# Stability of Rotation Pairs of Cycles for the Interval Maps

Authors: Taixiang Sun; Hongjian Xi; Hailan Liang; Qiuli He; Xiaofeng Peng
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

LetC0(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 (2mns,2mnt)⊣(γ,λ), then P(2mns,2mnt)⊂intP(γ,λ), where m≥0 is integer, n≥1 odd, 1≤s<t with s,t coprime, and 1≤γ<λ.


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## Body

## 1. Introduction

LetC0(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 (1.1)d(f,g)=sup⁡x∈I|f(x)-g(x)|.
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 fn(x)=x and fi(x)≠x for 1≤i≤n-1, and {fi(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 theSharkovskii theorem. To state it, let us first introduce the Sharkovskii ordering for positive integers: (1.2)3⊳5⊳7⊳⋯⊳2⋅3⊳2⋅5⊳2⋅7⊳⋯⊳2k⋅3⊳2k⋅5⊳2k⋅7⊳⋯⊳23⊳22⊳2⊳1.Theorem A (see [1]).
For any positive integersm 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 integersm 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)
Ifu/v≠1/2 and k/l∈[1/2,u/v) or k/l∈(u/v,1/2], then (u,v)⊣(k,l).(2)
Ifu/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]).
Letf∈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 integerv>u, write P(u,v)={f∈C0(I):f has a cycle with rotation pair (u,v)}.Theorem C (see [3]).
For any positive integersv>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(2mns,2mnt)⊣(γ,λ), then
(1.3)P(2mns,2mnt)⊂int⁡P(γ,λ),
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 Theorem1.2. To do this, we need the following definitions and lemmas.Lemma 2.1 (see [4, Lemma 1.4]).
Letf∈C0(I). If I0,I1,…,Im are compact subintervals of I with Im=I0 such that f(Ik-1)⊃Ik for 1≤k<m, then there exists a point y such that fm(y)=y and fk(y)∈Ik for every 0≤k<m.Lemma 2.2.
Letf∈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<⋯<yn} with rotation pair (m,n) satisfying f(yi)>yi for all 1≤i≤n-m and f(yi)<yi for all n-m+1≤i≤n.Proof.
We only prove the casef(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<⋯<en-2m+1<b such that f(ek)=ek+1 for every 1≤k≤n-2m and f(en-2m+1)=p. Let(2.1)Ik=[ek,ek+1]if1≤k≤n-2m,In-2m+2r+1=[en-2m+1,b]if0≤r≤m-2,In-2m+2r+2=[p,c]if0≤r≤m-1,In-1=[b,p].
Then f(Ii)⊃Ii+1 for i∈{1,2,…,n-1} and f(In)⊃I1. By Lemma 2.1, there exists a cycle Q={x1,x2,…,xn} such that xi∈Ii(1≤i≤n). Furthermore, Q can be renumbered so that Q={y1<y2<⋯<yn} with the desirable properties.Lemma 2.3.
Letf∈P(m,n); then f has a cycle Q={y1<y2<⋯<yn} with rotation pair (m,n) such that f(yi)>yi for all 1≤i≤n-m and f(yi)<yi for all n-m+1≤i≤n.Proof.
LetP={x1<x2<⋯<xn} 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(xk)<xk}; then s≥2,(xs-1,xs)∩F(f)≠∅, and f(xi)>xi for each 1≤i≤s-1. We may also assume that there exists some s<j≤n such that f(xj)>xj otherwise; let Q=P which completes the proof of Lemma 2.3.
Lett=min{k:k>s and f(xk)>xk} and p=max{(xs,xt)∩F(f)}. Then f(x)>x for all x∈(p,xt). Let j=min{k:fk+1(xt)≤p} and i=min{k:k≤j and fk+1(xt)≥fj(xt)}. Then fj+1(xt)<p<fi(xt)<fj(xt)≤fi+1(xt). It follows from Lemma 2.2 that f has a cycle Q={y1<y2<⋯<yn} such that Q with the desirable properties.Definition 2.4 (see [4]).
Letf∈C0(I). A cycle P of f with odd period n>1 is called a cycle of Stefan type if
(2.2)P={fn-1(c)<⋯<f2(c)<c<f(c)<⋯<fn-2(c)}
or
(2.3)P={fn-2(c)<⋯<f(c)<c<f2(c)<⋯<fn-1(c)}.Definition 2.5 (see [4, 5]).
Letf∈C0(I) and P={x1<x2<⋯<xn2m} be a cycle with period n2m, where n≥1 is odd and m≥0 is an integer. For each 0≤i≤m and each 1≤j≤2i, write A2ij={x(j-1)2m-in+1<x(j-1)2m-in+2<⋯<xj2m-in}. We call P a strongly simple cycle if one of the following three conditions hold.(1)
Ifm=0, then either n=1 or n>1 and P is a cycle of f of Stefan type, that is,
(2.4)P={fn-1(c)<⋯<f2(c)<c<f(c)<⋯<fn-2(c)}
or
(2.5)P={fn-2(c)<⋯<f(c)<c<f2(c)<⋯<fn-1(c)}.(2)
