NON-MONOTONICITY HEIGHT OF PM FUNCTIONS ON INTERVAL. 1. Introduction

Transcription

1 Acta Math. Univ. Comenianae Vol. LXXXVI, 2 (2017), pp NON-MONOTONICITY HEIGHT OF PM FUNCTIONS ON INTERVAL PINGPING ZHANG Abstract. Using the piecewise monotone property, we give a full description of non-monotonicity height of PM functions with a single fort on compact interval. This method is also available for general PM functions with finitely many forts, as well as those functions defined on the whole real line. Finally, we obtain a sufficient and necessary condition for the finite non-monotonicity height by characteristic interval. 1. Introduction Let I := [a, b] be a closed interval and f : I I be a continuous function. A point c (a, b) is called a fort if f is strictly monotonic in no neighborhood of c. As in [11], a continuous function f : I I is said to be piecewise monotonic (abbreviated as PM function) if the number N(f) of forts is finite. Then the sequence N(f n ) satisfies the ascending relation 0 = N(f 0 ) N(f) N(f n ) N(f n+1 ).... Denote P M(I, I) the set of all PM functions on I and let the least k N {0} satisfying N(f k ) = N(f k+1 ) if such k exists, and otherwise, be the nonmonotonicity height H(f) of f (cf. [13, 8]). It is known that non-monotonicity height H(f) is closely related to the problem of iterative roots. In 1961, Kuczma [2] gave a complete description of iterative roots for PM functions with H(f) = 0. In 1983, J. Zhang and L. Yang [11] put forward characteristic interval K(f) for PM function, and for the first time applied it to obtain monotonic iterative roots if those functions satisfy H(f) = 1. From [11, 13], we know that arbitrary PM functions satisfying H(f) 2 have no continuous iterative roots of order greater than N(f). Later, L. Liu and W. Zhang [9] proved that every continuous iterative root of a PM function with H(f) 1 is an extension an iterative root of f of the same order on the characteristic interval K(f). Further results on iterative roots of PM functions with H(f) 2 appeared in [6, 8]. Non-monotonicity height as assumption condition also appears in topological conjugacy between PM functions. In 2013, Y. Shi, L. Received September 5, 2015; revised July 14, Mathematics Subject Classification. Primary 39B12, 37E05, 26A18. Key words and phrases. iteration; fort; non-monotonicity height; piecewise monotonic function. Supported by ZR2014AL003, J12L59 and 2013Y04 grants.

2 288 PINGPING ZHANG Li and Z. Leśniak [10] constructed all homeomorphic solutions and continuously non-monotone solutions of the conjugacy equation ϕ f = g ϕ, where f : I I, g : J J are two given r-modal interval maps (special PM functions) with H(f) = H(g) = 1 and I, J are closed intervals. For a class of PM functions with H(f) 1, [3] gave a sufficient and necessary condition under which any two of these maps are topologically conjugate. Very recently, L. Li and W. Zhang [5, 7] gave a sufficient condition as well as a method of constructing the topological conjugacy between PM functions f and their iterative roots if H(f) = 1. As known from the previous work, H(f) plays an important role in studying PM functions. Thus, one problem was raised naturally: How to determine the H(f) for a given PM function f? This question was considered by Lin Li in [4]. It turns out that even very simple PM functions can become quite complicated under iteration. [4] investigated a class of polygonal functions with a unique vertex and determined the number of vertices under iteration by analyzing the slope of polygon. Those polygonal functions with opposite sign of slope are special kinds of PM functions, among which the particular case f(a) = a and f(b) = b also appeared in reference [1]. It seems that up to now there has not been a complete result of this problem (see [12]). In this paper, we are interested in non-monotonicity height H(f) of PM functions. In Section 2, we give a full description of H(f) for those functions with a single fort on compact interval. Our method using the piecewise monotone property is also available for general PM functions with finitely many forts, as well as those functions defined on the whole real line. Section 3 illustrates this method for those general PM functions by several examples. Finally, in Section 4, we discuss the relations between non-monotonicity height H(f) and characteristic interval K(f l ) for a given PM function f. 2. PM functions with a unique fort The main difficulties in finding H(f) of PM functions come from the sharply increasing number of forts under iteration. Using the piecewise monotonicity, we find that N(f n ) depends on the second order iterate of f. Furthermore, we prove that H(f) is uniquely determined by the second order iterate of all those critical points such as fort and two endpoints as well as fixed points of the given PM function. In order to determine H(f) by observing the change of N(f n ) and to avoid complicated computation, in this section, we only consider PM function on the compact interval I := [a, b], each of which has only one fort f 1 is increasing and f 2 is decreasing Let a function f P M(I, I) be of the form (2.1) f(t) = { f1 (t), a t t 0, f 2 (t), t 0 < t b,

