Proving Parseval's Theorem for Schwartz Functions with Compact Support

[Sonifa]

How to prove the following:

Suppose f is in the Schwartz Space ( smooth function with very fast decay). Its Fourier transform is smooth and has compact support contained in the interval (1/2,-1/2)

Show,

∫ (|f(x)|^2) dx = ∑ (|f(n)|^2) (where integral over R and sum up over n for all intergers)

 

[Opalg]

How to prove the following:

Suppose f is in the Schwartz Space ( smooth function with very fast decay). Its Fourier transform is smooth and has compact support contained in the interval (1/2,-1/2)

Show,

∫ (|f(x)|^2) dx = ∑ (|f(n)|^2) (where integral over R and sum up over n for all intergers)

I would start by using the Parseval theorem \(\displaystyle \int_{\mathbb{R}}|f(x)|^2dx = \int_{\mathbb{R}}|\hat{f}(\omega)|^2d\omega.\)

The sampling theorem (taking the sampling interval $\delta$ to be $\delta = 1$) says that \(\displaystyle f(x) = \sum_{n=-\infty}^\infty f(n)\phi(x-n),\) where $\phi$ is a smooth function whose Fourier transform $\hat{\phi}$ is identically $1$ on $\bigl(-\frac12,\frac12\bigr)$ and has support in $(-\pi,\pi)$. Then \(\displaystyle \hat{f}(\omega) = \sum_{n=-\infty}^\infty f(n)\hat{\phi}(\omega -n)\). Can you use that to show that \(\displaystyle \int_{\mathbb{R}}|\hat{f}(\omega)|^2d\omega = \sum_{n=-\infty}^\infty |f(n)|^2\)?

Edit. Correction: the Fourier transform of $\phi(x-n)$ is not $\hat{\phi}(\omega -n)$, but something like $e^{in\omega}\hat{\phi}(\omega) .$