Ifn=1 and m>0, then for each 1≤i≤m and each 1≤2k≤2i, f2i-1(A2i2k-1)=A2i2k and f2i-1(A2i2k)=A2i2k-1.(3)
Ifn>1 and m>0, then the following three conditions hold.
(i)
For each1≤i≤m and each 1≤2k≤2i, f2i-1(A2i2k-1)=A2i2k, and f2i-1(A2i2k)=A2i2k-1.(ii)
For each1≤j≤2m, A2mj is a cycle of f2m of Stefan type.(iii)
f maps each A2mi monotonically onto another A2mj, with one exception.Lemma 2.6 (see [4, 5]).
Iff∈C0(I) has a cycle with period n, then f has a strongly simple cycle with period n.
LetP={x1<x2<⋯<xn} be a cycle of f with period n>1. Then there is a unique map g:[x1,xn]→[x1,xn], which is called the linearization of P, satisfying
(1)
g(xi)=f(xi) 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.
Letf∈P(ks,kt), where s,t are coprime, k=n2m, n≥1 is odd, and m≥0 is an integer. Then f has a cycle P={z1<z2<⋯<zkt} with rotation pair (ks,kt) satisfying
(1)
f(y)<y if y∈Bi={z(i-1)2mn+1,…,zi2mn} for t-s+1≤i≤t and f(y)>y if y∈Bi={z(i-1)2mn+1,…,zi2mn} for 1≤i≤t-s;(2)
B1 is a strongly simple cycle of ft;(3)
f cyclically permutes the sets Bi (i=1,2,…,t).Proof.
By Lemma2.3, we may assume that R={x1<x2<⋯<xkt} is a cycle of f with rotation pair (ks,kt) satisfying
(2.6)f(y)>y∀y∈{x1,x2,…,xkt-ks},f(y)<y∀y∈{xkt-ks+1,…,xkt}.
Furthermore, we may assume that f is the linearization of R, I=[x1,xkt], 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,xkt].
We may assume thats/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<⋯<yt} (t>2) with rotation pair (s,t) satisfying f(y)>yforally∈{y1,…,yt-s} and f(y)<y for all y∈{yt-s+1…,yt}.
We can assumek≥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 ft(x)=x.
Claim 1.
We may assume that there exists a positive numberε>0 such that ft(y)>y for all y∈(x,x+ε).Proof.
Sincex∉R, there exists a positive number ε>0 such that (ft(y)-y)(y-x)>0 for all y∈(x-ε,x+ε)-{x} or (ft(y)-y)(y-x)<0 for all y∈(x-ε,x+ε)-{x}. If (ft(y)-y)(y-x)>0 for all y∈(x-ε,x+ε)-{x}, then Claim 1 holds. Now we assume (ft(y)-y)(y-x)<0 for all y∈(x-ε,x+ε)-{x}. Write u=max{y∈(p,x):ft(y)=y} since #{x∈Q:x>p}≥2.
We claim that for all1≤i≤t, (fi(u)-p)(fi(x)-p)>0. Indeed, if (fi(u)-p)(fi(x)-p)<0 for some 1≤i≤t, then there exists a point v∈(u,x) such that fi(v)=p; thus ft(v)=p, which implies (u,v)∩F(ft)≠∅. This is a contradiction.
We also claimu=max{fi(u):0≤i≤t}. Indeed, if fi(u)>u for some 1≤i≤t-1, then there exists a point v∈(u,x) such that fi(v)=v since x=maxQ. Let w=max{v∈(u,x):fi(v)=v}; then ft-i(w)=ft(w)>w. Since ft-i(x)<x, there exists a point e∈(w,x) such that ft-i(e)=e, which implies fi(e)=ft(e)>e and (e,x)∩F(fi)≠∅. This is a contradiction.
By usingu to replace x, we know that Claim 1 holds. Claim 1 is proven.
WriteS={y:ft(y)=x}∩(x,xkt]. Let T=minS if S≠∅ and T=xkt; otherwise. Put J=(x,T).
Claim 2.
J,f(J),…,ft-1(J) are pairwise disjoint and p∉⋃i=0t-1fi(J).Proof.
We first prove thatJ,f(J),…,ft-1(J) are pairwise disjoint. Suppose that there exist 0≤i<j≤t-1 and u,v∈J such that fi(u)=fj(v), then ft-j+i(u)=ft(v)>x. Since ft-j+i(x)<x, there exists a point y∈(x,u) such that ft-j+i(y)=x, which implies x>fj-i(x)=ft(y)>x. This is a contradiction.
Now we provep∉⋃i=0t-1fi(J). Suppose that there exist some 0≤i≤t-1 and u∈J such that fi(u)=p, then ft(u)=p, hence x∈ft((x,u)), which contradicts definition of T. Claim 2 is proven.
By definition ofT, it follows that R∩(⋃i=0t-1fi(J))≠∅ since otherwise we have ft(T)>T, which is impossible.
Ifft(J)⊂J, then ft|J has a cycle with period k. It follows from Claim 2 and Lemma 2.6 that f has a cycle P={z1<z2<⋯<zkt} with rotation pair (ks,kt) satisfying conditions (1), (2), and (3) of Lemma 2.7.
Ifft(J)⊄J, then ft(T)=x and there exists a point y∈J such that ft(y)≥T. Thus ft([x,y])∩ft([y,T])⊃[x,T]. By Lemma 2.3 of [4], ft 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<⋯<zkt} with rotation pair (ks,kt) satisfying conditions (1), (2), and (3) of Lemma 2.7. Lemma 2.7 is proven.


