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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.

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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 $\log|z|$.

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