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Fixed point theorem for weakly Chatterjeatype cyclic contractions
Fixed Point Theory and Applications volume 2013, Article number: 28 (2013)
Abstract
In this article, we introduce the notion of a Chatterjeatype cyclic weakly contraction and derive the existence of a fixed point for such mappings in the setup of complete metric spaces. Our result extends and improves some fixed point theorems in the literature. Example is given to support the usability of the result.
MSC:41A50, 47H10, 54H25.
1 Introduction and preliminaries
It is well known that the fixed point theorem of Banach, for contraction mappings, is one of the pivotal results in analysis. It has been used in many different fields of mathematics but suffers from one major drawback. More accurately, in order to use the contractive condition, a selfmapping T must be Lipschitz continuous, with the Lipschitz constant L<1. In particular, T must be continuous at all points of its domain.
A natural question arises:
Could we find contractive conditions which will imply the existence of a fixed point in a complete metric space but will not imply continuity?
Kannan [1, 2] proved the following result giving an affirmative answer to the above question.
Theorem 1.1 If (X,d) is a complete metric space and the mapping T:X\to X satisfies
where 0<k<\frac{1}{2} and x,y\in X, then T has a unique fixed point.
The mappings satisfying (1.1) are called Kannantype mappings.
A similar type of contractive condition has been studied by Chatterjea [3]. He proved the following result.
Theorem 1.2 If (X,d) is a complete metric space and T:X\to X satisfies
where 0<k<\frac{1}{2} and x,y\in X, then T has a unique fixed point.
In Theorems 1.1 and 1.2, there is no the requirement for the continuity of T.
Alber and GuerreDelabriere [4] introduced the concept of weakly contractive mappings and proved the existence of fixed points for singlevalued weakly contractive mappings in Hilbert spaces. Thereafter, in 2001, Rhoades [5] proved the fixed point theorem which is one of the generalizations of Banach’s contraction mapping principle because the weakly contractions contain contractions as a special case, and he also showed that some results of [4] are true for any Banach space. In fact, weakly contractive mappings are closely related to the mappings of Boyd and Wong [6] and of Reich types [7].
Fixed point problems involving different types of contractive type inequalities have been studied by many authors (see [1–24] and the references cited therein).
In [22], Kirk et al. introduced the following notion of a cyclic representation and characterized the Banach contraction principle in the context of a cyclic mapping.
Definition 1.1 [22]
Let X be a nonempty set and T:X\to X be an operator. By definition, X={\bigcup}_{i=1}^{m}{X}_{i} is a cyclic representation of X with respect to T if

(a)
{X}_{i}; i=1,\dots ,m are nonempty sets;

(b)
T({X}_{1})\subset {X}_{2},\dots ,T({X}_{m1})\subset {X}_{m},T({X}_{m})\subset {X}_{1}.
It is the aim of this paper to introduce the notion of a cyclic weakly Chatterjeatype contraction and then derive a fixed point theorem for such cyclic contractions in the framework of complete metric spaces.
2 Main results
To state and prove our main results, we will introduce our notion of a Chatterjeatype cyclic weakly contraction in a metric space. In this respect, let Φ denote the set of all monotone increasing continuous functions \mu :[0,\mathrm{\infty})\to [0,\mathrm{\infty}), with \mu (t)=0, if and only if t=0, and let Ψ denote the set of all lower semicontinuous functions {\psi :[0,\mathrm{\infty})}^{2}\to [0,\mathrm{\infty}), with \psi ({t}_{1},{t}_{2})>0, for {t}_{1},{t}_{2}\in (0,\mathrm{\infty}) and \psi (0,0)=0.
Definition 2.1 Let (X,d) be a metric space, m be a natural number, {A}_{1},{A}_{2},\dots ,{A}_{m} be nonempty subsets of X and Y={\bigcup}_{i=1}^{m}{A}_{i}. An operator T:Y\to Y is called a Chatterjeatype cyclic weakly contraction if

(1)
{\bigcup}_{i=1}^{m}{A}_{i} is a cyclic representation of Y with respect to T;

