# Which domain maximizes the energy of the Lebesgue measure?

This could be asked in more generality, but let me stick to a concrete case.

Usually one considers a fixed domain $E \subset \mathbb{C}$ and attaches to it the equilibrium probability measure $\nu_E$, the one that minimizes the energy integral $$I_E(\nu) := \int_E \int_E -\log{|z-w|} \, d\nu(z) \, d\nu(w).$$ I want to look at the opposite problem. Fix an infinite measure $\mu$ on $\mathbb{C}$, say the Lebesgue measure $dx\, dy$, and look for the measurable set $E \subset \mathbb{C}$ of unit measure $\mu(E) = 1$ for which the energy $I_E(\mu)$ is a maximum.

Is there a literature available on this type of question? In particular, what is the answer when $\mu$ is the Lebesgue measure? I am fine with assuming $E$ is a 'nice' domain of unit area rather than a general measurable set.

The same question could be posed in Euclidean space of any dimension, or a more general Riemannian manifold. For dimension one and the Lebesgue measure on $\mathbb{R}$ the maximum is clearly attained for a unit interval, with energy $$\int_0^1\int_0^1 -\log{|x-y|} \, dx \, dy = 3/2.$$

• It ought to be a circle (or generally a Euclidean sphere) of unit area (volume), right? – Noam D. Elkies May 21 '15 at 14:31
• Doesn't Steiner symmetrization prove this? – Boris Bukh May 21 '15 at 14:38
• @NoamD.Elkies: That is what I thought (the ball of unit volume), but I didn't see how to prove this. – Vesselin Dimitrov May 21 '15 at 16:31
• @BorisBukh: Thanks for the key word, I'll think about this. – Vesselin Dimitrov May 21 '15 at 17:28

## 1 Answer

That a circular disk maximizes the energy for the Lebesgue measure follows immediately from Riesz's inequality: $I(f,g,h)\le I(f^*,g^*,h^*)$,where $I(f,g,h) = \langle f, g*h\rangle$, and $f^*$ is the monotonically decreasing radial function whose superlevel sets have the same measures as those of $f$. In your case, take $f=g=1_E$ and $h=h^*=-log(|z|)$.

• This is probably only a small technical point, but the quoted result doesn't immediately apply because our $h$ is not non-negative. (I don't think this can be a real problem because it's silly to enter regions where $h<-100$, say, if we want to maximize, so for all practical purposes, $h$ is bounded below.) – Christian Remling May 22 '15 at 1:28
• @ChristianRemling, that's a good point. One way to work around it is, for a given $E$, to cut off $h$ at the diameter $d$ of $E$, $h(r>d)=h(d)$, and add a constant for positivity. You still get that the modified energy for $E$, which is the same as the real energy for $E$ plus a constant, is less than the modified energy for the disk, which is the same as the real energy for the disk plus the same constant. – Yoav Kallus May 22 '15 at 1:55