Can we extend a function $u \in H_0^1(\Omega)$ to $\overline{u} \in H_0^1(\widetilde{\Omega})$ with $\Omega \subset \widetilde{\Omega}$?

Question

Suppose we have $u \in H_0^1(\Omega)$. I want to know if it is always possible to extend it to an open set $\widetilde{\Omega}$ such that $\Omega \subset \tilde{\Omega}$ by using the extension: $$\overline{u}(x)=\left\{\begin{matrix} u(x) & if \ x \in \Omega \ \\ 0 & \quad \ \ \ if \ x \in \widetilde{\Omega}- \Omega \end{matrix}\right.$$ So that $\overline{u} \in H_0^1(\widetilde{\Omega})$.

To provide some aditional information, I need this to prove some result for which I need $\Omega \subset \subset \widetilde{\Omega}$ to be able to use Friedrichs Theorem.

Answer 1

If $f_n \in C^\infty_c(\Omega)$ satisfies

$$\|f_n - u\|_{H^1(\Omega)} \to 0,$$

then $\overline f_n \in C^\infty_c(\widetilde \Omega)$ satisfies

$$\| \overline f_n - \overline u \|_{H^1(\widetilde \Omega)} \to 0$$

Thus $\overline u \in H^1_0(\widetilde \Omega)$.

Answer 2

This is always true. In fact you may take $\tilde{\Omega}=\mathbb{R}^n $. The reason is that if $u_k\in C^\infty_c (\Omega) $ is an approximation of $u $, in $H^1 (\Omega) $, then the same sequence approximates the extension by zero in $H^1 (\tilde{\Omega}) $.