The goal is to proof for $u \in L^2([0,T];H^1_0)$ and $u' \in L^2([0,T];H^{-1})$ and $f \in L^2([0,T];L^2)$ that $$ (31) \qquad \langle u', v \rangle + B[u,v;t] = (f , v)_{L^2} $$ for all $v \in H^1_0$ and a.e $ t \in [0,T]$. $B$ is some specific bilinear form and $\langle u',v\rangle$ denotes the action of $u'$ on $v$ in $H^{-1}$.
By weak convergence in the proof of Theorem 3 the following equation is established for some $$v = \sum_{k=1}^N d^k(t) w_k \in C^1([0,T];H_0^1)$$ with $N$ fixed and $\{ w_k\}_{k=1}^{\infty} $ form an orthogonal basis of $H^1_0$ and $d^k(t)$ "given smooth functions".
$$ (30) \qquad \int_0^T \langle u', v \rangle + B[ u , v;t] dt = \int_0^T (f,v)_{L_2} dt$$
From this equation it is concluded that equation (31) holds and I don't understand this conclusion.
In the book it is argued that (30) holds for all $v \in L^2([0,T];H^1_0)$, since $v$'s in the above form are dense in the space. But for example since $N$ is fixed, $w_{N+1} \in L^2([0,T]; H^1_0)$ can not be expressed by such $v$'s. Am I missing that somehow also $N \to \infty$?
Then if we assume that (30) holds for all $v \in L^2([0,T];H^1_0)$, how would you go from an integral version (30) to a non integral version (31)?