A min-max formula for depth of the origin in a convex set Depth is defined as the distance to the boundary of a set, i.e., $\operatorname{depth}(x, C) = \operatorname{dist}(x, \mathbb{R}^n \backslash C)$. 
Let $C$ be a convex set that contains the origin. I believe that
$$\operatorname{depth}(0, C) = \min_{\|u\| = 1} \ \max_{v \in C} \ u \cdot v$$
I also believe the following lemma is true: given $u^\*$ that optimizes the above expression, $\max u^\* \cdot v$ is achieved at $v = \lambda u^\*$ for some scalar $\lambda$ (though not necessarily uniquely).
I don't know how to prove these things or where to look in the literature for these sorts of results.
 A: My office is quieter this morning, so let's try again.  I'll assume that $C$ is compact with $0 \in \operatorname{int} C$.  Let $B_r$ denote the ball of radius $r$ centered at $0$. Then
$\operatorname{depth} (0,C) = \max \{ r > 0 \mid B_r \subseteq C \}$.
The function $h_C(u) = \max_{v\in C} (u \cdot v)$ is called the support function of $C$.  Its crucial property here is that given two compact convex sets $C$ and $K$, $h_C \le h_K$ if and only if $C \subseteq K$.
On the one hand, if $B_r \subseteq C$ then for every unit vector $u$, $h_C(u) \ge h_{B_r}(u) = r$, so
$$\operatorname{depth} (0,C) \le \min_{\| u \| = 1} h_C(u).$$
On the other hand, if $h_C(u) \ge r$ for every unit vector $u$, then $h_C \ge h_{B_r}$ and so $B_r \subseteq C$, and thus
$$\operatorname{depth} (0,C) \ge \min_{\| u \| = 1} h_C(u).$$
Combining the previous two inequalities shows that the answer to your first question is "yes".
As for the second question, if $r = \operatorname{depth}(0,C)$ and $u^\ast$ achieves the minimum, then $r u^\ast \in \partial B_r \cap \partial C$. Since $u^\ast \cdot (r u^\ast) = r$, the maximum is achieved for $v = r u^\ast \in C$.
