Fix $u\_0\in H^1(\Omega)$ and $f=f(x,y,t)\in L^2(\Omega\times [0,T])$ where $\Omega$ is a sufficiently smooth bounded domain in $\mathbb{R}^2$.
Consider the problem of finding $u:\Omega\times[0,T]\to\mathbb{R}$ satisfying the following variational equation
$$
\begin{cases}
\langle \nabla u, \nabla v\rangle\_{L^2(\Omega)}+\langle u,v\rangle\_{L^2(\Omega)}=\langle f,v\rangle\_{L^2(\Omega)},\;\;\forall v\in H^1(\Omega),\;\forall t\in(0,T], \\\;u(\cdot,0)=u\_0, \;
\end{cases}
$$
It is clear that this variational problem comes from the equation 


$$\begin{cases}
-\Delta u+u=f, & \text{ on }\Omega\times(0,T],\\
u(0)=u\_0. 
\end{cases}$$


I did not put boundary conditions, but any boundary condition for me works as long as $\int\_{\partial\Omega}(\vec{n}\cdot\nabla u)v=0$ for all $v$.


**Note:** *All of the differential operators seen above are spacial operators, they involve no derivatives in time.*


**Questions:** 


* What are appropriate function spaces to look for solutions?
* Can we have that $u(\cdot,t)\in L^2([0,T])$ for all $t\in[0,T]$?
* Can we have that $\text{essup}\_{t\in[0,T]}\|u(\cdot,t)\|\_{L^2(\Omega)}\leq C$ for some constant $C\in\mathbb{R}$ independent of time and space?



A time dependent variational problem coming from a second order linear PDE
As you correctly pointed out, since the operators appearing here are only acting in space your problem amounts to solving an elliptic equation for each time. In particular assigning $u(0)=u\_0$ is meaningless: Even if the right-hand side $f$ is continuous in time, and without further compatibility assumptions between $f$ and $u\_0$, it is most likely false that the solution of the elliptic problem at time $t=0$ for the datum $f|\_{t=0}$ coincides with your "initial datum" $u\_0$.
Long story short: since time is not really a variable here, but rather a parameter, you need to figure out a framework to make sense of "regular dependence of the data on the parametere $t$", and then the dependence of the solution will follow. More details below.


In order to give a more specific answer to your question, let $K:=((-\Delta)+ 1)^{-1}$ be the inverse operator mapping the datum $f$ to the solution $u=Kf$ of $-\Delta u+u=f$.
By standard elliptic regularity $K$ is a continuous operator from $L^2$ to $H^2$. Here I'm assuming that your boundary conditions, which you apparently don't prescribe explicitly, are nice enough so that this standard regularity machinery works. If not, the just replace $H^2(\Omega)$ below by $H^1(\Omega)$.


**Answers**


* *Bochner space* is the key word here. In your first equation you implicitly consider $f(t)=f(.,.,t)$ as an element of $L^2(\Omega)$ for fixed time. (By the way: it seems that you work exclusively in spatial dimension 2, $(x,y)\in\Omega\subset \mathbb R^2$, if so it would be better to edit your question for the sake of clarity.) This is exactly what Bochner spaces do, typically $L^2((0,T);X)$ is the space of squared integrable functions $f:(0,T)\to X$ with values in the Banach space $X$, here you may want to consider of course $X=L^2(\Omega)$.
Just a quick comment, though: whether an $L^2$ function $f(x,y,t)$ in space-time (in the classical sense) can actually be considered as such a Bochner function $t\mapsto f(.,.,t)\in L^2(\Omega)$, and conversely, is actually slightly more delicate than what it may seem at first sight. But this is whole different topics so let me not elaborate too much here, just google "Bochner space" or look up this keyword here on Math.MO you'll find quasillions of related questions and answers.
* For your second question the answer is yes. But in order to define $u(t)$ you need to be able to evaluate $f(t)$ at (almost every) time $t$. This is my first answer to your question, so let's assume that indeed your datum $f\in L^2(0,T;L^2(\Omega))$. In this setup it makes sense to evaluate $f(t)\in L^2(\Omega)$ for a.e. $t\in (0,T)$, hence you can *define*
$$
u(t):= K[f(t)]=(-\Delta +1)^{-1}f(t) \in H^2(\Omega)
\qquad \mbox{for a.e. }t.
$$
So the answer is yes, and moreover you see that $u(t)$ lies in the better space $H^2(\Omega)$.
Actually what this shows is that $u$ can be properly defined as an element of the Bochner space $L^2(0,T;H^2(\Omega))$.
Note here that since $u$ is only defined for a.e. time clearly evaluating $u(0)$ does not make sense (you can always redefine $u$ on a negligible set of times and still get the same $L^2(0,T;H^2(\Omega))$ element)
* With these preliminaries out of the way, the answer to your question 3 should now be pretty obvious: No, you cannot hope for such $L^\infty(0,T;L^2)$, unless you improve the time regularity of $f$.
To see this, take any real-valued function of time $\eta=\eta(t)$ such that $\eta\in L^2(0,T)$. Take an arbitrary $F=F(x,y)\in L^2(\Omega)$ and set $f(x,y,t):=\eta(t)F(x,y)$.
Then clearly (the operator $-\Delta +1$ does not depend on time!) the solution of your problem is
$$
u(t)=\eta(t) KF
\qquad\mbox{hence}\qquad
\|u(t)\|\_{L^2(\Omega)}=|\eta(t)|\, \|KF\|\_{L^2(\Omega)}
$$
and therefore $\sup\limits\_{t\in (0,T)} \|u(t)\|\_{L^2(\Omega)} = +\infty$ unless $\eta\in L^\infty(0,T)$.
Note here that the same line of thoughts shows actually that the time regularity of $f$ gives the sharp time regularity of the solution, for example $f\in C^k([0,T];L^2)\Rightarrow u\in C^k([0,T];H^2)$ or whatever (you can substitute $C^k$ by whatever regularity scale you wish, roughly speaking).


