Research ArticleNon-Single-Valley Solutions for 119901-Order Feigenbaumrsquos TypeFunctional Equation 119891(120593(119909)) = 120593119901(119891(119909))

Min Zhang

College of Science China University of Petroleum Qingdao Shandong 266555 China

Correspondence should be addressed to Min Zhang zhangminmath163com

Received 19 March 2014 Accepted 16 June 2014 Published 10 July 2014

Academic Editor Cristina Marcelli

Copyright copy 2014 Min Zhang This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

This work deals with Feigenbaumrsquos functional equation 119891(120593(119909)) = 120593119901(119891(119909)) 120593(0) = 1 0 le 120593(119909) le 1 119909 isin [0 1] where 119901 ge 2

is an integer 120593119901 is the 119901-fold iteration of 120593 and 119891(119909) is a strictly increasing continuous function on [0 1] that satisfies 119891(0) = 0119891(119909) lt 119909 (119909 isin (0 1]) Using a constructive method we discuss the existence of non-single-valley continuous solutions of the aboveequation

1 Introduction

In 1978 Feigenbaum [1 2] and independently Couliet andTresser [3] introduced the notion of renormalization forreal dynamical systems In 1992 Sullivan [4] proved theuniqueness of the fixed point for the period-doubling renor-malization operator This fixed point of renormalizationsatisfies a functional equation known as the Cvitanovic-Feigenbaum equation

119892 (119909) = minus1

120582119892 (119892 (minus120582119909)) 0 lt 120582 lt 1

119892 (0) = 1 minus1 le 119892 (119909) le 1 119909 isin [minus1 1]

(1)

As mentioned above this equation and its solution playan important role in the theory initiated by Feigenbaum [1 2]However it is difficult to find an exact solution of the aboveequation in general This problem can be studied in classesof smooth functions or of continuous functions In classesof smooth functions the existence of smooth solutions for(1) has been established in [5ndash8] and references therein Asfar as we know continuous solutions of (1) in classes ofcontinuous functions have been relatively little researchedIn this direction we refer to [9 10] In particular Yang

and Zhang [9] demonstrated the existence of a single-valleycontinuous solution for the following equation

120593 (119909) =1

120582120593 (120593 (120582119909)) 0 lt 120582 lt 1

120593 (0) = 1 0 le 120593 (119909) le 1 119909 isin [0 1]

(2)

which is called the second type of Feigenbaumrsquos functionalequations In the last years a number of authors consideredthe more general equation

120593 (119909) =1

120582120593119901(120582119909) 0 lt 120582 lt 1

120593 (0) = 1 0 le 120593 (119909) le 1 119909 isin [0 1]

(3)

where 119901 ge 2 is an integer and 120593119901 is the 119901-fold iterationof 120593 It is easy to see that (2) is a special case of (3) For 119901large enough Eckmann et al [11] showed that there exists asolution of (3) similar to the function 120593(119909) = |1 minus 21199092| Forany 119901 ge 2 Zhang et al [12] and Liao et al [13] proved that (3)has single-valley-extended continuous solutions

In the present paper we will consider Feigenbaumrsquosfunctional equations

119891 (120593 (119909)) = 120593119901(119891 (119909))

120593 (0) = 1 0 le 120593 (119909) le 1 119909 isin [0 1] (4)

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 731863 8 pageshttpdxdoiorg1011552014731863

2 Abstract and Applied Analysis

where 119891(119909) is a strictly increasing continuous function on[0 1] that satisfies 119891(0) = 0 119891(119909) lt 119909 (119909 isin (0 1])We will prove the existence of single-valley-extended non-single-valley continuous solutions of (4) by the constructivemethod Obviously let 119891(119909) = 120582119909 then (4) is (3)

2 Basic Definitions and Lemmas

In this section we will give some characterizations of single-valley-extended non-single-valley continuous solutions of(4) they will be proved in the appendix

Definition 1 One calls 120593 a single-valley-extended continuoussolution of (4) if (i) 120593 is a continuous solution of (4) (ii)there exists 120572 isin (119891(1) 1) such that 120593 is strictly decreasingon [119891(1) 120572] and strictly increasing on [120572 1]

Definition 2 One calls 120593 a single-valley-extended non-single-valley continuous solution of (4) if (i) 120593 is a single-valley-extended continuous solution of (4) (ii) 120593 has at leastan extreme point on (0 119891(1))

In the following we always let 120582 = 119891(1) = 120593119901minus1(1) and

define the sets

Δ 119896 = [119891119896+1

(1) 119891119896(1)]

Δ1

119896= [119891119896(120572) 119891

119896(1)]

Δ2

119896= [119891119896+1

(1) 119891119896(120572)]

Δ120572

119896= [119891119896+1

(120572) 119891119896(120572)]

forall119896 ge 0

(5)

Obviously

Δ 119896 = Δ1

119896cup Δ2

119896 Δ

120572

119896= Δ1

119896+1cup Δ2

119896 (6)

and from the fact that 119891119896(1) and 119891119896(120572) are respectivelystrictly decreasing and lim119896rarr+infin119891

119896(1) = 0 lim119896rarr+infin119891

119896(120572) =

0 then

[0 1] =

+infin

⋃

119896=0

Δ 119896 =

+infin

⋃

119896=0

(Δ1

119896cup Δ2

119896)

[0 120572] =

+infin

⋃

119896=0

Δ120572

119896=

+infin

⋃

119896=0

(Δ1

119896+1cup Δ2

119896)

(7)

Lemma 3 Suppose that 120593(119909) is a single-valley-extended non-single-valley continuous solution of (4) and 120572 is the extremepoint of 120593 in (120582 1) Then the following conclusions hold

(i) 120593(119909) has a unique minimum point 120572 with 120593(120572) = 0(ii) 0 is a recurrent but not periodic point of 120593(iii) 120582 is an extreme point of 120593 and 120593(120582) gt 120582(iv) 120593(119909) has a unique fixed point 120573 = 120593(120573) on [0 1] and

120593119901minus1

(1) = 119891 (1) = 120582 lt 120573 lt 120572 (8)

(v) for 119909 isin [0 120582] and 0 le 119894 le 119901 minus 1 then 120593119894(119909) = 120572 if andonly if 119909 = 119891(120572) and 119894 = 119901 minus 1

Lemma 4 Suppose that 120593(119909) is a single-valley-extended non-single-valley continuous solution of (4) Let 119869 = [0 120582] 1198690 =120593(119869) and 119869119894 = 120593119894(1198690) then the following conclusions hold

(i) 1198690 1198691 119869119901minus2 sub (120582 1] are pairwise disjoint

(ii) for all 119894 = 0 1 119901 minus 2 then 120593119894 1198690 997891rarr 119869119894 is ahomeomorphism

Lemma 5 Suppose that 120593(119909) is a single-valley-extended non-single-valley continuous solution of (4) Then for all 119899 ge 1120593 has infinite many extreme points 119891119899(1) and 119891119899(120572) 119891119899(1) isthe maximum point of 120593 on [119891119899(120572) 120572] 119891119899(120572) is the minimumpoint of 120593 on [0 119891119899(1)]

Lemma 6 Suppose that 120593(119909) is a single-valley-extended non-single-valley continuous solution of (4) Then the equation120593119901minus1(119909) = 119891(119909) has only one solution 119909 = 1 on (120593(119891(120572)) 1]

Lemma7 Let12059311205932 be two single-valley-extended non-single-valley continuous solutions of (4) If

1205931 (119909) = 1205932 (119909) 119909 isin [120582 1] (9)

then 1205931(119909) = 1205932(119909) on [0 1]

3 Constructive Method of Solutions

In this section we will prove constructively the existenceof single-valley-extended non-single-valley continuous solu-tions of (4)

Theorem 8 Fix a strictly increasing continuous function 119891(119909)on [0 1] with 119891(0) = 0 119891(119909) lt 119909 (119909 isin (0 1]) Denote 119891(1) =120582 If 1205930(119909) is a continuous function on [120582 1] and satisfies thefollowing conditions

(i) there exists an 120572 isin (120582 1) such that 1205930(120572) = 0 and 1205930is strictly decreasing on [120582 120572] and strictly increasing on[120572 1]

(ii) 120593119901minus10(1) = 119891(1) = 120582 120593119901

0(120582) = 119891(1205930(1))

(iii) there exists an 1205720 isin (120582 1] with 120593119901minus1

0(1205720) = 0 1205930(120582) gt

1205720 denote 1198690 = [1205720 1] 119869119894 = 1205931198940(1198690) then

(a) 1198690 1198691 119869119901minus2 sub (120582 1] are pairwise disjoint(b) for all 119894 = 0 1 119901 minus 2 then 120593119894

0 1198690 997891rarr 119869119894 is a

homeomorphism(c) 120572 is an endpoint of 119869119901minus2 and 120572 = 120593

119901minus2

0(1205720)

(iv) the equation 120593119901minus10(119909) = 119891(119909) has only one solution 119909 =

1 on (1205720 1]

then there exists a uniquely single-valley-extended non-single-valley continuous function 120593(119909) satisfying the equation

119891 (120593 (119909)) = 120593119901(119891 (119909)) 119909 isin [0 1]

120593 (119909) = 1205930 (119909) 119909 isin [120582 1] (10)

Abstract and Applied Analysis 3

And 120593 has infinitely many extreme points Conversely if 1205930 isthe restriction on [120582 1] of a single-valley-extended non-single-valley continuous solution to (4) then conditions (119894)ndash(119894V) abovemust hold

Proof Suppose that 1205930 satisfies conditions (i)ndash(iv) Define120595+ = 120593

119901minus1

0|[1205720 1]

By condition (iii) we have that 120595+ is a ho-meomorphism And by 120595+(1205720) = 120593

119901minus1

0(1205720) = 0 le 120595+(1) we

know that 120595+ is strictly increasingFirstly we define 120593 on Δ 119896 by induction Obviously

120593 = 1205930 is well defined on Δ 0 Suppose that 120593(119909) is welldefined as120593119896(119909) onΔ 119896 and strictly increasing and decreasingrespectively on Δ1

119896and Δ2

119896for all 119896 le 119898 where 119898 ge 0 is a

certain integer Let

120593119898+1 (119909) = 120595minus1

+(119891 (120593119898 (119891

minus1(119909)))) (119909 isin Δ119898+1) (11)

then 120593(119909) is well defined as 120593119898+1(119909) on Δ119898+1 and strictlyincreasing and decreasing respectively on Δ1

119898+1and Δ2

119898+1

Thereby 120593(119909) is well defined as a continuous function 120593119896(119909)on Δ 119896 and strictly increasing and decreasing respectively onΔ1

119896and Δ2

119896for all 119896 ge 0 by induction And

120595+ (120593119896+1 (119891 (119909))) = 119891 (120593119896 (119909)) 119909 isin Δ 119896 (12)

Secondly we prove that 120593119896 and 120593119896+1 have the same valueon the common endpoint119891119896+1(1) ofΔ 119896 andΔ 119896+1 for all 119896 ge 0From condition (ii) we have

120595+ (1205930 (120582)) = 120593119901minus1

0(1205930 (120582)) = 120593

119901

0(120582) = 119891 (1205930 (1)) (13)

And letting119898 = 0 119909 = 120582 in (11) we get

1205931 (119891 (1)) = 120595minus1

+(119891 (1205930 (1))) = 1205930 (120582) = 1205930 (119891 (1)) (14)

That is 1205930 and 1205931 have the same value on the commonendpoint 119891(1) = 120582 of Δ 0 and Δ 1 Suppose that

120593119898 (119891119898(1)) = 120593119898minus1 (119891

119898(1)) (15)

where119898 ge 1 is a certain integer Let 119909 = 119891119898+1(1) in (11) thenwe have

120593119898+1 (119891119898+1

(1)) = 120595minus1

+(119891 (120593119898 (119891

119898(1))))

= 120595minus1

+(119891 (120593119898minus1 (119891

119898(1))))

= 120593119898 (119891119898+1

(1))

(16)

That is 120593119896 and 120593119896+1 have the same value on the commonendpoint 119891119896+1(1) of Δ 119896 and Δ 119896+1 for all 119896 ge 0 by inductionTherefore we can let

120593 (119909) = 1 (119909 = 0)

120593119896 (119909) (119909 isin Δ 119896) (17)

Since 120593119896 is continuous on Δ 119896 and increasing and decreasingrespectively on Δ1

119896and Δ2

119896for 119896 ge 0 and (14) (15) and (16)

we have that 120593 is a non-single-valley continuous function andhas infinitely many extreme points on (0 1]

Thirdly we prove that 120593 is continuous at 119909 = 0

as follows It is trivial that 119891119896(120572) is strictly decreasingand lim119896rarrinfin119891

119896(120572) = 0 We prove 120593119896(119891

119896(120572))|infin

119896=1is

strictly increasing on [1205720 1] by induction as follows Since119891(1205931(119891(120572))) gt 0 and from (11) we get

1205932 (1198912(120572)) = 120595

minus1

+(119891 (1205931 (119891 (120572))))

gt 120595minus1

+(0) = 120595

minus1

+(119891 (1205930 (120572))) = 1205931 (119891 (120572))

(18)

Suppose that 120593119898(119891119898(120572)) gt 120593119898minus1(119891

119898minus1(120572)) where 119898 ge 2 is a

certain integer Therefore by (11) and the fact that 120595minus1+∘ 119891 is

strictly increasing we have that

120593119898+1 (119891119898+1

(120572))

= 120595minus1

+(119891 (120593119898 (119891

119898(120572))))

gt 120595minus1

+(119891 (120593119898minus1 (119891

119898minus1(120572)))) = 120593119898 (119891

119898(120572))

(19)

Thereby 120593119896(119891119896(120572))|infin

119896=1is strictly increasing in [1205720 1] by

induction Let lim119896rarrinfin120593119896(119891119896(120572)) = 120574 then 120574 isin [1205720 1] From

(12) we have that

120593119901minus1

0(120593119896+1 (119891

119896+1(120572))) = 120595+ (120593119896+1 (119891

119896+1(120572)))

= 119891 (120593119896 (119891119896(120572)))

(20)

Let 119896 rarr infin we get 120593119901minus10(120574) = 119891(120574) By condition (iv) we

know 120574 = 1 = 120593(0) This proves that 120593 is continuous at 119909 = 0Thereby 120593 is a continuous function on [0 1] We have that120593(119909) satisfies (10) by (12) and 120593(119909) is unique from Lemma 7

Obviously if 1205930 is the restriction on [120582 1] of a single-valley-extended non-single-valley continuous solution to(4) then conditions (i)ndash(iv) must hold by the lemmas inSection 2

Example 9 Let 1205930(119909) [14 1] 997891rarr [0 1] be defined by

1205930 (119909) =

minus13

8119909 +

39

32 (

1

4le 119909 le

3

4)

119909 minus3

4 (

3

4le 119909 le 1)

(21)

Obviously 1205930 satisfies the conditions of Theorem 8 with 119901 =2 119891(119909) = 1199094 and 120582 = 119891(1) = 14 120572 = 34 Hence it is therestriction to [14 1] of a single-valley-extended continuoussolution 120593 to (4) Since 1205930 has the minimum point 120572 = 34

and 1205930(14) = 1316 gt 34 120593 is a single-valley-extendednon-single-valley continuous solution Its graph is depictedin Figure 1

Example 10 Let 1205930(119909) [15 1] 997891rarr [0 1] be defined by

1205930 (119909) =

minus109

25119909 +

218

125 (

1

5le 119909 le

2

5)

119909 minus2

5 (

2

5le 119909 le 1)

(22)

4 Abstract and Applied Analysis

x

120601(120582)

120601(f(120582))

120601(f(120572))

120601(x)

120601(1)

f(120572)f(120582)

Non-single-valley solution

1

1

0 120572120582

Figure 1 The graph of non-single-valley solution

Obviously 1205930 satisfies the conditions of Theorem 8 with 119901 =3 119891(119909) = 11990925 and 120582 = 119891(1) = 15 120572 = 25 Hence it is therestriction to [15 1] of a single-valley-extended continuoussolution 120593 to (4) Since 1205930 has the minimum point 120572 = 25

and 1205930(15) = 109125 gt 25 12059320(15) = 59125 gt 25

120593 is a single-valley-extended non-single-valley continuoussolution Its graph is similar to Figure 1

Appendix

Proof of Lemma 3 (i) Suppose that 120574 is a minimum point of120593 By (4) we have

119891 (120593 (120574)) = 120593119901(119891 (120574)) ge 120593 (120574) (A1)

And from 119891(0) = 0 119891(119909) lt 119909 (119909 isin (0 1]) we know 120593(120574) = 0If 120574 lt 120582 then for all 119894 = 1 2 we have 120593119894([0 120582]) = [0 1]Thus there exists 1199090 isin [0 120582] such that 120593119901minus1(1199090) = 0 And by(4) we have

119891 (120593 (119891minus1(1199090))) = 120593

119901(1199090) = 120593 (0) = 1 (A2)

This contradicts 119891(0) = 0 119891(119909) lt 119909 (119909 isin (0 1]) Thus 120574 gt 120582And from the definition of 120572 we know 120574 = 120572 and 120593(120572) =120593(120574) = 0

(ii) We now prove that for all 119899 ge 0 and each 119909 isin [0 1]we have

119891119899(120593 (119909)) = 120593

119901119899

(119891119899(119909)) (A3)

Obviously (A3) holds for 119899 = 1 by (4) Suppose that (A3)holds for 119899 le 119896 where 119896 is a certain integer Therefore byinduction and (4) we have that

120593119901119896+1

(119891119896+1

(119909))

= (120593119901119896

)119901

(119891119896+1

(119909)) = (120593119901119896

)119901minus1

∘ 120593119901119896

(119891119896+1

(119909))

= (120593119901119896

)119901minus1

(119891119896(120593 (119891 (119909))))

= (120593119901119896

)119901minus2

∘ 120593119901119896

(119891119896(120593 (119891 (119909))))

= (120593119901119896

)119901minus2

(119891119896(1205932(119891 (119909)))) = sdot sdot sdot

= (120593119901119896

)119901minus119894

(119891119896(120593119894(119891 (119909)))) = sdot sdot sdot

= 119891119896(120593119901(119891 (119909))) = 119891

119896(119891 (120593 (119909))) = 119891

119896+1(120593 (119909))

(A4)

that is (A3) holds for 119899 = 119896 + 1 Thereby (A3) is proved byinduction Let 119909 = 0 in (A3) we have that

119891119899(1) = 119891

119899(120593 (0)) = 120593

119901119899

(119891119899(0)) = 120593

119901119899

(0) (A5)

And it is trivial that 119891119899(1) is strictly decreasing andlim119899rarr+infin119891

119899(1) = 0 Thereby we have that

lim119899rarr+infin

120593119901119899

(0) = lim119899rarr+infin

119891119899(1) = 0 (A6)

that is we proved that 0 is a recurrent but not periodic pointof 120593

(iii) Firstly we prove that120582 is an extreme point of120593 By thefact that 120593 is strictly increasing on [120572 1] and (4) we have that120593119901 is strictly increasing in [119891(120572) 119891(1)] Thereby 120593 is strictly

monotone in [119891(120572) 119891(1)] Suppose that 120582 is not an extremepoint then 120593 is strictly decreasing on [119891(120572) 120572] We prove 120593is strictly decreasing on Δ120572

119896(119896 ge 0) respectively as follows

Obviously 120593 is strictly decreasing on Δ1205720 Suppose that 120593

is strictly decreasing on Δ120572119896for all 119896 le 119898 where 119898 ge 0 is a

certain integer By (4) we have that 120593119901 is strictly decreasingon Δ120572119898+1

Thereby 120593 is strictly monotone on Δ120572119898+1

Supposethat 120593 is strictly increasing on Δ120572

119898+1 then we claim that

120593 (119891119899(120572)) lt 120593 (119891

119898+1(120572)) forall119899 ge 119898 + 2 (A7)

We first prove that

120593 (119891119904(120572)) = 120593 (119891

119897(120572)) forall119904 = 119897 (A8)

Suppose that there exists 119904 = 119897 such that120593(119891119904(120572)) = 120593(119891119897(120572)) =120575 And from (A3) we get

120593119901119904minus1(120575) = 120593

119901119904

(119891119904(120572)) = 119891

119904(120593 (120572)) = 119891

119904(0) = 0 (A9)

Similarly we have 120593119901119897minus1(120575) = 0 If 119904 gt 119897 then

0 = 120593119901119904minus1(120575) = 120593

(119901119904minus119901119897)+(119901119897minus1)(120575) = 120593

(119901119904minus119901119897)(0) (A10)

Abstract and Applied Analysis 5

This contradicts that 0 is not a periodic point That iswe proved (A8) Thereby we have 120593(119891119899(120572)) = 120593(119891

119898+1(120572))

forall119899 ge 119898 + 2 And from the fact that 120593 is strictly increasingon Δ120572119898+1

we get 120593(119891119898+2(120572)) lt 120593(119891119898+1

(120572)) Suppose that120593(119891119899(120572)) lt 120593(119891

119898+1(120572)) where 119899 ge 119898 + 2 is a certain integer

If 120593(119891119899+1(120572)) gt 120593(119891119898+1(120572)) gt 120593(119891119899(120572)) and by the fact that120593119901119898+1minus1 is strictlymonotone on120593([119891119899+1(120572) 119891119899(120572)]) and (A3)

we get

120593119901119898+1minus1(120593 (119891

119899+1(120572)))

lt 120593119901119898+1minus1(120593 (119891

119898+1(120572)))

= 120593119901119898+1

(119891119898+1

(120572)) = 119891119898+1

(0) = 0

(A11)

or

120593119901119898+1minus1(120593 (119891119899(120572)))

lt 120593119901119898+1minus1(120593 (119891

119898+1(120572)))

= 120593119901119898+1

(119891119898+1

(120572)) = 119891119898+1

(0) = 0

(A12)

This contradicts that 120593119901119898+1minus1(119909) ge 0 That is 120593(119891119899+1(120572)) lt

120593(119891119898+1

(120572)) Thereby we proved (A7) by induction If120593(119891119898+1

(120572)) = 1 then by (A3) we have

120593119901119898+1

(1) = 120593119901119898+1

(120593 (119891119898+1

(120572))) = 120593 (120593119901119898+1

(119891119898+1

(120572)))

= 120593 (119891119898+1

(120593 (120572))) = 120593 (0) = 1

(A13)

This contradicts conclusion (ii) Thereby we get 120593(119891119899(120572)) lt120593(119891119898+1

(120572)) lt 1 forall119899 ge 119898 + 2 This contradicts that 120593(0) =1 Thereby 120593 is strictly decreasing on Δ120572

119898+1 Furthermore

we proved that 120593 is strictly decreasing on Δ120572

119896(119896 ge 0)

respectively by induction that is 120593 is single-valley solutionof (4) This contradicts the condition that 120593 is a non-single-valley solution of (4) That is we proved that 120582 is an extremepoint of 120593

Secondly we prove that 120593(120582) gt 120582 We claim that

120593 (119891119899(1)) = 120582 120593 (119891

119899(120572)) = 120582 forall119899 ge 1 (A14)

Suppose that 120593(119891119899(1)) = 120582 by (A5) we get 120593(119891119899(1)) =

120593119901119899+1(0) And from 120593

119901(0) = 120582 we have that 120593119901(0) = 120593119901

119899+1(0)

Let

119871 = 0 120593 (0) 120593119901(0) 120593

119901+1(0) 120593

119901119899

(0) (A15)

obviously 119871 is a limited set and 120593119894(0) isin 119871 forall119894 isin 119885

+By (A5) we have that 119891119894(1) = 120593

119901119894

(0) isin 119871 forall119894 isin 119885+

This contradicts that 119871 is limited Thereby we proved that120593(119891119899(1)) = 120582 Suppose that 120593(119891119899(120572)) = 120582 from (A3) we have

0 = 119891119899(120593 (120572)) = 120593

119901119899

(119891119899(120572))

= 120593119901119899minus1(120582) = 120593

119901119899+119901minus1

(0)

(A16)

This contradicts that 0 is not a periodic point of 120593 That is(A14) holds Suppose that 120593(120582) lt 120582 Since 120582 is an extremepoint of 120593 we have that 120593 is strictly increasing on [119891(120572) 120582]Furthermore we have 120593(119891(120572)) lt 120593(120582) lt 120582 We claim that

120593 (119891119899(1)) lt 120582 120593 (119891

119899(120572)) lt 120582 forall119899 ge 1 (A17)

It is trivial that (A17) holds for 119899 = 1 Suppose that (A17)holds for 119899 = 119898 where 119898 is a certain integer By (A3)we know 120593

119901119898

is strictly monotone on [119891119898+1(1) 119891119898(120572)] If120593(119891119898+1

(1)) gt 120582 then there exists 1199090 isin (119891119898+1

(1) 119891119898(120572))

such that 120593(1199090) = 120582 Thus 1199090 is an extreme point of1205932 on (119891

119898+1(1) 119891119898(120572)) That is 1205932 is not monotone on

[119891119898+1

(1) 119891119898(120572)] Furthermore 120593119901

119898

is not monotone on[119891119898+1

(1) 119891119898(120572)] This contradicts that 120593119901

119898

is monotone on[119891119898+1

(1) 119891119898(120572)] Thus 120593(119891119898+1(1)) lt 120582 Similarly we have

120593(119891119898+1

(120572)) lt 120582 That is (A17) holds for 119899 = 119898 + 1 Thereby(A17) holds by induction This contradicts that 120593(0) = 1Thus 120593(120582) gt 120582

(iv) Firstly suppose that 119902 is a fixed point of 120593 then by(A5) we have 119902 = 1 And by 120593(120572) = 0 we have 119902 = 120572 If 119902 isin(120572 1) then 119902 = 120593(119902) lt 120593(1) And by induction for all119898 ge 0we have 119902 = 120593119898(119902) lt 120593119898(1) Specially

