L’INSTITUTFOURIER DE ANNALES

A L

E S

E D L ’IN IT ST T U

F O U R

ANNALES

DE

L’INSTITUT FOURIER

Aihua FAN & Jörg SCHMELING

Everywhere divergence of one-sided ergodic Hilbert transform Tome 68, n^o6 (2018), p. 2477-2500.

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EVERYWHERE DIVERGENCE OF ONE-SIDED ERGODIC HILBERT TRANSFORM

by Aihua FAN & Jörg SCHMELING (*)

Abstract. — For a given number α ∈ (0,1) and a 1-periodic function f, we study the convergence of the series P∞

n=1 f(x+nα)

n , called one-sided Hilbert transform relative to the rotation x 7→ x+α mod 1. Among others, we prove that for any non-polynomial function of classC² having Taylor–Fourier series (i.e.

Fourier coefficients vanish onZ⁻), there exists an irrational numberα (actually a residual set ofα) such that the series diverges forallx. We also prove that for any irrational numberα, there exists a continuous functionf such that the series diverges for all x. The convergence of general seriesP^∞

n=1anf(x+nα) is also discussed in different cases involving the diophantine property of the number α and the regularity of the functionf.

Résumé. — Etant donné un nombreα ∈(0,1) et une fonction 1-périodique f, nous étudions la convergence de la sérieP∞

n=1 f(x+nα)

n , appelée la transformée de Hilbert latérale relative à la rotation x 7→x+α mod 1. Entre autres, nous démontrons que pour toute fonction non-polynomiale de classeC²admettant une série de Taylor–Fourier (i.e. les coefficients de Fourier sont nuls surZ⁻), il existe un αirrationnel (en réalité, un ensemble deαde deuxième catégorie au sens de Baire) tel que la série diverge pourtous lesx. Nous démontrons aussi que pour toutα irrationnel, il existe une fonction continuef telle que la série diverge pourtousles x. La convergence d’une série généraleP∞

n=1anf(x+nα) est aussi discutée pour divers cas où interviennent la propriété diophantienne du nombreαet la régularité de la fonctionf.

1. Introduction

Let f be a Lebesgue integrable function defined on the circle T=R/Z identified with [0,1) such thatR

Tf(x)dx= 0 and letα∈[0,1) be a given

Keywords:Ergodic Hilbert transform, Everywhere divergence, Irrational rotation.

2010Mathematics Subject Classification:37A30, 37A45.

(*) The authors’ thanks go to M. Björklund, M. Benedicks and C. Cuny for their valu- able remarks. The work was done during the visit in the Spring 2015 to Lund University of first author, who would like to thank the warm hospitality of Centre for Mathemat- ical Sciences, Lund University and to Knuth and Alice Wallenberg Foundation for its support. This work was also partially supported by NSFC 11471132.

number. We consider the following series (1.1)

∞

X

n=1

a_nf(x+nα)

where the coefficients{an}are complex numbers which are usually assumed square summable. LetT denote the rotation onTdefined by T x=x+α.

Then the series (1.1) takes the formP∞

n=1a_nf(Tⁿx), which may be called an ergodic series. Such ergodic series are studied for some hyperbolic sys- temsT in [9] and in many cases the almost everywhere (a.e.) convergence of P∞

n=1anf(Tⁿx) is ensured byP∞

n |an|²<∞(for the study of general ran- dom series of the formPanXn see [6], which is a continuation of [9]). The method used in [9] gives nothing about (1.1). It seems a delicate problem to study the pointwise convergence and the convergence in means of (1.1) in its generality.

Ifan= _n¹, the series (1.1) becomes the so-called one-sided ergodic Hilbert transform (EHT for short):

(1.2)

∞

X

n=1

f(x+nα)

n .

More generally, for any measure-preserving map T, the one-sided EHT takes the form P∞

n=1 f(Tⁿx)

n and was studied in the literature. In 1939, Izumi [13] raised the question of the a.e. convergence of the one-sided EHT.

In 1949, Halmos proved that for any non-atomic invariant measureµ, there exists a centered functionf ∈L²(µ) such that the one-sided EHT fails to converge in L²-norm. Later in 1959, Dowker and Erdős [8] constructed a centered functionf ∈L^∞(µ) which has the following stronger divergence

(1.3) lim sup

N→∞

N

X

n=1

f(Tⁿx) n

=∞ a.e.

(see also Del Junco and Rosenblatt [15] and see [1] for additional refer- ences). In 2009, Cuny [4] proved that for anyf ∈L¹(µ), theL¹-convergence of the one-sided EHT implies its a.e. convergence. This answered a question of Gaposhkin [11] who, in 1996, studied the one-sided EHT associated to a general unitary operatorU onL²(µ) and he gave an example of a unitary operator U and anf ∈ L²(µ) such that the one-sided EHT converges in L²-norm, but doesn’t converge a.e. ([11, p. 253-254]). It is still a question to find effective condition ensuring the a.e. convergence or theL^p-convergence for general one sided EHTs and even for (1.2).

