Parabolic time-dependent PDE notation/dual space confusion Suppose we have the PDE
$\frac{\partial u}{\partial t} - \Delta u = f(x,t)$  
with some boundary conditions.
I am confused about what it means to say that the weak time derivative $u_t \in L^2([O,T];H^{-1})$. It makes no sense to me. My teacher has written:
We have $u_t = \Delta u + f(x,t)$ so $u_t \in  L^2([O,T];H^{-1})$.
As far as I'm concerned, $u_t$ for a fixed $t$ is a function, not a functional that takes in a $H^1_0$ function and gives out a real number. What am I doing wrong?
 A: The reason for this is that $\Delta u \in L^2([0,T];H^{-1})$, as $u \in L^2([0,T];H^1_0)$. Since the right hand side the PDE is in $L^2([0,T];H^{-1})$, the left hand side must be too.
The functional associated to $u_t$ is given by integration by parts:
$$\langle u_t, v\rangle = \int u_t v dx = \int (\Delta u + f)v dx = \int -\nabla u \cdot \nabla v + fv dx,$$
for $v \in H^1_0$.  For each $v \in H^1_0$, $\langle u_t,v\rangle$ returns a real number.
EDIT:  In general, if $u \in L^2$, then $\partial_{x_i} u \in H^{-1}$ as follows:  for any $v \in H^1_0$, $\partial_{x_i} u$ acts on $v$ by the pairing
$$\langle \partial_{x_i} u,v\rangle = -\int u (\partial_{x_i} v) dx.$$
Note that when $u \in H^1$, this pairing agrees, by integration by parts, with the usual $L^2$ inner product
$$ \langle \partial_{x_i} u, v \rangle = \int (\partial_{x_i} u)v dx.$$
A: A function $u:[0,T]\times \Omega\to\mathbb{R}$ can be viewed from different angles. The most natural one is the one which I have just written: Put in $t$ and $x$ and obtain $u$. However, this is not always appropriate (as you should learn is PDE-classes) and often it makes much more sense, to view them as elements in function spaces, e.g. $u\in L^p([0,T]\times\Omega)$ (this makes a difference, e.g. this view does not allow for point evaluation of $u$).
Another view is the one you need here: $u$ can be viewed as a mapping from $[0,T]$ into a function space. For each $t$ you get a function depending on $x$: $u(t,\cdot):\Omega\to\mathbb{R}$. If you have this view, than you write, e.g. $u:[0,T]\to L^p(\Omega)$ or $u:[0,T]\to H^1(\Omega)$. If you now impose regularity for this mappings you end up with spaces like $L^p([0,T],H^1(\Omega))$. These spaces are called Bochner spaces.
