I already asked the question on MSE, and have tried to figure it out myself.
But the problem seems trickier than expected, so I guess MO is a better place to ask..
For the sake of completeness, I repeat the question.
For each $n \in \mathbb{N}$, let us consider the Cauchy problem \begin{equation} \partial_t u_n -\Delta u_n = f_n \text{ on } [0,T] \times \mathbb{T}^m \text{ with } u_n(t=0,x)=u^{0}_n(x) \end{equation} where $f_n(t,x) \in L^p_t\bigl( [0,T], L^q_x(\mathbb{T}^m) \bigr)$ for some fixed $p,q \in (1,\infty)$ and $u^{0}_n(x) \in W^{2,q}(\mathbb{T}^m)$. $T \in (0,\infty)$ is assumed to be fixed and $\mathbb{T}^m=(\mathbb{R}/\mathbb{Z})^m$ is the $m$-torus.
Then, it is well-known that there exists a unique weak solution, denoted as $u_n(t,x)$ for the above Cauchy problem.
Now, suppose that $u^{0}_n \to u^0$ in $W^{2,q}(\mathbb{T}^4)$ and $f_n \to f$ in $L^p_t\bigl( [0,T], L^q_x(\mathbb{T}^m) \bigr)$. Denote by $u(t,x)$ the unique weak solution of the Cauchy problem \begin{equation} \partial_t u -\Delta u = f \text{ on } [0,T] \times \mathbb{T}^m \text{ with } u(t=0,x)=u^0(x) \end{equation}
My question is that, do we have \begin{equation} u_n \to u \text{ in } W^{1,p}_t\bigl([0,T], L^q_x(\mathbb{T}^m) \bigr) \text{ as } n \to \infty? \end{equation}
I strongly suspect so, but cannot find a way to prove using maximal regularity or well-posedness. Could anyone please help me?