3
$\begingroup$

Does exists a short, simple proof of the inequality

$ \|u\|_{L^{2}(\Omega)} \leqslant C \|Du\| _{L^{2}(\Omega)} + \|u\| _{L^2{(\partial{\Omega})}} $ for $u\in H^{1}=W^{1,2}(\Omega) $

(Sobolev space with one weak derivative integrable in square), where $\Omega = \{ x\in\mathbb{R}^{n}:\ 1<|x|<2 \}$?

(we do not assume, that the trace of $u$ vanishes).

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ This question is perhaps not suitable for this website. Anyway, a quick sketch for your specific case: using a density argument it suffices to prove for $C^\infty\cap H^1$. For smooth functions, use the fundamental theorem of calculus to get $$ | u(r,\theta) | \leq |u(1,\theta)| + \int_1^2 |\partial_r u(s,\theta)| ds \leq |u(1,\theta)| + \left(\int_1^2 | DU(s,\theta) |^2 ds\right)^2 $$ Now square, integrate over $r,\theta$, and you are done. $\endgroup$
    – Willie Wong
    Commented Mar 8, 2011 at 15:24
  • $\begingroup$ Ack, the outer exponent in the far right of the displayed equation should be $1/2$, not $2$. $\endgroup$
    – Willie Wong
    Commented Mar 8, 2011 at 15:25
  • $\begingroup$ Granted this question has an easy answer, but is this really worth the votes to close? $\endgroup$
    – J.C. Ottem
    Commented Mar 12, 2011 at 3:45
  • $\begingroup$ By the way, the name is Friedrichs, not Friedrich. $\endgroup$
    – Michael Renardy
    Commented May 17, 2012 at 23:26

1 Answer 1

Reset to default
2
$\begingroup$

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.

Bazin.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .