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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 $\hat{f}^{(n)}(0)$.2012-08-28
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    @DavideGiraudo: Thanks for the suggestion.2013-01-15
  • 0
    Homework problem from ANU MATH3325, 2012.2013-09-20

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