Common Fixed Point Theorems for Four Mappings on Cone Metric Type Space

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Volume 2011, Article ID 589725,15pages doi:10.1155/2011/589725

Research Article

Common Fixed Point Theorems for Four Mappings

on Cone Metric Type Space

Aleksandar S. Cvetkovi´c,

1

_{Marija P. Stani´c,}

2

Sladjana Dimitrijevi´c,

2

_{and Suzana Simi´c}

2

1_{Department of Mathematics, Faculty of Mechanical Engineering, University of Belgrade,}

Kraljice Marije 16, 11120 Belgrade, Serbia

2_{Department of Mathematics and Informatics, Faculty of Science, University of Kragujevac,}

Radoja Domanovi´ca 12, 34000 Kragujevac, Serbia

Correspondence should be addressed to Aleksandar S. Cvetkovi´c,acvetkovic@mas.bg.ac.rs

Received 9 December 2010; Revised 26 January 2011; Accepted 3 February 2011

Academic Editor: Fabio Zanolin

Copyrightq2011 Aleksandar S. Cvetkovi´c 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.

In this paper we consider the so called a cone metric type space, which is a generalization of a cone metric space. We prove some common fixed point theorems for four mappings in those spaces. Obtained results extend and generalize well-known comparable results in the literature. All results are proved in the settings of a solid cone, without the assumption of continuity of mappings.

1. Introduction

Replacing the real numbers, as the codomain of a metric, by an ordered Banach space we obtain a generalization of metric space. Such a generalized space, called a cone metric space, was introduced by Huang and Zhang in1. They described the convergence in cone metric space, introduced their completeness, and proved some fixed point theorems for contractive mappings on cone metric space. Cones and ordered normed spaces have some applications in optimization theorysee2. The initial work of Huang and Zhang 1inspired many authors to prove fixed point theorems, as well as common fixed point theorems for two or more mappings on cone metric space, for example,3–14.

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All results are proved in the settings of a solid cone, without the assumption of continuity of mappings.

The paper is organized as follows. InSection 2we repeat some definitions and well-known results which will be needed in the sequel. InSection 3we prove common fixed point theorems. Also, we presented some corollaries which show that our results are generalization of some existing results in the literature.

2. Definitions and Notation

LetEbe a real Banach space andP a subset ofE. Byθwe denote zero element ofEand by intP the interior ofP. The subsetP is called a coneif and only if

iPis closed, nonempty andP /{θ};

iia, b∈Ê,a, b≥0, andx, y∈P implyaxby∈P;

iiiP∩−P {θ}.

For a given coneP, a partial orderingwith respect toPis introduced in the following way:x yif and only if y−x ∈ P. One writes x ≺ yto indicate thatx y, butx /y. If

y−x∈intP, one writesxy.

If intP /∅, the conePis called solid.

In the sequel we always suppose thatEis a real Banach space,Pis a solid cone inE, andis partial ordering with respect toP.

Analogously with definition of metric type space, given in 16, we consider cone metric type space.

Definition 2.1. Let X be a nonempty set and Ea real Banach space with coneP. A vector-valued functiond:X×X → Eis said to be a cone metric type function onX with constant

K≥1 if the following conditions are satisfied:

d1θdx, yfor allx, y∈Xanddx, y θif and only ifxy;

d2dx, y dy, xfor allx, y∈X;

d3dx, yKdx, z dz, yfor allx, y, z∈X.

The pairX, dis called a cone metric type spacein brief CMTS.

Remark 2.2. ForK1 inDefinition 2.1we obtain a cone metric space introduced in1. Definition 2.3. LetX, dbe a CMTS and{xn}a sequence inX.

c1{xn}converges tox∈Xif for everyc∈Ewithθcthere existsn0 ∈Æ such that dxn, xcfor alln > n0. We write limn→ ∞xnx, orxn → x,n → ∞.

c2If for everyc ∈ Ewithθ cthere existsn0 ∈ Æ such thatdxn, xm cfor all n, m > n0, then{xn}is called a Cauchy sequence inX.

