Capacity of a disc and its circular equilibrium measure

Put $\mu :={\omega }_{D(a,r)}^a$, the harmonic measure of the disc $K^{\circ }=D(a,r)$ at its centre. By [F5] the measure $\mu$ is a Borel probability measure carried by $\mathrm{\partial }K\subseteq K$, and for every Borel $f\ge 0$ one has $\int fd\mu =\frac{1}{2\pi }{\int }_0^{2\pi }f(a+r{e}^{it})dt$; in particular $\mu$ has no atoms, since for a single point $w$ the set $\{t\in [0,2\pi ):a+r{e}^{it}=w\}$ has at most two elements and Lebesgue measure zero, so $\mu (\{w\})=0$.

The circle average of the kernel. For $c\in \mathbb{C}$ put $M(c):=\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }c-r{e}^{it}\mathrm{\mid }dt\in [-\mathrm{\infty },\mathrm{\infty })$. If $\mathrm{\mid }c\mathrm{\mid }>r$, then $z\mapsto \log \mathrm{\mid }z-c\mathrm{\mid }$ is harmonic on an open set containing the closed disc $\overline{D(0,r)}$, so the circle mean-value property of [F6] gives $M(c)=\log \mathrm{\mid }0-c\mathrm{\mid }=\log \mathrm{\mid }c\mathrm{\mid }$.

If $0<\mathrm{\mid }c\mathrm{\mid }<r$, apply Jensen's formula [F7] on the unit disc to the entire function $f(\zeta ):=r\zeta -c$, which satisfies $f(0)=-c\mathrm{\ne }0$ and has the single zero ${\zeta }_0=c\mathrm{/}r$ of modulus $<1$: $\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }r{e}^{it}-c\mathrm{\mid }dt=\log \mathrm{\mid }f(0)\mathrm{\mid }+\log \frac{1}{\mathrm{\mid }{\zeta }_0\mathrm{\mid }}=\log \mathrm{\mid }c\mathrm{\mid }+\log \frac{r}{\mathrm{\mid }c\mathrm{\mid }}=\log r$, that is, $M(c)=\log r$.

If $c=0$, the integrand defining $M(c)$ is constantly $\log r$, so $M(0)=\log r$. If $\mathrm{\mid }c\mathrm{\mid }=r$, rotate the angle to write $c=r$ without changing the average. For $1\mathrm{/}2\le s<1$, step 1.3 gives $M(sr)=\log r$, and $\mathrm{\mid }{e}^{it}-s{\mathrm{\mid }}^2=(1-s{)}^2+4s{\sin }^2(t\mathrm{/}2)\ge 2{\sin }^2(t\mathrm{/}2)$. The positive part of $\log \mathrm{\mid }r(e^{it}-s)\mathrm{\mid }$ is bounded by ${\log }^{+}(2r)$; its negative part is bounded by $\mathrm{\mid }\log (r\sqrt{2})\mathrm{\mid }+{\log }^{-}\mathrm{\mid }\sin (t\mathrm{/}2)\mathrm{\mid }$. This last function is integrable on $[0,2\pi ]$: $(\sin {)}^{\mathrm{\prime }}(0)=1$ (The derivatives of sine and cosine are cosine and minus sine) gives $\sin v\ge v\mathrm{/}2$ for small positive $v$, the same bound applies near $t=2\pi$, and away from the endpoints the sine has a positive minimum. Thus its only singularities are bounded by constants plus $-\log t$ or $-\log (2\pi -t)$, both integrable. Dominated convergence (Dominated convergence) along $s\uparrow 1$ yields $M(r)=\log r$, and rotation gives $M(c)=\log r$ for every $\mathrm{\mid }c\mathrm{\mid }=r$.

Consequently, for $z\in \mathbb{C}$ and $c:=z-a$, the substitution $w=a+r{e}^{it}$ and [F5] give $U^{\mu }(z)=\int \log \frac{1}{\mathrm{\mid }z-w\mathrm{\mid }}d\mu (w)=\frac{1}{2\pi }{\int }_0^{2\pi }\log \frac{1}{\mathrm{\mid }z-a-r{e}^{it}\mathrm{\mid }}dt=-M(z-a)$; by steps 1.2, 1.3 and 2.1 this is $\log \frac{1}{r}$ when $\mathrm{\mid }z-a\mathrm{\mid }\le r$ and $\log \frac{1}{\mathrm{\mid }z-a\mathrm{\mid }}$ when $\mathrm{\mid }z-a\mathrm{\mid }\ge r$.

