Operators: Theory and Applications

Yuri G. Kondratiev

2018

OPERATOR THEORY

1 Introduction

Spaces

By IF we will always denote either the field IR of real numbers or the fieldC of complex numbers.

1.1. Definition A norm on a vector space X over IF is a map k · k : X ! [0,1) satisfying the conditions

(i) kxk = 0 () x = 0;(ii) k�xk = |�|kxk, 8x 2 X, 8� 2 IF;(iii) kx + yk kxk+ kyk, 8x, y 2 X (the triangle inequality).A vector space X equipped with a norm is called a normed space.

1.2. Definition A normed space X is called a Banach space if it is com-plete, i.e. if every Cauchy sequence in X is convergent.

1.3. Examples1. Any finite-dimensional normed space is a Banach space.2. Let lp, 1 p 1, denote the vector space of all sequences x =(xk)1k=1, xk 2 IF, such that

kxkp :=1X

k=1

|xk|p

!1/p

< 1, 1 p < 1,

kxk1 := supk2IN

|xk| < 1.

1

Then k · kp is a norm on lp and lp, 1 p 1, are Banach spaces.3. Let C([0, 1]) be the space of all continuous functions on [0, 1] and

kfkp :=✓Z 1

0|f(t)|pdt

◆1/p

, 1 p < 1,

kfk1 := sup0t1

|f(t)|.

Then each k · kp is a norm on C([0, 1]).The space C([0, 1]) with the norm k · k1 is a Banach space.The space C([0, 1]) with the norm k · kp, 1 p < 1, is not a Banachspace.Indeed, let us consider the sequence (fn) in C([0, 1]), where

fn(t) =

8>>>><

>>>>:

0 if 0 t 12 �

12n ,

nt� n�12 if 1

2 �12n < t < 1

2 + 12n ,

1 if 12 + 1

2n t 1 .

It is easily seen that each fn is a piece-wise linear function, 0 fn 1 and

kfn � fmkp <

Z 12+max{ 1

2n , 12m}

12�max{ 1

2n , 12m}

1pdt

!1/p

✓max

⇢1n

,1m

�◆1/p

! 0,

as m,n !1.

So, (fn) is a Cauchy sequence with respect to k ·kp. Now, suppose there exists f 2 C([0, 1])such that kf � fnkp ! 0. Taking into account that for an arbitrary � 2]0, 1/2[ there existsN such that

fn = 1 on [1/2 + �, 1], 8n > N,

we obtainZ 1

1/2+�|1� f(t)|pdt =

Z 1

1/2+�|fn(t)� f(t)|pdt kf � fnk

pp ! 0 as n !1.

Since the first integral is independent of n it has to be zero, which implies that f = 1 on[1/2 + �, 1] and hence on ]1/2, 1] since � was arbitrary. A similar argument shows thatf = 0 on [0, 1/2[. Thus f is discontinuous at t = 1/2 and we obtain a contradiction.

1.4. Definition Normed spaces X and Y are called isomorphic if thereexists a linear isometry from X onto Y . Isometry is a map which does notchange the norms of the corresponding points.

2

1.5. Theorem For any normed space (X, k · k) there exists a Banachspace (X 0, k · k0) and a linear isometry from X onto a dense linear subspaceof X 0. Two Banach spaces in which (X, k · k) can be so imbedded are iso-morphic .

1.6. Definition The space (X 0, k · k0) from Theorem 1.5 is called thecompletion of (X, k · k).

The completion of the space (C([0, 1]), k · kp), 1 p < 1, is the Banachspace Lp([0, 1]) from the theory of Lebesgue integral.

Operators

1.7. Definition Let X and Y be vector spaces. A map B : X ! Y iscalled a linear operator (map) if

B(�x + µz) = �Bx + µBz, 8x, z 2 X, 8�, µ 2 IF.

1.8. Theorem Let X and Y be normed spaces. For a linear operatorB : X ! Y the following statements are equivalent:(i) B is continuous;(ii) B is continuous at 0;(iii) there exists a constant C < +1 such that kBxk Ckxk, 8x 2 X.

1.9. Definition A liner operator B is bounded if it satisfies the last (andhence all) of the three conditions above. If B is bounded we define its normby the equality

kBk := inf{C : kBxk Ckxk, 8x 2 X}.

It is easy to see that

kBxk kBk kxk, 8x 2 X, (1.1)

kBk = inf{C : kBxk C, all x s.t. kxk 1} =

inf{C : kBxk C, all x s.t. kxk = 1} = sup

(kBxk

kxk: x 6= 0

)

=

sup{kBxk : kxk 1} = sup{kBxk : kxk = 1}

3

(prove these relations!).

1.10. Theorem Let X and Y be normed spaces. Then k · k is indeeda norm on the vector space B(X, Y ) of all bounded linear operators from Xinto Y, and

kABk kAkkBk, 8B 2 B(X, Y ), 8A 2 B(Y, Z). (1.2)

Moreover, if Y is complete then B(X, Y ) is a Banach space.

Let X be a Banach space. Theorem 1.10 says that

B(X) := B(X,X)

is actually what we call a Banach algebra.

A vector space E is called an algebra if for any pair (x, y) 2 E ⇥ E a unique productxy 2 E is defined with the properties

(xy)z = x(yz),

x(y + z) = xy + xz,

(x + y)z = xz + yz,

�(xy) = (�x)y = x(�y),

for all x, y, z 2 E and scalars �.E is called an algebra with identity if it contains an element e such that for all x 2 E ,

ex = xe = x, 8x 2 E .

A normed algebra is a normed space which is an algebra such that

kxyk kxkkyk, 8x, y 2 E ,

and if E has an identity e,kek = 1.

A Banach algebra is a normed algebra which is complete, considered as a normed space .

1.11. Definition Let X and Y be subspaces of X and Y . An operatorB : X ! Y is said to be an extension of B : X ! Y if Bx = Bx, 8x 2 X.

4

1.12. Theorem Let X and Y be Banach spaces and D be a dense lin-ear subspace of X. Let A be a bounded linear operator from D (equippedwith the X�norm) into Y. Then there exists a unique extension of A toa bounded linear operator A : X ! Y defined on the whole X; moreoverkAk = kAk.

1.13. Example Let X = Y be the space L2([a, b]), i.e. the completionof D = (C([a, b]), k · k2), (�1 < a < b < +1). Consider the operatorA : C([a, b]) ! C([a, b]) ⇢ L2([a, b]),

(Af)(t) =Z

b

a

k(t, ⌧)f(⌧)d⌧, where k 2 C([a, b]2).

Theorem 1.12 allows one to extend this operator to a bounded linear operatorA : L2([a, b]) ! L2([a, b]), since D is dense in its completion X = L2([a, b]).We only need to prove that A : D ! D is a bounded operator.Denote

K = max(t,⌧)2[a,b]2

|k(t, ⌧)|.

Using Cauchy–Schwarz inequality we obtain

|(Af)(t)| =

�����

Zb

a

k(t, ⌧)f(⌧)d⌧

�����

Zb

a

|k(t, ⌧)|2d⌧

!1/2 Zb

a

|f(⌧)|2d⌧

!1/2

Kp

b� akfk2, 8f 2 C([a, b]).

Therefore

kAfk2 =

Zb

a

|(Af)(t)|2dt

!1/2

K(b� a)kfk2, 8f 2 C([a, b]),

i.e. A is bounded and kAk K(b� a).(Let Y be the Banach space (C([a, b]), k · k1). Then the above proof showsthat A can be extended to a bounded linear operator A : L2([a, b]) !C([a, b]), kAk K

pb� a .)

1.14. Definition Let X be a normed space and xn 2 X, n 2 IN. Wesay that the series

P1

n=1 xn is convergent to a vector x 2 X if the sequence(Sj) of partial sums, Sj =

Pj

n=1 xn, converges to x. We then write1X

n=1

xn = x.

5

We say that the series converges absolutely if

1X

n=1

kxnk < 1.

1.15. Theorem Any absolutely convergent series in a Banach space isconvergent.

Proof: For m > k we have

kSm � Skk =

������

mX

n=k+1

xn

������

mX

n=k+1

kxnk ! 0, as k, m !1.

Hence the sequence (Sj) is Cauchy and therefore converges to some x 2 X. 2

2 Spectral theory of bounded linear opera-tors

Auxiliary results

Let us recall that for any normed spaces X and Y we denote by B(X, Y )the space of all bounded linear operators acting from X into Y . We use alsothe following notation B(X) = B(X, X).

2.1. Definition Let A be a linear operator from a vector space X intoa vector space Y. The kernel of A is the set

Ker(A) := {x 2 X : Ax = 0}.

The range of the operator A is the set

Ran(A) := {Ax : x 2 X}.

Ker(A) and Ran(A) are linear subspaces of X and Y correspondingly.(Why?) Moreover, if X and Y are normed spaces and A 2 B(X, Y ), thenKer(A) is closed (why?); this is not necessarily true for Ran(A).

6

2.2. Theorem (Banach) Let X and Y be Banach spaces and let B 2

B(X,Y ) be one-to-one and onto (i.e. Ker(B) = {0} and Ran(B) = Y ).Then the inverse operator B�1 : Y ! X is bounded, i.e. B�1

2 B(Y,X).

Proof: The proof of this fundamental result can be found in any textbook onfunctional analysis. 2

Let X and Y be vector spaces and let an operator B : X ! Y have aright inverse and a left inverse operators B�1

r, B�1

l: Y ! X,

B�1l

B = IX , BB�1r

= IY .

(By I we always denote the identity operator: Ix = x, 8x 2 X. Subscript(if any) indicates the space on which the identity operator acts.) Then Bhas a two-sided inverse operator B�1 = B�1

r= B�1

l. Indeed,

B�1r

= IXB�1r

= (B�1l

B)B�1r

= B�1l

(BB�1r

) = B�1l

IY = B�1l

.

2.3. Lemma Let X, Y and Z be vector spaces.(i) If operators B : X ! Y and T : Y ! Z are invertible, then TB : X ! Zis invertible too and (TB)�1 = B�1T�1.(ii) If operators B, T : X ! X commute: TB = BT , then TB : X ! X isinvertible if and only if both B and T are invertible.(iii) If operators B, T : X ! X commute and B is invertible, then B�1 andT also commute.

Proof: (i)

B�1T�1TB = B�1B = IX , TBB�1T�1 = TT�1 = IZ .

(ii) According to (i) we have to prove only that the invertibility of TB impliesthe invertibility of B and T . Let S : X ! X be the inverse of TB, i.e.STB = TBS = I. It is clear that ST is a left inverse of B. Since B andT commute, we have BTS = I, i.e. TS is a right inverse of B. Hence Bis invertible and its inverse equals ST = TS. Similarly we prove that T isinvertible.(iii)

B�1T = B�1TBB�1 = B�1BTB�1 = TB�1. 2

7

2.4. Lemma Let X be a Banach space, B 2 B(X) and kBk < 1. ThenI �B is invertible,

(I �B)�1 =1X

n=0

Bn (2.1)

and k(I �B)�1k 1/(1� kBk).

Proof: It follows from (1.2) that

kBnk kBn�1

kkBk · · · kBkn, 8n 2 IN.

Hence the series in the right-hand side of (2.1) is absolutely convergent and,consequently, convergent in B(X) (see Theorems 1.10 and 1.15). Let usdenote its sum by R 2 B(X). We have

(I �B)R = (I �B) limN!1

NX

n=0

Bn = limN!1

(I �BN+1) = I,

because kBN+1k kBkN+1

! 0 as N ! 1. Analogously we prove thatR(I �B) = I. Thus (I �B)�1 = R 2 B(X). Further,

k(I �B)�1k = lim

N!1

�����

NX

n=0

Bn

����� limN!1

NX

n=0

kBkn = limN!1

1� kBkN+1

1� kBk=

1

1� kBk. 2

2.5. Lemma Let X and Y be Banach spaces, A, B 2 B(X, Y ). Let A beinvertible and kBk < 1/kA�1

k. Then A + B is invertible,

(A + B)�1 =

1X

n=0

(�A�1B)n

!

A�1 = A�1

1X

n=0

(�BA�1)n

!

and

k(A + B)�1k

kA�1k

1� kBkkA�1k

Proof: We have

A + B = A(I + A�1B) = (I + BA�1)A

and kA�1Bk kA�1kkBk < 1, kBA�1

k kBkkA�1k < 1. Now it is left to

apply Lemmas 2.3(i) and 2.4. 2

8

The spectrum

In this subsection we will deal only with complex vector spaces, i.e. withthe case IF = C, if it is not stated otherwise.

2.6. Definition Let X be a Banach space and B 2 B(X). The resol-vent set ⇢(B) of the operator B is defined to be the set of all � 2 C suchthat B��I has an inverse operator (B��I)�1

2 B(X). The spectrum �(B)of the operator B is the complement of ⇢(B):

�(B) = C\⇢(B),

i.e. �(B) is the set of all � 2 C such that B � �I is not invertible on X.A complex number � is called an eigenvalue of B if there exists x 2 X\{0}

such that Bx = �x. In this case x is called an eigenvector of B correspondingto the eigenvalue �.

2.7. Lemma All eigenvalues of B belong to �(B).

Proof: Suppose � is an eigenvalue and x 6= 0 is a corresponding eigenvectorof B. Then (B � �I)x = 0. Since we have also (B � �I)0 = 0, the operatorB � �I is not one–to–one. So, B � �I is not invertible on X, i.e. � 2 �(B).2

On the other hand the set of all eigenvalues may be smaller than the spec-trum.

2.8. Examples1. If the space X is finite-dimensional then the spectrum of B 2 B(X) co-incides with the set of all its eigenvalues, i.e. with the set of zeros of thedeterminant of a matrix corresponding to B � �I (with respect to a givenbasis of X).2. Let X be one of the Banach spaces C([0, 1]), Lp([0, 1]), 1 p 1, andlet B be defined by the formula

Bf(t) = tf(t), t 2 [0, 1].

Then �(B) = [0, 1] but B does not have eigenvalues (exercise).

9

2.9. Lemma Let X be a Banach space and B 2 B(X). Then �(B) isa compact set and

�(B) ⇢ {� 2 C : |�| kBk}. (2.2)

Proof: Suppose |�| > kBk. Then

kBk < 1/|�|�1 = 1/k � ��1Ik = 1/k(��I)�1k.

