hilbert spaces spectra

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Spectral Theory in HilbertSpace

Lectures fall 2008

Christer Bennewitz


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Copyright c 19932008 by Christer Bennewitz


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Preface

The aim of these notes is to present a reasonably complete exposi-tion of Hilbert space theory, up to and including the spectral theoremfor the case of a (possibly unbounded) selfadjoint operator. As an ap-plication, eigenfunction expansions for regular and singular boundary

value problems of ordinary differential equations are discussed. Wefirst do this for the simplest Sturm-Liouville equation, and then, usingvery similar methods of proof, for a fairly general type of first ordersystems, which include so called Hamiltonian systems.

Prerequisites are modest, but a good understanding of Lebesgueintegration is assumed, including the concept of absolute continuity.Some previous exposure to linear algebra and basic functional analy-sis (uniform boundedness principle, closed graph theorem and maybeweak compactness of the unit ball in a (separable) Banach space) isexpected from the reader, but in the two places where we could haveused weak compactness, a direct proof has been given. The standard

proofs of the Banach-Steinhaus and closed graph theorems are givenin Appendix A. A brief exposition of the Riemann-Stieltjes integral,sufficient for our needs, is given in Appendix B. A few elementary factsabout ordinary linear differential equations are used. These are provedin Appendix C. In addition, some facts from elementary analytic func-tion theory are used. Apart from this the lectures are essentially self-contained.

Egevang, August 2008

Christer Bennewitz

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Contents

Preface i

Chapter 0. Introduction 1

Chapter 1. Linear spaces 5

Exercises for Chapter 1 8

Chapter 2. Spaces with scalar product 9Exercises for Chapter 2 13

Chapter 3. Hilbert space 15Exercises for Chapter 3 21

Chapter 4. Operators 23Exercises for Chapter 4 30

Chapter 5. Resolvents 31

Exercises for Chapter 5 33Chapter 6. Nevanlinna functions 35

Chapter 7. The spectral theorem 39Exercises for Chapter 7 44

Chapter 8. Compactness 45Exercises for Chapter 8 49

Chapter 9. Extension theory 511. Symmetric operators 512. Symmetric relations 53Exercises for Chapter 9 57

Chapter 10. Boundary conditions 59Exercises for Chapter 10 67

Chapter 11. Sturm-Liouville equations 69Exercises for Chapter 11 81

Chapter 12. Inverse spectral theory 831. Asymptotics of the m-function 842. Uniqueness theorems 87

Chapter 13. First order systems 91iii


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iv CONTENTS

Exercises for Chapter 13 98

Chapter 14. Eigenfunction expansions 101Exercises for Chapter 14 104

Chapter 15. Singular problems 105Exercises for Chapter 15 114

Appendix A. Functional analysis 117

Appendix B. Stieltjes integrals 121Exercises for Appendix B 127

Appendix C. Linear first order systems 129

Appendix. Bibliography 133


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CHAPTER 0

Introduction

Hilbert space is the most immediate generalization to the infinitedimensional case of finite dimensional Euclidean spaces (i.e., essentiallyRn for real, and Cn for complex vector spaces). Probably its most im-portant uses, and certainly its historical roots, are in spectral theory.

Spectral theory for differential equations originates with the method ofseparation of variables, used to solve many of the equations of math-ematical physics. This leads directly to the problem of expanding anarbitrary function in terms ofeigenfunctions of the reduced equation,which is the central problem of spectral theory. A simple exampleis that of a vibrating string. The string is supposed to be stretchedover an interval I R, be fixed at the endpoints a, b and vibratetransversally (i.e., in a direction perpendicular to the interval I) in aplane containing I. The string can then be described by a real-valuedfunction u(x, t) giving the location at time t of the point of the stringwhich moves on the normal to I through the point x

I. In appropri-

ate units the function u will then (for sufficiently small vibrations, i.e.,we are dealing with a linearization of a more accurate model) satisfythe following equation:

(0.1)

2u

x2=

2u

t2(wave equation)

u(a, t) = u(b, t) = 0 for t > 0 (boundary conditions)

u(x, 0) and ut(x, 0) given (initial conditions).

The idea in separating variables is first to disregard the initial condi-tions and try to find solutions to the differential equation that satisfythe boundary condition and are standing waves, i.e., of the special formu(x, t) = f(x)g(t). The linearity of the equation implies that sums ofsolutions are also solutions (the superposition principle), so if we canfind enough standing waves there is the possibility that any solutionmight be a superposition of standing waves. By substituting f(x)g(t)for u in (0.1) it follows that f(x)/f(x) = g(t)/g(t). Since the lefthand side does not depend on t, and the right hand side not on x, bothsides are in fact equal to a constant

. Since the general solution of

the equation g(t) + g(t) = 0 is a linear combination of sin( t) and1


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2 0. INTRODUCTION

cos(

t), it follows that

(0.2) f = f in If(a) = f(b) = 0,

and that g(t) = A sin(

t) + B cos(

t) for some constants A andB. As is easily seen, (0.2) has non-trivial solutions only when isan element of the sequence {j}1 , where j = ( baj)2. The numbers1, 2, . . . are the eigenvalues of (0.2), and the corresponding solutions(non-trivial multiples of sin(j

ba (x a))), are the eigenfunctions of(0.2). The set of eigenvalues is called the spectrum of (0.2). In general,a superposition of standing waves is therefore of the form u(x, t) =

(Aj sin(j t) + Bj cos(j t)) sin(j (x a)). If we assume thatwe may differentiate the sum term by term, the initial conditions of(0.1) therefore require that

Bj sin(

baj(x a)) and

Aj

baj sin(

baj(x a))are given functions. The question of whether (0.1) has a solution whichis a superposition of standing waves for arbitrary initial conditions, isthen clearly seen to amount to the question whether an arbitraryfunction may be written as a series

uj, where each term is an eigen-

function of (0.2), i.e., a solution for equal to one of the eigenvalues.We shall eventually show this to be possible for much more generaldifferential equations than (0.1).

The technique above was used systematically by Fourier in his The-orie analytique de la Chaleur (1822) to solve problems of heat conduc-tion, which in the simplest cases (like our example) lead to what arenow called Fourier series expansions. Fourier was never able to give asatisfactory proof of the completeness of the eigenfunctions, i.e., thefact that essentially arbitrary functions can be expanded in Fourierseries. This problem was solved by Dirichlet somewhat later, and atabout the same time (1830) Sturm and Liouville independently butsimultaneously showed weaker completeness results for more general

ordinary differential equations of the form (pu) + qu = u, withboundary conditions of the form Au + Bpu = 0, to be satisfied at theendpoints of the given interval. Here p and q are given, sufficiently reg-ular functions, and A, B given real constants, not both 0 and possiblydifferent in the two interval endpoints. The Fourier cases correspondto p 1, q 0 and A or B equal to 0.

For the Fourier equation, the distance between successive eigenval-ues decreases as the length of the base interval increases, and as thebase interval approaches the whole real line, the eigenvalues accumu-late everywhere on the positive real line. The Fourier series is thenreplaced by a continuous superposition, i.e., an integral, and we get

the classical Fourier transform. Thus a continuous spectrum appears,


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0. INTRODUCTION 3

and this is typical of problems where the basic domain is unbounded,or the coefficients of the equation have sufficiently bad singularities at

the boundary.In 1910 Hermann Weyl [12] gave the first rigorous treatment, in the

case of an equation of Sturm-Liouville type, of cases where continuousspectra can occur. Weyls treatment was based on the then recentlyproved spectral theorem by Hilbert. Hilberts theorem was a general-ization of the usual diagonalization of a quadratic form, to the case ofinfinitely many variables. Hilbert applied it to certain integral oper-ators, but it is not directly applicable to differential operators, sincethese are unbounded in a sense we will discuss in Chapter 4. Withthe creation of quantum mechanics in the late 1920s, these matters be-

came of basic importance to physics, and mathematicians, who had notadvanced much beyond the results of Weyl, took the matter up again.The outcome was the general spectral theorem, generally attributed toJohn von Neumann (1928), although essentially the same theorem hadbeen proved by Torsten Carleman in 1923, in a less abstract setting.Von Neumanns theorem is an abstract result, and detailed applica-tions to differential operators of reasonable generality had to wait untilthe early 1950s. In the meantime many independent results aboutexpansions in eigenfunctions had been given, particularly for ordinarydifferential equations.

In these lectures we will prove von Neumanns theorem. We will

then apply this theorem to differential equations, including those thatgive rise to the classical Fourier series and Fourier transform. Once onehas a result about expansion in eigenfunctions a host of other questionsappear, some of which we will discuss in these notes. Sample questionsare:

How do eigenvalues and eigenfunctions depend on the domainI and on the form of the equation (its order, coefficients etc.)?A partial answer is given if one can calculate the asymptoticdistributionof the eigenvalues, i.e., approximate the growth of

j as a function ofj. For simple ordinary differential operatorsthis can be done by fairly elementary means. The first suchresult for a partial differential equation was given by Weyl in1912, and his method was later improved and extended byCourant.

How well does the expansion converge when expanding dif-ferent classes of functions? Again, for ordinary differentialoperators some questions of this type can be handled by ele-mentary methods, but in general the answer lies in the explicitasymptotic behavior of the so called spectral projectors. Thefirst such asymptotic result was given by Carleman in 1934,

and his method has been the basis for most later results.


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4 0. INTRODUCTION

Can the equation be reconstructed if the spectrum is known? Ifnot, what else must one know? If different equations can have

the same spectrum, how many different equations? What dothey have in common? Questions like these are part of what iscalled inverse spectral theory. Really satisfactory answers haveonly been obtained for the equation u + qu = u, notablyby Gelfand and Levitan in the early 1950:s. Pioneering workwas done by Goran Borg in the 1940:s.

Another aspect of the first point is the following: Given a baseequation (corresponding to a free particle in quantum me-chanics) and another equation, which outside some boundedregion is close to the base equation (an obstacle has been in-

troduced), how can one relate the eigenfunctions for the twoequations? The main questions of so called scattering theoryare of this type.

