OPERATOR THEORY LECTURE NOTES

J. A. VIRTANEN

UNIVERSITY OF READING, SPRING 2016

Contents

1. Preliminaries 21.1. Banach spaces 21.2. Bounded linear operators on Banach spaces 31.3. Fundamental theorems 41.4. Hilbert space 42. Banach algebras and spectral theory 72.1. Banach algebras 72.2. Compact operators 122.3. Operators on Hilbert spaces 163. Fredholm theory 224. Weak convergence 25References 26

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1. Preliminaries

Here we recall some necessary results that are needed throughout. Proofs are omittedin most cases but can be easily found in the literature; those who have taken a coursein functional analysis are probably familiar with most of the results in this section. Westart with linear normed spaces and linear operators acting on Banach spaces, proceedto review the most fundamental results in the theory of Banach spaces and finish withsome basic Hilbert space theory.

1.1. Banach spaces. Throughout these notes F stands for either R or C. Let X be anormed linear space over F; that is, X is a vector space equipped with a norm ‖ · ‖ thatsatisfies the following properties:

(N1) ‖x+ y‖ ≤ ‖x|+ ‖y‖ for all x, y ∈ X;(N2) ‖αx‖ = |α|‖x‖ for all x ∈ X and α ∈ F;(N3) ‖x‖ > 0 if x 6= 0.

We say that a sequence (xn) in a normed space X is a Cauchy sequence if for everyε > 0 there is an N ∈ N such that ‖xn − xm‖ < ε whenever n > N and m > N . If everyCauchy sequence in X converges, then we say that X is complete. We say that X is aBanach space if it is a normed linear space that is complete.

Example 1.1. (a) The spaces Rn and Cn when equipped with the norm

‖x‖ =

(n∑k=1

|xk|2)1/2

are Banach spaces.(b) Let I be a set and 1 ≤ p <∞. Define lp(I) to be the set of all functions f : I → F

for which

‖f‖ =

(∑k∈I

|f(k)|p)1/p

<∞.

We also define l∞(I) to be the set of all functions f : I → F for which

‖f‖ = supk∈I|f(k)| <∞.

For 1 ≤ p ≤ ∞, lp(I) is a Banach space. We set lp = lp(N).(c) If X is a Hausdorff space, we define Cb(X) to be the set of all continuous functions

on X. The norm ‖f‖ of f ∈ Cb(X) is defined by

‖f‖ = supx∈X|f(x)|.

If X is a locally compact space, we define C0(X) to be the set of all continuous functionsf : X → F such that for every ε > 0 the set x ∈ X : |f(x)| ≥ ε is compact.

If X is compact, then C0(X) = Cb(X) and we write C(X) for it.(d) Let µ be a finite measure on a set Ω. For 0 < p < ∞, define Lp(Ω, dµ) to be the

space of all functions f on Ω with

‖f‖p =

(∫Ω

|f(x)|pdµ(x)

)1/p

<∞.

The space L∞(Ω, dµ) consists of all functions f on Ω with

‖f‖∞ = ess supx∈Ω |f(x)| <∞.2

It is not difficult to see that these spaces are linear normed spaces. For 1 ≤ p ≤ ∞,they are also Banach spaces, The spaces Lp(T) and Lp(D), where T = |z| = 1 andD = |z| < 1, play the most important role in our examples and also in connection withour study of some concrete linear operators acting on function spaces.

(e) For a set I, define c0(I) to be the set of all bounded functions f : I → F such thatfor every ε > 0 the set i ∈ I : |f(i) ≥ ε is finite. If I = N, we write c0 for c0(I). Notethat c0 consists of all sequences that converge to 0.

Exercise 1. Verify that the statements made in Example 1.1.

1.2. Bounded linear operators on Banach spaces. Let X and Y be Banach spacesand suppose that T : X → Y be a linear operator, i.e. for all x, y ∈ X and a ∈ F wehave T (x+ y) = Tx+ Ty and T (ax) = aTx. If there is a constant C such that

‖Tx‖ ≤ C‖x‖

for all x ∈ X, we say that T is a bounded linear operator. When Y = F, T is referred toas a linear functional. The space of all bounded linear operators T : X → Y is denotedby L(X, Y ) and we write L(X) for L(X,X).

Theorem 1.2. Let X and Y be Banach spaces. The class L(X, Y ) consisting of allbounded linear operators T : X → Y with the norm

‖T‖ = sup‖x‖≤1

‖Tx‖ = sup‖x‖=1

‖Tx‖ = supx 6=0

‖Tx‖‖x‖

is a Banach space. Also T is continuous if and only if it is bounded.

Proof. Exercise.

Let X be a Banach space. The space X∗ consisting of all bounded linear functionalson X is called the dual space of X.

Theorem 1.3. Suppose 1 ≤ p <∞ and 1/p+ 1/q = 1. Then

(Lp(Ω, dµ))∗ = Lq(Ω, dµ)

in the sense that for each g ∈ Lq(Ω, dµ)

Fgf =

∫Ω

f(x)g(x)dµ(x)

is a bounded linear functional on Lp(Ω, dµ) and each linear functional on Lp(Ω, dµ) is ofthis form.

Proof. See [18, Theorem 6.16].

Note that the preceding theorem fails when p =∞.We also need the concept of adjoint operators acting between two dual spaces. If T is

bounded linear operator between Banach spaces X and Y , then T ∗ : Y ∗ → X∗ definedby

T ∗(F )(x) = F (Tx), x ∈ X,F ∈ Y ∗

is called the adjoint of T .Let X be a Banach space. A bounded linear operator P on X is called a projection if

P 2 = P . Note also that P maps X onto a subspace and that on this subspace P acts asthe identity operator I.

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Proposition 1.4. Let X be a Banach space and let P be a projection on X. ThenQ = I − P is also a projection and

ranP = kerQ and kerP = ranQ.

Proof. Note that PQ = QP = 0 and Q2 = (I − P )Q = Q− PQ = Q.

We will later see that projections play an important role in operator theory, especiallyin connection with analytic function spaces.

1.3. Fundamental theorems. The proofs of the following four fundamental theoremscan be found in any text on functional analysis.

Theorem 1.5 (Open mapping theorem). Let X and Y be Banach spaces and let T be abounded linear operator from X onto Y . Then T maps open sets in X to open sets in Y .

The open mapping theorem has two important consequences: the inverse mappingtheorem and the closed graph theorem. The former states that if a bounded linearoperator T is both one-to-one and onto, then T−1 is also bounded. The latter consequenceis formulated in the following theorem.

Theorem 1.6 (Closed graph theorem). If T is a linear operator between two Banachspaces X and Y whose graph (x, Tx) : x ∈ X is closed in X × Y , then T is bounded.

In the infinite dimensional case, the following theorem depends on Zorn’s lemma, whichis equivalent to the axiom of choice.

Theorem 1.7 (Hahn-Banach theorem). If T is a bounded linear functional on a subspaceof a Banach space X, then T can be extended to a bounded linear functional on X withno increase in its norm.

A formulation of the last of the four fundamental results is given in the followingtheorem.

Theorem 1.8 (Principle of uniform boundedness). If Tα is a collection of boundedlinear operators between Banach spaces X and Y and if supα ‖Tαx‖ is finite for all x ∈ X,then supα ‖Tα‖ is finite.

1.4. Hilbert space. Hilbert spaces exhibit certain advantages over Banach spaces thatmake their study closer to that of finite dimensional Euclidean spaces. In particular themost important geometrical property is orthogonality and the fact that the dual space ofany Hilbert space can be identified with the space itself.

Definition 1.9. Let X be a linear space. An inner product is a function (· , ·) : X×X →C with the following properties:

(1) (ax+ by, z) = a(x, z) + b(y, z) for x, y, z ∈ X and a, b ∈ F;(2) (z, ax+ by) = a(z, x) + b(z, y) for x, y, z ∈ X and a, b ∈ F;

(3) (x, y) = (y, x) for x, y ∈ X;(4) (x, x) ≥ 0 for x ∈ X and (x, x) = 0 if and only if x = 0.

A linear space equipped with an inner product is called an inner product space. AHilbert space is a complete inner product space. The norm ‖·‖ on an inner product spaceis defined by ‖x‖ = (x, x)1/2.

Proposition 1.10. Let X be an inner product space. Then

|(x, y)| ≤ ‖x‖‖y‖for all x, y ∈ X. In particular, ‖ · ‖ as defined above is indeed a norm on X.

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Definition 1.11. Let X be an inner product space. We say that x, y ∈ X are orthogonaland write x⊥y if (x, y) = 0. A subset Y of X is said to be orthogonal if (x, y) = 0 for allx, y ∈ Y and further orthonormal if also ‖x‖ = 1 for each x ∈ Y .

Let us recall the Pythagorean theorem and the parallelogram law.

Proposition 1.12. If x1, . . . , xn is an orthogonal subset of an inner product space Xand x, y ∈ X, then ∥∥∥∑xk

∥∥∥2

=∑‖xk‖2 (1.1)

and‖x+ y‖2 + ‖x− y‖2 = 2‖x‖2 + 2‖y‖2. (1.2)

Proof. These are straightforward computations—see, for example, [4, Propositions 3.10and 3.11].

So far we have only dealt with inner product spaces and listed their basic properties.However, the deepest properties require that the norm be complete, that is, the space bea Hilbert space. Below are a few examples of familiar Hilbert spaces.

