# A Fixed Point Theorem in Orbitally Complete Partially Ordered Metric Spaces

Authors: G. V. R. Babu; P. D. Sailaja; K. T. Kidane
Journal: Journal of Operators
 (2013)

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## Abstract

Let(X,⪯) be a partially ordered set and T:X→X be a mapping. We prove a fixed point theorem for the map T satisfying a contractive condition in orbits, when X is T-orbitally complete. Our result extends and generalizes the results of Samet et al. (2013) to partially ordered sets. Also, we generalize the results of Ran and Reurings (2004).


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## Body

## 1. Introduction

In 1922, Banach [1] established a very fundamental result in fixed point theory, namely, the Banach contraction principle. Large number of generalizations have been made on this principle (we refer to Berinde and Vetro [2], Chatterjea [3], Ćirić [4], and Kannan [5, 6]). The Banach contraction principle forces the map T to be continuous. In 1968, Kannan [5] established a fixed point theorem for a map which need not be continuous. Later in 1971, Reich [7] proved a fixed point theorem through which the Banach contraction principle and Kannan's fixed point theorem are generalized.Recently, Samet et al. [8] established new fixed point theorems in complete metric spaces that generalize the Banach contraction principle and Reich’s and Kannan’s fixed point theorems.Theorem 1 (see [8]).
Let(X,d) be a complete metric space, φ:X→[0,∞) a lower semicontinuous function, and T:X→X a given mapping. Suppose that, for any 0<a<b<∞, there exists 0<γ(a,b)<1 such that a≤d(x,y)+φ(x)+φ(y)≤b implies
(1)d(Tx,Ty)+φ(Tx)+φ(Ty)≤γ(a,b)[d(x,y)+φ(x)+φ(y)],
for all x,y∈X. Then, T has a unique fixed point x*∈X. Moreover, one has φ(x*)=0.Theorem 2 (see [8]).
Let(X,d) be a complete metric space, φ:X→[0,∞) a lower semi-continuous function and T:X→X a given mapping. Suppose that there exists a constant γ∈(0,1/2) such that
(2)d(Tx,Ty)+φ(Tx)+φ(Ty)≤γ[d(x,Tx)+d(y,Ty)+φ(x)+φ(y)000000+φ(Tx)+φ(Ty)],
for all x,y∈X. Then, T has a unique fixed point x*∈X. Moreover, one has φ(x*)=0.Theorem 3 (see [8]).
Let(X,d) be a complete metric space, φ:X→[0,∞) a lower semi-continuous function and T:X→X a given mapping. Suppose that, there exist α,β,γ∈(0,1) with α+β+γ<1 such that for all x,y∈X,
(3)d(Tx,Ty)+φ(Tx)+φ(Ty)≤α[d(x,y)+φ(x)+φ(y)]00000+β[d(x,Tx)+φ(x)+φ(Tx)]00000+γ[d(y,Ty)+φ(y)+φ(Ty)].
Then, T has a unique fixed point x*∈X. Moreover, one has φ(x*)=0.Establishing the existence of fixed points in partially ordered sets is of recent interest in fixed point theory. In 2004, Ran and Reurings [9] established the existence of fixed points for contraction maps in partially ordered sets.Definition 4.
Let(X,⪯) be a partially ordered set. A map T:X→X is said to be non-decreasing if, for any x,y∈X with x⪯y, one has Tx⪯Ty.Theorem 5 (see [9]).
Let(X,⪯) be a partially ordered set and d be a metric on X such that (X,d) is a complete metric space. Suppose that T:X→X is a continuous map and there exists k∈(0,1) such that
(4)d(Tx,Ty)≤kd(x,y),
for each x,y∈X with x⪯y. If there exists x0∈X such that x0⪯Tx0, then T has a fixed point in X.Definition 6.
LetX be a nonempty set and T:X→X. Let x0∈X. The orbit of x0 is defined by 𝒪(x0)={x0,Tx0,T2x0,…}.Definition 7.