119902 = 120593119901119899minus1(119902) lt 120593

119901119899minus1(1) = 120593

119901119899minus1(120593 (0)) = 120593

119901119899

(0) (A18)

This contradicts (A5) Thereby 119902 lt 120572Secondly there exists at least one fixed point 120573 isin (0 120572) by

120593(0) = 1 120593(120572) = 0 We prove 120573 is the unique fixed point asfollows We claim that

120573 notin (119891119899+1

(1) 119891119899(1)] forall119899 ge 1 (A19)

Suppose that there exists119898 ge 1 such that120573 isin [119891119898(120572) 119891119898(1)]then by (A3) we have

119891119898(120593 (119891minus119898(120573))) = 120593

119901119898

(119891119898(119891minus119898(120573))) = 120593

119901119898

(120573) = 120573

(A20)

that is 120593(119891minus119898(120573)) = 119891minus119898(120573) And from 119891minus119898(120573) isin [120572 1] this

contradicts that 120593 does not have a fixed point on [120572 1] Thuswe have 120573 notin [119891119899(120572) 119891119899(1)] forall119899 ge 1 Suppose that there exists119898 ge 1 such that 120573 isin (119891119898+1(1) 119891119898(120572)) then by (A3) we have

119891119898minus1

(120593 (1198911minus119898

(120573))) = 120593119901119898minus1(119891119898minus1

(1198911minus119898

(120573)))

= 120593119901119898minus1(120573) = 120573

(A21)

Thereby 120593(1198911minus119898(120573)) = 1198911minus119898

(120573) That is 1198911minus119898(120573) isin (1198912(1)119891(120572)) is a fixed point of 120593 Since 120593 has no fixed pointon [119891(120572) 119891(1)] and 120593(120582) gt 120582 we have that 120593 must bestrictly increasing on [1198912(1) 119891(120572)] And 120593 has the fixed point1198911minus119898

(120573) isin (1198912(1) 119891(120572)) thus 120593119901 is strictly increasing on

[1198912(1) 119891(120572)] But by (4) and the fact that 120593 is decreasing on

[120582 120572] we have that 120593119901 is strictly decreasing on [1198912(1) 119891(120572)]This contradicts Thus we have 120573 notin [119891119899+1(1) 119891119899(120572)] forall119899 ge 1Thereby we prove (A19)Thus we have 120573 isin (120582 120572) And since120593 is decreasing on [120582 120572] we have that 120573 isin (120582 120572) is the uniquefixed point

6 Abstract and Applied Analysis

(v) By (4) we have 0 = 119891(120593(120572)) = 120593119901(119891(120572)) And

since 120572 is the unique minimum point of 120593 it follows that120593119901minus1(119891(120572)) = 120572 Thus the sufficiency is provedWe prove the necessity as follows Suppose that 120593119894(119909) = 120572

for some 119909 isin [0 120582] and 0 le 119894 le 119901 minus 1 Firstly we claim that

119909 notin (119891119899+1

(1) 119891119899(120572)) cup (119891

119899(120572) 119891

119899(1)) forall119899 ge 1

(A22)

Suppose that 119909 isin (119891119899(120572) 119891

119899(1)) then 120593119894+1 is not mono-

tone on (119891119899(120572) 119891

119899(1)) Thereby 120593119901 is not monotone on

(119891119899(120572) 119891

119899(1)) This contradicts that 120593119901 is monotone on

(119891119899(120572) 119891

119899(1)) by (A3) Thus 119909 notin (119891

119899(120572) 119891

119899(1)) We have

similarly that 119909 notin (119891119899+1(1) 119891119899(120572)) Thereby we prove (A22)that is 119909 = 119891119899(1) or 119909 = 119891119899(120572) forall119899 ge 1

Secondly we prove that

119909 = 119891119899(1) 119909 = 119891

119899+1(120572) forall119899 ge 1 (A23)

Suppose that 119909 = 119891119899(1) for some 119899 ge 1 then 120593119894(119891119899(1)) =

120572 And from (A5) we have 120572 = 120593119894(119891119899(1)) = 120593

119894(120593119901119899

(0)) =

120593119894+119901119899

(0) This contradicts that 0 is not a periodic point of 120593Thus we prove 119909 = 119891

119899(1) forall119899 ge 1 Suppose that 119909 = 119891119899(120572) for

some 119899 ge 2 then 120593119894(119891119899(120572)) = 120572 And from (A3) we have

120593119894(0) = 120593

119894(120593119901119899

(119891119899(120572))) = 120593

119901119899

(120593119894(119891119899(120572)))

= 120593119901119899

(120572) = 120593119901119899minus1(0)

(A24)

This contradicts that 0 is a recurrent point of120593Thuswe prove119909 = 119891119899(120572) forall119899 ge 2 Thereby (A23) holds That is we get 119909 =

119891(120572) and 120593119894(119891(120572)) = 120572Lastly we prove 119894 = 119901 minus 1 Suppose that there exists

119895 = 119894 0 le 119895 le 119901 minus 1 such that 120593119895(119891(120572)) = 120572 Then we cansuppose that 119895 lt 119894 and

0 = 120593119894+1(119891 (120572)) = 120593

119894minus119895(120593119895+1(119891 (120572))) = 120593

119894minus119895(0) (A25)

This contradicts that 0 is not a periodic point of 120593 Thus weprove 120593119895(119891(120572)) = 120572 forall119895 = 119894 Thereby we have 119894 = 119901 minus 1 and119909 = 119891(120572)

Proof of Lemma 4 (i) Firstly we prove that for all 119894 =

0 1 119901 minus 2 we have 119869119894 sub (120582 1] that is 119869119894 cap 119869 = 0 Weclaim that

120593119894+1(119891 (120572)) gt 120582 forall0 le 119894 le 119901 minus 2 (A26)

Suppose that there exists 1 le 119895 le 119901 minus 1 such that 120593119895(119891(120572)) =119909 le 120582 And from Lemma 3(v) we have 120572 = 120593

119901minus1(119891(120572)) =

120593119901minus1minus119895

(120593119895(119891(120572))) = 120593

119901minus1minus119895(119909) This contradicts Lemma 3(v)

Thus (A26) holds We claim that

120593119894+1(120582) gt 120582 forall0 le 119894 le 119901 minus 2 (A27)

It is trivial that 120593119894+1(120582) = 120582 by 0 is a recurrent point of120593 Suppose that there exists 1 le 119895 le 119901 minus 1 such that120593119895(120582) lt 120582 And from 120593

119895(119891(120572)) gt 120582 we have that there exists

119909 isin (119891(120572) 119891(1)) such that 120593119895(119909) = 120582 And from the fact that120582 is an extreme point of 120593 we get that 120593119895+1 is not monotoneon [119891(120572) 119891(1)] This contradicts that 120593119901 is strictly monotoneon [119891(120572) 119891(1)] Thus (A27) holds We claim that

120593119894+1(119909) gt 120582 forall0 le 119894 le 119901 minus 2 119909 isin [0 120582] (A28)

Suppose that there exists 119909 isin [0 120582) and 1 le 119895 le 119901 minus 1 suchthat 120593119895(119909) = 120582 And from (4) we get 120593119901minus119895(120582) = 120593

119901(119909) =

119891(120593(119891minus1(119909))) le 119891(1) = 120582 This contradicts (A27) Thus

120593119894+1(119909) = 120582 forall0 le 119894 le 119901 minus 2 119909 isin [0 120582] Suppose that there

exists 119909 isin [0 120582) and 1 le 119895 le 119901 minus 1 such that 120593119895(119909) lt 120582And from (A27) we have that there exists 119911 isin (119909 120582) suchthat 120593119895(119911) = 120582 And from (4) we get 120593119901minus119895(120582) = 120593

119901(119911) =

119891(120593(119891minus1(119911))) le 119891(1) = 120582 This contradicts (A27) Thus

(A28) holds That is 119869119894 cap 119869 = 0Secondly we prove that 119869119894 forall0 le 119894 le 119901 minus 2 are pairwise

disjoint Suppose that there exists 0 le 119894 lt 119895 le 119901minus 2 such that119869119894 cap 119869119895 = 119869119894119895 = 0 Let 119910 isin 119869119894119895 then there exist 119909119894 isin [0 120582] 119909119895 isin[0 120582] such that 120593119894+1(119909119894) = 119910 = 120593

119895+1(119909119895) Thereby we have

120593119901minus119895+119894

(119909119894) = 120593119901minus1minus119895

(120593119894+1(119909119894)) = 120593

119901minus1minus119895(120593119895+1(119909119895)) = 120593

119901(119909119895) =

119891(120593(119891minus1(119909119895))) le 119891(1) = 120582 This contradicts 119869119894 sub (120582 1] Thus

we proved that 1198690 1198691 119869119901minus2 are pairwise disjoint(ii) For all 119894 = 0 1 119901 minus 2 then 120593119894+1 119869 997891rarr 119869119894 is a

homeomorphismby Lemmas 3(i) and 3(v)Thereby120593119894 1198690 997891rarr119869119894 is also a homeomorphism

Proof of Lemma 5 120582 is a maximum point of 120593 on [119891(120572) 120572]by Lemma 3(iii) Suppose that 119891119898(1) is the maximum pointof 120593 on [119891119898(120572) 120572] where 119898 is some certain integer Firstlywe prove 119891119898(120572) is an extreme point of 120593 If otherwise 120593 isstrictly increasing on [119891119898+1(1) 119891119898(120572)] Thus 120593(119891119898+1(1)) lt120593(119891119898(120572)) and 120593 is strictly increasing on Δ119898 We claim that 120593

is strictly increasing on Δ 119896 for all 119896 ge 119898 It holds obviouslywhen 119896 = 119898 Suppose that 120593 is strictly increasing on Δ 119896 forall 119898 le 119896 le 119897 where 119897 is some certain integer 120593119901 is strictlyincreasing on Δ 119897+1 by (4) Thereby 120593 is strictly monotone onΔ 119897+1 If 120593 is strictly decreasing on Δ 119897+1 then we claim that

120593 (119891119904(1)) lt 120593 (119891

119897(120572)) forall119904 ge 119897 + 1 (A29)

Its proof is similar to (A7) and we omit it here Thereby120593(119891119904(1)) lt 120593(119891

119897(120572)) lt 1 forall119904 ge 119897 + 1 This contradicts

120593(0) = 1 Thus 120593 is strictly increasing on Δ 119897+1 Thereby 120593 isstrictly increasing on [0 119891119898(1)] by induction Furthermore120593(0) lt 120593(119891

119898(1)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is an extreme point of 120593Secondly we prove 119891119898(120572) is the minimum point of 120593 on

[0 119891119898(1)] Suppose that there exists 119897 ge 119898 + 1 such that

120593(119891119897(120572)) lt 120593(119891

119898(120572)) or 120593(119891119897(1)) lt 120593(119891119898(120572)) then we claim

that

120593 (119891119904(120572)) lt 120593 (119891

119898(120572)) forall119904 ge 119897 (A30)

or

120593 (119891119904(1)) lt 120593 (119891

119898(120572)) forall119904 ge 119897 (A31)

Abstract and Applied Analysis 7

The proof is similar to (A7) and we omit it here Thereby120593(0) lt 120593(119891

119898(120572)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is the minimum point of 120593 on [0 119891119898(1)]Thirdly we prove 119891119898+1(1) is an extreme point of 120593

120593119901minus1 is strictly increasing on [120593(119891119898(120572)) 120593(119891119898+1(1))] by (4)

and 120593 is strictly decreasing on [119891119898+1(1) 119891119898(120572)] And fromLemma 4(ii) we have that 120593119901minus1 is strictly increasing on 1198690 =[120593(119891(120572)) 1] Suppose that 119891119898+1(1) is not an extreme pointof 120593 then 120593 is strictly decreasing on [119891

119898+1(120572) 119891

119898+1(1)]

And from (4) and the fact that 120593 is strictly increasingon [119891

119898(120572) 119891

119898(1)] we get that 120593119901 is strictly increasing

on [119891119898+1

(120572) 119891119898+1

(1)] Thereby 120593119901minus1 is strictly decreasingon [120593(119891

119898+1(1)) 120593(119891

119898+1(120572))] This contradicts that 120593119901minus1 is

strictly increasing on 1198690 = [120593(119891(120572)) 1] Thus 119891119898+1(1) is theextreme point of 120593

Lastly we have that 119891119898+1(1) is the maximum point of 120593on [119891119898+1(120572) 120572] The proof is similar to 119891119898(120572) and we omit ithere Thereby 119891119899(1) is the maximum point of 120593 on [119891119899(120572) 120572]and 119891119899(120572) is the minimum point of 120593 on [0 119891119899(1)] for all 119899 ge1 by induction

Proof of Lemma 6 It is trivial that 119909 = 1 is a solution of theequation 120593119901minus1(119909) = 119891(119909) by (4) Suppose that 119909 = 1199090 isin

(120593(119891(120572)) 1] is an arbitrary solution of this equation that is120593119901minus1(1199090) = 119891(1199090) Since 120593([0 119891(120572)]) = [120593(119891(120572)) 1] we have

that there exists 119910 isin [0 119891(120572)] such that 120593(119910) = 1199090 We claimthat

120593 (119891119899(119910)) = 1199090 forall119899 ge 0 (A32)

Obviously (A32) holds for 119899 = 0 Suppose that (A32) holdsfor 119899 = 119896 where 119896 ge 0 is a certain integer Therefore by (4)we have that

120593119901minus1

(120593 (119891119896+1

(119910))) = 119891 (120593 (119891119896(119910))) = 119891 (1199090) = 120593

119901minus1(1199090)

(A33)

Since119891119896+1(119910) isin [0 119891(120572)) we have 120593(119891119896+1(119910)) isin (120593(119891(120572)) 1]And since 120593

119901minus1 is strictly increasing on [120593(119891(120572)) 1] by120593(119891(120572)) gt 120572 we have 120593(119891119896+1(119910)) = 1199090 That is (A32) holdsfor 119899 = 119896 + 1 Thereby (A32) is proved by induction By thefact that 119891119899(119910) is strictly decreasing and lim119899rarrinfin119891

119899(119910) = 0

we have 1199090 = lim119899rarr+infin120593(119891119899(119910)) = 120593(0) = 1

Proof of Lemma 7 There exist 120572 isin (120582 1) 120573 isin (120582 1) such that120593119894(120572) = 0 120593119894(120573) = 120573 (119894 = 1 2) by (8) Denote 1205930(119909) = 1205931(119909) =1205932(119909) (119909 isin [120582 1]) and 1205721 = 1205931(119891(120572)) lt 1205931(120582) = 1205930(120582) 1205722 =1205932(119891(120572)) lt 1205932(120582) = 1205930(120582) We prove that 1205721 = 1205722 as followsIt is trivial that 1205721 gt 120582 1205722 gt 120582 and

120593119901minus1

0(1205721) = 120593

119901minus1

1(1205721) = 120593

119901

1(119891 (120572)) = 0

120593119901minus1

0(1205722) = 120593

119901minus1

2(1205722) = 120593

119901

2(119891 (120572)) = 0

(A34)

by Lemmas 3(v) and 4 Since 120593119901minus10

is strictly monotone on[min1205721 1205722 1] we have that 1205721 = 1205722 Let 1205720 = 1205721 = 1205722 then1205720 lt 1205930(120582) and 120593

119901minus1

0(1205720) = 0 Define 120595+ = 120593

119901minus1

0|[1205720 1]

ByLemmas 3(v) and 4 we have that 120595+ is a homeomorphism

And by 120595+(1205720) = 0 le 120595+(1) we know that 120595+ is strictlyincreasing

We prove 1205931(119909) = 1205932(119909) on Δ 119896 for all 119896 ge 0 by inductionas follows

Obviously 1205931(119909) = 1205932(119909) holds on Δ 0 Suppose that1205931(119909) = 1205932(119909) holds on Δ 119896 for all 119896 le 119898 where 119898 ge 0 isa certain integer Let

120593 (119909) = 1205931 (119909) = 1205932 (119909) 119909 isin [119891119898+1

(1) 1] (A35)

Since 120593119894(119909) ge 120593119894(119891(120572)) = 1205720 gt 120582 for 119909 le 120582 and from (4) wehave

119891 (120593 (119891minus1(119909))) = 119891 (120593119894 (119891

minus1(119909))) = 120593

119901minus1

119894(120593119894 (119909))

= 120595+ (120593119894 (119909)) (119894 = 1 2 119909 isin Δ119898+1)

(A36)

Thereby

120593119894 (119909) = 120595minus1

+(119891 (120593 (119891

minus1(119909)))) (119894 = 1 2 119909 isin Δ119898+1)

(A37)

Thus we have 1205931(119909) = 1205932(119909) on Δ119898+1 Thus we get 1205931(119909) =1205932(119909) on Δ 119896 for all 119896 ge 0 by induction

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is partially supported by the National NaturalScience Foundation of China Tian Yuan Foundation (no11326129) and the Fundamental Research Funds for theCentral Universities (no 14CX02152A)

References

[1] M J Feigenbaum ldquoQuantitative universality for a class ofnonlinear transformationsrdquo Journal of Statistical Physics vol 19no 1 pp 25ndash52 1978

[2] M J Feigenbaum ldquoTheuniversalmetric properties of nonlineartransformationsrdquo Journal of Statistical Physics vol 21 no 6 pp669ndash706 1979

[3] P Couliet and C Tresser ldquoIterations drsquoendomorphismes etgroupe de renormalisationrdquo Journal de Physique Colloques vol39 no 5 pp C5-25ndashC5-28 1978

[4] D Sullivan ldquoBoubds quadratic differentials and renormaliza-tion conjecturesrdquo in Proceedings of the Mathematics into theTwenty-First Century Centennial Symposium F Browder Edpp 417ndash466 American Mathematical Society August 1992

[5] ADouady and JHHubbard ldquoOn the dynamics of polynomial-like mappingsrdquo Annales Scientifiques de lrsquoEcole NormaleSuperieure vol 18 no 2 pp 287ndash343 1985

[6] I Lanford ldquoA computer-assisted proof of the Feigenbaumconjecturesrdquo The American Mathematical Society vol 6 no 3pp 427ndash434 1982

8 Abstract and Applied Analysis

[7] J P Eckmann and P Wittwer ldquoA complete proof of theFeigenbaum conjecturesrdquo Journal of Statistical Physics vol 46no 3-4 pp 455ndash475 1987

[8] M Campanino and H Epstein ldquoOn the existence of Feigen-baumrsquos fixed pointrdquo Communications in Mathematical Physicsvol 79 no 2 pp 261ndash302 1981

[9] L Yang and J Z Zhang ldquoThe second type of Feigenbaumrsquosfunctional equationrdquo Science in China A vol 28 pp 1061ndash10691985

[10] P J McCarthy ldquoThe general exact bijective continuous solu-tion of Feigenbaumrsquos functional equationrdquo Communications inMathematical Physics vol 91 no 3 pp 431ndash443 1983

[11] J P Eckmann H Epstein and P Wittwer ldquoFixed pointsof Feigenbaumrsquos type for the equation 119891119901 (120582119909) = 120582119891(119909)rdquoCommunications in Mathematical Physics vol 93 no 4 pp495ndash516 1984

[12] A H Zhang L J Wang and W Song ldquoNonsingle-valley con-tinuous solutions of p-order Feigenbaums functional equationrdquoNortheastern Mathematical Journal vol 19 no 4 pp 375ndash3802003

[13] G Liao L Wang and Y Zhang ldquo119901-order Feigenbaumrsquos mapsrdquoNortheastern Mathematical Journal Dongbei Shuxue vol 20no 3 pp 284ndash290 2004

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Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

where 119891(119909) is a strictly increasing continuous function on[0 1] that satisfies 119891(0) = 0 119891(119909) lt 119909 (119909 isin (0 1])We will prove the existence of single-valley-extended non-single-valley continuous solutions of (4) by the constructivemethod Obviously let 119891(119909) = 120582119909 then (4) is (3)

2 Basic Definitions and Lemmas

In this section we will give some characterizations of single-valley-extended non-single-valley continuous solutions of(4) they will be proved in the appendix

Definition 1 One calls 120593 a single-valley-extended continuoussolution of (4) if (i) 120593 is a continuous solution of (4) (ii)there exists 120572 isin (119891(1) 1) such that 120593 is strictly decreasingon [119891(1) 120572] and strictly increasing on [120572 1]

Definition 2 One calls 120593 a single-valley-extended non-single-valley continuous solution of (4) if (i) 120593 is a single-valley-extended continuous solution of (4) (ii) 120593 has at leastan extreme point on (0 119891(1))

In the following we always let 120582 = 119891(1) = 120593119901minus1(1) and

define the sets

Δ 119896 = [119891119896+1

(1) 119891119896(1)]

Δ1

119896= [119891119896(120572) 119891

119896(1)]

Δ2

119896= [119891119896+1

(1) 119891119896(120572)]

Δ120572

119896= [119891119896+1

(120572) 119891119896(120572)]

forall119896 ge 0

(5)

Obviously

Δ 119896 = Δ1

119896cup Δ2

119896 Δ

120572

119896= Δ1

119896+1cup Δ2

119896 (6)

and from the fact that 119891119896(1) and 119891119896(120572) are respectivelystrictly decreasing and lim119896rarr+infin119891

119896(1) = 0 lim119896rarr+infin119891

119896(120572) =

0 then

[0 1] =

+infin

⋃

119896=0

Δ 119896 =

+infin

⋃

119896=0

(Δ1

119896cup Δ2

119896)

[0 120572] =

+infin

⋃

119896=0

Δ120572

119896=

+infin

⋃

119896=0

(Δ1

119896+1cup Δ2

119896)

(7)

Lemma 3 Suppose that 120593(119909) is a single-valley-extended non-single-valley continuous solution of (4) and 120572 is the extremepoint of 120593 in (120582 1) Then the following conclusions hold

(i) 120593(119909) has a unique minimum point 120572 with 120593(120572) = 0(ii) 0 is a recurrent but not periodic point of 120593(iii) 120582 is an extreme point of 120593 and 120593(120582) gt 120582(iv) 120593(119909) has a unique fixed point 120573 = 120593(120573) on [0 1] and

120593119901minus1

(1) = 119891 (1) = 120582 lt 120573 lt 120572 (8)

(v) for 119909 isin [0 120582] and 0 le 119894 le 119901 minus 1 then 120593119894(119909) = 120572 if andonly if 119909 = 119891(120572) and 119894 = 119901 minus 1

Lemma 4 Suppose that 120593(119909) is a single-valley-extended non-single-valley continuous solution of (4) Let 119869 = [0 120582] 1198690 =120593(119869) and 119869119894 = 120593119894(1198690) then the following conclusions hold

(i) 1198690 1198691 119869119901minus2 sub (120582 1] are pairwise disjoint

(ii) for all 119894 = 0 1 119901 minus 2 then 120593119894 1198690 997891rarr 119869119894 is ahomeomorphism

Lemma 5 Suppose that 120593(119909) is a single-valley-extended non-single-valley continuous solution of (4) Then for all 119899 ge 1120593 has infinite many extreme points 119891119899(1) and 119891119899(120572) 119891119899(1) isthe maximum point of 120593 on [119891119899(120572) 120572] 119891119899(120572) is the minimumpoint of 120593 on [0 119891119899(1)]

Lemma 6 Suppose that 120593(119909) is a single-valley-extended non-single-valley continuous solution of (4) Then the equation120593119901minus1(119909) = 119891(119909) has only one solution 119909 = 1 on (120593(119891(120572)) 1]

Lemma7 Let12059311205932 be two single-valley-extended non-single-valley continuous solutions of (4) If

1205931 (119909) = 1205932 (119909) 119909 isin [120582 1] (9)

then 1205931(119909) = 1205932(119909) on [0 1]

3 Constructive Method of Solutions

In this section we will prove constructively the existenceof single-valley-extended non-single-valley continuous solu-tions of (4)

Theorem 8 Fix a strictly increasing continuous function 119891(119909)on [0 1] with 119891(0) = 0 119891(119909) lt 119909 (119909 isin (0 1]) Denote 119891(1) =120582 If 1205930(119909) is a continuous function on [120582 1] and satisfies thefollowing conditions

(i) there exists an 120572 isin (120582 1) such that 1205930(120572) = 0 and 1205930is strictly decreasing on [120582 120572] and strictly increasing on[120572 1]

(ii) 120593119901minus10(1) = 119891(1) = 120582 120593119901

0(120582) = 119891(1205930(1))

(iii) there exists an 1205720 isin (120582 1] with 120593119901minus1

0(1205720) = 0 1205930(120582) gt

1205720 denote 1198690 = [1205720 1] 119869119894 = 1205931198940(1198690) then

(a) 1198690 1198691 119869119901minus2 sub (120582 1] are pairwise disjoint(b) for all 119894 = 0 1 119901 minus 2 then 120593119894

0 1198690 997891rarr 119869119894 is a

homeomorphism(c) 120572 is an endpoint of 119869119901minus2 and 120572 = 120593

119901minus2

0(1205720)

(iv) the equation 120593119901minus10(119909) = 119891(119909) has only one solution 119909 =

1 on (1205720 1]

then there exists a uniquely single-valley-extended non-single-valley continuous function 120593(119909) satisfying the equation

119891 (120593 (119909)) = 120593119901(119891 (119909)) 119909 isin [0 1]

120593 (119909) = 1205930 (119909) 119909 isin [120582 1] (10)

Abstract and Applied Analysis 3

And 120593 has infinitely many extreme points Conversely if 1205930 isthe restriction on [120582 1] of a single-valley-extended non-single-valley continuous solution to (4) then conditions (119894)ndash(119894V) abovemust hold