Related to the previous point is the problem of inverse scat-tering. Here one is given scattering data, i.e., the answer tothe question in the previous point, and the question is whetherthe equation is determined by scattering data, whether thereis a method for reconstructing the equation from the scatter-ing data, and similar questions. Many questions of this kindare of great importance to applications.


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CHAPTER 1

Linear spaces

This chapter is intended to be a quick review of the basic factsabout linear spaces. In the definition below the set K can be any field,although usually only the fields R of real numbers and C of complexnumbers are of interest.

Definition 1.1. A linear space or vector space over K is a set Lprovided with an addition +, which to every pair of elements u, v Lassociates an element u + v L, and a multiplication, which to every K and u L associates an element u L. The following rulesfor calculation hold:

(1) (u + v) + w = u + (v + w) for all u, v and w in L.(associativity)(2) There is an element 0 L such that u + 0 = 0 + u = u for

every u L. (existence of neutral element)(3) For every u L there exists v L such that u+v = v +u = 0.

One denotes v by

u. (existence of additive inverse)

(4) u + v = v + u for all u, v L. (commutativity)(5) (u + v) = u + v for all K and all u, v L.(6) ( + )u = u + u for all , K and all u L.(7) (u) = ()u for all , K and all u L.(8) 1u = u for all u L.

IfK = R we have a real linear space, ifK = C a complex linearspace. Axioms 13 above say that L is a group under addition, ax-iom 4 that the group is abelian (or commutative). Axioms 5 and 6 aredistributive laws and axiom 7 an associative law related to the multi-

plication by scalars, whereas axiom 8 gives a kind of normalization forthe multiplication by scalars.Note that by restricting oneself to multiplying only by real num-

bers, any complex space may also be viewed as a real linear space.Conversely, every real linear space can be extended to a complex lin-ear space (Exercise 1.1). We will therefore only consider complex linearspaces in the sequel.

Let M be an arbitrary set and let CM be the set of complex-valuedfunctions defined on M. Then CM, provided with the obvious defini-tions of the linear operations, is a complex linear space (Exercise 1.2).In the case when M =

{1, 2, . . . , n

}one writes Cn instead ofC{1,2,...,n}.

An element u Cn is of course given by the values u(1), u(2), . . . , u(n)5


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6 1. LINEAR SPACES

ofu so one may also regard Cn as the set of ordered n-tuples of complexnumbers. The corresponding real space is the usual Rn.

If L is a linear space and V a subset of L which is itself a linearspace, using the linear operations inherited from L, one says that V isa linear subspace ofL.

Proposition 1.2. A non-empty subset V ofL is a linear subspaceof L if and only if u + v V and u V for all u, v V and C.

The proof is left as an exercise (Exercise 1.3). If u1, u2, . . . , uk areelements of a linear space L we denote by [u1, u2, . . . , uk] the linearhull ofu1, u2, . . . , uk, i.e., the set of all linear combinations 1u1 + +kuk, where 1, . . . , k C. It is not hard to see that linear hulls arealways subspaces (Exercise 1.5). One says that u1, . . . , uk generatesL if L = [u1, . . . , uk], and any linear space which is the linear hullof a finite number of its elements is called finitely generated or finite-dimensional. A linear space which is not finitely generated is calledinfinite-dimensional. It is clear that if, for example, u1 is a linearcombination of u2, . . . , uk, then [u1, . . . , uk] = [u2, . . . , uk]. If none ofu1, . . . , uk is a linear combination of the others one says that u1, . . . , ukare linearly independent. It is clear that any finitely generated spacehas a set of linearly independent generators; one simply starts witha set of generators and goes through them one by one, at each stepdiscarding any generator which is a linear combination of those coming

before it. A set of linearly independent generators for L is called a basisfor L. A given finite-dimensional space L can of course be generatedby many different bases. However, a fundamental fact is that all suchbases ofL have the same number of elements, called the dimension ofL. This follows immediately from the following theorem.

Theorem 1.3. Suppose u1, . . . , uk generate L, and that v1, . . . , vjare linearly independent elements of L. Then j k.

Proof. Since u1, . . . , uk generate L we have v1 =k

s=1 x1sus,for some coefficients x11, . . . , x1k which are not all 0 since v1 = 0.By renumbering u1, . . . , uk we may assume x11

= 0. Then u1 =

1x11 v1 ks=2 x1sx11 us, and therefore v1, u2, . . . , uk generate L. In par-ticular, v2 = x21v1 +

ks=2 x2sus for some coefficients x21, . . . , x2k. We

can not have x22 = = x2k = 0 since v1, v2 are linearly independent.By renumbering u2, . . . , uk, if necessary, we may assume x22 = 0. Itfollows as before that v1, v2, u3, . . . , uk generate L. We can continue inthis way until we run out of either v:s (if j k) or u:s (if j > k). Butif j > k we would get that v1, . . . , vk generate L, in particular thatvj is a linear combination of v1, . . . , vk which contradicts the linearindependence of the v:s. Hence j k.

For a finite-dimensional space the existence and uniqueness of co-

ordinates for any vector with respect to an arbitrary basis now follows


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1. LINEAR SPACES 7

easily (Exercise 1.6). More importantly for us, it is also clear that Lis infinite dimensional if and only if every linearly independent subset

of L can be extended to a linearly independent subset of L with arbi-trarily many elements. This usually makes it quite easy to see that agiven space is infinite dimensional (Exercise 1.7).

If V and W are both linear subspaces of some larger linear spaceL, then the linear span [V, W] of V and W is the set

[V, W] = {u | u = v + w where v V and w W}.This is obviously a linear subspace of L. If in addition V W = {0},then for any u [V, W] there are unique elements v V and w Wsuch that u = v + w. In this case [V, W] is called the direct sum of Vand W and is denoted by V+W. The proof of these facts is left as an

exercise (Exercise 1.9).If V is a linear subspace of L we can create a new linear space

L/V, the quotient space of L by V, in the following way. We saythat two elements u and v of L are equivalent if u v V. It isimmediately seen that any u is equivalent to itself, that u is equivalentto v if v is equivalent to u, and that if u is equivalent to v, and v tow, then u is equivalent to w. It then easily follows that we may splitL into equivalence classes such that every vector is equivalent to allvectors in the same equivalence class, but not to any other vectors.The equivalence class containing u is denoted by u + V, and then

u + V = v + V precisely if u v V. We now define L/V as the set ofequivalence classes, where addition is defined by (u + V) + (v + V) =(u+v)+V and multiplication by scalar as (u+V) = u+V. It is easilyseen that these operations are well defined and that L/V becomes alinear space with neutral element 0 + V (Exercise 1.10). One definescodim V = dim L/V. We end the chapter by a fundamental fact aboutquotient spaces.

Theorem 1.4. dim V + codim V = dim L.We leave the proof for Exercise 1.11.


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8 1. LINEAR SPACES

Exercises for Chapter 1

Exercise1.1

.Let L be a real linear space, and let V be the setof ordered pairs (u, v) of elements of L with addition defined compo-

nentwise. Show that V becomes a complex linear space if one defines(x + iy)(u, v) = (xu yv,xv + yu) for real x, y. Also show that L canbe identified with the subset of elements of V of the form (u, 0), inthe sense that there is a one-to-one correspondence between the twosets preserving the linear operations (for real scalars).

Exercise 1.2. Let M be an arbitrary set and let CM be the setof complex-valued functions defined on M. Show that CM, providedwith the obvious definitions of the linear operations, is a complex linear

space.Exercise 1.3. Prove Proposition 1.2.

Exercise 1.4. Let M be a non-empty subset ofRn. Which of thefollowing choices ofL make it into a linear subspace ofCM?

(1) L = {u CM | |u(x)| < 1 for all x M}.(2) L = C(M) = {u CM | u is continuous in M}.(3) L = {u C(M) | u is bounded on M}.(4) L = L(M) = {u CM | u is Lebesgue integrable over M}.

Exercise 1.5. Let L be a linear space and uj L, j = 1, . . . , k.Show that [u1, u2, . . . , uk] is a linear subspace ofL.

Exercise 1.6. Show that if e1, . . . , en is a basis for L, then foreach u L there are uniquely determined complex numbers x1, . . . , xn,called coordinates for u, such that u = x1e1 + + xnen.

Exercise 1.7. Verify that L is infinite dimensional if and only ifevery linearly independent subset of L can be extended to a linearlyindependent subset of L with arbitrarily many elements. Then showthat u1, . . . , uk are linearly independent if and only if1u1+ +kuk =0 only for 1 = = k = 0. Also show that CM is finite-dimensionalif and only if the set M has finitely many elements.

Exercise 1.8. Let M be an open subset ofRn. Verify that L isinfinite-dimensional for each of the choices of L in Exercise 1.4 whichmake L into a linear space.

Exercise 1.9. Prove all statements in the penultimate paragraphof the chapter.

Exercise 1.10. Prove that ifL is a linear space and V a subspace,then L/V is a well defined linear space.

Exercise 1.11. Prove Theorem 1.4.


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CHAPTER 2

Spaces with scalar product

If one wants to do analysis in a linear space, some structure in ad-dition to the linearity is needed. This is because one needs some wayto define limits and continuity, and this requires an appropriate defini-tion of what a neighborhood of a point is. Thus one must introduce a

topology in the space. We will not deal with the general notion of topo-logical vector space here, but only the following particularly convenientway to introduce a topology in a linear space. This also covers mostcases of importance to analysis. A metric space is a set M providedwith a metric, which is a function d : M M R such that for anyx,y,z M the following holds.

(1) d(x, y) 0 and = 0 if and only if x = y. (positive definite)(2) d(x, y) = d(y, x). (symmetric)(3) d(x, y) d(x, z) + d(z, y). (triangle inequality)

A neighborhood of x M is then a subset O of M such that for some > 0 the set O contains all y M for which d(x, y) < . An openset is a set which is a neighborhood of all its points, and a closed setis one with an open complement. One says that a sequence x1, x2, . . .of elements in M converges to x M if d(xj , x) 0 as j .

