Bull Braz Math Soc, New Series https://doi.org/10.1007/s00574-018-0084-x

Equilibrium State for One-Dimensional Lorenz-Like Expanding Maps M. A. Bronzi1 · J. G. Oler1

Received: 19 May 2017 / Accepted: 12 March 2018 © Sociedade Brasileira de Matemática 2018

Abstract Let L : [0, 1]\{d} → [0, 1] be a one-dimensional Lorenz-like expanding map (d is the point of discontinuity), P = {(0, d), (d, 1)} and C α ([0, 1], P) the set of piecewise Hölder-continuous potentials of [0, 1] with the usual C 0 topology. In this context, applying a criteria by Buzzi and Sarig (Ergod Theory Dyn Syst 23(5):1383–1400, 2003, Th. 1.3), we prove that there exists an open and dense subset H of C α ([0, 1], P), such that each φ ∈ H admits exactly one equilibrium state. Keywords Equilibrium state · Lorenz Maps Mathematics Subject Classification Primary 37D25; Secondary 37D30 · 37D20

1 Introduction The thermodynamic formalism of piecewise monotonic maps on intervals was initiated (Hofbauer 1977a, b; Hofbauer and Keller 1982, 1993). In this context, being the potential a continuous function, the uniqueness and stochastic properties of the corresponding equilibrium state have been extensively studied in the works of these authors. In Denker et al. (1990) Denker, Keller and Urbanski extend Hofbauer’s results assuming supx∈[0,1] φ(x) < P(T, φ), where φ is a potential with bounded variation and just a finite number of discontinuities and P(T, φ) denotes the pressure. See also

B

M. A. Bronzi [email protected] J. G. Oler [email protected]

1

Faculdade de Matemática, FAMAT-UFU, Uberlândia, MG, Brazil

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M. A. Bronzi, J. G. Oler

(Denker and Urba´nski 1991; Denker et al. 1996; Haydn 1999; Przytycki 1990) for results on uniqueness in the complex context. Recently, in Inoquio-Renteria and Rivera-Letelier (2012), Li and Rivera-Letelier (2014a, b), using results of Keller (Keller 1985), the authors showed that, for a sufficiently regular interval map T , a Hölder-continuous potential φ admits a unique equilibrium state assuming that, for some integer n ≥ 1, supx∈[0,1] n1 Sn (φ) < P(T, φ)  j where Sn (φ) = n−1 j=0 φ ◦ T . Note that, in the case of a piecewise monotone interval map, the condition of Denker, Keller and Urbanski is not less general than the condition in Inoquio-Renteria and Rivera-Letelier (2012), Li and Rivera-Letelier (2014a), Li and Rivera-Letelier (2014b), since it is possible to get a unique equilibrium state for (T, φ) using transitivity and applying the condition supx∈[0,1] φ(x) < P(T, φ) in Denker et al. (1990) to an iterate of T and Sn φ. On the other hand, Buzzi and Sarig (Buzzi and Sarig 2003) proved that if a piecewise expanding map T : X → X is strongly topologically transitive, in the sense that for all nonempty open sets U , T (X ) ⊂ ∪k≥0 T k (U ), with a piecewise Hölder-continuous potential φ satisfying Ptop (φ, ∂P, T ) < Ptop (φ, T ), then φ supports a unique equilibrium state. Observe that the condition studied by Denker et al. (1990) can be replaced by the condition introduced by Buzzi and Sarig in Buzzi and Sarig (2003) since the condition Ptop (φ, ∂P, T ) < Ptop (φ, T ) can be used to reduce to checking sup Sn ϕ at 0 and 1 instead of everywhere interval. For the purpose of this work, as the model in Glendinning (1990), let L : [0, 1]\{d} → [0, 1] be a one-dimensional Lorenz-like expanding map where d is √ the point of discontinuity, L  (x) > 2 for any x ∈ [0, 1]\{d}, lim x→d + L(x) = 0, lim x→d − L(x) = 1, P = {(0, d), (d, 1)} formed by open intervals and C α ([0, 1], P) the set of piecewise Hölder-continuous potential of [0,1] with the usual C 0 topology. In this context, we present a mild improvement on Denker–Keller–Urbanski’s criterion for piecewise monotonic interval maps; see (Denker et al. 1990). Precisely, we prove that if φ is a piecewise Hölder-continuous potential in [0, 1]\{d} satisfying   1 1 max lim sup (Sn φ)(0), lim sup (Sn φ)(1) < Ptop (φ, L) n→∞ n n→∞ n

(*)

then φ admits a unique equilibrium state. Observe that Proposition 3.1 allows to check that (∗) is indeed equivalent to the Buzzi-Sarig criterion; see (Buzzi and Sarig 2003, Th. 1.3). By considering the work of Buzzi and Sarig (2003), this is expected since a one-dimensional Lorenz-like expanding map is understood as a strongly√topologically transitive piecewise expanding map. In fact, the hypothesis L  (x) > 2 for any x ∈ [0, 1]\{d} is fundamental to ensure that the condition L([0, 1]) ⊂ ∪k≥0 L k (U ) is verified for all one-dimensional Lorenz-like expanding map, where U is an open interval of [0, 1]; see (Williams 1979; Clark Robinson 2012). Moreover, this Lorenz map is conjugate to some βx + α mod 1 map; see (Rand 1978; Hofbauer and Keller 1993; Williams 1979). See (Buzzi 2004; Climenhaga and Thompson 2013; Climenhaga et al. 2017) for recent results about equilibrium states. The transitivity is an essential hypothesis to guarantees the unicity of equilibrium states. In Hofbauer (1981), Hofbauer studied whenever the measure of maximal entropy (which is unique) is fully supported or not for these linear mod 1 maps. In this

