Weak maximum principle - Schrödinger operator

Question

I think I'm missing something very basic here. I'm considering the problem given by

\begin{cases} Lu > 0, & \text{in } \Omega, \\ u \ge 0, & \text{on }\partial \Omega, \end{cases}

where $L = -\Delta + q(x)$, $\Omega$ is a bounded domain in $\mathbb{R}^n$, $q: \Omega \rightarrow \mathbb{R}$ is a bounded function such that $q(x) \ge 0$ in $\Omega$. I need to show that, for $u \in C^2(\Omega) \cap C(\overline{\Omega})$, $u \ge 0$ in $\Omega$. (This is the first step in showing the weak maximum principle holds, that is, the conclusion is true for $Lu \ge 0$). The proof goes like this: by contradiction, let's suppose there exists $\overline{x}$ such that $u(\overline{x}) < 0$ for some $\overline{x}$ in $\Omega$. Then the minimum point $x_0$ belongs to $\Omega$ and $\partial_{x_i x_i} u(x_0) \ge 0$ for all $i = 1, \dots, n$. Hence, $Lu(x_0) <0$, a contradiction.

I can't figure out why the "hence" holds. While true that the sum of those partial derivatives is non positive, the function $q$ is by hypotesis non negative. It would be true for the $-\Delta$ operator, I don't see why it holds for $L$.

Answer 1

We have

• $\partial_{x_i x_i} u(x_0) \ge 0$, hence $\Delta u(x_0) \ge 0$ and
• $u(x_0) \lt 0$, hence $q(x_0) u(x_0) \le 0$ (as $q \ge 0$ by assumption).

Therefore $-\Delta u(x_0) + q(x_0) u(x_0) \le 0$. This is already a contradiction, you do not need a strict inequality here.