Posted by: yglima | February 16, 2012

## CF1: uniform distribution and Denjoy-Koksma’s inequality

This first post discusses some uniform distribution results of the orbits of ${R_\alpha}$ on ${\mathbb T}$. It is a classical result that they are all uniformly distributed. Moreover, Denjoy-Koksma’s inequality states that, for iterates equal to the denominators of the continued fraction expansion of ${\alpha}$, the orbits are highly equidistributed. We give a proof of this result for the classical assumption of bounded variation and its strengthening for differentiable functions. The last part is devoted to describe the basic properties of a cylinder flow and how its ergodic properties are related to the uniform distribution of orbits of the respective irrational rotation. More specifically, we want to convince the reader that the ergodicity of a cylinder flow over ${R_\alpha}$ implies – on the contrary to Denjoy-Koksma’s inequality – that the orbits of the irrational rotation have discrepancy, in the sense that uniform distribution is not regular. This gives another reason, beyond their intrinsic ergodic theoretical importance, to investigate cylinder flows.

1. Uniform distribution

It is a classical result that the only ${R_\alpha}$-invariant probability measure is the Lebesgue measure ${m_{\mathbb T}}$ on ${\mathbb T}$. This is equivalent to

$\displaystyle \lim_{N\rightarrow\infty}\dfrac{1}{N}\sum_{n=1}^N f(x+n\alpha)=\int_{\mathbb T} fdm_{\mathbb T}\ \ \text{ for every }x\in\mathbb T$

for any continuous function ${f:\mathbb T\rightarrow\mathbb R}$. By approximation, this is equivalent to, for any interval ${I\subset\mathbb T}$, the frequency of visits to ${I}$ of any orbit of ${R_\alpha}$ being equal to ${m_{\mathbb T}(I)}$:

$\displaystyle \lim_{N\rightarrow\infty}\dfrac{\#\{1\le n\le N\,;\,x+n\alpha\in I\}}{N}=m_{\mathbb T}(I)\ \ \text{ for every }x\in\mathbb T.$

Definition 1 A sequence ${(x_n)}$ in ${\mathbb T}$ is uniformly distributed or equidistributed if

$\displaystyle \lim_{N\rightarrow\infty}\dfrac{\#\{1\le n\le N\,;\,x_n\in I\}}{N}=m_{\mathbb T}(I) \ \ \ \ \ (1)$

for any interval ${I\subset\mathbb T}$.

The unique ergodicity of ${R_\alpha}$ thus says that, for any ${x\in\mathbb T}$, the sequence ${x_n=x+n\alpha}$ is uniformly distributed. We can identify equidistribution in a quantitative fashion by considering the discrepancy of the sequence.

Definition 2 Let ${x_1,\ldots,x_N}$ be a finite sequence of real numbers. The number

$\displaystyle D_N=\sup_{0\le a

is called the discrepancy of the sequence. For an infinite sequence, the discrepancy ${D_N}$ is equal to the discrepancy of the first ${N}$ terms.

It is direct, and we leave the proof as an exercise to the reader, that

Proposition 3 The sequence ${(x_n)}$ is uniformly distributed iff

$\displaystyle \lim_{N\rightarrow\infty}D_N=0.$

In other words, the theorem states the following (in principle stronger yet equivalent) interesting fact: whenever the sequence is equidistributed, then the convergence in (1) is uniform along all intervals ${I}$.

The distinction between different types of equidistribution is made by the analysis of the non-averaged expressions

$\displaystyle N\cdot D_N=\sup_{0\le a

If a sequence has good equidistribution, one would hope for boundedness of ${N\cdot D_N}$. It turns out that, by the following result of W.M. Schmidt, this is never the case: ${\limsup_{N\rightarrow\infty}(N\cdot D_N)}$ is always infinite.

Theorem 4 (W.M. Schmidt) There is an absolute constant ${C>0}$ with the following property: for any infinite sequence ${(x_n)}$ of ${\mathbb T}$, we have the inequality

$\displaystyle D_N>C\cdot\dfrac{\log N}{N}$

for infinitely many positive integers ${N}$.

The interested reader may check the proof in the book Uniform distribution of sequences of Kuipers and Niederreiter.

