An irregular puncture does not force the Green kernel to vanish

As a subset of $\mathbb{C}$, the punctured disc $\mathrm{\Omega }$ is open, bounded and nonempty. It is path-connected: write $z=\rho e^{i\theta }$ with $0<\rho <1$. The radial segment $t\mapsto ((1-t)\rho +t\mathrm{/}2)e^{i\theta }$, $0\le t\le 1$, joins $z$ to $z\mathrm{/}(2\mathrm{\mid }z\mathrm{\mid })=e^{i\theta }\mathrm{/}2$ while staying at radii strictly between $0$ and $1$; a circular arc of radius $1\mathrm{/}2$ then joins that point to $1\mathrm{/}2$. Thus the path stays in $\mathrm{\Omega }$ and avoids $0$. Hence $\mathrm{\Omega }$ is a bounded complex domain in the sense of A complex domain is a nonempty connected open subset of $\mathbb{C}$.

Since $\mathrm{\Omega }$ is open, $\mathrm{\partial }\mathrm{\Omega }=\bar{\mathrm{\Omega }}\setminus \mathrm{\Omega }$, and $\bar{\mathrm{\Omega }}=\bar{\mathbb{D}}$: the unit disc is contained in the closure of $\mathrm{\Omega }$ and every point of $\mathrm{\partial }\mathbb{D}$ is a limit of points of $\mathrm{\Omega }$, while $0$ is not in $\mathrm{\Omega }$. Hence $\mathrm{\partial }\mathrm{\Omega }=\bar{\mathbb{D}}\setminus (\mathbb{D}\setminus \{0\})=(\bar{\mathbb{D}}\setminus \mathbb{D})\cup \{0\}=\{\mathrm{\mid }z\mathrm{\mid }=1\}\cup \{0\},$ so every boundary point of $\mathrm{\Omega }$ is either the puncture $0$ or a point of the unit circle.

The function $F_a(z)=-\log \mathrm{\mid }z-a\mathrm{\mid }$ is continuous on the boundary of $\mathrm{\Omega }$: $\mathrm{\mid }z-a\mathrm{\mid }\ge \min \{1-\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }a\mathrm{\mid }\}>0$ for $z\in \mathrm{\partial }\mathrm{\Omega }$ by step 2.1, since $a\mathrm{\ne }0$ and $\mathrm{\mid }a\mathrm{\mid }<1$. Define $h(z):=-\log \mathrm{\mid }1-\bar{a}z\mathrm{\mid }$ for $z\in \mathbb{D}$. The polynomial $1-\bar{a}z$ is holomorphic and nowhere zero on $\mathbb{D}$, because $\mathrm{\mid }\bar{a}z\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }<1$, so $h$ is harmonic on $\mathbb{D}$ by [F7], and in particular on $\mathrm{\Omega }$.

On the unit circle, $\mathrm{\mid }1-\bar{a}\xi \mathrm{\mid }=\mathrm{\mid }\xi -a\mathrm{\mid }$ for $\mathrm{\mid }\xi \mathrm{\mid }=1$: indeed $\mathrm{\mid }1-\bar{a}\xi \mathrm{\mid }=\mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }\bar{\xi }-\bar{a}\mathrm{\mid }=\mathrm{\mid }\xi -a\mathrm{\mid }$. Consequently $h(\xi )=-\log \mathrm{\mid }\xi -a\mathrm{\mid }=F_a(\xi )$ on $\mathrm{\mid }\xi \mathrm{\mid }=1$, while $h(0)=0$; the function $h$ is continuous on the closed unit disc, being a composition of continuous functions that is harmonic on the open disc.

For $z\in \mathbb{D}\setminus \{a\}$ the identity $F_a(z)-h(z)=-\log \mathrm{\mid }z-a\mathrm{\mid }+\log \mathrm{\mid }1-\bar{a}z\mathrm{\mid }=\log \mid \frac{1-\bar{a}z}{z-a}\mid =g_{\mathbb{D}}(z,a)$ holds by [F3]; so the difference of the singular term $F_a$ and the harmonic function $h$ is exactly the disc kernel.

Every Perron lower function for the datum $b:=F_a{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}$ is dominated by $h$. Let $v\in \mathcal{P}(b,\mathrm{\Omega })$ and $\epsilon >0$, and put $w:=v-h+\epsilon \log \mathrm{\mid }z\mathrm{\mid }$ on $\mathrm{\Omega }$. This $w$ is subharmonic on $\mathrm{\Omega }$: $v$ is subharmonic by [F4], $h$ is harmonic on $\mathrm{\Omega }$ by step 3.1, and $\log \mathrm{\mid }z\mathrm{\mid }$ is harmonic on $\mathrm{\Omega }\subseteq \mathbb{C}\setminus \{0\}$ by [F7], so [F8] applies to $v+(-h)+\epsilon \log \mathrm{\mid }z\mathrm{\mid }$. At a boundary point $\xi$ with $\mathrm{\mid }\xi \mathrm{\mid }=1$ one has ${\mathrm{lim\hspace{0.17em}sup}}_{z\to \xi }v(z)\le b(\xi )=F_a(\xi )$ by [F4] and step 3.1, while $-h(z)\to -F_a(\xi )$ by step 4.1 and $\epsilon \log \mathrm{\mid }z\mathrm{\mid }\to 0$, so ${\mathrm{lim\hspace{0.17em}sup}}_{z\to \xi }w(z)\le F_a(\xi )-F_a(\xi )+0=0$. At the puncture, [F4] applied at $0$ gives ${\mathrm{lim\hspace{0.17em}sup}}_{z\to 0}v(z)\le b(0)=-\log \mathrm{\mid }a\mathrm{\mid }$, and this value is finite, so $v$ is bounded above on a small punctured neighbourhood of $0$, $h$ is bounded there by step 4.1, and $\epsilon \log \mathrm{\mid }z\mathrm{\mid }\to -\mathrm{\infty }$; hence ${\mathrm{lim\hspace{0.17em}sup}}_{z\to 0}w(z)=-\mathrm{\infty }\le 0$. Thus $w$ is subharmonic on $\mathrm{\Omega }$ and has boundary limsup at most $0$ at every boundary point, over the two boundary cases and a general $\epsilon >0$.

