GENETICS OF ITERATIVE ROOTS FOR PM FUNCTIONS

. It is known that the time-one mapping of a ﬂow deﬁnes a discrete dynamical system with the same dynamical behaviors as the ﬂow, but conversely one wants to know whether a ﬂow embedded by a homeomorphism preserves the dynamical behaviors of the homeomorphism. In this paper we consider iterative roots, a weak version of embedded ﬂows, for the preservation. We refer an iterative root to be genetic if it is topologically conjugate to its parent function. We prove that none of PM functions with height being > 1 has a genetic root and none of iterative roots of height being > 1 is genetic even if the height of its parent function is equal to 1. This shows that most functions do not have a genetic iterative root. Further, we obtain a necessary and suﬃcient conditions under which a PM function f has a genetic iterative root in the case that f and the iterative root are both of height 1.

where φ n denotes the nth iterate of φ, i.e., φ n (x) := φ(φ n−1 (x)) and φ 0 (x) = x for all x ∈ I. Since Ch. Babbage [1] studied earlier in the 19-th century, the iterative root problem has attracted attentions of research and its basic theory was established (see [8,9,14]) for monontonic functions. More advances can be found in survey papers [2,17]. The non-monotonic case is much more complicated because those functions, not preserving orientation, possess complicated dynamical behaviors. Earlier results were given by Jingzhong Zhang and Lu Yang ( [16,19]) and by A. Blokh, E. Coven, M. Misiurewicz and Z. Nitecki ( [3]) for PM functions, a class of non-monotonic functions with finitely many forts (short name of non-monotonic points). In [16] iterative roots are obtained by extending those monotone iterative roots obtained on the "characteristic interval", the maximal monotone subinterval covering the range of f . Further progresses were made in [10,12,13] in recent years. This problem is important because it is a weak version of the embedding flow problem ( [4,6,15,18]), and defines fractional iterates. Moreover, it is related to information engineering ( [5,7]).
It is interesting to compare the dynamics of an iterative root φ with its parent function f . Actually not all iterative roots can inherit the dynamical properties of their parent function f . An iterative root φ of f is said to be genetic if φ is topologically conjugate to f on I, i.e., for each φ there exists a homeomorphism h : I → I such that So we need to identify which iterative roots are genetic or not. This problem was considered in [11] for a special class of PM functions of height 1, each of which has a characteristic interval bounded by an endpoint of the domain and is homeomorphic onto the characteristic interval but does not reach an endpoint of the characteristic interval outside. Theorem 2 of [12] indicates that all its iterative roots are in the simplest mode of 1-extension.
In this paper we generally discuss genetics of iterative roots for PM functions. In section 2 we introduce basic concepts about PM functions and a result on genetics. Our section 3 is devoted to non-genetic cases. We prove that none of PM functions f with height H(f ) > 1 has a genetic root, none of iterative roots φ of height H(φ) > 1 is genetic in the case that H(f ) = 1, and, in the case that both H(f ) = 1 and H(φ) = 1, the root φ is not genetic if its monotonicity on the characteristic interval of f is different from that of f . In section 4 we discuss in the case that H(f ) = H(φ) = 1 for necessary and sufficient conditions under which a PM function f has a genetic iterative root φ. We give examples to demonstrate our results in section 5.

2.
Preliminaries. Let a, b ∈ R such that a < b, and let f : I := [a, b] → I be a continuous function. A point c ∈ (a, b) is called a nonmonotonic point (or fort for short) of f if f is strictly monotonic in no neighborhood of c. The function f is said to be piecewise monotonic and called a PM functions if it has only finitely many forts. Let P M (I, I) denote the set of all piecewise monotonic self-mappings of I.
For f ∈ P M (I, I), we easily see that the cardinality of the set S(f ) of forts of f , denoted by N (f ), satisfies the ascending relation where the least positive integer k such that N (f k ) = N (f k+1 ) is referred to as the height H(f ) of f if such a k exists. Otherwise, the height is said to be ∞. The height is actually the short name of the concept nonmonotonicity height indicated in [13]. It is an important number to characterize the complexity of f .
The set S(f ) splits the whole interval I into subintervals such that f is strictly monotone on them, called monotone subinterval. As proved in [16,19], H(f ) = 1 if and only if the PM function f has a unique monotone subinterval, called characteristic interval, such that F is a self-mapping on the monotone subinterval and the monotone subinterval covers the range of F .
