Green Functions, Harmonic Measure, and Conformal Invariance: Examples and Counterexamples

Put $M(z):=(z-a)\mathrm{/}(1-\bar{a}z)=-({\phi }_a(z))$. Since $\mathrm{\mid }\bar{a}z\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }<1$ for $z\in \mathbb{D}$, the denominator does not vanish and $M$ is holomorphic on $\mathbb{D}$ by [F3]. For $z\in \mathbb{D}$ with $\mathrm{\mid }z\mathrm{\mid }=r$, $\mathrm{\mid }1-\bar{a}z{\mathrm{\mid }}^2-\mathrm{\mid }z-a{\mathrm{\mid }}^2=(1-\mathrm{\mid }a{\mathrm{\mid }}^2)(1-r^2)>0,$ so $\mathrm{\mid }M(z)\mathrm{\mid }<1$. No involution property of $M$ is needed.

Hence $G(z):=\log \mathrm{\mid }1-\bar{a}z\mathrm{\mid }-\log \mathrm{\mid }z-a\mathrm{\mid }=-\log \mathrm{\mid }M(z)\mathrm{\mid }$ satisfies $G(z)>0$ for $z\in \mathbb{D}\setminus \{a\}$ and $G(z)+\log \mathrm{\mid }z-a\mathrm{\mid }=\log \mathrm{\mid }1-\bar{a}z\mathrm{\mid }$ for $z\mathrm{\ne }a$.

On the circles $\mathrm{\mid }z\mathrm{\mid }=r$ with $\mathrm{\mid }a\mathrm{\mid }<r<1$ one has $1-\mathrm{\mid }M(z){\mathrm{\mid }}^2=(1-\mathrm{\mid }a{\mathrm{\mid }}^2)(1-r^2)\mathrm{/}\mathrm{\mid }1-\bar{a}z{\mathrm{\mid }}^2$, where $\mathrm{\mid }1-\bar{a}z\mathrm{\mid }\ge 1-\mathrm{\mid }a\mathrm{\mid }r\ge 1-\mathrm{\mid }a\mathrm{\mid }>0$; consequently $1-\mathrm{\mid }M(z){\mathrm{\mid }}^2\le (1-\mathrm{\mid }a{\mathrm{\mid }}^2)(1-r^2)\mathrm{/}(1-\mathrm{\mid }a\mathrm{\mid }{)}^2\to 0$ as $r\uparrow 1$, uniformly in the argument. For such $r$, $M(z)\mathrm{\ne }0$ on the circle, and $G(z)=-\log \mathrm{\mid }M(z)\mathrm{\mid }=-\frac{1}{2}\log \mathrm{\mid }M(z){\mathrm{\mid }}^2$; hence $G\to 0$ uniformly on the circles $\mathrm{\mid }z\mathrm{\mid }=r$ as $r\uparrow 1$, and in particular $G(z)\to 0$ as $\mathrm{\mid }z\mathrm{\mid }\to 1$.

$G$ is symmetric in its two arguments: writing $G(z,w)=\log \mathrm{\mid }1-\bar{w}z\mathrm{\mid }-\log \mathrm{\mid }z-w\mathrm{\mid }$ for the same formula in two variables, the identities $\mathrm{\mid }1-\bar{w}z\mathrm{\mid }=\mathrm{\mid }\overline{1-w\stackrel{‾}{z}}\mathrm{\mid }=\mathrm{\mid }1-w\bar{z}\mathrm{\mid }$ and $\mathrm{\mid }z-w\mathrm{\mid }=\mathrm{\mid }w-z\mathrm{\mid }$ give $G(z,w)=G(w,z)$: the two-variable formula is symmetric.

