Fixed Point Theorems in Partial 2-Metric Spaces

HASSAN M. ABU-DONIA, HANI A. ATIEMAN SAFAA

Department of Mathematics, Faculty of Science

Zagazig University

Zagazig 44519, Zagazig

E*<37

Abstract: - In this paper we aim to introduce the concept of partial 2-metric spaces and study some fixed point

theorems for self mappings defined on partial 2-metric spaces.

Key-Words: - Partial metric spaces, 2-metric spaces, partial 2-metric spaces, fixed point theorems, contractive

conditions.

Received: June 12, 2021. Revised: February 12, 2022. Accepted: February 28, 2022. Published: March 26, 2022.

1 Introduction

In recent years, several works on domain the-

ory are created so as to equip the semantics domain

with a notion of distance. Especially,the notion of

fixed point theory and contraction mapping was ex-

tended and elaborated with the introduction of con-

traction principle by Banach [5]. Definition of 2-

metric spaces was initiated by Gahler in a series of

papers ([6]-[8]). The 2-metric space have a unique

nonlinear structure, which is different from metric

spaces. More several of the authors studied and gen-

eralize some theorems in 2-metric spaces. Fixed point

theorem is an important tool in the theory of metric

spaces, it guarantees the existence and uniqueness of

fixed point of self maps of metric spaces. Iseki ([9]-

[11]) obtained basic results on fixed point of opera-

tors in 2-metric spaces. After this work for Iseki, sev-

eral authors studied and generalized fixed point the-

orems in 2-metric spaces. The notion of partial met-

ric space was introduced by Matthews ([14],[15]). A

partial metric space is obtained from metric space by

replacing the equality d(x, x) = 0 in the definition of

metric with the inequality d(x, x)≤d(x, y)for all

x, y. This notion features a big range of applications

not solely in several branches of mathematics, also

within the field of computer domain and semantics

([1]-[4],[12],[13]). Recently, authors have targeted

on partial metric spaces and its topological proper-

ties, and generalized fixed point theorems from the

category of metric spaces to the class of partial met-

ric spaces ([7]–[10]). In this paper we introduce the

concept of partial 2-metric spaces and study the fixed

point theorem under contraction self mapping on par-

tial 2-metric spaces.

1.1 Preliminaries

Definition 1 [6].A 2-metric space is a set Xwith a

non negative real valued function don X×X×Xsat-

isfying the following conditions: For every x, y, z, u

∈X, we have:

(M1) for two distinct point x, y in Xthere exist a

point zin Xsuch that d(x, y, z)



= 0,

(M2) d(x, y, z)=0when at least two of x, y and

zare equals,

(M3) d(x, y, z) = d(x, z, y) = d(z, y, x),

(M4) d(x, y, z)≤d(x, y, u) + d(x, u, z) +

d(u, y, z), and then the function dis called a 2-metric

function on X.

Example 2 [13].Let a mapping d:R3→[0,∞)be

defined by

d(x, y, z) = min{|x−y|,|y−z|,|z−x|}.

Then dis a 2-metric on R.

Definition 3 [6].A sequence {xn}in a 2-metric

space (X, d)is said to be convergent to a point x∈

X, that is limn→∞ xn=x, if limn→∞ d(xn, x, z) =

0for all z∈X, and the point xis called the limit of

the sequence {xn}in X.

Definition 4 [6].A sequence {xn}in a 2-metric

space (X, d)is called a Cauchy sequence if

limm,n→∞ d(xn, xm, a) = 0 for all a∈X.

Definition 5 [6].A 2-metric space (X, d)is

said to be complete if every Cauchy sequence in

Xis convergent.

Remark 6 [6].Every convergent sequence in a 2-

metric space is a Cauchy sequence .

