# 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.

• Yes. That is what I meant. @votes – Slm2004 Jul 14 '13 at 23:26
• Hmmm... the question is whether the mollification is super good: $\Vert u_\epsilon - u \Vert_\infty \to 0$ would allow us to bound $\left|\int_{\Omega}uv_{\epsilon}D\phi-\int_{\Omega}uvD\phi\right|$ by $\left\Vert u\right\Vert _{1}\left\Vert \left(v_{\epsilon}-v\right)\right\Vert _{\infty}\left\Vert D\phi\right\Vert _{\infty}$; but I am not sure when such good approximations exist. It's much easier when $u,v \in L^2$, where just $L^2$ approximation and Hoelder's theorem are enough. – Anthony Carapetis Aug 5 '13 at 14:10
• @Carapetis In general we don't have such good approximations. $L^1_{loc}$ really gives me much trouble. – Slm2004 Aug 11 '13 at 2:55

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?

Step 4: Consider the problem with additional assumption $u\geq 0$ and $v\geq 0$.

Firstly we assume $v\geq 1$. Define $\tilde{u}_n=\min\{u,\frac{n}{v}\}=u+(\frac{n}{v}-u)^+$. Since $u,\frac{n}{v} \in W^1(\Omega)$, then $\tilde{u}_n\in W^1(\Omega)$, satisfies $0\leq \tilde{u}_n v\leq n$, moreover $$D\tilde{u}_n=\begin{cases}Du,\quad uv<n\\-n\frac{Dv}{v^2},\quad uv\geq n\end{cases}\in L^1_{loc}(\Omega)$$ $$vD\tilde{u}_n+\tilde{u}_n Dv=\begin{cases}vDu+uDv,\quad uv<n\\0,\quad uv\geq n\end{cases}\in L^1_{loc}(\Omega)$$ Suppose $\displaystyle f_\epsilon(t)=\frac{t}{1+\epsilon t}$ defined on $\mathbb{R}_+^1$, where $\epsilon>0$. Then $f_\epsilon$ is a piecewise smooth function in $\mathbb{R}_+^1$ and $f'_\epsilon\in L^\infty(\mathbb{R}_+^1)$. By lemma 7.8 on Gilbarg and trudinger's book, $f_\epsilon(\tilde{u}_n)\in W^1(\Omega)$. Note that $f_\epsilon(u)\in L^\infty(\Omega)$ by the conclusion of step 3, $$\int_{\Omega}f_\epsilon(\tilde{u}_n) vD\phi=-\int_{\Omega}[Df_\epsilon(\tilde{u}_n)v+f_\epsilon(\tilde{u}_n)Dv]\phi$$ that is $$\int_{\Omega}\frac{\tilde{u}_n}{1+\epsilon \tilde{u}_n}vD\phi=-\int_{\Omega}\frac{\tilde{u}_nDv+vD\tilde{u}_n}{1+\epsilon \tilde{u}_n}\phi+\int_{\Omega}\frac{\epsilon \tilde{u}_nvD\tilde{u}_n}{(1+\epsilon \tilde{u}_n)^2}\phi\quad (1)$$ By dominating convergence theorem, as $\epsilon\to 0$ $$\int_{\Omega}f_\epsilon(\tilde{u}_n) vD\phi\to \int_{\Omega}\tilde{u}_nvD\phi$$ $$\int_{\Omega}\frac{\tilde{u}_nDv+vD\tilde{u}_n}{1+\epsilon\tilde{u}_n}\phi\to \int_{\Omega}(\tilde{u}_nDv+vD\tilde{u}_n)\phi$$ And $$\left|\int_{\Omega}\frac{\epsilon \tilde{u}_nvD\tilde{u}_n}{(1+\epsilon\tilde{u}_n)^2}\phi\right|\leq \epsilon n\int_{\Omega}|D\tilde{u}_n||\phi|\to 0\text{ as }\epsilon\to 0$$ Letting $\epsilon\to 0$, $(1)$ implies $$\int_{\Omega}\tilde{u}_nvD\phi=-\int_{\Omega}[vD\tilde{u}_n+\tilde{u}_nDv]\phi$$ Since for any $n>0$ $$|\tilde{u}_nv|\leq |uv|$$ $$|vD\tilde{u}_n+\tilde{u}_nDv|\leq |vDu+uDv|$$ by dominating convergence theorem, letting $n\to \infty$ $$\int_{\Omega}uvD\phi=-\int_{\Omega}[vDu+uDv]\phi$$

If we only know $v\geq 0$, consider $u$ and $v+1$, repeat the above proof $$\int_{\Omega}u(v+1)D\phi=-\int_{\Omega}[(v+1)Du+uD(v+1)]\phi$$ note that $u\in W^1(\Omega)$, this is equivalent to $$\int_{\Omega}uvD\phi=-\int_{\Omega}[vDu+uDv]\phi$$

