Dynamics in the integers - Floor function

Question

Let $\alpha$ be an irrational with $0<\alpha<1$. Consider the function given by \begin{align*} f: &\mathbb{N}\longrightarrow \mathbb{N}\\ &x\longmapsto [ \alpha\cdot x]\end{align*} where $[r]$ is the largest integer that is less than or equal to $r$ for $r\in\mathbb{R}$.

Let $N=\left[\frac{1}{\alpha}\right]$ be the integer part of $\frac{1}{\alpha}$, and consider the following numbers:

$A_1(n)=\#\{k\leq n : f(k)<f(k+1)\}$

$A_2(n)=\#\{k\leq n : f(k)=f(k+N)\}$

The questions are the following:

1) Is it true that $\displaystyle{\lim_{n\to\infty} \dfrac{A_1(n)}{n}=\alpha}$?

2) What is the value of $\displaystyle{\lim_{n\to\infty} \dfrac{A_2(n)}{n}}$, if it exists?

The problem above came out when I was trying to find examples of classes of finite ordered structures that satisfies strong conditions on their definable sets (for my Ph.D. Thesis). I translated the problem in simple terms, and I would really appreciate any help.

(If it is too easy, a hint will be good. If the problem is related with something in number theory, combinatorics, ergodic theory or anything else, I would also appreciate if you tell me how)

Answer 1

The answer to the first question is: yes. The function $f(x)$ jumps $0$ or $1$ at every integer (because $0\leq\alpha\leq 1$), and also $f(0)=0$, hence $A_1(n)=f(n+1)$. Therefore $$\frac{A_1(n)}{n}=\frac{[(n+1)\alpha]}{n}=\frac{n\alpha+O(1)}{n}=\alpha+O\left(\frac{1}{n}\right)=\alpha+o(1). $$

Regarding the second question: the limit exists and equals $1-N\alpha$. Indeed, $f(k+N)=[k\alpha+N\alpha]$, hence $f(k)=f(k+N)$ means that the fractional part of $k\alpha$ is less than $1-N\alpha$. Here we used that $0<N\alpha<1$. As $\alpha$ is irrational, the fractional parts of $k\alpha$ are equidistributed in $(0,1)$ by Weyl's theorem, hence the density of $k$'s with $f(k)=f(k+N)$ equals the length of the interval $(0,1-N\alpha)$ which is $1-N\alpha$.

More generally, if $1\leq M\leq N$ is fixed, then the density of $k$'s with $f(k)=f(k+M)$ equals $1-M\alpha$. In particular, the case $M=1$ yields an alternative proof for the first paragraph, since $1-(1-\alpha)=\alpha$.