OPERATORS Journal of Operators 2314-5072 2314-5064 Hindawi Publishing Corporation 532867 10.1155/2013/532867 532867 Research Article On n-Tupled Coincidence Point Results in Metric Spaces Imdad Mohammad 1 1 Soliman Ahmed H. 2, 3 3 Choudhury Binayak S. 4 Das Pradyut 4 Kong Qingkai 1 Department of Mathematics Aligarh Muslim University, Aligarh 202 002 India amu.ac.in 2 Department of Mathematics, Faculty of Science Al-Azhar University, Assiut 71524 Egypt azhar.edu.eg 3 Department of Mathematics, Faculty of Science King Khalid University, Abha 9004 Saudi Arabia kku.edu.sa 4 Department of Mathematics, Bengal Engineering and Science University Shibpur, Howrah, West Bengal 711103 India becs.ac.in 2013 17 3 2013 2013 12 11 2012 01 02 2013 05 02 2013 2013 Copyright © 2013 Mohammad Imdad et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We prove some n-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.

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 .

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., ). In this continuation, Lakshmikantham and Cirić  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 .

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 Section 2, 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 [<xref ref-type="bibr" rid="B11">22</xref>]).

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)xu,yv.

Definition 2 (see [<xref ref-type="bibr" rid="B11">22</xref>]).

Let (X,) be a partially ordered set and F:X×XX. 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,yX, (2)  x1,x2X,x1x2F(x1,y)F(x2,y),y1,y2X,y1y2F(x,y1)F(x,y2).

Definition 3 (see [<xref ref-type="bibr" rid="B11">22</xref>]).

Let (X,) be a partially ordered set and F:X×XX. 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 [<xref ref-type="bibr" rid="B11">22</xref>]).

Let (X,) be a partially ordered set equipped with a metric d such that (X,d) is a complete metric space. Let F:X×XX 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)]xu,yv. If there exist x0,y0X such that x0F(x0,y0) and y0F(y0,x0), then there exist x,yX such that x=F(x,y) and y=F(y,x).

Definition 5 (see [<xref ref-type="bibr" rid="B14">21</xref>]).

Let (X,) be a partially ordered set and F:X×XX and g:XX 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,x2X, g(x1)g(x2) implies F(x1,y)F(x2,y), for any yX, and, for all y1,y2X, g(y1)g(y2) implies F(x,y1)F(x,y2), for any xX.

Definition 6 (see [<xref ref-type="bibr" rid="B14">21</xref>]).

An element (x,y)X×X is called a coupled coincidence point of mappings F:X×XX and g:XX if (5)F(x,y)=g(x),F(y,x)=g(y).

Theorem 7 (see [<xref ref-type="bibr" rid="B14">21</xref>]).

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 limrt+ϕ(r)<t for each t>0. Let F:X×XX and g:XX 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,vX 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

F  is continuous,

X has the following properties:

if nondecreasing sequence {xn}x, then xnx for all n,

if nonincreasing sequence {yn}y, then yyn for all n.

if a nondecreasing sequence {xn}x, then xnx for all n0,

if a nonincreasing sequence {xn}x, then xnx for all n0.

If there exist x0,y0X such that (7)g(x0)F(x0,y0),g(y0)F(y0,x0), then there exist x,yX 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=1rXiX 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,

for all x11,x21X, x11x21 implies F(x11,x2,x3,,xr)F(x21,x2,x3,,xr)

for all x12,x22X, x12x22 implies F(x1,x12,x3,,xr)F(x1,x22,x3,,xr)

for all x13,x23X, x13x23 implies F(x1,x2,x13,,xr)F(x1,x2,x23,,xr)

for all x1r,x2rX, x1rx2r implies F(x1,x2,x3,,x1r)F(x1,x2,x3,,x2r).

Definition 9.

Let (X,) be a partially ordered set. Let F:i=1rXiX and g:XX 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,

for all x11,x21X, gx11gx21 implies F(x11,x2,x3,,xr)F(x21,x2,x3,,xr),

for all x12,x22X, gx12gx22 implies F(x1,x12,x3,,xr)F(x1,x22,x3,,xr),

for all x13,x23X, gx13gx23 implies F(x1,x2,x13,,xr)F(x1,x2,x23,,xr),

for all x1r,x2rX, gx1rgx2r implies F(x1,x2,x3,,x1r)F(x1,x2,x3,,x2r).

Definition 10.

Let X be a nonempty set. An element (x1,x2,x3,,xr)i=1rXi is called an r-tupled fixed point of the mapping F:i=1rXiX 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.

Let X be a nonempty set. An element (x1,x2,x3,,xr)i=1rXi is called an r-tupled coincidence point of the mappings F:i=1rXiX and g:XX 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.

Let X be a nonempty set. The mappings F:i=1rXiX and g:XX are said to be commutating if (11)g(F(x1,x2,,xr))=(F(g(x1),g(x2),,g(xr))   for all x1,x2,,xrX.