The most convenient, but not the only important, way of introduc-ing a metric in a linear space L is via a norm (Exercise 2.1). A normon L is a function : L R such that for any u, v L and C

(1) u 0 and = 0 if and only if u = 0. (positive definite)(2) u = ||u. (positive homogeneous)(3) u + v u + v. (triangle inequality)

The usual norm in the real space R3

is of course obtained from the dotproduct (x1, x2, x3) (y1, y2, y3) = x1y1 + x2y2 + x3y3 by setting x =x x. For an infinite-dimensional linear space L, it is sometimes

possible to define a norm similarly by setting u = u, u, where, is a scalar product on L. A scalar product is a function LL Csuch that for all u, v and w in L and all , C holds

(1) u + v,w = u, w + v, w. (linearity in first argument)(2) u, v = v, u. (Hermitian symmetry)(3) u, u 0 with equality only if u = 0. (positive definite)

If instead of (3) holds only

(3) u, u 0, (positive semi-definite)9


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10 2. SPACES WITH SCALAR PRODUCT

one speaks about a semi-scalar product. Note that (2) implies that

u, u

is real so that (3) makes sense. Also note that by combining (1)

and (2) we have w,u + v = w, u + w, v. One says that thescalar product is anti-linear in its second argument (Warning: In theso called Dirac formalism in quantum mechanics the scalar product isinstead anti-linear in the first argument, linear in the second). Togetherwith (1) this makes the scalar product into a sesqui-linear(=1 1

2-linear)

form. In words: A scalar product is a Hermitian, sesqui-linear andpositive definite form. We now assume that we have a scalar producton L and define u = u, u for any u L. To show that thisdefinition makes into a norm we need the following basic theorem.

Theorem 2.1. (Cauchy-Schwarz) If

,

is a semi-scalar product

on L, then for all u, v L holds |u, v|2 u, uv, v.Proof. For arbitrary complex we have 0 u + v,u + v =

||2u, u + u, v + v, u + v, v. For = rv, u with real rwe obtain 0 r2|u, v|2u, u 2r|u, v|2 + v, v. If u, u = 0 butu, v = 0 this expression becomes negative for r > 1

2v, v|u, v|2

which is a contradiction. Hence u, u = 0 implies that u, v = 0 sothat the theorem is true in the case when u, u = 0. If u, u = 0we set r = u, u1 and obtain, after multiplication by u, u, that0 |u, v|2 + u, uv, v which proves the theorem.

In the case of a scalar product, defining u = u, u, we maywrite the Cauchy-Schwarz inequality as |u, v| uv. In thiscase it is also easy to see when there is equality in Cauchy-Schwarzinequality. To see that is a norm on L the only non-trivial pointis to verify that the triangle inequality holds; but this follows fromCauchy-Schwarz inequality (Exercise 2.4).

Recall that in a finite dimensional space with scalar product it isparticularly convenient to use an orthonormal basis since this makesit very easy to calculate the coordinates of any vector. In fact, ifx1, . . . , xn are the coordinates of u in the orthonormal basis e1, . . . , en,then xj =

u, ej

(recall that e1, . . . , en is called orthonormal if all basis

elements have norm 1 and ej , ek = 0 for j = k). Given an arbitrarybasis it is easy to construct an orthonormal basis by use of the Gram-Schmidt method (see the proof of Lemma 2.2).

In an infinite-dimensional space one can not find a (finite) basis.The best one can hope for are infinitely many vectors e1, e2, . . . suchthat each finite subset is linearly independent, and any vector is thelimit in norm of a sequence of finite linear combinations of e1, e2, . . . .Again, it will turn out to be very convenient ife1, e2, . . . is an orthonor-mal sequence, i.e., ej = 1 for j = 1, 2, . . . and ej, ek = 0 for j = k.The following lemma is easily proved by use of the Gram-Schmidt pro-cedure.


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2. SPACES WITH SCALAR PRODUCT 11

Lemma 2.2. Any infinite-dimensional linear space L with scalarproduct contains an orthonormal sequence.

Proof. According to Chapter 1 we can find a linearly independentsequence in L, i.e., a sequence u1, u2, . . . such that u1, . . . , uk are lin-early independent for any k. Put e1 = u1/u1 and v2 = u2u2, e1e1.Next put e2 = v2/v2. If we have already found e1, . . . , ek, putvk+1 = uk+1

kj=1uk+1, ejej and ek+1 = vk+1/vk+1. I claim

that this procedure will lead to a well defined orthonormal sequencee1, e2, . . . . This is left for the reader to verify (Exercise 2.6).

Supposing we have an orthonormal sequence e1, e2, . . . in L a nat-ural question is: How well can one approximate (in the norm of L) anarbitrary vector u L by finite linear combinations of e1, e2, . . . . Hereis the answer:

Lemma 2.3. Suppose e1, e2, . . . is an orthonormal sequence in Land put, for any u L, uj = u, ej. Then we have

(2.1) u k

j=1

jej2 = u2 k

j=1

|uj|2 +k

j=1

|j uj |2

for any complex numbers 1, . . . , k.

The proof is by calculation (Exercise 2.7). The interpretation ofLemma 2.3 is very interesting. The identity (2.1) says that if we want

to choose a linear combination kj=1 jej of e1, . . . , ek which approxi-mates u well in norm, the best choice of coefficients is to take j = uj,j = 1, . . . , k. Furthermore, with this choice, the error is given exactlyby u kj=1 ujej2 = u2 kj=1|uj|2. One calls the coefficientsu1, u2, . . . the (generalized) Fourier coefficients ofu with respect to theorthonormal sequence e1, e2, . . . . The following theorem is an immedi-ate consequence of Lemma 2.3 (Exercise 2.8).

Theorem 2.4 (Bessels inequality). For any u the series

j=1|uj |2converges and one has

j=1

|uj |2 u2 .

Another immediate consequence of Lemma 2.3 is the next theorem(cf. Exercise 2.9).

Theorem 2.5 (Parsevals formula). The series

j=1 ujej converges

(in norm) to u if and only if

j=1|uj|2 = u2.There is also a slightly more general form of Parsevals formula.

Corollary 2.6. Suppose

j=1|uj |2 = u2 for someu L. Thenj=1 uj vj = u, v for any v L.


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Proof. Consider the following form on L.

[u, v] = u, v

j=1

uj vj .

Since |uj vj | 12 (|uj|2 + |vj |2) by the arithmetic-geometric inequality,Bessels inequality shows that the series is absolutely convergent. Itfollows that [, ] is a Hermitian, sesqui-linear form on L. Becauseof Bessels inequality it is also positive (but not positive definite).Thus [, ] is a semi-scalar product on L. Applying Cauchy-Schwarzinequality we obtain |[u, v]|2 [u, u][v, v]. By assumption [u, u] =u2

j=1|uj |2 = 0 so that the corollary follows.

It is now obvious that the closest analogy to an orthonormal basisin an infinite-dimensional space with scalar product is an orthonormalsequence with the additional property of the following definition.

Definition 2.7. An orthonormal sequence in L is called completeif the Parseval identity u2 = 1 |uj |2 holds for every u L.

It is by no means clear that we can always find complete orthonor-mal sequences in a given space. This requires the space to be separable.

Definition 2.8. A metric space M is called separable if it has adense, countable subset. This means a sequence u1, u2, . . . of elementsof M, such that for any u

M, and any > 0, there is an element uj

of the sequence for which d(u, uj ) < .

The vast majority of spaces used in analysis are separable (Exer-cise 2.10), but there are exceptions (Exercise 2.12).

Theorem 2.9. A infinite-dimensional linear space with scalar prod-uct is separable if and only if it contains a complete orthonormal se-quence.

The proof is left as an exercise (Exercise 2.11). Suppose e1, e2, . . . isa complete orthonormal sequence in L. We then know that any u Lmay be written as u =

j=1 ujej, where the series converges in norm.

Furthermore the numerical series j=1|uj|2 converges to u2. Thefollowing question now arises: Given a sequence 1, 2, . . . of complexnumbers for which

j=1|j|2 converges, does there exist an element

u L for which 1, 2, . . . are the Fourier coefficients? Equivalently,does

j=1 jej converge to an element u L in norm? As it turns

out, this is not always the case. The property required ofL is that itis complete. Warning: This is a totally different property from thecompleteness of orthonormal sequences we discussed earlier! To explainwhat it is, we need a few definitions.

Definition 2.10. A Cauchy sequence in a metric space M is a

sequence u1, u2, . . . of elements ofM such that d(uj, uk) 0 as j, k


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EXERCISES FOR CHAPTER 2 13

. More exactly: To every > 0 there exists a number such thatd(uj, uk) < if j > and k > .

It is clear by use of the triangle inequality that any convergentsequence is a Cauchy sequence. Far more interesting is the fact thatthis implication may sometimes be reversed.

Definition 2.11. A metric space M is called complete if everyCauchy sequence converges to an element in M.

A normed linear space which is complete is called a Banach space.If the norm derives from a scalar product,

j=1|j|2 converges and

e1, e2, . . . is an orthonormal sequence we put uk = kj=1 jej. If k < n

we then have (the second equality is a special case of Lemma 2.3)

un uk2 = n

j=k+1

jej2 =n

j=k+1

|j|2 =n

j=1

|j|2 k

j=1

|j|2 .

Since

j=1|j|2 converges the right hand side 0 as k, n . Henceu1, u2, . . . is a Cauchy sequence in L. It therefore follows that if L iscomplete, then

j=1 jej actually converges in norm to an element

of L. On the other hand, if L is not complete and e1, e2, . . . is anorthonormal sequence, then 1, 2, . . . may be chosen so that the series

j=1 jej does not converge in L although j=1|j|2 is convergent(Exercise 2.14).Exercises for Chapter 2

Exercise 2.1. Show that if is a norm on L, then d(x, y) =u v is a metric on L.

Exercise 2.2. Show that d(x, y) = arctan|x y| is a metric onR which can be extended to a metric on the set of extended realsR = R {}{}.

