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I've run into the problem of trying to evaluate the following:

$\max_{ \xi} \iint_{\partial B \times \partial B} \xi(x) \Phi(|x-y|) \xi(y) dS(x)dS(y)$

subject to $\int_{\partial B} \xi(y)dS(y) = 0$ and $\int_{\partial B}\xi^2(y)dS(y)=1$ where $B \subset \mathbb{R}^3$ is a ball of radius $1$ and $\Phi(|x-y|)=\frac{1}{|x-y|}$ is the Newtonian potential.

This seems to resemble an inverse fractional Soblev norm such as $H^{-1}$ and moreover appears to be related to the problem of finding an optimal Poincare constant.

My guess is that the maximum is obtained for $\xi=+1$ on the upper half and $\xi=-1$ on the lower half. Given this however, I still cannot do an explicit calculation to determine this quantity. Is there a standard reference for such problems arising in Potential Theory perhaps which will allow one to evaluate (even approximately) such expressions?

For instance I know I can rewrite the above as: $\int_{\partial B} |\nabla w|^2$ where $-\Delta w = \mu$ and $\mu(x) = \xi(x)dS(x)$ but I'm not sure how this can help me to evaluate such an expression.

To summarize, I would like to try to evaluate the above double integral for the particular function $\xi = +1$ on the upper half of the ball and $\xi=-1$ on the lower half. Being able to solve explicitly the above maximization problem would be a bonus.

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  • $\begingroup$ In terms of $w$, the solution of a Dirichlet problem with finte dimensional constraints is smooth, likely analytic. Thus the optimal $\xi$ must be smooth. It cannot be a sign function. $\endgroup$ Commented Nov 8, 2011 at 7:07
  • $\begingroup$ This is not a dirichlet energy, it is a non-local H^{-1} type energy and so solutions need not be smooth. $\endgroup$
    – Dorian
    Commented Nov 11, 2011 at 3:14

1 Answer 1

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The potential of $\xi$ is a harmonic function $u$ on the unit ball in the obvious way. $u(x) = \int_{\partial B} \frac{\xi(y)}{|x-y|} dS(y).$

Consider the spherical harmonics decomposition of $u$, given by

$u = \sum_{l=0}^\infty f^l r^l Y^l (\theta, \phi)$

The energy given above reduces to $C_n \int_{\partial B} u u_\nu dS = \sum l |f^l|^2$, while the normalization condition you have given reduces to $\int_{\partial B} |u_\nu|^2 dS = 1 = \sum l^2 |f^l|^2$. Ignoring the case $l=0$ (which has zero energy), you get the maximum of the two ratios at $l=1$ or $f^l = \delta_{1l}$.

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  • $\begingroup$ So are you saying that the maximum ration is $\sum l |f^l|^2 = 1$ since $f^l=\delta_{1l}$? $\endgroup$
    – Dorian
    Commented Dec 6, 2011 at 20:42
  • $\begingroup$ I find that strange as it should depend on $r$ but perhaps I'm missing something from your explanation. $\endgroup$
    – Dorian
    Commented Dec 6, 2011 at 20:43
  • $\begingroup$ Possibly involving some constant depending on dimension, yes. I just realized there may also be a factor of 2 involved, since my expression only captures the energy on the inside of the ball (in this case, $u$ on the outside of the ball is just the Kelvin transform of $u$ on the inside). $\endgroup$
    – Ray Yang
    Commented Dec 8, 2011 at 15:42

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