Example 1.13. (a) The space Fn equipped with the inner product

(x, y) = x1y1 + . . .+ xnyn

is a Hilbert space.(b) The space l2(I) equipped with the inner product

(x, y) =∑i∈I

x(i)y(i)

is a Hilbert space. If p 6= 2, then lp is not a Hilbert space.(c) The space C([a, b]) is not a Hilbert space; that is, the norm

‖x‖ = maxt∈[a,b]

|x(t)|

cannot be obtained from an inner product.(d) The space L2(µ) equipped with the inner product

(f, g) =

∫fgdµ

is a Hilbert space.

Exercise 2. Verify the statements made in the preceding example.

Theorem 1.14. Let L be a nonempty subset of a Hilbert space H. If L is closed andconvex, then L contains a unique element of smallest norm.

Proof. Let δ = inf ‖f‖ : f ∈ L. Then there is (fn) in L such that ‖fn‖ → δ as n→∞.According to (1.2) in Proposition 1.12,∥∥∥∥fn − fm2

∥∥∥∥2

= 2

∥∥∥∥fn2∥∥∥∥2

+ 2

∥∥∥∥fm2∥∥∥∥2

−∥∥∥∥fn + fm

2

∥∥∥∥2

≤ 2

∥∥∥∥fn2∥∥∥∥2

+ 2

∥∥∥∥fm2∥∥∥∥2

− δ2,

where the inequality follows from convexity. Therefore,

‖fn − fm‖2 ≤ 2 ‖fn‖2 + 2 ‖fm‖2 − 4δ2

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and solim supn,m→∞

‖fn − fm‖2 ≤ 2δ2 + 2δ2 − 4δ2 = 0,

which shows that (fn) is Cauchy. Since L is a closed subset of the complete space H, weconclude that there is f ∈ L such that fn → f , which implies ‖f‖ = limn→∞ ‖fn‖ = δ.

Definition 1.15. Let M be a subset of a Hilbert space H. The orthogonal complementM⊥ of M consists of all elements in H that are orthogonal to each element in M .

Note that M⊥ is a closed subspace of H; and M⊥ 6= H provided that M 6= 0.

Theorem 1.16. Let M be a closed subspace of a Hilbert space H. Then for f ∈ H thereare unique g ∈M and h ∈M⊥ such that f = g + h.

Proof. Let L = f−k : k ∈M, which is clearly nonempty, closed, and convex. Accordingto Theorem 1.14, there is a unique element h ∈ L of smallest norm. Let u ∈M be a unitvector. Then

h− (h, u)u ∈ Land so

‖h‖2 ≤ ‖h− (h, u)u‖2

= ‖h‖2 − (h, u)(h, u)− (h, u)(h, u) + (h, u)(h, u)‖u‖2,

which implies |(h, u)|2 ≤ 0, so that (h, u) = 0. Therefore h ∈ M⊥ and since also h ∈ L,there is g ∈M such that h = f − g.

For uniqueness, suppose that

f = g1 + h1 = g2 + h2

with gk ∈M and hk ∈M⊥. Then

M 3 g2 − g1 = h1 − h2 ∈M⊥,

which implies that g1 = g2 and h1 = h2.

Corollary 1.17. If M is a subspace of a Hilbert space H, then M⊥⊥ = M .

Proof. Since M ⊂M⊥⊥ and M⊥⊥ is closed, we have M ⊂M⊥⊥.If f ∈M⊥⊥, then by the previous theorem

f = g + h

with g ∈M and h ∈M⊥= M⊥. Since f ∈M⊥⊥,

0 = (f, h) = (g + h, h) = (h, h) = ‖h‖2

and so f = g ∈M .

Let H be a Hilbert space and let g ∈ H. Define ϕ : H → C by ϕ(f) = (f, g) for f ∈ H.Since

|ϕ(f)| = |(f, g)| ≤ ‖f‖‖g‖and, for g 6= 0,

‖ϕ‖ = supf 6=0

|(f, g)|‖f‖

≥ |(g, g)|‖g‖

= ‖g‖,

it follows that ϕ is a bounded linear functional on H with ‖ϕ‖ = ‖g‖.

Theorem 1.18 (Riesz representation theorem). Suppose that ϕ is a bounded linear func-tional on a Hilbert space H. Then there is a unique g ∈ H such that ϕ(f) = (f, g) forf ∈ H.

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Proof. Consider the kernel of ϕ

kerϕ = f ∈ H : ϕ(f) = 0 .

Since ϕ is continuous, kerϕ is a closed subspace of H. If kerϕ = H, we can choose g = 0.Otherwise, there is a unit vector h ∈ (kerϕ)⊥. In particular, ϕ(h) 6= 0.

Let f ∈ H. Then ϕ(f − ϕ(f)ϕ(h)

h) = ϕ(f)− ϕ(f)ϕ(h)

ϕ(h) = 0 and so f − ϕ(f)ϕ(h)

h ∈ kerϕ. Thus,

ϕ(f) = ϕ(f)(h, h) = ϕ(f)ϕ(h)

ϕ(h)(h, h)

= (ϕ(f)

ϕ(h)h, ϕ(h)h)

= (f − ϕ(f)

ϕ(h)h, ϕ(h)h) + (

ϕ(f)

ϕ(h)h, ϕ(h)h)

= (f, ϕ(f)h)

and we can put g = ϕ(h)h.If (f, g1) = (f, g2) for all f ∈ H, then (g1 − g2, g1 − g2) = 0 and so g1 = g2.

Corollary 1.19. Define L : H → H∗ by Lg = ϕg, where ϕg(f) = (f, g) for f ∈ H. Thenthe mapping L is an isometry and conjugate linear.

Definition 1.20. A set eii∈I is an orthonormal basis of a Hilbert space H if it isorthonormal and its span is dense in H.

Theorem 1.21. Each nontrivial Hilbert space has an orthonormal basis.

Proof. Most books on functional analysis prove this result, see e.g. [4, Theorem 3.29].

2. Banach algebras and spectral theory

2.1. Banach algebras. A vector space A over F which is equipped with a bilinear mapA2 → A satisfying a(bc) = (ab)c for all a, b, c ∈ A is called an algebra. A subspace Bof an algebra A is called a subalgebra if ab ∈ B for all a, b ∈ B. A norm ‖ · ‖ is calledsubmultiplicative if ‖ab‖ ≤ ‖a‖‖b‖ for all a, b ∈ A. The pair (A, ‖ · ‖) is called a normedalgebra. If A has an identity, 1, then we say A is unital and assume that ‖1‖ = 1. Acomplete unital normed algebra is called a Banach algebra. (Hence we only deal withunital Banach algebras.) The identity element is usually denoted by 1.

In this course the space of all bounded linear operators on a Banach space is the mostimportant example of Banach algebras.

Example 2.1. Let X be a Banach space. The space of all bounded linear operatorsL(X) is a Banach algebra. Indeed, the multiplication ST of two elements S, T ∈ L(X)is defined by

(ST )x = S(Tx)

for x ∈ X. Clearly, S(T +U) = ST +SU and (S+T )U = SU +TU for S, T ∈ L(X) andλ(ST ) = (λS)T = S(λT ), which shows bilinearity. Also, S(TU)x = S(TUx) = (ST )Ux,so S(TU) = (ST )U . It is not difficult to see that ‖TS‖ ≤ ‖T‖‖S‖ for all T, S ∈ L(X),so L(X) is a normed algebra. The identity operator I : X → X defined by Ix = x isclearly an identity element. That the norm is complete is left as an exercise. Thus, L(X)is a Banach algebra.

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Let A be a unital algebra. We say a ∈ A is invertible if there is a b ∈ A such thatab = ba = 1. The element b is unique and we write a−1 = b if a is invertible. Define

G(A) = G(A) = a ∈ A : a is invertible.It is not difficult to see that G(A) is a group. A subspace I of an algebra A is called anideal if

ab ∈ I and ba ∈ I for a ∈ A and b ∈ I.

Example 2.2. (a) If X is a compact space, then C(Ω) is a commutative Banach algebra.(b) The space of all n× n matrices Mn(F) is a noncommutative Banach algebra.(c) If (X,Ω, µ) is a σ-finite measure space, then L∞(X,Ω, µ) is a commutative Banach

algebra.(d) Let I be a closed ideal of a Banach algebra A. Then the quotient space A/I

equipped with the operations

(a+ I) + (b+ I) = (a+ b) + I and (a+ I)(b+ I) = ab+ I (a, b ∈ A)

and with the quotient norm

‖a+ I‖ = infb∈I‖a+ b‖

is a Banach algebra.

Let A be a unital algebra and a ∈ A. The spectrum σ(a) of a is defined by

σ(a) = λ ∈ C : λ− a /∈ G(A).

Remark 2.3. (a) If A is a unital algebra, and a, b ∈ A, then

1− ab ∈ G(A) if and only if 1− ba ∈ G(A).

Indeed, if c = (1− ab)−1, then 1 + bca is the inverse of 1− ba.(b) It follows that if a, b ∈ A,

σ(ab) \ 0 = σ(ba) \ 0.Note we may have σ(ab) 6= σ(ba). Indeed, let H be a Hilbert space with an orthonormalbasis enn≥1. Define shift operators u and v in the Banach algebra L(H) by u(ej) = ej+1,v(e1) = 0 and v(ej) = ej−1 if j > 1. Then 0 /∈ σ(vu) and 0 ∈ σ(uv).