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setting Hofbauer showed that there exists a unique measure of maximal entropy even though transitivity fails on a set of parameters of positive measure. The goal of this paper is checking the condition (∗) for an open and dense set of potentials, for a one-dimensional Lorenz-like expanding map (which, in particular, is √ strongly topologically transitive, because L  (x) > 2 ). We can now formulate our main result. Theorem A Let L : [0, 1]\{d} → [0, 1] be a one-dimensional Lorenz-like expanding map and C α ([0, 1], P) the set of piecewise Hölder-continuous potential of [0,1]. Then there exists an open and dense subset H of C α ([0, 1], P), in the C 0 topology, such that, each φ ∈ H admits a unique equilibrium state. In order to prove Theorem A we adapt a result of Buzzi and Sarig (Buzzi and Sarig 2003), which gives a sufficient condition for the existence and uniqueness of equilibrium state under a regularity condition on the pressure of the boundary of continuity domains of the dynamic. We express the pressure of the boundary of the corresponding partition for a one-dimensional Lorenz-like expanding map in terms of the Birkhoff average of the discontinuity and by a small perturbation of the potential along periodic points of sufficiently large period we guarantee the regularity condition of Buzzi-Sarig. In order to obtain adequate periodic points we use the conjugacy with a linear mod 1 transformation and the notion of cutting times in one dimensional dynamics; see (Bruin 1998; Hofbauer 1981). 1.1 Organization In Sect. 2 we review some standard facts on one-dimensional Lorenz-like expanding map and equilibrium states. Section 3 is dedicated to the proof of Theorem A. The construction of the set H is presented in Sect. 3.1 and in Sect. 3.2 it is shown that H is not empty. Finally the proof that each element of H admits a unique equilibrium state and that H is an open and dense set is in Sects. 3.3, 3.5 and 3.7.

2 Setting and Statements Lorenz maps originally arise from the study of geometric models for the Lorenz equations (Guckenheimer 1976; Guckenheimer and Williams 1979; Lorenz 1963; Sparrow 1982; Williams 1979). This model induces a one-dimensional Lorenz-like expanding map. The precise definition, as the model in Glendinning (Glendinning 1990), is as following: Definition 2.1 A one-dimensional Lorenz-like expanding map is a function L : [0, 1] → [0, 1] which has the following properties: (L.1) L has a unique discontinuity at x = d and L(d + ) = lim L(x) = 0, L(d − ) = lim L(x) = 1; x→d +

x→d −

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M. A. Bronzi, J. G. Oler Fig. 1 One-dimensional Lorenz-like expanding map

√ (L.2) For any x ∈ [0, 1]\{d}, L  (x) > 2 ; (L.3) Each inverse branch of L extends to a C 1+θ , θ > 0, function over [L(0), 1] or [0, L(1)] and if g denotes any of these inverse branches, g  (x) ≤ λ < 1 (See Fig. 1). We denote by P the natural partition of [0, 1]\{d}, i.e., P = {(0, d), (d, 1)}. The boundary of P is ∂P := {0, d, 1}. Also, define P (n−1) = {P0 ∩ L −1 (P1 ) ∩ · · · ∩ L −n+1 (Pn−1 ) = ∅ | Pi ∈ P}. Let φ : [0, 1] → R such that sup(φ) < ∞. This map φ is a piecewise Hölder continuous potential if the restriction to any elements of P is Hölder continuous, i.e., for all x, y in the same element of P, |φ(x)−φ(y)| ≤ K |x−y|α for some α > 0, K < ∞. Consider C α ([0, 1], P) := {φ : [0, 1] → R : φ is piecewise Hölder continuous potential}. According to Buzzi-Sarig (Buzzi and Sarig 2003), the pressure of subset S ⊂ [0, 1] and φ ∈ C α ([0, 1], P) is defined as ⎛ 1 Ptop (φ, S, L) = lim sup log ⎝ n→∞ n

⎞



sup e Sn φ(x) ⎠ ,

Cn ∈P (n−1) : S∩Cn =∅

x∈Cn

 n−1 (x)) where for x ∈ Cn , Cn ∈ P (n−1) , the Birkhoff average Sn φ(x) = n−1 j=1 φ(L α is well defined. The topological pressure of L for φ ∈ C ([0, 1], P) is defined by P(φ, L) = Ptop (φ, [0, 1], L). If the variational principle  Ptop (φ, L) =

sup

μ∈M L (X )

h μ (L) +

 φ dμ

where the supremum extends over the set M L (X ) of all ergodic L−invariant measure μ is attained for some μ, then μ is called an equilibrium state for φ. This theory was developed by Bowen (1975) and Ruelle (1978) in the context of Hölder-continuous potentials on hyperbolic dynamical systems, and has been applied to Axiom A systems,

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Anosov diffeomorphisms and other systems too, see e.g. (Chazottes and Keller 2012; Keller 1998) for more recent expositions. In this context, Buzzi and Sarig proved the following theorem: Theorem 2.1 (Buzzi and Sarig (2003)) Let (X, P, L) be a piecewise expanding map with φ ∈ C α ([0, 1], P). Assume that Ptop (φ, ∂P, L) < Ptop (φ, L). If M L (X ) denotes the set of invariant measures, then: 

(i) Ptop (φ, L) = supμ∈M L (X ) h μ (L) + φ dμ and this supremum is realized by at least one measure; (ii) there exist at most finitely many ergodic equilibrium states; and (iii) if, additionally, L is strongly topologically transitive in the sense that for all nonempty open sets U , ∪k≥0 L k (U ) ⊇ L(X ), then there exists a unique equilibrium state.

