Let $u\in\mathscr C^2(U)\cap\mathscr C(\bar U)$ be harmonic in the non-empty open and connected set $U\subset\mathbb R^n$. If there exists a Point $x_0\in U$, so that $u$ has a local Maximum at $x_0$, then $u$ is constant in $U$.
Proof:
Let $x_0$ be an point in $U$ with $u(x_0)\geq u(x)$ for all $x$ in some neighbourhood of $x_0$. Without loss of generality we assume the neighbourhood to be $\bar{B}_R(x_0)$ for some $R>0$. Due to the mean value property, we have: $$\frac1{\omega_nR^n}\int_{B_R(x_0)}u(x_0)dx=u(x_0)=\frac1{\omega_nR^n}\int_{B_R(x_0)}u(x)dx \Rightarrow \frac1{\omega_nR^n}\int_{B_R(x_0)}u(x_0)-u(x)dx=0$$. $u(x_0)-u(x)\geq0$ in $B_R(x_0)$ implies $u(x_0)-u(x)=0$ in $\bar B_R(x_0)$. We conclude that the set $V=\{x\in U|u(x)=u(x_0)\}$ is open in $U$. Since $V$ is clearly also closed, we have $U=V$ due to the connectedness.
I don't understand why $V$ is open in $U$. Help would be greatly appreciated as my colleage and I are on the verge of a mental breakdown.
Best Answer
In order to invoke the mean-value theorem, you need to assume that $B_{R}(x_{0}) \subset U$.
What you have shown is that for any $\mathbf{x}_{0} \in V$, the $B_{R}(x_{0})$ consists entirely of points where $u(\mathbf{x})=u(\mathbf{x}_{0})$. This means that $\mathbf{x} \in V$ for all $\mathbf{x} \in B_{R}(x_{0})$. Hence, $B_{R}(x_{0}) \subseteq V$.
You have hence shown that for any $\mathbf{x}_{0} \in V$ there exists an $R>0$ for which $B_{R}(x_{0}) \subseteq V$. By definition, this means that $V$ is an open set.