Abstract
Coupled fixed point results for nonlinear contraction mappings having a mixed monotone property in a partially ordered Gmetric space due to Choudhury and Maity are extended and unified. We also provide example to validate the main results in this article.
Mathematics Subject classification (2000): 47H10; 54H25.
Keywords:
coupled fixed points; mixed monotone property; Gmetric spaces; partially order set1. Introduction
One of the simplest and the most useful result in the fixed point theory is the BanachCaccioppoli contraction [1] mapping principle, a power tool in analysis. This principle has been generalized in different directions in different spaces by mathematicians over the years (see [210] and references mentioned therein). On the other hand, fixed point theory has received much attention in metric spaces endowed with a partial ordering. The first result in this direction was given by Ran and Reurings [11] and they presented applications of their results to matrix equations. Subsequently, Nieto and RodríguezLópez [12] extended the results in [11] for nondecreasing mappings and obtained a unique solution for a first order ordinary differential equation with periodic boundary conditions (see also, [1319]).
Bhaskar and Lakshmikantham [20] introduced the concept of a coupled fixed point and the mixed monotone property. Furthermore, they proved some coupled fixed point theorems for mappings which satisfy the mixed monotone property and gave some applications in the existence and uniqueness of a solution for a periodic boundary value problem. A number of articles in this topic have been dedicated to the improvement and generalization see in [2124] and reference therein.
Mustafa and Sims [25,26] introduced a new concept of generalized metric spaces, called Gmetric spaces. In such spaces every triplet of elements is assigned to a nonnegative real number. Based on the notion of Gmetric spaces, Mustafa et al. [27] established fixed point theorems in Gmetric spaces. Afterward, many fixed point results were proved in this space (see [2834]).
Recently, Choudhury and Maity [35] studied necessary conditions for existence of coupled fixed point in partially ordered Gmetric spaces. They obtained the following interesting result.
Theorem 1.1 ([35]). Let (X, ≼) be a partially ordered set such that X is a complete Gmetric space and F: X × X → X be a mapping having the mixed monotone property on X. Suppose there exists k ∈ [0,1) such that
for all x, y, z, u, v, w ∈ X for which x ≽ u ≽ w and y ≼ v ≼ z, where either u ≠ w or v ≠ z. If there exists x_{0}, y_{0 }∈ X such that
and either
(a) F is continuous or
(b) X has the following property:
(i) if a nondecreasing sequence {x_{n}} → x, then x_{n }≼ x for all n ∈ ℕ,
(ii) if a nonincreasing sequence {y_{n}} → y, then y_{n }≽ y for all n ∈ ℕ,
then F has a coupled fixed point.
The aim of this article is to extend and unify coupled fixed point results in [35] and to study necessary conditions to guarantee the uniqueness of coupled fixed point. We also provide illustrative example in support of our results.
2. Preliminaries
Throughout this article, (X, ≼) denotes a partially ordered set with the partial order ≼.
By x ≺ y, we mean x ≼ y but x ≠ y. If (X, ≼) is a partially ordered set. A mapping f: X → X is said to be nondecreasing (nonincreasing) if for all x, y ∈ X, x ≼ y implies f(x) ≼ f(y) (f(y) ≼ f(x), respectively).
Definition 2.1 ([20]). Let (X, ≼) be a partial ordered set. A mapping F: X × X → X is said to has the a mixed monotone property if F is monotone nondecreasing in its first argument and is monotone nonincreasing in its second argument, that is, for any x, y ∈ X
and
Definition 2.2 ([20]). An element (x, y) ∈ X × X is called a coupled fixed point of mapping F: X × X → X if
Consistent with Mustafa and Sims [25,26], the following definitions and results will be needed in the sequel.
