Stability of Rotation Pairs of Cycles for the Interval Maps

Abstract

Let 𝐶⁰(𝐼) be the set of all continuous self-maps of the closed interval 𝐼, and 𝐏(𝑢,𝑣)={𝑓∈𝐶⁰(𝐼)∶𝑓 has a cycle with rotation pair (𝑢,𝑣)} for any positive integer 𝑣>𝑢. In this paper, we prove that if (2^𝑚𝑛𝑠,2^𝑚𝑛𝑡)⊣(𝛾,𝜆), then 𝐏(2^𝑚𝑛𝑠,2^𝑚𝑛𝑡)⊂int𝐏(𝛾,𝜆), where 𝑚≥0 is integer, 𝑛≥1 odd, 1≤𝑠<𝑡 with 𝑠,𝑡 coprime, and 1≤𝛾<𝜆.

1. Introduction

Let 𝐶⁰(𝐼) be the set of all continuous self-maps of the closed interval 𝐼. For any 𝑓,𝑔∈𝐶⁰(𝐼), we define the distance between 𝑓 and 𝑔 by 𝑑𝑓,𝑔=sup_𝑥∈𝐼𝑓(𝑥)−𝑔(𝑥). (1.1) Then (𝐶⁰(𝐼),𝑑) becomes a metric space. For any subset 𝑀 of 𝐶⁰(𝐼), we use int𝑀 to denote the interior of 𝑀. A point 𝑥∈𝐼 is called a periodic point of 𝑓 with period 𝑛 if 𝑓^𝑛(𝑥)=𝑥 and 𝑓^𝑖(𝑥)≠𝑥 for 1≤𝑖≤𝑛−1, and {𝑓^𝑖(𝑥)∶0≤𝑖≤𝑛−1} is called a cycle with period 𝑛. Write 𝐹(𝑓)={𝑥∶𝑓(𝑥)=𝑥}, which is called the set of fixed points of 𝑓. For any subset 𝐴⊂𝐼, we use #𝐴 and [𝐴] to denote the cardinal number of 𝐴 and the smallest closed subinterval of 𝐼 containing 𝐴, respectively. Write [𝐴]=[𝑎;𝑏] if 𝐴={𝑎,𝑏}. For any positive integer 𝑛, write 𝐏(𝑛)={𝑓∈𝐶⁰(𝐼)∶𝑓 has a cycle with period 𝑛}.

One of the remarkable results in one-dimensional dynamics is the Sharkovskii theorem. To state it, let us first introduce the Sharkovskii ordering for positive integers: 3⊳5⊳7⊳⋯⊳2⋅3⊳2⋅5⊳2⋅7⊳⋯⊳2^𝑘⋅3⊳2^𝑘⋅5⊳2^𝑘⋅7⊳⋯⊳2³⊳2²⊳2⊳1. (1.2)

Theorem A (see [1]). For any positive integers 𝑚 and 𝑛, 𝐏(𝑛)⊂𝐏(𝑚) if 𝑛⊳𝑚.
Block [2] studied stability of cycles in the theorem of Sarkovskii and obtained the following theorem.

Theorem B (see [2]). For any positive integers 𝑚 and 𝑛, 𝐏(𝑛)⊂int𝐏(𝑚) if 𝑛⊳𝑚.
Blokh [3] introduced the following ordering among all pairs of positive integers (𝑘,𝑙) with 𝑘<𝑙.(1)If 𝑢/𝑣≠1/2 and 𝑘/𝑙∈[1/2,𝑢/𝑣) or 𝑘/𝑙∈(𝑢/𝑣,1/2], then (𝑢,𝑣)⊣(𝑘,𝑙).(2)If 𝑢/𝑣=𝑘/𝑙=𝑚/𝑛, where 𝑚,𝑛 are coprime, then (𝑢,𝑣)⊣(𝑘,𝑙) if and only if 𝑢/𝑚⊳𝑘/𝑚.He also defined the rotation pair and the rotation number of cycles with period 𝑛>1 for the interval maps.

Definition 1.1 (see [3]). Let 𝑓∈𝐶⁰(𝐼), 𝑃 be a cycle of 𝑓 with period 𝑛>1, and 𝑚=#{𝑦∈𝑃∶𝑓(𝑦)<𝑦}. Then (𝑚,𝑛) is called the rotation pair of 𝑃 and 𝑚/𝑛 the rotation number of 𝑃.
For any positive integer 𝑣>𝑢, write 𝐏(𝑢,𝑣)={𝑓∈𝐶⁰(𝐼)∶𝑓 has a cycle with rotation pair (𝑢,𝑣)}.

Theorem C (see [3]). For any positive integers 𝑣>𝑢 and 𝑙>𝑘, 𝐏(𝑢,𝑣)⊂𝐏(𝑘,𝑙)𝑖𝑓(𝑢,𝑣)⊣(𝑘,𝑙).
In this paper, we will study stability of rotation pairs of cycles for the interval maps. Our main result is the following theorem.

Theorem 1.2. If (2^𝑚𝑛𝑠,2^𝑚𝑛𝑡)⊣(𝛾,𝜆), then 𝐏2^𝑚𝑛𝑠,2^𝑚𝑛𝑡⊂int𝐏𝛾,𝜆, (1.3) where 𝑚≥0 is integer, 𝑛≥1 odd, 1≤𝑠<𝑡 with 𝑠, 𝑡 coprime, and 1≤𝛾<𝜆.