The dynamics of the rotationT_αx=x+α mod 1 depends strongly on the diophantine property of the numberα. Consequently, as we shall see,

the behavior of the associated one-sided EHTs are different for different α. We shall also see that the high order regularity (even the analyticity) off can not ensure the convergence of the one-sided EHT forα’s having very bad diophantine property (for long time, nα doesn’t come back to the neighborhood of 0). In some cases, the divergence of (1.2) takes place everywhere:

(1.4) lim sup

N→∞

N

X

n=1

f(x+nα) n

=∞ ∀x∈[0,1).

This reinforces the Dowker–Erdős’ result (1.3) for some Liouville rotations.

Forf ∈L¹(T) we denote byfb(n) then-th Fourier coefficient off defined byR

Tf(x)e^−2πinxdx. We adopt the notation kfk_A(T)=

∞

X

n=−∞

|fb(n)|.

For 0 < γ 6 1, Lip_γ will denote the space of functions on T such that

|f(x)−f(y)|6C|x−y|^γ. Forx∈Twe denote kxk= inf

n∈Z

|x−n|.

Notice that for all x, y ∈ T we have the triangle inequality kx+yk 6 kxk+kykand the estimate

2kxk6|sinπx|6πkxk.

In this note we are concentrated on the series (1.1) and (1.2). Our results are listed below.

(1) For any non-polynomial function f ∈ C²(T) with fb(n) = 0 for n <0, there exists a residual setRf depending onf such that for everyα∈ Rf the series (1.2) diverges for every x(Theorem 2.1).

(2) For any non-polynomial functionf ∈C²(T), there exists a residual setR_f depending onf such that for everyα∈ R_f the series (1.2) diverges for almost allx(Theorem 2.2).

(3) For any irrational number α, there exists a continuous function f such that the series (1.2) diverges for everyx(Theorem 2.3).

(4) For all f ∈ L² and for almost all α, the series (1.2) converges for almost allx(Theorem 3.1).

(5) If P∞ n∈Z\{0}

|^f(n)|b

knαk < ∞, the series (1.2) converges uniformly in x (Theorem 3.2).

(6) IfP∞ n∈Z\{0}

|^f(n)|b

2

knαk² <∞, the series (1.2) converges inL²-norm and for almost everyx(Theorem 3.5).

(7) Let f ∈ L²(T) with f(0) = 0 andb |fb(k)| 6C|k|^−β where C >0 andβ >1/2 are two constants. Letαbe an irrational number with convergents{p_n/q_n}. The series (1.2) converges inL²-mean and a.e.

if the following condition is satisfied

∞

X

m=1

log²q_m+1 qm^2β

<∞ (Theorem 3.6).

(8) Let α be an irrational number with convergents{pn/q_n}. For the functionf defined by the lacunary seriesP∞

m=1fb(q_m)e^2πiqmx with P

m>1|fb(q_m)|² <∞, the series (1.2) converges in L²-mean if and only if

∞

X

m=1

|fb(qm)|²log²qm+1<∞ (Proposition 3.7).

(9) Let α=p/qbe a rational number where p, qare coprime. For any f ∈L¹(T) withR

f(x)dx= 0, the seriesP

anf(x+nα) converges almost everywhere iff for anyj= 0,1, . . . , q−1, the numerical series P

ka_kq+j converges (Theorem 3.8).

Notice that for any polynomial f (of causefb(0) = 0) and any number α, the series (1.2) converges everywhere. But there are analytic functions f and irrational numbersαsuch that the series (1.2) diverges everywhere.

The behavior of the series (1.2) depends on that of partial sums of the seriesP∞

n=1n⁻¹e^2πinx. Notice that its real and imaginary parts are:

∞

X

n=1

1

ncos 2πnx= log 1 2|sinπx|,

∞

X

n=1

1

nsin 2πnx=π 1

2 −x

. These two series converge for all points x∈ (0,1). It is natural that the behavior of the series (1.1) will depend on that of partial sums of the series P∞

n=1ane^2πinx.

Section 2 will be devoted to the divergence of the one-sided EHT (1.2). Section 3 will be devoted to different convergences of the general ergodic series (1.1).

2. Divergence of one-sided ergodic Hilbert transform We first study the divergence of the series

∞

X

n=1

f(x+nα)

n .

We sayf ∈L¹(T) admits a Taylor–Fourier series if fb(n) = 0 forn6−1.

In the following,ζ(s) denotes the Riemannζ-functionP∞ n=1n^−s.

2.1. Statements on divergence

We first state three divergence statements that we will prove.

Theorem 2.1. — Letf ∈L¹(T)satisfy the following conditions (1) f(k) = 0b ifk60;fb(k)6= 0for infinitely many k.

(2) there existss >1such thatζ(s)<2andlim sup|k|^s|fb(k)|= 0.