Definition 2.3 ([26]). Let X be a nonempty set. Suppose that a mapping G: X × X × X → ℝ_{+ }satisfies:
(G_{1}) G(x,y,z) = 0 if x = y = z;
(G_{2}) G(x, x, y) > 0 for all x, y ∈ X with x ≠ y;
(G_{3}) G(x,x,y) ≤ G(x,y,z) for all x,y,z ∈ X with z ≠ y;
(G_{4}) G(x,y,z) = G(x,z,y) = G(y,z,x) = ..., (symmetry in all three variables);
(G_{5}) G(x,y,z) ≤ G(x,a,a) + G(a,y, z) for all x,y,z,a ∈ X (rectangle inequality).
Then G is called a Gmetric on X and (X, G) is called a Gmetric space.
Definition 2.4 ([26]). Let X be a Gmetric space and let {x_{n}} be a sequence of points of X, a point x ∈ X is said to be the limit of a sequence {x_{n}} if G(x,x_{n},x_{m}) → 0 as n, m → ∞ and sequence {x_{n}} is said to be Gconvergent to x.
From this definition, we obtain that if x_{n }→ x in a Gmetric space X, then for any ϵ > 0 there exists a positive integer N such that G(x,x_{n},x_{m}) < ϵ, for all n,m ≥ N.
It has been shown in [26] that the Gmetric induces a Hausdorff topology and the convergence described in the above definition is relative to this topology. So, a sequence can converge at the most to one point.
Definition 2.5 ([26]). Let X be a Gmetric space, a sequence {x_{n}} is called GCauchy if for every ϵ > 0 there is a positive integer N such that G(x_{n},x_{m},x_{l}) < ϵ for all n, m, l ≥ N, that is, if G(x_{n}, x_{m}, x_{l}) → 0, as n, m, l → ∞.
We next state the following lemmas.
Lemma 2.6 ([26]). If X is a Gmetric space, then the following are equivalent:
(1) {x_{n}} is Gconvergent to x.
(2) G(x_{n},x_{n},x) → 0 as n → ∞.
(3) G(x_{n},x,x) → 0 as n → ∞.
(4) G(x_{m},x_{n},x) → 0 as n,m → ∞.
Lemma 2.7 ([26]). If X is a Gmetric space, then the following are equivalent:
(a) The sequence {x_{n}} is GCauchy.
(b) For every ϵ > 0, there exists a positive integer N such that G(x_{n},x_{m},x_{m}) < ϵ, for all n,m ≥ N.
Lemma 2.8 ([26]). If X is a Gmetric space then G(x,y,y) ≤ 2G(y,x,x) for all x,y ∈ X.
Definition 2.9 ([26]). Let (X, G), (X', G') be two generalized metric spaces. A mapping f: X → X' is Gcontinuous at a point x ∈ X if and only if it is G sequentially continuous at x, that is, whenever {x_{n}} is Gconvergent to x, {f(x_{n})} is G'convergent to f(x).
Definition 2.10 ([26]). A Gmetric space X is called a symmetric Gmetric space if
for all x,y ∈ X.
Definition 2.11 ([26]). A Gmetric space X is said to be Gcomplete (or a complete Gmetric space) if every GCauchy sequence in X is convergent in X.
Definition 2.12 ([26]). Let X be a Gmetric space. A mapping F: X × X → X is said to be continuous if for any two Gconvergent sequences {x_{n}} and {y_{n}} converging to x and y, respectively, {F(x_{n},y_{n})} is Gconvergent to F(x,y).
3. Coupled fixed point in Gmetric spaces
Let Θ denotes the class of all functions θ: [0, ∞) × [0, ∞) → [0,1) which satisfies following condition:
For any two sequences {t_{n}} and {s_{n}} of nonnegative real numbers,
Following are examples of some function in Θ.
(1) θ_{1}(s,t) = k for s,t ∈ [0,∞), where k ∈ [0,1).
where k, l ∈ (0,1)
Now, we prove our main result.
Theorem 3.1. Let (X, ≼) be a partially ordered set such that there exists a complete Gmetric on X and F: X × X → X be a continuous mapping having the mixed monotone property. Suppose that there exists θ ∈ Θ such that
for all x, y, z, u, v, w ∈ X for which x ≽ u ≽ w and y ≼ v ≼ z where either u ≠ w or v ≠ z. If there exists x_{0}, y_{0 }∈ X such that
then F has a coupled fixed point.
