For some $\tau\in(0,1)$, let $f : (-\infty,0]\times [0,\tau] \rightarrow \mathbb{C}$ and $g:[0,\infty)\times[0,1]\rightarrow\mathbb{C}$ with $$f(0,t)=g(0,t)\text{ for }t\in [0,\tau]$$ be two smooth funtions which are periodic in the second variable, i.e. $f$ is $\tau$-periodic and $g$ is $1$-periodic. Further, assume that both $f$ and $g$ can be expressed as Fourier series, then above condition means that
$$\sum\limits_{k\in\mathbb{Z}} f_k e^{\frac{2\pi i k}{\tau}t} = \sum\limits_{r\in\mathbb{Z}} g_{r} e^{2\pi i rt} \text{ for }t\in [0,\tau]$$
Whenever $\frac{1}{\tau}\in\mathbb{Z}$, every function $t\mapsto e^{\frac{2 \pi i k}{\tau}t}$ appears on both the left and right hand side, and so $g_r = f_{k}$ whenever $r= \frac{k}{\tau}$ and $g_r=0$ else.
A naive thought would be that for $\tau\in\mathbb{Q}$, this generalizes to $$f_k \neq 0\Rightarrow \exists r\in\mathbb{Z} : \frac{k}{\tau}=r \text{ and } f_k = g_r$$ and $$g_k\neq 0 \Rightarrow \exists k\in\mathbb{Z} : r\tau = k \text{ and } f_k = g_r.$$
However, I am not sure if this can be proved, since now the basis funtions on the left and right do not have to share a period anymore. For example, if $\frac{l}{\tau}\not\in\mathbb{Z}$, then $r\tau\not\in\mathbb{Z}$ for any $r\in\mathbb{Z}$ \begin{align} f_l &= \frac{1}{\tau}\int\limits_0^\tau \left(\sum\limits_{k\in\mathbb{Z}} f_k e^{\frac{2\pi i k}{\tau} t} \right) e^{-\frac{2\pi i l}{\tau}t}dt\\\\ &= \frac{1}{\tau}\int\limits_0^\tau \left(\sum\limits_{r\in\mathbb{Z}} g_{r} e^{2\pi i rt}\right)e^{-\frac{2\pi i l}{\tau}t}dt\\\\ &=\frac{1}{\tau}\sum\limits_{r\in\mathbb{Z}}g_r\int_0^\tau e^{2\pi i(r-\frac{l}{\tau})t} dt\\\\ &= \frac{1}{\tau}\sum\limits_{r\in\mathbb{Z}} g_r \frac{e^{2\pi i \tau r}-1}{2\pi i(r-\frac{l}{\tau})} \end{align} which does not seem to neccessarily vanish. Is there some other way to get towards this "naive intuition"? I also have a feeling like I'm overlooking something obvious...