The two summands of $G$ are harmonic: $z\mapsto \log \mathrm{\mid }1-\bar{a}z\mathrm{\mid }$ is $(\log \mathrm{\mid }\cdot \mathrm{\mid })\circ (1-\bar{a}z)$ with $1-\bar{a}z$ holomorphic and nowhere zero on $\mathbb{D}$, and $z\mapsto \log \mathrm{\mid }z-a\mathrm{\mid }$ is $(\log \mathrm{\mid }\cdot \mathrm{\mid })\circ (z-a)$ with $z-a$ holomorphic and nowhere zero on $\mathbb{C}\setminus \{a\}$; hence both are harmonic on $\mathbb{D}\setminus \{a\}$ by [F2], and so is $G$ by [F3].

Leastness: let $u$ be any logarithmic-pole candidate at $a$. Then $u+\log \mathrm{\mid }z-a\mathrm{\mid }$ extends across $a$ to a harmonic function on $\mathbb{D}$ by [F1], and by step 2.1 the function $G+\log \mathrm{\mid }z-a\mathrm{\mid }$ agrees on $\mathbb{D}\setminus \{a\}$ with the harmonic function $\log \mathrm{\mid }1-\bar{a}z\mathrm{\mid }$ of step 3.2; hence $D:=u-G$ extends from $\mathbb{D}\setminus \{a\}$ to the difference of two harmonic functions on $\mathbb{D}$, which is harmonic by [F3]. Fix $\epsilon >0$ and let $r<1$ be so close to $1$ that $G<\epsilon$ on $\mathrm{\mid }z\mathrm{\mid }=r$, as step 2.2 permits. On that circle $D=u-G\ge -\epsilon$ because $u\ge 0$, so the infimum of $D$ over the closed disc $\{\mathrm{\mid }z\mathrm{\mid }\le r\}$ is at least $-\epsilon$ by [F4]. As $\epsilon \downarrow 0$ and $r\uparrow 1$ we get $D\ge 0$, that is $u\ge G$ on $\mathbb{D}\setminus \{a\}$. Hence $G$ is the pointwise least candidate, so $g_{\mathbb{D}}(\cdot ,a)=G$ by [F1].

By step 2.1 the function $G+\log \mathrm{\mid }z-a\mathrm{\mid }$ agrees on $\mathbb{D}\setminus \{a\}$ with the harmonic function $z\mapsto \log \mathrm{\mid }1-\bar{a}z\mathrm{\mid }$ of step 3.2, which is harmonic on all of $\mathbb{D}$. Together with steps 2.1 and 3.2 this shows that $G$ is a logarithmic-pole candidate at $a$ in the sense of [F1].

The kernel therefore has all the asserted properties: it is positive on $\mathbb{D}\setminus \{a\}$ by step 2.1, harmonic there with $G+\log \mathrm{\mid }z-a\mathrm{\mid }$ harmonic across $a$ by steps 3.2 and 4.2, symmetric by step 3.1 together with $g_{\mathbb{D}}=G$ from step 4.1, and it tends to zero at every boundary point of the unit circle by step 2.2; the logarithmic coefficient is one because $\log \mathrm{\mid }z-a\mathrm{\mid }$ is subtracted exactly once. No boundary datum was prescribed and no extension of $M$ beyond the disc was used, so irregular-boundary questions do not arise.

Let $A_E:=\{t\in [0,2\pi ):e^{it}\in E\}$. Choose the integer $m$ for which ${\alpha }^{\mathrm{\prime }}:=\alpha -2\pi m\in [0,2\pi )$, and put ${\beta }^{\mathrm{\prime }}:=\beta -2\pi m$, so ${\alpha }^{\mathrm{\prime }}<{\beta }^{\mathrm{\prime }}\le {\alpha }^{\mathrm{\prime }}+2\pi$. If ${\beta }^{\mathrm{\prime }}\le 2\pi$, then $A_E$ agrees with $[{\alpha }^{\mathrm{\prime }},{\beta }^{\mathrm{\prime }}]\cap [0,2\pi )$ up to the possible duplicate endpoint at $0$. If ${\beta }^{\mathrm{\prime }}>2\pi$, it agrees with $[{\alpha }^{\mathrm{\prime }},2\pi )\cup [0,{\beta }^{\mathrm{\prime }}-2\pi ]$ up to endpoints. In the full-circle case ${\beta }^{\mathrm{\prime }}-{\alpha }^{\mathrm{\prime }}=2\pi$, one has $A_E=[0,2\pi )$. In each case, splitting the integral at $2\pi$ if necessary and translating one part by $2\pi$, the periodicity in [F2] gives ${\int }_{A_E}{k}_z(t)dt={\int }_{\alpha }^{\beta }{k}_z(t)dt$. Endpoints have zero angular measure.

