# Onn-Tupled Coincidence Point Results in Metric Spaces

Authors: Mohammad Imdad; Ahmed H. Soliman; Binayak S. Choudhury; Pradyut Das
Journal: Journal of Operators
 (2013)

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## Abstract

We prove somen-tupled coincidence point results whenever n is even. We give here several new definitions like n-tupled fixed point, n-tupled coincidence point, and so forth. The main result is supported with the aid of an illustrative example.


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## Body

## 1. Introduction and Preliminaries

The classical Banach Contraction Principle proved in complete metric spaces continues to be an indispensable and effective tool in theory as well as applications which guarantees the existence and uniqueness of fixed points of contraction self-mappings besides offering a constructive procedure to compute the fixed point of the underlying map. There already exists an extensive literature on this topic, but keeping in view the relevance of this paper, we merely refer to [1–14].In 2006, Gnana Bhaskar and Lakshmikantham initiated the idea of coupled fixed point in partially ordered metric spaces and proved some interesting coupled fixed point theorems for mapping satisfying a mixed monotone property. In recent years, many authors obtained important coupled fixed point theorems (e.g., [15–20]). In this continuation, Lakshmikantham and Cirić [21] proved coupled common fixed point theorems for nonlinear ϕ-contraction mappings in partially ordered complete metric spaces which indeed generalize the corresponding fixed point theorems contained in Gnana Bhaskar and Lakshmikantham [22].As usual, this section is devoted to preliminaries which include basic definitions and results on coupled fixed point for nonlinear contraction mappings defined on partially ordered complete metric spaces. In Section2, we introduce the concepts of n-tupled coincidence point and n-tupled fixed point for mappings satisfying different contractive conditions and utilize these two definitions to obtain n-tupled coincidence point theorems for nonlinear ϕ-contraction mappings in partially ordered complete metric spaces.Now, we present some basic notions and results related to coupled fixed point in metric spaces.Definition 1 (see [22]).
Let(X,⪯) be a partially ordered set equipped with a metric d such that (X,d) is a metric space. Further, equip the product space X×X with the following partial ordering:
(1)for(x,y),(u,v)∈X×X,define(u,v)⪯(x,y)⟺x≽u,y⪯v.Definition 2 (see [22]).
Let(X,⪯) be a partially ordered set and F:X×X→X. One says that F enjoys the mixed monotone property if F(x,y) is monotonically nondecreasing in x and monotonically nonincreasing in y; that is, for any x,y∈X,
(2)x1,x2∈X,x1⪯x2⇒F(x1,y)⪯F(x2,y),y1,y2∈X,y1⪯y2⇒F(x,y1)≽F(x,y2).Definition 3 (see [22]).
Let(X,⪯) be a partially ordered set and F:X×X→X. One says that (x,y)∈X×X is a coupled fixed point of the mapping F if
(3)F(x,y)=x,F(y,x)=y.Theorem 4 (see [22]).
Let(X,⪯) be a partially ordered set equipped with a metric d such that (X,d) is a complete metric space. Let F:X×X→X be a continuous mapping having the mixed monotone property on X. Assume that there exists a constant k∈[0,1) with
(4)d(F(x,y),F(u,v))≤k2[d(x,u)+d(y,v)]∀x≽u,y⪯v.
If there exist x0,y0∈X such that x0⪯F(x0,y0) and y0≽F(y0,x0), then there exist x,y∈X such that x=F(x,y)  and  y=F(y,x).Definition 5 (see [21]).
Let(X,⪯) be a partially ordered set and F:X×X→X and g:X→X two mappings. The mapping F is said to have the mixed g-monotone property if F is monotone g-nondecreasing in its first argument and is monotone g-nonincreasing in its second argument, that is, if, for all x1,x2∈X, g(x1)⪯g(x2) implies F(x1,y)⪯F(x2,y), for any y∈X, and, for all y1,y2∈X, g(y1)⪯g(y2) implies F(x,y1)≽F(x,y2), for any x∈X.Definition 6 (see [21]).
