# Fixed Point Theorems on Ordered Metric Spaces through a Rational Contraction

Authors: Poom Kumam; Fayyaz Rouzkard; Mohammad Imdad; Dhananjay Gopal
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Ran and Reurings (2004) established an interesting analogue of Banach Contraction Principle in a complete metric space equipped with a partial ordering and also utilized the same oneto discuss the existence of solutions to matrix equations. Motivated by this paper, we prove results on coincidence points for a pair of weakly increasing mappings satisfying a nonlinear contraction condition described by a rational expression on an ordered complete metric space. The uniqueness of common fixed point is also discussed. Some examples are furnished to demonstrate the validity of the hypotheses of our results. As an application, we derive an existence theorem for the solution of an integral equation.


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## Body

## 1. Introduction with Preliminaries

A variety of generalizations of the Classical Banach Contraction Principle [1] are available in the existing literature of metric fixed point theory. The majority of these generalizations are obtained by improving the underlying contraction condition (e.g., [2]). Presently, there is vigorous research activity to prove existence results on complete metric spaces equipped with a partial ordering. In fact, various existence and uniqueness theorems on fixed and common fixed point for monotone mappings are of paramount importance in the study of nonlinear equations which generate natural interest to establish usable fixed point theorems in partial metric spaces (e.g., [1–24]).Very recently, Harjani et al. [25] proved a fixed point theorem in partially ordered metric spaces satisfying a contractive condition of rational type due to Jaggi [26]. The aim of this paper is to prove some results of Harjani et al. [25] type for a pair of self-mappings. We accomplish this using the concept of weakly increasing property due to Nashine and Samet [14] (also see [4, 27, 28]). Some examples are also furnished to demonstrate the validity of the hypotheses of our results. As an application, we establish the existence of solution to an integral equation (also see [2, 23, 29, 30]).Before presenting our results, we recall some notations, definitions, and examples required in our subsequent discussions.Definition 1.
Let𝒳 be a nonempty set. Then (𝒳,d,⪯) is called an ordered (partial) metric space if
(i)(𝒳,⪯) is a partially ordered set and (ii) (𝒳,d) is a metric space.Definition 2.
Let(𝒳,⪯) be a partially ordered set. Then(a)
elementsx,y∈𝒳 are called comparable with respect to “⪯” if either x⪯y or y⪯x;(b)
a mapping𝒯:𝒳→𝒳 is called nondecreasing with respect to “⪯” if x⪯y implies 𝒯x⪯𝒯y.Let𝒳 be a nonempty set and ℛ:𝒳→𝒳 be a given mapping. For every x∈𝒳, we denote by ℛ-1(x), the subset of 𝒳 defined by
(1)ℛ-1(x):={u∈𝒳∣ℛu=x}.Definition 3.
Let(𝒳,⪯) be a partially ordered set and let 𝒯,ℛ:𝒳→𝒳 be two mappings such that 𝒯𝒳⊆ℛ𝒳. We say that 𝒯 is weakly increasing with respect to ℛ if and only if for all x∈𝒳, one has:
(2)𝒯x⪯𝒯y,∀y∈ℛ-1(𝒯x).Remark 4.
Ifℛ:𝒳→𝒳 is the identity mapping on 𝒳, then 𝒯 is weakly increasing with respect to ℛ if and only if 𝒯(x)⪯𝒯𝒯(x) for all x∈𝒳.As mentioned earlier, the notion of weakly increasing mappings was introduced in [4] which is presently in use (e.g. [27, 28]). In what follows, we furnish a relatively new example to demonstrate the preceding definition.Example 5.
Consider𝒳=[0,∞) endowed with the natural order ≤. Define two mappings 𝒯,ℛ:𝒳→𝒳 as
(3)𝒯x={x2+1,0≤x<11,1≤x,ℛx={x3+1,0≤x<11,1≤x.
In order to show that the mapping𝒯 is weakly increasing with respect to mapping ℛ, we distinguish three cases.
Firstly, we consider the casex=0. Let y∈ℛ-1(𝒯(0)), that is, ℛ(y)=𝒯(0)=1 so that ℛ(y)=1 and henceforth y=0 or y≥1. By the definitions of 𝒯, we have
(4)1=𝒯(0)≤𝒯(y)=1.
