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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
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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
    
@Siminore : This is why I asked the question. I should have explained this in my question. I have a function in $H^1(\Omega)$ for a bounded smooth domain $\Omega$ and I want to approximate it by a smooth function, using mollifier(s) and convolution. If I use a single mollifier everywhere in $\Omega$, then values of the function outside of $\Omega$ will be used, and I don't want that. So I want to vary the supports of the mollifiers. –  Stefan Smith Feb 1 at 13:01
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2 Answers 2

up vote 2 down vote accepted

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 $\widetilde u^\varepsilon-\widetilde u^{\varepsilon'}$.

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Thanks. I posted this question ages ago and was not expecting to get another answer. The reason I asked it in the first place is that I have function $u \in H^1(\Omega)$ for a smooth bounded domain $\Omega$ that I want to approximate by a smooth function with the same trace (boundary values) on $\partial \Omega$. $u$ is a function of bounded variation. Do you know if there is a standard way of doing this? –  Stefan Smith Feb 1 at 13:05
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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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