3 Proof of Theorem A 3.1 Construction of set H As L is not continuous in d ∈ [0, 1] we make the following convention: Sn φ(d + ) is the right-hand limit of the function Sn φ(z) at d and Sn φ(d − ) is the left-hand limit of Sn φ(z) at d. More precisely, let φ ∈ C α ([0, 1], P) be a continuous function, then for any n ∈ N we define n−1  φ(L i (z)). Sn φ(d ± ) = lim z→d ±

By definition lim sup n→∞

L(d + )

= 0 and

L(d − )

i=0

= 1, so we conclude that:

1 1 1 1 Sn φ(d + ) = lim sup Sn φ(0) and lim sup Sn φ(d − ) = lim sup Sn φ(1). n n→∞ n n→∞ n n→∞ n

The following Lemma 3.1 guaranties that the above relations are well defined. Lemma 3.1 Let L : [0, 1]\{d} → [0, 1] be a one-dimensional Lorenz-like expanding map and consider φ ∈ C α ([0, 1], P). (i) If does not exist n 0 ∈ N such that L n 0 (0) = d, then lim sup n→∞

1 1 Sn φ(d + ) = lim sup Sn φ(0). n n→∞ n

(ii) If there exists n 0 ∈ N such that L n 0 (0) = d, then lim sup n→∞

1 1 Sn φ(d + ) = Sn φ(0). n n0 0

The same conclusion holds for d − replacing 0 for 1.

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Proof The first item comes easily from the definition of Birkhoff sum. Indeed, Sn φ(d + ) = lim Sn φ(x) = φ(d + ) + x→d +

n−2 

φ(L j (0)) = φ(d + ) + Sn−1 φ(0),

j=0

then lim sup n→∞

1 1 Sn φ(d + ) = lim sup Sn φ(0). n n n→∞

In order to prove (ii) let n = K n 0 + l, with 0 ≤ l < n 0 . Thus Sn φ(d + ) = φ(d + ) + S K n 0 +l−1 φ(0) = φ(d + ) + K Sn 0 φ(0) + Sl−1 φ(0),

where the second equality in the above equation occurs because L is piecewise increasing and φ is continuous. Indeed, for x close enough to d and x > d, one has L n 0 (x) > L n 0 (0) = d. Therefore, K 1 φ(d + ) l Sn φ(d + ) = + Sn 0 φ(0) + Sl−1 φ(0). n n n n Letting n → ∞ we have: lim sup n→∞

1 1 Sn φ(d + ) = Sn φ(0). n n0 0

The same argument gives similar results for d − replacing 0 by 1.

 

Remark 3.1 From now on we use lim supn→∞ n1 Sn φ(0) to refer one of the items in the above Lemma. The set H is defined as the set of φ ∈ C α ([0, 1], P) such that   1 1 Ptop (φ, L) > max lim sup (Sn φ)(0), lim sup (Sn φ)(1) . n→∞ n n→∞ n 3.2 H is Not Empty To show that H is not empty we show that it contains the null potential. Consider the potential φ ≡ 0, then Ptop (φ, L) = h top (L) and   1 1 Ptop (φ, L , ∂P) = max lim sup (Sn φ)(0), lim sup (Sn φ)(1) = 0. n→∞ n n→∞ n

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√ Since L  > 2 then h top (L) > 0 and L is strongly topologically transitive, so by Hofbauer (Hofbauer 1979) L admits a unique measure of maximal entropy. Thus Ptop (φ, L , ∂P) = 0 < h top (L) = Ptop (φ, L) and we conclude that φ ∈ H. 3.3 Every Element of H Admits a Unique Equilibrium State To show that every element of H admits a unique equilibrium state we first prove that the pressure of the boundary ∂P can be written in terms of asymptotic values of the Birkhoff average of the boundary point of the partition. We begin with the following simple result based on distortion argument. Lemma 3.2 Let L : [0, 1]\{d} → [0, 1] be a one-dimensional Lorenz-like expanding map and φ ∈ C α ([0, 1], P). Then for n large enough there exists a constant C > 0 such that |(Sn φ)(x) − (Sn φ)(y)| ≤ C, for all x, y ∈ Cn and Cn ∈ P (n−1) . Proof If φ is α-Hölder continuous on P there exists a constant K > 0 such that |(Sn φ)(x) − (Sn φ)(y)| ≤

n−1 

|φ(L (x)) − φ(L (y))| ≤ K i

i

i=0

n−1 

|L i (x) − L i (y)|α .

i=0

Since x, y ∈ Cn by property L.3 of Definition 2.1 there exists 0 < λ < 1 such that |(Sn φ)(x) − (Sn φ)(y)| ≤ K

∞ 

λiα .

i=0

Hence it is enough to take C = K

∞ 

λiα .

