Fast and slow points of Birkhoff sums
Fr\'ed\'eric Bayart (LMBP), Zoltan Buczolich (ELTE), Yanick Heurteaux, (LMBP)

TL;DR
This paper studies the growth behavior of Birkhoff sums for continuous zero-mean functions on the circle, revealing how typicality assumptions influence the growth rates in dynamical systems.
Contribution
It provides a detailed analysis of the growth rates of Birkhoff sums under different notions of typicality, extending the understanding to more general contexts.
Findings
Growth rates depend on the interpretation of 'typical'
Analysis applies to continuous functions with zero mean
Results extend to broader dynamical systems contexts
Abstract
We investigate the growth rate of the Birkhoff sums , where is a continuous function with zero mean defined on the unit circle and is a "typical" element of . The answer depends on the meaning given to the word "typical". Part of the work will be done in a more general context.
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Fast and slow points of Birkhoff sums
Frédéric Bayart, Zoltán Buczolich, Yanick Heurteaux
Université Clermont Auvergne, LMBP, UMR 6620 - CNRS, Campus des Cézeaux, 3 place Vasarely, TSA 60026, CS 60026 F-63178 Aubière Cedex, France.
Department of Analysis
ELTE Eötvös Loránd University
Pázmány Péter Sétány 1/c
1117 Budapest, Hungary
[email protected], [email protected], [email protected]
Abstract.
We investigate the growth rate of the Birkhoff sums , where is a continuous function with zero mean defined on the unit circle and is a “typical” element of . The answer depends on the meaning given to the word “typical”. Part of the work will be done in a more general context.
The first and the third author were partially supported by the grant ANR-17-CE40-0021 of the French National Research Agency ANR (project Front). The second author was supported by the Hungarian National Research, Development and Innovation Office–NKFIH, Grant 124003. He also thanks the Rényi Institute where he was a visiting researcher during the academic year 2017-18.
Mathematics Subject Classification: Primary : 37A05, Secondary : 11K55, 28A78, 60F15.
Keywords: Birkhoff sum, typical/generic properties, group rotation, coboundary, law of iterated logarithm.
1. Introduction
Let be the unit circle and let be irrational. Denote by , the set of continuous functions on with zero mean, and by the -th Birkhoff sum, . The rotation defines a uniquely ergodic transformation on with respect to the (normalized) Lebesgue measure . Hence for all we know that for all . The main purpose of this paper is to investigate the typical growth of .
There are several ways to understand this problem. We can fix (resp. ) and ask for the behaviour of for in a generic subset of and for a typical (resp. for a typical ). We can also consider it as a problem of two variables and ask for the behaviour of for in a generic subset of and for a typical . There are also several ways to understand the word “typical”. We can look for a residual set of the parameter space or for a set of full Lebesgue measure.
We shall try to put this in a general context. If we fix , then we consider the Birkhoff sums associated to a uniquely ergodic transformation on the compact metric space . Hence, let us fix an infinite compact metric space and an invertible continuous map such that is uniquely ergodic. Let be the ergodic measure, which is regular and continuous. We will also assume that it has full support (equivalenty, that all orbits of are dense). For and , the Birkhoff sum is now defined by . Using with for , let us define
[TABLE]
The set has already been studied by several authors. In particular, it was shown by Krengel [7] (when ) and later by Liardet and Volný [9] that, for all functions in a residual subset of , . We complete this result by showing that is also residual.
Theorem 1.1**.**
Suppose that satisfies . There exists a residual set such that for any , is residual and of full -measure in .
If we allow to vary in our initial problem, then the natural framework now is that of topological groups. Hence, we fix a compact and connected metric abelian group . By Corollary 4.4 in [8, Chapter 4], is a monothetic group, that is possesses a dense cyclic subgroup. Let be the Haar measure on . It is invariant under each translation, or group rotation . We define as the set of such that is ergodic. By well-known results of ergodic theory, belongs to if and only if is dense in ; in this case is uniquely ergodic, only the Haar measure is invariant with respect to . Moreover, is always nonempty, it is dense and its Haar measure is equal to 1 (see Theorem 4.5 in [8, Chapter 4]).
