My question relates to a translation property of Sobolev functions. Particularly, if $u\in W^{1,\,p}(\Omega)$, $\Omega\subset\mathbb{R}^{N}$, then there exists a constant $C$ such that for all $\omega\subset\subset\Omega$ ($\omega$ strongly included in $\Omega$), and all $h\in\mathbb{R}^{N}$ with $|h|<\text{dist}(\omega,\partial\Omega)$ we have $\|u(x+h)-u(x)\|_{L^{p}(\omega)}\leq C|h|$. Furthermore, we can take $C=\|\nabla u\|_{L^{p}(\Omega)}$.
Going through the proof of this in Brezis, we first consider $C_{c}^{\infty}(\mathbb{R}^{N})$ functions and the method is to parameterise the translation by $t\in[0,1]$ and using the fundamental theorem of calculus you can construct an integral representation of the translation difference and achieve the desired $p$ norms. That is, you get to,
\begin{align}
\int_{\omega}|u(x+h)-u(x)|^{p}dx\leq|h|^{p}\int_{0}^{1}dt\int_{\omega+th}|\nabla u(y)|^{p}dy.
\end{align}
From here Brezis continues with:
If $|h|<\text{dist}(\omega,\partial\Omega)$, there exists an open set $\omega'\subset\subset\Omega$ such that $\omega+th\subset\omega'$ for all $t\in[0,1]$ and thus,
\begin{align}
\|u(x+h)-u(x)\|_{L^{p}(\omega)}^{p}\leq|h|^{p}\int_{\omega'}|\nabla u|^{p}.
\end{align}
My question is, why does $|h|<\text{dist}(\omega,\partial\Omega)$ allows us to conclude there exists an open set $\omega'\subset\subset\Omega$ with the desired properties?
Best Answer
I believe I have achieved an answer to this. If we assume $\omega\subset\subset\Omega$ then for any $h\in\mathbb{R}^{N}$ and $t\in[0,1]$ such that $|h|<\text{dist}(\omega,\partial\Omega)$ we also have $\omega+th\subset\subset\Omega$. Now define $\omega'=\bigcup_{t\in[0,1]} \omega+th$, then clearly $\overline{\omega'}$ is compact in $\mathbb{R}^{N}$and $\overline{\omega'}\subset\Omega$. Hence, $\omega'\subset\subset\Omega$ and $\omega+th\subset\omega'$.
Would anyone care to let me know if this reasoning is correct?