For convenience we introduce the following notations: Let Fix(f ) consist of all fixed points of f : I → I. ξ ∈ Fix(f ) is said to be attractive (or repulsive) if there exists a neighbourhood O ξ of ξ such that lim n→+∞ f n (x) = ξ for every x ∈ O ξ (or if there exists a neighbourhood O ξ of ξ such that the sequence f n (x) does not tend to ξ for any x ∈ O ξ \{ξ}). As defined in [8, p.301], a reversing correspondence is a strictly increasing function f mapping I into itself with a fixed point ξ ∈ Fix(f ) and a strictly decreasing function ω mapping Fix(f ) onto itself such that ω(ξ) = ξ and the expression f (x) − x has opposite signs in the intervals (ξ 1 , ξ 2 ) and (ω(ξ 2 ), ω(ξ 1 )) for every ξ 1 , ξ 2 ∈ Fixf with ξ 1 < ξ 2 and (ξ 1 , ξ 2 )∩Fixf = ∅. More precisely, [17, Theorem 2.1] asserts the following results on genetics.
Proposition 1. In case (I,I) all strictly increasing iterative roots φ of every f ∈ CI(I, I) fixing both endpoints are genetic. In case (I,D) any decreasing iterative root of a reversing correspondence f ∈ CI(I, I) is not genetic.
In fact, the first result comes from the conjugation between iterative roots and f . The second results holds because of their opposite monotonicity.
In what follows, we discuss on genetics of iterative roots for PM functions generally, showing what iterative roots of a PM function are topologically conjugate to the PM function or not.
3. Non-genetic roots. In this section, we concentrate to non-genetic roots.
Lemma 3.1. Let f ∈ P M (I, I) and φ be a continuous iterative root of order n of f . If φ is genetic, then φ and f have the same forts, i.e., S(f ) = S(φ).
Proof. As shown in [19, p.121], one can see that On the other hand, there is a homeomorphism h such that since h is a homeomorphism, i.e., S(h) = ∅. Then, we have showing that #S(φ) = #{x ∈ I : h(x) ∈ S(f )} = #S(f ).
Lemma 3.1 enables us to obtain the following result. Proof. For an indirect proof to (i), assume that φ is a genetic iterative root of order n of f . We prove that Otherwise, N (φ k ) = N (φ k+1 ) for a certain integer 1 ≤ k ≤ n − 1. By Lemma 2.5 of [19], , a contradiction to the fact that H(f ) > 1. However, (3) contradicts to the fact which is implied by Lemma 3.1. Thus, result (i) is proved. The proof to the case of (ii) is similar. Assume that there is a continuous and genetic iterative root φ of order n of f with H(φ) ≥ 2. Then which contradicts to (4). This completes the proof of (ii).
In the case of (iii), we also have (4), i.e., S(φ) = S(f ). As shown in Lemma 3 of [12], φ has a characteristic interval, denoted by J. Since [min f, max f ] ⊂ [min φ, max φ] ⊂ J, we see that f and φ have the same characteristic interval J. Thus, this completes the proof of (iii) and therefore the proof is completed.
In what follows, our discussion is concentrated to the case that H(f ) = H(φ) = 1. In the case that H(f ) = 1, the mapping f has a unique characteristic interval as indicated in [16,19], denoted by K(f ), and all iterative roots φ of f with H(φ) = 1 are constructed in the mode of 1-extension The set S(f ) splits the interval I into N (f ) + 1 subintervals  Then, by the continuity of h, there are two interior points where I k and I k+1 are two consecutive sub-intervals in I(f ). It follows that f • h is non-monotone on the subinterval I i because the monotonicity of f on I k is different from the monotonicity of f on I k+1 . However, φ is a genetic iterative root of f , i.e., f • h = h • φ. It further implies that h • φ is non-monotone on the subinterval I i , but it contradicts to the fact that h • φ is monotone on every I j in I(f ), which holds because h is a homeomorphism on I and φ is monotone on I j . This proves result (i).
In order to prove (ii), let us consider h to be an orientation-preserving homeomorphism on I additionally. Then h(c 0 ) = c 0 . We claim that for all i = 1, 2, . . . , N (f ). In fact, by result (i) of this lemma and the continuity of h, we see that h(I 0 ) ⊆ I 0 . Assume that h(I 0 ) ⊂ I 0 but h(I 0 ) = I 0 . Then, by (6) and the continuity of h, h(I 1 ) ⊂ I 0 because I 1 is adjacent to I 0 . It follows that Note from result (i) that i=0 I i will be mapped by h into at most N (f ) sub-intervals in I(f ), which shows that h is not a surjection on I, a contradiction to the fact that h is a homeomorphism. This contradiction shows that showing h(c 1 ) = c 1 . Furthermore, we assume that (7) holds for a certain j ∈ {2, . . . , N (f ) − 1}, i.e., It follows that h is an orientation-preserving homeomorphism on I\ By result (i) of this lemma and the continuity of h, we see that h(I j+1 ) ⊆ , which contradicts to that h is a homeomorphism. Thus, h(I j+1 ) = I j+1 . This proves the claim inductively. Hence, it concludes that h(c i ) = c i for all i = 1, 2, . . . , N (f ) and completes the proof of (ii) since h is a homeomorphism on I. For result (iii), knowing from the fact that h is an orientationreserving homeomorphism on I, we have h(c 0 ) = c N (f )+1 . A similar discussion to the proof of (9) gives that h(I 0 ) = I N (f ) . Furthermore, similarly to (7), we can prove by induction that This completes the proof of (iii) and therefore the proof is completed.
Proof of Theorem 3.3. For an indirect proof, assume that φ is genetic. Then, there is a homeomorphism h : I → I such that (1) holds.
If h is an orientation-preserving homeomorphism on I, h . This gives a contradiction. Thus, there are no orientation-preserving homeomorphic solutions such that (1) holds.
If h is an orientation-reserving homeomorphism on I, then one can see from (iii) in Lemma 3.4 that since I is the characteristic interval of both f and φ.
On the other hand, notice that both h and φ are strictly increasing on I . Then, so is h • φ by (10). It follows that f • h is strictly increasing on I . By (11), we get that f is strictly decreasing on I N (f )+1− . Furthermore, it is assumed that f is strictly increasing on I . Then, we get that which contradicts to (12). Thus, there are no orientation-reserving homeomorphic solutions such that (1) holds. Therefore, the proof is completed.  Proof. For an indirect proof, assume that h is an orientation-reversing homeomorphic solution of equation (1). Then, one can see that for each However, by (12) and the results in (ii), where κ := (N (F ) + 1)/2. It leads to a contradiction and therefore the proof is completed.
By Lemma 4.1, the conjugation between a PM function and its an iterative root cannot be orientation-reversing in the cases (R-II) and (R-DD). In what follows, we find orientation-preserving homeomorphic solutions for equation (1) In fact, Fix(f ) ⊂ K(f ) and where 1 ≤ m ≤ N (f ) + 2 is a certain integer and α j , β j ∈ Fix(f ) such that Actually, (13) shows that for each θ ∈ A(f ) there exist uniquely two consecutive fixed points α and β of f , for which we assume α < β without loss of generality, such that θ ∈ (α, β). Since the cardinal #A(f ) is finite, we can find finitely many distinct subintervals (α j , β j ), This defines a decomposition of A as shown in (14). Furthermore, each A j (f ) can be grouped in different orbits, that is, Here Orb f (x 0 ) denotes the orbit of f starting from x 0 .  , s(j), is a singleton, denoted by f (c ji ) where c ji ∈ S * (f ), and there is an integer 1 ≤ k ≤ s(j) such that f (c ji ) lies between f (c jk ) and f 2 (c jk ) and φ(c ji ) lies between φ(c jk ) and φ 2 (c jk ), i.e., In order to prove this theorem, we need the following lemmas. Proof. We only consider that f has no fixed points in the interior of I. Otherwise, I is a union of closed intervals, each of which is bounded either by two consecutive fixed points of f or by one fixed point of f and a or b, one of the endpoints. Let U k denote such a closed interval. Clearly, f has no fixed points in the interior of U k . If there is an orientation-preserving homeomorphism h k : is an orientation-preserving homeomorphic solution of (1). When f has no fixed points in the interior of I, without loss of generality, we assume that f (x) < x for all x ∈ (a, b) since the proof is totally similar if f (x) > x for all x ∈ (a, b). Then, it suffices to discuss the following two cases: In case (I-1), it is easy to check that φ(x) < x and a = φ(a) < φ(b) = b. Choosing x 0 , y 0 ∈ (a, b) arbitrarily, one can define the sequences (x k ) and (y k ) such that Clearly, the sequence (x k ) is strictly decreasing and because φ is orientation-preserving on I and φ(x) < x. Similarly, the sequence (y k ) is also strictly decreasing and Let I i := [x i+1 , x i ] and J i := [y i+1 , y i ] for each i ∈ Z. Then, Obviously, there are infinitely many orientation-preserving homeomorphisms h 0 : Each h 0 defines a sequence (h i ) uniquely by the recursions Let which can be verified easily to satisfy equation (1). Further, we claim that h is continuous. In fact, by (15) and (16) we have For the same reason,

One can prove by induction that
implying that h i is an orientation-preserving homeomorphism from I i onto J i for all i ∈ Z. It follows from (18) that h is continuous at each point x i for all i ∈ Z. This proves the claimed continuity of h. Obviously, h is an orientation-preserving homeomorphism on I.
