# Common Fixed Point Results Using Generalized Altering Distances on Orbitally Complete Ordered Metric Spaces

Authors: Hemant Kumar Nashine; Zoran Kadelburg; Zorana Golubović
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We prove the existence of common fixed points for three relatively asymptotically regular mappings defined on an orbitally complete ordered metric space using orbital continuity of one of the involved maps. We furnish a suitable example to demonstrate the validity of the hypotheses of our results.


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## Body

## 1. Introduction and Preliminaries

Browder and Petryshyn introduced the concept of asymptotic regularity of a self-map at a point in a metric space.Definition 1.1 (see [1]).
A self-map𝒯 on a metric space (𝒳,d) is said to be asymptotically regular at a point x∈𝒳 if limn→∞d(𝒯nx,𝒯n+1x)=0.Recall that the set𝒪(x0;𝒯)={𝒯nx0:n=0,1,2,…} is called the orbit of the self-map 𝒯 at the point x0∈𝒳.Definition 1.2 (see [2]).
A metric space(𝒳,d) is said to be 𝒯-orbitally complete if every Cauchy sequence contained in 𝒪(x;𝒯) (for some x in 𝒳) converges in 𝒳.Here, it can be pointed out that every complete metric space is𝒯-orbitally complete for any 𝒯, but a 𝒯-orbitally  complete metric space need not be complete.Definition 1.3 (see [1]).
A self-map𝒯 defined on a metric space (𝒳,d) is said to be orbitally continuous at a point z in 𝒳 if for any sequence {xn}⊂𝒪(x;𝒯) (for some x∈𝒳), xn→z as n→∞ implies 𝒯xn→𝒯z as n→∞.Clearly, every continuous self-mapping of a metric space is orbitally continuous, but not conversely.Sastry et al. [3] extended the above concepts to two and three mappings and employed them to prove common fixed point results for commuting mappings. In what follows, we collect such definitions for three maps.Definition 1.4 (see [3]).
Let𝒮,𝒯,ℛ be three self-mappings defined on a metric space (𝒳,d).(1)
If for a pointx0∈𝒳, there exists a sequence {xn} in 𝒳 such that ℛx2n+1=𝒮x2n, ℛx2n+2=𝒯x2n+1, n=0,1,2,…, then the set 𝒪(x0;𝒮,𝒯,ℛ)={ℛxn:n=1,2,…} is called the orbit of (𝒮,𝒯,ℛ) at x0.(2)
The space(𝒳,d) is said to be (𝒮,𝒯,ℛ)-orbitally complete at x0 if every Cauchy sequence in 𝒪(x0;𝒮,𝒯,ℛ) converges in 𝒳.(3)
The mapℛ is said to be orbitally continuous at x0 if it is continuous on 𝒪(x0;𝒮,𝒯,ℛ).(4)
The pair(𝒮,𝒯) is said to be asymptotically regular (in short a.r.) with respect to ℛ at x0 if there exists a sequence {xn} in 𝒳 such that ℛx2n+1=𝒮x2n, ℛx2n+2=𝒯x2n+1, n=0,1,2,…, and d(ℛxn,ℛxn+1)→0 as n→∞.On the other side, Khan et al. [4] introduced the notion of an altering distance function, which is a control function that alters distance between two points in a metric space. This notion has been used by several authors to establish fixed point results in a number of subsequent works, some of which are noted in [5–9]. In [5], Choudhury introduced the concept of a generalized altering distance function in three variables which was further generalized by Rao et al. [10] to four variables and is defined as follows.Definition 1.5 (see [10]).
A functionψ:[0,+∞)4→[0,+∞) is said to be a generalized altering distance function if(i)
ψ is continuous,(ii)
ψ is nondecreasing in each variable,(iii)
ψ(t1,t2,t3,t4)=0⇔t1=t2=t3=t4=0.ℱ4 will denote the set of all functions ψ satisfying conditions (i)–(iii).Simple examples of generalized altering distance functions with four variables are(1.1)ψ(t1,t2,t3,t4)=kmax⁡{t1,t2,t3,t4},k>0;ψ(t1,t2,t3,t4)=max⁡{t1,t2,t3,t4}1+max⁡{t1,t2,t3,t4};ψ(t1,t2,t3,t4)=t1p+t2q+t3r+t4s,p,q,r,s≥1.On the other hand, fixed point theory has developed rapidly in metric spaces endowed with a partial ordering. The first result in this direction was given by Ran and Reurings [11] who presented its applications to matrix equations. Subsequently, Nieto and Rodríguez-López [12] extended this result for nondecreasing mappings and applied it to obtain a unique solution for a first order ordinary differential equation with periodic boundary conditions. Thereafter, several authors obtained many fixed point theorems in ordered metric spaces. For more details see [13–20] and the references cited therein.In this paper, an attempt has been made to derive some common fixed point theorems for three relatively asymptotically regular mappings defined on an orbitally complete ordered metric space, using orbital continuity of one of the involved maps and conditions involving a generalized altering distance function. The presented theorems generalize, extend, and improve some recent results given in [7, 14, 21, 22]. In the hypotheses, we have considered the space as not necessarily complete, the maps ℛ,𝒮, and 𝒯 as not necessarily continuous and the range of 𝒮 and 𝒯 may not be contained in the range of ℛ.


## 2. Results

### 2.1. Notations and Definitions

First, we introduce some further notations and definitions that will be used later.If(𝒳,≼) is a partially ordered set, then x,y∈𝒳 are called comparable if x≼y or y≼x holds. A subset 𝒦 of 𝒳 is said to be well ordered if every two elements of 𝒦 are comparable. If 𝒯:𝒳→𝒳 is such that, for x,y∈𝒳, x≼y implies 𝒯x≼𝒯y, then the mapping 𝒯 is said to be nondecreasing.Definition 2.1.
