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This is from my homework: given $f\in L^1(R)$ and $f$ is compactly supported, show that

  1. Fourier transform of $f$ , i.e $\hat{f(\zeta)}$ is $C^\infty$

  2. $\hat{f}(\zeta)$ is analytic and entire by showing that the taylor series of $\hat{f}$ at $\zeta = 0$ has infinite radius of convergence, and converges to $\hat{f}$

My attempt is:

For Part 1 since the domain is the real line, the support must be a closed interval and we can divide this interval into two parts: A where $|f|>0$ and B the rest. Then the integral over $|f|$ over A is less than $|f|^2$, while the integral over B is less that m(B), which is finite.

The I tried to show all derivatives are continuous by induction, but no success . For base case , I tried to show $\hat{f}(\zeta)$ is continuous. That is, $\lim_{ h\to \infty}|\hat{f}(\zeta+h) - \hat{f}(\zeta)|=0$. $|\hat{f}(\zeta+h) - \hat{f}(\zeta)|\leq \int|(e^{-i(\zeta+h)x}-e^{-i(\zeta)x})f(x)|dx \leq \int|e^{-ihx}-1||f(x)|dx$. I am stuck at this step and I guess power series would help to bound the RHS?

For induction step, assume $\hat{f}^n(\zeta)$ is continuous , let $|\hat{f}^{n+1}(\zeta+h) - \hat{f}^{n+1}(\zeta)|=D|\hat{f}^n(\zeta+h) - \hat{f}^n(\zeta)|$. I got trouble with putting D under integral sign and can't proceed..

For Part 2, I tried to bound $f$ with some $O(e^{ai})$ when $\zeta\to\infty$. Anyway I am totally lost from here. Could anyone show me how to do it and many thanks in advance.

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1 Answer 1

Let me treat in $\mathbb{R}^{n}$ instead.According to the fourier-laplace transform $$\hat{f}(\xi)=\int{e^{-ix\cdot\xi}f(x)dx},\quad \xi\in\mathbb{C}^{n}$$

Since now $f\in L^{1}$ with compact support,apply the Cauchy-Riemann operator to $\hat{f}(\xi)$ and exchange with integral(this is allowed since f is compact supported) to find that $\hat{f}(\xi)$ is a entire analytic function.

BTW,the result still hoids if f is only a distribution with compact support(and the method is essentially the same),which has many applications in linear PDE.

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