Proof. As F(X × X) ⊆ X, we can construct sequences {x_{n}} and {y_{n}} in X such that
Next, we show that
Since x_{0 }≼ F(x_{0},y_{0}) = x_{1 }and y_{0 }≽ F(y_{0},x_{0}) = y_{1}, therefore (3.3) holds for n = 0.
Suppose that (3.3) holds for some fixed n ≥ 0, that is,
Since F has a mixed monotone property, from (3.4) and (2.1), we have
for all x, y ∈ X and from (3.4) and (2.2), we have
for all x,y ∈ X. If we take y = y_{n }and x = x_{n }in (3.5), then we obtain
If we take y = y_{n+1 }and x = x_{n+1 }in (3.6), then
Now, from (3.7) and (3.8), we have
Therefore, by the mathematical induction, we conclude that (3.3) holds for all n ≥ 0, that is,
and
If there exists some integer k ≥ 0 such that
then G(x_{k+1},x_{k+1},x_{k}) = G(y_{k+1},y_{k+1},y_{k}) = 0 implies that x_{k }= x_{k+1 }and y_{k }= y_{k+1}. Therefore, x_{k }= F(x_{k},y_{k}) and y_{k }= F(y_{k},x_{k}) gives that (x_{k},y_{k}) is a coupled fixed point of F.
Now, we assume that G(x_{n+1},x_{n+1},x_{n}) + G(y_{n+1},y_{n+1},y_{n}) ≠ 0 for all n ≥ 0. Since x_{n }≼ x_{n+1 }and y_{n }≽ y_{n+1 }for all n ≥ 0 so from (3.1) and (3.2), we have
which implies that
Thus the sequence {G_{n+1 }:= G(x_{n+1}, x_{n+1}, x_{n}) + G(y_{n+1}, y_{n+1}, y_{n})} is monotone decreasing. It follows that G_{n }→ g as n → ∞ for some g ≥ 0. Next, we claim that g = 0. Assume on contrary that g > 0, then from (3.12), we obtain
On taking limit as n → ∞, we obtain
By property of function θ, we have G(x_{n}, x_{n}, x_{n1}) → 0, G(y_{n}, y_{n}, y_{n1}) → 0 as n → ∞ and we have
a contradiction. Therefore,
Similarly, we can prove that
Next, we show that {x_{n}} and {y_{n}} are Cauchy sequences. On contrary, assume that at least one of {x_{n}} or {y_{n}} is not a Cauchy sequence. By Lemma 2.7, there is an ϵ > 0 for which we can find subsequences {x_{n(k)}}, {x_{m(k)}} of {x_{n}} and {y_{n(k)}}, {y_{m(k)}} of {y_{n}} with m(k) > n(k) ≥ k such that
and
Using (3.16), (3.17) and the rectangle inequality, we have
On taking limit as k → ∞, we have
By the rectangle inequality, we get
Therefore, we have
This further implies that
On taking limit as k → ∞ and using (3.14), (3.15) and (3.18), we obtain
Since θ ∈ Θ, we have G(x_{n(k)}, x_{m(k)}, x_{m(k)}) → 0 and G(y_{n(k)}, y_{m(k)}, y_{m(k)}) → 0, that is
a contradiction. Therefore, {x_{n}} and {y_{n}} are GCauchy sequence. By Gcompleteness of X, there exists x,y ∈ X such that {x_{n}} and {y_{n}} Gconverges to x and y, respectively. Now, we show that F has a coupled fixed point. Since F is a continuous, taking n → ∞ in (3.2), we get
and
Therefore, x = F(x, y) and y = F(y, x), that is, F has a coupled fixed point.
