# Tripled Coincidence Point Theorems for Nonlinear Contractions in Partially Ordered Metric Spaces

Authors: Binayak S. Choudhury; Erdal Karapınar; Amaresh Kundu
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

Tripled fixed points are extensions of the idea of coupled fixed points introduced in a recent paper by Berinde and Borcut, 2011. Here using a separate methodology we extend this result to a triple coincidence point theorem in partially ordered metric spaces. We have defined several concepts pertaining to our results. The main results have several corollaries and an illustrative example. The example shows that the extension proved here is actual and also the main theorem properly contains all its corollaries.


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## Body

## 1. Introduction and Preliminaries

In recent times coupled fixed point theory has experienced a rapid growth in partially ordered metric spaces. The speciality of this line of research is that the problems herein utilize both order theoretic and analytic methods. References [1–19] are some instances of these works.Definition 1.1 (see [14]).
A functiong:R→R is said to be monotone nondecreasing (or increasing) if x≥y implies g(x)≥g(y).Definition 1.2 (see [14]).
LetX be a nonempty set. Let F:X×X→X be a mapping. An element (x,y) is called a coupled fixed point of F if
(1.1)F(x,y)=x,F(y,x)=y.Recently, Berinde and Borcut [20] extended the idea of coupled fixed points to tripled fixed points. The definition is as follows.Definition 1.3 (see [20]).
LetX be a nonempty set. Let F:X×X×X→X be a mapping. An element (x,y,z) is called a tripled fixed point of F if
(1.2)F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z.
They also extended the mixed monotone property to functions with three arguments.Definition 1.4 (see [20]).
Let(X,⪯) be a partially ordered set and F:X×X×X→X. The mapping F is said to have the mixed monotone property if for any x,y,z∈X(1.3)x1,x2∈X,x1⪯x2⟹F(x1,y,z)⪯F(x2,y,z),y1,y2∈X,y1⪯y2⟹F(x,y1,z)⪰F(x,y2,z),z1,z2∈X,z1⪯z2⟹F(x,y,z1)⪯F(x,y,z2).Our purpose here is to establish tripled coincidence point results in metric spaces with partial ordering. For that purpose we define mixedg-monotone property in the following. Mixed g-monotone property was already defined in the context of coupled fixed points [14]. Here in the spirit of Definition 1.4 we have made an extension of that.Definition 1.5.
Let(X,⪯) be a partially ordered set. Let g:X→X and F:X×X×X→X. The mapping F is said to have the mixed g-monotone property if for any x,y,z∈X. 							(1.4)x1,x2∈X,gx1⪯gx2⟹F(x1,y,z)⪯F(x2,y,z),y1,y2∈X,gy1⪯gy2⟹F(x,y1,z)⪰F(x,y2,z),z1,z2∈X,gz1⪯gz2⟹F(x,y,z1)⪯F(x,y,z2).Coupled coincidence point was defined by Lakshmikantham and Ćirić [14]. We also extend the concept of coupled coincidence point to tripled coincidence point in the following.Definition 1.6.
LetX be any nonempty set. Let g:X→X and F:X×X×X→X. An element (x,y,z) is called a tripled coincidence point of g and F if
(1.5)F(x,y,z)=gx,F(y,x,y)=gy,F(z,y,x)=gz.We extend the concept of commuting mappings given by Lakshmikantham and Ćirić [14], in the following definition.Definition 1.7.
LetX be a nonempty set. Then one says that the mappings g:X→X and F:X×X×X→X are commuting if for all x,y,z∈X(1.6)g(F(x,y,z))=F(gx,gy,gz).The following is the definition of compatible mappings which is an extension of the compatibility defined by Choudhury and Kundu in [8].Definition 1.8 (see [8]).
