Complexity of injective piecewise contracting interval maps
aa r X i v : . [ m a t h . D S ] F e b Complexity of injective piecewise contracting interval maps
E. Catsigeras , P. Guiraud and A. Meyroneinc September 3, 2018 Instituto de Matem´atica y Estad´ıstica Rafael Laguardia, Universidad de la Rep´ublica,Montevideo, Uruguay [email protected] Centro de Investigaci´on y Modelamiento de Fen´omenos Aleatorios Valpara´ıso, Facultad de Ingenier´ıa,Universidad de Valpara´ıso,Valpara´ıso, Chile [email protected] Departamento de Matem´aticas, Instituto Venezolano de Investigaciones Cient´ıficas,Apartado 20632, Caracas 1020A, Venezuela [email protected]
Abstract
We study the complexity of the itineraries of injective piecewise contracting maps on the interval. Weprove that for any such map the complexity function of any itinerary is eventually affine. We also provethat the growth rate of the complexity is bounded from above by the number N − N − n + 1. In these examples, the asymptotic dynamics takesplace in a minimal Cantor set containing all the discontinuities. Keywords: Interval map, Piecewise contraction, Complexity, Minimal Cantor set.MSC 2010: 37E05, 37E15, 37B10
Let (
X, d ) be a compact metric space and { X i } Ni =1 be a finite collection of N > X = S Ni =1 X i . Let f : X → X and assume f discontinuous on the set ∆ := { x ∈ X i ∩ X j , i = j ∈ { , . . . , N }} . If there exits a constant λ ∈ (0 ,
1) such that for any i ∈ { , . . . , N } the map f satisfies d ( f ( x ) , f ( y )) λ d ( x, y ) ∀ x, y ∈ X i , (1)then f is called a piecewise contracting map and each element of the collection { X i } Ni =1 is called a contractionpiece .In [5], we explored the diversity of asymptotic dynamics of these systems, and proved that a richdynamics can appear if the attractor contains discontinuity points. In particular, we exhibited three-dimensional examples with exponential complexity and positive topological entropy. On the other hand, ifthe attractor does not contain discontinuity points, then its dynamics is simple, just composed by a finitenumber of periodic orbits.In the present paper, we remain interested in the diversity of the dynamics but we restrict the study toa class of one-dimensional piecewise contracting maps. Our objective is to determine the range of all thepossible complexity functions in the whole considered class. In particular, we are interested in the relationbetween certain features of the discontinuity points and the complexity of the dynamics.1f a piecewise contracting map f is defined on a compact interval and each contraction piece is anopen interval, we say that f is a piecewise contracting interval map . For these systems, it has been shownthat generically the asymptotic dynamics is periodic, first for injective maps [2, 13, 14] and later for moregeneral one-dimensional maps [15]. In this paper, we are instead interested in the non-periodic asymptoticdynamics. In dimension one, there are few known examples of piecewise contracting maps with non-periodicattractors [4, 7, 9, 17], and none of them has orbits that accumulate in more than one discontinuity point.Therefore, little is known about the maximum complexity of the dynamics when the interval map has anarbitrary (finite) number of discontinuity points.By complexity of a map, we refer to the complexity function of the itineraries of its orbits. To definethe itineraries of a piecewise contracting map f , consider the set e X of those points of X whose orbit neverintersects ∆, that is e X := + ∞ \ n =0 f − n ( X \ ∆) , (2)and assume that e X is non-empty. We say that the sequence θ = { θ t } t ∈ N ∈ { , , . . . , N } N is the itinerary of x ∈ e X if for every t ∈ N and i ∈ { , . . . , N } we have θ t = i if and only if f t ( x ) ∈ X i . The complexityfunction of a sequence θ is the function defined for every n > p ( θ, n ) := L n ( θ ) where L n ( θ ) := { θ t . . . θ t + n − , t > } ∀ n > , that is, p ( θ, n ) gives the number of different words of length n contained in θ . Therefore, the complexityfunction of a sequence is a non-decreasing function of n . Also, if there exists n > p ( θ, n + 1) = p ( θ, n ), then p ( θ, n ) = p ( θ, n ) for all n > n . This implies that a symbolic sequence is eventually periodicif and only if its complexity function is eventually constant.In this paper we consider piecewise contracting maps satisfying a “separation property”. To define thisproperty, first note, from inequality (1), that for any i ∈ { , . . . , N } the restriction f | X i of f to the piece X i admits a continuous extension f i : X i → X which also satisfies inequality (1) on X i . Definition 1.1. (Separation)
We say that a piecewise contracting map f satisfies the separation property if for every i ∈ { , . . . , N } the continuous extension f i : X i → X is injective and f i ( X i ) ∩ f j ( X j ) = ∅ forany j ∈ { , . . . , N } such that j = i .A map f which satisfies the separation property is obviously injective on X \ ∆, but not necessarily onthe whole set X . A map f which is injective on X does not satisfy the separation property if and only ifthere are i and j in { , . . . , N } such that lim x → y f | X i ( x ) = lim x → z f | X j ( x ) for some y ∈ ∆ ∩ X i and z ∈ ∆ ∩ X j ,with y = z if i = j . It follows that not every injective map satisfies the separation property. Nevertheless,in dimension one, every injective map whose discontinuities are all of the first kind satisfies the separationproperty.Our main result is the following Theorem 1.2. We will later complement its statement with the additionalresults given by Theorem 2.10 and Theorem 3.1 about the relations between the complexity function andthe dynamical asymptotic behaviour of the orbits near the discontinuity points. Theorem 1.2.
1) Let θ be the itinerary of an orbit of a piecewise contracting interval map which has N contraction pieces and satisfies the separation property. Then, there exist α ∈ { , , . . . , N − } , β > and m > , such that the complexity function of θ satisfies p ( θ, n ) = αn + β ∀ n > m , (3) with β = 1 if α = N − .2) For every N > , there exists a piecewise affine contracting interval map f which has N contractionpieces and satisfies the separation property, such that p ( θ, n ) = ( N − n + 1 ∀ n > , (4) for every itinerary θ of f .
2s mention above, generically in the space of piecewise contracting interval maps, all the orbits areattracted by periodic orbits. Therefore, generically, the itinerary of any orbit is eventually periodic andhas an eventually constant complexity function. In other words, α = 0 in equality (3). Nevertheless, non-periodic attractors do appear when some orbits accumulate on discontinuity points [5]. In dimension twoor three, this can produce itineraries of polynomial or exponential complexity [5, 11, 12]. But in contrast,Theorem 1.2 proves that in dimension one the complexity of any non-periodic itinerary is affine.To prove Theorem 1.2, we will deduce equality (3) from precise results stated in Lemma 2.9 and Theo-rem 2.10, which relate the complexity of an itinerary with the recurrence properties of the correspondingorbit arbitrarily near the discontinuity points. In fact, the value of α equals the number of discontinuitypoints on which the orbit accumulates from both sides, and therefore is bounded above by the number ofdiscontinuities contained in the attractor (which is at most N − α = ( N − β depends on the transient behaviour of the dynamics and can be arbitrarily large; see relation(13).Part 1) of Theorem 1.2 states that the complexity of an itinerary is at most equal to ( N − n + 1. For N = 2, there are known examples of piecewise contracting maps whose itineraries have such a Sturmiancomplexity [3, 4, 6, 7, 9, 17]. In those examples, the attractor is a Cantor set supporting a minimal dynamics.Part 2) of Theorem 1.2 states that for any value of N > N − α of the complexity function, for any number N > N we construct for any N > N = 2 and a Sturmian complexity. As a consequence, we will prove with Theorem3.1 that the attractor of each of these maps inherit the Cantor structure and minimality of the attractorof the base case.The proof of part 2) of Theorem 1.2 provides for each N > In this subsection we give some preliminary results that are not specific to piecewise contracting intervalmaps. In fact, here X is not necessarily an interval and f : X → X may not satisfy the inequality (1),provided it admits continuous extensions on each continuity piece X i . Definition 2.1. (Atoms)
For every i ∈ { , . . . , N } let F i : P ( X ) → P ( X ) be defined by F i ( A ) = f ( A ∩ X i ) for all A ∈ P ( X ), where P ( X ) denotes the set of parts of X . Let n > i , . . . , i n ) ∈{ , . . . , N } n . We say that A i ,...i n := F i n ◦ F i n − ◦ · · · ◦ F i ( X ) is an atom of generation n if it is non-empty.We denote A n the set of all the atoms of generation n . Remark 2.2.
In the sequel we will often use the following basic properties of the atoms: by construction, A i i ...i n ⊂ A i ...i n ⊂ . . . ⊂ A i n , and if f is piecewise contracting then max A ∈A n +1 diam( A ) λ max A ∈A n diam( A )for all n >
1, where diam( A ) is the diameter of A .As shown by the following Lemma 2.3, the separation property implies that the atoms of a samegeneration are pairwise disjoint. Lemma 2.3.
Suppose that f satisfies the separation property. For every n >
1. if
A, B ∈ A n are such that A ∩ B = ∅ , then A = B ,2. if A i ...i n , A j ...j n ∈ A n and A i ...i n = A j ...j n , then ( i , . . . , i n ) = ( j , . . . , j n ) . roof. It is easy to show that 1) is true for n = 1. Suppose now that it is true for some n >
1. Let A , B ∈ A n +1 . Then there exists C and D ∈ A n and i , j ∈ { , . . . N } such that A = f i ( C ∩ X i ) and B = f j ( D ∩ X j ). Suppose that A ∩ B = ∅ . Since f i ( X i ) ∩ f j ( X j ) = ∅ for i = j , it follows that i = j . Now,since f i is injective, if A ∩ B = ∅ we have ( C ∩ X i ) ∩ ( D ∩ X i ) = ∅ , and C ∩ D = ∅ . Since C ∩ D = C ∩ D ,using the induction hypothesis we deduce that C = D and it follows that A = B .By the separation property 2) is true for n = 1. Suppose it is true for some n > A i ...i n +1 = A j ...j n +1 . Then, f i n +1 ( A i ...i n ∩ X i n +1 ) = f j n +1 ( A j ...j n ∩ X j n +1 ) and j n +1 = i n +1 , since f i n +1 ( X i n +1 ) ∩ f j n +1 ( X j n +1 ) = ∅ implies that j n +1 = i n +1 . On the other hand, if f i n +1 ( A i ...i n ∩ X i n +1 ) = f i n +1 ( A j ...j n ∩ X i n +1 ) then by injectiveness A i ...i n ∩ A j ...j n = ∅ , which implies by 1) that A i ...i n = A j ...j n .Using the induction hypothesis it follows that i k = j k for all k n .The following Lemma 2.4 and Lemma 2.5 give the relation between the itinerary of a point of e X andthe atoms visited by the orbit of that point. Lemma 2.4.
Let x ∈ e X and θ ∈ { , . . . , N } N be its itinerary. Then f t + n ( x ) ∈ A θ t θ t +1 ...θ t + n − for every t > and n > .Proof. Let t = 0. By the definitions of atom and itinerary we have that f ( x ) ∈ f ( X θ ) ⊂ A θ since x ∈ X θ .Assume that f n ( x ) ∈ A θ θ ...θ n − for some n >
1. Then f n +1 ( x ) = f ( f n ( x )) ∈ f ( A θ θ ...θ n − ∩ X θ n ) ⊂ A θ θ ...θ n . Now suppose t = 0, let y = f t ( x ) and ω be the itinerary of y . Then f t + n ( x ) = f n ( y ) ∈ A ω ...ω n − = A θ t θ t +1 ...θ t + n − . Lemma 2.5.
Suppose that f satisfies the separation property. Let x ∈ e X , t > , n > and θ be theitinerary of x . If f t + n ( x ) ∈ A i i ...i n then θ t θ t +1 . . . θ t + n − = i i . . . i n .Proof. Suppose t = 0. By Lemma 2.4 we have f n ( x ) ∈ A θ θ ...θ n − , therefore A i ...i n ∩ A θ ...θ n − = ∅ . ByLemma 2.3 we have A i ...i n = A θ ...θ n − and θ . . . θ n − = i . . . i n . Now suppose t = 0, let y = f t ( x ) and ω be the itinerary of y . Then f t + n ( x ) = f n ( y ) ∈ A i ...i n , which implies that ω . . . ω n − = i . . . i n , that is θ t . . . θ t + n − = i . . . i n . Corollary 2.6.
Suppose that f satisfies the separation property. Let x ∈ e X , t > , n > , θ be the itineraryof x , and ( i , i , . . . i n ) ∈ { . . . N } n . Then θ t θ t +1 . . . θ t + n − = i i . . . i n if and only if f t + n ( x ) ∈ A i ...i n . Proof.
It follows directly from Lemmas 2.4 and 2.5.Let x ∈ e X , I := { , . . . , N } and θ ∈ I N be the itinerary of x . Now, for any n > k ∈ I considerthe set L kn ( θ ) := { i . . . i n ∈ L n ( θ ) : { j ∈ I : ∃ t > f t + n ( x ) ∈ A i ...i n ∩ X j } = k } . A word of length n of θ belongs to L kn ( θ ) if it is the label of an atom that intersects at least k continuitypieces and if the orbit of f n ( x ) visits exactly k of these intersections. The following Lemma 2.7 puts inrelation the growth of the complexity function of an itinerary θ and the cardinality of the sets L kn ( θ ). Lemma 2.7.
Let x ∈ e X and θ ∈ I N be the itinerary of x . Then p ( θ, n + 1) p ( θ, n ) + N X k =2 ( k − L kn ( θ ) ∀ n > . (5) If moreover f satisfies the separation property, then (5) is an equality.Proof. Let n >
1, and observe that L n ( θ ) = S Nk =1 L kn ( θ ). First, we show the inclusion L n +1 ( θ ) ⊂ N [ k =1 B kn ( θ ) , (6)4here B kn ( θ ) := [ i ...i n ∈ L kn ( θ ) (cid:8) i . . . i n i n +1 ∈ I n +1 : ∃ t > f t + n ( x ) ∈ A i ...i n ∩ X i n +1 (cid:9) . Let i . . . i n +1 ∈ L n +1 ( θ ). Then, there exists t > i . . . i n +1 = θ t . . . θ t + n , which implies i . . . i n ∈ L kn ( θ ) for some k ∈ I (since i . . . i n ∈ L n ( θ ) and f t + n ( x ) ∈ X i n +1 ). On the other hand, byLemma 2.4, we have f t + n ( x ) ∈ A i ...i n . It follows that i . . . i n +1 ∈ B kn ( θ ), and thus (6) is true.If we suppose moreover that f satisfies the separation property, we can deduce that (6) is an equality.Indeed, if i . . . i n +1 ∈ S Nk =1 B kn ( θ ), then there exist k ∈ I and t > i . . . i n ∈ L kn ( θ ) and f t + n ( x ) ∈ A i ...i n ∩ X i n +1 . The latter implies that θ t + n = i n +1 , and i . . . i n = θ t . . . θ t + n − by Lemma 2.5.It follows that i . . . i n +1 ∈ L n +1 ( θ ).To finish the proof observe that for any k ∈ I the set B kn ( θ ) is defined by the union of disjoint sets thatsatisfy { i . . . i n i n +1 ∈ I n +1 : ∃ t > f t + n ( x ) ∈ A i ...i n ∩ X i n +1 } = k ∀ i . . . i n ∈ L kn ( θ ) , by definition of L kn ( θ ). So we have B kn ( θ ) = k L kn ( θ ). Moreover, since L kn ( θ ) ∩ L k ′ n ( θ ) = ∅ if k = k ′ , onthe one hand B kn ( θ ) ∩ B k ′ n ( θ ) = ∅ if k = k ′ , and on the other hand p ( θ, n ) = P Nk =1 L kn ( θ ). We deduce that N [ k =1 B kn ( θ ) = N X k =1 k L kn ( θ ) = N X k =1 L kn ( θ ) + N X k =1 ( k −
1) L kn ( θ ) = p ( θ, n ) + N X k =2 ( k −
1) L kn ( θ ) . Now, from (6) we conclude that (5) is true, and is an equality if f satisfies the separation property. From now on, we assume that the phase space X of f is a compact interval of R and that the contractionpieces are open intervals in X . This implies, in particular that the atoms are closed intervals. Also, sincethe number of pieces is finite, the map f has a finite number of discontinuities. We label the N contractionpieces { X i } Ni =1 of f , in such a way that X < X < · · · < X N . Definition 2.8.
Let x ∈ X and for any n > A n ( x ) := { A ∈ A n : ∃ t ∈ N : f t + n ( x ) ∈ A } . Let c ∈ ∆ and i ∈ { , . . . , N − } be such that c = X i ∩ X i +1 . Let n >
1, we say that c is n-left-right visited (in short nlr-visited ) by { f k ( x ) } k ∈ N if there exists A n ∈ A n ( x ) such that c ∈ A and { t ∈ N : f t + n ( x ) ∈ A n ∩ X i } 6 = ∅ and { t ∈ N : f t + n ( x ) ∈ A n ∩ X i +1 } 6 = ∅ . We denote ∆ nlr ( x ) the set of the discontinuities that are n lr-visited. We say that c is left-right recurrenlyvisited (in short lr-recurrently visited ) by { f k ( x ) } k ∈ N if c ∈ ∆ nlr ( x ) for all n >
1. We denote ∆ lr ( x ) the setof the discontinuities that are lr-recurrently visited.Note that for any x ∈ X and n > lr ( x ) ⊂ ∆ n +1 lr ( x ) ⊂ ∆ nlr ( x ) ⊂ ∆, because any atom ofgeneration n +1 is contained in an atom of generation n . Also, if c ∈ ∆ lr ( x ), then c is an accumulation point(by the left and by the right) of the orbit of x , since the diameter of the atoms of a piecewise contractingmap goes to 0 as their generation goes to infinity (see Remark 2.2). Lemma 2.9.
Let f be a piecewise contracting interval map satisfying the separation property. Let x ∈ e X and θ be its itinerary. Then, nlr ( x ) p ( θ, n + 1) − p ( θ, n ) ∀ n > . (7) Moreover, if n > is the smallest integer such that A ∩ ∆ for any A ∈ A n ( x ) with n > n , then p ( θ, n + 1) = p ( θ, n ) + nlr ( x ) ∀ n > n . (8)5ote that n exists and is bounded above by the smallest n > A ∈A n diam( A ) < min i ∈{ ,...,N } diam( X i ) , which in turn can be bounded above by a function of λ and the diameters of the continuity pieces (seeRemark 2.2). Proof.
Let n >
1. Suppose that c ∈ ∆ nlr ( x ), then there exists A ∈ A n ( x ) such that c ∈ A . Moreover thereexist i ∈ { , . . . , N } and t ∈ N such that f t + n ( x ) ∈ A ∩ X i . Therefore, according to Lemma 2.4, we have A ∩ A θ t ...θ t + n − = ∅ , and after Lemma 2.3 we have that A = A θ t ...θ t + n − . As there exists also t ′ ∈ N suchthat f t ′ + n ( x ) ∈ A ∩ X i +1 , we have that θ t . . . θ t + n − ∈ L kn ( θ ) for some k >
2. We deduce that∆ nlr ( x ) ⊂ N [ k =2 [ i ...i n ∈ L kn ( θ ) ( A i ...i n ∩ ∆ nlr ( x ))and it follows that nlr ( x ) N X k =2 X i ...i n ∈ L kn ( θ ) A i ...i n ∩ ∆ nlr ( x )) . Now, if A ∈ A n ( x ) and A ∩ ∆ nlr ( x ) = q , then A intersects at least q +1 continuity pieces that are visited bythe orbit of f n ( x ). It follows that for any k > i . . . i n ∈ L kn ( θ ) we have that A i ...i n ∩ ∆ nlr ( x )) k −
1. We deduce that nlr ( x ) N X k =2 ( k − L kn ( θ ) ∀ n > . (9)Now, let n > k >
2. If i . . . i n ∈ L kn ( θ ), then A i ...i n = ∅ and A i ...i n intersects at least k continuity pieces. As A i ...i n is a closed interval and the continuity pieces are open intervals, it follows that A i ...i n contains at least k − i . . . i n and i ′ . . . i ′ n are twodifferent words of L n ( θ ) then A i ...i n ∩ A i ′ ...i ′ n = ∅ . It follows that > N X k =2 ( k − L kn ( θ ) ∀ n > . (10)Then, inequalities (7) follow from (9), (10) and Lemma 2.7.Let n > n . Then, for any i . . . i n ∈ L n ( θ ) the atom A i ...i n intersects at most two continuity pieces,and therefore L kn ( θ ) = ∅ for all k >
3. Moreover, for any i . . . i n ∈ L n ( θ ) the discontinuity contained in A i ...i n belongs to ∆ nlr ( x ). We deduce that nlr ( x ) > L k ( θ ) = N X k =2 ( k − L kn ( θ ) , ∀ n > n , (11)which together with (9) and Lemma 2.7 implies (8). Theorem 2.10.
Let f be a piecewise contracting map satisfying the separation property. Let x ∈ e X and θ be its itinerary, then there exits m > such that p ( θ, n ) = n lr ( x ) + β ( x ) ∀ n > m , (12) with p ( θ, − lr ( x ) β ( x ) p ( θ, − m ( − lr ( x )) . (13) Proof.
For any c ∈ ∆, either c ∈ ∆ lr ( x ) or there exists ν ( c ) := min { n > c / ∈ ∆ nlr ( x ) } . As ∆ n +1 lr ( x ) ⊂ ∆ nlr ( x ) for all n >
1, it follows that for any c ∈ ∆ \ ∆ lr ( x ) we have that c / ∈ ∆ nlr ( x ) for all n > ν ( c ).Therefore, if n = max { ν ( c ) , c ∈ ∆ \ ∆ lr ( x ) } if ∆ = ∆ lr ( x ), and n = 1 otherwise, then ∆ nlr ( x ) = ∆ lr ( x )for all n > n . 6et m := max { n , n } . Then we can write (8) as p ( θ, n + 1) − p ( θ, n ) = lr ( x ) ∀ n > m , which implies p ( θ, n ) = p ( θ, m ) + ( n − m ) lr ( x ) for all n > m . It follows that (12) is true with β ( x ) = p ( θ, m ) − m lr ( x ) . (14)Recalling that nlr ( x ) > lr ( x ) for all n >
1, from (7) we obtain that p ( θ, − lr ( x ) p ( θ, n ) − n lr ( x ) p ( θ, − n ( − lr ( x )) ∀ n > , and setting n = m , we obtain (13) from (14). Proof of part 1) of Theorem 1.2.
For any x ∈ e X with itinerary θ we have p ( θ, − p ( θ, − lr ( x ) , (15)which implies, in particular, that 1 β ( x ). Together with Theorem 2.10, this proves equality (3) ofTheorem 1.2. In fact, equality (3) follows from equality (12) with α = lr ( x ) ∈ { , , . . . , N − } . Also,if α = N −
1, then lr ( x ) = β ( x ) = 1. Remark 2.11.
Now we give some direct consequences of Theorem 2.10 and we comment their relationswith other results. From Theorem 2.10 it follows that: There exists an itinerary with a complexity function which is not eventually constant if and only ifthere exists a discontinuity point c which is lr-recurrently visited by an orbit of e X . In particular, if thelimit set of f does not contain any discontinuity point, then the complexity function of every itinerary iseventually constant (recall that if c ∈ ∆ lr ( x ), then c belongs to the ω -limit set of x ). In this case, for any x ∈ e X with itinerary θ we have p ( θ, p ( θ, n ) p ( θ,
1) + ( m − ∀ n > , and p ( θ, n ) = β ( x ) is constant for any n > m . Moreover, when the limit set of f does not containdiscontinuity points, there exists a smallest integer m > m containsdiscontinuities. This integer m is an upper bound for m , which provides a uniform upper bound on β ( x )through inequalities (13). If lr ( x ) = 1 but ∆ lr ( x ) = ∆, then the itinerary θ of the orbit of x satisfies p ( θ, n ) = n + β ( x ) forall n large enough, where β ( x ) may be larger than 1. An example of a piecewise contracting map whoseitineraries have such a complexity can be found in [7]. In [8], it is shown that, up to a prefix of finite length,a sequence of complexity n + β is the image by a morphism of a Sturmian sequence. We conclude that, if lr ( x ) = 1 but ∆ lr ( x ) = ∆, the itinerary of any orbit, is Sturmian up to a morphism. Hence, up to amorphism, it is the itinerary of an irrational rotation, with respect to a suitable partition of the circle. If lr ( x ) >
1, the itinerary θ may be that of an irrational rotation: in fact, for some adequate valuesof α and β , the itineraries of an irrational rotation with respect to a suitable partition of the circle mayhave a complexity function of the form αn + β for all n large enough. However, not every sequence withsuch a complexity is itself an itinerary of an irrational rotation [1]. If all the discontinuities are lr-recurrently visited by the orbit of a point x ∈ e X , i.e. ∆ lr ( x ) = ∆, then m = n (see the definition of m in the proof of Theorem 2.10). Besides, from part 1 of Theorem 1.2, weknow that β ( x ) = 1 in this case. So, equality (12) becomes p ( θ, n ) = ( N − n + 1 ∀ n > n . (16)In the particular case where the map has two contraction pieces ( N = 2) and the (unique) discontinuitypoint is lr-recurrently visited by the orbit of x , then n = 1 and θ ( x ) is a Sturmian sequence. Therefore,it is also an itinerary of an irrational rotation. In general, if the itinerary θ ( x ) satisfies (16) for some N >
2, then it has the complexity of an itinerary of a N -interval exchange transformation satisfying theso-called Keane’s infinite distinct orbit condition [18, 19]. In fact it is proved in [16] the following result: if7 piecewise contracting map f has no periodic orbit, and is such that the image of any discontinuity andeach lateral limit of f belong to e X , then it is semi-conjugate to an interval exchange transformation. It ishowever not easy to exhibit examples satisfying these hypotheses, since generically a piecewise contractinginterval map has periodic points. In the next section, we will construct such examples for every N > θ . In the previous section we proved Theorem 2.10, which implies that the complexity of the itinerary of anyorbit of a piecewise contracting interval map satisfying the separation property is bounded from above by anaffine function whose slope is equal to the number of discontinuities of the map. However, as far as we know,there is still no example of piecewise contracting interval maps with more than one lr-recurrently visiteddiscontinuity. The purpose of this section is to construct such examples. Even more, we will constructexamples for which all the discontinuities are lr-recurrently visited by all the orbits. These maps generateitineraries with the maximal complexity for the fixed number N of contracting pieces. We say that theyhave “full” complexity.We are also interested in the asymptotic dynamics of such examples. It takes place in what we callthe attractor Λ of the piecewise contracting map f : X → X . To define the attractor we first recall thedefinition of the atoms A ∈ A n of generation n for any natural number n > ⊂ X as follows:Λ := ∞ \ n =1 Λ n where Λ n := [ A ∈A n A ∀ n > . (17)The sets Λ n can equivalently be recursively defined by Λ := f ( X \ ∆) and Λ n +1 := f (Λ n \ ∆) for all n > ω -limit points; see [5] for more details and examples.Precisely, in this section we prove the following theorem: Theorem 3.1.
For every N > , there exists a piecewise affine contracting map which has N contractionpieces and satisfies the separation property, whose attractor is a minimal Cantor set, and such that each ofits discontinuities is lr-recurrently visited by any orbit. Theorem 3.1, together with (16), proves immediately equation (4) of Theorem 1.2 for any n > n . Later,we will prove that it is always possible to construct the maps in such a way that n = 1 (see Lemma 3.8),to end the proof of part 2) of Theorem 1.2.Observe that the attractor of the piecewise contracting map of Theorem 3.1 contains all the disconti-nuities of the map. In fact any lr-recurrently visited discontinuity belongs to the ω -limit set of some orbit,and the attractor contains all the ω -limit sets.To prove Theorem 3.1, we will prove the following stronger statement: Assertion A:
For every N >
2, there exists N ordered disjoint open intervals X = [ c , c ) , X =( c , c ) , . . . , X N = ( c N − , c N ] of X := [ c , c N ] and f : X → X with all the following properties: P1)
The map f is piecewise contracting with contraction pieces X , . . . , X N and f | X i is affine with slope λ ∈ (0 , P2)
The map f satisfies the separation property. P3)
The attractor Λ of f is a Cantor set. P4)
The set S N − i =1 { f i ( c i ) , f i +1 ( c i ) } is a subset of e X . P5)
There exists i ∈ { , . . . , N − } such that { f n ( f i ( c i )) } n ∈ N or { f n ( f i +1 ( c i )) } n ∈ N is dense in Λ.8 For any x ∈ e X and i ∈ { , . . . , N − } we have that c i ∈ ∆ lr ( x ).To prove Assertion A and Theorem 3.1, we will need the following lemma. Lemma 3.2.
Let N > and c < c < · · · < c N in R . Let f : [ c , c N ] → [ c , c N ] be a piecewise contractingmap with contraction pieces X = [ c , c ) , X = ( c , c ) , . . . , X N = ( c N − , c N ] and which satisfies theseparation property. Suppose that there exists i ∈ { , . . . , N − } such that1) f j ( c i ) ∈ e X and { f n ( f j ( c i )) } n ∈ N is dense in Λ for some j ∈ { i, i + 1 } .2) c i ∈ ∆ lr ( x ) for some x ∈ e X .Then, for any ǫ > and y ∈ Λ such that Λ ∩ ( y, y + ν ) = ∅ (resp. Λ ∩ ( y − ν, y ) = ∅ ) for all ν > , thereexists l > such that f l ( x ) ∈ ( y, y + ǫ ) (resp. f l ( x ) ∈ ( y − ǫ, y ) ).Proof. We will make the proof for y ∈ Λ such that Λ ∩ ( y, y + ν ) = ∅ for all ν >
0, and without loss ofgenerality we will suppose that i = j = 1. Let ǫ > z ∈ Λ f ∩ ( y, y + ǫ ) and δ = min { z − y, y + ǫ − z } .Since { f n ( f ( c )) } n ∈ N is dense in Λ f , there exists n such that | f n ( f ( c )) − z | < δ . By injectivity of f on X \ ∆ the set P := ∪ n − l =0 f − l (∆) is finite, and for ρ := d ( f ( c ) , P ) > f n is continuouson ( f ( c ) − ρ, f ( c ) + ρ ). Using the continuity of f on [ c , c ], we obtain that there exists δ ′ > | f n ( f ( c )) − f n +1 ( x ) | < δ for all x ∈ ( c − δ ′ , c ). As c ∈ ∆ lr ( x ), there exists m such that f m ( x ) ∈ ( c − δ ′ , c ) and by the triangular inequality we deduce that | f l ( x ) − z | < δ for l = m + n + 1,that is f l ( x ) ∈ ( y, y + ǫ ).Before proving Assertion A, we show that together with Lemma 3.2 it implies Theorem 3.1: Proof of Theorem 3.1 as a Corollary of Assertion A.
Suppose that Assertion A is true and let f satisfying P1-6 for some N >
2. Then, f is a piecewise affine contracting interval map, it has the separationproperty, its attractor is a Cantor set and ∆ = ∆ lr ( x ) for any x ∈ e X . So, to prove Theorem 3.1 it remainsto prove that Λ is minimal and that ∆ = ∆ lr ( x ) for any x ∈ X \ e X . To this end, note that P1-6 donot impose any condition on the definition of f on ∆, and therefore f can be suitably defined on this set.So, we can assume that f ( c i ) ∈ { f i − ( c i ) , f i ( c i ) } for any i ∈ { , . . . , N − } , which implies by P4 that f (∆) ⊂ e X . Since f satisfies P1-6 , it satisfies the hypotheses of the Lemma 3.2. It follows that the orbit ofany point x ∈ Λ ∩ e X is dense in Λ. Now, since f ( c ) ∈ e X for all c ∈ ∆, the orbit of a point in Λ \ e X is alsodense in Λ. We deduce that Λ is minimal. Finally, P4 , P6 and f (∆) ⊂ e X imply that for any x ∈ X \ e X and c ∈ ∆ there exists p > c ∈ ∆ lr ( f p ( x )). Since ∆ lr ( f p ( x )) ⊂ ∆ lr ( x ), we conclude that anydiscontinuity of f is lr-recurrently visited by any orbit, ending the proof of Theorem 3.1.In the following subsections we will prove Assertion A. Let us describe briefly the route of the proof:The proof goes by induction on the number N > f . In Subsection 3.1, relyingon a known example, we prove that there exists a map satisfying P1-6 for N = 2. In Subsection 3.2,we construct a map g satisfying P1-6 with N + 1 contraction pieces, assuming the existence of a map f satisfying P1-6 with N contraction pieces and with contracting constant λ ∈ (0 , g from f ,we first choose an adequate point ξ ∈ e X , and its orbit { ξ r } r ∈ N , where ξ r = f r ( ξ ). Second, we “cut” theinterval X at each point ξ r with r > ξ ), and insert an interval G r substituting the point ξ r ,such that, for all r > G r is λ r . We define an affine map g | G r : G r → G r +1 for all r >
1. Inthis way, we have added a new discontinuity point of g at the point, say ξ . For all y S r > G r we definethe image g ( y ) from the image f ( x ) of the corresponding point x ∈ X \ { ξ r } r > . In particular g preservesthe old N − f . So g has N discontinuity points, hence N + 1 continuity pieces.Finally, in Proposition 3.7 we show that there exists a good choice of the cutting orbit { ξ r } r ∈ N , to make g inherit the properties P1-6 from f . P1-6 is a shorthand notation for “the properties P1 to P6 ”. .1 Full complexity with a single discontinuity In this subsection we prove that Assertion A holds for N = 2. We begin with a lemma about the sets Λ n (defined in (17)) for piecewise increasing maps with two contraction pieces. Lemma 3.3.
Let X := [0 , , c ∈ (0 , and f : X → X be a piecewise contracting map with contractionpieces X = [0 , c ) and X = ( c, . Suppose that the continuous extensions f and f of f are increasingand such that f ( c ) < f (1) < f (0) < f ( c ) = 1 .For every k ∈ N , let H k := f k (( f (1) , f (0))) . If c / ∈ H k for all k ∈ N , then H k = ( f k +1 (1) , f k +1 (0)) forall k ∈ N and Λ n = [0 , \ n − [ k =0 H k ∀ n > . (18) Moreover, { , } ⊂ e X , and for any n > and A ∈ A n there exists p and q in N such that A = [ f p (0) , f q (1)] .Proof. Assume that H k = ( f k +1 (1) , f k +1 (0)) for some k ∈ N . Then H k ⊂ [0 , c ) or H k ⊂ ( c, c / ∈ H k . Hence, f is continuous and increasing on H k , and H k +1 = ( f k +2 (1) , f k +2 (0)). As H = ( f (1) , f (0)),we have proved by induction that H k = ( f k +1 (1) , f k +1 (0)) for every k ∈ N .Now let us show (18) by induction. We have Λ := f ([0 , c ]) ∪ f ([ c, f (0) , ∪ [0 , f (1)] = [0 , \ H and hence (18) is true for n = 1. Now let n >
1, and assume that Λ n = [0 , \ ∪ n − k =0 H k . We shall provethat Λ n +1 = [0 , \ S nk =0 H k . First, observe thatΛ n +1 = f (cid:16) Λ n ∩ (0 , c ) (cid:17) ∪ f (cid:16) Λ n ∩ ( c, (cid:17) = f (Λ n \ { c } ) ∪ { f ( c ) , f ( c ) } , since c ∈ [0 , \ ∪ n − k =0 H k and therefore it belongs to the interior of Λ n . It follows that,Λ n +1 = f (cid:0) ([0 , \ { c } ) \ ∪ n − k =0 H k (cid:1) ∪ { , } . Besides, since f is injective on [0 , \ { c } , we have f (cid:0) ([0 , \ { c } ) \ ∪ n − k =0 H k (cid:1) = f ([0 , \ { c } ) \ f ( ∪ n − k =0 H k ) . On the other hand, we have f ( x ) / ∈ { , } for any x = c and c / ∈ H k for all k ∈ N , which implies thatΛ n +1 = ( f ([0 , \ { c } ) ∪ { , } ) \ ∪ n − k =0 f ( H k ) = ([0 , \ H ) \ ∪ nk =1 H k = [0 , \ ∪ nk =0 H k , as wanted.Let us prove that { , } ⊂ e X . Since c / ∈ f k ([ f (1) , f (0)]) for every k >
0, we have c / ∈ { f k (0) , f k (1) } forall k >
1. It follows that { , } ⊂ e X , because c / ∈ { , } .To end the proof, first observe that f ([0 , c ]) = [ f (0) ,
1] and f ([ c, , f (1)]. Therefore, the atomsof A are of the form [ f p (0) , f q (1)]. Now, as an induction hypothesis, assume that for some n > A ∈ A n there exists p and q in N such that A = [ f p (0) , f q (1)]. If B ∈ A n +1 , then, by definition ofatoms, there exist i ∈ { , } and A ∈ A n such that B = f i ( A ∩ X i ), where X := [0 , c ) and X := ( c, c / ∈ A , then A ⊂ X i and B = f ( A ) = [ f p +1 (0) , f q +1 (1)]. If c ∈ A , then either A ∩ X i = [ f p (0) , c ) or A ∩ X i = ( c, f q (1)]. In both cases A ∩ X i = ∅ , since { , } ⊂ e X . If follows that either B = [ f p +1 (0) ,
1] or B = [0 , f q +1 (1)]. Proposition 3.4.
Let λ and µ ∈ (0 , be such that λ + µ > and denote c := (1 − µ ) /λ . Let f : [0 , c ] → [0 , and f : [ c, → [0 , be defined by f ( x ) = λx + µ ∀ x ∈ [0 , c ] and f ( x ) = λx + µ − ∀ x ∈ [ c, . (19) Then, any map f : [0 , → [0 , such that f | [0 ,c ) = f | [0 ,c ) and f | ( c, = f | ( c, , and c / ∈ f k ([ f (1) , f (0)]) forall k ∈ N , satisfies P1-6 with X = [0 , c ) and X = ( c, . roof. Consider a map f : [0 , → [0 ,
1] which satisfies the hypotheses of Proposition 3.4. Then, it is easyto check that f satisfies P1-2 , with c = 0, c = c and c = 1. Also, f satisfies all the hypotheses of Lemma3.3, and it follows in particular that P4 holds.Let γ ∈ { , } and let us show that { f n ( γ ) } n ∈ N belongs to Λ. Since for any k ∈ N the set H k of Lemma3.3 is an open set of [0 , γ . Therefore, by (18) we have γ ∈ Λ n for all n ∈ N , that is γ ∈ Λ. Since γ ∈ e X and Λ ∩ e X is forward invariant, we deduce that { f n ( γ ) } n ∈ N ⊂ Λ.Now, let us prove that Λ has no isolated point. Let x ∈ Λ and ǫ >
0. Let n > A ) < ǫ for every A ∈ A n . Let A ∈ A n be such that x ∈ A , and let p and q ∈ N be such that A = [ f p (0) , f q (1)].If x ∈ ( f p (0) , f q (1)], then 0 < | f p (0) − x | < ǫ and if x ∈ [ f p (0) , f q (1)), then 0 < | f q (1) − x | < ǫ . Sinceboth points f p (0) and f q (1) ∈ Λ, we found a point in Λ \ { x } which is at a distance less than ǫ of x . Thisshows that Λ is a perfect set (recall that Λ is compact) and it proves at the same time that { f n (0) } n ∈ N and { f n (1) } n ∈ N are dense in Λ, i.e. f satisfies P5 . Now, Λ is totally disconnected because f satisfies theseparation property [5]. It follows that Λ is a Cantor set and f satisfies P3 .Now we show that f satisfies P6 . To this end, we prove that for every x ∈ e X and ǫ > l ∈ N and r ∈ N such that f l ( x ) ∈ ( c − ǫ, c ) and f r ( x ) ∈ ( c, c + ǫ ). Let x ∈ e X and ǫ >
0. Let n ∈ N be such that diam( A ) < ǫ/ A ∈ A n and let A ∈ A n be such that f n ( x ) ∈ A . Denote p and q the integers such that A = [ f p (0) , f q (1)] and let T = { t ∈ N : f n + t ( x ) ∈ ( c − ǫ, c + ǫ ) } . Arguing bycontradiction, assume that T = ∅ or that f n + t ( x ) ∈ ( c, c + ǫ ) for all t ∈ T . Then, by induction on t ∈ N ,we deduce that0 < ( f q + t (1) − c )( f n + t ( x ) − c ) and 0 f q + t (1) − f n + t ( x ) < ǫ/ ∀ t ∈ N . Therefore, for each t ∈ T we have f q + t (1) ∈ ( c,
1] and for each t / ∈ T we have | f q + t (1) − c | > | f n + t ( x ) − c | − | f q + t (1) − f n + t ( x ) | > ǫ/ . We deduce that f q + t (1) / ∈ ( c − ǫ/ , c ) for all t ∈ N . Now, let ν > f k (1) / ∈ ( c − ν, c ) forall k q . Then, f k (1) / ∈ ( c − ǫ , c ) for all k ∈ N , where ǫ = min { ν, ǫ/ } . On the other hand, there exit p ′ and q ′ such that [ f p ′ (0) , f q ′ (1)] is an atom of diameter strictly smaller than ǫ which contains c . Since f p ′ (0) ∈ Λ and f k (1) / ∈ ( c − ǫ , c ) for all k ∈ N , it follows that { f k (1) } k ∈ N is not dense in Λ, which is acontradiction. Therefore, T = ∅ and there exists l ∈ N such that f l ( x ) ∈ ( c − ǫ, c ). Now, if we assume that f n + t ( x ) ∈ ( c − ǫ, c ) for every t ∈ T , we deduce with an analogous proof that { f k (0) } k ∈ N is not dense in Λ.Therefore, there exists r ∈ N such that f r ( x ) ∈ ( c, c + ǫ ). Proof of Assertion A for N = 2 . Consider a map f : [0 , → [0 ,
1] defined by f ( x ) = λx + µ mod 1 ∀ x ∈ [0 , , where λ and µ ∈ (0 ,
1) are such that λ + µ >
1. Then, f is a piecewise contracting map which satisfiesthe hypothesis (19), with the particularity that f ( c ) = 0. Immediately, the “gap” between the atoms ofgeneration 1 is the interval ( f (1) , f (0)). It is standard to prove that if there exists a minimum naturalnumber k such that c ∈ f k ([ f (1) , f (0)]), then the attractor of f contains only periodic points. On theother hand, it has been proved using a rotation number approach that there is an uncountable set ofvalues of ( λ, µ ) such that f has no periodic points [3, 4, 6, 7, 17]. It follows that, for such values of ( λ, µ ), c / ∈ f k ([ f (1) , f (0)]) for all k ∈ N . Together with Proposition 3.4 this proves that Assertion A holds for N = 2. In the previous subsection we proved the existence of a map satisfying
P1-6 with N = 2 (a piecewisecontracting interval map with a single discontinuity). In this subsection we will complete the proof ofAssertion A, by induction on N .Let us assume that Assertion A holds for some N >
2. Then, there exists c < c < · · · < c N in R and f : X → X , where X = [ c , c N ], which satisfies P1-6 with the contraction pieces X = [ c , c ) , . . . , X N =11 c N − , c N ]. In the following, we denote ∆ f := { c i } i N − the set of the discontinuities of f , Λ f theattractor of f , and e X f the set defined by (2) where ∆ = ∆ f .Now, we construct a new map g with N + 1 contraction pieces and satisfying P1-6 , from the given map f that satisfies P1-6 and has N contraction pieces. The construction involves what we call a well-cutting orbit of f , defined as follows: Definition 3.5.
We say that an orbit { ξ r } r ∈ N of f is well-cutting , if ξ ∈ Λ f ∩ e X f and { ξ r } r ∈ N does notcontain any point of the following sets:1) the boundaries of the gaps of the Cantor set Λ f ,2) the orbits of c and c N ,3) the orbits of f i ( c i ) and f i +1 ( c i ) for all i ∈ { , . . . , N − } .Note that f has an uncountable number of well-cutting orbits. Indeed, ∆ f and the sets of items 1), 2),3) are countable. Therefore, the set P of all their pre-images is also countable. Since Λ f is uncountable,the complement of P in Λ f is uncountable and contains only well-cutting orbit of f . Also, a well-cuttingorbit is not eventually periodic. Indeed, by Theorem 2.10, the property P6 implies that the complexityfunction of the itinerary of ξ is not eventually constant.Let λ ∈ (0 ,
1) be the slope of f on any of its contraction pieces, and let { ξ r } r ∈ N be a well-cutting orbitof f . Consider the function φ : [ c , c N ] → R defined for any x ∈ [ c , c N ] by φ ( x ) = x + X n ∈N ( x ) λ n where N ( x ) := { n > ξ n < x } . (20)The following lemma gathers basic properties of φ that we will use in this section. Lemma 3.6.
