Julien Brémont

Planar random walk in a stratified quasi-periodic environment

Keywords: AMS 2020 subject classifications : 37E05, 37E10, 60G17, 60J10, 60K37 Markov chain, recurrence, stratified environment, quasi-periodic environment. 1

, we study the recurrence of inhomogeneous Markov chains in the plane, when the environment is horizontally stratified and the heterogeneity of quasi-periodic type.

Introduction

This article investigates the question of the recurrence of a class of inhomogeneous Markov chains in the plane, assuming the environment invariant under horizontal translations. This type of random walks were first considered by Matheron and de Marsily [START_REF] Matheron | Is transport in porous media always diffusive ? A counterexample[END_REF] around 1980, with a motivation coming from hydrology and the modelization of pollutants diffusion in a porous and stratified ground. In 2003, a discrete version was introduced by Campanino and Petritis in [START_REF] Campanino | Random walks on randomly oriented lattices[END_REF].

As in [START_REF] Brémont | On planar random walks in environments invariant by horizontal translations[END_REF][START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF][START_REF] Brémont | Random walk in a stratified independent random environment[END_REF], we consider an extension of the latter, restricting here to the plane and simplifying a little the hypotheses. We shall define a Markov chain (S k ) k≥0 in Z 2 , starting at the origin, such that the transition laws are constant on each stratum Z × {n}, n ∈ Z. The first and second coordinates will be respectively called "horizontal" and "vertical". For each (vertical) n ∈ Z, let positive reals α n , β n , γ n , with α n + β n + γ n = 1, and a probability measure µ n so that:

Hypothesis 1.1. ∃η > 0, ∀n ∈ Z, min{α n , β n , γ n } ≥ η, Supp(µ n ) ⊂ Z∩] -1/η, 1/η[, µ n (0) ≤ 1 -η.
The transition laws of (S k ) k≥0 are defined, for all (m, n) ∈ Z 2 and r ∈ Z, by: (m, n) αn -→ (m, n + 1), (m, n) βn -→ (m, n -1), (m, n)

γnµn(r) -→ (m + r, n).
Here is the corresponding picture:

m + r n n + 1 n -1 Z m Z 0 βn αn γnµn(r)
In [START_REF] Brémont | On planar random walks in environments invariant by horizontal translations[END_REF], for the general vertically flat case, a complete recurrence criterion was given. The asymptotic behaviour of the random walk is governed by the sums (γ -m ε -m /α -m + • • • + γ n-1 ε n-1 /α n-1 ) -m≤0≤n , associated with some horizontal flow defined by the environment and transverse to the vertical layer [-m, n). The central role is played by a two-variable function Φ(-m, n), introduced below, measuring the "horizontal dispersion" of the previous flow between vertical levels -m and n. The quantity deciding for the recurrence/transience of (S k ) k≥0 computes some "capacity of dispersion to infinity" of the environment. The abstract form of the criterion in [START_REF] Brémont | On planar random walks in environments invariant by horizontal translations[END_REF] is directly extracted from a Poisson kernel in a half-plane. It seems to be related to some notion of curvature at infinity of the level lines of the function Φ(-m, n). Several examples were next presented in [START_REF] Brémont | On planar random walks in environments invariant by horizontal translations[END_REF]. Roughly, a growth condition such as (log n) 1+δ on (γ 0 ε 0 /α 0 + • • • + γ n-1 ε n-1 /α n-1 ) is sufficient for transience, confirming the natural prevalence of transience results in the litterature on this model.

Extending [START_REF] Brémont | On planar random walks in environments invariant by horizontal translations[END_REF], for the general model where α n need not equal β n , a full characterization of the recurrence regime was shown in [START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF]. With some naturally generalized Φ(-m, n) (cf Definition 4.6 below), the form of the criterion is the same, emphasizing the fact that the environment defines a new metrization of Z 2 . The model appears notably more general than the vertically flat one. Several examples were next given in [START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF]. However, an empirical observation is that the methods employed for obtaining the structural results of [START_REF] Brémont | On planar random walks in environments invariant by horizontal translations[END_REF][START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF] are of very different nature than that used to treat examples. The analysis is in fact naturally divided in two parts, the second one never entering the mechanism of the random walk itself. The latter consists in studying fine properties of certain ergodic sums and is a source of interesting and difficult problems, for example closely related to temporal limit theorems and generalizations (cf Dolgopyat-Sarig [START_REF] Dolgopyat | Quenched and annealed temporal limit theorems for circle rotations[END_REF]). In [START_REF] Brémont | Random walk in a stratified independent random environment[END_REF], for the general model, the particular case when the transition laws are independent was studied in detail, with a precise quantification of the non-surprising fact that the transience regime largely prevails in the set of parameters.

The purpose of the present article is to complete [START_REF] Brémont | On planar random walks in environments invariant by horizontal translations[END_REF][START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF][START_REF] Brémont | Random walk in a stratified independent random environment[END_REF], by extending the applications of [START_REF] Brémont | On planar random walks in environments invariant by horizontal translations[END_REF][START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF]. We study for both the vertically flat and the general model the case when the transition laws are described by functions defined above an irrational rotation on the one-dimensional torus.

Preliminaries

Let T = R\Z be the one-dimensional torus. Unless otherwise stated, functions are defined on T, with arguments understood modulo one. We write x for the distance of a real x to Z.

We first recall classical facts about continued fractions. On this topic, one may consult Khinchin's book [START_REF] Khinchin | Continued fractions[END_REF]. Any irrational 0 < θ < 1 admits an infinite continued fraction expansion:

θ = 1 a 1 + 1 a 2 + • • • = [0, a 1 , a 2 , • • • ],
where the partial quotients (a i ) i≥1 are integers ≥ 1, obtained by successive iterations of the Gauss map x -→ {1/x}, starting from θ. The convergents (p n /q n ) n≥1 of θ are the truncations [0, a 1 , a 2 , • • • , a n ] = p n /q n , for n ≥ 1, of this continued fraction. The numerators (p n ) and denominators (q n ) check the same recurrence relation:

p n+1 = a n+1 p n + p n-1 , q n+1 = a n+1 q n + q n-1 , n ≥ 0,
with initial data p 0 = 0, p -1 = 1 and q 0 = 1, q -1 = 0. Classically (cf [START_REF] Khinchin | Continued fractions[END_REF], chap. 1):

1 2q n+1 ≤ 1 q n + q n+1 ≤ q n θ ≤ 1 q n+1 . (1)
Fixing θ ∈ Q, we consider the rotation T x = x + θ mod 1 on T and write T n f for f • T n , for any f : T → R. We also use cocycle notations, for x ∈ T:

f n (x) =    f (x) + • • • + f (T n-1 x), n ≥ 1, 0, n = 0, -f (T n x) -• • • -f (T -1 x), n ≤ -1. An important property is that f n+p (x) = f n (x) + T n f p (x), for any x ∈ T and n, p ∈ Z.
A function f : T -→ R with bounded variation will be said BV, with total variation written as V (f ). When f is BV, with T f (x) dx = 0, the Denjoy-Koksma inequality says that:

|f qn (x)| ≤ V (f ), n ≥ 1, x ∈ T. (2)
Let us now recall known facts concerning Ostrowski's expansions (cf Beck [START_REF] Beck | Probabilistic Diophantine approximation[END_REF], p. 23). Every integer q m ≤ n < q m+1 can be represented as:

n = 0≤k≤m b k q k , (3)
with 0 ≤ b 0 ≤ a 1 -1, 0 ≤ b j ≤ a j+1 , 1 ≤ j < m, and 1 ≤ b m ≤ a m+1 . Indeed, n = b m q m + r, for some 0 ≤ r < q m and 1 ≤ b m ≤ a m . Iterating this process furnishes the decomposition (3). Setting A -1 = 0 and A k = 0≤j≤k b j q j , for 0 ≤ k < m, by (3), we have for any function f :

f n (x) = m k=0 f b k q k (x + A k-1 θ).
When f is BV and centered, the Denjoy-Koksma inequality (2) furnishes the upper-bound:

|f n (x)| ≤ 0≤k≤m f q k ∞ b k ≤ V (f ) 0≤k≤m b k , x ∈ T. (4)
Set N = {0, 1, • • • }. For g : N → R + increasing to +∞ and x ≥ g(0), let g -1 (x) be the unique integer n ≥ 0 such that g(n) ≤ x < g(n + 1). By definition:

g(g -1 (x)) ≤ x < g(g -1 (x) + 1). (5)
Also, g -1 (g(n)) = n, for large n ∈ N. Finally, for f, g : N → R + , we write g f if there exists a constant C > 0 so that g(n) ≤ Cf (n), for large n ∈ N. We write f g if g f and f g.

The quasi-periodic vertically flat model

We first consider the vertically flat model, i.e.