(2)
\mu (d(Tx,Ty))\le \mu (\frac{1}{2}[d(x,Ty)+d(y,Tx)])\psi (d(x,Ty),d(y,Tx))
for any x\in {A}_{i}, y\in {A}_{i+1}, i=1,2,\dots ,m, where {A}_{m+1}={A}_{1}, \mu \in \mathrm{\Phi} and \psi \in \mathrm{\Psi}.
Theorem 2.1 Let (X,d) be a complete metric space, m\in \mathbb{N}, {A}_{1},{A}_{2},\dots ,{A}_{m} be nonempty closed subsets of X and Y={\bigcup}_{i=1}^{m}{A}_{i}. Suppose that T is a Chatterjeatype cyclic weakly contraction. Then T has a fixed point z\in {\bigcap}_{i=1}^{n}{A}_{i}.
Proof Let {x}_{0}\in X. We can construct a sequence {x}_{n+1}=T{x}_{n}, n=0,1,2,\dots .
If there exists {n}_{0}\in \mathbb{N} such that {x}_{{n}_{0}+1}={x}_{{n}_{0}}, hence the result. Indeed, we can see that T{x}_{{n}_{0}}={x}_{{n}_{0}+1}={x}_{{n}_{0}}.
Now, we assume that {x}_{n+1}\ne {x}_{n} for any n=0,1,2,\dots . As X={\bigcup}_{i=1}^{m}{A}_{i}, for any n>0, there exists {i}_{n}\in \{1,2,\dots ,m\} such that {x}_{n1}\in {A}_{{i}_{n}} and {x}_{n}\in {A}_{{i}_{n+1}}. Since T is a Chatterjeatype cyclic weakly contraction, we have
Since μ is a nondecreasing function, for all n=1,2,\dots , we have
This implies that d({x}_{n+1},{x}_{n})\le d({x}_{n},{x}_{n1}). Thus \{d({x}_{n+1},{x}_{n})\} is a monotone decreasing sequence of nonnegative real numbers and hence is convergent. Therefore, there exists r\ge 0 such that d({x}_{n+1},{x}_{n})\to r. Letting n\to \mathrm{\infty} in (2.2), we obtain that limd({x}_{n1},{x}_{n+1})=2r.
Letting n\to \mathrm{\infty} in (2.1) and using the continuity of μ and lower semicontinuity of ψ, we obtain that \mu (r)\le \mu (r)\psi (0,2r). This implies that \psi (2r,0)=0, hence r=0. Thus we have proved that
Now, we show that \{{x}_{n}\} is a Cauchy sequence. For this purpose, we prove the following result first.
Lemma 2.1 For every positive ϵ, there exists a natural number n such that if r,q\ge n with rq\equiv 1(modm), then d({x}_{r},{x}_{q})<\u03f5.
Proof Assume the contrary. Thus there exists \u03f5>0 such that for any n\in \mathbb{N}, we can find {r}_{n}>{q}_{n}\ge n with {r}_{n}{q}_{n}\equiv 1(modm) satisfying d({x}_{{r}_{n}},{x}_{{q}_{n}})\ge \u03f5.
Now, we take n>2m. Then, corresponding to {q}_{n}\ge n, we can choose {r}_{n} such that it is the smallest integer with {r}_{n}>{q}_{n} satisfying {r}_{n}{q}_{n}\equiv 1(modm) and d({x}_{{r}_{n}},{x}_{{q}_{n}})\ge \u03f5. Therefore, d({x}_{{r}_{nm}},{x}_{{q}_{n}})<\u03f5. By using the triangular inequality, we have
Letting n\to \mathrm{\infty} and using d({x}_{n+1},{x}_{n})\to 0, we obtain