The function φ is strictly increasing, left-continuous on [ c , c N ] , continuous on [ c , c N ] \{ ξ r } r > (in particular at ξ ), and lim x ց ξ r φ ( x ) = φ ( ξ r ) + λ r for all r > . Moreover, φ ([ c , c N ]) = [ φ ( c ) , φ ( c N )] \ ∞ [ r =1 G r where G r := ( φ ( ξ r ) , φ ( ξ r ) + λ r ] ∀ r > , and G r ∩ G l = ∅ for all r = l .Proof. Noting that N ( x ) ⊂ N ( x ′ ) for any x < x ′ , it is straightforward to show that φ is strictly increasing.Now we show that the left-hand limit of φ at any point x ∈ ( c , c N ] is equal to φ ( x ). Let x ∈ ( c , c N ]and ǫ >
0. Take m such that P ∞ n = m λ n < ǫ/ ρ > ξ n / ∈ ( x − ρ, x ) for all n < m .Now, if δ := min { ǫ/ , ρ } , then for any x ∈ ( x − δ, x ), we have | φ ( x ) − φ ( x ) | = x − x + X n ∈N ( x ) λ n − X n ∈N ( x ) λ n < ǫ X n ∈N ( x ) \N ( x ) λ n < ǫ, since min N ( x ) \ N ( x ) > m if ξ n / ∈ ( x − ρ, x ) for all n < m . With an analog proof, we can show thatthe right-hand limit of φ at x ∈ [ c , c N ) is equal to φ ( x ) if x = ξ r for any r >
1, and equal to φ ( ξ r ) + λ r if x = ξ r for some r > φ is strictly increasing, left-continuous and has a discontinuity jump of mag-nitude λ r at every point ξ r with r >
1, it is standard to check that φ ([ c , c N ]) = [ φ ( c ) , φ ( c N )] \ ∞ [ r =1 G r . Finally we show that G r ∩ G l = ∅ for all r = l . Let l and r > r = l and ξ r > ξ l . As φ ( ξ r ) ∈ φ ([ c , c N ]) we have that φ ( ξ r ) / ∈ G l , and as φ is injective we have that φ ( ξ r ) / ∈ G l . It follows that G r ∩ G l = ∅ , since φ ( ξ r ) > φ ( ξ l ). 12he following proposition shows how a well-cutting orbit of f and its associated function φ allow us toobtain a map which satisfies P1-6 with N + 1 contraction pieces. Therefore, it ends the proof of AssertionA by induction on N . Proposition 3.7.
Let { ξ r } r ∈ N be a well-cutting orbit of f and φ be defined according to (20) . Let ∆ g := φ (∆ f ∪ { ξ } ) . Then, any map g : [ φ ( c ) , φ ( c N )] → [ φ ( c ) , φ ( c N )] defined on [ φ ( c ) , φ ( c N )] \ ∆ g by g ( y ) = (cid:26) φ ◦ f ◦ φ − ( y ) if y ∈ φ ([ c , c N ]) \ ∆ g λ ( y − φ ( ξ r )) + φ ( ξ r +1 ) if y ∈ G r and r > satisfies P1-6 with N + 1 contraction pieces.Proof. Let d < d < · · · < d N +1 be such that { d , d , . . . , d N +1 } = φ (∆ f ∪ { c , ξ , c N } ). We denote j theinteger of { , . . . , N } such that d j = φ ( ξ ) . (22)Let Y := [ d , d ) , Y := ( d , d ) , . . . , Y N +1 := ( d N , d N +1 ] and Y := [ d , d N +1 ]. Since φ is strictly increasing,the sets Y j are all non-empty and pairwise disjoint. Let g : Y → Y be a map satisfying (21). We are goingto show that g satisfies P1-6 with Y , Y , . . . , Y N +1 . P1)
We first show that for any j ∈ { , , . . . , N } the map g is affine with slope λ on the interval ( d j , d j +1 ),that is g ( y ′ ) − g ( y ) = λ ( y ′ − y ) ∀ y, y ′ ∈ ( d j , d j +1 ) . (23)To prove (23), we fix j ∈ { , , . . . , N } and y, y ′ ∈ ( d j , d j +1 ), and we consider three cases:Case 1: Assume that y, y ′ ∈ φ ([ c , c N ]) and denote x := φ − ( y ) > φ − ( d j ) and x ′ := φ − ( y ′ ) < φ − ( d j +1 ).If we assume (with no loss of generality) y < y ′ , then x < x ′ and g ( y ′ ) − g ( y ) = f ( x ′ ) − f ( x ) + X n ∈N λ n where N := { n > f ( x ) ξ n < f ( x ′ ) } . Since f is affine on ( φ − ( d j ) , φ − ( d j +1 )) and has slope λ , we have that f ( x ′ ) − f ( x ) = λ ( x ′ − x ). On theother hand, since f is injective (separation property) and increasing on ( φ − ( d j ) , φ − ( d j +1 )) we have that n ∈ N if and only if n > n − ∈ N , where N := { n > x ξ n < x ′ } . It follows that g ( y ′ ) − g ( y ) = λ ( x ′ − x ) + X n ∈N λ n +1 = λ ( φ ( x ′ ) − φ ( x )) = λ ( y ′ − y ) . Case 2: Assume y ∈ G r for some r > y ′ ∈ φ ([ c , c N ]). Since G r is an interval and G r ∩ φ ([ c , c N ]) = ∅ ,we have G r ⊂ ( d j , d j +1 ). Therefore, φ ( ξ r ) ∈ ( d j , d j +1 ), because r = 0 and { ξ r } r ∈ N is a well-cutting orbit.So we can use Case 1 to obtain g ( y ′ ) − g ( φ ( ξ r )) = λ ( y ′ − φ ( ξ r )) . On the other hand, by definition of g on G r g ( φ ( ξ r )) − g ( y ) = φ ( ξ r +1 ) − λ ( y − φ ( ξ r )) − φ ( ξ r +1 ) , and the sum of these two equalities is (23).Case 3: Assume y ∈ G r and y ′ ∈ G ′ r for some r and r ′ > z ∈ φ ([ c , c N ]) ∩ ( d j , d j +1 ). Thenapplying Case 2 twice, we obtain g ( y ′ ) − g ( y ) = g ( y ′ ) − g ( z ) + g ( z ) − g ( y ) = λ ( y ′ − y ) , which ends the proof of (23).According to (23), we know that g is continuous on ∪ Nj =0 ( d j , d j +1 ). To study the continuity of g on the border of these intervals, we compute the left-hand and right-hand limits of g at the points of φ (∆ f ∪ { c , c N } ). Let i ∈ { , , . . . , N } and consider an increasing sequence { x n } n ∈ N in ( c i − , c i ) whichconverges to c i . Then, the sequence { φ ( x n ) } n ∈ N is increasing and converges to φ ( c i ), because φ is continuous13t any point that is not in the positive orbit of ξ , and { ξ r } r ∈ N is a well-cutting orbit. Using the definitionof g on φ ([ c , c N ]) \ ∆ g and the continuous extension f i of f | ( c i − ,c i ) to [ c i − , c i ] we obtain thatlim n →∞ g ( φ ( x n )) = lim n →∞ φ ( f ( x n )) = lim n →∞ φ ( f i ( x n )) = lim x → f i ( c i ) φ ( x ) . Once again, since { ξ r } r ∈ N is a well-cutting orbit, f i ( c i ) does not belong to { ξ r } r ∈ N and φ is continuous at f i ( c i ). It follows that lim y ր φ ( c i ) g ( y ) = φ ( f i ( c i )) . (24)Similarly, for any i ∈ { , , . . . , N − } we havelim y ց φ ( c i ) g ( y ) = φ ( f i +1 ( c i )) . (25)Since f i +1 ( c i ) = f i ( c i ) for any i ∈ { , . . . , N − } (separation property) and φ is injective, we deduce that g is discontinuous on φ (∆ f ). On the other hand, from (24) and (25) respectively, we obtain that g iscontinuous at φ ( c N ) and φ ( c ) respectively.It remains to study g at φ ( ξ ). As { ξ r } r ∈ N is a well-cutting orbit, there exists i ∈ { , , . . . , N } suchthat ξ ∈ ( c i − , c i ). Using the left continuity of φ , the monotonicity of f on ( c i − , c i ), and the continuityof f at ξ , we obtain that lim y ր φ ( ξ ) g ( y ) = φ ( f ( ξ )) = φ ( ξ ) . (26)On the other hand lim y ց φ ( ξ ) g ( y ) = lim x ց ξ φ ( x ) = φ ( ξ ) + λ. (27)Equalities (26) and (27) prove that g is discontinuous at φ ( ξ ). We conclude that the set of the discontinuitypoints of g is ∆ g , which together with (23) proves that g is a piecewise affine contracting map withcontractions pieces Y , Y , . . . , Y N +1 . In particular, the set of (2) writes for g as e X g = T + ∞ n =0 g − n ( X \ ∆ g ). P2)
Let us prove that g satisfies the separation property. For any j ∈ { , . . . , N + 1 } denote g j : Y j → [ φ ( c ) , φ ( c N )] the continuous extension of g | Y j to Y j . Then, Y j = [ φ ( c j − ) , φ ( ξ )] , Y j +1 = [ φ ( ξ ) , φ ( c j )] and Y j = (cid:26) [ φ ( c j − ) , φ ( c j )] if 1 j < j [ φ ( c j − ) , φ ( c j − )] if j + 1 < j N + 1 , where j is defined by (22). For every j ∈ { , . . . , N + 1 } the map g j is affine with slope λ >
0. Therefore, g j ( Y j ) is an interval whose boundaries are obtained using (24), (25), (26) and (27). Indeed, g j ( Y j ) = [ φ ( f j ( c j − )) , φ ( f j ( ξ ))] , g j +1 ( Y j +1 ) = [ φ ( f j ( ξ )) + λ, φ ( f j ( c j ))] , and g j ( Y j ) = (cid:26) [ φ ( f j ( c j − )) , φ ( f j ( c j ))] if 1 j < j [ φ ( f j − ( c j − )) , φ ( f j − ( c j − ))] if j + 1 < j N + 1 . On the one hand we have obtained that g j ( Y j ) ∩ g j +1 ( Y j +1 ) = ∅ . On the other hand, the monotonicityof φ and the separation property of f imply that the sets g ( Y ) , . . . , g j − ( Y j − ) , g j ( Y j ) ∪ g j +1 ( Y j +1 ) , g j +2 ( Y j +2 ) , . . . , g N +1 ( Y N +1 )are pairwise disjoint. It follows that g satisfies the separation property. P3)
Now we study the attractor of g , which can be written using the continuous extensions of g asΛ g = ∞ \ n =1 Λ g,n , where the sets Λ g,n are recursively defined asΛ g, = N +1 [ j =1 g j ( Y j ) and Λ g,n +1 = N +1 [ j =1 g j (Λ g,n ∩ Y j ) ∀ n > .
