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Bol. Soc. Paran. Mat. (3s.)v. 37 2(2019): 113–119.

c

SPM –ISSN-2175-1188 on line ISSN-00378712 in press

SPM: www.spm.uem.br/bspm doi:10.5269/bspm.v37i2.34690

A New Approach to the Study of Fixed Point Theorems for Simulation

Functions in G-Metric Spaces

Manoj Kumar and Rashmi Sharma

abstract: In this paper ﬁrst of all, we introduce the mapping ζ: [0,∞)×[0,∞)→

R, called the simulation function and the notion of Z-contraction with respect to

ζwhich generalize several known types of contractions. Secondly, we prove certain

ﬁxed point theorems using simulation functions in G-Metric spaces. An example is

also given to support our results.

Key Words: Simulation function; Contraction mapping; Z-contraction; Fixed

point; G-Metric spaces.

Contents

1 Introduction 113

2 Main Results 114

1. Introduction

Let (X, d) be a metric space and T:X→Xbe a mapping, then Tis called a

contraction(Banach Contraction) on Xif

d(T x, T y)≤λd(x, y)

for all x, y ∈X.

Where λis a real such that λ∈[0,1). A point x∈Xis called a ﬁxed point of T

if T x =x. The well-known Banach Contraction Principle[1] ensures the existence

and uniqueness of a ﬁxed point of a contraction on a complete metric space. After

this principle, several authors generalized this principle by introducing the various

contractions on metric spaces[2, 3-9]. In this work, we introduce a mapping namely

simulation funtion and the notion of Z-contraction. Among all the generalized

metric spaces, the notion of G-Metric spaces was introduced by Mustafa and Sims

in[10], where in the authors discuss the topological properties of this space and

proved the analog of the Banach Contraction Principle in the context of G-Metric

spaces.

Deﬁnition 1.1. AG-Metric space (X, G) is said to be symmetric if G(x, y, y) =

G(y, x, x) for all x, y ∈X.

Example 1.2. Let (X, d) be the usual metric space then the function G:

X×X×X→[0,∞) deﬁned by G(x, y, z) = max{d(x, y), d(y, z), d(z , x)}for all

x, y, z ∈Xis a G-Metric space.

2010 Mathematics Subject Classiﬁcation: 47H10, 54H25, 54C30.

Submitted December 30, 2016. Published March 06, 2017

113 Typeset by BSP

Mstyle.

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Soc. Paran. de Mat.

114 M. Kumar and R. Sharma

Deﬁnition 1.3. Let Xbe a nonempty set and G:X×X×X→[0,∞) be a

function satisfying the following properties:

(G1)G(x, y, z) = 0 if x=y=z,

(G2)0 < G(x, x, y) for all x, y ∈Xwith x6=y,

(G3)G(x, x, y)≤G(x, y, z ) for all x, y, z ∈X, with z6=y,

(G4)G(x, y, z) = G(x, z, y) = G(y, z , x) = ... (symmetry in all three variables),

(G5)G(x, y, z)≤G(x, a, a) + G(a, y, z), for all x, y, z , a ∈X(rectangle inequlity).

Then, the function Gis called a generalized metric, or, more speciﬁcally,a G-metric

on X, and the pair (X, G) is called a G-metric space.

Deﬁnition 1.4. Let (X, G),(X′, G′) be G-Metric spaces, then a function

f:X→X′is G-continuous at a point x∈Xif and only if it is G-sequentially con-

tinuous at x, that is, whenever {xn}is G-convergent to x,{f(xn)}is G′-convergent

to f(x).

Recently, Khojasteh et al. [11] introduced a new class of mappings called sim-

ulation functions. Later Argoubi et al. [12] slightly modiﬁed the deﬁnition of

simulation functions in the deﬁnition of simulation functions by withdrawing a

condition.

Let Z∗be the set of simulation functions in the sense of Argoubi et al.[12].

Deﬁnition 1.5. A simulation function is a mapping ζ: [0,∞)×[0,∞)→R

satisfying the following conditions:

(ζ1)ζ(t, s)< s −tfor all t, s > 0

(ζ2) if {tn}and {sn}are sequences in (0,∞) such that

limn→∞{tn}=limn→∞{sn}=l∈(0,∞),

then

limn→∞supζ(tn, sn)<0.