Definition 7 [14].Let Xbe a nonempty set. The map-

ping p:X×X→[0,∞)is said to be a partial met-

ric on Xif the following conditions are true. For any

x, y, z ∈X, we have:

(PM-1) x=yif and only if p(x, x) = p(y, y) =

p(x, y),

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(PM-2) p(x, x)≤p(x, y),

(PM-3) p(x, y) = p(y, x),

(PM-4) p(x, z)≤p(x, y) + p(y, z)−p(y, y),and

then the pair (X, p)is called a partial metric space,

(for short P MS).

2 Partial 2-metric spaces

In this section we have introduce the concept of

partial 2-metric spaces and some properties.

Definition 8 A mapping ρ:X3→R+where Xis a

non-empty set, is said to be a partial 2-metric on Xif

the following conditions are true. For every x, y, z, u

∈X, we have:

(P2M-1) ρ(x, x, x) = ρ(y, y, y) = ρ(z, z, z) =

ρ(x, y, z)when at least two of x, y and zare equals,

(P2M-2) ρ(x, x, x)≤ρ(x, y, z),

(P2M-3) ρ(x, y, z) = ρ(x, z, y) = ρ(z, y, x),

(P2M-4) ρ(x, y, z)≤ρ(x, y, u) + ρ(x, u, z) +

ρ(u, y, z)−ρ(u, u, u).Then the pair (X, ρ)is called

a partial 2-metric space; for short we write P2M-

space.

Example 9 Let X={0,1}, and let ρ(x, y, z) =

{2if x=y=z= 0

1otherwise , then (X, ρ)is a P2M-

space.

Theorem 10 Every 2-metric space is a P2M-space.

Proof. Let (X, d)be a 2-metric space, then from

the condition (M2) we obtain,

d(x, x, x) = d(y, y, y) = d(z, z, z) = 0,

when at least two point of any x, y, z are equals, that

is (P2M-1) is satisfied. Since d(x, y, z)≥0, and

d(x, x, x) = 0 ≤d(x, y, z),

so,

d(x, x, x)≤d(x, y, z),

which is the condition (P2M-2). Also from condition

(M2), we have

d(x, y, z) = d(x, z, y) = d(z, y, x),

which is the condition (P2M-3). From the condition

(M3), we have

d(x, y, z)≤d(x, y, u) + d(x, u, z)

+d(u, y, z),

and d(u, u, u) = 0,then we can write

d(x, y, z)≤d(x, y, u) + d(x, u, z)

+d(u, y, z)−d(u, u, u).

So (X, d)is a P2M-space.

From Example 9 shows that the inverse is not true,

then

we have (X, ρ)is a P2M-space but it is not 2-

metric space.

Definition 11 A modified P2M-space indefined by

replacing the condition (P2M-1) in Definition 8 by the

following:

(P2M-1)* x=y=z, if and only if ρ(x, x, x) =

ρ(y, y, y) = ρ(z, z, z) = ρ(x, y, z),then the pair

(X, ρ)is called a modified partial 2-metric space; for

short we write (P2M)*-space.

Example 12 Let ρ:R+×R+×R+→R+defined

by ρ(x, y, z) = max{x, y, z}for any x, y, z ∈R+.

Then the pair (R+, ρ)is a (P2M)*-space.

Example 13 Let R+= (0,∞)and R−= (−∞,0).

Consider the function ρ:R−×R−×R−→R+de-

fined by ρ(x, y, z) = −min{x, y, z}for any x, y, z ∈

R.Then the pair (R−, ρ)is a (P2M)*-space.

Example 14 Consider I={[a, b] : a≤b;a, b ∈

R}is the set of all closed intervals in R, let

the function ρ:I3→R+, which in defined by

ρ([a, b],[c, d], f, g]) = max{b, d, g} − min{a, c, f}.

Then the pair (I, ρ)is a (P2M)*-space.

Example 15 Let X= [0, a]and α≥a≥3.

Define the mapping ρ:X3→R+by ρ(x, y, z) =











x, if x=y=z= 0 ,

4α+x+y+z

3, if x, y, z ∈ {1,2,3}and x=y=z,

α+x+y+z

3,otherwise.

