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Fixed point theory for cyclic (ϕ  ψ)contractions
Fixed Point Theory and Applications volume 2011, Article number: 69 (2011)
Abstract
In this article, the concept of cyclic (ϕ  ψ)contraction and a fixed point theorem for this type of mappings in the context of complete metric spaces have been presented. The results of this study extend some fixed point theorems in literature.
2000 Mathematics Subject Classification: 47H10;46T99 54H25.
1. Introduction and preliminaries
One of the most important results used in nonlinear analysis is the wellknown Banach's contraction principle. Generalization of the above principle has been a heavily investigated branch research. Particularly, in [1] the authors introduced the following definition.
Definition 1. Let X be a nonempty set, m a positive integer, and T: X → X a mapping. X={\cup}_{i=1}^{m}{A}_{i} is said to be a cyclic representation of X with respect to T if

(i)
A_{ i } , i = 1, 2, ..., m are nonempty sets.

(ii)
T(A _{1}) ⊂ A _{2}, ..., T (A _{m1}) ⊂ A_{ m } , T (Am) ⊂ A _{1}.
Recently, 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., [2–5]). Now, we present the main result of [5]. Previously, we need the following definition.
Definition 2. Let (X, d) be a metric space, m a positive integer A_{1}, A_{2}, ..., A_{ m } nonempty closed subsets of X andX={\cup}_{i=1}^{m}{A}_{i}. An operator T: X → X is said to be a cyclic weak ϕcontraction if

(i)
X={\cup}_{i=1}^{m}{A}_{i} is a cyclic representation of X with respect to T.

(ii)
d(Tx, Ty) ≤ d(x, y)  ϕ(d(x, y)), for any X ∈ A_{ i }, y ∈ A _{i+1}, i = 1, 2, ..., m, where A _{m+1}= A _{1} and ϕ: [0, ∞) → [0, ∞) is a nondecreasing and continuous function satisfying ϕ(t) > 0 for t ∈ (0, ∞) and ϕ(0) = 0
Remark 3. For convenience, we denote by F the class of functions ϕ: [0, ∞) → [0, ∞) nondecreasing and continuous satisfying ϕ(t) > 0 for t ∈ (0, ∞) and ϕ(0) = 0.
The main result of [5] is the following.
Theorem 4. [[5], Theorem 6] Let (X, d) be a complete metric space, m a positive integer, A_{1}, A_{2}, ..., A_{ m } nonempty closed subsets of X andX={\cup}_{i=1}^{m}{A}_{i}. Let T: X → X be a cyclic weak ϕcontraction with ϕ ∈ F . Then T has a unique fixed point z\in {\cap}_{i=1}^{m}{A}_{i}.
The main purpose of this article is to present a generalization of Theorem 4.
2. Main results
First, we present the following definition.
Definition 5. Let (X, d) be a metric space, m a positive integer, A_{1}, A_{2}, ..., A_{ m } nonempty subsets of X andX={\cup}_{i=1}^{m}{A}_{i}. An operator T: X → X is a cyclic weak (ϕ  ψ)contraction if

