FAMILIES OF TRANSITIVE MAPS ON R WITH HORIZONTAL ASYMPTOTES

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1 REVISTA DE LA UNIÓN MATEMÁTICA ARGENTINA Vol. 59, No. 2, 2018, Pages Published online: May 28, 2018 FAMILIES OF TRANSITIVE MAPS ON R WITH HORIZONTAL ASYMPTOTES BLADISMIR LEAL, GUELVIS MATA, AND SERGIO MUÑOZ Abstract. We will prove the existence of a class of transitive maps on the real line R, with a discontinuity and horizontal asymptotes, whose set of periodic orbits is dense in R; that is, a class of chaotic families. In addition, we will show a rare phenomenon: the existence of periodic orbits of period three prevents the existence of transitivity. 1. Introduction The notion of transitivity is a fundamental tool in the study of maps with chaotic dynamics; see [19, 5, 3]. On boundaryless compact manifolds there is a well established theory about transitive diffeomorphisms; see [1, 15, 17, 18]. On compact intervals of R there exists a large class of examples and characterizations leading to such dynamic property; see [14, 16, 7] and references therein. In the non-compact cases the situation is very different, since there is no nown notion of hyperbolicity producing persistently chaotic dynamics and, regarding characterizations of transitivity, there are not many references. For example, in [9] it is proved that any Anosov diffeomorphism defined in the plane R 2 onto R 2 R 2 R 2 ) cannot be transitive; on the other hand, there are transitive diffeomorphisms of the plane R 2 minus a line of discontinuities see [4, 10]). In the case of non-bounded intervals the situation is similar to the case of non-compact manifolds; in this direction, in [12] it is shown that continuous transitive maps from R to R R R) must have infinite critical points; also in [13] there is a large class of examples of transitive maps on R. In any case, a characterization for transitivity is not shown and, in the context of such articles, small perturbations of those maps lose the transitive property. Recently, [11] shows the existence of a class of maps, leading to a geometric model of the well nown Boole transformation T x) = x 1 x see [2]), where a characterization for transitivity is possible; also, it is shown that, in the space of continuously differentiable maps from R \ {0} to R) the maps belonging to such geometric model are persistently transitive, with respect to the C 1 uniform topology. Such geometric model consists of maps with 2010 Mathematics Subject Classification. 37E99, 37D20. Key words and phrases. Increasing alternating systems, Horizontal asymptotes, Transitivity, Expansive alternating systems, Periodic orbit of period

2 376 B. LEAL, G. MATA, AND S. MUÑOZ domain R \ {0} onto R, increasing on each connected component of R \ {0} and non-bounded on R \ K, where K is a compact interval with 0 / K. In the attempt to extend the study of this type of maps and expand the space of transitive maps over unbounded intervals, we as ourselves: What is the dynamics of the maps bounded in R \ K, where K is a compact interval? Have these maps a non-empty intersection with the world of chaotic dynamics? The purpose of this article is to answer such questions positively, in addition to expanding the geometric model of the Boole maps, treated in [11], to a class with horizontal asymptotes; also to show a couple of characterizations of transitivity. Specifically, let f : R \ {0} R be a continuous map; we say that f is transitive if there is a point x R \ {0} such that its positive orbit with respect to f), that is O + f x) = {x, fx), f 2 x),... }, is dense in R. Definition 1. A continuous function f : R\{0} R is an increasing alternating system with asimptotes relative to its first pre-image if: 1a) f is strictly increasing in, 0) and 0, + ). 