# Fixed Point Theorem for Cyclic Chatterjea Type Contractions

Authors: Erdal Karapınar; Hemant Kumar Nashine
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We introduce the notion of cyclic weakly Chatterjea type contraction and generalized cyclic weakly Chatterjea type contraction in metric spaces. We discussed the existence of fixed point theorems of (generalized) cyclic weakly Chatterjea type contraction mappings in the context of complete metric spaces. Our main theorems extend and improve some fixed point theorems in the literature.


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## Body

## 1. Introduction and Preliminaries

Celebrated Banach’s contraction mapping principle is one of the cornerstones in the development of nonlinear analysis. Fixed point theorems have applications not only in the various branches of mathematics but also in economics, chemistry, biology, computer science, engineering, and others. In particular, such theorems are used to demonstrate the existence and uniqueness of a solution of differential equations, integral equations, functional equations, partial differential equations, and others. Due to the importance, generalizations of Banach fixed point theorem have been investigated heavily by many authors. (see, e.g., [1, 2]).Following this trend, in 1972, Chatterjea [3] introduced the following definition.Definition 1.1.
Let(X,d) be a metric space. A self-mapping T:X→X is called a Chatterjea type contraction if there exists α∈(0,1/2) such that for all x,y∈X the following inequality holds:
(1.1)d(Tx,Ty)≤α(d(x,Ty)+d(y,Tx)).In this interesting paper, Chatterjea [3] proved that every Chatterjea type contraction T on a complete metric space has a unique fixed point. Later, Choudhury [4] introduced a generalization of Chatterjea type contraction as follows.Definition 1.2.
A self-mappingT:X→X, on a metric space (X,d), is said to be a weakly C-contractive (or a weak Chatterjea type contraction) if for all x,y∈X,
(1.2)d(Tx,Ty)≤12[d(x,Ty)+d(y,Tx)]-ψ(d(x,Ty),d(y,Tx)),
where ψ:[0,+∞)2→[0,+∞) is a continuous function such that
(1.3)ψ(x,y)=0iffx=y=0.In [4], the author proves that every weak Chatterjea type contraction on a complete metric space has a unique fixed point.One of the interesting generalizations of a Banach’s contraction mapping principle was given by Kirk et al. [5] in 2003 by introducing the following notion of cyclic representation.Definition 1.3 (see [5]).
LetX be a nonempty set and let T:X→X be an operator. By definition, X=⋃i=1m‍Xi is a cyclic representation of X with respect to T if(a)
Xi,i=1,…,m are nonempty sets,(b)
T(X1)⊂X2,…,T(Xm-1)⊂Xm,
T(Xm)⊂X1.After the remarkable paper of Kirk et al. [5], some new fixed-point theorems for operators T defined on a complete metric space X with a cyclic representation of X with respect to T have appeared in the literature (see, e.g., [1, 2, 4–19]). Very recently, Karapınar [10] introduced the notion of the cyclic weak φ-contraction and proved fixed point theorems for these types of contractions.Definition 1.4 (see [10]).
Let(X,d) be a metric space, m∈ℕ,A1,A2,…,Am closed nonempty subsets of X, and Y=⋃i=1m‍Ai. An operator T:Y→Y is called a cyclic weak φ-contraction if(1)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(2)
there exists a continuous, nondecreasing functionφ:[0,∞)→[0,∞) with φ(t)>0 for t∈(0,∞) and φ(0)=0 such that
(1.4)d(Tx,Ty)≤d(x,y)-φ(d(x,y)),for any x∈Ai,y∈Ai+1,i=1,2,…,m where Am+1=A1.LetF denote all the continuous functions φ:[0,∞)→[0,∞) with φ(t)>0 for t∈(0,∞) and φ(0)=0.Theorem 1.5 (see [10]).
Let(X,d) be a complete metric space, m∈ℕ,A1,A2,…,Am nonempty subsets of X, and X=⋃i=1m‍Ai. Suppose that T is a cyclic weak φ-contraction with φ∈F. Then, T has a fixed point z∈⋂i=1n‍Ai.In this paper, we introduce the notions of cyclic weakly Chatterjea type contractions and generalized cyclic weakly Chatterjea type contractions and then derive fixed point theorems on these cyclic contractions in the setup of complete metric spaces. Our results generalize fixed point theorems [2, 5, 20] in the sense of metric spaces.


## 2. Main Results

### 2.1. Fixed Point Theory for the Cyclic Weakly Chatterjea Type Contractions in Complete Metric Space

First we introduce the notion of cyclic weakly Chatterjea type contraction in metric space.For convenience, we denote byF1 the class of functions ψ:[0,∞)2→[0,∞) lower semicontinuous satisfying ψ(x,y)=0 if and only if x=y=0.Definition 2.1.
Let(X,d) be a metric space, m∈ℕ,A1,A2,…,Am nonempty subsets of X, and Y=⋃i=1m‍Ai. An operator T:Y→Y is called a cyclic weakly Chatterjea type contraction if(1)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(2)
for anyx∈Ai,  y∈Ai+1,  i=1,2,…,m(2.1)d(Tx,Ty)≤α(d(x,Ty)+d(y,Tx))-ψ(d(x,Ty),d(y,Tx)),
where Am+1=A1, ψ∈F1 and α∈(0,1/2].Notice that the cyclic weak Chatterjea type contractions constitute a strictly larger class of mappings than cyclic weakφ-contractions.The main result of this section is the following.Theorem 2.2.
Let(X,d) be a complete metric space, m∈ℕ,  A1,A2,…,Am nonempty closed subsets of X, and X=⋃i=1m‍Ai. Suppose that T is a cyclic weakly Chatterjea type contraction. Then, T has a unique fixed point z∈⋂i=1n‍Ai.Proof.
Takex0∈X and consider the sequence given by xn+1=Txn,  n=0,1,2,…. If there exists n0∈ℕ such that xn0+1=xn0, then the existence of the fixed point is proved. Indeed, xn0+1=Txn0=xn0 and xn0 is the desired point. Suppose that xn+1≠xn for all n=0,1,2,…. Since X=⋃i=1m‍Ai, for any n>0 there exists in∈{1,2,…,m} such that xn-1∈Ain and xn∈Ain+1. Due to the fact that T is a cyclic weakly Chatterjea type contraction, we have(2.2)d(xn,xn+1)=d(Txn-1,Txn)≤α(d(xn-1,Txn)+d(xn,Txn-1))-ψ(d(xn-1,Txn),d(xn,Txn-1))=α(d(xn-1,xn+1)+d(xn,xn))-ψ(d(xn-1,xn+1),0)≤αd(xn-1,xn+1)≤α(d(xn-1,xn)+d(xn,xn+1)).
Therefore,
(2.3)d(xn,xn+1)≤α1-αd(xn-1,xn)foranyn=1,2,….
If we take k=α/(1-α)<1, then we have
(2.4)d(xn,xn+1)≤kd(xn-1,xn)foranyn=1,2,….
It follows that d(xn,xn+1)≤knd(x0,x1) and
(2.5)limn→∞d(xn,xn+1)=0.
In the sequel, we will prove that {xn} is a Cauchy sequence. First, we prove the following claim. For every ε>0 there exists n∈ℕ such that if r,q≥n with r-q≡1(m), then d(xr,xq)<ε.
Suppose, to the contrary, that there existsε>0 such that for any n∈ℕ we can find rn>qn≥n with rn-qn≡1(m) satisfying
(2.6)d(xqn,xrn)≥ε.