2. Some Lemmas

In this section, we prove Theorem 1.2. To do this, we need the following definitions and lemmas.

Lemma 2.1 (see [4, Lemma 1.4]). Let 𝑓∈𝐶⁰(𝐼). If 𝐼₀,𝐼₁,…,𝐼_𝑚 are compact subintervals of 𝐼 with 𝐼_𝑚=𝐼₀ such that 𝑓(𝐼_𝑘−1)⊃𝐼_𝑘 for 1≤𝑘<𝑚, then there exists a point 𝑦 such that 𝑓^𝑚(𝑦)=𝑦 and 𝑓^𝑘(𝑦)∈𝐼_𝑘 for every 0≤𝑘<𝑚.

Lemma 2.2. Let 𝑓∈𝐶⁰(𝐼). If there are points 𝑎, 𝑏, and 𝑐 such that 𝑓(𝑐)≤𝑎=𝑓(𝑎)<𝑏<𝑐≤𝑓(𝑏)  (resp., 𝑓(𝑐)≥𝑎=𝑓(𝑎)>𝑏>𝑐≥𝑓(𝑏)), then for any integers 𝑚 and 𝑛 with 𝑚/𝑛≤1/2 (resp. 1/2<𝑚/𝑛<1), 𝑓 has a cycle 𝑄={𝑦₁<𝑦₂<⋯<𝑦_𝑛} with rotation pair (𝑚,𝑛) satisfying 𝑓(𝑦_𝑖)>𝑦_𝑖 for all 1≤𝑖≤𝑛−𝑚 and 𝑓(𝑦_𝑖)<𝑦_𝑖 for all 𝑛−𝑚+1≤𝑖≤𝑛.

Proof. We only prove the case 𝑓(𝑐)≤𝑎=𝑓(𝑎)<𝑏<𝑐≤𝑓(𝑏) (the proof for the case 𝑓(𝑐)≥𝑎=𝑓(𝑎)>𝑏>𝑐≥𝑓(𝑏) is similar).
We may assume that (𝑎,𝑏)∩𝐹(𝑓)=∅, then 𝑓(𝑥)>𝑥 for all 𝑥∈(𝑎,𝑏). Choose 𝑝∈(𝑏,𝑐)∩𝐹(𝑓). Then there exist points 𝑎<𝑒₁<𝑒₂<⋯<𝑒_{𝑛−2𝑚+1}<𝑏 such that 𝑓(𝑒_𝑘)=𝑒_𝑘+1 for every 1≤𝑘≤𝑛−2𝑚 and 𝑓(𝑒_{𝑛−2𝑚+1})=𝑝. Let𝐼_𝑘=𝑒_𝑘,𝑒_𝑘+1𝐼if1≤𝑘≤𝑛−2𝑚,_{𝑛−2𝑚+2𝑟+1}=𝑒_{𝑛−2𝑚+1}𝐼,𝑏if0≤𝑟≤𝑚−2,_{𝑛−2𝑚+2𝑟+2}=𝐼𝑝,𝑐if0≤𝑟≤𝑚−1,_𝑛−1=𝑏,𝑝. (2.1) Then 𝑓(𝐼_𝑖)⊃𝐼_𝑖+1 for 𝑖∈{1,2,…,𝑛−1} and 𝑓(𝐼_𝑛)⊃𝐼₁. By Lemma 2.1, there exists a cycle 𝑄={𝑥₁,𝑥₂,…,𝑥_𝑛} such that 𝑥_𝑖∈𝐼_𝑖(1≤𝑖≤𝑛). Furthermore, 𝑄 can be renumbered so that 𝑄={𝑦₁<𝑦₂<⋯<𝑦_𝑛} with the desirable properties.

Lemma 2.3. Let 𝑓∈𝐏(𝑚,𝑛); then 𝑓 has a cycle 𝑄={𝑦₁<𝑦₂<⋯<𝑦_𝑛} with rotation pair (𝑚,𝑛) such that 𝑓(𝑦_𝑖)>𝑦_𝑖 for all 1≤𝑖≤𝑛−𝑚 and 𝑓(𝑦_𝑖)<𝑦_𝑖 for all 𝑛−𝑚+1≤𝑖≤𝑛.

Proof. Let 𝑃={𝑥₁<𝑥₂<⋯<𝑥_𝑛} be a cycle of 𝑓 with rotation pair (𝑚,𝑛). We may assume that 𝑚/𝑛≤1/2 (the proof for the case 1/2<𝑚/𝑛<1 is similar). Let 𝑠=min{𝑘∶𝑓(𝑥_𝑘)<𝑥_𝑘}; then 𝑠≥2,(𝑥_𝑠−1,𝑥_𝑠)∩𝐹(𝑓)≠∅, and 𝑓(𝑥_𝑖)>𝑥_𝑖 for each 1≤𝑖≤𝑠−1. We may also assume that there exists some 𝑠<𝑗≤𝑛 such that 𝑓(𝑥_𝑗)>𝑥_𝑗 otherwise; let 𝑄=𝑃 which completes the proof of Lemma 2.3.
Let 𝑡=min{𝑘∶𝑘>𝑠 and 𝑓(𝑥_𝑘)>𝑥_𝑘} and 𝑝=max{(𝑥_𝑠,𝑥_𝑡)∩𝐹(𝑓)}. Then 𝑓(𝑥)>𝑥 for all 𝑥∈(𝑝,𝑥_𝑡). Let 𝑗=min{𝑘∶𝑓^𝑘+1(𝑥_𝑡)≤𝑝} and 𝑖=min{𝑘∶𝑘≤𝑗 and 𝑓^𝑘+1(𝑥_𝑡)≥𝑓^𝑗(𝑥_𝑡)}. Then 𝑓^𝑗+1(𝑥_𝑡)<𝑝<𝑓^𝑖(𝑥_𝑡)<𝑓^𝑗(𝑥_𝑡)≤𝑓^𝑖+1(𝑥_𝑡). It follows from Lemma 2.2 that 𝑓 has a cycle 𝑄={𝑦₁<𝑦₂<⋯<𝑦_𝑛} such that 𝑄 with the desirable properties.