Then the pair (X, ρ)is a (P2M)*-space .

Remark 16 Every P2M-space is (P2M)*-space

and the inverse is not true.

Theorem 17 Let (X, ρ)be a P2M-space, and the

function dρ:X3→[0,∞), defined by dρ(x, y, z) =

3ρ(x, y, z)−ρ(x, x, x)−ρ(y, y, y)−ρ(z, z, z),then

(X, dρ)is a 2-metric space.

Proof. It’s clear that for all distinct elements

x, y, z ∈X, we have

dρ(x, y, z)= 0,

and from the condtion (P2M-1), we get

dρ(x, y, z) = 3ρ(x, y, z)−ρ(x, x, x)

−ρ(y, y, y)−ρ(z, z, z)

= 0,

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when at least two points x, y, z in Xare equals. From

the condtion (P2M-3), we obtain

dρ(x, y, z) = dρ(x, z, y) = dρ(z, y, x).

Also from the condtion (P2M-4), we

have

dρ(x, y, z)

= 3ρ(x, y, z)−ρ(x, x, x)−ρ(y, y, y)

−ρ(z, z, z).

≤3×[ρ(x, y, u) + ρ(x, u, z)

+ρ(u, y, z)−ρ(u, u, u)]

−ρ(x, x, x)−ρ(y, y, y)−ρ(z, z, z).

= 3ρ(x, y, u)−ρ(u, u, u)−ρ(x, x, x)

−ρ(y, y, y)+3ρ(x, u, z)−ρ(u, u, u)

−ρ(x, x, x)−ρ(z, z, z)+3ρ(u, y, z)

−ρ(u, u, u)−ρ(y, y, y)−ρ(z, z, z)

+ρ(x, x, x) + ρ(y, y, y) + ρ(z, z, z).

This implies,

dρ(x, y, z)≤dρ(x, y, u) + dρ(x, u, z) + dρ(u, y, z).

Then dρis a 2-metric on X, which the proof .

Definition 18 A sequence {xn}in a P2M-

space (X, ρ)converges to a point xin Xif

limn→∞ ρ(xn, x, z) = ρ(x, x, x)for all zin

Definition 19 A sequence {xn}in a P2M-space

(X, ρ)is said to be a Cauchy sequence if the

limn,m→∞ ρ(xm, xn, z)exists and finite, for all zin

Definition 20 AP2M-space (X, ρ)is said to be

complete if every Cauchy sequence in Xconverges to

an element xin Xsuch that limn,m→∞ ρ(xm, xn, z)

=ρ(x, x, x), for all zin X.

Definition 21 Let (X, ρ)be a P2M-space, a func-

tion f:X→Xis said to be a contraction if there

exist a constant c∈[0,1), such that ρ(fx, fy, z)≤

c×ρ(x, y, z)for all x, y, z ∈X.

Theorem 22 Let (X, ρ)be a complete P2M-space

and fbe a self-mapping on Xsatisfying the condi-

tion ρ(fx, fy, z)≤c×ρ(x, y, z)for all x, y, z ∈X

and c∈[0,1). Then fhas a unique fixed point.

Proof. Let x∈Xand for a fixed z∈

X, it is clear that from the condtion

(P2M-4) for each n, k ∈N, we have

ρ(fn+k+1(x), f n(x), z)

≤ρ(fn+k+1(x), f n(x), fn+k(x))

+ρ(fn+k+1(x), f n+k(x), z)

+ρ(fn+k(x), fn(x), z)

−ρ(fn+k(x), fn+k(x), fn+k(x)) .

This implies,

ρ(fn+k+1(x), f n(x), z)

≤ρ(fn+k+1(x), f n(x), fn+k(x))

+ρ(fn+k+1(x), f n+k(x), z)

+ρ(fn+k(x), fn(x), z).