Let(X,d) be a metric space and T:X→X. X is said to be T-orbitally complete if every Cauchy sequence in 𝒪(x), x∈X, converges to a point in X.Definition 8.
Let(X,d) be a metric space and T:X→X. T is said to be orbitally continuous at z∈X if Txn→Tz as n→∞ whenever xn→z as n→∞.IfT is continuous at z∈X, then T is orbitally continuous at z. But its converse need not be true.For more literature on these concepts, we refer to Türkoğlu et al. [10] and for works on ordered orbitally complete metric spaces we refer to Ding et al. [11] and Nashine et al. [12].Definition 9.
LetX be a nonempty set. Let {Ti/Ti:X→X,i=1,2,3,…} be a sequence of self-maps. An element x∈X is said to be a common fixed point of {Ti} if Ti(x)=x for each i=1,2,3,….In this paper, we prove a fixed point theorem in partially ordered metric spaceX, for a self-map T:X→X, when X is T-orbitally complete. Our result extends and generalizes the results of Samet et al. [8] to partially ordered sets. Our result also generalizes the results of Ran and Reurings [9].We use the following lemma in our main result.Lemma 10 (see [13]).
Suppose that(X,d) is a metric space. Let {xn} be a sequence in X such that d(xn,xn+1)→0 as n→∞. If {xn} is not a Cauchy sequence, then there exist ϵ>0 and sequences of the positive integers m(k) and n(k) with m(k)>n(k)>k such that d(xm(k),xn(k))≥ϵ, d(xm(k)-1,xn(k))<ϵ and limk→∞d(xm(k),xn(k))=ϵ and limk→∞d(xm(k)+1,xn(k)+1)=ϵ.


## 2. Main Result

Throughout this paper, we denote that(5)Ψ={ψ:[0,∞)⟶[0,∞):ψ00is  non-decreasing,  continuous  and00ψ(t)=0⇔t=0}.Theorem 11.
Let(X,⪯) be a partially ordered set and d a metric on X. Suppose that T:X→X is a non-decreasing map and x0∈X such that x0⪯Tx0. Suppose that there exist a lower semicontinuous function φ:X→[0,∞) and ψ∈Ψ such that the following condition holds.
“For each0≤a<b<∞, there exists γ(a,b)∈[0,1) such that a≤ψ(d(x,y))+φ(x)+φ(y)≤b implies
(6)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)≤γ(a,b)M(x,y),
where
(7)M(x,y)=max{ψ(d(x,y))+φ(x)+φ(y),0008.0ψ(d(x,Tx))+φ(x)+φ(Tx),00080ψ(d(y,Ty))+φ(y)+φ(Ty)}
for each x,y∈𝒪(x0)¯ with x⪯y.”
Assume thatX is T-orbitally complete. Then, the sequence {xn} defined by xn+1=Txn,n=0,1,2,…, is Cauchy in X. Let limn→∞xn=z,z∈X. Suppose that either(a)
T is orbitally continuous at z
or(b)
if{xn} is a nondecreasing sequence converging to x∈X, then xn⪯x, for all n.Then, z is a fixed point of T and φ(z)=0.Proof.
Letx0∈X such that x0⪯Tx0. Choose x1∈X such that x1=Tx0. Then, x0⪯x1. Since T is non-decreasing, we have x1=Tx0⪯Tx1. Now, we choose x2∈X such that x2=Tx1. Hence, x1=Tx0⪯Tx1=x2.
Continuing the same procedure, we obtain a sequence{xn}∈𝒪(x0)¯ such that
(8)xn+1=Txn,n=0,1,2,…,x0⪯x1⪯x2⪯⋯⪯xn⪯xn+1⪯⋯.
If there exists an integer n0 such that xn0=xn0+1, then, xn0 is a fixed point of T and hence we are through. Without loss of generality, we assume that xn≠xn+1 for each n.