Definition 2.4 (see [4]). Let 𝑓∈𝐶⁰(𝐼). A cycle 𝑃 of 𝑓 with odd period 𝑛>1 is called a cycle of Stefan type if 𝑃=𝑓^𝑛−1(𝑐)<⋯<𝑓²(𝑐)<𝑐<𝑓(𝑐)<⋯<𝑓^𝑛−2(𝑐) (2.2) or 𝑃=𝑓^𝑛−2(𝑐)<⋯<𝑓(𝑐)<𝑐<𝑓²(𝑐)<⋯<𝑓^𝑛−1(𝑐). (2.3)

Definition 2.5 (see [4, 5]). Let 𝑓∈𝐶⁰(𝐼) and 𝑃={𝑥₁<𝑥₂<⋯<𝑥_𝑛2^𝑚} be a cycle with period 𝑛2^𝑚, where 𝑛≥1 is odd and 𝑚≥0 is an integer. For each 0≤𝑖≤𝑚 and each 1≤𝑗≤2^𝑖, write 𝐴^𝑗_2𝑖={𝑥_(𝑗−1)2^𝑚−𝑖_𝑛+1<𝑥_(𝑗−1)2^𝑚−𝑖_𝑛+2<⋯<𝑥_𝑗2^𝑚−𝑖_𝑛}. We call 𝑃 a strongly simple cycle if one of the following three conditions hold.(1)If 𝑚=0, then either 𝑛=1 or 𝑛>1 and 𝑃 is a cycle of 𝑓 of Stefan type, that is, 𝑃=𝑓^𝑛−1(𝑐)<⋯<𝑓²(𝑐)<𝑐<𝑓(𝑐)<⋯<𝑓^𝑛−2(𝑐) (2.4) or 𝑃=𝑓^𝑛−2(𝑐)<⋯<𝑓(𝑐)<𝑐<𝑓²(𝑐)<⋯<𝑓^𝑛−1(𝑐). (2.5)(2)If 𝑛=1 and 𝑚>0, then for each 1≤𝑖≤𝑚 and each 1≤2𝑘≤2^𝑖, 𝑓^2𝑖−1(𝐴₂^2𝑘−1_𝑖)=𝐴₂^2𝑘_𝑖 and 𝑓^2𝑖−1(𝐴₂^2𝑘_𝑖)=𝐴₂^2𝑘−1_𝑖.(3)If 𝑛>1 and 𝑚>0, then the following three conditions hold.(i)For each 1≤𝑖≤𝑚 and each 1≤2𝑘≤2^𝑖, 𝑓^2𝑖−1(𝐴₂^2𝑘−1_𝑖)=𝐴₂^2𝑘_𝑖, and 𝑓^2𝑖−1(𝐴₂^2𝑘_𝑖)=𝐴₂^2𝑘−1_𝑖.(ii)For each 1≤𝑗≤2^𝑚, 𝐴^𝑗_2𝑚 is a cycle of 𝑓^2𝑚 of Stefan type.(iii)𝑓 maps each 𝐴^𝑖₂^𝑚 monotonically onto another 𝐴^𝑗_2𝑚, with one exception.

Lemma 2.6 (see [4, 5]). If 𝑓∈𝐶⁰(𝐼) has a cycle with period 𝑛, then 𝑓 has a strongly simple cycle with period 𝑛.
Let 𝑃={𝑥₁<𝑥₂<⋯<𝑥_𝑛} be a cycle of 𝑓 with period 𝑛>1. Then there is a unique map 𝑔∶[𝑥₁,𝑥_𝑛]→[𝑥₁,𝑥_𝑛], which is called the linearization of 𝑃, satisfying
(1)𝑔(𝑥_𝑖)=𝑓(𝑥_𝑖) for all 1≤𝑖≤𝑛,(2)𝑔|_{[𝑥𝑖,𝑥𝑖+1]} is linear for all 1≤𝑖≤𝑛−1.
By Theorem 7.5 of [4], we know that if 𝑔 has a strongly simple cycle with rotation pair (𝑝,𝑞), then 𝑓 has also a strongly simple cycle with rotation pair (𝑝,𝑞).