Let(𝒳,≼) be a partially ordered set and 𝒮,𝒯:𝒳→𝒳.(1)
[23] The pair (𝒮,𝒯) is called weakly increasing if 𝒮x≼𝒯𝒮x and 𝒯x≼𝒮𝒯x for all x∈𝒳.(2)
[24] The pair (𝒮,𝒯) is called partially weakly increasing if 𝒮x≼𝒯𝒮x for all x∈𝒳.(3)
[24] The mapping 𝒮 is called a weak annihilator of 𝒯 if 𝒮𝒯x≼x for all x∈𝒳.(4)
[24] The mapping 𝒮 is called dominating if x≼𝒮x for each x∈𝒳.Note that none of two weakly increasing mappings need to be nondecreasing. There exist some examples to illustrate this fact in [23]. Obviously, the pair (𝒮,𝒯) is weakly increasing if and only if the ordered pairs (𝒮,𝒯) and (𝒯,𝒮) are partially weakly increasing. Following is an example of an ordered pair (𝒮,𝒯) which is partially weakly increasing but not weakly increasing.Example 2.2 (see [24]).
Let𝒳=[0,1] be endowed with usual ordering.(1)
Let𝒮,𝒯:𝒳→𝒳 be defined by 𝒮x=x2 and 𝒯x=x. Clearly, (𝒮,𝒯) is partially weakly increasing. But 𝒯x=x≰x=𝒮𝒯x for x∈(0,1) implies that (𝒯,𝒮) is not partially weakly increasing.(2)
Let𝒮,𝒯:𝒳→𝒳 be defined by 𝒮x=x2 and 𝒯x=x3. Obviously, 𝒮𝒯x=x6≤x for all x∈𝒳. Thus 𝒮 is a weak annihilator of 𝒯.(3)
Let𝒮:𝒳→𝒳 be defined by 𝒮x=xn. Since x≤xn=𝒮x for all x∈𝒳, 𝒮 is a dominating map.Definition 2.3 (see [25, 26]).
Let(𝒳,d) be a metric space and f,g:𝒳→𝒳. The mappings f and g are said to be compatible if limn→∞d(fgxn,gfxn)=0, whenever {xn} is a sequence in 𝒳 such that limn→∞fxn=limn→∞gxn=t for some t∈𝒳.Definition 2.4.
Let𝒳 be a nonempty set. Then (𝒳,d,≼) is called an ordered metric space if(i)
(𝒳,d) is a metric space,(ii)
(𝒳,≼) is a partially ordered set.


### 2.2. Main Results

The first main result is as follows.Theorem 2.5.
Let(𝒳,d,≼) be an ordered metric space. Let 𝒮,𝒯,ℛ:𝒳→𝒳 be given mappings satisfying
(2.1)Ψ1(d(Sx,Ty))≤ψ1(M[S,T,R](x,y))-ψ2(M[S,T,R](x,y)),
for all x,y∈𝒪(x0;𝒮,𝒯,ℛ)¯ (for some x0) such that ℛx and ℛy are comparable, where
(2.2)M[S,T,R](x,y)=(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)]),
and  ψ1 and ψ2 are generalized altering distance functions (in ℱ4) and Ψ1(t)=ψ1(t,t,t,t). We assume the following hypotheses:(i)
(𝒮,𝒯) is a.r. with respect to ℛ at x0∈𝒳;(ii)
𝒳 is (𝒮,𝒯,ℛ)-orbitally complete at x0;(iii)
(ℛ,𝒮) and (ℛ,𝒯) are partially weakly increasing;(iv)
𝒮 and 𝒯 are dominating maps;(v)
𝒮 and 𝒯 are weak annihilators of ℛ;(vi)
for a nondecreasing sequence{xn}, xn≼yn for all n and yn→u as n→∞ imply that xn≼u for all n∈ℕ. Assume either(a)
𝒮 and ℛ are compatible; 𝒮 or ℛ is orbitally continuous at x0 or(b)
𝒯 and ℛ are compatible; 𝒯 or ℛ is orbitally continuous at x0.Then 𝒮,𝒯 and ℛ have a common fixed point. Moreover, the set of common fixed points of 𝒮,𝒯, and ℛ in 𝒪(x0;𝒮,𝒯,ℛ)¯ is well ordered if and only if it is a singleton.Proof.
Since(𝒮,𝒯) is a.r. with respect to ℛ at x0 in 𝒳, there exists a sequence {xn} in 𝒳 such that
(2.3)Sx2n-2=Rx2n-1,Tx2n-1=Rx2n,∀n∈N.
By the given assumptions, x2n-2≼𝒮x2n-2=ℛx2n-1≼𝒮ℛx2n-1≼x2n-1, and x2n-1≼𝒯x2n-1=ℛx2n≼𝒯ℛx2n≼x2n. Thus, for all n≥1, we have
(2.4)xn≼xn+1.
In view of (i), we have
(2.5)lim⁡n→∞d(Rxn,Rxn+1)=0.
Now, we assert that {ℛxn} is a Cauchy sequence in the metric space 𝒪(x0;𝒮,𝒯,ℛ).
From (2.5), it will be sufficient to prove that {ℛx2n} is a Cauchy sequence. We proceed by negation and suppose that {ℛx2n} is not a Cauchy sequence. Then, there exists ε>0 for which we can find two sequences of positive integers {m(i)} and {n(i)} such that for all positive integers i,
(2.6)n(i)>m(i)>i,d(Rx2m(i),Rx2n(i))≥ε,d(Rx2m(i),Rx2n(i)-2)<ε.