Theorem 3.2. Let (X, ≼) be a partially ordered set such that there exists a complete Gmetric on X and F: X × X → X be a mapping having the mixed monotone property. Suppose that there exists θ ∈ Θ such that
for all x, y, z, u, v, w ∈ X for which x ≽ u ≽ w and y ≼ v ≼ z where either u ≠ w or v ≠ z. If there exists x_{0},y_{0 }∈ X such that
and X has the following property:
(i) if a nondecreasing sequence {x_{n}} → x, then x_{n }≼ x for all n ∈ ℕ,
(ii) if a nonincreasing sequence {y_{n}} → y, then y_{n }≽ y for all n ∈ ℕ,
then F has a coupled fixed point.
Proof. Following arguments similar to those given in Theorem 3.1, we obtain a nondecreasing sequence {x_{n}} converges to x and a nonincreasing sequence {y_{n}} converges to y for some x,y ∈ X. By using (i) and (ii), we have x_{n }≼ x and y_{n }≽ y for all n.
If x_{n }= x and y_{n }= y for some n ≥ 0, then, by construction, x_{n+1 }= x and y_{n+1 }= y. Thus (x, y) is a coupled fixed point of F. So we may assume either x_{n }≠ x or y_{n }≠ y, for all n ≥ 0. Then by the rectangle inequality, we obtain
On taking limit as n → ∞, we have G(F(x,y),x,x) + G(F(y,x),y,y) = 0. Thus x = F(x,y) and y = F(x, y) and so (x, y) is a coupled fixed point of F.
Corollary 3.3. Let (X, ≼) be a partially ordered set such that there exists a complete Gmetric on X and F: X × X → X be a mapping having the mixed monotone property. Suppose that there exists η ∈ Θ such that
for all x, y, z, u, v, w ∈ X for which x ≽ u ≽ w and y ≼ v ≼ z, where either u ≠ w or v ≠ z. If there exists x_{0},y_{0 }∈ X such that
and either
(a) F is continuous or
(b) X has the following property:
(i) if a nondecreasing sequence {x_{n}} → x, then x_{n }≼ x for all n ∈ ℕ,
(ii) if a nonincreasing sequence {y_{n}} → y, then y_{n }≽ y for all n ∈ ℕ,
then F has a coupled fixed point.
Proof. For x,y,z,u,v,w ∈ X with x ≽ u ≽ w and y ≼ v ≼ z, where either u ≠ w or v ≠ z, from (3.20), we have
and
From (3.21) and (3.22), we have
for x,y,z,u,v,w ∈ X with x ≽ u ≽ w and y ≼ v ≼ z where either u ≠ w or v ≠ z, where
for all t_{1},t_{2 }∈ [0, ∞). It is easy to verify that θ ∈ Θ and we can apply Theorems 3.1 and 3.2. Hence F has a coupled fixed point.
Corollary 3.4. [[35], Theorems 3.1 and 3.2] Let (X,≼) be a partially ordered set such that there exists a complete Gmetric on X and F: X × X → X be a mapping having the mixed monotone property. Suppose that there exists a k ∈ [0,1) such that
for all x, y, z, u, v, w ∈ X for which x ≽ u ≽ w and y ≼ v ≼ z, where either u ≠ w or v ≠ z. If there exists x_{0}, y_{0 }∈ X such that
and either
(a) F is continuous or
(b) X has the following property:
(i) if a nondecreasing sequence {x_{n}} → x, then x_{n }≼ x for all n ∈ ℕ,
(ii) if a nonincreasing sequence {y_{n}} → y, then y_{n }≽ y for all n ∈ ℕ,
then F has a coupled fixed point.
Proof. Taking η(t_{1}, t_{2}) = k with k ∈ [0,1) for all t_{1}, t_{2 }∈ [0, ∞) in Theorems 3.1 and 3.2, result follows immediately.
Let Ω denotes the class of those functions ω: [0, ∞) → [0,1) which satisfies the condition: For any sequences {t_{n}} of nonnegative real numbers, ω(t_{n}) → 1 implies t_{n }→ 0.
