A characterization of zero topological entropy for a class of triangular mappings

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1 J. Math. Anal. Appl. 287 (2003) A characterization of zero topological entropy for a class of triangular mappings Juan Luis García Guirao a, and Jacek Chudziak b a Department of Mathematics, University of Castilla-La Mancha, Cuenca, Castilla-La Mancha, Spain b Department of Mathematics, University of Rzeszów, Rejtana 16 A, Rzeszów, Poland Received 16 April 2003 Submitted by U. Kirchgraber Abstract Sharkovskiĭ and Kolyada (1991) stated the problem of characterization triangular mappings having zero topological entropy. It is known that, even under some additional assumptions, this aim has not been reached. We solve this problem in the class of triangular mappings with basis map having closed set of periodic points Elsevier Inc. All rights reserved. Keywords: ECIT-89: European Conference on Iteration Theory held in Batschuns, Austria, 1989; Discrete dynamical system; Triangular mappings; Topological entropy; Chaos; Recurrent and uniformly recurrent points 1. Introduction and notation Let I =[0, 1] be the compact unit interval of the real line. We consider triangular mappings on the unit square, i.e., continuous transformations from I 2 into itself of the form F : (x, y) (f (x), g(x, y)). In this setting, the maps f and g are, respectively, called the basis and the fiber map of F.Foreveryx I, the maps g x defined by g x (y) = g(x,y) form a system of one-dimensionalmappingsdependingcontinuouslyon x. For more details, see, for instance, [1,3,4,12,13]. This paper has been partially supported by MCYT (Ministerio de Ciencia y Tecnología, Spain) and FEDER (Fondo Europeo de Desarrollo Regional), Grants BEC and BFM ; Fundación Séneca (Comunidad Autónoma de la Región de Murcia), Grant PI FS-01 and JCCM (Junta de Comunidades de Castilla-La Mancha), Grant PAC * Corresponding author. addresses: jlguirao@tel-cu.uclm.es (J.L.G. Guirao), chudziak@univ.rzeszow.pl (J. Chudziak) X/$ see front matter 2003 Elsevier Inc. All rights reserved. doi: /s x(03)

2 J.L.G. Guirao, J. Chudziak / J. Math. Anal. Appl. 287 (2003) By φ we denote a continuous map from a compact metric space X into itself. The pair (X, φ) is called the discrete dynamical system generated by φ on X. Foreveryx X and every integer n 1, we define φ n (x) = φ(φ n 1 (x)) and φ 0 as the identity map on X. A point x X is said to be periodic by φ if there exists a positive integer n such that φ n (x) = x. The smallest of the values m satisfying the previous condition is called the period of x. ByP(φ) we denote the set of all periodic points by φ. Forx X, wedefine the ω-limit set ω φ (x) of x by φ as the set of all accumulation points of the sequence {φ n (x)} n=0.letrec(φ) ={x x ω φ(x)} be the set of recurrent points of φ.asetm X is called minimal by φ if it is non-empty and it does not contain proper closed subsets N holding φ(n) = N. ByUR(φ) we denote the set of uniformly recurrent points of φ, i.e., all recurrent points with minimal ω-limit sets. A pair of points {x,y} X is said to be a Li Yorke pair of φ, if simultaneously holds lim inf d( φ n (x), φ n (y) ) = 0 and lim sup d ( φ n (x), φ n (y) ) > 0. Given a subset A X, we say that φ A is chaotic (in the sense of Li and Yorke, see [16]) if A contains a Li Yorke pair of φ. For the topic of the topological entropy we will use the definition introduced by Bowen (see [6]). Given ε>0 and a non-negative integer n, a subset E X is called an (n, ε)-separated set, ifforanyx y E there exists j {0, 1,...,n 1} with d(φ j (x), φ j (y)) > ε. Then the topological entropy h(φ) is defined by 1 ( ) h(φ) = lim lim sup ε 0 n log sup card(e), E where the supremum is taken over all (n, ε)-separated subsets E of X. Consider the following dynamical properties of a map φ: (P1) the topological entropy of φ is 0 (h(φ) = 0); (P2) the topological entropy of φ Rec(φ) is 0 (h(φ Rec(φ) ) = 0); (P3) φ Rec(φ) is non-chaotic; (P4) every recurrent point of φ is uniformly recurrent (Rec(φ) = UR(φ)); (P5) the period of every periodic point is power of two. To express (P5) we will use the typical terminology based on the Sharkovskiĭ s stratification, namely φ 2. Note that for triangular mappings the Sharkovskiĭ s order of coexisting cycles remains true (see [9]). It is well known that conditions (P1) (P5) are equivalent in the case of continuous maps of the unit interval (see [21]). Such equivalence establishes a useful characterization of the property of zero topological entropy which is a sign of dynamical simplicity. In 1989, Sharkovskiĭ and Kolyada (see [20]) formulated the problem of studying the relations between properties (P1) (P5) in the setting of triangular maps of the unit square. In the general case properties (P1) (P5) are not mutually equivalent (see [2,7,10,12,15]). Moreover, even under some additional assumptions on F, the equivalence is not reached. In [11] it is proved that in the case of triangular mappings non-decreasing on the fibres (i.e., on sets of the form I x ={x} I, x I ) conditions (P1), (P2) and (P5) are equivalent, (P3) implies (P4) and (P4) implies (P1). However (see [11, cf. Lemma 4.2]) there exists an example of

