4 Compact operators

4.1 Characterisation of the set of compact operators

Definition 4.1. T B(H) is called compact, if T maps bounded sets into pre-compactsets. Equivalently, T is compact if for every bounded sequence {xn} in H, the sequence{Txn} has a convergent subsequence.Example 4.2. It T is a finite rank operator, then T is compact. Indeed, let y1, . . . , yN bea basis in RanT . Then every Tx may be written uniquely as

Tx =Nn=1

nyn

with some coecients n. It is easy to see that n are linear functionals of x; thus n(x) =(x, zn) for some elements zn. So we obtain that T has the form

Tx =Nn=1

(x, zn)yn. (4.1)

Now if x(k) is a bounded sequence, then each of the sequences (x(k), z1), (x(k), z2),. . . isbounded, which allows to extract a convergent subsequence from Tx(k).

Theorem 4.3. An operator T B(H) is compact i it maps any weakly convergent se-quence into a norm convergent sequence.

Proof. 1. Let T be compact and let xn x weakly; we need to prove that Txn Tx in the operatornorm. It suces to consider the case x = 0; i.e. we assume xn 0 weakly and need to prove Txn 0.We have Txn 0 weakly; also, by the uniform boundedness principle, supnxn 0. By thecompactness of T , there exists a subsequence {xn} such that the norm limit z = limTxn exists. SinceTxn 0 weakly, we must have z = 0. This contradicts infTxn > 0.Remark. Here we have used a common trick: instead of proving that the sequence Txn converges, wehave proven that every subsequence has a convergent subsubsequence. Then necessarily the whole sequenceconverges.

2. Suppose T maps weakly convergent sequences into a norm convergent sequences. Let {xn} be abounded sequence. Then (Exercise!) there exists a weakly convergent subsequence xn. Then Txn convergesin the norm. Hence, T is compact.

Theorem 4.4. (i) A linear combination of compact operators is compact.

(ii) If A is compact and B is bounded, then both AB and BA are compact.

(iii) It Tn are compact and Tn T 0, then T is compact.(iv) If T is compact, then T is compact.

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Proof. (i), (ii) are straightforward consequences of the previous theorem. (iii) Let xk 0weakly; we need to prove that Txk 0. Without loss of generality assume xk 1.Fix > 0 and choose n suciently large so that Tn T /2. Since Tn is compact, wehave Tnxk 0 as k ; thus we can choose k0 such that Tnxk /2 for k k0. Sowe have

Txk Txk Tnxk+ Tnxk /2 + /2 = for k k0.

(iv) see Exercises.

The class of compact operators is denoted by S(H) (alternative notations: C(H),Comp(H)). The above theorem shows that S(H) is a closed two-sided ideal of the algebraB(H).Theorem 4.5. The set of finite rank operators is dense in S(H) in the operator norm.Proof. Let T S(H) and let > 0 be given; we need to construct a finite rank operatorT0 with T T0 . The image T [B1] of the unit ball B1 pre-compact, hence totallybounded. Let y1, . . . , yN be an -net for T [B1]. Let M be the span of y1, . . . , yN and let Pbe the orthogonal projection onto M . Then rankP N , and therefore rankPT N ; wewill choose T0 = PT .

Recall that for any z H, the element Pz M is the closest element of M to z:z Pz = inf

yMz y,

and therefore for any z T [B1] we havez Pz inf

kz yk .

Applying this to z = Tx, we get TxPTx for any x B1 and therefore TPT .

Corollary 4.6. The set S(H) is the norm closure of the set of all finite rank operatorsin H.

4.2 The Fredholm alternative

Let A be an N N matrix and let y CN . Consider the equationAx x = y (4.2)

for x CN and the corresponding homogeneous equationAx = x. (4.3)

It is a well-known fact studied in linear algebra that (4.2) has a solution for all y i (4.3)has no non-trivial solutions. Thus, we have an alternative: either (4.3) has a non-trivialsolution or the matrix IA is invertible. The following deep statement says that the sameis true for a compact operator A in an infinite dimensional space.

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Theorem 4.7 (The Fredholm alternative). Let A be a compact operator in H. Then eitherAx = x has a non-trivial solution, or the operator A I has a bounded inverse.Proof. We need to prove that if Ker(I A) = {0}, then the inverse (I A)1 exists.

