

It follows from (3.84) that

∥∥t2T (t)x− t20T (t0)x∥∥ ≤ ε‖x‖+ 2N‖x‖

πsup|s|≤N

∣∣eist − eist0∣∣ sups∈R

‖R(is, A)‖3,

so ∥∥t2T (t)− t20T (t0)∥∥ ≤ ε+

2N2

π|t− t0|

(∫ ∞

0

‖T (r)‖ dr)3

.

This shows that

lim supt→t0

∥∥t2T (t)− t20T (t0)∥∥ ≤ ε.

Since ε > 0 is arbitrary, t �→ t2T (t) is norm-continuous, and the result is proved.

3.14 The Second Order Cauchy Problem

Let A be a closed operator on a Banach space X . Given x, y ∈ X we consider theproblem

P 2(x, y)

⎧⎪⎨⎪⎩u′′(t) = Au(t) (t ≥ 0)

u(0) = x,

u′(0) = y.

3.14. THE SECOND ORDER CAUCHY PROBLEM 203

A classical solution of P 2(x, y) is a function u ∈ C2(R+, X) such that u(t) ∈ D(A)for all t ≥ 0 and P 2(x, y) holds. A mild solution is a function u ∈ C(R+, X) suchthat ∫ t

0

∫ s

0

u(r) dr ds =

∫ t

0

(t− s)u(s) ds ∈ D(A)

and

u(t) = x+ ty +A

∫ t

0

(t− s)u(s) ds (3.85)

for all t ≥ 0.If u is a classical solution, then integrating P 2(x, y) twice shows that u is a

mild solution. Conversely, if u is a mild solution and u ∈ C2(R+, X), then u is aclassical solution. This follows from (3.85) and the fact that A is closed.

Proposition 3.14.1. Let u ∈ C(R+, X) with abs(u) <∞. Let ω > max{abs(u), 0}.Then u is a mild solution of P 2(x, y) if and only if

u(λ) ∈ D(A) and λx+ y = (λ2 −A)u(λ) (3.86)

for all λ > ω.

Proof. There exists M ≥ 0 such that ‖v(t)‖ ≤ Meωt (t ≥ 0), where v(t) :=∫ t

0u(s) ds. Taking Laplace transforms, Corollary 1.7.6 shows that (3.85) holds if

and only ifu(λ)

λ2=

∫ ∞

0

e−λt

∫ t

0

v(s) ds dt ∈ D(A)

and

u(λ) =x

λ+

y

λ2+ A

u(λ)

λ2for all λ > ω.

Now let u ∈ C(R+, X). Assume that abs(u) < ∞, ω > max{abs(u), 0} and(ω,∞) ⊂ ρ(A). Then Proposition 3.14.1 shows that u is a mild solution of P 2(x, y)if and only if

u(λ) = λR(λ2, A)x+R(λ2, A)y (λ > ω). (3.87)

This relation will lead us to consider operators A such that λR(λ2, A) is a Laplacetransform.

Definition 3.14.2. A strongly continuous function Cos : R+ → L(X) is called acosine function if Cos(0) = I and

2Cos(t) Cos(s) = Cos(t+ s) + Cos(t− s) (t ≥ s ≥ 0). (3.88)

Lemma 3.14.3. Let Cos be a cosine function.

a) ω(Cos) <∞.

b) Cos(t)Cos(s) = Cos(s)Cos(t) for all s, t ≥ 0.

c) Define Cos(−t) = Cos(t) for t ≥ 0. Then Cos is strongly continuous on R,and

2Cos(t) Cos(s) = Cos(t+ s) + Cos(t− s) (s, t ∈ R). (3.89)

Proof. a) By the uniform boundedness principle,

M := sup0≤s≤2

‖Cos(s)‖ <∞.

Choose ω > 0 such that 2‖Cos(1)‖e−ω + e−2ω ≤ 1. We claim that ‖Cos(t)‖ ≤Meωt (t ≥ 0). This is certainly true for t ∈ [0, 2]. Assume that it holds for t ∈ [0, n],where n ∈ N, n ≥ 2. We claim that it holds for t ∈ [0, n + 1]. Let t ∈ (n − 1, n].Then

‖Cos(t+ 1)‖ = ‖2Cos(t) Cos(1)− Cos(t− 1)‖≤ 2‖Cos(1)‖Meωt +Meω(t−1)

=(2‖Cos(1)‖e−ω + e−2ω

)Meω(t+1)

≤ Meω(t+1).

b) Let t ≥ 0. Replacing s and t by t/2 in (3.88) gives Cos(t) = 2Cos(t/2)2−I.A simple induction shows that, for each n ≥ 1, Cos(t) is a polynomial in Cos(t/2n),so they commute.

Assume that Cos(t) commutes with Cos(rt/2n) for r = 1, . . . , k, for somek ≥ 1. Since

Cos

((k + 1)t

2n

)= 2Cos

(kt

2n

)Cos

(t

2n

)− Cos

((k − 1)t

2n

),

it follows that Cos(t) commutes with Cos((k + 1)t/2n).By induction, Cos(t) commutes with Cos(rt)/2n for all integers r, n ≥ 1. Now

b) holds by strong continuity of Cos.c) This follows easily from b).

We will frequently consider a cosine function to be extended to R as in Lemma3.14.3 c) without further notice.

In what follows, the Laplace integrals of operator-valued functions are inter-preted as in Sections 1.4 and 1.5.

Proposition 3.14.4. Let Cos : R+ → L(X) be strongly continuous. The followingassertions are equivalent:

(i) Cos is a cosine function.

(ii) One has abs(Cos) <∞, and there exist ω > max{abs(Cos), 0} and an oper-ator A such that (ω2,∞) ⊂ ρ(A) and

λR(λ2, A) =

∫ ∞

0

e−λt Cos(t) dt (λ > ω). (3.90)


In that case, we call A the generator of the cosine function Cos.

Proof. We extend Cos to an even function on R. Let ω > max{0, abs(Cos)}. Thenfor λ, μ > ω with λ �= μ, one has∫ ∞

0

∫ ∞

0

e−λte−μs(Cos(t+ s) + Cos(t− s)) ds dt =2

μ2 − λ2

(μQ(λ)− λQ(μ)

),

(3.91)where Q(λ) :=

∫∞0

e−λt Cos(t) dt. In fact,∫ ∞

0

∫ ∞

0

e−λte−μs(Cos(t+ s) + Cos(t− s)) ds dt

=

∫ ∞

0

e−λt

{∫ ∞

t

e−μ(r−t) Cos(r) dr +

∫ t

−∞e−μ(t−r) Cos(r) dr

}dt

=

∫ ∞

0

e−μr

∫ r

0

e(μ−λ)t dt Cos(r) dr +

∫ ∞

0

e−λt

∫ 0

−∞e−μ(t−r)Cos(r) dr dt

+

∫ ∞

0

e−λt

∫ t

0

e−μ(t−r) Cos(r) dr dt

=

∫ ∞

0

e−μr(μ− λ)−1(e(μ−λ)r − 1)Cos(r) dr

+

∫ 0

−∞

∫ ∞

0

e−(λ+μ)t dt eμr Cos(r) dr +

∫ ∞

0

∫ ∞

r

e−(λ+μ)t dt eμr Cos(r) dr

=1

μ− λ

(Q(λ)−Q(μ)

)+

1

λ+ μ

∫ ∞

0

e−μr Cos(r) dr

+1

λ+ μ

∫ ∞

0

e−(λ+μ)reμr Cos(r) dr

=1

μ− λ

(Q(λ)−Q(μ)

)+

1

λ+ μ

(Q(μ) +Q(λ)

)=

2

μ2 − λ2

(μQ(λ)− λQ(μ)

).

Now assume that Q(λ) = λR(λ2, A) (λ > ω). Then

2

μ2 − λ2

(μQ(λ)− λQ(μ)

)= 2λμ

R(λ2, A)−R(μ2, A)

μ2 − λ2

= 2λμR(λ2, A)R(μ2, A)

= 2

∫ ∞

0

e−λt Cos(t) dt

∫ ∞

0

e−μs Cos(s) ds.

Now (3.91) implies by the Uniqueness Theorem 1.7.3 that

Cos(t+ s) + Cos(t− s) = 2Cos(t) Cos(s) for all s, t ∈ R+. (3.92)


Conversely, assume that Cos is a cosine function. By Lemma 3.14.3, (3.92) holds,and by (3.91),

1

μ2 − λ2

(μQ(λ)− λQ(μ)

)=

∫ ∞

0

∫ ∞

0

e−λte−μs Cos(t) Cos(s) dt ds

= Q(λ)Q(μ).

Let R(λ) := 1√λQ(√λ) for λ > ω2. Then

R(λ)R(μ) =1√λ√μ

Q(√λ)Q(

√μ)

=1√λ√μ

1

μ− λ

(√μ Q(

√λ)−

√λ Q(

√μ))

=1

μ− λ

(R(λ)−R(μ)

)(λ, μ > ω2).

Thus, {R(λ) : λ > ω2} is a pseudo-resolvent. If R(λ)x = 0 for all λ > ω2, thenby the Uniqueness Theorem, Cos(t)x = 0 for all t ∈ R+. Since Cos(0) = I, thisimplies that x = 0. By Proposition B.6, there exists an operator A such that(ω2,∞) ⊂ ρ(A) and

λR(λ2, A) = λR(λ2) = Q(λ) =

∫ ∞

0

e−λt Cos(t) dt

for all λ > ω.

Let Cos be a cosine function on X with generator A. The sine functionSin : R→ L(X) associated with Cos is defined by

Sin(t) :=

∫ t

0

Cos(s) ds (t ∈ R). (3.93)

This means that Sin(t)x =∫ t

0Cos(s)x ds (x ∈ X), where the integral is a Bochner

integral. Then Sin is an odd function satisfying the functional equation

2 Sin(t) Sin(s) =

∫ t+s

t−s

Sin(r) dr (t, s ∈ R). (3.94)

This follows from integrating (3.89) twice.The following functional equation is also useful:

Sin(t+ s) = Cos(s) Sin(t) + Sin(s)Cos(t) (t, s ∈ R). (3.95)

To see this, differentiate (3.94) to obtain

2 Sin(t) Cos(s) = Sin(t+ s) + Sin(t− s).

Interchanging s and t gives

2 Sin(s)Cos(t) = Sin(s+ t) + Sin(s− t).

Since Sin is odd, adding these two equations gives (3.95).Moreover, we deduce from (3.90) that

R(λ2, A) =

∫ ∞

0

e−λt Sin(t) dt (λ > max{ω(Cos), 0}).

We collect some further properties of a cosine function Cos and its relationswith the generator A, and the associated sine function Sin.

Proposition 3.14.5. The following assertions hold:

a)∫ t

0(t− s)Cos(s)x ds ∈ D(A) and A

∫ t

0(t− s) Cos(s)x ds = Cos(t)x−x for all

x ∈ X, t ∈ R.

b) If x ∈ D(A), then Cos(t)x,Sin(t)x ∈ D(A) and ACos(t)x = Cos(t)Ax,ASin(t)x = Sin(t)Ax for all t ∈ R.

c) Let x, y ∈ X. Then x ∈ D(A) and Ax = y if and only if∫ t

0(t−s)Cos(s)y ds =

Cos(t)x− x for all t ∈ R.

d) D(A)={x∈X : limt↓0 2

t2(Cos(t)x−x) exists

}and Ax=limt↓0 2

t2(Cos(t)x−x).

e) A is densely defined.

f) For all x ∈ X, s, t ∈ R, one has Sin(t) Sin(s)x ∈ D(A) and

ASin(t) Sin(s)x = 12 (Cos(t+ s)x− Cos(t− s)x) .

Proof. a) It follows from Proposition 3.14.1 that Cos(·)x is a mild solution ofP 2(x, 0). This implies a) for t ≥ 0. It follows for t < 0 since Cos is even.

b) Let μ ∈ ρ(A). It follows from Proposition 3.1.5 that R(μ,A) Cos(t) =Cos(t)R(μ,A). This clearly implies b) (see Proposition B.7).

c) Assume that∫ t

0(t − s)Cos(s)y ds = Cos(t)x − x (t ≥ 0). Taking Laplace

transforms, we obtain

1

λR(λ2, A)y = λR(λ2, A)x− x

λ(λ > max{ω(Cos), 0}).

Hence, x ∈ D(A) and Ax = y. The converse follows from a) and b).d) Let x ∈ D(A), Ax = y. It follows from c) that

2

t2(Cos(t)x− x)− y =

2

t2

∫ t

0

(t− s) Cos(s)y ds− y

=2

t2

∫ t

0

(t− s)(Cos(s)y − y

)ds→ 0 as t ↓ 0,


since Cos(t)y → y as t ↓ 0. Conversely, let x, y ∈ X with limt↓0 2t2 (Cos(t)x−x) = y.

Then by a),

A2

t2

∫ t

0

(t− s)Cos(s)x ds =2

t2(Cos(t)x− x)→ y as t ↓ 0.

Since 2t2

∫ t

0(t − s) Cos(s)x ds → x as t ↓ 0 and since A is closed, it follows that

x ∈ D(A) and Ax = y.

e) Since x = limt↓0 2t2

∫ t

0(t − s)Cos(s)x ds for all x ∈ X, it follows from a)

that D(A) = X.f) This follows from (3.94) and a).

Using the functional equation (3.88), one sees that a cosine function neces-sarily has non-negative exponential type. More precisely, the following holds.

Proposition 3.14.6. a) Let Cos be a bounded cosine function and x ∈ X. Iflimt→∞ Cos(t)x = 0, then x = 0.

b) In particular, ω(Cos) ≥ 0 for each cosine function Cos.

Proof. a) Since Cos is bounded, it follows from the assumption that

x = limt→∞(Cos(2t)x+ x) = lim

t→∞ 2(Cos(t))2x = 0.

It is natural to transform the second order problem into a first order system.Let A be an operator on X . Consider the operator A on X ×X given by

D(A) := D(A)×X,

A(

xy

):=

(0 IA 0

)(xy

)=

(yAx

). (3.96)

Here, X ×X is considered with the norm ‖(x, y)‖X×X = ‖x‖X + ‖y‖X . Let λ ∈ Csuch that λ2 ∈ ρ(A). Then one easily verifies that λ ∈ ρ(A) and λ ∈ ρ(−A) and

R(λ,A) =

(λR(λ2, A) R(λ2, A)AR(λ2, A) λR(λ2, A)

), (3.97)

R(λ,−A) =

(λR(λ2, A) −R(λ2, A)−AR(λ2, A) λR(λ2, A)

). (3.98)

Using this we can prove the following theorem.

Theorem 3.14.7. The operator A generates a cosine function Cos on X if and onlyif A generates a once integrated semigroup S on X ×X. In that case, S is givenby

S(t) =(

Sin(t)∫ t

0Sin(s) ds

Cos(t)− I Sin(t)

), (3.99)

where Sin(t) =∫ t

0Cos(s) ds.


Proof. Assume that A generates a cosine function Cos. Then

AR(λ2, A)/λ = λR(λ2, A)− I/λ =

∫ ∞

0

e−λt(Cos(t)− I) dt.

It follows from (3.97) that∫ ∞

0

e−λtS(t) dt = R(λ,A)/λ (λ > ω(Cos)).

Thus, S is a once integrated semigroup and A is its generator.Conversely, assume that A generates a once integrated semigroup. Then

R(λ,A)/λ and hence AR(λ2, A)/λ = λR(λ2, A) − I/λ is a Laplace transform.Since 1/λ =

∫∞0

e−λt dt, λR(λ2, A) is also a Laplace transform; i.e., A is the gen-erator of a cosine function.

Corollary 3.14.8. Let A be the generator of a cosine function Cos and let Sin bethe associated sine function. Let x, y ∈ X and

u(t) := Cos(t)x+ Sin(t)y (t ≥ 0).

Then u is the unique mild solution of P 2(x, y).

Proof. It follows from (3.87) that u is a mild solution. Let v be another one anddefine

w(t) :=

∫ t

0

(t− s)(u(s)− v(s)) ds.

Then w is a classical solution of P 2(0, 0). Consider the function φ(t) :=

(w(t)w′(t)

).

Then φ is a classical solution of the homogeneous Cauchy problem associated withA with initial value φ(0) = 0. It follows from Lemma 3.2.9 that φ ≡ 0.

Let A be a bounded operator. Then

Cos(t) :=

∞∑n=0

t2n

(2n)!An

defines a continuous function from R+ into L(X), and for λ >√‖A‖ one has∫ ∞

0

e−λt Cos(t) dt =

∞∑n=0

∫ ∞

0

e−λt t2n

(2n)!dt An

=

∞∑n=0

An

λ2n+1

=1

λ

(I − A

λ2

)−1

= λR(λ2, A).


Thus, Cos is a cosine function and A is its generator. As a consequence of Theorem3.14.7, we may characterize cosine functions with bounded generators as follows.

Corollary 3.14.9. The following assertions are equivalent:

(i) The operator A generates a C0-semigroup on X ×X.

(ii) A generates a cosine function Cos such that limt↓0 ‖Cos(t)− I‖ = 0.

(iii) A is bounded.

Proof. (i) ⇒ (ii): If A generates a C0-semigroup, then limt↓0 ‖S(t)‖ = 0. Now (ii)follows from (3.99).

(ii)⇒ (iii): The Abelian Theorem 4.1.2 implies that limλ↓0 ‖λ2R(λ2, A)−I‖ =0. Hence, λ2R(λ2, A) is invertible for large λ. Consequently, D(A) = X.

(iii) ⇒ (i): If A is bounded, then A is also bounded.

We now deduce a simple perturbation result which will be considerably im-proved in Corollary 3.14.13.

Corollary 3.14.10. Let A be the generator of a cosine function and let B ∈ L(X).Then A+B generates a cosine function.

Proof. Consider the operator

B :=

(0 0B 0

)on X ×X. Then B(X ×X) ⊂ {0} ×X ⊂ D(A). It follows from Corollary 3.10.5

that A+ B =

(0 I

A+B 0

)generates a once integrated semigroup. Now the

claim follows from Theorem 3.14.7.

We have seen that A does not generate a C0-semigroup on X ×X unless Ais bounded. However, a semigroup exists on a natural “phase space”.

Theorem 3.14.11. The following assertions are equivalent:

(i) The operator A generates a cosine function.

(ii) There exists a Banach space V such that D(A) ↪→ V ↪→ X and such that thepart B of A in V ×X generates a C0-semigroup.

In that case, the Banach space V is uniquely determined by (ii). We call V ×Xthe phase space associated with A. Moreover, one has Sin(·)y ∈ C(R, V ) for ally ∈ X, Cos(·)x ∈ C1(R, X) ∩ C(R, V ) for all x ∈ V , Sin(·)x ∈ C(R, D(A)) for allx ∈ V , and B generates a C0-group J on V ×X given by

J (t) =

(Cos(t) Sin(t)Cos′(t) Cos(t)

)=

(Cos(t) Sin(t)ASin(t) Cos(t)

)(t ∈ R), (3.100)

where Sin is the sine function associated with Cos, and Cos′(t)x := ddt Cos(t)x =

ASin(t)x (x ∈ V ).


Here, V ×X is a Banach space for the norm ‖(x, y)‖V×X := ‖x‖V + ‖y‖X .Note that the operator B on V ×X is defined as follows:

D(B) = D(A)× V,

B(

xy

)=

(0 IA 0

)(xy

)=

(yAx

).

Proof of Theorem 3.14.11. (i) ⇒ (ii): Assume that A generates a cosine function.Then A generates a once integrated semigroup on X×X (Theorem 3.14.7). By theSandwich Theorem 3.10.4, there exists a Banach space Z such that D(A) ↪→ Z ↪→X ×X and such that the part B of A in Z generates a C0-semigroup J . DefineV := {x ∈ X : (x, 0) ∈ Z} with norm ‖x‖V := ‖(x, 0)‖Z . Then Z = V ×X . In fact,let (x, y) ∈ Z. Since (0, y) ∈ D(A) ⊂ Z, it follows that (x, 0) = (x, y)− (0, y) ∈ Z.Hence, x ∈ V and (x, y) ∈ V ×X . The converse inclusion is obvious.

Since Z is complete and Z ↪→ X×X, it follows that V is complete. It followsfrom the closed graph theorem that the embedding y �→ (0, y) from X into Z iscontinuous. So there exists β > 0 such that ‖(0, y)‖Z ≤ β‖y‖X (y ∈ X). Thus,

‖(x, y)‖Z ≤ ‖(x, 0)‖Z + ‖(0, y)‖Z ≤ ‖x‖V + β‖y‖X .

Now it follows from the closed graph theorem that ‖(x, y)‖V×X := ‖x‖V + ‖y‖Xdefines a norm on V ×X which is equivalent to ‖ · ‖Z . Thus, (ii) is proved.

(ii) ⇒ (i): Suppose that V is a Banach space such that D(A) ↪→ V ↪→ X andsuch that B generates a C0-semigroup J on V ×X. It follows from the SandwichTheorem 3.10.4 that A generates a once integrated semigroup S. Moreover,

S(t)z =

∫ t

0

J (s)z ds for all z ∈ V ×X, t ≥ 0.

In particular, S(·)z ∈ C(R+, D(A) × V ) ∩ C1(R+, V ×X) for all z ∈ V ×X. ByTheorem 3.14.7, the operator A generates a cosine function Cos and S is given by(3.99). Since Sin is odd and Cos is even, this implies that Sin(·)y ∈ C(R, V ) forall y ∈ X, Cos(·)x ∈ C1(R, X) ∩ C(R, V ) and Sin(·)x ∈ C(R, D(A)) for all x ∈ Vand that J is given by the first matrix in (3.100) for t ≥ 0. In order to proveuniqueness of V , we show that V is equal to the space

V := {x ∈ X : Sin(·)x ∈ C([0, 1], D(A))} .

Note that V is a Banach space for the norm

‖x‖˜V = ‖x‖X + sup

0≤s≤1‖ASin(s)x‖X ,

and we show that this norm is equivalent to ‖ ·‖V . Since V is completely describedby the operator A, this proves uniqueness.


As we noted above, one has V ⊂ V . It follows from the closed graph theoremthat the injection is continuous.

Next, we show that ‖x‖V ≤ c‖x‖˜V for all x ∈ V and some constant c ≥ 0.

First, since Sin(·)y ∈ C(R, V ) we note that, by the closed graph theorem again,

‖ Sin(t)y‖V ≤ c1‖y‖X (3.101)

for all y ∈ X, 0 ≤ t ≤ 2 and some constant c1 ≥ 0. Let x ∈ V . Then by (3.101)and Proposition 3.14.5 f),

‖x‖V =

∥∥∥∥12∫ 2

0

(x− Cos(t)x) dt+1

2Sin(2)x

∥∥∥∥V

≤ 1

2‖ Sin(2)x‖V +

1

2

∥∥∥∥∫ 2

0

(x− Cos(t)x) dt

∥∥∥∥V

≤ c1‖x‖X + sup0≤t≤2

‖x− Cos(t)x‖V≤ c1‖x‖X + sup

0≤t≤10≤s≤1

‖Cos(t+ s)x− Cos(t− s)x‖V

= c1‖x‖X + 2 sup0≤t≤10≤s≤1

‖ASin(t) Sin(s)x‖V

≤ c1‖x‖X + 2c1 sup0≤s≤1

‖A Sin(s)x‖X≤ 2c1‖x‖˜V .

We have shown that the norms ‖ · ‖V and ‖ · ‖˜V are equivalent on V . It remains

to show that V ⊂ V . Let x ∈ V . Proposition 3.14.5 b) implies that Cos(1)x ∈ V .Since 2 Sin(t) Cos(1)x = Sin(1 + t)x− Sin(1− t)x (by differentiation of (3.94)), itfollows that Sin(·)x ∈ C([0, 2], D(A)). Since the function is odd, one has Sin(·)x ∈C([−2, 2], D(A)). We have to show that x ∈ V . Since Sin(t)x ∈ V , it suffices toshow that limt↓0 ‖ 1t Sin(t)x− x‖

˜V = 0. By (3.94), we have∥∥∥∥ASin(s)

(1

tSin(t)x− x

)∥∥∥∥X

=

∥∥∥∥ 1

2t

∫ s+t

s−t

(ASin(r)x− ASin(s)x) dr

∥∥∥∥X

→ 0

as t ↓ 0, uniformly for s ∈ [0, 1]. This completes the proof of uniqueness of V .

Finally, we show that J extends to a C0-group. We have seen that the semi-group J is given by (3.100), and we can extend J to R by the same formulae. Forλ sufficiently large,

R(λ,A)z =

∫ ∞

0

e−λtJ (t)z dt


for z ∈ V ×X. Since Sin is odd and Cos is even, it follows from (3.98) that

R(λ,−A)z =

∫ ∞

0

e−λtJ (−t)z dt.

This shows that J defined by (3.100) is a C0-group on V ×X and its generator isthe part B of A in V ×X.

The phase space can be computed in many concrete cases (see Section 7.2and Examples 3.14.15 and 3.14.16). The phase space is also important in order toobtain classical solutions of the Cauchy problem of second order.

Corollary 3.14.12. Let A be the generator of a cosine function Cos on X. Denotethe associated phase space by V ×X, and the corresponding sine function by Sin.Let x ∈ D(A), y ∈ V . Then

u(t) := Cos(t)x+ Sin(t)y

defines a classical solution of P 2(x, y).

Proof. Since y ∈ V , one has Cos(·)y ∈ C1(R, X), and hence Sin(·)y ∈ C2(R, X).Since x ∈ D(A), it follows from Proposition 3.14.5 c) that Cos(·)x ∈ C2(R, X).Thus, u ∈ C2(R, X). Since u is a mild solution and A is closed, it follows that uis a classical solution.

The following perturbation result improves Corollary 3.14.10. It will be mostuseful for applications to elliptic operators given in Chapter 7.

Corollary 3.14.13. Let A be the generator of a cosine function with phase spaceV ×X. Let B ∈ L(V,X). Then A+B generates a cosine function with the samephase space.

Proof. The operator A =

(0 IA 0

)on V ×X with domain D(A)× V generates

a C0-semigroup on V × X. Since B =

(0 0B 0

)∈ L(V × X), it follows from

Corollary 3.5.6 that

(0 I

A+B 0

)= A+ B generates a C0-semigroup on V ×X.

It follows from Theorem 3.14.11 that A+B generates a cosine function.

The following corollary will have useful applications to hyperbolic equationsin Chapter 7.

Corollary 3.14.14. Let A be the generator of a cosine function on X with phasespace V ×X. Then the part AV of A in V generates a cosine function on V withphase space D(A)× V .


Proof. The operator A =

(0 IA 0

)on V ×X with domain D(A) = D(A) × V

generates a C0-semigroup on V ×X. Hence the part G of A in D(A) also generatesa C0-semigroup on D(A)×V (see the remarks preceding Proposition 3.10.3). Notethat

D(G) = {(x, y) ∈ D(A)× V : (y, Ax) ∈ D(A)× V } = D(A)×D(AV ).

Moreover, we have the continuous embeddings D(AV ) ↪→ D(A) ↪→ V . Now replac-ing X by V and V by D(A) in Theorem 3.14.11 we conclude that AV generates acosine function on V with phase space D(A)× V .

Next, we give two examples where the phase space can be determined easily.

Example 3.14.15. Let B be the generator of a C0-group U on X. Then A := B2

generates a cosine function Cos on X given by

Cos(t) :=1

2(U(t) + U(−t)) (t ∈ R).

The phase space is given by D(B)×X, where D(B) carries the graph norm.

Proof. Let M,ω ∈ R such that ‖U(t)‖ ≤ Meω|t| (t ∈ R). Let λ > ω. Thenλ ∈ ρ(B) ∩ ρ(−B) and (λ2 −A) = (λ−B)(λ+B). Hence, λ2 ∈ ρ(A) and

λR(λ2, A) = −λR(λ,B)R(−λ,B)

=1

2(R(λ,B)−R(−λ,B))

=1

2

∫ ∞

0

e−λt(U(t) + U(−t)) dt

=

∫ ∞

0

e−λt Cos(t) dt.

This shows that Cos is a cosine function with generator A.For x ∈ D(B) one has Cos′(t)x = 1

2 (BU(t)x− BU(−t)x). Thus, (3.100) de-fines a strongly continuous, exponentially bounded function J : R+ → L(D(B)×X) whose Laplace transform is the restriction of R(λ,A) to D(B)×X, by (3.97).Thus, J is a C0-semigroup whose generator is the part of A in D(B) × X . SoD(B)×X is the phase space, by Theorem 3.14.11.

Example 3.14.16. Let A be a selfadjoint operator on a Hilbert space H. Assumethat A is bounded above; i.e., (Ax|x)H ≤ ω‖x‖2H for all x ∈ D(A) and someω ∈ R. Then A generates a cosine function.

Proof. By the Spectral Theorem B.13, we can assume that H = L2(Ω, μ), Af =mf , D(A) = {f ∈ H : mf ∈ H}, where m : Ω → R is a measurable function.Moreover, the boundedness assumption implies that m(y) ≤ ω for μ-almost ally ∈ Ω.

First case: Assume that m ≤ 0 a.e. Let q(y) := i√−m(y), Bf := qf , D(B) :=

{f ∈ H : qf ∈ H}. Then B generates the C0-group U given by U(t)f := etqf .Since B2 = A, it follows from the preceding example that A generates a cosinefunction Cos. Moreover,

(Cos(t)f)(y) =1

2((U(t) + U(−t)) f)(y)

=1

2

(eit√−m(y) + e−it

√−m(y)

)f(y)

= cos(t√−m(y))f(y) (t ∈ R, f ∈ H, y ∈ Ω).

Second case: When m ≤ ω a.e., we can write m = m1 + m2 where m1 ≤ 0, 0 ≤m2 ≤ ω. Then A = A1 + B where A1f := m1f, D(A1) = D(A) and Bf := m2fdefines a bounded operator. It follows from Corollary 3.14.10 and the first casethat A generates a cosine function.

Next, we show that there is always a simple way to go from the second orderequation to the first order equation.

Theorem 3.14.17. Let A be the generator of a cosine function Cos. Then A gen-erates a holomorphic C0-semigroup T of angle π/2.

We give two different proofs of this theorem. The first uses the character-ization theorem for holomorphic semigroups. However, it does not give the bestpossible angle.

First proof. Let M,ω ≥ 0 such that ‖Cos(t)‖ ≤Meωt (t ≥ 0). Then by holomor-phic continuation (Proposition B.5) for Reλ > ω one has λ2 ∈ ρ(A) and

‖λR(λ2, A)‖ =∥∥∥∥∫ ∞

0

e−λt Cos(t) dt

∥∥∥∥ ≤ M

Reλ− ω.

Let ω1 := 2ω2. Let μ ∈ C such that Reμ > ω1. Write μ = reiθ, where −π/2 < θ <π/2, and let λ :=

√reiθ/2. Then

Re λ =√r cos

θ

2=√r

√cos θ + 1

2≥√

r cos θ

2≥√

ω1

2= ω.

Hence, μ = λ2 ∈ ρ(A) and

‖μR(μ,A)‖ ≤ |λ| M

Reλ− ω=

M√r√

r cos(θ/2)− ω=

M

cos(θ/2)− ω/√r

≤ M1√2− ω√

r

≤M2√2

if 1√2− ω√

r≥ 1

2√2; i.e., if r ≥ 8ω2.

We have shown that μ ∈ ρ(A) and ‖μR(μ,A)‖ ≤M2√2 whenever Reμ > ω1

and |μ| ≥ 8ω2. By Corollary 3.7.17, this implies that A generates a holomorphicsemigroup T . Since D(A) is dense, T is a C0-semigroup.

The second proof has the advantage of giving a formula which allows one tocompute the semigroup T from the cosine function Cos. In fact, we will prove theWeierstrass formula

T (t)x =

∫ ∞

0

e−s2/4t

√πt

Cos(s)x ds (t > 0). (3.102)

In the context of Example 3.14.15, this formula was established in Corollary 3.7.15.

Second proof of Theorem 3.14.17. Define T by (3.102). Then T (t) ∈ L(X) andT (·)x ∈ C∞((0,∞), X) for all x ∈ X. Let x ∈ X . We show that limt↓0 T (t)x = x.In fact, putting s = r

√t gives

T (t)x =

∫ ∞

0

e−r2/4

√π

Cos(r√t)x dr → x as t ↓ 0

by the dominated convergence theorem, since Cos(r√t)x → x, ‖Cos(r√t)‖ ≤

Meωr for all 0 < t < 1, for some M and ω, and 1√π

∫∞0

e−r2/4 dr = 1.

Let λ > ω. Then by Lemma 1.6.7,∫ ∞

0

e−λtT (t)x dt =

∫ ∞

0

∫ ∞

0

e−λt e−s2/4t

√πt

dt Cos(s)x ds

=

∫ ∞

0

1√λe−√λs Cos(s)x ds

= R(λ,A)x.

It follows from Definition 3.1.8 that T is a C0-semigroup and A is its generator.In order to show that T is holomorphic of angle π/2, we define

T (z)x =1√πz

∫ ∞

0

e−s2/4z Cos(s)x ds (Re z > 0).

Then T : C+ → L(X) is holomorphic. Let θ ∈ (0, π2). According to Definition

3.7.1, it remains to show that

supz∈Σθ|z|≤1

‖T (z)‖ <∞.

Let z ∈ Σθ, |z| ≤ 1. Let u := Re z/|z|2. Then∥∥e−s2/4z Cos(s)∥∥ ≤Me−us2/4eωs = Me−u(s−2ω/u)2/4eω

2/u.

3.15. SINE FUNCTIONS AND REAL CHARACTERIZATION 217

Hence,

‖T (z)‖ =

∥∥∥∥ 1

2√πz

∫R

e−s2/4z Cos(s) ds

∥∥∥∥≤ 1√|z| M

2√π

∫R

e−u(s−2ω/u)2/4 ds eω2/u

=1√|z| M√π

∫R

e−s2 ds1√ueω

2/u

=M√|z| |z|√

Re zeω

2|z|2/Re z

≤ M√cos θ

eω2|z|/ cos θ ≤ M√

cos θeω

2/ cos θ.


The converse of Theorem 3.14.17 is not true: a generator of a holomorphicC0-semigroup does not necessarily generate a cosine function. Indeed, generatorsof cosine functions satisfy a very restrictive spectral condition.

Proposition 3.14.18. Let A be the generator of a cosine function Cos. Then thereexists ω > 0 such that the spectrum σ(A) of A is contained in the parabola {ξ+ iη :η ∈ R, ξ ≤ ω2 − η2/4ω2}.Proof. There exist ω,M > 0 such that

‖Cos(t)‖ ≤Meωt (t ≥ 0).

Since λR(λ2, A) =∫∞0

e−λt Cos(t) dt for λ > ω, it follows from holomorphic con-tinuation (Proposition B.5) that λ2 ∈ ρ(A) whenever Reλ > ω. It is easy to seethat {

λ2 : λ ∈ C, Re λ > ω} ⊃ {

ξ + iη : η ∈ R, ξ > ω2 − η2

4ω2

}.

Example 3.14.19. Let H := L2(R), m(s) := −|s|+ is (s ∈ R),

(Af)(s) := m(s)f(s), D(A) := {f ∈ H : mf ∈ H}.Then A generates a holomorphic C0-semigroup on H. Since σ(A) = {−s ± is :s ∈ R+}, the spectrum of A is not contained in any parabola as described inProposition 3.14.18. Thus, A does not generate a cosine function.

3.15 Sine Functions and Real Characterization

In this section, we consider sine functions. These include the integrals of cosinefunctions as in Section 3.14, and elementary properties of sine functions will be


used to prove the generation theorem for cosine functions which is analogous tothe Hille-Yosida theorem. We shall also establish some useful perturbation results.Examples of sine functions occurring in applications will be given in the Notes ofChapter 7 and in Chapter 8.

Definition 3.15.1. An operator A on X generates a sine function if there existω,M ≥ 0 and a strongly continuous function Sin : R+ → L(X) such that thefollowing properties are satisfied:

a)∥∥∥∫ t

0Sin(s) ds

∥∥∥ ≤Meωt (t ≥ 0).

b) λ2 ∈ ρ(A) whenever λ > ω.

c) R(λ2, A) =∫∞0

e−λt Sin(t) dt (λ > ω).

We then call Sin the sine function generated by A.

If A generates a cosine function Cos, then

R(λ2, A) =1

λ

∫ ∞

0

e−λt Cos(t) dt =

∫ ∞

0

e−λt

∫ t

0

Cos(s) ds dt

for λ > ω(Cos). Thus A generates the sine function Sin given by Sin(t)x :=∫ t

0Cos(s)x ds. In other words, the definition is consistent with the previous notion

of Section 3.14.Next, we establish some relations between a sine function and its generator.

Proposition 3.15.2. Let Sin be a sine function and A be its generator. Then thefollowing hold:

a)∫ t

0(t− s) Sin(s)x ds ∈ D(A) and

A

∫ t

0

(t− s) Sin(s)x ds = Sin(t)x− tx (3.103)

for all x ∈ X.

b) If x ∈ D(A), then Sin(t)x ∈ D(A) and A Sin(t)x = Sin(t)Ax for all t ≥ 0.

c) Let x, y ∈ X. Then x ∈ D(A) and Ax = y if and only if∫ t

0

(t− s) Sin(s)y ds = Sin(t)x− tx (t ≥ 0). (3.104)

Proof. a) It follows from (3.87) that S(·)x is a mild solution of P 2(0, x). This isprecisely the claim.

b) This follows from Proposition 3.1.5 and Proposition B.7.


c) Let x, y ∈ X such that (3.104) holds. Taking Laplace transforms on bothsides gives

1

λ2R(λ2, A)y = R(λ2, A)x− x

λ2.

Hence, x ∈ D(A) and y = λ2x− (λ2−A)x = Ax. The converse implication followsfrom a) and b).

We now prove the characterization theorem for generators of cosine functions.It is mainly of theoretical interest since the condition (3.105) seems to be difficultto verify in concrete cases.

Theorem 3.15.3. Let A be a densely defined operator on a Banach space X. Thefollowing assertions are equivalent:

(i) A is the generator of a cosine function.

(ii) There exist ω,M ≥ 0 such that (ω2,∞) ⊂ ρ(A) and

1

k!

∥∥∥(λ− ω)k+1(λR(λ2, A)

)(k)∥∥∥ ≤M (3.105)

for all λ > ω and k ∈ N0.

Proof. (i)⇒ (ii): Assume that A generates a cosine function Cos. There exist M ≥0, ω ≥ 0 such that ‖Cos(t)‖ ≤ Meωt. Since λR(λ2, A)x =

∫∞0

e−λt Cos(t)x dt forall λ > ω and x ∈ X , the claim follows from Theorem 2.4.1.

(ii) ⇒ (i): Assume that (ii) is satisfied. By Theorem 2.4.1, there exists afunction S : R+ → L(X) satisfying

‖S(t+ h)− S(t)‖ ≤M

∫ t+h

t

eωs ds (t, h ≥ 0) (3.106)

such that

R(λ2, A) =

∫ ∞

0

e−λtS(t) dt (λ > ω).

Thus, S is a sine function and A is its generator. Let x ∈ D(A). Then by Propo-sition 3.15.2, one has

S(t)x− tx =

∫ t

0

(t− s)S(s)Axds (t ≥ 0).

It follows that S(·)x ∈ C1(R+, X). Since D(A) is dense in X, it follows fromLemma 3.3.3 that S(·)x ∈ C1(R+, X) for all x ∈ X. Let C(t)x := d

dtS(t)x (x ∈

X, t ≥ 0). It follows that C(t) is linear and by (3.106) that

‖C(t)‖ ≤Meωt (t ≥ 0).


Integration by parts shows that

λR(λ2, A)x =

∫ ∞

0

e−λtC(t)x dt (λ > ω)

for all x ∈ X . Thus, C is a cosine function and A is its generator.

Now we resume our investigation of sine functions. The following result par-allels Example 3.14.15.

Proposition 3.15.4. Let A be an operator such that A and −A generate once inte-grated semigroups. Then A2 generates a sine function. Moreover, the sine functionis exponentially bounded if both integrated semigroups are.

Proof. There exists ω ≥ 0 such that (ω,∞) ⊂ ρ(A) ∩ ρ(−A) and 1λ (λ− A)−1 and

1λ(λ+A)−1 are Laplace transforms (in the sense of Definition 3.1.4). Thus,

(λ2 − A2)−1 = (λ− A)−1(λ+ A)−1

exists for λ > ω, and by the resolvent equation we have

(λ2 − A2)−1 =1

2λ

((λ+ A)−1 + (λ− A)−1

).

Thus, (λ2 − A2)−1 is a Laplace transform. This shows that A2 generates a sinefunction Sin given by Sin(t) = 1

2 (S+(t) + S−(t)), where S+ and S− are the onceintegrated semigroups generated by A and −A, respectively.

The next example shows that the generator of a sine function is not neces-sarily densely defined.

Example 3.15.5. Let B be the generator of the translation group U on L1(R), givenby (U(t)f)(x) = f(x− t). Then B∗ and −B∗ generate once integrated semigroupson L∞(R) = L1(R)∗ (see Corollary 3.3.7 and Example 3.3.10). By Proposition3.15.4, (B∗)2 generates a sine function. However, D

((B∗)2

) ⊂ D(B∗) = W 1,∞(R),which is not dense in L∞(R) (cf. also Theorem 4.3.18).

Next, we establish a perturbation result for sine functions. Our proof is com-pletely different from the corresponding results on cosine functions (Corollary3.14.10 and Corollary 3.14.13).

Theorem 3.15.6. Let A be the generator of an exponentially bounded sine functionand let B ∈ L(D(A), X). Then A + B generates an exponentially bounded sinefunction.

Proof. Denote by Sin the sine function generated by A and let M,ω ≥ 0 such that‖Sin(t)‖ ≤Meωt (t ≥ 0). The idea of the proof is to solve the integral equation

SB(t) = Sin(t) +

∫ t

0

SB(s)B Sin(t− s) ds. (3.107)

First, we remark that Sin(t)x ∈ D(A) for all t ≥ 0, x ∈ X . In fact by Proposition

3.15.2,∫ t

0(t − s) Sin(t)x ds ∈ D(A) for all t > 0. Differentiating twice we obtain

that Sin(t)x ∈ D(A). Let α > 0. Multiplying (3.107) by e−(ω+α)t, we obtain theequivalent integral equation

U(t) = e−(ω+α)t Sin(t) +

∫ t

0

U(s)Be−(ω+α)(t−s) Sin(t− s) ds. (3.108)

Consider now the Banach space

C := {V : R+ → L(X) : V is strongly continuous and bounded}with norm ‖V ‖ := supt≥0 ‖V (t)‖. Suppose from now on that α > M‖B‖. Then

(JV )(t) :=

∫ t

0

V (s)Be−(ω+α)(t−s) Sin(t− s) ds

defines an operator J on C with norm ‖J‖L(C) ≤ M‖B‖α < 1. The integral equation

(3.108) can now be written in the form

(I − J)U = W

with W (t) := e−(ω+α)t Sin(t). Hence, it has a unique solution U (given by (I −J)−1W ). Therefore SB(t) := e(ω+α)tU(t) defines a solution of (3.107), which isexponentially bounded.

For λ > ω + α, let Q(λ) :=∫∞0

e−λtSB(t) dt. Then

Q(λ)−R(λ2, A) =

∫ ∞

0

e−λt(SB(t)− Sin(t)) dt

=

∫ ∞

0

e−λt

∫ t

0

SB(s)B Sin(t− s) ds dt

=

∫ ∞

0

∫ t

0

e−λsSB(s)Be−λ(t−s) Sin(t− s) ds dt

=

∫ ∞

0

∫ ∞

s

e−λsSB(s)Be−λ(t−s) Sin(t− s) dt ds

=

∫ ∞

0

e−λsSB(s)B

∫ ∞

0

e−λt Sin(t) dt ds

= Q(λ)BR(λ2, A).

Hence, Q(λ)(I −BR(λ2, A)) = R(λ2, A). The operator I −BR(λ2, A) is invertiblesince

‖BR(λ2, A)‖ ≤ ‖B‖∥∥∥∥∫ ∞

0

e−λt Sin(t) dt

∥∥∥∥≤ ‖B‖M(λ− ω)−1 < 1.


Since (λ2 −A−B) = (I −BR(λ2, A))(λ2 −A), it follows that λ2 ∈ ρ(A+B) and

R(λ2, A+B) = R(λ2, A)(I −BR(λ2, A))−1

= Q(λ)

=

∫ ∞

0

e−λtSB(t) dt

for all λ > ω+α. This shows that SB is a sine function and A+B is its generator.

3.16 Square Root Reduction for Cosine Functions

Let B be the generator of a C0-group and let ω ∈ R. Then by Corollary 3.14.10and Example 3.14.15, the operator A := B2 + ω generates a cosine function. Theaim of this section is to establish the following remarkable converse result: If A isthe generator of a cosine function on a UMD-space X, then there exist a generatorB of a C0-group and a number ω ≥ 0 such that A = B2+ω (see Corollary 3.16.8).By the results of Section 3.8, there is a square root B of A − ω for ω sufficientlylarge. We will show that B and −B always generate integrated semigroups, butthe UMD-property is needed in order to obtain a C0-group. We recall the followinglemma which is easy to prove.

Lemma 3.16.1. Let B be a closed operator on X, and A := B2. Let λ ∈ C. Ifλ2 ∈ ρ(A), then λ ∈ ρ(±B) and

R(λ,B) = (λ+B)R(λ2, A),

R(λ,−B) = (λ−B)R(λ2, A).

Proposition 3.16.2. Let B be a closed operator. Assume that A := B2 generates acosine function. Then B and −B generate exponentially bounded once integratedsemigroups.

Proof. Denote by Cos and Sin the cosine and the sine functions associated with A.Recall from Proposition 3.14.5 a) that

∫ t

0Sin(s)x ds ∈ D(A) and

A

∫ t

0

Sin(s)x ds = Cos(t)x− x

for all x ∈ X, t ≥ 0. Define

V+(t)x := Sin(t)x+B

∫ t

0

Sin(s)x ds (t ≥ 0). (3.109)

3.16. SQUARE ROOT REDUCTION FOR COSINE FUNCTIONS 223

Let λ ∈ ρ(A). Then

B

∫ t

0

Sin(s)x ds = BR(λ,A)(λ− A)

∫ t

0

Sin(s)x ds

= λBR(λ,A)

∫ t

0

Sin(s)x ds−BR(λ,A)(Cos(t)x− x)

for all x ∈ X. Since BR(λ,A) is bounded, it follows that V+ is strongly continuousand exponentially bounded. Moreover, for large λ > 0,∫ ∞

0

e−λtV+(t)x dt = R(λ2, A)x+1

λBR(λ2, A)x

=1

λ(λ+B)R(λ2, A)x =

1

λR(λ,B)x, (3.110)

by the preceding lemma. We have shown that V+ is a once integrated semigroupand B is its generator. Replacing B by −B shows that −B also generates a onceintegrated semigroup.

Now we assume that B is a closed operator such that B2 generates a cosinefunction. We want to investigate conditions under which this implies that B gen-erates a C0-group. Here is a characterization in terms of the behaviour of the sinefunction on D(B), which is considered as a Banach space for the graph norm.

Proposition 3.16.3. Let A be the generator of a cosine function Cos with associatedsine function Sin. Assume that B is a closed operator such that B2 = A. Then thefollowing are equivalent:

(i) B generates a C0-group U .

(ii) For all x ∈ X, Sin(t)x ∈ D(B) for almost all t ∈ (0,∞).

(iii) Sin(·)x ∈ C(R, D(B)) for all x ∈ X.

(iv) The phase space of Cos is D(B)×X.

In that case, the group U is given by

U(t)x = Cos(t)x+B Sin(t)x (3.111)

for all t ∈ R, x ∈ X.

For the proof we need the following results which are of independent interest.

Lemma 3.16.4. Let U : R→ L(X) be a mapping such that

a) U(t+ s) = U(t)U(s) (t, s ∈ R),

b) U(0) = I, and

c) U(·)x is measurable for all x ∈ X.

Then U is strongly continuous.

Proof. a) We show that M := sups∈[1,2] ‖U(s)‖ < ∞. Otherwise, by the uniformboundedness principle, there exist x ∈ X, sn ∈ [1, 2] such that ‖U(sn)x‖ ≥n (n ∈ N). Considering a subsequence if necessary, we can and do assume thatlimn→∞ sn =: γ exists. Since ‖U(·)x‖ is measurable, there exists a constantM1 ≥ 0and a measurable set F ⊂ [0, γ] with Lebesgue measure m(F ) > γ/2 such thatsupt∈F ‖U(t)x‖ ≤M1. Let

En :={(sn − t) : t ∈ F ∩ [0, sn]

}.

Then m(En) ≥ γ/2 for large n ∈ N. Now for t ∈ F ∩ [0, sn] we have

n ≤ ‖U(sn)x‖ ≤ ‖U(sn − t)‖ ‖U(t)x‖≤ ‖U(sn − t)‖M1.

Hence, ‖U(s)‖ ≥ n/M1 for all s ∈ En. Let E =⋂

k∈N⋃

n≥k En. Then m(E) ≥ γ/2and ‖U(s)‖ =∞ for all s ∈ E. This is a contradiction.

b) By the group property and a), one has for each t ∈ R,

sups∈[t+1,t+2]

‖U(s)‖ = sups∈[t+1,t+2]

‖U(s− t)U(t)‖ ≤M‖U(t)‖ (t ∈ R).

Thus U is locally bounded. It follows that for each x ∈ X , U(·)x is locally Bochnerintegrable.

c) Let x ∈ X, t0 ∈ R. Then

U(t0)x = U(t)U(t0 − t)x

and

U(t0 + h)x = U(t)U(t0 + h− t)x (t ∈ [1, 2], h ∈ R).

Hence,

‖U(t0)x− U(t0 + h)x‖ =

∥∥∥∥∫ 2

1

U(t)(U(t0 − t)x− U(t0 + h− t)x

)dt

∥∥∥∥≤ M

∫ 2

1

‖U(t0 − t)x− U(t0 + h− t)x‖ dt→ 0 as h→ 0,

using b) and the continuity of shifts on L1(R, X).

Lemma 3.16.5. Let F ⊂ R be a Lebesgue measurable set such that F + F ⊂ F . IfR \ F is a null set, then F = R.


Proof. Assume that N := R \ F �= ∅. Let a ∈ N . Then for x ∈ F , the assumptionimplies that a− x ∈ N (otherwise, a = x+ (a− x) ∈ F +F ⊂ F ). We have shownthat F ⊂ a−N , which is impossible since a−N is a null set.

Proof of Proposition 3.16.3. Assume that (i) holds. One has

Cos(t)x =1

2(U(t)x+ U(−t)x)

and

Sin(t)x =1

2

∫ t

0

(U(s) + U(−s))x ds.

It follows from Proposition 3.1.9 e) that Sin(t)x ∈ D(B) and

B Sin(t)x =1

2

(U(t)x− x− (U(−t)x− x)

)=

1

2(U(t)x− U(−t)x).

This proves (3.111). We have seen in Example 3.14.15 that (i) implies (iv). Theorem3.14.11 shows that (iv) implies (iii). The implication (iii) ⇒ (ii) is trivial.

It remains to show that (ii) ⇒ (i). It is obvious that for μ ∈ ρ(B) one has

R(μ,B)R(λ2, A) = R(λ2, A)R(μ,B) (λ > ω(Cos)).

Thus, by Proposition 3.1.5 and Proposition B.7, x ∈ D(B) implies that Sin(t)x,Cos(t)x ∈ D(B) and

B Sin(t)x = Sin(t)Bx, BCos(t)x = Cos(t)Bx.

Recall from (3.95) that

Sin(t+ s) = Cos(s) Sin(t) + Cos(t) Sin(s) (t, s ∈ R). (3.112)

Let x ∈ X and F := {t ∈ R : Sin(t)x ∈ D(B)}. By hypothesis, R \ F hasLebesgue measure zero. It follows from (3.112) that F +F ⊂ F . Now we concludefrom Lemma 3.16.5 that F = R. Define U(t)x := Cos(t)x + B Sin(t)x for allx ∈ X, t ∈ R. Since B is closed, U(t) is also closed. Hence, U(t) ∈ L(X). Ifx ∈ D(B), then U(·)x is continuous. Since D(B) ⊃ D(A) which is dense, it followsthat U(·)x is measurable for all x ∈ X (in fact, let x ∈ X, xn ∈ D(B) such thatlimn→∞ xn = x; then U(t)x = limn→∞ U(t)xn). It remains to show the groupproperty. Recall from Proposition 3.14.5 f) that Sin(t) Sin(s)x ∈ D(A) and

ASin(t) Sin(s)x =1

2

(Cos(t+ s)x− Cos(t− s)x

)(3.113)

for all t, s ∈ R, x ∈ X. Hence by (3.112),

U(t)U(s)

= Cos(t)Cos(s) + Cos(t)B Sin(s) +B Sin(t)Cos(s) + B Sin(t)B Sin(s)

= Cos(t)Cos(s) +B Sin(t+ s) +A Sin(t) Sin(s)

=1

2(Cos(t+ s) + Cos(t− s)) +B Sin(t+ s) +

1

2(Cos(t+ s)− Cos(t− s))

= Cos(t+ s) +B Sin(t+ s)

= U(t+ s).

It follows from Lemma 3.16.4 that U is a C0-group. Since V+(t) =∫ t

0U(s) ds for

t ≥ 0, where V+ is given by (3.109), we have by (3.110),∫ ∞

0

e−λtU(t)x dt = λ

∫ ∞

0

e−λtV+(t)x dt = R(λ,B)x

for large λ. This shows that B is the generator of U .

We will show that condition (ii) of Proposition 3.16.3 is automatically satis-fied if X is a UMD-space (as defined in Section 3.12), −A satisfies the hypothesesof Proposition 3.8.2, and B = i(−A)1/2 (as defined in Section 3.8). For this weneed some preparation concerning the Hilbert transform.

For ω ≥ 0 we define

L∞ω (R, X) :=

{f ∈ L1

loc(R, X) : ‖f‖ω,∞ := ess supt∈R

‖e−ω|t|f(t)‖ <∞}.

Then L∞ω (R, X) is a Banach space for the norm ‖ · ‖ω,∞.

Lemma 3.16.6. Let X be a UMD-space and let 0 ≤ ω < c. For f ∈ L∞ω (R, X) andε > 0 define the continuous function Hc

εf : R→ X by

(Hcεf)(t) =

∫|s|≥ε

e−c|s|

sf(t− s) ds.

Then for each τ > 0, limε↓0 Hcεf =: Hcf exists in L2((−τ, τ), X). Hence, Hc is a

bounded operator from L∞ω (R, X) into L2((−τ, τ), X) for each τ > 0.

Proof. Let τ ≥ 1, 0 < ε ≤ 1, |t| ≤ τ . Denote by χ the characteristic function of[−τ − 1, τ + 1]. Then

(Hcεf)(t) =

∫|s−t|≥ε

e−c|t−s|

t− sf(s) ds

= h1ε(t) + h2ε(t) + h3ε(t),

where

h1ε(t) :=

∫|s−t|≥ε

1

t− sχ(s)f(s) ds,

h2ε(t) :=

∫|s−t|≥ε

e−c|t−s| − 1

t− sχ(s)f(s) ds,

h3(t) :=

∫|s|≥τ+1

e−c|t−s|

t− sf(s) ds.

It is clear that ‖h3(t)‖ ≤ c1‖f‖ω,∞ for |t| ≤ τ , where c1 ≥ 0 is a constant. Next,observe that χf ∈ L2(R, X). If ψ(t) := 1

t(e−c|t| − 1), then ψ ∈ L2(R) and h2ε is

the convolution ((1− χ(−ε,ε))ψ) ∗ (χf). It follows from Proposition 1.3.2 that h2ε

converges uniformly on [−τ, τ ] to ψ ∗ (χf) as ε ↓ 0. Finally, let H be the Hilberttransform on L2(R, X) and HεR be as in Section 3.12. Then h1ε = πHεR(χf)|[−τ,τ ]

for R > 2τ + 1. Hence, h1ε → πH(χf)|[−τ,τ ] in L2((−τ, τ), X) as ε ↓ 0. Thusthe first assertion is proved. The second now follows from the Banach-Steinhaustheorem.

Now we are able to prove the main result.

Theorem 3.16.7 (Fattorini). Let A be the generator of a cosine function on a UMD-space X. Assume that (0,∞) ⊂ ρ(A) and supλ>0 ‖λR(λ,A)‖ <∞. Define (−A)1/2

as in Proposition 3.8.2 and let B = i(−A)1/2. Then B generates a C0-group andB2 = A.

Proof. It is immediate from Proposition 3.8.2 that B2 = A. We want to showcondition (ii) of Proposition 3.16.3. Let c > ω > ω(Cos). For t > 0 and y ∈ D(A),Proposition 3.8.2 gives

B Sin(t)y =1

πi

∫ ∞

0

λ−1/2R(λ,A) Sin(t)Ay dλ

= I1(t)y + I2(t)y,

where

I1(t)y :=1

πi

∫ c2

0

λ−1/2R(λ,A) Sin(t)Ay dλ

=1

πi

∫ c2

0

λ−1/2(λR(λ,A) Sin(t)y − Sin(t)y) dλ

and

I2(t)y :=1

πi

∫ ∞

c2λ−1/2R(λ,A) Sin(t)Ay dλ.

Since ω > ω(Cos), there exists M ≥ 0 such that ‖Cos(s)‖ ≤ Meωs (s ≥ 0).Consequently, ‖Sin(s)‖ ≤Mseωs (s ≥ 0) and∥∥∥∥∫ ∞

ε

e−√λs Sin(s) Sin(t)Ay ds

∥∥∥∥ ≤ c1

∫ ∞

0

e−√λsseωs ds = c1

1

(√λ− ω)2

for all ε > 0, where c1 := M‖ Sin(t)Ay‖. Since the function λ �→ λ−1/2(√λ−ω)−2

is in L1(c2,∞), and since

R(λ,A) Sin(t)Ay =

∫ ∞

0

e−√λs Sin(s) Sin(t)Ay ds,

we obtain from the dominated convergence theorem,

I2(t)y = limε↓0

1

πi

∫ ∞

c2λ−1/2

∫ ∞

ε

e−√λs Sin(s) Sin(t)Ay ds dλ.

Now we apply Fubini’s theorem and obtain

I2(t)y = limε↓0

2

πi

∫ ∞

ε

e−cs

sSin(s) Sin(t)Ay ds.

By (3.113), this gives

I2(t)y = limε↓0

1

πi

∫ ∞

ε

e−cs

s(Cos(t+ s)− Cos(t− s))y ds

= limε↓0

i

π

∫|s|≥ε

e−c|s|

sCos(t− s)y ds.

At this point we use that X is a UMD-space, so that we can apply Lemma 3.16.6.We have shown that I2(·)y = Hc

ε (Cos(·)y) for y ∈ D(A). Let x ∈ X, and letyn ∈ D(A) such that limn→∞ yn = x. Then Cos(·)yn converges to Cos(·)x inL∞ω (R, X). It follows from Lemma 3.16.6 that I2(·)yn converges in L2((−τ, τ), X)as n → ∞ for all τ > 0. Considering a subsequence if necessary, we obtain thatI2(t)yn converges a.e. in X as n→∞. Since I1(t) is a bounded operator it followsthat there exists a measurable subset F of R such that R\F has Lebesgue measure0 and B Sin(t)yn converges in X as n → ∞ for all t ∈ F . Since B is closed, thisimplies that Sin(t)x ∈ D(B) for all t ∈ F . This finishes the proof.

Now we obtain the following interesting characterization of generators ofcosine functions on UMD-spaces.

Corollary 3.16.8. Let A be an operator on a UMD-space. The following assertionsare equivalent:

(i) A generates a cosine function.

(ii) There exist a generator B of a C0-group and ω ≥ 0 such that A = B2 + ω.

Proof. (ii) ⇒ (i): This follows from Example 3.14.15 and Corollary 3.14.10.(i) ⇒ (ii): Assume that A generates a cosine function Cos. Then A generates

a C0-semigroup by Theorem 3.14.17. Hence, there exists ω ≥ 0 such that (ω,∞) ⊂ρ(A) and supλ>0 ‖λR(λ,A−ω)‖ <∞. The operator A−ω is also the generator of


a cosine function by Corollary 3.14.10. It follows from Theorem 3.16.7 that thereexists a C0-group with generator B such that A− ω = B2.

We conclude this section with some remarks. First, we mention that, givenan operator A, there may be infinitely many generators of groups whose square isA. We give an example.

Example 3.16.9. Let X = l2, and define A by

D(A) :={x = (xn)n∈N ∈ l2 : (n2xn)n∈N ∈ l2

},

(Ax)n := −n2xn.

Let (εn)n∈N be an arbitrary sequence in {−1, 1}. Define B on X by Bx :=(iεnnxn)n∈N with domain D(B) := {x : (nxn)n∈N ∈ l2}. Then B generatesthe C0-group U given by

U(t)x := (eiεnntxn)n∈N

and B2 = A.

It should be mentioned that even if the generator A of a cosine function hasa square root which generates a C0-group there may be other square roots whichdo not generate C0-groups. We give an example.

Example 3.16.10. Let A be the generator of a cosine function on a Banach spaceX. Assume that A is unbounded. Then we know from Theorem 3.14.7 that theoperator A on X ×X given by

D(A) = D(A)×X,

A =

(0 IA 0

),

does not generate a C0-semigroup. However, the operator A2 is given by

D(A2) = D(A)×D(A);

A2 =

(A 00 A

).

Thus, A2 generates a cosine function. If A has a square root which generates aC0-group, then so does A2, but this square root is different from A.

There exists an example of a generator A of a cosine function on a Banachspace X (which can even be chosen to be reflexive) such that

A− ω �= B2

for each ω ≥ 0 and each generator B of a C0-group. Of course in that case, X isnot a UMD-space. We refer to [Fat85, Section III.8] for such examples.


It is interesting to observe that the question whether a square root reductionexists for a cosine function is equivalent to the existence of a boundary group fora certain holomorphic semigroup.

To be more precise, let A be the generator of a cosine function. Then byTheorem 3.14.17, A generates a holomorphic C0-semigroup T of angle π/2. Letω > ω(T ). Then −(ω − A)1/2 generates a C0-semigroup V (see Theorem 3.8.3).Now it follows from Corollary 3.9.10 that B := i(ω − A)1/2 generates a C0-groupif and only if V is a holomorphic semigroup of angle π/2 and V has a boundarygroup (in the sense of Proposition 3.9.1). This is not always the case (if X failsto have the UMD property) as we pointed out above. However, we know fromProposition 3.16.2 that V always has an integrated boundary group (in the senseexplained in Theorem 3.9.13).

3.17 Notes

Section 3.1The characterization of mild solutions in terms of Laplace transforms given in Theorem3.1.3 appeared in [Neu94]. It is at the heart of the theory and it may have been known fora long time. Another way to define mild solutions by approximation schemes (“bonnessolutions”) has been given by Benilan, Crandall and Pazy [BCP88]; see also [BCP90].This way is most important in non-linear theory where Laplace transforms have not yetbeen used effectively.

The idea of defining semigroups and their generators directly by the property thatthe Laplace transform is a resolvent, is from [Are87b], but Theorem 3.1.7 is already con-tained in [DS59, Corollary VIII.1.16]. We should also mention the approach to semigroupsby Laplace transforms given by Kisynski [Kis76].

The characterization of generators in terms of well-posedness of the abstract Cauchyproblem for classical solutions is contained in [Kre71]. Related results are contained inthe lecture notes of van Casteren [Cas85]. Example 3.1.14 is taken from [Nag86, A-II,Example 1.4].

One may modify Theorem 3.1.12 by requiring a less restrictive existence assumptionbut a stronger uniqueness assumption by considering the Cauchy problem on a boundedinterval [0, τ ]:

C(τ)

{u′(t) = Au(t) (t ∈ [0, τ ]),

u(0) = x.

As before, we call u ∈ C([0, τ ], X) a mild solution if∫ t

0u(s) ds ∈ D(A) and u(t) − x =

A∫ t

0u(s) ds for all t ∈ [0, τ ]. Then the following holds.

Theorem 3.17.1. Let A be a closed operator, τ > 0. The following assertions are equiva-lent:

(i) For all x ∈ X there exists a unique mild solution of C(τ).

(ii) The operator A generates a C0-semigroup.

Other results on the local problem were obtained by Lyubich [Lyu66], Sova [Sov68],Oharu [Oha71] and Sanekata [San75].

3.17. NOTES 231

Section 3.2Integrated semigroups were introduced by Arendt in [Are84], [Are87a]. The systematictreatment based on techniques of Laplace transforms was developed by Arendt [Are87b],Neubrander [Neu88] and Kellermann [Kel86]. Theorem 3.2.13 is due to Arendt, Neubran-der and Schlotterbeck [ANS92] and Lemma 3.2.14 to Arendt, El-Mennaoui and Keyantuo[AEK94]. The theory of integrated semigroups has been developed in many directionsand we refer to the monographs of deLaubenfels [deL94] and Xiao and Liang [XL98] forfurther information. We do not aim to give an account of the many contributions, but wemention here a few extensions of the theory. First, one may consider α-times integratedsemigroups for some α ∈ R+; i.e., the power k in Definition 3.2.1 is replaced by someα ≥ 0 (see [Hie91a]). This leads to sharper regularity results for the solution of the as-sociated Cauchy problem than those which we have described. Moreover, a modificationof the Real Representation Theorem 2.2.1 shows that if a C∞-function r : (0,∞) → Xsatisfies

sup{‖λα+1r(n)(λ)/n!‖ : λ > 0, n ∈ N ∪ {0}

}< ∞

for some α > 0, then there exists a Holder continuous function F of exponent α satisfyingF (0) = 0 such that

r(λ) = λα

∫ ∞

0

e−λtF (t) dt (λ > 0)

(see [Hie91b]). Integrated Volterra equations were considered by Arendt and Kellermann[AK89] and deLaubenfels [deL90]. Second order problems in integrated form have been de-veloped by Kellermann and Hieber [KH89], Arendt and Kellermann [AK89] and Neubran-der [Neu89a].

For integrated solutions of implicit differential equations, see the papers of Arendtand Favini [AF93] and Knuckles and Neubrander [KN94]. Concerning the general theoryof degenerate differential equations in Banach spaces we refer to the monograph by Faviniand Yagi [FY99].

Many years before integrated semigroups were studied, Lions [Lio60] introduced dis-tribution semigroups. They have been further developed by Chazarain [Cha71] and Beals[Bea72] who also give applications to partial differential equations. Further informationcan be found for example in the monograph of Fattorini [Fat83].

In terms of integrated semigroups, Lions’s concept may now be described as follows:Let S be a k-times integrated semigroup on X. Denote by D(R+) the test functions onR with support in R+. For ϕ ∈ D(R+) define the operator T ∈ L(X) by

T (ϕ) := (−1)k∫ ∞

0

ϕ(k)(t)S(t) dt.

Then T : D(R+) → L(X) is a mapping satisfying the semigroup property

T (ϕ ∗ ψ) = T (ϕ)T (ψ).

It turns out that T defined as above is a distribution semigroup of exponential type andthat all distribution semigroups of exponential type as defined by Lions in [Lio60] are ofthis form (see [AK89] for a proof).

More generally, arbitrary distribution semigroups are defined as mappings U :D(R+) → L(X) such that

U(ϕ ∗ ψ) = U(ϕ)U(ψ)


and several other properties are satisfied. To each distribution semigroup one can as-sociate a generator A. These distribution semigroups are equivalent to local integratedsemigroups. More precisely, a closed operator A generates a distribution semigroup U ifand only if there exists k ∈ N such that the integrated Cauchy problem (ACP )k+1 (see(3.20)) is well-posed on a bounded interval; i.e., A generates a local integrated semigroupof order k. The generators of local integrated semigroups can be completely character-ized by spectral properties. See papers of Arendt, El-Mennaoui and Keyantuo [AEK94]or Okazawa and Tanaka [TanO90] for these and related results. These concepts have beenextended further in a series of papers by Lumer [Lum90], [Lum92], [Lum94], [Lum97],Cioranescu and Lumer [CL94] and Lumer and Neubrander [LN99], [LN97], and in themonograph of Melnikova and Filinkov [MF01].

A further concept, namely regularized semigroups, had been developed by Da Prato[DaP66], and was rediscovered by Davies and Pang [DP87]. Regularized semigroups wereextensively studied by deLaubenfels [deL94] and Miyadera and Tanaka [TM92], amongmany others and are now called C-semigroups, in general. Given a bounded operatorC one says, in the language of Laplace transforms, that an operator A generates a C-semigroup if CR(λ,A) is a Laplace transform (see [HHN92]). Thus, by Proposition 3.2.7,the generator of a k-times integrated semigroup is the same as a R(μ,A)k-semigroup,where μ ∈ ρ(A). Note, however, that in contrast to the situation of integrated semi-groups, the generator of a C-semigroup may have an empty resolvent set. Operators withempty resolvent set occur in particular in the context of Petrovskii correct systems ofpartial differential equations. Systems of this type can be treated by the theory of C-semigroups in a very efficient manner (see [HHN92]). Again, we refer to the monographsof deLaubenfels [deL94] and Xiao and Liang [XL98] for further information.

Another concept, which leads to a generalization of generators of bounded semi-groups and groups is the following.

A k-times integrated semigroup S is called tempered if ‖S(t)‖ ≤ ctk (t ≥ 0) forsome c ≥ 0. Arendt and Kellermann [AK89] showed that a densely defined operator Agenerates a k-times integrated tempered semigroup if and only if A generates a smoothdistribution semigroup of order k as introduced by Balabane and Emami-Rad [BE79],[BE85]. Smooth distribution groups allow a spectral calculus similar to the one knownfor bounded groups, as shown by Balabane, Emami-Rad and Jazar [BEJ93] and Jazar[Jaz95].

It was established in [KH89] that perturbation theory for integrated semigroups ismore complicated than for semigroups. We saw in Proposition 3.2.6 that perturbationby a scalar yields a quite complicated formula for the integrated semigroup. In general,generators of integrated semigroups are not even stable under bounded perturbations (incontrast to generators of C0-semigoups, by Corollary 3.5.6). For instance, let A be theunbounded generator of a cosine function (see Section 3.14), A be the operator given

by (3.96) and B be the bounded operator on X × X given by B(xy

)=

(−y0

). Then

A generates a once integrated semigroup, by Theorem 3.14.7. However, the range ofμ − (A + B) is contained in D(A)×X for any μ ∈ C, so ρ(A+ B) is empty and A+ Bdoes not generate a k-times integrated semigroup for any k ∈ N.

One perturbation theorem for integrated semigroups can be found in Corollary3.10.5, and we mention another one here. Using Theorem 3.2.8 and an argument similarto Proposition 3.11.12, Kaiser and Weis [KW03] have shown that A + B generates a

3.17. NOTES 233

(k + 2)-times integrated semigroup if A generates an exponentially bounded k-timesintegrated semigroup S and B : D(A) → X satisfies sup {‖BR(λ,A)‖ : Reλ = ω} < 1for some ω > ω(S).

Finally, we mention that, more generally, instead of Cauchy problems involvinga function and its derivatives at an instant of time, problems with memory may beconsidered. They lead to the theory of Volterra equations. Their solutions are governedby one-parameter families of operators, more general than semigroups or cosine functions.Vector-valued Laplace transforms also play a decisive role in this theory. We refer to themonograph of Pruss [Pru93] for a comprehensive presentation of Volterra equations. Astudy of regularized solutions of Volterra equations was initiated in [Liz00].

Section 3.3The Hille-Yosida theorem (in its general form due to Hille, Yosida, Feller, Miyadera andPhillips) was the starting point of the subsequent development of the theory of semi-groups. The classical approach of Yosida using what is now called the Yosida approx-imation and the classical method of Hille based on the convergence of the exponentialformula T (t)x = limn→∞(I − t

nA)−nx can be found in many textbooks on semigroup

theory (see, for example, those of Clement et al. [CHA87], Davies [Dav80], [Dav07], En-gel and Nagel [EN00], [EN06], Fattorini [Fat83], Goldstein [Gol85], Hille and Phillips[HP57], Kato [Kat66], Nagel et al. [Nag86], Pazy [Paz83] and Yosida [Yos80]). Hille andPhillips already mentioned explicitly on page 364 of [HP57] the problem of how to useWidder’s theorem in the proof of the Hille-Yosida theorem. The approach presented herebased on the real representation theorem for Laplace transforms solves this problem. Ourpresentation follows the lines of Arendt [Are87b].

A different condition which is sufficient for an operator to generate a C0-semigrouphas been given by Gomilko [Gom99] and Shi and Feng [SF00]. This condition has theadvantage of involving only the square of the resolvent and not higher powers.

Theorem 3.17.2. Let A be a densely defined operator on a Banach space X and supposethat there exist K and ω such that σ(A) ⊂ {λ ∈ C : Reλ ≤ ω} and∫ ∞

−∞

∣∣〈R(a+ is, A)2x, x∗〉∣∣ ds ≤ K‖x‖ ‖x∗‖a− ω

whenever a > ω, x ∈ X and x∗ ∈ X∗. Then A generates a C0-semigroup T satisfying‖T (t)‖ ≤ Meωt (t ≥ 0) for some M .

This theorem can be proved by using the Poisson representation of functions in a Hardyspace on a half-plane to establish the conditions of the Hille-Yosida theorem. Alterna-tively, a complex inversion formula can be used to define tT (t)x when x ∈ D(A2) andthen the operators T (t) extend by continuity to a C0-semigroup. The converse of the the-orem is true in Hilbert spaces. However, the derivative operator generates the C0-groupof translations on Lp(R), but it does not satisfy the hypotheses of Theorem 3.17.2 whenp �= 2.

Proposition 3.3.8 appeared first in a paper of Kato [Kat59]. We mention here thatProposition 3.3.8 is not true if merely the Radon-Nikodym property is assumed insteadof reflexivity (see [Are87a]). However, we have the following result [Are87b].

Proposition 3.17.3. Let X be a Banach space with the Radon-Nikodym property. Let Abe an operator on X satisfying the Hille-Yosida condition (3.23). Then A generates asemigroup in the sense of Definition 3.2.5.


The sun-dual T of a C0-semigroup T was already introduced by Hille and Phillips in[HP57]. Dual semigroups have been investigated systematically by Clement, Dieckmann,Gyllenberg, Thieme and van Neerven. For a comprehensive treatment and precise refer-ences, see the monograph by van Neerven [Nee92].

Section 3.4Dissipative operators and the Lumer-Phillips theorem are classical objects in semigrouptheory. Many further results in this direction can be found in the textbooks listed in theNotes of the previous section. It is also possible to characterize generators of positiveC0-semigroups of contractions by a similar notion, namely dispersiveness and a rangecondition. Even more generally, the norm may be replaced by a “half-norm” as studiedby Arendt, Chernoff and Kato [ACK82], Batty and Robinson [BR84] and Nagel et al.[Nag86, Section A-II.2]. If A is a densely defined dissipative operator and x ∈ D(A), thenRe〈Ax, x∗〉 ≤ 0 for all x∗ ∈ dN(x) (instead of merely some x∗ ∈ dN(x) as required by thedefinition). This is easy to see when A generates a contraction semigroup. For denselydefined operators, it was proved in [Bat78]; see also [ACK82, Theorem 2.5].

A very interesting class of contraction semigroups is the class of Ornstein-Uhlenbecksemigroups which arises naturally in the study of stochastic processes. For a thoroughstudy of this class of contraction semigroups, we refer to the book by Lorenzi and Bertoldi[LB07].

The Laplacian with Dirichlet boundary conditions is an easy example of an ellipticoperator in divergence form. Generalizations of the arguments used in Example 3.4.7lead to the theory of quadratic forms. For more information on this topic we refer to thebooks of Kato [Kat66], Dautray and Lions [DL90] and Davies [Dav80], [Dav95], and thesurvey article [Are04].

Section 3.5Hille-Yosida operators were studied by Kato already in [Kat59]. Later, these operatorswere investigated systematically by Sinestrari [Sin85] and Da Prato and Sinestrari [DS87],where Theorem 3.5.2 was first proved. A proof of the Da Prato-Sinestrari Theorem3.5.2 based on the theory of integrated semigroups was given by Kellermann and Hieber[KH89]. Theorem 3.5.2 may also be proved via the theory of non-linear semigroups, asshown by Benilan, Crandall and Pazy [BCP88]. In fact, operators which are not denselydefined are natural in the framework of the Crandall-Liggett-Benilan theorem which cor-responds to the Hille-Yosida theorem for non-linear operators. For a third, very different,approach using “abstract Sobolev towers”, see the work of Nagel and Sinestrari [NS94].The renorming Lemma 3.5.4 can be found in Pazy’s book [Paz83] and the bounded per-turbation Theorem 3.5.5 for Hille-Yosida operators is due to Kellermann and Hieber[KH89].

Section 3.6The Trotter-Kato theorem is a classical result in semigroup theory. It is frequently provedwith the help of the Hille-Yosida theorem (see, for example, [Paz83]). The proof which wegive here is based on the approximation theorem for Laplace transforms from Section 1.7,and is due to Jun Xiao and Liang [XL00] who also proved a general Trotter-Kato approx-imation theorem for integrated semigroups after previous work by Lizama [Liz94] andNicaise [Nic93]. For other variants of the Trotter-Kato theorem, we refer to Bobrowski’ssurvey article [Bob97b].

3.17. NOTES 235

Section 3.7The monographs by Amann [Ama95] and Lunardi [Lun95] are specialized texts on holo-morphic semigroups and parabolic problems. Much further information, in particular onmaximal regularity, interpolation and extrapolation scales of Banach spaces as well asmany applications to non-linear problems can be found there. Our proof of the charac-terization Theorem 3.7.11 uses only Laplace transform theory. Corollary 3.7.14 is takenfrom the lecture notes of Nagel et al. [Nag86]. The perturbation Theorem 3.7.25 is dueto Desch and Schappacher [DS88] with a different proof. If A generates a C0-semigroupon a reflexive space and B : D(A) → X is compact, then the estimate (3.49) holds (see[DS88]) or [EN00, p.179]). However, this is no longer true if X is not reflexive (see [Hes70]for an example). It was also shown in [DS88] (see also [AB06], [AB07], [DSS09]) that ifA+B generates a C0-semigroup for every B ∈ L(D(A), X) of rank 1, then A generatesa holomorphic C0-semigroup.

Section 3.8Fractional powers of sectorial operators are classical objects in semigroup theory, andthey now have a very extensive theory (see [MS01] and [Haa06]). Theorem 3.8.1 andmany further results in this direction can be found in the books of Amann [Ama95],Haase [Haa06] and others listed in the Notes on Section 3.3, and in the original papersof Balakrishnan [Bal60], Komatsu [Kom66] and others. The assumption that A is agenerator is not essential in Theorem 3.8.3. If A is densely defined and B = −A satisfiesthe assumptions of Proposition 3.8.2, then −(−A)1/2 generates a bounded holomorphicC0-semigroup; see for example, the books of Fattorini [Fat83] or Martınez and Sanz[MS01]. For a sectorial operator B and Re z > 0, one may define Bz to be the algebraicinverse of B−z. If A generates a bounded C0-semigroup and 0 < α < 1, then −(−A)α

generates a bounded holomorphic C0-semigroup.

Section 3.9Boundary values of holomorphic C0-semigroups already appeared in the book of Hille andPhillips [HP57]. The method which we use here is based on Laplace transform techniques.Theorem 3.9.4 was proved by Hormander [Hor60], but our approach follows the lines ofArendt, El-Mennaoui and Hieber [AEH97]. Basic information on the Riemann-Liouvillesemigroup and on fractional integration is in [HP57]; the idea of using the transferenceprinciple in this context is in [AEH97]. Results similar to Theorem 3.9.13 were firstobtained by Boyadzhiev and deLaubenfels [BdL93] and later improved by El-Mennaouiin [Elm92]. The proof of the implication (ii) ⇒ (i) of Theorem 3.9.13 follows the lines of[Elm92]. Corollary 3.9.14 is due to Hieber [Hie91a] with a different proof using techniquesfrom Fourier multipliers; see also [Lan68] and [Sjo70]. Further results in this directionsuch as behaviour of the critical exponent k = n|1/2 − 1/p| in Lp(Rn) and L1(Rn),results on more general types of operators such as pseudo-differential operators withsymbol a of the form a(ξ) = |ξ|α for some α > 0 can be found in [Sjo70], [Hie91a]and in Chapter 8. For example, if 1 < p < ∞, then iΔp generates a k-times integratedsemigroup on Lp(Rn) if and only if k ≥ n|1/2− 1/p|. Moreover, iΔ1 and iΔ∞ generate ak-times integrated semigroup on L1(Rn) and L∞(Rn), respectively, if and only if k > n/2.Further applications of boundaries of holomorphic semigroups are given by El-Mennaouiand Keyantuo, to Schrodinger operators in [EK96a] and to the wave equation in [EK96b].In particular, they show the remarkable result that the Schrodinger operator iΔ on Lp(Ω),where Ω = (−π, π)n, with Dirichlet or Neumann boundary conditions, generates a k-timesintegrated group for k > n

2| 12− 1

p|, and that this constant is optimal.

Section 3.10In this section we follow Arendt, Neubrander and Schlotterbeck [ANS92]. Extensions tofractional integrated semigroups have been given by Keyantuo [Key95a]. The perturba-tion result Corollary 3.10.5 is due to Kellermann and Hieber [KH89]. Applications of thesandwich theorem to differential operators have been given in [Are91].

Section 3.11Integrated semigroups were actually introduced first in the context of resolvent positiveoperators in [Are84]. Theorem 3.11.7 and Proposition 3.11.12 as well as Example 3.11.13are from [Are87a]. Theorem 3.11.5 and Example 3.11.6 are taken from [Are87b]. Theorem3.11.9 was proved by Arendt, Chernoff and Kato [ACK82]; it remains valid if X is anordered Banach space with normal cone X+ which has non-empty interior. Theorem3.11.10 is from [Are00], but our proof is different from the original one which is based onthe construction of an intermediate Banach lattice where Theorem 3.5.2 can be applied.Theorem 3.11.8 is due to Arendt and Benilan [AB92a]. It is no longer true on arbitraryBanach lattices: Grabosch and Nagel [GN89] showed that there exists a generator A ofa positive C0-semigroup such that D(A∗) is not a sublattice of X∗. Example 3.11.13also works on Lp(0, 1) if 1 < p < ∞, but not on L1(0, 1). In fact, the following result isdue to W. Desch (unpublished). It can also be seen in the framework of Miyadera–Voigtperturbation theory (see [Voi89]).

Theorem 3.17.4. Let A be the generator of a positive C0-semigroup on a space X of theform L1(Ω, μ). Let B : D(A) → X be linear and positive. If A+ B is resolvent positive,then A+B generates a C0-semigroup.

The analogous result also holds if X is an arbitrary Banach lattice, but A generatesa holomorphic, positive C0-semigroup, as shown by Arendt and Rhandi [AR91].

The theory of resolvent positive operators has been developed further by Thieme[Thi98a] and [Thi98b], where in particular spectral theory and perturbation theory arestudied. One may also study asymptotic behaviour. Assume that A is a resolvent positiveoperator generating a once integrated semigroup S. Then

lim supλ→∞

‖λR(λ,A)‖ < ∞

if and only if

lim supt↓0

1

t‖S(t)‖ < ∞.

Moreover, limt↓0 1tS(t) = I strongly if and only if limλ→∞ λR(λ,A) = I strongly (we refer

to [Are87a, Proposition 6.9]). Of course, an important class of resolvent positive operatorsare generators of positive C0-semigroups. In the spirit of this book, we have concentratedin this section on those results which are related to Laplace transform techniques. Werefer to the book [Nag86] edited by Nagel for the general theory of positive semigroups.

Finally, we should mention that there exist natural examples of operators which arenot resolvent positive even though the resolvent exists and is positive on some interval[λ1, λ2] where −∞ < λ1 < λ2 < ∞. Examples have been given by Greiner, Voigt andWolff [GVW81] and Ulm [Ulm99].

Section 3.12Theorem 3.12.2 and Example 3.12.3 are due to Driouich and El-Mennaoui [DE99]. Ex-tensions to resolvent families associated with Volterra equations are given in [CL03] and[Haa08].

3.17. NOTES 237

The acronym UMD-space stands for “unconditional martingale differences”, reflect-ing the original probabilistic definition of this class of Banach spaces. Burkholder [Bur81]showed that the probabilistic property implies that the Hilbert transform is bounded onLp(R, X) for 1 < p < ∞ and Bourgain [Bou83] established the converse. Every UMD-space is superreflexive (i.e., there is an equivalent norm which is uniformly convex), butthere are uniformly convex spaces which are not UMD-spaces. These and other propertiesof UMD-spaces are discussed in the survey article [Fra86] by Rubio de Francia.

The UMD-property has proved to be very important for the study of maximalregularity of solutions of inhomogeneous Cauchy problems. See, for example, [KW04],[Are04], [AB04].

Section 3.13A C0-semigroup T is said to be eventually norm-continuous if T is norm-continuous on(τ,∞) for some τ ≥ 0. It was shown in the book of Hille and Phillips [HP57] that if T iseventually norm-continuous, then for any real a, {λ ∈ σ(A) : Reλ ≥ a} is compact.

Theorem 3.13.2 was first proved by You [You92], but the simple proof given hereis due to El-Mennaoui and Engel [EE94]. With minor modifications, the proof showsthat if X is a Hilbert space and ‖T (τ)R(ω + is, A)n‖ → 0 as |s| → ∞ for some n ≥ 1and τ ≥ 0 then T is norm-continuous for t > τ . Other characterizations of immediatelyand eventually norm-continuous semigroups on Hilbert space have been given by Blascoand Martinez [BM96] and Blake [Bla01]. The latter characterization does not explicitlyinvolve any decay of the resolvent of A (see also [Ile07]).

Goersmeyer and Weis [GW99] have proved that if T is a positive semigroup on anLp-space and ‖R(ω+is,A)‖ → 0 as |s| → ∞, then T is norm-continuous for t > 0. On theother hand, Matrai [Mat08], and Chill and Tomilov [CT09], have constructed examplesshowing that Theorem 3.13.2 does not extend to all Banach spaces.

Section 3.14For further literature on cosine functions we refer to the monographs of Fattorini [Fat85]and Goldstein [Gol85] and the survey article of Bobrowski [Bob97b], and the referencesgiven there. Kellermann and Hieber [KH89] proved the relation between cosine functionsand integrated semigroups (Theorem 3.14.7). Uniqueness of the phase space (in Theorem3.14.11) is due to Kisynski [Kis72]. Further information about the phase space may befound in the Appendix of [Haa07a].

Corollary 3.14.13 seems to be a natural limit for perturbation results of that type.It is shown in [AB06, Theorem 2.2] (and stated in [AB07, Theorem 5.9]) that if A + Bgenerates a cosine function for every bounded operator B : D((ω −A)γ) → X of rank 1,where γ > 1/2, then A is bounded on X.

More general versions of well-posedness of the Cauchy problem of second order(k-times integrated cosine functions) are considered in [AK89] and, using a differentapproach, by Takenaka and Okazawa [TakO90]. Higher order problems are covered in[XL98].

Let X be a Hilbert space with inner product (·|·). The following result characterizesgenerators A of cosine functions X, up to choice of inner product, by a condition on thenumerical range W (A) := {(Ax|x) : x ∈ X, ‖x‖ = 1} related to the spectral conditionof Proposition 3.14.18. Here, part a) is due to Crouzeix [Cro04] who extended it to aremarkable piece of work on numerical range and functional calculus for matrices andoperators on Hilbert space [Cro07], [Cro08]. Part b) is due to Haase [Haa06, Corollary7.4.6].

Theorem 3.17.5. Let A be a densely defined operator on a Hilbert space X.

a) If W (A) is contained in the parabola Πω := {ξ + iη : η ∈ R, ξ ≤ ω2 − η2/4ω2} forsome ω > 0 and there is at least one point in ρ(A) ∩ (C \Πω), then A generates acosine function.

b) If A generates a cosine function, then there is an equivalent scalar product on Xwith respect to which the numerical range of A is contained in Πω for some ω > 0.

Section 3.15Sine functions were introduced by Arendt and Kellermann [AK89] where in particularthe perturbation result Theorem 3.15.6 is proved. Moreover, the following holds.

Theorem 3.17.6. Let A be a densely defined operator which generates a sine function Sinsatisfying lim supt↓0

1t‖Sin(t)‖ < ∞. Then A generates a C0-semigroup T . Moreover, T

has a holomorphic extension to the right half-plane.

The generation theorem for cosine functions (Theorem 3.15.3) is independently dueto Da Prato and Giusti [DG67] and Sova [Sov66].

Section 3.16Theorem 3.16.7 on square root reduction is due to Fattorini [Fat69]. At that timeUMD-spaces had not been investigated and Fattorini formulated the result for Lp-spaces(1 < p < ∞). But he clearly pointed out that boundedness of the Hilbert transform is thecrucial condition. Haase has given a different approach to Fattorini’s theorem via func-tional calculus [Haa07a], and he has shown that i(−A)1/2 generates a bounded C0-groupif A generates a bounded cosine function on a UMD-space [Haa09].

Part II

Tauberian Theorems andCauchy Problems

Again we consider the guide-line of this book, the characteristic equation

u(λ) = R(λ,A)x,

where u is a solution of the Cauchy problem

(CP )

{u′(t) = Au(t) (t ≥ 0),

u(0) = x,

and R(λ,A) is the resolvent of the operator A. The aim of this part is to studythe asymptotic behaviour of the solution u(t) as t→∞.

