On functions with Fourier transform having compact support I have another question from Stein & Shakarchi, Complex Analysis.
The problem is the following: Suppose $\hat{f}$ has compact support contained in $\left[-M,M\right]$ and let $f(z) = \sum_{n=0}^{\infty}{a_{n}z^{n}}$. Show that 
$$a_{n}= \frac{(2\pi i)^{n}}{n!} \int_{-M}^{M}{\hat{f}(\xi)\xi^{n} d \xi },\ \text{and so that}\ \ \lim_{n \rightarrow \infty}{\sup{(n! |a_{n}|)^{1/n}}} \leq 2 \pi M.$$ Conversely, if $f(z) = \sum_{n=0}^{\infty}{a_{n}z^{n}}$ is any power series with $\lim_{n \rightarrow \infty}{\sup{(n! |a_{n}|)^{1/n}} }\leq 2 \pi M$, then show that $f$ is entire and for every $\epsilon > 0$ there exists $A_{\epsilon} > 0$ such that $|f(z)| \leq A_{\epsilon}e^{2\pi (M+ \epsilon) |z|}.$
I managed to show that $a_{n}= \frac{(2\pi i)^{n}}{n!} \int_{-M}^{M}{\hat{f}(\xi)\xi^{n} d \xi }$ by using the inversion formula and by changing the order of summation, but I am kind of stuck at the next step and don't see how to deduce the inequality after simplyfing.
 A: As you say, the inversion formula
$$
\begin{align}
f(z)
&=\sum_{n=0}^\infty a_nz^n\\
&=\sum_{n=0}^\infty f^{(n)}(0)\frac{z^n}{n!}\\
&=\sum_{n=0}^\infty\left(\int_{-M}^M\hat{f}(\xi)(2\pi i\xi)^n\;\mathrm{d}\xi\right)\frac{z^n}{n!}\tag{1}
\end{align}
$$
yields that
$$
a_n=\frac{(2\pi i)^n}{n!}\int_{-M}^M\hat{f}(\xi)\;\xi^n\;\mathrm{d}\xi\tag{2}
$$
A simple estimate, using $|\xi|\le M$, gives
$$
|a_n|\le\frac{(2\pi)^n}{n!}\|\hat{f}\|_{L^1}M^n\tag{3}
$$
Therefore,
$$
|n!\:a_n|^{1/n}\le2\pi M\left(\|\hat{f}\|_{L^1}\right)^{1/n}\tag{4}
$$
Taking the $\limsup$ of $(4)$ yields the inequality.
Converse:
Supppose that $\limsup\limits_{n\to\infty}|n!\:a_n|^{1/n}\le2\pi M$. Then, because $n!>\sqrt{2\pi n}(n/e)^n$,
$$
\begin{align}
\limsup\limits_{n\to\infty}\;|a_n|^{1/n}&
\le\limsup\limits_{n\to\infty}\;2\pi M\frac{e}{n}(2\pi n)^{-\frac{1}{2n}}\\
&=0
\end{align}
$$
Thus, the radius of convergence of the series is $\infty$ and $f$ is entire. Furthermore, for any $\epsilon>0$, there is an $N$ so that if $n\ge N$,
$$
|a_n|\le\frac{(2\pi(M+\epsilon))^n}{n!}\tag{5}
$$
Thus,
$$
\begin{align}
\sum_{n=N}^\infty|a_nz^n|
&\le\sum_{n=N}^\infty\frac{(2\pi(M+\epsilon)|z|)^n}{n!}\\
&\le e^{2\pi(M+\epsilon)|z|}\tag{6}
\end{align}
$$
Since $e^{2\pi(M+\epsilon)|z|}$ grows faster than any polynomial in $|z|$, there is a constant, $A_\epsilon-1$ so that
$$
\sum_{n=0}^{N-1}|a_nz^n|\le(A_\epsilon-1)e^{2\pi(M+\epsilon)|z|}\tag{7}
$$
Adding $(6)$ and $(7)$, we get that
$$
\sum_{n=0}^\infty|a_nz^n|\le A_\epsilon e^{2\pi(M+\epsilon)|z|}\tag{8}
$$
A: Conversely:
We will first show that $f$ is entire.
We have for every $\varepsilon > 0$ that there exists an $N \in \mathbf N$ such that $|a_n| \leqslant \frac{[2 \pi (M + \epsilon)]^n}{n!}$ for all $n \geqslant N$.
From this we can conclude that $\limsup_{n \to \infty} |a_n|^{\frac1n} = 0$ and therefore $f$ is entire (root test).
So now we have
$$|f(z)| \leqslant \sum_{n = 0}^N |a_n| |z|^n + \sum_{n = N + 1}^\infty |a_n| |z|^n.$$
The last sum we can bound by $e^{2 \pi (M + \varepsilon) |z|}$ and the first one we can compare the coefficients of that polynomial with the coefficients of the function $C_{\varepsilon} e^{2 \pi (M + \varepsilon) |z|}$ to obtain the full bound.
Using Paley-Wiener we can now conclude something about the support.
A: By estimating directly we have
$$
|n! a_{n}|= \left|(2\pi i)^{n}\int_{-M}^{M}\hat{f}(\xi)\xi^{n} \,d \xi \right|\le
(2\pi)^{n} M^n\int_{-M}^{M}|\hat{f}(\xi)|\,d\xi=C (2\pi)^{n} M^n.
$$
