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).



Proof of Friedrichs inequality in a domain with simple geometry
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.


