$f \in H^1_0(\Omega)$ , $g \in W^{1,\infty}_{loc} \Rightarrow fg \in H_0^1$?

Question

As stated in the title I am facing this problem: proving that

$f \in H^1_0(\Omega)$ , $g \in W^{1,\infty}_{loc}(\Omega) \Rightarrow fg \in H_0^1$ .

$\Omega \subset \mathbb{R}^n $ is bounded and smooth.

If g were bounded it would be straightforward, but as it is I can't go on.

I have problems in both showing that $fg\in L^2$ and showing that the weak derivative exists and is in $L^2$. I tried showing that

$*) \ \ \partial_{x_i}(fg)=g\partial_{x_i}f +f\partial_{x_i}g $

since, if it were the case, one could write

$\int |\nabla(fg)|^2\leq 2\int|\nabla f|^2 +2C\int \frac{f^2}{d^2} \ \ $ where $d(x)=distance(x,\partial \Omega)$

but I can't neither conclude from here nor prove that *) holds.

Thanks

Answer 1

Are you sure this is true? If $f(x)=\sin x$ in $(0,\pi)$ and $g(x)=\frac1{x^6}$ then I don't think the product is in $H^1$. Note that $g$ is $C^1$ and so in $W^{1,\infty}_{loc}$.