Let(X,d) be a metric space. The mappings g and F, where g:X→X and F:X×X×X→X are said to be compatible if
(1.7)lim⁡n→∞d(gF(xn,yn,zn),F(gxn,gyn,gzn))=0,lim⁡n→∞d(gF(yn,xn,yn),F(gyn,gxn,gyn))=0,lim⁡n→∞d(gF(zn,yn,xn),F(gzn,gyn,gxn))=0,
whenever {xn},{yn},{zn} are sequences in X such that
(1.8)lim⁡n→∞F(xn,yn,zn)=gxn=x,lim⁡n→∞F(yn,xn,yn)=gyn=y,lim⁡n→∞F(zn,yn,xn)=gzn=z.


## 2. Main Results

Theorem 2.1.
Let(X,⪯) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose F:X×X×X→X and g:X→X are such that, g is monotone increasing, F has the mixed g-monotone property and
(2.1)d(F(x,y,z),F(u,v,w))≤ψ(max⁡{d(gx,gu),d(gy,gv),d(gz,gw)})
for all x,y,z∈X for which gx⪯gu, gy≽gv and gz⪯gw, where ψ:[0,+∞)→[0,+∞) is such that ψ(t) is monotone, ψ(t)<t and limr→t+ψ(r)<t for all t>0. Suppose F(X×X×X)⊆g(X), g is continuous, and {g,F} is a compatible pair. Suppose either  (a)
Fis continuous or(b)
X has the following properties:
(i)
if a nondecreasing sequence{αn}→α, then αn⪯α for all n,(ii)
if a nonincreasing sequence{βn}→β, then βn≽β for all n.
If there existx0,y0,z0∈X such that gx0⪯F(x0,y0,z0), gy0≽F(y0,x0,y0), and gz0⪯F(z0,y0,x0), then there exist x,y,z∈X such that
(2.2)F(x,y,z)=gx,F(y,x,y)=gy,F(z,y,x)=gz,
that is, g and F have a tripled coincidence point.Proof.
By a condition of the theorem, there existx0,y0,z0∈X such that gx0⪯F(x0,y0,z0), gy0≽F(y0,x0,y0), and gz0⪯F(z0,y0,x0). Since F(X×X×X)⊆g(X), we can choose x1,y1,z1∈X such that
(2.3)gx1=F(x0,y0,z0),gy1=F(y0,x0,y0),gz1=F(z0,y0,x0).
Continuing this process, we can construct sequences {xn},{yn}, and {zn} in X such that
(2.4)gxn+1=F(xn,yn,zn),gyn+1=F(yn,xn,yn),gzn+1=F(zn,yn,xn).
Next we will show that, for n≥0,
(2.5)gxn⪯gxn+1,gyn⪰gyn+1,gzn⪯gzn+1.
Since, gx0⪯F(x0,y0,z0),gy0≽F(y0,x0,y0), and gz0⪯F(z0,y0,x0), by (2.3), we get
(2.6)gx0⪯gx1,gy0⪰gy1,gz0⪯gz1,
that is, (2.5) holds for n=0.
We presume that (2.5) holds for some n=m>0. As F has the mixed g-monotone property and gxm⪯gxm+1,gym≽gym+1 and gzm⪯gzm+1, we obtain
(2.7)gxm+1=F(xm,ym,zm)⪯F(xm+1,ym,zm)⪯F(xm+1,ym,zm+1)⪯F(xm+1,ym+1,zm+1)=gxm+2,(2.8)gym+1=F(ym,xm,ym)⪰F(ym,xm,ym+1)⪰F(ym+1,xm,ym+1)⪰F(ym+1,xm+1,ym+1)=gym+2,(2.9)gzm+1=F(zm,ym,xm)⪯F(zm+1,ym,xm)⪯F(zm+1,ym+1,xm)⪯F(zm+1,ym+1,xm+1)=gzm+2.
Thus, (2.5) holds for n=m+1. Then, by induction, we conclude that (2.5) holds for n≥1.
If for somen∈ℕ,
(2.10)gxn=gxn+1,gyn=gyn+1,gzn=gzn+1,
then, by (2.4), (xn,yn,zn) is a tripled coincidence point of g and F. Therefore we assume, for any n∈ℕ,
(2.11)(gxn,gyn,gzn)≠(gxn+1,gyn+1,gzn+1).
