Green's function and the Riemann map

Remarks

Proof idea: since $\varphi$ is injective, $\varphi'(a) \ne 0$, so $\varphi(z) = (z - a) k(z)$ with $k$ holomorphic and zero-free on $\Omega$. Hence $-\log\abs{\varphi(z)} = -\log\abs{z - a} - \log\abs{k(z)}$, and $\log\abs{k}$ is harmonic on $\Omega$. Away from $a$, $\log\abs{\varphi}$ is harmonic because $\varphi$ is holomorphic and has no zero. A conformal map onto $\mathbb{D}$ is proper, so $\abs{\varphi(z)} \to 1$ as $z$ tends to the boundary, and $g(z, a) \to 0$ there. These are the defining properties, and the Green's function is unique by the maximum principle. Example: for $\Omega = \mathbb{D}$ and $\varphi(z) = (z - a)/(1 - \bar a z)$, one gets $g(z, a) = \log\abs{(1 - \bar a z)/(z - a)}$. Near the pole, $g(z, a) = -\log\abs{z - a} + \log\big(1/\abs{\varphi'(a)}\big) + o(1)$, and $1/\abs{\varphi'(a)}$ is the conformal radius of $\Omega$ at $a$. The hypothesis $\Omega \ne \C$ is needed: the plane has no Green's function, since a positive harmonic function on $\C \setminus \set{a}$ with a logarithmic pole at $a$ does not exist.

Going deeper

For a student

The formula can also be read backwards. If $g^*$ is a harmonic conjugate of $g(\cdot, a)$ on $\Omega \setminus \set{a}$, then $\exp(-(g + i g^*))$ is single-valued and gives the Riemann map. This is one classical route to the Riemann mapping theorem via potential theory, through the solution of a Dirichlet problem. On multiply connected domains the conjugate $g^*$ has periods, so $\exp(-(g + i g^*))$ is no longer single-valued.