Lemma 2.7. Let 𝑓∈𝐏(𝑘𝑠,𝑘𝑡), where 𝑠,𝑡 are coprime, 𝑘=𝑛2^𝑚, 𝑛≥1 is odd, and 𝑚≥0 is an integer. Then 𝑓 has a cycle 𝑃={𝑧₁<𝑧₂<⋯<𝑧_𝑘𝑡} with rotation pair (𝑘𝑠,𝑘𝑡) satisfying
(1)𝑓(𝑦)<𝑦 if 𝑦∈𝐵_𝑖={𝑧_(𝑖−1)2^𝑚_𝑛+1,…,𝑧_𝑖2^𝑚_𝑛} for 𝑡−𝑠+1≤𝑖≤𝑡 and 𝑓(𝑦)>𝑦 if 𝑦∈𝐵_𝑖={𝑧_(𝑖−1)2^𝑚_𝑛+1,…,𝑧_𝑖2^𝑚_𝑛} for 1≤𝑖≤𝑡−𝑠;(2)𝐵₁ is a strongly simple cycle of 𝑓^𝑡;(3)𝑓 cyclically permutes the sets 𝐵_𝑖 (𝑖=1,2,…,𝑡).

Proof. By Lemma 2.3, we may assume that 𝑅={𝑥₁<𝑥₂<⋯<𝑥_𝑘𝑡} is a cycle of 𝑓 with rotation pair (𝑘𝑠,𝑘𝑡) satisfying 𝑓𝑦>𝑦∀𝑦∈𝑥₁,𝑥₂,…,𝑥_{𝑘𝑡−𝑘𝑠},𝑓𝑦<𝑦∀𝑦∈𝑥_{𝑘𝑡−𝑘𝑠+1},…,𝑥_𝑘𝑡. (2.6) Furthermore, we may assume that 𝑓 is the linearization of 𝑅, 𝐼=[𝑥₁,𝑥_𝑘𝑡], and 𝑝 be the unique fixed point of 𝑓. Obviously, we have that 𝑓(𝑥)>𝑥 for all 𝑥∈[𝑥₁,𝑝) and 𝑓(𝑥)<𝑥 for all 𝑥∈(𝑝,𝑥_𝑘𝑡].
We may assume that 𝑠/𝑡≤1/2 (the proof for the case 1/2<𝑠/𝑡<1 is similar). If 𝑠/𝑡=1/2, then it follows from Theorem 7.18 of [4] that Lemma 2.7 holds. Now we assume 𝑠/𝑡<1/2.
By Theorem C, 𝑓 has a cycle 𝑄={𝑦₁<𝑦₂<⋯<𝑦_𝑡} (𝑡>2) with rotation pair (𝑠,𝑡) satisfying 𝑓(𝑦)>𝑦forall𝑦∈{𝑦₁,…,𝑦_𝑡−𝑠} and 𝑓(𝑦)<𝑦 for all 𝑦∈{𝑦_{𝑡−𝑠+1}…,𝑦_𝑡}.
We can assume 𝑘≥2 since otherwise there is nothing to prove. Furthermore, we may assume #{𝑥∈𝑄∶𝑥>𝑝}≥2 (the proof for the case #{𝑥∈𝑄∶𝑥<𝑝}≥2 is similar). Write 𝑥=max𝑄; then 𝑓^𝑡(𝑥)=𝑥.
Claim 1. We may assume that there exists a positive number 𝜀>0 such that 𝑓^𝑡(𝑦)>𝑦 for all 𝑦∈(𝑥,𝑥+𝜀).Proof. Since 𝑥∉𝑅, there exists a positive number 𝜀>0 such that (𝑓^𝑡(𝑦)−𝑦)(𝑦−𝑥)>0 for all 𝑦∈(𝑥−𝜀,𝑥+𝜀)−{𝑥} or (𝑓^𝑡(𝑦)−𝑦)(𝑦−𝑥)<0 for all 𝑦∈(𝑥−𝜀,𝑥+𝜀)−{𝑥}. If (𝑓^𝑡(𝑦)−𝑦)(𝑦−𝑥)>0 for all 𝑦∈(𝑥−𝜀,𝑥+𝜀)−{𝑥}, then Claim 1 holds. Now we assume (𝑓^𝑡(𝑦)−𝑦)(𝑦−𝑥)<0 for all 𝑦∈(𝑥−𝜀,𝑥+𝜀)−{𝑥}. Write 𝑢=max{𝑦∈(𝑝,𝑥)∶𝑓^𝑡(𝑦)=𝑦} since #{𝑥∈𝑄∶𝑥>𝑝}≥2.
We claim that for all 1≤𝑖≤𝑡, (𝑓^𝑖(𝑢)−𝑝)(𝑓^𝑖(𝑥)−𝑝)>0. Indeed, if (𝑓^𝑖(𝑢)−𝑝)(𝑓^𝑖(𝑥)−𝑝)<0 for some 1≤𝑖≤𝑡, then there exists a point 𝑣∈(𝑢,𝑥) such that 𝑓^𝑖(𝑣)=𝑝; thus 𝑓^𝑡(𝑣)=𝑝, which implies (𝑢,𝑣)∩𝐹(𝑓^𝑡)≠∅. This is a contradiction.
