Let $\Omega\subset\mathbb{R}^n$ be a bounded open set, let 
$$
C^1\_0(\overline\Omega) = \{u\in C^1(\Omega)\cap C(\overline\Omega):u|\_{\partial\Omega}=0\},
$$
and let $C^1\_c(\Omega)$ be the space of compactly supported $C^1$ functions in $\Omega$. We usually define $H^1\_0(\Omega)$ as the completion of $C^1\_c(\Omega)$ with respect to the $H^1$-norm. I want to define a new space, say $H^1\_\*(\Omega)$, as the completion of $C^1\_0(\overline\Omega)\cap H^1(\Omega)$ with respect to the $H^1$-norm. 
My question is, are these spaces the same? In other words, can any element of $C^1\_0(\overline\Omega)\cap H^1(\Omega)$ be approximated with arbitrary accuracy by compactly supported smooth functions in the $H^1$-norm? 


I can prove this under some assumptions on the boundary of $\Omega$, but the question is what happens if $\Omega$ is an arbitrary open set. The motivation for this question is that for solving boundary value problems with homogeneous Dirichlet conditions, it seems that the space $H^1\_\*$ is more natural, at least pedagogically. The introduction of $H^1\_0$ in a first lecture on Dirichlet problem appears "from nowhere" because it is not clear to the student why for instance, we have to take the completion of $C^1\_c$, which consists of functions with not only the values, but also the derivatives vanishing near the boundary. I thought it would be good to introduce $H^1\_\*$ first, and then later prove that it would have made no difference if we used $C^1\_c$ or $C^\infty\_c$ in the definition. However, I am stuck at proving this equivalence for general domains.



Variation on the Sobolev space $H^1_0$
Assume that $\Omega \subset \mathbb{R}^n$. I will prove that $C^1\_c (\Omega)$ is dense in $H^1 (\Omega) \cap C^1 (\Omega) \cap C\_0 (\Omega)$ endowed with the $H^1$ norm, where 
$$
 C\_0 (\Omega) = \{u: \Omega \to \mathbb{R} : \text{for every } \varepsilon > 0, \, u^{-1} (\mathbb{R} \setminus (-\varepsilon, \varepsilon)) \text{ is compact}\}. 
$$


Take $\theta \in C^1 (\mathbb{R})$ such that $\lim\_{\vert t \vert \to \infty} \frac{\theta (t)}{t} = 1$ and $\lim\_{\vert t \vert \to \infty} \theta' (t) = 1$ and $\theta = 0$ on $(-1, 1)$. For example,
$$
 \theta (t) = \begin{cases}
 \sqrt{1 + (t - 1)^2} & \text{if $t > 1$},\\
 0 & \text{if $- 1 \le t \le 1$},\\
 -\sqrt{1 + (t + 1)^2} & \text{if $t <- 1$},\\
\end{cases}
$$
In particular, $\theta'$ is bounded on $\mathbb{R}$. Set for $\lambda > 0$
$$
 \theta\_\lambda (t) = \lambda \theta (t/\lambda).
$$


For every $u \in H^1 (\Omega) \cap C^1 (\Omega) \cap C\_0 (\Omega)$ and $\lambda > 0$, define $u\_\lambda = \theta\_\lambda \circ u$. First observe that since $u \in C\_0 (\Omega)$ and $\theta\_\lambda = 0$ on $(-\lambda, \lambda)$, $\operatorname{supp} u\_\lambda$ is compact.
Next, for every $x \in \Omega$,
$$
 \lim\_{\lambda \to 0} u\_\lambda (x)
=u (x)
$$
and 
$$
 \lim\_{\lambda \to 0} \nabla u\_\lambda (x)
=\begin{cases}
\nabla u (x) & \text{if $u (x) \ne 0$},\\
0 & \text{if $u (x) = 0$}
\end{cases}
$$
By the implicit function theorem, $\{x \in \Omega : u(x) = 0 \text{ and } \nabla u (x) \ne 0\}$ is an $n-1$--dimensional embedded submanifold of $\Omega$. Hence for almost every $x \in \Omega$,
$$
 \lim\_{\lambda \to 0} u\_\lambda (x)
=u (x)
$$
and 
$$
 \lim\_{\lambda \to 0} \nabla u\_\lambda (x)
=
\nabla u (x).
$$


In order to conclude, observe that for every $x \in \Omega$,
$$
 \lvert u (x) - u\_\lambda (x) \rvert ^2
+ \lvert \nabla u (x) - \nabla u\_\lambda (x) \rvert ^2
\le \Bigl( \sup\_{t \in \mathbb{R}} \frac{\lvert \theta (t) - t\rvert}{\lvert t \rvert}\Bigr)^2 \lvert u (x) \rvert^2 + \Bigl( \sup\_{t \in \mathbb{R}} \lvert \theta' (t) - 1\rvert\Bigr)^2 \lvert \nabla u (x) \rvert^2.
$$
Hence, by Lebesgues' dominated convergence theorem, 
$$ \lim\_{\lambda \to 0} \int\_{\Omega} \lvert u - u\_\lambda \rvert ^2
+ \lvert \nabla u - \nabla u\_\lambda \rvert ^2=0.
$$