Contrary to what happens in Theorem 1.1, the growth of for a typical is not the same from the topological and from the probabilistic points of view. For the last one, the typical growth of has order .
Theorem 1.2**.**
- (i)
For all and all ,
[TABLE] 2. (ii)
There exists a residual subset such that, for all ,
[TABLE]
From a topological point of view, the typical growth of has order . Indeed, for with , let us introduce
[TABLE]
Theorem 1.3**.**
Suppose that satisfies . There exists a residual set such that for any we have .
We remark that, by the Kuratowski-Ulam theorem, Theorem 1.3 implies that there exists a residual set such that, for every , the set is residual in .
The last possibility is to fix and allow to vary. Without loss of generality, we may assume that . Again, topologically speaking, the typical growth of is not better than .
Corollary 1.4**.**
Suppose that satisfies . There exists a residual set such that for any , the set is residual in .
We finally come back to irrational rotations where we would like to get more precise statements. Let us fix and set
[TABLE]
When , , we simply denote by the set . We already know by the results mentioned before Theorem 1.1 that for in a residual subset of , where is the Lebesgue measure on . It turns out that a much stronger result is true: generically, these sets have zero Hausdorff dimension!
Theorem 1.5**.**
For any with , there exists a residual subset of such that, for any ,
We then do a similar study for Hölder functions , . Recall that a function belongs to if it has zero mean and if there exists a constant such that, for all ,
[TABLE]
The infimum of such constants is denoted by .
For a function , we have better bounds on depending on and on the arithmetical properties of . Indeed, it is known (see [8, Chapter 2, Theorem 5.4]) that where is the discrepancy of the sequence defined by
[TABLE]
For instance, if has type 1 (for example, if is an irrational algebraic number), using the well-known estimates of the discrepancy, we get that for all . In other words, for all , . We investigate the case and we show that the Hausdorff dimension of cannot always be large.
Theorem 1.6**.**
Let . There exists such that, for all ,
[TABLE]
This theorem is in stark contrast with Theorem 4.1 in [5]. In this last paper, a similar study of fast Birkhoff averages of subshifts is done. In this case, the sets which correspond to always have maximal dimension.
2. Useful lemmas
In this section, we provide lemmas which will be used several times for the proof of our main theorems. The first one allows to approximate step functions by continuous functions. In the statement of the theorem we use the standard notation for the function which equals if and equals 0 if not.
Lemma 2.1**.**
Let be a compact metric space, let be a continuous Borel probability measure on . Let be a step function such that and . Then there exists such that and except on a set of measure at most .
Proof.
Let be very small and be the finite set . We can write where . Since the measure is regular, we can find compact sets and open sets such that
[TABLE]
[TABLE]
By Urysohn’s lemma, one may find functions such that
[TABLE]
We then set . It is clear that
[TABLE]
Therefore,
[TABLE]
If , we now have and
[TABLE]
The function is continuous but is not necessarily in . Nevertheless, we observe that
[TABLE]
and we can modify to obtain a zero mean. Let and be such that and let with on the closed ball , outside and . We set
[TABLE]
Then , except on a set of measure at most and
[TABLE]
Choosing sufficiently small then gives the result. ∎
Our second lemma is a way to construct continuous functions in with large Birkhoff sums on large subsets. We give it in our general context of a uniquely ergodic transformation on an infinite compact metric space with non-atomic ergodic measure . As usual, satisfies . We denote by the complement of the set .
Lemma 2.2**.**
Let , , . Then there exist , and a compact set such that , and
[TABLE]
Proof.
Set We begin by fixing , any integer greater than , and such that . Let to be fixed later. We then consider a Rokhlin tower associated to , and (see for instance [3]). Namely, we consider such that the sets , , are pairwise disjoint and . We then consider a function equal to on , equal to on and equal to zero elsewhere.
We set
[TABLE]
Then, for any , for any , for any ,
[TABLE]
It follows that . In the same way, for any , for any , .