Now, we take n>2m. Then, corresponding to qn≥n we can choose rn in such a way that it is the smallest integer with rn>qn satisfying rn-qn≡1(m) and d(xqn,xrn)≥ε. Therefore, d(xqn,xrn-m)≤ε. Using the triangular inequality, we obtain
(2.7)ε≤d(xqn,xrn)≤d(xqn,xrn-m)+∑i=1m‍d(xrn-i,xrn-i+1)<ε+∑i=1m‍d(xrn-i,xrn-i+1).
Taking (2.5) into account and letting n→∞ in the inequality above, we find
(2.8)limn→∞d(xqn,xrn)=ε.
Again, by the triangular inequality, we have
(2.9)ε≤d(xqn,xrn)≤d(xqn,xqn+1)+d(xqn+1,xrn+1)+d(xrn+1,xrn)≤d(xqn,xqn+1)+d(xqn+1,xqn)+d(xqn,xrn)+d(xrn,xrn+1)+d(xrn+1,xrn)=2d(xqn,xqn+1)+d(xqn,xrn)+2d(xrn,xrn+1).
Passing to the limit as n→∞ in (2.8) and taking (2.5) and (2.8) into account, we get
(2.10)limn→∞d(xqn+1,xrn+1)=ε.
Since xqn and xrn lie in different adjacently labeled sets Ai and Ai+1 for certain 1≤i≤m, using the fact that T is a cyclic weakly Chatterjea type contraction, we have
(2.11)d(xqn+1,xrn+1)=d(Txqn,Txrn)≤α(d(xqn,Txrn)+d(xrn,Txqn))-ψ(d(xqn,Txrn),d(xrn,Txqn))≤α(d(xqn,xrn+1)+d(xrn,xqn+1))-ψ(d(xqn,xrn+1),d(xrn,xqn+1))≤α(d(xqn,xrn)+d(xrn,xrn+1)+d(xrn,xqn)+d(xqn,xqn+1))-ψ(d(xqn,xrn+1),d(xrn,xqn+1)).
Taking (2.5) and (2.8) into account together with the lower semicontinuity of ψ, the inequalities above yield
(2.12)ε≤α·ε-liminfn→+∞ψ(d(xqn,xrn+1),d(xrn,xqn+1))≤α·ε-ψ(ε,ε)≤α·ε
as n→∞. Hence, ψ(ε,ε)=0. From the fact that ψ(x,y)=0⇔x=y=0, we have ε=0, which is a contradiction. Therefore, our claim is proved.
We prove that(X,d) is a Cauchy sequence. Fix ε>0. By the claim, we find n0∈ℕ such that if r,q≥n0 with r-q≡1(m),
(2.13)d(xr,xq)≤ε2.
Since limn→∞d(xn,xn+1)=0, there exists n1∈ℕ such that
(2.14)d(xn,xn+1)≤ε2m
for any n≥n1. Suppose that r,s≥max{n0,n1} with s>r. Then, there exists k∈{1,2,…,m} such that s-r≡k(m). Therefore, we have s-r+φ≡1(m) for φ=m-k+1. So, we derive
(2.15)d(xr,xs)≤d(xr,xs+j)+d(xs+j,xs+j-1)+⋯+d(xs+1,xs).
By (2.13) and (2.14) and the inequality above, we get
(2.16)d(xr,xs)≤ε2+j×ε2m≤ε2+m×ε2m=ε.
This proves that (xn) is a Cauchy sequence. Since X is a complete metric space, there exists x∈X such that limn→∞xn=x.
Now, we prove thatx is a fixed point of T. Since limn→∞xn=x and, as X=⋃i=1m‍Ai is a cyclic representation of X with respect to T, the sequence (xn) has infinite terms in each Ai for i∈{1,2,…,m}. Suppose that x∈Ai,Tx∈Ai+1 and we take a subsequence xnk of (xn) with xnk∈Ai-1 (the existence of this subsequence is guaranteed by the previously mentioned comment). By using the contractive condition, we obtain
(2.17)d(xnk+1,Tx)=d(Txnk,Tx)≤α(d(xnk,Tx)+d(x,Txnk))-ψ(d(xnk,Tx),d(x,Txnk))=α(d(xnk,Tx)+d(x,xnk+1))-ψ(d(xnk,Tx),d(x,xnk+1)).
Letting n→∞ and by using xnk→x together with the lower semicontinuity of ψ, we have
(2.18)d(x,Tx)≤αd(x,Tx)-liminfn→+∞ψ(d(xnk,Tx),d(x,xnk+1))≤αd(x,Tx)-ψ(d(x,Tx),0)≤αd(x,Tx)
which is a contradiction (since α<1/2) unless d(x,Tx)=0. Therefore, x is a fixed point of T.
Finally, to prove the uniqueness of the fixed point, we havey,z∈X with y and z being fixed points of T. The cyclic character of T and the fact that y,z∈X are fixed points of T imply that y,z∈⋂i=1m‍Ai. Using the contractive condition we obtain
(2.19)d(y,z)=d(Ty,Tz)≤α(d(y,Tz)+d(z,Ty))-ψ(d(y,Tz),d(z,Ty)).
That is,
(2.20)d(y,z)≤2αd(y,z)-ψ(d(y,z),d(z,y))≤2αd(y,z).
This gives us ψ(d(y,z),d(z,y))=0, and, by our assumption about ψ, d(y,z)=0, that is, y=z. This finishes the proof.Corollary 2.3.
Let(X,d) be a complete metric space, m∈ℕ,A1,A2,…,Am nonempty subsets of X, and X=⋃i=1m‍Ai. Suppose that T:X→X be an operator such that(1)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(2)
there existsβ∈[0,1/2) such that
(2.21)d(Tx,Ty)≤β(d(x,Ty)+d(y,Tx))for any x∈Ai,y∈Ai+1,i=1,2,…,m where Am+1=A1. Then, T has a fixed point z∈⋂i=1n‍Ai.Proof.
Letβ∈[0,1/2). Here, it suffices to take the function ψ:[0,+∞)2→[0,+∞) as ψ(a,b)=(α-β)(a+b). It is clear that ψ satisfies the following conditions:(1)
ψ(a,b)=0 if and only if a=b=0,(2)
ψ(x,y)=(α-β)(x+y)=ψ(x+y,0). Hence, we apply Theorems 2.2 for α∈[1/4,1/2) and get the desired result.The following corollary gives us a fixed point theorem with a contractive condition of integral type for cyclic contractions.Corollary 2.4.
Let(X,d) be a complete metric space, m∈ℕ,  A1,A2,…,Am nonempty closed subsets of X, and X=⋃i=1m‍Ai. Suppose that T:X→X be an operator such that(i)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(ii)
there existsβ∈[0,1/2) such that
(2.22)∫0d(Tx,Ty)‍ρ(t)dt≤β∫0d(x,Ty)+d(y,Tx)‍ρ(t)dtfor any x∈Ai,y∈Ai+1,i=1,2,…,m where Am+1=A1 and ρ:[0,∞)→[0,∞) is a Lebesgue-integrable mapping satisfying ∫0ε‍ρ(t)dt for ε>0. Then T has unique fixed point z∈⋂i=1m‍Ai.Proof.
It is easily proved that the functionφ:[0,∞)→[0,∞) given by φ(t)=∫0t‍ρ(s)ds satisfies that φ∈F1. Therefore, Corollary 2.3 is obtained from Theorem 2.2, taking as φ the previously defined function and as ψ the function ψ(x,y)=(α-β)(x+y)=φ(x+y,0).If in Corollary2.4, we take Ai=X for i=1,2,…,m, we obtain the following result.Corollary 2.5.
