Product rule of weak derivatives

I am working on proving the following proposition: If $u,v\in {W^1(\Omega)}$ and $uv,uDv+vDu\in L^1_{\operatorname{loc}}(\Omega)$, then we have the product formula $$D(uv)=uDv+vDu.$$

The definition I use for weak derivative: A function $u\in L^1_{loc}(\Omega)$ has a $\alpha-$th weak derivative in $\Omega$ if there is a function $v\in L^1_{\operatorname{loc}}(\Omega)$ with: $$\int_{\Omega}uD^\alpha\phi=(-1)^{|\alpha|}\int_{\Omega}v\phi dx\quad \forall\,\phi\in C^\infty_0(\Omega)$$ $u\in W^1(\Omega)$ means it has first weak derivative.

I searched on the internet. Most of them just proved the case of $u\in W^1(\Omega)$ and $v\in C^1(\Omega)$. This is the first step of the proving. I want to prove the general case by approximation. Namely, let $v_\epsilon$ denote the regularization (or mollification) of $v\in W^1(\Omega)$, then it is true $$D(uv_\epsilon)=v_\epsilon Du+uDv_\epsilon$$ or equivalently $$\int_{\Omega}uv_\epsilon D\phi=-\int_{\Omega}(v_\epsilon Du+uDv_\epsilon)\phi dx\quad \forall\,\phi\in C^\infty_0(\Omega)$$ I am trying to prove $$\int_{\Omega}uv_\epsilon D\phi\to \int_{\Omega}uvD\phi\text{ as }\epsilon\to 0$$ $$\int_{\Omega}(v_\epsilon Du+uDv_\epsilon)\phi dx\to \int_{\Omega}(v Du+uDv)\phi dx\text{ as }\epsilon\to 0$$ But I can not accomplish, even for the first limit. Can anyone give me some idea? Maybe I am heading a wrong direction. But I am quite convinced that the mollification is so good that this must be true.

On the other hand, this formula is very elementary in Sobolev space. Probably you can prove this proposition by distribution theory(or generalized function), however, it sounds cheating for me somehow, because here our definition has noting with distribution.

Solution 1:

I still have no clue to prove $$\int_{\Omega}u v_\epsilon D\phi\to \int_{\Omega}uvD\phi\text{ as }\epsilon\to 0$$ So I prove it in another direction. I can solve the problem under the more restrictive assumption $uv, uDv,vDu\in L_{loc}^1(\Omega)$ respectively. The following is how I achieved

Step 1: prove the case $u\in W^1(\Omega)$, $v\in C^1(\Omega)$

Stpe 2: prove the case $u,v\in W^{1}(\Omega)\cap L^\infty_{loc}(\Omega)$ by step 1. Readers can also see the proof on page269 of Functional Analysis, Sobolev Spaces and Partial Differential Equations (Haim Brezis)

Step 3: Define $f\in C^0(\mathbb{R}^n)$ $$f_n(t)=\begin{cases}n,\quad t>n\\ t,\quad |t|\leq n\\ -n,\quad t<-n\end{cases}$$ then $f_n$ is piecewise smooth in $\mathbb{R}$ and $f_n'\in L^\infty (\mathbb{R})$. So $f_n(u)\in W^1(\Omega)$ by lemma 7.8 in Gilbarg and Trudinger's bk and $$D(f_n(u))(x)=\begin{cases}Du(x),\quad |u(x)|\leq n,\\0,\quad\quad |u(x)|>n.\end{cases}$$ Denote $u_n=f_n(u)\in W^1(\Omega)\cap L^\infty_{loc}(\Omega)$ and $v_n=f_n(v)\in W^1(\Omega)\cap L^\infty_{loc}(\Omega)$. By step 2 we have $u_nv_n\in W^1(\Omega)$ and $$\int_{\Omega} u_nv_n D\phi dx=-\int_{\Omega}(u_nDv_n+v_n Du_n)\phi dx\quad \forall\, \phi\in C^\infty_0(\Omega)$$ Note that by assumption $$|u_nv_n|\leq |uv|\in L^1_{loc}(\Omega) $$ $$|u_nDv_n|\leq |uDv|\in L^1_{loc}(\Omega) $$ $$|v_nDu_n|\leq |vDu|\in L^1_{loc}(\Omega) $$ By dominating convergence theorem, letting $n\to \infty$ $$\int_{\Omega} uv D\phi dx=-\int_{\Omega}(uDv+v Du)\phi dx\quad \forall\, \phi\in C^\infty_0(\Omega)$$ $Q.E.D.$

As I said before, we need $uDv$ and $vDu\in L^1_{loc}(\Omega)$ repectively, because we don't have $$|u_nDv_n+v_nDu_n|\leq |uDv+vDu|$$ This is very near to the original assumption. Can anyone give me more ideas?

Solution 2:

How about using the ACL characterisation of differentiability? I mean, $u$ has weak partial derivatives if and only if it is absolutely continuous on almost every line in any cardinal direction, and the classical partial derivatives (defined thus almost everywhere) are integrable.

Since the product of absolutely continuous functions is absolutely continuous, with $(fg)' = f' g + f g'$, it follows that $u v$ has the ACL property, and the classical partial derivatives of $u v$ obey the product rule almost everywhere. In order that $u v$ has weak partial derivatives it is therefore sufficient to assume that the classical partial derivatives of $u v$ are locally integrable.