1b) lim fx) = and lim fx) = +. x 0 + x 0 1c) There exist x 0 < 0 and x 1 > 0 such that f 1 0) = {x 0, x 1 }, x 1 and lim fx) = x 0. x 1d) fx) x for all x R \ {0}. If f : R\{0} R is as in Definition 1, we say that f is a SACAH. lim fx) = x + Theorem 2 Main Theorem). Let f : R\{0} R be a SACAH. Then f is transitive if and only if n 0 f n 0) is dense in R. Corollary 3. Let f : R\{0} R be a SACAH. If n 0 f n 0) is dense in R, then the set of periodic orbits of f is dense in R \ {0}. In contrast to Corollary 3, it is important to mention that in this wor we will show a particularity of this ind of family: The existence of a periodic period 3 orbit prevents the existence of transitivity. Corollary 4. The map B : R\{0} R defined by Bx) = x 1 is a transitive x SACAH. Curiously, n 0 B n 0) = Q. It is important to mention that the geometric model of the Boole transformation, called expansive increasing alternating systems, forms a set of transitive maps with non-empty interior see [11]); the alternating systems with asymptotes at the diagonal line y = x and those with horizontal asymptotes lie the models shown in this paper) are border elements of the set of transitive maps of R maps with a discontinuity). Another important point is that the type of maps shown in this article and those studied in [11] appear naturally as projections along invariant foliations or leaves) in the study of transitive diffeomorphisms of the plane R 2 minus a curve of discontinuities, and there is a relation between the plane dynamics and the projected one see [10] and recently [8]). These two latest wors are inspired Rev. Un. Mat. Argentina, Vol. 59, No )

3 TRANSITIVE MAPS ON R WITH HORIZONTAL ASYMPTOTES 377 by the wor of Devaney [4], who shows an example of a transitive diffeomorphism of the plane R 2 minus a line of discontinuity. Devaney s example was introduced by Hénon in [6] as a system associated with the movement of the three restricted bodies of classical mechanics. The article is organized in the following manner: In the first section, of notations and basic results, we grouped the points of the pre-images from zero to generate a partition, which will be very useful in proving our Main Theorem. In addition, we will show some properties with respect to the generated partition. Then, in the next section we will concentrate on the proof of the Main Theorem, together with the two corollaries enunciated above. 2. Notations and basic results In this section f denotes a SACAH. Let x 0 < 0 and x 1 > 0 as in Definition 1. On the other hand, R\{0} is the union of two connected components which we will denote by R 0 =, 0) and R 1 = 0, + ). Denote by f 0 = f R0 and f 1 = f R1, where f 0 x) = fx) for x R 0 and f 1 x) = fx) for x R 1. Since f 1 0), consider the following notation: n A n = f n 0) {0}. j=1 f j 0) {0}, for all n 1 and A f = n 1 From Definition 1 we obtain: Remar 1. a) f0, x 1 ) = R 0 and fx 0, 0) = R 1 ; b) fr 0 ) = x 0, + ) and fr 1 ) =, x 1 ); c) f 2 x 1, + ) = R 0 and f 2, x 0 ) = R 1. This remar tells us that for every point x R\A f, the orbit of x with respect to f visits each connected component R 0 and R 1 infinitely many times. The following definition is of great importance in the characterization of transitivity. Definition 5. Let f : R\{0} R be a SACAH. f is expansive if for each x, y R\A f, there exists N > 1 such that f N x).f N y) < 0. Next we show that the existence of a periodic orbit of period 3, for this type of transformations, prevents