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Consider $f_n(t)=\sum_{i=1}^{n}e^{ik_{i}t}$ with all $k_i$ some distinct integers for $t\in [-\pi,\pi)$. For $p>2$ I am interested in the maximum possible value of $$||f_n||_p,$$ where $f_n$ runs through all possible such sums with $n$ terms. Of particular interest is the case when $p$ is an even integer and $n\rightarrow \infty$ in case obtaining the sharp upper bound is too ambitious. Could it be that the Dirichlet kernel is the best such $f_n$?

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For even integer exponents, say $p=2k$ and $p \geq2$, the quantity is just the $k$-order additive energy of the set $S \subset \mathbb{Z}$ of non-zero Fourier coefficients. It is easy to see that this is maximized by any arithmetic progression of the desired length (which coincides with the $n$-order Dirichlet kernel).

In the case of $p=4$ this is just the statement that an arithmetic progression maximizes the quantity $$E(A) := \sum_{\substack{a,b,c,d \in A \\a+b=c+d}}1$$ of all integer subsets of cardinality $|A|$, which is easy to see.

For $p<2$ arithmetic progressions should minimize the quantity. This is a hard and difficult unsolved problem.

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  • $\begingroup$ Do you know by chance if the inequality $\| f\|_{1} \leq 0.9999 \sqrt{n}$ was ever proved/disproved somewhere? $\endgroup$ Commented Feb 21, 2021 at 0:07
  • $\begingroup$ under the assumption that $n\geq 2$. $\endgroup$ Commented Feb 21, 2021 at 0:10
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    $\begingroup$ Paata: Obviously you can get to a constant multiple of $\sqrt{n}$ by considering a dissociative set, but I'm not sure what is known about the absolute constants. I would suspect your inequality is true, but I'm not aware of any literature on the topic. $\endgroup$
    – Mark Lewko
    Commented Feb 21, 2021 at 0:41
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    $\begingroup$ Paata: This is discussed in Aistleitener's 2012 paper arxiv.org/pdf/1211.4640.pdf. At least as of the writing of that paper, the problem was open. Apparently, this problem has some important implications to rank one transforms. See the discussion there. $\endgroup$
    – Mark Lewko
    Commented Feb 21, 2021 at 1:13
  • $\begingroup$ Thank you for answering this. $\endgroup$
    – TOM
    Commented Feb 21, 2021 at 1:33

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