3 518 J.L.G. Guirao, J. Chudziak / J. Math. Anal. Appl. 287 (2003) a triangular mapping non-decreasing on the fibres with (P2) but neither (P3) nor (P4) (this example is based on ideas from [8]). The implication from (P4) to (P3) is still unknown. Therefore, even in such a particular case, properties (P1) (P5) are not equivalent. In the present paper, we consider the problem of the equivalence of (P1) (P5) in the class of triangular mappings with basis map having closed set of periodic points. We prove that for such a class of triangular mappings conditions (P1) (P5) are mutually equivalent. The assumption on the basis maps is a very easy to check requirement, held by a large class of maps and widely studied in the literature. Note that such maps necessarily have zero entropy but not conversely. There exist interval maps of type 2 (i.e., with periodic points with periods of all powers of two) and so with zero topological entropy (see [17]) such that the set of its periodic points is not closed (see, for instance, [21]). Moreover, we have the following Theorem 1 [18,19,21,23]. Let f be a continuous map from I into itself. Then the following statements are equivalent: (i) P(f) is closed, (ii) for any x I, ω f (x) is finite, (iii) Rec(f ) = P(f), (iv) UR(f ) = P(f). The statement of our main result is Main theorem. Let F be a triangular map with basis map f such that P(f) is closed. Then, conditions (P1) (P5) are mutually equivalent. 2. Partial and auxiliary results Before going to the proofs, let us make a brief comment about the tools that will be used in the following. As far as the topological entropy is concerned, if F is a triangular map there exists a formula and bounds by Bowen for h(f ) (see [6,22]), max { h(f ), h fib (F ) } h(f ) = sup h(f Ix ) h(f ) + h fib (F ), (1) x P(f) where h fib (F ) = sup x I h(f Ix ). Note that, for a triangular mapping F with basis f having closed set of periodic points, for computing h(f ) we can confine our attention to h(f Ix ),wherex P(f).Havingabasisf 2,wegeth(f ) = 0 (see [17]) and therefore h(f ) = h fib (F ). If all the maps g x are monotone (as in [11]), then for every x I is h(f Ix ) = 0 and consequently h(f ) = h(f ) (see [14]). Regarding the topological dynamics tools, between other results which will be conveniently quoted, we use the following weak version of the Kolyada s projection theorem (see [12, cf. Theorem 1]). Theorem 2. Let F be a triangular map with basis map f. Then, ( ) π 1 A(F ) = A(f ),

4 J.L.G. Guirao, J. Chudziak / J. Math. Anal. Appl. 287 (2003) where π 1 is the projection on the first coordinate and A(φ) denotes one of the following sets: P(φ), Rec(φ) or UR(φ), whereφ {F,f}. Theorem 3. Let F be a triangular map with basis map f such that P(f) is closed. Then (P1) and (P4) are equivalent. Proof. Assume that (P1) holds. Then, by (1), h(f ) = h(f Ix ) = 0foreveryx I. Fix an x P(f) and let m x be its period. Then h(f m x ) = m x h(f ) = 0(see[5, cf. Chapter VIII, Proposition 2] and therefore h(f m x Ix ) = 0. Let us consider the onedimensional map τ mx = g f mx 1 (x) g x.thenh(τ mx ) = 0, so Rec(τ mx ) = UR(τ mx ) (see [21, cf. Theorem 4.19]). Consequently, since Rec(F m x Ix ) ={x} Rec(τ mx ) and UR(F m x Ix ) ={x} UR(τ mx ),thenrec(f m x Ix ) = UR(F m x Ix ). Moreover, we have Rec(F m x ) = Rec(F ) and UR(F m x ) = UR(F ) (see [5]). Thus, Rec(F Ix ) = UR(F Ix ). Since x P(f) is arbitrarily fixed, Rec(F P(f) I ) = UR(F P(f) I ). On the other hand, the set P(f)is closed, which in view of Theorem 1, means that Rec(f ) = UR(f ) = P(f). Therefore, Rec(F ) = UR(F ). Now suppose that (P1) does not hold, i.e., h(f ) > 0. Since the set P(f) is closed, by (1) there exists x P(f) such that h(f Ix )>0. Let m x be the period of x by f. Thus, by [14, cf. Lemma 4.4], h(f m x Ix ) = m x h(f Ix )>0 (observe that this fact cannot be deduced from [5, cf. Chapter VIII, Proposition 2] because except for the case m x = 1, the fiber I x is not invariant by F ). Therefore, h(τ mx )>0 and using again [21, cf. Theorem 4.19] there is y Rec(τ m )\ UR(τ m ). Consequently, there exists a sequence of non-negative integers (n k ) k=0 such that lim k τ n k m x (y ) = y. Hence, (x, y ) = lim k (x, τ n k m x (y )) = lim k F n k m x (x, y ), i.e., (x, y ) Rec(F m x ) = Rec(F ). Observe that (x, y )/ UR(F ). Otherwise,y UR(τ mx ), which is not possible. Therefore, Rec(F ) UR(F ) ending the proof. For proving the equivalence between (P3) and (P4) (see Theorem 4) we develop the following lemma. Lemma 1. Let F be a triangular map with basis map f such that P(f) is closed. If F satisfies (P4), then for every (x, y) Rec(F ), themapf ωf (x,y) is non-chaotic. Proof. Let (x, y) Rec(F ). The case where ω F (x, y) is finite is trivial. Assume that ω F (x, y) is an infinite set. Since (P4) holds for F, by Theorem 3, h(f ) = 0. Therefore, by (1), h(f Ix ) = h(f ) = 0foreveryx P(f).Since(x, y) Rec(F ) and P(f)is closed, using Theorems 2 and 1, we get x P(f).Letm x be the period of x by f. Then, ω F (x, y) is contained in the fibers I xj,wherex j = f j (x) for j {0,...,m x }. Moreover, from (P4) it follows that ω F (x, y) is a minimal set by F. Thus, for every j {0,...,m x }, F m x has a minimal set M xj contained in I xj and ω F (x, y) = m x 1 j=0 M xj. By (P4) and [5, cf. Chapter VIII, Proposition 2] Rec(F m x ) = UR(F m x ). Hence, Rec(F m x Ixj ) = UR(F m x Ixj ). Since for interval maps (P4) implies (P3), for every j {0,...,m x } the map F m x Ixj is non-chaotic on the set of its recurrent points. Therefore, F m x Ixj is non-chaotic on M xj

