operator theory - università degli studi di...

Operator Theory

Stefano Meda

Universita di Milano-Bicocca

c© Stefano Meda 2009

Contents

1 Operators on finite dimensional spaces 1

1.1 Review of basic linear algebra . . . . . . . . . . . . . . . . . . . . 1

1.2 Functional calculus . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Diagonalisable operators . . . . . . . . . . . . . . . . . . . . . . . 7

1.4 First form of the spectral theorem . . . . . . . . . . . . . . . . . 13

1.5 Second form of the spectral theorem . . . . . . . . . . . . . . . . 16

2 Unbounded operators on Banach spaces 21

2.1 Basic definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.2 Bounded operators . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.3 Closed operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.4 Elementary properties of the spectrum: I . . . . . . . . . . . . . 31

2.5 Vector valued analytic functions . . . . . . . . . . . . . . . . . . 32

2.6 Elementary properties of the spectrum: II . . . . . . . . . . . . . 34

2.7 Closable operators . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3 Self adjoint operators 41

3.1 The adjoint operator I . . . . . . . . . . . . . . . . . . . . . . . . 41

3.2 The adjoint operator II . . . . . . . . . . . . . . . . . . . . . . . 44

3.3 Self adjoint operators . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.4 Extensions of symmetric operators . . . . . . . . . . . . . . . . . 54

4 One parameter semigroups 63

4.1 Exponential of a bounded operator . . . . . . . . . . . . . . . . . 63

4.2 Semigroups of operators . . . . . . . . . . . . . . . . . . . . . . . 64

4.3 Exponential of unbounded operators . . . . . . . . . . . . . . . . 71

4.4 Groups of unitary operators . . . . . . . . . . . . . . . . . . . . . 75

5 The spectral theorem 79

iii

iv CONTENTS

5.1 The functional calculus . . . . . . . . . . . . . . . . . . . . . . . 79

5.1.1 Background material . . . . . . . . . . . . . . . . . . . . . 79

5.1.2 The functional calculus on S(R) . . . . . . . . . . . . . . 80

5.1.3 The extended functional calculus . . . . . . . . . . . . . . 87

5.2 Properties of the extended functional calculus . . . . . . . . . . . 90

5.3 The spectral theorem I . . . . . . . . . . . . . . . . . . . . . . . . 95

5.4 Projection valued measures . . . . . . . . . . . . . . . . . . . . . 104

5.5 Spectral resolutions of the identity . . . . . . . . . . . . . . . . . 106

5.6 Operators associated to spectral resolutions . . . . . . . . . . . . 109

5.7 The spectral theorem II . . . . . . . . . . . . . . . . . . . . . . . 114

Chapter 1

Operators on finitedimensional spaces

In this chapter V will denote an n dimensional vector space over C and A alinear operator on V . We follow [DS, Ch VII]. We assume, though this is notnecessary for large parts of the theory we shall develop in this chapter, that Vis endowed with a Hilbertian norm.

1.1 Lecture I: review of basic linear algebra

Denote by v1, . . . , vn a basis of V ; then any v in V may be written in a uniqueway as

v = x1 v1 + · · ·+ xn vn,

where x1, . . . , xn are complex numbers. Denote by aij the ith component of Avjwith respect to v1, . . . , vn; then the vector Av has components given by a11 · · · a1n

......

an1 · · · ann

x1

...xn

.It is clear that the analysis of the operator A is easier when the associated matrixhas a simple form. It is known that two matrices M1 and M2 represent the samelinear operator A on V if and only if they are conjugate within GL(n,C), i.e.,if there exists an invertible matrix B with complex coefficients such that

M2 = BM1B−1.

If A admits a complete set of eigenvectors e1, . . . , en, associated to the (possiblyrepeated) eigenvalues λ1, . . . , λn, then its action on the vector v = x1 e1 + · · ·+xn en is given by

Av = x1 λ1 e1 + · · ·+ xn λn en,

and the matrix which represents A with respect to the basis e1, . . . , en is simplythe diagonal matrix diag(λ1, . . . , λn).

Here are some natural questions:

1

2 CHAPTER 1. OPERATORS ON FINITE DIMENSIONAL SPACES

(i) does any linear operator A admit a representation as a diagonal matrix?

(ii) If not, characterise the operators A for which this holds.

Complete answers to these questions are usually illustrated in basic coursesin linear algebra. The purpose of the first few lectures of this course is toreview the relevant results in a perhaps slightly different perspective, suitablefor generalisation to operators on infinite dimensional Hilbert spaces.

Exercise 1.1.1 Consider the operatorA on a two dimensional space V definedby

Av1 = v1 + v2 Av2 = v2,

where v1, v2 is a basis of V . Prove directly that A does not admit a represen-tation as a diagonal matrix.

The last exercise gives an answer to question (i) above. The answer to question(ii) above is more difficult, and will be the object of the next few lectures.

Definition 1.1.2 The resolvent set of A is

ρ(A) := ζ ∈ C : ζI −A is invertible.

The spectrum of A is the set

σ(A) := C \ ρ(A).

Exercise 1.1.3 Prove that ζ is in ρ(A) if and only if ζI −A is injective if andonly if ζI−A is surjective. Deduce that ζ is in σ(A) if and only if there exists vin V \0 such that Av = ζv, i.e., if and only if ζ is an eigenvalue of A. Recallthat any v in V \ 0 such that Av = ζv is called eigenvector associated tothe eigenvalue ζ.

Exercise 1.1.4 Prove that ζ is in σ(A) if and only if ζ solves the equation

det(ζI −A) = 0.

Hence σ(A) is a nonempty finite set, of cardinality at most dimV . For eachinteger k in 1, . . . ,dimV , give example of operators A for which the cardinalityof σ(A) is exactly k. We assume here that the reader is familiar with thedefinition of determinant of a linear operator.

Definition 1.1.5 For each ζ in C denote by N kζ the subspace of V defined by

N kζ = Ker(ζ −A)k.

If ζ is in σ(A), then N kζ is called the kth generalised eigenspace associated

to ζ.

Remark 1.1.6 The following are straightforward consequences of the defini-tion:

1.1. REVIEW OF BASIC LINEAR ALGEBRA 3

(i) N 0ζ ⊆ N 1

ζ ⊆ N 2ζ ⊆ · · · ;

(ii) ζ is in σ(A) if and only if N 1ζ 6= 0;

(iii) for each ζ in C there are at most dimV + 1 distinct subspaces N kζ of V ;

(iv) if A = ζI, then N 0ζ = 0 and N 1

ζ = N 2ζ = . . . = V .

A key role in what follows is played by the index of a complex number, whichwe shall define after the following preparatory lemma.

Lemma 1.1.7 For ζ in C, suppose that N kζ = N k+1

ζ . Then N kζ = N j

ζ for allj ≥ k.

Proof. Exercise for the reader. 2

Definition 1.1.8 For ζ in C, the index of ζ is the nonnegative integer ν(ζ),defined by

ν(ζ) := min k ∈ N : N kζ = N k+1

ζ .

Remark 1.1.9 The following are straightforward consequences of the defini-tion of the index of a complex number:

(i) ν(ζ) = 0 if and only if ζ is in ρ(A);

(ii) ν(ζ) ≤ dimV for every complex ζ.

Recall that if ζ is an eigenvalue of A, then its geometric multiplicity is thedimension of N 1

ζ . It is straightforward to prove that there is no relation betweenthe geometric multiplicity and the index of an eigenvalue of A.

Exercise 1.1.10 Show that the index and the spectrum of a linear transfor-mation A are invariant under conjugation with a nonsingular transformation Bof V .

Exercise 1.1.11 Compute the index of each of the following linear transfor-mations:

(i) λI, where λ is a nonzero complex number;

(ii) A, which acts on a basis v1, . . . , vn of V by

Av1 = v1Avk = vk−1 + vk k = 2, . . . , n;

(iii) A, which acts on a basis v1, v2 of V by

Av1 = −v2Av2 = v1.

Exercise 1.1.12 For each integer k ≤ dimV construct a linear operator withindex k.


1.2 Lecture II: functional calculus

The purpose of this lecture is to associate to any function f , holomorphic in aneighbourhood of the spectrum of a given operator A, a linear operator f(A)so that

(i) f(A) = A when f is the identity;

(ii) the map f 7→ f(A) is an homomorphism of algebras between the algebraof holomorphic functions in a neighbourhood of σ(A) and the algebra oflinear operators on V .

Requirements (i) and (ii) above force that we define

P (A) :=

N∑j=0

cj Aj

when P (ζ) =∑Nj=0 cj ζ

j . Clearly P and P (defined as P but with complexnumbers cj in place of cj) agree if and only if cj = cj for all j. But observe that

P 6= P does not imply P (A) 6= P (A).

Indeed, consider the operator A associated to the matrix[0 10 0

]with respect to the canonical basis of C2. It is straightforward to check thatA2 = A3 = 0, although the polynomials ζ 7→ ζ2 and ζ 7→ ζ3 are distinct.

This arises the following basic question: when do polynomials P and Qsatisfy the equation

P (A) = Q(A)?

Define the polynomial M by

M(x) =∏

ζ∈σ(A)

(x− ζ)ν(ζ). (1.2.1)

To answer the question above we shall make use of the following elementarylemma.

Lemma 1.2.1 Suppose that A is a linear operator on V . The following hold:

(i) there exists a polynomial Q such that Q(A) = 0;

(ii) if P (A) = 0, then M divides P ;

(iii) M(A) = 0, where M is defined in (1.2.1).

Proof. Part (i) is left as an exercise for the reader. Hint: given a basis v1, . . . , vnof V , the vectors v1, Av1, . . . , A

nv1 are linearly dependent, so that there existsa polynomial r1 such that r1(A)v1 = 0.

1.2. FUNCTIONAL CALCULUS 5

Now we prove (ii). Fix ζ in σ(A). By Taylor’s formula, we may write

P (x) =

deg(P )∑j=0

P (j)(ζ)

j!(x− ζ)j .

Pick v in N 1ζ \ 0. Then (A− ζI)jv = 0 for all positive integer j. Therefore

0 = P (A)v

=

deg(P )∑j=0

P (j)(ζ)

j!(A− ζI)jv

= P (ζ)v.

Then P (ζ) = 0. If ν(ζ) ≥ 2, choose v in N 2ζ \ N 1

ζ . Then (A− ζI)jv = 0 for allintegers j ≥ 2. Therefore

0 = P (A)v

=

deg(P )∑j=0

P (j)(ζ)

j!(A− ζI)jv

= P (ζ)v + P ′(ζ) (A− ζI)v.

We have already proved that P (ζ) = 0; furthermore (A − ζI)v 6= 0, becausev has been chosen outside N 1

ζ . Therefore, we may conclude that P ′(ζ) = 0.

By arguing inductively, we find that P (j)(ζ) = 0 for all j ≤ ν(ζ). This impliesthat the monomial (x− ζ)ν(ζ) divides P . Since ζ is a generic point in σ(A), therequired conlcusion follows.

Finally we prove (iii). The polynomial Q defined in (i) admits a factorisationof the form

Q(x) = aM(x)

N∏j=1

(x− λj)βj ,

where λ1, . . . , λN are distinct complex numbers, and βj are positive integers.Thus,

Q(A) = aM(A)

N∏j=1

(A− λjI)βj .

If λk is not in σ(A) for some k, then the operator (A−λkI)βk is invertible, andwe may write

(A− λkI)−βk Q(A) = aM(A)∏j 6=k

(A− λjI)βj .

Clearly the left hand side represents the null operators, and so does the righthand side. By iterating this procedure, we find that

0 = M(A)∏j

(A− λjI)βj ,

where now all the remaining λj ’s are in σ(A).


Now, consider any nonzero vector v in V . Fix ζ in σ(A). Since the righthand side of the formula above annihilates v,∏

w∈σ(A)\ζ

(A− wI)ν(w)+β(w)v

is in the kernel of (A− ζI)ν(ζ)+β(ζ) (here we have written β(ζ) instead of βj ifζ = λj and similarly for w). By definition of ν(ζ),∏

w∈σ(A)\ζ

(A− wI)ν(w)+β(w)v

is also in the kernel of (A− ζI)ν(ζ).

By iterating this procedure, we may conclude that M(A)v = 0 for all v inV , as required. 2

Theorem 1.2.2 Denote by P1 and P2 two polynomials. The following areequivalent:

(i) P1(A) = P2(A);

(ii) P1 − P2 has a zero of order ≥ ν(ζ) for each ζ in σ(A).

Proof. If (P1 − P2)(A) = 0, then M divides P1 − P2 by Lemma 1.2.1, and (ii)follows.

Conversely, if (ii) holds, then

(P1 − P2)(A) = M(A)P (A),

for some polynomial P . The required conclusion (i) then follows from the factthat M(A) = 0 (see Lemma 1.2.1 (iii)). 2

Given a linear operator A, denote by Φ the map defined on the algebra ofpolynomials by

Φ(p) = p(A).

Clearly Φ depends on the operator A; to avoid cumbersome notation we do notstress this dependence. An immediate consequence of Theorem 1.2.2 is that thekernel of Φ consists of all polynomials p such that M divides p, i.e.,

Djp(ζ) = 0 ∀ζ ∈ σ(A) ∀j ∈ 0, . . . , ν(ζ)− 1.

For reasons that will become clear later, it is convenient to extend the domainof definition of Φ from the algebra of polynomials to a larger algebra, i.e., thealgebra of holomorphic functions in a neighbourhood of the spectrum of A.

Definition 1.2.3 Given a finite subset F of C, we denote byH(F ) the space ofall functions which are holomorphic in a neighbourhood of F (the neighbourhoodmay depend on the function).

1.3. DIAGONALISABLE OPERATORS 7

Given a linear operator A, we define Φ : H(σ(A))→ L(V ) by

Φ(f) = p(A),

where p is any polynomial such that

Djp(ζ) = Djf(ζ) ∀ζ ∈ σ(A) ∀j ∈ 0, . . . , ν(ζ)− 1.

The map Φ is well defined by Theorem 1.2.2 above, and Φ(p) = p(A) for everypolynomial p. The following proposition summarises the main properties of themap Φ.

Proposition 1.2.4 For any linear operator A the associated map Φ definedabove possesses the following properties:

(i) Φ is an algebra homomorphism between H(σ(A)) and L(V ) that maps 1to the identity operator and the identity function to A;

(ii) the kernel of Φ is

f ∈ H(σ(A)) : Djf(ζ) = 0 ∀ζ ∈ σ(A) ∀j ∈ 0, . . . , ν(ζ)− 1;

(iii) Φ(H(σ(A))

)is a commutative subalgebra of L(V ).

Proof. Exercise. 2

Remark 1.2.5 Observe that if two functions f and g in H(σ(A)) agree onσ(A), then f(A) and g(A) may be different operators.

Exercise 1.2.6 Denote by A the linear operator on C3 which is representedby the matrix 1 0 3

2 1 20 0 2

.Find linear operators S and T such that

S4 = A3 A =

∞∑n=0

Tn

n!.

Exercise 1.2.7 How many different operators of the form f(A), with f inH(σ(A)), are there?

1.3 Lecture III: diagonalisable operators

The purpose of this lecture is to show that given a linear operator A on V ,there is a canonically associate finite number of mutually orthogonal projectionsEζ : ζ ∈ σ(A) of V such that

A =∑

ζ∈σ(A)

ζ Eζ +∑

ζ∈σ(A)

(A− ζI)Eζ .


As a consequence, we shall prove one of the forms of the spectral theorem forlinear operators on finite dimensional vector spaces (over C), i.e., the assertionthat

A is diagonalisable if and only if A has a basis of eigenvectors.

In particular, this holds when A is a normal operator (see Definition 1.3.6).

Definition 1.3.1 For each ζ in C, denote by eζ any function which is equalto 1 in a neighbourhood of ζ, whose closure excludes σ(A) \ ζ, and vanisheselsewhere. Set

Eζ := eζ(A)

This makes sense because eζ is in H(σ(A)). It is straightforward to check thatEζ does not depend on the choice of eζ .

The following proposition contains the main properties of the operators Eζ .

Proposition 1.3.2 The following hold:

(i) Eζ = 0 if and only if ζ is not in σ(A);

(ii) E2ζ = Eζ and Eζ1Eζ2 = 0 when ζ1 6= ζ2;

(iii) I =∑ζ∈σ(A)Eζ ;

(iv) if ζ is in σ(A), then (ζI −A)ν(ζ)Eζ = 0;

(v) RanEζ = N ν(ζ)ζ and the restriction of Eζ to N ν(ζ)

ζ is the identity operator;

(vi) if f is in H(σ(A)), then

f(A) =∑

ζ∈σ(A)

ν(ζ)−1∑j=0

Djf(ζ)

j!(A− ζI)jEζ ;

(vii) V =⊕

ζ∈σ(A)Nν(ζ)ζ ;

(viii) for each ζ in σ(A) the operator A leaves N ν(ζ)ζ invariant, i.e.,

AN ν(ζ)ζ ⊆ N ν(ζ)

ζ .

Proof. First we prove (i). Suppose that Eζ = 0. Take any polynomial Qζ suchthat Eζ = Qζ(A). Since Qζ(A) = 0, M divides Qζ by Lemma 1.2.1 (ii), henceQζ vanishes at every point of σ(A). If ζ were in σ(A), then we would haveQζ(ζ) = 1, by definition of eζ(A), so that ζ is not in σ(A).

Conversely, suppose that ζ is not in σ(A). Then, by the definition of Eζ ,any polynomial Pζ such that Pζ(A) = Eζ vanishes of order at least ν(w) for allw in σ(A). Hence Pζ(A) = 0 by Theorem 1.2.2, so that Eζ = 0.

Observe that (ii) is a straightforward consequence of the fact that Φ is analgebra homomorphism.


To prove (iii) observe that∑ζ∈σ(A) eζ is equal to 1 in a neighbourhood of

σ(A). Since Φ is an algebra homomorphism which maps 1 to I, the requiredconclusion follows.

Take any polynomial Qζ such that Eζ = Qζ(A). The polynomial

x 7→ (ζ − x)ν(ζ)Qζ(x)

vanishes at each point w of σ(A) of order at least ν(w). By Theorem 1.2.2 wemay conclude that (ζI −A)ν(ζ)Eζ = 0, and (iv) is proved.

The inclusion RanEζ ⊆ N ν(ζ)ζ is a direct consequence of (iv). To prove that

RanEζ is exactly N ν(ζ)ζ pick a polynomial Qζ such that Eζ = Qζ(A). Then

Qζ(A) = I +

deg(Qζ)∑j=ν(ζ)

Q(j)ζ (ζ)

j!(A− ζI)j .

Now, for y in N ν(ζ)ζ , we have (A− ζI)ν(ζ)y = 0, so that

Eζy = Qζ(A)y

= y +

deg(Qζ)∑j=ν(ζ)

Q(j)ζ (ζ)

j!(A− ζI)jy

= y.

Hence y is in RanEζ , as required. The last formula also shows that Eζ acts as

the identity on N ν(ζ)ζ , and (v) is proved.

To prove (vi) it suffices to show that the function f and the function g,defined by

g(x) =∑

ζ∈σ(A)

ν(ζ)−1∑j=0

Djf(ζ)

j!(x− ζ)jeζ(x)

satisfyf (j)(ζ) = g(j)(ζ) ∀j ∈ 0, . . . , ν(ζ)− 1 ∀ζ ∈ σ(A).

This is a routine verification, which is left to the reader.

To prove (vii), observe that, by (iii),

v =∑

ζ∈σ(A)

Eζv ∀v ∈ V,

so that V is the sum of the subspaces N ν(ζ)ζ , ζ in σ(A). To check that the sum

is direct, note that if v is in N ν(ζ)ζ ∩N ν(τ)

τ , then

v = Eζv = EζEτv

because the restriction of Eτ to N ν(τ)τ is the identity, and similarly for the

restriction of Eζ to N ν(ζ)ζ (see (v) above). But, if ζ 6= τ , then EζEτ = 0 by (ii),

so that v = 0, as required.


Finally, the proof of (viii) is a direct consequence of the fact that AEζ = EζA.2

Note that (vii) and (viii) above reduce the study of the operator A to the study

of its restrictions to the subspaces N ν(ζ)ζ , ζ in σ(A).

Exercise 1.3.3 Compute the spectral projections associated to the operatoron C2 which is represented, with respect to the canonical coordinates, by thematrix [

1 10 2

].

Proposition 1.3.4 Suppose that A is a linear operator on V . The followingare equivalent:

(i) A is a diagonalisable operator;

(ii) V admits a basis of eigenvectors of A;

(iii) ν(ζ) = 1 for every ζ in σ(A).

Proof. Clearly (i) implies (ii).

To prove that (ii) implies (iii), observe that⊕ζ∈σ(A)

N 1ζ ⊆

⊕ζ∈σ(A)

N ν(ζ)ζ = V.

The inclusion above follows from the elementary fact that N 1ζ ⊆ N

ν(ζ)ζ , and the

equality from Proposition 1.3.2 (vii). Since V has a basis of eigenvectors of A,the left hand side is equal to V , whence

N 1ζ = N ν(ζ)

ζ ∀ζ ∈ σ(A).

Thus, ν(ζ) = 1 for every ζ in σ(A), as required.

Finally, to prove that (iii) implies (i), observe that, by Proposition 1.3.2 (vi)and (v),

A =∑

ζ∈σ(A)

AEζ =∑

ζ∈σ(A)

ζEζ .

Hence A acts as multiplication by ζ on Ran(Eζ) = N 1ζ . 2

Note that, by Proposition 1.3.2 (vi), if ν(ζ) = 1 for all ζ in σ(A), then

f(A) =∑

ζ∈σ(A)

f(ζ)Eζ ∀f ∈ H(σ(A)),

whence f(A) depends only on the values of f at the points of σ(A).

Exercise 1.3.5 (Cayley–Hamilton) Prove that any operator A on V satis-fies its characteristic polynomial.


In view of the corollary above, it is worth finding necessary and/or sufficientconditions on A that imply that ν(ζ) = 1 for all ζ in σ(A). Theorem 1.3.13below is an interesting result in this direction.

Definition 1.3.6 A linear operator A on V is normal if

AA∗ = A∗A,

where A∗ is the adjoint operator of A, defined by

(Av,w) = (v,A∗w) ∀v, w ∈ V.

If A = A∗, then A is called self adjoint.

If A is self adjoint, then clearly A is normal. However, a normal operator maynot be self adjoint. For example consider the operator which acts on C2 by thediagonal matrix diag(i, 2i).

Proposition 1.3.7 Suppose that A is a linear operator on V . Then

Ker(A∗) = Ran(A)⊥ and Ker(A) = Ran(A∗)⊥.

Proof. Note that A∗y = 0 if and only if

(x,A∗y) = 0 ∀x ∈ V,

which is equivalent to(Ax, y) = 0 ∀x ∈ V,

which, of course, says that y is in the orthogonal complement of Ran(A)⊥. Thisproves the first assertion.

The second follows from the first applied to A∗ and the elementary fact thatA∗∗ = A. 2

This proposition and its proof hold almost verbatim for bounded operators onHilbert spaces.

Exercise 1.3.8 Prove that normality is invariant under conjugation with uni-tary operators, but not with nonsingular linear operators.

Exercise 1.3.9 Suppose that A is a linear operator such that (Av, v) = 0 forall v in V . Then A = 0. Note that this fails on real vector spaces (take, e.g., arotation of π/2 on R2).

Exercise 1.3.10 Suppose that A is a linear operator on V . Prove the follow-ing:

(i) A is normal if and only if

‖A∗v‖ = ‖Av‖ ∀v ∈ V ;


(ii) if A is normal, thenKer(A) = Ker(A∗),

so thatKer(ζ −A) = Ker(ζ −A∗),

and, more generally,

N kζ (A) = N k

ζ(A∗) ∀k ∈ N ∀ζ ∈ C;

(iii) the spectrum of A∗ is ζ : ζ ∈ σ(A);

(iv) ifA is normal and ζ, ω are distinct eigenvalues ofA, then the correspondingeigenspaces are orthogonal.

Definition 1.3.11 A linear operator E on V is a projection if E2 = E. Aprojection E is orthogonal when it is self adjoint

A similar definition holds in infinite dimensional Hilbert spaces; in that case werequire that E is a bounded operator.

Exercise 1.3.12 Suppose that E is a projection on V . The following areequivalent:

(i) E is self adjoint;

(ii) E is normal;

(iii) Ran(E) = Ker(E)⊥;

(iv) (Ev, v) = ‖Ev‖2 for every v in V .

Deduce that if A is normal then the projections Eζ , ζ in σ(A), are orthogonal.

Theorem 1.3.13 Suppose that A is a normal operator on V . Then ν(ζ) = 1for all ζ in σ(A).

Proof. Observe that for every v in V

‖Av‖2 = (Av,Av) = (A∗Av, v)

and that‖A∗v‖2 = (A∗v,A∗v) = (AA∗v, v).

Thus ‖Av‖ = ‖A∗v‖. By applying this formula to A−ζI, which is still a normaloperator, and to its adjoint A∗ − ζI, we see that

Ker(A− ζI) = Ker(A∗ − ζI). (1.3.1)

Now, suppose that ζ is in σ(A) and that y is in N 2ζ , i.e., (A− ζI)2y = 0. Then

(A− ζI)y is in the kernel of A− ζI, hence in the kernel of (A∗− ζI) by (1.3.1).Therefore

0 =((A∗ − ζI)(A− ζI)y, y

)=((A− ζI)y, (A− ζI)y

)= ‖(A− ζI)y‖2,

and (A− ζI)y = 0, i.e., y is in N 1ζ , as required. 2

Exercise 1.3.14 Does the converse of Theorem 1.3.13 hold?

1.4. FIRST FORM OF THE SPECTRAL THEOREM 13

1.4 Lecture IV: first form of the spectral theo-rem

The purpose of this lecture is to present a particular form of the spectral theoremfor self adjoint operators acting on a finite dimensional Hilbert space V . Inthis section we assume that A is self adjoint on V , i.e., A∗ = A. Then A mayrepresented with respect to any orthonormal basis of V by a hermitian matrix.We recall the following elementary facts from Section 1.3.

Proposition 1.4.1 Suppose that A is self adjoint. The following hold:

(i) σ(A) ⊂ R;

(ii) eigenvectors associated to distinct eigenvalues are orthogonal;

(iii) the operators Eζ : ζ ∈ σ(A) are mutually orthogonal projections.

Proof. To prove (i), observe that if ζ is in σ(A), and v is a nonzero vector inthe corresponding eigenspace N 1

ζ , then

ζ (v, v) = (Av, v) = (v,Av) = ζ (v, v),

so that ζ = ζ, i.e., ζ is real.

The proof of (ii) is similar and is left to the reader.

Finally, to prove (iii), recall that Eζ is the projection onto N 1ζ (for A is

self adjoint, hence ν(ζ) = 1 for all ζ in σ(A)). To conclude the proof of (iii), itremains to show that Eζ is self adjoint. Observe that a finite linear combinationof self adjoint operators with real coefficients is self adjoint. By definition

Eζ = p(A),

where p is any polynomial such that p(ζ) = 1 and p(ζ ′) = 0 for all ζ ′ in σ(A)\ζ.The polynomial

p(x) =∏

ζ′∈σ(A)\ζ

x− ζ ′

ζ − ζ ′

satisfies the requirements above and has real coefficients (remember that σ(A) ⊂R). This concludes the proof of the proposition. 2

Definition 1.4.2 A linear operator on V is said to be unitary if

U∗U = UU∗ = I.

Unitary operators are the natural isomorphisms of Hilbert spaces, as the follow-ing exercise shows.

Exercise 1.4.3 Prove that U is unitary if and only if U is an isometry, i.e.,

‖Uv‖ = ‖v‖ ∀v ∈ V.

Do you use the fact that V is finite dimensional? Similarly prove that U isunitary if and only if it preserves the inner product of V .


Now, we prove that any self adjoint operator A is unitarily equivalent to amultiplication operator on a direct sum of L2(R, µk) spaces with respect tosuitable measures µk.

For each ζ in σ(A) denote by n(ζ) the dimension of N 1ζ , i.e. the geometric

multiplicity of ζ, and by

v1ζ , . . . , vn(ζ)ζ

an orthonormal basis of N 1ζ . Since eigenspaces associated to distinct eigenvalues

are orthogonal (see Proposition 1.4.1 (ii)),

v1ζ , . . . , vn(ζ)ζ : ζ ∈ σ(A) (1.4.1)

is an orthonormal basis of V . Denote by ζ1, . . . , ζN1the distinct eigenvalues of

A, ordered so that n(ζj) ≥ n(ζk) whenever 1 ≤ j ≤ k ≤ N1. Set

n := maxn(ζ) : ζ ∈ σ(A)

andNk := #j : n(ζj) ≥ k.

Thus, the eigenvalues ζ1, . . . , ζNk are precisely those with multiplicity at leastk. Define measures µ1, . . . , µn by

µk =

Nk∑j=1

δζj ,

where δζj denotes the Dirac mass on the real line concentrated at ζj , and considerthe Hilbert space

H :=

n⊕k=1

L2(R, µk),

which consists of all n-tuples (f1, . . . , fn), where fk is in L2(R, µk), endowedwith the natural norm

‖(f1, . . . , fn)‖ :=[ n∑k=1

‖fk‖L2(R,µk)2]1/2

.

We define the multiplication operator M on H by the formula

M(f1, . . . , fn)(x) =(xf1(x), . . . , x fn(x)

)∀x ∈ R. (1.4.2)

Clearly M maps H into H.

Exercise 1.4.4 Define U : V → H by the formula

Uvkζj = (0, . . . , 0, fζj︸︷︷︸kth entry

, 0, . . . , 0) (1.4.3)

for every j in 1, . . . , Nk and for every k in 1, . . . , n (recall that Nk denotesthe cardinality of the set of the eigenvalues of geometric multiplicity at least k),and fζj is the function on R which is equal to 1 at ζj and vanishes elsewhere.Prove that U is a one-to-one isometry between V and H.