In applications, one typically has some information about the spectral be-haviour of A. This means that, in many cases, we know the Laplace transformu(λ) for Reλ > 0. This part is devoted to the problem of determining the relationbetween the asymptotic behaviour of u(t) as t → ∞ and that of u(λ) as λ ↓ 0.It contains two chapters. The first, Chapter 4, is devoted to the investigation ofarbitrary functions; in Chapter 5 the asymptotic behaviour of solutions of (CP )is investigated.

The Abelian theorems given in Chapter 4 show that convergence of u(t) ast → ∞ implies convergence of λu(λ) as λ ↓ 0. Much more interesting for the ap-plications which we have in mind, are Tauberian theorems, which give statementsabout the converse implication. However, they do need additional hypotheses, so-called Tauberian conditions.

Of particular importance are complex Tauberian theorems where the assump-tions involve the behaviour of u(λ) for λ close to the imaginary axis. Indeed, as-sumptions of this kind are directly related to spectral properties of the operatorvia the characteristic equation. We not only consider convergence of functions butwe also investigate their periodic behaviour. In fact, Chapter 4 contains an in-troduction to the theory of almost periodic and asymptotically almost periodicfunctions. The main results show that under suitable assumptions, countability ofthe spectrum implies asymptotic almost periodicity.

In the second chapter of this part, Chapter 5, the results of Chapter 4 areapplied to solutions of the Cauchy problem, and also a variety of new and specialresults are obtained. The most important framework is given by generators of C0-semigroups, but we also consider individual solutions in more general situations.This will be useful, for example, for the investigation of the heat equation withinhomogeneous boundary conditions which we give in Chapter 6.

In the first three sections of Chapter 5 the diverse abscissas of Laplace trans-form of the semigroup are related to spectral properties. Indeed, it turns out thatthe relation is not simple, in general. However there are very satisfying results inspecial cases; for example on Hilbert spaces (Section 5.2) or for positive semigroups(Section 5.3); both are important for applications.

The complex Tauberian theorems from Chapter 4 are applied to solutionsof (CP ) in many different situations. Asymptotically almost periodic solutions

241Part II

242

lead to splitting theorems which are considered in Section 5.4. Countability of thespectrum is a criterion for asymptotic periodicity. This is shown in Section 5.5 byexploiting the results of Chapter 4.

Typically, one has to assume boundedness of the solution for these appli-cations. This may be a difficult assumption to verify when the inhomogeneousCauchy problem is considered, but we show that this is automatic if the underly-ing semigroup is holomorphic and a certain spectral condition of non-resonance issatisfied. This last topic of Chapter 5 concludes our investigation of the interestinginterplay of spectrum and asymptotics.

Part II

Chapter 4

Asymptotics of LaplaceTransforms

Frequently, convergence of a function f : R+ → X for t→∞ implies convergenceof an average of this function. Assertions of this type are called Abelian theorems.A theorem is called Tauberian if, conversely, convergence of the function is deducedfrom the convergence of an average.

The Abelian theorems which we present in Section 4.1 are quite easy toprove. However, the Tauberian theorems corresponding to their converse versionsare much more delicate. They need additional hypotheses, so-called Tauberianconditions. Section 4.2 is devoted to Tauberian conditions of real type (for example,boundedness or positivity of f).

Interesting applications of these Abelian and real Tauberian theorems tosemigroups are given in Section 4.3 where mean ergodicity is discussed. This in-terrupts the general theme of this chapter, but the results will be useful in thesubsequent sections where the notion of mean ergodicity is needed.

In Section 4.4 a complex Tauberian theorem is proved with the help of anelegant contour argument. Here we make assumptions on holomorphic extensionsof f on the imaginary axis. We restrict ourselves to the case of one singularityin order to keep the ideas more transparent, but this case is already of specialinterest. For example, an immediate consequence is Gelfand’s theorem, sayingthat a bounded C0-group is trivial (i.e., the identity) if and only if the spectrumof its generator is reduced to {0}.

One interesting type of asymptotic behaviour for large time is almost pe-riodicity. The concept is introduced in Section 4.5 where elementary propertiesare proved for functions on R. In Section 4.6, Loomis’s theorem and its vector-valued version are proved by an elegant quotient method which allows one to ap-ply Gelfand’s theorem. The basic notion is the Carleman spectrum for a boundedmeasurable function defined on the line, and Loomis’s theorem states that any


244 4. ASYMPTOTICS OF LAPLACE TRANSFORMS

bounded uniformly continuous function f : R → C with countable spectrum isalmost periodic. There is one vector-valued version of Loomis’s theorem, whichis valid for every Banach space and involves an ergodicity condition. The othervector-valued version holds without further assumptions on the function but ageometric condition on the Banach space is needed.

Functions on the half-line are considered in Section 4.7. The naturally as-sociated “half-line” spectrum is discussed and the main theorem is a complexTauberian theorem for functions with countable spectrum, which is proved by thesame technique as we proved Loomis’s theorem.

In Section 4.8 we come back to functions defined on the line showing that theCarleman spectrum (defined by holomorphy) and the Beurling spectrum (definedby the Fourier transform) coincide. This allows us to prove a very general complexTauberian theorem for functions on the half-line in Section 4.9. Here we use Fouriertransform methods which allow one to reduce the problem to an application ofLoomis’s theorem (in the scalar case).

The structure of this chapter needs some explanation in view of our mainpurpose, namely the proof of a complex Tauberian theorem on the half-line. Wepresent three different methods by which we prove the result in increasing gen-erality (Theorems 4.4.8, 4.7.7 and 4.9.7); namely, the contour method, the quo-tient method and the Fourier method. If Loomis’s theorem (Corollary 4.6.4) isaccepted, the Fourier method of Section 4.9 is the most general. The contourmethod, presented as the first approach in Section 4.4, is the most elementary. Itgives us Gelfand’s theorem (Corollary 4.4.12) and other interesting consequences.The quotient method uses Gelfand’s theorem. It gives us Loomis’s theorem onthe line (Section 4.6), and in an elegant way the fairly general complex TauberianTheorem 4.7.7 on the half-line.

4.1 Abelian Theorems

Throughout this section, f denotes a function in L1loc(R+, X), where X is a Banach

space and R+ := [0,∞) is the right half-line.We consider the following three types of averages.

Definition 4.1.1. Let f∞ ∈ X. We say that

a) f(t) converges to f∞ in the sense of Abel as t → ∞ if abs(f) ≤ 0 and

A- limt→∞ f(t) := limλ↓0 λf(λ) = f∞;

b) f(t) is B-convergent to f∞ as t→∞, or simply write B- limt→∞ f(t) = f∞,

if for every δ > 0, limt→∞ 1δ

∫ t+δ

tf(s) ds = f∞;

c) f(t) converges to f∞ in the sense of Cesaro as t→∞ if

C- limt→∞ f(t) := lim

t→∞1

t

∫ t

0

f(s) ds = f∞.

4.1. ABELIAN THEOREMS 245

It will be convenient to consider the antiderivative of a function f given byF (t) :=

∫ t

0f(s) ds. Then the following Abelian theorem holds.

Theorem 4.1.2. Let f∞, F∞ ∈ X.

a) If limt→∞ f(t) = f∞, then B- limt→∞ f(t) = f∞.

b) If B- limt→∞ f(t) = f∞, then C- limt→∞ f(t) = f∞.

c) If C- limt→∞ f(t) = f∞, then A- limt→∞ f(t) = f∞.

d) If limt→∞ F (t) = F∞, then limλ↓0 f(λ) = F∞.

Proof. a) is obvious.b) Replacing f by f − f∞ we can assume that f∞ = 0. Then (taking δ = 1)

we have by assumption that limt→∞∫ t+1

tf(s) ds = 0. Let ε > 0. There exists

t0 ≥ 0 such that∥∥∥∫ t+1

tf(s) ds

∥∥∥ ≤ ε2 for all t ≥ t0. Moreover, there exists t1 ≥ t0

such that 1t

∫ t0+1

0‖ f(s) ‖ ds ≤ ε

2 for all t ≥ t1. Let t ≥ t1. Take n ∈ N such thatt− n ≤ t0 ≤ t− n+ 1. Then∥∥∥∥ 1

t

∫ t

0

f(s) ds

∥∥∥∥ ≤∥∥∥∥ 1

t

∫ t−n+1

0

f(s) ds

∥∥∥∥+

∥∥∥∥ 1

t

∫ t

t−n+1

f(s) ds

∥∥∥∥≤ 1

t

∫ t0+1

0

‖ f(s) ‖ ds+n−1∑k=1

∥∥∥∥∥ 1

t

∫ t−n+k+1

t−n+k

f(s) ds

∥∥∥∥∥≤ ε

2+

n− 1

t

ε

2≤ ε.

This proves the claim.c) Assume that C- limt→∞ f(t) = f∞. Then F (t) = O(t) as t → ∞, so f(λ)

exists for Reλ > 0 by Theorem 1.4.3.Let ε > 0. There exists τ > 0 such that ‖ 1

tF (t) − f∞ ‖ ≤ ε (t ≥ τ). Hence

by integration by parts,

lim supλ↓0

‖λf(λ)− f∞ ‖

= lim supλ↓0

∥∥∥∥λ ∫ ∞

0

e−λt (f(t)− f∞) dt

∥∥∥∥= lim sup

λ↓0

∥∥∥∥λ2

∫ ∞

0

e−λt (F (t)− tf∞) dt


λ↓0

∥∥∥∥λ2

∫ ∞

0

te−λt

(1

tF (t)− f∞

)dt


λ↓0λ2

∫ τ

0

te−λt

∥∥∥∥ 1

tF (t)− f∞

∥∥∥∥ dt+ limλ↓0

ελ2

∫ ∞

τ

te−λtdt

= limλ↓0

ελ2

∫ ∞

0

te−λtdt = ε.

d) Since f(λ) = F ′(λ) = λF (λ) (λ > 0) if abs(f) ≤ 0, this follows from theprevious parts since convergence implies A-convergence.

Concerning the behaviour for t→ 0, we note:

Proposition 4.1.3. Let x ∈ X. Assume that abs(f) <∞. Then

lim supλ→∞

‖λf(λ)− x ‖ ≤ lim supt↓0

‖ f(t)− x ‖.

In particular, limt↓0 f(t) = x implies limλ→∞ λf(λ) = x.

Proof. Since abs(f) <∞, there exist M,ω > 0 such that ‖F (t) ‖ ≤Meωt (t ≥ 0).Let b > 0, λ > ω. Then∥∥∥∥∫ ∞

b

λe−λtf(t) dt

∥∥∥∥ =

∥∥∥∥−λe−λbF (b) + λ2

∫ ∞

b

e−λtF (t) dt

∥∥∥∥≤ λe−λb‖F (b) ‖+ λ2 M

λ− ωe−(λ−ω)b.

Hence,

lim supλ→∞

‖λf(λ)− x ‖

= lim supλ→∞

∥∥∥∥ ∫ ∞

0

λe−λt (f(t)− x) dt

∥∥∥∥≤ lim

λ→∞

{sup

0≤t≤b‖ f(t)− x ‖+

∥∥∥∥∫ ∞

b

λe−λtf(t) dt

∥∥∥∥+

∫ ∞

b

λe−λt dt ‖x ‖}

= sup0≤t≤b

‖ f(t)− x ‖.

Here, we used throughout that λ∫∞0

e−λt dt = 1 (λ > 0). More generally, forα > −1 one has

λα+1

Γ(α+ 1)

∫ ∞

0

e−λttα dt = 1 (λ > 0).

One shows the following in a similar way as above.

Proposition 4.1.4. Let f∞ ∈ X, f0 ∈ X,α > −1.a) If C- limt→∞ f(t) = f∞, then limλ↓0 λα+1

Γ(α+1)

∫∞0

e−λttαf(t) dt = f∞.

b) If limt↓0 f(t) = f0, then limλ→∞ λα+1

Γ(α+1)

∫∞0

e−λttαf(t) dt = f0.

This allows one to compute the derivative of f at 0 from its Laplace transform:

Corollary 4.1.5. Let f ∈ L1loc(R+, X) such that abs(f) < ∞. Assume that

f ∈ C1 ([0, τ ], X) for some τ > 0. Then

limλ→∞

(λ2f(λ)− λf(0)

)= f ′(0).

4.2. REAL TAUBERIAN THEOREMS 247

Proof. Let α = 1. Applying Proposition 4.1.4 b) to g(t) := 1t (f(t)−f(0)), we obtain

that limλ→∞(λ2f(λ)−λf(0)) = limλ→∞ λ2∫∞0

e−λtt 1t(f(t)−f(0)) dt = f ′(0).

The following theorem gave the name to this type of result, and to this section.We deduce it from the corresponding result for Laplace transforms. Usually, oneuses Abel’s “summation by parts” for power series.

Theorem 4.1.6 (Abel’s Continuity Theorem). Let p(z) :=∑∞

n=0 anzn be a power

series, where an ∈ X. If∑∞

n=0 an = b, then limz↑1 p(z) = b.

Proof. Let f(t) = an if t ∈ [n, n + 1). Then limt→∞ F (t) = b. Hence,

abs(f) ≤ 0 and f(λ) = 1−e−λ

λ

∑∞n=0 an(e

−λ)n for all λ > 0. By Theorem 4.1.2d),

limz↑1

p(z) = limz↑1

∞∑n=0

anzn = lim

λ↓0

∞∑n=0

an(e−λ)n

= limλ↓0

λ

1− e−λf(λ) = lim

λ↓0f(λ)

= limt→∞F (t) = b.

4.2 Real Tauberian Theorems

Let f ∈ L1loc(R+, X). In the preceding section we established four notions of con-

vergence for f(t) (as t → ∞) of increasing generality: convergence in the usualsense, B-convergence, convergence in the sense of Cesaro, and convergence in thesense of Abel. None of the implications established in Theorem 4.1.2 is reversible.An additional condition, which allows one to reverse one of the implications iscalled a Tauberian condition, and the corresponding result a Tauberian theorem.Of particular interest are those results which allow one to deduce convergence inthe usual sense from convergence of a mean. In this section we consider assump-tions on f(t), i.e. real Tauberian conditions, in contrast to conditions on f(λ),known as complex Tauberian conditions, which will be the subject of the nextsection.

We first consider the reverse implication of Theorem 4.1.2 a).

A. Conditions under which B- limt→∞ f(t) = f∞ implies limt→∞ f(t) = f∞.

Let f ∈ L1loc(R+, X). In general, if B- limt→∞ f(t) = f∞, one cannot conclude

that limt→∞ f(t) = f∞ even if f is bounded. A simple example is the functionf =

∑∞n=1 χ[n,n+1/n] which has B-limit 0 (as t→∞) but does not converge.

Definition 4.2.1. A function g : R+ → X is called slowly oscillating if for all ε > 0there exist t0 ≥ 0, δ > 0 such that ‖ g(t)−g(s) ‖ ≤ ε whenever s, t ≥ t0, |t−s| ≤ δ.

A bounded continuous function is slowly oscillating if and only if it is uni-formly continuous. Indeed, one may think of Definition 4.2.1 as expressing thatthe function g is uniformly continuous at infinity. This is made more precise bythe following characterization.

Proposition 4.2.2. A function g : R+ → X is slowly oscillating if and only ifg = g0 + g1 where g1 : R+ → X is uniformly continuous and g0 : R→ X satisfieslimt→∞ g0(t) = 0.

Proof. It is clear that the condition is sufficient.In order to prove the converse, let g be slowly oscillating. Then we find a

decreasing sequence (δn)n∈N ⊂ (0,∞) such that limn→∞ δn = 0 and an increasingsequence (tn)n∈N ⊂ [0,∞) such that limn→∞ tn = ∞ and ‖ f(t + δ) − f(t) ‖ ≤ 1

nwhenever t ≥ tn, 0 < δ ≤ δn. For each n ∈ N, choose a partition of the in-terval [tn, tn+1] so that every subinterval has length smaller than δn. Define g1to be linear on each subinterval and equal to f at the endpoints of the subin-tervals. Extend g1 continuously to [0, t1]. Then g1 is uniformly continuous andlimt→∞ (f(t)− g1(t)) = 0. We put g0 := f − g1.

The following Tauberian theorem is easy to prove.

Theorem 4.2.3. Let f ∈ L1loc(R+, X) be slowly oscillating and f∞ ∈ X. If

B- limt→∞ f(t) = f∞, then lim

t→∞ f(t) = f∞.

Proof. For δ > 0 one has

lim supt→∞

‖ f(t)− f∞ ‖ ≤ lim supt→∞

∥∥∥∥∥ f(t)− 1

δ

∫ t+δ

t

f(s) ds

∥∥∥∥∥= lim sup

t→∞

∥∥∥∥∥ 1

δ

∫ t+δ

t

(f(t)− f(s)) ds

∥∥∥∥∥≤ lim sup

t→∞

(sup

t≤s≤t+δ‖ f(t)− f(s) ‖

)The last expression tends to 0 as δ ↓ 0 since f is slowly oscillating.

Next, we consider:

B. Conditions under which C- limt→∞ f(t) = f∞ implies limt→∞ f(t) = f∞.

First, we notice that, in general, convergence in the sense of Cesaro does notimply B-convergence even if f is bounded. For example, let f(t) := sin t. Then

C- limt→∞ f(t) = limt→∞ 1t

∫ t

0sin s ds = limt→∞ 1−cos t

t = 0. But 1δ

∫ t+δ

tsin s ds

= 1δ (cos(t+ δ)− cos t), and the B-limit does not exist.

A function f : R+ → X is called feebly oscillating (when t→∞) if

limt,s→∞t/s→1

‖ f(t)− f(s) ‖ = 0.

It is clear that every feebly oscillating function is slowly oscillating.

Example 4.2.4. Assume that there exist τ > 0,M ≥ 0 such that ‖tf(t)‖ ≤ M

(t ≥ τ). Then F (t) :=∫ t

0f(s) ds is feebly oscillating. In fact, ‖F (t) − F (s) ‖ ≤

M∫ t

sdrr = M log

(ts

)for t ≥ s ≥ τ .

Theorem 4.2.5. Let f ∈ L1loc(R+, X) and f∞ ∈ X. Assume that f is feebly oscil-

lating. If C- limt→∞ f(t) = f∞, then limt→∞ f(t) = f∞.

Proof. Let ε > 0. There exist δ > 0, t0 > 0 such that ‖ f(s)− f(t) ‖ < ε whenevert ≥ t0, s ∈ [t− δt, t+ δt]. Hence,∥∥∥∥∥ f(t)− 1

2δt

∫ t(1+δ)

t(1−δ)

f(s) ds

∥∥∥∥∥ =

∥∥∥∥∥ 1

2δt

∫ t(1+δ)

t(1−δ)

(f(t)− f(s)) ds

∥∥∥∥∥ ≤ ε if t ≥ t0.

The assumption implies that

1

2δt

∫ t(1+δ)

t(1−δ)

f(s) ds

=1 + δ

2δ

1

t(1 + δ)

∫ t(1+δ)

0

f(s) ds− 1− δ

2δ

1

t(1− δ)

∫ t(1−δ)

0

f(s) ds

→ f∞ as t→∞.

It follows that limt→∞ f(t) = f∞.

In the next theorem the Tauberian condition is of order-theoretic nature.

Theorem 4.2.6. Let X be an ordered Banach space with normal cone. Furthermorelet f : R+ → X+ be a function such that for some k ∈ N, t �→ tkf(t) is increasing.Then

C- limt→∞ f(t) = f∞ implies lim

t→∞ f(t) = f∞.

Proof. Since the cone is normal, there exists c ≥ 0 such that u ≤ x ≤ v implies‖x ‖ ≤ c(‖ u ‖+‖ v ‖) for u, x, v ∈ X (see Appendix C). Let ρ > 1. The assumptionimplies that

1

(ρ− 1)t

∫ ρt

t

f(s) ds =ρ

ρ− 1

1

ρt

∫ ρt

0

f(s) ds− 1

ρ− 1

1

t

∫ t

0

f(s) ds

→ f∞ (t→∞)

and, similarly, ρ(ρ−1)t

∫ t

t/ρf(s) ds→ f∞ (t→∞).

Since tkf(t) is increasing, one has∫ ρt

t

f(s) ds =

∫ ρt

t

skf(s)s−kds

≥ tkf(t)(ρt)−k(ρt− t) = f(t)tρ−k(ρ− 1)

and∫ t

t/ρf(s) ds ≤ f(t)tρk−1(ρ− 1). Thus,

ρ−k ρ

(ρ− 1)t

∫ t

t/ρ

f(s) ds− f∞ ≤ f(t)− f∞ ≤ ρk1

(ρ− 1)t

∫ ρt

t

f(s) ds− f∞.

Consequently,

lim supt→∞

‖ f(t)− f∞ ‖ ≤ c lim supt→∞

{∥∥∥∥∥ ρ−k ρ

(ρ− 1)t

∫ t

t/ρ

f(s) ds− f∞

∥∥∥∥∥+

∥∥∥∥ ρk 1

(ρ− 1)t

∫ ρt

t

f(s) ds− f∞

∥∥∥∥}≤ c ‖f∞‖ (1− ρ−k + ρk − 1).

Letting ρ ↓ 1 yields the claim.

C. Conditions under which A- limt→∞ f(t) = f∞ implies C- limt→∞ f(t) = f∞.

In general, Abel convergence does not imply Cesaro convergence. For example, letf(t) := t sin t (t ≥ 0). Then it is easy to see that

A- limt→∞ f(t) = lim

λ↓02λ2(1 + λ2)−2 = 0

but 1t

∫ t

0f(s) ds = sin t

t − cos t does not converge. Thus additional hypotheses areneeded.

First, we consider a boundedness condition on f .

Theorem 4.2.7. Let f∞ ∈ X. Assume that supt≥τ ‖f(t)‖ < ∞ for some τ ≥ 0. IfA- limt→∞ f(t) = f∞, then C- lim∞ f(t) = f∞.

Proof. a) We first assume that τ = 0. For β > 0 let eβ(t) := e−βt (t > 0). Thenspan{eβ : β > 0} is dense in L1(R+) (by Lemma 1.7.1). By hypothesis,

limα→∞

∫ ∞

0

e−sf(αs) ds = limλ↓0

∫ ∞

0

e−sf( s

λ

)ds

= limλ↓0

λ

∫ ∞

0

e−λsf(s) ds = f∞.

Hence,

limα→∞〈βeβ , f(α ·)〉 = lim

α→∞β

∫ ∞

0

e−βsf(αs) ds

= limα→∞

∫ ∞

0

e−sf

(α

βs

)ds

= f∞ = 〈βeβ , f∞〉 for all β > 0.

It follows that limα→∞〈h, f(α ·)〉 = f∞∫∞0

h(t) dt for all h ∈ L1(R+). Lettingh = χ(0,1], we obtain

limα→∞

1

α

∫ α

0

f(s) ds = limα→∞

∫ 1

0

f(αs) ds = limα→∞〈h, f(α ·)〉 = f∞.

b) If τ > 0 the result follows by applying a) to g(t) = f(t+ τ).

Next, we show that positivity is a Tauberian condition.

Theorem 4.2.8 (Karamata). Let X be an ordered Banach space with normal coneand let β > −1. Let f∞ ∈ X and f ∈ L1

loc(R+, X) such that f(t) ≥ 0 (t ∈ R+).Suppose that

a)∫∞0

e−λttβf(t) dt exists for all λ > 0; and

b) limλ↓0 λβ+1

Γ(β+1)

∫∞0

e−λttβf(t) dt = f∞.

Then C- limt→∞ f(t) = f∞.

This is a converse of Proposition 4.1.4 a). In particular, it follows from The-orem 4.2.8 that for positive f ∈ L1

loc(R+, X),

A- limt→∞ f(t) = f∞ implies C- lim

t→∞ f(t) = f∞.

Proof. It follows from the assumption that for n ∈ N0,

limλ↓0

λβ+1

Γ(β + 1)

∫ ∞

0

e−λt(e−λt)nf(t)tβdt

=1

(n+ 1)β+1limλ↓0

(λ(n+ 1))β+1

Γ(β + 1)

∫ ∞

0

e−λ(n+1)tf(t)tβdt

=1

(n+ 1)β+1f∞ = f∞

1

Γ(β + 1)

∫ ∞

0

tβe−t(e−t)ndt.

Consequently, for every polynomial p,

limλ↓0

λβ+1

Γ(β + 1)

∫ ∞

0

e−λtp(e−λt)tβf(t) dt = f∞1

Γ(β + 1)

∫ ∞

0

tβe−tp(e−t) dt.

Let q : [0, 1]→ R+ be given by q(x) := 0 if x < e−1, q(x) := x−1 if x ≥ e−1.Let δ > 0 and let q1, q2 be continuous functions such that 0 ≤ q1 ≤ q ≤ q2 on [0, 1]and

a) q1(x) = q(x) if x ≤ e−1 or x > e−1 + δ,

b) q2(x) = q(x) if x ≤ e−1 − δ or x > e−1,

c) sup0≤x≤1(q2(x)− q1(x)) ≤ e.

Now, let ε > 0. Choose δ > 0 such that eΓ(β+1)

∫ t2t1

tβe−tdt < ε, where e−t2 = e−1−δand e−t1 = e−1 + δ. By the Stone-Weierstrass theorem, there exist polynomials p1and p2 such that

q1 − ε ≤ p1 ≤ q1 ≤ q ≤ q2 ≤ p2 ≤ q2 + ε

on [0, 1]. Define kj : R+ → X (j = 1, 2) and h : (0,∞)→ X by:

kj(λ) :=λβ+1

Γ(β + 1)

∫ ∞

0

tβe−λtpj(e−λt)f(t) dt (λ > 0),

kj(0) :=f∞

Γ(β + 1)

∫ ∞

0

e−tpj(e−t)tβdt,

h(λ) :=λβ+1

Γ(β + 1)

∫ ∞

0

e−λtq(e−λt)tβf(t) dt (λ > 0).

Since 0 ≤ p2 − p1 ≤ q2 − q1 + 2ε, one has

0 ≤ k2(0)− k1(0) ≤ f∞Γ(β + 1)

e

∫ t2

t1

e−ttβdt+ 2εf∞

Γ(β + 1)

∫ ∞

0

e−ttβdt

≤ 3 εf∞.

The first part of the proof shows that limλ↓0 kj(λ) = kj(0). Let λ0 > 0 such that

‖ kj(λ)− kj(0) ‖ < ε for 0 < λ ≤ λ0, j = 1, 2. Let 0 < λ, λ ≤ λ0. Then

k1(λ)− k2(λ) ≤ h(λ)− h(λ) ≤ k2(λ)− k1(λ).

Hence, for a fixed constant c > 0,

‖h(λ)− h(λ) ‖ ≤ c{‖ k2(λ)− k1(λ) ‖+ ‖ k1(λ)− k2(λ) ‖

}≤ c

{‖ k2(λ)− k2(0) ‖+ ‖ k2(0)− k1(0) ‖+ ‖ k1(0)− k1(λ) ‖

+ ‖ k1(λ)− k1(0) ‖+ ‖ k1(0)− k2(0) ‖+ ‖ k2(0)− k2(λ) ‖}

≤ 4 c ε+ 2 · 3 c ε‖f∞‖ whenever 0 < λ, λ ≤ λ0.

This shows that h(λ) converges as λ ↓ 0. Since h(λ) = λβ+1

Γ(β+1)

∫ 1/λ

0sβf(s) ds, it

follows that limt→∞ t−(β+1)∫ t

0sβf(s) ds =: g exists. Hence,

f∞ = limλ↓0

λβ+1

Γ(β + 1)

∫ ∞

0

e−λtf(t)tβdt

= limλ↓0

(β + 1)λβ+2

Γ(β + 2)

∫ ∞

0

e−λttβ+1 1

tβ+1

∫ t

0

f(s)sβ ds dt

= (β + 1)g,

by the Abelian theorem mentioned above (Proposition 4.1.4 a)). Thus, g = f∞β+1 .

We have proved that t−(β+1)∫ t

1sβf(s) ds → f∞

β+1 as t → ∞. Since convergenceimplies Cesaro-convergence, we also have

limt→∞

1

t

∫ t

1

1

sβ+1

∫ s

1

rβf(r) dr ds =f∞

β + 1.

Hence, integration by parts yields

1

t

∫ t

1

f(s) ds =1

t

∫ t

1

1

sβd

ds

∫ s

1

rβf(r) dr ds

=1

tβ+1

∫ t

1

rβf(r) dr +β

t

∫ t

1

s−(β+1)

∫ s

1

rβf(r) dr ds

→ f∞β + 1

+ βf∞

β + 1= f∞.

D. Conditions under which limλ↓0 f(λ) = F∞ implies limt→∞ F (t) = F∞.

The following theorem is due to Hardy and Littlewood in the scalar-valued case.Let f ∈ L1

loc(R+, X) and let F (t) :=∫ t

0f(s) ds (t ≥ 0).

Theorem 4.2.9 (Hardy-Littlewood). Assume that M := supt≥τ t ‖ f(t) ‖ < ∞ for

some τ ≥ 0 and let F∞ ∈ X. If limλ↓0 f(λ) = F∞, then limt→∞ F (t) = F∞.

Proof. Replacing f by f(·+ τ) we can assume that τ = 0. For t > 0 we have∥∥∥F (t)− f(1t

) ∥∥∥ =

∥∥∥∥∫ t

0

f(s)(1− e−s/t) ds−∫ ∞

t

f(s)e−s/t ds

∥∥∥∥≤ M

(sup

0<s<tt1− e−s/t

s+

∫ ∞

1

e−r dr

r

)≤ M

(sup

0<r≤1

1− e−r

r+

∫ ∞

1

e−r dr

r

)<∞.

Since limt→∞ f( 1t) = F∞, it follows that F is bounded. But A- limt→∞ F (t) =

limλ↓0 f(λ) = F∞. It follows from Theorem 4.2.7 that C- limt→∞ F (t) = F∞. By


Example 4.2.4, F is feebly oscillating. Now Theorem 4.2.5 implies that limt→∞ F (t)= F∞.

The following classical example is a beautiful application of the Hardy-Littlewood theorem.

Example 4.2.10. Let f(t) := sin tt. Then f(λ) = − arctanλ+ π

2. In fact,

d

dλf(λ) =

∫ ∞

0

e−λt(−t)f(t) dt = −∫ ∞

0

e−λt sin t dt

= −∫ ∞

0

e−λt 1

2i(eit − e−it) dt

= − 1

2i

(1

λ− i− 1

λ+ i

)=

−11 + λ2

= − d

dλarctanλ.

Since limλ→∞ f(λ) = 0, the claim follows. Now by Theorem 4.2.9, it follows that∫∞0

sin tt

dt := limt→∞∫ t

0sin ss

ds = limλ↓0 f(λ) = π2.

If f is positive, then f(λ) is decreasing. So in the real-valued case it is clearthat

limλ↓0

f(λ) = supλ>0

f(λ) = supλ>0

supt>0

∫ t

0

e−λsf(s) ds

= supt>0

supλ>0

∫ t

0

e−λsf(s) ds = supt>0

F (t) = limt→∞F (t).

If X is an ordered Banach space we cannot argue like this (unless the norm isorder continuous). However, using the preceding results we obtain:

Theorem 4.2.11. Let X be an ordered Banach space with normal cone. Let f ∈L1loc(R+, X) and assume that f(t) ≥ 0 (t ≥ 0) and abs(f) ≤ 0. If limλ↓0 f(λ) = F∞

then limt→∞ F (t) = F∞.

Proof. We have A- limt→∞ F (t) = limλ↓0 λF (λ) = limλ↓0 f(λ) = F∞. It followsfrom Karamata’s theorem (Theorem 4.2.8) that C- limt→∞ F (t) = F∞. Since F isincreasing, the claim follows from Theorem 4.2.6.

E. Conditions under which A- limt→∞ f(t) = f∞ implies limt→∞ f(t) = f∞.

Let f ∈ L1loc(R+, X) and F (t) =

∫ t

0f(s) ds (t ≥ 0). Assume that abs(f) ≤ 0.

Since λF (λ) = f(λ), every Tauberian theorem of type E yields one of type D. Inorder to go the other way around, we apply Tauberian theorems of type D to thefunction fδ defined by

fδ(t) :=1

δ(f(t+ δ)− f(t)) (t ≥ 0).

Lemma 4.2.12. Let f∞ ∈ X, δ > 0. Consider the following assertions:

(i) limt→∞ f(t) = f∞.

(ii) limt→∞ 1δ

∫ t+δ

tf(s) ds = f∞.

(iii) limt→∞∫ t

0fδ(s) ds = f∞ − 1

δ

∫ δ

0f(s) ds.

(iv) limλ↓0 fδ(λ) = f∞ − 1δ

∫ δ

0f(s) ds.

(v) A- limt→∞ f(t) = f∞.

Then (i) ⇒ (ii) ⇔ (iii) ⇒ (iv) ⇔ (v).

Proof. The implication (i) ⇒ (ii) is obvious, and (ii) is equivalent to (iii) since∫ t

0

fδ(s) ds =1

δ

∫ t+δ

t

f(s) ds− 1

δ

∫ δ

0

f(s) ds.

Since

fδ(λ) =1

λδ(eλδ − 1)λf(λ)− eλδ

δ

∫ δ

0

e−λsf(s) ds, (4.1)

(iv) is equivalent to (v). By Theorem 4.1.2, (iii) implies (iv).

Remark 4.2.13. It is easy to see that f is B-convergent if (ii) or (iii) of Lemma4.2.12 holds for all δ ∈ (0, δ0] for some δ0 > 0. See also the Notes for more generalresults.

Now we can obtain the following Tauberian theorem by applying the Hardy-Littlewood Tauberian theorem to fδ.

Theorem 4.2.14. Let f ∈ L1loc(R+, X), f∞ ∈ X, δ0 > 0, τ ≥ 0,M ≥ 0. Assume that

t‖ f(t)− f(s) ‖ ≤M (4.2)

whenever t ≥ τ, |s− t| ≤ δ0. If A- limt→∞ f(t) = f∞, then limt→∞ f(t) = f∞.

Proof. By the assumption, lim supt→∞ t‖ fδ(t) ‖ < ∞ for all δ ∈ (0, δ0). More-over, f is slowly oscillating. Since A- limt→∞ f(t) = f∞, Lemma 4.2.12 gives us

limλ↓0 fδ(λ) = f∞− 1δ

∫ δ

0f(s) ds. Theorem 4.2.9 implies that limt→∞

∫ t

0fδ(s) ds =

f∞− 1δ

∫ δ

0f(s) ds for all δ ∈ (0, δ0). Hence, B- limt→∞ f(t) = f∞ by Lemma 4.2.12

and Remark 4.2.13. Now the claim follows from Theorem 4.2.3.

Now we can actually use Theorem 4.2.14 to prove a slight improvement of theHardy-Littlewood theorem; i.e., we deduce a Tauberian theorem of type D from aresult of type E as indicated above.

Theorem 4.2.15. Let f ∈ L1loc(R+, X), F∞ ∈ X. Assume that for some δ > 0,

lim supt→∞

∫ t+δ

t

r‖ f(r) ‖ dr <∞.

If limλ↓0 f(λ) = F∞, then limt→∞ F (t) = F∞, where F (t) :=∫ t

0f(s) ds.

Proof. We have

lim supt→∞

supt≤s≤t+δ

t‖F (t)− F (s) ‖ ≤ lim supt→∞

t

∫ t+δ

t

‖ f(s) ‖ ds

≤ lim supt→∞

∫ t+δ

t

s‖ f(s) ‖ ds <∞.

Thus, F satisfies (4.2). Since A- limt→∞ F (t) = limλ↓0 f(λ) = F∞, it follows fromTheorem 4.2.14 that limt→∞ F (t) = F∞.

Next, we consider an order condition.

Theorem 4.2.16. Let X be an ordered Banach space with normal cone, let f ∈L1loc(R+, X) such that tkf(t) is positive and increasing on [τ,∞) for some τ ≥ 0

and some k ∈ N. If A- limt→∞ f(t) = f∞, then limt→∞ f(t) = f∞.

Proof. Replacing f(t) by f(t+ τ) we can assume that τ = 0. Since f is positive, itfollows from Karamata’s Theorem 4.2.8 that C- limt→∞ f(t) = f∞. Now Theorem4.2.6 implies that limt→∞ f(t) = f∞.

F. Power series.

Let p(z) :=∑∞

n=0 anzn be a power series which converges for |z| < 1, where

an ∈ X . Defining f ∈ L1loc(R+, X) by

f(t) = an if t ∈ [n, n+ 1), (4.3)

the preceding results yield Tauberian theorems for p. In fact, one has abs(f) ≤ 0and

f(λ) =

(1− e−λ

λ

) ∞∑n=0

ane−λn (Reλ > 0). (4.4)

Thus, from Theorem 4.2.9 we obtain the following Tauberian counterpart of The-orem 4.1.6.

Theorem 4.2.17 (Hardy). Assume that supn∈N0n‖ an ‖ < ∞, and let b∞ ∈ X. If

limz↑1 p(z) = b∞, then∑∞

n=0 an = b∞.

The special case where limn→∞ nan = 0 had been proved by Tauber in 1897 andwas the starting point of Tauberian theory.

In the case of power series, theorems of types D and E are equivalent. Infact, let bn :=

∑nk=0 ak, or equivalently, a0 = b0, an = bn − bn−1 (n ∈ N). Then

q(z) :=∑∞

n=0 bnzn also converges for |z| < 1. Moreover,

∞∑k=0

akzk = (1− z)

∞∑k=0

bkzk (|z| < 1).

Thus, A- limn→∞ bn := limz↑1(1−z)∑∞

k=0 bkzk = limz↑1

∑∞k=0 akz

k whenever oneof the limits exists. So we obtain the following

Corollary 4.2.18. Let bn ∈ X such that supn∈N

n‖ bn−bn−1 ‖ <∞. If A- limn→∞ bn = b∞,

then limn→∞ bn = b∞.

G. Fourier series.

Let Y be a Banach space. Let f : R → Y be a continuous 2π-periodic function.By ck := 1

2π

∫ π

−πf(x)e−ikxdx ∈ Y (k ∈ Z), we denote the Fourier coefficients of f

and by

sm(x) :=m∑

k=−m

ckeikx (x ∈ R)

the Fourier sums. It is well known that, in general, sm(x) does not converge asm→∞. However, it converges in the sense of Cesaro.

Theorem 4.2.19 (Fejer). One has

limn→∞

1n+1

n∑m=0

sm(x) = f(x)

uniformly in x ∈ R.

Proof. a) We show that for m ∈ N, y ∈ R,∑m

k=−m eiky = Dm(y), where

Dm(y) :=

⎧⎨⎩cosmy − cos(m+ 1)y

1− cos y(y �∈ 2πZ)

2m+ 1 (y ∈ 2πZ)

(the so-called Dirichlet kernel). In fact,∑m

k=−m eiky = 1 +∑m

k=1(eiky + e−iky) is

real. Thus,

(1− cos y)m∑

k=−m

eiky = Re

((1− eiy)

m∑k=−m

eiky

)= Re(e−imy − ei(m+1)y)

= cosmy − cos(m+ 1)y.


b) We obtain from a) and periodicity,

sm(x) =

m∑k=−m

1

2π

∫ π

−π

f(t)e−iktdt eikx =1

2π

∫ π

−π

f(t)

m∑k=−m

eik(x−t)dt

=1

2π

∫ π

−π

f(t)Dm(x− t) dt =1

2π

∫ π

−π

f(x− t)Dm(t) dt.

c) Let

σn(x) :=1

n

n−1∑m=0

sm(x)

=1

2πn

∫ π

−π

f(x− t)1

1− cos t

n−1∑m=0

(cosmt− cos(m+ 1)t) dt

=1

2πn

∫ π

−π

f(x− t)1− cosnt

1− cos tdt

=1

2π

∫ π

−π

f(x− t)Fn(t) dt,

where

Fn(t) =

⎧⎪⎨⎪⎩1

n

(sin nt

2

sin t2

)2

(t �∈ 2πZ)

n (t ∈ 2πZ)

(the so-called Fejer kernel). Here, we used that cos a = 1− 2(sin a2 )

2 (a ∈ R).d) We have to show that σn(x) converges uniformly to f(x). If f ≡ 1, then

clearly σn(x) = 1. It follows from c) that 12π

∫ π

−πFn(t) dt = 1 (n ∈ N). Moreover,

Fn(t) ≥ 0 (t ∈ (−π, π)). Since Fn(t) = Fn(−t), we have

σn(x) =1

2π

∫ π

−π

f(x+ t)Fn(t) dt

=1

2π

∫ π

−π

f(x+ t) + f(x− t)

2Fn(t) dt.

Thus, σn(x)−f(x) = 12π

∫ π

−πFn(t)g(t) dt, where g(t) :=

12(f(x+t)+f(x−t))−f(x).

Let ε > 0. Since f is uniformly continuous, there exists δ > 0 such that ‖ g(t) ‖ ≤ εwhenever |t| ≤ δ. Thus,∥∥∥∥∥

∫ δ

−δ

Fn(t)g(t) dt

∥∥∥∥∥ ≤∫ δ

−δ

Fn(t) ‖ g(t) ‖ dt

≤ ε

∫ δ

−δ

Fn(t) dt ≤ ε · 2π

and ∥∥∥∥∫ π

δ

Fn(t)g(t) dt

∥∥∥∥ ≤ ‖g‖∞(π − δ)1

n

1

(sin δ/2)2< ‖g‖∞π

n(sin δ/2)−2.

Similarly, ‖ ∫ −δ

−πFn(t)g(t) dt ‖ ≤ ‖ g ‖∞ π

n(sin δ/2)−2. Hence, ‖ σn(x) − f(x) ‖ ≤

ε+ n−1‖ g ‖∞(sin δ/2)−2. Consequently, lim supn→∞ ‖σn − f ‖ ≤ ε.

Now we deduce the following from the Tauberian theorem Corollary 4.2.18.

Theorem 4.2.20. Assume that f : R → Y is a continuous 2π-periodic functionwhich is of bounded semivariation on [−π, π]. Then limn→∞ sn(x) = f(x) uni-formly in x ∈ R.

Proof. Let k ∈ Z, k �= 0. Then ck = 12πik

∫ π

−πe−ikxdf(x) by integration by parts.

Hence, there exists a constant c ≥ 0 such that |k| ‖ ck ‖ ≤ c (k ∈ Z). Consequently,m‖sm−sm−1‖∞ ≤ m(‖ cm ‖+‖ c−m ‖) ≤ 2c. Now the claim follows from Theorem4.2.9 and Corollary 4.2.18 by choosing as X the Banach space of all continuous2π-periodic functions on R with values in Y with the uniform norm ‖ · ‖∞.

H. Inversion of Laplace transforms.

Now we give a result for Laplace transforms which corresponds to Theorem 4.2.19and which leads to a proof of the Complex Inversion Theorem 2.3.4. Here X isany Banach space.

Theorem 4.2.21. Let f ∈ L1loc(R+, X) such that abs(‖f‖) <∞.

a) If ω > abs(‖f‖), then

limR→∞

∫ a

0

∥∥∥∥ 1

R

∫ R

0

1

2πi

∫ ω+ir

ω−ir

eλtf(λ) dλ− f(t)

∥∥∥∥ dt = 0

for all a > 0.

b) If f is continuous and exponentially bounded, f(0) = 0, and ω > ω(f), then

C- limr→∞

1

2πi

∫ ω+ir

ω−ir

eλtf(λ) dλ = f(t)

uniformly for t ∈ [0, a], for all a ≥ 0.

Proof. Let Φ(t) := 12π

(sin(t/2)

t/2

)2

(t ∈ R). Then Φ ∈ L1(R), and the inverse Fourier

transform of Φ is given by

(F−1Φ)(s) =

{12π

(1− |s|) (|s| ≤ 1),

0 (|s| > 1).

Let ΦR(t) := RΦ(Rt), so (F−1ΦR)(s) = (F−1Φ)(s/R), and {ΦR} acts an approx-imate unit on L1(R, X) and C0(R, X) as R→∞ (see Lemma 1.3.3).

Consider f as a function on R with f(s) = 0 (s < 0). Let ω > abs(‖f‖), t ≥0. Put

g(s) := e−ω(s+t)f(s+ t) (s ∈ R).

Then g ∈ L1(R, X) and

(Fg)(s) = eistf(ω + is).

By Fubini’s theorem and Theorem 1.8.1 b),

1

R

∫ R

0

1

2πi

∫ ω+ir

ω−ir

eλtf(λ) dλ dr

=1

2πeωt 1

R

∫ R

−R

∫ R

|s|eistf(ω + is) dr ds

= eωt

∫ R

−R

1

2π

(1− |s|

R

)(Fg)(s) ds

= eωt

∫ ∞

−∞ΦR(s)g(s) ds

= eωt(ΦR ∗ fω)(t),where fω(s) := e−ωsf(s) (note that ΦR(s) = ΦR(−s)). Now a) and b) follow, since‖ΦR∗fω−fω‖1 → 0 if ω > abs(‖f‖) and ‖ΦR∗fω−fω‖∞ → 0 if fω ∈ C0(R, X).

Now we are able to give the following proof of Theorem 2.3.4.

Proof of Theorem 2.3.4. Let F ∈ Lip0(R+, X). For Reλ = ω > 0, we have F (λ) =

dF (λ)/λ and

‖dF (λ)‖ ≤∫ ∞

0

e−ωt‖F‖Lip0(R+,X) dt = ‖F‖Lip0(R+,X)/ω.

For 0 ≤ t ≤ a and S > R > 0,∥∥∥∥∥ 1

2πi

∫ ω+iR

ω−iR

eλtF (λ) dλ− 1

2πi

∫ ω+iS

ω−iS

eλtF (λ) dλ

∥∥∥∥∥≤ eωa

π

∫ S

R

‖F‖Lip0(R+,X)

ω|ω + is| ds

≤ eωa

πω‖F‖Lip0(R+,X) log(S/R).

Hence, R �→ 12πi

∫ ω+iR

ω−iReλ·F (λ) dλ is feebly oscillating as a function from R+ to

C([0, a], X). It follows from Theorem 4.2.21 and Theorem 4.2.5 that

1

2πi

∫ ω+iR

ω−iR

eλtF (λ) dλ→ F (t)

uniformly for t ∈ [0, a].

4.3. ERGODIC SEMIGROUPS 261

4.3 Ergodic Semigroups

This section interrupts the general theme of this chapter: we consider convergencein mean of semigroups. This is an interesting illustration of some of the results inSections 4.1 and 4.2. Moreover, the notions introduced here will be useful in thecontext of almost periodic functions which form the subject of Section 4.5. Wealso prove a striking result due to Lotz: every C0-semigroup on an L∞-space hasa bounded generator (Corollary 4.3.19).

Let A be an operator on a Banach space X such that (0, λ0) ⊂ ρ(A) for someλ0 > 0 and

M := sup0<λ<λ0

‖λR(λ,A)‖ <∞. (4.5)

Note that (4.5) is satisfied if A generates a bounded C0-semigroup T . In fact, thenR(λ,A) =

∫∞0

e−λtT (t) dt (λ > 0) and (4.5) holds for M = supt≥0 ‖T (t)‖. Denoteby KerA := {x ∈ D(A) : Ax = 0} the kernel of A and by RanA := {Ax : x ∈D(A)} the range of A. Let x ∈ X. Since

AR(λ,A)x = λR(λ,A)x− x (0 < λ < λ0),

it follows from (4.5) that x ∈ RanA if and only if

limλ↓0

λR(λ,A)x = 0. (4.6)

Moreover,

λR(λ,A)x = x (0 < λ < λ0) if and only if x ∈ KerA. (4.7)

In particular,KerA ∩ RanA = {0}. (4.8)

In what follows, as elsewhere, limits in Banach spaces are norm-limits unlessspecified otherwise.

Proposition 4.3.1. Let A be an operator satisfying (4.5) and let x ∈ X.

a) The following assertions are equivalent:

(i) There exist λn ↓ 0 such that λnR(λn, A)x converges weakly as n→∞.

(ii) x0 := limλ↓0 λR(λ,A)x exists.

(iii) x ∈ KerA+RanA.

In that case, x0 ∈ KerA and x− x0 ∈ RanA.

b) If A generates a bounded C0-semigroup, then (i)–(iii) are equivalent to

(iv) x1 := limt→∞ 1t

∫ t

0T (s)x ds exists.


In that case, x1 = x0.

Proof. a) (i) ⇒ (iii): Assume that λnR(λn, A)x converges weakly to y as n→∞.By the resolvent equation, we have

μR(μ,A)λnR(λn, A)x =μ

μ− λn(λnR(λn, A)x)− μλn

μ− λnR(μ,A)x.

for all μ ∈ (0, λ0). Taking weak limits gives μR(μ,A)y = y. It follows from (4.7)that y ∈ KerA. Since λnR(λn, A)x − x = AR(λn, A)x ∈ RanA, it follows thaty − x is in the weak closure of RanA, which coincides with the norm closure.

(iii) ⇒ (ii): This follows from (4.6) and (4.7).(ii) ⇒ (i): This is trivial.b) Since the Laplace transform of u(t) := T (t)x is given by u(λ) = R(λ,A)x,

it follows from the Abelian Theorem 4.1.2 that (iv) implies (ii), and from theTauberian Theorem 4.2.7 that (ii) implies (iv).

Corollary 4.3.2. Let A be an operator satisfying (4.5).

a) The following assertions are equivalent:

(i) Px := limλ↓0 λR(λ,A)x exists for all x ∈ X.

(ii) X = KerA⊕ RanA.

In that case, P is the projection onto KerA along RanA.

b) If D(A) is dense, then (i) and (ii) are equivalent to

(iii) KerA separates KerA∗.

Note that (iii) means, by definition, that for all x∗ ∈ KerA∗ such that x∗ �= 0there exists x ∈ KerA such that 〈x, x∗〉 �= 0.

Proof. a) follows directly from Proposition 4.3.1 and (4.8).b) Assume that D(A) is dense.(ii) ⇒ (iii): Let x∗ ∈ KerA∗, x∗ �= 0. Then 〈Ax, x∗〉 = 0 for all x ∈ D(A).

Hence 〈y, x∗〉 = 0 for all y ∈ RanA. Since X = KerA ⊕ RanA, it follows that〈x, x∗〉 �= 0 for some x ∈ KerA.

(iii)⇒ (i): We first show that KerA+RanA is dense in X. In fact, otherwisethere exists x∗ ∈ X∗ \ {0} vanishing on KerA + RanA. Since x∗ vanishes onRanA, one has x∗ ∈ KerA∗. Thus, condition (iii) is violated, and the claim isproved. It follows from (4.6) and (4.7) that (λR(λ,A)x) converges as λ ↓ 0 forx ∈ KerA+RanA. This implies convergence for all x ∈ X by density.

Definition 4.3.3. A C0-semigroup T on X is called Cesaro-ergodic (or mean-ergo-dic) if

Qx := limt→∞

1

t

∫ t

0

T (s)x ds (4.9)


exists for all x ∈ X. The semigroup is called Abel-ergodic if its generator Asatisfies (4.5) and the equivalent conditions (i), (ii), (iii) of Corollary 4.3.2 aresatisfied.

Proposition 4.3.4. Let T be a C0-semigroup with generator A.

a) If T is Cesaro-ergodic, then T is Abel-ergodic and Q given by (4.9) is theprojection onto KerA along RanA.

b) Assume that T is bounded and Abel-ergodic. Then T is Cesaro-ergodic.

Proof. a) Assume that T is Cesaro-ergodic. It follows from the uniform bound-

edness principle that there exists M ≥ 0 such that 1t

∥∥ ∫ t

0T (s)x ds

∥∥ ≤ M‖x‖for all t > 0, x ∈ X. It follows from Theorem 3.1.7 that (0,∞) ⊂ ρ(A) andR(λ,A)x =

∫∞0

e−λtT (t)x dt for all λ > 0, x ∈ X . Integration by parts yields

‖λR(λ,A)x‖ =

∥∥∥∥∫ ∞

0

λ2e−λt

∫ t

0

T (s)x ds dt

∥∥∥∥≤ M

∫ ∞

0

λ2e−λtt dt ‖x‖= M‖x‖ (λ > 0).

Thus, condition (4.5) is satisfied. Since u(t) := T (t)x has Laplace tranform u(λ) =R(λ,A)x, the claim now follows from Theorem 4.1.2 and Proposition 4.3.1 a).

b) If T is bounded by M , then condition (4.5) is satisfied. So the claim followsfrom Proposition 4.3.1.

Corollary 4.3.5. Every bounded C0-semigroup on a reflexive Banach space X isCesaro-ergodic.

Proof. Let x ∈ X . It follows from reflexivity that condition (i) of Proposition4.3.1 is satisfied. Thus, T is Abel-ergodic and the claim follows from Proposition4.3.4.

Condition (iii) of Corollary 4.3.2 is frequently the most convenient way toverify Cesaro-ergodicity of a bounded C0-semigroup. Note that the dual assertionis always true: the space KerA∗ separates KerA whenever A generates a boundedC0-semigroup. More generally, the following holds.

Proposition 4.3.6. Let A be a densely defined operator satisfying (4.5). Let 0 �= x ∈KerA. Then there exists x∗ ∈ KerA∗ such that 〈x, x∗〉 �= 0.

Proof. Let y∗ ∈ X∗ such that 〈x, y∗〉 = 1. Let x∗ be a weak* limit point ofλR(λ,A)∗y∗ as λ ↓ 0. Since λR(λ,A)x = x, it follows that 〈x, x∗〉 = 1. Lety ∈ D(A). Then

limλ↓0

λR(λ,A)Ay = limλ↓0

λ(λR(λ,A)y − y) = 0.


Hence,〈Ay, x∗〉 = lim

λ↓0〈λR(λ,A)Ay, y∗〉 = 0.

Thus, x∗ ∈ KerA∗.

We have shown that for bounded C0-semigroups Abel- and Cesaro-ergodicityare equivalent. This is not always the case for arbitrary C0-semigroups as the

following easy example shows: Let X := C2, T (t) := eit(

1 t0 1

). Then T is

Abel-ergodic but not Cesaro-ergodic. However, by Karamata’s theorem (Theorem4.2.8) both notions do coincide for positive semigroups.

Theorem 4.3.7. Let T be a positive C0-semigroup on an ordered Banach space withnormal cone. Then T is Abel-ergodic if and only if T is Cesaro-ergodic.

Next, we extend the results on Cesaro-ergodicity to Lipschitz continuousintegrated semigroups.

Let A be an operator on X satisfying the Hille-Yosida condition

(0,∞) ⊂ ρ(A) and ‖(λR(λ,A))n‖ ≤M (4.10)

for all λ > 0, n ∈ N0 and some M ≥ 0. Then A generates a Lipschitz contin-uous once integrated semigroup S. More precisely, there exists S : R+ → L(X)satisfying

S(0) = 0, ‖S(t)− S(s)‖ ≤M |t− s| (s, t > 0)

and

R(λ,A) = λ

∫ ∞

0

e−λtS(t) dt (λ > 0);

see Theorem 3.3.1 and Section 3.5.If D(A) is dense, then A generates a bounded C0-semigroup T and 1

tS(t) =

1t

∫ t

0T (s) ds. If D(A) is not dense, then T does not exist on X, but the Cesaro

means 1tS(t) still make sense. Thus, the following theorem generalizes Proposition

4.3.4, but is more complicated to prove in this more general context.

Theorem 4.3.8. Let A be the generator of a Lipschitz continuous once integratedsemigroup S. The following conditions are equivalent:

(i) X = KerA⊕ RanA.

(ii) P := limt→∞ 1tS(t) exists in the strong operator topology.

In that case, P is the projection onto KerA along RanA.

Proof. (i) ⇒ (ii): If x ∈ KerA, then S(t)x = tx (t > 0). Next, let x = Ay ∈ RanA.We show that

limt→∞

1

tS(t)x = 0.


Note that ‖ 1tS(t)‖ ≤M . Since by Lemma 3.2.2,

S(t)y = ty +

∫ t

0

S(s)Ay ds,

one has

limt→∞

1

t2

∫ t

0

S(s)Ay ds = 0.

By the Abelian Theorem 4.1.2, it follows that

limt→∞

1

t

∫ t

0

1

s2

∫ s

0

S(r)Ay dr ds = C- limt→∞

1

t2

∫ t

0

S(s)Ay ds = 0.

Integration by parts yields

C- limt→∞

1

tS(t)Ay = lim

t→∞1

t

∫ t

0

1

sS(s)Ay ds

= limt→∞

1

t

∫ t

0

1

s

d

ds

∫ s

0

S(r)Ay dr ds

= limt→∞

{1

t2

∫ t

0

S(r)Ay dr +1

t

∫ t

0

1

s2

∫ s

0

S(r)Ay dr ds

}= 0.

Now observe that∥∥∥∥1t S(t)x− 1

sS(s)x

∥∥∥∥ ≤∥∥∥∥1t (S(t)x− S(s)x)

∥∥∥∥+

∣∣∣∣1t − 1

s

∣∣∣∣ ‖S(s)x‖≤ 1

t|t− s|2M‖x‖.

Thus, the function 1tS(t)Ay is feebly oscillating. It follows from the Tauberian

Theorem 4.2.5 that

limt→∞

1

tS(t)Ay = C- lim

t→∞1

tS(t)Ay = 0.

We have shown that 1tS(t)x converges to Px as t → ∞ for x ∈ KerA + RanA,

where P is the projection onto KerA along RanA. Since ‖1tS(t)‖ ≤M , it followsthat limt→∞ 1

tS(t)x = Px for all x ∈ X.

(ii) ⇒ (i): This is an Abelian theorem whose proof is analogous to Theorem4.1.2 c): Let x ∈ X. By Corollary 4.3.2, we have to show that

limλ↓0

λR(λ,A)x = limλ↓0

λ2

∫ ∞

0

e−λtS(t)x dt = Px.


Let ε > 0. There exists t0 such that ‖ 1tS(t)x− Px‖ ≤ ε for all t ≥ t0. Hence

lim supλ↓0

‖λR(λ,A)x− Px‖ = lim supλ↓0

∥∥∥∥∫ ∞

0

λ2e−λtt

(1

tS(t)x− Px

)dt


λ↓0

∫ t0

0

λ2e−λtt

∥∥∥∥1t S(t)x− Px

∥∥∥∥ dt+ ε

= ε.

Example 4.3.9. Consider the operator A on X := C[0, 1] given by

D(A) :={f ∈ C2[0, 1] : f(0) = f(1), f ′′(0) = 0

}, Af := f ′′.

It is not difficult to see that A is dissipative and (I −A)D(A) = X. By Corollary3.4.6, A generates a once integrated semigroup S satisfying

‖S(t)− S(s)‖ ≤ |t− s|.

Let f ∈ X. Then the constant function f(0) ∈ KerA and f − f(0) = Ag ∈ RanA,where

g(x) :=

∫ x

0

∫ t

0

(f(s)− f(0)) ds dt− x

∫ 1

0

∫ t

0

(f(t)− f(0)) ds dt.

It follows from Theorem 4.3.8 that

limt→∞

1

t(S(t)f)(x) = f(0)

uniformly for all x ∈ [0, 1], for all f ∈ X.

Next, we fix a bounded C0-semigroup T with generator A. Recall that theoperator A− iη generates the C0-semigroup (e−iηtT (t))t≥0, and we may considerergodicity for the rescaled semigroup.

Definition 4.3.10. a) Let η ∈ R. A vector x ∈ X is called ergodic at η (with respectto T ) if the mean

Mηx := limt→∞

1

t

∫ t

0

e−iηsT (s)x ds

converges in norm.

b) A vector x is called totally ergodic (with respect to T ) if x is ergodic at η forall η ∈ R.

c) The semigroup T is called totally ergodic if each vector x is totally ergodic; i.e.,if the semigroup (e−iηtT (t))t≥0 is Cesaro-ergodic for all η ∈ R.


By Xe we denote the space of all totally ergodic vectors in X and by Xe0 thespace of all vectors x ∈ Xe such that Mηx = 0 for all η ∈ R. Both spaces Xe andXe0 are closed and invariant under the semigroup.

It follows from Proposition 4.3.1 that a vector x ∈ X is totally ergodic if andonly if

xη := limα↓0

αR(α+ iη,A)x

exists for all η ∈ R. In that case, xη = Mηx. Moreover, Mηx ∈ Ker (A− iη) and

x−Mηx ∈ Ran(A− iη). In particular,

T (t)Mηx = eiηtMηx (t ≥ 0),

R(λ,A)Mηx =1

λ− iηMηx (λ ∈ ρ(A), λ �= iη).

If x ∈ X is totally ergodic we denote by

Freq(x) := {η ∈ R : Mηx �= 0}the set of all frequencies of X.

Proposition 4.3.11. Let x be totally ergodic (with respect to T ). Then Freq(x) iscountable.

Proof. We can assume that X is separable, as we may replace X by span{T (t)x :t ≥ 0}. Let M := supt≥0 ‖T (t)‖. Then

‖αR(α+ iη, A)‖ ≤M (α > 0, η ∈ R).

Let xη := Mηx = limt→∞ 1t

∫ t

0e−iηsT (s)x ds (η ∈ R). For η ∈ Freq(x), let

yη := xη/‖xη‖. Then ‖yη‖ = 1 and αR(α + iη, A)yη = yη (α > 0) and forμ ∈ Freq(x), μ �= η,

αR(α+ iη, A)yμ =α

α+ iη − iμyμ (α > 0).

Let μ, η ∈ Freq(x) such that η > μ. Then for α > 0,

M‖yη − yμ‖ ≥ ‖αR(α + iη,A)(yη − yμ)‖=

∥∥∥∥yη − α

α+ iη − iμyμ

∥∥∥∥≥ 1−

∣∣∣∣ α

α+ iη − iμ

∣∣∣∣= 1−

∣∣∣∣ 1

1 + i(η − μ)/α

∣∣∣∣ .Choosing α = η − μ, we obtain

M‖yη − yμ‖ ≥ 1−∣∣∣∣ 1

1 + i

∣∣∣∣ = 1−∣∣∣∣1− i

2

∣∣∣∣ = 1− 1√2> 0.

If Freq(x) is uncountable this contradicts the fact that X is separable.

It follows from Corollary 4.3.5 that every bounded C0-semigroup on a reflex-ive Banach space is totally ergodic. This can be generalized as follows.

Proposition 4.3.12. Assume that the vector x ∈ X has relatively weakly compactorbit {T (t)x : t ≥ 0}. Then x is totally ergodic with respect to T .

Proof. It follows from the assumption that {e−iηtT (t)x : t ≥ 0} is relativelyweakly compact for all η ∈ R. By Krein’s theorem [Meg98, Theorem 2.8.14], theclosed convex hull K := co{e−iηtT (t)x : t ≥ 0} is weakly compact. Since αR(α +iη,A)x =

∫∞0

αe−αte−iηtT (t)x dt ∈ K for all α > 0, condition (i) of Proposition

4.3.1 is satisfied and so 1t

∫ t

0e−iηtT (t)x dt converges as t → ∞, by Proposition

4.3.1.

Next we consider positive C0-semigroups on L1-spaces.

Proposition 4.3.13. Let X = L1(Ω, μ) where (Ω, μ) is a σ-finite measure space. LetT be a bounded positive C0-semigroup on X. If f ∈ X+ is ergodic at 0, then f istotally ergodic.

Proof. Since αR(α,A)f converges as α ↓ 0, the set K := {αR(α,A)f : 0 < α ≤ 1}is relatively compact. It follows that the solid hull so(K) := {g ∈ X : |g| ≤k for some k ∈ K} is relatively weakly compact (by [AB85, Theorem 13.8]). Letη ∈ R. Then

|αR(α+ iη,A)f | =

∣∣∣∣∫ ∞

0

αe−t(α+iη)T (t)f dt

∣∣∣∣≤

∫ ∞

0

αe−αtT (t)f dt = αR(α,A)f.

Thus, αR(α + iη, A)f ∈ so(K) for 0 < α ≤ 1. Consequently, condition (i) ofProposition 4.3.1 is satisfied.

In particular, if T is a positive bounded C0-semigroup on L1(Ω, μ), then T istotally ergodic whenever T is Cesaro-ergodic.

Here is a criterion which is sometimes convenient for verifying total ergodicity.

Proposition 4.3.14. Let X := L1(Ω, μ), where (Ω, μ) is a σ-finite measure space.Let T be a positive bounded C0-semigroup on X. Assume that there exists a func-tion u ∈ X such that u > 0 μ-a.e. and T (t)u ≤ u for all t ≥ 0. Then T is totallyergodic.

Proof. Suppose that f ∈ X and 0 ≤ f ≤ u. For α > 0,

0 ≤ αR(α,A)f ≤ αR(α,A)u ≤ u.

Since {g ∈ X : 0 ≤ g ≤ u} is weakly compact (see [Sch74, Theorem II.5.10 andProposition II.8.3]), it follows from Proposition 4.3.1 that limα↓0 αR(α,A)f exists.

Since {f ∈ X : 0 ≤ f ≤ u} is total in X, it follows that T is Abel-ergodic, andhence totally ergodic by Proposition 4.3.4 and Proposition 4.3.13.

In Proposition 4.3.14, the condition of a subinvariant strictly positive functioncannot be omitted. For example, the shift semigroup T on L1(R+) given by

(T (t)f)(x) = f(x+ t)

is not Cesaro-ergodic since KerA = {0}, but KerA∗ = C · 1 (where 1 denotes theconstant 1 function).

Next, we return to general Banach spaces, but we consider the special casewhere the Abel means in Proposition 4.3.1 converge with respect to the operatornorm.

Let A be an operator on X . We say that 0 is a simple pole of the resolvent if{λ ∈ C : 0 < |λ| < ε} ⊂ ρ(A) for some ε > 0 and there exists 0 �= P ∈ L(X) suchthat

R(λ,A)− P

λ(4.11)

has a holomorphic extension to the disc B(0, ε) := {λ ∈ C : |λ| < ε}.For example, if A generates a bounded C0-semigroup T , 0 ∈ σ(A) and A has

compact resolvent, then 0 is a simple pole. Indeed, if P is the spectral projectionof A associated with {0}, then AmP = 0 for some m ∈ N (see Proposition B.9 andthe subsequent remarks). It follows that

T (t)P =

m−1∑n=0

tnAnP

n!(t ≥ 0).

Since T is bounded, AP = 0. Hence,

R(λ,A) = R(λ,AZ)(I − P ) + P/λ,

where Z = (I − P )(X) and AZ is the part of A in Z. Since 0 �∈ σ(AZ), the claimfollows.

Proposition 4.3.15. Let A be an operator on X. The following assertions are equiv-alent:

(i) There exists λ0 > 0 such that (0, λ0) ⊂ ρ(A) and P := limλ↓0 λR(λ,A)converges in the operator norm.

(ii) 0 is a simple pole of the resolvent, or 0 ∈ ρ(A).

(iii) The range RanA is closed, (0, λ0) ⊂ ρ(A) for some λ0 > 0 and λR(λ,A)converges in the strong operator topology as λ ↓ 0.


Proof. (i)⇒ (ii): It follows from Proposition 4.3.2 thatX = KerA⊕RanA and thatP is the projection onto KerA along Y := RanA. Since P commutes with R(λ,A),we may consider the part AY of A in Y . Then (0, λ0) ⊂ ρ(AY ) and R(λ,AY ) =R(λ,A)|Y for λ ∈ (0, λ0) (see Proposition B.8). Since P = 0 on Y , it followsthat limλ↓0 ‖λR(λ,AY )‖ = 0. So there exists λ > 0 such that ‖λR(λ,AY )‖ ≤ 1

2 .Hence, dist(λ, σ(AY )) ≥ ‖R(λ,AY )‖−1 ≥ 2λ (see Corollary B.3). Thus 0 ∈ ρ(AY ),and so R(λ,A)|Y has a holomorphic extension to B(0, ε) for some ε > 0. SinceR(λ,A) = P/λ + R(λ,AY )(I − P ) (λ ∈ ρ(AY ), λ �= 0), this implies (ii). It alsoshows that RanA = Ran(AY ) = Y is closed.

(ii) ⇒ (i): It follows from (4.11) that limλ↓0 λR(λ,A) = P in L(X).

(i) ⇒ (iii): Strong convergence is trivial from (i). It has been shown abovethat RanA is closed.

(iii) ⇒ (ii): By assumption, Y := RanA is closed and X = KerA ⊕ Y byCorollary 4.3.2. Thus, the part AY in Y is invertible. Let P be the projection ontoKerA along Y . Since R(λ,A) = P/λ + R(λ,AY )(I − P ) (λ ∈ ρ(AY ), λ �= 0), theresult follows.

Next, we show a very peculiar phenomenon on the space X := L∞(Ω, μ),where (Ω, μ) is a measure space. Such a space has two remarkable propertiesconcerning convergence of sequences, namely

(DP )

{xn → 0 weakly in X and x∗n → 0 weakly in X∗

implies 〈xn, x∗n〉 → 0 as n→∞;

and

(G)

{x∗n → 0 weak* in X∗ implies

x∗n → 0 weakly in X∗ as n→∞.

The first property is called the Dunford-Pettis property. A space having the secondproperty is called a Grothendieck space. It is obvious that the properties (DP) and(G) are inherited by complemented subspaces. We refer to [Sch74, Theorems II.9.7and II.10.4] for a proof that L∞(Ω, μ) has these properties. The key argumentinvolving these two properties is expressed in the following lemma.

Lemma 4.3.16. Let X be a Banach space such that (G) and (DP) are satisfied.Suppose that Tn ∈ L(X) (n ∈ N) such that

limn→∞ ‖Tnx‖ = 0 for all x ∈ X

and

limn→∞ ‖T

∗nx∗‖ = 0 for all x∗ ∈ X∗.

Then

limn→∞ ‖T

2n‖ = 0.

Proof. Assume that ‖T 2n‖ does not converge to 0. Then there exist ε > 0 and a

subsequence such that ‖T 2nk‖ ≥ 2ε. By the Hahn-Banach theorem, we find xk ∈

X,x∗k ∈ X∗ such that ‖xk‖ = ‖x∗k‖ = 1 but∣∣〈Tnkxk, T

∗nkx∗k〉

∣∣ = ∣∣〈T 2nkxk, x

∗k〉∣∣ ≥ ε. (4.12)

Let yk := Tnkxk, y

∗k := T ∗nk

x∗k. Then for x∗ ∈ X∗,

|〈yk, x∗〉| = |〈xk, T∗nkx∗〉| ≤ ‖T ∗nk

x∗‖ → 0 as k →∞.

Thus, yk → 0 weakly.Let x ∈ X. Then

|〈x, y∗k〉| = |〈Tnkx, x∗k〉| ≤ ‖Tnk

x‖ → 0 as k →∞.

Thus, y∗k → 0 weak*. It follows from (G) that y∗k → 0 weakly. Now, (DP) impliesthat 〈yk, y∗k〉 → 0 as k →∞. This contradicts (4.12).

The following surprising result holds in particular on a space X := L∞(Ω, μ)for any measure space (Ω, μ).

Theorem 4.3.17. Let X be a Banach space satisfying (DP) and (G). Let A bea densely defined operator on X such that (0, λ0) ⊂ ρ(A) for some λ0 > 0. IfλR(λ,A)x converges weakly as λ ↓ 0 for all x ∈ X, then λR(λ,A) converges inL(X) as λ ↓ 0.Proof. It follows from Corollary 4.3.2 that Px := limλ↓0 λR(λ,A)x converges innorm for all x ∈ X , and that X = KerA ⊕ RanA. Replacing A by its part inRanA, we can assume that KerA = 0 and so P = 0.

Let x∗ ∈ X∗. Then for x ∈ X we have

〈x, λR(λ,A)∗x∗〉 = 〈λR(λ,A)x, x∗〉 → 0 as λ ↓ 0.Hence, λR(λ,A)∗x∗ → 0 weak* for all x∗ ∈ X∗. It follows from (G) thatλnR(λn, A)

∗x∗ → 0 weakly whenever λn ↓ 0, for all x∗ ∈ X∗. Now Corollary4.3.2 implies that ‖λR(λ,A)∗x∗‖ → 0 as λ ↓ 0, for all x∗ ∈ X∗.

We have shown that Tn := λnR(λn, A) satisfies the hypotheses of Lemma4.3.16 whenever λn ↓ 0. It follows that ‖(λR(λ,A))2‖ → 0 as λ ↓ 0. Since(λR(λ,A))2 = (I + AR(λ,A))2, it follows that

r(I +AR(λ,A)) ≤ ‖(I +AR(λ,A))2‖1/2 < 1

for λ > 0 small (where r(I +AR(λ,A)) denotes the spectral radius). This impliesthat AR(λ,A) is invertible. In particular, RanA = X . Now the claim follows fromProposition 4.3.15.

It is also interesting to consider convergence of λR(λ,A) for λ→∞. Again,the following result holds in particular on a space X = L∞(Ω, μ), where (Ω, μ) isa measure space.

Theorem 4.3.18. Let X be a Banach space satisfying (DP) and (G). Let A be adensely defined operator on X such that (λ0,∞) ⊂ ρ(A) and supλ>λ0

‖λR(λ,A)‖ <∞ for some λ0. Then A is bounded.

Proof. By Lemma 3.3.12, limn→∞(nR(n,A)x−x) = 0 for all x ∈ X . Let x∗ ∈ X∗.Then it follows that

〈x, nR(n,A)∗x∗ − x∗〉 = 〈nR(n,A)x− x, x∗〉 → 0 as n→∞,

for all x ∈ X. Thus, (nR(n,A)∗ − I)x∗ → 0 weak* as n→∞. It follows from (G)that (nR(n,A)∗ − I)x∗ → 0 weakly as n → ∞, for all x∗ ∈ X∗. But this impliesthat x∗ is in the weak closure of D(A∗) = R(λ,A∗)X∗ (λ ≥ λ0) for all x∗ ∈ X∗.Since the weak and norm closures coincide, A∗ is densely defined. It follows asabove that ‖(nR(n,A)∗ − I)x∗‖ → 0 as n → ∞, for all x∗ ∈ X∗. Now Lemma4.3.16 implies that ‖(nR(n,A)− I)2‖ → 0 as n→∞. In particular,

r((nR(n,A)− I)) ≤ ‖(nR(n,A)− I)2‖1/2 < 1

for n sufficiently large. This implies that nR(n,A) is invertible. Hence, D(A) = X ,so A is bounded by the closed graph theorem.

We deduce from Theorem 4.3.18 the following surprising and important re-sult.

Corollary 4.3.19 (Lotz). Let T be a C0-semigroup on the Banach space L∞(Ω, μ),where (Ω, μ) is a measure space. Then T has a bounded generator.

In other words, if a semigroup T defined onX := L∞(Ω, μ) converges stronglyto the identity as t ↓ 0, it converges already in the operator norm.

4.4 Complex Tauberian Theorems: the Contour Method

Let f ∈ L∞(R+, X) and let f(λ) =∫∞0

e−λtf(t) dt (Reλ > 0) be its Laplacetransform. We define the half-line spectrum sp(f) of f by

sp(f) :={η ∈ R : f does not have a holomorphic extension

to an open neighbourhood of iη in C}.

It turns out that countability of the spectrum with certain growth conditions is aTauberian hypothesis. Here we prove a Tauberian theorem of type D by completelyelementary contour arguments. For simplicity, we consider first the simplest casewhere the spectrum is empty, giving a qualitative result (Theorem 4.4.1) and thena quantified version (Theorem 4.4.6). Then we adapt the argument to the casewhen the spectrum consists of one point (Theorem 4.4.8). This suffices for manyinteresting applications (see Corollaries 4.4.12 and 4.4.13 and Theorems 4.4.14 and4.4.16) but we shall prove some more general results in Sections 4.7 and 4.9 by othermeans. In Theorem 4.4.18 we give a related result under different assumptions onf and f , and using a different contour method.

4.4. THE CONTOUR METHOD 273

Theorem 4.4.1. Let f ∈ L∞(R+, X), and assume that sp(f) is empty. Then

limt→∞

∫ t

0

f(s) ds = f(0).

Here, f denotes the holomorphic extension of f to a neighbourhood of 0. As usual,we will denote the open right half-plane {λ ∈ C : Reλ > 0} by C+.

Proof. We can assume that f(0) = 0. Otherwise, we replace f by f − χ(0,1)f(0).

There is a simply connected open set Ω containing C+ such that f has a holomor-phic extension (also denoted by f) to Ω.

Let gt(λ) :=∫ t

0e−λsf(s) ds (λ ∈ C, t ≥ 0). Let R > 1, and γ be a contour

consisting of the semi-circle {λ ∈ C : |λ| = R, Re λ > 0} and a path γ′ connectingiR with −iR and lying entirely in Ω ∩ {Reλ < 0} (except at the endpoints). Let

h(λ) := 1 +λ2

R2.

Since h(0) = 1, it follows from Cauchy’s theorem that

0 = −f(0) = − 1

2πi

∫γ

eλtf(λ)h(λ)dλ

λ;∫ t

0

f(s) ds = gt(0) =1

2πi

∫|λ|=R

eλtgt(λ)h(λ)dλ

λ.

Adding up, we have∫ t

0

f(s) ds =1

2πi

∫|λ|=RReλ>0

eλt(gt(λ)− f(λ))h(λ)dλ

λ

− 1

2πi

∫γ′eλtf(λ)h(λ)

dλ

λ

+1

2πi

∫|λ|=RReλ<0

eλtgt(λ)h(λ)dλ

λ

=: I1(t) + I2(t) + I3(t). (4.13)

We estimate these three integrals separately.I1(t): Let λ := Reiθ, θ ∈ (−π

2, π2). Then

‖ eλt(gt(λ)− f(λ)) ‖ =

∥∥∥∥∫ ∞

t

eλ(t−s)f(s) ds

∥∥∥∥=

∥∥∥∥∫ ∞

0

e−λsf(s+ t) ds

∥∥∥∥≤ ‖f‖∞

Reλ=‖f‖∞R cos θ

;∣∣∣∣h(λ)λ

∣∣∣∣ ≤ |1 + e2iθ| 1R

=2 cos θ

R.

Thus,

‖I1(t)‖ ≤ 1

2π

‖f‖∞R

2

RπR =

‖f‖∞R

(t ≥ 0).

I2(t): Note that limt→∞ ‖ I2(t) ‖ = 0 by the dominated convergence theorem.I3(t): Let λ := Reiθ, θ ∈ (π

2, 3π

2). Then

‖ eλtgt(λ) ‖ =∥∥∥∥ ∫ t

0

eλ(t−s)f(s) ds

∥∥∥∥ ≤ ‖f‖∞∫ t

0

eRs cos θds ≤ ‖f‖∞R | cos θ| ;∣∣∣∣h(λ)λ

∣∣∣∣ ≤ 2| cos θ|R

.

Hence,

‖ I3(t) ‖ ≤ 1

2π

‖f‖∞R

2

RπR =

‖f‖∞R

.

These estimates give

lim supt→∞

∥∥∥∥ ∫ t

0

f(s) ds

∥∥∥∥ ≤ 2‖f‖∞R

. (4.14)

Letting R→∞ proves the claim.

From the proof above we deduce the following.

Proposition 4.4.2. Let f ∈ L∞(R+, X). If R > 0 such that sp(f) ∩ [−R,R] = ∅,then

lim supt→∞

∥∥∥∥∫ t

0

f(s) ds− f(0)

∥∥∥∥ ≤ 2

Rlim supt→∞

‖ f(t) ‖.

Proof. Let c > lim supt→∞ ‖f(t)‖. Choose τ ≥ 0 such that ‖f(t + τ)‖ ≤ c for all

t ≥ 0 and apply the estimate (4.14) to f(·+ τ)− χ(0,1)

(f(0)− ∫ τ

0f(s) ds

).

The next two corollaries follow from Proposition 4.4.2 by replacing f(t) bye−iηtf(t).

Corollary 4.4.3. Let f ∈ L∞(R+, X). If η �∈ sp(f), then

supt≥0

∥∥∥∥∫ t

0

e−iηsf(s) ds

∥∥∥∥ <∞.

Corollary 4.4.4. Let f ∈ L∞(R+, X) such that limt→∞ ‖ f(t) ‖ = 0. Then

f(iη) = limt→∞

∫ t

0

e−iηsf(s) ds

for all η ∈ R\sp(f).


Next we give a quantified version of Theorem 4.4.1. When sp(f) is empty,

there is a continuous increasing function M : R+ → (0,∞) such that f has a

holomorphic extension (also denoted by f) to

ΩM :=

{λ ∈ C : Reλ > − 1

M(| Imλ|)}

and ‖f(λ)‖ ≤M(| Imλ|) for all λ ∈ ΩM . Define

Mlog(s) = M(s) [log(1 +M(s)) + log(1 + s)]

for s ∈ R+. The inverse M−1log of Mlog is an increasing function of (Mlog(0),∞)

onto (0,∞). Although we do not require M to be strictly increasing, we shalllet M−1 denote any increasing function from the range of M to R+ such thatM(M−1(t)) = t for all t in the range of M .

Example 4.4.5. a) If M is bounded, then M−1log (t) ∼ Cet as t → ∞, for some

constant C.b) If M(s) = β(1 + s)α where α, β > 0, then

M−1log (t) ∼ Cα,β

(t

log t

)1/α

as t→∞,

for some constant Cα,β .c) If M(s) = βeαs where α, β > 0, then M−1

log (t) ∼ 1αlog t as t→∞.

Theorem 4.4.6. Let f ∈ L∞(R+, X), and assume that sp(f) is empty. Let M andMlog be as above, and c ∈ (0, 1). Then there exist constants C and t0, dependingonly on ‖f‖∞, M and c, such that∥∥∥∥∫ t

0

f(s) ds− f(0)

∥∥∥∥ ≤ C

M−1log (ct)

for all t ≥ t0. (4.15)

Remark 4.4.7. In (4.15), the function M−1log cannot be replaced by M−1 in general.

For example, if X = C and f(t) = e−t, then the assumptions of Theorem 4.4.6 aresatisfied for any continuous increasing function M with M(0) = 1. So M−1 mayincrease arbitrarily fast, but∫ t

0

f(s) ds− f(0) = −e−t.

On the other hand, M−1 and M−1log behave similarly if M increases rapidly and

consistently (see Example 4.4.5). We shall see in Theorem 4.4.14 that, in thecontext of C0-semigroups, M−1

log (ct) can never be replaced in (4.15) by a function

increasing faster than M−1(Ct) if M is chosen in the optimal way. Example 4.4.15will show that it may not be possible to replace M−1

log by M−1 in (4.15) when Mincreases arbitrarily rapidly but inconsistently.

Proof of Theorem 4.4.6. We use the same method and notation as in the proof ofTheorem 4.4.1, taking Ω = ΩM and choosing γ′ and R as follows.

Let a = c−1/2 > 1 and take γ′ to be the union of γ0, γ+ and γ−, where

γ0(s) = − 1

aM(|s|) (−R ≤ s ≤ R) ,

γ±(τ) = τ ± iR (−(aM(R))−1 ≤ τ ≤ 0) .

On γ±, |h(λ)| ≤ C/R, so we can estimate the norm of the integral I2(t) over γ±by ∥∥∥∥∥

∫γ±

eλtf(λ)h(λ)dλ

λ

∥∥∥∥∥ ≤ C

∫ (aM(R))−1

0

1

RM(R)

e−tτ

Rdτ ≤ CM(R)

R2t.

Here and subsequently, C denotes a constant which may depend only on M , cand ‖f‖∞, although it may vary from place to place. On γ0, h(λ) is boundedindependently of R, and |etλ| ≤ e−t/aM(R). Hence, we can estimate the norm ofthe integral over γ0 by∥∥∥∥∫

γ0

eλtf(λ)h(λ)dλ

λ

∥∥∥∥ ≤ CM(R)e−t/aM(R)

∫γ0

∣∣∣∣dλλ∣∣∣∣ ≤ CM(R)(1+logR)e−t/aM(R).

With these estimates for I2(t) and those in the proof of Theorem 4.4.1 forI1(t) and I3(t), we obtain∥∥∥∥ ∫ t

0

f(s) ds

∥∥∥∥ ≤ 2‖f‖∞R

+ C

(M(R)

R2t+M(R)(1 + logR)e−t/aM(R)

).

Given t > c−1Mlog(1), choose R = M−1log (ct) > 1. Then

M(R)

R2t=

c

2R2 log ((1 +M(R))(1 +R))≤ C

R,

M(R)(1 + logR) ≤ C(1 +M(R))a(1 +R)a

R=

C

Ret/aM(R) .

Hence ∥∥∥∥ ∫ t

0

f(s) ds

∥∥∥∥ ≤ C

R=

C

M−1log (ct)

.

Now we consider the case when sp(f) is a single point, stating only thequalitative result.

Theorem 4.4.8. Let f ∈ L∞(R+, X), and assume that sp(f) = {η} and

supt≥0

∥∥∥∥∫ t

0

e−iηsf(s) ds

∥∥∥∥ <∞, (4.16)

where η ∈ R\{0}. Then limt→∞∫ t

0f(s) ds = f(0).

Remark 4.4.9. The growth condition (4.16) in Theorem 4.4.8 cannot be omitted.

For example, let f(t) := eit. Then f(λ) = (λ − i)−1, so sp(f) = {1}. However,∫ t

0f(s) ds = 1

i(eit − 1) does not converge as t→∞.

Proof of Theorem 4.4.8. The proof is similar to Theorem 4.4.1, with the followingchanges.

a) We take R > |η| and 0 < ε < min{|η|, R−|η|}. The path γ′ now consists oftwo separate paths connecting iR with i(η+ ε) and connecting i(η− ε) with −iR.

It lies entirely in a simply connected domain where f is defined holomorphically,and also in {Reλ < 0} (except at the four endpoints). The contour γ includes alsothe semi-circle {λ ∈ C : |λ− iη| = ε, Reλ > 0}.

b) Let

h(λ) :=

(1 +

ε2

(λ− iη)2

)η2

(η2 − ε2)

(1 +

λ2

R2

).

For λ = Reiθ, we have∣∣∣∣h(λ)λ

∣∣∣∣ ≤(1 +

ε2

(R− |η|)2)

η2

η2 − ε2|1 + e2iθ| 1

R

=

(1 +

ε2

(R− |η|)2)

η2

η2 − ε22| cos θ|

R. (4.17)

c) In (4.13) we have two additional integrals:

I4(t) :=1

2πi

∫|λ−iη|=εReλ>0

eλt(gt(λ)− f(λ))h(λ)dλ

λ;

I5(t) := − 1

2πi

∫|λ−iη|=εReλ>0

eλtgt(λ)h(λ)dλ

λ=

1

2πi

∫|λ−iη|=εReλ<0

eλtgt(λ)h(λ)dλ

λ.

We estimate these integrals as follows:I4(t): Let λ := iη + εeiθ, θ ∈ (−π

2, π2). Let F1(t) :=

∫ t

0e−iηsf(s) ds. Then by

assumption (4.16), K := supt≥0 ‖F1(t)‖ <∞. We have

‖ eλt(gt(λ)− f(λ)) ‖ =

∥∥∥∥ eλt ∫ ∞

t

e−λsf(s) ds

∥∥∥∥=

∥∥∥∥ eλt ∫ ∞

t

e−εseiθ d

dsF1(s) ds

∥∥∥∥=

∥∥∥∥−eλte−εteiθF1(t) + eλt∫ ∞

t

εeiθe−εseiθF1(s) ds

∥∥∥∥≤ K + ε

∥∥∥∥ ∫ ∞

t

e−εeiθ(s−t)F1(s) ds

∥∥∥∥≤ K + ε

K

ε cos θ= K

(1 +

1

cos θ

);

278 4. ASYMPTOTICS OF LAPLACE TRANSFORMS∣∣∣∣h(λ)λ

∣∣∣∣ ≤ |1 + e−2iθ| η2

η2 − ε22

1

|η| − ε= 2cos θ

η2

η2 − ε22

1

|η| − ε.

Consequently,

‖ I4(t) ‖ ≤ 1

2πK2 · 2 η2

η2 − ε22

1

|η| − εεπ = 4Kε

η2

η2 − ε21

|η| − ε.

I5(t): Let λ := iη + εeiθ, θ ∈ (π2 ,3π2 ). Then

‖ eλtgt(λ) ‖ =

∥∥∥∥ ∫ t

0

eλte−λsf(s) ds

∥∥∥∥=

∥∥∥∥ ∫ t

0

eλte−εseiθ d

dsF1(s) ds

∥∥∥∥=

∥∥∥∥ eλte−εteiθF1(t) + εeiθ∫ t

0

eλte−εseiθF1(s) ds

∥∥∥∥≤ K + εK

∫ t

0

e−ε(s−t) cos θds

≤ K

(1 +

1

| cos θ|)

.

Thus, as for I4(t) one obtains

‖ I5(t) ‖ ≤ 4Kεη2

η2 − ε21

|η| − ε.

d) The estimates for I1(t) and I3(t) now include additional factors appearingin (4.17). The dominated convergence theorem shows that limt→∞ ‖I2(t)‖ = 0.Thus, we obtain that

lim supt→∞

∥∥∥∥ ∫ t

0

f(s) ds

∥∥∥∥≤ 2‖f‖∞

R

(1 +

ε2

(R− |η|)2)

η2

η2 − ε2+ 8Kε

η2

η2 − ε21

|η| − ε.

Letting ε ↓ 0, we obtain

lim supt→∞

∥∥∥∥∫ t

0

f(s) ds

∥∥∥∥ ≤ 2‖f‖∞R

.

