R E S E A R C H
Open Access
n-tuplet fixed point theorems for contractive
type mappings in partially ordered metric
spaces
Müzeyyen Ertürk
1*and Vatan Karakaya
2A correction to this article has been published:
http://www.journalofinequalitiesandapplications.com/content/2013/1/368.
*Correspondence: merturk3263@gmail.com; merturk@yildiz.edu.tr
1Department of Mathematics, The Faculty of Arts and Sciences, Yildiz Technical University, Davutpasa Campus, Esenler, Istanbul, 34210, Turkey
Full list of author information is available at the end of the article
Abstract
In this paper, we study existence and uniquennes of fixed points of operator
F:Xn→Xwherenis an arbitrary positive integer andXis partially ordered complete
metric space.
MSC: Primary 47H10; 54H25; secondary 54E50
Keywords: fixed point theorems; nonlinear contraction; partially ordered metric space;n-tuplet fixed point; mixedg-monotone
1 Introduction
As it is known, fixed point theory is one of the oldest and most famous theory in math-ematics, and it has become an important tool for other areas of science such as approxi-mation theory, statistics, engineering and economics.
Among hundreds of fixed point theorems, the Banach contraction theorem [] is par-ticularly well known due to its simplicity and usefulness. It states that any contraction mapping of a complete metric space has a unique fixed point.
In , the Banach contraction principle were extended to metric space endowed with partial order by Ran and Reuring []. They pointed out that the contractivity condition on the nonlinear and monotone map is only assumed to hold on elements which are com-parable in the partial order. Afterward, Nieto and Rodriguez-Lopez [] extended results of Ran and Reuring for non-decreasing mapping and studied existence and uniqueness of first-order differential equations.
In , by following the above mentioned trend, Bhaskar and Lakshmikantham [] in-troduced mixed monotone property and gave their coupled fixed point theorem for map-pings with mixed monotone property. Also, they produced some applications related with the existence and uniqueness of solution for a periodic boundary value problem. This work of Bhaskar and Lakshmikantham has attracted the attention of many researchers. The con-cept of coupled fixed point for various contractive type mappings was studied by several authors [–]. Lakshmikantham and Ciric [] extended the results of [] for monotone non-linear contractive mapping and generalized mixed monotone concept. Berinde and Borcut [] introduced tripled fixed point theorem for non-linear mapping in partially
dered complete metric space as a generalization and extension of the coupled fixed point theorem.
Motivated by these studies, the quadruple fixed point theorem was given for different contractive type mappings [–].
In this paper, we generalize mentioned trend in the above for an arbitrary positive num-bern, that is, we introduce the concept ofn-tuplet fixed point theorem and prove some results.
2 Main results
Let us give new definitions for our aim.
Definition Let (X,≤) be partially ordered set andF :Xn→X. We say that F has
the mixed monotone property if F(x,x,x, . . . ,xn) is monotone non-decreasing in its
odd argument and it is monotone non-increasing in its even argument. That is, for any
x,x,x, . . . ,xn∈X y,z∈X, y≤z
⇒ F(y,x,x, . . . ,xn)≤F(z,x,x, . . . ,xn), y,z∈X, y≤z
⇒ F(x,y,x, . . . ,xn)≥F(x,z,x, . . . ,xn),
.. .
yn,zn∈X, yn≤zn
⇒ F(x,x,x, . . . ,yn)≤F(x,x,x, . . . ,zn) (ifnis odd), yn,zn∈X, yn≤zn
⇒ F(x,x,x, . . . ,yn)≥F(x,x,x, . . . ,zn) (ifnis even).
(.)
Definition LetXbe a nonempty set and F:Xn→Xa given mapping. An element
(x,x,x, . . . ,xn)∈Xnis called an-tuplet fixed point ofFif F(x,x,x, . . . ,xn) =x,
F(x,x, . . . ,xn,x) =x,
.. .
F(xn,x,x, . . . ,xn–) =xn.
(.)
Definition Let (X,≤) be partially ordered set andF:Xn→Xandg:X→X. We say that Fhas the mixedg-monotone property ifF(x,x,x, . . . ,xn) is monotoneg-non-decreasing
in its odd argument and it is monotoneg-non-increasing in its even argument. That is, for anyx,x,x, . . . ,xn∈X
y,z∈X, g(y)≤g(z)
⇒ F(y,x,x, . . . ,xn)≤F(z,x,x, . . . ,xn), y,z∈X, g(y)≤g(z)
..
