This was asked a long time ago on math.stackexchange with no answers.

Let $f:\mathbb{Z}\rightarrow \mathbb{C}.$ Assume that the support of $f$ is finite, say it is $[1,N],$ and that $\mid f\mid$ is bounded (as well as being nonzero at each point in its support if necessary). Define the fourier transform $$\widehat{f}:[0,1)\rightarrow \mathbb{C}$$ by $$\widehat{f(t)}=\sum_{n\in \mathbb{Z}} f(n)~e^{2i \pi n t}.$$

Parseval states that $$ \sum_{n \in \mathbb{Z}} \mid f(n) \mid^2 = \int_0^1 \mid\widehat{f(t)} \mid^2 \,dt $$ holds. Now let $v$ be a positive integer $\geq 2,$ and let the "sampled" function be $$ f_v(n)=\left\{ \begin{array}{ccc} f(n), & \quad\mathrm{if}\quad & v|n,\\ & & \\ 0 & & \mathrm{otherwise}. \end{array} \right. $$ Let $f=f_1$ for simplicity. What is the Parseval relationship for this function, as expressed in terms of the transform of the original function?

Due to the comb structure, one will have a `sinc' structure for the transform, and the comb can be truncated to the support $[1,N],$ which can presumably be used together with the convolution theorem in the frequency domain.

What I am really interested in, however, is estimating from below, the following quantity $$ \sum_{v=1}^m \sum_{n \in \mathbb{Z}} \mid f_v(n) \mid^2 = \sum_{v=1}^m \sum_{n \in \mathbb{Z}} \mid \mathbb{1}\{n\!\!\pmod v=0~\}|^2 \mid f(n)\mid^2= $$ $$ =\sum_{v=1}^m \int_0^1 \mid \widehat{\Phi_v(t-t')} \mid^2 \mid\widehat{f(t')} \mid^2 \,dt' \geq $$ $$ \stackrel{hopefully}{\geq} A(N,m) \int_0^1 \mid\widehat{f(t')} \mid^2 \,dt' $$ using some kind of uncertainty relation. Here, $\widehat{\Phi_v(t)}$ is the fourier transform of the comb, i.e., the sinc type function.

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

**Edit:** Let $|f|>0,$ on $[1,N]$. and can even take it to be essentially constant in magnitude if it helps.

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