 

i=0

Corollary 3.1 Lemma 3.2 is true on the closure of the cylinders Cn , denoted by Cn , i.e, if x, y ∈ Cn then there exists C > 0 such that |(Sn φ)(x) − (Sn φ)(y)| ≤ C. Proof Just write points in Cn as limits of points in Cn . Using Corollary 3.1, we can characterize the Ptop (φ, ∂P, L) as follows. Proposition 3.1 Let L : [0, 1]\{d} → [0, 1] be a one-dimensional Lorenz-like expanding map and φ ∈ C α ([0, 1], P), then   1 1 Ptop (φ, ∂P, L) = max lim sup (Sn φ)(0), lim sup (Sn φ)(1) . n→∞ n n→∞ n

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Proof Consider Cni , 1 ≤ i ≤ 4 the cylinder in P (n−1) such that {0, d, 1} ∩ ∂Cni = ∅, where ∂Cni is boundary of Cni . Since supx∈Cn e(Sn φ)(x) ≤ supx∈Cn e(Sn φ)(x) it follows that ⎛ Ptop (φ, ∂P, L) ≤ lim sup n→∞

1 log ⎝ n

⎞



sup e(Sn φ)(x) ⎠ .

Cn ∈P n−1 :∂ P ∩Cn =∅ x∈Cn

Using Lemma 3.2, for x ∈ Cni , we have (Sn φ)(x) ≤ (Sn φ)(bi ) + C, where bi ∈ {0, d − , d + , 1} depending on whether Cn is on the left or right hand side of the discontinuity d. It turns out that  4   1 e(Sn φ)(bi ) Ptop (φ, ∂P, L) ≤ lim sup log n→∞ n i=1   1 = max lim sup Sn φ(b j ) 1≤ j≤4   n→∞ n 1 1 ≤ max lim sup Sn φ(0), lim sup Sn φ(1) n→∞ n n→∞ n where in the above inequalities we used Lemma 3.1. Conversely, since supx∈Cn e(Sn φ)(x) ≥ inf x∈Cn e(Sn φ)(x) we obtain that ⎛ 1 Ptop (φ, ∂P, L) ≥ lim sup log ⎝ n→∞ n

Cn

⎞

 ∈P n−1 :∂ P ∩C

inf e(Sn φ)(x) ⎠ .

n  =∅

x∈Cn

Using Lemma 3.2 again, for x ∈ Cni , we have (Sn φ)(bi ) − C ≤ (Sn φ)(x). Thus Ptop (φ, ∂P, L) ≥ lim sup n→∞

    1 1 log emax1≤ j≤4 {Sn φ(b j )} ≥ max lim sup Sn φ(b j ) . 1≤ j≤4 n n→∞ n

Applying Lemma 3.1 again we have   1 1 Ptop (φ, ∂P, L) ≥ max lim sup Sn φ(0), lim sup Sn φ(1) . n→∞ n n→∞ n   Lemma 3.3 Let L : [0, 1]\{d} → [0, 1] be a one-dimensional Lorenz-like expanding map. Then the map Ptop ( · , ∂P, L) : C α ([0, 1], P) −→ R is continuous.

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Proof Note that for all ψ, φ ∈ C α ([0, 1], P) by Proposition 3.1 we have that    Ptop (ψ, ∂P, L) − Ptop (φ, ∂P, L)     1 1 = lim sup max {Sn ψ(0), Sn ψ(1)} − lim sup max {Sn φ(0), Sn φ(1)} n→∞ n  n→∞ n    1  Sn ψ(0) − Sn φ(0)  1  Sn ψ(1) − Sn φ(1)  + 2 lim sup ≤ 2 lim sup     2 2 n→∞ n n→∞ n ≤ ψ − φ  .   3.4 Proof that Every Element of H Admits a Unique Equilibrium State Consider φ ∈ H. By definition of H we have that   1 1 max lim sup Sn φ(0), lim sup Sn φ(1) < Ptop (φ, L). n→∞ n n→∞ n Thus by Proposition 3.1 we obtain   1 1 Ptop (φ, ∂P, L) = max lim sup Sn φ(0), lim sup Sn φ(1) < Ptop (φ, L). n→∞ n n→∞ n √ On the other hand, assuming the hypothesis L  > 2, the one-dimensional Lorenzlike expanding map is strongly topologically transitive (see Williams 1979; Clark Robinson 2012). Thus applying Theorem 2.1 we conclude that if φ ∈ H then φ admits a unique equilibrium state. 3.5 H is an Open Set in C α ([0, 1], P) First observe that H = H+ ∩ H− , where   1 H+ = φ : Ptop (φ, L) > lim sup (Sn φ)(0) n→∞ n   1 H− = φ : Ptop (φ, L) > lim sup (Sn φ)(1) . n→∞ n Consider φ ∈ C α ([0, 1], P). To prove that H is C 0 -open in C α ([0, 1], P) we need the following Lemmas: Lemma 3.4 Consider Ptop (·, L) : C α ([0, 1], P) → R defined by φ −→ Ptop (φ, L). Then for each φ, ψ ∈ C α ([0, 1], P) we have |Ptop (φ, L) − Ptop (ψ, L)| ≤ φ − ψ0 , in other words, Ptop (·, L) is a Lipschitz function.

