Abstract
We establish new coincidence point theorems for nonlinear contraction in ordered metric spaces. Also, we introduce an example to support our results. Some applications of our obtained results are given.
MSC: 54H25; 47H10; 54E50; 34B15.
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1. Introduction and Preliminaries
Generalization of the Banach principle [1] has been heavily investigated by many authors (see [2–14]). In particular, there has been a number of fixed point theorems involving altering distance functions. Such functions were introduced by Khan et al. [15].
Definition 1.1. [15]The function ϕ : [0, +∞) → [0, +∞) is called an altering distance function if the following properties are satisfied:
-
(1)
ϕ is continuous and nondecreasing.
-
(2)
ϕ(t) = 0 if and only if t = 0.
Khan et al. [15] proved the following theorem.
Theorem 1.1. Let (X, d) be a complete metric space, ψ an altering distance function and T : X → X satisfying
for x, y ∈ X and 0 < c < 1. Then, T has a unique fixed point.
Existence of fixed point in partially ordered sets has been considered by many authors. Ran and Reurings [14] studied a fixed point theorem in partially ordered sets and applied their result to matrix equations. While Nieto and Rodŕiguez-López [9] studied some contractive mapping theorems in partially ordered set and applied their main theorems to obtain a unique solution for a first order ordinary differential equation. For more works in partially ordered metric spaces, we refer the reader to [16–31].
Harjani and Sadarangani [7, 8] obtained some fixed point theorems in a complete ordered metric space using altering distance functions. They proved the following theorems.
Theorem 1.2. [8]Let (X, ≼) be a partially ordered set and suppose that there exists a metric d in X such that (X, d) is a complete metric space. Let f : X → X be a continuous and nondecreasing mapping such that
for comparable x, y ∈ X, where ψ and ϕ are altering distance functions. If there exists x0 ≼ f (x0), then f has a fixed point.
Theorem 1.3. [8]Let (X, ≼) be a partially ordered set and suppose that there exists a metric d in X such that (X, d) is a complete metric space. Assume that X satisfies if (x n ) is a nondecreasing sequence in X such that x n → x, then x n ≼ x for all n ∈ ℕ. Let f : X → X be a nondecreasing mapping such that
for comparable x, y ∈ X, where ψ and ϕ are altering distance functions. If there exists x0 ≼ f (x0), then f has a fixed point.
Altun and Simsek [3] introduced the concept of weakly increasing mappings as follows:
Definition 1.2. [3]Let (X, ≼) be a partially ordered set. Two mappings f, g : X → X are said to be weakly increasing if fx ≼ g(fx) and gx ≼ f(gx) for all x ∈ X.
Recently, Turkoglu [32] studied new common fixed point theorems for weakly compatible mappings on uniform spaces. While, Nashine and Samet [12] proved some new coincidence point theorems for a pair of weakly increasing mappings. Very recently, Shatanawi and Samet [33] proved some coincidence point theorems for a pair of weakly increasing mappings with respect to another map.
The aim of this article is to study new coincidence point theorems for a pair of weakly decreasing mappings satisfying (ψ, ϕ)-weakly contractive condition in an ordered metric space (X, d), where ϕ and ψ are altering distance functions.
2. Main Results
We start our study with the following definition:
Definition 2.1. Let (X, ≼) be a partially ordered set and T, f : X → X be two mappings. We say that f is weakly decreasing with respect to T if the following conditions hold:
-
(1)
fX ⊆ TX.
-
(2)
For all x ∈ X, we have fy ≼ fx for all y ∈ T -1(fx).
We need the following definition in our arguments.
Definition 2.2. [34]Let (X, d) be a metric space and f, g : X → X. If w = fx = gx for some x ∈ X, then x is called a coincidence point of f and g, and w is called a point of coincidence of f and g. The pair {f, g} is said to be compatible if and only if
whenever (x n ) is a sequence in X such that
for some t ∈ X.
Theorem 2.1. Let (X, ≼) be a partially ordered set and suppose that there exists a metric d on X such that (X, d) is a complete metric space. Let T, f : X → X be two maps such that for all x, y ∈ X with Tx and Ty are comparable, we have
where ϕ and ψ are altering distance functions. Assume that T and f satisfy the following hypotheses:
-
(i)
f is weakly decreasing with respect to T.
-
(ii)
The pair {T, f} is compatible.
-
(iii)
f and T are continuous.
Then, T and f have a coincidence point.
