Why does a compact Rieman surface not admit a Greens function? Here is my definition of a Greens function:
Definition: Let $X$ be a Riemann surface and $x_0$ a point in $X$.
(1) The Greens function of a relatively compact domain $G \subseteq X$ with regular boundary in $x_0$ is the unique continuous function
$$
g \colon \overline{G} \setminus \{x_0\} \to \mathbb{R},
$$
such that $g$ is harmonic on $G \setminus \{x_0\}$, vanishes on $\partial G$ and has a positive logarithmic singularity at $x_0$, that means for a chart $z$ around $x_0$ the function $x \mapsto g(x) + \log{|z(x)|}$ has a harmonic extension to $x_0$.
(2) The Greens function of $X$ in $x_0$ is the limit (only in case it is not $\infty$ everywhere!)
$$
g = \lim_{n \to \infty} g_n
$$
where $G_1 \subseteq G_2 \subseteq \ldots \subseteq \bigcup_{n=1}^\infty G_n = X$ are domains as above with Greens function $g_n$ in $x_0$.
Question: Why can a compact $X$ not have Greens functions?
My attempt: We can conclude that also the Greens function $g$ of a surface has a positive logarithmic singularity at $x_0$, and further that $\inf(g) = 0$. Now if $g$ were continuous on $X$, we could apply the maximum principle. But a positive logarithmic singularity means also that $\lim_{x \to x_0} g(x) = \infty$. So I don't know how to go on, I think I have to apply the maximum principle though.

Assume that if $g : X \setminus\{x_0\} \to \mathbb R$ is a harmonic function so that around $x_0$, $g(z) + \log |z|$ has a harmonic extension. Thus there is a small open disk $D$ around $x_0$ so that $g(z) \ge M > \inf g$ for all $z\in \overline D\setminus\{x_0\}$. In particular, minimum of $g$ when restricted to the compact set $X \setminus D$ is not attained at the boundary $\partial D$. So there is $y\in M\setminus D$ so that $g(x) = \inf_{x\in X\setminus D} g(x)$, but this is impossible by the strong maximum principle (unless $g$ is constant, which is also impossible in this case)
