A compact Riemann surface with a finite set of points removed is parabolic

Question

A Riemann surface $\mathcal{R}$ is called parabolic if it is not compact and doesn't carry a negative non-constant subharmonic function, and is called hyperbolic if it carries a negative non-constant subharmonic function.

Question: Let $\mathcal{X}$ be a compact Riemann surface with a finite set of points removed. Then how to show that $\mathcal{X}$ is parabolic? Can any one give me some references?

Remark: It might be more common to say that a Riemann surface is hyperbolic if it carries a complete hyperbolic metric. But I don't mean it here. The terminology here is called function-theoretic classification. The question above appears in many lectures, but none gives a proof. For example, see Page.22 of this lecture https://indico.ictp.it/event/a14295/session/1/contribution/5/material/1/0.pdf

Answer 1

There is a simple removability theorem for subharmonic functions: if it is bounded from above in a neighborhood of an isolated singularity then this singularity is removable.

Proof. Suppose our singularity is at 0, and our function is $u$ defined subharmonic and bounded in $|z|<R$. Let $$u_\epsilon=u+\epsilon\log|z|.$$ For any $r\in(0,R)$ let $v_r$ be the soluton of Dirichlet problem (Poisson integral) for $|z|<r$ with boundery values $u+\epsilon\log r$. Then by Maximum Principle, $u_r\leq v_r$ in $\delta<|z|<r$, where $\delta$ is sufficiently small. Letting $\epsilon\to 0$ we conclude that $u\leq v_r$. Now define $$\tilde{u} (0)=\limsup_{z\to 0} u(z),$$ and $\tilde u(z)=u(z)$ for $z\neq 0$. Then $\tilde{u}$ is subharmonic by the definition of a subharmonic function.