Finally, letting R→∞ proves the claim.

We derive from Theorem 4.4.8 a Tauberian theorem of type E for slowlyoscillating functions.

Corollary 4.4.10. Let f ∈ L∞(R+, X) be slowly oscillating. Assume that

a) sp(f) = ∅, orb) sp(f) = {η} and supt≥0 ‖

∫ t

0e−iηsf(s) ds ‖ <∞, where η ∈ R.

Then limt→∞ f(t) = 0.

Proof. We give the proof in the case when sp(f) = {η}.a) Assume that η �= 0. For δ > 0, consider fδ(t) := 1

δ(f(δ + t) − f(t)) as

before. It follows from (4.1) that sp(fδ) ⊂ {η}. Moreover,∫ t

0

e−iηsfδ(s) ds = δ−1

(∫ δ+t

δ

eiη(δ−s)f(s) ds−∫ t

0

e−iηsf(s) ds

),

which is bounded. It follows from Theorem 4.4.8 that limt→∞∫ t

0fδ(s) ds = fδ(0) =

−1δ

∫ δ

0f(s) ds (by (4.1)). Hence, B- limt→∞ f(t) = 0 by Lemma 4.2.12. Theorem

4.2.3 implies that limt→∞ f(t) = 0.b) Assume that η = 0. Let g(t) := eitf(t). Then g is bounded and slowly

oscillating. Moreover, sp(g) ⊂ {i} and supt≥0

∥∥ ∫ t

0e−isg(s) ds

∥∥ < ∞. It followsfrom a) that limt→∞ g(t) = 0. Hence, limt→∞ f(t) = 0.

Corollary 4.4.11. Let f : R+ → X be Lipschitz continuous such that

supt≥0

∥∥∥∥∫ t

0

f(s) ds

∥∥∥∥ <∞. (4.18)

Assume that sp(f) ⊂ {0}. Then limt→∞ f(t) = 0.

Note that (4.18) implies that C- limt→∞ f(t) = 0. Thus, Corollary 4.4.11 is aTauberian theorem of type A.

Proof of Corollary 4.4.11. We show that f is bounded. Then we can apply Corol-lary 4.4.10. Let F (t) :=

∫ t

0f(s) ds. By assumption, M := supt≥0 ‖F (t)‖ < ∞.

Moreover, there exists L ≥ 0 such that

‖f(t)− f(s)‖ ≤ L|t− s| (t, s ≥ 0).

Let x∗ ∈ X∗ such that ‖x∗‖ ≤ 1. Then x∗ ◦ f is differentiable a.e. and∣∣∣∣ ddt (x∗ ◦ f) (t)∣∣∣∣ ≤ L (t ≥ 0).

The Taylor expansion for x∗ ◦ F yields for s ≥ 0,

〈F (s+ 1), x∗〉 = 〈F (s), x∗〉+ 〈f(s), x∗〉+∫ s+1

s

(s+ 1− t)d

dt〈f(t), x∗〉 dt.


Hence,

|〈f(s), x∗〉| ≤ 2M +

∫ s+1

s

(s+ 1− t) · Ldt

= 2M +L

2.

The Hahn-Banach theorem implies that ‖f(s)‖ ≤ 2M + 12L (s ≥ 0).

Corollary 4.4.10 and 4.4.11 are first versions of a complex Tauberian theoremwhere the main hypothesis says that the spectrum is sufficiently small. General-izations will be given in Section 4.7 (Corollary 4.7.8) and in Section 4.9 (Theorem4.9.7). The proof of both generalizations will make use of Gelfand’s theorem whichwe obtain now as another immediate consequence of the contour method. In fact,applying Corollary 4.4.11 to bounded groups we obtain the following result.

Corollary 4.4.12 (Gelfand). Let A be the generator of a bounded C0-group U =(U(t))t∈R on X. If σ(A) ⊂ {0}, then U(t) = I for all t ∈ R. In particular, if σ(A)is empty, then X = {0}.

Proof. Let x ∈ D(A2), f(t) := AU(t)x. Then∫ t

0f(s) ds = U(t)x − x, which

is bounded. Since f(λ) = R(λ,A)Ax, we have sp(f) ⊂ {0}. Since ddtf(t) =

U(t)A2x, the function f is Lipschitz continuous. It follows from Corollary 4.4.11that limt→∞ f(t) = 0. Let M := supt∈R ‖U(t)‖. Since ‖Ax‖ = ‖U(−t)U(t)Ax‖ ≤M‖U(t)Ax‖, it follows that Ax = 0. Hence, U(t)x = x for all x ∈ D(A2), t ∈ R.Since R(1, A) has dense range D(A), the range D(A2) of R(1, A)2 is also dense.It follows that U(t) = I (t ∈ R). When σ(A) is empty, one can also apply theprevious case to A− iη for any η ∈ R.

Corollary 4.4.13. Let A be the generator of a bounded C0-group U . Then eachisolated point in σ(A) is an eigenvalue.

Proof. Let iη ∈ σ(A) be isolated. Consider the spectral projection P associatedwith iη, and let Y := PX �= {0} (see Proposition B.9). Then the group leaves thespace Y invariant. Consider the restricted group. The spectrum of its generatoris reduced to {iη}. Applying Corollary 4.4.12 to e−iηtU(t)|Y , one obtains thate−iηtU(t)y = y for all y ∈ Y .

Next, we prove a result on the asymptotic behaviour of a bounded C0-semigroup T with generator A, specifically the possible decay of ‖T (t)R(μ,A)‖for large t, where μ ∈ ρ(A) is fixed. Note that the rate of any such decay is inde-pendent of the choice of μ up to multiplicative constants, since (μ′−A)R(μ,A) isalways a bounded operator.

Let m : R+ → (0,∞) be a continuous decreasing function such that m(t)→ 0as t → ∞. We let m−1 denote any decreasing function from (0,∞) to R+ suchthat m(m−1(t)) = t for all t in the range of m.


Theorem 4.4.14. Let A be the generator of a bounded C0-semigroup T , and letμ ∈ ρ(A). The following are equivalent:

(i) σ(A) ∩ iR is empty.

(ii) limt→∞ ‖T (t)R(μ,A)‖ = 0.

More precisely:

a) Assume that σ(A) ∩ iR is empty, and let M : R+ → (0,∞) be a continuousincreasing function such that ‖R(is, A)‖ ≤ M(|s|) for all s ∈ R, and letc ∈ (0, 1). Then there exist C and t0 such that

‖T (t)R(μ,A)‖ ≤ C

M−1log (ct)

for all t ≥ t0. (4.19)

b) Assume that limt→∞ ‖T (t)R(μ,A)‖ = 0, and let m : R+ → (0,∞) be acontinuous decreasing function such that ‖T (t)R(μ,A)‖ ≤ m(t) for t ≥ 0and limt→∞m(t) = 0. Then σ(A)∩ iR is empty, and for each c ∈ (0, 1) thereexist C and s0 such that

‖R(is, A)‖ ≤ Cm−1

(c

|s|)

whenever |s| ≥ s0.

Proof. The implication from (1) to (2) can be seen from Theorem 4.4.1 with f(t) =T (t). Although that function is not measurable from R+ to L(X), its Laplace

transform f(λ) = R(λ,A) is holomorphic, and the proof of Theorem 4.4.1 remainsvalid. Alternatively, we give the following more precise argument for a). We letK = supt≥0 ‖T (t)‖.

a) First, by Corollary B.3, for any a > 1, ΩaM ⊂ ρ(A) and ‖R(λ,A)‖ ≤a

a−1M(| Imλ|) for all λ ∈ ΩaM .Let x ∈ X with ‖x‖ ≤ (a − 1)/a, and let f(t) = T (t)x (t ≥ 0). Then

‖f‖∞ ≤ K and f(λ) = R(λ,A)x, so ‖f(λ)‖ ≤ M(| Imλ|) for λ ∈ ΩaM . ByProposition 3.1.9 e),

T (t)R(μ,A)x = R(μ,A)x+ AR(μ,A)

∫ t

0

f(s) ds

= AR(μ,A)

(∫ t

0

f(s) ds− f(0)

).

Since AR(μ,A) is a bounded operator, the result follows from Theorem 4.4.8 andits proof, noting that C and t0 are independent of x for ‖x‖ ≤ (a− 1)/a.

b) Take s ∈ R, and consider x∗ ∈ D(A∗) with A∗x∗ = isx∗. Let x ∈ D(A),and put g(t) = e−ist〈T (t)x, x∗〉 (t ≥ 0). Then

g′(t) = e−ist〈AT (t)x− isT (t)x, x∗〉 = 0,


so g is constant. By the assumption (ii), limt→∞ ‖T (t)x‖ = 0, so limt→∞ g(t) = 0.Hence 0 = g(0) = 〈x, x∗〉. Since D(A) is dense, it follows that x∗ = 0. By theHahn-Banach theorem, A− is has dense range.

Let c′ ∈ (c, 1), and x ∈ D(A). By Proposition 3.1.9 f) applied to the rescaledsemigroup, ∫ t

0

e−isτT (τ)(A− is)x dτ = e−istT (t)x− x

for any t ≥ 0. Hence,

Kt‖(A− is)x‖ ≥∥∥∥∥∫ t

0

e−isτT (τ)(A− is)x dτ

∥∥∥∥≥ ‖x‖ − ‖T (t)x‖≥ ‖x‖ −m(t)‖(A− μ)x‖≥ ‖x‖ −m(t)

(‖(A− is)x‖+ (|μ|+ |s|)‖x‖).Thus,

‖(A− is)x‖ ≥ 1−m(t)(|μ|+ |s|)Kt+m(t)

‖x‖.

Let t = m−1 (c′/(|μ|+ |s|)), so m(t) ≤ c′/(|μ|+ |s|). Thus,

‖(A− is)x‖ ≥ (1− c′)‖x‖Km−1(c′/(|μ|+ |s|)) + c′/(|μ|+ |s|) .

This implies that A− is has closed range, and hence it is invertible. Moreover,

‖R(is, A)‖ ≤ K

1− c′m−1

(c′

|μ|+ |s|)+

c′

(1− c′)(|μ|+ |s|) ≤ Cm−1

(c

|s|)

whenever |s| is sufficiently large.

Part a) of Theorem 4.4.14 gives the sharpest result when M is chosen assmall as possible, that is,

M(s) = sup {‖R(is′, A)‖ : |s′| ≤ s} .Then part b) provides the lower bound

‖T (t)R(μ,A)‖ ≥ c

M−1(Ct)

for t sufficiently large and some constants c > 0 and C, and the sharpest estimateis given by taking

M−1(t) = inf{|s| : ‖R(is, A)‖ ≥ t}.Then the gap between this lower bound and the upper bound in (4.19) is small if Mgrows rapidly. To understand this more explicitly, we consider diagonal semigroupson Hilbert space.


Example 4.4.15. Let X = �2, (αn)n∈N be a strictly decreasing sequence in (0, 1],and λn = −αn + in. A C0-semigroup of contractions on �2 is defined by

T (t)x =(eλntxn

)(x = (xn) ∈ �2).

The generator A is given by

D(A) ={x ∈ �2 : (λnxn) ∈ �2

}={x ∈ �2 : (nxn) ∈ �2

},

Ax = (λnxn).

Moreover,

σ(A) = {λn : n ∈ N},‖R(is, A)‖ = sup

{|λn − is|−1 : n ∈ N}.

Take the optimal definitions of M and M−1 as above. Then M(k) = ‖R(ik, A)‖ =1/αk (k ∈ N). Let tk = 1/αk. Then

‖T (tk)A−1‖ = supn∈N

∣∣∣∣eλntk

λn

∣∣∣∣ ≥ e−1

|λk| ∼e−1

k=

e−1

M−1(tk)as k →∞.

When (αn) decays regularly and not too slowly, we may have

‖T (t)A−1‖ ∼ c

M−1(t)as t→∞. (4.20)

For example, when αn = n−γ where γ > 0,

‖T (t)A−1‖ ∼ cγt1/γ

∼ cγM−1(t)

.

On the other hand, (4.20) may fail if (αn) has relatively long periods of slowdecay. Suppose for example that (αn) is strictly decreasing and

α(k+1)! ≥ αk!+1

2(k ≥ 1).

Given c > 0, let tk = 1/(cα(k+1)!). Then

M−1(ctk) = (k + 1)!,

‖T (tk)A−1‖ ≥∣∣∣∣eλk!+1tkλk!+1

∣∣∣∣ ≥ e−2/c

|λk!+1| ∼ke−2/c

M−1(ctk).

Thus there is no upper bound of the form ‖T (t)A−1‖ ≤ C/M−1(ct). By arrangingthat the subsequence (αk!) decreases arbitrarily fast, this phenomenon can befound even when M increases arbitrarily fast.


The next result on asymptotic behaviour of C0-semigroups is analogous tothe Katznelson-Tzafriri theorem for contractions (see the Notes of this chapter).A C0-semigroup T = (T (t))t≥0 is called eventually differentiable if there existsτ ≥ 0 such that T (τ)X ⊂ D(A). Semigroup properties imply that T (t)X ⊂ D(A)for all t ≥ τ and T (t)X ⊂ D(A2) for all t ≥ 2τ . By the closed graph theorem,AT (t) ∈ L(X) for t ≥ τ and A2T (t) ∈ L(X) for t ≥ 2τ .

Theorem 4.4.16. Let A be the generator of an eventually differentiable, boundedC0-semigroup. The following are equivalent:

(i) σ(A) ∩ iR ⊂ {0}.(ii) limt→∞ ‖AT (t) ‖ = 0.

Proof. Let K := supt≥0 ‖T (t) ‖. Assume that (i) holds. Let f : R+ → L(X) bedefined by f(t) := AT (t+ 2τ). For x ∈ X,

‖f(t)x− f(s)x‖ =

∥∥∥∥∫ s

t

d

dr(f(r)x) dr

∥∥∥∥=

∥∥∥∥∫ s

t

A2T (r + 2τ)x dr

∥∥∥∥=

∥∥∥∥∫ s

t

T (r)A2T (2τ)x dr

∥∥∥∥≤ K‖A2T (2τ)‖ |t− s| ‖x‖ (s, t ≥ 0).

Thus f is Lipschitz continuous. Moreover, f(λ) = R(λ,A)AT (2τ). Hence, we have

sp(f) ⊂ {0}. Finally, ‖ ∫ t

0f(s) ds ‖ = ‖T (t+2τ)−T (2τ) ‖ ≤ 2K (t ≥ 0). It follows

from Corollary 4.4.11 that limt→∞ ‖ f(t) ‖ = 0.Conversely, assume that (ii) holds. Let iη ∈ σ(A). By Proposition B.2 d),

there exist xn ∈ D(A) with ‖xn ‖ = 1 such that limn→∞ ‖ (A− iη)xn ‖ = 0. Then(iηeiηt − AT (t)

)xn = iη

(eiηt − T (t)

)xn + T (t) (iη −A)xn

= iηeiηt∫ t

0

e−iηsT (s)(iη − A)xn ds+ T (t)(iη − A)xn

→ 0 as n→∞.

Thus, iηeiηt ∈ σ(AT (t)). Hence, |η| = |iηeiηt| ≤ ‖AT (t) ‖ for all t ≥ τ . Sincelimt→∞ ‖AT (t) ‖ = 0, we conclude that η = 0.

Theorem 4.4.1 shows that limt→∞∫ t

0f(s) ds = f(0) if f ∈ L∞(R+, X) and

sp(f) is empty. This is not true if the assumption that f is bounded is omitted(see Example 1.5.2). In the final part of this section we shall show that it is true

if f is exponentially bounded and f has a bounded holomorphic extension to ahalf-plane {λ : Reλ > −ε} for some ε > 0. Indeed, we shall show that

abs(f) ≤ hol0(f)

whenever f is exponentially bounded, where

hol0(f) := inf{ω ∈ R : f has a bounded holomorphic extension for Reλ > ω}

is the abscissa of boundedness of f . We begin with a general estimate.

Proposition 4.4.17. Let f ∈ L1loc(R+, X) be exponentially bounded and suppose that

hol(f) ≤ 0 and f is bounded on C+. Then there is a constant C such that∥∥∥∥∫ ∞

0

φ(t)f(t) dt

∥∥∥∥ ≤ C‖Fφ‖1

for all functions φ ∈ L1(R+) such that Fφ ∈ L1(R) and φf ∈ L1(R+, X).

Proof. First, we assume that φ has compact support and Fφ ∈ L1(R). The

Laplace transform φ is defined on C, and (Fφ)(s) = φ(is) = (Fφ)(−is). Takeω > max(0, ω(f)) and 0 < α < ω. The function t �→ e−ωtf(t) belongs to L1(R+, X)

and its Fourier transform is s �→ f(ω + is). Let ψ ∈ C∞c (R) with ψ(t) = e(ω−α)t

whenever t ∈ suppφ. Then (Fψ) ∗ (Fφ) ∈ L1(R). By the Fourier Inversion The-orem 1.8.1 d), F((Fψ) ∗ (Fφ))(t) = 4π2ψ(t)φ(t) = 4π2e(ω−α)tφ(t) for all t, and

((Fψ) ∗ (Fφ))(s) = 2πφ(α− ω − is) for all s. By Theorem 1.8.1 b),∫ ∞

0

e−αtf(t)φ(t) dt =

∫ ∞

0

e−ωtf(t)e(ω−α)tφ(t) dt

=1

2π

∫ ∞

−∞f(ω + is)φ(α− ω − is) ds. (4.21)

Now consider the contour integral∫f(z)φ(α− z) dz

around the rectangle with vertices α± ir, ω± ir, where r > 0. The integral alongthe bottom edge is ∫ ω

α

f(ξ − ir)φ(α − ξ + ir) dξ.

For α < ξ < ω,

φ(α− ξ + ir) =

∫ ∞

0

e−(α−ξ)tφ(t)e−irt dt→ 0

as r →∞, by the Riemann-Lebesgue lemma. Moreover,∣∣∣φ(α− ξ + ir)∣∣∣ ≤ 1√

2α

(∫ ∞

0

e2ωt|φ(t)|2 dt)1/2

,∥∥∥f(ξ − ir)∥∥∥ ≤ sup

z∈C+

‖f(z)‖,

whenever r > 0, α < ξ < ω. By the dominated convergence theorem,

limr→∞

∫ ω

α

f(ξ − ir)φ(α − ξ + ir) dξ = 0.

A similar argument shows that the integral along the top edge of the rectangletends to 0 as r →∞. By Cauchy’s theorem,

limr→∞

{∫ r

−r

f(ω + is)φ(α− ω − is) ds−∫ r

−r

f(α+ is)φ(−is) ds}

= 0.

From (4.21),∫ ∞

0

e−αtf(t)φ(t) dt = limr→∞

1

2π

∫ r

−r

f(α+ is)(Fφ)(−s) ds,

so ∥∥∥∥∫ ∞

0

e−αtφ(t)f(t) dt

∥∥∥∥ ≤ 1

2πsupz∈C+

∥∥∥f(z)∥∥∥ ‖Fφ‖1.Letting α ↓ 0 gives∥∥∥∥∫ ∞

0

φ(t)f(t) dt

∥∥∥∥ ≤(

1

2πsupz∈C+

‖f(z)‖)‖Fφ‖1.

Now, consider the case when φ is any function in L1(R+) such that Fφ ∈L1(R) and

∫∞0|φ(t)| ‖f(t)‖ dt < ∞. Let ψ ∈ C∞c (R) be any function satisfying

0 ≤ ψ ≤ 1, ψ(0) = 1 and∫∞−∞ ψ(t) dt = 1. Let ψn(t) = ψ(t/n) (t ∈ R) and

φn(t) = φ(t)ψn(t). Then (2π)−1(Fψn)(s) = (2π)−1n(Fψ)(ns), which forms amollifier. Hence, Fφn = (2π)−1Fφ ∗ Fψn → Fφ in L1(R) (see Lemma 1.3.3).Applying the previous result to the functions φn and taking the limit provides theresult.

Next, we show that the antiderivative can grow at most linearly if f isbounded on C+ and f is exponentially bounded. We shall apply this result toC0-semigroups in Theorem 5.1.8.

Theorem 4.4.18. Let f ∈ L1loc(R+, X) be exponentially bounded and suppose that

hol(f) ≤ 0 and f is bounded on C+. Then there is a constant c such that∥∥∥∥∫ t

0

f(s) ds

∥∥∥∥ ≤ c(1 + t) for all t ≥ 0.

Proof. Let C be as in Proposition 4.4.17 and let ω > max(0, ω(f)), so there existsM such that ‖f(s)‖ ≤ Meωs for all s ≥ 0. Take t > 0 and let α := t

2e−ωt and

φ := 1αχ(0,α) ∗ χ(0,t), so

φ(s) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩s/α (0 ≤ s < α),

1 (α ≤ s < t),t+ α− s

α(t ≤ s < t+ α),

0 (t+ α ≤ s).

Then α ≤ 1/(2ωe) and

(Fφ)(s) =1

α

(1− e−iαs

is

)(1− e−ist

is

)=

4e−isα/2e−ist/2

αs2sin

αs

2sin

st

2.

Hence,

‖Fφ‖1 =4

α

∫ ∞

0

| sin αs2 sin st

2 |s2

ds

≤ 2

∫ 1/α

0

| sin st2 |

sds+

4

α

∫ ∞

1/α

ds

s2

= 2

∫ t/2α

0

| sinu|u

du+ 4

≤∫ t/2α

1

du

u+ 6

= log (t/2α) + 6

= ωt+ 6.

Also,∫ t

0

f(s) ds =

∫ ∞

0

φ(s)f(s) ds+

∫ α

0

(1− s

α

)f(s) ds−

∫ t+α

t

t+ α− s

αf(s) ds,

so ∥∥∥∥∫ t

0

f(s) ds

∥∥∥∥ ≤∥∥∥∥∫ ∞

0

φ(s)f(s) ds

∥∥∥∥+

∫ α

0

‖f(s)‖ ds+∫ t+α

t

‖f(s)‖ ds

≤ C(ωt+ 6) +Mαeωα(1 + eωt)

≤ C(ωt+ 6) +Mt

2e−ωte1/2e2eωt

= 6C +(Cω +Me1/2e

)t.

Finally, we prove the result comparing abs(f) and hol0(f).

Theorem 4.4.19. Let f ∈ L1loc(R+, X) be exponentially bounded. Then abs(f) ≤

hol0(f).

Proof. Let ω > hol0(f) and let fω(t) := e−ωtf(t). Then fω(λ) = f(λ + ω), so

hol(fω) < 0 and fω is bounded on C+. By Theorem 4.4.18 and Theorem 1.4.3,

abs(fω) ≤ 0. Hence, abs(f) = abs(fω) + ω ≤ ω whenever ω > hol0(f).

Corollary 4.4.20. Let f ∈ L1loc(R+, X) be exponentially bounded. Assume that f

has a bounded holomorphic extension to a half-plane {λ ∈ C : Reλ > −ε} forsome ε > 0. Then

limt→∞

∫ t

0

f(s) ds = f(0).

4.5 Almost Periodic Functions

This section is divided into two parts: in the first we describe relatively compactorbits of C0-groups. The abstract results which we obtain are applied in the secondpart to characterize almost periodic functions on the real line.

A subset Q of R is called relatively dense if there exists a length l > 0 suchthat

[a, a+ l] ∩Q �= ∅for all a ∈ R.

Let U = (U(t))t∈R be a C0-group of isometries on a Banach space Y . Ouraim is to prove the following.

Theorem 4.5.1. Let y ∈ Y . The following assertions are equivalent:

(i) The set {U(t)y : t ∈ R} is relatively compact in Y .

(ii) The set {U(t)y : t ≥ 0} is relatively compact in Y .

(iii) For all ε > 0, the set Qε := {τ ∈ R : ‖U(τ)y − y ‖ ≤ ε} is relatively densein R.

(iv) y ∈ span{x ∈ Y : there exists η ∈ R such that U(t)x = eiηtx for all t ∈ R

}.

If these four equivalent conditions are satisfied we say that y has a relativelycompact orbit. Since Y is complete, a subsetK of Y is relatively compact if and onlyif K is precompact (i.e., for every ε > 0 it can be covered by a finite number of ε-balls). We shall use this equivalence frequently without comment, and we shall usethe terminology “relatively compact” except in some proofs where precompactnessis more relevant.

The equivalence of the conditions (i), (ii) and (iii) will be proved by simpledirect arguments. In order to show that they imply (iv) we need the following basicresult of harmonic analysis (see [Rud62, Section 1.5.2]).

Proposition 4.5.2. Let G be a compact abelian group with Haar measure dx. Letf : G→ C be continuous such that∫

G

f(x)γ(x) dx = 0

4.5. ALMOST PERIODIC FUNCTIONS 289

for every character γ (i.e., every continuous homomorphism γ : G → C \ {0}).Then f(x) = 0 for all x ∈ G.

Proof of Theorem 4.5.1. (i) ⇒ (ii): This is trivial.(ii) ⇒ (iii). Let ε > 0. By assumption, there exist t1, . . . , tm ≥ 0 such that

for all t ≥ 0 there exists j ∈ {1, . . . ,m} such that ‖U(t)y − U(tj)y ‖ ≤ ε. Letl := maxj=1,...,m tj . We show that [t, t + l] ∩ Qε �= ∅ for all t ∈ R. Let t ≥ 0.Choose j ∈ {1, . . . ,m} such that ‖U(t)y − U(tj)y‖ ≤ ε. Let τ := t − tj . Then‖U(τ)y − y ‖ = ‖U(−tj)(U(t)y − U(tj)y) ‖ = ‖U(t)y − U(tj)y ‖ ≤ ε. Thus, τ ∈Qε ∩ [t− l, t].

Let t < 0. Then there exists j ∈ {1, . . . ,m} such that ‖U(−t)y−U(tj)y ‖ ≤ ε.Let τ := t+ tj . Then ‖U(τ)y−y ‖ = ‖U(t)(U(tj)y−U(−t)y) ‖ ≤ ε. Thus, τ ∈ Qε

and τ ∈ [t, t+ l]. We have shown that [t, t+ l] ∩Qε �= ∅ for all t ∈ R.(iii) ⇒ (i): Let ε > 0. By assumption, there exists l > 0 such that for all

n ∈ Z, [ln, ln + l] ∩ Qε �= ∅. Since {U(t)y : t ∈ [0, 2l]} is compact, there existt1, . . . , tm ∈ [0, 2l] such that for all s ∈ [0, 2l] there exists j ∈ {1, . . . ,m} such that‖U(s)y−U(tj)y ‖ ≤ ε. Let t ∈ R. Take n ∈ Z such that t ∈ [ln, ln+ l] and chooseτ ∈ Qε ∩ [−ln,−ln+ l]. Then t+ τ ∈ [0, 2l]. There exists j ∈ {1, . . . ,m} such that‖U(t+ τ)y − U(tj)y ‖ ≤ ε. Thus,

‖U(t)y − U(tj)y ‖ ≤ ‖U(t)y − U(t)U(τ)y ‖+ ‖U(t+ τ)y − U(tj)y ‖ ≤ 2ε.

We have shown that the orbit {U(t)y : t ∈ R} is covered by balls B(U(tj)y, 2ε) ={z ∈ Y : ‖U(tj)y − z ‖ ≤ 2ε} (j = 1, . . . ,m). Thus, (i) holds.

(i) ⇒ (iv): Replacing Y by span{U(t)y : t ∈ R}, we can assume that everyorbit {U(t)x : t ∈ R} (x ∈ Y ) is relatively compact. Then U is bounded, bythe uniform boundedness principle. Denote by G the closure of {U(t) : t ∈ R}in Ls(Y ), the space L(Y ) with respect to the strong operator topology. ThenGx ⊂ K(x) := {U(t)x : t ∈ R}− for all x ∈ Y . By hypothesis, K(x) is compact.We consider the strong operator topology on G, and all limits in this proof willbe in that topology. Then G can be identified with a closed subset of

∏x∈Y K(x)

via S ∈ G �→ {Sx : x ∈ Y }. Hence, G is compact by Tychonov’s theorem. Sincemultiplication is jointly continuous on bounded subsets of Ls(X), S, T ∈ G impliesST ∈ G. Let T ∈ G. There exists a net (ti)i∈I in R such that T = limi∈I U(ti).Considering a subnet if necessary, we can assume that S := limi∈I U(−ti) existsas well. Then TS = ST = I. We show that inversion is continuous. In fact, letlimi∈I Ti = T in G. It follows from compactness that every subnet of (Ti)i∈I hasa subnet (Tj)j∈J such that limj T

−1j exists and joint continuity of multiplication

implies that limj T−1j = T−1. This implies that limi T

−1i = T−1. Thus, G is a

compact abelian group with continuous multiplication.Denote by G the dual group of G and by dS the Haar measure on G.

For γ ∈ G, x ∈ Y define Pγx :=∫Gγ(S)Sx dS ∈ Y . Then Pγ ∈ L(Y ) and

TPγx =∫Gγ(S)TSxdS = γ(T )

∫Gγ(TS)TSx dS = γ(T )Pγx (T ∈ G). The map-

ping φ(t) := γ(U(t)) is a continuous character on R. Hence, for each γ ∈ G, there


exists η ∈ R such that γ(U(t)) = eiηt (t ∈ R). Thus,

F := {Pγx : γ ∈ G, x ∈ Y }⊂ {

z ∈ Y : there exists η ∈ R such that U(t)z = eiηty for all t ∈ R}.

It remains to show that spanF is dense in Y . Let ψ ∈ Y ∗ such that 〈z, ψ〉 = 0 forall z ∈ F . Let x ∈ Y . Then

∫Gγ(S)〈Sx, ψ〉 dS = 0 for all γ ∈ G. Since S �→ 〈Sx, ψ〉

is a continuous mapping, it follows from Proposition 4.5.2 that 〈Sx, ψ〉 = 0 for allS ∈ G. In particular, 〈x, ψ〉 = 0. Thus, ψ = 0. It follows from the Hahn-Banachtheorem that spanF = Y .

(iv) ⇒ (i): It is obvious that the set

Y1 := {x ∈ Y : x has precompact orbit}is a closed subspace of Y . Since x ∈ Y1 whenever U(t)x = eiηtx for all t ∈ R, theimplication follows.

We recall the following facts from Section 4.3. An element x ∈ Y is calledtotally ergodic if

Mηx := limt→∞

1

t

∫ t

0

e−iηsU(s)x ds

converges for all η ∈ R. In that case,

U(t)Mηx = eiηtMηx (t ∈ R).

Moreover, the set of all frequencies

Freq(x) := {η ∈ R : Mηx �= 0}is countable (by Proposition 4.3.11).

It is obvious that the set Ye of all totally ergodic vectors is a closed subspaceof Y . We introduce the space Yap of all almost periodic vectors (with respect toU) defined by

Yap := {x ∈ Y : x has relatively compact orbit}= span{x ∈ Y : there exists η ∈ R such that U(t)x = eiηtx for all t ∈ R}.

Then Yap is a closed subspace of Y which is invariant under U . If x ∈ Y is aperiodic vector, i.e.,

U(t)x = eiξtx (t ∈ R)

for some ξ ∈ R, then Freq(x) ⊂ {ξ} and

Mηx =

{x if η = ξ,

0 if η �= ξ.(4.22)

It follows that all almost periodic vectors are totally ergodic.The following approximation result shows that every x ∈ Yap can be ap-

proximated by linear combinations of eigenvectors associated with frequencies ofx.


Proposition 4.5.3 (Spectral synthesis). Let x ∈ Yap. Then x ∈ span{y ∈ Y :there exists η ∈ Freq(x) such that U(t)y = eiηty for all t ∈ R}.Proof. The space Z := {y ∈ Yap : Mηy = 0 for all η ∈ R\Freq(x)} is closed andinvariant under the group. If η ∈ R\Freq(x) and y ∈ Z such that U(t)y = eiηty (t ∈R), then y = Mηy = 0. Now the claim follows from Theorem 4.5.1, applied to therestriction of U to Z.

Corollary 4.5.4. Let x ∈ Yap. Then

a) x = 0 if and only if Freq(x) = ∅.b) Freq(x) ⊂ {η} if and only if U(t)x = eiηtx (t ∈ R).

c) Freq(x) ⊂ {η1, . . . , ηm} if and only if x =∑m

j=1 xj with U(t)xj = eiηjtxj forall t ∈ R.

d) Let τ > 0. Then U(t+τ)x = U(t)x for all t ∈ R if and only if Freq(x) ⊂ 2πτZ.

Proof. a) and b) follow directly from Proposition 4.5.3. If Freq(x)⊂{η1, . . . , ηm},then by Proposition 4.5.3, x = limn→∞ xn, where xn =

∑mj=1 xnj for some xnj ∈ X

such that U(t)xnj = eiηjtxnj (j = 1, . . . ,m). We can assume that ηj �= ηk fork �= j. Then Mηj

x = limn→∞Mηjxn = limn→∞ xnj by (4.22). It follows that

x =∑m

j=1Mηjx. This proves one implication of c). The other follows from (4.22).We prove d). Assume that U(t + τ)x = U(t)x (t ∈ R). Recall that for z ∈

C, |z| = 1, one has

limn→∞

1

n

n−1∑k=0

zk =

{1 if z = 1,

0 if z �= 1.

Let η ∈ R. Then

Mηx = limn→∞

1

τn

∫ τn

0

U(s)xe−iηs ds

= limn→∞

1

τn

n−1∑k=0

∫ (k+1)τ

kτ

U(s)xe−iηs ds

= limn→∞

1

τn

n−1∑k=0

∫ τ

0

U(s)xe−iηs ds e−ikητ

=

{1/τ

∫ τ

0U(s)xe−iηs ds if ητ ∈ 2πZ,

0 if ητ �∈ 2πZ.

This proves one implication. The other follows directly from Proposition 4.5.3.

Now we come to the second part of this section where we consider a specialgroup of operators. Let X be a Banach space. By BUC(R, X) we denote the space


of all bounded uniformly continuous functions on R with values inX. It is a Banachspace for the uniform norm

‖ f ‖∞ = supt∈R

‖ f(t) ‖.

The shift group on BUC(R, X) defined by

(S(t)f)(r) := f(r + t) (r ∈ R, t ∈ R)

is a C0-group whose generator is denoted by B.

Lemma 4.5.5. For f ∈ BUC(R, X) one has f ∈ D(B) if and only if f is differen-tiable and f ′ ∈ BUC(R, X). In that case, Bf = f ′.

Proof. Let f ∈ D(B), g := Bf . Then S(t)f − f =∫ t

0S(s)g ds for all t ≥ 0. In

particular, f(t) = (S(t)f)(0) = f(0) +∫ t

0g(s) ds. This shows one implication of

the claim. Conversely, assume that f is differentiable and g := f ′ ∈ BUC(R, X).

Then (S(t)f − f)(r) = f(r + t) − f(r) =∫ r+t

rg(s) ds =

∫ t

0(S(s)g)(r) ds. Thus,

S(t)f − f =∫ t

0S(s)g ds. Hence, f ∈ D(B) and Bf = g.

Let η ∈ R, x ∈ X. By eiη ⊗ x we denote the periodic function given by

(eiη ⊗ x)(t) = eiηtx (t ∈ R).

Linear combinations of functions of the form eiη ⊗ x with η ∈ R, x ∈ X are calledtrigonometric polynomials.

Definition 4.5.6. A function f : R → X is called almost periodic if it can beapproximated uniformly on R by trigonometric polynomials. By

AP(R, X) := span{eiη ⊗ x : η ∈ R, x ∈ X}

we denote the space of all almost periodic functions on R with values in X.

Let η ∈ R. A function f ∈ BUC(R, X) satisfies

S(t)f = eiηtf for all t ∈ R

if and only if f = eiη ⊗ f(0). Thus, considering the group S on Y := BUC(R, X),we have Yap = AP(R, X). Now we can reformulate the results of the first part ofthis section for this special case.

Let f ∈ BUC(R, X). Let ε > 0. A real number τ > 0 is called an ε-period off if ‖f(τ + s)− f(s)‖ ≤ ε for all s ∈ R.

Theorem 4.5.7. Let f ∈ BUC(R, X). The following are equivalent:

(i) f is almost periodic.


(ii) For every ε > 0 the set of all ε-periods is relatively dense in R.

(iii) The orbit {S(t)f : t ∈ R} is relatively compact in BUC(R, X).

This is an immediate consequence of Theorem 4.5.1. It follows in particularthat for every f ∈ AP(R, X) there exist tn ∈ R such that limn→∞ tn =∞ and

‖ f(tn + s)− f(s) ‖ ≤ 1

nfor all s ∈ R. (4.23)

In particular, if f ∈ AP(R, X), then

‖ f ‖∞ = supt≥τ

‖ f(t) ‖ (4.24)

for all τ ∈ R.For f ∈ AP(R, X), η ∈ R, we define the mean

Mηf := limt→∞

1

t

∫ t

0

e−iηsS(s)f ds

= limα↓0

αR(α+ iη, B)f.

These limits exist in BUC(R, X) (i.e., with respect to the uniform norm) by the re-marks preceding Proposition 4.5.3 and by Proposition 4.3.1. Moreover, S(t)Mηf =eiηtMηf for all t ∈ R. Evaluating at 0, we deduce that

Mηf = eiη ⊗ (Mηf)(0). (4.25)

As before in the abstract setting, we let

Freq(f) := {η ∈ R : Mηf �= 0}

be the set of all frequencies of f . By Proposition 4.3.11, this is a countable set.Moreover the following property of spectral synthesis holds.

Proposition 4.5.8 (Spectral synthesis). Let f ∈ AP(R, X). Then

f ∈ span{eiη ⊗ x : η ∈ Freq(f), x ∈ f(R), }

where the closure is taken in BUC(R, X).

Proof. Let X0 := span{f(t) : t ∈ R} ⊂ X. Then by Theorem 4.5.7, f is also almostperiodic when it is considered as a function with values in X0. So we can assumethat X0 = X. Now the claim follows from Proposition 4.5.3.

From Corollary 4.5.4 we see the following.


Corollary 4.5.9. Let f ∈ AP(R, X). Then

a) f ≡ 0 if and only if Freq(f) = ∅.b) Freq(f) ⊂ {η} if and only if f = eiη ⊗ f(0).

c) f is a trigonometric polynomial if and only if Freq(f) is finite.

d) f is τ -periodic if and only if Freq(f) ⊂ 2πτ Z (where τ > 0).

We call assertion a) the uniqueness theorem for almost periodic functions.We also obtain from Theorem 4.5.7 that almost periodic functions have rel-

atively compact range.

Corollary 4.5.10. Let f ∈ AP(R, X). Then the set {f(t) : t ∈ R}− is compact.

Proof. By hypothesis, the set {S(t)f : t ∈ R} is relatively compact in BUC(R, X).Since evaluation at 0 is a continuous operator from BUC(R, X) into X , it followsthat the set {f(t) : t ∈ R} = {(S(t)f)(0) : t ∈ R} is relatively compact in X.

A function f ∈ BUC(R, X) is called weakly almost periodic if x∗ ◦ f is almostperiodic for all x∗ ∈ X∗.

Remark 4.5.11. There are various different definitions of weak almost periodicityin the literature. In particular, a function f ∈ BUC(R, X) is called weakly almostperiodic in the sense of Eberlein if the set

{S(t)f : t ∈ R}is relatively weakly compact in the Banach space BUC(R, X). This property isindependent of weak almost periodicity, in general. We refer to the Notes.

Proposition 4.5.12. Assume that f ∈ BUC(R, X) has relatively compact range. Iff is weakly almost periodic, then f is almost periodic.

Proof. Since f is separably valued, we can assume that X is separable. Then wecan consider X as a closed subspace of C[0, 1] (see [Woj91, p.36]). Let (Pn)n∈N bea bounded sequence of finite rank operators on C[0, 1] such that limn→∞ Png = gfor all g ∈ C[0, 1] (see [Woj91, pp.37,40]). Then Pn ◦ f is weakly almost periodic,hence almost periodic since it has values in a finite dimensional space. We havelimn→∞ Pn(f(t)) = f(t) for all t ∈ R. Since f has relatively compact range, thisconvergence is uniform in t ∈ R (see Proposition B.15). Thus, Pn ◦ f converges tof in BUC(R, C[0, 1]). Consequently, f is almost periodic.

Remark 4.5.13 (Almost periodic orbits). Let U be a bounded C0-group on a Ba-nach space Y and let x ∈ Y . Then the following are equivalent:

(i) x ∈ Yap (i.e., {U(t)x : t ∈ R} is relatively compact in Y ).

(ii) U(·)x ∈ AP(R, Y ).

4.6. COUNTABLE SPECTRUM AND ALMOST PERIODICITY 295

In fact, if U(·)x ∈ AP(R, Y ), then {U(t)x : t ∈ R} is relatively compact, byCorollary 4.5.10. Conversely, let x ∈ Yap. Let tn ∈ R. Then there exists a sub-sequence such that limk→∞ U(tnk

)x := y exists. This implies that U(tnk+ ·)x

converges to U(·)y in BUC(R, Y ). Thus, {U(t+ ·)y : t ∈ R} is relatively compactin BUC(R, Y ).

4.6 Countable Spectrum and Almost Periodicity

In this section we define the spectrum of a function f ∈ BUC(R, X) with thehelp of the Laplace transform (or more precisely, the Carleman transform). Themain result of this section says that, under suitable conditions on the space X, afunction f ∈ BUC(R, X) with countable spectrum is almost periodic.

Let f ∈ BUC(R, X). The Carleman transform f of f is defined by

f(λ) :=

{ ∫∞0

e−λtf(t) dt (Reλ > 0),

− ∫∞0

eλtf(−t) dt (Reλ < 0).

Thus, f is a holomorphic function defined on C\iR.Remark 4.6.1. Let f+ ∈ BUC(R+, X) be the restriction of f to R+ and f− ∈BUC(R+, X) be given by f−(t) = f(−t) (t ∈ R+). Then

f(λ) =

{f+(λ) (Reλ > 0),

−f−(−λ) (Reλ < 0),

where f+ and f− are the Laplace transforms of f+ and f−, respectively.

We use the same symbol for the Carleman transform and the Laplace trans-form. This will not lead to confusion.

A point iη ∈ iR is called regular for f if f has a holomorphic extension to aneighbourhood of iη (i.e., iη is regular if there exists an open neighbourhood V of

iη and a holomorphic function h : V → X such that h(λ) = f(λ) for all λ ∈ V \iR).The Carleman spectrum spC(f) is defined by

spC(f) = {η ∈ R : iη is not regular for f}. (4.26)

The following remark explains why this notion of spectrum is well adapted tospectral theory of C0-groups.

Remark 4.6.2 (Carleman spectrum and C0-groups). Let Y be a Banach space andU be a bounded C0-group on Y with generator A. Then σ(A) ⊂ iR, since A and−A generate bounded C0-semigroups. Let x ∈ Y and f(t) := U(t)x. Then the

Carleman transform f of f is given by

f(λ) = R(λ,A)x (λ ∈ C\iR). (4.27)

In particular,i spC(f) ⊂ σ(A). (4.28)

In fact, (4.27) is clear for Reλ > 0. If Re λ < 0, then f(λ) = −f−(−λ) =− ∫∞

0eλtU(−t)x dt = −R(−λ,−A)x = R(λ,A)x.

As in Section 4.5, we denote by S the shift group on BUC(R, X) and by Bits generator. For f ∈ BUC(R, X) and s ∈ R, let fs := S(s)f . An easy calculation(see Remark 4.6.2) shows that

(R(λ,B)f)(s) = fs(λ) = eλs(f(λ)−

∫ s

0

e−λtf(t) dt

)for λ ∈ C \ iR. We shall see in Lemma 4.6.8 that the singularities of R(·, B)

coincide with the singularities of f ; i.e., that the Carleman spectra of S(·)f andf(·) coincide.

Let η ∈ R. A function f ∈ BUC(R, X) is called uniformly ergodic at η if it isergodic at η with respect to S; i.e., if

Mηf := limt→∞

1

t

∫ t

0

e−iηsS(s)f ds (4.29)

exists in BUC(R, X). By Section 4.3, this is equivalent to saying that

Mηf = limα↓0

αR(α+ iη, B)f (4.30)

exists in BUC(R, X). Since (R(α+ iη,B)f)(s) = fs(α+ iη), this can be reformu-

lated by saying that αfs(α+ iη) converges as α ↓ 0 uniformly in s ∈ R.We say that f is totally ergodic if f is uniformly ergodic at each η ∈ R; i.e.,

if f is totally ergodic with respect to S in the sense of Sections 4.3 and 4.5.Next, we introduce a geometric condition of Banach spaces. Let c0 be the

Banach space of all complex sequences converging to 0 with the supremum norm(as in Example 1.1.5 b)). We say that X contains c0, and write briefly c0 ⊂ X,if there exists a closed subspace of X which is isomorphic to c0. Since closedsubspaces of reflexive spaces are reflexive, no reflexive Banach space contains c0.Also, any space of the form L1(Ω, μ) does not contain c0. See Appendix D forfurther information.

Now we can formulate the main theorem of this section.

Theorem 4.6.3. Let f ∈ BUC(R, X) have countable Carleman spectrum. Assumethat one of the following conditions is satisfied:

a) f is totally ergodic, or

b) f has relatively compact range, or

c) c0 �⊂ X.


Then f is almost periodic.

As a corollary, we note the important scalar case which is due to Loomis.Here, we let BUC(R) := BUC(R,C) and AP(R) := AP(R,C).

Corollary 4.6.4 (Loomis’s Theorem). Let f ∈ BUC(R) have countable Carlemanspectrum. Then f is almost periodic.

We have seen in the preceding section that every almost periodic function istotally ergodic and has relatively compact range. Thus, conditions a) and b) inTheorem 4.6.3 are necessary for the conclusion to hold. If the Banach space Xcontains c0, countability of the Carleman spectrum alone does not imply almostperiodicity. This is shown by the following example.

Example 4.6.5. Let X := c, the space of all convergent complex sequences x =(xn)n∈N with the supremum norm ‖x‖ = supn |xn|. Then X is isomorphic to c0.Let

f(t) :=(eit/n

)n∈N

=: (fn(t))n∈N .

Since f ′n(t) =ine

it/n, the function f is Lipschitz continuous and thus f∈BUC(R, c).

The Carleman transform f of f is given by

f(λ) =

(1

λ− i/n

)n∈N

for all λ ∈ C \iR. Consequently, spC(f) = {1/n : n ∈ N} ∪ {0}, which is count-able. However, f �∈ AP(R, c). In fact, f does not have relatively compact range.To see this, consider φn ∈ c∗ given by 〈x, φn〉 := xn for x = (xk)k∈N ∈ c, andφ∞ ∈ c∗ given by 〈x, φ∞〉 := limk→∞ xk. Then ‖φn ‖ ≤ 1 (n ∈ N ∪ {∞}) andlimn→∞〈x, φn〉 = 〈x, φ∞〉 for all x ∈ c. Suppose that K := {f(t) : t ∈ R} isrelatively compact in c. Then 〈x, φn〉 converges to 〈x, φ∞〉 uniformly on K (seeProposition B.15). In particular, limn→∞ Im〈f(t), φn〉 = limn→∞ sin t

n = 0 uni-formly in t ∈ R, which is absurd.

We need several auxiliary results for the proof of Theorem 4.6.3. The followingis a special kind of maximum principle for holomorphic functions.

Lemma 4.6.6. Let V be an open neighbourhood of iη such that V contains theclosed disc B(iη, 2r) = {z ∈ C : |z− iη| ≤ 2r}. Let h : V → X be holomorphic andc ≥ 0, k ∈ N0 such that

‖h(z) ‖ ≤ c

|Re z|k if |z − iη| = 2r, Re z �= 0.

Then ‖h(z) ‖ ≤ (4/3)kcr−k for all z ∈ B(iη, r).

Proof. We can assume that η = 0 (replacing h(z) by h(z + iη) otherwise). Define

g : V → X by g(z) :=

(1 +

z2

(2r)2

)k

h(z). Let |z| = 2r, z = 2reiθ. Then

‖ g(z) ‖ = |1 + ei2θ|k ‖h(z) ‖ = |eiθ(e−iθ + eiθ)|k ‖h(z) ‖= 2k | cos θ|k ‖h(z) ‖ ≤ c

rk.

It follows from the maximum principle that ‖ g(z) ‖ ≤ cr−k if |z| ≤ 2r. Let |z| ≤ r.

Then

∣∣∣∣∣(1 +

z2

(2r)2

)−1∣∣∣∣∣ =

∣∣∣∣ (2r)2

(2r)2 + z2

∣∣∣∣ ≤ 4r2

4r2 − r2=

4

3. Hence,

‖h(z) ‖ =∣∣∣∣∣(1 +

z2

(2r)2

)−k∣∣∣∣∣ ‖ g(z)‖ ≤ (4/3)kcr−k (|z| ≤ r).

We now describe the local spectrum associated with a bounded C0-groupand an individual vector. Let U be an isometric C0-group on a Banach space Ywith generator A. For x ∈ Y we define the space Yx := span{U(t)x : t ∈ R}. Itis invariant under the group U . We denote by Ax the generator of this restrictiongroup. Then σ(Ax) ⊂ σ(A) ⊂ iR, and R(λ,Ax) = R(λ,A)|Yx

(λ ∈ ρ(A)). Asbefore, C+ := {z ∈ C : Re z > 0} denotes the right half plane.

Lemma 4.6.7. Let x ∈ Y, η ∈ R. The following assertions are equivalent:

(i) iη ∈ ρ(Ax).

(ii) There exists an open neighbourhood V of iη and a holomorphic functionh : V → X such that h(λ) = R(λ,A)x for all λ ∈ C+ ∩ V .

In that case, h(λ) = R(λ,A)x for all λ ∈ V \iR.Proof. (i) ⇒ (ii): Let V := ρ(Ax), h(λ) := R(λ,Ax)x.

(ii) ⇒ (i): Assume (ii) and assume that V is connected. We first show thath(λ) = R(λ,A)x for all λ ∈ V \iR. Let μ ∈ ρ(A). Then k(λ) := (λ−A)R(μ,A)h(λ)defines a holomorphic function on V such that k(λ) = R(μ,A)x for all λ ∈ C+ ∩V . It follows from the uniqueness theorem that (λ − A)R(μ,A)h(λ) = k(λ) =R(μ,A)x for all λ ∈ V . This implies that R(μ,A)h(λ) = R(λ,A)R(μ,A)x =R(μ,A)R(λ,A)x for all λ ∈ V \iR. Since R(μ,A) is injective, the claim follows.

We now show that iη ∈ ρ(Ax). We have ‖R(λ,A) ‖ =∥∥ ∫∞

0e−λtU(t) dt

∥∥ ≤1/(Reλ) (Reλ > 0), and similarly, ‖R(λ,A) ‖ = ‖R(−λ,−A) ‖ ≤ −1/(Reλ) forReλ < 0. Thus ‖R(λ,A) ‖ ≤ 1/|Re λ| for all λ ∈ V \iR. Choose r > 0 such thatB(iη, 2r) ⊂ V . It follows from the assumption that R(·, A)z has a holomorphicextension to V for all z ∈ span{U(t)x : t ∈ R}. Now it follows from Lemma 4.6.6that

‖R(λ,A)z ‖ ≤ 4

3r‖ z ‖ for all λ ∈ B(iη, r)\iR.

Hence, ‖R(λ,Ax) ‖ ≤ 43r for all λ ∈ B(iη, r)\iR. This implies that iη ∈ ρ(Ax).

We define the local spectrum of A at x by

σ(A, x) := σ(Ax) = C\ρ(Ax).

Thus, iη ∈ σ(A, x) if and only if the equivalent conditions (i) and (ii) of Lemma4.6.7 are not satisfied.

Now we consider the shift group S on BUC(R, X) with generator B, as before.We show that the Carleman spectrum coincides with the local spectrum withrespect to B.

Lemma 4.6.8. Let f ∈ BUC(R, X). Then

i spC(f) = σ(B, f).

Proof. a) Let iη ∈ iR \σ(B, f). There exists an open neighbourhood V of iη and aholomorphic function h : V → BUC(R, X) such that h(λ) = R(λ,B)f (λ ∈ V \iR).Let k(λ) := h(λ)(0). Then k : V → X is holomorphic, k(λ) = (R(λ,B)f)(0) =∫∞0

e−λt(S(t)f)(0) dt = f(λ) for λ ∈ V ∩ C+, and k(λ) = (R(λ,B)f)(0) =

−(R(−λ,−B)f)(0) = − ∫∞0

eλt(S(−t)f)(0) dt = − ∫∞0

eλtf(−t) dt for λ ∈ V ,Reλ < 0. Thus, η �∈ spC(f).

b) Let η ∈ R\spC(f). By assumption, there exists an open neighbourhood Vof iη and a holomorphic function h : V → X such that

h(λ) =

{ ∫∞0

e−λtf(t) dt (λ ∈ V ∩ C+)

− ∫∞0

eλtf(−t) dt (λ ∈ V, Reλ < 0).

For λ ∈ V, s ∈ R, define

H(λ, s) := eλs(h(λ)−

∫ s

0

e−λtf(t) dt

).

Then for λ ∈ V, Reλ > 0,

(R(λ,B)f)(s) =

∫ ∞

0

e−λtf(t+ s) dt = H(λ, s) (s ∈ R)

and for λ ∈ V, Reλ < 0,

(R(λ,B)f)(s) = −(R(−λ,−B)f)(s) = −∫ ∞

0

eλtf(s− t) dt = H(λ, s).

Thus, ‖H(λ, s) ‖ ≤ ‖ f ‖∞/|Reλ| for λ ∈ V \ iR. By Lemma 4.6.6, this impliesthat for r > 0 such that B(iη, 2r) ⊂ V , one has ‖H(λ, s) ‖ ≤ 4‖ f ‖∞

/3r for all

λ ∈ B(iη, r), s ∈ R. We know that H(λ, ·) = R(λ,B)f ∈ BUC(R, X) if Reλ > 0.Since H(·, s) is holomorphic on B(iη, r) and bounded uniformly in s, it follows


from Corollary A.4 that λ �→ H(λ, ·) is holomorphic on B(iη, r) with values inBUC(R, X). Here, we choose linear functionals on BUC(R, X) of the form g �→〈g(s), x∗〉 for s ∈ R, x∗ ∈ X∗, ‖x∗‖ ≤ 1. Now, iη �∈ σ(B, f) by Lemma 4.6.7.

We deduce from Lemma 4.6.8 the following interesting observation.

Proposition 4.6.9. Let f ∈ BUC(R, X) and let η ∈ R\spC(f). Then f is uniformlyergodic at η and Mηf = 0.

Proof. It was shown in the proof of Lemma 4.6.8 that λ �→ R(λ,B)f has aholomorphic extension to a neighbourhood of iη. It follows in particular thatlimα↓0 αR(α+ iη, B)f = 0 in BUC(R+, X).

Proposition 4.6.10. Let f ∈ AP(R, X). Then spC(f) = Freq(f).

Proof. The inclusion Freq(f) ⊂ spC(f) is immediate from Proposition 4.6.9, al-though in this case the fact that f is uniformly ergodic at each point of R is alreadyknown (see Section 4.5). Since spC(f) is closed, it follows that Freq(f) ⊂ spC(f).

For the converse, let

Y0 = span {eiη ⊗ x : η ∈ Freq(f), x ∈ X}and Y := Y0. By Proposition 4.5.8, f ∈ Y . For g ∈ Y0, g has a holomorphicextension to C\ i Freq(f), and ‖g(λ)‖ ≤ ‖g‖∞/|Re λ|. By Lemma 4.6.6, ‖g(iη)‖ ≤8‖g‖∞

/(3 dist(η,Freq(f))

)if η ∈ R \ Freq(f). If (gn)n∈N is a sequence in Y0 such

that ‖gn− f‖∞ → 0, it follows that limn→∞ gn(λ) exists for λ ∈ C \ i Freq(f) andby Vitali’s Theorem A.5, this gives a holomorphic extension of f to C \ i Freq(f).Hence, spC(f) ⊂ Freq(f).

The condition that c0 �⊂ X will enter the proof of Theorem 4.6.3 in form ofthe following theorem due to Kadets.

Theorem 4.6.11 (Kadets). Assume that c0 �⊂ Xand let f ∈ BUC(R, X). If further-more S(t)f − f ∈ AP(R, X) for all t ∈ R, then f ∈ AP(R, X).

We will give the proof of Kadets’s theorem at the end of this section. ForX = C, Theorem 4.6.11 is due to H. Bohr. We will give a separate direct prooffor this easier case. For Theorem 4.6.3 we use the following spectral-theoreticreformulation of Kadets’s theorem.

Denote by π : BUC(R, X) → BUC(R, X)/AP(R, X) the quotient map. Re-call that the quotient space is a Banach space for the norm

‖π(f)‖ := inf{‖f + g‖ : g ∈ AP(R, X)}and π is contractive for this norm. Since S(t) leaves the space AP(R, X) invariant,

there exists a C0-group S on the quotient BUC(R, X)/AP(R, X) given by

S(t)π(f) := π(S(t)f).

We call its generator B. Now Kadets’s theorem can be rephrased in the followingform.


Corollary 4.6.12. The following assertions are equivalent:

(i) c0 �⊂ X.

(ii) B has empty point spectrum.

Proof. (i) ⇒ (ii): Theorem 4.6.11 says that 0 �∈ σp(B). So it suffices to show

that σp(B) = σp(B) + iR. Let η ∈ R, eiη(r) := eiηr. Then Qηf := eiη · f de-fines an isomorphism Qη on BUC(R, X) which leaves AP(R, X) invariant. More-

over, Q−ηS(t)Qη = eiηtS(t). Define Qη on BUC(R, X)/AP(R, X) by Qη(π(f)) :=

π(Qηf). Then Q−ηS(t)Qη = eiηtS(t). This implies the claim.(ii) ⇒ (i): We have to show that if c0 ⊂ X then there exists f ∈ BUC(R, X)

such that f �∈ AP(R, X) but S(t)f − f ∈ AP(R, X) for all t ≥ 0. Since thespace c0 is isomorphic to the space c, it suffices to consider X = c. Considerthe function f ∈ BUC(R, c) given by f(t) :=

(eit/n

)n∈N. Then f �∈ AP(R, c) by

Example 4.6.5. Let t ∈ R, g := S(t)f − f . Then g(s) =((eit/n − 1)eis/n

)n∈N =:

(gn(s))n∈N. Let hn(s) := (g1(s), . . . , gn(s), 0, 0, . . .). Since limn→∞(eit/n − 1) = 0,we have limn→∞ hn = g in BUC(R, c). But hn ∈ AP(R, c) for all n ∈ N. Thus,g ∈ AP(R, c).

Proof of Theorem 4.6.3. Let f ∈ BUC(R, X) such that spC(f) is countable.Case c): We assume that c0 �⊂ X. Let Y := BUC(R, X)/AP(R, X) and

consider the shift group S with generator B on Y . Assume that f �∈ AP(R, X);

i.e., f := π(f) �= 0. We consider the space Y˜f := span{S(t)f : t ∈ R} and the

induced operator B˜f (see the discussion before Lemma 4.6.7). Let η ∈ R\spC(f).

Then by Lemma 4.6.8, the function λ �→ R(λ,B)f has a holomorphic extensionh : V → BUC(R, X) where V is an open neighbourhood of iη. Then π ◦ h is a

holomorphic extension of λ �→ R(λ, B)f = π(R(λ,B)f). It follows from Lemma

4.6.7 that iη �∈ σ(B˜f). We have proved that σ(B

˜f) ⊂ spC(f). Thus, σ(B˜f

) is

countable, closed and non-empty (by Corollary 4.4.12), so it contains an isolated

point. It follows from Corollary 4.4.13 that B˜fhas non-empty point spectrum.

Hence, B also has non-empty point spectrum. This contradicts Corollary 4.6.12.The proof is finished in this case.

Case a): Now assume that f is totally ergodic. Consider the space E := {g ∈BUC(R, X) : g is totally ergodic} which is closed and invariant under the shiftgroup. We denote the shift group on E also by S and its generator by B. Considerthe quotient space E := E/AP(R, X), the quotient map π : E → E and the induced

group S on E given by S(t)π(g) := π(S(t)g) with generator B. It suffices to show

that B has empty point spectrum. Then the proof given in Case a) carries over.

Assume that g ∈ E , η ∈ R such that S(t)π(g) = eiηtπ(g) (t ∈ R). Thene−iηtS(t)g − g ∈ AP(R, X) (t ∈ R). It follows that

Mηg − g = limt→∞

1

t

∫ t

0

(e−iηsS(s)g − g

)ds ∈ AP(R, X).


But Mηg = eiη ⊗ (Mηg)(0) ∈ AP(R, X). Thus g ∈ AP(R, X).Case b): Assume that f has relatively compact range. It follows from Case c)

applied in the scalar case that f is weakly almost periodic. Hence, f ∈ AP(R, X)by Proposition 4.5.12.

Now Theorem 4.6.3 is completely proved in the case a) where f is assumed tobe uniformly ergodic at each η ∈ spC(f). The proof in the other cases is completeadmitting Kadets’s theorem. Note however that for the case b), where f is assumedto have relatively compact range, we merely need Kadets’s theorem in the scalarcase. The scalar case is much easier to prove and will be particularly important forthe general complex Tauberian theorem presented in Section 4.9. For this reason,we first give a direct proof of Kadets’s theorem in the scalar case.

The following reformulation will be useful.

Lemma 4.6.13. Let X be a Banach space. The following assertions are equivalent:

(i) If f ∈ AP(R, X) such that F (t) :=∫ t

0f(s) ds (t ∈ R) is bounded, then

F ∈ AP(R, X).

(ii) If f ∈ BUC(R, X) such that S(t)f − f ∈ AP(R, X) for all t ∈ R, thenf ∈ AP(R, X).

Proof. (ii) ⇒ (i): Let f ∈ AP(R, X) such that F (t) =∫ t

0f(s) ds is bounded. Then

F ∈ BUC(R, X) and S(t)F − F =∫ t

0S(s)f ds ∈ AP(R, X) for all t ∈ R. Hence,

F ∈ AP(R, X) by (ii).(i) ⇒ (ii): Let f ∈ BUC(R, X) such that S(t)f − f ∈ AP(R, X) for all t ∈ R.

Let λ > 0 and g = R(λ,B)f . Then S(t)g−g ∈ AP(R, X) and g ∈ D(B). It followsthat

g′ = limt↓0

1/t (S(t)g − g) ∈ AP(R, X).

Thus by (i), g(t)− g(0) =∫ t

0g′(s) ds defines an almost periodic function. We have

shown that R(λ,B)f ∈ AP(R, X) for all λ > 0. Since f = limλ→∞ λR(λ,B)f inBUC(R, X) (see Proposition 3.1.9 a)), it follows that f ∈ AP(R, X).

Proof of Theorem 4.6.11 in the scalar case. Let f ∈ AP(R) such that F (t) :=∫ t

0f(s) ds is bounded. We show that F ∈ AP(R). Then the result follows from

Lemma 4.6.13. It follows easily from Definition 4.5.6 that we can assume that fis real-valued. It is clear that F ∈ BUC(R). Let sk ∈ R (k ∈ N). There existsa subsequence tm := skm

such that g := limm→∞ S(tm)f exists in BUC(R). Wehave to show that S(tm)F has a convergent subsequence. Then F ∈ AP(R) byTheorem 4.5.7.

a) Since (S(tm)F ) (t) =∫ tm+t

0f(s) ds =

∫ t

0S(tm)f(s) ds + F (tm), and

(F (tm))m∈N is bounded, it follows that S(tm)F has a limit point G for compactconvergence (by which we mean the topology of uniform convergence on bounded

intervals), where G(t) =∫ t

0g(s) ds+ d for some d ∈ R. Clearly, G ∈ BUC(R) and

supG := supt∈R G(t) ≤ supF, inf G ≥ inf F . Since S(−tm)g → f as m → ∞,

for the same reason S(−tm)G has a limit point H for compact convergence, H isdifferentiable, H ′ = f, supH ≤ supG, infH ≥ inf G. Hence, H(t) = F (t) + c forsome c ∈ R and c+ supF ≤ supG ≤ supF and inf F + c ≥ inf G ≥ inf F . Hence,c = 0. Thus, supG = supF and inf G = inf F . This determines d uniquely. So thelimit point G is unique. Hence, S(tm)F converges to G ∈ BUC(R) uniformly onbounded intervals.

b) Now it remains to show that S(tm)F converges uniformly to G. If not,passing to a subsequence we can assume that there exist ε > 0 and sm ∈ R suchthat |F (tm+ sm)−G(sm)| ≥ ε > 0 (m ∈ N). Taking subsequences, we can assumethat h := limm→∞ S(tm+sm)f and h1 := limm→∞ S(sm)g = h1 exist in BUC(R).Then h = h1, since

|h(t)− h1(t)| ≤ |h(t)− f(t+ tm + sm)|+ |f(t+ tm + sm)− g(sm + t)|+ |g(sm + t)− h1(t)|

→ 0 as m→∞.

Let H(t) :=∫ t

0h(s) ds. By a), there exist constants c1, c2 such that

limm→∞ (S(tm + sm)F ) (t) = H(t) + c1 and lim

m→∞ (S(sm)G) (t) = H(t) + c2

uniformly on bounded intervals, and inf F = infH + c1, and inf G = infH + c2.Since inf F = inf G, it follows that c1 = c2. Thus ε ≤ limm→∞ |F (tm + sm) −G(sm)| = |c1 − c2| = 0, a contradiction.

Now we have proved Kadets’s theorem in the scalar case, and hence also theproof of Theorem 4.6.3 in the case b), and in particular of Loomis’s theorem, iscomplete.

Before proving Kadets’s theorem in full generality, we consider another specialcase which can be deduced from the scalar case by a short but clever argument.

Recall that a Banach space X is called uniformly convex if for all ε > 0 thereexists δ > 0 such that for ‖x‖ ≤ 1, ‖y‖ ≤ 1,

‖x− y‖ ≥ ε⇒∥∥∥∥x+ y

2

∥∥∥∥ ≤ 1− δ. (4.31)

Every uniformly convex space is reflexive, and Lp-spaces are uniformly convex for1 < p <∞.

Proof of Theorem 4.6.11 for uniformly convex spaces.Let f ∈ AP(R, X) such that F (t) =

∫ t

0f(s) ds is bounded. We claim that F is

almost periodic. We can assume that

‖F‖∞ := supt∈R

‖F (t)‖ = 1.

Since by the scalar case, F is weakly almost periodic, in view of Proposition 4.5.12it suffices to show that F has precompact range.

Assume that this is false. Then there exist ε > 0, tn ∈ R such that

‖F (tn)− F (tm)‖ ≥ ε (n �= m).

Choose δ > 0 according to (4.31). Let x∗ ∈ X∗ such that ‖x∗‖ = 1 and ‖x∗◦F‖∞ >1− δ

4 . Replacing (tn)n∈N by a subsequence if necessary, we can assume that

S(tn)f → f in BUC(R, X)

S(tn)(x∗ ◦ F ) → g in BUC(R).

Then ‖g‖∞ ≥ 1 − δ4 , since S(−tn)g → x∗ ◦ F in BUC(R). Let t ∈ R such that

|g(t)| ≥ 1− δ2 . Then

lim supn,m→∞n�=m

‖F (tn + t)− F (tm + t)‖

= lim supn,m→∞n�=m

∥∥∥∥F (tn)− F (tm) +

∫ t

0

(f(tn + s)− f(tm + s)) ds


n,m→∞n�=m

‖F (tn)− F (tm)‖ ≥ ε.

Hence by (4.31),

1− δ

2≤ |g(t)| = lim

n,m→∞n�=m

|〈F (t+ tn) + F (t+ tm), x∗〉|2

≤ lim supn,m→∞n �=m

‖F (t+ tn) + F (t+ tm)‖2

≤ 1− δ.

This is a contradiction.

For the proof of Kadets’s theorem in the general case we use the followingcharacterization which is proved in Appendix D.

Theorem 4.6.14. Let X be a Banach space. The following are equivalent:

(i) c0 ⊂ X.

(ii) there is a divergent series∑∞

n=1 xn in X which is unconditionally bounded;i.e., there exists M ≥ 0 such that∥∥∥∥ m∑

j=1

xnj

∥∥∥∥ ≤M

whenever nj ∈ N (j = 1, 2, . . . ,m) such that n1 < n2 < . . . < nm.

Proof of Kadets’s Theorem 4.6.11.Assume that there exists f ∈ AP(R, X) such that F (t) =

∫ t

0f(s) ds is bounded,

but F �∈ AP(R, X). We show that this implies that c0 ⊂ X. In view of Lemma4.6.13, this proves Kadets’s theorem.

By the scalar case already proved, F is weakly almost periodic.a) We show that there exists ε > 0 such that the set

U(ε) := {t ∈ R : ‖F (t)‖ < ε}is not relatively dense. Assume that, on the contrary, U(ε) is relatively dense forall ε > 0.

Let ε > 0. We show that R =⋃n

j=0(U(ε) + sj) for suitable s0, . . . , sn ∈ R. Infact, there exists l > 0 such that U(ε/2) ∩ [a − l, a] �= ∅ for all a ∈ R. Since F isLipschitz continuous, there exists δ > 0 such that ‖F (z)‖ ≤ ε/2 implies ‖F (r)‖ ≤ εfor all r ∈ [z, z + δ]. Let s0 := 0, s1 := δ, s2 := 2δ, . . . , sn := nδ, where nδ > l. Lett ∈ R. Take s ∈ [t− l, t] ∩ U(ε/2). Then

[s, s+ δ] ⊂ U(ε) = U(ε) + s0,

[s+ δ, s+ 2δ] = [s, s+ δ] + δ ⊂ U(ε) + s1,

...

[s+ (n− 1)δ, s+ nδ] ⊂ U(ε) + sn.

Thus, t ∈ [s, s+ l] ⊂ ⋃nj=0(U(ε) + sj).

Since f ∈ AP(R, X), the function S(t)F − F =∫ t

0S(r)f dr ∈ AP(R, X) for

all t ∈ R. In particular, S(sj)F − F has precompact range Kj . Now let t ∈ R.Then t = s+ sj for some s ∈ U(ε), j ∈ {0, 1, 2, . . . , n}. Thus, F (t) = F (s+ sj) =(S(sj)F − F ) (s) + F (s) ∈ Kj + B(0, ε). Thus, the range of F is contained in⋃n

j=0Kj + B(0, ε). Since ε > 0 is arbitrary, this implies that F has precompactrange. This is impossible by Proposition 4.5.12.

b) For γ ∈ R, δ > 0, let

Vγ(δ) :={t ∈ R : ‖F (t+ γ)− F (t)− F (γ)‖ < δ

}.

We show that⋂n

j=1 Vγj(δ) is relatively dense in R for all δ > 0, γ1, γ2, . . . , γn ∈ R.

In fact, let γ := maxj=1,...,n |γj |. Then

‖F (t+ γj)− F (t)− F (γj)‖ =

∥∥∥∥∫ γj

0

(f(t+ s)− f(s)) ds

∥∥∥∥≤ γ. sup

0≤s≤γ‖f(t+ s)− f(s)‖.

Thus,

n⋂j=1

Vγj(δ) ⊃

{t ∈ R : ‖f(t+ s)− f(s)‖ ≤ δ

γfor all s ∈ [0, γ]

}.

The last set is relatively dense since f ∈ AP(R, X).c) By a) and b), there exists ε > 0 such that the following holds:

If δ > 0, γ1, γ2, . . . , γr ∈ R, then U(ε) �⊃ ∩rj=1Vγj

(δ). (4.32)

We construct tn ∈ R such that ‖F (tn)‖ ≥ ε (n ∈ N) and

‖F (tn + tj1 + . . .+ tjr)− F (tn)− F (tj1 + . . .+ tjr)‖ ≤ 2−n (4.33)

whenever j1, j2, . . . , jr ∈ N, 1 ≤ j1 < j2 < . . . < jr ≤ n− 1. Choose t1 ∈ R \U(ε).Assume that t1, . . . , tn−1 are constructed. By (4.32), there exists tn ∈ R\U(ε) suchthat tn ∈ Vγ(2

−n) whenever γ = tj1 + . . .+ tjr with 1 ≤ j1 < j2 < . . . < jr ≤ n−1.This means that (4.33) holds.

d) Let xn := F (tn). Then the series∑∞

n=1 xn diverges. We show that∥∥∥∥∥m∑

k=1

xik

∥∥∥∥∥ ≤ ‖F‖∞ + 1 (4.34)

whenever i1 < i2 < . . . < im. In fact,∥∥∥∥∥F(

m∑k=1

tik

)−

m∑k=1

F (tik)

∥∥∥∥∥ ≤∥∥∥∥∥F

(m∑

k=1

tik

)− F (tim)− F

(m−1∑k=1

tik

)∥∥∥∥∥+

∥∥∥∥∥F(

m−1∑k=1

tik

)− F (tim−1)− F

(m−2∑k=1

tik

)∥∥∥∥∥+ . . .+

∥∥∥F (ti2 + ti1)− F (ti2)− F (ti1)∥∥∥

≤ 2−im + . . .+ 2−i2 ≤ 1.

Thus, ∥∥∥∥∥m∑

k=1

xik

∥∥∥∥∥ =

∥∥∥∥∥m∑

k=1

F (tik)

∥∥∥∥∥ ≤ 1 + ‖F‖∞.

Now it follows from Theorem 4.6.14 that c0 ⊂ X .

4.7 Asymptotically Almost Periodic Functions

In this section we study bounded uniformly continuous functions on the half-line.The main result is a Tauberian theorem (Theorem 4.7.7) which says that, underadditional assumptions, such a function with countable spectrum is asymptoticallyalmost periodic. Here we again use a quotient method similar to the precedingsection. A similar result, in the case when the spectrum has at most one point,was given in Theorem 4.4.8 with a proof by elementary contour integrals.

4.7. ASYMPTOTICALLY ALMOST PERIODIC FUNCTIONS 307

LetX be a Banach space and denote by BUC(R+, X) the space of all boundeduniformly continuous functions defined on R+ with values in X . It is a Banachspace for the uniform norm

‖ f ‖∞ = supt≥0

‖ f(t) ‖ (f ∈ BUC(R+, X)).

By C0(R+, X) we denote the closed subspace consisting of all f ∈ BUC(R+, X)such that limt→∞ ‖ f(t) ‖ = 0.

For x ∈ X, η ∈ R, we now consider the function eiη ⊗ x : t �→ eiηtx on R+

and denote byAP(R+, X) := span{eiη ⊗ x : η ∈ R, x ∈ X}

the space of all almost periodic functions on the half-line (the closure being takenin BUC(R+, X)).

Proposition 4.7.1. Every f ∈ AP(R+, X) has a unique extension f ∈ AP(R, X)

and ‖ f ‖∞ = ‖ f ‖∞. Moreover,

‖ f ‖∞ = supt≥τ

‖ f(t) ‖ for all τ ≥ 0. (4.35)

Proof. Let f ∈ AP(R+, X). There exist trigonometric polynomials fn ∈ span{eiη⊗x : η ∈ R, x ∈ X} such that fn → f in BUC(R+, X) as n → ∞. It follows from

(4.24) that (fn)n∈N is a Cauchy sequence in BUC(R, X). Let f be the limit of

(fn)n∈N in BUC(R, X). Then f ∈ AP(R, X) and (4.35) follows from (4.24).


AP(R+, X) ∩ C0(R+, X) = {0}. (4.36)

ByAAP(R+, X) := C0(R+, X)⊕ AP(R+, X),

we denote the space of all asymptotically almost periodic functions on the half-line.It follows from (4.35) that for f = f0 + f1 with f0 ∈ C0(R+, X), f1 ∈ AP(R+, X),one has ‖f1‖∞ ≤ ‖f‖∞. Thus, AAP(R+, X) is a closed subspace of BUC(R+, X).

The following observation is useful for later purposes. By (4.23), there existtn ∈ R+ such that limn→∞ tn =∞ and

‖f(tn + s)− f1(s)‖ ≤ 1

nfor all s ∈ R+. (4.37)

In the following, we consider the shift semigroup S on BUC(R+, X) given by

(S(t)f) (s) = f(t+ s) (s, t ≥ 0, f ∈ BUC(R+, X)).

It is a C0-semigroup of contractions, and its generator will be denoted by B.

Similarly to Section 4.6, for f ∈ BUC(R+, X), Reλ > 0 and s ≥ 0,

(R(λ,B)f)(s) = fs(λ) = eλs(f(λ)−

∫ s

0

e−λtf(t) dt

),

where fs := S(s)f . We shall see in Lemma 4.7.9 that the singularities of R(·, B)f

and f(·) in iR coincide, and we shall exploit this to prove an analogue of part ofTheorem 4.6.3 (Theorem 4.7.7).

Let f ∈ BUC(R+, X). Let η ∈ R. We say that f is uniformly ergodic at η iff is ergodic at η with respect to S; i.e., if

Mηf := limt→∞

1

t

∫ t

0

e−iηsS(s)f ds

converges in BUC(R+, X). This is equivalent to the convergence of

Mηf = limα↓0

αR(α+ iη, B)f.

In that case,Mηf = eiη ⊗ (Mηf)(0). (4.38)

A function f ∈ BUC(R+, X) is called totally ergodic if f is ergodic at all η ∈R. From now on, we denote by E(R+, X) the set of all totally ergodic functions inBUC(R+, X). This is a closed subspace of BUC(R+, X) containing AAP(R+, X).

In the following we will consider the quotient space

E := E(R+, X)/AAP(R+, X)

with quotient map π : E(R+, X)→ E . Then E is a Banach space for the norm

‖π(f) ‖ := inf{‖f − g‖∞ : g ∈ AAP(R+, X)} (f ∈ E(R+, X)).

Since E(R+, X) and AAP(R+, X) are invariant under the shift semigroup, we can

define a C0-semigroup S on E by

S(t)π(f) := π (S(t)f) (t ≥ 0, f ∈ E(R+, X)).

We denote by B the generator of S.The interesting fact about this construction is the following.

Proposition 4.7.2. Each S(t) is isometric and surjective. Thus, S extends to anisometric C0-group on E . Moreover, B has empty point spectrum.

Proof. It is immediate that ‖S(t)‖ ≤ ‖S(t)‖ = 1 for all t ≥ 0. We show that S(t)is isometric. Let f ∈ E(R+, X), t ≥ 0 and g ∈ AAP(R+, X). Define h : R+ → Xby

h(s) :=

{g(s− t) (s ≥ t)

g(0) + f(s)− f(t) (s < t).


Then h ∈ AAP(R+, X) and

‖f − h‖∞ = sups≥t

‖f(s)− g(s− t)‖ = sups≥0

‖f(s+ t)− g(s)‖

= ‖S(t)f − g‖∞.

Hence, ‖π(f)‖ ≤ ‖π(S(t)f)‖ = ‖S(t)π(f)‖.Since S(t) maps E(R+, X) onto E(R+, X), it follows that S(t) is surjective.

By Proposition 3.1.23, S extends to a C0-group on E .It remains to show that the point spectrum of B is empty. Let f ∈ E(R+, X)

and η ∈ R such that S(t)π(f) = eiηtπ(f) (t ≥ 0). Then 1t

∫ t

0e−iηsS(s)π(f) ds =

π(f) (t ≥ 0). On the other hand, limt→∞ 1t

∫ t

0e−iηsS(s)f ds = Mηf ∈ AP(R+, X).

Applying π on both sides, we conclude that limt→∞ 1t

∫ t

0e−iηsS(s)π(f) ds =

π(Mηf) = 0. Hence, π(f) = 0.

The following simple result is essentially a reformulation of part of Proposi-tion 4.7.2 (see Lemma 4.6.13).

Proposition 4.7.3. Let f ∈ AAP(R+, X) and F (t) :=∫ t

0f(s) ds. Suppose that F

is bounded and uniformly ergodic at 0. Then F ∈ AAP(R+, X).

Proof. Note first that

S(s)F − F =

∫ s

0

S(r)f dr ∈ AAP(R+, X).

Hence,

M0F − F = limt→∞

1

t

∫ t

0

(S(s)F − F ) ds ∈ AAP(R+, X).

Since M0F is a constant function, it follows that F ∈ AAP(R+, X).

Next, we characterize asymptotically almost periodic functions by relativecompactness of the orbits under S. This is analogous to the characterization ofalmost periodic functions on the line proved in Section 4.5 (see Theorem 4.5.7).

Theorem 4.7.4. Let f ∈ BUC(R+, X). The following are equivalent:

(i) f ∈ AAP(R+, X).

(ii) The orbit {S(t)f : t ≥ 0} is relatively compact in BUC(R+, X).

Proof. (i)⇒ (ii): If h ∈ C0(R+, X), then limt→∞ S(t)h = 0. Thus, {S(t)h : t ≥ 0} isprecompact. If f = eiη⊗x, then S(t)f = eiηtf , so f has precompact orbit. It followsthat every f ∈ span ({eiη ⊗ x : η ∈ R, x ∈ X} ∪ C0(R+, X)) = AAP(R+, X) hasprecompact orbit.

(ii) ⇒ (i): Assume that Of := {S(t)f : t ≥ 0} is relatively compact inBUC(R+, X). Then f ∈ E(R+, X) by Proposition 4.3.12, and Oπ(f) := π(Of ) =


{S(t)π(f) : t ≥ 0} is relatively compact in E . It follows from Theorem 4.5.1 that

π(f) ∈ span{g ∈ E : S(t)g = eiηtg (t ∈ R) for some η ∈ R

}. By Proposition 4.7.2,

this set is reduced to {0}. Thus, π(f) = 0; i.e., f ∈ AAP(R+, X).

Similarly to the results on the line in Section 4.5, one can also characterizeasymptotically almost periodic functions by “the relative density of asymptoticε-periods”, but the definition on the half-line is slightly different.

Let f ∈ BUC(R+, X). For ε > 0, λ > 0, we let

Qε,λ(f) :={τ ∈ R+ : ‖f(t+ τ)− f(t)‖ ≤ ε whenever t ≥ λ

}.

We say that a subset Q of R+ is relatively dense in R+ if there exists a lengthl > 0 such that

[a, a+ l] ∩Q �= ∅ for all a ∈ R+.

Theorem 4.7.5. Let f ∈ BUC(R+, X). The following assertions are equivalent:

(i) f ∈ AAP(R+, X).

(ii) For all ε > 0 there exists a λ > 0 such that Qε,λ(f) is relatively dense inR+.

Proof. (ii) ⇒ (i): By Theorem 4.7.4, we have to show that the set {S(t)f : t ≥ 0}is precompact in BUC(R+, X). Let ε > 0. By assumption, there exists λ > 0 suchthat Qε,λ(f) is relatively dense in R+. Choose a length l such that [a, a + l] ∩Qε,λ(f) �= ∅ for all a ≥ 0. Since {S(t)f : 0 ≤ t ≤ 2λ+ l} is precompact, it sufficesto cover {S(t)f : t ≥ 2λ + l} by finitely many balls of radius 2ε. There existt1, . . . , tm ∈ [λ, λ+ l] such that for all t ∈ [λ, λ+ l] there exists j ∈ {1, . . . ,m} suchthat

‖S(t)f − S(tj)f‖∞ ≤ ε.

Now let t ≥ 2λ+l. There exists τ ∈ Qε,λ(f)∩ [t−λ−l, t−λ]. Then t−τ ∈ [λ, λ+l].There exists j ∈ {1, . . . ,m} such that

‖S(t− τ)f − S(tj)f‖∞ ≤ ε.

Hence,

‖S(t)f − S(tj)f‖∞ ≤ ‖S(t)f − S(t− τ)f‖∞ + ‖S(t− τ)f − S(tj)f‖∞ ≤ 2ε

since τ ∈ Qε,λ(f) and t− τ ≥ λ.(i) ⇒ (ii): Let f ∈ AAP(R+, X). Let ε > 0. There exist λ ≥ 0 and h ∈

AP(R, X) such that ‖f(t)− h(t)‖ ≤ ε/3 for all t ≥ λ. Let τ > 0 be an ε/3-periodfor h. Then τ ∈ Qε,λ(f). Since ε/3-periods for h are relatively dense in R, Qε, λ(f)is relatively dense in R+.

Recall from Section 4.4 that the half-line spectrum sp(f) of a function f ∈BUC(R+, X) is defined in the following way: call iη ∈ iR regular for f if f has aholomorphic extension to a neighbourhood of iη. Then we set

sp(f) ={η ∈ R : iη is not regular for f

}(4.39)


The half-line spectrum should not be confused with the Carleman spectrum in-troduced in the preceding section. Indeed, if f is the restriction of a functiong ∈ BUC(R, X), then the half-line spectrum of f is much smaller than the Carle-man spectrum of g in general. We make this more precise in the following remark.

Remark 4.7.6 (Comparison of half-line spectrum and Carleman spectrum).a) Let f ∈ BUC(R+, X). Assume that there exists η ∈ R \ sp(f). Then f has atmost one extension g ∈ BUC(R, X) such that η �∈ spC(g).

In fact, assume that there are two extensions g, h ∈ BUC(R, X) such thatη �∈ spC(g) ∪ spC(h). Consider k := g − h. Then k|R+

= 0. It follows that the

Carleman transform k of k is 0 on the right half-plane. Since k is a holomorphicfunction on C \ iR with a holomorphic extension to a neighbourhood of iη, it

follows from the uniqueness theorem for holomorphic functions that k(λ) = 0 alsoon the left half-plane. This implies that g − h = 0 by the uniqueness theorem forLaplace transforms.

b) Let f ∈ BUC(R+, X) be exponentially decreasing, but f �= 0. Then spC(g) = Rfor all extensions g ∈ BUC(R, X) of f , but sp(f) = ∅.

In fact, there existM ≥ 0, ε > 0 such that ‖f(t)‖ ≤Me−εt. Thus the Laplace

transform f(λ) of f has a holomorphic extension to the region {Reλ > −ε}and f(λ) is bounded for Reλ ≥ −ε/2. Now assume that g ∈ BUC(R, X) is anextension of f such that spC(g) �= R. Then the Carleman transform g(λ) of g

agrees with f(λ) for Reλ > 0, and hence for Reλ > −ε by the uniqueness theoremfor holomorphic functions. Since g(λ) is bounded for Reλ ≤ −ε/2, it follows thatg extends to a bounded entire function. So g is constant by Liouville’s theorem.Since limλ→∞ f(λ) = 0, it follows that f = 0. Hence, f = 0.

It is clear from the preceding remark that the assumption that the half-linespectrum be countable is much less restrictive than countability of the Carlemanspectrum. In addition, we have the following observation about countability of apart of the spectrum.

Let f ∈ E(R+, X). As in Section 4.3, we define the set of all frequencies of fby

Freq(f) := {η ∈ R : Mηf �= 0} .Since (Mηf)(0) = limα↓0 αf(α + iη), it follows from (4.38) that Freq(f) ⊂ sp(f).It follows from Proposition 4.3.11 that Freq(f) is countable.

Now we can formulate the main result of this section which is a complexTauberian theorem. With the help of Proposition 4.7.2, we are able to reduce theproof to an application of Gelfand’s theorem (Corollary 4.4.12).

Theorem 4.7.7. Let f be a totally ergodic function in BUC(R+, X) with countablehalf-line spectrum sp(f). Then f is asymptotically almost periodic.

The following corollary illustrates particularly well the Tauberian characterof this theorem.

Corollary 4.7.8. Let f ∈ E(R+, X) with countable half-line spectrum.

a) If Freq(f) = ∅, then f ∈ C0(R+, X).

b) If Freq(f) = {0}, then limt→∞ f(t) exists.

c) If Freq(f) ⊂ 2πτZ, where τ > 0, then there exists a τ -periodic, continuous

function g : R+ → X such that limt→∞ ‖ f(t)− g(t) ‖ = 0.

Proof. By Theorem 4.7.7, one has f = g + h ∈ AP(R+, X) + C0(R+, X). SinceFreq(g) = Freq(f), the claim follows from Corollary 4.5.9.

For the proof of Theorem 4.7.7 we need the following lemma which is analo-gous to Lemma 4.6.8.

Lemma 4.7.9. Let f ∈ BUC(R+, X). Assume that the Laplace transform f of f hasa holomorphic extension to a neighbourhood of iη, where η ∈ R. Then the functionR(·, B)f : C+ → BUC(R+, X) has a holomorphic extension to a neighbourhood ofiη.

Proof. We can assume that η = 0. Let V be a connected neighbourhood of 0 andg : V → X be holomorphic such that g(λ) = f(λ) for λ ∈ V ∩ C+. For λ ∈ V let

G(λ, s) := eλs(g(λ)−

∫ s

0

e−λtf(t) dt

)(s ∈ R+).

Then for Reλ > 0,

G(λ, s) =

∫ ∞

0

e−λtf(t+ s) dt = (R(λ,B)f)(s).

It is clear that G(·, s) : V → X is holomorphic for all s ∈ R+. Choose r > 0 suchthat B(0, 2r) ⊂ V . We show that

sups∈R+, |λ|≤r

‖G(λ, s) ‖ <∞.

In fact, let M := sup|λ|≤2r ‖ g(λ) ‖. If Reλ > 0, then

‖G(λ, s) ‖ = ‖ (R(λ,B)f)(s) ‖ ≤ ‖R(λ,B)f‖∞ ≤ ‖f‖∞Reλ

.

If Re λ < 0, |λ| = 2r, then |eλs| ≤ 1 and so

‖G(λ, s) ‖ ≤ M +

∫ s

0

e−Reλ (t−s) dt ‖f‖∞ ≤M +‖f‖∞|Reλ|

≤ 2rM + ‖ f ‖∞|Reλ| .


