-1

Suppose f is of moderate decrease and that its fourier transform $\hat{f}$ is supported in $I=[-1/2,1/2]$. If $\chi$ is the characteristic function of $I$,then show $\hat{f}(\xi)=\chi(\xi)\sum_{-\infty}^\infty f(n)e^{-2\pi in \xi}$.

I just do not see how we can have a sum instead of an integral. Any ideas on how to solve this?

Recall that $f(x)$ is of moderate decrease if there exists a constant $A>0$ such that $|f(x)| \leq \frac{A}{1+x^2}$

Epsilon
  • 69

2 Answers2

2

If $\hat{f}(\xi)$ has compact support it means you can write it as

$$ \hat{f}(\xi) = \chi(\xi) \hat{g}(\xi) $$

where $\hat{g}(\xi)$ is a periodic function, with period $f_0 = 1$. Since $\hat{g}(\xi)$ is periodic in $\left[-1/2,1/2 \right]$ you can write the Fourier series, whose coefficients are

$$ f(n) = \int_{-1/2}^{1/2} \hat{g}(\xi)e^{-i2\pi n \xi}d\xi = \int_{-1/2}^{1/2} \hat{f}(\xi)e^{-i2\pi n \xi}d\xi $$

therefore

$$ \hat{f}(\xi) = \chi(\xi) \sum_{j=-\infty}^{\infty}f(n)e^{i2\pi n \xi} $$

user8469759
  • 5,285
  • The "moderate decrease" tells you the Fourier transform exists, because at least $f$ is in $L^1$ and therefore it's Fourier transform exists. – user8469759 Mar 19 '18 at 15:13
  • What did you use in your last step? I thought the fourier inversion formula used only integrals, not sums.. – Epsilon Mar 19 '18 at 15:33
  • Do you agree with me that $\hat{f}(\xi) = \chi(\xi) \hat{g}(\xi)$ where $\hat{g}$ is some periodic function in $[-1/2,1/2]$? you can easily construct such a function. – user8469759 Mar 19 '18 at 15:39
  • Assuming "yes" as answer, do you agree with me that since $\hat{g}$ is periodic then you can write $\hat{g}$ as $$\hat{g}(\xi) = \sum_{j\in \mathbb{Z}} c_n e^{i 2\pi n \xi}$$? – user8469759 Mar 19 '18 at 15:43
  • Therefore you can substitute that expression in the one I gave for $\hat{f}$ in terms of $\hat{g}$, there's no inversion involved. The only subtlety to be observed is that since $|f(x)| < A/(1+x^2)$, then $f(x) \in L^2$, therefore by Parseval $\hat{f}$ is in $L^2$, and therefore $\hat{g}$ is in $L^2(I)$ therefore you can write the Fourier series. – user8469759 Mar 19 '18 at 15:46
  • Thanks a lot. This will be the accepted answer. – Epsilon Mar 19 '18 at 15:46
0

Use the [Whittaker-Shannon interpolation formula] (https://en.wikipedia.org/wiki/Whittaker%E2%80%93Shannon_interpolation_formula)

under the form you can find in the upsaid article :

$$f(t)=\underbrace{\sum_{k=-\infty}^{\infty}f(n)\delta(t-nT)}_{f(t) . III_T(t)}\ \star \text{sinc}(\frac{t}{T})$$

(III means Dirac comb) with $T=1$ and then take the Fourier Transform of both sides using the facts that the Fourier Transform of

  • a convolution product is an ordinary product.

  • Dirac $\delta(t-n)$ is (by translation rule) $FT(\delta(t-n))=\underbrace{FT(\delta(t))}_{=1} e^{-2i\pi n\xi}.$

  • $t \mapsto $ sinc $(t)$ function is $\chi \mapsto \chi(\xi)$ function.

Jean Marie
  • 81,803
  • Doesn't this approach require to know a little bit about distributions? – user8469759 Mar 19 '18 at 15:27
  • Yes, of course, but only the most elementary one, the Dirac distribution which is usually met rather soon in lectures on Fourier Transform, at least for students that are in engineering studies. Your answer [+1] has the advantage to use only Fourier series. – Jean Marie Mar 19 '18 at 15:33
  • @user8469759 Yes, but only because of the needlessly abstruse notation! Juust write $\text{sinc}((t-nT)/T)$ in place of that convolution and the need for distributions goes away. – David C. Ullrich Mar 19 '18 at 15:57
  • @David C. Ullrich You are perfectly right. But sooner or later using Dirac's distribution and the Dirac comb is conceptualy a simplification. – Jean Marie Mar 19 '18 at 17:25