Distributional Laplacian of a compact logarithmic potential

By [F11] the support $S\mathrm{\ne }\mathrm{\varnothing }$ is compact. Fix $R>0$ and put $S_0:={\max }_{w\in S}\mathrm{\mid }w\mathrm{\mid }<\mathrm{\infty }$. For $z\in B(0,R)$ one has $\log \mathrm{\mid }z-w\mathrm{\mid }\le \log (R+S_0)$, so $p_{\mu }^{+}(z)\le M{\log }^{+}(R+S_0)<\mathrm{\infty }$, where $M=\mu (\mathbb{C})$. Also $p_{\mu }^{-}(z)\le \int {\log }^{+}(1\mathrm{/}\mathrm{\mid }z-w\mathrm{\mid })d\mu (w)$. Tonelli and the radial computation ${\int }_{\mathrm{\mid }u\mathrm{\mid }<1}\log (1\mathrm{/}\mathrm{\mid }u\mathrm{\mid })dA(u)=2\pi {\int }_0^{1}r\log (1\mathrm{/}r)dr=\pi \mathrm{/}2$ give ${\int }_{B(0,R)}{p}_{\mu }^{-}(z)dA(z)\le M{\int }_{\mathrm{\mid }u\mathrm{\mid }<1}\log (1\mathrm{/}\mathrm{\mid }u\mathrm{\mid })dA(u)=\pi M\mathrm{/}2<\mathrm{\infty }$. Thus $p_{\mu }^{-}<\mathrm{\infty }$ outside an area-null subset of $B(0,R)$ and $p_{\mu }\in (-\mathrm{\infty },\mathrm{\infty })$ almost everywhere; in particular $p_{\mu }$ is not identically $-\mathrm{\infty }$ on the connected domain $\mathbb{C}$.

Let $z_n\to z$ and choose $C$ with $\log \mathrm{\mid }{z}_n-w\mathrm{\mid }\le C$ for all $w\in S$ and all $n$; the functions $h_n(w):=C-\log \mathrm{\mid }{z}_n-w\mathrm{\mid }$ are nonnegative and measurable, so [F9] gives ${\mathrm{lim\hspace{0.17em}inf}}_n\int h_{n}d\mu \ge \int {\mathrm{lim\hspace{0.17em}inf}}_n{h}_{n}d\mu$, that is, ${\mathrm{lim\hspace{0.17em}sup}}_n{p}_{\mu }(z_n)\le \int {\mathrm{lim\hspace{0.17em}sup}}_n\log \mathrm{\mid }{z}_n-w\mathrm{\mid }d\mu (w)=p_{\mu }(z)$, because $\log \mathrm{\mid }{z}_n-w\mathrm{\mid }\to \log \mathrm{\mid }z-w\mathrm{\mid }$ for $w\mathrm{\ne }z$ and $\to -\mathrm{\infty }$ for $w=z$. Hence $p_{\mu }$ is upper semicontinuous.

For every $w\in \mathbb{C}$ the function $z\mapsto \log \mathrm{\mid }z-w\mathrm{\mid }$ is subharmonic by [F2] applied to the holomorphic function $z\mapsto z-w$; by [F5] it therefore satisfies the circle mean inequality $\log \mathrm{\mid }a-w\mathrm{\mid }\le \frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }a+r{e}^{it}-w\mathrm{\mid }dt$ for all $a\in \mathbb{C}$ and $r>0$.

Fix $a\in \mathbb{C}$, $r>0$ and put $G(t,w):=\log \mathrm{\mid }a+r{e}^{it}-w\mathrm{\mid }$; since $G^{+}\le \log (\mathrm{\mid }a\mathrm{\mid }+r+S_0+1)$ on $S$, the extended integral ${\int }_0^{2\pi }{\int }_{S}Gd\mu dt$ and its reversed iterated integral both equal $\int G^{+}-\int G^{-}$ by two applications of [F7], the difference being well defined because $\int \int G^{+}<\mathrm{\infty }$. Integrating the inequality of step 1.3 over $\mu$ and using this identity gives $p_{\mu }(a)\le \frac{1}{2\pi }{\int }_0^{2\pi }{\int }_{S}Gd\mu dt=\frac{1}{2\pi }{\int }_0^{2\pi }{p}_{\mu }(a+r{e}^{it})dt$.

