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?

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

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