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Example 2.4. LetB{ei|i1, . . . , n}be orthonormal basis ofÊ

n _{with inner product}_·_,_·_{. Let}

p >0, and define

Xp

x|x:0,1−→Ê

n_,

₁

0

|xt, ek|pdt∈Ê, k1, . . . , n

, 2.1

wherexrepresents class of elementxwith respect to equivalence relation of functions equal almost everywhere. We chooseEÊ

n _and

PB

y∈Ê

n _|_{y, e} i

≥0, i1, . . . , n. 2.2

We show thatPBis a solid cone. Letyk∈PB,k∈Æ, with property limk→∞yky. Since scalar

product is continuous, we get limk→∞yk, ei limk→∞yk, ei y, ei,i1, . . . , n. Clearly,

it must bey, ei ≥ 0,i 1, . . . , n, and, hence,y ∈PB, that is,PB is closed. It is obvious that

θ /e1∈PB/{θ}, and fora, b≥0, and allz, y∈PB, we haveazby, ei az, eiby, ei≥0,

i1, . . . , n. Finally, ifz ∈PB∩−PBwe havez, ei ≥0 and−z, ei ≥0,i1, . . . , n, and it

follows thatz, ei 0,i 1, . . . , n, and, since Bis complete, we getz 0. Let us choose

z n_i₁ei. It is obvious thatz∈intPB, since if not, for everyε >0 there existsy /∈ PBsuch

that|1−y, ei| ≤ n_i₁|1−y, ei|21/2z−y< ε. If we chooseε1/4, we conclude that it must bey, ei>1−1/4>0, hencey∈PB, which is contradiction.

Finally, defined:Xp×Xp → PBby

df, g

n

i1

ei

₁

0

_f₋_g

t, eipdt, f, g∈Xp. 2.3

Then it is obvious that Xp, d is CMTS withK 2p−1. Letf, g,hbe functions such that f, e1 1,g, e1 −2,h, e1 0, andf, ei g, ei h, ei 0,i2, . . . , n, withp2 give

df, g 9e1,df, h e1, anddh, g 4e1, which proves 5e1 df, h dh, gdf, g

9e1, but 9e1df, g2df, h dh, g 10e1.

The following properties are well known in the case of a cone metric space, and it is easy to see that they hold also in the case of a CMTS.

Lemma 2.5. LetX, dbe a CMTS over-ordered real Banach spaceEwith a coneP. The following properties holda, b, c∈E.

p1Ifabandbc, thenac.

p2Ifθacfor allc∈intP, thenaθ.

p3Ifaλa, wherea∈Pand0≤λ <1, thenaθ.

p4Letxn → θinEand letθc. Then there exists positive integern0such thatxncfor eachn > n0.

Definition 2.6see17. LetF, G:X → Xbe mappings of a setX. IfyFxGxfor some

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Definition 2.7see17. LetFandGbe self-mappings of setXandCF, G {x∈X :Fx Gx}. The pair{F, G}is called weakly compatible if mappingsFandGcommute at all their coincidence points, that is, ifFGxGFxfor allx∈ CF, G.

Lemma 2.8see5. LetFandGbe weakly compatible self-mappings of a setX. IfFandGhave a unique point of coincidenceyFxGx, thenyis the unique common fixed point ofFandG.

3. Main Results

Theorem 3.1. LetX, dbe a CMTS with constant1 ≤ K ≤ 2 andP a solid cone. Suppose that self-mappingsF, G, S, T :X → X are such thatSX ⊂ GX,TX ⊂ FXand that for some constant

λ∈0,1/Kfor allx, y∈Xthere exists

ux, y∈

KdFx, Gy, KdFx, Sx, KdGy, Ty, Kd

Fx, TydGy, Sx

2

, 3.1

such that the following inequality

dSx, Ty λ Ku

x, y, 3.2

holds. If one ofSX,TX,FX, orGXis complete subspace ofX, then{S, F}and{T, G}have a unique point of coincidence inX. Moreover, if{S, F}and{T, G}are weakly compatible pairs, thenF,G,S, andThave a unique common fixed point.