1.4. FIRST FORM OF THE SPECTRAL THEOREM 15

Theorem 1.4.5 Suppose that A is self adjoint on the N dimensional Hilbertspace V . Then the operator U : V → H defined in (1.4.3) is a one-to-oneisometry such that

UAU∗(f1, . . . , fn) = M(f1, . . . , fn) ∀(f1, . . . , fn) ∈ H.

Proof. By Exercise 1.4.4, the map U is a one-to-one isometry between V andH. It remains to prove that UAU∗ = M . Pick an element (f1, . . . , fn) of H.Then there exist constants ckj : j = 1, . . . , Nk, k = 1, . . . , n such that

fk =

Nk∑j=1

ckj fζj

µk almost everywhere (recall that µk is a measure concentrated in the setζ1, . . . ζNk). Note also that

‖fk‖L2(R,µk)2

=

Nk∑j=1

|ckj |2,

because the functions fζ1 , . . . , fζNk are orthonormal in L2(R, µk).

Furthermore,

U∗(f1, . . . , fn) =

N1∑j=1

c1j v1ζj + · · ·+

Nn∑j=1

cnj vnζj ,

so that

UAU∗(f1, . . . , fn) = U( N1∑j=1

c1j ζj v1ζj + · · ·+

Nn∑j=1

cnj ζj vnζj

)

=( N1∑j=1

c1j ζj fζj , · · · ,Nn∑j=1

cnj ζj fζj

)= M(f1, . . . , fn),

as required. 2

This is one of the forms of the spectral theorem that can be generalised to (pos-sibly unbounded) self adjoint operators on infinite dimensional Hilbert spaces.

To follow the proof of the theorem above, it may be helpful to consider thefollowing arrays

ζ1 · · · · · · · · · · · · · · · ζN1

ζ1 · · · · · · · · · ζN2

· · · · · · · · · · · ·ζ1 · · · ζNn

andv1ζ1 · · · · · · · · · · · · · · · v1ζN1

v2ζ1 · · · · · · · · · v2ζN2

· · · · · · · · · · · ·vnζ1 · · · vnζNn


All the eigenvalues with multiplicity at least j are listed in the ith row for alli ≥ j. The last line contains the eigenvalues which have multiplicity exactlyequal to n.

In the case where the eigenvalues of A are all distinct the construction abovesimplifies. We need only consider the measure µ1, and now N1 is just thedimension of V and H is just L2(R, µ1).

1.5 Lecture V: second form of the spectral the-orem

Definition 1.5.1 A family Pλ : λ ∈ R of orthogonal projections on Vis called a spectral resolution of the identity if it possesses the followingproperties:

(i) PλPλ′ = Pmin(λ,λ′);

(ii) there exists real numbers α and β such that

Pλ = 0 ∀λ ∈ (−∞, α) and Pλ = I ∀λ ∈ [β,∞);

(iii) for every v in Vlimλ↓λ0

Pλv = Pλ0v.

Thus, λ 7→ Pλ is a “increasing” projection valued function on the real line.

Exercise 1.5.2 Suppose that P1 and P2 are orthogonal projections of V . Thefollowing are equivalent:

(i) P1 ≤ P2 (i.e., (P2v − P1v, v) ≥ 0 for all v in V );

(ii) ‖P1v‖ ≤ ‖P2v‖ for all v in V ;

(iii) Ran(P1) ⊆ Ran(P2);

(iv) P2P1 = P1;

(v) P1P2 = P1.

This result holds verbatim also when V is an infinite dimensional Hibert space.

Exercise 1.5.3 Suppose that A and B are linear operators on V and that Ais self adjoint. Then AB = 0 if and only if Ran(A) and Ran(B) are orthogonal.

The second form of the spectral theorem uses the concept of spectral resolutionof the identity associated to the self adjoint operator A, which we now define.

Definition 1.5.4 For every λ in R, define the operator Pλ by

Pλ :=∑

ζ∈σ(A), ζ≤λ

Eζ ,

1.5. SECOND FORM OF THE SPECTRAL THEOREM 17

where Eζ are the spectral projections defined in Lecture 1.2. The family Pλλ∈Ris called the spectral resolution of the identity associated to the self ad-joint operator A.

In the next proposition we shall prove that the spectral resolution of the identityassociated to A is, indeed, a spectral resolution of the identity in the sense ofDefinition 1.5.1.

Proposition 1.5.5 Let Pλ be the spectral resolution of the identity associatedto the self adjoint operator A. The following hold:

(i) for every λ in R, the operator Pλ is the orthogonal projection onto

⊕ζ∈σ(A): ζ≤λ

N 1ζ ;

(ii) if λ < ζmin, then Pλ = 0, and if λ ≥ ζmax, then Pλ = I;

(iii) PλPµ = Pmin(λ,µ);

(iv) for each λ0 in R, the map λ 7→ Pλ is constant in a right neighbourhood ofλ0. Hence

limλ↓λ0

Pλv = Pλ0v ∀v ∈ V ;

(v) for each pair v and w in V , the function λ 7→ (Pλv, w) is a step function(hence of bounded variation). In particular, if ζ1 < · · · < ζN1 are thedistinct eigenvalues of A, then

(Pλv, w) =

N1∑j=1

(Eζjv, w) 1[ζj ,∞)(λ);

(vi) for each v and w in V

(v, w) =

∫ ∞−∞

d(Pλv, w) and ‖v‖2 =

∫ ∞−∞

d(Pλv, v),

where these integrals are (Lebesgue–) Stieltjes integrals with respect to themeasures associated to the distribution functions λ 7→ (Pλv, w) and λ 7→(Pλv, v) respectively.


6

-........‖Pζ1v‖

2

ζ1

r.................r

ζ2

‖Pζ2v‖2 ........

The graph of λ 7→ (Pλv, v)

Proof. To prove (i), first observe that Pλ is a projection. Indeed,

P2λ =

∑ζ,ζ′∈σ(A); ζ,ζ′≤λ

EζEζ′ =∑

ζ∈σ(A); ζ≤λ

Eζ = Pλ,

because Eζ is a projection and EζEζ′ = 0 when ζ 6= ζ ′. To show that Pλ isorthogonal, it suffices to prove that Pλ is self adjoint. But this is clear, becausePλ is a finite sum of self adjoint operators, by Proposition 1.4.1 (iii). Recall thatEζ is the orthogonal projection onto N 1

ζ , and that N 1ζ and N 1

ζ′ are orthogonalsubspaces whenever ζ 6= ζ ′, by Proposition 1.4.1. On the one hand, clearly,

Ran(Pλ) ⊆⊕

ζ∈σ(A): ζ≤λ

N 1ζ .

On the other hand, if

v =∑

ζ∈σ(A): ζ≤λ

vζ ,

where vζ is in N 1ζ , then

Pλv =∑

ζ∈σ(A): ζ≤λ

Pλvζ

=∑

ζ∈σ(A): ζ≤λ

Eζvζ

=∑

ζ∈σ(A): ζ≤λ

vζ

= v,

as required to conclude the proof of (i).

Statement (ii) is obvious, and (iii) follows from the fact that the projectionsEζ satisfy EζEζ′ = 0 when ζ 6= ζ ′.

To prove the first statement in (iv), fix λ0 in R and denote by ζ the leasteigenvalue of A greater than λ0. By definition of Pλ the operator λ 7→ Pλ isconstant for all λ in [λ0, ζ), as required. The second statement follows directlyfrom the first.

1.5. SECOND FORM OF THE SPECTRAL THEOREM 19

To prove (v) note that, by definition of Pλ,

(Pλv, w) =∑ζ≤λ

(Eζv, w),

which is constant in each of the connected components of R \ σ(A), hence it isa step function. A straightforward verification shows that the right hand sideis equal to

N1∑j=1

(Eζjv, w) 1[ζj ,∞)(λ),

as required.

Finally, to prove (vi), observe that the Lebesgue–Stjelties measure associatedto the distribution function

λ 7→N1∑j=1

(Eζjv, w) 1[ζj ,∞)(λ)

is justN1∑j=1

(Eζjv, w) δζj .

Therefore ∫ ∞−∞

d(Pλv, w) =

N1∑j=1

(Eζjv, w)

∫ ∞−∞

dδζj

=( N1∑j=1

Eζjv, w)

= (v, w),

threby proving the first formula. The second formula follows immediately fromthe first. 2

Since A is self adjoint, the index of each of its eigenvalues is equal to 1; thus,the definition of the operator f(A) depends only on the values of f on σ(A),and not on its derivatives. We may extend the functional calculus for A definedin Lecture II to all complex valued functions on the real line, by just defining

f(A) =∑

ζ∈σ(A)

f(ζ)Eζ . (1.5.1)

Theorem 1.5.6 (Spectral theorem: II form) Given a self adjoint operatorA, the following representation holds

(Av,w) =

∫ ∞−∞

λ d(Pλv, w) ∀v, w ∈ V,

where Pλ is the spectral resolution of the identity associated to A. Moregenerally, for every function f on σ(A)(

f(A)v, w)

=

∫ ∞−∞

f(λ) d(Pλv, w) ∀v, w ∈ V,

where f(A) is defined in (1.5.1).


Proof. We give a proof of the second formula, the first being a special case ofthe second. By arguing as in the proof of Proposition 1.5.5 (v), we see that∫ ∞

−∞f(λ) d(Pλv, w) =

N1∑j=1

(Eζjv, w)

∫ ∞−∞

f(λ) dδζj

=( N1∑j=1

f(ζj)Eζjv, w)

= (f(A)v, w),

as required. 2

Chapter 2

Unbounded operators onBanach spaces

In this chapter, B will denote a (possibly infinite dimensional) Banach spaceover C.

2.1 Lecture VI: basic definitions

Definition 2.1.1 A linear operator on B is a pair (X,A), where X is a(possibly not closed) linear subspace of B and A : X → B is a linear operator:X is called the domain of A and will be denoted by Dom(A).

Note that we do not assume that Dom(A) is dense in B, unless explicitly stated.

Definition 2.1.2 Given two linear operators A and B on B, we say that Bis an extension of A, and we write B ⊃ A, if Dom(B) ⊃ Dom(A) and therestriction of B to Dom(A) agrees with A.

Consider the operators A and B on L2(R) defined by

Dom(A) = C∞c (R) Af = i f ′ ∀f ∈ Dom(A)

Dom(B) = C1c (R) Bf = i f ′ ∀f ∈ Dom(B).

Clearly B ⊃ A.

We consider two more examples of linear operators, which are one dimensionalmodels of the Dirichlet and Neumann Laplacian. Suppose that −∞ < a <b <∞. Let HD and HN be the operators on L2((a, b)) defined by

Dom(HD) = f ∈ C2([a, b]) : f(a) = 0 = f(b) HDf = −f ′′

Dom(HN ) = f ∈ C2([a, b]) : f ′(a) = 0 = f ′(b) HNf = −f ′′.

It is not hard to check that the domains of HD and HN are dense in L2((a, b)).

21

22 CHAPTER 2. UNBOUNDED OPERATORS ON BANACH SPACES

Definition 2.1.3 A linear operator on B is bounded if Dom(A) = B and

supf 6=0

‖Af‖‖f‖

<∞.

The supremum above is called the operator norm of A and it is denoted by|||A|||.

Exercise 2.1.4 Prove that for a linear operator A on B the following areequivalent:

(i) A is a bounded operator on B;

(ii) A is continuous at every point of B;

(iii) A is continuous at 0.

Exercise 2.1.5 Suppose that A is a linear operator on B with dense domain.Assume that

supf∈Dom(A)\0

‖Af‖‖f‖

<∞.

Prove that there exists a unique bounded linear operator A that extends A.

Exercise 2.1.6 Prove that the operators A and B defined just below Defini-tion 2.1.2 do not admit bounded extensions to L2(R).

Now we introduce a class of linear operators, which will play a key role in whatfollows.

Definition 2.1.7 Suppose that m is a measurable function on the measurespace (M,M, µ). Define the multiplication operator Am by

Dom(Am) := f ∈ L2(µ) : mf ∈ L2(µ)Amf := mf ∀f ∈ Dom(Am).

Explicitly, the domain of Am is the space of all measurable functions on M suchthat ∫

M

|f |2[1 + |m|2

]dµ <∞.

Definition 2.1.8 Suppose that m is a complex valued measurable function ona measure space (M,M, µ). We say that a complex number z is in the essentialrange of m if for every ε > 0 the inverse image of the ball B(z, ε) has positivemeasure. The essential range of m will be denoted by Raness(m).

Exercise 2.1.9 Prove that the essential range of a measurable function is aclosed set.

2.1. BASIC DEFINITIONS 23

Definition 2.1.10 Suppose that m is a complex valued measurable functionon a measure space (M,M, µ). The essential supremum of m is the number‖m‖∞, defined by

‖m‖∞ = sup|z| : z ∈ Raness(m).

Theorem 2.1.11 Suppose that the measure space (M,M, µ) possesses the fol-lowing property: for every measurable set E such that µ(E) =∞, there exists ameasurable subset E′ of E such that 0 < µ(E′) <∞. Prove the following:

(i) the multiplication operator Am is bounded on L2(µ) if and only if m is inL∞(µ); in this case |||Am||| = ‖m‖∞;

(ii) Dom(Am) is dense in L2(µ) if and only if m is finite a.e..

Proof. First we prove (i). If m is in L∞(M), then

‖Amf‖22 ≤ ‖m‖∞2 ‖f‖22,

so that |||Am|||2 ≤ ‖m‖∞.

Conversely, suppose that Am is bounded. For each α < ‖m‖∞, the set Eα :=|m| > α has positive finite measure (in case µ(Eα) = ∞, by assumption, wemay choose E′α ⊂ Eα such that 0 < µ(E′α) <∞). Therefore∥∥∥Am 1Eα

‖1Eα‖2

∥∥∥2

2

=

∫Eα

|m|2

µ(Eα)dµ ≥ α2

This implies |||Am|||2 ≥ α. By taking the supremum with respect to α < ‖m‖∞,we may conclude that ‖m‖∞ ≤ |||Am|||2. Thus m must be essentially bounded.The estimates proved so far also imply that |||Am|||2 = ‖m‖∞.

Next we prove (ii). If m were not finite almost everywhere, there should exista measurable set E such that µ(E) > 0 and m =∞ on E. Then either µ(E) <∞, or µ(E) = ∞. In the latter case by assumption there exists a measurableset E′ ⊂ E such that 0 < µ(E′) < ∞, and we may proceed using E′ insteadof E. Then a function f in Dom(Am) must be equal to 0 almost everywhereon E, and 1E is a function in L2(M) which is orthogonal to Dom(Am). HenceDom(Am) cannot be dense in L2(M).

Conversely, suppose that m is finite almost everywhere. Given g in L2(M)we show that we can approximate g by a sequence of function in Dom(Am). Foreach positive integer N , define

gN = g 1EN ,

where EN := |m| ≤ N. Observe that EN ↑ M , because m is finite almosteverywhere. Furthermore gN is in Dom(Am). Indeed,∫

M

|mgN |2 dµ ≤ N2

∫EN

|g|2 dµ = N2 ‖g‖22 <∞.

Finally, gN → g pointwise almost everywhere, and |gN − g| ≤ 2|g|. Hence∫M

|gN − g|2 dµ→ 0

by the Dominated Convergence Theorem. 2


Exercise 2.1.12 Suppose that A is a densely defined linear operator in aHilbert space H such that (Af, f) = 0 for all f in Dom(A). Prove that A = 0.

2.2 Lecture VII: bounded operators

It is straightforward to check that the set of bounded operators on B is a vectorspace, which we denote by L(B), and that the operator norm is a norm on L(B).

Proposition 2.2.1 The space L(B) is a Banach space with respect to the op-erator norm.

Proof. Suppose that An is a Cauchy sequence of operators in L(B). Then forevery ε > 0 there exists an integer ν such that

|||Am −An||| ≤ ε ∀m,n ≥ ν.

Consequently,

‖Amf −Anf‖ ≤ ε ‖f‖ ∀m,n ≥ ν ∀f ∈ B. (2.2.1)

Thus, for every f in B, Anf is a Cauchy sequence in B. Since B is complete,there exists an element in B that we denote Af such that

limn→∞

Anf = Af, (2.2.2)

where the convergence is in the norm of B. It is straightforward to check thatthe operator A thus defined is linear. By letting m tend to ∞ in (2.2.1), weobtain

‖Af −Anf‖ ≤ ε ‖f‖ ∀n ≥ ν ∀f ∈ B.

This implies that A−An is a bounded operator on B, and that

|||A−An||| ≤ ε ∀n ≥ ν.

Thus, A is bounded (because it is the sum of the bounded operators An andA−An), and limn→∞ |||An −A||| = 0, as required. 2

Exercise 2.2.2 Prove that the vector space K(B) of all compact linear oper-ators on B is a two sided closed ideal in L(B).

Exercise 2.2.3 Suppose that H is a Hilbert space and that J is a nonnullnorm closed ideal in L(H). Prove that K(H) is contained in J , by showing thatany finite rank operator is in J . Hint : follow the following steps: First step:Given v, w in H, define

Rv,wf := (f, v)w.

Show that Rv,w is a rank one operator with Ran(Rv,m) = 〈w〉, and that thefollowing formulae hold:

Rv,wRv′,w′ = (w′, v)Rv′,w αRv,w = Rv,αw, Rv,v = ‖v‖2 P〈v〉,

2.2. BOUNDED OPERATORS 25

where P〈v〉 denotes the orthogonal projection onto the subspace generated by v.Furthermore, if T is in L(H), then

Rv,wT = RT∗v,w and TRv,w = Rv,Tw.

Finally, if T 6= 0 and Th 6= 0, then

1

‖Th‖2RTh,wTRw,h = Rw,w.

Second step: Show that J contains all finite rank operators. Third step: showthat the norm closure of the space of finite rank operators is K(H).

In the case where B is the space L(X ) of bounded operators on the Banachspace X , there are at least four topologies which deserve consideration:

(i) the uniform topology, i.e., the norm topology on L(X ) induced by theoperator norm. In this topology, An → A if limn→∞ |||An −A||| = 0;

(ii) the weak topology, i.e., the weakest locally convex topology for which allthe functionals in L(X )′ are continuous. In this topology, An → A iflimn→∞ |V (An)− V (A)| = 0 for every V in L(X )′;

(iii) the strong operator topology, i.e., the locally convex topology generatedby the seminorms A 7→ ‖Af‖, where f is in X . In this topology, An → Aif and only if limn→∞ ‖Anf −Af‖ = 0 for every f in X ;

(iv) the weak operator topology, i.e., the locally convex topology generated bythe seminorms A 7→ |〈F,Af〉|, where f is in X and F is in X ′. In thistopology, An → A if and only if limn→∞ 〈F,Anf −Af〉 = 0 for every fin X and F in X ′.

It is clear that the uniform topology is stronger than the weak topology andthat the strong operator topology is stronger than the weak operator topology.

Moreover, the weak topology is stronger than the weak operator topology.Indeed, the weak topology is the weakest topology for which all functionalsin L(X ) are continuous, whereas the weak operator topology is the weakesttopology for which all the functionals of the form A 7→ 〈F,Af〉 are continuous.

Exercise 2.2.4 Consider the operators Tn, Sn and Wn on `2, defined by

Tn(x1, x2, . . .) =(x1/n, x2/n, . . .

)Sn(x1, x2, . . .) =

(0, . . . , 0, xn+1, xn+2, . . .

)n zeroes

Wn(x1, x2, . . .) =(0, . . . , 0, x1, x2, . . .

)n zeroes.

Prove that

(i) Tn tends to 0 in the uniform topology;

(ii) Sn tends to 0 in the strong operator topology, but not in the uniformtopology;


(iii) Wn tends to 0 weakly but not in the strong operator topology.

Exercise 2.2.5 For each t > 0 consider the operator Tt defined on L2(R) by

Ttϕ(x) = ϕ(x+ t) ∀x ∈ R.

(i) Compute the norm of Tt.

(ii) In what topology and to what limit does Tt converge as t tends to ∞?

(iii) Answer a similar question in the Hilbert space L2(R, γ), where γ denotesthe Gaussian measure defined by

dγ(x)

dx= e−x

2

∀x ∈ R.

2.3 Lecture VIII: closed operators

Definition 2.3.1 The graph of a linear operator A on B is the subset GA ofB × B defined by

GA := (f,Af) : f ∈ Dom(A).

We endow B × B with the norm

‖(f, g)‖ :=(‖f‖2 + ‖g‖2

)1/2. (2.3.1)

The motivation for this choice is that in the case where B is a Hilbert space, thisnorm is the Hilbertian norm associated to the inner product on B × B definedby (

(f1, g1), (f2, g2))

:= (f1, f2) + (g1, g2).

Exercise 2.3.2 Prove that B ⊃ A if and only if GB ⊃ GA.

Definition 2.3.3 A linear operator A on B is closed if GA is closed in B×B.

Exercise 2.3.4 Prove that if A is bounded on B, then A is closed.

The vector space Dom(A), endowed with the norm

‖f‖1 =(‖f‖2 + ‖Af‖2

)1/2, (2.3.2)

will be denoted by B1.

Proposition 2.3.5 Suppose that A is a linear operator on B. The followingare equivalent:

(i) A is closed;

(ii) for every sequence fn in Dom(A) such that there exist f and g in B forwhich

limn→∞

‖fn − f‖ = 0 and limn→∞

‖Afn − g‖ = 0,

we have f is in Dom(A) and Af = g;

2.3. CLOSED OPERATORS 27

(iii) B1 is a Banach space.

Proof. First we prove that (i) implies (ii). Suppose that fn is a sequence inDom(A) such that there exist f and g in B for which

limn→∞


‖Afn − g‖ = 0.

Then clearly (fn, Afn) converges to (f, g) in the graph norm defined in (2.3.1)above. Since A is closed by assumption, GA is a closed subset of B × B, hence(f, g) is in GA, so that f is in Dom(A) and g = Af .

Next we prove that (ii) implies (iii). Suppose that fn is a Cauchy sequencein B1. We need to prove that there exists f in Dom(A) such that

limn→∞

‖fn − f‖1 = 0.

Since fn is a Cauchy sequence in B1, both fn and Afn are Cauchy se-quences in B. Since B is complete, there exist f and g in B such that

limn→∞


‖Afn − g‖ = 0.

By assumption, f is in Dom(A) and g = Af . It follows that fn is convergentto f in B1, so that B1 is complete, as required.

Finally, we prove that (iii) implies (i). Suppose that (f, g) is in GA. Weneed to prove that (f, g) is in GA. Since (f, g) is in GA, there exist a sequencefn in Dom(A) such that

limn→∞


‖Afn − g‖ = 0.

Then fn is a Cauchy sequence in B1. Since B1 is complete by assumption,there exists h in Dom(A) such that

limn→∞

‖fn − h‖1 = 0.

Therefore

limn→∞

‖fn − h‖ = 0 and limn→∞

‖Afn −Ah‖ = 0.

By the uniqueness of the limit, we may conclude that f = h and g = Ah, sothat (f, g) is in GA, as required. 2

The norm ‖ ‖1 defined in (2.3.2) is clearly equivalent to the norm ‖ ‖1′ definedby

‖f‖1′ = ‖f‖ + ‖Af‖. (2.3.3)

Sometimes it is convenient to use this norm instead of ‖ ‖1.

Corollary 2.3.6 Suppose that A and B are linear operators on B, and that Bis bounded. Then A is closed if and only of A+B is closed.


Proof. Clearly Dom(A + B) = Dom(A). We consider on Dom(A) two norms:the norm ‖ ‖1′, defined in (2.3.3), and the norm ‖ ‖2′, defined by

‖f‖2′ = ‖f‖ + ‖(A+B)f‖.

We claim that

(1 + |||B|||)−1 ‖f‖1′ ≤ ‖f‖2′ ≤ (1 + |||B|||) ‖f‖1′

The required conclusion follows directly from the claim, because these inequal-ities state that ‖ ‖1 and ‖ ‖2 are equivalent norms on Dom(A), hence Dom(A)is a Banach space with respect to the first norm if and only if it is a Banachspace with respect to the second norm, and this condition is equivalent to sayingthat A and A+B are closed, respectively, by the proposition above.

To prove the right inequality, observe that

‖f‖ + ‖(A+B)f‖ ≤ ‖f‖ + ‖Af‖ + |||B||| ‖f‖≤ (1 + |||B|||) (‖f‖ + ‖Af‖).

To prove the left inequality, observe that, by the triangle inequality,

‖f‖ + ‖(A+B)f‖ ≥ ‖f‖ +∣∣‖Af‖ − |||B||| ‖f‖∣∣

≥ ‖f‖ + ‖Af‖ − |||B||| ‖f‖.

Therefore‖f‖ + ‖Af‖ ≤ (1 + |||B|||) ‖f‖ + ‖(A+B)f‖,

which implies the required estimate. 2

We give an example of a closed operator which is unbounded. Define Af :=f ′ on Dom(A) = C1

([0, 1]

). We think of A as a densely defined operator on

C([0, 1]

), endowed with the uniform norm. We leave the reader the proof that

A is a closed operator.

Given a nonnull function f in C1([0, 1]

)such that f(1) = f ′(1) = 0, define

fε(x) := f(x/ε) ∀x ∈ [0, ε],

and fε(x) := 0 in [1/ε, 1]. Clearly fε is in C1([0, 1]

)and ‖fε‖∞ = ‖f‖∞.

Furthermore,

‖f ′ε‖∞ =1

ε‖f ′‖∞,

which tends to ∞ as ε tends to 0. Thus, A cannot possibly be bounded, asrequired.

Another example of a closed unbounded operator is the following. Define alinear operator on L2((0, 1)) by

Dom(A) = f ∈W 1,2((0, 1)) : f(0) = 0 = f(1)Af = i f ′ ∀f ∈ Dom(A).

Here f ′ is the distributional derivative of f , which belongs to L2((0, 1)), becausef is in W 1,2((0, 1)). Note that the condition f(0) = 0 = f(1), which is a priorimeaningless for a generic function in L2((0, 1)), makes sense for functions inW 1,2((0, 1)). Indeed, every such function admits a unique continuous represen-tative, which extends to a continuous function on [0, 1]. For this extension thevalues at 0 and 1 are of course well defined.

2.3. CLOSED OPERATORS 29

Exercise 2.3.7 Prove that A is unbounded.

The proof of the closedness of A uses the following elementary fact, which westate as a lemma for future reference.

Lemma 2.3.8 If fn is a sequence in W 1,2((0, 1)), and there exist f and g inL2((0, 1)) such that

limn→∞

‖fn − f‖2 = 0 and that limn→∞

‖f ′n − g‖2 = 0,

then f is in W 1,2((0, 1)), and g = f ′.

Proof. Since f is in L2((0, 1)), it suffices to show that the distributional deriva-tive of f is in L2((0, 1)). Since fn is in W 1,2((0, 1)),

∫ 1

0

ϕf ′n dλ = −∫ 1

0

ϕ′ fn dλ, ∀ϕ ∈ C∞c ((0, 1)).

When n tends to∞ the left and the right hand sides converge to −∫ 1

0ϕg dλ and

to −∫ 1

0ϕ′ f dλ, respectively. Hence f is in W 1,2((0, 1)) and g = f ′, as required.

2

We go back to the example above. Suppose that fn is in Dom(A), and thatthere exist f and g in L2((0, 1)) such that

limn→∞

‖fn − f‖2 = 0 and that limn→∞

‖Afn − g‖2 = 0.

We need to prove that f is in Dom(A) and that Af = g. By the lemma, g is inW 1,2((0, 1)) and i f ′ = g.

To conclude the proof of the closedness of A it remains to show that f is inDom(A), i.e., that f(0) = 0 = f(1). Recall that we may assume that both fnand f are absolutely continuous. Observe that

fn(t) =

∫ t

0

f ′n dλ ∀t ∈ [0, 1]

(here we have used the fact that fn(0) = 0, because fn is in Dom(A)) andsimilarly that

f(t)− f(0) =

∫ t

0

f ′ dλ ∀t ∈ [0, 1].

Since f ′n is convergent to f ′ in L2((0, 1)),

limn→∞

fn(t) =

∫ t

0

f ′(s) ds ∀t ∈ [0, 1],

so that f(t) − f(0) = limn→∞ fn(t) for every t in [0, 1]. In particular, f(1) −f(0) = 0. Since fn converges to f in L2((0, 1)) and it converges pointwise, itmust converge to f almost everywhere. This forces f(0) = 0, hence f(1) = 0,and f is in Dom(A), as required.

Observe that the domain of the operator A in the example above is strictlycontained in L2((0, 1)).


Exercise 2.3.9 Prove that the Laplace operator ∆ on Rn with domain C∞c (Rn)is not closed.

Theorem 2.3.10 (Closed graph) Suppose that A is a closed linear operatoron a Banach space B with Dom(A) = B. Then A is bounded.

Proof. Since A is closed, the graph GA of A is a closed subspace of B×B, hencea Banach space with respect to the norm defined above. Consider the linearoperator from GA to B defined by

(f,Af) 7→ f ∀f ∈ B.

This operator is bounded, because

‖f‖B ≤(‖f‖B2 + ‖Af‖B2

)1/2 ∀f ∈ B,

and it is onto, because A is defined on all of B. Furthermore, it is clearlyinjective. By the Open Mapping Theorem the inverse operator, mapping f to(f,Af) is bounded, i.e., there exists a constant C such that(

‖f‖B2 + ‖Af‖B2)1/2 ≤ C ‖f‖B ∀f ∈ B,

which implies the estimate

‖Af‖B ≤ C ‖f‖B ∀f ∈ B,

i.e., the boundedness of A, as required. 2

Corollary 2.3.11 Suppose that A is a closed operator which maps Dom(A)onto B in a one-to-one fashion. Then A−1 is bounded.