Using the contraction condition it follows that

ρ(fn+k+1(x), f n(x), fn+k(x))

=ρ(f(fn+k(x)), fn+k(x), fn(x))

≤cn+k×ρ(f(x), x, fn(x));

and

ρ(fn+k+1(x), f n+k(x), z)

=ρ(f(fn+k)(x), fn+k(x), z)

≤cn+k×ρ(f(x), x, z).

From (2) and (3) into (1), we get

ρ(fn+k+1(x), f n(x), z)

≤cn+k×ρ(f(x), x, fn(x)) + cn+k×ρ(f(x), x, z)

+ρ(fn+k(x), fn(x), z).

Also we have,

ρ(fn+k(x), fn(x), z)

≤ρ(fn+k(x), fn+k−1(x), z)

+ρ(fn+k(x), fn(x), fn+k−1(x))

+ρ(fn+k−1(x), fn(x), z)

−ρ(fn+k−1(x), fn+k−1(x), fn+k−1(x)).

So,

ρ(fn+k(x), fn(x), z)

≤ρ(fn+k(x), fn+k−1(x), z)

+ρ(fn+k(x), fn(x), fn+k−1(x))

+ρ(fn+k−1(x), fn(x), z).

≤cn+k−1×ρ(f(x), x, fn(x))

+cn+k−1×ρ(f(x), x, z)

+ρ(fn+k−1(x), fn(x), z).

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Thus,

ρ(fn+k+1(x), f n+k(x), z)

≤cn+k×ρ(f(x), x, z) + cn+k−1×ρ(f(x), x, z) + ...

+cn×ρ(f(x), x, z) + cn+k×ρ(f(x), x, f n(x))

+cn+k−1×ρ(f(x), x, fn(x)) + ....

+cn×ρ(f(x), x, fn(x)) + ρ(fn(x), fn(x), z).

Then,

ρ(fn+k+1(x), f n+k(x), z)

≤[cn+k+cn+k−1+.... +cn]

×[ρ(f(x), x, z) + ρ(f(x), x, f n(x))]

+cn×ρ(f(x), x, z).

≤[cn+k+cn+k−1+.... +cn]

×[ρ(f(x), x, z) + ρ(f(x), x, f n(x))]

+cn×ρ(f(x), x, z).

Which implies,

ρ(fn+k+1(x), f n+k(x), z)

≤cn×1−ck+1

1−c

×[ρ(f(x), x, z) + ρ(f(x), x, f n(x))]

+cn×ρ(x, x, z).

≤cn×cn

1−c

×[ρ(f(x), x, z) + ρ(f(x), x, f n(x))]

+cn×ρ(x, x, z).

Consequently {fn(x)}is a Cauchy sequence in the

P2M-space (X, ρ), and

lim

n,m→∞

ρ(fm(x), fn(x), z) = 0 ∀z∈X.

Since Xis Complete P2M-space. Then we can

choose a∈Xsuch that fn(x)converges to a, so,

lim

n,m→∞

ρ(fm(x), fn(x), z) = lim

n,m→∞

ρ(fn(x), a, z)

=ρ(a, a, a) = 0 ∀z∈X .

In the following we will show that a∈Xis a fixed

point for f. We have,

ρ(f(a), a, z)

≤ρ(f(a), fn+1(x), z) + ρ(f(a), a, f n+1(x))

+ρ(fn+1(x), a, z).

=ρ(fn+1(x), f (a), z) + ρ(fn+1(x), f (a), a)

+ρ(fn+1(x), a, z).

≤c×[ρ(fn(x), a, z) + ρ(fn(x), a, a)]

+ρ(fn+1(x), a, z).

As n→ ∞,we get

ρ(f(a), a, z) = 0,∀z∈X,

so, f(a) = a. For the uniqueness proof, we assume

that there exist another fixed point b∈X, so, f(b) =

b. Now,

ρ(b, a, z) = ρ(f(b), f (a), z)≤c×ρ(b, a, z).