Exercise 2.3. Consider the linear space C1

[0, 1], consisting ofcomplex-valued, differentiable functions with continuous derivative, de-fined in [0, 1]. Show that the following are all norms on C1[0, 1].

u = sup0x1|u(x)| , u1 =

10|u| ,

u1, = u + u.Invent some more norms in the same spirit!

Exercise 2.4. Find all cases of equality in Cauchy-Schwarz in-equality for a scalar product! Then show that , defined by u =u, u, where , is a scalar product, is a norm.


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Exercise 2.5. Show that u, v =

1

0u(x)v(x) dx is a scalar prod-

uct on the space C[0, 1] of continuous, complex-valued functions definedon [0, 1].

Exercise 2.6. Finish the proof of Lemma 2.2.

Exercise 2.7. Prove Lemma 2.3.

Exercise 2.8. Prove Bessels inequality!

Exercise 2.9. Prove Parsevals formula!

Exercise 2.10. It is well known that the set of step functions whichare identically 0 outside a compact subinterval of an interval I are densein L2(I). Use this to show that L2(I) is separable.

Exercise 2.11. Prove Theorem 2.9.

Hint: Use Gram-Schmidt!

Exercise 2.12. Let L be the set of complex-valued functions u ofthe form u(x) =

kj=1 je

ijx where 1, . . . , k are (a finite number of)different real numbers and 1, . . . , k are complex numbers. Show thatL is a linear subspace of C(R) (the functions continuous on the realline) on which u, v = limT 12T

TT uv serves as a scalar product.

Then show that the norm of eix is 1 for any R and that eix isorthogonal to eix as soon as

= . Conclude that

Lis not separable.

Exercise 2.13. Show that as metric spaces the set Q of rationalnumbers is not complete but the set R of reals is.

Exercise 2.14. Suppose L is a space with scalar product which isnot complete, and that e1, e2, . . . is a complete orthonormal sequencein L. Show that there exists a sequence 1, 2, . . . of complex numbers,such that

|j |2 < but jej does not converge to any elementofL.


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CHAPTER 3

Hilbert space

A Hilbert space is a linear space H (we will as always assume thatthe scalars are complex numbers) provided with a scalar product suchthat the space is also complete, i.e., any Cauchy sequence (with respectto the norm induced by the scalar product) converges to an element

of H. We denote the scalar product of u and v H by u, v andthe norm of u by u = u, u. It is usually required, and we willfollow this convention, that the space be separable as well, i.e., thereis a countable, dense subset. Recall that this means that any elementcan be arbitrarily well approximated in norm by elements of this densesubset. In the present case this means that H has a complete orthonor-mal sequence, and conversely, if the space has a complete orthonormalsequence it is separable (Theorem 2.9). As is usual we will also assumethat H is infinite-dimensional.

Example 3.1. The space 2 consists of all infinite sequences u =

(u1, u2, . . . ) of complex numbers for which |uj|2 < , i.e., whichare square summable. The scalar product of u with v = (v1, v2, . . . ) isdefined as u, v = ujvj . This series is absolutely convergent since|ujvj| (|uj|2 + |vj|2)/2 and u, v are square summable. Show that 2is a Hilbert space (Exercise 3.1)!

The space Hilbert himself dealt with was 2. Actually, any Hilbertspace is isometrically isomorphic to 2, i.e., there is a bijective (one-to-one and onto) linear map H u u 2 such that u, v = u, vfor any u and v in H (Exercise 3.2). This is the reason any complete,separable and infinite-dimensional space with scalar product is called

a Hilbert space. However, there are infinitely many isomorphisms thatwill serve, and none of them is natural, i.e., in general to be preferredto any other, so the fact that all Hilbert spaces are isomorphic is notparticularly useful in practice.

Example 3.2. The most important example of a Hilbert space isL2(, ) where is some domain in Rn and is a (Radon) measuredefined there; often is simply Lebesgue measure. The space consistsof (equivalence classes of) complex-valued functions on , measurablewith respect to and with integrable square over with respect to .That this space is separable and complete is proved in courses on the

theory of integration.15


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16 3. HILBERT SPACE

Given a normed space one may of course ask whether there is ascalar product on the space which gives rise to the given norm in the

usual way. Here is a simple criterion.

Lemma 3.3. (parallelogram identity) If u and v are elements ofH,then

u + v2 + u v2 = 2u2 + 2v2.Proof. A simple calculation gives uv2 = uv, uv = u2

(u, v + v, u) +v2. Adding this for the two signs the parallelogramidentity follows.

The name parallelogram identity comes from the fact that thelemma can be interpreted geometrically, as saying that the sum of the

squares of the lengths of the sides in a parallelogram equals the sumof the squares of the lengths of the diagonals. This is a theorem thatcan be found in Euclids Elements. Given a normed space, Lemma 3.3shows that a necessary condition for the norm to be associated witha scalar product is that the parallelogram identity holds for all vec-tors in the space. It was proved by von Neumann in 1929 that this isalso sufficient (Exercise 3.3). We shall soon have another use for theparallelogram identity.

In practice it is quite common that one has a space with scalar prod-uct which is not complete (such a space is often called a pre-Hilbertspace). In order to use Hilbert space theory, one must then embedthe space in a larger space which is complete. The process is calledcompletion and is fully analogous to the extension of the rational num-bers to the reals, which is also done to make the Cauchy convergenceprinciple valid. In very brief outline the process is as follows. Startingwith a (not complete) normed linear space L let Lc be the set of allCauchy sequences in L. The set Lc is made into a linear space in theobvious way. We may embed L in Lc by identifying u L with thesequence (u , u , u , . . . ). In Lc we may introduce a semi-norm (i.e.,a norm except that there may be non-zero elements u in the space forwhich u = 0) by setting (u1, u2, . . . ) = limuj. Now letNc be thesubspace of Lc consisting of all elements with semi-norm 0, and putH = Lc/Nc, i.e., elements in Lc are identified whenever the distancebetween them is 0. One may now prove that induces a norm on Hunder which H is complete, and that through the identification abovewe may consider the original space L as a dense subset of H. If theoriginal norm came from a scalar product, then so will the norm of H.We leave to the reader to verify the details, using the hints provided(Exercise 3.4).

The process above is satisfactory in that it shows that any normedspace may be completed (in fact, the same process works in any metricspace). Equivalence classes of Cauchy sequences are of course rather ab-

stract objects, but in concrete cases one can often identify the elements


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3. HILBERT SPACE 17

of the completion of a given space with more concrete objects. So, forexample, one may view L2(, ) as the completion, in the appropriate

norm, of the linear space C0() of functions which are continuous in and 0 outside a compact subset of .

In the sequel H is always assumed to be a Hilbert space. There aretwo properties which make Hilbert spaces far more convenient to dealwith than more general spaces. The first is that any closed, linear sub-space has a topological complement which can be chosen in a canonicalway (Theorem 3.7). The second, a Hilbert space can be identified withits topological dual (Theorem 3.8). Both these properties are actuallytrue even if the space is not assumed separable (and of course if thespace is finite-dimensional), as our proofs will show. To prove them we

start with the following definition.Definition 3.4. A set M is called convex if it contains all line-

segments connecting two elements of the set, i.e., ifu and v M, thentu + (1 t)v M for all t [0, 1].

A subset of a metric space is of course called closed if all limitsof convergent sequences contained in the subset are themselves in thesubset. It is easily seen that this is equivalent to the complement ofthe subset being open, in the sense that it is a neighborhood of all itspoints (check this!).

Lemma3.5

.Any closed, convex subsetK ofH has a unique elementof smallest norm.

Proof. Put d = inf{u | u K}. Let u1, u2, . . . be a minimizingsequence, i.e., uj K and uj d. By the parallelogram identitywe then have

uj uk2 = 2uj2 + 2uk2 4(uj + uk)/22 .On the right hand side the two first terms both tend to 2d2 as j, k .By convexity (uj + uk)/2 K so the last term is 4d2. Thereforeu1, u2, . . . is a Cauchy sequence, and has a limit u which obviously has

norm d and is in K, since K is closed. If u and v are both minimizingelements, replacing uj by u and uk by v in the calculation above im-mediately shows that u = v, so the minimizing element is unique.

Lemma 3.6. Suppose M is a proper (i.e., M = H) closed, linearsubspace of H. Then there is a non-trivial normal to M, i.e., anelement u = 0 in H such that u, v = 0 for all v M.

Proof. Let w / M and put K = w + M. Then K is obviouslyclosed and convex so it has a smallest element u which is non-zerosince 0 / K. Let v = 0 be in M so that u + av K for any scalara. Hence

u

2

u + av

2 =

u

2 + 2 Re(a

v, u

) +

|a

|2

v

2. Setting

a = u, v/v2 we obtain (|u, v|/v)2 0 so that u, v = 0.


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18 3. HILBERT SPACE

Two subspaces M and N are said to be orthogonal if every elementin M is orthogonal to every element in N. Then clearly M

N =

{0

}so the direct sum of M and N is defined. In the case at hand this iscalled the orthogonal sum of M and N and denoted by M N. ThusM N is the set of all sums u + v with u M and v N. If M andN are closed, orthogonal subspaces of H, then their orthogonal sum isalso a closed subspace of H (Exercise 3.5). If A is an arbitrary subsetofH we define

A = {u H | u, v = 0 for all v A}.This is called the orthogonal complement of A. It is easy to see thatA is a closed linear subspace ofH, that A B implies B A andthat A

(A) (Exercise 3.6).

When M is a linear subspace of H an alternative way of writingM is H M. This makes sense because of the following theorem ofcentral importance.

Theorem 3.7. Suppose M is a closed linear subspace of H. ThenM M = H.

Proof. M M is a closed linear subspace of H so if it is notall ofH, then it has a non-trivial normal u by Lemma 3.6. But if u isorthogonal to both M and M, then u M (M) which showsthat u cannot be = 0. The theorem follows.

A nearly obvious consequence of Theorem 3.7 is that M = Mfor any closed linear subspace M ofH (Exercise 3.7).

