No temporal distributional limit theorem for a.e. irrational translation

D M I T R Y D O L G O P Y AT O M R I S A R IG

N O T E M P O R A L DIS T R I B U T IO N A L L I M I T T H E O R E M F O R A. E.

IR R AT IO N A L T R A N S L AT IO N

U N E T R A N S L AT IO N IR R AT IO N N E L L E T Y PIQ U E N E S AT IS F A I T PA S D E T H É O R È M E L I M I T E T E M P O R E L

Abstract. — Bromberg and Ulcigrai constructed piecewise smooth functions on the circle such that the set ofαfor which the sumPn−1

k=0f(x+kα mod 1) satisfies a temporal distributional limit theorem along the orbit of a.e.xhas Hausdorff dimension one. We show that the Lebesgue measure of this set is equal to zero.

Résumé. — Bromberg et Ulcigrai ont construit des fonctions lisses par morceaux sur le cercle pour lesquelles l’ensemble desαtels que la sommePn−1

k=0f(x+kα mod 1) satisfait un théorème limite temporel le long de l’orbite de presque toutxest un ensemble de dimension de Hausdorff 1. Nous montrons que cet ensemble est de mesure nulle.

1. Introduction and statement of main result

1.1. Background

SupposeT :X →Xis a map,f :X →Ris a function, andx₀ ∈Xis afixed initial condition. We say that theT–ergodic sums Sn =f(x₀) +f(T x₀) +· · ·+f(Tⁿ⁻¹x₀)

2010Mathematics Subject Classification:37D25, 37D35.

DOI:https://doi.org/10.5802/ahl.4

(*) This work was partially supported by the ISF grants 199/14, 1149/18 and by the BSF grant 2016105. The authors thank Adam Kanigowski for simplifying the statement and the proof of Lemma 3.2.

satisfy a temporal distributional limit theorem (TDLT) on the orbit of x₀, if there exists a non-constant real valued random variable Y, centering constants AN ∈ R and scaling constantsBN → ∞s.t.

(1.1) S_n−A_N

B_N −−−→

N→∞ Y in distribution,

whennis sampled uniformly from {1, . . . , N}andx₀ is fixed. Equivalently, for every Borel set E ⊂Rs.t. P(Y ∈∂E) = 0,

1 N Card

16n6N : S_n−A_N

B_N ∈E

−−−→N→∞ P(Y ∈E).

We allow and expect A_N, B_N, Y to depend on T, f, x₀.

Such limit theorems have been discovered for several zero entropy uniquely ergodic transformations, including systems where the more traditional spatial limit theorems, withx₀ is sampled from a measure onX, fail [Bec10, Bec11, ADDS15, DS17b, PS17, DS18]. Of particular interest are TDLT for

R_α : [0,1]→[0,1], R_α(x) =x+α mod 1, f_β(x) := 1_[0,β)(x)−β,

because the R_α–ergodic sums of f_β along the orbit of x represent the discrepancy of the sequence x+nα mod 1 with respect to [0, β) [Sch78, CK76, Bec10]. Another source of interest is the connection to the “deterministic random walk” [AK82, ADDS15].

The validity of the TDLT forR_α andf_β depends on the diophantine properties ofα andβ. Recall thatα∈(0,1) is badly approximable if for some c >0,|qα−p|>c/|q|

for all irreducible fractions p/q. Equivalently, the digits in the continued fraction expansion ofα are bounded [Khi63]. Say that β ∈(0,1) is badly approximable with respect to α if for some C > 0, |qα−β−p| > C/|q| for all p, q ∈ Z, q 6= 0. If α is badly approximable then everyβ ∈Q∩(0,1) is badly approximable with respect to α. The recent paper [BU18] shows:

Theorem 1.1 (Bromberg–Ulcigrai [BU18]). — Suppose α is badly approximable and β is badly approximable with respect to α, e.g. β ∈ Q∩(0,1). Then the R_α- ergodic sums of f_β satisfy a temporal distributional limit theorem with Gaussian limit on the orbit of every initial condition.

The set of badly approximableαhas Hausdorff dimension one [Jar29], but Lebesgue measure zero [Khi24]. This leads to the following question:Is there a β s.t. the R_α– ergodic sums of f_β satisfy a temporal distributional limit theorem for a.e. α and a.e.

initial condition?

In this paper we answer this question negatively.

1.2. Main result

To state our result in its most general form, we need the following terminology.