Proof Suppose that 1205930 satisfies conditions (i)ndash(iv) Define120595+ = 120593

119901minus1

0|[1205720 1]

By condition (iii) we have that 120595+ is a ho-meomorphism And by 120595+(1205720) = 120593

119901minus1

0(1205720) = 0 le 120595+(1) we

know that 120595+ is strictly increasingFirstly we define 120593 on Δ 119896 by induction Obviously

120593 = 1205930 is well defined on Δ 0 Suppose that 120593(119909) is welldefined as120593119896(119909) onΔ 119896 and strictly increasing and decreasingrespectively on Δ1

119896and Δ2

119896for all 119896 le 119898 where 119898 ge 0 is a

certain integer Let

120593119898+1 (119909) = 120595minus1

+(119891 (120593119898 (119891

minus1(119909)))) (119909 isin Δ119898+1) (11)

then 120593(119909) is well defined as 120593119898+1(119909) on Δ119898+1 and strictlyincreasing and decreasing respectively on Δ1

119898+1and Δ2

119898+1

Thereby 120593(119909) is well defined as a continuous function 120593119896(119909)on Δ 119896 and strictly increasing and decreasing respectively onΔ1

119896and Δ2

119896for all 119896 ge 0 by induction And

120595+ (120593119896+1 (119891 (119909))) = 119891 (120593119896 (119909)) 119909 isin Δ 119896 (12)

Secondly we prove that 120593119896 and 120593119896+1 have the same valueon the common endpoint119891119896+1(1) ofΔ 119896 andΔ 119896+1 for all 119896 ge 0From condition (ii) we have

120595+ (1205930 (120582)) = 120593119901minus1

0(1205930 (120582)) = 120593

119901

0(120582) = 119891 (1205930 (1)) (13)

And letting119898 = 0 119909 = 120582 in (11) we get

1205931 (119891 (1)) = 120595minus1

+(119891 (1205930 (1))) = 1205930 (120582) = 1205930 (119891 (1)) (14)

That is 1205930 and 1205931 have the same value on the commonendpoint 119891(1) = 120582 of Δ 0 and Δ 1 Suppose that

120593119898 (119891119898(1)) = 120593119898minus1 (119891

119898(1)) (15)

where119898 ge 1 is a certain integer Let 119909 = 119891119898+1(1) in (11) thenwe have

120593119898+1 (119891119898+1

(1)) = 120595minus1

+(119891 (120593119898 (119891

119898(1))))

= 120595minus1

+(119891 (120593119898minus1 (119891

119898(1))))

= 120593119898 (119891119898+1

(1))

(16)

That is 120593119896 and 120593119896+1 have the same value on the commonendpoint 119891119896+1(1) of Δ 119896 and Δ 119896+1 for all 119896 ge 0 by inductionTherefore we can let

120593 (119909) = 1 (119909 = 0)

120593119896 (119909) (119909 isin Δ 119896) (17)

Since 120593119896 is continuous on Δ 119896 and increasing and decreasingrespectively on Δ1

119896and Δ2

119896for 119896 ge 0 and (14) (15) and (16)

we have that 120593 is a non-single-valley continuous function andhas infinitely many extreme points on (0 1]

Thirdly we prove that 120593 is continuous at 119909 = 0

as follows It is trivial that 119891119896(120572) is strictly decreasingand lim119896rarrinfin119891

119896(120572) = 0 We prove 120593119896(119891

119896(120572))|infin

119896=1is

strictly increasing on [1205720 1] by induction as follows Since119891(1205931(119891(120572))) gt 0 and from (11) we get

1205932 (1198912(120572)) = 120595

minus1

+(119891 (1205931 (119891 (120572))))

gt 120595minus1

+(0) = 120595

minus1

+(119891 (1205930 (120572))) = 1205931 (119891 (120572))

(18)

Suppose that 120593119898(119891119898(120572)) gt 120593119898minus1(119891

119898minus1(120572)) where 119898 ge 2 is a

certain integer Therefore by (11) and the fact that 120595minus1+∘ 119891 is

strictly increasing we have that

120593119898+1 (119891119898+1

(120572))

= 120595minus1

+(119891 (120593119898 (119891

119898(120572))))

gt 120595minus1

+(119891 (120593119898minus1 (119891

119898minus1(120572)))) = 120593119898 (119891

119898(120572))

(19)

Thereby 120593119896(119891119896(120572))|infin

119896=1is strictly increasing in [1205720 1] by

induction Let lim119896rarrinfin120593119896(119891119896(120572)) = 120574 then 120574 isin [1205720 1] From

(12) we have that

120593119901minus1

0(120593119896+1 (119891

119896+1(120572))) = 120595+ (120593119896+1 (119891

119896+1(120572)))

= 119891 (120593119896 (119891119896(120572)))

(20)

Let 119896 rarr infin we get 120593119901minus10(120574) = 119891(120574) By condition (iv) we

know 120574 = 1 = 120593(0) This proves that 120593 is continuous at 119909 = 0Thereby 120593 is a continuous function on [0 1] We have that120593(119909) satisfies (10) by (12) and 120593(119909) is unique from Lemma 7

Obviously if 1205930 is the restriction on [120582 1] of a single-valley-extended non-single-valley continuous solution to(4) then conditions (i)ndash(iv) must hold by the lemmas inSection 2

Example 9 Let 1205930(119909) [14 1] 997891rarr [0 1] be defined by

1205930 (119909) =

minus13

8119909 +

39

32 (

1

4le 119909 le

3

4)

119909 minus3

4 (

3

4le 119909 le 1)

(21)

Obviously 1205930 satisfies the conditions of Theorem 8 with 119901 =2 119891(119909) = 1199094 and 120582 = 119891(1) = 14 120572 = 34 Hence it is therestriction to [14 1] of a single-valley-extended continuoussolution 120593 to (4) Since 1205930 has the minimum point 120572 = 34

and 1205930(14) = 1316 gt 34 120593 is a single-valley-extendednon-single-valley continuous solution Its graph is depictedin Figure 1

Example 10 Let 1205930(119909) [15 1] 997891rarr [0 1] be defined by

1205930 (119909) =

minus109

25119909 +

218

125 (

1

5le 119909 le

2

5)

119909 minus2

5 (

2

5le 119909 le 1)

(22)

4 Abstract and Applied Analysis

x

120601(120582)

120601(f(120582))

120601(f(120572))

120601(x)

120601(1)

f(120572)f(120582)

Non-single-valley solution

1

1

0 120572120582

Figure 1 The graph of non-single-valley solution

Obviously 1205930 satisfies the conditions of Theorem 8 with 119901 =3 119891(119909) = 11990925 and 120582 = 119891(1) = 15 120572 = 25 Hence it is therestriction to [15 1] of a single-valley-extended continuoussolution 120593 to (4) Since 1205930 has the minimum point 120572 = 25

and 1205930(15) = 109125 gt 25 12059320(15) = 59125 gt 25

120593 is a single-valley-extended non-single-valley continuoussolution Its graph is similar to Figure 1

Appendix

Proof of Lemma 3 (i) Suppose that 120574 is a minimum point of120593 By (4) we have

119891 (120593 (120574)) = 120593119901(119891 (120574)) ge 120593 (120574) (A1)

And from 119891(0) = 0 119891(119909) lt 119909 (119909 isin (0 1]) we know 120593(120574) = 0If 120574 lt 120582 then for all 119894 = 1 2 we have 120593119894([0 120582]) = [0 1]Thus there exists 1199090 isin [0 120582] such that 120593119901minus1(1199090) = 0 And by(4) we have

119891 (120593 (119891minus1(1199090))) = 120593

119901(1199090) = 120593 (0) = 1 (A2)

This contradicts 119891(0) = 0 119891(119909) lt 119909 (119909 isin (0 1]) Thus 120574 gt 120582And from the definition of 120572 we know 120574 = 120572 and 120593(120572) =120593(120574) = 0

(ii) We now prove that for all 119899 ge 0 and each 119909 isin [0 1]we have

119891119899(120593 (119909)) = 120593

119901119899

(119891119899(119909)) (A3)

Obviously (A3) holds for 119899 = 1 by (4) Suppose that (A3)holds for 119899 le 119896 where 119896 is a certain integer Therefore byinduction and (4) we have that

120593119901119896+1

(119891119896+1

(119909))

= (120593119901119896

)119901

(119891119896+1

(119909)) = (120593119901119896

)119901minus1

∘ 120593119901119896

(119891119896+1

(119909))

= (120593119901119896

)119901minus1

(119891119896(120593 (119891 (119909))))

= (120593119901119896

)119901minus2

∘ 120593119901119896

(119891119896(120593 (119891 (119909))))

= (120593119901119896

)119901minus2

(119891119896(1205932(119891 (119909)))) = sdot sdot sdot

= (120593119901119896

)119901minus119894

(119891119896(120593119894(119891 (119909)))) = sdot sdot sdot

= 119891119896(120593119901(119891 (119909))) = 119891

119896(119891 (120593 (119909))) = 119891

119896+1(120593 (119909))

(A4)

that is (A3) holds for 119899 = 119896 + 1 Thereby (A3) is proved byinduction Let 119909 = 0 in (A3) we have that

119891119899(1) = 119891

119899(120593 (0)) = 120593

119901119899

(119891119899(0)) = 120593

119901119899

(0) (A5)

And it is trivial that 119891119899(1) is strictly decreasing andlim119899rarr+infin119891

119899(1) = 0 Thereby we have that

lim119899rarr+infin

120593119901119899

(0) = lim119899rarr+infin

119891119899(1) = 0 (A6)

that is we proved that 0 is a recurrent but not periodic pointof 120593

(iii) Firstly we prove that120582 is an extreme point of120593 By thefact that 120593 is strictly increasing on [120572 1] and (4) we have that120593119901 is strictly increasing in [119891(120572) 119891(1)] Thereby 120593 is strictly

monotone in [119891(120572) 119891(1)] Suppose that 120582 is not an extremepoint then 120593 is strictly decreasing on [119891(120572) 120572] We prove 120593is strictly decreasing on Δ120572

119896(119896 ge 0) respectively as follows

Obviously 120593 is strictly decreasing on Δ1205720 Suppose that 120593

is strictly decreasing on Δ120572119896for all 119896 le 119898 where 119898 ge 0 is a

certain integer By (4) we have that 120593119901 is strictly decreasingon Δ120572119898+1

Thereby 120593 is strictly monotone on Δ120572119898+1

Supposethat 120593 is strictly increasing on Δ120572

119898+1 then we claim that

120593 (119891119899(120572)) lt 120593 (119891

119898+1(120572)) forall119899 ge 119898 + 2 (A7)

We first prove that

120593 (119891119904(120572)) = 120593 (119891

119897(120572)) forall119904 = 119897 (A8)

Suppose that there exists 119904 = 119897 such that120593(119891119904(120572)) = 120593(119891119897(120572)) =120575 And from (A3) we get

120593119901119904minus1(120575) = 120593

119901119904

(119891119904(120572)) = 119891

119904(120593 (120572)) = 119891

119904(0) = 0 (A9)

Similarly we have 120593119901119897minus1(120575) = 0 If 119904 gt 119897 then

0 = 120593119901119904minus1(120575) = 120593

(119901119904minus119901119897)+(119901119897minus1)(120575) = 120593

(119901119904minus119901119897)(0) (A10)

Abstract and Applied Analysis 5

This contradicts that 0 is not a periodic point That iswe proved (A8) Thereby we have 120593(119891119899(120572)) = 120593(119891

119898+1(120572))

forall119899 ge 119898 + 2 And from the fact that 120593 is strictly increasingon Δ120572119898+1

we get 120593(119891119898+2(120572)) lt 120593(119891119898+1

(120572)) Suppose that120593(119891119899(120572)) lt 120593(119891

119898+1(120572)) where 119899 ge 119898 + 2 is a certain integer

If 120593(119891119899+1(120572)) gt 120593(119891119898+1(120572)) gt 120593(119891119899(120572)) and by the fact that120593119901119898+1minus1 is strictlymonotone on120593([119891119899+1(120572) 119891119899(120572)]) and (A3)

we get

120593119901119898+1minus1(120593 (119891

119899+1(120572)))

lt 120593119901119898+1minus1(120593 (119891

119898+1(120572)))

= 120593119901119898+1

(119891119898+1

(120572)) = 119891119898+1

(0) = 0

(A11)

or

120593119901119898+1minus1(120593 (119891119899(120572)))

lt 120593119901119898+1minus1(120593 (119891

119898+1(120572)))

= 120593119901119898+1

(119891119898+1

(120572)) = 119891119898+1

(0) = 0

(A12)

This contradicts that 120593119901119898+1minus1(119909) ge 0 That is 120593(119891119899+1(120572)) lt

120593(119891119898+1

(120572)) Thereby we proved (A7) by induction If120593(119891119898+1

(120572)) = 1 then by (A3) we have

120593119901119898+1

(1) = 120593119901119898+1

(120593 (119891119898+1

(120572))) = 120593 (120593119901119898+1

(119891119898+1

(120572)))

= 120593 (119891119898+1

(120593 (120572))) = 120593 (0) = 1

(A13)

This contradicts conclusion (ii) Thereby we get 120593(119891119899(120572)) lt120593(119891119898+1

(120572)) lt 1 forall119899 ge 119898 + 2 This contradicts that 120593(0) =1 Thereby 120593 is strictly decreasing on Δ120572

119898+1 Furthermore

we proved that 120593 is strictly decreasing on Δ120572

119896(119896 ge 0)

respectively by induction that is 120593 is single-valley solutionof (4) This contradicts the condition that 120593 is a non-single-valley solution of (4) That is we proved that 120582 is an extremepoint of 120593

Secondly we prove that 120593(120582) gt 120582 We claim that

120593 (119891119899(1)) = 120582 120593 (119891

119899(120572)) = 120582 forall119899 ge 1 (A14)

Suppose that 120593(119891119899(1)) = 120582 by (A5) we get 120593(119891119899(1)) =

120593119901119899+1(0) And from 120593

119901(0) = 120582 we have that 120593119901(0) = 120593119901

119899+1(0)

Let

119871 = 0 120593 (0) 120593119901(0) 120593

119901+1(0) 120593

119901119899

(0) (A15)

obviously 119871 is a limited set and 120593119894(0) isin 119871 forall119894 isin 119885

+By (A5) we have that 119891119894(1) = 120593

119901119894

(0) isin 119871 forall119894 isin 119885+

This contradicts that 119871 is limited Thereby we proved that120593(119891119899(1)) = 120582 Suppose that 120593(119891119899(120572)) = 120582 from (A3) we have

0 = 119891119899(120593 (120572)) = 120593

119901119899

(119891119899(120572))

= 120593119901119899minus1(120582) = 120593

119901119899+119901minus1

(0)

(A16)

This contradicts that 0 is not a periodic point of 120593 That is(A14) holds Suppose that 120593(120582) lt 120582 Since 120582 is an extremepoint of 120593 we have that 120593 is strictly increasing on [119891(120572) 120582]Furthermore we have 120593(119891(120572)) lt 120593(120582) lt 120582 We claim that

120593 (119891119899(1)) lt 120582 120593 (119891

119899(120572)) lt 120582 forall119899 ge 1 (A17)

It is trivial that (A17) holds for 119899 = 1 Suppose that (A17)holds for 119899 = 119898 where 119898 is a certain integer By (A3)we know 120593

119901119898

is strictly monotone on [119891119898+1(1) 119891119898(120572)] If120593(119891119898+1

(1)) gt 120582 then there exists 1199090 isin (119891119898+1

(1) 119891119898(120572))

such that 120593(1199090) = 120582 Thus 1199090 is an extreme point of1205932 on (119891

119898+1(1) 119891119898(120572)) That is 1205932 is not monotone on

[119891119898+1

(1) 119891119898(120572)] Furthermore 120593119901

119898

is not monotone on[119891119898+1

(1) 119891119898(120572)] This contradicts that 120593119901

119898

is monotone on[119891119898+1

(1) 119891119898(120572)] Thus 120593(119891119898+1(1)) lt 120582 Similarly we have

120593(119891119898+1

(120572)) lt 120582 That is (A17) holds for 119899 = 119898 + 1 Thereby(A17) holds by induction This contradicts that 120593(0) = 1Thus 120593(120582) gt 120582

(iv) Firstly suppose that 119902 is a fixed point of 120593 then by(A5) we have 119902 = 1 And by 120593(120572) = 0 we have 119902 = 120572 If 119902 isin(120572 1) then 119902 = 120593(119902) lt 120593(1) And by induction for all119898 ge 0we have 119902 = 120593119898(119902) lt 120593119898(1) Specially

119902 = 120593119901119899minus1(119902) lt 120593

119901119899minus1(1) = 120593

119901119899minus1(120593 (0)) = 120593

119901119899

(0) (A18)

This contradicts (A5) Thereby 119902 lt 120572Secondly there exists at least one fixed point 120573 isin (0 120572) by

120593(0) = 1 120593(120572) = 0 We prove 120573 is the unique fixed point asfollows We claim that

120573 notin (119891119899+1

(1) 119891119899(1)] forall119899 ge 1 (A19)

Suppose that there exists119898 ge 1 such that120573 isin [119891119898(120572) 119891119898(1)]then by (A3) we have

119891119898(120593 (119891minus119898(120573))) = 120593

119901119898

(119891119898(119891minus119898(120573))) = 120593

119901119898

(120573) = 120573

(A20)

that is 120593(119891minus119898(120573)) = 119891minus119898(120573) And from 119891minus119898(120573) isin [120572 1] this

contradicts that 120593 does not have a fixed point on [120572 1] Thuswe have 120573 notin [119891119899(120572) 119891119899(1)] forall119899 ge 1 Suppose that there exists119898 ge 1 such that 120573 isin (119891119898+1(1) 119891119898(120572)) then by (A3) we have

119891119898minus1

(120593 (1198911minus119898

(120573))) = 120593119901119898minus1(119891119898minus1

(1198911minus119898

(120573)))

= 120593119901119898minus1(120573) = 120573

(A21)

Thereby 120593(1198911minus119898(120573)) = 1198911minus119898

(120573) That is 1198911minus119898(120573) isin (1198912(1)119891(120572)) is a fixed point of 120593 Since 120593 has no fixed pointon [119891(120572) 119891(1)] and 120593(120582) gt 120582 we have that 120593 must bestrictly increasing on [1198912(1) 119891(120572)] And 120593 has the fixed point1198911minus119898

(120573) isin (1198912(1) 119891(120572)) thus 120593119901 is strictly increasing on

[1198912(1) 119891(120572)] But by (4) and the fact that 120593 is decreasing on

[120582 120572] we have that 120593119901 is strictly decreasing on [1198912(1) 119891(120572)]This contradicts Thus we have 120573 notin [119891119899+1(1) 119891119899(120572)] forall119899 ge 1Thereby we prove (A19)Thus we have 120573 isin (120582 120572) And since120593 is decreasing on [120582 120572] we have that 120573 isin (120582 120572) is the uniquefixed point

6 Abstract and Applied Analysis

(v) By (4) we have 0 = 119891(120593(120572)) = 120593119901(119891(120572)) And

since 120572 is the unique minimum point of 120593 it follows that120593119901minus1(119891(120572)) = 120572 Thus the sufficiency is provedWe prove the necessity as follows Suppose that 120593119894(119909) = 120572

for some 119909 isin [0 120582] and 0 le 119894 le 119901 minus 1 Firstly we claim that

119909 notin (119891119899+1

(1) 119891119899(120572)) cup (119891

119899(120572) 119891

119899(1)) forall119899 ge 1

(A22)

Suppose that 119909 isin (119891119899(120572) 119891

119899(1)) then 120593119894+1 is not mono-

tone on (119891119899(120572) 119891

119899(1)) Thereby 120593119901 is not monotone on

(119891119899(120572) 119891

119899(1)) This contradicts that 120593119901 is monotone on

(119891119899(120572) 119891

119899(1)) by (A3) Thus 119909 notin (119891

119899(120572) 119891

119899(1)) We have

similarly that 119909 notin (119891119899+1(1) 119891119899(120572)) Thereby we prove (A22)that is 119909 = 119891119899(1) or 119909 = 119891119899(120572) forall119899 ge 1

Secondly we prove that

119909 = 119891119899(1) 119909 = 119891

119899+1(120572) forall119899 ge 1 (A23)

Suppose that 119909 = 119891119899(1) for some 119899 ge 1 then 120593119894(119891119899(1)) =

120572 And from (A5) we have 120572 = 120593119894(119891119899(1)) = 120593

119894(120593119901119899

(0)) =

120593119894+119901119899

(0) This contradicts that 0 is not a periodic point of 120593Thus we prove 119909 = 119891

119899(1) forall119899 ge 1 Suppose that 119909 = 119891119899(120572) for

some 119899 ge 2 then 120593119894(119891119899(120572)) = 120572 And from (A3) we have

120593119894(0) = 120593

119894(120593119901119899

(119891119899(120572))) = 120593

119901119899

(120593119894(119891119899(120572)))

= 120593119901119899

(120572) = 120593119901119899minus1(0)

(A24)

This contradicts that 0 is a recurrent point of120593Thuswe prove119909 = 119891119899(120572) forall119899 ge 2 Thereby (A23) holds That is we get 119909 =

119891(120572) and 120593119894(119891(120572)) = 120572Lastly we prove 119894 = 119901 minus 1 Suppose that there exists

119895 = 119894 0 le 119895 le 119901 minus 1 such that 120593119895(119891(120572)) = 120572 Then we cansuppose that 119895 lt 119894 and

0 = 120593119894+1(119891 (120572)) = 120593

119894minus119895(120593119895+1(119891 (120572))) = 120593

119894minus119895(0) (A25)

This contradicts that 0 is not a periodic point of 120593 Thus weprove 120593119895(119891(120572)) = 120572 forall119895 = 119894 Thereby we have 119894 = 119901 minus 1 and119909 = 119891(120572)

Proof of Lemma 4 (i) Firstly we prove that for all 119894 =

0 1 119901 minus 2 we have 119869119894 sub (120582 1] that is 119869119894 cap 119869 = 0 Weclaim that

120593119894+1(119891 (120572)) gt 120582 forall0 le 119894 le 119901 minus 2 (A26)

Suppose that there exists 1 le 119895 le 119901 minus 1 such that 120593119895(119891(120572)) =119909 le 120582 And from Lemma 3(v) we have 120572 = 120593

119901minus1(119891(120572)) =

120593119901minus1minus119895

(120593119895(119891(120572))) = 120593

119901minus1minus119895(119909) This contradicts Lemma 3(v)

Thus (A26) holds We claim that

120593119894+1(120582) gt 120582 forall0 le 119894 le 119901 minus 2 (A27)

It is trivial that 120593119894+1(120582) = 120582 by 0 is a recurrent point of120593 Suppose that there exists 1 le 119895 le 119901 minus 1 such that120593119895(120582) lt 120582 And from 120593

119895(119891(120572)) gt 120582 we have that there exists

119909 isin (119891(120572) 119891(1)) such that 120593119895(119909) = 120582 And from the fact that120582 is an extreme point of 120593 we get that 120593119895+1 is not monotoneon [119891(120572) 119891(1)] This contradicts that 120593119901 is strictly monotoneon [119891(120572) 119891(1)] Thus (A27) holds We claim that

120593119894+1(119909) gt 120582 forall0 le 119894 le 119901 minus 2 119909 isin [0 120582] (A28)

Suppose that there exists 119909 isin [0 120582) and 1 le 119895 le 119901 minus 1 suchthat 120593119895(119909) = 120582 And from (4) we get 120593119901minus119895(120582) = 120593

119901(119909) =

119891(120593(119891minus1(119909))) le 119891(1) = 120582 This contradicts (A27) Thus

120593119894+1(119909) = 120582 forall0 le 119894 le 119901 minus 2 119909 isin [0 120582] Suppose that there

exists 119909 isin [0 120582) and 1 le 119895 le 119901 minus 1 such that 120593119895(119909) lt 120582And from (A27) we have that there exists 119911 isin (119909 120582) suchthat 120593119895(119911) = 120582 And from (4) we get 120593119901minus119895(120582) = 120593

119901(119911) =

119891(120593(119891minus1(119911))) le 119891(1) = 120582 This contradicts (A27) Thus

(A28) holds That is 119869119894 cap 119869 = 0Secondly we prove that 119869119894 forall0 le 119894 le 119901 minus 2 are pairwise

disjoint Suppose that there exists 0 le 119894 lt 119895 le 119901minus 2 such that119869119894 cap 119869119895 = 119869119894119895 = 0 Let 119910 isin 119869119894119895 then there exist 119909119894 isin [0 120582] 119909119895 isin[0 120582] such that 120593119894+1(119909119894) = 119910 = 120593

119895+1(119909119895) Thereby we have

120593119901minus119895+119894

(119909119894) = 120593119901minus1minus119895

(120593119894+1(119909119894)) = 120593

119901minus1minus119895(120593119895+1(119909119895)) = 120593

119901(119909119895) =

119891(120593(119891minus1(119909119895))) le 119891(1) = 120582 This contradicts 119869119894 sub (120582 1] Thus

we proved that 1198690 1198691 119869119901minus2 are pairwise disjoint(ii) For all 119894 = 0 1 119901 minus 2 then 120593119894+1 119869 997891rarr 119869119894 is a

homeomorphismby Lemmas 3(i) and 3(v)Thereby120593119894 1198690 997891rarr119869119894 is also a homeomorphism

