Fourier series for general $f \in L^1(\mathbb{R})$ with compact support Let $f \in L^1(\mathbb{R})$ , which has a fourier transform with compact support in $[-T,T]$. 
I am trying to show that for the function $f^{\land}  : [-T,T]\rightarrow \mathbb{R}$ the equality
$$f^{\land} (\omega) = \frac{1}{\sqrt2 T}\sum_{n\in \mathbb{Z}}c_n\exp\left(in\frac{\pi}T\omega\right)$$
holds, with $c_n = \sqrt{\frac\pi T}f(-n\frac{\pi}T)$, where $f^\land$ denotes the fourier transform of $f$ on $\mathbb{R}$ restricted to $[-T,T]$.
I am not quite sure how to find this series representation. Basically, what I did is I defined the function $F(x) := f(\frac T \pi x)* 1_{[-\pi,\pi]}$. Then by integration and substitution I got that the fourier transform of $F$ is given by: $F^\land (x) = \frac{\pi}T f^\land(\frac \pi T x)* 1_{[-T,T]}.$
This is where I get stuck. How do I get the desired series above, or is my approach totally wrong?
Any help would be greatly appreciated!
 A: I'm going to assume $T=1$. And I'm not going to bother getting the $\pi$'s straight - recall the Litttlewood Convention, to the effect that $2\pi=1$.
Let's write $F=\hat f$.
As noted already, since $f$ and $\hat f$ are integrable the Inversion Theorem shows that $$f(t)=\int_{-1}^1e^{it\xi}F(\xi)\,d\xi.$$This shows that the Fourier coefficients of $F$ are given by values of $f$, as in that formula.
It also shows that $f$ extends to an entire function in the plane, defined by $$f(z)=\int_{-1}^1e^{iz\xi}F(\xi)\,d\xi.$$Define $f_y(t)=f(t+iy)$. Then $f_y$ is the inverse Fourier transform of $e^{-y\xi}F(\xi)$. There is a smooth function with compact support which equals $e^{-y\xi}$ on $[-1,1]$. Hence there is a Schwarz function $\phi_y$ with $\hat\phi_y(\xi)=e^{-y\xi}$ on $[-1,1]$, and hence $$f_y=f*\phi_y.$$So $||f_y||_1\le||f||_1||\phi_y||_1$. We can certainly obtain $||\phi_y||_1$ bounded for $|y|\le1$, hence $$||f_y||_1\le c||f||_1\quad(|y|\le 1).$$
In particular $$\int_{-1}^1\int_{-\infty}^\infty|f(t+iy)\,dtdy<\infty.$$
So $$\sum_{n\in\mathbb Z}\iint_{|z-n|<1}|f(z)| <\infty,$$which implies by the mean-value property of holomorphic functions that $$\sum_{n\in\mathbb Z}|f(n)|<\infty.$$So the Fourier series for $F$ converges uniformly, hence converges to $F$.
