Bapurao C. Dhage
Kasubai, Gurukul Colony, Ahmedpur-413 515, Dist: Latur, Maharashtra, India
E-mail: [email protected]
Abstract
We present two basic hybrid fixed point theorems for contractive mappings in par- tially ordered metric spaces and derive some interesting special cases. Our main result is illustrated with an example.
(2010) Mathematics Subject Classifications: 47H10, 54G25 Keywords and phrases: Partially ordered set; Metric space;
Contractive mapping; Hybrid fixed point theorem.
1 Introduction
The study of hybrid fixed point theorems for the contraction mappings in partially ordered metric spaces is initiated in Ran and Reuring [5] which are further continued by Nieto and Rodringuez-Lopez [4] and applied to boundary value problems of nonlinear first order ordinary differential equations for proving the existence results under certain monotonic conditions. Since then many mathematicians have established several hybrid fixed point theorems (FPTs) for different classes of contraction mappings in partially ordered metric spaces. For details, a reader is referred to a recent paper of Dhage et al. [2] and the references cited therein. The main advantage of these hybrid fixed point theorems is that the uniqueness of the fixed point of the mappings in question may be obtained under certain additional conditions. So the hybrid FPTs are useful for proving the existence as well as uniqueness theorems for some nonlinear problems under certain monotonic conditions (see Dhage [1] and the references therein). In the present paper we generalize the class of contraction mappings to contractive mappings and prove some basic hybrid FPTs in partially ordered metric spaces. In the following section we prove our main results of this paper.
2 Basic Hybrid FPTs
An order relation on a non-empty set X is a reflexive, antisymmetric and transitive relation and the non-empty setXtogether with the order relationis a partially ordered set and denoted by (X,). We need the following definitions in what follows.
0Corresponding author∗: [email protected] (Bapurao C. Dhage) Tel. +91 2381 262826 (India)
Definition 2.1. A mapping T : X → X is called monotone nondecreasing if it preserves the order relation, that is, if x y then T x T y for all x, y ∈ X. Similarly, T is called monotone nonincreasing it preserves the order relation reversely, i.e., x y implies T x T y for all x, y ∈ X. T is simply called monotonic if it is either monotone nondecreasing or monotone nonincreasing on X.
Definition 2.2. A mapping T : X → X is said to dominate a point x ∈ X from above (resp. from below) if x T x (resp. T x x). T is called dominating from above (resp. dominating from below) on X if it dominates every point of X from above (resp.
from below). T is called dominating onX if it is either dominating above or dominating below on X.
Definition 2.3. A mapping T : X → X is called mixed dominating if it dominates every point x∈ X from above or from below i.e. if one of x T x and x T x holds for every x∈X.
It is clear that mixed dominating is a weaker property of a mapping than dominating above and dominating below in a partially ordered metric space.
Example 2.1. LetX =R and letf :R→R be a mapping defined as f(x) = 2x.
Then f is mixed dominating on R because x≤T xif x≥0 and x≥T xif x≤0.
We introduce a metric d on X so that (X,, d) is now becomes a partially ordered metric space. An orbit O(x;T) of a mapping T : X → X at a point x ∈ X is a set of points in X defined as
O(x;T) ={x, T x, T2x, . . . .}
The mapping T is called T-orbitally continuous if for any sequence {xn} in O(x;T), xn→x∗ implies T xn →T x∗ for all x∈X.
Definition 2.4 (Dhage [1]). A mapping T : X → X is called partially compact if T(C) is relatively compact subset of the metric space X for every totally ordered set or chain C in X.
Note that every compact set is partially compact, but the converse is not necessarily true. Now we are well equipped with all necessary details to prove the main results of this paper.
Theorem 2.1. Let (X,, d) be a partially ordered metric space and let T :X →X be a nondecreasing mapping satisfying the contractive condition
d(T x, T y)< d(x, y) (2.1)
for all comparable elements x, y ∈ X with x 6=y. Suppose that there is an x0 ∈ X such thatx0 T x0 and the sequence {Tnx0} of successive iterates of T at x0 has a convergent subsequence {Tnkx0} converging to a point u∈X. If T is continuous atu and u and T u are comparable, then uis a fixed point ofT which is further unique if X is totally ordered.