1. First consider the case when A has a finite rank. The question basically reducesto linear algebra. Let e1,. . . ,eN be a basis in RanA. Then for each x RanA we havex =N

i=1 xiei and A acts on elements from RanA according to

Ax =Ni=1

(Ax)iei, (Ax)i =Nj=1

Aijxj

with some matrix Aij. Condition Ker(I A) = {0} means that det(I A) = 0. Now givenz H, let us construct a solution x to x Ax = z by setting

x = z +Ni=1

iei

with some coecients i to be determined below. Our equation reduces to

Ni=1

iei Az Ni=1

Nj=1

Aijj

ei = 0.

Let Az =

i iei, then the above equation reduces to the linear system

i Nj=1

Aijj = i, i = 1, . . . , N.

This system has a solution since det(I A) = 0.2. Consider the general case. Write A = A0+A1, where A0 is finite rank and A1 < 1/2.

Then I A1 is invertible and we haveI A = I A0 A1 = (I A1)(I (I A1)1A0) = (I A1)(I A2),

whereA2 = (IA1)1A0 has a finite rank. Since Ker(IA) = {0}, we also get Ker(IA2) ={0} and so, by the first part of the proof, I A2 is invertible. It follows that I A isinvertible.

Corollary 4.8. Let A be a compact operator in H and = 0. Then either Ax = x has anon-trivial solution, or (A I)1 exists.

Volterra operator shows that the Fredholm alternative is false for = 0.Further, compactness is crucial for the Fredholm alternative. Consider, for example,

the operator A of multiplication by x in L2(0, 1). Then, as we have already seen, A has noeigenvalues, yet (A I) is not invertible if (A) = [0, 1].

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4.3 The spectra of compact operators

Here our aim is to describe the spectrum of compact operators and to prove a spectraltheorem for compact self-adjoint operators.

Theorem 4.9. Let A be a compact operator. Then (A) \ {0} consists of eigenvalues offinite multiplicities (i.e. dimKer(A I) 0; let us prove that the number of eigenvalues of A with || > is finite.For simplicity, here we will assume that A is normal; for the general case, see, for example,Kreyszig, Theorem 8.3-1. Suppose, to get a contradiction, that there is an infinite sequenceof eigenvalues n of A with |n| > . Let xn be the corresponding sequence of normalisedeigenvectors. Then by Theorem 2.23, xn are mutually orthogonal (it is here that we usethat A is normal). It follows that {xn} is an orthonormal system. Then, as in part 1 of theproof, we get that Axn 0, but on the other hand

Axn = |n|xn = |n| > ,which is a contradiction.

3. Finally, if = 0 is not an eigenvalue of A, then by the Fredholm alternative, is notin the spectrum of A. Thus, the non-zero spectrum of A consists of eigenvalues.

Theorem 4.10. Let A be compact. Then 0 (A).Proof. Suppose to the contrary that 0 (A). Then, for some C > 0, we have

A1x Cx, x H,or, taking Ax = y, we get

Ay C1y, x H.Now let {yn}n=1 be a sequence such that yn = 1 and yn 0 weakly. Then by Theo-rem 4.3, we have Ayn 0. This contradicts to Ayn C1.

The dimension of KerA may be zero, finite, or infinite.The above analysis completes the description of the spectrum of an arbitrary compact

operator. In fact, it is straightforward to see any sequence {n}n=1 (with prescribed finitemultiplicities) such that |n| 0 as n may be a spectrum of a compact operator:Theorem 4.11. Let {n}Nn=1, N , be a finite or infinite sequence of complex numbers;if N = , assume that |n| 0 as n . Assume that for each n, a positive integer

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(n) is prescribed; let also (0) be a prescribed number, which may be zero, positive integer,or infinity. Then there exists a Hilbert space H and a normal operator A in H such that

(A) = Nn=1{n} {0},dimKer(A nI) = (n) for all n, and dimKerA = (0).Proof. Let

k = (0) +Nn=1

(n) .