Let(X,d) be a complete metric space and let T:X→X be a mapping such that for any x,y∈X,
(2.23)∫0d(Tx,Ty)‍ρ(t)dt≤β∫0d(x,Ty)+d(y,Tx)‍ρ(t)dt,
where ρ:[0,∞)→[0,∞) is a Lebesgue-integrable mapping satisfying ∫0ε‍ρ(t)dt for ε>0 and the constant β∈[0,1/2). Then T has unique fixed point.If in Theorem2.2 we put Ai=X for i=1,2,…,m we get the main result of [4].Corollary 2.6.
Let(X,d) be a complete metric space and T:X→X a mapping such that for any x,y∈X,
(2.24)d(Tx,Ty)≤α(d(x,Ty)+d(y,Tx))-ψ(d(x,Ty)+d(y,Tx)),
where ψ∈F1 and α∈(0,1/2). Then T has unique fixed point.Example 2.7.
LetX=ℝ with the usual metric. Suppose A1=[0,1],A2=[0,1/2], and Y=⋃i=12‍Ai. Define T:Y→Y such that Tx=x/9 for all x∈Y. It is clear that ⋃i=12‍Ai is a cyclic representation of Y with respect to T. Furthermore, if ψ:[0,+∞)2→[0,+∞) is defined by ψ(x,y)=(x+y)/9, then ψ∈F2. Here T is a generalized cyclic weakly Chatterjea type contraction for α=1/2. To see this fact we examine three cases.(i)
Suppose thatx≥y. Then, the inequality (2.1) turns into
(2.25)d(Tx,Ty)=|x7-y7|=x-y7≤12[d(x,Ty)+d(y,Tx)]-ψ(d(x,Ty),d(y,Tx))=12[|x-y7|+|y-x7|]-19[|x-y7|+|y-x7|]=12[x-y7+|y-x7|]-19[x-y7+|y-x7|]=718[x-y7+|y-x7|].
If y<x/7, then (2.25) becomes
(2.26)x-y7≤718[x-y7+x7-y]=49(x-y).
Hence, (2.1) is satisfied. If y≥x/7, then (2.25) becomes
(2.27)x-y7≤718[x-y7+y-x7]=13(x+y).
Thus, (2.1) is true.(ii)
Suppose thaty/7≤x≤y. Then, the inequality (2.1) yields that
(2.28)d(Tx,Ty)=|x7-y7|=y-x7≤12[d(x,Ty)+d(y,Tx)]-ψ(d(x,Ty),d(y,Tx))=12[|x-y7|+|y-x7|]-19[|x-y7|+|y-x7|]=718[|x-y7|+|y-x7|]=718[x-y7+y-x7]=13(x+y).
So, (2.1) holds.(iii)
Finally, suppose thaty/7≥x. Then, the inequality (2.1) yields that
(2.29)d(Tx,Ty)=|x7-y7|=y-x7≤12[d(x,Ty)+d(y,Tx)]-ψ(d(x,Ty),d(y,Tx))=12[|x-y7|+|y-x7|]-19[|x-y7|+|y-x7|]=718[|x-y7|+|y-x7|]=718[x-y7+y-x7]=13(y+x).
Hence, (2.1) is true. Therefore, all conditions of Theorem 1.5 are satisfied, and so T has a fixed point (which is z=0∈⋂i=12‍Ai.)


### 2.2. Fixed Point Theory for the Generalized Cyclic Weakly Chatterjea Type Contractions in Complete Metric Space

In this section we derive fixed point theorems for self-maps satisfying certain generalized cyclic weakly Chatterjea type contractions in a complete metric space.For convenience, we denote byF2 the class of functions ψ:[0,∞)4→[0,∞) lower semicontinuous satisfying ψ(x,y,z,t)=0 if and only if x=y=z=t=0.We introduce the notion of generalized cyclic weakly Chatterjea type contraction in metric space.Definition 2.8.
Let(X,d) be a metric space, m∈ℕ,A1,A2,…,Am nonempty subsets of X and Y=⋃i=1m‍Ai. An operator T:Y→Y is called a generalized cyclic weakly Chatterjea type contraction if(1)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(2)
(2.30)d(Tx,Ty)≤α[d(x,Tx)+d(y,Ty)+d(x,Ty)+d(y,Tx)]-ψ(d(x,Tx),d(x,Ty),d(x,Ty),d(y,Tx)),
for any x∈Ai,y∈Ai+1,i=1,2,…,m where Am+1=A1 and
(2.31)ψ(d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))∈F2andα∈(0,14].We state the main result of this section as follows.Theorem 2.9.
Let(X,d) be a complete metric space, m∈ℕ,  A1,A2,…,Am nonempty closed subsets of X and X=⋃i=1m‍Ai. Suppose that T is a generalized cyclic weakly C-contraction. Then, T has a unique fixed point z∈⋂i=1n‍Ai.Proof.
As in the proof of Theorem2.2, we take x0∈X and construct a sequence by defining xn+1=Txn,n=0,1,2,…. Suppose that there exists n0∈ℕ such that xn0+1=xn0. Then, since xn0+1=Txn0=xn0, the existence of the fixed point is proved. So, we assume that xn+1≠xn for any n=0,1,2,…. Since X=⋃i=1m‍Ai, for any n>0, there exists in∈{1,2,…,m} such that xn-1∈Ain and xn∈Ain+1. Regarding that T is a generalized cyclic weakly Chatterjea type contraction, we have
(2.32)d(xn,xn+1)=d(Txn-1,Txn)≤α[d(xn-1,Txn-1)+d(xn,Txn)+d(xn-1,Txn)+d(xn,Txn-1)]-ψ(d(xn-1,Txn-1),d(xn,Txn),d(xn-1,Txn),d(xn,Txn-1))=α[d(xn-1,xn)+d(xn,xn+1)+d(xn-1,xn+1)]-ψ(d(xn-1,xn),d(xn,xn+1),d(xn-1,xn+1),0)≤α[d(xn-1,xn)+d(xn,xn+1)+d(xn-1,xn+1)]≤2α[d(xn-1,xn)+d(xn,xn+1)].
Therefore,
(2.33)d(xn,xn+1)≤2α1-2αd(xn-1,xn)foranyn=1,2,….
If we take k=α/(1-α)<1, then we have
(2.34)d(xn,xn+1)≤kd(xn-1,xn)foranyn=1,2,….
It follows that d(xn,xn+1)≤knd(x0,x1) and
(2.35)limn→∞d(xn,xn+1)=0.
We show that {xn} is a Cauchy sequence. For this purpose, first, we prove the following claim. For every ε>0 there exists n∈ℕ such that if r,q≥n with r-q≡1(m), then d(xr,xq)<ε.
Assume the contrary. So there existsε>0 such that for any n∈N we can find rn>qn≥n with rn-qn≡1(m) satisfying
(2.36)d(xqn,xrn)≥ε.
Now, we take n>2m. Then, corresponding to qn≥n we can choose rn in such a way that it is the smallest integer with rn>qn satisfying rn-qn≡1(m) and d(xqn,xrn)≥ε. Therefore, d(xqn,xrn-m)≤ε by using the triangular inequality
(2.37)ε≤d(xqn,xrn)≤d(xqn,xrn-m)+∑i=1m‍d(xrn-i,xrn-i+1)<ε+∑i=1m‍d(xrn-i,xrn-i+1).