Proof. Define a sequence {xn}={Tnx0} of iterates ofT atx0 as
xn+1 =Tn+1x0 =T xn, n= 0,1,2, . . . (2.2) Since T is nondecreasing, one has
x0 x1 · · · xn · · ·, (2.3) or
Tmx0 Tnx0∀m≥n ∈N. Denote
Cn =d(xn, xn+1)>0 for each n = 0,1,2, . . .. Then from (2.1) we obtain
C0 > C1 >· · ·> Cn>· · · . (2.4) Thus{Cn}is a decreasing sequence of positive real numbers. Hence{Cn}is convergent, and there is a positive real number C > 0 such that
n→∞lim Cn =C. (2.5)
Now every subsequence of {Cn}also converges to the same limit point. Therefore,
k→∞lim Cnk =C = lim
k→∞Cnk+1. (2.6)
Suppose that a subsequence {Tnkx0} of {Tnx0} is convergent and converges to the point, say u∈X, i.e.limk→∞Tnkx0 =u. Since T is continuous at u, one has
k→∞lim Tnk+1x0 = lim
k→∞T Tnkx0 =T
k→∞lim Tnkx0
=T u and
k→∞lim Tnk+2x0 = lim
k→∞T2Tnkx0 =T2
k→∞lim Tnkx0
=T2u.
Therefore,
k→∞lim d Tnkx0, Tnk+1x0
=d(u, T u) and
k→∞lim d Tnk+1x0, Tnk+2x0
=d(T u, T2u).
Now from (2.6) it follows that
d(T u, T2u) =d(u, T u). (2.7)
As u and T uare comparable, either uT u orT uu. If u6=T u, in both the cases, by (2.1) we obtain
d(T u, T2u)< d(u, T u) which is a contraction to (2.7). Hence u=T u.
To prove uniqueness, assume thatX is totally ordered and let v(6=u) be another fixed point of T. Since X is totally ordered, either u v or v u. Then in both the cases, from (2.1) we obtain a contradiction. This completes the proof.
Theorem 2.2. Let (X,, d) be a partially ordered metric space and let T : X → X be a nondecreasing mapping satisfying the contractive condition (2.1). Suppose that there is an x0 ∈X such that x0 T x0 and the sequence {Tnx0} of successive iterates of T at x0 has a convergent subsequence{Tnkx0} converging to a point u∈X. If T is continuous at u and u and T uare comparable, then u is a fixed point of T which is further unique if X is totally ordered.
Remark 2.1. We note the continuity of the mappingT in above Theorems 2.1 and 2.1 at every cluster point ofO(x;T) may be replaced with the condition ofT-orbitally continuity of T on the metric space X.
Corollary 2.1. Let(X,)be a partially ordered set and there exists a metric spacedsuch that (X, d) is partially compact. Let T : X → X be a nondecreasing mapping satisfying (2.1). Suppose that T is continuous and there exists an x0 ∈ X such that x0 T x0 or x0 T x0. If T is mixed dominating at every cluster point of O(x0, T), then T has a fixed point which is further unique if X is totally ordered.
Proof. Define a sequence {xn} = {Tnx0} of successive iteration of T at x0. Since T is nondecreasing and x0 T x0, we have that {Tnx0} is totally ordered set in X. As X is partially compact, the sequence {Tnx0} has a convergent subsequence, say {Tnkx0} converging to some point u ∈ X. Further since T dominates every cluster point of {Tnx0}, it follows that u and T u are comparable. Now the desired conclusion follows by
an application of Theorem 2.1.
Theorem 2.3. Let (X,, d) be a totally ordered metric space and let T : X → X be a nondecreasing mapping. Suppose that there exists a positive integer p such that
d(Tpx, Tpy)< d(x, y) (2.8) for all comparable x, y ∈ X with x 6= y. Suppose that there exists a x0 ∈ X such that x0 T x0 and the sequence {Tpnx0} of iteration of Tp atx0 has a convergent subsequence converging to the point u. If T is continuous at u, then u is a unique fixed point of T. Proof. Let S = Tp. Suppose that a subsequence {Tpnkx0} of the sequence {Tpnx0} is convergent and converges to u. Then S is continuous at u, because T is continuous at u. Now, an application of Theorem 2.1 yields thatS has a unique fixed point, that is, it is a point u ∈ X such that S(u) = Tp(u) = u. Now T(u) = T(Tpu) = S(T u), showing that T uis again a fixed point of S. By the uniqueness of u,we get T u=u.The proof is
complete.