If k = , take H = 2, and if k < , take H = Ck. The operator A is constructed asan operator of multiplication by an appropriate sequence {tn}. This sequence consists ofeigenvalues {n}Nn=1 (each eigenvalue n is repeated (n) times); also it contains a finiteor infinite number (0) of zeros.

Remark 4.12. Using the Volterra operator (see Section 2.5), in the case k < one canalso construct A as a (non-normal) operator in an infinite dimensional space.

4.4 The spectral decomposition for compact self-adjoint opera-tors

Theorem 4.13 (The Hilbert-Schmidt theorem). Let A be a compact self-adjoint operator inH. Then (A) \ {0} consists of real eigenvalues of finite multiplicities, that can accumulateonly to 0. There exists a complete orthonormal basis in H which consists of eigenvalues ofA.

First we need a lemma.

Lemma 4.14. Let A = A B(H). Then for the spectral radius of A we have r(A) = A.Proof. For any operator T B(H) we have T T T2; on the other hand,

T2 = supx1

Tx2 = supx1

(T Tx, x) T T,

and therefore T T = T2 (weve already seen variants of this calculation). In particular,for a self-adjoint operator A we have A2 = A2. It follows that Ak = Ak for k = 2nand so, taking the limit over a subsequence in the spectral radius formula, we get

r(A) = limn

An1/n = limn

A2n1/2n = A.

Corollary 4.15. Let A be a bounded self-adjoint operator such that (A) = {0}; thenA = 0.

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Proof of the Hilbert-Schmidt theorem. The description of the spectrum of A is a conse-quence of Theorem 4.9 and the elementary fact that all eigenvalues of self-adjoint operatorsare real. Let us prove the existence of the basis of eigenvalues. For each non-zero eigenvalue of A choose an orthonormal basis in Ker(A I). The collection of all elements of thesebases is an orthonormal set in H. Let M be the closed span of this orthonormal set. ThenM is invariant for A; since A is self-adjoint, it follows that M reduces A, i.e. M is alsoinvariant for A.

Consider the restriction A0 = A|M ; we would like to prove that A0 = 0. The operatorA0 is self-adjoint and compact inM (because A is self-adjoint and compact inM). SupposeA0 has a non-zero eigenvalue with a non-zero eigenvector x M; then x is also aneigenvector of A, so x M . This contradiction shows that A0 has no non-zero eigenvectors.Then, by Corollary 4.15, we have A0 = 0.

Thus, appending the basis of eigenvectors of A in M by any orthonormal basis in M,we obtain an orthonormal eigenbasis of A in H.

Let A be a compact self-adjoint operator. Let n be the sequence of the non-zeroeigenvalues of A. It is convenient to enumerate these eigenvalues with multiplicities takeninto account, i.e. for example, if dimKer(A 5I) = 17, then the number 5 is repeated17 times in the list of eigenvalues of A. Then we can enumerate the eigenvalues and theeigenvalues in a coordinated way, i.e. we have a list of real numbers 1,2, . . . and a list ofelements x1, x2, . . . of H such that Axn = nxn for all n.

If KerA is trivial, then the above list of eigenvectors is {xn} forms a complete set. IfKerA is non-trivial, it is convenient to add to the list of eigenvalues a sequence of zeros(finite or infinite, depending on whether dimKerA is finite or infinite), and to add to thelist of eigenvectors a sequence which forms a basis in KerA. The above theorem says thatif we form a list of eigenvectors in this way, then the orthonormal system {xn} is complete.

Clearly, the operator A in this case can be written as

A =n=1

n(, xn)xn. (4.4)

This is the spectral representation for a compact self-adjoint operator. Since xn are or-thogonal, and n 0 as n , the above infinite sum converges in the operator norm(Exercise 4.21). Of course, the terms with n = 0 can be omitted from the sum.

A similar representation is valid for any compact normal operator; the only dierenceis that n are in general complex numbers.

The spectral representation (4.4) allows one to easily compute functions of A. Indeed,we have

Am =n=1

mn (, xn)xn

for any integer m. This is easily extended to

p(A) =n=1

p(n)(, xn)xn

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for any polynomial p with p(0) = 0 (we need the last condition to ensure the norm conver-gence of the series). Further, if A 0 is a non-negative compact self-adjoint operator, wecan set

A =n=1

n(, xn)xn; (4.5)

then it is straightforward to check thatA agrees with the definition given earlier (see

Reed-Simon, Theorem VI.9). Other functions (exponential, resolvent etc) may be definedin this way. You can read further on this functional calculus for self-adjoint operators inReed-Simon, Section VII.1 (this is optional).

