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.


1 Answer 1


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 $h(z)<o(\log|z|),\; z\to\infty$. 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 $h(z)<o(|z|^\alpha)$.

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 $\log|z|$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.