Let $$D(f,g):=\int_{\mathbb{R}^3\times\mathbb{R}^3}\frac{1}{|x-y|}\overline{f(x)}g(y)~dxdy$$ with $f,g$ real valued and sufficiently integrable be the usual Coulomb energy. Under the assumption $D(|f|,|f|)<\infty$ it can be seen that $D(f,f)\geq 0$ (see for example Lieb-Loss, Analysis 9.8). The reason for that is basically that the Fourier transform of $\frac{1}{|\cdot|}$ is non-negative.

In two dimensions the Newton kernel is given by $-\log|\cdot|$ which does not have a positive Fourier transform anymore. I saw the claim that under further assumptions the positivity of the Coulomb energy however does still hold true. Precisely:

If $f\in L^1(\mathbb{R}^2)\cap L^{1+\epsilon}(\mathbb{R}^2)$ for some $\epsilon>0$ such that $$\int_{\mathbb{R}^2}\log(2+|x|)|f(x)|~dx<\infty$$ and $$\int f=0,$$ then $$D(f,f):=-\int_{\mathbb{R}^2\times\mathbb{R}^2}\log{|x-y|}\overline{f(x)}f(y)~dxdy\geq0.$$

The reference given is Carlen, Loss: Competing symmetries, the logarithmic HLS inequalitiy and Onofir's inequality on $S^n$. I don't see how the claim follows from that paper.

Do you have any other reference for the claim above or see a reason why it should be true?


1 Answer 1


This is true when the support of $f$ is contained in the unit disc. If the support is contained in a disc $|z|<R$, then $(f,f)$ is bounded from below by a constant that depends on $R$. This minor nuisance makes the logarithmic potential somewhat different from the Newtonian potential, however most statements of potential theory are similar for these two cases, or can be easily modified. For the details, the standard reference is MR0350027 Landkof, N. S. Foundations of modern potential theory. Springer-Verlag, New York-Heidelberg, 1972.

(This is copied from my ans on math.overflow).

  • $\begingroup$ Under the assumption $\int f=0$ the size of support does not matter: the integral scales. $\endgroup$ Dec 3, 2013 at 3:19
  • $\begingroup$ With log as a kernel, it does not scale. A constant is added to the log. $\endgroup$ Dec 3, 2013 at 5:55

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