Proof of Lemma 5 120582 is a maximum point of 120593 on [119891(120572) 120572]by Lemma 3(iii) Suppose that 119891119898(1) is the maximum pointof 120593 on [119891119898(120572) 120572] where 119898 is some certain integer Firstlywe prove 119891119898(120572) is an extreme point of 120593 If otherwise 120593 isstrictly increasing on [119891119898+1(1) 119891119898(120572)] Thus 120593(119891119898+1(1)) lt120593(119891119898(120572)) and 120593 is strictly increasing on Δ119898 We claim that 120593

is strictly increasing on Δ 119896 for all 119896 ge 119898 It holds obviouslywhen 119896 = 119898 Suppose that 120593 is strictly increasing on Δ 119896 forall 119898 le 119896 le 119897 where 119897 is some certain integer 120593119901 is strictlyincreasing on Δ 119897+1 by (4) Thereby 120593 is strictly monotone onΔ 119897+1 If 120593 is strictly decreasing on Δ 119897+1 then we claim that

120593 (119891119904(1)) lt 120593 (119891

119897(120572)) forall119904 ge 119897 + 1 (A29)

Its proof is similar to (A7) and we omit it here Thereby120593(119891119904(1)) lt 120593(119891

119897(120572)) lt 1 forall119904 ge 119897 + 1 This contradicts

120593(0) = 1 Thus 120593 is strictly increasing on Δ 119897+1 Thereby 120593 isstrictly increasing on [0 119891119898(1)] by induction Furthermore120593(0) lt 120593(119891

119898(1)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is an extreme point of 120593Secondly we prove 119891119898(120572) is the minimum point of 120593 on

[0 119891119898(1)] Suppose that there exists 119897 ge 119898 + 1 such that

120593(119891119897(120572)) lt 120593(119891

119898(120572)) or 120593(119891119897(1)) lt 120593(119891119898(120572)) then we claim

that

120593 (119891119904(120572)) lt 120593 (119891

119898(120572)) forall119904 ge 119897 (A30)

or

120593 (119891119904(1)) lt 120593 (119891

119898(120572)) forall119904 ge 119897 (A31)

Abstract and Applied Analysis 7

The proof is similar to (A7) and we omit it here Thereby120593(0) lt 120593(119891

119898(120572)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is the minimum point of 120593 on [0 119891119898(1)]Thirdly we prove 119891119898+1(1) is an extreme point of 120593

120593119901minus1 is strictly increasing on [120593(119891119898(120572)) 120593(119891119898+1(1))] by (4)

and 120593 is strictly decreasing on [119891119898+1(1) 119891119898(120572)] And fromLemma 4(ii) we have that 120593119901minus1 is strictly increasing on 1198690 =[120593(119891(120572)) 1] Suppose that 119891119898+1(1) is not an extreme pointof 120593 then 120593 is strictly decreasing on [119891

119898+1(120572) 119891

119898+1(1)]

And from (4) and the fact that 120593 is strictly increasingon [119891

119898(120572) 119891

119898(1)] we get that 120593119901 is strictly increasing

on [119891119898+1

(120572) 119891119898+1

(1)] Thereby 120593119901minus1 is strictly decreasingon [120593(119891

119898+1(1)) 120593(119891

119898+1(120572))] This contradicts that 120593119901minus1 is

strictly increasing on 1198690 = [120593(119891(120572)) 1] Thus 119891119898+1(1) is theextreme point of 120593

Lastly we have that 119891119898+1(1) is the maximum point of 120593on [119891119898+1(120572) 120572] The proof is similar to 119891119898(120572) and we omit ithere Thereby 119891119899(1) is the maximum point of 120593 on [119891119899(120572) 120572]and 119891119899(120572) is the minimum point of 120593 on [0 119891119899(1)] for all 119899 ge1 by induction

Proof of Lemma 6 It is trivial that 119909 = 1 is a solution of theequation 120593119901minus1(119909) = 119891(119909) by (4) Suppose that 119909 = 1199090 isin

(120593(119891(120572)) 1] is an arbitrary solution of this equation that is120593119901minus1(1199090) = 119891(1199090) Since 120593([0 119891(120572)]) = [120593(119891(120572)) 1] we have

that there exists 119910 isin [0 119891(120572)] such that 120593(119910) = 1199090 We claimthat

120593 (119891119899(119910)) = 1199090 forall119899 ge 0 (A32)

Obviously (A32) holds for 119899 = 0 Suppose that (A32) holdsfor 119899 = 119896 where 119896 ge 0 is a certain integer Therefore by (4)we have that

120593119901minus1

(120593 (119891119896+1

(119910))) = 119891 (120593 (119891119896(119910))) = 119891 (1199090) = 120593

119901minus1(1199090)

(A33)

Since119891119896+1(119910) isin [0 119891(120572)) we have 120593(119891119896+1(119910)) isin (120593(119891(120572)) 1]And since 120593

119901minus1 is strictly increasing on [120593(119891(120572)) 1] by120593(119891(120572)) gt 120572 we have 120593(119891119896+1(119910)) = 1199090 That is (A32) holdsfor 119899 = 119896 + 1 Thereby (A32) is proved by induction By thefact that 119891119899(119910) is strictly decreasing and lim119899rarrinfin119891

119899(119910) = 0

we have 1199090 = lim119899rarr+infin120593(119891119899(119910)) = 120593(0) = 1

Proof of Lemma 7 There exist 120572 isin (120582 1) 120573 isin (120582 1) such that120593119894(120572) = 0 120593119894(120573) = 120573 (119894 = 1 2) by (8) Denote 1205930(119909) = 1205931(119909) =1205932(119909) (119909 isin [120582 1]) and 1205721 = 1205931(119891(120572)) lt 1205931(120582) = 1205930(120582) 1205722 =1205932(119891(120572)) lt 1205932(120582) = 1205930(120582) We prove that 1205721 = 1205722 as followsIt is trivial that 1205721 gt 120582 1205722 gt 120582 and

120593119901minus1

0(1205721) = 120593

119901minus1

1(1205721) = 120593

119901

1(119891 (120572)) = 0

120593119901minus1

0(1205722) = 120593

119901minus1

2(1205722) = 120593

119901

2(119891 (120572)) = 0

(A34)

by Lemmas 3(v) and 4 Since 120593119901minus10

is strictly monotone on[min1205721 1205722 1] we have that 1205721 = 1205722 Let 1205720 = 1205721 = 1205722 then1205720 lt 1205930(120582) and 120593

119901minus1

0(1205720) = 0 Define 120595+ = 120593

119901minus1

0|[1205720 1]

ByLemmas 3(v) and 4 we have that 120595+ is a homeomorphism

And by 120595+(1205720) = 0 le 120595+(1) we know that 120595+ is strictlyincreasing

We prove 1205931(119909) = 1205932(119909) on Δ 119896 for all 119896 ge 0 by inductionas follows

Obviously 1205931(119909) = 1205932(119909) holds on Δ 0 Suppose that1205931(119909) = 1205932(119909) holds on Δ 119896 for all 119896 le 119898 where 119898 ge 0 isa certain integer Let

120593 (119909) = 1205931 (119909) = 1205932 (119909) 119909 isin [119891119898+1

(1) 1] (A35)

Since 120593119894(119909) ge 120593119894(119891(120572)) = 1205720 gt 120582 for 119909 le 120582 and from (4) wehave

119891 (120593 (119891minus1(119909))) = 119891 (120593119894 (119891

minus1(119909))) = 120593

119901minus1

119894(120593119894 (119909))

= 120595+ (120593119894 (119909)) (119894 = 1 2 119909 isin Δ119898+1)

(A36)

Thereby

120593119894 (119909) = 120595minus1

+(119891 (120593 (119891

minus1(119909)))) (119894 = 1 2 119909 isin Δ119898+1)

(A37)

Thus we have 1205931(119909) = 1205932(119909) on Δ119898+1 Thus we get 1205931(119909) =1205932(119909) on Δ 119896 for all 119896 ge 0 by induction

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is partially supported by the National NaturalScience Foundation of China Tian Yuan Foundation (no11326129) and the Fundamental Research Funds for theCentral Universities (no 14CX02152A)

References

[1] M J Feigenbaum ldquoQuantitative universality for a class ofnonlinear transformationsrdquo Journal of Statistical Physics vol 19no 1 pp 25ndash52 1978

[2] M J Feigenbaum ldquoTheuniversalmetric properties of nonlineartransformationsrdquo Journal of Statistical Physics vol 21 no 6 pp669ndash706 1979

[3] P Couliet and C Tresser ldquoIterations drsquoendomorphismes etgroupe de renormalisationrdquo Journal de Physique Colloques vol39 no 5 pp C5-25ndashC5-28 1978

[4] D Sullivan ldquoBoubds quadratic differentials and renormaliza-tion conjecturesrdquo in Proceedings of the Mathematics into theTwenty-First Century Centennial Symposium F Browder Edpp 417ndash466 American Mathematical Society August 1992

[5] ADouady and JHHubbard ldquoOn the dynamics of polynomial-like mappingsrdquo Annales Scientifiques de lrsquoEcole NormaleSuperieure vol 18 no 2 pp 287ndash343 1985

[6] I Lanford ldquoA computer-assisted proof of the Feigenbaumconjecturesrdquo The American Mathematical Society vol 6 no 3pp 427ndash434 1982

8 Abstract and Applied Analysis

[7] J P Eckmann and P Wittwer ldquoA complete proof of theFeigenbaum conjecturesrdquo Journal of Statistical Physics vol 46no 3-4 pp 455ndash475 1987

[8] M Campanino and H Epstein ldquoOn the existence of Feigen-baumrsquos fixed pointrdquo Communications in Mathematical Physicsvol 79 no 2 pp 261ndash302 1981

[9] L Yang and J Z Zhang ldquoThe second type of Feigenbaumrsquosfunctional equationrdquo Science in China A vol 28 pp 1061ndash10691985

[10] P J McCarthy ldquoThe general exact bijective continuous solu-tion of Feigenbaumrsquos functional equationrdquo Communications inMathematical Physics vol 91 no 3 pp 431ndash443 1983

[11] J P Eckmann H Epstein and P Wittwer ldquoFixed pointsof Feigenbaumrsquos type for the equation 119891119901 (120582119909) = 120582119891(119909)rdquoCommunications in Mathematical Physics vol 93 no 4 pp495ndash516 1984

[12] A H Zhang L J Wang and W Song ldquoNonsingle-valley con-tinuous solutions of p-order Feigenbaums functional equationrdquoNortheastern Mathematical Journal vol 19 no 4 pp 375ndash3802003

[13] G Liao L Wang and Y Zhang ldquo119901-order Feigenbaumrsquos mapsrdquoNortheastern Mathematical Journal Dongbei Shuxue vol 20no 3 pp 284ndash290 2004

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Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

And 120593 has infinitely many extreme points Conversely if 1205930 isthe restriction on [120582 1] of a single-valley-extended non-single-valley continuous solution to (4) then conditions (119894)ndash(119894V) abovemust hold

Proof Suppose that 1205930 satisfies conditions (i)ndash(iv) Define120595+ = 120593

119901minus1

0|[1205720 1]

By condition (iii) we have that 120595+ is a ho-meomorphism And by 120595+(1205720) = 120593

119901minus1

0(1205720) = 0 le 120595+(1) we

know that 120595+ is strictly increasingFirstly we define 120593 on Δ 119896 by induction Obviously

120593 = 1205930 is well defined on Δ 0 Suppose that 120593(119909) is welldefined as120593119896(119909) onΔ 119896 and strictly increasing and decreasingrespectively on Δ1

119896and Δ2

119896for all 119896 le 119898 where 119898 ge 0 is a

certain integer Let

120593119898+1 (119909) = 120595minus1

+(119891 (120593119898 (119891

minus1(119909)))) (119909 isin Δ119898+1) (11)

then 120593(119909) is well defined as 120593119898+1(119909) on Δ119898+1 and strictlyincreasing and decreasing respectively on Δ1

119898+1and Δ2

119898+1

Thereby 120593(119909) is well defined as a continuous function 120593119896(119909)on Δ 119896 and strictly increasing and decreasing respectively onΔ1

119896and Δ2

119896for all 119896 ge 0 by induction And

120595+ (120593119896+1 (119891 (119909))) = 119891 (120593119896 (119909)) 119909 isin Δ 119896 (12)

Secondly we prove that 120593119896 and 120593119896+1 have the same valueon the common endpoint119891119896+1(1) ofΔ 119896 andΔ 119896+1 for all 119896 ge 0From condition (ii) we have

120595+ (1205930 (120582)) = 120593119901minus1

0(1205930 (120582)) = 120593

119901

0(120582) = 119891 (1205930 (1)) (13)

And letting119898 = 0 119909 = 120582 in (11) we get

1205931 (119891 (1)) = 120595minus1

+(119891 (1205930 (1))) = 1205930 (120582) = 1205930 (119891 (1)) (14)

That is 1205930 and 1205931 have the same value on the commonendpoint 119891(1) = 120582 of Δ 0 and Δ 1 Suppose that

120593119898 (119891119898(1)) = 120593119898minus1 (119891

119898(1)) (15)

where119898 ge 1 is a certain integer Let 119909 = 119891119898+1(1) in (11) thenwe have

120593119898+1 (119891119898+1

(1)) = 120595minus1

+(119891 (120593119898 (119891

119898(1))))

= 120595minus1

+(119891 (120593119898minus1 (119891

119898(1))))

= 120593119898 (119891119898+1

(1))

(16)

That is 120593119896 and 120593119896+1 have the same value on the commonendpoint 119891119896+1(1) of Δ 119896 and Δ 119896+1 for all 119896 ge 0 by inductionTherefore we can let

120593 (119909) = 1 (119909 = 0)

120593119896 (119909) (119909 isin Δ 119896) (17)

Since 120593119896 is continuous on Δ 119896 and increasing and decreasingrespectively on Δ1

119896and Δ2

119896for 119896 ge 0 and (14) (15) and (16)

we have that 120593 is a non-single-valley continuous function andhas infinitely many extreme points on (0 1]

Thirdly we prove that 120593 is continuous at 119909 = 0

as follows It is trivial that 119891119896(120572) is strictly decreasingand lim119896rarrinfin119891

119896(120572) = 0 We prove 120593119896(119891

119896(120572))|infin

119896=1is

strictly increasing on [1205720 1] by induction as follows Since119891(1205931(119891(120572))) gt 0 and from (11) we get

1205932 (1198912(120572)) = 120595

minus1

+(119891 (1205931 (119891 (120572))))

gt 120595minus1

+(0) = 120595

minus1

+(119891 (1205930 (120572))) = 1205931 (119891 (120572))

(18)

Suppose that 120593119898(119891119898(120572)) gt 120593119898minus1(119891

119898minus1(120572)) where 119898 ge 2 is a

certain integer Therefore by (11) and the fact that 120595minus1+∘ 119891 is

strictly increasing we have that

120593119898+1 (119891119898+1

(120572))

= 120595minus1

+(119891 (120593119898 (119891

119898(120572))))

gt 120595minus1

+(119891 (120593119898minus1 (119891

119898minus1(120572)))) = 120593119898 (119891

119898(120572))

(19)

Thereby 120593119896(119891119896(120572))|infin

119896=1is strictly increasing in [1205720 1] by

induction Let lim119896rarrinfin120593119896(119891119896(120572)) = 120574 then 120574 isin [1205720 1] From

(12) we have that

120593119901minus1

0(120593119896+1 (119891

119896+1(120572))) = 120595+ (120593119896+1 (119891

119896+1(120572)))

= 119891 (120593119896 (119891119896(120572)))

(20)

Let 119896 rarr infin we get 120593119901minus10(120574) = 119891(120574) By condition (iv) we

know 120574 = 1 = 120593(0) This proves that 120593 is continuous at 119909 = 0Thereby 120593 is a continuous function on [0 1] We have that120593(119909) satisfies (10) by (12) and 120593(119909) is unique from Lemma 7

Obviously if 1205930 is the restriction on [120582 1] of a single-valley-extended non-single-valley continuous solution to(4) then conditions (i)ndash(iv) must hold by the lemmas inSection 2

Example 9 Let 1205930(119909) [14 1] 997891rarr [0 1] be defined by

1205930 (119909) =

minus13

8119909 +

39

32 (

1

4le 119909 le

3

4)

119909 minus3

4 (

3

4le 119909 le 1)

(21)

Obviously 1205930 satisfies the conditions of Theorem 8 with 119901 =2 119891(119909) = 1199094 and 120582 = 119891(1) = 14 120572 = 34 Hence it is therestriction to [14 1] of a single-valley-extended continuoussolution 120593 to (4) Since 1205930 has the minimum point 120572 = 34

and 1205930(14) = 1316 gt 34 120593 is a single-valley-extendednon-single-valley continuous solution Its graph is depictedin Figure 1

Example 10 Let 1205930(119909) [15 1] 997891rarr [0 1] be defined by

1205930 (119909) =

minus109

25119909 +

218

125 (

1

5le 119909 le

2

5)

119909 minus2

5 (

2

5le 119909 le 1)

(22)

4 Abstract and Applied Analysis

x

120601(120582)

120601(f(120582))

120601(f(120572))

120601(x)

120601(1)

f(120572)f(120582)

Non-single-valley solution

1

1

0 120572120582

Figure 1 The graph of non-single-valley solution

Obviously 1205930 satisfies the conditions of Theorem 8 with 119901 =3 119891(119909) = 11990925 and 120582 = 119891(1) = 15 120572 = 25 Hence it is therestriction to [15 1] of a single-valley-extended continuoussolution 120593 to (4) Since 1205930 has the minimum point 120572 = 25

and 1205930(15) = 109125 gt 25 12059320(15) = 59125 gt 25

120593 is a single-valley-extended non-single-valley continuoussolution Its graph is similar to Figure 1

Appendix

Proof of Lemma 3 (i) Suppose that 120574 is a minimum point of120593 By (4) we have

119891 (120593 (120574)) = 120593119901(119891 (120574)) ge 120593 (120574) (A1)

And from 119891(0) = 0 119891(119909) lt 119909 (119909 isin (0 1]) we know 120593(120574) = 0If 120574 lt 120582 then for all 119894 = 1 2 we have 120593119894([0 120582]) = [0 1]Thus there exists 1199090 isin [0 120582] such that 120593119901minus1(1199090) = 0 And by(4) we have

119891 (120593 (119891minus1(1199090))) = 120593

119901(1199090) = 120593 (0) = 1 (A2)

This contradicts 119891(0) = 0 119891(119909) lt 119909 (119909 isin (0 1]) Thus 120574 gt 120582And from the definition of 120572 we know 120574 = 120572 and 120593(120572) =120593(120574) = 0

(ii) We now prove that for all 119899 ge 0 and each 119909 isin [0 1]we have

119891119899(120593 (119909)) = 120593

119901119899

(119891119899(119909)) (A3)

Obviously (A3) holds for 119899 = 1 by (4) Suppose that (A3)holds for 119899 le 119896 where 119896 is a certain integer Therefore byinduction and (4) we have that

120593119901119896+1

(119891119896+1

(119909))

= (120593119901119896

)119901

(119891119896+1

(119909)) = (120593119901119896

)119901minus1

∘ 120593119901119896

(119891119896+1

(119909))

= (120593119901119896

)119901minus1

(119891119896(120593 (119891 (119909))))

= (120593119901119896

)119901minus2

∘ 120593119901119896

(119891119896(120593 (119891 (119909))))

= (120593119901119896

)119901minus2

(119891119896(1205932(119891 (119909)))) = sdot sdot sdot

= (120593119901119896

)119901minus119894

(119891119896(120593119894(119891 (119909)))) = sdot sdot sdot

= 119891119896(120593119901(119891 (119909))) = 119891

119896(119891 (120593 (119909))) = 119891

119896+1(120593 (119909))

(A4)

that is (A3) holds for 119899 = 119896 + 1 Thereby (A3) is proved byinduction Let 119909 = 0 in (A3) we have that

119891119899(1) = 119891

119899(120593 (0)) = 120593

119901119899

(119891119899(0)) = 120593

119901119899

(0) (A5)

And it is trivial that 119891119899(1) is strictly decreasing andlim119899rarr+infin119891

119899(1) = 0 Thereby we have that

lim119899rarr+infin

120593119901119899

(0) = lim119899rarr+infin

119891119899(1) = 0 (A6)

that is we proved that 0 is a recurrent but not periodic pointof 120593

(iii) Firstly we prove that120582 is an extreme point of120593 By thefact that 120593 is strictly increasing on [120572 1] and (4) we have that120593119901 is strictly increasing in [119891(120572) 119891(1)] Thereby 120593 is strictly

monotone in [119891(120572) 119891(1)] Suppose that 120582 is not an extremepoint then 120593 is strictly decreasing on [119891(120572) 120572] We prove 120593is strictly decreasing on Δ120572

119896(119896 ge 0) respectively as follows

Obviously 120593 is strictly decreasing on Δ1205720 Suppose that 120593

is strictly decreasing on Δ120572119896for all 119896 le 119898 where 119898 ge 0 is a

certain integer By (4) we have that 120593119901 is strictly decreasingon Δ120572119898+1

Thereby 120593 is strictly monotone on Δ120572119898+1

Supposethat 120593 is strictly increasing on Δ120572

119898+1 then we claim that

120593 (119891119899(120572)) lt 120593 (119891

119898+1(120572)) forall119899 ge 119898 + 2 (A7)

We first prove that

120593 (119891119904(120572)) = 120593 (119891

119897(120572)) forall119904 = 119897 (A8)

Suppose that there exists 119904 = 119897 such that120593(119891119904(120572)) = 120593(119891119897(120572)) =120575 And from (A3) we get

120593119901119904minus1(120575) = 120593

119901119904

(119891119904(120572)) = 119891

119904(120593 (120572)) = 119891

119904(0) = 0 (A9)

Similarly we have 120593119901119897minus1(120575) = 0 If 119904 gt 119897 then

0 = 120593119901119904minus1(120575) = 120593

(119901119904minus119901119897)+(119901119897minus1)(120575) = 120593

(119901119904minus119901119897)(0) (A10)

Abstract and Applied Analysis 5

This contradicts that 0 is not a periodic point That iswe proved (A8) Thereby we have 120593(119891119899(120572)) = 120593(119891

119898+1(120572))

forall119899 ge 119898 + 2 And from the fact that 120593 is strictly increasingon Δ120572119898+1

we get 120593(119891119898+2(120572)) lt 120593(119891119898+1

(120572)) Suppose that120593(119891119899(120572)) lt 120593(119891

119898+1(120572)) where 119899 ge 119898 + 2 is a certain integer

If 120593(119891119899+1(120572)) gt 120593(119891119898+1(120572)) gt 120593(119891119899(120572)) and by the fact that120593119901119898+1minus1 is strictlymonotone on120593([119891119899+1(120572) 119891119899(120572)]) and (A3)

we get

120593119901119898+1minus1(120593 (119891

119899+1(120572)))

lt 120593119901119898+1minus1(120593 (119891

119898+1(120572)))

= 120593119901119898+1

(119891119898+1

(120572)) = 119891119898+1

(0) = 0

(A11)

or

120593119901119898+1minus1(120593 (119891119899(120572)))

lt 120593119901119898+1minus1(120593 (119891

119898+1(120572)))

= 120593119901119898+1

(119891119898+1

(120572)) = 119891119898+1

(0) = 0

(A12)

This contradicts that 120593119901119898+1minus1(119909) ge 0 That is 120593(119891119899+1(120572)) lt

120593(119891119898+1

(120572)) Thereby we proved (A7) by induction If120593(119891119898+1

(120572)) = 1 then by (A3) we have

120593119901119898+1

(1) = 120593119901119898+1

(120593 (119891119898+1

(120572))) = 120593 (120593119901119898+1

(119891119898+1

(120572)))

= 120593 (119891119898+1

(120593 (120572))) = 120593 (0) = 1

(A13)

This contradicts conclusion (ii) Thereby we get 120593(119891119899(120572)) lt120593(119891119898+1

(120572)) lt 1 forall119899 ge 119898 + 2 This contradicts that 120593(0) =1 Thereby 120593 is strictly decreasing on Δ120572

119898+1 Furthermore

we proved that 120593 is strictly decreasing on Δ120572

119896(119896 ge 0)

respectively by induction that is 120593 is single-valley solutionof (4) This contradicts the condition that 120593 is a non-single-valley solution of (4) That is we proved that 120582 is an extremepoint of 120593

Secondly we prove that 120593(120582) gt 120582 We claim that

120593 (119891119899(1)) = 120582 120593 (119891

119899(120572)) = 120582 forall119899 ge 1 (A14)

Suppose that 120593(119891119899(1)) = 120582 by (A5) we get 120593(119891119899(1)) =

120593119901119899+1(0) And from 120593

119901(0) = 120582 we have that 120593119901(0) = 120593119901

119899+1(0)

Let

119871 = 0 120593 (0) 120593119901(0) 120593

119901+1(0) 120593

119901119899

(0) (A15)

obviously 119871 is a limited set and 120593119894(0) isin 119871 forall119894 isin 119885

+By (A5) we have that 119891119894(1) = 120593

119901119894

(0) isin 119871 forall119894 isin 119885+

This contradicts that 119871 is limited Thereby we proved that120593(119891119899(1)) = 120582 Suppose that 120593(119891119899(120572)) = 120582 from (A3) we have

0 = 119891119899(120593 (120572)) = 120593

119901119899

(119891119899(120572))

= 120593119901119899minus1(120582) = 120593

119901119899+119901minus1

(0)

(A16)

This contradicts that 0 is not a periodic point of 120593 That is(A14) holds Suppose that 120593(120582) lt 120582 Since 120582 is an extremepoint of 120593 we have that 120593 is strictly increasing on [119891(120572) 120582]Furthermore we have 120593(119891(120572)) lt 120593(120582) lt 120582 We claim that

120593 (119891119899(1)) lt 120582 120593 (119891

119899(120572)) lt 120582 forall119899 ge 1 (A17)