Then there exists a residual setR ⊂[0,1]of irrational numbers such that for eachα∈ R, we have

lim sup

n→+∞

N

X

n=1

f(x+nα) n

= +∞, ∀x∈[0,1).

The solution of ζ(s₀) = 2 verifies 1< s₀= 1.72865. . . <2. Thesin the condition (2) must verifys > s0 >1. So the condition (2) implies thatf admits an absolutely convergent Fourier series. All non polynomial func- tions of classC²admitting Taylor–Fourier series satisfies the conditions (1) and (2). The following analytic functions are examples

f(x) =

∞

X

n=1

rⁿe^2πinx= re^2πix

1−re^2πix = re^2πix−r²

1−2rcos(2πx) +r² (0< r <1).

Theorem 2.2. — Let f : T → R be an integrable function whose Fourier coefficients verify the following conditions

(1) f(0) = 0,b fb(k)6= 0for infinitely manyk.

(2) there existss >1such thatζ(s)<2andlim sup|k|^s|fb(k)|= 0.

Then there exists a residual setR ⊂[0,1]of irrational numbers such that for eachα∈ R, we have

lim inf

N→+∞

N

X

n=1

f(x+nα)

n =−∞, lim sup

N→+∞

N

X

n=1

f(x+nα)

n = +∞,

for almost everyx.

For the last theorem, we have succeeded in proving the a.e. divergence.

We wonder if the everywhere divergence is still true.

Theorem 2.3. — For any irrational number α∈ (0,1), there exists a continuous function f : T → Cwith R

Tf(x)dx= 0 having an absolutely convergent Fourier series such that

lim sup

N→∞

N

X

n=1

f(x+nα) n

= +∞ ∀x∈[0,1).

In order to prove these three theorems, we developfinto its Fourier series and we shall see that the behavior of the one-sided EHT relies heavily on that of the following trigonometric polynomials

G_N(x) =

N

X

n=1

e^2πinx n .

We shall also need a result due to Jacobsthal which concerns the biggest gap between natural numbers coprime with a given natural number. We get together such preliminaries as several lemmas before we prove the theorems.

2.2. Some lemmas

Lemma 2.4. — Assume 0< c <1/2. Then sup

N>1

sup

kxk>c

|GN(x)|6 π c. Proof. — Notice thatGN(1/2) =PN

n=1 (−1)ⁿ

n so that sup_N>1|GN(1/2)|6 1. Also notice that

|G⁰_N(x)|= 2π

N

X

n=1

e^2πinx

6 2π

|sinπx| 6 π c if 1/2>|x|>c. Then, by the Newton–Leibniz formula we get

|G_N(x)|6|G_N(1/2)|+

Z x 1/2

G⁰_N(y)dy

61 + π 2c 6π

c.

Lemma 2.5.

GN(x) = logN−2

N

X

n=1

sin²πnx n +O(1)

where the constant in O(1) is uniform in x and in N. In particular, if

|xN|6C for some constantC >0, then

GN(x) = logN+O(1)

where the constant inO(1)doesn’t depend onxandN, but onC.

Proof.

GN(x)−GN(0) =

N

X

n=1

e^2πinx−1

n .

Its imaginary part isPN n=1

sin(2πnx)

n which is uniformly bounded inxand inN (see [16], p. 4). Its real part is equal to

N

X

n=1

cos(2πnx)−1

n =−2

N

X

n=1

sin²(πnx)

n .

We conclude for the first assertion by observing thatGN(0) = logN+O(1).

Suppose|xN|6C. Just using|sinx|6|x|, we get

N

X

n=1

sin²πnx

n 6π²x²

N

X

n=1

n=π²x²N(N+ 1)/26π²C². A corollary is that if |N x|6C, then we have.

sup

m>2

|GN(mx)|6|GN(x)|+O(1) = logN+O(1).

Lemma 2.6. — Let (φk)⊂[0,1) be an arbitrary sequence of numbers and let(nk)⊂Na sequence of increasing positive integers. For any interval I⊂[0,1)of positive length, the limsup set

lim sup

k→∞

{x∈[0,1) :nkx+φk ∈I mod 1}

has full Lebesgue measure.

Proof. — The space [0,1) identified with the circle is compact. The se- quence (φk) has a limit point, say φ. Without loss of generality, we can assume thatφ_k tends toφas ktends to infinity. So, the intervals−φ_k+I contains a common intervalI⁰ with positive length when k is sufficiently large. We can also assume thatI⁰ ⊂I−φk for allk. Sincenk is increasing, for almost all pointsx, the sequencen_kx (mod 1) is uniformly distributed.

So, for almost every point x, nkx ∈ I⁰ mod 1 for infinitely many k. A fortiori,n_kx+φ_k ∈I mod 1 for infinitely many k.