(i)
X={\cup}_{i=1}^{m}{A}_{i} is a cyclic representation of X with respect to T,

(ii)
ϕ(d(Tx, Ty)) ≤ ϕ(d(x, y))  ψ(d(x, y)), for any X ∈ A_{ i }, y ∈ A _{i+1}, i = 1, 2, ..., m, where A _{m+1}= A _{1} and ϕ, ψ ∈ F .
Our main result is the following.
Theorem 6. Let (X, d) be a complete metric space, m a positive integer, A_{1}, A_{2}, ..., A_{ m } nonempty subsets of X andX={\cup}_{i=1}^{m}{A}_{i}. Let T: X → X be a cyclic (ϕ  ψ)contraction. Then T has a unique fixed pointz\in {\cap}_{i=1}^{m}{A}_{i}.
Proof. Take x_{0} ∈ X and consider the sequence given by
If there exists n_{0} ∈ ℕ such that {x}_{{n}_{0}+1}={x}_{{n}_{0}} then, since {x}_{{n}_{0}+1}=T{x}_{{n}_{0}}={x}_{{n}_{0}}, the part of existence of the fixed point is proved. Suppose that x_{n+1}≠ x_{ n } for any n = 0, 1, 2, .... Then, since X={\cup}_{i=1}^{m}{A}_{i}, for any n > 0 there exists i_{ n } ∈ {1, 2, ..., m} such that {x}_{n1}\in {A}_{{i}_{n}} and {x}_{n}\in {A}_{{i}_{n+1}}. Since T is a cyclic (ϕ  ψ)contraction, we have
From 2.1 and taking into account that ϕ is nondecreasing we obtain
Thus {d(x_{ n } , x_{n+1})} is a nondecreasing sequence of nonnegative real numbers. Consequently, there exists γ ≥ 0 such that \underset{n\to \infty}{lim}d\left({x}_{n},{x}_{n+1}\right)=\gamma. Taking n → ∞ in (2.1) and using the continuity of ϕ and ψ, we have
and, therefore, ψ(γ) = 0. Since ψ ∈ F, γ = 0, that is,
In the sequel, we will prove that {x_{ n } } is a Cauchy sequence.
First, we prove the following claim.
Claim: For every ε > 0 there exists n ∈ ℕ such that if p, q ≥ n with p  q ≡ 1(m) then d(x_{ p } , x_{ q } ) < ε.
In fact, suppose the contrary case. This means that there exists ε > 0 such that for any n ∈ ℕ we can find p_{ n } > q_{ n } ≥ n with p_{ n }  q_{ n } ≡ 1(m) satisfying
Now, we take n > 2m. Then, corresponding to q_{ n } ≥ n use can choose p_{ n } in such a way that it is the smallest integer with p_{ n } > q_{ n } satisfying p_{ n }  q_{ n } ≡ 1(m) and d\left({x}_{{q}_{n}},{x}_{{p}_{n}}\right)\ge \epsilon. Therefore, d\left({x}_{{q}_{n}},{x}_{{p}_{nm}}\right)\le \epsilon. Using the triangular inequality
Letting n → ∞ in the last inequality and taking into account that lim_{n→∞}d(x_{ n } , x_{n+1}) = 0, we obtain
Again, by the triangular inequality
Letting n → ∞ in (2.4) and taking into account that \underset{n\to \infty}{lim}d\left({x}_{n},{x}_{n+1}\right)=0 and (2.4), we get
Since {x}_{{q}_{n}} and {x}_{{p}_{n}} lie in different adjacently labelled sets A_{ i } and A_{i+1}for certain 1 ≤ i ≤ m, using the fact that T is a cyclic (ϕ  ψ)contraction, we have
Taking into account (2.4) and (2.6) and the continuity of ϕ and ψ, letting n → ∞ in the last inequality, we obtain
and consequently, ψ(ε) = 0. Since ψ ∈ F, then ε = 0 which is 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 n_{0} ∈ ℕ such that if p, q ≥ n_{0} with p  q ≡ 1(m)