An element(x,y)∈X×X is called a coupled coincidence point of mappings F:X×X→X and g:X→X if
(5)F(x,y)=g(x),F(y,x)=g(y).Theorem 7 (see [21]).
Let(X,⪯) be a partially ordered set equipped with a metric d such that (X,d) is a complete metric space. Assume that there is a function ϕ:[0,∞)→[0,∞) with ϕ(t)<t and limr→t+ϕ(r)<t for each t>0. Let F:X×X→X and g:X→X be maps such that F has the mixed g-monotone property and
(6)d(F(x,y),F(u,v))≤ϕ(d(g(x),g(u))+d(g(y),g(v))2)
for all x,y,u,v∈X for which g(x)⪯g(u) and g(y)≽g(v). Suppose that F(X×X)⊆g(X),g is continuous and commutes with F besides(a)
Fis continuous,(b)
X has the following properties:
(i)
if nondecreasing sequence{xn}→x, then xn⪯x for all n,(ii)
if nonincreasing sequence{yn}→y, then y⪯yn for all n.(iii)
if a nondecreasing sequence{xn}→x, then xn⪯x for all n≥0,(iv)
if a nonincreasing sequence{xn}→x, then xn≽x for all n≥0.If there exist x0,y0∈X such that
(7)g(x0)⪯F(x0,y0),g(y0)≽F(y0,x0),
then there exist x,y∈X such that
(8)g(x)=F(x,y),g(y)=F(y,x).
That is, F and g have a coupled coincidence point.


## 2. Main Results

Throughout the paper,r stands for a general even natural number.Definition 8.
Let(X,⪯) be a partially ordered set and F:∏i=1r‍Xi→X a mapping. The mapping F is said to have the mixed monotone property if F is nondecreasing in its odd position arguments and nonincreasing in its even position arguments, that is, if, (i)
for allx11,x21∈X, x11⪯x21 implies F(x11,x2,x3,…,xr)⪯F(x21,x2,x3,…,xr)(ii)
for allx12,x22∈X, x12⪯x22 implies F(x1,x12,x3,…,xr)≽F(x1,x22,x3,…,xr)(iii)
for allx13,x23∈X, x13⪯x23 implies F(x1,x2,x13,…,xr)⪯F(x1,x2,x23,…,xr)
for allx1r,x2r∈X, x1r⪯x2r implies F(x1,x2,x3,…,x1r)≽F(x1,x2,x3,…,x2r).Definition 9.
Let(X,⪯) be a partially ordered set. Let F:∏i=1r‍Xi→X and g:X→X be two mappings. Then the mapping F is said to have the mixed g-monotone property if F is g-nondecreasing in its odd position arguments and g-nonincreasing in its even position arguments, that is, if,(i)
for allx11,x21∈X, gx11⪯gx21 implies F(x11,x2,x3,…,xr)⪯F(x21,x2,x3,…,xr),(ii)
for allx12,x22∈X, gx12⪯gx22 implies F(x1,x12,x3,…,xr)≽F(x1,x22,x3,…,xr),(iii)
for allx13,x23∈X, gx13⪯gx23 implies F(x1,x2,x13,…,xr)⪯F(x1,x2,x23,…,xr),for all x1r,x2r∈X, gx1r⪯gx2r implies F(x1,x2,x3,…,x1r)≽F(x1,x2,x3,…,x2r).Definition 10.
LetX be a nonempty set. An element (x1,x2,x3,…,xr)∈∏i=1r‍Xi is called an r-tupled fixed point of the mapping F:∏i=1r‍Xi→X if
(9)x1=F(x1,x2,x3,…,xr),x2=F(x2,x3,…,xr,x1),x3=F(x3,…,xr,x1,x2),⋮xr=F(xr,x1,x2,…,xr-1).Definition 11.