α n = β n = (1-γ n )/2, n ∈ Z.
As a preliminary remark, we discuss the case when the sequence (ε

n γ n /(1 -γ n )) n∈Z is periodic. Proposition 3.1.
For the vertically flat model, let (ε n γ n /(1 -γ n )) n∈Z be periodic with period N ≥ 1. Then the random walk is either recurrent or transient, according to whether:

0≤n<N ε n γ n /(1 -γ n ) = 0 or = 0.
This follows from [START_REF] Brémont | On planar random walks in environments invariant by horizontal translations[END_REF], respectively Prop 1.4 i) and Corollary 1.3 i). This extends the case of the Campanino-Petritis model in [START_REF] Campanino | Random walks on randomly oriented lattices[END_REF], when

µ n = δ xn with x n = ε n = (-1) n , α n = β n = γ n = 1/3.
Turning to quasi-periodic situations, we shall generalize [START_REF] Brémont | On planar random walks in environments invariant by horizontal translations[END_REF], Prop. 1.5, giving in particular a better understanding of the Campanino-Petritis model in this quasi-periodic context.

Theorem 3.2. Let θ = [0, a 1 , a 2 , • • • ] ∈ Q, with n≥1 log(1 + a n )/(a 1 + • • • + a n ) = +∞, and T x = x + θ mod 1 on T. Let f : T → R be piecewise K-Lipschitz, with zero mean. Under Hypothesis 1.1, let α n = β n and ε n γ n /(1 -γ n ) = f (nθ), n ∈ Z. Then the random walk is recurrent. Remark. -This is shown in [4], Prop. 1.5, for f = (1 [0,1/2) -1 [1/2,1) )/2, corresponding to the Campanino-Petritis model with µ n = δ xn and x n = 1 [0,1/2) (nθ) -1 [1/2,1) (nθ).
As a consequence of the theorem, the random walk is recurrent when

x n = 1 [0,1/2) (x + nθ) -1 [1/2,1) (x + nθ), n ∈ Z,
for any x ∈ T, taking f (. + x). As noticed in [START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF], Prop. 7.1, the condition on θ is generic in measure, since n≥1 1/(a 1 + • • • + a n ) = +∞, for a.-e. θ, cf Khinchin [START_REF] Khinchin | Metrische Kettenbruchprobleme[END_REF].

The other direction is in general more delicate, since requiring lower bounds on the ergodic sums. We just give an example.

Proposition 3.3. Let θ = [0, a 1 , a 2 , • • • ] ∈ Q,
with a 1 odd and a n even for n ≥ 2. We suppose that for some δ > 1, a n+1 ≥ (a n ) δ , for large n. Let f = 1 [0,1/2) -1 [1/2,1) . Under Hypothesis 1.1, let α n = β n and ε n γ n /(1 -γ n ) = f (x + nθ), n ∈ Z, for some x ∈ T. Then, for Lebesgue almost-every x ∈ T, the random walk is transient.

Remark. -In the last proposition, one may take for example the angle θ ∈ (0, 1) defined by the partial quotients a n = 2 2 n-1 -1 , n ≥ 1.

3.1 Proof of Theorem 3.2

Fix θ = [0, a 1 , a 2 , • • • , • • • ] ∈ Q, T x = x +
θ mod 1 on T and f , as in the statement of the theorem. Using cocycle notations (f n (x)) n∈Z , introduce for n ≥ 1 the following positive functions ϕ(n) and ϕ + (n) such that:

ϕ 2 (n) = n 2 + -n≤k< ≤n (f (x) -f k (x)) 2 and ϕ 2 + (n) = n 2 + -n≤k< ≤n,k ≥0 (f (x) -f k (x)) 2 .
The dependence on x of ϕ(n) and ϕ

+ (n) is implicit. Obviously, n ≤ ϕ + (n) ≤ ϕ(n).
The next lemma gives some control in the other direction.

Lemma 3.4.

There exists a constant C 0 > 0, uniform on x ∈ T, such that for all n ≥ 1 and 1 ≤ m ≤ 4a n+1 :

ϕ 2 (mq n ) ≤ 2ϕ 2 + (mq n ) + C 0 m 4 q 2 n .

Proof of the lemma:

Step 1. In the sequel, we simplify

f n (x) into f n , n ∈ Z. Setting A = -n≤k≤-1,1≤ ≤n (f -f k ) 2 , we have ϕ 2 (n) = ϕ 2 + (n) + A. Then: A = n 1≤ ≤n f 2 + n -n≤k≤-1 f 2 k -2 -n≤k≤-1 f k 1≤ ≤n f . (6)
We next have:

-2 -n≤k≤-1 f k 1≤ ≤n f =   1≤ ≤n (f -f -)   2 -   1≤ ≤n f   2 -   1≤ ≤n f -   2 . (7)
Now, classically:

1≤k< ≤n (f -f k ) 2 = 2≤ ≤n ( -1)f 2 + 1≤ ≤n-1 (n -)f 2 -2 1≤k< ≤n f k f = 1≤ ≤n ( -1)f 2 + 1≤ ≤n (n -)f 2 -2 1≤k< ≤n f k f = n 1≤ ≤n f 2 -   1≤ ≤n f   2 . ( 8
)
Proceeding symmetrically for the other part of A, we obtain from ( 6), ( 7) and ( 8):

A = 1≤k< ≤n (f -f k ) 2 + -n≤k< ≤-1 (f -f k ) 2 +   1≤ ≤n (f -f -)   2 .
Consequently:

ϕ 2 (n) ≤ 2ϕ 2 + (n) +   1≤ ≤n (f -f -)   2 .
Step 2. Let n ≥ 1 and 1 ≤ m ≤ 4a n+1 . Setting B = 1≤ ≤mqn (f -f -), we have:

B = 0≤u<m 1≤ ≤qn (f uqn+ -f -uqn-) = 0≤u<m 1≤ ≤qn (f uqn -f -uqn + T uqn f -T -uqn f -).
Using Denjoy-Koksma's inequality (2), for any integer 0 ≤ u < m, we have |f uqn (x)| ≤ uV (f ), idem for f -uqn (x). As a result:

B = O(m 2 q n ) + 0≤u<m 1≤ ≤qn (T uqn f -T -uqn f -).
Fixing any integer 0 ≤ u < m, we have :

1≤ ≤qn (T uqn f -T -uqn f -) = qn-1 k=0 (q n -k)f (x + uq n θ + kθ) + qn k=1 (q n + 1 -k)f (x -uq n θ -kθ).
Using the Denjoy-Koksma inequality (2) for the (q n + 1)-term in the second sum on the right hand side and making a change of variable in the first one, we get:

1≤ ≤qn (T uqn f -T -uqn f -) = qn k=1 k(f (x + uq n θ + (q n -k)θ) -f (x -uq n θ -kθ)) + O(q n ) = qn k=1 k(f (x u + -kθ) -f (x u --kθ)) + O(q n ),
when setting x u + = x + (u + 1)q n θ and x u -= x -uq n θ. By (1) and the hypothesis m ≤ 4a n+1 :

x u + -x u -≤ (2u + 1) q n θ ≤ (8a n+1 + 1)/q n+1 ≤ 9/q n .
Denote by [x u -, x u + ] the short interval on T between x u -and x u + and by D the number of discontinuities of f . Recall also that there is exactly one kθ mod 1, 1 ≤ k ≤ q n , in each interval [ /q n , ( + 1)/q n ), 0 ≤ < q n , on T. As a result, for a given discontinuity of f , there at most 10 values of 1 ≤ k ≤ q n such that [x u -, x u + ] -kθ contains this discontinuity. Hence, using that f is K-Lipschitz on intervals containing no discontinuity, an upper-bound for the last sum is:

D × 10q n × 2 f ∞ + q 2 n × K × 9 q n = O(q n ),
As a result, B = O(m 2 q n ) + O(mq n ) = O(m 2 q n ), which ends the proof of the lemma.

We turn to the proof of Theorem 3.2. We assume that α n = β n and that µ n and thus ε n are such that ε n γ n /(1 -γ n ) = f (x + nθ), n ∈ Z, for some x ∈ T. The statement of the theorem corresponds to x = 0. Introduce the following definition, due to Feller (1969): Definition 3.5. A non-decreasing function g : R + → R + satisfies dominated variation, if there exists a constant C > 0 so that for large x > 0, g(2x) ≤ Cg(x). Hence, iterating, for all K > 0, there exists C K so that for large x > 0, g(Kx) ≤ C K g(x).

In [START_REF] Brémont | On planar random walks in environments invariant by horizontal translations[END_REF], setting R k = k≤i≤ ε i γ i /(1 -γ i ), for integers k ≤ , we considered the two functions:

Φ 2 (n) = n 2 + -n≤k≤ ≤n (R k ) 2 , Φ 2 + (n) = n 2 + -n≤k≤ ≤n,k >0 (R k ) 2 ,
The following results were then established:

Theorem 3.6. ([4], Lemma 6.1, Theorem 1.2. and Corollary 1.3 i))
1) The functions Φ and Φ + satisfy dominated variation with a constant C = C(η) depending only on η, where η is introduced in Hypothesis 1.1.

