operator theory on hilbert space class...

OPERATOR THEORY ON HILBERT SPACE

Class notes

John Petrovic

Contents

Chapter 1. Hilbert space 1

1.1. Definition and Properties 1

1.2. Orthogonality 3

1.3. Subspaces 7

1.4. Weak topology 9

Chapter 2. Operators on Hilbert Space 13

2.1. Definition and Examples 13

2.2. Adjoint 15

2.3. Operator topologies 17

2.4. Invariant and Reducing Subspaces 20

2.5. Finite rank operators 22

2.6. Compact Operators 23

2.7. Normal operators 27

Chapter 3. Spectrum 31

3.1. Invertibility 31

3.2. Spectrum 34

3.3. Parts of the spectrum 38

3.4. Spectrum of a compact operator 40

3.5. Spectrum of a normal operator 43

iii

iv CONTENTS

Chapter 4. Invariant subspaces 47

4.1. Compact operators 47

4.2. Line integrals 49

4.3. Invariant subspaces for compact operators 52

4.4. Normal operators 56

Chapter 5. Spectral radius algebras 64

5.1. Compact operators 64

CHAPTER 1

Hilbert space

1.1. Definition and Properties

In order to define Hilbert space H we need to specify several of its features. First, it is a complex vector

space — the field of scalars is C (complex numbers). [See Royden, p. 217.] Second, it is an inner product

space. This means that there is a complex valued function 〈x, y〉 defined on H×H with the properties that, for

all x, y, z ∈ H and α, β ∈ C:

(a) 〈αx+ βy, z〉 = α〈x, z〉+ β〈y, z〉; it is linear in the first argument;

(b) 〈x, y〉 = 〈y, x〉; it is Hermitian symmetric;

(c) 〈x, x〉 ≥ 0; it is non-negative;

(d) 〈x, x〉 = 0 iff x = 0; it is positive.

In every inner product space it is possible to define a norm as ‖x‖ = 〈x, x〉1/2.

Exercise 1.1.1. Prove that this is indeed a norm.

Finally, Hilbert space is complete in this norm (meaning: in the topology induced by this norm).

Example 1.1.1. Cn is an inner product space with 〈x, y〉 =∑nk=1 xkyk and, consequently, the norm ‖x‖ =√∑n

k=1 |xk|2. Completeness: if x(k)∞k=1 is a Cauchy sequence in Cn (here x(k) = (x(k)1 , x

(k)2 , . . . , x

(k)n )) then so

is x(k)m for any fixed m, 1 ≤ m ≤ n, and C is complete.

Example 1.1.2. LetH0 denote the collection of all complex sequences, i.e. functions a : N→ C, characterized

by the fact that an 6= 0 for a finite number of positive integers n. Define the inner product on H0 by 〈a, b〉 =∑∞n=0 anbn. The space H0 is not complete in the induced norm. Indeed, the sequence a(k)k∈N, defined by

a(k)n = 1/2n if n ≤ k and a

(k)n = 0 if n > k is a Cauchy sequence, but not convergent.

1

2 1. HILBERT SPACE

Example 1.1.3. Let `2 denote the collection of all complex sequences a = an∞n=1 such that∑∞n=1 |an|2

converges. Define the inner product on `2 by 〈a, b〉 =∑∞n=1 anbn. Suppose that a(k)∞k=1 is a Cauchy sequence

in `2. Then so is a(k)n ∞k=1 for each n, hence there exists an = limk→∞ a

(k)n . First we show that a ∈ `2. Indeed,

choose K so that for k ≥ K we have ‖a(k) − a(K)‖ ≤ 1. Then, using Minkowski’s Inequality for sequences (see

Royden, p. 122), for any N ∈ N,N∑n=1

|an|21/2

≤

N∑n=1

|an − a(K)n |2

1/2

+

N∑

1=0

|a(K)n |2

1/2

= limk→∞

N∑n=1

|a(k)n − a(K)

n |21/2

+

N∑n=1

|a(K)n |2

1/2

≤ lim supk→∞

‖a(k) − a(K)‖+ ‖a(K)‖ ≤ 1 + ‖a(K)‖.

Thus a = an ∈ `2. Moreover, a(k) converges to a, i.e. limk→∞ ‖a − a(k)‖ = 0. Let ε > 0 and choose M so

that k, j ≥M implies that ‖a(k) − a(j)‖ < ε. For such k ≥M and any N , we have

N∑n=1

|an − a(k)n |2 = lim

j→∞

N∑n=1

|a(j)n − a(k)

n |2 ≤ lim supj→∞

‖a(j) − a(k)‖2 ≤ ε2.

Since N is arbitrary, it follows that ‖a− a(k)‖ ≤ ε and, therefore, `2 is Hilbert space.

Example 1.1.4. The space L2 of functions f : X → C, such that∫X|f |2dµ <∞ (where X is usually [0, 1] and

µ Lebesgue measure). The inner product is defined by 〈f, g〉 =∫Xfg dµ and L2 is complete by the Riesz–Fisher

Theorem (see Royden, p. 125).

Example 1.1.5. The space H2. Let X = T (the unit circle) and µ the normalized Lebesgue measure on T.

The Hardy space H2 consists of those functions in L2(T) such that 〈f, eint〉 = 0 for n = −1,−2, . . . .

Some important facts.

Proposition 1.1.1 (Parallelogram Law). ‖x+ y‖2 + ‖x− y‖2 = 2‖x‖2 + 2‖y‖2.

Proposition 1.1.2 (Polarization Identity). 4〈x, y〉 = 〈x+ y, x+ y〉 − 〈x− y, x− y〉+ i〈x+ iy, x+ iy〉 − i〈x−

iy, x− iy〉.

Exercise 1.1.2. Prove Propositions 1.1.1 and 1.1.2.

1.2. ORTHOGONALITY 3

Problem 1. Let ‖ · ‖ be a norm on Banach space X , and define 〈x, y〉 as in Polarization Identity. Assuming

that the norm satisfies the Parallelogram Law, prove that 〈x, y〉 defines an inner product.

1.2. Orthogonality

In Linear Algebra a basis of a vector space is defined as a minimal spanning set. In Hilbert space such a

definition is not very practical. It is hard to speak of minimality when a basis can be infinite. In fact, a basis can

be uncountable, so if eii∈I is such a basis, what is the meaning of∑i∈I xiei?

Definition 1.2.1. An orthonormal subset of Hilbert space H is a set E such that (a) ‖e‖ = 1, for all e ∈ E ;

(b) if e1, e2 ∈ E and e1 6= e2 then 〈e1, e2〉 = 0. An orthonormal basis in H is a maximal orthonormal set. We use

abbreviations o.n.s. and o.n.b. for orthonormal set and orthonormal basis, respectively.

Theorem 1.2.1. Every Hilbert space has an orthonormal basis.

Proof. Let e be a unit vector in H. Then E = e is an orthonormal set. Let M be the collection of all

orthonormal sets in H that contain E . By the Hausdorff Maximal Principle (Royden, p.25) there exists a maximal

chain C of such orthonormal sets, partially ordered by inclusion. Let N be the union of all elements of C. Then

N is a maximal orthonormal set, hence a basis of H.

If the set e is replaced by any orthonormal set, the same proof yields a stronger result.

Theorem 1.2.2. Every orthonormal set in Hilbert space can be extended to an orthonormal basis.

Example 1.2.1. For k ∈ N, let ek denote the sequence with only one non-zero entry, lying in the kth position

and equal to 1. The set ekk∈N is an o.n.b. for `2. (If a vector x ∈ `2 is orthogonal to all ek, then each of its

components is zero, so x = 0.)

Example 1.2.2. The set e1, e3, e5, . . . is an orthonormal set in `2 but not a basis.

Example 1.2.3. The set

1√2π,

cos t√π,

sin t√π,

cos 2t√π,

sin 2t√π, . . .

is an o.n.b. in L2(−π, π).

4 1. HILBERT SPACE

Example 1.2.4. The set

1√2π

eint : n ∈ Z

is another o.n.b. in L2(−π, π).

In Linear Algebra, if eii∈I is an o.n.b. then every vector x can be written as∑i∈I〈x, ei〉ei. In Hilbert space

our first task is to make sense of this sum since the index set I need not be countable.

Theorem 1.2.3 (Bessel’s Inequality). Let eiki=1 be an o.n.s. in H, and let x ∈ H. Then∑ki=1 |〈x, ei〉|2 ≤

‖x‖2.

Proof. If we write xi = 〈x, ei〉, then

0 ≤ ‖x−k∑i=1

xiei‖2 = 〈x−k∑i=1

xiei, x−k∑i=1

xiei〉 = ‖x‖2 − 2Re〈x,k∑i=1

xiei〉+ 〈k∑i=1

xiei,

k∑j=1

xjej〉

= ‖x‖2 − 2Rek∑i=1

xi〈x, ei〉+k∑i=1

k∑j=1

xixj〈ei, ej〉 = ‖x‖2 − 2Rek∑i=1

xixi +k∑i=1

xixi = ‖x‖2 −k∑i=1

|xi|2.

Corollary 1.2.4. Let E = eii∈I be an o.n.s. in H, and let x ∈ H. Then 〈x, ei〉 6= 0 for at most a countable

number of i ∈ I.

Proof. Let x ∈ H be fixed and let En = ei : |xi| ≥ 1/n. If ei1 , ei2 , . . . , eik ∈ En then

‖x‖2 ≥k∑j=1

|xij |2 ≥ k(1/n2).

So, for each n ∈ N, En is a finite set, and E = ∪nEn.

In view of Corollary 1.2.4 the expressions like∑〈x, ei〉ei turn out to be the usual infinite series. Our next

task is to establish their convergence. The following Lemma will be helpful in this direction.

Lemma 1.2.5. If xii∈N is a sequence of complex numbers and eii∈N is an o.n.s. in H, then the series∑i∈N xiei and

∑i∈N |xi|2 are equiconvergent.

1.2. ORTHOGONALITY 5

Proof. Let sm and σm denote the partials sums of∑i∈N xiei and

∑i∈N |xi|2, respectively. Then

‖sm − sn‖2 = ‖∑mi=n+1 xiei‖2 = 〈

∑mi=n+1 xiei,

∑mj=n+1 xjej〉 =

∑mi=n+1 |xi|2 = |σm − σn|

so the series are equiconvergent.

Now we can establish the convergence of∑i∈I〈x, ei〉ei. We will use notation xi = 〈x, ei〉 for the Fourier

coefficients of x ∈ H relative to the fixed basis eii∈I .

Corollary 1.2.6 (Parseval’s Identity). Let eii∈I be an o.n.s. in H, and let x ∈ H. Then the series∑i∈I xiei and

∑i∈I |xi|2 converge and ‖

∑i∈I xiei‖2 =

∑i∈I |xi|2.

Proof. Since only a countable number of terms in each series is non-zero, we can rearrange them and consider

the series∑∞i=1 xiei and

∑∞i=1 |xi|2. The latter series converges by the Bessel’s Inequality and Lemma 1.2.5 implies

that the former series converges too. Moreover, their partial sums sm and σm satisfy ‖sm‖ = σm, so the last

assertion of the corollary follows by letting m go to ∞.

Now we are in the position to show that, in Hilbert space, every o.n.b. indeed spans H. Of course, the

minimality is a direct consequence of the definition.

Theorem 1.2.7. Let E = eii∈I be an o.n.b. in H. Then, for each x ∈ H, x =∑i∈I xiei, where xi = 〈x, ei〉.

Proof. Let xi = 〈x, ei〉 and y = x −∑i∈I xiei. (Well defined since the series converges.) Then 〈y, ek〉 =

〈x, ek〉 − 〈∑i∈I xiei, ek〉 = 0, for each k ∈ I, so y ⊥ E . If y 6= 0, then E ∪ y/‖y‖ is an o.n.s., contradiciting the

maximality of E , so y = 0.

The following is the analogue of a well known Linear Algebra fact. We use notation card I for the cardinal

number of the set I.

Theorem 1.2.8. Any two orthonormal bases eii∈I and fjj∈J in H have the same cardinal number.

Proof. We will assume that both cardinal numbers are infinite. If either of them is finite, one knows from

Linear Algebra that the other one is finite and equal to the first. Let j ∈ J be fixed and let Ij = i ∈ I : 〈fj , ei〉 6=

6 1. HILBERT SPACE

0. By Corollary 1.2.4, Ij is at most countable. Further, ∪j∈JIj = I. Indeed, if i0 ∈ I \ ∪j∈JIj then 〈fj , ei0〉 = 0

for all j ∈ J so it would follow that ei0 = 0. Since card Ij ≤ ℵ0 we see that card I ≤ card J ·ℵ0 = card J . Similarly,

card J ≤ card I. By Cantor–Bernstein Theorem, (see, e.g., “Proofs from the book”, p.90) card I = card J .

Definition 1.2.2. The dimension of Hilbert space H, denoted by dimH, is the cardinal number of a basis

of H.

In this course we will assume that dimH ≤ ℵ0.

Exercise 1.2.1. If H is an infinite dimensional Hilbert space, then H is separable iff dimH = ℵ0. [Given a

countable basis, use rational coefficients. Given a countable dense set, approximate each element of a basis close

enough to exclude all other basis elements.]

Next, we want to address the question: when can we identify two Hilbert spaces? We need a vector space

isomorphism (i.e., a linear bijection) that preserves the inner product.

Definition 1.2.3. If H and K are Hilbert spaces, an isomorphism is a linear surjection U : H → K such

that, for all x, y ∈ H, 〈Ux,Uy〉 = 〈x, y〉. In this situation we say that H and K are isomorphic.

Exercise 1.2.2. Prove that 〈Ux,Uy〉 = 〈x, y〉 for all x, y ∈ H iff ‖Ux‖ = ‖x‖ for all x ∈ H. Conclude that a

Hilbert space isomorphism is injective.

Theorem 1.2.9. Every separable Hilbert space of infinite dimension is isomorphic to `2. Every Hilbert space

of finite dimension n is isomorphic to Cn.

Proof. We will assume that H is an infinite dimensional Hilbert space and leave the finite dimensional case

as an exercise. Since H is separable, there exists an o.n.b. en∞n=1. For x ∈ H, let xi = 〈x, ei〉 and U(x) =

(x1, x2, x3, . . . ). By Parseval’s Identity, the series∑∞i=1 |xi|2 converges, so the sequence (x1, x2, x3, . . . ) belongs

to `2. Thus U is well-defined, linear (because the inner product is linear in the first argument), and isometric:

1.3. SUBSPACES 7

‖Ux‖2 =∑∞i=1 |xi|2 = ‖x‖2. Finally, if (y1, y2, y3, . . . ) ∈ `2 then

∑∞i=1 |yi|2 converges so, by Lemma 1.2.5,∑∞

n=1 ynen converges and U(∑∞n=1 ynen) = (y1, y1, y1, . . . ). Thus, U is surjective and the theorem is proved.

Exercise 1.2.3. Prove that every Hilbert space of finite dimension n is isomorphic to Cn.

Problem 2. Let H ne a separable Hilbert space and M a subspace of H. Prove that M is a separable

Hilbert space.

Problem 3. The Haar system ϕm,n, m ∈ N, 1 ≤ n ≤ 2m, is defined as:

ϕm,n(x) =

2m/2, ifn− 12m

≤ x ≤ n− 1/22m

,

−2m/2, ifn− 1/2

2m≤ x ≤ n

2m,

0, if x /∈[n− 12m

,n

2m

).

Prove that this system is an o.n.b. of L2[0, 1].

1.3. Subspaces

Example 1.3.1. Let H = L2[0, 1] and let G be a measurable subset of [0, 1]. Denote by L2(G) the set of

functions in L2 that vanish outside of G. Then L2(G) is a closed subspace of H. Further, if f ∈ L2(G) and

g ∈ L2(Gc), then 〈f, g〉 = 0.

Definition 1.3.1. If M is a closed subspace of the Hilbert space H, then the orthogonal complement of M,

denoted M⊥, is the set of vectors in H orthogonal to every vector in M.

Exercise 1.3.1. Prove that M⊥ is a closed subspace of H.

Theorem 1.3.1. Let M be a closed subspace of Hilbert space H, and let x ∈ H. Then there exist unique

vectors y in M and z in M⊥ so that x = y + z.

8 1. HILBERT SPACE

Proof. Let eii∈I and fjj∈J be orthonormal bases forM andM⊥, respectively. Their union is an o.n.b.

of H so x =∑i∈I〈x, ei〉ei +

∑j∈J〈x, fj〉fj and we define y =

∑i∈I〈x, ei〉ei, z =

∑j∈J〈x, fj〉fj . Then y ∈ M,

z ∈M⊥, and x = y + z.

Suppose now that x = y1 + z1 = y2 + z2, where y1, y2 ∈M and z1, z2 ∈M⊥. Then y1− y2 = z2− z1 belongs

to both M and M⊥, so 〈y1 − y2, y1 − y2〉 = 0 and it follows that y1 = y2, and consequently z1 = z2.

Definition 1.3.2. In the situation described in Theorem 1.3.1 we say that H is the orthogonal direct sum of

M and M⊥, and we write H =M⊕M⊥. When z = x+ y with x ∈M and y ∈M⊥ we often write z = x⊕ y.

Theorem 1.3.1 allows us to define a map P : H →M by Px = y. It is called the orthogonal projection of H

onto M, and it is denoted by PM. Here are some of its properties.

Theorem 1.3.2. Let M be a closed subspace of Hilbert space H and let P be the orthogonal projection on

M. Then:

(a) P is a linear transformation;

(b) ‖Px‖ ≤ ‖x‖, for all x ∈ H;

(c) P 2 = P ;

(d) KerP =M⊥ and RanP =M.

Proof. Let eii∈I and fjj∈J be orthonormal bases for M and M⊥, respectively, and let Q = I − P be

the orthonormal projection onM⊥. If x′, x′′ ∈ H and α′, α′′ ∈ C, then P (α′x′+α′′x′′) =∑i∈I〈α′x′+α′′x′′, ei〉 =

α′Px′ + α′′Px′′, so (a) holds.

(b) If x ∈ H, then x = Px+Qx and Px ⊥ Qx. Therefore, ‖x‖2 = ‖Px‖2 + ‖Qx‖2 ≥ ‖Px‖2.

(c) If y ∈M then Py = y. Now, for any x ∈ H, Px ∈M so P 2x = P (Px) = Px.

(d) If Px = 0 then x = Qx ∈ M⊥. If x ∈ M⊥ then Qx = x by (c), so Px = 0. The other assertion is

obvious.

Problem 4. Prove that PMx is the unique point in M that is nearest to x, meaning that ‖x − PMx‖ =

inf‖x− h‖ : h ∈M.

1.4. WEAK TOPOLOGY 9

Problem 5. In L2[0, 1] find the orthogonal complement to the subspace consisting of:

(a) all polynomials in x;

(b) all polynomials in x2;

(c) all polynomials in x with the free term equal to 0;

(d) all polynomials in x with the sum of coefficients equal to 0.