Our next goal is to prove that the spectrum of an element in a Banach algebra is alwaysnonempty, closed and bounded. The following auxiliary result, despite its simplicity, is asurprisingly powerful tool in basic spectral theory.

Theorem 2.4. Let A be a unital Banach algebra and let a ∈ A be such that ‖a‖ < 1.Then 1− a ∈ G(A) and

(1− a)−1 =∞∑n=0

an, (2.1)

which is referred to as the Neumann series for (1− a)−1.

Proof. Since A is submultiplicative, the series∑∞

n=0 an is absolutely convergent, and hence

convergent in A by the completeness of A. Let us denote its sum by b ∈ A. Then

(1− a)b = (1− a) limm→∞

m∑n=0

an = limm→∞

(1− am+1

)= 1

because ‖an+1‖ ≤ ‖a‖n+1 → 0. Similarly, b(1− a) = 1. Thus, (1− a)−1 = b. 8

Theorem 2.5. If A is a Banach algebra, then G(A) is open in A and the mapping

f : G(A)→ A, f(a) = a−1

is continuous.

Proof. Let a ∈ G(A). If b ∈ A satisfies ‖b− a‖ < ‖a−1‖−1, then ‖ba−1 − 1‖ = ‖(b− a)a−1‖ ≤

‖b− a‖ ‖a−1‖ < 1. So, ba−1 ∈ G(A), and therefore b = (ba−1)a ∈ G(A). Hence, G(A) isopen in A.

For all b ∈ A, we define lb : A → A and rb : A → A (left and right multiplications byb) by setting

lb(a) = ba and rb(a) = ab.

If we define f : G(A)→ A such that f(a) = a−1, then

f(a) =(la−1 f ra−1

)(a)

= a−1

for all a ∈ G(A). Since ra−1(a) = 1, f(1) = 1, la−1(1) = a−1 and for all a ∈ G(A) themaps la−1 , ra−1 are continuous at 1 and a, respectively, it is enough to show that f iscontinuous at 1. Indeed, if ‖1− b‖ < 1 and a = 1− b, then by Theorem 2.4,

∞∑n=0

an = (1− a)−1 = b−1.

Let ‖1− b‖ < ε < 1/2. If a = 1− b, then

‖f(b)− f(1)‖ =∥∥b−1 − 1

∥∥ ≤ ‖1− b‖∥∥b−1∥∥

≤ ‖a‖∞∑n=0

‖a‖n = ‖a‖(1− ‖a‖

)−1 ≤ 2ε.

Thus, f is continuous at 1.

We can now prove that the spectrum is always bounded and closed, and hence acompact set.

Theorem 2.6. Let A be a Banach algebra and let a ∈ A. Then σ(a) is closed and

σ(a) ⊂ λ ∈ C : |λ| ≤ ‖a‖ .

Proof. If |λ| > ‖a‖, then ‖λ−1a‖ < 1. So, 1− λ−1a ∈ G(A) by Theorem 2.4. Therefore,λ− a = λ(1− λ−1a) ∈ G(A), and hence λ /∈ σ(a).

Define f : C→ A by setting f(λ) = λ−a. Then f is continuous, and thus f−1(G(A)

)=

C \ σ(a) is open since G(A) is open. Therefore, σ(a) is closed.

It is not difficult to give examples of Banach algebras over R and elements in themwhose spectra are empty. Hence, from now on, we only consider Banach algebras A overC and show that the spectrum of a in A is always nonempty.

Theorem 2.7 (Gelfand). If A is a unital Banach algebra and a ∈ A, then σ(a) 6= ∅.

Proof. Let λ ∈ C \ σ(a). Since G(A) is open, there is an r > 0 such that µ − a ∈ G(A)whenever |µ− λ| < r. Let φ ∈ A∗. Recall the following resolvent identity:

(µ− a)−1 − (λ− a)−1 = (µ− a)−1((λ− a)(λ− a)−1

)−((µ− a)−1(µ− a)

)(λ− a)−1

= (µ− a)−1((λ− a)− (µ− a)

)(λ− a)−1

= (λ− µ)(µ− a)−1(λ− a)−1.

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Hence, we conclude that

φ((µ− a)−1

)− φ((λ− a)−1

)µ− λ

=φ((µ− a)−1 − (λ− a)−1

)µ− λ

=φ(−(µ− λ)(µ− a)−1(λ− a)−1

)µ− λ

= φ(−(µ− a)−1(λ− a)−1

)−→ φ

(−(λ− a)−2

)as µ→ λ by the continuity of inverse (see Theorem 2.5) and the continuity of φ. So themap λ 7→ φ

((λ− a)−1

)is analytic on C \ σ(a). Also, by Theorem 2.5,

φ((λ− a)−1

)= λ−1φ

((1− λ−1a)−1

)−→ 0

as |λ| → ∞ since (1 − λ−1a)−1 → 1. Assume that σ(a) = ∅. Then the map λ 7→φ((λ− a)−1

)is entire, i.e. analytic in the whole plane, and bounded. Thus the map λ 7→

φ((λ−a)−1

)must reduce to 0 by Liouville’s theorem [18, Theorem 10.23]. Consequently,

φ(a−1) = 0 for all φ ∈ A∗. The Hahn-Banach theorem implies that a−1 = 0, which is acontradiction. Thus, σ(a) 6= ∅.

In order to gain more information about the spectrum σ(a), we may wish to ask whathappens to it when a is subjected to an elementary transformation. We state the answerin the case of polynomial transformations. The proof, which is left as an exercise, is anapplication of the fundamental theorem of algebra. Denote by C[z] the algebra of allpolynomials in an indeterminate z with complex coefficients. Let A be an algebra andlet a ∈ A. If p ∈ C[z] is the polynomial

p = λ0 + λ1z + λ2z2 + · · ·+ λnz

n,

then we set

p(a) = λ0 + λ1a+ λ2a2 + · · ·+ λna

n.

The spectral mapping theorem states that

σ (p(a)) = p (σ(a)) . (2.2)

A fundamental concept in the study of operators is the spectral radius, which is ofsignificant importance both in the theory and computation. Let A be a Banach algebraand a ∈ A. The spectral radius of a is defined by

r(a) = supλ∈σ(a)

|λ|. (2.3)

Since σ(a) is bounded and nonempty, r(a) is well defined. Notice that the supremum isattained because σ(a) is compact. Also, r(ab) = r(ba) since σ(ab) \ 0 = σ(ba) \ 0.

Clearly r(0) = 0. Is it possible to have r(a) = 0 for some a 6= 0? Define

a =

(0 01 0

).

Then a2 = 0, and since r(an) = (r(a))n, which follows from the spectral mapping theoremfor polynomials, we have r(a) = 0.

While finding the spectrum can often be a very hard task, the spectral radius can becomputed more easily. A useful tool for this is the spectral radius formula, which weprove next. We start with an auxiliary lemma.

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Lemma 2.8. Let X be a normed space and let x ∈ X \ 0. Define ϕ : X∗ → C bysetting ϕ(x∗) = x∗(x). Then ϕ ∈ X∗∗ and ‖ϕ‖ = ‖x‖.

Proof. It is clear that ϕ is linear. Let x∗ ∈ X∗. Then

|ϕ(x∗)| = |x∗(x)| ≤ ‖x‖ ‖x∗‖ ,so ϕ ∈ X∗∗ and ‖ϕ‖ ≤ ‖x‖. It is a consequence of the Hahn-Banach theorem thatthere exists x∗ ∈ X∗ such that ‖x∗‖ = 1 and x∗(x) = ‖x‖. Therefore, ‖ϕ‖ ≥ |ϕ(x∗)| =|x∗(x)| = ‖x‖.

Theorem 2.9 (Beurling). Let A be a unital Banach algebra and let a ∈ A. Then

r(a) = infn≥1‖an‖

1n = lim

n→∞‖an‖

1n .

Proof. Let λ ∈ σ(a). Then λn ∈ σ(an), n = 1, 2, . . ., by the spectral mapping theoremfor polynomials. Therefore, by Lemma 2.6, |λn| = |λ|n ≤ ‖an‖, n = 1, 2, . . ., and so

|λ| ≤ infn≥1‖an‖1/n ≤ lim inf

n→∞‖an‖1/n .

Consequently,

r(a) ≤ infn≥1‖an‖1/n ≤ lim inf

n→∞‖an‖1/n . (2.4)

Let G = λ ∈ C : 0 < |λ| < 1/r(a). (Note that G = C \ 0 if r(a) = 0.) If λ ∈ G,then |λ−1| > r(a), and so λ−1−a ∈ G(A), which implies that (1−λa) = λ(λ−1−a) is alsoinvertible. For a∗ ∈ A∗, we define f : G → C by setting f(λ) = a∗

((1 − λa)−1

). Let us

show that the map f is analytic. Define g(λ) = λ−1 on C \ 0 and h(λ) = a∗((λ− a)−1

)on C \ σ(a). Then it is easy to see that g is analytic. Also, by the proof of Theorem 2.7,the map h is analytic. Moreover, if λ ∈ G, then |g(λ)| = |λ−1| > r(a), and henceg(λ) ∈ C\σ(a). Therefore, since f(λ) = g(λ)h

(g(λ)

), the function f is analytic. Clearly,

f is bounded in the open ball B(0, r) where r < 1/r(a). Since f has an extension whichis analytic in B(0, 1/r(a)) (see [18, Theorem 10.20]), it can be represented by a uniquepower series in B(0, 1/r(a)) (see [18, Theorem 10.16]). Consequently, there are complexnumbers cn such that

f(λ) =∞∑n=0

cnλn (2.5)

for all λ ∈ G. Moreover, if |λ| < 1/ ‖a‖(≤ 1/r(a)

), then ‖λa‖ < 1, so by Theorem 2.4,

(1− λa)−1 =∞∑n=0

λnan,

and hence

f(λ) =∞∑n=0

a∗(an)λn. (2.6)

This implies that cn = a∗(an) for all n by the uniqueness of the numbers cn in the powerseries (2.5), and therefore the power series (2.6) converges for all λ ∈ G.

