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.

  • $\begingroup$ Why are you using a different symbol for the area -- $A^*$ looks like some dual object. Also, is $L$ the perimeter of the boundary? $\endgroup$
    – Igor Rivin
    Jun 3, 2012 at 19:08
  • $\begingroup$ What is the angle $\theta$? $\endgroup$
    – Deane Yang
    Jun 4, 2012 at 10:39
  • $\begingroup$ Hopefully clarified this. I mean that the curve is $(r(\theta),\theta)$ in polar coordinates. $\endgroup$
    – Dorian
    Jun 4, 2012 at 11:12
  • $\begingroup$ 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? $\endgroup$
    – Deane Yang
    Jun 4, 2012 at 13:02
  • $\begingroup$ Ok I have hopefully clarified the question. Yes I believe that the name for what I'm saying is star shaped. $\endgroup$
    – Dorian
    Jun 4, 2012 at 13:06

1 Answer 1


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.

  • $\begingroup$ Thanks for your answer. Do you think the result will hold if one assumes that the curve is star shaped and symmetric with respect to reflections across the origin? This seems quite reasonable. $\endgroup$
    – Dorian
    Jun 5, 2012 at 22:53

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