We also claim 𝑢=max{𝑓^𝑖(𝑢)∶0≤𝑖≤𝑡}. Indeed, if 𝑓^𝑖(𝑢)>𝑢 for some 1≤𝑖≤𝑡−1, then there exists a point 𝑣∈(𝑢,𝑥) such that 𝑓^𝑖(𝑣)=𝑣 since 𝑥=max𝑄. Let 𝑤=max{𝑣∈(𝑢,𝑥)∶𝑓^𝑖(𝑣)=𝑣}; then 𝑓^𝑡−𝑖(𝑤)=𝑓^𝑡(𝑤)>𝑤. Since 𝑓^𝑡−𝑖(𝑥)<𝑥, there exists a point 𝑒∈(𝑤,𝑥) such that 𝑓^𝑡−𝑖(𝑒)=𝑒, which implies 𝑓^𝑖(𝑒)=𝑓^𝑡(𝑒)>𝑒 and (𝑒,𝑥)∩𝐹(𝑓^𝑖)≠∅. This is a contradiction.
By using 𝑢 to replace 𝑥, we know that Claim 1 holds. Claim 1 is proven.
Write 𝑆={𝑦∶𝑓^𝑡(𝑦)=𝑥}∩(𝑥,𝑥_𝑘𝑡]. Let 𝑇=min𝑆 if 𝑆≠∅ and 𝑇=𝑥_𝑘𝑡; otherwise. Put 𝐽=(𝑥,𝑇).
Claim 2. 𝐽,𝑓(𝐽),…,𝑓^𝑡−1(𝐽) are pairwise disjoint and 𝑝∉⋃^𝑡−1_𝑖=0𝑓^𝑖(𝐽).Proof. We first prove that 𝐽,𝑓(𝐽),…,𝑓^𝑡−1(𝐽) are pairwise disjoint. Suppose that there exist 0≤𝑖<𝑗≤𝑡−1 and 𝑢,𝑣∈𝐽 such that 𝑓^𝑖(𝑢)=𝑓^𝑗(𝑣), then 𝑓^{𝑡−𝑗+𝑖}(𝑢)=𝑓^𝑡(𝑣)>𝑥. Since 𝑓^{𝑡−𝑗+𝑖}(𝑥)<𝑥, there exists a point 𝑦∈(𝑥,𝑢) such that 𝑓^{𝑡−𝑗+𝑖}(𝑦)=𝑥, which implies 𝑥>𝑓^𝑗−𝑖(𝑥)=𝑓^𝑡(𝑦)>𝑥. This is a contradiction.
Now we prove 𝑝∉⋃^𝑡−1_𝑖=0𝑓^𝑖(𝐽). Suppose that there exist some 0≤𝑖≤𝑡−1 and 𝑢∈𝐽 such that 𝑓^𝑖(𝑢)=𝑝, then 𝑓^𝑡(𝑢)=𝑝, hence 𝑥∈𝑓^𝑡((𝑥,𝑢)), which contradicts definition of 𝑇. Claim 2 is proven.
By definition of 𝑇, it follows that 𝑅∩(⋃^𝑡−1_𝑖=0𝑓^𝑖(𝐽))≠∅ since otherwise we have 𝑓^𝑡(𝑇)>𝑇, which is impossible.
If 𝑓^𝑡(𝐽)⊂𝐽, then 𝑓^𝑡|_𝐽 has a cycle with period 𝑘. It follows from Claim 2 and Lemma 2.6 that 𝑓 has a cycle 𝑃={𝑧₁<𝑧₂<⋯<𝑧_𝑘𝑡} with rotation pair (𝑘𝑠,𝑘𝑡) satisfying conditions (1), (2), and (3) of Lemma 2.7.
If 𝑓^𝑡(𝐽)⊄𝐽, then 𝑓^𝑡(𝑇)=𝑥 and there exists a point 𝑦∈𝐽 such that 𝑓^𝑡(𝑦)≥𝑇. Thus 𝑓^𝑡([𝑥,𝑦])∩𝑓^𝑡([𝑦,𝑇])⊃[𝑥,𝑇]. By Lemma 2.3 of [4], 𝑓^𝑡 has a cycle of period 3 on 𝐽. It follows from Claim 2, Theorem A, and Lemma 2.6 that 𝑓 has a cycle 𝑃={𝑧₁<𝑧₂<⋯<𝑧_𝑘𝑡} with rotation pair (𝑘𝑠,𝑘𝑡) satisfying conditions (1), (2), and (3) of Lemma 2.7. Lemma 2.7 is proven.