LetT :=R/Z. We say that f :T→R is piecewise smooth if there exists a finite set S ⊂ T s.t. f is continuously differentiable on T\S and ∃ ψ : T → R with bounded variation s.t. f⁰ = ψ on T\S. For example: f_β(x) = 1_[0,β)(x)−β (take S={0, β},ψ ≡0). We show:

Theorem 1.2. — Letf be a piecewise smooth function of zero mean. Then there is a set of full measureE ⊂ T×T s.t. if (α, x)∈ E then the Rα–ergodic sums of f do not satisfy a TDLT on the orbit of x.

The condition^R_Tf = 0 is necessary: By Weyl’s equidistribution theorem, for every α 6∈ Q, f Riemann integrable s.t. ^R

Tf = 1, and x₀ ∈ T, S_n/N −−−→^dist

N→∞ U[0,1] as n∼U(1, . . . , N). See Section 1.4 for the notation.

This paper has a companion [DS17a] which gives a different proof of Theorem 1.2, in the special case f(x) ={x} − ¹₂. Unlike the proof given below, [DS17a] does not identify the set ofαwhere the TDLT fails, but it does give more information on the different scaling limits for the distributions of S_n, n ∼U(1, . . . , N_k) along different subsequences N_k → ∞. [DS17a] also shows that if we randomize both n and α by sampling (n, α) uniformly from {1, . . . , N} ×T, then S_n− _N^{1 PN}_k=1S_k/√

lnN converges in distribution to the Cauchy distribution.

The methods of [DS17a] are specific for f(x) ={x} −¹₂, and we do not know how to apply them to other functions such asf_β(x) = 1_[0,β)(x)−β.

1.3. The structure of the proof Supposef is piecewise smooth and has mean zero.

We shall see below that iff is continuous, then for a.e. α, f is an R_α-coboundary, therefore S_n are bounded, hence (1.1) cannot hold with B_N → ∞, Y non-constant.

We remark that (1.1) does hold withBN ≡1,AN =f(x0),Y = distribution of minus the transfer function, but this is not a TDLT since no actual scaling is involved.

The heart of the proof is to show that if f is discontinuous, then for a.e. α, the temporal distributions of the ergodic sums have different asymptotic scaling behavior on different subsequences. The proof of this has three independent parts:

(1) A reduction to the case f(x) = ^Pd_m=1b_mh(x+β_m),h(x) := {x} −¹₂.

(2) A proof that if N ⊂ N has positive lower density, then there exists M > 1 s.t. the following set has full Lebesgue measure in (0,1):

A(N, M) :=

(

α ∈(0,1) : ∃n_k ↑ ∞, r_k6M s.t. for all k:

r_kq_nk ∈ N, a_nk+1/(a₁+· · ·+a_nk)→ ∞

)

.

Here a_n andq_n are the partial quotients and principal denominators of α, see Section 3.1.

(3) Construction of N =N(b₁, . . . , bd;β1, . . . , βd)⊆N with positive density, s.t.

for everyα∈ A(N, M) and a.e.x, one can analyze the temporal distributions of the Birkhoff sums of ^Pd_m=1b_mh(x+β_m).

1.4. Notation

n ∼U(1, . . . , N) means that n is a random variable taking values in {1, . . . , N}, each with probability _N¹. U[a, b] is the uniform distribution on [a, b]. Lebesgue’s measure is denoted by mes. N = {1,2,3, . . .} and N0 = N∪ {0}. If x ∈ R, then kxk:= dist(x,Z) and {x} is the unique number in [0,1) s.t.x ∈ {x}+Z. Card(·) is the cardinality. Ifε >0, then a=b±ε means that |a−b|6ε.

2. Reduction to the case f (x) =

b

h(x + β

)

Let h(x) = {x} − ¹₂, and let G denote the collection of all non-identically zero functions of the formf(x) = ^Pd_m=1b_mh(x+β_m), whered∈N, b_i, β_i ∈R.We explain how to reduce the proof of Theorem 1.2 from the case of a general piecewise smooth f(x) to the casef ∈ G.

The following proposition was proved in [DS17a]. Let C(T) denote the space of continuous real-valued functions on T with the sup norm.

Proposition 2.1. — Iff(t) is differentiable on T\ {β₁, . . . , β_d} and f⁰ extends to a function with bounded variation on T, then there ared ∈N0,b₁, . . . , b_d∈R s.t.

for a.e. α∈T there isϕα ∈C(T) s.t.

f(x) =

d

X

i=1

b_ih(x+β_i) +

Z

T

f(t) dt+ϕ_α(x)−ϕ_α(x+α) (x6=β₁, . . . , β_d).

The following proposition was proved in [DS17b]. Let (Ω,B, µ) be a probability space, and let T : Ω→Ω be a probability preserving map.

