Positive part of $y$ with $y\in L^2(0,T; H_0^1(\Omega))$ and $y'\in L^2(0,T; H^{-1}(\Omega))$ 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 
$\chi_{Q^+} y' \in L^2(0,T;H^{-1}(\Omega))$.
 A: 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 $(\theta^+)'\not\in L^1(V^*)$.