Finally, for any , for any ,
[TABLE]
Moreover,
[TABLE]
provided is large enough.
Thanks to Lemma 2.1, we approximate by a continuous function with and except on a set of measure , with . Fix . Then except if . Let . Then . Moreover, for all and all . Clearly, . We conclude by taking for the closure of . ∎
3. Fast and slow points of Birkhoff sums - I
In this section, we prove Theorems 1.1 and 1.3. Their proofs share many similarities and depend heavily on Lemma 2.2 applied in suitable situations. We will also need that if is a uniquely ergodic transformation on , then the set of -coboundaries for , namely the set of functions for some , is dense in (see for instance [9, Lemma 1]). It is convenient to work with a coboundary since its Birkhoff sums are uniformly bounded.
Proof of Theorem 1.1.
Let be a dense sequence of coboundaries in and let be such that Let , and be given by Lemma 2.2 for , , , . We set and we observe that, for ,
[TABLE]
Since is compact and is continuous, we can choose and an open set containing such that, for any , for any ,
[TABLE]
Let which is a residual set in and pick . There exists an increasing sequence going to such that for all . We set . Since the set has full measure. Moreover, since has full support and for all , is also residual in . Finally if belongs to , then (1) is true for infinitely many , which shows Theorem 1.1. ∎
In the next proof is replaced by the compact connected metric abelian group and we consider uniquely ergodic translations . We recall that for these translations, all non-constant characters are -coboundaries: they can be written as , where .
Proof of Theorem 1.3.
Since is compact we can choose a sequence of trigonometric polynomials which is dense in (see [10, Section 1.5.2]). Let , that is is ergodic. Since is a -coboundary for all , , there exists such that
[TABLE]
Let , and be given by Lemma 2.2 for , , , , . Set and observe that, for , ,
[TABLE]
Since is compact in and is continuous, we can choose and an open set such that and, for any ,
[TABLE]
We now observe that is dense in for any . Hence, is a residual subset of and any satisfies that belongs to since (2) is true for infinitely many integers . ∎
Proof of Corollary 1.4.
This corollary follows easily from Theorem 1.3 and from the Kuratowski-Ulam theorem. Indeed, we know that there exist a residual set and such that, for all , is residual. Now, setting , for any , is residual. ∎
4. Fast and slow points of Birkhoff sums - II
We turn to the proof of Theorem 1.2. Its first part heavily depends on the following Menshov-Rademacher inequality (see for instance [2, Chapter 4]).
Lemma 4.1**.**
Let be a sequence of orthonormal random variables and be a sequence of real numbers. Then
[TABLE]
Proof of Theorem 1.2 part (i).
Recall that . Without loss of generality, we suppose and we consider as a random variable on the probability space . Next we show that is an orthonormal sequence. Indeed, let be the Fourier expansion of . Then, for ,
[TABLE]
Now, is zero provided and is equal to 1 otherwise. Moreover, let us fix and set , . Then except if , namely except if . If , using that is torsion-free since is compact and connected, this can only happen if . Therefore, we have shown that
[TABLE]
Applying Lemma 4.1 with yields
[TABLE]
Let and for ,
[TABLE]
Using Markov’s inequality and (3), we get
[TABLE]
Since , the Borel-Cantelli lemma implies that and the conclusion follows. ∎
Remark 4.2**.**
In fact, the same proof shows that, for any ,
[TABLE]
To prove the second part of Theorem 1.2, we shall use both a Baire category and a probabilistic argument. The probabilistic part is based on the the following lemma, which is a consequence of the proof of the law of the iterated logarithm done in [1] (the important point here is that we need a choice of which does not depend on the particular choice of the sequence).
We recall that a random variable has a Rademacher distribution if .
Lemma 4.3**.**
Let and . There exists such that, for any sequence of independent Rademacher variables defined on the same probability space ,
[TABLE]
The following lemma is the key point of our proof.
Lemma 4.4**.**
Let , and . There exist , and with , and
[TABLE]
Proof.