Lemma 2.7. — Suppose that{ck}k>1is a sequence of numbers such that ck 6= 0for infinitely manyk’s andlim sup|k|^s|ck|= 0for somes >1. Then there exists a strictly increasing subsequence{k`}`>1 of positive integers such that for any`>1, we have

(ζ(s)−1)|ck_`|>

∞

X

m=2

|cmk_`| whereζ(s) =P∞

n=1n^−s.

Proof. — Letk<sub>`</sub> (`>1) be defined inductively in the following way. Let A1be the set of maximizing points of maxk>1|k|^s|ck|. Since lim sup|k|^s|ck|= 0 and there are infinitely manyc_k 6= 0,A₁ is non-empty and finite. Let

k1= maxA1.

Now letA₂ be the set of maximizing points of max_k>k1|k|^s|c_k|, which is also non-empty and finite. Let

k₂= maxA₂. It is clear thatk1< k2. Inductively, we define

k`+1= max

m > k`:|m|^s|cm|= max

k>k`

|k|^s|ck|

. By the definition ofk`, we have

∀m>2, k^s_{`</sub>|c<sub>k`}|>(mk_{`</sub>)<sup>s</sup>|c<sub>mk`}|, i.e. m^−s|c_{k`</sub>|>|c<sub>mk`}|.

Taking sum overm>2, we get the desired result.

Letqn(α) denote the denominator ofn-th convergent ofα. Letϕ:N→N be an increasing function. Define

Bϕ(α) ={qn(α) :ϕ(qn(α))< qn+1(α)}.

Usually ϕ increases very fast. So, we asked that for q_n(α) ∈ Bϕ(α) the denominatorqn+1(α) next toqn(α) is much larger thanqn(α).

Lemma 2.8. — Let Λ ⊂ N be an arbitrary infinite subset of natural numbers. For genericα, we have

#(Λ∩ Bϕ(α)) =∞.

We will apply Lemma 2.8 to Λ ={k`}, the sequence appearing in Lem- ma 2.7, withϕ(n) =e^∆n/c(n)(∆>1 being a large number andc(n) being a sequence tending to 0). In order to prove Lemma 2.8 we need a result due to Jacobsthal.

2.3. An estimate on Jacobsthal’s function

Let N = p^α₁¹. . . p^α_k^k be the prime factorization of a natural number N∈N. Assume that

1 =m1< m2<· · ·< mi< mi+1<· · ·

are the integers which are coprime withN. Jacobsthal’s function is defined as

g(N) = max

16i<∞(mi+1−mi), (N ∈N).

What we will need is g(N) = o(N) as N → ∞. The estimate on g(N) below was known to Jacobsthal [14]. But for completeness we include a proof. There are much better estimates known (see for example [12]), but the one presented here suffices for our purpose.

Theorem 2.9. — LetN =p^α₁¹. . . p^α_k^k. Theng(N)6(k+1) 2^k−1 +1.

Proof. — Since the definition ofg(N) implies that any interval of length g(N) contains at least one number coprime toN, we need to find a lower bound onm∈Nsuch that for any integernthe intervalI:= [n, n+m−1]

contains at least one integer coprime withN. Any such a lower bound will be an upper bound ofg(N).

Let 1 6j 6 k and let 1 6i1 < · · · < ij 6k be given. We denote by K(i₁, . . . , i_j) the number of integersl ∈I that are divisible by p_i1. . . p_ij. These integersl are the following ones

n6p_i1. . . p_ij <2p_i1. . . p_ij <· · ·< K(i₁, . . . , i_j)p_i1. . . p_ij 6n+m−1.

The numberK(i1, . . . , ij) depends on n. But it has the following bounds independent ofn:

m p_i1. . . p_ij

−16K(i₁, . . . , i_j)6 m

p_i1. . . p_ij

+ 1.

By the inclusion-exclusion principle, the numberLof natural numbersl∈I with gcd(l, N)>1 is given by

L= X

16i6k

K(i)− X

16i₁<i₂6k

K(i1, i2) +· · ·+ (−1)^k+1K(1, . . . , k).

Hence the numberM of natural numbers l ∈ I that are coprime with N verifies

M =m−L

=m− X

0<i6k

K(i) + X

0<i1<i26k

K(i1, i2) +· · ·+ (−1)^k+2K(1, . . . , k)

>m·

k

Y

i=1

1− 1

p_i

−

k

X

i=1

k i

>m·

k

Y

i=1

i i+ 1−

k

X

i=1

k i

>m· 1

k+ 1− 2^k−1 .

ThereforeM >0 ifm>(k+ 1) 2^k−1

+ 1.

Since N>2·3^k(N)−1, we have k(N)6log₃

3 2N

=δlog₂ 3

2N

whereδ= 1/log₂3<1. We conclude

N→∞lim g(N)

N 6 lim

N→∞

k(N)·2^k(N)

N 6 lim

N→∞

δlog₂ ³₂N

· ³₂Nδ

N = 0.

