## 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).

## The geometry of situation: diameter vs radius

Mathematicians and engineers are disinclined to agree about anything in public: should the area of a circle be described using the neat formula $\pi r^2$ or in terms of the more easily measured diameter as $\frac{1}{4} \pi d^2$, for example?
J. Bryant and C. Sangwin, How Round is Your Circle?

I will argue on behalf of the diameter but from a mathematician’s perspective. The diameter of a nonempty set $A\subset \mathbb R^2$ is

$\displaystyle \mathrm{diam}\, A = \sup_{a,b\in A} |a-b|$

Whether $\mathrm{diam}\, \varnothing$ should be $0$ or $-\infty$ I’ll leave for you to decide. The radius of $A$ can be defined as

$\displaystyle \mathrm{rad}\, A = \inf_{x\in \mathbb R^2}\sup_{a\in A}|x-a|$

For a circle — whether this word means $\mathbb S^1$ or $\mathbb D^2$ — these definitions indeed agree with the diameter and radius. The example of $\mathbb S^1$ shows that in the definition of the radius we should not require $x\in A$.

The problem of determining the radius of a given set was posed in 1857 by J.J.Sylvester in Quarterly Journal of Pure and Applied Mathematics. Thanks to Google Books, I can reproduce his article in its entirety:

Suppose that $f\colon A\to \mathbb R^2$ is a map of $A$ that is nonexpanding/short/metric/1-Lipschitz or whatyoucallit: $| f(a)- f(b) | \le |a-b|$ for all $a,b\in A$. Clearly, the diameter does not increase: $\mathrm{diam}\, f(A)\le \mathrm{diam}\,A$. What happens to the radius is not nearly as obvious…

It turns out that the radius does not increase either. Indeed, by Kirszbraun’s theorem $f$ can be extended to a 1-Lipschitz map of the entire plane, and the extended map tells us where the center of a bounding circle should go. Kirszbraun’s theorem is valid in $\mathbb R^n$ for every $n$, as well as in a Hilbert space. Hence, nonexpanding maps do not increase the radius of any subset of a Hilbert space.

However, general normed vector spaces are different…

The example given below is wrong; the map is not 1-Lipschitz. I keep it for historical record.