If $\varphi \in C^\infty_0(\Omega)$ and $u \in H^1(\Omega)$ then $\varphi u \in W^{1, 2}_0(\Omega)$? Let $\Omega \subset \mathbb R^n$, $\varphi \in C^\infty_0(\Omega)$ and $u \in W^{1, 2}(\Omega)$. In my functionnal analysis course, we use the fact that $\varphi u \in W^{1, 2}_0 (\Omega)$ and I am not very sur why this is true.
If $\Omega$ is sufficiently smooth, then I think I see why: in this case $C^\infty(\overline{\Omega})$ is dense in $W^{1, 2}(\Omega)$ so we can find $(u_n)_n \subset C^\infty(\overline{\Omega})$ with $u_n \to u$ and so $\varphi u_n \to \varphi u$. As $(\varphi u_n)_n \subset C^\infty_0(\Omega)$, we can conclude that $\varphi u \in W^{1, 2}_0(\Omega)$. Is this right? Is the statement still true when $\Omega$ is any open set in $\mathbb R^n$ ?
 A: The $H=W$ result in [1] mentioned by gerw, if I understand correctly, is Evans' Theorem 2 of Section 5.3.2 (page 265 in the second edition):

THEOREM 2 (Global approximation by smooth functions). Assume U is bounded, and suppose as well that $u \in W^{k, p}(U)$ for some $1 \leq p<\infty$. Then there exist functions $u_{m} \in C^{\infty}(U) \cap W^{k, p}(U)$ such that
$$
u_{m} \rightarrow u \quad \text { in } W^{k, p}(U) .
$$
Note carefully that we do not assert $u_{m} \in C^{\infty}(\overline{U})$ (but see Theorem 3 below).

It is Theorem 3 that requires some regularity of the boundary (approximation by functions in $C^{\infty}(\overline{U})$).
edit It was pointed out in the comments that Evans' formulation is only for bounded sets. The extension to unbounded sets is not too hard using cutoff functions, but for the particular result you're after, you can proceed directly (see edit below).
For clarity I repeat the proof sketch:

*

*Let $u_n\in C^\infty(\Omega) \cap W^{k,p}(\Omega)$ such that $u_n\xrightarrow{W^{k,p}(\Omega)} u$ from Theorem 2.

*Verify $\phi u_n \xrightarrow{W^{k,p}(\Omega)} \phi u$ with product rule

*Conclude, since $W^{k,p}_0(\Omega) := \overline{C^\infty_c(\Omega)}^{W^{k,p}(\Omega)}$ and $\phi u_n\in C^\infty_c(\Omega)$.

So the result is true even for $k\neq1$ and $p\neq 2$.
edit In the case of an unbounded $\Omega$, it is enough to show that $\phi u \in W^{k,p}_0(K)$ for some bounded subset $K\subset \Omega$. Such a $K$ is provided by the support of $\phi$ which is compact (hence bounded); since $u|_K\in W^{k,p}(K)$, the proof works without other modifications.
I imagine you could also fashion a different proof, being careful with the supports, using Theorem 1.

 [1] Meyers, Norman G.; Serrin, James, (H = W), Proc. Natl. Acad. Sci. USA 51, 1055-1056 (1964). ZBL0123.30501. 