Hence, according to Lemma 2.5 the operator B � �I is invertible, i.e. � /2�(B). So, we have proved (2.2).Let us take an arbitrary �0 2 ⇢(B). Then for any � 2 C such that |���0| <1/k(B � �0I)�1

k we conclude from Lemma 2.5 and the representation

B � �I = (B � �0I) + (�0 � �)I (2.3)

that the operator B � �I is invertible, i.e. � 2 ⇢(B). Hence ⇢(B) is an openset and its complement �(B) is closed. So, �(B) is a bounded (see (2.2))closed set, i.e. a compact set. 2

2.10. Definition Let X be a Banach space and B 2 B(X). The operator-valued function

⇢(B) 3 � ! R(B; �) := (B � �I)�12 B(X)

is called the resolvent of the operator B.

2.11. Lemma (the resolvent equation)

R(B; �)�R(B; �0) = (�� �0)R(B; �)R(B; �0), 8�, �0 2 ⇢(B).

Proof:

R(B; �)�R(B; �0) = (B � �I)�1� (B � �0I)�1 =

(B � �I)�1((B � �0I)� (B � �I))(B � �0I)�1 =

(�� �0)R(B; �)R(B; �0). 2

2.12. Lemma If operators A, B 2 B(X) commute: AB = BA, then forany � 2 ⇢(B) and µ 2 ⇢(A) the operators A, B, R(A; µ) and R(B; �) all

10

commute with each other.Proof: It is clear that the operators A, B, A�µI and B��I commute. Nowthe desired result follows from Lemma 2.3(iii). 2

Taking A = B in the last lemma we obtain the following result.2.13. Corollary For any �, µ 2 ⇢(B) the operators B, R(B; �) and R(B; µ)commute. 2

For any normed space X we denote by X⇤ its dual (= adjoint) space, i.e.the space of all bounded linear functionals on X, i.e. B(X, IF).

2.14. Theorem Let Z be a complex Banach space, ⌦ ⇢ C be an opensubset of the complex plane and F : ⌦ ! Z be a Z–valued function. Thenthe following statements are equivalent:(i) for any �0 2 ⌦ there exists the derivative

dF

d�(�0) = F 0(�0) := lim

�!�0

1

�� �0(F (�)� F (�0)) 2 Z,

i.e. ����1

�� �0(F (�)� F (�0))� F 0(�0)

����! 0 as � ! �0;

(ii) any �0 2 ⌦ has a neighbourhood where F (�) can be represented by anabsolutely convergent series

F (�) =1X

n=0

(�� �0)nFn(�0), Fn(�0) 2 Z;

(iii) for any G 2 Z⇤ the complex–valued function

⌦ 3 � 7! G(F (�)) 2 C

is holomorphic (= analytic) in ⌦.If Z = B(X,Y ) for some Banach spaces X and Y , then for the operator–valued function F the above statements are equivalent to(iv) for any x 2 X and g 2 Y ⇤ the complex–valued function

⌦ 3 � 7! g(F (�)x) 2 C

11

is holomorphic (= analytic) in ⌦.

Proof: It is clear that each of the statements (i) and (ii) implies (iii). (Why?)It is also obvious that (iii) implies (iv). Indeed, for any x 2 X and g 2 Y ⇤

the mappingB(X, Y ) 3 B 7! g(Bx) 2 C

is a bounded linear functional on B(X, Y ). The opposite implications arevery non–trivial. We will not give the proof here. It can be found, e.g., in E.Hille & R.S. Phillips, Functional Analysis and Semi–Groups, Ch. III, Sect.2. 2

2.15. Definition A vector–valued (operator–valued) function is called holo-morphic (= analytic) in ⌦ if it satisfies the conditions (i)–(iii) (conditions(i)–(iv)) of the above theorem.

2.16. Remark We will not use the non–trivial part of Theorem 2.14. Wewill only use the fact that each of the statements (i) and (ii) implies (iii) and(iv). In the proof of the following theorem we will check that the resolventsatisfies both (i) and (ii).

2.17. Theorem The B(X)–valued function R(B; ·) is analytic in ⇢(B) andhas the following properties

dR(B; �)

d�

!

�=�0

= R2(B; �0), 8�0 2 ⇢(B), (2.4)

��R(B; �) ! I as |�|!1, (2.5)

kR(B; �)k �1

d(�, �(B)), � 2 ⇢(B), (2.6)

where d(�, K) := inf{|�� µ| : µ 2 K} denotes the distance of � from K.

Proof: Let us take an arbitrary �0 2 ⇢(B). It follows from Lemma 2.5 and theproof of Lemma 2.9 that the function R(B; ·) is bounded in a neighbourhoodof �0. From the resolvent equation (Lemma 2.11) we obtain

R(B; �) ! R(B; �0) as � ! �0

12

and

lim�!�0

R(B; �)�R(B; �0)

�� �0= lim

�!�0

R(B; �)R(B; �0) = R2(B; �0).

Thus (2.4) is valid for any point �0 2 ⇢(B), i.e R(B; ·) is analytic in ⇢(B).Note that (2.3) and Lemma 2.5 imply the following Taylor expansion

R(B; �) =1X

n=0

(�� �0)nRn+1(B; �0), if |�� �0| < 1/k(B � �0I)�1

k. (2.7)

Further,

k � �R(B; �)� Ik = k(I � ��1B)�1� Ik =

�����

1X

n=1

��nBn

�����

1X

n=1

|�|�nkBkn =

|�|�1kBk

1� |�|�1kBk=

kBk

|�|� kBk! 0, as |�|!1

(see Lemma 2.4) and (2.5) is proved.Let us take an arbitrary �0 2 ⇢(B). The proof of Lemma 2.9 implies that� 2 ⇢(B) if |���0| < 1/kR(B; �0)k. Hence d(�0, �(B)) � 1/kR(B; �0)k, i.e.

kR(B; �0)k �1

d(�0, �(B)), 8�0 2 ⇢(B). 2

2.18. Lemma Let X be a normed space. Then for any z 2 X there existsg 2 X⇤ such that g(z) = kzk and kgk = 1.

Proof: Let X0 = lin{z} be the one–dimensional subspace of X spannedby z:

X0 := {↵z| ↵ 2 IF}.

We define a functional g0 : X0 ! IF by the equality g0(↵z) = ↵kzk. It is clearthat g0(z) = kzk and kg0k = 1. (Why?) Due to the Hahn–Banach theorem g0

can be extended to a functional g 2 X⇤ such that g(z) = kzk and kgk = 1. 2

2.19. Lemma �(B) 6= ; for any B 2 B(X).

13

Proof: Let us suppose that �(B) = ; and take arbitrary x 2 X\{0}, g 2 X⇤.It follows from Theorem 2.17 that the function

f(�) := g(R(B; �)x)

is analytic in ⇢(B) = C and f(�) ! 0 as |�| ! 1 (see (2.5) and alsoTheorem 2.14 and Remark 2.16). Liouville’s theorem implies f ⌘ 0. Let ustake g 2 X⇤ such that g(R(B; �0)x) = kR(B; �0)xk for the given x 2 X\{0}and some �0 2 C (see Lemma 2.18). Then we obtain

0 = f(�0) = g(R(B; �0)x) = kR(B; �0)xk,

i.e. R(B; �0)x = 0. Consequently x = (B � �0I)R(B; �0)x = 0 for x 2X\{0}. The obtain contradiction shows that �(B) cannot be empty. 2

Combining Lemmas 2.9, 2.19 and Theorem 2.17 we obtain the following re-sult.

2.20. Theorem Let X be a Banach space and B 2 B(X). The spec-trum of B is a non-empty compact set contained in the closed disk {� 2C : |�| kBk} and the resolvent (B � �I)�1 is an analytic B(X)–valuedfunction on C\�(B). 2

2.21. Remark Let (ak) be a sequence of real numbers. We will use thefollowing notation:

lim infn!1

an := limn!1

infk�n

ak, lim supn!1

an := limn!1

supk�n

ak.

These limits exist (finite or infinite) because bn := infk�n ak and cn :=sup

k�nak are monotone. The Sandwich Theorem implies that the limit

limn!1 an exists iflim infn!1

an = lim supn!1

an.

2.22. Definition Let B 2 B(X). The spectral radius r(B) of the operatorB is defined by the equality

r(B) := sup{|�| : � 2 �(B)}.

14

This is the radius of the minimal disk centred at 0 and containing �(B).

2.23. Theorem (the spectral radius formula) Let X be a Banach spaceand B 2 B(X). Then

r(B) = limn!1

kBnk

1/n. (2.8)

Proof: Let us take an arbitrary � 2 �(B). We have

Bn� �nI = (B � �I)(�n�1I + �n�2B + · · ·+ �Bn�2 + Bn�1).

Since B��I is not invertible and the operators in the RHS commute, Bn��nI

is not invertible (see Lemma 2.3(ii)). So, � 2 �(B) implies �n2 �(Bn).

Consequently

r(B)n = (sup{|�| : � 2 �(B)})n = sup{|�|n : � 2 �(B)} r(Bn).

Hence, according to (2.2) (with Bn instead of B)

r(B) (r(Bn))1/n kBn

k1/n,

and thereforer(B) lim inf

n!1kBn

k1/n.

Now it is su�cient to prove that

lim supn!1

kBnk

1/n r(B). (2.9)

If � 2 C is such that |�| > kBk, then � /2 �(B) and

(B � �I)�1 = ���1(I � ��1B)�1 = �

1X

n=0

��n�1Bn (2.10)

(see Lemma 2.4). Let us take arbitrary x 2 X, g 2 X⇤ and consider thefunction

f(�) := g(R(B; �)x).

If |�| > kBk then (2.10) implies the following Laurent expansion

f(�) = �

1X

n=0

��n�1g(Bnx). (2.11)

15

Since R(B; ·) is analytic in C\�(B) (see Theorem 2.17), so is f (see Theorem2.14 and Remark 2.16). Hence, (2.11) is valid if |�| > r(B). Taking � =aei✓, a > r(B) and integrating the series for �m+1f(�) term by term withrespect to ✓ we have

Z 2⇡

0am+1ei(m+1)✓f(aei✓)d✓ = �

Z 2⇡

0

1X

n=0

am�nei(m�n)✓g(Bnx)d✓ =

�

1X

n=0

am�ng(Bnx)Z 2⇡

0ei(m�n)✓d✓ = �2⇡g(Bmx).

Thus

|g(Bmx)| 1

2⇡

Z 2⇡

0am+1

|g(R(B; aei✓)x)|d✓ am+1M(a)kgkkxk,

whereM(a) := sup

0✓2⇡

kR(B; aei✓)k

(see (1.1)). Taking g 2 X⇤ such that g(Bmx) = kBmxk, kgk = 1 (see Lemma2.18) we obtain

kBmxk am+1M(a)kxk, 8x 2 X,

i.e.kBm

k am+1M(a), 8m 2 IN.

Consequentlylim supm!1

kBmk

1/m a, if a > r(B),

i.e. we have proved (2.9). 2

The functional calculus

Let p(�) =P

N

n=0 an�n be a polynomial and let B 2 B(X). It is naturalto define p(B) by the equality p(B) =

PN

n=0 anBn. It turns out that onecan define f(B) for any complex–valued function f which is analytic in someneighbourhood of �(B).

Let f be a complex–valued function which is analytic in a disk {� 2 C :

16

|�| < rf}, where rf > 0 is some given number. Then we have the followingTaylor expansion

f(�) =1X

n=0

an�n, |�| < rf ,

where an = f (n)(0)/n! and the series is absolutely convergent.

2.24. Definition Let X be a Banach space, B 2 B(X) and r(B) < rf ,i.e. �(B) ⇢ {� 2 C : |�| < rf}. Then we define f(B) by the equality

f(B) :=1X

n=0

anBn. (2.12)

We need to check that this definition makes sense. Let us take " > 0 suchthat r(B) + " < rf . It follows from the spectral radius formula that for su�-ciently large n we have kBn

k < (r(B)+")n. Therefore the power series (2.12)is absolutely convergent. Hence, f(B) 2 B(X) (see Theorems 1.10 and 1.15).

A disadvantage of the above definition is that f has to be analytic in somedisk containing �(B). This does not seem fair if �(B) is far from being adisk, e.g. is a curve. So, it is natural to look for an alternative definition off(B).

2.25. Definition A bounded open set ⌦ ⇢ C is called admissible if itsboundary @⌦ consists of a finite number of smooth closed pairwise disjointcurves. The orientation of @⌦ is chosen in such a way that ⌦ remains on theleft as we move along @⌦ in the positive direction.

Let ⌦ be an admissible set and let f be complex–valued function whichis analytic in some neighbourhood of the closure Cl(⌦) of ⌦. Then we canwrite the Cauchy integral formula from Complex Analysis:

f(�0) =1

2⇡i

Z

@⌦

f(�)

�� �0d�, �0 2 ⌦.

The formal substitution �0 = B gives

f(B) =1

2⇡i

Z

@⌦f(�)(�I �B)�1d�.

17

The RHS is well defined and belongs to B(X) if @⌦T

�(B) = ;. Here andbelow the integrals are understood as norm limits of the corresponding Rie-mann sums. This motivates the following definition.

2.26. Definition Let X be a Banach space, B 2 B(X). Suppose f isanalytic in some (open) neighbourhood �f of �(B) and ⌦ is an admissibleset such that

�(B) ⇢ ⌦ ⇢ Cl(⌦) ⇢ �f . (2.13)

Then

f(B) := �1

2⇡i

Z

@⌦f(�)R(B; �)d�. (2.14)

It is always possible to find an admissible set satisfying (2.13). Indeed, �(B)is a closed bounded subset of the open set �f . Therefore d(�(B), @�f ) > 0.Here d(K1, K2) := inf{|�1 � �2| : �1 2 K1, �2 2 K2} denotes the distancebetween two sets.

2.27. Proposition The RHS of (2.14) does not depend on the choice ofan admissible set satisfying (2.13).

Proof: Let ⌦ and ⌦1 be admissible sets satisfying (2.13). Then there ex-ists an admissible set ⌦0 such that

�(B) ⇢ ⌦0 ⇢ Cl(⌦0) ⇢ ⌦\

⌦1 ⇢ �f .

