If $u$ and $v$ have weak derivatives,what about $uv$?

Question

$\Omega$ is a domain in $R^n$, Let $u\in L^1_{\text{loc}}(\Omega)$. If there exists $g_i \in L^1_{\text{loc}}(\Omega)$ such that $$\int_\Omega g_i \phi \, dx=-\int_\Omega u \frac{\partial \phi_i}{\partial x},\phi \in C_0^\infty(\Omega)$$ Then we say $u$ has weak partial derivatives $g_i$.

If $u$ and $v$ have weak partial derivatives, does $uv$ have weak partial derivatives? Or what conditions should we add to $u$ and $v$?

Answer 1

If $\def\loc{\text{loc}}u,v\in L^\infty_{\loc}(\Omega)$, then $uv\in L^1_{\loc}(\Omega)$ and $u\frac{\partial v}{\partial x_i}+v\frac{\partial u}{\partial x_i}\in L^1_{\loc}(\Omega)$. It remains to show that if $U\subset \Omega$ is a open set with $\overline{U}\subset\Omega$ compact, then $$\int_U (uv)\frac{\partial\varphi}{\partial x_i}=-\int_U \left(u\frac{\partial v}{\partial x_i}+v\frac{\partial u}{\partial x_i}\right)\varphi,\ \forall\ \varphi\in C_0^\infty(U)$$

To this end I refer your to Brezis page 269, Proposition 9.4.

Answer 2

You can also ask that $uv$, $u'v$ and $uv'$ are in $L^1_{loc}$ : then the usual product rule holds. (Use the sequential characterization of weak derivatives with $\mathcal C^\infty$ functions and pass to the $L^1_{loc}$ limit in the usual product rule).