From (2.6) and using the triangular inequality, we get
(2.7)ε≤d(Rx2m(i),Rx2n(i))≤d(Rx2m(i),Rx2n(i)-2)+d(Rx2n(i)-2,Rx2n(i)-1)+d(Rx2n(i)-1,Rx2n(i))<ε+d(Rx2n(i)-2,Rx2n(i)-1)+d(Rx2n(i)-1,Rx2n(i)).
Letting i→∞ in the above inequality and using (2.5), we obtain
(2.8)lim⁡i→∞d(Rx2m(i),Rx2n(i))=ε.
Again, the triangular inequality gives us
(2.9)|d(Rx2n(i),Rx2m(i)-1)-d(Rx2n(i),Rx2m(i))|≤d(Rx2m(i)-1,Rx2m(i)).
Letting i→∞ in the above inequality and using (2.5) and (2.8), we get
(2.10)lim⁡i→∞d(Rx2n(i),Rx2m(i)-1)=ε.
Similarly, we have
(2.11)lim⁡i→∞d(Rx2n(i)+1,Rx2m(i)-1)=ε.
On the other hand, we have
(2.12)d(Rx2n(i),Rx2m(i))≤d(Rx2n(i),Rx2n(i)+1)+d(Rx2n(i)+1,Rx2m(i))=d(Rx2n(i),Rx2n(i)+1)+d(Tx2n(i),Sx2m(i)-1).
Then, from (2.5), (2.8), and the continuity of Ψ1, we get by letting i→∞ in (2.1)
(2.13)Ψ1(ε)≤lim⁡i→∞Ψ1(d(Sx2m(i)-1,Tx2n(i))).
Now, using the considered contractive condition (2.1) for x=x2m(i)-1 and y=x2n(i), we have
(2.14)Ψ1(d(Sx2m(i)-1,Tx2n(i)))≤ψ1(12d(Rx2m(i)-1,Rx2n(i)),d(Rx2m(i)-1,Rx2m(i)),d(Rx2n(i),Rx2n(i)+1),12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))])-ψ2(12d(Rx2m(i)-1,Rx2n(i)),d(Rx2m(i)-1,Rx2m(i)),d(Rx2n(i),Rx2n(i)+1),12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))]).
Then, from (2.5), (2.10), (2.11), and the continuity of ψ1 and ψ2, we get by letting i→∞ in the above inequality
(2.15)lim⁡i→∞Ψ1(d(Sx2m(i)-1,Tx2n(i)))≤ψ1(ε,0,0,ϵ)-ψ2(ε,0,0,ϵ)≤Ψ1(ε)-ψ2(ε,0,0,ϵ).
Now, combining (2.13) with the above inequality, we get
(2.16)Ψ1(ε)≤Ψ1(ε)-ψ2(ε,0,0,ϵ),
which implies that ψ2(ε,0,0,ϵ)=0, which is a contradiction since ε>0. Hence {ℛxn} is a Cauchy sequence in 𝒪(x0;𝒮,𝒯,ℛ). Since 𝒳 is (𝒮,𝒯,ℛ)-orbitally complete at x0, there exists some z∈𝒳 such that ℛxn→z as n→∞.
Finally, we prove the existence of a common fixed point of the three mappings𝒮,𝒯, and ℛ.
We have(2.17)Rx2n+1=Sx2n⟶zasn⟶∞,Rx2n+2=Tx2n+1⟶zasn⟶∞.
Suppose that (a) holds. Since {𝒮,ℛ} are compatible, we have
(2.18)lim⁡n→∞SRx2n+2=lim⁡n→∞RSx2n+2=Rz.
Also, x2n+1≼𝒯x2n+1=ℛx2n+2. Now
(2.19)Ψ1(d(SRx2n+2,Tx2n+1))≤ψ1(12d(RRx2n+2,Rx2n+1),d(RRx2n+2,SRx2n+2),d(Rx2n+1,Tx2n+1),12[d(RRx2n+2,Tx2n+1)+d(SRx2n+2,Rx2n+1)])-ψ2(12d(RRx2n+2,Rx2n+1),d(RRx2n+2,SRx2n+2),d(Rx2n+1,Tx2n+1),12[d(RRx2n+2,Tx2n+1)+d(SRx2n+2,Rx2n+1)]).
Assume that ℛ is orbitally continuous. Passing to the limit as n→∞, we obtain
(2.20)Ψ1(d(Rz,z))≤ψ1(d(Rz,z),0,0,d(Rz,z))-ψ2(d(Rz,z),0,0,d(Rz,z))≤Ψ1(d(Rz,z))-ψ2(d(Rz,z),0,0,d(Rz,z)),
so ψ2(d(ℛz,z),0,0,d(ℛz,z))=0, which implies that
(2.21)Rz=z.
Now, x2n+1≼𝒯x2n+1 and 𝒯x2n+1→z as n→∞, so by the assumption we have x2n+1≼z and (2.1) becomes
(2.22)Ψ1(d(Sz,Tx2n+1))≤ψ1(12d(Rz,Rx2n+1),d(Sz,Rz),d(Tx2n+1,Rx2n+1),12[d(Rz,Tx2n+1)+d(Sz,Rx2n+1)])-ψ2(12d(Rz,Rx2n+1),d(Sz,Rz),d(Tx2n+1,Rx2n+1),12[d(Rz,Tx2n+1)+d(Sz,Rx2n+1)]).
Passing to the limit as n→∞ in the above inequality and using (2.21), it follows that
(2.23)Ψ1(d(Sz,z))≤ψ1(0,d(Sz,z),0,12d(Sz,z))-ψ2(0,d(Sz,z),0,12d(Sz,z))≤Ψ1(d(Sz,z))-ψ2(0,d(Sz,z),0,12d(Sz,z)),
which holds unless ψ2(0,d(𝒮z,z),0,(1/2)d(𝒮z,z))=0, so
(2.24)Sz=z.
