ANALYSIS TOOLS WITH APPLICATIONS 579 35. Compact and Fredholm Operators and the Spectral Theorem In this section H and B will be Hilbert spaces. Typically H and B will be separable, but we will not assume this until it is needed later. 35.1. Compact Operators. Proposition 35.1. Let M be a finite dimensional subspace of a Hilbert space H then (1) M is complete (hence closed). (2) Closed bounded subsets of M are compact. Proof. Using the Gram-Schmidt procedure, we may choose an orthonormal n basis φ1,...,φn of M. Define U : M C to be the unique unitary map such { } th → n that Uφi = ei where ei is the i standard basis vector in C . It now follows that M is complete and that closed bounded subsets of M are compact since the same is true for Cn. Definition 35.2. A bounded operator K : H B is compact if K maps bounded sets into precompact sets, i.e. K(U) is compact→ in B, where U := x H : x < 1 { ∈ k k } is the unit ball in H. Equivalently, for all bounded sequences xn ∞ H, the { }n=1 ⊂ sequence Kxn ∞ has a convergent subsequence in B. { }n=1 Notice that if dim(H)= and T : H B is invertible, then T is not compact. ∞ → Definition 35.3. K : H B is said to have finite rank if Ran(K) B is finite dimensional. → ⊂ Corollary 35.4. If K : H B is a finite rank operator, then K is compact. In particular if either dim(H→) < or dim(B) < then any bounded operator K : H B is finite rank and hence∞ compact. ∞ → Example 35.5. Let (X, µ) be a measure space, H = L2(X, µ) and n k(x, y) fi(x)gi(y) ≡ i=1 X where 2 fi,gi L (X, µ) for i =1,...,n. ∈ 2 2 Define (Kf)(x)= X k(x, y)f(y)dµ(y), then K : L (X, µ) L (X, µ) is a finite rank operator and hence compact. → R Lemma 35.6. Let := (H, B) denote the compact operators from H to B. Then (H, B) isanormclosedsubspaceofK K L(H, B). K Proof. The fact that is a vector subspace of L(H, B) will be left to the reader. K Now let Kn : H B be compact operators and K : H B be a bounded operator → → such that limn Kn K op =0. We will now show K is compact. →∞ k − k First Proof. Given >0, choose N = N() such that KN K <. k − k Using the fact that KN U is precompact, choose a finite subset Λ U such that ⊂ minx Λ y KN x <for all y KN (U) . Then for z = Kx0 K(U) and x Λ, ∈ k − k ∈ ∈ ∈ z Kx = (K KN )x0 + KN (x0 x)+(KN K)x k − k k − − − k 2 + KN x0 KN x . ≤ k − k 580 BRUCE K. DRIVER† Therefore minx Λ z KN x < 3, which shows K(U) is 3 bounded for all >0, K(U) is totally∈ boundedk − andk hence precompact. Second Proof. Suppose xn n∞=1 is a bounded sequence in H. By compactness, 1 { } 1 ∞ ∞ ∞ there is a subsequence xn n=1 of xn n=1 such that K1xn n=1 is convergent in B. Working inductively, we may construct{ } subsequences © ª © ª 1 2 m xn ∞ x ∞ x ∞ x ∞ ... { }n=1 ⊃ n n=1 ⊃ n n=1 ···⊃ { n }n=1 ⊃ m such that Kmxn n∞=1 is convergent© ª in©B ªfor each m. By the usual Cantor’s diago- { } n nalization procedure, let yn := x , then yn ∞ is a subsequence of xn ∞ such n { }n=1 { }n=1 that Kmyn ∞ is convergent for all m. Since { }n=1 Kyn Kyl (K Km) yn + Km(yn yl) + (Km K) yl) k − k ≤ k − k k − k k − k 2 K Km + Km(yn yl) , ≤ k − k k − k lim sup Kyn Kyl 2 K Km 0 as m , n,l k − k ≤ k − k → →∞ →∞ which shows Kyn ∞ is Cauchy and hence convergent. { }n=1 Proposition 35.7. A bounded operator K : H B is compact iff there exists → finite rank operators, Kn : H B, such that K Kn 0 as n . → k − k → →∞ Proof. Since K(U) is compact it contains a countable dense subset and from this it follows that K (H) is a separable subspace of B. Let φn be an orthonormal N { } basis for K (H) B and PN y = (y,φn)φn be the orthogonal projection of y ⊂ n=1 N onto span φn n=1. Then limN PPN y y =0for all y K(H). { } →∞ k − k ∈ Define Kn PnK —afinite rank operator on H. For sake of contradiction ≡ suppose that lim supn K Kn = >0, in which case there exists xnk U →∞ k − k ∈ such that (K Kn )xn for all nk. Since K is compact, by passing to a k − k k k ≥ subsequence if necessary, we may assume Kxn ∞ is convergent in B. Letting { k }nk=1 y limk Kxnk , ≡ →∞ (K Kn )xn = (1 Pn )Kxn (1 Pn )(Kxn y) + (1 Pn )y k − k k k k − k k k ≤ k − k k − k k − k k Kxn y + (1 Pn )y 0 as k . ≤ k k − k k − k k → →∞ But this contradicts the assumption that is positive and hence we must have limn K Kn =0, i.e. K is an operator norm limit of finite rank operators. The→∞ conversek − directionk follows from Corollary 35.4 and Lemma 35.6. Corollary 35.8. If K is compact then so is K∗. Proof. Let Kn = PnK be as in the proof of Proposition 35.7, then Kn∗ = K∗Pn is still finite rank. Furthermore, using Proposition 12.16, K∗ K∗ = K Kn 0 as n k − nk k − k → →∞ showing K∗ is a limit of finite rank operators and hence compact. 35.2. Hilbert Schmidt Operators. Proposition 35.9. Let H and B be a separable Hilbert spaces, K : H B be a → bounded linear operator, en n∞=1 and um m∞=1 be orthonormal basis for H and B respectively. Then: { } { } ANALYSIS TOOLS WITH APPLICATIONS 581 2 2 (1) n∞=1 Ken = m∞=1 K∗um allowing for the possibility that the sums are infiknite.k In particulark the Hilbertk Schmidt norm of K, P P 2 ∞ 2 K HS := Ken , k k n=1 k k X is well defined independent of the choice of orthonormal basis en ∞ . We { }n=1 say K : H B is a Hilbert Schmidt operator if K HS < and let HS(H, B) denote→ the space of Hilbert Schmidt operatorsk k from H∞to B. (2) For al l K L(H, B), K = K∗ and ∈ k kHS k kHS K K := sup Kh : h H h =1 . k kHS ≥ k kop {k k ∈ 3kk } (3) The set HS(H, B) isasubspaceof (H, B) and isanormon K k·kHS HS(H, B) for which (HS(H, B), HS) is a Hilbert space. The inner prod- uct on HS(H, B) is given by k·k ∞ (35.1) (K1,K2)HS = (K1en,K2en). n=1 X N (4) Let PN x := (x, en)en be orthogonal projection onto span ei : i N n=1 { ≤ } ⊂ H and for K HS(H, B), let Kn := KPn. Then P∈ 2 2 K KN K KN 0 as N , k − kop ≤ k − kHS → →∞ whichshowsthatfinite rank operators are dense in (HS(H, B), HS) . (5) If L is another Hilbert space and A : L H and C : B L arek·k bounded operators, then → → KA K A and CK K C . k kHS ≤ k kHS k kop k kHS ≤ k kHS k kop Proof. Items 1. and 2. By Parsaval’s equality and Fubini’s theorem for sums, ∞ 2 ∞ ∞ 2 ∞ ∞ 2 ∞ 2 Ken = (Ken,um) = (en,K∗um) = K∗um . n=1 k k n=1 m=1 | | m=1 n=1 | | m=1 k k X X X X X X This proves K is well defined independent of basis and that K = k kHS k kHS K∗ HS . For x H 0 ,x/ x maybetakentobethefirst element in an orthonormalk k basis∈ for H\{ and} hencek k x K K . x ≤ k kHS ° ° ° k k° Multiplying this inequality by °x shows° Kx K x and hence K k° k ° k k ≤ k kHS k k k kop ≤ K . k kHS Item 3. For K1,K2 L(H, B), ∈ ∞ 2 K1 + K2 = K1en + K2en k kHS v k k un=1 uX t ∞ 2 [ K1en + K2en ] = K1en + K2en ∞ ≤ v k k k k k{k k k k}n=1k2 un=1 uX t K1en ∞ + K2en ∞ = K1 + K2 . ≤ k{k k}n=1k2 k{k k}n=1k2 k kHS k kHS 582 BRUCE K. DRIVER† From this triangle inequality and the homogeneity properties of , we now eas- k·kHS ily see that HS(H, B) is a subspace of (H, B) and HS is a norm on HS(H, B). Since K k·k ∞ ∞ (K1en,K2en) K1en K2en n=1 | | ≤ n=1 k kk k X X ∞ 2 ∞ 2 K1en K2en = K1 K2 , ≤ v k k v k k k kHS k kHS un=1 un=1 uX uX t t the sum in Eq. (35.1) is well defined and is easily checked to define an inner product 2 on HS(H, B) such that K =(K1,K2) . To see that HS(H, B) is complete k kHS HS in this inner product suppose Km ∞ is a — Cauchy sequence in HS(H, B).