Proof. It suffices to show that A−1 is closed and then apply the Closed GraphTheorem. Observe that

GA−1 = (Af, f) : f ∈ Dom(A).

Hence GA−1 is the image of GA under the isomorphism (x, y) 7→ (y, x) of B×B.Since this isomorphism preserves closed sets, and GA is closed, GA−1 is closed,as required. 2

Exercise 2.3.12 Suppose that B is a Banach space with respect two differentnorms ‖ ‖1 and ‖ ‖2, and that there exists a constant C such that

‖f‖1 ≤ C‖f‖2 ∀f ∈ B.

Prove that ‖ ‖1 and ‖ ‖2 are equivalent norms.

Exercise 2.3.13 Suppose that F is a subset of B∗ that separates the pointsof B (i.e., F (f) = 0 for all F in F implies f = 0). Prove that if T is a linearoperator on B with Dom(T ) = B and such that F T is continuous for all F inF , then T is continuous.

2.4. ELEMENTARY PROPERTIES OF THE SPECTRUM: I 31

2.4 Lecture IX: elementary properties of thespectrum: I

Definition 2.4.1 Suppose that A is a linear operator on B. We say that thecomplex number ζ is in the resolvent set of A if ζI − A maps Dom(A) in aone-to-one fashion onto B and the resolvent operator

R(ζ;A) := (ζI −A)−1

is a bounded operator on B. The resolvent set of A will be denoted by %(A).The spectrum σ(A) of A is defined to be C \ %(A).

One of the reasons for which spectral theory on infinite dimensional Banachspaces is far more difficult than in the finite dimensional case is that the followingresult, whose proof we leave as an exercise, is false in the infinite dimensionalcase.

Proposition 2.4.2 Suppose that B is finite dimensional, and that A is a linearoperator defined on B. The following hold:

(i) A is bounded;

(ii) A is invertible if and only if A is injective if and only if A is surjective;

(iii) Ran(A) is dense if and only if Ran(A) = B;

(iv) if A is invertible, then A−1 is bounded.

Proof. Exercise. 2

Consider the operator A defined just below Definition 2.1.2. There exists a(purely algebraic) extension of A to a linear operator on B1. Note that thisextension cannot be bounded for we have proved in Exercise 2.1.6 that A doesnot admit a bounded extension to all of B.

This is an example of an unbounded linear operator which is defined every-where.

Exercise 2.4.3 Find counterexamples of statements (ii)-(iv) in the case ofoperators on infinite dimensional Banach spaces, by looking at operators L, R,defined on `2 by

L(x1, x2, x3, . . .) = (x2, x3, . . .) R(x1, x2, x3, . . .) = (0, x1, x2, x3, . . .),

and C, defined on `1 by

C(x1, x2, x3, . . .) =(x1,

x22,x33, . . .

).

1See, for instance, p. 38 of A.E. Taylor, D.C. Lay, Introduction to functional analysis, JohnWiley & Sons, 1980.


Definition 2.4.4 Suppose that A is a closed operator on B. We say that apoint ζ in σ(A) is in the point spectrum of A if ζ is an eigenvalue of A. Wesay that a point ζ in σ(A) is in the residual spectrum of A if it is not aneigenvalue and Ran(ζI −A) is not dense in B.

The main reason for introducing the residual spectrum of an operator is thatself adjoint operators have empty residual spectrum.

The spectrum of an operator is very sensitive to the domain of the operator asthe following examples show.

Denote by I the interval (0, 1) and consider the following four operators alldefined on dense subspaces of L2(I):

Dom(A1) = W 1,2(I) A1f = i f ′

Dom(A2) = f ∈W 1,2(I) : f(0) = 0 A2f = i f ′

Dom(A3) = f ∈W 1,2(I) : f(0) = f(1) A3f = i f ′

Dom(A4) = f ∈W 1,2(I) : f(0) = 0 = f(1) A4f = i f ′.

Exercise 2.4.5 Prove the following:

(i) the operators A1, . . . , A4 are closed;

(ii) the spectrum of A1 is C and it is entirely point spectrum.

(iii) the spectrum of A2 is empty;

(iv) the spectrum of A3 is the set 2πk : k ∈ Z and it is entirely pointspectrum;

(v) the spectrum of A4 is C and it is entirely residual spectrum. In particular,Ran(ζI −A) has codimension one for each ζ in C.

Hint: the following result may be helpful.

Proposition 2.4.6 Let Ω be an open interval in R, a in C∞(Ω), f in C(Ω).Suppose that u is in D′(Ω) and solves u′ + au = f in D′(Ω), i.e., in the senseof distributions. Then u is in C1(Ω), hence u is a classical solution.

A similar result holds for higher order linear equations with smooth coefficientsand continuous data. If the datum is in L2(I), then it is not hard to show thatthe unique solution to u′ + au = f is given by the usual variation of constantsformula.

2.5 Lecture X: vector valued analytic functions

Complex numbers appear naturally in the spectral theory of linear operators.The purpose of this lecture is to illustrate the basic definitions and propertiesof analytic functions with values in a Banach space. The main difference withthe scalar case is that there may be several interesting topologies on a singleBanach space; this reflects in the fact that there are several a priori differentdefinitions of analytic function. We denote by B a complex Banach space.

2.5. VECTOR VALUED ANALYTIC FUNCTIONS 33

Definition 2.5.1 Suppose that D is an open subset of C. We say that afunction f : D → B is strongly analytic in D if for every ζ in D there existsan element, which we denote by f ′(ζ), of B such that

limh→0

∥∥∥f(ζ + h)− f(ζ)

h− f ′(ζ)

∥∥∥ = 0.

In this case f ′(ζ) is called the strong derivative of f at ζ.

Definition 2.5.2 Suppose that D is an open subset of C. We say that afunction f : D → B is weakly analytic in D if for every F in B′ the functionF f is an analytic function in D.

It is straightforward to check that a strongly analytic function with values inB is weakly analytic. It is a surprising fact, proved by N. Dunford, that theconverse statement hold.

Theorem 2.5.3 Suppose that D is an open subset of C, and that f : D → B.The following are equivalent:

(i) f is strongly analytic in D;

(ii) f is weakly analytic in D.

Proof. The proof that (i) implies (ii) is straightforward, and is left to the reader.

Now we prove that (ii) implies (i). Fix ζ in D. For every z in B(ζ, r), themap F 7→ F (f(z)) is a linear functional `z on B′ of norm ‖f(z)‖. For a fixed Fin B′

|〈F, `z〉| = |F (f(z))|≤ supz∈B(ζ,r)

|F (f(z))|

<∞,because the function F f is continuous in D and bounded on B(ζ, r), for r <d(ζ,Dc). By the Banach–Steinhaus Theorem, the functionals `z : z ∈ B(ζ, r)have uniformly bounded norms, i.e.,

M := supz∈B(ζ,r)

‖f(z)‖ <∞.

We have proved that f is locally bounded.

Next we prove that f is continuous. By the Cauchy integral formula

(F f)(z)− (F f)(ζ) =1

2πi

∫γ

(F f)(w)[ 1

w − z− 1

w − ζ

]dw

=z − ζ2πi

∫γ

(F f)(w)

(w − z) (w − ζ)dw,

where z is in B(ζ, r/2) and γ denotes the positively oriented boundary of B(ζ, r).Therefore

|(F f)(z)− (F f)(ζ)| ≤ |z − ζ|2π

supw∈γ|(F f)(w)| 2πr

r2/4

≤ 4

r|z − ζ|M ‖F‖ ∀F ∈ B′.


Now, choose F such that ‖F‖ = 1 and∣∣F (f(z)− f(ζ))∣∣ = ‖f(z)− f(ζ)‖.

Then we conclude that f is continuous.

We claim that f is representable via a vector valued Cauchy integral formula.Observe that the integral ∫

γ

f(w)

w − ζdw

makes sense, because f is continuous, and that for each F in B′

F(∫

γ

f(w)

w − ζdw)

=

∫γ

(F f)(w)

w − ζdw.

Now, the right hand side is equal to 2πi (F f)(ζ), because F f is analytic, sothat

F( 1

2πi

∫γ

f(w)

w − ζdw − f(ζ)

)= 0 ∀F ∈ B′.

This implies1

2πi

∫γ

f(w)

w − ζdw − f(ζ) = 0,

as claimed.

From this formula it is straightforward to deduce, as in the classical case,that

f ′(ζ) =1

2πi

∫γ

f(w)

(w − ζ)2dw,

which proves the required strong analyticity of f in D. 2

Proposition 2.5.4 Suppose that X is a complex Banach space, that D is anopen subset of C, and that f : D → L(X ). The following are equivalent:

(i) for each x in X and each X in X ′ the function ζ 7→ X(f(ζ)x) is analyticin D;

(ii) for each x in X the map ζ 7→ f(ζ)x is strongly analytic from D to X ;

(iii) f is strongly analytic from D to L(X ).

Proof. We leave the proof as an exercise for the reader. 2

2.6 Lecture XI: elementary properties of thespectrum: II

In this lecture we prove some elementary properties of the spectrum of a linearoperator. In particular, we shall prove that the resolvent set is open, whence thespectrum is closed, and that the resolvent operator is analytic on the spectrumwith values in L(B).

2.6. ELEMENTARY PROPERTIES OF THE SPECTRUM: II 35

Theorem 2.6.1 Suppose that A is a linear operator on B. The following hold:

(i) if σ(A) is not all of C, then A is closed;

(ii) the resolvent set %(A) is open. Furthermore, if ζ is in %(A) , then the ballB(ζ, |||R(ζ;A)|||−1) is contained in %(A), and

R(ζ + w;A) =

∞∑k=0

(−w)k R(ζ;A)k+1 (2.6.1)

for each w in B(0, |||R(ζ;A)|||−1);

(iii) the map ζ 7→ R(ζ;A) is a L(B) valued analytic function;

(iv) the resolvent operators satisfies the following identities:

R(ζ;A)−R(ω;A) = (ω − ζ)R(ζ;A)R(ω;A)

R(ζ;A)R(ω;A) = R(ω;A)R(ζ;A)

d

dζR(ζ;A) = −R(ζ;A)2 ∀ζ, ω ∈ %(A);

(v) if A is bounded, then σ(A) is contained in B(0, |||A|||).

Proof. For notational simplicity, in this proof we shall write Rz instead ofR(z;A).

First we prove (i). Suppose that ζ is in %(A). The operator ζI − A isclosed, because it is the inverse of the continuous, hence closed operator, R(ζ;A).Clearly A = A− ζI + ζI, so A is the sum of a closed operator and a continuousoperator, hence it is closed by Corollary 2.3.6.

To prove (ii), suppose that ζ is in %(A), and consider the series

∞∑k=0

(−w)k Rk+1ζ .

We claim that this series is convergent in the uniform topology of operators forall w in B(0, |||Rζ |||−1). Indeed,

∞∑k=0

∣∣∣∣∣∣(−w)k Rk+1ζ

∣∣∣∣∣∣ ≤ ∞∑k=0

|w|k∣∣∣∣∣∣Rζ∣∣∣∣∣∣k+1

,

which is clearly convergent in B(0, |||Rζ |||−1).

Denote by Cw the sum of the series above. We shall prove that

Cw (ζ + w −A) = IDom(A) and (ζ + w −A)Cw = I. (2.6.2)

The first equality implies that ζ+w−A is injective, and the second that ζ+w−Ais surjective. Thus, ζ + w − A is a one to one mapping from Dom(A) onto B,


with inverse Cw. Therefore ζ + w is in %(A) and Rζ+w = Cw. It remains toprove the equalities above. The first is straightforward:

Cw (ζ + w −A) = Cw (ζ −A) + wCw

=

∞∑k=0

(−w)k Rk+1ζ (ζ −A) +

∞∑k=0

w (−w)k Rk+1ζ

= IDom(A).

To prove the second, we first prove that

(ζ −A)

∞∑k=0

(−w)k Rk+1ζ =

∞∑k=0

(−w)k (ζ −A)Rk+1ζ .

This is not obvious, because A is not assumed to be continuous. Since ζ is in%(A), A is closed by (i), hence ζ − A is closed. Fix f in B, and denote by fNthe element of B, defined by

fN =

N∑k=0

(−w)k Rk+1ζ f.

Clearly fN is in Dom(A), limN→∞ fN = f∞, and

limN→∞

(ζ −A)fN = limN→∞

N∑k=0

(−w)k (ζ −A)Rk+1ζ

=

∞∑k=0

(−w)k Rkζ .

It follows that f∞ is in Dom(A) and

(ζ −A)f∞ =

∞∑k=0

(−w)k Rkζ ,

as required.

Now we prove the second equality in (2.6.2). Observe that

(ζ + w −A)Cw = (ζ −A)Cw + wCw

=

∞∑k=0

(−w)k (ζ −A)Rk+1ζ +

∞∑k=0

w (−w)k Rk+1ζ

= I.

The proof of (ii) is complete.

From (2.6.1), we deduce that the mapping ζ 7→ Rζ is weakly analytic, hencestrongly analytic, and (iii) is proved.

To prove (iv), observe that

Rζ −Rw = Rζ (w −A)Rw −Rw=[Rζ (w −A)− I

]Rw

=[Rζ (w − ζ + ζ −A)− I

]Rw

= (w − ζ)Rζ Rw,

(2.6.3)

2.7. CLOSABLE OPERATORS 37

and the first required formula is proved. By exchanging the roles of ζ and w wefind

Rw −Rζ = (ζ − w)Rw Rζ ,

which, together the equality above, yields

0 = (w − ζ)[RζRw −RwRζ

],

which, for ζ 6= w, implies that Rζ and Rw commute. Finally, (2.6.3) impliesthat

Rw −Rζw − ζ

= −Rw Rζ .

Now we let w tend to ζ and obtain the third of the required formulae.

Finally, to prove (v), we first consider the Neumann series

∞∑k=0

Ak

zk+1.

It is straightforward to check that if |z| > |||A|||, then this series is convergent inthe uniform topology of operators. By arguing much as in the proof of (ii), wemay show that the sum of the Neumann series is Rz, thereby proving that z isin %(A). Clearly this implies that σ(A) is contained in B(0, |||A|||), as required.2

By Theorem 2.6.1 (i), the spectral theory of linear operators that are not closedis not very interesting. This is the reason for which we dedicate so much careto closed operators.

2.7 Lecture XII: closable operators

In practice, differential operators are defined on quite small subspaces of a Ba-nach space B, and, usually, they are not closed operators. This raises thequestion whether such operators admit a closed extension. Those which do arecalled closable and their smallest possible extension is called their closure. Thepurpose of this lecture is to study the closability of operators and the closure ofclosable operators.

Definition 2.7.1 An operator A is called closable if it admits at least oneclosed extension.

There are operators which are not closable. This is due to the fact that GA isnot always the graph of an operator: see Proposition 3.2.8 below.

Proposition 2.7.2 Suppose that A is a closable linear operator. Then thereexists a closed extension A which is minimal, i.e., if B is a closed extension ofA, then B ⊃ A.

Proof. Suppose that A1 is a closed extension of A. We show that GA1⊇ GA.


Indeed, let (f, g) be in GA. Then there exists a sequence fn in Dom(A)such that

fn → f and Afn → g

as n tends to ∞. Since A1 extends A and A1 is closed, f si in Dom(A1) andg = Af , so that (f, g) is in GA1 , as required.

Define A as the restriction of A1 to the completion B1 of Dom(A) withrespect to the norm ∥∥f∥∥

1

′=∥∥f∥∥ +

∥∥A1f∥∥ .

Then its graph is GA and it is the minimal closed extension of A. We leave theverification of these facts to the reader.

2

Observe that given a linear operator A, the space Dom(A), endowed with thenorm

‖f‖1′ = ‖f‖ + ‖Af‖,

always admits an abstract completion Dom(A). The point is that this abstractcompletion can be embedded in B if and only if A is closable.

Definition 2.7.3 Suppose that A is a closable linear operator. Its minimalclosed extension, constructed in Proposition 2.7.2, is called the closure of Aand is denoted by A.

Suppose that A is closable. The proof of Proposition 2.7.2 shows that A mayalso be characterised as follows: an element f in B is in the domain of A if andonly if there exists a sequence fn in Dom(A) such that

limn→∞

fn = f

and Afn is convergent, to g say. In this case Af = g.

Proposition 2.7.4 If A is closable, then GA = GA.

Proof. On the one hand GA ⊃ GA, because A is an extension of A. HenceGA ⊃ GA, for GA is closed. On the other hand, if (f,Af) is in GA, then thereexists a sequence fn in Dom(A) such that

limn→∞

fn = f and limn→∞

Afn = Af.

Then clearly (f,Af) is the limit in B × B of the sequence (fn, Afn), whichbelongs to GA. Hence (f,Af) is in GA, as required. 2

Exercise 2.7.5 Suppose that A is an operator on B. The following are equiv-alent:

(i) A is closable;

(ii) GA is a graph;

2.7. CLOSABLE OPERATORS 39

(iii) if (0, g) is in GA, then g = 0.

A very important class of closable operators is that of symmetric operators onHilbert spaces.

Definition 2.7.6 Suppose that A is a linear operator on a Hilbert space H.We say that A is symmetric if

(Af, g) = (f,Ag) ∀f, g ∈ Dom(A).

Exercise 2.7.7 Suppose that A is a symmetric operator on H. Prove thefollowing:

(i) for every f in Dom(A)

‖Af ± if‖2 = ‖Af‖2 + ‖f‖2;

(ii) prove that the operators A± iI are one to one.

Proposition 2.7.8 Suppose that A is a symmetric densely defined operator ona Hilbert space H. Then A is closable.

Proof. By Exercise 2.7.5 it suffices to prove that if (0, g) is in GA, then g = 0.Suppose that (0, g) is in GA. Then there exists a sequence fn in Dom(A) suchthat

limn→∞

fn = 0 and limn→∞

Afn = g.

We prove that g is orthogonal to Dom(A). Indeed,

(ϕ, g) = limn→∞

(ϕ,Afn)

= limn→∞

(Aϕ, fn)

= 0 ∀ϕ ∈ Dom(A),

where we have used the symmetry of A in the penultimate equality. Then g isorthogonal to the closure of Dom(A), hence to H. Therefore g = 0, as required.2

Exercise 2.7.9 Determine the closure of the Laplacian ∆ with domain C∞c (Rn).

Exercise 2.7.10 Prove that the operator HD, defined just above Defini-tion 2.1.3, is not closed. Prove that HD is closable, and determine its closure.

Chapter 3

Self adjoint operators

In this chapter we shall consider only linear operators on Hilbert spaces.

3.1 Lecture XIII: the adjoint operator I

The purpose of this lecture is to introduce the concept of (Hilbert space) adjointof a densely defined linear operator, and study some of its basic properties.In this lecture we always assume that A is a densely defined linear operator ona complex Hilbert space H.

Definition 3.1.1 The adjoint of a densely defined linear operator A on H isthe operator A∗, defined by

Dom(A∗) = g ∈ H : ∃γ in H s.t. (Af, g) = (f, γ) for all f in Dom(A)A∗g = γ ∀g ∈ Dom(A∗).

The definition ofA∗g makes sense only if γ is uniquely determined by the elementg in Dom(A). Suppose that γ1 and γ2 are elements in H such that

(f, γ1) = (Af, g) = (f, γ2) ∀f ∈ Dom(A).

Then (f, γ1 − γ2) = 0 for all f in Dom(A). Since Dom(A) is dense in H, wemay conclude that γ1 − γ2 = 0, as required.

On the contrary, if Dom(A) is not dense, and there exists a constant C suchthat

|(Af, g)| ≤ C ‖f‖ ∀f ∈ Dom(A),

then the functional f 7→ (Af, g) admits many extensions to bounded functionalson H, and there is no way to define A∗g.

Note that if f is in Dom(A) and g is in Dom(A∗), then

(Af, g) = (f,A∗g).

Remark 3.1.2 The following are equivalent:

41

42 CHAPTER 3. SELF ADJOINT OPERATORS

(i) g is in Dom(A∗);

(ii) there exists a constant Cg such that∣∣(Af, g)∣∣ ≤ Cg ‖f‖ ∀f ∈ Dom(A).

We note that g is in Dom(A∗) if and only if the linear functional f 7→ (Af, g),initially defined on Dom(A), extends to a bounded linear functional on H. Inthis case, the functional will be represented by an element γ in H, by the RieszRepresentation Theorem.

Exercise 3.1.3 Suppose that H is finite dimensional, and that MA is thematrix that represents the linear operator A with respect to some basis. Thenthe matrix MA∗ which represents the adjoint operator A∗ with respect to the

same basis is the conjugate transpose matrix MT

A.

Denote by I the interval [0, 1]. Suppose that K is in L2(I × I), and considerthe integral operator A on L2(I) with kernel K, i.e.,

Af(x) =

∫ 1

0

K(x, y) f(y) dy ∀f ∈ L2(I).

It is not hard to check that A is bounded on L2(I) with norm ≤ ‖K‖L2(I×I),and that the adjoint operator A∗ is the integral operator with kernel (x, y) 7→K(y, x).

Exercise 3.1.4 Prove that the operator A defined above is compact. Hint:prove that A is the limit in the operator norm of a sequence of operators offinite rank.

Exercise 3.1.5 Consider the convolution operator T on L2(R), defined by

Tf = f ∗ ϕ ∀f ∈ L2(R),

where ϕ is in L1(R). Prove that T is bounded on L2(R) and find T ∗.

Proposition 3.1.6 Suppose that the multiplication operator Dom(Am) (see Def-inition 2.1.7 for the notation) is densely defined in L2(M). Then the adjoint ofAm is the operator Am.

Proof. First we prove that Dom(A∗m) = Dom(Am). A function g is in Dom(A∗m)if and only if the linear functional f 7→

∫Mmf g dµ, initially defined on Dom(Am),

extends to a bounded linear functional on L2(M). By the Riesz RepresentationTheorem, this happens if and only if there exists γ in L2(M) such that∫

M

mf g dµ =

∫M

f γ dµ ∀f ∈ Dom(Am).

Since Dom(Am) is dense in L2(M), γ = mg, whence g must be in Dom(Am).

Now suppose that g is in Dom(A∗). Then

(Amf, g) =

∫M

mf g dµ =

∫M

f mg dµ = (f,Amg) ∀f, g ∈ Dom(Am).

Therefore A∗m = Am, as required. 2

3.1. THE ADJOINT OPERATOR I 43

Exercise 3.1.7 Compute the adjoint of ∆ on L2(Rn); here ∆ is the standardLaplacian, defined on C∞c (Rn). Then compute the adjoint of ∆.

Exercise 3.1.8 Consider the left shift operator L on `2, defined by

L(x1, x2, . . .) = (x2, x3, . . .)

for every sequence xn in `2. Compute L∗. Do the same for the right shiftoperator R, defined by

R(x1, x2, . . .) = (0, x1, x2, . . .).

Exercise 3.1.9 Prove that if A is bounded on H, then A∗ is bounded on Hand |||A||| = |||A∗|||. Prove also that (A∗)∗ = A.

In view of the exercise above, ∗ is an isometry of L(H). The following propositionsummarises the main properties of the ∗ operator.


(i) ∗ is a conjugate linear isometric isomorphism of L(H);

(ii) ∗ is involutive;

(iii) (ST )∗ = T ∗ S∗ for all S and T in L(H);

(iv) if A and A−1 are in L(H), then A∗ is invertible in L(H) and (A∗)−1 =(A−1)∗;

(v) |||A∗A||| = |||A|||2 for every A in L(H).

Proof. The proof of (i)-(iv) is left to the reader.

To prove (v) observe that, on the one hand,

|||A∗A||| ≤ |||A∗||| |||A||| = |||A|||2,

where we have used the Exercise 3.1.9. On the other hand

|||A|||2 = sup‖f‖=1

‖Af‖2

= sup‖f‖=1

|(Af,Af)|

= sup‖f‖=1

|(f,A∗Af)|

≤ |||A∗A|||,

and the required equality follows. 2

The properties of the involution ∗ contained in this proposition and the fact thatL(H) is complete are usually summarised by saying that L(H) is a C∗-algebra(with involution ∗).


Definition 3.1.11 A linear operator on H is said to be unitary if

U∗U = UU∗ = I.

Exercise 3.1.12 Prove that the following are equivalent:

(i) U is unitary;

(ii) Ran(U) = H and (Uf,Ug) = (f, g) for every f and g in H.

(iii) Ran(U) = H and ‖Uf‖ = ‖f‖ for every f in H;

Compare with the analogous result in the finite dimensional case (Exercise 1.4.3).An operator U such that ‖Uf‖ = ‖f‖ for every f in H is called an isometry.Construct an isometry on `2 which is not onto.

3.2 Lecture XIV: the adjoint operator II

The purpose of this lecture is to study some additional properties of the adjointoperator, and to introduce the notion of self adjoint operator. The propertiesof self adjoint operators will be the object of subsequent lectures.

Definition 3.2.1 Suppose that A is a densely defined operator on a complexHilbert space H, and that Dom(A∗) is dense in H. We may then define theoperator (A∗)∗, which we denote by A∗∗. We may then proceed recursively todefine the operators A∗∗∗, A∗∗∗∗ and so forth.

The following theorem contains information concerning the relationship betweenthe closure of an operator and its adjoint. We shall make repeated use of theoperator V : H×H → H×H, defined by

V (f, g) = (−g, f) ∀f, g ∈ H. (3.2.1)

It is straightforward to check that V is a surjective isometry of H × H. As aconsequence, for every subspace E of H×H the following holds:

V (E⊥) =(V (E)

)⊥. (3.2.2)

We leave the easy verification of this fact to the reader.

We shall also need the following property.

Exercise 3.2.2 Suppose that U is a unitary operator on the Hilbert space H,and that M is a (possibly nonclosed) subspace of H. Then U(M) = (UM).

Theorem 3.2.3 Suppose that A is a densely defined operator on H. The fol-lowing hold:

(i) GA∗ = (V GA)⊥;

(ii) A∗ is a closed operator;

3.2. THE ADJOINT OPERATOR II 45

(iii) A is closable if and only if Dom(A∗) is dense in H. In this case A = A∗∗;

(iv) if A is closable, then (A)∗ = A∗.

Proof. First we prove (i). The point (f, g) is in V (GA)⊥ if and only if

(Aϕ, f) = (ϕ, g) ∀ϕ ∈ Dom(A).

This is equivalent to the fact that f is in Dom(A∗) and that A∗f = g, i.e., that(f, g) is in GA∗ , as required.

Now (ii) follows directly from (i) and the fact that the orthogonal comple-ment of a subspace of an Hilbert space is closed.

Next we prove (iii). First we shall prove that if Dom(A∗) is dense in H, thenA is closable. By Exercise 2.7.5, it suffices to prove that GA is a graph; we shallprove that it is the graph of A∗∗, which is well defined by the hypothesis thatDom(A∗) be dense in H. Observe that

GA =(G⊥A)⊥

(because V 2 = −I) =((V 2GA)⊥)⊥

=((V (V GA))⊥)⊥

(because V is unitary) =(V (V GA)⊥)⊥

=(V GA∗)

⊥

(A∗∗ is defined for A∗ is densely defined) = GA∗∗ ,

as required.

To prove that if A is closable, then Dom(A∗) is dense, suppose that ψ is inDom(A∗)⊥. Observe that (ψ, 0) is in (GA∗)

⊥. Indeed, for every ϕ in Dom(A∗)((ϕ,A∗ϕ), (ψ, 0)

)= (ϕ,ψ) = 0,

as required.

But then (ψ, 0) is in V GA by (i). Now,

V GA = V GA = V GA,

(the first equality follows from Exercise 3.2.2 and the last equality from theformula GA = GA) i.e., (0, ψ) is in GA. Since A is a linear operator, ψ = 0,hence Dom(A∗) is dense, as required.

Finally, we prove (iv). By (i), GA∗ = (V GA)⊥ and G(A)∗ = (V GA)⊥. Thus,

it suffices to prove that (V GA)⊥ = (V GA)⊥. But

(V GA)⊥ = (V GA)⊥ = (V GA)⊥ = (V GA)⊥ = GA∗ .

Since A∗ is closed by (ii), GA∗ = GA∗ , as required. 2

Exercise 3.2.4 Suppose that A is densely defined. Prove the following:

(i) Ker(A∗) = Ran(A)⊥ ∩Dom(A∗);


(ii) if A is closed, then Ker(A) = Ran(A∗)⊥ ∩Dom(A).

Proposition 3.2.5 Suppose that A is a densely defined operator in H. ThenA is symmetric if and only if A∗ ⊃ A.

Proof. If A∗ ⊃ A, then A is clearly symmetric.

Conversely, if A is symmetric, then Dom(A) ⊂ Dom(A∗) and the restrictionof A∗ to Dom(A) agrees with A. 2

Definition 3.2.6 Suppose that A is a densely defined operator in H. We saythat A is self adjoint if A∗ = A.

Observe that if A is self adjoint, then A is symmetric. The converse is false. Itmay very well happen that A∗ is a proper extension of A, as examples belowwill show.

Note that A is self adjoint if and only if A is symmetric and Dom(A∗) =Dom(A).

Corollary 3.2.7 Suppose that A is a densely defined operator on H. The fol-lowing hold:

(i) if A is symmetric, then A is closable and A = A∗∗;

(ii) if A is self adjoint, then A is closed.

Proof. The proof is left as an exercise. 2

We conclude this lecture with an example of a densely defined operator A suchthat Dom(A∗) = 0.

Proposition 3.2.8 Denote by en and orthonormal basis of the separableHilbert space H and by xn a sequence, which is dense in H. Define A onen by

Aen = xn ∀n ∈ N,

and extend A on Dom(A) := line1, e2, e3, . . . by linearity. The following hold:

(i) the graph of A is dense in H×H;

(ii) Dom(A∗) = 0.