Problem 6. If M and N are subspaces of Hilbert space that are orthogonal to each other, then the sum

M + N = x + y : x ∈ M, y ∈ N is a subspace. Show that the theorem is not true if M and N are either:

closed but not orthogonal or orthogonal but not closed.

1.4. Weak topology

Read Royden, page 236–238.

Example 1.4.1. Consider the sequence of functions cosntn∈N in L1[0, 2π]. It is easy to see that this

sequence is not convergent. However, for any function f ∈ L∞,∫ 1

0f(t) cosnt dt→ 0 as n→∞. Since L∞ is the

dual space of L1, we say that cosnt→ 0 weakly, and we write w − limn cosnt = 0.

Example 1.4.2. Consider the sequence of functions cosntn∈N in L∞[0, 2π]. Notice that, while not a

convergent sequence, if f ∈ L1 then∫ 1

0f(t) cosnt dt → 0 as n → ∞. Since L∞ is the dual space of L1, we say

that cosnt→ 0 in the weak∗ topology.

In a Banach space X it is useful to consider three topologies: the norm topology, induced by the norm; weak

topology — the smallest topology in which all bounded linear functionals on X are continuous; weak∗ topology

(meaningful when X is the dual space of Y so that Y ⊂ X ∗) — the smallest topology in which some bounded

linear functionals on X are continuous (those that can be identified as elements of Y). In order to dicuss these

topologies (and understand their role), we need to find out what bounded linear functionals on Hilbert space H

look like.

10 1. HILBERT SPACE

Theorem 1.4.1 (Riesz Representation Theorem). If L is a bounded linear functional on H, then there is a

unique vector y ∈ H such that L(x) = 〈x, y〉 for every x ∈ H. Moreover, ‖L‖ = ‖y‖.

Proof. Assuming that such y exists, we can write it as y =∑i∈N yiei relative to a fixed o.n.b. eii∈N.

Then yi = 〈y, ei〉 = 〈ei, y〉 = L(ei). Therefore, we define y =∑i∈N L(ei) ei, and all it remains to prove is the

convergence of the series. Let sn =∑ni=1 L(ei) ei. Then L(sn) =

∑ni=1 L(ei)Lei = ‖sn‖2, so ‖sn‖2 ≤ ‖L‖‖sn‖

from which it follows that ‖sn‖ ≤ ‖L‖. Thus the series∑ni=1 L(ei) ei converges and the result follows from

Lemma 1.2.5.

We see that if L ∈ H∗, the dual space of H, then L = Ly. The mapping Φ : H → H∗ defined by Φ(y) = Ly is

a norm preserving surjection. It is conjugate linear: Φ(α1y1 +α2y2) = α1y1 +α2y2. Nevertheless, we identify H∗

with H. Consequently, H is reflexive (i.e., H∗∗ = H) so the weak∗ and weak topologies on H coincide. Therefore,

we will work with 2 topologies: weak and norm induced. The absence of a qualifier will always mean that it is

the latter.

Exercise 1.4.1. Prove that the weak topology is weaker than the norm toplogy, i.e., if G is a weakly open

set then G is an open set.

Example 1.4.3. If enn∈N is an orthonormal sequence in H then w − lim en = 0 but the sequence is not

convergent.

Exercise 1.4.2. Prove that the Hilbert space norm is continuous but not weakly continuous.

The following result shows why weak topology is important. [See Royden, p. 237]

Theorem 1.4.2 (Banach-Alaoglu). The unit ball x ∈ H : ‖x‖ ≤ 1 in Hilbert space H is weakly compact.

Remark 1.4.1. The unit ball B1 of H is NOT compact (assuming that H is infinite dimensional). Reason:

if enn∈N is an o.n.b. then the set e1, e2, e3, . . . is closed but not totally bounded, hence not compact.

Exercise 1.4.3. Prove that if a bounded set in H is weakly closed then it is weakly compact.

1.4. WEAK TOPOLOGY 11

In spite of the fact that the weak topology is weaker then the norm topology, some of the standard results

remain true.

Theorem 1.4.3. A weakly convergent sequence is bounded.

Proof. Suppose that xn is a weakly convergent sequence. Then, for any y ∈ H, the sequence 〈xn, y〉 is a

convergent sequence of complex numbers, which implies that it is bounded. In other words, for any y ∈ H there

exists C = C(y) > 0 such that |〈xn, y〉| ≤ C. This means that, for each n ∈ N, xn can be viewed as a bounded

linear functional on H. By the Uniform Bounded Principle (Royden, p. 232), these functionals are uniformly

bounded, i.e., there exists M > 0 such that, for all n ∈ N, ‖xn‖ ≤M .

Although weakly convergent sequence need not be convergent there are situation when it does.

Theorem 1.4.4. If xnn∈N is a weakly convergent sequence in a compact set K then it is convergent.

Proof. Since xnn∈N ⊂ K, it has an accumulation point x′ and a subsequence x′n converging to z. If xn

had another accumulation point x′′, then there would be another subsequence x′′n converging to w. It would follow

that w − limx′n = x′ and w − limx′′n = x′′. Since xn is weakly convergent this implies that x′ = x′′, so it has

only one accumulation point, namely the limit.

By definition, the weak topology W is the smallest one in which every bounded linear functional L on H is

continuous. This means that, for any such L and any open set G in the complex plane, L−1(G) ∈ W. Since open

disks form a base of the usual topology in C it suffices to require that L−1(G) ∈ W for each open disk G. Notice

that x ∈ L−1(G) iff L(x) ∈ G, so if G = z : |z − z0| < r and z0 = L(x0) then x ∈ L−1(G) iff |L(x − x0)| < r.

Now Riesz Representation Theorem implies that L−1(G) = x ∈ H : |〈x − x0, y〉 < r for some y ∈ H. We

conclude that a subbase of W consists of the sets W = W (x0; y, r) = x ∈ H : |〈x− x0, y〉 < r.

Exercise 1.4.4. Prove that a bounded linear functional L is continuous in a topology T iff L−1(G) ∈ T for

every open disk G.

12 1. HILBERT SPACE

Problem 7. Prove that a subspace of Hilbert space is closed iff it is weakly closed.

Problem 8. Prove that Hilbert space is weakly complete.

Problem 9. Let xnn∈N be a sequence in Hilbert space with the property that ‖xn‖ = 1, for all n, and

〈xm, xn〉 = c, if m 6= n. Prove that xnn∈N is weakly convergent.

Problem 10. Find the weak closure of the unit sphere in Hilbert space.

CHAPTER 2

Operators on Hilbert Space

“Nobody, except topologists, is interested in problems about Hilbert space; the people who work in Hilbert

space are interested in problems about operators”.

Paul Halmos

2.1. Definition and Examples

Read Section 10.2 in Royden’s book. Operator always means linear and bounded. The algebra of all bounded

linear operators on H is denoted by L(H).

Example 2.1.1. Let H = Cn and A = [aij ] an n × n matrix. The operator of multiplication by A is linear

and bounded. Indeed, for x = (x1, x2, . . . , xn) and M = sup1≤i≤n

(∑nj=1 |aij |2

)1/2

,

‖Ax‖ = sup1≤i≤n

|n∑j=1

aijxj | ≤ sup1≤i≤n

n∑j=1

|aij |21/2 n∑

j=1

|xj |21/2

= M‖x‖

so ‖A‖ ≤M .

Example 2.1.2. Let H = `2 and A = [aij ]∞i,j=1, where aij = ci if i = j and aij = 0 if i 6= j. We call

such matrix diagonal and denote it by diag(c1, c2, . . . ), or diag(cn). The operator A (or, more precisely, the

operator of multiplication by A) is bounded iff c = (c1, c2, . . . ) ∈ `∞ (i.e., when c is a bounded sequence).

Indeed, let x = (x1, x2, . . . ) ∈ `2, so Ax = (c1x1, c2x2, . . . ) and ‖Ax‖2 =∑∞i=1 |cixi|2. If |ci| ≤ M , i ∈ N, then

‖Ax‖2 ≤M2∑∞i=1 |xi|2 = ‖x‖2 so A is bounded. On the other hand, if c /∈ `∞, then for each n there exists in so

that |cin | ≥ n. Then ‖Aein‖ = ‖cinein‖ ≥ n→∞ and A is unbounded.

Remark 2.1.1. It is extremely hard to decide, in general, whether an operator A is bounded just by studying

its matrix [〈Aej , ei〉]∞i,j=1.

13

14 2. OPERATORS ON HILBERT SPACE

Example 2.1.3. Let H = `2 and let S be the unilateral shift, defined by S(x1, x2, . . . ) = (0, x1, x2, . . . ).

Notice that ‖S(x1, x2, . . . )‖2 = 02 + |x1|2 + |x2|2 + · · · = ‖x‖2 so ‖S‖ = 1. In fact, S is an isometry, hence

injective, but it is not surjective!

Example 2.1.4 (Multiplication on L2). Let h be a measurable function and define Mhf , for f ∈ L2, by

(Mhf)(t) = h(t)f(t). If h ∈ L∞ (essentially bounded functions — see Royden, p. 118), then

‖Mhf‖2 =∫|hf |2 ≤ ‖h‖2∞

∫|f |2 = ‖h‖2∞‖f‖2

so Mh is a bounded operator on L2 and ‖Mh‖ ≤ ‖h‖∞. On the other hand, for ε > 0, there exists a set C ⊂ [0, 1]

of positive measure so that |h(t)| ≥ ‖h‖∞ − ε for t ∈ C. If f = χC then

‖Mhf‖2 =∫|hf |2 =

∫C

|h|2 ≥ (‖h‖∞ − ε)2µ(C) = (‖h‖∞ − ε)2‖f‖2,

and it follows that ‖Mh‖ ≥ ‖h‖∞ − ε. We conclude that ‖Mh‖ = ‖h‖∞ and Mh is bounded iff h ∈ L∞.

Example 2.1.5 (Integral operators on L2). Let K : [0, 1] × [0, 1] → C be measurable and square integrable

with respect to planar Lebesgue measure. We define the operator TK by (Tkf)(x) =∫ 1

0K(x, y)f(y) dy. Now

‖TKf‖2 =∫ 1

0

|Tkf(x)|2 dx =∫ 1

0

∣∣∣∣∫ 1

0

K(x, y)f(y) dy∣∣∣∣2 dx ≤ ∫ 1

0

(∫ 1

0

|K(x, y)f(y)| dy)2

dx

≤∫ 1

0

∫ 1

0

|K(x, y)|2 dy∫ 1

0

|f(y)|2 dydx = ‖f‖2

∫ 1

0

∫ 1

0

|K(x, y)|2 dydx.

Therefore, TK is bounded and ‖TK‖ ≤∫ 1

0

∫ 1

0|K(x, y)|2 dydx

1/2

.

Example 2.1.6 (Weighted shifts). Let H = `2 and let cnn∈N be a bounded sequence of complex numbers.

A weighted shift W on `2 is defined by W (x1, x2, . . . ) = (0, c1x1, c2x2, . . . ). It can be written as W = S diag(cn)

so it is a bounded operator and ‖W‖ = ‖diag(cn)‖.

In some situations it is useful to have an alternate formula for the operator norm. In what follows we will

use notation B1 for the closed unit ball of H, i.e. B1 = x ∈ H : ‖x‖ ≤ 1.

Proposition 2.1.1. Let T be linear operator on Hilbert space. Then ‖T‖ = sup|〈Tx, y〉| : x, y ∈ B1.

2.2. ADJOINT 15

Proof. Let α denote the supremum above, and let us assume that T 6= 0 (otherwise there is nothing to

prove). Clearly, for x, y ∈ B1, |〈Tx, y〉| ≤ ‖T‖, so α ≤ ‖T‖. In the other direction,

α ≥ sup|〈Tx, y〉| : x, y ∈ B1, Tx 6= 0, y =Tx

‖Tx‖

= sup|〈Tx, Tx

‖Tx‖〉| : x ∈ B1, Tx 6= 0

= sup‖Tx‖ : x ∈ B1, Tx 6= 0

= ‖T‖,

and the proof is complete.

2.2. Adjoint

In Linear Algebra we learn that the column space of matrix A = [aij ]ni,j=1 and the null space of its transpose

AT are orthogonal complements in Rn. In Cn, AT needs to be replaced by A∗ = [aji]ni,j=1. In this situation,

(2.1) 〈Ax, y〉 = 〈x,A∗y〉.

Exercise 2.2.1. Prove that, if A is an n× n matrix and x, y ∈ Cn, then 〈Ax, y〉 = 〈x,A∗y〉.

Example 2.2.1. Let h ∈ L∞ and let Mh be the operator of multiplication on L2. Then (Mh)∗ = Mh.

The following result will show that a relation (2.1) is available for any operator.

Proposition 2.2.1. If T is an operator on H then there exists a unique operator S on H such that 〈Tx, y〉 =

〈x, Sy〉, for all x, y ∈ H.

Proof. Let y ∈ H be fixed. Then ϕ(x) = 〈Tx, y〉 is a bounded linear functional on H. By Riesz Repre-

sentation Theorem there exists a unique z ∈ H such that ϕ(x) = 〈x, z〉, for all x ∈ H. Define Sy = z. Then


〈Tx, y〉 = 〈x, Sy〉. To show that S is linear, let Sy1 = z1, Sy2 = z2, and let x ∈ H. Then

〈x, S(α1y1 + α2y2)〉 = 〈Tx, α1y1 + α2y2〉 = α1〈Tx, y1〉+ α2〈Tx, y2〉

= α1〈x, Sy1〉+ α2〈x, Sy2〉 = 〈x, α1Sy1 + α2Sy2〉.

By the uniqueness part of Riesz Representation Theorem S is linear. That S is unique can be deduced by

contradiction: if 〈x, Sy〉 = 〈x, S′y〉 for all x, y ∈ H then 〈x, Sy−S′y〉 = 0 for all x which implies that Sy−S′y = 0

for all y, hence S = S′. Finally, S is bounded: ‖Sy‖2 = 〈Sy, Sy〉 = 〈TSy, y〉 ≤ ‖TSy‖‖y‖ ≤ ‖T‖‖Sy‖‖y‖ so

‖Sy‖ ≤ ‖T‖‖y‖ and ‖S‖ ≤ ‖T‖.

Definition 2.2.1. If T ∈ L(H) then the adjoint of T , denoted T ∗, is the unique operator on H satisfying

〈Tx, y〉 = 〈x, T ∗y〉, for all x, y ∈ H.

Here are some of the basic properties of the involution T 7→ T ∗.

Proposition 2.2.2.

(a) I∗ = I

(b) T ∗∗ = (T ∗)∗ = T ;

(c) ‖T ∗‖ = ‖T‖;

(d) (α1T1 + α2T2)∗ = α1T∗1 + α2T

∗2 ;

(e) (T1T2)∗ = T ∗2 T∗1 ;

(f) if T is invertible then so is T ∗ and (T ∗)−1 = (T−1)∗;

(g) ‖T 2‖ = ‖T ∗T‖.

Proof. The assertion (a) is obvious and (b) follows from 〈x, T ∗∗y〉 = 〈T ∗x, y〉 = 〈y, T ∗x〉 = 〈Ty, x〉 = 〈x, Ty〉.

It was shown in the proof of Proposition 2.2.1 that ‖T ∗‖ ≤ ‖T‖ so ‖T ∗∗‖ ≤ ‖T ∗‖ ≤ ‖T‖ and (c) follows from

(b). We leave (d) as an exercise and notice that 〈x, (T1T2)∗y〉 = 〈T1T2x, y〉 = 〈T2x, (T1)∗y〉 = 〈x, (T2)∗(T1)∗y〉

establishes (e). As a consequence of (a) and (e), T ∗(T−1)∗ = (T−1T )∗ = I and (T−1)∗T ∗ = (TT−1)∗ = I which

2.3. OPERATOR TOPOLOGIES 17

is (f). Finally, ‖T ∗T‖ ≤ ‖T ∗‖‖T‖ = ‖T‖2 and to prove the opposite inequality let ε > 0 and let x be a unit vector

such that ‖Tx‖ ≥ ‖T‖ − ε. Then ‖T ∗T‖ ≥ ‖T ∗Tx‖ ≥ 〈T ∗Tx, x〉 = ‖Tx‖2 > (‖T‖ − ε)2, and (g) is proved.

Example 2.2.2. Let H = `2 and the let S be the unilateral shift (see Example 2.1.3). Then S∗(x1, x2, . . . ) =

(x2, x3, . . . ). The operator S∗ is called the backward shift.

Example 2.2.3. Let TK be the integral operator on L2 (see Example 2.1.5). Then (TK)∗ = TK∗ , where

K∗(x, y) = K(y, x).

We now give the Hilbert space formulation of the relation with which we have opened this section.

Theorem 2.2.3. If T is an operator on Hilbert space H then KerT = (RanT ∗)⊥.

Proof. Let x ∈ KerT and let y ∈ RanT ∗. Then there exists z ∈ H such that y = T ∗z. Therefore

〈x, y〉 = 〈x, T ∗z〉 = 〈Tx, z〉 = 0 so x ∈ (RanT ∗)⊥. In the other direction, if x ∈ (RanT ∗)⊥ and z ∈ H, then

〈Tx, z〉 = 〈x, T ∗z〉 = 0. Taking z = Tx we see that Tx = 0, and the proof is complete.

We notice that, for T ∈ L(H) and x, y ∈ H, the expression 〈Tx, y〉 is a form that is linear in the first and

conjugate linear in the second argument. It turns out that this is sufficient for a polarization identity.

Proposition 2.2.4 (Second Polarization Identity).

4〈Tx, y〉 = 〈T (x+ y), x+ y〉 − 〈T (x− y), x− y〉+ i〈T (x+ iy), x+ iy〉 − i〈T (x− iy), x− iy〉.

Exercise 2.2.2. Prove Second Polarization Identity.

2.3. Operator topologies

In this section we take a look at the algebra L(H). It has three useful topologies which lead to 3 different

types of convergence.

Definition 2.3.1. A sequence of operators Tn ∈ L(H) converges uniformly (or in norm) to an operator T if

‖Tn−T‖ → 0, n→∞. A sequence of operators Tn ∈ L(H) converges strongly to an operator T if ‖Tnx−Tx‖ → 0,


n→∞, for all x ∈ H. A sequence of operators Tn ∈ L(H) converges weakly to an operator T if 〈Tnx−Tx, y〉 → 0,

n→∞, for any x, y ∈ H.

It follows from the definition that the weak topology is the weakest of the three, while then norm topology

(a.k.a. the uniform topology) is the strongest. Are they different?

Proposition 2.3.1. The operator norm is continuous with respect to the uniform topology but discontinuous

with respect to the strong and weak topologies.

Proof. The first assertion is a consequence of the inequality |‖A‖ − ‖B‖| ≤ ‖A − B‖. To prove the other

two, let enn∈N be an o.n.b. of H, Hn = ∨∞k=nek, Pn = PHn. Then Pn → 0 strongly (hence weakly) since

‖Pnx‖2 =∑∞k=n+1 |xk|2 → 0. However, ‖Pn‖ = 1 which does not converge to 0.

Example 2.3.1. We say that an operator T is a rank one operator if there exist u, v ∈ H so that Tx = 〈x, v〉u.

