Conformal covariance of the canonical planar Green kernel

Since $f$ is injective and holomorphic on the domain $\mathrm{\Omega }$, [F2] gives $f^{\mathrm{\prime }}(a)\mathrm{\ne }0$; thus the holomorphic function $w\mapsto f(w)-f(a)$ has a zero of order one at $a$, so by [F2] there is a disc $D(a,\rho )\subseteq \mathrm{\Omega }$ and a holomorphic $q$ on $D(a,\rho )$ with $f(z)-f(a)=(z-a)q(z)$ and $q(a)=f^{\mathrm{\prime }}(a)\mathrm{\ne }0$. Making $\rho$ smaller, $q$ is nowhere zero on $D(a,\rho )$, so [F3] provides a holomorphic $L$ on $D(a,\rho )$ with $\exp L=q$; then both components of $L$ are $C^2$ by [F2], so $\mathrm{Re}L$ is harmonic on $D(a,\rho )$ by [F3], and for $0<\mathrm{\mid }z-a\mathrm{\mid }<\rho$ $\log \mathrm{\mid }f(z)-f(a)\mathrm{\mid }=\log \mathrm{\mid }z-a\mathrm{\mid }+\mathrm{Re}L(z),$ because $\mathrm{\mid }f(z)-f(a)\mathrm{\mid }=\mathrm{\mid }z-a\mathrm{\mid }\cdot \mathrm{\mid }q(z)\mathrm{\mid }=\mathrm{\mid }z-a\mathrm{\mid }{e}^{\mathrm{Re}L(z)}$.

The inverse $f^{-1}:{\mathrm{\Omega }}^{\mathrm{\prime }}\to \mathrm{\Omega }$ is holomorphic by [F2], and by [F4] its derivative satisfies $(f^{-1}{)}^{\mathrm{\prime }}(f(a))=1\mathrm{/}{f}^{\mathrm{\prime }}(a)\mathrm{\ne }0$; the same computation as in step 1.1, applied to $f^{-1}$ at the pole $f(a)$, therefore supplies a disc $D(f(a),{\rho }^{\mathrm{\prime }})\subseteq {\mathrm{\Omega }}^{\mathrm{\prime }}$ and a harmonic function $R$ on it with $\log \mathrm{\mid }{f}^{-1}(w)-a\mathrm{\mid }=\log \mathrm{\mid }w-f(a)\mathrm{\mid }+R(w)(0<\mathrm{\mid }w-f(a)\mathrm{\mid }<{\rho }^{\mathrm{\prime }})\mathrm{.}$

Define $G(z):=g_{{\mathrm{\Omega }}^{\mathrm{\prime }}}(f(z),f(a))$ for $z\in \mathrm{\Omega }\setminus \{a\}$. Then $G$ is a logarithmic-pole candidate at $a$ on $\mathrm{\Omega }$: it is nonnegative because $g_{{\mathrm{\Omega }}^{\mathrm{\prime }}}$ is, it is harmonic by [F4] because $g_{{\mathrm{\Omega }}^{\mathrm{\prime }}}(\cdot ,f(a))$ is harmonic on ${\mathrm{\Omega }}^{\mathrm{\prime }}\setminus \{f(a)\}$ and $f$ is holomorphic on $\mathrm{\Omega }\setminus \{a\}$ with image ${\mathrm{\Omega }}^{\mathrm{\prime }}\setminus \{f(a)\}$, and $G(z)+\log \mathrm{\mid }z-a\mathrm{\mid }=[g_{{\mathrm{\Omega }}^{\mathrm{\prime }}}(f(z),f(a))+\log \mathrm{\mid }f(z)-f(a)\mathrm{\mid }]-\mathrm{Re}L(z)$ extends across $a$ to a harmonic function, because the bracket is the harmonic corrector of $g_{{\mathrm{\Omega }}^{\mathrm{\prime }}}$ composed with $f$ and $\mathrm{Re}L$ is harmonic by step 1.1.

Symmetrically, $H(w):=g_{\mathrm{\Omega }}(f^{-1}(w),a)$ for $w\in {\mathrm{\Omega }}^{\mathrm{\prime }}\setminus \{f(a)\}$ is a logarithmic-pole candidate at $f(a)$ on ${\mathrm{\Omega }}^{\mathrm{\prime }}$: nonnegativity, harmonicity and the logarithmic pole follow as in step 2.2 with the roles of $f$ and $f^{-1}$ exchanged, the local harmonic remainder being supplied by step 2.1.

By the least-candidate characterization in [F1] applied on $\mathrm{\Omega }$, the candidate $G$ of step 2.2 dominates the canonical kernel: $g_{\mathrm{\Omega }}(z,a)\le g_{{\mathrm{\Omega }}^{\mathrm{\prime }}}(f(z),f(a))$ for all $z\in \mathrm{\Omega }\setminus \{a\}$.

Applying [F1] on ${\mathrm{\Omega }}^{\mathrm{\prime }}$ to the candidate $H$ of step 3.1 gives $g_{{\mathrm{\Omega }}^{\mathrm{\prime }}}(w,f(a))\le g_{\mathrm{\Omega }}(f^{-1}(w),a)$ for all $w\in {\mathrm{\Omega }}^{\mathrm{\prime }}\setminus \{f(a)\}$; substituting $w=f(z)$ for $z\in \mathrm{\Omega }\setminus \{a\}$ yields the reverse inequality $g_{{\mathrm{\Omega }}^{\mathrm{\prime }}}(f(z),f(a))\le g_{\mathrm{\Omega }}(z,a)$.

The two inequalities of steps 3.2 and 4.1 are opposite, so $g_{\mathrm{\Omega }}(z,a)=g_{{\mathrm{\Omega }}^{\mathrm{\prime }}}(f(z),f(a))$ for every $z\in \mathrm{\Omega }\setminus \{a\}$. Only the least-candidate order of [F1] and the local behaviour of the two biholomorphisms were used: neither $f$ nor $f^{-1}$ was extended to a boundary point.