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)<M\}}. 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.

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