. (.)
yn,zn∈X, g(yn)≤g(zn)
⇒ F(x,x,x, . . . ,yn)≤F(x,x,x, . . . ,zn) (ifnis odd), yn,zn∈X, g(xn)≤g(yn)
⇒ F(x,x,x, . . . ,yn)≥F(x,x,x, . . . ,zn) (ifnis even).
Note that ifgis the identity mapping, this definition reduces to Definition .
Definition LetX be a nonempty set andF:Xn→Xa given mapping. An element
(x,x,x, . . . ,xn)∈Xnis called an-tuplet coincidence point ofF:Xn→Xandg:X→Xif F(x,x,x, . . . ,xn) =g(x),
F(x,x, . . . ,xn,x) =g(x),
.. .
F(xn,x,x, . . . ,xn–) =g(xn).
(.)
Note that ifgis the identity mapping, this definition reduces to Definition .
Definition Let (X,≤) be partially ordered set andF:Xn→Xandg:X→X.Fandg
called commutative if
gF(x,x,x, . . . ,xn)
=Fg(x),g(x),g(x), . . . ,g(xn)
(.)
for allx,x,x, . . . ,xn∈X.
Letdenote the all functionsφ : [,∞)→[,∞), which are continuous and satisfy that
(i) φ(t) <t,
(ii) limr→t+φ(r) <tfor eachr> .
Since we want to shorten expressions in the following theorem, consider Condition in the following forXanF.
Condition Suppose either
(i) Fis continuous, or
(ii) Xhas the following property:
(a) if non-decreasing sequencexk→x, thenxk≤xfor allk, (b) if non-increasing sequenceyk→y, thenyk≥yfor allk.
Theorem Let(X,≤)be partially ordered set and suppose that(X,d)is complete met-ric space.Assume F:Xn→X and g:X→X are such that F has the mixed g-monotone property and
dF(x,x,x, . . . ,xn),F(y,y,y, . . . ,yn)
≤φ
d(g(x),g(y)) +d(g(x),g(y)) +· · ·+d(g(xn),g(yn)) n
for all x,x,x, . . . ,xn∈X for which g(x)≤g(y),g(x)≥g(y), . . . ,g(xn)≤g(yn) (if n is odd),g(xn)≥g(yn) (if n is even).Assume that F(Xn)⊂g(X)and g commutes with F.Also, suppose that Conditionis satisfied.If there exist x
,x,x, . . . ,xn∈X such that
gx≤Fx,x,x, . . . ,xn,
gx≥Fx,x, . . . ,xn,x, ..
.
gxn≤Fxn,x,x, . . . ,xn– (if n is odd),
gxn≥Fxn,x,x, . . . ,xn– (if n is even)
(.)
then there exist x,x,x, . . . ,xn∈X such that F(x,x,x, . . . ,xn) =g(x),
F(x,x, . . . ,xn,x) =g(x),
.. .
F(xn,x,x, . . . ,xn–) =g(xn),
that is,F and g have a n-tuplet coincidence point.
Proof Letx
,x,x, . . . ,xn∈Xbe such that (.). SinceF(Xn)⊂g(X), we construct the
sequence (x
k), (xk), . . . , (xnk) as follows:
gx
k
=Fx
k–,xk–, . . . ,xnk–
,
gxk=Fxk–, . . . ,xnk–,xk–,
.. .
gxnk=Fxnk–,xk–, . . . ,xnk––
(.)
fork= , , , . . . . We claim that
gxk–≤gxk,
gxk–≥gxk, ..
.
gxnk–≤gxnk (ifnis odd),
gxnk–≥gxnk (ifnis even)
(.)
for allk≥. For this, we will use the mathematical induction. The inequalities in (.) holdk= because of (.), that is, we have
gx≤Fx,x, . . . ,xn=gx,
.. .
gxn≤Fxn,x,x, . . . ,xn–=gxn (ifnis odd),
gxn≥Fxn,x,x, . . . ,xn–=gxn (ifnis even).
Thus, our claim is true fork= . Now, suppose that the inequalities in (.) holdk=m. In this case,
gxm–≤gxm,
gxm–≥gxm, ..
.
gxnm–≤gxnm (ifnis odd),
gxnm–≥gxnm (ifnis even).
(.)