LetX be a nonempty set. An element (x1,x2,x3,…,xr)∈∏i=1r‍Xi is called an r-tupled coincidence point of the mappings F:∏i=1r‍Xi→X and g:X→X if
(10)gx1=F(x1,x2,x3,…,xr),gx2=F(x2,x3,…,xr,x1),gx3=F(x3,…,xr,x1,x2),⋮gxr=F(xr,x1,x2,…,xr-1).Definition 12.
LetX be a nonempty set. The mappings F:∏i=1r‍Xi→X and g:X→X are said to be commutating if
(11)g(F(x1,x2,…,xr))=(F(g(x1),g(x2),…,g(xr))
for all x1,x2,…,xr∈X.Now, we are equipped to prove our main result as follows.Theorem 13.
Let(X,⪯) be a partially ordered set equipped with a metric d such that (X,d) is a complete metric space. Assume that there is a function ϕ:[0,+∞)→[0,+∞) with ϕ(t)<t and limr→t+ϕ(r)<t for each t>0. Further, suppose that F:∏i=1r‍Xi→X and g:X→X are two maps such that F has the mixed g-monotone property satisfying the following conditions:(i)
F(∏i=1r‍Xi)⊆g(X),(ii)
g is continuous and monotonically increasing,(iii)
(g,F) is a commutating pair,(iv)
(12)d(F(x1,x2,…,xr),F(y1,y2,…,yr))≤ϕ(1r∑n=1r‍d(g(xn),g(yn))),for all x1,x2,x3,…,xr,y1,y2,y3,…,yr∈X,with gx1⪯gy1,gx2≽gy2,gx3⪯gy3,…,gxr≽gyr. Also, suppose that either(a)
F is continuous or(b)
X has the following properties:
(13)(i)ifanondecreasingsequence{xn}→x,thenxn⪯x∀n≥0,(ii)ifanonincreasingsequence{xn}→x,thenxn≽x∀n≥0.If there exist x01,x02,x03,…,x0r∈X such that
(14)gx01⪯F(x01,x02,x03,…,x0r),gx02≽F(x02,x03,…,x0r,x01),gx03⪯F(x03,…,x0r,x01,x02),⋮gx0r≽F(x0r,x01,x02,…,x0r-1),
then F and g have a r-tupled coincidence point; that is, there exist x1,x2,x3,…,xr∈X such that
(15)gx1=F(x1,x2,x3,…,xr),gx2=F(x2,x3,…,xr,x1),gx3=F(x3,…,xr,x1,x2),⋮gxr=F(xr,x1,x2,x3,…,xr-1).Proof.
Starting withx01,x02,x03,…,x0r in X, we define the sequences {xn1},{xn2},{xn3},…,{xnr} in X as follows:
(16)gxn+11=F(xn1,xn2,xn3,…,xnr),gxn+12=F(xn2,xn3,…,xnr,xn1),gxn+13=F(xn3,…,xnr,xn1,xn2),⋮gxn+1r=F(xnr,xn1,xn2,xn3,…,xnr-1).
Now, we prove that for all n≥0,
(17)gxn1⪯gxn+11,gxn2≽gxn+12,gxn3⪯gxn+13,…,gxnr≽gxn+1r.(18)gx01⪯F(x01,x02,x03,…,x0r)=x11,gx02≽F(x02,x03,…,x0r,x01)=x12,gx03⪯F(x03,…,x0r,x01,x02)=x13,⋮gx0r≽F(x0r,x01,x02,x03,…,x0r-1)=x1r.
