Some Fixed Point Theorem for Mapping on Complete G-Metric Spaces

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Volume 2008, Article ID 189870,12pages doi:10.1155/2008/189870

Research Article

Some Fixed Point Theorem for Mapping on

Complete

G-Metric Spaces

Zead Mustafa, Hamed Obiedat, and Fadi Awawdeh

Department of Mathematics, The Hashemite University, P.O. Box 330127, Zarqa 13115, Jordan

Correspondence should be addressed to Zead Mustafa,[email protected]

Received 1 April 2008; Accepted 10 July 2008

Recommended by Brailey Sims

We prove some fixed point results for mapping satisfying suﬃcient conditions on complete G-metric space, also we showed that if theG-metric spaceX, Gis symmetric, then the existence and uniqueness of these fixed point results follow from well-known theorems in usual metric space

X, dG, whereX, dGis the usual metric space which defined from theG-metric spaceX, G.

Copyrightq2008 Zead Mustafa 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.

1. Introduction

During the sixties, the notion of 2-metric space introduced by G¨ahler see 1, 2 as a generalization of usual notion of metric spaceX, d. But diﬀerent authors proved that there is no relation between these two functions, for instance, Ha et al. in3show that 2-metric need not be continuous function, further there is no easy relationship between results obtained in the two settings.

In 1992, Bapure Dhage in his Ph.D. thesis introduce a new class of generalized metric space calledD-metric spaces4,5.

In a subsequent series of papers, Dhage attempted to develop topological structures in such spacessee5–7. He claimed thatD-metrics provide a generalization of ordinary metric functions and went on to present several fixed point results.

But in 2003 in collaboration with Brailey Sims, we demonstrated in8that most of the claims concerning the fundamental topological structure ofD-metric space are incorrect, so, we introduced more appropriate notion of generalized metric space as follows.

Definition 1.1see9. Let X be a nonempty set, and letG :X×X×X → Rbe a function satisfying the following properties:

G1Gx, y, z 0 ifx y z;

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G3Gx, x, y≤Gx, y, z,for allx, y, z∈X,with_{z /}y;

G4Gx, y, z Gx, z, y Gy, z, x · · ·,symmetry in all three variables;

G5Gx, y, z≤Gx, a, a Ga, y, z, for allx, y, z, a∈X,rectangle inequality.

Then the function G is called ageneralized metric, or, more specifically, aG-metriconX, and the pairX, Gis called aG-metric space.

Definition 1.2see9. LetX, Gbe aG-metric space, and letxnbe sequence of points of X, a pointx∈Xis said to be thelimitof the sequencexn, if limn,m→∞Gx, xn, xm 0, and one says that the sequencexnisG-convergenttox.

Thus, that ifxn→xin aG-metric spaceX, G, then for any >0,there existsN∈N such thatGx, xn, xm< , for alln, m≥N.

Proposition 1.3see9. LetX, Gbe aG-metric space, then the following are equivalent.

1 xnisG-convergent tox.

2Gxn, xn, x→0, asn→ ∞.

3Gxn, x, x→0, asn→ ∞.

4Gxm, xn, x→0, asm, n→ ∞.

Definition 1.4see9. LetX, Gbe aG-metric space, a sequencexnis calledG-Cauchy if for every > 0, there isN ∈ N such thatGxn, xm, xl < ,for alln, m, l ≥ N; that is, if Gxn, xmxl→0 asn, m, l→ ∞.

Proposition 1.5see8. IfX, Gis aG-metric space, then the following are equivalent.

1The sequencexnisG-Cauchy.

2For every >0,there existsN∈Nsuch thatGxn, xm, xm< ,for alln, m≥N.

Definition 1.6see9. LetX, GandX, Gbe twoG-metric spaces, and letf :X, G →

X, Gbe a function, thenf is said to beG-continuous at a pointa∈ Xif and only if, given >0, there existsδ >0 such thatx, y∈X; andGa, x, y< δimpliesGfa, fx, fy< . A functionfisG-continuous atXif and only if it isG-continuous at alla∈X.

Proposition 1.7see9. LetX, G,X, Gbe twoG-metric spaces. Then a functionf:X→X isG-continuous at a pointx∈Xif and only if it isGsequentially continuous atx; that is, whenever

xnisG-convergent tox,fxnisG-convergent tofx.