It is su�cient to prove that

1

2⇡i

Z

@⌦f(�)R(B; �)d� =

1

2⇡i

Z

@⌦0

f(�)R(B; �)d�, (2.15)

because the same argument will apply to the pair ⌦1 and ⌦0.Let us take arbitrary x 2 X, g 2 X⇤ and prove that

1

2⇡i

Z

@⌦f(�)g(R(B; �)x)d� =

1

2⇡i

Z

@⌦0

f(�)g(R(B; �)x)d�. (2.16)

The function � 7! f(�)g(R(B; �)x) 2 C is analytic in a neighbourhood ofCl(⌦)\⌦0. Since the boundary of this set equals @⌦

S@⌦0, (2.16) follows

from the Cauchy theorem.

18

Thus,

g✓

1

2⇡i

Z

@⌦f(�)R(B; �)xd��

1

2⇡i

Z

@⌦0

f(�)R(B; �)xd�◆

= 0, 8g 2 X⇤.

Consequently

1

2⇡i

Z

@⌦f(�)R(B; �)xd� =

1

2⇡i

Z

@⌦0

f(�)R(B; �)xd�, 8x 2 X,

(see Lemma 2.18), i.e. (2.15) holds. 2

Our aim now is to show that Definition 2.26 does not contradict the “com-mon sense”, i.e. that in the situations where we now what f(B) is, (2.14)gives the same result.

2.28. Lemma Let ⌦ be an admissible neighbourhood of �(B). Then forany �0 62 Cl(⌦) we have

(B � �0I)m = �1

2⇡i

Z

@⌦(�� �0)

mR(B; �)d�, 8m 2 Z. (2.17)

In particular

R(B; �0) = (B � �0I)�1 = �1

2⇡i

Z

@⌦(�� �0)

�1R(B; �)d�.

Proof: Let us denote the RHS of (2.17) by Am. Since �0 62 Cl(⌦), the functionf(�) := (�� �0)m is analytic in some neighbourhood of Cl(⌦) and

1

2⇡i

Z

@⌦(�� �0)

md� = 0.

Using this equality and the resolvent equation (Lemma 2.11)

R(B; �) = R(B; �0) + (�� �0)R(B; �0)R(B; �),

we obtain

Am = �R(B; �0)1

2⇡i

Z

@⌦(�� �0)

md��

R(B; �0)1

2⇡i

Z

@⌦(�� �0)

m+1R(B; �)d� = R(B; �0)Am+1.

19

HenceAm = (B � �0I)�1Am+1, m = 0,±1,±2, . . .

This recursion formula shows that (2.17) follows from the case m = 0. Wethus have to prove that

�1

2⇡i

Z

@⌦R(B; �)d� = I.

According to Proposition 2.27 it will su�ce to prove that

�1

2⇡i

Z

|�|=r

R(B; �)d� = I

for su�ciently large r > 0. If |�| = r > kBk we have the following absolutelyand uniformly convergent expansion

(B � �I)�1 = �

1X

n=0

��n�1Bn

(see (2.10)). Termwise integration gives

�1

2⇡i

Z

|�|=r

R(B; �)d� =1X

n=0

Bn1

2⇡i

Z

|�|=r

��n�1d� = B0 = I. 2

2.29. Corollary Let ⌦ be an admissible neighbourhood of �(B). Then forany polynomial p we have

p(B) = �1

2⇡i

Z

@⌦p(�)R(B; �)d�.

In particular

Bn = �1

2⇡i

Z

@⌦�nR(B; �)d�, 8n 2 IN

[{0}.

Proof: It is su�cient to represent p(�) in the form p(�) =P

N

m=0 cm(���0)m,�0 62 Cl(⌦), and apply the last lemma. 2

2.30. Theorem Let f be analytic in a disk �f = {� 2 C : |�| < rf},

20

where rf > r(B). Then Definitions 2.24 and 2.26 give the same result.

Proof: Termwise integration gives

�1

2⇡i

Z

@⌦f(�)R(B; �)d� = �

1

2⇡i

Z

@⌦

1X

n=0

an�n

!

R(B; �)d� =

1X

n=0

an

✓�

1

2⇡i

Z

@⌦�nR(B; �)d�

◆=

1X

n=0

anBn

(see Corollary 2.29). 2

Let f and g be functions analytic in some (open) neighbourhoods �f and�g of �(B). It follows directly from Definition 2.26 that

(↵f + �g)(B) = ↵f(B) + �g(B), 8↵, � 2 C. (2.18)

2.31. Theorem (fg)(B) = f(B)g(B).

Proof: Let us take admissible sets ⌦f and ⌦g such that

�(B) ⇢ ⌦f ⇢ Cl(⌦f ) ⇢ ⌦g ⇢ Cl(⌦g) ⇢ �f

\�g. (2.19)

Using the resolvent equation (Lemma 2.11)

R(B; �)R(B; µ) = �R(B; �)

µ� ��

R(B; µ)

�� µ, � 6= µ,

we obtain

f(B)g(B) =

�1

2⇡i

Z

@⌦f

f(�)R(B; �)d�

!

�1

2⇡i

Z

@⌦g

g(µ)R(B; µ)dµ

!

=

�1

2⇡i

Z

@⌦f

f(�)

�1

2⇡i

Z

@⌦g

g(µ)R(B; �)R(B; µ)dµ

!

d� =

�1

2⇡i

Z

@⌦f

f(�)R(B; �)

1

2⇡i

Z

@⌦g

g(µ)

µ� �dµ

!

d��

�1

2⇡i

Z

@⌦f

f(�)

1

2⇡i

Z

@⌦g

g(µ)

�� µR(B; µ)dµ

!

d�.

21

Since � 2 ⌦g for any � 2 @⌦f and µ 62 Cl(⌦f ) for any µ 2 @⌦g (see (2.19)),the Cauchy theorem implies

1

2⇡i

Z

@⌦g

g(µ)

µ� �dµ = g(�),

1

2⇡i

Z

@⌦f

f(�)

�� µd� = 0.

Therefore

f(B)g(B) = �1

2⇡i

Z

@⌦f

f(�)g(�)R(B; �)d��

1

2⇡i

Z

@⌦g

1

2⇡i

Z

@⌦f

f(�)

�� µd�

!

g(µ)R(B; µ)dµ = (fg)(B)� 0 = (fg)(B). 2

2.32. Corollary f(B)g(B) = g(B)f(B).

Proof: f(B)g(B) = (fg)(B) = (gf)(B) = g(B)f(B). 2

2.33. Theorem Let A, B 2 B(X) and AB = BA. Suppose f and g areanalytic in some neighbourhoods �f and �g of �(A) and �(B) correspond-ingly. Then f(A)g(B) = g(B)f(A).

Proof: Let us take admissible sets ⌦f and ⌦g such that

�(A) ⇢ ⌦f ⇢ Cl(⌦f ) ⇢ �f , �(B) ⇢ ⌦g ⇢ Cl(⌦g) ⇢ �g.

Since AB = BA implies R(A; µ)R(B; �) = R(B; �)R(A; µ) for any � 2 ⇢(B)and µ 2 ⇢(A) (see Lemma 2.12), we have

f(A)g(B) =

�1

2⇡i

Z

@⌦f

f(µ)R(A; µ)dµ

!

�1

2⇡i

Z

@⌦g

g(�)R(B; �)d�

!

=

�1

4⇡2

Z

@⌦f

Z

@⌦g

f(µ)g(�)R(A; µ)R(B; �)d�dµ =

�1

4⇡2

Z

@⌦g

Z

@⌦f

f(µ)g(�)R(B; �)R(A; µ)dµd� =

�1

2⇡i

Z

@⌦g

g(�)R(B; �)d�

!

�1

2⇡i

Z

@⌦f

f(µ)R(A; µ)dµ

!

= g(B)f(A). 2

22

2.34. Theorem (the spectral mapping theorem) Let f be analyticin some neighbourhood of �(B). Then

�(f(B)) = f(�(B)) := {f(�) : � 2 �(B)}.

Proof: Let us prove that

�(f(B)) ⇢ f(�(B)). (2.20)

Let take any ⇣ 2 C\f(�(B)). Then f � ⇣ 6= 0 in some neighbourhood of�(B). Hence the function g := 1/(f � ⇣) is analytic in a neighbourhood of�(B). Now Theorem 2.31 implies

g(B)(f(B)� ⇣I) =

1

f � ⇣(f � ⇣)

!

(B) = 1(B) = I.

Similarly (f(B) � ⇣I)g(B) = I. Therefore g(B) = R(f(B); ⇣) and ⇣ 62

�(f(B)). Thus⇣ 62 f(�(B)) =) ⇣ 62 �(f(B)),

i.e. (2.20) holds.It is left to prove that

f(�(B)) ⇢ �(f(B)). (2.21)

Let us take an arbitrary µ 2 �(B) and consider the function

h(�) :=

((f(�)� f(µ))/(�� µ) if � 6= µ,f 0(µ) if � = µ.

It is clear that h is analytic in some neighbourhood of �(B) and f(�)�f(µ) =(�� µ)h(�). It follows from Theorem 2.31 that

f(B)� f(µ)I = (B � µI)h(B).

The operator B�µI is not invertible, because µ 2 �(B). Since the operatorsB � µI and h(B) commute (see Corollary 2.32), the operator f(B)� f(µ)Icannot be invertible (see Theorem 2.3(ii)). Thus

µ 2 �(B) =) f(µ) 2 �(f(B)),

i.e. (2.21) holds. 2

23

2.35. Theorem Let f be analytic in some neighbourhood �f of �(B) and gbe analytic in some neighbourhood �g of f(�(B)). Then for the compositiong � f (defined by (g � f)(�) := g(f(�))) we have

(g � f)(B) = g(f(B)).

Proof: The RHS of the last equality is well defined due to the spectral map-ping theorem.

Let us take admissible sets ⌦f and ⌦g such that

�(B) ⇢ ⌦f ⇢ Cl(⌦f ) ⇢ �f ,

�(f(B)) = f(�(B)) ⇢ f(Cl(⌦f )) ⇢ ⌦g ⇢ Cl(⌦g) ⇢ �g.

For any ⇣ 2 @⌦g the function h⇣ := 1/(f � ⇣) is analytic in some neigh-bourhood of Cl(⌦f ). Exactly as in the proof of Theorem 2.34 we prove thath⇣(B) = R(f(B); ⇣). Using the Cauchy theorem we obtain

g(f(B)) = �1

2⇡i

Z

@⌦g

g(⇣)R(f(B); ⇣)d⇣ = �1

2⇡i

Z

@⌦g

g(⇣)h⇣(B)d⇣ =

�1

2⇡i

Z

@⌦g

g(⇣)

�1

2⇡i

Z

@⌦f

1

f(�)� ⇣R(B; �)d�

!

d⇣ =

�1

2⇡i

Z

@⌦f

R(B; �)

1

2⇡i

Z

@⌦g

g(⇣)

⇣ � f(�)d⇣

!

d� =

�1

2⇡i

Z

@⌦f

g(f(�))R(B; �)d� = �1

2⇡i

Z

@⌦f

(g � f)(�)R(B; �)d� =

(g � f)(B). 2

24

Summary

Suppose X is a Banach space and B 2 B(X). Let A(B) be the set ofall functions f which are analytic in some neighbourhood of �(B). (Theneighbourhood can depend on f 2 A(B).) The map

A(B) 3 f 7! f(B) 2 B(X)

defined by (2.14) has the following properties:(i) if f, g 2 A(B), ↵, � 2 C, then ↵f + �g, fg 2 A(B) and (↵f + �g)(B) =↵f(B) + �g(B), (fg)(B) = f(B)g(B);(ii) if f has the power series expansion f(�) =

P1

n=0 cn�n, valid in a neigh-bourhood of �(B), then f(B) =

P1

n=0 cnBn;(iii) �(f(B)) = f(�(B)) (the spectral mapping theorem);(iv) if f 2 A(B) and g 2 A(f(B)), then (g � f)(B) = g(f(B)).

Linear operator equations

Let X and Y be normed spaces and A 2 B(X, Y ). The adjoint operatorA⇤ : Y ⇤

! X⇤ is defined by the equality

(A⇤g)(x) := g(Ax), 8g 2 Y ⇤, 8x 2 X.

It is not di�cult to prove that A⇤2 B(Y ⇤, X⇤) and kA⇤

k = kAk (FunctionalAnalysis I). It turns out that A⇤ plays a very important role in the study ofthe following linear operator equation

Ax = y, y 2 Y. (2.22)

2.36. Lemma For the closure of the range of A we have

Ran(A) ⇢ {y 2 Y : g(y) = 0, 8g 2 Ker(A⇤)}. (2.23)

Proof: Let us take an arbitrary y 2 Ran(A) and g 2 Ker(A⇤). There existxn 2 X such that yn := Axn converge to y as n !1. Hence

g(y) = g( limn!1

Axn) = limn!1

g(Axn) = limn!1

(A⇤g)(xn) = limn!1

0(xn) = 0. 2

25

2.37. Corollary (a necessary condition of solvability) If (2.22) has asolution then g(y) = 0, 8g 2 Ker(A⇤).

Proof: (2.22) is solvable if and only if y 2 Ran(A). 2

Now we are going to prove that (2.23) is in fact an equality. We start withthe following corollary of the Hahn–Banach theorem.

2.38. Lemma Let Y0 be a closed linear subspace of Y and y0 2 Y \Y0.Then there exists g 2 Y ⇤ such that g(y0) = 1, g(y) = 0 for all y 2 Y0 andkgk = 1/d, where d = d(y0, Y0) = infy2Y0 ky0 � yk. (d > 0 because Y0 isclosed.)

Proof: LetY1 := {�y0 + y : � 2 IF, y 2 Y0}.

Define a functional g1 : Y1 ! IF by the equality

g1(�y0 + y) = �.

It is clear that g1 is linear and g1(y0) = 1, g1(y) = 0 for all y 2 Y0. Further,

kg1k = sup�y0+y 6=0

|g1(�y0 + y)|

k�y0 + yk= sup

�y0+y 6=0

|�|

k�y0 + yk=

sup�6=0, y2Y0

|�|

k�y0 + yk= sup

�6=0, y2Y0

1

ky0 + ��1yk= sup

z2Y0

1

ky0 + zk=

1

infz2Y0 ky0 + zk=

1

infw2Y0 ky0 � wk=

1

d.

