# 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 $H^1({\mathbb R}^n)$?

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Fixed it. Using the equation environment is really bad TeXing. In this site you can just use . If you're TeXing your own stuff, I suggest either the align environment or the gather environment. – Patrick Da Silva Dec 28 '11 at 3:18
I do not know the answer. May I ask you why you are considering this strange "regularization"? – Siminore Apr 23 '12 at 17:09
If you formally differentiate $\tilde{u}$, are you sure you can avoid troubles if $\varphi$ approaches zero? It goes in the denominator, and, in principle, this may produce a singularity. – Siminore Apr 23 '12 at 17:13

## 1 Answer

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 $\widetilde u\in H^1(\Bbb R^n)$.

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