2. Denjoy-Koksma’s inequality

The above theorem guarantees the existence of infinitely many times for which the uniform distribution behaves in an irregular fashion. Nevertheless, for the special case of the irrational rotation, there are indeed iterates ${N}$ for which the sequence ${0,\alpha,\ldots,N\alpha}$ is highly equidistributed in the sense that ${N\cdot D_N}$ is bounded, and this is given by Denjoy-Koksma’s inequality. The iterates exhibiting such behavior are the denominators of the continued fraction expansion of ${\alpha}$.

Given ${\alpha}$ irrational, consider its continued fraction expansion

$\displaystyle \alpha\ =\ a_0+\dfrac{1}{a_1+\dfrac{1}{a_2+\dfrac{\ \ \ 1\ \ \ }{\ddots}}}\ \doteq\ [a_0;a_1,a_2,\ldots]\ ,$

whose ${n^{th}}$convergentis

$\displaystyle \alpha_n=\dfrac{p_n}{q_n}=[a_0;a_1,a_2,\ldots,a_n],\ n\ge 0.$

The ${q_n}$ are called the partial quotient denominators or just denominators of ${\alpha}$. They give the best rational approximations of ${\alpha}$. More precisely, the approximation is equal to

$\displaystyle \|q_n\alpha\|=q_n\cdot\left|\alpha-\dfrac{p_n}{q_n}\right|$

and is always at most ${1/q_n}$. On these denominators, the dynamics of the rotation ${x\mapsto x+\alpha}$ is partially described by the rational rotation ${x\mapsto x+p_n/q_n}$, and that’s why the sequence is highly equidistributed. This is the content of the next

Lemma 5 For each ${k=0,1,\ldots,q_n-1}$, there is one and exactly one integer ${i=0,1,\ldots,q_n-1}$ for which

$\displaystyle \{k\alpha\}\in\left[\dfrac{i}{q_n}\,,\,\dfrac{i+1}{q_n}\right)\,\cdot$

Proof: The argument is based on the article Sur la conjugaison différentiable des difféomorphismes du cercle à des rotations of Michel Herman. Let ${p/q=p_n/q_n}$. Firstly, assume

$\displaystyle 0<\alpha-\dfrac{p}{q}<\dfrac{1}{q^2}\,\cdot$

Then, for ${k=0,1,\ldots,q-1}$,

$\displaystyle 0

and so

$\displaystyle 0<\{k\alpha\}-\dfrac{i}{q}<\dfrac{1}{q}\ \ \Longrightarrow\ \ \{k\alpha\}\in\left[\dfrac{i}{q}\,,\,\dfrac{i+1}{q}\right),$

where ${i}$ is the residue of ${k p}$ modulo ${q}$. As ${k p}$ runs overs all residues modulo ${q}$, every such interval contains at least (and then exactly) one element of the set ${\{\{k\alpha\}\,;k=0,1,\ldots,q-1\}}$.

The case ${0 is similar, in which case ${\{k\alpha\}\in[(i-1)/q,i/q)}$. $\Box$

Exercise 1 Show that the above lemma implies that ${q_n\cdot D_{q_n}\le 1}$ for every ${n\ge 1}$.

Translating to the setup of functions, we arrive at the

Theorem 6 (Denjoy-Koksma’s inequality) If ${\phi:\mathbb T\rightarrow\mathbb R}$ has bounded variation, then

$\displaystyle \left|\sum_{k=0}^{q_n-1}\phi(x+k\alpha)-q_n\cdot\int_{\mathbb T}\phi dm_{\mathbb T}\right|\le {\rm Var}(\phi)$

for every ${x\in\mathbb T}$.