2. Main Results

In this section, we deﬁne the simulation function, give some examples and prove

a related ﬁxed point result.

Deﬁnition 2.1. Let (X, G) be a G-Metric space, f:X→Xa mapping and

ζ∈Z. Then fis called a Z-contraction with respect to ζif the following condition

is satisﬁed

ζ(G(f x, fy, f z), G(x, y, z)) ≥0 for all x, y, z ∈X. (2.1)

Lemma 2.2. Let (X, G) be a G-Metric space and f:X→Xbe a Z-

contraction with respect to ζ∈Z. Then, fis asymptotically regular at every

x∈X.

Proof: Let x∈Xbe arbitrary. If for some p∈N

we have fpx=fp+1x, that is f y =y, where y=fp−1x, that is fz =z, where

z=fp−1x

Study of Fixed Point Theorems for Simulation Functions in G-Metric Spaces115

then, fny=fn−1fy =fn−1y=... =f y =yfor all n∈N. Now for suﬃcient large

n∈N, we obtain

G(fnx, f n+1x, fn+1x) = G(fn−p+1 fp−1x, f n−p+2fp−1x, f n−p+2fp−1x)

=G(fn−p+1y, f n−p+2y, f n−p+2y)

=G(y, y, y ) = 0

Therefore, limn→∞ G(fnx, f n+1x, f n+1x) = 0

Suppose, fnx6=fn−1xfor all n∈N, then it follows from (1) that

0≤ζ(G(fn+1x, f nx, f nx), G(fnx, f n−1x, fn−1x))

=ζ(G(ffnx, f fn−1x, f f n−1x), G(fnx, f n−1x, fn−1x))

≤G(fnx, f n−1x, fn−1x)−G(fn+1 x, f nx, fnx)

The above inequality show that {G(fnx, f n−1x, fn−1x)}is a monotonically de-

creasing sequence of non-negative reals and so it must be convergent.

Let limn→∞ G(fnx, f n+1x, fn+1x) = r≥0. If r > 0 then since fis Z-

contraction with respect to ζ∈Ztherefore, we have

0≤limn→∞supζ(G(fn+1 x, f nx, fnx), G(fnx, f n−1x, fn−1x)) <0.

This, contradiction shows that r= 0, that is, limn→∞ G(fnx, f n+1x, f n+1x) = 0.

Thus, fis an asymptotically regular mapping at x.

Lemma 2.3. Let (X, G) be a G-Metric space and f:X→Xbe a Z-

contraction with respect to ζ. Then the Picard sequence {xn}generated by f

with initial value x0∈Xis a bounded sequence, where xn=f xn−1for all n∈N.

Proof: Let x0∈Xbe arbitrary and {xn}be the Picard sequence, that is,

xn=fxn−1for all n∈N. On the contrary, assume that {xn}is not bounded.

Without loss of generality we can assume that xn+p6=xnfor all n, p ∈N. Since

{xn}is not bounded, there exists a subsequence {xn}such that n1= 1 and each

k∈N,nk+1 is the minimum integer such that

G(xn(k)+1, xn(k), xn(k))>1

and

G(xm, xn(k), xn(k))≤1

for nk≤m≤n(k)+1 −1. Therefore, by the triangular inequality, we have

1<G(xn(k)+1, xn(k), xn(k))

≤G(xn(k)+1, xn(k)+1 −1, xn(k)+1 −1) + G(xn(k)+1 −1, xn(k), xn(k))

≤G(xn(k)+1, xn(k)+1 −1, xn(k)+1 −1) + 1.