## 3. Proof of Theorem1.2

In this section, we will give the proof of Theorem1.2.Proof of Theorem1.2.
We may assumes/t≤1/2 (the proof for the case 1/2<s/t<1 is similar). Let f∈P(2mns,2mnt). 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.
Ifm≥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 (2m(n+2)s,2m(n+2)t).Proof.
By Lemma2.7, we know that f has a cycle {z1<z2<⋯<z2mnt} with rotation pair (2mns,2mnt) satisfying
(1)
f(y)<y if y∈Bi={z(i-1)2mn+1,…,zi2mn} for t-s+1≤i≤t and f(y)>y if y∈Bi={z(i-1)2mn+1,…,zi2mn} for 1≤i≤t-s;(2)
B1 is a strongly simple cycle of ft;(3)
f cyclically permutes the sets Bi (i=1,2,…,t).
For each1≤l≤2m, let z1(l) denote the midpoint of the n points in Cl={x(l-1)n+1<⋯<xln} and zj(l)=f2mt(j-1)(z1(l))(1<j≤n). Then for each 1≤l≤2m, we have either(3.1)zn(l)<zn-2(l)<⋯<z3(l)<z1(l)<z2(l)<⋯<zn-3(l)<zn-1(l)
or
(3.2)zn(l)>zn-2(l)>⋯>z3(l)>z1(l)>z2(l)>⋯>zn-3(l)>zn-1(l).
Furthermore, the blocks Cl can be renumbered so that ft(z1(l))=z1(l+1) for 1≤l<2m. Then
(3.3)ft(zj(l))=zj(l+1)if1≤l<2m,1≤j≤n,ft(zj(2m))=zj+1(1)if1≤j<n,ft(zn(2m))=z1(1).
Since ft([z1(l);z2(l)])⊃[z1(l+1);z2(l+1)] for 1≤l<2m and ft([z1(2m);z2(2m)])⊃[z3(1);z2(1)], there exist points z-3,z-2,z-1,z0 such that f2mt(z-i)=z-i+1(i=0,1,2,3) satisfying(1)
for1≤l≤2m either
(3.4)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)
or
(3.5)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);(2)
f(l-1)t+i(z-j)∈[fi(B1)] for 1≤l≤2m and 0≤i<t and j=0,1,2,3.Let
(3.6)ε=min⁡{|fi(z-3(1))-fj(z-3(1))|:0≤i<j<2m(n+2)t}10
and U={g∈C0(I):d(fi,gi)<ε for all 1≤i≤2m(n+2)t}. Then for every g∈U and 0≤i<j≤2m(n+2)t-1, we have fi(z-3(1))<fj(z-3(1)) if and only if gi(z-3(1))<gj(z-3(1)), and g2m(n+4)t(z-3(1))∈[g2m2t(z-3(1));g2m6t(z-3(1))]. Put
(3.7)Il=[gl-1(z-3(1));g2m2t+l-1(z-3(1))]for1≤l≤2m(n+2)t.
Then we have
(3.8)g(Il)⊃Il+1,if1≤l≤2m(n+2)t-1,g(I2m(n+2)t)⊃I1.
This yields a cycle O={y,g(y),…,g2m(n+2)t(y)} such that gi-1(y)∈Ii for i=1,2,…,2m(n+2)t. it is easy to verify that the rotation pair of O is (2m(n+2)s,2m(n+2)t). Claim 3 is proven.Claim 4.
Ifm≥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 (2m-1s,2m-1t).Proof.
By Lemma2.7, we know that f has a cycle {x1<x2<⋯<x2mt} with rotation pair (2ms,2mt) satisfying
(1)
