MIT OpenCourseWarehttp://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.104 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 18. Tuesday April 14: Compact operators Last time we considered invertible elemenets of B(H), the algebra of bounded operators on a separable Hilbert space, and also the finite rank operators. The latter form an ideal which is closed under taking adjoints. We also showed the the closure of this ideal, the elements in B(H) which are the limits of (norm-convergent) sequences of finite rank operators, also form an ideal which is closed under taking adjoints and also norm, i.e. metrically, closed. Definition 8. An element K ∈ B(H), the bounded operators on a separable Hilbert space, is said to be compact (the old terminology was ‘totally bounded’ and you might still see this) if the image of the unit ball is precompact, i.e. has compact closure – that is if the closure of K{u ∈ H; �u�H ≤ 1} is compact in H. Lemma 12. An operator K ∈ B(H) is compact if and only if the image {Kun} of any weakly convergent sequence {un} in H is strongly, ie. norm, convergent. Proof. First suppose that un � u is a weakly convergent sequence in H and that K is compact. We know that �un� < C is bounded so the sequence Kun is contained in CK(B(0, 1)) and hence in a compact set (clearly if D is compact then so is cD for any constant c.) Thus, any subsequence of Kun has a convergent subseqeunce and the limit is necessarily Ku since Kun � Ku (true for any bounded operator by computing (18.1) (Kun, v) = (un, K∗v) (u, K∗v) = (Ku, v).)→ But the condition on a sequence in a metric space that every subsequence of it has a subsequence which converges to a fixed limit implies convergence. (If you don’t remember this, reconstruct the proof: To say a sequence vn does not converge to v is to say that for some � > 0 there is a subsequence along which d(vnk , v) ≥ �. This is impossible given the subsequence of subsequence condition (coverging to the fixed limit v.) Conversely, suppose that K has this property of turning weakly convergent into strongly convergent sequences. We want to show that K(B(0, 1)) has compact closure. This just means that any sequence in K(B(0, 1)) has a (strongly) con-vergent subsequence – where we do not have to worry about whether the limit is in the set or not. Such a sequence is of the form Kun where un is a sequence in B(0, 1). However we know that the ball is weakly compact, that is we can pass to a subsequence which converges weakly, unj � u. Then, by the assumption of the Lemma, Kunj Ku converges strongly. Thus un does indeed have a convergent →subsequence and hence K(B(0, 1)) must have compact closure. � Proposition 25. An operator K ∈ B(H), bounded on a separable Hilbert space, is compact if and only if it is the limit of a norm-convergent sequence of finite rank operators, i.e. the ideal of compact operators K(H) is the norm closure of the ideal of finite rank operators. Proof. So, we need to show that a compact operators is the limit of a convergent sequence of finite rank operators. To do this we use one of the characterizations of compact subsets of a separable Hilbert space discussed earlier. Namely, if ei is an orthonormal basis of H then a subset I ⊂ H is compact if and only if it is closed and bounded and has equi-small tails with respec to {ei}, meaning given � > 0� � � � 105 LECTURE NOTES FOR 18.102, SPRING 2009 there exits N such that (18.2) |(v, ei)|2 < �2 ∀ v ∈ I. i>N Now we shall apply this to the set K(B(0, 1)) where we assume that K is compact – so this set is contained in a compact set. Hence (18.2) applies to it. Namely this means that for any � > 0 there exists n such that (18.3) |(Ku, ei)|2 < �2 ∀ u ∈ H, �u�H ≤ 1. i>n For each n consider the first part of these sequences and define (18.4) Knu = (Ku, ei)ei. k≤n This is clearly a linear operator and has finite rank – since its range is contained in the span of the first n elements of {ei}. Since this is an orthonormal basis, (18.5) �Ku − Knu�2 = (Ku, ei)2 H | |i>n Thus (18.3) shows that �Ku − Knu�H ≤ �. Now, increasing n makes �Ku − Knu�smaller, so given � > 0 there exits n such that for all N ≥ n, (18.6) �K − KN �B = sup �Ku − Knu�H ≤ �. �u�≤1 Thus indeed, Kn K in norm and we have shown that the compact operators are →contained in the norm closure of the finite rank operators. For the converse we assume that Tn → K is a norm convergent sequence in B(H) where each of the Tn is of finite rank – of course we know nothing about the rank except that it is finite. We want to conclude that K is compact, so we need to show that K(B(0, 1)) is precompact. It is certainly bounded, by the norm of K. By one of the results on compactness of sets in a separable Hilbert space we know that it suffices to prove that every weakly convergent sequence in K(B(0, 1)) has a strongly convergent subsequence – meaning norm convergent. The limit need not be in K(B(0, 1)) but must exist of the set is to have compact closure. So, suppose vk is a weakly convergent sequence in K(B(0, 1)). Well then, it is of the form Kuk where uk is a sequence in the unit ball. Of necessity this has a weakly convergent subsequence, so we can assume that uk � u is weakly convergent, by passing to a subsequence of the original sequence. Now, each Tn is of finite rank so the sequences Tnvk are each strongly convergent as k → ∞ – namely they are weakly convergent because (Tnvk, w) = (vk, T n ∗w), and in a finite dimensional space. Use the triangle inequality and definition of the norm of an operator to see that (18.7) �Kvk − Kvl� ≤ �Kvk − Tnvk� + �Tnvk − Tnvl� + �Tnvl − Kvl� ≤ 2�K − Tn�B + �Tnvk − Tnvl�. Now, given � > 0 first choose n so large that �K − Tn� < �/3. Then, having fixed n, use the fact that Tnvk is Cauchy to choose p such that k, l > p implies �Tnvk − Tnvl� < �/3. It follows that Kvk is Cauchy and hence convergent by the completeness of Hilbert space. Thus K is compact. �106 LECTURE NOTES FOR 18.102, SPRING 2009 Notice that this