Leibniz's rule in $W^{1,\,p}(\Omega)\cap L^\infty(\Omega)$

Question

Is it true?

If $\Omega\subset\mathbb{R}^n$ is bounded and $u,v\in W^{1,\,p}(\Omega)\cap L^\infty(\Omega)$, then $uv\in W^{1,\,p}(\Omega)\cap L^\infty(\Omega)$ and $$\nabla(uv)=u\nabla v+v\nabla u.$$

I proved for $2\leq p$.

Answer 1

This is true for $p\geqslant 1$. The case $p=1$, being a limiting one, requires more efforts. It will be sufficient to establish the rule $$ \partial_j(uv)=u\partial_j v+v\partial_j u,\quad j=,\dots,n,\tag{1} $$ for the weak derivatives $\partial_j$ just in case $p=1$, with the equality $(1)$ understood in a weak sense, i.e., as an integral identity $$ \int\limits_{\Omega}uv\partial_j\varphi\,dx= -\int\limits_{\Omega}(u\partial_j v+v\partial_j u)\varphi\,dx\quad \forall\, \varphi\in C_0^{\infty}(\Omega)\tag{2} $$ where $C_0^{\infty}(\Omega)$ denotes the space of infinitely differentiable functions compactly supported in the domain $\Omega\subset\mathbb{R}^n$, $n\geqslant 2$. The basic idea underlying the proof is to approximate $u,v$ by smooth functions. While smooth functions are dense in the Sobolev space $W^{1,1}(\Omega)$ w.r.t. its norm, it is not the case with the Lebesgue space $L^{\infty}(\Omega)$. However, essentially bounded functions can be approximated in $L^{\infty}(\Omega)$ by smooth functions in the weak$^\ast$- topology which is enough to establish the identity $(2)$.

Denote by $\omega_{\varepsilon}$ a mollifier, i.e., let $\omega_{\varepsilon}=\varepsilon^{-n}\omega(x/\varepsilon)$, $\varepsilon>0$, with some $\omega\in C_0^{\infty}(\Omega)$ such that $$ \int\limits_{\mathbb{R}^n}\omega(x)\,dx=1,\quad {\rm supp\,}\omega\subset B_1=\{x\in\mathbb{R}^n\colon\,|x|<1\}. $$ Take and fix an arbitrary test function $\varphi\in C_0^{\infty}(\Omega)$ to be substituted into identity $(2)$, and notice there is some $\delta\overset{\rm def}{=}{\rm dist}({\rm supp\,}\varphi,\partial\Omega)>0$. Denote $\Omega_{\delta}=\{x\in\Omega\,\colon\,{\rm dist}(x,\partial\Omega)>\delta\}$ and notice that ${\rm supp\,}\varphi\subset\overline{\Omega}_{\delta}$. Extensions of $u,v$ by zero outside $\Omega$ denote by $\tilde{u},\tilde{v}$, and let $u_{\varepsilon}=\omega_{\varepsilon}\ast\tilde{u},\; v_{\varepsilon}=\omega_{\varepsilon}\ast\tilde{v}$. It is clear that $u_{\varepsilon}\in C^{\infty}(\mathbb{R}^n)$ while $$ \lim_{\varepsilon\to 0}\|u-u_{\varepsilon}\|_{W^{1,1} (\Omega_{\delta})}=0.\tag{3} $$ Convergence of $u_{\varepsilon}$ in the norm of $L^{\infty}(\Omega_{\delta})$ requires $u\in C(\Omega)$ which may not be the case, but it is clear that $$ \lim_{\varepsilon\to 0}\int\limits_{\Omega_{\delta}}u_{\varepsilon}\psi\,dx= \int\limits_{\Omega_{\delta}}u\,\psi\,dx\quad \forall\, \psi\in L^1(\Omega_{\delta}),\tag{4} $$ i.e., smooth functions $u_{\varepsilon}$ converge to $u$ in the weak$^\ast$- topology on $L^{\infty}(\Omega_{\delta})$ being dual to $L^1(\Omega_{\delta})$.

The fact that $u_{\varepsilon},v_{\varepsilon}\in C^{\infty}(\Omega_{\delta})$ implies $$ \int\limits_{\Omega_{\delta}}u_{\varepsilon}v_{\varepsilon}\partial_j\varphi\,dx= -\int\limits_{\Omega_{\delta}}(u_{\varepsilon}\partial_j v_{\varepsilon}+ v_{\varepsilon}\partial_j u_{\varepsilon})\varphi\,dx, $$ whence by $(3),(4)$ follows $(2)$.