Without loss of generality, we may assume that . Lemma 4.3 gives us a value of associated to and . We then consider a sequence of independent Rademacher variables defined on the same probability space . We select a neighbourhood of so that, setting
[TABLE]
we have . This is possible since, denoting by a basis of neighbourhoods of [math] in , we have
[TABLE]
By compactness of , is contained in a finite union . We set and, for , . The sets provide a Borelian partition of .
We then split each into a disjoint sum with . For define by . We finally put
[TABLE]
so that
[TABLE]
Let us fix . For all and all , there exists exactly one integer , that we will denote by , such that . Hence, for and ,
[TABLE]
Moreover, for , the integers and are different: otherwise, would belong to .
Applying Lemma 4.3 to the sequence \big{(}X_{k(j,u,x)}\varphi_{k(j,u,x)}(x+ju)\big{)}_{0\leq j\leq N-1} which is a sequence of independent Rademacher variables, we get the existence of such that and
[TABLE]
Hence
[TABLE]
Keeping in mind that holds as well, by Fubini’s theorem we can select and fix such that
[TABLE]
Given , according to Lemma 2.1, the function can be approximated by a continuous function such that and which coincides with except in a set of measure less than . It follows that for every and for any , except in a set of measure less than . Finally, if is sufficiently small, inequality (4) is still satisfied if we replace by . ∎
Proof of Theorem 1.2, part (ii).
Let be a sequence of trigonometric polynomials dense in . For all and all , since is a -coboundary for , we know that . We then find with and such that, for all , . We apply Lemma 4.4 with , and . We get a function , an integer and a set . We define and so that . The way we constructed all these objects ensures that, for any ,
[TABLE]
This yields the existence of a such that, for any and any ,
[TABLE]
We finally consider the residual set and we pick . There exists an increasing sequence such that . Let which has full measure and pick . There exists a subsequence of such that for all . We then have
[TABLE]
which allows us to conclude. ∎
Remark 4.5**.**
The proof gives slightly more than announced: there exists a residual set such that, for all and all ,
[TABLE]
5. Fast and slow points for irrational rotations on the circle
Throughout this section, we fix .
5.1. A partition of
To get an estimate of the Hausdorff dimension of , which is more precise than the result already obtained on its measure, we will need a refinement of Rokhlin towers specific to irrational rotations. We shall use the following system of partitions of associated to the irrational rotation , as it is described for instance in [11, Lecture 9, Theorem 1]. Let be the -th convergent of in its continued fraction expansion. Define
[TABLE]
Denote also \Delta_{j}^{(n)}=R_{\alpha}^{j}\big{(}\Delta_{0}^{(n)}\big{)}. For any , the intervals , and , , are pairwise disjoint and their union is the whole . We shall denote by the length of . It is well known that
[TABLE]
5.2. Continuous functions
The main step towards the proof of Theorem 1.5 is the following lemma which improves partly Lemma 2.2.
Lemma 5.1**.**
Let , , , and . Then there exist with , a compact set , and an integer such that
[TABLE]
[TABLE]
Proof.
Let be such that . Let also be a large integer and consider the partition of described in Section 5.1:
[TABLE]
where the convergents of are . Since it will be easier to deal with even numbers we put , which is the greatest even integer less than . Hence and are even. We define a continuous function with zero mean such that
- •
on , and on , , except on two very small intervals of size where is affine to ensure that vanishes at the boundary of and .
- •
on , and on , , except on two very small intervals of size where is affine to ensure that vanishes at the boundary of and .
- •
if we set .
We set (resp. ) the (largest) subinterval of (resp. ) such that and we let
[TABLE]
If belongs to , then for all and . Therefore, we have . On the other hand, is the union of at most
- •
intervals of size ;
- •
intervals of size ;
- •
intervals of size .
Hence we have
[TABLE]
if we choose sufficiently large and then sufficiently small. ∎
Proof of Theorem 1.5.
