MathOverflow is a question and answer site for professional mathematicians. 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

I am studying the following article by Benoit Perthame:

Somewhere in the middle of it, I'm stuck at proving a certain limit equality. Maybe it's obvious and I can't get it.

$$ \int_{(\Bbb{R})} \left(\chi(\xi,u)\star \varphi_\varepsilon \right)^2d \xi \to |u| \text{ in } {L}^1_{loc} $$

where $\varphi_\varepsilon(t,x)$ is a regularizing kernel, $u$ satisfies $$\partial_t u +\text{div}A(u)=0 \text{ and }\text{ in }\mathcal{D}^\prime((o,\infty)\times \Bbb{R}^d) $$ and

$$ \chi(\xi,u)=\begin{cases} 1 & {0\leq \xi\leq u} \newline -1 & u \leq \xi \leq 0 \newline 0 & \text{otherwise} \end{cases}$$

Thank you.

[edit] Sorry. I forgot to mention that $u \in L^1_{loc}$.

share|cite|improve this question
(a) I think if the equality were to hold, it will hold regardless of the equation. For continuous functions it is easy to check that the limit holds even pointwise, since $|u| = \int_{\mathbb{R}} \chi(\xi,u)^2 d\xi$ and it is just pushing integrals around and justifying a few Fubini's. (b) If $u\in D'$, you don't necessarily have that $|u|$ is some well-defined object in $L^1_{loc}$. For example, what is $|\delta'|$? So perhaps what the equation is doing is giving $u$ a priori some regularity so that the absolute value of $u$ is well-defined. – Willie Wong Feb 29 '12 at 14:16
Ah, your desired expression appear in the "Proof for (2.2)" step. In (2.2), which is part of the statement of Theorem 2.1. So by hypothesis $u$ is $L^1_{loc}$. So you can probably use the result for continuous functions and Luzin's theorem to get convergence. – Willie Wong Feb 29 '12 at 14:29
@Willie Wong: Thank you for your comment. I'm starting to understand now. – Beni Bogosel Feb 29 '12 at 16:38

Your Answer


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

Browse other questions tagged or ask your own question.