Suppose \(\Omega\subseteq \mathbb{R}^d\) is a bounded domain and \(u\in C^2(\Omega)\cap C(\bar{\Omega})\) is subharmonic . Then either for every \(x\in \Omega\)

\[ u(x)<\max_{x\in \partial\Omega} u(x) \]

or \(u\) is constant.

Proof

According to the weak maximum principle we have \(u(x)\le \max_{\partial\Omega}u\). Consider the set \(U=\{x\mid u(x)=M\}\). Since \(u\) is continuous \(U\) is closed.

Assume there is point \(x\) with \(u(x)=\max_{\partial\Omega}u=:M\), i.e. \(U\) is not empty. Using the mean value property there is a suitable ball \(B_r(x)\subseteq \Omega\) with

\[ M=u(x)\le \frac{1}{\lvert B_r(x)\rvert}\int_{B_r(x)} u\le M. \]

Therefore, \(U\) is open, since \(B_r(x)\subseteq U\).

Since \(\Omega\) is path connected and therefore connected the only open, closed and nonempty subset of \(\Omega\) is \(\Omega\) itself. In conclusion, \(u\) is constant.

Remarks

See also Link to heading