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Proof The proof of the Lemma follows from Theorem 9.7 of Walters (1982), with appropriate modifications.   Lemma 3.5 The functionals P+ , P− :C α ([0, 1], P) → R, defined by φ −→ P− (φ) = lim supn→∞ n1 (Sn φ)(1) and φ −→ P+ (φ) = lim supn→∞ n1 (Sn φ)(0) are continuous. Proof We give the proof only for the map P+ , the other case is analogous. As in Lemma 3.1 we consider two cases. If there is n 0 ∈ N such that L n 0 (0) = d, then Lemma 3.5 is trivial, since P+ is constant. If there is not n 0 such that L n 0 (0) = d, then by the definition of Birkhoff sum, we have that     1 1 |P+ (φ) − P+ (ψ)| = lim sup (Sn φ)(0) − lim sup (Sn ψ)(0) n→∞ n n→∞ n 1 ≤ lim sup |(Sn φ(0) − Sn ψ(0))| n→∞ n (3.1) ≤ φ − ψ0 . Thus the inequality (3.1) proves the Lemma.

 

3.6 Proof that H is C 0 -Open in C α ([0, 1], P) Now, let us prove the first part of Theorem A. Note that to prove that H is open in C α ([0, 1], P) it is sufficient to show that H+ and H− are open in C α ([0, 1], P). ˜ = Ptop (φ, L) − P+ (φ). Consider the map P˜ : H+ → R defined by P(φ) ˜ Combining Lemmas 3.4 and

3.5 we obtain that P is continuous and H+ =  φ : Ptop (φ, L) > P+ (φ) = P˜ −1 (A), with A = (0, ∞). Thus H+ is an open set of C([0, 1], R). Similarly we get H+ is open and we conclude that H is open in C α ([0, 1], P). 3.7 H is a Dense Set in C α ([0, 1], P) Pick φ ∈ C α ([0, 1], P), our proof consists in the construction of a sequence of potentials φ ,k,l ∈ H such that φ ,k,l → φ in C 0 -topology. To this end, firstly using properties of linear mod 1 transformation we construct an auxiliary family {An }n∈N of subsets of [0,1] generated by the partition F = {(0, (1 − α)/β) , ((1 − α)/β, 1)} (see below). Then we find periodic points with large enough period in the cylinders that are the right and left hand side of the discontinuity. Finally we construct φ ,k,l by perturbing φ along the orbit of the above mentioned periodic points in order to obtain higher pressure and dominate the pressure of the boundary. 3.8 Conjugation to a Linear mod 1 Transformation Recall that one-dimensional Lorenz-like maps are topologically conjugate to a linear mod 1 transformation as in Faller and Pfister (2009), Glendinning (1990), Hofbauer

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Equilibrium State for One-Dimensional... Fig. 2 Linear mod1 transformation: T (x) = βx + αmod1

(1981). Consider 1 < β ≤ 2 and α ≥ 0 such that α + β ≤ 2, then the map T (x) = βx + α mod 1 is called linear mod 1 transformation (See Fig. 2). Viewed as a map of the interval a linear mod 1 transformation has a point of discontinuity at D = (1−α)/β. In addition T (D + ) = 0, T (D − ) = 1, T (1) = β + α − 1, T (0) = α and T (x) =

⎧ ⎨ βx + α, ⎩

if x ∈ [0, (1 − α)/β)

βx + α − 1, if x ∈ ((1 − α)/β, 1].

Observe that linear mod 1 transformations are strongly topologically transitive if √ β > 2. More details and properties about this class of maps can be found in Hofbauer (1981), Palmer (1979), Parry (1966), Parry (1979). In this context, Glendinning proved the following: Theorem 3.1 (Glendinning (1990)) Let L : [0, 1]\{d} → [0, 1] be a one-dimensional Lorenz-like expanding map. Then L is topologically conjugate to a linear mod 1 transformation T and h top (L) = ln(β). 3.9 Construction of Auxiliary Sets { An }n∈N Note that by Theorem 3.1 to find a periodic point for a one-dimensional Lorenz-like expanding map is equivalent to find a periodic point for a linear mod 1 transformation. Therefore, the construction of periodic points will be done using linear mod 1 transformation T . The boundary of such a system ([0, 1], F, T ), where F = {(0, D) , (D, 1)}, is ∂F = {0, D, 1}. Let F (n) be the collection of n-cylinders dynamically defined by the transformation T , i.e., the non empty intersections Cn = ∩nj=1 T − j (F j ), where F j ∈ F. We define Cn+ and Cn− being the cylinders respectively at the right and left

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hand side of the discontinuity D, i.e., D ∈ ∂Cn± , where + or − represent the cylinders on the right or left side of D, respectively. We introduce an auxiliary family An by induction as follows: put A0 := (D, 1) and for n ≥ 0 we write An+1 =

⎧ / An ⎨ T (An ), if D ∈ ⎩

T (A∗n ), if D ∈ An ,

where A∗n is the connected component of An \{D} containing T n (D + ). Definition 3.1 An integer N is a cutting time for T if D ∈ A N . This notion can be found e.g. in Graczyk and Swipolhk atek (1998). 3.10 Construction of Periodic Points In this section we find periodic points with large enough periods in the cylinders that are on right and left hand side of the discontinuity. More precisely, we will prove using properties of linear mod 1 transformations that there exists sequences of integers {Nk± }k∈N and sequences of periodic points such that pk± → d ± , d ± ∈ ∂C N ± and ±