2.4. Proof of Lemma 2.8

Let n1 < n2 < · · · < nk < . . . be the elements of Λ. For k, l ∈ N we consider the sets

B_k,l:={α∈R : q_l(α) =n_k andq_l+1(α)> ϕ(q_l(α)}, Bk:=[

l>1

Bk,l. These sets are open. Moreover we have

Bϕ:=\

N

[

k>N

Bk={α : #(Λ∩ Bϕ(α)) =∞}. This set is aGδ–set and it is left to prove that it is dense.

We observe first that if p∈Nand gcd(p, n_k) = 1, thenn_k is an approx- imant forp/nk. Moreover if

p nk

=a0+ 1

a₁+ 1

a2+. .. + 1 a_l ,

thenp=pl(p/nk),nk =ql(p/nk). Furthermore, for any integera`+1, let pl+1

ql+1

:=a0+ 1

a₁+ 1

a2+. .. + 1 al+ 1

al+1

.

Then we havepl(pl+1/ql+1) =pl and

(2.1) ql(pl+1/ql+1) =nk.

Moreover we have gcd(p_l+1, q_l+1) = 1, p_l+1 = a_l+1p_l+p_l−1 and q_l+1 = al+1ql+q_l−1.Hence ifal+1 is sufficiently large,

(2.2) pl+1

q_l+1 ∈B_k Now

(2.3)

p nk

−p_l+1 ql+1

=

p_l ql

−p_l+1 ql+1

= 1

nkql+1

< 1 n²_k.

It follows from (2.1), (2.2) and (2.3) that it remains to show that the reduced fractionsp/n_k are getting more and more dense askincreases. In fact, by Theorem 2.9, g(nk) = o(nk). This implies that two consecutive reduced fraction of the formp/nk have a distance of ordero(1) as ktends to infinity, which completes the proof of Lemma 2.8.

We finish our preliminaries with two facts on continued fractions which will be frequently used later:

∀n>1, 1

2q_n+1 6kqnαk6 1 q_n+1 (2.4)

∀m < q_n, kmαk>kqnαk.

(2.5)

We refer to Khinchin ([17, Theorem 9 and Theorem 13, Theorem 16]).

2.5. Proofs of Theorem 2.1 and of Theorem 2.2

We first prove Theorem 2.2. Letck =fb(k). The sequence{ck}k>1satisfies the condition of Lemma 2.7. Take the sequence Λ = {k`} in Lemma 2.7.

Apply Lemma 2.8 to Λ and ϕ(n) = e^∆n/c(n), where the constant ∆ >1 will be determined later and

c(n) = min{|ck<sub>`</sub>|:k`6n}, (n>1).

Then we get a residual setRf such that for each α∈ Rf there exists a subsequence of {k_{`</sub>} which is a subsequence {q<sub>n`}(α)} of {q_n(α)} (which depends onα!) such that

(2.6) ∀`>1, c(qn<sub>`</sub>(α)) logqn_{`</sub>+1(α)>∆qn<sub>`}(α).

The number αbeing fixed for the discussion below, we will simply write qn_{`</sub> andqn<sub>`}+1forqn_{`</sub>(α) andqn<sub>`}+1(α). Recall thatqn_{`</sub>(α) andqn<sub>`}+1(α) are the denominators of two consecutive convergents ofα.

TheN-th partial sum of the series in question can be written as SN(x) =

N

X

n=1

f(x+nα)

n = X

k∈Z\{0}

cke^2πikxGN(kα).

Let 0< <1/4 be a fixed small number. For any fixed`, we will consider the partial sum withN = [qn<sub>`</sub>+1]. We will cut the sum over k into four subsums:

Sq<sub>n`+1</sub>(x) =S`,A(x) +S`,B(x) +S`,C(x) +S`,D(x) where

S_`,A(x) = X

|k|<q_n`

c_ke^2πikxG_{qn`+1</sub>(kα) S<sub>`,B}(x) = X

|k|>q_n`+1

c_ke^2πikxG<sub>qn`+1</sub>(kα) S`,C(x) = X0

q_{n`</sub><|k|<q<sub>n`}+1

cke^2πikxGq_n`+1(kα).

S`,D(x) = X

16|m|6q_{n`</sub>+1/q<sub>n`}

cmq<sub>n`</sub>e<sup>2πimqn`x</sup>Gq_{n`+1</sub>(mqn<sub>`}α)

whereP0

means that the sum is taken overk’s which are not multiples of q_{n`</sub>. As we shall see,S<sub>`,D}(x) will be the principal term.

Since f is real, c_−k =ck and consequently all the four sums above are real.

For|k|< qn_{`</sub>, we havekkαk>1/qn<sub>`}. So, by Lemma 2.4, we have (2.7) |S`,A(x)|6 X

|k|<q_n`

|ck| ·πqn_{`</sub> 6πkfk<sub>A(T)</sub>qn<sub>`}.

