# Bounded Subharmonic functions in the Plane are constant.

I would like to prove this result without any appeals to theorems like, say the three-lines theorem, but only through the use of the fundamental solution, $$\log |z - w|$$ in the plane.

The proof I have seen goes as follows:

Consider $$u : \mathbb{C} \to \mathbb{R}$$ a subharmonic, and bounded function. Consider a logarithmic perturbation of $$u$$, given by: $$v_\epsilon(z) = u(z) - \epsilon \log |z|$$ Observe that when $$|z| > 1$$, we have that $$\log |z| > 0$$, so that $$v_\epsilon(z) < u(z)$$ here. Moreover (here is the portion of the proof I take issue with): $$\sup_{|z| > 1} v_\epsilon = \max_{S(0,1)}v_\epsilon = \max_{S(0,1)} u = \max_{\overline{D}(0,1)}u$$ Thus, for any $$z$$ such that $$|z| > 1$$, we have that: $$u(z) = v_\epsilon(z) + \epsilon \log |z| \leq \max_{\overline{D(0,1)}} u + \epsilon \log|z|$$ As the LHS is independent of $$\epsilon$$, we may send $$\epsilon$$ to $$0$$, and conclude that $$u(z) \leq \max_{\overline{D(0,1)}}u$$, and taking suprema, get that $$\sup_{\mathbb{C}}u = \max_{\overline{D(0,1)}}u$$. Thus, $$u$$ attains an interior maximum, and hence is constant.

Question: Why are we allowed to apply the maximum principle with $$v_\epsilon$$ ? I though that the maximum principle only applied on bounded domains?

• The maximum principle holds on any domain (meaning open, connected subset of the plane). See, for example, the statement here – Brevan Ellefsen Jun 11 '19 at 0:53
• You can essentially think of this as partitioning your region into a collection of bounded domains and running the result on each – Brevan Ellefsen Jun 11 '19 at 1:03