We know that if $u$ has a weak derivative $u'$, and if $f \in C^1$ with $f'$ bounded then $(f(u))'= f'(u)u'$.
But how interpret the duality pairing $$\langle (f(u))', v \rangle = \langle f'(u)u', v \rangle?$$
Is it equal to $\langle u', f'(u)v \rangle$. I don't think so, I think we need smoothness to say so. But is there any way to write the equality above as one involving the weak derivative $u'$?
Let us for example here take the duality pairing of $H^{-1}$ and $H^1$.