Definition 1.8 see 9. A G-metric space X, G is called symmetric G-metric space if Gx, y, y Gy, x, xfor allx, y∈X.

Proposition 1.9 see9. LetX, Gbe a G-metric space, then the function Gx, y, zis jointly continuous in all three of its variables.

Proposition 1.10see8. EveryG-metric spaceX, Gwill define a metric spaceX, dGby

dGx, y Gx, y, y Gy, x, x, ∀x, y∈X. 1.1

Note that ifX, Gis a symmetricG-metric space, then

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However, ifX, Gis not symmetric, then it holds by theG-metric properties that

3

2Gx, y, y≤dGx, y≤3Gx, y, y, ∀x, y∈X, 1.3 and that in general these inequalities cannot be improved.

Definition 1.11see9. A G-metric spaceX, Gis said to beG-complete or completeG -metricif everyG-Cauchy sequence inX, GisG-convergent inX, G.

Proposition 1.12 see 9. A G-metric space X, G is G-complete if and only ifX, dG is a complete metric space.

2. Main results

Here we start our work with the following theorem.

Theorem 2.1. LetX, Gbe a completeG-metric space, and letT :X →Xbe a mapping satisfying one of the following conditions:

GTx, Ty, Tz≤{aGx, y, zbGx, Tx, TxcGy, Ty, TydGz, Tz, Tz}

2.1

or

GTx, Ty, Tz≤ {aGx, y, z bGx, x, Tx cGy, y, Ty dGz, z, Tz}

2.2

for allx, y, z∈Xwhere0≤abcd <1, thenThas a unique fixed point (sayu, i.e.,Tu u), andT isG-continuous atu.

Proof. Suppose thatTsatisfies condition2.1, then for allx, y∈X, we have

GTx, Ty, Ty≤aGx, y, y bGx, Tx, Tx cdGy, Ty, Ty,

GTy, Tx, Tx≤aGy, x, x bGy, Ty, Ty cdGx, Tx, Tx. 2.3

Suppose thatX, Gis symmetric, then by definition of metricX, dGand1.2, we get

dGTx, Ty≤adGx, y cdb

2 dGx, Tx

cdb

2 dGy, Ty, ∀x, y∈X. 2.4 In this line, since 0 < abcd <1, then the existence and uniqueness of the fixed point follows from well-known theorem in metric spaceX, dG see10.

However, ifX, Gis not symmetric then by definition of metricX, dGand1.3, we get

dGTx, Ty≤adGx, y 2cdb

3 dGx, Tx

2cdb

3 dGy, Ty, 2.5

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Letx0∈Xbe an arbitrary point, and define the sequencexnbyxn Tnx0. By2.1,

we have

Gxn, xn1, xn1≤aGxn−1, xn, xn bGxn−1, xn, xn cdGxn, xn1, xn1, 2.6

then

Gxn, xn1, xn1≤ ₁₋ab

cdGxn−1, xn, xn. 2.7

Letq ab/1−cd, then 0≤q <1 since 0≤abcd <1. So,

Gxn, xn1, xn1≤qGxn−1, xn, xn. 2.8

Continuing in the same argument, we will get

Gxn, xn1, xn1≤qnGx0, x1, x1. 2.9

Moreover, for alln, m∈N; n < m, we have by rectangle inequality that

Gxn, xm, xm≤Gxn, xn1, xn1Gxn1, xn2, xn2Gxn2, xn3, xn3· · ·Gxm−1, xm, xm ≤qn_qn1_{· · ·}_qm−1_G_x

0, x1, x1

≤ qn

1−qGx0, x1, x1,

2.10

and so limGxn, xm, xm 0, as n, m → ∞. Thus xn isG-Cauchy sequence. Due to the completeness ofX, G, there existsu∈Xsuch thatxnisG-converge tou.

Suppose thatTu/u, then

Gxn, Tu, Tu≤aGxn−1, u, u bGxn−1, xn, xn cdGu, Tu, Tu, 2.11

taking the limit as n → ∞, and using the fact that the function G is continuous, then Gu, Tu, Tu≤cdGu, Tu, Tu. This contradiction implies thatu Tu.

To prove uniqueness, suppose that_{u /}vsuch thatTv v, then

Gu, v, v≤aGu, v, v bGu, Tu, Tu cdGv, Tv, Tv aGu, v, v, 2.12

which implies thatu v.

