# Proof of $uv\in W^{1,p}(\Omega )$ and $\partial_i(uv)=u\partial _i v+v\partial _iu$.

Let $\Omega \subset \mathbb R^d$ an open set and $u,v\in W^{1,p}(\Omega )\cap L^\infty (\Omega )$. Prove that $uv\in W^{1,p}(\Omega )$ and $$\partial _i(uv)=u\partial _iv+v\partial _iu.$$

I in fact don't really understand why we want $u,v\in L^\infty (\Omega )$. But here is the proof :

Let $K\subset \subset \Omega$ and $\varphi_n$ a standard mollifier. We denote $$u_n=\varphi_n*u\quad \text{and}\quad v_n=\varphi_n*v.$$ We know that $u_n,v_n\in W^{1,p}(\Omega )$ and $\partial_i u_n=\varphi_n*\partial_i u$ and $\partial _i v_n=\varphi_n*\partial _iv,$ and thus $$u_n\to u\quad \text{and}\quad v_n\to v$$ in $W^{1,p}(\Omega )$.

Moreover $$\|u_n\|_{L^\infty (K)}\leq \|u\|_{L^\infty (\Omega )}\quad \text{and}\quad \|v_n\|_{L^\infty (K)}\leq \|v\|_{L^\infty (\Omega )}.$$

Supposons WLOG that $u_n\to u$ and $\partial _i u_n\to \partial u$ a.e. Then $$\partial _i(u_nv_n)=u_n\partial _i v_n+v_n\partial _i u_n\to u\partial _iv+v\partial _iu \in L^p(\Omega ).$$

Q1) Is it the limit in $L^p(\Omega )$ are pointwise ?

Then if $\zeta \in \mathcal C^1_0(\Omega )$, $$-\int_\Omega uv\partial _i \zeta =\lim_{n\to \infty }-\int_\Omega u_nv_n\partial _i\zeta =\lim_{n\to \infty }\int \partial _i(u_nv_n)\zeta =\lim_{n\to \infty }\int_\Omega (u_n\partial _iv_n+v_n\partial _iu_n)\zeta =\int_\Omega (u\partial_i v+v\partial _i u)\zeta$$ and thus $uv\in W^{1,p}(\Omega )$ and $\partial_i (uv)=u\partial_i v+v\partial_i u$.

Q2) Where did we use the fact that $u,v\in L^\infty$ and $\partial _i(u_nv_n)=u_n\partial _i v_n+v_n\partial _i u_n\to u\partial _iv+v\partial _iu \in L^p(\Omega )$ ?

Q3)In what the compact $K$ is important ? I don't have the impression we used it.

• It is because they chose the mollifiered of u and v which are bounded and converge in w^{1,p} – Guy Fsone Dec 30 '17 at 13:57
• To answer Q2, remember that $uv \in W^{1,p}$ requires all derivatives of $uv$ to be in $L^p$. Now look at the product rule and think about why the right hand side is in $L^p$ and why we need $L^\infty$ for this. This should also help you with the convergence. – mlk Dec 30 '17 at 14:02