±

±

k

±

L Nk ( pk ) = pk for some pk ∈ C N ± , where + or − represent the cylinders on the k right or left side of d, respectively. Lastly, we show that P(φ, ∂P, L) can be calculated by the average of pk± . Lemma 3.6 Let N + be a cutting time for T and C N + = (D + , B + ) a cylinder. Suppose + that T N (B + ) > B + , then there exists a periodic point pk+ of period N + for T such + that pk ∈ C N + . Analogous results are obtained to the cylinders at the left hand side of D. +

Proof We first show that if T N (B + ) > B + then there exists a periodic point of period − + N + for T in C N + . The case for T N (B − ) < B − is similar. Observe that T N (C N + ) = + + A N + . As N + is a cutting time we obtain that A N + = (T N (D + ), T N (B)) and + + + + D ∈ A N + . So T N (C N + ) = (T N (D + ), T N (B)). As T N (B + ) > B + and D is a cutting time we obtain +

+

+

T N (C N + ) = A N + = (T N (D + ), T N (B + )) ⊃ (D, B + ) = C N + , +

∈ C N + such that T N ( p + ) = p+ . so there exists p + N+ N+ N+

 

Lemma 3.7 There exists infinitely many cutting times N0+ < N1+ < · · · < Nk+ < · · · +

such that T Nk (B + ) > B + , for k ∈ N. Similar results are get to the cylinders at the left hand side of D. Proof We first prove that there exists infinitely many cutting times N0+ < N1+ < + · · · < Nk+ < · · · such that T Nk (B + ) > B + , for k ∈ N. We can proceed similarly

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to prove the second case. To this end, fix N0+ > 0. By contradiction suppose that + for all cutting times N + ≥ N0+ , we have that T N (B + ) < B + . Thus there exists + cutting times N0+ < N1+ < · · · < Nk+ < · · · such that T Nk (B + ) < B + , for k ∈ N. Therefore, |C N + | k+1

|C N + |

=

k

+

|A N + |

β

=

k+1 + Nk+1 −Nk+

|A N + | k

|T Nk (D + ) − D| +

+

|T Nk (D + ) − T Nk (B + )|

.

(3.2)

The second equality above comes from the fact that there is no cutting time between + and consequently Nk+ and Nk+1 +

+

+

|A N + | = β Nk+1 −Nk |T Nk (D + ) − D|. k+1

Now , +

+

|T Nk (D + ) − D| |T

Nk+

(D + ) − T

Nk+

(B + )|

≥

+

+

|T Nk (D + ) − T Nk (B + )| − |T Nk (B + ) − D| |T

= 1−

|T |T

Nk+

Nk+

Nk+

(D + ) − T

Nk+

(B + )| +

(B + ) − D|

(D + ) − T

Nk+

(B + )|

=1−

1 |T Nk (B + ) − D| + |D − B + | β Nk

(3.3) +

Since we are considering T Nk (B + ) < B + we have +

1 |T Nk (B + ) − D| 1 − >− +. Nk+ |D − T (B + )| β β Nk

(3.4)

Thus, using (3.2), (3.3) and (3.4) we have that |C N + | k+1

|C N + |

>1−

k

1 +

β Nk

.

As a consequence we have that |C N + | k+1

|C N + |

=

k0

|C N + | k+1

|C N + | k

>

·

|C N + | k

|C N + |

· ··· ·

|C N + |

k−1

k0 +1

|C N + | k0

∞   1  1 − j =: γ > 0. β j=1

Therefore, |C N + | > γ · |C N + |, for all k ≥ k0 . This gives us a contradiction since k

k0

by construction |C N + | → 0 when k → ∞. k

 

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M. A. Bronzi, J. G. Oler +

Corollary 3.2 Consider C N + belonging the collection P (Nk ) of Nk+ -cylinder for a k one-dimensional Lorenz-like expanding map L. Assuming that d ∈ ∂C N + , then there k

+

exists pk+ ∈ C N + such that L Nk ( pk+ ) = pk+ . Analogously it is possible to obtain a k

periodic point pk− for the cylinder C N − . k

Proof Since a one-dimensional Lorenz-like expanding map L is topologically conjugate to a linear mod 1 transformation T (Theorem 3.1) it follows immediately from   Lemma 3.6 that periodic point pk± there exists. Lemma 3.8 Let L : [0, 1]\{d} → [0, 1] be a one-dimensional Lorenz-like expanding map and consider φ ∈ C α ([0, 1], P). If Nk+ ∈ N such that d ∈ ∂C N + , pk+ ∈ C N + k

+

and L Nk ( pk+ ) = pk+ , then

   1 1 + + lim sup lim Sn φ( pk ) = lim sup lim Sn φ( pk ) , n→∞ n k→∞ n n→∞ k→∞

k



(3.5)

Replacing pk+ by pk− we get similar results. Proof We give the proof only for pk+ . The case pk− is similar. To see this, we first observe that as sup(φ) < ∞ and then   n−1  1   Sn φ( p + ) ≤ 1 |φ(L j ( pk+ ))| ≤ sup(φ), k  n n

for all n, k ∈ N.