We use the notation T = u⊗v. Let Tn = en⊗e1. Then 〈Tnx, y〉 = x1yn → 0 while Tnx = x1en is not a convergent

sequence. Thus, the weak and strong topologies are different.

Example 2.3.2. The involution T 7→ T ∗ is continuous in uniform topology. (‖T ∗n − T ∗‖ = ‖Tn − T‖). Also,

it is continuous in the weak topology, because

|〈(T ∗n − T ∗)x, y〉| = |〈x, (Tn − T )y〉| = |〈(Tn − T )y, x〉| .

However, it is not continuous in the strong topologies. Counterexample: let S be the unilateral shift, and

Tn = (S∗)n. Then Tn → 0 strongly but T ∗n is not a strongly convergent sequence. Indeed, for any x =

(x1, x2, . . . ) ∈ H, ‖Tnx‖2 = ‖(xn+1, xn+2, . . . )‖2 =∑∞k=n |xk|2 → 0, as n → ∞. On the other hand, for x = e1,

T ∗nx = Sne1 = en, which is not a convergent sequence.

An operator T ∈ L(H) is a continuous mapping when H is given the strong topology. We will write, following

Halmos, (s→s). One may ask about the other types of continuity.

Theorem 2.3.2. The three types of continuity (s→s), (w→w), and (s→w) are all equivalent.

2.3. OPERATOR TOPOLOGIES 19

Proof. Suppose that T is continuous, and let W be a weakly open neighborhood of Tx0 in H. We will show

that T−1(W ) is weakly open. It suffices to prove this assertion in the case when W belongs to the subbase of the

weak topology. To that end, let W = W (Tx0, y, r) = x ∈ H : |〈x − Tx0, y〉| < r. Then z ∈ T−1(W ) ⇔ Tz ∈

W ⇔ |〈Tz − Tx0, y〉| < ε ⇔ |〈z − x0, T∗y〉| < ε. We see that z ∈ T−1(W ) iff z ∈ V (x0, T

∗y, ε) so T−1(W ) = V

which is a weakly open set.

The implication (w→w)⇒(s→w) is trivial, so we concentrate on the implication (s→w)⇒(s→s). To that end,

suppose that T is not continuous. Then it is unbounded, so there exists a sequence xnn∈N of unit vectors such

that ‖Txn‖ ≥ n2, n ∈ N. Clearly, xn/n→ 0 and the assumption (s→w) implies that Txn/n weakly converges to

0. By Theorem 1.4.3 the sequence Txn/n is bounded which contradicts the fact that ‖Txn/n‖ ≥ n.

The fact that every operator in L(H) is weakly continuous has an interesting consequence.

Corollary 2.3.3. If T is a linear operator on H then T (B1) is closed.

Proof. Banach-Alaoglu Theorem established that B1 is weakly compact so, by Theorem 2.3.2, T (B1) is

weakly compact, hence weakly closed, hence norm closed.

Exercise 2.3.1. Prove that if F is a closed and bounded set in H then T (F ) is closed.

At the end of this section we consider a situation that occurs quite frequently.

Theorem 2.3.4. Let M be a linear manifold that is dense in Hilbert space H. Every bounded linear trans-

formation T :M→H can be uniquely extended to a bounded linear transformation T : H → H. In addition, the

operator norm of T equals ‖T‖.

Proof. Let x ∈ H. Then there exists a sequence xnn∈N ⊂ M converging to x. Since xnn∈N is also a

Cauchy sequence, for every ε > 0 there exists N ∈ N such that, m,n ≥ N ⇒ ‖xm − xn‖ < ε/‖T‖. It follows

that, for m,n ≥ N , ‖Txm − Txn‖ < ε, so Txnn∈N is a Cauchy sequence, hence convergent, and there exists

y = limn→∞ Txn. We will define T x = y, i.e., T (limxn) = limTxn.


First we need to establish that the definition is independent of the sequence xnn∈N. If x′nn∈N is another

sequence converging to x, we form the sequence (x1, x′1, x2, x

′2, . . . ) which also converges to x. By the previous,

the sequence (Tx1, Tx′1, Tx2, Tx

′2, . . . ) must converge, and therefore, both of the subsequences Txnn∈N and

Tx′nn∈N must have the same limit.

Notice that, if xn → x, the continuity of the norm implies that ‖T x‖ = ‖ limTxn‖ = lim ‖Txn‖ ≤

lim ‖T‖‖xn‖ = ‖T‖‖x‖ so ‖T‖ ≤ ‖T‖. Since the other inequality is obvious we see that ‖T‖ = ‖T‖. In particular,

T is a bounded operator. Also, T (αx+ βy) = T (α limxn + β lim yn) = T (lim(αxn + βyn)) = limT (αxn + βyn) =

lim(αTxn + βTyn) = α limTxn + β limTyn = αTx+ βT y, so T is linear.

Finally, suppose that T1 and T2 are two continuous extensions of T , and let x ∈ H. If xn → x, the continuity

implies that both T1xn → T1x and T2xn → T2x. If xn ∈ M then T1xn = T2xn, so T1x = T2x. Therefore, the

extension is unique, and the proof is complete.

Need an example

2.4. Invariant and Reducing Subspaces

WhenM is a closed subspace of H, we can always write H =M⊕M⊥. Relative to this decomposition, any

operator T acting on H can be written as a 2× 2 matrix with operator entries

(2.2) T =

X Y

Z W

.

It is sometimes convenient to consider only the initial space or the target space as a direct sum. In such a situation

we will use a 1 × 2 or 2 × 1 matrix. Thus[X Y

]will describe an operator T :M⊕M⊥ → H; if f ∈ M and

g ∈M⊥ then [X Y ][fg

]= Xf + Y g.

A subspace M is invariant for T if, for any x ∈ M, Tx ∈ M. It is reducing for T if both M and M⊥ are

invariant for T .

2.4. INVARIANT AND REDUCING SUBSPACES 21

Example 2.4.1. The subspace (0) consisting of zero vector only is an invariant subspace for any operator T .

Also, H is an invariant subspace for any operator T . Because they are invariant for every operator they are called

trivial. A big open problem in Operator theory is whether every operator has a non-trivial invariant subspace.

Example 2.4.2. If M is a closed subspace of H and T1 is an operator on M with values in M, then the

operator T = T1 ⊕ 0, defined by Tx = T1x if x ∈ M and Tx = 0 if x ∈ M⊥ is an operator in L(H). However, if

M is not invariant for T1, the same definition (Tx = T1x for x ∈M, Tx = 0 for x ∈M⊥) describes the operator[T1 0

].

Proposition 2.4.1. If T is an operator on Hilbert space H, and P = PM is the projection onto the closed

subspace M, then the following are equivalent:

(a) M is invariant for T ;

(b) PTP = TP ;

(c) Z = 0 in (2.2).

Proof. It is not hard to see that the matrix for P is [ I 00 0 ] so PTP − TP =

[0 0−Z 0

]. This establishes (b) ⇔

(c). Since[fg

]∈M iff g = 0, we see that T

[f0

]=[XfZf

]∈M for all x ∈ H iff Z = 0 so (a) ⇔ (c).

Example 2.4.3. Let S be the unilateral shift, n ∈ N, and M = ∨k≥nek. Then SM = ∨k≥n+1ek ⊂M.

Proposition 2.4.2. If T is an operator on Hilbert space H, and P = PM then the following are equivalent:

(a) M is reducing for T ;

(b) PT = TP ;

(c) Y, Z = 0 in (2.2);

(d) M is invariant for T and T ∗.

Proof. Since PT − TP =[

0 Y−Z 0

]we see that (b) ⇔ (c). Further, the matrix for T ∗ is

[X∗ Z∗

Y ∗ W∗]

so, by

Proposition 2.4.1, M is invariant for T and T ∗ iff Z = Y ∗ = 0 and (c) ⇔ (d). In order to prove that (a) ⇔ (d)


it suffices to show that M is invariant for T ∗ iff M⊥ is invariant for T . By Proposition 2.4.1, M is invariant for

T ∗ iff Y ∗ = 0 (iff Y = 0). On the other hand T[

0g

]=[Y gWg

]∈M⊥ iff Y g = 0 for all g.

Exercise 2.4.1. Prove that the matrix for T ∗ is[X∗ Z∗

Y ∗ W∗].

Example 2.4.4. Let T = Mh, let E ⊂ [0, 1], m(E) > 0, and let M = L2(E). If f ∈ M then Tf = hf ∈ M.

Also, T ∗ = Mh and T ∗f = hf ∈M, so M is reducing for T .

Example 2.4.5. Let S be the unilateral shift, n ∈ N, and M = ∨k≥nek. Then M is invariant for S but not

reducing, since en ∈M but S∗en = en−1 /∈M.

2.5. Finite rank operators

The closest relatives of finite matrices are the finite rank operators.

Definition 2.5.1. An operator T is a finite rank operator if its range is finite dimensional. We denote the

set of finite rank operators by F.

Example 2.5.1. If T is a rank one operator u ⊗ v (see Example 2.3.1) then the range of u ⊗ v is the one

dimensional subspace spanned by u, so u⊗ v ∈ F.

The rank one operators turn out to be the building blocks out of which finite rank operators are made.

Proposition 2.5.1. If T is a linear operator on H then T belongs to F iff there exist vectors u1,u2,. . . ,un,

and v1,v2,. . . ,vn such that Tx =∑ni=1〈x, vi〉ui.

Proof. Suppose that RanT is of finite dimension n, and let e1, e2, . . . , en be an o.n.b. of RanT . Then

Tx =∑ni=1〈Tx, ei〉ei =

∑ni=1〈x, T ∗ei〉ei. We leave the converse as an exercise.

Exercise 2.5.1. Prove that if there exist vectors u1, u2, . . . , un, v1, v2, . . . , vn such that Tx =∑ni=1〈x, vi〉ui,

for all x ∈ H, then RanT is of dimension at most n.

2.6. COMPACT OPERATORS 23

Exercise 2.5.2. Prove that if T =∑ui ⊗ vi then T ∗ =

∑vi ⊗ ui.

The next theorem summarizes some very important properties of the class F.

Theorem 2.5.2. The set F is a minimal ∗-ideal in L(H).

Here the star means that F is closed under the operation of taking adjoints.

Proof. It is obvious that F is a subspace of L(H). Furthermore, if T ∈ F and A ∈ L(H), then RanTA ⊂

RanT so TA ∈ F. Also, if T is of finite rank, then according to Proposition 2.5.1, T =∑ni=1 ui ⊗ vi so

T ∗ =∑ni=1 vi ⊗ ui. It follows that T ∗ ∈ F and the same is true of T ∗A∗, for any A ∈ L(H). Consequently, AT

is of finite rank, and F is a ∗-ideal. To see that it is minimal, it suffices to show that, if J is a non-zero ideal,

then J contains all rank one operators. Let T ∈ J , T 6= 0. Then there exists vectors x, y, such that ‖y‖ = 1 and

y = Tx. Let u ⊗ v be a rank one operator. Since J is an ideal, it contains the product (u ⊗ y)T (x ⊗ v) which

equals u⊗ v.

A finite rank operator is a generalization of a finite matrix. What happens when we take the closure of F in

some topology?

Exercise 2.5.3. Prove that the strong closure of F is L(H). [Hint: Prove that Pn → I strongly.] Conclude

that the weak closure of F is also L(H).

2.6. Compact Operators

Exercise 2.5.3 established that the strong closure of F is L(H). Therefore, we consider the norm topology.

Definition 2.6.1. An operator T in L(H) is compact if it is the limit of a sequence of finite rank operators.

We denote the set of compact operators by K.

Example 2.6.1. Let T = diag(cn) as in Example 2.1.2, with limn→∞ cn = 0. Then T is compact. Reason:

take Tn = diag(c1, c2, . . . , cn, 0, 0, . . . ). Then Tn ∈ F and ‖T − Tn‖ = sup|ck| : k ≥ n+ 1 → 0. It follows that T

is compact.


Example 2.6.2. Let T = TK as in Example 2.1.5. If K ∈ L2([0, 1] × [0, 1]) then TK is compact. We will

point out at several different sequences in F that all converge to TK

We start with a function theoretic approach: simple functions are dense in L2 (Royden, p. 128), and a

similar proof establishes that simple functions are dense in L2([0, 1] × [0, 1]). Since a simple function is a linear

combination of the characteristic functions of rectangles χ[a,b]×[c,d](x, y) = χ[a,b](x)χ[c,d](y) it follows that K(x, y)

is the L2 limit of functions of the form Kn(x, y) =∑ni=1 fi(x)gi(y), so TK is the norm limit of TKn

, which are all

finite rank operators.

Exercise 2.6.1. Verify that TKn∈ F, if Kn(x, y) =

∑ni=1 fi(x)gi(y).

Our second approach is exploiting the fact that L2 is Hilbert space. If ejj∈N is an o.n.b. of L2 we can, for

a fixed y, write K(x, y) =∑∞j=1 kj(y)ej(x). Now define KN (x, y) =

∑Nj=1 kj(y)ej(x) and notice that TKN

→ TK

as N →∞.

Exercise 2.6.2. Verify that TKN∈ F and that limN→∞ TKN

= TK , if KN (x, y) is as above.

Our last method is based on the matrix for TK . Let kij = 〈TKej , ei〉, with enn∈N an o.n.b. of L2([0, 1]).

First we notice that, for any f ∈ L2,∑k |〈f, ek〉|2 = ‖

∑k〈f, ek〉ek‖2 = ‖f‖2. Therefore,

∞∑j=1

|〈TKej , ei〉|2 =∞∑j=1

|〈ej , T ∗Kei〉|2 =∞∑j=1

|〈T ∗Kei, ej , 〉|2 = ‖T ∗Kei‖2

=

1∫0

∣∣∣∣∣∣1∫

0

K∗(y, x)ei(x) dx

∣∣∣∣∣∣2

dy =

1∫0

∣∣∣∣∣∣1∫

0

K(x, y)ei(x) dx

∣∣∣∣∣∣2

dy.

It follows that, for any n ∈ N,

n∑i=1

∞∑j=1

|kij |2 =n∑i=1

1∫0

∣∣∣∣∣∣1∫

0

K(x, y)ei(x) dx

∣∣∣∣∣∣2

dy =

1∫0

n∑i=1

∣∣∣∣∣∣1∫

0

K(x, y)ei(x) dx

∣∣∣∣∣∣2

dy

≤1∫

0

∞∑i=1

∣∣∣∣∣∣1∫

0

K(x, y)ei(x) dx

∣∣∣∣∣∣2

dy =

1∫0

1∫0

|K(x, y)|2 dxdy


so the series∞∑i=1

∞∑j=1

|kij |2 converges. Operators whose matrices satisfy this condition are called Hilbert–Schmidt

operators. The Hilbert-Scmidt norm is defined as ‖TK‖2 = ∞∑i=1

∞∑j=1

|kij |21/2, and it satisfies the inequality

‖A‖ ≤ ‖A‖2. Hilbert-Scmidt operators are compact because we can define Tn to be the matrix consisting of the

first n rows of the matrix of TK and having the remaining entries 0. Then each Tn ∈ F and ‖Tn − TK‖ → 0.

Indeed, RanTn ⊂ ∨e1, e2, . . . , en, and ‖TK − Tn‖2 ≤ ‖TK − Tn‖22 =∞∑

i=n+1

∞∑j=1

|kij |2 → 0, n→∞.

Exercise 2.6.3. Prove that the Hilbert-Scmidt norm is indeed a norm and, for any T ∈ L(H), ‖T‖ ≤ ‖T‖2.

Next we consider some of the properties of compact operators. The first one follows directly from the

definition.

Theorem 2.6.1. The set K is the smallest closed ∗-ideal in L(H).

The following result reveals the motivation for calling these operators compact.

Theorem 2.6.2. An operator T in L(H) is compact iff it maps the closed unit ball of H into a compact set.

Proof. Suppose that K is compact and let ynn∈N be a sequence in K(B1). We will show that there exists

a subsequence of yn that converges to an element of K(B1). Notice that, for every n ∈ N, yn = Kxn, and xn

belongs to the weakly compact set B1. Thus, there exists a subsequence xnk converging weakly to x ∈ B1.

Thus, it suffices to show that Kxnkconverges to Kx. Let Kn be a sequence in F thaty converges to K. For

any m ∈ N, Km(B1) is a bounded and closed set (by Corollary 2.3.3) that is contained in a finite dimensional

subspace of H, so it is compact. By Theorem 1.4.4, Kmxnkk∈N converges to Kmx. Now, let ε > 0. Then

there exists N ∈ N such that ‖K −KN‖ < ε/3. Further, with N fixed, there exists k0 ∈ N so that, for k ≥ k0,

‖KNxnk−KNx‖ < ε/3. Therefore, for k ≥ k0,

‖Kxnk−Kx‖ ≤ ‖(K −KN )xnk

‖+ ‖KN (xnk− x)‖+ ‖(KN −K)x‖ < ε

3+ε

3+ε

3= ε.

Thus, ynk= Kxnk

is a convergent subsequence converging to Kx ∈ K(B1) so K(B1) is a compact set.


Suppose now that K(B1) is compact and let n ∈ N. Notice that ∪y∈K(B1)B(y, 1/n) is an open covering of

the compact set K(B1), so there exist vectors x(n)1 , x

(n)2 , . . . , x

(n)k ∈ H so that ∪ki=1B(Kx(n)

i , 1/n) is a covering of

K(B1). LetHn be the span of Kx(n)1 ,Kx

(n)2 , . . . ,Kx

(n)k and Pn the orthogonal projection onHn. Finally, let Kn =

PnK. Clearly, Kn ∈ F. Let ε > 0, and choose N > 1/ε. If n ≥ N , and ‖x‖ ≤ 1, then ‖Kx−Knx‖ = ‖Kx−PnKx‖.

Since PnKx is the point in Hn closest to Kx, it follows that ‖Kx−Knx‖ ≤ inf1≤i≤n ‖Kx−Kx(n)i ‖ < 1/n < ε.

Thus Kn → K and the proof is complete.

Remark 2.6.1. In many texts the characterization of compact operators, established in Theorem 2.6.2, is

taken to be the definition of a compact operator.

Exercise 2.6.4. Prove that if F is a closed and bounded set in H and T is a compact operator in L(H) then

T (F ) is a compact set.

There is another characterization of compact operators:

Proposition 2.6.3. If T is a linear operator on H then T is compact iff it maps every weakly convergent

sequence into a convergent sequence. In this situation, if w − limxn = x then limTxn = Tx.

Proof. Suppose first that T is compact and let w − limxn = x. By Proposition 1.4.3, there exists M > 0

such that, for all n ∈ N, ‖xn‖ ≤ M . Therefore, Txn/M ∈ T (B1), which is compact by Theorem 2.6.2. Now

Theorem 1.4.4 implies that limxn = x.

