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.
 A: 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 $a_n = \frac{1}{n!} \hat{f}^{(n)}(0)$.