Now, we must show that the inequalities in (.) holdk=m+ . If we consider (.) and mixedg-monotone property ofFtogether with (.), we have
gxm=Fxm–,xm–, . . . ,xnm–
≤Fxm,xm–, . . . ,xnm–
≤Fxm,xm,xm–, . . . ,xnm––,xnm–
.. .
≤Fxm,xm, . . . ,xnm–,xnm–
≤Fxm,xm, . . . ,xnm=gxm+,
gxm=Fxm–, . . . ,xnm–,xm–
≥Fxm,xm–, . . . ,xnm–,xm–
≥Fxm,xm, . . . ,xnm–,xm–
.. .
≥Fxm,xm, . . . ,xnm,xm–
≥Fxm,xm, . . . ,xnm,xm=gxm+, ..
.
gxnm=Fxnm–,xm–, . . . ,xnm––
≤Fxnm,xm–, . . . ,xnm––
≤Fxnm,xm,xm–, . . . ,xnm––
.. .
≤Fxnm,xm, . . . ,xmn–=gxnm+ (ifnis odd),
gxnm=Fxnm–,xm–, . . . ,xnm––
≥Fxnm,xm–, . . . ,xnm––
≥Fxnm,xm,xm–, . . . ,xnm––
.. .
≥Fxnm,xm,xm, . . . ,xnm–,xnm––
≥Fxnm,xm, . . . ,xmn–=gxnm+ (ifnis even).
Thus, (.) is satisfied for allk≥. So, we have,
· · · ≥gxk≥gxk–≥ · · · ≥gx≥gx,
· · · ≤gxk≤gxk–≤ · · · ≤gx≤gx, ..
.
· · · ≥gxnk≥gxnk–≥ · · · ≥gxn≥gxn (ifnis odd),
· · · ≤gxnk≤gxnk–≤ · · · ≤gxn≤gxn (ifnis even).
(.)
For the simplicity, we define
δk=d
gxk,gxk++dgxk,gxk++· · ·+dgxnk,gxnk+. We will show that
δk+≤n·φ
δk n
. (.)
By (.), (.) and (.), we get
dgxk+,gxk+
=dFxk,xk, . . . ,xnk,Fxk+,xk+, . . . ,xnk+
≤φ
d(g((xk),g(xk+))) +d(g(xk),g(xk+)) +· · ·+d(g(xnk),g(xnk+))
n
=φ
δk n
, (.)
dgxk+,gxk+
=dFxk, . . . ,xnk,xk,Fxk+, . . . ,xnk+,xk+
≤φ
d(g(x
k),g(xk+)) +· · ·+d(g(xnk),g(xnk+)) +d(g(xk),g(xk+))
n
=φ
δk n
, (.)
dgxnk,gxnk+
=dFxnk,xk, . . . ,xnk–,Fxnk+,xk+, . . . ,xnk–+
≤φ
d(g((xnk),g(xkn+))) +d(g(xk),g(xk+)) +· · ·+d(g(xkn–),g(xnk–+))
n
=φ
δk n
. (.)
Due to (.)-(.), we conclude that
dgxk+,gxk++dgxk+,gxk++· · · +dgxnk+,gxnk+≤nφ
δk n
. (.)
Hence, we get (.).
Sinceφ(t) <tfor allt> , thenδk+≤nφ(δnk) <n·δnk =δkfor allk∈N. So, (δk) is
mono-tone decreasing. Since it is bounded below, there is someδ≥ such that
lim
k→∞δk=δ+. (.)
We want to show thatδ= . Suppose thatδ> . Then taking the limit asδk→δ+ of both
sides of (.) and keeping in mind that we assume thatlimr→t+φ(r) <tfor allt> , we
have
δ= lim
k→∞δk+≤klim→∞n·φ
δk n
=n lim
δk→δ+
φ
δk n
<nδ
n=δ (.)
which is a contradiction. Thus,δ= , that is
lim
k→∞
dgxk,gxk++dgxk,gxk++· · ·+dgxnk,gxnk+= . (.)
Now we prove thatg(x
k),g(xk), . . . ,g(xnk) are Cauchy sequences. Suppose that at least one
ofg(x
k),g(xk), . . . ,g(xnk) is not Cauchy. So, there exists anε> for which we can find
sub-sequence of integerj(k),l(k) withj(k) >l(k)≥ksuch that
tk=d
gxj(k),gxl(k)+dgxj(k),gxl(k)+· · ·+dgxnj(k),gxnl(k)≥ε. (.)
