If $u \in L^2(0,T;H^1)$ has a distributional derivative $u' \in L^2(0,T;H^{-1})$, does $(u^+)' \in L^2(0,T;H^{-1})$ make sense?

Question

If $u \in L^2(0,T;H^1)$ has a distributional derivative $u' \in L^2(0,T;H^{-1})$, does $(u^+)' \in L^2(0,T;H^{-1})$ exist, i.e., is $u^+$ weakly differentiable in time? By $u^+$ I mean the positive part of $u$. It is tempting to saying that $$(u^+)' = \begin{cases} u' &: u \geq 0\\ 0 &: u < 0 \end{cases} $$ however this is not particularly rigorous. Does anyone know how to treat such objects?

Answer 1

This answer is probably wrong. Please disregard.

I have a note that Wloka's Partial Differential Equations, Section 25, is a good place to look, but I don't have a copy handy to check right now.

A possible outline is like this. Let $\mathcal{H}$ be your space of $L^2(0,T; H^1)$ functions with a weak derivative in $L^2(0,T; H^{-1})$. First you verify the chain rule that for $\Phi : \mathbb{R} \to \mathbb{R}$ which is, say, $C^1$ and has a bounded derivative, we have $(\Phi \circ u)' = (\Phi' \circ u) u'$. (It may help to verify it first on a dense subspace of $\mathcal{H}$ consisting of appropriately smooth functions.) Then you choose a sequence $\Phi_n(x)$ converging pointwise to $x^+$ and with uniformly bounded derivatives, apply the chain rule to $\Phi_n$, and pass to the limit with appropriate justification.

This doesn't work because even though $\Phi_n' \circ u$ converges pointwise and boundedly, it does not follow that $(\Phi_n' \circ u) u'$ converges in $H^{-1}$.