Let $W \subset \mathcal{L}^2(\mathbb{R})$ be a linear subspace. I want to show that for a certain $\phi \in W$, $\{ \phi_m : m \in \mathbb{Z}\}$ is a basis for $W$. Here $\phi_m$ describes a translation of $\phi$ by $m$, i.e. $\phi_m(x):= \phi(x-m)$. For this we normally have to prove that there exists a sequence $(c_m)_{m \in \mathbb{Z}} \subset \mathcal{l}^2(\mathbb{Z})$ such that \begin{equation} f = \sum_{m \in \mathbb{Z}} c_m \phi_m \end{equation} holds.
Now there is stated that this representation is equivalent to \begin{equation} \hat{f} (\zeta) = c(\zeta) \hat{\phi}(\zeta) \end{equation} with $c$ being $2\pi$-periodic in $\mathcal{L}^2([0,2\pi])$, and the $\hat{}$-operator describing the Fourier transform.
The question ist, why does this equivalence hold?
$\hat {\phi_m} (\zeta)=e^{im\zeta}\hat {\phi} (\zeta)$ so $\hat {{\sum c_m\phi_m} (t)}=\sum c_m e^{im\zeta} \hat {\phi} (\zeta)=c(\zeta)\hat {\phi} (\zeta)$. For the converse part expand $c(\zeta)$ in its Fourier series and use the fact $f=\sum c_n\phi_m$ if the two sides have the same FT.