In dimensions $d\geq 3$ the Coulomb energy is always nonnegative (since the Fourier transform of $\frac{1}{\\cdot\^{d2}}$ is nonnegative). What can one say about positivity properties of the Coulomb energy in $d=2$? $$D(f,g):=\displaystyle\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}\overline{f(x)}g(y)\log{\xy\}~dxdy$$ Is $\int f=0$, $f\in L^1(\mathbb{R}^2)$ sufficient for $D(f,f)\geq 0$? References would be appreciated.

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. SpringerVerlag, New YorkHeidelberg, 1972. 


To deal with this singular integral is rather subtle. If you think about it, it may be possible that your integral is actually not welldefined 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), 265276), 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 selfadvertising ;) 


$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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, $$\color{#00f}{\large% \varepsilon = {1 \over 2\pi}\int_{{\mathbb R}^{2}} \verts{\nabla\Phi\pars{\vec{r}}}^{2}\,\dd^{2}\vec{r}\ \geq\ 0} $$ 

