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 min-max formula for depth of the origin in a convex set
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 $v = r u^\ast \in C$.


