# Positivity of the Coulomb energy in two dimensions

In dimensions $d\geq 3$ the Coulomb energy is always non-negative (since the Fourier transform of $\frac{1}{\|\cdot\|^{d-2}}$ is non-negative). What can one say about positivity properties of the Coulomb energy in $d=2$? $$D(f,g):=-\displaystyle\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\overline{f(x)}g(y)\log{\|x-y\|}~dxdy$$ Is $\int f=0$, $f\in L^1(\mathbb{R}^2)$ sufficient for $D(f,f)\geq 0$? References would be appreciated.

• shouldn't the domain of each of the integrals be $\mathbb{R}$? Dec 1, 2013 at 18:42
• No, $f,g\in L^1(\mathbb{R}^2)$ or $L^1_{\text{loc}}(\mathbb{R}^2)$.
– whz
Dec 1, 2013 at 18:51
• Asked 2 days ago on math.stackexchange Dec 1, 2013 at 19:20
• Didn't receive any answers over there so I figured, I should try here. I'm sorry if this is not allowed.
– whz
Dec 1, 2013 at 19:23
• Does the same hold if we replace the Lebesgue measure with a signed measure with compact support?(for $\mathbb{R}^{n\geq 3}$)
– BigM
Jun 17, 2015 at 23:46

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.

• I copied my answer there. Someone has to vote it up to give me the bounty:-) Dec 1, 2013 at 20:49
• Thank you very much for the reference. One more question: Does Theorem 1.16 from your reference not also imply that $\inf f=0$ is sufficient for $D(f,f)\geq 0$? I mean if $\nu$ is the measure with density $f$, we have $\nu(1)=\int 1\times f(x)~dx=0$ and 1.16 implies that in this case the mutual energy is non-negative. I'm not entirely sure what the notation $\nu(g)$ for a function $g$ is supposed to mean, I guess $\nu(g)=\int g(x) \nu(dx)$.
– whz
Dec 2, 2013 at 7:28
• @whz: Your guess about $\nu(g)$ is correct, but your conjecture is not: let the support of $f$ consist of two small pieces very far away from each other. Then the energy is negative. Apr 8, 2015 at 3:11

To deal with this singular integral is rather subtle. If you think about it, it may be possible that your integral is actually not well-defined under the only assumptions you suggest.

One way to deal properly with it, is to assume your function $f$ integrates the $\log$ at infinity, and that $D(|f|,|f|)<+\infty$. Then, your statement has been proved in "Two problems on potential theory for unbounded sets", by Cegrell, Kolodziej and Levenberg (Math. Scand. 83 (1998), 265-276), cf. Theorem 2.5. Here $f$ is actually allowed to be a signed measure.

You may also want to restate everything in terms of the energy over the Riemann sphere (i.e. the one point compactification of $\mathbb R^2$). This is a strategy I used in "http://ecp.ejpecp.org/article/view/1818" and "http://www.sciencedirect.com/science/article/pii/S0021904512000573". Sorry for the self-advertising ;)

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \varepsilon&=-\int_{{\mathbb R}^{2}}\int_{{\mathbb R}^{2}} \fermi\pars{\vec{r}}\fermi\pars{\vec{r}'}\ln\pars{\verts{\vec{r} - \vec{r}'}}\,\dd^{2}\vec{r}\,\dd^{2}\vec{r}' = -\int_{{\mathbb R}^{2}} \int_{{\mathbb R}^{2}}\fermi\pars{\vec{r}}\Phi\pars{\vec{r}}\,\dd^{2}\vec{r} \tag{1} \end{align} where $\ds{% \Phi\pars{\vec{r}} \equiv \int_{{\mathbb R}^{2}} \fermi\pars{\vec{r}'}\ln\pars{\verts{\vec{r} -\vec{r}'}}\,\dd^{2}\vec{r}'}$. Also

$$\nabla^{2}\Phi\pars{\vec{r}} =\int_{{\mathbb R}^{2}}\fermi\pars{\vec{r}'} \overbrace{\braces{\nabla^{2}\ln\pars{\verts{\vec{r} -\vec{r}'}}}}^{\ds{=\ 2\pi\,\delta\pars{\vec{r} - \vec{r}'}}}\,\dd^{2}\vec{r}' =2\pi\,\fermi\pars{\vec{r}}$$

$\pars{1}$ is reduced to \begin{align} \varepsilon&= -\,{1 \over 2\pi} \int_{{\mathbb R}^{2}}\Phi\pars{\vec{r}}\nabla^{2}\Phi\pars{\vec{r}}\,\dd^{2}\vec{r} = -\,{1 \over 2\pi} \int_{{\mathbb R}^{2}}\braces{% \nabla\cdot\bracks{\Phi\pars{\vec{r}}\nabla\Phi\pars{\vec{r}}}\,\dd^{2}\vec{r} - \verts{\nabla\Phi\pars{\vec{r}}}^{2}}\,\dd^{2}\vec{r} \\[3mm]&= -\,{1 \over 2\pi} \int_{{\mathbb R}^{2}} \nabla\cdot\bracks{\Phi\pars{\vec{r}}\nabla\Phi\pars{\vec{r}}}\,\dd^{2}\vec{r} + {1 \over 2\pi}\int_{{\mathbb R}^{2}} \verts{\nabla\Phi\pars{\vec{r}}}^{2}\,\dd^{2}\vec{r} \end{align} The first integral is reduced to a 'line integration', with boundaries with goes '$\to \infty$', via Stokes Theorem and it goes to zero. Then, $$\color{#00f}{\large% \varepsilon = {1 \over 2\pi}\int_{{\mathbb R}^{2}} \verts{\nabla\Phi\pars{\vec{r}}}^{2}\,\dd^{2}\vec{r}\ \geq\ 0}$$