2) The random walk is recurrent if and only

if n≥1 n -2 (Φ -1 (n)) 2 /Φ -1 + (n) = +∞. 3) The condition n≥1 1/Φ + (n) < +∞ is sufficient for transience.
Using that f is bounded and that

|R k | -|f (x) -f k (x)| ≤ f ∞ , k ≤ , it
is immediate that for some constant C > 0 depending only on η (hence uniform on x ∈ T), for all n ≥ 1:

Φ(n)/C ≤ ϕ(n) ≤ CΦ(n) and Φ + (n)/C ≤ ϕ + (n) ≤ CΦ + (n). Corollary 3.7.
1) The functions ϕ -1 and ϕ -1 + satisfy dominated variation, i.e. for any K > 0, there exists a constant C K > 0, independent of x ∈ T, so that for large y > 0:

ϕ -1 (Ky) ≤ C K ϕ -1 (y) and ϕ -1 + (Ky) ≤ C K ϕ -1 + (y). (9)
2) The random walk is recurrent if and only

if n≥1 n -2 (ϕ -1 (n)) 2 /(ϕ -1 + (n)) = +∞. 3) The condition n≥1 1/ϕ + (n) < +∞ is sufficient for transience.
Let us reprove a concrete version of dominated variation of ϕ -1 and ϕ -1 + in the following lemma. Concerning for example ϕ + , we essentially show that n

-→ ϕ 2 + (n)/n, n > 0, is non- decreasing. For a < b in Z and x ∈ T, let: ψ(a, b) = a≤k< ≤b (f (x) -f k (x)) 2 , ( 10
)
where the dependence on x is implicit on the left hand side.

Lemma 3.8. Let integers a < b < c and x ∈ T. Then:

ψ(a, c) c -a ≥ ψ(a, b -1) b -a + ψ(b + 1, c) c -b .
Also, for large n, uniformly in x ∈ T:

ϕ + (2n) ≥ 2 1/4 ϕ + (n). ( 11
)
Proof of the lemma:

We write f k in place of f k (x). Decompose ψ(a, c) = ψ(a, b) + ψ(b, c) + a≤k<b< ≤c (f -f k ) 2
and then expend:

a≤k<b< ≤c (f -f k ) 2 = (b -a) b< ≤c f 2 + (c -b) a≤k<b f 2 k -2 b< ≤c f a≤k<b f k . As before, ψ(b + 1, c) = (c -b) b< ≤c f 2 -( b< ≤c f ) 2 and ψ(a, b -1) = (b -a) a≤k<b f 2 k - ( a≤k<b f k ) 2 . When substituting: a≤k<b< ≤c (f -f k ) 2 = b -a c -b   ψ(b + 1, c) + ( b< ≤c f ) 2   + c -b b -a   ψ(a, b -1) + ( a≤k<b f k ) 2   -2 b< ≤c f a≤k<b f k .
As a consequence:

ψ(a, c) ≥ c -a c -b ψ(b + 1, c) + c -a b -a ψ(a, b -1) +   b -a c -b b< ≤c f - c -b b -a a≤k<b f k   2 , (12)
giving the result. Concerning the last part of the lemma, for any integer n ≥ 1, we have: [START_REF] Devulder | Random walk in random environment in a two-dimensional stratified medium with orientations[END_REF].

ψ(0, 2n) ≥ 2n n + 1 ψ(0, n), ψ(-2n, 0) ≥ 2n n + 1 ψ(-n, 0). Since ϕ 2 + (n) = n 2 + ψ(-n, 0) + ψ(0, n), we get
Remark. -As a complement, let us observe that:

c -a c -b ψ(b + 1, c) + c -a b -a ψ(a, b -1) ≥ ( ψ(a, b -1) + ψ(b + 1, c)) 2 , ( 13
)
as this is equivalent to the true relation:

c -b b -a ψ(a, b -1) + b -a c -b ψ(b + 1, c) ≥ 2 ψ(a, b -1) ψ(b + 1, c).
Using ( 13) in [START_REF] Dolgopyat | Quenched and annealed temporal limit theorems for circle rotations[END_REF], this thus implies some reverse triangular inequality:

ψ(a, c) ≥ ψ(a, b -1) + ψ(b + 1, c).
We start the proof of Theorem 3.2. A corollary of Lemma 3.4 is that there exists a constant C 0 > 0, independent of x ∈ T, such that for all n ≥ 1 and 1 ≤ m ≤ 4a n+1 :

ϕ(mq n ) ≤ C 0 (ϕ + (mq n ) + m 2 q n ). ( 14
)
Let now n ≥ 1 and ≥ 0 be such that 2 ≤ 4a n+1 . We make the following discussion:

-Case 1. If ϕ + (2 q n ) ≥ 2 2
q n , then, using ( 14) and next ( 9) at the end:

ϕ -1 (ϕ + (2 q n )) ≥ ϕ -1 ((ϕ + (2 q n ) + 2 2 q n )/2) ≥ ϕ -1 (ϕ(2 q n )/(2C 0 )) ≥ ϕ -1 (ϕ(2 q n )) C 2C0 = 2 q n C 2C0 . By Ostrowski's expansion (3), for |k| ≤ 2 q n ≤ 4q n+1 , |f k | ≤ 4 × V (f )(a 1 + • • • + a n+1 ). Hence, for some C > 0, ϕ + (2 q n ) ≤ C2 q n (a 1 + • • • + a n+1 ). Thus, with C 1 = 1/(CC 2C0
), for large n:

(ϕ -1 (ϕ + (2 q n ))) 2 2 q n ϕ + (2 q n ) ≥ 1 C 2C0 2 q n ϕ + (2 q n ) ≥ C 1 a 1 + • • • + a n+1 . (15)
-Case 2. Suppose that ϕ + (2 q n ) < 2 2 q n . As ϕ + (2 q n ) ≥ 2 q n , there exists 0 ≤ ≤ so that 2 2 q n ≤ ϕ + (2 q n ) < 2 2( +1) q n . We get in this case, using ( 14) and ( 9):

ϕ -1 (ϕ + (2 q n )) ≥ ϕ -1 (2 2 q n ) ≥ ϕ -1 ((ϕ + (2 q n ) + 2 2 q n )/5) ≥ ϕ -1 (ϕ(2 q n )/(5C 0 )) ≥ ϕ -1 (ϕ(2 q n )) C 5C0 = 2 q n C 5C0 .
As a result, we can write in this case, redefining

C 1 := min{C 1 , 1/(16C 5C0 )} > 0: (ϕ -1 (ϕ + (2 q n ))) 2 2 q n ϕ + (2 q n ) ≥ 1 C 5C0 2 2 q 2 n 2 q n ϕ + (2 q n ) ≥ 1 4C 5C0 2 ≥ C 1 a 1 + • • • + a n+1 .
Hence, ( 15) is true for any ≥ 0 such that 2 ≤ 4a n+1 . This now gives, for large n > 0:

ϕ+(qn)≤k<ϕ+(4an+1qn) 1 k 2 (ϕ -1 (k)) 2 ϕ -1 + (k) ≥ 0≤ ≤1+log 2 an+1 ϕ+(2 qn)≤k<ϕ+(2 +1 qn) 1 k 2 (ϕ -1 (k)) 2 ϕ -1 + (k) ≥ 0≤ ≤1+log 2 an+1 (ϕ -1 (ϕ + (2 q n ))) 2 ϕ -1 + (ϕ + (2 +1 q n )) ϕ+(2 qn)≤k<ϕ+(2 +1 qn) 1 k 2 ≥ 0≤ ≤1+log 2 an+1 (ϕ -1 (ϕ + (2 q n ))) 2 2 +1 q n ϕ+(2 qn)≤k<ϕ+(2 +1 qn) 1 k 2 .
Using now relation [START_REF] Khinchin | Metrische Kettenbruchprobleme[END_REF], we arrive at:

ϕ+(qn)≤k<ϕ+(4an+1qn) 1 k 2 (ϕ -1 (k)) 2 ϕ -1 + (k) ≥ C 1 2(a 1 + • • • + a n+1 ) 0≤ ≤1+log 2 an+1 ϕ + (2 q n ) ϕ+(2 qn)≤k<ϕ+(2 +1 qn) 1 k 2 .
Using that 1/k 2 ≥ 1/k -1/(k + 1), we obtain:

ϕ+(2 qn)≤k<ϕ+(2 +1 qn) 1 k 2 ≥ 1 ϕ + (2 q n ) + 1 - 1 ϕ + (2 +1 q n ) -1 .
Now, for large n (uniformly in ≥ 0), applying [START_REF] Devulder | Random walk in random environment in a two-dimensional stratified medium with orientations[END_REF], we have ϕ

+ (2 +1 q n ) ≥ 2 1/4 ϕ + (2 q n ).
As a result, for large n, we obtain:

ϕ+(2 qn)≤k<ϕ+(2 +1 qn) 1 k 2 ≥ 1 -2 -1/4 2 1 ϕ + (2 q n ) .
This thus furnishes, for large n > 0:

ϕ+(qn)≤k<ϕ+(4an+1qn) 1 k 2 (ϕ -1 (k)) 2 ϕ -1 + (k) ≥ C 1 (1 -2 -1/4 ) 4(a 1 + • • • + a n+1 ) (1 + log 2 a n+1 ).
Finally, notice that 4a n+1 q n ≤ q n+5 , for n ≥ 0. Hence, for large N > 0:

k≥ϕ+(q N ) (ϕ -1 (k)) 2 ϕ -1 + (k) ≥ 1 5 n≥N ϕ+(qn)≤k<ϕ+(qn+5) (ϕ -1 (k)) 2 ϕ -1 + (k) ≥ C 1 (1 -2 -1/4 ) 5 n≥N (1 + log 2 a n+1 ) 4(a 1 + • • • + a n+1 )
.

