MIT18_102s09_lec12

MIT18_102s09_lec12 - MIT OpenCourseWare http://ocw.mit.edu...

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Unformatted text preview: MIT OpenCourseWare http://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, R n or C n (and hence in any finite dimensional normed space) any closed and bounded set is compact. Also recall that the convergence of a sequence in C n 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 C n . 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 e k is an othonormal sequence and u n u is a convergent sequence then given > there exists N such that (12.1) | ( u n ,e k ) | 2 < 2 n....
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MIT18_102s09_lec12 - MIT OpenCourseWare http://ocw.mit.edu...

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