3 NON-MONOTONICITY HEIGHT OF PM FUNCTIONS ON INTERVAL 289 where f 1, f 2 are continuous, strictly monotonic with different monotonicities. In this subsection, we investigate that f 1 is increasing and f 2 is decreasing, the main result on H(f) reads as follows. Theorem 1. Suppose that f P M(I, I) is of the form (2.1). (i) If t 0 f(t 0 ), then H(f) = 1 and (2.2) lim n f n (t 0 ) = t, where t is a fixed point of f. (ii) If min{f(a), f(b)} t 0, then H(f) = 1. Moreover, (2.2) holds in the case where the unique fixed point t of f is attracting. Proof. In case (i), it follows from the condition t 0 f(t 0 ) that f(t) [a, t 0 ] for all t [a, b], which implies the iterate f n for n 1, proceeds on the monotonic subinterval [a, t 0 ], and its number of forts is invariant under iteration, i.e., N(f n ) N(f) = 1 for all n 2. Thus H(f) = 1 is proved. Furthermore, the monotonicity of f 1 yields (2.3) f(t 0 ) f 2 (t 0 ) f 3 (t 0 )... f n (t 0 ) a, that is, {f n (t 0 )} is a decreasing sequence which tends to t [a, t 0 ). Moreover, by continuity of f, f(t ) = t and (2.2) is proved. In case (ii), inequality min{f(a), f(b)} t 0 shows that f(t) [min{f(a), f(b)}, f(t 0 )] [t 0, b], t [a, b]. Thus the iterate f n proceeds on the monotonic subinterval [t 0, b], i.e., N(f n ) N(f) = 1 for all n N, which implies that H(f) = 1. Moreover, we assert that (2.4) f 2 (t 0 ) f 4 (t 0 )... f 2n (t 0 )... t... f 2n 1... f 3 (t 0 ) f(t 0 ). In fact, f([t 0, b]) [t 0, b] yields f n (t 0 ) [t 0, b] for any positive integer n. Since f 2 is strictly increasing on [t 0, b] and t 0 f 2 (t 0 ), there exists t (t 0, b] such that (2.5) f 2n (t 0 ) < t and f 2n (t 0 ) < f 2n+2 (t 0 ). By the same argument, we also have (2.6) f 2n 1 (t 0 ) > t and f 2n 1 (t 0 ) > f 2n+1 (t 0 ). Then (2.4) is proved by (2.5) (2.6), which gives the assertion of (ii). This completes the proof. Remark 1. Under the assumption of case (ii), the function f can have a point z of period 2 such that t 0 < z < t < f(z) < f(t 0 ). Theorem 2. Suppose that f P M(I, I) is of the form (2.1). (i) If min{f(a), f 2 (t 0 )} t 0 > f(b), then H(f) = 2. (ii) If min{f(b), f 2 (a)} t 0 > f(a), then H(f) = 2. (iii) If f(t 0 ) > t 0 > max{f(a), f(b)} and min{f 2 (t 0 ), f 2 (a), f 2 (b)} t 0, then H(f) = 2.

