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Consider a compact, convex domain $\Omega \subset \mathbb{R}^3$ with $|\Omega|=1$ with smooth boundary $\partial \Omega$.

Now consider the electric potential generated by this uniform mass distribution: $\phi = \int_{\Omega} \frac{1}{|x-y|} dx$. Question: I would like to know if there is a relationship between mean curvature $H$ and $\phi$. What I have in mind is an inequality of the form $\|\phi - \bar \phi\|_{L^p(\partial \Omega)} \leq C \|H - \bar H\|_{L^p(\partial \Omega)}$ where $\bar \phi$ and $\bar H$ denote the averages over $\partial \Omega$ of $\phi$ and $H$ respectively and $p$ can be anything for the time being.

Although one term is much more 'non-local' than the other, for a convex body it seems reasonable that the closer to being an surface of constant mean curvature, the closer one is to being an equipotential surface. This is related to a previous question of mine which ended up going unanswered regarding whether the only convex, compact equipotential surface in $\mathbb{R}^3$ was a sphere or not.

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I have no time to look for a precise reference, but this belongs to the general theory of rearrangement in PDEs. Reliable authors are Talenti, Bandle and perhaps Mossino. – Denis Serre Sep 8 2011 at 15:46
Isn't $\phi$ infinite along the boundary of $\Omega$? – Deane Yang Sep 8 2011 at 16:11
No. It's like integrating 1/r in R^3 which will give you something of order 1 since surface measure is r^2 – Dorian Sep 8 2011 at 16:40
A more elaborate way of seeing this is that the characteristic function of $\Omega$ is an $L^p$ function for all $p \in [1,\infty]$ and standard elliptic theory says that $\phi$ must be $H^2$ so in fact continuous. – Dorian Sep 8 2011 at 16:42
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Isn't this problem related to the capacitor problem? Have a look at Sections 11.15-11.17 of the book "Analysis" of E. H. Lieb and M. Loss (2nd. edition), or the book "Function Spaces and Potential Theory" by D. R. Adams and L. I. Hedberg. – Pedro Lauridsen Ribeiro Sep 8 2011 at 21:23
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