3. Proof of Theorem 1.2

In this section, we will give the proof of Theorem 1.2.

Proof of Theorem 1.2. We may assume 𝑠/𝑡≤1/2 (the proof for the case 1/2<𝑠/𝑡<1 is similar). Let 𝑓∈𝐏(2^𝑚𝑛𝑠,2^𝑚𝑛𝑡). We wish to show that there exists a neighbourhood 𝑈 of 𝑓 in 𝐶⁰(𝐼) such that every 𝑔∈𝑈 has a cycle with rotation pair (𝛾,𝜆). The proof will be carried out in a number of stages.

Claim 3. If 𝑚≥0 and 𝑛≥3, then there exists a neighourhood 𝑈 of 𝑓 in 𝐶⁰(𝐼) such that every 𝑔∈𝑈 has a cycle with rotation pair (2^𝑚(𝑛+2)𝑠,2^𝑚(𝑛+2)𝑡).

Proof. By Lemma 2.7, we know that 𝑓 has a cycle {𝑧₁<𝑧₂<⋯<𝑧₂^𝑚_𝑛𝑡} with rotation pair (2^𝑚𝑛𝑠,2^𝑚𝑛𝑡) satisfying
(1)𝑓(𝑦)<𝑦 if 𝑦∈𝐵_𝑖={𝑧_(𝑖−1)2^𝑚_𝑛+1,…,𝑧_𝑖2^𝑚_𝑛} for 𝑡−𝑠+1≤𝑖≤𝑡 and 𝑓(𝑦)>𝑦 if 𝑦∈𝐵_𝑖={𝑧_(𝑖−1)2^𝑚_𝑛+1,…,𝑧_𝑖2^𝑚_𝑛} for 1≤𝑖≤𝑡−𝑠;(2)𝐵₁ is a strongly simple cycle of 𝑓^𝑡;(3)𝑓 cyclically permutes the sets 𝐵_𝑖 (𝑖=1,2,…,𝑡).
For each 1≤𝑙≤2^𝑚, let 𝑧₁(𝑙) denote the midpoint of the 𝑛 points in 𝐶_𝑙={𝑥_{(𝑙−1)𝑛+1}<⋯<𝑥_𝑙𝑛} and 𝑧_𝑗(𝑙)=𝑓^{2𝑚𝑡(𝑗−1)}(𝑧₁(𝑙))(1<𝑗≤𝑛). Then for each 1≤𝑙≤2^𝑚, we have either𝑧_𝑛(𝑙)<𝑧_𝑛−2(𝑙)<⋯<𝑧₃(𝑙)<𝑧₁(𝑙)<𝑧₂(𝑙)<⋯<𝑧_𝑛−3(𝑙)<𝑧_𝑛−1(𝑙) (3.1) or 𝑧_𝑛(𝑙)>𝑧_𝑛−2(𝑙)>⋯>𝑧₃(𝑙)>𝑧₁(𝑙)>𝑧₂(𝑙)>⋯>𝑧_𝑛−3(𝑙)>𝑧_𝑛−1(𝑙). (3.2) Furthermore, the blocks 𝐶_𝑙 can be renumbered so that 𝑓^𝑡(𝑧₁(𝑙))=𝑧₁(𝑙+1) for 1≤𝑙<2^𝑚. Then 𝑓^𝑡𝑧_𝑗(𝑙)=𝑧_𝑗(𝑙+1)if1≤𝑙<2^𝑚𝑓,1≤𝑗≤𝑛,_𝑡𝑧_𝑗2^𝑚=𝑧_𝑗+1(1)𝑓if1≤𝑗<𝑛,_𝑡𝑧_𝑛2^𝑚=𝑧₁(1). (3.3) Since 𝑓^𝑡([𝑧₁(𝑙);𝑧₂(𝑙)])⊃[𝑧₁(𝑙+1);𝑧₂(𝑙+1)] for 1≤𝑙<2^𝑚 and 𝑓^𝑡([𝑧₁(2^𝑚);𝑧₂(2^𝑚)])⊃[𝑧₃(1);𝑧₂(1)], there exist points 𝑧₋₃,𝑧₋₂,𝑧₋₁,𝑧₀ such that 𝑓^2𝑚𝑡(𝑧_−𝑖)=𝑧_−𝑖+1(𝑖=0,1,2,3) satisfying(1)for 1≤𝑙≤2^𝑚 either 𝑧₁(𝑙)<𝑓^{(𝑙−1)𝑡}𝑧₋₁<𝑓^{(𝑙−1)𝑡}𝑧₋₃<𝑓^{(𝑙−1)𝑡}𝑧₋₂<𝑓^{(𝑙−1)𝑡}𝑧₀<𝑧₂(𝑙) (3.4) or 𝑧₁(𝑙)>𝑓^{(𝑙−1)𝑡}𝑧₋₁>𝑓^{(𝑙−1)𝑡}𝑧₋₃>𝑓^{(𝑙−1)𝑡}𝑧₋₂>𝑓^{(𝑙−1)𝑡}𝑧₀>𝑧₂(𝑙); (3.5)(2)𝑓^{(𝑙−1)𝑡+𝑖}(𝑧_−𝑗)∈[𝑓^𝑖(𝐵₁)] for 1≤𝑙≤2^𝑚 and 0≤𝑖<𝑡 and 𝑗=0,1,2,3.Let 𝑓𝜀=min^𝑖𝑧₋₃(1)−𝑓^𝑗𝑧₋₃(1)∶0≤𝑖<𝑗<2^𝑚(𝑛+2)𝑡10 (3.6) and 𝑈={𝑔∈𝐶⁰(𝐼)∶𝑑(𝑓^𝑖,𝑔^𝑖)<𝜀 for all 1≤𝑖≤2^𝑚(𝑛+2)𝑡}. Then for every 𝑔∈𝑈 and 0≤𝑖<𝑗≤2^𝑚(𝑛+2)𝑡−1, we have 𝑓^𝑖(𝑧₋₃(1))<𝑓^𝑗(𝑧₋₃(1)) if and only if 𝑔^𝑖(𝑧₋₃(1))<𝑔^𝑗(𝑧₋₃(1)), and 𝑔^{2𝑚(𝑛+4)𝑡}(𝑧₋₃(1))∈[𝑔^2𝑚2𝑡(𝑧₋₃(1));𝑔^2𝑚6𝑡(𝑧₋₃(1))]. Put 𝐼_𝑙=𝑔^𝑙−1𝑧₋₃(1);𝑔^{2𝑚2𝑡+𝑙−1}𝑧₋₃(1)for1≤𝑙≤2^𝑚(𝑛+2)𝑡. (3.7) Then we have 𝑔𝐼_𝑙⊃𝐼_𝑙+1,if1≤𝑙≤2^𝑚()𝑔𝐼𝑛+2𝑡−1,_{2𝑚(𝑛+2)𝑡}⊃𝐼₁. (3.8) This yields a cycle 𝑂={𝑦,𝑔(𝑦),…,𝑔^{2𝑚(𝑛+2)𝑡}(𝑦)} such that 𝑔^𝑖−1(𝑦)∈𝐼_𝑖 for 𝑖=1,2,…,2^𝑚(𝑛+2)𝑡. it is easy to verify that the rotation pair of 𝑂 is (2^𝑚(𝑛+2)𝑠,2^𝑚(𝑛+2)𝑡). Claim 3 is proven.

