Abstract
In this paper, we introduce the notion of a cyclic -contraction for the pair of self-mappings on the set X. We utilize our definition to introduce some common fixed point theorems for the two mappings f and T under a set of conditions. Also, we introduce an example to support the validity of our results. As application of our results, we derive some common fixed point theorems of integral type.
MSC:54H25, 47H10.
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1 Introduction
In recent years many authors established interesting results in fixed point theory in (ordered) metric spaces. One of the popular topics in the fixed point theory is the cyclic contraction. Kirk et al. [1] established the first result in this interesting area. Meantime, other authors obtained important results in this area (see [1–12]).
We begin with the definition of a cyclic map.
Definition 1.1 Let A and B be non-empty subsets of a metric space and . Then T is called a cyclic map if and .
In 2003, Kirk et al. [1] gave the following interesting theorem in fixed point theory for a cyclic map.
Theorem 1.1 ([1])
Let A and B be nonempty closed subsets of a complete metric space . Suppose that is a cyclic map such that
If , then T has a unique fixed point in .
Recently, several authors proved many results in fixed point theory for cyclic mappings, satisfying various (nonlinear) contractive conditions (see [1–12]). Some of contractive conditions are based on functions called control functions which alter the distance between two points in a metric space. Such functions were introduced by Khan et al. [13].
Definition 1.2 (altering distance function, [13])
The function is called an altering distance function if the following properties are satisfied:
-
(1)
ϕ is continuous and nondecreasing;
-
(2)
if and only if .
For some fixed point theorems based on an altering distance function, we refer the reader to [14–20].
Let X be a nonempty set. Then is called an ordered metric space if and only if is a metric space and is a partially ordered set. Two elements are called comparable if or .
Altun et al. [21, 22] introduced the notion of weakly increasing mappings and proved some existing theorems. For some works in the theory of weakly increasing mappings, we refer the reader to [23, 24].
Definition 1.3 ([21])
Let be a partially ordered set. Two mappings are said to be weakly increasing if and for all .
The purpose of this paper is to obtain common fixed point results for mappings satisfying nonlinear contractive conditions of a cyclic form based on the notion of an altering distance function.
2 Main result
We start with the following definition.
Definition 2.1 Let be an ordered metric space and A, B be nonempty closed subsets of X. Let be two mappings. The pair is called a cyclic -contraction if
-
(1)
ψ is an altering distance function;
-
(2)
has a cyclic representation w.r.t. the pair ; that is, , and ;
-
(3)
There exists such that for any comparable elements with and , we have
(2.1)
Definition 2.2 Let be a partially ordered set and A, B be closed subsets of X with . Let be two mappings. The pair is said to be -weakly increasing if for all and for all .
From now on, by ψ we mean altering distance functions unless otherwise stated.
In the rest of this paper, ℕ stands for the set of nonnegative integer numbers.
Theorem 2.1 Let be an ordered complete metric space and A, B be nonempty closed subsets of X. Let be two mappings such that the pair is -weakly increasing. Assume the following:
-
(1)
The pair is a cyclic -contraction;
-
(2)
f or T is continuous.
Then f and T have a common fixed point.
Proof Choose . Let . Since , we have . Also, let . Since , we have . Continuing this process, we can construct a sequence in X such , , and .
Since f and T are -weakly increasing, we have
We divide our proof into the following steps.
Step 1: We will show that is a Cauchy sequence in .
Subcase 1: Suppose that for some . Since and are comparable elements in X with and , we have
Since , we have and hence . Similarly, we may show that . Hence is a constant sequence in X, so it is a Cauchy sequence in .
Subcase 2: for all . Given . If n is even, then for some . Since , and , are comparable, we have
If
then
which is a contradiction. Thus
therefore
If n is odd, then for some . Since and are comparable with and , we have
If
then
which is a contradiction. Therefore
and hence
From (2.3) and (2.5), we have
Since ψ is an altering distance function, we have is a bounded nonincreasing sequence. Thus there exists such that
On letting in (2.6), we have
Since , we have and hence . Thus
Next, we show that is a Cauchy sequence in the metric space . It is sufficient to show that is a Cauchy sequence in . Suppose to the contrary; that is, is not a Cauchy sequence in . Then there exists for which we can find two subsequences and of such that is the smallest index for which
This means that
From (2.8), (2.9) and the triangular inequality, we get that
On letting in the above inequalities and using (2.7), we have
Again, from (2.8) and the triangular inequality, we get that
Letting in the above inequalities and using (2.7) and (2.10), we get that
Since and are comparable with and , we have
Letting and using the continuity of ψ, we get that
Since , we have and hence , a contradiction. Thus is a Cauchy sequence in .
Step 2: Existence of a common fixed point.
Since is complete and is a Cauchy sequence in X, we have converges to some , that is, . Therefore
Since is a sequence in A, A is closed and , we have . Also, since is a sequence in B, B is closed and , we have . Now, we show that u is a fixed point of f and T. Without loss of generality, we may assume that f is continuous, since , we get . By the uniqueness of limit, we have . Now, we show that . Since with and , we have
Since , we get that and hence . □
Theorem 2.1 can be proved without assuming the continuity of f or the continuity of T. For this instance, we assume that X satisfies the following property:
-
(P)
If is a nondecreasing sequence in X with , then .
Now, we state and prove the following result.
Theorem 2.2 Let be an ordered complete metric space and A, B be nonempty closed subsets of X. Let be two mappings such that the pair is -weakly increasing. Assume the following:
-
(1)
The pair is a cyclic -contraction;
-
(2)
X satisfies property (P).