The kernel $k_z(t):=(1-\mathrm{\mid }z{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }{e}^{it}-z{\mathrm{\mid }}^2$ is continuous and positive on $\mathbb{R}$ by [F2], and $1-\mathrm{\mid }z{\mathrm{\mid }}^2>0$ for $z\in \mathbb{D}$, so $k_z$ is a nonnegative measurable function on the interval $[\alpha ,\beta ]$ of finite length $\beta -\alpha \le 2\pi$.

Combining [F1] with step 1.1 expresses the harmonic measure of the arc as an integral of $k_z$ over $[\alpha ,\beta ]$: ${\omega }_{\mathbb{D}}^z(E)={\int }_{A_E}{k}_z(t)\frac{dt}{2\pi }=\frac{1}{2\pi }{\int }_{\alpha }^{\beta }{k}_z(t)dt\mathrm{.}$

The right-hand integral is an ordinary integral of a continuous function on a compact interval: by [F2] $k_z$ is Riemann integrable on $[\alpha ,\beta ]$ and its Riemann and Lebesgue integrals over that interval coincide, so the value in step 2.1 is well defined and equals the displayed Riemann integral.

At $z=0$ the kernel is identically one, because $\mathrm{\mid }{e}^{it}-0\mathrm{\mid }=1$ and $1-\mathrm{\mid }0{\mathrm{\mid }}^2=1$; the identity of step 2.1 therefore gives ${\omega }_{\mathbb{D}}^0(E)=\frac{1}{2\pi }{\int }_{\alpha }^{\beta }1dt=\frac{\beta -\alpha }{2\pi },$ a number in $[0,1]$, equal to $1$ exactly when the arc is the full circle $\beta =\alpha +2\pi$ and equal to the normalized angular length otherwise.

The calculation used only the explicitly given Poisson density and the elementary integration of a continuous periodic kernel; Dependent Choice was used only through [F1]. In particular no harmonicity of $z\mapsto {\omega }_{\mathbb{D}}^z(E)$, and no regularity of an indicator of $E$ as a boundary datum, was used or asserted here.

Define the Cayley map $C(w):=(w-i)\mathrm{/}(w+i)$, which is holomorphic on $\mathbb{C}\setminus \{-i\}$ and in particular on $\mathbb{H}$, and satisfies $\mathrm{\mid }C(w)\mathrm{\mid }<1$ for $w\in \mathbb{H}$ because $\mathrm{\mid }w+i{\mathrm{\mid }}^2-\mathrm{\mid }w-i{\mathrm{\mid }}^2=4\mathrm{Im}w>0$. For real $t$ one computes $C(t)=\frac{(t-i{)}^2}{t^2+1}=\frac{t^2-1-2ti}{t^2+1}=e^{i\theta (t)},\theta (t):=2\arctan t-\pi \in (-2\pi ,0),$ so $\mathrm{\mid }C(t)\mathrm{\mid }=1$, $C(t)\mathrm{\ne }1$, and by the chain rule and [F3] $C^{\mathrm{\prime }}(t)=\frac{2i}{(t+i{)}^2},\mathrm{\mid }{C}^{\mathrm{\prime }}(t)\mathrm{\mid }=\frac{2}{t^2+1}={\theta }^{\mathrm{\prime }}(t)\mathrm{.}$

For every $R>0$ the substitution $s=(t-x)\mathrm{/}y$ on $[-R,R]$, which is admissible by [F4] with $\phi (t)=(t-x)\mathrm{/}y$ and $g(s)=1\mathrm{/}(\pi (1+s^2))$, gives ${\int }_{-R}^R{k}_z(t)dt=\frac{1}{\pi }(\arctan \frac{R-x}{y}-\arctan \frac{-R-x}{y}),$ and letting $R\to \mathrm{\infty }$ with [F3] yields the improper Riemann integral ${\int }_{\mathbb{R}}{k}_{z}dt=1$. Since $k_z$ is continuous and nonnegative, [F5] makes $k_z$ Lebesgue integrable with ${\int }_{\mathbb{R}}{k}_{z}d{\lambda }_1=1$.