Proof. Let us choosex0 ∈ Xarbitrary. SinceSX ⊂ GX, there existsx1 ∈X such that Gx1

Sx0 z0. SinceTX ⊂FX, there existsx2 ∈X such thatFx2 Tx1 z1. We continue in this

manner. In general,x2n1 ∈Xis chosen such thatGx2n1 Sx2nz2n, andx2n2∈Xis chosen

such thatFx2n2 Tx2n1z2n1.

First we prove that

dzn, zn1αdzn−1, zn, n≥1, 3.3

whereαmax{λ, λK/2−λK}, which will lead us to the conclusion that{zn}is a Cauchy

sequence, sinceα ∈ 0,1 it is easy to see that 0 < λK/2−λK < 1. To prove this, it is necessary to consider the cases of an odd integernand of an evenn.

Forn21,∈Æ0, we havedz21, z22 dSx22, Tx21, and from3.2there

exists

ux22, x21∈

KdFx22, Gx21, KdFx22, Sx22,

KdGx21, Tx21, KdFx22, Tx21 ₂dGx21, Sx22

Kdz21, z2, Kdz21, z22,Kdz2₂, z22

,

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such thatdz21, z22λ/Kux22, x21. Thus we have the following three cases: idz21, z22λdz21, z2;

iidz21, z22 λdz21, z22, which, because of property p3, implies

dz21, z22 θ;

iiidz21, z22λ/2dz2, z22, that is, by usingd3,

dz21, z22 λK₂ dz2, z21 λK₂ dz21, z22, 3.5

which impliesdz21, z22λK/2−λKdz2, z21.

Thus, inequality3.3holds in this case. Forn2,∈Æ0, we have

dz2, z21 dSx2, Tx21

λ

Kux2, x21, 3.6

where

ux2, x21∈

KdFx2, Gx21, KdFx2, Sx2,

KdGx21, Tx21, KdFx2, T21 ₂dGx21, Sx2

Kdz2−1, z2, Kdz2, z21,Kdz2−₂1, z21

.

3.7

Thus we have the following three cases:

idz2, z21λdz2−1, z2;

iidz2, z21λdz2, z21, which impliesdz2, z21 θ;

iiidz2, z21 λ/2dz2−1, z21 λK/2dz2−1, z2 λK/2dz2, z21, which impliesdz2, z21λK/2−λKdz2, z2−1.

So, inequality3.3is satisfied in this case, too.

Therefore,3.3is satisfied for alln∈Æ0, and by iterating we get

dzn, zn1αndz0, z1. 3.8

SinceK≥1, form > nwe have

dzn, zmKdzn, zn1 K2dzn1, zn2 · · ·Km−n−1dzm−1, zm

Kαn_K2_αn1_{· · ·}_Km−n_αm−1_d_z 0, z1

Kαn

1−Kαdz0, z1−→θ, asn−→ ∞.

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Now, byp4andp1, it follows that for everyc∈intPthere exists positive integern0such

thatdzn, zmcfor everym > n > n0, so{zn}is a Cauchy sequence.

Let us suppose that SX is complete subspace of X. Completeness of SX implies existence ofz∈SXsuch that limn→ ∞z2n limn→ ∞Sx2nz. Then, we have

lim

n→ ∞Gx2n1nlim→ ∞Sx2nnlim→ ∞Fx2nnlim→ ∞Tx2n1 z, 3.10

that is, for anyθc, for suﬃciently largenwe havedzn, zc. Sincez∈SX⊂GX, there

existsy∈Xsuch thatzGy. Let us prove thatzTy. Fromd3and3.2, we have

dTy, zKdTy, Sx2n

KdSx2n, zλu

x2n, y

Kdz2n, z, 3.11

where

ux2n, y

∈

KdFx2n, Gy

, KdFx2n, Sx2n, Kd

Gy, Ty, Kd

Fx2n, Ty

dGy, Sx2n

2

Kdz2n−1, z, Kdz2n−1, z2n, Kd

z, Ty, Kd

z2n−1, Ty

dz, z2n

2

.