4 290 PINGPING ZHANG Proof. In case (i), the assumption f(a) t 0 > f(b) leads to the result f 2 f 1 (t), a t t 0, (2.7) f 2 (t) = f 2 f 2 (t), t 0 < t f 1 (t 0 ), f 1 f 2 (t), f 1 (t 0 ) < t b, which implies that t 0, f 1 (t 0 ) are two forts of f 2. Using the assumption min{f(a), f 2 (t 0 )} t 0, we obtain that (2.8) f 2 (t) [min{f(a), f 2 (t 0 ), f 2 (b)}, f(t 0 )] [t 0, f(t 0 )] [t 0, f 1 (t 0 )], t [a, b]. Noticing f([t 0, f 1 (t 0 )]) [t 0, f(t 0 )] [t 0, f 1 (t 0 )], it follows that the iterate f n (n 3) proceeds on the monotonic subinterval [t 0, f 1 (t 0 )]. Thus N(f n ) 2 (n 2) is a consequence of the monotonicity of f 2, and then H(f) = 2. Regarding case (ii), with a similar proof as that of (i), we have f([a, b]) [f(a), f(t 0 )] from the assumption f(a) < t 0 f(b). Then f 1 f 1 (t), a t f 1 (t 0 ), (2.9) f 2 (t) = f 2 f 1 (t), f 1 (t 0 ) < t t 0, f 2 f 2 (t), t 0 < t b, implying f 1 (t 0 ), t 0 are two forts of f 2. The formula (2.9) also shows that max f 2 = max{f 1 f 1 (f 1 (t 0 )), f 2 f 2 (b)} f(t 0 ) b, min f 2 = min{f 1 f 1 (a), f 2 f 1 (t 0 )} t 0, that is, f 2 (t) [t 0, b] for all t [a, b]. Hence, the iterate f n (n 3) proceeds on the monotonic subinterval [t 0, b] and N(f n ) N(f 2 ) = 2 for all integers n 2. Therefore, H(f) = 2. In case (iii), the assumption f(t 0 ) > t 0 > max{f(a), f(b)} shows that f([a, b]) [min{f(a), f(b)}, f(t 0 )], and then f 1 f 1 (t), a t f 1 1 (t 0 ), (2.10) f 2 f 2 f 1 (t), f 1 1 (t 0 ) < t t 0, (t) = f 2 f 2 (t), t 0 < t f 2 1 (t 0 ), f 1 f 2 (t), f 1 2 (t 0 ) < t b, which follows that f 1 1 (t 0 ), t 0, f 2 1 (t 0 ) are three forts of f 2. Now (2.10) jointly with the assumption t 0 min{f 2 (t 0 ), f 2 (a), f 2 (b)}, imply max f 2 = max{f 1 f 1 (f 1 1 (t 0 )), f 2 f 2 (f 2 1 (t 0 ))} = f(t 0 ) f 2 1 (t 0 ), min f 2 = min{f 1 f 1 (a), f 2 f 1 (t 0 ), f 1 f 2 (b)} t 0, that is, f 2 ([a, b]) [t 0, f 1 2 (t 0 )], and the iterate f n (n 3) proceeds on the monotone subinterval [t 0, f 1 2 (t 0 )]. Thus, N(f n ) N(f 2 ) = 3 for all n 2 and H(f) = 2. This completes the proof.