A linear form on H is complex-valued linear function on H. Nat-urally is said to be continuous if(uj ) (u) whenever uj u. Theset of continuous linear forms on a Banach space B (or a more generaltopological vector space) is made into a linear space in an obvious way.This space is called the dual ofB, and is denoted by B. A continuouslinear form on a Banach space B has to be bounded in the sense thatthere is a constant C such that |(u)| Cu for any u B. Forsuppose not. Then there exists a sequence of elements u1, u2, . . . of

B for which |(uj)|/uj . Setting vj = uj/(uj) we then havevj 0 but |(vj)| = 1 0, so can not be continuous. Conversely, if is bounded by C then |(uj) (u)| = |(uj u)| Cuj u 0 ifuj u, so a bounded linear form is continuous. The smallest possiblebound of a linear form is called the norm of , denoted .

It is easy to see that provided with this norm B is complete, so thedual of a Banach space is a Banach space (Exercise 3.8). A familiarexample is given by the space Lp(, ) for 1 p < , where isa domain in Rn and a Radon measure defined in . The dual ofthis space is Lq(, ), where q is the conjugate exponent to p, in thesense that 1

p+ 1

q= 1. A simple example of a bounded linear form on

a Hilbert space H is (u) = u, v, where v is some fixed element of


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3. HILBERT SPACE 19

H. By Cauchy-Schwarz inequality |(u)| vu so v. But(v) =

v

2 so actually

=

v

. The following theorem, which has

far-reaching consequences for many applications to analysis, says thatthis is the only kind of bounded linear form there is on a Hilbert space.In other words, the theorem allows us to identify the dual of a Hilbertspace with the space itself.

Theorem 3.8 (Riesz representation theorem). For any boundedlinear form on H there is a unique element v H such that (u) =u, v for all u H. The norm of is then = v.

Proof. The uniqueness of v is clear, since the difference of twopossible choices of v must be orthogonal to all of H (for example toitself). If (u) = 0 for all u then we may take v = 0. Otherwise we setM = {u H | (u) = 0} which is obviously linear because is, andclosed since is continuous. Since M is not all ofH it has a normal w =0 by Lemma 3.6, and we may assume w = 1. If now u is arbitrary inH we put u1 = u ((u)/(w))w so that (u1) = (u) (u) = 0, i.e.,u1 M so u1, w = 0. Hence u, w = ((u)/(w))w, w = (u)/(w)so (u) = u, v where v = (w)w. We have already proved that =v.

So far we have tacitly assumed that convergence in a Hilbert spacemeans convergence in norm, i.e., uj u means uj u 0. Thisis called strong convergence; one writes s-lim uj = u or uj

u. There

is also another notion of convergence which is very important. By def-inition uj tends to u weakly, in symbols w-lim uj = u or uj u, ifuj, v u, v for every v H. It is obvious that strong convergenceimplies weak convergence to the same limit (the scalar product is con-tinuous in its arguments by Cauchy-Schwarz), but the converse is nottrue (Exercise 3.9). We have the following important theorem.

Theorem 3.9. Every bounded sequence in H has a weakly con-vergent subsequence. Conversely, every weakly convergent sequence isbounded.

Proof.

The first claim is a consequence of the weak compact-ness of the unit ball of the dual of a Banach space. Since we do notwant to assume knowledge of this, we will give a direct proof. To thisend, suppose v1, v2, . . . is the given sequence, bounded by C, and lete1, e2, . . . be a complete orthonormal sequence in H. The numericalsequence {vj, e1}j=1 is then bounded and so has a convergent sub-sequence, corresponding to a subsequence {v1j}j=1 of the v:s, by theBolzano-Weierstrass theorem. The numerical sequence {v1j, e2}j=1is again bounded, so it has a convergent subsequence, correspondingto a subsequence {v2j}j=1 of {v1j}j=1. Proceeding in this manner weget a sequence of sequences

{vkj

}j=1, k = 1, 2, . . . , each element of

which is a subsequence of those preceding it, and with the property that


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20 3. HILBERT SPACE

vn = limjvnj , en exists. I claim that {vjj}j=1 converges weakly tov = vnen. Note that {vjj , en}j=1 converges to vn since it is a subse-quence of {vnj , en}j=1 from j = n on. Furthermore Nn=1|vn|2 C2for all N since it is the limit as j of Nn=1|vNj , en|2 whichby Bessels inequality is bounded by vN j2 C2. It follows that

n=1|vn|2 C2 so that v is actually an element ofH.To show the weak convergence, let u =

unen be arbitrary in

H. Suppose > 0 given arbitrarily. Writing u = u + u whereu =

Nn=1 unen we may now choose N so large that u < so that|vjj , u| < C. Furthermore |v, u| < C and vjj , u v, u

so limj|vjj , u v, u| 2C. Since > 0 is arbitrary the weakconvergence follows.

The converse is an immediate consequence of the Banach-Steinhausprinciple of uniform boundedness.

Theorem 3.10 (Banach-Steinhaus). Let1, 2, . . . be a sequence ofbounded linear forms on a Banach spaceB which is pointwise bounded,i.e., such that for each u B the sequence 1(u), 2(u), . . . is bounded.Then 1, 2, . . . is uniformly bounded, i.e., there is a constant C suchthat |j(u)| Cu for every u B and j = 1, 2, . . . .

Assuming Theorem 3.10 (for a proof, see Appendix A), we cancomplete the proof of Theorem 3.9, since a weakly convergent sequencev

1, v

2, . . . can be identified with a sequence of linear forms

1,

2, . . .

by setting j(u) = u, vj. Since a convergent sequence of numbersis bounded it follows that we have a pointwise bounded sequence oflinear functionals. By Theorem 3.10 there is a constant C such that|u, vj| Cu for every u H and j = 1, 2, . . . . In particular,setting u = vj gives vj C for every j.


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Exercise3.1

.Prove the completeness of

2

!Hint: Given a Cauchy sequence show first that each coordinate con-verges.

Exercise 3.2. Prove that any Hilbert space is isometrically iso-morphic to 2, i.e., there is a bijective (one-to-one and onto) linearmap H u u 2 such that u, v = u, v for any u and v in H.

Exercise 3.3. Suppose L is a linear space with norm whichsatisfies the parallelogram identity for all u, v L. Show that u, v =14

3k=0 i

ku + ikv2 is a scalar product on L.Hint: Show first that u, u = u2, that v, u = u, v and thatiu,v = iu, v. Then show that u + v, w u, w v, w = 0 andfrom that u,v = u, v for any rational number . Finally usecontinuity.

Exercise 3.4. Show that the semi-norm on the space Lc definedin the text is well-defined, i.e., that the limit limuj exists for anyelement (u1, u2, . . . ) Lc. Then verify that H = Lc/Nc can be given anorm under which it is complete, that L may be viewed as isometricallyand densely embedded in H, and that H is a Euclidean space (a spacewith scalar product) if

Lis.

Exercise 3.5. Show that if M and N are closed, orthogonal sub-spaces ofH, then also M N is closed.

Exercise 3.6. Show that is A H, then A is a closed linearsubspace ofH, that A B implies B A and that A (A).

Exercise 3.7. Verify that M = M for any closed linear subspaceM of H, and also that for an arbitrary set A H the smallest closedlinear subspace containing A is A.

Exercise 3.8. Show that a bounded linear form on a Banach space

Bhas a least bound, which is a norm on

B, and that

B is complete

under this norm.

Exercise 3.9. Show that an orthonormal sequence does not con-verge strongly to anything but tends weakly to 0. Conclude that if in aEuclidean space every weakly convergent sequence is convergent, thenthe space is finite-dimensional.

Hint: Show that the distance between two arbitrary elements in thesequence is

2 and use Bessels inequality to show weak convergence

to 0.


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CHAPTER 4

Operators

A bounded linear operator from a Banach space B1 to another Ba-nach space B2 is a linear mapping T : B1 B2 such that for someconstant C we have T u2 Cu1 for every u B1. The smallestsuch constant C is called the norm of the operator T and denoted by

T. Like in the discussion of linear forms in the last chapter it followsthat the boundedness ofT is equivalent to continuity, in the sense thatT uj T u2 0 ifuj u1 0 (Exercise 4.1). IfB1 = B2 = B onesays that T is an operator on B. The operator-norm defined above hasthe following properties (Here T : B1 B2 and S are bounded linearoperators, and B1, B2 and B3 Banach spaces).

(1) T 0, equality only if T = 0,(2) T = ||T for any C,(3) S+ T S + T if S : B1 B2,(4) ST ST if S : B2 B3.

We leave the proof to the reader (Exercise 4.1). Thus we have made theset of bounded operators from B1 to B2 into a normed space B(B1, B2).In fact, B(B1, B2) is a Banach space (Exercise 4.2). We write B(B)for the bounded operators on B. Because of the property (4) B(B) iscalled a Banach algebra.

Now let H1 and H2 be Hilbert spaces. Then every bounded operatorT : H1 H2 has an adjoint1 T : H2 H1 defined as follows. Con-sider a fixed element v H2 and the linear form H1 u Tu , v2which is obviously bounded by Tv2. By the Riesz representa-tion theorem there is therefore a unique element v H1, such thatTu , v2 = u, v1. By the uniqueness, and since Tu , v2 dependsanti-linearly on v, it follows that T : v v is a linear operator fromH2 to H1. It is also bounded, since v21 = T v, v2 Tv1v2,so that T T. The adjoint has the following properties.

Proposition 4.1. The adjoint operation B(H1, H2) T T B(H2, H1) has the properties:

(1) (T1 + T2) = T1 + T

2 ,

(2) (T) = T for any complex number ,(3) (T2T1)

= T1 T2 if T2 : H2 H3,

(4) T = T,

1Also operators between general Banach spaces, or even more general topolog-

ical vector spaces, have adjoints, but they will not concern us here.

23


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24 4. OPERATORS

(5) T = T,(6)

TT

=

T

2.

Proof. The first four properties are very easy to show and areleft as exercises for the reader. To prove (5), note that we alreadyhave shown that T T and combining this with (4) gives theopposite inequality. Use of (5) shows that TT TT =T2 and the opposite inequality follows from T u22 = TTu , u1 TT u1u1 TTu21 so (6) follows. The reader is asked to fillin the details missing in the proof (Exercise 4.3).

