Distance between the boundaries of a convex set and its shrunk version

Question

Let $S$ be a compact convex set and $S'=\{x\in S: d(x, \partial S)\ge \delta\}$, where $d$ is the Euclidean distance and $\partial S$ is the boundary of $S$. Assume that we choose $\delta>0$ such that $S'$ is non-empty. I was wondering how $$\sup_{x\in S}\inf_{y\in S'}d(x,y)$$ is related to $\delta$. I guess the curvature may be involved here.

Answer 1

Let $m=\sup_{x\in S}\inf_{y\in S'} \|x-y\|$. Since $S$ is compact and $d$ is continuous, we see that $S'$ is closed and hence compact. Hence $m=\max_{x\in S}\min_{y\in S'} \|x-y\|$.

Note that $m \ge \delta$. Choose $x \in S, y \in S'$ such that $\|x-y\| = m$. Note that $[s',s] \subset S$ and then $s \in \partial S$ (otherwise there is some $t>1$ such that $s'+t(s-s') \in S$, and then $\|x-y\| - s'\| = t \|x-y\| > m$). Since $\|x-y\| \ge \delta$ we have $m \ge \delta$.

Note that $m$ can be strictly larger than $\delta$. Take the $S$ to be the subset of the plane given by the convex hull of $\overline{B}(0,\delta) $ and the point $(p,0)$. Then $S'=\{0\}$ and $m = |p|$.