Due to the Hahn–Banach theorem g1 can be extended to a functional g 2 Y ⇤

such that g(y0) = 1, g(y) = 0 for all y 2 Y0 and kgk = 1/d. 2

2.39. Theorem

Ran(A) = {y 2 Y : g(y) = 0, 8g 2 Ker(A⇤)}.

Proof: According to Lemma 2.36 it is su�cient to prove that

N (A) := {y 2 Y : g(y) = 0, 8g 2 Ker(A⇤)} ⇢ Ran(A).

26

Suppose the contrary: there exists y0 2 N (A), y0 62 Ran(A). Lemma 2.38implies the existence of g 2 Y ⇤ such that g(y0) = 1, g(y) = 0 for all y 2Ran(A). Then

(A⇤g)(x) = g(Ax) = 0, 8x 2 X,

i.e A⇤g = 0, i.e. g 2 Ker(A⇤). Since y0 2 N (A), we have g(y0) = 0. Contra-diction! 2

2.40. Corollary Suppose Ran(A) is closed. Then (2.22) is solvable if andonly if g(y) = 0, 8g 2 Ker(A⇤).

Proof: (2.22) is solvable if and only if y 2 Ran(A) = Ran(A). 2

Projections

2.41. Definition Let X be a normed space. An operator B 2 B(X) iscalled a projection if it is idempotent, i.e. B2 = B.

2.42. Lemma Let P 2 B(X) be a projection. Then Q := I � P is aprojection, PQ = QP = 0 and

Ran(P ) = Ker(Q), Ran(Q) = Ker(P ).

Proof: Q2 = (I � P )2 = I � 2P + P 2 = I � 2P + P = I � P = Q, i.e. Q is aprojection.

PQ = P (I � P ) = P � P 2 = 0 = (I � P )P = QP.

It follows from the equality QP = 0 that Ran(P ) ⇢ Ker(Q) (why?). Onthe other hand for any x 2 Ker(Q) we have x � Px = 0, i.e. x = Px,hence x 2 Ran(P ). Thus Ker(Q) ⇢ Ran(P ). So, Ran(P ) = Ker(Q). Theequality Ran(Q) = Ker(P ) follows from the last one if P is replaced by Q. 2

2.43. Definition Let L, L1, L2 be subspaces of a vector space X. Wesay that L is the direct sum of L1, L2 and write L = L1 � L2 if

L1 \ L2 = {0} and L = L1 + L2 := {y1 + y2 : y1 2 L1, y2 2 L2}.

27

2.44. Proposition Let L, L1, L2 be subspaces of a vector space X. ThenL = L1 � L2 i↵ every element y of L can be uniquely written as y = y1 + y2

with yk 2 Lk.

Proof: Exercise. 2

2.45. Lemma If P 2 B(X) is a projection, then Ran(P ) is closed andX = Ran(P )�Ker(P ).

Proof: Ran(P ) = Ker(I � P ), so Ran(P ) is closed. The equality I =P + (I � P ) and Lemma 2.42 imply

Ran(P ) + Ker(P ) = Ran(P ) + Ran(I � P ) = X.

So, it is left to prove that Ran(P ) \ Ker(P ) = {0}. Let us take an arbi-trary x 2 Ran(P )\Ker(P ). Since x 2 Ran(P ), there exists y 2 X such thatx = Py. Consequently Px = P 2y = Py = x. On the other hand x 2 Ker(P ),i.e. Px = 0. Thus x = 0. 2

2.46. Theorem Let X be a Banach space and P 2 B(X) be a non–trivialprojection, i.e. P 6= 0, I. Then �(P ) = {0, 1}.

Proof: Using the spectral mapping theorem (Theorem 2.34) we obtain

{0} = �(0) = �(P � P 2) = {�� �2 : � 2 �(P )}.

So, � 2 �(P ) =) � = 0 or 1, i.e. �(P ) ⇢ {0, 1}.On the other hand, Ran(P ) 6= {0} and Ran(I � P ) 6= {0} (why?), i.e.Ker(I � P ) 6= {0} and Ker(P ) 6= {0} (see Lemma 2.42). Consequently theoperators P and P � I are not invertible, i.e. {0, 1} ⇢ �(P ). 2

Suppose B 2 B(X) and ⌦ is an admissible set such that @⌦T

�(B) = ;. Considerthe operator

P := �1

2⇡i

Z

@⌦R(B;�)d�. (2.24)

If ⌦T

�(B) = ; then P = 0 (follows from the Cauchy theorem). If �(B) ⇢ ⌦ thenP = 1(B) = I (Lemma 2.28). The only non–trivial case is when there exists a non–emptysubset � of �(B) such that � 6= �(B), � ⇢ ⌦, ⌦

T(�(B)\�) = ; and both � and �(B)\�

are closed. This can happen only if �(B) is not connected. Let an admissible set ⌦0 andopen sets �, �0 be such that

� ⇢ ⌦ ⇢ Cl(⌦) ⇢ �, �(B)\� ⇢ ⌦0 ⇢ Cl(⌦0) ⇢ �0

28

and �T

�0 = ;. Let us consider the function f : �1 := �S

�0 ! C which equals 1 in� and 0 in �0. It is analytic in �1. Since �(B) ⇢ ⌦1 := ⌦

S⌦0 and @⌦1 = @⌦

S@⌦0,

we haveP = �

12⇡i

Z

@⌦R(B;�)d� = �

12⇡i

Z

@⌦1

f(�)R(B;�)d� = f(B).

It is clear that f2 = f . Hence P 2 = P (Theorem 2.31). It follows from the spectralmapping theorem (Theorem 2.34) that �(P ) = �(f(B)) = {0, 1}. Thus P is a non–trivialprojection which commutes with g(B) for any function g analytic in some neighbourhoodof �(B) (Corollary 2.32). P is called the spectral projection corresponding to the spectralset �. Since

g(B)Px = Pg(B)x 2 Ran(P ), 8x 2 X,

andPg(B)x0 = g(B)Px0 = g(B)0 = 0, 8x0 2 Ker(P ),

Ran(P ) and Ker(P ) are invariant under g(B), i.e.

g(B)Ran(P ) ⇢ Ran(P ), g(B)Ker(P ) ⇢ Ker(P ).

Compact operators

2.47. Definition Let X be a normed space. A set K ⇢ X is relatively com-pact if every sequence in K has a Cauchy subsequence. K is called compactif every sequence in K has a subsequence which converges to an element of K.

2.48. Proposition K relatively compact =) K bounded.K compact =) K closed and bounded.Any subset of a relatively compact set is relatively compact.Any closed subset of a compact set is compact.

2.49. Theorem K is relatively compact if and only if it is totally bounded,i.e. for every " > 0, X contains a finite set, called "�net for K, such thatthe finite set of open balls of radius " and centres in the "�net covers K.

Proof: See e.g. C. Go↵man & G. Pedrick, First Course in Functional Anal-ysis, Section 1.11, Lemma 1. 2

29

2.50. Proposition Let X be finite–dimensional. K ⇢ X is relativelycompact i↵ K is bounded. K ⇢ X is compact i↵ K is closed and bounded.

2.51. Theorem On a finite–dimensional vector space X, any norm k · k

is equivalent to any other norm k ·k0, i.e. there exist constants c1, c2 > 0 such

thatc1kxk kxk

0 c2kxk, 8x 2 X.

Proof : See e.g. E. Kreyszig, Introductory Functional Analysis with Appli-cations, Theorem 2.4-5. 2

2.52. Definition Let X and Y be normed spaces. A linear operatorT : X ! Y is called compact (or completely continuous) if it maps boundedsets of X into relatively compact sets of Y . We will denote the set of allcompact linear operators acting from X into Y by Com(X,Y ).

Since any relatively compact set is bounded, T 2 Com(X, Y ) maps boundedsets into bounded sets. In particular the image T (SX) of the unit ball

SX := {x 2 X : kxk 1} (2.25)

is bounded, i.e T 2 B(X, Y ). Thus

Com(X, Y ) ⇢ B(X, Y ).

It follows from Definition 2.47, that a linear operator T : X ! Y is compacti↵ for any bounded sequence xn 2 X, n 2 IN, the sequence (Txn)n2IN has aCauchy subsequence.

2.53. Lemma Let X and Y be normed spaces. A linear operator T :X ! Y is compact i↵ T (SX) is relatively compact.

Proof: Since SX is bounded, T 2 Com(X, Y ) implies that T (SX) is rela-tively compact (see Definition 2.52).

Hence we need to prove that if T (SX) is relatively compact then T 2

Com(X, Y ). It is clear that for any t > 0 the set tT (SX) = T (tSX) is rel-atively compact. (Why?) Let W ⇢ X be an arbitrary bounded set. ThenW ⇢ tSX if t � sup{kxk : x 2 W}. Using the obvious fact that any subset

30

of a relatively compact set is relatively compact (see Proposition 2.48) weconclude from T (W ) ⇢ T (tSX) that T (W ) is relatively compact. 2

2.54. Theorem Let X, Y and Z be normed spaces.(i) If T1, T2 2 Com(X, Y ) and ↵, � 2 IF, then ↵T1 + �T2 2 Com(X, Y ).(ii) If T 2 Com(X,Y ), A 2 B(Z,X) and B 2 B(Y, Z), then TA 2 Com(Z, Y )and BT 2 Com(X,Z).(iii) If Tk 2 Com(X, Y ), k 2 IN, and kT � Tkk ! 0 as k ! 1, thenT 2 Com(X,Y ).

Proof: (i). Let xn 2 X, n 2 IN be an arbitrary bounded sequence. Itfollows from the compactness of T1 that (T1xn) has a Cauchy subsequence(T1x(1)

n). Using the compactness of T2 we can extract a Cauchy subsequence

(T2x(2)n

) from the sequence (T2x(1)n

). It is clear that (T1x(2)n

) is a Cauchysequence. Consequently any bounded sequence xn 2 X, n 2 IN has a sub-sequence (x(2)

n) such that

⇣(↵T1 + �T2)x(2)

n

⌘is a Cauchy sequence. Thus

↵T1 + �T2 2 Com(X, Y ).(ii) Let zn 2 Z, n 2 IN be an arbitrary bounded sequence. Then (Azn)is a bounded sequence in X and it follows from the compactness of T that(TAzn) has a Cauchy subsequence (TAz(1)

n). So, TA 2 Com(Z, Y ). For any

bounded sequence xn 2 X, n 2 IN, the sequence (Txn) has a Cauchy subse-quence (Tx(1)

n). It is easy to see that (BTx(1)

n) is a Cauchy sequence. Hence

BT 2 Com(X, Z).(iii) Let xn 2 X, n 2 IN be an arbitrary bounded sequence. Since T1 is com-pact (T1xn) has a Cauchy subsequence (T1x(1)

n). Using the compactness of T2

we can extract a Cauchy subsequence (T2x(2)n

) from the sequence (T2x(1)n

). Itis clear that (T1x(2)

n) is a Cauchy sequence. Repeating this process we obtain

sequences (x(m)n

)n2IN in X such that (x(m1)n

) is a subsequence of (x(m0)n

) ifm1 > m0, and (Tkx(m)

n)n2IN is a Cauchy sequence if k m. Let us consider

the diagonal sequence (x(n)n

). It is easily seen that (x(n)n

) is a subsequence of(xn). (Tkx(n)

n)n2IN is a Cauchy sequence for any k 2 IN, because (Tkx(n)

n)n�k is

a subsequence of the Cauchy sequence (Tkx(k)n

)n2IN. Let us prove that (Tx(n)n

)is also Cauchy. We have

kTx(n)n� Tx(m)

mk kTx(n)

n� Tkx

(n)nk+ kTkx

(n)n� Tkx

(m)mk+

kTkx(m)m� Tx(m)

mk kT � Tkk

⇣kx(n)

nk+ kx(m)

mk

⌘+ kTkx

(n)n� Tkx

(m)mk.

31

For a given " > 0, we choose k so large that

kT � Tkk <"

3M,

where M := supn2IN kxnk. Then we determine n0 so that

kTkx(n)n� Tkx

(m)mk <

"

3, 8n, m � n0.

Now we havekTx(n)

n� Tx(m)

mk < ", 8n, m � n0.

Hence (Tx(n)n

) is a Cauchy sequence, i.e. T 2 Com(X, Y ). 2

Propositions (i) and (iii) of the last theorem mean that Com(X, Y ) is aclosed linear subspace of B(X, Y ). Let Y = X. It follows from (ii) that

Com(X) := Com(X, X)

is actually what we call an ideal in the Banach algebra B(X).

2.55. Definition We say that T 2 B(X, Y ) is a finite rank operator ifRan(T ) is finite–dimensional.

It is clear that any finite rank operator is compact (see Proposition 2.50).Theorem 2.54(iii) implies that a limit of a convergent sequence of finite rankoperators is compact.

2.56. Theorem Every bounded subset of a normed space X is relativelycompact i↵ X is finite–dimensional.

Proof: See e.g. L.A. Liusternik & V.J. Sobolev, Elements of FunctionalAnalysis, §16). 2

2.57. Corollary The identity operator IX is compact i↵ X is finite–dimensional. 2

2.58. Theorem Let T 2 Com(X,Y ) and let at least one of the spacesX and Y be infinite–dimensional. Then T is not invertible.

Proof: Suppose T is invertible: T�12 B(Y, X). Then operators T�1T = IX

and TT�1 = IY are compact by Theorem 2.54(ii). This however contradictsCorollary 2.57. 2

32

2.59. Corollary Let X be an infinite–dimensional Banach space andT 2 Com(X). Then 0 2 �(T ). 2

2.60. Theorem Any non-zero eigenvalue � of a compact operator T 2

Com(X) has finite multiplicity, i.e. the subspace X� spanned by eigenvec-tors of T corresponding to � is finite–dimensional.

Proof: Any bounded subset ⌦ of X� is relatively compact because Tx =�x, 8x 2 ⌦, i.e. ⌦ = ��1T (⌦), and T is a compact operator. Hence X� isfinite–dimensional by Theorem 2.56. 2

2.61. Theorem Let T 2 Com(X). Then �(T ) is at most countableand has at most one limit point, namely, 0. Any non-zero point of �(T )is an eigenvalue of T , which has finite multiplicity according to Theorem2.60. In a neighbourhood of �0 2 �(T )\{0} the resolvent R(T ;�) admits the followingrepresentation:

R(T ;�) =kX

n=0

(�� �0)�n�1(T � �0I)nP + R(T ;�)

where R(T ;�) is analytic in some neighbourhood of �0, P is the spectral projection cor-responding to �0. P and hence (T � �0I)nP are finite rank operators.

