A limit involving a regularizing kernel - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T03:23:07Z http://mathoverflow.net/feeds/question/89801 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89801/a-limit-involving-a-regularizing-kernel A limit involving a regularizing kernel Beni Bogosel 2012-02-28T21:37:09Z 2012-02-29T16:49:03Z <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>$$\chi(\xi,u)=\begin{cases} 1 &amp; {0\leq \xi\leq u} \newline -1 &amp; u \leq \xi \leq 0 \newline 0 &amp; \text{otherwise} \end{cases}$$</p> <p>Thank you.</p> <p> Sorry. I forgot to mention that $u \in L^1_{loc}$.</p>