Secondly, we argue the casex≥1. Let y∈ℛ-1(𝒯(x)), that is, ℛ(y)=𝒯(x)=1, which amounts to say (in view of definition R) that y=0 or y≥1, implying thereby
(5)1=𝒯(x)≤𝒯(y)=1.
Finally, we need to consider the case0<x<1. Let y∈ℛ-1(𝒯(x)), that is, ℛ(y)=𝒯(x). By the definition of 𝒯, we have 𝒯(x)=x2+1, so that ℛ(y)=x2+1. Now, in view of definition of ℛ, we have ℛ(y)=y3+1, so that y3+1=x2+1, yielding thereby y=(x)2/3. Now, we have
(6)x2+1=𝒯(x)≤𝒯(y)=𝒯((x)2/3)=(x)4/3+1.
Thus we have shown that𝒯 is weakly increasing with respect to ℛ.Definition 6.
Let(𝒳,⪯) be an ordered metric space. We say that 𝒳 is regular if and only if the following hypothesis holds:
if{zn} is a nondecreasing sequence in 𝒳 with respect to ⪯ such that limn→∞zn=z∈𝒳, then zn⪯z for all n∈ℕ.Definition 7.
A pair(ℛ,𝒯) of self-mappings of a metric space (𝒳,d) is said to be compatible if and only if limn→+∞d(𝒯ℛxn,ℛ𝒯xn)=0, whenever {xn} is a sequence in 𝒳 such that limn→+∞ℛxn=limn→+∞𝒯xn=z∈𝒳.Definition 8 (see [18]).
A pair(ℛ,𝒯) of self-mappings of a metric space (𝒳,d) is said to be reciprocally continuous if and only if limn→∞ℛ𝒯xn=ℛz and limn→∞𝒯ℛxn=𝒯z for every sequence {xn} in 𝒳 satisfying
(7)limn→∞ℛxn=limn→∞𝒯xn=z
for some z∈𝒳.
Notice that a pair of continuous mappings is always reciprocally continuous but not conversely as substantiated by examples in [18].Definition 9 (see [19]).
A pair(ℛ,𝒯) of self mappings of a metric space (𝒳,d) is said to be weakly reciprocally continuous if and only if limn→∞ℛ𝒯xn=ℛz, for every sequence {xn} in 𝒳 satisfying
(8)limn→∞ℛxn=limn→∞𝒯xn=z
for some z∈𝒳.
Evidently, every pair of reciprocally continuous mappings is always weakly reciprocally but not conversely as demonstrated in Pant et al. [19].


## 2. Results

The main result of this paper runs as followsTheorem 10.
Let(𝒳,⪯) be a partially ordered set equipped with a metric d on 𝒳 such that (𝒳,d) is a complete metric space. Let 𝒯,ℛ:𝒳→𝒳 be two mappings satisfying (for pairs (x,y)∈𝒳×𝒳 wherein ℛx and ℛy are comparable)
(9)d(𝒯x,𝒯y)≤αd(ℛx,𝒯x)·d(ℛy,𝒯y)1+d(ℛx,ℛy)d(𝒯x,𝒯y)≤+βd(ℛx,ℛy),
where α, β are nonnegative real numbers with α+β<1. Suppose that (a)
𝒳 is regular and 𝒯 is weakly increasing with ℛ,(b)
the pair(ℛ,𝒯) is commuting as well as weakly reciprocally continuous.Then 𝒯 and ℛ have a coincidence point. That is, there exists u∈𝒳 such that ℛu=𝒯u.Proof.
Letx0 be an arbitrary point in 𝒳. Since 𝒯𝒳⊆ℛ𝒳, one can inductively construct a sequence {xn} in 𝒳 defined by
(10)ℛxn=𝒯xn-1,∀n∈ℕ.
Asx1∈ℛ-1(𝒯x0) and x2∈ℛ-1(𝒯x1), using weakly increasing property of 𝒯 with respect to ℛ, we obtain
(11)ℛx1=𝒯x0⪯𝒯x1=ℛx2⪯𝒯x2=ℛx3.
