Proof of Friedrichs inequality in a domain with simple geometry 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).
 A: 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.