5 NON-MONOTONICITY HEIGHT OF PM FUNCTIONS ON INTERVAL 291 Theorem 3. Suppose that f P M(I, I) is of the form (2.1). (i) If f has no fixed points on [a, t 0 ) and f(b) t 0 > f 2 (a), then H(f) (2, ) is finite. (ii) If f has no fixed points on [a, t 0 ) and f(t 0 ) > t 0 > max{f(a), f(b)}, f 2 (t 0 ) t 0 > min{f 2 (a), f 2 (b)}, then H(f) (2, ) is finite. Proof. In case (i), we first claim that t 0 > f(a). Otherwise, the assumption f(b) t 0 implies that f(t) t 0 for all t [a, b]. Then f 2 (t) t 0 is a contradiction to the condition t 0 > f 2 (a). Hence, f([a, b]) [f(a), f(t 0 )] and (2.11) f 2 (t) = f 1 f 1 (t), a t f 1 (t 0 ), f 2 f 1 (t), f 1 (t 0 ) < t t 0, f 2 f 2 (t), t 0 < t b. We conclude f 1 (t 0 ), t 0 are two forts of f 2. Note that f is strictly increasing on subinterval [a, t 0 ] with no fixed point, it is true that f(t) > t for t [a, t 0 ] and there exists a finite positive integer i such that f i (a) t 0 > f i 1 (a) > > f(a). Moreover, the assumption f(b) t 0 yields f n (t 0 ) t 0 for any integer n 1. Thus we get f i ([a, t 0 ]) [t 0, b]. On the other hand, by the facts that f(b) t 0 and f is strictly decreasing on [t 0, b], f is a self-mapping on [t 0, b], and then f n ((t 0, b]) [t 0, b] for any n N. Hence, f i ([a, b]) [t 0, b] and thus the number of the forts of f n (n i) is identical to f i, that is, N(f n ) is bounded for n. Therefore, H(f) = A < for an integer A > 2. In case (ii), by using the assumptions f 2 (t 0 ) t 0 and the monotonicity of f on subinterval [t 0, b], for an arbitrary integer n 1 we get (2.12) t 0 f 2 (t 0 ) f 2n (t 0 ) f 2n 1 (t 0 ) f 3 (t 0 ) f(t 0 ), inductively. The formula (2.12) means that f has a fixed point t (t 0, f(t 0 )). Note that f has no fixed points and is strictly increasing on subinterval [a, t 0 ], it is true that f(t) > t for all t [a, t 0 ]. Consequently, f 2 (b) > f(b) and f 2 (a) > a since f(t 0 ) > t 0 > max{f(a), f(b)}. Hence, min{f 2 (a), f 2 (b)} > min{f(a), f(b)}. As in case (i), repeating this process, we get a strictly increasing sequence {min{f n (a), f n (b)}} fulfilling (2.13) min{f k (a), f k (b)} t 0 for certain positive integer k. Then by using the results f(t 0 ) f 1 2 (t 0 ), (2.12) and (2.13), we get f k ([a, b]) [t 0, f 1 2 (t 0 )], and then N(f n ) N(f k ) for all integers n k + 1. Therefore, N(f n ) is finite for n and H(f) = B < for an integer B > 2. This completes the proof.

6 292 PINGPING ZHANG Theorem 4. Suppose that f P M(I, I) is of the form (2.1). (i) If f(a) t 0 > max{f(b), f 2 (t 0 )}, then. (ii) If f(t 0 )>t 0 >max{f(a), f(b), f 2 (t 0 )}, then. (iii) If f has fixed points on [a, t 0 ) and f(t 0 )>t 0, then. Proof. In case (i), from the assumption f(a) t 0 > f(b), we have f([a, b]) = [f(b), f(t 0 )] = [f(b), t 0 ] (t 0, f(t 0 )]. Then f(f[a, b]) = [f 2 (b), f(t 0 )] [f 2 (t 0 ), f(t 0 )] [f 2 (t 0 ), f(t 0 )], which covers the point t 0 since f 2 (t 0 ) < t 0 < f(t 0 ). Hence, f 2 has two forts t 0, f 2 1 (t 0 ). Following the same process, one checks that f 3 has at least three forts t 0, f 2 1 (t 0 ), f 2 2 (t 0 ) and {N(f n )} is strictly increasing for n. Therefore, N(f n ) is unbounded as n, and then. In case (ii), the assumption f(t 0 ) > t 0 > max{f(a), f(b)} implies that the points f 1 1 (t 0 ), f 1 2 (t 0 ), t 0 are forts of f 2. The remaining proof is similar as that of (i). That is, by the condition t 0 > f 2 (t 0 ), every range f n ([a, b]) (n 2) covers the interval [f 2 (t 0 ), f(t 0 )] containing t 0 as its interior point. Then new forts appear after every iteration and lim n N(f n ) =. Hence,. In case (iii), without loss of generality, assume that t 1 [a, t 0 ) is a fixed point of f. Note that f(t 0 ) > t 0, then as discussed for case (i)-(ii), every f n ([a, b]) covers the interval [t 1, f(t 0 )] for any integer n 1, which contains t 0 as its interior point. Therefore, new forts appear continually under every iteration and lim n N(f n ) =. Therefore,. This completes the proof. The main result of Theorems 1 4 is presented in Table 1 as follows. Table 1. Results on nonmonotonicity height H(f). Nonmonotonicity Conditions height t 0 f(t 0 ) H(f) = 1 t 0 < f(t 0 ) min{f(a), f(b)} t 0 H(f) = 1 f(b) t 0 > f(a); f 2 (a) t 0 H(f) = 2 f(t) > t, t [a, t 0 ) f 2 (a) < t 0 2 < H(f) < f(a) t 0 > f(b) f 2 (t 0 ) t 0 H(f) = 2 f 2 (t 0 ) < t 0 max{f(a), f(b)} < t 0 ; f 2 (t 0 ) t 0 min{f 2 (a), f 2 (b)} t 0 H(f) = 2 f(t) > t, t [a, t 0 ) min{f 2 (a), f 2 (b)} < t 0 2 < H(f) < f 2 (t 0 ) < t 0 f has fixed points on [a, t 0 )