Claim 4. If 𝑚≥1 and 𝑛=1, then there exists a neighourhood 𝑈 of 𝑓 in 𝐶⁰(𝐼) such that every 𝑔∈𝑈 has a cycle with rotation pair (2^𝑚−1𝑠,2^𝑚−1𝑡).

Proof. By Lemma 2.7, we know that 𝑓 has a cycle {𝑥₁<𝑥₂<⋯<𝑥₂^𝑚_𝑡} with rotation pair (2^𝑚𝑠,2^𝑚𝑡) satisfying
(1)if 𝑦∈𝐵_𝑖={𝑥_(𝑖−1)2^𝑚₊₁,…,𝑥_𝑖2^𝑚} and 𝑡−𝑠+1≤𝑖≤𝑡, then 𝑓(𝑦)<𝑦; if 𝑦∈𝐵_𝑖={𝑥_(𝑖−1)2^𝑚₊₁,…,𝑥_𝑖2^𝑚} and 1≤𝑖≤𝑡−𝑠, then 𝑓(𝑦)>𝑦;(2)𝐵₁ is a strongly simple cycle of 𝑓^𝑡;(3)𝑓 cyclically permutes the sets 𝐵_𝑖 (𝑖=1,2,…,𝑡).
Since 𝑓^{2𝑚−1𝑡}(𝑥₁)=𝑥₂ and 𝑓^{2𝑚−1𝑡}(𝑥₂)=𝑥₁, there exist points 𝑥₁≤𝑎<𝑏≤𝑥₂ such that 𝑓^{2𝑚−1𝑡}(𝑏)=𝑥₁<𝑥₂=𝑓^{2𝑚−1𝑡}(𝑎) and 𝑓^𝑖(𝑎),𝑓^𝑖(𝑏)∈[𝑓^𝑖(𝑥₁);𝑓^𝑖(𝑥₂)] for 0≤𝑖≤2^𝑚−1𝑡−1. Let𝑓𝜀=min𝑏−𝑎,min^𝑖𝑥₁−𝑓^𝑗𝑥₁∶0≤𝑖<𝑗<2^𝑚𝑡10 (3.9) and 𝑈={𝑔∈𝐶⁰(𝐼)∶𝑑(𝑓^𝑖,𝑔^𝑖)<𝜀 for all 1≤𝑖≤2^𝑚−1𝑡}. Then for every 𝑔∈𝑈, we have 𝑔^{2𝑚−1𝑡}(𝑎)>𝑎,𝑔^{2𝑚−1𝑡}(𝑏)<𝑏. (3.10) This yields a cycle 𝑂={𝑦,𝑓(𝑦),…,𝑓^{2𝑚−1𝑡−1}(𝑦)} such that 𝑔^𝑖(𝑦)∈[𝑔^𝑖(𝑎);𝑔^𝑖(𝑏)] for 𝑖=0,1,…,2^𝑚−1𝑡−1. it is easy to verify that the rotation pair of 𝑂 is (2^𝑚−1𝑠,2^𝑚−1𝑡). Claim 4 is proven.

Claim 5. If 𝑚=0 and 𝑛=1, then there exists a neighourhood 𝑈 of 𝑓 in 𝐶⁰(𝐼) such that every 𝑔∈𝑈 has a cycle with rotation pair (𝛾,𝜆).

Proof. By Lemma 2.7, 𝑓, has a cycle 𝑃={𝑥₁<𝑥₂<⋯<𝑥_𝑡} with rotation pair (𝑠,𝑡) such that 𝑓(𝑥_𝑖)>𝑥_𝑖 for all 1≤𝑖≤𝑡−𝑠 and 𝑓(𝑥_𝑖)<𝑥_𝑖 for all 𝑡−𝑠+1≤𝑖≤𝑡.
Choose two integers 𝑢,𝑣 with 𝑢,𝑣 coprime such that 𝑠/𝑡<𝑢/𝑣<𝛾/𝜆. Without loss of generality, we can assume 𝑓(𝑥_{𝑡−𝑠+1})<𝑥_𝑡−𝑠. Take 𝑤∈𝐹(𝑓)∩(𝑥_𝑡−𝑠,𝑥_{𝑡−𝑠+1}). Put 1≤𝑙=𝑡𝑢−𝑠𝑣, then there exist points 𝑦_𝑖∈(𝑥_𝑡−𝑠,𝑥_{𝑡−𝑠+1}) (𝑖=0,1,…,2𝑙−1) such that 𝑥_𝑡−𝑠<𝑦_2𝑙−2<𝑦_2𝑙−4<⋯<𝑦₂<𝑦₀<𝑤<𝑦₁<𝑦₃<⋯<𝑦_2𝑙−3<𝑦_2𝑙−1<𝑥_{𝑡−𝑠+1} with 𝑓(𝑦_𝑖)=𝑦_𝑖+1 (𝑖=0,1,…,2𝑙−2) and 𝑓(𝑦_2𝑙−1)=𝑥_𝑡−𝑠. Let 𝑓𝜀=min^𝑖𝑦₀−𝑓^𝑗𝑦₀∶0≤𝑖<𝑗<2𝑙+𝑡10 (3.11) and 𝑈={𝑔∈𝐶⁰(𝐼)∶𝑑(𝑓^𝑖,𝑔^𝑖)<𝜀 for all 1≤𝑖≤(𝑡−2𝑠)𝑣}. Then for every 𝑔∈𝑈, we have 𝑔^{(𝑡−2𝑠)𝑣}(𝑦₀)<𝑦₀<𝑔(𝑦₀) and 𝑔²(𝑦₀)<𝑔(𝑦₀).
Let 𝑧=max{𝑥∈𝐼∶𝑥<𝑦₀ and 𝑔^{(𝑡−2𝑠)𝑣}(𝑥)=𝑥}, 𝛼=min{𝑧,𝑔(𝑧),…,𝑔^{(𝑡−2𝑠)𝑣−1}(𝑧)}, and 𝑤₁∈(𝑦₀,𝑔(𝑦₀))∩𝐹(𝑔).