Let us show that lim supn→∞ ‖an‖1/n ≤ r(a). Let λ ∈ G. If a∗ ∈ A∗, then the

convergence of the power series (2.6) implies that there is Ma∗ > 0 such that

supn∈N|a∗(λnan)| < Ma∗ .

Define ϕn : A∗ → C by setting ϕn(a∗) = a∗(λnan). Then

|ϕn(a∗)| = |a∗(λnan)| < Ma∗ , n = 1, 2, . . .11

and therefore by the Banach-Steinhaus theorem, there exists Cλ > 0 such that

‖λnan‖ = ‖ϕn‖ ≤ Cλ, n = 1, 2, . . .

(where the equality is justified by the preceding lemma), which implies that ‖an‖1/n ≤(Cλ)

1/n/ |λ|, and hence

lim supn→∞

‖an‖1/n ≤∣∣λ−1

∣∣ .Consequently, each complex number z with |z| > r(a) satisfies

lim supn→∞

‖an‖1/n ≤ |z| ,

and so

lim supn→∞

‖an‖1/n ≤ r(a).

This, together with (2.4), completes the proof.

2.2. Compact operators. We say that a topological space X is compact if every opencover of K has a finite subcover. If X is a topological space with topology T , and if Yis a subset of X, then

TY = Y ∩ U : U ∈ T is a topology on Y and Y with this topology is called a subspace of X.

Example 2.10. (a) The space R is not compact because (n, n + 2) : n ∈ Z has nofinite subcover.

(b) The space X = 0 ∪ 1/n : n ∈ N is a compact subspace of R because everyopen cover has an element which contains 0 and all but finite number of points of X.

(c) A finite subset of a topological space is compact.(d) The set (0, 1] is not compact because the open cover (1/n, 2) : n ∈ N has no

finite subcover. Similarly, (0, 1) is not compact. To show that [0, 1] is compact is a lotmore difficult.

Theorem 2.11. (a) A closed subspace of a compact space is compact.(b) A compact subspace of a Hausdorff space is closed.

Proof. Exercise∗.

We say X is limit point compact if each infinite subset of X has a limit point. We sayX is sequentially compact if every sequence of X has a convergent subsequence.

Theorem 2.12. If X is metrizable, then the following are equivalent:

(i) X is compact;(ii) X is limit point compact;

(iii) X is sequentially compact.

Proof. Exercise∗.

We say X is totally bounded if for every ε > 0, there is a finite cover of X by ε-ballsB(xi, ε) with xi ∈ X for i = 1, . . . , n.

Theorem 2.13. A subset of a metric space is compact if and only if it is complete andtotally bounded.

Proof. Exercise∗. 12

Let X and Y be Banach spaces, and T : X → Y be linear. We say T is a compactoperator and write T ∈ K(X, Y ) if T (SX) is compact, where SX = x ∈ X : ‖x‖ ≤ 1.Notice that compact operators are bounded.

Theorem 2.14. Let X and Y be Banach spaces and T ∈ L(X, Y ). The following areequivalent:

(i) T ∈ K(X, Y );

(ii) T (U) is compact for every bounded U ⊂ X;(iii) (Txn) has a convergent subsequence for every bounded sequence (xn) in X;(iv) T (SX) is totally bounded.

Proof. Exercise.

Theorem 2.15. Let X, Y and Z be Banach spaces. Then

(i) if K ∈ K(X, Y ) and T ∈ L(Y, Z), TK ∈ K(X,Z);(ii) if K ∈ K(Y, Z) and T ∈ L(X, Y ), KT ∈ K(X,Z).

In particular, K(X) is a closed ideal of the Banach algebra L(X).

Proof. (i) If (xn) is bounded, then there is a subsequence (xnk) of (xn) such that (Kxnk

)converges. Because T is bounded, (TKxnk

) converges.(ii) Exercise.

The first two assertions imply that K(X) is an ideal of L(X). Suppose that T ∈ K(X).If ε > 0, there is an S ∈ K(X) such that ‖T − S‖ < ε. Since S is totally bounded, thereare x1, . . . , xn such that if x ∈ SX , then ‖Sx − Sxi‖ < ε for some i. Thus, if x ∈ SX ,

‖Tx− Txi‖ ≤ ‖Tx− Sx‖+ ‖Sx− Sxi‖+ ‖Sxi − Txi‖ ≤ 3ε,

so T is totally bounded. Alternatively, we can use Theorem 2.14 (iii) to prove that K(X)is closed.

Later when we analyze the spectra of compact operators, the properties of their adjointoperators play an important role. The following theorem shows that the adjoint of acompact operator is always compact.

Theorem 2.16 (Schauder). Let X and Y be Banach spaces and let T ∈ K(X, Y ). ThenT ∗ ∈ K(Y ∗, X∗).

Proof. Let ε > 0. Since T (SX) is totally bounded, there are x1, . . . , xn ∈ SX such that ifx ∈ SX , then

‖T (x)− T (xi)‖ <ε

3for some i. Define S ∈ L(Y ∗,Cn) by setting

S(τ) = (τ(T (x1)), . . . , τ(T (xn))).

Since S is of finite rank, S ∈ K(Y ∗,Cn). Therefore, there are τ1, . . . , τm ∈ SY ∗ such thatfor all τ ∈ SY ∗ (note the following norm in Cn)

‖S(τ)− S(τk)‖ = ‖(τ(T (x1))− τk(T (x1)), . . . , τ(T (xn))− τk(T (xn)))‖

= max1≤j≤n

|T ∗(τ)(xj)− T ∗(τk)(xj)| <ε

3

for some k. Let x ∈ SX . Then ‖T (x)− T (xi)‖ < ε/3 for some i and

|T ∗(τ)(xi)− T ∗(τk)(xi)| <ε

313

by the above equality. So, if τ ∈ SY ∗ , then for some k ∈ 1, . . . ,m, we have

|T ∗(τ)(x)− T ∗(τk)(x)| ≤ |T ∗(τ)(x)− T ∗(τ)(xi)|+ |T ∗(τ)(xi)− T ∗(τk)(xi)|+ |T ∗(τk)(xi)− T ∗(τk)(x)| ≤ ε,

hence ‖T ∗(τ)− T ∗(τk)‖ = sup‖x‖≤1 |T ∗(τ)(x)− T ∗(τk)(x)| ≤ ε. Therefore, T ∗(SY ∗) istotally bounded.

We use the following result to show that all bounded operators on a finite-dimensionalnormed space are compact.

Theorem 2.17 (Riesz). Let X be a normed vector space, and suppose that Y is a properclosed vector subspace of X. Then for any ε > 0, there exists x ∈ X such that ‖x‖ = 1and

‖x+ Y ‖ = infy∈Y‖x+ y‖ > 1− ε.

Proof. Let s ∈ X \ Y , and let t = ‖s+ Y ‖ > 0. Let δ ∈ (0, 1). Then there exists y0 ∈ Ysuch that

t ≤ ‖s− y0‖ ≤t

δ.

Let x = c(s− y0), where c = ‖s− y0‖−1. Then ‖x‖ = 1 and if y ∈ Y ,

‖x− y‖ = ‖c(s− y0)− y‖ = c∥∥s− y0 − c−1y

∥∥ = c∥∥s− (y0 + c−1y)

∥∥≥ ct = ‖s− y0‖−1 t ≥ δ

tt = δ.

Therefore, ‖x+ Y ‖ ≥ δ. So, if ε > 0, there exists δ ∈ (1− ε, 1) and a unit vector x ∈ Xsuch that ‖x+ Y ‖ ≥ δ > 1− ε. This completes the proof.

Theorem 2.18. Let X be a normed space. If SX is compact, then X is finite dimensional.

Proof. Suppose that SX is compact but dimX =∞. Choose x1 ∈ X such that ‖x1‖ = 1.Then spanx1 is a proper closed subspace of X. By Riesz’s Lemma 2.17, there is x2 ∈ Xsuch that ‖x2‖ = 1 and ‖x2 − x1‖ ≥ 1/2. Now spanx1, x2 is a proper closed subspaceof X, and so again by Riesz’s Lemma 2.17, there is x3 ∈ X such that ‖x3‖ = 1 and‖x3 − xk‖ ≥ 1/2, k = 1, 2. Proceeding by induction, we get a sequence (xn) in SX suchthat ‖xm − xn‖ ≥ 1/2 for m 6= n, and hence (xn) has no convergent subsequence, whichcontradicts the compactness of SX .

Corollary 2.19. Let X be a Banach space. Then L(X) = K(X) if and only if X isfinite dimensional.