This means that c≥1, which contradicts that c∈

[0,1). So, we must have ρ(b, a, z) = 0,∀z∈X.

Then a=b, which the proof.

Lemma 23 Assume that xn→xas n→ ∞

in a P2M-space such that ρ(x, x, x)=0then

limn→∞ ρ(xn, y, z) = ρ(x, y, z)for every x, y and

zin X.

Proof. First note that limn→∞ ρ(xn, x, z) =

ρ(x, x, x) = 0 ∀z∈X.

From the condition (P2M-4) we find that

ρ(xn, y, z)

≤ρ(xn, y, x) + ρ(xn, x, z) + ρ(x, y, z)

−ρ(x, x, x),

≤ρ(xn, y, x) + ρ(xn, x, z) + ρ(x, y, z).

ρ(xn, y, z)−ρ(x, y, z)≤ρ(xn, y, x) + ρ(xn, x, z).

Also,

ρ(x, y, z)

≤ρ(x, y, xn) + ρ(x, xn, z)

+ρ(xn, y, z)−ρ(xn, xn, xn),

≤ρ(x, y, xn) + ρ(x, xn, z) + ρ(xn, y, z).

ρ(x, y, z)−ρ(xn, y, z)≤ρ(x, y, xn) + ρ(x, xn, z).

Hence,

ρ(xn, y, x) + ρ(xn, x, z)

≤ |ρ(x, y, z)−ρ(xn, y, z)|

≤ρ(x, y, xn) + ρ(x, xn, z).

Let n→ ∞, and from the condition of (P2M-1) we

conclude the claim.

Lemma 24 (1) A sequence {xn}is a Cauchy se-

quence in the a P2M-space (X, ρ)if and only if

{xn}is also a Cauchy sequence in the 2-metric space

(X, dρ).

(2) (X, ρ)is complete if and only if (X, dρ)is also

complete. Moreover

lim

n→∞

ρ(x, xn, z) = lim

n,m→∞

ρ(xn, xm, z) = ρ(x, x, x)

⇔lim

n→∞

dρ(x, xn, z) = 0.

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Proof. (1) Let {xn}is a Cauchy sequence in the

P2M-space (X, ρ),then there exist α∈Rsuch that

ε > 0,there is n(ε)∈N, we have

|ρ(xn, xm, z)−α| ≤ ε

3for all n, m ≥n(ε)and z∈X.

Since,

dρ(xn, xm, z)

= 3ρ(xn, xm, z)−ρ(xn, xn, xn)

−ρ(xm, xm, xm)−ρ(z, z, z)

= [3ρ(xn, xm, z)−3α]−[ρ(xn, xn, xn)−α]

−[ρ(xm, xm, xm)−α]−[ρ(z, z, z)−α].

From P2M-2, we get ρ(xn, xn, xn)≤ρ(xn, xm, z),

then,

|ρ(xn, xn, xn)−α| ≤ ε

Then,

dρ(xn, xm, z)

≤3|ρ(xn, xm, z)−α|+|ρ(xn, xn, xn)−α|

+|ρ(xm, xm, xm)−α|+|ρ(z, z, z)−α|

dρ(xn, xm, z)≤ε∀n, m ≥n(ε).

Then {xn}is a Cauchy sequence in (X, dρ).Now, we

prove that every Cauchy sequence{xn}in (X, dρ)is

a Cauchy sequence in (X, ρ).Since {xn}is a Cauchy

sequence in (X, dρ),then

dρ(xn, xm, z)≤ε∀n, m ≥n0(ε).

Take ε=1

2then there exist n0(ε)∈Nsuch that

dρ(xn, xm, z)≤1

2∀n, m ≥n0(ε).

Since,

dρ(xn, xn0, z) + ρ(xn, xn, z)

=dρ(xn0, xn, z) + ρ(xn0, xn0, z),

then we have,

|ρ(xn, xn, z)|

=|dρ(xn0, xn, z)−dρ(xn, xn0, z) + ρ(xn0, xn0, z)|

≤2dρ(xn0, xn, z) + |ρ(xn0, xn0, z)|

<1 + |ρ(xn0, xn0, z)|.