Proof: See e.g. W. Rudin, Functional Analysis, Theorem 4.25 and T. Kato,Perturbation Theory For Linear Operators, Ch. III, §6, Section 5. 2

2.62. Theorem Let X and Y be Banach spaces and T 2 B(X,Y ). ThenT is compact if and only if T ⇤ is compact.

Proof: See W. Rudin, Functional Analysis, Theorem 4.19. 2

Let (X, k · kX) be a normed space and Z be its linear subspace. SupposeZ is equipped with a norm k · kZ . We will say that (Z, k · kZ) is continuouslyembedded in (X, k · kX) if the operator I : (Z, k · kZ) ! (X, k · kX), Ix = x,is continuous, i.e.

kxkX constkxkZ , 8x 2 Z.

If I : (Z, k · kZ) ! (X, k · kX) is compact we say that (Z, k · kZ) is compactlyembedded in (X, k · kX).

33

2.63. Proposition Let (Z, k · kZ) be compactly embedded in (X, k · kX),Y be a normed space and let A : Y ! (Z, k · kZ) be bounded. ThenA 2 Com(Y,X).

Proof: Apply Theorem 2.54(ii) to the operator A = IA. 2

2.64. Example Let m 2 INS{0}, �1 < a < b < 1. We denote by

Cm([a, b]) the space of all m times continuously di↵erentiable functions on[a, b] equipped with the norm

kuk(m) :=mX

j=0

maxt2[a,b]

�����dju(t)

dtj

����� .

Cm([a, b]) is a Banach space. Cm([a, b]) is compactly embedded in Cn([a, b])if m > n. (Let X and Z be some spaces of functions defined on a compactset K. Normally it is reasonable to expect Z to be compactly embedded inX if Z consists of functions which are “more smooth” than functions fromX.) It follows from Proposition 2.63 that if a linear operator T is continuousfrom Cn([a, b]) to Cm([a, b]), m > n, then it is compact in Cn([a, b]). Forexample the operator defined by the formula

(Tu)(t) :=Z

b

a

k(t, s)u(s)ds, k 2 C1([a, b]2),

is bounded from C0([a, b]) = C([a, b]) to C1([a, b]) and hence compact inC([a, b]).

3 The geometry of Hilbert spaces

3.1. Definition An inner (scalar) product space is a vector space H

together with a map (· , ·) : H⇥H! IF such that

(�x + µy, z) = �(x, z) + µ(y, z), (3.1)

(x, y) = (y, x), (3.2)

(x, x) � 0, and (x, x) = 0 () x = 0, (3.3)

for all x, y, z 2 H and �, µ 2 IF.

34

(3.1)–(3.3) imply

(x, �y + µz) = �(x, y) + µ(x, z), (3.4)

(x, 0) = (0, x) = 0. (3.5)

3.2. Examples1. CN , (x, y) :=

PN

k=1 wkxkyk, wk > 0. (For wk = 1 we have the standardinner product on CN .)2. l2, (x, y) :=

P1

k=1 xkyk. (The series converges because of the inequality|xkyk| (|xk|

2 + |yk|2)/2.)

3. C([0, 1]), (f, g) :=R 10 f(t)g(t)dt.

4. C1([0, 1]), (f, g) :=R 10

⇣f(t)g(t) + f 0(t)g0(t)

⌘dt.

3.3. Theorem Every inner product space becomes a normed space by set-ting kxk := (x, x)1/2.

Proof: Linear Analysis. 2

3.4. Theorem (Cauchy–Schwarz inequality)

|(x, y)| kxkkyk, 8x, y 2 H. (3.6)

Proof: Linear Analysis. 2

3.5. Definition A system {x↵}↵2J ⇢ H is called orthogonal if (x↵, x�) = 0,for all ↵, � 2 J such that ↵ 6= �.

3.6. Proposition (Pythagoras’ theorem) If a system of vectors x1, . . . , xn 2

H is orthogonal, then

kx1 + x2 + · · ·+ xnk2 = kx1k

2 + kx2k2 + · · ·+ kxnk

2.

Proof: Express both sides in terms of inner products. 2

3.7. Proposition (polarization identity) If IF = IR then

4(x, y) = kx + yk2� kx� yk2, 8x, y 2 H.

35

If IF = C then

4(x, y) = kx + yk2� kx� yk2 + ikx + iyk2

� ikx� iyk2, 8x, y 2 H.

Proof: The same. 2

3.8. Proposition (parallelogram law)

kx + yk2 + kx� yk2 = 2kxk2 + 2kyk2, 8x, y 2 H.

Proof: The same. 2

According to Proposition 3.8 for the norm induced by an inner productthe parallelogram law holds. Now the following result implies that a norm isinduced by an inner product i↵ the parallelogram law holds.

3.9. Theorem (Jordan – Von Neumann) A norm satisfying the par-allelogram law is derived from an inner product. 2

3.10. Definition A complete inner product space is called a Hilbert space.

So, every Hilbert Space is a Banach space and the whole Banach space theorycan be applied to Hilbert spaces.

3.11. Examples 1. CN with any of the norms of Example 3.2-1 is aHilbert space.2. l2 is a Hilbert space.3. The inner product space from Example 3.2-3 is not a Hilbert space.

Orthogonal complements

3.12. Theorem Let L be a closed linear subspace of a Hilbert space H.For each x 2 H there exists a unique point y 2 L such that

kx� yk = d(x, L) := inf{kx� zk : z 2 L}.

This point satisfies the equality

(x� y, z) = 0, 8z 2 L.

36

Proof: Step I. Let yn 2 L be such that kx�ynk ! d := d(x, L). Applying the parallelogramlaw to the vectors x� yn and x� ym and using the fact that (yn + ym)/2 2 L we have

kyn � ymk2 = 2kx� ynk

2 + 2kx� ymk2� kx� yn + x� ymk

2 =

2kx� ynk2 + 2kx� ymk

2� 4kx�

12(yn + ym)k2

2kx� ynk2 + 2kx� ymk

2� 4d2

! 2d2 + 2d2� 4d2 = 0 as n, m !1.

Hence (yn) is a Cauchy sequence, its limit y belongs to L and

kx� yk = lim kx� ynk = d(x, L).

Step II. Let us take an arbitrary z 2 L such that kzk = 1. Then w := y+�z 2 L, 8� 2 IF(why?). For � = (x� y, z) we have

d2 kx� wk2 = (x� y � �z, x� y � �z) = kx� yk2 � �(z, x� y)�

�(x� y, z) + |�|2 = kx� yk2 � |�|2 = d2� |�|2.

Hence(x� y, z) = � = 0, 8z 2 L s.t. kzk = 1.

Consequently (x� y, z) = 0, 8z 2 L\{0} (why?). It is clear that the last equality is validfor z = 0 as well.Step III. Let y0 2 L be such that kx � y0k = d(x, L). Then we obtain from Step II:(x� y0, z) = 0, 8z 2 L. Hence (y � y0, z) = 0, 8z 2 L. Taking z = y � y0 2 L we obtainky � y0k

2 = 0, i.e. y = y0. This proves the uniqueness. 2

3.13. Remark The last theorem and its proof remain true in the casewhen H is an inner product space and L is its complete linear subspace.

3.14. Definition The orthogonal complement M? of a set M ⇢ H isthe set

M? := {x 2 H : (x, y) = 0, 8y 2 M}.

3.15. Proposition Let M be an arbitrary subset of H. Then(i) M? is a closed linear subspace of H;(ii) M ⇢ M??;(iii) M is dense in H =) M? = {0}.

Proof: Exercise. 2

3.16. Theorem For any closed linear subspace M of a Hilbert space H we

37

have H = M �M?.

Proof: According to Theorem 3.12 for any x 2 H there exists y 2 M suchthat

kx� yk = d(x, M) and (x� y, z) = 0, 8z 2 M.

So, u := x� y 2 M? andx = y + u

is a decomposition of the required form, i.e. H = M + M?. Let us provenow that M \M? = {0}. For any w 2 M \M? we have

kwk2 = (w,w) = 0, i.e. w = 0. 2

Complete orthonormal sets

3.17. Definition The linear span of a subset A of a vector space X is thelinear subspace

linA =

(NX

n=1

�nxn : xn 2 A, �n 2 IF, n = 1, . . . , N, N 2 IN

)

.

If X is a normed space than the closed linear span of A is the closure of linA.We will use the following notation for the closed linear span of A

spanA := Cl(linA).

3.18. Proposition If A is a finite set then spanA = linA.

Proof: Cf. Example 1.3–1. 2

3.19. Definition A system {e↵}↵2J ⇢ H is called orthonormal if

(e↵, e�) = ��

↵=

(1 if � = ↵ ,0 if � 6= ↵ .

38

3.20. Definition A system {e↵}↵2J ⇢ H is called linearly independent if forany finite subset {e↵1 , . . . , e↵N},

PN

n=1 cne↵n = 0 =) c1 = · · · = cN = 0.

3.21. Proposition Any orthonormal system is linearly independent.

Proof: SupposeP

N

n=1 cne↵n = 0. Then

0 =

NX

n=1

cne↵n , e↵m

!

=NX

n=1

cn(e↵n , e↵m) = cm

for any m = 1, . . . , N . 2

Let us consider the Gauss approximation problem: suppose {e1, . . . , eN}

is an orthonormal system in an inner product space H. For a given x 2 H

find c1, . . . , cN such that kx�P

N

n=1 cnenk is minimal.

3.22. Theorem The Gauss approximation problem has a unique solu-tion cn = (x, en). The vector w := x �

PN

n=1(x, en)en is orthogonal toL := lin{e1, . . . , eN}. Moreover, kwk2 = kxk2

�P

N

n=1 |(x, en)|2 (Bessel’sequality) and, consequently,

PN

n=1 |(x, en)|2 kxk2 (Bessel’s inequality).

Proof: For arbitrary c1, . . . , cN we have

�����x�NX

n=1

cnen

�����

2

=

x�NX

n=1

cnen, x�NX

m=1

cmem

!

=

kxk2�

NX

n=1

cn(en, x)�NX

m=1

cm(x, em) +

NX

n=1

cnen,NX

m=1

cmem

!

=

kxk2�

NX

n=1

|(x, en)|2 +

NX

n=1

(x, en)en �

NX

n=1

cnen,NX

m=1

(x, em)em �

NX

m=1

cmem

!

=

kxk2�

NX

n=1

|(x, en)|2 +

�����

NX

n=1

(x, en)en �

NX

n=1

cnen

�����

2

.

Therefore kx�P

N

n=1 cnenk is minimal when cn = (x, en) and kwk2 = kxk2�P

N

n=1 |(x, en)|2. The fact that w is orthogonal to L follows from Theorem

39

3.12 (see also Remark 3.13). Let us give an independent direct proof of thisfact:

(w, em) =

x�NX

n=1

(x, en)en, em

!

= (x, em)�NX

n=1

(x, en)(en, em) =

(x, em)�NX

n=1

(x, en)�m

n= (x, em)� (x, em) = 0

for any m = 1, . . . , N . So, w is orthogonal to e1, . . . , eN , i.e. to L. 2

3.23. Lemma Let H be a Hilbert space and {en} be an orthonormalsystem, cn 2 IF, n 2 IN and

P1

n=1 |cn|2 < 1. Then the series

P1

n=1 cnen isconvergent and for its sum y we have

kyk =

1X

n=1

|cn|2

!1/2

.

Proof: Let ym =P

m

n=1 cnen. Then, using Pythagoras’ theorem (see Proposi-tion 3.6) and the convergence of the series

P1

n=1 |cn|2, we obtain

kym � ykk2 =

������

mX

n=k+1

cnen

������

2

=mX

n=k+1

|cn|2! 0, as k, m !1, (m > k).

Thus (ym) is a Cauchy sequence, has a limit y and

kyk2 = limm!1

kymk2 = lim

m!1

mX

n=1

|cn|2 =

1X

n=1

|cn|2. 2

3.24. Corollary Let H be a Hilbert space and {en} be an orthonormal set.For any x 2 H the series

P1

n=1(x, en)en is convergent and for its sum y wehave

kyk =

1X

n=1

|(x, en)|2!1/2

kxk.

Proof: It follows from Theorem 3.22 that

NX

n=1

|(x, en)|2 kxk2

40

for any N 2 IN. Therefore1X

n=1

|(x, en)|2 kxk2. 2

3.25. Definition The Fourier coe�cients of an element x 2 H with respectto an orthonormal set {en} are the numbers (x, en).

3.26. Theorem Let {en} be an orthonormal set in a Hilbert space H.Then the following statements are equivalent(i) x =

P1

n=1(x, en)en, 8x 2 H;(ii) kxk2 =

P1

n=1 |(x, en)|2, 8x 2 H (Parseval’s identity);(iii) (x, en) = 0, 8n 2 IN =) x = 0;(iv) lin{en} is dense in H.

Proof: (i) =) (ii) Follows from Corollary 3.24.

(ii) =) (iii) Immediate.

(iii) =) (iv) For an arbitrary x 2 H and y :=P1

n=1(x, en)en (see Corol-lary 3.24) we have (x � y, em) = 0, 8m 2 IN (see the proof of Theorem3.22). Then x = y by (iii). It is clear that y 2 span{en}. Consequentlyx 2 span{en} for any x 2 H, i.e. Cl(lin{en}) = H.

(iv) =) (i) For an arbitrary x 2 H and y =P1

n=1(x, en)en we have (x �y, en) = 0, 8n 2 IN. Therefore (x � y, z) = 0, 8z 2 lin{en}, i.e. x � y 2(lin{en})?. But (lin{en})? = {0} since lin{en} is dense (see Proposition3.15). Hence x = y. 2

3.27. Definition The orthonormal set {en} is called complete if it sat-isfies any (and hence all) of the conditions of Theorem 3.26.

3.28. Examples1. An orthonormal subset of a finite-dimensional Hilbert space is completei↵ it is a basis.2. Let H = l2,

en = (0, . . . , 0, 1| {z }n

, 0, . . .).