By hypothesis, the series on the right hand side diverges. From Corollary 3.7, we conclude that the random walk is recurrent.

Proof of Proposition 3.3

Let us place in the context of this proposition, namely

α n = β n and ε n γ n /(1-γ n ) = f (x+nθ), n ∈ Z, for some x ∈ T. Here f = 1 [0,1/2) -1 [1/2,1) and the angle θ = [0, a 1 , a 2 , • • • ] is such that a 1 is
odd and a n is even for n ≥ 2, together with a n+1 ≥ (a n ) δ , for large n, for some fixed δ > 1.

From the relation q n+1 = a n+1 q n + q n-1 , n ≥ 0, and q 0 = 1, q -1 = 0, we recursively obtain that q n is odd, n ≥ 1. This implies that for all n ≥ 1 and x ∈ T, |f qn (x)| ≥ 1, since the number of (T k x) 0≤k<qn that fall in the intervals [0, 1/2) and [1/2, 1) are different.

Let 2/3 < β < 1 and define m k = (a k+1 ) β , k ≥ 1. Introduce :

A k = {x ∈ T, f mq k (x) = mf q k (x), 0 ≤ m ≤ m k }.
Write L T for Lebesgue measure on T and denote by [x, x+q k θ] the short interval on T determined by x and x + q k θ on T.

If T r [x, x + q k θ] does not contain either 0 or 1/2, for 0 ≤ r < m k q k , then x ∈ A k , because f (T r x) = f (T r+q k x), for 0 ≤ r < m k q k . Hence, if x ∈ T\A k , there exist 0 ≤ r < m k q k such that either 0 or 1/2 belongs to T r [x, x + q k θ]. As a result, T\A k ⊂ ∪ 0≤r<m k q k ,y∈{0,1/2} ([y -q k θ, y] -rθ). Using (1): L T (T\A k ) ≤ 2m k q k q k+1 ≤ 2m k a k+1 ≤ 2(a k+1 ) -(1-β) .
Since (a k ) grows at least geometrically, k a

-(1-β) k < +∞.
By the first lemma of Borel-Cantelli, we deduce that for Lebesgue a.-e. x, x ∈ A k for large k.

Let N k = a 1 + • • • + a k and observe that N k ∼ k→+∞ a k , since for large k, a k ≥ (a k-1 ) δ , with δ > 1. As m k = (a k+1 ) β , we now choose β < 1 close enough to 1, so that m k ≥ 1000N k , for large k. Let now m k ≥ m ≥ 100N k . For 0 ≤ , ≤ mq k ,
we make the Euclidean divisions of , by q k : = aq k + b and = a q k + b , 0 ≤ b, b < q k and where 0 ≤ a, a ≤ m k . Then, almost-surely for large k, since x ∈ A k , we have f aq k (x) = af q k (x), f a q k (x) = a f q k (x). Thus:

f (x) -f (x) = (a -a )f q k (x) + T aq k f b (x) -T a q k f b (x).
Using the upper-bound (4), coming from Ostrowski's expansion (3), for f b ∞ and f b ∞ , we have (since

V (f ) = 2), |T aq k f b (x)| ≤ 2N k and |T a q k f b (x)| ≤ 2N k . Hence, a.-e., for large k, using the fact that |f q k (x)| ≥ 1: |f (x) -f (x)| ≥ |a -a ||f q k (x)| -|T aq k f b (x) -T a q k f b (x)| ≥ |a -a | -4N k . (16)
Consequently, for m k ≥ m ≥ 100N k , by ( 16):

ϕ 2 + (mq k ) ≥ 0≤ ≤ ≤mq k (f -f ) 2 ≥ 0≤a <m/4;m/2<a<m;0≤b,b <q k (f aq k +b (x) -f a q k +b (x)) 2 ≥ 0≤a <m/4;m/2<a<m;0≤b,b <q k (m/4 -4N k ) 2 ≥ (m/4) 2 q 2 k (m/5) 2 ≥ m 4 q 2 k /400. (17)
In order to conclude the argument we shall apply Corollary 3.7, 3), and show the convergence of n≥1 1/ϕ + (n). Let us write for a.-e. x and large K > 0:

n≥ K 1 ϕ + (n) ≤ k≥K 1≤m≤a k+1 mq k ≤n<(m+1)q k 1 ϕ + (n) ≤ k≥K 1≤m≤a k+1 q k ϕ + (mq k ) ≤ k≥K   1≤m≤100N k + 100N k <m≤m k + m k <m≤a k+1   q k ϕ + (mq k ) = k≥K [ U k + V k + W k ] . 1) Considering k≥K U k , we fix k ≥ K and 1 ≤ m ≤ 100N k .
Using first the function ψ introduced in (10), before Lemma 3.8, we have:

ϕ 2 + (mq k ) = (mq k ) 2 + ψ(-mq k , 0) + ψ(0, mq k ). ( 18
)
By Lemma 3.8 and next [START_REF] Kim | The recurrence time for interval exchange maps[END_REF], for some (next generic) constant c > 0, a.-e. for large k > 0:

ϕ + (mq k ) ≥ c mq k /(m k-1 q k-1 )ϕ + (m k-1 q k-1 ) ≥ c mq k /(m k-1 q k-1 )m 2 k-1 q k-1 .
As a result:

ϕ + (mq k ) ≥ c √ ma k m 3/2 k-1 q k-1 ≥ c √ ma 1/2+3β/2 k q k-1 ≥ c √ ma 3β/2-1/2 k q k . (19)
We obtain, a.-e., for large K > 0, via [START_REF] Pène | Transient random walk in Z 2 with stationary orientations[END_REF], still for a generic c > 0, using at the end that N k ∼ k→+∞ a k , β > 2/3 and that (a k ) grows at least geometrically:

k≥K U k ≤ c k≥K 1≤m≤100N k 1 √ ma 3β/2-1/2 k ≤ c k≥K √ N k a 3β/2-1/2 k ≤ c k≥K √ a k a 3β/2-1/2 k ≤ c k≥1 a 1-3β/2 k < +∞.
2) Considering now k≥K V k , let k ≥ K and 100N k < m ≤ m k . Using [START_REF] Kim | The recurrence time for interval exchange maps[END_REF], we have a.-e. for large K > 0 that ϕ + (mq k ) ≥ m 2 q k /20. So, for a generic constant c > 0:

k≥K V k ≤ c k≥K 100N k <m≤m k 1 m 2 ≤ c k≥K 1 N k ≤ c k≥K a -1 k < +∞.
3) For k≥K W k , let k ≥ K and m k < m ≤ a k+1 . Again, using (18), Lemma 3.8 and finally [START_REF] Kim | The recurrence time for interval exchange maps[END_REF], for m k , we obtain, a.-e., for large K > 0, for some generic constant c > 0:

ϕ + (mq k ) ≥ c m/m k ϕ + (m k q k ) ≥ c m/m k m 2 k q k ≥ c √ mm 3/2 k q k .
As a consequence, we can write, for some generic c > 0, using at the end that β > 2/3 > 1/3:

k≥K W k ≤ c k≥K m k <m≤a k+1 1 √ mm 3/2 k ≤ c k≥K √ a k+1 m 3/2 k ≤ c k≥K a 1/2-3β/2 k+1 < +∞.
This completes the proof of the proposition.

The quasi-periodic general case

Let again T x = x + θ mod 1 be an irrational rotation on T. The basic assumption in this section will be that for some BV functions f, g : T → R, with f centered, for some x ∈ T:

β n /α n = e f (T n-1 x) , γ n ε n /α n = g(T n x), n ∈ Z.
Implicitly, Hypothesis 1.1 will always be realized, uniformly in x ∈ T. Introduce some definitions.

ρ n =    β1 α1 • • • βn αn , n ≥ 1, 1, n = 0, αn+1 βn+1 • • • α0 β0 , n ≤ -1.
For n ≥ 0, let:

v + (n) = 0≤k≤n ρ k and v -(n) = (q 0 /p 0 ) -n-1≤k≤-1 ρ k .
In the same way, introduce for n ≥ 0:

w + (n) = 0≤k≤n 1/ρ k and w -(n) = (p 0 /q 0 ) -n-1≤k≤-1 1/ρ k .
As already mentioned in the Introduction, for the random walk to be recurrent, the vertical random walk has first to be. The necessary and sufficient condition for this (cf [START_REF] Karlin | The classification of birth and death processes[END_REF]) is:

lim n→+∞ v + (n) = +∞ and lim n→+∞ v -(n) = +∞.
Since f is BV and centered, the Denjoy-Koksma inequality (2) says that |f ±qn (x)| ≤ V (f ), for any x ∈ T. As a result, ρ n = e fn(x) does not go to zero, neither as n → +∞, nor as n → -∞, implying that the two previous conditions hold.

