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I asked a question recently about generalizing an inequality due to Gage. This inequality asserts that given a convex domain $\Omega$ in $\mathbb{R}^2$ with support function $p(X) = \langle X, \nu \rangle$ where $X$ is the position vector with respect to the origin and $\nu$ is the normal on $\partial \Omega$ that

$\int_{\partial \Omega} p^2 dS \leq \frac{LA}{\pi}$, holds for some choice of origin where $L$ is the length of the curve and $A$ the area.

I have been trying to prove the following generalization for star shaped sets:

$\int_{\partial \Omega} p^2 dS \leq \int_{\partial \Omega^*} (p^*)^2 dS^* + C(L,A)(A^*-A)$,

where here $\Omega^*$ is the convex hull of $\Omega$, $p^*$ is the support function of the convex hull, $A$ is the area of the set and $A^*$ the area of the convex hull. Using Green's theorem I have been able to deduce the following: $\int_0^{2\pi} p(\theta)^2 \frac{dS}{d\theta}{d\theta} = \int_0^{2\pi} p^*(\theta)^2 \frac{dS^*}{d\theta} dS^* + \int_0^{2\pi} p^*(\theta)p(\theta)( \frac{dS^*}{d\theta} -\frac{dS}{d\theta}) d\theta + \frac{1}{2} \int_{0}^{2\pi} [p(\theta)^*-p^(\theta)][r^*(\theta)^2 - r(\theta)^2] d\theta$,

where $r^*$ and $r$ are the lengths to $\partial \Omega^*$ and $\partial \Omega$ respectively and we can paramateterize by $\theta$ since the set is star shaped. I thus need only control the term

$\int_0^{2\pi} p^*(\theta) p(\theta)(\frac{dS^*}{d\theta} - \frac{dS}{d\theta}) d\theta$

but this is not clear at all to me. My intuition suggests it should be negative.

I would appreciate any ideas or suggestions for alternative approaches to this.

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Your inequality is not true in general. You can have $\int_{\partial\Omega} p^2 dS >\int_{\partial\Omega^*} (p^*)^2 dS^* + C(A^*-A)$ for any $C$ while keeping $L$ and $A$ a priori bounded.

Indeed, $$ \int_{\partial\Omega} p^2 dS = \int_0^{2\pi} \frac{r^4}{\sqrt{r^2+{r'}^2}} d\theta $$ where $r=r(\theta)$ is the radial function and $\theta$ is the angular coordinate. Begin with $\Omega^*$ being a polygon approximating a long and narrow ellipse. Its radial function $r^*(\theta)$ is piecewise smooth. Approximate $r^*$from below by a piecewise constant function so that the difference is bounded by a small $\varepsilon$. Near points of discuntinuity, connect the constant pieces by linear functions with almost infinite derivative. The resulting function is $r(\theta)$. If the approximation is chosen so that $r=r^*$ at points corresponding to vertices of the original polygon $\Omega^*$, then $\Omega^*$ is indeed the convex hull of the set $\Omega$ defined by $r(\theta)$. However, $$ \int_{\partial\Omega} p^2 dS \approx \int_0^{2\pi} r^3 d\theta\ge \int_0^{2\pi} (r^*-\varepsilon)^3 d\theta > \int_0^{2\pi} \frac{(r^*)^4}{\sqrt{(r^*)^2+{(r^*)'}^2}} = \int_{\partial\Omega^*} (p^*)^2 dS^* $$ for $\varepsilon$ sufficiently small, since the term ${(r*)'}^2$ in the denominator is not negligible.

Moreover, $\int p^2 dS - \int(p^*)^2$ is bounded away from zero, but $A^*-A\to 0$ as $\varepsilon\to 0$ and $L$ stays bounded, so the difference $\int p^2 dS - \int(p^*)^2$ is not majored by $C(L,A)(A^*-A)$.

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Thank you for this interesting example, but if $C(L,A) > 0$ this improves the inequality no? If you see the last term that I added in my calculation, the term $\int_0^{2\pi} [p^*(\theta)-p(\theta)][(r^*)^2(\theta)-r(\theta)] d\theta$, this term indeed is controlled by $C(A^*-A)$ where $C>0$. Thus your example shows that what I want is certainly false for $C(L,A)=0$ but the question remains if adding such a term compensates. – Dorian Jun 7 '12 at 9:55
By improves, is a strictly weaker inequality. Sorry. – Dorian Jun 7 '12 at 10:18
Sorry, I misread the question. But the example works anyway. I edited the answer so that it answers the correct question. – Sergei Ivanov Jun 7 '12 at 20:19

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