To show thatTisG-continuous atu, letyn⊆Xbe a sequence such that limyn u. we can deduce that

Gu, Tyn, Tyn≤aGu, yn, yn bGu, Tu, Tu cdGyn, Tyn, Tyn

aGu, yn, yn cdGyn, Tyn, Tyn, 2.13

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Taking the limit asn→ ∞, from which we see thatGu, Tyn, Tyn→0 and so, by

Proposition 1.7,Tyn→u Tu. It is proved thatT isG-continuous atu.

IfT satisfies condition2.2, then the argument is similar to that above. However, to show that the sequencexnisG-Cauchy, we start with

Gxn, xn, xn1≤aGxn−1, xn−1, xn bcGxn−1, xn−1, xn dGxn, xn, xn1, 2.14

then

Gxn, xn, xn1≤ a₁₋bc

d Gxn−1, xn−1, xn. 2.15 Letq abc/1−d, then 0≤q <1 since 0≤abcd <1.

Continuing in the same way, we find that

Gxn, xn, xn1≤qnGx0, x0, x1. 2.16

Then for all n, m ∈ N; n < m, we have by repeated use of the rectangle inequality Gxn, xn, xm≤qn_/₁₋_q_G_x

0, x0, x1.

Corollary 2.2. LetX, Gbe a completeG-metric space and letT :X →Xbe a mapping satisfying one of the following conditions:

GTmx, Tmy, Tmz

≤aGx, y, y bGx, Tmx, Tmx cGy, Tmy, Tmy dGz, Tmz, Tmz

2.17 or

GTmx, Tmy, Tmz≤aGx, y, ybGx, x, TmxcGy, y, TmydGz, z, Tmz,

2.18

for allx, y, z∈X, where0 ≤abcd <1. ThenT has a unique fixed point (sayu), andTm_is G-continuous atu.

Proof. From the previous theorem, we see thatTm_{has a unique fixed point}_say_u_{, that is,} Tm_u_u_{. But}_T_u_T_Tm_u_Tm1_u_Tm_T_u_{, so}_T_u_{is another fixed point for}_Tm and by uniquenessTu u.

GTx, Ty, Tz≤kmax ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

Gx, Tx, Tx, Gy, Ty, Ty, Gz, Tz, Tz

⎫ ⎪ ⎪ ⎬ ⎪ ⎪

⎭ 2.19

or

GTx, Ty, Tz≤kmax ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

Gx, x, Tx, Gy, y, Ty, Gz, z, Tz

⎫ ⎪ ⎪ ⎬ ⎪ ⎪

⎭, 2.20

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Proof. Suppose thatTsatisfies condition2.19, then for allx, y∈X, GTx, Ty, Ty≤kmax{Gx, Tx, Tx, Gy, Ty, Ty},

GTy, Tx, Tx≤kmax{Gy, Ty, Ty, Gx, Tx, Tx}. 2.21

Suppose thatX, Gis symmetric, then by definition of the metricX, dGand1.2we get dGTx, Ty≤kmax{dGx, Tx, dGy, Ty}, ∀x, y∈X. 2.22 Sincek < 1, then the existence and uniqueness of the fixed point follows from a theorem in metric spaceX, dG see11.

However, ifX, Gis not symmetric, then by definition of the metricX, dGand1.3, we get

dGTx, Ty≤ 4k

3 max{dGx, Tx, dGy, Ty}, ∀x, y∈X. 2.23 The metric condition gives no information about this map since 4k/3 need not be less than 1, but we will proof it byG-metric.

Letx0 ∈ X be an arbitrary point, and define the sequencexnbyxn Tnx0. By 2.19, we can verify that

Gxn, xn1, xn1≤kmax{Gxn−1, xn, xn, Gxn, xn1, xn1}

kGxn−1, xn, xn since 0≤k <1.

2.24

Continuing in the same argument, we will find

Gxn, xn1, xn1≤knGx0, x1, x1. 2.25

For alln, m∈N; n < m, we have by rectangle inequality that

Gxn, xm, xm≤Gxn, xn1, xn1Gxn1, xn2, xn2Gxn2, xn3, xn3· · ·Gxm−1, xm, xm ≤kn_kn1_{· · ·}_km−1_G_x

0, x1, x1

≤ kn

1−kGx0, x1, x1.

