Maximum principle for a compact logarithmic potential

Set $S:=\mathrm{supp}\mu$, $m:=\mu (\mathbb{C})>0$ and $p:=p_{\mu }=-U^{\mu }$, so that $p(z)={\int }_{\mathbb{C}}\log \mathrm{\mid }z-w\mathrm{\mid }d\mu (w)\in [-\mathrm{\infty },\mathrm{\infty })$ for every $z\in \mathbb{C}$.

By [F2] the set $S$ is closed, carries $\mu$ and is contained in every closed carrier; since $\mu \mathrm{\ne }0$ is carried by a compact set, $S$ is a nonempty compact subset of that carrier and $\mu (\mathbb{C}\setminus S)=0$, so the function $p$ of step 1.1 satisfies $p(z)={\int }_S\log \mathrm{\mid }z-w\mathrm{\mid }d\mu (w)$ for every $z$.

For $x\in S$ the hypothesis gives $p(x)\ge -M>-\mathrm{\infty }$ for the function $p$ of step 1.1; were $\mu (\{x\})>0$, the diagonal contributes $+\mathrm{\infty }$ to the integral for $U^{\mu }(x)$, while the kernel on the compact support is bounded below, so $U^{\mu }(x)=+\mathrm{\infty }$, that is $p(x)=-\mathrm{\infty }$, so $\mu (\{x\})=0$.

Suppose, for contradiction, that $p(z_0)<-M$ for some $z_0\in \mathbb{C}$; then $z_0\mathrm{\notin }S$ by the hypothesis at the points of $S$, and with $p$ as in step 1.1 one can choose $c$ with $p(z_0)<c<-M$ (if $p(z_0)$ is finite take $c=(p(z_0)-M)\mathrm{/}2$ and if $p(z_0)=-\mathrm{\infty }$ take $c=-M-1$) and set $A:=\{z\in \mathbb{C}\setminus S:p(z)<c\}$.

Hence for every $x\in S$ the decreasing measurable sets $B(x,1\mathrm{/}j)$ satisfy $\mu (B(x,1\mathrm{/}j))\downarrow \mu (\{x\})=0$ as $j\to \mathrm{\infty }$: continuity from above applies to the finite measure $\mu$, and $\mu (\{x\})=0$ is step 2.2.

Since $S$ carries $\mu$ by step 2.1, for $z_0\mathrm{\notin }S$ with $\delta :=d(z_0,S)>0$ the kernel $\log \mathrm{\mid }z-w\mathrm{\mid }$ and its partial derivatives in $z$ of order at most two are continuous and uniformly bounded on $B(z_0,\delta \mathrm{/}2)\times S$. These constant bounds are $\mu$-integrable because $\mu$ is finite. Apply [F4] successively along coordinate intervals to the kernel and its first derivatives: the measurable differentiated integrands obey these bounds, so $\mathrm{\Delta }p(z)={\int }_S{\mathrm{\Delta }}_z\log \mathrm{\mid }z-w\mathrm{\mid }d\mu (w)=0$ by [F5]. Dominated convergence in [F4] makes the resulting derivatives continuous. Thus $p\in C^2$ locally off $S$ and is harmonic there by [F6].

Fix $x\in \mathrm{\partial }S$, $\epsilon \in (0,1)$ and $z\in \mathbb{C}\setminus S$ with $\mathrm{\mid }z-x\mathrm{\mid }<\epsilon \mathrm{/}8$; the continuous function $w\mapsto \mathrm{\mid }z-w\mathrm{\mid }$ attains a least value over the nonempty compact $S$ of step 2.1 at some $y\in S$, so $\mathrm{\mid }y-z\mathrm{\mid }=d(z,S)\le \mathrm{\mid }z-x\mathrm{\mid }$ and hence $\mathrm{\mid }y-x\mathrm{\mid }\le 2\mathrm{\mid }z-x\mathrm{\mid }$ and $\mathrm{\mid }y-w\mathrm{\mid }\le \mathrm{\mid }y-z\mathrm{\mid }+\mathrm{\mid }z-w\mathrm{\mid }\le 2\mathrm{\mid }z-w\mathrm{\mid }$ for every $w\in S$; the estimates below hold for every nearest point $y$, so only its existence is used.