Set δn=max{d(gxn,gxn+1),d(gyn,gyn+1),d(gzn,gzn+1)}.
Then(2.12)δn>0∀n≥0.
Then, by (2.1), (2.4) and (2.5), we have
(2.13)d(gxn,gxn+1)=d(F(xn-1,yn-1,zn-1),F(xn,yn,zn))≤ψ(max⁡{d(gxn-1,gxn),d(gyn-1,gyn),d(gzn-1,gzn)}),d(gyn,gyn+1)=d(F(yn-1,xn-1,yn-1),F(yn,xn,yn))≤ψ(max⁡{d(gyn-1,gyn),d(gxn-1,gxn),d(gyn-1,gyn)}),d(gzn,gzn+1)=d(F(zn-1,yn-1,xn-1),F(zn,yn,xn))≤ψ(max⁡{d(gzn-1,gzn),d(gyn-1,gyn),d(gxn-1,gxn)}).
Thus, from (2.13) we obtain that
(2.14)δn=max⁡{d(gxn,gxn+1),d(gyn,gyn+1),d(gzn,gzn+1)}≤ψ(max⁡{d(gxn-1,gxn),d(gyn-1,gyn),d(gzn-1,gzn)}).
It then follows from (2.12) and a property ψ, that for all n≥1,
(2.15)δn≤ψ(δn-1)<δn-1.
Thus, {δn} is a monotone decreasing sequence of nonnegative real numbers. So, there exist a δ≥0 such that
(2.16)lim⁡n→∞δn=δ.
Suppose δ>0. Letting n→∞ in (2.14),  using (2.15), (2.16), and a property of ψ, we get
(2.17)δ≤ψ(δ)<δ,
which is a contradiction. Thus δ=0, or
(2.18)lim⁡n→∞δn=0,
or
(2.19)lim⁡n→∞d(gxn+1,gxn)=0,lim⁡n→∞d(gyn+1,gyn)=0,lim⁡n→∞d(gzn+1,gzn)=0.
Now, we will prove that {gxn}, {gyn}, and {gzn} are Cauchy sequences. Suppose, to the contrary, that at least one of {gxn}, {gyn}, and {gzn} is not a Cauchy sequence. So, there exists an ɛ>0 for which we can find subsequences {gxn(k)} of {gxn}, {gyn(k)} of {gyn}, and {gzn(k)} of {gzn} with n(k)>m(k)≥k such that
(2.20)αk=max⁡{d(gxn(k),gxm(k)),d(gyn(k),gym(k)),d(gzn(k),gzm(k))}≥ɛ.
Additionally, corresponding to m(k), we may choose n(k) such that it is the smallest integer satisfying (2.20). Then, for all k≥0,
(2.21)max⁡{d(gxn(k)-1,gxm(k)),d(gyn(k)-1,gym(k)),d(gzn(k)-1,gzm(k))}<ɛ.
By using (2.20) and (2.21) we have for k≥0,
(2.22)ɛ≤αk=max⁡{d(gxn(k),gxm(k)),d(gyn(k),gym(k)),d(gzn(k),gzm(k))}≤max⁡{d(gxn(k),gxn(k)-1)+d(gxn(k)-1,gxm(k)),d(gyn(k),gyn(k)-1)+d(gyn(k)-1,gym(k)),d(gzn(k),gzn(k)-1)+d(gzn(k)-1,gzm(k))}≤max⁡{d(gxn(k),gxn(k)-1),d(gyn(k),gyn(k)-1),d(gzn(k),gzn(k)-1)}+ɛ≤δn(k)-1+ɛ.
Letting k→∞ in (2.22), and using (2.19), we get
(2.23)lim⁡k→∞αk=lim⁡k→∞max⁡{d(gxn(k),gxm(k)),d(gyn(k),gym(k)),d(gzn(k),gzm(k))}=ɛ.
Let, for k≥0,
(2.24)βk=max⁡{d(gxn(k)+1,gxm(k)+1),d(gyn(k)+1,gym(k)+1),d(gzn(k)+1,gzm(k)+1)}.