Proof. To prove (i) suppose that (ϕ,ψ) is in G⊥A. We shall prove that ϕ = 0 = ψ.

Indeed, the statement (ϕ,ψ) in G⊥A implies((ϕ,ψ), (ej , Aej)

)= 0 ∀j ∈ N,

i.e.,(ϕ, ej) = −(ψ,Aej).

3.3. SELF ADJOINT OPERATORS 47

By definition of A the right hand side is equal to −(ψ, xj). Choose a subsequencejk such that xjk tends to ψ: then −(ψ, xjk) = ϕ(jk), and a limiting argumentshows that

limk→∞

ϕ(jk) = −‖ψ‖2.

By the Riemann–Lebesgue lemma we may conclude that ‖ψ‖ = 0, i.e., thatψ = 0. But then (ϕ, ej) = 0 for all j, so that ϕ = 0.

To prove (ii) we use the fact, proved above, that GA∗ =(V (GA)

)⊥. Since

GA is dense, so is V (GA), hence its orthogonal complement is (0, 0), i.e., theonly point in GA∗ is the origin. Hence Dom(A∗) = 0, as required. 2

3.3 Lecture XV: self adjoint operators

The purpose of this lecture is to study the basic properties of self adjoint oper-ators.

The definition of self adjoint operator is very subtle, and, in general, it is hardto check directly from the definition that a given operator is self adjoint. Thefollowing characterisation is often useful.

Theorem 3.3.1 (Basic criterion of self adjointness) Suppose that A is adensely defined symmetric operator on H. The following are equivalent:

(i) A is self adjoint;

(ii) A is a closed operator and Ker(A∗ + iI) = 0 = Ker(A∗ − iI);

(iii) Ran(A+ iI) = H = Ran(A− iI).

Proof. First we prove that (i) implies (ii). Since A = A∗ by definition of selfadjoint operator, A is closed, because so is A∗. Suppose that f is in Ker(A∗−iI).Then f is in Dom(A∗), and

A∗f = i f.

Thus,(f,A∗f) = (f, i f),

so that(Af, f) = (−i f, f)

by the definition of the adjoint operator and properties of inner products. ButA = A∗, so that (Af, f) = i (f, f). This, combined with the previous formula,implies that (f, f) = 0, i.e., f = 0, as required.

Next we prove that (ii) implies (iii). We shall prove that Ker(A∗ + iI) = 0implies that Ran(A − iI) = H. The proof that Ker(A∗ − iI) = 0 implies thatRan(A+ iI) = H is similar and is omitted.

Step I: Ran(A− iI) is dense. Suppose that g is in Ran(A− iI)⊥. We mustshow that g = 0. Indeed, g is in Ran(A− iI)⊥ if and only if

(Aϕ, g) = i (ϕ, g) ∀ϕ ∈ Dom(A).


This implies that g is in Dom(A∗) and that

(ϕ,A∗g) = i (ϕ, g) ∀ϕ ∈ Dom(A).

Since Dom(A) is dense in H, we may conclude that A∗g = −i g, so that g is inKer(A∗ + iI). Then g = 0 by assumption, as required.

Step II: Ran(A− iI) is closed. Since A is a symmetric operator,

‖Af − if‖2 = ‖Af‖2 + ‖f‖2

≥ ‖f‖2.(3.3.1)

Suppose that fn is a sequence in Dom(A) such that (A−iI)fn is convergent.We shall prove that its limit, g say, is still in Ran(A − iI). By (3.3.1), thesequence fn is a Cauchy sequence; therefore fn is convergent, say to f .Since A is closed, f is in Dom(A), and Afn tends to Af , so that

g = limn→∞

(A− iI)fn = Af − if,

i.e., g is in Ran(A− iI), hence Ran(A− iI) is closed.

Step I and Step II imply that Ran(A− iI) = H, as required to conclude theproof of the implication (ii) implies (iii).

Finally, we prove that (iii) implies (i). Working backwords in Step I above,we may easily check that

Ran(A∓ iI) dense =⇒ Ker(A∗ ± iI) = 0.

Since A is symmetric, A∗ ⊃ A. Thus, we need to show that Dom(A∗) ⊆Dom(A). Suppose that g is in Dom(A∗). Observe that Dom(A∗) = Dom(A∗ −iI). Since Ran(A− iI) = H, there exists f in Dom(A) such that

(A− iI)f = (A∗ − iI)g.

We shall prove that f = g. Observe that f is in Dom(A∗), hence f − g is inDom(A∗), and

(A∗ − iI)(f − g) = 0.

This implies f − g = 0, because Ker(A∗ − iI) = 0, as required. 2

Exercise 3.3.2 Prove that if A is a symmetric densely defined operator suchthat %(A) ∩ R 6= ∅, then A is self adjoint.

Suppose that s is in %(A) ∩ R 6= ∅. Then for ε small enough, the disc B(s, 2ε)is contained in %(A) (because %(A) is open). The operator

As,ε :=2

ε(A− sI)

satisfies Theorem 3.3.1 (iii), hence it is self adjoint.

A direct consequence of Theorem 3.3.1 is that ±i are in the resolvent set of anyself adjoint operator. A slight variation of some of the arguments used in theproof of Theorem 3.3.1 give the following sharp result.


Proposition 3.3.3 Suppose that A is a self adjoint operator. Then every com-plex number ζ with Im ζ 6= 0 is in the resolvent set of A and

|||R(ζ;A)||| ≤ |Im(ζ)|−1.

Proof. First we prove that if Im(ζ) 6= 0, then ζI −A is injective. Indeed,

‖Af − ζf‖2 = ‖Af‖2 − 2(Re(ζ)

)(Af, f) + |ζ|2 ‖f‖2

≥ ‖Af‖2 − 2|Re(ζ)| ‖Af‖ ‖f‖ + |ζ|2 ‖f‖2

≥ −|Re(ζ)|2 ‖f‖2 + |ζ|2 ‖f‖2

≥ |Im(ζ)|2 ‖f‖2 ∀f ∈ Dom(A).

(3.3.2)

By arguing as in the proof of the implication (ii) implies (iii), we may showthat ζI −A injective implies Ran(ζI −A) = H. Note that ζI −A is closed, forA is self adjoint, hence closed, and ζI is bounded (see Corollary 2.3.6). ThenζI − A is invertible, with continuous inverse (recall that A is closed). Then(3.3.2) implies the required norm estimate. 2

Corollary 3.3.4 A closed symmetric densely defined operator A is self adjointif and only if σ(A) ⊂ R.

Proof. First suppose that A is self adjoint. By Proposition 3.3.3 every point inC \ R is contained in the resolvent set, hence σ(A) ⊂ R, as required.

Conversely, suppose that σ(A) is contained in R. Then the points ±i are inthe resolvent set of A, so that A is self adjoint by Theorem 3.3.1. 2

The next result was announced in Lecture 2.4, and states that self adjointoperators have empty residual spectrum, i.e., if ζ is in σ(A), and it is not aneigenvalue, then Ran(ζI −A) is dense in H.

Proposition 3.3.5 Suppose that A is a self adjoint operator on H. Then theresidual spectrum of A is empty.

Proof. Suppose that ζ is in σ(A) and it is not an eigenvalue, and pick ψ in theorthogonal complement of Ran(ζI −A). Then

(ζϕ−Aϕ,ψ) = 0 ∀ϕ ∈ Dom(A).

This is equivalent to saying that

ζ (ϕ,ψ) = (Aϕ,ψ) ∀ϕ ∈ Dom(A).

By the definition of the adjoint operator, we see that ψ is in Dom(A∗), whichagrees with Dom(A), because A is self adjoint. Then

ζ (ϕ,ψ) = (ϕ,Aψ) ∀ϕ ∈ Dom(A).

Since Dom(A) is dense in H, we may conclude that Aψ = ζ ψ (recall that ζmust be real). Since ζ is not an eigenvalue by assumption, ψ = 0, as required.2


Proposition 3.3.6 Suppose that A is a self adjoint operator, and that ζ is acomplex number with Im(ζ) 6= 0. Then

R(ζ;A)∗ = R(ζ;A).

Proof. First we show that R(ζ;A)∗ maps H into Dom(A). Indeed, for every ηin Dom(A)

(R(ζ;A)∗ξ, (ζI −A)η) = (ξ,R(ζ;A)(ζI −A)η) = (ξ, η),

because R(ζ;A) is a left inverse of ζI − A. Thus, the linear functional η 7→(R(ζ;A)∗ξ, (ζI − A)η) extends to a continuous linear functional on H. HenceR(ζ;A)∗ξ is in Dom((ζI −A)∗), i.e., in Dom(A), and

((ζI −A)R(ζ;A)∗ξ, η) = (ξ, η) ∀ξ ∈ H ∀η ∈ Dom(A).

Therefore (ζI −A)R(ζ;A)∗ = IH.

Next, observe that for every ξ in Dom(A) and every η in H

(R(ζ;A)∗(ζI −A)ξ, η) = ((ζI −A)ξ,R(ζ;A)η)

= (ξ, (ζI −A)R(ζ;A)η)

= (ξ, η),

whence R(ζ;A)∗(ζI − A) = IDom(A). Thus, R(ζ;A)∗ is a two sided inverse of

ζI −A, hence it agrees with R(ζ;A), as required. 2

By using Proposition 3.3.3 it is easy to find closed symmetric densely definedoperators which are not self adjoint. The operator A4 in Exercise 2.4.5 is a closedsymmetric densely defined operator, but it is not self adjoint because σ(A4) = C.Note that the operator A1 in the same exercise is even not symmetric.

Exercise 3.3.7 Are the operators A2 and A3 in Exercise 2.4.5 self adjoint?Prove also that A∗1 = A4 and that A∗4 = A1. This gives another proof of thefact that A1 and A4 are not self adjoint. For which ζ in C the equation

if ′ − ζf = g

has a unique solution ϕ in W 2,1(I) with ϕ(0) = ϕ(1) for any g in L2(I)?

Theorem 3.3.8 Consider the multiplication operator Am on a σ-finite measurespace (M,M, µ). Prove the following:

(i) Am is symmetric if and only if m is real;

(ii) if m is real, then Am is self adjoint;

(iii) the spectrum of Am is the essential range of m. Furthermore

R(ζ;Am) = A1/(ζ−m).


Proof. To prove (i), note that (Amf, g) = (f,Amg) is equivalent to∫M

(m−m) f g dµ = 0 ∀f, g ∈ Dom(Am).

If m is real, the integral above vanishes for every f and g is Dom(Am), whenceAm is symmetric.

Conversely, if Am is symmetric, the integral above vanishes, for every f andg in Dom(Am). Set

EN := |m| ≤ N.

Note that the functions 1En and m−m1EN are in Dom(Am). Hence∫EN

|m−m|2 dµ =

∫M

(m−m) 1EN (m−m) 1EN dµ = 0.

Thus, Im(m) = 0 almost everywhere on EN . Sincem is finite almost everywhere,EN ↑M , so that Imm = 0 almost everywhere, as required.

To prove (ii), recall that Dom(A∗m) = Dom(Am) by Exercise 3.1.6. If m isreal, then Am is symmetric by (i), and the required conclusion follows.

Finally suppose that ζ is not in Raness(m). Since Raness(m) is a closed set,the function

rζ :=1

ζ −m

is in L∞(M). Denote by Tζ the operator Arζ . Clearly Tζ is bounded on L2(M)and

Tζ (ζ −Am) = IDom(Am) and (ζ −Am)Tζ = I.

Therefore ζ−Am is invertible and the inverse is Tζ . Thus ζ is in %(Am), whencethe spectrum of Am is contained in Raness(m).

Suppose now that λ is in Raness(m). We shall prove that λ is in σ(Am).For every positive integer n the set

Sn := x : |m(x)− λ| < 1/n

has positive measure. In the case where µ(Sn) =∞, we take S′n ⊂ Sn such that0 < µ(S′n) <∞. Observe that 1Sn is in Dom(Am), and that

‖(λ−Am) 1Sn‖22

=

∫Sn

|λ−m|2 dµ

≤ 1

n2µ(Sn)

=1

n2‖1Sn‖2

2.

This implies that λ − Am cannot have a bounded inverse, for otherwise 1Sn =(λ−Am)−1f for some f in L2(M) and

‖(λ−Am)−1f‖2 ≥ n ‖f‖2 ∀n.

2


It is not hard to prove that, in fact, if ζ is not in Raness(m), then

|||(ζ −Am)−1||| =1

d(ζ,Raness(m)).

Exercise 3.3.9 Prove that any nonempty closed subset E of C is the spectrumof some normal operator on a Hilbert space H. Hint: look at multiplicationoperators on L2(C) with m(z) := z1E(z) + z01Ec(z), where z0 is an arbitrarilychosen point in E.

Definition 3.3.10 Suppose that H1 and H2 are complex Hilbert spaces, andthat A1 and A2 are two operators on H1 and H2 respectively. We say that A1

and A2 are unitarily equivalent if there exists a unitary operator U : H2 → H1

such thatA1 = UA2U

−1.

One of the interesting features of unitary equivalence of operators is that itpreserves the spectrum, the symmetry, hence the self adjointness, as proved inthe following proposition.

Proposition 3.3.11 Suppose that H1 and H2 are complex Hilbert spaces, andthat A1 and A2 are two unitarily equivalent operators on H1 and H2 respectively.The following hold:

(i) A1 is symmetric if and only if A2 is;

(ii) ζ is in σ(A1) if and only if ζ is in σ(A2);

(iii) A1 is self adjoint if and only if A2 is.

Proof. First we prove (i). Suppose that A1 is symmetric. We shall prove thatA2 is symmetric. Note that ϕ is in Dom(A2) if and only if Uϕ is in Dom(A1).Pick ϕ and ψ in Dom(A2). Clearly

(A2ϕ,ψ) = (U−1A1Uϕ,ψ)

= (A1Uϕ, (U−1)∗ψ)

= (A1Uϕ,Uψ)

(A1 is symmetric) = (Uϕ,A1Uψ)

= (ϕ,U∗A1Uψ)

= (ϕ,U−1A1Uψ)

= (ϕ,A2ψ),

so A2 is symmetric. By reversing the role of A1 and A2 we obtain (i).

To prove (ii), suppose that ζ is in %(A1). We shall prove that ζ is in %(A2).Consider the operator Tζ := U−1R(ζ;A1)U (recall that R(ζ;A1) = (ζI−A1)−1).Observe that

Tζ(ζI −A2) = UR(ζ;A1)U−1(ζI −A2)UU−1 = UR(ζ;A1)(ζI −A1)U−1

= IDom(A1),


and that

(ζI −A2)Tζ = UU−1(ζI −A2)UR(ζ;A1)U−1 = U(ζI −A1)R(ζ;A1)U−1

= I,

so that ζ is in %(A2).

Finally, to prove (iii), we suppose that A1 is self adjoint. Then, A1 is closed,densely defined and symmetric. Therefore so is A2. Since A1 is self adjoint, itsspectrum is real. By (ii), so is σ(A2), whence A2 is self adjoint by Corollary 3.3.4.2

This proposition is very useful in all cases in which a unitary operator U whichintertwines A1 and A2 is easy to construct, and it is easy to show (or well known)that either A1 or A2 is self adjoint. We give an example, which is a particularcase of a class of examples studied in one of the subsequent lectures. Recall thefollowing characterisation of the Sobolev space W 1,2(R).

Proposition 3.3.12 A function f is in W 1,2(R) if and only if its Fourier trans-form satisfies the following∫ ∞

−∞|f(ξ)|2 (1 + |ξ|2) dξ <∞.

Consider the differential operator A defined by

Dom(A) = W 1,2(R), Af = i f ′ ∀f ∈ Dom(A).

It is straightforward to check that A is closed and symmetric. Note that

FAF−1 = M,

where F denotes the Fourier transform (see the beginning of Chapter 5 for itsdefinition) and M is the multiplication operator defined by

Mf(x) = −x f(x)

for all f in L2(R) such that Mf is in L2(R). Since (−x) is real, M is symmetric.From Theorem 3.3.8 we know that M is self adjoint, with spectrum equal to R.Hence A is self adjoint, with spectrum equal to R, by Proposition 3.3.11.

Exercise 3.3.13 Prove that the operator A above is self adjoint by using theBasic Criterion of Self Adjointness. Then prove the same result by showingdirectly that A∗ = A.

One of the forms of the Spectral Theorem asserts that every self adjoint operatoris unitarily equivalent to a multiplication operator on a suitable measure space.However, this is not very useful for practical purposes, and, in general, it iseasier to prove self adjointness directly.


3.4 Lecture XVI: extensions of symmetric op-erators

In this section we consider a densely defined symmetric operator A. An impor-tant issue is whether A admits any self adjoint extensions.

There are symmetric operators which do not admit self adjoint extensions. Forinstance, consider the operator A, defined by

Dom(A) = C∞c (R+) Af = i f ′ ∀f ∈ Dom(A).

It is straightforward to check that A is a symmetric operator, and that thedomain of its closure is

Dom(A) = f ∈W 1,2(R+) : f(0) = 0.

Further, A is not self adjoint, for A∗ is the operator defined by

Dom(A∗) = W 1,2(R+) A∗f = i f ′ ∀f ∈ Dom(A∗).

By using the fact that distributional solutions of the equations

f ′ − f = 0 and f ′ + f = 0

are classical solutions, we find that

Ker(A∗ − iI) = 0 and Ker(A∗ + iI) = spane−x.

By the Basic Criterion of Self Adjointness, A is not self adjoint.

Suppose that A is a self adjoint extension of A (hence of A). Then

Dom(A) ⊂ Dom(A) ⊂ Dom(A∗), (3.4.1)

with proper inclusions. Therefore, Dom(A) contains at least one function, g say,in W 1,2(R+) such that g(0) 6= 0. By symmetry, we must have

(Ag, g) = (g, Ag),

i.e., ∫ ∞0

[ig′ g − g ig′

]dλ = 0.

A direct calculation (write g = u+ iv with u and v real) shows that the integralabove is equal to −i |g(0)|2. This implies g(0) = 0, i.e., g is in Dom(A), whichcontradicts (3.4.1). Thus, no self adjoint extensions of A exist.

Definition 3.4.1 Suppose that A is a symmetric densely defined operator inH. We say that A is semibounded from below if there exists c in R suchthat

(Af, f) ≥ c (f, f) ∀f ∈ Dom(A).

Similarly, we say that A is semibounded from above if −A is semiboundedfrom below, i.e., there exists c in R such that

(Af, f) ≤ c (f, f) ∀f ∈ Dom(A).

3.4. EXTENSIONS OF SYMMETRIC OPERATORS 55

Exercise 3.4.2 Suppose that A is a self adjoint operator such that

(Af, f) ≥ 0 ∀f ∈ Dom(A).

Prove that σ(A) is contained in [0,∞) and that

|||(ζI −A)−1||| ≤ |ζ|−1 ∀ζ ∈ (−∞, 0).

Theorem 3.4.3 Suppose that A is a symmetric densely defined operator in H.If A is semibounded, then A admits at least one self adjoint extension.

Proof. Since A is symmetric, it is closable by Proposition 2.7.8, so we mayassume that A is closed.

We prove the theorem in the case where A is semibounded from below. Byreplacing A with A − (c − 1)I, we may assume that A is semibounded frombelow by 1, i.e.,

(Af, f) ≥ (f, f) ∀f ∈ Dom(A). (3.4.2)

We endow Dom(A) with the norm

‖f‖′ = (Af, f)1/2.

It is indeed a norm by (3.4.2). We denote the normed space (Dom(A), ‖ ‖′)simply by HF and its abstract completion by HF .

Step I. We claim thatHF may be identified with a subspaceH′ ofH. Indeed,the inclusion map J : HF → H satisfies the estimate

‖Jf‖ = ‖f‖ ≤ (Af, f)1/2 = ‖f‖′ ∀f ∈ Dom(A). (3.4.3)

By Exercise 2.1.5, J extends to a bounded linear operator J from HF to H. Toconclude the proof of the claim, it suffices to prove that J is injective. Supposethat J f = 0 for some f in HF . We need to show that f = 0. Pick a sequencefn in Dom(A) such that

limn→∞

∥∥fn − f∥∥HF = 0.

By (3.4.3) and the fact that J restricted to Dom(A) is the identity,

limn→∞

∥∥fn∥∥H = limn→∞

∥∥Jfn − J f∥∥H = 0. (3.4.4)

Now, let F in HF such that

(f, F )HF =∥∥f∥∥HF ,

where (·, ·)HF denotes the inner product on HF . Since Dom(A) is dense in HF ,

we may approximate F by a sequence Fm in Dom(A). Choose ν such that∣∣(f, Fν)HF

∣∣ ≥ 1

2

∥∥f∥∥HF ,whence

limn→∞

∣∣(fn, Fν)HF

∣∣ ≥ 1

2

∥∥f∥∥HF ,


Now observe that∣∣(fn, Fν)HF

∣∣ =∣∣(fn, AFν)

∣∣ ≤ ∥∥fn∥∥H ∥∥AFν∥∥H ,

which is convergent to 0 by (3.4.4). This implies that ‖f‖HF = 0, so that f = 0,

as required to conclude the proof of the injectivity of J , and of the claim. Set

H′ := J HF ‖f‖H′ = ‖J−1f‖HF .

Clearly H′ is an isometric copy of HF .

Step II. Define the operator AF as follows

Dom(AF ) = Dom(A∗) ∩H′ AF f = A∗f ∀f ∈ Dom(AF ). (3.4.5)

Now we prove that AF is symmetric. We need to show that for every ϕ and ψin Dom(AF ) we have

(AFϕ,ψ) = (ϕ,AFψ),

i.e.,(A∗ϕ,ψ) = (ϕ,A∗ψ),

Since Dom(A) is dense inH′ with respect to the norm ‖ ‖′, there exist sequencesϕn and ψn in Dom(A) that converge in the H norm to ϕ and ψ respectively.Then

(A∗ϕ,ψ) = limn→∞

(A∗ϕ,ψn)

(by the definition of A∗) = limn→∞

(ϕ,Aψn)

= limn→∞

limm→∞

(ϕm, Aψn)

(A is symmetric) = limn→∞

limm→∞

(Aϕm, ψn).

Note that we may exchange the order of the limits in the iterated limit above,because both the iterated limits are equal to lim(n,m)→∞(ϕm, ψn)′. Therefore

(A∗ϕ,ψ) = limm→∞

limn→∞

(Aϕm, ψn)

= limm→∞

(Aϕm, ψ)

= limm→∞

(ϕm, A∗ψ)

= (ϕ,A∗ψ),

as required.

Step III. We show that AF is self adjoint. Since AF is symmetric, AF ⊆ A∗F .Thus, it remains to prove that A∗F ⊆ AF . Suppose that ϕ is in Dom(A∗F ). Weneed to show that ϕ is in Dom(A∗) ∩H′. Since ϕ is in Dom(A∗F ),

(AFψ,ϕ) = (ψ,A∗Fϕ) ∀ψ ∈ Dom(AF ).

In particular,(Aψ,ϕ) = (ψ,A∗Fϕ) ∀ψ ∈ Dom(A),

for Dom(A) is contained in Dom(AF ) and AF restricted to Dom(A) agrees withA. The formula above shows that ϕ is in Dom(A∗).


It remains to show that ϕ is in H′. Note that

|(AFψ,ϕ)| = |(ψ,A∗Fϕ)|≤ ‖ψ‖H ‖A∗Fϕ‖H≤ ‖ψ‖H′ ‖A∗Fϕ‖H ∀ψ ∈ Dom(AF ).

Therefore ψ 7→ (AFψ,ϕ) extends to a continuous linear functional onH′. There-fore, there exists γ′ in H′ such that

(AFψ,ϕ) = (ψ, γ′)′ ∀ψ ∈ Dom(AF ).

A straightforward limiting argument shows that (ψ, γ′)′ = (Aψ, γ′) for every ψin Dom(A). Thus,

(Aψ,ϕ) = (Aψ, γ′) ∀ψ ∈ Dom(A).

This implies that γ′ is in Dom(A∗), hence in Dom(AF ) (because γ′ is in H′ byconstruction), and the formula above may be rewritten as follows

(ψ,A∗Fϕ) = (ψ,AF γ′) ∀ψ ∈ Dom(A).

A density argument then shows that this equality holds for every ψ in H. Inparticular, for every ψ in Dom(AF ).

(AFψ,ϕ) = (AFψ, γ′) ∀ψ ∈ Dom(AF ); (3.4.6)

we have used the symmetry of AF and the fact that both ψ and γ′ belong toDom(AF ). We now prove that Ran(AF ) is dense in H. Then (3.4.6) will implythat ϕ = γ′, i.e., that ϕ is in H′, as required to conclude the proof.

Suppose that ψ is in Dom(A) and that η is in H. The linear functionalψ 7→ (ψ, η) extends to a bounded linear functional on H′. Indeed,

|(ψ, η)| ≤ ‖ψ‖ ‖η‖≤ ‖ψ‖H′ ‖η‖.

Since Dom(A) is dense, the required extension follows. By the Riesz Represen-tation Theorem, there exists η′ in H′ such that

(ψ, η) = (ψ, η′)′ ∀ψ ∈ Dom(A).

Furthermore,(ψ, η′)′ = (Aψ, η′) ∀ψ ∈ Dom(A).

These two formulae show that η′ is in Dom(A∗), hence in Dom(AF ). Therefore

(ψ, η) = (ψ,AF η′) ∀ψ ∈ Dom(A).

Since Dom(AF ) is dense in H, we may conclude that AF η′ = η. Therefore

Ran(A) = H, as required to conclude the proof of the theorem. 2

Definition 3.4.4 Suppose that A is a semibounded symmetric operator in H(recall that A is densely defined). The self adjoint extension AF of A constructedin the proof of Theorem 3.4.3 is called the Friedrichs extension of A.


Consider the operator HD, defined just above Definition 2.1.3. Observe thatHD is semibounded from below. Indeed,

(HDf, f) =

∫ b

a

|f ′|2 dλ ≥ 0 ∀f ∈ Dom(HD).

A more refined estimate, which uses Poincare’s inequality, is the following

(HDf, f) =

∫ b

a

|f ′|2 dλ ≥ c ‖f‖22 ∀f ∈ Dom(HD),

where c depends on b−a. In particular, c = 1 when b−a = 1. Now we determinethe Friedrichs extension of HD.

We claim that Dom(H∗D) = W 2,2(I)∩W 1,20 (I), where I denotes the interval

(a, b). Indeed, if ϕ is in Dom(H∗D), then there exists a constant C such that

|(HDψ,ϕ)| ≤ C ‖ψ‖2,

so that the functional ψ 7→ (HDψ,ϕ) extends to a bounded linear functional onL2(I). Therefore, by the Riesz Representation Theorem, there exists γ in L2(I)such that

(HDψ,ϕ) = (ψ, γ) ∀ψ ∈ Dom(HD);

in particular this holds for ψ in C∞c (I). Thus, ϕ has a second order derivativein L2(I), which is equal to γ, and ϕ is in W 2,2(I). Observe that an integrationby parts argument shows that

(HDψ,ϕ) = ψ′(a)ϕ(a)− ψ′(b)ϕ(b) +

∫ b

a

ψ′ ϕ′ dλ

= ψ′(a)ϕ(a)− ψ′(b)ϕ(b) +

∫ b

a

ψ (−ϕ′′) dλ

= ψ′(a)ϕ(a)− ψ′(b)ϕ(b) + (ψ, γ).

The right hand side is equal to (ψ, γ) for every ψ in Dom(HD) provided thatϕ(a) = 0 = ϕ(b). Thus, ϕ is also in W 1,2

0 (I).

Conversely, if ϕ is in W 2,2(I)∩W 1,20 (I), then its distributional derivative ϕ′′

is in L2(I), and an integration by parts argument similar to that above gives

(ψ,ϕ′′) = (ψ′′, ϕ) ∀ψ ∈ C∞c (I).

Thus, ψ 7→ (HDψ,ϕ) extends to a bounded functional on H, hence ϕ is inDom(H∗D), as required.

Recall that the space H′ in the proof of the theorem above is the completionof Dom(HD) with respect to the norm

‖f‖′ = (HDf, f)1/2.

It is well known that this space is W 1,20 (I), so that

Dom((HD)F

)= W 2,2(I) ∩W 1,2

0 (I) and (HD)Fϕ = −ϕ′′.


Note that the closure of HD is the operator

HDf = −f ′′ ∀f ∈W 2,2(I) ∩W 1,20 (I)

(see Exercise 2.7.10), so that HD = H∗D, i.e., the closure HD of HD is selfadjoint, and agrees with the Friedrichs extension of HD.

It is instructive to discuss a variant of the operator HD above. Denote by H ′Dthe operator obtained by restricting HD to C∞c (I) (all functions in this spacehave vanishing derivatives at both ends of the interval I). The operator H ′D isclearly symmetric, hence closable. The domain of its closure is the completionof C∞c (I) with respect to the graph norm. By Sobolev space theory, this com-pletion is just W 2,2

0 (I). We emphasise that if f is the continuous representativeof an element in W 2,2

0 (I), then both f and f ′ vanish at the ends of I, whereasif g is in W 2,2(I)∩W 1,2

0 (I), then g vanishes at ∂I, but g′ may not vanish at ∂I.

It may be shown that H ′D has several self adjoint extensions. For example,

there is an extension H ′D of H ′D “with boundary conditions” ϕ′(0) = 0 = ϕ′(1).This extension is not the Friedrichs extension, as we now show. Indeed, byarguing much as above (for HD), it is straightforward to check that

Dom((H ′D)∗) = W 2,2(I) and (H ′D)∗f = −f ′′

(we do not need that f be inW 1,20 (I), because all the functions in Dom(H ′D) have

vanishing derivatives at ∂I). Since the completion of Dom(H ′D) with respect tothe norm

‖f‖′ = (HDf, f)1/2

is W 1,20 (I),

Dom((H ′D)F

)= W 2,2(I) ∩W 1,2

0 (I) and (H ′D)Fϕ = −ϕ′′.

