# On construction of a $\mathbb{Q}$ periodic function with Fourier series

Taking $f$ a function decreasing exponentially at infinity we can consider the periodic function given by following Fourier series:

$$F(x)= \sum\limits_{n =1}^{\infty} f(n) e^{2 i \pi n x}$$

Using this function, let's consider following function:

$$G(x)=\sum\limits_{q =1}^{\infty} \sum\limits_{a =0}^{q-1} F(x+\frac{a}{q}) = \sum\limits_{q =1}^{\infty} \sum\limits_{n =1}^{\infty} \sum\limits_{a =0}^{q-1} f(n) e^{2 i \pi n (x+\frac{a}{q})}$$

(see below the simple calculation showing that this sum is well defined)

The question is: Is this function "$\mathbb{Q}$ periodic" ? If not why ?

We see (with some effort due to the fact that $a$ and $q$ are not always prime together...) that adding a fraction $\frac{c}{d}$ to the $\frac{a}{q}$ (modulo 1) will not change the elements of the infinite sum.

The sum is done on the elements of the set $\{(a,q); q\in \mathbb{N^*}; 0\le a<q\}=\{(kp,kq); \frac{p}{q} \in ]0,1[ \cap \mathbb{Q}, p\land q=1\} \cup \{(0,q); q\in \mathbb{N^*}\}$

Function $G(x)$ is well defined:

The sum on $a$ is independant and we have immediatly, posing $n=mq$:

$$G(x)=\sum\limits_{q =1}^{\infty} \sum\limits_{m=1}^{\infty} q \; f(qm) e^{2 i \pi mqx}$$

(as $\sum\limits_{a =0}^{q-1} e^{2 i \pi n \frac{a}{q}}=0$ if q does not divide $n$ and $=q$ if $q$ divides $n$)

Under this form it is clear that the sum converges and that $G(x)$ is well defined. It seems also that this forms shows that $G(x)$ is not $\mathbb{Q}$ periodic (if it was then $G(x)$ would be the constant function and all its Fourier coefficients would be zero - so where is the problem ?)

Any reference on $\mathbb{Q}$ periodic function is welcome.

This function $G(x)$ is not periodic, even if the terms of the sum are the same. The fact to add a rational number $\frac{c}{d}$ changes the order of summation of the terms and finally the sum defining $G(x+\frac{c}{d})$ exists but is different from $G(x)$.
Suppose $G(x)$ is $\mathbb{Q}$ periodic then as $G(x)$ is also equal to:
$$G(x)=\sum\limits_{q =1}^{\infty} \sum\limits_{m=1}^{\infty} q \; f(qm) e^{2 i \pi mqx} = \sum\limits_{n =1}^{\infty} A_n e^{2 i \pi nx}$$
(here we can change the order of summation as the sum is absolutely convergent!), with the $A_n$ such that $\sum\limits_{n =1}^{N} A_n$ converges (as $f$ has been chosen fast decreasing ). It means $\sum\limits_{n =1}^{\infty} A_n e^{2 i \pi nx}$ converges uniformly so that $G(x)$ is continuous. Therefore the function $G(x)$ should be constant and its Fourier coefficient should be zero except the constant term and this is not the case, so $G(x)$ cannot be periodic.