For $w\in \mathbb{H}$ and real $t$, direct expansion gives $1-\mathrm{\mid }C(w){\mathrm{\mid }}^2=\frac{4\mathrm{Im}w}{\mathrm{\mid }w+i{\mathrm{\mid }}^2},\mathrm{\mid }C(t)-C(w)\mathrm{\mid }=\frac{2\mathrm{\mid }t-w\mathrm{\mid }}{\mathrm{\mid }t+i\mathrm{\mid }\mathrm{\mid }w+i\mathrm{\mid }},$ the second identity because $(t-i)(w+i)-(w-i)(t+i)=2i(t-w)$. Multiplying, and using [F1], [F3] and step 1.1, $\frac{P(C(w),C(t))\mathrm{\mid }{C}^{\mathrm{\prime }}(t)\mathrm{\mid }}{2\pi }=\frac{4y}{\mathrm{\mid }w+i{\mathrm{\mid }}^2}\cdot \frac{\mathrm{\mid }t+i{\mathrm{\mid }}^2\mathrm{\mid }w+i{\mathrm{\mid }}^2}{4\mathrm{\mid }t-w{\mathrm{\mid }}^2}\cdot \frac{2}{2\pi (t^2+1)}=\frac{y}{\pi ((t-x{)}^2+y^2)}=k_z(t)$ whenever $w=z=x+iy$.

By [F5] the set function ${\omega }^z(A):={\int }_A{k}_{z}d{\lambda }_1$, defined on the Lebesgue sigma-algebra and hence on the Borel sets of $\mathbb{R}$, is a measure, and ${\omega }^z(\mathbb{R})=1$ by step 1.2. So $d{\omega }_{\mathbb{H}}^z=k_{z}d{\lambda }_1$ is a Borel probability measure on $\mathbb{R}$.

Representation for compactly supported data. Let $f$ be continuous with compact support and define $\psi$ on $\mathrm{\partial }\mathbb{D}$ by $\psi (\zeta ):=f(i(1+\zeta )\mathrm{/}(1-\zeta ))$ for $\zeta \mathrm{\ne }1$ and $\psi (1):=0$; this is well defined with $C$-values on the unit circle, and it is continuous: near $\zeta =1$ one has $\mathrm{\mid }1+\zeta \mathrm{\mid }\ge 2-\mathrm{\mid }1-\zeta \mathrm{\mid }$, so the parameter $i(1+\zeta )\mathrm{/}(1-\zeta )$ tends to infinity and $\psi$ vanishes there because $f$ does, while continuity away from $1$ follows from continuity of $f$ and of $\zeta \mapsto i(1+\zeta )\mathrm{/}(1-\zeta )$. Let $u_D$ be the Poisson integral of $\psi$ and $u:=u_D\circ C$ on $\mathbb{H}$. Then $u$ is harmonic by [F2], and for $z\in \mathbb{H}$ the definition of the Poisson integral, the identity $\psi (C(t))=f(t)$ and step 2.1 give $u(z)=\frac{1}{2\pi }{\int }_0^{2\pi }\psi (e^{i\theta })P(C(z),e^{i\theta })d\theta =\frac{1}{2\pi }{\int }_{\mathbb{R}}f(t)P(C(z),C(t))\mathrm{\mid }{C}^{\mathrm{\prime }}(t)\mathrm{\mid }dt={\int }_{\mathbb{R}}fd{\omega }_{\mathbb{H}}^z,$ the middle equality being the substitution $\theta =\theta (t)$ of step 1.1 on a compact parameter interval, legitimate by [F4] and the vanishing of the integrand near the ends $0$ and $-2\pi$ of the period interval, together with $2\pi$-periodicity.