Some quasi-invariant measures on T will play a role. In the sequel, we consider the space of Borel probability measures on T, equipped with its usual (metrizable) weak- * topology (using continuous w : T → R, as test functions), for which this space is compact. We denote by T ν the image by T of a Borel probability measure ν on T. By definition, w dT ν = T w dν, for any bounded measurable w : T → R. Let us recall the following folklore result. Theorem 4.2. ([9], Prop. 5.8, or [START_REF] Aaronson | Invariant measures and asymptotics for some skew products[END_REF], Prop. 1.1.) Let h : T → R be BV and centered. There exists a unique Borel probability measure ν h on T such that dT ν h = e T -1 h dν h . This measure has no atom.

The proof of existence, relying on non-atomicity, is incomplete in [START_REF] Conze | Marches en milieu aléatoire et mesures quasi-invariantes pour un système dynamique[END_REF] and too abstract in [START_REF] Aaronson | Invariant measures and asymptotics for some skew products[END_REF]. We choose to reprove existence and atomicity in an elementary way below. Also, it is well-known (cf [START_REF] Conze | Marches en milieu aléatoire et mesures quasi-invariantes pour un système dynamique[END_REF]) that ν h is absolutely continuous with respect to Lebesgue measure L T if and only if h = log u -log T u, for some L T -integrable u > 0, otherwise it is singular. When ν h is as in Theorem 4.2, notice the following relation, for any bounded measurable w : T → R:

T w dν h = T e -h T w dν h . (20)
We shall show in this section:

Theorem 4.3.
Let θ ∈ Q and T x = x+θ mod 1 on T. Let BV functions f, g : T → R, with f centered. Suppose that β n /α n = e f (T n-1 x) and γ n ε n /α n = g(T n x), n ∈ Z, for some x ∈ T.

i) Suppose that T gdν f = 0. Then for all x ∈ T, the random walk is transient. ii) Assume that g = h -e -f T h, for some bounded h. Introduce:

1) f = u -T u, with e u ∈ L 1 (L T ). 2) f (x 0 + x) = f (x 0 -x), for some x 0 ∈ T and L T -a.-e. x ∈ T.
Then, under either condition 1) or 2), for L T -a.-e. x, the random walk is recurrent.

Remark. -As soon as f is not identically zero (i.e. ν f = L T ), it is possible that T gdν f = 0, while T g(x)dx = 0. Indeed, there exist an interval I and a real t such that ν f (I) = ν f (I + t), so g = 1 I -1 I+t convenes. In item ii) of the theorem, the condition g = h -e -f T h, for some bounded h, implies T gdν f = 0. Reciprocally, when T gdν f = 0 and supposing a Diophantine condition on θ together with a regularity condition on both f and g, one can find a bounded h so that g = h -e -f T h. For instance, one has the following statement:

η(θ) = sup{r ∈ R * + , lim inf q→+∞ q r qθ = 0} ≥ 1. (21)
Let m > η(θ) be an integer. Assume that f ∈ C 2m (T, R) is centered and g ∈ C m (T, R) verifies T g dν f = 0. Then g = h -e -f T h, for some continuous h on T. Proof of the lemma: By Arnold [START_REF] Arnold | Small denominators I. Mapping the circle onto itself. (Russian)[END_REF], cf also Conze-Marco [START_REF] Conze | Remarks on step cocycles over rotations, centralizers and coboundaries[END_REF] (Thm 2.1), since f (m) is C m and m > η(θ), one has f (m) = v -T v, for some continuous v. By successive integrations, we have f = u -T u, with u of class C m and zero mean. Hence e -f = e T u /e u and so ν f is the measure with density e u with respect to L T . The hypothesis on g is thus (ge u )(x)dx = 0. As ge u is of class C m , using one more time [START_REF] Arnold | Small denominators I. Mapping the circle onto itself. (Russian)[END_REF], we have ge u = H -T H, for a continuous H. Finally, h = e -u H is bounded, as continuous on T, and satisfies g = h -e -f T h.

In the context of Theorem 4.3 ii), when

T gdν f = 0, for instance when g = h -e -f T h with h bounded (and even simply when g = 0), transience requires some strongly dissymmetric behaviour between v + (n) and w + (n) or between v -(n) and w -(n), as n → +∞. We build an example in the next proposition. Condition 1) of Theorem 4.3 ii), for example satisfied for f = 1 [0,1/2) -1 [1/2,1) with x 0 = 1/4, prevents this dissymmetry to occur. Proposition 4.5. In the context of Theorem 4.3, there exists θ ∈ Q and some BV centered f so that f = u -T u, with u ≥ 0, such that for any bounded g, for L T -a.-e. x, the random walk is transient.

Preliminaries

As in [START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF], we introduce functions Φ str (n), Φ(n) and Φ + (n) describing the average horizontal macrodispersion of the environment. The last two respectively correspond to Φ u (n) and Φ u,+ (n) in [START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF], Definition 2.3, with d = 1, u = 1 ∈ R + and ε s = m s , with the notations of [START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF]. Definition 4.6. i) The structure function, depending only on the vertical, is defined for n ≥ 0 by:

Φ str (n) =   n -v -1 -(n)≤k≤v -1 + (n) 1 ρ k    1/2
.

2) For m, n ≥ 0, introduce:

Φ(-m, n) =    -v -1 -(m)≤k≤ ≤v -1 + (n) ρ k ρ   1 ρ 2 k + 1 ρ 2 + s=k γ s ε s α s ρ s 2      1/2 . For n ≥ 0, set Φ(n) = Φ(-n, n) and Φ + (n) = Φ 2 (-n, 0) + Φ 2 (0, n).
As in [START_REF] Brémont | Random walk in a stratified independent random environment[END_REF], we rectify a misleading point appearing in [START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF], Definition 2.3 1), where the term "standard Lebesgue measure" on the half Euclidean ball S d-1

+ = {x ∈ R d | x = 1, x 1 ≥ 0}
has to be understood as "uniform probability measure". The following result is extracted from [START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF], Theorem 2.4, Proposition 2.5 1) and Lemma 6.11. Theorem 4.7.

i) The random walk is recurrent if and only if

n≥1 1 n 2 (Φ -1 (n)) 2 Φ -1 + (n) = +∞.
ii) The condition n≥1 1/Φ(n) < +∞ is sufficient for the transience of the random walk. It is equivalent to transience whenever Φ Φ + .

As a general fact, it is rather directly verified that Φ str Φ + Φ. Other general results, fully detailed in [START_REF] Brémont | Random walk in a stratified independent random environment[END_REF], section 3.1, are:

Φ + (n) Φ str (n) +    -v -1 -(n)≤k≤ ≤0 or 0≤k≤ ≤v -1 + (n) ρ k ρ s=k γ s ε s α s ρ s 2    1/2 , ( 22
)
as well as:

Φ(n) Φ str (n) +    -v -1 -(n)≤k≤ ≤v -1 + (n) ρ k ρ s=k γ s ε s α s ρ s 2    1/2 . ( 23
)
An essential point (recalled in detail in [START_REF] Brémont | Random walk in a stratified independent random environment[END_REF], end of Section 3.1) is that the inverse functions Φ -1 str , Φ -1 + and Φ -1 check dominated variation in the sense of Definition 3.5.

Proof of Theorem 4.3 i)

Let us start with a lemma, inspired from [START_REF] Conze | Marches en milieu aléatoire et mesures quasi-invariantes pour un système dynamique[END_REF] (Proposition 4.2). Introduce the notation :

A(n, g, x) = n k=0 g(T k x)/ρ k (x) n k=0 1/ρ k (x)
.

Recall that f is fixed as in Theorem 4.3.

Lemma 4.8. i) (Partial reproof of Theorem 4.2) Let (x n ) n≥1 in T and (N n ) be an increasing sequence of integers. For x ∈ T, write ρ n (x) = e fn(x) , n ∈ Z. Then any cluster point µ, for the weak- * topology, of the sequence of probability measures on T:

Nn k=0 δ T k xn /ρ k (x n ) Nn k=0 1/ρ k (x n ) n≥1 ( 24
)
is non-atomic and verifies dT µ = e T -1 f dµ. The solution of this last equation is unique.

ii) Let g : T → R be BV. Then A(n, g, x) → n→+∞ T gdν f , uniformly in x.

Proof of the lemma:

As a preliminary remark, for any (x n ) and (N n ), we have, as n → +∞:

Nn k=0 1/ρ k (x n ) → +∞ and (1/ρ Nn (x n ))/( Nn k=0 1/ρ k (x n )) → 0. ( 25
)
The first point comes from the observation that, independently of x n , ρ q (x n ) ≥ e -V (f ) , for any ≥ 1, as follows from the Denjoy-Koksma inequality [START_REF] Arnold | Small denominators I. Mapping the circle onto itself. (Russian)[END_REF]. Next, the second point can be equivalently rewritten as: i) Call (µ n ) n≥1 the sequence in (24) and consider a cluster point µ of it for the weak- * topology.