Proof: Let ${q=q_n}$ and ${x\in\mathbb T}$. By Lemma 5, each interval

$\displaystyle I_i=\left[x+\dfrac{i}{q}\,,\,x+\dfrac{i+1}{q}\right)\subset\mathbb T\,, \ \ i=0,1,\ldots,q-1,$

has exactly one element ${x_i=x+k_i\alpha}$ of the set ${\{\{x+k\alpha\}\,;\,k=0,1,\ldots,q-1\}}$. Thus

$\begin{array}{rcl} \left|\displaystyle\sum_{k=0}^{q-1}\phi(x+k\alpha)-q\displaystyle\int_{\mathbb T}\phi dm_{\mathbb T}\right|&=& \left|\displaystyle\sum_{k=0}^{q-1}\dfrac{1}{m_{\mathbb T}(I_k)}\int_{I_k}\{\phi(x_k)-\phi(x)\}dm_{\mathbb T}\right|\\ &&\\&\le&\displaystyle\sum_{k=0}^{q-1}\dfrac{1}{m_{\mathbb T}(I_k)}\displaystyle\int_{I_k}|\phi(x_k)-\phi(x)|dm_{\mathbb T}\\ &&\\&\le&\displaystyle\sum_{k=0}^{q-1}\sup_{y,z\in I_k}|\phi(y)-\phi(z)|\\ &&\\&\le&\displaystyle\sum_{k=0}^{q-1}{\rm Var}(\phi|_{I_k})\\ &&\\&\le&{\rm Var}(\phi). \end{array}$

$\Box$

Denjoy-Koksma’s inequality can be improved if ${\phi}$ is of class ${C^1}$. Let ${\|\cdot\|_{C^0}}$ denote the ${C^0}$-norm.

Theorem 7 If ${\phi:\mathbb T\rightarrow\mathbb R}$ is of class ${C^1}$, then

$\displaystyle \left\|\sum_{k=0}^{q_n-1}\phi(x+k\alpha)-q_n\cdot\int_{\mathbb T}\phi dm_{\mathbb T}\right\|_{C^0}\rightarrow 0\ \ \ \text{ as }n\rightarrow\infty.$

Proof: Again, we can assume that ${\phi}$ belongs to the set ${C^1_0(\mathbb T)}$ of ${C^1}$ functions of zero mean. The main property we’ll use is that the ${C^1}$-closure of the set

$\displaystyle B(R_\alpha,C^1)=\left\{u\circ R_{\alpha}-u\,;\,u\in C^1(\mathbb T)\right\}$

of ${C^1}$ coboundary functions is equal to ${C^1_0(\mathbb T)}$. For this, recall the set of trigonometric polynomials

$\displaystyle \left\{\psi(x)=\sum_{k=-n}^n \hat\psi(k)\cdot e^{2\pi ikx}\,;\,\hat\psi(0)=0\right\}$

of zero mean is ${C^1}$-dense in ${C^1_0(\mathbb T)}$. Hence, it suffices to prove that each trigonometric polynomial is a coboundary. This is a simple task: given ${\phi(x)= \sum_{k=-n}^{n} \hat{\phi}(k)\cdot e^{2\pi ikx}}$,

$\displaystyle u(x)=\sum_{k=-n}^{n}\dfrac{\hat\phi (n)}{e^{2\pi ik\alpha}-1}\cdot e^{2\pi ikx}$

satisfies ${\phi(x)=u(x+\alpha)-u(x)}$.

The theorem now follows: given ${\phi\in C^1_0(\mathbb T)}$ and ${\varepsilon>0}$, let ${\tilde\phi=u\circ R_{\alpha}-u}$ be a coboundary for which ${\|\phi-\tilde\phi\|_{C^1}<\varepsilon}$. In particular, ${{\rm Var}(\phi-\tilde\phi)<\varepsilon}$. Then

$\displaystyle \begin{array}{rcl} \left\|\sum_{k=0}^{q_n-1}\phi(x+k\alpha)\right\|_{C^0} &\le&\left\|\displaystyle \sum_{k=0}^{q_n-1}(\phi-\tilde\phi)(x+k\alpha)\right\|_{C^0}\\ &&\\ &&+\left\|\displaystyle \sum_{k=0}^{q_n-1}\tilde\phi(x+k\alpha)\right\|_{C^0}\\ &&\\ &\le&{\rm Var}(\phi-\tilde{\phi})+\left\|u\circ R_\alpha^{q_n}-u\right\|_{C^0}\\ &&\\ &<&\varepsilon+\|u'\|_{C^0}\cdot \|q_n\alpha\|\,, \end{array}$

where in the second equality we used Denjoy-Koksma’s inequality and in the third the mean value inequality. For large ${n}$, the last expression is smaller than ${2\varepsilon}$. $\Box$

Remark 1 Given any ${r\in{\mathbb N}\cup\{\infty\}}$, the trigonometric polynomials are also dense in ${C^r(\mathbb T)}$ and then the same argument above proves that the ${C^r}$-closure of the set

$\displaystyle B(R_\alpha,C^r)=\left\{u\circ R_{\alpha}-u\,;\,u\in C^r(\mathbb T)\right\}$

is equal to ${C^r_0(\mathbb T)}$, the set of ${C^r}$ functions of zero mean.