Proof. Let x0 ∈ X. Since fX ⊆ TX, we choose x1 ∈ X such that fx0 = Tx1. Also, since fX ⊆ TX, we choose x2 ∈ X such that fx1 = Tx2. Continuing this process, we can construct a sequences (x n ) in X such that Txn+1= fx n . Now, since x1 ∈ T-1(fx0) and x2 ∈ T-1(fx1), by using the assumption that f is weakly decreasing with respect to T, we obtain
By induction on n, we conclude that
Hence,
If for some n0 ∈ X, then . Thus, is a coincidence point of T and f. Hence, we may assume that Txn+1≠ Tx n for all n ∈ ℕ.
Since Tx n and Txn+1are comparable, then by (1), we have
If
then
So, ϕ(d(Txn+1, Txn+2)) = 0 and hence d(Txn+1, Txn+2) = 0, a contradiction.
Thus,
Therefore, we have
Since ψ is a nondecreasing function, we get that {d(Txn+1, Tx n ): n ∈ ℕ} is a nonincreasing sequence. Hence, there is r ≥ 0 such that
Letting n → +∞ in (2) and using the continuity of ψ and ϕ, we get that
Thus, ϕ(r) = 0 and hence r = 0. Therefore,
Now, we prove that (Tx n ) is a Cauchy sequence in X. Suppose to the contrary; that is, (Tx n ) is not a Cauchy sequence. Then, there exists ε > 0 for which we can find two subsequences of positive integers (Txm(i)) and (Txn(i)) such that n(i) is the smallest index for which
This means that
From (4), (5) and the triangular inequality, we have
On letting i → +∞ in above inequality and using (3), we have
Also,
Letting i → +∞ in the above inequalities and using (3), we get that
Since Txn(i)-1and Txm(i)are comparable, by (1), we have
Letting i → +∞ in the above inequalities, and using (3), (6) and (7), we get that
Therefore, ϕ(ε) = 0 and hence ε = 0, a contradiction. Thus, {Tx n } is a Cauchy sequence in the complete metric space X. Therefore, there exists u ∈ X such that
By the continuity of T, we have
Since Txn+1= fx n → u, Tx n → u, and the pair {T, f} is compatible, we have
By the triangular inequality, we have
Letting n → +∞ and using the fact that T and f are continuous, we get that d(fu, Tu) = 0. Hence, fu = Tu, that is, u is a coincidence point of T and f.
Theorem 2.2. Let (X, ≼) be a partially ordered set and suppose that there exists a metric d on X. Let T, f : X → X be two maps such that for all x, y ∈ X with Tx and Ty are comparable, we have
where ϕ and ψ are altering distance functions. Suppose that the following hypotheses are satisfied:
-
(i)
If (x n ) is a nonincreasing sequence in X with respect to ≼ such that x n → x ∈ X as n → +∞, then x n ≽ x for all n ∈ ℕ.
-
(ii)
f is weakly decreasing with respect to T.
-
(iii)
TX is a complete subspace of X.
Then, T and f have a coincidence point.
Proof. Following the proof of Theorem 2.1, we have (Tx n ) is a Cauchy sequence in (TX, d). Since TX is complete, there is v ∈ X such that
Since {Tx n } is a nonincreasing sequence in X. By hypotheses, we have Tx n ≽ Tv for all n ∈ ℕ. Thus, by (8), we have
Letting n → +∞ in the above inequalities, we get that
Hence, ϕ(d(Tv, fv)) = 0. Since ϕ is an altering distance function, we get that d(Tv, fv) = 0. Therefore, Tv = fv. Thus, v is a coincidence point of T and f.
By taking ψ(t) = t and ϕ(t) = (1 - k)t, k ∈ [0, 1) in Theorems 2.1 and 2.2, we have the following two results.
Corollary 2.1. Let (X, ≼) be a partially ordered set and suppose that there exists a metric d on X such that (X, d) is a complete metric space. Let T, f : X → X be two maps such that for all x, y ∈ X with Tx and Ty are comparable, we have
Assume that T and f satisfy the following hypotheses:
-
(i)
f is weakly decreasing with respect to T.
-
(ii)
The pair {T, f} is compatible.
-
(iii)
f and T are continuous.
If k ∈ [0, 1), then T and f have a coincidence point.