Again, for allk≥0,
(2.25)αk=max⁡⁡{d(gxn(k),gxm(k)),d(gyn(k),gym(k)),d(gzn(k),gzm(k))}≤max⁡{d(gxn(k),gxn(k)+1)+d(gxn(k)+1,gxm(k)+1)+d(gxm(k)+1,gxm(k)),d(gyn(k),gyn(k)+1)+d(gyn(k)+1,gym(k)+1)+d(gym(k)+1,gym(k)),d(gzn(k),gzn(k)+1)+d(gzn(k)+1,gzm(k)+1)+d(gzm(k)+1,gzm(k))}≤max⁡⁡{d(gxn(k),gxn(k)+1),d(gyn(k),gyn(k)+1),d(gzn(k),gzn(k)+1)}+max⁡⁡{d(gxn(k)+1,gxm(k)+1),d(gyn(k)+1,gym(k)+1),d(gzn(k)+1,gzm(k)+1)}+max⁡⁡{d(gxm(k),gxm(k)+1),d(gym(k),gym(k)+1),d(gzm(k),gzm(k)+1)}≤δn(k)+1+βk+δm(k)+1.
Analogously we have for k≥0,(2.26)βk=max⁡{d(gxn(k)+1,gxm(k)+1),d(gyn(k)+,gym(k)+1),d(gzn(k)+1,gzm(k)+1)}≤max⁡⁡{d(gxn(k)+1,gxn(k))+d(gxn(k),gxm(k))+d(gxm(k),gxm(k)+1),d(gyn(k)+1,gyn(k))+d(gyn(k),gym(k))+d(gym(k),gym(k)+1),d(gzn(k)+1,gzn(k))+d(gzn(k),gzm(k))+d(gzm(k),gzm(k)+1)}≤max⁡{d(gxn(k),gxn(k)+1),d(gyn(k),gyn(k)+1),d(gzn(k),gzn(k)+1)}+max⁡{d(gxn(k),gxm(k)),d(gyn(k),gym(k)),d(gzn(k),gzm(k))}+max⁡{d(gxm(k),gxm(k)+1),d(gym(k),gym(k)+1),d(gzm(k),gzm(k)+1)}≤δn(k)+1+αk+δm(k)+1.
Letting k→∞ in (2.25) and (2.26), we get that
(2.27)lim⁡k→∞max⁡{d(gxn(k)+1,gxm(k)+1),d(gyn(k)+1,gym(k)+1),d(gzn(k)+1,gzm(k)+1)}=lim⁡k→∞βk=ɛ=lim⁡k→∞αk.
Since n(k)>m(k), for k≥0, we have
(2.28)gxn(k)⪰gxm(k),gyn(k)⪯gym(k),gzn(k)⪰gzm(k).
Then from (2.1), (2.4), and (2.28), we have for k≥0,
(2.29)d(gxn(k)+1,gxm(k)+1)=d(F(xn(k),yn(k),zn(k)),F(xm(k),ym(k),zm(k)))≤ψ(max⁡⁡{d(gxn(k),gxm(k)),d(gyn(k),gym(k)),d(gzn(k),gzm(k))}),d(gyn(k)+1,gym(k)+1)=d(F(yn(k),xn(k),yn(k)),F(ym(k),xm(k),ym(k)))≤ψ(max⁡⁡{d(gyn(k),gym(k)),d(gxn(k),gxm(k)),d(gyn(k),gym(k))}),d(gzn(k)+1,gzm(k)+1)=d(F(zn(k),yn(k),xn(k)),F(zm(k),ym(k),xm(k)))≤ψ(max⁡⁡{d(gzn(k),gzm(k)),d(gyn(k),gym(k)),d(gxn(k),gxm(k))}).
From (2.29) for k≥0, we get
(2.30)βk≤ψ(max⁡{d(gxn(k),gxm(k)),d(gyn(k),gym(k)),d(gzn(k),gzm(k))})=ψ(αk).
Letting k→∞ in (2.30), using (2.20), (2.27), and a property of ψ, we get
(2.31)ɛ≤ψ(ɛ)<ɛ,
which is a contradiction. This shows that {gxn}, {gyn}, and {gzn} are Cauchy sequences.
