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Let $G$ be a finite cyclic group of order $n$ ($n$ need not be prime) and $\mathcal{F}$ the normalized discrete Fourier transform defined on $G$.

Is there a canonical way to construct an eigenfunction $f:G\rightarrow [0,1]$ of $\mathcal{F}$ such that $\mathcal{F}(f)=f$ and $f(0)=1$?

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    $\begingroup$ The average of the (suitably normalized) delta function and constant function. $\endgroup$
    – Terry Tao
    Commented Aug 15 at 14:32
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    $\begingroup$ @TerryTao Yes, I should have written that as a trivial example. You can actually take the average of every subgroup and its dual and get the same. Are there any other known constructions beyond convex combinations of such? $\endgroup$
    – Itay
    Commented Aug 15 at 15:35
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    $\begingroup$ Since ${\mathcal F}^4 = I$, you can average any function $f$ with its first three Fourier transforms ${\mathcal F}(f)$, ${\mathcal F}^2(f)$, ${\mathcal F}^3(f)$ and normalize, and this is a complete description of the eigenfunctions of eigenvalue 1 (the eigenfunctions for the other three eigenvalues $i, -1, -i$ are produced similarly). If $f$ is even then ${\mathcal F}^2(f)=f$ and one just needs to average $f$ and ${\mathcal F}(f)$ as before. $\endgroup$
    – Terry Tao
    Commented Aug 15 at 17:13
  • $\begingroup$ Getting the result to be non-negative and maximized at the origin may require some further tricks and normalizations, e.g., restricting $f$ to first be non-negative and positive definite. $\endgroup$
    – Terry Tao
    Commented Aug 15 at 17:15
  • $\begingroup$ @TerryTao Thank you! $\endgroup$
    – Itay
    Commented Aug 15 at 18:17

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