An equality in $L^2(0,T;V')$!? Weak solution to PDE via Galerkin approximations I have the heat equation
$$u' - \Delta u = f$$
as equality in $L^2(0,T;V')$,i.e.,
$$(u',v) + (\nabla u, \nabla v) = (f,v)$$
for all $v \in L^2(0,T;V)$, where I used the same brackets for duality pairing and inner product for succinctness.
Let $w_j$ be the basis in $V$ and $H$. I read in a book that the finite dimensional (Galerkin) approximations to the PDE 
$$(u_n',w_j) + (\nabla u_n, \nabla w_j) = (f, w_j), \quad\text{for $j=1,...,n$}$$
can be written as
$$\frac{d}{dt}u_n - \Delta u_n = P_nf\tag{1}$$
where $P_n$ is a projection operator.
In what sense is this latter equation an equality? Presumable not an equality in $L^2(0,T;V')$ because the weak form I wrote above only holds for the basis functions. However, the author later takes an inner product of (1) with an element on $L^2(0,T;V)$, how can he do that?
 A: Well, it is indeed an abuse of notation though, by no means you could say (1) is an equality in $L^2(0,T;V')$, because (1) only holds in a finite dimensional subspace. Multiply (1) by the basis of the finite dimensional subspace
$$
(u_n',w_j) - (\Delta u_n,w_j) = (P_n f,w_j).
$$
Proper assumption about the boundary condition and continuity leads to:
$$
(u_n',w_j) - (\Delta u_n,w_j) = (u_n',w_j) + (\nabla u_n, \nabla w_j).
$$
Hence:
$$
(P_n f,w_j) = (f,w_j), \quad \text{for } j=1,\ldots,n,
$$
which is to say, $P_n f$ is the $L^2$-projection of $f$ onto this finite dimensional subspace.
I didn't see the paper, but I guess the reason to multiply a function in $L^2(0,T;V)$ is to get some a priori error estimate for this approximation by exploiting the fact that
$$
(u'-u_n',w_j) + \big(\nabla (u-u_n), \nabla w_j\big) = (f-P_n f,w_j).
$$

EDIT: question in the comment, instead of claiming (1) holds, rather what we can do is the following
$$
\begin{aligned}
&(u_n',\phi) + a(u_n,\phi)
\\
=& (u_n',P_n\phi) + a(u_n,P_n\phi) + (u_n',\phi - P_n\phi) + a(u_n,\phi-P_n\phi) 
\\
=& (P_n f,P_n \phi) + (u_n',\phi - P_n\phi) + a(u_n,\phi-P_n\phi) 
\\
=& (P_n f,\phi) + \color{blue}{(u_n',\phi - P_n\phi)} + \color{red}{a(u_n,\phi-P_n\phi) }.
\end{aligned}
$$
Blue term could vanish if we integrate w.r.t. time and integration by parts to move the derivative to $\phi-P_n \phi$. However for the red term to vanish, we must have something like $\Delta u_n = 0$...

EDIT2: If we say $f = g$ in $L^2(0,T;V')$, I don't know what the author did, but according to the Evans' book (see Section 7.1), it is 
$$\newcommand{\lsub}[2]{{\vphantom{#2}}_{#1}{#2}}
\lsub{V'}{\langle f(t,\cdot),v \rangle}_V = \lsub{V'}{\langle g(t,\cdot),v \rangle}_V,
$$
for any $v\in V$, at a.e. time $t\in [0,T]$. This is the same with what you wrote in the comment though.