If H1 = H2 = H3 = H, then the properties (1)(4) above are theproperties required for the star operation to be called an involution

on the algebra B(H), and a Banach algebra with an involution, alsosatisfying (5) and (6), is called a B algebra. There are no less thanthree different useful notions of convergence for operators in B(H1, H2).We say that Tj tends to T

uniformly ifTj T 0, denoted by Tj T, strongly ifTjuT u2 0 for every u H1, denoted Tj T,

and weakly if Tju, v2 Tu , v2 for all u H1 and v H2,

denoted Tj T.

It is clear that uniform convergence implies strong convergence andstrong convergence implies weak convergence, and it is also easy to seethat neither of these implications can be reversed.

Of particular interest are so called projection operators. A pro-jection P on H is an operator in B(H) for which P2 = P. If P isa projection then so is I P, where I is the identity on H, since(I P)(I P) = I P P + P2 = I P. Setting M = PH andN = (I P)H it follows that M is the null-space of I P since Mclearly consist of those elements u H for which P u = u. SimilarlyN is the null-space of P. Since P and I P are bounded (i.e., con-tinuous) it therefore follows that M and N are closed. It also followsthat M N = {0} and the direct sum M+N of M and N is H (thismeans that any element of H can be written uniquely as u + v withu M and v N). Conversely, ifM and N are linear subspaces ofH,M N = {0} and M+N = H, then we may define a linear map P sat-isfying P2 = P by setting P w = u ifw = u + v with u M and v N.As we have seen P can not be bounded unless M and N are closed.There is a converse to this: IfM and N are closed, then P is bounded.This follows immediately from the closed graph theorem (Exercise 4.4).In the case when the projection P, and thus also I P, is bounded,the direct sum MN is called topological. If M and N happen to beorthogonal subspaces P is called an orthogonal projection. ObviouslyN = M then, since the direct sum of M and N is all of

H. We have

the following characterization of orthogonal projections.


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4. OPERATORS 25

Proposition 4.2. A projection P is orthogonal if and only if itsatisfies P = P.

Proof. If P = P and u M, v N, then u, v = Pu , v =u, Pv = u , Pv = u, 0 = 0 so M and N are orthogonal. Con-versely, suppose M and N orthogonal. For arbitrary u, v H wethen have Pu , v = P u , P v + P u, (I P)v = P u , P v so thatalso u , Pv = P u , P v. Hence Pu , v = u , Pv holds generally, i.e.,P = P.

An operator T for which T = T is called selfadjoint. Hence anorthogonal projection is the same as a selfadjoint projection. We willhave much more to say about selfadjoint operators in a more general

context later. Another class of operators of great interest are the uni-tary operators. This is an operator U : H1 H2 for which U = U1.Since Uu,Uv2 = UUu,v1 = u, v1 the operator U preserves thescalar product; such an operator is called isometric. If U is isomet-ric we have u, v1 = Uu , Uv2 = UUu,v1, so that U is a leftinverse of U for any isometric operator. If dim H1 = dim H2 < ,then a left inverse of a linear operator is also a right inverse, so inthis case isometric and unitary (orthogonal in the case of a real space)are the same thing. If dimH1 = dim H2 or both spaces are infinite-dimensional, however, this is not the case. For example, in the space2 we may define U(x1, x2, . . . ) = (0, x1, x2, . . . ), which is obviously

isometric (this is a so called shift operator), but the vector (1, 0, 0, . . . )is not the image of anything, so the operator is not unitary. Its ad-joint is U(x1, x2, . . . ) = (x2, x3, . . . ), which is only a partial isometry,namely an isometry on the vectors orthogonal to (1, 0, 0, . . . ). See alsoExercise 4.8.

It is never possible to interpret a differential operator as a boundedoperator on some Hilbert space of functions. We therefore need todiscuss unbounded operators as well. Similarly, we will need to discussoperators that are not defined on all of H. Thus we now consider alinear operator T : D(T) H2, where the domain D(T) of T is somelinear subset of

H1. T is not supposed bounded. Another such operator

S is said to be an extension of T if D(T) D(S) and Su = T u forevery u D(T). We then write T S. We must discuss the conceptof adjoint. The form u Tu , v2 is, for fixed v H2, only defined foru D(T), and though linear not necessarily bounded, so there may notbe any v H1 such that Tu , v2 = u, v1 for all u D(T). Evenif there is, it may not be uniquely determined, since if w D(T) wecould replace v by v + w with no change in u, v. We thereforemake the basic assumption that D(T) = {0}, i.e., D(T) is dense inH1. T is then said to be densely defined2. In this case v H1 is

2We will discuss the case of an operator which is not densely defined in Chap-

ter 9.


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26 4. OPERATORS

clearly uniquely determined by v H2, if it exists. It is also obviousthat v depends linearly on v, so we define

D(T) to be those v

H2

for which we can find a v H1, and set Tv = v. There is no reasonto expect the adjoint T to be densely defined. In fact, we may haveD(T) = {0}, so T may not itself have an adjoint. To understand thisrather confusing situation it turns out to be useful to consider graphsof operators.

The graph of T is the set GT = {(u , Tu) | u D(T)}. This set isclearly linear and may be considered a linear subset of the orthogonaldirect sum H1 H2, consisting of all pairs (u1, u2) with u1 H1 andu2 H2 with the natural linear operations and provided with the scalarproduct (u1, u2), (v1, v2) = u1, v11 + u2, v22. This makes H1 H2into a Hilbert space (Exercise 4.6).We now define the boundary operator U : H1 H2 H2 H1 by

U(u1, u2) = (iu2, iu1) (the terminology is explained in Chapter 9). Itis clear that U is isometric and surjective (onto H2 H1). It followsthat Uis unitary. IfH1 = H2 = H it is clear that Uis selfadjoint andinvolutary (i.e., U2 is the identity). Now put(4.1) (GT)

:=U((H1 H2) GT) = (H2 H1) UGT.The second equality is left to the reader to verify who should alsoverify that (GT)

is a graph of an operator (i.e., the second componentof each element in (GT)

is uniquely determined by the first) if and

only if T is densely defined. If T is densely defined we now define Tto be the operator whose graph is (GT)

. This means that T is theoperator whose graph consists of all pairs (v, v) H2 H1 such thatTu , v2 = u, v1 for all u D(T), i.e., our original definition. Animmediate consequence of (4.1) is that T S implies S T.

We say that an operator is closed if its graph is closed as a sub-space of H1 H2. This is an important property; in many ways theproperty of being closed is almost as good as being bounded. An ev-erywhere defined operator is actually closed if and only if it is bounded(Exercise 4.7). It is clear that all adjoints, having graphs that are or-thogonal complements, are closed. Not all operators are closeable, i.e.,have closed extensions; for this is required that the closure GT of GTis a graph. But it is clear from (4.1) that the closure of the graph is(GT)

. So, we have proved the following proposition.

Proposition 4.3. Suppose T is a densely defined operator in aHilbert space H. ThenT is closeable if and only if the adjoint T isdensely defined. The smallest closed extension (the closure) T of T isthen T.

The proof is left to Exercise 4.9. Note that if T is closed, its do-main

D(T) becomes a Hilbert space if provided by the scalar product

u, vT = u, v1 + T u , T v2.


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4. OPERATORS 27

In the rest of this chapter we assume that H1 = H2 = H. A denselydefined operator T is then said to be symmetric if T

T. In other

words, if Tu , v = u , Tv for all u, v D(T). Thus Tu , u is alwaysreal for a symmetric operator. It therefore makes sense to say thata symmetric operator is positive if Tu , u 0 for all u D(T). Adensely defined symmetric operator is always closeable since T is auto-matically densely defined, being an extension of T. If actually T = T

the operator is said to be selfadjoint. This is an important propertybecause these are the operators for which we will prove the spectraltheorem. In practice it is usually quite easy to see if an operator issymmetric, but much more difficult to decide whether a symmetric op-erator is selfadjoint. When one wants to interpret a differential operator

as a Hilbert space operator one has to choose a domain of definition;in many cases it is clear how one may choose a dense domain so thatthe operator becomes symmetric. With luck this operator may havea selfadjoint closure3, in which case the operator is said to be essen-tially selfadjoint. Otherwise, given a symmetric T, one will look forselfadjoint extensions of T. If S is a symmetric extension of T, we getT S S T so that any selfadjoint extension of T is a restric-tion of the adjoint T. There is now obviously a need for a theory ofsymmetric extensions of a symmetric operator. We will postpone thediscussion of this until Chapter 9. Right now we will instead studysome very simple, but typical, examples.

Example 4.4. Consider the differential operator ddx

on some openinterval I. We want to interpret it as a densely defined operator inthe Hilbert space L2(I) and so must choose a suitable domain. A con-venient choice, which would work for any differential operator withsmooth coefficients, is the set C0 (I) of infinitely differentiable func-tions on I with compact support, i.e., each function is 0 outside somecompact subset of I. It is well known that C0 (I) is dense in L

2(I).Let us denote the corresponding operator T0; it is usually called theminimal operator for d

dx. Sometimes it is the closure of this operator

which is called the minimal operator, but this will make no difference

to the calculations in the sequel. We now need to calculate the adjointof the minimal operator.

Let v D(T0 ). This means that there is an element v L2(I)such that

I

v =

Iv for all C0 (I) and that T0 v = v. In-

tegrating by parts we have

Iv =

I(

v) since the boundaryterms vanish. Here

v denotes any integral function of v. Thus we

have

I(v +

v) = 0 for all C0 (I). We need the following

lemma.

3This is the same as T being selfadjoint. Show this!


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28 4. OPERATORS

Lemma 4.5 (du Bois Reymond). Suppose u is locally square inte-grable on R, i.e., u

L2(I) for every bounded real interval I. Also

suppose that u = 0 for every C0 (R). Then u is (almost ev-erywhere) equal to a constant.

