Fixed Point Theorems for Contractive Mappings in Complete G-Metric Spaces

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Volume 2009, Article ID 917175,10pages doi:10.1155/2009/917175

Research Article

Fixed Point Theorems for Contractive Mappings in Complete G-Metric Spaces

Zead Mustafa

¹

and Brailey Sims

²

1Department of Mathematics, The Hashemite University, P.O. Box 330127, Zarqa 13115, Jordan

2School of Mathematical and Physical Sciences, The University of Newcastle, NSW 2308, Australia

Correspondence should be addressed to Zead Mustafa,[email protected] Received 31 December 2008; Accepted 7 April 2009

Recommended by H´el`ene Frankowska

We prove some fixed point results for mappings satisfying various contractive conditions on CompleteG-metric Spaces. Also the Uniqueness of such fixed point are proved, as well as we showed these mappings areG-continuous on such fixed points.

Copyrightq2009 Z. Mustafa and B. Sims. 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

Metric spaces are playing an increasing role in mathematics and the applied sciences.

Over the past two decades the development of fixed point theory in metric spaces has attracted considerable attention due to numerous applications in areas such as variational and linear inequalities, optimization, and approximation theory.

Diﬀerent generalizations of the notion of a metric space have been proposed by Gahler 1,2and by Dhage3,4. However, HA et al.5have pointed out that the results obtained by Gahler for his 2 metrics are independent, rather than generalizations, of the corresponding results in metric spaces, while in6the current authors have pointed out that Dhage’s notion of aD-metric space is fundamentally flawed and most of the results claimed by Dhage and others are invalid.

In 2003 we introduced a more appropriate and robust notion of a generalized metric space as follows.

Definition 1.1see7. Let X be a nonempty set, and letG :X×X×X → R be a function satisfying the following axioms:

G1Gx, y, z 0 ifxyz,

G20< Gx, x, y, forall x, y∈X, with x /y,

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G3Gx, x, y≤Gx, y, z, forall x, 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 functionGis called a generalized metric, or, more specifically aG-metric onX, and the pairX, Gis called aG-metric space.

Example 1.2see7. LetX, dbe a usual metric space, thenX, GsandX, GmareG-metric space, where

Gs

x, y, z d

x, y d

y, z

dx, z, ∀x, y, z∈X, Gm

x, y, z

max d

x, y , d

y, z

, dx, z

, ∀x, y, z∈X. 1.1

We now recall some of the basic concepts and results forG-metric spaces that were introduced in7.

Definition 1.3. LetX, Gbe aG-metric space, let xnbe a sequence of points ofX, we say thatxnisG-convergent toxif lim_{n,m→ ∞}Gx, xn, xm 0; that is, for any >0,there exists N∈Nsuch thatGx, xn, xm< , for alln, m≥Nthroughout this paper we mean byNthe set of all natural numbers. We refer toxas the limit of the sequencexnand writexn −−−→G x.

Proposition 1.4. LetX, Gbe aG-metric space then the following are equivalent.

1 xnisG-convergent tox. 2Gxn, xn, x → 0, asn → ∞.

3Gxn, x, x → 0, asn → ∞.

Definition 1.5. LetX, Gbe a G-metric space, a sequence xn is calledG-Cauchy if given >0, there isN∈Nsuch thatGxn, xm, xl< ,for alln, m, l≥Nthat is ifGxn, xm, xl → 0 asn, m, l → ∞.

Proposition 1.6. In aG-metric spaceX, G, the following are equivalent.

1The sequencexnisG-Cauchy.

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

Definition 1.7. LetX, GandX, GbeG-metric spaces and letf :X, G → X, Gbe a function, thenfis said to beG-continuous at a pointa∈Xif given >0, there existsδ >0 such thatx, y ∈ X;Ga, x, y< δimpliesGfa, fx, fy< . A functionfisG-continuous onXif and only if it isG-continuous at alla∈X.

Proposition 1.8. Let X, G, X, G be G-metric spaces, then a function f : X → X is G- continuous at a pointx ∈ X if and only if it isG-sequentially continuous atx; that is, whenever xnisG-convergent tox,fxnisG-convergent tofx.

Proposition 1.9. LetX, Gbe aG-metric space, then the functionGx, y, zis jointly continuous in all three of its variables.