3.12

Therefore we have the following four cases:

idTy, zKλdz2n−1, z Kdz2n, zKλ·c/2Kλ K·c/2K c, asn → ∞; iidTy, zKλdz2n−1, z2nKdz2n, zKλ·c/2KλK·c/2K c, asn → ∞; iiidTy, zKλdz, Ty Kdz2n, z, that is,

dTy, z K

1−Kλdz2n, z K

1−Kλ·

1−Kλ

K ·cc, asn−→ ∞; 3.13

ivdTy, zKλ/2dz2n−1, Ty dz, z2n Kdz2n, z, that is, because ofd3,

dTy, z Kλ

2

Kdz2n−1, z Kd

z, Tydz, z2nKdz2n, z, 3.14

which implies

dTy, z 1

1−K2_λ/₂

K2_λ

2 dz2n−1, z

Kλ

2 K

dz2n, z

K2λ

2−K2_λ

K2_λ

c

2

Kλ2 2−K2_λ

2−K2_λ

Kλ2

c

2 c, asn−→ ∞,

3.15

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Therefore,dTy, z cfor eachc ∈ intP. So, byp2we havedTy, z θ, that is,

TyGyz,yis a coincidence point, andzis a point of coincidence ofTandG.

SinceTX ⊂ FX, there existsv ∈Xsuch thatz Fv. Let us prove thatSv z. From

d3and3.2, we have

dSv, zKdSv, Tx2n1 KdTx2n1, zλuv, x2n1 Kdz2n1, z, 3.16

where

uv, x2n1

∈

KdFv, Gx2n1, KdFv, Sv, KdGx2n1, Tx2n1, K

dFv, Tx2n1 dGx2n1, Sv

2

Kdz, z2n, Kdz, Sv, Kdz2n, z2n1, Kdz, z2n1 ₂dz2n, Sv

.

3.17

Therefore we have the following four cases:

idSv, zKλdz, z2n Kdz2n1, z; iidSv, zKλdz, Sv Kdz2n1, z; iiidSv, zKλdz2n, z2n1 Kdz2n1, z;

ivdSv, zKλ/2dz, z2n1 dz2n, Sv Kdz2n1, z.

By the same arguments as above, we conclude thatdSv, z θ, that is,SvFvz. So,zis a point of coincidence ofSandF, too.

Now we prove thatzis unique point of coincidence of pairs{S, F}and{T, G}. Suppose that there existsz∗which is also a point of coincidence of these four mappings, that is,Fv∗ Gy∗ Sv∗Ty∗z∗. From3.2,

dz, z∗ dSv, Ty∗ λ Ku

v, y∗, 3.18

where

uv, y∗∈

KdFv, Gy∗, KdFv, Sv, dGy∗, Ty∗, Kd

Fv, Ty∗dGy∗, Sv

2

{Kdz, z∗, θ}.

3.19

Usingp3we getdz, z∗ θ, that is,zz∗. Therefore,zis the unique point of coincidence of pairs{S, F}and{T, G}. If these pairs are weakly compatible, thenzis the unique common fixed point ofS,F,T, andG, byLemma 2.8.

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We give one simple, but illustrative, example.

Example 3.2. LetX Ê,EÊ, andP 0,∞. Let us definedx, y |x−y|

2_{for all}_{x, y}_∈_X_.

ThenX, dis a CMTS, but it is not a cone metric space since the triangle inequality is not satisfied. Starting with Minkowski inequalitysee18forp2, by using the inequality of arithmetic and geometric means, we get

|x−z|2≤x−y2y−z22x−y|x−z| ≤2x−y2y−z2. 3.20

Here,K2.

Let us define four mappingsS, F, T, G:X → Xas follows:

SxMaxb, Fxaxb, TxMcxd, Gxcxd, 3.21

where x ∈ X,a /0, c /0, and M < 1/√2. Since SX FX TX GX X we have trivially SX ⊂ GX and TX ⊂ FX. Also, X is a complete space. Further, dSx, Ty

|Maxb−Mcyd|2 M2dFx, Gy, that is, there existsλ M2 < 1/2 1/K such that3.2is satisfied.