Corollary 2.2. Let (X, ≼) be a partially ordered set and suppose that there exists a metric d on X. Let T, f : X → X be two maps such that for all x, y ∈ X with Tx and Ty are comparable, we have
Suppose that the following hypotheses are satisfied:
-
(i)
If (x n ) is a nonincreasing sequence in X with respect to ≼ such that x n → x ∈ X as n → +∞, then x n ≽ x for all n ∈ ℕ.
-
(ii)
f is weakly decreasing with respect to T.
-
(iii)
TX is a complete subspace of X.
If k ∈ [0, 1), then T and f have a coincidence point.
Corollary 2.3. Let (X, ≼) be a partially ordered set and suppose that there exists a metric d on X such that (X, d) is a complete metric space. Let T, f : X → X be two maps such that for all x, y ∈ X with Tx and Ty are comparable, we have
Assume that T and f satisfy the following hypotheses:
-
(i)
f is weakly decreasing with respect to T.
-
(ii)
The pair {T, f} is compatible.
-
(iii)
f and T are continuous.
If a1 + a2 + a3 + a4 ∈ [0, 1), then T and f have a coincidence point.
Proof. Follows from Corollary 2.1 by noting that
□
Corollary 2.4. Let (X, ≼) be a partially ordered set and suppose that there exists a metric d on X such that (X, d) is a complete metric space. Let f : X → X be a map such that for all comparable x, y ∈ X, we have
where ϕ and ψ are altering distance functions. Assume that f satisfies the following hypotheses:
-
(i)
f(fx) ≼ fx for all x ∈ X.
-
(ii)
f is continuous.
Then, f has a fixed point.
Proof. Follows from Theorem 2.1 by taking T = i X (the identity map).
Corollary 2.5. Let (X, ≼) be a partially ordered set and suppose that there exists a metric d on X such that (X, d) is complete. Let f : X → X be a map such that for all comparable x, y ∈ X, we have
where ϕ and ψ are altering distance functions. Suppose that the following hypotheses are satisfied:
-
(i)
If (x n ) is a nonincreasing sequence in X with respect to ≼ such that x n → x ∈ X as n → +∞, then x n ≽ x for all n ∈ ℕ.
-
(ii)
f(fx) ≼ fx for all x ∈ X.
Then, f has a fixed point.
Proof. Follows from Theorem 2.2 by taking T = i X (the identity map).
By taking ψ(t) = t and ϕ(t) = (1 - k)t, k ∈ [0, 1) in Corollaries 2.4 and 2.5, we have the following results.
Corollary 2.6. Let (X, ≼) be a partially ordered set and suppose that there exists a metric d on X such that (X, d) is a complete metric space. Let f : X → X be a map such that for all comparable x, y ∈ X, we have
Assume f satisfies the following hypotheses:
-
(i)
f(fx) ≼ fx for all x ∈ X.
-
(ii)
f is continuous.
If k ∈ [0, 1), then f has a fixed point.
Corollary 2.7. Let (X, ≼) be a partially ordered set and suppose that there exists a metric d on X such that (X, d) is complete. Let f : X → X be a map such that for all comparable x, y ∈ X, we have
Suppose that the following hypotheses are satisfied:
-
(i)
If (x n ) is a nonincreasing sequence in X with respect to ≼ such that x n → x ∈ X as n → +∞, then x n ≽ x for all n ∈ ℕ.
-
(ii)
f(fx) ≼ fx for all x ∈ X.
If k ∈ [0, 1), then f has a fixed point.
Now, we introduce an example to support our results.
Example 2.1. Let X = [0, +∞). Define d : X × X → ℝ by d(x, y) = |x - y|. Define f, T : X → X by
and
Then,
-
(1)
fX ⊆ TX.
-
(2)
f and T are continuous.
-
(3)
The pair {f, T} is compatible.
-
(4)
f is weakly decreasing with respect to T.
-
(5)
For all x, y ∈ X, we have
Proof. The proof of (1) and (2) is clear.
To prove (3), let (x n ) be any sequence in X such that
for some t ∈ X. Since , we have . Since Tx n → t as n → +∞, we have (x n ) has at most only finitely many elements greater than 1. Thus, and for all n ∈ ℕ except at most for finitely many elements. Thus, we have and as n → +∞. By uniqueness of limit, we get that and hence t = 0. Thus, x n → 0 as n → +∞. Since f and T are continuous, we have fx n → f 0 = 0 and Tx n → T 0 = 0 as n → +∞. Therefore,
Thus, the pair {f, T} is compatible.