SinceX is complete, there exist x,y,z∈X such that
(2.32)lim⁡n→∞gxn=x,lim⁡n→∞gyn=y,lim⁡n→∞gzn=z.
From (2.4) and (2.32), using the continuity of g, we have
(2.33)gx=lim⁡n→∞g(gxn+1)=lim⁡n→∞g(F(xn,yn,zn)),(2.34)gy=lim⁡n→∞g(gyn+1)=lim⁡n→∞g(F(yn,xn,yn)),(2.35)gz=lim⁡n→∞g(gzn+1)=lim⁡n→∞g(F(zn,yn,xn)).
Now we will show that gx=F(x,y,z),gy=F(y,x,y), and gz=F(z,y,x).
Sinceg and F are compatible,  in addition with (2.33), (2.34), and (2.35), respectively imply
(2.36)lim⁡n→∞d(g(F(xn,yn,zn)),F(g(xn),g(yn),g(zn)))=0,(2.37)lim⁡n→∞d(g(F(yn,xn,yn)),F(g(yn),g(xn),g(yn)))=0,(2.38)lim⁡n→∞d(g(F(zn,yn,xn)),F(g(zn),g(yn),g(xn)))=0.
Suppose now the assumption (a) holds, that is, F is continuous.
For alln≥0, we have
(2.39)d(gx,F(gxn,gyn,gzn))≤d(gx,g(F(xn,yn,zn)))+d(g(F(xn,yn,zn)),F(gxn,gyn,gzn)).
Taking the limit as n→∞, using (2.32), (2.33), (2.36), and the facts that g and F are continuous, we have d(gx,F(x,y,z))=0.
Similarly, by using (2.32), (2.34), and (2.37) and (2.32), (2.35), and (2.38), respectively,  and also the facts that g and F are continuous, we have d(gy,F(y,x,y))=0 and d(gz,F(z,y,x))=0.
Thus we have proved thatg and F have a tripled coincidence point.
Suppose that the assumption(b) holds. Since {gxn},{gzn} are nondecreasing and gxn→x with gzn→z and also {gyn} is nonincreasing with gyn→y, by assumption (b) we have for all n(2.40)gxn⪯x,gyn⪰y,gzn⪯z.
By virtue of monotone increasing property of g we have
(2.41)ggxn⪯gx,ggyn⪰gy,ggzn⪯gz.
Now using (2.4) we have
(2.42)d(gx,F(x,y,z))≤d(gx,g(gxn+1))+d(g(g(xn+1)),F(x,y,z))≤d(gx,g(gxn+1))+d(g(F(xn,yn,zn)),F(gxn,gyn,gzn))+d(F(gxn,gyn,gzn),F(x,y,z))≤d(gx,g(gxn+1))+d(g(F(xn,yn,zn)),F(gxn,gyn,gzn))+ψ(max⁡{d(ggxn,gx),d(ggyn,gy),d(ggzn,gz)}),(by(2.1),(2.41)).
Taking the limit as n→∞ in the above inequality, using (2.33), (2.36), and (2.41) we have
(2.43)d(gx,F(x,y,z))≤lim⁡n→∞ψ(max⁡{d(ggxn,gx),d(ggyn,gy),d(ggzn,gz)}).
By (2.33), (2.34), (2.35), and the property of ψ, we have
(2.44)d(gx,F(x,y,z))≤ψ(0)=0,
that is
(2.45)gx=F(x,y,z).
In a similar manner  using (2.33), (2.34), (2.35), and (2.36), (2.37), (2.38), respectively, we obtain
(2.46)gy=F(y,x,y),gz=F(z,y,x).
Thus, we proved that g and F have a tripled coincidence point.
This completes the proof of the theorem.Corollary 2.2.
