# 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.

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

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I see that my proof of the converse is just a fleshing out of Jonas's, but I will leave it for the sake of completeness. – robjohn Oct 24 '11 at 17:23
oops. I forgot to mention entirety. – robjohn Oct 24 '11 at 17:33

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.

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Good proof (+1). I just fleshed it out a bit. :-) – robjohn Oct 24 '11 at 17:45
@robjohn Thanks! But I'm not so sure if the extra detail is useful, if you're up to the point that you're reading Stein and Shakarchi, you should be able to fill in the gaps, I think? – Jonas Teuwen Oct 24 '11 at 18:36

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

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