We denote(9)un=ψ(d(xn,xn+1))+φ(xn)+φ(xn+1)foreachn=0,1,2,….
We show that un+1≤un for each n=0,1,2,…. Suppose, if possible, there exists n0 such that un0<un0+1. That is,
(10)0<un0=ψ(d(xn0,xn0+1))+φ(xn0)+φ(xn0+1)<ψ(d(xn0+1,xn0+2))+φ(xn0+1)+φ(xn0+2)=un0+1.
Then, from (6), there exists γ(un0,un0+1)∈[0,1) such that
(11)ψ(d(Txn0,Txn0+1))+φ(Txn0)+φ(Txn0+1)≤γ(un0,un0+1)M(xn0,xn0+1),
where
(12)M(xn0,xn0+1)=max{ψ(d(xn0,xn0+1))+φ(xn0)+φ(xn0+1),22222222.2ψ(d(xn0,xn0+1))+φ(xn0)+φ(xn0+1),222222222ψ(d(xn0+1,xn0+2))+φ(xn0+1)+φ(xn0+2)}=max{un0,un0+1}=un0+1.
Hence, from (11), we get
(13)ψ(d(Txn0,Txn0+1))+φ(Txn0)+φ(Txn0+1)≤γ(un0,un0+1)un0+1;
that is,
(14)un0+1≤γ(un0,un0+1)un0+1,
a contradiction since γ(un0,un0+1)∈[0,1). Therefore, un+1≤un for each n=0,1,2,…. That is, the sequence {un} is a decreasing sequence of real numbers. Hence, there exists u≥0 such that limn→∞un=u and u≤un for each n=0,1,2,….
We now show thatu=0. Suppose that u>0. Since 0<u≤un≤u0 for all n, from (6), there exists γ(0,u0)∈[0,1) such that
(15)un+1=ψ(d(Txn,Txn+1))+φ(Txn)+φ(Txn+1)≤γ(0,u0)M(xn,xn+1),
where
(16)M(xn,xn+1)=max{ψ(d(xn,xn+1))+φ(xn)+φ(xn+1),0000002.00ψ(d(xn,xn+1))+φ(xn)+φ(xn+1),000002000ψ(d(xn+1,xn+2))+φ(xn+1)+φ(xn+2)}=max{un,un+1}=un.
Thus, from (15), we get
(17)un+1≤γ(0,u0)un.
Letting n→∞, we get
(18)u≤γ(0,u0)u,
a contradiction. Hence, u=0.
Therefore(19)limn→∞un=limn→∞[ψ(d(xn,xn+1))+φ(xn)+φ(xn+1)]=0.
Thus, we have limn→∞ψ(d(xn,xn+1))=0 and limn→∞φ(xn)=0.
By the property ofψ, we have
(20)limn→∞d(xn,xn+1)=0.
We now show that{xn}⊆𝒪(x0) is a Cauchy sequence in X. Suppose if possible that {xn} is not a Cauchy sequence in X. Then, there exist ϵ>0 and sequences of the positive integers {m(k)} and {n(k)} with m(k)>n(k)>k such that
(21)d(xm(k),xn(k))≥ϵ.
We choose m(k), the least positive integer satisfying (21). Then, we have m(k)>n(k)>k with
(22)d(xm(k),xn(k))≥ϵ,d(xm(k)-1,xn(k))<ϵ.
Then, by Lemma 10, we have
(23)limk→∞d(xm(k),xn(k))=ϵ;limk→∞d(xm(k)+1,xn(k)+1)=ϵ.
Now, since limk→∞d(xm(k),xm(k)-1)=0, there exists a number K such that
(24)d(xm(k),xm(k)-1)<ϵforeachk≥K.
Hence,
(25)d(xm(k),xn(k))≤d(xm(k),xm(k)-1)+d(xm(k)-1,xn(k))<ϵ+ϵforallk≥K=2ϵforallk≥K.
Since ψ is non-decreasing, we have
(26)ψ(d(xm(k),xn(k)))≤ψ(2ϵ)forallk≥K.