3. Cylinder flows

We now focus our attention on cylinder flows. As defined in CF0, for any irrational number ${\alpha}$ and function ${\phi:\mathbb T\rightarrow G}$, where ${G=\mathbb R}$ or ${\mathbb Z}$, we define the cylinder flow ${F=F(\alpha,\phi):\mathbb T\times G\rightarrow\mathbb T\times G}$ by

$\displaystyle F(x,y)=(x+\alpha,y+\phi(x)).$

The dynamics of ${F}$ is intimately connected to the Birkhoff sums of ${\phi}$ under ${R_\alpha}$: denoting these by ${S_n\phi}$, we have

$\displaystyle F^n(x,y)=(x+n\alpha,y+S_n\phi(x)).$

By Birkhoff’s theorem, ${S_n\phi/n}$ converges to ${\int_{\mathbb T}\phi dm_{\mathbb T}}$ almost surely. In particular, if ${\phi}$ has drift, i.e. ${\int_{\mathbb T}\phi dm_{\mathbb T}\not=0}$, then

$\displaystyle |S_n\phi(x)|\rightarrow\infty\ \text{ for almost every }x\in\mathbb T$

and so almost every orbit of ${F}$ diverges to infinity. From now on, we assume ${\phi}$ has no drift.

There is a natural invariant measure for ${F}$: it is ${m_{\mathbb T}\times m_G}$, where ${m_{\mathbb T}}$ and ${m_G}$ are the Lebesgue measure of ${\mathbb T}$ and ${G}$, respectively. It is an infinite sigma-finite measure. Measures of this type often behave in very strange ways and fail to comply with rules forming the basis of finite ergodic theory but, regardless this, are worth studying and open the possibility for very different statistical behaviors.

Let ${(X,\mathcal A,\mu,F)}$ be a measure-preserving system: ${(X,\mathcal A,\mu)}$ is a measure space, ${\mu}$ a (possibly infinite) ${\sigma}$-finite measure and ${F}$ is a measurable map on ${X}$ invariant under ${\mu}$.

Definition 8 We say ${\mu}$ is conservative if ${\mu(A)=0}$ whenever ${A\in\mathcal A}$ is such that ${\{F^{-n}A\}_{n\ge 0}}$ are pairwise disjoint. We say ${F}$ is ergodic if it has only trivial invariant sets, that is, if ${\mu(A)=0}$ or ${\mu(X\backslash A)=0}$ whenever ${A}$ is a measurable set invariant under ${F}$.

If ${\mu}$ is finite, then it is necessarily conservative. This is Poincaré’s recurrence theorem (see ERT1). In the infinite measure situation, this is not generally the case. For example, the translation ${x\mapsto x+1}$, ${x\in\mathbb Z}$, is ${m_{\mathbb Z}}$-invariant but non-conservative. Even more weird, it is ergodic! This is a pathological example that indeed is global in the sense of the

Exercise 2 Let ${(X,\mathcal A,\mu,F)}$ be an invertible measure-preserving system. If it is ergodic and non-conservative, then it is isomorphic to the translation ${x\mapsto x+1}$, ${x\in\mathbb Z}$.

In the non-invertible situation, there are many other examples. See the end of this paper of J. Aaronson and T. Meyerovitch. As cylinder flows are invertible and clearly not isomorphic to the translation in ${\mathbb Z}$, conservativity follows if we prove ergodicity.