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Definition 1.10. AG-metric spaceX, Gis said to beG-completeor a completeG-metric space if everyG-Cauchy sequence inX, GisG-convergent inX, G.

2. The Main Results

We begin with the following theorem.

Theorem 2.1. LetX, Gbe a completeG-metric space and letT : X → X be a mapping which satisfies the following condition, for allx, y, z∈X,

G

Tx, T y

, Tz

≤kmax G

x, y, z

, Gx, Tx, Tx, G y, T

y , T

y , Gz, Tz, Tz, G

x, T y

, T y

, G

y, Tz, Tz

, Gz, Tx, Tx , 2.1

wherek∈0,1/2. ThenT has a unique fixed point (sayu) andT isG-continuous atu.

Proof. Suppose thatT satisfies condition2.1, letx0 ∈Xbe an arbitrary point, and define the sequencexnbyxnTⁿx0, then by2.1, we have

Gxn, xn1, xn1≤kmax{Gxn−1, xn, xn, Gxn−1, xn1, xn1, Gxn, xn1, xn1} 2.2

so,

Gxn, xn1, xn1≤kmax{Gxn−1, xn1, xn1, Gxn−1, xn, xn}. 2.3

But, byG5, we have

Gxn−1, xn1, xn1≤Gxn−1, xn, xn Gxn, xn1, xn1. 2.4 So,2.3becomes

Gxn,x_n1,x_n1≤kmax{Gxn−1, xn, xn Gxn, xn1, xn1, Gxn−1, xn, xn}. 2.5

So, it must be the case that

Gxn, x_n1, x_n1≤k{Gx_n−1, xn, xn Gxn, x_n1, x_n1}, 2.6

which implies

Gxn, xn1, xn1≤ k

1−kGxn−1, xn, xn. 2.7

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Letqk/1−k, thenq <1 and by repeated application of2.7, we have

Gxn, xn1, xn1≤qⁿGx0, x1, x1. 2.8 Then, for alln, m∈N, n < m,we have by repeated use of the rectangle inequality and2.8 that

Gxn, xm, xm≤Gxn, xn1, xn1 Gxn1, xn2, xn2

Gx_n2, x_n3, x_n3 · · ·Gx_m−1, xm, xm

≤

qⁿ qⁿ¹ · · ·q^m−1

Gx0, x1, x1

≤ qⁿ

1−q Gx0, x1, x1.

2.9

Then, lim Gxn, xm, xm 0, asn, m → ∞, since limqⁿ/1−q Gx0, x1, x1 0, asn, m →

∞. Forn, m, l∈NG5implies thatGxn, xm, xl≤Gxn, xm, xm Gxl, xm, xm, taking limit asn, m, l → ∞, we getGxn, xm, xl → 0. SoxnisG-Cauchy a sequence. By completeness ofX, G, there existsu∈Xsuch thatxnisG-converges tou. Suppose thatTu/u, then

Gxn, Tu, Tu≤kmax

Gx_n−1, u, u, Gx_n−1, xn, xn, Gu, Tu, Tu

Gx_n−1, Tu, Tu, Gu, xn, xn , 2.10 taking the limit as n → ∞, and using the fact that the function G is continuous on its variables, we haveGu, Tu, Tu ≤ k Gu, Tu, Tu, which is a contradiction since 0 ≤ k < 1/2. So,u Tu. To prove uniqueness, suppose thatv /uis such thatTv v, then 2.1implies thatGu, v, v≤kmax{Gu, v, v, Gv, u, u}, thusGu, v, v≤kGv, u, uagain by the same argument we will findGv, u, u≤kGu, v, v, thus

Gu, v, v≤k²Gu, v, v 2.11

which implies thatuv, since 0≤k <1/2. To see thatT isG-continuous atu, letyn⊆Xbe a sequence such that limyn u, then

G T

yn

, Tu, T yn

≤kmax

⎧⎪

⎪⎨

⎪⎪

⎩ G

yn, u, yn

, G yn, T

yn

, T yn

, Gu, Tu, Tu, G

yn, Tu, Tu , G

u, T yn

, T yn

⎫⎪

⎪⎬

⎪⎪

⎭, 2.12

and we deduce that

G T

yn

, u, T yn

≤kmax G

yn, u, yn

, G yn, T

yn

, T yn

, G

yn, u, u 2.13

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butG5implies that

G yn, T

yn

, T yn

≤G

yn, u, u G

u, T yn

, T yn

2.14

and2.13leads to the following cases, 1GTyn, u, Tyn≤kGyn, yn, u, 2GTyn, u, Tyn≤kGyn, u, u, 3GTyn, u, Tyn≤qGyn, u, u.

