TITLE: Convex subset of Hilbert space as intersection of closed balls
QUESTION [8 upvotes]: How does one prove that any closed, convex, and bounded subset of a Hilbert space is the intersection of the closed balls that contain it?

REPLY [9 votes]: Let $K$ be the convex, closed and bounded set and let $L$ be the intersection of the closed balls containing $K$. Clearly $K\subseteq L$. 
If this inclusion is proper, then by translation we may assume that $0\in L\setminus K$.
$K$ is convex and closed, so we can choose a supporting hyperplane $\Pi$ orthogonal to $k\in \Pi\cap K$  so that $K\subseteq\Pi_+=k+\{h\in H:\mbox{Re}(h,k)\ge0\}$. By scaling, we may assume that $\|k\|=1$. 
For $r>0$, consider the closed balls $B_r$ of radius $r$ and center $(\frac12+r)k$. We have $\bigcup_{r>0}B_r\supseteq\Pi_+\supseteq K$. Since $K$ is bounded, there is $r>0$ with $B_r\supseteq K$. Since $0\not\in B_r$ by construction, $0\not\in L$, a contradiction.