In order to establish the converse, we will demonstrate that T (B1) is compact by showing that every sequence

in T (B1) has a convergent subsequence. Let ynn∈N ⊂ T (B1). Then yn = Txn, for xn ∈ B1, so the Banach–

Alaoglu Theorem implies that xn has a weakly convergent subsequence xnk and, by assumption, Txnk

is

a (strongly) convergent subsequence of Txn.

Example 2.6.3. We have seen in Example 2.6.1 that if T − diag(cn) and cn → 0, then T is compact. The

converse is also true: if en is the o.n.b. which makes T diagonal, then Ten → 0 (because w− lim en = 0 and T

is compact) so ‖cnen‖ → 0.

2.7. NORMAL OPERATORS 27

It is useful to know that compactness is inherited by the parts of an operator.

Theorem 2.6.4. Suppose that T is a compact operator on Hilbert space H =M⊕M⊥ and that, relative to

this decomposition, T = [X YZ W ]. Then each of the operators X,Y, Z,W is compact.

Proof. Let Tn be a sequence of finite rank operators that converges to T . Write, for each n ∈ N,

Tn =[Xn Yn

Zn Wn

]. Then all the operators Xn, Yn, Zn,Wn ∈ F and they converge to X,Y, Z,W , respectively.

Exercise 2.6.5. Prove that Xn, Yn, Zn,Wn ∈ F and that they converge to X,Y, Z,W , respectively. [Consider

the projections P1 = PM and P2 = PM⊥ and notice that, for example P1TP2 = [ 0 Y0 0 ], so ‖Yn − Y ‖ ≤ ‖Tn − T‖

and RanYn ⊂ RanP1TnP2 the later being finite dimensional.]

2.7. Normal operators

Definition 2.7.1. If T is an operator on Hilbert space H then:

(a) T is normal if TT ∗ = T ∗T ;

(b) T is self-adjoint (or Hermitian) if T = T ∗;

(c) T is positive if 〈Tx, x〉 ≥ 0 for all x ∈ H;

(d) T is unitary if TT ∗ = T ∗T = I.

Example 2.7.1. Let T = diag(cn). Then T ∗ = diag(cn) so T is normal. Also, T = T ∗ iff cn ∈ R, n ∈ N, and

T is positive iff cn ≥ 0, n ∈ N. Finally, T ∗T = diag(|cn|2) so T is unitary iff |cn| = 1, n ∈ N.

Exercise 2.7.1. Let T = Mh on L2. Prove that T is normal and that it is: self-adjoint iff h(x) ∈ R, a.e.;

positive iff h(x) ≥ 0 a.e.; unitary iff |h(x)| = 1 a.e..

The relationship between T and T ∗ that defines each of these classes allows us to establish some of their

significant properties.

Proposition 2.7.1. An operator T on Hilbert space H is self-adjoint iff 〈Tx, x〉 is real for any x ∈ H.


Proof. If T = T ∗ then 〈Tx, x〉 = 〈x, T ∗x〉 = 〈x, Tx〉 = 〈Tx, x〉 so 〈Tx, x〉 ∈ R. On the other hand, if 〈Tx, x〉

is real for any x ∈ H then Second Polarization Identity implies that 〈Tx, y〉 = 〈Ty, x〉 so T = T ∗.

Exercise 2.7.2. Prove that 〈Tx, x〉 ∈ R implies that 〈Tx, y〉 = 〈Ty, x〉.

Corollary 2.7.2. If P is a positive operator on Hilbert space H then P is self-adjoint.

Example 2.7.2. If P is the orthogonal projection on a subspace M of Hilbert space H, then P is a positive

operator. Indeed, if z ∈ H write z = x+ y relative to H =M⊕M⊥. By Theorem 1.3.2, Pz = x and Py = 0, so

〈Pz, z〉 = 〈x, x+ y〉 = ‖x‖2 ≥ 0.

Combining Theorem 1.3.2 and Example 2.7.2 we see that every projection is a positive idempotent. In fact,

the converse is also true.

Theorem 2.7.3. If T is an idempotent self-adjoint operator then T is a projection onM = x ∈ H : Tx = x.

Proof. Let z ∈ H and write it as z = Tz + (z − Tz). Now T (Tz) = Tz so Tz ∈ M. Also, z − Tz ∈ M⊥.

Indeed, if x ∈M, then 〈x, z − Tz〉 = 〈x, z〉 − 〈x, Tz〉 = 〈x, z〉 − 〈Tx, z〉 = 0.

By Proposition 2.1.1, the norm of every operator T in L(H) can be computed by considering the supremum

of the values of its bilinear form 〈Tx, y〉. The next result shows that, when T is self adjoint, it suffices to consider

only some pairs of x, y ∈ B1.

Proposition 2.7.4. If T is a self-adjoint operator on Hilbert space H then ‖T‖ = sup|〈Tx, x〉| : ‖x‖ = 1.

Proof. Clearly, |〈Tx, x〉| ≤ ‖T‖‖x‖2, so if we denote by α the supremum above, we have that α ≤ ‖T‖. To

prove that α = ‖T‖, we use the Second Polarization Identity, and we notice that, in view of the assumption T = T ∗

and Proposition 2.7.1, 4Re〈Tx, y〉 = 〈T (x+y), x+y〉−〈T (x−y), x−y〉. Moreover, using Parallelogram Law, and

assuming that x and y are unit vectors, we obtain that 4Re〈Tx, y〉 ≤ α‖x+y‖2 +α‖x−y‖2 = α(2‖x‖2 +2‖y‖2) =

4α. When x is selected so that ‖Tx‖ 6= 0, y = Tx/‖Tx‖ we obtain Re‖Tx‖ ≤ α so ‖T‖ ≤ α.


Exercise 2.7.3. Prove that two product of two self-adjoint operators is self-adjoint iff the operators commute.

Remark 2.7.1. If we write A = (T + T ∗)/2 and B = (T − T ∗)/2i then the operators A,B are self-adjoint

and T = A+ iB. We call them the real part and the imaginary part of T .

Proposition 2.7.5. If T is an operator on Hilbert space H then the following are equivalent.

(a) T is a normal operator;

(b) ‖Tx‖ = ‖T ∗x‖ for all x ∈ H;

(c) the real and imaginary part of T commute.

Proof. Notice that ‖Tx‖2 − ‖T ∗x‖2 = 〈(T ∗T − TT ∗)x, x〉. If T is normal then the right side is 0, so (a)

implies (b). If (b) is true, then the left side is 0, for all x. Since T ∗T − TT ∗ is self-adjoint, Proposition 2.7.4

implies that its norm is 0, so (b) implies (a). A calculation shows that, if A and B are the real and imaginary

part of T , resp., then AB −BA = (T ∗T − TT ∗)/2i so (a) is equivalent to (c).

In Definition 1.2.3 we have introduced the concept of the Hilbert space isomorphsim. Since it preserves the

inner product (〈Ux,Uy〉 = 〈x, y〉), it preserves the norm, and hence both weak and strong toplogies. Therefore,

if U : H → K, we do not distinguish between an operator T ∈ L(H) and UTU−1 ∈ L(H), and we say that they

are unitarily equivalent. Since, by Definition 2.7.1, an operator T is unitary iff TT ∗ = T ∗T = I, we should check

that UU∗ = U∗U = I.

Exercise 2.7.4. Verify that UU∗ = IK and U∗U = IH.

Notice that both equalities need to be verified, because it is quite possible for one to hold but not the other.

Example: the unilateral shift S satisfies S∗S = I 6= SS∗.

Exercise 2.7.5. Prove that T is an isometry iff T ∗T = I.

Exercise 2.7.3 asserts that the product of two self-adjoint operators is itself self-adjoint iff the operators

commute. What if self-adjoint is replaced by normal? If M,N are commuting normal operators, their product


is normal if MN commutes with N∗M∗ and it looks like we need the additional assumption that M commutes

with N∗ (which also gives that M∗ commutes with N). When an operator T commutes with both N and N∗ we

say that T doubly commutes with N . When N is normal we can establish even a stronger result.

Theorem 2.7.6 (Fuglede–Putnam Theorem). Suppose that M , N are normal operators and T ∈ L(H)

intertwines M and N : MT = TN . Then M∗T = TN∗.

Proof. Let λ be a complex number, and denote A = λM , B = λN . Notice that AT = TB, so A2T =

A(AT ) = A(TB) = (AT )B = (TB)B = TB2, and inductively AkT = TBK , for k ∈ N. It follows that, if we

denote the exponential function by exp(z), exp(A)T = T exp(B). It is not hard to see that exp(−A) exp(A) = I

so

T = exp(−A)T exp(B).

If we denote by U1 = exp(A∗ − A), U2 = exp(B − B∗), then both U1, U2 are unitary operators. Indeed,

U∗1 = [∑

(A∗ − A)n/n!]∗ =∑

(A − A∗)n/n! = exp(A − A∗) = U−11 , and similarly for U2. Now we have that

exp(A∗)T exp(−B∗) = U1TU2 and ‖ exp(A∗)T exp(−B∗)‖ = ‖T‖. We conclude that

‖ exp(λM∗)T exp(−λN∗)‖ = ‖T‖

for all λ ∈ C. Now f(λ) = exp(λM∗)T exp(−λN∗) is an entire bounded function, hence a constant. Therefore,

f ′(0) = 0. On the other hand, f ′(λ) = M∗ exp(λM∗)T exp(−λN∗) + exp(λM∗)T exp(−λN∗)(−N∗) so f ′(0) =

M∗T − TN∗, and the theorem is proved.

Exercise 2.7.6. Prove that exp(−T ) exp(T ) = I for any operator T ∈ L(H).

Corollary 2.7.7. The product of two normal operators is itself normal iff the operators commute.

Exercise 2.7.7. Prove Corollary 2.7.7.

CHAPTER 3

Spectrum

3.1. Invertibility

In Linear Algebra we learn that each the properties of being invertible, injective, or surjective implies the

other two. Things are very different in infinite dimesional Hilbert space.

Example 3.1.1. Let T = diag(1/n). It is easy to see that KerT = (0) so T is injective. However, it is not

surjective, because its range does not contain the sequence (1, 1/2, 1/3, . . . ) ∈ `2.

Exercise 3.1.1. Prove that T = diag(1/n) is injective but (1, 1/2, 1/3, . . . ) /∈ RanT .

Example 3.1.2. The backward shift S∗ (see Example 2.2.2) is surjective: given (y1, y2, . . . ) ∈ `2 we have that

S∗(0, y1, y2, . . . ) = (y1, y2, . . . ). On the other hand S∗e1 = 0 so S∗ is not injective. Also, S∗S(x1, x2, x3, . . . ) =

S∗(0, x1, x2, . . . ) = (x1, x2, x3, . . . ), so S∗S = I. However, SS∗(x1, x2, x3, . . . ) = S(x2, x3, . . . ) = (0, x2, x3, . . . )

so SS∗ 6= I.

We say that an operator T is left invertible if there exists an operator L ∈ L(H) such that LT = I. It is right

invertible if there exists an operator R such that TR = I. Therefore, the unilateral shift S is left invertible, while

S∗ is right invertible. Since S is injective, it is tempting to jump to the conclusion that an operator is injective

iff it is left invertible.

Example 3.1.3. The Volterra integral operator V is defined on L2 by V f(x) =∫ x

0f(t) dt. Since this is an

integral operator TK with K = χE(x, y) where E = (x, y) : y ≤ x and χE ∈ L2, V is a compact operator so it

cannot be left invertible. Yet, V is injective since V f = 0 implies that f = 0.

Exercise 3.1.2. Prove that the Volterra integral operator V is injective.

31

32 3. SPECTRUM

Exercise 3.1.3. Prove that the range of the Volterra integral operator V is a dense linear manifold in H.

Instead of injectivity, another condition plays a major role in the questions about invertibility.

Definition 3.1.1. An operator T ∈ L(H) is bounded below if there exists α > 0 such that ‖Tx‖ ≥ α‖x‖, for

all x ∈ H.

Example 3.1.4. Let T = diag(cn). Then T is bounded below iff |cn| ≥ α > 0, n ∈ N.

An immediate consequence of this property concerns the range of the operator.

Theorem 3.1.1. If an operator T on Hilbert space H is bounded below then its range is a closed subset of H.

Proof. Let yn be a sequence of vectors in RanT converging to y. Then yn = Txn for some xn ∈ H, so

‖yn − ym‖ = ‖Txn − Txm‖ ≥ α‖xn − xm‖. Since yn is a Cauchy sequence, the same is true of xn. Let

x = limxn. Then Txn → Tx, i.e. yn → Tx. Thus y = Tx ∈ RanT , and RanT is closed.

Example 3.1.3 shows that the injectivity is not sufficient to guarantee the left invertibility. The next result

gives the correct necessary and sufficient conditions.

Theorem 3.1.2. Let T be an operator in L(H). The following are equivalent:

(a) T is left invertible;

(b) KerT = (0) and RanT is closed;

(c) T is bounded below.

Proof. If LT = I then ‖x‖ = ‖LTx‖ ≤ ‖L‖‖Tx‖, so T is bounded below with α = 1/‖L‖, and (a) ⇒ (c).

Clearly, if T is bounded below it must be injective, and the fact that its range is closed is Theorem 3.1.1, so (c)

implies (b). If (b) is true then, by the Open Mapping Theorem (Royden, p.230), there exists a bounded linear

operator L1 : RanT → H, such that L1T = I. If we define L = [ L1 0 ] relative to H = RanT ⊕ (RanT )⊥ (see

Example 2.4.2), then L ∈ L(H) and LT = I.

3.1. INVERTIBILITY 33

Exercise 3.1.4. Prove that T = [ I A0 I ] is bounded below for any operator A.

A similar characterization is available for surjectivity. The most efficient approach seems to be based on the

observation that T is right invertible iff T ∗ is left invertible. In order to continue in this direction we need the

following result, which is significant on its own.

Theorem 3.1.3. The operator T has closed range iff the range of T ∗ is closed.

Proof. Since T ∗∗ = T it suffices to prove one of the two implications. To that end, let RanT be closed, and

let xn be a sequence of vectors such that T ∗xn converges to y. We will show that y ∈ RanT ∗. Since RanT is closed

we can write H = RanT ⊕KerT ∗. If relative to this decomposition xn = x′n⊕x′′n, then T ∗xn = T ∗x′n so, without

loss of generality, we may assume that the sequence xn belongs to RanT . The convergence of T ∗xn implies

the weak convergence so, for any z ∈ H, 〈T ∗xn, z〉 → 〈y, z〉. It follows that 〈xn, T z〉 → 〈y, z〉 and, moreover,

that 〈xn, w〉 converges for any w ∈ H. Indeed, if we write w = w1 ⊕ w2, where w1 ∈ RanT (so w1 = Tz1)

and w2 ∈ KerT ∗, (so 〈xn, w2〉 = 0), we see that xn is a weakly convergent sequence. If w − limxn = x then

w − limT ∗xn = T ∗x. On the other hand, T ∗xn converges to y, so y = T ∗x ∈ RanT ∗.

Now we can deliver the promised characterizations of surjectivity.

Theorem 3.1.4. Let T be an operator in L(H). The following are equivalent:

(a) T is right invertible;

(b) T ∗ is bounded below.

(c) T is surjective.

Proof. The equivalence of (a) and (b) follows from Theorem 3.1.2 applied to T ∗. Further, TR = I implies

that TR is surjective. Since RanTR ⊂ RanT , T is surjective and (a) implies (c). Finally, let T be surjective.

This implies that KerT ∗ = (0) and also, via Theorem 3.1.3, that RanT ∗ is closed. Applying Theorem 3.1.2 we

see that T ∗ is left invertible and the result follows by taking adjoints.

We close this section with a sufficient condition for invertibility that is of quite a different nature.

34 3. SPECTRUM

Theorem 3.1.5. If T is an operator on Hilbert space H and ‖I − T‖ < 1 then T is invertible.

Proof. Let α = 1 − ‖I − T‖ ∈ (0, 1]. If x ∈ H, then ‖Tx‖ = ‖x − (1 − T )x‖ ≥ ‖x‖ − ‖1 − T‖‖x‖ = α‖x‖

so T is bounded below. Suppose now that the range of T is not dense in H. Then there exists y ∈ H such that

d = inf‖y − x‖ : x ∈ RanT > 0. It follows that there exists x ∈ RanT such that (1− α)‖y − x‖ < d. (Obvious

if α = 1, otherwise β = 1/(1−α) > 1 so there exists x such that ‖y−x‖ < βd.) Notice that x+T (y−x) ∈ RanT

so d ≤ ‖y − x− T (y − x)‖ ≤ ‖I − T‖‖y − x‖ < d, which is a contradiction, so T has dense range.

Second proof: The series I + T + T 2 + T 3 + . . . converges in the operator norm, and it is easy to verify

that (I − T )(I + T + T 2 + T 3 + . . . ) = I.

Exercise 3.1.5. Prove that, if ‖T‖ < 1, the series∑∞n=0 T

n converges uniformly.

Exercise 3.1.6. Verify that, if ‖T‖ < 1, (I − T )−1 =∑∞n=0 T

n.

3.2. Spectrum

A complex number λ belongs to the spectrum of an operator T (notation: λ ∈ σ(T )) if T − λI is not

invertible. The complement of σ(T ) is called the resolvent set of T and is denoted by ρ(T ). The spectral radius

of T , r(T ) = sup|λ| : λ ∈ σ(T ). While it is more pedantic to write λI, it is customary to omit the identity and

write just λ for the operator λI. As usual, the interest in the spectrum of a linear operator T is motivated by the

finite dimensional case. In that situation, λ ∈ σ(T ) iff λ is an eigenvalue of T , and eigenvalues play an essential

role in the structure theory via the Jordan form. As we will see, the situation is quite different in the infinite

dimensional Hilbert space.

Example 3.2.1. Let T = diag(cn). If λ = cn for some n, then T −λ has non-trivial kernel (containing en) so

the spectrum contains the whole diagonal. Is there more? If T = diag(1/n) then T is not invertible so 0 belongs

to the spectrum of T , although it is not one of the diagonal entries and not an eigenvalue. What about the

sequence cn = (1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 4/5, . . . )? The operator T = diag(cn) is not invertible, but neither

3.2. SPECTRUM 35

is T − 1, so both 0 and 1 belong to the spectrum of T . Should we include limit points of the sequence as well?

The truth is, we cannot address the problem before we establish some essential properties of the spectrum.

Proposition 3.2.1. If T is an operator on Hilbert space H then r(T ) ≤ ‖T‖.

Proof. If |λ| > ‖T‖ then ‖T/λ‖ < 1. By Theorem 3.1.5, the operator I−T/λ is invertible, so λ /∈ σ(T ).

Example 3.2.2. Let S∗ be the backward shift on `2 (see Example 2.2.2). If |λ| < 1 then the sequence

u = λn is in `2 and it is an eigenvector of S∗, i.e., S∗u = λu so S∗ − λ has non-trivial kernel and is not

invertible. Consequently, the spectrum of S∗ contains the open unit disk. On the other hand, ‖S∗‖ = ‖S‖ = 1

so, by Proposition 3.2.1, σ(S∗) is contained in the closed unit disk.

Example 3.2.2 raises once again the question whether the spectrum must contain its boundary points.