Now, using (22), we get
(27)0<a=ψ(ϵ)≤ψ(d(xm(k),xn(k)))≤ψ(d(xm(k),xn(k)))+φ(xm(k))+φ(xn(k))<ψ(2ϵ)+um(k)+un(k)foreachk≥K≤ψ(2ϵ)+u0+u0foreachk≥K=ψ(2ϵ)+2u0=b(say)foreachk≥K.
That is, 0<a≤ψ(d(xm(k),xn(k)))+φ(xm(k))+φ(xn(k))<b for each k≥K.
Therefore, from (6), there exists γ(a,b)∈[0,1) such that
(28)ψ(d(xm(k)+1,xn(k)+1))+φ(xm(k)+1)+φ(xn(k)+1)≤γ(a,b)M(xm(k),xn(k)),
where
(29)M(xm(k),xn(k))=max{ψ(d(xm(k),xn(k)))+φ(xm(k))+φ(xn(k)),00002.00ψ(d(xm(k),xm(k)+1))+φ(xm(k))+φ(xm(k)+1),0000200ψ(d(xn(k),xn(k)+1))+φ(xn(k))+φ(xn(k)+1)}.
Letting k→∞, we get
(30)limk→∞M(xm(k),xn(k))=ψ(ϵ).
Now, letting k→∞ in (28), we get
(31)ψ(ϵ)≤γ(a,b)ψ(ϵ),
a contradiction. Hence, {xn}⊆𝒪(x0) is a Cauchy sequence in X. Since X is T-orbitally complete, there exists z∈X such that limn→∞xn=z.
Sinceφ is lower semi-continuous, we have
(32)φ(z)≤liminfn→∞φ(xn)=0.
Hence, φ(z)=0.
Now, suppose that(a) holds. That is, T is orbitally continuous at z. Then, xn+1=Txn→Tz. Hence,Tz=z.
Now, suppose that(b) holds. Then, we have xn⪯z for all n.
Ifxn=z for infinitely many n, then xn+1=Txn=Tz. But xn+1→z. Hence, Tz=z. So, without loss of generality, we assume that d(xn,z)>0 for infinitely many n. That is, there exists a subsequence {xn(k)} of {xn} such that d(xn(k),z)>0 for each k=0,1,2,….
As the sequence{d(xn(k),z)} is convergent, it is bounded. Hence, there exists l≥0 such that d(xn(k),z)≤l for each k=0,1,2,….
Sinceψ is non-decreasing, we have
(33)ψ(d(xn(k),z))≤ψ(l)foreachk=0,1,2,….
Now, for each k=0,1,2,…, we have
(34)0≤ψ(d(xn(k),z))≤ψ(d(xn(k),z))+φ(xn(k))+φ(z)<ψ(d(xn(k),z))+un(k)≤ψ(l)+u0=b(say).
Hence, from (6), there exists γ(0,b)∈[0,1) such that
(35)ψ(d(xn(k)+1,Tz))+φ(xn(k)+1)+φ(Tz)≤γ(0,b)M(xn(k),z),
where
(36)M(xn(k),z)=max{ψ(d(xn(k),z))+φ(xn(k))+φ(z),mmmmmllψ(d(xn(k),xn(k)+1))+φ(xn(k))+φ(xn(k)+1),mmmmmlψ(d(z,Tz))+φ(z)+φ(Tz)}
and limk→∞M(xn(k),z)=ψ(d(z,Tz))+φ(Tz), since φ(z)=0.
Lettingk→∞ in (35), we get
(37)ψ(d(z,Tz))+φ(Tz)≤γ(a,b)[ψ(d(z,Tz))+φ(Tz)].
Thus, ψ(d(z,Tz))+φ(Tz)=0. That is, Tz=z.


## 3. Corollaries and Examples

The following are the possible corollaries of Theorem11.Corollary 12.
