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Thursday, October 23, 2014

Record a problem

Let H be a Hilbert space. It is well-known that if Z is a convex and closed subset of H, then for every xH, there is a unique element zZ such that xz=d(x,Z). Let's denote this unique element z by PZx.
Problem. Let Z1Z2Z3 be a chain of closed convex subsets of a Hilbert space H, show that:
(a) If n=1Zn, then xPZnxxPZx.
(b) If n=1Zn=, then xPZnx.
Since Zn's are not subspace of H, PZn itself is not a linear map, and therefore standard tools from functional analysis cannot be used. Nevertheless, by assuming Zn's are further a closed subspace, it is instructive to see a weak-convergence argument to conclude the result in (a) --- a slightly complicated solution.

Some also asked me can n=1Zn= happen? Below is an example to show there is a chain of closed convex subsets with empty intersection:

Example. Let Zn={(1,12,,1n,a1,a2,):(a1,a2,)2}, then Zn is convex, a closed subset of 2, and descending in n. However, if there is (x1,x2,)n=1Zn, then necessarily xn=1/n for every n, and this element is not in 2, a contradiction. So n=1Zn=.
  Next what is d(x,Zn) in this particular example? By definition for zZn we have xz2=nk=1|xk1k|2+k>n|xkzk|2, hence for fixed x2 the smallest possible xz is d(x,Zn)=nk=1|xk1k|2, of course this diverges to infinity as n for any fixed x2. What's more, PZnx=(1,12,,1n,xn+1,xn+2,).

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