Maximum distance between a point in the convex hull of $A$ and its closest neighbour in $A$

Question

I have a bounded set $A \subset \mathbb{R}^{n}$. I'm trying to compute the maximum distance (or an upper bound to it) between a point belonging to the convex hull of $A$ and its closest neighbour in $A$: \begin{equation*} \max_{x \in \mathcal{C}(A)} d(x, \Pi_{A}(x)), \end{equation*} where $\mathcal{C}(A)$ is the convex hull of $A$, and $\Pi_{A}(a) = \mathop{\mathrm{argmax}}_{b \in A} d(a, b)$. A trivial upper bound would be $\mathrm{diam}(A)$, since $\mathrm{diam}(A) = \mathrm{diam}(\mathcal{C}(A))$. My initial guess for a better bound is $\frac{1}{2} \mathrm{diam}(A)$, but I'm unable to prove it formally.

Answer 1

If $x$ belongs to a convex hull, then it lies on some line segment $[y,z]$ where both $y,z\in A$, i.e. $x\in [y,z]$, so $dist(y,z)=dist(y,x)+dist(x,z)$. Hence $dist(x,A)\le \min\{ (dist(x,y),dist(x,z)\}\le \frac{dist(y,z)}2\le \frac12 \mathrm{diam}(A)$.