Then f and T have a common fixed point.
Proof We follow the proof of Theorem 2.1 step by step to construct a nondecreasing sequence in X with , and for some . Since , , A and B are closed subsets of X, we get . Using property (P), we get for all . Since and , we have
Letting in the above inequality, we get . Since , we get , hence . Similarly, we may show that . Thus u is a common fixed point of f and T. □
Taking (the identity function) in Theorem 2.1, we have the following result.
Corollary 2.1 Let be an ordered complete metric space and A, B be nonempty closed subsets of X. Let be two mappings such that the pair is -weakly increasing and has a cyclic representation with respect to the pair . Suppose that there exists such that for any two comparable elements with and , we have
If f or T is continuous, then f and T have a common fixed point.
The continuity of f or T in Corollary 2.1 can be dropped.
Corollary 2.2 Let be an ordered complete metric space and A, B be nonempty closed subsets of X. Let be two mappings such that the pair is -weakly increasing and such that has a cyclic representation with respect to the pair . Suppose that there exists such that for any two comparable elements with and , we have
If X satisfies property (P), then f and T have a common fixed point.
By taking in Theorem 2.1, we have the following result.
Corollary 2.3 Let be an ordered complete metric space and A, B be nonempty closed subsets of X with . Let be a mapping such that for all . Suppose that there exists such that for all and , we have
Assume the following:
-
(1)
f is a cyclic map;
-
(2)
f is continuous.
Then f has a fixed point.
The continuity of f in Corollary 2.3 can be dropped.
Corollary 2.4 Let be an ordered complete metric space and A, B be nonempty closed subsets of X with . Let be a mapping such that for all . Suppose that there exists such that for all and , we have
Assume the following:
-
(1)
f is a cyclic map;
-
(2)
X satisfies property (P).
Then f has a fixed point.
Taking in Theorem 2.1, we have the following result.
Corollary 2.5 Let be an ordered complete metric space. Let be two weakly increasing mappings. Suppose that there exists such that for any two comparable elements , we have
If f or T is continuous, then f and T have a common fixed point.
The continuity of f or T in Corollary 2.5 can be dropped.
Corollary 2.6 Let be an ordered complete metric space. Let be two weakly increasing mappings. suppose that there exists such that for any two comparable elements , we have
If X satisfies property (P), then f and T have a common fixed point.
To support the validity of our results, we introduce the following nontrivial example.
Example 2.1 On , consider
We introduce a relation on X by if and only if . Define by the formulae
and
Also, define by . Let and . Then
-
(1)
is a complete ordered metric space;
-
(2)
has a cyclic representation with respect to the pair ;
-
(3)
The pair is weakly -increasing;
-
(4)
X satisfies property (P);
-
(5)
For every two comparable elements with and , we have
Proof The proof of part (1) is clear. Since and , we conclude that has a cyclic representation with respect to the pair . To prove part (3), given . If , then . Thus and hence . If , then and . Thus and hence . Therefore for all . Similarly, we may show that for all . So, the pair is weakly -increasing. To prove part (4), let be a nondecreasing sequence such that . Then . So, for all n except for finitely many. Since is a nondecreasing with respect to ⪯, we have . Since for all but finitely many, then there exists such that for all . So, for all and hence for all . Thus X satisfies property (P). To prove part (5), given two comparable elements with and . We divide the proof into the following cases:
-
Case one: and . Here, we have and hence . Thus
-
Case two: and . Here and . Since and , then and for some .
If , then and hence . Thus , which is impossible.
If , then . Thus
If , then . Thus
-
Case three: and . Here and . Thus
-
Case four: and . Here and .
Note that f and T satisfy all the hypotheses of Theorem 2.1. Hence f and T have a fixed point. Here 0 is the fixed point of f and T. □
3 Applications
Denote by Λ the set of functions satisfying the following hypotheses:
-
(h1)
μ is a Lebesgue-integrable mapping on each compact of ;
-
(h2)
For every , we have
Theorem 3.1 Let be an ordered complete metric space and A, B be nonempty closed subsets of X. Let be two mappings such that the pair is -weakly increasing and has a cyclic representation w.r.t. the pair . Suppose that there exist and such that for any two comparable elements with and , we have
If f or T is continuous, then f and T have a common fixed point.
Proof Follows from Theorem 2.1 by defining via and noting that ψ is an altering distance function. □
The continuity of f or T in Theorem 3.1 can be dropped.
Theorem 3.2 Let be an ordered complete metric space and A, B be nonempty closed subsets of X. Let be two mappings such that is -weakly increasing and has a cyclic representation w.r.t. the pair . Suppose that there exist and such that for any two comparable elements with and , we have
If X satisfies property (P), then f and T have a common fixed point.
By taking in Theorems 3.1 and 3.2, we have the following results.
Corollary 3.1 Let be an ordered complete metric space. Let be two weakly increasing mappings. Suppose that there exist and such that for any two comparable elements , we have
If f or T is continuous, then f and T have a common fixed point.
Corollary 3.2 Let be an ordered complete metric space. Let be two weakly increasing mappings. Suppose that there exist and such that for any two comparable elements , we have
If X satisfies property (P), then f and T have a common fixed point.
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Shatanawi, W., Postolache, M. Common fixed point results for mappings under nonlinear contraction of cyclic form in ordered metric spaces. Fixed Point Theory Appl 2013, 60 (2013). https://doi.org/10.1186/1687-1812-2013-60
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DOI: https://doi.org/10.1186/1687-1812-2013-60