Claim 6. (𝑔^𝑖(𝑧)−𝑤₁)(𝑔^𝑖(𝑦₀)−𝑤₁)>0 for any 𝑖∈{0,1,…,(𝑡−2𝑠)𝑣}.

Proof. Assume on the contrary that (𝑔^𝑖(𝑧)−𝑤₁)(𝑔^𝑖(𝑦₀)−𝑤₁)≤0 for some 𝑖∈{0,1,…,(𝑡−2𝑠)𝑣}; then there exists a point 𝑐∈[𝑧,𝑦₀) such that 𝑔^𝑖(𝑐)=𝑤₁; thus 𝑔^{(𝑡−2𝑠)𝑣}(𝑐)=𝑤₁, which implies (𝑧,𝑦₀)∩𝐹(𝑔^{(𝑡−2𝑠)𝑣})≠∅, a contradiction.

Claim 7. If [𝛼,𝑦₀]∩𝐹(𝑔)≠∅, then 𝑔 has a cycle with rotation pair (𝛾,𝜆).

Proof. Indeed, if [𝛼,𝑦₀]∩𝐹(𝑔)≠∅, let 𝑧₀=max{[𝛼,𝑦₀]∩𝐹(𝑔)}, then 𝑧₀∈[𝛼,𝑧). Since 𝑔(𝑦₀)>𝑦₀ and [𝑧,𝑦₀]∩𝐹(𝑔)=∅, we have 𝑔(𝑧)>𝑧. Let 𝑗=min{𝑘∶𝑓^𝑘+1(𝑧)≤𝑧₀} and 𝑖=min{𝑘∶𝑘≤𝑗 and 𝑓^𝑘+1(𝑧)≥𝑓^𝑗(𝑧)}. Then 𝑓^𝑗+1(𝑧)≤𝑧₀<𝑓^𝑖(𝑧)<𝑓^𝑗(𝑧)≤𝑓^𝑖+1(𝑧). It follows from Lemma 2.2 that 𝑔 has a cycle with rotation pair (𝛾,𝜆).

In the following, we assume that [𝛼,𝑦₀]∩𝐹(𝑔)=∅.

Claim 8. 𝑔^𝑖+1(𝑧)∈[𝑔^𝑖(𝑧);𝑤₁] or 𝑤₁∈[𝑔^𝑖+1(𝑧);𝑔^𝑖(𝑧)] for any 𝑖∈{0,1,…,(𝑡−2𝑠)𝑣}.

Proof. Assume on the contrary that 𝑔^𝑖(𝑧)∈[𝑔^𝑖+1(𝑧);𝑤₁] for some 𝑖∈{0,1,…,(𝑡−2𝑠)𝑣}; then 𝑔^𝑖(𝑧)>𝑦₀. Since 𝑔^𝑖+1(𝑦₀)∈[𝑔^𝑖(𝑦₀);𝑤₁], we have [𝑔^𝑖(𝑧);𝑔^𝑖(𝑦₀)]∩𝐹(𝑔)≠∅. Let 𝑤₂∈[𝑔^𝑖(𝑧);g^𝑖(𝑦₀)]∩𝐹(𝑔); then 𝑤₂>𝑦₀ and there exists a point 𝑑∈[𝑧,𝑦₀] such that 𝑔^{(𝑡−2𝑠)𝑣}(𝑑)=𝑤₂; thus, (𝑑,𝑦₀]∩𝐹(𝑔^{(𝑡−2𝑠)𝑣})≠∅, a contradiction.
By Claims 7 and 8, we know that 𝑔 has a cycle with rotation number 𝑢/𝑣. It follows from Theorem C that 𝑔 has a cycle with rotation pair (𝛾,𝜆), which completes the proof of Claim 5.

Theorem 1.2 now follows immediately from Claim 3, Claim 4, Claim 5, and Theorem C.

Acknowledgments

The project was supported by NSF of China (10861002), NSF of Guangxi (2010GXNSFA013106, 2011GXNSFA014781), SF of Education Department of Guangxi (200911MS212), and Innovation Project of Guangxi Graduate Education 105931003060.