2.26 Then, limGxn, xm, xm 0, asn, m → ∞, and thusxnisG-Cauchy sequence. Due to the completeness ofX, G, there existsu∈Xsuch thatxn→u.

Suppose thatTu/u, thenGxn1, Tu, Tu≤kmax{Gxn1, xn2, xn2, Gu, Tu,

Tu}and by taking the limit asn→ ∞, and using the fact that the functionGis continuous, we get thatGu, Tu, Tu≤kGu, Tu, Tu. This contradiction implies thatu Tu.

To prove uniqueness, suppose that _{u /}v such that Tv v, then Gu, v, v ≤ kmax{Gv, v, v, Gu, u, u} 0 which implies thatu v.

To show thatTisG-continuous atu, letyn⊆Xbe a sequence such that limyn u, then

Gu, Tyn, Tyn≤kmax{Gu, Tu, Tu, Gyn, Tyn, Tyn} kGyn, Tyn, Tyn.

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Corollary 2.4. LetX, Gbe a completeG-metric space and letT :X →Xbe a mapping satisfying one of the following conditions for somem∈N:

GTmx, Tmy, Tmz≤kmax ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

Gx, Tm_x_{, T}m_x_, Gy, Tm_y_{, T}m_y_, Gz, Tm_z_{, T}m_z

⎫ ⎪ ⎪ ⎬ ⎪ ⎪

⎭ 2.28

or

GTm_x_{, T}m_y_{, T}m_z_≤_k_max ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

Gx, x, Tm_x_, Gy, y, Tm_y_, Gz, z, Tmz

⎫ ⎪ ⎪ ⎬ ⎪ ⎪

⎭, 2.29

for allx, y, z∈X, thenThas a unique fixed point (sayu) andTmisG-continuous atu.

Proof. We use the same argument as inCorollary 2.2.

GTx, Ty, Ty≤kmax{Gx, Ty, Ty, Gy, Tx, Tx, Gy, Ty, Ty} 2.30

or

GTx, Ty, Ty≤kmax{Gx, x, Ty, Gy, y, Tx, Gy, y, Ty}, 2.31

for allx, y∈X, wherek ∈0,1. ThenThas a unique fixed point (sayu), andTisG-continuous at u.

Proof. Suppose thatTsatisfies condition2.30, then for allx, y∈X,

GTx, Ty, Ty≤kmax{Gx, Ty, Ty, Gy, Tx, Tx, Gy, Ty, Ty},

GTy, Tx, Tx≤kmax{Gx, Ty, Ty, Gy, Tx, Tx, Gx, Tx, Tx}. 2.32

Suppose thatX, Gis symmetric, then by definition of the metricX, dGand1.2, we have.

dGTx, Ty≤ k 2max ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

dGx, Ty, dGy, Tx, dGy, Ty

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ k 2max ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

dGx, Ty, dGy, Tx, dGx, Tx

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

≤kmax{dGx, Ty, dGy, Tx, dGx, Tx, dGy, Ty}, ∀x, y∈X.

2.33

Since 0≤k <1, then the existence and uniqueness of the fixed point follows from a theorem in metric spaceX, dG see12.

However, ifX, Gis not symmetric, then by definition of the metricX, dGand1.3, we have

dGTx, Ty≤ 2k 3 max ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

dGx, Ty, dGy, Tx, dGy, Ty

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ 2k 3 max ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

dGx, Ty, dGy, Tx, dGx, Tx

⎫ ⎪ ⎪ ⎬ ⎪ ⎪

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for allx, y∈X, then the metric spaceX, dGgives no information about this map since 4k/3 need not be less than 1. But we will proof it byG-metric.

Letx0 ∈ X be arbitrary point, and define the sequencexnbyxn Tnx0, then by 2.30and usingk <1, we deduce that

Gxn, xn1, xn1≤kmax{Gxn−1, xn1, xn1, Gxn, xn1, xn1} kGxn−1, xn1, xn1. 2.35

So,

Gxn, xn1, xn1≤kGxn−1, xn1, xn1, 2.36

and using

Gxn−1, xn1, xn1≤kmax{Gxn−2, xn1, xn1, Gxn, xn−1, xn−1, Gxn, xn1, xn1}, 2.37

then,

Gxn, xn1, xn1≤k2max{Gxn−2, xn1, xn1, Gxn, xn−1, xn−1}. 2.38

Continuing in this procedure, we will have

Gxn, xn1, xn1≤knΓn, 2.39

whereΓn max{Gxi, xj, xj; for alli, j∈ {0,1, . . . , n1}}.

