# Generalization of an inequality due to Gage for curve shortening

There is a well known inequality due to Gage which asserts the following. Let $\Omega$ be a smooth, convex set in $\mathbb{R}^2$ and let $p = \langle X, \nu \rangle$ be the support function of $\Omega$, where $X = \langle x, y \rangle$ with respect to some origin $O$ and $\nu$ is the normal to the boundary.

Denoting $A$ and $L$ and the area and length of the curve, then it holds that $\int_{\partial \Omega} p^2 dS \leq \frac{AL}{\pi}$ for some particular choice of origin $O \in \Omega$.

Question/Conjecture: Given an arbitrary simply connected, smooth set $\Omega$, does it hold that $\int_{\partial \Omega} p^2 dS \leq \frac{LA^*}{\pi}$ where $A^*$ denotes the area of the convex hull of $\Omega$ and $L$ is the length of the original boundary $\partial \Omega$.

Update June 04/2012: There has been an answer to my original question so I would like to ask if a related although weaker assertion is true. Let $\partial \Omega$ be paramaterizable by the angle $\theta$ in polar coordinates, so that the curve is represented by $(r(\theta),\theta)$. Then $p = p(\theta)$ is obviously single valued. This means precisely that the domain $\Omega$ is star shaped. Let $p^*$ be the support function of the convex hull $\Omega^*$. Does it hold that $\int_{\partial \Omega} p^2 \leq \int_{\partial \Omega^*} (p^*)^2 dS$?

Any direction to references on related questions would also be greatly appreciated.

• Why are you using a different symbol for the area -- $A^*$ looks like some dual object. Also, is $L$ the perimeter of the boundary? Jun 3, 2012 at 19:08
• What is the angle $\theta$? Jun 4, 2012 at 10:39
• Hopefully clarified this. I mean that the curve is $(r(\theta),\theta)$ in polar coordinates. Jun 4, 2012 at 11:12
• You should distinguish a bit more carefully between the domain $\Omega$ and its boundary $\partial\Omega$. And if the boundary is parameterizable by the polar angle, then isn't the support function automatically single-valued? And $\Omega$ is called a star-shaped domain? Jun 4, 2012 at 13:02
• Ok I have hopefully clarified the question. Yes I believe that the name for what I'm saying is star shaped. Jun 4, 2012 at 13:06

The conjecture is false. Let $ABCD$ be a square inscribed in the unit circle. Consider the following piecewise smooth loop in the plane. First it starts from $A$ and moves between $A$ and $B$ along the circle back and forth $N$ times where $N$ is a large odd number. Then it goes from $B$ to $C$ along the the straight line segment $[BC]$. Then it moves between $C$ and $D$ along the circle back and forth $N$ times, and finally returns from $D$ to $A$ along the segment $[DA]$.
This loop is neither smooth nor simple, but it can be approximated by a smooth simple loop which has almost the same derivative most of the time. This approximation bounds a domain $\Omega$ for which we have $L\approx \pi N+2\sqrt2$, $A^*\approx \frac\pi2+1<\pi$. By symmetry, the best choice of the origin is the center of the circle, and then $\int_{\partial\Omega} p^2\approx \pi N +2$. Hence $$\lim_{N\to\infty}\int_{\partial\Omega} p^2/LA^* = \frac1{\pi/2+1} > \frac1\pi ,$$ thus the desired inequality fails if $N$ is large enough.