Common Fixed Point Theorems for Weak Contraction Mapping of Integral Type in Modular Spaces

Common Fixed Point Theorems for Weak Contraction

Mapping of Integral Type in Modular Spaces

R. A. Rashwan

1,∗

_,

_{H. A. Hammad}

2

1_{Department of Mathematics, Faculty of Science, Assuit University, Assuit 71516, Egypt}

2_{Department of Mathematics, Faculty of Science, Sohag University, Sohag 82524, Egypt}

∗_{Corresponding Author: rr [email protected]}

Copyright c⃝2014 Horizon Research Publishing All rights reserved.

Abstract

Inthispaper,weprovethreecommonfixedpointtheoremsforweakcontractionmappingsofintegraltypein modularspaces.Inthefirsttheoremweproveacommonfixedpointofρ−compatiblemappingssatisfyinga(ϕ−Ψ)−weak contraction.Thesecondtheorem is anotherversionofthefirsttheorem.Inthethirdtheoremwestudyacommonfixedpoint of ρ−compatible mappings inmodular spaces involving alteringdistances of integraltype. The results extendedseveral similarresultsinmetricandBanachspaces.

Keywords

Common Fixed Point, Modular Spaces,ρ−Compatible Maps, Comparsion Function, Lebesgue-stieltjes Inte-grable Mapping, Altering Distance Function

2010 Mathematics Subject Classification:Primary 47H10, Secondary 45H25

1

Introduction

Let(X, d)be a complete metric space andf :X →Xa self mapping ofX.Suppose thatFf ={x∈X :F(x) =x}

is the set of fixed points off.The classical Banach’s fixed point theorem is established in Banach [5] by using the following contractive defination: there existsk∈[0,1)such that∀x, y∈X,we have

d(f x, f y)≤kd(x, y). (1.1) Rhoades [19] proved the following theorem:

Theorem 1.1

LetT be a mapping from a complete metric space(X, d)into itself satisfying

d(T x, T y)≤d(x, y)−ϕ(d(x, y))∀x, y∈X, (1.2) whereϕ : R+ _→R+_{is continuous and nondecreasing such thatϕis positive onR}+_{\{0}, ϕ(0) = 0and lim}

t→∞ϕ(t) = ∞.

ThenT has a unique fixed point inX.

Branciari [8] and Rhoades [20] proved the following theorems for contraction mapping and weakly contractive mapping of integral type, respectively which are generalization of the Banach fixed point theorem.

Theorem 1.2

[8] LetT be a mapping from complete metric space(X, d)into itself satisfying

d(T x,T y_∫ )

0

φ(t)dt≤k d_∫(x,y)

0

φ(t)dt∀x, y∈X, (1.3)

wherek∈[0,1)is a constant andφ: R+ →R+be a Lebesgue-integrable mapping which is summable, nonnegative, and such that for eachϵ >0,

ϵ

∫

0

φ(t)dt >0.ThenThas a unique fixed pointz∈X,such that for eachx∈X, limn_→∞Tn_x=z.

Theorem 1.3

[20] LetTbe a mapping from complete metric space(X, d)into itself satisfying

d(T x,T y∫ )

0

φ(t)dt≤k m∫(x,y)

0

φ(t)dt∀x, y∈X, (1.4)

wherem(x, y) = max{d(x, y), d(x, T x), d(y, T y),d(x,T y)+₂d(y,T x)},and

d(T x,T y∫ )

0

φ(t)dt≤k M∫(x,y)

0

φ(t)dt∀x, y∈X, (1.5)

withM(x, y) = max{d(x, y), d(x, T x), d(y, T y), d(x, T y), d(y, T x)},

wherek∈[0,1)andφ:R+→R+in both cases is as defined in Theorem 1.1. ThenT has a unique fixed pointz∈X,such that for eachx∈X, limn_→∞Tn_x=z.

Afterward, many authors extended this work to more general contractive conditions. The works noted in [21, 3, 1, 11, 17].

The following definition is taken from Breind [6]

Definition 1.1

A single valued mappingf on a metric spaceX is called a weak contraction or(δ, L)−weak con-traction if and only if there exists two constantsL≥0andδ∈[0,1)such that

d(f x, f y)≤δd(x, y) +L(d(y, f x), ∀x, y∈X. (1.6) We shall employ the following definitions in the sequal to obtain our results.

