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We are in trouble with the following exercise, we are almost sure about it's related to Dirichlet's energy and Dirichlet's principle, but we don't know how to prove it. The exercise says:

Let $u\in C^2(\Omega)$, $u=0$ in $\partial \Omega$, being $\Omega$ a regular domain. Prove that:

$$ \int_\Omega |\nabla u|^2 dx\leq \epsilon \int_\Omega |\Delta u|^2 dx + \frac{1}{4\epsilon}\int_\Omega u^2 dx \text{ } \forall \epsilon>0$$

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    Are you sure you've transcribed the inequality correctly? Both integrals look the same to me... – 2017-01-23
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    Ups! The equation accidentally had lost the laplacian, thanks! – 2017-01-23
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    OK, that looks better. My hint would be to integrate by parts, and be aware of the [Peter-Paul inequality](https://en.wikipedia.org/wiki/Young's_inequality#Elementary_case). – 2017-01-23

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Let $u\in C^2_0(\Omega)$ then

$$ \int_\Omega |\nabla u|^2 dx\leq \epsilon \int_\Omega |\Delta u|^2 dx + \frac{1}{4\epsilon}\int_\Omega u^2 dx \text{ } \forall \epsilon>0$$

Using that $u=0$ on the boundary we get with integration by parts

$$\int_Ω \langle \nabla u, \nabla u \rangle \ dx=-\int_Ω u\ \Delta u \ dx \leq \int_Ω |u \ \Delta u| \ dx $$

Now using Young's inequality we get $$|u \cdot \Delta u|=|\frac{1}{\sqrt{2\epsilon}}u | \cdot|\sqrt{2\epsilon}\Delta u | \leq \frac{1}{2\epsilon} \frac{1}{2}|u|^2 +2\epsilon \frac{1}{2} |\Delta u|^2=\frac{1}{4\epsilon}u^2+\epsilon|\Delta u|^2$$

Just insert this into the integral above and the proof is completed.

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    Thanks a lot to both of you!!! It was easier than we thought, simply using Peter-Paul inequality... – 2017-01-23