Additionally, corresponding toj(k), we can choosel(k) such that it is the smallest integer satisfying (.) andj(k) >l(k)≥k. Thus,
dgxj(k)–,gxl(k)+dgxj(k)–,gxl(k)+· · ·+dgxnj(k)–,gxln(k)<ε. (.)
By using triangle inequality and having (.) and (.) in mind
ε≤tk
≤dgxj(k),gxj(k)–+dgxj(k)–,gxl(k)
+dgxj(k),gxj(k)–+dgxj(k)–,gxl(k)
+· · ·+dgxnj(k),gxnj(k)–+dgxnj(k)–,gxnl(k)
< dgxj(k),gxj(k)–+dgxj(k),gxj(k)–
+· · ·+dgxnj(k),gxnj(k)–+ε
< δj(k)–+ε. (.)
Lettingk→ ∞in (.) and using (.)
lim
k→∞tk=klim→∞
dgxj(k),gxl(k)+dgxj(k),gxl(k)+· · ·+dgxnj(k),gxnl(k)
=ε+. (.)
We apply triangle inequality to (.) as the following.
tk =d
gxj(k),gxl(k)+dgxj(k),gxl(k)
+· · ·+dgxnj(k),gxnl(k)
≤dgxj(k),gxj(k)++dgxj(k)+,gxl(k)++dgxl(k)+,gxl(k)
+dgxj(k),gxj(k)++dgxj(k)+,gxl(k)++dgxl(k)+,gxl(k)
+· · ·+dgxnj(k),gxnj(k)++dgxnj(k)+,gxnl(k)++dgxnl(k)+,gxnl(k)
≤δj(k)++δl(k)++d
gxj(k)+,gxl(k)++dgxj(k)+,gxl(k)+
+· · ·+dgxnj(k)+,gxnl(k)+. (.) Sincej(k) >l(k), then
gxj(k)≥gxl(k),
gxj(k)≤gxl(k), ..
.
gxnj(k)≥gxnl(k) (ifnis odd),
gxnj(k)≤gxnl(k) (ifnis even).
(.)
So, from (.), (.) and (.), we get
dgxj(k)+,gxl(k)+
=dFxj(k),xj(k), . . . ,xnj(k),Fxl(k),xl(k), . . . ,xnl(k)
≤φ
n
dgxj(k),gxl(k)+dgxj(k),gxl(k)+· · ·+dgxnj(k),gxnl(k)
=φ
tk n
dgxj(k)+,gxl(k)+
=dFxj(k), . . . ,xnj(k),xj(k),Fxl(k), . . . ,xln(k),xl(k)
≤φ
n
dgxj(k),gxl(k)+· · ·+dgxnj(k),gxnl(k)+dgxj(k),gxl(k)
=φ tk n , (.) .. .
dgxnj(k)+,gxnl(k)+
=dFxnj(k),xj(k), . . . ,xnj(–k),Fxnl(k),xl(k), . . . ,xnl(–k)
≤φ
n
dgxnj(k),gxnl(k)+dgxj(k),gxl(k)+· · ·+dgxnj(–k),gxnl(–k)
=φ tk n . (.)
Combining (.) with (.)-(.), we get
tk≤δj(k)++δl(k)+
+dgx
j(k)+
,gx
l(k)+
+dgx
k(k)+
,gx
l(k)+
+· · ·+dgxnj(k)+,gxnl(k)+
≤δj(k)++δl(k)++tk+n·φ
tk n
<δj(k)++δl(k)++n·
tk
n. (.)
Lettingk→ ∞, we obtain a contradiction. This show thatg(x
k),g(xk), . . . ,g(xnk) are Cauchy
sequences. SinceXis complete metric space, there existx,x, . . . ,xn∈Xsuch that
lim
k→∞g
xk=x, lim
k→∞g
xk=x, . . . , lim
k→∞g
xnk=xn. (.)
Sincegis continuous, (.) implies that
lim
k→∞g
gxk=gx, lim
k→∞g
gxk=gx, . . . ,
lim
k→∞g
gxnk=gxn.
(.)
From (.) and by regarding commutativity ofFandg
ggxk+=gFxk,xk, . . . ,xnk=Fgxk,gxk, . . . ,gxnk,
ggxk+=gFxk, . . . ,xnk,xk=Fgxk, . . . ,gxnk,gxk,
.. .
ggxnk+=gFxnk,xk, . . . ,xnk–=Fgxnk,gxk, . . . ,gxnk–.