It is trivial that (A17) holds for 119899 = 1 Suppose that (A17)holds for 119899 = 119898 where 119898 is a certain integer By (A3)we know 120593

119901119898

is strictly monotone on [119891119898+1(1) 119891119898(120572)] If120593(119891119898+1

(1)) gt 120582 then there exists 1199090 isin (119891119898+1

(1) 119891119898(120572))

such that 120593(1199090) = 120582 Thus 1199090 is an extreme point of1205932 on (119891

119898+1(1) 119891119898(120572)) That is 1205932 is not monotone on

[119891119898+1

(1) 119891119898(120572)] Furthermore 120593119901

119898

is not monotone on[119891119898+1

(1) 119891119898(120572)] This contradicts that 120593119901

119898

is monotone on[119891119898+1

(1) 119891119898(120572)] Thus 120593(119891119898+1(1)) lt 120582 Similarly we have

120593(119891119898+1

(120572)) lt 120582 That is (A17) holds for 119899 = 119898 + 1 Thereby(A17) holds by induction This contradicts that 120593(0) = 1Thus 120593(120582) gt 120582

(iv) Firstly suppose that 119902 is a fixed point of 120593 then by(A5) we have 119902 = 1 And by 120593(120572) = 0 we have 119902 = 120572 If 119902 isin(120572 1) then 119902 = 120593(119902) lt 120593(1) And by induction for all119898 ge 0we have 119902 = 120593119898(119902) lt 120593119898(1) Specially

119902 = 120593119901119899minus1(119902) lt 120593

119901119899minus1(1) = 120593

119901119899minus1(120593 (0)) = 120593

119901119899

(0) (A18)

This contradicts (A5) Thereby 119902 lt 120572Secondly there exists at least one fixed point 120573 isin (0 120572) by

120593(0) = 1 120593(120572) = 0 We prove 120573 is the unique fixed point asfollows We claim that

120573 notin (119891119899+1

(1) 119891119899(1)] forall119899 ge 1 (A19)

Suppose that there exists119898 ge 1 such that120573 isin [119891119898(120572) 119891119898(1)]then by (A3) we have

119891119898(120593 (119891minus119898(120573))) = 120593

119901119898

(119891119898(119891minus119898(120573))) = 120593

119901119898

(120573) = 120573

(A20)

that is 120593(119891minus119898(120573)) = 119891minus119898(120573) And from 119891minus119898(120573) isin [120572 1] this

contradicts that 120593 does not have a fixed point on [120572 1] Thuswe have 120573 notin [119891119899(120572) 119891119899(1)] forall119899 ge 1 Suppose that there exists119898 ge 1 such that 120573 isin (119891119898+1(1) 119891119898(120572)) then by (A3) we have

119891119898minus1

(120593 (1198911minus119898

(120573))) = 120593119901119898minus1(119891119898minus1

(1198911minus119898

(120573)))

= 120593119901119898minus1(120573) = 120573

(A21)

Thereby 120593(1198911minus119898(120573)) = 1198911minus119898

(120573) That is 1198911minus119898(120573) isin (1198912(1)119891(120572)) is a fixed point of 120593 Since 120593 has no fixed pointon [119891(120572) 119891(1)] and 120593(120582) gt 120582 we have that 120593 must bestrictly increasing on [1198912(1) 119891(120572)] And 120593 has the fixed point1198911minus119898

(120573) isin (1198912(1) 119891(120572)) thus 120593119901 is strictly increasing on

[1198912(1) 119891(120572)] But by (4) and the fact that 120593 is decreasing on

[120582 120572] we have that 120593119901 is strictly decreasing on [1198912(1) 119891(120572)]This contradicts Thus we have 120573 notin [119891119899+1(1) 119891119899(120572)] forall119899 ge 1Thereby we prove (A19)Thus we have 120573 isin (120582 120572) And since120593 is decreasing on [120582 120572] we have that 120573 isin (120582 120572) is the uniquefixed point

6 Abstract and Applied Analysis

(v) By (4) we have 0 = 119891(120593(120572)) = 120593119901(119891(120572)) And

since 120572 is the unique minimum point of 120593 it follows that120593119901minus1(119891(120572)) = 120572 Thus the sufficiency is provedWe prove the necessity as follows Suppose that 120593119894(119909) = 120572

for some 119909 isin [0 120582] and 0 le 119894 le 119901 minus 1 Firstly we claim that

119909 notin (119891119899+1

(1) 119891119899(120572)) cup (119891

119899(120572) 119891

119899(1)) forall119899 ge 1

(A22)

Suppose that 119909 isin (119891119899(120572) 119891

119899(1)) then 120593119894+1 is not mono-

tone on (119891119899(120572) 119891

119899(1)) Thereby 120593119901 is not monotone on

(119891119899(120572) 119891

119899(1)) This contradicts that 120593119901 is monotone on

(119891119899(120572) 119891

119899(1)) by (A3) Thus 119909 notin (119891

119899(120572) 119891

119899(1)) We have

similarly that 119909 notin (119891119899+1(1) 119891119899(120572)) Thereby we prove (A22)that is 119909 = 119891119899(1) or 119909 = 119891119899(120572) forall119899 ge 1

Secondly we prove that

119909 = 119891119899(1) 119909 = 119891

119899+1(120572) forall119899 ge 1 (A23)

Suppose that 119909 = 119891119899(1) for some 119899 ge 1 then 120593119894(119891119899(1)) =

120572 And from (A5) we have 120572 = 120593119894(119891119899(1)) = 120593

119894(120593119901119899

(0)) =

120593119894+119901119899

(0) This contradicts that 0 is not a periodic point of 120593Thus we prove 119909 = 119891

119899(1) forall119899 ge 1 Suppose that 119909 = 119891119899(120572) for

some 119899 ge 2 then 120593119894(119891119899(120572)) = 120572 And from (A3) we have

120593119894(0) = 120593

119894(120593119901119899

(119891119899(120572))) = 120593

119901119899

(120593119894(119891119899(120572)))

= 120593119901119899

(120572) = 120593119901119899minus1(0)

(A24)

This contradicts that 0 is a recurrent point of120593Thuswe prove119909 = 119891119899(120572) forall119899 ge 2 Thereby (A23) holds That is we get 119909 =

119891(120572) and 120593119894(119891(120572)) = 120572Lastly we prove 119894 = 119901 minus 1 Suppose that there exists

119895 = 119894 0 le 119895 le 119901 minus 1 such that 120593119895(119891(120572)) = 120572 Then we cansuppose that 119895 lt 119894 and

0 = 120593119894+1(119891 (120572)) = 120593

119894minus119895(120593119895+1(119891 (120572))) = 120593

119894minus119895(0) (A25)

This contradicts that 0 is not a periodic point of 120593 Thus weprove 120593119895(119891(120572)) = 120572 forall119895 = 119894 Thereby we have 119894 = 119901 minus 1 and119909 = 119891(120572)

Proof of Lemma 4 (i) Firstly we prove that for all 119894 =

0 1 119901 minus 2 we have 119869119894 sub (120582 1] that is 119869119894 cap 119869 = 0 Weclaim that

120593119894+1(119891 (120572)) gt 120582 forall0 le 119894 le 119901 minus 2 (A26)

Suppose that there exists 1 le 119895 le 119901 minus 1 such that 120593119895(119891(120572)) =119909 le 120582 And from Lemma 3(v) we have 120572 = 120593

119901minus1(119891(120572)) =

120593119901minus1minus119895

(120593119895(119891(120572))) = 120593

119901minus1minus119895(119909) This contradicts Lemma 3(v)

Thus (A26) holds We claim that

120593119894+1(120582) gt 120582 forall0 le 119894 le 119901 minus 2 (A27)

It is trivial that 120593119894+1(120582) = 120582 by 0 is a recurrent point of120593 Suppose that there exists 1 le 119895 le 119901 minus 1 such that120593119895(120582) lt 120582 And from 120593

119895(119891(120572)) gt 120582 we have that there exists

119909 isin (119891(120572) 119891(1)) such that 120593119895(119909) = 120582 And from the fact that120582 is an extreme point of 120593 we get that 120593119895+1 is not monotoneon [119891(120572) 119891(1)] This contradicts that 120593119901 is strictly monotoneon [119891(120572) 119891(1)] Thus (A27) holds We claim that

120593119894+1(119909) gt 120582 forall0 le 119894 le 119901 minus 2 119909 isin [0 120582] (A28)

Suppose that there exists 119909 isin [0 120582) and 1 le 119895 le 119901 minus 1 suchthat 120593119895(119909) = 120582 And from (4) we get 120593119901minus119895(120582) = 120593

119901(119909) =

119891(120593(119891minus1(119909))) le 119891(1) = 120582 This contradicts (A27) Thus

120593119894+1(119909) = 120582 forall0 le 119894 le 119901 minus 2 119909 isin [0 120582] Suppose that there

exists 119909 isin [0 120582) and 1 le 119895 le 119901 minus 1 such that 120593119895(119909) lt 120582And from (A27) we have that there exists 119911 isin (119909 120582) suchthat 120593119895(119911) = 120582 And from (4) we get 120593119901minus119895(120582) = 120593

119901(119911) =

119891(120593(119891minus1(119911))) le 119891(1) = 120582 This contradicts (A27) Thus

(A28) holds That is 119869119894 cap 119869 = 0Secondly we prove that 119869119894 forall0 le 119894 le 119901 minus 2 are pairwise

disjoint Suppose that there exists 0 le 119894 lt 119895 le 119901minus 2 such that119869119894 cap 119869119895 = 119869119894119895 = 0 Let 119910 isin 119869119894119895 then there exist 119909119894 isin [0 120582] 119909119895 isin[0 120582] such that 120593119894+1(119909119894) = 119910 = 120593

119895+1(119909119895) Thereby we have

120593119901minus119895+119894

(119909119894) = 120593119901minus1minus119895

(120593119894+1(119909119894)) = 120593

119901minus1minus119895(120593119895+1(119909119895)) = 120593

119901(119909119895) =

119891(120593(119891minus1(119909119895))) le 119891(1) = 120582 This contradicts 119869119894 sub (120582 1] Thus

we proved that 1198690 1198691 119869119901minus2 are pairwise disjoint(ii) For all 119894 = 0 1 119901 minus 2 then 120593119894+1 119869 997891rarr 119869119894 is a

homeomorphismby Lemmas 3(i) and 3(v)Thereby120593119894 1198690 997891rarr119869119894 is also a homeomorphism

Proof of Lemma 5 120582 is a maximum point of 120593 on [119891(120572) 120572]by Lemma 3(iii) Suppose that 119891119898(1) is the maximum pointof 120593 on [119891119898(120572) 120572] where 119898 is some certain integer Firstlywe prove 119891119898(120572) is an extreme point of 120593 If otherwise 120593 isstrictly increasing on [119891119898+1(1) 119891119898(120572)] Thus 120593(119891119898+1(1)) lt120593(119891119898(120572)) and 120593 is strictly increasing on Δ119898 We claim that 120593

is strictly increasing on Δ 119896 for all 119896 ge 119898 It holds obviouslywhen 119896 = 119898 Suppose that 120593 is strictly increasing on Δ 119896 forall 119898 le 119896 le 119897 where 119897 is some certain integer 120593119901 is strictlyincreasing on Δ 119897+1 by (4) Thereby 120593 is strictly monotone onΔ 119897+1 If 120593 is strictly decreasing on Δ 119897+1 then we claim that

120593 (119891119904(1)) lt 120593 (119891

119897(120572)) forall119904 ge 119897 + 1 (A29)

Its proof is similar to (A7) and we omit it here Thereby120593(119891119904(1)) lt 120593(119891

119897(120572)) lt 1 forall119904 ge 119897 + 1 This contradicts

120593(0) = 1 Thus 120593 is strictly increasing on Δ 119897+1 Thereby 120593 isstrictly increasing on [0 119891119898(1)] by induction Furthermore120593(0) lt 120593(119891

119898(1)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is an extreme point of 120593Secondly we prove 119891119898(120572) is the minimum point of 120593 on

[0 119891119898(1)] Suppose that there exists 119897 ge 119898 + 1 such that

120593(119891119897(120572)) lt 120593(119891

119898(120572)) or 120593(119891119897(1)) lt 120593(119891119898(120572)) then we claim

that

120593 (119891119904(120572)) lt 120593 (119891

119898(120572)) forall119904 ge 119897 (A30)

or

120593 (119891119904(1)) lt 120593 (119891

119898(120572)) forall119904 ge 119897 (A31)

Abstract and Applied Analysis 7

The proof is similar to (A7) and we omit it here Thereby120593(0) lt 120593(119891

119898(120572)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is the minimum point of 120593 on [0 119891119898(1)]Thirdly we prove 119891119898+1(1) is an extreme point of 120593

120593119901minus1 is strictly increasing on [120593(119891119898(120572)) 120593(119891119898+1(1))] by (4)

and 120593 is strictly decreasing on [119891119898+1(1) 119891119898(120572)] And fromLemma 4(ii) we have that 120593119901minus1 is strictly increasing on 1198690 =[120593(119891(120572)) 1] Suppose that 119891119898+1(1) is not an extreme pointof 120593 then 120593 is strictly decreasing on [119891

119898+1(120572) 119891

119898+1(1)]

And from (4) and the fact that 120593 is strictly increasingon [119891

119898(120572) 119891

119898(1)] we get that 120593119901 is strictly increasing

on [119891119898+1

(120572) 119891119898+1

(1)] Thereby 120593119901minus1 is strictly decreasingon [120593(119891

119898+1(1)) 120593(119891

119898+1(120572))] This contradicts that 120593119901minus1 is

strictly increasing on 1198690 = [120593(119891(120572)) 1] Thus 119891119898+1(1) is theextreme point of 120593

Lastly we have that 119891119898+1(1) is the maximum point of 120593on [119891119898+1(120572) 120572] The proof is similar to 119891119898(120572) and we omit ithere Thereby 119891119899(1) is the maximum point of 120593 on [119891119899(120572) 120572]and 119891119899(120572) is the minimum point of 120593 on [0 119891119899(1)] for all 119899 ge1 by induction

Proof of Lemma 6 It is trivial that 119909 = 1 is a solution of theequation 120593119901minus1(119909) = 119891(119909) by (4) Suppose that 119909 = 1199090 isin

(120593(119891(120572)) 1] is an arbitrary solution of this equation that is120593119901minus1(1199090) = 119891(1199090) Since 120593([0 119891(120572)]) = [120593(119891(120572)) 1] we have

that there exists 119910 isin [0 119891(120572)] such that 120593(119910) = 1199090 We claimthat

120593 (119891119899(119910)) = 1199090 forall119899 ge 0 (A32)

Obviously (A32) holds for 119899 = 0 Suppose that (A32) holdsfor 119899 = 119896 where 119896 ge 0 is a certain integer Therefore by (4)we have that

120593119901minus1

(120593 (119891119896+1

(119910))) = 119891 (120593 (119891119896(119910))) = 119891 (1199090) = 120593

119901minus1(1199090)

(A33)

Since119891119896+1(119910) isin [0 119891(120572)) we have 120593(119891119896+1(119910)) isin (120593(119891(120572)) 1]And since 120593

119901minus1 is strictly increasing on [120593(119891(120572)) 1] by120593(119891(120572)) gt 120572 we have 120593(119891119896+1(119910)) = 1199090 That is (A32) holdsfor 119899 = 119896 + 1 Thereby (A32) is proved by induction By thefact that 119891119899(119910) is strictly decreasing and lim119899rarrinfin119891

119899(119910) = 0

we have 1199090 = lim119899rarr+infin120593(119891119899(119910)) = 120593(0) = 1

Proof of Lemma 7 There exist 120572 isin (120582 1) 120573 isin (120582 1) such that120593119894(120572) = 0 120593119894(120573) = 120573 (119894 = 1 2) by (8) Denote 1205930(119909) = 1205931(119909) =1205932(119909) (119909 isin [120582 1]) and 1205721 = 1205931(119891(120572)) lt 1205931(120582) = 1205930(120582) 1205722 =1205932(119891(120572)) lt 1205932(120582) = 1205930(120582) We prove that 1205721 = 1205722 as followsIt is trivial that 1205721 gt 120582 1205722 gt 120582 and

120593119901minus1

0(1205721) = 120593

119901minus1

1(1205721) = 120593

119901

1(119891 (120572)) = 0

120593119901minus1

0(1205722) = 120593

119901minus1

2(1205722) = 120593

119901

2(119891 (120572)) = 0

(A34)

by Lemmas 3(v) and 4 Since 120593119901minus10

is strictly monotone on[min1205721 1205722 1] we have that 1205721 = 1205722 Let 1205720 = 1205721 = 1205722 then1205720 lt 1205930(120582) and 120593

119901minus1

0(1205720) = 0 Define 120595+ = 120593

119901minus1

0|[1205720 1]

ByLemmas 3(v) and 4 we have that 120595+ is a homeomorphism

And by 120595+(1205720) = 0 le 120595+(1) we know that 120595+ is strictlyincreasing

We prove 1205931(119909) = 1205932(119909) on Δ 119896 for all 119896 ge 0 by inductionas follows

Obviously 1205931(119909) = 1205932(119909) holds on Δ 0 Suppose that1205931(119909) = 1205932(119909) holds on Δ 119896 for all 119896 le 119898 where 119898 ge 0 isa certain integer Let

120593 (119909) = 1205931 (119909) = 1205932 (119909) 119909 isin [119891119898+1

(1) 1] (A35)

Since 120593119894(119909) ge 120593119894(119891(120572)) = 1205720 gt 120582 for 119909 le 120582 and from (4) wehave

119891 (120593 (119891minus1(119909))) = 119891 (120593119894 (119891

minus1(119909))) = 120593

119901minus1

119894(120593119894 (119909))

= 120595+ (120593119894 (119909)) (119894 = 1 2 119909 isin Δ119898+1)

(A36)

Thereby

120593119894 (119909) = 120595minus1

+(119891 (120593 (119891

minus1(119909)))) (119894 = 1 2 119909 isin Δ119898+1)

(A37)

Thus we have 1205931(119909) = 1205932(119909) on Δ119898+1 Thus we get 1205931(119909) =1205932(119909) on Δ 119896 for all 119896 ge 0 by induction

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is partially supported by the National NaturalScience Foundation of China Tian Yuan Foundation (no11326129) and the Fundamental Research Funds for theCentral Universities (no 14CX02152A)

References

[1] M J Feigenbaum ldquoQuantitative universality for a class ofnonlinear transformationsrdquo Journal of Statistical Physics vol 19no 1 pp 25ndash52 1978

[2] M J Feigenbaum ldquoTheuniversalmetric properties of nonlineartransformationsrdquo Journal of Statistical Physics vol 21 no 6 pp669ndash706 1979

[3] P Couliet and C Tresser ldquoIterations drsquoendomorphismes etgroupe de renormalisationrdquo Journal de Physique Colloques vol39 no 5 pp C5-25ndashC5-28 1978

[4] D Sullivan ldquoBoubds quadratic differentials and renormaliza-tion conjecturesrdquo in Proceedings of the Mathematics into theTwenty-First Century Centennial Symposium F Browder Edpp 417ndash466 American Mathematical Society August 1992

[5] ADouady and JHHubbard ldquoOn the dynamics of polynomial-like mappingsrdquo Annales Scientifiques de lrsquoEcole NormaleSuperieure vol 18 no 2 pp 287ndash343 1985

[6] I Lanford ldquoA computer-assisted proof of the Feigenbaumconjecturesrdquo The American Mathematical Society vol 6 no 3pp 427ndash434 1982

8 Abstract and Applied Analysis

[7] J P Eckmann and P Wittwer ldquoA complete proof of theFeigenbaum conjecturesrdquo Journal of Statistical Physics vol 46no 3-4 pp 455ndash475 1987

[8] M Campanino and H Epstein ldquoOn the existence of Feigen-baumrsquos fixed pointrdquo Communications in Mathematical Physicsvol 79 no 2 pp 261ndash302 1981

[9] L Yang and J Z Zhang ldquoThe second type of Feigenbaumrsquosfunctional equationrdquo Science in China A vol 28 pp 1061ndash10691985

[10] P J McCarthy ldquoThe general exact bijective continuous solu-tion of Feigenbaumrsquos functional equationrdquo Communications inMathematical Physics vol 91 no 3 pp 431ndash443 1983

[11] J P Eckmann H Epstein and P Wittwer ldquoFixed pointsof Feigenbaumrsquos type for the equation 119891119901 (120582119909) = 120582119891(119909)rdquoCommunications in Mathematical Physics vol 93 no 4 pp495ndash516 1984

[12] A H Zhang L J Wang and W Song ldquoNonsingle-valley con-tinuous solutions of p-order Feigenbaums functional equationrdquoNortheastern Mathematical Journal vol 19 no 4 pp 375ndash3802003

[13] G Liao L Wang and Y Zhang ldquo119901-order Feigenbaumrsquos mapsrdquoNortheastern Mathematical Journal Dongbei Shuxue vol 20no 3 pp 284ndash290 2004

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

x

120601(120582)

120601(f(120582))

120601(f(120572))

120601(x)

120601(1)

f(120572)f(120582)

Non-single-valley solution

1

1

0 120572120582

Figure 1 The graph of non-single-valley solution

Obviously 1205930 satisfies the conditions of Theorem 8 with 119901 =3 119891(119909) = 11990925 and 120582 = 119891(1) = 15 120572 = 25 Hence it is therestriction to [15 1] of a single-valley-extended continuoussolution 120593 to (4) Since 1205930 has the minimum point 120572 = 25

and 1205930(15) = 109125 gt 25 12059320(15) = 59125 gt 25

120593 is a single-valley-extended non-single-valley continuoussolution Its graph is similar to Figure 1

Appendix

Proof of Lemma 3 (i) Suppose that 120574 is a minimum point of120593 By (4) we have

119891 (120593 (120574)) = 120593119901(119891 (120574)) ge 120593 (120574) (A1)

And from 119891(0) = 0 119891(119909) lt 119909 (119909 isin (0 1]) we know 120593(120574) = 0If 120574 lt 120582 then for all 119894 = 1 2 we have 120593119894([0 120582]) = [0 1]Thus there exists 1199090 isin [0 120582] such that 120593119901minus1(1199090) = 0 And by(4) we have

119891 (120593 (119891minus1(1199090))) = 120593

119901(1199090) = 120593 (0) = 1 (A2)

This contradicts 119891(0) = 0 119891(119909) lt 119909 (119909 isin (0 1]) Thus 120574 gt 120582And from the definition of 120572 we know 120574 = 120572 and 120593(120572) =120593(120574) = 0

(ii) We now prove that for all 119899 ge 0 and each 119909 isin [0 1]we have

119891119899(120593 (119909)) = 120593

119901119899

(119891119899(119909)) (A3)

Obviously (A3) holds for 119899 = 1 by (4) Suppose that (A3)holds for 119899 le 119896 where 119896 is a certain integer Therefore byinduction and (4) we have that

120593119901119896+1

(119891119896+1

(119909))

= (120593119901119896

)119901

(119891119896+1

(119909)) = (120593119901119896

)119901minus1

∘ 120593119901119896

(119891119896+1

(119909))

= (120593119901119896

)119901minus1

(119891119896(120593 (119891 (119909))))

= (120593119901119896

)119901minus2

∘ 120593119901119896

(119891119896(120593 (119891 (119909))))

= (120593119901119896

)119901minus2

(119891119896(1205932(119891 (119909)))) = sdot sdot sdot

= (120593119901119896

)119901minus119894

(119891119896(120593119894(119891 (119909)))) = sdot sdot sdot

= 119891119896(120593119901(119891 (119909))) = 119891

119896(119891 (120593 (119909))) = 119891

119896+1(120593 (119909))

(A4)

that is (A3) holds for 119899 = 119896 + 1 Thereby (A3) is proved byinduction Let 119909 = 0 in (A3) we have that

119891119899(1) = 119891

119899(120593 (0)) = 120593

119901119899

(119891119899(0)) = 120593

119901119899

(0) (A5)

And it is trivial that 119891119899(1) is strictly decreasing andlim119899rarr+infin119891

119899(1) = 0 Thereby we have that

lim119899rarr+infin

120593119901119899

(0) = lim119899rarr+infin

119891119899(1) = 0 (A6)

that is we proved that 0 is a recurrent but not periodic pointof 120593

(iii) Firstly we prove that120582 is an extreme point of120593 By thefact that 120593 is strictly increasing on [120572 1] and (4) we have that120593119901 is strictly increasing in [119891(120572) 119891(1)] Thereby 120593 is strictly

monotone in [119891(120572) 119891(1)] Suppose that 120582 is not an extremepoint then 120593 is strictly decreasing on [119891(120572) 120572] We prove 120593is strictly decreasing on Δ120572

119896(119896 ge 0) respectively as follows

Obviously 120593 is strictly decreasing on Δ1205720 Suppose that 120593

is strictly decreasing on Δ120572119896for all 119896 le 119898 where 119898 ge 0 is a

certain integer By (4) we have that 120593119901 is strictly decreasingon Δ120572119898+1

Thereby 120593 is strictly monotone on Δ120572119898+1

Supposethat 120593 is strictly increasing on Δ120572

119898+1 then we claim that

120593 (119891119899(120572)) lt 120593 (119891

119898+1(120572)) forall119899 ge 119898 + 2 (A7)

We first prove that

120593 (119891119904(120572)) = 120593 (119891

119897(120572)) forall119904 = 119897 (A8)

Suppose that there exists 119904 = 119897 such that120593(119891119904(120572)) = 120593(119891119897(120572)) =120575 And from (A3) we get

120593119901119904minus1(120575) = 120593

119901119904

(119891119904(120572)) = 119891

119904(120593 (120572)) = 119891

119904(0) = 0 (A9)