Using the trivial estimate|GN(x)|6logN+γ+o(1) (γbeing the Euler constant) and the hypothesis|ck||k|^s=o(1), we get

(2.8) |S`,B(x)|6 X

|k|>q_n`+1

1

k^s·log(qn_{`</sub>+1) =O logqn<sub>`}+1

q_n^s−1_`+1

!

=O(1) For any ksuch thatq_{n`</sub> < k < q<sub>n`+1} andq_n`6 |k, we have

k=`qn<sub>`</sub>+r (16`6qn<sub>`</sub>+1/qn_{`</sub>, 16r < qn<sub>`}).

Then

kkαk>krαk − k`qn<sub>`</sub>αk> 1 qn_`

−qn_`+1

qn_`

· 1 qn_`+1

=1− qn_`

. By Lemma 2.4, for suchkwe have

|Gq_{n`+1</sub>(kα)|6 π 1−qn<sub>`}

so that

(2.9) |S`,C(x)|6 X⁰

q_{n`</sub><|k|<q<sub>n`}+1

|ck| · π

1−qn_` 6 π

1−kfk_A(T)qn_`. Since c_−k =ck, we have

|S`,D(x)|>2|cq<sub>n`</sub>||Gq_{n`</sub>+1(qn<sub>`}α)||cos(2πqn_{`</sub>x+φq<sub>n`})|

−2

∞

X

m=2

|cmq_{n`</sub>||Gq<sub>n`+1}(mqn_`α)|

where φq_{n`</sub> is the sum of the argument of cq<sub>n`} and the argument of Gq_{n`</sub>+1(qn<sub>`}α). Sincekqn_{`</sub>αk ·qn<sub>`}+16, by Lemma 2.5, we have

|Gq_{n`+1</sub>(qn<sub>`}α)|= logqn_`+1+O(1);

|Gq_{n`</sub>+1(mq<sub>n`}α)|6logq_n`+1+O(1) (∀m>2).

So,

|S`,D(x)|

(2.10)

>2 |cq_{n`</sub>||cos(2πqn<sub>`}x+φq_n`)| −

∞

X

m=2

|cmq_n`|

!

(logqn_`+1+O(1)).

When cos(2πq_{n`</sub>x+φ<sub>qn`}) is positive and when the difference on the right hand side of (2.10) is positive, we will have S`,D(x)>0 and we can take off the absolute value on the left hand side of (2.10).

Takeδ >0 such thatζ(s) +δ <2. Apply Lemma 2.6 to a small interval I= (−η, η) centered at zero such that cos 2πη > ζ(s)−1 +δ. For almost allx, there exist infinitely manyqn_` depending onxsuch that

cos(2πqn_{`</sub>x+φq<sub>n`})>ζ(s)−1 +δ.

For such`, if we use Lemma 2.7 we get

(2.11) S_{`,D</sub>(x)>2δ|c<sub>qn`}|(logq_n`+1+O(1)).

Combining (2.7), (2.8), (2.9) and (2.11), we obtain that for almost everyx we have

lim sup

N→+∞

N

X

n=1

f(x+nα)

n = +∞.

We choose

∆ = π

2δkfkA(T)

1 + 1

1−

.

We can also prove that for almost every xwe have lim inf

n→∞

N

X

n=1

f(x+nα)

n =−∞.

The only change to do is to take a small interval centered at 1/2 instead ofI= (−η, η). Thus we have proved Theorem 2.2.

The proof of Theorem 2.1 is easier. Because, in this case, f admits a Taylor–Fourier series and in the place of (2.10) we have directly the esti- mate

|S`,D(x)|> |cq<sub>n`</sub>| −

∞

X

m=2

|cmq_n`|

!

(logq_n`+1+O(1)).

2.6. Proof of Theorem 2.3

The idea of proof is the same as above. Take a summable sequence of positive numbers{c`}`>1such that

∀`>1, c`>

∞

X

j=`+1

cj.

For example,c` =r<sup>`</sup> with 0 < r <1/2. Take a very sparse subsequence {qn_`}from the denominators{qn}of the convergentspn/qn ofαsuch that

lim

`→∞

R_{`</sub>logq<sub>n`+1} qn`−1+1

= +∞, where R_{`</sub>=c<sub>`}−

∞

X

j=`+1

c_j.

Then define

f(x) =

∞

X

j=1

cje^2πiqnjx.

It is a continuous function withkfkA(T)<∞. Notice that it is a lacunary series in the sense thatf(n) = 0 forb n6=qn_j. We can write

q_n`+1

X

k=1

f(x+kα)

k =

∞

X

j=1

cje^2πiqnjxGq_n`+1(qnjα).

Cut the sum into

q_n`+1

X

k=1

f(x+kα)

k =S`,A(x) +S`,B(x) +S`,D(x)

where

S_`,A(x) =

`−1

X

j=1

c_je^2πiqnjxG_qn`+1(q_njα)

S`,B(x) =

∞

X

j=`+1

cje^2πiqnjxGq_n`+1(qnjα)

S_{`,D</sub>(x) =c<sub>`}e<sup>2πiqn`x</sup>G<sub>qn`+1</sub>(q_n`α).

