Function monotony between [0,T] and $L^2$

Question

Let $\Omega\subseteq\mathbb{R}^N$ be a bounded and smooth domain. If $z:[0,T]\to L^2(\Omega)$ is a function in $H^1([0,T],L^2(\Omega))$ with the property that $z'(t)(x):=z'(t,x)>0$ a.e. on $\Omega$ for any $t\in [0,T]$ is it true that $z$ is an incresing function wrt the order of $L^2(\Omega)$, i.e. $z(t_1,x)>z(t_2,x)$ for almost all $x\in\Omega$ if $t_1>t_2$?

I did not succeed in finding a counterexample. Moreover I find this type of result being used in research articles on reaction-diffusion systems. For the definition and some basic properties of $H^1([0,T],L^2(\Omega))$ see here https://www.uni-ulm.de/fileadmin/website_uni_ulm/mawi.inst.020/abschlussarbeiten/MA_Marcel_Kreuter.pdf

Is there any concept of Dini derivative for such functions?

Answer 1

First, since you have $H^1(0,T)$ imbedds in $\mathscr{C}^0([0,T])$, $z$ can be seen as an element of $\mathscr{C}^0([0,T];L^2(\Omega))$ and you can speak without ambiguity of $z(t_1)$ and $z(t_2)$. Now, for $t_1>t_2$, the formula (that has to be understood in $L^2(\Omega)$) \begin{align*} z(t_1) = z(t_2) + \int_{t_2}^{t_1} \partial_t z (s)\,\mathrm{d} s, \end{align*} is true when $z$ is smooth and remains true in your setting by a density argument. You recover your inequality because equality in $L^2(\Omega)$ implies equality a.e.