Passing to the limit as n→∞ in the last inequality and taking into account that limn→∞d(xn,xn+1)=0, we obtain
(2.38)limn→∞d(xqn,xrn)=ε.
Again, by the triangular inequality
(2.39)ε≤d(xqn,xrn)≤d(xqn,xqn+1)+d(xqn+1,xrn+1)+d(xrn+1,xrn)≤d(xqn,xqn+1)+d(xqn+1,xqn)+d(xqn,xrn)+d(xrn,xrn+1)+d(xrn+1,xrn)=2d(xqn,xqn+1)+d(xqn,xrn)+2d(xrn,xrn+1).
Taking (2.35) and (2.38) into account, we get
(2.40)limn→∞d(xqn+1,xrn+1)=ε
as n→∞ in (2.38).
Sincexqn and xrn lie in different adjacently labeled sets Ai and Ai+1 for certain 1≤i≤m, using the fact that T is a generalized cyclic weakly Chatterjea type contraction, we have
(2.41)d(xqn+1,xrn+1)=d(Txqn,Txrn)≤α[d(xqn,Txqn)+d(xrn,Txrn)+d(xqn,Txrn)+d(xrn,Txqn)]-ψ(d(xqn,Txqn),d(xrn,Txrn),d(xqn,Txrn),d(xrn,Txqn))≤α[d(xqn,xqn+1)+d(xrn,xrn+1)+d(xqn,xrn+1)+d(xrn,xqn+1)]-ψ(d(xqn,xqn+1),d(xrn,xrn+1),d(xqn,xrn+1),d(xrn,xqn+1)).
Taking into account (2.35) and (2.38) and the lower semicontinuity of ψ, letting n→∞ in the last inequality, we obtain
(2.42)ε≤α[ε+0+ε+0]-liminfn→+∞ψ(d(xqn,xqn+1),d(xrn,xrn+1),d(xqn,xrn+1),d(xrn,xqn+1))≤α[ε+0+ε+0]-ψ(ε,0,ε,0)≤2α·ε
and from the last inequality, ψ(ε,0,ε,0)≤-2αε≤0. Therefore φ(ε,0,ε,0)=0. From the fact that ψ(x,y,z,t)=0⇔x=y=z=t=0, we have ε=0, which is a contradiction. Therefore, our claim is proved.
In the sequel, we will prove that(X,d) is a Cauchy sequence. Fix ε>0. By the claim, we find n0∈ℕ such that if r,q≥n0 with r-q≡1(m)(2.43)d(xr,xq)≤ε2.
Since limn→∞d(xn,xn+1)=0, we also find n1∈ℕ such that
(2.44)d(xn,xn+1)≤ε2m
for any n≥n1. Suppose that r,s≥max{n0,n1} and s>r. Then there exists k∈{1,2,…,m} such that s-r≡k(m). Therefore, s-r+φ≡1(m) for φ=m-k+1. So, we have
(2.45)d(xr,xs)≤d(xr,xs+j)+d(xs+j,xs+j-1)+⋯+d(xs+1,xs).
By (2.43) and (2.44) and from the last inequality, we get
(2.46)d(xr,xs)≤ε2+j×ε2m≤ε2+m×ε2m=ε.
This proves that (xn) is a Cauchy sequence. Since X is a complete metric space, there exists x∈X such that limn→∞xn=x. In what follows, we prove that x is a fixed point of T. In fact, since limn→∞xn=x and as X=⋃i=1m‍Ai is a cyclic representation of X with respect to T, the sequence (xn) has infinite terms in each Ai for i∈{1,2,…,m}. Suppose that x∈Ai,Tx∈Ai+1 and we take a subsequence xnk of (xn) with xnk∈Ai-1 (the existence of this subsequence is guaranteed by the previously mentioned comment). By using the contractive condition, we can obtain
(2.47)d(xnk+1,Tx)=d(Txnk,Tx)≤α[d(xnk,Txnk)+d(x,Tx)+d(xnk,Tx)+d(x,Txnk))-ψ(d(xnk,Txnk),d(x,Tx),d(xnk,Tx),d(x,Txnk))=α[d(xnk,xnk+1)+d(x,Tx)+d(xnk,Tx)+d(x,xnk+1))-ψ(d(xnk,xnk+1),d(x,Tx),d(xnk,Tx),d(x,xnk+1)).
Passing to the limit as n→∞ and using xnk→x, lower semicontinuity of ψ, we have
(2.48)d(x,Tx)≤12d(x,Tx)-liminfn→+∞ψ(d(xnk,xnk+1),d(x,Tx),d(xnk,Tx),d(x,xnk+1))≤12d(x,Tx)-ψ(0,d(x,Tx),d(x,Tx),0)≤12d(x,Tx),
which is a contradiction unless d(x,Tx)=0, and, therefore, x is a fixed point of T. Finally, to prove the uniqueness of the fixed point, we have y,z∈X with y and z being fixed points of T. The cyclic character of T and the fact that y,z∈X are fixed points of T imply that y,z∈⋂i=1m‍Ai. Using the contractive condition we obtain
(2.49)d(y,z)=d(Ty,Tz)≤α[d(y,Ty)+d(z,Tz)+d(y,Tz)+d(z,Ty)]-ψ(d(y,Ty),d(z,Tz),d(y,Tz),d(z,Ty)).
that is,
(2.50)d(y,z)≤2αd(y,z)-ψ(0,0,d(y,z),d(z,y))≤d(y,z).
This gives us ψ(0,0,d(y,z),d(z,y))≤(1-2α)d(y,z)≤0, and, by our assumption about ψ, d(y,z)=0, that is, y=z. This finishes the proof.Corollary 2.10.
Let(X,d) be a complete metric space, m∈ℕ, A1,A2,…,Am nonempty subsets of X, and X=⋃i=1m‍Ai. Suppose that T:X→X be an operator such that(1)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(2)
there existsβ∈[0,1/4) such that
(2.51)d(Tx,Ty)≤β[d(x,Tx)+d(y,Ty)+d(x,Ty)+d(y,Tx)]
for any x∈Ai,y∈Ai+1,i=1,2,…,m, where Am+1=A1. Then, T has a fixed point z∈⋂i=1n‍Ai.Proof.
Letβ∈[0,1/4). Here, it suffices to take the function ψ:[0,+∞)4→[0,+∞) defined by ψ(a,b,c,e)=(α-β)(a+b+c+e). Obviously, ψ satisfies that ψ(a,b,c,e)=0 if and only if a=b=c=e=0, and ψ(x,y,z,t)=(α-β)(x+y+z+t)=ψ(x+y+z+t,0). Then, we can apply Theorems 2.9.The following corollary gives us a fixed point theorem with a contractive condition of integral type for cyclic contractions.Corollary 2.11.
Let(X,d) be a complete metric space, m∈ℕ,A1,A2,…,Am nonempty closed subsets of X and X=⋃i=1m‍Ai. Suppose that T:X→X be an operator such that(1)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(2)
there existsβ∈[0,1/4) such that
(2.52)∫0d(Tx,Ty)‍ρ(t)dt≤β∫0d(x,Tx)+d(y,Ty)+d(x,Ty)+d(y,Tx)‍ρ(t)dtfor any x∈Ai,y∈Ai+1,i=1,2,…,m, where Am+1=A1 and ρ:[0,∞)→[0,∞) is a Lebesgue-integrable mapping satisfying ∫0ε‍ρ(t)dt for ε>0. Then T has unique fixed point z∈⋂i=1m‍Ai.Proof.