Now, since x2n≼𝒮x2n and 𝒮x2n→z as n→∞  implies that x2n≼z, from (2.1)
(2.25)Ψ1(d(Sx2n,Tz))≤ψ1(d(Rx2n,Rz),d(Rx2n,Sx2n),d(Rz,Tz),12(d(Rx2n,Tz)+d(Sx2n,Rz)))-ψ2(d(Rx2n,Rz),d(Rx2n,Sx2n),d(Rz,Tz),12(d(Rx2n,Tz)+d(Sx2n,Rz))).
Passing to the limit as n→∞, we have
(2.26)Ψ1(d(z,Tz))≤ψ1(0,0,d(z,Tz),d(z,Tz))-ψ2(d(z,Tz),0,0,d(z,Tz))≤Ψ1(d(z,Tz))-ψ2(0,0,d(z,Tz),d(z,Tz)),
which gives that
(2.27)z=Tz.
Therefore, 𝒮z=𝒯z=ℛz=z, hence z is a common fixed point of ℛ,𝒮, and 𝒯. The proof is similar when 𝒮 is orbitally continuous.
Similarly, the result follows when condition (b) holds.
Now, suppose that the set of common fixed points of𝒮,𝒯, and ℛ in 𝒪(x0;𝒮,𝒯,ℛ)¯ is well ordered. We claim that it cannot contain more than one point. Assume to the contrary that 𝒮u=𝒯u=ℛu=u and 𝒮v=𝒯v=ℛv=v but u≠v. By supposition, we can replace x by u and y by v in (2.1) to obtain
(2.28)Ψ1(d(u,v))=Ψ1(d(Su,Tv))≤ψ1(d(Ru,Rv),d(Ru,Su),d(Rv,Tv),12[d(Ru,Tv)+d(Su,Rv)])-ψ2(d(Ru,Rv),d(Ru,Su),d(Rv,Tv),12[d(Ru,Tv)+d(Su,Rv)])=ψ1(d(u,v),0,0,d(u,v))-ψ2(d(u,v),0,0,d(u,v))<Ψ1(d(u,v)),
a contradiction. Hence, u=v. The converse is trivial.Now, it is easy to state a corollary of Theorem2.5 involving a contraction of integral type.Corollary 2.6.
Let𝒮,𝒯, and ℛ satisfy the conditions of Theorem 2.5, except that condition (2.1) is replaced by the following: there exists a positive Lebesgue integrable function u on ℝ+ such that ∫0εu(t)dt>0 for each ε>0 and that
(2.29)∫0Ψ1(d(Sx,Ty))u(t)dt≤∫0ψ1(M[S,T,R](x,y))u(t)dt-∫0ψ2(M[S,T,R](x,y))u(t)dt.
Then, 𝒮,𝒯, and ℛ have a common fixed point. Moreover, the set of common fixed points of 𝒮,𝒯, and ℛ in 𝒪(x0;𝒮,𝒯,ℛ)¯ is well ordered if and only if it is a singleton.Remark 2.7.
If we take(2.30)ψ1(t1,t2,t3,t4)=max⁡{t1,t2,t3,t4},ψ2(t1,t2,t3,t4)=(1-k)max⁡{t1,t2,t3,t4},
for k∈(0,1), then Ψ1(t)=t for all t≥0, and the contractive condition (2.1) becomes
(2.31)d(Sx,Ty)≤kmax⁡{d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)]},
which corresponds to the contraction given by Theorem  2.1 in [24] by taking ψ(t)=t and φ(t)=(1-k)t. Hence, the result of Abbas et al. [24] is covered by Theorem 2.5 for three maps.
Other results could be derived for other choices ofψ1 and ψ2.As consequences of Theorem2.5, we may state the following corollaries.Corollary 2.8.
Let(𝒳,d,≼) be an ordered metric space. Let 𝒯,ℛ:𝒳→𝒳 be given mappings satisfying
(2.32)Ψ1(d(Tx,Ty))≤ψ1(d(Rx,Ry),d(Rx,Tx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Tx)])-ψ2(d(Rx,Ry),d(Rx,Tx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Tx)]),
for all x,y∈𝒪(x0;𝒯,𝒯,ℛ)¯ such that ℛx and ℛy are comparable, where ψ1 and ψ2 are generalized altering distance functions (in ℱ4) and Ψ1(t)=ψ1(t,t,t,t). We assume the following hypotheses:(i)
𝒯 is a.r. with respect to ℛ at x0;(ii)
𝒳 is (𝒯,ℛ)-orbitally complete at x0;(iii)
𝒯 or ℛ is orbitally continuous at x0;(iv)
(𝒯,ℛ) is partially weakly increasing;(v)
𝒯 is a dominating map;(vi)
𝒯 is a weak annihilator of ℛ;(vii)
𝒯 and ℛ are compatible.Let for a nondecreasing sequence {xn} with xn≼yn  for all n, yn→u as n→∞ imply that xn≼u for all n∈ℕ.
Then𝒯 and ℛ have a common fixed point. Moreover, the set of common fixed points of 𝒯 and ℛ in 𝒪(x0;𝒯,𝒯,ℛ)¯ is well ordered if and only if it is a singleton.Corollary 2.9.