Let(X,⪯) be a partially ordered set. Suppose that T:X→X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T-orbitally complete. Assume that there exists x0∈X such that x0⪯Tx0. Suppose that there exist a lower semi-continuous function φ:X→[0,∞), ψ∈Ψ, and k∈[0,1) such that
(38)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)≤kM(x,y),
where
(39)M(x,y)=max{ψ(d(x,y))+φ(x)+φ(y),mmmψ(d(x,Tx))+φ(x)+φ(Tx),mmllψ(d(y,Ty))+φ(y)+φ(Ty)}
for each x,y∈𝒪(x0)¯ with x⪯y.
Then, the sequence{xn} defined by xn+1=Txn,n=0,1,2,…, is Cauchy in X. Let limn→∞xn=z. Suppose that either (a) or (b) of Theorem 11 holds, then z is a fixed point of T and φ(z)=0.Corollary 13.
Let(X,⪯) be a partially ordered set. Suppose that T:X→X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T-orbitally complete metric space. Assume that there exists x0∈X such that x0⪯Tx0. Suppose that there exist a lower semi-continuous function φ:X→[0,∞), ψ∈Ψ such that the following condition holds.
“For each0≤a<b<∞, there exists γ(a,b)∈[0,1) such that a≤ψ(d(x,y))+φ(x)+φ(y)≤b implies
(40)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)≤γ(a,b)[ψ(d(x,y))+φ(x)+φ(y)]
for each x,y∈𝒪(x0)¯ with x⪯y.”
Then, the sequence{xn} defined by xn+1=Txn,n=0,1,2,…, is Cauchy in X. Let limn→∞xn=z. Suppose that either (a) or (b) of Theorem 11 holds, then z is a fixed point of T and φ(z)=0.Proof.
Since inequality (40) implies inequality (6); the conclusion of the corollary follows from Theorem 11.Corollary 14.
Let(X,⪯) be a partially ordered set. Suppose that T:X→X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T-orbitally complete. Assume that there exists x0∈X such that x0⪯Tx0. Suppose that there exist a lower semi-continuous function φ:X→[0,∞), ψ∈Ψ such that the following condition holds.
“For each0≤a<b<∞, there exist α(a,b),β(a,b)∈[0,1) with α(a,b)+β(a,b)<1 such that a≤ψ(d(x,y))+φ(x)+φ(y)≤b implies
(41)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)≤α(a,b)[ψ(d(x,Tx))+φ(x)+φ(Tx)]+β(a,b)[ψ(d(y,Ty))+φ(y)+φ(Ty)]
for each x,y∈𝒪(x0)¯ with x⪯y.”
Then, the sequence{xn} defined by xn+1=Txn,n=0,1,2,…, is Cauchy in X. Let limn→∞xn=z. Suppose that either (a) or (b) of Theorem 11 holds, then z is a fixed point of T and φ(z)=0.Proof.
Inequality (41) implies inequality (6). Hence, the conclusion follows from Theorem 11.As an immediate consequence, we obtain the following corollary.Corollary 15.
Let(X,⪯) be a partially ordered set. Suppose that T:X→X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T-orbitally complete. Assume that there exists x0∈X such that x0⪯Tx0. Suppose that there exist a lower semi-continuous function φ:X→[0,∞), ψ∈Ψ, and α,β∈[0,1) with α+β<1 such that
(42)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)≤α[ψ(d(x,Tx))+φ(x)+φ(Tx)]+β[ψ(d(y,Ty))+φ(y)+φ(Ty)]
for each x,y∈𝒪(x0)¯ with x⪯y.
Then, the sequence{xn} defined by xn+1=Txn,n=0,1,2,…, is Cauchy in X. Let limn→∞xn=z. Suppose that either (a) or (b) of Theorem 11 holds, then z is a fixed point of T and φ(z)=0.Corollary 16.
Let(X,⪯) be a partially ordered set. Suppose that T:X→X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T-orbitally complete. Assume that there exists x0∈X such that x0⪯Tx0. Suppose that there exist a lower semi-continuous function φ:X→[0,∞), ψ∈Ψ such that the following condition holds.
