Smoothing a Sobolev function Let $u \in H^1({\mathbb R}^n)$, $n \geq 2$.   Let $\varphi \in C^\infty_0({\mathbb R}^n)$ with $\varphi \geq 0$.  Let $\eta$ be a smoothing kernel with $\eta \in C^\infty_0({\mathbb R}^n)$, $\eta \geq 0$, $\int \eta \,dx = 1$.  For $t > 0$, define $\eta_t$ by $\eta_t(x)=\frac{1}{t^n}\eta(\frac{x}{t})$.  Define ${\tilde u}$ by 
$$
{\tilde u}(x)=
   \begin{cases}
    u(x); &\text{if } \varphi(x)=0, \\
\\
   \int_{{\mathbb R}^n} \eta_{\varphi(x)}(y-x) u(y)\, dy; & \text{if } \varphi(x) > 0.
   \end{cases}
$$
My question is, is ${\tilde u}$ in $H^1({\mathbb R}^n)$?  
 A: Let
$$
{\tilde u}(x)=
   \begin{cases}
    u(x); &\text{if } \varphi(x)=0, \\
\\
   \int_{{\mathbb R}^n} \eta_{\varphi(x)}(y-x) u(y)\, dy; & \text{if } \varphi(x) > 0.
   \end{cases}
$$
We shall first show that $\widetilde u$ is $L^2(\mathbb R^n)$. First we observe that $\widetilde u$ in $\mathbb R^n\smallsetminus\mathrm{supp}\,\varphi$ is identical to $u$ and for every $x\in\mathrm{supp}\,\varphi$ 
$$
|\widetilde u(x)| \le \int_{\mathbb{R}^n}|\eta_{\varphi(x)}(y-x)|\, |u(y)|\, dy\le
\|u\|_{L^2(\mathbb R)^n}\|\eta\|_{L^2(\mathbb R)^n}=M.
$$
Hence $\widetilde u$ is a sum of an $L^2$-function and a bounded and compactly supported function, and hence also an $L^2$-function. Thus $\widetilde u\in L^2(\mathbb R^n)$.
Next, following the idea of Davide Giraudo.
It is clear that $\widetilde u$ can be written as
$$
J_\varphi[u](x)=\widetilde u(x)=\int_{\Bbb R^n}\eta(t)\,u\big(x+\varphi(x)t\big)\,dt. \tag{1}
$$
and if $u$ is sufficiently smooth, then 
\begin{align}
\partial_j\widetilde u(x)&=\int_{\Bbb R^n}\eta(t)\,\Big(1+\sum_{k=1}^n\partial_k\varphi(x)t\Big)\partial_ju(x+\varphi(x)t)dt\\
&=\widetilde{\partial_j u}(x)+\sum_{k=1}^n \partial_k\varphi(x)\int_{\Bbb R^n}t\,\eta(t)\,\partial_ju\big(x+\varphi(x)t\big)\,dt,
\end{align}
and
$$
\int_{\Bbb R^n}t\,\eta(t)\,\partial_ju\big(x+\varphi(x)t\big)\,dt=
   \begin{cases}
    u(x)\int_{\mathbb R^n}t\,\eta(t)\,dt; &\text{if } \varphi(x)=0, \\
\\
   \int_{{\mathbb R}^n} (y-x)\,\eta_{\varphi(x)}(y-x) \partial_j u(y)\, dy; & \text{if } \varphi(x) > 0.
   \end{cases}
$$
It is not hard to see (using similar arguments) that $\widetilde{\partial_j u}\in L^2(\mathbb R^n)$.
It remains to explain what happens if $u$ is not smooth. 
In such casse we can find smooth $u^\varepsilon$, such that $u^\varepsilon\to u$, in the $H^1$-norm, and simply check that is $u^\varepsilon-u^{\varepsilon'}$ tends to zero, as $\varepsilon,\,\varepsilon'\to 0$, so does $\widetilde u^\varepsilon-\widetilde u^{\varepsilon'}$.
A: It's not an answer, but there are just some ideas. Maybe it will help. 
We can write 
$$\widetilde u(x)=\int_{\Bbb R^n}\eta(t)u(x+\varphi(x)t)dt,$$
since it's true when $\varphi(x)=0$, and when it's not the case we use a substitution. 
When $u$ is a test functions, it appears that $\widetilde u\in L^2(\Bbb R^n)$ and 
$$\partial_j\widetilde u(x)=\int_{\Bbb R^n}\eta(t)\sum_{k=1}^n\left(1+\partial_k\varphi(x)t\right)\partial_ju(x+\varphi(x)t)dt,$$
which proves that $\widetilde u\in H^1(\Bbb R^n)$.
