Geometry of numbers argument: counting integers with some linear condition I am interested in the proof of the following result:
Suppose that $A > 1$, $\lambda \in \mathbb{R}$, and for $0 < Z \leq 1$, let $U(Z)$ be the number of integer solutions $v$ of 
\begin{eqnarray}
|v| < ZA \ \  \  \text{ and } \ \ \   \| \lambda v \| < Z A^{-1}.
\end{eqnarray}
Then, if $0 < Z_1 < Z_2 \leq 1$, we have 
$$
U(Z_1) \gg (Z_1/Z_2) \ U(Z_2).
$$
I would greatly appreciate any comments or hints on this! Thank you very much!
PS Here $\|x\|$ denotes the distance the closest integer. 
 A: If you don't mind, I'll reformulate your problem slightly. Let $X = ZA$, $B = A^2$. Then $XB^{-1} = ZA^{-1}$. We would like to know the number of integer solutions $U'(X)$ to the system of inequalities
$$
\begin{cases}|v| < X;\\\|\lambda v\|<XB^{-1}.\end{cases}
$$
Let $\lfloor x \rfloor$ denote the largest integer not exceeding $x$, and put $\{x\} = x - \lfloor x \rfloor$. Then
$$
\|\lambda v\| = \|(\lfloor \lambda \rfloor + \{\lambda\})v\| = \|\lfloor \lambda\rfloor v + \{\lambda\}v\| = \|\{\lambda\}v\|.
$$
Since $0 \leq \{\lambda\} < 1$, without loss of generality we may assume that $\lambda$ satisfies $0 \leq \lambda < 1$.
Suppose that $XB^{-1} > 1/2$. Then $0 \leq \|\lambda v\| \leq 1/2 < XB^{-1}$ is true for any choice of $\lambda$ or $v$, which means that $U(X)$ is equal to $2\lfloor X\rfloor + 1$. The same observation appplies to the case when $\lambda = 0$.
Suppose that $XB^{-1} \leq 1/2$ and $0 < \lambda < 1$. Denote the largest integer in the interval $(-\lambda X-1, \lambda X+1)$ by $k$. The fact that $\|\lambda v\| < XB^{-1}$ simply means that there exists some integer $n \in \{-k, -k+1, \ldots, k-1, k\}$ such that
$$
\left|n - \lambda v\right| < XB^{-1}.
$$
Verify that there are at least $\lambda X/2$ and at most $2 \lambda X +3$ possible values of $n$. 
Further, each $\lambda v$ is contained in exactly one interval of the form $(n - XB^{-1},n + XB^{-1})$, whose length is $2XB^{-1}$. Verify that this interval contains at least $X(\lambda B)^{-1}$ and at most $2X(\lambda B)^{-1} + 1$ numbers of the form $\lambda v$, where $v$ is an integer.
We conclude that, when $X/B \leq 1/2$ and $0 < \lambda < 1$,
$$
\frac{\lambda X}{2}\cdot \frac{X}{\lambda B} \leq U'(X) \leq (2 \lambda X + 3)(2 X(\lambda B)^{-1} + 1).
$$
Otherwise,
$$
U'(X) = 2\lfloor X\rfloor + 1.
$$
Finally, note that $U(Z) = U'(ZA)$. Plug in $B = A^2$ and $X=ZA$ inside the above expressions and deduce $U(Z_1) \gg (Z_1/Z_2)U(Z_2)$.
A: Is it a reference you want? Check chapter 12 (if I remember correctly) in Davenport's book on diophantine equations and inequalities.
A: I can show the (exact) inequality
$$(*) \ \ \  \ \ \ V(Z_1) \geq \left(\frac{Z_1}{Z_2} \right)^2 V(Z_2) \ \ \quad (\frac{2}{A} \leq Z_1 \leq Z_2 \leq \frac{A}{2}), $$
for a smoothed version of $U$ defined by
$$
V(Z) = \sum_{\nu \in \mathbb{Z}} \mathrm{sinc}^2\left( \frac{\nu}{2ZA} \right) \left( 1 - \frac{A ||\lambda \nu||}{Z} \right)_{+}.
$$
Indeed, the Fourier transform of the function $f(x) =\mathrm{sinc}^2\left( \frac{x}{2} \right)$ is given by the formula $\hat{f}(x)= (1- |x|)_+$, so that we have 
$$V(Z) =  \sum_{\nu \in \mathbb{Z}} \  f \left( \frac{\nu}{ZA} \right) \hat{f}\left( \frac{A ||\lambda \nu||}{Z} \right).$$
Now, if $Z^{-1} A \geq 2$ and $ZA \geq 2$, then Poisson summation formula yields
$$
\hat{f}\left( \frac{A ||\lambda \nu||}{Z} \right) = \sum_{n \in \mathbb{Z}} \hat{f}\left( \frac{A (\lambda \nu + n )}{Z} \right) = \frac{Z}{A} \sum_{\mu \in \mathbb{Z}} f \left( \frac{Z \mu}{A} \right) e(\lambda \mu \nu), \\
\text{and similarly} \ \ \ \sum_{\nu \in \mathbb{Z}} f \left( \frac{\nu}{ZA} \right) e(\lambda \mu \nu) = ZA \ \hat{f}\left( ZA ||\lambda \nu||\right),
$$
where $e(x) = e^{2 i \pi x}$.
We thus get 
$$V(Z) = \frac{Z}{A} \sum_{\nu \in \mathbb{Z}} \  f \left( \frac{\nu}{ZA} \right) \left(  \sum_{\mu \in \mathbb{Z}} f \left( \frac{Z \mu}{A} \right) e(\lambda \mu \nu)\right) \\
= \frac{Z}{A} \sum_{\mu \in \mathbb{Z}} f \left( \frac{Z \mu}{A} \right) \left(  \sum_{\nu \in \mathbb{Z}} \  f \left( \frac{\nu}{ZA} \right)  e(\lambda \mu \nu)\right) \\
=Z^2  \sum_{\mu \in \mathbb{Z}} f \left( \frac{Z \mu}{A} \right) \hat{f}\left( ZA ||\lambda \nu||\right) \\
= Z^2 V(Z^{-1}).
$$
Since $V(Z^{-1})$ is a decreasing function of $Z$, this yields $(*)$.
Note that $(*)$ implies an inequality of the form
$$ U'(Z_1) \gg \left(\frac{Z_1}{Z_2} \right)^2 U(Z_2),$$
with
$$
U'(Z_1) = \sum_{\substack{\nu \in \mathbb{Z} \\ ||\lambda \nu|| < Z_1 A^{-1}}} \min \left( 1, \left(\frac{Z_1 A}{|\nu|}\right)^2 \right).
$$
