This is a followup to this earlier question

Let $f:\mathbb{Z}\rightarrow \{\pm 1\}.$ Assume that the support of $f$ is finite, say it is contained in $[1,N],$ it can even be taken to be $[1,N]$ if it helps. To avoid divisibility issues, assume that $N$ is is not divisible by any integer in $[1,m]$. One could, for example, take $N=m\#-1,$ where $m\#$ is the $m^{th}$ primorial.

Define the fourier transform $\widehat{f}$ on $[0,1)$ by $\widehat{f(t)}=\sum_{n\in \mathbb{Z}} f(n)~e^{-2i \pi n t}.$

Now let $v$ be a positive integer $\geq 2,$ and let the "projected" function be $$ f_v(n)=\left\{ \begin{array}{ccc} f(n), & \quad\mathrm{if}\quad & v|n,\\ & & \\ 0 & & \mathrm{otherwise}. \end{array} \right. $$ Of course $f_1$ is simply $f.$ Can one obtain a nontrivial bound of the form $$ \sum_{v=1}^m \mid \sum_{n \in \mathbb{Z}} f_v(n) \mid^2 {\geq} A(N,m) \int_0^1 \mid\widehat{f(t')} \mid^2 \,dt' $$ or similar, using some kind of uncertainty relation.

We can take $m\ll N,$ a fractional power of $N$ or even a power of $\log N.$

  • $\begingroup$ What prevents all sums on the left from being $0$? I mean, $N=m!$, $f(n)=e^{2\pi i n/N}$, say. $\endgroup$
    – fedja
    Jun 21 '19 at 0:34
  • $\begingroup$ @fedja, please see edit. $\endgroup$
    – kodlu
    Jun 21 '19 at 0:38
  • 1
    $\begingroup$ The new condition is not scale-invariant and, thereby meaningless. Just perturb all values in my example by $c$ where $c$ is extremely small. $\endgroup$
    – fedja
    Jun 21 '19 at 0:44
  • $\begingroup$ @fedja now I see. Thanks. I will update with a meaningful constraint, when I have time to think carefully. $\endgroup$
    – kodlu
    Jun 21 '19 at 3:42
  • $\begingroup$ @fedja please see new edit $\endgroup$
    – kodlu
    Jun 22 '19 at 12:52

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