It follows from Lemma 4.6.6 that ‖G(λ, s) ‖ ≤ 43

(2rM+‖ f ‖∞

r

)=: c for all λ ∈

B(0, r), s ∈ R+.Now the claim follows from Corollary A.4 if we choose linear functionals ψ

on BUC(R+, X) which are of the form 〈f, ψ〉 := 〈f(s), x∗〉 where s ∈ R+, x∗ ∈

X∗, ‖x∗ ‖ ≤ 1.

Proof of Theorem 4.7.7. We keep the notation introduced before Proposition 4.7.2.Let f ∈ E(R+, X) have countable spectrum. Assume that f := π(f) �= 0. Then

E˜f:= span{S(t)f : t ∈ R} �= {0}.

Denote by B˜fthe generator of the group S restricted to E

˜f. Let η ∈ R\sp(f).

By Lemma 4.7.9, there exists an open neighbourhood V of iη and a holomorphicfunction H : V → BUC(R+, X) such that H(λ) = R(λ,B)f for λ ∈ V, Reλ >0. Since H(λ) ∈ E(R+, X) for Reλ > 0, it follows from the identity theorem

(Proposition A.2) that H(λ) ∈ E(R+, X) for all λ ∈ V . The function π◦H : V → Eis holomorphic and for λ ∈ V ∩ C+, (π ◦ H)(λ) = R(λ, B)π(f). It follows from

Lemma 4.6.7 that iη �∈ σ(B˜f ). Thus, σ(B˜f ) is countable. But then σ(B

˜f ) contains

an isolated point which is an eigenvalue of B˜fby Corollary 4.4.13. This contradicts

Proposition 4.7.2. Thus, π(f) = 0; i.e., f ∈ AAP(R+, X).

An even more general version of Theorem 4.7.7 will be given in Section 4.9.However, before that some further preparation concerning harmonic analysis isgiven in the following section.

We note an immediate corollary of Lemma 4.7.9.

Corollary 4.7.10. Let f ∈ BUC(R+, X). If η ∈ R \ sp(f), then f is uniformlyergodic at η and Mηf = 0.

We now give an example which shows that the condition of total ergodicityis crucial in Theorem 4.7.7, even in the scalar case. This contrasts dramaticallywith the situation of the entire line where total ergodicity is a consequence ofcountability of the Carleman spectrum if c0 �⊂ X (Theorem 4.6.3.

The example also shows that ergodicity has to be uniform with respect totranslates of the function. Recall that total ergodicity means by definition that thefunction is uniformly ergodic at each η ∈ R.

Example 4.7.11. The function f(t) := sin√t has half-line spectrum sp(f) = {0}

and 0 �∈ Freq(f). Since f(t) does not converge to 0 as t→∞, it follows from The-

orem 4.7.7 that f is not uniformly ergodic at 0. However, limt→∞ 1t

∫ t

0f(s) ds = 0.

Proof. a) We show that

f(λ) =

√πe−1/4λ

2λ3/2(Reλ > 0).

In fact, f(t) = sin√t =

∑∞n=0

(−1)ntn+1/2

(2n+1)! . Thus, for λ > 0, integrating term by

term, we obtain

f(λ) =

∫ ∞

0

sin√t e−λt dt

=

∞∑n=0

(−1)n(2n+ 1)!

∫ ∞

0

tn+1/2e−λt dt

=∞∑

n=0

(−1)n(2n+ 1)!

1

λn+3/2

∫ ∞

0

un+1/2e−u du

=

∞∑n=0

(−1)n(2n+ 1)!

λ−n−3/2Γ(n+ 3/2).

Standard formulas for the Gamma function give

Γ(n+ 3/2) = (n+ 1/2)(n− 1/2) . . .1

2Γ(1/2)

=(2n+ 1)(2n− 1) . . . 1

2n+1

√π =

(2n+ 1)!

n!22n+1

√π.

Thus,

f(λ) =∞∑

n=0

λ−n−3/2 (−1)nn!22n+1

√π

=

√π

2λ3/2

∞∑n=0

1

n!

(−1)n(4λ)n

=

√π

2λ3/2e−1/4λ.

By uniqueness of holomorphic extensions, this formula is true for Reλ > 0. Itfollows that sp(f) = {0}.

b) For t > 0,

1

t

∫ t

0

f(s) ds =1

t

∫ √t

0

2u sinu du

=1

t[−2u cosu]

√t

0 +2

t

∫ √t

0

cosu du

= −2 cos√t√

t+

2

tsin√t→ 0

as t→∞.c) One can directly see that f is not uniformly ergodic at 0. In fact, given

t > 0 we may choose s such that

(2n+ 1/4)π <√s <

√s+ t < (2n+ 3/4)π


for some integer n (since√s+ t−√s→ 0 as s→∞). Then f(u) > 1√

2whenever

s ≤ u ≤ s+ t, so 1t

∫ s+t

sf(u) du > 1√

2. Thus, f is not uniformly ergodic at 0.

We want to extend Theorem 4.7.7 to bounded measurable functions. Thenwe can only obtain assertions for the means of f , for example B-convergence, butthis is sufficient for interesting applications to power series.

Let f ∈ L∞(R+, X). We define the half-line spectrum sp(f) as in Section 4.4;viz.,

sp(f) :=

{η ∈ R : f(λ) does not have a holomorphic extension

to a neighbourhood of iη

}.

Moreover, we say that f is totally ergodic if for each η ∈ R,

(Mηf)(t) := limα↓0

α

∫ ∞

0

e−(α+iη)sf(t+ s) ds

converges uniformly in t ∈ R+. In that case, it follows as in (4.25) that

Mηf = eiη ⊗ x,

where eiη(t) = eiηt, x = (Mηf)(0) ∈ X . We set

Freq(f) := {η ∈ R : Mηf �= 0}.The following is a Tauberian theorem where B-convergence is deduced.

Theorem 4.7.12. Let f ∈ L∞(R+, X) be totally ergodic. Assume that sp(f) iscountable and Freq(f) ⊂ {0}. Then B-limt→∞ f(t) = (M0f)(0).

Proof. Let δ > 0, g(t) := 1δ

∫ t+δ

tf(s) ds. Then g ∈ BUC(R+, X). Integrating by

parts we obtain for Reλ > 0,

g(λ) = − 1

δλ

∫ ∞

0

d

dt

(e−λt

) ∫ t+δ

t

f(s) ds dt

=1

δλ

(∫ δ

0

f(s) ds+

∫ ∞

0

e−λt(f(t+ δ)− f(t)) dt

)

=1

δλ

(∫ δ

0

f(s) ds+ eλδ∫ ∞

δ

e−λtf(t) dt− f(λ)

)

=1

δλ(eλδ − 1)f(λ) +

1

δλ

(∫ δ

0

f(s) ds− eλδ∫ δ

0

e−λsf(s) ds

)

=1

δλ(eλδ − 1)f(λ) +

(1− eλδ

λδ

∫ δ

0

f(s) ds+ eλδ∫ δ

0

f(s)1− e−λs

λδds

).

The right-hand summand defines an entire function of λ. Since 1δλ (e

λδ−1) is entire,it follows that sp(g) ⊂ sp(f).

Next, we show that g is totally ergodic. Let r ∈ R, gr(s) := g(s+r), fr(s) :=

f(s + r). Then gr(t) = 1δ

∫ t+δ

tfr(s) ds. By the computation above, we have for

Reλ > 0,

gr(λ) =1

δλ(eλδ − 1)fr(λ) +

1− eλδ

λδ

∫ δ

0

fr(s) ds+ eλδ∫ δ

0

fr(s)1− e−λδ

λδds.

Let λ = α + iη (α > 0). Then αfr(α + iη) converges uniformly in r ∈ R+ asα ↓ 0. Consequently, so does αgr(α+ iη). We have shown that g is totally ergodic.

Moreover, (M0g)(0) = limα↓0 αg(α) = limα↓0 αf(α) = (M0f)(0), whereas forη �= 0, (Mηg)(0) = 0. It follows from Corollary 4.7.8 that g(t) converges as t→∞;i.e., f(t) is B-convergent as t→∞. It follows from Theorem 4.1.2 that

B- limt→∞ f(t) = lim

t→∞ g(t) = (M0g)(0) = (M0f)(0).

Lemma 4.7.13. Let f ∈ L∞(R+, X). Let η ∈ R such that

supt≥0

∥∥∥∥∫ t

0

e−iηsf(s) ds

∥∥∥∥ <∞.

Then f is uniformly ergodic at η and Mηf = 0.

Proof. Let M be such that∥∥∥∫ t

0e−iηsf(s) ds

∥∥∥ ≤M (t ≥ 0). Let r ≥ 0. Then∥∥∥∥∫ t

0

e−iηsf(s+ r) ds

∥∥∥∥ =

∥∥∥∥eiηr ∫ t+r

t

e−iηsf(s) ds

∥∥∥∥ ≤ 2M.

As an application of Theorem 4.7.12 we obtain the following result on powerseries. We let T := {z ∈ C : |z| = 1}.Proposition 4.7.14. Let an ∈ X such that M := supm∈N ‖

∑mn=0 an‖ <∞. Consider

the power series

p(z) :=

∞∑n=0

anzn (|z| < 1).

Assume that for each z ∈ T \ {1} the function p has a holomorphic extension to aneighbourhood of z. Then limn→∞ an = 0.

Proof. Let f(t) := an for t ∈ [n, n + 1), n ∈ N0. Then f ∈ L∞(R+, X) and‖f‖∞ ≤ 2M . For Reλ > 0, one has

f(λ) =

∞∑m=0

am

∫ m+1

m

e−λt dt =

∞∑m=0

ame−λm

∫ 1

0

e−λt dt

=1− e−λ

λ

∞∑m=0

ame−λm.

Thus, sp(f) ⊂ 2πZ. We will show that supt≥0

∥∥∥∫ t

0e−iηsf(s) ds

∥∥∥ < ∞ for all η ∈2πZ. Then Lemma 4.7.13 and Corollary 4.7.10 imply that f is totally ergodic andMηf = 0 for all η ∈ R. Finally, Theorem 4.7.12 implies that

limn→∞ an = lim

n→∞

∫ n+1

n

f(s) ds = B- limn→∞ f(t) = 0.

Let η = 0. For t ≥ 0, let n ∈ N such that t ∈ [n, n+ 1). Then

∫ t

0

f(s) ds =n−1∑m=0

am + an(t− n).

Thus,∥∥∥∫ t

0f(s) ds

∥∥∥ ≤ 3M .

Let η = 2πk, k ∈ Z \ {0}. Let t ≥ 0, t ∈ [n, n+ 1). Then

∫ t

0

e−iηsf(s) ds =n−1∑m=0

am

∫ m+1

m

e−i2πkt dt+ an

∫ t

n

e−i2πks ds

= an

∫ t

n

e−i2πks ds.

Hence,∥∥∥∫ t

0e−iηsf(s) ds

∥∥∥ ≤ 2M .

Corollary 4.7.15 (Katznelson-Tzafriri). Let T ∈ L(X) be an operator such thatM := supn∈N ‖Tn‖ < ∞. Then σ(T ) ∩ T ⊂ {1} if and only if limn→∞ ‖Tn(I −T )‖ = 0.

Proof. a) Assume that σ(T ) ∩ T ⊂ {1}. Let

p(z) :=

∞∑n=0

(T − I)Tnzn = (T − I)(I − zT )−1 for |z| < 1.

Then ∥∥∥∥∥m∑

n=0

(T − I)Tn

∥∥∥∥∥ = ‖Tm+1 − I‖ ≤M + 1.

Thus the claim follows from Proposition 4.7.14.b) Suppose that limn→∞ ‖Tn(I −T )‖ = 0 and μ ∈ σ(T )∩T. By the spectral

mapping theorem, μn(1− μ) ∈ σ(Tn(I − T )). Then

|1− μ| = |μn(1− μ)| ≤ ‖Tn(I − T )‖ → 0.

Hence, μ = 1.

4.8 Carleman Spectrum and Fourier Transform

For functions on the line, the Carleman spectrum is the natural notion if weare interested in the properties of the Laplace transform (or more precisely, theCarleman transform). However, the Fourier transform is also a powerful tool. Inorder to make it available we show in this section that the Carleman spectrumcoincides with the support of the Fourier transform of the given function and alsowith the Beurling spectrum.

As applications we give several results which allow us to deduce asymptoticproperties of a function from the nature of its spectrum. For example, we show thata function f ∈ BUC(R, X) is almost periodic whenever it has discrete spectrum.A function f ∈ L∞(R, X) is τ -periodic if and only if its spectrum is contained in2πτZ.

Let f ∈ L1(R). As in Section 1.8 and Appendix E, we denote by Ff ∈ C0(R)the Fourier transform given by

Ff(t) :=∫R

e−istf(s) ds (t ∈ R) (4.40)

and we let

Ff(t) :=∫R

eitsf(s) ds (t ∈ R). (4.41)

By S(R) we denote the Schwartz space; i.e., the space of all infinitely dif-ferentiable functions f : R → C such that supt∈R |f (m)(t)|(1 + |t|)k < ∞ for allm, k ∈ N0. Then F is a bijective mapping from S(R) into S(R) with inverse(2π)−1F . See Appendix E for further information.

Let X be a Banach space. By L1(R, (1 + |t|)−k dt;X) we denote the space ofall functions f ∈ L1

loc(R, X) such that∫R‖ f(t) ‖ (1+ |t|)−k dt <∞, where k ∈ N0.

Note that L1(R, X) ⊂ L1(R, (1 + |t|)−k dt;X) for all k ∈ N0.Let f ∈ L1(R, (1+ |t|)−k dt;X), where k ∈ N0. Then we define Ff as a linear

mapping from S(R) into X by

〈ϕ,Ff〉 =

∫R

f(t)(Fϕ)(t) dt (ϕ ∈ S(R)).

The support of Ff is defined by

suppFf :=

{η ∈ R : for all ε > 0 there exists ϕ ∈ S(R) such that

suppϕ ⊂ (η − ε, η + ε) and 〈ϕ,Ff〉, �= 0

}.

(4.42)

In the particular case when k = 0, i.e. f ∈ L1(R, X), Fubini’s theorem shows that〈ϕ,Ff〉 =

∫R

∫Rf(t)e−ist dt ϕ(s) ds. Hence,

suppFf = {s ∈ R : (Ff)(s) �= 0}−, (4.43)

4.8. CARLEMAN SPECTRUM AND FOURIER TRANSFORM 319

and our notation is consistent with (4.40) and the identification of absolutely regu-lar functions with distributions when X = C (see Theorem 1.8.1 b) and AppendixE).

If f ∈ L1(R, (1 + |t|)−k dt;X), then one can define the Carleman transform

f : C \ (iR) → X and the Carleman spectrum spC(f) exactly as in Section 4.6.We are going to show that suppFf and spC(f) coincide.

We recall the notion of mollifier (ρn)n∈N from Section 1.3, but here we shallassume that ρ1 ∈ S(R). The function ρ1 ∈ S(R) satisfies

∫Rρ1(t) dt = 1. Then

ρn ∈ S(R) is given by ρn(t) = nρ1(nt) (t ∈ R, n ∈ N). Thus,

Fρn ∈ S(R) (n ∈ N), (4.44)

and Fρn(t) = Fρ1( tn), so

limn→∞(Fρn)(t) = 1 (t ∈ R); (4.45)

|(Fρn)(t)| ≤∫R

|ρ1(s)| ds (t ∈ R, n ∈ N). (4.46)

Moreover, if supp ρ1 ⊂ (−1, 1), thensupp ρn ⊂

(− 1n ,

1n

)(n ∈ N); (4.47)

and if suppFρ1 ⊂ (−1, 1), thensuppFρn ⊂ (−n, n) (n ∈ N). (4.48)

Theorem 4.8.1. Let k ∈ N0, f ∈ L1(R, (1 + |t|)−k dt;X). Then spC(f) = suppFf .

Proof. Let η ∈ R. Then by the definition of the Carleman transform, for α > 0,one has

f(α+ iη) =

∫ ∞

0

e−αte−iηtf(t) dt and

f(−α+ iη) = −∫ 0

−∞eαte−iηtf(t) dt.

Hence,

f(α + iη)− f(−α+ iη) =

∫ ∞

−∞e−α|t|e−iηtf(t) dt. (4.49)

Let ϕ ∈ S(R). Then by Fubini’s theorem,

〈ϕ,Ff〉 =

∫R

f(t)(Fϕ)(t) dt

= limα↓0

∫R

e−α|t|f(t)(Fϕ)(t) dt

= limα↓0

∫R

e−α|t|f(t)∫R

e−istϕ(s) ds dt

= limα↓0

∫R

∫R

e−α|t|e−istf(t) dt ϕ(s) ds.

Thus by (4.49), we obtain

〈ϕ,Ff〉 = limα↓0

∫R

(f(α+ is)− f(−α+ is)

)ϕ(s) ds (4.50)

for all ϕ ∈ S(R).From this, the inclusion suppFf ⊂ spC(f) follows immediately: Assume

that η �∈ spC(f). Then there exists ε > 0 such that f has a bounded holomorphic

extension to B(iη, ε). This implies that limα↓0(f(α+ is)− f(−α+ is)

)= 0 when-

ever s ∈ (η − ε, η + ε). Thus, by (4.50) and the dominated convergence theorem,〈ϕ,Ff〉 = 0 whenever ϕ ∈ S(R) and suppϕ ⊂ (η − ε, η + ε).

Next, we prove the other inclusion: spC(f) ⊂ suppFf .a) We first deal with the special case where k = 0; i.e., f ∈ L1(R). Then Ff

is a continuous function. Let η ∈ R \ suppFf . Then there exists ε > 0 such thatFf vanishes on (η − ε, η + ε). Putting

f(iβ) :=

∫ ∞

0

e−iβtf(t) dt = −∫ 0

−∞e−iβtf(t) dt

for β ∈ (η − ε, η + ε), f has a continuous extension to a neighbourhood of iη.It follows from Morera’s theorem that this extension is holomorphic. Thus, η �∈spC(f).

b) Now let k ∈ N0 be arbitrary. We first give an estimate of f . Let M :=∫R(1 + |t|)−k ‖ f(t) ‖ dt. Then

‖ f(λ) ‖ ≤Mkke1−k|Reλ|−k (4.51)

whenever 0 < |Reλ| < 1. In fact, if 0 < Re λ < 1, then

‖ f(λ) ‖ ≤∫ ∞

0

e−Reλt‖ f(t) ‖ dt

≤ M supt>0

((1 + t)ke−Reλt

)≤ Mkke1−k(Reλ)−k.

The estimation is similar for −1 < Re λ < 0.Now let η ∈ R \ suppFf . Then there exists ε > 0 such that 〈ϕ,Ff〉 = 0 for

all ϕ ∈ S(R) with suppϕ ⊂ (η − 2ε, η + 2ε). Let (ρn)n∈N be a mollifier in S(R),such that supp ρn ⊂ (− 1

n ,1n ). Let fn := (Fρn) · f . Since Fρn ∈ S(R), we have

fn ∈ L1(R, X).For n > 1/ε, we have suppFfn ∩ (η − ε, η + ε) = ∅. In fact, let ϕ ∈ S(R)

such that suppϕ ⊂ (η − ε, η + ε). Then 〈Ffn, ϕ〉 =∫RfnFϕ =

∫RfFρn · Fϕ =∫

fF(ρn ∗ ϕ) = 0 since ρn ∗ ϕ ∈ S(R) and supp ρn ∗ ϕ ⊂ supp ρn + suppϕ ⊂(− 1

n ,1n ) + (η − ε, η + ε) ⊂ (η − 2ε, η + 2ε).

From part a) of the proof we conclude that fn has a holomorphic extension toB(iη, ε). Since ‖ fn(t) ‖ ≤ ‖f(t)‖ ‖ρ1‖1, we obtain from (4.51) (replacing f by fn)

that ‖ fn(λ) ‖ ≤ M‖ρ‖1kke1−k|Re λ|−k whenever 0 < |Reλ| < 1. Now it followsfrom Lemma 4.6.6 that

‖ fn(λ) ‖ ≤ c (λ ∈ B(iη, ε/2)) (4.52)

for all n ∈ N, where c is a constant independent of n ∈ N. Since fn(t) → f(t) asn→∞ for all t ∈ R (see Lemma 1.3.3), it follows from the dominated convergence

theorem that limn→∞ fn(λ) = f(λ) if Reλ > 0. It follows from Vitali’s theorem

(Theorem A.5) that f has a holomorphic extension toB(iη, ε/2). Thus, η �∈ spC(f).

The following are consequences of Theorem 4.8.1, and extensions of parts ofCorollary 4.5.9 (see Proposition 4.6.10).

Theorem 4.8.2. Let f ∈ L1(R, (1 + |t|)−k dt;X) for some k ∈ N0.

a) spC(f) = ∅ if and only if f(t) = 0 a.e.

b) spC(f) = {0} if and only if f is a polynomial.

c) spC(f) ⊂ {η1 . . . ηm} if and only if there exist polynomials p1, . . . , pm suchthat f(t) =

∑mj=1 pj(t)e

iηjt (t ∈ R).

Proof. a) Assume that spC(f) = ∅. Then it follows from the proof of Theorem4.8.1 that

∫Rf(t)Fϕ(t) dt = 0 for all ϕ ∈ C∞c (R). Since C∞c (R) is dense in S(R)

for the canonical topology of S(R) and since F is continuous (see Appendix E), itfollows that

∫Rf(t)(Fϕ)(t) dt = 0 for all ϕ ∈ S(R). Hence,

∫Rf(t)ψ(t) dt = 0 for

all ψ ∈ S(R). This implies that f(t) = 0 a.e.b) Assume that spC(f) = {0}. Then the Laurent series

f(λ) =

∞∑n=−∞

anλn

converges for λ ∈ C \ {0}, where

an :=1

2πi

∫|z|=r

f(z)z−n−1 dz (n ∈ Z)

independently of r > 0.From (4.51) we obtain a constant c > 0 such that

‖ f(λ) ‖ ≤ c|Reλ|−k (4.53)

for 0 < |Reλ| < 1. This implies that∥∥∥∥∥ 1

2πi

∫|z|=r

zn+k−1(1 +z2

r2)kf(z) dz

∥∥∥∥∥ ≤ c 2krn (n ∈ Z, r ∈ (0, 1))

(observe that for z = reiθ, ‖ (1 + z2

r2)kf(z) ‖ ≤ |1 + ei2θ|kc |r cos θ|−k = 2kcr−k).

Hence,

c 2krn ≥∥∥∥∥∥∥

k∑j=0

(k

j

)r−2j · 1

2πi

∫|z|=r

zn+k−1+2j f(z) dz

∥∥∥∥∥∥=

∥∥∥∥∥∥k∑

j=0

(k

j

)r−2ja−n−k−2j

∥∥∥∥∥∥for all r ∈ (0, 1). Multiplying by r2k, we obtain

c2krn+2k ≥∥∥∥∥∥∥

k∑j=0

(k

j

)r2k−2ja−n−k−2j

∥∥∥∥∥∥ .Now let n ≥ 1−2k. Then the left-hand term, as well as all terms on the right exceptfor j = k, converge to 0 as r ↓ 0. Consequently, a−n−3k = 0 whenever n ≥ 1− 2k,

i.e., am = 0 if m < −k, so f(λ) =∑∞

n=−k anλn. Let f0(t) :=

∑kj=0 a−j−1t

j/j!.

Then f0(λ) =∑−1

n=−k anλn. Thus f− f0 is entire and so spC(f−f0) = ∅. It follows

from a) that f(t) = f0(t) a.e.c) We prove the assertion by induction. Let m = 1. Then spC(f) = {η1}. Let

g(t) := e−iη1tf(t). Then spC(g) = 0. It follows from b) that g is a polynomial.

Now assume that the assertion holds form and that spC(f) = {η1, . . . , ηm+1},where ηj �= ηl if j �= l. Replacing f(t) by e−iηm+1tf(t) if necessary, we can assume

that ηm+1 = 0. It follows from the proof of b) that 0 is a pole of f . Moreover, thereexists a polynomial pm+1 such that (f − pm+1) has a holomorphic extension toa neighbourhood of 0. Thus, spC(f − pm+1) ⊂ {η1, . . . , ηm}. Now it follows fromthe inductive assumption that (f − pm+1)(t) =

∑mn=1 pn(t)e

iηnt (t ∈ R).

Corollary 4.8.3. Let f ∈ L∞(R, X).

a) If spC(f) = {0}, then f is constant.

b) If spC(f) is finite, then f is a trigonometric polynomial.

If f ∈ L∞(R, X), there is an alternative way to describe the spectrum of f .We define the Beurling spectrum by

spB(f) :=

{η ∈ R : for all ε > 0 there exists g ∈ L1(R)

such that suppFg ⊂ (η − ε, η + ε) and f ∗ g �= 0

}.

Proposition 4.8.4. Let f ∈ L∞(R, X). Then spC(f) = spB(f).


Proof. By Theorem 4.8.1, we have

spC(f) = suppFf

={η ∈ R : for all ε > 0 there exists ϕ ∈ S(R) such that

suppϕ ⊂ (η − ε, η + ε) and

∫R

f(t)(Fϕ)(t) dt �= 0}

={η ∈ R : for all ε > 0 there exists h ∈ S(R) such that

suppFh ⊂ (η − ε, η + ε) and

∫R

f(t)h(t) dt �= 0}


suppFh ⊂ (η − ε, η + ε) and

∫R

f(t)h(−t) dt �= 0}


suppFh ⊂ (η − ε, η + ε) and (f ∗ h)(0) �= 0}.

This shows that spC(f) ⊂ spB(f).In order to show the converse, let η ∈ spB(f). Let ε > 0. Then there exists

g ∈ L1(R) such that suppFg ⊂ (η− ε, η+ ε) and f ∗ g �= 0. Replacing g by g(·− t)if necessary, we can assume that (f ∗ g)(0) �= 0.

Take a mollifier (ρn)n∈N in S(R) such that supp ρn ⊂ (− 1n ,

1n ), and de-

fine gn := Fρn · g. By (4.45), (4.46) and the dominated convergence theorem,limn→∞ gn = g in L1(R). It follows from Proposition 1.3.2 that limn→∞(gn ∗f)(0) = (g ∗ f)(0) �= 0.

We have ρn ∗ Fg ∈ C∞(R) (see Proposition 1.3.6) and supp(ρn ∗ Fg) ⊂(− 1

n, 1n) + (η − ε, η + ε) ⊂ (η − 2ε, η + 2ε) if n > 1/ε. Moreover, (ρn ∗ Fg)(s) =∫

ρn(s− r)Fg(r) dr =∫ρn(s− r)

∫g(t)e−irtdt dr =

∫ρn(r)

∫g(t)e−i(s−r)tdt dr =

F((Fρn)·g)(s). Hence, Fgn = F((Fρn)·g) ∈ C∞c (R) ⊂ S(R). Since F is a bijectionfrom S(R) onto S(R), it follows that gn ∈ S(R). We have shown that there existsn > 1/ε such that gn ∈ S(R), suppFgn ⊂ (η − 2ε, η + 2ε) and (f ∗ gn)(0) �= 0.Since ε > 0 is arbitrary, it follows that η ∈ suppFf = spC(f).

So far, we have established the identity of three different notions of spectrumof a function f ∈ L∞(R, X): the Carleman spectrum, the support of its Fouriertransform, and the Beurling spectrum. Next, we consider two situations in whicha special form of the spectrum of a function tells us a lot about the nature of thefunction itself (Theorems 4.8.7 and 4.8.8).

We need the following auxiliary result.

Lemma 4.8.5. Let f ∈ L∞(R, X) and g ∈ L1(R). Then spC(f ∗ g) ⊂ spC(f) ∩suppFg.Proof. a) Let η ∈ R \ spC(f). By Proposition 4.8.4, there exists ε > 0 such that


f ∗ h = 0 whenever h ∈ L1(R), suppFh ⊂ (η − ε, η + ε). Hence, (f ∗ g) ∗ h =(f ∗ h) ∗ g = 0. Thus, η �∈ spB(f ∗ g) = spC(f ∗ g).

b) Let η �∈ suppFg. We show that η �∈ spB(f ∗ g). There exists ε > 0 suchthat (Fg)(r) = 0 for all r ∈ (η − ε, η + ε). Let h ∈ L1(R) such that suppFh ⊂(η− ε, η+ ε). Then F(g ∗h) = Fg · Fh = 0. Thus, g ∗h = 0. Hence, (f ∗ g) ∗h = 0.This proves the claim.

Remark 4.8.6. The same property is true if f ∈ L1(R, (1 + |t|)−k dt;X) for somek ∈ N0 and g ∈ S(R).

Now we can prove that a function f ∈ BUC(R, X) is almost periodic wheneverit has discrete Carleman spectrum. Thus, the geometric condition “c0 �⊂ X”,which appeared in Theorem 4.6.3, is not needed if the spectrum does not have anyaccumulation point.

Theorem 4.8.7. Let f ∈ BUC(R, X). If the spectrum spC(f) of f is discrete, thenf is almost periodic.

Proof. Let (ρn)n∈N be a mollifier in S(R) such that suppFρ1 ⊂ (−1, 1). ByProposition 1.3.2 and Lemma 4.8.5, f ∗ ρn ∈ BUC(R, X) and spC(f ∗ ρn) ⊂spC(f) ∩ suppFρn ⊂ spC(f) ∩ (−n, n). Thus, f ∗ ρn has finite spectrum, andit follows from Corollary 4.8.3 that f ∗ρn ∈ AP(R, X). By Lemma 1.3.3, it followsthat f = limn→∞ f ∗ ρn ∈ AP(R, X).

Next, we give a spectral characterization of periodic functions extending partof Corollary 4.5.9. If g is a function defined on R we let g(t) = g(−t) (t ∈ R).

Theorem 4.8.8. Let f ∈ L∞(R, X) and let τ > 0. Then f is τ-periodic (i.e.,f(t+ τ) = f(t) t-a.e.) if and only if spC(f) ⊂ 2π

τ Z.

Proof. a) Let f be τ -periodic. Then

f(λ) = (1− e−λτ )−1

∫ τ

0

e−λtf(t) dt

for λ ∈ C \ iR. Thus, f extends to a meromorphic function with at most simplepoles at λn := 2πin/τ (n ∈ Z). The residue at λn is given by

cn :=1

τ

∫ τ

0

e−2πint/τf(t) dt (n ∈ Z).

Therefore we have

spC(f) = {2πn/τ : n ∈ Z, cn �= 0} .b) Assume that spC(f) ⊂ 2π

τ Z. Let (ρn)n∈N be a mollifier in S(R) such thatsuppFρn ⊂ (−n, n) and ρn = ρn (n ∈ N). Then ρn ∗ f ∈ BUC(R, X) andspC(ρn ∗ f) ⊂ suppFρn ∩ spC(f) ⊂ (−n, n) ∩ spC(f) ⊂ (−n, n) ∩ 2π

τ Z. It follows

4.9. COMPLEX TAUBERIAN THEOREMS: THE FOURIER METHOD 325

from Corollary 4.8.3 and Proposition 4.6.10 that ρn∗f is a τ -periodic trigonometricpolynomial.

Let ϕ ∈ Cc(R). Then limn→∞ ρn ∗ϕ = ϕ in L1(R). Thus,∫f(t+ τ)ϕ(t) dt =

limn→∞∫f(t+ τ) (ρn ∗ϕ)(t) dt = limn→∞

∫(f ∗ρn)(t+ τ)ϕ(t) dt = limn→∞

∫(f ∗

ρn)(t)ϕ(t) dt = limn→∞∫f(t) (ρn ∗ ϕ)(t) dt =

∫f(t)ϕ(t) dt. Since ϕ ∈ Cc(R) is

arbitrary, it follows that f(t+ τ) = f(t) t-a.e.

4.9 Complex Tauberian Theorems: the Fourier Method

In this section we present an approach via Fourier transforms to complex Tauberiantheorems for Laplace transforms. This method was already used in a restrictedform by Ingham in 1935 (see [Ing35]). It will eventually lead to the most generalcomplex Tauberian theorem presented in this book (Theorem 4.9.7).

It is not such a great difference to consider bounded measurable functionswhich are slowly oscillating, instead of uniformly continuous functions as we didbefore, but the point is that the notion of spectrum is changed. Instead of consid-ering a point iη as regular if the Laplace transform has a holomorphic extensionin a neighbourhood of iη, we merely assume that a locally integrable extension tothe imaginary axis exists. This leads to a smaller spectrum which we call the weakhalf-line spectrum. Thus, asking that this small spectrum be countable is a weakerhypothesis. It turns out that this weaker hypothesis is more natural or easier toverify in some applications (see the Notes of this section).

In the case where the weak half-line spectrum is empty, the proof is completelyelementary and leads to a slight generalization of Ingham’s Tauberian theorem.We consider this case first (Theorem 4.9.5). The general case will be proved bya Hahn-Banach argument which allows us to apply Loomis’s theorem in a quitetricky way.

Let f ∈ L∞(R+, X). We define the weak half-line spectrum spw(f) of f asfollows.

Definition 4.9.1. Let η ∈ R. We say that iη is a weakly regular point for f if thereexist ε > 0 and h ∈ L1((η − ε, η + ε), X) such that

f(α + i·)→ h in the distributional sense on (η − ε, η + ε) as α ↓ 0. (4.54)

Then the weak half-line spectrum spw(f) of f is defined as the set of all real

numbers which are not weakly regular for f .

Of course, (4.54) means by definition that

limα↓0

∫R

f(α + is)ϕ(s) ds =

∫ η+ε

η−ε

h(s)ϕ(s) ds (4.55)

for all test functions ϕ ∈ D(η − ε, η − ε) = C∞c (η − ε, η − ε).


It is clear that spw(f) is a closed subset of R. Moreover, if f(λ) has a con-tinuous extension to C+ ∪ (i(η − ε), i(η + ε)), where η ∈ R, ε > 0, then clearlyη �∈ spw(f). In particular, we have spw(f) ⊂ sp(f). The inclusion is strict ingeneral as the following example shows.

Example 4.9.2. Let X := l2,

f(t) :=(n−1e−t/n

)n∈N

.

Then f has a continuous extension to C+ ∪ iR, but not a holomorphic extensionto a neighbourhood of 0.

Lemma 4.9.3. Let f ∈ L∞(R+, X). Then there exists h ∈ L1loc(R\ spw(f), X) such

that

f(α+ i·)→ h in the distributional sense on R \ spw(f) as α ↓ 0. (4.56)

We set f(is) := h(s) (s ∈ R \ spw(f)).Proof. For all x ∈ R \ spw(f), we find an open neighbourhood Ux of x and hx ∈L1(Ux, X) such that f(α + i·) → hx in the distributional sense on Ux as α ↓ 0.Clearly, hx(t) = hy(t) almost everywhere on Ux ∩ Uy whenever x, y ∈ R \ spw(f).Hence, there exists a function h ∈ L1

loc(R \ spw(f), X) such that

h|Ux= hx (x ∈ R \ spw(f)).

It remains to show (4.56). Let ϕ ∈ D(R \ spw(f)). Let K := suppϕ. There existx1, x2, . . . , xn such that the sets Uj := Uxj

(j = 1, 2, . . . , n) cover K. Let ϕj ∈D(R\spw(f)) (j = 1, 2, . . . , n) be a partition of unity subordinate to this covering;i.e., 0 ≤ ϕj ≤ 1, suppϕj ⊂ Uj (j = 1, 2, . . . , n),

∑nj=1 ϕj(x) = 1 for all x ∈ K.

Then ϕ =∑n

j=1 ψj , where ψj := ϕϕj . Since suppψj ⊂ Uj ,

limα↓0

∫R

f(α+ is)ψj(s) ds =

∫Uj

h(s)ψj(s) ds

for all j = 1, 2, . . . , n. Hence,

limα↓0

∫R

f(α+ is)ϕ(s) ds =

∫R\spw(f)

h(s)ϕ(s) ds. (4.57)

Since ϕ ∈ D(R \ spw(f)) is arbitrary, this is precisely the meaning of (4.56).

Lemma 4.9.4. Let f ∈ L∞(R+, X). Then

ϕ ∗ f ∈ C0(R+, X)

for all ϕ ∈ S(R) such that Fϕ ∈ C∞c (R) and suppFϕ ∩ spw(f) = ∅.

4.9. COMPLEX TAUBERIAN THEOREMS: THE FOURIER METHOD 327

Here as elsewhere, we identify a function defined on R+ with its extensionby 0 on R. In particular,

(ϕ ∗ f)(t) =∫ ∞

0

f(s)ϕ(t− s) ds (t ∈ R).

Proof of Lemma 4.9.4. Let ϕ ∈ S(R) such that Fϕ ∈ C∞c (R) and suppFϕ ∩spw(f) = ∅. Then Fϕ · f(i·) ∈ L1(R, X), where f(i·) was defined in Lemma 4.9.3.Let t ≥ 0. Then the inverse Fourier transform of the function s �→ ϕ(t− s) is thefunction η �→ (2π)−1eiηtFϕ(η). Thus by Theorem 1.8.1 b),

(ϕ ∗ f)(t) =

∫ ∞

0

f(s)ϕ(t− s) ds

= limα↓0

∫ ∞

0

e−αsf(s)ϕ(t− s) ds

= limα↓0

(2π)−1

∫R

f(α + iη)eiηtFϕ(η) dη

= (2π)−1

∫R

f(iη)Fϕ(η)eiηt dη.

It follows from the Riemann-Lebesgue lemma (Theorem 1.8.1) that limt→∞(ϕ ∗f)(t) = 0.

Now we prove the complex Tauberian theorem in the case where the weakhalf-line spectrum is empty.

Theorem 4.9.5 (Ingham). Let f ∈ L∞(R+, X) be slowly oscillating. If spw(f) = ∅,then

limt→∞ f(t) = 0.

Proof. a) We show that g ∗ f ∈ C0(R+, X) for all g ∈ L1(R). By Proposition 1.3.2,Tg := (g ∗ f)|R+

defines a bounded linear operator from L1(R) into BUC(R+, X).Let φ be a continuous linear functional on BUC(R+, X) vanishing on C0(R+, X).By the Hahn-Banach theorem, it suffices to show that 〈Tg, φ〉 = 0 for all g ∈ L1(R).Let h := T ∗φ ∈ L∞(R). Then

〈g ∗ f, φ〉 =∫R

hg dt

for all g ∈ L1(R). It follows from Lemma 4.9.4 that∫Rhϕdt = 0 if ϕ ∈ S(R) and

Fϕ ∈ C∞c (R). Hence, suppFh = ∅ (see (4.42)). Thus, spC(h) = ∅ by Theorem4.8.1 which implies that h = 0 by Theorem 4.8.2. This proves the claim. We haveshown that g ∗ f ∈ C0(R+, X) for all g ∈ L1(R).

b) Taking in particular g := 1δχ[0,δ], it follows from a) that B-limt→∞ f(t) = 0.

Now it follows from Theorem 4.2.3 that limt→∞ f(t) = 0.


We will see in Example 5.5.7 that in general the hypothesis that g be slowlyoscillating cannot be omitted.

Next, we consider the case of countable weak spectrum and we extend Theo-rem 4.7.12. We need the notion of uniform ergodicity on a subset of R, for boundedmeasurable functions, similar to the total ergodicity of Section 4.7.

Definition 4.9.6. Let f ∈ L∞(R+, X) and let E be a subset of R. We say that f isuniformly ergodic on E if for all η ∈ E the limit

(Mηf) (t) = limα↓0

α

∫ ∞

0

e−(α+iη)sf(t+ s) ds

exists uniformly in t ∈ R+.

As in (4.25) it follows thatMηf = eiη ⊗ x

where eiη(t) = eiηt, x = (Mηf) (0) ∈ X.

The following complex Tauberian theorem is the main result of this section.Its proof is based on the same idea as the one of Theorem 4.9.5, but it is lesselementary since Loomis’s theorem is used.

Theorem 4.9.7. Let f ∈ L∞(R+, X) be slowly oscillating. Assume that spw(f) iscountable and that f is uniformly ergodic on spw(f). Then

f = f0 + f1,

where f1 ∈ AP(R+, X) and limt→∞ f0(t) = 0. In particular, if f ∈ BUC(R+, X)then f ∈ AAP(R+, X).

Proof. a) We show that g ∗ f ∈ AAP(R+, X) for all g ∈ L1(R) (and for thiswe do not need the assumption that f is slowly oscillating). As in the proof ofTheorem 4.9.5, we consider the operator T : L1(R) → BUC(R+, X) given byTg := (g ∗ f)|R+

. Let φ ∈ BUC(R+, X)∗ be a continuous linear functional onBUC(R+, X) which vanishes on AAP(R+, X). We have to show that 〈g ∗ f, φ〉 = 0for all g ∈ L1(R). Then the claim follows from the Hahn-Banach theorem.

Let h := T ∗φ. As in the proof of Theorem 4.9.5 it follows from Lemma4.9.4 that spC(h) ⊂ spw(f). Hence, spC(h) is countable. Let ϕ ∈ L1(R). SincespC(ϕ ∗ h) ⊂ spC(h) (by Lemma 4.8.5), spC(ϕ ∗ h) is also countable. Moreover,ϕ∗h ∈ BUC(R) by Proposition 1.3.2. It follows from Loomis’s theorem (Corollary4.6.4) that ϕ ∗ h ∈ AP(R) for all ϕ ∈ L1(R).

Next, let ϕ ∈ C∞c (0,∞). We show that Mη(ϕ ∗ h) = 0 for all η ∈ R in orderto conclude that ϕ ∗ h = 0 by Corollary 4.5.9.

First, since Freq(ϕ ∗h) ⊂ spC(ϕ ∗ h) ⊂ spw(f), it is clear that Mη(ϕ ∗ h) = 0for all η ∈ R \ spw(f).

Now, let η ∈ spw(f). Let cη := Mη(ϕ ∗ h)(0). ThenMη(ϕ ∗ h)(t) = cηe

iηt (see (4.25)).

4.10. NOTES 329

We have to show that cη = 0.For λ ∈ R, we define eλ(t) := eλt (t ∈ R). Note that

cη = limα↓0

α

∫ ∞

0

e−(α+iη)t (ϕ ∗ h) (t) dt= lim

α↓0α〈χR+

e−(α+iη), ϕ ∗ h〉= lim

α↓0α〈h, χR−eα+iη ∗ ϕ〉

= limα↓0

α〈(f ∗ χR−eα+iη ∗ ϕ)|R+ , φ〉.

However,

kα(t) := α

∫ 0

−∞f(t− s)ei(α+iη)s ds = α

∫ ∞

0

f(t+ s)e−i(α+iη)s ds

converges to the function eiη⊗x as α ↓ 0 uniformly in t ≥ 0. Since ϕ ∈ C∞c (0,∞),this implies that (kα ∗ ϕ)(t) converges to

eiηt∫ ∞

0

e−iηsϕ(s) ds x =: k(t)

uniformly in t ≥ 0 as α ↓ 0. Thus, cη = 〈k, φ〉. Since k ∈ AP(R+, X), it followsthat cη = 0.

We have shown that h ∗ ϕ = 0 for all ϕ ∈ C∞c (0,∞). Choosing a mollifier(ρn)n∈N ⊂ C∞c (−∞, 0), we deduce that for all k ∈ L1(R),

〈k, h〉 = limn→∞〈k ∗ ρn, h〉 = lim

n→∞〈k, ρn ∗ h〉 = 0.

Thus, T ∗φ = h = 0. Here, for v : R → X we let v(t) = v(−t). It follows that〈g ∗ f, φ〉 = 0 for all φ ∈ BUC(R+, X)∗ vanishing on AAP(R+, X). Hence, g ∗ f ∈AAP(R+, X).

b) Since f is slowly oscillating, f = g0 + g1 where g1 ∈ BUC(R+, X) andlimt→∞ g0(t) = 0, by Proposition 4.2.2. Let (ρn)n∈N ⊂ C∞c (0,∞) be a mollifierwith supp ρn ⊂ (0, 1/n). Then ρn ∗ g0 ∈ C0(R+, X), and ρn ∗ f ∈ AAP(R+, X)by a). Hence, ρn ∗ g1 = ρn ∗ f − ρn ∗ g0 ∈ AAP(R+, X). Since g1 ∈ BUC(R+, X),limn→∞ ρn ∗ g1 = g1 in BUC(R+, X). Thus, g1 ∈ AAP(R+, X). Now, g1 = f1 + f2with f1 ∈ AP(R+, X), f2 ∈ C0(R+, X). Thus, f = f1 + (f2 + g0).

4.10 Notes

Sections 4.1 and 4.2The prototype for Abelian theorems was Abel’s classical continuity theorem (Theorem4.1.6) which was proved by N. Abel [Abe26] in 1826. Tauberian theorems form theircounterpart and the first result of this type is Tauber’s classical theorem from 1897


[Tau97] which is mentioned in connection with Hardy’s Theorem 4.2.17. There is anenormous amount of literature on Tauberian theorems. We refer to the monographsby Widder [Wid71] and by Doetsch [Doe50, Volume I] for the classical results. Themonographs of Pitt [Pit58] and van de Lune [Lun86], and Chatterji’s historical account[Cha84], cover some of the subsequent developments, and Korevaar’s book [Kor04] is awide-ranging account which provides much historical information up to recent times.

The notion of B-limit was introduced by Arendt and Pruss [AP92] following ideas ofBatty [Bat90] showing how to pass from theorem of type D to results of type E. Theorem4.2.15 illustrates in a surprising way how this strategy can be used. The presentationgiven here is close to [AP92].

With the help of Wiener’s Tauberian theorem the following clarification of B-convergence was given by Arendt and Batty [AB00].

Theorem 4.10.1. Let u ∈ L∞(R, X), u∞ ∈ X, and suppose that

limt→∞

1

δ

∫ t+δ

t

u(s) ds = u∞

holds for δ = δ1 and δ = δ2, where δ1 and δ2 are rationally independent. Then

limt→∞

(ρ ∗ u)(t) =(∫

R

ρ

)u∞

for all ρ ∈ L1(R). In particular, B-limt→∞ u(t) = u∞.

Karamata’s theorem has important applications to the study of eigenvalue distri-butions (see e.g. [Sim79, Theorem 10.6]). An elegant short proof of Karamata’s theoremin the scalar case has been given by Konig [Kon60]. Further Abelian and Tauberiantheorems for positive vector-valued functions were given by El-Mennaoui [Elm94].

Another direction of Tauberian theorems for Laplace transforms occurs in theframework of limitation theory. A final result in this direction has been obtained byStadtmuller and Trautner [ST81] and by Kratz and Stadtmuller [KS90] in the discretecase.

Section 4.3The results at the beginning of the section are quite standard.

A systematic study of mean convergence of integrated semigroups was carried outby El-Mennaoui [Elm92]. In particular, Theorem 4.3.7, saying that for positive C0-semigroups Cesaro-ergodicity and Abel-ergodicity are equivalent, can be extended tointegrated semigroups. We state a special case explicitly as follows:

Let E be a Banach lattice with order continuous norm. Let A be a resolvent positiveoperator, so A generates a once integrated semigroup S (by Theorem 3.11.7). Then

X = KerA⊕ RanA

if and only if

Px = limt→∞

1

tS(t)x

exists for all x ∈ E. In that case, P is the projection onto KerA along RanA. See [Elm92,Theorem 4.1].

4.10. NOTES 331

Theorem 4.3.7 and Theorem 4.3.8 are explicitly proved in Arendt and Pruss [AP92].More generally, El-Mennaoui [Elm92] studied strong convergence of 1

tkS(t) (as t → ∞),

where S is a k-times integrated semigroup and k ∈ N. This is a very natural problem.In fact, if A is the generator of S, then for x ∈ D(Ak+1), S(·)x ∈ Ck+1((0,∞), X) andu(t) := S(k)(t)x defines a classical solution of the Cauchy problem{

u′(t) = Au(t) (t ≥ 0),

u(0) = x.

Thus, 1tkS(t)x = 1

tk

∫ t

0

(t−s)k−1

(k−1)!u(s) ds is the kth Cesaro mean of the solution u and

describes its asymptotic behaviour for t → ∞.Proposition 4.3.13 is due to Arendt and Batty [AB92a] who also showed that the

result does not hold on every Banach lattice. In fact, an example is given in [AB92a]which shows that a positive C0-semigroup on a space C(K), where K is compact, maywell be Cesaro-ergodic without being totally ergodic. However, in Proposition 4.3.13 theL1-space may be replaced by any Banach lattice with order continuous norm.

The striking properties concerning C0-semigroups on L∞-spaces are due to Lotz[Lot85]. In fact, it had been proved before by Kishimoto and Robinson [KR81] that everygenerator of a positive C0-semigroup on an L∞-space is bounded, and independently ofLotz, Coulhon [Cou84] had proved Corollary 4.3.19 for contraction semigroups. However,it was Lotz who discovered the interesting interplay of the geometric properties (G) and(DP). He proved in particular Lemma 4.3.16 and also proved ergodic theorems in thediscrete case (i.e., for power bounded operators).

Other examples of Banach spaces having both properties (DP) and (G) are thefollowing:

a) C(K), where K is an F -space (cf. the Notes of Section 2.7).

b) H∞(D), the space of all bounded holomorphic functions on the unit disc D := {z ∈C : |z| < 1} with the supremum norm.

We refer to [Nag86, Section A-II.3] and the references given there.

Section 4.4.A version of the complex Tauberian Theorem 4.4.1 was proved by Ingham [Ing35]. Theingenious contour argument proof of a case of Ingham’s theorem was given in Korevaar’sbeautiful article [Kor82] which also gives an elementary proof of the prime number theo-rem based on Ingham’s Tauberian theorem. Korevaar was inspired by Newman [New80]who proved the corresponding result for Dirichlet series; see also the book by Newman[New98]. The quantified version presented in Theorem 4.4.6 is due to Batty and Duy-ckaerts [BD08]. Remark 4.4.7 shows that the estimate (4.15) is quite sharp when M isbounded or grows very slowly, and Borichev and Tomilov [BT10] have shown that it isoptimal for polynomial growth.

Theorem 4.4.8, as stated here, is due to Arendt and Batty [AB88], who also gaveother versions allowing a countable number or even a null set of singularities. Here,merely the case of one singularity is presented which allows several interesting applicationsincluding those given in this section and later in the chapter. A quantified version is givenin [BD08]. Further versions are contained in the work of Arendt and Pruss [AP92]. Cesaroconvergence is investigated in [AB95] with the help of contour arguments similar to thoseof Theorem 4.4.8. For example, it is shown that a function f ∈ L1

loc(R+, X) is Cesaro


convergent if ‖f(t)‖ = O(t) as t → ∞ and every point of iR is either regular for f or apole of order 1.

Corollary 4.4.12 is contained in Arveson’s work on spectral subspaces [Arv82], butit is usually associated with Gelfand’s name. More precisely, it is the corresponding resulton bounded operators which is due to Gelfand (saying that an isometry whose spectrumis {1} is necessarily the identity). Since the weak spectral mapping theorem holds forbounded C0-groups [Nag86, A-III, Theorem 7.4], Corollary 4.4.12 follows immediatelyfrom Gelfand’s classical theorem on isometries. Another elegant proof of Gelfand’s theo-rem is due to Allan and Ransford [AR89], see also the survey article of Zemanek [Zem94].Extensions of Gelfand’s theorem have been obtained by Zarrabi [Zar93]. He proved inparticular that an invertible contraction T on a Banach space X with countable spectrumis already an isometry if it satisfies the growth condition

limn→∞

log ‖T−n‖√n

= 0.

Theorem 4.4.14 is taken from [BD08]. Remarkably the implication (ii) ⇒ (i) doesnot seem to have been in the earlier literature of operator semigroups. The implication(i) ⇒ (ii) was already implicit in [AB88] (see [Bat94, p.41]). Quantified versions wereobtained for polynomial growth of M on arbitrary Banach spaces [BEPS06], and forpolynomial or exponential growth of M [Bur98], [Leb96], [LR05], using methods whichgenerally give slightly slower rates of decay than (4.19). For M(s) = β(1 + s)α, it isshown in [BT10] that the rate in (4.19) is optimal on general Banach spaces, but M−1

log

can be replaced by M−1 in the case of Hilbert spaces. A version of Theorem 4.4.14 whenσ(A) ∩ iR is finite is given in [Mar11].

These results have applications to energy decay of damped wave equations of theform ⎧⎪⎪⎪⎨⎪⎪⎪⎩

∂2u

∂t2= Δu− 2a(x)

∂u

∂t(t > 0, x ∈ Ω),

u(x, t) = 0 (t > 0, x ∈ ∂Ω),

u(·, 0) = u0 ∈ H10 (Ω),

∂u

∂t(·, 0) = u1 ∈ L2(Ω).

(4.58)

Here, Ω is a smooth or convex open subset of Rn with boundary ∂Ω, and a : Ω → [0,∞)is a bounded continuous function representing the damping at each point. Let X be theHilbert space H1

0 (Ω)× L2(Ω), and let A be the operator on X defined by

D(A) :=(H2(Ω) ∩H1

0 (Ω))×H1

0 (Ω),

A :=

(0 IΔ −2a(x)

).

Then A generates a C0-semigroup of contractions on X , and the energy of the solutionu of (4.58) is given by E(u, t) := 1

2‖T (t)(u0, u1)‖2X . The following result was proved by

Lebeau [Leb96] with a slightly slower rate of decay in b), and the optimal rate given herewas established in [Bur98]. Once a) is proved, b) follows directly from Theorem 4.4.14 a)with M(s) = βeαs.

Theorem 4.10.2. Consider the damped wave equation (4.58), where Ω is a bounded openset and a is not identically zero. We assume that Ω is smooth or convex. Let A andE(u, t) be as above. Then the following hold.

4.10. NOTES 333

a) σ(A) ⊂ {λ ∈ C : −2‖a‖∞ ≤ Reλ < 0}, and there exist α, β > 0 such that‖R(is, A)‖ ≤ βeα|s| for all s ∈ R.

b) There is a constant C such that E(u, t) ≤ C‖(u0, u1)‖2H2×H1

(log(2 + t))2for all u0 ∈ H2(Ω)∩

H10 (Ω), u1 ∈ H1

0 (Ω), and t ≥ 0.

When the data is sufficiently smooth that (u0, u1) ∈ D(Ak), the rate of energydecay in b) can readily be improved to O

((log t)−k

). In many cases, the rate is much

faster than logarithmic. When the domain of damping {x ∈ Ω : a(x) > 0} satisfies the“geometric optics condition”, the energy decay is exponential (and uniform with respectto ‖(u0, u1)‖X), i.e., ω(T ) < 0 and ‖R(is, A)‖ is bounded [BLR89]. When Ω is a rectanglein R2 and the domain of damping is a strip across Ω, the geometric optics condition isnot satisfied but ‖R(is,A)‖ = O(s2) as |s| → ∞ [LR05, Example 3]. In this case, theoptimal rate of decay E(u, t) = O(t−1) follows from [BT10, Theorem 2.4].

Laplace transform methods can also be applied to local energy decay of wave equa-tions on exterior domains, a subject which already featured prominently in the 1967monograph by Lax and Phillips [LP67]. In this context, Ω = Rn \K where K is a com-pact obstacle, and a = 0 in (4.58). The quantity of study is now the local energy ofthe solutions, and one restricts attention to initial data supported in a fixed compactset. Thus one is interested in the asymptotic behaviour of ‖P1T (t)R(μ,A)P2‖ where P1

and P2 are multiplications by characteristic functions of balls. The semigroup propertydoes not hold for P1T (t)P2, but Laplace transform results are still applicable. HoweverP1R(λ,A)P2 has a singularity at the origin when n is even, so a quantified version ofTheorem 4.4.8 is used (see [BD08]). The general result, originally obtained in [Bur98], isthat the local energy decays at least logarithmically. For the case when the obstacle is“not trapping”, earlier results had shown that the decay occurs at an exponential rate ifn is odd and at a polynomial rate if n is even (see [MRS77], [MS78]).

Theorem 4.4.16 is due to Arendt and Pruss [AP92]; it is a continuous analogueof the Katznelson-Tzafriri theorem [KT86] (see Corollary 4.7.15). Most of these resultswere originally proved by means of harmonic analysis; the link with Korevaar’s contourmethods for complex Tauberian theorems was first established by Allan, Ransford andO’Farrell [AOR87] whose work inspired [AB88].

Theorem 4.4.18 and Theorem 4.4.19 were discovered by Blake [Bla99] (see also[BB00]) who developed a method used by van Neerven [Nee96b] in the case of orbits ofC0-semigroups (see the Notes on Section 5.1). These results are valid not only when fis exponentially bounded, but even when e−ωtf(t) ∈ Lp(R+, X) for some ω and somep > 1. However, they are not valid in the case p = 1. Bloch [Blo49] gave an examplewhere e−ωtf(t) ∈ L1(R+) whenever ω > 0, abs(f) = 0 and hol0(f) = −∞. His examplecan be adapted to show that the estimate c(1+ t) in the conclusion of Theorem 4.4.18 issharp for the class of exponentially bounded functions f [Bat03].

Section 4.5The material presented here is standard. Almost periodic functions were introduced byHarald Bohr [Boh25] in 1925, and the first edition of his book [Boh47] was published in1934. Further textbooks are those of Levitan and Zhikov [LZ82], Fink [Fin74] and Ame-rio and Prouse [AP71]. Concerning the role of almost periodic functions for dynamicalsystems governed by partial differential equations, we refer to Haraux [Har91].


Some authors use the terminology “scalarly almost periodic” functions instead ofweakly almost periodic functions in order to distinguish from the notion of “weakly almostperiodic in the sense of Eberlein”. We refer to Milnes [Mil80] and to the Notes of Section4.7 for more information and comparison of these different notions.

Section 4.6It was Loomis [Loo60] who proved that bounded uniformly continuous functions withcountable spectrum are almost periodic in the scalar case. The extension to the vector-valued case when c0 �⊂ X was included in the book of Levitan and Zhikov [LZ82] afterKadets had proved his striking result (Theorem 4.6.11) in [Kad69]. For the proof ofKadets’s theorem we follow [LZ82].

The condition of total ergodicity was used by Levitan [Lev66] in the context ofantiderivatives of almost periodic functions. Theorem 4.6.3 in case a) is due to Ruessand Vu [RV95], but with a different proof. The condition of total ergodicity turned outto be crucial in later developments on the half-line (see Sections 4.7, 4.9). The proofof Theorem 4.6.3 which we give here, is due to Arendt and Batty [AB97], but Lemma4.6.6 and Lemma 4.6.7 are taken from Batty, van Neerven and Rabiger [BNR98a]. Severalresults presented here can be extended to measurable functions. In particular, a version ofTheorem 4.6.3 remains true if f ∈ L∞(R, X) is slowly oscillating at infinity and a priorinot continuous. Then the result says that f is equal almost everywhere to an almostperiodic function (see the paper of Arendt and Batty [AB00, Corollary 3.3]).

A harmonic analytic approach to countable spectrum on the line and almost pe-riodicity was taken by Baskakov [Bas78], [Bas85] and Basit [Bas95], [Bas97], using theBeurling spectrum instead of the Carleman spectrum (c.f. Section 4.8).

Section 4.7The main result of this section, Theorem 4.7.7 is due to Batty, van Neerven and Rabiger[BNR98a], [BNR98b] who also proved Lemma 4.7.9. However, their proof is more compli-cated and based on the countable spectrum theorem (Theorem 5.5.4). The direct proofvia the quotient method given here is due to Arendt and Batty [AB99] who proved Propo-sition 4.7.2 in particular. Example 4.7.11 was considered by Staffans [Sta81, p.608], Ruessand Vu [RV95, Example 3.12] and Batty, van Neerven and Rabiger [BNR98b]. The proofof the Katznelson-Tzafriri theorem (Corollary 4.7.15) presented here is similar to the onegiven by Arendt and Pruss [AP92].

Here our emphasis is on (strong) asymptotic almost periodicity, but there are manyinteresting results on weak versions of these notions. A function f ∈ BUC(R+, X) iscalled weakly asymptotically almost periodic in the sense of Eberlein (in short, Eberlein-w.a.a.p.) if the set {S(t)f : t ≥ 0} is relatively weakly compact in BUC(R+, X). Ruessand Summers have investigated this notion in a series of articles [RS86], [RS87], [RS88a],[RS88b], [RS89], [RS90a], [RS90b], [RS92a], [RS92b]; see also papers of Ruess [Rue91],[Rue95] and Rosenblatt, Ruess and Sentilles [RRS91]. They show in a convincing waythat this is the right notion for evolution equations. It is different from weak asymptoticalmost periodicity (w.a.a.p.) in the sense that x∗ ◦ f is asymptotically almost periodicfor all x∗ ∈ X∗. An example of an Eberlein-w.a.a.p. orbit of a bounded C0-semigroup,which is not w.a.a.p. is given in [RS90b, p.180].

The notion of Eberlein-w.a.a.p. functions is particularly useful in the context ofthe mean ergodic theorem for non-linear semigroups; see [RS87], [RS88a], [RS90a] and[RS92a]. An Eberlein-w.a.a.p. function f splits, f = f0 + f1, where f1 is almost periodicand f0 is such that 0 is in the weak closure of {S(t)f0 : t ≥ 0} in BUC(R+, X) (see

4.10. NOTES 335

Theorem 5.4.11). This implies that Mηf0 = 0 for all η ∈ R, but otherwise the asymptoticbehaviour as t → ∞ of the function f0 is very weak. It can still happen that for somesequence tn → ∞, ‖S(tn)f0 − f0‖∞ → 0 (see [RRS91, Section 3]).

Section 4.8In this section we closely follow the book of Pruss [Pru93] where in particular Proposition4.8.4 is proved with the help of our favourite fudge factor. A different proof is given inDavies’s book [Dav80, Chapter 8]. Theorem 4.8.7 on discrete spectrum is contained in apaper of Arendt and Schweiker [AS99] with a slightly different proof. In more abstractcontexts the result appeared already in work of Baskakov [Bas85], Beurling [Beu47] andReiter [Rei52].

Section 4.9In this section we follow closely Chill’s thesis [Chi98a] (see also [Chi98b]). In particular,Theorem 4.9.7 is due to Chill with the proof which we give here. This theorem seemsto be a definitive complex Tauberian theorem involving countable spectrum. It is worthmentioning that uniform ergodicity is automatic outside the weak half-line spectrum.More precisely, Chill [Chi98a, Lemma 1.16] proved the following.

Lemma 4.10.3. Let f ∈ L∞(R+, X) and let η ∈ R \ spw(f). Then f is uniformly ergodicat η.

The more general weak half-line spectrum seems to be more natural in the context ofVolterra equations. A first investigation of asymptotic behaviour of the correspondingsolution operators (which are called resolvents in the theory of Volterra equations) hasbeen carried out by Arendt and Pruss [AP92] (see also [Pru93]). Theorem 4.9.7 can nowbe more directly applied, as shown by Chill and Pruss [CP01] and Fasangova and Pruss[FP01].

Chapter 5

Asymptotics of Solutions ofCauchy Problems

In this chapter, we give various results concerning the long-time asymptotic be-haviour of mild solutions of homogeneous and inhomogeneous Cauchy problems onR+ (see Section 3.1 for the definitions and basic properties). For the most part, weshall assume that the homogeneous problem is well-posed, so that the operator Agenerates a C0-semigroup T , mild solutions of the homogeneous problem (ACP0)are given by u(t) = T (t)x =: ux(t) (Theorem 3.1.12), and mild solutions of theinhomogeneous problem (ACPf ) are given by u(t) = T (t)x + (T ∗ f)(t), whereT ∗ f is the convolution of T and f (Proposition 3.1.16). In typical applications,the operator A and its spectral properties will be known, but solutions u will notbe known explicitly, so the objective is to obtain information about the behaviourof u from the spectral properties of A. To achieve this, we shall apply the results ofearlier chapters, making use of the fact that the Laplace transform of u can easilybe described in terms of the resolvent of A.

In Section 5.1, we obtain general relations between spectral bounds of A,abscissas associated with the Laplace transform of T , growth bounds of T andits associated integrated semigroup, and the behaviour of convolutions T ∗ f forgeneral f .

In Sections 5.2 and 5.3, we give more precise relations between spectral andgrowth bounds in the cases of semigroups on Hilbert spaces and positive semi-groups on Banach lattices.

In Section 5.4, we use the general theory of ergodicity and asymptotic almostperiodicity (Sections 4.3 and 4.7) to obtain splitting theorems (Glicksberg-deLeeuwtheorems) for C0-semigroups with relatively (weakly) compact orbits. In Section5.5, we apply the complex Tauberian theorem (Theorem 4.7.7 or Theorem 4.9.7)to the case when the imaginary part of the spectrum of A is countable.


338 5. ASYMPTOTICS OF SOLUTIONS OF CAUCHY PROBLEMS

In Section 5.6, we consider the asymptotic behaviour of T ∗ f , showing inparticular that T ∗ f is bounded when T is a bounded holomorphic C0-semigroup,f is bounded, and T and f are out of phase in a sense described by their Laplacetransforms.

5.1 Growth Bounds and Spectral Bounds

Let T be a C0-semigroup on X with generator A. Recall from Section 3.1 that Tis exponentially bounded, and the exponential growth bound ω(T ) is defined by:

ω(T ) = inf

{ω ∈ R : there exists Mω such that ‖T (t)‖ ≤Mωe

ωt for all t ≥ 0

}.

By the uniform boundedness principle applied to {e−ωtT (t) : t ≥ 0},

ω(T ) = inf

{ω ∈ R : for each x ∈ X, there exists Mω,x

such that ‖T (t)x‖ ≤Mω,xeωt for all t ≥ 0

}= sup

x∈Xω(ux), (5.1)

where ux(t) := T (t)x. This suggests several possible ways of defining other bounds,for example by replacing ω(ux) by hol(ux) or abs(ux), and/or taking the supremumin (5.1) not over all x ∈ X, but only over x ∈ D(A) (i.e., considering classicalsolutions ux of the homogeneous Cauchy problem rather than mild solutions).Later in this section, we shall consider such bounds, and also bounds associatedwith the spectrum and resolvent of the generator A, but first we establish someproperties of ω(T ).

The first elementary result exploits the semigroup property of T to obtainsome simple properties of the growth bound.

Proposition 5.1.1. Let T be a C0-semigroup on X. Then

a) ω(T ) = limt→∞ t−1 log ‖T (t)‖ = inf

t>0t−1 log ‖T (t)‖.

b) The spectral radius r(T (t)) of T (t) is etω(T ).

c) Let x ∈ X and ω ∈ R, and suppose that∫∞0

e−ωt‖T (t)x‖ dt < ∞. Thenω(ux) ≤ ω.

Proof. a) For ω > ω(T ),

log ‖T (t)‖t

≤ logMω

t+ ω,

solim supt→∞

t−1 log ‖T (t)‖ ≤ ω.

5.1. GROWTH BOUNDS AND SPECTRAL BOUNDS 339

Hence,lim supt→∞

t−1 log ‖T (t)‖ ≤ ω(T ).

For the reverse inequality we may assume that ‖T (t)‖ > 0 for all t ≥ 0. Forτ > 0 and nτ ≤ t < (n+ 1)τ ,

‖T (t)‖ = ‖T (τ)nT (t− nτ)‖ ≤ Cτ‖T (τ)‖n≤ C′τ‖T (τ)‖t/τ = C ′τ exp

((τ−1 log ‖T (τ)‖)t),

where

Cτ = sup0≤s≤τ

‖T (s)‖, C ′τ =

{Cτ if ‖T (τ)‖ ≥ 1,

Cτ

‖T (τ)‖ if ‖T (τ)‖ < 1.

Thus,ω(T ) ≤ τ−1 log ‖T (τ)‖

for all τ > 0. This proves a).b) By the spectral radius formula and a),

r(T (t)) = limn→∞ ‖T (t)

n‖1/n = limn→∞ ‖T (nt)‖

1/n = limn→∞ exp

(t

ntlog ‖T (nt)‖

)= etω(T ).

c) Take τ ≥ 1. For τ − 1 ≤ t ≤ τ ,

‖ux(τ)‖ = ‖T (τ)x‖ = ‖T (τ − t)T (t)x‖ ≤ C1‖T (t)x‖ = C1‖ux(t)‖.

Hence,

e−ωτ‖ux(τ)‖ ≤ C1

∫ τ

τ−1

e−ωτ‖ux(t)‖ dt ≤ C1e|ω|

∫ ∞

0

e−ωt‖ux(t)‖ dt <∞.

This shows that τ �→ e−ωτ‖ux(τ)‖ is bounded on [1,∞) and hence on R+, soω(ux) ≤ ω.

It follows from Proposition 5.1.1 c) that

ω(T ) = inf

{ω ∈ R :

∫ ∞

0

e−ωt‖T (t)x‖ dt <∞ for all x ∈ X

}(5.2)

= sup {abs(‖ux‖) : x ∈ X}= abs(‖T‖).

The following result (which we call Datko’s theorem, although there wereother contributions; see the Notes) shows that the infimum in (5.2) is never at-tained. Indeed, when applied to e−ωtT (t) with p = 1, condition b) shows thatω(T ) < ω if

∫∞0

e−ωt‖T (t)x‖ dt <∞ for all x ∈ X .

Recall from Section 1.3 that T ∗ f denotes the convolution of T with f , so

(T ∗ f)(t) :=∫ t

0

T (t− s)f(s) ds =

∫ t

0

T (s)f(t− s) ds

when f ∈ L1loc(R+, X).

Theorem 5.1.2 (Datko’s Theorem). Let T be a C0-semigroup on X, and let 1 ≤p <∞. The following are equivalent:

(i) ω(T ) < 0.

(ii) For all x ∈ X, ux ∈ Lp(R+, X).

(iii) For all f ∈ Lp(R+, X), T ∗ f ∈ Lp(R+, X).

(iv) For all f ∈ L∞(R+, X), T ∗ f ∈ L∞(R+, X).

(v) For all f ∈ C0(R+, X), T ∗ f ∈ C0(R+, X).

(vi) For all f ∈ C0(R+, X),

supt≥0

∥∥∥∥∫ t

0

T (s)f(s) ds

∥∥∥∥ <∞.

(vii) For all f ∈ AP(R+, X),

supt≥0

∥∥∥∥∫ t

0

T (s)f(s) ds

∥∥∥∥ <∞.

(viii) There is a constant C such that∥∥∥∥∫ t

0

T (s)f(s) ds

∥∥∥∥ ≤ C sup0≤s≤t

‖f(s)‖ (5.3)

for all f ∈ C([0, t], X) and all t ≥ 0.

Proof. First, assume that (i) holds. Then there exist M and α > 0 such that‖T (t)‖ ≤Me−αt for all t ≥ 0. Since ‖ux(t)‖ ≤Me−αt‖x‖, ux ∈ Lp(R+, X), so (ii)holds.

Proposition 1.3.5 shows that (iii), (iv) and (v) hold. The proofs of (vi), (vii)and (viii) all follow from the estimate∥∥∥∥∫ t

0

T (s)f(s) ds

∥∥∥∥ ≤ ∫ t

0

Me−αs sup0≤r≤t

‖f(r)‖ ds ≤ M

αsup

0≤s≤t‖f(s)‖.

(ii) ⇒ (i): By hypothesis, x �→ ux maps X into Lp(R+, X), and it is easy tocheck that this map has closed graph. Hence, there is a constant C such that∫ ∞

0

‖T (t)x‖p dt ≤ C‖x‖p (5.4)

for all x ∈ X .Suppose that ω(T ) ≥ 0. By Proposition 5.1.1 b), there exists λ ∈ σ(T (1))

with |λ| = eω(T ) ≥ 1, and λ is in the topological boundary of σ(T (1)). Now λ is anapproximate eigenvalue of T (1) (Proposition B.2), so there is a sequence (xk) inX such that ‖xk‖ = 1 and limk→∞ ‖T (1)xk−λxk‖ = 0. Then limk→∞ ‖T (1)nxk−λnxk‖ = 0 for n ∈ N. Passing to a subsequence of (xk), we may assume that

‖xk‖ = 1, ‖T (1)nxk − λnxk‖ ≤ 1

2(n = 1, 2, . . . , k).

Hence, ‖T (n)xk‖ ≥ 12 (n = 1, 2, . . . , k). If n− 1 ≤ t ≤ n ≤ k, then

1

2≤ ‖T (n)xk‖ = ‖T (n− t)T (t)xk‖ ≤ C1‖T (t)xk‖,

where C1 = sup0≤s≤1 ‖T (s)‖. Thus, ‖T (t)xk‖ ≥ 1/(2C1) whenever 0 ≤ t ≤ k, so∫ ∞

0

‖T (t)xk‖p dt ≥(

1

2C1

)p

k‖xk‖p (k = 1, 2, . . .).

This contradicts (5.4). It follows that ω(T ) < 0.(iii)⇒ (ii): Choose ω > max(0, ω(T )). Take x ∈ X , and let f(t) := e−ωtT (t)x.

Then f ∈ Lp(R+, X), so (iii) implies that T ∗ f ∈ Lp(R+, X). But

(T ∗ f)(t) =∫ t

0

T (s)(e−ω(t−s)T (t− s)x

)ds =

(1− e−ωt

ω

)ux(t).

Thus, ‖ux(t)‖ ≤(

ω

1− e−ω

)‖(T ∗ f)(t)‖ for t ≥ 1, and ux is bounded on [0, 1], so

ux ∈ Lp(R+, X).(iv) or (v) ⇒ (viii): Define Vt : C0(R+, X)→ X by Vtg := (T ∗ g)(t). Either

(iv) or (v) implies that supt≥0 ‖Vtg‖ <∞ for each g ∈ C0(R+, X). By the uniformboundedness principle, there is a constant C such that∥∥∥∥∫ t

0

T (s)g(t− s) ds

∥∥∥∥ ≤ C‖g‖∞

for all g ∈ C0(R+, X).Given t ≥ 0 and f ∈ C([0, t], X), choose g ∈ C0(R+, X) such that g(s) =

f(t− s) whenever 0 ≤ s ≤ t and ‖g‖∞ = sup0≤s≤t ‖f(s)‖. Then∥∥∥∥∫ t

0

T (s)f(s) ds

∥∥∥∥ =

∥∥∥∥∫ t

0

T (s)g(t− s) ds

∥∥∥∥ ≤ C‖g‖∞ = C sup0≤s≤t

‖f(s)‖.

(vi) or (vii)⇒ (viii): As in the proof of (iv) or (v)⇒ (viii), (vi) (respectively,(vii)) implies that there is a constant C such that∥∥∥∥∫ t

0

T (s)g(s) ds

∥∥∥∥ ≤ C‖g‖∞

for all g ∈ C0(R+, X) (respectively, g ∈ AP(R+, X)). Given t ≥ 0 and f ∈C([0, t], X) there exists an extension g ∈ C0(R+, X) (respectively, a periodic ex-tension g) such that ‖g‖∞ = sup0≤s≤t ‖f(s)‖. It follows that∥∥∥∥∫ t

0

T (s)f(s) ds

∥∥∥∥ ≤ C sup0≤s≤t

‖f(s)‖.

(viii) ⇒ (i): Take ω > max(0, ω(T )), so there exists M such that ‖T (s)‖ ≤Meωs for all s ≥ 0. For t ≥ 0 and x ∈ X, let f(s) := eωsT (t − s)x (0 ≤ s ≤ t).Then (5.3) gives (

eωt − 1

ω

)‖T (t)x‖ ≤ CMeωt‖x‖.

Thus,

‖T (t)x‖ ≤ CMω‖x‖1− e−ω

for all t ≥ 1. It follows that M0 := supt≥0 ‖T (t)‖ <∞. Putting f(s) := T (t− s)xin (5.3) gives

t‖T (t)x‖ ≤ CM0‖x‖for all t ≥ 0 and all x ∈ X. Thus, ‖T (t)‖ ≤ CM0/t < 1 for sufficiently large t, soω(T ) = inft>0 t

−1 log ‖T (t)‖ < 0.

The following corollary of Datko’s theorem can also be proved by a variationof the method of Lemma 3.2.14, without assuming that T is a semigroup.

Corollary 5.1.3. Let T be a C0-semigroup on X. Then there exists x ∈ X suchthat ω(ux) = ω(T ). In particular, if for each x ∈ X there exist Mx and αx > 0such that ‖T (t)x‖ ≤ Mxe

−αxt for all t ≥ 0, then there exist M and α > 0 suchthat ‖T (t)‖ ≤Me−αt for all t ≥ 0.

Proof. The result is trivial if ω(T ) = −∞, so we assume that ω(T ) > −∞. Replac-ing T (t) by e−ω(T )tT (t), we may assume that ω(T ) = 0. Then the result followsimmediately from Theorem 5.1.2, (ii) ⇒ (i).

Recall from Sections 1.4 and 1.5 that

abs(T ) := sup {abs(ux) : x ∈ X}= inf

{ω ∈ R : for all x ∈ X, lim

τ→∞

∫ τ

0

e−ωtT (t)x dt exists

},

T (λ)x := limt→∞

∫ t

0

e−λsT (s)x ds = ux(λ) (λ > abs(T ), x ∈ X),

hol(T ) := inf

{ω ∈ R : T extends to a holomorphic

function from {Reλ > ω} to L(X)

}= sup {hol(ux) : x ∈ X}= inf

{ω ∈ R : for all x ∈ X, ux has a holomorphic

extension to {λ ∈ C : Reλ > ω}}.

We now define

ω1(T ) := sup {ω(ux) : x ∈ D(A)}= inf

{ω ∈ R :

∫ ∞

0

e−ωt‖T (t)x‖ dt <∞ for all x ∈ D(A)

}= inf

{ω ∈ R : for all x ∈ D(A), there exists Mω,x

such that ‖T (t)x‖ ≤Mω,xeωt for all t ≥ 0

}= inf

{ω ∈ R : there exists Mω such that

‖T (t)R(λ,A)‖ ≤Mωeωt for all t ≥ 0

}.

Here, λ is any point in ρ(A). The equalities follow from the definition of ω(ux),Proposition 5.1.1 c) and the uniform boundedness principle.

It is clear from (1.10), (1.14) and the definitions that hol(T ) ≤ abs(T ) ≤ ω(T )and ω1(T ) ≤ ω(T ).

The spectral bound s(A) of the generator A of T is defined by

s(A) := sup {Re λ : λ ∈ σ(A)} ,with the convention that s(A) = −∞ if σ(A) is empty.

The following results make precise the relation between the spectrum andresolvent of A and abscissas associated with the Laplace transform of T . In par-ticular, Proposition 5.1.4 shows that s(A) <∞.

Proposition 5.1.4. Let T be a C0-semigroup on X with generator A. Then

s(A) = hol(T ). (5.5)

Moreover, for x ∈ X,ux(λ) = R(λ,A)x (5.6)

whenever Reλ > s(A),

R(λ,A)x = limτ→∞

∫ τ

0

e−λtT (t)x dt (5.7)

whenever Reλ > abs(T ), and

sup {‖R(λ,A)‖ : Reλ > ω} <∞ (5.8)

whenever ω > ω(T ).

Proof. The functions T (λ) and R(λ,A) are holomorphic for Reλ > hol(T ) and forλ ∈ ρ(A), respectively, and they coincide for λ > ω(T ). Thus T has a holomorphicextension for Reλ > min(s(A), hol(T )) and this implies that hol(T ) ≤ s(A) and(5.6) and (5.7) hold. The equality (5.5) now follows from Theorem 3.1.7.

For ω > ω′ > ω(T ), there exists M such that ‖T (t)‖ ≤ Meω′t for all t. For

Reλ > ω,

‖R(λ,A)x‖ ≤∫ ∞

0

e−Reλt‖T (t)x‖ dt ≤ M‖x‖Reλ− ω′

,

so

sup {‖R(λ,A)‖ : Reλ > ω} ≤ M

ω − ω′<∞.

When Reλ = abs(T ), the existence of the limit in (5.7) for all x ∈ X doesnot correspond exactly to the existence of R(λ,A) (consider, for example, T (t) = Iwith λ ∈ iR, λ �= 0). The following result describes the relation between these twoproperties and stability of classical solutions of the homogeneous problem.

Proposition 5.1.5. Let T be a C0-semigroup on X, and let λ ∈ C. The followingare equivalent:

(i) limt→∞

∫ t

0

e−λsT (s)x ds exists for all x ∈ X.

(ii) λ ∈ ρ(A) and limt→∞ ‖e

−λtT (t)x‖ = 0 for all x ∈ D(A).

In that case,

R(λ,A)x = limt→∞

∫ t

0

e−λsT (s)x ds

for all x ∈ X.

Proof. Replacing T (t) by e−λtT (t), we may assume that λ = 0.

(i)⇒ (ii): Suppose that Bx := limt→∞∫ t

0T (s)x ds exists for all x ∈ X . Then

1

h(T (h)Bx−Bx) = − 1

h

∫ h

0

T (s)x ds→ −x

as h ↓ 0. Thus, Bx ∈ D(A) and ABx = −x for all x ∈ X .Now suppose that x ∈ D(A). Then

BAx = limt→∞

∫ t

0

T (s)Axds = limt→∞T (t)x− x,

by Proposition 3.1.9. Thus, y := limt→∞ T (t)x exists. Since limt→∞∫ t

0T (s)x ds

also exists, it follows that y = 0 and BAx = −x (for all x ∈ D(A)). Thus A hasan algebraic inverse −B. By Proposition B.1, 0 ∈ ρ(A) and B = −A−1 = R(0, A).

(ii) ⇒ (i): Suppose that 0 ∈ ρ(A) and limt→∞ ‖T (t)y‖ = 0 for all y ∈ D(A).Then, for x ∈ X, ∫ t

0

T (s)x ds = −∫ t

0

T (s)AR(0, A)x ds

= R(0, A)x− T (t)R(0, A)x

→ R(0, A)x

as t→∞.

The following result is the analogue of Theorem 1.4.3 for semigroups.

Proposition 5.1.6. Let T be a C0-semigroup on X with generator A, let S be theassociated integrated semigroup:

S(t)x :=

∫ t

0

T (s)x ds (x ∈ X),

and let

S(t) :=

{S(t)−R(0, A) if 0 ∈ ρ(A),

S(t) if 0 ∈ σ(A).

Thenabs(T ) = ω1(T ) = ω(S).

Proof. The fact that abs(T ) = ω(S) follows from Proposition 1.4.5, Remark 1.4.6and Proposition 5.1.5. To prove that abs(T ) = ω1(T ), we may assume that ω(T ) <0 (replacing T (t) by e−ωtT (t)). Then 0 ∈ ρ(A). By Proposition 3.1.9, S(t)x ∈ D(A)and AS(t)x = T (t)x− x for all x ∈ X. Hence,

S(t)x = S(t)x−R(0, A)x = T (t)A−1x.

Thus,abs(T ) = ω(S) = sup {ω(uA−1x) : x ∈ X} = ω1(T ).

It turns out that the spectral bound s(A) is of limited use in the study ofasymptotic behaviour—the spectrum of an operator may be unstable under smallperturbations. However, such instability can only occur when the norm of theresolvent is large, so it is more useful to consider the pseudo-spectral bound definedby

s0(A) := inf

{ω > s(A) : there exists Cω such that

‖R(λ,A)‖ ≤ Cω whenever Reλ > ω

}.


It is clear from (5.8) that

s(A) ≤ s0(A) ≤ ω(T ).

It follows from (5.5) and the uniform boundedness principle that

s0(A) = sup {hol0(ux) : x ∈ X} ,

where hol0(f) is the abscissa of boundedness of f defined in Sections 1.5 and 4.4.Theorem 4.4.19 provides the following result.

Theorem 5.1.7. Let T be a C0-semigroup on X with generator A. Then abs(T ) ≤s0(A).

Proof. By Theorem 4.4.19, abs(ux) ≤ hol0(ux) for each x ∈ X . Hence, abs(T ) =supx abs(ux) ≤ supx hol0(ux) = s0(A).

Proceeding as in the proof of Theorem 4.4.14 a), we can deduce from Theo-rem 4.4.18 the following more precise information about asymptotic behaviour forindividual vectors.

Theorem 5.1.8. Let T be a C0-semigroup on X with generator A, and let S be theassociated integrated semigroup. Let x ∈ X, and suppose that hol(ux) ≤ 0 and ux

is bounded on C+. Then there is a constant c (depending on x) such that

‖S(t)x‖ ≤ c(1 + t)

for all t ≥ 0. Moreover, for each μ ∈ ρ(A), there is a constant cμ (depending on xand μ) such that

‖T (t)R(μ,A)x‖ ≤ cμ(1 + t)

for all t ≥ 0.

Proof. The first statement is immediate from Theorem 4.4.18. By Proposition 3.1.9,

T (t)R(μ,A)x = R(μ,A)x+

∫ t

0

T (s)AR(μ,A)x ds

= R(μ,A)x+ (μR(μ,A)− I)S(t)x,

and the second statement follows.

We now summarise the general relations between spectral bounds, abscissasand growth bounds associated with semigroups, obtained in Proposition 5.1.4,Proposition 5.1.6 and Theorem 5.1.7.

Theorem 5.1.9. Let T be a C0-semigroup on X with generator A. Then

s(A) = hol(T ) ≤ ω1(T ) = abs(T ) ≤ s0(A) ≤ ω(T ).

Now, we give two examples which show that none of the inequalities in The-orem 5.1.9 can be replaced by an equality, and we shall give a further example inSection 5.3. In Theorem 5.1.12 and the next two sections of this chapter, we shallsee that further equalities are valid under various additional assumptions on Xand/or T .

Example 5.1.10. There is a C0-semigroup T on a Hilbert space X such that s(A) <ω1(T ) < s0(A) = ω(T ).

Let X be the Hilbert space

X :=

{x = (xn)n∈N : xn ∈ Cn,

∞∑n=1

‖xn‖2 <∞},

‖x‖ :=

( ∞∑n=1

‖xn‖2)1/2

,

where the norm on Cn is the Euclidean norm. Let Bn := (β(n)i,j )1≤i,j≤n be the

n × n complex matrix with β(n)i,i+1 = 1 for 1 ≤ i < n, β

(n)i,j = 0 otherwise, and let

An := i2nIn +Bn. Let A be the operator on X defined by

D(A) :=

{x ∈ X :

∞∑n=1

22n‖xn‖2 <∞},

Ax := (Anxn)n∈N.

Since ‖Bn‖ = 1 and Bnn = 0,

∥∥etAn∥∥ =

∥∥etBn∥∥ ≤ n−1∑

j=0

tj

j!≤ et.

On the other hand, if xn := n−1/2(1, 1, 1, . . . , 1)T ∈ Cn, then ‖xn‖ = 1 and

∥∥etBnxn − etxn∥∥2 =

1

n

n−1∑m=0

⎛⎝ m∑j=0

tj

j!− et

⎞⎠2

→ 0

as n→∞. Thus, supn∥∥etAn

∥∥ = et.

We may define T (t) : X → X by

T (t)x :=(etAnxn

)n∈N .

Then ‖T (t)‖ = et, and T is a C0-semigroup with generator A and ω(T ) = 1.

For x ∈ D(A),

‖T (t)x‖2 =

∞∑n=1

∥∥etAnxn

∥∥2 ≤∞∑

n=1

⎛⎝n−1∑j=0

tj

j!

⎞⎠2

‖xn‖2

≤∞∑

n=1

⎛⎝n−1∑j=0

tj

2jj!

⎞⎠2

22n‖xn‖2 ≤ et∞∑

n=1

22n‖xn‖2.

Thus, ω1(T ) ≤ 12 .

On the other hand, if 0 < α < 1/2 and xn = αn(1, 1, . . . , 1)T ∈ Cn, thenx = (xn) ∈ D(A) and

‖T (t)x‖2 =

∞∑n=1

n−1∑m=0

⎛⎝ m∑j=0

tj

j!

⎞⎠2

α2n

≥∞∑

n=1

2n−2∑r=0

⎛⎜⎝ ∑j+k=r

0≤j,k≤n−1

r!

j!k!

⎞⎟⎠ tr

r!α2n

≥∞∑

n=1

2n−2∑r=0

2r

r + 1

tr

r!α2n

=

∞∑r=0

∑n≥ r

2+1

α2n 2rtr

(r + 1)!

≥∞∑r=0

αr+3

1− α2

2rtr

(r + 1)!=

α2

1− α2

e2αt − 1

2t.

Thus, ω(ux) ≥ α. It follows that ω1(T ) =12 .

To calculate the spectral bounds, note that σ(An) = {i2n} and

‖R(λ,An)‖ = ‖R(λ− i2n, Bn)‖ ≤ 1

|λ− i2n| − 1(5.9)

if |λ− i2n| > 1. It follows that supn ‖R(λ,An)‖ <∞ whenever λ /∈ {i2n : n ∈ N}.Hence, σ(A) = {i2n : n = 1, 2, . . .}, so s(A) = 0.

It also follows from (5.9) that s0(A) ≤ 1. On the other hand,

‖R(1 + i2n, A)‖ ≥ ‖R(1, Bn)‖ =∥∥∥∥ n−1∑

j=0

Bjn

∥∥∥∥ ≥ n1/2.

Hence, s0(A) = 1.

We shall see in Section 5.2 that the equality s0(A) = ω(T ) holds for allC0-semigroups on Hilbert space.

Example 5.1.11. There is a positive C0-semigroup on a (reflexive) Banach latticeX such that s(A) = ω1(T ) = s0(A) < ω(T ).

Let X := Lp(1,∞) ∩ Lq(1,∞), where 1 ≤ p ≤ q < ∞. Then X is a Banachlattice with the natural ordering and norm:

‖f‖ := max(‖f‖p, ‖f‖q).

