I want to show the following are equivalent on a Riemann surface $W$:
- Green's function $g_W (p, q)$ exists.
- There exists a bounded harmonic function on $W$.
- There exists a positive harmonic function on $W$.
Clearly 2. implies 3. I've shown 3. implies 1. via a Perron family argument, controlling the Perron family to which the Green's function is a solution. I suspect 1. implies 2. is similar, e.g. constructing a Perron family which is controlled by the Green's function, forcing the Perron solution to be bounded. However, I can't seem to get this to work, and I'm wondering if 1. implies 2. is necessarily true, since equivalence of 2. and 3. seems like a very strong statement.
It definitely holds for $W$ simply connected, as by the uniformization theorem $W$ is conformally equivalent with the disk, so we simply take the real/imaginary part of that conformal map. Along these lines, I was thinking of somehow using the fact that the covering space of $W$ is the disk and constructing a holomorphic right inverse $W \to \mathbb D$ to the covering map.
Some definitions for clarity: Let $W$ be a connected Riemann surface and $q \in W$. The map $g_W (p, q)$ is defined as the Perron solution to the family of compactly supported subharmonic functions $v: W \setminus \{q\} \to [-\infty, \infty)$ satisfying \begin{equation} \limsup_{p \to q} [v(p) + \log |z(p)|] < \infty \end{equation} for some coordinate chart $z : U \to \mathbb D$ sending $p$ to the origin. We say the Green's function exists if
Just a general remark about what happens with references but no proofs.
The result is not true in general; it is true if $W \subset \mathbb C$ by capacity theory, but otherwise both inclusions are strict -
(Riemann surfaces satisfying 1 are also called hyperbolic, while those that do not satisfy 2 are called $O_{HB}$, and those that do not satisfy 3 are called $O_{HP}$ and there a few more classes coming from the existence or not of bounded analytic functions, harmonic functions with bounded Dirichlet integral etc - in the plane the complement of the usual Cantor set is hyperbolic - positive capacity - hence admits bounded harmonic functions, but the Cantor set has Lebesgue measure zero, so its complement doesn't admit bounded analytic functions).
Counterexamples are complicated and for example, the book by Ahlfors and Sario on Riemann Surfaces has them in chapter 4, while the proof that $1$ implies $2$ for planar surfaces is classic and appears in Ahlfors and Sario but also in Conway Functions of One Complex Variable II