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 $H^1({\mathbb R}^n)$?
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 $\widetilde u\in H^1(\Bbb R^n)$.