The energy of $\mu$. Choose $R>2r=\mathrm{diam}(\mathrm{\partial }K)$; by [F1], $k_R=k+\log R\ge 0$ on $\mathrm{\partial }K\times \mathrm{\partial }K$ and $k_R=k+\log R$ pointwise, so the iterated integral of $k_R$ against $\mu \otimes \mu$ equals $\int U^{\mu }d\mu +\log R$; since $\mathrm{supp}\mu =\mathrm{\partial }K$ and $U^{\mu }=\log \frac{1}{r}$ there by step 3.1 with $\mathrm{\mid }z-a\mathrm{\mid }=r$, [F1] gives $I(\mu )=\int U^{\mu }d\mu =\log \frac{1}{r}$, a finite real number.

Let $\sigma \in P(K)$ be a Borel probability measure on $K$ with $I(\sigma )<+\mathrm{\infty }$; then $\sigma$ is a finite positive compactly supported measure of total mass $1$, and since $\mathrm{supp}\sigma \subseteq K$ step 3.1 gives $U^{\mu }=\log \frac{1}{r}$ on $\mathrm{supp}\sigma$, so by [F1] the mixed energy is $I(\mu ,\sigma )=\int U^{\mu }d\sigma =\log \frac{1}{r}=I(\mu )$.

By [F4], whose Countable Choice hypothesis is supplied by the Axiom of Choice of the statement, the pair $\mu ,\sigma$ of step 5.1 satisfies $I(\sigma -\mu )=I(\sigma )-2I(\mu ,\sigma )+I(\mu )=I(\sigma )-\log \frac{1}{r}\ge 0$, with equality if and only if $\sigma =\mu$; hence every $\sigma \in P(K)$ with finite energy has $I(\sigma )\ge \log \frac{1}{r}=I(\mu )$, with equality only for $\sigma =\mu$, while $\sigma \in P(K)$ with $I(\sigma )=+\mathrm{\infty }$ also satisfies $I(\sigma )\ge I(\mu )$ since $I(\mu )$ is finite. Therefore $V_K={\inf }_{\sigma \in P(K)}I(\sigma )=I(\mu )=\log \frac{1}{r}$, and $\mu$ is the unique minimizer.

By step 6.1 the unique minimizer of the energy over $P(K)$ is $\mu$, so [F3] identifies $\mu$ as the equilibrium measure of $K$ and shows it is the only one; the capacity is $\mathrm{cap}(K)=e^{-V_K}=e^{-\log (1\mathrm{/}r)}=r$.

The boundary circle. The circle $\mathrm{\partial }K$ is compact and nonempty and $P(\mathrm{\partial }K)\subseteq P(K)$, so its Robin constant satisfies $V_{\mathrm{\partial }K}\ge V_K$; conversely every $\sigma \in P(\mathrm{\partial }K)$ is a probability carried by $K$ with $\mathrm{supp}\sigma \subseteq \mathrm{\partial }K\subseteq K$, so step 3.1 gives $U^{\mu }=\log \frac{1}{r}$ on $\mathrm{supp}\sigma$ and the argument of steps 5.1 and 6.1 applies verbatim to yield $I(\sigma )\ge I(\mu )$ with equality only for $\sigma =\mu$; hence $V_{\mathrm{\partial }K}=V_K=\log \frac{1}{r}$ and $\mathrm{cap}(\mathrm{\partial }K)=r$, with unique equilibrium measure $\mu$, which is carried by $\mathrm{\partial }K$.

Combining steps 3.1, 7.1 and 8.1 gives the displayed potential, the capacity $r$ of both $K$ and $\mathrm{\partial }K$, the Robin constant $V_K=\log (1\mathrm{/}r)$, and the identification of normalized arclength as the unique equilibrium measure of each of the two compact sets. Two choice principles are spent in the calculation, both supplied by the Axiom of Choice of the statement: Countable Choice in step 6.1 through [F4], and Dependent Choice in steps 1.1, 3.1 and 8.1 through [F5].