j=0

Thus the double superior limit lim supn,k→∞ n1 Sn φ( pk+ ) exists. Now, for fixed k ∈ N, we can be write n = qn+ Nk+ + rn+ , 0 ≤ rn+ ≤ Nk+ − 1. Thus, the following limit exists   qn+ 1 1 lim Sn φ( pk + ) = lim S + φ( pk+ ) + + + S + φ( pk+ ) n→∞ n n→∞ qn+ N + + rn+ Nk qn Nk + rn+ rn k =

1 S + φ( pk+ ), Nk+ Nk

(3.6) where we get the last equality because, as 0 ≤ rn+ ≤ Nk+ − 1, then Srn+ φ( pk+ ) is limited. On the other hand, for fixed n ∈ N, as pk+ → d + the following limit exists by continuity, i.e, 1 1 (3.7) lim Sn φ( pk+ ) = Sn φ(d + ). k→∞ n n Combining (3.6) and (3.7) we get (3.5).  

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Equilibrium State for One-Dimensional...

Corollary 3.3 Let L : [0, 1]\{d} → [0, 1] be a one-dimensional Lorenz-like expanding map and consider φ ∈ C α ([0, 1], P). If pk+ is a sequence of periodic points satisfying the conditions of Lemma 3.8 then 1 1 + + S Nk+ φ( pk ) = lim sup n Sn φ(0). Nk n→∞

lim sup k→∞

Replacing 0 by 1 and pk+ by pk− we get similar results. Proof For each n, as limk→∞ pk+ = d + , we obtain lim

k→∞

1 1 Sn φ( pk ) = Sn φ(d + ). n n

(3.8)

Letting n → ∞ we can rewrite (3.8) as 

 1 1 lim sup lim Sn φ( pk ) = lim sup Sn φ(d + ). k→∞ n n→∞ n→∞ n By Lemma 3.8 we obtain  lim sup k→∞

 1 1 Sn φ( pk ) = lim sup Sn φ(d + ). n→∞ n n→∞ n lim

As lim

n→∞

1 1 Sn φ( pk+ ) = S N φ( pk ) n Nk k  

this finishes the proof. 3.11 Construction of Potential φ,k,l

Recall that by Corollary 3.3, there exists a subsequence Nk+ → ∞ such that d ∈ ∂C N + ±

±

±

±

k

and pk± ∈ C N ± such that L Nk ( pk ) = pk± . Let I j = (L j ( pk ) − δk± , L j ( pk ) + δk± ) k

be intervals, where 0 ≤ j ≤ Nk± − 1. Since the orbit of pk± is a finite set, there exists ± ± δk± > 0 such that Ii ∩ I j = ∅, for all 0 ≤ j, i ≤ Nk± − 1 with i = j. Here and δ± ± subsequently, we denote δk,l = k . Consider l ⎧ ± Nk± −1 ⎪ k −1 ⎪ N  ± ± ⎪ ⎪ ⎨ B ,k, j,l (x) , x ∈ Ij ± B ,k,l (x) = j=0 j=0 ⎪ ⎪ ⎪ ⎪ ⎩ 0 , other wise.

123

M. A. Bronzi, J. G. Oler

±

Fig. 3 Bump function B ,k, j,l ±

where B ,k, j,l is the bump function defined by Fig. 3. ±

±

Lemma 3.9 Consider φ ,k,l (x) = φ(x) + B ,k,l (x), where φ ∈ C α ([0, 1], P). Then we have the following properties: ±

(1) φ ,k,l is Holder continuous; ±

(2) φ ,k,l − φC 0 < , for all φ ∈ C α ([0, 1], P); ±

(3) As φ ,k,l is built on the orbit of the periodic point pk± , we have that ±

±

±

S Nk φ ,k,l ( pk ) = S Nk φ( pk ) + . ±

 

Proof The proof follows of the construction of potential φ ,k,l . ±

±

Lemma 3.10 Let O( pk ) be the orbit of pk and χO( p± ) the characteristic function of k ± ± O( pk ). Then liml→∞ B ,k,l (x) = · χO( p± ) (x), for all x ∈ [0, 1]. k

±

±

Proof If x ∈ O( pk ), then there exists j0 ∈ {0, 1, . . . , Nk± −1} such that x = L j0 ( pk ).  Nk± −1 ± ± ± j0 ± Then B ,k,l (x) = B ,k,l (L j0 ( pk± )) = j=0 B ,k,l, j (L ( pk )) = , since by the ±

±

±

±

construction B ,k,l, j (L j0 ( pk )) = 0, whenever j = j0 , and B ,k,l, j0 (L j0 ( pk )) = . ±

Therefore, liml→∞ B ,k,l (x) = .

