A limit involving a regularizing kernel - MathOverflow most recent 30 from http://mathoverflow.net2013-05-20T03:23:07Zhttp://mathoverflow.net/feeds/question/89801http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/89801/a-limit-involving-a-regularizing-kernelA limit involving a regularizing kernelBeni Bogosel2012-02-28T21:37:09Z2012-02-29T16:49:03Z
<p>I am studying the following article by Benoit Perthame: <a href="http://www.mendeley.com/research/uniqueness-error-estimates-first-order-quasilinear-conservation-laws-via-kinetic-entropy-defect-measure/#" rel="nofollow">http://www.mendeley.com/research/uniqueness-error-estimates-first-order-quasilinear-conservation-laws-via-kinetic-entropy-defect-measure/#</a></p>
<p>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. </p>
<p>$$ \int_{(\Bbb{R})} \left(\chi(\xi,u)\star \varphi_\varepsilon \right)^2d \xi \to |u| \text{ in } {L}^1_{loc} $$</p>
<p>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</p>
<p>$$ \chi(\xi,u)=\begin{cases} 1 & {0\leq \xi\leq u} \newline
-1 & u \leq \xi \leq 0 \newline
0 & \text{otherwise} \end{cases}$$</p>
<p>Thank you.</p>
<p>[edit] Sorry. I forgot to mention that $u \in L^1_{loc}$.</p>