Proposition 2.2. — Suppose f = g +ϕ−ϕ◦T µ-a.e. with f, g, ϕ : Ω → R measurable. If the ergodic sums of g satisfy a TDLT along the orbit of a.e.x, then so do the ergodic sums of f.

These results show that if Theorem 1.2 holds for every f ∈ G, then Theorem 1.2 holds for anydiscontinuous piecewise smooth function with zero mean. As for con- tinuous piecewise smooth functions with zero mean, these are R_α-cohomologous to g ≡ 0 for a.e. α because the b_i in Proposition 2.1 must all vanish. Since the zero function does not satisfy the TDLT, continuous piecewise smooth functions do not satisfy a TDLT.

3. The set A has full measure

3.1. Statement and plan of proof

Let α be an irrational number, with continued fraction expansion denoted by [a₀;a₁, a₂, a₃, . . .] :=a₀ + 1

a₁ +. . ., a₀ ∈ Z, a_i ∈N (i> 1). We call a_n the quotients of α. Let p_n/q_n denote the principal convergents of α, determined recursively by

q_n+1 =a_n+1q_n+qn−1, p_n+1 =a_n+1p_n+pn−1

andp0 =a0, q0 = 1;p1 = 1+a₁a0,q1 =a1. We callqntheprincipal denominatorsand a_i the partial quotients ofα. Sometimes – but not always! – we will writeq_k=q_k(α), p_k =p_k(α), a_k =a_k(α).

Given N ⊂ Nand M >1, let A =A(N, M)⊂(0,1) denote the set of irrational α∈(0,1) s.t. for some subsequence n_k↑ ∞,

(3.1) ∃r_k 6M s.t. r_kq_nk ∈ N, a_nk+1

(a₀+· · ·+a_nk) −−−→

k→∞ ∞.

Thelower density of N is d(N) := lim inf _N¹ Card(N ∩[1, N]). The purpose of this section is to prove:

Theorem 3.1. — If a setN has positive lower density, then there exists M such that A(N, M) has full Lebesgue measure in (0,1).

The proof consists of the following three lemmas:

Lemma 3.2. — For almost all α there is n₀ = n₀(α) s.t. if k > n₀ and a_k+1 >

1

4k(lnk)(ln lnk), then a_k+1/(a₁+· · ·+a_k)> ¹₈ln lnk.

Lemma 3.3. — Supposeα ∈(0,1)\Q and(p, q) ∈N0×N satisfy gcd(p, q) = 1 and|qα−p|6 _qL¹ whereL>4. Then there existsk s.t.q=q_k(α)anda_k+1(α)> ¹₂L.

Lemma 3.4. — Suppose ψ :R+ →Ris a non-decreasing function s.t.

(3.2) ^X

n

1

nψ(n) =∞.

Suppose N ⊂ N has positive lower density. For all M sufficiently large, for a.e.

α∈(0,1)there are infinitely many pairs(m, n)∈N0×Ns.t.n∈ N,gcd(m, n)6M, and |nα−m|6 _nψ(n)¹ .

Remark 3.5. — By the monotonicity of ψ, if e^k−1 < n < e^k then ψe^k−1 6 ψ(n)6ψe^k. Hence (3.2) holds iff ^P_kψ(e¹k) =∞.

Remark 3.6. — If N = N, then Lemma 3.4 holds with M = 1 by the classical Khinchine Theorem. We do not know if Lemma 3.4 holds withM = 1 for every set N with positive lower density.

Proof of Theorem 3.1 given Lemmas 3.2–3.4. — We apply these lemmas with ψ(t) =c(lnt) (ln lnt) (ln ln lnt) and c >1/ln(¹⁺

√5 2 ).

Fix M > 1 as in Lemma 3.4. Then ∃ Ω ⊂ (0,1) of full measure s.t. for every α∈Ω there are infinitely many (m, n)∈N0×Nas follows. Let m^∗ :=m/gcd(m, n), n^∗ :=n/gcd(m, n),p:= gcd(m, n), then

(1) pn^∗ ∈ N, p6M, |n^∗α−m^∗|= ^|nα−m|_p 6 _n^∗_ψ(n^{1 ∗}₎ (∵n^∗ 6n);

(2) ∃ k s.t. n^∗ = q_k(α) and a_k+1(α) > ¹₂ψ(q_k) (∵ Lemma 3.3). By its recursive definition, q_k> k-th Fibonacci number> ¹₃(¹⁺

√ 5

3 )^k. So for all k large enough, a_k+1(α)> ¹₂ψ(q_k)> ¹₄k(lnk)(ln lnk);

(3) a_k+1/(a₁+· · ·+a_k)> ¹₈ln lnk → ∞ (∵Lemma 3.2).