For simplicity, we keep the same notations (x n ),(N n ) and assume that (µ n ) converges to µ. Let a ∈ T. For any δ > 0, we have µ((a -δ, a + δ)) ≤ lim inf µ n ((a -δ, a + δ)). It is therefore enough to show that µ n ((a -δ, a + δ)) is arbitrary small for large n, for a well-chosen δ > 0.

Fix an integer K ≥ 1 and take δ > 0 so that the intervals (a -δ, a + δ) -kθ, 0 ≤ k ≤ 2q K , on T are disjoint. Then:

µ n ((a -δ, a + δ)) = Nn k=0 1 (a-δ,a+δ)-kθ (x n )/ρ k (x n ) Nn k=0 1/ρ k (x n ) . Let 0 ≤ τ 1,n < • • • < τ Ln,n ≤ N n , for some L n ≥ 0, be the subsequence of 0 ≤ k ≤ N n such that x n ∈ (a -δ, a + δ) -kθ. If L n = 0, we have µ n ((a -δ, a + δ)) = 0. When L n ≥ 2,
by hypothesis on δ, we have τ k,n + q K < τ k+1,n , for 1 ≤ k < L n , and τ Ln-1,n + 2q K < τ Ln,n . Using the Denjoy-Koksma inequality (2), giving 1/ρ k±q l (x n ) ≥ e -V (f ) /ρ k (x n ), we obtain, when L n ≥ 2:

Nn k=0 1 ρ k (x n ) ≥ 1≤k≤Ln-1 0≤ ≤K 1 ρ τ k,n +q (x n ) + 0≤ ≤K 1 ρ τ Ln ,n -q (x n ) ≥ 1≤k≤Ln (1/ρ τ k,n (x n )) 0≤ ≤K e -V (f ) = (K + 1)e -V (f ) 1≤k≤Ln (1/ρ τ k,n (x n )).
When L n = 1, noticing that either τ Ln,n > q K or τ Ln,n + q K < N n , as soon as n is large enough, the same reasoning holds and we obtain the same equality. Hence, always:

µ n ((a -δ, a + δ)) ≤ 1≤k≤Ln 1/ρ τ k,n (x n ) (K + 1)e -V (f ) 1≤k≤Ln (1/ρ τ k,n (x n )) = e V (f ) K + 1 .
This can be made arbitrary small, when choosing K large enough. Hence µ is non-atomic.

Next, for any continuous h :

T → R, A(N n , h, x n ) → T hdµ.
Since µ is non-atomic, this holds for any h continuous except at countably many points and in particular if h is BV. Since f is BV, e -f is also BV. Thus for any continuous h, A(N n , e -f T h, x n ) → T e -f T hdµ. It now follows from (25), that for any continuous h:

T e -f T hdµ = T hdµ,
giving dT µ = e T -1 f dµ. Thus µ solves the equation dT µ = e T -1 f dµ. For unicity of the solution, see [START_REF] Conze | Marches en milieu aléatoire et mesures quasi-invariantes pour un système dynamique[END_REF] (Theorem 5.6). This solution is hence non-atomic. This completes the proof of this point.

ii) If the result is not true, using that |A(n, g, x)| ≤ g ∞ , there exists a ∈ R, (x n ) in T and

N n → +∞, such that A(N n , g, x n ) → a =
T gdν f . By compacity of the weak- * topology, for some subsequence (N ζ(n) ) of (N n ) and (x ζ(n) ) of (x n ), the sequence of measures:

N ζ(n) k=0 δ T k x ζ(n) /ρ k (x ζ(n) ) N ζ(n) k=0 1/ρ k (x ζ(n) )
, n ≥ 1, converges to some probability µ on T. By i), dT µ = e T -1 f dµ and so µ = ν f , by unicity. As

ν f is non-atomic, A(N ζ(n) , g, x ζ(n) ) → T gdν f , contradicting A(N n , g, x n ) → a = T gdν f .
This ends the proof of the lemma.

We turn to the proof of Theorem 4.3 i), fixing BV functions f and g, with f centered and T gdν f = 0. Let x ∈ T and ρ n = ρ n (x) = e fn(x) , n ∈ Z, as before. Observe first that (cf [START_REF] Brémont | Random walk in a stratified independent random environment[END_REF], section 3.1, 2), for the first inequality):

Φ str (n)    -v -1 -(n)≤k≤ ≤v -1 + (n) ρ k ρ + ρ ρ k    1/2    -v -1 -(n)≤k≤ ≤v -1 + (n) ρ k ρ s=k 1 ρ s 2    1/2
, considering, for the second inequality, only in the last inside sum the terms for s = k and s = .

Introduce now the following function Ψ, essentially corresponding to Φ when g = 1 ("essentially", because, by (23) and the previous inequality, the definition can be simplified when g = 1):

Ψ(n) =    -v -1 -(n)≤k≤ ≤v -1 + (n) ρ k ρ s=k 1 ρ s 2    1/2 . ( 26
)
Notice now that (cf again [START_REF] Brémont | Random walk in a stratified independent random environment[END_REF], section 3.1, 2), for the first line), using at the end that g is bounded:

Φ(n) Φ str (n) +    -v -1 -(n)≤k≤ ≤v -1 + (n) ρ k ρ s=k γ s ε s α s ρ s 2    1/2 = Φ str (n) +    -v -1 -(n)≤k≤ ≤v -1 + (n) ρ k ρ s=k g(T s x) ρ s 2    1/2 Ψ(n).
We now prove the reverse inequality. Using Lemma 4.8 ii), let first M ≥ 1 be such that for n ≥ M and all x ∈ T:

n k=0 g(T k x)/ρ k (x) n k=0 1/ρ k (x) - T gdν f ≤ T gdν f /2.
This gives the inequality

| s=k g(T s x)/ρ s (x)| ≥ (1/2)| T gdν f |× s=k 1/ρ s (x), whenever -k > M . Consequently: Φ(n) ≥    -v -1 -(n)≤k≤ ≤v -1 + (n), -k>M ρ k ρ s=k g(T s x) ρ s 2    1/2 ≥ 1 2 T gdν f    -v -1 -(n)≤k≤ ≤v -1 + (n), -k>M ρ k ρ s=k 1 ρ s 2    1/2 . ( 27
)
Observe now that:

   -v -1 -(n)≤k≤ ≤v -1 + (n), -k≤M ρ k ρ s=k g(T s x) ρ s 2    1/2 ≤ g ∞    -v -1 -(n)≤k≤ ≤v -1 + (n), -k≤M ρ k ρ s=k 1 ρ s 2    1/2 .
As M is fixed and η ≤ ρ k (y)/ρ k+1 (y) ≤ 1/η, for any k ∈ Z and y ∈ T, where η comes from Hypothesis 1.1, we get when 0 ≤ -k ≤ M :

ρ k ρ s=k 1 ρ s 2 1.
Therefore, for some constant C > 0, depending on M :

   -v -1 -(n)≤k≤ ≤v -1 + (n), -k≤M ρ k ρ s=k 1 ρ s 2    1/2 ≤ C    -v -1 -(n)≤k≤v -1 + (n) 1    1/2 v -1 -(n) + v -1 + (n). (28)
We now show that v -1

+ (n)/Ψ 2 (n) → 0. Indeed, by (26)
:

Ψ 2 (n) ≥ 0≤k≤ ≤v -1 + (n) ρ k ρ s=k 1 ρ s 2 0≤k≤ ≤v -1 + (n) ρ k ρ + ρ ρ k 0≤k≤v -1 + (n) ρ k 0≤k≤v -1 + (n) 1/ρ k (v -1 + (n)) 2 ,
by the Cauchy-Schwarz inequality in the final step. Thus Ψ 2 (n

)/v -1 + (n) v -1 + (n) → +∞, as n → +∞. In the same way, Ψ 2 (n)/v -1 -(n) → +∞, as n → +∞. By (28):    -v -1 -(n)≤k≤ ≤v -1 + (n), -k≤M ρ k ρ s=k 1 ρ s 2    1/2 = o(Ψ(n)).
From ( 27) and (28), we deduce that Ψ Φ and thus finally Ψ Φ. The same argumentation shows that Φ + Ψ + , where Ψ "essentially" corresponds to g = 1 and is defined by:

Ψ + (n) =    -v -1 -(n)≤k≤ ≤v -1 + (n),k ≥0 ρ k ρ s=k 1 ρ s 2    1/2 . ( 29
)
By Theorem 4.7 i), we are therefore left to proving:

n≥1 1 n 2 (Ψ -1 (n)) 2 Ψ -1 + (n) < +∞. (30)
We shall give two proofs of (30). The first one just a reinterpretation. Consider another random walk, this time defined by (changing only g and keeping the same BV function f and x ∈ T):

µ n = δ +1 , ε n = 1, α n = γ n , β n /α n = e f (T n-1 x) , n ∈ Z.
Since γ n ε n /α n = 1, this case corresponds to g = 1. As previously indicated (before (26)), the functions Φ and Φ+ in this case verify Φ Ψ and Φ+ Ψ + . By Theorem 4.7 i), condition (30) is just the transience criterion of this new random walk. The latter being obviously transient (a.-s., the horizontal coordinate tends monotonically to +∞), the condition in (30) is verified and this ends the first proof.