41

Then {en}n2IN is a complete orthonormal set (why?).3. The set {en}n2Z,

en(t) = ei2⇡nt,

is orthonormal in the inner product space C([0, 1]) of Examples 3.2-3, 3.11-3(check this!). It is one of the basic results of the classical Fourier analysisthat it has also the properties (ii)–(iv) of Theorem 3.26 (with Z and

P1

n=�1

instead of IN andP1

n=1 correspondingly). The Fourier seriesP1

n=�1(f, en)en

of a function f 2 C([0, 1]) is convergent to f with respect to the norm k · k2,but may be not uniformly convergent, i.e. not convergent with respect to thenorm k · k1.

3.29. Theorem (Gram–Schmidt orthogonalization process) For anarbitrary countable or finite set {yn} of vectors of an inner product space Hthere exists a countable or finite orthonormal set {en} such that

lin{yn} = lin{en}.

Proof: Let us first construct a set {zn} such that lin{yn} = lin{zn} and zn

are linearly independent. We define zn inductively:

z1 = yn1 ,

where yn1 is the first non-zero yn,

z2 = yn2 ,

where yn2 is the first yn not in the linear span of z1, and, generally,

zk = ynk,

where ynkis the first yn not in the linear span of z1, . . . , zk�1. It is clear that

lin{yn} = lin{zn} and zn are linearly independent (prove this!).Now we define inductively vectors en such that for any N the set {en}

N

n=1

is orthonormal and lin{en}N

n=1 = lin{zn}N

n=1. By our construction z1 is non-zero, so

e1 =z1

kz1k

42

is well defined and has all the necessary properties in the case N = 1. Assumethat vectors e1, . . . , eN�1 have been defined. By hypothesis zN /2 LN�1 :=lin{en}

N�1n=1 = lin{zn}

N�1n=1 , so the vector

uN = zN �

N�1X

k=1

(zN , ek)ek

must be non-zero and hence

eN =uN

kuNk

is well defined. We have

lin{en}N

n=1 = lin(eN ,LN�1) = lin(uN ,LN�1) = lin(zN ,LN�1) = lin{zn}N

n=1.

Further, uN 2

⇣lin{en}

N�1n=1

⌘?

by Theorem 3.22. Consequently eN is orthogo-

nal to any element of the orthonormal set {en}N�1n=1 . It is clear that keNk = 1.

So the set {en}N

n=1 is orthonormal. This completes the induction step.

Thus we have constructed an orthonormal set {en} such that

lin{en} = lin{zn} = lin{yn}. 2

3.30. Definition A normed space is called separable if it contains a count-able dense subset.

3.31. Theorem A Hilbert space H is separable i↵ it contains a count-able (or finite) complete orthonormal set.

Proof: Let H be a separable Hilbert space, {yn}n2IN be a dense subset andlet {en} be the orthonormal set of the last theorem. Then lin{en} = lin{yn}

is dense in H, i.e. {en} is complete.Conversely, assume that H contains a countable (or finite) complete or-

thonormal set {en}. Let

QN =

(NX

n=1

�nen : �1, . . . ,�N 2 Q + iQ

)

,

43

where Q is the field of rational numbers, and let

Q = [QN .

Then (exercise) Q is a countable dense subset of H. 2

Let U : H1 ! H2 be an isomorphism of inner product spaces H1 and H2

(see Definition 1.4). It follows from the polarization identity (see Proposition3.7) that U preserves the inner products

(Ux, Uy)2 = (x, y)1, 8x, y 2 H1.

3.32. Theorem All infinite dimensional separable Hilbert spaces are iso-morphic to l2 and hence to each other.

Proof: Let H be an infinite dimensional separable Hilbert space and let{en}n2IN be a complete orthonormal set in it. Let us define U : l2 ! H bythe formula U(�n)n2IN =

P1

n=1 �nen (see Lemma 3.23). It is clear that U islinear. Lemma 3.23 and Theorem 3.26 (Definition 3.27) imply that U is ontoand an isometry. 2

3.33. Remark Suppose {e↵} is a not necessarily countable orthonormal set. It iscalled complete if

(x, e↵) = 0, 8↵ =) x = 0.

Any two complete orthonormal subsets of a Hilbert space H have the same cardinality.This cardinality is called the dimension of H. Two Hilbert spaces are isomorphic i↵ theyhave the same dimension. (For proofs see e.g. C. Go↵man & G. Pedrick, First Course inFunctional Analysis, Section 4.7.).

Bounded linear functionals on a Hilbert space

3.34. Theorem (Riesz) f is a bounded linear functional on a Hilbert spaceH, i.e. a bounded linear operator from H to IF if and only if there exists aunique z 2 H such that

f(x) = (x, z), 8x 2 H. (3.7)

44

Moreover, kfk = kzk.

Proof: It is clear that the equality (3.7) defines a bounded linear functional for an ar-bitrary z 2 H. So, we have to prove that any bounded linear functional on H has a uniquerepresentation of the form (3.7).Uniqueness. If z1, z2 satisfy (3.7), then

(x, z1) = (x, z2), 8x 2 H,

and therefore z1 � z2 2 H? = {0}.

Existence. If f = 0, take z = 0. Suppose f 6= 0. Then Ker(f) is a closed linear subspaceof H (see Definition 2.1) and Ker(f) 6= H. Let y 2 Ker(f)?, y 6= 0. We have

f(f(x)y � f(y)x) = f(x)f(y)� f(x)f(y) = 0, 8x 2 H,

i.e.f(x)y � f(y)x 2 Ker(f), 8x 2 H.

Consequently(f(x)y � f(y)x, y) = 0, 8x 2 H.

The left-hand side of the last equality equals f(x)kyk2�f(y)(x, y) = f(x)kyk2�(x, f(y)y).Thus

f(x) = (x, z), 8x 2 H, where z :=f(y)kyk2

y.

The equality kfk = kzk follows from the relations

|f(x)| = |(x, z)| kzkkxk,8x 2 H, and f(z) = kzk2. 2

4 Spectral theory in Hilbert spaces

The adjoint operator

4.1. Theorem Let H1 and H2 be Hilbert spaces and B 2 B(H1,H2). Thereexists a unique operator B⇤

2 B(H2,H1) such that

(Bx, y) = (x, B⇤y), 8x 2 H1, 8y 2 H2. (4.1)

Proof: Let us take an arbitrary y 2 H2. Then

H1 3 x 7! f(x) := (Bx, y) 2 IF

45

is a linear functional and

|f(x)| kBxkkyk kBkkxkkyk = (kBkkyk)kxk, 8x 2 H1.

So, f is a bounded linear functional on H1 and kfk kBkkyk. According to the Riesztheorem (Theorem 3.34) there exists a unique z = z(B, y) 2 H1 such that

(Bx, y) = f(x) = (x, z), 8x 2 H1,

and kzk = kfk kBkkyk. It is easy to see that the map

H2 3 y 7! B⇤y := z 2 H1

is linear (check this!). We have already proved that

kB⇤yk kBkkyk, 8y 2 H2,

i.e. B⇤ is bounded andkB⇤

k kBk. (4.2)

It is clear that (4.1) is satisfied and that the constructed operator B⇤ is the unique oper-ator satisfying this equality. 2

4.2. Definition The operator B⇤ from the previous theorem is calledthe adjoint of B.

4.3. Theorem LetH1, H2 andH3 be Hilbert spaces, B, B1, B2 2 B(H1,H2)and T 2 B(H2,H3). Then(i) (↵B1 + �B2)⇤ = ↵B⇤

1 + �B⇤

2 , 8↵, � 2 IF;(ii) (TB)⇤ = B⇤T ⇤;(iii) B⇤⇤ = B;(iv) kB⇤

k = kBk;(v) kB⇤Bk = kBB⇤

k = kBk2;(vi) if B is invertible then so is B⇤ and (B⇤)�1 = (B�1)⇤.

Proof: (i)

((↵B1 + �B2)x, y) = ↵(B1x, y) + �(B2x, y) = ↵(x,B⇤

1y) + �(x, B⇤

2y) =(x, (↵B⇤

1 + �B⇤

2)y), 8x 2 H1, 8y 2 H2.

(ii)(TBx, z) = (Bx, T ⇤z) = (x, B⇤T ⇤z), 8x 2 H1, 8z 2 H3.

(iii)

(B⇤⇤x, y) = (y, B⇤⇤x) = (y, (B⇤)⇤x) = (B⇤y, x) = (x,B⇤y) = (Bx, y),8x 2 H1, 8y 2 H2.

46

Consequently((B �B⇤⇤)x, y) = 0, 8x 2 H1, 8y 2 H2.

Taking y = (B �B⇤⇤)x we obtain (B �B⇤⇤)x = 0, 8x 2 H1.(iv) follows from (4.2) and (iii):

kBk = k(B⇤)⇤k kB⇤k kBk.

(v)kB⇤Bk kB⇤

kkBk = kBk2

(see (iv)). On the other hand

kBk2 = supkxk=1

kBxk2 = supkxk=1

(Bx,Bx) = supkxk=1

(x, B⇤Bx)

supkxk=1

kxkkB⇤Bxk = kB⇤Bk.

Thus kB⇤Bk = kBk2. Using this equality with B⇤ instead of B we derivefrom (iii) and (iv) that kBB⇤

k = kBk2.(vi) It is su�cient to take the adjoint operators in the equality

BB�1 = IH2 , B�1B = IH1 ,

(see (ii)). 2

Parts (i)–(iii) of Theorem 4.3 show that B(H) is a Banach algebra withan involution. Part (v) shows that it is, in fact, a C⇤–algebra.

A mapping x 7! x⇤ of an algebra E into itself is called an involution if it has the fol-lowing properties

(↵x + �y)⇤ = ↵x⇤ + �y⇤,

(xy)⇤ = y⇤x⇤,

x⇤⇤ = x,

for all x, y 2 E and ↵, � 2 IF.A Banach algebra E with an involution satisfying the identity

kx⇤xk = kxk2, 8x 2 E ,

is called a C⇤–algebra.

The analogue of Theorem 4.3(i) for operators acting on normed spaceshas the following form

(↵B1 + �B2)⇤ = ↵B⇤

1 + �B⇤

2 , 8↵, � 2 IF. (4.3)

47

Let us explain why this equality is di↵erent from that of Theorem 4.3(i). IfH1 and H2 are Hilbert spaces, then we have two definitions of the adjointof B 2 B(H1,H2): Definition 4.2 and that for operators acting on normedspaces. B⇤

2 B(H2,H1) according to the first one, while the second one givesus an operator B⇤

2 B(H⇤

2,H⇤

1). In order to connect them to each other wehave to use the identification H

⇤

1 ⇠ H1, H⇤

2 ⇠ H2 given by Theorem 3.34.This identification is conjugate linear (anti-linear):

fj(x) = (x, zj), j = 1, 2, 8x 2 H =) ↵f1(x)+�f2(x) = (x, ↵z1+�z2), 8x 2 H.

This is why we have the complex conjugation in Theorem 4.3(i) and do nothave it in (4.3).Note also that we use the following definition of linear operations on X⇤ (foran arbitrary normed space X)

(f1 + f2)(x) := f1(x) + f2(x), 8f1, f2 2 X⇤, 8x 2 X,

(↵f)(x) := ↵f(x), 8f 2 X⇤, 8x 2 X, 8↵ 2 IF.

One can replace the last definition by

(↵f)(x) := ↵f(x), 8f 2 X⇤, 8x 2 X, 8↵ 2 IF.

In this case Theorem 4.3(i) is valid for operators acting on normed spaces.

4.4. Theorem Let H1 and H2 be Hilbert spaces, B 2 B(H1,H2). Then

Ker(B⇤) = Ran(B)?, Ker(B) = Ran(B⇤)?.

Proof:

B⇤y = 0 () (x, B⇤y) = 0, 8x 2 H1 () (Bx, y) = 0, 8x 2 H1 ()

y 2 Ran(B)?.

Thus Ker(B⇤) = Ran(B)?. Since B⇤⇤ = B, the second assertion follows fromthe first if B is replaced by B⇤. 2

4.5. Corollary (cf. Theorem 2.39). Let H1 and H2 be Hilbert spaces,B 2 B(H1,H2). Then

Cl(Ran(B)) = Ker(B⇤)?, Cl(Ran(B⇤)) = Ker(B)?.

48

Proof: Note that for any linear subspace M of a Hilbert space H we havethe equality Cl(M) = M??. (Prove this!) 2

4.6. Definition Let H,H1 and H2 be Hilbert spaces. An operator B 2

B(H) is said to be(i) normal if BB⇤ = B⇤B,(ii) self-adjoint if B⇤ = B, i.e. if

(Bx, y) = (x, By), 8x, y 2 H.

An operator U 2 B(H1,H2) is called unitary if U⇤U = IH1 , UU⇤ = IH2 , i.e.if U�1 = U⇤.

Note that any self-adjoint operator is normal. Any unitary operator U 2

B(H) is normal as well.

4.7. Theorem An operator B 2 B(H) is normal if and only if

kBxk = kB⇤xk, 8x 2 H. (4.4)

Proof: We havekBzk2 = (Bz,Bz) = (B⇤Bz, z),

kB⇤zk2 = (B⇤z, B⇤z) = (BB⇤z, z)

for any z 2 H. So, (4.4) holds if B is normal.If kBzk = kB⇤zk, 8z 2 H, then using the polarization identity (see Propo-sition 3.7) we obtain

(Bx,By) = (B⇤x, B⇤y), 8x, y 2 H,

i.e.((B⇤B �BB⇤)x, y) = 0, 8x, y 2 H.

Taking y = (B⇤B�BB⇤)x we deduce that (B⇤B�BB⇤)x = 0, 8x 2 H. 2

4.8. Theorem Let B 2 B(H) be a normal operator. Then(i) Ran(B⇤)? = Ker(B) = Ker(B⇤) = Ran(B)?,(ii) Bx = ↵x =) B⇤x = ↵x,(iii) eigenvectors corresponding to di↵erent eigenvalues of B are orthogonal

49

to each other.

Proof: (i) follows from Theorems 4.4 and 4.7. Applying (i) to B � ↵I inplace of B we obtain (ii). (Exercise: prove that B � ↵I is normal.)Finally, suppose Bx = ↵x, By = �y and ↵ 6= �. We have from (ii)

↵(x, y) = (↵x, y) = (Bx, y) = (x, B⇤y) = (x, �y) = �(x, y).