So everyα ∈Ω belongs to A =A(N, M), andA has full measure.

Next we prove Lemmas 3.2–3.4.

3.2. Proof of Lemma 3.2 By [DV86], for almost everyα

(a₁+· · ·+ak+1)−max_j6k+1aj

klnk → 1

ln 2 <2.

So ifk is large enough, and ak+1 > ¹₄k(lnk)(ln lnk) then

j6k+1maxa_j =a_k+1, a₁+· · ·+a_k

klnk 62, and a_k+1

a₁+· · ·+a_k > 1

8ln lnk.

3.3. Proof of Lemma 3.3

For every (p, q) as in the lemma, |qα−p|< _2q¹. A classical result in the theory of continued fractions [Khi63, Theorem 19] says that in this case∃k s.t.q=q_k(α), p= p_k(α).

To estimate a_k+1 = a_k+1(α) we recall the following facts, valid for the principal denominators of any irrationalα∈(0,1) [Khi63]:

(1) |q_kα−p_k|> _q ¹

k+qk+1;

(2) q_k+1+q_k <(a_k+1+ 2)q_k, whence by (a) a_k+1 > _q ¹

k|q_kα−p_k|−2.

In our case, |q_kα−p_k|=|qα−p|6 _qk¹_L, so a_k+1 > L−2> ^L₂. 3.4. Preparations for the proof of Lemma 3.4

Let (Ω,F,P) be a probability space, and A_k ∈ F be measurable events. Given D >1, we say that Ak are D-quasi-independent, if

(3.3) P(A_k1 ∩A_k2)6DP(A_k1)P(A_k2) for all k₁ 6=k₂.

The following proposition is a slight variation on Sullivan’s Borel–Cantelli Lemma from ([Sul82]):

Proposition 3.7. — For every D > 1 there exists a constant δ(D) > 0 such that the following holds in any probability space:

(a) If A_k are D-quasi-independent measurable events s.t. limk→∞P(A_k) = 0 but

P

kP(A_k) = ∞, thenP(A_k occurs infinitely often)>δ(D).

(b) The quasi-independence assumption in(a)can be weakened to the assumption that for somer ∈N,P(A_k1 ∩A_k2)6DP(A_k1)P(A_k2) for all |k₂ −k₁|>r.

(c) One can take δ(D) = _2D¹ .

Proof. — Since P(A_k)→0 but ^PP(A_k) =∞, there is an increasing sequence N_j such that lim_j→∞^PN_k=N^j+1

j+1P(A_k) = _D¹.

Let B_j be the event that at least one of events {A_k}^N_k=N^j+1_j+1 occurs. Since B_j =

UNj+1 k=Nj+1

A_k\^Sk−1_j=N

j+1A_j, P(B_j)>

Nj+1

X

k=Nj+1

P(A_k)− ^X

Nj+16k1<k26Nj+1

P(A_k1 ∩Ak2)

>

Nj+1

X

k=Nj+1

P(A_k)−D ^X

Nj+16k1<k26Nj+1

P(A_k1)P(A_k2)

>

Nj+1

X

k=Nj+1

P(A_k)− D 2





Nj+1

X

k=Nj+1

P(A_k)





2

.

Since lim_j→∞^PN_k=N^j+1

j+1P(A_k) = _D¹ and D>1,lim infP(B_j)> _2D¹ .

LetE denote the event that A_j happens infinitely often. E is also the event that B_j happens infinitely often, therefore E = ^T∞_n=1^S∞_j=n+1B_j. In a probability space,

the measure of a decreasing intersection of sets is the limit of the measure of these sets. SoP(E)>lim infP(B_j)> _2D¹ , proving (a) and (c).

Part (b) follows from part (a) by applying it to the sets{A<sub>kr+`</sub>}where 06`6r−1

is chosen to get^P_kP(Akr+`) = ∞.

Themultiplicity of a collection of measurable sets{E_k}is defined to be the largest K s.t. there areK different k_i with P(^TK_i=1E_ki)>0.

Proposition 3.8. — Let E_k be measurable sets in a finite measure space. If the multiplicity of {E_k} is less than K, then

mes ^[

k

Ek

!

> 1 K

X

k

mes(E_k).

Proof. — 1^S

iEi > _K^{1 P}i1_Ei almost everywhere.

Proposition 3.9. — For every non-empty open intervalI ⊂[0,1], Card

(m, n)∈ {0, . . . , N}² : m

n ∈I, gcd(m, n) = 1

∼3 mes(I)N²/π², as N→ ∞.

Proof. — This classical fact due to Dirichlet follows from the inclusion-exclusion principle and the identityζ(2) =π²/6, see [HW08, Theorem 459].