It is easily proved that the functionφ:[0,∞)→[0,∞) given by φ(t)=∫0t‍ρ(s)ds satisfies that φ∈F1. Therefore, Corollary 2.10 is obtained from Theorem 2.9, taking as φ the perviously defined function and as ψ the function ψ(x,y,z,t)=(1/4-α)(x+y+z+t)=φ(x+y+z+t,0).If in Corollary2.11, we take Ai=X for i=1,2,…,m, we obtain the following result.Corollary 2.12.
Let(X,d) be a complete metric space and let T:X→X be a mapping such that for any x,y∈X,
(2.53)∫0d(Tx,Ty)‍ρ(t)dt≤β∫0d(x,Tx)+d(y,Ty)+d(x,Ty)+d(y,Tx)‍ρ(t)dt,
where ρ:[0,∞)→[0,∞) is a Lebesgue-integrable mapping satisfying ∫0ε‍ρ(t)dt for ε>0 and the constant β∈[0,1/4). Then T has unique fixed point.If in Theorem2.9 we put Ai=X for i=1,2,…,m, we have the generalized result of [4].Corollary 2.13.
Let(X,d) be a complete metric space and let T:X→X be a mapping such that for any x,y∈X,
(2.54)d(Tx,Ty)≤α[d(x,Tx)+d(y,Ty)+d(x,Ty)+d(y,Tx)]-ψ(d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)),
where ψ∈F2 and α∈(0,1/4]. Then T has unique fixed point.Example 2.14.
LetX=ℝ with the usual metric. Suppose A1=[0,1], A2=[0,1/3], A3=[0,1/9], A4=[0,1/27] and Y=⋃i=14‍Ai. Define T:Y→Y such that Tx=x/4 for all x∈Y. It is clear that ⋃i=14‍Ai is a cyclic representation of Y with respect to T. Furthermore, if ψ:[0,+∞)4→[0,+∞) is defined by ψ(x,y,z,t)=(x+y+z+t)/16, then ψ∈F2. As in Example 2.7, it can be easily shown that T is a generalized cyclic weakly Chatterjea type contraction α=1/4. Therefore, all conditions of Theorem 2.9 are satisfied, and so T has a fixed point (which is z=0∈⋂i=14‍Ai).


## 2.1. Fixed Point Theory for the Cyclic Weakly Chatterjea Type Contractions in Complete Metric Space

First we introduce the notion of cyclic weakly Chatterjea type contraction in metric space.For convenience, we denote byF1 the class of functions ψ:[0,∞)2→[0,∞) lower semicontinuous satisfying ψ(x,y)=0 if and only if x=y=0.Definition 2.1.
Let(X,d) be a metric space, m∈ℕ,A1,A2,…,Am nonempty subsets of X, and Y=⋃i=1m‍Ai. An operator T:Y→Y is called a cyclic weakly Chatterjea type contraction if(1)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(2)
for anyx∈Ai,  y∈Ai+1,  i=1,2,…,m(2.1)d(Tx,Ty)≤α(d(x,Ty)+d(y,Tx))-ψ(d(x,Ty),d(y,Tx)),
where Am+1=A1, ψ∈F1 and α∈(0,1/2].Notice that the cyclic weak Chatterjea type contractions constitute a strictly larger class of mappings than cyclic weakφ-contractions.The main result of this section is the following.Theorem 2.2.
Let(X,d) be a complete metric space, m∈ℕ,  A1,A2,…,Am nonempty closed subsets of X, and X=⋃i=1m‍Ai. Suppose that T is a cyclic weakly Chatterjea type contraction. Then, T has a unique fixed point z∈⋂i=1n‍Ai.Proof.
Takex0∈X and consider the sequence given by xn+1=Txn,  n=0,1,2,…. If there exists n0∈ℕ such that xn0+1=xn0, then the existence of the fixed point is proved. Indeed, xn0+1=Txn0=xn0 and xn0 is the desired point. Suppose that xn+1≠xn for all n=0,1,2,…. Since X=⋃i=1m‍Ai, for any n>0 there exists in∈{1,2,…,m} such that xn-1∈Ain and xn∈Ain+1. Due to the fact that T is a cyclic weakly Chatterjea type contraction, we have(2.2)d(xn,xn+1)=d(Txn-1,Txn)≤α(d(xn-1,Txn)+d(xn,Txn-1))-ψ(d(xn-1,Txn),d(xn,Txn-1))=α(d(xn-1,xn+1)+d(xn,xn))-ψ(d(xn-1,xn+1),0)≤αd(xn-1,xn+1)≤α(d(xn-1,xn)+d(xn,xn+1)).
Therefore,
(2.3)d(xn,xn+1)≤α1-αd(xn-1,xn)foranyn=1,2,….
If we take k=α/(1-α)<1, then we have
(2.4)d(xn,xn+1)≤kd(xn-1,xn)foranyn=1,2,….
It follows that d(xn,xn+1)≤knd(x0,x1) and
(2.5)limn→∞d(xn,xn+1)=0.
In the sequel, we will prove that {xn} is a Cauchy sequence. First, we prove the following claim. For every ε>0 there exists n∈ℕ such that if r,q≥n with r-q≡1(m), then d(xr,xq)<ε.
Suppose, to the contrary, that there existsε>0 such that for any n∈ℕ we can find rn>qn≥n with rn-qn≡1(m) satisfying
(2.6)d(xqn,xrn)≥ε.
Now, we take n>2m. Then, corresponding to qn≥n we can choose rn in such a way that it is the smallest integer with rn>qn satisfying rn-qn≡1(m) and d(xqn,xrn)≥ε. Therefore, d(xqn,xrn-m)≤ε. Using the triangular inequality, we obtain
(2.7)ε≤d(xqn,xrn)≤d(xqn,xrn-m)+∑i=1m‍d(xrn-i,xrn-i+1)<ε+∑i=1m‍d(xrn-i,xrn-i+1).
Taking (2.5) into account and letting n→∞ in the inequality above, we find
(2.8)limn→∞d(xqn,xrn)=ε.
Again, by the triangular inequality, we have
(2.9)ε≤d(xqn,xrn)≤d(xqn,xqn+1)+d(xqn+1,xrn+1)+d(xrn+1,xrn)≤d(xqn,xqn+1)+d(xqn+1,xqn)+d(xqn,xrn)+d(xrn,xrn+1)+d(xrn+1,xrn)=2d(xqn,xqn+1)+d(xqn,xrn)+2d(xrn,xrn+1).
Passing to the limit as n→∞ in (2.8) and taking (2.5) and (2.8) into account, we get
(2.10)limn→∞d(xqn+1,xrn+1)=ε.
Since xqn and xrn lie in different adjacently labeled sets Ai and Ai+1 for certain 1≤i≤m, using the fact that T is a cyclic weakly Chatterjea type contraction, we have
(2.11)d(xqn+1,xrn+1)=d(Txqn,Txrn)≤α(d(xqn,Txrn)+d(xrn,Txqn))-ψ(d(xqn,Txrn),d(xrn,Txqn))≤α(d(xqn,xrn+1)+d(xrn,xqn+1))-ψ(d(xqn,xrn+1),d(xrn,xqn+1))≤α(d(xqn,xrn)+d(xrn,xrn+1)+d(xrn,xqn)+d(xqn,xqn+1))-ψ(d(xqn,xrn+1),d(xrn,xqn+1)).
Taking (2.5) and (2.8) into account together with the lower semicontinuity of ψ, the inequalities above yield
(2.12)ε≤α·ε-liminfn→+∞ψ(d(xqn,xrn+1),d(xrn,xqn+1))≤α·ε-ψ(ε,ε)≤α·ε
as n→∞. Hence, ψ(ε,ε)=0. From the fact that ψ(x,y)=0⇔x=y=0, we have ε=0, which is a contradiction. Therefore, our claim is proved.
