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Let $Ω ⊂ R^d$ be a bounded domain for which the divergence theorem holds. Assume $u ∈ C^2(\bar{Ω} ), u = 0$ on $∂Ω$. Show that for every $ε > 0,$

$2\int_Ω |∇u(x)|^2 dx ≤ ε\int_Ω(Δu(x))^2 dx +\frac{1}{ε}\int_Ω u^2(x) dx.$

I tried using Divergence theorem for some $w=\epsilon u^2$, but no success. At least a hint would be appreciated.

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    If you integrate the LHS by parts (using the boundary condition) and use Cauchy's inequality on the product $\int |\Delta u||u|$ it should get you there! – 2017-01-31
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    But how do I include an $\epsilon$? – 2017-01-31
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    There's a neat trick to do this: use that $\int |\Delta u||u| = \int |\varepsilon^{1/2} \Delta u||\varepsilon^{-1/2}u|$ and again apply Cauchy. I think in general people call this the Peter-Paul inequality – 2017-01-31
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    @Matt It's actually CS followed by the AM-GM. – 2017-01-31
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    @Dr.MV I believe both methods should work, though perhaps I misspoke: Wikipedia calls it Young's inequality, though I remembered learning it as Cauchy's inequality (the Peter-Paul inequality still stands). In fact, it seems that Young's inequality is simply a special case of AM-GM! – 2017-01-31
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    Young is, more precisely, a special case of the weighted AM-GM inequality. – 2017-01-31

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Let $\phi=\frac1{\sqrt \epsilon} u$ and $\psi =\sqrt{\epsilon}\, \nabla^2 u$. Using the divergence theorem along with the condition that $u=0$ for $x\in \partial \Omega$, we find

$$\begin{align} \int_\Omega \left|\nabla u\right|^2\,dx&=-\int_\Omega u \nabla^2 u\,dx\\\\ &=-\int_\Omega \phi \psi \,dx \tag 1 \end{align}$$

Now, applying the Cauchy-Schwarz inequality to the right-hand side of $(1)$ reveals

$$\left(\int_\Omega \left|\nabla u\right|^2\,dx\right)^2\le \left(\int_\Omega \phi^2\,dx\right)\left(\int_\Omega \psi^2\,dx\right) \tag 2$$

Then, applying the AM-GM inequality to the right-hand side of $(2)$ yields

$$\left(\int_\Omega \phi^2\,dx\right)\left(\int_\Omega \psi^2\,dx\right)\le \frac14 \left(\int_\Omega \phi^2\,dx+\int_\Omega \psi^2\right)^2 \tag 3$$

whence taking square roots and using $(2)$, we obtain the coveted inequality

$$\int_\Omega |\nabla u^2|\,dx\le \frac12 \int_\Omega \left(\frac1{\epsilon}u^2+\epsilon(\nabla^2 u)^2\right)$$