Theorem 2.4. Let (X,, d) be a totally ordered metric space and let T : X → X be a nondecreasing mapping. Suppose that there exists a positive integer p such that the contraction condition (2.8) holds. Suppose further that there exists an x0 ∈X such that x0 T x0 and the sequence {Tpnx0} of iteration of Tp atx0 has a convergent subsequence converging to the point u. If T is continuous at u, then u is a unique fixed point of T. Corollary 2.2. Let (X,) be a totally ordered set and there exists a metric d such that (X, d) is partially compact. Let T : X → X be a nondecreasing mapping satisfying (2.8) and there exists an x0 ∈ X such that x0 T x0 or x0 T x0. If T is mixed dominating and continuous at every cluster point of O(x0, T), then T has a unique fixed point.
Remark 2.2. The conclusion of Theorems 2.1 and 2.3 also remains true if we replace the contractive conditions (2.1) and (2.8) respectively by
d(T x, T y)<max
d(x, y), d(x, T x), d(y, T y),1 2 h
d(x, T y) +d(y, T x)i
(2.9) and
d(Tpx, Tpy)<max
d(x, y)d(x, Tpx), d(y, Tpy),1 2 h
d(x, Tpy) +d(y, Tpx)i
(2.10) for all comparable elementsx, y ∈X provided the right hand side of the inequalities (2.9) and (2.10) is not zero. The proofs of these results may be obtained by closely observing the proofs of Theorem 2.1 and 2.3 with appropriate modifications.
3 An Example
Let J = [a, b] be a closed and bounded interval in R, the set of real numbers, where a, b ∈ R, a < b. Define the usual order relation ≤ and usual standard metric d on J defined as
d(x, y) = |x−y|.
Consider a continuous and nondecreasing functionf :J →J satisfying|f0(x)|<1 for allx∈(a, b). Clearly, J is compact and f is a partially contractive on J. To see this, let x, y ∈J be such thatx < y. Then
d(f x, f y) = |f x−f y|
=
Z y
x
|f0(x)|ds
< |x−y|
= d(x, y).
Further if there is an element x0 ∈ J such that x0 ≤f x0 or x0 ≥f x0, then f has a unique fixed point in view of the fact that J is totally ordered set in R.
Example 3.1. Given a closed and bounded intervalJ = [0,1]in R, consider the mapping f on J defined by f(x) = x2
2 for x∈[0,1].
Clearly, f is continuous and nondecreasing on J into itself. Now, |f0(x)| =x < 1 for eachx∈(0,1). Furthermore, there is an elementx0 = 1 in [0,1] such that f(1) = 12 <1.
Hence an application of above result yields that f has a unique fixed point in [0,1]. In this case the unique fixed point of f is x∗ = 0.
Remark 3.1. The fixed point results of this paper may be extended to commuting and non-commuting pairs of mappings in partially ordered metric spaces satisfying the gen- eralized contrastive type conditions of the form of the inequalities (2.9) and (2.10) for proving the common fixed point results. These and other similar results form further scope and open problems in the area of hybrid fixed point theory.
References
[1] B. C. Dhage,Hybrid fixed point theory in partially ordered normed linear spaces and applications to fractional integral equations, Diff. Equ. Appl. 5 (2013), 155-184.
[2] B. C. Dhage H. K. Nashine and V. S Patil, Common fixed points for some variants of weakly contraction mappings in partially ordered metric spaces, Adv. Fixed Point Theory 3 (2013), 29-48.
[3] M. Edelstein,On fixed points and periodic points under contractive mappings, J Lond.
Math. Soc 37 (1962), 74-79.
[4] J. J. Nieto and R. Rodriguez-Lopez, Contractive mappings theorems in partially or- dered sets and applications to ordinary differential equations, Order 22 (2005), 223- 239.
[5] A. C. M. Ran, M.C. R. Reurings,A fixed point theorem in partially ordered sets and some applications to matrix equations, Proc. Amer. Math. Soc.132(2003), 14351443.
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