Boundary limits for compactly supported data. If $z_j\to t_0$ with $t_0\in \mathbb{R}$ and $z_j\in \mathbb{H}$, then $C(z_j)\to C(t_0)\in \mathrm{\partial }\mathbb{D}\setminus \{1\}$ by continuity of $C$ at $t_0$, and $u_D$ is continuous on $\bar{\mathbb{D}}$ with boundary values $\psi$ by [F1]; hence $u(z_j)=u_D(C(z_j))\to \psi (C(t_0))=f(t_0)$ by step 3.1.

General bounded continuous data. Let $f$ be bounded and continuous and let ${\chi }_N:\mathbb{R}\to [0,1]$ be continuous with ${\chi }_N=1$ on $[-N,N]$ and ${\chi }_N=0$ outside $[-N-1,N+1]$ (for instance ${\chi }_N(t)=\max \{0,\min \{1,N+1-\mathrm{\mid }t\mathrm{\mid }\}\}$). Each $f_N:=f{\chi }_N$ is continuous with compact support, so $u_{f_N}$ is harmonic on $\mathbb{H}$ by step 3.1. If $K\subseteq \mathbb{H}$ is compact and $R$ is such that $\mathrm{\mid }w\mathrm{\mid }\le R$ on $K$, then for $N>2R$ and $w=x+iy\in K$ the tail bound ${\int }_{\mathrm{\mid }t\mathrm{\mid }>N}{k}_w(t)dt\le \frac{2R}{\pi }{\int }_N^{\mathrm{\infty }}\frac{4}{t^2}dt=\frac{8R}{\pi N}$ holds, because $\mathrm{\mid }t-x\mathrm{\mid }\ge \mathrm{\mid }t\mathrm{\mid }-R\ge \mathrm{\mid }t\mathrm{\mid }\mathrm{/}2$ and $(t-x{)}^2+y^2\ge (t-x{)}^2$ there; hence $\mathrm{\mid }{u}_f-u_{f_N}\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\cdot 8R\mathrm{/}(\pi N)$ on $K$ and $u_{f_N}\to u_f$ locally uniformly on $\mathbb{H}$. Therefore $u_f$ is harmonic on $\mathbb{H}$ by [F6].

Boundary limits for general data. Fix $t_0\in \mathbb{R}$ and $\eta >0$. Continuity of $f$ at $t_0$ gives $\delta >0$ with $\mathrm{\mid }f(t)-f(t_0)\mathrm{\mid }<\eta$ for $\mathrm{\mid }t-t_0\mathrm{\mid }<\delta$. Splitting the defining integral at $\mathrm{\mid }t-t_0\mathrm{\mid }=\delta$ and using ${\int }_{\mathbb{R}}{k}_z=1$ from step 1.2, $\mid u_f(z)-f(t_0)\mid \le \eta +2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\mathrm{\mid }t-t_0\mathrm{\mid }\ge \delta }{k}_z(t)dt,$ and the last integral equals $\frac{1}{\pi }(\pi -\arctan \frac{t_0+\delta -x}{y}+\arctan \frac{t_0-\delta -x}{y})$, which tends to $0$ as $x\to t_0$, $y\downarrow 0$ by [F3], since then $(t_0\pm \delta -x)\mathrm{/}y\to \pm \mathrm{\infty }$. Hence $u_f(z)\to f(t_0)$ as $z\to t_0$ inside $\mathbb{H}$.

The density $k_z$ therefore defines a Borel probability measure of total mass one on $\mathbb{R}$ (step 2.2) whose integrals produce, for every bounded continuous boundary function, a harmonic function with the prescribed limit at every finite boundary point of $\mathbb{H}$ (steps 4.2 and 5.1). Steps 1.2 and 2.2 use ${\mathrm{A}\mathrm{C}}_{\omega }$ through the measure and integral interface of [F5], as does the integral interpretation in step 3.1; step 4.2 also uses it through the locally uniform harmonic-limit theorem [F6], while steps 1.1, 2.1 and the arctangent limits are choice-free calculations.

