Orthonormal system Let $\varphi\in L^2(\mathbb{R})$, prove that $\{e^{2\pi i m x}\varphi(x)\}$ is an orthonormal system iff $$\sum_{n\in\mathbb{Z}}|\varphi(x-n)|^2=1 \ \ a.e  \ x$$
How do you prove this. The hint is the LHS is an $L^1([0,1))$ function. How would you approche this? I know you have to Parseval somewhere. 
Thanks
 A: This is essentially the Poisson summation formula. The scalar products are the Fourier coefficients of $F=|\varphi|^2$, so we have an ONS if and only if $\widehat{F}(n)=0$ for $n\in\mathbb Z$, $n\not=0$, and $\widehat{F}(0)=1$. If we now use the version quoted near the end of the wikipedia article, $\sum F(x+n)=\sum\widehat{F}(k) e^{2\pi i kx}$, then the claim is an immediate consequence.
However, that was a bit formal since the Poisson summation formula needs interpretation in situations when the series are not obviously convergent.
For a rigorous proof, we simply forget about all this and start from scratch. As you already pointed out, $G(x):=\sum |\varphi(x-n)|^2\in L^1(0,1)$, and, by Fubini,
$$
\widehat{G}_k = \int_0^1 G(x)e^{-2\pi ikx}\, dx = \sum \int_0^1 \ldots = \int_{-\infty}^{\infty} F(x)^{-2\pi ikx}\, dx =\widehat{F}(k) .
$$
So we have an ONS if and only if $\widehat{G}_k=\delta_{k0}$, or, equivalently $G\equiv 1$ (on $x\in (0,1)$ originally, but then everywhere by periodicity), by the uniqueness of the Fourier coefficients of $L^1$ functions.