Let Tp be the C0-semigroup on Lp(1,∞) defined by

(Tp(t)g) (s) := g(set),

and let T be the positive C0-semigroup on X obtained by restricting Tp to X . LetA and Ap be the generators of T and Tp, respectively.

For f ∈ X ,

‖T (t)f‖ = max

{(∫ ∞

1

|f(set)|p ds)1/p

,

(∫ ∞

1

|f(set)|q ds)1/q

}

= max

{e−t/p

(∫ ∞

et|f(r)|p dr

)1/p

, e−t/q

(∫ ∞

et|f(r)|q dr

)1/q}

≤ max{e−t/p‖f‖p, e−t/q‖f‖q

}≤ e−t/q‖f‖.

On the other hand, if

f(s) :=

{1 (et ≤ s ≤ et + 1),

0 otherwise,

then ‖f‖ = 1 and ‖T (t)f‖ = e−t/q. Thus, ‖T (t)‖ = e−t/q and ω(T ) = −1/q.For Reλ < −1/p, let fλ(s) := sλ. Then fλ ∈ X and T (t)fλ = eλtfλ, so

fλ ∈ D(A) and Afλ = λfλ. Hence,

σ(A) ⊇{λ ∈ C : Reλ ≤ − 1

p

},

and s(A) ≥ −1/p.In the case p = q, we now know that s(Ap) = ω(Tp) = −1/p, and for

Reλ := α > −1/p and f ∈ Lp(1,∞),

(R(λ,Ap)f) (s) =

∫ ∞

0

e−λtf(set) dt = sλ∫ ∞

s

f(r)

rλ+1dr.

For 1 < p < q <∞ and p′ such that 1p + 1

p′ = 1, we have∫ ∞

1

|(R(λ,Ap)f)(s)|q ds =

∫ ∞

1

sαq∣∣∣∣∫ ∞

s

f(r)

rλ+1dr

∣∣∣∣q ds

≤∫ ∞

1

sαq‖f‖qp(∫ ∞

s

dr

r(α+1)p′

)q/p′

ds

=‖f‖qp

((α+ 1)p′ − 1)q/p′

∫ ∞

1

s−q/p ds

=‖f‖qpp

((α+ 1)p′ − 1)q/p′(q − p).

If 1 = p < q <∞, then ∫ ∞

1

|(R(λ,Ap)f)(s)|q ds ≤ ‖f‖q1q − 1

.

In each case, R(λ,Ap) maps Lp(1,∞) into X , so D(Ap) ⊂ X . Since

R(λ,A) = T (λ) = Tp(λ)|X = R(λ,Ap)|Xfor λ > −1/q, it follows that A is the part of Ap in X, σ(A) ⊂ σ(Ap) ⊂ {Reλ ≤−1/p}, and R(λ,A) = R(λ,Ap)|X for Reλ > −1/p (see Proposition B.8). Forf ∈ X and Reλ = α > −1/p,

‖R(λ,A)f‖ ≤ max

{‖R(λ,A)f‖p, ‖f‖pp1/q

((α+ 1)p′ − 1)1/p′(q − p)1/q

}≤ max

{p

αp+ 1,

p1/q

((α+ 1)p′ − 1)1/p′(q − p)1/q

}‖f‖

if 1 < p < q <∞;

‖R(λ,A)f‖ ≤ max

{1

α + 1,

1

(q − 1)1/q

}‖f‖

if 1 = p < q < ∞. Thus, s0(A) = −1/p. It follows from Theorem 5.1.9 thatω1(T ) = −1/p.

We shall see in Section 5.3 that the equality s(A) = s0(A) holds for allpositive semigroups on Banach lattices, while the equality s(A) = ω(T ) holds forall positive semigroups on Lp-spaces. We conclude this section by showing thats(A) = ω(T ) for all holomorphic semigroups.

Theorem 5.1.12. Let T be a holomorphic C0-semigroup on X with generator A.Then ω(T ) = s(A). Moreover, there exists λ ∈ σ(A) such that Reλ = ω(T ).

5.2. SEMIGROUPS ON HILBERT SPACES 351

Proof. For each x ∈ X, ux has a holomorphic extension to a sector Σθ, given byux(z) = T (z)x. By Theorem 2.6.2,

ω(ux) = inf

{ω ∈ R : λ �→ R(λ,A)x has a holomorphic

extension to {λ : Reλ > ω}}

≤ s(A).

Hence,ω(T ) = sup

x∈Xω(ux) ≤ s(A).

The final statement follows from the fact that {λ ∈ σ(A) : Reλ ≥ s(A) − 1} isnonempty and compact (see Theorem 3.7.11 and Corollary 3.7.17).

5.2 Semigroups on Hilbert Spaces

Example 5.1.10 shows that there are C0-semigroups on Hilbert spaces such thats(A) < ω1(T ) < s0(A). In that example, s0(A) = ω(T ), and we now show thatthis equality always holds on Hilbert spaces.

Theorem 5.2.1. Let T be a C0-semigroup on a Hilbert space X with generator A.Then s0(A) = ω(T ).

Proof. Let x ∈ X . For ω > ω(T ), the function s �→ R(ω + is, A)x on R is theFourier transform of the function t �→ e−ωtT (t)x on R+. By Plancherel’s Theorem1.8.2, ∫ ∞

0

e−2ωt‖T (t)x‖2 dt = 1

2π

∫ ∞

−∞‖R(ω + is, A)x‖2 ds. (5.10)

Suppose that s0(A) < ω(T ) and let C := supReλ>ω(T ) ‖R(λ,A)‖ < ∞. Forω(T ) < ω1 < ω2,

R(ω1 + is, A)x = R(ω2 + is, A)x+ (ω2 − ω1)R(ω1 + is, A)R(ω2 + is, A)x,

so‖R(ω1 + is, A)x‖ ≤ (1 + C(ω2 − ω1)) ‖R(ω2 + is, A)x‖.

By (5.10),∫ ∞

0

e−2ω1t‖T (t)x‖2 dt ≤ (1 + C(ω2 − ω1))2∫ ∞

0

e−2ω2t‖T (t)x‖2 dt.

Letting ω1 ↓ ω(T ) gives∫ ∞

0

e−2ω(T )t‖T (t)x‖2 dt ≤ (1 + C(ω2 − ω(T )))2∫ ∞

0

e−2ω2t‖T (t)x‖2 dt <∞

for all x ∈ X. The implication (ii) ⇒ (i) of Theorem 5.1.2, with p = 2 and T (t)replaced by e−ω(T )tT (t), gives a contradiction.

Theorem 5.2.1 is not valid for X = Lp(0, 1) (1 < p <∞, p �= 2).

Example 5.2.2. Let 2 < q <∞. Example 5.1.11 shows that there is a C0-semigroupTq on L2(1,∞) ∩ Lq(1,∞) whose generator Aq satisfies s0(Aq) = − 1

2< −1

q=

ω(Tq). There is a linear homeomorphism Jq of Lq(0, 1) onto L2(1,∞) ∩ Lq(1,∞)[LT77, Corollary II.2.e.8]. Let Sq(t) := J−1

q Tq(t)Jq. Then Sq is a C0-semigroup on

Lq(0, 1) whose generator Bq satisfies s0(Bq) = − 12 < − 1

q = ω(Sq).

Let 1 < p < 2, and let q be the conjugate index, so that 1p + 1

q = 1 and

Lp(0, 1) = Lq(0, 1)∗. Let Sp(t) := Sq(t)∗. By Corollary 3.3.9, Sp is a C0-semigroup

on Lp(0, 1), whose generator Bp = B∗q satisfies s0(Bp) = −12< −1

q= ω(Sp).

We shall see in the next section that ω(T ) = s0(A) = s(A) for all positivesemigroups on Lp-spaces.

The analogue of Theorem 5.2.1 for individual orbits is not true.

Example 5.2.3. There is a C0-semigroup T on a Hilbert space X with a vectorx ∈ X and a real number a < ω(ux) such that ux = R(·, A)x has a boundedholomorphic extension to the half-plane {λ ∈ C : Reλ > a}.

Let X := L2(1,∞) and (T (t)f)(s) := f(set). By Example 5.1.11, s(A) =ω(T ) = ω(A) = −1/2. Let A1 be the generator of the C0-semigroup on L1(1,∞)∩L2(1,∞) obtained by restricting T , so s0(A1) = −1, again by Example 5.1.11.For f ∈ L1(1,∞) ∩ L2(1,∞), R(·, A1)f has an extension to a bounded holomor-phic map of {λ ∈ C : Reλ > a} into L1(1,∞) ∩ L2(1,∞) whenever a > −1.But R(λ,A)f = R(λ,A1)f when Reλ is large, so R(·, A)f has an extension to abounded holomorphic map of {λ ∈ C : Reλ > a} into L2(1,∞). However, it ispossible to choose f ∈ L1(1,∞)∩L2(1,∞) such that ω(uf ) = −1/2 (where ω(uf )is calculated in L2(1,∞)). For example, let

f(s) =

{1 (en ≤ s ≤ en + n−2; n ∈ N),

0 otherwise.

Then (T (n)f)(s) = 1 for 1 ≤ s ≤ 1 + n−2e−n, so ‖T (n)f‖2 > n−1e−n/2. Hence,ω(uf ) ≥ −1/2. On the other hand, ω(uf ) ≤ ω(T ) = −1/2.

5.3 Positive Semigroups

Let T be a C0-semigroup on an ordered Banach space X , with generator A. Werecall from Section 3.11 that T is positive if and only if A is resolvent positive.

Example 5.1.11 shows that there are positive C0-semigroups T on spaces ofthe form Lp(Ω, μ)∩Lq(Ω, μ) (1 ≤ p < q <∞) such that s(A) < ω(T ). On the otherhand, we shall show in this section that s(A) = abs(T ) = s0(A) for all positive

5.3. POSITIVE SEMIGROUPS 353

semigroups on any ordered Banach space with normal cone, and that s(A) = ω(T )for all positive semigroups on Lp(Ω, μ). Note that Proposition 3.11.2 shows thats(A) ∈ σ(A) if A generates a positive semigroup and σ(A) is non-empty. Thefollowing result makes this more precise. We give a proof that abs(T ) ∈ σ(A) basedon Theorem 1.5.3, whereas Proposition 3.11.2 was proved by means of Bernstein’sTheorem 2.7.7.

Theorem 5.3.1. Let X be an ordered Banach space with normal cone and let T bea positive C0-semigroup on X with generator A. Then

s(A) = ω1(T ) = abs(T ) = s0(A).

Moreover, s(A) ∈ σ(A) if s(A) > −∞.

Proof. By Theorem 5.1.9, s(A) ≤ ω1(T ) = abs(T ) ≤ s0(A). It suffices to provethat abs(T ) ∈ σ(A) if abs(T ) > −∞, and that supReλ>ω ‖R(λ,A)‖ <∞ wheneverω > abs(T ), so abs(T ) = s0(A).

Suppose that abs(T ) > −∞ and abs(T ) ∈ ρ(A). Let ε > 0 such that the ballB(abs(T ), ε) ⊂ ρ(A). For x ∈ X+, the function ux is positive with Laplace trans-form R(λ,A)x (Reλ > ω1(T )), which has a holomorphic extension to B(abs(T ), ε).By Theorem 1.5.3, abs(ux) ≤ abs(T ) − ε. By linearity, abs(ux) ≤ abs(T ) − εfor all x ∈ X. But this contradicts the definition of abs(T ). This proves thatabs(T ) ∈ σ(A) if abs(T ) > −∞, and hence s(A) = abs(T ). By Corollary 3.11.3,supReλ>ω ‖R(λ,A)‖ <∞.

Example 5.3.2. There exist a Hilbert space X which is a vector lattice with con-tinuous lattice operations, and a positive C0-semigroup T on X such that s(A) =ω1(T ) < s0(A) = ω(T ).

Let X be the Sobolev space

H1(1,∞) := {f ∈ L2(1,∞) : f ′ ∈ L2(1,∞)},

(see Appendix E) with

‖f‖H1(1,∞) :=(‖f‖22 + ‖f ′‖22)1/2 .

Then X is a Hilbert space, and it is a vector lattice with the properties that∥∥ |f |∥∥ = ‖f‖ [DL90, Chapter IV, Section 7, Proposition 6] and lattice operationsare continuous (see [BY84, p.219]), but it is not a Banach lattice and the positivecone is not normal.

Let T be the C0-semigroup on H1(1,∞) given by (T (t)f)(s) := f(set). Thegenerator A of T is given by

D(A) ={f ∈ H1(1,∞) : s �→ sf ′(s) ∈ H1(1,∞)

},

(Af)(s) = sf ′(s).

The semigroup governs the following very natural partial differential equation:

∂u

∂t= s

∂u

∂s, (t > 0, s > 1),

u(0, s) = u0(s), (s > 1),

where u(t, s) := (T (t)u0)(s).For α < −1/2, the function fα(s) := sα lies in X and T (t)fα = eαtfα. Hence,

s(A) ≥ −1/2. For f ∈ H1(1,∞),

‖T (t)f‖2H1(1,∞) =

∫ ∞

1

∣∣f(set)∣∣2 ds+

∫ ∞

1

e2t∣∣f ′(set)∣∣2 ds

= e−t

∫ ∞

et|f(r)|2 dr + et

∫ ∞

et|f ′(r)|2 dr

≤ et‖f‖2H1(1,∞).

Thus, ω(T ) ≤ 1/2. Choose non-zero g ∈ C∞c (R) with support in R+. Givent ≥ 0, let f(s) := g(s − et). Then ‖f‖2H1(1,∞) =

∫∞0

(|g(s)|2 + |g′(s)|2) ds and

‖T (t)f‖2H1(1,∞) ≥ et∫∞1|g′(r)|2 dr. It follows that ω(T ) ≥ 1/2, so s0(A) = 1/2 by

Theorem 5.2.1. We shall show that abs(T ) ≤ −1/2. It then follows from Theorem5.1.9 that s(A) = ω1(T ) = −1/2.

Let S be the corresponding C0-semigroup on L2(1,∞) with generator B,so abs(S) = ω1(S) = ω(S) = −1/2 (Example 5.1.11). Let ω > −1/2 and f ∈H1(1,∞). For t > 0, let

gt :=

∫ t

0

e−ωrT (r)f dr ∈ H1(1,∞),

g := R(ω,B)f ∈ L2(1,∞).

Then limt→∞ ||gt−g‖2 = 0, since ω > abs(S). We have to show that limt→∞ ||gt−g‖H1(1,∞) = 0.

By Proposition 3.1.9, gt ∈ D(A− ω) = D(A) ⊂ D(B), and

(A− ω)gt = (B − ω)gt = e−ωtS(t)f − f.

But (Agt)(s) = sg′t(s), so

g′t = h · (ωgt − f + e−ωtS(t)f),

where h(s) := s−1. Since |h(s)| ≤ 1 for all s ∈ (1,∞), it is clear that

limt→∞ ‖g

′t − h(ωg − f)‖2 = 0,

so limt→∞ gt exists in H1(1,∞). Thus, abs(uf ) ≤ −1/2 for all f ∈ H1(1,∞), soabs(T ) ≤ −1/2.

Now we consider positive semigroups on Lp(Ω, μ), where we aim to show thats(A) = ω(T ). For p = 2, this result is immediate from Theorems 5.3.1 and 5.2.1.There is also an easy proof for p = 1, which we present in Proposition 5.3.7. Thegeneral case needs some preliminaries. We work with the product space R × Ω,and we use a vector-valued norm on Lp(R× Ω). We begin by defining this norm,and establishing its properties.

Let (Ω, μ) be a σ-finite measure space, and consider R×Ω to be equipped withthe product of Lebesgue measure m on R and the given measure μ on Ω. Let 1 ≤p <∞. We write Lp(Ω) for Lp(Ω, μ) and we identify Lp(R×Ω) with Lp(R, Lp(Ω)),so that the notations g(t, y) and g(t)(y) (t ∈ R, y ∈ Ω) are interchangeable. Weconsider the non-linear map Φ : Lp(R, Lp(Ω))→ Lp(Ω) given by

Φ(g) :=

(∫ ∞

−∞|g(t)|p dt

)1/p

.

This is the composition of three maps:

Φ1 : Lp(R× Ω)→ L1(R× Ω), Φ1(g) := |g|p,

Φ2 : L1(R× Ω)→ L1(Ω), Φ2(h)(y) :=

∫ ∞

−∞h(t, y) dt,

Φ3 : L1(Ω)→ Lp(Ω), Φ3(k) := |k|1/p.This makes it clear that Φ is well defined.

Lemma 5.3.3. Let g, h ∈ Lp(R, Lp(Ω)), f ∈ L∞(Ω), s ∈ R. Then

a) ‖Φ(g)‖Lp(Ω) = ‖g‖Lp(R×Ω).

b) Φ(gs) = Φ(g), where gs(t) = g(t+ s).

c) Φ(f · g) = |f |Φ(g), where (f · g)(t, y) = f(y)g(t, y).

d) Φ(g + h) ≤ Φ(g) + Φ(h) in Lp(Ω).

e) Φ is continuous.

Proof. a), b) and c) are all trivial. To prove d), let Gy(t) := g(t, y), Hy(t) :=h(t, y) (t ∈ R, y ∈ Ω). For μ-almost all y, Gy ∈ Lp(R) and Hy ∈ Lp(R), soMinkowski’s inequality gives

‖Gy +Hy‖Lp(R) ≤ ‖Gy‖Lp(R) + ‖Hy‖Lp(R). (5.11)

But

‖Gy‖Lp(R) =

(∫ ∞

−∞|g(t, y)|p dt

)1/p

= Φ(g)(y),

etc., so (5.11) givesΦ(g + h)(y) ≤ Φ(g)(y) + Φ(h)(y).


This holds μ-a.e., so Φ(g + h) ≤ Φ(g) + Φ(h) in Lp(Ω).Now, e) follows from a) and d). By d),

Φ(g) ≤ Φ(g − h) + Φ(h),

soΦ(g)− Φ(h) ≤ Φ(g − h).

Similarly,Φ(h)− Φ(g) ≤ Φ(h− g) = Φ(g − h).

Since Φ(g) etc. are real-valued, this shows that

|Φ(g)− Φ(h)| ≤ Φ(g − h).

Hence,

‖Φ(g)− Φ(h)‖Lp(Ω) ≤ ‖Φ(g − h)‖Lp(Ω) = ‖g − h‖Lp(R,Lp(Ω)).

Lemma 5.3.3 shows that Φ is a convex, vector-valued function. The nextlemma is a vector-valued instance of Jensen’s inequality.

Lemma 5.3.4. Let G : [a, b]→ Lp(R, Lp(Ω)) be continuous. Then

Φ

(∫ b

a

G(t) dt

)≤∫ b

a

Φ(G(t)) dt in Lp(Ω).

Proof. By Lemma 5.3.3, c) and d),

Φ

(b− a

2n

2n−1∑r=0

G

(rb+ (2n − r)a

2n

))≤ b− a

2n

2n−1∑r=0

Φ

(G

(rb+ (2n − 1)a

2n

)).

Letting n→∞ and using the continuity of Φ (Lemma 5.3.3 e)) gives the result.

For g ∈ Lp(R, Lp(Ω)) and a bounded operator T on Lp(Ω), we may defineT ◦ g ∈ Lp(R, Lp(Ω)) by

(T ◦ g)(t) = T (g(t)).

Proposition 5.3.5. Let T be a positive bounded linear operator on Lp(Ω), and g ∈Lp(R, Lp(Ω)). Then

Φ(T ◦ g) ≤ T (Φ(g)). (5.12)

Proof. Both sides of (5.12) depend continuously on g, so it suffices to assume thatg is a simple function

g(t)(y) =n∑

k=1

χAk(t)gk(y),

where A1, . . . , An are disjoint Borel subsets of R, and g1, . . . , gn ∈ Lp(Ω). Lethk := m(Ak)

1/pgk (k = 1, . . . , n). Then

Φ(T ◦ g) =(

n∑k=1

m(Ak)|Tgk|p)1/p

=

(n∑

k=1

|Thk|p)1/p

,

T (Φ(g)) = T

(n∑

k=1

m(Ak)|gk|p)1/p

= T

(n∑

k=1

|hk|p)1/p

.

Take αk ∈ Q + iQ with∑n

k=1 |αk|p′ ≤ 1, where p′ is the conjugate index of p(maxk |αk| ≤ 1 if p = 1). By Holder’s inequality,

Re

(n∑

k=1

αkhk

)≤(

n∑k=1

|hk|p)1/p

= Φ(g).

Applying T gives

Re

(n∑

k=1

αkThk

)= T

(Re

n∑k=1

αkhk

)≤ T (Φ(g)).

Now (n∑

k=1

|(Thk)(y)|p)1/p

= sup

{Re

(n∑

k=1

αk(Thk)(y)

): αk ∈ Q+ iQ,

n∑k=1

|αk|p′ ≤ 1

}≤ T (Φ(g))(y) μ-a.e.

Thus, Φ(T ◦ g) ≤ T (Φ(g)).

Theorem 5.3.6. Let (Ω, μ) be a σ-finite measure space, 1 ≤ p < ∞, and T be apositive C0-semigroup on Lp(Ω), with generator A. Then s(A) = ω(T ).

Proof. First, assume that s(A) < 0. Fix α > max(0, ω(T )). Take f ∈ Lp(Ω) anddefine g ∈ Lp(R, Lp(Ω)) by

g(t) :=

{e−αtT (t)f (t ≥ 0),

0 (t < 0).

Define G : R+ → Lp(R, Lp(Ω)) by

G(s) := T (s) ◦ g−s,

where g−s(t) := g(t− s). Thus,

G(s)(t) =

{e−α(t−s)T (t)f (0 ≤ s ≤ t),

0 (−∞ < t < s).

Then

Φ

(∫ m

0

G(s) ds

)=

(∫ ∞

0

∣∣∣∣∣∫ min(m,t)

0

e−α(t−s)T (t)f ds

∣∣∣∣∣p

dt

)1/p

=1

α

{∫ ∞

0

(e−αmax(0,t−m) − e−αt

)p

|T (t)f |p dt}1/p

.

Thus,

0 ≤ 1

α

{∫ ∞

0

(e−αmax(0,t−m) − e−αt

)p

|T (t)f |p dt}1/p

= Φ

(∫ m

0

G(s) ds

)≤

∫ m

0

Φ(G(s)) ds

=

∫ m

0

Φ (T (s) ◦ g−s) ds

≤∫ m

0

T (s) (Φ(g−s)) ds

=

∫ m

0

T (s)(Φ(g)) ds,

where we have used Lemma 5.3.4, Proposition 5.3.5 and Lemma 5.3.3 b) in thethird, fifth and sixth lines respectively. Since abs(T ) = s(A) < 0 (Theorem 5.3.1),∫ m

0

T (s)(Φ(g)) ds→ R(0, A)(Φ(g)) in Lp(Ω),

as m→∞ (Proposition 5.1.4). By the monotone convergence theorem,

0 ≤ 1

α

{∫ ∞

0

(1− e−αt

)p |T (t)f |p dt}1/p

≤ R(0, A)(Φ(g)).

Hence, (1− e−α

α

)(∫ ∞

1

|T (t)f |p dt)1/p

≤ R(0, A)(Φ(g)).

Taking norms in Lp(Ω) gives∫Ω

∫ ∞

1

|(T (t)f)(y)|p dt dμ(y) ≤(

α

1− e−α

)p

‖R(0, A)‖p‖Φ(g)‖pLp(Ω),

so ∫ ∞

1

‖T (t)f‖pLp(Ω) dt <∞.

It follows from Theorem 5.1.2 (ii) ⇒ (i), that ω(T ) < 0.In the general case, we may apply the case above to the semigroup e−ωtT (t)

for ω > s(A), and we deduce that ω(T ) < ω whenever ω > s(A). Thus ω(T ) ≤s(A).

For p = 1, there is a much simpler proof of Theorem 5.3.6. A Banach latticeX is said to be an L-space if

‖x+ y‖ = ‖x‖+ ‖y‖ for all x, y ∈ X+. (5.13)

If (Ω, μ) is any measure space, then L1(Ω, μ) is an L-space. If Ω is a locallycompact, Hausdorff space, then C0(Ω)

∗ (which can be identified with the space ofall regular Borel measures on Ω) is an L-space. On the other hand, any L-space isisomorphic as a Banach lattice to a space of the form L1(Ω, μ) [Sch74, TheoremII.8.5].

Proposition 5.3.7. Let T be a positive C0-semigroup on an L-space X, with gen-erator A. Then s(A) = ω(T ).

Proof. By (5.13), there exists x∗ ∈ X∗+ such that 〈x, x∗〉 = ‖x‖ for all x ∈ X+. Let

ω > abs(T ) = s(A) (Theorem 5.3.1). For x ∈ X+ and τ ≥ 0,∫ τ

0

e−ωt‖T (t)x‖ dt =⟨∫ τ

0

e−ωtT (t)x dt, x∗⟩≤ 〈R(ω,A)x, x∗〉.

Hence, ∫ ∞

0

e−ωt‖T (t)x‖ dt <∞

for all x ∈ X+, and so for all x ∈ X. It follows from Datko’s theorem (Theorem5.1.2) that ω(T ) < ω whenever ω > s(A), and therefore ω(T ) = s(A).

Theorem 5.3.6 is also true in the case p = ∞ (even without the assumptionthat the semigroup is positive), but for the trivial reason that the semigroup isnorm-continuous (Corollary 4.3.19). A more interesting case is that of spaces ofthe form C0(Ω), and this can be deduced from Proposition 5.3.7 by duality.

Theorem 5.3.8. Let Ω be a locally compact, Hausdorff space, and T be a positiveC0-semigroup on C0(Ω), with generator A. Then s(A) = ω(T ).

Proof. Let X := C0(Ω)∗, which is an L-space and therefore has order-continuous

norm. Let Y := D(A∗). By Theorem 3.11.8, Y is a closed ideal in X , so Y is alsoan L-space. Let T (t) := T (t)∗|Y . By Proposition 3.3.14, T is a C0-semigroupon Y whose generator A is the part of A∗ in Y . Moreover, σ(A ) = σ(A) (see

Propositions B.8 and B.11). Let ω > s(A) = s(A ). By Proposition 5.3.7, thereexists M such that ‖T (t)‖ ≤Meωt for all t ≥ 0. For x ∈ X and y∗ ∈ Y ,

|〈T (t)x, y∗〉| ≤Meωt‖x‖ ‖y∗‖.For x∗ ∈ X∗ and λ > ω(T ), R(λ,A)∗x∗ ∈ D(A∗) ⊂ Y . Furthermore, x =limλ→∞ λR(λ,A)x (Proposition 3.1.9), and c := lim supλ→∞ λ‖R(λ,A)‖ < ∞since T is exponentially bounded. Hence,

|〈T (t)x, x∗〉| = limλ→∞

|〈T (t)x, λR(λ,A)∗x∗〉|≤ lim sup

λ→∞Meωt‖x‖λ‖R(λ,A)∗x∗‖

≤ cMeωt‖x‖ ‖x∗‖.It follows that ‖T (t)‖ ≤ cMeωt, so ω(T ) < ω whenever ω > s(A).

5.4 Splitting Theorems

Let T be a bounded C0-semigroup on X, and let x ∈ X . In this section, we shallapply the theory of asymptotic behaviour of functions on R+, as developed inChapter 4, to the special case of the orbit ux, where

ux(t) := T (t)x.

We shall see that ergodicity and (asymptotic) almost periodicity of ux correspondto natural semigroup properties of x, and also to compactness properties of theorbit. In particular, the main results will be two splitting theorems. We saw alreadyin Proposition 4.3.12 that a vector x is totally ergodic with respect to T if the orbit{T (t)x : t ≥ 0} is relatively weakly compact. The weak splitting theorem showsthat x can be uniquely decomposed as x = x0 + x1, where x0 is totally ergodicwith all means Mηx0 = 0, and x1 is in the closed linear span of the unimodulareigenvectors of T . The strong splitting theorem states that if the orbit of x isrelatively compact in the norm topology, then limt→∞ ‖T (t)x0‖ = 0.

As in Section 4.3, we let Xe denote the space of all vectors x ∈ X which aretotally ergodic with respect to T . Thus, x ∈ Xe if and only if

Mηx := limt→∞

1

t

∫ t

0

e−iηsT (s)x ds

exists (in the norm topology of X), for each η ∈ R. We also let

Xe0 := {x ∈ Xe : Mηx = 0 for all η ∈ R} ,X0 := {x ∈ X : ‖T (t)x‖ → 0 as t→∞} .

Since T is bounded, all these are closed T -invariant subspaces of X, and

X0 ⊂ Xe0 ⊂ Xe.

5.4. SPLITTING THEOREMS 361

In Example 5.4.3, we shall exhibit these subspaces in a very fundamental exampleof multiplier semigroups, but we first give two very simple general results.

Proposition 5.4.1. Let T be a bounded C0-semigroup on X. Then ux∈BUC(R+, X)for each x ∈ X. Moreover, the map x �→ ux is bounded and linear from X intoBUC(R+, X).

Proof. Uniform continuity of ux follows from the strong continuity of T and theestimate

‖ux(t+ h)− ux(t)‖ = ‖T (t)(T (h)x− x)‖ ≤M‖T (h)x− x‖,

where M = sups≥0 ‖T (s)‖. The other properties are immediate.

Recall from Section 4.7 that a function f ∈ BUC(R+, X) is totally ergodic ifit is totally ergodic with respect to the shift semigroup on BUC(R+, X); i.e., foreach η ∈ R,

(Mηf)(t) := limτ→∞

1

τ

∫ τ

0

e−iηsf(t+ s) ds

exists inX , uniformly for t ≥ 0. The space E(R+, X) of all totally ergodic functionsis a closed subspace of BUC(R+, X).

Proposition 5.4.2. Let T be a bounded C0-semigroup on X, and let x ∈ X. Thenux ∈ E(R+, X) if and only if x ∈ Xe. In that case,

Mη(ux)(t) = T (t)Mηx = eiηtMηx

for all t ≥ 0.

Proof. Observe that

1

τ

∫ τ

0

e−iηsux(t+ s) ds = T (t)

(1

τ

∫ τ

0

e−iηsT (s)x ds

).

It follows easily that ux ∈ E(R+, X) if and only if x ∈ Xe, and that thenMη(ux)(t) = T (t)Mηx. By (4.38), Mη(ux)(t) = eiηtMη(ux)(0) = eiηtMηx.

Let T be a bounded C0-semigroup on X with generator A. A vector x ∈ Xis a unimodular eigenvector of T if there exists η ∈ R such that T (t)x = eiηtx forall t ≥ 0 (equivalently, x ∈ D(A) and Ax = iηx). Let

Xap = span {unimodular eigenvectors of T}

be the space of almost periodic vectors of T . Then Xap is a closed T -invariantsubspace of X. Proposition 5.4.2 shows that Mηx ∈ Xap whenever x ∈ Xe andη ∈ R. When T is the restriction of a C0-group of isometries on X , this definitionof Xap is consistent with the definition given in Section 4.5.

Example 5.4.3. Let μ be a Borel measure on C− := {λ ∈ C : Reλ ≤ 0}, and letX := Lp(μ) := Lp(C−, μ) for some 1 ≤ p <∞. Let T be the multiplier semigroupgiven by

(T (t)f)(λ) := eλtf(λ) (f ∈ X,λ ∈ C−, t ≥ 0).

The generator A is given by

D(A) =

{f ∈ X :

∫C−|λf(λ)|p dμ(λ) <∞

},

(Af)(λ) = λf(λ).

Furthermore,

σ(A) = suppμ,

σp(A) = σp(A∗) = {λ : λ atom of μ}.

The measure μ can be decomposed as μ = μ−+ ν = μ−+ νa + νn, where μ−is the restriction of μ to C− := {λ ∈ C : Reλ < 0}, ν is the restriction of μ to iR,and νa and νn are the atomic and non-atomic parts of ν, respectively. Since μ−,νa and νn are carried by disjoint sets, X splits in a natural way as

X = Lp(μ−)⊕ Lp(ν) = Lp(μ−)⊕ Lp(νa)⊕ Lp(νn).

It is easy to verify that T is totally ergodic (this is automatic for 1 < p <∞, sinceX is then reflexive), and

Xe = X,

Xe0 = Lp(μ−)⊕ Lp(νn),

X0 = Lp(μ−),Xap = Lp(νa).

For f ∈ X, uf is asymptotically almost periodic if and only if f ∈ Lp(μ−)⊕Lp(νa).Moreover, there is a bounded C0-group U on Lp(ν) such that U(t)f = T (t)f forall f ∈ Lp(ν) and all t ≥ 0.

Although multiplier semigroups are very special in some ways, we shall seein the results which follow that some of the features of Example 5.4.3 hold verygenerally.

Proposition 5.4.4. Let T be a bounded C0-semigroup on X. Then

a) Xap ⊂ Xe.

b) Xap = {x ∈ X : ux ∈ AP(R+, X)}.c) Xe0 ∩Xap = {0}.


d) There is a bounded C0-group U on Xap such that T (t)x = U(t)x for allx ∈ Xap and all t ≥ 0.

Proof. Suppose that y is a unimodular eigenvector, with uy(t) = T (t)y = eiηty forall t ≥ 0. Then uy ∈ AP(R+, X) and y = T (t)(e−iηty) ∈ T (t)(Xap). By linearityand continuity (Proposition 5.4.1), ux ∈ AP(R+, X) for all x ∈ Xap, and T (t)(Xap)is a dense subspace of Xap.

Let x ∈ Xap. Then ux ∈ E(R+, X), so x ∈ Xe by Proposition 5.4.2. Let t ≥ 0.By (4.23), there exist sn ∈ R+ such that

limn→∞ ‖x− T (sn + t)x‖ = lim

n→∞ ‖ux(0)− ux(sn + t)‖ = 0.

Hence,‖x‖ = lim

n→∞ ‖T (sn)T (t)x‖ ≤M‖T (t)x‖,where M := sups≥0 ‖T (s)‖. Thus,

M−1‖x‖ ≤ ‖T (t)x‖ ≤M‖x‖, (5.14)

for all x ∈ Xap and all t ≥ 0. This implies that T (t)(Xap) is closed, and we sawabove that it is a dense subspace of Xap. It follows from this and (5.14) that T |Xap

extends to a bounded C0-group U on Xap.If x ∈ Xe0∩Xap, thenMη(ux)(0) = Mηx = 0 for all η, so x = 0 by Proposition

4.7.1 and Corollary 4.5.9.Finally, suppose that ux ∈ AP(R+, X) and let π : X → X/Xap =: X be the

quotient map. Then π ◦ ux ∈ AP(R+, X) and for each η ∈ R,

Mη(π ◦ ux)(0) = π(Mη(ux)(0)) = π(Mηx) = 0,

since Mηx is a unimodular eigenvector (Proposition 5.4.2). By Proposition 4.7.1and Corollary 4.5.9, π ◦ ux = 0, so ux(t) ∈ Xap for all t ≥ 0. In particular,x = ux(0) ∈ Xap.

In Proposition 5.4.15, we shall extend part b) of Proposition 5.4.4 to indi-vidual solutions of homogeneous Cauchy problems, and in particular to individualorbits of (unbounded) semigroups. On the other hand, the following example showsthat the assumption that T is bounded is important for many aspects of Proposi-tion 5.4.4.

Example 5.4.5. There is an (unbounded) C0-semigroup T on a Banach space Xsuch that ω(T ) = 0, the set of vectors x such that limt→∞ ‖T (t)x‖ = 0 is dense inX, and the span of the unimodular eigenvectors of T is dense in X.

Let w : R+ → R+ be a continuous function with the following properties:

a) w(0) = 1,

b) w is strictly decreasing,

c) limt→∞ w(t) = 0,

d) w(s+ t) ≥ w(s)w(t) (s, t ≥ 0).

Let

X :={f : R+ → C : f continuous, lim

t→∞ f(t)w(t) = 0},

‖f‖ := supt≥0

|f(t)|w(t),

(T (t)f) (s) := f(s+ t).

Then T is a C0-semigroup on X , with ‖T (t)‖ = 1/w(t), and we may choose w suchthat ω(T ) = 0 (for example, w(t) = (1 + t)−1). For η ∈ R, eiη is a unimodulareigenvalue of T , where eiη(t) := eiηt.

Suppose that f ∈ X has compact support. Then limt→∞ ‖T (t)f‖ = 0. Given

ε > 0, choose τ such that supp f ⊂ [0, τ ] and w(τ) < ε(supt≥0 |f(t)|+ ε

)−1. By

Fejer’s Theorem (Theorem 4.2.19), there is a trigonometric polynomial p of periodτ such that |f(t) − p(t)| < ε whenever 0 ≤ t ≤ τ . For t > τ , f(t) = 0 and thereexists t′ ∈ [0, τ ] such that p(t) = p(t′), so

|f(t)− p(t)|w(t) = |p(t′)|w(t) ≤ (|f(t′)|+ ε)w(τ) < ε.

Thus, ‖f − p‖ < ε. Since the functions of compact support are dense in X, itfollows that the unimodular eigenvectors span a dense subspace of X , and alsothat the vectors f with limt→∞ ‖T (t)f‖ = 0 are dense.

The following theorem is one of the main results of this section.

Theorem 5.4.6 (Strong Splitting Theorem). Let T be a bounded C0-semigroup onX, and let x ∈ X. The following are equivalent:

(i) x ∈ X0 ⊕Xap.

(ii) ux is asymptotically almost periodic.

(iii) {T (t)x : t ≥ 0} is relatively compact.

Moreover, if x = x0 + x1 where x0 ∈ X0 and x1 ∈ Xap, then ‖x1‖ ≤M‖x‖, whereM = supt≥0 ‖T (t)‖.Proof. The implication (i) ⇒ (ii) follows immediately from the fact thatAAP(R+, X) = C0(R+, X)⊕ AP(R+, X), together with Proposition 5.4.4 b).

Suppose that ux is asymptotically almost periodic. Then ux = f +g for somef ∈ C0(R+, X) and g ∈ AP(R+, X). By (4.37), there is a sequence (tn) in R+ suchthat

limn→∞ ‖ux(tn + s)− g(s)‖ = 0,


uniformly for s ≥ 0. Let x1 := g(0) = limn→∞ ux(tn). For s ≥ 0,

g(s) = limn→∞ ux(tn + s) = lim

n→∞T (s)ux(tn) = T (s)x1.

Thus, g = ux1. By Proposition 5.4.4 b), x1 ∈ Xap. Moreover, ux−x1

= f ∈C0(R+, X), so x − x1 ∈ X0. By Proposition 5.4.4 c), x1 is unique. Moreover,‖x1‖ = limn→∞ ‖T (tn)x‖ ≤M‖x‖. This proves the implication (ii) ⇒ (i) and thefinal statement of the theorem.

The implication (ii) ⇒ (iii) follows from the fact that any asymptoticallyalmost periodic function has relatively compact range (see Theorem 4.7.4).

Now suppose that {T (t)x : t ≥ 0} is relatively compact. Let (tn) be a sequencein R+. By assumption, there is a subsequence (tnk

) such that T (tnk)x converges

to a limit x1 in X . Then (S(tnk)ux) (t) = T (t + tnk

)x → T (t)x1 uniformly fort ≥ 0. Thus, {S(t)ux : t ≥ 0} is relatively compact in BUC(R+, X), so Theorem4.7.4 shows that ux is asymptotically almost periodic.

A bounded C0-semigroup T on X is said to be asymptotically almost periodicif the equivalent conditions of Theorem 5.4.6 are satisfied for every x ∈ X . Inother words, T is asymptotically almost periodic if and only if X = X0 ⊕ Xap

(as a topological direct sum). In the literature, such semigroups are often called“almost periodic”, but we will not use this loose terminology.

We shall see in Section 5.5 that a totally ergodic semigroup with generatorA is asymptotically almost periodic if σ(A) ∩ iR is countable. We can see thisimmediately in the special case when A has compact resolvent, i.e., R(λ,A) iscompact for λ ∈ ρ(A) (see Appendix B).

Proposition 5.4.7. Let T be a bounded C0-semigroup on X such that the generatorA of T has compact resolvent. Then X = X0 ⊕Xap.

Proof. Choose λ ∈ ρ(A). For x ∈ D(A),

{T (t)x : t ≥ 0} = {R(λ,A)T (t)(λI − A)x : t ≥ 0} ,which is relatively compact. It follows by density of D(A) that {T (t)x : t ≥ 0} isrelatively compact for all x ∈ X , so the result follows from Theorem 5.4.6.

Now we turn towards the weak splitting theorem. First, we require some pre-liminary results of a very classical nature. Recall from Section 4.5 that a complex-valued trigonometric polynomial is a function p : R→ C of the form

p =

m∑j=1

λjeiηj

for some m ∈ N, λj ∈ C and distinct ηj ∈ R, where eiη(t) = eiηt. Then p is totallyergodic with means

Mη(p)(0) =

{λj if η = ηj for some j,

0 otherwise.


We shall write μη(p) for Mη(p)(0).

Proposition 5.4.8. Let η1, η2, . . . , ηk ∈ R, and ε > 0. There is a trigonometricpolynomial p : R→ C such that

a) p(t) ≥ 0 for all t,

b) μ0(p) = 1,

c) 1 ≥ μηj(p) > 1− ε for j = 1, 2, . . . , k.

Proof. Let {η′1, η′2, . . . , η′m} be a basis overQ of the Q-linear span of {η1, η2, . . . , ηk},so that ηj =

∑mr=1 βjrη

′r for some unique βjr ∈ Q. Replacing η′r by η′r/nr and βjr

by βjrnr where nr is a common multiple of the denominators of β1r, β2r, . . . , βmr,we may assume that βjr ∈ Z.

For j = 1, 2, . . . ,m and N ∈ N, let FjN be the Fejer kernel corresponding tothe frequency η′j (see the proof of Theorem 4.2.19):

FjN (t) :=

N∑n=−N

(1− |n|

N

)eiη

′jt =

1

N

⎛⎝ sinNη′jt

2

sinη′jt2

⎞⎠2

.

Let

pN (t) :=

m∏j=1

FjN (t)

=∑

1≤n1,...,nm≤N

(1− |n1|

N

). . .

(1− |nm|

N

)exp (i(n1η

′1 + . . .+ nmη′m)t) .

Then pN (t) ≥ 0 and the Q-independence of {η′1, . . . , η′m} implies that

μ0(pN ) = 1,

μηj (pN ) =

(1− |βj1|

N

). . .

(1− |βjm|

N

)for j = 1, 2, . . . , k. The result follows by choosing N sufficiently large.

Corollary 5.4.9. There is a net (pα)α∈Λ of complex-valued trigonometric polyno-mials such that

a) pα(t) ≥ 0 for all t ∈ R and all α ∈ Λ;

b) μ0(pα) = 1 for all α ∈ Λ; and

c) limα μη(pα) = 1 for all η ∈ R.


Proof. Let P be the set of all trigonometric polynomials p such that p(t) ≥ 0 forall t and μ0(p) = 1. For p ∈ P , define νp : R→ C by

νp(η) := μη(p).

By Proposition 5.4.8, the constant function 1 is in the closure of {νp : p ∈ P} inCR for the topology of pointwise convergence, and the result follows.

Proposition 5.4.10. Let T be a bounded C0-semigroup on X, and let x ∈ Xe. Thenthere is a net (xα)α∈Λ in X such that

a) For each α, xα ∈ Xap;

b) For each α, xα is in the closed convex hull of {T (t)x : t ≥ 0}; andc) limα Mη(xα) = Mη(x) for all η ∈ R.

Proof. Given a complex-valued trigonometric polynomial p(t) =∑

η μη(p)eiηt

(where μη(p) = 0 for all except finitely many η), let

p · x :=∑η

μ−η(p)Mη(x)

= limt→∞

1

t

∫ t

0

∑η

μ−η(p)e−iηsT (s)x ds

= limt→∞

1

t

∫ t

0

p(s)T (s)x ds.

Since Mη(x) is a unimodular eigenvector of T , p · x ∈ Xap and Mη(p · x) =μ−η(p)Mη(x).

Suppose that p(t) ≥ 0 for all t and μ0(p) = 1. Then

p · x = limt→∞

1t

∫ t

0p(s)T (s)x ds

1t

∫ t

0p(s) ds

.

For each t > 0,1t

∫ t

0p(s)T (s)x ds

1t

∫ t

0p(s) ds

is the mean value of ux with respect to a probability measure on [0, t], and thereforeit is in the closed convex hull of the orbit of x. It follows that p · x also lies in thisclosed set.

The result now follows by taking (pα)α∈Λ as in Corollary 5.4.9 and puttingxα := pα · x.

Now we can give the second main theorem of this section, which is the ana-logue of Theorem 5.4.6 for the weak topology.

Theorem 5.4.11 (Weak Splitting Theorem). Let T be a bounded C0-semigroup onX, let x ∈ X, and suppose that {T (t)x : t ≥ 0} is relatively weakly compact. Thenx = x0 + x1 for some unique x0 ∈ Xe0 and x1 ∈ Xap. Moreover, ‖x1‖ ≤ M‖x‖,where M := supt≥0 ‖T (t)‖.Proof. The uniqueness follows from Proposition 5.4.4. For existence, we may replaceX by the closed linear span of the orbit of x, so we may assume that every vectorin X is totally ergodic with respect to T . Let (xα)α∈Λ be as in Proposition 5.4.10.The property b) of Proposition 5.4.10 shows that ‖xα‖ ≤M‖x‖ and {xα : α ∈ Λ}is relatively weakly compact, by Krein’s theorem [Meg98, Theorem 2.8.14]. Hence,there is a subnet which is weakly convergent to a limit x1 ∈ X with ‖x1‖ ≤M‖x‖.Since {xα} ⊂ Xap and Xap is norm closed and hence weakly closed, x1 ∈ Xap.Let x0 := x − x1. Since limα Mη(x − xα) = 0 and Mη is bounded, hence weaklycontinuous on X, Mη(x0) = 0.

Corollary 5.4.12. Let T be a bounded C0-semigroup on a reflexive space X. ThenX = Xe0 ⊕Xap as a topological direct sum.

Comparison of Theorems 5.4.11 and 5.4.6 provides the following corollary.

Corollary 5.4.13. Let T be a bounded C0-semigroup on X, let x ∈ X, and supposethat x is totally ergodic with respect to T with all means Mηx = 0 and that theorbit {T (t)x : t ≥ 0} is relatively compact. Then limt→∞ ‖T (t)x‖ = 0.

Corollary 5.4.14. Let T be a bounded C0-semigroup on X, let x ∈ X, and supposethat x is totally ergodic with respect to T and that the orbit {T (t)x : t ≥ 0} isrelatively compact. If Mηx = 0 for all η ∈ R \ {0}, then limt→∞ T (t)x exists.

Proof. This follows by applying Corollary 5.4.13 with x replaced by x−M0x.

Next, we consider mild solutions of Cauchy problems, without assumingthe existence of a semigroup. For some purposes, it is not natural to specify aninitial value, and we therefore extend our previous terminology by saying thatu ∈ C(R+, X) is a mild solution of the abstract Cauchy problem

(ACPf ) u′(t) = Au(t) + f(t) (t ≥ 0),

where f ∈ L1loc(R+, X), if

∫ t

0u(r) dr ∈ D(A) and u(t) = u(0) + A

∫ t

0u(r) dr +∫ t

0f(r) dr for all t ≥ 0. It is easy to see that this implies that u(t) = u(s) +

A∫ t

su(r) dr +

∫ t

sf(r) dr for t ≥ s ≥ 0.

We begin by describing the almost periodic solutions of the homogeneousCauchy problem, thereby extending Proposition 5.4.4 c).

Proposition 5.4.15. Let A be a closed linear operator on X, and let u ∈ AP(R+, X)be a mild solution of (ACP0). For each ε > 0, there exist n ∈ N, x1, . . . , xn ∈ D(A),and η1, . . . , ηn ∈ R such that Axj = iηjxj and∥∥∥∥u(t)− n∑

j=1

eiηjtxj

∥∥∥∥ < ε (5.15)


for all t ≥ 0.

Proof. Let Z be the set of all u ∈ BUC(R+, X) which are mild solutions of (ACP0).Let (un) be a sequence in Z, u ∈ BUC(R+, X) and suppose that ‖un − u‖∞ → 0.

For t ≥ 0,∫ t

0un(s) ds ∈ D(A) and

un(t) = un(0) + A

∫ t

0

un(s) ds.

Since A is closed, it follows on letting n→∞ that

u(t) = u(0) + A

∫ t

0

u(s) ds.

Thus, Z is a closed subspace of BUC(R+, X). Moreover, if S is the shift semigroupon BUC(R+, X) and u ∈ Z, then S(t)u ∈ Z for all t ≥ 0.

Now, suppose that u ∈ Z ∩ AP(R+, X). By Proposition 4.7.1, u = g|R+for

some g ∈ AP(R, X). Let

Y := span{SR(t)g : t ∈ R} ⊂ BUC(R, X),

where SR is the shift group on BUC(R, X). Applying Theorem 4.5.1 to the restric-tion of SR to Y shows that

g ∈ span {eiη ⊗ x : η ∈ R, x ∈ X, eiη ⊗ x ∈ Y } . (5.16)

Let t ∈ R. By (4.23) there exist τn →∞ such that

limn→∞ ‖SR(t+ τn)g − SR(t)g‖∞ = 0. (5.17)

For t + τn > 0, (SR(t + τn)g)|R+= S(t + τn)u, so it follows from (5.17) and the

first paragraph that (SR(t)g)|R+∈ Z for every t ∈ R, and hence that h|R+

∈ Z forevery h ∈ Y .

Now suppose that h := eiη ⊗ x ∈ Y . Take t > 0 and let

ζ :=

⎧⎨⎩eiηt − 1

iηif η �= 0,

t if η = 0.

We can choose t in such a way that ζ �= 0. Then

x =1

ζ

∫ t

0

h(s) ds ∈ D(A),

and

eiηtx = h(t) = h(0) +A

∫ t

0

h(s) ds = x+ ζAx.

Hence, Ax = iηx. The result follows from this and (5.16).

Now we show that the splitting of AAP(R+, X) as C0(R+, X)⊕AP(R+, X)respects mild solutions of inhomogeneous Cauchy problems.


Proposition 5.4.16. Let A be a closed linear operator on X, let u0 ∈ C0(R+, X),u1 ∈ AP(R+, X), f0 ∈ C0(R+, X), f1 ∈ AP(R+, X), u = u0+u1, and f = f0+f1.Suppose that u is a mild solution of (ACPf ). Then u0 and u1 are mild solutionsof (ACPf0) and (ACPf1), respectively.

Proof. It is easy to verify that the function t �→ (u1(t), f1(t)) is almost periodic(with values in X ×X), and hence that t �→ (u(t), f(t)) is asymptotically almostperiodic. By (4.37), there is a sequence (τn)n≥1 in R+ such that τn →∞ and

supt≥0

‖u(t+ τn)− u1(t)‖ → 0, supt≥0

‖f(t+ τn)− f1(t)‖ → 0

as n→∞. Since u is a mild solution of (ACPf ),

u(t+ τn) = A

(∫ t

0

u(s+ τn) ds

)+

∫ t

0

f(s+ τn) ds+ u(τn).

Since A is closed, it follows on letting n→∞ that

u1(t) = A

(∫ t

0

u1(s) ds

)+

∫ t

0

f1(s) ds+ u1(0),

so u1 is a mild solution of (ACPf1). By linearity, u0 = u− u1 is a mild solution of(ACPf0).

The following corollary is a generalization of part of the strong splittingtheorem (Theorem 5.4.6, (i) ⇔ (ii)).

Corollary 5.4.17. Let T be a C0-semigroup on X, let x ∈ X, and suppose that ux

is asymptotically almost periodic. Then there exist unique x0 and x1 in X suchthat

a) x = x0 + x1;

b) limt→∞ ‖T (t)x0‖ = 0; and

c) There is a sequence (yn) in the linear span of the unimodular eigenvectorsof T such that limn→∞ T (t)yn = T (t)x1 uniformly for t ≥ 0.

Conversely, if a), b) and c) hold, then ux is asymptotically almost periodic.

Proof. Let ux = v0 + v1, where v0 ∈ C0(R+, X) and v1 ∈ AP(R+, X). By Propo-sition 5.4.16 (with A equal to the generator of T and f0 = f1 = 0), v0 = ux0

and v1 = ux1for some x0, x1 ∈ X . Now, a) and b) are immediate, c) follows from

Proposition 5.4.15, and the uniqueness follows from (4.36). The converse statementis immediate, as in Theorem 5.4.6.

5.5. COUNTABLE SPECTRAL CONDITIONS 371

5.5 Countable Spectral Conditions

In this section, we shall give various results showing that solutions of well-posedhomogeneous Cauchy problems are asymptotically almost periodic under assump-tions including boundedness of the solution and countability of the purely imagi-nary part of the (local) spectrum of A, using the results and methods of Chapter4. In Section 5.6, we shall extend some of the results to inhomogeneous Cauchyproblems which are not necessarily associated with C0-semigroups.

Suppose that A generates a C0-semigroup T on a complex Banach space X.For x ∈ X , we again put ux(t) := T (t)x (t ≥ 0). Then ux is an exponentiallybounded function and its Laplace transform ux(λ) coincides with R(λ,A)x forlarge real λ. We shall assume that ux is bounded, and our first step is to relatethe half-line spectrum sp(ux) of ux in the sense of Section 4.7 to the operator A.

The imaginary local resolvent set ρu(A, x) of A at x is defined to be the setof all points iη ∈ iR such that there exist an open set U containing C+ ∪{iη} anda holomorphic function g : U → X such that g(λ) ∈ D(A) and (λ − A)g(λ) = xfor all λ ∈ C+. We shall see in Proposition 5.5.1 that g(λ) = R(λ,A)x wheneverλ ∈ C+ ∩ ρ(A), so g is uniquely determined if U is connected.

The imaginary local spectrum σu(A, x) of A at x is:

σu(A, x) := iR \ ρu(A, x).

It is clear from the definition that ρu(A, x) is open in iR, so σu(A, x) is closed.WhenA generates a bounded C0-group, Lemma 4.6.7 shows that σu(A, x) coincideswith the local spectrum σ(A, x) defined in Section 4.6.

The following is a local version of Proposition B.5.

Proposition 5.5.1. Let T be a C0-semigroup on X with generator A, and let x ∈ X.

a) Let V be a connected open subset of C, let g : V → X be holomorphic, andsuppose that there is a subset U of V , with a limit point in V , such thatg(λ) ∈ D(A) and (λ− A)g(λ) = x whenever λ ∈ U . Then g(λ) ∈ D(A) and(λ− A)g(λ) = x whenever λ ∈ V .

b) If Reλ > hol(ux), then ux(λ) ∈ D(A) and (λ−A)ux(λ) = x.

c) If Re λ > hol(ux) and λ ∈ ρ(A), then ux(λ) = R(λ,A)x.

Proof. a) Fix μ ∈ ρ(A). For λ ∈ U ,

(λ− A)R(μ,A)g(λ) = R(μ,A)x.

Since AR(μ,A) = μR(μ,A)− I is a bounded operator on X, the left-hand side isholomorphic on V . By uniqueness of holomorphic extensions, this formula is truewhenever λ ∈ V . Hence,

g(λ) = R(μ,A)x− (λ− μ)R(μ,A)g(λ) ∈ D(A),


and(λ− A)g(λ) = x,

since R(μ,A) is injective.Statement b) follows by applying a) with g := ux, and c) is then immediate.

The following result identifies σu(A, x) with sp(ux) when ux is bounded.

Proposition 5.5.2. Let T be a C0-semigroup on X with generator A. Let x ∈ X,and suppose that ux is bounded. Then

a) σu(A, x) ⊂ σ(A) ∩ iR.

b) sp(ux) = {η ∈ R : iη ∈ σu(A, x)}.Proof. a) By Proposition 5.5.1 c), ux(λ) = R(λ,A)x whenever λ ∈ C+ ∩ ρ(A).Thus, we may define a holomorphic function g : C+ ∪ ρ(A)→ X by

g(λ) :=

{ux(λ) (λ ∈ C+),

R(λ,A)x (λ ∈ ρ(A)).

This shows that σu(A, x) ⊂ σ(A) ∩ iR.b) Suppose that iη /∈ σu(A, x), so there exist an open set U containing C+ ∪

{iη} and a holomorphic function g : U → X such that g(λ) ∈ D(A) and (λ −A)g(λ) = x for all λ ∈ C+. Then g(λ) = R(λ,A)x = ux(λ) whenever λ ∈ ρ(A) ∩C+. By uniqueness of holomorphic extensions, g(λ) = ux(λ) whenever λ ∈ C+, soη /∈ sp(ux).

Conversely, suppose that η /∈ sp(ux), so there exists a connected open setV containing iη and a holomorphic function g : V → X such that g(λ) = ux(λ)for all λ ∈ V ∩ C+. By Proposition 5.5.1 b), g(λ) ∈ D(A) and (λ − A)g(λ) = xfor all x ∈ V ∩ C+. By Proposition 5.5.1 a), this holds for all λ ∈ V . Thus,iη /∈ σu(A, x).

The next four results appear in decreasing order of generality, with the as-sumptions becoming stronger but less technical.

Theorem 5.5.3. Let T be a C0-semigroup on X with generator A. Let x ∈ X, andsuppose that the following conditions are satisfied:

a) ux : t �→ T (t)x is bounded and uniformly continuous;

b) σu(A, x) is countable; and

c) For each iη ∈ σu(A, x), limα↓0 αT (s)ux(α+ iη) exists, uniformly for s ≥ 0.

Then ux is asymptotically almost periodic. If all the limits in c) are zero, then‖T (t)x‖ → 0 as t→∞.


Proof. By Proposition 5.5.2, condition b) is equivalent to sp(ux) being countable.Moreover, c) is equivalent to total ergodicity of ux. Indeed, if S denotes the shiftsemigroup on BUC(R+, X), then

(S(s)ux) (t) = T (s)ux(t), so(S(s)ux

)(λ) = T (s)ux(λ).

The result follows from Theorem 4.7.7 or 4.9.7.

Theorem 5.5.4. Let T be a bounded C0-semigroup on X, with generator A. Letx ∈ X, and suppose that the following conditions are satisfied:

a) σu(A, x) is countable; and

b) x ∈ Ker(A− iη) + Ran(A− iη) for each iη ∈ σu(A, x).

Then x ∈ X0 ⊕Xap. If x ∈ Ran(A− iη) for each iη ∈ σu(A, x), then x ∈ X0.

Proof. By Proposition 5.4.1, ux is uniformly continuous and bounded. Moreover,ux(α + iη) = R(α + iη,A)x for α > 0, limα↓0 αT (s)ux(α + iη) exists, uniformlyfor s ≥ 0, by Proposition 4.3.1 and the boundedness of T , and the limit is 0 ifx ∈ Ran(A− iη). Now the results follow from Theorem 5.5.3.

The results of Section 4.3 (applied with T (t) replaced by e−iηtT (t)) showthat condition b) of Theorem 5.5.4 is equivalent to any of the following:

(i) If x∗ ∈ D(A∗), A∗x∗ = iηx∗ for some iη ∈ σu(A, x) and 〈y, x∗〉 = 0 for ally ∈ Ker(A− iη), then 〈x, x∗〉 = 0;

(ii) limα↓0 αR(α+ iη, A)x exists in X whenever iη ∈ σu(A, x);

(iii) limt→∞ t−1∫ t

0e−iηsT (s)x ds exists whenever iη ∈ σu(A, x).

Since the Abel limit exists (and equals 0) when iη ∈ ρu(A, x), the restriction ineach condition that iη ∈ σu(A, x) is redundant.

Recall from Definition 4.3.10 that a bounded C0-semigroup T on X is saidto be totally ergodic if the condition (iii) above is satisfied for all x ∈ X (and allη ∈ R). By Proposition 5.4.2, T is totally ergodic if and only if each orbit ux is atotally ergodic function. As in the case of individual η and x discussed above, thisproperty can be characterized in several ways in terms of the generator A.

Recall also from Section 5.4 that T is said to be asymptotically almost pe-riodic if each orbit ux is asymptotically almost periodic (see Theorem 5.4.6 forequivalent properties). Thus, every asymptotically almost periodic semigroup istotally ergodic, but the converse is not true (see Example 5.5.9). The followingresult shows that the converse is true under a countable spectral condition. Notethat Ran(A− iη) is dense in X if and only if iη does not belong to the point spec-trum σp(A

∗) of A∗, and that σp(A) ∩ iR ⊂ σp(A∗) when A generates a bounded

semigroup (see Proposition 4.3.6).


Theorem 5.5.5 (Countable Spectrum). Let T be a bounded C0-semigroup on Xwith generator A, and suppose that σ(A) ∩ iR is countable.

a) If T is totally ergodic, then T is asymptotically almost periodic.

b) If σp(A∗) ∩ iR is empty, then ‖T (t)x‖ → 0 as t→∞, for each x ∈ X.

Proof. This is immediate from Proposition 5.5.2, Theorem 5.5.4 and the re-marks above.

Theorem 5.5.6. Let T be a bounded C0-semigroup on a reflexive space X withgenerator A, and suppose that σ(A) ∩ iR is countable. Then the following hold:

a) T is asymptotically almost periodic.

b) If σp(A)∩ iR ⊂ {0}, then there is a projection P such that limt→∞ T (t) = Pin the strong operator topology.

Proof. a) This is immediate from Corollary 4.3.5 and Theorem 5.5.5 a).b) By a), X = X0 ⊕Xap. Let P be the projection onto Xap along X0. The

assumption on σp(A) ∩ iR implies that Xap = {x ∈ X : T (t)x = x (t ≥ 0)}, andthe result follows.

Theorem 5.5.6 is also true if T is a positive, bounded, Cesaro-ergodic C0-semigroup on L1(Ω, μ) and σ(A) ∩ iR is countable. This follows from Proposition4.3.14.

We saw in Proposition 5.4.7 that Theorem 5.5.5 can be proved more directlywhen A has compact resolvent. There is also a simple proof when T is a boundedholomorphic C0-semigroup in the sense of Definition 3.7.1. Then σ(A) ∩ iR ⊂ {0}by Theorem 3.7.11, and there is a constant c such that ‖AT (t)‖ ≤ c/t for allt > 0, by Theorem 3.7.19. Hence, limt→∞ T (t)Ay = 0 for all y ∈ D(A), andlimt→∞ T (t)(x1 + x2) = x1 whenever x1 ∈ KerA and x2 ∈ RanA. If A is totallyergodic, then KerA+RanA is dense in X, so limt→∞ T (t)x exists for all x ∈ X .

The following example shows that the assumption of uniform continuity can-not be omitted from Theorem 5.5.3.

Example 5.5.7. There is a C0-semigroup T on a Hilbert space X and a vector x ∈ Xsuch that σ(A)∩iR is empty and ux is bounded, but ux is not asymptotically almostperiodic (and not uniformly continuous, by Theorem 5.5.3).

Let X := �2 and T be the C0-semigroup defined by:

(T (t)x)2n−1 := eλnt(x2n−1 + tx2n),

(T (t)x)2n := eλntx2n,

where λn := in− 1/n. The generator A is given by:

D(A) ={x ∈ �2 : (nxn) ∈ �2

},

(Ax)2n−1 = x2n + λnx2n−1,

(Ax)2n = λnx2n,

σ(A) = {λn : n ≥ 1}.


Now take x ∈ �2 given by:

x2n−1 := 0, x2n := n−3/2.

Then

(T (t)x)2n−1 =teλnt

n3/2,

(T (t)x)2n =eλnt

n3/2.

A simple argument involving Riemann sums of s−3e−2/s shows that

‖T (t)x‖2 =∞∑

n=1

1 + t2

n3e−2t/n →

∫ ∞

0

s−3e−2/s ds =

∫ ∞

0

ue−2u du =1

4

as t→∞. Thus, all the assumptions of Theorem 5.5.3, except uniform continuity,are satisfied, with c) vacuous, and T has no unimodular eigenvectors since A has noimaginary eigenvalues. However, limt→∞ ‖T (t)x‖ �= 0, so ux is not asymptoticallyalmost periodic, by Corollary 5.4.17.

Theorem 5.5.4 was obtained from Theorem 5.5.3 by observing that the con-vergence in condition c) of Theorem 5.5.3 is automatically uniform when T isuniformly bounded. The next example shows that this may not be valid for indi-vidual bounded orbits of unbounded semigroups.

Example 5.5.8. There is a (norm-continuous, unbounded) C0-semigroup T on aHilbert space X and a vector x ∈ X such that σ(A)∩iR = {0}, 0 /∈ σp(A)∪σp(A∗),and ux is bounded and uniformly continuous, but ux is not asymptotically almostperiodic (and not totally ergodic).

Let X := �2, and T be the norm-continuous semigroup given by

(T (t)x)2n−1 := e−t/n(x2n−1 + tx2n),

(T (t)x)2n := e−t/nx2n.

The generator A is the bounded operator given by:

(Ax)2n−1 = x2n − x2n−1

n,

(Ax)2n = −x2n

n,

σ(A) =

{− 1

n: n ≥ 1

}∪ {0}.

Moreover, 0 /∈ σp(A) ∪ σp(A∗), so RanA = X, and Xap = {0}.

Now takex2n−1 := 0, x2n := n−3/2.

As in Example 5.5.7, ‖T (t)x‖ → 12 as t → ∞. Thus, ux : t �→ T (t)x is bounded.

Since A is bounded, it follows that t �→ AT (t)x = u′x(t) is bounded, so ux isuniformly continuous. Since T has no unimodular eigenvectors and limt→∞ ‖T (t)x‖�= 0, it follows from Corollary 5.4.17 that ux is not asymptotically almost periodic.By Theorem 4.7.7 or 4.9.7 (see also Theorem 5.5.3), ux is not totally ergodic.

It is not difficult to verify directly that limα↓0 αR(α,A)x does not exist, whichalso shows that ux is not totally ergodic, and therefore not asymptotically almostperiodic.

The next example shows that the countability condition in Theorem 5.5.5(and hence in other results of this section) is best possible in a certain sense.

Example 5.5.9. Let E be any uncountable closed subset of R. There is a C0-groupT of isometries on a Banach space X (even a Hilbert space) such that σ(A) ⊂ iEand σp(A

∗) is empty, but T is not (asymptotically) almost periodic.Choose [a, b] so that E ∩ [a, b] is uncountable, and let

E′ := {x ∈ E ∩ [a, b] : for all ε > 0, E ∩ [a, b] ∩ (x− ε, x+ ε) is uncountable}.Then (E ∩ [a, b]) \E′ is countable, and E′ is compact with no isolated points. If Econtains no interval, then E′ is homeomorphic to the Cantor set [Wil70, Theorem30.3]. If E contains an interval, then E clearly contains a subset homeomorphicto the Cantor set. There is a non-zero non-atomic Borel measure on the Cantorset (for example, the Lebesgue-Stieltjes measure associated with the Lebesgue-Cantor function [Tay73, Sections 2.7,4.5]). Hence, there is a non-zero non-atomicmeasure μ supported by iE. Let T be the multiplier (semi)group on Lp(μ), where1 ≤ p <∞ (see Example 5.4.3). Then

σ(A) = suppμ ⊂ iE, σp(A∗) = ∅,

but T is not asymptotically almost periodic.

Suppose that T is a bounded, totally ergodic semigroup on X and thereis a dense subspace Y of X such that σu(A, y) is countable for each y ∈ Y .From Theorem 5.5.4 and the fact that X0 ⊕ Xap is closed, it follows that T isasymptotically almost periodic. The following example shows that the conversedoes not hold.

Example 5.5.10. There is a C0-semigroup T on a Banach space X such thatlimt→∞ ‖T (t)x‖ = 0 for all x ∈ X, but σu(A, x) = iR whenever x �= 0.

Let X := L1(R+, w(t)dt), where w : R+ → R+ satisfies:

a) w is non-increasing,

b) limt→∞ w(t) = 0,

c) For each a > 0, there is a constant ca > 0 such that w(t) ≥ cae−at for all

t ≥ 0.

Let T be the C0-semigroup of contractions on X defined by

(T (t)f) (s) :=

{f(s− t) (s ≥ t ≥ 0),

0 (t > s ≥ 0),

and let A be the generator. For f ∈ X,

‖T (t)f || =∫ ∞

t

|f(s− t)|w(s) ds =∫ ∞

0

|f(s)|w(s+ t) ds→ 0

as t→∞.For Reλ < 0, let

gλ(s) :=eλs

w(s)(s ≥ 0).

Then gλ is bounded, by c), so eλ(s) := eλs defines an element of X∗. For f ∈ Xand t ≥ 0,

〈T (t)f, eλ〉 =

∫ ∞

t

f(s− t)eλs ds =

∫ ∞

0

f(s)eλ(s+t) ds

= eλt∫ ∞

0

f(s)eλsds = eλt〈f, eλ〉.

Thus,T ∗(t)eλ = eλteλ (t ≥ 0),

so eλ ∈ D(A∗) and A∗eλ = λeλ.Suppose that f ∈ X and λ �→ R(λ,A)f has a holomorphic extension F to a

connected neighbourhood V of some point iη0 in iR. We shall show that f = 0.For λ ∈ V ∩ C+,

R(1, A)f = F (λ) + (λ− 1)R(1, A)F (λ).

By uniqueness of holomorphic extensions,

R(1, A)f = F (λ) + (λ− 1)R(1, A)F (λ) = (λ−A)R(1, A)F (λ)

for all λ ∈ V . Hence, for λ ∈ V ∩ C−,∫ ∞

0

eλs (R(1, A)f) (s) ds = 〈R(1, A)f, hλ〉 = 〈R(1, A)F (λ), (λ− A∗)hλ〉 = 0.

As a function of λ, the left-hand side is holomorphic on C− and vanishes on C−∩V .Therefore, ∫ ∞

0

eλs (R(1, A)f) (s) ds = 0 (λ ∈ C−).

By uniqueness of Laplace transforms, this implies that R(1, A)f = 0. Hence, f = 0,since R(1, A) is injective.

5.6 Solutions of Inhomogeneous Cauchy Problems

In this section, we consider the asymptotic behaviour of mild solutions of inho-mogeneous Cauchy problems. In practice, once one knows that the solution isbounded and uniformly continuous, it often follows that further asymptotic prop-erties are inherited by u from asymptotic properties of a semigroup generated by A(or spectral properties of A) and of the inhomogeneity f . For example, Theorems5.6.6 and 5.6.8 are results of this type under assumptions of non-resonance andcountable spectrum, respectively. Thus, we are interested first in conditions whichensure that a mild solution is bounded and uniformly continuous.

We first assume that A generates a bounded C0-semigroup T on X, and thatf ∈ L1

loc(R+, X) is given. Recall from Proposition 3.1.16 that the unique solutionof the inhomogeneous Cauchy problem

(ACPf )

{u′(t) = Au(t) + f(t),

u(0) = x,

is given by

u(t) = T (t)x+ (T ∗ f)(t),where

(T ∗ f)(t) =∫ t

0

T (t− s)f(s) ds.

Thus, we are seeking conditions which ensure that T ∗f is bounded and uniformlycontinuous.

We consider first the simple case when ω(T ) < 0.

Proposition 5.6.1. Let T be a semigroup on X with generator A, and suppose thatω(T ) < 0.

a) If f ∈ L∞(R+, X), then T ∗ f is bounded.

b) If f ∈ BUC(R+, X), then T ∗ f ∈ BUC(R+, X).

c) If f ∈ AAP(R+, X), then T ∗ f ∈ AAP(R+, X).

d) If f∞ := limt→∞ f(t) exists, then limt→∞(T ∗ f)(t) = R(0, A)f∞.

e) If f ∈ AP(R+, X), then there exist unique x ∈ X and g ∈ AP(R+, X) suchthat (T ∗ f)(t) = T (t)x + g(t) for all t ≥ 0. If f is τ-periodic, then g isτ -periodic.

Proof. Parts a), b) and the case f∞ = 0 of d) all follow from Proposition 1.3.5.Moreover, the map f �→ T ∗ f is bounded on BUC(R+, X).

Let Y := {ux : x ∈ X}, where ux(t) = T (t)x. Since ω(T ) < 0, Y is aclosed subspace of C0(R+, X). Let Z := Y ⊕ AP(R+, X), a closed subspace of

5.6. SOLUTIONS OF INHOMOGENEOUS CAUCHY PROBLEMS 379

AAP(R+, X). Suppose that f = eiη⊗y for some η ∈ R and y ∈ X , so f(t) = eiηty.Then

(T ∗ f)(t) = eiηt∫ t

0

e−iηsT (s)y ds

= eiηt∫ t

0

d

ds

(−e−iηsT (s)R(iη,A)y)ds

= T (t) (−R(iη,A)y) + eiηtR(iη,A)y. (5.18)

Thus, T ∗ f ∈ Z, and the almost periodic part of T ∗ f has the same period asf . By linearity and continuity, T ∗ f ∈ Z for all f ∈ AP(R+, X), and the almostperiodic part of T ∗ f is τ -periodic when f is τ -periodic. This proves e), and d)follows from the case f∞ = 0 and the case η = 0 of (5.18). Finally, c) follows fromd) and e).

In the context of Proposition 5.6.1 a), we cannot conclude that T ∗ f isuniformly continuous.

Example 5.6.2. There exist a C0-semigroup S on a Hilbert space X with ω(S) < 0and a function f ∈ L∞(R+, X) such that S ∗ f is not uniformly continuous.

Let T , X and x be as in Example 5.5.7. Take ω > ω(T ), and let S(t) :=e−ωtT (t) and f(t) := T (t)x. Then f is bounded and continuous, and

(S ∗ f)(t) =∫ t

0

e−ωsT (t)x ds =1− e−ωt

ωT (t)x.

Since T (·)x is not uniformly continuous, S ∗ f is not uniformly continuous.

Remark 5.6.3. In part e) of Proposition 5.6.1, T ∗ f is an asymptotically almostperiodic mild solution of the inhomogeneous Cauchy problem (ACPf ), with initialvalue 0. The decompositions f = 0 + f and T ∗ f = ux + g, where ux(t) =T (t)x, correspond to the splitting AAP(R+, X) = C0(R+, X) ⊕ AP(R+, X). Aspredicted by Proposition 5.4.16, ux is a mild solution of (ACP )0 with initial valuex, and g is a mild solution of (ACP )f with initial value −x. When f is τ -periodic,x = R(1, T (τ))((T ∗ f)(τ)) and it is easy to verify directly that T ∗ f − ux isτ -periodic.

Now suppose that T is bounded but ω(T ) = 0. Then Datko’s theorem (The-orem 5.1.2) shows that there exist bounded f such that T ∗ f is unbounded.The simplest examples in which T ∗ f is unbounded arise from eigenvalues. Iff(t) = T (t)x = eiηtx for all t ≥ 0, then (T ∗ f)(t) = teiηtx, and this is unboundedif x �= 0. Note that there is resonance between T and f , reflected in the fact thatiη ∈ σ(A) ∩ i sp(f).

In order to obtain positive results showing that T ∗f is bounded when ω(T ) =0, we have to place some constraints on f and we may also impose assumptionson T . The first possibility (which is somewhat dual to the case considered inProposition 5.6.1) is to assume that f ∈ L1(R+, X).

Proposition 5.6.4. Let T be a bounded C0-semigroup on X, and let f ∈ L1(R+, X).Then

a) T ∗ f ∈ BUC(R+, X).

b) If T is asymptotically almost periodic, then T ∗ f ∈ AAP(R+, X).

c) If limt→∞ T (t) exists in the strong operator topology, then limt→∞(T ∗ f)(t)exists.

d) If limt→∞ T (t) = 0 in the strong operator topology, then limt→∞(T ∗ f)(t) =0.

Proof. First, note that ‖(T ∗ f)(t)‖ ≤ M‖f‖1, where M := supt≥0 ‖T (t)‖. Thus,the map f �→ T ∗f is bounded, and by density it suffices to prove the results whenf has support in [0, τ ] for some τ . But then

(T ∗ f)(t) = T (t− τ)((T ∗ f)(τ))

for t ≥ τ , and all the results are immediate.

Another way to obtain results that T ∗ f is bounded is to impose a conditionof non-resonance by assuming that σ(A) ∩ i sp(f) is empty. However, there areexamples (see Examples 5.1.10 and 5.1.11 after rescaling) of bounded semigroupswhere σ(A) ∩ iR is empty (so non-resonance occurs for all f), but ω(T ) = 0, andthen by Datko’s theorem there exist bounded f such that T ∗ f is unbounded.Thus, non-resonance is not sufficient on its own to obtain positive results aboutboundedness of T ∗ f ; we need to impose further assumptions on T , or f , or both.If T is holomorphic and ω(T ) = 0, then σ(A)∩ iR is nonempty (Theorem 5.1.12),and we now establish that non-resonance implies boundedness of T ∗ f when T isholomorphic.

Theorem 5.6.5 (Non-resonance Theorem). Let T be a holomorphic C0-semigroupon X with generator A such that supt≥0 ‖T (t)‖ < ∞. Let f ∈ L∞(R+, X), andsuppose that σ(A) ∩ i sp(f) is empty. Then T ∗ f ∈ BUC(R+, X).

Proof. Note first that T is norm-continuous on (0,∞), and that −iσ(A) ∩ R iscompact (by Corollary 3.7.18) and disjoint from sp(f) by assumption. Let ψ ∈C∞c (R) be such that ψ = 1 on a neighbourhood of −iσ(A) ∩R in R and ψ = 0 ona neighbourhood of sp(f). Define G,H : R→ L(X) by

G(t) :=

{T (t)− ∫∞

0(F−1ψ)(t− s)T (s) ds (t ≥ 0),

− ∫∞0

(F−1ψ)(t− s)T (s) ds (t < 0),

H(η) :=

{(1− ψ(η))R(iη,A) (iη ∈ ρ(A)),

0 (iη ∈ σ(A)).

Then G ∈ L∞(R,L(X)) and G is continuous on R \ {0}. Also, H ∈ C∞(R,L(X))and, for all large |η|, H(η) = R(iη,A), so H ′′(η) = −2R(iη,A)3. By Corollary3.7.18, there is a constant C such that ‖H ′′(η)‖ ≤ C|η|−3 for all large |η|. Hence,H ′′ ∈ L1(R,L(X)) and F−1H ′′ : R→ L(X) is bounded. We shall use this to showthat G ∈ L1(R,L(X)).

Let ρ ∈ C∞c (R). By two applications of Theorem 1.8.1 b), the dominatedconvergence theorem and integration by parts,∫ ∞

−∞(Fρ)(t)(F−1H ′′)(t) dt =

∫ ∞

−∞ρ(η)H ′′(η) dη

=

∫ ∞

−∞ρ′′(η)H(η) dη

= limξ↓0

∫ ∞

−∞ρ′′(η)(1− ψ(η))R(ξ + iη,A) dη

= limξ↓0

∫ ∞

−∞(Fρ′′)(t)Gξ(t) dt

= limξ↓0

∫ ∞

−∞(−t2)(Fρ)(t)Gξ(t) dt,

where

Gξ(t) :=

{e−ξtT (t)− ∫∞

0(F−1ψ)(t− s)e−ξsT (s) ds (t ≥ 0),

− ∫∞0

(F−1ψ)(t− s)e−ξsT (s) ds (t < 0).

By the dominated convergence theorem, ‖Gξ(t)−G(t)‖ → 0 as ξ ↓ 0. Moreover,

‖Gξ(t)‖ ≤(1 + ‖F−1ψ‖1

)M

for all ξ > 0 and t ∈ R, where M := supt≥0 ‖T (t)‖. By the dominated convergencetheorem again,∫ ∞

−∞(Fρ)(t)(F−1H ′′)(t) dt =

∫ ∞

−∞(Fρ)(t)(−t2)G(t) dt.

Since this holds for all ρ ∈ C∞c (R), it follows that

(F−1H ′′)(t) = −t2G(t)

a.e. (in fact, everywhere, since both sides are continuous). Hence,G ∈ L1(R,L(X)).Now define g, h : R→ X by

g(t) :=

∫ ∞

0

(F−1ψ)(t− s)f(s) ds,

h(η) :=

{ψ(η)f(iη) (η /∈ sp(f)),

0 (η ∈ sp(f)).

Then g ∈ BUC(R, X) (by Proposition 1.3.2 c)) and h ∈ C∞c (R, X). A similarargument to the previous paragraph shows that

(F−1h′′)(t) = −t2g(t),

so that g ∈ L1(R, X). For t ≥ 0,

(T ∗ f)(t) =

∫ ∞

0

G(t− s)f(s) ds+

∫ ∞

0

∫ ∞

0

(F−1ψ)(t− s− r)T (r)f(s) dr ds

=

∫ ∞

0

G(t− s)f(s) ds+

∫ ∞

0

T (r)g(t− r) dr.

Both of these terms may be regarded as being convolutions of functions on R,where f(s) = 0 for s < 0 and T (r) = 0 for r < 0. Since T is bounded andg ∈ L1(R, X), and f is bounded and G ∈ L1(R,L(X)), both terms are boundedand uniformly continuous on R and hence on R+ (see Propositions 1.3.2, 1.3.5 andRemark 1.3.8).

Next, we give a result which shows that asymptotic properties of T ∗f gener-ally follow from those of T and f when non-resonance holds and T ∗ f is boundedand uniformly continuous.

Theorem 5.6.6. Let T be a bounded C0-semigroup on X, and let f ∈ BUC(R+, X).Suppose that σ(A) ∩ i sp(f) is empty and T ∗ f ∈ BUC(R+, X).

a) If T is asymptotically almost periodic and f ∈ AAP(R+, X), then T ∗ f ∈AAP(R+, X).

b) If limt→∞ T (t) exists in the strong operator topology and limt→∞ f(t) exists,then limt→∞(T ∗ f)(t) exists.

c) If limt→∞ T (t) = 0 in the strong operator topology and limt→∞ f(t) = 0,then limt→∞(T ∗ f)(t) = 0.

Proof. First, let Y be any of the spaces BUC(R+, X), AAP(R+, X), C0(R+, X)or the space of continuous functions f : R+ → X such that limt→∞ f(t) exists.We assume that Y contains all orbits ux of T (x ∈ X), f ∈ Y and σ(A) ∩ i sp(f)is empty, and we shall prove that (T ∗ ϕ ∗ f)|R+

∈ Y whenever ϕ ∈ L1(R) andFϕ ∈ C∞c (R). Here we are regarding T and f as being defined on R with T (t) = 0and f(t) = 0 for t < 0, and the convolutions are defined accordingly (see Section1.3). We shall also consider the convolution T ∗ ϕ : R→ L(X) defined by

(T ∗ ϕ)(t)x =

∫ ∞

0

ϕ(t− s)T (s)x ds.

Note that T ∗ϕ is bounded and uniformly norm-continuous (see Proposition 1.3.5c)).


We proceed in a similar way as in Theorem 5.6.5. Let ψ ∈ C∞c (R) be such thatψ = 1 on a neighbourhood of −iσ(A) ∩ supp(Fϕ) and ψ = 0 on a neighbourhoodof sp(f). Define G,H : R→ L(X) by:

G := T ∗ ϕ− T ∗ ϕ ∗ F−1ψ,

H(η) :=

{(Fϕ)(η)(1− ψ(η))R(iη,A) (iη ∈ ρ(A)),

0 (iη ∈ σ(A)).

Then G ∈ BUC(R, X), H ∈ C∞c (R,L(X)), and as in the proof of Theorem 5.6.5,(F−1H ′′)(t) = −t2G(t). Hence, G ∈ L1(R,L(X)). It follows that (G ∗ f)|R+

∈ Y ,(see Remark 5.6.3 b)).

Let g := ϕ ∗ F−1ψ ∗ f , and let

h(η) :=

{(Fϕ)(η)ψ(η)f(iη) (η �∈ sp(f)),

0 (η ∈ sp(f)).

Then g ∈ BUC(R, X), h ∈ C∞c (R, X) and (F−1h′′)(t) = −t2g(t). Hence, g ∈L1(R, X). We claim that this implies that (T∗g)|R+

∈ Y . Note that T∗(χ(a,b)⊗x) =χ(a,b)∗ux ∈ Y (see Remark 5.6.3 b)). Since the step functions are dense in L1(R, X)and convolution with T is a bounded map from L1(R, X) into L∞(R, X), it followsthat (T ∗ g)|R+

∈ Y , as claimed.Since

T ∗ ϕ ∗ f = G ∗ f + T ∗ ϕ ∗ F−1ψ ∗ f= G ∗ f + T ∗ g,

it follows that (T ∗ ϕ ∗ f)|R+∈ Y , as claimed.

Now suppose that T ∗ f ∈ BUC(R+, X). Let (ϕn) be an approximate unit inL1(R) such that Fϕn ∈ C∞c (R) for every n (see Lemma 1.3.3). Since (T ∗f)(0) = 0,we can consider T ∗f as an element of BUC(R, X) and we obtain that T ∗ϕn ∗f =ϕn ∗ (T ∗ f)→ T ∗ f uniformly as n→∞. It follows that (T ∗ f)|R+

∈ Y , and thiscompletes the proof.

Finally in this section, we consider the situation when the assumption ofnon-resonance is replaced by a countable spectral condition. As in Theorem 5.6.6,we assume that the inhomogeneity f and the given mild solution u of (ACPf )are bounded and uniformly continuous (and totally ergodic), and we aim to showthat asymptotic properties of f are transferred to u. In contrast to Section 5.5 andthe earlier part of this section, we consider problems which may not be associatedwith C0-semigroups; i.e., A is not assumed to be a generator. Thus we shall extendTheorem 5.5.3 in several directions.

Let f, u ∈ BUC(R+, X). Recall from Section 5.4 that u is said to be a mildsolution of the inhomogeneous Cauchy problem

(ACPf ) u′(t) = Au(t) + f(t) (t ≥ 0),


if∫ t


u(t) = u(0) +A

∫ t

0

u(s) ds+

∫ t

0

f(s) ds

for all t ≥ 0. As in Section 4.7, we let S be the shift semigroup on BUC(R+, X),(S(t)f)(s) := f(s+ t), and B be the generator of S, so

(R(λ,B)f)(t) =

∫ ∞

0

e−λsf(s+ t) ds = ft(λ) (λ ∈ C+, t ≥ 0),

where ft(s) := f(s + t). The following proposition relates the Laplace transform,spectrum and ergodicity of u to the properties of f .

Proposition 5.6.7. Let A be a closed operator on X, let f ∈ BUC(R+, X), and letu be a bounded, uniformly continuous, mild solution of the inhomogeneous Cauchyproblem (ACPf ). Then

a) For t ≥ 0 and λ ∈ C+ ∩ ρ(A),

(R(λ,B)u)(t) = R(λ,A) ((R(λ,B)f)(t)) +R(λ,A)(u(t)).

b) sp(u) ⊂ sp(f) ∪ {η ∈ R : iη ∈ σ(A)}.c) If f is uniformly ergodic at η where iη ∈ ρ(A) ∩ iR, then u is uniformly

ergodic at η and Mηu(t) = R(iη,A)(Mηf(t)).

Proof. Note first that

u(s+ t) = u(t) +A(vt(s)) + gt(s), (5.19)

where vt(s) :=∫ s+t

tu(r) dr and gt(s) :=

∫ s+t

tf(r) dr. Moreover, for Reλ > 0,

(R(λ,B)u)(t) =

∫ ∞

0

e−λsu(s+ t) ds = λvt(λ),

and similarly(R(λ,B)f)(t) = λgt(λ).

For λ ∈ C+, the integral∫∞0

e−λsvt(s) ds is absolutely convergent, and it follows

from (5.19) that∫∞0

e−λsA(vt(s)) ds is also absolutely convergent. By Proposition1.6.3, vt(λ) ∈ D(A) and

A(vt(λ)) =

∫ ∞

0

e−λsu(s+ t) ds− gt(λ)− λ−1u(t).

Hence,(λ− A)((R(λ,B)u)(t)) = (R(λ,B)f)(t) + u(t),


and a) follows.

Since f(λ) = (R(λ,B)f)(0) and u(λ) = (R(λ,B)u)(0), it follows from a) that

u(λ) = R(λ,A)(f(λ)

)+R(λ,A)(u(0))

for λ ∈ C+∩ρ(A). If iη ∈ ρ(A) and η /∈ sp(f), then f has a holomorphic extensionnear iη, and then u also has a holomorphic extension. This proves b).

Since Mηf = limα↓0 αR(α+ iη,B)f , c) follows from a).

We now give the extension of Theorem 5.5.3 to inhomogeneous Cauchy prob-lems. Although this case is not a corollary of the Tauberian theorem (Theorem4.7.7), the idea of the proof is similar.

Theorem 5.6.8. Let A be a closed operator on X such that σ(A)∩ iR is countable.Let f : R+ → X be asymptotically almost periodic, and u be a bounded, uni-formly continuous, mild solution of the inhomogeneous Cauchy problem (ACPf ),and suppose that u is uniformly ergodic at η whenever iη ∈ σ(A) ∩ iR. Then

a) u is asymptotically almost periodic, and

iFreq(u) ∩ ρ(A) = iFreq(f) ∩ ρ(A).

b) If τ > 0, f is τ -periodic and iFreq(u) ⊂ ρ(A) ∪ (2πiτ

)Z, then u = u0 + u1,

where u0 is a mild solution of (ACP0), limt→∞ u0(t) = 0, and u1 is a τ-periodic solution of (ACPf ).

c) If iFreq(f) ⊂ σ(A) ∪ {0} and iFreq(u) ⊂ ρ(A) ∪ {0}, then limt→∞ u(t) =R(0, A)(M0f(0)).

d) If iFreq(f) ⊂ σ(A) and iFreq(u) ⊂ ρ(A), then limt→∞ u(t) = 0.