Continuing this process indefinitely, we get(12)ℛx1⪯ℛx2⪯ℛx3⪯⋯⪯ℛxn⪯ℛxn+1⪯⋯.
Now, we proceed to show that forn∈ℕ,
(13)d(ℛ(xn+1),ℛ(xn+2))≤(β1-α)nd(ℛ(x1),ℛ(x2)).
Asℛ(x2)≽ℛ(x1), using (9), we have
(14)d(ℛ(x2),ℛ(x3))=d(𝒯(x1),𝒯(x2))≤αd(ℛ(x1),𝒯(x1))·d(ℛ(x2),𝒯(x2))1+d(ℛ(x1),ℛ(x2))≤+βd(ℛ(x1),ℛ(x2)),≤(owingtod(ℛ(x1),ℛ(x2))≤1+d(ℛ(x1),ℛ(x2)))≤αd(ℛ(x2),ℛ(x3))+βd(ℛ(x1),ℛ(x2))
which implies that
(15)(1-α)d(ℛ(x2),ℛ(x3))≤βd(ℛ(x1),ℛ(x2))
so that
(16)d(ℛ(x2),ℛ(x3))≤(β1-α)d(ℛ(x1),ℛ(x2)).
Let (13) holds for all n>0. As ℛ(xn+1)≽ℛ(xn), using (9), we have
(17)d(ℛ(xn+2),ℛ(xn+3))=d(𝒯(xn+1),𝒯(xn+2))≤αd(ℛ(xn+1),𝒯(xn+1))d(ℛ(xn+2),𝒯(xn+2))1+d(ℛ(xn+1),ℛ(xn+2))≤+βd(ℛ(xn+1),ℛ(xn+2)),≤(owingtod(ℛ(xn+1),ℛ(xn+2)))≤≤1+d(ℛ(xn+1),ℛ(xn+2))≤≤αd(ℛ(xn+2),ℛ(xn+3))≤+βd(ℛ(xn+1),ℛ(xn+2)),
so that(18)d(ℛ(xn+2),ℛ(xn+3))≤(β1-α)d(ℛ(xn+1),ℛ(xn+2))≤(β1-α)n+1d(ℛx1,ℛx2).
Now withh=β/(1-α), for any m>n, we have
(19)d(ℛ(xn),ℛ(xm))≤d(ℛ(xn),ℛ(xn+1))+d(ℛ(xn+1),ℛ(xn+2))+⋯d(ℛ(xm-1),ℛ(xm))≤[hn+hn+1+⋯hm-1]d(ℛx1,ℛx2)≤[hn1-h]d(ℛx1,ℛx2).
As0<h=β/(1-α)<1, we have limn→+∞d(ℛ(xn),ℛ(xm))=0, that is, {ℛxn} is Cauchy sequences in the complete metric space (𝒳,d) and hence there exists some x∈𝒳 such that
(20)limn→+∞𝒯(xn)=limn→+∞ℛ(xn)=x.
Owing to commutativity of𝒯 with ℛ and by using (10) (for each n≥1), we have
(21)ℛ(ℛxn+1)=ℛ(𝒯xn)=𝒯(ℛxn),thatisℛ-1(𝒯(ℛxn))=ℛxn+1.
Since𝒯 is weakly increasing with ℛ, we can write
(22)ℛ(ℛxn+1)=𝒯(ℛxn)⪯𝒯(ℛxn+1)=ℛ(ℛxn+2)
so that ℛ(ℛxn) is nondecreasing. As the maps ℛ and 𝒯 are weakly reciprocally continuous, limn→∞ℛ(ℛxn)=limn→∞ℛ(𝒯xn-1)=ℛx, which together with regularity of 𝒳 gives rise to ℛ(ℛxn)⪯ℛx; that is, ℛ(ℛxn) and ℛx are comparable. On using condition (9), we have
(23)d(ℛx,𝒯x)≤d(ℛx,ℛ(ℛxn+1))+d(ℛ(ℛxn+1),𝒯x)=d(ℛx,ℛ(ℛxn+1))+d(ℛ(𝒯xn),𝒯x)=d(ℛx,ℛ(ℛxn+1))+d(𝒯(ℛxn),𝒯x)=d(ℛx,ℛ(ℛxn+1))+d(𝒯x,𝒯(ℛxn))≤d(ℛx,ℛ(ℛxn+1))+αd(ℛx,𝒯x)d(ℛ(ℛxn),𝒯(ℛxn))1+d(ℛx,ℛ(ℛxn))+βd(ℛx,ℛ(ℛxn)).
