Which convex subsets of a normed space are intersections of balls? Let $(V, \lVert \cdot \rVert)$ be a normed space. For any $A \subseteq V$, let $O(A)$ be the intersection of all closed balls containing $A$, or more precisely, let $O \colon 2^V \to 2^V$ be defined by the formula:
\begin{equation}
\forall A \in 2^V \quad O(A) = \bigcap_{x \in V} \overline{B}(x, r_A(x)),
\end{equation}
where $\overline{B}(x, r) = \{ y \in V \colon \lVert y - x \rVert \le r\}$ and $r_A(x) = \sup \{ \lVert a - x \rVert \colon a \in A\}$. Clearly, for every $A \subseteq V$ set $O(A)$ is convex. In the case when $A$ is (metrically) unbounded, $O(A) = V$, because one can allow for $r_A(x)$ to equal $\infty$. In the case when $A = \{x\}$, then $O(A) = \{x\}$.

Question:
I am interested in whether there is some characterization of what types of sets must be in the range of $O$.

For example, if $V = \mathbb{R}^2$, I suspect that regardless of the used norm, the range of $O$ will contain all line segments which are parallel to one of at least two noncollinear directions. Here, the reason being that:

*

*if a given direction is parallel to the supporting line of the unit ball at some point and the supporting line at that point is unique, then that direction has the mentioned property,

*if at a given point of the unit ball we have some family of supporting lines, then the directions from the boundary of this family, here, lines with the highest and lowest leading coefficient in the slope-intercept form, have this property

 A: According to the comment in my previous answer, let met try a more general approach. I assume that $V$ is a normed vector space such that the unit sphere has a unique supporting hyperplane for each of its boundary point.
Warning: this is just a preliminary work that I am posting not to lose it. I intend to finish it when I have more time...
The key point is to prove that larger and larger spheres ressemble more and more to half-spaces. A precise statement could be the following: Choose a unit vector $e \in V$ and define $B_\lambda = B(-\lambda e, \lambda)$ for all $\lambda > 0$.

*

*Claim 1: If $\lambda' \geq \lambda$, $B_\lambda \subset B_{\lambda'}$.

Indeed, if $x \in B_\lambda$, $\|x + \lambda e\| \leq \lambda$. Thus
$$
\|x+ \lambda' e \| = \|x+ \lambda e + (\lambda' - \lambda)e\| \leq
\|x+ \lambda e\| + (\lambda' - \lambda)\|e\| \leq \lambda',
$$
showing that $x \in B_{\lambda'}$.

*

*Claim 2: $C = \bigcup_{\lambda > 0} B_\lambda$ is a convex cone.

It is convex as if $x, y \in C$, there exists $\lambda > 0$ so that $x, y \in B_\lambda$, so the segment $[x,y]$ is in $B_\lambda \subset C$. Now, if $x \in C$ and $\alpha \geq 0$, let $\lambda > 0$ be such that $x \in B_\lambda$. Then we have
$$
\|\alpha x + \alpha \lambda e\| \leq \alpha \|x + \lambda e\| \leq \alpha \lambda
$$
so $\alpha x \in B_{\alpha \lambda} \subset C$.
It should be noted that $C$ is not closed. On $\mathbb{R}^n$, if we choose $e = e_1$, then $C = \{x \in \mathbb{R}^n, x_1 < 0\} \cup \{0\}$.

*

*Claim 3: $\overline{C}$ is a half-space.

As $e \not\in B_\lambda$ for all $\lambda$, $e \not\in C$ so, from then Hahn-Banach theorem, there exists $\phi \in V^*$ such that $\phi(e) > 0$, $\phi(x) \leq 0$ for all $x \in \overline{C}$. If $\overline{C} \neq \{x \in V, \phi(x) \leq 0\}$, there exists $x_0 \in V$ such that $\phi(x_0) < 0$ (as the complement of $\overline{C}$ is open). From the Hahn-Banach theorem, there exists $\psi \in V^*$ such that $\psi(x_0) > 0$ and $\psi(x) \leq 0$ for all $x \in \overline{C}$. $\phi$ and $\psi$ cannot be proportional so  $\ker(\phi)$ and $\ker(\psi)$ are two distinct supporting hyperplanes for $\overline{C}$ and thus for (say) $B_1$. This contradicts the assumption showing that $\overline{C}^is a half-space.
Assume now that $A$ is a non-empty bounded convex set. As is well-known, $A$ is the intersection of all its supporting half-spaces, so we want to prove that $O(A)$ is contained in all these half-spaces. The idea is as in my previous answer: we construct a sequence of balls whose center goes to infinity in some direction so they have bigger and bigger radius and ressemble near infinity to the half-space we chose. Our first task is to find the right direction in which to move our centers:

*

*Claim 4: (assume that $V$ is reflexive) For any (vector) hyperplane $H$, there exists a vector $e \in \overline{B}(0, 1)$ such that $H$ is the supporting hyperplane of $\overline{B}(-e, 1)$ at $0$.

Let $\phi \in V^*$ be such that $H = \ker(\phi)$ and $\|\phi\|= 1$. If $\dim(V)$ is finite, there exists $e \in \overline{B}(0, 1)$ such that $\phi(e) = 1$. The affine hyperplace $\{x \in V, \phi(x) = 1\}$ is then trivially a supporting hyperplane for $\overline{B}(0, 1)$. The same conclusion holds if $V$ is reflexive as the unit ball is weakly compact and $\phi$ is weakly continuous.
A: Let me assume that $V$ is a Hilbert space.
I claim that all closed bounded convex subsets $A$ satisfy $A = O(A)$.
Assume that $A = \bigcap_i \Omega_i$ for some family of convex subsets $\Omega_i$. As $A \subset \Omega_i$, we have $O(A) \subset O(\Omega_i)$. Taking intersections, we have
$$
A \subset O(A) \subset \bigcap_i O(\Omega_i).
$$
If we find a family of subsets $B_i$ such that $C_i \supset O(B_i)$ satisfies $A = \bigcap_i C_i$, we obtain that $A = O(A)$.
As is well-known, $A$ is the intersection of its supporting half-spaces. As half spaces $\Omega$ cannot satisfy $\Omega = O(\Omega)$, we need to choose some better candidates for our $\Omega$. So we choose cylinders: given a vector $e \in V$ and $x_0 \in A$, let
$$
a = \inf_{x \in A} \langle e, x\rangle, b = \sup_{x \in A} \langle e, x\rangle\text{ and } r = \sup_{x \in A} \|p_{e^\perp}(x-x_0)\|.
$$
Let $C_{e, x_0}$ be the cylinder defined by
$$
C_{e, x_0} = \{x \in V | \langle e, x\rangle \in [a, b] \text{ and } \|p_{e^\perp}(x-x_0)\| \leq r \}.
$$
For any $e$ and $x_0$, $C_{e, x_0}$ is contained in the supporting half-space $H_{e, a} = \{x\in V | \langle x, e\rangle \geq a\}$ so $A = \bigcap_{e, x_0} C_{e, x_0}$.
All that remains to do is to prove that $O(C_{e, x_0}) \subset H_{e, a}$.
The idea is just to see that $O(C_{e, x_0})$ is contained in the intersection of the balls centered at $x_0 + \lambda e$ for $\lambda >> 0$ and of radius $r_{C_{e, x_0}}(x_0 + \lambda e)$. Details are painful to write down but the idea is geometrically obvious: the balls will ressemble more and more to the half-space $H_{e, a}$.