We mimic the proof of Theorem 1.1. Recall that
[TABLE]
Let be a sequence of coboundaries which is dense in . Then for any , there exists such that . Let , and be given by Lemma 5.1 for , and . We set and observe that, for ,
[TABLE]
There exists such that, for any and any ,
[TABLE]
Since the sequence is dense in , is a residual subset of . Pick . There exists an increasing sequence such that . We set and observe that, for any ,
[TABLE]
Moreover, . For any , the properties of the sets ensure that . Since , we conclude that and therefore . ∎
5.3. Hölder functions
We now modify the previous construction to adapt it to Hölder continuous functions.
Lemma 5.2**.**
Let , , with , , , , . There exist a continuous function with , , an integer , and a compact set such that
[TABLE]
[TABLE]
Proof.
The construction of will be more or less difficult depending on the arithmetical properties of . Let be the th convergent of in its continued fraction expansion. For each , there exists such that . We define
[TABLE]
We then fix such that , and
[TABLE]
If moreover , we also require that .
Let be a large integer and consider the partition of described in Section 5.1:
[TABLE]
Again for ease of notation we suppose that and are even; if not, a modification similar to the one used in the proof of Lemma 5.1 can be used.
First case: . Then, for large enough, for some fixed . We fix such an and we then define as follows:
- •
on , , is equal to on , equal to on .
- •
On , , is equal to on , equal to on .
- •
is equal to [math] otherwise.
It is then clear that , and . Recalling that for we then set
[TABLE]
[TABLE]
[TABLE]
[TABLE]
Observe that if , then and that , . It follows that, for with constants which do not depend on and may change from line to line
[TABLE]
provided is large enough. Thus (7) is satisfied with and for large values of . Moreover, is contained in the union of
- •
intervals of size (the intervals which are not considered);
- •
intervals of size (the extreme parts of the intervals );
- •
intervals of size (the intervals of the following generation ).
Hence, for large enough,
[TABLE]
Since , and , (8) is also satisfied provided is large enough.
Second case: . This time, the intervals coming from are too long to be neglected with respect to the -measure. By the choice of , we know that there exist integers as large as we want such that
[TABLE]
we will fix such an later. We keep the same values for , , and and the same definition for on as in the first case. On the other hand, we define on by imposing on , on if and on , on if . We then set
[TABLE]
[TABLE]
[TABLE]
[TABLE]
and . Remember that . We can still use (5.3) and can deduce analogously for any ,
[TABLE]
provided is large enough. From now on we can fix a sufficiently large . The set consists of at most
- •
intervals of size ;
- •
intervals of size ;
- •
intervals of size ;
- •
intervals of size .
Thus,
[TABLE]
By using (9) and (11) we conclude exactly as before since
[TABLE]
and
[TABLE]
∎
Proof of Theorem 1.6.
We will prove slightly more than announced. Let be the closed subspace of defined by
[TABLE]
The space , equipped with the norm of the uniform convergence is now again a separable complete metric space. We will prove that, for all functions in a residual subset of , for all , . Since provided , it is sufficient to prove this inequality for belonging to a sequence which is dense in . Now, the countable intersection of residual sets remaining residual, we just have to prove that, for a fixed , all functions in a residual subset of satisfy .
Let be a sequence of -coboundaries which is dense in and with . For any , there exists such that . Let , and be given by Lemma 5.2 with , , and . We set so that , is dense in and, for any in the compact set , there exists with
[TABLE]
We can then find such that, for all and all , there exists with
[TABLE]
We set which is a residual subset of . Pick . There exists an increasing sequence such that . We set and observe that, for any ,
[TABLE]
so that . Now the construction of the sets ensures that
[TABLE]
∎
Question 5.3**.**
Is the value optimal? In particular, it does not depend on the type of , which may look surprizing.
6. Miscellaneous remarks
6.1. Open questions
Our study suggests further questions. The first one is related to Corollary 1.4.
Question 6.1**.**
Does there exist such that
- (i)
for all , for all ,
[TABLE] 2. (ii)
for all , there exists a residual subset of such that, for all ,
[TABLE]
It can be shown that works for (ii). Indeed, Lemma 4.4 and Fubini’s theorem imply that, for all , all and all , there exist , , and with , and for . Translating if necessary, we may assume that . We then conclude exactly as in the proof of Theorem 1.2.