On makingn→∞ in the preceeding inequality, one gets
(24)d(ℛx,𝒯x)=0,
so that𝒯x=ℛx. Thus, we have shown that 𝒯 and ℛ have a coincidence point. This completes the proof.Settingα=0 in Theorem 10, we deduce the following.Corollary 11.
Let(𝒳,⪯) be a partially ordered set on which there is a metric d on 𝒳 such that (𝒳,d) is a complete metric space. Let 𝒯,ℛ:𝒳→𝒳 be given mappings satisfying (for pairs (x,y)∈𝒳×𝒳 wherein ℛx and ℛy are comparable)
(25)d(𝒯x,𝒯y)≤βd(ℛx,ℛy),
where β is a nonnegative real number with β<1. Suppose that (a)
𝒳 is regular and 𝒯 is weakly increasing with ℛ,(b)
the pair (ℛ,𝒯) is commuting as well as weakly reciprocally continuous. Then 𝒯 and ℛ have a coincidence point. That is, there exists u∈𝒳 such that ℛu=𝒯u.Theorem 12.
Let(𝒳,⪯) be a partially ordered set equipped with a metric d on 𝒳 such that (𝒳,d) is a complete metric space. Let 𝒯,ℛ:𝒳→𝒳 be two mappings satisfying (for pairs (x,y)∈𝒳×𝒳 wherein ℛx and ℛy are comparable)
(26)d(𝒯x,𝒯y)≤αd(ℛx,𝒯x)·d(ℛy,𝒯y)1+d(ℛx,ℛy)d(𝒯x,𝒯y)≤+βd(ℛx,ℛy),
where α, β are non-negative real numbers with α+β<1. Suppose that (a)
𝒯 is weakly increasing with ℛ,(b)
the pair{𝒯,ℛ} is compatible and reciprocally continuous. Then 𝒯 and ℛ have a coincidence point. That is, there exists u∈𝒳 such that ℛu=𝒯u.Proof.
Proceeding on the lines of the proof of Theorem10, one can furnish a sequence {xn} such that
(27)limn→+∞𝒯(xn)=limn→+∞ℛ(xn)=x.
Now, it remains to show thatx is a coincidence point of 𝒯 and ℛ. To accomplish this, as the pair {𝒯,ℛ} is compatible as well as reciprocally continuous,
(28)limn→∞d(ℛ(𝒯(xn)),𝒯(ℛ(xn)))=0,(29)ℛ(x)=limn→∞ℛ(𝒯(xn)),𝒯(x)=limn→∞𝒯(ℛ(xn)),
whenever
(30)limn→+∞𝒯(xn)=limn→+∞ℛ(xn)=x.
By using (29) in (28), we have d(𝒯x,ℛx)=0, so that 𝒯x=ℛx.By appealing Theorems10 and 12, one can also have the following natural theorem.Theorem 13.
Let(𝒳,⪯) be a partially ordered set on which there is a metric d on 𝒳 such that (𝒳,d) is a complete metric space. Let 𝒯:𝒳→𝒳 be nondecreasing mapping satisfying (for pairs (x,y)∈𝒳×𝒳 wherein ℛx and ℛy are comparable)
(31)d(𝒯x,𝒯y)≤αd(x,𝒯x)·d(y,𝒯y)1+d(x,y)+βd(x,y),
where α, β are non-negative real numbers such that α+β<1. Suppose that (I)
𝒯x⪯𝒯(𝒯x) for all x∈𝒳,(II)
either𝒯 is continuous or 𝒳 is regular. Then, 𝒯 has a fixed point.Proof.
The proof of this theorem can be outlined on the lines of the proof of Theorem10 realizing ℛ to be the identity mapping on 𝒳.Now, we introduce the following property which will be utilized in our next theorem.Property (A).