Let(X,⪯) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose F:X×X×X→X and g:X→X are such that F has the mixed g-monotone property and
(2.47)d(F(x,y,z),F(u,v,w))≤ψ(max⁡{d(gx,gu),d(gy,gv),d(gz,gw)})
for any x,y,z∈X for which gx⪯gu, gy≽gv and gz⪯gw, where ψ:[0,+∞)→[0,+∞) be such that ψ(t) is monotone, ψ(t)<t and limr→t+ψ(r)<t for all t>0. Suppose F(X×X×X)⊆g(X), g is continuous, and F and g are commuting. Suppose either  (a)
F is continuous, or(b)
X has the  following property:
(i)
if a nondecreasing sequence{xn}→x,  then xn⪯x for all n,(ii)
if a nonincreasing sequence{yn}→y, then yn≽y for all n.
If there existx0,y0,z0∈X such that gx0⪯F(x0,y0,z0), gy0≽F(y0,x0,y0), and gz0⪯F(z0,y0,x0), then there exist x,y,z∈X such that
(2.48)F(x,y,z)=gx,F(y,x,y)=gy,F(z,y,x)=gz,
that is, F and g have a tripled coincidence point.Proof.
Since a commuting pair is also a compatible pair, the result of the Corollary2.2 follows from Theorem 2.1.Later, by an example, we will show that the Corollary2.2 is properly contained in Theorem 2.1.Corollary 2.3.
Let(X,⪯) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose F:X×X×X→X be such that F has the mixed monotone property and
(2.49)d(F(x,y,z),F(u,v,w))≤ψ(max⁡{d(x,u),d(y,v),d(z,w)})
for any x,y,z∈X for which x⪯u,y≽v and z⪯w, where ψ:[0,+∞)→[0,+∞) be such that ψ(t) is monotone, ψ(t)<t and limr→t+ψ(r)<t for all t>0. Suppose(a)
F is continuous, or(b)
X has the following property:
(i)
if a nondecreasing sequence{xn}→x, then xn⪯x for all n,(ii)
if a nonincreasing sequence{yn}→y, then yn≽y for all n.
If there existx0,y0,z0∈X such that x0⪯F(x0,y0,z0), y0≽F(y0,x0,y0), and z0⪯F(z0,y0,x0), then there exist x,y,z∈X such that
(2.50)F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z,
that is, F has a tripled fixed point.Proof.
Takingg(x)=x in Theorem 2.1  we obtain Corollary 2.3.Corollary 2.4.
Let(X,⪯) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose F:X×X×X→X and g:X→X are such that F has the mixed monotone property and
(2.51)d(F(x,y,z),F(u,v,w))≤kmax⁡{d(x,u),d(y,v),d(z,w)}
for any x,y,z∈X for which x⪯u,y≽v and z⪯w, where 0<k<1. Suppose either(a)
F is continuous,  or(b)
X has the following  property:
(i)
if a nondecreasing sequence{xn}→x, then  xn⪯x for all n,(ii)
if a nonincreasing sequence{yn}→y, then  yn≽y for all n.
If there existx0,y0,z0∈X such that x0⪯F(x0,y0,z0), y0≽F(y0,x0,y0), and z0⪯F(z0,y0,x0), then there exist x,y,z∈X such that
(2.52)F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z,
that is, F has a tripled coincidence point.Proof.
Takingψ(t)=kt, t>0 where 0<k<1, in Corollary 2.3 we obtain Corollary 2.4.The following corollary is the result of Berinde and Borcut in [20].Corollary 2.5.
Let(X,⪯) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose F:X×X×X→X be such that F has the mixed monotone property and
(2.53)d(F(x,y,z),F(u,v,w))≤a1d(x,u)+a2d(y,v)+a3d(z,w)
for any x,y,z∈X for which x⪯u,y≽v and z⪯w, where a1+a2+a3<1. Suppose either(a)
F is continuous, or(b)
X has the following property:
(i)
if a nondecreasing sequence{xn}→x, then xn⪯x for all n(ii)
if a nonincreasing sequence{yn}→y, then yn≽y for all n.
If there existx0,y0,z0∈X such that x0⪯F(x0,y0,z0), y0≽F(y0,x0,y0), and z0⪯F(z0,y0,x0), then there exist x,y,z∈X such that
(2.54)F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z,
that is, F has a tripled fixed point.Proof.