 N±+ ± ± ± / j=0 I j then liml→∞ B ,k,l (x) = On the other hand, let x ∈ [0, 1]\O( pk ). If x ∈ ± ± 0. Otherwise x ∈ I ± j for some j ∈ {0, 1, . . . , Nk − 1}. In this case, since δk,l → 0 as +  N± ± ± l → ∞, it follows that for l0 large enough we have x ∈ / j=0 I j and lim B ,k,l (x) = ±

±

l→∞

0. Therefore, for all x ∈ [0, 1]\O( pk ), we get that liml→∞ B ,k,l (x) = 0 this complete the proof.   ± , ∂P, L) = 0. Lemma 3.11 Under the above assumptions liml→∞ Ptop (B ,k,l

123

Equilibrium State for One-Dimensional... ± Proof By Lemma 3.10, we have that liml→∞ B ,k,l (x) = · χO( p± ) (x). As O( pk± ) k is a closed set we obtain that χO( p± ) : [0, 1] → R is upper semi-continuous. Thus, k applying Lemma 3.3 we have that ± , ∂P, L) = Ptop ( · χO( p± ) , ∂P, L) lim Ptop (B ,k,l k

l→∞

   1 = · lim sup max Sn χO( p± ) (0), Sn χO( p± ) (1) k k n→∞ n = 0, / O( pk+ ) and χO( p+ ) (L j (x)) = 0, for all 0 ≤ j ≤ n − 1. since L j (x) ∈

 

k

+ − Corollary 3.4 Define φ ,k,l (x) = max{φ ,k,l (x), φ ,k,l (x)}, for x ∈ [0, 1]. Then

lim Ptop (φ ,k,l , ∂P, L) = Ptop (φ, ∂P, L).

l→∞

3.12 Proof that H is C 0 -Dense in C α ([0, 1], P) Our proof starts by recalling the follows result proved by Buzzi and Sarig in Buzzi and Sarig (2003), where L is the Lorenz map. Proposition 3.2 (Buzzi and Sarig 2003) Consider φ a piecewise uniformly continuous potential and let ν be an ergodic probability measure. If ν(S) > 0, then Ptop (φ, S, L) ≥ h ν (L) + φ dν, where h ν (L) is the metric entropy of ν. Fix φ ∈ C α ([0, 1], P). Remember that our proof consists in the construction of a potential φ ,k,l such that φ ,k,l − φC 0 < and Ptop (φ ,k,l , L) > Ptop (φ ,k,l , ∂P, L),

(3.9)

i.e, φ ,k,l ∈ H. To this end, fix > 0 and consider + − φ ,k,l (x) = max{φ ,k,l (x), φ ,k,l (x)}, for x ∈ [0, 1].

By Lemma 3.9 the potential φ ,k,l satisfies the condition φ ,k,l − φC 0 < , for all k, l. Thus we just need to show that the inequality (3.9) is true. By definition of Ptop (φ ,k,l , L), there exists k0 , l0 such that Ptop (φ ,k,l , L) ≥ Ptop (φ ,k,l , ∂P, L) for all k ≥ k0 and l ≥ l0 . To obtain a contradiction, suppose that for all k, l, Ptop (φ ,k,l , L) = Ptop (φ ,k,l , ∂P, L).

(3.10)

123

M. A. Bronzi, J. G. Oler +

Consider C N + ∈ P (Nk k

−1)

+

as in Proposition 3.6, then there exists pk+ ∈

C N + such that L Nk ( pk+ ) = pk+ . Furthermore one can construct a measure k   + + 1  Nk −1 μk (·) = j=0 δ L j ( p + ) (·), where δ L j ( p + ) is the Dirac measure with N+ k

δ L j ( p+ ) (L j ( pk+ )) k

k

k

= 1, j ∈ {0, 1, . . . , Nk − 1}. As μ+ k (C Nk ) > 0 by Proposition

3.2 we have

+ + Ptop (φ ,k,l , L) ≥ Ptop (φ ,k,l , L) ≥ Ptop (φ ,k,l , C Nk , L)

≥ h μ+ (L) +



+ φ ,k,l dμ+ k

k

Nk −1 1  1 + + φ ,k,l (L j ( pk+ )) = (S Nk φ ,k,l )( pk+ ) Nk Nk j=0 1 = (S Nk φ)( pk+ ) + , Nk

=

where in the last equality we are applying Lemma 3.9. We get the same conclusion for pk− ∈ C N − , i.e., k

Ptop (φ ,k,l , L) ≥

1 (S − φ)( pk− ) + . Nk− Nk

Thus  Ptop (φ ,k,l , L) ≥ max

 1 1 + − (S + φ)( pk ), − (S N − φ)( pk ) + . k Nk+ Nk Nk

(3.11)

Letting l → ∞ in the inequation (3.11) and combining (3.10) with Corollary 3.4 we get  Ptop (φ, ∂P, L) ≥ max

 1 1 + − (S + φ)( pk ), − (S N − φ)( pk ) + . k Nk+ Nk Nk

Indeed letting k → ∞ and combining Corollary 3.3 with Proposition 3.1 we obtain 

 1 1 Ptop (φ, ∂P, L) ≥ max lim sup + (S N + φ)( pk+ ), lim sup − (S N − φ)( pk− ) +

k k t→∞ Nk k→∞ Nk   1 1 = max lim sup (Sn φ)(0), lim sup (Sn φ)(1) +

n→∞ n n→∞ n = Ptop (φ, ∂P, L) + ,

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Equilibrium State for One-Dimensional...

which contradicts (3.10). Thus we show that the inequality (3.9) is true. Acknowledgements We thank Ali Tahzibi for proposing the problem and for many helpful suggestions during the preparation of the paper. Also, we thank Daniel Smania and Krerley Oliveira for some helpful conversations and comments on the problem.

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