Poisson density of harmonic measure on a disc

Every boundary point of a disc is regular. Fix $\xi \in \mathrm{\partial }D(c,R)$ and put ${\zeta }_{\ast }:=c+2(\xi -c)$, so that $\mathrm{\mid }{\zeta }_{\ast }-c\mathrm{\mid }=2R$ and $\mathrm{\mid }{\zeta }_{\ast }-\xi \mathrm{\mid }=R$. Define $b(w):=\log R-\log \mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }$ for $w\in D(c,R)$. Then $b$ is harmonic on $D(c,R)$ by [F4], since $w\mapsto \log \mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }$ is the composition of $\log \mathrm{\mid }\cdot \mathrm{\mid }$ with the translation $w\mapsto w-{\zeta }_{\ast }$, which is holomorphic and nowhere zero on the disc. Moreover $\mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }>R$ for $w\in D(c,R)$, so $b<0$ there; $b(\xi )=\log R-\log R=0$, so $b\to 0$ at $\xi$ by continuity of $b$ at $\xi$. For every neighbourhood $V$ of $\xi$, choose an open neighbourhood $W$ with $\xi \in W\subseteq V$. If $\mathrm{\partial }D(c,R)\setminus W$ is empty, the uniform separation condition in [F3] is vacuous and any $c_V<0$ works. Otherwise the continuous boundary extension of $b$ attains a strictly negative maximum on the nonempty compact set $\mathrm{\partial }D(c,R)\setminus W$; choosing this maximum as $c_V$ gives the required bound at every point of $\mathrm{\partial }D(c,R)\setminus V$. Hence $b$ is a barrier at $\xi$ and $\xi$ is regular by [F3]; since $\xi$ was arbitrary, $D(c,R)$ is a bounded regular plane domain.

For a continuous $\phi :\mathrm{\partial }D(c,R)\to \mathbb{R}$ put $\psi (t):=\phi (c+R{e}^{it})$ and $u(\zeta ):=P[\psi ]((\zeta -c)\mathrm{/}R)$ for $\zeta \in D(c,R)$. By [F1] applied to the unit disc, $u$ is harmonic on $D(c,R)$ and continuous on its closure with boundary values $\phi$. For $\zeta =z$ the definition of the Poisson kernel gives, with $\xi =c+R{e}^{it}$, $u(z)=\frac{1}{2\pi }{\int }_0^{2\pi }\phi (\xi )\frac{R^2-\mathrm{\mid }z-c{\mathrm{\mid }}^2}{\mathrm{\mid }\xi -z{\mathrm{\mid }}^2}dt,$ because $1-\mathrm{\mid }(z-c)\mathrm{/}R{\mathrm{\mid }}^2=(R^2-\mathrm{\mid }z-c{\mathrm{\mid }}^2)\mathrm{/}{R}^2$ and $\mathrm{\mid }{e}^{it}-(z-c)\mathrm{/}R\mathrm{\mid }=\mathrm{\mid }\xi -z\mathrm{\mid }\mathrm{/}R$.

Since all boundary points of $D(c,R)$ are regular by step 1.1, the regularized Perron envelope $H_{\phi }$ of a continuous datum $\phi$ is harmonic on $D(c,R)$ and has the boundary limit $\phi$ at every boundary point; it is therefore a continuous harmonic extension of $\phi$ to the closure, and so is $u$ by step 1.2. Uniqueness [F2] gives $H_{\phi }=u$, hence by step 1.2 $H_{\phi }(z)=\frac{1}{2\pi }{\int }_0^{2\pi }\phi (c+R{e}^{it})\frac{R^2-\mathrm{\mid }z-c{\mathrm{\mid }}^2}{\mathrm{\mid }c+R{e}^{it}-z{\mathrm{\mid }}^2}dt\mathrm{.}$

Define the measure $\nu$ on $\mathrm{\partial }D(c,R)$ by $\nu (E):=\frac{1}{2\pi }{\int }_{\{t\in [0,2\pi ):c+R{e}^{it}\in E\}}\frac{R^2-\mathrm{\mid }z-c{\mathrm{\mid }}^2}{\mathrm{\mid }c+R{e}^{it}-z{\mathrm{\mid }}^2}dt$. The integrand is continuous and positive, so $\nu$ is a finite Borel measure on the compact circle, and step 2.1 says exactly that $\int \phi d\nu =H_{\phi }(z)$ for every continuous $\phi$; taking $\phi \equiv 1$ and using the representation theorem of [F2], $\nu$ is a probability measure. Hence $\nu$ is a harmonic measure for $D(c,R)$ at $z$, and by the uniqueness in [F2], $\nu ={\omega }_{D(c,R)}^z$. Finally, the parametrization $t\mapsto c+R{e}^{it}$ is arclength measured in units $ds=Rdt$, so the density of ${\omega }_{D(c,R)}^z$ with respect to $s$ is $(R^2-\mathrm{\mid }z-c{\mathrm{\mid }}^2)\mathrm{/}(2\pi R\mathrm{\mid }\xi -z{\mathrm{\mid }}^2)$, which is the displayed second form.

Consequently the harmonic measure of the disc $D(c,R)$ at $z$ has the Poisson density of the statement, in both its angle form and its arclength form, and it is a probability measure on the boundary circle. The kernel computation of steps 1.2 and 3.1 is choice-free; $\mathrm{D}\mathrm{C}$ was used only in step 2.1 through the uniqueness theorem [F2] and in the identification of $\nu$.