Letting k→ ∞ and using Lemma 2.2 we get

limk→∞G(xn(k)+1, xn(k), xn(k)) = 1

116 M. Kumar and R. Sharma

By (1), we get G(xn(k)+1, xn(k), xn(k))≤G(xn(k)+1 −1, xn(k)−1, xn(k)−1), therefore

using the triangular inequality we obtain

1< G(xn(k)+1, xn(k), xn(k))≤G(xn(k)+1 −1, xn(k)−1, xn(k)−1)

≤G(xn(k)+1 −1, xn(k), xn(k)) + G(xn(k), xn(k)−1, xn(k)−1)

≤1 + G(xn(k), xn(k)−1, xn(k)−1)

Letting k→ ∞ and using Lemma 2.2, we obtain

limk→∞ G(xn(k)+1 −1, xn(k)−1, xn(k)−1) = 1

Now, since fis a Z-contraction with respect to ζ∈Ztherefore, we have

0≤limk→∞supζ(G(f xn(k)+1 −1, f xn(k)−1, fxn(k)−1))

=limk→∞supζ(G(xn(k)+1 , xn(k), xn(k)), G(xn(k)+1 −1, xn(k)−1, xn(k)−1)) <0

This contradiction proves result.

Theorem 2.4. Let (X, G) be a complete G-Metric space and f:X→Xbe a

Z-contraction with respect to ζ. Then, fhas a unique ﬁxed point uin Xand for

every x0∈Xthe Picard sequence {xn}where xn=f xn−1for all n∈Nconverges

to the ﬁxed point of f.

Proof: Let x0∈Xbe arbitrary and {xn}be the Picard sequence, that is,

xn=fxn−1for all n∈N. We shall show that this sequence is a Cauchy sequence.

For this, let

Cn=sup{G(xi, xj, xj) : i, j ≥n}

Note that the sequence {xn}is a monotonically decreasing sequence of positive

reals and by Lemma 2.3 the sequence {xn}is bounded, threrefore Cn<∞for all

n∈N. Thus, {Cn}is monotonic bounded sequence, therefore convergent, that is,

there exists C≥0 such that limn→∞Cn=C. We shall show that C= 0. If C > 0

then by the deﬁnition Cn, for every k∈Nthere exists mk> nk≥kand

Ck−1

k< G(xm(k), xn(k), xn(k))≤Ck

Hence,

limk→∞G(xm(k), xn(k), xn(k))≤Ck(2.2)

Using (1) and the triangular inequality, we obtain

G(xm(k), xn(k), xn(k))≤G(xm(k)−1, xn(k)−1, xn(k)−1)

≤G(xm(k)−1, xm(k), xm(k)) + G(xm(k), xn(k), xn(k))

+G(xn(k), xn(k)−1, xn(k)−1)

Study of Fixed Point Theorems for Simulation Functions in G-Metric Spaces117

Using Lemma 2.2, (2) and letting k→ ∞ in the above inequality we get

limk→∞ G(xm(k)−1, xn(k)−1, xn(k)−1) = C. (2.3)

Since Tis a Z-contraction with respect to ζ∈Ztherefore using (1), (2), (3) and

(ζ2), we get

0≤limk→∞supζ(G(xm(k)−1, xn(k)−1, xn(k)−1), G(xm(k), xn(k), xn(k))) <0

This contradiction proves that C= 0 and so {xn}is a Cauchy sequence. Since Xis

a complete G-Metric space, there exists u∈Xsuch that limn→∞xn=u. We shall

show that the point uis a ﬁxed point of f. Suppose f u 6=uthen G(u, f u, fu)>0.

Again, using (1), ζ1,ζ2, we have

0≤limn→∞supζ(G(f xn, f u, f u), G(xn, u, u))

≤limn→∞supζ[G(xn, u, u)−G(xn=1 , f u, fu)]

=−G(u, f u, fu)

This contradiction shows that G(u, f u, f u) = 0, that is, f u =u. Thus, uis a ﬁxed

point of f.

Example 2.5. Let X= [0,1] and G:X×X→Rbe deﬁned by G(x, y, z) =

max{|x−y|,|y−z|,|z−x|}. Then, (X, G) is a complete G-Metric space. Deﬁne a

mapping f:X→Xas f x =x

x+1 for all x∈X.fis a continuous function but it

is not a Banach contraction. But it is a Z-contraction with respect to ζ∈Z, where

ζ(t, s) = s

s+ 1 −tfor all t, s ∈[0,∞).

