Proof of Friedrich inequality in a domain with simple geometry - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T11:36:48Z http://mathoverflow.net/feeds/question/57826 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57826/proof-of-friedrich-inequality-in-a-domain-with-simple-geometry Proof of Friedrich inequality in a domain with simple geometry Maciej S. 2011-03-08T13:51:54Z 2012-06-14T23:22:00Z <p>Does exists a short, simple proof of the inequality</p> <p>$\|u\|_{L^{2}(\Omega)} \leqslant C \|Du\| _{L^{2}(\Omega)} + \|u\| _{L^2{(\partial{\Omega})}}$ for $u\in H^{1}=W^{1,2}(\Omega)$ </p> <p>(Sobolev space with one weak derivative integrable in square), where $\Omega = \{ x\in\mathbb{R}^{n}:\ 1&lt;|x|&lt;2 \}$?</p> <p>(we do not assume, that the trace of $u$ vanishes).</p> http://mathoverflow.net/questions/57826/proof-of-friedrich-inequality-in-a-domain-with-simple-geometry/94565#94565 Answer by Bazin for Proof of Friedrich inequality in a domain with simple geometry Bazin 2012-04-19T20:17:28Z 2012-04-19T20:17:28Z <p>Let $\Omega$ be an open subset of $\mathbb R^n$ with a $C^1$ boundary and $u\in H^1(\Omega)$. We compute with $D_{x_1}=-i\partial_{x_1}$, $$2\Re\langle D_{x_1}u, i x_1u\rangle=-2\Re\int_\Omega x_1(\partial_{x_1}u)\ \overline{u} dx =-\int_\Omega x_1\partial_1(\vert u\vert^2) dx= -\int_\Omega \partial_1(x_1\vert u\vert^2) dx+ \int_\Omega \vert u\vert^2 dx,$$ so that with Green's formula $$2\Re\langle D_{x_1}u, i x_1u\rangle=\Vert u\Vert_{L^2(\Omega)}^2 -\int_{\partial \Omega} x_1\vert u\vert^2\nu_1 d\sigma,$$ and thus (Cauchy-Schwarz) $\Vert u\Vert_{L^2(\Omega)}^2\le \sup_{x\in \partial \Omega}{\vert x_1\vert} \Vert u\Vert_{L^2(\partial\Omega)}^2 +2\sup_{x\in \partial \Omega}{\vert x_1\vert}\Vert u\Vert_{L^2(\Omega)} \Vert D_{x_1}u\Vert_{L^2(\Omega)}$ implying $$\Vert u\Vert_{L^2(\Omega)}^2\le \sup_{x\in \partial \Omega}{\vert x_1\vert} \Vert u\Vert_{L^2(\partial\Omega)}^2 +\frac 12 \Vert u\Vert_{L^2(\Omega)}^2 +2 \sup_{x\in \partial \Omega}{\vert x_1\vert}^2 \Vert D_{x_1}u\Vert_{L^2(\Omega)}^2.$$ The term $\frac 12\Vert u\Vert_{L^2(\Omega)}^2$ in the rhs can be absorbed in the lhs, yielding the sought inequality. One could also fiddle with the choice of the multiplier $x_1$ and get better constants by replacing $x_1$ by another function and $\partial _1$ by another vector field.</p> <p>Bazin.</p>