In each case take the limit as n → ∞ to see that Gu, Tyn, Tyn → 0 and so, by Proposition 1.4, we have that the sequence Tyn is G-convergent to u Tu, therefor Proposition 1.8implies thatTisG-continuous atu.

Remark 2.2. If theG-metric space is boundedthat is, for someM >0 we haveGx, y, z≤M for allx, y, z ∈ Xthen an argument similar to that used above establishes the result for 0≤k <1.

Corollary 2.3. LetX, Gbe a completeG-metric spaces and letT : X → X be a mapping which satisfies the following condition for somem ∈ N and for allx, y, z∈X:

G

T^mx, T^m y

, T^mz

≤kmax

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

G x, y, z

, Gx, T^mx, T^mx, G

y, T^m y

, T^m y

, Gz, T^mz, T^mz, G

x, T^m y

, T^m y

, G

y, T^mz, T^mz , Gz, T^mx, T^mx,

⎫⎪

⎪⎪

⎬

⎪⎪

⎪⎭

, 2.15

wherek∈0,1/2, thenT has a unique fixed point (sayu), andT^misG-continuous atu.

Proof. From the previous theorem, we have thatT^mhas a unique fixed pointsay u, that is, T^mu u. ButTu TT^mu T^m1u T^mTu, soTuis another fixed point forT^m and by uniquenessTuu.

Theorem 2.4. LetX, Gbe a completeG-metric space, and letT : X → X be a mapping which satisfies the following condition for allx, y, z∈X:

G

Tx, T y

, Tz

≤kmax

⎧⎪

⎪⎨

⎪⎪

⎩ G

x, T y

, T y

G

y, Tx, Tx ,

G

y, Tz, Tz G

z, T y

, T

y

, Gx, Tz, Tz Gz, Tx, Tx

⎫⎪

⎪⎬

⎪⎪

⎭, 2.16

wherek∈0,1/2, thenT has a unique fixed point (sayu), andTisG-continuous atu.

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Proof. Suppose thatTsatisfies the condition2.16, letx0∈Xbe an arbitrary point, and define the sequencexnbyxnTⁿx0, then by2.16we get

Gxn, xn1, xn1≤kmax

⎧⎪

⎪⎨

⎪⎪

⎩

Gxn−1, xn1, xn1 Gxn, xn, xn, Gxn, xn1, xn1 Gxn, xn1, xn1,

Gx_n−1, x_n1, x_n1 Gxn, xn, xn

⎫⎪

⎪⎬

⎪⎪

⎭ kmax{Gx_n−1, x_n1, x_n1, 2Gxn, x_n1, x_n1},

2.17

since 0≤k <1/2, then it must be the case that

Gxn, xn1, xn1≤kGxn−1, xn1, xn1 2.18

but fromG5, we have

Gxn−1, xn1, xn1≤Gxn−1, xn, xn Gxn, xn1, xn1, 2.19

so2.18implies that

Gxn, xn1, xn1≤ k

1−k Gxn−1, xn, xn, 2.20

letqk/1−k, thenq <1 and by repeated application of2.20, we have

Gxn, x_n1, x_n1≤qⁿGx0, x1, x1. 2.21

Then, for all n, m ∈ N, n < m, we have, by repeated use of the rectangle inequality,Gxn, xm, xm ≤ Gxn, xn1, xn1 Gxn1, xn2, xn2 Gxn2, xn3, xn3 · · · Gx_m−1, xm, xm ≤ qⁿ qⁿ¹ · · · q^m−1Gx0, x1, x1 ≤ qⁿ/1 − q Gx0, x1, x1. So, lim Gxn, xm, xm 0, asn, m → ∞andxnisG-Cauchy sequence. By the completeness ofX, G, there existsu∈Xsuch thaxnisG-convergent tou.Suppose thatTu/u, then

