Take the tour ×  Let $\Omega \subset R^n$ be bounded and open ,
$u\in C^2(\Omega)\cap C(\bar \Omega)$ be a solution of
$-\triangle u=f$ in $\Omega$ , $u=0$ on $\partial \Omega$. Prove
that there exists a constant $C$, depending only on $n$ and
$diam(\Omega )$ such that $||u||_{L^\infty} \le C
||f||_{L^\infty}$ . I have got some idea like comparing $u$ with
parabolas and using maximum principle or so . Some how i am not
able do anything on this problem . Any kind of help or solution
is appreciated. Thanks a lot.  share|improve this question add
comment  1 Answer  up vote 3 down vote accepted  Your basic idea
is correct.  Let $M = \|f\|_\infty$, and let $x_0\in\Omega$. The
functions  $$ \tilde{u}^\pm(x) = u(x) \pm \frac{M}{2n} (x -
x_0)^2 $$  are seen to have the property that  $$ -\triangle
\tilde{u}^+ = -\triangle u - M = f - M \leq 0 $$  is subharmonic
and  $$ -\triangle \tilde{u}^- = - \triangle u + M = f + M \geq
0 $$  is superharmonic.  So by the maximal/minimal principles
for sub/super harmonic functions