Proposition 3.10. — Suppose α = [0;a₁, a₂, . . .] and α = [0;a_{`+1</sub>, a<sub>`+2}, . . .].

Then the principal convergents p_{`</sub>/q<sub>`} of α and the principal convergents p_{`</sub>/q<sub>`} of α are related by

p_l+l p_l+l+1 q_l+l q_l+l+1

!

= pl−1 p_l ql−1 q_l

! p_l p_l+1 q_l q_l+1

!

. Proof. — Since a₀ = 0, the recurrence relations forp_n/q_n imply

p_n p_n+1 q_n q_n+1

!

= pn−1 p_n qn−1 q_n

! 0 1 1 a_n+1

!

, p₀ p₁ q₀ q₁

!

= 0 1

1 a₁

!

. So

p_n p_n+1 q_n q_n+1

!

= 0 1

1 a₁

!

· · · 0 1 1 a_n+1

!

. It follows that

p_l+l p_l+l+1 q_l+l q_l+l+1

!

= pl−1 p_l ql−1 q_l

! p_l p_l+1 q_l q_l+1

!

,

where p_i/q_i are the principal convergents ofα:= [0;al+1, al+2, . . .].

3.5. Proof of Lemma 3.4

Without loss of generality, limt→∞ψ(t) = ∞, otherwise replace ψ(t) by the bigger monotone functionψ(t) + lnt.

FixM >1, to be determined later. Let

Ω_k:={(m, n)∈N² :n∈ N, n∈[e^k−1, e^k],0< m < n,gcd(m, n)6M}, Am,n,k:=

(

α∈T:|nα−m|6 1 e^kψ(e^k)

)

, Ak:= ^[

(m,n)∈Ω_k

Am,n,k,

A:={α∈T:α belongs to infinitely many A_k}.

The lemma is equivalent to saying that A has full Lebesgue measure for a suitable choice of M.

We will prove a slightly different statement. Fixε >0 small. Given an non-empty intervalI ⊂[ε,1−ε], let

Ω_k(I) :=

(m, n)∈Ω_k : m n ∈I

, Ak(I) := ^[

(m,n)∈Ω_k(I)

Am,n,k,

A(I) := {α∈T:α belongs to infinitely many A_k(I)}.

We will prove that there exists a positive constantδ =δ(ε, M) s.t. for all intervals I ⊂[ε,1−ε], mes(A(I)∩I)>δmes(I). It then follows by a standard density point argument (see below) that A ∩[ε,1−ε] has full measure. Since ε is arbitrary, the lemma is proved.

Claim 1. — There existsK =K(ε) such that for every k > K, the multiplicity of {Am,n,k}(m,n)∈Ω_k(I) is uniformly bounded byM.

Proof. — Suppose (m_i, n_i) ∈ Ω_k(I) (i = 1,2) and A_m1,n1,k∩A_m2,n2,k 6= ∅. Then there is α s.t. |n_iα−m_i| 6δ_k := _ekψ(e^{1 k}) (i = 1,2). Choose K =K(ε) so large that k > K ⇒δ_k< _4e^εk.

Ifk > K, thenα> ^m_niⁱ −δk >minI−^ε₂ > ^ε₂. Let ri := gcd(m_i, ni) and (n^∗_i, m^∗_i) :=

1

ri(n_i, m_i). Then |n^∗_iα −m^∗_i| 6 δ_k and m^∗_i 6 n^∗_i 6 n_i 6 e^k, so |n^∗₂m^∗₁ −n^∗₁m^∗₂| =

1

α|m^∗₁(n^∗₂α−m^∗₂)−m^∗₂(n^∗₁α−m^∗₁)|6 ^2e_ε/2^kδk <1. Son^∗₂m^∗₁ =n^∗₁m^∗₂. Since gcd(n^∗_i, m^∗_i) = 1, (n^∗₁, m^∗₁) = (n^∗₂, m^∗₂). It follows that (n₂, m₂)∈ {(rn^∗₁, rm^∗₁) :r= 1, . . . , M}. So the multiplicity of {A_m,n,k}_{(m,n)∈Ωk(I)} is uniformly bounded by M. Claim 2. — Let d(N) := lim inf _N¹ Card(N ∩ [1, N]) > 0, then there exists M =M(N) and K^f=K(ε,^f N,|I|) s.t. for allk > K,^f

(3.4) d(N) mes(I)

4M ψ(e^k) 6mes(Ak(I))6 6 mes(I) ψ(e^k) . In particular,mes(Ak(I))−−−→

k→∞ 0 and ^Pmes(Ak(I)) =∞.