Thus, the Friedrichs extension of H ′D agrees with that of HD.

In the case of a general symmetric densely defined operator the problem ofexistence of self adjoint extensions of A may be reduced to a purely algebraicproblem.

Definition 3.4.5 Suppose that A is a symmetric densely defined operator.The deficiency subspaces D− and D+ of A are defined by

D± := f ∈ Dom(A∗) : A∗f = ±if.

The (possibly infinite) dimensions d− and d+ of D− and D+ are called defi-ciency indices of A.

Exercise 3.4.6 Suppose that A is a symmetric densely defined operator.Prove that

D± = f ∈ H : (Ah, f) = ∓i(h, f) for all h ∈ Dom(A).

The following criterion, due to von Neumann, is of fundamental importance.


Theorem 3.4.7 Suppose that A is a symmetric densely defined operator. ThenA admits at least one self adjoint extensions if and only if its deficiency indicesare equal. Furthermore, the following are equivalent:

(i) the closure of A is self adjoint:

(ii) d− = 0 = d+;

(iii) A has exactly one self adjoint extension.

Theorem 3.4.8 Suppose that A is a symmetric densely defined operator inL2(M), where (M,M, µ) is a measure space. If for every f in Dom(A) also fis in Dom(A) and

Af = Af,

then A admits at least one self adjoint extension.

Proof. We show that complex conjugation is a one to one mapping from D+

onto D−. Indeed, suppose that f is in D+. By Exercise 3.4.6

(Ah, f) = −i (h, f) ∀h ∈ Dom(A).

Observe that (Ah, f

)=(Ah, f

)=

∫M

Ah f dµ

=(Ah, f

)(by Exercise 3.4.6) = −i

(h, f

)= i(h, f

),

i.e., f is in D−. Working backwards, we see that f in D− implies that f is inD+.

This implies that d− = d+, and the required result follows from Theo-rem 3.4.7. 2

Consider the operator A, defined by

Dom(A) = C∞c (R+) Af = i f ′.

Integration by parts shows that A is symmetric. It is not hard to check that f isin D±, i.e., A∗f = ±i f , if and only if f is in W 1,2(R+) and f ′ = ±f . The onlydistributional solutions to these equations are the exponentials f±(x) = e±x, ofwhich only f− is in W 1,2(R+). Thus, d+ = 0 and d− = 1, so that A does notadmit self adjoint extensions.

If, instead, we consider the operator A, defined by

Dom(A) = C∞c (I) Af = i f ′,

where I denotes the interval (0, 1), then we find that d+ = 1 = d−, and Aadmits self adjoint extensions, according to Theorem 3.4.7.


Exercise 3.4.9 Work out all the details of the example above.

We conclude this chapter by showing that the operator A4 defined just aboveExercise 2.4.5 has infinitely many self adjoint extensions. From Exercise 3.3.7We know that A∗4 is the operator A1 given by

Dom(A1) = W 1,2(I) and A1f = if ′

It is straighforward to check that Ker(A∗4±iI) are the one dimensional subspacesof L2(I) generated by the exponential functions e∓·. Thus, d± = 1, and A4 hasself adjoint extensions. For each complex α such that |α| = 1 consider theoperator Aα4 defined by

Dom(Aα4 ) = f ∈W 1,2(I) : f(0) = α f(1), and Aα4 f = if ′.

Clearly Aα4 ⊃ A4. It is straightforward to check that Aα4 is self adjoint (by usingthe Basic Criterion of Self Adjointness).

Exercise 3.4.10 Prove that Aα4 is self adjoint for all complex α such that|α| = 1.

Chapter 4

One parameter semigroups

The main issue of this chapter is to define the exponential of a (possibly un-bounded) operator A on a complex Banach space B. If A is bounded, then theexponential of A may be defined as for matrices. The case where A is unboundedis more delicate, and occupies most of this chapter.

4.1 Lecture XVII: exponential of a bounded op-erator

Consider a bounded operator A on the complex Banach space B. The purposeof this lecture is to define and study the properties of the operators e−tA, wheret is in R.

Definition 4.1.1 Suppose that A is a bounded operator on the complex Ba-nach space B. For each ζ in C the exponential e−ζA of A by

e−ζA =

∞∑k=0

(−1)k

k!(ζA)k

(the series is convergent in the uniform topology).

Exercise 4.1.2 Compute the exponential of the following matrices:

A1 =

[0 10 0

]A2 =

[0 1−1 0

]A3 =

[a b0 −a

]a 6= 0, b ∈ R.

Exercise 4.1.3 Compute the exponential of the multiplication operator Am(see Definition 2.1.7) on L2(µ) when m is bounded.

Exercise 4.1.4 Compute the exponential of the operator A, defined on L2(Z)by

Af(x) = f(x+ 1)− f(x) ∀x ∈ Z.

63

64 CHAPTER 4. ONE PARAMETER SEMIGROUPS

The main result concerning exponential operators associated to bounded oper-ators is the following.

Theorem 4.1.5 Suppose that A is bounded on B. The following hold:

(i) the map ζ 7→ e−ζA is analytic in the uniform topology of operators;

(ii) for every ζ and ω in C

e−ζA e−ωA = e−(ζ+ω)A;

(iii) for every ζ in C

d

dζe−ζA = −A e−ζA = −e−ζAA,

where the derivative is taken in the uniform topology of operators;

(iv) for every f in B, the B valued function e−tAf solves the Cauchy problemu′(t) +Au(t) = 0

u(0) = f ;

(v) in the case where B is a Hilbert space, the adjoint of eζA is eζA∗. In

particular, if A is self adjoint, then(eitA

)∗= e−itA ∀t ∈ R.

Consequently, eitA is unitary for every t in R.

Proof. The proof is left as an exercise for the reader. 2

Exercise 4.1.6 Give a bound for |||e−tA||| when t in R.

4.2 Lecture XVIII: semigroups of operators

It is well known that the unique distribution solution of the Cauchy problemu′(t) + a u(t) = 0 ∀t ∈ Ru(0) = 1

is the exponential function u(t) = e−at. Given a complex Banach space B anda linear operator A on B, it is natural to speculate whether, given f in B, theBanach valued Cauchy problem

u′(t) +Au(t) = 0 ∀t ∈ Ru(0) = f

(4.2.1)

has any solution u : R→ B in an appropriate sense. We have already seen howto construct a solution to this problem in the case where A is bounded.

In this lecture, we shall focus on the case where A is unbounded. We assumethat Dom(A) is dense in B.

4.2. SEMIGROUPS OF OPERATORS 65

Definition 4.2.1 A one parameter family Tt : t ∈ [0,∞) of bounded oper-ators on B is called a strongly continuous semigroup of operators if thefollowing hold:

(i) T0 = I and TsTt = Ts+t for all s and t in R+;

(ii) Tt tends to the identity in the strong operator topology as t tends to 0+,i.e., for every f in B

limt→0+

‖Ttf − f‖ = 0.

The semigroup Tt is said to be a contraction semigroup if

|||Tt||| ≤ 1 ∀t ∈ R+.

Proposition 4.2.2 Suppose that Tt is a strongly continuous semigroup ofoperators on B. The following hold:

(i) the map t 7→ Tt is continuous from [0,∞) to L(B), with respect to thestrong operator topology;

(ii) |||Tt||| is bounded on compacta of [0,∞).

Proof. Fix t > 0. Clearly,

Tt+hf − Ttf = Tt(Thf − f).

If h > 0, then Thf − f tends to 0 in B by the strong continuity of Tt at 0.Since Tt is bounded, we conclude that Tt+hf−Ttf tends to 0 in B. Thus, t 7→ Ttis strongly right continuous on (0,∞). The right continuity at 0 is part of thedefinition of a strongly continuous semigroup.

Suppose now that h < 0. Then

Tt+hf − Ttf = Tt+h(f − T−hf).

To conclude, we need to prove that |||Tt+h||| is uniformly bounded for small h.It suffices to prove that there exists T > 0 such that

sups∈[0,T ]

|||Ts||| <∞.

We argue by contradiction. If such a T does not exist, then there exists asequence tk such that tk → 0 and

supk|||Ttk ||| =∞. (4.2.2)

By strong continuity at 0, for every f in B

supk‖Ttkf‖ <∞.

By the Uniform Boundedness Principle,

supk|||Ttk ||| <∞,


which contradicts (4.2.2).

This implies the required boundedness on compacta and implies the strongleft continuity of Tt. 2

An example wheresupt>0|||Tt||| =∞

is given by the multiplication by et on L2(R).

Definition 4.2.3 The infinitesimal generator of the strongly continuoussemigroup Tt is the operator A defined by

Af = limt→0+

f − Ttft

for all f for which the above limit exists (in B).

Notice that the domain of the infinitesimal generator A of Tt is precisely theset of all f in B such that the function t 7→ Ttf is strongly right differentiableat t = 0: for such f we have

Af = − d

dt

+

Ttf |t=0.

Proposition 4.2.4 Suppose that Tt is a strongly continuous one parametersemigroup of operators with infinitesimal generator A. The following hold:

(i) for each f in Dom(A) the map t 7→ Ttf is strongly differentiable on (0,∞)and strongly right differentiable at 0. Furthermore, Ttf is in Dom(A) forall t > 0 and

ATtf = TtAf = − d

dt

+

Ttf ;

(ii) Dom(A) is dense in B

(iii) A is a closed operator.

Proof. First we prove (i). Note that t 7→ Ttf is right differentiable at 0 because fis in Dom(A). To prove the right differentiability of t 7→ Ttf on (0,∞), observefirst that t 7→ Ttf is continuous, by Proposition 4.2.2. Then note that

Tt+hf − Ttfh

=Th − Ih

(Ttf)

by the semigroup property, so that

limh→0+

Tt+hf − Ttfh

= limh→0+

Th − Ih

(Ttf) = limh→0+

TtThf − f

h. (4.2.3)

Since f is in Dom(A),

limh→0+

Thf − fh

= −Af,


whence

limh→0+

TtThf − f

h= −TtAf,

because Tt is continuous. Therefore the limit in the middle term of (4.2.3)exists, so that Ttf is in Dom(A), by the definition of Dom(A), and it is equalto −A(Ttf). We have proved that Ttf is right differentiable at t and that

d

dt

+

Ttf = −ATtf = −TtAf ∀f ∈ Dom(A).

To prove that Tt is left differentiable at t and that

d

dt

−Ttf = −ATtf = −TtAf ∀f ∈ Dom(A),

writeTt−hf − Ttf

−h= −Tt−h

f − Thfh

;

the required conclusion follows much as above (here we use the local uniformboundedness of |||Tt−h||| with respecto to h).

Next we prove (ii). For f in B and t > 0, define

ft :=1

t

∫ t

0

Tsf ds,

where the integral is the strong limit of the corresponding Riemann sums. Weclaim that ft is in Dom(A), and that limt→0+ ‖f − ft‖ = 0. To prove that ft isin Dom(A), observe that

Tεft =1

t

∫ t+ε

ε

Tsf ds.

Therefored

dεTεft|ε=0

=Ttf − f

t,

as required. To prove that ft tends to f as t tends to 0, we may proceed as inthe proof of the fundamental theorem of calculus. We omit the details.

Finally we prove (iii). Suppose that fn is a sequence in Dom(A) such thatfn → f and Afn → g. Since fn is in Dom(A),

fn − Ttfn =

∫ t

0

TsAfn ds.

Now we let n tend to ∞, and get that

f − Ttf =

∫ t

0

Tsg ds.

Therefore

limt→0

f − Ttft

= limt→0

1

t

∫ t

0

Tsg ds = g,

whence f is in Dom(A) and Af = g, as required. 2


Since T0 = I, we have shown that if A is the infinitesimal generator of a stronglycontinuous semigroup Tt, then for f in Dom(A) the function Ttf solves theCauchy problem (4.2.1). It may be shown that Ttf is the unique solution of(4.2.1). The situation in the case where A is bounded is much simpler.

Exercise 4.2.5 Prove by “elementary methods” that if A is bounded, thene−tAf is the unique solution to the Cauchy problem (4.2.1).

It would be desirable to find a solution of the Cauchy problem (4.2.1) for allinitial data f in B. The argument above shows that if f is in Dom(A), thent 7→ Ttf is differentiable at t > 0, and solves the Cauchy problem (4.2.1).However, by Exercise 4.2.7 below, there may exist f in B such that t 7→ Ttf isnot differentiable at any point t in (0,∞). We shall see that this phenomenoncannot happen when A is a self adjoint operator on a Hilbert space. Thus,in this case, for any initial datum f , Ttf is in Dom(A) and solves the Cauchyproblem (see Corollary 5.7.5).

Given an unbounded operator A on B, a natural issue is to construct a stronglycontinuous semigroup of operators such that A is its infinitesimal generator.This problem will be considered in the next lecture.

Exercise 4.2.6 For each p in [1,∞] consider the family Tt of operators onLp(R) defined by

Ttf(x) = f(x− t) ∀f ∈ Lp(R) ∀x ∈ R.

Prove the following:

(i) Tt is a contraction on Lp(R) for all p in [1,∞];

(ii) Tt is a strongly continuous semigroup of operators on Lp(R) for all p in[1,∞);

(iii) Tt is not a strongly continuous semigroup of operators on L∞(R);

(iv) for p in [1,∞) compute the infinitesimal generator of Tt on Lp(R);

(v) for p = 2 check that the infinitesimal generator of Tt is of the form iA,where A is a self adjoint operator.

Exercise 4.2.7 Consider the translation semigroup of Exercise 4.2.6 as actingon BC(R), the Banach space of all bounded uniformly continuous functions onR with the uniform norm. Prove that Tt is a strongly continuous contractionsemigroup. Prove also that for a generic f in BC(R), the function Ttf need notbe in the domain of the infinitesimal generator of the semigroup.

For each p in [1,∞] consider the family Tt of operators on Lp(Rn) defined by

Ttf(x) =

f ∗ wt(x) if t > 0

f(x) if t = 0 ∀f ∈ Lp(Rn) ∀x ∈ Rn,


where wt denotes the Gauss–Weierstrass kernel

wt(x) = (4πt)−n/2 e−|x|2/(4t) ∀t ∈ R+ ∀x ∈ Rn.

Define the Fourier transform F of a function f in the Schwartz class S(Rn) by

Ff(ξ) =1

(2π)n/2

∫ ∞−∞

f(x) e−iξx dx ∀ξ ∈ Rn.

It is known that

Fwt(ξ) = (2π)−n/2 e−t|ξ|2

∀t ∈ R+ ∀ξ ∈ Rn.

Observe that‖wt‖1 = (2π)n/2 Fwt(0) = 1. (4.2.4)


(i) Tt is a contraction on Lp(Rn) for all p in [1,∞];

(ii) Tt is not a strongly continuous semigroup of operators on L∞(Rn);

(iii) Tt is a strongly continuous semigroup of operators on Lp(Rn) for all p in[1,∞);

(iv) the infinitesimal generator of Tt on L2(Rn) is the operator −∆ on theSobolev space W 2,d(Rn).

Proof. To prove (i) observe that

‖Ttf‖p ≤ ‖f‖p ‖wt‖1,

and the required estimate follows from (4.2.4).

If Tt were strongly continuous on L∞(Rn), then Ttf would converge uni-formly to f as t tends to 0+. Take f = 1[0,∞); then Ttf is continuous (in fact,it is smooth), hence its uniform limit, which is f , must be continuous. Thiscontradiction proves (ii).

To prove (iii), Suppose that δ is in R+. Observe that∫B(0,δ)c

|wt(y)|dy =

∫B(0,δ/

√t)c|w1(y)|dy

(because w1 is in L1(Rn)) → 0

as t tends to 0+. Since∫Rn w1(y) dy = 1, we may write

f ∗ wt(x)− f(x) =

∫Rn

(τyf − f)(x)wt(y) dy

=

∫B(0,δ)

(τyf − f)(x)wt(y) dy +

∫B(0,δ)c

(τyf − f)(x)wt(y) dy.

Then ∥∥f ∗ wt − f∥∥p ≤ supy∈B(0,δ)

∥∥τyf − f∥∥p + 2∥∥f∥∥

p

∫B(0,δ)c

|wt(y)|dy.


Given ε > 0 there exists δ in R+ such that∥∥τyf − f∥∥p < ε/2 for every y such

that |y| < δ. Once we have chosen δ, we may choose t small enough so that∫B(0,δ)c

|wt(y)|dy < ε/(4∥∥f∥∥

p). Thus,∥∥f ∗ wt − f∥∥p < ε

for t small enough, as required.

Now we prove (iv). Suppose first that f is a function in L2(Rn) such that

limt→0+

f − Ttft

exists in L2(Rn). We denote by g this limit; we must show that f is in W 2,2(Rn)and that g = −∆f .

Observe that

1− wt(ξ)t

− |ξ|2 =1

t

∫ t

0

|ξ|2 e−s|ξ|2

ds− |ξ|2

= |ξ|2[1

t

∫ t

0

e−s|ξ|2

ds− 1]

→ 0

(4.2.5)

pointwise as t tends to 0+. By Plancherel’s theorem,

limt→0+

∥∥∥f 1− wtt− g∥∥∥2

= 0.

By Fatou’s theorem

∞ > lim inft→0+

∫Rn

∣∣∣f(ξ)1− wt(ξ)

t

∣∣∣2 dξ

≥∫Rn

∣∣f(ξ)∣∣2 lim inf

t→0+

∣∣∣1− wt(ξ)t

∣∣∣2 dξ

=

∫Rn

∣∣f(ξ)∣∣2 |ξ|4 dξ,

whence f is in W 2,2(Rn). We have used (4.2.5) in the last equality. Furthermore,from (4.2.5) we see that ∣∣∣1− wt(ξ)

t− |ξ|2

∣∣∣ ≤ |ξ|2. (4.2.6)

By the Dominated Convergence Theorem

0 = limt→0+

∫Rn

∣∣f(ξ)∣∣2 ∣∣∣1− wt(ξ)

t− |ξ|2

∣∣∣2 dξ

= limt→0+

∥∥∥f − Ttft

+ ∆f∥∥∥2

2

.

(4.2.7)

Therefore g = −∆f , as required.

4.3. EXPONENTIAL OF UNBOUNDED OPERATORS 71

Conversely, suppose that f is in W 2,2(Rn). We must show that f is in thedomain of the infinitesimal generator of Tt and that −( d/ dt)Ttf|t=0 = −∆f .Observe that

limt→0+

∥∥∥f − Ttft

+ ∆f∥∥∥2

2

= limt→0+

∫Rn

∣∣f(ξ)∣∣2 ∣∣∣1− wt(ξ)

t− |ξ|2

∣∣∣2 dξ

By (4.2.5), (4.2.6) and the Dominated Convergence Theorem, the right handside tends to 0 as t tends to 0+. This gives the required conclusion. 2

Exercise 4.2.9 Suppose that Tt is a one parameter semigroup on B thatis continuous with respect to the uniform topology. Prove that its infinitesimalgenerator is bounded. Hint : Define the operators

Ah :=I − Thh

and Mh :=1

h

∫ h

0

Ts ds.

Prove that Mt is invertible for t small enough, and that

MtAε = AtMε ∀ε, t ∈ R+.

4.3 Lecture XIX: exponential of unbounded op-erators

Suppose that A is unbounded. Then the series∑∞k=0

(−1)kk! (tA)k cannot con-

verge in the uniform topology. Moreover, if the series

∞∑k=0

(−1)k

k!(tA)kf

converges for some f in B, then f must be in⋂∞k=0 Dom(Ak). Since we want the

putative e−tA to be bounded on B, we must find an alternative way to define theexponential. The purpose of this lecture is to define e−tA for a class of closeddensely defined operators A.

Theorem 4.3.1 (Hille–Yosida) Suppose that A is a densely defined operator.The following are equivalent:

(i) A is the infinitesimal generator of a (strongly continuous) contractionsemigroup;

(ii) A is closed, (−∞, 0) is in %(A) and

|||(ζI +A)−1||| ≤ 1

ζ∀ζ ∈ (0,∞).

Proof. First we prove that (i) implies (ii). In fact, we prove more, i.e., that forevery ζ such that Re(ζ) > 0

|||(ζI +A)−1||| ≤ 1

Re(ζ).


Suppose that Re(ζ) > 0, and consider the operator

Sζf =

∫ ∞0

e−ζt Ttf dt ∀f ∈ B.

Observe that the integrand is a continuous function on [0,∞) with values in B.Thus, the integral may be interpreted as a generalised Riemann integral. SinceTt is contractive,

‖Sζf‖ ≤∫ ∞0

e−Re(ζ)t dt ‖f‖

=‖f‖

Re(ζ)∀f ∈ B.

Hence Sζ is bounded with

|||Sζ ||| ≤1

Re(ζ).

It remains to prove that Sζ satisfies

(ζI +A)Sζ = Sζ(ζI +A) = I.

We prove that Sζf is in Dom(A) for every f in B, and that ASζf = f−ζ Sζf .Indeed,

Sζf − TsSζfs

= −1

s

∫ ∞0

e−ζt[Ts+tf − Ttf

]dt

=1− eζs

s

∫ ∞0

e−ζt Ttf dt+eζs

s

∫ s

0

e−ζt Ttf dt.

The right hand side tends to f − ζSζf as s tends to 0. Thus Sζf is in Dom(A)and

ASζf = f − ζSζf,i.e., (ζI +A)Sζ = I, as required.

Now we show that if f is in Dom(A), then Sζ(ζI +A)f = f . Indeed,

Sζ(ζI +A)f = ζ Sζf +

∫ ∞0

e−ζt TtAf dt

= ζ Sζf −∫ ∞0

e−ζtd

dtTtf dt

(by parts) = ζ Sζf + f − ζ∫ ∞0

e−ζt Ttf dt

= f,

as required.

Next we prove that (ii) implies (i). For each λ > 0, define the operator A(λ)

byA(λ) = λ

[I − λ(λI +A)−1

].

Clearly A(λ) is a bounded operator on B. We claim that A(λ) converges stronglyto A on Dom(A) as λ tends to ∞, i.e.

limλ→∞

‖A(λ)ξ −Aξ‖ = 0 ∀ξ ∈ Dom(A).

4.3. EXPONENTIAL OF UNBOUNDED OPERATORS 73

First, we observe that for ξ in Dom(A)

A(λ)ξ = λ (λI +A)−1Aξ. (4.3.1)

Thus, to prove the claim it suffices to show that λ (λI+A)−1 converges stronglyto the identity on B. If ξ is in Dom(A), then∥∥λ (λI +A)−1ξ − ξ

∥∥ =∥∥(λI +A)−1Aξ

∥∥≤∣∣∣∣∣∣(λI +A)−1

∣∣∣∣∣∣ ‖Aξ‖≤ ‖Aξ‖

λ,

which tends to 0 as λ tends to ∞. If ξ is in B, there exists a sequence ξn inDom(A) such that limn→∞ ‖ξ − ξn‖ = 0. Then∥∥λ (λI +A)−1ξ − ξ

∥∥≤∥∥λ (λI +A)−1ξ − λ (λI +A)−1ξn

∥∥ +∥∥λ (λI +A)−1ξn − ξn

∥∥ +∥∥ξn − ξ∥∥

≤∥∥λ (λI +A)−1ξn − ξn

∥∥ + 2∥∥ξn − ξ∥∥ .

Now, given ε > 0, choose n large enough so that∥∥ξn − ξ∥∥ < ε. Then choose λ

such that ‖Aξn‖/λ < ε; hence∥∥λ (λI +A)−1ξ − ξ∥∥ ≤ 3ε

for λ large enough, and the claim is proved.

Since A(λ) is bounded, we may construct the semigroup

e−tA(λ)

:=

∞∑k=0

(−t)k

k!(A(λ))k.

Note that e−tA(λ)

is a contraction semigroup, and that e−tA(λ)

is strongly con-vergent as λ tends to ∞. Indeed,

‖e−tA(λ)

‖ ≤ e−λt∞∑k=0

(tλ2)k

k!|||(λI +A)−1|||k

≤ 1,

so that e−tA(λ)

is a contraction semigroup.

Now suppose that ξ is in Dom(A), and that λ1 and λ2 and t are positive.Then

e−tA(λ1)

ξ − e−tA(λ2)

ξ =

∫ t

0

d

ds

[e−sA

(λ1)

e−(t−s)A(λ2)

ξ]

ds.

Note that

d

ds

[e−sA

(λ1)

e−(t−s)A(λ2)

ξ]

= e−sA(λ1)

e−(t−s)A(λ2)(

A(λ2) −A(λ1))ξ;

we have used the fact that A(λ1) and A(λ2) are permutable operators to write

e−sA(λ1)

e−(t−s)A(λ2)

= es(A(λ2)−A(λ1))−tA(λ2)

.


Thus, we obtain

∥∥e−tA(λ1)

ξ − e−tA(λ2)

ξ∥∥ ≤ ∫ t

0

∣∣∣∣∣∣e−sA(λ1)

e−(t−s)A(λ2) ∣∣∣∣∣∣ ‖A(λ1)ξ −A(λ2)ξ‖ ds

≤ t ‖A(λ1)ξ −A(λ2)ξ‖.(4.3.2)

Therefore, for every ξ in Dom(A) there exists an element in B, which we denoteby Ttξ, such that

limλ→∞

∥∥e−tA(λ)

ξ − Ttξ∥∥ = 0,

i.e., e−tA(λ)

is strongly convergent on Dom(A). Suppose now that ξ is in B.Then there exists a sequence ξn in Dom(A) such that limn→∞ ‖ξn − ξ‖ = 0,so that∥∥e−tA

(λ1)

ξ − e−tA(λ2)

ξ∥∥

≤∥∥e−tA

(λ1)

ξ − e−tA(λ1)

ξn∥∥ +

∥∥e−tA(λ1)

ξn − e−tA(λ2)

ξn∥∥ +

∥∥e−tA(λ2)

ξn − e−tA(λ2)

ξ∥∥

≤ 2 ‖ξ − ξn‖ +∥∥e−tA

(λ1)

ξn − e−tA(λ2)

ξn∥∥ ,

which is clearly small for n, λ1 and λ2 large. Hence e−tA(λ)

is strongly convergent

on all of B. We shall denote by Ttξ the strong limit of e−tA(λ)

ξ.

We remark for further use that we have, in fact, proved that for every T > 0

limλ→∞

sup0≤t≤T

‖e−tA(λ)

ξ − Ttξ‖ = 0 ∀ξ ∈ B. (4.3.3)

It remains to prove that Tt is a (strongly continuous) contraction semi-group. Note that ∥∥Ttξ∥∥ = lim

λ→∞

∥∥e−tA(λ)

ξ∥∥

≤ ‖ξ‖ ∀ξ ∈ B ∀t ∈ R+,

so that Tt is a contraction semigroup.

To prove that Tt is strongly continuous, we argue as follows. By thetriangle inequality

‖Ttξ − ξ‖ ≤ ‖Ttξ − e−tA(λ)

ξ‖ + ‖e−tA(λ)

ξ − ξ‖

Fix ε > 0 and choose λ so large that the first summand is < ε for all t in [0, T ](this can be done by (4.3.3)). Then choose δ so small that the second summandis < ε for all t in [0, δ) (and for λ fixed chosen above). This can be done because

the semigroup e−tA(λ)

is strongly continuous.

Finally, we prove that the infinitesimal generator of Tt, that we provision-

ally call A′, agrees with A. Since A(λ) is the infinitesimal generator of e−tA(λ),

e−tA(λ)

ξ − ξ = −∫ t

0

e−sA(λ)

A(λ)ξ ds ∀ξ ∈ Dom(A).

4.4. GROUPS OF UNITARY OPERATORS 75

Since A(λ) tends to A strongly on Dom(A) and e−tA(λ) converges to Ttuniformly on compacta in the strong operator topology by (4.3.3),

Ttξ − ξ = −∫ t

0

TsAξ ds ∀ξ ∈ Dom(A).

Therefore, for every ξ in Dom(A)

d

dtTtξ|t=0

= −Aξ,

so that A′ is an extension of A. Then λI +A′ is an extension of λI +A. SinceλI + A is surjective, so is, a fortiori, λI + A′. If A′ were a proper extensionof A, there would exist η in Dom(A′) \ Dom(A). But then λI + A′ would notbe injective, which contradicts the fact that A′ generates a strongly continuoussemigroup (hence λI + A is invertible by the implication (i) =⇒ (ii) provedabove). Hence A′ = A, as required. 2

For later purposes, we need the following.

Proposition 4.3.2 Suppose that T (1)t and T (2)

t are strongly continuous

semigroups on B. If their infinitesimal generators agree, then T(1)t = T

(2)t for

every t in [0,∞).

Proof. This follows easily from the injectivity of the Laplace transform. Weomit the details. 2

4.4 Lecture XX: groups of unitary operators

Definition 4.4.1 A one parameter family Ut : t ∈ R of bounded operatorson H is called a strongly continuous group of operators if the followinghold:

(i) U0 = I and UsUt = Us+t for all s and t in R;

(ii) Ut tends to the identity in the strong operator topology as t tends to 0,i.e., for every ξ in H

limt→0

∥∥Utξ − ξ∥∥ = 0.

The group Ut is said to be unitary if for every t in R the operator Ut isunitary.

Note that if Ut is a group of operators, then U−t = U−1t . If Ut is unitary,then U−t = U∗t .

Exercise 4.4.2 Suppose that Ut is a one parameter group of unitary opera-tors on H. Prove that if t 7→ Ut is continuous with respect to the weak operatortopology, then it is continuous with respect to the strong operator topology.


