Phragmén-Lindelöf theorem

First some notation. Let ${\partial }_{\mathrm{\infty }}G$ be the extended boundary of $G$. That is, the boundary of $G$, plus optionally the point at infinity if in fact$G$ is unbounded.

Theorem.

Let $G$ be a simply connected region and let $f\mathrm{:}G\mathrm{\to }\mathrm{C}$and $\phi \mathrm{:}G\mathrm{\to }\mathrm{C}$ be analytic functions. Furthermoresuppose that $\phi$ never vanishes and is bounded on $G$. If $M$ is a constantand ${\mathrm{\partial }}_{\mathrm{\infty }}\mathit{}G\mathrm{=}A\mathrm{\cup }B$ such that

1. for every $a\in A$, ${\overline{\lim }}_{z\to a}|f(z)|\le M$, and

2. for every $b\in B$, and $\eta >0$, ${\overline{\lim }}_{z\to b}|f(z)|{|\phi (z)|}^{\eta }\le M$,

then $\mathrm{|}f\mathit{}\mathrm{(}z\mathrm{)}\mathrm{|}\mathrm{\le }M$ for all $z\mathrm{\in }G$.

This theorem is a generalization of the maximal modulus principle, but instead of requiring that the function is bounded as we approach the boundary, we only need a restriction on its growth to it to in fact be bounded in all of $G$.

If you let $A={\partial }_{\mathrm{\infty }}G$ (and $\phi \equiv 1$ perhaps), then you get almost exactly one version of the maximal modulus principle. In this case it turns out that $G$ need not be simply connected since that is only needed todefine $z\mapsto \phi {(z)}^{\eta }$.

In fact the requirement that $G$ be simply connected can be eased up a bit in this theorem since it is only needed locally. So the theorem is still true if for every point$z\in {\partial }_{\mathrm{\infty }}G$ there exists an open neighbourhood $N$ of $z$ suchthat $N\cap G$ is simply connected.