We prove that(X,d) is a Cauchy sequence. Fix ε>0. By the claim, we find n0∈ℕ such that if r,q≥n0 with r-q≡1(m),
(2.13)d(xr,xq)≤ε2.
Since limn→∞d(xn,xn+1)=0, there exists n1∈ℕ such that
(2.14)d(xn,xn+1)≤ε2m
for any n≥n1. Suppose that r,s≥max{n0,n1} with s>r. Then, there exists k∈{1,2,…,m} such that s-r≡k(m). Therefore, we have s-r+φ≡1(m) for φ=m-k+1. So, we derive
(2.15)d(xr,xs)≤d(xr,xs+j)+d(xs+j,xs+j-1)+⋯+d(xs+1,xs).
By (2.13) and (2.14) and the inequality above, we get
(2.16)d(xr,xs)≤ε2+j×ε2m≤ε2+m×ε2m=ε.
This proves that (xn) is a Cauchy sequence. Since X is a complete metric space, there exists x∈X such that limn→∞xn=x.
Now, we prove thatx is a fixed point of T. Since limn→∞xn=x and, as X=⋃i=1m‍Ai is a cyclic representation of X with respect to T, the sequence (xn) has infinite terms in each Ai for i∈{1,2,…,m}. Suppose that x∈Ai,Tx∈Ai+1 and we take a subsequence xnk of (xn) with xnk∈Ai-1 (the existence of this subsequence is guaranteed by the previously mentioned comment). By using the contractive condition, we obtain
(2.17)d(xnk+1,Tx)=d(Txnk,Tx)≤α(d(xnk,Tx)+d(x,Txnk))-ψ(d(xnk,Tx),d(x,Txnk))=α(d(xnk,Tx)+d(x,xnk+1))-ψ(d(xnk,Tx),d(x,xnk+1)).
Letting n→∞ and by using xnk→x together with the lower semicontinuity of ψ, we have
(2.18)d(x,Tx)≤αd(x,Tx)-liminfn→+∞ψ(d(xnk,Tx),d(x,xnk+1))≤αd(x,Tx)-ψ(d(x,Tx),0)≤αd(x,Tx)
which is a contradiction (since α<1/2) unless d(x,Tx)=0. Therefore, x is a fixed point of T.
Finally, to prove the uniqueness of the fixed point, we havey,z∈X with y and z being fixed points of T. The cyclic character of T and the fact that y,z∈X are fixed points of T imply that y,z∈⋂i=1m‍Ai. Using the contractive condition we obtain
(2.19)d(y,z)=d(Ty,Tz)≤α(d(y,Tz)+d(z,Ty))-ψ(d(y,Tz),d(z,Ty)).
That is,
(2.20)d(y,z)≤2αd(y,z)-ψ(d(y,z),d(z,y))≤2αd(y,z).
This gives us ψ(d(y,z),d(z,y))=0, and, by our assumption about ψ, d(y,z)=0, that is, y=z. This finishes the proof.Corollary 2.3.
Let(X,d) be a complete metric space, m∈ℕ,A1,A2,…,Am nonempty subsets of X, and X=⋃i=1m‍Ai. Suppose that T:X→X be an operator such that(1)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(2)
there existsβ∈[0,1/2) such that
(2.21)d(Tx,Ty)≤β(d(x,Ty)+d(y,Tx))for any x∈Ai,y∈Ai+1,i=1,2,…,m where Am+1=A1. Then, T has a fixed point z∈⋂i=1n‍Ai.Proof.
Letβ∈[0,1/2). Here, it suffices to take the function ψ:[0,+∞)2→[0,+∞) as ψ(a,b)=(α-β)(a+b). It is clear that ψ satisfies the following conditions:(1)
ψ(a,b)=0 if and only if a=b=0,(2)
ψ(x,y)=(α-β)(x+y)=ψ(x+y,0). Hence, we apply Theorems 2.2 for α∈[1/4,1/2) and get the desired result.The following corollary gives us a fixed point theorem with a contractive condition of integral type for cyclic contractions.Corollary 2.4.
Let(X,d) be a complete metric space, m∈ℕ,  A1,A2,…,Am nonempty closed subsets of X, and X=⋃i=1m‍Ai. Suppose that T:X→X be an operator such that(i)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(ii)
there existsβ∈[0,1/2) such that
(2.22)∫0d(Tx,Ty)‍ρ(t)dt≤β∫0d(x,Ty)+d(y,Tx)‍ρ(t)dtfor any x∈Ai,y∈Ai+1,i=1,2,…,m where Am+1=A1 and ρ:[0,∞)→[0,∞) is a Lebesgue-integrable mapping satisfying ∫0ε‍ρ(t)dt for ε>0. Then T has unique fixed point z∈⋂i=1m‍Ai.Proof.
It is easily proved that the functionφ:[0,∞)→[0,∞) given by φ(t)=∫0t‍ρ(s)ds satisfies that φ∈F1. Therefore, Corollary 2.3 is obtained from Theorem 2.2, taking as φ the previously defined function and as ψ the function ψ(x,y)=(α-β)(x+y)=φ(x+y,0).If in Corollary2.4, we take Ai=X for i=1,2,…,m, we obtain the following result.Corollary 2.5.
Let(X,d) be a complete metric space and let T:X→X be a mapping such that for any x,y∈X,
(2.23)∫0d(Tx,Ty)‍ρ(t)dt≤β∫0d(x,Ty)+d(y,Tx)‍ρ(t)dt,
where ρ:[0,∞)→[0,∞) is a Lebesgue-integrable mapping satisfying ∫0ε‍ρ(t)dt for ε>0 and the constant β∈[0,1/2). Then T has unique fixed point.If in Theorem2.2 we put Ai=X for i=1,2,…,m we get the main result of [4].Corollary 2.6.
Let(X,d) be a complete metric space and T:X→X a mapping such that for any x,y∈X,
(2.24)d(Tx,Ty)≤α(d(x,Ty)+d(y,Tx))-ψ(d(x,Ty)+d(y,Tx)),
where ψ∈F1 and α∈(0,1/2). Then T has unique fixed point.Example 2.7.
LetX=ℝ with the usual metric. Suppose A1=[0,1],A2=[0,1/2], and Y=⋃i=12‍Ai. Define T:Y→Y such that Tx=x/9 for all x∈Y. It is clear that ⋃i=12‍Ai is a cyclic representation of Y with respect to T. Furthermore, if ψ:[0,+∞)2→[0,+∞) is defined by ψ(x,y)=(x+y)/9, then ψ∈F2. Here T is a generalized cyclic weakly Chatterjea type contraction for α=1/2. To see this fact we examine three cases.(i)
Suppose thatx≥y. Then, the inequality (2.1) turns into
(2.25)d(Tx,Ty)=|x7-y7|=x-y7≤12[d(x,Ty)+d(y,Tx)]-ψ(d(x,Ty),d(y,Tx))=12[|x-y7|+|y-x7|]-19[|x-y7|+|y-x7|]=12[x-y7+|y-x7|]-19[x-y7+|y-x7|]=718[x-y7+|y-x7|].
If y<x/7, then (2.25) becomes
(2.26)x-y7≤718[x-y7+x7-y]=49(x-y).
Hence, (2.1) is satisfied. If y≥x/7, then (2.25) becomes
(2.27)x-y7≤718[x-y7+y-x7]=13(x+y).