Definition 1.2

[16] A function Ψ : R+ _{→ R}+ _{is called a comparison function if it satisfies the following}

condi-tions

(i)Ψis monotone increasing,Ψ(t)< tfor somet >0,

(ii)Ψ(0) = 0,

(iii) lim

n_→∞Ψ

n_{(t) = 0, ∀t≥0.}

Definition 1.3

[14, 4] The function ψ : [0,∞) → [0,∞) is called an altering distance function if and only if the following properties are satisfied

1.ψis continuous and non-decreasing. 2.ψ(t) = 0⇔t= 0.

Theorem 1.4

[16] Let(X, d)be a complete metric space and f : X → X satisfies a(L, ψ)-weak contraction of integral type

d(f x,f y∫ )

0

φ(t)dυ(t)≤L(

d(∫x,f x)

0

φ(t)dυ(t))r(

d(∫y,f x)

0

φ(t)dυ(t)) +ψ(

d∫(x,y)

0

φ(t)dυ(t))∀x, y∈X, (1.7)

whereL≥0, r≥0.Suppose that

(i)ψ:R+→R+is a continuous comparison function andυ:R+→R+is a monotone increasing functions,

(ii) φ : R+ _{→ R}+ _{be a Lebesgue-Stieltjes integrable mapping which is summable, nonnegative and for each ϵ > 0,}

ϵ

∫

0

φ(t)dυ(t)>0.

Thenf has a unique fixed pointx∗∈Xsuch that for eachx∈X, limn_→∞fn_x=x∗_.

Theorem 1.5

[16] Let (X, d)be a complete metric space andf : X → X satisfies a(ϕ, ψ)-weak contraction of integral type

d(f x,f y∫ )

0

φ(t)dυ(t)≤ϕ(

d(∫x,f x)

0

φ(t)dυ(t))(

d(∫y,f x)

0

φ(t)dυ(t)) +ψ(

d∫(x,y)

0

φ(t)dυ(t))∀x, y∈X. (1.8)

Suppose that

(i)ψ:R+_→R+_{is a continuous comparison function,}

(ii)Φ, υ:R+_→R+_{are monotone increasing functions such thatΦis continuous andΦ(0) = 0,}

(iii) φ : R+ → R+ be a Lebesgue-Stieltjes integrable mapping which is summable, nonnegative and for eachϵ > 0, ϵ

∫

0

φ(t)dυ(t)>0andυ :R+ →R+is also an increasing function. Thenf has a unique fixed pointx∗ ∈X such that for eachx∈X, limn_→∞fnx=x∗.

In 2012, H. Aydi [4] presented the following definition and fixed point theorem for contractive condition of integral type involving altering distances as the following

Definition 1.4

[4]u: [0,+∞)→[0,+∞)is subadditive on each[a, b]⊂[0,+∞)if,

a+b

∫

0

φ(t)dt≤ a

∫

0

φ(t)dt+

b

∫

0

φ(t)dt.

Theorem 1.6

[4] Let(X, d)be a complete metric space andf :X →Xsuch that

ψ(

d(f x,f y_∫ ))

0

u(t)dt)≤ψ(θ(x, y))−Φ(θ(x, y)), (1.9)

for eachx, y∈Xwith non−negative real numbersα, β, γsuch that

2α+β+ 2γ <1,whereψ,Φare altering distances, and

θ(x, y) =α

d(x,f x)+_∫d(y,f y)

0

u(t)dt

+β d_∫(x,y)

0

u(t)dt+γ

max_{d(x,f y_∫ ),d(y,f x)_}

0

u(t)dt, (1.10)

whereu(t) : [0,+∞) → [0,+∞)be a Lebesgue−integrable mapping which is summable, subadditive on each subset of

R+_{,non-negative such that for each}

ϵ >0, ϵ

∫

0

Thenf has a unique fixed point inX.

The notion of modular space, as a generalization of a metric space, was introduced by Nakano[13] in 1950 and redefined and generalized by Musielak and Orlicz[12] in 1959. In the existence of fixed point theory and Banach contraction principle occupies a prominent place in the study of metric spaces, it become a most popular tool in solving problems in mathematical analysis and construct methods in mathematics to solve problems in applied mathematics and sciences. In this article we study and prove the existence of fixed point theorems for mappings in modular spaces.

