# Irrational rotation - recurrence times

I consider the irrational rotation $T_\alpha(x) = x + \alpha \text{ mod } 1$ for given irrational $\alpha \in [0,1]$. For a given open interval $A \subset [0,1]$ with length $|A|>0$, I consider the recurrence times $I = \{n\in \mathbb{N}: T^n(0) \in A \}$. I want to show that $\sum_{i \in I} p\cdot(1-p)^i \to |A|$ as $p \to 0$.

My very informal motivation for this is that the above sum should be equal to $\sum_{n\in \mathbb{N}} p \cdot (1-p)^{n\cdot\frac{1}{|A|}}$ "give or take" "a few" $(1-p)$-factors (which tend to $1$ as $p \to 0$), and the latter sum can be shown to converge to $|A|$ as $p \to 0$.

I have obtained a somewhat similar (but obviously not identical) result for the case of rational $\alpha$ (happy to add details, but I'm not sure if it's useful), and tried to derive the above using a rational series converging to $\alpha$, but wasn't successful.

Unfortunately, I have virtually no background in ergodic theory, numbers theory and similar subjects which apparently treat irrational rotations, and thus don't quite know where to start.

You can recover this result in two steps:

• a variation on Birkhoff's ergodic theorem yields the Cesàro convergence of the sums;
• Cesàro convergence implies Abel convergence.

First step: $T_\alpha$ preserves the Lebesgue measure and is uniquely ergodic. Hence, for all $f \in \mathcal{C} (\mathbb{S}_1, \mathbb{R})$,

$$\lim_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} f \circ T_\alpha^k = \int_{\mathbb{S}_1} f(x) \ dx,$$

where the limit is uniform. By approximating $\mathbf{1}_A$ from above and from below by continuous functions, we get that:

$$\lim_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} \mathbf{1}_A \circ T_\alpha^k = |A|,$$

where the limit is again uniform. I had to use uniform ergodicity to be able to tell something about a specific starting point ($0$) instead of a generic one.

Second step: let $p \in (0,1)$. Let $(a_k)_{k \geq 0}$ be a bounded real sequence. Then:

$$\sum_{n=0}^{+ \infty} (1-p)p^n a_n = \sum_{n=1}^{+ \infty} \frac{n(1-p)^2 p^n}{p} \frac{1}{n} \sum_{k=0}^{n-1} a_k$$

Now, putting $b_n^p := n(1-p)^2p^{n-1}$, for all $p \in (0,1)$, the sequence $(b_n^p)_{n \geq 1}$ defines a probability measure on the positive integers. In addition, for all $n$, $b_n^p$ converges to $0$ as $p$ goes to $1$. Hence,

$$\lim_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} a_k = \ell \quad \implies \quad \lim_{p \to 1} \sum_{n=0}^{+ \infty} (1-p)p^n a_n = \ell.$$

Finally, take $a_n := \mathbf{1}_A \circ T_\alpha^n (0)$, and apply the first step.

In the range $n \le jp^{1/2} < n+1$, the number of $j$'s contributing to the sum is $|A|p^{-1/2} +o(p^{-1/2})$. So that portion of the sum satisfies $$\sum_{j=np^{-1/2}}^{(n+1)p^{-1/2}} p(1-p)^j \chi_I(j)= (|A|p^{1/2}+o(p^{1/2})) (1-p)^{np^{-1/2}} ;$$ here, I can treat $(1-p)^j$ as a constant when doing the sum, at the expense of a multiplicative error $(1-p)^{p^{-1/2}}$, which can be absorbed into the other error term, from the ergodic theorem.

So the whole sum equals $$|A|p^{1/2}(1+o(1)) \sum_{n\ge 0} (1-p)^{np^{-1/2}} = (1+o(1))|A| \frac{p^{1/2}}{1-(1-p)^{p^{-1/2}}} \to |A| ,$$ as desired.