# Diameter vs radius, part II

A set $A$ in a metric space $X$ has diameter $\mathrm{diam}\, A=\sup_{a,b\in A} |a-b|$ and radius $\mathrm{rad}\, A = \inf_{x\in X}\sup_{a\in A} |a-x|$. It’s easier to find the radius by expressing it in a different form: the smallest (or infimal) value of $r$ such that the $\bigcap_{a\in A} \overline{B}(a,r)\ne\varnothing$, where $\overline{B}(a,r)$ is the closed ball of radius $r$ with center $a$.

Suppose $X$ is a normed linear space and $f\colon A\to X$ is a map that does not increase distances (hence does not increase the diameter of any set). I already said that the radius may increase under $f$, but my example was incorrect. Here is a correct one: the set $A$ consists of 3 points in red.

The blue hexagon is the unit sphere in this space. The three points in red have distance 2 from one another. So do their images under $f$, but the radius increases from $1$ to $2/\sqrt{3}$. The set $f(A)$ consists of three vertices of the regular hexagon (in green) circumscribed about the blue one.

I think this example is as bad as it gets in two dimensions: that is, we should have $\mathrm{rad}\, f(A) \le \frac{2}{\sqrt{3}} \mathrm{rad}\, A$ in any 2-dimensional normed space. Informally, the worst case is when the unit ball is triangular, which it can’t be because of the symmetry requirement. The hexagon is the next worst thing.

In higher dimensions the constant cannot be smaller than $\frac{2}{\sqrt{3}}$, since the above construction can be implemented in a subspace. I don’t know whether the constant grows with dimension or not (either way it can’t exceed 2, as remarked before).