Since \underset{n\to \infty}{lim}d\left({x}_{n},{x}_{n+1}\right)=0 we also find n_{1} ∈ ℕ such that
for any n ≥ n_{1}.
Suppose that r, s ≥ max{n_{0}, n_{1}} and s > r. Then there exists k ∈ {1, 2, ..., m} such that s  r ≡ k(m). Therefore, s  r + j ≡ 1(m) for j = m  k + 1. So, we have d(x_{ r } , x_{ s } ) ≤ d(x_{ r } , x_{s+j})+ d(x_{s+j}, x_{s+j1})+ ⋯ + d(x_{s+1}, x_{ s } ). By (2.7) and (2.8) and from the last inequality, we get
This proves that (x_{ n } ) is a Cauchy sequence. Since X is a complete metric space, there exists x ∈ X such that lim_{n→∞}x_{ n } = x. In what follows, we prove that x is a fixed point of T. In fact, since \underset{n\to \infty}{lim}{x}_{n}=x and, as X={\cup}_{i=1}^{m}{A}_{i} is a cyclic representation of X with respect to T, the sequence (x_{ n } ) has infinite terms in each A_{ i } for i ∈ {∈ 1, 2, ..., m}.
Suppose that x ∈ A_{ i } , Tx ∈ A_{i+1}and we take a subsequence {x}_{{n}_{k}} of (x_{ n } ) with {x}_{{n}_{k}}\in {A}_{i1} (the existence of this subsequence is guaranteed by the abovementioned comment). Using the contractive condition, we can obtain
and since {x}_{{n}_{k}}\to x and ϕ and ψ belong to F, letting k → ∞ in the last inequality, we have
or, equivalently, ϕ(d(x, Tx)) = 0. Since ϕ ∈ F, then 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 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\in {\cap}_{i=1}^{m}{A}_{i}. Using the contractive condition we obtain
and from the last inequality
Since ψ ∈ F, d(y, z) = 0 and, consequently, y = z. This finishes the proof.
In the sequel, we will show that Theorem 6 extends some recent results.
If in Theorem 6 we take as ϕ the identity mapping on [0, ∞) (which we denote by Id_{[0, ∞)}), we obtain the following corollary.
Corollary 7. Let (X, d) be a complete metric space m a positive integer, A_{1}, A_{2}, ..., A_{ m } nonempty subsets of X andX={\cup}_{i=1}^{m}{A}_{i}. Let T: X → X be a cyclic (Id_{[0, ∞)}  ψ) contraction. Then T has a unique fixed pointz\in {\cap}_{i=1}^{m}{A}_{i}.
Corollary 7 is a generalization of the main result of [5] (see [[5], Theorem 6]) because we do not impose that the sets A_{ i } are closed.
If in Theorem 6 we consider ϕ = Id_{[0, ∞)} and ψ = (1  k)Id_{[0, ∞)} for k ∈ [0, 1) (obviously, ϕ, ψ ∈ F), we have the following corollary.
Corollary 8. Let (X, d) be a complete metric space m a positive integer, A_{1}, A_{2}, ..., A_{ m } nonempty subsets of X andX={\cup}_{i=1}^{m}{A}_{i}. Let T: X → X be a cyclic (Id_{[0, ∞)}  (1  k)Id_{[0, ∞)}) contraction, where k ∈ [0, 1). Then T has a unique fixed pointz\in {\cap}_{i=1}^{m}{A}_{i}.
Corollary 8 is Theorem 1.3 of [1].
The following corollary gives us a fixed point theorem with a contractive condition of integral type for cyclic contractions.
Corollary 9. Let (X, d) be a complete metric space, m a positive integer, A_{1}, A_{2}, ..., A_{ m } nonempty closed subsets of X andX={\cup}_{i=1}^{m}{A}_{i}. Let T: X → X be an operator such that

