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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.

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If $0$ is not in $C$, then its depth is zero, right? but your minmax formula won't give zero. – Gerry Myerson Apr 12 '11 at 6:57
Sounds like a homework problem... Do you understand the case when $\partial C$ is $C^2$-smooth? If NO take some calculus course. If YES, then you can take $\epsilon$-nbhd of $C$, its boundary is $C^{1,1}$ and is is as $C^2$ for all practical purposes. – Anton Petrunin Apr 12 '11 at 13:22
Gerry: right, I should have specified that C contains the origin. Anton: it's a problem that came up in my research, not a homework problem. I'm not a mathematician, though. – John Schulman Apr 12 '11 at 13:52
Both statements are true (with the assumption that $C$ contains the origin in its interior; you probably also mean to assume that $C$ is compact so that you really can write min/max instead of inf/sup). Since someone is sawing through a brick wall two floors above me, I can't concentrate enough to write out a proof right now, so here are some key words to help you get started: support function, polar body, and gauge/Minkowski functional. These are discussed in any book on convex geometry. – Mark Meckes Apr 12 '11 at 14:15
up vote 5 down vote accepted

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 $v = r u^\ast \in C$.

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