the existence of transitivity and expansiveness. Lemma 6. Let f : R\{0} R be a SACAH. If there exists x a periodic point of period 3, then f is neither expansive nor transitive. Proof. Let x be a periodic point of period 3. Denote the orbit of x by x = x 0, fx) = x 1 and f 2 x) = x 2. From Remar 1 x i, x 0 ) or x i x 1, + ), for some i {0, 1, 2}. If x i, x 0 ), then B =, x i ] x 0, fx i )] 0, f 2 x i )] is f-invariant, that is, fb) B. On the other hand, if x i x 1, + ), then we have that B = [x i, + ) [fx i ), x 1 ) [f 2 x i, 0) is f-invariant. In any case f is neither expansive nor transitive. Rev. Un. Mat. Argentina, Vol. 59, No )

4 378 B. LEAL, G. MATA, AND S. MUÑOZ In the proof of the Main Theorem we will need to mae use of the symbolic dynamics. For it, consider the following space of sequences: Σ 2 = {a = a 0, a 1,... ) : a j {0, 1} and j 0}, with the usual topology induced by the metric d 2 a, b) = Σ 2. Let us consider the subset + n=0 Σ2, 2) = {a Σ 2 : 1, 1, 1) a and 0, 0, 0) a}, a n b n 2 n, for a, b where 1, 1, 1) a and 0, 0, 0) a mean that 1, 1, 1) a j, a j+1, a j+2 ) and 0, 0, 0) a j, a j+1, a j+2 ), for all j 0, respectively. It is well nown that Σ 2 and Σ2, 2) are compact and the shift σ : Σ 2 Σ 2 defined by σa 0, a 1, a 2,...) = a 1, a 2, a 3,...) is continuous and transitive. Also, Σ2, 2) is invariant with respect to σ and σ is transitive restricted to Σ2, 2). Let us denote by Perσ, 3) the set of all periodic orbits of period 3 belonging to σ. With this notation in mind consider the set Σ 2, 2) = {a Σ2, 2) : σ j a) Perσ, 3), for all j 0}. Observe that Σ 2, 2) is not compact and Σ 2, 2) is invariant with respect to σ. Also, σ restricted to Σ 2, 2) is a transitive function; denote this restriction by σ. Definition 7. Let K be a finite set of R. P K shall denote a partition of R generated by K, if its elements are intervals of R satisfying: a) I K and I K = for all I P K. b) I = R \ K; I P K c) If I, J P K with I J, then I J =. P K is the set of all the connected components of R \ K. Consider the following lemma of set theory. Lemma 8. Let {K n }, n 1, be a sequence of finite sets of R such that K n K n+1 and n 1 K n is dense in R. If for each sequence I n of bounded atoms of P Kn, I n P Kn, I n+1 I n, then diami n ) 0. Observe that if A B, then for each J P B there is a single I P A such that J I. Lemma 9. Let n > 1 and P An be the partition generated by A n. If I is an atom of P An, then f n I) {R 0, x 0, 0), 0, x 1 ), R 1 }. Proof. By induction. The case n = 1 is clear from Remar 1 and the fact that P A1 = {R 0, x 0, 0), 0, x 1 ), R 1 }. Let us suppose that the lemma is true for n 1. Let J P An+1 ; since A n A n+1, there exists I P An such that J I. By the inductive hypothesis we have four options. The first: If f n I) = R 0, then from Remar 1, f n+1 I) = x 0, + ); this means that there exists y I such that f n+1 y) = 0, that is, Rev. Un. Mat. Argentina, Vol. 59, No )

5 TRANSITIVE MAPS ON R WITH HORIZONTAL ASYMPTOTES 379 from Definition 7, there exist J 1 and J 2 in P An+1 such that J 1 J 2 {y} = I, {y} = J 1 J 2, f n+1 J 1 ) = x 0, 0) and f n+1 J 2 ) = R 1. From this together with Definition 7 it follows that J = I 1 or J = J 2 ; in any case we have that f n+1 J) {R 0, x 0, 0), 0, x 1 ), R 1 }. The second option: If f n I) = x 0, 0), then f n+1 I) = R 1, therefore I A n+1 =, so by Definition 7, I P An+1 and consequently J = I. This shows that f n+1 J) = x 0, 0). The other two options are similar. So f n+1 J) {R 0, x 0, 0), 0, x 1 ), R 1 }. Lemma 10. For each n > 0, we have that f n R 0 ) f n R 1 ) = I P An I. Proof. By induction. For n = 1 the