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 $|f(x)| \leq \frac{A}{1+x^2}$
On
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.
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
$$ \hat{f}(\xi) = \chi(\xi) \sum_{j=-\infty}^{\infty}f(n)e^{i2\pi n \xi} $$