Similarly we have 120593119901119897minus1(120575) = 0 If 119904 gt 119897 then

0 = 120593119901119904minus1(120575) = 120593

(119901119904minus119901119897)+(119901119897minus1)(120575) = 120593

(119901119904minus119901119897)(0) (A10)

Abstract and Applied Analysis 5

This contradicts that 0 is not a periodic point That iswe proved (A8) Thereby we have 120593(119891119899(120572)) = 120593(119891

119898+1(120572))

forall119899 ge 119898 + 2 And from the fact that 120593 is strictly increasingon Δ120572119898+1

we get 120593(119891119898+2(120572)) lt 120593(119891119898+1

(120572)) Suppose that120593(119891119899(120572)) lt 120593(119891

119898+1(120572)) where 119899 ge 119898 + 2 is a certain integer

If 120593(119891119899+1(120572)) gt 120593(119891119898+1(120572)) gt 120593(119891119899(120572)) and by the fact that120593119901119898+1minus1 is strictlymonotone on120593([119891119899+1(120572) 119891119899(120572)]) and (A3)

we get

120593119901119898+1minus1(120593 (119891

119899+1(120572)))

lt 120593119901119898+1minus1(120593 (119891

119898+1(120572)))

= 120593119901119898+1

(119891119898+1

(120572)) = 119891119898+1

(0) = 0

(A11)

or

120593119901119898+1minus1(120593 (119891119899(120572)))

lt 120593119901119898+1minus1(120593 (119891

119898+1(120572)))

= 120593119901119898+1

(119891119898+1

(120572)) = 119891119898+1

(0) = 0

(A12)

This contradicts that 120593119901119898+1minus1(119909) ge 0 That is 120593(119891119899+1(120572)) lt

120593(119891119898+1

(120572)) Thereby we proved (A7) by induction If120593(119891119898+1

(120572)) = 1 then by (A3) we have

120593119901119898+1

(1) = 120593119901119898+1

(120593 (119891119898+1

(120572))) = 120593 (120593119901119898+1

(119891119898+1

(120572)))

= 120593 (119891119898+1

(120593 (120572))) = 120593 (0) = 1

(A13)

This contradicts conclusion (ii) Thereby we get 120593(119891119899(120572)) lt120593(119891119898+1

(120572)) lt 1 forall119899 ge 119898 + 2 This contradicts that 120593(0) =1 Thereby 120593 is strictly decreasing on Δ120572

119898+1 Furthermore

we proved that 120593 is strictly decreasing on Δ120572

119896(119896 ge 0)

respectively by induction that is 120593 is single-valley solutionof (4) This contradicts the condition that 120593 is a non-single-valley solution of (4) That is we proved that 120582 is an extremepoint of 120593

Secondly we prove that 120593(120582) gt 120582 We claim that

120593 (119891119899(1)) = 120582 120593 (119891

119899(120572)) = 120582 forall119899 ge 1 (A14)

Suppose that 120593(119891119899(1)) = 120582 by (A5) we get 120593(119891119899(1)) =

120593119901119899+1(0) And from 120593

119901(0) = 120582 we have that 120593119901(0) = 120593119901

119899+1(0)

Let

119871 = 0 120593 (0) 120593119901(0) 120593

119901+1(0) 120593

119901119899

(0) (A15)

obviously 119871 is a limited set and 120593119894(0) isin 119871 forall119894 isin 119885

+By (A5) we have that 119891119894(1) = 120593

119901119894

(0) isin 119871 forall119894 isin 119885+

This contradicts that 119871 is limited Thereby we proved that120593(119891119899(1)) = 120582 Suppose that 120593(119891119899(120572)) = 120582 from (A3) we have

0 = 119891119899(120593 (120572)) = 120593

119901119899

(119891119899(120572))

= 120593119901119899minus1(120582) = 120593

119901119899+119901minus1

(0)

(A16)

This contradicts that 0 is not a periodic point of 120593 That is(A14) holds Suppose that 120593(120582) lt 120582 Since 120582 is an extremepoint of 120593 we have that 120593 is strictly increasing on [119891(120572) 120582]Furthermore we have 120593(119891(120572)) lt 120593(120582) lt 120582 We claim that

120593 (119891119899(1)) lt 120582 120593 (119891

119899(120572)) lt 120582 forall119899 ge 1 (A17)

It is trivial that (A17) holds for 119899 = 1 Suppose that (A17)holds for 119899 = 119898 where 119898 is a certain integer By (A3)we know 120593

119901119898

is strictly monotone on [119891119898+1(1) 119891119898(120572)] If120593(119891119898+1

(1)) gt 120582 then there exists 1199090 isin (119891119898+1

(1) 119891119898(120572))

such that 120593(1199090) = 120582 Thus 1199090 is an extreme point of1205932 on (119891

119898+1(1) 119891119898(120572)) That is 1205932 is not monotone on

[119891119898+1

(1) 119891119898(120572)] Furthermore 120593119901

119898

is not monotone on[119891119898+1

(1) 119891119898(120572)] This contradicts that 120593119901

119898

is monotone on[119891119898+1

(1) 119891119898(120572)] Thus 120593(119891119898+1(1)) lt 120582 Similarly we have

120593(119891119898+1

(120572)) lt 120582 That is (A17) holds for 119899 = 119898 + 1 Thereby(A17) holds by induction This contradicts that 120593(0) = 1Thus 120593(120582) gt 120582

(iv) Firstly suppose that 119902 is a fixed point of 120593 then by(A5) we have 119902 = 1 And by 120593(120572) = 0 we have 119902 = 120572 If 119902 isin(120572 1) then 119902 = 120593(119902) lt 120593(1) And by induction for all119898 ge 0we have 119902 = 120593119898(119902) lt 120593119898(1) Specially

119902 = 120593119901119899minus1(119902) lt 120593

119901119899minus1(1) = 120593

119901119899minus1(120593 (0)) = 120593

119901119899

(0) (A18)

This contradicts (A5) Thereby 119902 lt 120572Secondly there exists at least one fixed point 120573 isin (0 120572) by

120593(0) = 1 120593(120572) = 0 We prove 120573 is the unique fixed point asfollows We claim that

120573 notin (119891119899+1

(1) 119891119899(1)] forall119899 ge 1 (A19)

Suppose that there exists119898 ge 1 such that120573 isin [119891119898(120572) 119891119898(1)]then by (A3) we have

119891119898(120593 (119891minus119898(120573))) = 120593

119901119898

(119891119898(119891minus119898(120573))) = 120593

119901119898

(120573) = 120573

(A20)

that is 120593(119891minus119898(120573)) = 119891minus119898(120573) And from 119891minus119898(120573) isin [120572 1] this

contradicts that 120593 does not have a fixed point on [120572 1] Thuswe have 120573 notin [119891119899(120572) 119891119899(1)] forall119899 ge 1 Suppose that there exists119898 ge 1 such that 120573 isin (119891119898+1(1) 119891119898(120572)) then by (A3) we have

119891119898minus1

(120593 (1198911minus119898

(120573))) = 120593119901119898minus1(119891119898minus1

(1198911minus119898

(120573)))

= 120593119901119898minus1(120573) = 120573

(A21)

Thereby 120593(1198911minus119898(120573)) = 1198911minus119898

(120573) That is 1198911minus119898(120573) isin (1198912(1)119891(120572)) is a fixed point of 120593 Since 120593 has no fixed pointon [119891(120572) 119891(1)] and 120593(120582) gt 120582 we have that 120593 must bestrictly increasing on [1198912(1) 119891(120572)] And 120593 has the fixed point1198911minus119898

(120573) isin (1198912(1) 119891(120572)) thus 120593119901 is strictly increasing on

[1198912(1) 119891(120572)] But by (4) and the fact that 120593 is decreasing on

[120582 120572] we have that 120593119901 is strictly decreasing on [1198912(1) 119891(120572)]This contradicts Thus we have 120573 notin [119891119899+1(1) 119891119899(120572)] forall119899 ge 1Thereby we prove (A19)Thus we have 120573 isin (120582 120572) And since120593 is decreasing on [120582 120572] we have that 120573 isin (120582 120572) is the uniquefixed point

6 Abstract and Applied Analysis

(v) By (4) we have 0 = 119891(120593(120572)) = 120593119901(119891(120572)) And

since 120572 is the unique minimum point of 120593 it follows that120593119901minus1(119891(120572)) = 120572 Thus the sufficiency is provedWe prove the necessity as follows Suppose that 120593119894(119909) = 120572

for some 119909 isin [0 120582] and 0 le 119894 le 119901 minus 1 Firstly we claim that

119909 notin (119891119899+1

(1) 119891119899(120572)) cup (119891

119899(120572) 119891

119899(1)) forall119899 ge 1

(A22)

Suppose that 119909 isin (119891119899(120572) 119891

119899(1)) then 120593119894+1 is not mono-

tone on (119891119899(120572) 119891

119899(1)) Thereby 120593119901 is not monotone on

(119891119899(120572) 119891

119899(1)) This contradicts that 120593119901 is monotone on

(119891119899(120572) 119891

119899(1)) by (A3) Thus 119909 notin (119891

119899(120572) 119891

119899(1)) We have

similarly that 119909 notin (119891119899+1(1) 119891119899(120572)) Thereby we prove (A22)that is 119909 = 119891119899(1) or 119909 = 119891119899(120572) forall119899 ge 1

Secondly we prove that

119909 = 119891119899(1) 119909 = 119891

119899+1(120572) forall119899 ge 1 (A23)

Suppose that 119909 = 119891119899(1) for some 119899 ge 1 then 120593119894(119891119899(1)) =

120572 And from (A5) we have 120572 = 120593119894(119891119899(1)) = 120593

119894(120593119901119899

(0)) =

120593119894+119901119899

(0) This contradicts that 0 is not a periodic point of 120593Thus we prove 119909 = 119891

119899(1) forall119899 ge 1 Suppose that 119909 = 119891119899(120572) for

some 119899 ge 2 then 120593119894(119891119899(120572)) = 120572 And from (A3) we have

120593119894(0) = 120593

119894(120593119901119899

(119891119899(120572))) = 120593

119901119899

(120593119894(119891119899(120572)))

= 120593119901119899

(120572) = 120593119901119899minus1(0)

(A24)

This contradicts that 0 is a recurrent point of120593Thuswe prove119909 = 119891119899(120572) forall119899 ge 2 Thereby (A23) holds That is we get 119909 =

119891(120572) and 120593119894(119891(120572)) = 120572Lastly we prove 119894 = 119901 minus 1 Suppose that there exists

119895 = 119894 0 le 119895 le 119901 minus 1 such that 120593119895(119891(120572)) = 120572 Then we cansuppose that 119895 lt 119894 and

0 = 120593119894+1(119891 (120572)) = 120593

119894minus119895(120593119895+1(119891 (120572))) = 120593

119894minus119895(0) (A25)

This contradicts that 0 is not a periodic point of 120593 Thus weprove 120593119895(119891(120572)) = 120572 forall119895 = 119894 Thereby we have 119894 = 119901 minus 1 and119909 = 119891(120572)

Proof of Lemma 4 (i) Firstly we prove that for all 119894 =

0 1 119901 minus 2 we have 119869119894 sub (120582 1] that is 119869119894 cap 119869 = 0 Weclaim that

120593119894+1(119891 (120572)) gt 120582 forall0 le 119894 le 119901 minus 2 (A26)

Suppose that there exists 1 le 119895 le 119901 minus 1 such that 120593119895(119891(120572)) =119909 le 120582 And from Lemma 3(v) we have 120572 = 120593

119901minus1(119891(120572)) =

120593119901minus1minus119895

(120593119895(119891(120572))) = 120593

119901minus1minus119895(119909) This contradicts Lemma 3(v)

Thus (A26) holds We claim that

120593119894+1(120582) gt 120582 forall0 le 119894 le 119901 minus 2 (A27)

It is trivial that 120593119894+1(120582) = 120582 by 0 is a recurrent point of120593 Suppose that there exists 1 le 119895 le 119901 minus 1 such that120593119895(120582) lt 120582 And from 120593

119895(119891(120572)) gt 120582 we have that there exists

119909 isin (119891(120572) 119891(1)) such that 120593119895(119909) = 120582 And from the fact that120582 is an extreme point of 120593 we get that 120593119895+1 is not monotoneon [119891(120572) 119891(1)] This contradicts that 120593119901 is strictly monotoneon [119891(120572) 119891(1)] Thus (A27) holds We claim that

120593119894+1(119909) gt 120582 forall0 le 119894 le 119901 minus 2 119909 isin [0 120582] (A28)

Suppose that there exists 119909 isin [0 120582) and 1 le 119895 le 119901 minus 1 suchthat 120593119895(119909) = 120582 And from (4) we get 120593119901minus119895(120582) = 120593

119901(119909) =

119891(120593(119891minus1(119909))) le 119891(1) = 120582 This contradicts (A27) Thus

120593119894+1(119909) = 120582 forall0 le 119894 le 119901 minus 2 119909 isin [0 120582] Suppose that there

exists 119909 isin [0 120582) and 1 le 119895 le 119901 minus 1 such that 120593119895(119909) lt 120582And from (A27) we have that there exists 119911 isin (119909 120582) suchthat 120593119895(119911) = 120582 And from (4) we get 120593119901minus119895(120582) = 120593

119901(119911) =

119891(120593(119891minus1(119911))) le 119891(1) = 120582 This contradicts (A27) Thus

(A28) holds That is 119869119894 cap 119869 = 0Secondly we prove that 119869119894 forall0 le 119894 le 119901 minus 2 are pairwise

disjoint Suppose that there exists 0 le 119894 lt 119895 le 119901minus 2 such that119869119894 cap 119869119895 = 119869119894119895 = 0 Let 119910 isin 119869119894119895 then there exist 119909119894 isin [0 120582] 119909119895 isin[0 120582] such that 120593119894+1(119909119894) = 119910 = 120593

119895+1(119909119895) Thereby we have

120593119901minus119895+119894

(119909119894) = 120593119901minus1minus119895

(120593119894+1(119909119894)) = 120593

119901minus1minus119895(120593119895+1(119909119895)) = 120593

119901(119909119895) =

119891(120593(119891minus1(119909119895))) le 119891(1) = 120582 This contradicts 119869119894 sub (120582 1] Thus

we proved that 1198690 1198691 119869119901minus2 are pairwise disjoint(ii) For all 119894 = 0 1 119901 minus 2 then 120593119894+1 119869 997891rarr 119869119894 is a

homeomorphismby Lemmas 3(i) and 3(v)Thereby120593119894 1198690 997891rarr119869119894 is also a homeomorphism

Proof of Lemma 5 120582 is a maximum point of 120593 on [119891(120572) 120572]by Lemma 3(iii) Suppose that 119891119898(1) is the maximum pointof 120593 on [119891119898(120572) 120572] where 119898 is some certain integer Firstlywe prove 119891119898(120572) is an extreme point of 120593 If otherwise 120593 isstrictly increasing on [119891119898+1(1) 119891119898(120572)] Thus 120593(119891119898+1(1)) lt120593(119891119898(120572)) and 120593 is strictly increasing on Δ119898 We claim that 120593

is strictly increasing on Δ 119896 for all 119896 ge 119898 It holds obviouslywhen 119896 = 119898 Suppose that 120593 is strictly increasing on Δ 119896 forall 119898 le 119896 le 119897 where 119897 is some certain integer 120593119901 is strictlyincreasing on Δ 119897+1 by (4) Thereby 120593 is strictly monotone onΔ 119897+1 If 120593 is strictly decreasing on Δ 119897+1 then we claim that

120593 (119891119904(1)) lt 120593 (119891

119897(120572)) forall119904 ge 119897 + 1 (A29)

Its proof is similar to (A7) and we omit it here Thereby120593(119891119904(1)) lt 120593(119891

119897(120572)) lt 1 forall119904 ge 119897 + 1 This contradicts

120593(0) = 1 Thus 120593 is strictly increasing on Δ 119897+1 Thereby 120593 isstrictly increasing on [0 119891119898(1)] by induction Furthermore120593(0) lt 120593(119891

119898(1)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is an extreme point of 120593Secondly we prove 119891119898(120572) is the minimum point of 120593 on

[0 119891119898(1)] Suppose that there exists 119897 ge 119898 + 1 such that

120593(119891119897(120572)) lt 120593(119891

119898(120572)) or 120593(119891119897(1)) lt 120593(119891119898(120572)) then we claim

that

120593 (119891119904(120572)) lt 120593 (119891

119898(120572)) forall119904 ge 119897 (A30)

or

120593 (119891119904(1)) lt 120593 (119891

119898(120572)) forall119904 ge 119897 (A31)

Abstract and Applied Analysis 7

The proof is similar to (A7) and we omit it here Thereby120593(0) lt 120593(119891

119898(120572)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is the minimum point of 120593 on [0 119891119898(1)]Thirdly we prove 119891119898+1(1) is an extreme point of 120593

120593119901minus1 is strictly increasing on [120593(119891119898(120572)) 120593(119891119898+1(1))] by (4)

and 120593 is strictly decreasing on [119891119898+1(1) 119891119898(120572)] And fromLemma 4(ii) we have that 120593119901minus1 is strictly increasing on 1198690 =[120593(119891(120572)) 1] Suppose that 119891119898+1(1) is not an extreme pointof 120593 then 120593 is strictly decreasing on [119891

119898+1(120572) 119891

119898+1(1)]

And from (4) and the fact that 120593 is strictly increasingon [119891

119898(120572) 119891

119898(1)] we get that 120593119901 is strictly increasing

on [119891119898+1

(120572) 119891119898+1

(1)] Thereby 120593119901minus1 is strictly decreasingon [120593(119891

119898+1(1)) 120593(119891

119898+1(120572))] This contradicts that 120593119901minus1 is

strictly increasing on 1198690 = [120593(119891(120572)) 1] Thus 119891119898+1(1) is theextreme point of 120593

Lastly we have that 119891119898+1(1) is the maximum point of 120593on [119891119898+1(120572) 120572] The proof is similar to 119891119898(120572) and we omit ithere Thereby 119891119899(1) is the maximum point of 120593 on [119891119899(120572) 120572]and 119891119899(120572) is the minimum point of 120593 on [0 119891119899(1)] for all 119899 ge1 by induction

Proof of Lemma 6 It is trivial that 119909 = 1 is a solution of theequation 120593119901minus1(119909) = 119891(119909) by (4) Suppose that 119909 = 1199090 isin

(120593(119891(120572)) 1] is an arbitrary solution of this equation that is120593119901minus1(1199090) = 119891(1199090) Since 120593([0 119891(120572)]) = [120593(119891(120572)) 1] we have

that there exists 119910 isin [0 119891(120572)] such that 120593(119910) = 1199090 We claimthat

120593 (119891119899(119910)) = 1199090 forall119899 ge 0 (A32)

Obviously (A32) holds for 119899 = 0 Suppose that (A32) holdsfor 119899 = 119896 where 119896 ge 0 is a certain integer Therefore by (4)we have that

120593119901minus1

(120593 (119891119896+1

(119910))) = 119891 (120593 (119891119896(119910))) = 119891 (1199090) = 120593

119901minus1(1199090)

(A33)

Since119891119896+1(119910) isin [0 119891(120572)) we have 120593(119891119896+1(119910)) isin (120593(119891(120572)) 1]And since 120593

119901minus1 is strictly increasing on [120593(119891(120572)) 1] by120593(119891(120572)) gt 120572 we have 120593(119891119896+1(119910)) = 1199090 That is (A32) holdsfor 119899 = 119896 + 1 Thereby (A32) is proved by induction By thefact that 119891119899(119910) is strictly decreasing and lim119899rarrinfin119891

119899(119910) = 0

we have 1199090 = lim119899rarr+infin120593(119891119899(119910)) = 120593(0) = 1

Proof of Lemma 7 There exist 120572 isin (120582 1) 120573 isin (120582 1) such that120593119894(120572) = 0 120593119894(120573) = 120573 (119894 = 1 2) by (8) Denote 1205930(119909) = 1205931(119909) =1205932(119909) (119909 isin [120582 1]) and 1205721 = 1205931(119891(120572)) lt 1205931(120582) = 1205930(120582) 1205722 =1205932(119891(120572)) lt 1205932(120582) = 1205930(120582) We prove that 1205721 = 1205722 as followsIt is trivial that 1205721 gt 120582 1205722 gt 120582 and

120593119901minus1

0(1205721) = 120593

119901minus1

1(1205721) = 120593

119901

1(119891 (120572)) = 0

120593119901minus1

0(1205722) = 120593

119901minus1

2(1205722) = 120593

119901

2(119891 (120572)) = 0

(A34)

by Lemmas 3(v) and 4 Since 120593119901minus10

is strictly monotone on[min1205721 1205722 1] we have that 1205721 = 1205722 Let 1205720 = 1205721 = 1205722 then1205720 lt 1205930(120582) and 120593

119901minus1

0(1205720) = 0 Define 120595+ = 120593

119901minus1

0|[1205720 1]

ByLemmas 3(v) and 4 we have that 120595+ is a homeomorphism

And by 120595+(1205720) = 0 le 120595+(1) we know that 120595+ is strictlyincreasing

We prove 1205931(119909) = 1205932(119909) on Δ 119896 for all 119896 ge 0 by inductionas follows

Obviously 1205931(119909) = 1205932(119909) holds on Δ 0 Suppose that1205931(119909) = 1205932(119909) holds on Δ 119896 for all 119896 le 119898 where 119898 ge 0 isa certain integer Let

120593 (119909) = 1205931 (119909) = 1205932 (119909) 119909 isin [119891119898+1

(1) 1] (A35)

Since 120593119894(119909) ge 120593119894(119891(120572)) = 1205720 gt 120582 for 119909 le 120582 and from (4) wehave

119891 (120593 (119891minus1(119909))) = 119891 (120593119894 (119891

minus1(119909))) = 120593

119901minus1

119894(120593119894 (119909))

= 120595+ (120593119894 (119909)) (119894 = 1 2 119909 isin Δ119898+1)

(A36)

Thereby

120593119894 (119909) = 120595minus1

+(119891 (120593 (119891

minus1(119909)))) (119894 = 1 2 119909 isin Δ119898+1)

(A37)

Thus we have 1205931(119909) = 1205932(119909) on Δ119898+1 Thus we get 1205931(119909) =1205932(119909) on Δ 119896 for all 119896 ge 0 by induction

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is partially supported by the National NaturalScience Foundation of China Tian Yuan Foundation (no11326129) and the Fundamental Research Funds for theCentral Universities (no 14CX02152A)

References

[1] M J Feigenbaum ldquoQuantitative universality for a class ofnonlinear transformationsrdquo Journal of Statistical Physics vol 19no 1 pp 25ndash52 1978

[2] M J Feigenbaum ldquoTheuniversalmetric properties of nonlineartransformationsrdquo Journal of Statistical Physics vol 21 no 6 pp669ndash706 1979

[3] P Couliet and C Tresser ldquoIterations drsquoendomorphismes etgroupe de renormalisationrdquo Journal de Physique Colloques vol39 no 5 pp C5-25ndashC5-28 1978

[4] D Sullivan ldquoBoubds quadratic differentials and renormaliza-tion conjecturesrdquo in Proceedings of the Mathematics into theTwenty-First Century Centennial Symposium F Browder Edpp 417ndash466 American Mathematical Society August 1992

[5] ADouady and JHHubbard ldquoOn the dynamics of polynomial-like mappingsrdquo Annales Scientifiques de lrsquoEcole NormaleSuperieure vol 18 no 2 pp 287ndash343 1985

[6] I Lanford ldquoA computer-assisted proof of the Feigenbaumconjecturesrdquo The American Mathematical Society vol 6 no 3pp 427ndash434 1982

8 Abstract and Applied Analysis

[7] J P Eckmann and P Wittwer ldquoA complete proof of theFeigenbaum conjecturesrdquo Journal of Statistical Physics vol 46no 3-4 pp 455ndash475 1987

[8] M Campanino and H Epstein ldquoOn the existence of Feigen-baumrsquos fixed pointrdquo Communications in Mathematical Physicsvol 79 no 2 pp 261ndash302 1981

[9] L Yang and J Z Zhang ldquoThe second type of Feigenbaumrsquosfunctional equationrdquo Science in China A vol 28 pp 1061ndash10691985

[10] P J McCarthy ldquoThe general exact bijective continuous solu-tion of Feigenbaumrsquos functional equationrdquo Communications inMathematical Physics vol 91 no 3 pp 431ndash443 1983

[11] J P Eckmann H Epstein and P Wittwer ldquoFixed pointsof Feigenbaumrsquos type for the equation 119891119901 (120582119909) = 120582119891(119909)rdquoCommunications in Mathematical Physics vol 93 no 4 pp495ndash516 1984

[12] A H Zhang L J Wang and W Song ldquoNonsingle-valley con-tinuous solutions of p-order Feigenbaums functional equationrdquoNortheastern Mathematical Journal vol 19 no 4 pp 375ndash3802003

[13] G Liao L Wang and Y Zhang ldquo119901-order Feigenbaumrsquos mapsrdquoNortheastern Mathematical Journal Dongbei Shuxue vol 20no 3 pp 284ndash290 2004

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Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

This contradicts that 0 is not a periodic point That iswe proved (A8) Thereby we have 120593(119891119899(120572)) = 120593(119891

119898+1(120572))

forall119899 ge 119898 + 2 And from the fact that 120593 is strictly increasingon Δ120572119898+1

we get 120593(119891119898+2(120572)) lt 120593(119891119898+1

(120572)) Suppose that120593(119891119899(120572)) lt 120593(119891

119898+1(120572)) where 119899 ge 119898 + 2 is a certain integer

If 120593(119891119899+1(120572)) gt 120593(119891119898+1(120572)) gt 120593(119891119899(120572)) and by the fact that120593119901119898+1minus1 is strictlymonotone on120593([119891119899+1(120572) 119891119899(120572)]) and (A3)

we get

120593119901119898+1minus1(120593 (119891

119899+1(120572)))

lt 120593119901119898+1minus1(120593 (119891

119898+1(120572)))