In case (I-2), it is easy to check that φ(x) < x for all x ∈ (a, b) and a = φ(a) < φ(b) < b. Choose initial points x 0 = φ(b) and y 0 = f (b), which define sequences (x i ) and (y i ) respectively by (15). Similarly to case (I-1), one can find infinitely many orientation-preserving homeomorphismsh : Finally, let which satisfies equation (1). Furthermore, By (19) we can check that h is an orientation-preserving homeomorphism. Therefore, the proof is completed. Proof. In order to prove the necessity, consider a nonempty branch A j (f ), in which every point can be presented as f (c i ) for a certain for a certain s(i) ∈ Z. It follows from (5) that On the other hand, we assumed that there are orientation-preserving homeomorphisms h : which implies that #A j (f ) = 1. This completes the proof of the necessity in case (ii).
In what follows, we prove the sufficiency. We only consider the situation j = 1, i.e., a singleton, where c ∈ S(f ) ∪ {c 0 , c N (f )+1 }. From (24) we see that As shown in the proof of Theorem 1 of [12], the function φ, being an iterative root of f of order n with H(φ) = 1, can be presented as where φ is an iterative root of f | K(f ) . Notice from Theorem 2 of [19] that because both f and φ are increasing on K(f ). It follows from (25) and (26) that where . Therefore, we are ready to find conjugations between f and φ: First, by Lemma 4.3, we can find infinitely many orientation-preserving homeomorphisms h : then, we extend the above h from K(f ) to the whole I, defining i.e., h(c i ) = c i for all i = 0, 1, . . . , N (f ) + 1. This proves that h is an orientationpreserving homeomorphism on I and completes the proof of sufficiency. Therefore, the whole proof is completed.
Proof. We first prove the necessity. Without loss of generality, consider a branch ) and both f (d 1 ) and f (d 2 ) are not contained in an orbit of f . Thus there are two subcases to be considered: We only consider subcase (C-up) since subcase (C-down) can be discussed similarly. In subcase (C-up), it can be seen from (26) that since x < f (x) for all x ∈ (α d , β d ). It means that φ j (x) < φ k (x) for any integers j < k on the subinterval (α d , β d ). Moreover, we assumed that there are orientationpreserving homeomorphisms h : This implies that since h is an orientation-preserving homeomorphism. In order to guarantee the following inequalities i.e., Since n ≥ 2 and k ∈ Z + , we get k = 1, implying that Furthermore, the above two inequalities can be changed into . This completes the proof of the necessity.
In what follows, we prove the sufficiency. We only consider the situation j = 1, i.e., As shown in the proof of Theorem 1 of [12], the function φ, being an iterative root of f of order n with H(φ) = 1, can be presented as in (26). Notice from Theorem 2 of [19] that Fix(φ) = Fix(f ) because both f and φ are increasing on K(f ). It follows from (26) that where . Therefore, we are ready to find conjugations between f and φ: First, by Lemma 4.3, we can find infinitely many orientation-preserving homeomorphisms h : Moreover, the proof of Lemma 4.3 shows that the conjugation mapping h only requires the initial oneh 0 : [x 0 , x 1 ] → [y 0 , y 1 ] to be an orientation-preserving homeomorphism. Note from the assumption that and that for iterative roots φ It leads us to choose a suitableh 0 which satisfies by (17). Then, we extend the above h from K(f ) to the whole I, defining , ∀ x ∈ I i , for every i ∈ {0, 1, . . . , N (f )}\{ }; finally, we check that the function satisfies equation (1). Additionally, from the proof of Lemma 4.3 and the fact Fix(φ) = Fix(f ) we see that It follows from (31-34) that , h(c i ) = c i for all i = 0, 1, . . . , N (f ) + 1. This proves that h is an orientationpreserving homeomorphism on I and completes the proof of sufficiency. Therefore, the whole proof is completed.