The set $A$ of step 2.3 is bounded: with $r_0:=\max \{1,{\sup }_{w\in S}\mathrm{\mid }w\mathrm{\mid }\}$ (finite by step 2.1) and $\mathrm{\mid }z\mathrm{\mid }\ge 2r_0$ one has $\mathrm{\mid }z-w\mathrm{\mid }\ge \mathrm{\mid }z\mathrm{\mid }\mathrm{/}2$ for every $w\in S$, so by step 2.1 $p(z)\ge m\log (\mathrm{\mid }z\mathrm{\mid }\mathrm{/}2)\to +\mathrm{\infty }$ and $p(z)\ge c$ for all large $\mathrm{\mid }z\mathrm{\mid }$; therefore $\bar{A}$ is closed and bounded, hence compact.

On the region $F:=\{w:\mathrm{\mid }x-w\mathrm{\mid }\ge \epsilon \}$ both $\mathrm{\mid }z-w\mathrm{\mid }$ and $\mathrm{\mid }y-w\mathrm{\mid }$ exceed $3\epsilon \mathrm{/}4$ for the points $z,y$ of step 3.3, so the mean value estimate for the logarithm gives $\mathrm{\mid }\log \mathrm{\mid }z-w\mathrm{\mid }-\log \mathrm{\mid }y-w\mathrm{\mid }\mathrm{\mid }\le 8\mathrm{\mid }z-x\mathrm{\mid }\mathrm{/}(3\epsilon )$, while on the region $N:=\{w:\mathrm{\mid }x-w\mathrm{\mid }<\epsilon \}$ the nearest-point inequality of step 3.3 gives $\log \mathrm{\mid }z-w\mathrm{\mid }\ge \log \mathrm{\mid }y-w\mathrm{\mid }-\log 2$.

Splitting the integral of step 2.1 at $N$ and $F$ and integrating the two pointwise bounds of step 4.1 for the nearest point $y$ of step 3.3 gives $p(z)\ge p(y)-\mu (B(x,\epsilon ))\log 2-8m\mathrm{\mid }z-x\mathrm{\mid }\mathrm{/}(3\epsilon )\ge -M-\mu (B(x,\epsilon ))\log 2-8m\mathrm{\mid }z-x\mathrm{\mid }\mathrm{/}(3\epsilon )$, the last inequality by the hypothesis at $y\in S$; the near part of $\int \log \mathrm{\mid }y-w\mathrm{\mid }d\mu$ is finite because $p(y)\ge -M$ is finite and the integrand on $F$ is bounded, so no difference of infinities occurs.

Consequently, for every $\eta >0$ there is $r>0$ with $p(z^{\mathrm{\prime }})\ge -M-\eta$ whenever $z^{\mathrm{\prime }}\in \mathbb{C}\setminus S$ and $\mathrm{\mid }{z}^{\mathrm{\prime }}-x\mathrm{\mid }<r$: by step 3.1 choose $j\ge 2$ with $\mu (B(x,1\mathrm{/}j))\log 2\le \eta \mathrm{/}2$, and put $\epsilon :=1\mathrm{/}j$ and $r:=\min \{\epsilon \mathrm{/}8,3\epsilon \eta \mathrm{/}(16m)\}>0$, so that the last term of step 5.1 is less than $\eta \mathrm{/}2$; thus ${\mathrm{lim\hspace{0.17em}sup}}_{z^{\mathrm{\prime }}\to x,z^{\mathrm{\prime }}\mathrm{\notin }S}{U}^{\mu }(z^{\mathrm{\prime }})\le M$ at every $x\in \mathrm{\partial }S$.

The set $A$ of step 2.3 is nonempty and open, and $\bar{A}\cap \mathrm{\partial }S=\mathrm{\varnothing }$: since $c<-M$ and step 6.1 applies at every $x\in \mathrm{\partial }S$ with $\eta :=(-M-c)\mathrm{/}2>0$, giving $-M-\eta =(-M+c)\mathrm{/}2>c$, the set $A$ misses a whole ball about each boundary point, so no limit point of $A$ lies in $\mathrm{\partial }S$; hence $\bar{A}\subseteq \mathbb{C}\setminus S$, because a point of $\bar{A}$ lying in $S$ would have every ball about it meeting both $S$ and $\mathbb{C}\setminus S$, that is, would lie in $\mathrm{\partial }S$.