Definition 4.4.3 Suppose that Ut is a one parameter group of unitaryoperators. The operator A, defined by

Dom(A) = ξ ∈ H : t 7→ Utξ is strongly differentiable at 0

Aξ = − d

dt(Utξ)|t=0

∀ξ ∈ Dom(A),

is called the infinitesimal generator of Ut.

Theorem 4.4.4 (Stone) The following hold:

(i) if L is self adjoint, then iL generates a strongly continuous group of uni-tary operators;

(ii) the infinitesimal generator of a strongly continuous group of unitary oper-ators is of the form iL for some self adjoint operator L.

Proof. First we prove (i). Since σ(L) is contained in R, the spectra of iL and−iL are contained in iR. Furthermore, the estimate

|||(λI ± iL)−1||| ≤ 1

λ∀λ ∈ R+

is a direct consequence of Proposition 3.3.3. Thus, both iL and −iL satisfythe hypotheses of the Hille–Yosida Theorem 4.3.1, so that iL and −iL gener-ate strongly continuous contraction semigroups, which we denote by T+

t andT−t , respectively. Set

Ut :=

T+t ∀t ∈ (0,∞)

I if t = 0

T−−t ∀t ∈ (−∞, 0).

Clearly |||Ut||| ≤ 1 for every t in R. We shall prove that Ut is a one parametergroup of unitary operators. We retain the notation of the proof of the Hille–Yosida Theorem. By arguing as in its proof, we see that

T+t = lim

λ→∞e−t(iL)

(λ)

and T−t = limλ→∞

e−t(−iL)(λ)

∀t ∈ R+,

where the convergence is in the strong operator topology. Define U(λ)t by

U(λ)t :=

e−t(iL)

(λ) ∀t ∈ (0,∞)

I if t = 0

et(−iL)(λ) ∀t ∈ (−∞, 0).

(Warning : U(λ)t need not be unitary!) It is straightforward to check that(

(iL)(λ))∗

= (−iL)(λ), whence((−iL)(λ)

)∗= (iL)(λ).

Indeed, it suffices to prove that the adjoint of (λI + iL)−1 is (λI − iL)−1. Thisis equivalent to R(iλ;L)∗ = R(−iλ;L), which holds by Proposition 3.3.6. FromTheorem 4.1.5 (v) we then deduce that(

e−t(iL)(λ))∗

= e−t((iL)(λ))∗ = e−t(−iL)

(λ)

∀t ∈ R+

4.4. GROUPS OF UNITARY OPERATORS 77

and (e−t(−iL)

(λ))∗= e−t((−iL)

(λ))∗ = e−t(iL)(λ)

∀t ∈ R+.

By taking the limit as λ tends to∞ in the strong operator topology, we see that

U∗t = (T+t )∗ = T−t = U−t ∀t ∈ R+.

Now, observe that if t > 0

UtU∗t = lim

λ→∞e−t(iL)

(λ)−t(−iL)(λ) = U∗t Ut

in the strong operator topology; here we have used the fact that the operators(iL)(λ) and (−iL)(λ) commute. As in the proof of the Hille–Yosida Theorem(iL)(λ) and (−iL)(λ) tend strongly to iL and −iL respectively, whence theirsum tends strongly to 0 (on Dom(L)).

We claim that e−t(iL)(λ)−t(−iL)(λ) tends strongly to the identity. Note that

the claim implies the relation

UtU∗t = I = U∗t Ut ∀t > 0.

A similar formula may be proved for t < 0, thereby completing the proof thatUt is unitary.

To prove the claim, write A(λ) instead of (iL)(λ) + (−iL)(λ). By arguingmuch as in the proof of the Hille–Yosida Theorem, it is straightforward to check

that |||e−t(iL)(λ) ||| ≤ 1 and |||e−t(−iL)(λ) ||| ≤ 1 for all t > 0. Thus |||e−tA(λ) ||| ≤ 1.Then

e−tA(λ)

ξ =

∫ t

0

d

dse−sA

(λ)

ξ ds+ ξ

= −∫ t

0

e−sA(λ)

A(λ)ξ ds+ ξ ∀ξ ∈ H.

Hence

‖e−tA(λ)

ξ − ξ‖ ≤∫ t

0

‖e−sA(λ)

A(λ)ξ‖ ds

≤ t ‖A(λ)ξ‖ ∀ξ ∈ H.

SinceA(λ)ξ tends to 0 as λ tends to∞ for every ξ in Dom(L), so does ‖e−tA(λ)

ξ − ξ‖,as required. A three epsilon argument based on the density of Dom(L) in Hproves that

limλ→+∞

‖e−tA(λ)

ξ − ξ‖ = 0 ∀ξ ∈ H.

To complete the proof of (i) it remains to prove that the infinitesimal gen-erator A′ of Ut is equal to iL. On the one hand, if ξ is in Dom(L), then bothT+t ξ and T−t ξ are strongly right differentiable at the origin and

d

dt

+

T+t ξ|t=0 = −iLξ and

d

dt

+

T−t ξ|t=0 = iLξ,

so that Utξ is strongly differentiable at 0 and

A′ξ = − d

dtUtξ|t=0 = iLξ.


Hence A′ is an extension of iL. On the other hand, if ξ is in Dom(A′), thenT+t ξ is right differentiable at 0, so that ξ is in Dom(L) and

A′ξ = iLξ.

Thus, A′ = iL, as required.

Next we prove (ii). Suppose that Ut is a strongly continuous one parametergroup of unitary operators. Its restriction to [0,∞) is a strongly continuouscontraction semigroup. Hence the spectrum of its infinitesimal generator A iscontained in ζ : Re ζ ≥ 0. Similarly, the formula

T−t = U−t ∀t ∈ [0,∞)

defines a strongly continuous semigroup of contractions. It is straightforwardto check that its infinitesimal generator is −A (with the same domain as A).Thus, its spectrum is contained in ζ : Re ζ ≥ 0. Therefore the spectrum ofA is contained in iR. Set L := iA. Then σ(L) ⊂ R. Furthermore, L is closedand densely defined, because, iL generates a strongly continuous semigroup.By Corollary 3.3.4, to prove that L is self adjoint it suffices to show that L issymmetric. Suppose that ξ and η are in Dom(L). Then

(Lξ, η) = −i d

dt(Utξ, η)|t=0

= −i d

dt(ξ, U−tη)|t=0

= −i (ξ,−iLη)

= (ξ,Lη),

as required. 2

Chapter 5

The spectral theorem

In this chapter H will denote a separable Hilbert space over C and L a selfadjoint operator with domain Dom(L) in H. Recall that Dom(L) is dense in Hby definition of self adjoint operator.

5.1 Lecture XXI: the functional calculus

5.1.1 Background material

The Fourier transform F of a function f in the Schwartz class S(R) is de-fined by

Ff(t) =1√2π

∫ ∞−∞

f(s) e−ist ds ∀t ∈ R.

It is known that F is a topological isomorphism of S(R) (endowed with itsstandard topology generated by a natural sequence of seminorms), whose inverseis

F−1f(s) =1√2π

∫ ∞−∞

f(t) eist dt.

Furthermore

F(fg) =1√2π

(Ff) ∗ (Fg). (5.1.1)

Recall that the convolution between two integrable functions ϕ and ψ is de-fined by

ϕ ∗ ψ(t) =

∫ ∞−∞

ϕ(t− s)ψ(s) ds ∀t ∈ R.

We shall write also f instead of Ff and f instead of F−1f . It may be shown thatF extends to a surjective isometry of L2(R). If T is a tempered distribution,i.e., an element of S ′(R), then its Fourier transform is defined by

〈f,FT 〉 = 〈Ff, T 〉 ∀f ∈ S(R).

It is well known that F is a topological isomorphism of S ′(R).

79

80 CHAPTER 5. THE SPECTRAL THEOREM

We need more notation. For every z in C we denote by ez : R → C theexponential

ez(x) = ezx.

Also, for every t in R let τt : S(R) → S(R) denote the translation operatordefined by

τtf(s) = f(s− t) ∀s ∈ R.

The reflection operator ˜ on S(R) is defined by

f(s) = f(−s) ∀s ∈ R.

We extend τt to S ′(R) by duality, thus

〈ϕ, τtT 〉 = 〈τ−tϕ, T 〉 ∀T ∈ S ′(R) ∀ϕ ∈ S(R).

We extend the reflection operator to S ′(R) similarly. Here 〈·, ·〉 denotes thepairing between S(R) and S ′(R) (with respect to the Lebesgue measure). Notethat the reflection commutes with the Fourier transform, i.e.,

(Ff)˜ = F(f) ∀f ∈ S(R). (5.1.2)

Furthermore, ⟨f , T

⟩=⟨f , T

⟩∀f ∈ S(R) ∀T ∈ S ′(R). (5.1.3)

Remark 5.1.1 A finite Borel measure µ is a tempered distribution. Denote byµ the Fourier transform of µ as a tempered distribution, i.e., µ is the tempereddistribution such that

〈ϕ, µ〉 = 〈ϕ, µ〉 ∀ϕ ∈ S(R).

We claim that µ is the continuous function, defined by

µ(t) =1√2π

∫ ∞−∞

e−its dµ(s) ∀t ∈ R.

Indeed,

〈ϕ, µ〉 =

∫ ∞−∞

dµ(s)1√2π

∫ ∞−∞

ϕ(t) e−ist dt

(by Fubini’s Theorem) =

∫ ∞−∞

dt ϕ(t)1√2π

∫ ∞−∞

e−ist dµ(s).

5.1.2 The functional calculus on S(R)

Suppose that L is a self adjoint operator on H. By Stone’s Theorem (seeTheorem 4.4.4), the operator −iL generates a strongly continuous group ofunitary operators, which we denote by eitL.

5.1. THE FUNCTIONAL CALCULUS 81

Definition 5.1.2 (functional calculus in S(R)) For each f in S(R), definethe operator f(L) by

f(L)η =1√2π

∫ ∞−∞

f(t) eitLη dt ∀η ∈ H. (5.1.4)

We shall also write Φ(f) instead of f(L).

Note that the integrand is a continuous H-valued function on R. Hence theabove integral may be interpreted as a generalised Riemann integral with valuesin H. The integral is convergent, because ‖eitLη‖ = ‖η‖ and f is a Schwartzfunction. Note that f(L) is a bounded operator on H. Indeed,

‖f(L)ξ‖ ≤ 1√2π

∫ ∞−∞|f(t)| ‖eitLξ‖ dt

(eitL is contractive) ≤ ‖ξ‖ 1√2π

∫ ∞−∞|f(t)|dt.

The last integral is convergent, because f is a Schwartz function. Therefore

|||f(L)||| ≤ 1√2π

∫ ∞−∞|f(t)|dt. (5.1.5)

A better estimate of the norm of f(L) will be obtained later.

It is tempting to define f(L) directly as

1√2π

∫ ∞−∞

f(t) eitL dt.

Unfortunately, in most cases this integral does not make sense as a L(H) valuedintegral. Indeed, the function t 7→ eitL may fail to be strongly Borel measurable.Recall1 that t 7→ eitL is strongly measurable if and only if it is weakly measurable(i.e., for every bounded linear functional F on L(H) the scalar valued functiont 7→ F (eitL) is measurable) and its range is a separable subspace of L(H). Thelast requirement is rarely satisfied. Consider, for instance, the unitary group Utof translations in L2(R) (see Exercise 4.2.6), defined by

Utf(x) = f(x− t) ∀t, x ∈ R ∀f ∈ L2(R).

From Exercise 4.2.6 we know that its infinitesimal generator is the operator T ,defined by

T f = f ′ ∀f ∈W 1,2(R).

Denote by L the operator iT (which is self adjoint). If eitL : t ∈ R wereseparable, there would exist a sequence tn in R such that eitnL is dense ineitL : t ∈ R. But this is impossible, as the following exercise shows.

Exercise 5.1.3 In the example above, prove that for every t in R \ tn√

2 ≤ |||eitL − eitnL||| ≤ 2.

1See Pettis’ Theorem on p. 131 of K. Yosida, Functional Analysis, VI Edition, SpringerVerlag, 1980.


Observe that for every ξ and η in H the function t 7→ (e−itLξ, η) is bounded, for

|(e−itLξ, η)| ≤ ‖ξ‖ ‖η‖ ∀t ∈ R.

Definition 5.1.4 For every ξ and η in H, denote by µξ,η the tempereddistribution on R such that

µξ,η(t) =1√2π

(e−itLξ, η) ∀t ∈ R.

For the sake of brevity, we shall write µξ instead of µξ,ξ. Furthermore, definethe map Jξ : S(R)→ H, by

Jξf = f(L)ξ,

where f(L) is defined in (5.1.4).

Remark 5.1.5 A straightforward computation shows that

4 µξ,η = µξ+η − µξ−η + i µξ+iη − i µξ−iη.

Hence

4µξ,η = µξ+η − µξ−η + i µξ+iη − i µξ−iη.

The Schwartz space, endowed with the uniform norm, is a normed ∗-algebrawith respect to the pointwise product of functions, and with involution given bycomplex conjugation. It is not a Banach algebra, because S(R) is not completewith respect to ‖ ‖∞. We already know that L(H) is a Banach ∗-algebra withrespect to the composition of operators and with involution given by the adjointmap.

We define the map Φ : S(R)→ L(H) by

Φ(f) = f(L) ∀f ∈ S(R).

The next proposition contains a few basic properties of Φ.

Proposition 5.1.6 The map Φ is a ∗-homomorphism from the normed ∗-algebra S(R) (endowed with the uniform norm) to the Banach ∗-algebra L(H)(endowed with the operator norm).

The proof of Proposition 5.1.6 hinges on the following lemma, which is of inde-pendent interest.

Lemma 5.1.7 Suppose that T is a tempered distribution such that T is a boundedcontinuous function and ⟨

|f |2, T⟩≥ 0 ∀f ∈ S(R).

Then T is the restriction to S(R) of a (nonnegative) regular Borel measure with

total variation√

2πT (0).


Proof. First we prove that〈ϕ, T 〉 ≥ 0

for every nonnegative function ϕ in C∞c (R). Given such a function ϕ and ε > 0,consider the function ψε =

√ϕ+ εG, where G denotes the Gaussian

G(x) = e−x2

∀x ∈ R.

It is straightforward to check that ψε is in S(R). By assumption,

〈ϕ+ εG, T 〉 =⟨ψ2ε , T

⟩≥ 0.

It is straightforward to check that ϕ+ εG is convergent to ϕ in S(R), as ε tendsto 0. Therefore

〈ϕ, T 〉 = limε→0〈ϕ+ εG, T 〉

≥ 0,

as required. Observe also that 〈ϕ, T 〉 is nonnegative whenever ϕ is a nonnegativeSchwartz function. Indeed, ϕ is the limit in the topology of S(R) of a sequenceof nonnegative functions with compact support.

Next we prove that

〈ϕ, T 〉 ≤√

2π T (0) ‖ϕ‖∞

for every nonnegative function ϕ in C∞c (R). Observe that for every c > 0

ec−cs2/R2

≥ 1 ∀s ∈ [−R,R].

Therefore for every ϕ in C∞c (R) with support contained in [−R,R]

ecG√c/R ‖ϕ‖∞ ≥ ϕ and − ecG

√c/R ‖ϕ‖∞ ≤ ϕ.

Here G√c/R(s) = G(

√cs/R) = e−cs

2/R2

. Since G is a Schwartz function⟨G√c/R, T

⟩≥ 0. The two inequalities above imply that for every real valued

Schwartz function ϕ,

|〈ϕ, T 〉| ≤ ec⟨G√c/R, T

⟩‖ϕ‖∞.

For c fixed, the family G√c/R : R > 0 is increasing with R, whence so is

⟨G√c/R, T

⟩: R > 0. Therefore⟨

G√c/R, T

⟩≤ lims→∞

⟨G√c/s, T

⟩∀R > 0.

We now prove that the limit above is equal to√

2πT (0). Set ε =√c/R. Since

G is even,〈Gε, T 〉 = 〈(Gε), (T )〉

=1√2ε

∫ ∞−∞

e−|v|2/(4ε) T (t) dt

=√

2

∫ ∞−∞

e−|t|2

T (2√εt) dt

→√

2π T (0),


as required. Here we have used the Lebesgue Dominated Convergence Theorem.We have proved that for every c > 0 and for every real valued ϕ in C∞c (R) withcompact support

|〈ϕ, T 〉| ≤ ec√

2π T (0) ‖ϕ‖∞.

By taking the infimum with respect to c > 0 we obtain that

|〈ϕ, T 〉| ≤√

2π T (0) ‖ϕ‖∞ ∀ϕ ∈ C∞c (R)

(ϕ real valued). Clearly this estimate implies a similar estimate for complex val-ued functions in C∞c (R). Thus, T is a bounded linear functional on C∞c (R) withrespect to the supremum norm. Hence it extends to a bounded linear functionalon C0(R)2, which is the closure of C∞c (R) with respect to the supremum norm.By a celebrated result of Riesz, T is a (positive) finite regular Borel measure.

Finally observe that

T (R) =√

2π T (0),

i.e. the total variation of T is√

2π T (0), as required. 2

Proof.[of Proposition 5.1.6] Observe preliminarily that

(f(L)ξ, η) =1√2π

∫ ∞−∞

f(t) (eitLξ, η) dt

=⟨f , (µξ,η˜)⟩

= 〈f, µξ,η〉 ∀ξ, η ∈ H.

(5.1.6)

By (5.1.5) we know that Φ(f) is in L(H) for every Schwartz function f . Weneed to prove that

(i) f(L)∗ = f(L);

(ii) (fg)(L) = f(L) g(L);

(iii) |||f(L)||| ≤ ‖f‖∞.

First we prove (i). Let ξ and η be in H. Then

(f(L)∗ξ, η) = (ξ, f(L)η) = (f(L)η, ξ) = 〈f, µη,ξ〉 =⟨f, µη,ξ

⟩=⟨f, µξ,η

⟩= (f(L)ξ, η),

as required. We have used the fact that µη,ξ = µξ,η in the fifth equality above.To prove this fact, note that

µη,ξ(t) = µη,ξ(−t) =1√2π

(eitLη, ξ

)= µξ,η.

2The Banach space of all continuous functions vanishing at infinity, endowed with thesupremum norm


Next we prove (ii). Notice that

(fg)(L)η =1√2π

∫ ∞−∞

(fg)(t) eitLη dt

(by (5.1.1)) =1

2π

∫ ∞−∞

(f ∗ g)(t) eitLη dt

=1

2π

∫ ∞−∞

∫ ∞−∞

f(t− s) ei(t−s)L[eisLη

]g(s) dsdt

(t− s = v) =1

2π

∫ ∞−∞

dv f(v) eivL(∫ ∞−∞

eisLη g(s) ds)

= f(L)(g(L)η

),

as required.

To prove (iii), we observe that (5.1.6) and (ii) (with f in place of g) implythat ⟨

|f |2, µξ⟩

= (|f |2(L)ξ, ξ) = (f(L)∗f(L)ξ, ξ) = ‖f(L)ξ‖2

≥ 0 ∀f ∈ S(R).(5.1.7)

By Lemma 5.1.7, µξ is a nonnegative regular Borel measure, with total variation

µξ(R) =√

2π µξ(0) = (e−itLξ, ξ)|t=0 = ‖ξ‖2.

By (5.1.7) and the formula above

‖f(L)ξ‖2 =

∫R|f |2 dµξ

≤ ‖f‖∞2µξ(R)

= ‖f‖∞2 ‖ξ‖2 ∀ξ ∈ H.

Hence |||f(L)||| ≤ ‖f‖∞, as required to conclude the proof of (iii) and of theproposition. 2

It is well known that the completion of (S(R), ‖ ‖∞) is isomorphic to the Banachalgebra (C0(R), ‖ ‖∞), where C0(R) denotes the space of continuous functionson R vanishing at infinity. By Proposition 5.1.6, Φ admits a unique continuousextension to a ∗-homomorphism from C0(R) into L(H).

We shall describe further extensions of the functional calculus Φ. First weestablish some of the properties of Φ(f) when f in the Schwartz class.

Proposition 5.1.8 Suppose that L is a self adjoint operator on H, f and g arein S(R) and ξ and η are in H. Then

(i) µξ is a positive regular Borel measure, with total variation ‖ξ‖2 and µξ,ηis a complex regular Borel measure

(ii) the following hold:(f(L)ξ, g(L)η

)=

∫Rf g dµξ,η and

(f(L)ξ, η

)=

∫Rf dµξ,η.


In particular,

‖f(L)ξ‖2 =

∫R|f |2 dµξ.

Proof. In Proposition 5.1.6, we have already proved µξ is a positive regular

Borel measure, with total variation ‖ξ‖2. By Remark 5.1.5 and Fourier inversionformula µξ,η is a linear combination of four finite positive Borel measures, henceit is a complex measure. By (5.1.6) and the fact that µξ,η is a complex measure,we may conclude that (

f(L)ξ, η)

= 〈f, µξ,η〉

=

∫ ∞−∞

f dµξ,η,(5.1.8)

and the second formula in (ii) is proved. The first follows from the second (withfg in place of f), because(

f(L)ξ, g(L)η)

=((fg)(L)ξ, η

).

The third formula in (ii) follows from the first with g = f and η = ξ. 2

Exercise 5.1.9 Suppose that L is a self adjoint operator on H and that f andg are in S(R). The following hold:

(i) if f is real valued, then f(L) is self adjoint;

(ii) if f is nonnegative, then f(L) is a nonnegative operator on H, i.e.

(f(L)ξ, ξ) ≥ 0 ∀ξ ∈ H;

(iii) if Lψ = λψ for some complex number λ, then f(L)ψ = f(λ)ψ;

(iv) for every ζ in C and f in S(R) define fζ by

fζ(t) = (ζ − t) f(t) ∀t ∈ R.

Then fζ is in S(R) and

(ζI − L) f(L)η = fζ(L) η = f(L) (ζI − L)η ∀η ∈ Dom(L).

Exercise 5.1.10 Prove that µξ,η is a complex measure with total variation atmost ‖ξ‖ ‖η‖. (Hint : use the Radon–Nikodym Theorem.)

Exercise 5.1.11 Prove that if λ is an eigenvalue with normalised eigenvectorξλ, then µξλ = δλ, the Dirac mass at λ. (Hint : look at the proof of Exer-cise 5.1.9 (iii).)


5.1.3 The extended functional calculus

In this section we extend the Schwartz functional calculus to a fairly wide classof functions. First, we remark that the Schwartz class S(R) is dense inL2(µξ) for every ξ in H. Indeed, by a well known result [Ru, Thm 3.14] Cc(R)is dense in L2(µξ); moreover, C∞c (R) is dense in Cc(R) in the uniform topology,whence in the L2(µξ) norm, because µξ is a finite measure.

Proposition 5.1.12 The map Jξ (see Definition 5.1.4) extends uniquely to anisometry, which we still denote by Jξ, from L2(µξ) to H. For all f and g inL2(µξ) (∫

R|f |2 dµξ

)1/2= ‖Jξf‖,

(Jξ(f),Jξ(g)

)=

∫Rf g dµξ

and(Jξ(f), ξ

)=∫R f dµξ.

Proof. By Proposition 5.1.8 (ii), Jξ is an isometry from the normed (non-complete) space S(R), endowed with the L2(µξ) norm, into H. By abstractnonsense, Jξ extends uniquely to an isometry, still denoted Jξ, from the com-pletion of S(R), i.e. L2(µξ), to H. Explicitly, given f in L2(µξ), there exists asequence fn in S(R) such that

limn→∞

‖f − fn‖2 = 0.

Then Jξf := limn→∞ Jξ(fn). It is straightforward to check that Jξ(f) does notdepend on the sequence fn in S(R) chosen to approximate f in L2(µξ).

Since Jξ is an isometry, the first and the second formula in the statementhold. The third formula follows from the second formula in Proposition 5.1.8 (ii)(with η = ξ), for Jξ is an isometry. 2

The isometry Jξ constructed in Proposition 5.1.12 need not be surjective.An important case in which Jξ is onto is when ξ is a cyclic vector (see Defini-tion 5.3.1 below). This fact will play a key role in the proof of the multiplicativeform of the spectral theorem (see Theorem 5.3.3).

Definition 5.1.13 Let f be a (possibly unbounded) Borel function on R.Define the operator Φ(f) as follows:

Dom(Φ(f)

):= ξ ∈ H : f ∈ L2(µξ)

andΦ(f)ξ := Jξf ∀ξ ∈ Dom

(Φ(f)

)(µξ and Jξ are defined in Definition 5.1.4).

If f is in S(R), then Φ(f) agrees with the operator f(L) defined in (5.1.4),because when f is in S(R), Jξf is precisely defined to be f(L)ξ (see Defini-tion 5.1.4).

The functional calculus introduced in Definition 5.1.13 applies to a large class offunctions. A natural question is whether L, the resolvent operator R(ζ;L)


and the unitary group eitL may be obtained as Φ(f) for suitable f(see Definition 5.1.13). This is the content of the next proposition. First weneed more notation. We denote by eit, j and rζ the functions defined by

eit(s) := eist, j(s) := s and rζ(s) :=1

ζ − s∀s ∈ R.

Here ζ is in C\R. We denote by Φ(eit), Φ(j) and Φ(rζ) the operators associatedto eit, j and rζ , respectively, as in Definition 5.1.13.


Φ(1R) = I, Φ(eit) = eitL, Φ(j) = L and Φ(rζ) = R(ζ;L);

here R(ζ;L) denotes the resolvent operator of L with pole ζ.

Proof. We prove the first formula. For every ξ in H, the measure µξ is finite,so that 1R is in L2(µξ), whence Dom(Φ(1R)) = H. Furthermore, by Proposi-tion 5.1.12,

(Φ(1R)ξ, ξ) =

∫ ∞−∞

1R dµξ

= µξ(R)

(by Proposition 5.1.8) = ‖ξ‖2,which implies the required conclusion.

Now we prove the second formula. By definition Φ(eit)ξ = Jξ(eit), so thatthe third formula in Proposition 5.1.12 gives(

Φ(eit)ξ, ξ)

=

∫ ∞−∞

eits dµξ(s)

(by Remark 5.1.1) =√

2π µξ(−t)=(eitLξ, ξ

)∀ξ ∈ H.

This implies that Φ(eit) = eitL, as required.

Next we prove the third formula. By the second formula, the infinitesimalgenerator of Φ(eit) is −iL. We shall prove that the infinitesimal generator ofΦ(eit) is −iΦ(j), and the required conclusion will follow.

Suppose that ξ is in Dom(L). Observe that

d

dt(Φ(eit)ξ,Lξ) = i(Φ(eit)Lξ,Lξ)

= i

∫ ∞−∞

eit dµLξ

(by Lemma 5.2.3) = i

∫ ∞−∞

eit |j|2 dµξ.

Now we evaluate both sides at t = 0 and obtain

i‖Lξ‖2 = i

∫ ∞−∞|j|2 dµξ,


whence ξ is in Dom(Φ(j)). Furthermore

i(Lξ, ξ) =d

dt(Φ(eit)ξ, ξ)|t=0 =

d

dt

∫ ∞−∞

eit dµξ|t=0.

Observe that ∣∣∣eits − 1

t− is

∣∣∣ =∣∣∣i ∫ s

0

(eitu − 1) du∣∣∣

≤ 2|s| ∀s ∈ R.

Then, by the Lebesgue Dominated Convergence Theorem,

d

dt

∫ ∞−∞

eit dµξ|t=0 = i

∫ ∞−∞

j dµξ = i(Φ(j)ξ, ξ).

Thus, Φ(j) is an extension of L. Since L is self adjoint, iI+L is one to one ontoH. If Φ(j) were a proper extension of L, iI + Φ(j) would be a proper extensionof iI+L, so that iI+Φ(j) would not be injective. However, if ξ is in the kernelof iI + Φ(j), then

0 =((iI + Φ(j)

)ξ, ξ)

= (Φ(i+ j)ξ, ξ) =

∫ ∞−∞

(i+ j) dµξ

= i‖ξ‖2 +

∫ ∞−∞

j dµξ,

which forces ξ = 0, a contradiction. Thus, Φ(j) = L, as required.

Finally, we prove the fourth formula. Recall (from the proof of the Hille–Yosida Theorem) that for every ζ in C with Re(ζ) > 0

(R(ζ; iL)ξ, ξ

)=

∫ ∞0

e−ζt(eitLξ, ξ

)dt

(by the second formula) =

∫ ∞0

dt

∫ ∞−∞

e−ζt eits dµξ(s)

(by Fubini’s Theorem) =

∫ ∞−∞

dµξ(s)

∫ ∞0

e−ζt eits dt

=

∫ ∞−∞

1

ζ − isdµξ(s)

=(rζ(iL)ξ, ξ

)∀ξ ∈ H.

This implies that R(ζ; iL) = rζ(iL), hence that R(−iζ;L) = r−iζ(L) for all ζwith Re(ζ) > 0, i.e., R(w;L) = rw(L) for all w with Re(w) < 0. If Re(ζ) < 0,then we may argue similarly, starting with the formula

(R(ζ; iL)ξ, ξ

)=

∫ ∞0

eζt(e−itLξ, ξ

)dt,

and conclude that R(w;L) = rw(L) for all w with Re(w) > 0.

This concludes the proof of (iii) and of the proposition. 2


5.2 Lecture XXII: properties of the extendedfunctional calculus

Denote by BB(R) the Banach ∗-algebra of bounded Borel functions on Rwith respect to the uniform norm. Clearly C0(R) is a closed ∗-subalgebra ofBB(R).

Proposition 5.2.1 (Bounded functional calculus) The functional calculusΦ (see Definition 5.1.13) satisfies the following:

(i) |||Φ(f)||| ≤ ‖f‖∞ for all f in BB(R);

(ii) if fn is a sequence in BB(R) such that

fn → f pointwise and supn ‖fn‖∞ <∞,

then Φ(fn) tends to Φ(f) in the strong operator topology;

(iii) Φ is a continuous ∗-homomorphism from BB(R) to L(H);

(iv) for every ξ and η in H, and every f in BB(R)

(Φ(f)ξ, η

)=

∫ ∞−∞

f dµξ,η.