Assuming the truth of the lemma for the moment it follows that,choosing the appropriate representative in the equivalence class of v,v +

v is constant. Hence v is locally absolutely continuous with

derivative v. It follows that D(T0 ) consists of functions in L2(I)which are locally absolutely continuous in I with derivative in L2(I),and that T0 v = v. Conversely, all such functions are in D(T0 ),as follows immediately by partial integration in

I

v =

I

v. Theoperator T0 is therefore also a differential operator, generated by

d

dx.

The differential operator ddx is called the formal adjoint of ddx andthe operator T0 is called the maximal operator belonging to ddx . Inthe same way any linear differential operator (with sufficiently smoothcoefficients) has a formal adjoint, obtained by integration by parts.For ordinary differential operators with smooth coefficients one canalways calculate adjoints in essentially the way we just did; for partialdifferential operators matters are more subtle and one needs to use thelanguage of distribution theory.

Proof of Lemma 4.5. Let C0 (R) and assume that

= 1.

Given

C0 (R) we put 0(x) = (x) and (x) = x(0).It is clear that is infinitely differentiable. It also has compact support

(why?), sou

= 0 by assumption. Butu

=u

u so that

(u K) = 0 where K =

u does not

depend on . Since C0 (R) is dense in L2(R) this proves that u = K

a.e., so that u is constant.

For the minimal operator of a differential operator to be symmetricit is clear that the differential operator has to be formally symmet-ric, i.e., the formal adjoint has to coincide with the original operator.In Example 4.4 D(T0) D(T0 ) but there is a minus sign preventingT0 from being symmetric. However, it is clear that had we startedwith the differential operator i d

dxinstead, then the minimal opera-

tor would have been symmetric, but the domains of the minimal andmaximal operators unchanged. One may then ask for possible selfad-joint extensions of the minimal operator, or equivalently for selfadjointrestrictions of the maximal operator.

Example 4.6. Let T1 be the maximal operator of i ddx on theinterval I. Let u, v D(T1) and a, b I. Then

ba

T1uv b

auT1v =

i

b

a(uv + uv) = iu(a)v(a) iu(b)v(b). Since u, v, T1u and T1v are all

in L2(I) the limit ofuv exists in both endpoints ofI. Consider the case

I = R. Since |u(x)|2 has limits as x and is integrable, the limits


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4. OPERATORS 29

must both be 0. Hence T1u, v u, T1v = 0 for any u, v D(T1), sothe maximal operator is symmetric and therefore selfadjoint (how does

this follow?). It also follows that the maximal operator is the closure ofthe minimal operator so the minimal operator is essentially selfadjoint.

Example 4.7. Consider the same operator as in Example 4.6 butfor the interval (0, ). If u D(T1) we obtain T1u, u u, T1u =i|u(0)|2. To have a symmetric restriction ofT1 we must therefore requireu(0) = 0, and with this restriction on the domain of T1 we obtain amaximal symmetric operator T. If now u D(T) and v D(T1) weobtain Tu , vu, T1v = iu(0)v(0) = 0 so that T = T1. T is thereforenot selfadjoint so no matter how we choose the domain the differentialoperator

i d

dx, though formally symmetric, will not be selfadjoint in

L2(0, ). One says that i ddx has no selfadjoint realization in L2(0, ).Example 4.8. We finally consider the operator of Example 4.6 for

the interval (, ). We now have(4.2) T1u, v u, T1v = i(u()v() u()v()).In particular, for u = v it follows that for u to be in the domain of asymmetric restriction ofT1 we must require |u()| = |u()| so that usatisfies the boundary condition u() = eiu() for some real . From(4.2) then follows that if v is in the domain of the adjoint, then v willhave to satisfy the same boundary condition. On the other hand, if we

impose this condition, then the resulting operator will be selfadjoint(because its adjoint will be symmetric). It follows that restricting thedomain ofT1 by such a boundary condition is exactly what is requiredto obtain a selfadjoint restriction. Each in [0, 2) gives a differentselfadjoint realization, but there are no others.

The examples show that there may be a unique selfadjoint real-ization of our formally symmetric differential operator, none at all, orinfinitely many depending on circumstances. It can be a very difficultproblem to decide which of these possibilities occur in a given case.In particular, much effort has been devoted to decide whether a given

differential operator on a given domain has a unique selfadjoint real-ization.


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30 4. OPERATORS


Exercise4.1

.Prove that boundedness is equivalent to continuityfor a linear operator between normed spaces. Then prove the properties

of the operator norm listed at the beginning of the chapter.

Exercise 4.2. Suppose B1 and B2 are Banach spaces. Show thatso is B(B1, B2).

Exercise 4.3. Fill in the details of the proof of Proposition 4.1.

Exercise 4.4. Show that if M and N are closed subspaces of Hwith MN = {0} and M+N = H, then the corresponding projectionsonto M and N are bounded operators.

Hint: The closed graph theorem!

Exercise 4.5. Show that a non-trivial (i.e., the range is not {0})projection is orthogonal if and only if its operator norm is 1.

Exercise 4.6. Suppose H1 and H2 are Hilbert spaces. Show thatthe orthogonal direct sum H1 H2 is also a Hilbert space.

Exercise 4.7. Show that a bounded, everywhere defined operatoris automatically closed. Conversely, that an everywhere defined, closedoperator is bounded.

Hint: The closed graph theorem!

Exercise 4.8. Show that if U is unitary, then all eigen-values ofU have absolute value || = 1. Also show that ife1 and e2 are eigen-vectors corresponding to eigen-values 1 and 2 respectively, then e1and e2 are orthogonal if 1 = 2.

Exercise 4.9. Show that ifT is densely defined and closeable, thenthe closure of T is T.


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CHAPTER 5

Resolvents

We now consider a closed, densely defined operator T in the Hilbertspace H. We define the solvability and deficiency spaces of T at by

S = {u H | (T )v = u for some v D(T)}

D = {u D(T) | Tu = u}.The following basic lemma is valid.

Lemma 5.1. Supposed T is closed and densely defined. Then

(1) D = H S.(2) If T is symmetric and Im = 0, then S is closed and H =

S D(3) If T is selfadjoint and Im = 0, then(T)v = u is uniquely

solvable for any u H (i.e., S = H), T has no non-realeigen-values (i.e., D = {0}), andv 1|Im |u.

Proof. Any element of the graph of T is of the form (v,v + u),where u S. To see this, simply put u = T v v for any v D(T).Now T v , w v,w = u + v,w v,w = u, w, so it followsthat (w,w) GT , i.e., w D, if and only if w is orthogonal to S.This proves (1).

IfT is symmetric and (v,v+u) GT, then v+u, v = v,v+u,i.e., Im v2 = Imv, u, which is vu by Cauchy-Schwarz in-equality. If Im = 0 we obtain v 1|Im |u, so that v is uniquelydetermined by u; in particular T has no non-real eigen-values. Further-more, suppose that u1, u2, . . . is a sequence in S converging to u, and

that (vj, vj + uj) GT. Then v1, v2, . . . is also a Cauchy sequence,since vj vk 1|Im |uj uk. Thus vj tends to some limit v, andsince T is closed we have (v,v + u) GT. Hence u S, so that Sis closed and (2) follows.

Finally, if T is self-adjoint, then T = T is symmetric so it has nonon-real eigen-values. If Im = 0 it follows that D = {0} so that (3)follows and the proof is complete.

In the rest of this chapter we assume that T is a selfadjoint operator.We define the resolvent set of T as

(T) = { C | T has a bounded, everywhere defined inverse} ,31


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32 5. RESOLVENTS

and the spectrum (T) ofT as the complement of(T). By Lemma 5.1.3the spectrum is a subset of the real line. For every

(T) we now

define the resolvent of T at as the operator R = (T )1. Theresolvent has the following properties.

Theorem 5.2. The resolvent of a selfadjoint operator T has theproperties:

(1) R 1/|Im | if Im = 0.(2) (R)

= R for (T).(3) R R = ( )RR for and (T).

The last statement is called the (first) resolvent relation.

Proof.The first claim is simply a re-statement of Lemma 5.1.3.Note that all elements of GT are of the form (Ru,Ru + u). Now

w = (R)w precisely if Ru, w = u, w for all u H. AddingRu, w to both sides we obtain Ru,w + w = Ru + u, w,so that (w, w + w) GT, i.e., w = Rw. This proves (2). Finally,suppose (w,w+u) GT. Since GT is linear it follows that (v,v+u) GT if and only if (v,v+u)(w,w+u) = (vw, (vw)+()w) GT. But this means exactly that (3) holds.

Theorem 5.3. The resolvent set (T) is open, and the function R is analytic in the uniform operator topology as a B(H)-valuedfunction. This means (by definition) that R can be expanded in apower series with respect to around any point in (T) and that theseries converges in operator norm in a neighborhood of the point. Infact, if (T), then (T) for | | < 1/R and

R =

k=0

( )kRk+1 for | | < 1/R .

Finally, the function (T) Ru, v is analytic for all u, v H,and for u = v it maps the upper and lower half-planes into themselves.

Proof.The series is norm convergent if | | < 1/R since()kRk+1 R(| |R)k, which is a term in a convergent

geometric series. Writing T = T ( ) and applying thisto the series from the left and right, one immediately sees that theseries represents the inverse of T . We have verified the formulafor R and it also follows that (T) is open. Now by Theorem 5.2we have 2i ImRu, u = Ru, u u, Ru = (R R)u, u =2i Im RRu, u = 2i Im Ru2. It follows that ImRu, u hasthe same sign as Im . The analyticity of Ru, v follows since wehave a power series expansion of it around any point in (T), by theseries for R. Alternatively, from Theorem 5.2.3 it easily follows that

ddRu, v = R2u, v (Exercise 5.1).


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Analytic functions that map the upper and lower halfplanes intothemselves have particularly nice properties. Our proof of the gen-

eral spectral theorem will be based on the fact that Ru, u is such afunction, so we will make a detailed study of them in the next chapter.

That (T) is open means of course that the spectrum is alwaysa closed subset ofR. It is customary to divide the spectrum into (atleast) two disjoint subsets, the point spectrum p(T) and the continuousspectrum c(T), defined as follows.

p(T) = { C | T is not one-to-one}c(T) = (T) \ p(T).

