The question is summarized in the title.
More precisely, let $f(x,t) \in L^2\bigl([0,1]^2 \times [0,1] \bigr)$ with $\int_{[0,1]^2} f(x,t) d^2x=0$ for all $t \in [0,1]$.
Then, we can apply the inverse Laplacian $(-\Delta)^{-1}_x$ to this $f(x,t)$ to see that \begin{equation} (-\Delta)^{-1}_xf(x,t) \in H^2_xL^2_t ([0,1]^2 \times [0,1] \bigr) \end{equation} That is, $\int_0^1 \lVert (-\Delta)^{-1}f(\cdot, t) \rVert_{H^2_x}^2 dt < \infty$.
Now, let us further assume that $(-\Delta)^{-1}_xf(x,t)$ is in fact $\textit{jointly}$ continuous with respect to $(x,t)$ on $[0,1]^2 \times [0,1]$.
Then, I wonder if \begin{equation} \lVert f(\cdot, t) \rVert_{L^2_x}^2= \int_{[0,1]^2} \lvert f(x,t) \rvert^2 d^2x \end{equation} is still continuous as a function of $t \in [0,1]$.
Since, $x$ and $t$ are independent variables, I guess the answer would be yes, but I cannot justify it. Could anyone please help me?
Not true. Here is a counterexample on the spatial domain $[0,1]$, it can be easily extended to $[0,1]^2$. Define $$ f_n(x) := \sin( n \pi x ). $$ Then $\|f_n\|_{L^2(0,1)}^2 = \frac12$, $f_n \rightharpoonup 0$, and $f_n \not \to 0$ in $L^2(0,1)$. For $t\in [\frac1{n+1}, \frac1n]$ define $$ f(t,x) = \lambda f_n(x) + (1-\lambda) f_{n+1}(x), $$ where $\lambda\in [0,1]$ is that $$ t = \lambda \frac1n + (1-\lambda) \frac1{n+1}. $$ Set $f(0,x)=0$. Then $t \mapsto f(\cdot,t)$ is continuous for $t>0$, but the limit $\lim_{t\searrow0} f(\cdot,t)$ does not exist in $L^2(0,1)$. (The map $t\mapsto \lambda\mapsto f$ is continuous. And $f(\cdot, \frac1n) = f_n$.) Moreover, since $f_n$ and $f_{n+1}$ are orthogonal, $\|f(t,\cdot)\|_{L^2}^2 \ge \frac14$ for all $t>0$, so that $t\mapsto \|f(t,\cdot)\|_{L^2}^2$ is not continuous at $t=0$.
Now $(-\Delta_x)^{-1}f$ is (assuming homogeneous boundary conditions) $$\begin{split} (-\Delta_x)^{-1}f(t,\cdot) &= \lambda (-\Delta_x)^{-1} f_n + (1-\lambda) (-\Delta_x)^{-1}f_{n+1}\\ &= \lambda \frac1{n^2\pi^2}f_n + \lambda \frac1{(n+1)^2\pi^2}f_{n+1} \end{split}, $$ which is continuous at $t=0$, i.e., $(-\Delta_x)^{-1}f(\cdot,0) \to 0$ for $t\to 0$.
For spatial domain $[0,1]^d$, $d>1$, define $$ f_n(x) := \prod_{k=1}^d \sin( n \pi x_k ). $$ And the same reasoning as above applies. The example can also be modified for different boundary conditions: set $f_n$ the be the $n$-th eigenfunction of $-\Delta$.