## Strong maximum principle

Strong maximum principle: Suppose ${Lu\geq 0}$ with ${c=0}$, ${u\in C^2(\Omega)\cap C^1(\overline{\Omega})}$ where ${\Omega}$ is connected and open in ${\mathbb{R}^n}$. If ${u}$ achieves maximum over ${\overline{\Omega}}$ at an interior point, then ${u}$ must be a constant.

Proof: Suppose ${M=\sup\limits_{\overline{\Omega}}u}$. Set ${V=\{x\in \Omega|u(x). If ${u}$ achieves maximum at an interior pt and not constant over all ${\Omega}$, then ${\partial V\cap \Omega\neq \emptyset}$. choose a point ${x_0}$ such that ${\text{dist}(x_0,\partial V)< \text{dist}(x_0,\partial \Omega)}$. Consider the largest ball ${B_r(x_0)\subset V}$ with radius ${r}$ centering at ${x_0}$. Suppose ${y\in \partial B_r\cap \partial V}$, then ${u(y)=M}$ and ${D_u(y)=0}$, because ${M}$ is the maximum. However, applying the hopf lemma to ${u}$ in ${V}$, we know ${\displaystyle \frac{\partial u}{\partial n}>0}$ where ${\nu}$ is the outer normal vector of ${B_r(x_0)}$. Contradiction.

Remark: If ${c\leq 0}$ and ${u}$ achieves nonngative maximum in the interior of ${\Omega}$. Then ${u}$ must be a constant.