Let(𝒳,d,≼) be an ordered metric space. Let  𝒯:𝒳→𝒳 be a mapping satisfying
(2.33)Ψ1(d(Tx,Ty))≤ψ1(d(x,y),d(x,Tx),d(y,Ty),12[d(x,Ty)+d(y,Tx)])-ψ2(d(x,y),d(x,Tx),d(y,Ty),12[d(x,Ty)+d(y,Tx)]),
for all (x,y)∈𝒪(x0;𝒯)¯ such that x and y are comparable, where ψ1 and ψ2 are generalized altering distance functions (in ℱ4) and Ψ1(t)=ψ1(t,t,t,t). We assume the following hypotheses:(i)
𝒯 is a.r. at some point x0 of 𝒳;(ii)
𝒳 is 𝒯-orbitally complete at x0;(iii)
𝒯 is a dominating map. Let for a nondecreasing sequence {xn} with xn≼yn  for all n, yn→u as n→∞ imply that xn≼u for all n∈ℕ.
Then𝒯 has a fixed point. Moreover, the set of fixed points of 𝒯 in 𝒪(x0;𝒯)¯ is well ordered if and only if it is a singleton.We present an example showing the usage of our results.Example 2.10.
Let the set𝒳=[0,+∞) be equipped with the usual metric d and the order defined by
(2.34)x≼y⟺x≥y.
Consider the following self-mappings on 𝒳:
(2.35)Rx=6x,Sx={12x,0≤x≤12,x,x>12,Tx={13x,0≤x≤13,x,x>13.
Take x0=1/2. Then it is easy to show that
(2.36)O(x0;S,T,R)⊂{12k⋅3l:k,l∈N},O(x0;S,T,R)¯=O(x0;S,T,R)∪{0}
and all the conditions (i)–(vi) and (a)-(b) of Theorem 2.5 are fulfilled. Take ψ1(t1,t2,t3,t4)=max{t1,t2,t3,t4} and ψ2(t1,t2,t3,t4)=(5/6)max{t1,t2,t3,t4}. Then contractive condition (2.1) takes the form
(2.37)|12x-13y|≤16max⁡{|6x-6y|,112x,173y,12[|6x-13y|+|6y-12x|]},
for x,y∈𝒪(x0;𝒮,𝒯,ℛ). Using substitution y=tx, t>0, the last inequality reduces to
(2.38)|3-2t|≤max⁡{6|1-t|,112,173t,12[|6-13t|+|6t-12|]},
and can be checked by discussion on possible values for t>0. Hence, all the conditions of Theorem 2.5 are satisfied and 𝒮,𝒯,ℛ have a common fixed point (which is 0).Remark 2.11.
It was shown by examples in [22] that (in similar situations)(1)
if the contractive condition is satisfied just on𝒪(x0;𝒮,𝒯,ℛ), there might not exist a (common) fixed point;(2)
under the given hypotheses (common), fixed point might not be unique in the whole space𝒳.


## 2.1. Notations and Definitions

First, we introduce some further notations and definitions that will be used later.If(𝒳,≼) is a partially ordered set, then x,y∈𝒳 are called comparable if x≼y or y≼x holds. A subset 𝒦 of 𝒳 is said to be well ordered if every two elements of 𝒦 are comparable. If 𝒯:𝒳→𝒳 is such that, for x,y∈𝒳, x≼y implies 𝒯x≼𝒯y, then the mapping 𝒯 is said to be nondecreasing.Definition 2.1.
Let(𝒳,≼) be a partially ordered set and 𝒮,𝒯:𝒳→𝒳.(1)
[23] The pair (𝒮,𝒯) is called weakly increasing if 𝒮x≼𝒯𝒮x and 𝒯x≼𝒮𝒯x for all x∈𝒳.(2)
[24] The pair (𝒮,𝒯) is called partially weakly increasing if 𝒮x≼𝒯𝒮x for all x∈𝒳.(3)
[24] The mapping 𝒮 is called a weak annihilator of 𝒯 if 𝒮𝒯x≼x for all x∈𝒳.(4)
[24] The mapping 𝒮 is called dominating if x≼𝒮x for each x∈𝒳.Note that none of two weakly increasing mappings need to be nondecreasing. There exist some examples to illustrate this fact in [23]. Obviously, the pair (𝒮,𝒯) is weakly increasing if and only if the ordered pairs (𝒮,𝒯) and (𝒯,𝒮) are partially weakly increasing. Following is an example of an ordered pair (𝒮,𝒯) which is partially weakly increasing but not weakly increasing.Example 2.2 (see [24]).
Let𝒳=[0,1] be endowed with usual ordering.(1)
Let𝒮,𝒯:𝒳→𝒳 be defined by 𝒮x=x2 and 𝒯x=x. Clearly, (𝒮,𝒯) is partially weakly increasing. But 𝒯x=x≰x=𝒮𝒯x for x∈(0,1) implies that (𝒯,𝒮) is not partially weakly increasing.(2)
Let𝒮,𝒯:𝒳→𝒳 be defined by 𝒮x=x2 and 𝒯x=x3. Obviously, 𝒮𝒯x=x6≤x for all x∈𝒳. Thus 𝒮 is a weak annihilator of 𝒯.(3)
Let𝒮:𝒳→𝒳 be defined by 𝒮x=xn. Since x≤xn=𝒮x for all x∈𝒳, 𝒮 is a dominating map.Definition 2.3 (see [25, 26]).
Let(𝒳,d) be a metric space and f,g:𝒳→𝒳. The mappings f and g are said to be compatible if limn→∞d(fgxn,gfxn)=0, whenever {xn} is a sequence in 𝒳 such that limn→∞fxn=limn→∞gxn=t for some t∈𝒳.Definition 2.4.
Let𝒳 be a nonempty set. Then (𝒳,d,≼) is called an ordered metric space if(i)
(𝒳,d) is a metric space,(ii)
(𝒳,≼) is a partially ordered set.


## 2.2. Main Results

The first main result is as follows.Theorem 2.5.
