Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
@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 '14 at 13:01
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 $\widetilde u^\varepsilon-\widetilde u^{\varepsilon'}$.

share|cite|improve this answer
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 '14 at 13:05

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.