equality is clear from Remar 1. Let us suppose that the lemma is true for n 1. First, using the inductive hypothesis, note that f n 1 R 0 ) f n 1 R 1 ) = f n 1 R \ {0}) = f 1 I). I P An Let x f n 1 R 0 ) f n 1 R 1 ); then x I P An f 1 I), that is, there exists I P An such that x f 1 I); since I A n =, it follows that x R \ A n+1 ; therefore, x I P An+1 I and consequently f n 1 R 0 ) f n 1 R 1 ) I P An+1 I. Let y I P An+1 I, then there exists J P An+1 such that y J. So, fy) A n =, and therefore there exists I P Aa such that fy) I. Consequently y I P An f 1 I), which completes the proof. Lemma 11. Let f be a SACAH. f is expansive if and only if A f = is dense in R \ {0}. A n {0} Proof. ) Suppose that f is expansive, and that A f = n=1 A n {0} is not dense in R \ {0}. Then, there is an interval J such that J f n 0) = for all n 0; thus, f n J) R 0 or f n J) R 1 for all n 0. Let x y be in J, then since f is expansive there exists N > 0 such that f N x) f N y) < 0; from this, f N J) R 0 and f N J) R 1. This is a contradiction. ) Suppose that A f = n=1 A n {0} is dense in R \ {0}. Let a b be in R f and consider, without loss of generality, that a < b. Since A f is dense there exists x A f such that x a, b). Let N 1 be the minimum such that f N 0) a, b). Then, f N a) f N b) < Proof of the main results Main Theorem. ) Suppose that f is expansive. Then, from Lemma 11, R f is totally disconnected and fr f ) = R f. Consider the function h : R f Σ 2, 2) n=1 Rev. Un. Mat. Argentina, Vol. 59, No )

6 380 B. LEAL, G. MATA, AND S. MUÑOZ defined by hx) = a, where a n = { 0 if f n x) < 0, 1 if f n x) > 0. Note that from Remar 1, h is well defined. Our objective is to show that h is a homeomorphism and that σ h = h f, that is, h is a topological conjugation between f and σ. That means f is transitive since σ is, so the proof follows. h is one to one. Indeed, let x y in R f and let us denote a = hx) and b = hy). Since f is expansive, there exists N > 1 such that f N x) f N y) < 0, therefore a N b N and so a b. h is continuous. Indeed, let x R f and let us denote a = hx). Let ɛ > 0, then there exists M > 1 such that, if b Σ 2, 2) with a j = b j for all 0 j M, then d 2 a, b) < ɛ. On the other hand, there exists I P AM+1 such that x I. From this follows that f j I) {0} = for all 0 j M +1, then from definition of h, f j I) R aj for all 0 j M. Now, let δ = dx, I); then for y x δ, x + δ) R f I it follows that f j y) R aj for all 0 j M. Then, calling b = hy) we conclude that d 2 hx), hy)) < ɛ. h is onto. Indeed, let a = a 0, a 1, a 2,... ) Σ 2, 2) and remember that R 0 =, 0), R 1 = 0, + ). Consider the sets H 0 = R a0 ; H 1 = R a0 f 1 R a1 );. H n = R a0 f 1 R a1 ) f n R an ), for all n 1. Observe that if x H n, then f j x) R aj for all 0 j n. To continue with the proof, we need to prove the following claims. Claim 1: H n and H n P An, for all n 1. Proof. By induction. Let us show that for n = 1 the result is true. Note that H 0 = R 0 or R 1 and f 1 0) = {x 0, x 1 }. Also, f 1 R 0 ) =, x 0 ) 0, x 1 ) and f 1 R 1 ) = x 0, 0) x 1, + ). Then, intersecting it follows that H 1 and H 1 P A1. Suppose that H n and H n P An for n 1. Suppose that H n+1 =, that is, H n f n 1 R 0 ) =, and suppose also that a n+1 = 0. From the inductive hypothesis and Lemma 9 it follows that f n H n ) {R 0, x 0, 0), 0, x 1 ), R 1 } and since H n+1 = then f n H n ) = x 0, 0). So, a n = 0. Now, since H n 1 and from Lemma 9, f n 1 H n 1 ) {R 0, x 0, 0), 0, x 1 ), R 1 }. From Remar 1 it follows that f n 1 H n 1 ) = R 0 or f n 1 H n 1 ) = x 0, 0). This proves that a n 1 = 0. Thus, a n 1, a n, a n+1 ) = 0, 0, 0) but this is a contradiction with the fact a Σ 2, 2). In the case a n+1 = 1 the proof is similar obtaining a n 1, a n, a n+1 ) = 1, 1, 1), which is a contradiction for the same reason as before. Consequently, H n+1. It remains to prove that H n+1 P An+1, indeed if f n 1 0) H n =, then H n P An+1. From Lemma 10 and the fact that H n+1 there exists I P An+1 Rev. Un. Mat. Argentina, Vol. 59, No )

