Let $f$ be in $L^{2}(\mathbb{R})$, and let $\hat f$ be its Fourier transform, defined by
$$ \hat f(\omega) = \int_{\mathbb{R}} f(x)e^{-i \omega x} dx$$
Prove that the system $\{f(t-n)\}_{n \in \mathbb{Z}}$ is orthonormal in $L^{2}(\mathbb{R})$ if and only if
$$\sum_{k \in \mathbb{Z}} |\hat f (\omega + 2 \pi k)|^2 \equiv 1$$
for almost all $\omega$.
Thank you.
Define $g_n(t) = f(t-n)$. From the definition of Fourier transform,
$\hat{g}_n(\omega) = \hat{f}(\omega) e^{in \omega}$.
Plancherel theorem states:
$\langle g_n, g_m \rangle = \langle \hat{g}_n, \hat{g}_m \rangle $,
but it also holds that
$\langle \hat{g}_n, \hat{g}_m \rangle = \langle e^{i \omega (n-m)}, |\hat{f}|^2 \rangle$ (this follows by writing down the integrals).
Define
$$h(\omega) = \sum_{k=-\infty}^\infty |\hat{f}(\omega+2\pi k)|^2 $$
$h(\omega) \in L^1([0,2\pi])$, because
$\int_0^{2\pi}|h(\omega)|d\omega=\sum_{k=-\infty}^\infty \int_0^{2\pi}|\hat{f}(\omega+2\pi k)|^2=\sum_{k=-\infty}^\infty \int_{-2\pi k}^{-2\pi (k+1)}|\hat{f}(\omega)|^2=||\hat f||_{L^2}=2\pi ||f||_{L^2}<\infty$
Now, $h$ has a Fourier series in $[0, 2\pi]$, the coefficients being
$c_{m} = \frac{1}{2\pi}\int_0^{2\pi} h(\omega) e^{-i \omega m} d\omega = \frac{1}{2\pi}\sum_{k=-\infty}^\infty \int_0^{2\pi}|\hat{f}(\omega+2\pi k)|^2 e^{-i\omega m}d\omega=\frac{1}{2\pi}\sum_{k=-\infty}^\infty \int_{-2\pi k}^{-2\pi (k+1)}|\hat{f}(\omega)|^2e^{-i\omega m}e^{i2\pi km}d\omega=\frac{1}{2\pi}\sum_{k=-\infty}^\infty \int_{-2\pi k}^{-2\pi (k+1)}|\hat{f}(\omega)|^2e^{-i\omega m}d\omega=\frac{1}{2\pi} \int_{\mathbb{R}}|\hat{f}(\omega)|^2e^{-i\omega m}d\omega=\langle e^{-i \omega m}, |\hat{f}|^2 \rangle$
If $\hat{g}_n$ are orthonormal, then by the above, $c_m=1$ if $m=0$, $0$ otherwise. This gives $h(\omega) \equiv 1$, by the completeness of the $\{e^{inx}\}_{n \in \mathbb{Z}}$ system. In the other direction, if $h(\omega) \equiv 1$, its Fourier coefficients are $c_m=1$ if $m=0$, $0$ otherwise, which gives the orthogonality of the $\hat{g}_n$.
Thanks!