Theorem 3.2.2. If T is an operator on Hilbert space H then σ(T ) is a non-empty compact set.

Proof. Proposition 3.2.1 shows that the spectrum of T is bounded. To show that it is closed, we will show

that ρ(T ) is open. Let λ0 ∈ ρ(T ) so that T − λ0 is invertible. Since

‖1− (T − λ0)−1(T − λ)‖ = ‖(T − λ0)−1 [T − λ0 − (T − λ)] ‖ = ‖(T − λ0)−1‖|λ− λ0|

we see that ‖1− (T −λ0)−1(T −λ)‖ < 1 if |λ−λ0| is sufficiently small. By Theorem 3.1.5, for such λ the operator

(T − λ0)−1(T − λ) is invertible so the same is true of T − λ. Consequently ρ(T ) is open.

Our next goal is to show that the spectrum of a bounded operator cannot be empty. In order to do that, let

x, y ∈ H, and consider the complex-valued function F (λ) = 〈(T − λ)−1x, y〉 defined for λ ∈ ρ(T ).

Proposition 3.2.3. The function F is analytic in ρ(T ) ∪ ∞.

Proof. Let λ0 ∈ ρ(T ). Write

T − λ = (T − λ0)− (λ− λ0) = (T − λ0)[1− (T − λ0)−1(λ− λ0)

]

36 3. SPECTRUM

and notice that if |λ− λ0| is sufficiently small, then ‖(T − λ0)−1(λ− λ0)‖ < 1. By Exercise 3.1.6, we can write

(T − λ)−1 = (T − λ0)−1∞∑n=0

(T − λ0)−n(λ− λ0)n.

Therefore, the function F (λ) =∞∑n=0〈(T − λ0)−n−1x, y〉(λ − λ0)n is analytic in a neighborhood of λ0. As for

λ0 =∞, we consider the function

(3.1) G(λ) = F (1/λ) = 〈(T − 1/λ)−1x, y〉

at λ = 0. Since T − 1/λ = −(1− λT )/λ, for λ 6= 0, Theorem 3.1.5 and Exercise 3.1.6 show that, for λ sufficiently

small (but different from 0), the operator T − 1/λ is invertible and G(λ) = −λ∞∑n=0〈Tnx, y〉λn is analytic at

0. Furthermore, F (∞) = G(0) = 0. If the spectrum of T were empty, F would be an entire function that is

bounded, hence by Liouville’s Theorem, a constant. Since F (∞) = 0 it would follow that F is a zero function for

any x, y ∈ H, which is impossible. (Take x = (T − λI)y, y 6= 0.) Thus σ(T ) is non-empty.

Now we can return to Example 3.2.2 and conclude that the spectrum of S∗ is the closed unit disk. What

about σ(S)?

Exercise 3.2.1. A complex number λ belongs to σ(T ) iff λ ∈ σ(T ∗).

Exercise 3.2.2. Given a non-empty compact set F ⊂ C, show that there exists an operator T ∈ L(H) such

that σ(T ) = F .

Example 3.2.3. The spectrum of the unilateral shift S is the closed unit disk. However, S has no eigenvalues.

Theorem 3.2.4 (Spectral mapping theorem). Let T ∈ L(H) and let p be a polynomial. Then σ(p(T )) =

p(σ(T )).

Proof. Suppose that λ0 ∈ σ(T ), and write p(λ)− p(λ0) = (λ− λ0)q(λ). Then p(T )− p(λ0) = (T − λ0)q(T )

and it is not hard to see that the operator A = p(T )− p(λ0) cannot be invertible. Otherwise, we would have that

3.2. SPECTRUM 37

T − λ0 has both the left inverse A−1q(T ) and the right inverse q(T )A−1. Thus p(λ0) ∈ σ(p(T )), and we obtain

that p(σ(T )) ⊂ σ(p(T )).

To prove the converse, let λ0 ∈ σ(p(T )), and let λ1, λ2, . . . , λn be the roots of p(λ) = λ0. Then p(T )− λ0 =

α(T − λ1)(T − λ2) . . . (T − λn) for some non-zero complex number α. Since p(T ) − λ0 is not invertible there

exists j, 1 ≤ j ≤ n, such that T − λj is not invertible. For this j, λj ∈ σ(T ) and p(λj) = λ0 so λ0 ∈ (σ(T )).

Consequently, σ(p(T )) ⊂ p(σ(T )) and the theorem is proved.

Exercise 3.2.3. Let X and T be operators in L(H), and suppose that X is invertible. Then σ(X−1TX) =

σ(T ).

In many instances it is quite hard to determine the spectrum of an operator. However, it may be possible to

determine its spectral radius, using the next result.

Theorem 3.2.5 (Spectral Radius Formula). Let T ∈ L(H). Then r(T ) = limn→∞ ‖Tn‖1/n.

Proof. By the Spectral mapping theorem, σ(Tn) = [σ(T )]n so [r(A)]n = r(An) ≤ ‖An‖. Thus, r(A) ≤

‖An‖1/n and r(A) ≤ lim infn→∞ ‖An‖1/n. In order to prove the converse we consider the function G(λ) defined

by (3.1) for λ 6= 0 and 1/λ ∈ ρ(T ). For such λ, G is analytic by Proposition 3.2.3 and it can be represented

by the convergent series −λ∑∞n=0 λ

n〈Tnx, y〉. Thus, the sequence λn〈Tnx, y〉 must be bounded. That means

that for each y, the sequence of bounded linear functionals λnTnx is bounded at y, i.e., there exists C(y) such

that |〈λnTnx, y〉| ≤ C(y). By the Uniform Boundedness Principle, the sequence λnTnx is uniformly bounded.

This means that, for each x, there exists C(x), such that ‖λnTnx‖ ≤ C(x). Applying the Uniform Boundedness

Principle once again, we obtain M > 0 such that |λ|n‖Tn‖ ≤M , n ∈ N. It follows that |λ|‖Tn‖1/n ≤M1/n and

|λ| lim supn→∞ ‖Tn‖1/n ≤ 1. Since this is true for any λ such that 1/λ ∈ ρ(T ) it holds all the more whenever

1/|λ| > r(T ). It follows that lim supn→∞ ‖Tn‖1/n ≤ r(T ) and the theorem is proved.

38 3. SPECTRUM

3.3. Parts of the spectrum

A combination of Theorems 3.1.2 and 3.1.4 established that an operator is invertible iff it is bounded below

and has closed range.

Definition 3.3.1. A complex number λ belongs to the approximate point spectrum σapp(T ) of a linear

operator T if T − λ is not bounded below. It belongs to the compression spectrum σcomp(T ) of T if the closure

of Ran (T − λ) is a proper subspace of H. Finally, it belongs to σp(T ) — the point spectrum of T , if it is an

eigenvalue of T .

Remark 3.3.1. There is more than one classification of the parts of the spectrum. The residual spectrum is

σcomp(T )− σp(T ), and the continuous spectrum is σ(T )− (σcomp(T ) ∪ σp(T )). The left spectrum consists of those

complex numbers λ such that T − λ is not left invertible, and similarly for the right spectrum.

Example 3.3.1. Let T = diag(cn). First we notice that T is invertible iff the sequence cn is invertible.

Indeed, if cndn = 1, and dn ∈ `∞, define T−1 = diag(dn). If T is invertible, then T−1en = en/cn so 1/|cn| =

‖T−1en‖ ≤ ‖T−1‖ shows that 1/cn ∈ `∞. Therefore, λ ∈ σ(T ) iff cn − λ is not invertible, which is true iff there

exists a subsequence cnk such that cnk

− λ → 0. In other words, if and only if λ is an accumulation point of

cn. Thus σ(T ) is the closure of the diagonal.

What are the parts of the spectrum of diag(cn)? Suppose that ‖(T−λ)x‖ ≥ α‖x‖. Then∑∞i=1 |(cn−λ)xn|2 ≥

α∑∞i=1 |xn|2. By taking x = en we obtain that |cn−λ| ≥

√α, for all n ∈ N, which means that cn−λ is invertible

and, hence, λ /∈ σ(T ). This shows that σ(T ) ⊂ σapp(T ) and therefore σ(T ) = σapp(T ).

The previous example is a special case of a more general result.

Theorem 3.3.1. If T is a normal operator then σ(T ) = σapp(T ).

Proof. By Proposition 2.7.5, taking into account that T − λ is normal, for any x ∈ H, ‖(T − λ)x‖ =

‖(T ∗−λ)x‖ so σp(T ) = σp(T ∗). Also, λ ∈ σp(T ∗)⇔ Ker (T ∗−λ) 6= (0)⇔ Ran (T ∗−λ)∗ is not dense⇔ Ran (T−λ)

3.3. PARTS OF THE SPECTRUM 39

is not dense ⇔ λ ∈ σcomp(T ). Conclusion: σcomp(T ) ⊂ σp(T ) ⊂ σapp(T ). Since σ(T ) = σapp(T ) ∪ σcomp(T ) the

result follows.

Remark 3.3.2. The proof of Theorem 3.3.1 established that a complex number λ belongs to σp(T ∗) iff

λ ∈ σcomp(T ).

Exercise 3.3.1. If T ∈ L(H) then σ(T ) = σapp(T ) ∪ σcomp(T ).

Since the spectrum is the union of two parts, it is interesting that its boundary is always in the same one.

Theorem 3.3.2. The boundary of the spectrum is included in the approximate point spectrum.

Proof. Let λ ∈ ∂σ(T ). The spectrum of T is closed so λ ∈ σ(T ), which means that either λ ∈ σapp(T )

(in which case there is nothing to prove) or λ ∈ σcomp(T ). In the latter case there exists a non-zero vector x

orthogonal to Ran (T − λ). Let λn ⊂ ρ(T ) such that λn → λ. Since T − λn is invertible, we can define unit

vectors fn = (T − λn)−1f/‖(T − λn)−1f‖. Now

‖(T − λ)fn‖2 ≤ ‖(T − λ)fn‖2 + ‖(T − λn)fn‖2 = ‖(λ− λn)fn‖2 = |λ− λn| → 0

where we have used the fact that (T − λ)fn is a multiple of f , hence orthogonal to (T − λn)fn. Consequently,

λ ∈ σapp(T ).

Example 3.3.2. We have seen in Example 3.2.3 that the spectrum of the unilateral shift S is D−. By

Exercise 3.2.1 the same is true of σ(S∗). Since S is an isometry, 0 canot be an eigenvalue of S. (Sx = 0 imnplies

‖x‖ = ‖Sx‖ = 0.) If λ 6= 0 then S(x1, x2, . . . ) = λ(x1, x2, . . . ) leads to 0 = λx1 and xn = λxn+1, n ∈ N, and we

see that x = 0. Therefore, σp(S) is empty and, by Exercise 3.3.2, so is σcomp(S∗).

The equation S∗x = λx leads to xn+1 = λxn, n ∈ N, and thus to x = x1(1, λ, λ2, . . . ). Therefore, x is a

non-zero vector in `2 iff |λ| < 1. Consequently, σp(S∗) = σcomp(S) = D.

By Theorem 3.3.2, the approximate point spectra of S and S∗ include the unit circle T. For S that is all

because, if |λ| < 1 then ‖Sx − λx‖ ≥ |‖Sx‖ − ‖λx‖| = (1 − |λ|)‖x‖ so S − λ is bounded below. On the other

hand, the approximate point spectrum always includes the eigenvalues, so σapp(S∗) = D−.

40 3. SPECTRUM

Theorem 3.3.3. Suppose thatM is a closed subspace of Hilbert space H, and that, relative to H =M⊕M⊥,

T =[T1 00 T2

]. Then σ(T ) = σ(T1) ∪ σ(T2).

Proof. If T −λ is not invertible then T1−λ and T2−λ cannot both be invertible, so σ(T ) ⊂ σ(T1)∪σ(T2).

On the other hand, if either T1 or T2 is not bounded below, say ‖T1xn‖ → 0, then ‖T (xn ⊕ 0)‖ → 0, so

σapp(T1)∪σapp(T2) ⊂ σ(T ). The corresponding inclusion for the compression spectra can be obtained by switching

to the adjoints and using Exercise 3.3.2.

Problem 11. Suppose that H =M1 ⊕M2 ⊕ . . . and that relative to this decomposition T = diag(Tn) is a

diagonal matrix with operator entries T1, T2, . . . . Is it true that σ(T ) = (∪σ(Tn))−?

3.4. Spectrum of a compact operator

In this section we take a more detailed look at compact operators and their spectra.

Theorem 3.4.1. Let T be a compact operator, let λ be a non-zero complex number, and suppose that T − λ

is not bounded below. Then λ ∈ σp(T ).

Proof. Let xn be a sequence of unit vectors such that ‖(T − λ)xn‖ → 0, n → ∞. Since B1 is weakly

compact, xn has a weakly convergent subsequence xnk, so the compactness of T implies that Txnk

is a

convergent sequence. Let x = limk Txnk. Notice that ‖x‖ ≥ ‖λxnk

‖ − ‖(T − λ)xnk‖ → |λ| so x is a non-zero

vector. Moreover, ‖(T − λ)x‖ ≤ ‖(T − λ)(Txnk− x)‖+ ‖(T − λ)Txnk

‖ → 0 so λ ∈ σp(T ).

Theorem 3.4.1 established that the non-zero points in the approximate point spectrum are eigenvalues. Our

goal is to prove a similar inclusion for the compression spectrum. We start with the following result.

Theorem 3.4.2. Let T be a compact operator and let λ be a non-zero complex number. Then Ran (T − λ) is

closed.

Proof. First we show that, if RanT is closed, it must be finite dimensional. Indeed, if we denote by T1 the

restriction of T to its initial space (KerT )⊥, then T1 is an injective linear transformation from (KerT )⊥ onto

3.4. SPECTRUM OF A COMPACT OPERATOR 41

RanT , hence invertible. Let B be the intersection of the closed ball of radius ‖T−11 ‖ and RanT . Now, if y ∈ B

then y = T1x, for some x ∈ (KerT )⊥, so x = T−11 y. Since ‖y‖ ≤ ‖T−1

1 ‖ it follows that x ∈ B1 ∩ (KerT )⊥. We

conclude that B is contained in the compact set T (B1∩(KerT )⊥) so B must be compact, hence finite dimensional.

Next we observe that Ker (T − λ) must be finite dimensional. Reason: Ker (T − λ) is invariant for T and

the restriction of T to Ker (T − λ) is a compact operator with range Ker (T − λ). (If x ∈ Ker (T − λ) write

x = 1λ [Tx− (T − λ)x] = T (x/λ) ∈ T (Ker (T − λ)).)

Finally, we prove the theorem. Let S be the restriction of T − λ to Ker (T − λ)⊥. Notice that RanS =

Ran (T − λ) so it suffices to show that RanS is closed. By Theorem 3.1.2 we will accomplish this goal by

establishing that S is bounded below. However, if S is not bounded below then Theorem 3.4.1 shows that

(T − λ)x = 0 for some nonzero vector x in Ker (T − λ)⊥. This is impossible, so RanS is closed and the proof is

complete.

Before we can proceed we need this technical result.

Lemma 3.4.3. Let T be a compact operator and let λn be a sequence of complex numbers. Suppose that

there exists a nested sequence of distinct subspaces M1 ( M2 ( M3 ( . . . such that (T − λn)Mn+1 ⊂ Mn.

Then λn converges to 0.

Proof. Let en be an sequence of unit vectors such that e1 ∈ M1 and en+1 ∈ Mn+1 Mn. Clearly, this

is an orthonormal system. Moreover, for n ≥ 2, 〈(T − λn)en, en〉 = 0 which implies that ‖Ten‖ ≥ |〈Ten, en〉| =

|〈(T − λn)en, en〉 + 〈λnen, en〉| = |λn|. Since T is compact and w − lim en = 0 it follows that limn Ten = 0 so

limn λn = 0.

Theorem 3.4.1 shows that if λ ∈ σ(T ) then either λ = 0, or λ ∈ σp(T ), or T − λ is bounded below (hence

injective) but not surjective. By Theorem 3.1.4, T − λ not being surjective is the same as (T − λ)∗ not being

bounded below. Since T ∗ is also compact, another application of Theorem 3.4.1 allows us to conclude that

λ ∈ σp(T ∗). The next result shows that there is even less variation in the spectrum of a compact operator.

42 3. SPECTRUM

Theorem 3.4.4. Let T be a compact operator and let λ be a non-zero complex number. Then λ ∈ σp(T ) iff

λ ∈ σp(T ∗).

Proof. Clearly, it suffices to prove either direction. Suppose that λ ∈ σp(T ). By Theorem 3.4.2, the range

of T − λ is closed. We will show that it must be a proper subspace of H. Suppose to the contrary that T − λ is

surjective, and denote Mn = Ker (T − λ)n. Since λ is an eigenvalue of T we can inductively define a sequence

xn of nonzero vectors such that (T − λ)xn = xn−1, with x0 = 0. Clearly xn belongs to Mn but not to Mn−1,

and (T − λ)Mn+1 ⊂ Mn, so Lemma 3.4.3 implies that the constant sequence λ, λ, λ, . . . converges to 0, which

contradicts the assumption that λ 6= 0. Therefore, Ran (T − λ) (which coincides with Ker (T − λ)∗) is a proper

subspace of H and λ ∈ σp(T ∗).

To summarize, the spectrum of a compact operator consists of the point spectrum and, possibly, 0. On the

infinite dimensional Hilbert space, 0 must be in the spectrum because if a compact operator T were invertible,

then so would be the identity (a product of TT−1), contradicting the conclusions of Example 2.6.1. Thus we have

a corollary.

Corollary 3.4.5. The spectrum of a compact operator consists of 0 and its eigenvalues.

It is reasonable to ask about the location of the eigenvalues.

Theorem 3.4.6. For any C > 0 there is a finite number of linearly independent eigenvectors of a compact

operator corresponding to eigenvalues λ such that |λ| ≥ C.

Proof. Suppose to the contrary that there is an infinite sequence xn of unit vectors, and a sequence of

eigenvalues λn of T , |λn| ≥ C, so that Txn = λnxn. Let Mn = ∨nk=1xk. If x ∈ Mn then x =∑nk=1 ckxk so

(T − λn)x = (T − λn)∑nk=1 ckxk =

∑nk=1 ck(T − λn)xk =

∑nk=1 ck(λk − λn)xk ∈Mn−1. Applying Lemma 3.4.3

we obtain that λn → 0, which contradicts |λn| ≥ C.

Corollary 3.4.7. If λ is a non-zero eigenvalue of a compact operator T , then the nullspace of T − λ is a

finite dimensional subspace.

3.5. SPECTRUM OF A NORMAL OPERATOR 43

Corollary 3.4.8. The spectrum of a compact operator T is at most countable, and the only accumulation

point of it can be zero.

Remark 3.4.1. If T = diag(cn) where c1 = 1 and cn = 0 for n ≥ 2, then T is compact, and σ(T ) = 0, 1 so

it has no accumulation points.