= 120593119901119898+1

(119891119898+1

(120572)) = 119891119898+1

(0) = 0

(A11)

or

120593119901119898+1minus1(120593 (119891119899(120572)))

lt 120593119901119898+1minus1(120593 (119891

119898+1(120572)))

= 120593119901119898+1

(119891119898+1

(120572)) = 119891119898+1

(0) = 0

(A12)

This contradicts that 120593119901119898+1minus1(119909) ge 0 That is 120593(119891119899+1(120572)) lt

120593(119891119898+1

(120572)) Thereby we proved (A7) by induction If120593(119891119898+1

(120572)) = 1 then by (A3) we have

120593119901119898+1

(1) = 120593119901119898+1

(120593 (119891119898+1

(120572))) = 120593 (120593119901119898+1

(119891119898+1

(120572)))

= 120593 (119891119898+1

(120593 (120572))) = 120593 (0) = 1

(A13)

This contradicts conclusion (ii) Thereby we get 120593(119891119899(120572)) lt120593(119891119898+1

(120572)) lt 1 forall119899 ge 119898 + 2 This contradicts that 120593(0) =1 Thereby 120593 is strictly decreasing on Δ120572

119898+1 Furthermore

we proved that 120593 is strictly decreasing on Δ120572

119896(119896 ge 0)

respectively by induction that is 120593 is single-valley solutionof (4) This contradicts the condition that 120593 is a non-single-valley solution of (4) That is we proved that 120582 is an extremepoint of 120593

Secondly we prove that 120593(120582) gt 120582 We claim that

120593 (119891119899(1)) = 120582 120593 (119891

119899(120572)) = 120582 forall119899 ge 1 (A14)

Suppose that 120593(119891119899(1)) = 120582 by (A5) we get 120593(119891119899(1)) =

120593119901119899+1(0) And from 120593

119901(0) = 120582 we have that 120593119901(0) = 120593119901

119899+1(0)

Let

119871 = 0 120593 (0) 120593119901(0) 120593

119901+1(0) 120593

119901119899

(0) (A15)

obviously 119871 is a limited set and 120593119894(0) isin 119871 forall119894 isin 119885

+By (A5) we have that 119891119894(1) = 120593

119901119894

(0) isin 119871 forall119894 isin 119885+

This contradicts that 119871 is limited Thereby we proved that120593(119891119899(1)) = 120582 Suppose that 120593(119891119899(120572)) = 120582 from (A3) we have

0 = 119891119899(120593 (120572)) = 120593

119901119899

(119891119899(120572))

= 120593119901119899minus1(120582) = 120593

119901119899+119901minus1

(0)

(A16)

This contradicts that 0 is not a periodic point of 120593 That is(A14) holds Suppose that 120593(120582) lt 120582 Since 120582 is an extremepoint of 120593 we have that 120593 is strictly increasing on [119891(120572) 120582]Furthermore we have 120593(119891(120572)) lt 120593(120582) lt 120582 We claim that

120593 (119891119899(1)) lt 120582 120593 (119891

119899(120572)) lt 120582 forall119899 ge 1 (A17)

It is trivial that (A17) holds for 119899 = 1 Suppose that (A17)holds for 119899 = 119898 where 119898 is a certain integer By (A3)we know 120593

119901119898

is strictly monotone on [119891119898+1(1) 119891119898(120572)] If120593(119891119898+1

(1)) gt 120582 then there exists 1199090 isin (119891119898+1

(1) 119891119898(120572))

such that 120593(1199090) = 120582 Thus 1199090 is an extreme point of1205932 on (119891

119898+1(1) 119891119898(120572)) That is 1205932 is not monotone on

[119891119898+1

(1) 119891119898(120572)] Furthermore 120593119901

119898

is not monotone on[119891119898+1

(1) 119891119898(120572)] This contradicts that 120593119901

119898

is monotone on[119891119898+1

(1) 119891119898(120572)] Thus 120593(119891119898+1(1)) lt 120582 Similarly we have

120593(119891119898+1

(120572)) lt 120582 That is (A17) holds for 119899 = 119898 + 1 Thereby(A17) holds by induction This contradicts that 120593(0) = 1Thus 120593(120582) gt 120582

(iv) Firstly suppose that 119902 is a fixed point of 120593 then by(A5) we have 119902 = 1 And by 120593(120572) = 0 we have 119902 = 120572 If 119902 isin(120572 1) then 119902 = 120593(119902) lt 120593(1) And by induction for all119898 ge 0we have 119902 = 120593119898(119902) lt 120593119898(1) Specially

119902 = 120593119901119899minus1(119902) lt 120593

119901119899minus1(1) = 120593

119901119899minus1(120593 (0)) = 120593

119901119899

(0) (A18)

This contradicts (A5) Thereby 119902 lt 120572Secondly there exists at least one fixed point 120573 isin (0 120572) by

120593(0) = 1 120593(120572) = 0 We prove 120573 is the unique fixed point asfollows We claim that

120573 notin (119891119899+1

(1) 119891119899(1)] forall119899 ge 1 (A19)

Suppose that there exists119898 ge 1 such that120573 isin [119891119898(120572) 119891119898(1)]then by (A3) we have

119891119898(120593 (119891minus119898(120573))) = 120593

119901119898

(119891119898(119891minus119898(120573))) = 120593

119901119898

(120573) = 120573

(A20)

that is 120593(119891minus119898(120573)) = 119891minus119898(120573) And from 119891minus119898(120573) isin [120572 1] this

contradicts that 120593 does not have a fixed point on [120572 1] Thuswe have 120573 notin [119891119899(120572) 119891119899(1)] forall119899 ge 1 Suppose that there exists119898 ge 1 such that 120573 isin (119891119898+1(1) 119891119898(120572)) then by (A3) we have

119891119898minus1

(120593 (1198911minus119898

(120573))) = 120593119901119898minus1(119891119898minus1

(1198911minus119898

(120573)))

= 120593119901119898minus1(120573) = 120573

(A21)

Thereby 120593(1198911minus119898(120573)) = 1198911minus119898

(120573) That is 1198911minus119898(120573) isin (1198912(1)119891(120572)) is a fixed point of 120593 Since 120593 has no fixed pointon [119891(120572) 119891(1)] and 120593(120582) gt 120582 we have that 120593 must bestrictly increasing on [1198912(1) 119891(120572)] And 120593 has the fixed point1198911minus119898

(120573) isin (1198912(1) 119891(120572)) thus 120593119901 is strictly increasing on

[1198912(1) 119891(120572)] But by (4) and the fact that 120593 is decreasing on

[120582 120572] we have that 120593119901 is strictly decreasing on [1198912(1) 119891(120572)]This contradicts Thus we have 120573 notin [119891119899+1(1) 119891119899(120572)] forall119899 ge 1Thereby we prove (A19)Thus we have 120573 isin (120582 120572) And since120593 is decreasing on [120582 120572] we have that 120573 isin (120582 120572) is the uniquefixed point

6 Abstract and Applied Analysis

(v) By (4) we have 0 = 119891(120593(120572)) = 120593119901(119891(120572)) And

since 120572 is the unique minimum point of 120593 it follows that120593119901minus1(119891(120572)) = 120572 Thus the sufficiency is provedWe prove the necessity as follows Suppose that 120593119894(119909) = 120572

for some 119909 isin [0 120582] and 0 le 119894 le 119901 minus 1 Firstly we claim that

119909 notin (119891119899+1

(1) 119891119899(120572)) cup (119891

119899(120572) 119891

119899(1)) forall119899 ge 1

(A22)

Suppose that 119909 isin (119891119899(120572) 119891

119899(1)) then 120593119894+1 is not mono-

tone on (119891119899(120572) 119891

119899(1)) Thereby 120593119901 is not monotone on

(119891119899(120572) 119891

119899(1)) This contradicts that 120593119901 is monotone on

(119891119899(120572) 119891

119899(1)) by (A3) Thus 119909 notin (119891

119899(120572) 119891

119899(1)) We have

similarly that 119909 notin (119891119899+1(1) 119891119899(120572)) Thereby we prove (A22)that is 119909 = 119891119899(1) or 119909 = 119891119899(120572) forall119899 ge 1

Secondly we prove that

119909 = 119891119899(1) 119909 = 119891

119899+1(120572) forall119899 ge 1 (A23)

Suppose that 119909 = 119891119899(1) for some 119899 ge 1 then 120593119894(119891119899(1)) =

120572 And from (A5) we have 120572 = 120593119894(119891119899(1)) = 120593

119894(120593119901119899

(0)) =

120593119894+119901119899

(0) This contradicts that 0 is not a periodic point of 120593Thus we prove 119909 = 119891

119899(1) forall119899 ge 1 Suppose that 119909 = 119891119899(120572) for

some 119899 ge 2 then 120593119894(119891119899(120572)) = 120572 And from (A3) we have

120593119894(0) = 120593

119894(120593119901119899

(119891119899(120572))) = 120593

119901119899

(120593119894(119891119899(120572)))

= 120593119901119899

(120572) = 120593119901119899minus1(0)

(A24)

This contradicts that 0 is a recurrent point of120593Thuswe prove119909 = 119891119899(120572) forall119899 ge 2 Thereby (A23) holds That is we get 119909 =

119891(120572) and 120593119894(119891(120572)) = 120572Lastly we prove 119894 = 119901 minus 1 Suppose that there exists

119895 = 119894 0 le 119895 le 119901 minus 1 such that 120593119895(119891(120572)) = 120572 Then we cansuppose that 119895 lt 119894 and

0 = 120593119894+1(119891 (120572)) = 120593

119894minus119895(120593119895+1(119891 (120572))) = 120593

119894minus119895(0) (A25)

This contradicts that 0 is not a periodic point of 120593 Thus weprove 120593119895(119891(120572)) = 120572 forall119895 = 119894 Thereby we have 119894 = 119901 minus 1 and119909 = 119891(120572)

Proof of Lemma 4 (i) Firstly we prove that for all 119894 =

0 1 119901 minus 2 we have 119869119894 sub (120582 1] that is 119869119894 cap 119869 = 0 Weclaim that

120593119894+1(119891 (120572)) gt 120582 forall0 le 119894 le 119901 minus 2 (A26)

Suppose that there exists 1 le 119895 le 119901 minus 1 such that 120593119895(119891(120572)) =119909 le 120582 And from Lemma 3(v) we have 120572 = 120593

119901minus1(119891(120572)) =

120593119901minus1minus119895

(120593119895(119891(120572))) = 120593

119901minus1minus119895(119909) This contradicts Lemma 3(v)

Thus (A26) holds We claim that

120593119894+1(120582) gt 120582 forall0 le 119894 le 119901 minus 2 (A27)

It is trivial that 120593119894+1(120582) = 120582 by 0 is a recurrent point of120593 Suppose that there exists 1 le 119895 le 119901 minus 1 such that120593119895(120582) lt 120582 And from 120593

119895(119891(120572)) gt 120582 we have that there exists

119909 isin (119891(120572) 119891(1)) such that 120593119895(119909) = 120582 And from the fact that120582 is an extreme point of 120593 we get that 120593119895+1 is not monotoneon [119891(120572) 119891(1)] This contradicts that 120593119901 is strictly monotoneon [119891(120572) 119891(1)] Thus (A27) holds We claim that

120593119894+1(119909) gt 120582 forall0 le 119894 le 119901 minus 2 119909 isin [0 120582] (A28)

Suppose that there exists 119909 isin [0 120582) and 1 le 119895 le 119901 minus 1 suchthat 120593119895(119909) = 120582 And from (4) we get 120593119901minus119895(120582) = 120593

119901(119909) =

119891(120593(119891minus1(119909))) le 119891(1) = 120582 This contradicts (A27) Thus

120593119894+1(119909) = 120582 forall0 le 119894 le 119901 minus 2 119909 isin [0 120582] Suppose that there

exists 119909 isin [0 120582) and 1 le 119895 le 119901 minus 1 such that 120593119895(119909) lt 120582And from (A27) we have that there exists 119911 isin (119909 120582) suchthat 120593119895(119911) = 120582 And from (4) we get 120593119901minus119895(120582) = 120593

119901(119911) =

119891(120593(119891minus1(119911))) le 119891(1) = 120582 This contradicts (A27) Thus

(A28) holds That is 119869119894 cap 119869 = 0Secondly we prove that 119869119894 forall0 le 119894 le 119901 minus 2 are pairwise

disjoint Suppose that there exists 0 le 119894 lt 119895 le 119901minus 2 such that119869119894 cap 119869119895 = 119869119894119895 = 0 Let 119910 isin 119869119894119895 then there exist 119909119894 isin [0 120582] 119909119895 isin[0 120582] such that 120593119894+1(119909119894) = 119910 = 120593

119895+1(119909119895) Thereby we have

120593119901minus119895+119894

(119909119894) = 120593119901minus1minus119895

(120593119894+1(119909119894)) = 120593

119901minus1minus119895(120593119895+1(119909119895)) = 120593

119901(119909119895) =

119891(120593(119891minus1(119909119895))) le 119891(1) = 120582 This contradicts 119869119894 sub (120582 1] Thus

we proved that 1198690 1198691 119869119901minus2 are pairwise disjoint(ii) For all 119894 = 0 1 119901 minus 2 then 120593119894+1 119869 997891rarr 119869119894 is a

homeomorphismby Lemmas 3(i) and 3(v)Thereby120593119894 1198690 997891rarr119869119894 is also a homeomorphism

Proof of Lemma 5 120582 is a maximum point of 120593 on [119891(120572) 120572]by Lemma 3(iii) Suppose that 119891119898(1) is the maximum pointof 120593 on [119891119898(120572) 120572] where 119898 is some certain integer Firstlywe prove 119891119898(120572) is an extreme point of 120593 If otherwise 120593 isstrictly increasing on [119891119898+1(1) 119891119898(120572)] Thus 120593(119891119898+1(1)) lt120593(119891119898(120572)) and 120593 is strictly increasing on Δ119898 We claim that 120593

is strictly increasing on Δ 119896 for all 119896 ge 119898 It holds obviouslywhen 119896 = 119898 Suppose that 120593 is strictly increasing on Δ 119896 forall 119898 le 119896 le 119897 where 119897 is some certain integer 120593119901 is strictlyincreasing on Δ 119897+1 by (4) Thereby 120593 is strictly monotone onΔ 119897+1 If 120593 is strictly decreasing on Δ 119897+1 then we claim that

120593 (119891119904(1)) lt 120593 (119891

119897(120572)) forall119904 ge 119897 + 1 (A29)

Its proof is similar to (A7) and we omit it here Thereby120593(119891119904(1)) lt 120593(119891

119897(120572)) lt 1 forall119904 ge 119897 + 1 This contradicts

120593(0) = 1 Thus 120593 is strictly increasing on Δ 119897+1 Thereby 120593 isstrictly increasing on [0 119891119898(1)] by induction Furthermore120593(0) lt 120593(119891

119898(1)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is an extreme point of 120593Secondly we prove 119891119898(120572) is the minimum point of 120593 on

[0 119891119898(1)] Suppose that there exists 119897 ge 119898 + 1 such that

120593(119891119897(120572)) lt 120593(119891

119898(120572)) or 120593(119891119897(1)) lt 120593(119891119898(120572)) then we claim

that

120593 (119891119904(120572)) lt 120593 (119891

119898(120572)) forall119904 ge 119897 (A30)

or

120593 (119891119904(1)) lt 120593 (119891

119898(120572)) forall119904 ge 119897 (A31)

Abstract and Applied Analysis 7

The proof is similar to (A7) and we omit it here Thereby120593(0) lt 120593(119891

119898(120572)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is the minimum point of 120593 on [0 119891119898(1)]Thirdly we prove 119891119898+1(1) is an extreme point of 120593

120593119901minus1 is strictly increasing on [120593(119891119898(120572)) 120593(119891119898+1(1))] by (4)

and 120593 is strictly decreasing on [119891119898+1(1) 119891119898(120572)] And fromLemma 4(ii) we have that 120593119901minus1 is strictly increasing on 1198690 =[120593(119891(120572)) 1] Suppose that 119891119898+1(1) is not an extreme pointof 120593 then 120593 is strictly decreasing on [119891

119898+1(120572) 119891

119898+1(1)]

And from (4) and the fact that 120593 is strictly increasingon [119891

119898(120572) 119891

119898(1)] we get that 120593119901 is strictly increasing

on [119891119898+1

(120572) 119891119898+1

(1)] Thereby 120593119901minus1 is strictly decreasingon [120593(119891

119898+1(1)) 120593(119891

119898+1(120572))] This contradicts that 120593119901minus1 is

strictly increasing on 1198690 = [120593(119891(120572)) 1] Thus 119891119898+1(1) is theextreme point of 120593

Lastly we have that 119891119898+1(1) is the maximum point of 120593on [119891119898+1(120572) 120572] The proof is similar to 119891119898(120572) and we omit ithere Thereby 119891119899(1) is the maximum point of 120593 on [119891119899(120572) 120572]and 119891119899(120572) is the minimum point of 120593 on [0 119891119899(1)] for all 119899 ge1 by induction

Proof of Lemma 6 It is trivial that 119909 = 1 is a solution of theequation 120593119901minus1(119909) = 119891(119909) by (4) Suppose that 119909 = 1199090 isin

(120593(119891(120572)) 1] is an arbitrary solution of this equation that is120593119901minus1(1199090) = 119891(1199090) Since 120593([0 119891(120572)]) = [120593(119891(120572)) 1] we have

that there exists 119910 isin [0 119891(120572)] such that 120593(119910) = 1199090 We claimthat

120593 (119891119899(119910)) = 1199090 forall119899 ge 0 (A32)

Obviously (A32) holds for 119899 = 0 Suppose that (A32) holdsfor 119899 = 119896 where 119896 ge 0 is a certain integer Therefore by (4)we have that

120593119901minus1

(120593 (119891119896+1

(119910))) = 119891 (120593 (119891119896(119910))) = 119891 (1199090) = 120593

119901minus1(1199090)

(A33)

Since119891119896+1(119910) isin [0 119891(120572)) we have 120593(119891119896+1(119910)) isin (120593(119891(120572)) 1]And since 120593

119901minus1 is strictly increasing on [120593(119891(120572)) 1] by120593(119891(120572)) gt 120572 we have 120593(119891119896+1(119910)) = 1199090 That is (A32) holdsfor 119899 = 119896 + 1 Thereby (A32) is proved by induction By thefact that 119891119899(119910) is strictly decreasing and lim119899rarrinfin119891

119899(119910) = 0

we have 1199090 = lim119899rarr+infin120593(119891119899(119910)) = 120593(0) = 1

Proof of Lemma 7 There exist 120572 isin (120582 1) 120573 isin (120582 1) such that120593119894(120572) = 0 120593119894(120573) = 120573 (119894 = 1 2) by (8) Denote 1205930(119909) = 1205931(119909) =1205932(119909) (119909 isin [120582 1]) and 1205721 = 1205931(119891(120572)) lt 1205931(120582) = 1205930(120582) 1205722 =1205932(119891(120572)) lt 1205932(120582) = 1205930(120582) We prove that 1205721 = 1205722 as followsIt is trivial that 1205721 gt 120582 1205722 gt 120582 and

120593119901minus1

0(1205721) = 120593

119901minus1

1(1205721) = 120593

119901

1(119891 (120572)) = 0

120593119901minus1

0(1205722) = 120593

119901minus1

2(1205722) = 120593

119901

2(119891 (120572)) = 0

(A34)

by Lemmas 3(v) and 4 Since 120593119901minus10

is strictly monotone on[min1205721 1205722 1] we have that 1205721 = 1205722 Let 1205720 = 1205721 = 1205722 then1205720 lt 1205930(120582) and 120593

119901minus1

0(1205720) = 0 Define 120595+ = 120593

119901minus1

0|[1205720 1]

ByLemmas 3(v) and 4 we have that 120595+ is a homeomorphism

And by 120595+(1205720) = 0 le 120595+(1) we know that 120595+ is strictlyincreasing

We prove 1205931(119909) = 1205932(119909) on Δ 119896 for all 119896 ge 0 by inductionas follows

Obviously 1205931(119909) = 1205932(119909) holds on Δ 0 Suppose that1205931(119909) = 1205932(119909) holds on Δ 119896 for all 119896 le 119898 where 119898 ge 0 isa certain integer Let

120593 (119909) = 1205931 (119909) = 1205932 (119909) 119909 isin [119891119898+1

(1) 1] (A35)

Since 120593119894(119909) ge 120593119894(119891(120572)) = 1205720 gt 120582 for 119909 le 120582 and from (4) wehave

119891 (120593 (119891minus1(119909))) = 119891 (120593119894 (119891

minus1(119909))) = 120593

119901minus1

119894(120593119894 (119909))

= 120595+ (120593119894 (119909)) (119894 = 1 2 119909 isin Δ119898+1)

(A36)

Thereby

120593119894 (119909) = 120595minus1

+(119891 (120593 (119891

minus1(119909)))) (119894 = 1 2 119909 isin Δ119898+1)

(A37)

Thus we have 1205931(119909) = 1205932(119909) on Δ119898+1 Thus we get 1205931(119909) =1205932(119909) on Δ 119896 for all 119896 ge 0 by induction

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is partially supported by the National NaturalScience Foundation of China Tian Yuan Foundation (no11326129) and the Fundamental Research Funds for theCentral Universities (no 14CX02152A)

References

[1] M J Feigenbaum ldquoQuantitative universality for a class ofnonlinear transformationsrdquo Journal of Statistical Physics vol 19no 1 pp 25ndash52 1978

[2] M J Feigenbaum ldquoTheuniversalmetric properties of nonlineartransformationsrdquo Journal of Statistical Physics vol 21 no 6 pp669ndash706 1979

[3] P Couliet and C Tresser ldquoIterations drsquoendomorphismes etgroupe de renormalisationrdquo Journal de Physique Colloques vol39 no 5 pp C5-25ndashC5-28 1978

[4] D Sullivan ldquoBoubds quadratic differentials and renormaliza-tion conjecturesrdquo in Proceedings of the Mathematics into theTwenty-First Century Centennial Symposium F Browder Edpp 417ndash466 American Mathematical Society August 1992

[5] ADouady and JHHubbard ldquoOn the dynamics of polynomial-like mappingsrdquo Annales Scientifiques de lrsquoEcole NormaleSuperieure vol 18 no 2 pp 287ndash343 1985

[6] I Lanford ldquoA computer-assisted proof of the Feigenbaumconjecturesrdquo The American Mathematical Society vol 6 no 3pp 427ndash434 1982

8 Abstract and Applied Analysis

[7] J P Eckmann and P Wittwer ldquoA complete proof of theFeigenbaum conjecturesrdquo Journal of Statistical Physics vol 46no 3-4 pp 455ndash475 1987

[8] M Campanino and H Epstein ldquoOn the existence of Feigen-baumrsquos fixed pointrdquo Communications in Mathematical Physicsvol 79 no 2 pp 261ndash302 1981

[9] L Yang and J Z Zhang ldquoThe second type of Feigenbaumrsquosfunctional equationrdquo Science in China A vol 28 pp 1061ndash10691985

[10] P J McCarthy ldquoThe general exact bijective continuous solu-tion of Feigenbaumrsquos functional equationrdquo Communications inMathematical Physics vol 91 no 3 pp 431ndash443 1983

[11] J P Eckmann H Epstein and P Wittwer ldquoFixed pointsof Feigenbaumrsquos type for the equation 119891119901 (120582119909) = 120582119891(119909)rdquoCommunications in Mathematical Physics vol 93 no 4 pp495ndash516 1984

[12] A H Zhang L J Wang and W Song ldquoNonsingle-valley con-tinuous solutions of p-order Feigenbaums functional equationrdquoNortheastern Mathematical Journal vol 19 no 4 pp 375ndash3802003

[13] G Liao L Wang and Y Zhang ldquo119901-order Feigenbaumrsquos mapsrdquoNortheastern Mathematical Journal Dongbei Shuxue vol 20no 3 pp 284ndash290 2004

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Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

(v) By (4) we have 0 = 119891(120593(120572)) = 120593119901(119891(120572)) And

since 120572 is the unique minimum point of 120593 it follows that120593119901minus1(119891(120572)) = 120572 Thus the sufficiency is provedWe prove the necessity as follows Suppose that 120593119894(119909) = 120572

for some 119909 isin [0 120582] and 0 le 119894 le 119901 minus 1 Firstly we claim that

119909 notin (119891119899+1

(1) 119891119899(120572)) cup (119891

119899(120572) 119891

119899(1)) forall119899 ge 1

(A22)

Suppose that 119909 isin (119891119899(120572) 119891

119899(1)) then 120593119894+1 is not mono-

tone on (119891119899(120572) 119891

119899(1)) Thereby 120593119901 is not monotone on

(119891119899(120572) 119891

119899(1)) This contradicts that 120593119901 is monotone on

(119891119899(120572) 119891

119899(1)) by (A3) Thus 119909 notin (119891

119899(120572) 119891

119899(1)) We have

similarly that 119909 notin (119891119899+1(1) 119891119899(120572)) Thereby we prove (A22)that is 119909 = 119891119899(1) or 119909 = 119891119899(120572) forall119899 ge 1

Secondly we prove that

119909 = 119891119899(1) 119909 = 119891

119899+1(120572) forall119899 ge 1 (A23)

Suppose that 119909 = 119891119899(1) for some 119899 ge 1 then 120593119894(119891119899(1)) =

