# Parseval type lower bound on sum of squares of function projections

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 $$\log N.$$

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