Ifℛ(xn) is a nondecreasing sequence in X such that limn→+∞ℛ(xn)=x, then ℛ(xn) is comparable to ℛ(x), for all n∈N.Theorem 14.
Let(𝒳,⪯) be a partially ordered set on which there exists a metric d on 𝒳 such that (𝒳,d) is a complete metric space. Let 𝒯,ℛ:𝒳→𝒳 be given mappings satisfying (for pairs (x,y)∈𝒳×𝒳 wherein ℛx and ℛy are comparable)
(32)d(𝒯x,𝒯y)≤αd(ℛx,𝒯x)·d(ℛy,𝒯y)1+d(ℛx,ℛy)d(𝒯x,𝒯y)≤+βd(ℛx,ℛy),
where α, β are non-negative real numbers with α+β<1. Suppose that (a)
𝒳 is regular and 𝒯 is weakly increasing with ℛ,(b)
the pair (ℛ,𝒯) is commuting as well as weakly reciprocally continuous,(c)
ℛ satisfies Property (A). Then ℛ and 𝒯 have a common fixed point.Proof.
Proceeding on the lines of the proof of Theorem10, one can inductively construct nondecreasing sequence {ℛxn}, such that limn→∞𝒯xn=limn→∞ℛxn+1=x, and ℛ(x)=𝒯(x). Since ℛ(xn) and ℛ(x) are comparable (for all n∈N), by using condition (32) we have
(33)d(ℛx,ℛxn+1)=d(𝒯x,𝒯xn)d(ℛx,ℛxn+1)≤αd(ℛx,𝒯x)d(ℛxn,𝒯xn)1+d(ℛx,ℛxn)d(ℛx,ℛxn+1)≤+βd(ℛx,ℛxn).
Taking the limit asn→∞, one gets
(34)x=ℛx=𝒯x.
This completes the proof.Theorem 15.
Let(𝒳,⪯) be a partially ordered set on which there exists a metric d on 𝒳 such that (𝒳,d) is a complete metric space. Let 𝒯,ℛ:𝒳→𝒳 be given mappings satisfying (for pairs (x,y)∈𝒳×𝒳 wherein ℛx and ℛy are comparable)
(35)d(𝒯x,𝒯y)≤αd(ℛx,𝒯x)·d(ℛy,𝒯y)1+d(ℛx,ℛy)d(𝒯x,𝒯y)≤+βd(ℛx,ℛy),
where α, β are non-negative real numbers with α+β<1. Suppose that (a)
𝒯 is weakly increasing with ℛ,(b)
the pair{𝒯,ℛ} is compatible and reciprocally continuous,(c)
ℛ satisfies Property (A).
Thenℛ and 𝒯 have a common fixed point.Proof.
Proof is obvious in view of Theorems12 and 14.


## 3. Uniqueness Results

In what follows, we investigate the conditions under which Theorem10 ensures the uniqueness of common fixed point.Theorem 16.
If, in addition to the hypotheses of Theorem10, every (x,x*)∈𝒳×𝒳, there exists a u∈𝒳 such that 𝒯u is upper bound of 𝒯x and 𝒯x*, then 𝒯 and ℛ have a unique common fixed point.Proof.
In view of Theorem10, the set of coincidence points of the maps 𝒯 and ℛ is non-empty. If x and x* are two coincidence points of the maps 𝒯 and ℛ (i.e., ℛx=𝒯x, ℛx*=𝒯x*), then we proceed to show that
(36)ℛx=ℛx*.
In view of the additional hypothesis of this theorem, there existsu∈𝒳 such that 𝒯u is upper bound of 𝒯x and 𝒯x*. Put u0=u, and choose u1∈𝒳 such that ℛu1=𝒯u0. Now, proceeding on the lines of the proof of Theorem 10, one can inductively define sequence {ℛun} such that
(37)ℛun=𝒯un-1∀n,
wherein
(38)ℛu1⪯ℛu2⪯ℛx3⪯⋯⪯ℛun⪯ℛun+1⪯⋯.