Last remark raises a question: can a compact operator have a one-point spectrum? Since compact operators

are never invertible, the single point is necessarily 0, so the question can be reformulated as: are there compact

quasinilpotent operators? (An operator T is quasinilpotent if σ(T ) = 0.) In finite dimensions, a quasinilpotent

operator is nilpotent, i.e. there exists a positive integer N such that TN = 0. This need not be the case in infinite

dimensional Hilbert space.

Example 3.4.1. Let T be a weighted shift (see Example 2.1.6) with weight sequence 1/nn∈N. It is compact

following Example 2.6.1. Since Wen = (1/n)en+1 it follows that W ken = 1n(n+1)...(n+k−1)en+k. This shows that

W k is a product of Sk and a diag( 1n(n+1)...(n+k−1) ). Since Sk is an isometry, ‖W k‖ = supn 1

n(n+1)...(n+k−1) =

1/k!. Now r(W ) = limk ‖W k‖1/k = limk(1/k!)1/k = 0. Therefore, W is a compact quasinilpotent operator.

3.5. Spectrum of a normal operator

On the first glance, normal operators appear to be too diverse to fit one description. Before we can correct

this misconception, we will need to make a thorough study of this class, and some of its prominent subclasses.

Theorem 3.5.1. (a) If T is a unitary operator then σ(T ) is a subset of the unit circle. (b) If T is a self-

adjoint operator then σ(T ) is a subset of the real axis. (c) If T is a positive operator then σ(T ) is a subset of the

non-negative real axis. (d) If T is a non-trivial projection then σ(T ) = 0, 1.

Proof. All operators listed are normal, so by Theorem 3.3.1, it suffices to prove assertions (a) – (d) with

σapp(T ) instead of σ(T ). To that end, we will prove that, if λ does not belong to the appropriate set, then T − λ

is bounded below.

44 3. SPECTRUM

(a) If T is unitary and |λ| 6= 1, then ‖Tx− λx‖ ≥ |‖Tx‖ − ‖λx‖| = |(1− λ)| ‖x‖ so T is bounded below.

(b) Let λ = α+ iβ. Then ‖Tx− λx‖2 = ‖Tx−αx‖2 − 2Re〈Tx−αx, iβx〉+ ‖iβx‖2. If α, β are real numbers

and T = T ∗ we have that 〈T − αx, x〉 ∈ R by Proposition 2.7.1, and it follows that Re〈Tx − αx, iβx〉 = 0.

Therefore, ‖Tx− λx‖2 ≥ |β|2‖x‖2, so β 6= 0 implies that T − λ is bounded below.

(c) If T ≥ 0 then T is self-adjoint, so σ(T ) ⊂ R. Notice that ‖Tx− λx‖2 = ‖Tx‖2 − 2Re〈Tx, λx〉+ ‖λx‖2. If

λ < 0 then 〈Tx, λx〉 < 0 (by definition of a positive operator) so ‖Tx − λx‖2 ≥ |λ|2‖x‖2 and T − λ is bounded

below.

(d) If T is a non-trivial projection then neither T nor I − T (the projection on the orthogonal complement

of the range of T ) can be invertible, so 0, 1 ⊂ σ(T ). If λ /∈ 0, 1, a calculation shows that 1λ(1−λ)T −

1λ is the

inverse of T .

Exercise 2.7.1 asserts that the operator of multiplication by an L∞ function is a normal operator. In addition,

it showed that Mh belongs to one of the important subclasses iff its (essential) range belonged to a specific subset

of the complex plane. On the other hand, Theorem 3.5.1 showed that for a general normal operator, a membership

in each of the mentioned subclasses implies the analogous behavior of its spectrum. This is no coincidence. First

we need a proposition.

Proposition 3.5.2. Let T = Mh on L2. Then the following are equivalent:

(a) RanT is dense;

(b) h(x) 6= 0 a.e.;

(c) T is injective;

(d) T ∗ is injective.

Proof. Let A = x : h(x) = 0. Suppose that µ(A) 6= 0 and let f = χA. For any g ∈ L2, 〈Tg, f〉 =∫hgf =∫

Ahg = 0 so f is a non-zero function that is orthogonal to RanT . Thus (a) implies (b). Next, if Tf = 0 then

h(x)f(x) = 0 a.e., so assuming (b) we see that f = 0, and (c) follows. Notice that if T ∗f = 0 then h(x)f(x) = 0

so Tf = 0 and (c) implies (d). Finally, the implication (d) ⇒ (a) is a direct consequence of Theorem 2.2.3.

3.5. SPECTRUM OF A NORMAL OPERATOR 45

Recall that the essential range of a function h ∈ L∞(X,µ) is the set of all complex numbers z such that the

measure of Eε = x ∈ X : |h(x)− z| < ε is different from zero for all ε > 0.

Theorem 3.5.3. Let T = Mh on L2. Then σ(T ) is the essential range of h.

Proof. Notice that Mh − λ is a multiplication by h − λ. Let us denote by A = x : h(x) 6= λ, B = x :

h(x) = λ, and define a function g(x) = 1/(h(x)− λ) if x ∈ A and g(x) = 0 if x ∈ B.

Suppose first that λ ∈ ρ(T ). By Proposition 3.5.2, µ(B) = 0. Thus, g(x) = 1/(h(x)−λ) a.e. and MgMh−λ =

Mh−λMg = I. Since the assumption is that Mh−λ is invertible, the operator Mg is bounded, and by Example 2.1.4,

g ∈ L∞. The estimate |g(x)| ≤ M a.e. implies that |h(x) = λ| ≥ 1/M a.e. so µ(E1/M ) = 0 and λ is not in the

essential range of h.

Conversely, if λ is not in the essential range of h, then there exists ε0 > 0 such that µ(Eε0) = 0. Consequently,

|h(x)−λ| ≥ ε0 a.e., whence |g(x)| ≤ 1/ε0 a.e., and Mg is a bounded operator. This shows that Mh−λ is invertible

and the proof is complete.

Proposition 3.2.1 established that r(T ) ≤ ‖T‖. For normal operators more can be said, and the following

result paves the way to that goal.

Proposition 3.5.4. If T is a normal operator then ‖Tn‖ = ‖T‖n, n ∈ N.

Proof. First we notice that, in view of Proposition 2.7.5, for n ∈ N,

‖Tnx‖2 = 〈Tnx, Tnx〉 = 〈T ∗Tnx, Tn−1x〉 ≤ ‖T ∗Tnx‖‖Tn−1x‖ = ‖Tn+1x‖‖Tn−1x‖ ≤ ‖Tn+1‖‖Tn−1‖‖x‖2

so ‖Tn‖2 ≤ ‖Tn+1‖‖Tn−1‖.

Now we prove the assertion of the proposition using induction. We will assume that ‖T‖ 6= 0, otherwise the

theorem is trivially correct. It is easy to see that the statement is valid for n = 0 and n = 1. Suppose that it is

true for n. Then

‖T‖2n = (‖T‖n)2 = ‖Tn‖2 ≤ ‖Tn+1‖‖Tn−1‖ ≤ ‖Tn+1‖‖T‖n−1

46 3. SPECTRUM

and, dividing both sides by ‖T‖n−1, it follows that ‖T‖n+1 ≤ ‖Tn+1‖. Since the opposite inequality is obvious,

the theorem is proved.

Corollary 3.5.5. If T is a normal operator then r(T ) = ‖T‖.

Proof. By Theorems 3.2.5 and 3.5.4, ‖T‖ = n√‖Tn‖ → r(T ).

CHAPTER 4

Invariant subspaces

4.1. Compact operators

We have seen that the spectrum of a compact operator consists of the eigenvalues and 0 which may be but is

not necessarily an eigenvalue. Furthermore, each of the eigenspaces E(λ) = Ker (T − λ), corresponding to λ 6= 0,

is finite dimensional. The situation is especially pleasant when T is self-adjoint, in addition to being compact.

One of the benefits of this additional hypothesis concerns the eigenspaces.

Proposition 4.1.1. If T is a compact, self-adjoint operator on Hilbert space, and if λ, µ are two different

eigenvalues of T , then the corresponding eigenspaces E(λ), E(µ) are mutually orthogonal.

Proof. If Tx = λx and Ty = µy, then λ〈x, y〉 = 〈Tx, y〉 = 〈x, Ty〉 = µ〈x, y〉, since µ ∈ R. Given that λ 6= µ

it follows that 〈x, y〉 = 0.

Proposition 4.1.1 shows that H can be written as a direct sum M⊕M⊥, where M = ⊕n∈NE(λn), the

orthogonal direct sum of all eigenspaces. When T is self-adjoint, the subspace M⊥ is just a mirage.

Theorem 4.1.2. If T is a compact, self-adjoint operator on H space, and σp(T ) = λii∈I , then H =

⊕i∈IE(λi).

Proof. Let M = ⊕i∈IE(λi) and suppose that M 6= H. Notice that M is invariant for T = T ∗, so M⊥ is

also reducing for T . Let T1 be the restriction of T to M⊥. Then σ(T1) ⊂ σ(T ) by Theorem 3.3.3. Since T1 is

compact, if λ 6= 0 is in its spectrum it must be an eigenvalue. However, the corresponding eigenvectors would

also be eigenvectors of T and, as such, would belong to M. It follows that T1 must be quasinilpotent. On the

other hand T1 is normal which would necessitate that its norm and spectral radius are equal, so T1 = 0 which

means that M⊥ ⊂ E(0) ⊂M. The obtained contradiction shows that H = ⊕i∈IE(λi).

47

48 4. INVARIANT SUBSPACES

Remark 4.1.1. Each eigenspace E(λ) is reducing for a self-adjoint operator so, relative to the decomposition

H = ⊕i∈IE(λi), T can be represented as diag(Ti), where Ti is an operator mapping E(λi) into itself, and σ(Ti)

is a singleton λi. In addition, regardless of whether T is self-adjoint or not, each eigenspace is hyperinvariant

for T . This means that it is invariant for any operator that commutes with T . Indeed, if A commutes with T ,

then T − λ annihilates Ax together with x.

When T is not self-adjoint, the situation is much more complicated. The eigenspaces need not be mutually

orthogonal any more. The eigenvectors do not necessarily span H. In fact, there are compact operators without

eigenvalues, (so they are necessarily quasinilpotent). Still, we can see some of the structure remaining. The

eigenspaces are hyperinvariant (if there are any), although they need not be reducing. Since all operators on Cn

are compact, it is instructive to look at finite matrices.

Example 4.1.1. Let T = [ 1 10 1 ] acting on C2. Then σ(T ) = 1 and E(1) = C⊕ (0) which is neither invariant

for T ∗, nor is the span of eigenvectors of T equal to C2.

Example 4.1.2. Let T = [ 2 10 3 ] acting on C2. The eigenvalues of T are 2 and 3, with corresponding eigenvectors

[ 10 ] and [ 1

1 ], and they are not mutually orthogonal.

When T has eigenvalues, it must have a non-trivial invariant subspace. What about the case of a compact

quasinilpotent operator?

Example 4.1.3. Let T be the Volterra-type integral operator with kernel K, i.e., Tf(x) =∫ x

0K(x, y)f(y) dy.

It is compact (Example 2.6.2) and has no eigenvalues different from 0. Indeed, let λ ∈ σ(T ), λ 6= 0 and let f ∈ L2

be the appropriate eigenfunction. Define g(x) =∫ x

0|f(y)|2 dy. Clearly, g is a monotone differentiable function

and g′(x) = |f(x)|2 a.e. Let a = supx ∈ [0, 1] : g(x) = 0. (Since g(0) = 0 such a number exists.) Now, for

a.e. x,

|λf(x)|2 = |Tf(x)|2 =

∣∣∣∣∣∣x∫

0

K(x, y)f(y) dy

∣∣∣∣∣∣2

≤x∫

0

|K(x, y)|2 dyx∫

0

|f(y)|2 dy,

4.2. LINE INTEGRALS 49

so |λ|2g′(x)/g(x) ≤∫ x

0|K(x, y)|2 dy for a.e. x ∈ (a, 1). By integrating the last inequality we obtain

|λ|2 ln g(x) |1a ≤1∫a

x∫0

|K(x, y)|2 dy ≤ ‖T‖2

which is a contradiction since ln g(1) = ln ‖f‖2 and ‖T‖ are finite, but ln g(a) is not.

This example shows that there are many compact quasinilpotent operators. For the Volterra-type integral

operators we can exhibit some invariant subspaces.

Theorem 4.1.3. Let T be a Volterra-type integral operator with kernel K, let a ∈ [0, 1], and let Ma = f ∈

L2 : f(x) = 0 when x ≤ a. Then Ma is a subspace of L2 that is invariant for T .

Exercise 4.1.1. Prove Theorem 4.1.3.

A deep result in the theory of integral operators is that every compact quasinilpotent operator is unitarily

equivalent to an operator of the form as in Example 4.1.3. Consequently every compact operator (quasinilpotent

or not) has an invariant subspace. As we will demonstrate, there is a way to prove an even stronger theorem.

(See Theorem 4.3.2 below.)

4.2. Line integrals

In this section we make a brief detour, by considering line integrals of functions of a complex variable with

values in L(H).

Example 4.2.1. Let T ∈ L(H) and consider the function ρ(λ) = (T − λ)−1 defined for λ ∈ ρ(T ). This

function is known as the resolvent of T .

Let C be a curve in the complex plane. We will assume that it is parametrized by a continuous function

γ : [0, 1]→ C and that it is rectifiable, which means that γ is a function of bounded variation. Suppose that S is

a function defined and continuous on C, with values in L(H). Let P be a partition of [0, 1]: 0 = t0 < t1 < t2 <


· · · < tn = 1 and, for 1 ≤ k ≤ n let t∗k ∈ [tk−1, tk]. Then we have a partition of C with points γi = γ(ti) and

intermediate points γ∗i = γ(t∗i ). Let us denote ∆γi = γi − γi−1 and consider the sum

n∑k=1

S(γ∗k) ∆γk.

It can be shown that these sums converge to a unique operator S which we denote as S =∫CS(γ) dγ. Moreover,

if T is an operator that commutes with each S(γ), then T commutes with S.

Example 4.2.2. Let T ∈ L(H), and let C be a curve in ρ(T ) defined by γ = γ(t). For every λ ∈ ρ(T ), the

function ρ(λ) is a continuous function (in the uniform topology), so we can consider∫Cρ(γ) dγ.

What happens when the curve C is replaced by a curve C ′ that is not far from C?

Theorem 4.2.1. Let C0 be a rectifiable curve in the resolvent set of T , and let C1 be a curve homotopic to

C0. Then∫C0ρ(γ) dγ =

∫C1ρ(γ) dγ.

Remark 4.2.1. All these facts can be established following the same procedures as in the case when the

integrand is a complex-valued function. [See Conway.]

Now we turn to operators. Example 4.2.2 showed that the operator∫Cρ(γ) dγ is well defined. It turns out

that this operator has some interesting properties.

Theorem 4.2.2. Let C be a simple closed rectifiable curve in ρ(T ). Then the operator

(4.1) P = − 12πi

∫C

ρ(λ) dλ

is a projection (not necessarily orthogonal) that commutes with every operator that commutes with T . Conse-

quently, the subspaces RanP and KerP are both invariant for T .

Proof. Let C ′ be a simple closed rectifiable curve in ρ(T ) that lies inside C and is homotopic to C. Then

(2πi)2P 2 =∫C

ρ(γ) dγ∫C′ρ(λ) dλ =

∫C

∫C′ρ(γ)ρ(λ) dγdλ.

4.2. LINE INTEGRALS 51

A calculation shows that ρ(γ)ρ(λ) = [ρ(γ)− ρ(λ)](γ − λ)−1. Thus we have that

(2πi)2P 2 =∫C′ρ(γ)

∫C

(γ − λ)−1 dλdγ −∫C

ρ(λ)∫C′

(γ − λ)−1 dγdλ = −2πi∫C′ρ(γ) dγ − 0 = (2πi)2P.

So, P 2 = P , and it follows from the definition of the integral and ρ(λ), that if A commutes with T then A

commutes with P .

Finally, if y ∈ RanP , then Ty = TPy = PTy so Ty ∈ RanP . Similarly, if x ∈ KerP , then 0 = TPx = PTx

so Tx ∈ KerP .

Exercise 4.2.1. Verify that ρ(γ)ρ(λ) = [ρ(γ)− ρ(λ)](γ − λ)−1.

Theorem 4.2.2 required that the closed curve C lies in ρ(T ), but made no reference to the spectrum of

T . Consequently, we may have a part of the spectrum inside C and a part outside. In that case we obtain a

decomposition of T .

Theorem 4.2.3. Let T be an operator in L(H), let C be a simple closed rectifiable curve in ρ(T ), let P be

the projection defined in (4.1), and let T ′ and T ′′ be the restrictions of T to RanP and KerP , respectively. Then

T = T ′ + T ′′, the spectrum of T ′ is precisely the subset of σ(T ) inside C, and the spectrum of T ′′ is precisely the

subset of σ(T ) outside C.

Proof. Since ρ(λ) commutes with P , for any λ ∈ ρ(T ), the subspaces RanP and KerP are invariant for

ρ(λ). Let ρ′(λ) and ρ′′(λ) denote the restrictions of ρ(λ) to these subspaces. If we denote by I ′ and I ′′ the

identity operators on these subspaces, then ρ′(λ)(λI ′−T ′) = I ′ and ρ′′(λ)(λI ′′−T ′′) = I ′′. Therefore, if λ ∈ ρ(T )

then λ must belong to both ρ(T ′) and ρ(T ′′). In the other direction, if λ ∈ ρ(T ′) ∩ ρ(T ′′) then there exist

operators A′ and A′′ such that A′(λI ′ − T ′) = I ′ and A′′(λI ′′ − T ′′) = I ′′. Now we can define, for any x ∈ H,

Ax = A′Px + A′′(I − P )x. It is not hard to see that the restricitons of A to RanP and KerP are precisely A′

and A′′, and that A(λI − T )x = x when x belongs to either RanP or KerP . It follows that A(λI − T )x = x

holds for all x ∈ H, so λ ∈ ρ(T ). We conclude that λ ∈ σ(T ) iff λ ∈ σ(T ′) or λ ∈ σ(T ′′).


Suppose now that λ lies outside of C. We will show that λ ∈ ρ(T ′), which is true iff there exists an operator A′

acting on RanP and satsifying A′(λI ′ − T ′) = I ′. Actually, we will show that there exists an operator A ∈ L(H)

that commutes with T and A(λI − T ) = P . To that end, we notice that

(T − λI)ρ(γ) = (T − λI)(T − γI)−1 = (T − γI)(T − γI)−1 + (γ − λ)(T − γI)−1 = I + (γ − λ)(T − γI)−1.

Therefore,

(4.2) (T − λI)1

2πi

∫C

ρ(γ)(γ − λ)−1 dγ =1

2πi

∫C

(γ − λ)−1 dγ I +1

2πi

∫C

ρ(γ) dγ = 0− P = −P.

On the other hand, if λ lies inside of C, then the integral in (4.2) equals I − P , so the restriction to KerP yields

I ′′. Once again, this shows that λI ′′ − T ′′ is invertible.

