From: bruck@math.usc.edu (Ronald Bruck) Subject: Re: convex sets in Hilbert space Date: Mon, 24 Jan 2000 13:12:06 -0800 Newsgroups: sci.math Summary: [missing] In article , Gabor Szekelyhidi wrote: :It is well known, that if H is a Hilbert space, and E is a closed convex :set in H, then for every point x in H there is a unique point in E that is :of minimal distance from x. : :I read in a book, that the converse of this is unknown, but the book was :written in 1979 (Kirillov, Gvisiani : Problems and Theorems in Functional :Analysis), and I can't seem to find any more recent information. : :Does anyone know what has been done about this problem? I'm not quite sure what the converse of this is. If it's supposed to be, "If X is a Banach space with the property that for every nonempty closed convex subset C of X, and for every point x of X, there exists a unique point of C which is closest to x, then X must be a Hilbert space", then it's trivially false; every reflexive rotund (= strictly convex) space (e.g., L^p, has this property. Every reflexive space satisfies existence of a proximinal point (because the norm is sequentially weakly lower semi-continuous), and the strict convexity guarantees uniqueness. If it's supposed to be, "If X is a Banach space with the property that every closed nonempty subset E with the property that each x in X has a unique proximinal point in E ==> E must be convex then X is a Hilbert space", then yes, I believe this is still open. --Ron Bruck ============================================================================== From: Simon Subject: Re: convex sets in Hilbert space Date: Tue, 25 Jan 2000 12:57:29 GMT Newsgroups: sci.math Yes. The converse (by which I suppose you mean "every Chebyshev set in a Hilbert space in convex", where E is a Chebyshev set in H provided for every point x in H there is a unique point in E that is of minimal distance from x) is still unsolved. (I think anyone who solved it would send me a preprint with a suitably triumphant message attached). It may well depend on whether the Hilbert space is separable. It is true if the Hilbert space is finite dimensional, or under a variety of assumptions about the Chebyshev set. Note that $E=\emptyset$ is a counterexample to your "well known" statement. But you didn't want to know that. I probably state a few false theorems per year in my classes because of the empty set, and my students never seem to notice. Cheers Simon. [quote of original message deleted --djr] -- Dr Simon Fitzpatrick, Shenton Park, Western Australia Mathematician and International Correspondence Chess Master http://www.q-net.net.au/~dsf/Simon.html