Again, by the triangular inequality,
Letting n\to \mathrm{\infty} and using d({x}_{n+1},{x}_{n})\to 0, we get
Consider
and
On taking n\to \mathrm{\infty} in inequalities (2.5) and (2.6), we have
and
As {x}_{{q}_{n}} and {x}_{{r}_{n}} lie in different adjacently labeled sets {A}_{i} and {A}_{i+1} for certain 1\le i\le m, using the fact that T is a Chatterjeatype cyclic weakly contraction, we obtain
On taking n\to \mathrm{\infty} in (2.9), using (2.7) and (2.8), the continuity of μ and lower semicontinuity of ψ, we get that
Consequently, \psi (\epsilon ,\epsilon )\le 0, which is contradiction with \epsilon >0. Hence the result is proved. □
Now, using Lemma 2.1, we will show that \{{x}_{n}\} is a Cauchy sequence in Y. Fix \u03f5>0. By Lemma 2.1, we can find {n}_{0}\in \mathbb{N} such that r,q\ge {n}_{0} with rq\equiv 1(modm)
Since limd({x}_{n},{x}_{n+1})=0, we can also find {n}_{1}\in \mathbb{N} such that
for any n\ge {n}_{1}.
Assume that r,s\ge max\{{n}_{0},{n}_{1}\} and s>r. Then there exists k\in \{1,2,\dots ,m\} such that sr\equiv k(modm). Hence sr+t=1(modm) for t=mk+1. So, we have
Using (2.10), (2.11) and (2.12), we obtain
Hence \{{x}_{n}\} is a Cauchy sequence in Y. Since Y is closed in X, then Y is also complete and there exists x\in Y such that lim{x}_{n}=x.
Now, we will prove that x is a fixed point of T.
As Y={\bigcup}_{i=1}^{m}{A}_{i} is a cyclic representation of Y with respect to T, the sequence \{{x}_{n}\} has infinite terms in each {A}_{i} for i=\{1,2,\dots ,m\}. Suppose that x\in {A}_{i}, Tx\in {A}_{i+1} and we take a subsequence \{{x}_{{n}_{k}}\} of \{{x}_{n}\} with {x}_{{n}_{k}}\in {A}_{i}. By using the contractive condition, we can obtain
Letting n\to \mathrm{\infty} and using the continuity of μ and lower semicontinuity of ψ, we have
which is a contradiction unless d(x,Tx)=0. Hence x is a fixed point of T.
Now, we will prove the uniqueness of the fixed point.
Suppose that {x}_{1} and {x}_{2} ({x}_{1}\ne {x}_{2}) are two fixed points of T. Using the contractive condition and the continuity of μ and lower semi continuity of ψ, we have
which is a contradiction unless {x}_{1}={x}_{2}. Hence the main result is proved. □
If \mu (a)=a, then we have the following result.
Corrollary 2.1 Let (X,d) be a complete metric space, m\in \mathbb{N}, {A}_{1},{A}_{2},\dots ,{A}_{m} be nonempty closed subsets of X and Y={\bigcup}_{i=1}^{m}{A}_{i}. Suppose that T:Y\to Y is an operator such that