Proof. — mes(A_m,n,k) = mes^hm_n −r_m,n,^m_n +r_m,nⁱ= 2r_m,nwherer_m,n = _nekψ(e^{1 k}). Sincen ∈[e^k−1, e^k],

(3.5) Card(Ω_k(I))

M e^2kψ(e^k) 6mesA_k(I)6 e Card(Ω_k(I)) e^2kψ(e^k) ,

where the lower bound uses Claim 1 and Proposition 3.8.

Card(Ω_k(I)) satisfies the boundsA−B 6Card(Ω_k(I))6A where A:= Card

(m, n) :n∈ N, n ∈[e^k−1, e^k], m n ∈I

B := Card

(m, n) :n∈ N, n ∈[e^k−1, e^k], m

n ∈I,gcd(m, n)>M

. Choose K^f=K^f(ε,N,|I|)> K(ε) s.t. for all k >K^f

(1) Card{n ∈ N : 06n 6e^k}> ^√1₂d(N)

(2) Card{n ∈[e^k−1, e^k]∩N:p|n}62(e^k−e^k−1)/p for all p>1;

(3) For all n > e^K−1e , p>1, n

p√

2mes(I)6Card

m∈N: m

n ∈I, p|m

6 2n

p mes(I).

Ifk >K^f, then ¹₂d(N)e^2kmes(I)6A62e^2kmes(I) and B 6

∞

X

p=M

Card

(m, n) :n ∈[e^k−1, e^k], m

n ∈I, p|m, p|n

6

∞

X

p=M

2(e^k−e^k−1)

p · 2e^kmes(I)

p <4e^2kmes(I)

∞

X

p=M

1 p² 6 1

4d(N)e^2kmes(I), provided we choose M s.t.

∞

X

p=M

p⁻² < 1

16d(N).

Together we get ¹₄d(N)e^2kmes(I) 6 Card(Ω_k(I)) 6 2e^2kmes(I). The claim now

follows from (3.5).

Claim 3. — There exists D=D(N, M), r=r(M), andKˆ = ˆK(ε,N, I) s.t. for all k₁, k₂ >Kˆ s.t.|k₁−k₂|> r(M),

(3.6) mes(A_k1(I)∩ A_k2(I)|I)6Dmes(A_k1(I)|I) mes(A_k2(I)|I) Proof. — By Claim 2, if k1, k2 are large enough, then

(3.7) mes(A_k1(I)|I) mes(A_k2(I)|I)> d(N) 5M

!2

1 ψ(e^k1)ψ(e^k2),

where we put 5 instead of 4 in the denominator to deal with edge effects arising from mes(A_k(I)\I) =O_ekψ(e^{1 k})

.

To prove the claim, it remains to bound mes (A_k1(I)∩ A_k2(I)|I) from above by

const

R1R2, whereR_i :=ψ(e^ki).

A cylinder is a set of the form

[[a₁, . . . , a_n]] ={α∈(0,1)\Q:a_i(α) = a_i (16i6n)}.

Equivalently, α ∈[[a₁, . . . , a_n]] iff α has an infinite continued fraction expansion of the formα = [0;a₁, . . . , a_n,∗,∗, . . .].

Our plan is to cover A_ki(I) by unions of cylinders of total measure O(1/R_i), and then use the following well-known fact: There is a constant G > 1 s.t. for any (a₁, . . . , an, b1, . . . , bm)∈N^n+m,

(3.8) G⁻¹ 6 mes[[a₁, . . . , a_n;b₁, . . . , b_m]]

mes[[a₁, . . . , a_n]] mes[[b₁, . . . , b_m]] 6G.

This is because the invariant measure _{ln 2}¹ _1+x^dx of T : (0,1)→(0,1), T(x) = {¹_x} (the Gauss map) is a Gibbs–Markov measure, thanks to the bounded distortion of T, see [ADU93, Section 2].

To cover A_ki(I) by cylinders, it is enough to cover Am,n,ki by cylinders for every (m, n) ∈ Ω_ki(I). Suppose α ∈ A_m,n,ki. Then r := gcd(m, n) 6 M and (m^∗, n^∗) :=

1

r(m, n) satisfies

gcd(m^∗, n^∗) = 1, n^∗ ∈ ^[

|k^∗_i−k_i|6lnM

[e^k∗i−1, e^k∗i], |n^∗α−m^∗|< 1 n^∗R_i.