One may be interested in showing directly the convergence of the series in (30). We now furnish the argument. This may help in the future to manipulate the recurrence criterion for studying other examples. Introduce the following functions:

Ψ ++ (n) =    0≤k≤ ≤v -1 + (n) ρ k ρ s=k 1 ρ s 2    1/2 , (31)
as well as:

Ψ +-(n) =    -v -1 -(n)≤k≤ ≤0 ρ k ρ s=k 1 ρ s 2    1/2
.

We shall next use repeatedly properties like max(a, b) a + b, √ a + b √ a + √ b, etc, for a, b ≥ 0. From the definition of Ψ + given in (29), we have:

Ψ + (n) Ψ ++ (n) + Ψ +-(n). Hence, Ψ -1 + (n) min{Ψ -1 ++ (n), Ψ -1 +-(n)}, thus furnishing: 1 Ψ -1 + (n) 1 Ψ -1 ++ (n) + 1 Ψ -1 +- (n)
.

In order to prove (30), we thus have to show the two convergences:

n≥1 1 n 2 (Ψ -1 (n)) 2 Ψ -1 ++ (n) < +∞ and n≥1 1 n 2 (Ψ -1 (n)) 2 Ψ -1 +-(n) < +∞. ( 32
)
We establish the first one, the case of the second one being similar.

First, using Hypothesis 1.1, for some c > 1 depending only on η, we have 0≤k≤n+1 ρ k ≤ c 0≤k≤n ρ k , for all n ≥ 0. Hence, for large n > 0, uniformly on x ∈ T, using [START_REF] Brémont | Markov chains in a stratified environment, ALEA[END_REF]:

n ≥ 0≤ ≤v -1 + (n) ρ ≥ 0≤ ≤v -1 + (n)+1 ρ /c ≥ n/c. (33)
Thus, for large n > 0 (uniformly on x ∈ T):

v -1 + (n/c 2 )≤ ≤v -1 + (n) ρ n, (34)
We now have, using (26):

Ψ(n) ≥    -v -1 -(n)≤k≤0≤ ≤v -1 + (n) ρ k ρ s=0 1 ρ s 2    1/2 . ( 35
)
The variables k and are now independent on the right hand side. Let us define:

ζ(n) = √ n    0≤ ≤v -1 + (n) ρ s=0 1 ρ s 2    1/2 .
Obviously, by (33):

ζ(n) n v -1 + (n) s=0 1 ρ s . ( 36
)
Analogously to (33), we have -v -1 -(n)≤k≤0 ρ k n, so we get, by ( 35) and (34):

Ψ(n) ζ(n) ≥ √ n    v -1 + (n/c 2 )≤ ≤v -1 + (n) ρ s=0 1 ρ s 2    1/2 n v -1 + (n/c 2 ) s=0 1 ρ s ≥ (n/c 2 ) v -1 + (n/c 2 ) s=0 1 ρ s . ( 37
) Set F (n) = v -1 + (n) k=0 1/ρ k and G(n) = nF (n).
The first inequality in (37) gives Ψ -1 (n) ζ -1 (n). Moreover, the last inequalities in (37) also provide, together with (36):

ζ -1 (n) G -1 (n).
In order to establish the first part of (32), it is thus sufficient to show that:

n≥1 G -1 (n) n 2 1 Ψ -1 ++ (n) < +∞. ( 38
)
For the analysis of Ψ ++ , we fix an integer K > c 2 , where the constant c > 1 appears in (33). Define now for any integer u ≥ 0 the quantity:

A u = v -1 + (K u )<k≤v -1 + (K u+1 ) 1/ρ k . (39)
Starting from the definition (31) of Ψ ++ :

Ψ ++ (K n ) =    0≤k≤ ≤v -1 + (K n ) ρ k ρ l   k≤s≤ 1/ρ s   2    1/2    v -1 + (K 0 )<k≤ ≤v -1 + (K n ) ρ k ρ l   k≤s≤ 1/ρ s   2    1/2 .
Proceeding as for (34), we also have v -1

+ (K u )<k≤v -1 + (K u+1 ) ρ k K u .
Using this, we continue:

Ψ ++ (K n )       0≤u≤v≤n-1 v -1 + (K u )<k≤v -1 + (K u+1 ) v -1 + (K v )< ≤v -1 + (K v+1 ),k≤ ρ k ρ l    v -1 + (K u )<s≤v -1 + (K v+1 ) 1/ρ s    2       1/2   0≤u≤v≤n-1 K u K v (A u + • • • + A v ) 2   1/2
.

We arrive at:

Ψ ++ (K n ) 0≤u≤v<n K u/2 K v/2 (A u + • • • + A v ) 0≤l<n A l 0≤u≤ K u/2 ≤v<n K v/2 K n/2 0≤ <n A K /2 . ( 40
)
Let N ≥ 1 be an integer. We have, by (40):

N n=0 Ψ ++ (K n ) K n 0≤n≤N 1 K n/2 0≤ <n A K /2 = 0≤ <N A <n≤N K /2 K n/2 0≤ <N A F (K N ). (41)
Now, (41) furnishes, still for N ≥ 1:

0≤k≤Ψ++(K N ) 1 Ψ -1 ++ (k) 0≤n<N Ψ++(K n )<k≤Ψ++(K n+1 ) 1 Ψ -1 ++ (k) 0≤n<N Ψ ++ (K n+1 ) K n F (K N ). (42)
Let us define:

Z(n) = 0≤k≤n 1/Ψ -1 ++ (k).
We extend the notation to real x > 0, by Z(x) = Z( x ), using the floor function. Idem for Ψ ++ . The last inequality (42) thus says that for n ≥ 1, taking N so that K

N -1 ≤ n < K N : Z(Ψ ++ (n/K)) F (n) = G(n) n .
In particular, using (5):

Z(Ψ ++ (G -1 (n)/K)) G(G -1 (n)) G -1 (n) n G -1 (n) . Next, if n ≤ Ψ ++ (G -1 (n)/K), then Z(n) n/G -1 (n).
Otherwise, the last inequality gives:

Z(n) ≤ Z(Ψ ++ (G -1 (n)/K)) + Ψ++(G -1 (n)/K)<k≤n 1/Ψ -1 ++ (k) ≤ n G -1 (n) + n Ψ -1 ++ (Ψ ++ (G -1 (n)/K)) n G -1 (n) .
Finally, using the last inequality and the definition of Z(n), we show (38):

n≥1 G -1 (n) n 2 1 Ψ -1 ++ (n) n≥1 1 Z(n) 2 1 Ψ -1 ++ (n) ≤ n≥1 1 Z(n -1)Z(n) 1 Ψ -1 ++ (n) ≤ n≥1 1 Z(n -1) - 1 Z(n) < +∞.
This concludes the second proof of (30) and of Theorem 4.3 i).

Remark. -The previous proof in fact shows that, in complete generality, the condition Φ 2 str Φ implies the transience of the random walk.

Proof of Theorem ii)

In this section, f and g are BV functions, with f centered. We suppose that g = h -e -f T h, for a bounded h. As T s g(x) = γ s ε s /α s and T s (e -f T h)/ρ s = (T s+1 h)/ρ s+1 , we first have:

   -v -1 -(n)≤k≤ ≤v -1 + (n) ρ k ρ s=k γ s ε s α s ρ s 2    1/2 =    -v -1 -(n)≤k≤ ≤v -1 + (n) ρ k ρ s=k T s g ρ s 2    1/2    -v -1 -(n)≤k≤ ≤v -1 + (n) ρ k ρ s=k T s h ρ s - T s+1 h ρ s+1 2    1/2    -v -1 -(n)≤k≤ ≤v -1 + (n) ρ k ρ 1/ρ 2 k + 1ρ 2 l    1/2
Φ str (n).

See for example [START_REF] Brémont | Random walk in a stratified independent random environment[END_REF], section 3.1, 2), for details of the last step. Next, we deduce by (23) that Φ(n) Φ + (n) Φ str (n). Using Theorem 4.7 ii), the recurrence of the random walk is now equivalent to the divergence of n≥1 1/Φ str (n), or, using (34) and the analogous version for v -, of (cf Definition 4.1):

n≥1 1 n(w + • v -1 + (n) + w -• v -1 -(n))
.

Because of the monotonicity of the general term of the previous series, by usual condensation, this is equivalent, for any fixed K > 1, to showing the divergence of:

n≥1 √ K n w + • v -1 + (K n ) + w -• v -1 -(K n ) . (43)
1) Suppose first that f = u -T u, with e u ∈ L 1 (L T ). Then, by the Law of Large Numbers (for the second step), a.-e., as n → +∞:

w + (n)(x) = n k=0 e -u(x)+T k u(x) ∼ ne -u(x)
T e u(y) dy.