(1)
{\bigcup}_{i=1}^{m}{A}_{i} is a cyclic representation of Y with respect to T;

(2)
d(Tx,Ty)\le \frac{1}{2}[d(x,Ty)+d(y,Tx)]\psi (d(x,Ty),d(y,Tx))
for any x\in {A}_{i}, y\in {A}_{i+1}, i=1,2,\dots ,m, where {A}_{m+1}={A}_{1} and \psi \in \mathrm{\Psi}. Then T has a fixed point z\in {\bigcap}_{i=1}^{n}{A}_{i}.
If \psi (a,b)=(\frac{1}{2}k)(a+b), where k\in [0,\frac{1}{2}), we have the following result.
Corrollary 2.2 Let (X,d) be a complete metric space, m\in \mathbb{N}, {A}_{1},{A}_{2},\dots ,{A}_{m} be nonempty closed subsets of X and Y={\bigcup}_{i=1}^{m}{A}_{i}. Suppose that T:Y\to Y is an operator such that

(1)
{\bigcup}_{i=1}^{m}{A}_{i} is a cyclic representation of Y with respect to T;

(2)
there exists k\in [0,\frac{1}{2}) such that d(Tx,Ty)\le k[d(x,Ty)+d(y,Tx)]
for any x\in {A}_{i}, y\in {A}_{i+1}, i=1,2,\dots ,m, where {A}_{m+1}={A}_{1}. Then T has a fixed point z\in {\bigcap}_{i=1}^{n}{A}_{i}.
3 Applications
Other consequences of our results, for mappings involving contractions of integral type, are given in the following. In this respect, denote by Λ the set of functions \mu :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) satisfying the following hypotheses:

(h1)
μ is a Lebesgueintegrable mapping on each compact of [0,\mathrm{\infty});

(h2)
for any \u03f5>0, we have {\int}_{0}^{\u03f5}\mu (t)>0.
Corrollary 3.1 Let (X,d) be a complete metric space, m\in \mathbb{N}, {A}_{1},{A}_{2},\dots ,{A}_{m} be nonempty closed subsets of X and Y={\bigcup}_{i=1}^{m}{A}_{i}. Suppose that T:Y\to Y is an operator such that

(1)
{\bigcup}_{i=1}^{m}{A}_{i} is a cyclic representation of Y with respect to T;

(2)
there exists k\in [0,\frac{1}{2}) such that
{\int}_{0}^{d(Tx,Ty)}\alpha (s)\phantom{\rule{0.2em}{0ex}}ds\le k{\int}_{0}^{d(x,Ty)+d(y,Tx)}\alpha (s)\phantom{\rule{0.2em}{0ex}}ds
for any x\in {A}_{i}, y\in {A}_{i+1}, i=1,2,\dots ,m, where {A}_{m+1}={A}_{1} and \alpha \in \mathrm{\Lambda}. Then T has a fixed point z\in {\bigcap}_{i=1}^{n}{A}_{i}.
If we take {A}_{i}=X, i=1,2,\dots ,m, we obtain the following result.
Corrollary 3.2 Let (X,d) be a complete metric space and T:X\to X be a mapping such that
for any xy\in X, k\in [0,\frac{1}{2}) and \alpha \in \mathrm{\Lambda}. Then T has a fixed point z\in {\bigcap}_{i=1}^{n}{A}_{i}.
Example 3.1 Let X be a subset in ℝ endowed with the usual metric. Suppose {A}_{1}=[0,1], {A}_{2}=[0,\frac{1}{2}] and Y={\bigcup}_{i=1}^{2}{A}_{i}. Define T:Y\to Y such that Tx=\frac{x}{5} for all x\in Y. It is clear that {\bigcup}_{i=1}^{2}{A}_{i} is a cyclic representation of Y with respect to T. Furthermore, if \mu :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is given as \mu (t)=t and {\psi :[0,\mathrm{\infty})}^{2}\to [0,\mathrm{\infty}) is given by \psi (x,y)=\frac{1}{7}(x+y), then \psi \in \mathrm{\Psi}.
Now, we prove that T satisfies the inequality of Chatterjeatype cyclic weakly contraction, i.e., \mu (d(Tx,Ty))\le \mu (\frac{1}{2}[d(x,Ty)+d(y,Tx)])\psi (d(x,Ty),d(y,Tx)). To see this fact, we examine three cases.
Case 1. Suppose that x\ge y. Then
and
If y<\frac{x}{5}, then
Hence, the given inequality is satisfied.
If y\ge \frac{x}{5}, then
Hence the given inequality is satisfied.
Case 2. Suppose that \frac{y}{5}\le x\le y. Then from (3.1) and (3.2), we have
Hence the given inequality is satisfied.
Case 3. Finally, suppose that \frac{y}{5}\ge x. Then from (3.1) and (3.2), we have
and
Hence the given inequality is satisfied.
Therefore, all the conditions of Theorem 2.1 are satisfied, and so T has a fixed point (which is z=0\in {\bigcap}_{i=1}^{2}{A}_{i}).
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Chandok, S., Postolache, M. Fixed point theorem for weakly Chatterjeatype cyclic contractions. Fixed Point Theory Appl 2013, 28 (2013). https://doi.org/10.1186/16871812201328
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DOI: https://doi.org/10.1186/16871812201328