Thus, (2.1) is true.(ii)
Suppose thaty/7≤x≤y. Then, the inequality (2.1) yields that
(2.28)d(Tx,Ty)=|x7-y7|=y-x7≤12[d(x,Ty)+d(y,Tx)]-ψ(d(x,Ty),d(y,Tx))=12[|x-y7|+|y-x7|]-19[|x-y7|+|y-x7|]=718[|x-y7|+|y-x7|]=718[x-y7+y-x7]=13(x+y).
So, (2.1) holds.(iii)
Finally, suppose thaty/7≥x. Then, the inequality (2.1) yields that
(2.29)d(Tx,Ty)=|x7-y7|=y-x7≤12[d(x,Ty)+d(y,Tx)]-ψ(d(x,Ty),d(y,Tx))=12[|x-y7|+|y-x7|]-19[|x-y7|+|y-x7|]=718[|x-y7|+|y-x7|]=718[x-y7+y-x7]=13(y+x).
Hence, (2.1) is true. Therefore, all conditions of Theorem 1.5 are satisfied, and so T has a fixed point (which is z=0∈⋂i=12‍Ai.)


## 2.2. Fixed Point Theory for the Generalized Cyclic Weakly Chatterjea Type Contractions in Complete Metric Space

In this section we derive fixed point theorems for self-maps satisfying certain generalized cyclic weakly Chatterjea type contractions in a complete metric space.For convenience, we denote byF2 the class of functions ψ:[0,∞)4→[0,∞) lower semicontinuous satisfying ψ(x,y,z,t)=0 if and only if x=y=z=t=0.We introduce the notion of generalized cyclic weakly Chatterjea type contraction in metric space.Definition 2.8.
Let(X,d) be a metric space, m∈ℕ,A1,A2,…,Am nonempty subsets of X and Y=⋃i=1m‍Ai. An operator T:Y→Y is called a generalized cyclic weakly Chatterjea type contraction if(1)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(2)
(2.30)d(Tx,Ty)≤α[d(x,Tx)+d(y,Ty)+d(x,Ty)+d(y,Tx)]-ψ(d(x,Tx),d(x,Ty),d(x,Ty),d(y,Tx)),
for any x∈Ai,y∈Ai+1,i=1,2,…,m where Am+1=A1 and
(2.31)ψ(d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))∈F2andα∈(0,14].We state the main result of this section as follows.Theorem 2.9.
Let(X,d) be a complete metric space, m∈ℕ,  A1,A2,…,Am nonempty closed subsets of X and X=⋃i=1m‍Ai. Suppose that T is a generalized cyclic weakly C-contraction. Then, T has a unique fixed point z∈⋂i=1n‍Ai.Proof.
As in the proof of Theorem2.2, we take x0∈X and construct a sequence by defining xn+1=Txn,n=0,1,2,…. Suppose that there exists n0∈ℕ such that xn0+1=xn0. Then, since xn0+1=Txn0=xn0, the existence of the fixed point is proved. So, we assume that xn+1≠xn for any n=0,1,2,…. Since X=⋃i=1m‍Ai, for any n>0, there exists in∈{1,2,…,m} such that xn-1∈Ain and xn∈Ain+1. Regarding that T is a generalized cyclic weakly Chatterjea type contraction, we have
(2.32)d(xn,xn+1)=d(Txn-1,Txn)≤α[d(xn-1,Txn-1)+d(xn,Txn)+d(xn-1,Txn)+d(xn,Txn-1)]-ψ(d(xn-1,Txn-1),d(xn,Txn),d(xn-1,Txn),d(xn,Txn-1))=α[d(xn-1,xn)+d(xn,xn+1)+d(xn-1,xn+1)]-ψ(d(xn-1,xn),d(xn,xn+1),d(xn-1,xn+1),0)≤α[d(xn-1,xn)+d(xn,xn+1)+d(xn-1,xn+1)]≤2α[d(xn-1,xn)+d(xn,xn+1)].
Therefore,
(2.33)d(xn,xn+1)≤2α1-2αd(xn-1,xn)foranyn=1,2,….
If we take k=α/(1-α)<1, then we have
(2.34)d(xn,xn+1)≤kd(xn-1,xn)foranyn=1,2,….
It follows that d(xn,xn+1)≤knd(x0,x1) and
(2.35)limn→∞d(xn,xn+1)=0.
We show that {xn} is a Cauchy sequence. For this purpose, first, we prove the following claim. For every ε>0 there exists n∈ℕ such that if r,q≥n with r-q≡1(m), then d(xr,xq)<ε.
Assume the contrary. So there existsε>0 such that for any n∈N we can find rn>qn≥n with rn-qn≡1(m) satisfying
(2.36)d(xqn,xrn)≥ε.
Now, we take n>2m. Then, corresponding to qn≥n we can choose rn in such a way that it is the smallest integer with rn>qn satisfying rn-qn≡1(m) and d(xqn,xrn)≥ε. Therefore, d(xqn,xrn-m)≤ε by using the triangular inequality
(2.37)ε≤d(xqn,xrn)≤d(xqn,xrn-m)+∑i=1m‍d(xrn-i,xrn-i+1)<ε+∑i=1m‍d(xrn-i,xrn-i+1).
Passing to the limit as n→∞ in the last inequality and taking into account that limn→∞d(xn,xn+1)=0, we obtain
(2.38)limn→∞d(xqn,xrn)=ε.
Again, by the triangular inequality
(2.39)ε≤d(xqn,xrn)≤d(xqn,xqn+1)+d(xqn+1,xrn+1)+d(xrn+1,xrn)≤d(xqn,xqn+1)+d(xqn+1,xqn)+d(xqn,xrn)+d(xrn,xrn+1)+d(xrn+1,xrn)=2d(xqn,xqn+1)+d(xqn,xrn)+2d(xrn,xrn+1).
Taking (2.35) and (2.38) into account, we get
(2.40)limn→∞d(xqn+1,xrn+1)=ε
as n→∞ in (2.38).
Sincexqn and xrn lie in different adjacently labeled sets Ai and Ai+1 for certain 1≤i≤m, using the fact that T is a generalized cyclic weakly Chatterjea type contraction, we have
(2.41)d(xqn+1,xrn+1)=d(Txqn,Txrn)≤α[d(xqn,Txqn)+d(xrn,Txrn)+d(xqn,Txrn)+d(xrn,Txqn)]-ψ(d(xqn,Txqn),d(xrn,Txrn),d(xqn,Txrn),d(xrn,Txqn))≤α[d(xqn,xqn+1)+d(xrn,xrn+1)+d(xqn,xrn+1)+d(xrn,xqn+1)]-ψ(d(xqn,xqn+1),d(xrn,xrn+1),d(xqn,xrn+1),d(xrn,xqn+1)).
Taking into account (2.35) and (2.38) and the lower semicontinuity of ψ, letting n→∞ in the last inequality, we obtain
(2.42)ε≤α[ε+0+ε+0]-liminfn→+∞ψ(d(xqn,xqn+1),d(xrn,xrn+1),d(xqn,xrn+1),d(xrn,xqn+1))≤α[ε+0+ε+0]-ψ(ε,0,ε,0)≤2α·ε
and from the last inequality, ψ(ε,0,ε,0)≤-2αε≤0. Therefore φ(ε,0,ε,0)=0. From the fact that ψ(x,y,z,t)=0⇔x=y=z=t=0, we have ε=0, which is a contradiction. Therefore, our claim is proved.