14e denote G := S ∞ r =1 G r and we recall that φ ( X ) = Y \ G by Lemma 3.6.Let us show by induction that Λ g,n ∩ φ ( X ) = φ (Λ f,n ) ∀ n > . (28)First note thatΛ g, ∩ φ ( X ) = N +1 [ j =1 ( g j ( Y j ∩ φ ( X )) ∪ g j ( Y j ∩ G )) ∩ φ ( X ) = N +1 [ j =1 g j ( Y j ∩ φ ( X )) ∩ φ ( X ) , since for any j ∈ { , . . . , N + 1 } we have Y j ∩ G = Y j ∩ G and therefore g j ( Y j ∩ G ) = g ( Y j ∩ G ) ⊂ G .Besides, if y ∈ Y j ∩ φ ( X ) and y = d j then g j ( y ) = (cid:26) φ ◦ f j ◦ φ − ( y ) if 1 j < j + 1 φ ◦ f j − ◦ φ − ( y ) if j + 1 j N + 1 , (29)and if y = d j then g j ( y ) = φ ◦ f j ◦ φ − ( y ) and g j +1 ( y ) = φ ( f j ( ξ )) + λ = φ ( ξ ) + λ ∈ G. (30)Denote Z , Z , . . . , Z N the sets defined by Z j = Y j ∪ Y j +1 and Z j = (cid:26) Y j if 1 j < j Y j +1 if j < j N , then Λ g, ∩ φ ( X ) = N [ j =1 φ ◦ f j ◦ φ − ( Z j ∩ φ ( X )) . As Z j ∩ φ ( X ) = φ ( X j ) for any j ∈ { , . . . , N } , we deduce thatΛ g, ∩ φ ( X ) = N [ j =1 φ ( f j ( X j )) = φ (Λ f, ) . Now assume that Λ g,n ∩ φ ( X ) = φ (Λ f,n ) for some n >
1. As before, we haveΛ g,n +1 ∩ φ ( X ) = N +1 [ j =1 g j (Λ g,n ∩ Y j ∩ φ ( X )) ∩ φ ( X ) = N [ j =1 φ ◦ f j ◦ φ − (Λ g,n ∩ Z j ∩ φ ( X )) . To obtain that Λ g,n +1 ∩ φ ( X ) = φ (Λ f,n +1 ) and complete the induction, it is enough to show thatΛ g,n ∩ Z j ∩ φ ( X ) = φ (Λ f,n ∩ X j ) . (31)On the one hand, using the induction hypothesis, we obtain Λ g,n ∩ φ ( X ) = φ (Λ f,n ). On the other hand wehave Z j ∩ φ ( X ) = φ ( X j ). As φ is injective, it follows thatΛ g,n ∩ Z j ∩ φ ( X ) = φ (Λ f,n ∩ X j ) ∀ j ∈ { , . . . , N } . Let y ∈ φ (Λ f,n ∩ X j \ (Λ f,n ∩ X j )) and x ∈ Λ f,n ∩ X j \ (Λ f,n ∩ X j ) be such that y = φ ( x ). Then, x ∈ { c j − , c j } and φ is continuous at x , since x / ∈ { ξ r } r ∈ N . Let { x n } n ∈ N be a sequence of Λ f,n ∩ X j which converges to x . Then, the sequence { φ ( x n ) } n ∈ N belongs to Λ g,n ∩ Z j and converges to y = φ ( x ). Itfollows that y ∈ Λ g,n ∩ Z j ∩ φ ( X ) and we have proved that φ (Λ f,n ∩ X j ) ⊂ Λ g,n ∩ Z j ∩ φ ( X ). To show theconverse inclusion, take y ∈ Λ g,n ∩ Z j ∩ φ ( X ) \ (Λ g,n ∩ Z j ). Then, y ∈ Z j \ Z j ⊂ { φ ( c j − ) , φ ( c j ) , φ ( ξ ) } .If y = φ ( ξ ), then j = j and y ∈ φ (Λ f,n ∩ X j ), since ξ ∈ Λ f ∩ X j . As c j − and c j are lr-recurrentlyvisited, there exist two atoms A j − and A j in the set of the atoms of generation n of f such that c j − and c j belong to the interior of A j − and A j , respectively. As { c j − , c j } ⊂ X j , it follows that c j − ∈ A j − ∩ X j and c j ∈ A j ∩ X j . Recalling that Λ f,n is the union of all the atoms of generation n of f , we deduce that15 c j − , c j } ⊂ Λ f,n ∩ X j and y ∈ φ (Λ f,n ∩ X j ). This ends the proof of (31) and completes the proof byinduction of (28).Using the injectivity of φ and (28) we obtain thatΛ g ∩ φ ( X ) = φ (Λ f ) . (32)Now we show that Λ g ∩ G = { φ ( ξ r ) + λ r } r > . To this end, we prove by induction thatΛ g,n ∩ G = { φ ( ξ r ) + λ r , r n } ∪ ∞ [ r = n +1 G r ∀ n > . (33)Using equations (29) and (30) it follows thatΛ g, ∩ G = N +1 [ j =1 ( g j ( Y j ∩ φ ( X )) ∪ g j ( Y j ∩ G )) ∩ G = { g j +1 ( d j ) } ∪ N +1 [ j =1 g j ( Y j ∩ G ) ∩ G Recalling the inclusion of G in the union of the sets Y , Y , . . . , Y N +1 , and the definition of g in G , we obtainΛ g, ∩ G = { φ ( ξ ) + λ } ∪ N +1 [ r =2 G r , which proves (33) for n = 1.Now, assume that (33) holds for some n >
1. As d j = φ ( ξ ) ∈ φ (Λ f ) = Λ g ∩ φ ( X ) we have that d j ∈ Λ g ⊂ Λ g,n . On the other hand, there exits a decreasing sequence in Λ f which converges to ξ ,because Λ f is a Cantor set and ξ is not a border of a gap of Λ f . By (32), the image of this sequence by φ belongs to Λ g and it is decreasing by monotonicity of φ . From the continuity of φ at ξ , it follows that d j ∈ Λ g ∩ Y j +1 ⊂ Λ g,n ∩ Y j +1 . Now, using once again equations (29) and (30), we deduceΛ g,n +1 ∩ G = N +1 [ j =1 ( g j (Λ g,n ∩ Y j ∩ φ ( X )) ∪ g j (Λ g,n ∩ Y j ∩ G )) ∩ G = { g j +1 ( d j ) } ∪ N +1 [ j =1 g j (Λ g,n ∩ Y j ∩ G ) ∩ G. As Λ g,n ∩ Y j \ (Λ g,n ∩ Y j ) ⊂ { d j − , d j } , and d , d , . . . , d N +1 do not belong to G , we have that g j (Λ g,n ∩ Y j ∩ G ) = g (Λ g,n ∩ Y j ∩ G ) for all j ∈ { , , . . . , N + 1 } . Therefore,Λ g,n +1 ∩ G = { φ ( ξ ) + λ } ∪ g (Λ g,n ∩ G ) . Using the induction hypothesis, we obtainΛ g,n +1 ∩ G = { φ ( ξ r ) + λ r , r n + 1 } ∪ ∞ [ r = n +2 G r , which complete the proof of (33).Note that (33) can also be written asΛ g,n ∩ G = { φ ( ξ r ) + λ r } r > ∪ ∞ [ r = n +1 Int( G r ) ∀ n > , and recall that the sets G r are pairwise disjoint. This implies that Λ g ∩ G = { φ ( ξ r )+ λ r } r > , which togetherwith (32) gives Λ g = φ (Λ f ) ∪ { φ ( ξ r ) + λ r } r > . (34)Now we show that Λ g is a Cantor set. By definition of attractor Λ g is compact. Besides, it is totallydisconnected because g satisfies the separation property (see Theorem 5.2 of [5]). It remains to show thatΛ g has no isolated point. Let y ∈ φ (Λ f ) and x ∈ Λ f be such that y = φ ( x ). As Λ f is a Cantor set,16here exists a sequence { x n } n ∈ N in Λ f \ { x } which converges to x and { φ ( x n ) } n ∈ N belongs to Λ g \ { y } . If y / ∈ { φ ( ξ r ) } r > , then φ is continuous at x and { φ ( x n ) } n ∈ N converges to y . If y = φ ( ξ r ) for some r > { x n } n ∈ N is increasing, since ξ r does not belong to the boundaries of the gaps ofthe Cantor set Λ f . Using the left-continuity of φ , we obtain that { φ ( x n ) } n ∈ N converges to y . Now, let y = φ ( ξ r ) + λ r for some r > { x n } n ∈ N be a decreasing sequence in Λ f which converges to ξ r . Then, { φ ( x n ) } n ∈ N converges to y and belongs to Λ g \ { y } , since y / ∈ φ ( X ). P4)
Let us prove that g j ( d j ) and g j +1 ( d j ) belong to e X g for every j ∈ { , , . . . , N } . First note that forany x such that f n ( x ) / ∈ ∆ f ∪ { ξ } for all n ∈ N , we have g n ◦ φ ( x ) = φ ◦ f n ( x ) ∀ n ∈ N and φ ( x ) ∈ e X g . (35)Let j ∈ { , , . . . , N } and y ∈ { g j ( d j ) , g j +1 ( d j ) } . If y = g j +1 ( d j ), then using (29) and (30) we obtain that y = φ ( x ) for some point x ∈ { f j ( c j ) , f j +1 ( c j ) , f j − ( c j − ) , f j ( c j − ) , f ( ξ ) } . Since { ξ n } n ∈ N is a well-cuttingorbit and f i ( c i ) and f i +1 ( c i ) belong to e X f for any i ∈ { , , . . . , N − } , it follows from (35) that y ∈ e X g .Now, if y = g j +1 ( d j ) then by (30) we have that g n ( y ) ∈ G for all n ∈ N , and therefore y ∈ e X g . P5)