Further, settingx0=x, and x0*=x*, one can also define the sequences {ℛxn} and {ℛxn*} such that
(39)ℛxn=𝒯xn-1,ℛxn*=𝒯xn-1*.
For everyn≥1, we have
(40)ℛxn=𝒯x,ℛxn*=𝒯x*∀n≥1.
Since𝒯u=ℛu1 is upper bound of 𝒯x=ℛx1 and 𝒯x*=ℛx1*, then
(41)ℛx1=ℛx⪯ℛu1,ℛx1*=ℛx*⪯ℛu1.
It is easy to show thatℛx⪯ℛun and ℛx*⪯ℛun for all n≥1, then ℛx and ℛx* are comparable with ℛun. On using (9), we have
(42)d(ℛx,ℛun+1)=d(𝒯x,𝒯un)d(ℛx,ℛun+1)≤αd(ℛx,𝒯x)d(ℛun,𝒯un)1+d(ℛx,ℛun)d(ℛx,ℛun+1)≤+βd(ℛx,ℛun)
or
(43)d(ℛx,ℛun+1)≤βd(ℛx,ℛun).
Owing to (43), we can write
(44)d(ℛx,ℛun+1)≤βd(ℛx,ℛun)≤β2d(ℛx,ℛun-1)≤⋯≤βn+1d(ℛx,ℛu0).
Taking the limit asn→∞ in (44), we get
(45)limn→∞d(ℛx,ℛun)=0,
as 0<β<1.
Similarly, one can also show that(46)limn→∞d(ℛx*,ℛun)=0.
On using (45) and (46), we can have
(47)d(ℛx,ℛx*)≤d(ℛx,ℛun+1)+d(ℛun+1,ℛx*)
so that limn→∞d(ℛx,ℛx*)0, that is, ℛx=ℛx*.  Thus, we have proved (36).
Sinceℛx=𝒯x, owing to commutativity of 𝒯 and ℛ, one can write
(48)ℛ(ℛx)=ℛ(𝒯x)=𝒯(ℛx)
which on inserting ℛx=z gives rise to
(49)ℛz=𝒯z.
Thus,z is another coincidence point of the pair. Now, due to ℛx=ℛx* (for every coincidence point x and x*) and owing to the fact that z is coincidence point of the pair {ℛ,𝒯}, it follows that ℛz=ℛx, that is,
(50)ℛz=z.
On making use of (49) and (50), we can have
(51)z=ℛz=𝒯z
which shows that z is a common fixed point of 𝒯 and ℛ.
To prove the uniqueness, letp be another common fixed point of the pair {ℛ,𝒯}. Since p and z are coincidence points, then ℛp=ℛz.
Owing to thatp and z are common fixed point of the pairs {ℛ,𝒯}, one can have
(52)p=ℛp=ℛz=z.
This completes the proof.The following simple example demonstrates Theorem16.Example 17.
Consider𝒳={(1,0),(0,1)}⊂R2 equipped with natural order (x,y)≤(z,t)⇔x≤z and y≤t. Thus (𝒳,≤), is a partially ordered set wherein the two elements are not comparable to each other. Also, (𝒳,≤) is a complete metric space under Euclidean metric. Defined mappings 𝒯 and ℛ:𝒳→𝒳 as
(53)𝒯(x,y)={(1,0),(x,y)=(1,0)(1,0),(x,y)=(0,1),(54)ℛ(x,y)={(1,0),(x,y)=(1,0)(0,1),(x,y)=(0,1).
Asℛ(x,y) and ℛ(z,t) in 𝒳 are merely comparable to themselves, inequality (9) is vacuously satisfied for every α,β.
Notice that𝒯(𝒳)=(1,0)⊂ℛ(𝒳)=𝒳. Also, for (x,y)=(1,0),
(55)ℛ(z,t)=𝒯(x,y)=𝒯(1,0)=(1,0)⟹(z,t)=(1,0)
so that
(56)𝒯(1,0)=𝒯(x,y)≤𝒯(z,t)=𝒯(1,0).
Otherwise, for (z,t)=(0,1),
(57)ℛ(z,t)=𝒯(x,y)=𝒯(0,1)=(1,0)⟹(z,t)=(1,0)
so that
(58)(1,0)=𝒯(0,1)=𝒯(x,y)≤𝒯(z,t)=𝒯(1,0)=(1,0).
