Maximum principle for harmonic functions

Remarks

Proof idea for part 1: the set $A = \set{x \in U : u(x) = \sup_U u}$ is closed in $U$ by continuity. It is open: if $x \in A$ and $\overline{B}(x, r) \subseteq U$, the mean value property over the ball makes $u(x)$ the average of values $\le u(x)$, which forces $u = u(x)$ on the ball. As $U$ is connected and $A$ is not empty, $A = U$. Part 2 follows: the maximum of $u$ on the compact set $\overline{U}$ is attained, and if it were attained only inside, part 1 would make $u$ constant. Example: $u(x, y) = x^2 - y^2$ on the unit disc takes its maximum $1$ on the boundary, at $(\pm 1, 0)$, and the value $0$ at the centre. Boundedness is needed in part 2: on the half-plane $\set{x > 0}$, $u = x$ is $0$ on the boundary and positive inside. Connectedness is needed in part 1. On two disjoint discs, the function equal to $0$ on one and $1$ on the other is harmonic and attains its maximum without being constant. This is the harmonic case of the general maximum principle for elliptic operators.

Going deeper

For a student

The weak maximum principle gives uniqueness for the Dirichlet problem. Two harmonic functions with the same boundary values differ by a harmonic function that is $0$ on the boundary, hence $0$ inside. It also gives stability: if the boundary values of two solutions differ by at most $\varepsilon$, then the solutions differ by at most $\varepsilon$ everywhere. Subharmonic functions satisfy the same maximum principle, and Perron's method builds solutions of the Dirichlet problem from it.