Suppose \(\Omega\subseteq \mathbb{R}^d\) is a bounded and open subset and \(u\in C^2(\Omega)\cap C(\bar{\Omega})\) is subharmonic . Then
\[ \max_{x\in \bar{\Omega}} u(x) = \max_{x\in \partial\Omega} u(x). \]
Proof
Consider two cases. First \(\Delta u>0\). If there is a local maximum in \(x_0 \in \Omega\) the Hessian \(\nabla^2 u(x_0)\) is semi negative definite. Since \(\Delta u\) is the trace of the Hessian we obtain \(\Delta u\le 0\), which is a contradiction.
If \(\Delta u\ge 0\) we consider \(v(x)=\lvert x\rvert^2\). Then \(\Delta (u+\varepsilon v)=\Delta u + \varepsilon 2d >0\) for all \(\varepsilon>0\). The first case implies
\[ \max_{x\in \Omega}u(x)\le \max_{x\in \Omega} \bigl(u(x)+\varepsilon v(x)\bigr) = \max_{x\in \partial\Omega}\bigl(u(x) + \varepsilon v(x)\bigr) \le \max_{x\in \partial\Omega}u(x)+\varepsilon\max_{x\in \Omega}v(x), \]where we used that \(v\ge 0\). Since \(\varepsilon>0\) was arbitrary we obtain \(\max_{x\in \Omega} u(x)\le \max_{x\in \partial\Omega}u(x)\).
Remarks
- One can show, that for a smooth enough harmonic function \(u\) the function \(\lvert \nabla u\rvert^2\) is subharmonic, and therefore \[ \max_{\Omega} \lvert \nabla u\rvert^2 = \max_{\partial\Omega} \lvert \nabla u\rvert^2. \]