Since ↵ 6= �, we conclude (x, y) = 0. 2

4.9. Theorem If U 2 B(H1,H2), the following statements are equiva-lent.(i) U is unitary.(ii) Ran(U) = H2 and (Ux, Uy) = (x, y), 8x, y 2 H1.(iii) Ran(U) = H2 and kUxk = kxk, 8x 2 H1.

Proof: If U is unitary, then Ran(U) = H2 because UU⇤ = IH2 . Also, U⇤U = IH1 , sothat

(Ux, Uy) = (x,U⇤Uy) = (x, y), 8x, y 2 H1.

Thus (i) implies (ii). It is obvious that (ii) implies (iii).It follows from the polarization identity (see Proposition 3.7) that (iii) implies (ii). So, if(iii) holds, then

(U⇤Ux, y) = (Ux, Uy) = (x, y), 8x, y 2 H1.

Consequently U⇤U = IH1 (why?). On the other hand (iii) implies that U is a linear isom-etry of H1 onto H2. Hence U is invertible. Since U⇤U = IH1 , we have U�1 = U⇤ (why?),i.e. U is unitary. 2

Theorem 4.9 shows that the notion of a unitary operator coincides with thenotion of an isomorphism of Hilbert spaces (see Definition 1.4) and that theseisomorphisms preserve the inner products (see also the equality in the para-graph between Theorems 3.31 and 3.32).

4.10. Theorem(i) B is self-adjoint, � 2 IR =) �B is self-adjoint.(ii) B1, B2 are self-adjoint =) B1 + B2 is self-adjoint.(iii) Let B1, B2 be self-adjoint. Then B1B2 is self-adjoint if and only if B1

and B2 commute.(iv) If Bn, n 2 IN are self-adjoint and kB �Bnk ! 0, then B is self-adjoint.

50

Proof: Exercise. 2

4.11. Theorem Let H be a complex Hilbert space (i.e. IF = C) andB 2 B(H). Then B is self-adjoint i↵ (Bx, x) is real for all x 2 H.

Proof: If B is self-adjoint, then

(Bx, x) = (x, Bx) = (Bx, x), 8x 2 H,

i.e. (Bx, x) is real.Let us prove the converse. For any x, y 2 H we have

4(Bx, y) = (B(x + y), x + y)� (B(x� y), x� y) +

i(B(x + iy), x + iy)� i(B(x� iy), x� iy) (4.5)

(check this expanding the right-hand side; this is the polarization identity foroperators) and similarly

4(x, By) = (x + y, B(x + y))� (x� y, B(x� y)) +

i(x + iy, B(x + iy))� i(x� iy, B(x� iy)).

Since(Bz, z) = (Bz, z) = (z, Bz), 8z 2 H,

we have(Bx, y) = (x, By), 8x, y 2 H,

i.e. B is self-adjoint. 2

4.12. Theorem All eigenvalues of a self-adjoint operator B 2 B(H) arereal and eigenvectors corresponding to di↵erent eigenvalues of B are orthog-onal to each other.

Proof: According to Theorem 4.8(iii) it is su�cient to prove the first state-ment. Let � be an eigenvalue and x 2 H\{0} be a corresponding eigenvector:Bx = �x. Using Theorem 4.11 we obtain

� =(Bx, x)

kxk22 IR. 2

51

4.13. Theorem Let H be a Hilbert space. Each of the following fourproperties of a projection P 2 B(H) implies the other three:(i) P is self-adjoint.(ii) P is normal.(iii) Ran(P ) = Ker(P )?.(iv) (Px, x) = kPxk2, 8x 2 H.

Proof: It is trivial that (i) implies (ii).Theorem 4.8(i) shows that Ker(P ) = Ran(P )? if P is normal. Since P is a projectionRan(P ) is closed (see Lemma 2.45). So, Ran(P ) = Ran(P )?? = Ker(P )? (see the proofof Corollary 4.5). Thus (ii) implies (iii).Let us prove that (iii) implies (i). Indeed, if (iii) holds, then

(Px, (I � P )y) = 0, ((I � P )x, Py) = 0, 8x, y 2 H,

(see Lemma 2.42). Therefore,

(Px, y) = (Px, Py + (I � P )y) = (Px, Py) = (Px + (I � P )x, Py) = (x, Py),

i.e. P is self-adjoint. So, we have proved that the properties (i)–(iii) are equivalent to eachother.Now it is su�cient to prove that (iii) is equivalent to (iv). We have seen above that (iii)implies the equality (Px, (I � P )x) = 0, 8x 2 H. Hence

(Px, x) = (Px, Px) + (Px, (I � P )x) = (Px, Px) = kPxk2, 8x 2 H.

Finally, assume (iv) holds. Let us take arbitrary y 2 Ran(P ) and z 2 Ker(P ) and considerx = y + tz, t 2 IR. It is clear that Px = y. Consequently

0 kPxk2 = (Px, x) = (y, y + tz) = kyk2 + t(y, z).

Thust(y, z) � �kyk2, 8t 2 IR,

i.e.(y, z) = 0, 8y 2 Ran(P ), 8z 2 Ker(P ),

i.e. (iii) holds.(In the case when H is a complex Hilbert space the proof can be slightly simplified: afterproving the implications (i) =) (ii) =) (iii) =) (iv) as above, we obtain directly fromTheorem 4.11 that (iv) =) (i).) 2

4.14. Definition Property (iii) of the last theorem is usually expressed

52

by saying that P is an orthogonal projection.

Numerical range

4.15. Theorem Let H be a Hilbert space and B 2 B(H). Suppose thereexists c > 0 such that

|(Bx, x)| � ckxk2, 8x 2 H. (4.6)

Then B is invertible and kB�1k 1/c.

Proof: It follows from (4.6) that Ker(B) = {0} and

ckxk2 kBxkkxk, 8x 2 H,

i.e.kxk c�1

kBxk, 8x 2 H. (4.7)

Let us prove that Ran(B) is closed. For any y 2 Cl(Ran(B)) there existxn 2 H such that Bxn ! y as n !1. It follows from (4.7) that

kxn � xmk c�1kBxn �Bxmk ! 0 as n,m !1.

Hence (xn) is a Cauchy sequence in the Hilbert space H. Let us denote itslimit by z. We have

Bz = B✓

limn!1

xn

◆= lim

n!1Bxn = y.

Thus y 2 Ran(B), i.e. Ran(B) is closed.(4.6) implies that if x 2 Ran(B)?, then x = 0. So, Ran(B)? = {0}, i.e.Ran(B) is dense in H (why?). Consequently Ran(B) = H and the operatorB is invertible. The inequality kB�1

k 1/c follows from (4.7). 2

4.16. Definition Let H be a Hilbert space and let B 2 B(H). Thenthe set

Num(B) := {(Bx, x) : kxk = 1, x 2 H} (4.8)

53

is called the numerical range of the operator B.

It is clear that

|(Bx, x)| kBxkkxk kBkkxk2, 8x 2 H,

so,Num(B) ⇢ {µ 2 C : |µ| kBk}. (4.9)

The numerical range of any bounded linear operator is convex (see e.g. G.Bachman & L. Narici, Functional Analysis, 21.4).

4.17. Theorem For any B 2 B(H) we have

�(B) ⇢ Cl(Num(B)). (4.10)

Proof: Let us take an arbitrary � 2 C\Cl(Num(B)) and z 2 H such thatkzk = 1. It is clear that (Bz, z) 2 Num(B). Hence

|((B � �I)z, z)| = |(Bz, z)� �| � d > 0,

where d is the distance from � to Cl(Num(B)). Now for arbitrary x 2 H\{0}we have

|((B � �I)x, x)| = kxk2

�����

(B � �I)x

kxk,

x

kxk

!����� � dkxk2.

According to Theorem 4.15 B � �I is invertible, i.e. � /2 �(B). Hence�(B) ⇢ Cl(Num(B)). 2

54

Spectra of self-adjoint and unitary operators

It follows from Theorem 4.11 that if B 2 B(H) is self-adjoint then Num(B) ⇢IR. Therefore Theorem 4.17 implies the following result.

4.18. Theorem Let B 2 B(H) be self-adjoint and

m := infkxk=1

(Bx, x), M := supkxk=1

(Bx, x).

Then �(B) ⇢ [m,M ] ⇢ IR. 2

4.19. Theorem If U 2 B(H) is unitary, then �(U) ⇢ {� 2 C : |�| = 1}.

Proof: It follows from Theorem 4.9(iii) that kUk = 1 = kU�1k. If |�| > 1

then � /2 �(U) by Lemma 2.9. Suppose |�| < 1. We have

U � �I = U(I � �U�1)

and k�U�1k = |�| < 1. Hence U ��I is invertible by Lemmas 2.3(i) and 2.4.

So, � /2 �(U), i.e. �(U) ⇢ {� 2 C : |�| = 1}. 2

Spectra of normal operators

4.20. Theorem The spectral radius of an arbitrary normal operatorB 2 B(H) equals its norm: r(B) = kBk (cf. Theorem 2.23).

Proof: Replacing x by Bx in (4.4) we obtain

kB2xk = kB⇤Bxk, 8x 2 H.

So, kB2k = kB⇤Bk. Now Theorem 4.3(v) implies kB2

k = kB⇤Bk = kBk2.Hence

kB2k = kBk2.

Since (Bn)⇤ = (B⇤)n (see Theorem 4.3(ii)), if B is normal then so is Bn

(why?). Hence it follows by induction that kBmk = kBkm for all integers m

of the form 2k. Using Theorem 2.23 we can write

r(B) = limn!1

kBnk

1/n = limk!1

kB2kk

1/2k= lim

k!1

kBk = kBk. 2

55

4.21. Theorem Let B 2 B(H) be a normal operator. Then

kBk = supkxk=1

|(Bx, x)|. (4.11)

Proof: Theorems 4.17 and 4.20 imply

kBk = r(B) = sup{|�| : � 2 �(B)} sup{|�| : � 2 Cl(Num(B))} =

sup{|�| : � 2 Num(B)} = supkxk=1

|(Bx, x)| kBk. 2

Compact operators in Hilbert spaces

4.22. Theorem Let X be a normed space, H be a Hilbert space andlet T 2 Com(X,H). Then there exists a sequence of finite rank operatorsTn 2 Com(X,H), such that kT � Tnk ! 0 as n !1 (see Definition 2.55).

Proof: Let y(n)1 , . . . , y(n)

knbe a 1/n–net of the relatively compact set T (SX)

(see (2.25) and Theorem 2.49) and let Ln := span{y(n)1 , . . . , y(n)

kn}. Let Pn be

the orthogonal projection onto Ln, i.e.

Pny = z, where y = z + w, z 2 Ln, w 2 L?

n,

(see Theorem 3.16). It is easy to see that Tn := PnT is a finite rank operatorand

k(T � Tn)xk = kTx� PnTxk = dist(Tx,Ln) <1

n, 8x 2 SX ,

(see Theorem 3.12). Thus kT � Tnk 1/n. 2

4.23. Remark A subset {en}n2IN of a Banach space X is called a Schauderbasis if for any x 2 X there exists unique representation of the form

x =1X

n=1

�nen, �n 2 IF.

It is clear that any Banach space having a Schauder basis is separable (cf.the proof of Theorem 3.31). Theorem 4.22 remains valid if we replace H by

56

a Banach space having a Schauder basis. In 1973 P. Enflo gave a negativesolution to the long standing problem on existing of a Schauder basis in anyseparable Banach space: he constructed a separable Banach space withoutSchauder basis. He also proved that there exist a separable Banach spaceand a compact operator acting in it such that this operator is not a limit ofa convergent sequence of finite rank operators.

Hilbert–Schmidt operators

Let H1 and H2 be separable Hilbert spaces and {en}n2IN be a complete or-thonormal set in H1. Then for any operator T 2 B(H1,H2) the quantityP1

n=1 kTenk2 is independent of the choice of a complete orthonormal set {en}

(this quantity may be infinite). Indeed, if {fn}n2IN is a complete orthonormalset in H2 then

1X

n=1

kTenk2 =

1X

n=1

1X

k=1

|(Ten, fk)|2 =

1X

n=1

1X

k=1

|(en, T⇤fk)|

2 =1X

k=1

kT ⇤fkk2.

So,P1

n=1 kTenk2 =

P1

n=1 kTe0nk

2 for any two complete orthonormal sets {en}

and {e0n} in H1.

4.24. Definition The Hilbert–Schmidt norm of an operator T 2 B(H1,H2)is defined as

kTkHS :=

1X

n=1

kTenk2

!1/2

and the operators for which it is finite are called Hilbert–Schmidt operators.

Note thatkTk kTkHS. (4.12)

Indeed,

kTxk =

�����T

1X

n=1

(x, en)en

!����� =

�����

1X

n=1

(x, en)Ten

����� 1X

n=1

|(x, en)|kTenk

1X

n=1

|(x, en)|2!1/2 1X

n=1

kTenk2

!1/2

= kTkHSkxk, 8x 2 H1,

57

(see Theorem 3.26).

4.25. Exercise Prove that the set of all Hilbert–Schmidt operators is asubspace of B(H1,H2) and that k · kHS is indeed a norm on that subspace.

4.26. Theorem Hilbert–Schmidt operators are compact.

Proof: Let T be a Hilbert–Schmidt operator and let {en}n2IN be a completeorthonormal set in H1. The operator TN defined by

TN

1X

n=1

anen

!

=NX

n=1

anTen

is a finite rank operator (why?). Moreover for any x =P1

n=1 anen 2 H1 wehave

kTx� TNxk =

������

1X

n=N+1

anTen

������

1X

n=N+1

|an|kTenk

0

@1X

n=N+1

|an|2

1

A1/20

@1X

n=N+1

kTenk2

1

A1/2

kxk

0

@1X

n=N+1

kTenk2

1

A1/2

and therefore

kT � TNk

0

@1X

n=N+1

kTenk2

1

A1/2

! 0 as N !1.

Hence T is compact being the limit of a sequence of compact operators (seeTheorem 2.54(iii)). 2

4.27. Example (Integral operators) Let H = L2([0, 1]) and let a mea-surable function

k : [0, 1]⇥ [0, 1] ! IF

be such that Z 1

0

Z 1

0|k(t, ⌧)|2d⌧dt < 1.

Define the operator K by

(Kf)(t) :=Z 1

0k(t, ⌧)f(⌧)d⌧.