4.5 The canonical form of a compact operator

One can write a certain series representation even for non-normal operators. It is lessecient than in the normal case but still is very useful.

Theorem 4.16 (Canonical form of a compact operator). Let A be a compact operator inH. Then there exists (not necessarily complete) orthonormal sets {xn}Nn=1 and {yn}Nn=1,N , and positive real numbers {sn}Nn=1 with sn 0 so that

A =Nn=1

sn(, xn)yn, (4.6)

where the sum is norm convergent.

Proof. (this is Theorem VI.17 from Reed-Simon). Since A is compact, so is AA. Thus AAis compact and self-adjoint. By the Hilbert-Schmidt theorem, there exists an orthonormalset {xn}Nn=1 so that AAxn = nxn with n = 0 and so that AA is the zero operator onthe subspace orthogonal to {xn}Nn=1. Then we can write

AA =n=1

n(, xn)xn,

and so

|A| = AA =n=1

n(, xn)xn. (4.7)

Now recall that A = U |A|, where U is a partial isometry with the initial subspace Ran|A|and the final subspace RanA; we have UUx = x for all x in the initial subspace. Setsn =

n and yn = Uxn. For every n we have xn Ran|A| and so UUxn = xn. Thus, we

have(yn, ym) = (Uxn, Uxm) = (U

Uxn, xm) = (xn, xm) = nm,

i.e. {yn} is an orthonormal set. Applying U to (4.7), we obtain the required decompositionof A. Convergence is discussed in Exercise 4.21.

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Exercises

Exercise 4.17. Let {xn} be an orthonormal sequence in H. Prove that xn 0 weakly inH.Exercise 4.18. Prove that the identity operator in an infinite dimensional space is notcompact.

Exercise 4.19. Let A be a compact operator in H and let p be a polynomial. Prove thatp(A) is compact i p(0) = 0.

Exercise 4.20. Prove that the operator of multiplication by x in L2(0, 1) is not compact.

Exercise 4.21. Consider a finite sum

S =Mm=1

am(, xm)ym,

where {xm}Mm=1 and {ym}Mm=1 are orthonormal sets. Prove that S = maxm=1...M

|am|. Use thisto prove that the sequence of partial sums

AN =Nn=1

sn(, xn)yn

of the series (4.6) is a Cauchy sequence in B(H). Conclude that the series (4.6) convergesin the operator norm. Note that the series may not be absolutely convergent in B(H)!Exercise 4.22. By using (4.5), prove that if A 0 is compact, then

A is compact.

Exercise 4.23. Let T be a finite rank one operator on H. Using the representation (4.1),prove that T is also a finite rank operator.Exercise 4.24. Let T B(H) be compact. Prove that T is also compact. Hint: use theprevious exercise and Theorems 4.4, 4.5.

Exercise 4.25. (An alternative way of proving that T is compact i T is compact). LetT B(H).(i) Using Theorem 4.3(ii), prove that if T is compact, then T T is compact.

(ii) Using Theorem 4.3, prove that if T T is compact, then T is compact.

(iii) Conclude that T is compact i T is compact.

Exercise 4.26. Let {xn}n=1 be a bounded sequence in a separable Hilbert space H. Provethat there exists a weakly convergent subsequence of {xn}. Proceed as follows. Let {yk}k=1be a dense set in H. Consider the sequence {(xn, y1)}n=1. Prove that one can select aconvergent subsequence of this sequence. Next, consider {(xn, y2)} over this subsequence.Select a convergent subsequence of this, etc. By using a diagonal subsequence, constructa subsequence {xn} such that {(xn, yk)}n=1 converges for all k. Prove that xn convergesweakly in H.Exercise 4.27. Let An be a sequence of bounded operators which converges to zerostrongly, and let K be a compact operator. Prove that AnK 0 as n . Hint:consider the case of a finite rank K first. Then use Theorem 4.5 and Exercise 1.40.

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