Consequentely, the sequence {ρ(xn, xn, z)}nis

bounded in R,and so there exist a∈R, such that

a subsequence {ρ(xnk, xnk, z)}kis convergent to a,

i.e limk→∞ ρ(xnk, xnk, z) = a. It remins to prove

that {ρ(xn, xn, z)}nis a Cauchy sequence in R.Since

{xn}is Cauchy sequence (X, dρ)given ε > 0∃

n(ε)∈Nsuch that dρ(xn, xm, z)<3ε

2∀n, m ≥

n(ε).Thus,

|ρ(xn, xn, z)−ρ(xm, xm, z)|

3|dρ(xm, xn, z)−dρ(xn, xm, z)|

≤2

3dρ(xm, xn, z) = ε.

Therefore,

lim

n→∞

ρ(xn, xn, z) = a, and we get

|ρ(xn, xm, z)−α|

=|ρ(xn, xm, z)−ρ(xn, xn, z) + ρ(xn, xn, z)−α|

≤ |ρ(xn, xm, z)−ρ(xn, xn, z)|

+|ρ(xn, xn, z)−α|

< dρ(xn, xm, z) + |ρ(xn, xn, z)−α|

< ε, ∀n, m ≥n(ε).

Then {xn}is a Cauchy sequence in (X, ρ).

(2) Now, we prove completeness of (X, dρ)im-

plies completeness of (X, ρ).If {xn}is a Cauchy se-

quence in (X, ρ)then it is also a Cauchy sequence in

(X, dρ).Since (X, dρ)is a complete 2-metric space,

then there exist y∈Xsuch that

lim

n→∞

dρ(y, xn, z) = lim

n→∞ 3ρ(y, xn, z)−ρ(y, y, y)

−ρ(xn, xn, xn)−ρ(z, z, z).

Since {xn}is a Cauchy sequence in (X, ρ),let ε > 0

then ∃n0∈N,such that dρ(xm, xn, z)≤3ε

2. Then,

|ρ(xn, xn, z)−ρ(xm, xm, z)|

3|dρ(xm, xn, z)−dρ(xn, xm, z)|

≤2

3dρ(xm, xn, z) = ε∀n, m > n0.

This shows that (X, ρ)is complete.

Conversely, we prove (X, dρ)is complete. Let

{xn}is a Cauchy sequence in (X, dρ).Then {xn}is

a Cauchy sequence in (X, ρ)and it is convergent to

y∈Xwith

lim

n,m→∞

ρ(xn, xm, z) = lim

n→∞

ρ(y, xn, z) = ρ(y, y, y).

Then,

ρ(y, xn, z)−ρ(y, y, y)< ε;n≥n(ε).

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Since,

dρ(y, xn, z)

= 3ρ(y, xn, z)−ρ(y, y, y)

−ρ(xn, xn, xn)−ρ(z, z, z)

<|ρ(y, xn, z)−ρ(y, y, y)|

+|ρ(y, xn, z)−ρ(xn, xn, xn)|

+|ρ(y, xn, z)−ρ(z, z, z)|

= 3ε < ε.

Then dρ(y, xn, z)< ε ∀n≥n(ε).Then (X, dρ)is

complete. Finally, it’s a simple to check that limn→∞

dρ(x, xn, z) = 0 if and only if

lim

n→∞

ρ(x, xn, z) = lim

n,m→∞

ρ(xn, xm, z) = ρ(x, x, x).

Theorem 25 Let (X, ρ)be a P2M-space, let T:

X→Xbe a map for which the inequality

aρ(T x, T y, z) + b[ρ(x, T x, z) + ρ(y, T y, z)

+c[ρ(x, T y, z) + ρ(y, T y, z)(1)

≤sρ(x, y, z) + rρ(x, T 2x, z),(4)

holds for all x, y in Xwhere the constants a, b, c, r

and ssatsisfy

0≤s−b

a+b<1,

a+b= 0, a +b+c > 0, c −r > 0, c > 0.