We shall show that
Fx,x, . . . ,xn=gx,
Fx, . . . ,xn,x=gx,
.. .
Fxn,x, . . . ,xn–=gxn.
Suppose now, (i) holds. Then by (.), (.) and (.), we have
gx= lim
k→∞g
gxk+= lim
k→∞g
Fxk,xk, . . . ,xnk
= lim
k→∞F
gxk,gxk, . . . ,gxnk
=F
lim
k→∞g
xk, lim
k→∞g
xk, . . . , lim
k→∞g
xnk
=Fx,x, . . . ,xn. (.)
Analogously,
gx=ggxk+= lim
k→∞g
Fxk, . . . ,xnk,xk
= lim
k→∞F
gxk, . . . ,gxnk,gxk
=F
lim
k→∞g
xk, . . . ,lim
k→∞g
xnk, lim
k→∞g
xk
=Fx, . . . ,xn,x, (.)
.. .
gxn=ggxnk+= lim
k→∞g
Fxn,x, . . . ,xn–
= lim
k→∞F
gxnk,gxk, . . . ,gxnk–
=Flim
k→∞g
xnk, lim
k→∞g
xk, . . . ,lim
k→∞g
xnk–
=Fxn,x, . . . ,xn–. (.)
Thus, we have
gx=Fx,x, . . . ,xn,
gx=Fx, . . . ,xn,x,
.. .
gxn=Fxn,x, . . . ,xn–.
g(xnk–) are non-decreasing andg(x
k),g(xk), . . . ,g(xnk) are non-increasing and by considering g(x
k)→x,g(xk)→x, . . . ,g(xnk)→xnwe have
gxk≥x, gxk≤x, . . . , gxnk≤xn (ifnis odd),
gxnk≥xn (ifnis even)
(.)
for allk. Thus, by triangle inequality and (.)
dgx,Fx,x, . . . ,xn
≤dgx,ggxk+
+dggxk+,Fx,x, . . . ,xn
≤dgx,ggxk+
+φ
n
dggxk,gx+dggxk,gx
+· · ·+dggxnk,gxn. (.)
Lettingk→ ∞implies thatd(g(x),F(x,x, . . . ,xn))≤. Hence,g(x) =F(x,x, . . . ,xn). Analogously, we can get that
gx=Fx, . . . ,xn,x, . . . , gxn=Fxn,x, . . . ,xn–.
Thus, we proved thatFandghave an-tuplet coincidence point.
Corollary The above theorem reduces to Theorem.of[]for n= and g(x) =x if(i)is
satisfied andφ(t) =ct where c∈(, ).
The following corollary is a generalization of Corollary . in [] and Theorem . in [].
Corollary Let(X,≤)be a partially ordered set and suppose that(X,d)is complete metric space.Suppose F:Xn→X and there existφ∈such that F has the mixed g-monotone property and there exist a m∈[, )with
φFx,x, . . . ,xn,Fy,y, . . . ,yn
≤m
n
dgx,gy+dgx,gy+· · ·+dgxn,gyn (.)
for all x,x, . . . ,xn,y,y, . . . ,yn∈X for which g(x)≤g(y),g(x)≥g(y), . . . ,g(xn)≤g(yn)
that
gz≤Fz,z, . . . ,zn,
gz≥Fz, . . . ,zn,z, ..
.
gzn≤Fzn,z, . . . ,zn– (if n is odd),
gzn≥Fzn,z, . . . ,zn– (if n is even)
(.)
then there exist x,x, . . . ,xn∈X such that
gx=Fx,x, . . . ,xn,
gx=Fx, . . . ,xn,x,
.. .
gxn=Fxn,x, . . . ,xn–,
that is,F and g have a n-tuplet coincidence point.
Proof It is sufficient to takeφ=mtwithm∈[, ) in previous theorem.
3 Uniqueness ofn-tuplet fixed point
For all (x,x, . . . ,xn), (y,y, . . . ,yn)∈Xn,
x,x, . . . ,xn≤y,y, . . . ,yn
⇐⇒ x≤y, x≥y, . . . , xn≤yn (ifnis odd),
xn≥yn (ifnis even). (.)
We say that (x,x, . . . ,xn) is equal to (y,y, . . . ,yn) if and only ifx=y,x=y, . . . ,xn=yn.