Indeed, if x, y ∈X, then by a simple calculation it can be shown that

ζ(G(f x, fy, f y), G(x, y, y)) ≥0.

Clearly, 0 is the ﬁxed point of f.

Corollary 2.6. Let (X, G) be a complete G-Metric space and f:X→Xbe

a mapping satisfying the following condition: G(f x, f y, f y)≤λG(x, y, y) for all

x, y, y ∈X, where λ∈[0,1]. Then, fhas a unique ﬁxed point in X.

Proof: Deﬁne ζB: [0,∞)×[0,∞)→Rby ζB(t, s, s) = λs−tfor all s, t ∈[0,∞).

Note that, the mapping fis a Z-contraction with respect to ζB∈Z. Therefore,

the result follows by taking ζ=ζBin Theorem 2.4.

Corollary 2.7. Let (X, G) be a complete G-Metric space and f:X→X

be a mapping satisfying the following condition: G(f x, f y, f y)≤G(x, y, y)−

ϕ(G(x, y, y)) for all x, y, y ∈X, where ϕ: [0,∞)→[0,∞) is lower semi continuous

function and ϕ−1(0) = {0}. Then, fhas a unique ﬁxed point in X.

Proof: Deﬁne ζR: [0,∞)×[0,∞)→Rby ζR(t, s, s) = s−ϕ(s)−tfor all

s, t ∈[0,∞). Note that, the mapping fis a Z-contraction with respect to ζR∈Z.

Therefore, the result follows by taking ζ=ζRin Theorem 2.4.

Corollary 2.8. Let Let (X, G) be a complete G-Metric space and f:X→X

be a mapping satisfying the following condition: G(f x, fy , fy)≤ϕ(G(x, y, y))×

118 M. Kumar and R. Sharma

×G(x, y, y) for all x, y, y ∈X, where ϕ: [0,+∞)→[0,1) be a mapping such that

limsupt→r+ϕ(t)<1, for all r > 0. Then, fhas a unique ﬁxed point.

Proof: Deﬁne ζR: [0,∞)×[0,∞)→Rby ζR(t, s, s) = sϕ(s)−tfor all

s, t ∈[0,∞). Note that, the mapping fis a Z-contraction with respect to ζR∈Z.

Therefore, the result follows by taking ζ=ζRin Theorem 2.4.

Corollary 2.9. Let Let (X, G) be a complete G-Metric space and f:X→X

be a mapping satisfying the following condition: G(f x, f y, f y)≤η(G(x, y, y)) for

all x, y, y ∈X, where η: [0,+∞)→[0,+∞) be an upper semi continuous mapping

such that η(t)< t for all t > 0 and η(0) = 0. Then, fhas a unique ﬁxed point.

Proof: Deﬁne ζBW: [0,∞)×[0,∞)→Rby ζBW(t, s, s) = sη(s)−tfor all

s, t ∈[0,∞). Note that, the mapping fis a Z-contraction with respect to ζBW∈Z.

Therefore, the result follows by taking ζ=ζBWin Theorem 2.4.

Corollary 2.10. Let Let (X, G) be a complete G-Metric space and f:X→X

be a mapping satisfying the following condition: RG(f x,f y ,f y

0φ(t)dt ≤G(x, y, y)

for all x, y ∈X, where φ: [0,∞)→[0,∞) is a function such that Rt

0φ(t)dt exists

and Rc

0φ(t)dt > ǫ, for each ǫ > 0. Then, fhas a unique ﬁxed point.

Proof: Deﬁne ζK: [0,∞)×[0,∞)→Rby ζK(t, s, s) = s−Rt

0φ(u)du for

all s, t ∈[0,∞). Then, ζK∈Z. Therefore, the result follows by taking ζ=ζKin

Theorem 2.4.

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Manoj Kumar(Corresponding Author),

Rashmi Sharma,

Department of Mathematics,

Lovely Professional University,

Phagwara, Punjab,

India.

E-mail address: manojantil18@gmail.com, manoj.19564@lpu.co.in

E-mail address: rashmisharma.lpu@gmail.com