In the same way, a.-e., w -(n)(x) is linear. For v + , using again the Law of Large Numbers, for a positive, but maybe non-integrable, function, there is a.-e. some κ(x) > 0 such that:

v + (n)(x) = e u(x) n k=0 e -T k u(x) ≥ nκ(x), for n ≥ 1.
The same property is true, a.-e., for v -(n), as n → +∞. Hence, a.-e., there is some c(x) > 0 so that

v -1 + (n) ≤ c(x)n and v -1 -(n) ≤ c(x)n, for large n.
We obtain, a.-e., for large n > 0:

w + • v -1 + (n) ≤ w + (c(x)n) n. Idem, w -•v -1 -(n) n.
These make the general term in (43) not go to zero, so the series diverges. 2) Suppose that (instead of the L 1 -condition) for some x 0 ∈ T, then f (x + x 0 ) = f (x 0 -x), for a.-e. x ∈ T. Using the denominators (q n ) of the convergents of the angle θ, one has:

0≤k≤qn e f k (x) = 0≤k≤qn e f qn -k (x) = 0≤k≤qn e f -k (x)+T -k fq n (x) 0≤k≤qn e f -k (x) , (45)
using the Denjoy-Koksma's inequality (2). As a result, one obtains that v + (q n ) v -(q n ), as n → +∞, uniformly on x ∈ T. For the same reason:

w + (q n ) w -(q n ), (46)
as n → +∞, also uniformly on x ∈ T.

From v + (q n ) v -(q n ), independently on x ∈ T, we can now fix some large K > 1 and p 0 such that for any n ≥ 1, there exists p with

K p ≤ v + (q n ) ≤ K p+p0 and K p ≤ v -(q n ) ≤ K p+p0 . This gives v -1 + (K p ), v -1 -(K p ) ≤ q n .
In (43), the term corresponding to p verifies (uniformly on x ∈ T):

√ K p w + • v -1 + (K p ) + w -• v -1 -(K p ) ≥ K -p0 v + (q n ) w + (q n ) + w -(q n ) v + (q n ) w + (q n ) , (47)
using (46) for the last step. Next, immediately from the definition of the model, the set {x ∈ T, the random walk is transient} is measurable and T -invariant, hence has Lebesgue measure zero or one, by ergodicity of (T, T, L T ). If the random walk were transient for a.-e. x, then, by (47) and the convergence of the series in (43), for a.-e. x:

(v + (q n )/w + (q n ))(x 0 + x) → 0 and (v + (q n )/w + (q n ))(x 0 -x) → 0, (48)

as n → +∞. However, using (46) and the symmetry assumption in the final step, we can write:

v + (q n ) w + (q n ) (x 0 + x) v + (q n ) w -(q n ) (x 0 + x) = 0≤k≤qn e f k (x0+x) 0≤k≤qn e -f -k (x0+x) 0≤k≤qn e f k (x0+x)
0≤k≤qn e f k (x0-x) .

As a result, for a.-e. x ∈ T:

v + (q n ) w + (q n ) (x 0 + x) w + (q n ) v + (q n ) (x 0 -x).
By (48), the left hand side goes to 0, as n → +∞, whereas the right hand side goes to +∞. This contradiction completes the proof of Theorem 4.3 ii).

Remark. -In a similar way, but without the symmetry assumption, suppose that f (x) = u(x) -u(x + y), for some BV function u and some parameter y ∈ T. Let us show that for a.-e. (x, y) ∈ T 2 the random walk is recurrent. Indeed, in the previous proof, part 2), if ever transience holds for some (x, y), then, by (47) and the convergence of the series in (43): This is impossible again, when reversing the roles of x and y. We thus have recurrence for a.-e. (x, y) ∈ T 2 . Rather generally, in Theorem 4.3 ii), it would be interesting if the symmetry assumption 2) could be dropped. This raises the question, for f : T → R, BV and centered, of understanding the a.-e. behaviour, as n → ∞, of ratios of the form n k=0 e f k (x) / n k=0 e -f k (x) . Also in Theorem 4.3 ii), supposing only T gdν f = 0 for g (in place of g = h -e -f T h, with h bounded) requires to find sharp upper-bounds on sums of the form: n k=0 e -f k (x) T k g(x).

v + (n)/w + (n) =

Proof of Proposition 4.5

By Theorem 4.7 ii), to prove transience for the random walk, it is enough to show the convergence of n≥1 1/Φ str (n). By definition of Φ str , it is sufficient to establish that:

n≥1 1/ nw + • v -1 + (n) < +∞. ( 49
)
We choose f in the form f = u -T u, with u ≥ 0. Then e -u is integrable, so v + (n) is a.-e. linear, as n → +∞, by the Law of Large Numbers, as for (44). As a consequence, we obtain that for any x ∈ T, v -1 + (n) n, as n → +∞. Setting U = e u , we have w + (n) ∼ e -u(x) U n (x), so in order to obtain (49), it is enough to show that for a.-e. x:

n≥1 1/ nU n < +∞. ( 50
)
Let us build u (and f = u -T u). Let the rotation angle θ ∈ Q be defined by the partial quotients a m = m 6 , m ≥ 1. Introduce h B,∆ (x) = B(1 -|x|/∆) + , for ∆ > 0, B > 0. It is a piecewise linear pick function of height B and width ∆, centered at zero. Let (q m ) be the denominators of the convergents of θ. For m ≥ 1, set h m = h Bm,∆m , with:

∆ m = 1/(m 2 q m ), B m = m 2 /q m . (51)
We first define f = m≥1 f m , where:

f m = qm-1 k=0 T -k (h m -T qm h m ).
For large m, the (kθ) 0≤k<qm are approximately equally spaced and the sum in the definition of f m involves functions with disjoint supports. For m ≥ 1, f m is centered and, using (51) and (1):

V (f m ) ≤ q m V (h m -T qm h m ) ≤ Cq m (B m /∆ m ) q m θ ≤ C/m 2 .
As a result, f is BV and centered. We now check that f is a.-e. equal to some u -T u. For m ≥ 1, f m = u m -T u m , with:

u m = qm-1 k=0 T -k qm-1 =0 T h m = | |<qm (q m -| |)T h m . (52)
The Lebesgue measure of the support of u m is ≤ 2q m ∆ m . As m≥1 q m ∆ m < +∞, by the first lemma of Borel-Cantelli, a.-e. x ∈ T belongs to the support of u m , for only finitely many m. Hence u = m≥1 u m is well-defined a.-e. and f = u -T u, a.-e..

The Diophantine type η(θ) of θ, defined (21), also equal lim sup log q n+1 / log q n . Hence, here η(θ) = 1, as a m = m 6 , m ≥ 1. For x ∈ T and r > 0, let τ r (x) = min{n ≥ 1, T n x < r}. By a result of Kim and Marmi [START_REF] Kim | The recurrence time for interval exchange maps[END_REF], for a.-e. x:

Definition 4. 1 .

 1 Fixing x ∈ T, let ρ n = ρ n (x) = e fn(x) , n ∈ Z. In other words:

Lemma 4. 4 .

 4 Introduce the Diophantine type of θ:

e

  -f k (xn)+f Nn (xn) = Nn k=0 e T k f Nn -k (xn) = Nn k=0 e T Nn -k f k (xn) → +∞, for the same reason.

e

  u k (x)-u k (x+y) / qn k=0 e -u k (x)+u k (x+y) → 0. Now, the set of (x, y) ∈ T 2 verifying this property is clearly invariant by the joint action of T ×Id and Id × T on T 2 , which is ergodic. Hence this set has measure 0 or 1. If this is 1, one obtains that for a.-e. (x, y), as n → +∞: qn k=0 e u k (x)-u k (y) / qn k=0 e -u k (x)+u k (y) → 0.

Acknowledgments. We thank Yves Derriennic for inviting to revisit these questions and the referee for detailed comments.

Recall that U = e u and let s m = τ ∆m/2 (x), m ≥ 1. For a.-e. x ∈ T, decompose now:

We next have, using (52) and h m (T sm x) ≥ B m /2:

Also, a.-e., using (53), for large m,

Next, the denominators of the convergents of θ verify q m = m 6 q m-1 +q m-2 ≤ (2m) 6 q m-1 , so brutally q m ≤ (2m) 6m ≤ e cm log m . Hence s m+1 = O(e 3cm log m ). Together with (55), we deduce that the series in (54) is finite. This ends the proof of the proposition.

As a final remark, rather generally, when f is not an additive coboundary, then the asymptotic behaviour of (log(v + (n)/w + (n))) n≥1 is somehow that of a classical random walk in R. Lemma 4.9. Let f ∈ L 1 (T, R), centered, and not a.-e. equal to u -T u, for some measurable u. Define

as in Definition 4.1. Then one of the following three situations occur:

Proof of the lemma :

As a first point, v + (n) → +∞, a.-e., since, a.-e., the random walk (f k ) k≥0 is recurrent, as k → +∞, since f is integrable and centered. We have v + (n)(x) ∼ e f (x) v + (n -1)(T x) and w + (n)(x) ∼ e -f (x) w + (n -1)(T x), so the following set is T -invariant:

It hence has Lebesgue measure 0 or 1, by ergodicity. If this measure is 1, we can a.-e. define ψ(x) = lim sup n→+∞ (v + (n)(x)/w + (n)(x)). The opening remark on equivalents yields: