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.



Averaged Parseval Relation for Sampling a Function on Integers
If one were to assume $|f|$ is almost constant on its support, then a rough estimate is available. 


Assume for now $|f| = 1$ on its support. Similar argument can be made if we know that there exists some $C>0$ such that $C \inf\_{n\in \{1, \ldots, N\}} |f(n)| \geq \sup\_{n \in \{1, \ldots, N\}} |f(n)|$. 


In the case $|f| = 1$ on its support, the sum of $\ell\_2$ norms that the OP is interested becomes 


$$ \sum\_{\nu = 1}^m \underbrace{\lfloor \frac{N}{\nu} \rfloor}\_{= \|f\_\nu\|\_2^2} $$


which by elementary inequalities gives


$$ N \sum\_{\nu = 1}^m \frac{1}{\nu} \geq \sum\_{\nu = 1}^m \lfloor \frac{N}{\nu} \rfloor \geq N \sum\_{\nu = 1}^m \frac{1}{\nu} - m $$


The sum $\sum\_{\nu = 1}^m \frac{1}{\nu}$ is the $n$th [Harmonic number](https://en.wikipedia.org/wiki/Harmonic_number) about which much is known. For a very rough estimate we have it is bounded


$$ \ln(m) + 1 \geq \sum\_{\nu = 1}^m \frac{1}{\nu} \geq \ln(m) $$


So you get


$$ N \ln(m) + N \geq \sum\_{\nu = 1}^m \lfloor \frac{N}{\nu} \rfloor \geq N \ln(m) - m $$


As you want to compare this to $\| f\|\_{2}^2 = N$ in this case, you see that a very naive bound of 


$$ A(N,m) \geq \ln(m) - \frac{m}{N} $$


would work. This can of course be sharpened using better estimates of the Harmonic numbers. 


In the case you have $C > 1$, you see that the same argument leads to $A(N,m) \geq \frac1{C^2} (\ln(m) - \frac{m}{N})$ as a rough estimate. Of course, as I mentioned in the comments, there is the trivial lower bound of $A(N,m) \geq 1$, which seems sharp as $C \nearrow \infty$, at least when $m \ll N$. 