Thus 𝒯 is weakly increasing with respect to ℛ.
If{(xn,yn)}⊂𝒳 is a nondecreasing sequence converging to (x,y)∈𝒳, then necessarily {(xn,yn)} must be constant sequence, that is, (xn,yn)=(x,y), for all n∈N, so that limit (x,y) is an upper bond for all the terms in the sequence which shows that 𝒳 is regular.
By using definitions of the maps𝒯 and ℛ, we have
(59)(x,y)=ℛ(x,y)=limn→∞𝒯(xn,yn)=limn→∞ℛ𝒯(xn,yn)
wherein limn→∞ℛ(xn,yn)=limn→∞𝒯(xn,yn)=(x,y) so that
(60)ℛ(x,y)=limn→∞ℛ𝒯(xn,yn),
which shows that the pair {ℛ,𝒯} is weakly reciprocally continuous. Also, the pair {ℛ,𝒯} is clearly commuting.
Now, we show that for everyx and x* in 𝒳, there exists a u∈𝒳 such that 𝒯u is upper bound of 𝒯x and 𝒯x*. If x=x*, then choice of u is obvious. Otherwise, if x=(1,0) and x*=(0,1), we can choose u=(0,1) such that 𝒯u=𝒯(0,1)=(1,0) is upper bound of 𝒯x=(1,0) and 𝒯x*=(1,0).
Thus, we have shown that all the conditions of Theorem16 are satisfied and (x,y)=(1,0) is the unique common fixed point of 𝒯 and ℛ.Theorem 18.
If, in addition to the hypotheses of Theorem12, for pairs (x,x*)∈𝒳×𝒳, there exists a u∈𝒳 such that 𝒯u is upper bound of 𝒯x and 𝒯x*, then 𝒯 and ℛ have a unique common fixed point.Proof.
Proof is obvious in view of Theorems12 and 16.Corollary 19.
In addition to the hypotheses of Theorem16 (or Theorem 18), suppose that for every (x,x*)∈𝒳×𝒳 there exists a u∈𝒳 such that 𝒯u is upper bound of 𝒯x and 𝒯x*. Then 𝒯 has a unique fixed point; that is, there exists a unique x∈𝒳 such that x=𝒯x.Proof.
In Theorem16, if ℛ=I (the identity mapping on X), we have the result.The following example demonstrates Theorem18.Example 20.
Consider𝒳=[0,∞) equipped with the usual metric and natural order ≤. Define two mappings 𝒯,ℛ:𝒳→𝒳 by
(61)𝒯x={x3+2,0≤x<13,1≤x,ℛx={2x3+1,0≤x<13,1≤x.
Then, evidently (𝒳,≤) is a partially ordered set and the maps 𝒯 and ℛ satisfy inequality (9) with β=1/2 and α=1/3, for x,y∈[0,∞).
To show that𝒯 is weakly increasing with respect to ℛ, notice that 𝒯(𝒳)=[2,3]⊂[1,3]=ℛ(𝒳). Firstly, we argue the case x=0. Let y∈ℛ-1(T0), that is,ℛ(y)=𝒯(0)=2 so that 2y3+1=2 or y=1/21/3. Using definitions of 𝒯 and ℛ, we have
(62)2=𝒯(0)≤𝒯(y)=𝒯(121/3)=12+2.
Secondly, if x≥1, then we have ℛ(y)=𝒯(x)=3. By using definition of ℛ, we have y≥1, so that
(63)3=𝒯(x)≤𝒯(y)=3.
Finally, we consider the case 0<x<1. Let y∈ℛ-1(𝒯(x)), that is, ℛ(y)=𝒯(x). In view of definition of 𝒯, we have 𝒯(x)=x3+2, so that ℛ(y)=x3+2. By using the definition of ℛ, we have ℛ(y)=2y3+1, so that 2y3+1=x3+2 or y=((x3+1)/2)1/3. Thus, we have
(64)x3+2=𝒯(x)≤𝒯(y)=𝒯((x3+12)1/3)=x3+12+2.