Proof. First we prove (i). Since µξ is a finite measure, every f in BB(R) is inL2(µξ). Hence Dom

(Φ(f)

)is all of H. Furthermore, for every ξ in H

‖Φ(f)ξ‖ = ‖Jξf‖(Jξ is an isometry) = ‖f‖L2(µξ)

≤ ‖f‖∞ µξ(R)1/2

(by Proposition 5.1.8) = ‖f‖∞ ‖ξ‖,

so that

|||Φ(f)||| ≤ ‖f‖∞,

as required.

Next we prove (ii). Suppose that fn tends to f pointwise, and supn ‖fn‖∞ <∞. Since µξ is a finite measure, ‖f − fn‖L2(µξ) → 0 by the Dominated Conver-gence Theorem. Then

limn→∞

Φ(fn)ξ = limn→∞

Jξfn

(because ‖f − fn‖L2(µξ) → 0) = Jξf= Φ(f)ξ ∀ξ ∈ H,

as required.

5.2. PROPERTIES OF THE EXTENDED FUNCTIONAL CALCULUS 91

Now, we prove (iii). The continuity of Φ has already been proved in (i). Toprove that Φ is a ∗-homomorphism observe that(

Φ(f)∗ξ, ξ)

=(Φ(f)ξ, ξ

)(by Proposition 5.1.12) =

∫ ∞−∞

f dµξ

=

∫ ∞−∞

f dµξ

(by Proposition 5.1.12) =(Φ(f)ξ, ξ

)∀ξ ∈ H,

so that Φ(f)∗ = Φ(f). Furthermore(Φ(f)Φ(g)ξ, ξ

)=(Φ(g)ξ,Φ(f)∗ξ

)=(Φ(g)ξ,Φ(f)ξ

)(by Proposition 5.1.12) =

∫ ∞−∞

f g dµξ

=(Φ(fg)ξ, ξ

)∀ξ ∈ H,

so that Φ(fg) = Φ(f) Φ(g), as required to conclude the proof of (iii).

Finally, (iv) follows by polarisation from Remark 5.1.5 and the third formulain Proposition 5.1.12. 2

Exercise 5.2.2 For every ε > 0, denote by Gε the function

Gε(t) = e−ε2t2 ∀t ∈ R.

The operator Gε(L) tends to the identity operator I in the strong operatortopology, i.e.,

limε→0+

‖Gε(L)ξ − ξ‖ = 0 ∀ξ ∈ H.

Proposition 5.1.14 shows that if f is unbounded, then the operator Φ(f)(see Definition 5.1.13) need not be bounded. It is natural to ask when Φ(f)is, at least, densely defined. We shall need the following result, which is ofindependent interest.

Lemma 5.2.3 Suppose that f is a (possibly unbounded) Borel function on Rand that ξ is in Dom(f(L)). Then

dµf(L)ξ = |f |2 dµξ

(i.e., µf(L)ξ is absolutely continuous with respect to µξ and the Radon–Nikodymderivative of µf(L)ξ with respect to µξ is equal to |f |2).

Proof. Suppose that E is a Borel set in R. We compute

µf(L)ξ(E) =

∫ ∞−∞

1E dµf(L)ξ

(by Proposition 5.1.12) =(Φ(1E)Φ(f)ξ,Φ(f)ξ

).


Choose a sequence fn in S(R) such that limn→∞ ‖fn − f‖L2(µξ) = 0 (recallthat f is in L2(µξ), because ξ is in Dom(Φ(f)) by assumption). We claim that

limn→∞

(Φ(1E)Φ(fn)ξ,Φ(fn)ξ

)=(Φ(1E)Φ(f)ξ,Φ(f)ξ

).

Indeed,∣∣(Φ(1E)Φ(fn)ξ,Φ(fn)ξ

)−(Φ(1E)Φ(f)ξ,Φ(f)ξ

)∣∣ may be estimated by

the sum of∣∣(Φ(1E)Φ(fn − f)ξ,Φ(fn)ξ

)∣∣ and∣∣(Φ(1E)Φ(f)ξ,Φ(fn − f)ξ

)∣∣. Eachof these two term is dominated by

|||Φ(1E)||| ‖Jξ(fn − f)‖ max(‖Jξfn‖, ‖Jξf‖

),

which tends to 0 as n tends to ∞.

By the properties of the bounded functional calculus(Φ(1E)Φ(fn)ξ,Φ(fn)ξ

)=(Φ(1E |fn|2)ξ, ξ

)=

∫ ∞−∞

1E |fn|2 dµξ

→∫ ∞−∞

1E |f |2 dµξ

as n tends to ∞. By combining the formulae above, we obtain that

µf(L)ξ(E) =

∫ ∞−∞

1E |f |2 dµξ =

∫E

|f |2 dµξ,

which is the required conclusion. 2

Corollary 5.2.4 Suppose that f is a Borel function on R that is finite every-where. Then Dom(f(L)) is dense in H.

Proof. Set En := |f | ≤ n. Since f is finite everywhere,⋃nEn = R. Suppose

that ξ is in Ran(1En(L)). We claim that ξ is in Dom(f(L)). Observe that

ξ = 1En(L)ξ,

for 1En(L) is idempotent by Proposition 5.2.1. Indeed, if ξ is in Ran(1En(L)

),

then there exists η in H such that ξ = 1En(L)η. Then

1En(L)ξ = 1En(L)2η

= 1En(L)η

= ξ,

as required. Therefore ∫ ∞−∞|f |2 dµξ =

∫ ∞−∞|f |2 dµ1En (L)ξ

(by Lemma 5.2.3) =

∫ ∞−∞

1En |f |2 dµξ

≤ n2 µξ(En)

≤ n2 ‖ξ‖2,

5.2. PROPERTIES OF THE EXTENDED FUNCTIONAL CALCULUS 93

hence ξ is in Dom(f(L)), as claimed.

Now, for any η in H

limn→∞

‖η − 1Enη‖ = limn→∞

[∫ ∞−∞|1− 1En(s)|2 dµη(s)

]1/2= 0

by the Dominated Convergence Theorem, so that Dom(f(L)) is dense, as re-quired. 2

Proposition 5.2.5 For each ξ in H the support of µξ is contained in the spec-trum of L.

Proof. Since L is self adjoint, its spectrum is contained in R. Since the spectrumis closed, R\σ(L) is the union of at most countably many open intervals. Weshow that µξ(I) = 0 for each such interval I.

Indeed, let B be any open disc in the complex plane such that B ∩ R = I.Let Bn be a strictly increasing sequence of discs contained in B and with thesame centre as B such that B =

⋃nBn. Set

Im := Bm ∩ R.

It suffices to show that µ(Im) = 0 for all m. Choose n > m. Then

µξ(Im) =

∫ ∞−∞

1Bm∩R(t) dµξ(t)

=

∫Bm∩R

dµξ(t)1

2πi

∫∂B+

n

dz

z − t;

the last inequality follows from the Cauchy integral formula. It is straightfor-ward to check that we may interchange the order of integration in the formulaabove by Fubini’s Theorem (note that |z − t| ≥ rn − rm, where rn and rm arethe radii of Bn and Bm, respectively). Thus,

µξ(Im) =1

2πi

∫∂B+

n

dz

∫Bm∩R

dµξ(t)

z − t

(by Lemma 5.2.3) =1

2πi

∫∂B+

n

dz

∫ ∞−∞

1

z − tdµ1Bm∩R(L)ξ(t)

=1

2πi

∫∂B+

n

(Rzξ

′m, ξ

′m) dz,

where Rz denote the resolvent of L with pole z and ξ′m stands for 1Bm∩R(L)ξ.Since z 7→ Rz is holomorphic outside the spectrum of L, the integral on the righthand side of the formula above vanishes, whence so does µξ(Im), as required. 2


Lemma 5.2.6 Suppose that ξ is in H, and f is in L2(µξ). Then for every ηin H the function f is in L1(

∣∣µξ,η∣∣), where |µξ,η| denotes the total variation ofthe complex measure µξ,η,

‖f‖L1(|µξ,η|) ≤ ‖f‖L2(µξ) ‖η‖

and (f(L)ξ, η

)=

∫ ∞−∞

f dµξ,η. (5.2.1)

Proof. By the Radon–Nikodym Theorem, there exists a measurable function hsuch that |h| = 1 and

dµξ,η = h d|µξ,η|.Suppose now that f is bounded. Then

‖f‖L1(|µξ,η|) =

∫ ∞−∞|f |d|µξ,η| =

∫ ∞−∞|f |hdµξ,η

(by Proposition 5.2.1 (iv)) =((|f |h)(L)ξ, η

)≤∥∥(|f |h)(L)ξ

∥∥ ‖η‖=(∫ ∞−∞|f |2 |h|2 dµξ

)1/2‖η‖

= ‖f‖L2(µξ) ‖η‖,

as required. The general case follows by approximating a (possibly unbounded)function f by a sequence fn of bounded functions converging to f pointwiseand satisfying |fn| ≤ |f |, and using Fatou’s Theorem on the left hand side andthe Dominated Convergence Theorem on the right hand side.

To prove (5.2.1), observe that both sides of (5.2.1) define a bounded conju-gate linear functional on H (as a function of η), and that the two sides agree if ηis in Dom(f(L)), by polarisation. Now, if Dom(f(L)) is dense, then a straight-forward density argument completes the proof. If Dom(f(L)) is not dense, thenwe may approximate f by a sequence fn of bounded functions converging tof pointwise and satisfying |fn| ≤ |f |. Then for each n(

fn(L)ξ, η) =

∫ ∞−∞

fn dµξ,η.

Furthermore, ∣∣∣∫ ∞−∞

fn dµξ,η −∫ ∞−∞

f dµξ,η

∣∣∣ ≤ ∫ ∞−∞|fn − f |d|µξ,η|

(by the first part of the proof) ≤ ‖fn − f‖L2(µξ) ‖η‖(by the DCT) → 0

and ∣∣(fn(L)ξ, η)−(f(L)ξ, η

)∣∣ ≤ ‖fn(L)ξ − f(L)ξ‖ ‖η‖

(by Proposition 5.1.12) =(∫ ∞−∞|fn − f |2 dµξ

)1/2‖η‖

(by the DCT) → 0.

This proves (5.2.1), and completes the proof of the proposition. 2

5.3. THE SPECTRAL THEOREM I 95

Exercise 5.2.7 Prove that if f and g are Borel functions on R, f is boundedand g is finite everywhere, then

f(L)g(L) = (fg)(L)

on Dom(g(L)).

Exercise 5.2.8 Prove that ξ is in Dom(L) if and only if µξ is twice differen-tiable at 0.

5.3 Lecture XXIII: The spectral theorem I

In this section we prove the multiplicative form of the spectral theorem, due tovon Neumann. We need some notation.

Suppose that µj is a sequence of positive Borel measures on R. We denote by`2(L2(µj)

)the space of all sequences fj, where fj is in L2(µj), such that

‖fj‖`2(L2) :=( ∞∑j=1

‖fj‖L2(µj)2)1/2

<∞.

Definition 5.3.1 Let L be a self adjoint operator in H. A vector ξ in H issaid to be a cyclic vector for L if

spaneitLξ : t ∈ R = H.

Exercise 5.3.2 Suppose that A is a self adjoint operator on a finite dimen-sional space. Prove the following:

(i) if A has multiple eigenvalues, then it does not admit cyclic vectors;

(ii) if all the eigenvalues of A are simple, then A has cyclic vectors.

Theorem 5.3.3 (von Neumann) Suppose that L is a self adjoint operator on aseparable Hilbert space H. Then there exist a σ-compact measure space (X,M, µ),where µ is a Borel measure, a real-valued measurable function a and a surjectiveisometry J : L2(µ)→ H such that

J−1LJ f = a f ∀f ∈ L2(µ) such that J f ∈ Dom(L).

Proof. First we prove the result in the case where L has a cyclic vector, ξsay. We shall show that in this case X = R, a(x) = x and µ is the measure µξdefined in Definition 5.1.4. Recall that µξ is a positive Borel measure on R byProposition 5.1.8 (i), and that the map Jξ : S(R)→ H defined by

Jξf = f(L)ξ,

extends to an isometry, also denoted by Jξ, from L2(µξ) to H.

We claim that Jξ is surjective. First observe that Ran(Jξ) is dense inH, because it contains spaneitLξ : t ∈ R, and ξ is cyclic.


Now we prove that Ran(Jξ) is closed. This will complete the proof of theclaim. Suppose that η is in H. Since Ran(Jξ) is dense, there exists a sequenceηk in Ran(Jξ) such that ηk → η as k tends to ∞. Since Jξ is isometric, Jξ isinvertible, and J−1ξ ηk is a Cauchy sequence in L2(µξ). Therefore there exists

g in L2(µξ) such that ‖J−1ξ ηk − g‖L2(µξ) tends to 0. Since Jξ is continuous

ηk = JξJ−1ξ ηk

→ Jξg.

Since ‖ηk − η‖ → 0, η = Jξg, so that η is in Ran(Jξ). This concludes the proofof the claim.

Next we show that

J−1ξ eitLJξf = Meit f ∀f ∈ L2(µξ). (5.3.1)

Note thateitLJξf =

(eitf

)(L)ξ = Jξ

(eitf

)∀f ∈ S(R)

by the bounded functional calculus (Proposition 5.2.1 (iii)), so that

J−1ξ eitLJξf = eit f

= Meit f ∀f ∈ S(R).

Since the Schwartz space is norm dense in L2(µξ), there exists a sequence fnin S(R) such that limn→∞ ‖fn − f‖L2(µξ) = 0. Since J−1ξ eitLJξ is bounded on

L2(µξ),limn→∞

‖J−1ξ eitLJξfn − J−1ξ eitLJξf‖L2(µξ) = 0.

Furthermore,limn→∞

‖eitfn − eitf‖L2(µξ) = 0.

ThusJ−1ξ eitLJξf = lim

n→∞J−1ξ eitLJξfn

= limn→∞

eit fn

= Meit f ∀f ∈ L2(µξ)

and (5.3.1) is proved.

Now suppose that f is in Dom(Ma). Note that∣∣∣ei(t+h) f − eit f

h− iaeit f

∣∣∣ = |f |∣∣∣eih − 1

h− ia

∣∣∣≤ 2 |f | |a|.

We have use the formula

eihx − 1

h− ix = i

∫ x

0

(eihu − 1) du.

Thus, the right hand side of (5.3.1) is strongly differentiable in L2(µξ) at everypoint and its derivative at t = 0 is equal to iaf , i.e., to iMaf . Now observe thatif f in Dom(Ma), then ∫ ∞

−∞s2 |f(s)|2 dµξ(s) <∞.


Furthermore, j is in L2(µJξf ). Indeed,∫ ∞−∞

s2 dµJξf (s) =

∫ ∞−∞

s2 |f(s)|2 dµξ(s)

by Lemma 5.2.3. Therefore Jξf is in the domain of −iL, i.e., in the domainof the infinitesimal generator of the group eitL. Then t 7→ eitLJξf is differ-entiable at every t and its derivative at t = 0 is equal to iLJξf . Since J−1ξ iscontinuous, the left hand side of (5.3.1) is differentiable at t = 0 with derivativeiJ−1ξ LJξf . We have proved that

J−1ξ LJξf = Ma f ∀f ∈ Dom(Ma),

as required. This concludes the proof of the theorem in the case where L admitsa cyclic vector.

Next we consider the case where L does not admit a cyclic vector. Forevery ξ in H we denote by Hξ the closed subspace spaneitLξ : t ∈ R. Observethat Hξ is eisLs∈R invariant. Indeed, suppose that η is in Hξ. Then thereexists a sequence ηk in spaneitLξ : t ∈ R which tends to η as k tends to∞. Since eisL is continuous, eisLηk tends to eisLη. Furthermore, eisLηk is againin spaneitLξ : t ∈ R by the linearity of eisL and the group property. Asa consequence, the restriction of eisL to Hξ is a group of unitary operatorson the Hilbert space Hξ, and its infinitesimal generator, which, by abstractnonsense, is densely defined in Hξ, agrees with the restriction of −iL to Hξ,i.e., it is the operator −iL on Dom(L) ∩Hξ.

Clearly, ξ is a cyclic vector of Hξ. Then by the first part of the theorem toL ↑ Hξ, we see that

Jξ : L2(µξ)→ Hξ

is a surjective isometry,

J−1ξ eitLJξf = Meit f ∀f ∈ L2(µξ),

and

J−1ξ LJξf = Ma f ∀f ∈ Dom(L ↑ Hξ).

Furthermore, H⊥ξ is eitLt∈R invariant because eitL is unitary.

We shall proceed iteratively. Let ξ1 be any nonzero vector in H, and considerHξ1 . Then take ξ2 in H⊥ξ1 , and consider Hξ2 . Then take ξ3 in (Hξ1 ⊕ Hξ2)⊥,and consider Hξ3 . By iterating this construction, we obtain a direct orthogonaldecomposition

H =

N⊕j=1

Hξj ,

where N is either a positive integer or ∞, each Hξj is eitLt∈R invariant, andξj is a cyclic vector for L ↑ Hξj , which is self adjoint on Dom(L) ∩Hξj .

We consider the Hilbert space `2(L2(µξj )) defined above. Let

U : `2(L2(µξj ))→ H


be defined by

U(f1, f2, f3, . . .) =

N∑j=1

Uξjfj =

N∑j=1

fj(L)ξj ;

note that fj(L)ξj is inHξj and that the sum above is an orthogonal sum. Clearly

‖U(f1, f2, f3, . . .)‖ =( N∑j=1

‖Uξjfj‖2)1/2

(because Uξj is an isometry) =( N∑j=1

∫R|fj |2 dµξj

)1/2= ‖(f1, f2, f3, . . .)‖`2(L2(µξj )).

Thus, U is an isometry and

U−1LU(f1, f2, f3, . . .) = (af1, af2, af3, . . .)

where a(s) = s, and U(f1, f2, f3, . . .) is in Dom(L). It is straightforward tocheck that U is surjective. Indeed, given η in H, there is a (unique) orthogonaldecomposition of η as

∑j ηj , where ηj is inHξj , and ηj is the image of a (unique)

function fj in L2(µξj ). Therefore U(f1, f2, f3, . . .) = η.

Now, set X =⋃Nj=1 Rj (the union of N copies of R), and consider the natural

topology on X: we say that a subset A of X is open if and only of there existopen sets Aj ⊂ Rj such that A =

⋃Nj=1Aj . It is straightforward to check that

X is σ-compact. Indeed, we may write X =⋃Nh=1K

(h), where

K(h) =

h⋃j=1

[−h, h]j ,

where [−h, h]j ⊂ Rj . We define a Borel measure on X as follows. Suppose that

Ej is a Borel set in Rj and consider E =⋃Nj=1Ej . We set

µ(E) :=

N∑j=1

µξj (Ej).

It is straightforward to check that µ is a Borel measure on X. To prove thatL2(µ) is isometrically isomorphic to `2

(L2(µξj )

), we associate to each function

f on X the sequence T f := (f1, f2, f3, . . .) of its restrictions to (R1,R2,R3, . . .).We show that T is the required isometry. Note that if

f =

H∑h=1

ch 1Eh , Eh ∩ Ek = ∅ if h 6= k, and Eh =⋃Nj=1E

hj with Ehj ⊂ Rj


is a simple function in L2(µ), then

‖f‖L2(µ)2

=

H∑h=1

|ch|2 µ(Eh)

(by definition of µ) =

N∑j=1

H∑h=1

|ch|2 µξj (Ehj )

=

N∑j=1

∫Rj

∣∣∣ H∑h=1

ch 1Ehj

∣∣∣2 dµξj

=

N∑j=1

∫Rj|fj |2 dµξj

= ‖(f1, f2, f3, . . .)‖`2(L2(µξ))2,

where we have denoted by fj the function∑Hh=1 c

h 1Ehj .

If now f is a nonnegative function in L2(µ), we may approximate f by asequence of simple functions converging to f from below. A straightforwardapplication of the monotone convergence theorem then gives

‖f‖L2(µ)2

= ‖(f1, f2, f3, . . .)‖`2(L2(µξ))2.

The general case follows by decomposing a generic complex valued function onX as the sum of u+ − u− + iv+ − iv−, where u and v denote the real and theimaginary part of f . We leave the details to the reader.

Finally, we show that T is surjective. Indeed, if (f1, f2, f3, . . .) is a functionin `2(L2(µξj )), denote by f the function on X whose restriction to Rj agreeswith fj . Then clearly T f = (f1, f2, f3, . . .), so that T is surjective.

Now the map J := U T is the required isometry of L2(µ) onto H. Thefunction a in the statement of the theorem is the function whose restriction toRj is equal to the identity. Thus, if f is a measurable function on X, then Mafis the function on X such that

Maf(x) = xRRjf(x) ∀x ∈ Rj ,

where RRj denotes the restriction operator from X to Rj .This concludes the proof of the theorem. 2

Remark 5.3.4 von Neumann’s result asserts that each self adjoint operatoron a separable Hilbert space is unitarily equivalent to a multiplication operatorwith a suitable real-valued function on a space L2(X,µ), where X is σ-compactand µ is σ-finite. Note, however, that there is no canonical choice of (X,M, µ),for the structure of the measure space depends on the subspaces Hξj , and thereis freedom in chosing them.

Exercise 5.3.5 Suppose that ξ is cyclic for L. Is it true that the support ofµξ is equal to σ(L)?


Exercise 5.3.6 Prove that λ is an eigenvalue of L if and only if λ is aneigenvalue of the multiplication operator Ma, where a is as in the statement ofvon Neumann’s theorem.

Definition 5.3.7 A self adjoint operator which admits cyclic vectors is saidto have simple spectrum.

Example 5.3.8 Let L denote the operator iD on

Dom(L) = W 1,2(R).

We already know that L is self adjoint. Furthermore, it is not hard to show that(eitLf

)(s) = f(s− t) ∀s, t ∈ R.

Let ξ be a function in L2(R) such that ξ is continuous and ξ(τ) 6= 0 for all τ inR. We prove that ξ is a cyclic vector for L.

We need to show that the linear span of translates of ξ is dense in L2(R),equivalently (by passing to the Fourier transform), that functions of the form

( N∑j=1

cj eitjt)ξ(τ)

are norm dense in L2(R).

Suppose that f is in L2(R) and let ε > 0. Choose N such that

‖f − fχ[−N,N ]‖2 < ε.

Let P be a trigonometric polynomial on [−N,N ] such that∥∥P − fχ[−N,N ]

ξ

∥∥2< ε;

note thatfχ[−N,N ]

ξis in L2([−N,N ]). Then

‖P ξ − fχ[−N,N ]‖2 ≤ ‖ξ‖∞∥∥P − fχ[−N,N ]

ξ

∥∥2

≤ ε ‖ξ‖∞.

Therefore

‖f − P ξ‖2 ≤ ‖f − fχ[−N,N ]‖2 + ‖fχ[−N,N ] − P ξ‖2≤ ε

(1 + ‖ξ‖∞

),

which proves that ξ is a cyclic vector.

Next, we construct the spectral measure. Let ζ be defined by

ζ(t) =1√2π

(eitLξ, ξ) ∀t ∈ R.


Since the group eitL is just the translation group, we find that

ζ(t) =1√2π

ξ ? ξ(t),

where ξ(t) = ξ(−t). Thus,

dµ(τ) =1√2π|ξ(τ)|2 dτ.

The associated isometry is then given by

J f = f(L)ξ

=

∫ ∞−∞

f(t) eitLξ dt

=

∫ ∞−∞

f(t) τtξ dt

= f ∗ ξ.

Note that

J−1eitLJ f = J−1τt(f ∗ ξ)

= J−1[(τtf) ∗ ξ

]= J−1

[(Meitf) ∗ ξ]

= Meit(f).

For every ε > 0 let hε be defined by

hε(t) = (4πε)−1/2 e−t2/(4ε) ∀t ∈ R.

By taking hε instead of ξ in the above argument we obtain isometries Jε :L2(R, e−ε|τ |2 dτ)→ H, defined by

Jεf = f ∗ hε.

By passing to the limit as ε tends to 0, we obtain the isometry J0 : L2(R) →L2(R) given by

J0f = f ,

and the corresponding spectral measure

dµ0(t) =1√2π

dτ.

Von Neumann’s spectral theorem extends to finite sets of commuting operators.

Definition 5.3.9 Let L1, . . . ,Lk be self adjoint operators on a separableHilbert space H. We say that L1, . . . ,Lk commute, or that they are per-mutable, if the unitary groups eitL1 , . . . , eitLk commute.


Definition 5.3.10 Let L1, . . . ,Lk be commuting self adjoint operators ona separable Hilbert space H. We say that a vector ξ in H is a cyclic for(L1, . . . ,Lk) if

H = spaneit1L1 · · · eitkLk : (t1, . . . , tk) ∈ Rk.

In this case we say that (L1, . . . ,Lk) has simple spectrum.

Theorem 5.3.11 (von Neumann) Suppose that L1, . . . ,Lk are permutableself adjoint operators on a separable Hilbert space H, and that ξ is cyclic for(L1, . . . ,Lk). Then there exist a Borel measure µ on Rk and a unitary mapJ : L2(µ)→ H such that

J−1eitLjJ f(τ1, . . . , τk) = eitτj f(τ1, . . . , τk) ∀f ∈ L2(µ) ∀j ∈ 1, . . . , k.

Proof. The proof is similar to the proof of Theorem 5.3.3 and is omitted. 2

A result similar to von Neumann’s theorem 5.3.3 holds also in the case where(L1, . . . ,Lk) does not admit a cyclic vector. We leave the reader the task ofstating and proving such result.

We conclude this section with a remark about the extended functional calculusconstructed in Definition 5.1.13. Suppose that m is a Borel function on R. Wemay consider the operator Φ(m) on the domain

η ∈ H : m ∈ L2(µη),

and the operator J−1Φ(m)J on L2(X,µ), where the notation is as in thestatement of Theorem 5.3.3. It is natural to speculate whether the opera-tors J−1 Φ(m)J and Mma agree; here Mma is the multiplication operatorby m a on L2(X,µ).

Proposition 5.3.12 Suppose that m is a Borel function on R. The isometryJ (constructed in Theorem 5.3.3) is a one to one map between Dom(Mma)and Dom(m(L)). Furthermore

J−1m(L)J f = Mmaf ∀f ∈ Dom(Mma).

Proof. First we assume that L has a cyclic vector, ξ say, in which case Jis just Jξ. Furthermore, since a(s) = s, we write m instead of m a.

Note that f is in Dom(Mm) if and only if∫ ∞−∞|mf |2 dµξ <∞.

Since Jξf = f(L)ξ, and dµf(L)ξ = |f |2 dµξ by Lemma 5.2.3, this is equivalentto ∫ ∞

−∞|m|2 dµJξf <∞.

Thus, f is in Dom(Mm) if and only if m is in L2(µJξf ), i.e., if and only if Jξfis in Dom(m(L)).


Furthermore, there exist sequences fn and mn of bounded measurablefunctions, such that |fn| ≤ |f |, |mn| ≤ |m|,

fn → f and mn → m pointwise.

Since Jf(L)ξ is an isometry from L2(µJξf ) to H and mn → m in the L2(µJξf )norm (by the Dominated Convergence Theorem), Jf(L)ξ(mn) tends to Jf(L)ξ(m)in H, so that

(m(L)f(L)ξ, η) = limn→∞

(mn(L)f(L)ξ, η)

(because mn is in BB(R)) = limn→∞

(f(L)ξ,mn(L)η) ∀η ∈ H.

Similarly, since Jξ is an isometry from L2(µξ) to H and fn → f in the L2(µξ)norm (by the Dominated Convergence Theorem), Jξ(fn) tends to Jξ(f) in H,so that

(m(L)f(L)ξ, η) = limn→∞

limj→∞

(fj(L)ξ,mn(L)η)

(by the BB(R) functional calculus) = limn→∞

limj→∞

((mnfj)(L)ξ, η)

(by Proposition 5.2.1 (iv)) = limn→∞

limj→∞

∫ ∞−∞

mn fj dµξ,η ∀η ∈ H.

Now, we know that mf is in L2(µξ), whence, by Lemma 5.2.6, mf is in L1(|µξ,η|)for every η in H. By the Dominated Convergence Theorem (applied twice)

limn→∞

limj→∞

∫ ∞−∞|mn fj −mf |d|µξ,η| = 0,

so that

limn→∞

limj→∞

∫ ∞−∞

mn fj dµξ,η =

∫ ∞−∞

mf dµξ,η

= ((mf)(L)ξ, η) ∀η ∈ H,

where the last equality follows from Lemma 5.2.6 (because mf is in L2(µξ)).Then m(L)Jξf = Jξ(mf), and the required formula follows.

Next we consider the general case (i.e., L does not admit a cyclicvector). Suppose that f is in L2(µ). Denote by fj and mj the restrictions off and m to Rj , respectively (see the proof of Theorem 5.3.3 for the notation).Note that mj is also the restriction of m a to Rj , for the restriciton of a to Rjis the identity (see the proof of Theorem 5.3.3).

Note that f is in Dom(Mma) if and only if∫X

|(m a) f |2 dµ =

N∑j=1

∫ ∞−∞|mjfj |2 dµξj <∞.

Since |fj |2 dµξj = dµJξj fj , f is in Dom(Mma) if and only if

N∑j=1

∫ ∞−∞|mj |2 dµJξj fj <∞,


i.e., if and only ifm is in L2(µJ f ), equivalently, if and only if J f is in Dom(m(L)).