Second, Theorem 1.5 improves Theorem 1.1 for rotations of the circle by replacing nowhere dense sets with the more precise notion of sets with zero Hausdorff dimension. There are also enhancements of meager sets, for instance -porous sets (see [12])
Question 6.2**.**
Does there exist a residual subset of such that, for any , is -porous?
In the spirit of Theorem 1.2, the next step would be to perform a multifractal analysis of the exceptional sets. Precisely, let and . Let us set
[TABLE]
These sets have Lebesgue measure zero.
Question 6.3**.**
Can we majorize the Hausdorff dimension of ?
We could also replace everywhere the by .
Question 6.4**.**
Let with . Does there exist such that is residual? has full measure?
6.2. Other sums
The study of is a particular case of the series . In the particular case this series, also called the one-sided ergodic Hilbert transform, was thoroughly investigated in [4].
In [4], the authors show that for any non-polynomial function with values in , there exists a residual set of irrational numbers depending on such that, for every ,
[TABLE]
for almost every and they ask if this holds for every (they show that this is the case if when ). We provide a counterexample.
Example 6.5**.**
Let and be defined by its Fourier coefficients , for , for . A small computation shows that
[TABLE]
We shall prove that the one-sided ergodic Hilbert transform of is bounded at . Indeed, setting
[TABLE]
it is easy to show that
[TABLE]
Now, it is well-known that the imaginary part of , namely is uniformly bounded in and (see e.g. [6, p.4]).
Question 6.6**.**
Can we investigate, in the spirit of this paper and of [4], the case , with ?
6.3. Coboundaries in
The natural norm in is given by
[TABLE]
One may wonder whether, in Theorem 1.6, we have residuality in instead of in . A natural way to do that would be to prove that the coboundaries are dense in . This is not the case, which shows again that is a weird space.
In we denote the ball of radius centered at by , that is if and only if . We shall prove the following precise statement.
Theorem 6.7**.**
For any for any there exists such that for any the function is not a (and hence not a )-coboundary, that is there is no such that . Hence -coboundaries are not dense in .
Proof.
By induction we select , , with the following properties. If we let \displaystyle h_{k}=\Big{(}\frac{k}{n_{k+1}}\Big{)}^{1/\xi} then the intervals
[TABLE]
(all these intervals are considered on ),
[TABLE]
[TABLE]
For this property we can use that is a union of intervals, which by (15) are of total measure less than and the sequence is uniformly distributed on , especially if we suppose that the s are denominators of suitable convergents of and recall Subsection 5.1. We also suppose that is maximal possible, by this we mean that if and then
[TABLE]
By the definition of and (16) we have
[TABLE]
Next we define . On an interval , , we define in the following way: and
[TABLE]
otherwise is linear on each with . If then we set .
It is obvious that with .
Suppose that and proceeding towards a contradiction suppose that with a Then there exists such that .
Clearly, for any and any , we have
[TABLE]
We will prove in (28) and (29) that for any function , its Birkhoff sums are not bounded and this will provide a contradiction.
Suppose is fixed. Since we have for any
[TABLE]
This and (19) imply that for
[TABLE]
Next we consider the cases when , . Then (17) applies. Suppose first that there exists , , such that . The construction of on ensures that
[TABLE]
provided was choosen sufficiently large.
If then either , or . In this latter case with and we can suppose by the inductive definition of the that . Thus
[TABLE]
Similarly if we can suppose that
[TABLE]
In case one of belongs to a , , and the other is not an element of any such interval then at some in and a combination of (23) and (24), or (25) is applicable.
Summarizing, we have finally shown that for all ,
[TABLE]
Since by (21) and (26) we obtain
[TABLE]
We claim that either
[TABLE]
(see (18) as well), or
[TABLE]
It is clear that for large this will contradict (20).
Next suppose that the negation of (28) and the negation of (29) hold.
This implies
[TABLE]
[TABLE]
On the other hand, by (16) and (27)
[TABLE]
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