So (17) holds for n=0. Suppose (17) holds for some n>0. Consider
(19)gxn+11=F(xn1,xn2,xn3,…,xnr)⪯F(xn+11,xn2,xn3,…,xnr)⪯F(xn+11,xn+12,xn3,…,xnr)⪯F(xn+11,xn+12,xn+13,…,xnr)⪯F(xn+11,xn+12,xn+13,…,xn+1r)=gxn+21,gxn+12=F(xn2,xn3,…,xnr,xn1)≽F(xn+12,xn3,…,xnr,xn1)≽F(xn+12,xn+13,…,xnr,xn1)≽F(xn+12,xn+13,…,xn+1r,xn1)≽F(xn+12,xn+13,…,xn+1r,xn+11)=gxn+22,gxn+13=F(xn3,…,xnr,xn1,xn2)⪯F(xn+13,…,xnr,xn1,xn2)⪯F(xn+13,…,xn+1r,xn1,xn2)⪯F(xn+13,…,xn+1r,xn+11,xn2)⪯F(xn+13,…,xn+1r,xn+11,xn+12)=gxn+23,⋮gxn+1r=F(xnr,xn1,xn2,xn3,…,xnr-1)≽F(xn+1r,xn1,xn2,xn3,…,xnr-1)≽F(xn+1r,xn+11,xn2,xn3,…,xnr-1)≽F(xn+1r,xn+11,xn+12,xn3,…,xnr-1)≽F(xn+1r,xn+11,xn+12,xn+13,…,xnr-1)≽F(xn+1r,xn+11,xn+12,xn+13,…,xn+1r-1)=gxn+2r.
Then, by induction, (17) holds for all n≥0.
Using (16) and (17), we have
(20)d(g(xm1),g(xm+11))=d(F(xm-11,xm-12,…,xm-1r),F(xm1,xm2,…,xmr))≤ϕ(1r∑n=1r‍d(g(xm-1n),g(xmn))).
Similarly, we can inductively write
(21)d(g(xm2),g(xm+12))≤ϕ(1r∑n=1r‍d(g(xm-1n),g(xmn))),⋮d(g(xmr),g(xm+1r))≤ϕ(1r∑n=1r‍d(g(xm-1n),g(xmn))).
Therefore, by putting
(22)δm=d(g(xm1),g(xm+11))+d(g(xm2),g(xm+12))+⋯+d(g(xmr),g(xm+1r)),
we have
(23)δm=d(g(xm1),g(xm+11))+d(g(xm2),g(xm+12))+⋯+d(g(xmr),g(xm+1r))≤rϕ(1r∑n=1r‍d(g(xm-1n),g(xmn)))=rϕ(1rδm-1).
Since ϕ(t)<t for all t>0, therefore, δm≤δm-1 for all m so that {δm} is a nonincreasing sequence. Since it is bounded below, there is some δ≥0 such that
(24)limm→∞δm=+δ.
We shall show that δ=0. Suppose, on the contrary that δ>0. Taking the limits as m→+∞ of both the sides of (23) and keeping in mind our supposition that limr→t+ϕ(r)<t for all t>0, we have
(25)δ=limm→∞δm≤limm→∞rϕ(1rδm-1)=rϕ(1rδ)<rδr=δ,
which is a contradiction so that δ=0 yielding thereby
(26)limm→∞d(g(xm1),g(xm+11))+d(g(xm2),g(xm+12))+⋯+d(g(xmr),g(xm+1r))=0.
Next we show that all the sequences {g(xm1)},{g(xm2)},…, and {g(xmr)} are Cauchy sequences. If possible, suppose that at least one of {g(xm1)},{g(xm2)},… and {g(xmr)} is not a Cauchy sequence. Then there exists  ϵ>0 and sequences of positive integers {l(k)} and {m(k)} such that for all positive integers k,
(27)m(k)>l(k)>k,d(gxl(k)1,gxm(k)1)+d(gxl(k)2,gxm(k)2)+⋯+d(gxl(k)r,gxm(k)r)≥ϵ,d(gxl(k)1,gxm(k)-11)+d(gxl(k)2,gxm(k)-12)+⋯+d(gxl(k)r,gxm(k)-1r)<ϵ.