Example 2.20. Let (X,Ω, µ) be a measure space, and k : X ×X → F be a measurablefunction such that ∫

X

|k(x, y)|dµ(y) ≤ c1 for a.e. x (2.7)

and ∫X

|k(x, y)|dµ(x) ≤ c1 for a.e. y (2.8)

for some constants cj. Define K : L2(µ)→ L2(µ) by

Kf(x) =

∫X

k(x, y)f(y)dµ(y).

The mapping K is called an integral operator with kernel k. There are many other func-tion spaces where integral operators can be defined, e.g. the space of continuous functions

14

and other Lp spaces, and their properties depend on those of their kernel functions. Weshow that (2.7) and (2.8) are sufficient for K to be bounded on L2(µ) and ‖K‖ ≤ (c1c2)1/2.It is left as an exercise to show that K is compact if k ∈ L2(X ×X).

If f ∈ L2, then

|K(f(x)| ≤∫|k(x, y)|1/2|k(x, y)|1/2|f(y)|dµ(y)

≤(∫|k(x, y)|dµ(y)

)1/2(∫|k(x, y)‖f(y)|2dµ(y)

)1/2

,

and so ∫|Kf(x)|2dµ(x) ≤ c1

∫ ∫|k(x, y)||f(y)|2dµ(y)dµ(x) ≤ c1c2‖f‖2

2,

which shows that Kf ∈ L2 and ‖K‖ ≤ (c1c2)1/2.

To deal with integral operators on spaces of continuous functions, we introduce theconcept of equicontinuity. Let X be compact and F ⊂ C(X). We say that F is equicon-tinuous if for each ε > 0 and x0 ∈ X, there is a neighborhood U of x0 such that|f(x)− f(x0)| < ε for all x ∈ U and f ∈ F . Clearly, if f ∈ C(X), then f is equicontin-uous. In general, if we have more than one continuous function in our set, we may haveto choose a different U for each function unless the set is equicontinuous.

Example 2.21. Suppose that k ∈ C([0, 1]2). We show that the integral operator K withkernel k is compact on C([0, 1]). For f ∈ C([0, 1]) and s, s′ ∈ [0, 1], we have

|Kf(s)−Kf(s′)| ≤∫ 1

0

|k(s, t)− k(s′, t)||f(t)|dt ≤ supt∈[0,1]

|k(s, t)− k(s′, t)|‖f‖∞.

Therefore, since k is continuous on a compact set [0, 1]2, given ε > 0 and x0 ∈ [0, 1], thereis a δ > 0 such that

|Kf(x)−Kf(x0)| ≤ ε

for all x ∈ B(x0, δ) and f ∈ SC([0,1]). Thus, f is equicontinuous. Also, ‖Kf‖∞ ≤ ‖k‖ forall f ∈ SC([0,1]). It follows from the following theorem that K is compact on C([0, 1]).

Theorem 2.22 (Arzela-Ascoli). Let X be compact and F ⊂ C(X). If F is equicontinu-ous and pointwise bounded; that is, sup|f(x)| : f ∈ F <∞ for every x ∈ X, then F istotally bounded.

Proof. Let ε > 0. For each x ∈ X, there is an open neighborhood Ux of x such that

|f(x)− f(y)| < ε for all y ∈ Ux and all f ∈ F . (2.9)

Since Ux : x ∈ X is an open cover of the compact set X, there are x1, . . . , xn ∈ X suchthat X =

⋃nj=1 Uxj . Since F is pointwise bounded, (2.9) implies that there is an M > 0

such that sup|f(x)| : f ∈ F , x ∈ X ≤M .Let D = |z| ≤ M. Define p : F → Dn by p(f) = (f(x1), . . . , f(xn)). Since

Dn is compact and p(F) ⊂ Dn, there are f1, . . . , fm ∈ F such that if f ∈ F , then‖p(f) − p(fk)‖ = max1≤j≤n |f(xj) − fk(xj)| ≤ ε for some k. Thus, if f ∈ F and x ∈ X,then there is a k such that |f(xj) − fk(xj)| for j = 1, . . . , n, and x ∈ Uxi for some i, sothat

|f(x)− fk(x)| ≤ |f(x)− f(xi)|+ |f(xi)− fk(xi)|+ |fk(xi)− fk(x)| < 3ε.

15

Example 2.23. (a) Define T : l2 → l2 by

Tx =(xkk

)for x = (x1, x2, . . .) ∈ l2.

Note that T is linear. Define Tn : l2 → l2 by

Tnx =(x1,

x2

2,x3

3, . . . ,

xnn, 0, 0, . . .

).

Then each Tn is finite rank and hence compact. Also, a direct computation shows that

‖(T − Tn)x‖2 ≤ ‖x‖2

(n+ 1)2,

and so ‖T − Tn‖ ≤ (n+ 1)−1 → 0. Thus, T is compact on l2.(b) Note that if we only know that there are compact operators Tn such that ‖(T −

Tn)x‖ → 0 for every x, then we cannot conclude that T is compact. Indeed, consider theoperators Tn on l2 defined by

Tnx = (x1, . . . , xn, 0, 0, . . .).

Then

‖x− Tnx‖ =∞∑

k=n+1

|xk|2 → 0

for every x ∈ l2, but I is not compact because l2 is infinite dimensional.

2.3. Operators on Hilbert spaces. Throughout this section H stands for Hilbertspaces. Let T ∈ L(H). Then there is a unique operator S ∈ L(H) such that

(Tf, g) = (f, Sg) for all f, g ∈ H.We call S the adjoint of T and write T ∗ = S. To show uniqueness, suppose thereis another operator S ′ ∈ L(H) such that (Tf, g) = (f, S ′g) for all f, g ∈ H. Then((S − S ′)g, f) = 0 for all f, g ∈ H, which implies that S = S ′. To show existence, wedefine ϕ : H → C by

ϕ(f) = (Tf, g),

where g ∈ H. Then ϕ is a bounded linear functional, and so by the Riesz representationtheorem, there is an h ∈ H such that

ϕ(f) = (f, h)

for all f ∈ H. For g ∈ H, we set Sg = h. Then S is a bounded linear operator thatsatisfies (Tf, g) = (f, Sg) for all f, g ∈ H.

Proposition 2.24. If S, T ∈ L(H), then

(i) T ∗∗ = (T ∗)∗ = T ;(ii) ‖T‖ = ‖T ∗‖;

(iii) (T ∗)−1 = (T−1)∗ provided that T is invertible;(iv) ‖T‖2 = ‖T ∗T‖.

Proof. (i) If f, g ∈ H, then (f, T ∗∗g) = (T ∗f, g) = (g, T ∗f) = (Tg, f) = (f, Tg).(ii) Note that ‖T ∗f‖2 = (TT ∗f, f) ≤ ‖T‖‖T ∗f‖‖f‖ so that ‖T ∗∗‖ ≤ ‖T ∗‖ ≤ ‖T‖, and

apply (i).The other two assertions are left as an exercise.

Proposition 2.25. If T ∈ L(H), then

kerT = (ranT ∗)⊥ and kerT ∗ = (ranT )⊥.16

Proof. If f ∈ kerT , then(f, T ∗g) = (Tf, g) = 0

for all g ∈ H, and so f ∈ (ranT ∗)⊥. If f ∈ (ranT ∗)⊥, then

(Tf, g) = (f, T ∗g) = 0

implies that f ∈ kerT . The other assertion now follows from Proposition 2.24.

Let T ∈ L(H). Then

(i) T is normal if TT ∗ = T ∗T ;(ii) T is self-adjoint if T = T ∗;(iii) T is positive if (Tf, f) ≥ 0 for all f ∈ H;(iv) T is unitary if T ∗T = TT ∗ = I.

Proposition 2.26. An operator T ∈ L(H) is self-adjoint if and only if (Tf, f) is realfor all f ∈ H.

Proof. If T is self-adjoint, then (Tf, f) = (f, Tf) = (Tf, f). On the other hand, if (Tf, f)

is real for all f ∈ H, then (Tf, f) = (Tf, f) = (f, T ∗f) = (T ∗f, f).

Corollary 2.27. If P is positive, then P is self-adjoint.

Proposition 2.28. If T ∈ L(H), then T ∗T is positive.

Proof. For f ∈ H,(T ∗Tf, f) = (Tf, Tf) = ‖Tf‖2 ≥ 0.

We say that a linear operator T : X → Y between two Banach spaces is bounded belowif there is an ε > 0 such that ‖Tf‖ ≥ ε‖f‖ for all f ∈ X.

Proposition 2.29. Let T ∈ L(H). Then T is invertible if and only if T is bounded belowand ranT is dense.

Proof. Suppose T is invertible. Then ranT = H and hence the range of T is dense. Also,

‖Tf‖ ≥ 1

‖T−1‖‖T−1Tf‖ =

1

‖T−1‖‖f‖

for all f ∈ H, and so T is bounded below.Conversely, if T is bounded below, there is an ε > 0 such that ‖Tf‖ ≥ ε‖f‖ for all

f ∈ H. Suppose that Tfn → y. Then

‖fn − fm‖ ≤1

ε‖Tfn − Tfm‖,

so (fn) is Cauchy, which implies that there is an f ∈ H such that fn → f . Since T iscontinuous, Tfn → Tf , and hence y = Tf ∈ ranT . Thus, ranT is closed.