Proof. By Proposition 5.6.7 c), u is uniformly ergodic at η whenever iη ∈ ρ(A)∩ iRso u is totally ergodic. Let B be the generator of the shift semigroup on the spaceE(R+, X) (see Section 4.7). By Proposition 5.6.7 a),

(R(λ,B)u)(t) = (R(λ,B)(R(λ,A) ◦ f))(t) + R(λ,A)(u(t))

for λ ∈ C+ ∩ ρ(A). The space AAP(R+, X) is invariant under the operation ofcomposition with a fixed member of L(X) and under R(λ,B), so R(λ,B)(R(λ,A)◦f) ∈ AAP(R+, X). Hence,

R(λ, B)π(u) = π(R(λ,B)u) = π(R(λ,A) ◦ u)

whenever λ ∈ C+ ∩ ρ(A), where

π : E(R+, X)→ Y := E(R+, X)/AAP(R+, X)


is the quotient map and B is the generator of the C0-group S on E induced by theshift semigroup (see Proposition 4.7.2). This shows that the map λ �→ R(λ, B)π(u)

has a holomorphic extension to a map h : ρ(A)→ E , given by h(λ) = π(R(λ,A)◦u).It now follows as in the proof of Theorem 4.7.7 that π(u) = 0, so u is asymptoticallyalmost periodic.

The fact that iFreq(u) ∩ ρ(A) = iFreq(f) ∩ ρ(A) follows from Proposition5.6.7 c), and the remaining statements follow from Corollary 4.7.8 and Proposition5.4.16.

The following corollary generalizes several parts of Proposition 5.6.1.

Corollary 5.6.9. Let A be a closed operator on X such that σ(A) ∩ iR is empty.Let f ∈ BUC(R+, X), and u be a bounded, uniformly continuous, mild solution ofthe inhomogeneous Cauchy problem (ACPf ).

a) If f is totally ergodic, then u is totally ergodic.

b) If f is asymptotically almost periodic, then u is asymptotically almost peri-odic, and Freq(u) = Freq(f).

c) If f is τ-periodic, then u = u0 + u1, where u0 is a mild solution of (ACP0),limt→∞ u0(t) = 0, and u1 is a τ -periodic solution of (ACPf ).

d) If f∞ := limt→∞ f(t) exists, then limt→∞ u(t) = R(0, A)f∞.

e) If limt→∞ f(t) = 0, then limt→∞ u(t) = 0.

Proof. a) follows from Proposition 5.6.7 c), and the remaining statements fromTheorem 5.6.8.

5.7 Notes

Accounts of the asymptotic behaviour of C0-semigroups have appeared in the books ofDaletskii and Krein [DK74], Levitan and Zhikov [LZ82], Nagel et al. [Nag86], van Neerven[Nee96c], Chicone and Latushkin [CL99] and Eisner [Eis10].

Section 5.1The growth bound ω(T ) appeared in the book of Hille and Phillips [HP57], while ω1(T )arose in papers of D’Jacenko [Jac76] and Zabczyk [Zab79]. It is possible to define higherorder and fractional growth bounds in the following way.

For μ > ω(T ), μ−A is a sectorial operator, and the fractional powers (μ−A)α andR(μ,A)α are defined whenever α ≥ 0 (see the Notes on Section 3.8). Define

ωα(T ) := ω(‖T (·)R(μ,A)α‖)= inf {ω(ux) : x ∈ D((μ−A)α)} .

This is independent of μ > ω(T ), since D((μ−A)α) is independent of μ (see Proposition3.8.2). With only minor modifications, the proof of Proposition 5.1.5 shows that

ωα+1(T ) = sup {abs(ux) : x ∈ D((μ− A)α)} .

5.7. NOTES 387

Moreover, the resolvent identity may be used to show that

hol(uR(μ,A)x) = hol(ux),

from which it follows that

s(A) = inf {hol(ux) : x ∈ D(An)} ≤ ωn(T )

for all positive integers n, and hence that s(A) ≤ ωα(T ) for all α ≥ 0.Theorem 5.1.2 is mostly due to Datko [Dat70], [Dat72], with contributions also from

Pazy [Paz72], van Neerven [Nee96a] and Schuler and Vu [SV98]. The equality of ω1(T )and abs(T ) (Proposition 5.1.6) and Proposition 5.1.5 were established by Neubrander[Neu86]. Theorem 5.1.7 was first proved by Weis and Wrobel [WW96], following prelim-inary results of Slemrod [Sle76] (showing that ω2(T ) ≤ s0(A)) and van Neerven, Strauband Weis [NSW95] (showing that ωα(T ) ≤ s0(A) whenever α > 1). In [WW96], inter-polation theory was used to show that ωα(T ) is a convex function of α and therefore itis continuous for α > 0. Thus, Theorem 5.1.7 followed from the result of [NSW95]. ABanach space X is said to have Fourier type p (where 1 ≤ p ≤ 2) if the Fourier trans-

form defines a bounded linear map from Lp(R, X) into Lp′(R, X), where 1/p+ 1/p′ = 1.Every Banach space has Fourier type 1; every superreflexive space has Fourier type p forsome p > 1; X has Fourier type 2 if and only if X is (isomorphic to) a Hilbert space(see also the Notes on Section 1.8). van Neerven, Straub and Weis [NSW95] showed thatωα(T ) ≤ s0(A) if X has Fourier p and α > 1/p − 1/p′, and Weis and Wrobel [WW96]extended this to the case when α = 1/p−1/p′ and p < 2 (for the case p = 2, see the noteson Section 5.2). van Neerven [Nee09] has given another variation of the result. Assumethat X has (Rademacher) type q1 and cotype q2, where 1 ≤ q1 ≤ 2 and q2 ≥ 2, and let

α = 1/q1−1/2; for example, X may be any Lp-space (1 ≤ p < ∞) and then α =∣∣∣ 1p − 1

2

∣∣∣.Then ωα(T ) is bounded above by an abscissa for the resolvent of A at least as largeas, but often equal to, s0(A). For the definitions and properties of type and cotype, see[Woj91] for example.

Trefethen and Embree [TE05] provide many examples of the behaviour of thepseudo-spectrum (sets where the norm of the resolvent of an operator is large) includingstrong evidence that the pseudo-spectrum is much more stable than the spectrum underperturbations. For α ≥ 0, there is an associated pseudo-spectral bound defined by

sα(A) = inf

{ω > s(A) : there exists Cω such that

‖R(λ,A)‖ ≤ Cω(1 + |λ|)α whenever Re λ > ω

}.

The proofs of Theorem 5.1.7 can be modified to show that ωα+1(T ) ≤ sα(A) for anysemigroup on any Banach space (earlier, Slemrod [Sle76] proved that ωn+2(T ) ≤ sn(A),and Wrobel [Wro89] showed that ωn+1(T ) ≤ sn(A) if X has non-trivial Fourier type).

van Neerven [Nee96b] proved Theorem 5.1.8 by means of Laplace inversion along awell-chosen contour. Both conclusions that growth is at most linear are sharp.

Suppose that x ∈ X, and ux is defined and bounded on C+, and let α > 1. Undercertain additional assumptions, ‖T (t)R(μ,A)αx‖ → 0 as t → ∞. This was established byHuang and van Neerven [HN99] when X has the analytic Radon-Nikodym property, andby Batty, Chill and van Neerven [BCN98] when T is sun-reflexive in the sense of [Nee92];i.e., when R(μ,A) is weakly compact, by a theorem of de Pagter [Pag89]. In particular,

it is true if X is reflexive (the Hilbert space case was first covered by Huang [Hua99]).However, if X = C0(R), T is the shift group: (T (t)f)(s) = f(s + t), and g ∈ Cc(R),then ug exists and is bounded on C+, but ‖T (t)R(μ,A)αg‖ = ‖R(μ,A)αg‖ for all t ≥ 0.Blake [Bla99] has obtained further results related to Theorem 5.1.8. A detailed accountof many of these refinements of Theorems 5.1.7 and 5.1.8 is given in the book of vanNeerven [Nee96c, Chapter 4].

Example 5.1.10 is a modification, due to Wrobel [Wro89], of an example of Zabczyk[Zab75]. Example 5.1.11 is due to Arendt [Are94b], and van Neerven [Nee96d] has ana-lyzed it for many rearrangement-invariant function spaces. The first example of a positiveC0-semigroup with s(A) < ω(T ) was given by Greiner, Voigt and Wolff [GVW81], andthe first example of a positive C0-group by Wolff [Wol81].

Theorem 5.1.12 is a consequence of the spectral mapping theorem:

σ(T (t)) \ {0} ={eμt : μ ∈ σ(A)

},

which is valid for eventually norm-continuous semigroups. It appeared in the book of Hilleand Phillips [HP57] and it can be proved either by product space techniques (see [Nag86])or by Banach algebra methods (see [HP57] or [Dav80]), but we do not know of a proofby Laplace transform methods of either the spectral mapping theorem for holomorphicsemigroups, or Theorem 5.1.12 for eventually norm-continuous semigroups. Greiner andMuller [GM93] obtained a spectral mapping theorem for all exponentially bounded inte-grated semigroups. Martinez and Mazon [MM96], Blake [Bla01] and Nagel and Poland[NP00] have established a version of the spectral mapping theorem for asymptoticallynorm-continuous semigroups (see the Notes on Section 5.2).

Section 5.2Theorem 5.2.1 originated in the work of Gearhart [Gea78] who considered the case whenT is a contraction semigroup on a Hilbert space X. He showed that eμt ∈ ρ(T (t)) if andonly μ + (2πi/t)Z ⊂ ρ(A) and supn∈Z ‖R(A,μ + 2πin/t)‖ < ∞. This was extended tothe non-contractive case with new and simpler proofs, independently by Herbst [Her83],Howland [How84], Huang [Hua85] and Pruss [Pru84]. A consequence of these results isthat if the resolvent of A is bounded on the imaginary axis, then T is “hyperbolic”; i.e.,X splits as a topological direct sum of closed, T -invariant subspaces, X = X− ⊕ X+,such that ω(T |X−) < 0 and there is a C0-group U on X+ with T (t)|X+ = U(−t) andω(U) < 0. A particularly beautiful proof of Theorem 5.2.1 was given by Latushkinand Montgomery-Smith [LM95] which is based on the theory of “evolution semigroups”.See also the monograph by Chicone and Latushkin [CL99] where many far-reachingconsequences are given.

Our short proof of Theorem 5.2.1 is a simplified version of a proof given by Weiss[Wei88]. For higher order growth and spectral bounds, Weiss [Wei90] and Wrobel [Wro89]showed that ωn(T ) = sn(A) for semigroups on Hilbert space. For Banach spaces, Greiner[Gre84] gave a characterisation of σ(T (t)) based on Fejer’s theorem, from which Theorem5.2.1 can be deduced in the case of Hilbert spaces. Examples 5.2.2 and 5.2.3 are due toWeis [Wei98] and Arendt [Are94b], respectively.

Building on preliminary work of Martinez and Mazon [MM96], Blake [Bla01] provedthe following variant of Theorem 5.2.1, showing that the absence of norm-continuity forlarge t is reflected in a rather precise way in the shape of the spectrum and the growthof the resolvent (i.e., the shape of the pseudo-spectrum).

5.7. NOTES 389

For a C0-semigroup T with generator A, let

s∞0 (A) := inf

{ω ∈ R : there exist bω > 0 and Cω > 0 such that λ ∈ ρ(A) and

‖R(μ,A)‖ ≤ Cω whenever Re λ > ω and | Imλ| > bω

},

δ(T ) := inf

{ω ∈ R : there exists Mω > 0 such that

lim suph→0

‖T (t+ h)− T (t)‖ ≤ Mωeωt for all t ≥ 0

}.

We say that T is asymptotically norm-continuous if δ(T ) < ω(T ).

Theorem 5.7.1. Let T be a C0-semigroup on a Hilbert space, with generator A. Thens∞0 (A) = δ(T ).

Section 5.3Part of Theorem 5.3.1 was proved by Greiner, Voigt and Wolff [GVW81], and the remain-der by Neubrander [Neu86]. Example 5.3.2 is due to Arendt [Are94b]. Theorem 5.3.6 wasfirst proved by Weis [Wei95], answering a question which had been open for some time.The proof in [Wei95] used interpolation theory and the theory of evolution semigroupsdeveloped by Latushkin and Montgomery-Smith [LM95]. The simplified proof given herefollows a later method of Weis [Wei96]. Proposition 5.3.5 appeared in a proof given byMontgomery-Smith [Mon96] but it can alternatively be established by methods shown in[Haa07b]. Another variant of the proof of Theorem 5.3.6 is given in [Wei98]. The specialcases of p = 1 (Proposition 5.3.7) and p = 2 had been proved by Derndinger [Der80]and Greiner and Nagel [GN83], respectively. Our proof of Proposition 5.3.7 is taken from[Der80]. Theorem 5.3.8 was proved by Derndinger [Der80] for compact Ω and by Battyand Davies [BD82] for locally compact Ω. Note that the proofs of Proposition 5.3.7 andTheorem 5.3.8 do not use the lattice properties, and the equality s(A) = ω(T ) holds forpositive semigroups on ordered Banach spaces where either the norm on X+ or the normon X∗+ is additive in the sense of (5.13) (see [BD82]). In particular, this is true for C∗-algebras (the result for positive semigroups on unital C∗-algebras was first establishedby Groh and Neubrander [GN81]). For a positive semigroup on a Banach lattice with(Rademacher) type p and cotype q, one has ωα(T ) ≤ s0(A) for α = 1/p − 1/q [Nee09](see the Notes on Section 5.1).

There are many aspects of asymptotic behaviour of positive semigroups which arenot covered in this book. An early account of the theory was given in [Nag86]. See alsothe Notes on Section 5.5.

Section 5.4Splitting theorems for relatively compact orbits of semigroups, such as Theorems 5.4.6and 5.4.11, are often associated with the names of Glicksberg and de Leeuw. In [GL61],they were the first to obtain such a theorem for general Banach spaces, following specialcases due to Jacobs [Jac56, etc]. In those papers, the splitting theory was carried out forvery general semigroups of operators, and the methods were algebraic and topological.For one-parameter semigroups, the Glicksberg-deLeeuw theorem is the following variantof Theorem 5.4.11.

Theorem 5.7.2 (Glicksberg-deLeeuw Theorem). Let T be a bounded C0-semigroup on X,and suppose that {T (t)x : t ≥ 0} is weakly relatively compact for each x ∈ X. Then

X = Xw0 ⊕Xap,


whereXw0 := {x ∈ X : 0 is in the weak closure of {T (t)x : t ≥ 0}} .

Accounts of the general Glicksberg-deLeeuw theory can be found in the books ofKrengel [Kre85, Section 2.4], van Neerven [Nee96c, Section 5.7], Engel and Nagel [EN00,Section V.2] and Eisner [Eis10, Section I.1].

It is easy to deduce the strong splitting theorem (Theorem 5.4.6) from Theorem5.7.2, but comparison with the weak splitting theorem (Theorem 5.4.11) is more delicate.A priori, it is not clear that the spaces Xe0 and Xw0 are contained in each other. However,it is easy to verify that the following properties are equivalent for a totally ergodicsemigroup T on X:

(i) There is a closed T -invariant subspace Y of X such that X = Y ⊕ Xap as atopological direct sum.

(ii) There is a bounded projection P of X onto Xap such that PT (t) = T (t)P for allt ≥ 0.

(iii) X = Xe0 ⊕Xap.

When these properties hold, Y = KerP = Xe0. The weak and strong splitting theoremsand the Glicksberg-deLeeuw theorem provide important special cases when the propertieshold, and they show in particular that Xe0 = Xw0 if the orbits of T are weakly relativelycompact.

There are some totally ergodic C0-semigroups for which the properties (i), (ii) and(iii) above do not hold. In particular, Woodward [Woo74] showed that the space E(R)of all totally ergodic functions in BUC(R) is strictly larger than E0(R) ⊕ AP(R), whereE0(R) is the space of totally ergodic functions whose means are all zero, and therefore(iii) does not hold for the C0-group of shifts on E(R). The weak splitting theorem showsthat there is such a splitting of the space of all Eberlein-w.a.a.p. functions (see the Noteson Section 4.7).

The descriptions of the spaces Xe0 and Xw0 are both rather weak, and the strongestresult has been obtained by Ruess and Summers [RS90b].

Theorem 5.7.3. Let T be a bounded C0-semigroup on X, let x ∈ X, and suppose that{T (t)x : t ≥ 0} is relatively weakly compact. Then ux is Eberlein-w.a.a.p. Moreover,x = x0 + x1, where x1 ∈ Xap and ux0 is Eberlein-w.a.a.p. with means Mηux0 = 0 for allη ∈ R.

Ruess and Summers [RS87], [RS88a], [RS90a], [RS92a], [RS92b] have also carriedout detailed investigations of orbits, and almost-orbits, of non-linear semigroups andsolutions of non-autonomous Cauchy problems. Their results show that Eberlein’s notionof weak asymptotic almost periodicity is important even in those cases.

The construction of trigonometric polynomials as in Proposition 5.4.8 by means ofFejer kernels occurred in Bohr’s book [Boh47] where it was used in a proof (attributedto De La Vallee Poussin) of a fundamental property of almost periodic functions. Othertechniques in the proof of Theorem 5.4.11 appear in the work of Datry and Muraz [DM95],[DM96], who have considered splittings in a very abstract situation.

Section 5.5There is a very large literature on the subject of the local spectrum for a boundedoperator, or a commuting family of bounded operators. The book of Erdelyi and Wang

5.7. NOTES 391

[EW85] includes an account of the theory for unbounded operators. In the literature,the operators are usually assumed to satisfy the “single-valued extension property”, butin the context of Section 5.5, we consider only a peripheral part of the local spectrum,for which we have chosen a definition which makes this property hold automatically (seeProposition 5.5.1) and which is consistent with the notion of spectrum for functions (seeProposition 5.5.2).

An alternative notion of imaginary local spectrum, σu(T, x), of a C0-semigroup Thas been introduced by Batty and Yeates [BY00], using ideas of Albrecht [Alb81]. WhenT is bounded, the definition of σu(T, x) is as follows.

For a bounded semigroup T and f ∈ L1(R+), define f(T ) ∈ L(X) by f(T )x =∫∞0

f(t)T (t)x dt. A point iη ∈ iR is in ρu(T, x) if there exist n ∈ N, f1, . . . , fn ∈ L1(R+),a neighbourhood V of the point ((Ff1)(−η), . . . , (Ffn)(−η)) in Cn and holomorphicfunctions gi : V → X for each i = 1, . . . , n such that

n∑i=1

(zi − fi(T )) gi(z) = x (z = (z1, . . . , zn) ∈ V ).

Then σu(T, x) := iR \ ρu(T, x).While it is easy to see that σu(T, x) is contained in σu(A, x), it remains open whether

equality holds. Theorem 5.5.3 remains valid when σu(A, x) is replaced by σu(T, x) (see[BY00]). Part b) of Theorem 5.5.5 was proved independently by Arendt and Batty [AB88]and Lyubich and Vu [LV88]. The proof in [AB88] was based on the contour integralmethod of Section 4.4, combined with an unusual argument by transfinite induction. Thefunctional analytic proof in [LV88] is related to the quotient method of Sections 4.7 and4.8. Part a) of Theorem 5.5.5 was proved by Lyubich and Vu [LV90a]. Prototypes ofTheorem 5.5.5 for norm-continuous semigroups, some discrete semigroups, and the caseof empty peripheral spectrum had been obtained by Sklyar and Shirman [SS82], Atzmon[Atz84], and Huang [Hua83] (see also [Hua93a], [Hua93b]), respectively.

A different proof of Theorem 5.5.5 b) was subsequently given by Esterle, Strouseand Zouakia [ESZ92]. We summarize their method in the next few paragraphs.

Let E be a closed subset of R. A function f ∈ L1(R+) is said to be of spectralsynthesis with respect to E if there is a sequence (gn) in L1(R) such that limn→∞ ‖gn −f‖1 = 0 and, for each n, Fgn vanishes on a neighbourhood of E. (Here, we are regardingf as a member of L1(R) with f(t) = 0 for t < 0.) If f is of spectral synthesis with respectto E, then Ff vanishes on E. If the boundary of E is countable and Ff vanishes on E,then f is of spectral synthesis with respect to E [Kat68, p.230].

For a bounded semigroup T such that σ(A) ∩ iR is countable, it was shown byEsterle, Strouse, and Zouakia [ESZ92], using an abstract Mittag-Leffler theorem (a gen-eralization of Baire’s category theorem), that the linear span of the union of the rangesof f(T ) for all f which are of spectral synthesis with respect to iσ(A)∩R is dense in X.Then, Theorem 5.5.5 b) follows from the following analogue of the Katznelson-Tzafriritheorem [KT86].

Theorem 5.7.4. Let T be a bounded C0-semigroup on X and let f ∈ L1(R+) be of spectralsynthesis with respect to iσ(A) ∩ R. Then ‖T (t)f(T )‖ → 0 as t → ∞.

Theorem 5.7.4 was proved in [ESZ92] using methods of harmonic analysis, and in[Vu92] using a functional analytic method. In the next paragraph, we sketch a proof


which uses the ideas of Section 4.9, in particular a version of Ingham’s Theorem 4.9.5 forfunctions on R.

Let g ∈ L1(R) be such that Fg vanishes in a neighbourhood of iσ(A) ∩ R, and let(g ∗ T )(t)x =

∫∞0

g(s − t)T (s)x ds. Then g ∗ T ∈ BUC(R, X), and it is not difficult toshow that g ∗ T has distributional Fourier transform

H(s) =

{(Fg)(−s)R(is, A) (is ∈ ρ(A)),

0 (is ∈ σ(A)),

in the sense that ∫ ∞

−∞(g ∗ T )(t)(Fϕ)(t) dt =

∫ ∞

−∞H(s)ϕ(s) ds

for all ϕ ∈ C∞c (R). If ρ ∈ S(R) is such that Fρ has compact support, then (Fρ) ·H ∈ L1(R,L(X)) with F−1((Fρ) · H) = ρ ∗ (g ∗ T ) ∈ C0(R,L(X)) by the Riemann-Lebesgue lemma. Using a mollifier (ρn), it follows that g ∗ T ∈ C0(R,L(X)). Choosing gto approximate f , it follows that T (t)f(T ) = (f ∗ T )(t) → 0 as t → ∞.

Another proof of Theorem 5.5.5 b) using the abstract Mittag-Leffler theorem hasbeen given in [BCT02]. This was one of a series of papers, also including [Tom01], [CT03],[CT04] and [BCT07], in which Chill and Tomilov investigated stability of semigroups inthe sense of strong operator convergence to 0. An excellent survey is given in [CT07]. Forgeneral Banach spaces the main results are as follows.

Theorem 5.7.5. Let T be a bounded C0-semigroup T on X, with generator A. Thenlimt→∞ ‖T (t)x‖ = 0 for all x ∈ X if any of the following conditions holds:

a)

{x ∈ X : lim

α↓0

∥∥αR(α+ is, A)2x∥∥ = 0 for all s ∈ R

}is dense in X.

b) For some γ > 1,

{x ∈ X : lim

α↓0

∫R

∥∥αγ−1R(α+ is,A)γx∥∥ ds = 0

}is dense in X.

c) For each s0 ∈ R, there exists a neighbourhood U of s0 in R such that{x ∈ X : lim

α↓0

∫U

∥∥αR(α+ is, A)2x∥∥ ds = 0

}is dense in X.

The assumptions of Theorem 5.5.5 b) imply that⋂

s∈R Ran(A−is) is dense in X, bythe abstract Mittag-Leffler theorem. If x ∈ Ran(A−is), then limα↓0

∥∥αR(α+ is, A)2x∥∥ =

0. So Theorem 5.5.5 b) follows from case a) of Theorem 5.7.5, which is proved by meansof an edge-of-the-wedge theorem.

Theorem 5.7.5 can be improved when X has non-trivial Fourier type. For example,for a Hilbert space X , density of{

x ∈ X : limα↓0

∥∥√αR(α+ is,A)x∥∥ = 0 for all s ∈ R

}is sufficient for stability. Moreover there is a variant of the integral condition b) of The-orem 5.7.5 which is both necessary and sufficient on Hilbert spaces. We refer the readerto the survey [CT07], or the original articles, for the details.

Greenfield [Gre94] (see also [BBG96]) proved the following quantitative version ofTheorem 5.5.5 b).

5.7. NOTES 393

Theorem 5.7.6. Let T be a C0-semigroup of contractions on X, and suppose that σ(A)∩iRis countable. Let X∗1 be the weak*-closed linear span of the unimodular eigenvectors ofthe dual semigroup T ∗ on X∗. Then

limt→∞

‖T (t)x‖ = inf {‖x− y‖ : y ∈ X0}= sup {|〈x, x∗〉| : x∗ ∈ X∗1 , ‖x∗‖ ≤ 1}

for each x ∈ X.

A consequence of Theorem 5.5.6 is that if X is superreflexive (i.e., if there is anequivalent uniformly convex norm on X), and A generates a bounded C0-semigroup Ton X and σ(A) ∩ iR is countable, then every ultrapower of T is asymptotically almostperiodic. Huang and Rabiger [HR94] proved a converse result, that σ(A)∩iR is countableif every ultrapower of T is asymptotically almost periodic (the discrete version was givenearlier by Nagel and Rabiger [NR93]; see also the work of Rabiger and Wolff [RW95]).

There was a considerable interval before the global Theorem 5.5.5 was improved tothe local Theorem 5.5.3. Batty and Vu [BV90] gave a few results on individual orbits,and an intermediate stage between the global and the local was considered by Huang[Hua93a], [Hua93b] and Batty [Bat96]. Theorem 5.5.3 was first proved by Batty, vanNeerven and Rabiger [BNR98b], and the method given here is from [AB99].

Theorems 5.5.5 and 5.5.3, and their discrete analogues, have been extended in var-ious directions: to weighted results (showing that ‖T (t)x‖/w(t) → 0 where w is a weighton R+ and ‖T (t)‖ ≤ w(t)), by Allan and Ransford [AR89], Vu [Vu93], Kerchy [Ker97]and Batty and Yeates [BY00]; to representations of subsemigroups of locally compactabelian groups, by Lyubich and Vu [LV90b], Batty and Vu [BV92], Batty and Yeates[BY00] and Kerchy [Ker99]; to once integrated semigroups, by El-Mennaoui [Elm94] (us-ing Theorem 5.5.5 and the extrapolation construction of Section 3.10); and to Volterraequations, by Arendt and Pruss [AP92].

Examples 5.5.7–5.5.10 are taken from papers of Arendt and Batty [AB88], Battyand Vu [BV90], and Batty, van Neerven and Rabiger [BNR98b]. Using a direct sum ofweighted shifts, van Neerven [Nee00] has extended Example 4.7.11 to give an exampleof a C0-semigroup T with a vector x such that ux is bounded and uniformly continuous,σu(A, x) = {0}, and limα↓0 αT (s)ux(α) exists for each s ≥ 0, but ux is not asymptoticallyalmost periodic. Thus, the assumption of uniform convergence cannot be omitted fromcondition c) of Theorem 5.5.3.

Perron-Frobenius theory provides a special interplay between spectral propertiesof the generator and asymptotic behaviour of a positive semigroup. The lecture notes[Nag86] give a complete account of the state of the art in 1986. They were written beforethe results of Section 5.5 on countable spectrum were discovered. Consequently someresults can now be formulated or proved more easily, and we describe some of them here.

A basic result in Perron-Frobenius theory is the cyclicity of the boundary spectrum.If A generates a bounded positive C0-semigroup and is ∈ σ(A) for some s ∈ R, then ins ∈σ(A) for all n ∈ Z. If the semigroup is eventually norm-continuous (i.e., norm-continuouson (t0,∞) for some t0 ≥ 0), then it follows that σ(A) ∩ iR ⊂ {0}. The following resultwas proved in [Nag86] by means of the discrete Katznelson-Tzafriri theorem (Corollary4.7.15), but it now has the same proof as Theorem 5.5.6 b).

Theorem 5.7.7. Let T be a bounded, positive, eventually norm-continuous, Cesaro-ergodicC0-semigroup on a Banach lattice. Then there is a projection P such that limt→∞ T (t) =P in the strong operator topology.

A positive C0-semigroup on a Banach lattice X is said to be irreducible if the onlyclosed T–invariant ideals of X are {0} and X (see Appendix C for the definition ofideals). If A generates a bounded, positive, irreducible, C0-semigroup and A has compactresolvent, then Perron-Frobenius theory establishes that σ(A) ∩ iR is either empty orequal to 2πiτZ for some τ ≥ 0. The following is then an easy consequence of Proposition5.4.7 (see [Are04, Sections 3.4, 3.5]).

Theorem 5.7.8. Let T be a bounded, positive, irreducible C0-semigroup on a Banach latticeX, and assume that the generator A has compact resolvent. Then there is a periodic C0-group U on a closed subspace Y of X such that, for each x ∈ X there exists y ∈ Y suchthat limt→∞ ‖T (t)x− U(t)y‖ = 0.

Let X = Lp(Ω, μ), where (Ω, μ) is a σ-finite measure space and 1 ≤ p < ∞. Abounded operator S onX is a kernel operator if there is a measurable function k : Ω×Ω →C such that (Sf)(y) =

∫Ωk(y, z) dμ(z) for μ-almost all y ∈ Ω and all f ∈ X. For abstract

characterizations of kernel operators, see [Sch74, Proposition IV.9.8] or [Mey91, Theorem3.3.11]. All bounded operators on �p are kernel operators. Many elliptic operators generateirreducible C0-semigroups of kernel operators, and the properties of the kernels have beenmuch studied (see [Dav90], [Ouh05], [Rob91]).

The following result is due to Greiner [Gre82] (see also [Are08]).

Theorem 5.7.9. Let T be a positive, bounded, irreducible C0-semigroup on Lp(Ω, μ), withgenerator A. Suppose that T (t0) is a kernel operator for some t0 > 0 and that 0 ∈ σp(A).Then there is a projection P of rank one such that limt→∞ T (t) = P in the strong operatortopology.

It is remarkable that any positive, bounded C0-semigroup of kernel operators onLp(Ω, μ) satisfies σp(A) ∩ iR ⊂ {0}. This is also due to Greiner [Gre82] and it wasdiscussed more recently in [Dav05], [Kei06], [Wol07] and [Are08]. The Gaussian semigroupon L1(Rn) is positive, contractive and irreducible, and it consists of kernel operators (seeExample 3.7.6). However 0 is not an eigenvalue of its generator, and the semigroup doesnot converge.

It should be mentioned that for cosine functions countable spectrum also gives theexpected asymptotic behaviour. The following result is due to Arendt and Batty [AB97,Proposition 4.9] (see also [Bas85, Theorem 10]).

Theorem 5.7.10. Let Cos be a bounded cosine function on a Banach space X with gener-ator A. Assume that the following conditions hold:

a) c0 �⊂ X;

b) σ(A) is countable; and

c) 0 �∈ σ(A).

Then for each x, y ∈ X the function

u(t) := Cos(t)x+ Sin(t)y (t ∈ R)

is almost periodic.

5.7. NOTES 395

Note that u is the unique mild solution of P 2(x, y) as defined in Section 3.14. Here,

Sin(t)y :=

∫ t

0

Cos(s)y ds (t ∈ R).

Since Cos is bounded, one knows that σ(A) ⊂ (−∞, 0]. An equivalent formulation of thetheorem is to say that, under the conditions a), b) and c), one has

X = span{x ∈ D(A) : there exists η ∈ R such that Ax = −η2x

}.

Surveys of the topics of this section have previously been written by Batty [Bat94],Vu [Vu97], van Neerven [Nee96c, Chapter 5], and Chill and Tomilov [CT07].

Section 5.6A version of Theorem 5.6.5 was first given by Basit [Bas97] under the stronger assump-tion that f has an extension g ∈ BUC(R, X) such that σ(A) ∩ i spC(g) is empty (seeSection 4.6). Basit’s method originated in Lyapunov’s finite-dimensional theory, and itwas developed and applied to infinite-dimensional Cauchy problems on the line by Vu[Vu91], Ruess and Vu [RV95] and Schuler and Vu [SV98]. It involves solving operatorequations of the form AY − Y B = C (Lyapunov equations) for an operator Y from asubspace of BUC(R, X) into X. The proof of Theorem 5.6.5 given here is due to Battyand Chill [BC99] who carried out the basic argument in a more general situation inwhich T is not necessarily a semigroup. It shows that the conclusion of boundedness inTheorem 5.6.5 is valid, not only for holomorphic semigroups, but also for many even-tually norm-continuous semigroups and for asymptotically norm-continuous semigroupson Hilbert space. Blake [Bla99] has further refined the method and extended Theorem5.6.5 to various classes of asymptotically norm-continuous semigroups on Banach spaces,including eventually differentiable semigroups (for the definition of asymptotically norm-continuous semigroups, see the Notes on Section 5.2 above). Theorem 5.6.6 also appearedin [BC99]. The role of smooth functions in the proof also leads to another result of [BC99]that T ∗ f is bounded and uniformly continuous if T is bounded, R(iη, A) exists and isbounded for large |η|, and f is bounded with bounded derivatives of first and secondorder.

Proposition 5.6.7 and Theorem 5.6.8 are due to Arendt and Batty [AB99], followingresults of Ruess and Vu [RV95] for inhomogeneous Cauchy problems on R. Batty, Hutterand Rabiger [BHR99] obtained a version for periodic Cauchy problems. Applications toinhomogeneous Volterra equations have been given by Arendt and Batty [AB00], Chilland Pruss [CP01] and Fasangova and Pruss [FP01].

Part III

Applications and Examples

In this part of the book we present some applications and examples which illustratehow the theory developed in Parts I and II can be used. There are three chapterswhich are independent of each other; they all use basic concepts from distributiontheory which can be found in Appendix E.

In Chapter 6 the heat equation with inhomogeneous boundary conditions isinvestigated. The idea of the approach presented here is to work entirely in spacesof continuous functions. We assume that Ω is a bounded open set on which theDirichlet problem is well-posed. This is a very weak regularity assumption on theboundary, and it is well known from potential theory. Based on this assumption,the results of this chapter rely on the methods developed in Parts I and II anddo not use complicated results of partial differential equations. Resolvent positiveoperators (as developed in Section 3.11) play an important role giving the transi-tion from the elliptic problem to a parabolic problem. Results of Part II will beused to show how the asymptotic behaviour of the given function on the boundarydetermines the asymptotic behaviour of the solution.

In the approach of Chapter 6 we do not use Hilbert space techniques at all.This is different in Chapter 7 where we prove well-posedness of a fairly general hy-perbolic equation in L2(Ω). The results are based on the theory of cosine functionsas they are presented in Section 3.14. Most important is the role of the phase spaceintroduced there. We need a brief introduction to quadratic form methods whichis given in Section 7.1. Here we only consider the most simple case; the spectraltheorem for selfadjoint operators (as stated in Appendix B) plays a major role inthis introduction. Then the results of Chapter 7 follow from the general theory ofcosine functions given in Section 3.14.

In Chapter 8 differential operators with constant coefficients on Rn, and moregenerally pseudo-differential operators, are considered. With the help of the notionof integrated semigroups, precise results on well-posedness and regularity of thecorresponding parabolic problem in Lp-spaces are obtained. In particular, it will beshown that the wave equation is not well-posed in the semigroup sense on Lp(Rn)for p �= 2. This explains why Chapter 7 is restricted to L2(Ω). However, it will beshown that the wave equation as well as some other equations from mathematicalphysics lead to k-times integrated semigroups on Lp-spaces. An important issue isto determine the best possible value of k depending on p. This tells us somethingabout the regularity properties of the equations which we consider. A principaltool in Chapter 8 is the theory of Fourier multipliers. A resume of some of therequired results, including Mikhlin’s theorem about Fourier multipliers on Lp(Rn)for 1 < p < ∞, is given in Appendix E without proofs. Some further results onFourier multipliers are proved in Section 8.2, including a weak form of Mikhlin’stheorem which is valid for L1(Rn).

Part III 399

Chapter 6

The Heat Equation

In this chapter we consider the Laplacian on spaces of continuous functions. IfΩ ⊂ Rn is an open, bounded set with boundary ∂Ω which is Dirichlet regular, wewill show that the Laplacian generates a holomorphic semigroup on the space

C0(Ω) := {u ∈ C(Ω) : u|∂Ω = 0}.

Furthermore, using the theory of resolvent positive operators developed in Section3.11 we show that the heat equation with inhomogeneous boundary conditions iswell-posed. We use the results of Chapter 5 to study the asymptotic behaviour ofits solutions.

6.1 The Laplacian with Dirichlet Boundary Conditions

Let Ω ⊂ RN be an open, bounded set with boundary ∂Ω =: Γ. Given ϕ ∈ C(Γ),we consider the Dirichlet problem

D(ϕ)

⎧⎪⎨⎪⎩u ∈ C(Ω),

u|Γ = ϕ,

Δu = 0 in D(Ω)′.Here and throughout this chapter, D(Ω)′ is the space of all distributions on Ω, andwe identify functions in C(Ω) with their restrictions to Ω and locally integrablefunctions on Ω with distributions on Ω. Thus, C(Ω) ⊂ C(Ω) ⊂ L1

loc(Ω) ⊂ D(Ω)′,and the third line of D(ϕ) says that

∫ΩuΔψ dx = 0 for all ψ ∈ D(Ω) (see Appendix

E). Although we allow complex-valued functions ϕ and u, D(ϕ) is essentially areal problem; u is a solution of D(ϕ) if and only if Reu and Imu are solutions ofD(Reϕ) and D(Imϕ) respectively. It is well known that a function satisfying D(ϕ)is in C∞(Ω) (see [Rud91, p.220], for example). So we look for harmonic functionsin Ω having a continuous extension to the boundary with prescribed boundary


402 6. THE HEAT EQUATION

values. We will not talk about methods to solve the Dirichlet problem. For us,it serves as a reference problem. In fact, this problem is well studied in potentialtheory (see [DL90, Chapter II], [Hel69], [GT83], [Kel67], [Lan72]).

Many geometric properties of the boundary are known to be sufficient forwell-posedness of the Dirichlet problem.

Definition 6.1.1. The set Ω is called Dirichlet regular if for all ϕ ∈ C(Γ) thereexists a solution of D(ϕ).

Examples 6.1.2. a) If n = 1, then each bounded open set is Dirichlet regular (see[DL90, Chapter II, Section 4, Example 6]). On the other hand, if n ≥ 2 and Ω ⊂ Rn

is open, then Ω \ {z} is not Dirichlet regular for any z ∈ Ω (see [DL90, ChapterII, Section 4, Remark 1]).

b) If the boundary of Ω is C1, or more generally, Lipschitz continuous, then Ω isDirichlet regular (see [DL90, Chapter II, Section 4, Proposition 4]).

c) For n = 2, any simply connected Ω ⊂ R2 is Dirichlet regular (see [Con73,Chapter X, Corollary 4.18]).

d) For n = 3, Lebesgue’s cusp is a simply connected set which is not Dirichletregular (see [Lan72, p.287], [AD08]).

Next, we establish the elliptic maximum principle. It will be important forus to consider distributional inequalities which are easy to define. Let f ∈ D(Ω)′.We write

f ≥ 0 if 〈ϕ, f〉 ≥ 0 for all ϕ ∈ D(Ω)+, (6.1)

where D(Ω)+ := {ϕ ∈ D(Ω) : ϕ(x) ≥ 0 for all x ∈ Ω}. If f ∈ L1loc(Ω) is identified

with a distribution in D(Ω)′, then f ≥ 0 as a distribution if and only if f(x) ≥ 0a.e. in Ω.

Theorem 6.1.3 (Elliptic maximum principle). Let M ≥ 0, λ ≥ 0, u ∈ C(Ω) suchthat

a) λu−Δu ≤ 0 in D(Ω)′; andb) u|Γ ≤M .

Then u ≤M on Ω.

Proof. By considering the real and imaginary parts of u separately, we may assumethat u is real-valued. Let c := maxx∈Ω u(x).

First case: We assume that u ∈ C2(Ω). Assume that c > M . Let γ >√λ and

δ := supx∈Ω eγx1 (where x = (x1, . . . , xn)). Choose ε > 0 such that M + εδ < c.Let v(x) := u(x) + εeγx1 . Then v ∈ C2(Ω) ∩ C(Ω) and v ≤ M + εδ < c on Γ,but maxx∈Ω v(x) ≥ c. Thus, there exists x0 ∈ Ω such that v(x0) = maxΩ v(x). It

6.1. THE LAPLACIAN WITH DIRICHLET BOUNDARY CONDITIONS 403

follows that D2jv(x0) =

d2

dt2v(x0 + tej) ≤ 0 (where ej = (0, 0, . . . , 1, 0, . . . , 0)), and

so Δv(x0) ≤ 0. Hence,

0 ≤ λv(x0)−Δv(x0) = λu(x0)−Δu(x0) + εeγx01(λ− γ2)

≤ εeγx01(λ− γ2),

which is a contradiction.Second case: Now let u be arbitrary and assume that c > M . Let

K := {x ∈ Ω : u(x) = c}.Then assumption b) implies that K is a non-empty compact subset of Ω. LetΩ′ ⊂ Ω be open such that K ⊂ Ω′ ⊂ Ω′ ⊂ Ω. Then c1 := sup∂Ω′ u(x) < c. Letc1 < c2 < c, c2 > M . Denote by (ρk)k∈N a mollifier in C∞c (Rn) with ρk ≥ 0 andsupp ρk ⊂ {y ∈ Rn : |y| < 1/k} (see Section 1.3). Then

vk(x) := (ρk ∗ u)(x) =∫|y|<1/k

u(x− y)ρk(y) dy

is defined for x ∈ Ωk := {y ∈ Ω : dist(y,Γ) > 1/k} and vk ∈ C∞(Ωk). Moreover,vk converges to u uniformly on compact subsets of Ω as k → ∞. Hence, thereexists k ∈ N such that Ω′ ⊂ Ωk, supΩ′ vk > c2 and sup∂Ω′ vk < c2. But

λvk(x)−Δvk(x) = 〈ρk(x− ·), λu−Δu〉 ≤ 0

for all x ∈ Ω′. This is impossible by the first case.

It follows immediately from Theorem 6.1.3 thatD(ϕ) has at most one solutionfor all each ϕ ∈ C(Γ) and the solution is real if ϕ is real-valued.

Next, we consider the space X := C(Ω)×C(Γ) which is a Banach lattice forthe ordering

(u, ϕ) ≥ 0⇐⇒ u ≥ 0 and ϕ ≥ 0

(u ∈ C(Ω), ϕ ∈ C(Γ)) and the norm

‖(u, ϕ)‖ := max{‖u‖C(Ω), ‖ϕ‖C(Γ)

},

with

‖u‖C(Ω) := maxx∈Ω

|u(x)|,‖ϕ‖C(Γ) := max

z∈Γ|ϕ(z)|.

On C(Ω) we consider the Laplacian Δmax with maximal domain; i.e.,

D(Δmax) := {u ∈ C(Ω) : Δu ∈ C(Ω)},Δmaxu := Δu in D(Ω)′.

It is obvious that Δmax is a closed operator.


Remark 6.1.4. It is known that D(Δmax) �⊂ C2(Ω) whenever Ω is a non-emptyopen set in Rn (n ≥ 2). However, we will see in Lemma 6.1.5 that D(Δmax)is contained in C1(Ω) (cf. Remark 3.7.7 b)). Thus, there always exist functionsu ∈ C1(Ω) such that Δu ∈ C(Ω) but for some i, j, the distribution DiDju is nota function in C(Ω). This fact may be considered as unpleasant. However, for ourpurposes it does not matter. We will see that solutions of the heat equation arealways of class C∞.

We consider the operator A on X given by

D(A) := D(Δmax)× {0},A(u, 0) := (Δu,−u|Γ).

Thus, for u ∈ D(Δmax), f ∈ C(Ω), ϕ ∈ C(Γ), we have −A(u, 0) = (f, ϕ) if andonly if {

−Δu = f in D(Ω)′,u|Γ = ϕ ;

(6.2)

i.e., if and only if u solves Poisson’s equation. For this reason we call A the Poissonoperator. Since Δmax is closed, it follows that A is also closed.

By En we denote the Newtonian potential; i.e., En : Rn \ {0} → R is givenby

En(x) :=

⎧⎪⎪⎪⎨⎪⎪⎪⎩|x|/2 if n = 1,log |x|2π

if n = 2,

− 1

n(n− 2)ωn

1

|x|n−2if n ≥ 3,

where ωn := |B(0, 1)| is the volume of the unit ball in Rn. Then En ∈ C∞(Rn\{0})and En, DjEn ∈ L1

loc(Rn) (j = 1, . . . , n), as is easy to see.

Let f ∈ Cc(Rn). Then one has v := En ∗ f ∈ C1(Rn). Moreover,

Δv = f in D(Rn)′. (6.3)

We refer to ([DL90, Chapter II, Section 3]) for this standard fact of distributiontheory. Frequently, v = En∗f is called the Newtonian potential of f . Note however,that v �∈ C2(Rn) in general.

We deduce the following regularity result which will be useful. In the proofand elsewhere in this chapter, we do not distinguish notationally between functionson Ω and their restrictions to Ω′ ⊂ Ω.

Lemma 6.1.5. Let u, f ∈ C(Ω) such that Δu = f in D(Ω)′. Then u ∈ C1(Ω). Iff ∈ Ck(Ω) for some k ∈ N, then u ∈ Ck+1(Ω).

Proof. a) Let Ω′ be open such that Ω′ ⊂ Ω. Let ρ ∈ D(Ω) such that ρ(x) = 1 onΩ′. Consider ρf ∈ Cc(Rn) and v := En ∗ (ρf) ∈ C1(Rn). Then Δv = ρf in D(Rn)′

(see (6.3)). Hence,Δ(u− v) = f − ρf = 0 in D(Ω′)′.

Thus, u−v is harmonic and so in C∞(Ω′). Consequently, u = (u−v)+v ∈ C1(Ω′).Since Ω′ is arbitrary, the first assertion is proved.

b) We prove the second assertion. It is true for k = 0 by a). Assume thatit holds for k ∈ N0. Assume that f ∈ Ck+1(Ω). Then u ∈ C1(Ω) by a) andΔDju = DjΔu = Djf in D(Ω)′. Since Djf ∈ Ck(Ω), it follows that Dju ∈Ck+1(Ω) (j = 1, . . . , n) by the inductive hypothesis. Hence, u ∈ Ck+2(Ω).

Theorem 6.1.6. Assume that Ω is Dirichlet regular. The Poisson operator A isresolvent positive and s(A) < 0.

Proof. a) Let λ ≥ 0 and suppose that λ ∈ ρ(A). Then R(λ,A) ≥ 0. In fact, letf ∈ C(Ω), ϕ ∈ C(Γ), (u, 0) = R(λ,A)(f, ϕ). Then λu − Δu = f in D(Ω)′ andu|Γ := ϕ. If f ≤ 0 and ϕ ≤ 0, it follows from Theorem 6.1.3 that u ≤ 0 in Ω.

b) We show that 0 ∈ ρ(A). Let f ∈ C(Ω) and ϕ ∈ C(Γ). Let f ∈ Cc(Rn) bean extension of f . Let w := −En ∗ f . Then w ∈ C(Rn) and −Δw = f in D(Ω)′.Let v be the solution of the Dirichlet problem D(ϕ − ψ), where ψ = w|Γ. Thenu = v + w ∈ C(Ω), u|Γ = ϕ and

Δu = Δv +Δw = Δw = −f in D(Ω)′.We have shown that A is surjective. It follows from Theorem 6.1.3 that A isinjective. Hence, 0 ∈ ρ(A) since A is closed.

c) Let Q := R+ ∩ ρ(A). Then by a), R(λ,A) ≥ 0 for all λ ∈ Q and

R(0, A)−R(λ,A) = λR(λ,A)R(0, A) ≥ 0,

hence 0 ≤ R(λ,A) ≤ R(0, A) for all λ ∈ Q. Thus, ‖R(λ,A)‖ ≤ ‖R(0, A)‖ for allλ ∈ Q. By Corollary B.3, it follows that |λ − μ| ≥ ‖R(λ,A)‖−1 ≥ ‖R(0, A)‖−1

for all λ ∈ Q, μ ∈ σ(A). Since 0 ∈ Q, this implies that Q = R+. It follows fromProposition 3.11.2 that s(A) < 0.

In the remainder of this chapter we assume that Ω is Dirichlet regular.The Poisson operator is not densely defined and is not a Hille-Yosida operator

since‖λR(λ,A)‖ ≥ λ (λ > 0).

In fact, let (u, 0) = R(λ,A)(0, 1Γ). Then u|Γ = 1Γ. Hence by Theorem 6.1.3,

‖λR(λ,A)‖ ≥ ‖λ(u, 0)‖ = λ.

Moreover, since the polynomials are dense in C(Ω) by the Stone-Weierstrass the-orem, it follows that

D(A) = C(Ω)× {0}. (6.4)

If we consider the part Ac of A in D(A) = C(Ω) × {0}, then we obtain aHille-Yosida operator as we shall see in the next theorem. The operator Ac is givenby

D(Ac) = {(u, 0) : u ∈ D(Δmax), u|Γ = 0} ,Ac(u, 0) = (Δmaxu, 0).

Since R(λ,A)X ⊂ D(A), it follows from Proposition B.8 that (s(A),∞) ⊂ ρ(Ac)and R(λ,Ac) = R(λ,A)|

D(A)for all λ > s(A). Thus, Ac is a resolvent positive

operator and s(Ac) ≤ s(A) < 0.Identifying C(Ω) × {0} with C(Ω), Ac is identified with the operator Δc on

C(Ω) given by

D(Δc) = C0(Ω) ∩D(Δmax),

Δcu = Δu in D(Ω)′.Here, C0(Ω) = {u ∈ C(Ω) : u|Γ = 0}.Theorem 6.1.7. Assume that Ω is Dirichlet regular. Then the operator Δc on C(Ω)is dissipative and resolvent positive, and s(Δc) < 0.

Proof. We have established above that Δc is resolvent positive and s(Δc) < 0. Itremains to show that Δc is dissipative. Let t > 0, u ∈ D(Δc), u − tΔu = f . Wehave to show that ‖u‖C(Ω) ≤ ‖f‖C(Ω). Let M := ‖f‖C(Ω). Let θ ∈ [0, 2π] and

v := Re(eiθu). Then (v−M)− tΔ(v−M) = Re(eiθf)−M ≤ 0 in D(Ω)′. It followsfrom Theorem 6.1.3 that v −M ≤ 0; i.e., Re(eiθu) ≤M for all θ. We deduce that‖u‖C(Ω) ≤M .

Note that the operator Δc is also not densely defined. So we consider thepart Δ0 of Δc in C0(Ω) = D(Δc). Then Δ0 is given by

D(Δ0) = {u ∈ C0(Ω) : Δu ∈ C0(Ω)} ,Δ0u = Δu in D(Ω)′.

Since D(Ω) ⊂ D(Δ0), the operator Δ0 is densely defined. Moreover, for λ >s(Δc), R(λ,Δc)C0(Ω) ⊂ D(Δc) ⊂ C0(Ω). Consequently, (s(Δc),∞) ⊂ ρ(Δ0) andR(λ,Δ0) = R(λ,Δc)|C0(Ω) for all λ > s(Δc). Hence, ‖λR(λ,Δ0)‖ ≤ ‖λR(λ,Δc)‖≤ 1 for λ > 0. Applying the Hille-Yosida theorem we obtain the following result.

Theorem 6.1.8. Assume that Ω is Dirichlet regular. Then the operator Δ0 generatesa positive contractive C0-semigroup T0 on C0(Ω).

Next, we prove holomorphy of T0. More generally, the following holds.

Theorem 6.1.9. Assume that Ω is Dirichlet regular. Then Δc generates a boundedholomorphic semigroup Tc on C(Ω). The operator Δ0 generates a bounded holo-morphic C0-semigroup on C0(Ω).

Proof. We recall from Example 3.7.6 that the Laplacian generates a bounded holo-morphic C0-semigroup on C0(Rn); i.e., defining the operator L := ΔC0(Rn) onC0(Rn) by

D(L) := {f ∈ C0(Rn) : Δf ∈ C0(R

n)} ,Lu := Δu in D(Rn)′,

there exists M ≥ 0 such that

λ ∈ ρ(L) and ‖λR(λ,L)‖ ≤M (6.5)

whenever Reλ > 0.We show a similar estimate for Δc. Let f ∈ C(Ω), Re λ > 0, g := R(λ,Δc)f .

Let f ∈ C0(Rn) be an extension of f with ‖f‖C0(Rn) = ‖f‖C(Ω) and let g :=

R(λ,L)f . Let ϕ := g|Γ and (w, 0) = R(λ,A)(0, ϕ), where A is the Poisson operatoron X = C(Ω)×C(Γ). Then w ∈ C(Ω), λw−Δw = 0 in D(Ω)′ and w|Γ = ϕ = g|Γ.Thus, h := g − w ∈ C0(Ω) and λh −Δh = λg −Δg = f in D(Ω)′. It follows thath = g. Observe that

‖R(λ,A)‖ ≤ ‖R(0, A)‖by Corollary 3.11.3 (since A is resolvent positive and s(A) < 0). Hence,

‖w‖C(Ω) ≤ ‖R(0, A)‖ ‖ϕ‖C(Γ) ≤ ‖R(0, A)‖ ‖g‖C(Ω).

Hence, setting c := 1 + ‖R(0, A)‖ we have,

‖g‖C(Ω) = ‖g − w‖C(Ω) ≤ c‖g‖C(Ω) ≤cM

|λ| ‖f‖C0(Rn) =cM

|λ| ‖f‖C(Ω)

by (6.5).We have shown that

‖λR(λ,Δc)‖ ≤ cM (Reλ > 0). (6.6)

It follows from Corollary 3.7.12 that Δc generates a bounded holomorphic semi-group Tc on C(Ω). Since Δ0 is the part of Δc in D(Δc) = C0(Ω), the secondassertion is an immediate consequence (see Remark 3.7.13).

We should mention that

‖Tc(t)‖ ≤ 1 for all t > 0. (6.7)

This follows from Proposition 3.7.16. In fact, one can show that Δc is dissipative(regularizing as in Theorem 6.1.3). We will not use (6.7) and we omit the proof.Moreover,

T0(t) = Tc(t)|C0(Ω) (t > 0). (6.8)

This follows from (3.46). It follows from Theorem 5.1.12 or Theorem 5.3.8 thats(Δ0) = ω(T0). Since s(Δ0) < 0, we conclude that ω(T0) < 0. Since Tc is holo-morphic we have Tc(t)C(Ω) ⊂ D(Δc) ⊂ C0(Ω) for all t > 0, hence Tc(t + s) =T0(t)Tc(s) for all t > 0, s > 0. We conclude that

‖Tc(t)‖ ≤Me−εt (t ≥ 0) (6.9)

for some M > 0 and ε > 0.Finally, as a consequence of Theorem 6.1.9 we note the following regularity

result.


Proposition 6.1.10. Assume that Ω is Dirichlet regular. Let u0 ∈ C0(Ω), and letu(t, x) = (T0(t)u0)(x) (t ≥ 0, x ∈ Ω). Then⎧⎪⎨⎪⎩

u ∈ C∞((0,∞)× Ω) ∩ C(R+ × Ω),

ut(t, x) = Δu(t, x) (t > 0, x ∈ Ω),

u(0, x) = u0(x) (x ∈ Ω).

(6.10)

Proof. Since T0 is holomorphic, for f ∈ C0(Ω) we have T0(·)f ∈ C∞((0,∞), D(Δk0))

for all k ∈ N. It follows from Lemma 6.1.5 that D(Δk0) ⊂ Ck(Ω). The closed graph

theorem implies that for each open set Ω′ such that Ω′ ⊂ Ω the restriction map ofD(Δk

0) into Ck(Ω′) is continuous where D(Δk0) carries the graph norm and Ck(Ω′)

the norm

‖u‖Ck(Ω′) = max{‖Dαu‖C(Ω′) : |α| ≤ k},where α = (α1, . . . , αn) is a multi-index, |α| = ∑n

j=1 αi, Dα = Dα11 . . . Dαn

n and

D0u = u. In particular, T (·)f ∈ C∞((0,∞), Ck(Ω′)) for all k ∈ N. This impliesthat the function satisfies (6.10).

6.2 Inhomogeneous Boundary Conditions

In Section 6.1 we solved the elliptic problem showing that the Poisson operatoris resolvent positive. Now we prove well-posedness of an evolutionary problemwith time-dependent boundary conditions, by converting it into an inhomogeneousCauchy problem and using the results of Chapter 3. We keep the notation of Section6.1.

Let τ > 0. Given u0 ∈ C(Ω) and ϕ ∈ C([0, τ ], C(Γ)), we consider theparabolic problem

Pτ (u0, ϕ)

⎧⎪⎨⎪⎩u′(t) = Δu(t) (t ∈ [0, τ ]),

u(t)|Γ = ϕ(t) (t ∈ [0, τ ]),

u(0) = u0.

Remark 6.2.1. Let n = 3. Then Ω is a solid body and the solution u of Pτ (u0, ϕ)describes the heat flow in Ω. More precisely, for x ∈ Ω, u0(x) is the given initialtemperature at the point x ∈ Ω. For t ∈ [0, τ ], z ∈ Γ, the quantity ϕ(t)(z) is thegiven temperature at z at the time t. We may imagine that the boundary is heatedby some resistance surrounding Ω. Then the solution u(t)(x) is the temperatureat the point x ∈ Ω at the time t ∈ [0, τ ].

Definition 6.2.2. A mild solution of Pτ (u0, ϕ) is a function u ∈ C([0, τ ], C(Ω))such that

Δ

∫ t

0

u(s) ds = u(t)− u0 in D(Ω)′ (6.11)

6.2. INHOMOGENEOUS BOUNDARY CONDITIONS 409

andu(t)|Γ = ϕ(t) (6.12)

for all t ∈ [0, τ ].

Note that (6.11) implies in particular that∫ t

0u(s) ds ∈ D(Δmax) for all t ∈

[0, τ ].Consider the Poisson operator A on X = C(Ω) × C(Γ) given by A(u, 0) =

(Δu,−u|Γ) on D(A) = D(Δmax) × {0}. Given U0 ∈ X and Φ ∈ C([0, τ ], X), weconsider the Cauchy problem{

U ′(t) = AU(t) + Φ(t) (t ∈ [0, τ ]),

U(0) = U0.(6.13)

Proposition 6.2.3. Let u0 ∈ C(Ω), U0 := (u0, 0), and let ϕ ∈ C([0, τ ], C(Γ)),Φ(t) := (0, ϕ(t)). Let U ∈ C([0, τ ], X). Then U is a mild solution of (6.13) if andonly if U is of the form U(t) = (u(t), 0) (t ∈ [0, τ ]) where u is a mild solution ofPτ (u0, ϕ).

Proof. If U is a mild solution of (6.13), then

U(t) =d

dt

∫ t

0

U(s) ds ∈ D(A) = C(Ω)× {0}

for all t ∈ [0, τ ]. Thus, U(t) = (u(t), 0) for some u ∈ C([0, τ ], C(Ω)). Now the claimis immediate from the definition of A and Definition 6.2.2.

Now let u0 ∈ D(Δmax), U0 = (u0, 0) ∈ D(A). Let ϕ ∈ C([0, τ ], C(Γ)), Φ(t) =(0, ϕ(t)). Then

AU0 +Φ(0) = (Δu0,−u0|Γ + ϕ(0)).

Thus, the consistency condition

AU0 +Φ(0) ∈ D(A) = C(Ω)× {0} (6.14)

from Theorem 3.11.10 becomes

u0|Γ = ϕ(0). (6.15)

This is obviously a necessary condition for the existence of a mild solution ofPτ (u0, ϕ). Now we obtain the following from Theorem 3.11.10.

Proposition 6.2.4. Assume that Ω is Dirichlet regular. Let ϕ0 ∈ C(Γ), ϕ′ ∈L1((0, τ), C(Γ)), ϕ(t) := ϕ0 +

∫ t

0ϕ′(s) ds (t ∈ [0, τ ]). Let u0 ∈ D(Δmax). If condi-

tion (6.15) is satisfied, then there exists a unique mild solution of Pτ (u0, ϕ).

Next, we obtain the weak parabolic maximum principle as a direct conse-quence of Theorem 3.11.11. It will serve as an a priori estimate for solutions ofPτ (u0, ϕ).

Proposition 6.2.5 (Parabolic maximum principle). Assume that Ω is Dirichlet reg-ular. Let u be a mild solution of Pτ (u0, ϕ), where u0 and ϕ are real-valued. Letc+, c− ∈ R be constants such that

c− ≤ u0 ≤ c+ and

c− ≤ ϕ(t) ≤ c+ (t ∈ [0, τ ]).

Then c− ≤ u(t) ≤ c+ (t ∈ [0, τ ]).

Proof. Note that e(t) := c+ defines a mild solution of Pτ (c+, c+). Let v(t) :=c+ − u(t). Then v is a mild solution of Pτ (c+ − u0, c+ − ϕ). Since c+ − u0 ≥ 0and c+−ϕ(t) ≥ 0, it follows from Theorem 3.11.11, applied to (6.13) with U(t) =(v(t), 0), U0 = (c+− u0, 0), Φ(t) = (0, c+−ϕ(t)), that c+− u(t) = v(t) ≥ 0 for allt ∈ [0, τ ]. The other inequality is proved in a similar way.

Let u be a mild solution of Pτ (u0, ϕ), where u0 and ϕmay be complex-valued.By considering Re(eiθu), which is a mild solution of Pτ (Re(e

iθu0),Re(eiθϕ)), it

follows from Proposition 6.2.5 that

‖u‖C([0,τ ],C(Ω)) ≤ max{‖ϕ‖C([0,τ ],C(Γ)) , ‖u0‖C(Ω)

}. (6.16)

Here, we consider C([0, τ ], C(Ω)) and C([0, τ ], C(Γ)) as Banach spaces for thenorms

‖u‖C([0,τ ],C(Ω)) = sup0≤t≤τ

‖u(t)‖C(Ω) and ‖ϕ‖C([0,τ ],C(Γ)) = sup0≤t≤τ

‖ϕ(t)‖C(Γ),

respectively.Now we can prove well-posedness of Pτ (u0, ϕ).

Theorem 6.2.6. Assume that Ω is Dirichlet regular. Let u0 ∈ C(Ω) and ϕ ∈C([0, τ ], C(Γ)) such that u0|Γ = ϕ(0). Then there exists a unique mild solutionof Pτ (u0, ϕ).

Proof. Uniqueness follows from Lemma 3.2.9. For existence, choose u0n ∈ D(Δmax)such that limn→∞ u0n = u0 in C(Ω). Choose ϕn ∈ C1([0, τ ], C(Γ)) such thatϕn(0) = u0n|Γ and ϕn → ϕ as n → ∞ in C([0, τ ], C(Γ)). For example, onemay let ϕn(t) := (1 − λ(nt))ψn(t) + λ(nt)u0n|Γ, where λ(s) := (1 − min(s, 1))2,ψn ∈ C1([0, τ ], C(Γ)) and ‖ψn − ϕ‖C([0,τ ],C(Γ)) < 1/n. By Proposition 6.2.4, thereexists a unique mild solution un of Pτ (u0n, ϕn). By (6.16), we have

‖un − um‖C([0,τ ],C(Ω)) ≤ max{‖ϕn − ϕm‖C([0,τ ],C(Γ)) , ‖u0n − u0m‖C(Ω)

}.

Hence, (un)n∈N is a Cauchy sequence in C([0, τ ], C(Ω)). Let u := limn→∞ un in

C([0, τ ], C(Ω)). Then u(t)|Γ = limn→∞ ϕn(t) = ϕ(t). Since Δmax

∫ t

0un(s) ds =

un(t)− u0n and Δmax is closed, it follows that∫ t

0

u(s) ds = limn→∞

∫ t

0

un(s) ds ∈ D(Δmax) and Δmax

∫ t

0

u(s) ds = u(t)− u0

6.2. INHOMOGENEOUS BOUNDARY CONDITIONS 411

for all t ∈ [0, τ ]. We have shown that u is a mild solution of Pτ (u0, ϕ).

So far, we have seen that for each u0 ∈ C(Ω) and ϕ ∈ C([0, τ ], C(Γ)) satisfyingϕ(0) = u0|Γ there exists a unique mild solution. We now show that the mildsolution u is always of class C∞ on (0, τ ] × Ω. In fact, we may identify u witha continuous function defined on [0, τ ] × Ω with values in R by letting u(t, x) :=u(t)(x) (t ∈ [0, τ ], x ∈ Ω). Then the following holds.

Theorem 6.2.7. Assume that Ω is Dirichlet regular. Let u0 ∈ C(Ω) and ϕ ∈C([0, τ ], C(Γ)) such that u0|Γ = ϕ(0). Let u be the mild solution of Pτ (u0, ϕ).Then

u ∈ C∞((0, τ ]× Ω).

Proof. a) Assume that u0 = 0. Let v(t) :=∫ t

0u(s) ds. Then

v ∈ C1([0, τ ], C(Ω)), v(t) ∈ D(Δmax) and v′(t) = Δv(t) for t ∈ [0, τ ].

Let 0 < t0 ≤ τ, x0 ∈ Ω. Choose r > 0 such that B(x0, r) ⊂ Ω, and let C :=[0, τ ]×B(x0, r) and C′ := [t0/2, τ ]×B(x0, r/2). Choose ξ ∈ C∞([0, τ ]×Rn) suchthat ξ ≡ 1 on C′, ξ ≡ 0 on ([0, τ ] × Rn) \ C, and ξ ≡ 0 on [0, t0/4] × Rn. Letw := ξ · v on [0, τ ] × Ω, w := 0 on [0, τ ] × (Rn \ Ω). Then w ∈ C1([0, τ ], C0(Rn)),and

w′(t) = ξ′(t)v(t) + ξ(t)v′(t) on [0, τ ]× Ω,

with w′(t) = 0 on [0, τ ] × (Rn \ Ω). It follows from Lemma 6.1.5 and the closedgraph theorem that D(Δmax) ↪→ C1(Ω) when D(Δmax) carries the graph normand C1(Ω) has the natural Frechet topology. In particular, ∇v is continuous on[0, τ ]× C. We have

Δw(t) = ξ(t)Δv(t) + 2∇ξ(t) · ∇v(t) + Δξ(t)v(t) on [0, τ ]× Ω.

Let

f(t) :=

{ξ′(t)v(t)− 2∇ξ(t) · ∇v(t)−Δξ(t)v(t) on [0, τ ]× Ω,

0 on [0, τ ]× (Rn \Ω).

Then f ∈ C([0, τ ], C0(Rn)) and w′(t) = Δw(t) + f(t) on [0, τ ]× Rn. Denote by Gthe Gaussian semigroup on C0(Rn), i.e.

G(t)g := kt ∗ g, where kt(x) := (4πt)−n/2e−|x|2/4t.

Since w(0) = 0, it follows from Proposition 3.1.16 that

w(t) =

∫ t

0

G(t− s)f(s) ds, i.e.,

w(t, x) =

∫ t

0

∫Rn

(4π(t− s))−n/2e−|x−y|2/4(t−s)f(s, y) dy ds

for all 0 < t ≤ t0, x ∈ Rn. Since f ≡ 0 on C ′ and outside C, the integrandhas no singularities for (t, x) in the interior of C ′. Thus, w is of class C∞ in aneighbourhood of (t0, x0) in (0, τ ]×Ω. Since v = w in C′ and (t0, x0) is arbitrary,it follows that v, and hence also u, belong to C∞((0, τ ]× Ω).

b) Now consider the general case when u0|Γ = ϕ(0). Let w0 be the solutionof the Dirichlet problem D(ϕ(0)). Consider v(t) := u(t) − w0. Then v is a mildsolution of

v′(t) = Δv(t) (t ∈ [0, τ ])

and v(0)|Γ = 0. Denote by T0 the C0-semigroup generated by Δ0 on C0(Ω). Letw(t) := v(t) − T0(t)v(0). Then w is a mild solution of Pτ (0, ϕ − ϕ(0)). Hence,w ∈ C∞((0, τ ] × Ω) by a). Since T0(·)v(0) ∈ C∞((0,∞) × Ω) by Proposition6.1.10, the proof is complete.

Now we can reformulate the results. For this, we consider the parabolic do-main

Ωτ := (0, τ ]× Ω

with parabolic boundary

Γτ = ({0} × Ω) ∪ ((0, τ ]× Γ),

where τ > 0. Thus, Ωτ is a cylinder and Γτ is the topological boundary withoutthe top.

Theorem 6.2.8. Assume that Ω is Dirichlet regular. Then for every ψ ∈ C(Γτ )there exists a unique function u ∈ C(Ωτ ) ∩ C∞(Ωτ ) such that{

ut −Δu = 0 in Ωτ , and

u|Γτ= ψ.

(6.17)

Thus, (6.17) is formulated exactly as the Dirichlet problem, the Laplacianbeing replaced by the parabolic operator d

dt −Δ, Ω by the parabolic domain Ωτ

and Γ by the parabolic boundary Γτ .

6.3 Asymptotic Behaviour

We keep the notation of the preceding section. But now we consider the problemon the half-line

P∞(u0, ϕ)

⎧⎪⎨⎪⎩u′(t) = Δu(t) (t ≥ 0),

u(t)|Γ = ϕ(t) (t ≥ 0),

u(0) = u0,

6.3. ASYMPTOTIC BEHAVIOUR 413

where u0 ∈ C(Ω) and ϕ ∈ C(R+, C(Γ)) are given functions. As before, by a mildsolution of P∞(u0, ϕ) we understand a function u ∈ C(R+, C(Ω)) such that

Δ

∫ t

0

u(s) ds = u(t)− u0 in D(Ω)′ and

u(t)|Γ = ϕ(t)

for all t ∈ R+.We assume throughout this section that Ω is Dirichlet regular.It follows from Theorem 6.2.6 that for each u0 ∈ C(Ω) and ϕ ∈ C(R+, C(Γ))

such that u0|Γ = ϕ(0) there exists a unique mild solution u of P∞(u0, ϕ). Inthis section we study the asymptotic behaviour of u(t) as t → ∞. The resultsare analogous to (and in some cases, consequences of) abstract results given inSections 5.4 and 5.6. We start with Cesaro convergence.

Proposition 6.3.1. Let ϕ : R+ → C(Γ) be continuous and bounded. Assume that

limt→∞

1

t

∫ t

0

ϕ(s) ds = ϕ∞ in C(Γ).

Let u0 ∈ C(Ω) satisfying u0|Γ = ϕ(0) and let u be the mild solution of P∞(u0, ϕ).Then

limt→∞

1

t

∫ t

0

u(s) ds = u∞ exists in C(Ω).

Moreover, Δu∞ = 0 in D(Ω)′ and u∞|Γ = ϕ∞.

Proof. By (6.16), u is bounded. Taking Laplace transforms, we have λu(λ) −Δu(λ) = u0 and u(λ)|Γ = ϕ(λ) (λ > 0). Denote by w(λ) the solution of theDirichlet problem D(ϕ(λ)). Then u(λ)− w(λ) ∈ C0(Ω) and

λ(u(λ)− w(λ))−Δ(u(λ)− w(λ)) = u0 − λw(λ).

Thus, u(λ) − w(λ) = R(λ,Δc)(u0 − λw(λ)). Let u∞ be the solution of D(ϕ∞).By Theorem 4.1.2, limλ↓0 λϕ(λ) = C − limt→∞ ϕ(t) = ϕ∞. It follows from themaximum principle (Theorem 6.1.3) that λw(λ) converges to a function u∞ inC(Ω) as λ ↓ 0. Clearly, u∞ solves D(ϕ∞). Thus, u(λ)−w(λ)→ R(0,Δc)(u0−u∞)in C(Ω) as λ ↓ 0. Consequently, λ(u(λ)−w(λ))→ 0 in C(Ω) as λ ↓ 0. This impliesthat limλ↓0 λu(λ) = u∞ in C(Ω). By Theorem 4.2.7, this implies the claim.

Next we consider uniform continuity.

Proposition 6.3.2. Let ϕ ∈ BUC(R+, C(Γ)) and u0 ∈ C(Ω) such that u0|Γ = ϕ(0).Let u be the mild solution of P∞(u0, ϕ). Then u ∈ BUC(R+, C(Ω)).

Proof. By (6.16), u is bounded. For δ > 0 let uδ(t) := u(t + δ) − u(t), ϕδ(t) :=ϕ(t + δ) − ϕ(t) (t ≥ 0). Then uδ is the mild solution of P∞(uδ(0), ϕδ). Since

ϕδ → 0 in BUC(R+, C(Γ)) and uδ → 0 in C(Ω) as δ ↓ 0, it follows from (6.16) thatuδ(t)→ 0 as δ ↓ 0 uniformly on R+. This means that u is uniformly continuous.

Using Propositions 6.2.3 and 6.3.2, the results of Chapter 5 on inhomogeneousCauchy problems give the following.

Theorem 6.3.3. Let ϕ ∈ AAP(R+, C(Γ)) and u0 ∈ C(Ω) such that u0|Γ = ϕ(0).Denote by u the mild solution of P∞(u0, ϕ). Then

a) u ∈ AAP(R+, C(Ω)) and Freq(u) = Freq(ϕ).

b) If ϕ = ϕ1 + ϕ2 where ϕ1 ∈ AP(R+, C(Γ)), ϕ2 ∈ C0(R+, C(Γ)), and u =u1 + u2 where u1 ∈ AP(R+, C(Ω)) and u2 ∈ C0(R+, C(Ω)), then u1 is themild solution of P∞(u1(0), ϕ1) and u2 is the mild solution of P∞(u2(0), ϕ2).

c) If limt→∞ ϕ(t) = ϕ∞ exists in C(Γ), then limt→∞ u(t) = u∞ where u∞ isthe solution of the Dirichlet problem D(ϕ∞).

Proof. a) Let X := C(Ω) × C(Γ). Consider the function U : R+ → X definedby U(t) := (u(t), 0). Then by Proposition 6.3.2, U ∈ BUC(R+, X). Let Φ(t) :=(0, ϕ(t)). Then Φ ∈ AAP(R+, X) and Proposition 6.3.2 shows that U is a mildsolution of {

U ′(t) = AU(t) + Φ(t) (t ≥ 0),

U(0) = (u0, 0),(6.18)

where A is the Poisson operator. Since s(A) < 0, it follows from Corollary 5.6.9that U ∈ AAP(R+, X), hence u ∈ AAP(R+, C(Ω)). Moreover, it also follows thatFreq(u) = Freq(U) = Freq(Φ) = Freq(ϕ).

b) This is a direct consequence of Propositions 5.4.16 and 6.2.3.c) By Corollary 5.6.9, limt→∞ U(t) exists and equals R(0, A)(0, ϕ∞) =

(u∞, 0), so that u∞ is the solution of D(ϕ∞).

Corollary 6.3.4. Let ϕ ∈ AP(R+, C(Γ)). Then there exists a unique u0 ∈ C(Ω)satisfying u0|Γ = ϕ(0) such that the mild solution u of P∞(u0, ϕ) is almost periodic.

Proof. Existence: Let v0 ∈ C(Ω) such that v0|Γ = ϕ(0). Let v be the mild solutionof P∞(v0, ϕ). Then v = v1+v2 where v1 ∈ AP(R+, C(Ω)) and v2 ∈ C0(R+, C(Ω)),by Theorem 6.3.3. Moreover, v1 is the mild solution of P∞(v1(0), ϕ). So we maychoose u0 = v1(0).

Uniqueness: Assume that the mild solution u of P∞(u0, ϕ) is almost periodic.Then v = u− u ∈ AP(R+, C(Ω)) and v is the mild solution of P∞(u0 − u0, 0). Itfollows from Theorem 6.3.3 c) that limt→∞ v(t) = 0. Hence, v(t) ≡ 0 by (4.36).

In the situation of Corollary 6.3.4, if ϕ is τ -periodic, then u is also τ -periodic.This follows from Corollary 4.5.4, since Freq(u) = Freq(ϕ) ⊂ 2π

τZ (see also Corol-

lary 5.6.9 c).Finally, we consider the inhomogeneous heat equation with inhomogeneous

boundary conditions.

6.4. NOTES 415

Given u0 ∈ C(Ω), ϕ ∈ C(R+, C(Γ)) such that u0|Γ = ϕ(0) and f ∈C(R+, C(Ω)), we consider the problem

P∞(u0, ϕ, f)

⎧⎪⎨⎪⎩u′(t) = Δu(t) + f(t) (t ≥ 0),

u(t)|Γ = ϕ(t) (t ≥ 0),

u(0) = u0.

A mild solution is a continuous function u : R+ → C(Ω) such that u(0) =

u0, u(t)|Γ = ϕ(t),∫ t

0u(s) ds ∈ D(Δmax) and

Δ

∫ t

0

u(s) ds+

∫ t

0

f(s) ds = u(t)− u0 in D(Ω)′ (6.19)

for all t ≥ 0. By Theorem 6.2.6, there is at most one mild solution of P∞(u0, ϕ, f).By Proposition 3.7.22, the function

v(t) :=

∫ t

0

Tc(t− s)f(s) ds (t ≥ 0) (6.20)

defines a mild solution of P∞(0, 0, f). Since Ω is Dirichlet regular, by Theorem6.2.6, there exists a unique mild solution w of P∞(u0, ϕ, 0). Hence, u = v +w is amild solution of P∞(u0, ϕ, f). We have shown the following.

Theorem 6.3.5. Let f ∈ C(R+, C(Ω)), ϕ ∈ C(R+, C(Γ)) and u0 ∈ C(Ω) such thatu0|Γ = ϕ(0). Then there exists a unique mild solution of P∞(u0, ϕ, f).

Recall that ω(Tc) < ∞. Thus, v = Tc ∗ f ∈ BUC(R+, C(Ω)) (respectively,AAP(R+, C(Ω))) if f ∈ BUC(R+, C(Ω)) (respectively, f ∈ AAP(R+, C(Ω))), byProposition 5.6.1. So we obtain the following from Theorem 6.3.3.

Theorem 6.3.6. Let f ∈ AAP(R+, C(Ω)), ϕ ∈ AAP(R+, C(Γ)) and u0 ∈ C(Ω).Assume that u0|Γ = ϕ(0). Let u be the mild solution of P∞(u0, ϕ, f). Then u ∈AAP(R+, C(Ω)).

Corollary 6.3.7. Let f : R+ → C(Ω) and ϕ : R+ → C(Γ) be continuous. Assumethat limt→∞ f(t) = f∞ exists in C(Ω) and limt→∞ ϕ(t) = ϕ∞ in C(Γ). Let u0 ∈C(Ω) such that u0|Γ = ϕ(0). Let u be the mild solution of P∞(u0, ϕ, f). Thenlimt→∞ u(t) = u∞ exists in C(Ω) and{

u∞|Γ = ϕ∞,

−Δu∞ = f∞ in D(Ω)′. (6.21)

6.4 Notes

The approach to solving the heat equation with the help of the Poisson operator istaken from [Are00], where general strongly elliptic operators in divergence form with


bounded measurable coefficients are also considered. Greiner [Gre87] developed an ab-stract perturbation theory for boundary conditions. Theorems 6.1.7 and 6.1.8 are provedin [AB98], where it is also shown that Dirichlet regularity is a necessary condition. Lumerand Schnaubelt [LS99] consider also non-cylindrical domains. Lumer gave a proof of theholomorphy of the semigroup generated by Δ0 which is based on the maximum principle(cf. [LP79]). In Theorem 6.1.9 we use the properties of resolvent positive operators toprove holomorphy.

Theorem 6.2.7 is an adaptation of Evans’s proof [Eva98, Section 2.3, Theorem 8] tothe solutions defined here. Theorem 6.2.8 was probably first proved by Tychonoff [Tyc38]in 1938 with the help of integral equations. Other proofs were given by Fulks [Ful56],[Ful57] and Babuska and Vyborny [BV62] (see also the work of Lumer [Lum75]).

Concerning an Lp-approach to boundary value problems via holomorphic semi-groups we refer to the monograph by Taira [Tai95].

Chapter 7

The Wave Equation

In this chapter we study the wave equation

utt = Δu

on an open subset Ω of Rn. We will use the theory of cosine functions and work onL2(Ω). We first consider the Laplacian with Dirichlet boundary conditions. This isa selfadjoint operator and well-posedness is a consequence of the spectral theorem.A further aim is to replace the Laplace operator by a general elliptic operator. Thiswill be done by a perturbation theorem for selfadjoint operators which we provein Section 7.1.

We give a brief introduction to symmetric sesquilinear forms which are thenatural tool for proving selfadjointness. However, we restrict ourselves to the min-imum needed to show that quite general equations can be solved in a simple wayby functional analytical methods. The restriction to Dirichlet boundary conditionsand to second order operators is not essential; we choose these in order to presentthe simplest case.

7.1 Perturbation of Selfadjoint Operators

A selfadjoint operator which is bounded above generates a cosine function (seeExample 3.14.16). We will present a perturbation result in terms of the formdomain which again leads to a generator of a cosine function with the same phasespace. It is important to know this since the phase space yields the natural domainfor initial data for classical solutions (Corollary 3.14.12).

We will see in the following section that the abstract setting which we presenthere is very well adapted to elliptic operators.

Let H be a Hilbert space with scalar product (·|·)H and norm ‖ · ‖H . Weconsider another Hilbert space V with scalar product (·|·)V and norm ‖ · ‖V .Moreover, we assume that V is continuously embedded into H with dense image.


418 7. THE WAVE EQUATION

This means, we assume that V ⊂ H, that V is dense in H and that there exists aconstant ω > 0 such that

ω‖u‖2H ≤ ‖u‖2V (7.1)

for all u ∈ V . We use the abbreviation

Vd↪→ H

for these three properties.

In this situation, we associate to V an operator AH on H by postulating:

D(AH) :=

{u ∈ V : there exists f ∈ H such that

(u|v)V = −(f |v)H for all v ∈ V

},

AHu := f.

Note that f is uniquely determined by u since V is dense in H.

We call AH the operator on H associated with V .

Proposition 7.1.1. The operator AH is selfadjoint and bounded above by −ω. Inparticular, AH generates a holomorphic C0-semigroup T on H of angle π/2 satis-fying

‖T (z)‖ ≤ e−ωRe z (Re z > 0).

Moreover, AH generates a cosine function.

Proof. a) AH is symmetric. In fact, let u, v ∈ D(AH). Then by the definition ofAH ,

(AHu|v)H = −(u|v)V = −(v|u)V = (AHv|u)H = (u|AHv)H .

b) For u ∈ D(AH) one has

(AHu|u)H = −(u|u)V ≤ −ω‖u‖2H ,

by (7.1). Thus, AH is bounded above by −ω.c) We show that −AH is surjective. Let f ∈ H. Then F (v) := (v|f)H defines

a continuous linear form on V . By the Riesz-Frechet lemma, there exists a uniqueu ∈ V such that (v|f)H = (v|u)V for all v ∈ V . Hence by the definition of AH ,one has u ∈ D(AH) and −AHu = f . It follows from Theorem B.14 that AH isselfadjoint. The remaining two assertions follow from Example 3.7.5 and Example3.14.16.

Let us consider multiplication operators as a first example. It is very simple;nevertheless, it is a generic example by the spectral theorem (Theorem B.13).

7.1. PERTURBATION OF SELFADJOINT OPERATORS 419

Example 7.1.2 (Multiplication operators). Let (Y, μ) be a measure space, ω > 0and m : Y → [ω,∞) be measurable. Let H := L2(Y, μ) and let

V := L2(Y,mdμ) :=

{u ∈ H :

∫Y

|u|2mdμ <∞}.

Then V is a Hilbert space for the scalar product

(u|v)V :=

∫Y

u(x)v(x)m(x) dμ(x)

and Vd↪→ H.

It is easy to see that the operator AH on H associated to V is the multipli-cation operator given by

D(AH) = {u ∈ H : mu ∈ H},AHu = −mu.

(cf. Example B.12). The cosine function Cos generated by AH is given by

(Cos(t)f)(x) = (cos t√

m(x))f(x)

(cf. Example 3.14.16). Thus,

Cos(t) =1

2(U(t) + U(−t)),

where U is the C0-group on H given by

(U(t)f)(x) = eit√

m(x)f(x)

(cf. Example 3.14.16). The generator B of U is given by

D(B) = {f ∈ H :√m · f ∈ H},

Bf = i√m · f.

Thus, V = D(B). It follows from Example 3.14.15 that the phase space of Cos isV ×H.

In Example 7.1.2, the phase space of the cosine function associated with

V is V × H. In fact, this is the case whenever Vd↪→ H. The spectral theorem

applied to the selfadjoint operator AH shows that there is a measure space (Y, μ),a measurable function m : Y → [ω,∞) and a unitary equivalence U taking H ontoL2(Y, μ) and AH onto the operator appearing in Example 7.1.2. The next result,which is the converse of Proposition 7.1.1, shows that V is uniquely determinedby AH . Hence, it follows from Example 7.1.2 that U takes V onto L2(Y,mdμ) andthat the phase space is V ×H.


Proposition 7.1.3. Let B be a selfadjoint operator on H which is bounded above by

−ω, where ω > 0. Then there exists a unique Hilbert space V such that Vd↪→ H and

such that B is the operator associated with V . Moreover, the phase space associatedwith the cosine function generated by B is V ×H.

Proof. Uniqueness: Assume that B is associated with V where Vd↪→ H. We show

that D(B) is dense in V . In fact, let v ∈ V such that (u|v)V = 0 for all u ∈ D(B).Then

(Bu|v)H = −(u|v)V = 0

for all u ∈ D(B). Since B is selfadjoint, it follows that v ∈ D(B) and Bv = 0. SinceB is invertible, we conclude that v = 0. This proves the claim. Now we observethat

‖u‖2V = (u|u)V = −(Bu|u)Hfor all u ∈ D(B). Thus, V is the completion of D(B) for the norm ‖u‖V :=√−(Bu|u)H . Moreover, (u|v)V = −(Bu|v)H for all u, v ∈ D(B). Thus, the scalarproduct is also determined by B.

Existence: Using the spectral theorem, we may assume that B is a multipli-cation operator. Then the assertion is proved in Example 7.1.2.

Propositions 7.1.1 and 7.1.3 establish a bijective correspondence between self-adjoint operators B on H which are bounded above by −ω (where ω > 0) and

Hilbert spaces V such that Vd↪→ H and (7.1) is satisfied. One frequently calls V

the form domain of B, and (·|·)V the sesquilinear form associated with B.From Corollary 3.14.13 we now deduce our first perturbation result for self-

adjoint operators.

Corollary 7.1.4. Let V be a Hilbert space such that Vd↪→ H and let AH be the

operator associated with V on H. Let C ∈ L(V,H). Then AH + C generates acosine function on H with phase space V ×H.

Our next aim is to introduce another kind of perturbation. For this we needsome preparation.

Let Vd↪→ H. A mapping ϕ : V → C is called antilinear if

ϕ(u+ v) = ϕ(u) + ϕ(v) (u, v ∈ V ), and

ϕ(λu) = λϕ(u) (u ∈ V, λ ∈ C).

ByV ′ := {ϕ : V → C : ϕ antilinear and continuous}

we denote the antidual of V . It is a Banach space for the norm

‖ϕ‖V ′ := sup‖u‖V ≤1

|ϕ(u)|.

7.1. PERTURBATION OF SELFADJOINT OPERATORS 421

We embed H into V ′ in the following way. For f ∈ H we define ϕf ∈ V ′ by

ϕf (u) := (f |u)H .

It is clear that the mapping f �→ ϕf : H → V ′ is linear, injective and continuous.This is the desired embedding.

For ϕ ∈ V ′ we use the notation

(ϕ|u) := ϕ(u) (u ∈ V ).

Thus, if f ∈ H we have

(ϕf |u) = (f |u)H (u ∈ V ).

By the Riesz-Frechet lemma, the mapping A : V → V ′ given by

(Au|v) := −(u|v)V (u, v ∈ V )

is an isometric isomorphism from V onto V ′. In particular, V ′ is itself a Hilbertspace for the scalar product

(f |g)V ′ := (A−1f |A−1g)V (f, g ∈ V ′).


ω‖ϕf‖2V ′ ≤ ‖f‖2H (7.2)

for all f ∈ H. In fact,

ω‖ϕf‖2V ′ = ω supu∈V

‖u‖V ≤1

|(f |u)H |2

≤ supu∈V√

ω‖u‖H≤1

|(f |√ωu)H |2

≤ ‖f‖2H ,

by the Cauchy-Schwarz inequality.

Now we identify H with a subspace of V ′, and we identify f and ϕf forf ∈ H. In particular, for f ∈ H we write

(f |v) = (f |v)H (v ∈ V ).

Having this in mind, the proof of the following proposition is easy.

Proposition 7.1.5. The operator A on V ′ is selfadjoint with upper bound −ω. More-over, H × V ′ is the phase space associated with the cosine function generated byA on V ′. The operator AH associated with V is the part of A in H.


Proof. We first prove that

(Au|f)V ′ = −(u|f)H (7.3)

for all u ∈ V, f ∈ H.In fact, let w = A−1f ∈ V . Then

(w|u)V = −(Aw|u) = −(f |u)H .

Hence,(Au|f)V ′ = (u|A−1f)V = (u|w)V = −(u|f)H .