By steps 1.1 and 2.1 the hypotheses of [F4] and [F6] apply to the bounded complex domain $\mathrm{\Omega }$ with the continuous zero datum, so step 5.1 gives $w\in \mathcal{P}(0,\mathrm{\Omega })$ and [F6] gives $w\le 0$, that is $v\le h-\epsilon \log \mathrm{\mid }z\mathrm{\mid }$ on $\mathrm{\Omega }$. Since $-\log \mathrm{\mid }z\mathrm{\mid }>0$ on $\mathrm{\Omega }$ and $\epsilon >0$ is arbitrary, letting $\epsilon \downarrow 0$ yields $v\le h$ on $\mathrm{\Omega }$.

Conversely, each function $h+\epsilon \log \mathrm{\mid }z\mathrm{\mid }$ with $\epsilon >0$ is a Perron lower function for the datum $b$. It is subharmonic on $\mathrm{\Omega }$ because $h$ is harmonic there by step 3.1 and $\epsilon \log \mathrm{\mid }z\mathrm{\mid }$ is harmonic there by [F7], and at every boundary point the limsup condition of [F4] holds: at $\mathrm{\mid }\xi \mathrm{\mid }=1$ the limit is $h(\xi )+\epsilon \log 1=F_a(\xi )=b(\xi )$ by step 4.1, and at $0$ the function tends to $-\mathrm{\infty }\le b(0)$ because $h$ stays bounded near $0$ by step 4.1 while $\epsilon \log \mathrm{\mid }z\mathrm{\mid }\to -\mathrm{\infty }$. Hence [F5] gives $U_b\ge h+\epsilon \log \mathrm{\mid }z\mathrm{\mid }$ on $\mathrm{\Omega }$ for every $\epsilon >0$, and letting $\epsilon \downarrow 0$ gives $U_b\ge h$ on $\mathrm{\Omega }$; step 5.1 gave $U_b\le h$ because the supremum of a family all of whose members are at most $h$ is at most $h$.

Therefore $U_b=h$ on $\mathrm{\Omega }$. Since $h$ is continuous on $\mathrm{\Omega }$ by step 3.1, the regularized envelope of [F5] is $H_b(z)={\lim }_{\rho \downarrow 0}\sup \{h(w):w\in \mathrm{\Omega },\mathrm{\mid }w-z\mathrm{\mid }<\rho \}=h(z)(z\in \mathrm{\Omega })\mathrm{.}$

Step 1.1 makes $\mathrm{\Omega }$ a bounded complex domain and $a\in \mathrm{\Omega }$, so [F2] applies with $b_a=b$ and $h_a=H_b$: the canonical Green kernel of $\mathrm{\Omega }$ at $a$ exists and equals $g_{\mathrm{\Omega }}(z,a)=F_a(z)-H_b(z)=F_a(z)-h(z)=g_{\mathbb{D}}(z,a)(z\in \mathrm{\Omega }\setminus \{a\}),$ the last equality by step 4.2. In particular $\mathrm{\Omega }$ is Greenian at $a$ and the canonical kernel is the restriction of the disc kernel.

Since $a\mathrm{\ne }0$, the point $0$ lies in $\mathbb{D}\setminus \{a\}$ and the formula of [F3] extends continuously to it, giving $g_{\mathbb{D}}(0,a)=\log \mathrm{\mid }1\mathrm{/}(0-a)\mathrm{\mid }=\log (1\mathrm{/}\mathrm{\mid }a\mathrm{\mid })$. By step 8.1 the same formula represents $g_{\mathrm{\Omega }}(\cdot ,a)$ on $\mathrm{\Omega }\setminus \{a\}$, so ${\lim }_{z\to 0}{g}_{\mathrm{\Omega }}(z,a)=\log \frac{1}{\mathrm{\mid }a\mathrm{\mid }},$ and this value is strictly positive because $0<\mathrm{\mid }a\mathrm{\mid }<1$. By step 2.1 the puncture $0$ is a boundary point of $\mathrm{\Omega }$, so the canonical Green kernel does not vanish at this Euclidean boundary point, and a definition requiring vanishing at every Euclidean boundary point would exclude it.

Countable Choice is used exactly through the cited existence theorem [F2], which supplies both the existence of the canonical kernel on the bounded domain $\mathrm{\Omega }$ and its identification with $F_a-H_b$; the disc formula of [F3], the Perron comparisons of steps 5.1, 6.2 and 7.1, and the puncture limit of step 9.1 use no choice principle.