Vanishing periodizations $\sum_{k \in \mathbb Z} f(t+ak)$ of a function $f$ for different values of $a$ implies $f=0$? Consider a continuous function $f : \mathbb R \to \mathbb C$ with rapid decay (e.g. $|f(t)| < e^{-t^2}$). For a constant $a>0$ let
$$
F_a(t) = \sum_{k \in \mathbb Z} f(t+ak)
$$
be the periodization of $f$ with period $a$. Assume that $F_a = 0$, that is, $F_a$ vanishes identically. In general, this does not imply that $f$ vanishes identically. How about $F_a = 0$ for different values of $a$? Formally, assume that
$$
F_{a_1} = F_{a_2} = \dotsb =  F_{a_n} = 0\tag{$*$}\label{star}
$$
for different $a_j >0$. Can we find conditions on the $a_j$'s so that \eqref{star} implies $f=0$?
 A: The Fourier coefficients of the periodic function $F_a$ are (up to a factor)
\begin{align*}
\int_0^a F_a(t) e^{2\pi in t/a}\, dt & =\sum_k \int_0^a f(t+ak)e^{2\pi int/a}\, dt\\
& = \sum_k \int_{ka}^{(k+1)a} f(s) e^{2\pi ins/a}\, ds \\
& = \int_{-\infty}^{\infty} f(s)e^{2\pi i ns/a}\, ds .
\end{align*}
So the condition $F_a=0$ is equivalent to $\widehat{f}$ vanishing on $(1/a)\mathbb Z$, and no finite set of such conditions can imply that $f=0$.
A: From Poisson summation, you have that
$$F_a(t) = \frac{1}{a} \sum_{n \in Z} \widehat{f}(n/a)e(tn/a).$$
Further we know that $F_a(t)$ will identically vanish if and only if the Fourier coefficients identically vanish. This leaves you with a question about zeros of the Fourier transform of a rapidly decaying function.
It follows from this that such a set of $a_i$'s, say $A$ will suffice as long a you can guarantee that for some $\xi$ such that $\widehat{f}(\xi) \neq 0$ there is an $a \in A$ such that $\xi a \in Z$. Clearly, for instance, answering the question from the comments, $A=R^{+}$ will suffice.
A: The answer is no.
Proof by induction :
Let $f_n \neq 0$ with compact support such that the periodizations of $a_1,...,a_n$ are zero.
Then there exists an integer $m$ such that the function $f_{n+1}(x) = f_n(x) - f_n(x + m a_{n+1})$ is non zero.
But for $f_{n+1}$ the periodizations of $a_1,...,a_{n+1}$ are zero and $f_{n+1}$ has also compact support.