Then

q_n`+1

X

k=1

f(x+kα) k

>|S`,D(x)| − |S`,A(x)| − |S`,B(x)|.

For j < `, we have kqn<sub>j</sub>αk >kqn<sub>`−1</sub>αk > 1/(2(qn_`−1+1)). Then by Lem- ma 2.4,

|S`,A(x)|62πkfk<sub>A(T)</sub>(qn<sub>`−1</sub>+1).

By the trivial estimate|GN(x)|6logN+O(1), we get

|S`,B(x)|6 X

j>`+1

|cj|(log(qn_`+1) +O(1)).

Sincekqn_{`</sub>αk61/qn<sub>`}+1, we havekqn_{`</sub>αk ·qn<sub>`}+16. So, by Lemma 2.5,

|S`,D(x)|>|c`|(log(qn`+1) +O(1)).

Thus

q_n`+1

X

k=1

f(x+kα) k

>R`(logqn<sub>`</sub>+1+O(1))−2πkfk_A(T)(qn`−1+1).

The right hand side tends to infinity.

3. Convergences of ^P∞_n=1anf(x+nα)

Now we present some results on the convergence (a.e. convergence, L²- convergence or uniform convergence) of the series

∞

X

n=1

anf(x+nα).

3.1. Almost everywhere convergence for almost all α

Theorem 3.1. — Assumef ∈L²(T)withR

f(x)dx= 0. For almost all α∈(0,1) and almost allx∈(0,1), the series P

anf(x+nα)converges if one of the following conditions is satisfied:

(1) P|an|²log²n <∞;

(2) P

|an|²logn < ∞ and f ∈ A(T) (in particular f ∈ Lip_γ with γ >1/2).

Proof.

(1). — On the product space T×T, the product measure dα⊗dx is considered as a probability measure. Then we consider the random variables

X_n=x+nα (mod 1), (n>0).

We claim that any coupleXn andXmwith n6=mare P-independent. In fact, take any two bounded Borel functionsg₁ andg₂ onT. We can prove that

Eg₁(X_n)g₂(X_m) =Eg₁(X_n)Eg₂(X_m)

where E refers to the expectation with respect to dα⊗dx. In fact, by developingg1andg2 into their Fourier series, we get

Eg₁(Xn)g₂(Xm) =X

bg₁(k₁)bg₂(k₂)Ee^{2πi(k1+k2)x+(nk1+mk2)α}

= X

k₁+k₂=0 nk₁+mk₂=0

bg₁(k₁)bg₂(k₂)

=bg1(0)bg2(0) =Eg1(Xn)Eg2(Xm).

SinceR

f(x)dx= 0, the above independence implies the orthogonality of the system{f(X_n)} in L²(dα⊗dx). Then, by the Menshov–Rademacher theorem and the hypothesis P

|an|²log²n < ∞, the random series Panf(Xn) converges dα⊗dx-almost everywhere. Hence, we conclude by using the Fubini theorem.

(2). — Assume thatf ∈A(T) andP∞

n=1|an|²logn < ∞. By a result of Gaposhkin [10] which is a consequence of the Carleson theorem on the a.e. convergence of Fourier series, for any given x, the seriesP

anf(x+ nα) converges for almost everyα. So, by the Fubini theorem, we conclude that for almost every α, the series P

anf(x+nα) converges for almost

everyx.

Notice that no tripleX`, Xm, Xn areP-independent.

3.2. Uniform convergence when αis diophantine

Theorem 3.2. — Let α be an irrational number, and let f ∈ L¹(T) withR

f(x)dx= 0, and (a_n)⊂C. Suppose X

n∈Z\{0}

|f(n)|b

knαk <∞; lim

n→∞an= 0,

∞

X

n=0

|an−an+1|<∞.

Then the seriesP∞

n=0anf(x+nα)converges everywhere, even uniformly onx.

Proof. — Under the first condition, the following cocycle equation ad- mits a unique solutiong∈A(T):

g(x+α)−g(x) =f(x).

Actually, by taking the Fourier transform, we find the solution g(x) = X

n∈Z\{0}

fb(n)

e^2πinα−1e^2πinα. Thus

X

n>0

anf(x+nα) =X

n>0

an[g(x+ (n+ 1)α)−g(x+nα)].

Sincean →0, by a summation by parts, we get X

n>0

anf(x+nα) =X

k>0

(ak−1−ak)g(x+kα) (with convention a−1 = 0). So P

|an−an+1| < ∞ implies the uniform

convergence of the series in question.