The proof follows from Corollary2.4, since the inequality in Corollary 2.5 implies that Corollary 2.4.Remark 2.6.
The method used in the proof of Corollary2.5 is different from that used by Berinde and Borcut [20].Next we discuss an example.Example 2.7.
LetX=ℜ. Then (X,⪯) is a partially ordered set with the partial ordering defined by x⪯y if and only if |x|≤|y| and x·y≥0.Letd(x,y)=|x-y| for x,y∈ℜ. Then (X,d) is a complete metric space.Letg:X→X be defined as g(x)=x2/10,for all x∈X.LetF:X×X×X→X be defined as
(2.55)F(x,y,z)=x2-y2+z29,∀x,y,z∈X.
Then F obeys the mixed g-monotone property.Letψ:[0,∞)→[0,∞) be defined as ψ(t)=(1/3)t for all t∈[0,∞).Let,{xn},{yn}, and {zn} be three sequences in X such that
(2.56)lim⁡n→∞F(xn,yn,zn)=lim⁡n→∞g(xn)=a,lim⁡n→∞F(yn,xn,yn)=lim⁡n→∞g(yn)=b,lim⁡n→∞F(zn,yn,xn)=lim⁡n→∞g(zn)=c.
Then explicitly,
(2.57)lim⁡n→∞⁡xn2-yn2+zn29=lim⁡n→∞xn210,∀x,y,z∈X,or,10a-10b+10c9=aimplya-10b+10c=0.
Again,
(2.58)lim⁡n→∞yn2-xn2+yn29=lim⁡n→∞yn210,∀x,y,z∈X,or,10b-10a+10b9=bimply11b-10a=0.
And
(2.59)lim⁡n→∞zn2-yn2+xn29=lim⁡n→∞zn210,∀x,y,z∈X,or,10c-10b+10a9=cimplyc-10b+10a=0.
Then from the above relations we have, a=0,b=0, and c=0.Therefore,(2.60)d(g(F(xn,yn,zn)),F(gxn,gyn,gzn))⟶0asn⟶∞,d(g(F(yn,xn,yn)),F(gyn,gxn,gyn))⟶0asn⟶∞,d(g(F(zn,yn,xn)),F(gzn,gyn,gxn))⟶0asn⟶∞.
Hence, the pair (g,F) is compatible in X.Also,x0=0,z0=c(>0), and y0=0 are three points in X such that g(x0)=g(0)=0<c2/9=F(0,0,c)=F(x0,y0,z0),g(y0)=g(0)=0=F(0,0,0)=F(y0,x0,y0), and g(z0)=g(c)=c2/10<c2/9=F(c,0,0)=F(z0,y0,x0).We next verify inequality (2.1) of Theorem 2.1. We take x,y,z,u,v,w∈X, such that gx⪯gu,gz⪯gw and gy≽gv, that is, x2≤u2,z2≤w2, and y2≥v2.LetA=max{d(gx,gu),d(gy,gv),d(gz,gw)}=max{|(x2-u2)|,|(y2-v2)|,|(z2-w2)|}.Thend(F(x,y,z), F(u,v,w))=d((x2-y2+z2)/9,(u2-v2+w2)/9)=(|(x2-u2)-(y2-v2)+(z2-w2))/3|≤(|(x2-u2)|+|(y2-v2)|+|(z2-w2)|)/9≤3A/9=A/3=ψ(A)=ψ(max{d(gx,gu),d(gy,gv),d(gz,gw)}).Thus it is verified that the functionsg,F, and ψ satisfy all the conditions of Theorem 2.1. Here (0,0,0) is the tripled coincidence point of g and F in X.Remark 2.8.
It is observed that in Example2.7 the function F and g do not commute, but they are compatible. Hence Corollary 2.2 cannot be applied to this example. This shows that Theorem 2.1 properly contains Corollary 2.2. Also g≠I, so the results of Berinde and Borcut [20] cannot be applied to this example. This shows that result in [20] is effectively generalised.



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*Source: 329298-2012-07-19.xml*