Let H be an infinite dimensional and separable Hilbert Space. Do there exist disjoint, closed and bounded subsets A,B of H which satisfy the following conditions? (1) Each of A,B is convex and has a non-empty interior with respect to H. (2) Given any positive real number e, there is a point of A and a point of B whose distance apart is less than e....If the answer to this question is "YES", can A and B be congruent?...Any information about this subject will be much appreciated.
1 Answer
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No. Convex closed subsets are weakly closed, and since $A,B$ are also assumed bounded, we can make $x_n,y_n$ converge weakly to limits $x\in A$, $y\in B$ on a subsequence if $x_n\in A$, $y_n\in B$. If now also $\|x_n -y_n\|\to 0$, then $x=y\in A\cap B$.
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$\begingroup$ Many thanks, Christian, for this information. $\endgroup$ Aug 15, 2014 at 20:09
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2$\begingroup$ If you replace $H$ by an arbitrary non reflexive space $X$, then the answer is "yes". James' theorem gives a norm one functional $F$ on $X$ which does not achieve its norm on the unit ball $B_X$ of $X$. Let $A = \{x\in X: 1 \le F(x)\} \cap 2B_X$ and set $B= B_X$. $\endgroup$ Aug 15, 2014 at 21:06