I want to show that, for a function $f$ of moderate decrease and $\hat{f}(\xi)$ supported in $I=[-\frac{1}{2},\frac{1}{2}]$,
$$f(x) = \sum^{\infty}_{n=-\infty} f(n)K(x-n)$$
where $K(y) = \frac{sin( \pi y)}{\pi y}$
The book provides the hint to show that $\hat{f(\xi)} = \chi_I(\xi) \sum^{\infty}_{n=-\infty} f(n)e^{-2 \pi i n \xi}$ where $\chi$ is the characteristic function
I know that $${K}(\xi) = \hat{\chi}_I(\xi) $$ and I can see Poisson's summation formula inside there somewhere, but I can't proceed because I can't come up with justification.
If I multiply the inside of the first sum by $1 = e^{-2 \pi i n \xi}e^{-2 \pi i(x-n) \xi}$ I have $$\sum^{\infty}_{n=-\infty} f(n)K(x-n)e^{-2 \pi i n \xi}e^{-2 \pi i(x-n) \xi}$$ and I thought about applying Poisson to get this to be equal to: $$\sum^{\infty}_{n=-\infty} \hat{f} \ast \chi_I (\xi)$$but this doesn't seem right. However, I have the suspicion that convolutions are involved somewhere otherwise I don't know how to justify taking the FT/inverse FT of a product of functions. I've been stuck for a while, so any help is appreciated.