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 limrt+ϕ(r)<t for each t>0. Further, suppose that F:i=1rXiX and g:XX are two maps such that F has the mixed g-monotone property satisfying the following conditions:

F(i=1rXi)g(X),

g is continuous and monotonically increasing,

(g,F) is a commutating pair,

(12)d(F(x1,x2,,xr),F(y1,y2,,yr))ϕ(1rn=1rd(g(xn),g(yn))),

for all x1,x2,x3,,xr,y1,y2,y3,,yrX,with gx1gy1,gx2gy2,gx3gy3,,gxrgyr. Also, suppose that either

F is continuous or

X has the following properties: (13)(i)if  a  nondecreasing  sequence  {xn}x,thenxnxn0,(ii)if  a  nonincreasing  sequence  {xn}x,thenxnxn0.

If there exist x01,x02,x03,,x0rX such that (14)gx01F(x01,x02,x03,,x0r),gx02F(x02,x03,,x0r,x01),gx03F(x03,,x0r,x01,x02),gx0rF(x0r,x01,x02,,x0r-1), then F and g have a r-tupled coincidence point; that is, there exist x1,x2,x3,,xrX 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 with x01,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 n0, (17)gxn1gxn+11,gxn2gxn+12,gxn3gxn+13,,gxnrgxn+1r.(18)gx01F(x01,x02,x03,,x0r)=x11,gx02F(x02,x03,,x0r,x01)=x12,gx03F(x03,,x0r,x01,x02)=x13,gx0rF(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 n0.

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(1rn=1rd(g(xm-1n),g(xmn))). Similarly, we can inductively write (21)d(g(xm2),g(xm+12))ϕ(1rn=1rd(g(xm-1n),g(xmn))),d(g(xmr),g(xm+1r))ϕ(1rn=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(1rn=1rd(g(xm-1n),g(xmn)))=rϕ(1rδm-1). Since ϕ(t)<t for all t>0, therefore, δmδm-1 for all m so that {δm} is a nonincreasing sequence. Since it is bounded below, there is some δ0 such that (24)limmδm=+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 limrt+ϕ(r)<t for all t>0, we have (25)δ=limmδmlimmrϕ(1rδm-1)=rϕ(1rδ)<rδr=δ, which is a contradiction so that δ=0 yielding thereby (26)limmd(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)ϵd(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)ϵd(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)limkd(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(1rn=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)ϵrϕ(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,,xrX such that (36)limmg(xm1)=x1,limmg(xm2)=x2,,limmg(xmr)=xr. By the continuity of g and (36), we can have (37)limmg(g(xm1))=g(x1),limmg(g(xm2))=g(x2),,limmg(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)=limmg(g(xm+11))=limmg(F(xm1,xm2,,xmr))=F(limmg(xm1),limmg(xm2),,limmg(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.

Since g is monotonically increasing, therefore (41)g(g(xmi))g(xi)when  i  is  odd,g(g(xmi))g(xi)when  i  is  even.

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(1rn=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 limrt+ϕ(r)<t for each t>0. Further, suppose that F:i=1rXiX 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(1rn=1rd(xn,yn)), for allx1,x2,x3,,xr,y1,y2,y3,,yrX with x1y1,x2y2,x3y3,,xryr. Also, suppose that either

F is continuous or

X has the following properties:

if a nondecreasing sequence {xn}x, then xnx for all n0,

if a nonincreasing sequence {xn}x, then xnx for all n0.

If there exist x01,x02,x03,,x0rX such that (45)x01F(x01,x02,x03,,x0r),x02F(x02,x03,,x0r,x01),x03F(x03,,x0r,x01,x02),x0rF(x0r,x01,x02,,x0r-1), then F has an r-tupled fixed point in X; that is, there exist x1,x2,x3,,xrX 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.

Setting g=I, the identity mapping, in Theorem 13, we obtain Corollary 14.

Also, Theorem 13 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=1rXiX and g:XX are two maps such that F has the mixed g-monotone property satisfying the following conditions:

F(i=1rXi)g(X),

g is continuous and monotonically increasing,

(g,F) is a commutating pair,

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,,yrX with gx1gy1,gx2gy2,gx3gy3,,gxrgyr. Also, suppose that either

F is continuous or

X has the following properties:

if a nondecreasing sequence {xn}x, then xnx for all n0,

if a nonincreasing sequence {xn}x, then xnx for all n0.

If there exist x01,x02,x03,,x0rX such that (47)gx01F(x01,x02,x03,,x0r),gx02F(x02,x03,,x0r,x01),gx03F(x03,,x0r,x01,x02),gx0rF(x0r,x01,x02,,x0r-1), then F and g have an r-tupled coincidence point in X;

that is, there exist x1,x2,x3,,xrX 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.

Setting ϕ(t)=k·t with k[0,1) in Theorem 13, we obtain Corollary 15.

The following example illustrates Theorem 13.

Example 16.

Let X=[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,yX. Define g:XX as g(x)=x/(r-1) wherein r is fixed and r>1 for all xX. Also, define F:i=1rXiX by (49)F(x1,x2,,xr)=x1-x2+x3-+xr-1-xrr2-1, for all x1,x2,,xrX. 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(1rn=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.

Acknowledgments

All the authors are grateful to both the learned referees for their fruitful suggestions and remarks towards the improvement of this paper.

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