$\|x_0 − y_0\| = \operatorname{dist}(A, B)$

Question

Let $X$ be a reflexive Banach space and let $A, B$ be non-empty, closed, and convex subsets of $X$. If $B$ is bounded, prove that there exist $x_0$ in $A$ and $y_0$ in $B$ with $\|x_0 − y_0\| = \operatorname{dist}(A, B)$.

Answer 1

One may proceed in the following steps:

1. In a Banach space $E$, for convex subsets, weakly closed=strongly closed. (HINT: use Hahn--Banach theorem.)

Proof. Let $C$ be a nonempty, convex, strongly closed subset of $E$. We prove that $C^c$ is open in the weak topology. Let $x_0\notin C$. By Hahn--Banach theorem, there exists a closed hyperplane strictly separating $\{x_0\}$ and $C$. Thus, there exist some $f\in E^\star$ and some $\alpha\in \mathbb{R}$ such that \begin{equation} f(x_0) <\alpha <f(y), \qquad \forall y\in C. \end{equation} Set \begin{equation} V= \{x\in E:f(x) <\alpha\}, \end{equation} so that $x_0\in V$, $V\cap C=\varnothing$ (i.e., $V\subseteq C^c$) and $V$ is open in the weak topology.

2. Show that, given any element $x\in X$, there is an $a\in A$ such that $\|x-a\|=\text{dist}(x,A)$. (HINT: consider an infimizing sequence and use Eberlein--Smulian theorem.)

Proof. Let $\{a_n\}_{n=1}^\infty\subseteq A$ be an infimizing sequence, i.e., $\|x-a_n\|\to \text{dist}(x,A)$. So, the sequence $\{a_n\}_{n=1}^\infty$ is bounded for the norm topology. By Eberlein--Smulian theorem, there is a subsequence $\{a_{n_k}\}_{k=1}^\infty$ converges to some $a\in X$ weakly $\sigma(X,X^\star)$. Since $A$ is weakly closed, $a\in A$.

3. Let $\varphi(b)=\text{dist}(b,A), \ b\in B$. Check that $\varphi$ is convex and sequentially lower semicontinuous for the weak topology. (HINT: in a Banach space, a convex function that is continuous for the norm topology is sequentially lower semicontinuous for the weak topology.)

Definition. (sequentially lower semicontinuous). Let $E$ be a topological sppace. A function $f:E\to (-\infty,+\infty]$ is said to be sequentially lower semicontinuous at some $x\in E$, if \begin{equation} \lim_{n\to \infty}x_k =x \ \ \text{in $E$}\quad \text{implies}\ \ f(x) \leq \liminf_{n\to\infty} f(x_n).\tag{$\star$} \end{equation} If ($\star$) holds for every $x\in E$, then we say $f$ is a sequentially lower semicontinuous function.

Proof. (1) We claim that: in a Banach space $E$, a convex function $f$ that is continuous for the norm topology is sequentially lower semicontinuous for the weak topology.

Suppose, by contradiction, that $f$ is not sequentially lower semicontinuous at some $x_0\in E$ for the weak topology. Then there exist a sequence $\{x_n\}_{n=1}^\infty\subseteq X$, such that $x_n\rightharpoonup x_0$, but $f(x_0)>\liminf_{n\to\infty} f(x_n)$. So there is an $\varepsilon_0>0$ such that $f(x_0)-\varepsilon_0 >\liminf_{n\to\infty}f(x_n)$. Thus there exists a subsequence of $\{x_n\}_{n=1}^\infty$ (still denoted by $\{x_n\}_{n=1}^\infty$) such that $f(x_n)\leq f(x_0)-\varepsilon_0$. By Mazur's theorem, $x_n \rightharpoonup x_0 \in \overline{\text{conv}\left(\{x_n\}_{n=1}^\infty\right)}$, i.e., there exists a sequence $\{y_k\}_{k=1}^\infty$ made up of convex combinations of the $x_n$'s that converges strongly to $x_0$, i.e., \begin{equation} y_k =\sum_{n=1}^{n_k}\alpha_n^kx_n, \ \ 0\leq \alpha_n^k\leq 1, \ \ \sum_{n=1}^{n_k} \alpha_n^k =1, \ \ y_k\to x_0. \end{equation} Since $f$ is convex and continuous for the norm topology, \begin{equation} f(x_0)\leftarrow f(y_k) \leq \sum_{n=1}^{n_k} \alpha_n^k f(x_n)\leq (f(x_0)-\varepsilon_0)\cdot \sum_{n=1}^{n_k} \alpha_n^k =f(x_0)-\varepsilon_0, \end{equation} a contradiction!

(2) We check that $\varphi$ is convex and continuous for the norm topology.

Let $b_1,b_2\in B$ and $t\in [0,1]$ be fixed. Given $\varepsilon>0$ there is some $a_1\in A$ and some $a_2\in A$ such that \begin{equation} \|b_1-a_1\| \leq \varphi (b_1) +\varepsilon, \ \ \|b_2-a_2\| \leq \varphi (b_2) +\varepsilon. \end{equation} Hence, \begin{equation} \|tb_1+(1-t) b_2 -\left[ta_1+(1-t)a_2\right]\| \leq t\varphi(b_1) +(1-t) \varphi(b_2) +\varepsilon. \end{equation} But $ta_1+(1-t)a_2 \in A$, so that \begin{equation} \varphi (tb_1 +(1-t)b_2) \leq t\varphi(b_1) +(1-t) \varphi(b_2) +\varepsilon, \ \forall \varepsilon>0. \end{equation} So, $\varphi$ is convex.

Let $b_1,b_2\in B$ be arbitrary, write that \begin{equation} \|b_1-a\| \leq \|b_1-b_2\| +\|b_2-a\| . \end{equation} Taking $\inf_{a\in A}$ leads to $\varphi(b_1)\leq \|b_1-b_2\| +\varphi(b_2)$. Then exchange $b_1,b_2$, we can conclude $|\varphi(b_1)-\varphi(b_2)|\leq \|b_1-b_2\|$. So $\varphi$ is continuous for the norm topology.

4. Let $E$ be a reflexive Banach space, $K$ is a nonempty, strongly bounded, weakly closed subset, $f:K\to (-\infty,+\infty]$ is a sequentially lower semicontinuous function for the weak topology. Then there exists some $x_0\in K$ such that \begin{equation} f(x_0) =\inf_{x\in K}f(x). \end{equation} Proof. There is a sequence $\{x_n\}_{n=1}^\infty\subseteq K$ such that \begin{equation} f(x_n) \to \inf_{x\in K}f(x). \end{equation} By Eberlein--Smulian theorem, there is a subsequence $\{x_{n_k}\}_{k=1}^\infty$ and some $x_0\in K$ such that $x_{n_k} \rightharpoonup x_0$. Thus, \begin{equation} f(x_0) \leq \liminf_{k\to \infty} f(x_{n_k}) =\inf_{x\in K} f(x) \leq f(x_0), \end{equation} i.e., $f(x_0)= \inf_{x\in K}f(x)$.