According toTheorem 3.1,{S, F}and{T, G}have a unique point of coincidence inX, that is, there exists uniquez∈Xand there existx, y∈Xsuch thatzSxFxTy Gy. It is easy to see thatx−b/a,y−d/c, andz0.

If{S, F}is weakly compatible pair, we haveSFx FSx, which impliesMb b, that is,b 0. Similarly, if{T, G}is weakly compatible pair, we haveTGy GTy, which implies

Mdd, that is,d0. Thenxy 0, andz0 is the unique common fixed point of these four mappings.

The following two theorems can be proved in the same way asTheorem 3.1, so we omit the proofs.

Theorem 3.3. LetX, dbe a CMTS with constantK ≥ 2 andP a solid cone. Suppose that self-mappings F, G, S, T : X → X are such thatSX ⊂ GX,TX ⊂ FX and that for some constant

λ∈0,2/K2_{for all}_{x, y}_∈_X_{there exists}

ux, y∈

KdFx, Gy, KdFx, Sx, KdGy, Ty, Kd

Fx, TydGy, Sx

2

, 3.22

dSx, Ty λ Ku

x, y, 3.23

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Theorem 3.4. LetX, dbe a CMTS with constantK ≥ 1 andP a solid cone. Suppose that self-mappings F, G, S, T : X → X are such thatSX ⊂ GX,TX ⊂ FX and that for some constant

λ∈0,1/Kfor allx, y∈Xthere exists

ux, y∈

KdFx, Gy, KdFx, Sx, KdGy, Ty,d

Fx, TydGy, Sx

2

, 3.24

dSx, Ty λ Ku

x, y, 3.25

Theorems3.1and3.4are generalizations of13, Theorem 2.2. As a matter of fact, for

K1, from Theorems3.1and3.4, we get13, Theorem 2.2.

If we chooseT SandG F, from Theorems3.1,3.3, and3.4we get the following results for two mappings on CMTS.

Corollary 3.5. LetX, dbe a CMTS with constant1 ≤ K ≤ 2 andP a solid cone. Suppose that self-mappingsF, S:X → Xare such thatSX⊂FXand that for some constantλ∈0,1/Kfor all

x, y∈Xthere exists

ux, y∈

KdFx, Fy, KdFx, Sx, KdFy, Sy, Kd

Fx, SydFy, Sx

2

, 3.26

dSx, Sy λ Ku

x, y, 3.27

holds. IfFXorSXis complete subspace ofX, thenFandShave a unique point of coincidence inX. Moreover, if{F, S}is a weakly compatible pair, thenFandShave a unique common fixed point.

Corollary 3.6. LetX, dbe a CMTS with constantK ≥ 2andP a solid cone. Suppose that self-mappingsF, S :X → Xare such thatSX⊂ FXand that for some constantλ ∈0,2/K2_{for all}

ux, y∈

KdFx, Fy, KdFx, Sx, KdFy, Sy, Kd

Fx, SydFy, Sx

2

, 3.28

dSx, Sy λ Ku

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Corollary 3.7. LetX, dbe a CMTS with constantK ≥ 1andP a solid cone. Suppose that self-mappingsF, S :X → X are such thatSX ⊂ FXand that for some constantλ ∈ 0,1/Kfor all

ux, y∈

KdFx, Fy, KdFx, Sx, KdFy, Sy,d

Fx, SydFy, Sx

2

, 3.30

dSx, Sy λ Ku

x, y, 3.31

Theorem 3.8. LetX, dbe a CMTS with constantK ≥ 1 andP a solid cone. Suppose that self-mappingsF, G, S, T :X → X are such thatSX ⊂GX,TX ⊂FXand that there exist nonnegative constantsai,i1, . . . ,5, satisfying

a1a2a32Kmax{a4, a5}<1, a3Ka4K2<1, a2Ka5K2<1, 3.32

such that for allx, y∈Xinequality

dSx, Tya1d

Fx, Gya2dFx, Sx a3d

Gy, Tya4d

Fx, Tya5d

Gy, Sx,

3.33

Proof. We define sequences{xn}and{zn}as in the proof ofTheorem 3.1. First we prove that

dzn, zn1αdzn−1, zn, n≥1, 3.34

where

αmax

a1a3a5K

1−a2−a5K

,a1a2a4K

1−a3−a4K

, 3.35

which implies that{zn}is a Cauchy sequence, since, because of3.32, it is easy to check that

