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.� � � 71 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 12. Tuesay, Mar 17: Compactness and weak convergence A subset in a general metric space is one with the property that any sequence in it has a convergent subsequence, with its limit in the set. You will recall with pleasure no doubt the equivalence of this condition to the (more general since it makes good sense in an arbitrary topological space) equivalence of this with the covering condition, that any open cover of the set has a finite subcover. So, in a separable Hilbert space the notion of a compact set is already fixed. We want to characterize it – in the problems this week you will be asked to prove several characterizations. A general result in a metric space is that any compact set is both closed and bounded, so this must be true in a Hilbert space. The Heine-Borel theorem gives a converse to this, Rn or Cn (and hence in any finite dimensional normed space) any closed and bounded set is compact. Also recall that the convergence of a sequence in Cn is equivalent to the convergence of the n sequences given by its components and this is what is used to pass first from R to C and then to Cn . All of this fails in infinite dimensions and we need some condition in addition to being bounded and closed for a set to be compact. To see where this might come from, observe that a set, S, consisting of the points of a convergent sequence, s : N −→ M, together with its limit, s, in any metric space is always compact. The set here is the image of the sequence, thought of as a map from the integers into the metric space, together with the limit (which might of course already be in the image). Certainly this set is bounded, since the distance from the intial point is certainly bounded. Moreover it is closed, although you might need to think about this for a minute. A sequence in the set which is the image of another sequence consists of elements of the original sequence in any order and maybe repeated at will. Since the original sequence may itself have reapeated points, the labelling of points is by no means unique. However S is closed since M \ S is open – a point in p ∈ M \ S is at a finite no-zero distance, d(p, s) from the limit so B(p, d(p, s)/2) can contain only finitely many elements of S hence a smaller open ball does not meet it. Lemma 6. The image of a convergent sequence in a Hilbert space is a set with equi-small tails with respect to any orthonormal sequence, i.e. if ek is an othonormal sequence and un u is a convergent sequence then given � > 0 there exists N such→that (12.1) |(un, ek)|2 < �2 ∀ n. k>N Proof. Bessel’s inequality shows that for any u ∈ H, (12.2) |(u, ek)|2 ≤ �u�2 . k The convergence of this series means that (12.1) can be arranged for any single element un or the limit u by choosing N large enough, thus given � > 0 we can choose N� so that (12.3) |(u, ek)|2 < �2/2. k>N�� � � 72 LECTURE NOTES FOR 18.102, SPRING 2009 In fact, for any orthonormal sequence such as ek – whether complete or not, (12.4) P : H � u �−→ P u = (u, ek)ek ∈ H k is continuous and in fact has norm at most one. Indeed from Bessel’s inequality, �P u�2 ≤ �u�2 . Now, applying this to � (12.5) PN u = (u, ek)ek k>N the convergence un → u implies the convergence in norm �PN un� → �PN u� and so (12.6) |(u, ek)|2 < �2 . k>N� So, we have arranged (12.1) for n > n� with N = N�. Of course, this estimate remains valid if N is increased, and we may arrange it for n ≤ n� by chossing N large enough. Thus indeed (12.1) holds for all n if N is chosen large enough. � This suggest one useful characterization of compact sets in a separable Hilbert space. Proposition 19. A set K ⊂ H in a separabel Hilbert space is compact if and only if it is bounded, closed and has equi-small tails with respect to any one orthonormal basis. Proof. We already know that a compact set is closed and bounded. Suppose the equi-smallness of tails condition fails with respect to some orthonormal basis ek. This means that for some � > 0 and all N there is an element uN ∈ K such that (12.7) |(uN , ek)|2 ≥ �2 . k>N Then the sequence {uN } can have no convergent subsequence, since this would contradict the Lemma we have just proved, hence K is not compact in this case. Thus we have proved the equi-smallness of tails condition to be necessary for the compactness of a closed, bounded set. So, it remains to show that it is sufficient. So, suppose K is closed, bounded and satisfies the equi-small tails condition with respect to an orthonormal basis ek and {un} is a sequence in K. We only need show that {un} has a Cauchy subsequence, since this will converge (H being complete) and the limit will be in K (since it is closed). Now, consider each of the sequences of coefficients (un, ek) in C. Here k is fixed. This sequence is bounded: (12.8) |(un, ek)| ≤ �un� ≤ C by the boundedness of K. So, by the Heine-Borel theorem, there is a subsequence of unl such that (unl , ek) converges as l → ∞. We can apply this argument for each k = 1, 2, . . . . First extracting a subsequence of {un} so that the sequence (un, e1) converges ‘along this subsequence’. Then extract a subsequence of this subsequence so that (un, e2) also converges along this sparser subsequence, and continue inductively. Then pass to the ‘diagonal’ subsequence of {un} which has kth entry the kth term in the kth subsequence. It is ‘eventually’ a subsequence of each of the subsequences previously constructed – meaning it coincides with a subsequence for some point onward (namely the� � � � 73 LECTURE NOTES FOR 18.102, SPRING 2009 kth term onward for the kth subsquence). Thus, for this subsequence each of the (unl , ek) converges. Now, let’s relabel this subsequence vn for simplicity of notation and consider Bessel’s identity (the orthonormal set ek is complete by assumption) for the differ-ence �vn − vn+l�2 = (vn − vn+l, ek)2