Theorem 3.5. Let (X, ≼) be a partially ordered set such that there exists a complete Gmetric on X and F: X × X → X be a mapping having the mixed monotone property. Suppose that there exists ω ∈ Ω such that
for all x, y, z, u, v, w ∈ X for which x ≽ u ≽ w and y ≼ v ≼ z where either u ≠ w or v ≠ z. If there exists x_{0},y_{0 }∈ X such that
and either
(a) F is continuous or
(b) X has the following property:
(i) if a nondecreasing sequence {x_{n}} → x, then x_{n }≼ x for all n ∈ ℕ,
(ii) if a nonincreasing sequence {y_{n}} → y, then y_{n }≽ y for all n ∈ ℕ,
then F has a coupled fixed point.
Proof. Taking θ(t_{1},t_{2}) = ω(t_{1 }+ t_{2}) for all t_{1},t_{2 }∈ [0,∞) in Theorems 3.1 and 3.2, result follows.
Taking ω(t) = k with k ∈ [0,1) for all t ∈ [0, ∞) in Theorem 3.5, we obtain the following corollary.
Corollary 3.6. Let (X, ≼) be a partially ordered set such that there exists a complete Gmetric on X and F: X × X → X be a mapping having the mixed monotone property. Suppose that there exists ∈ [0,1) such that
for all x, y, z, u, v, w ∈ X for which x ≽ u ≽ w and y ≼ v ≼ z where either u ≠ w or v ≠ z. If there exists x_{0},y_{0 }∈ X such that
and either
(a) F is continuous or
(b) X has the following property:
(i) if a nondecreasing sequence {x_{n}} → x, then x_{n }≼ x for all n ∈ ℕ,
(ii) if a nonincreasing sequence {y_{n}} → y, then y_{n }≽ y for all n ∈ ℕ,
then F has a coupled fixed point.
Theorem 3.7. Let (X, ≼) be a partially ordered set such that there exists a complete Gmetric on X and F: X × X → X be a mapping having the mixed monotone property and F(x,y) ≼ F(y,x), whenever x ≼ y. Suppose that there exists θ ∈ Θ such that
for all x, y, z, u, v, w ∈ X for which w ≼ u ≼ x≼ y ≼ v ≼ z, where either u ≠ w or v ≠ z. If there exists x_{0},y_{0 }∈ X such that
and either
(a) F is continuous or
(b) X has the following property:
(i) if a nondecreasing sequence {x_{n}} → x, then x_{n }≼ x for all n ∈ ℕ,
(ii) if a nonincreasing sequence {y_{n}} → y, then y_{n }≽ y for all n ∈ ℕ,
then F has a coupled fixed point.
Proof. By given hypothesis, there exist x_{0},y_{0 }∈ X such that
We define x_{1},y_{1 }∈ X by
Since x_{0 }≼ y_{0}, by given assumptions, we have F(x_{0},y_{0}) ≼ F(y_{0},x_{0}). Hence
Continuing the above process, we have two sequences {x_{n}} and {y_{n}} such that
and
for all n ≥ 0. If there is k ∈ ℕ such that x_{k }= y_{k }= α (say), then we have
that is, α = F(α, α). Therefore, (α, α) is a coupled fixed point of F. Next, assume that
for all n ∈ ℕ. Further, using similar arguments as stated in Theorem 3.1, we may assume that (x_{n},y_{n}) ≠ (x_{n+1},y_{n+1}). Then, in view of (3.27), for all n ≥ 0, the inequality (3.26) holds with
The rest of the proof follows by following the same steps as given in Theorem 3.1 for case (a). For case (b), we follow the same steps as given in Theorem 3.2.
Example 3.8. Let X = ℕ ∪ {0} and G: X × X × X → X be define by
Then X is a complete Gmetric space. Let partial order ≼ on X be defined as follows: For x,y ∈ X,
Let F: X × X → X be defined by
If w ≼ u ≼ x ≼ y ≼ v ≼ z holds, then we have w ≥ u ≥ x > y ≥ v ≥ z. Therefore F(x,y) = F(u,v) = F(w,z) = 1 and F(y,x) = F(v,u) = F(z,w) = 0. So the left side of (3.26) becomes
and (3.26) is satisfied for all θ ∈ Θ. Thus Theorem 3.7 is applicable to this example with x_{0 }= 0 and y_{0 }= 81. Moreover, F has coupled fixed points (0,0) and (1,0).