120572 And from (A5) we have 120572 = 120593119894(119891119899(1)) = 120593

119894(120593119901119899

(0)) =

120593119894+119901119899

(0) This contradicts that 0 is not a periodic point of 120593Thus we prove 119909 = 119891

119899(1) forall119899 ge 1 Suppose that 119909 = 119891119899(120572) for

some 119899 ge 2 then 120593119894(119891119899(120572)) = 120572 And from (A3) we have

120593119894(0) = 120593

119894(120593119901119899

(119891119899(120572))) = 120593

119901119899

(120593119894(119891119899(120572)))

= 120593119901119899

(120572) = 120593119901119899minus1(0)

(A24)

This contradicts that 0 is a recurrent point of120593Thuswe prove119909 = 119891119899(120572) forall119899 ge 2 Thereby (A23) holds That is we get 119909 =

119891(120572) and 120593119894(119891(120572)) = 120572Lastly we prove 119894 = 119901 minus 1 Suppose that there exists

119895 = 119894 0 le 119895 le 119901 minus 1 such that 120593119895(119891(120572)) = 120572 Then we cansuppose that 119895 lt 119894 and

0 = 120593119894+1(119891 (120572)) = 120593

119894minus119895(120593119895+1(119891 (120572))) = 120593

119894minus119895(0) (A25)

This contradicts that 0 is not a periodic point of 120593 Thus weprove 120593119895(119891(120572)) = 120572 forall119895 = 119894 Thereby we have 119894 = 119901 minus 1 and119909 = 119891(120572)

Proof of Lemma 4 (i) Firstly we prove that for all 119894 =

0 1 119901 minus 2 we have 119869119894 sub (120582 1] that is 119869119894 cap 119869 = 0 Weclaim that

120593119894+1(119891 (120572)) gt 120582 forall0 le 119894 le 119901 minus 2 (A26)

Suppose that there exists 1 le 119895 le 119901 minus 1 such that 120593119895(119891(120572)) =119909 le 120582 And from Lemma 3(v) we have 120572 = 120593

119901minus1(119891(120572)) =

120593119901minus1minus119895

(120593119895(119891(120572))) = 120593

119901minus1minus119895(119909) This contradicts Lemma 3(v)

Thus (A26) holds We claim that

120593119894+1(120582) gt 120582 forall0 le 119894 le 119901 minus 2 (A27)

It is trivial that 120593119894+1(120582) = 120582 by 0 is a recurrent point of120593 Suppose that there exists 1 le 119895 le 119901 minus 1 such that120593119895(120582) lt 120582 And from 120593

119895(119891(120572)) gt 120582 we have that there exists

119909 isin (119891(120572) 119891(1)) such that 120593119895(119909) = 120582 And from the fact that120582 is an extreme point of 120593 we get that 120593119895+1 is not monotoneon [119891(120572) 119891(1)] This contradicts that 120593119901 is strictly monotoneon [119891(120572) 119891(1)] Thus (A27) holds We claim that

120593119894+1(119909) gt 120582 forall0 le 119894 le 119901 minus 2 119909 isin [0 120582] (A28)

Suppose that there exists 119909 isin [0 120582) and 1 le 119895 le 119901 minus 1 suchthat 120593119895(119909) = 120582 And from (4) we get 120593119901minus119895(120582) = 120593

119901(119909) =

119891(120593(119891minus1(119909))) le 119891(1) = 120582 This contradicts (A27) Thus

120593119894+1(119909) = 120582 forall0 le 119894 le 119901 minus 2 119909 isin [0 120582] Suppose that there

exists 119909 isin [0 120582) and 1 le 119895 le 119901 minus 1 such that 120593119895(119909) lt 120582And from (A27) we have that there exists 119911 isin (119909 120582) suchthat 120593119895(119911) = 120582 And from (4) we get 120593119901minus119895(120582) = 120593

119901(119911) =

119891(120593(119891minus1(119911))) le 119891(1) = 120582 This contradicts (A27) Thus

(A28) holds That is 119869119894 cap 119869 = 0Secondly we prove that 119869119894 forall0 le 119894 le 119901 minus 2 are pairwise

disjoint Suppose that there exists 0 le 119894 lt 119895 le 119901minus 2 such that119869119894 cap 119869119895 = 119869119894119895 = 0 Let 119910 isin 119869119894119895 then there exist 119909119894 isin [0 120582] 119909119895 isin[0 120582] such that 120593119894+1(119909119894) = 119910 = 120593

119895+1(119909119895) Thereby we have

120593119901minus119895+119894

(119909119894) = 120593119901minus1minus119895

(120593119894+1(119909119894)) = 120593

119901minus1minus119895(120593119895+1(119909119895)) = 120593

119901(119909119895) =

119891(120593(119891minus1(119909119895))) le 119891(1) = 120582 This contradicts 119869119894 sub (120582 1] Thus

we proved that 1198690 1198691 119869119901minus2 are pairwise disjoint(ii) For all 119894 = 0 1 119901 minus 2 then 120593119894+1 119869 997891rarr 119869119894 is a

homeomorphismby Lemmas 3(i) and 3(v)Thereby120593119894 1198690 997891rarr119869119894 is also a homeomorphism

Proof of Lemma 5 120582 is a maximum point of 120593 on [119891(120572) 120572]by Lemma 3(iii) Suppose that 119891119898(1) is the maximum pointof 120593 on [119891119898(120572) 120572] where 119898 is some certain integer Firstlywe prove 119891119898(120572) is an extreme point of 120593 If otherwise 120593 isstrictly increasing on [119891119898+1(1) 119891119898(120572)] Thus 120593(119891119898+1(1)) lt120593(119891119898(120572)) and 120593 is strictly increasing on Δ119898 We claim that 120593

is strictly increasing on Δ 119896 for all 119896 ge 119898 It holds obviouslywhen 119896 = 119898 Suppose that 120593 is strictly increasing on Δ 119896 forall 119898 le 119896 le 119897 where 119897 is some certain integer 120593119901 is strictlyincreasing on Δ 119897+1 by (4) Thereby 120593 is strictly monotone onΔ 119897+1 If 120593 is strictly decreasing on Δ 119897+1 then we claim that

120593 (119891119904(1)) lt 120593 (119891

119897(120572)) forall119904 ge 119897 + 1 (A29)

Its proof is similar to (A7) and we omit it here Thereby120593(119891119904(1)) lt 120593(119891

119897(120572)) lt 1 forall119904 ge 119897 + 1 This contradicts

120593(0) = 1 Thus 120593 is strictly increasing on Δ 119897+1 Thereby 120593 isstrictly increasing on [0 119891119898(1)] by induction Furthermore120593(0) lt 120593(119891

119898(1)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is an extreme point of 120593Secondly we prove 119891119898(120572) is the minimum point of 120593 on

[0 119891119898(1)] Suppose that there exists 119897 ge 119898 + 1 such that

120593(119891119897(120572)) lt 120593(119891

119898(120572)) or 120593(119891119897(1)) lt 120593(119891119898(120572)) then we claim

that

120593 (119891119904(120572)) lt 120593 (119891

119898(120572)) forall119904 ge 119897 (A30)

or

120593 (119891119904(1)) lt 120593 (119891

119898(120572)) forall119904 ge 119897 (A31)

Abstract and Applied Analysis 7

The proof is similar to (A7) and we omit it here Thereby120593(0) lt 120593(119891

119898(120572)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is the minimum point of 120593 on [0 119891119898(1)]Thirdly we prove 119891119898+1(1) is an extreme point of 120593

120593119901minus1 is strictly increasing on [120593(119891119898(120572)) 120593(119891119898+1(1))] by (4)

and 120593 is strictly decreasing on [119891119898+1(1) 119891119898(120572)] And fromLemma 4(ii) we have that 120593119901minus1 is strictly increasing on 1198690 =[120593(119891(120572)) 1] Suppose that 119891119898+1(1) is not an extreme pointof 120593 then 120593 is strictly decreasing on [119891

119898+1(120572) 119891

119898+1(1)]

And from (4) and the fact that 120593 is strictly increasingon [119891

119898(120572) 119891

119898(1)] we get that 120593119901 is strictly increasing

on [119891119898+1

(120572) 119891119898+1

(1)] Thereby 120593119901minus1 is strictly decreasingon [120593(119891

119898+1(1)) 120593(119891

119898+1(120572))] This contradicts that 120593119901minus1 is

strictly increasing on 1198690 = [120593(119891(120572)) 1] Thus 119891119898+1(1) is theextreme point of 120593

Lastly we have that 119891119898+1(1) is the maximum point of 120593on [119891119898+1(120572) 120572] The proof is similar to 119891119898(120572) and we omit ithere Thereby 119891119899(1) is the maximum point of 120593 on [119891119899(120572) 120572]and 119891119899(120572) is the minimum point of 120593 on [0 119891119899(1)] for all 119899 ge1 by induction

Proof of Lemma 6 It is trivial that 119909 = 1 is a solution of theequation 120593119901minus1(119909) = 119891(119909) by (4) Suppose that 119909 = 1199090 isin

(120593(119891(120572)) 1] is an arbitrary solution of this equation that is120593119901minus1(1199090) = 119891(1199090) Since 120593([0 119891(120572)]) = [120593(119891(120572)) 1] we have

that there exists 119910 isin [0 119891(120572)] such that 120593(119910) = 1199090 We claimthat

120593 (119891119899(119910)) = 1199090 forall119899 ge 0 (A32)

Obviously (A32) holds for 119899 = 0 Suppose that (A32) holdsfor 119899 = 119896 where 119896 ge 0 is a certain integer Therefore by (4)we have that

120593119901minus1

(120593 (119891119896+1

(119910))) = 119891 (120593 (119891119896(119910))) = 119891 (1199090) = 120593

119901minus1(1199090)

(A33)

Since119891119896+1(119910) isin [0 119891(120572)) we have 120593(119891119896+1(119910)) isin (120593(119891(120572)) 1]And since 120593

119901minus1 is strictly increasing on [120593(119891(120572)) 1] by120593(119891(120572)) gt 120572 we have 120593(119891119896+1(119910)) = 1199090 That is (A32) holdsfor 119899 = 119896 + 1 Thereby (A32) is proved by induction By thefact that 119891119899(119910) is strictly decreasing and lim119899rarrinfin119891

119899(119910) = 0

we have 1199090 = lim119899rarr+infin120593(119891119899(119910)) = 120593(0) = 1

Proof of Lemma 7 There exist 120572 isin (120582 1) 120573 isin (120582 1) such that120593119894(120572) = 0 120593119894(120573) = 120573 (119894 = 1 2) by (8) Denote 1205930(119909) = 1205931(119909) =1205932(119909) (119909 isin [120582 1]) and 1205721 = 1205931(119891(120572)) lt 1205931(120582) = 1205930(120582) 1205722 =1205932(119891(120572)) lt 1205932(120582) = 1205930(120582) We prove that 1205721 = 1205722 as followsIt is trivial that 1205721 gt 120582 1205722 gt 120582 and

120593119901minus1

0(1205721) = 120593

119901minus1

1(1205721) = 120593

119901

1(119891 (120572)) = 0

120593119901minus1

0(1205722) = 120593

119901minus1

2(1205722) = 120593

119901

2(119891 (120572)) = 0

(A34)

by Lemmas 3(v) and 4 Since 120593119901minus10

is strictly monotone on[min1205721 1205722 1] we have that 1205721 = 1205722 Let 1205720 = 1205721 = 1205722 then1205720 lt 1205930(120582) and 120593

119901minus1

0(1205720) = 0 Define 120595+ = 120593

119901minus1

0|[1205720 1]

ByLemmas 3(v) and 4 we have that 120595+ is a homeomorphism

And by 120595+(1205720) = 0 le 120595+(1) we know that 120595+ is strictlyincreasing

We prove 1205931(119909) = 1205932(119909) on Δ 119896 for all 119896 ge 0 by inductionas follows

Obviously 1205931(119909) = 1205932(119909) holds on Δ 0 Suppose that1205931(119909) = 1205932(119909) holds on Δ 119896 for all 119896 le 119898 where 119898 ge 0 isa certain integer Let

120593 (119909) = 1205931 (119909) = 1205932 (119909) 119909 isin [119891119898+1

(1) 1] (A35)

Since 120593119894(119909) ge 120593119894(119891(120572)) = 1205720 gt 120582 for 119909 le 120582 and from (4) wehave

119891 (120593 (119891minus1(119909))) = 119891 (120593119894 (119891

minus1(119909))) = 120593

119901minus1

119894(120593119894 (119909))

= 120595+ (120593119894 (119909)) (119894 = 1 2 119909 isin Δ119898+1)

(A36)

Thereby

120593119894 (119909) = 120595minus1

+(119891 (120593 (119891

minus1(119909)))) (119894 = 1 2 119909 isin Δ119898+1)

(A37)

Thus we have 1205931(119909) = 1205932(119909) on Δ119898+1 Thus we get 1205931(119909) =1205932(119909) on Δ 119896 for all 119896 ge 0 by induction

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is partially supported by the National NaturalScience Foundation of China Tian Yuan Foundation (no11326129) and the Fundamental Research Funds for theCentral Universities (no 14CX02152A)

References

[1] M J Feigenbaum ldquoQuantitative universality for a class ofnonlinear transformationsrdquo Journal of Statistical Physics vol 19no 1 pp 25ndash52 1978

[2] M J Feigenbaum ldquoTheuniversalmetric properties of nonlineartransformationsrdquo Journal of Statistical Physics vol 21 no 6 pp669ndash706 1979

[3] P Couliet and C Tresser ldquoIterations drsquoendomorphismes etgroupe de renormalisationrdquo Journal de Physique Colloques vol39 no 5 pp C5-25ndashC5-28 1978

[4] D Sullivan ldquoBoubds quadratic differentials and renormaliza-tion conjecturesrdquo in Proceedings of the Mathematics into theTwenty-First Century Centennial Symposium F Browder Edpp 417ndash466 American Mathematical Society August 1992

[5] ADouady and JHHubbard ldquoOn the dynamics of polynomial-like mappingsrdquo Annales Scientifiques de lrsquoEcole NormaleSuperieure vol 18 no 2 pp 287ndash343 1985

[6] I Lanford ldquoA computer-assisted proof of the Feigenbaumconjecturesrdquo The American Mathematical Society vol 6 no 3pp 427ndash434 1982

8 Abstract and Applied Analysis

[7] J P Eckmann and P Wittwer ldquoA complete proof of theFeigenbaum conjecturesrdquo Journal of Statistical Physics vol 46no 3-4 pp 455ndash475 1987

[8] M Campanino and H Epstein ldquoOn the existence of Feigen-baumrsquos fixed pointrdquo Communications in Mathematical Physicsvol 79 no 2 pp 261ndash302 1981

[9] L Yang and J Z Zhang ldquoThe second type of Feigenbaumrsquosfunctional equationrdquo Science in China A vol 28 pp 1061ndash10691985

[10] P J McCarthy ldquoThe general exact bijective continuous solu-tion of Feigenbaumrsquos functional equationrdquo Communications inMathematical Physics vol 91 no 3 pp 431ndash443 1983

[11] J P Eckmann H Epstein and P Wittwer ldquoFixed pointsof Feigenbaumrsquos type for the equation 119891119901 (120582119909) = 120582119891(119909)rdquoCommunications in Mathematical Physics vol 93 no 4 pp495ndash516 1984

[12] A H Zhang L J Wang and W Song ldquoNonsingle-valley con-tinuous solutions of p-order Feigenbaums functional equationrdquoNortheastern Mathematical Journal vol 19 no 4 pp 375ndash3802003

[13] G Liao L Wang and Y Zhang ldquo119901-order Feigenbaumrsquos mapsrdquoNortheastern Mathematical Journal Dongbei Shuxue vol 20no 3 pp 284ndash290 2004

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

The proof is similar to (A7) and we omit it here Thereby120593(0) lt 120593(119891

119898(120572)) lt 1 This contradicts 120593(0) = 1 Thus 119891119898(120572)

is the minimum point of 120593 on [0 119891119898(1)]Thirdly we prove 119891119898+1(1) is an extreme point of 120593

120593119901minus1 is strictly increasing on [120593(119891119898(120572)) 120593(119891119898+1(1))] by (4)

and 120593 is strictly decreasing on [119891119898+1(1) 119891119898(120572)] And fromLemma 4(ii) we have that 120593119901minus1 is strictly increasing on 1198690 =[120593(119891(120572)) 1] Suppose that 119891119898+1(1) is not an extreme pointof 120593 then 120593 is strictly decreasing on [119891

119898+1(120572) 119891

119898+1(1)]

And from (4) and the fact that 120593 is strictly increasingon [119891

119898(120572) 119891

119898(1)] we get that 120593119901 is strictly increasing

on [119891119898+1

(120572) 119891119898+1

(1)] Thereby 120593119901minus1 is strictly decreasingon [120593(119891

119898+1(1)) 120593(119891

119898+1(120572))] This contradicts that 120593119901minus1 is

strictly increasing on 1198690 = [120593(119891(120572)) 1] Thus 119891119898+1(1) is theextreme point of 120593

Lastly we have that 119891119898+1(1) is the maximum point of 120593on [119891119898+1(120572) 120572] The proof is similar to 119891119898(120572) and we omit ithere Thereby 119891119899(1) is the maximum point of 120593 on [119891119899(120572) 120572]and 119891119899(120572) is the minimum point of 120593 on [0 119891119899(1)] for all 119899 ge1 by induction

Proof of Lemma 6 It is trivial that 119909 = 1 is a solution of theequation 120593119901minus1(119909) = 119891(119909) by (4) Suppose that 119909 = 1199090 isin

(120593(119891(120572)) 1] is an arbitrary solution of this equation that is120593119901minus1(1199090) = 119891(1199090) Since 120593([0 119891(120572)]) = [120593(119891(120572)) 1] we have

that there exists 119910 isin [0 119891(120572)] such that 120593(119910) = 1199090 We claimthat

120593 (119891119899(119910)) = 1199090 forall119899 ge 0 (A32)

Obviously (A32) holds for 119899 = 0 Suppose that (A32) holdsfor 119899 = 119896 where 119896 ge 0 is a certain integer Therefore by (4)we have that

120593119901minus1

(120593 (119891119896+1

(119910))) = 119891 (120593 (119891119896(119910))) = 119891 (1199090) = 120593

119901minus1(1199090)

(A33)

Since119891119896+1(119910) isin [0 119891(120572)) we have 120593(119891119896+1(119910)) isin (120593(119891(120572)) 1]And since 120593

119901minus1 is strictly increasing on [120593(119891(120572)) 1] by120593(119891(120572)) gt 120572 we have 120593(119891119896+1(119910)) = 1199090 That is (A32) holdsfor 119899 = 119896 + 1 Thereby (A32) is proved by induction By thefact that 119891119899(119910) is strictly decreasing and lim119899rarrinfin119891

119899(119910) = 0

we have 1199090 = lim119899rarr+infin120593(119891119899(119910)) = 120593(0) = 1

Proof of Lemma 7 There exist 120572 isin (120582 1) 120573 isin (120582 1) such that120593119894(120572) = 0 120593119894(120573) = 120573 (119894 = 1 2) by (8) Denote 1205930(119909) = 1205931(119909) =1205932(119909) (119909 isin [120582 1]) and 1205721 = 1205931(119891(120572)) lt 1205931(120582) = 1205930(120582) 1205722 =1205932(119891(120572)) lt 1205932(120582) = 1205930(120582) We prove that 1205721 = 1205722 as followsIt is trivial that 1205721 gt 120582 1205722 gt 120582 and

120593119901minus1

0(1205721) = 120593

119901minus1

1(1205721) = 120593

119901

1(119891 (120572)) = 0

120593119901minus1

0(1205722) = 120593

119901minus1

2(1205722) = 120593

119901

2(119891 (120572)) = 0

(A34)

by Lemmas 3(v) and 4 Since 120593119901minus10

is strictly monotone on[min1205721 1205722 1] we have that 1205721 = 1205722 Let 1205720 = 1205721 = 1205722 then1205720 lt 1205930(120582) and 120593

119901minus1

0(1205720) = 0 Define 120595+ = 120593

119901minus1

0|[1205720 1]

ByLemmas 3(v) and 4 we have that 120595+ is a homeomorphism

And by 120595+(1205720) = 0 le 120595+(1) we know that 120595+ is strictlyincreasing

We prove 1205931(119909) = 1205932(119909) on Δ 119896 for all 119896 ge 0 by inductionas follows

Obviously 1205931(119909) = 1205932(119909) holds on Δ 0 Suppose that1205931(119909) = 1205932(119909) holds on Δ 119896 for all 119896 le 119898 where 119898 ge 0 isa certain integer Let

120593 (119909) = 1205931 (119909) = 1205932 (119909) 119909 isin [119891119898+1

(1) 1] (A35)

Since 120593119894(119909) ge 120593119894(119891(120572)) = 1205720 gt 120582 for 119909 le 120582 and from (4) wehave

119891 (120593 (119891minus1(119909))) = 119891 (120593119894 (119891

minus1(119909))) = 120593

119901minus1

119894(120593119894 (119909))

= 120595+ (120593119894 (119909)) (119894 = 1 2 119909 isin Δ119898+1)

(A36)

Thereby

120593119894 (119909) = 120595minus1

+(119891 (120593 (119891

minus1(119909)))) (119894 = 1 2 119909 isin Δ119898+1)

(A37)

Thus we have 1205931(119909) = 1205932(119909) on Δ119898+1 Thus we get 1205931(119909) =1205932(119909) on Δ 119896 for all 119896 ge 0 by induction

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is partially supported by the National NaturalScience Foundation of China Tian Yuan Foundation (no11326129) and the Fundamental Research Funds for theCentral Universities (no 14CX02152A)

References

[1] M J Feigenbaum ldquoQuantitative universality for a class ofnonlinear transformationsrdquo Journal of Statistical Physics vol 19no 1 pp 25ndash52 1978

[2] M J Feigenbaum ldquoTheuniversalmetric properties of nonlineartransformationsrdquo Journal of Statistical Physics vol 21 no 6 pp669ndash706 1979

[3] P Couliet and C Tresser ldquoIterations drsquoendomorphismes etgroupe de renormalisationrdquo Journal de Physique Colloques vol39 no 5 pp C5-25ndashC5-28 1978

[4] D Sullivan ldquoBoubds quadratic differentials and renormaliza-tion conjecturesrdquo in Proceedings of the Mathematics into theTwenty-First Century Centennial Symposium F Browder Edpp 417ndash466 American Mathematical Society August 1992

[5] ADouady and JHHubbard ldquoOn the dynamics of polynomial-like mappingsrdquo Annales Scientifiques de lrsquoEcole NormaleSuperieure vol 18 no 2 pp 287ndash343 1985

[6] I Lanford ldquoA computer-assisted proof of the Feigenbaumconjecturesrdquo The American Mathematical Society vol 6 no 3pp 427ndash434 1982

8 Abstract and Applied Analysis

[7] J P Eckmann and P Wittwer ldquoA complete proof of theFeigenbaum conjecturesrdquo Journal of Statistical Physics vol 46no 3-4 pp 455ndash475 1987

[8] M Campanino and H Epstein ldquoOn the existence of Feigen-baumrsquos fixed pointrdquo Communications in Mathematical Physicsvol 79 no 2 pp 261ndash302 1981

[9] L Yang and J Z Zhang ldquoThe second type of Feigenbaumrsquosfunctional equationrdquo Science in China A vol 28 pp 1061ndash10691985

[10] P J McCarthy ldquoThe general exact bijective continuous solu-tion of Feigenbaumrsquos functional equationrdquo Communications inMathematical Physics vol 91 no 3 pp 431ndash443 1983

[11] J P Eckmann H Epstein and P Wittwer ldquoFixed pointsof Feigenbaumrsquos type for the equation 119891119901 (120582119909) = 120582119891(119909)rdquoCommunications in Mathematical Physics vol 93 no 4 pp495ndash516 1984

[12] A H Zhang L J Wang and W Song ldquoNonsingle-valley con-tinuous solutions of p-order Feigenbaums functional equationrdquoNortheastern Mathematical Journal vol 19 no 4 pp 375ndash3802003

[13] G Liao L Wang and Y Zhang ldquo119901-order Feigenbaumrsquos mapsrdquoNortheastern Mathematical Journal Dongbei Shuxue vol 20no 3 pp 284ndash290 2004

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

[7] J P Eckmann and P Wittwer ldquoA complete proof of theFeigenbaum conjecturesrdquo Journal of Statistical Physics vol 46no 3-4 pp 455ndash475 1987

[8] M Campanino and H Epstein ldquoOn the existence of Feigen-baumrsquos fixed pointrdquo Communications in Mathematical Physicsvol 79 no 2 pp 261ndash302 1981

[9] L Yang and J Z Zhang ldquoThe second type of Feigenbaumrsquosfunctional equationrdquo Science in China A vol 28 pp 1061ndash10691985

[10] P J McCarthy ldquoThe general exact bijective continuous solu-tion of Feigenbaumrsquos functional equationrdquo Communications inMathematical Physics vol 91 no 3 pp 431ndash443 1983

[11] J P Eckmann H Epstein and P Wittwer ldquoFixed pointsof Feigenbaumrsquos type for the equation 119891119901 (120582119909) = 120582119891(119909)rdquoCommunications in Mathematical Physics vol 93 no 4 pp495ndash516 1984

[12] A H Zhang L J Wang and W Song ldquoNonsingle-valley con-tinuous solutions of p-order Feigenbaums functional equationrdquoNortheastern Mathematical Journal vol 19 no 4 pp 375ndash3802003

[13] G Liao L Wang and Y Zhang ldquo119901-order Feigenbaumrsquos mapsrdquoNortheastern Mathematical Journal Dongbei Shuxue vol 20no 3 pp 284ndash290 2004

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of