Assume k_i is so large that R_i = ψ(e^ki) > 4. Then Lemma 3.3 gives a_li+1 > ^R₂ⁱ. ThusA_ki(I)⊂ C_ki(I, R_i) where

C_k(I, R) := ^[

k^∗∈[k−lnM,k]







α∈(0,1)\Q:∃` s.t.

q_{`</sub>(α)∈[e<sup>k∗−1</sup>, e<sup>k∗</sup>], a`+1(α)>R/2 p`(α)/q<sub>`}(α)∈I







. This is a union of cylinders, becauseq`(α), p<sub>`</sub>(α), a<sub>`+1</sub>(α) are constant on cylinders of length`+ 1.

We claim that for somec^∗(M) which only depends on M, for all k_i large enough, (3.9) mes(C_ki(I, R_i))6 c^∗(M) mes(I)

R_i .

Every rational ^m_n ∈(0,1) has two finite continued fraction expansions: [0;a₁, . . . , a<sub>`</sub>] and [0;a<sub>1</sub>, . . . , a`−1,1] with a` >1. We write `=`(^m_n) andai =ai(^m_n). With this notation

C_ki(I, R_i) = ^[

k_i^∗∈[ki−lnM,ki] n∈[e^k∗i−1,e^k∗i]

[

gcd(m,n)=1 m/n∈I

[

b>Ri/2

a₁

m n

, . . . , a_`(^m

n)

m n

, b

∪

a₁

m n

, . . . , a_`(^m

n)

m n

−1,1, b

. We have [[a₁, . . . , a<sub>`</sub>]] = (<sup>p</sup><sub>q</sub><sup>`+p`−1</sup>

`+q`−1,^p_q^`

`) or (<sup>p</sup><sub>q</sub><sup>`</sup>

`,<sup>p</sup><sub>q</sub><sup>`+p`−1</sup>

`+q`−1), depending on the parity of`, see for instance [Khi63]. Since |p<sub>`</sub>q`−1−p`−1q_{`</sub>|= 1 and q<sub>`+1} =a_{`+1</sub>q<sub>`}+q`−1, we have mes([[a<sub>1</sub>, . . . , a<sub>`</sub>, b]]) = _q ¹

`+1(q`+1+q`) = _(bq ¹

`+q`−1)((b+1)q`+q`−1) 6 _b(b+1)q^{1 2}

`

, leading to mes(Cki(I, Ri))6 ^X

k^∗_i∈[k_i−lnM,k_i]

X

n∈[e^k∗i−1,e^k

∗ i]

X

gcd(m,n)=1 m/n∈I

X

b>Ri/2

2 n²b(b+ 1) 6 8 lnM

e^{2(ki−1−lnM)}R_i

e^kiM

X

n=1

#

m ∈N: m

n ∈I,gcd(m, n) = 1

6 c^∗(M)

R_i mes(I) where c^∗(M) only depends onM. The last step uses Proposition 3.9.

Next we cover A_k1(I)∩ A_k2(I) by cylinders. Suppose without loss of generality that k2 > k1. Arguing as before one sees that if

(3.10) k₂ > k₁+ lnM + 1,

then Ak1(I)∩ Ak2(I) can be covered by sets [[a₁, . . . , a`, b, a₁, . . . , a_l, b]] as follows:

The convergents p_i/q_i of (every) α in [[a₁, . . . , a_`, b, a₁, . . . , a_l, b]], (16i6l+l+ 2), satisfy

(1) q_l ∈[e^k1∗−1, e^k∗1], k₁^∗ ∈[k₁−lnM, k₁], p_l/q_l ∈I, b>R₁/2;

(2) q_l+l+1 ∈[e^k2∗−1, e^k∗2], k₂^∗ ∈[k₂−lnM, k₂], p_l/q_l ∈I, b>R₂/2 (3) k₂^∗ > k^∗₁ (this is where (3.10) is used).

We claim that

(3.11) [[a₁, . . . , a_`, b]]⊂ C_k1(I),

(3.12) b 6e^{k∗2−k∗1+1},

(3.13) [[a₁, . . . , a_`, b]]⊂ ^[

|r|63

C_k2−k₁+r−lnb([0,1], R₂).

(3.11) follows from (1). Next,e^k2∗>q_l+1>bq_l>be^k∗1−1 proving (3.12). To prove (3.13), letp_i/q_i, 16i6`+ 2, be the principal convergents of (every) α∈[[b, a<sub>1</sub>, . . . , a<sub>`</sub>, b]].

By Proposition 3.10, q_l+1+l = ql−1p_l+1 +q_lq_l+1, whence q_lq_l+1 6 q_l+1+l 6 2q_lq_l+1. Sinceql ∈[e^k∗1−1, e^k∗1] and q_l+l+1 ∈[e^k2∗−1, e^k∗2],

(3.14) e^{k2∗−k∗1−2} 6 q_l+1+l

2q_l 6q_l+1 6 q_l+1+l

q_l 6e^{k2∗−k∗1+1}.