The function $p$ is continuous on the nonempty compact set $\bar{A}$ of step 3.4 and attains there a minimum at some $a\in \bar{A}$; every point $b\in \mathrm{\partial }A$ satisfies $p(b)=c$, because $b\in \bar{A}\subseteq \mathbb{C}\setminus S$ by step 7.1 and $p$ is continuous at $b$, points of $A$ approach $b$ with $p<c$ and points outside $A$ approach $b$ with $p\ge c$ (on $S$ because $p\ge -M>c$ by the hypothesis, on $\mathbb{C}\setminus S$ by the definition of $A$ in step 2.3); since $p(a)\le p(z_0)<c$, the minimiser $a$ lies in $A$, and $p(z)\ge c>p(a)$ for every $z\in \mathbb{C}\setminus S$ outside $\bar{A}$, so $p$ attains a global minimum over $\mathbb{C}\setminus S$ at the interior point $a$.

Let $\mathrm{\Omega }$ be the connected component of $\mathbb{C}\setminus S$ containing $a$; it is open and path-connected, hence a domain, $p{\mathrm{\mid }}_{\mathrm{\Omega }}$ is harmonic by step 3.2, and $a\in \mathrm{\Omega }$ is an interior local minimum of $p{\mathrm{\mid }}_{\mathrm{\Omega }}$, so [F12] forces $p\equiv p(a)$ on $\mathrm{\Omega }$, with $p(a)<c<-M$ by step 8.1.

The component $\mathrm{\Omega }$ is a proper subset of $\mathbb{C}$, because $\mathrm{\Omega }\subseteq \mathbb{C}\setminus S$ and $S\mathrm{\ne }\mathrm{\varnothing }$ by step 2.1; hence $\mathrm{\partial }\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$, since otherwise $\bar{\mathrm{\Omega }}=\mathrm{\Omega }\cup \mathrm{\partial }\mathrm{\Omega }=\mathrm{\Omega }$ would make the nonempty proper subset $\mathrm{\Omega }$ both open and closed in the connected space $\mathbb{C}$.

Every point $\zeta \in \mathrm{\partial }\mathrm{\Omega }$ lies in $\mathrm{\partial }S$: it lies in $\bar{\mathrm{\Omega }}\subseteq \overline{\mathbb{C}\setminus S}$, and if $\zeta \mathrm{\notin }S$ then $\zeta \in \mathbb{C}\setminus S$, whose component ${\mathrm{\Omega }}^{\mathrm{\prime }}$ is open by [F10] and disjoint from $\mathrm{\Omega }$, so $\bar{\mathrm{\Omega }}$ is contained in the closed set $\mathbb{C}\setminus {\mathrm{\Omega }}^{\mathrm{\prime }}$, which does not contain $\zeta$; hence $\zeta \in S$, and every ball about $\zeta$ also meets $\mathbb{C}\setminus S$ because $\zeta$ is a boundary point of the domain $\mathrm{\Omega }$ of step 9.1, so $\zeta \in \mathrm{\partial }S$.

If $\mathrm{\Omega }$ were unbounded, choose $R>2r_0$ with $m\log (R\mathrm{/}2)>p(a)$ and a point $z\in \mathrm{\Omega }$ with $\mathrm{\mid }z\mathrm{\mid }\ge R$; then step 3.4 gives $p(z)\ge m\log (\mathrm{\mid }z\mathrm{\mid }\mathrm{/}2)\ge m\log (R\mathrm{/}2)>p(a)=p(z)$ by step 9.1, a contradiction.

If $\mathrm{\Omega }$ were bounded, step 10.1 gives a point $\zeta \in \mathrm{\partial }\mathrm{\Omega }$, hence $\zeta \in \mathrm{\partial }S$ by step 10.2, and step 6.1 with $\eta :=(-M-p(a))\mathrm{/}2>0$ gives a ball $B(\zeta ,r)$ with $p(z)\ge -M-\eta =(-M+p(a))\mathrm{/}2>p(a)$ for all $z\in (\mathbb{C}\setminus S)\cap B(\zeta ,r)$, using $p(a)<-M$ from step 9.1; but $\zeta \in \mathrm{\partial }\mathrm{\Omega }$ makes $B(\zeta ,r)$ meet $\mathrm{\Omega }$, and any $z$ in that intersection satisfies $p(z)=p(a)$ by step 9.1, a contradiction.

Both cases of steps 11.1 and 10.3 are impossible, so the supposition of step 2.3 is false: $p\ge -M$ holds on $\mathbb{C}\setminus S$, and on $S$ it is exactly the hypothesis, so $p\ge -M$ on all of $\mathbb{C}$ and therefore $U^{\mu }=-p\le M$ everywhere.