# Derivation of a product : Sobolev spaces.

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 $\|u\|_\infty$

• thanks for your answer. For Q1), how can I be sure that $\rho_n$ really exist ? And for 2), we have that $u_n\to u$ in $W^{1,p}$ not in $L^p$... is it the same ? For 3) how do you know that $u_n$ and $v_n$ are $C^\infty$ function ? – MSE Dec 27 '16 at 11:45
• @MSE : there exists explicit exemples of such mollifiers such as this one : en.wikipedia.org/wiki/Mollifier#Concrete_example – Tryss Dec 27 '16 at 11:47
• @MSE And for your second question, I really suggest that you learn the definition of $W^{1,p}$ before going further. Basically, $u\to u$ in $W^{1,p}$ means that $u_n \to u$ in $L^p$ and that $\partial u_n \to \partial u$ in $L^p$ too – Tryss Dec 27 '16 at 11:51
• So, for 1), if $f\in L^p$, then we alway have an approximation of identity s.t. $\rho_n*f\to f$ in $L^p$ ? (it's not written explicitely in your link) – MSE Dec 27 '16 at 13:46
• @MSE: to complete the answer for your question 2, notice that $$\|u\|_{W^{1,p}}=\|u\|_{L^p}+\sum_{i=1}^n\|\partial _i u\|.$$ Therefore, $u_n\to u\in W^{1,p}$ $\iff$ $\|u_n-u\|_{L^p}\longrightarrow 0$ and $\|\partial _iu_n-\partial_i u\|_{L^p}\longrightarrow 0$ for all $i$. For 1), yes, there is always such a $\rho_n$. – Surb Dec 27 '16 at 13:57