α∈0,1. To prove this, it is necessary to consider the cases of an odd and of an even integer

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Forn21,∈Æ0, we havedz21, z22 dSx22, Tx21, and from3.33we

have

dSx22, Tx21a1dFx22, Gx21 a2dFx22, Sx22

a3dGx21, Tx21 a4dFx22, Tx21 a5dGx21, Sx22, 3.36

that is,

dz21, z22a1dz21, z2 a2dz21, z22 a3dz2, z21 a4dz21, z21 a5dz2, z22

a1a3dz2, z21 a2dz21, z22 a5dz2, z22

a1a3dz2, z21 a2dz21, z22 a5Kdz2, z21 a5Kdz21, z22

a1a3a5Kdz2, z21 a2a5Kdz21, z22.

3.37

Therefore,

dz21, z22

a1a3a5K

1−a2−a5K

dz2, z21, 3.38

that is, inequality3.34holds in this case.

Similarly, forn 2, ∈Æ0, we havedz2, z21 dSx2, Tx21, and from3.33

we have

dSx2, Tx21a1dFx2, Gx21 a2dFx2, Sx2 a3dGx21, Tx21 a4dFx2, Tx21 a5dGx21, Sx2,

3.39

that is,

dz2, z21a1dz2−1, z2 a2dz2−1, z2 a3dz2, z21 a4dz2−1, z21 a5dz2, z2

a1a2dz2−1, z2 a3dz2, z21 a4dz2−1, z21

a1a2dz2−1, z2 a3dz2, z21 a4Kdz2−1, z2 a4Kdz2, z21

a1a2a4Kdz2−1, z2 a3a4Kdz2, z21.

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Thus,

dz2, z21 a₁1₋_aa2a4K 3−a4K

dz2−1, z2, 3.41

and inequality3.34holds in this case, too.

By the same arguments as inTheorem 3.1we conclude that{zn}is a Cauchy sequence.

Let us suppose that SX is complete subspace of X. Completeness of SX implies existence ofz∈SXsuch that limn→ ∞z2n limn→ ∞Sx2nz. Then, we have

lim

n→ ∞Gx2n1nlim→ ∞Sx2nnlim→ ∞Fx2nnlim→ ∞Tx2n1 z, 3.42

that is, for anyθc, for suﬃciently largenwe havedzn, zc. Sincez∈SX⊂GX, there

existsy∈Xsuch thatzGy. Let us prove thatzTy. Fromd3and3.33, we have

dTy, zKdTy, Sx2n

KdSx2n, z

a1Kd

Fx2n, Gy

a2KdFx2n, Sx2n a3Kd

Gy, Ty

a4Kd

Fx2n, Ty

a5Kd

Gy, Sx2n

KdSx2n, z

a1Kdz2n−1, z a2Kdz2n−1, z2n a3Kd

z, Ty

a4Kd

z2n−1, Ty

a5Kdz, z2n Kdz2n, z

a1Kdz2n−1, z a2Kdz2n−1, z2n a3Kd

z, Ty

a4K2dz2n−1, z a4K2d

z, Tya5Kdz, z2n Kdz2n, z.

3.43

The sequence{zn}converges toz, so for eachc∈intPthere existsn0 ∈Æsuch that for every n > n0

dTy, z 1

1−a3K−a4K2

a1Kdz2n−1, z a2Kdz2n−1, z2n a4K2dz2n−1, z a5Kdz, z2n Kdz2n, z

a1K

1−a3K−a4K2 ·

1−a3K−a4K2

a1K ·

c

5

a2K

1−a3K−a4K2 ·

1−a3K−a4K2

a2K ·

c

5

a4K2

1−a3K−a4K2

·1−a3K−a4K2

a4K2

·c

5

a5K

1−a3K−a4K2

·1−a3K−a4K2

a5K ·

c

5

K

1−a3K−a4K2 ·

1−a3K−a4K2

K · c

5

c,

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because of3.32. Now, byp2 it follows thatdTy, z θ, that is, Ty z. So, we have

Ty Gyz, that is,yis a coincidence point, andzis a point of coincidence of mappingsT

andG.

