# Subharmonic function in unbounded regions

The harmonic majorization for a subharmonic function $$h$$ is well-known for bounded regions $$\Omega \subset \mathbb{C}$$: $$h \le 0 \text{ in }\partial \Omega \Longrightarrow h \le 0 \text{ in }\Omega.$$ I know this is related to maximum principle.

I need a reference for the same result for unbounded regions of the complex plane. I think one should further assume that the function is bounded on $$\overline{\Omega}$$. I saw this somewhere but I am not able to find a reference.

Yes, this is called the Phragmen-Lindelof Principle: For every region on the Riemann sphere, if $$h$$ is subharmonic and bounded from above, and $$\limsup_{z\to\zeta}h(z)\leq 0$$ for all $$\zeta\in\partial\Omega$$, except finitely many points, then $$h\leq 0$$ in $$\Omega$$. If your domain $$\Omega$$ is an unbounded domain in $$C$$, just include $$\infty$$ to this finite exceptional set.

There are many improvements of this, for example, finite exceptional set can be replaced by a set of zero capacity. Boundedness from above can be replaced by a weaker condition $$h(z). This can be replaced by a weaker growth condition, if something is known about the shape of the unbounded domain near infinity. For example, if the portion of $$\Omega$$ near $$\infty$$ is contained in a sector of opening $$<\pi/\alpha$$, then instead of boundedness one can impose the growth condition $$h(z).

Refs. Ransford, Potential theory in the plane,

Levin, Lectures on entire functions,

Hayman, Kennedy, Subharmonic functions.

In fact, the proof is very simple. Suppose $$h$$ is bounded from above and $$h(z)\leq 0$$ on $$\partial\Omega$$, where $$\Omega$$ is an unbounded domain. Here $$\partial$$ is with respect to $$C$$, so it does not include $$\infty$$. Suppose for simplicity that $$\Omega$$ does not intersect the unit disk. Consider $$u(z)=h(z)-\epsilon\log|z|$$, where $$\epsilon>0$$. Then $$\limsup_{z\to\zeta}u(z)\leq 0$$ for $$\zeta\in \partial^*\Omega$$, the boundary with respect to the Riemann sphere, so it includes $$\infty$$. By the usual Maximum principle we conclude that $$u(z)\leq 0$$ on $$\Omega$$. Passing to the limit for fixed $$z$$ as $$\epsilon\to 0$$, we obtain $$h(z)\leq 0$$.

To obtain the result under other conditions, you use other auxilliary functions in place of $$\log|z|$$.