If $\Delta f\geq0$ for $f\in C^2(D)$ where $D\subset\mathbb{R}^n$ is convex, show that $f$ has no local maximum in $D$ unless it is constant.

Question

So the Laplacian is the trace of the Hessian matrix and also if $f$ attains local maximum at for example $x_0\in D$ then $\nabla^2 f(x_0)\leq0$. This concludes that all eigenvalues of $\nabla^2 f(x_0)$ are $0$. How should we proceed from here?

Answer 1

Hint. If $\Delta f\ge 0$, then $$ f(\xi)\le \frac{1}{|B_r(\xi)|}\int_{B_r(\xi)}f(x)\,dx. \qquad\text{Gauss Law of Arithmetic Mean} $$ Hence, if $\xi$ local maximum, then $f$ is constant in the whole ball $B_r(\xi)\subset D$.

However, $f$ does not have to be constant on the whole of $D$. For example $$ f(x)=\left\{ \begin{array}{ccc} 0 & \text{if} & |x|\le 1,\\ (|x|-1)^4 & |x|>1. \end{array} \right. $$