Now,
(28)ϵ≤d(gxl(k)1,gxm(k)1)+d(gxl(k)2,gxm(k)2)+⋯+d(gxl(k)r,gxm(k)r)≤d(gxl(k)1,gxm(k)-11)+d(gxl(k)2,gxm(k)-12)+⋯+d(gxl(k)r,gxm(k)-1r)+d(gxm(k)-11,gxm(k)1)+d(gxm(k)-12,gxm(k)2)+⋯+d(gxm(k)-1r,gxm(k)r),
that is,
(29)ϵ≤d(gxl(k)1,gxm(k)1)+d(gxl(k)2,gxm(k)2)+⋯+d(gxl(k)r,gxm(k)r)≤ϵ+d(gxm(k)-11,gxm(k)1)+d(gxm(k)-12,gxm(k)2)+⋯+d(gxm(k)-1r,gxm(k)r).
Letting k→∞ in the above inequality and using (26), we have
(30)limk→∞d(gxl(k)1,gxm(k)1)+d(gxl(k)2,gxm(k)2)+⋯+d(gxl(k)r,gxm(k)r)=ϵ.
Again,
(31)d(gxl(k)+11,gxm(k)+11)+d(gxl(k)+12,gxm(k)+12)+⋯+d(gxl(k)+1r,gxm(k)+1r)≤d(gxl(k)+11,gxl(k)1)+d(gxl(k)+12,gxl(k)2)+⋯+d(gxl(k)+1r,gxl(k)r)+d(gxl(k)1,gxm(k)1)+d(gxl(k)2,gxm(k)2)+⋯+d(gxl(k)r,gxm(k)r)+d(gxm(k)1,gxm(k)+11)+d(gxm(k)2,gxm(k)+12)+⋯+d(gxm(k)r,gxm(k)+1r),(32)d(gxl(k)1,gxm(k)1)+d(gxl(k)2,gxm(k)2)+⋯+d(gxl(k)r,gxm(k)r)≤d(gxl(k)+11,gxl(k)1)+d(gxl(k)+12,gxl(k)2)+⋯+d(gxl(k)+1r,gxl(k)r)+d(gxl(k)+11,gxm(k)+11)+d(gxl(k)+12,gxm(k)+12)+⋯+d(gxl(k)+1r,gxm(k)+1r)+d(gxm(k)1,gxm(k)+11)+d(gxm(k)2,gxm(k)+12)+⋯+d(gxm(k)r,gxm(k)+1r).
Letting k→∞ in the above inequalities, using (26) and (30), we have
(33)limk→∞{d(gxl(k)+11,gxm(k)+11)+d(gxl(k)+12,gxm(k)+12)+⋯+d(gxl(k)+1r,gxm(k)+1r)}=ϵ.
Now,
(34)d(gxl(k)+11,gxm(k)+11)+d(gxl(k)+12,gxm(k)+12)+⋯+d(gxl(k)+1r,gxm(k)+1r)=d(F(xl(k)1,xl(k)2,…,xl(k)r),F(xm(k)1,xm(k)2,…,xm(k)r))+d(F(xl(k)2,xl(k)3,…,xl(k)r,xl(k)1),F(xm(k)2,xm(k)3,…,xm(k)r,xl(k)1))…+d(F(xl(k)r,xl(k)1,…,xl(k)r-1),F(xm(k)r,xm(k)1,…,xm(k)r-1))≤rϕ(1r∑n=1r‍d(g(xl(k)n),g(xm(k)n))).
Letting k→∞ in the above inequality, using (30), (33), and the property of ϕ, we have
(35)ϵ≤rϕ(ϵr)<rϵr=ϵ,
which is a contradiction. Therefore, {g(xm1)},{g(xm2)},…, and {g(xmr)} are Cauchy sequences in (X,d). Since the metric space (X,d) is complete, so there exist x1,x2,…,xr∈X such that
(36)limm→∞g(xm1)=x1,limm→∞g(xm2)=x2,…,limm→∞g(xmr)=xr.