Theorem In addition to hypothesis Theorem , assume that for all (x,x, . . . ,xn),
(y,y, . . . ,yn)∈Xnthere exist(z,z, . . . ,zn)∈Xnsuch that
Fz,z, . . . ,zn,Fz, . . . ,zn,z, . . . ,Fzn,z, . . . ,zn–
is comparable to
Fx,x, . . . ,xn,Fx, . . . ,xn,x, . . . ,Fxn,x, . . . ,xn–
and
Then F and g have a unique n-tuplet common fixed point,that is,there exist(x,x, . . . ,xn)∈ Xnsuch that
x=gx=Fx,x, . . . ,xn,
x=gx=Fx, . . . ,xn,x,
.. .
xn=gxn=Fxn,x, . . . ,xn–.
Proof From the Theorem , the set ofn-tuplet coincidences is non-empty. We will show that if (x,x, . . . ,xn) and (y,y, . . . ,yn) aren-tuplet coincidence points, that is, if
gx=Fx,x, . . . ,xn,
gx=Fx, . . . ,xn,x,
.. .
gxn=Fxn,x, . . . ,xn–
and
gy=Fy,y, . . . ,yn,
gy=Fy, . . . ,yn,y,
.. .
gyn=Fyn,y, . . . ,yn–,
then
gx=gy,
gx=gy,
.. .
gxn=gyn.
(.)
By assumption there is (z,z, . . . ,zn)∈Xnsuch that
Fz,z, . . . ,zn,Fz, . . . ,zn,z, . . . ,Fzn,z, . . . ,zn– (.)
is comparable with
Fx,x, . . . ,xn,Fx, . . . ,xn,x, . . . ,Fxn,x, . . . ,xn– (.)
and
Define sequences (g(z
k)), (g(zk)), . . . , (g(znk)) such thatz=z,z=z, . . . ,zn=znand
gzk=Fzk–,zk–, . . . ,znk–,
gzk=Fzk–, . . . ,znk–,zk–,
.. .
gzkn=Fznk–,zk–, . . . ,zkn––.
(.)
Since (.) and (.) comparable with (.), we may assume that
gx,gx, . . . ,gxn≥gz,gz, . . . ,gzn=gz,gz, . . . ,gzn.
By using (.), we get that
gx,gx, . . . ,gxn≥gzk,gzk, . . . ,gznk
for allk. From (.), we have
gx≤gzk,
gx≥gzk, ..
.
gxn≤gznk (ifnis odd),
gxn≥gznk (ifnis even).
(.)
By (.) and (.), we have
dgx,gzk+
=dFx,x, . . . ,xn,Fzk,zk, . . . ,znk
≤φ
n
dgx,gzk+dgx,gzk+· · ·+dgxn,gzn k
, (.)
dgx,gzk+
=dFx, . . . ,xn,x,Fzk, . . . ,znk,zk
≤φ
n
dgx,gzk+· · ·+dgxn,gznk+dgx,gzk, (.) ..
.
dgxn,gznk+
=dFxn,x, . . . ,xn–,Fznk,zk, . . . ,znk–
≤φ
n
Adding (.)-(.), we get
d(g(x),g(z
k+)) +d(g(x),g(zk+)) +· · ·+d(g(xn),g(znk+))
n
≤φ
n
dgx,gzk+· · ·+dgxn,gznk+dgx,gzk. (.)
Hence, it follows
d(g(x),g(z
k+)) +d(g(x),g(zk+)) +· · ·+d(g(xn),g(znk+))
n
≤φk
n
dgx,gzk+· · ·+dgxn,gznk+dgx,gzk (.)
for eachk≥. It is known thatφ(t) <tandlimr→t+φ(r) <timplylimk→∞φk(t) = for
eacht> . Thus, from (.)
lim
k→∞d
gx,gzk+= ,
lim
k→∞d
gx,gzk+= ,
.. .
lim
k→∞d
gx,gzk+= .
(.)
Analogously, we can show that
lim
k→∞d
gy,gzk+= ,
lim
k→∞d
gy,gzk+= ,
.. .
lim
k→∞d
gy,gzk+= .
(.)
Combining (.) and (.) and by using the triangle inequality
dgx,gy≤dgx,gzk++dgy,gzk+→ ask→ ∞,
dgx,gy≤dgx,gzk++dgy,gzk+→ ask→ ∞, ..
.
dgxn,gyn≤dgxn,gznk++dgyn,gznk+→ ask→ ∞.
(.)