ify∈Bi={x(i-1)2m+1,…,xi2m} and t-s+1≤i≤t, then f(y)<y; if y∈Bi={x(i-1)2m+1,…,xi2m} and 1≤i≤t-s, then f(y)>y;(2)
B1 is a strongly simple cycle of ft;(3)
f cyclically permutes the sets Bi (i=1,2,…,t).
Sincef2m-1t(x1)=x2 and f2m-1t(x2)=x1, there exist points x1≤a<b≤x2 such that f2m-1t(b)=x1<x2=f2m-1t(a) and fi(a),fi(b)∈[fi(x1);fi(x2)] for 0≤i≤2m-1t-1. Let(3.9)ε=min⁡{b-a,min⁡{|fi(x1)-fj(x1)|:0≤i<j<2mt}}10
and U={g∈C0(I):d(fi,gi)<ε for all 1≤i≤2m-1t}. Then for every g∈U, we have
(3.10)g2m-1t(a)>a,g2m-1t(b)<b.
This yields a cycle O={y,f(y),…,f2m-1t-1(y)} such that gi(y)∈[gi(a);gi(b)] for i=0,1,…,2m-1t-1. it is easy to verify that the rotation pair of O is (2m-1s,2m-1t). Claim 4 is proven.Claim 5.
Ifm=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 Lemma2.7, f, has a cycle P={x1<x2<⋯<xt} with rotation pair (s,t) such that f(xi)>xi for all 1≤i≤t-s and f(xi)<xi for all t-s+1≤i≤t.
Choose two integersu,v with u,v coprime such that s/t<u/v<γ/λ. Without loss of generality, we can assume f(xt-s+1)<xt-s. Take w∈F(f)∩(xt-s,xt-s+1). Put 1≤l=tu-sv, then there exist points yi∈(xt-s,xt-s+1) (i=0,1,…,2l-1) such that xt-s<y2l-2<y2l-4<⋯<y2<y0<w<y1<y3<⋯<y2l-3<y2l-1<xt-s+1 with f(yi)=yi+1 (i=0,1,…,2l-2) and f(y2l-1)=xt-s. Let
(3.11)ε=min⁡{|fi(y0)-fj(y0)|:0≤i<j<2l+t}10
and U={g∈C0(I):d(fi,gi)<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).
Letz=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.
(gi(z)-w1)(gi(y0)-w1)>0 for any i∈{0,1,…,(t-2s)v}.Proof.
Assume on the contrary that(gi(z)-w1)(gi(y0)-w1)≤0 for some i∈{0,1,…,(t-2s)v}; then there exists a point c∈[z,y0) such that gi(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:fk+1(z)≤z0} and i=min{k:k≤j and fk+1(z)≥fj(z)}. Then fj+1(z)≤z0<fi(z)<fj(z)≤fi+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.
gi+1(z)∈[gi(z);w1] or w1∈[gi+1(z);gi(z)] for any i∈{0,1,…,(t-2s)v}.Proof.
Assume on the contrary thatgi(z)∈[gi+1(z);w1] for some i∈{0,1,…,(t-2s)v}; then gi(z)>y0. Since gi+1(y0)∈[gi(y0);w1], we have [gi(z);gi(y0)]∩F(g)≠∅. Let w2∈[gi(z);gi(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 Claims7 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.Theorem1.2 now follows immediately from Claim 3, Claim 4, Claim 5, and Theorem C.



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