It follows from (7.3) that for all u, v ∈ V ,

(Au|v)V ′ = −(u|v)H = −(v|u)H = (Av|u)V ′ = (u|Av)V ′ .Thus, A is symmetric. Moreover,

(Au|u)V ′ = −‖u‖2H ≤ −ω‖u‖2V ′for all u ∈ V by (7.2). Thus, A is bounded above by −ω. Since A is surjective, itfollows from Theorem B.14 that A is selfadjoint. In particular, V = D(A) is dense

in V ′. Thus, Hd↪→ V ′. Let B be the operator on V ′ associated with H. It follows

from (7.3) that B is an extension of A. Since both operators are invertible, theyare equal. It follows from Proposition 7.1.3 that the phase space associated withthe cosine function generated by B on V ′ is H × V ′. The final assertion is easy toverify.

We will illustrate Proposition 7.1.5 by considering multiplication operatorsin the following example. The discussion before Proposition 7.1.3 shows that theydescribe the most general situation; in fact, the example gives an alternative proofof Proposition 7.1.5.

Example 7.1.6 (Antidual associated with multiplication operator). Let H :=L2(Y, μ), V := L2(Y,mdμ) where m : Y → [ω,∞) is measurable, ω > 0. Then wemay identify V ′ with L2(Y, 1

m dμ) by letting

(w|v) :=∫Y

wv dμ

for all w ∈ L2(Y, 1mdμ), v ∈ V . Since for f ∈ H, one has

(ϕf |v) =∫Y

f v dμ,

the embeddingH ↪→ V ′ : f �→ ϕf corresponds to the identity mapping L2(Y, μ) ↪→L2(Y, 1

mdμ). Now the mapping A : V → V ′ is given by Au = −mu. In fact, for

u ∈ V one has

(Au|v) = −∫Y

uvmdμ

7.2. THE WAVE EQUATION IN L2(Ω) 423

for all v ∈ V .Note that A, considered as an operator on V ′, is associated to the subspace

H = L2(Y, dμ) of V ′ = L2(Y, 1m dμ). Thus, A generates a cosine function on V ′

and the associated phase space is H × V ′.

Now we are in the position to prove the following general perturbation result.

Theorem 7.1.7. Let Vd↪→ H and identify H with a subspace of V ′ in the canonical

way. Denote by A : V → V ′ the isomorphism given by the Riesz-Frechet lemma.Let C ∈ L(V,H), B ∈ L(H,V ′). Then the part (A+B +C)H of A+B +C in Hgenerates a cosine function Cos on H whose phase space is V ×H.

Note that

D((A+B + C)H) = {u ∈ V : Au+Bu+ Cu ∈ H}.

Proof. We consider A as an unbounded operator on V ′ with domain V . Thenwe know from Proposition 7.1.5 that A generates a cosine function on V ′ withassociated phase space H ×V ′. Since B ∈ L(H,V ′), A+B also generates a cosinefunction on V ′ with phase space H × V ′ (by Corollary 3.14.13). Now we applyCorollary 3.14.14 to deduce that the part (A + B)H of A + B in H generatesa cosine function with associated phase space D(A + B) × H = V × H. SinceC ∈ L(V,H), one has (A +B + C)H = (A +B)H + C. Now the claim follows byanother application of Corollary 3.14.13.

Corollary 7.1.8. Under the conditions of Theorem 7.1.7, the operator (A+B+C)Hgenerates a holomorphic C0-semigroup on H of angle π/2.

Proof. This follows from Theorem 3.14.17.

7.2 The Wave Equation in L2(Ω)

Let Ω ⊂ Rn be an open set. In this section we will consider the wave equation onΩ with Dirichlet boundary conditions. We consider first the Laplacian and thenmore general elliptic operators.

For this, we recall some distributional notions (see Appendix E). We denotethe first Sobolev space in L2(Ω) by H1(Ω), i.e., H1(Ω) = W 1,2(Ω), and we define

H10 (Ω) := D(Ω)H

1(Ω). This allows us to give a meaning to Dirichlet boundary

conditions: for f ∈ H1(Ω) we say that f |∂Ω = 0 weakly if f ∈ H10 (Ω).

As usual, we consider L2(Ω) as a subspace of D(Ω)′. In particular, if f ∈H1(Ω), then Δf ∈ D(Ω)′ is defined by

〈ϕ,Δf〉 = 〈Δϕ, f〉 =n∑

j=1

∫Ω

fD2jϕdx = −

n∑j=1

∫Ω

DjfDjϕdx


for all ϕ ∈ D(Ω). Thus, to say that Δf ∈ L2(Ω) means that there exists a functiong ∈ L2(Ω) such that

−n∑

j=1

∫Ω

DjfDjϕdx =

∫Ω

gϕdx

for all ϕ ∈ D(Ω). We then identify g and Δf .Next, we define the Laplacian with Dirichlet boundary conditions. This exam-

ple has already been given in Chapter 3 (Example 3.4.7). Here we show how it fitsinto the setting of the preceding section. After that, it will be easy to investigatemore general elliptic operators than the Laplacian.

Example 7.2.1 (The Dirichlet Laplacian). Define the operator ΔL2(Ω) on L2(Ω) by

D(ΔL2(Ω)) :={f ∈ H1

0 (Ω) : Δf ∈ L2(Ω)},

ΔL2(Ω)f := Δf.

Then ΔL2(Ω) is selfadjoint and bounded above by 0. Moreover, ΔL2(Ω) generates acosine function with phase space H1

0 (Ω)× L2(Ω).

Proof. Let H := L2(Ω) and V := H10 (Ω) with scalar product

(u|v)V :=

∫Ω

uv dx+

n∑j=1

∫Ω

DjuDj v dx.

Then clearly, Vd↪→ H. Let B be the operator on H which is associated with V .

We show that ΔL2(Ω) = B + I. In fact, let u ∈ D(B), Bu =: f . Then u ∈ H10 (Ω)

and

−n∑

j=1

∫Ω

DjuDjϕdx−∫Ω

uϕdx = −(u|ϕ)V =

∫Ω

fϕdx

for all ϕ ∈ H10 (Ω). Taking ϕ ∈ D(Ω), we obtain

〈ϕ,Δu〉 = 〈Δϕ, u〉 =n∑

j=1

∫Ω

uD2jϕdx

= −n∑

j=1

∫Ω

DjuDjϕdx =

∫Ω

fϕ dx+

∫Ω

uϕdx.

Hence, Δu = f + u. This shows that u ∈ D(ΔL2(Ω)) and Bu = ΔL2(Ω)u − u.Conversely, let u ∈ D(ΔL2(Ω)). Then u ∈ H1

0 (Ω) and for ϕ ∈ D(Ω),

−(u|ϕ)V = −n∑

j=1

∫Ω

DjuDjϕ dx−∫Ω

uϕ dx = 〈ϕ,Δu〉 −∫Ω

uϕ dx.

7.2. THE WAVE EQUATION IN L2(Ω) 425

Since D(Ω) is dense in H10 (Ω), it follows that −(u|ϕ)V = (Δu− u | ϕ)L2(Ω) for all

ϕ ∈ H10 (Ω). Thus, u ∈ D(B) and Bu = Δu− u = ΔL2(Ω)u− u.We have shown that B = ΔL2(Ω)−I. Hence, ΔL2(Ω) = B+I is also selfadjoint.

Since B generates a cosine function with phase space H10 (Ω) × L2(Ω), so does

ΔL2(Ω), by Corollary 3.14.13.

Now we obtain the following well-posedness result for the wave equation.

Theorem 7.2.2 (Wave equation). Let f ∈ H10 (Ω) such that Δf ∈ L2(Ω). Let

g ∈ H10 (Ω). Then there exists a unique function u ∈ C2(R+, L

2(Ω)) such that

a) Δu(t) ∈ L2(Ω) for t ≥ 0;

b) u(t) ∈ H10 (Ω) for t ≥ 0;

c) u′′(t) = Δu(t) for t ≥ 0;

d) u(0) = f, u′(0) = g.

Proof. Denote by ΔL2(Ω) the Dirichlet Laplacian and by Cos the cosine functiongenerated by ΔL2(Ω) on L2(Ω) (see Example 7.2.1). Let Sin be the associated sinefunction. Then u(t) = Cos(t)f + Sin(t)g is a solution of⎧⎪⎪⎪⎨⎪⎪⎪⎩

u ∈ C2(R+, L2(Ω)),

u(t) ∈ D(ΔL2(Ω)) (t ≥ 0)

u′′(t) = ΔL2(Ω)u(t) (t ≥ 0)

u(0) = f, u′(0) = g,

by Corollary 3.14.12. Uniqueness follows from Corollary 3.14.8.

Next, we consider general uniformly elliptic operators of second order. Letaij ∈ L∞(Ω) be complex-valued coefficients such that aij = aji and

n∑i,j=1

aij(x)ξiξj ≥ α|ξ|2

for all ξ ∈ Rn, x-a.e. on Ω, where α > 0 is fixed. This last condition is calleduniform ellipticity and is equivalent to saying that the smallest eigenvalue of thehermitian matrix (aij(x)) is at least α for almost all x ∈ Ω.

Let bi, ci, d ∈ L∞(Ω) be complex-valued functions (i = 1, 2, . . . , n). We con-sider the formal elliptic second order operator

Lu :=

n∑i,j=1

Di(aijDju) +

n∑j=1

(Dj(bju) + cjDju) + du. (7.4)

It is possible to give a sense to L by multiplying (7.4) by a test function andintegration by parts. More precisely, we define L as follows.


Let u ∈ H10 (Ω). Define the distribution Lu ∈ D(Ω)′ by

〈ϕ, Lu〉 = −n∑

i,j=1

∫Ω

aij(x)(Dju)(x)(Diϕ)(x) dx

−n∑

j=1

∫Ω

bj(x)u(x)(Djϕ)(x) dx+

n∑j=1

∫Ω

cj(x)(Dju)(x)ϕ(x) dx

+

∫Ω

d(x)u(x)ϕ(x) dx

for all ϕ ∈ D(Ω). Then L : H10 (Ω) → D(Ω)′ is linear. Consider the part LH of L

in H := L2(Ω); i.e., LH is the operator on H given by

D(LH) = {u ∈ H10 (Ω) : Lu ∈ L2(Ω)}

LHu = Lu.

Then the following holds.

Theorem 7.2.3. The operator LH generates a cosine function on L2(Ω) with phasespace H1

0 (Ω)× L2(Ω).

Proof. Let α > 0 and consider V := H10 (Ω) with the scalar product

(u|v)V =

∫Ω

n∑i,j=1

aij(x)Diu(x)Dj v(x) dx+ α

∫Ω

u(x)v(x) dx.

Then ‖u‖V =√

(u|u)V is equivalent to the given norm on H10 (Ω). Thus, V is a

Hilbert space and Vd↪→ L2(Ω). We identify L2(Ω) with a subspace of V ′. Define

B ∈ L(H,V ′) by

(Bu|ϕ) := −∫Ω

n∑j=1

bjuDjϕdx (ϕ ∈ V = H10 (Ω))

and C ∈ L(V,H) by

Cu :=n∑

j=1

cjDju+ du− αu.

Let A : V → V ′ be the isomorphism given by the Riesz-Frechet lemma. It followsfrom Theorem 7.1.7 that the operator (A+B + C)H given by

D((A+B + C)H) :={u ∈ H1

0 (Ω) : Au+Bu+ Cu ∈ L2(Ω)},

(A+B + C)Hu := Au+Bu+ Cu,

generates a cosine function on L2(Ω) with associated phase space H10 (Ω)×L2(Ω).

Since D(Ω) is dense in H10 (Ω), the restriction mapping V ′ → D(Ω)′ is injective;

7.3. NOTES 427

thus, we may identify V ′ with a subspace of D(Ω)′. With this identification onehas LH = (A+B + C)H .

Now we deduce from Corollary 3.14.12 well-posedness of the following hy-perbolic problem. As before, L : H1

0 (Ω) → D(Ω)′ denotes the elliptic operatorassociated with the coefficients aij , bi, ci, d.

Corollary 7.2.4 (Hyperbolic equation). Let f ∈ H10 (Ω) such that Lf ∈ L2(Ω) and

let g ∈ H10 (Ω). Then there exists a unique function u ∈ C2(R+, L

2(Ω)) satisfying

a) u(t) ∈ H10 (Ω), Lu(t) ∈ L2(Ω) (t ≥ 0);

b) u′′(t) = Lu(t) (t ≥ 0);

c) u(0) = f, u′(0) = g.

Proof. This follows from Theorem 7.2.3 and Corollary 3.14.12 as in the proof ofTheorem 7.2.2.

Since each generator of a cosine function is also the generator of a holomor-phic C0-semigroup (by Theorem 3.14.17), we also obtain well-posedness of thecorresponding parabolic equation. The result is an extension of Example 3.7.24.

Corollary 7.2.5 (Parabolic equation). Let f ∈ L2(Ω). Then there exists a uniquefunction u ∈ C∞((0,∞), L2(Ω)) ∩ C(R+, L

2(Ω)) such that

a) u(t) ∈ H10 (Ω), Lu(t) ∈ L2(Ω) (t > 0);

b) u′(t) = Lu(t) (t > 0);

c) u(0) = f .

Proof. Denote by T the holomorphic C0-semigroup generated by LH . Then u(t) =T (t)f is the unique solution of a), b) and c), by Theorem 3.7.19.

7.3 Notes

Section 7.1 merely contains a direct approach to constructing selfadjoint operators byscalar products. We refer to the textbooks [Dav80], [Dav95], [Kat66] and [RS72] for asystematic treatment of quadratic form methods.

Elliptic operators generating cosine functions are described in the monographs byFattorini [Fat83], [Fat85] and Goldstein [Gol85], but the perturbation arguments leadingto Theorem 7.1.7 and Corollary 7.2.4 may be new (in the case B �= 0; i.e., when thecoefficients bj do not vanish).

Section 7.2 gives a fairly general well-posedness result on L2(Ω), and the restrictionto Dirichlet boundary conditions has been chosen merely for convenience. However, theseresults are definitely restricted to L2(Ω) and no longer valid on Lp(Ω) (p �= 2). This willbe made precise in Example 8.4.9. On L1(Rn) or C0(Rn) the Laplacian does not generatea cosine function. However, the following holds for n = 3.


Theorem 7.3.1. Let X := L1(R3) or C0(R3), and{D(ΔX) := {f ∈ X : Δf ∈ X},ΔXf := Δf.

Then ΔX generates a sine function Sin on X given by

(Sin(t)f)(x) =1

tσ2

∫∂B(x,t)

f(z) dσ(z)

where σ denotes the surface measure on ∂B(x, t) := {z ∈ R3 : |x − z| = t} and σ2 :=2|B(0, 1)| is the surface area of the 2-dimensional sphere. Thus, (Sin(t)f)(x) is t-timesthe mean of f over the sphere ∂B(x, t).

This can be seen by inspecting the proofs given in [Eva98, Section 2.4].Generalizing the method of spherical means, a systematic treatment of an “abstract

Laplacian” (given as the closure of the sum of n generators of commuting cosine functions)is given by Keyantuo [Key95b].

Chapter 8

Translation Invariant Operatorson Lp(Rn)

In this chapter we consider differential operators with constant coefficients, andmore generally pseudo-differential operators on Lp(Rn). The realization Opp(a) inLp(Rn) of such an operator is translation invariant. We assume that the “symbol”a satisfies certain smoothness and growth assumptions. In particular, when a isa polynomial, then Opp(a) is a differential operator. In the following sections weinvestigate the question under which conditions on the symbol a the operatorOpp(a) generates a C0-semigroup or an integrated semigroup on Lp(Rn). This

question is closely related to the problem whether eta or∫ t

0(t−s)k−1

k! esa ds is aFourier multiplier for Lp(Rn). In Section 8.2 (see also Appendix E) we considerFourier multipliers in some detail. Since we are interested in the case p = 1 aswell as 1 < p <∞, we need to include Fourier multipliers on L1(Rn). Bernstein’slemma and a partition of unity argument are our main tools. Proposition 8.2.3gives a simple criterion for a function to belong to the Fourier algebra FL1(Rn)and hence to be a Fourier multiplier on L1(Rn). These techniques are essential forour main results.

Assuming a suitable growth condition on the symbol a, which is in particularfulfilled for elliptic and even hypoelliptic polynomials, we prove in Section 8.3 thatthe operator Opp(a) associated to a generates a k-times integrated semigroup onLp(Rn) for some k ∈ N provided ρ(Opp(a)) �= ∅ and the range of a lies in a lefthalf-plane. Observe that the result covers the case of the operator iΔ which hasalready been considered in Section 3.9 (Theorem 3.9.4 and Corollary 3.9.14). It isinteresting to note that the order of integration stated in Theorem 8.3.6 is in factoptimal for homogeneous symbols of the form a(ξ) = i|ξ|m. These results are alsoclosely related to the existence of the boundary group of the Poisson semigroupon Lp(Rn) (see Corollary 8.3.11) and to Littman’s result on the cosine functiongenerated by the Laplacian on Lp(Rn) (see Theorem 8.3.12).

430 8. TRANSLATION INVARIANT OPERATORS ON LP (RN )

In the final section of this chapter, we consider systems of differential op-erators with constant coefficients on Lp-spaces. Brenner’s result (Theorem 8.4.3)states that first order symmetric, hyperbolic systems generate C0-semigroups onLp(Rn)N for p �= 2 if and only if the matrices commute. This means that thesolutions of the wave equation, Maxwell’s equation or Dirac’s equation are notgoverned by a C0-semigroup on Lp-spaces if p �= 2. We prove, however, that thesolutions are given by integrated semigroups on these spaces.

8.1 Translation Invariant Operators and C0-semigroups

In this section we consider Cauchy problems{u′(t) = Au(t) (t ≥ 0),

u(0) = u0,

where A is the realization of a pseudo-differential operator in a function space Xof the form X = Lp(Rn) (1 ≤ p <∞) or C0(Rn), and u0 ∈ X. More precisely, letm > 0 and ρ ∈ [0, 1]. We define Sm

ρ,0 to be the set of all functions a ∈ C∞(Rn)such that for each multi-index α ∈ Nn

0 there exists a constant Cα such that

|Dαa(ξ)| ≤ Cα(1 + |ξ|)m−ρ|α| (ξ ∈ Rn).

Obviously, a polynomial of order m belongs to Sm1,0. We call a ∈ C∞(Rn) a symbol

if a ∈ Smρ,0 for some m > 0 and some ρ ∈ [0, 1]. For a symbol a we define the

pseudo-differential operator Op(a) associated to a by

Op(a)u(x) :=

∫Rn

eix·ξa(ξ)Fu(ξ)dξ (x ∈ Rn, u ∈ S(Rn)),

where x · ξ is the scalar product of x and ξ, and Fu denotes the Fourier transformof u. The operator OpX(a) defined by

OpX(a)f := F−1(aFf),

D(OpX(a)) := {f ∈ X : F−1(aFf) ∈ X}, (8.1)

is called the realization of Op(a) in X, or the X-realization of Op(a). When X =Lp(Rn), we may write Opp(a) for OpX(a). Here, F−1(aFf) is interpreted in thesense of distributions (see Appendix E) as follows: as usual, we identify f ∈ Xwith Tf ∈ S(Rn)′ given by

〈ϕ, Tf 〉 =∫Rn

fϕ (ϕ ∈ S(Rn)).

The Fourier transform is an isomorphism of S(Rn)′ (see (E.10)) which implies thatFf ∈ S(Rn)′. Since a and all its derivatives are polynomially bounded, a · Ff is a

8.1. TRANSLATION INVARIANT OPERATORS AND C0-SEMIGROUPS 431

well defined element of S(Rn)′ (see (E.3)). Hence, F−1(aFf) ∈ S(Rn)′ since F isan isomorphism of S(Rn)′.

It is not difficult to verify that OpX(a) is a closed operator in X whenevera is a symbol. In addition, OpX(a) is densely defined since S(Rn) ⊂ D(OpX(a)).Moreover, by (E.12), OpX(a) is a differential operator of order m with constantcoefficients aα ∈ C, i.e.,

OpX(a)f =∑|α|≤m

aαDαf (f ∈ D(OpX(a))), (8.2)

when a is the polynomial of order m of the form

a(ξ) =∑|α|≤m

aα(iξ)α (ξ ∈ Rn). (8.3)

A polynomial a of the form (8.3) is called elliptic if its principal part am, definedby

am(ξ) :=∑|α|=m

aα(iξ)α (ξ ∈ Rn),

vanishes only at ξ = 0. We call OpX(a) an elliptic operator on X if a is an ellipticpolynomial. Moreover, a polynomial a is called hypoelliptic if

Dαa(ξ)

a(ξ)→ 0 as |ξ| → ∞

whenever |α| �= 0.Our first lemma in this section shows that for operators OpX(a) under con-

sideration there is a close relationship between the resolvent set of OpX(a) andFourier multipliers for X. For the definition of Fourier multipliers and the spaceMX(Rn) we refer to Appendix E, but we note that a symbol a is a Fourier multi-plier for X if and only if OpX(a) is a bounded operator on X. In order to simplifyour notation, we also write AX for OpX(a) if no confusion seems likely. Finally,given a symbol a we set

a(Rn) := {a(ξ) : ξ ∈ Rn}.Lemma 8.1.1. Let X be one of the spaces Lp(Rn) (1 ≤ p < ∞) or C0(Rn). Let abe a symbol and let λ ∈ C. Then λ ∈ ρ(AX) if and only if (λ− a) is nowhere zeroand (λ− a)−1 ∈MX(Rn). In particular, a(Rn) ⊂ σ(AX).

Proof. Assume that (λ− a)−1 ∈MX(Rn). For f ∈ X set

Trλf := F−1((λ− a)−1Ff).

If f ∈ D(AX), then Trλ(λ−AX)f = F−1((λ− a)−1(λ− a)Ff) = f . Moreover, iff ∈ X, then Trλf ∈ D(AX) since F−1((λ− a)(λ− a)−1Ff) = f ∈ X. Hence,

(λ−AX)Trλf = F−1((λ− a)(λ− a)−1Ff) = f.

We have therefore proved that (λ−AX) is invertible with inverse operator Trλ .Conversely, let λ ∈ ρ(AX). If f ∈ D(AX), then τaf ∈ D(AX) and AXτaf =

τaAXf , where (τaf)(x) := f(x−a). Hence, R(λ,AX)τa = τaR(λ,AX). PropositionE.1 shows that there exists rλ ∈ MX(Rn) satisfying R(λ,AX)f = F−1(rλFf)(f ∈ X). Therefore,

f = (λ−AX)R(λ,AX)f = F−1((λ− a)rλFf) (f ∈ X),

which implies by the uniqueness theorem for Fourier transforms that (λ−a)rλ = 1a.e. Since a is continuous and rλ is bounded, it follows that λ �∈ a(Rn) and rλ =(λ− a)−1 a.e.

Remark 8.1.2. The above proof shows that, given λ ∈ ρ(AX), we have

R(λ,AX)f = F−1(rλFf) (f ∈ X),

where rλ = (λ− a)−1.

Proposition 8.1.3. Let X be one of the spaces Lp(Rn) (1 ≤ p <∞) or C0(Rn). Leta be a symbol. Then the following assertions are equivalent:

(i) eta ∈MX(Rn) for all t ≥ 0 and there exist constants M,ω ≥ 0 such that

‖eta‖MX (Rn) ≤Meωt (t ≥ 0).

(ii) AX generates a C0-semigroup on X.

Proof. (i) ⇒ (ii): By replacing a(ξ) by a(ξ) − ω, we may assume without loss ofgenerality that ω = 0. It follows from Proposition E.2 and the assumptions that

supξ∈Rn

etRe a(ξ) = ‖eta‖M2(Rn) ≤ ‖eta‖MX(Rn) ≤ M (t ≥ 0). (8.4)

Hence, Re a ≤ 0. For λ > 0 and f ∈ X, the integral∫∞0

e−λtF−1(etaF)f dtconverges in X and it is easy to see that it coincides in S(Rn)′ with F−1((λ −a)−1)Ff . Hence, (λ− a)−1 ∈MX(Rn) and

‖(λ− a)−1‖MX(Rn) ≤∫ ∞

0

e−λt‖eta‖MX(Rn) dt ≤M

∫ ∞

0

e−λtdt ≤ M

λ.

By Lemma 8.1.1, we conclude that (0,∞) ⊂ ρ(AX).For f ∈ X set

T (t)f :=

{F−1(etaFf) (t > 0),

f (t = 0).

By assumption, T (t) ∈ L(X) for all t ≥ 0. In order to prove that the mapping T :R+ → L(X) is strongly continuous, assume, for the time being, that f ∈ D(AX).Since

a

∫ t

0

esads = eta − 1 (t ≥ 0)

8.1. TRANSLATION INVARIANT OPERATORS AND C0-SEMIGROUPS 433

it follows easily that

‖T (t)f − f‖X =

∥∥∥∥∫ t

0

F−1(esaF)AXf ds

∥∥∥∥X

≤Mt‖AXf‖X (t ≥ 0)

for f ∈ D(AX). Since D(AX) is dense in X and since ‖T (t)‖L(X) ≤ M by as-sumption, it follows that T is strongly continuous.

Finally, let f ∈ S(Rn). By Fubini’s theorem,∫ ∞

0

e−λtF−1(etaFf) dt = F−1

(∫ ∞

0

e−λtetaFf dt

)= F−1((λ− a)−1Ff) (λ > 0).

Since (λ− a)−1 ∈MX(Rn), it follows from Remark 8.1.2 that

R(λ,AX) =

∫ ∞

0

e−λtT (t) dt (λ > 0).

Thus, the assertion follows from Theorem 3.1.7.(ii) ⇒ (i): Denote by TX the C0-semigroup generated by AX . Since TX(t)

commutes with translations for all t ≥ 0, it follows from Proposition E.1 that thereexists ut ∈MX(Rn) such that

TX(t)f = F−1(utFf) (f ∈ S(Rn), t ≥ 0).

By Theorem 3.1.7, we have ‖TX(t)‖L(X) ≤ Meωt (t ≥ 0) for some M,ω ≥ 0.Let f ∈ S(Rn), ϕ ∈ C∞c (Rn). For λ sufficiently large, we have by Remark 8.1.2,Definition 3.1.6 and Fubini’s theorem that∫ ∞

0

e−λt〈ϕ, etaFf〉 dt = 〈ϕ, (λ− a)−1Ff〉

= 〈Fϕ,F−1(λ− a)−1Ff〉= 〈Fϕ,R(λ,AX)f〉=

∫ ∞

0

e−λt〈Fϕ,F−1utFf〉 dt.

The uniqueness theorem for Laplace transforms implies that

〈ϕ, etaFf〉 = 〈ϕ, utFf〉for all t ≥ 0, ϕ ∈ C∞c (Rn) and f ∈ S(Rn). This implies that ut = eta a.e. for eacht ≥ 0. Thus, eta ∈MX(Rn) and

‖eta‖MX(Rn) = ‖ut‖MX(Rn) = ‖TX(t)‖L(X) ≤Meωt (t ≥ 0).

Since M2(Rn) = L∞(Rn) (see Proposition E.2 b)), the following corollary isobvious.

Corollary 8.1.4. Let a be a symbol. Then AL2(Rn) generates a C0-semigroup onL2(Rn) if and only if there exists ω ∈ R such that

Re a(ξ) ≤ ω for all ξ ∈ Rn.

A necessary condition for eta to belong to MX(Rn) for t > 0 is given in thenext lemma.

Lemma 8.1.5. Let X be one of the spaces Lp(Rn) (1 ≤ p <∞) or C0(Rn). Considera polynomial a of order m with principal part am. Suppose that eta ∈MX(Rn) forall t ≥ 0 and that ‖eta‖MX (Rn) ≤ Meωt (t ≥ 0) for suitable constants M,ω ≥ 0.Then eam ∈MX(Rn).

Proof. Let a = am+am−1+ · · ·+a0, where each term aj is homogeneous of degreej (j = 0, . . . ,m). The change of variables ξ �→ t−1/mξ implies by Proposition E.2e) that ut, defined by

ut(ξ) := eam(ξ)etam−1(t−1/mξ) . . . eta0 ,

belongs toMX(Rn) and ‖eta‖MX(Rn) = ‖ut‖MX(Rn) for all t > 0. By assumption,there exists C > 0 such that ‖ut‖MX(Rn) ≤ C for all t ∈ (0, 1). Since limt↓0 ut(ξ) =

eam(ξ) for all ξ ∈ Rn, it follows from Proposition E.2 f) that eam ∈MX(Rn).

When AX is a first order differential operator of the form

AXf =

n∑j=1

ajDjf + a0f,

where aj ∈ C (j = 0, 1, . . . , n), then AX generates a C0-semigroup given by

(T (t)f)(x) = ea0tf(x+ ta),

where a = (a1, . . . , an), X = Lp(Rn) (1 ≤ p <∞) or X = C0(Rn). In the followingproposition a converse assertion is proved.

Proposition 8.1.6. Let X be one of the spaces Lp(Rn) (1 ≤ p < ∞, p �= 2) orC0(Rn). Assume that AX is a differential operator of the form (8.2) on X suchthat the symbol of the principal part am satisfies Re am = 0. Then AX generatesa C0-semigroup on X if and only if the order m of AX is 1.

Proof. It follows from Proposition 8.1.3 and Lemma 8.1.5 that eam ∈Mp(Rn) forX = Lp(Rn) and eam ∈M∞(Rn) for X = C0(Rn). Since Ream = 0 and p �= 2 byassumption, it follows from Theorem E.4 a) that m = 1.

Conversely, if m = 1, then ‖eta‖MX(Rn) = eta0 for all t ≥ 0 and the assertionfollows from Proposition 8.1.3.

Note that the special case of the symbol a(ξ) = −i|ξ|2 was already consideredin Theorem 3.9.4.

8.2. FOURIER MULTIPLIERS 435

8.2 Fourier Multipliers

In this section on Fourier multipliers we give several sufficient conditions for afunction to be a Fourier multiplier for Lp(Rn). The results presented in the fol-lowing are the basis of our subsequent analysis of Cauchy problems in Lp(Rn)corresponding to operators of the form (8.1).

We start with a classical result due to Bernstein. Recall from Appendix E thatfor j ∈ N0 the space Hj(Rn) is defined to be the space of all functions f ∈ L2(Rn)whose distributional derivatives Dαf belong to L2(Rn) for |α| ≤ j. Plancherel’stheorem implies that f ∈ Hj(Rn) if and only if ξ �→ ξαFf(ξ) belong to L2(Rn)for all |α| ≤ j. It is not hard to verify that there exist constants C1, C2 > 0 suchthat

C1(1 + |ξ|2)j ≤∑|α|≤j

|ξα|2 ≤ C2(1 + |ξ|2)j (ξ ∈ Rn),

from which it follows that f ∈ Hj(Rn) if and only if ξ �→ (1 + |ξ|2)j/2Ff(ξ)

belongs to L2(Rn), and that the norm(∑

|α|≤j ‖Dαf‖2L2(Rn)

)1/2is equivalent to

‖(1 + | · |2)j/2Ff(·)‖L2(Rn). Thus,

Hj(Rn) ={f ∈ L2(Rn) : F−1((1 + | · |2)j/2Ff(·)) ∈ L2(Rn)

}.

Lemma 8.2.1 (Bernstein). Let u ∈ Hj(Rn) for some j > n2 . Then Fu ∈ L1(Rn)

and there exists a constant C (depending only on n and j) such that

‖Fu‖L1(Rn) ≤ C‖u‖1−(n/2j)L2(Rn)

( ∑|α|=j

‖Dαu‖L2(Rn)

)n/2j

(u ∈ Hj(Rn)).

Proof. For R > 0, we obtain by the Cauchy-Schwarz inequality and Plancherel’stheorem,

‖Fu‖L1(Rn) =

∫|ξ|≤R

1 · |Fu(ξ)| dξ +∫|ξ|≥R

|ξ|−j |ξ|j |Fu(ξ)| dξ

≤(∫

|ξ|≤R

1 dξ

)1/2

‖u‖L2(Rn)

+

(∫|ξ|≥R

|ξ|−2j dξ

)1/2 (∫|ξ|≥R

|ξ|2j |Fu(ξ)|2 dξ)1/2

≤ CRn/2‖u‖L2(Rn) + CR(n/2)−j∑|α|=j

‖Dαu‖L2(Rn)

for some constant C (depending on n and j). The assertion follows by choosing

R := ‖u‖−1/jL2(Rn)

(∑|α|=j ‖Dαu‖L2(Rn)

)1/j

.

Our next result on a “partition of unity” will be very useful in the sequel.

Lemma 8.2.2. There exists ϕ ∈ C∞c (Rn) satisfying ϕ ≥ 0, suppϕ ⊂ {ξ ∈ Rn :12< |ξ| < 2

}, and ∑

k∈Zϕ(2−kξ) = 1 (ξ �= 0).

Proof. Choose f ∈ C∞c (Rn) such that supp f ⊂ {ξ ∈ Rn : 12< |ξ| < 2}, f ≥ 0

and f(ξ) > 0 if 1√2≤ |ξ| ≤√2. For ξ ∈ Rn, set f0(ξ) :=

∑k∈Z f(2

−kξ). Then

f0 ∈ C∞(Rn\{0}), f0(ξ) > 0 for all ξ ∈ Rn \ {0} and f0(2−kξ) = f0(ξ) for all

k ∈ Z and all ξ �= 0. Hence, the function ϕ defined by

ϕ(ξ) :=

⎧⎨⎩0 (ξ = 0),f(ξ)

f0(ξ)(ξ �= 0),

satisfies the desired assertions.

A very efficient sufficient condition for a function to belong to Mp(Rn) isgiven by Mikhlin’s theorem (see Theorem E.3). In fact, let j := min{k ∈ N : k >n2} and define MM as

MM :={m : Rn → C : m ∈ Cj(Rn\{0}), |m|M <∞}

,

where the norm | · |M is defined as

|m|M := max|α|≤j

supξ∈Rn\{0}

|ξ||α||Dαm(ξ)|.

Mikhlin’s theorem then states that MM ↪→ MLp(Rn) provided 1 < p < ∞ (seeTheorem E.3). Note that Mikhlin’s theorem does not hold for p = 1. In the fol-lowing we give a simple criterion for a function to belong to ML1(Rn). We setFL1(Rn) := {Fg : g ∈ L1(Rn)} = {f ∈ C0(Rn) : Ff ∈ L1(Rn)}. This space is aBanach space for the norm inherited from L1(Rn); i.e.,

‖f‖FL1(Rn) := ‖F−1f‖L1(Rn) = (2π)−n‖Ff‖L1(Rn).

The convolution theorem for Fourier transforms shows that FL1(Rn) ⊂ M1(Rn)isometrically; i.e., ‖f‖FL1(Rn) = ‖f‖M1(Rn) for f ∈ FL1(Rn). Moreover, by Propo-sition E.2, FL1(Rn) ↪→Mp(Rn) for 1 ≤ p ≤ ∞. Let ε > 0 and put j := min{k ∈N : k > n

2 }. Define

Mε := {m ∈ Cj(Rn) : |m|Mε<∞},

where|m|Mε

:= max|α|≤j

supξ∈Rn

|ξ||α|+ε|Dαm(ξ)|.

Then (Mε, | · |Mε) is a Banach space and the following holds true.

Proposition 8.2.3. Let ε > 0. Then Mε ↪→ FL1(Rn).

Proof. Let m ∈ Mε. Choose ψ ∈ C∞c (Rn) such that ψ(ξ) = 1 whenever |ξ| ≤ 2and write m = ψm + (1 − ψ)m. It follows from Bernstein’s Lemma 8.2.1 thatF(ψm) ∈ L1(Rn). We may therefore assume that m(ξ) = 0 whenever |ξ| ≤ 2.

Let ϕ be a function as in Lemma 8.2.2, and for k ∈ Z set

mk := mϕk, ϕk(ξ) := ϕ(2−kξ) (ξ ∈ Rn).

Then m =∑∞

k=1 mk. By Leibniz’s rule, we obtain for α with |α| ≤ j,

|Dαmk(ξ)| =

∣∣∣∣ ∑β≤α

(α

β

)Dα−βm(ξ)2−k|β|(Dβψ)(2−kξ)

∣∣∣∣≤ C

∑β≤α

(α

β

)|ξ|−(|α−β|+ε)2−k|β|‖Dβψ‖∞

≤ C∑β≤α

(α

β

)2−k|β|

2k(|α−β|+ε)

≤ C2−k(|α|+ε),

where C denotes a constant (which may differ from line to line). Here, we usedthe fact that 2k−1 < |ξ| for ξ ∈ supp mk. The L2-norm of Dαmk may hence beestimated as follows:

‖Dαmk‖L2(Rn) ≤ C2−k(|α|+ε)

(∫2k−1<|ξ|<2k+1

1 dξ

)1/2

≤ C2−k(|α|+ε)(2kn

)1/2(|α| ≤ j).

This implies that

‖Dαmk‖L2(Rn) ≤ C2−k(|α|+ε−(n/2)) (k ≥ 1, |α| ≤ j).

Hence, mk ∈ Hj(Rn) and it follows from Lemma 8.2.1 that

‖Fmk‖L1(Rn) ≤ C(2−k(ε−(n/2))

)1−(n/2j) (2−k(j+ε−(n/2))

)n/2j

= C2−kε.

Therefore, ‖Fm‖L1(Rn) ≤ ∑∞k=1 ‖Fmk‖L1(Rn) < ∞ and it follows that m ∈

FL1(Rn). The closed graph theorem implies that the embeddingMε ↪→ FL1(Rn)is continuous (in fact, the constants above are proportional to |m|Mε

).

Lemma 8.2.4. Let a : Rn → C be continuous such that a ∈ Cj(Rn\{0}), wherej = min{k ∈ N : k > n

2 }. Assume that there exist constants m > 0 and Cα > 0such that

|Dαa(ξ)| ≤ Cα|ξ|m−|α| (0 < |α| ≤ j, |ξ| ≤ 1, ξ �= 0).

Let ψ ∈ C∞c (Rn) such that ψ(ξ) = 1 for all ξ ∈ Rn with |ξ| ≤ 1. Then aψ ∈FL1(Rn).

Proof. Replacing a(ξ) by a(ξ)− a(0), we may assume that a(0) = 0. Then

|a(ξ)| =∣∣∣∣∫ 1

0

ξ · ∇a(tξ) dt

∣∣∣∣ ≤ C0|ξ|m

for some constant C0.For k ∈ Z−, set vk(ξ) := a(ξ)ψ(ξ)ϕ(2−kξ), where ϕ is a function as in

Lemma 8.2.2. The L2-norm of Dαvk may be estimated exactly as in the proofof Proposition 8.2.3, giving

‖Dαvk‖L2(Rn) ≤ C2k(m−|α|+(n/2)) (k ≤ −1, |α| ≤ j).

It follows from Lemma 8.2.1 that

‖Fvk‖L1(Rn) ≤ C2kmn/2j.

Thus,−1∑

k=−∞‖Fvk‖L1(Rn) ≤ C

∞∑k=1

2−kmn/2j <∞,

and it follows that∑−1

k=−∞ vk ∈ FL1(Rn). Since aψ −∑−1k=−∞ vk ∈ Hj(Rn),

Lemma 8.2.1 gives the result.

Examples 8.2.5. The assumptions of Lemma 8.2.4 are in particular satisfied forthe functions a : Rn → C defined by

a) a(ξ) := c|ξ|m (m > 0, c ∈ C);

b) a(ξ) := ei|ξ|m

(m > 0);

c) a(ξ) :=

∫ 1

0

(1− s)k−1

(k − 1)!eis|ξ|

m

ds (m > 0, k ∈ N).

Proposition 8.2.6. Let 1 ≤ p ≤ ∞ and let m ∈ Cj(Rn) for some j > n2. Suppose

that m(ξ) = 0 whenever |ξ| ≤ 1. Let ε > 0 and ρ ∈ (−∞, 1]. Assume that thereexists a constant M ≥ 1 such that

sup0<|α|≤j

(sup|ξ|≥1

|Dαm(ξ)| |ξ|ε+ρ|α|)1/|α|

≤ M,

sup|ξ|≥1

|m(ξ)‖ξ|ε ≤ M.

If ε > n∣∣12− 1

p

∣∣(1−ρ), then m ∈Mp(Rn) and there exists a constant C (depending

on n, p, ρ and ε but otherwise independent of m and M) such that

‖m‖Mp(Rn) ≤ CM1+n| 12− 1p |.

Proof. By Proposition E.2 c), we may assume without loss of generality that 1 ≤p ≤ 2. Let ϕ be a function as in Lemma 8.2.2. For k ∈ Z, put mk := mϕk, whereϕk(ξ) = ϕ(2−kξ) for ξ ∈ Rn. We claim that

‖m‖Mp(Rn) ≤∞∑k=0

‖mk‖Mp(Rn) <∞.

Observe that the first inequality follows from the assumption that m(ξ) = 0 for|ξ| ≤ 1. In order to estimate ‖mk‖Mp(Rn), note that |ξ| > 2k−1 for ξ ∈ supp mk.By Leibniz’s rule, we have

|Dαmk(ξ)| =

∣∣∣∣∑β≤α

(α

β

)Dα−βm(ξ)2−k|β|(Dβϕ)(2−kξ)

∣∣∣∣≤

{C0M2−kε (|α| = 0),

CαM|α|2k(−ε−ρ|α|) (|α| �= 0),

for suitable constants C0, Cα > 0. Consequently, there exist constants Cαn suchthat

‖Dαmk‖L2(Rn) ≤{C0nM2−kε2kn/2 (|α| = 0),

CαnM|α|2k(−ε−ρ|α|)2kn/2 (|α| �= 0).

Choosing now j > n2, we conclude by Lemma 8.2.1 that

‖mk‖M1(Rn) = ‖mk‖FL1(Rn)

≤ C(M2−kε2kn/2

)1−(n/2j) (M j2k(−ε−ρj)2kn/2

)n/2j

≤ CM1−(n/2j)Mn/22k(−ε+n(1−ρ)/2),

for a suitable constant C > 0. Setting θ := 2(1− 1

p

)for p ∈ (1, 2), it follows from

Proposition E.2 d) that

‖mk‖Mp(Rn) ≤ ‖mk‖1−θM1(Rn)‖mk‖θM2(Rn)

≤ CM1+n| 12− 1p |2k(−ε+(1−ρ)n| 12− 1

p |).

Thus,∑∞

k=0 ‖mk‖Mp(Rn) <∞ and the proof is complete.

For a symbol a ∈ Smρ,0 and r > 0, we consider the following growth hypothesis:


(Hr): There exist constants C,L > 0 such that

|a(ξ)| ≥ C|ξ|r

for all ξ ∈ Rn with |ξ| ≥ L.

It is clear that if a ∈ Smρ,0 satisfies (Hr) then r ≤ m.

Remark 8.2.7. We note that by the Seidenberg-Tarski theorem (see [Hor83, The-orem 11.1.3]), Hypothesis (Hr) is in particular satisfied for all polynomials a sat-isfying |a(ξ)| → ∞ as |ξ| → ∞. Hence, assumption (Hr) holds for hypoellipticpolynomials. If a is an elliptic polynomial of order m, then (Hr) is satisfied withr = m.

Lemma 8.2.8. Let 1 ≤ p ≤ ∞, N ∈ N,m ∈ (0,∞), ρ ∈ [0, 1] and r > 0. Supposethat a ∈ Sm

ρ,0 satisfies (Hr) and that 0 �∈ a(Rn). If N > n∣∣12 − 1

p

∣∣ (m−ρ−r+1r

), then

a−N ∈Mp(Rn).

Proof. Let ψ ∈ C∞c (Rn) such that

ψ(ξ) :=

{1 (|ξ| ≤ max (L, 1)),

0 (|ξ| ≥ L+ 1),

where L is the constant arising in Hypothesis (Hr). Then, writing a−N = ψa−N +(1−ψ)a−N , we conclude by Lemma 8.2.1 that it suffices to prove that (1−ψ)a−N ∈Mp(Rn). Now, Dα((1 − ψ)a−N )(ξ) = Dα(a−N )(ξ) for |ξ| ≥ L + 1. Using theassumption that a ∈ Sm

ρ,0 and a satisfies (Hr), and noting that r ≤ m, one seesthat

|Dα(a−N )(ξ)| ≤ Cα|ξ|−rN+(m−r−ρ)|α| (|ξ| ≥ L+ 1),

for suitable constants Cα. Hence, the assertion follows from Proposition 8.2.6.

Lemma 8.2.9. Let 1 ≤ p ≤ ∞, N ∈ N, m ∈ (0,∞), ρ ∈ [0, 1], r > 0 and leta ∈ Sm

ρ,0. Assume that supξ∈Rn Re a(ξ) ≤ −1 and that Hypothesis (Hr) is satisfied.

If N > n∣∣ 12 − 1

p

∣∣ (1+m−ρr

), then etaa−N ∈ Mp(Rn) and there exists a constant C

(depending on N,n, ρ, p,m and r but otherwise independent of a) such that

‖etaa−N‖Mp(Rn) ≤ C(1 + t)n| 12− 1p | (t ≥ 0).

Proof. By Proposition E.2 c), we may restrict ourselves to the case 1 ≤ p ≤ 2. Letψ ∈ C∞c (Rn) such that 0 ≤ ψ ≤ 1 and

ψ(ξ) :=

{1 (|ξ| ≤ L1),

0 (|ξ| ≥ L1 + 1),

8.3. LP -SPECTRA AND INTEGRATED SEMIGROUPS 441

where L1 := max(L,C−1/r, 1) and C,L are the constants appearing in (Hr). Fort ≥ 0, we set ut := etaa−N . By Lemma 8.2.1, we conclude that ψut ∈ M1(Rn)and that

‖ψut‖M1(Rn) ≤ Cn(1 + t)n/2 (t ≥ 0),

for some constant Cn. Since

‖ψut‖M2(Rn) = ‖ψut‖L∞(Rn) ≤ 1

for all t ≥ 0, it follows from Proposition E.2 d) that

‖ψut‖Mp(Rn) ≤ Cn(1 + t)n| 12− 1p | (t ≥ 0),

for a suitable constant Cn. Writing ut = ψut + (1 − ψ)ut, we conclude that itremains to prove the assertion for (1− ψ)ut instead of ut. Now, by Leibniz’s rule,

Dαut =∑

β+γ=α

α!

β!γ!Dβ(eta)Dγ(a−N ) (t ≥ 0).

Since a ∈ Smρ,0, we have

|(Dβeta)(ξ)| ≤ Cβ(1 + t)|β||ξ||β|(m−ρ) (|ξ| ≥ L).

As in the proof of Lemma 8.2.8,

|(Dγa−N )(ξ)| ≤ Cγ |ξ|−rN+|γ|(m−r−ρ) (|ξ| ≥ L),

and it follows that there exists a constant C > 1 such that

sup0<|α|≤j

sup|ξ|≥1

(|Dα [(1− ψ)(ξ)ut(ξ)] ||ξ|rN+|α|(ρ−m)

)1/|α|≤ C(1 + t) (t > 0).

Since L ≥ C−1/r, we see that there exists a constant C > 1 such that

sup|ξ|≥1

((1− ψ)(ξ) ut)(ξ)|ξ|rN ≤ C

for t ≥ 0. Hence, the assertion follows from Proposition 8.2.6.

8.3 Lp-spectra and Integrated Semigroups

For a symbol a ∈ Smρ,0 and r > 0, consider again the Hypothesis (Hr) introduced

in the previous Section 8.2:

(Hr): There exist constants C,L > 0 such that

|a(ξ)| ≥ C|ξ|r

for all ξ ∈ Rn with |ξ| ≥ L.

In order to obtain a precise description of σ(Opp(a)), Lemma 8.1.1 shows thatwe need to decide whether or not the function (λ − a)−1 is an Lp-multiplier. Ingeneral, this is a difficult matter. However, if the symbol a satisfies Hypothesis(Hr), the situation is much simpler. Indeed, we have the following result. Recallthat a(Rn) was defined as a(Rn) = {a(ξ) : ξ ∈ Rn}.Proposition 8.3.1. Let 1 ≤ p <∞, m ∈ (0,∞), ρ ∈ [0, 1] and r > 0. Suppose thata ∈ Sm

ρ,0 satisfies (Hr). If ρ(Opp(a)) �= ∅, then σ(Opp(a)) = σ(Op2(a)) = a(Rn).

The following lemma will be useful in the proof of Proposition 8.3.1.

Lemma 8.3.2. Let 1 ≤ p < ∞,m ∈ (0,∞) and ρ ∈ [0, 1]. Suppose that a ∈ Smρ,0

and ρ(Opp(a)) �= ∅. Let q be a polynomial of order k of the form q(t) = cktk +

. . .+ c0 (t ∈ R), with coefficients c0, . . . , ck ∈ C. Then

Opp(q(a)) = q(Opp(a)).

Proof. It is clear that Opp(q(a)) is an extension of q(Opp(a)). Moreover, we have

D(q(Opp(a))) = D(Opp(a)k). We are claiming that D(Opp(q(a))) = D(Opp(a)

k),and we shall prove this by induction on k. For k = 1, this is trivial. Let μ ∈ρ(Opp(a)). Then, there exist d0 ∈ C, dk ∈ C\{0} and polynomials q1, q2 of degreek − 1 such that

q(t) = (μ− t)q1(t) + d0 = dktk + q2(t) (t ∈ R).

For f ∈ D(Opp(q(a))) ⊂ Lp(Rn), we have F−1((μ− a)q1(a)Ff) + d0f ∈ Lp(Rn).Hence, F−1((μ − a)q1(a)Ff) ∈ Lp(Rn). Since μ ∈ ρ(Opp(a)), we have (μ − a)−1

∈ Mp(Rn) by Lemma 8.1.1. Thus, F−1(q1(a)Ff) ∈ Lp(Rn). Therefore, we havef ∈ D(Opp(q1(a))) = D(Opp(a)

k−1) = D(Opp(q2(a))) by the induction hypoth-

esis. Moreover, dkF−1(akFf) = F−1(q(a)Ff) − F−1(q2(a)Ff) ∈ Lp(Rn). Thus,Opp(a)

k−1f ∈ D(Opp(a)) and f ∈ D(Opp(a)k) as required.

Proof of Proposition 8.3.1. Note first that Lemma 8.1.1 together with the fact thatM2(Rn) = L∞(Rn) implies that σ(Op2(a)) coincides with a(Rn). Since a(Rn) ⊂σ(Opp(a)) by Lemma 8.1.1, we only need to prove that σ(Opp(a)) ⊂ a(Rn). Choose

λ ∈ C\a(Rn). The assumption (Hr) and Lemma 8.2.8 imply that (λ − a)−N ∈Mp(Rn) ifN is sufficiently large. Therefore, by Lemma 8.1.1, 0 ∈ ρ(Opp((λ−a)N )).It follows from Lemma 8.3.2 that

Opp((λ− a)N ) =(Opp(λ− a)

)N=(λ−Opp(a)

)N. (8.5)

This implies that (λ−Opp(a))N is invertible. It follows that λ−Opp(a) is invertible

with inverse (λ−Opp(a))N−1((λ−Opp(a))

N )−1, which is bounded by the closedgraph theorem.

A quantitative version of Proposition 8.3.1 is given in the following theorem.

Theorem 8.3.3. Let 1 ≤ p < ∞, m ∈ (0,∞), ρ ∈ [0, 1] and r > 0. Suppose thata ∈ Sm

ρ,0 satisfies (Hr). Then the following assertions hold true:

a) If n∣∣12− 1

p

∣∣ (m−ρ−r+1r

)< 1, then σ(Opp(a)) = σ(Op2(a)).

b) If ρ �= 1, then the bound given in a) is optimal; i.e., if n∣∣12− 1

p

∣∣ (1−ρm

)> 1,

there exists a ∈ Smρ,0, satisfying (Hr) with r = m, such that σ(Opp(a)) �=

σ(Op2(a)).

Proof. The assertion a) follows by combining Proposition 8.3.1 with Lemma 8.2.8

and Lemma 8.1.1. In order to prove assertion b), let m ∈(0, n(1−ρ)

2

)and let

a : Rn → C be a C∞-function such that

a(ξ) :=

{|ξ|mei|ξ|

1−ρ

(|ξ| ≥ 2),

1 (|ξ| ≤ 1),

and |a(ξ)| ≥ 1 for all ξ ∈ Rn. Then a ∈ Smρ,0 and (Hr) is satisfied with r = m. It

follows from Theorem E.4 b) that a−1 ∈Mp(Rn) only if n∣∣12− 1

p

∣∣ ≤ m1−ρ . Therefore,

by Lemma 8.1.1, 0 ∈ σ(Opp(a)) if n∣∣12 − 1

p

∣∣1−ρm > 1. Since 0 �∈ a(Rn) = σ(Op2(a)),

the assertion follows.

Observing that for elliptic polynomials of degree m we have ρ = 1 and (Hr)is satisfied with r = m, we immediately have the following corollary.

Corollary 8.3.4. Let 1 ≤ p <∞ and let a be an elliptic polynomial. Then

σ(Opp(a)) = a(Rn).

Remark 8.3.5. It is worthwhile noticing that assertion a) of Theorem 8.3.3 is nolonger true if Hypothesis (Hr) is not satisfied. In fact, consider the symbol a givenby

a(ξ) := −i(ξ1 + ξ22 + ξ23 − i) (ξ ∈ R3).

Then σ(Op2(a)) = {z ∈ C : Re z = −1}, but, by Theorem E.4 c)(i), we havea−1 �∈ Mp(Rn) if p �= 2. Hence, by Lemma 8.1.1, 0 ∈ σ(Opp(a)) whenever p �= 2.

We now consider the question whether operators associated to symbols a ∈Smρ,0 satisfying Hypothesis (Hr) are generators of integrated semigroups on Lp-

spaces. To this end, let Np be the smallest integer such that

Np > n

∣∣∣∣12 − 1

p

∣∣∣∣ (1 +m− ρ

r

)(1 ≤ p <∞).

We then have the following result (see Corollary 3.9.14 for the special case wherea(ξ) = −i|ξ|2).

Theorem 8.3.6. Let 1 ≤ p < ∞,m ∈ (0,∞), ρ ∈ [0, 1] and r > 0. Suppose thata ∈ Sm

ρ,0 satisfies (Hr). Then the following assertions are equivalent:

(i) ρ(Opp(a)) �= ∅ and supξ∈Rn Re a(ξ) ≤ ω for some ω ∈ R.

(ii) The operator Opp(a) generates an Np-times integrated semigroup on Lp(Rn).

(iii) σ(Opp(a)) ⊂ {z ∈ C : Re z ≤ ω} for some ω ∈ R.

Proof. (i)⇒ (ii): By rescaling we may assume that ω = −1 (see Proposition 3.2.6).It follows from Proposition 8.3.1 that 0 ∈ ρ(Opp(a)). For t ≥ 0 and k ∈ N define

the function ukt : Rn → C by

ukt :=

∫ t

0

(t− s)k−1

(k − 1)!esads. (8.6)

Integrating by parts we obtain

ukt =

eta

ak−

k∑j=1

1

(k − j)!

tk−j

aj. (8.7)

We conclude from Lemma 8.1.1 and from the fact thatMp(Rn) is a Banach algebrathat there exists a constant C such that∥∥∥∥ k∑

j=1

1

(k − j)!

tk−j

aj

∥∥∥∥Mp(Rn)

≤ C(1 + t)k−1 (t ≥ 0). (8.8)

By assumption, the symbol a satisfies (Hr). It thus follows from Lemma 8.2.9 that∥∥∥∥ eta

aNp

∥∥∥∥Mp(Rn)

≤ C(1 + t)n| 12− 1p | (t ≥ 0), (8.9)

for some constant C. Combining (8.8) with (8.9) it follows that uNp

t ∈ Mp(Rn)for all t ≥ 0 and that

‖uNp

t ‖Mp(Rn) ≤ C(1 + t)α (t ≥ 0),

for some constants C,α.For f ∈ Lp(Rn) and t ≥ 0 set

S(t)f := F−1(uNp

t Ff).

Since

a

∫ t

0

uNps ds = u

Np

t − tNp

(Np)!(t ≥ 0),

it follows that for f ∈ S(Rn) and r > t ≥ 0 we have

‖S(t)f − S(r)f‖Lp(Rn)

≤ ‖Opp(a)f‖Lp(Rn)

∫ r

t

‖uNps ‖Mp(Rn) ds+ (rNp − tNp)‖f‖Lp(Rn).

Thus, S(·)f : R+ → Lp(Rn) is continuous. Since S(Rn) is dense in Lp(Rn) and Sis locally bounded, it follows that S : R+ → L(Lp(Rn)) is strongly continuous.

It remains to show that the generator of S is Opp(a). To this end, fix λ > 0and let f ∈ S(Rn). It follows from Fubini’s theorem that∫ ∞

0

e−λtS(t)f dt =

∫ ∞

0

e−λtF−1(uNp

t Ff) dt

= F−1

(∫ ∞

0

e−λtuNp

t dtFf

)= F−1

(1

λNp(λ− a)−1Ff

)=

1

λNpR(λ,Opp(a))f.

Since S(Rn) is dense in Lp(Rn), the assertion follows by Definition 3.2.1.(ii) ⇒ (iii): This follows from the definition of an integrated semigroup.(iii) ⇒ (i): This is a consequence of Proposition 8.3.1.

The following is immediate from Theorem 8.3.3 and Theorem 8.3.6.

Corollary 8.3.7. Let 1 ≤ p < ∞, m ∈ (0,∞), ρ ∈ [0, 1] and r > 0. Assume thata ∈ Sm

ρ,0 satisfies (Hr) and supξ∈Rn Re a(ξ) ≤ ω for some ω ∈ R. For N ∈ N,there exists a constant δN > 0 such that Opp(a) generates an N -times integrated

semigroup on Lp(Rn) provided∣∣12 − 1

p

∣∣ < δN .

Example 8.3.8. The example of the symbol a given by

a(ξ) := (−i)(ξ1 − ξ22 − ξ23 − i)(ξ1 + ξ22 + ξ23 + i) (ξ ∈ R3)

shows that Opp(a) generates an integrated semigroup on Lp(R3) only for certainvalues of p. Indeed, we verify that supξ Re a(ξ) = 0 and (Hr) is satisfied with

r = 1. Hence, by Theorem 8.3.3, we see that ρ(Opp(a)) �= ∅ provided∣∣12− 1

p

∣∣ <19. Therefore, Opp(a) generates a once integrated semigroup on Lp(R3) provided∣∣12 − 1

p

∣∣ < 112 . However, by Theorem E.4 c)(ii), σ(Opp(a)) �= a(Rn) if

∣∣12 − 1

p

∣∣ > 38 .

Proposition 8.3.1 implies that σ(Opp(a)) = C if∣∣12− 1

p

∣∣ > 38. Thus, Opp(a) does not

generate an N -times integrated semigroup on Lp(R3) for any N for those valuesof p.

We consider now the case where a is no longer a symbol belonging to Smρ,0

but a is a homogeneous function of the form

a(ξ) = i|ξ|m or a(ξ) = −i|ξ|m (ξ ∈ Rn) (8.10)

for some m > 0. In that case, the realization of the pseudo-differential operatorassociated to a in function spaces X of the form Lp(Rn) (1 ≤ p < ∞) or C0(Rn)is defined as follows. For f ∈ X and a of the form (8.10), define aFf ∈ S(Rn)′ asthe mapping

ϕ �→∫Rn

fF(aϕ) dx (ϕ ∈ S(Rn)). (8.11)

Notice that F(aϕ) ∈ L2(Rn)∩L∞(Rn) by Plancherel’s theorem and the Riemann-Lebesgue lemma. We even have F(aϕ) ∈ L1(Rn) by Lemma 8.2.4 and Example8.2.5. Together with the inequality

‖f‖q ≤ ‖f‖θ1‖f‖1−θ∞ (1 < q <∞, θ := 1/q),

this implies that F(aϕ) ∈ Lq(Rn) for 1 ≤ q ≤ ∞. It follows that the mappinggiven in (8.11) is well defined. It is also not difficult to verify that aFf ∈ S(Rn)′.We thus define, for a of the form (8.10),

OpX(a) := F−1(aFf),

D(OpX(a)) := {f ∈ X : F−1(aFf) ∈ X}. (8.12)

We note that the assertions of Lemma 8.1.1 and Remark 8.1.2 remain true if thesymbol a is replaced by a homogeneous function a of the form (8.10).

By the proof of Theorem 8.3.6, the operator Opp(a) := OpLp(Rn)(a), definedas in (8.12), generates a k-times integrated semigroup on Lp(Rn) (1 ≤ p <∞) forsome k > 0 if and only if uk

t ∈Mp(Rn), where ukt is defined by

ukt (ξ) :=

∫ t

0

(t− s)k−1

(k − 1!)esa(ξ) ds (ξ ∈ Rn).

Notice first that

ukt (ξ) = tk

∫ 1

0

(1− s)k−1

(k − 1)!esta(ξ) ds (ξ ∈ Rn).

The change of variables ξ �→ t−1/mξ implies by Proposition E.2 e) that ukt ∈

Mp(Rn) if and only if uk1 ∈Mp(Rn), and then

‖ukt ‖Mp(Rn) = tk‖uk

1‖Mp(Rn).

In order to determine whether uk1 ∈ Mp(Rn), let ψ ∈ C∞c (Rn) such that

ψ(ξ) = 1 for |ξ| ≤ 1. It follows from Lemma 8.2.4 and from Example 8.2.5 thatψuk

1 ∈ FL1(Rn) ⊂Mp(Rn) (1 ≤ p <∞). Furthermore,

(1− ψ)uk1 = (1− ψ)

ea

ak− (1− ψ)

k∑j=1

1

(k − j)!

1

aj.

Now, since∣∣Dα((1 − ψ)

∑kj=1

1(k−j)!

1aj )(ξ)

∣∣ ≤ Cα

|ξ|m+|α| for |α| ≤ l with l > n2 , it

follows from Proposition 8.2.3 that the second term on the right-hand side abovebelongs to Mp(Rn). By Theorem E.4 b), we conclude that uk

1 ∈ Mp(Rn) when1 n2 )

in the case m �= 1; and uk1 ∈Mp(Rn) when 1 n−12 ) for the case m = 1.

Consider now the case k = 0. It follows from the proof of Proposition 8.1.3that Opp(a) := OpLp(Rn)(a), defined as in (8.12), generates a C0-semigroup on

Lp(Rn) (1 ≤ p < ∞) if and only if u0t : ξ �→ eit|ξ|

m ∈ Mp(Rn) and ‖u0t‖Mp(Rn)

is exponentially bounded in t. The change of variables ξ �→ t−1/mξ implies byProposition E.2 e) that ‖u0

t‖Mp(Rn) = ‖u01‖Mp(Rn) for all t > 0. Let now ψ ∈

C∞c (Rn) such that ψ(ξ) = 1 for ξ ∈ Rn with |ξ| ≤ 1, and write u01 = u0

1ψ+u01(1−ψ).

Then u01ψ ∈Mp(Rn) by Lemma 8.2.4 and Example 8.2.5. It follows from Theorem

E.4 b) that u01 ∈ Mp(Rn) when 1 < p < ∞ (respectively, p = 1) if and only if

n∣∣12 − 1

p

∣∣ ≤ 0 (respectively, n2 < 0) in the case m �= 1; and u0

1 ∈ Mp(Rn) when

1 < p < ∞ (respectively, p = 1) if and only if (n − 1)∣∣ 12− 1

p

∣∣ ≤ 0 (respectively,n−12

< 0) for the case m = 1. We have therefore proved the following result.

Theorem 8.3.9. Let 1 ≤ p < ∞, k ∈ N0 and m > 0. Define a : Rn → C bya(ξ) := i|ξ|m.

a) If m �= 1 and 1 n

2).

b) If m = 1 and 1 n−1

2).

Remark 8.3.10. a) We note that the assertions of Theorem 8.3.9 remain true forthe homogeneous function a given by a(ξ) := −i|ξ|m.

b) Let OpC0(a) be the operator on C0(Rn) defined as in (8.12) with Lp(Rn) re-

placed by C0(Rn). Then the assertions of Theorem 8.3.9 remain true if Opp(a) inTheorem 8.3.9 is replaced by OpC0

(a) and 1/p by 0.

Theorem 8.3.9 has interesting consequences for boundary values of holomor-phic semigroups as discussed in Section 3.9. In fact, consider the Poisson semigroupT defined as in Example 3.7.9. There we proved that T is a bounded holomorphicC0-semigroup of angle π/2 on Lp(Rn) for 1 ≤ p < ∞. Its generator is given byApf = F−1(−| · |Ff) for f ∈ D(Ap) = {f ∈ Lp(Rn) : F−1(−| · |Ff) ∈ Lp(Rn)}.Theorem 8.3.9 and Remark 8.3.10 imply the following corollary.

Corollary 8.3.11. a) Let 1 < p < ∞ and let Tp be the Poisson semigroup onLp(Rn). Then Tp admits a boundary group on Lp(Rn) in the sense of Propo-sition 3.9.1 if and only if (n − 1)

∣∣12 − 1

p

∣∣ ≤ 0, i.e., if and only if n = 1 orp = 2.

b) Let T1 be the Poisson semigroup on L1(Rn). Then T1 does not admit aboundary group on L1(Rn) in the sense of Proposition 3.9.1.

Another interesting consequence of Theorem 8.3.9 concerns the questionwhether the Laplacian Δp generates a cosine function on Lp(Rn) (see Example3.7.6, Section 3.9 and Section 3.14). Indeed, let 1 ≤ p <∞ and recall that Δp onLp(Rn) may be written as

Δpf = F−1(−| · |2Ff),

D(Δp) ={f ∈ Lp(Rn) : F−1(−| · |2Ff) ∈ Lp(Rn)

}.

Since Δp generates a bounded C0-semigroup on Lp(Rn) (see Example 3.7.6), itfollows that (0,∞) ⊂ ρ(Δp) and supλ>0 ‖λR(λ,Δp)‖ <∞. Moreover, by Remark8.1.2 we have

R(λ,Δp)f = F−1

(1

λ+ | · |2Ff

)(λ > 0, f ∈ S(Rn).

For the time being, let n = 1. Then Δp = A2p, where Ap is the generator of the C0-

group Tp of shifts considered in Example 3.3.10. By Example 3.14.15, Δp generatesa cosine function Cos on Lp(R) given by

Cos(t) =1

2(Tp(t) + Tp(−t)) (t ∈ R).

For n > 1, the situation is different. Indeed, suppose that Δp generates acosine function Cos on Lp(Rn) for n > 1. For the time being, let 1 < p < ∞.Then Lp(Rn) is a UMD-space and it follows from Theorem 3.16.7 that i(−Δp)

1/2,defined as in Proposition 3.8.2, generates a C0-group U on Lp(Rn). By Example

3.8.5 or by noting that |ξ| = ∫∞0

λ−12

|ξ|λ+|ξ|2 dλ (ξ ∈ Rn), we see that −(−Δp)

1/2

coincides with the generator of the Poisson semigroup T on Lp(Rn); i.e.,

−(−Δp)1/2f = F−1(−|ξ|Ff) (f ∈ S(Rn)).

Since the Poisson semigroup T is a bounded holomorphic C0-semigroup of angleπ/2 on Lp(Rn), i(−Δp)

1/2 generates the boundary semigroup of T in the sense ofProposition 3.9.1. By Corollary 8.3.11, we conclude that this implies that p = 2,since we assumed that n > 1.

Finally, consider the case p = 1 and assume that Δ1 generates a cosinefunction Cos on L1(Rn) for n > 1. Observe that Δ2 generates a cosine functionCos on L2(Rn) given by

Cos(t)f = F−1(cos(t| · |)Ff) (t ∈ R, f ∈ L2(Rn)).

The Riesz-Thorin interpolation theorem [Hor83, Theorem 7.1.12] implies that Δp

generates a cosine function on Lp(Rn) for 1 < p < 2. This contradicts the assertionproved above and we have therefore proved the following result.

Theorem 8.3.12. Let 1 ≤ p < ∞ and assume that the Laplacian Δp generates acosine function on Lp(Rn). Then n = 1 or p = 2.

8.4. SYSTEMS OF DIFFERENTIAL OPERATORS ON LP -SPACES 449

8.4 Systems of Differential Operators on Lp-spaces

In this section we consider initial value problems for systems of the form⎧⎨⎩∂u

∂t= Au (t ≥ 0, x ∈ Rn),

u(0, x) = u0(x) (x ∈ Rn),

where u : R+ × Rn → CN and A is an N ×N -matrix whose entries (Aij)1≤i,j≤N

are differential operators with constant coefficients of order mij in the sense of(8.2). The realization of A in Lp(Rn)N (1 ≤ p < ∞) is defined as follows: leta : Rn → L(CN ) be of the form

a(ξ) :=

⎛⎜⎝ a11(ξ) . . . a1N (ξ)...

...aN1(ξ) . . . aNN (ξ)

⎞⎟⎠ (ξ ∈ Rn), (8.13)

where aij(ξ) :=∑|α|≤mij

aijα(iξ)α. Let m := max{mi,j : 1 ≤ i, j ≤ N}. Then

a(ξ) = a0(ξ)+a1(ξ)+· · ·+am(ξ), where each term aj (0 ≤ j ≤ m), is homogeneousof degree j. The term am is called the principal part of a. For 1 ≤ p <∞ we define

Apf := F−1(aFf),D(Ap) :=

{f ∈ Lp(Rn)N : F−1(aFf) ∈ Lp(Rn)N

}, (8.14)

where the Fourier transform of vector-valued functions is defined by applying thetransform elementwise. The proof of Lemma 8.1.1 and Proposition 8.1.3 imply thefollowing result.

Lemma 8.4.1. The operator A2 generates a C0-semigroup on L2(Rn)N if and onlyif there exists ω ∈ R such that

sup{∥∥et(a(ξ)−ωI)

∥∥ : ξ ∈ Rn, t ≥ 0}<∞. (8.15)

Matrix-valued symbols a satisfying (8.15) have been completely characterizedby Kreiss. The following consequence of his result will be very useful in the sequel.

Proposition 8.4.2. Suppose that a of the form (8.13) satisfies (8.15). Assume thatσ(am(ξ)) ⊂ iR for all ξ ∈ Rn. Then there exists S ∈ L∞(Rn,L(CN )) such thatS(ξ) is invertible and S(ξ)−1a(ξ)S(ξ) is diagonal for all ξ ∈ Rn.

For a proof of Proposition 8.4.2 we refer to [Kre59].In the following we examine in detail the special case of symmetric hyperbolic

systems on Lp(Rn)N . To this end, let a : Rn → L(CN ) be of the form

a(ξ) :=

n∑j=1

Mj(iξj),

where M1, . . . ,Mn are hermitian N ×N -matrices. Then the following holds true.

Theorem 8.4.3 (Brenner). Let 1 ≤ p < ∞ such that p �= 2. Let a : Rn → L(CN )be given by

a(ξ) =n∑

j=1

Mj(iξj) (8.16)

where M1, . . . ,Mn are hermitian matrices. Then Ap generates a C0-semigroup onLp(Rn)N if and only if the matrices M1, . . . ,Mn commute.

We base the proof of Theorem 8.4.3 on the following two lemmas. Here,MN

p (Rn) is the space of all (N × N)-matrices m = (mij) where mij ∈ Mp(Rn)for i, j = 1, 2, . . . , N (see Appendix E).

Lemma 8.4.4. Let 1 ≤ p <∞ such that p �= 2. Let a be of the form (8.16) and setb = −ia. Suppose that ea ∈ MN

p (Rn). Then the eigenvalues λk(·) of b(·) can bechosen in such a way that

λk(ξ) =n∑

j=1

λkjξj (ξ ∈ Rn), (8.17)

where λkj ∈ R for 1 ≤ j ≤ n, 1 ≤ k ≤ N .

Proof. The implicit function theorem implies that there is an open ball U ⊂ Rn

and C∞-functions λk : U → R and uk : U → CN (k = 1, . . . , N) such that, for allξ ∈ U, ‖uk(ξ)‖ ≥ 1, b(ξ)uk(ξ) = λk(ξ)uk(ξ) and {uk(ξ) : k = 1, . . . , N} is a basisof CN . For the time being, we fix k and write λ and u for λk and uk.

Let ξ0 ∈ U and let ψ ∈ C∞c (U) such that ψ(ξ0) = 1. Choose v ∈ C∞c (Rn,CN )such that u(ξ) · v(ξ) = 1 (ξ ∈ U).

For t > 0, we have

ψ(ξ)eitλ(ξ)u(ξ) = ψ(ξ)eta(ξ)u(ξ) (ξ ∈ U),

and therefore

ψ(ξ)eitλ(ξ) = eta(ξ)ψ(ξ)u(ξ) · v(ξ) (ξ ∈ U).

Since ψu, v ∈ C∞c (Rn,CN ), each of their coordinates belongs toMp(Rn). It followsfrom the homogeneity of a and Proposition E.2 e) that there exists a constant Csuch that

‖ψeitλ‖Mp(Rn) ≤ C‖eta‖MNp (Rn) = C‖ea‖MN

p (Rn). (8.18)

Define now μ(ξ) := λ(ξ0 + ξ)− λ(ξ0)− ξ · ∇λ(ξ0) if ξ0 + ξ ∈ U , and set

ft(ξ) :=

{ψ(ξ0 + t−1/2ξ)eitμ(t

−1/2ξ) if ξ0 + t−1/2ξ ∈ U,

0 otherwise.


For g ∈Mp(Rn) and x ∈ Rn, let (τxg)(ξ) := g(ξ − x) and let hx(ξ) := eiξ·x. ThenF−1(τxg · F)ϕ = hx · F−1(gF)(h−x · ϕ) and F−1((hx · g)F)ϕ = τ−x(F−1(gF)ϕ)for all ϕ ∈ S(Rn). It follows that τxg, hx · g ∈Mp(Rn) and

‖τxg‖Mp(Rn) = ‖hx · g‖Mp(Rn) = ‖g‖Mp(Rn).

Using these relations, (8.18) and Proposition E.2 e), we have that ‖ft‖Mp(Rn) ≤C‖ea‖MN

p (Rn) for all t > 0. Moreover, limt→∞ ft(ξ) = eiP (ξ) uniformly on compact

subsets of Rn, where P (ξ) := 12

∑ni,j=1 DiDjλ(ξ0)ξiξj . Proposition E.2 f) implies

that eiP ∈ Mp(Rn). However, since we assumed that p �= 2, Theorem E.4 a)implies that P ≡ 0. Hence, all the second derivatives of λ vanish at an arbitrarypoint ξ0 ∈ U , which implies that λ is linear on U .

We have shown that

λk(ξ) = λk0 +n∑

j=1

λkjξj (ξ ∈ U, k = 1, . . . , N),

where λk0, λkj ∈ R (k = 1, . . . , N, j = 1, . . . , n). It follows that

det(zI − b(ξ)) =

N∏k=1

(z − λk0 −

n∑j=1

λkjξj

)(ξ ∈ U, z ∈ C). (8.19)

Since both sides of the equation above are polynomials in ξ, it follows that (8.19)holds for ξ ∈ Rn and z ∈ C. By homogeneity, λk0 = 0 for all k = 1, . . . , N . Thus,we can choose

λk(ξ) =

n∑j=1

λkjξj (ξ ∈ Rn).

Lemma 8.4.5. Let a : Rn → L(CN ) be of the form (8.16) and let b = −ia. As-sume that the eigenvalues λ(·) of b(·) are of the form (8.17). Then the matricesM1, . . . ,Mn commute.

Proof. Let λj(·), j = 1, . . . , r be the distinct linear functions representing theeigenvalues of b(·) for ξ ∈ Rn. Denote by V the set where two or more eigenvaluescoincide. Then

b(ξ) =

r∑j=1

λj(ξ)Pj(ξ) (ξ ∈ Rn\V ),

where Pj(ξ) are orthogonal projections given by

Pj(ξ) =

∏k �=j(b(ξ)− λk(ξ))∏k �=j(λj(ξ)− λk(ξ))

=:Nj(ξ)

Dj(ξ).

Since ‖Pj(ξ)‖ = 1 for ξ ∈ Rn\V , we have ‖Nj(ξ)‖ = |Dj(ξ)| for ξ ∈ Rn \ V , andhence for all ξ ∈ Rn by continuity. The entries of Nj(ξ) are polynomials, so this

implies that they are divisible by each of the linear factors λj(ξ)−λk(ξ) of Dj(ξ).Since Nj(ξ) and Dj(ξ) both have degree r− 1, it follows that Pj(ξ) is constant forall ξ ∈ Rn\V . Set Pj := Pj(ξ). Then

b(ξ) =

r∑j=1

λj(ξ)Pj (ξ ∈ Rn\V ), (8.20)

and by continuity (8.20) holds for all ξ ∈ Rn. Obviously, the projections Pj com-mute and it thus follows that the matrices M1, . . . ,Mn also commute.

Proof of Theorem 8.4.3. The proof of Theorem 8.4.3 may now be completed as fol-lows. Suppose that Ap generates a C0-semigroup on Lp(Rn). Then, by the proof ofProposition 8.1.3, eta ∈MN

p (Rn) and ‖eta‖MNp (Rn) = ‖ea‖MN

p (Rn) by Proposition

E.2 e). Lemma 8.4.4 and Lemma 8.4.5 imply that the matrices commute.The converse implication is easy to prove. In fact, if M1, . . . ,Mn commute,

then the Mj may be simultaneously diagonalised by a unitary matrix U so thatDj = UMjU

∗, where Dj = diag(λj1, . . . , λjn). Hence,

eta(ξ) = exp

(it

n∑j=1

U∗DjUξj

)= U∗ exp

(it

n∑j=1

Djξj

)U

and it follows that eta ∈ MNp (Rn) and supt≥0 ‖eta‖MN

p (Rn) < ∞. The assertionnow follows as in the proof of Proposition 8.1.3. The C0-semigroup generated byAp may be written explicitly in terms of U and N translation semigroups onLp(Rn).

The following result describes the generalisation of Theorem 8.4.3 to thesituation of systems of arbitrary order m.

Theorem 8.4.6 (Brenner). Let 1 ≤ p < ∞ such that p �= 2. Assume that a :Rn → L(CN ) of the form (8.13) satisfies σ(am(ξ)) ⊂ iR for all ξ ∈ Rn. Then Ap

(defined as in (8.14)) generates a C0-semigroup on Lp(Rn)N if and only if thereexist commuting diagonalisable matrices M1, . . . ,Mn, with real eigenvalues suchthat

am(ξ) =

n∑j=1

Mj(iξj) (ξ ∈ Rn). (8.21)

We do not aim to give here a detailed proof of Theorem 8.4.6. For this werefer to [Bre73]. We only notice that Proposition 8.1.5 generalizes to the situationof systems discussed in Theorem 8.4.6. Hence, if Ap generates a C0-semigroupon Lp(Rn)N , then eta ∈ MN

p (Rn) which implies that eam ∈ MNp (Rn), where am

denotes the principal part of a. One can now show that the order m is necessarily1 and that am is of the form (8.21).

Starting from this situation, we show in the following that the operator Ap

generates a k-times integrated semigroup on Lp(Rn)N for suitable k > 0 provided

A2 generates a C0-semigroup on L2(Rn)N and further additional assumptions onthe symbol a are satisfied. More precisely, the following holds true.

Theorem 8.4.7. Let 1 < p <∞. Assume that a of the form (8.13) is homogeneousof degree m for some m ≥ 1. Suppose that σ(a(ξ)) ⊂ iR for all ξ ∈ Rn and thatthe number of distinct eigenvalues of a(ξ) is constant, and the rank of a(ξ) is alsoconstant, for ξ ∈ Rn\{0}.

a) If A2 generates a C0-semigroup on L2(Rn)N , then Ap generates a k-timesintegrated semigroup on Lp(Rn)N provided k > n|12 − 1

p |.b) If A2 generates a C0-semigroup on L2(Rn)N and in addition σ(a(ξ)) ={iα1|ξ|, . . . , iαN |ξ|} for all ξ ∈ Rn, where α1, . . . , αN ∈ R, then Ap generatesa k-times integrated semigroup on Lp(Rn)N provided k ≥ (n− 1)

∣∣ 12− 1

p

∣∣.The proof of Theorem 8.4.7 is based on the following lemma.

Lemma 8.4.8. Let the assumption of Theorem 8.4.7 a) be satisfied. Then the func-tion uk

t : Rn → L(CN ) defined by

ukt (ξ) :=

∫ t

0

(t− s)k−1

(k − 1)!esa(ξ) ds (t ≥ 0, ξ ∈ Rn)

belongs to MNp (Rn) provided k > n

∣∣12− 1

p

∣∣. If the assumption b) is also satisfied,

then ut ∈MNp (Rn) provided k ≥ (n− 1)

∣∣12 − 1

p

∣∣.Proof. By assumption, a is homogeneous of order m and σ(a(ξ)) ⊂ iR for all ξ ∈Rn. It follows from Lemma 8.4.1 and Proposition 8.4.2 that a(ξ) is diagonalisablefor all ξ ∈ Rn. Therefore and by virtue of our assumptions, the minimal polynomialof the matrix −ia(ξ) has only (K+1) simple roots λl(ξ) (l = 0, . . . ,K) for some Ksatisfying 0 ≤ K ≤ N−1. Observe next that the eigenvalues λl(·) are homogeneousfunctions of degree m since a(·) is homogeneous of degree m.

For t ≥ 0 and ξ ∈ Rn \ {0}, let q be a polynomial of degree K in one variablesuch that q(λl(ξ)) = eitλl(ξ) for l = 0, . . . ,K. Then

eta(ξ) = q(a(ξ)).

We now examine the form of the coefficients Cj(t, ξ) of q.Denote by L(ξ) the (K + 1) × (K + 1)-matrix whose l-th row is given by

(λl(ξ)K , λl(ξ)

K−1, . . . , λl(ξ), 1) (l = 0, . . . ,K). Since

detL(ξ) =∏

l<j≤K

(λl(ξ)− λj(ξ)) �= 0

for all ξ �= 0, we have

Cj(t, ξ) = (detL(ξ))−1K∑l=0

(detClj(ξ))e

itλl(ξ),


where C lj(ξ) is the (K + 1) × (K + 1)-matrix defined by replacing the element

λl(ξ)K−j of L(ξ) by 1 and all other elements of L(ξ) of row l and column j by 0.

Therefore,

eta(ξ) =

K∑j=0

Cj(ξ)(a(ξ))K−j

=

K∑l=0

eitλl(ξ)1

detL(ξ)

K∑j=0

(detClj(ξ))a(ξ)

K−j (t ≥ 0, ξ �= 0).

Moreover, since ξ �→ detL(ξ) is homogeneous of degree h := (K + 1)Km/2 andξ �→ detC l

j(ξ) is homogeneous of degree h− (K − j)m for all l ∈ {0, . . . ,K}, thefunctions Φl : Rn \ {0} → L(CN ) given by

Φl(ξ) :=1

detL(ξ)

K∑j=0

(detC lj(ξ))(a(ξ))

K−j (l = 0, . . . ,K)

are homogeneous of degree 0.Since by assumption the number of distinct eigenvalues of a(ξ) is constant for

ξ ∈ Rn\{0}, it follows from [Kat82, Theorem II.5.13a] that the eigenvalues λl(·)are C∞-functions on Rn\{0}. Hence,

eta(ξ) =

K∑l=0

eitλl(ξ)Φl(ξ), (8.22)

where Φl ∈ C∞(Rn\{0},L(CN )) is homogeneous of degree 0 for l = 0, . . . ,K.It follows from Mikhlin’s theorem (see Theorem E.3) that Φl ∈ MN

p (Rn). Byassumption, either λl is identically zero on Rn \{0} or it is homogeneous of degreem on Rn \ {0} and therefore satisfies (Hr) with r = m. It follows from the proofof Theorem 8.3.9, Proposition 8.2.6 and Lemma 8.2.1 that

ξ �→∫ t

0

(t− s)k−1

(k − 1)!eisλl(ξ)ds

belongs to Mp(Rn) provided k > n∣∣12 − 1

p

∣∣. Since max1≤i,j≤N{‖aij‖Mp(Rn)} is an

equivalent norm to ‖a‖MNp (Rn), we conclude from (8.22) that uk

t ∈ MNp (Rn) if

k > n∣∣ 12 − 1

p

∣∣.If σ(a(ξ)) = {iα1|ξ|, . . . , iαN |ξ|} for α1, . . . , αN ∈ R and all ξ ∈ Rn, then the

representation (8.22) together with Theorem 8.3.9 implies that ukt ∈ MN

p (Rn) if

k ≥ (n− 1)∣∣ 12 − 1

p

∣∣.Proof of Theorem 8.4.7. Thanks to Lemma 8.4.8 it is now no longer difficult toextend the proof of Theorem 8.3.6 to the present situation of systems and to show


that Ap is the generator of a k-times integrated semigroup S on Lp(Rn)N givenby

S(t)f := F−1(uktFf) (t ≥ 0, f ∈ Lp(Rn)N ).

We finish this section by applying Theorem 8.4.7 to certain systems arisingin mathematical physics. We start with the wave equation on Rn.

Example 8.4.9 (Wave equation on Rn). Consider the classical wave equation

wtt = !w (t ∈ R, x ∈ Rn).

Introducing the variable u := (∇w,wt)T , the wave equation can be written as a

symmetric, hyperbolic system with

a(ξ) = i

⎛⎜⎜⎜⎝0 . . . . ξ1...

.... ξnξ1 ξ2 . . ξn 0

⎞⎟⎟⎟⎠(n+1)×(n+1)

(ξ ∈ Rn).

It follows from Theorem 8.4.3 that the operator Ap in Lp(Rn)n+1 associated with adoes not generate a C0-semigroup on Lp(Rn)n+1 if p �= 2 and n > 1. This propertyof the wave equation was observed first by Littman [Lit63]. The eigenvalues of a(ξ)are λ0(ξ) = 0 of multiplicity (n−1) and λ1,2(ξ) = ±i|ξ| each of multiplicity 1 (ξ ∈Rn). Hence, given p ∈ (1,∞), by Theorem 8.4.7, the operator Ap on Lp(Rn)n+1

associated with a generates a k-times integrated semigroup on Lp(Rn)n+1 if k ≥(n− 1)

∣∣12 − 1

p

∣∣.Example 8.4.10 (Maxwell’s equations). We consider Maxwell’s equations in thecase where current and charge densities are zero and units are chosen so that thespeed of light is one. Then Maxwell’s equations can be written as

∂

∂t

(uv

)=

(0 − rotrot 0

)(uv

),

(u(0)v(0)

)=

(u0v0

),

where u, v : R3 → C3. Note that Maxwell’s equations may be rewritten as asymmetric, hyperbolic system satisfying the assumptions of Theorem 8.4.7 b). Infact, a : R3 → L(C6) given by

a(ξ) = i

⎛⎜⎜⎜⎜⎜⎜⎝0 0 0 0 −ξ3 ξ20 0 0 ξ3 0 −ξ10 0 0 −ξ2 ξ1 00 ξ3 −ξ2 0 0 0−ξ3 0 ξ1 0 0 0ξ2 −ξ1 0 0 0 0

⎞⎟⎟⎟⎟⎟⎟⎠has eigenvalues λ0(ξ) = 0, λ1,2(ξ) = ±i|ξ| (ξ ∈ R3), each of multiplicity 2. Hence,by Theorem 8.4.7 b), the Maxwell operator Ap associated to a generates a once

integrated semigroup on Lp(R3)6. Note that, by Theorem 8.4.3, the Maxwell op-erator Ap does not generate a C0-semigroup on Lp(R3)6 if p �= 2.

Example 8.4.11 (Dirac’s equation). The relativistic description of the motion of aparticle of mass m with spin 1/2 is provided by Dirac’s equation

∂u

∂t(x, t) = c

3∑j=1

Mj∂u

∂xj(x, t)−M4

mc2

ihu(x, t) + V (x, t) (x ∈ R3, t ≥ 0).

Here, u is a function defined on R3 × R+ which takes values in C4, c is the speedof light, h is Planck’s constant, and M1,M2,M3,M4 are 4× 4 matrices given by

M1 :=

⎛⎜⎜⎝0 0 0 10 0 1 00 1 0 01 0 0 0

⎞⎟⎟⎠ , M2 :=

⎛⎜⎜⎝0 0 0 −i0 0 i 00 −i 0 0i 0 0 0

⎞⎟⎟⎠ ,

M3 :=

⎛⎜⎜⎝0 0 1 00 0 0 −11 0 0 00 −1 0 0

⎞⎟⎟⎠ , M4 :=

⎛⎜⎜⎝1 0 0 00 1 0 00 0 −1 00 0 0 −1

⎞⎟⎟⎠.

If V ≡ 0 and units are chosen so that all constants are equal to 1, then Dirac’sequation may be written as a symmetric, hyperbolic system on X := Lp(R3)4

(1 < p <∞) of the form

d

dt

(vw

)=

(0 Ap

Ap 0

)(vw

)+ i

(I 00 −I

)(vw

)(t ≥ 0),(

v(0)w(0)

)=

(v0w0

),

where

Ap :=

⎛⎜⎝ ∂

∂x3

∂

∂x1− i

∂

∂x2∂

∂x1+ i

∂

∂x2− ∂

∂x3

⎞⎟⎠with domain D(Ap) := {f ∈ Lp(R3)2 : Apf ∈ Lp(R3)2} in Lp(R3)2.