Let(𝒳,d,≼) be an ordered metric space. Let 𝒮,𝒯,ℛ:𝒳→𝒳 be given mappings satisfying
(2.1)Ψ1(d(Sx,Ty))≤ψ1(M[S,T,R](x,y))-ψ2(M[S,T,R](x,y)),
for all x,y∈𝒪(x0;𝒮,𝒯,ℛ)¯ (for some x0) such that ℛx and ℛy are comparable, where
(2.2)M[S,T,R](x,y)=(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)]),
and  ψ1 and ψ2 are generalized altering distance functions (in ℱ4) and Ψ1(t)=ψ1(t,t,t,t). We assume the following hypotheses:(i)
(𝒮,𝒯) is a.r. with respect to ℛ at x0∈𝒳;(ii)
𝒳 is (𝒮,𝒯,ℛ)-orbitally complete at x0;(iii)
(ℛ,𝒮) and (ℛ,𝒯) are partially weakly increasing;(iv)
𝒮 and 𝒯 are dominating maps;(v)
𝒮 and 𝒯 are weak annihilators of ℛ;(vi)
for a nondecreasing sequence{xn}, xn≼yn for all n and yn→u as n→∞ imply that xn≼u for all n∈ℕ. Assume either(a)
𝒮 and ℛ are compatible; 𝒮 or ℛ is orbitally continuous at x0 or(b)
𝒯 and ℛ are compatible; 𝒯 or ℛ is orbitally continuous at x0.Then 𝒮,𝒯 and ℛ have a common fixed point. Moreover, the set of common fixed points of 𝒮,𝒯, and ℛ in 𝒪(x0;𝒮,𝒯,ℛ)¯ is well ordered if and only if it is a singleton.Proof.
Since(𝒮,𝒯) is a.r. with respect to ℛ at x0 in 𝒳, there exists a sequence {xn} in 𝒳 such that
(2.3)Sx2n-2=Rx2n-1,Tx2n-1=Rx2n,∀n∈N.
By the given assumptions, x2n-2≼𝒮x2n-2=ℛx2n-1≼𝒮ℛx2n-1≼x2n-1, and x2n-1≼𝒯x2n-1=ℛx2n≼𝒯ℛx2n≼x2n. Thus, for all n≥1, we have
(2.4)xn≼xn+1.
In view of (i), we have
(2.5)lim⁡n→∞d(Rxn,Rxn+1)=0.
Now, we assert that {ℛxn} is a Cauchy sequence in the metric space 𝒪(x0;𝒮,𝒯,ℛ).
From (2.5), it will be sufficient to prove that {ℛx2n} is a Cauchy sequence. We proceed by negation and suppose that {ℛx2n} is not a Cauchy sequence. Then, there exists ε>0 for which we can find two sequences of positive integers {m(i)} and {n(i)} such that for all positive integers i,
(2.6)n(i)>m(i)>i,d(Rx2m(i),Rx2n(i))≥ε,d(Rx2m(i),Rx2n(i)-2)<ε.
From (2.6) and using the triangular inequality, we get
(2.7)ε≤d(Rx2m(i),Rx2n(i))≤d(Rx2m(i),Rx2n(i)-2)+d(Rx2n(i)-2,Rx2n(i)-1)+d(Rx2n(i)-1,Rx2n(i))<ε+d(Rx2n(i)-2,Rx2n(i)-1)+d(Rx2n(i)-1,Rx2n(i)).
Letting i→∞ in the above inequality and using (2.5), we obtain
(2.8)lim⁡i→∞d(Rx2m(i),Rx2n(i))=ε.
Again, the triangular inequality gives us
(2.9)|d(Rx2n(i),Rx2m(i)-1)-d(Rx2n(i),Rx2m(i))|≤d(Rx2m(i)-1,Rx2m(i)).
Letting i→∞ in the above inequality and using (2.5) and (2.8), we get
(2.10)lim⁡i→∞d(Rx2n(i),Rx2m(i)-1)=ε.
Similarly, we have
(2.11)lim⁡i→∞d(Rx2n(i)+1,Rx2m(i)-1)=ε.
On the other hand, we have
(2.12)d(Rx2n(i),Rx2m(i))≤d(Rx2n(i),Rx2n(i)+1)+d(Rx2n(i)+1,Rx2m(i))=d(Rx2n(i),Rx2n(i)+1)+d(Tx2n(i),Sx2m(i)-1).
Then, from (2.5), (2.8), and the continuity of Ψ1, we get by letting i→∞ in (2.1)
(2.13)Ψ1(ε)≤lim⁡i→∞Ψ1(d(Sx2m(i)-1,Tx2n(i))).
Now, using the considered contractive condition (2.1) for x=x2m(i)-1 and y=x2n(i), we have
(2.14)Ψ1(d(Sx2m(i)-1,Tx2n(i)))≤ψ1(12d(Rx2m(i)-1,Rx2n(i)),d(Rx2m(i)-1,Rx2m(i)),d(Rx2n(i),Rx2n(i)+1),12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))])-ψ2(12d(Rx2m(i)-1,Rx2n(i)),d(Rx2m(i)-1,Rx2m(i)),d(Rx2n(i),Rx2n(i)+1),12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))]).
Then, from (2.5), (2.10), (2.11), and the continuity of ψ1 and ψ2, we get by letting i→∞ in the above inequality
(2.15)lim⁡i→∞Ψ1(d(Sx2m(i)-1,Tx2n(i)))≤ψ1(ε,0,0,ϵ)-ψ2(ε,0,0,ϵ)≤Ψ1(ε)-ψ2(ε,0,0,ϵ).