Next, letp_ei/qei(16i6l) denote the principal convergents of (every)α_e∈[[a₁, . . . , a_`, b]].

Then ^p_q^l+1

l+1

= 1/(b + ^epl

eq_l), so q_l+1 = bq_el +p_el, whence bq_el 6 q_l+1 6 (b + 1)q_el. Thus qe_l ∈ [(b+ 1)⁻¹q_l+1, b⁻¹q_l+1]. It follows that the l-th principal convergent of every αe ∈[[a₁, . . . , a_`, b]] satisfies

(3.15) qe_l ∈[e^{k2∗−k∗1−3−lnb}, e^{k∗2−k1∗+1−lnb}].

It is now easy to see (3.13).

By (3.13),A_k1(I)∩ A_k2(I)∩I ⊂^S_|r|63^U_[a,b]⊂Ck

1(I)

U

[a⁰,b⁰]⊂C_k

2−k1+r−lnb([0,1])[[a, b, a⁰, b]].

Now arguing as in the proof of (3.9) and using (3.8) we obtain (3.16) mes(A_k1(I)∩ A_k2(I)∩I)

6 ^X

k^∗_i∈[ki−lnM,ki] i=1,2;|r|63

X

n∈[e^k∗1−1

,e^k∗1]

X

gcd(m,n)=1 m/n∈I

[exp(k^∗₂−k₁^∗+1)]

X

b=[R1/2]

2Gmes(Ck2−k1+r−lnb([0,1],R2)) n²b(b+1)

6 const mes(I) R₁R₂ .

(3.16) uses the estimate mes(C_k2−k₁+r−lnb([0,1], R₂)) =O(1/R₂) which is also valid when k₂ −k₁ +r−lnb is small, provided we choose M large enough so that the asymptotic in Proposition 3.9 holds for all N > M with I = [0,1]. See the proof of (3.9).

Combining (3.16) with (3.7), we find that under (3.10) A_ki(I) are D-quasi-indep-

endent for sufficiently largeD, proving Claim 3.

Claims 2 and 3 allow us to apply Sullivan’s Borel–Cantelli Lemma (Proposition 3.7).

We obtain δ = δ(M) s.t. for every interval I ⊂ [ε,1−ε], mes(A ∩I) > δmes(I).

This means that [ε,1−ε]\ Ahas no Lebesgue density points, and therefore must have measure zero. So A has full measure in [ε,1−ε]. Since ε is arbitrary, A has

full measure.

4. Proof of Theorem 1.2

As explained in Section 2, it is enough to prove Theorem 1.2 forf(x) of the form

Pd

m=1bmh(x+βm) 6≡ 0 with h(x) = {x} − ¹₂. Without loss of generality, βi are different andb_i 6= 0. Notice that

f(x) = −

d

X

m=1

b_m

∞

X

j=1

sin(2πj(x+β_m))

πj .

Thereforekfk²₂ = _2π¹2

P

n 1

n²D(β₁n, . . . , β_dn), where D:T^d→R is D(γ₁, . . . , γ_d) :=

Z 1 0

" _d X

m=1

b_msin(2π(y+γ_m))

#²

dy.

(4.1)

Since f 6≡ 0, D(β₁n, . . . , β_dn) > 0 for some n. Let T denote the closure in T^d of O:={(β₁n, . . . , β_dn) modZ:n ∈Z}.This is a minimal set for the translation by (β₁, . . . , β_d) onT^d, so a standard compactness argument shows that for everyε₀ >0,

the set

(4.2) N :={n∈N: D(β₁n, . . . , β_dn)> ε₀}

is syndetic: its gaps are bounded. Thus N has positive lower density.

By Theorem 3.1, if M is sufficiently large then the set A := A(N, M) has full measure inT. Let

S_n(α, x) :=

n−1

X

k=0

f(x+kα).

The proof of Theorem 1.2 for f(x) above consists of two parts:

Theorem 4.1. — Supposeα∈ A, then for a.e.x∈[0,1), there exist A_k(x)∈R and B_k(x), N_k(x)→ ∞ such that

S_n(α, x)−A_k(x) B_k(x)

−−−→dist

k→∞ U[0,1], as n∼U(0, . . . , N_k(x)).

Theorem 4.2. — Supposeα∈ A, then for a.e.x∈[0,1), there are noA_N(x)∈R and B_N(x)→ ∞ such that

S_n(α, x)−A_N(x) B_N(x)

−−−→dist

N→∞ U[0,1], as n∼U(0, . . . , N).