7 TRANSITIVE MAPS ON R WITH HORIZONTAL ASYMPTOTES 381 such that H n+1 = H n I. Then, from Definition 7, H n = I, from where it follows that H n+1 P An+1. Now, if f n 1 0) H n from Lemma 9 there is a single y H n such that f n+1 y) = 0. Then, from Definition 7 and the fact that A n A n+1, there exist I 1 and I 2 in P An+1 disjoint with I 1 I 2 = H n \ {y}. From this and Lemma 9, if f n+1 I 1 ) R 1 then f n+1 I 2 ) R 0 or conversely. It means that I 1 or I 2 is contained in f n 1 R an+1 ). So from Lemma 10 H n+1 = H n I 1 or H n+1 = H n I 2 ; in any case, H n+1 P An+1. This completes the proof of Claim 1. Claim 2: There exists N 1 such that H N is a bounded atom of P AN. Proof. Suppose that H N is not bounded for all N 1. Case 1. If a 0 = 0, then a 3 = 0, a 3+1 = 0 and a 3+2 = 1, for all 0. By induction. Since H 0 = R 0 and H 1 is not bounded it follows that H 1 =, x 0 ). From Remar 1 fh 1 ) = x 0, 0) and f 2 H 1 ) = R 1, then a 1 = 0 and a 2 = 1. So, this proves the case = 0. Suppose that a 3 = 0, a 3+1 = 0 and a 3+2 = 1. Then, from this and Lemma 9, it follows that f 3 H 3 ) = R 0 and f 3+1 H 3+1 ) = x 0, 0); therefore, f 3+2 H 3+2 ) = R 1. Since f 3+3 H 3+2 ) =, x 1 ), there exists y H 3+2 such that f 3+3 y) = 0. Since H 3+3 is not bounded, we have that H 3+3 =, y) and f 3+3 H 3+3 ) = R 0, and consequently a 3+3 = 0. Then, f 3+1)+1 H 3+3 ) = x 0, + ). So there exists y 1 H 3+3 such that f 3+1)+1 y 1 ) = 0. From Lemma 9 and the fact that H 3+1)+1 H 3+3 is not bounded, H 3+1)+1 =, y 1 ). Therefore f 3+1)+1 H 3+1)+1 ) = x 0, 0) and f 3+1)+2 H 3+1)+1 = R 1, so this shows that a 3+1)+1 = 0 and a 3+1)+2 = 1. Case 2. If a 0 = 1, then a 3 = 1, a 3+1 = 1 and a 3+2 = 0, for all 0. The proof follows similarly. Note that in both cases 1 and 2 it is shown that a Σ 2, 2) is a periodic orbit of period 3 with respect to the shift σ, but this is a contradiction from the supposition that H N is not bounded for all N 1. Claim 3: If x n 1 H n, then x R f. Proof. Suppose that x A f. From Claim 2, there exists N > 1 such that H n is bounded for n N and from Claim 1 there exists y A N such that H = x, y) or H = y, x). We have the following possible cases. A1) Suppose that H N = x, y) and a N = 1. Then a N+3 = 1, a N+3+1 = 0, and a N+3+2 = 0, for all 0. We will prove the claim A1) by induction on. For = 0 the claim is valid. Indeed, from our hypothesis and Lemma 9, f N+1 H N+1 ) = R 0 ; consequently, a N+1 = 0. Then, f N+2 H N+1 ) = x 0, + ). From this there exists y 1 H N+1 such that f N+2 y 1 ) = 0. So, from Claim 1 and the fact that x H N+2 and H N+2 x, y), it follows that f N+2 H N+2 ) = x 0, 0). Therefore a N+2 = 0. Rev. Un. Mat. Argentina, Vol. 59, No )

8 382 B. LEAL, G. MATA, AND S. MUÑOZ Suppose that the claim is valid for some 0. Since x H N+3+2 [x, y] and H N+3+2 H N+3+1 x, y), it follows that f N+3+2 H N+3+2 ) = R 1. Denote by m = N Then, applying f, f m+1 H m ) =, x 1 ) and we obtain that there exists y 1 H m such that f m+1 y 1 ) = 0. Then, from Claim 1 and the fact that x H m+1, it follows that H m+1 = x, y 1 ) and f m+1 H m+1 ) = R 0 ; consequently, a m+1 = 0. Applying f we have that f m+2 H m+1 ) = x 0, + ). Since x H m+2 and using the same last argument, f m+2 H m+2 ) = x 0, 0) and f m+3 H m+2 ) = R 1. So a m+2 = 0 and a