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Suppose that $A$ and $B$ are nonempty closed convex subsets of the Hilbert space $H$.

Case I: If $A$ or $B$ is bounded, I can prove that there exist $a\in A$ and $b\in B$ such that $$d(A,B) = d(a,b),$$ where $d(A,B) := \inf\{d(x,y)\mid x\in A,\, y\in B\}$.

Case II: If both $A$ and $B$ are unbounded and assume $\lim_{\|x\|,\|y\|\to\infty} d(x,y) = \infty$, how to prove the same result as in Case I?

The proof of Case I:

Without loss of generality, we can assume that $A$ is bounded. For any $n\geq 1$, there exists some $x_n\in A$ such that $$d(x_n, B) < d(A, B) + \frac1n.$$ Since $(x_n)$ is bounded, there exists some $a\in A$ such that $x_n\rightharpoonup a$ in the weak topology of $H$, i.e. $$\lim_{n\to\infty} \langle x_n, y\rangle = \langle a, y\rangle\quad \forall y\in H.$$

Since $$\begin{align} \|x_n - P_B(x_n)\|^2 & = \|x_n-a+a-P_B(a)+P_B(a)-P_B(x_n)\|^2 \\ & = \|a-P_B(a)\|^2 + \|x_n-a+P_B(a)-P_B(x_n)\|^2 \\ & \quad + 2\mathrm{Re}\langle a-P_B(a), x_n-a\rangle + 2\mathrm{Re}\langle a-P_B(a), P_B(a)-P_B(x_n)\rangle \\ & \geq \|a-P_B(a)\|^2 + 2\mathrm{Re}\langle a-P_B(a), x_n-a\rangle. \end{align}$$ We have \begin{align*} (d(A,B))^2 & \leq \|a-P_B(a)\|^2\leq\|x_n-P_B(x_n)\|^2 - 2\mathrm{Re}\langle a-P_B(a),x_n-a\rangle\\ & < \biggl(d(A,B)+\frac{1}{n}\biggr)^2 - 2\mathrm{Re}\langle a-P_B(a),x_n-a\rangle \\ & \quad \to (d(A,B))^2 \qquad \text{as } n\to\infty. \end{align*}

It follows that $d(a,b) = d(A,B)$ where $b = P_B(a)$.

Stephen
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1 Answers1

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It seems to me that case II follows from case I: Let $K[x,r]$ denote the closed ball in $H$ with center $x$ and radius $r$. Under the assumptions of case II let $\alpha:= \inf \{\|x-y\| \mid x \in A, y \in B\}$. Choose $M:= \alpha +1$. By assumption there is some $r_1 > 0$ such that $$ \forall x \in A, ~ \forall y \in B: ~ (\|x\|\ge r_1 \wedge \|y\|\ge r_1 ~ \Rightarrow ~ \|x-y\| \ge M). $$ Set $A_0:=A \cap K[0,r_2]$ and $B_0:= B\cap K[0,r_2]$ with $r_2:= r_1 + M$. Let $x \in A,y \in B$ such that $x \in A\setminus A_0$ or $y \in B\setminus B_0$. If $\|x\|,\|y\| \ge r_1$ then $\|x-y\| \ge M$. If $\|x\| < r_1$ then $x \in A_0$ (since $r_2 > r_1$), hence $y \notin B_0$, so $\|y\| > r_2$. Therefore $$ \|x-y\| \ge \|y\|-\|x\| > r_2-r_1 = M. $$ The same way $\|y\| < r_1$ implies $\|x-y\| > M$. Thus $$ \alpha= \inf \{\|x-y\| \mid x \in A, y \in B\} = \inf \{\|x-y\| \mid x \in A_0, y \in B_0\}. $$ Now $A_0,B_0$ are bounded, nonempty, closed and convex. From case I you get some $a \in A_0$ and $b \in B_0$ with $\|a-b\|= \alpha$.

Gerd
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