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If $f \in L^2[0,1]$ then it follows from the uniqueness-theorem for Fourier series that if $$ \int_0^1 f(x)e^{-2\pi i x n} dx = 0, \quad n \in \mathbb Z $$ then $f=0$ almost everywhere. Now let $F \subset \mathbb R$ be a general set of frequencies that is (a) periodic and (b) has density one, i.e. $$ F = \alpha \mathbb Z + \{ c_1, \dots c_N \} = \{ an+c_j : n \in \mathbb Z, j \in \{ 1,\dots, N\} \}, \quad c_j \neq c_i, \ i \neq j $$ and $$ \frac Na=1. $$ I'm wondering whether one can still conclude that if $$ \int_0^1 f(x)e^{-2\pi i x \ell} \, dx = 0, \quad \ell \in F $$ then $f=0$ almost everywhere.

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    $\begingroup$ you've got $\alpha,$ then $a$ in your expressions for $F$ $\endgroup$
    – kodlu
    Commented Nov 10, 2023 at 13:26

1 Answer 1

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Under your conditions, the system $e^{2\pi i\ell_nx}$ is complete. Moreover, it is a Riesz basis for $L^2[0,1]$. This follows from the result of B. Ya. Levin in

Interpolation by entire functions of exponential type (Russian) in: "Mathematical physics and functional analysis", 1, 136-146, Kharkiv 1969.

Indeed the canonical product with zeros $\ell_n$ is a periodic function, therefore it is a function of the "sine-type" in the sense of Levin, and he proved that zeros of such functions give a Riesz basis. For a modern reference for this see

G. Semmler, Complete interpolating sequences, the discrete Muckenhoupt condition, and conformal mapping. Ann. Acad. Sci. Fenn. Math. 35 (2010), no. 1, 23–46.

(This journal is reely available in https://www.acadsci.fi/mathematica/)

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