Let's consider the following problem.
Let $A,C$ be two nonempty sets in a normed space $X$. Let $R(A;C)$ denote the set of real numbers $r \ge 0$ such that $A$ is contained in a closed ball of radius $r$ centered at some point of $C$ (a closed ball of radius $r = 0$ and center $c$ is just the singleton $\{c\}$).
Show that if X is a reflexive Banach space, $A$ is bounded, and $C$ is closed and convex, then $R(A;C)$ has a minimum (and hence it is a closed unbounded interval).
I tried to solve it by using the following theorem.
Theorem. Let $X$ be a reflexive Banach space, $C \subset X$ convex and closed, $f \colon C \to \mathbb{R} \cup \{+\infty\}$ convex and continuous. Suppose that one of the following two conditions is satisfied: $C$ is bounded; $C$ is unbounded and $f$ is coercive ($\lim_{\lVert x \rVert \to + \infty} f(x) = +\infty$). Then $f$ has a minimum on $C$ ($\exists \min_C f$).
In particular, I defined $f \colon C \to \mathbb{R}$ as $f(c) = \inf \{ r \ge 0 \colon A \subset B_r(c) \}$, where $B_r(c)$ is the closed ball of radius $r$ centered at $c$. My idea is that it is enough to show $f$ has a minimum on $C$ (by using the theorem above) to get that $R(A;C)$ has a minimum. But I'm not sure about my idea.
How would you solve the problem?
Let $f(c) = \sup_{a \in A} \|a-c\|$. Since $A$ is bounded, $f$ is finite, Furthermore $f$ is Lipschitz hence continuous. In addition, it is straightforward to show that $\lim_{\|c\| \to \infty} f(c) = +\infty$.
Pick some $c'\in C$ and define $K= \{ (c, f(c)) | c \in C, f(c) \le f(c') \}$. $K$ is closed, convex and bounded. Since $K$ is (norm)closed and convex, it is weakly closed.
Since $X \times \mathbb{R}$ is reflexive, Banach Alaogu tells us that the closed ball is weakly compact. Since $K$ is a weakly closed subset of some closed ball, it is weakly compact.
Define the (weakly continuous) functional $\phi((x,\alpha)) = \alpha$ and note that since $K$ is weakly compact that $\phi$ has a minimiser $(c^*, \alpha^*) \in K$. It follows from this and the above that $f(c) \ge f(c^*)$ for all $c \in C$.