Find the fourier coefficient of given function

Question

I'ven been reading Dimitris Koukoulopouls' analytic number theory book.

I'm stuck at p.63, the following part.

Let $f : \mathbb{R} \to \mathbb{C}$ be twice-continuously differentiable function of $\mathbb{R}$ and $f^{(j)} << \frac{1}{x^2}$ for $|x| \ge 1$ and $j \in \{0,1,2\}$, so that

$$\hat{f}(\xi)= \int_{\mathbb{R}} f(x) e^{-2\pi i \xi x} dx$$

and he defines $$S_f(x) := \sum_{n \ge 1} f(nx)$$

Next, he uses Poisson's summation formula to deduce that $$\sum_{n\in \mathbb{Z}} f(nx) = \frac{1}{x} \sum_{n \in \mathbb{Z}} \hat{f} (n/x)$$

I don't see why this is true.

Here is what I've tried to justify the above equation but failed:

Set $$g(t) = \sum_{n \in \mathbb{Z}} f(t+nw)$$

Then this $g$ is a periodic function with period $w$.

So its $m$-th Fourier coefficient is: $$c_m = \frac{1}{w} \int_{0}^{w} g(t) e^{-2\pi i m t} dt$$

Then
$\begin{align*} c_m &= \frac{1}{w} \int_{0}^{w} \sum_{n \in \mathbb{Z}} f(t+nw) e^{-2\pi i m t} dt \\ & = \sum_{n \in \mathbb{Z}} \frac{1}{w} \int_{0}^{w} f(t+nw) e^{-2 \pi i mt} dt \end{align*}$

And Substituting $t+nw= y$ in the equation gives $$c_m = \sum_{n \in \mathbb{Z}} \frac{1}{w}\int_{wn}^{w(n+1)} f(y) e^{-2 \pi i m (y-nw)} dy $$

I can't see how to proceed, could you give me any advice?

Answer 1

Try defining $g$ in terms of $f$ (no sum, just $f$).

Fix $x$ and set $g(t)=f(tx)$. Then \[ \hat g(\xi )=\int _{\mathbb R}g(t)e(-\xi t)dt=\int _{\mathbb R}f(tx)e(-\xi t)dt=\frac {1}{x}\int _{\mathbb R}f(t)e(-\xi t)dt=\frac {\hat f(\xi )}{x}\] so Poisson on $g$ says \[ \sum _{n}f(nx)=\sum _{n}g(n)=\sum _n\hat g(n)=\frac {1}{x}\sum _n\hat f(n)\]