Let $u,v\in W^{1,p}(\Omega )\cap L^\infty (\Omega )$, $p\in[1,\infty ]$. Then, $u,v\in W^{1,p}(\Omega )$ and $$\partial _i(uv)=u\partial _iv+v\partial _iu.$$
I have problem to understand the proof. Let $p\in [1,\infty )$ and let $D\subset \subset \Omega $ an open. Let $\rho_n$ ba a standard mollifier. Define for $n$ large enough $$u_n=\rho_n* u\quad \text{and}\quad v_n=\rho_n*v.$$ Then, $$u_n\longrightarrow u\text{ in }W^{1,p}(\Omega )\quad \text{and}\quad v_n\longrightarrow v\text{ in }W^{1,p}(\Omega ),$$ and $$\|u_n\|_{L^\infty (\Omega )}\leq \|u\|_{L^\infty (\Omega )}\quad \text{and}\quad \|v_n\|_{L^\infty (\Omega )}\leq \|v\|_{L^\infty (\Omega )}.$$
Quest 1 : Why such $\rho_n$ exist and why do we have the previous convergence in $W^{1,p}(\Omega )$ and the inequality $\|u_n\|_{L^\infty (\Omega )}\leq \|u\|_{L^\infty (\Omega )}$ (and same with $v_n$). So as I see $\rho_n$ is more an approximation of identity, but still, why can I do that ?
WLOG, one may assume the $u_n\to u$ a.e. in $D$ and $\partial _i u_n\to \partial _i u$ a.e. in $D$.
Quest 2 : Why can we assume that ?
We have in $D$ that $$\partial _i(u_nv_n)=u_n\partial _i v_n+v_n\partial _i u_n\to v\partial _i u+u\partial _i v\in L(D).$$
Quest 3: We do we have this relation ? Isn't it what we wanted to prove at the beginning ? I really don't understand why $$\partial _i(u_nv_n)=u_n\partial _i v_n+v_n\partial _i u_n,$$ I have the impression that it's what we want to prove, no ? I neither don't understand why it converge to $v\partial _i u+u\partial _i v$... This proof looks so weird...
If I can understand all what happen before, the conclusion will be fine.
Some help :
Question 1)
It is indeed an approximation of the identity that is $C^\infty$. The norm inequality come directly from the Young's inequality :
$$\| \rho_n \ast u \|_{\infty} \leq \| \rho_n\|_1 \|u\|_{\infty}$$
Question 2)
If $u_n \to u$ in $L^p$, there exists a subsequence $u_{\phi(n)}$ that converge a.e to $u$. So we can just consider this subsequence
Question 3)
The $u_n$ and $v_n$ are $C^\infty$ functions, so it's the usual product rule. For the convergence, split the epsilons in two :
$$| u_n \partial_i v_n - u_n \partial_i v + u_n \partial_i v - u\partial_i v| \leq |u_n \partial_i (v_n - v) | + | (u_n-u) \partial_i v|$$
It works because the $u_n$ are bounded by $\|u\|_\infty$