Green kernel of the disc at a nonzero pole

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.