The density $k_z$ is continuous and nonnegative on $\mathbb{R}$ by [F1], so its restriction to the compact interval $[a,b]$ is Riemann integrable by [F3]; the interval is Borel, so the integral ${\int }_{[a,b]}{k}_{z}d{\lambda }_1$ defining ${\omega }_{\mathbb{H}}^z([a,b])$ is well posed.

Substituting $s=(t-x)\mathrm{/}y$ on $[a,b]$, admissible by [F3] with $\phi (t)=(t-x)\mathrm{/}y$ (so ${\phi }^{\mathrm{\prime }}=1\mathrm{/}y\mathrm{\ne }0$) and $g(s)=1\mathrm{/}(\pi (1+s^2))$, gives for the Riemann integral ${\int }_a^b{k}_z(t)dt=\frac{1}{\pi }{\int }_{(a-x)\mathrm{/}y}^{(b-x)\mathrm{/}y}\frac{ds}{1+s^2}=\frac{1}{\pi }(\arctan \frac{b-x}{y}-\arctan \frac{a-x}{y}),$ the last equality by the antiderivative in [F2].

By the Riemann-Lebesgue agreement of [F3] the same value is the Lebesgue integral of $k_z$ over $[a,b]$, hence ${\omega }_{\mathbb{H}}^z([a,b])=\frac{1}{\pi }(\arctan \frac{b-x}{y}-\arctan \frac{a-x}{y})\mathrm{.}$

The displayed value lies in $[0,1]$: it is nonnegative because $\arctan$ is strictly increasing and $a<b$ imply $\arctan ((a-x)\mathrm{/}y)<\arctan ((b-x)\mathrm{/}y)$, and it is at most $(\pi \mathrm{/}2-(-\pi \mathrm{/}2))\mathrm{/}\pi =1$ because $\arctan$ takes values in $(-\pi \mathrm{/}2,\pi \mathrm{/}2)$.

Boundary values. If $t_0\in (a,b)$ and $z\to t_0$ with $z\in \mathbb{H}$, then $(b-x)\mathrm{/}y\to +\mathrm{\infty }$ and $(a-x)\mathrm{/}y\to -\mathrm{\infty }$, so the value tends to $(\pi \mathrm{/}2+\pi \mathrm{/}2)\mathrm{/}\pi =1$. If $t_0\mathrm{\notin }[a,b]$, then $(b-x)\mathrm{/}y$ and $(a-x)\mathrm{/}y$ tend to the same signed infinity, so their arctangents have the same limit and the value tends to $0$. If $t_0=a$ and $z$ approaches $a$ with $x=a$, then $(a-x)\mathrm{/}y=0$ and $(b-x)\mathrm{/}y\to +\mathrm{\infty }$, so the value tends to $(\pi \mathrm{/}2-0)\mathrm{/}\pi =1\mathrm{/}2$; the same computation at $t_0=b$ gives $1\mathrm{/}2$.

Therefore the harmonic measure of a real interval for the upper half-plane is the explicit arctangent expression of step 3.1, bounded between zero and one by step 4.1 and attaining the boundary values one, zero and one half in the three configurations of step 4.2. The use of ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through the Borel-measure and Lebesgue-integral interface in steps 1.1 and 3.1, including their Riemann-to-Lebesgue comparison.

The annulus $A$ is connected: for $w_k={\rho }_k{e}^{i{\theta }_k}\in A$, $k=1,2$, the path $w(t)=((1-t){\rho }_1+t{\rho }_2)\exp \left(i\right((1-t){\theta }_1+t{\theta }_2\left)\right)$, $t\in [0,1]$, has modulus between ${\rho }_1$ and ${\rho }_2$ and hence values in $A$, and it joins $w_1$ to $w_2$. Together with openness and boundedness from [F1] this makes $A$ a bounded complex domain.