Remark 3.9. A Gmetric naturally induces a metric d_{G }given by d_{G}(x,y) = G(x,y,y) + G(x,x,y) [25]. From the condition that either u ≠ w or v ≠ z, the inequality (3.1), (3.19), (3.25) and (3.26) do not reduce to any metric inequality with the metric d_{G}. Therefore, the corresponding metric space (X, d_{G}) results are not applicable to Example 3.8.
Remark 3.10. Example 3.8 is not supported by Theorems 3.1, 3.2 and 3.5. This is evident by the fact that the inequality (3.1), (3.19) and (3.25) are not satisfied when w = u = x = y = 3, v = 0 and z = 1. Moreover, the coupled fixed point is not unique.
4. Uniqueness of coupled fixed point in Gmetric spaces
In this section, we study necessary conditions to obtain the uniqueness of a coupled fixed point in the setting of partially ordered Gmetric spaces. If (X, ≼) is a partially ordered set, then we endow the product of X × X with the following partial order:
For (x, y), (u, v) ∈ X × X, (x, y) ⊴ (u, v) if and only if x ≼ u and y ≽ v.
Theorem 4.1. In addition to the hypotheses in Theorem 3.1, suppose that for every (x, y), (z, t) ∈ X × X, there exists a point (u,v) ∈ X × X that is comparable to (x,y) and (z,t). Then F has a unique coupled fixed point.
Proof. From Theorem 3.1, F has a coupled fixed points. Suppose (x, y) and (z, t) are coupled fixed points of F, that is, x = F(x,y),y = F(y,x),z = F(z,t) and t = F(t,z). Next, we claim that x = z and y = t. By given hypothesis, there exists (u,v) ∈ X × X that is comparable to (x,y) and (z,t). We put u_{0 }= u and v_{0 }= v and construct sequences {u_{n}} and {v_{n}} by
Since (u,v) is comparable with (x,y), we assume that (u_{0},v_{0}) = (u,v) ⊴ (x,y). Using the mathematical induction, it is straight forward to prove that
From (3.1), we have
Consequently, sequence {G(x,x,u_{n}) + G(y,y,v_{n})} is nonnegative and decreasing, so
for some g ≥ 0. We claim that g = 0. Indeed, if g > 0 then following similar arguments to those given in the proof of Theorem 3.1, we conclude that
Since θ ∈ Θ, we obtain G(x,x,u_{n1}) → 0 and G(v_{n1},y,y) → 0. Therefore,
which is a contradiction. Hence
Similarly, one can prove that
and
From rectangular inequality, we have
and
Combine (4.6) and (4.7), we have
Taking n → ∞, by (4.2), (4.3), (4.4) and (4.5), we have G(z,x,x) + G(y,t,t) ≤ 0. So G(z,x,x) = 0 and G(y,t,t) = 0, that is, z = x and y = t. Therefore, F has a unique coupled fixed point. This completes the proof.
Theorem 4.2. In addition to the hypotheses in Theorem 3.2, suppose that for every (x, y), (z, t) ∈ X × X, there exists a point (u,v) ∈ X × X that is comparable to (x,y) and (z,t). Then F has a unique coupled fixed point.
Proof. Proof is similar to the one given in Theorems 4.1 and 3.2.
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
All authors read and approved the final manuscript.
5. Acknowledgements
This study was supported by the Higher Education Research Promotion and National Research University Project of Thailand, Office of the Higher Education Commission under the Computational Science and Engineering Research Cluster (CSEC Grant No. 54000267).
W. Sintunavarat would like to thank the Research Professional Development Project under the Science Achievement Scholarship of Thailand (SAST). The authors thank the referee for comments and suggestions on this manuscript.
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