Setup: Let $p$ be a large prime number and $F = \{0, \dots, p-1\}$ be the field of order $p$. Let $I$ denote the discrete interval $I = \{1, \dots, M\}$ for some $M < p$. Regard both $I$ and $F$ as subsets of the real line, and let $f : [0,p) \to R$ be a smooth indicator function of $I$ with support contained in $\tilde I = \{0, \dots, M+1\}$.

Extend $f$ to be periodic on the line with period $p$ and (abusing notation) call this function $f$ as well. Now sample $f$ at the integers to obtain a discrete function (abusing notation still) called $f$ with period $p$.

This last $f$ has a finite fourier series $$f(x) = \sum_{\xi=0}^{p-1} \hat f(\xi) \exp(2 \pi i \xi x / p)$$ with fourier coefficients $$\hat f(\xi) = \frac1p \sum_{z=0}^{p-1} f(z) \exp(- 2 \pi i \xi z / p).$$ and since f is smooth its fourier coefficients decay quickly, $\sum_{\xi=0}^{p-1} \hat f(\xi) = O(1).$

Problem:

One should be able to show (1) The coefficients $\hat f(\xi)$ can be dropped for $\xi > p/M$, and (2) For $\xi = 1, \dots, p/M$ the bound $|\hat f(\xi)| \leq M/p$ holds.

I think this can be accomplished with an appropriate form of the Poisson summation formula.

`$\varphi_1(x) = \exp( ( |x|^2 - 1 )^{-1} )$`

for`$|x| < 1$`

and zero otherwise. Observe`$\text{supp} \varphi_1 \subset [-1,1]$`

. Define`$\varphi_2(x) = \varphi_1(2x/(M_1+1)-1)$`

, so that`$\text{supp} \varphi_2 \subset [0,M_1+1]$`

. Set`$c^{-1} = \int_0^{M_1+1} \varphi_2(x) \ dx$`

and put`$\varphi_3 = c \varphi_2 * \mathbb I_{[0,M_1+1]},$`

where $\mathbb I_{[0,M_1+1]}$ is the indicator function of $[0,M_1+1]$. This $\varphi_3$ is manifestly smooth on $\mathbb F_p \simeq \{0,\dots,p-1\}$ and satisfies $\varphi_3|_{[1,M]} \equiv 1$,`$\text{supp} \varphi_3 \subset [0,M_1+1]$`

. $\endgroup$ – John Dusel Nov 16 '11 at 16:19