# Nonexpanding Jordan curves

A simple closed curve ${\Gamma}$ on the plane can be parameterized by a homeomorphism ${f\colon \mathbb T\to \Gamma}$ in infinitely many ways. It is natural to look for “nice” parameterizations: smooth ones, for example. I do not want to require smooth ${\Gamma}$ here, so let us try to find ${f}$ that is nonexpanding, that is ${|f(a)-f(b)|\le |a-b|}$ for all ${a, b\in \mathbb T}$. Note that Euclidean distance is used here, not arclength.

What are some necessary conditions for the existence of a nonexpanding parameterization?

1. The curve must have length at most ${2\pi}$, since nonexpanding maps do not increase length. But this is not sufficient: an equilateral triangle of sidelength ${2\pi/3}$ has no nonexpanding parameterization, despite its length being ${2\pi}$.
2. The curve must have diameter at most 2 (which the triangle in item 1 fails). Indeed, nonexpanding maps do not increase the diameter either. However, ${\mathrm{diam}\,\Gamma\le 2}$ is not sufficient either: an equilateral triangle of sidelength ${2}$ has no nonexpanding parameterization, despite its diameter being 2 (and length ${6 < 2\pi}$).
3. The curve must be contained in some closed disk of radius 1. This is not as obvious as the previous two items. We need Kirszbraun’s theorem here: any nonexpanding map ${f\colon \mathbb T\to \Gamma}$ extends to a nonexpanding map ${F\colon \mathbb R^2\to\mathbb R^2}$, and therefore ${f(\mathbb T)}$ is contained in the closed disk of radius 1 centered at ${F(0)}$. (This property fails for the triangle in item 2.)

The combination of 1 and 3 (with 2 being superseded by 3) still is not sufficient. A counterexample is given by any polygon that has length ${2\pi}$ but is small enough to fit in a unit disk, for example:

Indeed, since the length is exactly ${2\pi}$, a nonexpanding parametrization must have constant speed 1. But mapping a circular arc onto a line segment with speed 1 increases pairwise Euclidean distances, since we are straightening out the arc.
1. It is sufficient for ${\Gamma}$ to be a convex curve contained in the unit disk. Indeed, the nearest-point projection onto a convex set is a nonexpanding map, and projecting the unit circle onto the curve in this way gives the desired parameterization.
2. It is sufficient for the curve to have length at most 4. Indeed, in this case there exists a parameterization ${f\colon \mathbb T\to\Gamma}$ with ${|f'|\le 4/(2\pi) = 2/\pi}$. For any ${a, b\in \mathbb T}$ the length of the shorter subarc ${\gamma}$ between ${a}$ and ${b}$ is at most ${(\pi/2)|a-b|}$. Therefore, the length of ${f(\gamma)}$ is at most ${|a-b|}$, which implies ${|f(a)-f(b)|\le |a-b|}$.
Clearly, neither of two sufficient conditions is necessary for the existence of a nonexpanding parameterization. But one can consider “length ${\le 2\pi}$ is necessary, length ${\le 4}$ is sufficient” a reasonable resolution of the problem: up to some constant, the length of a curve can decide the existence one way or the other.