m+3 = 1. Replacing m we have that a N+3+1) = 0, a N+3+1)+1 = 0, and a N+3+1)+2 = 1, and that is what we wanted to prove. The demonstrations of the following claims follow similarly as in the case A1). In each case we indicate the components of the sequence a. A2) Suppose that H N = x, y) and a N = 0. Then, 2.1) if f N H N ) = x 0, 0), then a N+3 = 0, a N+3+1 = 1 and a N+3+2 = 0 for all 0; 2.2) if f N H N ) = R 0, then a N+3 = 0, a N+3+1 = 0 and a N+3+2 = 1 for all 0. A3) Suppose that H N = y, x) and a N = 0. Then a N+3 = 0, a N+3+1 = 1, and a N+3+2 = 1 for all 0. A4) Suppose that H N = y, x) and a N = 1. Then, 4.1) if f N H N ) = 0, x 1 ), then a N+3 = 1, a N+3+1 = 0, and a N+3+2 = 1 for all 0; 4.2) if f N H N ) = R 1, then a N+3 = 1, a N+3+1 = 1, and a N+3+2 = 0 for all 0. In all cases we proved that the point σ ) N a) is periodic of period 3 and this is a contradiction. Consequently, x R f. Let a Σ 2, 2). From Claim 2, n 1 H n ; from the injectivity of h and Claim 2, there exists a single x n 1 H n, and from Claim 3, x R f. So, from definitions of h and the sets H n it follows that hx) = a. This proves that h is onto Σ 2, 2). h 1 is continuous. Let a Σ 2, 2) and ɛ > 0. Let us denote x = h 1 a). In the same way of constructing the H n s in the proof that h is onto, we have that x H n for all n 1. Then, from Lemma 8, diamh n ) 0 when n +. Therefore, there exists N > 1 such that H N x ɛ, x + ɛ). On the other hand, there exists δ > 0 such that if b Σ 2, 2) with d 2 a, b) < δ then a j = b j for all 0 j N. Then, y = h 1 b) H j for 0 j N, that is, y = h 1 b) H N x ɛ, x + ɛ). This proves the continuity of h 1. Finally, it remains to prove that h f = σ h. Indeed, given x R f and denoting a = hfx)) we have that a n = 0 if f n+1 x) < 0, or a n = 1 if f n+1 x) > 0 for all n 0. On the other hand, if b = hx) we have that σ hx)) = b 1, b 2, b 3,...). Rev. Un. Mat. Argentina, Vol. 59, No )

9 TRANSITIVE MAPS ON R WITH HORIZONTAL ASYMPTOTES 383 Note that b 1, b 2, b 3,..., ) = a 0, a 1, a 2,...). This shows that h fx) = σ hx), for all x R f. ) Suppose that f is transitive and suppose by contradiction) that f is not expansive. Then, there exists x y such that f j x).f j y) > 0, for all j 0. Suppose, without loss of generality, that x < y. That means f j [x, y]) A f = for all j 0; from this we conclude that f j [x, y]) does not have points of the set n 0 f n 0). Denote by I = [x, y]. Since f is transitive there exists 1 such that f I) I. Since f is continuous, increasing in each connected component of R\{0} and f j I) R aj, j 0, it follows that f I) I is an interval. Note that f 2 I) f I) I), so f 2 I) f I) or I is an interval and also contained in R 0 or R 1. Given J = n 1 f n I), then J is invariant by f, but this is a contradiction since f is transitive. Proof of Corollary 3. In the proof of the Main Theorem, it is shown that f is topologically conjugated with σ : Σ 2, 2) Σ 2, 2) and the set of periodic points of σ is dense in Σ 2, 2); from this the result follows. Corollary 12. All maps f : R\{0} R which are SACAH and transitive are topologically conjugated with each other. Proof. Let f and g be SACAH maps. Then, from the Main Theorem there exist homeomorphisms h 1 : R f Σ 2, 2) and h 2 : R g Σ 2, 2) such that h 1 f = σ h 1 and h 2 g = σ h 2. Taing h = h 1 2 h 1 : R f R g we have that h f = g h. Let us now consider an interesting example of this ind of maps. Let B : R\{0} R be defined by Bx) = x x, x > 0, = x 1 1 x, x < 0. Lemma 13. B n 0) = Q. n 0 Proof of Corollary 4. The proof follows from Lemma 13 and the Main Theorem. Proof of Lemma 13. Observe that if a Q, then B 1 a) Q. From this we conclude that n 0 B n 0) Q. Now, we will show the other inclusion, that is, Q n 0 B n 0). First observe that: Also, note that for all x > 1, it follows that Bx) < 1. 1) if B x) A B, then B j x) A B, for 0 j. 2) Also note that B x) = Bx), for all x 0. In general, for all n 1 and x R B it follows that B n x) = B n x), 3) Rev. Un. Mat. Argentina, Vol. 59, No )

