Let $u$ be a $C^2$ solution for $-\Delta u=f$ in a bounded set $\Omega$. Show that $$sup_{\bar \Omega} \vert u\vert\le sup_{\bar \Omega}\,f+sup_{\partial\Omega} \vert u\vert$$
If the maximum of $u$ is attained on the boundary $\partial\Omega$, then it's trivial. So I consider the case that the maximum of $u$ is attained in $\Omega$, denoting the maximum point by $x_0$.
What I know is that $-\Delta u(x_0)\le 0$ and so $f(x_0)\le 0$. And I don't know how to go on. I guess we need to use the comparison principle(or equvilantly, the maximum principle). But I don't know how to proceed.
Any help will be appreciated.
Best Answer
Here are two hints. First, I don't believe the estimate you're after is true as stated. You need a constant on the $f$ term on the right, i.e. your target estimate should be $$ \sup_{\bar{\Omega}} |u| \le C \sup_{\bar{\Omega}} |f| + \sup_{\partial \Omega} |u|. $$ This is needed because there are nontrivial solutions to $-\Delta u = \lambda u$ with $u=0$ on $\partial \Omega$ and $\lambda>0$ arbitrarily large (the eigenfunctions of the Laplacian). If we normalize so that $\sup_{\bar{\Omega}} |u| =1$, then it's clear that we need some constant on the $f$ term on the right in order to compensate.
The second hint is to indeed use the maximum principle, but not on $u$ directly. Define the function $$ v(x) = u(x) + \alpha |x|^2 $$ for some constant $\alpha >0$ and study $-\Delta v$.