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For $f:\mathbb{Z}/p^k \mathbb{Z}\to \mathbb{C}$, define the Fourier transform $\widehat{f}:\mathbb{Z}/p^k \mathbb{Z}\to \mathbb{C}$ in the usual way, viz., $\widehat{f}(\xi) = \sum_x f(x) e(-\xi x/p^k)$, where $e(t) = e^{2\pi i t}$. What is an example of a function $f:\mathbb{Z}/p^k \mathbb{Z}\to \{-1,1\}$ such that $|\widehat{f}(\xi)|\ll p^{k/2}$ for all $\xi$?

For $k=1$, there are lots of choices: we can let $f(\xi) = \left(\frac{\xi}{p}\right)$ for $\xi\ne 0$, and $f(\xi) = 1$ (say) for $\xi = 0$, but other things (such as $f(\xi) = \left(\frac{\xi^3 + a \xi + b}{p}\right)$, or other algebraic-geometrical choices) also work.

What I want is an explicit construction (preferrably one that is as "clean" as possible) for $k>1$. Existence is not hard.

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  • $\begingroup$ Does $x$ run over $\mathbb Z/p\mathbb Z$ or $\mathbb Z/p^k\mathbb Z$ in $\hat f(\xi)$? And do you really mean $e(-\xi x/p)$, or should it be $e(-\xi x/p^k)$, with $x$ running over $\mathbb Z/p^k\mathbb Z$? $\endgroup$
    – Joe Silverman
    Commented Mar 1, 2023 at 11:10
  • $\begingroup$ A construction based on Rudin-Shapiro polynomials should work, see the introduction to Flat Littlewood polynomials exist by Paul Balister, Bela Bollobas, Robert Morris, Julian Sahasrabudhe and Marius Tiba. By the way, why is "existence not hard"? A random construction would miss by a log factor. $\endgroup$
    – Ben Green
    Commented Mar 1, 2023 at 11:56
  • $\begingroup$ I meant $p^k$ in the denominator - thanks. $\endgroup$
    – H A Helfgott
    Commented Mar 1, 2023 at 12:47
  • $\begingroup$ Not sure I understand your comment Harald. If one uses the results of Balister arxiv.org/abs/1909.08777 then you can take $n = p^k$ and this will give an example; the exponential sum is actually uniformly small at all $\theta$, not just $\theta = r/p^k$. I'll have to think whether there is an example that is more natural as regards the $p$-structure on $\mathbf{Z}/p^k \mathbf{Z}$. $\endgroup$
    – Ben Green
    Commented Mar 1, 2023 at 13:59
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    $\begingroup$ I was hoping for a simpler answer, but technically it does. $\endgroup$
    – H A Helfgott
    Commented Mar 2, 2023 at 21:29

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