“For each0≤a<b<∞ there exist α(a,b),β(a,b), and γ(a,b)∈[0,1) with α(a,b)+β(a,b)+γ(a,b)<1 such that a≤ψ(d(x,y))+φ(x)+φ(y)≤b implies
(43)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)≤α(a,b)[ψ(d(x,y))+φ(x)+φ(y)]+β(a,b)[ψ(d(x,Tx))+φ(x)+φ(Tx)]+γ(a,b)[ψ(d(y,Ty))+φ(y)+φ(Ty)]
for each x,y∈𝒪(x0)¯ with x⪯y.”
Then, the sequence{xn} defined by xn+1=Txn,n=0,1,2,…, is Cauchy in X. Let limn→∞xn=z. Suppose that either (a) or (b) of Theorem 11 holds, then z is a fixed point of T and φ(z)=0.Proof.
The conclusion follows from Theorem11, since inequality (43) implies inequality (6).We now obtain the following corollary from Corollary16.Corollary 17.
Let(X,⪯) be a partially ordered set and d a metric on X. Suppose that T:X→X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T-orbitally complete. Assume that there exists x0∈X such that x0⪯Tx0. Suppose that there exist a lower semi-continuous function φ:X→[0,∞), ψ∈Ψ and there exist α,β, and γ∈[0,1) with α+β+γ<1 such that
(44)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)≤α[ψ(d(x,y))+φ(x)+φ(y)]+β[ψ(d(x,Tx))+φ(x)+φ(Tx)]+γ[ψ(d(y,Ty))+φ(y)+φ(Ty)]
for each x,y∈𝒪(x0)¯ with x⪯y.
Then, the sequence{xn} defined by xn+1=Txn,n=0,1,2,…, is Cauchy in X. Let limn→∞xn=z. Suppose that either (a) or (b) of Theorem 11 holds, then z is a fixed point of T and φ(z)=0.Remark 18.
By choosingψ(t)=t in Corollaries 13, 15, and 17, we obtain partially ordered versions of Theorems 1, 2, and 3, respectively.Remark 19.
By choosingφ≡0 and ψ(t)=t, t≥0, in Corollary 12, we obtain Theorem 5 as a corollary to Corollary 12, which in turn Theorem 5 follows as a corollary to Theorem 11.Theorem 20.
Let(X,⪯) be a partially ordered set. Suppose that T1:X→X is a non-decreasing selfmap on X. Assume that there is a metric d on X such that X is T1-orbitally complete. Assume that there exists x0∈X such that x0⪯T1x0. Let {Ti/Ti:X→X,i=1,2,3,…} be a sequence of maps and suppose that there exist a lower semi-continuous function φ:X→[0,∞), ψ∈Ψ, and k∈[0,1) such that
(45)ψ(d(T1x,Tjy))+φ(T1x)+φ(Tjy)≤kM(x,y),
where
(46)M(x,y)=max{(Tjy)ψ(d(x,y))+φ(x)+φ(y),mmmmmllψ(d(x,T1x))+φ(x)+φ(T1x),mmmmmlψ(d(y,Tjy))+φ(y)+φ(Tjy)}
for each x,y∈𝒪(x0)¯ with x⪯y and j=1,2,3,….
If either(a) or (b) of Theorem 11 holds, then the family of maps {Tj/j=1,2,3,…} has a common fixed point.Proof.
Ifj=1, then T1 has a fixed point by Corollary 12. Let that fixed point be u (say). Then, φ(u)=0. We show that Tju=u for each j=2,3,4,…. Let j>1 be fixed such that Tju≠u.
Now, using (45), we get
(47)ψ(d(T1u,Tju))+φ(T1u)+φ(Tju)≤kM(u,u),
where
(48)M(u,u)=max{(Tjy)ψ(d(u,u))+φ(u)+φ(u),0000002.00ψ(d(u,T1u))+φ(u)+φ(T1u),000000200ψ(d(u,Tju))+φ(u)+φ(Tju)}=ψ(d(u,Tju))+φ(Tju).
