Let $\Omega \subset \mathbb R^n$ be a domain, sufficiently smooth. Let $T>0$. Define the space $W(0,T)$ by $$ W(0,T) = \{ y \in L^2(0,T; H^1_0(\Omega)): \ y'\in L^2(0,T;H^{-1}(\Omega)),\ $$ where $y'$ is the time derivative of $y$, $H^{-1}(\Omega)$ the dual space of $H^1_0(\Omega)$.
The question is: Let $y\in W(0,T)$. Does the positive part $y^+$ of $y$, $$ y^+(x,t) = \max(0, y(x,t)) $$ belong to $W(0,T)$ as well?
My intuition is, that the answer is 'no', however, I am not able to find a counter-example.
The difficulty here, is to prove that $(y^+)' \in L^2(0,T;H^{-1}(\Omega))$.
If one knows that in addition $y' \in L^2(0,T;L^2(\Omega))$, then one can show that $$ (y^+)' = \chi_{Q^+} y', $$ where $\chi_{Q^+}$ is the characteristic function of the set $$ Q^+=\{(x,t): \ y(x,t)>0\}. $$ If $y'\in L^2(0,T;H^{-1}(\Omega))$ only, then I think it is not true that $\chi_{Q^+} y' \in L^2(0,T;H^{-1}(\Omega))$.
It appears that the answer to the question is indeed negative. Here is a counterexample.
Let $\Omega = I = (0,1)$. Define $V:=H_0^1(\Omega)$, $H:=L^2(\Omega)$ with the scalar products $(u,v)_V = \int_\Omega \nabla u\cdot \nabla v $ and $(u,v)_H=\int_\Omega u v $.
The example exploits the fact that the mapping $u\mapsto u^+$ is unbounded from $V^*$ to $V^*$ (that is, there is no constant $c>0$ such that $\|u^+\|_{V^*}\le c \|u\|_{V^*}$ holds for all $u\in H$).
Functions in space. Define the functions $\psi_n(x):=\sin(n \pi x)$. Then by elementary calculations one obtains $$ \|\psi_n\|_{V} = \frac{n \pi}{\sqrt2}, \quad \|\psi_n\|_{H} = \frac{1}{\sqrt2}, \quad \|\psi_n\|_{V^*} = \frac{1}{\sqrt2 n \pi}. $$ Now let us estimate the positive part of $\psi_n$. First, one computes $$ \int_\Omega \psi_n^+ = \left\lfloor \frac{n+1}2\right\rfloor \int_0^{1/n} \sin(n\pi x) \ge \frac 1\pi. $$ The function $e(x)=1$ is not in $V$, so we approximate it by $v_e(x):=\min(4x,1,4(1-x))$. Then $$ \|e-v_e\|_H^2 = 2\int_0^{1/4}(4x)^2 = \frac16. $$ Then we get $$ \langle \psi_n^+ , v_e\rangle_{V^*,V} =( \psi_n^+ , v_e)_H \ge (\psi_n^+ , e)_H - \|\psi_n^+\|_H\|v_e-e\|_H \ge \frac1\pi - \frac1{\sqrt{12}} = 0.0296\dots \ge \frac15. $$ Hence it holds $\|\psi_n^+\|_{V^*} \ge c$ for some constant $c>0$. We will exploit the different scaling of $\|\psi_n\|_{V^*} $ and $\|\psi_n^+\|_{V^*} $ in the construction of the counterexample.
Functions in time. Let $\phi\in H_0^1(I)$ be given with $\phi\ne0$ and $\phi\ge0$. Define $\phi_n(t):=n(n+1)\cdot \phi( n(n+1)t -n)$. These functions have support on $(\frac1{n+1},\frac1n)$. It holds $\|\phi_n\|_{L^1(I)} =\|\phi\|_{L^1(I)}$ and $$ \|\phi_n\|_{L^2(I)} \le \sqrt2 n\|\phi\|_{L^2(I)},\quad \|\phi_n'\|_{L^2(I)} \le \sqrt2 n^3\|\phi'\|_{L^2(I)},\quad \|\phi_n'\|_{L^1(I)} \ge n^2\|\phi'\|_{L^1(I)} $$
Series definition. Define $$ \theta(x,t):=\sum_{i=1}^\infty n^{-3} \phi_n(t)\psi_n(x). $$ Note that the supports of $\phi_n$ are distinct. Due to the estimates above we obtain $$ \|\theta\|_{L^2(I;V)}^2 = \sum_{i=1}^\infty n^{-6} \|\phi_n\|_{L^2(I)}^2\|\psi_n\|_V^2 \le c \sum_{i=1}^\infty n^{-2} < \infty, $$ $$ \|\theta'\|_{L^2(I;V^*)}^2 = \sum_{i=1}^\infty n^{-6} \|\phi_n'\|_{L^2(I)}^2\|\psi_n\|_{V^*}^2 \le c \sum_{i=1}^\infty n^{-2} < \infty, $$ hence the series converges in $W(0,T)$ and $\theta \in W(0,T)$. The time derivative of $\theta^+$ exists and belongs to $L^1_{loc}(I;V^*)$. However it is not in $L^1(V^*)$: Assume this would be true, then by continuity of the integral it follows $$ \|(\theta^+)'\|_{L^1(V^*)} =\lim_{N\to\infty} \int_{1/(N+1)}^1 \|(\theta^+)'\|_{V^*} =\lim_{N\to\infty} \sum_{i=1}^N n^{-3}\|\phi_n'\|_{L^1(I)}\|\psi_n^+\|_{V^*}\\ \ge c\,{\lim\sup\,}_{N\to\infty} \sum_{i=1}^N n^{-1} = +\infty, $$ and it follows $(\theta^+)'\not\in L^1(V^*)$.