SinceTX ⊂ FX, there existsv ∈ X such thatz Fv. Let us prove thatSv z, too. Fromd3and3.33, we have

dSv, zKdSv, Tx2n1 KdTx2n1, z

a1KdFv, Gx2n1 a2KdFv, Sv a3KdGx2n1, Tx2n1

a4KdFv, Tx2n1 a5KdGx2n1, Sv KdTx2n1, z

a1Kdz, z2n a2Kdz, Sv a3Kdz2n, z2n1

a4Kdz, z2n1 a5Kdz2n, Sv KdTx2n1, z

a1Kdz, z2n a2Kdz, Sv a3Kdz2n, z2n1

a4Kdz, z2n1 a5K2dz2n, z a5K2dSv, z KdTx2n1, z,

3.45

and by the same arguments as above, we conclude thatdSv, z θ, that is,Sv Fv z. Thus,zis a point of coincidence of mappingsSandF, too.

Suppose that there exists z∗ which is also a point of coincidence of these four mappings, that is,Fv∗Gy∗Sv∗Ty∗z∗. From3.33we have

dz, z∗ dSv, Ty∗

a1Kd

Fv, Gy∗a2KdFv, Sv a3Kd

Gy∗, Ty∗

a4Kd

Fv, Ty∗a4Kd

Gy∗, Sv

a1Kdz, z∗ a2Kdz, z a3Kdz∗, z∗ a4Kdz, z∗ a5Kdz∗, z

a1a4a5Kdz, z∗,

3.46

andbecause ofp3it follows thatzz∗. Therefore,zis the unique point of coincidence of pairs{S, F}and{T, G}, and we havezSvFvGyTy. If{S, F}and{T, G}are weakly compatible pairs, thenzis the unique common fixed point ofS,F,T, andG, byLemma 2.8.

The proofs for the cases in whichFX,GX, orTXis complete are similar.

Theorem 3.8 is a generalization of 13, Theorem 2.8. Choosing K 1 from

Theorem 3.8we get the following corollary.

Corollary 3.9. Let X, d be cone metric space and P a solid cone. Suppose that self-mappings

F, G, S, T :X → X are such thatSX⊂ GX,TX ⊂ FXand that there exist nonnegative constants

ai,i1, . . . ,5, satisfyinga1a2a32 max{a4, a5}<1, such that for allx, y∈Xinequality

dSx, Tya1d

Fx, Gya2dFx, Sx a3d

Gy, Tya4d

Fx, Tya5d

Gy, Sx,

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If we chooseT SandGF, fromTheorem 3.8, we get the following result for two mappings on CMTS.

Corollary 3.10. LetX, dbe a CMTS with constantK ≥ 1andP a solid cone. Suppose that self-mappingsF, S : X → X are such thatSX ⊂ FXand that there exist nonnegative constants ai,

i1, . . . ,5, satisfying

a1a2a32Kmax{a4, a5}<1, a3Ka4K2<1, a2Ka5K2<1, 3.48

such that for allx, y∈Xinequality

dSx, Sya1d

Fx, Fya2dFx, Sx a3d

Fy, Sya4d

Fx, Sya5d

Fy, Sx,

3.49

holds. If one ofSXorFXis complete subspace ofX, thenSandFhave a unique point of coincidence inX. Moreover, if{F, S}is a weakly compatible pair, thenFandShave a unique common fixed point.

Acknowledgments

The authors are indebted to the referees for their valuable suggestions, which have contributed to improve the presentation of the paper. The first two authors were supported in part by the Serbian Ministry of Science and Technological DevelopmentsGrant no. 174015.

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