This means that the point spectrum consists of the eigen-values of T,

and the continuous spectrum of those for which S is dense in Hbut not closed. This follows since (T )1 is automatically boundedif S = H, by the closed graph theorem (Exercise 5.2). For non-selfadjoint operators there is a further possibility; one may have Snon-dense even if is not an eigenvalue. Such values of constitutethe residual spectrum which by Lemma 5.1 is empty for selfadjointoperators.

An eigenvalue for a selfadjoint operator is said to have finite mul-tiplicity if the eigenspace is finite-dimensional. Removing from thespectrum all isolated points which are eigenvalues of finite multiplic-ity leaves one with the essential spectrum. The name comes from the

fact that the essential spectrum is quite stable under perturbations(changes) of the operator T, but we will not discuss such matters here.


Exercise 5.1. Suppose that R is the resolvent of a self-adjointoperator T in a Hilbert space H. Show directly from Theorem 5.2.3 thatif u, v H, then Ru, v is analytic (has a complex derivative)for (T), and find an expression for the derivative. Also show thatif u H, then Ru, u is increasing in every point of R.

Exercise 5.2. Show that if T is a closed operator with S =

Hand / p(T), then (T).Hint: The closed graph theorem!

Exercise 5.3. Show that if T is a self-adjoint operator, then U =(T+i)(Ti)1 = I+2iRi is unitary. Conversely, ifU is unitary and 1is not an eigen-value, then T = i(U+ I)(U I)1 is selfadjoint. Whatcan one do if 1 is an eigen-value? This transform, reminiscent of aMobius transform, is called the Cayley transform and was the basis forvon Neumanns proof of the spectral theorem for unbounded operators.


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41/139

CHAPTER 6

Nevanlinna functions

Our proof of the spectral theorem is based on the following repre-sentation theorem.

Theorem 6.1. Suppose F is analytic in C \ R, F() = F(),and F maps each of the upper and lower half-planes into themselves.

Then there exists a unique, left-continuous, increasing function with(0) = 0 and

d(t)1+t2

< , and unique real constants and 0,such that

(6.1) F() = + +

1

t t

1 + t2

d(t),

where the integral is absolutely convergent.

For the meaning of such an integral, see Appendix B. FunctionsF with the properties in the theorem are usually called Nevanlinna,

Herglotz or Pick functions. I am not sure who first proved the theorem,but results of this type play an important role in the classical bookEindeutige analytische Funktionen by Rolf Nevanlinna (1930). We willtackle the proof through a sequence of lemmas.

Lemma 6.2 (H. A. Schwarz). Let G be analytic in the unit disk,and put u(R, ) = Re G(Rei). For |z| < R < 1 we then have:

(6.2) G(z) = i Im G(0) +1

2

Rei + z

Rei z u(R, ) d.

Proof. According to Poissons integral formula (see e.g. Chapter 6of Ahlfors: Complex Analysis (McGraw-Hill 1966)), we have

Re G(z) =1

2

R2 |z|2|Rei z|2 u(R, ) d .

The integral here is easily seen to be the real part of the integral in(6.2). The latter is obviously analytic in z for |z| < R < 1, so the twosides of (6.2) can only differ by an imaginary constant. However, forz = 0 the integral is real, so (6.2) follows.

The formula (6.2) is not applicable for R = 1, since we do not

know whether Re G has reasonable boundary values on the unit circle.35


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36 6. NEVANLINNA FUNCTIONS

However, if one assumes that Re G 0 the boundary values exist atleast in the sense of measure, and one has the following theorem.

Theorem 6.3 (Riesz-Herglotz). LetG be analytic in the unit circlewith positive real part. Then there exists an increasing function on[0, 2] such that

G(z) = i Im G(0) +1

2

ei + z

ei z d() .

With a suitable normalization the function will also be unique,but we will not use this. To prove Theorem 6.3 we need some kindof compactness result, so that we can obtain the theorem as a limit-

ing case of Lemma 6.2. What is needed is weak compactness in thedual of the continuous functions on a compact interval, provided withthe maximum norm. This is the classical Helly theorem. Since we as-sume minimal knowledge of functional analysis we will give the classicalproof.

Lemma 6.4 (Helly).

(1) Suppose {j}1 is a uniformly bounded1 sequence of increas-ing functions on an interval I. Then there is a subsequenceconverging pointwise to an increasing function.

(2) Suppose

{j

}1 is a uniformly bounded sequence of increasing

functions on a compact interval I, converging pointwise to .Then

(6.3)

I

f dj I

f d as j ,

for any function f continuous on I.

Proof. Let r1, r2, . . . be a dense sequence in I, for example an enu-meration of the rational numbers in I. By Bolzano-Weierstrass theo-rem we may choose a subsequence {1j}1 of{j}1 so that 1j(r1) con-verges. Similarly, we may choose a subsequence

{2j

}1 of

{1j

}1 such

that 2j(r2) converges; as a subsequence of 1j(r1) the sequence 2j (r1)still converges. Continuing in this fashion, we obtain a sequence of se-quences {kj}j=1, k = 1, 2, . . . such that each sequence is a subsequenceof those coming before it, and such that (rn) = limj kj (rn) existsfor n k. Thus jj (rn) (rn) as j for every n, since jj (rn) isa subsequence of nj(rn) from j = n on. Clearly is increasing, so ifx I but = rn for all n, we may choose an increasing subsequence rjk ,k = 1, 2, . . . , converging to x, and define (x) = limk (rjk).

Suppose x is a point of continuity of . If rk < x < rn we getjj (rk) (rn) jj (x) (x) jj (rn) (rk). Given > 0 we may

1i.e., all the functions are bounded by a fixed constant


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6. NEVANLINNA FUNCTIONS 37

choose k and n such that (rn) (rk) < . We then obtain

limj(jj (x) (x)) limj(jj (x) (x)) .Hence {jj}1 converges pointwise to , except possibly in points ofdiscontinuity of . But there are at most countably many such discon-tinuities, being increasing. Hence repeating the trick of extractingsubsequences, and then using the diagonal sequence, we get a subse-quence of the original sequence which converges everywhere in I. Wenow obtain (1).

If f is the characteristic function of a compact interval whose end-points are points of continuity for and all j it is obvious that (6.3)holds. It follows that (6.3) holds if f is a stepfunction with all discon-

tinuities at points where and all j are continuous. Iff is continuousand > 0 we may, by uniform continuity, choose such a stepfunctiong so that supI|f g| < . If C is a common bound for all j we thenobtain |

I(f g) d| < 2C and similarly with replaced by j. It

follows that limj|

If dj

I

f d| 4C and since is arbitrarypositive (2) follows.

Proof of Theorem 6.3. According to Lemma 6.2 we have, for|z| < 1,

G(Rz) = i Im G(0) +1

2

ei + z

ei

zdR() ,

where R() = Re G(Re

i) d. Hence R is increasing, 0 andbounded from above by R(). Now Re G is a harmonic function so ithas the mean value property, which means that R() = 2 Re G(0).This is independent of R, so by Hellys theorem we may choose a se-quence Rj 1 such that R converges to an increasing function . Useof the second part of Hellys theorem completes the proof.

To prove the uniqueness of the function of Theorem 6.1 we needthe following simple, but important, lemma.

Lemma 6.5 (Stieltjes inversion formula). Let be complex-valuedof locally bounded variation, and such that

d(t)t2+1

is absolutely con-

vergent. Suppose F() is given by (6.1). Then if y < x are points ofcontinuity of we have

(x) (y) = lim0

1

2i

xy

(F(s + i) F(s i) ds

= lim0

1

x

y

d(t)

(t

s)2 + 2

ds .


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38 6. NEVANLINNA FUNCTIONS

Proof. By absolute convergence we may change the order of inte-gration in the last integral. The inner integral is then easily calculated

to be1

(arctan((x t)/) arctan((y t)/)).

This is bounded by 1, and also by a constant multiple of 1/t2 if isbounded (verify this!). Furthermore it converges pointwise to 0 outside[y, x], and to 1 in (y, x) (and to 1

2for t = x and t = y). The theorem

follows by dominated convergence.

Proof of Theorem 6.1. The uniqueness of follows immedi-ately on applying the Stieltjes inversion formula to the imaginary partof (6.1) for = s + i.

We obtain (6.1) from the Riesz-Herglotz theorem by a change ofvariable. The mapping z = 1+i1i maps the upper half plane bijectively

to the unit disk, so G(z) = iF() is defined for z in the unit diskand has positive real part. Applying Theorem 6.3 we obtain, aftersimplification,

F() = Re F(i) +1

2

1 + tan(/2)

tan(/2) d() .

Setting t = tan(/2) maps the open interval (, ) onto the real axis.For = the integrand equals , so any mass of at gives riseto a term with 0. After the change of variable we get

F() = + +

1 + t

t d(t) ,

where we have set = Re F(i) and (t) = ()/(2). Since

1 + t

t =

1

t t

1 + t2

(1 + t2)

we now obtain (6.1) by setting (t) =t

0(1 + s2) d(s).

It remains to show the uniqueness of and . However, setting

= i, it is clear that = Re F(i), and since we already know that is unique, so is .

Actually one can calculate directly from F since by dominatedconvergence Im F(i)/ as . It is usual to refer to as themass at infinity, an expression explained by our proof. Note, however,that it is the mass of at infinity and not that of !


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CHAPTER 7

The spectral theorem

Theorem 7.1. (Spectral theorem) Suppose T is selfadjoint. Thenthere exists a unique, increasing and left-continuous family {Et}tR oforthogonal projections with the following properties:

Et commutes with T, in the sense that T Et is the closure ofEtT. Et 0 as t and Et I (= identity on H) as t

(strong convergence). T = t dEt in the following sense: u D(T) if and only if

t2 dEtu, u < , Tu , v =

t dEtu, v and T u2 =

t2 dEtu, u.

The family {Et}tR of projections is called the resolution of the iden-tity for T. The formula T =

t dEt can be made sense of directly by

introducing Stieltjes inte

hilbert spaces spectra

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