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Aug 15 at 18:17 comment added Itay @TerryTao Thank you!
Aug 15 at 17:15 comment added Terry Tao 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.
Aug 15 at 17:13 comment added Terry Tao 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.
Aug 15 at 15:35 comment added Itay @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?
Aug 15 at 14:32 comment added Terry Tao The average of the (suitably normalized) delta function and constant function.
Aug 15 at 9:47 history asked Itay CC BY-SA 4.0