Let's consider the function$F(x,y)$ defined by:
$$F(x,y)= \sum_{n=1}^{\infty} f(nx) e^{-2 i \pi ny}$$
with $f(x)$ decreasing exponentially at infinity, $f(x)=0$ and $\int_{-\infty}^{\infty} f(|t|)dt=0$
It seems complex to understand the behavior of $F(x,y)$ (especially for $x$ near zero with $y$ fixed !).
Applying the Poisson sommation formula to $f(|t|x) e^{-2 i\pi |t|y}$we have:
$$F(x,y)= \sum_{n=-\infty}^{\infty} f(|n|x) e^{-2 i \pi |n|y}= \frac{1}{x} \sum_{n=-\infty}^{\infty} \hat{f}(\frac{n+y}{x})$$
Noting: $\hat{f}(t)=\int_{-\infty}^{\infty} f(|t|) e^{-2 i\pi |t|y} dt$
So it seems that, if for example $\hat{f}(t) \sim \frac{1}{t^2}$ at infinity then we have for all y fixed, $ F(x,y) \to 0$ for $x \to 0$, so that we should have $\int_{0}^{1} F(x,y) \overline{F(x,y)} dy \to 0$ for $x \to 0$ (all functions are continuous on $y$ variable on $[0,1]$)
But when I calculate directly $\int_{0}^{1} F(x,y) \overline{F(x,y)} dy$, I find:
$$\int_{0}^{1} \sum_{n=1}^{\infty} f(nx) e^{-2 i \pi ny} .\overline{\sum_{k=1}^{\infty} f(kx) e^{-2 i \pi ky}} dy = \sum_{n=1}^{\infty} f(nx) \overline{f(nx)} $$
which obviously is in $\frac{1}{x}$ (by Poisson Summation formula). So where is the incoherence ?
Any reference on this type of sums?