Step 5: Consider the most general case with no extra assumption. Since $$u^+v^+, u^+v^-, u^-v^+,u^-v^-\in L^\infty_{loc}(\Omega)$$ $$v^{\pm}Du^\pm+u^\pm Dv^\pm\in L^\infty_{loc}(\Omega)\text{ respectively }$$ step 4 will imply $$u^\pm v^\pm\in W^1(\Omega),\quad D(u^\pm v^\pm)=v^{\pm}Du^\pm+u^\pm Dv^\pm$$ So $$uv= u^+v^+-u^+v^-- u^-v^++u^-v^-\in W^1(\Omega)$$ and $$D(uv)=uDv+vDu$$

• I use too many steps. Maybe there is an easier way to solve it. Any comment is welcome. – Slm2004 Aug 12 '13 at 18:15

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.

I am stuck in this problem tonight, I found this answer helpful, but it may be simplified as follows.

• Step 2 can be done if we only assume $u$ or $v \in W^1(\Omega) \cap L^\infty(\Omega)$.

• Step 4 can be easier if one notes the monotonicity and hence Step 3 is unnecessary. (The key is still to define $U_n := \min\{u,\frac{n}{v}\}$.)

• I also do not understand why $u^+ v^+ \in L^\infty_\mathit{loc}$ in Step 5. But I found an alternative (easier?) way to prove the general case.

I post my answer here and hope it's correct !!

Step 3: Consider the problem under the assumption $u,v \geq 0$

we first assume $v \geq 1$ and define $U_n := \min\{u,\frac{n}{v}\}$. By Lemma 7.6 of Gilbarg-Trudinger's book, $U_n \in W^1(\Omega) \cap L^\infty(\Omega)$, and $D(U_n v) = Du \chi_{\{uv < n\}} + (- \frac{n Dv}{v^2})\chi_{\{uv \geq n\}}$. Hence $U_n Dv + v DU_n = (u Dv + v Du)\chi_{\{uv < n\}} \in L^1_{\mathit{loc}}(\Omega).$ Step 2 implies that, for each $\phi \in C^1_0 (\Omega), \int_\Omega U_n v D\phi = - \int_\Omega [U_n Dv + v DU_n] \phi.$

I note the monotonicity $U_n v = \min \{ uv,n \} \nearrow uv$

and $U_n Dv + v DU_n\to u Dv + v Du$ a.e. and $|U_n Dv + v DU_n| \leq |u Dv + v Du| \in L^1(\mathit{supp} \phi)$, MCT and LDCT imply that \begin{align*} \int_\Omega u v D\phi & = \Big( \int_{\{ D\phi \geq 0\}} + \int_{\{ D\phi < 0 \}} \Big) uv D \phi = \lim_{n \to \infty} \Big( \int_{\{ D\phi \geq 0\}} + \int_{\{ D\phi < 0 \}} \Big) U_n v D \phi \\ = & \lim_{n \to \infty} \int_\Omega U_n v D \phi = \lim_{n \to \infty} - \int_\Omega [U_n Dv + v DU_n] \phi = - \int_\Omega (u Dv + v Du)\phi. \end{align*}

The above formula is also true for general $v \geq 0$, since \begin{align*} \int_\Omega u v D\phi + \int_\Omega u D\phi & = \int_\Omega u (v+1) D\phi = - \int_\Omega (u D(v+1) + (v+1) Du)\phi \\ & = - \int_\Omega (u Dv + v Du)\phi - \int_\Omega Du \phi. \end{align*}

Finally, we decompose $u = u^+ - u^-, v=v^+ - v^-$. Since $u^+v^+ = uv \chi_{\{ u>0,v>0 \}}$ and $u^+ Dv^+ + v^+ Du^+ = (uDv + vDu) \chi_{\{u>0,v>0 \}}$, they are in $L^1_{\mathit{loc}}(\Omega)$. Other similar functions also in $L^1_{\mathit{loc}}(\Omega)$ without proof. Then by Step 3, for each $\phi \in C^1_0(\Omega)$,

\begin{align*} & \int_\Omega u v D\phi = \int_\Omega (u^+ v^+ - u^+ v^- - u^- v^+ + u^- v^-) D\phi \\ &= - \int_\Omega \Big( (u^+ Dv^+ + v^+ Du^+) - (u^+ Dv^- + v^- Du^+) -(u^- Dv^+ + v^+ Du^-) + (u^- Dv^- + v^- Du^-) \Big)\phi \\& = - \int_\Omega (u Dv + v Du)\phi. \end{align*}