5 520 J.L.G. Guirao, J. Chudziak / J. Math. Anal. Appl. 287 (2003) for every j {0,...,m x }. Consequently, since x P(f) the map F m x, and thus F,is non-chaotic on the union of M xj which is equal to ω F (x, y). Theorem 4. Let F be a triangular map with basis map f such that P(f) is closed. Then (P3) and (P4) are equivalent. Proof. Suppose that F Rec(F ) is chaotic, so there exist points z, t Rec(F ) holding lim inf d( F n (z), F n (t) ) = 0, lim sup d ( F n (z), F n (t) ) > 0. (2) Thus, there exists an increasing sequence of positive integers (n k ) k=0 such that lim k d(f n k (z), F n k(t)) = 0. Since (F n k (z)) k=0 and (F n k(t)) k=0 have convergent subsequences, without loss of generality we can assume that lim k F n k (z) = u and lim k F n k (t) = v. Then, u = v and u ω F (z) ω F (t). IfRec(F ) and UR(F )were equal, then ω F (z) and ω F (t) will be minimal sets and therefore, ω F (u) = ω F (z) = ω F (t). Thus, F ωf (u) will be chaotic, which in view of Lemma 1 is not possible. So, it is proved that Rec(F ) UR(F ). Now, suppose that Rec(F ) UR(F ). By Theorem 3, h(f ) > 0. Thus, using the fact that the set P(f)is closed, by (1) there exists a point x P(f)such that h(f Ix )>0. Let m x be the period of x by f and τ mx be the one-dimensional map defined in the following way g f mx 1 (x) g x. Then, h(τ mx )>0 and therefore τ mx is chaotic on the set of its recurrent points (see [21, cf. Theorem 4.19]). Hence, there exist z 1,t 1 Rec(τ mx ) such that lim inf d( τ n m x (z 1 ), τ n m x (t 1 ) ) = 0, lim sup d ( τm n x (z 1 ), τm n x (t 1 ) ) > 0. (3) For every integer n 1andy I is F m x n (x, y) = (x, τ n m x (y)). Therefore, if we define z = (x, z 1 ) and t = (x, t 1 ),thenz, t Rec(F ) andby(3)weget lim inf d( F m x n (z), F mx n (t) ) = 0, Hence, and lim sup d ( F m x n (z), F mx n (t) ) > 0. 0 lim inf d( F n (z), F n (t) ) lim inf d( F m x n (z), F m x n (t) ) = 0 0 < lim sup obtaining that F Rec(F ) is chaotic. d ( F m x n (z), F mx n (t) ) lim sup d ( F n (z), F n (t) ), 3. Proof of the main result Main theorem. Let F be a triangular map with basis map f such that P(f) is closed. Then, conditions (P1) (P5) are mutually equivalent. Proof. By Theorems 3 and 4, we get the mutually equivalence between (P1), (P3) and (P4). On other hand, by [10] and without additional assumptions (P5) is always equivalent to h(f P(F) ) = 0. Therefore, since (P2) implies h(f P(F) ) = 0, we obtain that (P2) implies

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