Let $\Omega\subseteq \mathbb{R}^n$ be an open set and show that
$$
\lVert{\nabla u\rVert}_{L^2}^2 \leq \epsilon\lVert{\Delta u\rVert}_{L_2}^2 + \frac{1}{4\epsilon}\lVert{u\rVert}_{L^2}^2
$$
for any $\epsilon>0$ and $u\in H_0^2(\Omega)$.
It would suffice to porve the inequality for smooth compactly supported functions since these are dense in $H_0^2(\Omega)$. Other than the fact that I will probably need Green's identity, I am not sure how to proceed. Where does the $\epsilon$ come into play?
Thanks!
Best Answer
By Green's identity and the Cachy-Schwarz inequality $$ \int_\Omega|\nabla u|^2=-\int_\Omega u\,\Delta u \le \int_\Omega |u\,\Delta u|\le\Bigl(\int_\Omega|u|^2\Bigr)^{1/2}\Bigl(\int_\Omega|\Delta u|^2\Bigr)^{1/2}, $$ that is, $$ \|\nabla u\|_{L^2}^2\le\|u\|_{L^2}\|\Delta u\|_{L^2}. $$ From the inequality $$ a\,b\le\frac12(a^2+b^2), $$ we get for any $\epsilon>0$ $$ a\,b=(\sqrt{2\,\epsilon}\,a)\Bigl(\frac{1}{\sqrt{2\,\epsilon}}\,b\Bigr)\le\epsilon\,a^2+\frac{1}{4\,\epsilon}\,b^2. $$