Forn, m∈N; n < m,letΓ max{Γi; for alli n, . . . , m−1}. Then, for alln, m∈N; n < m, we have by rectangle inequality

Gxn, xm, xm≤Gxn, xn1, xn1Gxn1, xn2, xn2Gxn2, xn3, xn3· · ·Gxm−1, xm, xm ≤knΓnkn1Γn1· · ·km−1Γm−1

≤knkn1· · ·km−1Γ

≤ kn

1−kΓ.

2.40

This prove that limGxn, xm, xm 0, asn, m → ∞, and thusxnisG-Cauchy sequence. SinceX, GisG-complete then there existsu∈Xsuch thatxnisG-converge tou.

Gxn, Tu, Tu≤kmax{Gxn−1, Tu, Tu, Gu, xn1, xn1, Gu, Tu, Tu}. 2.41

Taking the limit as n → ∞, and using the fact that the function G is continuous, we get Gu, Tu, Tu≤kGu, Tu, Tu, this contradiction implies thatu Tu.

To prove the uniqueness, suppose that_{u /}vsuch thatTv v. So, by2.30, we have that

Gu, v, v≤kmax{Gu, v, v, Gv, u, u} ⇒Gu, v, v≤kGv, u, u. 2.42

Again we will findGv, u, u≤kGu, v, v, so

Gu, v, v≤k2Gu, v, v; 2.43

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Gu, Tyn, Tyn≤kmax{Gu, Tyn, Tyn, Gyn, Tu, Tu, Gyn, Tyn, Tyn}.

2.44

But,Gyn, Tyn, Tyn ≤ Gyn, u, u Gu, Tyn, Tyn, so,Gu, Tyn, Tyn ≤ k/1− kGyn, u, u.

Taking the limit asn→ ∞, from which we see thatGu, Tyn, Tyn→0 and so, by

Proposition 1.7, we haveTyn→u Tuwhich implies thatT isG-continuous atu.

Corollary 2.6. LetX, Gbe a completeG-metric space, and letT :X→X, be a mapping satisfying one of the following conditions:

GTx, Ty, Tz≤kmax ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

Gx, Ty, Ty, Gx, Tz, Tz, Gy, Tx, Tx, Gy, Tz, Tz, Gz, Tx, Tx, Gz, Ty, Ty

⎫ ⎪ ⎪ ⎬ ⎪ ⎪

⎭ 2.45

or

GTx, Ty, Tz≤kmax ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

Gx, x, Ty, Gx, x, Tz, Gy, y, Tx, Gy, y, Tz, Gz, z, Tx, Gz, z, Ty

⎫ ⎪ ⎪ ⎬ ⎪ ⎪

⎭, 2.46

for allx, y, z∈Xwherek∈0,1, thenThas a unique fixed point (sayu) andT isG-continuous at u.

Proof. If we letz yin conditions2.45and2.46, then they become conditions2.30and

2.31, respectively, inTheorem 2.5; so the proof follows fromTheorem 2.5.

Corollary 2.7. LetX, Gbe a completeG-metric space and letT :X →Xbe a mapping satisfying one of the following conditions:

GTmx, Tmy, Tmz≤kmax ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

Gx, Tm_y_{, T}m_y_{, G}_{x, T}m_z_{, T}m_z_, Gy, Tm_x_{, T}m_x_{, G}_{y, T}m_z_{, T}m_z_, Gz, Tm_x_{, T}m_x_{, G}_{z, T}m_y_{, T}m_y

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭,

GTm_x_{, T}m_y_{, T}m_z_≤_k_max ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

Gx, x, Tm_y_{, G}_{x, x, T}m_z_, Gy, y, Tm_x_{, G}_{y, y, T}m_z_, Gz, z, Tmx, Gz, z, Tmy

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭,

GTmx, Tmy, Tmy≤kmax{Gx,Tmy,Tmy, Gy,Tmx,Tmx, Gy,Tmy,Tmy},

2.47 or,

GTm_x_{, T}m_y_{, T}m_y_≤_k_max_{_G_{x, x, T}m_y_{, G}_{y, y, T}m_x_{, G}_{y, y, T}m_y_}_, _2.48

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Proof. The proof follows fromTheorem 2.5,Corollary 2.6, and from an argument similar to that used inCorollary 2.2.