By step 1.2 $p_{\mu }$ is upper semicontinuous, by step 1.1 it is not identically $-\mathrm{\infty }$ on $\mathbb{C}$, and by step 2.1 it satisfies the circle mean inequality at every closed disc in $\mathbb{C}$; each disc lies in some $B(0,R)$ and the inequality of step 2.1 is exactly the one required by [F5], so $p_{\mu }$ is subharmonic on $\mathbb{C}$, and [F4] makes it locally integrable.

Let $z_0\mathrm{\notin }S$ and $\delta :=\frac{1}{2}d(z_0,S)>0$. On $B(z_0,\delta )$ every $w\in S$ satisfies $\mathrm{\mid }z-w\mathrm{\mid }\ge \delta$, so every partial derivative of order $1$ or $2$ in $z$ of $(z,w)\mapsto \log \mathrm{\mid }z-w\mathrm{\mid }$ is bounded on $B(z_0,\delta )\times S$ by a constant depending only on $\delta$; since $\mu$ is finite, [F10] applied to $x$ and $y$ derivatives lets the Laplacian pass inside the integral, and [F3] gives $\mathrm{\Delta }{p}_{\mu }(z)=\int {\mathrm{\Delta }}_z\log \mathrm{\mid }z-w\mathrm{\mid }d\mu (w)=0$ for $z\in B(z_0,\delta )$. The resulting first and second derivatives are continuous by Dominated convergence, since the kernel derivatives are continuous away from the uniformly separated support and obey the same integrable constant bounds. Hence $p_{\mu }$ is harmonic on the open set $\mathbb{C}\setminus S$.

Let $\phi \in C_c^{\mathrm{\infty }}(\mathbb{C})$. If $\phi =0$ the identity is immediate; otherwise take a nonempty compact set $L$ containing $\mathrm{supp}\phi$ and choose $A>\max \{\mathrm{\mid }z-w\mathrm{\mid }:z\in L,w\in S\}$. Then ${\int }_L\mathrm{\mid }\log \mathrm{\mid }z-w\mathrm{\mid }\mathrm{\mid }dA(z)\le {\int }_{\mathrm{\mid }u\mathrm{\mid }<A}\mathrm{\mid }\log \mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }dA(u)<\mathrm{\infty }$ uniformly for $w\in S$. Since $\mathrm{\Delta }\phi$ is bounded on $L$, the integrand $\log \mathrm{\mid }z-w\mathrm{\mid }\mathrm{\Delta }\phi (z)$ is product-integrable on $L\times S$, and [F8] gives $\int p_{\mu }\mathrm{\Delta }\phi dA=\int (\int \log \mathrm{\mid }z-w\mathrm{\mid }\mathrm{\Delta }\phi (z)dA(z))d\mu (w)$. The inner integral is the distributional pairing $⟨{\mathrm{\Delta }}_x\log \mathrm{\mid }x-w\mathrm{\mid },\phi ⟩$, which by [F6] equals $2\pi \phi (w)$; hence $\frac{1}{2\pi }\int p_{\mu }\mathrm{\Delta }\phi dA=\int \phi d\mu$ for every test function, that is, ${\mu }_{p_{\mu }}=\mu$ in the normalization of Distributional Riesz measure of a plane subharmonic function, equivalently $\mathrm{\Delta }{p}_{\mu }=2\pi \mu$ and $\mathrm{\Delta }{U}^{\mu }=-\mathrm{\Delta }{p}_{\mu }=-2\pi \mu$.