4.3. Invariant subspaces for compact operators

In Section 4.1 we have discovered that every compact operator on Hilbert space has an invariant subspace.

What more is there to say? For one thing, if λ is an eigenvalue of T , then E(λ) is hyperinvariant. Thus, it is

natural to ask whether a compact quasinilpotent operator always has a hyperinvariant subspace.

Before we address this question, let us take a look at the set of all operators that commute with T . It is

called the commutant of T , it is denoted by T′, and it is an algebra. The last statement means that T′ is

closed under sums, products, and multiplication by scalars.

Exercise 4.3.1. Prove that T′ is an algebra.

Definition 4.3.1. A subalgebra of L(H) is transitive if it is weakly closed, unital (containing the identity

operator), and has only the trivial invariant subspaces.

Example 4.3.1. The algebra L(H) is transitive. It is clearly weakly closed and unital. If L(H) had a non-

trivial invariant subspace M, then we could pick non-zero vectors x ∈ M⊥ and y ∈ M, and consider the rank

one operator T = x⊗ y. This would lead to a contradiciton, since y ∈M but Ty = (x⊗ y)y = 〈y, y〉x ∈M⊥.

4.3. INVARIANT SUBSPACES FOR COMPACT OPERATORS 53

A big open problem in operator theory is whether L(H) is the only transitive algebra. This is true when H

is finite dimensional.

Theorem 4.3.1 (Burnside’s Theorem). Let H be a finite dimensional vector space of dimension larger than

1. If A is a transitive algebra of linear transformations on H, then A = L(H).

Proof. We will show that A contains a rank one operator. Let T0 be an operator with minimal non-zero

rank d. If d > 1, choose x1 and x2 so that vectors T0x1, T0x2 are linearly independent, and then choose A ∈ A so

that AT0x1 = x2. (Such an operator A exists, otherwise AT0x1 : A ∈ A would be a subspace of H, invariant for

A.) Then T0AT0x1 (= T0x2) and T0x1 are linearly independent, and T0AT0−λT0 is not a zero transformation for

any λ ∈ C. On the other hand, there exists a complex number λ0 such that the restriction of T0A− λ0 to RanT0

is not invertible. Therefore, T0AT0 − λ0T0 has rank less than d and greater than 0, contradicitng the minimality

of d. Hence d = 1.

If T0 = x⊗ y, we will show that A contains all rank one operators. Let u⊗ v be a rank one operator. Once

again, there must be an operator A1 ∈ A such that A1x = u. Notice that the algebra A∗ = A∗ : A ∈ A is also

transitive. Therefore, there exists an operator A2 ∈ A such that A∗2y = v. Then A1T0A2 = u⊗ v so A contains

all rank one operators and, hence, all finite rank operators, i.e. L(H).

Exercise 4.3.2. Prove that if A is a subalgebra of L(H) and x ∈ H, then Ax = Ax : A ∈ A is a subspace

of H, invariant for A.

Exercise 4.3.3. Prove that A is transitive iff A∗ is transitive.

Theorem 4.3.2 (Lomonosov’s Theorem). Let A be a non-scalar operator on Hilbert space that commutes

with a compact operator. Then A has a nontrivial hyperinvariant subspace.

The proof of this result uses a fixed point theorem.

Theorem 4.3.3. Let F be a compact and convex subset of Hilbert space H, and let T be a linear operator in

L(H) with the property that T (F ) ⊂ F . Then there exists p ∈ H such that Tp = p.


Proof. For every n ∈ N, let Tn = (1 + T + T 2 + · · ·+ Tn−1)/n. The set Tn(F ) is convex, (Exercise 4.3.4),

and compact, as the image of a compact set under a continuous map. Also, Tn(F ) ⊂ F , because if x ∈ F then

T kx ∈ F , 0 ≤ k ≤ n − 1, and K is convex. Further, for any m,n ∈ N, Tm(F )Tn(F ) ⊂ Tm(F ) ∩ Tn(F ) which

shows that the family Tn(F )n∈N has a finite intersection property. Since they are all subset of a compact set

F , they all have a non-empty intersection, i.e., there exists p ∈ ∩Tn(F ) : n ∈ N. We will show that Tp = p.

Suppose, to the contrary, that Tp 6= p. Then there exists α > 0 such that ‖Tp−p‖ ≥ α. Since F is a bounded

set, there exists M > 0 such that ‖x‖ ≤ M , for x ∈ F . Let n be a positive integer satisfying n > 2M/α. Since

p ∈ Tn(F ), there exists xn ∈ F such that p = Tnxn and, therefore,

Tp− p = (T − 1)Tnxn = (T − 1)1 + T + T 2 + · · ·+ Tn−1

nxn =

Tn − 1n

xn.

Then α ≤ ‖Tp− p‖ = ‖(Tn − 1)/nxn‖ ≤ (‖Tnxn‖+ ‖xn‖)/n ≤ 2M/n which contradicts the choice of n.

Exercise 4.3.4. Prove that if C is a convex set in Hilbert space H and T ∈ L(H), then T (C) is a convex set.

Now we can prove the result which is frequently referred to as the Lomonosov’s Lemma.

Theorem 4.3.4. If A is a transitive subalgebra of L(H) and if K is a non-zero compact operator in L(H),

then there exists an operator A ∈ A and a non-zero vector x ∈ H such that AKx = x.

Proof. Without loss of generality we will assume that ‖K‖ = 1. As we have already noticed, it suffices to

consider the case when K is quasinilpotent. Let x0 be a vector in H such that ‖Kx0‖ > 1 and notice that this

implies that ‖x0‖ > 1, so the closed ball B(x0, 1) does not contain 0. Let D be the image under K of the closed

ball B(x0, 1). By Exercise 2.6.4, D is a compact set. In addition, it is convex, by Exercise 4.3.4 and it does not

contain 0. Indeed, for any x ∈ B(x0, 1), ‖Kx‖ ≥ ‖Kx0‖ − ‖K(x− x0)‖ > 1− ‖x− x0‖ ≥ 0.

For an operator T ∈ A, consider the set UT = y ∈ H : ‖Ty − x0‖ < 1. Notice that UT = T−1(z :

‖z − x0‖ < 1 so it is an open set. Moreover, every non-zero vector y belongs to UT , for some T ∈ A. Indeed, A

is transitive so the linear manifold Ty : T ∈ A must be dense in H and, hence, there exists T ∈ A such that

‖Ty − x0‖ < 1, which means that y ∈ UT . Thus, ∪T∈AUT is a covering of H − 0, and all the more of D. As

4.3. INVARIANT SUBSPACES FOR COMPACT OPERATORS 55

established earlier, D is a compact set, so there exist operators T1, T2, . . . , Tn ∈ A such that D ⊂ ∪ni=1UTi. This

means that, for any y ∈ D there exists Ti, 1 ≤ i ≤ n, such that ‖Tiy − x0, ‖ < 1.

Now, for each j, 1 ≤ j ≤ n, and y ∈ D, we define αj(y) = max0, 1 − ‖Tjy − x0‖. Notice that each αj is

continuous on D, 0 ≤ αj ≤ 1, and∑nj=1 αj(y) > 0, for all y ∈ D. Furthermore, αj(y) 6= 0 iff ‖Tjy − x0‖ < 1.

Define, for y ∈ D and 1 ≤ j ≤ n,

βj(y) =αj(y)n∑i=1

αi(y),

and notice that each βj is continuous on D, 0 ≤ βj ≤ 1, and∑nj=1 βj(y) = 1, for all y ∈ D. Also, βj(y) 6= 0 iff

αj(y) 6= 0 iff ‖Tjy − x0‖ < 1. Finally, let Ψ : D → H be defined by Ψ(y) =∑nj=1 βj(y)Tjy. It is easy to see that

Ψ is continuous on D. We will show that Ψ(D) ⊂ B(x0, 1). Let y ∈ D. Then

‖Ψ(y)− x0‖ = ‖n∑j=1

βj(y)Tjy −n∑j=1

βj(y)x0‖ ≤n∑j=1

|βj(y)|‖Tjy − x0‖ ≤ 1

so Ψ(y) ∈ B(x0, 1) and Ψ(D) ⊂ B(x0, 1). If we define Φ : B(x0, 1)→ H by Φ(y) = Ψ(Ky), then Φ is a continuous

map of B(x0, 1) into itself. Since B(x0, 1) is a compact, convex set, Theorem 4.3.3 shows that Φ has a fixed

point p ∈ B(x0, 1), hence non-zero. Now we define the operator A =n∑j=1

βj(Kp)Tj which is in A. Finally,

AKp =n∑j=1

βj(Kp)TjKp = Ψ(Kp) = Φ(p) = p.

Now we can prove Lomonosov’s theorem.

Proof of Lomonosov’s Theorem. Let A = A′ and suppose, to the contrary, that A is transitive. By

Theorem 4.3.4, there exists an operator T ∈ A′ such that TKx = x. In other words, a compact operators AK

has 1 as an eigenvalue. Let E(1) denote the appropriate eigenspace which is finite dimensional. Since A commutes

with TK, the subspace E(1) is invariant for A as well. The restriction of A to E(1) must have an eigenvalue λ

and, since E(1) is invariant for A, we see that λ is an eigenvalue for A (not just the restriction). Let M denote

the eigenspace of A corresponding to λ, i.e., M = x ∈ H : Ax = λx. Being an eigenspace, it is hyperinvariant

for A. It is not (0), so it remains to notice that it is not H because A 6= λ.


4.4. Normal operators

We have seen in Exercise 2.7.1 that a multiplication operator Mh on L2 is a normal operator. In this section

we will show that, in a sense, every normal operator is a multiplication by an essentially bounded function.

Example 4.4.1. Let T = [ a 00 b ], with a, b ∈ C. Then TT ∗ = T ∗T . Let X = 1, 2 and let µ be a counting

measure on X. Notice that L2(X,µ) is the collection of all functions f : X → C with norm(∫X|f |2 dµ

)1/2 =(|f(1)|2 + |f(2)|2

)1/2. Since this is the Euclidean norm, we see that L2(X,µ) is just L(C2). Finally, let h be a

function on X, h(1) = a, h(2) = b. Then T can be identified with Mh on L(C2).

Remark 4.4.1. A similar construction can be made for the case when T is an n × n diagonal matrix,

T = diag(cn).

Example 4.4.2. Let T = diag(cn), with cn ∈ C for all n ∈ N. Let X = N and µ(n) = 1/2n. Then (X,µ)

is a finite measure space. Further, let h : X → C be defined by h(n) = cn. Then T can be identified with the

operator Mh on L2(X,µ).

The last example shows the danger of going through the motions. What does it mean “can be identified”?

While it is easy to see that Tf = Mhf for any sequence f , their domains are not the same. Namely, T acts on

`2 but Mh acts on L2(X,µ), and these 2 spaces are not the same. For example, the sequence (1, 1, 1, . . . ) belongs

to L2(X,µ) but not to `2. However, these two spaces are isomorphic. Let U : L2(X,µ) → `2 be defined by

U(f) = (f(1)/√

2, f(2)/√

22, f(3)/√

23, . . . ). It is easy to verify that U is injective and surjective linear map so,

by the Open Mapping Principle, it is an isomorphism. Moreover, if f ∈ L2(X,µ), then

U−1TU(f) = U−1T (f(1)√

2,f(2)√

22,f(3)√

23, . . . ) = U−1(

c1f(1)√2

,c2f(2)√

22,c3f(3)√

23, . . . )

= U−1(h(1)f(1)√

2,h(2)f(2)√

22,h(3)f(3)√

23, . . . ) = hf,

so T is unitarily equivalent to Mh.


Exercise 4.4.1. Prove that the map U : L2(X,µ) → `2, constructed in Example 4.4.2, is an isometric

isomorphism.

Notice that in Examples 4.4.1 and 4.4.2 the measure was defined on each of the pieces. What happens if

pieces are not that obvious? How do we define a piece?

Definition 4.4.1. A vector ξ is cyclic for an operator T if the set p(T )ξ : p is a polynomial is dense in H.

An operator T is cyclic if it has a cyclic vector.

Example 4.4.3. Let T = S, the unilateral shift. The vector ξ = e1 is cyclic for S. If x ∈ `2, x = (x1, x2, . . . )

then x can be approximated by truncated sequences (x1, x2, . . . , xn, 0, 0, . . . ) =∑nk=1 xkek =

∑nk=1 T

ke1.

Example 4.4.4. Let . . . , e−2, e−1, e0, e1, e2, . . . be an o.n.b. of H, and let T be the bilateral shift : Ten =

en+1, n ∈ Z. Then ξ = e0 is not a cyclic vector for T , because p(T )e0− = ∨∞k=0ek. However, T ∗en = en−1,

n ∈ Z, so we need to replace polynomials in T by polynomials in T and T ∗, i.e., f(T ) =∑ni,j=1 T

iT ∗j . If the set

f(T )ξ : f is a polynomial in T, T ∗ is dense in H, we say that e0 is a star-cyclic vector for T .

Before we proceed, we revisit the Stone–Weierstrass Theorem [Bartle, p. 184]. Although it is proved under

the assumption that K is a compact subset of Rp, the same proof is valid when K is a compact set in C. Also,

we will rephrase it using the following terinology. We will say that an algebra A of functions separates points on

K if, for any two distinct points x, y ∈ K there is a function f ∈ A such that f(x) 6= f(y). If for each x ∈ K

there is a function g ∈ A such that g(x) 6= 0, we say that A vanishes at no point of K.

Theorem 4.4.1 (Stone–Weierstrass Theorem). Let A be an algebra of continuous, real-valued functions on

a compact set K in C. If A separates points on K and if A vanishes at no point of K, then the uniform closure

of B of A consists of all real-valued continuous functions on K.

The Stone–Weierstrass Theorem deals only with real-valued functions of complex variable. Now we extend

it to complex-valued functions. We will require that A be self-adjoint, meaning that if f ∈ A the f ∈ A.


Theorem 4.4.2. Let A be a self-adjoint algebra of continuous, complex functions on a compact set K in C.

If A separates points on K and if A vanishes at no point of K, then the uniform closure of B of A consists of all

complex continuous functions on K.

Proof. Let f = u+iv be a continuous function on K, and let AR denote the set of all real-valued functions in

A. Since u, v are continuous real-valued continuous function on K, it suffice to show that every such function lies

in the closure of AR. Since AR is clearly an algebra, the result will follow from the Stone–Weierstrass Theorem,

once we show that AR separates points on K and vanishes at no point of K.

Suppose that z1, z2 are distinct points in K. By assumption, A separates points on K so it contains a function

f such that f(z1) 6= f(z2). Also, A vanishes at no point of K, so it contains two functions g, h such that g(z1) 6= 0,

h(z2) 6= 0. Then, the function

F (z) =f(z)g(z)− f(z2)g(z)f(z1)g(z1)− f(z2)g(z1)

belongs to A and has the property that F (z1) = 1, F (z2) = 0. Notice that, if F = u + iv ∈ A, then F ∈ A and

u = (F + F )/2 ∈ AR. Clearly, u(z1) = 1, u(z2) = 0 so AR separates points on K.

Let z0 ∈ K. Then there exists a function G ∈ A such that G(z0) 6= 0. Let λ be a complex number such that

λG(z0) > 0 and notice that H = Re(λG) is a function in AR such that H(z0) > 0. Thus, AR vanishes at no point

of K and the proof is complete.

Now we are ready to establish a stronger connection between normal operators and operators of multiplication.

Theorem 4.4.3. Let T be a normal operator in L(H) with a star-cyclic vector ξ. Then there exists a finite

measure µ on σ(T ), a bounded function h : σ(T ) → R, and an isomporphism U : L2(σ(T ), µ) → H such that

U−1TUf(x) = h(x)f(x) for a.e. x ∈ σ(T ) and all f ∈ L2(σ(T ), µ).

Proof. Let A be the algebra of complex-valued polynomials in z, z. For f ∈ A we define L(f) = 〈f(T )ξ, ξ〉.

Clearly, L is a linear functional and it is bounded onA. Indeed, |L(f)| = |〈f(T )ξ, ξ〉| ≤ ‖f(T )ξ‖‖ξ‖ ≤ ‖f(T )‖‖ξ‖2.

Further, T is normal, so f(T ) is also normal and, by Corollary 3.5.5, ‖f(T )‖ = r(f(T )) = sup|λ| : λ ∈ σ(f(T )).


Finally, by the Spectral Mapping Theorem, λ ∈ σ(f(T )) iff λ = f(µ), for some µ ∈ σ(T ). Thus, ‖f(T )‖ =

sup|f(µ)| : µ ∈ σ(T ) = ‖f‖∞. We conclude that |L(f)| ≤ ‖f‖∞‖ξ‖2, so L is bounded on A. By Theorem 4.4.2,

A is dense in C(σ(T )) so we can extend L to a bounded linear functional on C(σ(T )). If f is a non-negative function

in C(σ(T )), then so is√f and it can be approximated by a sequence fn ∈ A. It follows that f can be approximated

by the sequence fnfn and, by the continuity of L, L(f)f = limL(fnfn) = 〈fn(T )fn(T )ξ, ξ〉 = ‖fn(T )ξ‖2 ≥ 0.

Thus, L is positive, and by Riesz Representation Theorem [Royden, p. 352] there exists a finite positive measure

µ on σ(T ) such that 〈f(T )ξ, ξ〉 =∫f dµ. Now define the operator U on A by U(f) = f(T )ξ. Since |f |2 = ff

we have that∫|f |2 dµ = 〈f(T )f(T )ξ, ξ〉 = ‖f(T )ξ‖2 = ‖U(f)‖2. That way, U is an isometry on A. Further,

A is dense in L2(µ) because it is dense in C(σ(T )), and the latter set is dense in L2 ([Rudin, Theorem 3.14]).

Therefore, by Theorem 2.3.4, U can be extended to an isometry U : L2(σ(T ), µ)→ H. Since ξ is star-cyclic, the

set f(T )ξ : f ∈ A is dense in H so the range of U is dense. Since U is bounded below its range is closed so U

is surjective.

Finally, if we denote by f(z) the function zf(z), then U−1TU(f) = U−1Tf(T )ξ = U−1f(T )ξ = f so T can

be identified with Mz on L2(σ(T ), µ).

What if T does not have a star-cyclic vector?

Theorem 4.4.4. Let T be a normal operator in L(H). Then there exists a compact set X, a finite measure µ

on X, a bounded function h : X → R, and an isomporphism U : L2(X,µ)→ H such that U−1TUf(x) = h(x)f(x)

for a.e. x ∈ X and all f ∈ L2(X,µ).

Proof. Let x1 be a non-zero vector and let M1 be the closed linear span of f(T )x1 : f ∈ A. If M1 = H

then x1 is a star-cyclic vector for T and Theorem 4.4.3 applies. IfM1 6= H there exists a non-zero vector x2 ∈M⊥1 .