By the continuity of g and (36), we can have
(37)limm→∞g(g(xm1))=g(x1),limm→∞g(g(xm2))=g(x2),…,limm→∞g(g(xmr))=g(xr).
Using (16) and the commutativity of F with g, we get
(38)g(g(xm+11))=g(F(xm1,xm2,…,xmr))=F(g(xm1),g(xm2),…,g(xmr)),g(g(xm+12))=g(F(xm2,xm3,…,xmr))=F(g(xm2),g(xm3),…,g(xm1)),⋮g(g(xm+1r))=g(F(xmr,xm1,…,xmr-1))=F(g(xmr),g(xm1),…,g(xmr-1)).
Now, we show that F and g have an r-tupled coincidence point. To accomplish this, suppose (a) holds, then using (16), (37), and the continuities of F and g, we obtain
(39)g(x1)=limm→∞g(g(xm+11))=limm→∞g(F(xm1,xm2,…,xmr))=F(limm→∞g(xm1),limm→∞g(xm2),…,limm→∞g(xmr))=F(x1,x2,…,xr).
Similarly, we can also show that
(40)g(x2)=F(x2,x3,…,xr,x1),g(x3)=F(x3,…,xr,x1,x2),⋮g(xr)=F(xr,x1,…,xr-1).
Hence the element (x1,x2,…,xr)∈∏i=1r‍Xi is a r-tupled coincidence point of the mappings F and g. Next, assume that (b) holds. Since {g(xmi)} is nondecreasing or nonincreasing as i is odd or even and g(xmi)→xi as m→∞, we have g(xmi)⪯xi, when  i  is odd while g(xmi)≽xi, when i is even.
Sinceg is monotonically increasing, therefore
(41)g(g(xmi))⪯g(xi)wheniisodd,g(g(xmi))≽g(xi)wheniiseven.
On using triangle inequality together with (16), we get
(42)d(g(x1),F(x1,…,xr))≤d(g(x1),g(g(xm+11)))+d(g(g(xm+11)),F(x1,…,xr))≤d(g(x1),g(g(xm+11)))+ϕ(1r∑n=1r‍d(g(g(xmn)),g(xn))).
Letting m→∞ in the above inequality and using (37), we have g(x1)=F(x1,x2…,xr). Similarly, we can also show that
(43)g(x2)=F(x2,x3…,x1),…,g(xr)=F(xr,x1…,xr-1)
which shows that F and g have an r-tupled coincidence point. This completes the proof.Corollary 14.
Let(X,⪯) be a partially ordered set equipped with a metric d such that (X,d) is a complete metric space. Assume that there is a function ϕ:[0,+∞)→[0,+∞) with ϕ(t)<t and limr→t+ϕ(r)<t for each t>0. Further, suppose that F:∏i=1r‍Xi→X is a mapping such that F has the mixed monotone property satisfying the following conditions:
(44)d(F(x1,x2,…,xr),F(y1,y2,…,yr))≤ϕ(1r∑n=1r‍d(xn,yn)),
for allx1,x2,x3,…,xr,y1,y2,y3,…,yr∈X with x1⪯y1,x2≽y2,x3⪯y3,…,xr≽yr. Also, suppose that either(a)
F is continuous or(b)
X has the following properties:
(i)
if a nondecreasing sequence{xn}→x, then xn⪯x for all n≥0,(ii)
if a nonincreasing sequence{xn}→x, then xn≽x for all n≥0.
If there existx01,x02,x03,…,x0r∈X such that
(45)x01⪯F(x01,x02,x03,…,x0r),x02≽F(x02,x03,…,x0r,x01),x03⪯F(x03,…,x0r,x01,x02),⋮x0r≽F(x0r,x01,x02,…,x0r-1),
then F has an r-tupled fixed point in X; that is, there exist x1,x2,x3,…,xr∈X such that
(46)x1=F(x1,x2,x3,…,xr),x2=F(x2,x3,…,xr,x1),x3=F(x3,…,xr,x1,x2),⋮xr=F(xr,x1,x2,x3,…,xr-1).Proof.