Let i ∈ { , . . . , N − } and l ∈ { i, i + 1 } be such that { f n ( f l ( c i )) } n ∈ N is dense in Λ f . Let us denote x := f l ( c i ) and y := φ ( x ). Using (29) we obtain that there exists j = j such that y ∈ { g j ( d j ) , g j +1 ( d j ) } .Therefore, to prove P5 for g it is enough to show that { g n ( y ) } n ∈ N is dense in Λ g . Applying (35) to x ,we obtain that y ∈ e X g and that g n ( y ) = φ ( f n ( x )) for all n ∈ N . Besides, according to (32) we have { φ ( f n ( x )) } n ∈ N ⊂ Λ g , since { f n ( x ) } n ∈ N ⊂ Λ f .First let y ∈ Λ g \ { φ ( ξ r ) , φ ( ξ r ) + λ r } r > . Then, y ∈ φ (Λ f ) and there exists { n k } k ∈ N such that { f n k ( x ) } k ∈ N converges to x := φ − ( y ) ∈ Λ f . Since x / ∈ { ξ r } r > , it follows that φ is continuous at x and { g n k ( y ) } k ∈ N converges to y .Now, let y = φ ( ξ r ) for some r >
1. Since c i is lr-recurrently visited by { f k ( x ) } k ∈ N , and Λ f ∩ ( ξ r − ν, ξ r ) = ∅ for all ν > ξ r is not border of gap) by Lemma 3.2 we have that the orbit of x accumulatesfrom the left on ξ r . Using the left-continuity of φ we obtain that the orbit of y accumulates on y = φ ( ξ r ).Using now Λ f ∩ ( ξ r , ξ r + ν ) = ∅ for all ν >
0, we obtain that there exists a subsequence of { f n ( x ) } n ∈ N which converges to ξ r from the right-hand side. The image by φ of this subsequence converges to φ ( ξ r )+ λ r .It follows that φ ( ξ r ) + λ r is also a limit point of { g n ( y ) } n ∈ N . P6)
Now we prove that all the discontinuities of g are lr-recurrently visited by the orbits of the points of e X g . Let j ∈ { , . . . , N } and y ∈ e X g . We are going to show that there exist two sequences { n k } k ∈ N and { m k } k ∈ N going to infinity such that g n k ( y ) < d j < g m k ( y ) for all k ∈ N and { g n k ( y ) } k ∈ N and { g m k ( y ) } k ∈ N converge to d j .First, let us show the above assertion for j = j . We denote c the point of ∆ f such that φ ( c ) = d j . If y ∈ e X g ∩ φ ( X ), then g n ( y ) = φ ◦ f n ◦ φ − ( y ) ∀ n ∈ N , (36)since e X g ∩ φ ( X ) is forward invariant by g . It follows that x := φ − ( y ) belongs to e X f and c is lr-recurrentlyvisited by { f k ( x ) } k ∈ N by property P6 of f . Therefore, there exist two sequences { n k } k ∈ N and { m k } k ∈ N going to infinity such that f n k ( x ) < c < f m k ( x ) for all k ∈ N and { f n k ( x ) } k ∈ N and { f m k ( x ) } k ∈ N convergeto c . Using (36), the monotonicity of φ and its continuity at c , we deduce that { g n k ( y ) } k ∈ N and { g m k ( y ) } k ∈ N satisfy the required properties.Now assume y ∈ G ⊂ f X g and let r > y ∈ G r . Then, g n ( y ) ∈ G r + n for all n ∈ N , that is φ ( f n ( ξ r )) < g n ( y ) φ ( f n ( ξ r )) + λ n + r ∀ n ∈ N . (37)As ξ r ∈ e X f , we have that c is lr-recurrently visited by { f k ( ξ r ) } k ∈ N . Therefore, there exists two sequences { n k } k ∈ N and { m k } k ∈ N going to infinity such that f n k ( ξ r ) < c < f m k ( ξ r ) for all k ∈ N and { f n k ( ξ r ) } k ∈ N and { f m k ( ξ r ) } k ∈ N converge to c . Using (37) and the continuity of φ at c we have that both { g n k ( y ) } k ∈ N and { g m k ( y ) } k ∈ N converge to φ ( c ) = d j . On the other hand, by monotonicity of φ and the left-hand side of(37) we have that d j < g m k ( y ) for all k ∈ N . Again by monotonicity of φ , we have φ ( f n k ( ξ r )) < d j , whichimplies that φ ( f n k ( ξ r )) + λ n k + r < d j because d j / ∈ G . We deduce from (37) that g n k ( y ) < d j .17inally, we show that d j = φ ( ξ ) is lr-recurrently visited by the orbit of any point of e X g . Let d j ∈ ∆ g \ { d j } be a discontinuity of g such that { g n ( g l ( d j )) } n ∈ N is dense in Λ g for some l ∈ { j, j + 1 } (weproved in P6 that it exists). Since d j = d j we already know that d j is lr-recurrently visited by { g k ( y ) } k ∈ N for any y ∈ e X g . Now, since φ ( ξ ) is not a border of gap of Λ g (because ξ is not a border of gap of Λ f and φ is continuous at this point) we can apply Lemma 3.2 to deduce that d j is lr-recurrently visited by { g k ( y ) } k ∈ N for any y ∈ e X g . Lemma 3.8.
If any atom of generation of f contains at most one point of ∆ f , then there exists awell-cutting orbit of f such that any atom of generation of g contains at most one point of ∆ g .Proof. As f satisfies P6 , any discontinuity of f is contained in an atom of generation 1 of f and this atomis unique because of P2 . It follows that one of the N atoms of generation 1 of f does not contain anydiscontinuity. Let us denote A i this atom, where i ∈ { , . . . , N } is such that A i = f (( c i − , c i )). Then, A i ∩ Λ f,n = ∅ for all n >
1, since by P6 for any n > n − c i in its interior. It follows that A i ∩ Λ f = ∅ . Moreover, A i ∩ Λ f is compact, totally disconnectedand any point of Int( A i ) ∩ Λ f is not isolated, because Λ f is a Cantor set. Now, as the atoms are compactand disjoint, if x ∈ ( A i \ Int( A i )) ∩ Λ f then x is a border of gap of Λ f and there exists a sequence inInt( A i ) ∩ Λ f which converges to x . We deduce that A i ∩ Λ f is a Cantor set. Therefore, there exists ξ ∈ A i such that { f k ( ξ ) } k ∈ N is a well-cutting orbit of f .For every i ∈ { , . . . , N } let B i = g (( φ ( c i − ) , φ ( c i ))), where φ is the function defined in (20) with a well-cutting orbit of f such that ξ ∈ A i . Then, for any atom B of generation 1 of g there exists i ∈ { , . . . , N } such that B ⊆ B i . Moreover, we can show that B i ∩ φ ( X ) = φ ( A i ) ∀ i ∈ { , . . . , N } , where A i := f (( c i − , c i )) is an atom of generation 1 of f . Now, let d = d ′ ∈ ∆ g and assume by contradictionthat there exists an atom of generation 1 of g which contains d and d ′ . Then, there exits an atom ofgeneration 1 of f which contains two elements of ∆ f ∪ { ξ } , which is a contradiction. Proof of part 2) of Theorem 1.2.
From Theorem 3.1 we deduce that equality (4) of Theorem 1.2 holds forany n > n , where n is defined in Lemma 2.9 and is not necessarily equal to 1. For any piecewisecontracting map f , the integer n is bounded above by n ,f := min { n > A ∩ ∆ f ∀ A ∈ A f,n } ,where A f,n is the set of the atoms of generation n of f . Obviously, if f has only two contraction piecesthen n ,f = 1. Lemma 3.8 proves that if f satisfies P1-6 and n ,f = 1, then for a suitable choice of thewell-cutting orbit of f , we have n ,g = 1 for the map g of Proposition 3.7. Acknowledgements.
EC and PG have been supported by the Math-Amsud Regional Program 16-MATH-06 (PHYSECO). EC thanks the invitations of Instituto Venezolano de Investigaciones Cient´ıficas (IVIC-Venezuela) and Universidad de Valpara´ıso (CIMFAV Chile), and the partial financial support of CSIC(Universidad de la Rep´ublica, Uruguay) and ANII (Uruguay). AM and PG thank the invitation of Uni-versidad de la Rep´ublica and the partial financial support of CSIC (Uruguay). AM thanks the financialsupport of the CIMFAV CID 04/03, and the members of the CIMFAV for their kindness during his stay.
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