GTx, Ty, Ty≤kmax{Gx, Ty, Ty, Gy, Tx, Tx} 2.49

or

GTx, Ty, Ty≤kmax{Gx, x, Ty, Gy, y, Tx}, 2.50

for allx, y∈X, wherek∈0,1, thenThas a unique fixed point (sayu), andTisG-continuous atu.

Proof. Since whenever the mapping satisfies condition 2.49, or 2.50, then it satisfies condition 2.45, or 2.46, respectively, in Theorem 2.5. Then the proof follows from

Theorem 2.5.

Theorem 2.9. LetX, Gbe a completeG-metric space, and letT :X→X, be a mapping satisfying one of these conditions

GTx, Ty, Ty≤a{Gx, Ty, Ty Gy, Tx, Tx} 2.51

or

GTx, Ty, Ty≤a{Gx, x, Ty Gy, y, Tx}, 2.52

for allx, y∈X, wherea∈0,1/2, thenThas a unique fixed point (sayu), andT isG-continuous atu.

Proof. Suppose thatTsatisfies condition2.51, then we have

GTx, Ty, Ty≤a{Gy, Tx, Tx Gx, Ty, Ty},

GTy, Tx, Tx≤a{Gx, Ty, Ty Gy, Tx, Tx}, 2.53

for allx, y∈X.

Suppose thatX, Gis symmetric, then by definition of the metricX, dGand1.2, we get

dGTx, Ty≤a{dGx, Ty dGy, Tx} ∀x, y∈X. 2.54

Since 0≤2a <1, then the existence and uniqueness of the fixed point follow from a theorem in metric spaceX, dG see13.

However, ifX, Gis not symmetric, then by definition of the metricX, dGand1.3, we have

dGTx, Ty≤ 4a

3 dGx, Ty 4a

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Letx0 ∈ X be arbitrary point, and define the sequencexnbyxn Tnx0, then by 2.51, we have

Gxn, xn1, xn1≤a{Gxn−1, xn1, xn1 Gxn, xn, xn} aGxn−1, xn1, xn1. 2.56

But

Gxn−1, xn1, xn1≤aGxn−1, xn, xn aGxn, xn1, xn1, 2.57

thus we have

Gxn, xn1, xn1≤ ₁₋a

aGxn−1, xn, xn. 2.58

Let k a/1 −a, hence 0 ≤ k < 1 then continue in this procedure, we will get that Gxn, xn1, xn1≤knGx0, x1, x1.

For alln, m∈N; n < m, we have by rectangle inequality

Gxn, xm, xm≤Gxn, xn1, xn1Gxn1, xn2, xn2Gxn2, xn3, xn3· · ·Gxm−1, xm, xm

≤knkn1· · ·km−1Gx0, x1, x1

≤ kn

1−kGx0, x1, x1.

2.59

Then, limGxn, xm, xm 0, as n, m → ∞, and so, xn is G-Cauchy sequence. By completeness ofX, G, there existsu∈Xsuch thatxnisG-converge tou.

Gxn, Tu, Tu≤a{Gxn−1, Tu, Tu Gu, xn, xn}. 2.60

Taking the limit as n → ∞, and using the fact that the function G is continuous, then Gu, Tu, Tu≤aGu, Tu, Tu. This contradiction implies thatu Tu.

To prove uniqueness, suppose that _{u /}v such that Tv v, then Gu, v, v ≤ a{Gu, v, v Gv, u, u}, so

Gu, v, v≤

k a

1−a

Gv, u, u 2.61

again by the same argument, we can verify thatGu, v, v≤k2Gu, v, v,which implies that u v.

Gu, Tyn, Tyn≤a{Gu, Tyn, Tyn Gyn, Tu, Tu}, 2.62

and soGu, Tyn, Tyn≤a/1−aGyn, Tu, Tu.

Taking the limit as n → ∞, from which we see that Gu, Tyn, Tyn → 0. By

(12)

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