7 NON-MONOTONICITY HEIGHT OF PM FUNCTIONS ON INTERVAL f 1 is decreasing and f 2 is increasing In this subsection, we investigate the function (2.1) in which f 1 is decreasing and f 2 is increasing. The discussion is similar to that of the Subsection 2.1 since g(t) := a + b f(t) for all t [a, b]. Therefore, we only give the result and omit their proofs. Theorem 5. Suppose that f P M(I, I) is of the form (2.1). (i) If f(t 0 ) t 0, then H(f) = 1 and (2.2) holds, where t is a fixed point of f. (ii) If t 0 max{f(a), f(b)}, then H(f) = 1 and (2.2) holds, where t is the unique fixed point of f. Theorem 6. Suppose that f P M(I, I) is of the form (2.1). (i) If f(b) t 0 > f(a) and t 0 f 2 (b), then H(f) = 2. (ii) If f(a) t 0 > f(b) and t 0 f 2 (t 0 ), then H(f) = 2. (iii) If min{f(a), f(b)} > t 0 > f(t 0 ) and t 0 max{f 2 (t 0 ), f 2 (a), f 2 (b)}, then H(f) = 2. Theorem 7. Suppose that f P M(I, I) is of the form (2.1). (i) If f has no fixed points on (t 0, b] and f 2 (b) > t 0 > f(a), then H(f) (2, ) is finite. (ii) If f has no fixed points on (t 0, b] and min{f(a), f(b)} > t 0 > f(t 0 ), max{f 2 (a), f 2 (b)} > t 0 f 2 (t 0 ), then H(f) (2, ) is finite. Theorem 8. Suppose that f P M(I, I) is of the form (2.1). (i) If f(a) t 0 > f(b) and f 2 (t 0 ) > t 0, then. (ii) If min{f(a), f(b), f 2 (t 0 )} > t 0 > f(t 0 ), then. (iii) If f has fixed points on (t 0, b] and t 0 > f(t 0 ), then. Table 2 presents all cases listed in Theorems 5 8 as follows. Table 2. Results on nonmonotonicity height H(f). Nonmonotonicity Conditions height f(t 0 ) t 0 H(f) = 1 f(t 0 ) < t 0 max{f(a), f(b)} t 0 H(f) = 1 f(a) < t 0 f(b); f 2 (b) t 0 H(f) = 2 f(t) < t, t [t 0, b) f 2 (a) > t 0 2 < H(f) < f(b) < t 0 f(a) f 2 (t 0 ) t 0 H(f) = 2 f 2 (t 0 ) > t 0 min{f(a), f(b)} > t 0 ; f 2 (t 0 ) t 0 max{f 2 (a), f 2 (b)} t 0 H(f) = 2 f(t) < t, t [t 0, b) max{f 2 (a), f 2 (b)} > t 0 2 < H(f) < f 2 (t 0 ) > t 0 f has fixed points on [t 0, b)