Now if ranT is closed, then ranT = ranT = H, so T is onto. Since T is boundedbelow, kerT = 0. Thus, T−1 exists. It is linear since T is linear. If g ∈ H, then g = Tffor some f ∈ H, and

‖T−1g‖ = ‖f‖ ≤ 1

ε‖Tf‖ =

1

ε‖g‖,

which shows that T−1 is bounded.

Remark 2.30. The preceding theorem is also true for operators acting on Banach spacesand can be proved the same way.

Corollary 2.31. Let T ∈ L(H). If T and T ∗ are both bounded below, then T is invertible.17

Proof. We need to show that the range of T is dense. Indeed,

(ranT )⊥ = kerT ∗ = 0

and, according to Corollary 1.17,

ranT = (ranT )⊥⊥ = 0⊥ = H.

Therefore, the range is dense in H.

Proposition 2.32. Let T ∈ L(H).

(i) If T is self-adjoint, then σ(T ) is real.(ii) If T is positive, then σ(T ) is nonnegative.

Proof. Let λ = α + iβ, α, β ∈ R, β 6= 0. To prove the first assertion, we need to showthat T − λ is invertible. Note that K = (T − α)/β is self-adjoint, and T − λ is invertibleif and only if K − i is invertible. According to Corollary 2.31 above, it suffices to showthat both K − i and (K − i)∗ = K + i are bounded below.

Let f ∈ H, then

‖(K ± i)f‖ = ((K ± i)f, (K ± i)f)

= ‖Kf‖2 ∓ i(Kf, f)± i(f,Kf) + ‖f‖2

= ‖Kf‖2 + ‖f‖2 ≥ ‖f‖2.

Thus, T − λ is invertible and σ(T ) ⊂ R.To prove the second statement, suppose in addition that T is positive and λ < 0. Then

‖(T − λ)f‖2 = ‖Tf‖2 − 2λ(Tf, f) + λ2‖f‖2 ≥ λ2‖f‖2

since (Tf, f) ≥ 0. On the other hand, (T − λ)∗ = T − λ. Thus, Corollary 2.31 impliesthat T − λ is invertible.

We say P ∈ L(H) is a projection if P 2 = P and P is self-adjoint. Let M be a closedsubspace of H and define PM on H by

PMf = g with f = g + h,

where g ∈M and h ∈M⊥ are unique—see Theorem 1.16.

Theorem 2.33. If M is a closed subspace of H, then PM is a projection of H onto M .If P is another projection on H with ranP = M , then P = PM .

Proof. Let f1, f2 ∈ H, and λ1, λ2 ∈ C. Then fi = gi + hi, i = 1, 2 with gi ∈ M andhi ∈M⊥. Write

λ1f1 + λ2f2 = (λ1g1 + λ2g2) + (λ1h1 + λ2h2)

where λ1g1 + λ2g2 ∈M and λ1h1 + λ2h2 ∈M⊥. Now, by uniqueness, we have

PM(λ1f1 + λ2f2) = λ1g1 + λ2g2 = λ1PMf1 + λ2PMf2

so PM is a linear operator on H.On the other hand, since g1 and h1 are orthogonal, ‖PMf1‖2 = ‖g1‖2 ≤ ‖g1‖2 +‖h1‖2 =‖f1‖2, which implies that ‖PM‖ ≤ 1 and PM ∈ L(H). Moreover, since (PMf1, f2) =(g1, g2 + h2) = (g1, g2) = (g1 + h1, g2) = (f1, PMf2), PM is self-adjoint.

If f ∈ M , then PMf = f ∈ M , so ranPM = M . Also, if f = g + h ∈ H, thenPMf = g = PMg = PM(PMf) = P 2

Mf . Thus, PM is a projection.The uniqueness of PM is left as an exercise.

18

Example 2.34. Consider

L2 = L2(∂D) = f : ∂D→ C : ‖f‖22 =

∫ 2π

0

|f(eiθ)|2 dθ2π

<∞,

which is a Hilbert space with inner product

(f, g) =

∫ 2π

0

f(eiθ)g(eiθ)dθ

2π.

Put en(eiθ) = einθ. Then ‖en‖ = 1 and (en, em) = 0 if n 6= m. The Stone-Weierstrassapproximation theorem states that for every continuous function on f : ∂D → C, thereare trigonometric polynomials pn : ∂D→ C defined by

pn(eiθ) =n∑

k=−n

αkek(eiθ)

such that pn → f uniformly on ∂D (see [14, Theorem 3.1]). Since the space of allcontinuous functions on ∂D is dense in L2, it follows that the set en : n ∈ Z is anorthonormal basis of L2.

Recall that if f ∈ L1(∂D), then the kth Fourier coefficient fk of f is defined by

fk(eiθ) =

∫ 2π

0

f(eiθ)e−ikθdθ

2π.

If f ∈ L2(∂D), then

‖f −n∑

k=−n

fkek‖L2 → 0 (2.10)

(see [14, Theorem 5.5]).Define the Hardy H2 by

H2 = f ∈ L2 : fk = 0 for k < 0.

Then H2 is a closed subspace of L2 and hence there exists the orthogonal projection P ofL2 onto H2 (see Theorem 2.33). We call P the Riesz projection. If follows that if f ∈ L2,then f ∼

∑∞k=−∞ fkek as in (2.10), and so

(Pf)(eiθ) =∞∑k=0

fkeikθ.

Let a ∈ L∞ and define the Toeplitz operator Ta on H2 by

Taf = P (af).

The function a is called the symbol of Ta. Clearly Ta is linear. It is also bounded because

‖Ta‖2 ≤ ‖a‖∞‖f‖2

for all f ∈ H2. In fact, ‖Ta‖ = ‖a‖∞, which is left as an exercise.

Recall that if T is bounded on a Banach space X and if Tn are finite rank such that‖T − Tn‖ → 0, then T is compact. The converse statement is not true in Banach spacesin general (see [9]) but is true in Hilbert spaces.

Theorem 2.35. A linear operator T on a Hilbert space H is compact if and only if thereis a sequence of finite rank operators Tn such that ‖T − Tn‖ → 0.

19

Proof. Suppose that T be compact. Since the closure T (SH) is compact, it is separable.Thus we can find a basis e1, e2, . . . for ranT . Let Pn be the orthogonal projection ofH onto the closed linear span of e1, . . . , en (see Theorem 2.33 below). Put Tn = PnT .Then Tn has finite rank.

Let h ∈ H and put k = Th. Then

Pnk =n∑k=1

(k, ek)ek and k =∞∑k=1

(k, ek)ek.

Therefore,‖Tnh− Th‖ = ‖Pnk − k‖ → 0. (2.11)

Let ε > 0. Since T is compact, there are h1, . . . , hm such that if ‖h‖ ≤ 1, ‖Th−Thj‖ <ε/3 for some j. For n ≥ 1, we have

‖Th− Tnh‖ ≤ ‖Th− Thj‖+ ‖Thj − Tnhj‖+ ‖Pn(Thj − Th)‖≤ 2‖Th− Thj‖+ ‖Thj − Tnhj‖ ≤ 2ε/3 + ‖Thj − Tnhj‖.

According to (2.11), there is n0 such that ‖Thj − Tnhj‖ < ε/3 for all hj and n > n0.

The next example deals with a very important class of linear operators, namely mul-tiplication operators.

Example 2.36. Let ϕ ∈ L∞ := L∞(X,µ) and define Mϕ on L2 := L2(X,µ) by Mϕf =ϕf , where µ is a probability measure; that is, µ(X) = 1. Clearly, Mϕ is linear and‖Mϕf‖2 ≤ ‖ϕ‖∞‖f‖2, which shows that Mϕ ∈ L(L2) and ‖Mϕ‖ ≤ ‖ϕ‖∞.

To show that ‖Mϕ‖ = ‖ϕ‖∞, let

En =

x ∈ X : |ϕ(x)| ≥ ‖ϕ‖∞ −

1

n

.

Then χEn ∈ L2 and

‖MϕχEn‖2 =

(∫X

|ϕχEn|2dµ)1/2

≥(∫

X

(‖ϕ‖∞ −1

n)2|χEn|2dµ

)1/2

=

(‖ϕ‖∞ −

1

n

)‖χEn‖2,

which implies that ‖Mϕ‖ ≥ ‖ϕ‖∞ − 1n

for all n ∈ N. Thus, ‖Mϕ‖ = ‖ϕ‖∞.In order to find the adjoint of Mϕ, observe that

(Mϕf, g) =

∫X

ϕfgdµ =

∫X

fϕgdµ = (f,Mϕg).

Therefore, M∗ϕ = Mϕ.

Since L∞ is commutative, Mϕ commutes with Mϕ and hence is a normal operator. Ifϕ ∈ L∞,Mϕ = M∗

ϕ if and only if ϕ = ϕ, that is Mϕ is self-adjoint if and only if ϕ is real.

Moreover, Mϕ is a projection if and only if ϕ2 = ϕ, that is, ϕ is a characteristic function.

As in the previous example, many classes of concrete operators acting on functionspaces are generated by functions, often referred to as symbols, and their operator-theoretic properties are often encoded in the properties of their symbols. To demonstratethis further, we characterize the spectrum of Ma in terms of the essential range of thefunction a. Let f be measurable on X. The essential range R(f) of f is defined by

R(f) =λ ∈ C : µ(x : |f(x)− λ| < ε) > 0 for all ε > 0

.

