Exponential decay of Fourier coefficients correspond to analytic functions I am looking for a proof of the Payley-Wiener theorem saying that an analytic function on $\mathbb{T}$ (the torus) has exponentially decaying Fourier coefficients. I also found a proof of this fact in this book, but I do not get what they are doing there, i.e. how they apply Cauchy's formula. 
If somebody could elaborate or prove this from scratch, I would be very grateful.

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*If $f(t)$ is $2\pi$ periodic and $C^\infty$, with $c_n = \frac{1}{2\pi} \int_0^{2\pi} f(t)e^{-i n t}dt$ then for any $k \in \mathbb{N}$ : $f^{(k)}(t) = \sum_{n= - \infty}^\infty (i n)^k c_n e^{i n t}$ and hence $n^k c_n \to 0$ as $n \to \pm \infty$, so that the coefficients $c_n$  are more than polynomially decreasing.

*Now if $f(t) = g(e^{it})$ where $g(z)$ is analytic on $1-\epsilon < |z| < 1+\epsilon$ then for those $z$ : $\ g(z) = \sum_{n=-\infty}^\infty c_n z^n$ (the same coefficients $c_n$ as before (*)) and hence $c_n (1+\frac{\epsilon}{2})^{|n|} \to 0$ as $n \to \pm \infty$, so that the coefficients $c_n$ are exponentially decreasing.
(*) means that the coefficients of the Fourier series of $h(t)  =g(re^{it})$ are $r^n c_n$, which you can show from the Cauchy integral theorem
