Let $u \in L^2(0,T;H^1)$ have weak derivative $u' \in L^2(0,T;H^{-1}).$ Let $c$ be a constant.
I want to show that $$\langle ((u-c)^+)', (u-c)^+ \rangle = \langle u', (u-c)^+ \rangle$$ where $(f)^+ = \max\{0, f\}$ (the brackets denote the duality pairing).
Is it as easy as recognising $((u-c)^+)' = u'\chi_{\{u-c > 0\}}$ by linearity of the weak derivative? Then it is equal to the RHS when $\{u-c > 0\}$, and when $\{u-c \leq 0\}$, the argument on the RHS is $0$ so again it is equal.