Does the weak derivative satisfy the product rule? For a locally integrable function $f$, a weak derivative $f'$ satisfies the following relation: $\int{f\phi^{\alpha}}=(-1)^{\alpha}\int{f'\phi}$. Does a weak derivative also satisfy the product rule? For instance, is $(fg)'=f'g+fg'$?
 A: Suppose that $f$ is weakly differentiable and that $\psi$ is bounded and smooth.  If $\phi$ is smooth and compactly supported so is $\psi \phi$ and
$$\int (f' \psi) \phi \, dx = \int f' (\psi \phi) \, dx = - \int f (\psi \phi)' \, dx = - \int f \psi' \phi \, dx - \int f \psi \phi' \, dx$$
which rearranges to
$$\int (f \psi) \phi' \, dx = - \int( f'\psi + f \psi') \phi \, dx.$$
This means that  $f\psi$ is weakly differentiable and $(f \psi)' = f'\psi + f \psi'$.
As pointed out in the comments in general there is going to be an issue of $fg$ being non-locally integrable. If you can approximate $g$ in the right way by smooth functions you can circumvent that difficulty but that will require some use of Sobolev spaces.
A: Counterexample
Let 
$H(x) = \begin{cases}1&\text{if }x>0,\\0&\text{otherwise.}\end{cases}$
This function is locally integrable and so is its $n$:th power since $H^n = H$ for $n=1,2,3,\ldots$
The weak derivative $H'$ is a distribution given by
$$
\langle H', \phi \rangle
= -\langle H, \phi' \rangle
= -\int_{-\infty}^{\infty} H(x) \, \phi'(x) \, dx
= -\int_{0}^{\infty} \phi'(x) \, dx
= -[\phi(x)]_{0}^{\infty} 
= \phi(0)
= \langle \delta, \phi \rangle
.
$$
Thus we should also have $(H^n)' = H' = \delta.$ But according to the power rule, which follows from the product rule, it should be $(H^n)' = n H^{n-1} H' = n H^{n-1} \delta.$ There is not even any way to redefine $H(0)$ so that $n H^{n-1} \delta = \delta$ for all $n.$