10 384 B. LEAL, G. MATA, AND S. MUÑOZ where R B = R \ A B and A B = n 1 B n 0) {0}. Case 1. For all n Z, we have that n A B. From 2) it is sufficient to show that n A B for all n 1. The proof is by induction. For n = 1 it is clear since B 1 0) = { 1, 1}. Suppose that j A B for 1 j n. Then, Bn + 1) = n 1 n and applying B we have B 2 n + 1) = 1 n. Applying B again we obtain B 3 n + 1) = n 1. Therefore, by our inductive hypothesis n 1 A B ; from this and 2) it follows that n + 1) A B. This proves Case 1. From the proof of Case 1, from 1) and 2), we have 1 n A B for all n Z \ {0}. 4) Case 2. Consider > 0, n > such that n = q +r, where r {0, 1,..., 1} with q 3. Then, for t 1 and 2t + 1 q we have B 3t 1 ) n 2t = n n 2t 1) B 3t ) = n n 2t B 3t+1 ) n 2t + 1) =. n We will do the proof of Case 2 by finite induction. Let us see that the equations are valid for t = 1: B 2 ) = 1 n n = n 2 n since q 2, n 2 > 0, it follows that n 2 > 0. Therefore, n B 3 ) = < 0 and B4 ) = n 3. n n 2 n Suppose that the equations are valid for t and suppose that 2t + 1) + 1 q. Let us see that the equations are valid for t + 1. Note that 3t + 2 = 3t + 1) 1 and n 2t + 1) 3t+3 = 3t+1). On the other hand, since 2t+1)+1 q, then > 0. So applying B to B 3t+1 ) we obtain n B 3t+2 ) = 1 n n 2t + 2) = n 2t + 1) n 2t + 1), where n 2t + 2) > 0 because 2t + 1) + 1 q. Applying B again, B 3t+3 ) = n n 2t + 2) < 0 and ) n B 3t+1)+1 n 2t + 1) + 1) =, Rev. Un. Mat. Argentina, Vol. 59, No )

11 TRANSITIVE MAPS ON R WITH HORIZONTAL ASYMPTOTES 385 which is what we wanted to prove. n Case 3. For each n 1, m A B, m n. Let us show this by induction on n. For n = 1 it is valid from 4). Suppose j that it is valid for n, that is, m A B, for all m j and for 1 j n. Denote = n + 1. If m =, then the proof follows since 1 A B. Suppose that m >, then there exists q 1 such that m = q. + r, where r {0, 1,..., n}. Observe that B ) = 1 + m m = m. If q = 1, we have that m = + r. Then, B ) = + r = r m. Since 0 r 1, by the inductive hypothesis we have r A B. Then, from 2) and 3) it follows that m A B, m {,..., 2 1}. Now, suppose that q = 2, then m = 2 + r. Therefore m > 0 and B 2 ) m = m 2 m = r + r. r By the inductive hypothesis, + r A B. Then, from 2) and 3), m A B for all m {2, 2 + 1,..., 3 1}. Finally, suppose that q 3, then from Case 2, for t 1, 2t + 1 q and m = q. + r, B 3t 1 ) m 2t = m m 2t 1) B 3t ) = m m 2t B 3t+1 ) m 2t + 1) =. m Tae t 0 1 such that 2t = q 1 or 2t = q. If 2t = q 1, then m 2t + 1) = q. + r q 1) = + r. B 3t0+1 ) = + r and remember that = n + 1, m B 3t0+2 n + 1 ) = B 3t0+2 ) = 1 m m + r = r = r n + 1 A B, by our inductive hypothesis since r {0, 1,..., n}. From 2) and 3) it follows that n + 1 m A B, for all m 3n + 1). Now, if 2t = q, B 3t0+1 n + 1 ) q. + r q. = = r m n + 1 n + 1 A B Rev. Un. Mat. Argentina, Vol. 59, No )