It follows from Theorem 8.4.3 that the Dirac operator

Dp :=

(0 Ap

Ap 0

)+ i

(I 00 −I

),

with domain D(Dp) := D(Ap) ×D(Ap), generates a C0-semigroup on Lp(R3)4 ifand only if p = 2. In the following, we show that the Dirac operator Dp generatesa twice integrated semigroup on Lp(R3)4 if 1 < p <∞.


In order to do so, let A be a linear operator on a Banach space Y , let A onY × Y be given by

A :=

(0 AA 0

), D(A) := D(A)×D(A),

and let B be the bounded operator on Y ×Y defined by B := i

(I 00 −I

). Then

the following holds true.

Lemma 8.4.12. Let k ∈ N0. Then, the operator A generates an (exponentiallybounded) k-times integrated semigroup on Y ×Y if and only if A and −A generate(exponentially bounded) k-times integrated semigroups on Y .

Proof. Define U ∈ L(Y × Y ) by U := 1√2

(I II −I

). Then A = UDU−1, where

D :=

(A 00 −A

). Since D generates a k-times integrated semigroup if and only

if A and −A both do so, the result follows from the remarks before Theorem3.5.7.

Lemma 8.4.13. Assume that A generates an exponentially bounded once integratedsemigroup on Y × Y . Then A+ B with domain D(A) generates an exponentiallybounded twice integrated semigroup on Y × Y .

Proof. It follows from Lemma 8.4.12 that A and −A generate once integratedsemigroups on Y which are exponentially bounded. Moreover, Proposition 3.15.4implies that A2 generates an exponentially bounded sine function (Sin(t))t≥0 onY . Thus,

R(λ2, A2) =

∫ ∞

0

e−λt Sin(t) dt (λ > abs(Sin)).

We conclude that (λ−A− B) is invertible for λ > abs(Sin) and that

R(λ,A+ B) =(

λ+ i AA λ− i

)R(λ2 + 1, A2) (λ > abs(Sin)).

It follows from Theorem 3.15.6 that A2 − I generates a sine function (SinI(t))t≥0

on Y . Thus,

R(λ2 + 1, A2) = R(λ2, A2 − I) =

∫ ∞

0

e−λt SinI(t) dt,

for λ sufficiently large. For t ≥ 0, we set

S11(t) :=

∫ t

0

SinI(s) ds+ i

∫ t

0

(t− s) SinI(s) ds,

S22(t) :=

∫ t

0

SinI(s) ds− i

∫ t

0

(t− s) SinI(s) ds,

S12(t) := S21(t) := A

∫ t

0

(t− s) SinI(s) ds.

We note that S12(t) and S21(t) are well defined for t ≥ 0, since by Proposition

3.15.2,∫ t

0(t− s) SinI(s)x ds ∈ D(A2) for all x ∈ Y and all t ≥ 0. We now set

S(t) :=

(S11(t) S12(t)S21(t) S22(t)

)(t ≥ 0).

Then abs(S) <∞ and we verify that

R(λ,A+ B) = λ2

∫ ∞

0

e−λtS(t) dt (λ > abs(S)).

Thus, A+ B generates a twice integrated semigroup on Y × Y .

Finally, consider again the situation of the Dirac equation. The eigenvaluesof the symbol of Ap may be computed to be λ1,2(ξ) := ±i|ξ|. Hence, by Theorem8.4.7 it follows that Ap and −Ap generate exponentially bounded once integrated

semigroups on Lp(R3)2. By Lemma 8.4.12, Ap :=

(0 Ap

Ap 0

)generates an expo-

nentially bounded once integrated semigroup on Lp(R3)2 ×Lp(R3)2. Furthermoreby Lemma 8.4.13, Dp = Ap+B generates a twice integrated semigroup on Lp(R3)4.We have thus proved the following result.

Theorem 8.4.14. Let 1 < p < ∞. Then the Dirac operator Dp = Ap + B onLp(R3)4, with domain D(Ap)×D(Ap), generates a twice integrated semigroup onLp(R3)4.

8.5 Notes

Section 8.1Most of the content of this section is more or less standard. The results given in Propo-sition 8.1.3 and Proposition 8.1.6 are based on the well known properties of Fouriermultipliers listed in Appendix E.

Section 8.2An excellent reference for more information on Fourier multipliers is [Ste93]. A proof ofBernstein’s result (Lemma 8.2.1) can be found for instance in [Hor83]. Proposition 8.2.3is due to Hieber [Hie91a]. The remaining part of this section follows the lines of [Hie95].

Section 8.3The result described in Theorem 8.3.3 on Lp-spectral independence for pseudo-differentialoperators is due to Hieber [Hie95]. Corollary 8.3.4 was first shown by Iha and Schu-bert [IS70]. For further results on invariance of the Lp-spectrum of certain classes ofpseudo-differential operators, see [Sch71] and [LS97]. Theorem 8.3.6 and Theorem 8.3.9on integrated semigroups generated by operators associated to symbols a ∈ Sm

ρ,0 or tohomogeneous symbols a of the form a(ξ) = |ξ|m are due to Hieber [Hie91a], [Hie95]. Forrelated results see also [Lan68], [Sjo70], [BE85], [deL94]. It seems that Corollary 8.3.11on the boundary group of the Poisson semigroup does not exist in the literature. It is

8.5. NOTES 459

however strongly related to Lp estimates of the wave equation; see [Per80]. The resultdescribed in Theorem 8.3.12 on the cosine function generated by the Laplacian in Lp(Rn)was first proved by Littman [Lit63] by direct calculations (not using the theory of cosinefunctions).

Section 8.4The Cauchy problem for systems of differential operators with constant coefficients ofthe form described in Theorem 8.4.3 and Theorem 8.4.6 was investigated in detail byBrenner (see [Bre66] and [Bre73]). Theorem 8.4.3 and Theorem 8.4.6 are due to him.Our proof follows essentially the lines of [Bre66]. For the rest of the section we followclosely [Hie91c]. Example 8.4.11 and Theorem 8.4.14 can be found in [Hie91d]. For furtherinformation on the systems discussed in Section 8.4, see also Chapter 1 of [Fat83].

Appendix A

Vector-valued HolomorphicFunctions

Let X be a Banach space and let Ω ⊂ C be an open set. A function f : Ω→ X isholomorphic if

f ′(z0) := limh→0

h∈C\{0}

f(z0 + h)− f(z0)

h(A.1)

exists for all z0 ∈ Ω.If f is holomorphic, then f is continuous and weakly holomorphic (i.e. x∗ ◦ f

is holomorphic for all x∗ ∈ X∗). If Γ := {γ(t) : t ∈ [a, b]} is a finite, piecewisesmooth contour in Ω, we can form the contour integral

∫Γf(z) dz. This coincides

with the Bochner integral∫ b

af(γ(t))γ′(t) dt (see Section 1.1). Similarly we can

define integrals over infinite contours when the corresponding Bochner integral isabsolutely convergent. Since⟨∫

Γ

f(z) dz, x∗⟩=

∫Γ

〈f(z), x∗〉 dz,

many properties of holomorphic functions and contour integrals may be extendedfrom the scalar to the vector-valued case, by applying the Hahn-Banach theorem.For example, Cauchy’s theorem is valid, and also Cauchy’s integral formula:

f(w) =1

2πi

∫|z−z0|=r

f(z)

z − wdz (A.2)

whenever f is holomorphic in Ω, the closed ball B(z0, r) is contained in Ω andw ∈ B(z0, r). As in the scalar case one deduces Taylor’s theorem from this.

Proposition A.1. Let f : Ω → X be holomorphic, where Ω ⊂ C is open. Letz0 ∈ Ω, r > 0 such that B(z0, r) ⊂ Ω. Then

W. Arendt et al., Vector-valued Laplace Transforms and Cauchy Problems: Second Edition, 461Monographs in Mathematics 96, DOI 10.1007/978-3-0348-0087-7, © Springer Basel AG 2011

462 A. VECTOR-VALUED HOLOMORPHIC FUNCTIONS

f(z) =

∞∑n=0

an(z − z0)n

converges absolutely for |z − z0| < r, where

an :=1

2πi

∫|z−z0|=r

f(z)

(z − z0)n+1dz.

We also mention a special form of the identity theorem.

Proposition A.2 (Identity theorem for holomorphic functions). Let Y be a closedsubspace of a Banach space X. Let Ω be a connected open set in C and f : Ω→ Xbe holomorphic. Assume that there exists a convergent sequence (zn)n∈N ⊂ Ω suchthat limn→∞ zn ∈ Ω and f(zn) ∈ Y for all n ∈ N. Then f(z) ∈ Y for all z ∈ Ω.

Note that for Y = {0}, we obtain the usual form of the identity theorem.

Proof. Let x∗ ∈ Y 0 := {y∗ ∈ X∗ : 〈y, y∗〉 = 0 (y ∈ Y )}. Then x∗ ◦f(zn) = 0 for alln ∈ N. It follows from the scalar identity theorem that x∗ ◦ f(z) = 0 for all z ∈ Ω.Hence, f(z) ∈ Y 00 = Y for all z ∈ Ω.

In the following we show that every weakly holomorphic function is holomor-phic. Actually, we will prove a slightly more general assertion which turns out tobe useful. A subset N of X∗ is called norming if

‖x‖1 := supx∗∈N

|〈x, x∗〉|

defines an equivalent norm on X . A function f : Ω → X is called locally boundedif supK ‖f(z)‖ <∞ for all compact subsets K of Ω.

Proposition A.3. Let Ω ⊂ C be open and let f : Ω → X be locally bounded suchthat x∗ ◦ f is holomorphic for all x∗ ∈ N , where N is a norming subset of X∗.Then f is holomorphic.

In particular, if X = L(Y, Z), where Y, Z are Banach spaces, and if f : Ω→X is locally bounded, then the following are equivalent:

(i) f is holomorphic.

(ii) f(·)y is holomorphic for all y ∈ Y .

(iii) 〈f(·)y, z∗〉 is holomorphic for all y ∈ Y, z∗ ∈ Z∗.

Proof. We can assume that ‖x‖1 = ‖x‖ for all x ∈ X. In order to show holomorphyat z0 ∈ Ω we can assume that z0 = 0, replacing Ω by Ω− z0 otherwise. For smallh, k ∈ C\{0}, let

u(h, k) :=f(h)− f(0)

h− f(k)− f(0)

k.

463

We have to show that for ε > 0 there exists δ > 0 such that ‖ u(h, k) ‖ ≤ εwhenever |h| ≤ δ and |k| ≤ δ. Let r > 0 such that B(0, 2r) ⊂ Ω and

M := supz∈B(0,2r)

‖ f(z) ‖ <∞.

Then by Cauchy’s integral formula, for |z| < r, |h| ≤ r, |k| ≤ r, h, k �= 0, x∗ ∈ N ,

〈u(h, k), x∗〉 =1

2πi

∫|z|=2r

〈f(z), x∗〉{1

h

(1

z − h− 1

z

)− 1

k

(1

z − k− 1

z

)}dz

=h− k

2πi

∫|z|=2r

〈f(z), x∗〉z(z − h)(z − k)

dz.

Hence, |〈u(h, k), x∗〉| ≤ |h− k|M/r2. Since N is norming, we deduce that

‖u(h, k) ‖ ≤ |h− k| Mr2

.


Corollary A.4. Let Ω ⊂ C be a connected open set and Ω0 ⊂ Ω be open. Leth : Ω0 → X be holomorphic. Assume that there exists a norming subset N ofX∗ such that for all x∗ ∈ N there exists a holomorphic extension Hx∗ : Ω → Cof x∗ ◦ h. If supx∗∈N

z∈Ω|Hx∗(z)| < ∞, then h has a unique holomorphic extension

H : Ω→ X.

Proof. Again we assume that ‖ · ‖1 = ‖ · ‖. Let

Y :={y = (yx∗)x∗∈N ⊂ C : ‖y‖∞ := sup

x∗∈N|yx∗ | <∞

},

and let H : Ω→ Y be given by H(z) := (Hx∗(z))x∗∈N . It follows from PropositionA.3 that H is holomorphic. By x ∈ X �→ (〈x, x∗〉)x∗∈N , one defines an isometricinjection from X into Y . Since H(z) ∈ X for z ∈ Ω0, it follows from the identitytheorem (Proposition A.2) that H(z) ∈ X for all z ∈ Ω.

We will extend Proposition A.3 considerably in Theorem A.7. Before that weprove Vitali’s theorem.

Theorem A.5 (Vitali). Let Ω ⊂ C be open and connected. Let fn : Ω → X beholomorphic (n ∈ N) such that

supn∈N

z∈B(z0,r)

‖fn(z)‖ <∞

whenever B(z0, r) ⊂ Ω. Assume that the set

Ω0 :={z ∈ Ω : lim

n→∞ fn(z) exists}

A. VECTOR-VALUED HOLOMORPHIC FUNCTIONS

464 A. VECTOR-VALUED HOLOMORPHIC FUNCTIONS

has a limit point in Ω. Then there exists a holomorphic function f : Ω→ X suchthat

f (k)(z) = limn→∞ f (k)

n (z)

uniformly on all compact subsets of Ω for all k ∈ N0.

Proof. Let l∞(X) := {x = (xn)n∈N ⊂ X : ‖x‖∞ := sup ‖xn‖ < ∞}. Then l∞(X)is a Banach space for the norm ‖·‖∞ and the space c(X) of all convergent sequencesis a closed subspace of l∞(X). Consider the function F : Ω → l∞(X) given byF (z) = (fn(z))n∈N. It follows from Proposition A.3 that F is holomorphic. (Onemay take N to be the space of all functionals on l∞(X) of the form (xn)n∈N �→〈xk, x

∗〉 where k ∈ N, x∗ ∈ X∗, ‖x∗‖ ≤ 1). Since F (z) ∈ c(X) for all z ∈ Ω0,it follows from the identity theorem (Proposition A.2) that F (z) ∈ c(X) for allz ∈ Ω. Consider the mapping φ ∈ L(c(X), X) given by φ((xn)n∈N) = limn→∞ xn.Then f = limn→∞ fn = φ ◦ F : Ω→ X is holomorphic.

Finally, we prove uniform convergence on compact sets. Let B(z0, r) ⊂ Ω andk ∈ N0. It follows from (A.2) that

1

k!f (k)n (z) =

1

2πi

∫|w−z0|=r

fn(w)

(w − z)k+1dw.

Now the dominated convergence theorem implies that f(k)n (z) converges uniformly

on B(z0, r/2) to f (k)(z). Since every compact subset of Ω can be covered by afinite number of discs, the claim follows.

If in Vitali’s theorem (fn) is a net instead of a sequence, the proof showsthat f(z) = lim fn(z) exists for all z ∈ Ω and defines a holomorphic functionf : Ω→ X .

Next we recall a well known theorem from functional analysis.

Theorem A.6 (Krein-Smulyan). Let X be a Banach space and W be a subspaceof the dual space X∗. Denote by B∗ the closed unit ball of X∗. Then W is weak*closed if and only if W ∩B∗ is weak* closed.

For a proof, see [Meg98, Theorem 2.7.11].

Now we obtain the following convenient criterion for holomorphy.

Theorem A.7. Let Ω ⊂ C be open and connected, and let f : Ω → X be a locallybounded function. Assume that W ⊂ X∗ is a separating subspace such that x∗ ◦ fis holomorphic for all x∗ ∈W . Then f is holomorphic.

Here, W is called separating if 〈x, x∗〉 = 0 for all x∗ ∈ W implies x = 0(x ∈ X).

Proof. Let Y := {x∗ ∈ X∗ : x∗ ◦ f is holomorphic}. Since W ⊂ Y , the subspaceY is weak* dense. It follows from Vitali’s theorem (applied to nets if X is not

465

separable) that Y ∩ B∗ is weak* closed. Now it follows from the Krein-Smulyantheorem that Y = X∗. Hence, f is holomorphic by Proposition A.3.

Notes: Usually, Vitali’s theorem is proved with the help of Montel’s theorem which isonly valid in finite dimensions. A vector-valued version is proved in the book of Hille andPhillips [HP57] by a quite complicated power-series argument going back to Liouville.The very simple proof given here is due to Arendt and Nikolski [AN00] who also provedTheorem A.7 (see also [AN06]).

A. VECTOR-VALUED HOLOMORPHIC FUNCTIONS

Appendix B

Closed Operators

Let X be a complex Banach space. An operator on X is a linear map A : D(A)→X, where D(A) is a linear subspace of X , known as the domain of A. The rangeRanA, and the kernel KerA, of A are defined by:

RanA := {Ax : x ∈ D(A)},KerA := {x ∈ D(A) : Ax = 0}.

The operator A is densely defined if D(A) is dense in X .An operator A is closed if its graph G(A) is closed in X ×X, where

G(A) := {(x,Ax) : x ∈ D(A)}.Thus, A is closed if and only if

Whenever (xn) is a sequence in D(A), x, y ∈ X ,‖xn − x‖ → 0 and ‖Axn − y‖ → 0, then x ∈ D(A) and Ax = y.

It is immediate from this that if A is closed and α, β ∈ C with α �= 0, then theoperator αA+ β with D(αA+ β) = D(A) is closed.

An operator A is said to be closable if there is an operator A (known as theclosure of A) such that G(A) is the closure of G(A) in X ×X. Thus A is closableif and only if

Whenever (xn) is a sequence in D(A), y ∈ X,‖xn‖ → 0 and ‖Axn − y‖ → 0, then y = 0.

When A is closable,

D(A) =

{x ∈ X : there exist xn ∈ D(A) and y ∈ X

such that ‖xn − x‖ → 0 and ‖Axn − y‖ → 0

},

Ax = y.

468 B. CLOSED OPERATORS

For an operator A, D(A) becomes a normed space with the graph norm

‖x‖D(A) := ‖x‖+ ‖Ax‖.

The operator A : D(A) → X is always bounded with respect to the graph norm,and A is closed if and only if D(A) is a Banach space in the graph norm. Notethat if A is replaced by αA + β where α �= 0, then the space D(A) is unchangedand the graph norm is replaced by an equivalent norm.

Let A be a closed operator on X. A subspace D of D(A) is said to be a coreof A if D is dense in D(A) with respect to the graph norm. Thus, D is a core ofA if and only if A is the closure of A|D, or equivalently for each x ∈ D(A) there isa sequence (xn) in D such that ‖xn − x‖ → 0 and ‖Axn −Ax‖ → 0.

An operator A on X is said to be invertible if there is a bounded operator A−1

on X such that A−1Ax = x for all x ∈ D(A) and A−1y ∈ D(A) and AA−1y = yfor all y ∈ X.

Proposition B.1. Let A be an operator on X. The following assertions are equiv-alent:

(i) A is invertible.

(ii) RanA = X and there exists δ > 0 such that ‖Ax‖ ≥ δ‖x‖ for all x ∈ D(A).

(iii) A is closed, RanA is dense in X, and there exists δ > 0 such that ‖Ax‖ ≥δ‖x‖ for all x ∈ D(A).

(iv) A is closed, RanA = X and KerA = {0}.Proof. The equivalence of (i) and (ii) is an easy consequence of the definition. Sinceany bounded operator has closed graph, and since

G(A−1) = {(y, x) : (x, y) ∈ G(A)},

any invertible operator is closed. Thus, (i) and (ii) imply (iii) and (iv). When (iii)holds, G(A) is complete, and the map (x,Ax) �→ Ax is an isomorphism of G(A)onto RanA, so RanA is complete and (ii) follows. When (iv) holds, the inversemapping theorem can be applied to the map A from D(A) (with the graph norm)to X , showing that A−1 exists as a bounded map from X to D(A) and hence toX.

Let λ ∈ C. Then λ is said to be in the resolvent set ρ(A) of A if λ − Ais invertible, and we write R(λ,A) := (λ − A)−1. The remarks in the previousparagraphs show that if ρ(A) is non-empty, then A is closed. The function R(·, A) :ρ(A)→ L(X) is the resolvent of A. The spectrum of A is defined to be:

σ(A) := C \ ρ(A),

469

and the spectral bound is:

s(A) := sup{Reλ : λ ∈ σ(A)}if the supremum exists (s(A) := −∞ if σ(A) is empty). The point spectrum σp(A),and approximate point spectrum σap(A), of A are defined by:

σp(A) := {λ ∈ C : Ker(λ−A) �= {0}} ,σap(A) :=

{λ ∈ C : there exist xn ∈ D(A) such that

‖xn‖ = 1 and limn→∞ ‖(λ− A)xn‖ = 0

}.

Thus, σp(A) and σap(A) consist of the eigenvalues and approximate eigenvalues ofA, respectively. It is clear that σp(A) ⊂ σap(A) ⊂ σ(A).

Proposition B.2. Suppose that A has non-empty resolvent set, and let μ ∈ ρ(A).Let λ ∈ C, λ �= μ. Then

a) λ ∈ ρ(A) if and only if (μ− λ)−1 ∈ ρ(R(μ,A)). In that case,

R(λ,A) = (μ− λ)−1((μ− λ)−1 −R(μ,A)

)−1R(μ,A). (B.1)

b) λ ∈ σp(A) if and only if (μ− λ)−1 ∈ σp(R(μ,A)).

c) λ ∈ σap(A) if and only if (μ− λ)−1 ∈ σap(R(μ,A)).

d) The topological boundary of σ(A) is contained in σap(A).

Proof. Parts a), b) and c) follow immediately from the identity

λ−A = (μ− λ)((μ− λ)−1 −R(μ,A)

)(μ−A).

Part d) follows from a), c) and the corresponding result for bounded operators.Alternatively, d) may be proved directly in exactly the same way as for boundedoperators.

Corollary B.3. For any operator A, ρ(A) is open and σ(A) is closed in C. Moreover,if μ ∈ ρ(A), λ ∈ C and |λ− μ| < ‖R(μ,A)‖−1, then λ ∈ ρ(A), and

R(λ,A) =∞∑

n=0

(μ− λ)nR(μ,A)n+1,

where the series is norm-convergent. Hence,

‖R(λ,A)‖ ≤ ‖R(μ,A)‖1− |λ− μ| ‖R(μ,A)‖ .

Moreover, R(·, A) is holomorphic on ρ(A) with values in L(X) and

R(μ,A)(n)

n!= (−1)nR(μ,A)n+1 (n ∈ N).

B. CLOSED OPERATORS

Proof. This is immediate from (B.1) and the Neumann expansion, (I − T )−1 =∑∞n=0 T

n, when T is a bounded operator with ‖T‖ < 1.

Proposition B.4. Let A be an operator on X, and let λ, μ ∈ ρ(A). Then

R(λ,A)−R(μ,A) = (μ− λ)R(λ,A)R(μ,A). (B.2)

Proof. The identity (B.2) follows by rearranging (B.1).

Proposition B.5. Let A be an operator on X, and U be a connected open subset ofC. Suppose that U ∩ ρ(A) is nonempty and that there is a holomorphic functionF : U → L(X) such that {λ ∈ U ∩ ρ(A) : F (λ) = R(λ,A)} has a limit point inU . Then U ⊂ ρ(A) and F (λ) = R(λ,A) for all λ ∈ U .

Proof. Let V = {λ ∈ U ∩ ρ(A) : F (λ) = R(λ,A)}, μ ∈ ρ(A), x ∈ D(A), y ∈ X.For λ ∈ V ,

F (λ)(λ−A)x = x, (B.3)

F (λ)y = R(μ,A)y − (λ− μ)R(μ,A)F (λ)y, (B.4)

using (B.2). By uniqueness of holomorphic extensions (Proposition A.2), (B.3) and(B.4) are valid for all λ ∈ U . Now, (B.4) implies that F (λ)y ∈ D(A) and

R(μ,A)(λ− A)F (λ)y = F (λ)y + (λ− μ)R(μ,A)F (λ)y

= R(μ,A)y.

Since R(μ,A) is injective, (λ − A)F (λ)y = y for all λ ∈ U . This and (B.3) implythat λ ∈ ρ(A) and F (λ) = R(λ,A).

The equation (B.2) is known as the resolvent equation or resolvent identity. Afunction R : U → L(X), defined on a subset U of C, is said to be a pseudo-resolventif it satisfies the resolvent equation; i.e., if

R(λ)−R(μ) = (μ− λ)R(λ)R(μ) (λ, μ ∈ U).

The following proposition is easy to prove.

Proposition B.6. Let R : U → L(X) be a pseudo-resolvent. Then

a) KerR(λ) and RanR(λ) are independent of λ ∈ U .

b) There is an operator A on X such that R(λ) = R(λ,A) for all λ ∈ U if andonly if KerR(λ) = {0}.An operator A is said to have compact resolvent if ρ(A) �= ∅ and R(λ,A) is

a compact operator on X . Since the compact operators form an ideal of L(X), itis immediate from (B.2) that this property is independent of λ ∈ ρ(A). When Ahas compact resolvent, then σ(A) is a discrete subset of C. This follows from (B.1)and the fact that the spectrum of a compact operator has 0 as its only limit point.

The following is easy to prove.

471

Proposition B.7. Let A be an operator on X with non-empty resolvent set, and letT ∈ L(X). The following are equivalent:

(i) R(λ,A)T = TR(λ,A) for all λ ∈ ρ(A).

(ii) R(λ,A)T = TR(λ,A) for some λ ∈ ρ(A).

(iii) For all x ∈ D(A), Tx ∈ D(A) and ATx = TAx.

For an operator A, the powers An (n ≥ 2) are defined recursively:

D(An) :={x ∈ D(An−1) : An−1x ∈ D(A)

},

Anx := A(An−1x).

Note that D((λ−A)n) = D(An) for all λ ∈ C, n ∈ N. It is easy to see that An isinvertible if and only if A is invertible, and then (An)−1 = (A−1)n.

If A is densely defined and ρ(A) �= ∅, then D(An) is a core for A, for eachn ∈ N. To see this, let λ ∈ ρ(A). Then R(λ,A) has dense range D(A). It followsthat the range D(An−1) of R(λ,A)n−1 is dense in X . Let x ∈ D(A). There isa sequence (ym)m∈N in D(An−1) converging to (λ − A)x. Let xm := R(λ,A)ym.Then xm ∈ D(An), ‖xm − x‖ → 0 and ‖Axm −Ax‖ → 0.

Let A be an operator on X , and let Y be a closed subspace of X. The partof A in Y is the operator AY on Y defined by

D(AY ) := {y ∈ D(A) ∩ Y : Ay ∈ Y },AY y := Ay.

The following results are easy to prove.

Proposition B.8. Let A be an operator on X, and let Y be a closed subspace of X.

a) If D(A) ⊂ Y , then ρ(A) ⊂ ρ(AY ) and R(λ,AY ) = R(λ,A)|Y for all λ ∈ρ(A).

b) Suppose that ρ(A) �= ∅ and there is a projection P of X onto Y such thatPR(λ,A) = R(λ,A)P for some λ ∈ ρ(A). Then A maps D(A)∩Y into Y , AY

is the restriction of A to D(A) ∩ Y , λ ∈ ρ(AY ) and R(λ,AY ) = R(λ,A)|Y .One situation where the conditions of Proposition B.8 b) are satisfied is

described in the following.

Proposition B.9. Let A be a closed operator on X with ρ(A) �= ∅, and suppose thatthere are a compact subset E1 and a closed subset E2 of C such that E1∩E2 = ∅ andE1∪E2 = σ(A). Then there is a bounded projection P on X such that R(λ,A)P =PR(λ,A) for all λ ∈ ρ(A), P (X) ⊂ D(A), σ(AY ) = E1 and σ(AZ) = E2, whereY := P (X), Z := (I − P )(X). Moreover, P is unique, and A|Y ∈ L(Y ).

B. CLOSED OPERATORS


The projection P is known as the spectral projection of A associated with E1.

Proof. Take μ ∈ ρ(A) and consider R(μ,A) ∈ L(X). Then σ(R(μ,A)) = E′1 ∪ E′2,where

E′1 := {(μ− λ)−1 : λ ∈ E1}and

E′2 :=

{{(μ− λ)−1 : λ ∈ E2} if D(A) = X,

{(μ− λ)−1 : λ ∈ E2} ∪ {0} otherwise.

Then E′1 and E′2 are compact and disjoint. By the functional calculus for boundedoperators (see [DS59, p.573]), there is a unique bounded projection P on Xsuch that R(λ,A)P = PR(λ,A) for all λ ∈ ρ(A), σ(R(μ,A)|Y ) = E′1 andσ(R(μ,A)|Z) = E′2. Since 0 �∈ σ(R(μ,A)|Y ), Y ⊂ D(A) and A|Y is bounded bythe closed graph theorem. The remaining properties follow easily from PropositionB.2 a).

Suppose that A has compact resolvent, let λ ∈ ρ(A) and μ ∈ σ(A). Let Pbe the spectral projection of A associated with {μ}. Then there exists m ∈ Nsuch that (R(λ,A)P − (λ − μ)−1P )m = 0 (see [DS59, Theorem VII.4.5]). Hence,(A− μ)mP = 0.

Given an operator A on X , let

G(A∗) := {(x∗, y∗) ∈ X∗ ×X∗ : 〈Ax, x∗〉 = 〈x, y∗〉 for all x ∈ D(A)} ,which is a weak* closed subspace of X∗×X∗. If (and only if) A is densely defined,then G(A∗) is the graph of an operator A∗ in X∗, known as the adjoint of A. Forthe remainder of this appendix, we shall assume that A is densely defined, and weshall consider properties of A∗.

When A is closed, the operator A can be recovered from A∗ in the followingway.

Proposition B.10. Let A be a closed, densely defined operator on X, and let x, y ∈X. The following assertions are equivalent:

(i) x ∈ D(A) and Ax = y.

(ii) 〈x,A∗x∗〉 = 〈y, x∗〉 for all x∗ ∈ D(A∗).

Hence, D(A∗) is weak* dense in X∗.

Proof. The implication (i) ⇒ (ii) is immediate from the definition of A∗. For theconverse, suppose that (x, y) /∈ G(A). By the Hahn-Banach theorem, there exists(x∗, y∗) ∈ X∗×X∗ such that 〈x, x∗〉+ 〈y, y∗〉 �= 0 but 〈u, x∗〉+ 〈Au, y∗〉 = 0 for allu ∈ D(A). The latter condition implies that y∗ ∈ D(A∗) and A∗y∗ = −x∗. Thus,〈x,A∗y∗〉 = −〈x, x∗〉 �= 〈y, y∗〉, so (ii) is violated.

If D(A∗) is not weak* dense in X∗, then by the Hahn-Banach theorem thereexists y ∈ X such that y �= 0 and 〈y, x∗〉 = 0 for all x∗ ∈ D(A∗). By the previouspart, A0 = y, which is absurd.

473

If A is closable (and densely defined), it is easy to see that (A)∗ = A∗, soD(A∗) is weak* dense by Proposition B.10. Conversely, it is easy to see that ifD(A∗) is weak* dense, then A is closable.

Proposition B.11. Let A be a closed, densely defined operator on X. Then

a) A∗ is invertible if and only if A is invertible, and then (A∗)−1 = (A−1)∗.

b) σ(A∗) = σ(A), and R(λ,A∗) = R(λ,A)∗ for all λ ∈ ρ(A).

c) σ(A) = σap(A) ∪ σp(A∗).

Proof. a) If A is invertible, it is easy to verify that (A−1)∗A∗x∗ = x∗ for allx∗ ∈ D(A∗) and A∗(A−1)∗y∗ = y∗ for all y∗ ∈ X∗. Thus, A∗ is invertible and(A∗)−1 = (A−1)∗.

Now suppose that A∗ is invertible, and let δ =∥∥(A∗)−1

∥∥−1. Since KerA∗ =

{0}, RanA is dense in X, by a simple application of the Hahn-Banach theorem.For x ∈ X , there exists x∗ ∈ X∗ such that ‖x∗‖ = 1 and 〈x, x∗〉 = ‖x‖. Lety∗ = (A∗)−1x∗ ∈ D(A∗), so that ‖y∗‖ ≤ δ−1 and A∗y∗ = x∗. Hence,

‖Ax‖ ≥ δ |〈Ax, y∗〉| = δ |〈x,A∗y∗〉| = δ‖x‖.

It follows from Proposition B.1 that A is invertible.b) This follows from a) by replacing A by λ−A.c) This follows from applying Proposition B.1 and the fact that RanA is

dense in X if and only if KerA∗ = {0} (by the Hahn-Banach theorem), with Areplaced by λ−A.

Now, let H be a Hilbert space with inner product (·|·)H . Identifying H∗ withH by means of the Riesz-Frechet lemma, we obtain the following. If A is a denselydefined operator on H, the adjoint A∗ of A is defined by

D(A∗) :=

{x ∈ H : there exists y ∈ H such that

(Au|x)H = (u|y)H for all u ∈ D(A)

},

A∗x = y.

We say that A is selfadjoint if A = A∗.

Example B.12 (Multiplication operators). Let (Ω, μ) be a measure space, H :=L2(Ω, μ), m : Ω→ R a measurable function. Define the operator Mm on H by

D(Mm) := {f ∈ H : mf ∈ H},Mmf := mf.

It is easy to see that Mm is selfadjoint.

B. CLOSED OPERATORS


Let H, H be Hilbert spaces. Two operators A on H and A on H are calledunitarily equivalent if there exists a unitary operator U : H → H such that

D(A) = U−1D(A),

Ax = U−1AUx.

It is easy to see that, in that case, A is selfadjoint whenever A is.Now we can formulate the spectral theorem as follows; we refer to [RS72,

Theorem VIII.4] for a proof.

Theorem B.13 (Spectral Theorem). Each selfadjoint operator is unitarily equiva-lent to a real multiplication operator.

Thus, selfadjoint and real multiplication operators are effectively the samething. In proofs, we frequently regard an arbitrary selfadjoint operator as being areal multiplication operator.

A selfadjoint operator A is always symmetric; i.e., (Ax|y)H = (x|Ay)H forall x, y ∈ D(A). In particular, (Ax|x)H ∈ R for all x ∈ D(A). We say that A isbounded above if there exists ω ∈ R such that

(Ax|x)H ≤ ω(x|x)H (x ∈ D(A)).

In that case, ω is called an upper bound of A. If A is a multiplication operator Mm,then this is equivalent to saying that

m(y) ≤ ω for almost all y ∈ Ω.

It is easy to see (for example, from the spectral theorem) that for any selfadjointoperator A, we have σ(A) ⊂ R and ω is an upper bound for A if and only ifσ(A) ⊂ (−∞, ω]; i.e., ω ≥ s(A). Similarly, we say that A is bounded below by ω if

(Ax|x)H ≥ ω(x|x)H (x ∈ D(A)).

The definition of selfadjointness is not easy to verify in practice. Here is ahandy criterion, for a proof of which we refer to [RS72, Theorem X.1].

Theorem B.14. Let A be an operator on H and let ω ∈ R. The following areequivalent:

(i) A is selfadjoint with upper bound ω.

(ii) a) (Ax|y)H = (x|Ay)H (x, y ∈ D(A)),b) (Ax|x)H ≤ ω(x|x)H (x ∈ D(A)), andc) there exists λ > ω such that Ran(λ− A) = X.

Finally, we mention one or two topics concerning bounded operators. By theclosed graph theorem, an operator T on a Banach space X is bounded if T isclosed and D(T ) = X. Conversely, a densely defined, closed, bounded operator is

475

everywhere defined. By convention, a bounded operator T on a Banach space Xwill be assumed to be defined on the whole of X . The spectral radius of T will bedenoted by r(T ), so

r(T ) = sup{|λ| : λ ∈ σ(T )} = inf{‖Tn‖1/n : n ∈ N

}.

In order to allow a convenient citation in the book, we state the followingstandard fact whose proof is straightforward. Note that a family of bounded linearoperators is equicontinuous if and only if it is bounded.

Proposition B.15. Let X,Y be Banach spaces, Tn ∈ L(X,Y ) (n ∈ N) such thatsupn∈N ‖Tn‖ <∞. The following are equivalent:

(i) (Tnx)n∈N converges for all x in a dense subspace of X.

(ii) (Tnx)n∈N converges for all x ∈ X.

(iii) (Tnx)n∈N converges uniformly in x ∈ K for all compact subsets K of X.

Notes: The material of this appendix is standard, and can be found in various books, forexample [Kat66, Chapter 3].

B. CLOSED OPERATORS

Appendix C

Ordered Banach Spaces

Let X be a real Banach space. By a positive cone in X we understand a closedsubset X+ of X such that

X+ +X+ ⊂ X+; (C.1)

R+ ·X+ ⊂ X+; (C.2)

X+ ∩ (−X+) = {0}; and (C.3)

X+ −X+ = X. (C.4)

Then an ordering on X is introduced by setting

x ≤ y ⇐⇒ y − x ∈ X+.

The space X together with the positive cone is called a real ordered Banach space.The elements of X+ are called positive.

Remark C.1. Property (C.3) is frequently expressed by saying that X+ is a propercone, and (C.4) says thatX+ is generating. We assume these properties throughoutwithout further notice.

If x∗ ∈ X∗, then we say that x∗ is positive and write x∗ ≥ 0 if

〈x, x∗〉 ≥ 0 for all x ∈ X+.

The set X∗+ := {x∗ ∈ X∗ : x∗ ≥ 0} is closed and satisfies (C.1), (C.2) and (C.3).For x, y ∈ X such that x ≤ y we denote by

[x, y] := {z ∈ X : x ≤ z ≤ y}

the order interval defined by x and y. One says that the cone X+ is normal if allorder intervals are bounded.

478 C. ORDERED BANACH SPACES

Proposition C.2. The cone X∗+ is normal. The cone X+ is normal if and only if

X∗+ −X∗+ = X∗.

Thus, if X+ is normal then (X∗, X∗+) is also an ordered Banach space with

normal cone. We call X∗+ the dual cone of X+.

If the cone X+ is normal then there is a constant c ≥ 0 such that

y ≤ x ≤ z =⇒ ‖x‖ ≤ cmax(‖y‖, ‖z‖). (C.5)

Indeed, passing to an equivalent norm one can even arrange that c = 1.If X is a real ordered Banach space we tacitly consider the complexification

of X . So in this book an ordered Banach space is always the complexification of areal ordered Banach space. Thus, any C∗-algebra is an ordered Banach space withnormal cone.

Let X be an ordered Banach space. A linear mapping T : X → X is calledpositive if

Tx ∈ X+ for all x ∈ X+.

Then we write T ≥ 0. If S, T : X → X are linear, we write S ≤ T if T − S ≥ 0.If X+ is normal, every positive linear mapping T : X → X is continuous.

Moreover, there is a constant k ≥ 0 such that

±S ≤ T =⇒ ‖S‖ ≤ k‖T‖ (C.6)

if S, T : X → X are linear.A real ordered Banach space X is a lattice if for all x, y ∈ X there exists a

least upper bound x∨y of x and y (i.e., x∨y ∈ X, x∨y ≥ x, x∨y ≥ y and w ≥ x, yimplies w ≥ x ∨ y). In that case, there also exists a largest lower bound x ∧ y =−((−x) ∨ (−y)). One sets x+ = x ∨ 0, x− = (−x)+, |x| = x ∨ (−x) = x+ + x−.Then X is called a real Banach lattice if in addition the following compatibilitycondition is satisfied:

|x| ≤ |y| =⇒ ‖x‖ ≤ ‖y‖ (C.7)

for all x, y ∈ X. Thus, the cone of a Banach lattice is always normal.In this book, a Banach lattice is the complexification of a real Banach lattice.

Important examples of Banach lattices are the spaces Lp(Ω, μ) (1 ≤ p ≤ ∞), where(Ω, μ) is a measure space, and

C(K) := {f : K → C : f continuous},where K is a compact space.

Let X be a real Banach lattice. A subspace Y of X is called a sublattice if

x ∈ Y implies |x| ∈ Y.

The space Y is called an ideal if

x ∈ Y, y ∈ X, |y| ≤ |x| implies y ∈ Y.

479

Let (Ω, μ) be a σ-finite measure space and X = Lp(Ω, μ), where 1 ≤ p <∞.Then Y is a closed ideal of X if and only if

Y = {f ∈ X : f |S = 0 a.e.}

for some measurable subset S of Ω.If M ⊂ X is a subset, then

Md := {x ∈ X : |x| ∨ |y| = 0 for all y ∈M}

is a closed ideal of X . One says that M is a band if M = Mdd. In that case,M ⊕Md = X.

IfX is a complex Banach lattice, then a subspace Y of X is called a sublattice(ideal, band) if

a) x ∈ Y =⇒ Rex ∈ X, and

b) Y ∩XR is a sublattice (ideal, band) of XR,

where XR denotes the underlying real Banach lattice.An ordered Banach space has order continuous norm if each decreasing pos-

itive sequence (xn)n∈N converges; i.e.,

If xn ≥ xn+1 ≥ 0 (n ∈ N), then limn→∞ xn exists.

The spaces Lp(Ω, μ) (1 ≤ p <∞) have order continuous norm, but L∞(Ω, μ) andC(K) do not if they have infinite dimension. Also, the dual of a C∗-algebra hasorder continuous norm.

Let X be a Banach lattice. Then the following assertions are equivalent:

(i) If 0 ≤ xn ≤ xn+1 and supn∈N ‖xn‖ <∞, then (xn)n∈N converges.

(ii) X is a band in X∗∗.

(iii) c0 is not isomorphic to a closed subspace of X.

In assertion (ii), we identify X with a closed subspace of X∗∗ via the canonicalevaluation mapping.

A Banach lattice X satisfying the equivalent conditions (i), (ii), and (iii)is called a KB-space. Every reflexive Banach lattice and every space of the formL1(Ω, μ) are KB-spaces. Moreover, if X is a KB-space then X has order continuousnorm. The space c0 does have order continuous norm but is not a KB-space. Eachclosed ideal of a KB-space is a band.

Notes: We refer to the monograph [Sch74] by Schaefer and to the survey article [BR84]for all this and for further information.

C. ORDERED BANACH SPACES

Appendix D

Banach Spaces which Contain c0

We let c0 be the Banach space of all complex sequences a = (ar)r≥1 such thatlimr→∞ ar = 0, with ‖a‖ = supr |ar|. For n ≥ 1, let en := (δnr)r≥1, so ‖en‖ = 1and ∥∥∥∥∥

m∑n=1

αnen

∥∥∥∥∥ = maxn|αn|

for all m ∈ N and α1, . . . , αm ∈ C.A complex Banach space X is said to contain c0 if there is a closed linear

subspace Y of X which is isomorphic (linearly homeomorphic) to c0. This is equiv-alent to the existence of a sequence (xn)n≥1 in X and strictly positive constantsc1 and c2 such that

c1 maxn|αn| ≤

∥∥∥∥∥m∑

n=1

αnxn

∥∥∥∥∥ ≤ c2 maxn|αn| (D.1)

for all m ∈ N and α1, . . . , αm ∈ C. Then the map∑m

n=1 αnxn �→∑m

n=1 αnenextends to an isomorphism of the closed linear span of {xn} onto c0.

Since c0 is not reflexive, a reflexive Banach space cannot contain c0. Moreover,for any measure space (Ω, μ), the space L1(Ω, μ) does not contain c0.

A formal series∑∞

n=1 xn in X is said to be unconditionally bounded if thereis a constant M such that ∥∥∥∥ m∑

j=1

xnj

∥∥∥∥ ≤M (D.2)

whenever m ∈ N and 1 ≤ n1 < n2 < · · · < nm. The series∑

n en in c0 isunconditionally bounded (with M = 1), but it is divergent. It follows that anyBanach space which contains c0 has a divergent, unconditionally bounded series.In this appendix, we shall give a converse result showing that if X contains adivergent, unconditionally bounded series, then X contains c0.

482 D. BANACH SPACES WHICH CONTAIN C0

Lemma D.1. Suppose that∑

n xn is a divergent, unconditionally bounded series ina complex Banach space X, and let M be as in (D.2). Then∥∥∥∥∥

m∑n=1

αnxn

∥∥∥∥∥ ≤ 4M max1≤n≤m

|αn|

for all m ∈ N and α1, . . . , αm ∈ C.

Proof. First suppose that αn ≥ 0 for n = 1, 2, . . . ,m. By rearranging x1, x2, . . . , xm,we may suppose that 0 ≤ α1 ≤ α2 ≤ · · · ≤ αm. Then

m∑n=1

αnxn = α1

m∑n=1

xn + (α2 − α1)

m∑n=2

xn + · · ·+ (αm − αm−1)xm.

Hence,∥∥∥∥∥m∑

n=1

αnxn

∥∥∥∥∥ ≤ α1M + (α2 − α1)M + · · ·+ (αm − αm−1)M = αmM.

The general case follows by decomposing each complex number αn as∑3

j=0 αnjij

where αnj ≥ 0 and |αnj | ≤ |αn|.Lemma D.2. Suppose that X contains a divergent, unconditionally bounded series.Then there is a sequence (yj)j≥1 in X such that ‖yj‖ = 1 for all j and∥∥∥∥∥∥

m∑j=1

βjyj

∥∥∥∥∥∥ ≤ 3

2max

1≤j≤m|βj |

for all m ∈ N and β1, . . . , βm ∈ C.

Proof. Let∑

n xn be a divergent, unconditionally bounded series, and let

γk := sup

{∥∥∥∥∥m∑

n=k+1

αnxn

∥∥∥∥∥ : m > k,αn ∈ C, |αn| ≤ 1

}.

By Lemma D.1, γk is finite, and clearly (γk) is a decreasing sequence. Let γ :=limk→∞ γk. Then γ > 0, since

γk ≥ sup

{∥∥∥∥∥m∑

n=k+1

xn

∥∥∥∥∥ : m > k

}

and∑

n xn is divergent. Replacing xn by (5/4γ)xn, we may assume that γ = 5/4.

483

Choose k1 ≥ 1 such that γk1< 3/2. Since γk1

> 1, there exist k2 > k1 andαn ∈ C (k1 < n ≤ k2) such that |αn| ≤ 1 and

ν1 :=

∥∥∥∥∥k2∑

n=k1+1

αnxn

∥∥∥∥∥ > 1.

Iterating this, we may choose k1 < k2 < . . . and αn ∈ C (n > k1) such that|αn| ≤ 1 and

νj :=

∥∥∥∥∥∥kj+1∑

n=kj+1

αnxn

∥∥∥∥∥∥ > 1.

Let

yj := ν−1j

kj+1∑n=kj+1

αnxn.

Then ‖yj‖ = 1. Moreover, if m ∈ N and βj ∈ C (j = 1, . . . ,m) and jn is chosen sothat kjn < n ≤ kjn+1 (n > k1), then∥∥∥∥∥∥

m∑j=1

βjyj

∥∥∥∥∥∥ =

∥∥∥∥∥∥km+1∑

n=k1+1

βjnν−1jn

αnxn

∥∥∥∥∥∥ ≤ 3

2max

k1<n≤km+1

∣∣βjnν−1jn

αn

∣∣ ≤ 3

2max

1≤j≤m|βj |.

Theorem D.3. Suppose that X contains a divergent, unconditionally bounded series∑n xn. Then X contains c0.

Proof. Let (yj) be as in Lemma D.2. Let m ∈ N and βj ∈ C (j = 1, . . . ,m). Then∥∥∥∑mj=1 βjyj

∥∥∥ ≤ 32maxj |βj |. We shall establish that

∥∥∥∑mj=1 βjyj

∥∥∥ ≥ 12maxj |βj |,

so that (yj) satisfies the condition (D.1), and therefore X contains c0.Choose k such that |βk| = maxj |βj |, and choose x∗ ∈ X∗ such that ‖x∗‖ = 1

and βk〈yk, x∗〉 = |βk|. Let

β′j :=

{βj (j �= k),

−βk (j = k).

Then∥∥∥∥∥∥m∑j=1

βjyj

∥∥∥∥∥∥ ≥ Re 〈m∑j=1

βjyj , x∗〉 = 2|βk|+Re 〈

m∑j=1

β′jyj , x∗〉

≥ 2|βk| −∥∥∥∥∥∥

m∑j=1

β′jyj

∥∥∥∥∥∥ ≥ 2|βk| − 3

2max

j|βj | = 1

2max

j|βj |.

D. BANACH SPACES WHICH CONTAIN C0

484 D. BANACH SPACES WHICH CONTAIN C0

This completes the proof.

Notes: Theorem D.3 is due to Bessaga and Pelczynski [BP58]. They also showed thatX contains c0 if (and only if) there is a sequence of unit vectors (yj) in X such that∑

j |〈yj , x∗〉| < ∞ for all x∗ ∈ X∗. Our proof, which is adapted from [LZ82], establishessuch a property but in a specific way which eliminates some of the cases considered in[BP58]. Moreover, this proof shows (when constants 5/4 and 3/2 are replaced by constantsarbitrarily close to 1) that X contains c0 “almost isometrically”, thereby establishinga positive solution to the “distortion problem” in c0. This was first proved by James[Jam64].

Another direct proof of Theorem D.3 is given in a paper of Eberhardt and Greiner[EG92]. There are numerous other characterizations of Banach spaces which contain c0,some of which may be found in the books of Guerre-Delabriere [Gue92], Lindenstraussand Tzafriri [LT77] and Megginson [Meg98]. Note in particular that a Banach lattice Xdoes not contain c0 (as a subspace, or equivalently as a sublattice) if and only if X isa KB-space, that is, every bounded increasing sequence in X converges [LT77, TheoremII.1.c.4], [Mey91, Theorem 2.4.12].

Appendix E

Distributions and FourierMultipliers

In this appendix we collect basic facts on distributions and Fourier multipliers.They are needed at various places in the book; those which are essential to un-derstanding Parts I and II are also explained at the appropriate point in the text,while other results from this appendix are needed only for examples in Chapter 3or for the applications in Part III.

First, we consider distributions on Rn. A multi-index is an element α =(α1, . . . , αn) ∈ Nn

0 . We write |α| for ∑nj=1 αj , Dj for ∂

∂xjand Dα for Dα1

1 · · ·Dαnn .

We denote by D(Rn) (or by C∞c (Rn) in other contexts) the space of all complex-valued C∞-functions on Rn with compact supports (the test functions), and byS(Rn) the Schwartz space of all smooth, rapidly decreasing functions on Rn, i.e.

S(Rn) := {ϕ ∈ C∞(Rn) : ‖ϕ‖m,α <∞ for all m ∈ N0, α ∈ Nn0} ,

where

‖ϕ‖m,α := supx∈Rn

(1 + |x|)m|Dαϕ(x)|.

When equipped with the topology defined by the family of all norms ‖·‖m,α, S(Rn)is a Frechet space, and D(Rn) is a dense subspace of S(Rn).

We denote by D(Rn)′ the space of all distributions, i.e., linear maps f : ϕ �→〈ϕ, f〉 of D(Rn) into C such that for each compact K ⊂ Rn there exist m ∈ N andC > 0 such that

|〈ϕ, f〉| ≤ C sup|α|≤m

supx∈Rn

|Dαϕ(x)|

for all ϕ ∈ D(Rn) with suppϕ ⊂ K. Let S(Rn)′ be the space of all temperate distri-butions, i.e., continuous linear maps from S(Rn) into C. Then S(Rn)′ is embeddedin D(Rn)′ in a natural way.

486 E. DISTRIBUTIONS AND FOURIER MULTIPLIERS

We considerD(Rn)′ to have the topology arising from the duality with D(Rn),so a net (fα) of distributions converges to 0 in D(Rn)′ if and only if 〈ϕ, fα〉 → 0for all ϕ ∈ D(Rn).

Any locally integrable f : Rn → C can be identified with a distribution by

〈ϕ, f〉 :=∫Rn

ϕ(x)f(x) dx (ϕ ∈ S(Rn)). (E.1)

We shall make such identifications freely.A function f : Rn → C is said to be absolutely regular if there exists k ∈ N0

such that x �→ (1 + |x|)−kf(x) is Lebesgue integrable on Rn. For an absolutelyregular function f , the corresponding distribution is temperate.

Any continuous linear map T : S(Rn) → S(Rn) induces an adjoint. Wenow describe how this enables operators of multiplication, differentiation, Fouriertransform and convolution to be extended from functions to distributions.

Let g : Rn → C be a C∞-function. Then ϕ · g ∈ D(Rn) for all ϕ ∈ D(Rn).Given a distribution f ∈ D(Rn)′, we can define g · f by

〈ϕ, g · f〉 := 〈ϕ · g, f〉 (ϕ ∈ D(Rn)). (E.2)

If, for each multi-index α, there exists mα ∈ N and cα > 0 such that

|(Dαg)(x)| ≤ cα(1 + |x|)mα (x ∈ Rn), (E.3)

then the map ϕ �→ ϕ · g is continuous from S(Rn) into S(Rn), and thereforeg · f ∈ S(Rn)′ whenever f ∈ S(Rn)′.

Given a distribution f ∈ D(Rn)′, the derivatives Djf (j = 1, . . . , n) aredefined in D(Rn)′ by

〈ϕ,Djf〉 := −〈Djϕ, f〉 (ϕ ∈ D(Rn)). (E.4)

Then Dj maps S(Rn)′ into itself. Integration by parts shows that this notationis consistent when differentiable functions are identified with distributions, andthe product law extends to derivatives of products of differentiable functions anddistributions as discussed above. For higher order derivatives, (E.4) becomes

〈ϕ,Dαf〉 = (−1)|α|〈Dαϕ, f〉 (ϕ ∈ D(Rn)). (E.5)

Next, we consider convolutions. For functions f and g, the convolution f ∗ gis defined by

(f ∗ g)(x) :=∫Rn

f(x− y)g(y) dy

whenever the integral exists. For ϕ,ψ ∈ S(Rn), ψ ∗ ϕ ∈ S(Rn) and the mapψ �→ ψ∗ϕ is continuous. Hence, the convolution ϕ∗f of ϕ ∈ S(Rn) and f ∈ S(Rn)′

can be defined by

〈ψ, ϕ ∗ f〉 := 〈ψ ∗ ϕ, f〉 (ψ ∈ S(Rn)), (E.6)

487

where ϕ(x) := ϕ(−x), and then ϕ ∗ f ∈ S(Rn)′. An easy calculation shows thatthis is consistent when functions are identified with distributions.

An alternative way to define ϕ ∗ f is as follows. For x ∈ Rn and ψ ∈ S(Rn),let τxψ(y) := ψ(y − x). For ϕ ∈ S(Rn), τxϕ ∈ S(Rn) and the map x �→ τxϕ iscontinuous on S(Rn). For f ∈ S(Rn)′, let

(ϕ ∗ f)(x) := 〈τxϕ, f〉 (x ∈ Rn). (E.7)

Then ϕ ∗ f is a continuous, bounded function.These two definitions of ϕ ∗ f are consistent when functions are identified

with distributions. Moreover,

Dj(ϕ ∗ f) = (Djϕ) ∗ f. (E.8)

For f ∈ L1(Rn), the Fourier transform Ff of f is defined by:

(Ff)(ξ) =

∫Rn

e−ix·ξf(x) dx (ξ ∈ Rn), (E.9)

where x · ξ :=∑n

j=1 xjξj . The Fourier inversion theorem [Hor83, Theorem 7.1.5]shows that F is a linear and topological isomorphism of S(Rn), and

(F−1ϕ)(ξ) = (2π)−n(Fϕ)(−ξ) (ϕ ∈ S(Rn), ξ ∈ Rn).

The Fourier transform therefore induces an isomorphism of S(Rn)′, also denotedby F :

〈ϕ,Ff〉 := 〈Fϕ, f〉 (ϕ ∈ S(Rn), f ∈ S(Rn)′). (E.10)

A simple application of Fubini’s theorem shows that this notation is consistentwhen f ∈ L1(Rn) and f is identified with a distribution in S(Rn)′.

The following relations, which are elementary for functions, extend to distri-butions f :

F−1f = (2π)−n(Ff ) = (2π)−nF f , where 〈ϕ, f〉 := 〈ϕ, f〉, (E.11)

FDjf = iξj · Ff, (E.12)

F(ϕ ∗ f) = (Fϕ) · (Ff) (ϕ ∈ S(Rn)). (E.13)

Plancherel’s theorem states that

〈Fϕ,F ψ〉 = (2π)n〈ϕ, ψ〉 (ϕ,ψ ∈ S(Rn)), (E.14)

where ψ is the complex conjugate of ψ, and hence F extends by continuity to alinear operator F on the Hilbert space L2(Rn) such that (2π)−n/2F is unitary.This also says that, for each f ∈ L2(Rn), the distribution Ff belongs to L2(Rn).

Many of the concepts above can be extended to the case of distributions onan open subset Ω of Rn. Let D(Ω) be the space of test functions on Ω, i.e., C∞-functions of compact support in Ω, and D(Ω)′ be the space of distributions on Ω,

E. DISTRIBUTIONS AND FOURIER MULTIPLIERS


i.e., linear functionals f on D(Ω) such that for each compact K ⊂ Ω there existm ∈ N and C > 0 such that

|〈ϕ, f〉| ≤ C sup|α|≤m

supx∈Ω

|Dαϕ(x)|

for all ϕ ∈ D(Ω) with suppϕ ⊂ K. Locally integrable functions on Ω can beidentified with distributions, and the derivatives Dj of a distribution f are definedby

〈ϕ,Djf〉 := −〈Djϕ, f〉 (ϕ ∈ D(Ω)).For m ∈ N and 1 ≤ p ≤ ∞, the Sobolev space Wm,p(Ω) is defined by

Wm,p(Ω) := {f ∈ Lp(Ω) : Dαf ∈ Lp(Ω) for all α ∈ Nn0 with |α| ≤ m},

where Dαf is understood in the sense of distributions. Thus, f ∈Wm,p(Ω) if andonly if for each α ∈ Nn

0 with |α| ≤ m there exists fα ∈ Lp(Ω) such that∫Ω

ϕfα dx = (−1)|α|∫Ω

(Dαϕ)f dx (ϕ ∈ D(Ω)).

In the special case when n = 1, f ∈Wm,p(Ω) if and only if f ∈ Cm−1(Ω), f (m−1)

is absolutely continuous, and f (j) ∈ Lp(Ω) for j = 0, 1, . . . ,m. Equipped with thenorm

‖f‖Wm,p(Ω) :=∑|α|≤m

‖Dαf‖p,

Wm,p(Ω) becomes a Banach space. The closure of D(Ω) in Wm,p(Ω) is denotedby Wm,p

0 (Ω). For p = 2, we also use the notation

Hm(Ω) := Wm,2(Ω) and Hm0 (Ω) := Wm,2

0 (Ω).

Equipped with the equivalent norm

‖f‖Hm(Ω) :=

⎛⎝ ∑|α|≤m

‖Dαf‖22

⎞⎠1/2

,

Hm(Ω) is a Hilbert space with the inner product

(f |g)Hm(Ω) =∑|α|≤m

∫Ω

DαfDαg dx.

Note that Plancherel’s theorem and (E.12) show that

Hm(Rn) = {f ∈ L2(Rn) : ξα · Ff ∈ L2(Rn) for all α ∈ Nn0 with |α| ≤ m},

where ξα is the function ξ �→ ξα11 ξα2

2 · · · ξαnn . Hence, f ∈ Hm(Rn) if and only if

ξ �→ (1 + |ξ|2)m/2(Ff)(ξ) belongs to L2(Rn).

489

Now we consider Fourier multipliers. If g is a C∞-function satisfying theestimates (E.3), then the map ϕ �→ F−1gFϕ := F−1(g · (Fϕ)) is a continuouslinear map on S(Rn). It is a classical problem to seek conditions on a function suchthat such a map becomes continuous on a function space X = Lp(Rn) (1 ≤ p ≤ ∞)or C0(Rn). Let m : Rn → C be an absolutely regular function. For ϕ ∈ S(Rn), wedefine m · (Fϕ) ∈ S(Rn)′ by

〈ψ,m · (Fϕ)〉 :=∫Rn

ψm · (Fϕ) dx (ψ ∈ S(Rn)).

Then we consider the distribution F−1(m · (Fϕ)) ∈ S(Rn)′. We call m a Fouriermultiplier for X if F−1(m·(Fϕ)) ∈ X for all ϕ ∈ S(Rn) and there exists a constantC such that

‖F−1(m · (Fϕ))‖X ≤ C‖ϕ‖X (ϕ ∈ S(Rn)).

Then the map ϕ �→ F−1(m · (Fϕ)) extends to a bounded linear operator Tm :f �→ F−1mFf on X (in the case when X = L∞(Rn), the extension is weak*continuous). When m is a C∞-function, Tmf agrees with the distribution F−1(m ·Ff) defined earlier.

We denote the space of all Fourier multipliers for X by MX(Rn), or byMp(Rn) when X = Lp(Rn), with the usual identification of functions which coin-cide a.e. We put

‖m‖MX(Rn) := ‖Tm‖L(X).

Fourier multipliers are bounded functions, and ‖m‖MX (Rn) ≥ ‖m‖∞ (see Propo-sition E.2). It follows easily that MX(Rn) is a Banach space. Note also thatMC0

(Rn) ⊂M∞(Rn).

For a ∈ Rn, define τa ∈ L(X) by τaf(x) := f(x − a). If m ∈ MX(Rn), it iseasy to see that

Tmτa = τaTm (a ∈ Rn). (E.15)

Conversely, we have the following result.

Proposition E.1. Let X = Lp(Rn) (1 ≤ p ≤ ∞) or C0(Rn), and assume thatT ∈ L(X) satisfies (E.15). Then there exists m ∈MX(Rn) such that

Tf = F−1mFf (f ∈ X).

For a proof of Proposition E.1, see [Hor60].

For N ∈ N, we let MNp (Rn) be the space of all matrices m = (mij)1≤i,j≤N ,

where mij ∈ Mp(Rn). Each such matrix m defines a bounded operator F−1mFon Lp(Rn)N , where F : Lp(Rn)N → Lp(Rn)N acts on each coordinate function,and matrix multiplication operates as usual. The norm onMN

p (Rn) is taken to be

the norm of the operator F−1mF when Lp(Rn)N is given the norm of Lp(Rn ×{1, . . . , N}). Note that M1

p(Rn) =Mp(Rn).


Proposition E.2. Let 1 ≤ p ≤ ∞, N ∈ N. Then the following hold true:

a) MNp (Rn) is a Banach algebra.

b) MN2 (Rn) = L∞(Rn,L(CN )) := {(mij)1≤i,j≤N : mij ∈ L∞(Rn)N}.

c) MNp (Rn) =MN

p′ (Rn), where 1/p+ 1/p′ = 1.

d) MN1 (Rn) ⊂MN

p (Rn) ⊂MN2 (Rn). Moreover, for m ∈MN

1 (Rn),

‖m‖MNp (Rn) ≤ ‖m‖θMN

1 (Rn)‖m‖1−θMN

2 (Rn), (E.16)

where θ := 2∣∣ 1p − 1

2

∣∣.e) Given a ∈ MN

p (Rn) define at by at(ξ) := a(tξ) for t > 0, ξ ∈ Rn. Then

at ∈MNp (Rn) for all t > 0 and

‖at‖MNp (Rn) = ‖a‖MN

p (Rn) (t > 0).

f) Let (aj)j∈N ⊂ MNp (Rn). Assume that there exists a constant C > 0 such

that ‖aj‖MNp (Rn) ≤ C for j ∈ N. Let a ∈ L∞(Rn) such that aj(x) → a(x)

for almost all x ∈ Rn as j →∞. Then a ∈MNp (Rn) and ‖a‖MN

p (Rn) ≤ C.

Proof. We give only sketches of the proofs; details may be found in [Hor60] or[Ste93].

a) follows from the formal identity F−1(m1m2)F = (F−1m1F)(F−1m2F),b) is an easy consequence of Plancherel’s theorem and c) is easily proved by duality,showing even that the equalities are isometric.

For d), we can assume by c) that 1 ≤ p ≤ 2. Let m ∈ MNp (Rn). By c),

F−1mF is bounded on Lp(Rn)N and on Lp′(Rn)N . Moreover the two versionsof the map agree on Lp(Rn)N ∩ Lp′(Rn)N . By the Riesz-Thorin theorem [Hor83,Theorem 7.1.12], F−1mF extends to a bounded linear operator on L2(Rn)N . Thisshows that MN

p (Rn) ⊂ MN2 (Rn). The inclusion MN

1 (Rn) ⊂ MNp (Rn) and the

inequality (E.16) also follows from the Riesz-Thorin theorem.e) follows from the fact that F−1atF = J−1

t F−1aFJt, where Jt is the isom-etry, (Jtf)(ξ) := t−n/pf(tξ), on Lp(Rn), and f) is proved by taking limits throughthe definitions of Fourier multipliers.

An extremely useful sufficient condition for a functionm to belong toM1p(R

n)for 1 n

2 }. Define the Banach space MM by

MM :={m : Rn → K : m ∈ Cj(Rn \ {0}), |m|M <∞}

, (E.17)

where the norm | · |M is defined by

|m|M := max|α|≤j

supξ∈Rn\{0}

|ξ||α||Dαm(ξ)|. (E.18)

We then have the following result.

491

Theorem E.3 (Mikhlin). Let 1 < p <∞. Then MM ↪→Mp(Rn).

For a proof of Mikhlin’s theorem, we refer to [Ste93, Theorem VI.4.4].The following results on Fourier multipliers will be useful in Chapter 8.

Theorem E.4. Let 1 ≤ p ≤ ∞. Then the following hold true:

a) Let a be a real homogeneous polynomial on Rn of degree m > 1. Then eia ∈Mp(Rn) if and only if p = 2.

b) Let a ∈ C∞(Rn) satisfy

a(ξ) :=

{|ξ|−βe−i|ξ|α (|ξ| ≥ 2),

0 (|ξ| ≤ 1),

where α > 0 and β ≥ 0.

(i) If α �= 1 and 1 < p <∞ (respectively, p = 1), then a ∈ Mp(Rn) if and

only if n∣∣12− 1

p

∣∣ ≤ βα(respectively, n

2< β

α).

(ii) If α = 1 and 1 < p < ∞ (respectively, p = 1) then a ∈ Mp(Rn) if andonly if (n− 1)

∣∣12 − 1

p

∣∣ ≤ β (respectively, n−12 < β).

c) Define a1 : R3 → C and a2 : R3 → C by

a1(ξ) := (−i)(ξ1 + ξ22 + ξ23 − i),

a2(ξ) := ξ1 + ξ22 + ξ23 + i.

(i) If p �= 2, then a−11 �∈ Mp(R3).

(ii) Let a := a1a2. Define the operator Ap on Lp(R3) by Apf := F−1(aFf)with D(Ap) := {f ∈ Lp(R3) : F−1(aFf) ∈ Lp(R3)}. If ∣∣12 − 1

p

∣∣ > 38 ,

then σ(Ap) = C.

For a proof of the assertions of Theorem E.4 we refer to [Hor60] (assertiona)), [FS72], [Miy81] and [Per80] (assertion b)), [KT80] (assertion c)i)) and [IS70](assertion c)ii)).

Finally, we note one instance of Mikhlin’s Theorem.For x ∈ R, define

signx :=

⎧⎪⎨⎪⎩1 (x > 0),

0 (x = 0),

−1 (x < 0).

Then sign ∈MM . By Mikhlin’s theorem, sign ∈Mp(R) for 1 < p <∞.For ϕ ∈ S(R), one finds that F−1(−i sign)Fϕ is a function given by

(F−1(−i sign)Fϕ)(x) = limε↓0

1

π

∫|x−y|≥ε

ϕ(y)

x− ydy. (E.19)

This is known as the Hilbert transform of ϕ. Thus, we have the following.

Proposition E.5. Let 1 < p < ∞. Then the Hilbert transform is a bounded linearoperator on Lp(R).

Notes: The material on distributions is very standard and can be found in many books,for example [Hor83]. The basic material on Fourier multipliers can be found in [Ste93].

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Notation

Function and Distribution Spaces

AAP(R+, X) space of asymptotically almost periodic functions . . . . . 307

AP(I,X),AP(I) space of almost periodic functions . . . . . . . . . . . 292, 297, 307

BSV([a, b], X) space of functions of bounded semivariation . . . . . . . . . . . . 48

BSVloc(R+, X) space of functions of locally bounded semivariation . . . . . 48

BUC(I,X),BUC(I) space of bounded, uniformly continuous functions . . . . . 15

Lipω(R+, X) space of Lipschitz continuous functions . . . . . . . . . . . . . 63, 77

D(Ω)′ space of distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485, 488

D(Ω) space of test functions . . . . . . . . . . . . . . . . . . . . . . . . 15, 485, 487

E , E(R+, X) space of totally ergodic functions . . . . . . . . . . . . . . . . 301, 308

FL1(Rn) Fourier algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436

MM Mikhlin space of Fourier multipliers . . . . . . . . . . . . . . 436, 490

MNp (Rn) space of matrices of Fourier multipliers . . . . . . . . . . . . . . . 489

MX(Rn),Mp(Rn) space of Fourier multipliers on X or Lp(Rn) . . . . . . . . . . 489

Mε strong Mikhlin space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436

S(Rn)′ space of temperate distributions . . . . . . . . . . . . . . . . . . . . . . 486

S(Rn) Schwartz space of rapidly decreasing functions . . . 319, 485

E quotient of space of totally ergodic functions . . . . . 301, 308

C(I,X), C(Ω) space of continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . 15

c(X) space of convergent sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 41

526 NOTATION

Ck(I,X), Ck(Ω) space of k-times continuously differentiable functions . . . 15

C∞(I,X), C∞(Ω) space of infinitely differentiable functions . . . . . . . . . . . . . . 15

C0(I,X), C0(Ω) space of continuous functions vanishing at infinity . . . . . . 15

c0 space of null sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10, 481

Cc(I,X), Cc(Ω) space of continuous functions with compact support . . . 15

C∞c (I,X), C∞c (Ω) space of infinitely differentiable functions with compact sup-port . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

C∞W ((ω,∞), X) Widder space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64, 78

C1ω(R+, X) space of functions with continuous, exponentially bounded

derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

H2(C+, X), H2(C+) Hardy spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Hm(Ω), Hm0 (Ω) Sobolev space of order (m, 2) . . . . . . . . . . . . . . . . . . . . . . . . . 488

Lp(Ω, μ) space of p-integrable functions on a measure space . . . . 175

Lp(I) space of Lebesgue p-integrable functions . . . . . . . . . . . . . . . 14

Lp(I,X) space of Bochner p-integrable functions . . . . . . . . . . . . 13, 14

lp space of p-summable sequences . . . . . . . . . . . . . . . . . . . . 10, 132

L∞(I,X), L∞(I) space of bounded measurable functions . . . . . . . . . . . . . . . . 14

l∞(X) space of bounded sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

L∞ω (I,X) space of exponentially bounded functions . . . . . . . . . 77, 226

L1loc(R+, X) space of locally Bochner integrable functions . . . . . . . . . . . 13

Smρ,0 space of symbols of pseudo-differential operators . . . . . . 430

Wm,p(Ω),Wm,p0 (Ω) Sobolev space of order (m, p) . . . . . . . . . . . . . . . . . . . . . . . . . 488

Dual Spaces and Subspaces

V ′ antidual of V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420

X∗ dual space of X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

X sun-dual of X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

X0 space of vectors converging to 0 . . . . . . . . . . . . . . . . . . . . . . . 360

NOTATION 527

Xe space of totally ergodic vector in X . . . . . . . . . . . . . . . . . . . . 267

Xap space of almost periodic vectors in X . . . . . . . . . . . . 290, 361

Xe0 space of totally ergodic vectors with means 0 . . . . . . . . . 267

Norms and Dualities

(·|·)H inner product on a Hilbert space H . . . . . . . . . . . . . . . . . . . . 45

(·|·) duality between a space and its antidual . . . . . . . . . . . . . . 421

〈·, ·〉 duality between a space and its dual . . . . . . . . . . . . . . . 7, 485

‖ · ‖D(A) graph norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468

‖ · ‖W Widder norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64, 78

|α| norm of multi-index α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485

| · |M Mikhlin norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436, 490

‖ · ‖p Lebesgue-Bochner norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13, 14

| · |Mεstrong Mikhlin norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436

‖ · ‖ω,∞ exponentially bounded norm . . . . . . . . . . . . . . . . . . . . . . 77, 226

|π| norm of partition π . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Functions, Integrals and Abscissas

abs(f), abs(dF ) abscissa of convergence . . . . . . . . . . . . . . . . . . . . . . 27, 30, 56, 57

χΩ characteristic function of Ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Cos cosine function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

hol(f), hol(T ) abscissa of holomorphy of f or T . . . . . . . . . . . . . . . . . . . 33, 35

hol0(f) abscissa of boundedness of f . . . . . . . . . . . . . . . . . . . . . . . . . . . 33∫ b

ag(t) dF (t) Riemann-Stieltjes integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49∫ b

ag(t) dt Riemann integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50∫

If(t) dt Bochner integral of f over I . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

ω(f), ω(T ) exponential growth bound of f or T . . . . . . . . . . . . . . . . 29, 30

ω1(T ) exponential growth bound of classical solutions . . . . . . . 343

528 NOTATION

sign signum function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138, 491

Sin sine function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206, 218

En Newtonian potential on Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404

eλ exponential function t �→ eλt . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

eλ ⊗ x the function t �→ eλtx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292

f ∗ g, T ∗ f convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21, 24, 487

kt, kz Gaussian kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

S(g, π) Riemann sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

S(g, F, π) Riemann-Stieltjes sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

ux orbit of T through x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30, 337

V (π, F ) variation of F over π . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

V (F ), V[a,b](F ) total variation of F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Operators

Δ distributional Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

Δ0 Laplacian on C0(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406

Δp Laplacian on Lp(Rn) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

ΔX Laplacian on X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

Δmax Laplacian on C(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403

ΔL2(Ω) Dirichlet-Laplacian on L2(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . 140

Ap system of differential operators on Lp(Rn) . . . . . . . . . . . . 449

AX pseudo-differential operator on X . . . . . . . . . . . . . . . . . . . . . 431

K(X) space of compact operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

L(X,Y ),L(X) space of bounded linear operators . . . . . . . . . . . . . . . . . . . . . . 24

OpX(a),Opp(a) pseudo-differential operator on X or Lp(Rn) . . . . . . . . . . 430

KerA kernel of A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261, 467

RanA range of A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261, 467

NOTATION 529

A closure of an operator A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467

Φ Riesz operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

ΦS Riesz-Stieltjes operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

A∗ adjoint of an operator A . . . . . . . . . . . . . . . . . . . . . . . . . . 472, 473

AH operator associated with quadratic form . . . . . . . . . . . . . . . 419

AY part of an operator A in Y . . . . . . . . . . . . . . . . . . . . . . . 136, 471

B−z fractional power of B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

B1/2 square root of B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

D(A) domain of an operator A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467

Dα higher order partial derivative . . . . . . . . . . . . . . . . . . . 485, 486

Dj partial derivative ∂/∂xj . . . . . . . . . . . . . . . . . . . . . . . . . . 485, 486

R(λ,A) resolvent of an operator A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468

Spectrum and Resolvent Set

spB(f) Beurling spectrum of f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322

spC(f) Carleman spectrum of f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295

sp(f) half-line spectrum of f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272

spw(f) weak half-line spectrum of f . . . . . . . . . . . . . . . . . . . . . . . . . . 325

ρ(A) resolvent set of an operator A . . . . . . . . . . . . . . . . . . . . . . . . . 468

ρu(A, x) imaginary local resolvent set . . . . . . . . . . . . . . . . . . . . . . . . . . 371

σ(A, x) local spectrum of A at x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299

σ(A) spectrum of A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468

σp(A) point spectrum of A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469

σu(A, x) imaginary local spectrum of A at x . . . . . . . . . . . . . . . . . . . 371

σap(A) approximate point spectrum of A . . . . . . . . . . . . . . . . . . . . . 469

r(T ) spectral radius of an operator T . . . . . . . . . . . . . . . . . . . . . . . 475

s(A) spectral bound of A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188, 469

530 NOTATION

s0(A) pseudo-spectral bound of A . . . . . . . . . . . . . . . . . . . . . . . . . . . 345

Subsets of Rn or C

C+ open right half-plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

C− open left half-plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362

T unit circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316

N set of positive integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

N0 set of non-negative integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

∂Ω topological boundary of Ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401

R+ set of non-negative real numbers . . . . . . . . . . . . . . . . . . . . . . . . 13

Σα sector of angle α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

Transformations

f reflection of f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

f , T Laplace or Carleman transform of f or T . . . . . . 27, 32, 295

F Fourier transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44, 487

L Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

LS Laplace-Stieltjes transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

F conjugate Fourier transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

dF Laplace-Stieltjes transform of F . . . . . . . . . . . . . . . . . . . . . . . 55

Cauchy Problems

(ACP0) homogeneous abstract Cauchy problem . . . . . . . . . . . . . . . . 108

(ACPf ) inhomogeneous abstract Cauchy problem . . . . . . . . . . . . . 117

(ACPk+1) (k + 1)-times integrated abstract Cauchy problem . . . . 129

ACP0(R) abstract Cauchy problem on the line . . . . . . . . . . . . . . . . . . 118

D(ϕ) Dirichlet problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401

P∞(u0, ϕ, f) inhomogeneous heat equation . . . . . . . . . . . . . . . . . . . . . . . . . 415

Pτ (u0, ϕ) heat equation with inhomogeneous boundary conditions 408,412

NOTATION 531

Other Notation

(Hr) growth hypothesis for symbols . . . . . . . . . . . . . . . . . . . . . . . . . 439

Freq(x),Freq(f) set of frequencies of vector x or function f . . 267, 293, 315

dN(x) subdifferential of the norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

supp support of a function or distribution . . . . . . . . . . . . . . . . . . 318

B(x, ε) closed ball, centre x, radius ε . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

D closure of a set D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

B(x, ε) open ball, centre x, radius ε . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

m(Ω) Lebesgue measure of Ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Mηx,Mηf mean of vector x or function f at η . . . . 266, 293, 308, 315

x · ξ scalar product of x and ξ in Rn . . . . . . . . . . . . . . . . . . 430, 487

x ≤ y ordering in a Banach space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477

X+ positive cone in X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477

Zd↪→ X continuous dense embedding . . . . . . . . . . . . . . . . . . . . . 184, 418

Z ↪→ X continuous embedding of Z in X . . . . . . . . . . . . . . . . . . . . . 184

Index

AAbel-convergence, 244, 257Abel-ergodic, 263Abelian theorem, 243, 245abscissa

of boundedness, 33, 285of convergence, 27, 31, 56, 58of holomorphy, 33

absolutelycontinuous, 15, 18convergent, 14regular, 486

adjoint, 472, 473almost periodic

function on the half-line, 307function on the line, 292orbits, 294vector, 290, 361

almost separably valued, 7analytic

Radon-Nikodym property, 61representation, 84

antiderivative, 15antidual, 420, 422antilinear, 420approximate

eigenvalue, 469point spectrum, 469unit, 23

approximation theorem, 41, 67asymptotically

almost periodic, 307, 365norm-continuous, 389

BB-convergence, 244Banach lattice, 478band, 479Bernstein, 90, 100, 435Beurling spectrum, 322

Bochner, 9integrable, 9integral, 9

boundarygroup, 172semigroup, 171

boundedabove, 214, 474holomorphic semigroup, 150semivariation, 48variation, 15, 48

Brenner, 450, 452

CCarleman

spectrum, 295spectrum and C0-groups, 295transform, 295

Cauchy problemabstract, 108inhomogeneous, 117on the line, 118

Cesaro-convergence, 244Cesaro-ergodic, 262character, 289classical solution, 108, 117, 203closable operator, 467closed operator, 467closure, 467Coifman-Weiss, 175compact resolvent, 470complete orbit, 120completely monotonic, 90, 106complex

inversion, 75, 259representation, 81Tauberian condition, 247

cone, 477converges

in the sense of Abel, 244in the sense of Cesaro, 244

534 INDEX

convex, 91convolution, 21, 24, 26, 486core, 468cosine function, 203countable spectrum, 374, 385countably valued, 6

DDa Prato-Sinestrari, 142Datko, 340densely defined, 467Desch-Schappacher, 161Dirac’s equation, 456Dirichlet

boundary conditions, 140, 423kernel, 257Laplacian, 424problem, 401regular, 402

dissipative, 137distribution, 485, 487

semigroup, 231dominated convergence, 11dual cone, 478Dunford-Pettis

property, 270theorem, 19

Eeigenvalue, 469elliptic

equation, 170maximum principle, 402operator, 425, 431polynomial, 431

ergodic vector, 266eventually differentiable, 284exponential growth bound, 29, 30, 338

FFattorini, 227feebly oscillating, 249Fejer, 257

kernel, 258

first order perturbation, 160form domain, 420Fourier

coefficients, 257inversion theorem, 45, 487multiplier, 489sums, 257transform, 44, 487type, 61, 387

fractional powers, 163frequencies, 267, 293, 311Fubini, 12function

absolutely continuous, 15, 18absolutely regular, 486almost separably valued, 7asymptotically almost periodic,

307Bochner integrable, 9completely monotonic, 90convex, 91countably valued, 6feebly oscillating, 249holomorphic, 461Laplace transformable, 28Lipschitz continuous, 15locally bounded, 462measurable, 6normalized, 100of bounded semivariation, 48of bounded variation, 15, 48of weak bounded variation, 48Riemann integrable, 50Riemann-Stieltjes integrable, 49simple, 6slowly oscillating, 247step, 6strongly continuous, 24test, 15, 485, 487totally ergodic, 296, 308, 315uniformly ergodic, 296, 308, 328weakly measurable, 7

fundamental theorem of calculus, 18

INDEX 535

GGaussian semigroup, 150, 153, 156,

170, 172, 183Gelfand, 280generating cone, 477generator

infinitesimal, 114of C0-group, 119of C0-semigroup, 112of cosine function, 205of integrated semigroup, 122of semigroup, 126of sine function, 218

Glicksberg-deLeeuw, 389graph norm, 468Grothendieck space, 270group

C0, 119, 295boundary, 172integrated, 179

HHormander, 173half-line spectrum, 272, 310, 315Hardy, 256Hardy-Littlewood, 253Hilbert transform, 198, 491Hille-Yosida

operator, 141theorem, 134

holomorphicfunction, 461semigroup, 148

hyperbolicequation, 427semigroup, 388system, 449

hypoelliptic, 431

Iideal, 192, 478identity theorem, 462

imaginary localresolvent set, 371spectrum, 371

improper integral, 13infinitesimal generator, 114Ingham, 327integral

absolutely convergent, 14Bochner, 9improper, 13Laplace, 27Laplace-Stieltjes, 55Riemann, 50Riemann-Stieljes, 49

integrated semigroup, 122integration by parts, 50intermediate points, 49interpolation property, 90inversion

complex, 75, 259Phragmen-Doetsch, 73Post-Widder, 42, 73

invertible, 468irreducible, 394

Jjump, 99

KKadets, 300Karamata, 251Katznelson-Tzafriri, 317, 391KB-space, 479kernel, 261, 394, 467Krein-Smulyan, 464

LL-space, 359Laplace

integral, 27, 32transform, 63, 110transformable, 28

536 INDEX

Laplace-Stieltjesintegral, 55transform, 63

Laplacianand boundary group, 173and boundary integrated group,

183and cosine functions, 448first order perturbation, 160generates Gaussian semigroup, 150on continuous functions, 403square root, 170with Dirichlet boundary condi-

tions, 140, 424with inhomogeneous boundary con-

ditions, 408largest lower bound, 478lattice, 478least upper bound, 478Lebesgue point, 16Lipschitz continuous, 15local

integrated semigroup, 232spectrum, 299, 371

locally bounded, 462Loomis, 297Lotz, 272Lumer-Phillips, 139

MMaxwell’s equations, 455mean-ergodic, 262measurable, 6Mikhlin, 491mild solution, 108, 117, 119, 203, 368,

408, 413, 415mollifier, 23, 319multi-index, 485multiplication operator, 419, 473

NNewtonian potential, 404non-resonance, 380normal cone, 477

normalization, 100normalized

antiderivative, 28function, 100

norming, 462

Ooperator

adjoint, 472, 473associated with form, 418closable, 467closed, 467elliptic, 425, 431Hille-Yosida, 141invertible, 468kernel, 394multiplication, 419, 473Poisson, 404positive, 478pseudo-differential, 430resolvent positive, 188Riesz, 72Riesz-Stieltjes, 66sectorial, 162selfadjoint, 150, 473symmetric, 474

ordercontinuous norm, 479interval, 477

ordered Banach space, 477

PPaley-Wiener, 46parabolic

domain, 412equation, 427maximum principle, 410problem, 408

part, 471partitioning points, 49period, 292periodic vector, 290

INDEX 537

perturbationcompact, 161first order, 160of C0-semigroup, 144of cosine function, 210, 213of Hille-Yosida operator, 143, 144of integrated semigroup, 187, 232of resolvent positive operator, 195of selfadjoint operator, 420, 423of sine function, 220relatively bounded, 159

Petrovskii correct systems, 232Pettis, 7phase space, 210Phragmen-Doetsch, 73Phragmen-Lindelof, 176Plancherel, 45point spectrum, 469Poisson

equation, 404operator, 404semigroup, 152, 170, 447

positivecone, 477element, 477functional, 477operator, 478

Post-Widder, 42, 73Pruss, 82primitive, 15principal

part, 431, 449value, 13

proper cone, 477pseudo-differential operator, 430pseudo-resolvent, 470pseudo-spectral bound, 345

RRadon-Nikodym property, 19, 72range, 261, 467real

Banach lattice, 478ordered Banach space, 477

representation, 69, 78Tauberian condition, 247

realization, 430regular point, 295, 310regularized semigroup, 232relatively compact orbit, 288relatively dense, 288, 310representation

analytic, 84complex, 81Paley-Wiener, 46real, 69, 78Riesz-Stieltjes, 66

resolvent, 335, 468compact, 470equation, 470identity, 470positive, 188set, 468

Riemannintegrable, 50integral, 50sum, 50

Riemann-Lebesgue, 45Riemann-Liouville semigroup, 175Riemann-Stieltjes

integrable, 49integral, 49sum, 49

Riesz operator, 72Riesz-Stieltjes

operator, 66representation, 66

Ssandwich theorem, 185Schwartz space, 318, 485sectorial operator, 162selfadjoint operator, 150, 473semigroup, 126

C-, 232C0, 111Abel-ergodic, 263

538 INDEX

asymptotically almost periodic,365

asymptotically norm-continuous,388

boundary, 171bounded holomorphic, 150Cesaro-ergodic, 262distribution, 231eventually differentiable, 284Gaussian, 150, 153, 156, 170, 172,

183holomorphic, 148hyperbolic, 388irreducible, 394k-times integrated, 122local integrated, 232norm-continuous, 201once integrated, 122Poisson, 152, 170, 447regularized, 232Riemann-Liouville, 175smooth distribution, 232totally ergodic, 266, 373

separating, 8, 262, 464sesquilinear form, 420similar operators, 144simple

function, 6pole, 269

sine function, 206, 218slowly oscillating, 247smooth distribution semigroup, 232smoothing effect, 158Sobolev space, 488spectral

projection, 472bound, 188, 343, 469radius, 475synthesis, 291, 293, 391theorem, 474

spectrum, 468approximate point, 469Beurling, 322Carleman, 295

half-line, 272, 310, 315imaginary local, 371local, 299point, 469weak half-line, 325

square root, 164step function, 6strong convergence, 31strong splitting theorem, 364strongly continuous, 24subdifferential, 137sublattice, 478sun-dual, 137support, 318symbol, 430symmetric, 474

TTauberian

condition, 243, 247theorem, 88, 243, 247

temperate distribution, 485tempered integrated semigroup, 232test function, 15, 485, 487Theorem

Abel, 247Analytic Representation, 84Approximation, 41, 67Bernstein, 100Bochner, 9Brenner, 450, 452Coifman-Weiss, 175Complex Inversion, 75Complex Representation, 81Countable spectrum, 374Da Prato-Sinestrari, 142Datko, 340Desch-Schappacher, 161Dominated Convergence, 11Dunford-Pettis, 19Fattorini, 227Fejer, 257Fubini, 12Gelfand, 280

INDEX 539

Glicksberg-deLeeuw, 389Hormander, 173Hardy, 256Hardy-Littlewood, 253Hille-Yosida, 134Identity, 462Ingham, 327Kadets, 300Karamata, 251Katznelson-Tzafriri, 317, 391Krein-Smulyan, 464Loomis, 297Lotz, 272Lumer-Phillips, 139Mikhlin, 491Non-resonance, 380Paley-Wiener, 46Pettis, 7Phragmen-Doetsch Inversion, 73Phragmen-Lindelof, 176Plancherel, 45Post-Widder Inversion, 42, 73Real Representation, 69, 78Riesz-Stieltjes Representation, 66Sandwich, 185Spectral, 474Splitting, 364, 368, 389Tauberian, 88Trotter-Kato, 146Uniqueness, 40, 294Vitali, 463

totally ergodicfunction, 296, 308, 315semigroup, 266, 373vector, 266, 290, 361

transference principle, 175trigonometric polynomial, 292, 365Trotter-Kato, 146

UUMD-space, 198unconditionally bounded, 304, 481uniform ellipticity, 425

uniformlyconvex, 303ergodic, 296, 308, 328

unimodular eigenvector, 361uniqueness

sequence, 40theorem, 40, 294

unitarily equivalent, 474

Vvariation of constants formula, 118,

158Vitali, 463

Wwave equation, 170, 332, 425, 455weak

bounded variation, 48half-line spectrum, 325splitting theorem, 368

weaklyalmost periodic, 294almost periodic in the sense of

Eberlein, 294asymptotically almost periodic,

334holomorphic, 461measurable, 7regular point, 325

Weierstrass formula, 216well-posed, 115

YYoung’s inequality, 22, 24

vector-valued laplace transforms and cauchy problems: second edition

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