Gxn, Tu, Tu≤kmax

⎧⎪

⎪⎨

⎪⎪

⎩

Gx_n−1, Tu, Tu Gu, xn, xn, Gu, Tu, Tu Gu, Tu, Tu,

Gxn−1, Tu, Tu Gu, xn, xn

⎫⎪

⎪⎬

⎪⎪

⎭. 2.22

Taking the limit as n → ∞, and using the fact that the function G is continuous in its variables, we get

Gu, Tu, Tu≤kmax{2Gu, Tu, Tu, Gu, Tu, Tu}, 2.23

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since 0≤k <1/2, this contradiction implies thatuTu.To prove uniqueness, suppose that v /usuch thatTv v, then

Gu, v, v≤kmax

⎧⎪

⎪⎨

⎪⎪

⎩

Gu, v, v Gv, u, u, Gv, v, v Gv, v, v, Gu, v, v Gv, u, u

⎫⎪

⎪⎬

⎪⎪

⎭, 2.24

so we deduce that Gu, v, v ≤ kGu, v, v Gv, u, u. This implies that Gu, v, v ≤ k/1− kGv, u, u and by repeated use of the same argument we will find Gv, u, u ≤ k/1−kGu, v, v. Therefor we getGu, v, v≤ k/1−k²Gv, u, u, since 0< k/1−k <1, this contradiction implies thatu v. To show thatT isG-continuous atuletyn ⊆ Xbe a sequence such that limyn uinX, G, then

G T

yn

, Tu, Tu

≤kmax

⎧⎪

⎪⎨

⎪⎪

⎩ G

yn, Tu, Tu G

u, T yn

, T yn

, Gu, Tu, Tu, Gu, Tu, Tu G

yn, Tu, Tu G

u, T yn

, T yn

⎫⎪

⎪⎬

⎪⎪

⎭. 2.25

Thus,2.25becomes

G T

yn

, u, u

≤k G

yn, u, u G

u, T yn

, T yn

2.26

but by G5 we have Gu, Tyn, Tyn ≤ 2GTyn, u, u, therefor 2.26 implies that GTyn, u, u≤kGyn, u, u 2kGTyn, u, uand we deduce that

G T

yn

, u, u

≤ k 1−2kG

yn, u, u

. 2.27

Taking the limit of 2.27 as n → ∞, we see that GTyn, u, u → 0 and so, by Proposition 1.8, we haveTyn → uTuwhich implies thatT isG-continuous atu.

Corollary 2.5. LetX, Gbe a completeG-metric space, and letT : X → X be a mapping which satisfies the following condition for somem∈ Nand for allx, y, z∈X:

G

T^mx, T^m y

, T^mz

≤kmax

⎧⎪

⎪⎨

⎪⎪

⎩ G

x, T^m y

, T^m y

G

y, T^mx, T^mx ,

G

y, T^mz, T^mz G

z, T^m y

, T^m

y

, Gx, T^mz, T^mz Gz, T^mx, T^mx

⎫⎪

⎪⎬

⎪⎪

⎭, 2.28

wherek∈0,1/2, thenT has a unique fixed point (sayu), andT^misG-continuous atu.

Proof. The proof follows from the previous theorem and the same argument used in Corollary 2.3.

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Theorem 2.6. LetX, Gbe a completeG-metric space, and letT : X → X be a mapping which satisfies the following condition, for allx, y∈X,

G

Tx, T y

, T y

≤kmax G

y, T y

, T y

G x, T

y , T

y

, 2G

y, Tx, Tx , 2.29

wherek ∈0,1/3, thenT has a unique fixed point, sayu, andTisG-continuous atu.

