Extending by zero a Sobolev function

Let $\Omega\subset\mathbb{R}^n$ be a bounded open set and $u\in W_0^{1,2}(\Omega)$. Define $B_R=B(x_0,R)$ for $x_0\in\partial\Omega$ and consider $\tilde{u}=u\chi_{\Omega\cap B_{2R}}$. Do we need some extra assumptions in order to prove that $\tilde{u}\in W_0^{1,2}(B_{3R})$?

I am doing this because I want to bound

$$\Big(\int_{\Omega\cap B_{2R}}u^2\Big)^{1/2}\leq C\Big(\int_{\Omega\cap B_{2R}}|\nabla u|^{2_*}\Big)^{1/2_*}$$ using Sobolev-Poincaré's inequality, but it is not true that $u\in W_0^{1,2}(\Omega\cap B_{2R})$.

Thank you very much for your help.

• The function $\tilde {u}$ defined in $B_{3R}$ (I'm assuming that you are extending it by zero outside $\Omega$) does not have weak derivatives in general. Every Sobolev function is absolutely continuous on almost every segment of line parallel to the coordinate axis. – Tomás Mar 21 '14 at 1:01
• I understand, thanks :) – Peter C. Mar 21 '14 at 18:09
• In case $\,\partial\Omega=\{x_0\}$, a closure of of the subspace $\{u\in C_0^{\infty}(\mathbb{R}^n)\,\colon {\rm supp\,}u\in\Omega\}$ in the Sobolev space $\,W^{1,2}(\mathbb{R}^n)\,$ does coincide with that of the subspace $\,C_0^{\infty}(\mathbb{R}^n)\,$ when $n\geqslant 2$, and hence there is no problem at all. In case $\,\partial\Omega\,$ is Lipschitz, your assumption $u\notin W_0^{1,2}(\Omega\cap B_{2R})$ implies that $\tilde{u}$ is not weakly differentiable in $(B_{3R})$, and hence $\tilde{u}\notin W_0^{1,2}(B_{3R})$. But your bound holds for all $u\in W_0^{1,2}(\Omega)$. – mkl314 Mar 23 '14 at 15:59