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:

77 letters, 119 citations on Google Scholar

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.

For example, consider the taxicab distance \|(x,y)\|=|x|+|y| in the plane. The 4-point set in red is mapped by f isometrically in such a way that \mathrm{rad}\, f(A)=2\,\mathrm{rad}\, A.

Taxicab distance

This example is as bad as it gets: for any subset A of a metric space X we have

\mathrm{rad}\, f(A) \le \mathrm{diam}\, f(A)\le \mathrm{diam}\,A \le 2\,\mathrm{rad}\,A

provided that f \colon A\to Y is a nonexpanding map into a metric space Y.

The extra factor of 2 in \mathrm{rad}\, f(A) \le 2\,\mathrm{rad}\,A causes problems when one wants control some quantity \gamma(A) under iteration of f. There’s a big difference between \gamma(f(A))\le 0.8\, \gamma (A) and \gamma(f(A)) \le 1.6\, \gamma (A).

One thought on “The geometry of situation: diameter vs radius

  1. I thought the definition of radius is half of the diameter.. never thought it was so complicated to compute:)

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