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.

Solution 1:

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 $$ \sum_{n=0}^\infty|a_nz^n|\le A_\epsilon e^{2\pi(M+\epsilon)|z|}\tag{8} $$

Solution 2:

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.

Solution 3:

By estimating directly we have $$ |n! a_{n}|= \left|(2\pi i)^{n}\int_{-M}^{M}\hat{f}(\xi)\xi^{n} \,d \xi \right|\le (2\pi)^{n} M^n\int_{-M}^{M}|\hat{f}(\xi)|\,d\xi=C (2\pi)^{n} M^n. $$