Coming back to our specific skew product, the mere assumption of zero drift guarantees that ${m_{\mathbb T}\times m_G}$ is conservative, according to the following

Theorem 9 (Atkinson) Let ${(X,\mathcal A,\mu,T)}$ be an ergodic probability measure-preserving system and ${\phi:X\rightarrow{\mathbb R}}$ a measurable function such that ${\int_X\phi d\mu=0}$. Then ${\mu}$-almost every ${x\in X}$ has the following property: for any set ${A\in\mathcal A}$ containing ${x}$ with ${\mu(A)>0}$ and any ${\varepsilon>0}$, the set

$\displaystyle \left\{n>0\,;\,T^nx\in A\text{ and }|S_n\phi(x)|<\varepsilon\right\}$

is infinite.

The proof is in the paper Recurrence of co-cycles and random walks. A classical result of W. Gottschalk and G. Hedlund characterizes transitivity.

Theorem 10 (Gottschalk – Hedlund) Let ${X}$ be a compact metric space, ${T:X\rightarrow X}$ a minimal homeomorphism and ${\phi:X\rightarrow\mathbb R}$ continuous. Then ${F:X\times\mathbb R\rightarrow X\times\mathbb R}$ defined by

$\displaystyle F(x,y)=(Tx,y+\phi(x))$

is transitive if and only if

• ${\int_X\phi d\mu=0}$ and
• ${\sup_{n\ge 1}\|S_n\phi\|_{C^0}=\infty}$.

Another theorem of Gottschalk and Hedlund indeed shows that, under the assumptions of Theorem 10, the two conditions above are equivalent to ${\phi}$ not being a coboundary. See the book Topological dynamics for a proof.

Not much can be obtained in the topological perspective. By this we mean that, when ${\phi:\mathbb T\rightarrow\mathbb R}$ is continuous, ${F}$ is never minimal, whose proof is the content of this paper of P. Le Calvez and J.-C. Yoccoz.

Theorem 11 (Le Calvez – Yoccoz) There are no minimal homeomorphisms on an open annulus.

The next post will discuss ergodicity of cylinder flows. Before finishing the present one, let’s emphasize the relation between ergodicity and irregularity of equidistribution: Denjoy-Koksma’s inequality only gives information on the partial quotient denominators of ${\alpha}$. It might happen that, for different iterates, the Birkhoff sums become unbounded. The analysis of this, at least in the qualitative perspective, can be made via cylinder flows: if ${F}$ is ergodic, then ${S_n\phi}$ must assume arbitrarily large values, because the orbit of a generic point ${(x,y)}$ has to visit sets far from the fiber ${\mathbb T\times\{y\}}$. In particular no such high equidistribution as that in Denjoy-Koksma’s inequality might happen. It is because of this that

“Ergodicity of cylinder flows are related to the irregularity of equidistribution”.

Previous posts: CF0.

## Responses

1. Hi Yuri,

As a side remark, it is worth to mention that there are no good analogues of the Denjoy-Koksma inequality in “higher dimensions”: for instance, in “Appendice 1” of this article here (http://www.ams.org/mathscinet-getitem?mr=1367354), Yoccoz shows that, given $\theta$ an increasing non-bounded function from $\mathbb{R}_+$ to itself, there are a minimal rotation of the two-torus $\mathbb{T}^2$, a real-analytic function $\phi:\mathbb{T}^2\to\mathbb{C}$ of zero mean, and a full (Lebesgue) measure subset $\Omega$ such that, for every $p\in\Omega$, one has $|S_n\phi(p)|\geq n/\theta(n)$.

Best, Matheus

2. Hi,

A remark: you should assume that phi is not a coboundary in theorem 10. As otherwise, the map F is bounded on the fibers and you do not have transitivity… Moreover, it enlights your last remark as Birkhoff sum of a coboundary is definitely regular!

Theorem 11 is definitely geometric. I suspect that there exists a theorem which says that a skew-product with fiber Z is never minimal. Do you know if this is true ?

• Hi Vincent,

You’re absolutely right. As it is, this is just equivalent to conservativity. I forgot to add that
1. the basis transformation $T$ is minimal, and
2. the Birkhoff sums are unbounded:
$\sup_{n\ge 1}\|S_n\phi\|_{C^0}=\infty$.

Another theorem of Gottschalk and Hedlund guarantees that $(1+2)\iff\phi$ is not a coboundary, as you remarked.