Settingg=I, the identity mapping, in Theorem 13, we obtain Corollary 14.
Also, Theorem13 immediately yields the following corollary.Corollary 15.
Let(X,⪯) be a partially ordered set equipped with a metric d such that (X,d) is a complete metric space. Suppose that F:∏i=1r‍Xi→X and g:X→X are two maps such that F has the mixed g-monotone property satisfying the following conditions:(i)
F(∏i=1r‍Xi)⊆g(X),(ii)
g is continuous and monotonically increasing,(iii)
(g,F) is a commutating pair,(iv)
d(F(x1,x2,…,xr),F(y1,y2,…,yr))≤(k/r)∑n=1rd(g(xn),g(yn)),k∈[0,1)for all x1,x2,x3,…,xr,y1,y2,y3,…,yr∈X with gx1⪯gy1,gx2≽gy2,gx3⪯gy3,…,gxr≽gyr. Also, suppose that either(a)
F is continuous or(b)
X has the following properties:
(i)
if a nondecreasing sequence{xn}→x, then xn⪯x for all n≥0,(ii)
if a nonincreasing sequence{xn}→x, then xn≽x for all n≥0.If there exist x01,x02,x03,…,x0r∈X such that
(47)gx01⪯F(x01,x02,x03,…,x0r),gx02≽F(x02,x03,…,x0r,x01),gx03⪯F(x03,…,x0r,x01,x02),⋮gx0r≽F(x0r,x01,x02,…,x0r-1),
then F and g have an r-tupled coincidence point in X;
that is, there existx1,x2,x3,…,xr∈X such that
(48)gx1=F(x1,x2,x3,…,xr),gx2=F(x2,x3,…,xr,x1),gx3=F(x3,…,xr,x1,x2),⋮gxr=F(xr,x1,x2,x3,…,xr-1).Proof.
Settingϕ(t)=k·t with k∈[0,1) in Theorem 13, we obtain Corollary 15.
The following example illustrates Theorem13.Example 16.
LetX=[0,1]. Then X is a complete metric space under natural ordering ⪯ of real numbers and natural metric d(x,y)=|x-y| for all x,y∈X. Define g:X→X as g(x)=x/(r-1) wherein r is fixed and r>1 for all x∈X. Also, define F:∏i=1r‍Xi→X by
(49)F(x1,x2,…,xr)=x1-x2+x3-⋯+xr-1-xrr2-1,
for all x1,x2,…,xr∈X. Define ϕ:[0,∞)→[0,∞) as ϕ(t)=(r/(r+1))t, where r is fixed as earlier. Then ϕ has all the properties mentioned in Theorem 13. Also F and g are commutating mapping in X.
Next, we verify inequality (12) (of Theorem 13)
(50)d(F(x1,x2,…,xr),F(y1,y2,…,yr))=d(x1-x2+x3-x4+⋯+xr-1-xrr2-1,y1-y2+y3-y4+⋯+yr-1-yrr2-1)=1r+1|x1-x2+x3-x4+⋯+xr-1-xrr-1-y1-y2+y3-y4+⋯+yr-1-yrr-1|≤rr+11r{|x1-y1|+|x2-y2|+⋯+|xr-yr|r-1}=rr+11r((gxr,gyr)d(gx1,gy1)+d(gx2,gy2)+⋯+d(gxr,gyr)(gx2,gy2))=ϕ(1r∑n=1rd(gxn,gyn)).
Thus all the conditions of Theorem 13 (without order) are satisfied and (0,0,…,0) is a r-tupled coincidence point of F and g.



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*Source: 532867-2013-03-17.xml*