Therefore 𝒯 is weakly increasing with respect to ℛ.
Since𝒯 and ℛ, are continuous, therefore this pair of maps is reciprocally continuous.
Now, we show that maps𝒯 and ℛ, are compatible. If limn→∞xn=k, and k≥1, then limn→∞𝒯xn=limn→∞ℛxn=3 and henceforth
(65)limn→∞𝒯ℛxn=limn→∞ℛ𝒯xn=3
implying thereby limn→∞d(ℛ𝒯xn,𝒯ℛxn)=0 so that 𝒯 and ℛ, are compatible.
Thus, we have shown that all the conditions of Theorem18 are satisfied and x=3 is the unique fixed point of 𝒯 and ℛ.


## 4. An Application

In this section, we present an application of Theorem13 and used the idea of Ćirić et al. [29] to define a partial order and prove an existence theorem for the solution of an integral equation.Theorem 21.
Consider the integral equation(66)x(t)=∫0TK(t,s,x(s))ds+g(t),t∈[0,T]
with T>0 wherein(i)
the functionsK:[0,T]×[0,T]×ℝn→ℝn and g:ℝn→ℝnare continuous,(ii)
for eacht,s∈[0,T],
(67)K(t,s,x(s))≪K(t,s,∫0TK(s,τ,x(τ))dτ+g(s)),where ≪ denotes a partial order relation on ℝn,(iii)
there exists a continuous functionp:[0,T]×[0,T]→ℝ+ such that
(68)|K(t,s,u)-K(t,s,v)|≤p(t,s)|u-v|
for each t,s∈[0,T] and is also comparable u,v∈ℝn,(iv)
supt∈[0,T]∫0Tp(t,s)ds≤β<1.Then the integral equation (66) has a unique solution x* in C([0,T],ℝn).Proof.
Consider𝒳:=C([0,T],ℝn) with the usual supremum norm, that is,
(69)∥x∥=maxt∈[0,T]|x(t)|,
for x,y∈C([0,T]ℝn). Define on 𝒳 a partial order as follows (x,y∈C([0,T],ℝn)):
(70)x⪯yiffx(t)≪y(t)foranyt∈[0,T].
Then(𝒳,⪯) is a partially ordered set and (𝒳,∥·∥) is a complete metric space.
Moreover for any increasing sequence{xn} in 𝒳 converging to x*∈𝒳, we have xn(t)≪x*(t) for any t∈[0,T]. Also for every x,y∈𝒳 there exists c∈𝒳 which depends on x and y and is also comparable to x and y (cf. [21]).
Define𝒯:C([0,T])→C([0,T]) by
(71)𝒯u(t)=∫0TK(t,s,u(s))ds+g(t),hhhhhhhhht∈[0,T],u∈C([0,T]).
Let us prove that(72)𝒯u⪯𝒯(𝒯u)∀u∈C([0,T]).
Letu∈C([0,T]). From (ii), for all t∈[0,T], we have
(73)𝒯u(t)=∫0TK(t,s,u(s))ds+g(t)≪∫0TK(t,s,∫0T‍K(s,τ,u(τ))dτ+g(s))ds+g(t)=∫0TK(t,s,𝒯u(s))ds+g(t)=𝒯(𝒯u)(t).
Then (72) holds.
Now, for allx,y∈C([0,T]) with y⪯x, we have (by (iii))
(74)|𝒯x(t)-𝒯y(t)|≤∫0T|K(t,s,x(s))-K(t,s,y(s))|ds≤∫0Tp(t,s)(|x(s)-y(s)|)ds≤∥x-y∥∫0Tp(t,s)ds.
Hence(75)∥𝒯x(t)-𝒯y(t)∥≤∥x-y∥supt∈[0,T]∫0Tp(t,s)ds≤β∥x-y∥.
On the other hand, it is demonstrated in [16] that condition of regularity is satisfied for 𝒳=C([0,T];ℝn).
Thus, all the hypotheses of Theorem13 are satisfied for α=0, and then 𝒯 has a fixed point u*∈C([0,T];ℝn); that is, u* is a solution of the integral equation (66).



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*Source: 206515-2013-08-22.xml*