$C_{0}^{\infty}(\mathbb{R}^{n})$ is dense in $H^{1}(\Omega)$, for a Lipschitz domain? Let $\Omega$ be an open, bounded domain with Lipschitz boundary, for example the unit cube $(0,1)\times...\times(0,1)$. 
Is it true that for any $u\in H^{1}(\Omega)$, there exists a sequence $u_{j}\in C_{0}^{\infty}(\mathbb{R}^{n})$ converging to $u$ in $H^{1}(\Omega)$?
If the boundary is smooth then the result can be found in many textbooks.
Any references are welcome.
Thanks.

This boils down to the existence of an extension operator: Take $\Omega'=B_r(0)$ with $r$ large such that $\bar\Omega \subset \Omega'$. Then there is a continuous extension operator $E:H^1(\Omega)\to H^1_0(\Omega')$ with $(Eu)|_\Omega = u$. Now $C_c^\infty(\Omega') $ is dense in $ H^1_0(\Omega')$.
Now let $u\in H^1(\Omega)$, then $Eu\in H^1_0(\Omega')$. For every $\epsilon>0$ there is $\phi\in C^\infty_c(\Omega')$ with $\| Eu - \phi\|_{H^1(\Omega')}\le \epsilon$. This implies $\| u - \phi\|_{H^1(\Omega)}\le \epsilon$. Since $\phi\in C_c^\infty(\mathbb R^d)$ as well, we get the denseness result.
