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 $f=0$?
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2$\begingroup$ Is it obvious that $F_a = 0$ for all $a > 0$ implies $F = 0$? $\endgroup$– LSpiceCommented Feb 19, 2023 at 16:40
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$\begingroup$ No for now that's also not obvious to me. Actually I'm also wondering what happens if the $a$'s belong to an interval, $a \in (x,y)$ where $x<y$. Would this be enough to conclude that $f=0$? $\endgroup$– user975628Commented Feb 19, 2023 at 16:44
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$\begingroup$ For $n = 2$ the ratio $a_1 / a_2$ should be irrational, right ? $\endgroup$– jjcaleCommented Feb 19, 2023 at 18:16
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3 Answers
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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 $f=0$.
answered Feb 19, 2023 at 18:43
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$\begingroup$ It's not clear to me if the last conclusion follows or not. It might be the case that the inequality $|f(t)|<e^{-t^2}$ prevents its Fourier transform from vanishing on a large set of lattices (which will imply, for instance, lots of very closely spaced zeros). $\endgroup$ Commented Feb 19, 2023 at 19:02
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1$\begingroup$ @MarkLewko The OP did not specify what they meant by "rapid decay". Certainly in the Schwartz space $\mathcal{S}(\mathbb{R})$ there is a nonzero function whose Fourier transform vanishes at a prescribed discrete subset of $\mathbb{R}$. $\endgroup$ Commented Feb 19, 2023 at 19:22
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2$\begingroup$ For the decay in "e.g." we may choose $\widehat{f}(t) = e^{-b t^2} \prod_{i=1}^n sin(\pi t a_i)$ with sufficient small $b > 0$ . $\endgroup$– jjcaleCommented Feb 19, 2023 at 19:54
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1$\begingroup$ @GiorgioMetafune Take any closed interval disjoint from the sequence, and then take a (not identically zero) smooth function supported in that interval. This function will be in the Schwartz class and vanish at the given sequence. $\endgroup$ Commented Feb 19, 2023 at 22:18
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1$\begingroup$ @GHfromMO True, thank you! $\endgroup$ Commented Feb 19, 2023 at 22:20
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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.
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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.