Now, combining (2.13) with the above inequality, we get
(2.16)Ψ1(ε)≤Ψ1(ε)-ψ2(ε,0,0,ϵ),
which implies that ψ2(ε,0,0,ϵ)=0, which is a contradiction since ε>0. Hence {ℛxn} is a Cauchy sequence in 𝒪(x0;𝒮,𝒯,ℛ). Since 𝒳 is (𝒮,𝒯,ℛ)-orbitally complete at x0, there exists some z∈𝒳 such that ℛxn→z as n→∞.
Finally, we prove the existence of a common fixed point of the three mappings𝒮,𝒯, and ℛ.
We have(2.17)Rx2n+1=Sx2n⟶zasn⟶∞,Rx2n+2=Tx2n+1⟶zasn⟶∞.
Suppose that (a) holds. Since {𝒮,ℛ} are compatible, we have
(2.18)lim⁡n→∞SRx2n+2=lim⁡n→∞RSx2n+2=Rz.
Also, x2n+1≼𝒯x2n+1=ℛx2n+2. Now
(2.19)Ψ1(d(SRx2n+2,Tx2n+1))≤ψ1(12d(RRx2n+2,Rx2n+1),d(RRx2n+2,SRx2n+2),d(Rx2n+1,Tx2n+1),12[d(RRx2n+2,Tx2n+1)+d(SRx2n+2,Rx2n+1)])-ψ2(12d(RRx2n+2,Rx2n+1),d(RRx2n+2,SRx2n+2),d(Rx2n+1,Tx2n+1),12[d(RRx2n+2,Tx2n+1)+d(SRx2n+2,Rx2n+1)]).
Assume that ℛ is orbitally continuous. Passing to the limit as n→∞, we obtain
(2.20)Ψ1(d(Rz,z))≤ψ1(d(Rz,z),0,0,d(Rz,z))-ψ2(d(Rz,z),0,0,d(Rz,z))≤Ψ1(d(Rz,z))-ψ2(d(Rz,z),0,0,d(Rz,z)),
so ψ2(d(ℛz,z),0,0,d(ℛz,z))=0, which implies that
(2.21)Rz=z.
Now, x2n+1≼𝒯x2n+1 and 𝒯x2n+1→z as n→∞, so by the assumption we have x2n+1≼z and (2.1) becomes
(2.22)Ψ1(d(Sz,Tx2n+1))≤ψ1(12d(Rz,Rx2n+1),d(Sz,Rz),d(Tx2n+1,Rx2n+1),12[d(Rz,Tx2n+1)+d(Sz,Rx2n+1)])-ψ2(12d(Rz,Rx2n+1),d(Sz,Rz),d(Tx2n+1,Rx2n+1),12[d(Rz,Tx2n+1)+d(Sz,Rx2n+1)]).
Passing to the limit as n→∞ in the above inequality and using (2.21), it follows that
(2.23)Ψ1(d(Sz,z))≤ψ1(0,d(Sz,z),0,12d(Sz,z))-ψ2(0,d(Sz,z),0,12d(Sz,z))≤Ψ1(d(Sz,z))-ψ2(0,d(Sz,z),0,12d(Sz,z)),
which holds unless ψ2(0,d(𝒮z,z),0,(1/2)d(𝒮z,z))=0, so
(2.24)Sz=z.
Now, since x2n≼𝒮x2n and 𝒮x2n→z as n→∞  implies that x2n≼z, from (2.1)
(2.25)Ψ1(d(Sx2n,Tz))≤ψ1(d(Rx2n,Rz),d(Rx2n,Sx2n),d(Rz,Tz),12(d(Rx2n,Tz)+d(Sx2n,Rz)))-ψ2(d(Rx2n,Rz),d(Rx2n,Sx2n),d(Rz,Tz),12(d(Rx2n,Tz)+d(Sx2n,Rz))).
Passing to the limit as n→∞, we have
(2.26)Ψ1(d(z,Tz))≤ψ1(0,0,d(z,Tz),d(z,Tz))-ψ2(d(z,Tz),0,0,d(z,Tz))≤Ψ1(d(z,Tz))-ψ2(0,0,d(z,Tz),d(z,Tz)),
which gives that
(2.27)z=Tz.
Therefore, 𝒮z=𝒯z=ℛz=z, hence z is a common fixed point of ℛ,𝒮, and 𝒯. The proof is similar when 𝒮 is orbitally continuous.
Similarly, the result follows when condition (b) holds.
Now, suppose that the set of common fixed points of𝒮,𝒯, and ℛ in 𝒪(x0;𝒮,𝒯,ℛ)¯ is well ordered. We claim that it cannot contain more than one point. Assume to the contrary that 𝒮u=𝒯u=ℛu=u and 𝒮v=𝒯v=ℛv=v but u≠v. By supposition, we can replace x by u and y by v in (2.1) to obtain
(2.28)Ψ1(d(u,v))=Ψ1(d(Su,Tv))≤ψ1(d(Ru,Rv),d(Ru,Su),d(Rv,Tv),12[d(Ru,Tv)+d(Su,Rv)])-ψ2(d(Ru,Rv),d(Ru,Su),d(Rv,Tv),12[d(Ru,Tv)+d(Su,Rv)])=ψ1(d(u,v),0,0,d(u,v))-ψ2(d(u,v),0,0,d(u,v))<Ψ1(d(u,v)),
a contradiction. Hence, u=v. The converse is trivial.Now, it is easy to state a corollary of Theorem2.5 involving a contraction of integral type.Corollary 2.6.
Let𝒮,𝒯, and ℛ satisfy the conditions of Theorem 2.5, except that condition (2.1) is replaced by the following: there exists a positive Lebesgue integrable function u on ℝ+ such that ∫0εu(t)dt>0 for each ε>0 and that
(2.29)∫0Ψ1(d(Sx,Ty))u(t)dt≤∫0ψ1(M[S,T,R](x,y))u(t)dt-∫0ψ2(M[S,T,R](x,y))u(t)dt.