(i)
X={\cup}_{i=1}^{m}{A}_{i} is a cyclic representation of X with respect to T .

(ii)
There exists k ∈ [0, 1) such that
{\int}_{0}^{d\left(Tx,Ty\right)}\rho \left(t\right)dt\le k{\int}_{0}^{d\left(x,y\right)}\rho \left(t\right)dt
for any X ∈ A_{ i }, y ∈ A_{i+1}, i = 1, 2, ..., m where A_{m+1}= A_{1}, and ρ: [0, ∞) → [0, ∞) is a Lebesgueintegrable mapping satisfying {\int}_{0}^{\epsilon}\rho \left(t\right)dt for ε> 0.
Then T has unique fixed pointz\in {\cap}_{i=1}^{m}{A}_{i}.
Proof. It is easily proved that the function ϕ: [0, ∞) → [0, ∞) given by \phi \left(t\right)={\int}_{0}^{t}p\left(s\right)ds satisfies that ϕ ∈ F. Therefore, Corollary 9 is obtained from Theorem 6, taking as ϕ the abovedefined function and as ψ the function ψ(t) = (1  k)ϕ(t).
If in Corollary 9, we take A_{ i } = X for i = 1, 2, ..., m we obtain the following result.
Corollary 10. Let (X, d) be a complete metric space and T: X → X a mapping such that for x, y ∈ X,
where ρ: [0, ∞) → [0, ∞) is a Lebesgueintegrable mapping satisfying{\int}_{0}^{\epsilon}\rho \left(t\right)dtfor ε > 0 and the constant k ∈ [0, 1). Then T has a unique fixed point.
Notice that this is the main result of [6]. If in Theorem 6 we put A_{ i } = X for i = 1, 2, ..., m we have the result.
Corollary 11. Let (X, d) be a complete metric space and T: X → X an operator such that for x, y ∈ X,
where ϕ, ψ ∈ F . Then T has a unique fixed point.
This result appears in [7].
3. Example and remark
In this section, we present an example which illustrates our results. Throughout the article, we let ℕ* = ℕ\{0}.
Example 12. Consider X=\left\{\frac{1}{n}:n\in {\mathbb{N}}^{*}\right\}\cup \left\{0\right\} with the metric induced by the usual distance in ℝ, i.e., d(x, y) = x  y. Since X is a closed subset of ℝ, it is a complete metric space. We consider the following subsets of X:
Obviously, X = A_{1} ∪ A_{2}. Let T: X → X be the mapping defined by
It is easily seen that X = A_{1} ∪ A_{2}is a cyclic representation of X with respect to T. Now we consider the function ρ: [0, ∞) → [0, ∞) defined by
It is easily proved that{\int}_{0}^{t}\rho \left(s\right)ds={t}^{1\u2215t}for t ≤ 1.
In what follows, we prove that T satisfies condition (ii) of Corollary 9.
In fact, notice that the function ρ(t) is a Lebesgueintegrable mapping satisfying{\int}_{0}^{\epsilon}\rho \left(t\right)dt>0for ε > 0. We take m, n ∈ ℕ* with m ≥ n and we will prove
Since{\int}_{0}^{t}\rho \left(s\right)ds={t}^{1\u2215t}for t ≤ 1 and, as diam(X) ≤ 1, the last inequality can be written as
or equivalently,
or equivalently,
or equivalently,
In order to prove that this last inequality is true, notice that
and, therefore,
On the other hand, from
we obtain
and, thus,
Since\frac{n+m+1}{mn}\ge 1,
Finally, (3.2) and (3.3) give us (3.1).
Now we take x=\frac{1}{n}, n ∈ ℕ* and y = 0. In this case, condition (ii) of Corollary 9 for k=\frac{1}{2} has the form
The last inequality is true since
and, then,
Consequently, since assumptions of Corollary 9 are satisfied, this corollary gives us the existence of a unique fixed point (which is obviously x = 0).
This example appears in [6].
Now, we connect our results with the ones appearing in [3]. Previously, we need the following definition.
Definition 13. A function φ: [0, ∞) → [0, ∞) is a (c)comparison function if \sum _{k=0}^{\infty}{\phi}^{k}\left(t\right) converges for any t∈ [0, ∞). The main result of[3]is the following.
Theorem 14. Let (X, d) be a complete metric space, m a positive integer, A_{1}, A_{2}, ..., A_{ m } nonempty subsets of X, X={\cup}_{i=1}^{m}{A}_{i}and φ: [0, ∞) → [0, ∞) a (c)comparison function. Let T: X → X be an operator and we assume that

(i)
X={\cup}_{i=1}^{m}{A}_{i} is a cyclic representation of X with respect to T.

(ii)
d(Tx, Ty) ≤ φ(d(x, y)), for any X ∈ A_{ i } and y ∈ A _{i+1}, where A _{m+1}= A.
Then T has a unique fixed pointz\in {\cap}_{i=1}^{m}{A}_{i}.
Now, the contractive condition of Theorem 6 can be written as
for any x ∈ A_{ i } , y ∈ A_{i+1}, where A_{m+1}= A_{1}, and ϕ, φ ∈ F.
Particularly, if we take ϕ = Id_{[0, ∞)} and \phi \left(t\right)=\frac{{t}^{2}}{1+t}, it is easily seen that ϕ, φ ∈ F. On the other hand,
and
Moreover, for every t ∈ (0, ∞), \sum _{k=0}^{\infty}{\left(\varphi \phi \right)}^{\left(k\right)}\left(t\right) diverges. Therefore, ϕ  φ is not a (c)comparison function. Consequently, our Theorem 6 can be applied to cases which cannot treated by Theorem 14.
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Acknowledgements
KS was partially supported by the "Ministerio de Education y Ciencia", Project MTM 2007/65706.
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The authors declare that they have no competing interests.
5. Authors' contributions
The authors have contributed in obtaining the new results presented in this article. All authors read and approved the final manuscript.
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Karapinar, E., Sadarangani, K. Fixed point theory for cyclic (ϕ  ψ)contractions. Fixed Point Theory Appl 2011, 69 (2011). https://doi.org/10.1186/16871812201169
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DOI: https://doi.org/10.1186/16871812201169
Keywords
 cyclic (ϕ  ψ)contraction
 fixed point theory.