Banach fixed point theorem and Picard-Lindelöf applied to this equation (explanation needed) Consider the following equation which holds for all $w$ in some space,
$$\langle v(t), w \rangle = \langle v(0), w \rangle - \int_0^t \langle F(s,v(s)), w \rangle$$
where $\langle F(s,v),w \rangle$ is locally Lipschitz continuous in $v$. 

This integral equation has a solution by help of Banach's fixed point theorem in analogy to the existence theorem of Picard-Lindelöf. 

I don't see how this is similar to the equation $y'(t) = f(t, y(t))$ or its integral, because we have the duality pairing $\langle v(t), w \rangle$ here. How to overcome this?
Source is Section 2.1 of http://link.springer.com/article/10.1007%2Fs10492-012-0004-0.
edit: Maybe I put the integral inside the duality pairing and then apply the ODE existence theory to the first component of the duality pairing??
EDIT: some text from the paper:
http://imgur.com/1r2Ijil
http://imgur.com/Frw3IsK
http://imgur.com/drlwRgv
 A: In the article it is written $w_j$ is a finite dimensional basis of a finite dimensional space $A := span(w_1...w_k)$ and that the equation you mention has a solution $[0, T]\to A$ (only, it is written just before (2.1))
The author showed that $\|\langle F(s,u),w\rangle - F(s,v),w\rangle\|\le C\|u-v\|$ for all $u,v$ in a bigger space than $A$. Then, as Alex has written, let 
$$
F_j(y) = \langle v(0),w_j\rangle + \int_0^t \langle F(s,y), w_j\rangle ds.
$$
If you assume that $F_j\colon A\to C([0,T])$; let $y_n:\mathbb R\to A$ be given, and $(F_j(y_n))_{j\le k}$ is a family of $k$ elements, there exists then a unique $y_{n+1}(t)\in A$ s.t.
$$
\forall j, t,\qquad F_j(y_n)(t) = \langle y_{n+1}(t), w_j\rangle
$$
because the family $(w_j)_j$ is a basis!! This transition $y_n\to y_{n+1}$ is now clear, and with the relation
$$
\forall j, \langle y_{n+1}, w_j\rangle = F_j(y_n)
$$
you prove that there exists a fixed point for the sequence $(y_k)$...
Note: It seems for me that the finite basis hypothesis is the key for the proof... Moreover, i'm not totally sure about the spaces above, but you will correct me if needed...
EDIT: how to get the fixed point? As usually: Let $\|\cdot\|$ be the norm as explained in the article (like $L^{m'}$) norm and define the new norm in time
$$
\|f\|_T = sup_{t\in[-T,T]}(e^{-Lt}\|f(t)\|)
$$
Moreover, as the deimension is finite and all norm are equivalent, the $\|\cdot\|$ norm in $A$ is equivalent to the norm defined by $f\in A$, $\sum_j |\langle f, w_j\rangle|$. Therefore, exists $C'$ such that
$$
\|y_{n+1}(t) - y_n(t)\| \le C' \int_0^t \|y_n(s)-y_{n-1}(s)\|ds
$$
Then,
$$
\begin{split}
\|y_{n+1}-y_n\|_T &\le C'\sup_{[-T,T]} e^{-Lt}\int_0^t\|y_n(s)-y_{n-1}(s)\|e^{-Ls}e^{Ls}ds \\
&\le C'\|y_n-y_{n-1}\|_T \sup_{[-T,T]} \frac 1Le^{-Lt}[e^{Lt}-1] \\
&\le \frac{C'}{L}\|y_n - y_{n-1}\|_T
\end{split}
$$
Thus for $L<C'$ the sequence is contracting. We just have to check that the space $\mathcal C([0, T]; A)$ with $\|\cdot\|_T$ is a Banach space. $A$ with $\|\cdot\|$ is a Banach (it is $L^{m'}$ projected on a closed space), thus $C([0,T],A)$ with this sup norm is also a Banach: Everything is fine and you can apply Banach fixed point theorem!
A: I cannot read the linked paper, I would have to buy the paper :). If I understand right, the formulation you give is a weak formulation for the integral equation
$$v(t)=v(0)+\int_{0}^{t}F(s,v(s))d s,$$ which is again a formulation for the ode you mentioned. In my opinion, it is essential that you name the test space for $w$ as well as what they (the authors) understand by a weak solution of this type of ode. Now let $w$ be fixed, then, we have to show that the mapping
$$
C_{w}:A\rightarrow A,\ C_{w}(x)=\langle v_{0},w\rangle+\int_{0}^{t}\langle F(s,v(s)),w\rangle ds
$$
is a contraction for $A$ a proper Banach space (you should know, which space to pick, or the paper should mention this fact). Let $\|\cdot\|_{A}$ be the norm of the space $A$, we have for $x,y\in A$ and $w$ arbitrary but fixed:
$$\|C_{w}(x)-C_{w}(y)\|_{A}\leq \int_{0}^{t}L\|x-y\|_{A}\ d s \|w\|\leq Lt\|x-y\|_{A}\|w\|,$$
applying Cauchy Schwarz (if we have an inner product) and using the local Lipschitz continuity. So for sufficiently small $t$ this is a contraction also depending on $w$. Now one might be able to assume $\|w\|=1$ (depends on the definition of the weak solution) and we have the contraction, and thus a solution in the - still to be specified - space $A$. 
Of course, first of all you have to show, that the $C_{w}$ is mapping $A$ to $A$, i.e. $C_{w}$ is a self-map. But I cannot show this, since I do not know $A$....
Maybe this is a bit helpful or gives some ideas for the solution...
Yours,
Alex