Then Thas at least one fixed point.

Proof. Take an arbitrary point x0∈X, define the

sequence xn+1 =T xn,n= 0,1,2,3... . Sustituting

x=xn,y=xn+1 into equation (4), we have

aρ(T xn, T xn+1, z) + b[ρ(xn, T xn, z)

+ρ(xn+1, T xn+1, z)] + c[ρ(xn, T xn+1, z)

+ρ(xn+1, T xn+1, z)]

≤sρ(xn, xn+1, z) + rρ(xn, T 2xn, z),

which implies

aρ(xn+1, xn+2, z) + b[ρ(xn, xn+1, z)

+ρ(xn+1, xn+2, z)] + c[ρ(xn, xn+2, z)

+ρ(xn+1, xn+1, z)]

≤sρ(xn, xn+1, z) + rρ(xn, xn+2, z).

Rewriting this inequality as

(a+b)ρ(xn+1, xn+2, z)+(c−r)ρ(xn, xn+2, z)

+cρ(xn+1, xn+1, z)

≤(s−b)ρ(xn, xn+1, z),

and using the fact

(c−r)ρ(xn, xn+2, z) + cρ(xn+1, xn+1, z)≥0

where c−r > 0, c > 0.Then we obtain

ρ(xn+1, xn+2, z)≤αρ(xn, xn+1, z)

α=(s−b)

(a+b),(a+b)= 0,0≤α≤1.Thus,

ρ(xn+1, xn+2, z)≤αρ(xn, xn+1, z)

≤α2ρ(xn−1, xn, z)

≤α3ρ(xn−2, xn−1, z)

≤... ≤αn+1ρ(x0, x1, z).

We will show {xn}is a cauchy sequence. Since

ρ(xn, xn+1, z)≤αn+1ρ(x0, x1, z).

Taking n→ ∞,0≤α≤1,

then lim

n→∞

ρ(xn, xm, z)→0(exist and finite).

Then {xn}is a Cauchy sequence in (X, ρ).By

Lemma 24, {xn}is also a Cauchy sequence in

(X, dρ).Since (X, ρ)is complete then (X, dρ)is also

complete. Thus there exists x∈Xsuch that xn→x

in (X, dρ); moreover, by Lemma 24,

lim

n→∞

ρ(x, xn, z)

=lim

n,m→∞

ρ(xn, xm, z)

=ρ(x, x, x) = 0 ⇔lim

n→∞

dρ(x, xn, z) = 0.

Now, we will show that xbe a fixed point of T. Sub-

stituting x=xnand y=xinto (4), we obtain

aρ(T xn, T x, z) + b[ρ(xn, T xn, z)

+ρ(x, T x, z)]

+c[ρ(xn, T x, z) + ρ(x, T xn, z)]

≤sρ(xn, x, z) + rρ(x, T 2xn, z),

which implies

aρ(xn+1, T x, z) + b[ρ(xn, xn+1, z)

+ρ(x, T x, z)] + c[ρ(xn, T x, z)

+ρ(xn, T xn, z)]

≤sρ(xn, x, z) + rρ(x, xn+2, z).

Taking the limit as n→ ∞ and using Lemma 23, we

get

(a+b+c)ρ(x, T x, z)≤0.

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Since (a+b+c)>0,

then we have 0≤(a+b+c)ρ(x, T x, z)≤0,

this means that ρ(x, T x, z) = 0.From theorem 17,

we get

0≤dρ(x, T x, z) = 3 ρ(x, T x, z)

−ρ(x, x, x)−ρ(T x, T x, T x)−ρ(z, z, z)

=−ρ(T x, T x, T x)−ρ(z, z, z)≤0,

hence dρ(x, T x, z) = 0,that is x=T x, which the

proof.

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