By commutativity ofFandg,
ggx=gFx,x, . . . ,xn=Fgx,gx, . . . ,gxn,
ggx=gFx, . . . ,xn,x=Fgx, . . . ,gxn,gx,
.. .
ggxn=gFxn,x, . . . ,xn–=Fgxn,gx, . . . ,gxn–.
(.)
Denoteg(x) =w,g(x) =w, . . . ,g(xn) =wn. Since (.), we get
gw=Fw,w, . . . ,wn,
gw=Fw, . . . ,wn,w,
.. .
gwn=Fwn,w, . . . ,wn–.
(.)
Thus, (w,w, . . . ,wn) is an-tuplet coincidence point. Then from assumption in theorem
withy=w, . . . ,yn=wnit followsg(w) =g(x),g(w) =g(x), . . . ,g(wn) =g(xn), that is
gw=w, . . . , gwn=wn. (.)
From (.) and (.),
w=gw=Fw,w, . . . ,wn,
w=gw=Fw, . . . ,wn,w,
.. .
wn=gwn=Fwn,w, . . . ,wn–.
Therefore, (w,w, . . . ,wn) isn-tuplet common fixed point ofFandg. To prove the
unique-ness, assume that (u,u, . . . ,un) is anothern-tuplet common fixed point. Then by
assump-tion in theorem we have
u=gu=gw,
u=gu=gw,
.. .
un=gun=gwn.
property and
dFx,x,x, . . . ,xn,Fy,y,y, . . . ,yn
≤φ
d(g(x),g(y)) +d(g(x),g(y)) +· · ·+d(g(xn),g(yn)) n
(.)
for all x,x,x, . . . ,xn,y,y,y, . . . ,yn∈X for which x≤y,x≥y, . . . ,xn≤yn(if n is odd),
xn≥yn(if n is even).Suppose there exist x
,x,x, . . . ,xn∈X such that
x≤Fx,x,x, . . . ,xn,
x≥Fx,x, . . . ,xn,x,
xn≤Fxn,x,x, . . . ,xn– (if n is odd), ..
.
xn≥Fxn,x,x, . . . ,xn– (if n is even).
(.)
Assume also that Conditionholds.Then there exist x,x,x, . . . ,xnsuch that
x=Fx,x,x, . . . ,xn,
x=Fx,x, . . . ,xn,x,
.. .
xn=Fxn,x,x, . . . ,xn–.
(.)
That is F has a n-tuplet fixed point.
Proof Takeg(x) =x, then the assumption in Theorem are satisfied. Thus, we get the
result.
Corollary Let(X,≤)be partially ordered set and suppose that(X,d)is complete metric space.Suppose F:Xn→X and there existφ∈such that F has the mixed g-monotone property and there exist m∈[, )with
φFx,x, . . . ,xn,Fy,y, . . . ,yn
≤m
n
dgx,gy+dgx,gy+· · ·+dgxn,gyn (.)
for all x,x, . . . ,xn,y,y, . . . ,yn∈X for which x≤y, x≥y, . . . , xn≤yn(if n is odd), xn≥yn(if n is even).Suppose there exist x,x,x, . . . ,xn∈X such that
x≤Fx,x,x, . . . ,xn,
x≥Fx,x, . . . ,xn,x, ..
xn≤Fxn,x,x, . . . ,xn– (if n is odd),
xn≥Fxn,x,x, . . . ,xn– (if n is even).
Assume also that Conditionholds.Then there exist x,x,x, . . . ,xnsuch that
x=Fx,x,x, . . . ,xn,
x=Fx,x, . . . ,xn,x,
.. .
xn=Fxn,x,x, . . . ,xn–.
(.)
That is F and g have n-tuplet coincidence point.
Proof Takingφ(t) =m·twithm∈[, ) in above corollary we obtain this corollary.
Competing interests
The authors declare that they have no competing interests.
Author details
1Department of Mathematics, The Faculty of Arts and Sciences, Yildiz Technical University, Davutpasa Campus, Esenler, Istanbul, 34210, Turkey.2Current address: Department of Mathematical Engineering, Yildiz Technical University, Davutpasa Campus, Esenler, Istanbul, 34220, Turkey.
Acknowledgements
This work is supported by Yildiz Technical University Scientific Research Projects Coordination Unit under the project number BAPK 2012-07-03-DOP03.
Received: 4 January 2013 Accepted: 5 April 2013 Published: 22 April 2013
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doi:10.1186/1029-242X-2013-196