20

Notice that R(f) = f(X) if f is continuous on X. However, in general, a point in therange of f need not be in its essential range and there may be points in the complementof the range that are contained in the essential range.

Lemma 2.37. If f ∈ L∞, then R(f) is compact and

‖f‖∞ = supλ∈R(f)

|λ|.

Proof. If λ0 /∈ R(f), there exists ε > 0 such that X1 := x ∈ X : |f(x) − λ0| < ε hasmeasure zero. Let λ ∈ B(λ0, ε/2). If λ ∈ R(f), then X2 := x ∈ X : |f(x) − λ| < ε/2has positive measure but for x ∈ X2 we have |f(x) − λ0| ≤ |f(x) − λ| + |λ − λ0| < ε,that is, X2 ⊂ X1, which is a contradiction. Thus, B(λ0, ε/2) ⊂ C \ R(f), and so R(f) isclosed.

Let us show that sup|λ| : λ ∈ R(f) ≤ ‖f‖∞. If |λ| > ‖f‖∞, then there is a δ > 0such that

|f(x)|+ δ ≤ |λ|for almost every x, and so the set of all x for which |f(x)− λ| < δ/2 is of measure zero.Thus, λ /∈ R(f). It follows that R(f) is bounded and closed, and hence compact. Thereverse inequality is left as an exercise.

Lemma 2.38. If f ∈ L∞, σ(f) = R(f).

Proof. If λ /∈ σ(f), (f − λ)−1 is essentially bounded, and so there is an M such thatM−1 ≤ |f(x) − λ| for almost every x ∈ X, that is, λ /∈ R(f). Conversely, if λ /∈ R(f),then there is a δ > 0 such that

x ∈ X : |f(x)− λ| < δhas measure zero. Thus, |(f − λ)−1| ≤ δ−1 almost everywhere, and so λ /∈ σ(f).

A subalgebraM of L(H) is said to be maximal abelian provided that it is commutativeand is not properly contained in any commutative subalgebra of L(H).

Proposition 2.39. The algebra M = Mϕ : ϕ ∈ L∞ is maximal abelian.

Proof. Clearly, M is commutative. Suppose that T ∈ L(L2) commutes with M. Ifψ = T1, then ψ ∈ L2 and

Tϕ = TMϕ1 = MϕT1 = ϕψ (2.12)

for ϕ ∈ L∞. In addition, if

En = x ∈ X : |ψ(x)| ≥ ‖T‖+ 1/n,then it follows that

‖T‖‖χEn‖2 ≥ (‖T‖+ 1/n) ‖χEn‖2.

Therefore, ‖χEn‖2 = 0 and x ∈ X : |ψ(x)| ≥ ‖T‖ has measure zero. Thus, ψ ∈ L∞and Tϕ = Mψϕ for ϕ ∈ L∞. Since L∞ is dense in L2, T = Mψ ∈M.

Proposition 2.40. If H is a Hilbert space and if A is a maximal abelian subalgebra ofL(H), then σ(T ) = σA(T ) for T in A.

Proof. Clearly, σ(T ) ⊂ σA(T ) for T ∈ A. If λ /∈ σ(T ), then (T − λ)−1 ∈ L(H). Since(T − λ)−1 commutes with A and A is maximal abelian, we have (T − λ)−1 ∈ A. Thusλ /∈ σA(T ).

Corollary 2.41. If ϕ ∈ L∞, then σ(Mϕ) = R(ϕ).

Proof. Since M = Mψ : ψ ∈ L∞ is maximal abelian, σM(T ) = σ(T ) for T ∈ M. ThespacesM and L∞ are isomorphic and so σM(Mϕ) = σL∞(ϕ) = R(ϕ) by Lemma 2.38.

21

3. Fredholm theory

Let X be a Banach space and T ∈ L(X). We say T is Fredholm and write T ∈ Φ(X)if

dim kerT <∞ and dim(X/ ranT ) <∞.We first show that the range of a Fredholm operator is always closed. The concept of thedirect sum Y ⊕Z plays a key role in this. Let Y and Z be subspaces of X. If Y ∩Z = 0and X = Y + Z, we write

X = Y ⊕ Z.

Theorem 3.1. Let Y be a subspace of a Banach space X.

(i) If Y has finite dimension, there exists a closed subspace Z of X such that X = Y ⊕Z.(ii) If Y has finite codimension, there exists a finite-dimensional, and therefore closed,

subspace Z of X such that X = Y ⊕ Z.

Proof. (i) Suppose that dimY < ∞. Let ϕ1, . . . , ϕn be a basis for Y ∗. By the Hahn-Banach theorem, for k = 1, . . . , n, there exists φk ∈ X∗ such that φk is an extension ofϕk. Let φ : X → Cn be such that φ(x) =

(φ1(x), . . . , φn(x)

), and let Z = ker(φ). Then

Z is closed. If x ∈ Y ∩ Z, then

0 = φ(x) =(ϕ1(x), . . . , ϕn(x)

).

So, if ϕ ∈ Y ∗, there exist α1, . . . , αn ∈ C such that

ϕ(x) = α1ϕ1(x) + · · ·+ αnϕn(x) = 0.

Hence, ϕ(x) = 0 for all ϕ ∈ Y ∗ which implies that x = 0 by the Hahn-Banach theorem.Thus, Y ∩ Z = 0.

Let x ∈ X. Since the restriction φ Y : Y → Cn is an isomorphism, there existsy ∈ Y such that φ(y) = φ(x). If z = x − y, then φ(z) = φ(x) − φ(y) = 0, and hence,z ∈ ker(φ) = Z. Therefore, we have x = y + z with y ∈ Y and z ∈ Z, hence X = Y ⊕Z.

(ii) Suppose that dimX/Y < ∞. Let x1 + Y, . . . , xn + Y be a basis for X/Y . LetZ = spanx1, . . . , xn. Then Z is finite-dimensional. If x ∈ Y ∩ Z, then there existα1, . . . , αn ∈ C such that x = α1x1 + · · ·+ αnxn and

0 = x+ Y = α1x1 + · · ·+ αnxn + Y = α1(x1 + Y ) + · · ·+ αn(xn + Y )

which implies that α1 = . . . = αn = 0, and hence Y ∩ Z = 0. If x ∈ X, thenx+Y = α1(x1 +Y )+ · · ·+αn(xn+Y ) for some α1, . . . , αn ∈ C. Let z = α1x1 + · · ·+αnxn,and let y = x−z. Then it is straightforward to see that y ∈ Y . Therefore, X = Y ⊕Z.

Theorem 3.2. Let X, Y be Banach spaces and let T ∈ L(X, Y ). If there exists a closedvector subspace Z of Y such that ranT ⊕ Z = Y , then ranT is closed in Y .

Proof. The mapX/ kerT → Y, x+ kerT 7→ Tx

is bounded, has the same range as T and is one-to-one. Thus we can assume that T isone-to-one.

Define S : X × Z → Y such that S(x, z) = Tx + z. Since T (X)⊕a Z = Y , S is onto.Also, S is one-to-one since T is one-to-one and T (X) ⊕a Z = Y . Moreover, since T iscontinuous, ‖S(x, z)‖ ≤ ‖Tx‖ + ‖z‖ ≤ ‖T‖‖x‖ + ‖z‖ ≤ M(‖x‖ + ‖z‖) = M‖(x, z)‖,where M > max‖T‖, 1, and hence S is continuous. Therefore, S−1 is continuous bythe open mapping theorem. Let x ∈ X. Then (x, 0) = S−1

(Tx)

and it follows that

‖Tx‖ ≥ ‖S−1‖−1‖x‖. Hence T is bounded below, and by the proof of Proposition 2.29,the set T (X) is closed.

22

Theorem 3.3. If T is Fredholm on a Banach space, then ranT is closed.

Proof. Apply the previous two theorems.

Our next goal is a characterization of Fredholm operators in terms of invertibility inthe Calkin algebra L(X)/K(X) (see Example 2.2). The following Riesz’s theorem forcompact operators plays an important role in it.

Theorem 3.4. Let T be a compact operator on a Banach space X. Then T − λ isFredholm for all λ 6= 0.

Proof. Let Z = ker(T − λ). If x ∈ Z, then Tx − λx = 0, and so Tx = λx ∈ Z. Thisshows that T (Z) ⊂ Z. Therefore the restriction T Z is in K(Z). Since λ 6= 0, it followsthat the identity operator is compact on Z and hence the set Z is finite-dimensional byTheorem 2.18.

Next we show that ran(T − λ) is closed. Since Z is finite-dimensional, there exists aclosed vector space Y ⊂ X such that X = Y ⊕ Z. If z ∈ Z, then (T − λ)z = 0. So,(T − λ)(X) = (T − λ)(Y ). If (T − λ) Y is not bounded below, there exists a sequence(xn) in Y such that ‖xn‖ = 1 and (T−λ)xn → 0. Since T ∈ K(X), there is a subsequence(x1

n) of (xn) such that(T (x1

n))

is convergent. Therefore,

x1n =

1

λ

(T (x1

n)− (T − λ)(x1n))→ x.

for some x ∈ Y = Y . Also,

0 = limn→∞

(T − λ)(xn),

hence (T − λ)x = 0, and so x ∈ ker(T − λ) = Z. Since Y ∩ Z = 0, x = 0, which isa contradiction. Therefore, (T − λ)(X) = (T − λ)(Y ) is closed by the proof of Proposi-tion 2.29.