Boundary points of the outer circle are regular. Fix $\xi \in C_R$ and set ${\zeta }_{\ast }:=2\xi$, so that $\mathrm{\mid }{\zeta }_{\ast }\mathrm{\mid }=2R$ and $\mathrm{\mid }{\zeta }_{\ast }-\xi \mathrm{\mid }=R$. Define $b(w):=\log R-\log \mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }$ for $w\in A$. Since ${\zeta }_{\ast }\mathrm{\notin }\bar{A}$, the map $w\mapsto w-{\zeta }_{\ast }$ is a holomorphic map with values in $\mathbb{C}\setminus \{0\}$ on the open set containing $\bar{A}$, so $w\mapsto \log \mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }$ is harmonic on $A$ by [F4] and $b$ is harmonic, hence subharmonic, on $A$ by [F4]. For $w\in A$ the reverse triangle inequality gives $\mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }\ge \mathrm{\mid }{\zeta }_{\ast }\mathrm{\mid }-\mathrm{\mid }w\mathrm{\mid }>2R-R=R$, so $b(w)<0$; and $b(\xi )=\log R-\log R=0$ with $b$ continuous at $\xi$, so $b(w)\to 0$ as $w\to \xi$. Finally $b\le 0$ on $\mathrm{\partial }A$ with equality only at $\xi$: on $C_R$ the identity $\mathrm{\mid }\eta -{\zeta }_{\ast }\mathrm{\mid }=R$ holds exactly when $\eta =\xi$, and on $C_r$ one has $\mathrm{\mid }\eta -{\zeta }_{\ast }\mathrm{\mid }\ge 2R-r>R$. Hence for every neighbourhood $V$ of $\xi$ the continuous function $b$ is strictly negative on the compact set $\mathrm{\partial }A\setminus V$ (if that set is empty there is nothing to bound), so its maximum there is some $c_V<0$; thus $b$ is a barrier at $\xi$, and $\xi$ is regular by [F3].

Boundary points of the inner circle are regular. Fix $\xi \in C_r$ and set ${\zeta }_{\ast }:=\xi \mathrm{/}2$, so that $\mathrm{\mid }{\zeta }_{\ast }\mathrm{\mid }=r\mathrm{/}2$ and $\mathrm{\mid }\xi -{\zeta }_{\ast }\mathrm{\mid }=r\mathrm{/}2$. Define $b(w):=\log (r\mathrm{/}2)-\log \mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }$ for $w\in A$. Here ${\zeta }_{\ast }\mathrm{\notin }\bar{A}$, so $b$ is harmonic on $A$ by [F4] and hence subharmonic. For $w\in A$ the reverse triangle inequality gives $\mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }\ge \mathrm{\mid }w\mathrm{\mid }-\mathrm{\mid }{\zeta }_{\ast }\mathrm{\mid }>r-r\mathrm{/}2=r\mathrm{/}2$, so $b(w)<0$; and $b(\xi )=0$ with $b$ continuous at $\xi$, so $b(w)\to 0$ as $w\to \xi$. On $\mathrm{\partial }A$ one has $b\le 0$ with equality only at $\xi$: the equality $\mathrm{\mid }\eta -{\zeta }_{\ast }\mathrm{\mid }=r\mathrm{/}2$ forces $\mathrm{\mid }\eta \mathrm{\mid }=r$ and $\eta$ on the ray through ${\zeta }_{\ast }$, that is $\eta =\xi$, while for $\mathrm{\mid }\eta \mathrm{\mid }=R$ one has $\mathrm{\mid }\eta -{\zeta }_{\ast }\mathrm{\mid }\ge R-r\mathrm{/}2>r\mathrm{/}2$. So for every neighbourhood $V$ of $\xi$ the maximum of $b$ over the compact set $\mathrm{\partial }A\setminus V$ is a $c_V<0$, and $b$ is a barrier at $\xi$; by [F3] the point $\xi$ is regular.