58

Then K : L2([0, 1]) ! L2([0, 1]) is a Hilbert–Schmidt operator and

kKkHS =✓Z 1

0

Z 1

0|k(t, ⌧)|2d⌧dt

◆1/2

.

Proof: For t 2 [0, 1] let kt(⌧) := k(t, ⌧). Let {en}n2IN be a complete orthonor-mal set in L2([0, 1]). Then

Ken(t) =Z 1

0k(t, ⌧)en(⌧)d⌧ = (kt, en).

Hence

kKenk2 =

Z 1

0|Ken(t)|2dt =

Z 1

0|(kt, en)|2dt

and therefore

1X

n=1

kKenk2 =

1X

n=1

Z 1

0|(kt, en)|2dt =

Z 1

0

1X

n=1

|(kt, en)|2dt =

Z 1

0kktk

2dt =Z 1

0

Z 1

0|k(t, ⌧)|2d⌧dt < 1,

since {en} is a complete orthonormal set in L2([0, 1]).

5 Spectral theory of compact normal opera-tors

In this chapter H will always denote a Hilbert space.

5.1. Theorem Let T 2 Com(H) be a normal operator. Then T hasan eigenvalue � such that |�| = kTk.

Proof: This result follows from Theorems 2.61 and 4.20. Since we havenot proved Theorem 2.61, we give an independent proof of Theorem 5.1.

There is nothing to prove if T = 0. So, let us suppose that T 6= 0.It follows from Theorem 4.21 that there exist xn 2 H, n 2 IN such thatkxnk = 1 and |(Txn, xn)| ! kTk. We can suppose that the sequence ofnumbers (Txn, xn) is convergent (otherwise we could take a subsequence of

59

(xn)). Let � be the limit of this sequence. It is clear that |�| = kTk 6= 0. Wehave

kTxn � �xnk2 = (Txn � �xn, Txn � �xn) = kTxnk

2� �(xn, Txn)�

�(Txn, xn) + |�|2kxnk2 = kTxnk

2� 2Re(�(Txn, xn)) + |�|2

2|�|2 � 2Re(�(Txn, xn)) ! 2|�|2 � 2|�|2 = 0 as n !1.

Hence kTxn � �xnk ! 0 as n !1.Since T is a compact operator the sequence (Txn) has a Cauchy subsequence(Txnk

)k2IN. The last sequence is convergent because the space H is complete.Let us denote its limit by y. It follows from the formula

xnk=

1

�(Txnk

� (Txnk� �xnk

))

that the sequence (xnk)k2IN converges to x := ��1y. Consequently

Tx = T✓

limk!1

xnk

◆= lim

k!1

Txnk= y = �x

and

kxk =���� lim

k!1

xnk

���� = limk!1

kxnkk = 1.

Thus x 6= 0 is an eigenvector of T corresponding to the eigenvalue �. 2

5.2. Theorem (Spectral theorem for a compact normal operator)Let T 2 Com(H) be a normal operator. Then there exists a finite or a count-able orthonormal set {en}

N

n=1, N 2 IN[ {1}, of eigenvectors of T such thatany x 2 H has a unique representation of the form

x =NX

n=1

cnen + y, y 2 Ker(T ), cn 2 C. (5.1)

One then has

Tx =NX

n=1

�ncnen, (5.2)

where �n 6= 0 is the eigenvalue of T corresponding to the eigenvector en.Moreover,

�(T )\{0} = {�n}N

n=1, (5.3)

60

|�1| � |�2| � · · · (5.4)

andlim

n!1�n = 0, if N = 1. (5.5)

Proof: Step I We will use an inductive process. Let �1 be an eigenvalue of Tsuch that |�1| = kTk (see Theorem 5.1) and let e1 be the corresponding eigen-vector such that ke1k = 1. Suppose we have already constructed non-zeroeigenvalues �1, . . . ,�k of T satisfying (5.4) and corresponding eigenvectorse1, . . . , ek such that the set {en}

k

n=1 is orthonormal. Let

Lk := lin{e1, . . . , ek}, Hk := L?

k.

It is clear that

L1 ⇢ L2 ⇢ · · · ⇢ Lk, H1 � H2 � · · · � Hk. (5.6)

Any x 2 Lk has a unique representation of the form x =P

k

n=1 cnen, cn 2 C.Consequently

Tx =kX

n=1

�ncnen 2 Lk, T ⇤x =kX

n=1

�ncnen 2 Lk

(see Theorem 4.8(ii)). Hence

TLk ⇢ Lk, T ⇤Lk ⇢ Lk.

For any z 2 Hk = L?

kwe have

(x, Tz) = (T ⇤x, z) = 0, (x, T ⇤z) = (Tx, z) = 0, 8x 2 Lk,

because T ⇤x, Tx 2 Lk. Thus Tz, T ⇤z 2 Hk. So,

THk ⇢ Hk, T ⇤Hk ⇢ Hk.

Let us consider the restriction of T to the Hilbert space Hk:

Tk := T |Hk2 B(Hk).

It is easy to see that Tk is a compact normal operator (why?). There are twopossibilities.

61

Case I. Tk = 0. In this case the construction terminates.Case II. Tk 6= 0. In this case there exists an eigenvalue �k+1 6= 0 of Tk

such that |�k+1| = kTkk = kT |Hkk (see Theorem 5.1). It follows from the

construction and from (5.6) that |�k+1| |�k| · · · |�1|. Let ek+1 2 Hk =L?

kbe an eigenvector of Tk corresponding to �k+1 and such that kek+1k =

1. It is clear that ek+1 is an eigenvector of T and that the set {en}k+1n=1 is

orthonormal.This construction gives us a finite or a countable orthonormal set {en}

N

n=1,N 2 IN [ {1}, of eigenvectors of T and corresponding non-zero eigenvalues�n satisfying (5.4). This set is finite if we have Case I for some k and infiniteif we always have Case II.Step II Let us prove (5.5). Suppose (5.5) does not hold. Then there exists� > 0 such that |�n| � �, 8n 2 IN (see (5.4)). Consequently

kTen � Temk2 = k�nen � �memk

2 = |�n|2 + |�m|

2� 2�2 > 0 if n 6= m.

Hence (Ten) cannot have a Cauchy subsequence. However, this contradictsthe compactness of T . So, we have proved (5.5).

Step III Let us consider the subspace

HN := \N

n=1Hn.

If N is finite, then according to our construction T |HN = 0, i.e. HN ⇢

Ker(T ). If N = 1, then for any x 2 HN we have

kTxk kT |Hkkkxk = |�k+1|kxk ! 0 as k !1.

Thus kTxk = 0, i.e. Tx = 0, 8x 2 HN . Consequently HN ⇢ Ker(T ). Onthe other hand Ker(T ) ⇢ L

?

k= Hk for any k by Theorem 4.8(iii). Hence

HN = Ker(T ). (5.7)

Step IV For an arbitrary x 2 H we have

x�NX

n=1

(x, en)en, ek

!

= (x, ek)�NX

n=1

(x, en)(en, ek) = (x, ek)� (x, ek) = 0,

8k = 1, . . . , N,

62

(cf. the proof of Theorem 3.22). Thus x has a representation of the form(5.1), where cn = (x, en) and

y := x�NX

n=1

(x, en)en 2 HN = Ker(T ).

Taking the inner products of both sides of (5.1) with ek and using (5.7) weobtain ck = (x, ek), i.e. a representation of the form (5.1) is unique.

Step V (5.2) follows immediately from (5.1) and the continuity of T . It

is left to prove (5.3). Let us take an arbitrary � 2 C\⇣{�n}

N

n=1 [ {0}⌘. It is

clear that the distance d from � to the closed set {�n}N

n=1 [ {0} is positive.(5.1), (5.2) imply

(T � �I)x =NX

n=1

(�n � �)(x, en)en � �y.

It is easy to see that the operator T � �I is invertible and

(T � �I)�1x =NX

n=1

(�n � �)�1(x, en)en � ��1y. (5.8)

Note that

k(T � �I)�1xk2 =NX

n=1

|�n � �|�2|(x, en)|2 + |�|�2

kyk2

d�2

NX

n=1

|(x, en)|2 + kyk2

!

= d�2kxk2 < 1

(see Proposition 3.6, Lemma 3.23 and (5.7)). So, we have proved that � /2�(T ), i.e. �(T ) ⇢ {�n}

N

n=1[{0}. On the other hand we have {�n}N

n=1 ⇢ �(T ).2

5.3. Remark Since any self-adjoint operator is normal (see Definition4.6), the previous theorem is valid for self-adjoint operators. In this case alleigenvalues �n are real (see Theorem 4.12 or Theorem 4.18). This variant ofTheorem 5.2 is known as the Hilbert–Schmidt theorem.

63

Let Pn be the orthogonal projection onto the one–dimensional subspace gen-erated by en, i.e. Pn := (·, en)en and let PKer (T) be the orthogonal projectiononto Ker(T ). Then (5.1), (5.2) and (5.8) can be rewritten in the followingway:

I =NX

n=1

Pn + PKer (T), (5.9)

T =NX

n=1

�nPn, (5.10)

R(T ; �) = (T � �I)�1 =NX

n=1

(�n � �)�1Pn � ��1PKer (T) (5.11)

(see Step IV of the proof of Theorem 5.2), where the series are stronglyconvergent. (A series

P1

n=1 An, An 2 B(X, Y ), is called strongly convergentif the series

P1

n=1 Anx converges in Y for any x 2 X.) On the other hand itis easy to see that in the case N = 1 the series (5.9) and (5.11) do notconverge in the B(H)–norm, while (5.10) does (prove this!).

Suppose a function f is analytic in some neighbourhood �f of �(T ) and⌦ is an admissible set such that

�(T ) ⇢ ⌦ ⇢ Cl(⌦) ⇢ �f .

Then we obtain from Definition 2.26, (5.11) and the Cauchy theorem

f(T )x =✓�

1

2⇡i

Z

@⌦f(�)R(T ; �)d�

◆x =

�1

2⇡i

Z

@⌦f(�)

NX

n=1

(�n � �)�1Pnx� ��1PKer (T)x

!

d� =

�

NX

n=1

✓1

2⇡i

Z

@⌦f(�)(�n � �)�1d�

◆Pnx +

✓1

2⇡i

Z

@⌦f(�)��1d�

◆PKer (T)x =

NX

n=1

f(�n)Pnx + f(0)PKer (T)x, 8x 2 H. (5.12)

The series here are convergent because (Pnx, Pmx) = 0 if m 6= n andPN

n=1 kPnxk2 =P

N

n=1 |(x, en)|2 kxk2 (see Lemma 3.23 and Corollary 3.24).

64

Note that if dim(H) < +1 it may happen that 0 62 �(T ) and 0 62 ⌦. If inthis case f(0) 6= 0 then

1

2⇡i

Z

@⌦f(�)��1d� = 0 6= f(0).

Nevertheless (5.12) holds because PKer (T) = 0, since Ker (T) = 0.

It is clear that the RHS of (5.12) is well defined for any bounded on �(T )not necessarily analytic function f . This motivates the following definition.5.4. Definition Let T 2 Com(H) be a normal operator and f be a boundedfunction on �(T ). Then

f(T )x :=NX

n=1

f(�n)Pnx + f(0)PKer (T)x, 8x 2 H.

Note that if dim(H) < +1 and 0 62 �(T ), then f may be not definedat 0. The above definition, however, still makes sense because in this casePKer (T) = 0, since Ker (T) = 0, and we assume that f(0)PKer (T) = 0.

The operator f(T ) is well defined because

�����

NX

n=1

f(�n)Pnx + f(0)PKer (T)x

�����

2

=

NX

n=1

|f(�n)(x, en)|2 + |f(0)|2kPKer (T)xk2

sup�2�(T )

|f(�)|

!2

kxk2 < 1, 8x 2 H, (5.13)

(cf. Step V of the proof of Theorem 5.2). It follows from (5.13) that

kf(T )k sup�2�(T )

|f(�)|.

It is easy to see that if Ker(T ) = {0}, then

kf(T )k sup�2{�n}

|f(�)| = sup�2�(T )\{0}

|f(�)|.

65

Since kf(T )enk = kf(�k)enk = |f(�n)| and kf(T )yk = kf(0)yk = |f(0)|kykfor any y 2 Ker(T ), we have

kf(T )k �

(sup

�2�(T ) |f(�)| if Ker (T) 6= {0},sup

�2�(T )\{0} |f(�)| if Ker (T) = {0}.

Thus

kf(T )k =

(sup

�2�(T ) |f(�)| if Ker (T) 6= {0},sup

�2�(T )\{0} |f(�)| if Ker (T) = {0}.(5.14)

If dim(H) < +1 and Ker(T ) = {0}, then 0 62 �(T ) and �(T ) \ {0} =�(T ). If dim(H) = +1 and Ker(T ) = {0}, then 0 2 �(T ) is the only limitpoint of the set �(T ) \ {0}. Therefore

sup�2�(T )

|f(�)| = sup�2�(T )\{0}

|f(�)|, if f is continuous at 0.

However, if f is not continuous at 0, the above equality is not necessarilytrue.

It is easy to see that the functional calculus f 7! f(T ) from Definition5.4 has the same properties as that from Definition 2.26 (see Theorems 2.30,2.31, 2.34 and 2.35). The advantages of Definition 5.4 are that f has to bedefined only on �(T ) and may be non-analytic. The disadvantage is thatT has to be a compact normal operator acting on a Hilbert space, whileDefinition 2.26 deals with an arbitrary bounded linear operator acting on aBanach space.

5.5. Theorem Let T 2 Com(H) be a normal operator and f be a boundedfunction on �(T ). The operator f(T ) is normal. It is compact if and only ifone of the following statements is true:(i) N < +1, dim(Ker(T )) < +1, i.e. dim(H) < +1,(ii) N < +1, dim(Ker(T )) = +1 and f(0) = 0,(iii) N = +1, dim(Ker(T )) < +1 and f(�n) ! 0 as n !1,(iv) N = +1, dim(Ker(T )) = +1, f(0) = 0 and f(�n) ! 0 as n !1.The operator f(T ) is self–adjoint if and only if one of the following state-ments is true:(a) Ker(T ) = {0} and f(�n) 2 IR for each n,(b) Ker(T ) 6= {0}, f(0) 2 IR and f(�n) 2 IR for each n.

Proof: Exercise. 2

66