Let $D$ be a bounded domain, with boundary $\partial D$.
Suppose $u \in C^{2}(D)\cap C^{0}(D\cup\partial D)$ satisfies $\Delta u - u \geq 0$ in all of $D$.
I am supposed to prove then that $u\leq \max\{\ \max\limits_{\partial D}(u), \ 0 \ \}$.
MY ATTEMPT:
I think that:
$\max \{ \max\limits_{C} u , 0 \} = \max\limits_{C} u \ \ \ \ $ when $\max\limits_{C} u > 0$
$\max \{ \max\limits_{C} u , 0 \} = 0 \ \ \ \ $ when $\max\limits_{C} u \leq 0$
So I thought about treating these cases seperately, however I get stuck.
$\ $
The maximum of $u$ must occur in $D\cup \partial D$.
Suppose that $\mathbf{p} \in D$ is a maximum of $u$. Then $u_{x}(\mathbf{p})=u_{y}(\mathbf{p})=0$, and also, $u_{xx}(\mathbf{p})\leq 0$ and $u_{yy}(\mathbf{p})\leq 0$ (since $\mathbf{p}$ is a maximum).
Then I find that $u_{xx}(\mathbf{p})+ u_{yy}(\mathbf{p}) \leq 0$, which means that $\Delta u(\mathbf{p}) \leq 0$.
This is where I get stuck.
I'm trying to get a contradiction using this method. If I can only say that $u$ is positive then $\Delta u(\mathbf{p})- u \leq 0$ is immediate, and we have a contradiction (meaning that $\mathbf{p} \in D$ cannot be a maximum and the maximum must occur on the boundary $\partial D$.). But I can't figure out why.
Can somebody help me out?
We don't need to worry about the set where $u\le 0$ because the inequality is automatic there. On the open set $U=\{x : u>0 \}$ (possibly empty), $\Delta u\ge 0$, so $u$ is subharmonic on $U$ and thus satisfies the maximum principle there. So $u\le\max_{\partial U}u$ on $U$, but the boundary of $U$ is contained in $\partial D\cup U^c$, so we obtain the desired estimate.