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Let $\Omega \subset R^n$ be bounded and open, $u\in C^2(\Omega)\cap C(\bar \Omega)$ be a solution of $-\Delta u=f$ 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.

1 Answers 1

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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 you have that

$ u(x) \leq \tilde{u}^+(x) \leq \sup_{\partial\Omega} \tilde{u}^+ \leq \frac{M}{2n} \mathrm{diam}(\Omega)^2 $

and

$ u(x) \geq \tilde{u}^-(x) \geq \inf_{\partial\Omega}\tilde{u}^- \geq - \frac{M}{2n} \mathrm{diam}(\Omega)^2 $

And so you get

$ \|u\|_\infty \leq \frac{\mathrm{diam}(\Omega)^2}{2n} \|f\|_\infty $

and the constant indeed depends only on the dimension and the diameter of the set $\Omega$.

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    Okay. Glad to hear.2012-07-23