With respect to your second remark, I don’t know and, up to my knowledge, there aren’t results in this direction (but I have to check with more care).

Thanks, Yuri.

3. Hi Matheus,

what is the precise (original) reference for Theorem (Remark ?) 7 ?
Is it due to Koksma ?

This is closely related to my work with M. Triestino http://arxiv.org/abs/1207.1481, where we extend this to any C^2 circle diffeomorphism of irrational rotation number as the base dynamics. Indeed, if you read our proof carefully in the context of an irrational rotation, it is the very same as yours…

Best,

Andrés

• Hi Andrés,

as far as I know, the original reference for “Theorem 7” in this post is Michel Herman’s article “Sur la conjugaison différentiable des difféomorphismes du cercle à des rotations”.

Indeed, by looking in this article at the proposition at the paragraph (3.2) at page 74 (after the proof of Denjoy-Koksma inequality), you can find a slightly more general version of “Theorem 7” based on the modulus of continuity of $\phi$.

Best,

Matheus

4. Nice, many thanks Matheus.

Here is a closely related question for which affirmative answer would trivialize everything (though I’m afraid the answer should be negative): assume you have a C^2 circle diffeomorphism of irrational rotation number, as well as a C^1 test function u on the circle. Is it true that the derivatives of the Birkhoff means of u at the approximating times of the rotation number uniformly converge to zero ? Is this true at least for a subsequence of these times ?

All the best,

Andrés

• Hi Andrés,

I’ve not thought seriously about your question, but I think that the following remark gives some reason to think that a positive answer in some cases is not completely impossible.

Let me oversimplify the situation by considering $f$ a smooth circle diffeomorphism with Diophantine rotation number and let the test function $u$ be a coboundary of a $C^1$ function $\phi$, i.e., $u = \phi\circ f - \phi$. Then, we know that $f$ is smooth ($C^1$) conjugated to the rotation $R_{\alpha}$, i.e., $f = h\circ R_{\alpha}\circ h^{-1}$ with $h$ a $C^1$ diffeomorphism. On the other hand, the $n$th Birkhoff sum of $u$ is $u\circ f^n - u$, and hence its first derivative is $u'(f^n)\cdot (f^n)' - u' = u'(f^n)\cdot ((f^n)'-1)+(u'(f^n)-u')$. Now, we note that $f^n=h\circ R_{n\alpha}\circ h^{-1}$, so that $(f^n)' = h'(R_{n\alpha}\circ h^{-1})\cdot (h^{-1})'$ is close to $1$ and $f^n$ is close to the identity when $n=q_k$. In particular, in this very special case ($f$ smooth with Diophantine rotation number and $u$ a coboundary of a $C^1$ function, the question has a positive answer (if I’m not mistaken).

Best,

Matheus

5. Hi again Matheus,

yes, actually, if you assume only that the differ is C^1 conjugate to the rotation one can still show the desired convergence without the assumption on u. The problem are Liouvillean rotation numbers. I’m afraid that this shouldn’t be true in such case. Anyway, some kind of convergence can be established (ion L^1, if I remember well).

Best,

Andrés

• Hi Andrés,

Ah, ok, now I see your point! I’ll try to keep your question in mind and if I find something interesting to tell you, I will add it as a reply here.

Best,

Matheus

6. Hi Matheus. Very quickly just to say that the answer to the question above is negative. This arose in a discussion with S. Crovisier who carried out the construction of a counter-example using the Anosov-Katok method (counter-examples are indeed generic…)

7. Hi,

If I am not mistaken, the argument in the proof of Denjoy-Koksma does not work: you can not necessarily find a value y such that the sum is zero (zero mean does not imply the existence of a value 0… unless you assume that the function is continuous). The proof is still valid for continuous with bounded variation. It is not hard to fix to make it work for general BV functions but need some Abel summation.

Vincent

• Hi Vincent,

You’re absolutely right. I will ask the owner of the post (Yuri) to make the necessary changes.

Best,

Matheus

8. Hi Vincent,

Thanks for the observation. I just posted a different proof (without the use of Abel summation) which follows Herman’s paper Sur la conjugaison différentiable des difféomorphismes du cercle à des rotations. I hope this time the proof is correct.

Best, Yuri.

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