Then, from (47), we obtain
(49)ψ(d(T1u,Tju))+φ(Tju)≤k[ψ(d(T1u,Tju))+φ(Tju)],
a contradiction. Hence, Tju=u for each j=1,2,3,….
Thus, the conclusion of the theorem follows.The following is an example in support of Theorem11.Example 21.
LetX=[0,2) with the usual metric. We define a partial order ⪯ on X by
(50)⪯∶={(x,y):x,y∈[0,1),x≥y}∪{(x,y):x,y∈[1,2),x=y}.
We define T:X→X, ψ:[0,∞)→[0,∞), and φ:X→[0,∞) by
(51)T(x)={x2,ifx∈[0,1),1+x2,ifx∈[1,2),ψ(t)=t2,t≥0,φ(x)={x2,if0≤x≤1,34x,if1<x<2.
Let x0=1/4; then, 𝒪(x0)={(1/2n)/n≥2,n∈Z+} and 𝒪(x0)¯=𝒪(x0)∪{0}.
AndTx0=1/8 so that x0⪯Tx0.
We now verify inequality (6).
In this example, we chooseγ(a,b)=3/4 for each a,b∈[0,∞), 0≤a<b satisfying a≤ψ(d(x,y))+φ(x)+φ(y)≤b for all x,y∈𝒪(x0)¯.
Letx,y∈𝒪(x0)¯.
Case (i) (x=1/2n, y=1/2m, and n≥m≥2). Suppose that a≤ψ(d(x,y))+φ(x)+φ(y)≤b. That is, a≤(1/2)[(1/2m)-(1/2n)]+(1/2n+1)+(1/2m+1)≤b. That is, a≤1/2m≤b. Now,
(52)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)=12m+1≤3412m=γ(a,b)[ψ(d(x,y))+φ(x)+φ(y)]≤γ(a,b)M(x,y).
Case (ii) (x=1/2n, n≥2, and y=0). Suppose that a≤ψ(d(x,y))+φ(x)+φ(y)≤b; that is, a≤1/2n≤b. Now,
(53)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)=12n+1≤3412n=γ(a,b)[ψ(d(x,y))+φ(x)+φ(y)]≤γ(a,b)M(x,y).
Case (iii) (x=y=1/2n,n≥2). Suppose that a≤ψ(d(x,y))+φ(x)+φ(y)≤b; that is, a≤1/2n≤b. Now,
(54)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)=12n+1.
Hence the verification of inequality (6) is the same as in Case (ii).
Case (iv) (x=y=0). In this case, ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)=0 so that inequality (6) holds trivially.From all the above cases, it is clear that inequality (6) holds for all x,y∈𝒪(x0)¯. Hence, all the hypotheses of Theorem 11 hold and 0 is the fixed point of T in 𝒪(x0)¯. Also, φ(0)=0.Now, forx=0 and y=1, suppose that there exists 0≤a<b<∞ such that a≤ψ(d(x,y))+φ(x)+φ(y)=ψ(1)+φ(0)+φ(1)<b. Then, we have
(55)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)=ψ(1)+φ(0)+φ(1)≰γ(a,b)[ψ(d(x,y))+φ(x)+φ(y)]
so that inequality (1) fails to hold for any φ and γ(a,b)∈[0,1). Hence Theorem 1 is not applicable.Example 22.
LetX=[0,1) with the usual metric. We define a partial order ⪯ on X by
“x⪯y if and only if x≥y for each x,y∈X.”
We defineT:X→X, ψ:[0,∞)→[0,∞), and φ:X→[0,∞) by
(56)T(x)={0,ifx=0,14,if12≤x≤1,316,if14≤x<12,3212n+1,if12n+1≤x<12n,n≥2,ψ(t)=2t,t≥0,φ(x)={x6,if0≤x<316,x-316,if316≤x<1.