8 294 PINGPING ZHANG 3. PM functions with finitely many forts From the proofs of Theorems 1 8 and Tables 1 2, we note that H(f) is independent of concrete route, and is uniquely determined by the second order iterate of the unique fort and two endpoints as well as fixed points of the given PM function. Our method using the piecewise monotone property is also available for general PM functions with finitely many forts, as well as those functions defined on the whole real line. In the following, we present several examples illustrating that H(f) of the general PM functions is also determined by the second iterate of its critical points such as forts, two endpoints and fixed points. Example 1. Consider the mapping φ 1 : [0, 1] [0, 1] defined by 2 3 t + 4 5, t [0, 1 5 ], φ 1 (t) := 1 3 t + 3 5, t ( 1 5, 3 10 ], t + 1, t ( 3 10, 2 5 ], 1 2 t + 2 5, t ( 2 5, 1]. Clearly, φ 1 has three forts t 1 := 1 5, t 2 := 3 10, t 3 := 2 5. Note that then H(φ 1 ) = 1 by (i) of Theorem 5. min{φ 1 (t 1 ), φ 1 (t 3 )} > t 3,

9 NON-MONOTONICITY HEIGHT OF PM FUNCTIONS ON INTERVAL 295 Example 2. Consider the mapping φ 2 : (, 4] (, 4] defined by 2 3t, t (, 1], φ 2 (t) := 2 3 t + 4 3, t (1, 2], 3 2t 3, t (2, 3], 3 2t + 6, t (3, 4]. It is easy to see that t 1 := 1, t 2 := 2, t 3 := 3 are forts of φ 2. Furthermore, φ 2 (t 3 ) > t 1 > implying that H(φ 2 ) = 2 by (i) of Theorem 6. lim φ 2(x) and t 1 > φ 2 2(t 3 ), x Example 3. Consider the mapping φ 3 : R R defined by 4t, t (, 1], 4t + 8, t (1, 2], φ 3 (t) := t 2, t (2, 3], 4 t, (3, + ). Clearly, t 1 := 1,t 2 := 2 and t 3 := 3 are forts of φ 3. Since lim t φ 3 (t) = and φ 3 (t 1 ) > t 1 > φ 2 3(t 1 ), we obtain that H(φ 3 ) = similar to (ii) of Theorem Relations between H(f) and K(f l ) In this subsection, we give a sufficient and necessary condition under which H(f) of a PM function f is finite.

10 296 PINGPING ZHANG Theorem 9. If f P M(I, I), then H(f) is finite if and only if there exists a characteristic interval K(f l ) for an l N. Proof. Sufficiency. If there exists a characteristic interval K(f l ) of the PM function f l for an l N, then the iterate (f l ) n of f l proceeds on the strictly monotonic subinterval K(f l ) I and the number N(f nl ) of forts is invariant under iteration, i.e., N(f nl ) N(f l ) for all n 1. Thus H(f l ) is finite. Consequently, H(f) is finite by using N(f) N(f l ) and the finity of H(f l ). Necessarity. If H(f) is finite, then N(f H(f)+i ) = N(f H(f) ) for all i N. Consequently, we obtain (4.1) N((f H(f) ) n ) = N(f H(f) ), n N. Let l = H(f). The formula (4.1) changes into N(f nl ) = N(f l ), which shows that the iterate of f l is invariant on forts. According to the definition of characteristic interval, there exists a strictly monotonic subinterval K(f l ) I as the characteristic interval of f l such that f l (I) K(f l ). Theorem 9 implies an essential difference between H(f) < and. That is, H(f) < guarantees a characteristic interval of f l for l N, which describes some invariance of f, in turn those properties determine its dynamical behavior. While for, there is no such a strictly monotonic subinterval since the range of f n (n = 1, 2,... ) covers at least one fort under each iteration. Acknowledgment. The author is grateful to the referees for their careful reading and comments.

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