2

Preliminaries

We will start with a brief recollection of basic concepts and facts in modular spaces and modular metric spaces (see [9], [10], [18], [2] and [7]).

Definition 2.1

[18] LetXbe an arbitrary vector space overK= (RorC). a) A functionalρ:X →[0,∞]is called modular if:

(i)ρ(x) = 0iffx= 0.

(ii)ρ(αx) =ρ(x)forα∈Kwith|α|= 1, for allx∈X.

(iii)ρ(αx+βy)≤ρ(x) +ρ(y)ifα, β≥0, α+β = 1, for allx, y∈X. If (iii) is replaced by:

(iv)ρ(αx+βy)≤αρ(x) +βρ(y)forα, β≥0,α+β= 1, for allx, y∈X, then the modular is called convex modular. b) Ifρa modular inX,then the setXρ={x∈X :ρ(αx)→0asα→0}is called a modular space.

Remark 2.1

[18] Note thatρ is an increasing function as the following, suppose 0 < a < b, then, property (iii) withy= 0shows that

ρ(ax) =ρ(a

b(bx))≤ρ(bx).

Definition 2.2

[18] LetXρbe a modular space.

a) A sequence(xn)n∈N inXρis said to be:

(i)ρ−convergent toxifρ(xn−x)→0asn→ ∞.

(ii)ρ−Cauchy ifρ(xn−xm)→0asn, m→ ∞.

b)Xρisρ−complete if everyρ−Cauchy sequence isρ−convergent. c) A subsetB⊂Xρis said to beρ−closed if for any sequence

(xn)n∈N ⊂Bandxn→xwe havex∈B.

d) A subsetB⊂Xρis calledρ−bounded ifδρ(B) =sup ρ(x−y)<∞

for allx, y∈B, whereδρ(B)is called theρ−diameter ofB.

e)ρhas the Fatou property if:

ρ(x−y)≤lim inf ρ(xn−yn),

wheneverxn→xandyn→yasn→ ∞.

f)ρis said to satisfy the∆2−condition ifρ(2xn)→0, whenever(xn)→0asn→ ∞.

Remark 2.2

[7] Note that sinceρdoes not satisfy a priori the triangle inequality, we cannot expect that if {xn}and

Definition 2.3

[18] LetXρ be a modular space, where ρsatisfies the∆2−condition. Two self−mappingsT andh

ofX are calledρ−compatible ifρ(T hxn−hT xn)→0, whenever(xn)n∈N is a sequence inXρsuch thathxn → zand T xn →zfor some pointz∈Xρ.

3

Main Results

In this section, we study the existence of a common fixed point forρ−compatible mappings satisfying a(ϕ−Ψ)−weak contraction of integral type in modular spaces

Theorem 3.1

Let Xρ be a ρ−complete modular space, where ρsatisfies the ∆2−condition. Suppose c, l ∈ R+,

c > landT, h:Xρ→Xρare twoρ−compatible mappings such thatT(Xρ)⊆h(Xρ)and

ρ(c(T x∫−T y)

0

φ(t)dν(t)≤Ψ(

ρ(l(hx∫−hy))

0

φ(t)dν(t))

+ϕ(

ρ(l(hx∫−T x))

0

φ(t)dν(t))(

ρ(l(hy∫−T x))

0

φ(t)dν(t)), (3.1)

for eachx, y∈Xρ,Ψ :R+→R+is a continuous compasion function andν, ϕ:R+ →R+are montone increasing func-tions, such thatϕ(0) = 0.Letφ:R+_→R+_{be a Lebesgue−Stieltjes integrable mapping which is summable, nonnegative,}

and such that for each

ϵ >0, ϵ

∫

0

φ(t)dν(t)>0. (3.2)

If one ofhorTis continuous, then there exists a unique common fixed point ofhandT.

