Maximum principle for subharmonic functions

Remarks

Proof idea for part 1: let $M = \sup_\Omega u$ and $A = \set{x \in \Omega : u(x) = M}$. The set $A = \set{u \ge M}$ is closed in $\Omega$ because $u$ is upper semicontinuous. Let $b \in A$. The sub-mean inequality on small spheres about $b$ and $u \le M$ force $u = M$ almost everywhere on a small ball about $b$. Since $A$ is closed, $u = M$ on that whole ball, so $A$ is open. Connectedness gives $A = \Omega$ when $A$ is nonempty. For part 2, extend $u$ to $\overline{\Omega}$ by its limit superior at boundary points. The extension is upper semicontinuous on a compact set, so it attains its maximum, and part 1 rules out a positive maximum inside.

Boundedness of $\Omega$ is needed in part 2. On the half-space $\set{x \in \R^n : x_1 > 0}$, the function $u(x) = x_1$ is harmonic, hence subharmonic, and tends to $0$ at every boundary point, but it is positive everywhere. Connectedness is needed in part 1. On two disjoint discs, the function equal to $0$ on one disc and $1$ on the other is subharmonic and attains its supremum, but it is not constant. Applied to $u$ and $-u$ for a harmonic $u$, it gives the maximum and minimum principles for harmonic functions.

Going deeper

For a student

This is the tool behind uniqueness for the Dirichlet problem and behind the Perron method, where subharmonic functions are compared with harmonic ones. It also gives the comparison principle: if $u$ is subharmonic, $v$ is superharmonic, and $\limsup (u - v) \le 0$ at the boundary of a bounded domain, then $u \le v$ inside. Unbounded domains need extra growth conditions, which is the subject of Phragmén–Lindelöf type results.