Proof of Theorem 4.2. We first consider case (i), i.e., A(f ) = ∅. In this case, where P 2 (f ) denotes the set of all 2-periodic points of f .
Clearly, B j ∩ B k = ∅ for all j = k in {1, . . . , m}. Furthermore, each B j (f ) can be grouped in different orbits, that is, with an value f (c ji ) ∈ B ji for each i = 1, .., s(j) and Orb f (f (c ji )) ∩ Orb f (f (c jk )) = ∅ for i = k. Here Orb f (x 0 ) denotes the orbit of f initiated from x 0 . where c ji ∈ S * (f ), and there is an integer 1 ≤ k ≤ s(j) such that f (c ji ) lies between f (c jk ) and f 2 (c jk ) and φ(c ji ) lies between φ(c jk ) and φ 2 (c jk ), i.e., Proof. We only consider the situation that f has no 2-periodic points in the interior of I. Otherwise, I is a union of closed intervals, each of which is bounded either by two consecutive 2-period points or by one 2-period point and an endpoint. Let U k denote such a closed interval. If we can find an orientation-preserving homeomorphism h k : is an orientation-preserving homeomorphic solution of (1).
Since we consider f has no 2-period points in the interior of I, we only need to discuss in the two cases: In case (I-1), let η be the unique fixed point of f . We only prove that η is an attractive fixed point because the case that η is a repulsive fixed point can be proved similarly. It is easy to check that η is also an attractive fixed point of φ and a = φ(b) < φ(a) = b. Choose initial points x 0 , y 0 ∈ (a, b) arbitrarily and extend them to two infinite-sequences (x k ) and (y k ) respectively as Clearly, because φ is orientation-reversing on I and the fixed point η is attractive. Similarly, we have Let I i denote the compact interval between x i and x i+2 for each i ∈ Z. Then I = +∞ i=−∞ I i . Moreover, Let J i denote the compact interval between y i and y i+2 for each i ∈ Z. We similarly have I = Each h 0 defines a sequence (h i ) uniquely by the recursions Let which can be verified easily by equation (1). In fact, by (36) and (37) we have For the same reason, One can prove by induction that implying that h i is an orientation-preserving homeomorphism from I i onto J i for all i ∈ Z. It follows from (18) that h is continuous at each point x i for all i ∈ Z. This proves the claimed continuity of h. Obviously, h is an orientation-preserving homeomorphism on I.
In case (I-2), i.e., a < f (b) < f (a) < b, we construct orientation-preserving homeomorphisms h : I → I by (38), (39) and (40) as done in the same manner as in the above case (I-1) with either x 0 = f (a) and y 0 = g(a) or x 0 = f (b) and y 0 = g(b) for the starting points in (36). Therefore, this completes the proof.
Proof of Theorem 4.6. We first consider case (i), i.e., B(f ) = ∅. In this case By Theorem 2 of [19], all periodic points of f lie in the characteristic interval K(f ) = I := [c , c +1 ], which implies that because f is decreasing on K(f ). Note that the function φ, being an iterative root of f of order n with H(φ) = 1, can be presented as in (26), i.e., where φ is an iterative root of f | K(f ) . Further, by Theorem 2 of [19], because both f and φ are decreasing on K(f ). Let Then by (41) and (43) we have for each i = 0, 1, . . . , N (f ) + 1 since n is odd. Therefore, we are ready to find conjugations between f and φ: First, by Lemma 4.7, we can find infinitely many orientation-preserving homeomorphisms h : Then, we extend the above h from K(f ) to the whole I, defining , h(c i ) = c i for all i = 0, 1, . . . , N (f ) + 1. This proves that h is an orientationpreserving homeomorphism on I and completes the proof in case (i).
Remark that for the case H(f ) > 1 all iterative roots of such a PM function f are not conjugate to f and for the case H(f ) = 1 all iterative roots φ satisfying H(φ) ≥ 2 are not conjugate to f . It shows that most PM functions have nongenetic iterative roots, which do not possess the same dynamical behaviors as the corresponding PM functions.