By Theorem 2.16, T ∗ is compact, and so dim ker(T ∗−λ) <∞. Let W = X/ ran(T −λ)and define π : X → W by π(x) = x + ran(T − λ). Then ranπ∗ ⊂ ker(T ∗ − λ). Since π∗

is one-to-one, we have

dimW = dimW ∗ = dimπ∗(W ∗) ≤ ker(T ∗ − λ) <∞.

Let T ∈ L(X). An operator S ∈ L(X) for which I −TS and I −ST are both compactis referred to as a regularizer of T .

Theorem 3.5. Let T ∈ L(X). If T is Fredholm, then there is an S ∈ L(X) such thatI − TS and I − ST are both compact. Conversely, if there are operators S1, S2 ∈ L(X)such that I − TS1 and I − S2T are compact, then T is Fredholm.

Proof. Suppose that I − TS1 and I − S2T are compact operators. Then

TS1 = I +W1 and S2T = I +W2

for some W1 ∈ K(X) and W2 ∈ K(X). Since kerT ⊂ ker(S2T ) = ker(I + W2),dim kerT ≤ dim ker(I + W2) < ∞ (see Theorem 3.4). Similarly, since (I + W1)(X) =TS1(X) ⊂ T (X), dim(X/ ranT ) ≤ dim(X/ ran(I+W2)) <∞ (see Theorem 3.4). There-fore, T is Fredholm.

Suppose that T is Fredholm. Then

kerT ⊕X1 = X = ranT ⊕X2,23

where X1 is closed and dimX2 < ∞ (see Theorem 3.1). Define T1 : X1 → ranT byT1x = Tx. Then T1 is onto and since kerT ∩X1 = 0, T1 is also one-to-one. Thus, T−1

1

is continuous by the open mapping theorem. Define S : X → X by

Sx =

T−1

1 x if x ∈ ranT,

0 if x ∈ X2 and indT 6= 0,

Wx if x ∈ X2 and indT = 0,

where W : X2 → kerT is a linear isomorphism (such W exists when indT = 0). ThenS is continuous and TST = T , so kerST = kerT and ranTS = ranT and ST, TSare both projections on X. Thus, ran(I − ST ) = kerST = kerT , which implies thatI − ST is of finite rank. Similarly, if x2 = (I − TS)x = x − TSx for some x ∈ X, thenx = TSx+ x2 ∈ ranT ⊕X2, which implies that x ∈ X2. Thus, ran(I − TS) ⊂ X2 and soI − TS is of finite rank.

Remark 3.6. In the previous theorem, the operator S can be showed to be Fredholm.

Theorem 3.7 (Atkinson). A bounded linear operator T on X is Fredholm if and only ifT +K(X) is invertible in the Calkin algebra L(X)/K(X).

Proof. If T is Fredholm, then there is an S ∈ L(X) such that I − ST and I − TS arecompact, and hence

(T +K(X))(S +K(X)) = I +K(X) = (S +K(X))(T +K(X)). (3.1)

Conversely, if T + K(X) is invertible, then there is an operator S ∈ L(X) such that(3.1) holds, and so ST − I = W1 and TS − I = W2 for some compact operators Wj.Thus, T is Fredholm by the previous theorem.

Theorem 3.8 (Index theorem). If T and S are Fredholm on X, then ST is Fredholmand

ind(ST ) = ind(S) + ind(T ).

Proof. TBD.

Theorem 3.9. The space of all Fredholm operators Φ(X) is open in L(X) and the indexfunction

ind : Φ(X)→ Z, T 7→ indT

is continuous.

Proof. Let T ∈ Φ(X, Y ). By Theorem 3.5 and Remark 3.6, there is a V ∈ Φ(X) suchthat V T = 1 + W1 and TV = 1 + W2 for some W1 ∈ K(X) and W2 ∈ K(X). Setδ < ‖V ‖−1. Let S ∈ L(X) be such that ‖S‖ < δ. Then ‖V S‖ < 1 and ‖SV ‖ < 1, andso (I + V S)−1 ∈ L(X) and (I + SV )−1 ∈ L(Y ) by Theorem 2.4. Let V1 = V (I + SV )−1

and V2 = (I + V S)−1V . Then

(T + S)V1 = (T + S)V (I + SV )−1 = (I +W2 + SV )(I + SV )−1

=[(I +W2(I + SV )−1

)(I + SV )

](I + SV )−1

= I +W2(I + SV )−1

and

V2(T + S) = (I + V S)−1V (T + S) = (I + V S)−1(I +W1 + V S)

= (I + V S)−1[(I + V S)

(I + (I + V S)−1W1

)]= I + (I + V S)−1W1

24

Therefore, since (W +V S)−1W1 and W2(I+SV )−1 are compact operators, I− (T +S)V1

and I − V2(T + S) are compact operators, and so T + S is Fredholm by Theorem 3.5.Consequently, Φ(X) is open in L(X).

To prove continuity, let W ∈ Φ(X) be such that ‖T −W‖ < δ. Then ‖TV −WV ‖ < 1,and hence S = I − TV +WV is invertible in L(X) by Theorem 2.4. Since T +WV T =TV T + T − TV T +WV T = TV T + ST = T + ST and so WV T = ST ,

ind(W ) + ind(V ) + ind(T ) = ind(S) + ind(T )

by Theorem 3.8, and hence ind(W ) = − ind(V ) because S is invertible.

Let X be a Banach space and T ∈ L(X). The essential spectrum σess(T ) of T is definedby

σess(T ) = λ ∈ C : T − λ /∈ Φ(X),where Φ(X) is the space of all Fredholm operators on X.

Proposition 3.10. If X is a Banach space and T ∈ L(X), then σess(T ) is nonemptyand compact.

Proof. Exercise.

Proposition 3.11. If K is a compact operator on a Banach space, then σess(K) = 0.

Proof. Exercise.

4. Weak convergence

If X is a normed space, the weak topology on X is the topology defined by the collectionof seminorms px∗ : x∗ ∈ X∗, where

px∗(x) = |x∗(x)|.A sequence (xn) in X is said to be weakly convergent if there is an x ∈ X such that

x∗(xn)→ x∗(x)

for all x∗ ∈ X∗.

Proposition 4.1. Let X be a Banach space, (xn) be a sequence in X and x ∈ X.

(i) If ‖xn − x‖ → 0, then xn → x weakly.(ii) If xn → x weakly, then supn ‖xn‖ <∞.

Proof. Suppose that ‖xn − x‖ → 0. If x∗ ∈ X∗, then

|x∗(xn)− x∗(x)| = |x∗(xn − x)| ≤ ‖x∗‖‖xn − x‖ → 0.

Thus, xn → x weakly.Suppose that xn → x weakly. Since X ⊂ X∗∗, each xn generates a bounded linear

functional on X∗ defined by xn(x∗) = x∗(xn) with ‖xn‖ = sup|x∗(xn)| : ‖x∗‖ ≤ 1.Since xn → x weakly, we have x∗(xn) → x∗(x) for all x∗, and so for each x∗ ∈ X∗,supn |xn(x∗)| is bounded. Thus, by Theorem 1.8, supn ‖xn‖ <∞.

It is left as an exercise to show that the converse of (i) is not true in general.We say that (x∗n) is weak-star convergent in X∗ if (x∗n(x)) converges for all x ∈ X. This

defines the weak-star topology on X∗.

Theorem 4.2 (Alaoglu). If X is a normed space, then SX∗ is compact in the weak-startopology.

Proof. See [2, Theorem 3.1]. 25

Proposition 4.3. Let H be a separable Hilbert space. Then xn → x weakly in H if andonly if (xn, y)→ (x, y) for all y ∈ H.

Proof. Suppose xn → x weakly. Let y ∈ H and define x∗(z) = (z, y) for z ∈ H. Thenx∗ ∈ H∗ and hence (xn, y) = x∗(xn)→ x∗(x) = (x, y).

Conversely, suppose that (xn, y) → (x, y) for all y ∈ H. Let x∗ ∈ H∗. Theorem 1.18implies that there is an h ∈ H such that x∗(z) = (z, h) for all z ∈ H. Since x∗(xn) =(xn, h)→ (x, h) = x∗(x), we have xn → x weakly.

Lemma 4.4. Let H be a separable Hilbert space and let T ∈ L(H). If fn → f weakly inH, then Tfn → Tf weakly.

Proof. Suppose that fn → f weakly. If g ∈ H, then

limn→∞

(Tfn, g) = limn→∞

(fn, T∗g) = (f, T ∗g) = (Tf, g).

Therefore, Tfn → Tf weakly by Proposition 4.3.

Theorem 4.5. A linear operator T on a separable Hilbert space H is compact if and onlyif ‖Tfn‖ → 0 whenever fn → 0 weakly in H.

Proof. Suppose that T is compact and fn → 0 weakly. Then supn ‖fn‖ < ∞ by Propo-sition 4.1 and hence there is a subsequence (fnk

) and g ∈ H such that Tfnk→ g. By

Lemma 4.4, Tfnk→ 0 weakly. Thus, g = 0.

Conversely, suppose that Tfn → 0 whenever fn → 0 weakly. Let fn ∈ SH . By Theo-rem 4.2, (fn) has a weakly convergent subsequence (fnk

). Therefore, (Tfnk) converges in

the norm.

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