Then, 𝒮,𝒯, and ℛ have a common fixed point. Moreover, the set of common fixed points of 𝒮,𝒯, and ℛ in 𝒪(x0;𝒮,𝒯,ℛ)¯ is well ordered if and only if it is a singleton.Remark 2.7.
If we take(2.30)ψ1(t1,t2,t3,t4)=max⁡{t1,t2,t3,t4},ψ2(t1,t2,t3,t4)=(1-k)max⁡{t1,t2,t3,t4},
for k∈(0,1), then Ψ1(t)=t for all t≥0, and the contractive condition (2.1) becomes
(2.31)d(Sx,Ty)≤kmax⁡{d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)]},
which corresponds to the contraction given by Theorem  2.1 in [24] by taking ψ(t)=t and φ(t)=(1-k)t. Hence, the result of Abbas et al. [24] is covered by Theorem 2.5 for three maps.
Other results could be derived for other choices ofψ1 and ψ2.As consequences of Theorem2.5, we may state the following corollaries.Corollary 2.8.
Let(𝒳,d,≼) be an ordered metric space. Let 𝒯,ℛ:𝒳→𝒳 be given mappings satisfying
(2.32)Ψ1(d(Tx,Ty))≤ψ1(d(Rx,Ry),d(Rx,Tx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Tx)])-ψ2(d(Rx,Ry),d(Rx,Tx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Tx)]),
for all x,y∈𝒪(x0;𝒯,𝒯,ℛ)¯ such that ℛx and ℛy are comparable, where ψ1 and ψ2 are generalized altering distance functions (in ℱ4) and Ψ1(t)=ψ1(t,t,t,t). We assume the following hypotheses:(i)
𝒯 is a.r. with respect to ℛ at x0;(ii)
𝒳 is (𝒯,ℛ)-orbitally complete at x0;(iii)
𝒯 or ℛ is orbitally continuous at x0;(iv)
(𝒯,ℛ) is partially weakly increasing;(v)
𝒯 is a dominating map;(vi)
𝒯 is a weak annihilator of ℛ;(vii)
𝒯 and ℛ are compatible.Let for a nondecreasing sequence {xn} with xn≼yn  for all n, yn→u as n→∞ imply that xn≼u for all n∈ℕ.
Then𝒯 and ℛ have a common fixed point. Moreover, the set of common fixed points of 𝒯 and ℛ in 𝒪(x0;𝒯,𝒯,ℛ)¯ is well ordered if and only if it is a singleton.Corollary 2.9.
Let(𝒳,d,≼) be an ordered metric space. Let  𝒯:𝒳→𝒳 be a mapping satisfying
(2.33)Ψ1(d(Tx,Ty))≤ψ1(d(x,y),d(x,Tx),d(y,Ty),12[d(x,Ty)+d(y,Tx)])-ψ2(d(x,y),d(x,Tx),d(y,Ty),12[d(x,Ty)+d(y,Tx)]),
for all (x,y)∈𝒪(x0;𝒯)¯ such that x and y are comparable, where ψ1 and ψ2 are generalized altering distance functions (in ℱ4) and Ψ1(t)=ψ1(t,t,t,t). We assume the following hypotheses:(i)
𝒯 is a.r. at some point x0 of 𝒳;(ii)
𝒳 is 𝒯-orbitally complete at x0;(iii)
𝒯 is a dominating map. Let for a nondecreasing sequence {xn} with xn≼yn  for all n, yn→u as n→∞ imply that xn≼u for all n∈ℕ.
Then𝒯 has a fixed point. Moreover, the set of fixed points of 𝒯 in 𝒪(x0;𝒯)¯ is well ordered if and only if it is a singleton.We present an example showing the usage of our results.Example 2.10.
Let the set𝒳=[0,+∞) be equipped with the usual metric d and the order defined by
(2.34)x≼y⟺x≥y.
Consider the following self-mappings on 𝒳:
(2.35)Rx=6x,Sx={12x,0≤x≤12,x,x>12,Tx={13x,0≤x≤13,x,x>13.
Take x0=1/2. Then it is easy to show that
(2.36)O(x0;S,T,R)⊂{12k⋅3l:k,l∈N},O(x0;S,T,R)¯=O(x0;S,T,R)∪{0}
and all the conditions (i)–(vi) and (a)-(b) of Theorem 2.5 are fulfilled. Take ψ1(t1,t2,t3,t4)=max{t1,t2,t3,t4} and ψ2(t1,t2,t3,t4)=(5/6)max{t1,t2,t3,t4}. Then contractive condition (2.1) takes the form
(2.37)|12x-13y|≤16max⁡{|6x-6y|,112x,173y,12[|6x-13y|+|6y-12x|]},
for x,y∈𝒪(x0;𝒮,𝒯,ℛ). Using substitution y=tx, t>0, the last inequality reduces to
(2.38)|3-2t|≤max⁡{6|1-t|,112,173t,12[|6-13t|+|6t-12|]},
and can be checked by discussion on possible values for t>0. Hence, all the conditions of Theorem 2.5 are satisfied and 𝒮,𝒯,ℛ have a common fixed point (which is 0).Remark 2.11.
It was shown by examples in [22] that (in similar situations)(1)
if the contractive condition is satisfied just on𝒪(x0;𝒮,𝒯,ℛ), there might not exist a (common) fixed point;(2)
under the given hypotheses (common), fixed point might not be unique in the whole space𝒳.



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*Source: 382094-2012-05-22.xml*