To prove f is weakly decreasing with respect to T, let x, y ∈ X be such that y ∈ T-1(fx). If x ∈ [0, 1], then
In this case, we must have Ty = y2. Thus, . Hence, . Therefore,
If x > 1, then . Thus, . In this case, we have Ty = y2. Thus,
So,
Therefore,
Therefore, f is weakly decreasing with respect to T.
To prove (5), let x, y ∈ X.
Case 1: If x, y ∈ [0, 1], then
Case 2: If x, y ∈ (1, +∞), then
Case 3: (x ∈ [0, 1] and y ∈ (1, +∞)) or (y ∈ [0, 1] and x ∈ (1, +∞)).
Without loss of generality, we may assume that x ∈ [0, 1] and y ∈ (1, +∞). Then,
If
then
If
then
Thus, f and T satisfy all the hypotheses of Corollary 2.1. Therefore, T and f have a coincidence point. Here (0, 0) is the coincidence point of f and T.
3. Applications
Denote by Λ the set of functions λ : [0, +∞) → [0, +∞) satisfying the following hypotheses:
-
(1)
λ is a Lebesgue-integrable mapping on each compact of [0, +∞).
-
(2)
For every ε > 0, we have .
It is an easy matter to see that the mapping ψ : [0, +∞) → [0, +∞) defined by
is an altering distance function. Now, we have the following results:
Theorem 3.1. Let (X, ≼) be a partially ordered set and suppose that there exists a metric d on X such that (X, d) is a complete metric space. Let T, f : X → X be two maps such that for all x, y ∈ X with Tx and Ty are comparable, we have
where λ, μ ∈ Λ. Assume that T and f satisfy the following hypotheses:
-
(1)
f is weakly decreasing with respect to T.
-
(2)
The pair {T, f} is compatible.
-
(3)
f and T are continuous.
Then, T and f have a coincidence point.
Proof. Follows from Theorem 2.1 by taking and . □
Theorem 3.2. Let (X, ≼) be a partially ordered set and suppose that there exists a metric d on X. Let T, f : X → X be two maps such that for all x, y ∈ X with Tx and Ty are comparable, we have
where λ, μ ∈ Λ. Suppose that the following hypotheses are satisfied:
-
(1)
If (x n ) is a nonincreasing sequence in X with respect to ≼ such that x n → x ∈ X as n → +∞, then x n ≽ x for all n ∈ ℕ.
-
(2)
f is weakly decreasing with respect to T.
-
(3)
TX is a complete subspace of X.
Then, T and f have a coincidence point.
Proof. Follows from Theorem 2.2 by taking and . □
Now, our aim is to give an existence theorem for a solution of the following integral equation:
where T > 0. Let X = C([0, T]) be the set of all continuous functions defined on [0, T]. Define
by
Then, (X, d) is a complete metric space. Define an ordered relation ≤ on X by
Then, (X, ≤) is a partially ordered set. Now, we prove the following result.
Theorem 3.3. Suppose the following hypotheses hold:
-
(1)
K : [0, T] × [0, T] × ℝ+ → ℝ+ and g : ℝ → ℝ are continuous.
-
(2)
For each t, s ∈ [0, T], we have
-
(3)
There exist a continuous function G : [0, T] × [0, T] → [0, +∞] such that
for each comparable u, v ∈ ℝ and each t, s ∈ [0, T].
-
(4)
for some r < 1.
Then, the integral equation (9) has a solution u ∈ C([0, T]).
Proof. Define f : C([0, T]) → C([0, T]) by
Now, we have
Thus, we have f(fx) ≤ fx for all x ∈ C([0, T]).
For x, y ∈ C([0, T]) with x ≼ y, we have
Moreover, if (f n ) is a nonincreasing sequence in C([0, T]) such that f n → f as n → +∞, then f n ≥ f for all n ∈ ℕ (see [9]). Thus, all the required hypotheses of Corollary 2.7 are satisfied. Thus, there exist a solution u ∈ C([0, T]) of the integral equation (9).
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Acknowledgements
The authors thank the editor and the referees for their useful comments and suggestions. Special thank goes to the Referee #3 for his suggestion to formulate Definition 2.1 in more suitable and validity form.
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Shatanawi, W., Mustafa, Z. & Tahat, N. Some Coincidence Point Theorems for Nonlinear Contraction in Ordered Metric Spaces. Fixed Point Theory Appl 2011, 68 (2011). https://doi.org/10.1186/1687-1812-2011-68
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DOI: https://doi.org/10.1186/1687-1812-2011-68