Notice that M1 is invariant (hence reducing) for T and T ∗, so the same is true of M⊥1 . Now, either the closed

linear span of f(T )x2 : f ∈ A equals M⊥1 , in which case T = T1 ⊕ T2 and both T1 and T2 are star-cyclic, or

we continue the process. Applying the Hausdorff Maximal Principle, we obtain a decomposition of H relative to

which T = diag(Ti) and each of the operators on the diagonal is star-cyclic. By Theorem 4.4.3, for each i there


exists a finite measure space (Xi, µi), a function hi ∈ L2(Xi, µi), and unitary operator Ui : L2(Xi, µi) → Mi,

such that U−1i TiUi = Mhi . Next we define X to be the union of Xi and µ a measure on X so that µ = µi on

Xi. Finally, we define a function h so that h = hi on Xi and a unitary operator U = diag(Ui). Then T can be

identified with Mh on L2(X,µ), i.e., U−1TU = Mh.

We will now introduce a very important concept.

Definition 4.4.2. If X is a set, Ω a σ-algebra of subsets of X, and H is Hilbert space, a spectral measure

for (X,Ω,H) is a function E : Ω→ L(H) such that

(a) for each ∆ in Ω, E(∆) is a projection;

(b) E(∅) = 0 and E(X) = 1;

(c) E(∆1 ∩∆2) = E(∆1)E(∆2);

(d) if ∆ii∈I are pairwise disjoint sets in Ω, then E(∪i∈I∆i) =∑i∈I E(∆i).

Example 4.4.5. Let X = N, let Ω be the set of all subsets of N, and let enn∈N be an o.n.b. ofH. For ∆ ⊂ N,

define E(∆) to be the projection onto the span ∨n∈∆en. Properties (a) and (b) of Definition 4.4.2 are obvious.

Since E(∆)ei is either ei or 0, depending on whether i belongs to ∆ or not, we see that E(∆1)E(∆2)ei = 0 unless

i ∈ ∆1 ∩∆2, in which case it equals ei. Thus, for x =∑xiei, E(∆1)E(∆2)x =

∑i∈∆1∩∆2

xiei = E(∆1 ∩∆2)x,

and (c) holds as well. Finally, if ∆ii∈I are pairwise disjoint sets in Ω, and ∆ = ∪i∈I∆i, writing x =∑n∈N xnen,

we have that E(∆)x =∑i∈∆1

xiei +∑i∈∆2

xiei + · · · = E(∆1)x+ E(∆2)x+ . . . .

Example 4.4.6. If X is a set, Ω a σ-algebra of subsets of X, and µ a measure on Ω, let H = L2(X,µ), and

define, for ∆ ∈ Ω and f ∈ L2, E(∆)f = χ∆f . Then, E is a spectral measure.

Exercise 4.4.2. Verify that E in Example 4.4.6 is a spectral measure.

We will now show that the equality U−1TU = Mh, established in Theorem 4.4.3, can be extended in the

following manner. Suppose that F is a bounded function on σ(T ). Then we can define F (T ) = UMFhU−1 since,

for x ∈ H, U−1x ∈ L2 and MFhU−1x is also in L2, so UMFhU

−1x is well defined.


Theorem 4.4.5. Let T be a bounded linear operator on Hilbert space H. The mapping F 7→ F (T ) is an

algebra homomorphism from L∞(σ(T ), µ) to L(H).


Remark 4.4.2. The homomorphism F 7→ F (T ) is called a functional calculus.

Example 4.4.6 shows that a spectral measure can be defined using multiplication by characteristic functions.

We present a variation on this theme.

Theorem 4.4.6. If T is a normal operator on Hilbert space, ∆ is a measurable subset of σ(T ), and F = χ∆,

then the mapping E defined by E(∆) = F (T ) = UMFhU−1 is a spectral measure.


Exercise 4.4.5. What is E when T = diag(cn)?

Let x, y ∈ H and denote by f = U−1x and g = U−1y. Since U is a surjective isometry, U−1 = U∗ so f = U∗x

and g = U∗y. If F = χ∆ then, by definition, 〈E(∆)x, y〉 = 〈F (T )x, y〉 = 〈UMFhU−1x, y〉 = 〈MFhf, g〉 =∫

F h fg dµ. On the other hand, E is the spectral measure of T , so 〈E(∆)x, y〉 also defines a measure ν(∆). It

is often called the scalar spectral measure of T .

Exercise 4.4.6. Verify that ν is a measure.

Now, 〈E(∆)x, y〉 is equal to∫χ∆ dν as well as to

∫F hfg dµ, so we have the equality

(4.3)∫F hfg dµ =

∫F dν

whenever F is a characteristic function. Since every simple function is a linear combination of characteristic

functions, it is not hard to see that (4.3) remains true when F is a simple function. Further, every bounded

function can be approximated by simple functions so, by relying on Lebesgue Dominated Convergence Theorem,


we obtain that (4.3) holds for any bounded function F . In particular, if F (λ) = λ, we obtain that 〈Tx, y〉 =∫λ dν =

∫λ d〈E(λ)x, y〉. Since this is true for all x, y ∈ H, we can write T =

∫λ dE(λ) or

(4.4) T =∫λ dE.

More generally, since (4.3) holds for any bounded function F , it follows that, for any such function,

(4.5) F (T ) =∫F (λ) dE.

Theorem 4.4.6 established that to every normal operator there corresponds a spectral measure. The following

result shows how essential this measure is for the operator.

Theorem 4.4.7. If T is a normal operator and E the associated spectral measure, then an operator A com-

mutes with T iff A commutes with E(∆) for every Borel set ∆ ⊂ σ(T ).

Proof. Let x, y ∈ H, and let F be a bounded function on σ(T ). Then

〈AF (T )x, y〉 = 〈F (T )x,A∗y〉 =∫F (λ) d〈E(λ)x,A∗y〉, and

〈F (T )Ax, y〉 =∫F (λ) d〈E(λ)Ax, y〉.

If A and T commute, Fuglede–Putnam Theorem implies that A commutes with T ∗, hence with F (T ), for any

bounded function F . In particular, by taking F = χ∆, we obtain that 〈E(∆)x,A∗y〉 = 〈E(∆)Ax, y〉 or, equiv-

alently that 〈AE(∆)x, y〉 = 〈E(∆)Ax, y〉. Since this holds for all x, y ∈ H it follows that A commutes with

E(∆).

Conversely, if A commutes with E(∆), then 〈E(∆)x,A∗y〉 = 〈AE(∆)x, y〉 = 〈E(∆)Ax, y〉. Since 〈ATx, y〉 =∫λ d〈E(λ)x,A∗y〉 and 〈TAx, y〉 =

∫λ d〈E(λ)Ax, y〉, we obtain that 〈ATx, y〉 = 〈TAx, y〉 for all x, y ∈ H. Thus

AT = TA and the proof is complete.

Theorem 4.4.7 has an important consequence that concerns the existence of hyperinvariant subspaces.


Corollary 4.4.8. If T is a normal operator in L(H), and E is its spectral measure, then E(∆) is a hyper-

invariant subspace for T , for any Borel set ∆ ⊂ σ(T ). Consequently, if T is not a scalar multiple of the identity,

then T has a non-trivial hyperinvariant subspace.

Exercise 4.4.7. Prove Corollary 4.4.8.

CHAPTER 5

Spectral radius algebras

5.1. Compact operators

In Section 4.3 we have shown that every compact operator is contained in an algebra, namely its commutant,

that is not transitive. Are there other algebras that would contain a given operator and still have an invariant

subspace? We will show that the answer is affirmative. Let us denote the class of quasinilpotent operators as Q.

The following is a direct consequence of Theorem 4.3.4.

Proposition 5.1.1. Let A be a unital subalgebra of L(H) and let K be a compact operator in L(H). If

AK ∈ Q for each A ∈ A, then A has a n. i. s.

Proof. If A is transitive, by Theorem 4.3.4 there exists A ∈ A such that 1 ∈ σp(AK), so AK /∈ Q.

Our goal is to find an algebra A with the property stated in Proposition 5.1.1. Let A ∈ L(H). For m ∈ N,

define

(5.1) dm =m

1 +mr(A), and Rm =

( ∞∑n=0

d2nm A

∗nAn

)1/2

.

Exercise 5.1.1. Prove that the series in (5.1) converges uniformly and, for each m ∈ N, Rm is invertible

with ||R−1m || ≤ 1.

If A is an operator in L(H) and Rm is as in (5.1), we associate with A the collection

BA =T ∈ L(H) : sup

m||RmAR−1

m || <∞.

Exercise 5.1.2. Show that BA is an algebra.

We will show that BA contains all operators that commute with A. In fact, we can prove a stronger result.

64


Proposition 5.1.2. Suppose A is a nonzero operator, B is a power bounded operator commuting with A,

and T is an operator for which AT = BTA. Then T ∈ BA.

An operator T is power bounded if there exists C > 0 such that ‖Tn‖ ≤ C, for all n ∈ N. For example, if

‖T‖ ≤ 1, then T is power bounded.

Proof. It is easy to verify that A2T = B2TA2. Using induction one can prove that AnT = BnTAn, for

every n ∈ N. The operator B is power bounded so there is a constant C such that ‖Bn‖ ≤ C, for each n ∈ N.

For any vector x ∈ H and any positive integer m, we have that

(5.2) ‖Rmx‖2 = 〈Rmx,Rmx〉 = 〈R2mx, x〉 =

∞∑n=0

d2nm 〈A∗

nAnx, x〉 =∞∑n=0

d2nm 〈Anx,Anx〉 =

∞∑n=0

d2nm ‖Anx‖2.

On the other hand, ‖AnTR−1m x‖ = ‖BnTAnR−1

m x‖ ≤ C‖T‖‖AnR−1m x‖ so we obtain that

‖RmTR−1m x‖2 =

∞∑n=0

d2nm ‖AnTR−1

m x‖2

≤ C2‖T‖2∞∑n=0

d2nm ‖AnR−1

m x‖2

= C2‖T‖2‖RmR−1m x‖2

= C2‖T‖2‖x‖2.

Thus T ∈ BA.

From this we deduce an easy consequence.

Corollary 5.1.3. Let T be an operator such that AT = λTA for some complex number λ with |λ| ≤ 1.

Then T ∈ BA. In particular BA contains the commutant of A.

Example 5.1.1. If u and v are unit vectors then Bu⊗v = T ∈ L(H) : v is an eigenvector for T ∗. Let

A = u ⊗ v be a rank one operator, with u and v are unit vectors. One knows that r(u ⊗ v) = |〈u, v〉|. A

66 5. SPECTRAL RADIUS ALGEBRAS

calculation shows that, for n ∈ N, An = 〈u, v〉n−1 u⊗ v and A∗nAn = r2n−2 v ⊗ v. Therefore,

R2m = I +

( ∞∑n=1

d2nm r2n−2

)v ⊗ v = I +

d2m

1− d2mr

2v ⊗ v.

Let λm =√

1 + d2m/(1− d2

mr2) for every m ∈ N. Notice that λm →∞ as m→∞. Indeed, either dm → 1/r

or, if A is quasinilpotent, λm =√

1 +m2. If we denote by M the one dimensional space spanned by v then,

relative to H =M⊕M⊥, the matrix of Rm is Rm =[λm 00 1

]and R−1

m =[

1/λm 00 1

]. If T is an arbitrary operator,

say T = [X YZ W ], then

RmTR−1m =

λm 0

0 1

X Y

Z W

1/λm 0

0 1

X Y λm

Z/λm W

and it is easy to see that supm ‖RmTR−1

m ‖ <∞ if and only if Y = 0. This means that M⊥ is invariant for T or,

equivalently, that M is invariant for T ∗, and this is true iff v is an eigenvector for T ∗.

Exercise 5.1.3. Prove that r(u⊗ v) = |〈u, v〉|.

Now we define QA = T ∈ L(H) : ‖RmTR−1m ‖ → 0.

Theorem 5.1.4. QA is a two sided ideal in BA and every operator in QA is quasinilpotent. Furthermore, if

A is quasinilpotent, then A ∈ QA.

Proof. Let T ∈ QA and let X ∈ BA. Then ‖RmTXR−1m ‖ ≤ ‖RmTR−1

m ‖ ‖RmXR−1m ‖ → 0 so QA is a right

ideal. Since the same estimate holds for XT we see that QA is a two sided ideal in BA. On the other hand

r(T ) = r(RmTR−1m ) ≤ ‖RmTR−1

m ‖ which shows that if T ∈ QA then it must be quasinilpotent. Finally, if A ∈ Q

then r(A) = 0 and dm = m. Using (5.2) we see that

‖RmAR−1m x‖2 =

∞∑n=0

m2n‖An+1R−1m x‖2 =

1m2

∞∑n=0

m2n+2‖An+1R−1m x‖2

=1m2

[−‖R−1

m x‖2 +∞∑n=0

m2n‖AnR−1m x‖2

]=

1m2

[‖x‖2 − ‖R−1

m x‖2]≤ ‖x‖

2

m2

from which it follows that ‖RmAR−1m ‖ ≤ 1/m→ 0, m→∞.


Remark 5.1.1. The ideal QA need not contain every quasinilpotent operator in BA. Indeed, if A is the

unilateral forward shift a calculation shows that R2m = 1/(1− d2

m). Since every operator commutes with a scalar

multiple of the identity it follows that BA = L(H). On the other hand, ‖RmTR−1m ‖ = ‖T‖ for any T in L(H), so

QA = (0).

The following result justifies our interest in QA.

Theorem 5.1.5. If QA 6= (0) and there exists a nonzero compact operator in BA, then BA has a n. i. s.

Proof. Let K be a nonzero compact operator in BA. Without loss of generality we may assume that QK = 0

for every Q ∈ QA. Indeed, if QK 6= 0 for some Q ∈ QA, then QK is a compact quasinilpotent operator with the

property that BAQK ⊂ Q and the result follows from Proposition 5.1.1.

Let Q be a fixed nonzero operator in QA and let T be an arbitrary operator in BA. Then QT ∈ QA and,

hence, QTK = 0. Since K 6= 0 there is a nonzero vector z in the range of K. Clearly, QTz = 0 so Tz ∈ kerQ for

all T ∈ BA. Naturally, the closure of the subspace Tz : T ∈ BA is an invariant subspace for BA. It is nonzero

since z 6= 0 and the identity operator is in BA. Finally, it is not H since it is contained in the kernel of a nonzero

operator Q.

From Theorem 5.1.5 we deduce some easy consequences.

Corollary 5.1.6. Suppose that A is a quasinilpotent operator, B is a power bounded operator commuting

with A, and K is a nonzero compact operator satisfying AK = BKA. Then BA has a n. i. s.

Proof. By Proposition 5.1.2, K is in BA. Since A ∈ Q, Theorem 5.1.4 shows that A ∈ QA. The result then

follows from Theorem 5.1.5.

Corollary 5.1.7. Suppose that A is a quasinilpotent operator, λ is a complex number, and K is a nonzero

compact operator satisfying AK = λKA. Then either BA or BA∗ has a n. i. s. In any case, A has a proper

hyperinvariant subspace.


Proof. If |λ| ≤ 1 Corollary 5.1.6 implies that BA has a n. i. s. For |λ| > 1, we have A∗K∗ = (1/λ)K∗A∗

so the same argument shows that BA∗ has a n. i. s. If M is such a subspace then it is hyperinvariant for A∗. It

follows that M⊥ is a proper hyperinvariant subspace for A.

Now we arrive to the main result of this section.

Theorem 5.1.8. Let K be a nonzero compact operator on the separable, infinite dimensional Hilbert space

H. Then BK has a n. i. s.

Proof. We will show that QK 6= (0). The result will then follow from Theorem 5.1.5. Of course, if K is

quasinilpotent, Theorem 5.1.4 shows that K ∈ QK . Therefore, for the rest of the proof, we will assume that

r(K) > 0.

Notice that x⊗ y ∈ QA iff ‖Rm(x⊗ y)R−1m ‖ → 0. However, ‖Rm(x⊗ y)R−1

m ‖ = ‖Rmx‖‖R−1m y‖ so it suffices

to exhibit a rank one operator x⊗ y with supm ‖Rmx‖ <∞ and limm ‖R−1m y‖ = 0. A vector y with the desired

property is supplied by the following lemma.

Lemma 5.1.9. Suppose that K is a compact operator and r(K) > 0. Then there exists a unit vector v such

that∥∥R−1

m v∥∥→ 0, m→∞.

Proof. Let λ be a complex number in σ(K) such that |λ| = r(K). Then λ ∈ σ(K∗) so there are unit vectors

u and v for which Ku = λu and K∗v = λv. An easy calculation shows that K (u⊗ v) = (u⊗ v)K so that u⊗v ∈

K′ ⊂ BA. It then follows that supm ||Rmu|| ||R−1m v|| < ∞. On the other hand, a straightforward calculation

shows that ||Rmu|| → ∞, m→∞. Since supm ||Rmu|| ||R−1m v|| <∞ it must follow that ||R−1

m v|| → 0.

Exercise 5.1.4. Prove that ‖Rmu‖ → ∞, m→∞.

So it remains to provide a nonzero vector x with the property that

(5.3) supm‖Rmx‖ <∞.


To that end, it suffices for x to satisfy

(5.4) lim supn‖Knx‖1/n < r(K).

Indeed, (5.4) implies that the power series∑∞n=0 ||Knx||2 zn has radius of convergence bigger than 1/r2 and,

consequently, the series∑n ‖Knx‖2/r2n converges. Since

||Rmx||2 =∞∑n=0

(m

1 +mr

)2n

||Knx||2

and m/(1 +mr) is an increasing sequence converging to 1/r, we see that (5.4) implies (5.3).

It is not hard to see that, if K has an eigenvalue λ with the property that |λ| < r(K), then any eigenvector

corresponding to λ satisfies (5.4). Thus we may assume that 0 is an isolated point of σ(K). Let Γ be a positively

oriented circle around the origin such that 0 is the only element of σ(K) inside the circle, and let

P = − 12πi

∫Γ

(K − λI)−1 dλ.

By Theorem 4.2.2, P is a projection that commutes with K, and the restriction K0 of K to the invariant subspace

RanP is quasinilpotent. It follows that, if x is a unit vector in RanP , then ||Knx||1/n = ||Kn0 x||

1/n ≤ ‖Kn0 ‖1/n →

0. This completes the proof of the theorem.

Exercise 5.1.5. Prove that if u⊗ v is a rank one operator, then ‖u⊗ v‖ = ‖u‖‖v‖.

As mentioned earlier, the presence of proper invariant subspaces for BK (K compact) is an advancement in

invariant subspace theory only if BK differs from K′ . We do not know at the present time if BK can equal

K′ for a compact nonzero operator K on an infinite dimensional space. We do know that the answer is no if

K has positive spectral radius.

Proposition 5.1.10. Let K be a compact operator on an infinte dimensional Hilbert space such that r(K) > 0.

Then BK 6= K′ .


Proof. Notice that the vectors x and y obtained in the proof of Theorem 5.1.8 satisfy (5.3) and K∗y = λy,

with |λ| = r(K). Since it was established that x ⊗ y ∈ BK it suffices to prove that K(u ⊗ v) 6= (u ⊗ v)K. This

follows from the fact that Kx 6= λx which is a simple consequence of (5.3).

operator theory on hilbert space class...

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