# To show Taylor series of a Fourier transform $\hat{f }$ converges to $\hat{f}$

I got some trouble with the following question.

Say $f$ is in $L^1(R)$ with compact support . I need to show (1) $\hat{f(\zeta)}$ is infinitely differentiable and all derivatives are continuous. (2) show Taylor series of $\hat{f(\zeta)}$ at $\zeta$ = 0 has infinitely radius of convergence , and converges to $\hat{f}$ (i.e. $\hat{f}$ is analytic and entire).

I did part (1) and showed that radius of convergence of the Taylor series is infinity. But stuck at the part to show the TS does converge to $\hat{f}$.

by calculation TS = $\sum_{n=0}^{\infty}\hat{f^n(0)}\zeta^n/n! = (-i)^n\int_{-M}^Mx^nf(x)dx\zeta^n/n!$.

So $a_n= (-i)^n\int_{-M}^Mx^nf(x)/n!$. I used root test to show R is infinity.

I then managed to show the error term $R(N):=\hat{f}-TS(N)=\hat{f}-\sum_{n=0}^{N}\hat{f^n(0)}\zeta^n/n! \to 0$ as $N\to\infty$ but kind of stuck at here. Any thoughts ? Thanks in advance.

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Expand $e^{-i \zeta t} = \sum_{k=0}^\infty \frac{(-i \zeta t)^n}{n!}$ and note that the convergence is uniform on compact sets ($\mathbb{supp} f$). Then express $\int \frac{(-i \zeta t)^n}{n!} f(t) dt$ in terms of $\hat{f}^{(n)}(0)$. –  copper.hat Aug 28 '12 at 16:10
@DavideGiraudo: Thanks for the suggestion. –  copper.hat Jan 15 '13 at 7:32
Homework problem from ANU MATH3325, 2012. –  Scott Morrison Sep 20 '13 at 6:20

I'm not exactly sure where you got stuck, hopefully the following clarifies. If you can show that $\hat{f}$ is equal pointwise to a power series with infinite radius of convergence, then you are finished.
Let $B$ be such that $\text{supp} f \subset [-B,B]$. Note that for any $z \in \mathbb{C}$, $\int |(z t)^n f(t)| dt \leq (|z| B)^n \|f\|_1$, and hence $\sum_{n=0}^\infty \int |\frac{(z t)^n}{n!} f(t)| dt \leq e^{|z|B} \|f\|_1$.
It follows from the dominated convergence theorem that \begin{eqnarray} \hat{f}(\omega) &=& \int e^{-i\omega t} f(t) dt \\ &=& \int \sum_{n=0}^\infty \frac{(-i\omega t)^n}{n!} f(t) dt \\ &=& \sum_{n=0}^\infty \int \frac{(-i\omega t)^n}{n!} f(t) dt \\ &=& \sum_{n=0}^\infty \frac{(-i\omega )^n}{n!} \int t^n f(t) dt \\ &=& \sum_{n=0}^\infty \omega^n (-i)^n \int \frac{t^n}{n!} f(t) dt \\ &=& \sum_{n=0}^\infty a_n \omega^n \end{eqnarray} where $a_n = (-i)^n \int \frac{t^n}{n!} f(t) dt$. The power series has an infinite radius of convergence, hence $\hat{f}$ is entire, and it follows that $a_n = \frac{1}{n!} \hat{f}^{(n)}(0)$.