Recall that the irrationality measure (also called Liouville–Roth con- stant) of an irrational numberα, denoted byµ(α), is defined by

µ(α) = inf

µ:∃A >0,∀p∈Z,∀q∈N^∗,

α−p q > A

q^µ

. It is well known thatµ(α) = 2 for almost all irrational numbersα(Khint- chine), µ(α) = 2 for all irrational algebraic numbers (Roth), µ(e) = 2, µ(π) < 7,60630853, µ(log 2) < 3,57455391. If µ(α) = ∞, α is called a Liouville number. The set of Liouville numbers is aGδ dense set, but its Hausdorff dimension is zero.

Corollary 3.3. — Supposef ∈C^β(T)withβ > µ(α)andR

f(x)dx= 0. ThenP

anf(x+nα)converges uniformly (onx) if lima_n = 0, X

|an−a_n+1|<∞.

Proof. — By the hypothesis on f, we have |fb(n)| 6 B|n|^−β. By the definition ofµ(α) we have knαk>A|n|^−µ+1 forµ > µ(α). Thus

X|fb(n)|

knαk 6B A

X 1

|n|^β−µ+1 <∞.

Corollary 3.4. — For almost allα, for anyf ∈C²⁺withR

f(x)dx= 0, the seriesP

anf(x+nα)converges uniformly (onx) if lima_n = 0, X

|an−a_n+1|<∞.

3.3. L²-convergence and a.e. convergence when αis diophantine

Theorem 3.5. — Let f ∈ L²(T) with R

f(x)dx = 0. The series Panf(x+nα)converges inL²-norm if and only if

(3.1) lim

p,q→∞

X

k∈Z

|fb(k)|²

q

X

n=p

ane^2πinkα

2

= 0.

The condition(3.1) is satisfied when the series P∞

n=1a_ne^2πinx converges uniformly (onx). The condition(3.1)is also satisfied when

(3.2) X

n∈Z\{0}

|fb(n)|²

knαk² <∞; lim

n→∞an = 0,

∞

X

n=1

|an−an+1|<∞.

Proof. — We have Z

T

f(x+nα)f(x)dx=X

k∈Z

|f(k)|b ²e^2πiknα.

It follows that the spectral measure off is the following discrete measure σ_f =X

k∈Z

|fb(k)|²δ_kα. By the spectral lemma, we have

Z

q

X

p

a_nf(x+nα)

2

dx= Z

q

X

p

a_ne^2πint

2

σ_f(t).

Now we can conclude for the first assertion by the Cauchy criterion for L²-convergence.

The second assertion is an immediate consequence.

For the third assertion, we check the condition (3.1) by an Abel summa- tion and the fact thatPn

k=0e^2πikx=O(kxk⁻¹) and obtain

q

X

n=p

a_ne^2πinkα

6C |ap|+|aq|+ 1 kkαk

q−1

X

n=p

|an−a_n+1|

!

for some constantC >0.

In particular, a sufficient condition for P∞ n=1

f(x+nα)

n to converge inL²- norm is P

n∈Z\{0}

|^f(n)|b

2

knαk² < ∞. By Cuny’s result, P

n∈Z\{0}

|^f(n)|b

2

knαk² < ∞ implies the a.e. convergence ofPf(x+nα)

n . Similarly, a sufficient condition forP

n=1 (−1)ⁿ

n f(x+nα) to converge in L²-norm isP

n∈Z\{0}

|^f(n)|b

2

knα−1/2k² <∞. We should only notice that

q

X

p

(−1)ⁿ

n e^2πinkα=

q

X

p

1

ne2πin(kα−1/2)

and then make an Abel summation. It seems that the oscillation of (−1)ⁿ= e^2πin·1/2doesn’t promote the convergence. But for a fixedαand for almost allβ ∈(0,1), the series Pe^2iπnβ

n f(x+nα) converges in L² and a.e. This is a consequence of the result of Cuny [4] applied to the Dunford–Schwartz operator

T f(x) =e^2iπβf(x+α).

The size of the exceptional set ofβ was studied by Chevallier, Cohen and Conze [2]. Another oscillation sequence is the Möbius functionµ(n). It can be deduced from Cuny and Weber [7] that for any f ∈ L^p (p > 1), the seriesPµ(n)

n f(x+nα) converge inL^p and a.e.

The sufficient condition (3.2) is not so satisfactory, because P|^f(n)|b

2

knαk² <

∞is not so transparent. If we assumefb(k) =O(|k|^−β). ThenP|^f(n)|b

2

knαk² <∞ is ensured byβ > µ(α)−1/2 (>3/2). This can be improved toβ >1/2 in the case of the one-sided EHT (1.2). More precisely, we have the following theorem.

Theorem 3.6. — Let f ∈ L²(T) with f(0) = 0b and |fb(k)| 6 C|k|^−β whereC >0andβ >1/2are two constants. Letαbe an irrational number with convergents{p_n/q_n}. The one-sided EHTP∞

n=1

f(x+nα)

n converges in L²-mean and a.e. if the following condition is satisfied

(3.3)

∞

X

m=1

log²qm+1

q^2βm

<∞.