Let x0=2/3. Then, Tx0=1/4 so that x0⪯Tx0. Consider
(57)O(x0)={23,14,316,316,316,…},O(x0)¯=O(x0).
We now verify inequality (6).
We chooseγ(a,b)=12/69 for each 0≤a<b<∞ satisfying a≤ψ(d(x,y))+φ(x)+φ(y)≤b for all x,y∈𝒪(x0)¯.
Letx,y∈𝒪(x0)¯.
Case (i) (x=2/3,y=1/4). Suppose that a≤ψ(d(x,y))+φ(x)+φ(y)≤b; that is, a≤11/8≤b. Now,
(58)M(x,y)=max{ψ(d(x,y))+φ(x)+φ(y),mmmmmmψ(d(x,Tx))+φ(x)+φ(Tx),mmmmmllψ(d(y,Ty))+φ(y)+φ(Ty)}=max{118,118,316}=118,ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)=316<1269118=γ(a,b)M(x,y).
Case (ii) (x=2/3,y=3/16). Suppose that a≤ψ(d(x,y))+φ(x)+φ(y)≤b; that is, a≤69/48≤b. Now,
(59)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)=ψ(116)+φ(14)+φ(316)=316,M(x,y)=max{ψ(d(x,y))+φ(x)+φ(y),mmmmmm.ψ(d(x,Tx))+φ(x)+φ(Tx),mmmmmllψ(d(y,Ty))+φ(y)+φ(Ty)}=max{6948,118,0}=6948
so that
(60)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)≤γ(a,b)M(x,y).
Case (iii) (x=1/4, y=3/16). Suppose that a≤ψ(d(x,y))+φ(x)+φ(y)≤b; that is, a≤3/16≤b. Now,
(61)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)=ψ(0)+φ(316)+φ(316)=0.
Hence, (6) holds trivially.
Case (iv) (x=y=2/3). Suppose that a≤ψ(d(x,y))+φ(x)+φ(y)≤b; that is, a≤23/24≤b. Now,
(62)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)=ψ(0)+φ(14)+φ(14)=18,M(x,y)=max{ψ(d(x,y))+φ(x)+φ(y),mmmmmllψ(d(x,Tx))+φ(x)+φ(Tx),mmmmmlψ(d(y,Ty))+φ(y)+φ(Ty)}=max{2324,118,118}=118.
Then,
(63)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)=18<1269118=γ(a,b)M(x,y).
Case (v). (let x=y=1/4). Suppose that a≤ψ(d(x,y))+φ(x)+φ(y)≤b; that is, a≤1/8≤b. Now,
(64)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)=ψ(0)+φ(316)+φ(316)=0.
Hence, (6) holds trivially.Hence, all the hypotheses of Theorem11 hold and 3/16 is the fixed point of T in 𝒪(x0)¯. Also, φ(3/16)=0.We observe that atx=0 and y=3/16,
(65)ψ(d(Tx,Ty))+φ(Tx)+φ(Ty)=ψ(316)+φ(0)+φ(316),ψ(d(x,y))+φ(x)+φ(y)=ψ(316)+φ(0)+φ(316),ψ(d(x,Tx))+φ(x)+φ(Tx)+ψ(d(y,Ty))+φ(y)+φ(Ty)=2[φ(0)+φ(316)].
Hence, inequality (2) does not hold for any φ,ψ, and γ∈[0,1/2). Therefore, Theorem 2 is not applicable.Also,ψ(d(Tx,Ty))=ψ(3/16) and ψ(d(x,y))=3/16. Hence, ψ(d(Tx,Ty))≤kψ(d(x,y)) does not hold for any ψ∈Ψ and k∈[0,1). Thus, inequality (4) fails to hold, and hence Theorem 5 is not applicable.Thus, by Remark19, we conclude that Theorem 11 is a generalization of Theorem 5.



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*Source: 404573-2013-12-18.xml*