Proof. Suppose thatT satisfies the condition2.29. Let x0 ∈ X be an arbitrary point, and define the sequencexnbyxnTⁿx0, then by2.29, we have

Gxn, xn1, xn1≤kmax

Gxn, xn1, xn1 Gxn−1, xn1, xn1,

2Gxn, xn, xn, , 2.30

thusGxn, x_n1, x_n1≤kGxn, x_n1, x_n1 kGx_n−1, x_n1, x_n1and so

Gxn, x_n1, x_n1≤ k

1−k Gx_n−1, x_n1, x_n1. 2.31 But byG5we have

Gxn−1, xn1, xn1≤Gxn−1, xn, xn Gxn, xn1, xn1. 2.32

Letpk/1−2k, thenp∈0,1sincek∈0,1/3and from2.31we deduce that

Gxn, xn1, xn1≤pGxn−1, xn, xn. 2.33 Continuing this procedure we get Gxn, x_n1, x_n1 ≤ pⁿGx0, x1, x1. Then, for all n, m ∈ N, n < m, we have by repeated use of the rectangle inequality that Gxn, xm, xm ≤ Gxn, xn1, xn1 Gxn1, xn2, xn2 Gxn2, xn3, xn3 · · ·Gxm−1, xm, xm≤pⁿpⁿ¹

· · ·p^m−1Gx0, x1, x1≤pⁿ/1−p Gx0, x1, x1.Thus, lim Gxn, xm, xm 0, asn, m → ∞, so,xnisG-Cauchy a sequence. By completeness ofX, G, there existsu∈Xsuch thatxn isG-convergent tou. Suppose thatTu/u, then

Gxn, Tu, Tu≤kmax

Gu, Tu, Tu Gxn−1, Tu, Tu,

2Gu, xn, xn , 2.34 taking the limit as n → ∞, and using the fact that the function G is continuous in its variables, we obtain Gu, Tu, Tu ≤ 2kGu, Tu, Tu. Since 0 < k < 1/3 this is a contradiction so,u Tu. To prove uniqueness, suppose thatv /uis such thatTv v, then

Gu, v, v≤kmax

Gv, v, v Gu, v, v,

2Gv, u, u, , 2.35

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thusGu, v, v≤kmax{Gu, v, v,2Gv, u, u}and we deduce that

Gu, v, v≤2kGv, u, u. 2.36

By the same argument we get

Gv, u, u≤2k Gu, v, v, 2.37

hence,Gu, v, v≤4k² Gu, v, vwhich implies thatuvsince 0≤k <1/3⇒0≤4k²<1.

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

G

Tu, T yn

, T yn

≤kmax G

yn, T yn

, T yn

G u, T

yn

, T yn

, 2G

yn, Tu, Tu , 2.38

therefore,2.38implies two cases.

Case 1. Gu, Tyn, Tyn≤2kGyn, u, u.

Case 2. Gu, Tyn, Tyn≤k/1−kGyn, Tyn, Tyn.

But, byG5we haveGyn, Tyn, Tyn≤Gyn, u, u Gu, Tyn, Tyn, so case 2 implies thatGu, Tyn, Tyn≤ p Gyn, u, u.In each case taking the limit as n → ∞, we see thatGu, Tyn, Tyn → 0 and so, byProposition 1.8, we haveTyn → uTuwhich implies thatT isG-continuous atu.

Corollary 2.7. LetX, Gbe a completeG-metric spaces, and letT :X → X be a mapping which satisfies the following condition for somem∈ Nand for allx, y∈X :

G

T^mx, T^m y

, T^m y

≤kmax G

y, T^m y

, T^m y

G x, T^m

y , T^m

y

, 2G

y, T^mx, T^mx , 2.39

wherek ∈0,1/3, thenT has a unique fixed point, sayu, andT^misG-continuous atu.

Proof. The proof follows from the previous theorem and the same argument used in Corollary 2.3. The following theorem has been stated in 8 without proof, but this can be proved by using Theorem2.6as follows.

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Theorem 2.8see8. LetX, Gbe a completeG-metric space and letT :X → Xbe a mapping which satisfies the following condition, for allx, y, z∈X,

G

Tx, T y

, Tz

≤kmax

⎧⎪

⎪⎨

⎪⎪

⎩

Gz, Tx, Tx G

y, Tx, Tx ,

G

y, Tz, Tz

Gx, Tz, Tz ,

G x, T

y , T

y G

z, T y

, T

y

⎫⎪

⎪⎬

⎪⎪

⎭, 2.40

wherek∈0,1/3, thenT has a unique fixed point, sayu, andTisG-continuous atu.

Proof. Settingz y in condition 2.40, then it reduced to condition 2.29, and the proof follows from Theorem2.6.

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