12 386 B. LEAL, G. MATA, AND S. MUÑOZ by the inductive hypothesis since r {0, 1,..., n}. Therefore, from 2) and 3) it follows that n + 1 m A B, for all m 3n + 1). The proof of Case 3 is complete. To finish the proof of Lemma 13, let p Q \ {0}. If 0 < p 1, from Case 3 we have p A B. If p > 1, then from 1) Bp) 0, 1) and Bp) Q; consequently, there exist integers n 1 and m > n such that Bp) = n m. From Case 3 and 2), p A B. Finally, if p < 0, by 1) and 2) we have that p A B. This completes the proof of Lemma 13. Acnowledgments The authors express their gratitude to the Vice-Rector for Postgraduate Studies and Research and the Research Group Clean Energy and Enviroment Research of the National University of Chimborazo UNACH) for the support received during the development of this research. The author Bladismir Leal is grateful to Neptalí Romero for his good suggestions and conversations. References [1] D. V. Anosov. Geodesic flows on closed Riemann manifolds with negative curvature. Proceedings of the Stelov Institute of Mathematics, no ). American Mathematical Society, MR [2] G. Boole. On the comparison of transcendents, with certain applications to the theory of definite integrals. Phil. Trans. R. Soc. Lond ), [3] A. Crannell. The role of transitivity in Devaney s definition of chaos. Amer. Math. Monthly ), no. 9, MR [4] R. L. Devaney. The baer transformation and a mapping associated to the restricted threebody problem. Comm. Math. Phys ), no. 4, MR [5] R. L. Devaney. An introduction to chaotic dynamical systems. Second edition. Addison- Wesley, MR [6] M. Hénon. Generating families in the restricted three-body problem, II: Quantitative study of bifurcations. Lecture Notes in Physics, New Series m: Monographs, 65. Springer, Berlin, MR [7] S. Kolyada and L. Snoha. Some aspects of topological transitivity a survey. In: Iteration theory ECIT 94) Opava), Grazer Math. Ber., 334, Karl-Franzens-Univ. Graz, Graz, MR [8] F. Lenarduzzi. Generalized Hénon-Devaney maps of the plane. Doctoral thesis, IMPA, [9] P Mendes. On Anosov diffeomorphisms on the plane. Proc. Amer. Math. Soc ), no. 2, MR [10] S. Muñoz. Robust transitivity and ergodicity of transformations of the real line and real plane. Doctoral thesis, IMPA, [11] S. Muñoz. Robust transitivity of maps of the real line. Discrete Contin. Dyn. Syst ), no. 3, MR [12] A. Nagar and S. P. Sesha Sai. Some classes of transitive maps on R. J. Anal ), MR [13] A. Nagar, V. Kannan and S. P. Sesha Sai. Properties of topologically transitive maps on the real line. Real Anal. Exchange /02), no. 1, MR [14] W. Parry. Symbolic dynamics and transformations of the unit interval. Trans. Amer. Math. Soc ), MR [15] E. Pujals. From hyperbolicity to dominated splitting. In: Partially hyperbolic dynamics, laminations, and Teichmüller flow, , Fields Inst. Commun., 51, Amer. Math. Soc., Providence, RI, MR Rev. Un. Mat. Argentina, Vol. 59, No )

13 TRANSITIVE MAPS ON R WITH HORIZONTAL ASYMPTOTES 387 [16] W. Reddy. Expanding maps on compact metric spaces. Topology Appl ), no. 3, MR [17] M. Shub. Topologically transitive diffeomorphisms of T 4. In: Symposium on Differential Equations and Dynamical Systems University of Warwic, ), 39 40, Lecture Notes in Mathematics, 206, Springer-Verlag, Berlin, New Yor, [18] S. Smale. Differentiable dynamical systems. Bull. Amer. Math. Soc ), MR [19] M. Velleoop and R. Berglund. On intervals, transitivity = chaos. Amer. Math. Monthly ), no. 4, MR B. Leal Facultad de Ingeniería, Universidad Nacional de Chimborazo, Riobamba, Ecuador and Departamento de Matemática, Facultad de Ciencias, La Hechicera, Universidad de los Andes, Mérida 5101, Venezuela bladismir@ula.ve G. Mata Facultad de Ingeniería, Universidad Nacional de Chimborazo, Riobamba, Ecuador and Departamento de Matemática, Facultad de Ciencias, La Hechicera, Universidad de los Andes, Mérida 5101, Venezuela gmata@ula.ve S. Muñoz Faculdade de Tecnologia, Universidade do Estado do Rio de Janeiro UERJ), Resende, RJ, Brasil sergiomunoz056@gmail.com Received: December 9, 2017 Accepted: May 3, 2018 Rev. Un. Mat. Argentina, Vol. 59, No )