# Central Limit Theorem(s) for irrational rotation

Let $\alpha$ be irrational and $T: S^1 \rightarrow S^1$ be the rotation by $\alpha$. I'm interested in what type of Central Limit Theorem (if any) can hold for sums $Y_n = \frac{1}{\sqrt{n}}\sum_{k=1}^{n} f(T^k x)$.

I've done some googling and found statements like "generic smooth functions $f$ do not obey a CLT for irrational rotation", but was unable to find a definitive cohesive reference. The specific example of sums I'm interested in are of the form $\approx \sum_{k}\log \left\vert \frac{f(T^{2k}x)}{f(T^{2k+1}x)}\right\vert$ for some smooth $f$.

• What sort of conditions do you impose on $f$? You mention smooth functions but your example is not smooth. It looks like you are assuming that $\int_{S^1} f(z) ds = 0$. Jun 29, 2014 at 20:59
• @DouglasZare: edited so it's clear the relevant r.v. have mean zero. Jun 29, 2014 at 21:13
• Maybe you're more looking for something along the lines of "error estimates in the ergodic theorem." See for example this question: mathoverflow.net/questions/4411/… Jun 30, 2014 at 0:51
• There is a paper of Harry Kesten (Uniform Distribution Mod 1) published in the Annals in 1960 -- also a follow-up paper a couple of years later, dealing with the case where $f$ is a characteristic function minus its expectation, and showing the limit distribution is Cauchy Jun 30, 2014 at 1:00
• @MarcinKotowski I am aware of that. What I was trying to say was that a circle rotation is as non-random as possible (given that it's ergodic); for example, it has pure point spectrum. Of course, I said it in a maximally misleading way. Jun 30, 2014 at 19:45

The result depends on the approximation properties of $\alpha$.

Of course one has to assume $\int_{S^1} f(z)dz=0$. A rotation by $\alpha$ has the effect that the $k$-th Fourier coefficient of $f$ is multiplied by $\exp(2\pi i \cdot k \alpha)$. Hence, the $k$-th Fourier coefficient (for $k \neq 0$) of $T_n(f)$ is just $$\frac1{\sqrt{n}} \sum_{l=1}^n \exp(2\pi i \cdot k l \cdot\alpha) \cdot \hat f(k) =\frac1{\sqrt{n}} \cdot \exp(2\pi i \cdot k \cdot\alpha) \cdot \frac{1 - \exp(2\pi i \cdot k n \cdot\alpha)}{1 - \exp(2\pi i \cdot k \cdot\alpha)} \cdot \hat f(k).$$

If $\alpha$ is algebraic (or diophantine generic), then $|\alpha - p/k| \geq C/k^M$ for some constants $C$ and $M$. Hence, $|1 - \exp(2\pi i \cdot k \cdot\alpha)|^{-1}$ grows at most like a polynomial in $k$. If $f$ is smooth, then $k \mapsto \hat f(k)$ decays rapidly, so that $k \mapsto |1 - \exp(2\pi i \cdot k \cdot\alpha)|^{-1} \hat f(k)$ is still in $\ell^1(\mathbb Z)$. This altogether implies that $T_n(f)$ converges to zero uniformly on $S^1$.

On the other side, if $\alpha$ is some well-chosen Liouville number and $f$ some special constructed smooth function, then I believe that $T_n(f)$ need not converge pointwise (or uniformly).

• So $T_n(f)\to0$ as $n\to\infty$. My function is $z\mapsto\Re z$. What power of $n$ should I divide $\sum_{l=1}^n \exp(2\pi \cdot kl\cdot\alpha)\cdot \hat{f](k)$ to get something positive and finite? Thanks in advance. Nov 3, 2023 at 3:58

May I suggest Michael Lacey's more-or-less definitive paper on this topic, On central limit theorems, modulus of continuity and Diophantine type for irrational rotations (Journal d'Analyse Mathematique 61 (1993) 47-59). In that paper, Lacey proves that if $T$ is a rotation by an irrational number with irrationality measure $\mu$, then:

1. For $\beta>\frac{1}{2(\mu-1)}$ there does not exist a $\beta$-Hoelder continuous function which satisfies a central limit theorem with respect to $T$, and furthermore the limit distribution of $\frac{1}{\sqrt{n}}\sum_{k=0}^{n-1}f \circ T^k$ is trivial.
2. For $\beta<\frac{1}{2(\mu-1)}$ there does exist a $\beta$-Hoelder continuous function which satisfies a central limit theorem with respect to $T$.

Lacey's result is in fact more general: it considers functional central limit theorems with convergence to fractional Brownian motion with a general exponent $H \in (0,1)$. The case of an ordinary (functional) central limit theorem corresponds to $H=1/2$.

Let $\mathcal B$ be a Banach space of functions on $S^1$ containing the nonconstant trigonometric polynomials as a dense subset, such that $\|f\| \ge \|f\|_\infty$ and all members of $\mathcal B$ have mean $0$. Then there is a dense $G_\delta$ set of pairs $(\alpha, f) \in S^1 \times \mathcal B$ such that $Y_n(f)(x)$ are unbounded for all $x \in S^1$.

In fact, let $U_m$ be the set of $(\alpha, f)$ for which there exist a positive integer $n$ such that $|Y_{n}(f)(x)| > m$ for all $x \in S^1$. Note that $U_m$ is open, and any $(\alpha, f) \in \bigcup_m U_m$ satisfies the requirements. So it suffices to show $U_m$ is dense. Consider any $f_0 \in \mathcal B$, $\alpha_0 \in S^1$ and $\epsilon > 0$. Take a trigonometric polynomial $f_1(x) = \sum_{j=-K}^K c_j \exp(ij x)$ with $\|f_1 - f_0\| < \epsilon$. Let $\alpha = 2 \pi M/N$ for coprime positive integers $M$ and $N$ with $|\alpha - \alpha_1|< \epsilon$ and $N > K$. Note that $Y_n(f_1)(x)$ is periodic in $n$ with period $N$ and $Y_N(f_1)(x) = 0$, so $Y_n(f_1)(x) = O(1/\sqrt{n})$ (uniformly in $x$). On the other hand, $f_2(x) = \exp(i N x)$ has $f_2(n \alpha + x) = f_2(x)$ for all integers $n$, so that $Y_n(f_2)(x) = \sqrt{n} f_2(x)$. Thus we can take $f = f_1 + c f_2$ with $|c|$ small enough that $\|f - f_0\| < \epsilon$, and for sufficiently large $n$ we have $|Y_n(f)(x)| > m$ for all $x \in S^1$.