# Is the distance between disjoint closed convex subsets of a Hilbert space positive? Is it attained?

Let $H$ be an infinite dimensional and separable Hilbert space. Let $A,B$ be infinite, closed and convex subsets of $H$. If $A$ and $B$ are disjoint and if at least one of them is bounded, is the greatest lower bound of the distances between a point of $A$ and a point of $B$ always equal to a positive real number $\epsilon$? If the answer to this qustion is "yes", does there always exist a point of $A$ and a point of $B$ whose distance apart is $\epsilon$?

• What properties of $A,B$, and Hilbert spaces look like they might be useful in giving an affirmative answer? Commented Aug 23, 2014 at 18:58

Assume that $A$ is bounded. Take $x_n\in A, y_n\in B$ such that $\lim||x_n-y_n||=e$. Since $A$ is bounded - sequence $x_n$ is bounded and so is $y_n$. Thus using Banach-Alaoglu theorem one can choose subsequences $x_{n_k},y_{n_k}$ converging in the weak topology to $x,y$. Now since closed and convex sets are weakly closed (Mazur theorem) one gets that $x\in A$ and $y\in B$. Moreover $e\leq||x-y||\leq \liminf ||x_{n_k}-y_{n_k}||=e$ (norm is weakly lower semicontinouos) so that $||x-y||=e$. Since $A$ and $B$ are disjoint thus $e>0$.

• Oh, you beat me to it. I guess I'll leave my answer anyway. +1. Commented Aug 23, 2014 at 19:18

Yes to both questions.

First, recall that a closed convex set in a Hilbert space is weakly closed, and that a weakly closed and norm bounded set is weakly compact (Alaoglu's theorem).

Let $\epsilon := \inf\{\|x-y\| : x \in A, y \in B\}$. We will show there exist $x \in A$, $y \in B$ with $\|x-y\| = \epsilon$ which proves both statements. Choose a sequence $x_n \in A$, $y_n \in B$ such that $\|x_n - y_n\| \to \epsilon$. Now $A$ is weakly compact, so we may pass to a subsequence and assume that $x_n$ converges weakly to some $x \in A$. Also, we may intersect $B$ with a sufficiently large closed ball to produce a closed, bounded, convex $B_1 \subset B$ containing all the $y_n$. $B_1$ is likewise weakly compact so we can pass to a further subsequence and assume that $y_n$ converges weakly to some $y \in B_1 \subset B$.

Clearly $\|x-y\| \ge \epsilon$. On the other hand, by Cauchy-Schwarz, $$|\langle x_n - y_n, x-y \rangle| \le \|x_n - y_n\| \|x-y\|.$$ By weak convergence, the left side converges to $|\langle x-y, x-y \rangle| = \|x-y\|^2$ and the right side converges to $\epsilon \|x-y\|$ by assumption. We conclude $\|x-y\|^2 \le \epsilon \|x-y\|$, or in other words $\|x-y\| \le \epsilon$. (If it turns out that $\|x-y\|=0$ the conclusion is still true.)

• Many thanks for your responses and the detailed proofs that justify them. Commented Aug 24, 2014 at 20:12