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 $\mathrm{rad}\, f(A)=2\,\mathrm{rad}\, A$ .

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”

lianxin says:

2012-04-18 at 15:06

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

The geometry of situation: diameter vs radius

Related

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

Leave a Reply Cancel reply