In the sequel, we will prove that(X,d) is a Cauchy sequence. Fix ε>0. By the claim, we find n0∈ℕ such that if r,q≥n0 with r-q≡1(m)(2.43)d(xr,xq)≤ε2.
Since limn→∞d(xn,xn+1)=0, we also find n1∈ℕ such that
(2.44)d(xn,xn+1)≤ε2m
for any n≥n1. Suppose that r,s≥max{n0,n1} and s>r. Then there exists k∈{1,2,…,m} such that s-r≡k(m). Therefore, s-r+φ≡1(m) for φ=m-k+1. So, we have
(2.45)d(xr,xs)≤d(xr,xs+j)+d(xs+j,xs+j-1)+⋯+d(xs+1,xs).
By (2.43) and (2.44) and from the last inequality, we get
(2.46)d(xr,xs)≤ε2+j×ε2m≤ε2+m×ε2m=ε.
This proves that (xn) is a Cauchy sequence. Since X is a complete metric space, there exists x∈X such that limn→∞xn=x. In what follows, we prove that x is a fixed point of T. In fact, since limn→∞xn=x and as X=⋃i=1m‍Ai is a cyclic representation of X with respect to T, the sequence (xn) has infinite terms in each Ai for i∈{1,2,…,m}. Suppose that x∈Ai,Tx∈Ai+1 and we take a subsequence xnk of (xn) with xnk∈Ai-1 (the existence of this subsequence is guaranteed by the previously mentioned comment). By using the contractive condition, we can obtain
(2.47)d(xnk+1,Tx)=d(Txnk,Tx)≤α[d(xnk,Txnk)+d(x,Tx)+d(xnk,Tx)+d(x,Txnk))-ψ(d(xnk,Txnk),d(x,Tx),d(xnk,Tx),d(x,Txnk))=α[d(xnk,xnk+1)+d(x,Tx)+d(xnk,Tx)+d(x,xnk+1))-ψ(d(xnk,xnk+1),d(x,Tx),d(xnk,Tx),d(x,xnk+1)).
Passing to the limit as n→∞ and using xnk→x, lower semicontinuity of ψ, we have
(2.48)d(x,Tx)≤12d(x,Tx)-liminfn→+∞ψ(d(xnk,xnk+1),d(x,Tx),d(xnk,Tx),d(x,xnk+1))≤12d(x,Tx)-ψ(0,d(x,Tx),d(x,Tx),0)≤12d(x,Tx),
which is a contradiction unless d(x,Tx)=0, and, therefore, x is a fixed point of T. Finally, to prove the uniqueness of the fixed point, we have y,z∈X with y and z being fixed points of T. The cyclic character of T and the fact that y,z∈X are fixed points of T imply that y,z∈⋂i=1m‍Ai. Using the contractive condition we obtain
(2.49)d(y,z)=d(Ty,Tz)≤α[d(y,Ty)+d(z,Tz)+d(y,Tz)+d(z,Ty)]-ψ(d(y,Ty),d(z,Tz),d(y,Tz),d(z,Ty)).
that is,
(2.50)d(y,z)≤2αd(y,z)-ψ(0,0,d(y,z),d(z,y))≤d(y,z).
This gives us ψ(0,0,d(y,z),d(z,y))≤(1-2α)d(y,z)≤0, and, by our assumption about ψ, d(y,z)=0, that is, y=z. This finishes the proof.Corollary 2.10.
Let(X,d) be a complete metric space, m∈ℕ, A1,A2,…,Am nonempty subsets of X, and X=⋃i=1m‍Ai. Suppose that T:X→X be an operator such that(1)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(2)
there existsβ∈[0,1/4) such that
(2.51)d(Tx,Ty)≤β[d(x,Tx)+d(y,Ty)+d(x,Ty)+d(y,Tx)]
for any x∈Ai,y∈Ai+1,i=1,2,…,m, where Am+1=A1. Then, T has a fixed point z∈⋂i=1n‍Ai.Proof.
Letβ∈[0,1/4). Here, it suffices to take the function ψ:[0,+∞)4→[0,+∞) defined by ψ(a,b,c,e)=(α-β)(a+b+c+e). Obviously, ψ satisfies that ψ(a,b,c,e)=0 if and only if a=b=c=e=0, and ψ(x,y,z,t)=(α-β)(x+y+z+t)=ψ(x+y+z+t,0). Then, we can apply Theorems 2.9.The following corollary gives us a fixed point theorem with a contractive condition of integral type for cyclic contractions.Corollary 2.11.
Let(X,d) be a complete metric space, m∈ℕ,A1,A2,…,Am nonempty closed subsets of X and X=⋃i=1m‍Ai. Suppose that T:X→X be an operator such that(1)
X=⋃i=1m‍Ai is a cyclic representation of X with respect to T;(2)
there existsβ∈[0,1/4) such that
(2.52)∫0d(Tx,Ty)‍ρ(t)dt≤β∫0d(x,Tx)+d(y,Ty)+d(x,Ty)+d(y,Tx)‍ρ(t)dtfor any x∈Ai,y∈Ai+1,i=1,2,…,m, where Am+1=A1 and ρ:[0,∞)→[0,∞) is a Lebesgue-integrable mapping satisfying ∫0ε‍ρ(t)dt for ε>0. Then T has unique fixed point z∈⋂i=1m‍Ai.Proof.
It is easily proved that the functionφ:[0,∞)→[0,∞) given by φ(t)=∫0t‍ρ(s)ds satisfies that φ∈F1. Therefore, Corollary 2.10 is obtained from Theorem 2.9, taking as φ the perviously defined function and as ψ the function ψ(x,y,z,t)=(1/4-α)(x+y+z+t)=φ(x+y+z+t,0).If in Corollary2.11, we take Ai=X for i=1,2,…,m, we obtain the following result.Corollary 2.12.
Let(X,d) be a complete metric space and let T:X→X be a mapping such that for any x,y∈X,
(2.53)∫0d(Tx,Ty)‍ρ(t)dt≤β∫0d(x,Tx)+d(y,Ty)+d(x,Ty)+d(y,Tx)‍ρ(t)dt,
where ρ:[0,∞)→[0,∞) is a Lebesgue-integrable mapping satisfying ∫0ε‍ρ(t)dt for ε>0 and the constant β∈[0,1/4). Then T has unique fixed point.If in Theorem2.9 we put Ai=X for i=1,2,…,m, we have the generalized result of [4].Corollary 2.13.
Let(X,d) be a complete metric space and let T:X→X be a mapping such that for any x,y∈X,
(2.54)d(Tx,Ty)≤α[d(x,Tx)+d(y,Ty)+d(x,Ty)+d(y,Tx)]-ψ(d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)),
where ψ∈F2 and α∈(0,1/4]. Then T has unique fixed point.Example 2.14.
LetX=ℝ with the usual metric. Suppose A1=[0,1], A2=[0,1/3], A3=[0,1/9], A4=[0,1/27] and Y=⋃i=14‍Ai. Define T:Y→Y such that Tx=x/4 for all x∈Y. It is clear that ⋃i=14‍Ai is a cyclic representation of Y with respect to T. Furthermore, if ψ:[0,+∞)4→[0,+∞) is defined by ψ(x,y,z,t)=(x+y+z+t)/16, then ψ∈F2. As in Example 2.7, it can be easily shown that T is a generalized cyclic weakly Chatterjea type contraction α=1/4. Therefore, all conditions of Theorem 2.9 are satisfied, and so T has a fixed point (which is z=0∈⋂i=14‍Ai).



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*Source: 165698-2012-08-23.xml*