The proof that m(L)J f = J (mf) in the general case is similar to that inthe case where L admits a cyclic vector. We omit the details. 2

5.4 Lecture XXIV: projection valued measures

Definition 5.4.1 Let (M,M) be a measurable space. A projection valuedmeasure on M is a map P : M→ L(H) such that

(i) P(∅) = 0 and P(M) = I;

(ii) P(E) is an orthogonal projection for every E in M;

(iii) P(E ∩ F ) = P(E)P(F );

(iv) if E ∩ F = ∅, then P(E ∪ F ) = P(E) + P(F );

(v) for every ξ and η in H the map Pξ,η, defined by

Pξ,η(E) = (P(E)ξ, η),

is a complex measure.

In the case where M is the Borel σ-algebra of a locally compact Hausdorff space,we require also that Pξ,η be regular.

We shall write Pξ instead of Pξ,ξ. Note the following direct consequences of thedefinition:

(i) Pξ is a positive measure and Pξ(M) = ‖ξ‖2;

(ii) P(E)P(F ) = P(F )P(E) for every E and F in M;

(iii) if E ∩ F = ∅, then P(E)P(F ) = P(∅) = 0. Since P(E) and P(F ) areorthogonal projections, it follows that

Ran(P(E)) ⊥ Ran(P(F ));

(iv) P is finitely additive but not countably additive in the uniform topologyof operators.

The finite additivity is an immediate consequence of Definition 5.4.1 (iv).To check that P is not in general countably additive, it suffices to observethat if P(En) is not 0, then |||P(En)||| = 1. Therefore if only finitely manyprojections P(En) vanish, the series

∑n P(En) cannot converge in the

uniform topology;

(v) for every ξ in H, the map from M to H, defined by E 7→ P(E)ξ, iscountably additive.

5.4. PROJECTION VALUED MEASURES 105

Indeed, suppose that En are pairwise disjoint and set E =⋃∞n=1En.

Since for every η in H, Pξ,η is a complex measure,

(P(E)ξ, η

)=

∞∑n=1

(P(En)ξ, η

)= limN→∞

( N∑n=1

P(En)ξ, η). (5.4.1)

Now, the series∑∞n=1 P(En)ξ is convergent in H. Indeed, since the pro-

jections P(En) are mutually orthogonal,

∞∑n=1

‖P(En)ξ‖2 ≤ ‖ξ‖2,

by Bessel’s inequality. Hence

limN→∞

∞∑n=N

‖P(En)ξ‖2 = 0.

Furthermore, for every Q ≥ N

[ Q∑n=N

‖P(En)ξ‖2]1/2

=∥∥∥ Q∑n=N

P(En)ξ∥∥∥ ,

so that

limN→∞

∥∥∥ Q∑n=N

P(En)ξ∥∥∥ = 0,

and the series∑∞n=1 P(En)ξ is convergent by Cauchy’s criterion. Now

(5.4.1) implies that the sum of the series is P(E)ξ, as required.

(vi) If P(En) = 0 for every n, then P(E) = 0.

Indeed, for ξ in H

‖P(E)ξ‖2 =(P(E)ξ, ξ) = Pξ(E) =

∞∑n=1

Pξ(En) = 0,

as required.

An important family of functions of the operator L is the following. To eachBorel set E in R, we associate the bounded operator 1E(L). Since ‖1E‖∞ = 1,we have |||1E(L))||| ≤ 1, by Proposition 5.2.1 (i).

Proposition 5.4.2 Suppose that L is a self adjoint operator on H. The map-ping P : B → L(H), defined by

P(E) = 1E(L)

is a projection valued measure.


Proof. First we prove that for every E in R the operator 1E(L) is a self adjointprojection on H. We need to show that 1E(L)∗ = 1E(L), and that 1E(L)2 =1E(L). This follows directly from Proposition 5.2.1, and the formulae 1E = 1Eand 12

E = 1E .

Next we show that E 7→ 1E(L) is a measure. Clearly P(∅) = 0 and P(R) = I.Furthermore, the formula 1E∩F (L) = 1E(L) 1F (L) follows from 1E∩F = 1E 1Fand Proposition 5.2.1. The proof of the other properties of P is similar and isomitted. 2

5.5 Lecture XXV: spectral resolutions of theidentity

The purpose of this lecture is to introduce the so called spectral resolutionsof the identity, which are important tools in the spectral theory of self adjointoperators. The relationship between a spectral measure and the associatedspectral resolution of the identity is similar to that between a measure and itsdistribution function.

Definition 5.5.1 Suppose that I is a possibly unbounded interval of the realline and that F : I → C. We say that F is of bounded variation if thereexists a constant C such that for every positive integer N and finite number ofpoints

x0 < x1 < . . . < xN−1 < xN

we have ∑j

|F (xj)− F (xj−1)| ≤ C.

The infimum of all constants C for which this holds is called the total variationof F and is denoted by ‖F‖BV .

We recall that a complex valued function F on R is of bounded variation ifand only if it may be written as

F1 − F2 + iF3 − iF4,

where the Fj are real increasing functions such that

limx→∞

Fj(x)− limx→−∞

Fj(x) <∞.

If F is a function of bounded variation which is right continuous at all points,then each of the functions Fj is increasing and right continuous at every point.We associate to Fj the corresponding Lebesgue–Stjeltjes measure µj , and define

µ(E) := µ1(E)− µ2(E) + iµ3(E)− iµ4(E)

for every Borel measurable subset E of the real line. Since Fj has finite totalvariation, µj is a finite Borel measure and µ is a complex Borel measure on R.

5.5. SPECTRAL RESOLUTIONS OF THE IDENTITY 107

Definition 5.5.2 A family Ps : s ∈ R of orthogonal projections on His called a spectral resolution of the identity if it possesses the followingproperties:

(i) PsPs′ = Pmin(s,s′);

(ii) for every ξ in H

lims↓−∞

Psξ = 0 and lims↑∞Psξ = I;

(iii) for every ξ in Hlims↓s0Psξ = Ps0ξ.


(i) if s < s′, then Ps − Ps′ is an orthogonal projection;

(ii) if s < t < u, then Ran(Pt − Ps) and Ran(Pu − Pt) are orthogonal;

(iii) for every ξ and η in H the function s 7→ (Psξ, η) is a function of boundedvariation, with total variation ≤ ‖ξ‖ ‖η‖.

Proof. First we prove (i). Since

[Ps′ − Ps

]2=[Ps − Ps′

] [Ps − Ps′

]= P2

s − PsPs′ − Ps′Ps + Ps′= Ps − Ps − Ps + Ps′= −Ps + Ps′ ,

Ps − Ps′ is a projection. We have repeatedly used Definition 5.5.2 (i). ByExercise 1.3.12, to prove that Ps − Ps′ is an orthogonal projection it remainsto show that Ps − Ps′ is self adjoint. This is clear, for Ps′ − Ps is a linearcombination with real coefficients of self adjoint operators.

Now we prove (ii). Given ξ and η in H,((Pt − Ps)ξ, (Pu − Pt)η

)=(ξ, (Pt − Ps)(Pu − Pt)η

)= 0,

because

(Pt − Ps) (Pu − Pt) = PtPu − P2t − PsPu + PsPt

= Pt − Pt − Ps + Ps= 0,

as required.


Finally we prove (iii). Suppose that s0 < s1 < s2 < sN . Since Psj − Psj−1

is an orthogonal projection,

N∑j=1

∣∣(Psjξ, η)− (Psj−1ξ, η)

∣∣=

N∑j=1

∣∣((Psj − Psj−1)ξ, (Psj − Psj−1

)η)∣∣

≤N∑j=1

∥∥(Psj − Psj−1)ξ∥∥ ∥∥(Psj − Psj−1

)η∥∥

≤( N∑j=1

∥∥(Psj − Psj−1)ξ∥∥ 2)1/2 ( N∑

j=1

∥∥(Psj − Psj−1)η∥∥ 2)1/2

;

the last inequality follows from Schwarz’s inequality. Since (Psj − Psj−1)ξ and

(Psk − Psk−1)ξ are orthogonal for j 6= k,

N∑j=1

∥∥(Psj − Psj−1)ξ∥∥ 2

=∥∥(PsN − Ps0)ξ

∥∥ 2 ≤ ‖ξ‖2.

SimilarlyN∑j=1

∥∥(Psj − Psj−1)η∥∥ 2

=∥∥(PsN − Ps0)η

∥∥ 2 ≤ ‖η‖2

The required estimate follows. 2

Suppose that f is a continuous function on R, and that ξ is in H. For eachpartition P := α = s0 < s1 < . . . < sM = β of [α, β], we consider thecorresponding Riemann–Stjelties sums

S(f ;P ; ξ) :=

M∑j=1

f(s′j)[Psjξ − Psj−1

ξ],

where s′j is in [sj−1, sj ]. We set δ(P ) := maxsj − sj−1 : j = 1, . . . ,M.

Proposition 5.5.4 Suppose that f is a continuous function on R, and that ξis in H. Then the limit of S(f ;P ; ξ) as δ(P ) tends to 0 exists.

Proof. Given ε > 0, let δ > 0 be such that |s− s′| < δ implies |f(s)− f(s′)| < ε.Suppose that P is as above and that δ(P ) < δ. Consider another partitionQ := α = t0 < t1 < . . . < tN = β of [α, β] such that δ(Q) < δ and theassociated Riemann sums

S(f ;Q; ξ) :=

N∑j=1

f(t′j)[Ptjξ − Ptj−1

ξ],

where t′j is in [tj−1, tj ]. Let P ∪ Q := α = u0 < u1 < . . . < uO = β be thesuperposition of the two partitions. Then

S(f ;P ; ξ)− S(f ;Q; ξ) =

O∑`=1

ε`[Pujξ − Puj−1ξ

],

5.6. OPERATORS ASSOCIATED TO SPECTRAL RESOLUTIONS 109

where |ε`| ≤ 2ε for all `. Since the summands on the right hand side are pairwiseorthogonal,

‖S(f ;P ; ξ)− S(f ;Q; ξ)‖ ≤ ε( O∑`=1

∥∥Pujξ − Ptu−1ξ∥∥ 2)1/2

≤ ε ‖ξ‖.

Thus, the required limit exists. 2

Definition 5.5.5 Suppose that f is a continuous function on R, and that ξis in H. Then the limit of S(f ;P ; ξ) as δ(P ) tends to 0 is denoted by∫ β

α

f(s) dPsξ.

We define ∫ ∞−∞

f(s) dPsξ

to be the limit in norm, whenever it exists, of∫ βαf(s) dPsξ as α tends to −∞

and β tends to ∞.

5.6 Lecture XXVI: Operators associated to spec-tral resolutions

We now show how to associate a self adjoint operator to a given spectralresolution of the identity Ps.For each ξ and η in H, we denote by νξ,η the Lebesgue–Stieltjes measureassociated to the distribution function

s 7→(Psξ, η

)∀s ∈ R.

Note that this function is right continuous at every point (a consequenceof Definition 5.5.2 (iii)). We write νξ instead of νξ,ξ. A simple calculation givesthe following polarisation formula

4 νξ,η = νξ+η − νξ−η + iνξ+iη − iνξ−iη. (5.6.1)

Note that this formula implies that

4 |νξ,η| ≤ νξ+η + νξ−η + νξ+iη + νξ−iη. (5.6.2)

Note that νξ is a positive measure, because s 7→(Psξ, η

)is an increasing

function on R, and it is a regular Borel measure.

Lemma 5.6.1 The set ξ ∈ H :

∫ ∞−∞

s2 dνξ <∞

is a vector space, dense in H.


Proof. Denote by E the set in the statement of the lemma. To prove that E isa vector space, suppose that ξ1 and ξ2 are in E. Observe that for −∞ ≤ r ≤s ≤ ∞

νξ1+ξ2((r, s]

)=((Ps − Pr)(ξ1 + ξ2), (ξ1 + ξ2)

)= νξ1

((r, s]

)+ 2 Re

((Ps − Pr)ξ1, ξ2

)+ νξ2

((r, s]

)≤ 2 νξ1

((r, s]

)+ 2 νξ2

((r, s]

).

Similarly, we can prove that

νξ1+ξ2(I) ≤ 2 νξ1(I) + 2 νξ2(I)

for every interval I. Now the assumption j2 in L1(νξ1)∩L1(νξ2) implies that j2

is in L1(νξ1+ξ2), i.e., that ξ1 + ξ2 is in E. A similar, perhaps easier, argumentshows that if j2 is in L1(νξ) and α is a complex number, then j2 is in L1(ναξ),so that αξ is in E. Thus, E is a vector space, as required.

Now we prove that E is dense in H. Suppose that η is in H. Observepreliminarily that if −∞ < s0 < s1 < ∞, then Ps1η − Ps0η is in E. We needto show that the function s 7→ s2 is integrable with respect to the measure νη′

where, for the sake of brevity, we have denoted by η′ the vector Ps1η − Ps0η.We shall prove that νη′ is supported in the interval [s0, s1], and the requiredintegrability property will follow. Observe that

(Ptη′, η′) =

0 if t ≤ s0‖Ptη − Ps0η‖

2if s0 < t ≤ s1

‖Ps1η − Ps0η‖2

if s1 < t.

Hence the support of νη′ is contained in [s0, s1] as claimed.

Now we prove that η can be approximated by elements of the form Ps1η −Ps0η. Indeed,∥∥η − (Ps1η − Ps0η)∥∥ 2

=∥∥Ps0η∥∥ 2

+∥∥(I − Ps1)η∥∥ 2

,

which tends to 0 as s0 tends to −∞ and s1 tends to∞, because of the propertiesof spectral resolutions of the identity. 2

Definition 5.6.2 Define the operator T by

Dom(T ) :=ξ ∈ H :

∫ ∞−∞

s2 dνξ(s) <∞

and

(T ξ, ξ) =

∫ ∞−∞

sdνξ(s) ∀ξ ∈ Dom(T ).

Since Dom(T ) is dense inH by Lemma 5.6.1, the formula above defines uniquelyT ξ for all ξ in Dom(T ).

In the next lemma we prove some properties of the operator T that will be usedbelow.


Lemma 5.6.3 The following hold:

(i) if ξ is in Dom(T ), then Ptξ is in Dom(T ) for every t in R;

(ii) for every ξ and η in Dom(T )

(T ξ, η) =

∫ ∞−∞

sdνξ,η(s);

(iii) T is symmetric;

(iv) for every ξ in Dom(T )T Ptξ = PtT ξ; (5.6.3)

(v) if ξ is in Dom(T ), then

‖T ξ‖2 =

∫ ∞−∞

s2 dνξ(s);

(vi) T is a closed operator.

Proof. First we prove (i). By Proposition 5.5.3

νPtξ((−∞, s]

)=

νξ((−∞, s]

)∀s ∈ (−∞, t]

‖Ptξ‖2 ∀s ∈ (t,∞)

for every s in R. Hence∫ ∞−∞

s2 dνPtξ(s) =

∫ t

−∞s2 dνξ(s)

≤∫ ∞−∞

s2 dνξ(s),

which is finite by assumption.

To prove (ii) note that, by the polarisation formula (5.6.2), if ξ and η are inDom(T ), then j2 is in L1(|νξ,η|), and then (5.6.1) implies that

(T ξ, η) =

∫ ∞−∞

sdνξ,η.

We now prove (iii). Note that if ξ and η are in Dom(T ), then νξ,η = νη,ξ.Indeed,

νξ,η((−∞, s]

)=(Psξ, η

)=(ξ,Psη

)=(Psη, ξ

)= νη,ξ

((−∞, s]

)∀s ∈ R.

Therefore

(T ξ, η) =

∫ ∞−∞

sdνη,ξ(s)

= (T η, ξ)= (ξ, T η),


as required to prove the symmetry of T .

Next we prove (iv). Observe that

(PtT ξ, ξ) = (T ξ,Ptξ)(T is symmetric and Ptξ is in Dom(T )) = (ξ, T Ptξ)

= (T Ptξ, ξ)

(by definition of T ) =

∫ ∞−∞

sdνPtξ,ξ

(νPtξ,ξ is real) =

∫ ∞−∞

sdνPtξ,ξ

(by definition of T ) = (T Ptξ, ξ) ∀ξ ∈ Dom(T ).

Since Dom(T ) is dense in H, the required property follows.

We prove (v). We determine the distribution function of νξ,PtT ξ for any ξin Dom(T ):

νξ,PtT ξ((−∞, s]

)= (Psξ,PtT ξ)= (Pmin(s,t)ξ, T ξ)

(by (i) and (ii)) =

∫ ∞−∞

udνξ,Pmin(s,t)ξ(u) ∀s ∈ R.

Now, observe that

νξ,Pmin(s,t)ξ

((−∞, u]

)= (Puξ,Pmin(s,t)ξ)

= (Pmin(s,t,u)ξ, ξ)

=

‖Puξ‖2 ∀u ≤ min(s, t)

‖Pmin(s,t)ξ‖2 ∀u > min(s, t).

Hencedνξ,Pmin(s,t)ξ(u) = 1(−∞,min(s,t)](u) dνξ(u).

By combining the formulae above, we obtain

νξ,PtT ξ((−∞, s]

)=

∫ min(s,t)

−∞udνξ(u),

whencedνξ,PtT ξ(s) = s1(−∞,t](s) dνξ(s). (5.6.4)

Then

‖T ξ‖2 = limt→∞

(T ξ,PtT ξ)

(by Lemma 5.6.3 (ii)) = limt→∞

∫ ∞−∞

sdνξ,PtT ξ(s)

(by (5.6.4)) = limt→∞

∫ t

−∞s2 dνξ(s)

=

∫ ∞−∞

s2 dνξ(s) ∀ξ ∈ Dom(T ),


as required.

Finally, we prove (vi). Suppose that ξn is a sequence in Dom(T ), suchthat

ξn → ξ and T ξn → η.

We need to prove that ξ is in Dom(T ) and that T ξ = η. Note that the distri-bution function of νξn tends pointwise to the distribution function of νξ as ntends to ∞. Indeed,∣∣(Psξn, ξn)− (Psξ, ξ)

∣∣ ≤ ∣∣(Ps(ξn − ξ), ξn)∣∣+∣∣(Psξn, ξn − ξ)∣∣

≤ ‖ξn − ξ‖ supn‖ξn‖

which tends to 0 as n tends to ∞. Then νξn tends vaguely (i.e., with respectto the weak∗ topology of M(R), the Banach space of all complex measures onR) to νξ as n tends to ∞ [Fo, Prop. 7.19]. Denote by j the identity function(j(s) = s for every s in R). Then∫ t

−tj2 dνξ = lim

n→∞

∫ t

−tj2 dνξn

≤ limn→∞

∫ ∞−∞

j2 dνξn

= limn→∞

‖T ξn‖2

= ‖η‖2.

By taking the supremum with respect to t, we may conclude that ξ is in Dom(T ).Observe that T is closable, because it is symmetric and densely defined (seeProposition 2.7.8). Then η must be T ξ, for otherwise the points (ξ, η) and(ξ, T ξ) would both belong to GT , which would contradict the closability of T .2

Proposition 5.6.4 The operator T of Definition 5.6.2 is self adjoint.

Proof. Observe that T is a densely defined operator by Lemma 5.6.1 and itis symmetric and closed by Lemma 5.6.3 (iii) and (vi). Therefore the range ofT ± iI is closed (see the proof after formula (3.3.1)). By the Basic Criterionof Self Adjointness, to prove that T is self adjoint, it remains to show that therange of T ± iI is dense in H. Suppose that η is orthogonal to Ran(T + iI).Then

(T ξ + iξ, η) = 0 ∀ξ ∈ Dom(T ).

and(T ξ + iξ, η) = lim

t→∞

((Pt − P−t)(T ξ + iξ), η

)= limt→∞

((Pt − P−t)(T ξ + iξ), (Pt − P−t)η

)= limt→∞

((T + iI)(Pt − P−t)ξ, (Pt − P−t)η

)= limt→∞

∫ t

−t(s+ i) dνξ,η(s).


Recall that(Pt−P−t

)η is in Dom(T ) for every t > 0 (see the proof of Lemma 5.6.1).

Now we choose ξ =(Pt − P−t

)η, and obtain

0 = limt→∞

∫ t

−t(s+ i) dνη(s).

Since νη is a positive measure, this implies that νη(R) = 0, hence νη = 0. Since

νη(R) = ‖η‖2, we may conclude that η = 0. Thus, T + iI is onto. Similarly, wemay prove that T − iI is onto.

The proof of the proposition is complete. 2

5.7 Lecture XXVII: the spectral theorem II

Suppose that L is a self adjoint operator. We may associate to L a projectionvalued measure P, defined by

P(E) = 1E(L) ∀E ∈ B,

where the operator on the right hand side is obtained from the extended func-tional calculus developed in Proposition 5.2.1. We associate to the p.v.m. P itsdistribution function s 7→ Ps, defined by

Ps := P((−∞, s]

)= 1(−∞,s](L) ∀s ∈ R.

It is straightforward to check that Ps is a spectral resolution of the identity,which we call the spectral resolution of the identity associated to L.

We claim that for every ξ and η in H, the Lebesgue–Stieltjes measureνξ,η associated to the spectral resolution Ps defined above is precisely themeasure µξ,η.

Indeed, by the definition of spectral resolution of the identity associated to L(Psξ, η

)=(1(−∞,s](L)ξ, η

)(by (5.2.1)) =

∫ ∞−∞

1(−∞,s] dµξ,η

= µξ,η((−∞, s]

)∀s ∈ R,

so that µξ,η is the Lebesgue–Stieltjes measure associated to the distributionfunction s 7→ (Psξ, η).

Exercise 5.7.1 Determine the spectral resolution of the identity associatedto the multplication operator Mm, where m is real.

Note that the spectral resolution of the identity associated to L is constant oneach interval contained in %(L). Indeed, µξ vanishes in R \ σ(L) for every ξ inH. Hence so does µξ,η, for all ξ and η in H.

5.7. THE SPECTRAL THEOREM II 115

Theorem 5.7.2 (Spectral theorem: II form) Suppose that L is a self ad-joint operator and that Ps is the associated spectral resolution of the identity.Then

Dom(L) =ξ ∈ H :

∫ ∞−∞

s2 d(Psξ, ξ) <∞

and

(Lξ, η) =

∫ ∞−∞

sd(Psξ, η) ∀ξ ∈ Dom(L) ∀η ∈ H.

More generally, for every continuous function f on R

(f(L)ξ, η

)=

∫ ∞−∞

f(s) d(Psξ, η) ∀ξ ∈ Dom(f(L)) ∀η ∈ H,


Proof. From the definition of the extended functional calculus (see Defini-tion 5.1.13) and Proposition 5.1.14, we see that

Dom(L) =ξ ∈ H :

∫ ∞−∞

s2 dµξ(s) <∞

and that

(Lξ, ξ) =

∫ ∞−∞

sdµξ(s) ∀ξ ∈ Dom(L).

More generally, for every Borel function on R

(f(L)ξ, η) =

∫ ∞−∞

f dµξ,η ∀ξ ∈ Dom(f(L)) ∀η ∈ H

by Lemma 5.2.6. We have proved that µξ,η = νξ,η. It is known that gener-alised Riemann–Stieltjes integrals and the associated Lebesgue–Stieltjes inte-grals agree on the vector space of continuous integrable functions. The threeformulae in the statement of the theorem follow. 2

In the following corollary, we prove that for every ξ in Dom(L), the element Lξis representable as a H valued Riemann–Stieltjes integral.

Corollary 5.7.3 Suppose that L is a self adjoint operator and that Ps is theassociated spectral resolution of the identity. Then

Lξ =

∫ ∞−∞

sdPsξ ∀ξ ∈ Dom(L) ∀η ∈ H

(the integral above is as in Definition 5.5.5). More generally, for every contin-uous function f on R

f(L)ξ =

∫ ∞−∞

f(s) dPsξ ∀ξ ∈ Dom(f(L)) ∀η ∈ H,



Proof. We prove the first formula. The proof of the second is similar, and isomitted.

Denote by j the identity function. If ξ is in Dom(L), then j is in L2(µξ),hence in L1(µξ), for |j| ≤ j2 at infinity and j is continuous. Therefore,

(Lξ, ξ) =

∫ ∞−∞

j dνξ

(by the DCT) = limα→∞β→−∞

∫ β

α

j dνξ.

(5.7.1)

Since j is continuous and s 7→ (Psξ, ξ) is the distribution function of the mea-sure νξ, ∫ β

α

j dνξ =

∫ β

α

j(s) d(Psξ, ξ)

(by definition of RS integral) = limδ(P )↓0

S(j;P ).

Now,

S(j;P ) =

N∑k=1

sk−1((Psk − Psk−1

)ξ, ξ)

=( N∑k=1

sk−1 (Psk − Psk−1)ξ, ξ

)→(∫ β

α

j(s) dPsξ, ξ)

as δ(P ) ↓ 0 (by the continuity of the inner product). We have proved that∫ β

α

j dνξ =(∫ β

α

j(s) dPsξ, ξ)

∀ξ ∈ Dom(L).

By taking the limit of both sides as α and β tend to −∞ and ∞ respectively,and using (5.7.1), we may conclude that

(Lξ, ξ) =(∫ ∞−∞

j(s) dPsξ, ξ)

∀ξ ∈ Dom(L).

Since Dom(L) is dense in H,

Lξ =

∫ ∞−∞

j(s) dPsξ ∀ξ ∈ Dom(L),

as required. 2

One may wonder whether the spectral resolution of the identity associated toa given self adjoint operator L is the unique spectral resolution of the identityfor which the formulae in the statement of Theorem 5.7.2 hold. The answer iscontained in the following proposition.

5.7. THE SPECTRAL THEOREM II 117

Proposition 5.7.4 Suppose that Ps and P ′s are two spectral resolutions ofthe identity such that∫ ∞

−∞sd(Psξ, ξ) = (Lξ, ξ) =

∫ ∞−∞

sd(P ′sξ, ξ) ∀ξ ∈ Dom(L).

Then Ps = P ′s for every s in R.

Proof. Recall that

(eitLξ, ξ) =

∫ ∞−∞

eits d(Psξ, ξ) ∀ξ ∈ H.

In the rest of this proof we shall denote the operator eitL by Ut. For every t inR define the operator U ′t by

(U ′tξ, ξ) =

∫ ∞−∞

eits d(P ′sξ, ξ) ∀ξ ∈ H.

It is straightforward, though tedious, to check that for every ξ in Dom(L) botht 7→ Utξ and t 7→ U ′tξ are solutions of the Cauchy problem

u′(t) + iLu(t) = 0 ∀t ∈ R+

u(0) = ξ.

It is well known that this problem has a unique solution, hence Ut = U ′t for allt > 0. A similar argument proves that Ut = U ′t for all t < 0, so that Ut andU ′t agree. Then the Fourier transform of the measures νξ and ν′ξ agree; by theinjectivity of the Fourier transform, νξ = ν′ξ, hence

(Psξ, ξ) = (P ′sξ, ξ) ∀s ∈ R ∀ξ ∈ Dom(L).

Since Dom(L) is dense in H, we may conclude that Ps = P ′s for every s in R,as required. 2

We conclude this chapter by proving an interesting consequence of the spectraltheorem. We consider the one parameter family e−tL : t ≥ 0 of operators,spectrally defined (i.e., e−tL := Φ(e−t)).

Corollary 5.7.5 Suppose that L is a nonnegative self adjoint operator (i.e.(Lf, f) ≥ 0 for every f in H). For each f in H, the H valued function t 7→ e−tLfsolves the Cauchy problem

u′(t) + Lu(t) = 0 ∀t ∈ R+

u(0) = f.

Proof. By Exercise 3.4.2, σ(L) is contained in [0,∞). The properties of thebounded functional calculus together the fact that for every g inH the support ofthe measure µg is contained in σ(L) (see Proposition 5.2.5) imply that e−tL isa strongly continuous semigroup of contractions on H, and that its infinitesimalgenerator is L.


We claim that for every f in H and for every t > 0 the function e−tLf is inDom(L). Indeed, by the definition of the extended functional calculus, e−tLf isin Dom(L) if and only if ∫ ∞

0

s2 dµe−tLf (s) <∞.

We have used the facts that σ(L) is contained in [0,∞) (see Exercise 3.4.2) andthe support of µe−tLf is contained in the spectrum of L (see Proposition 5.2.5).Now, by Lemma 5.2.3, the left hand side is equal to∫ ∞

0

s2 e−2ts dµf (s) ≤ C2

t2‖f‖2 ∀t ∈ R+,

where C = maxu>0 u e−u, and the claim is proved.

By Proposition 4.2.4 (i), the H valued function t 7→ e−tLf is strongly dif-ferentiable at t > 0 with derivative equal to −Le−tLf . Thus, e−tLf solves theequation. Clearly, it satisfies the initial condition. The proof is complete. 2

Remark 5.7.6 Notice the difference with the case of Banach spaces, where wemust assume that the initial datum f is in Dom(L).

Bibliography

[DS] N. Dunford and J.T. Schwartz, Linear Operators. Part I. GeneralTheory, Wiley Classic Library Edition, 1988.

[EG] L. C. Evans, R. F. Gariepy, Measure Theory and Fine Propertiesof Functions, Studies in Advanced Mathematics, CRC Press, BocaRaton, FL, 1992.

[Fo] G.B. Folland, Real analysis. Modern techniques and their applica-tions, Second edition, J. Wiley & Sons, 1999.

[RS1] M. Reed and B. Simon, Methods of modern Mathematical Physics.Vol. I, Academic Press.

[RS2] M. Reed and B. Simon, Methods of modern Mathematical Physics.Vol. II, Academic Press.

[Ru] W. Rudin, Real and complex analysis, Third Edition, McGraw-Hill,1987.

[Y] K. Yosida, Functional Analysis, Grundlehren der mathematischenWissenschaften 123, Springer Verlag, Berlin Heidelberg New York,1985.

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