Proof: Letα ∈ R+ _{be the conjugate of} c

l such that l c +

1

α = 1.Letx◦be an arbitrary point of Xρ and generate

induc-tively the sequence(T xn)n∈N asT xn=hxn+1.For each integern≥1,inequality (3.1) shows that

ρ(c(T xn−∫1−T xn))

0

φ(t)dν(t)≤Ψ(

ρ(l(hxn−∫1−hxn))

0

φ(t)dν(t))

+ϕ(

ρ(l(hxn−∫1−T xn−1))

0

φ(t)dν(t))(

ρ(l(hxn∫−T xn−1))

0

φ(t)dν(t))

≤Ψ(

ρ(c(T xn−∫2−T xn−1))

0

φ(t)dν(t)) +ϕ(

ρ(c(T xn−∫2−T xn−1))

0

φ(t)dν(t))(

ρ(c(T xn−∫1−T xn−1))

0

φ(t)dν(t))

≤Ψ(

ρ(c(T xn−∫2−T xn−1))

0

φ(t)dν(t))≤...≤Ψn(

ρ(c(∫x−T x))

0

φ(t)dν(t)). (3.3) Taking the limit in (3.3) asn→ ∞yields

lim

n_→∞

ρ(c(T xn∫−1−T xn))

0

φ(t)dν(t) = 0, (3.4)

since lim

n→∞Ψ n₍

ρ(c(x_∫−T x))

0

lim

n_→∞ρ(c(T xn−1−T xn)) = 0. (3.5)

Now we show that(T xn)n∈N isρ−Cauchy. If not, then there exists anϵ >0and two sequences of integers{n(s)},{m(s)},

withn(s)> m(s)≥s,such that

ρ(l(T xn(s)−T xm(s)))≥ϵfors= 1,2, ... (3.6)

We can assume that

ρ(l(T xn(s)₋1−T xm(s)))< ϵ. (3.7)

To prove inequality (3.7) we can see [18]. Again from (3.1), we get

ρ(c(T xm(s)∫−T xn(s)))

0

φ(t)dν(t) ≤ Ψ(

ρ(l(hxm(s)∫−hxn(s)))

0

φ(t)dν(t))

+ϕ(

ρ(l(hxm(s)∫−T xm(s)))

0

φ(t)dν(t))(

ρ(l(hxn(s)∫−T xm(s)))

0

φ(t)dν(t))

= Ψ(

ρ(l(T xm(s)−∫1−T xn(s)−1))

0

φ(t)dν(t))+

ϕ(

ρ(l(T xm(s)∫−1−T xm(s)))

0

φ(t)dν(t))(

ρ(l(T xn(s)∫−1−T xm(s)))

0

φ(t)dν(t)), (3.8)

moreover,

ρ(l(T xm(s)₋1−T xn(s)₋1)) =ρ(

αl

α(T xm(s)−1−T xm(s)) +ρ( lc

c(T xm(s)−T xn(s)−1)

≤ρ(αl(T xm(s)−1−T xm(s))) +ρ(c(T xm(s)−T xn(s)−1)), (3.9)

taking the limit ass→ ∞in(3.9) and using (3.5), (3.7) and using∆2−condition, we have

ρ(l(T xm(s)−∫1−T xn(s)−1))

0

φ(t)dν(t))≤

ϵ

∫

0

φ(t)dν(t)), (3.10)

in (3.8), taking the limit ass→ ∞and using (3.10), (3.7) and (3.6), we get

ϵ

∫

0

φ(t)dν(t)) ≤

ρ(c(T xn(s)∫−T xm(s)))

0

φ(t)dν(t))≤Ψ(

ρ(l(T xn(s)−∫1−T xm(s)−1))

0

φ(t)dν(t))

≤ Ψ(

ϵ

∫

0

φ(t)dν(t))< ϵ

∫

0

φ(t)dν(t)),

SinceXρisρ−complete, then there existsz∈Xρsuch that

ρ(c(T xn−z))→0asn→ ∞.

If T is continuous, thenT2_x

n → T z and T hxn → T z.Since ρ(c(hT xn −T hxn)) → 0, then byρ−compatibility, hT xn→T z.We now prove thatzis a fixed point ofT, if not, we have from (3.1)

ρ(c(T2x∫n−T xn))

0

φ(t)dν(t)≤Ψ(

ρ(l(hT x∫n−hxn))

0

φ(t)dν(t))

+ϕ(

ρ(l(hT x∫n−T2xn))

0

φ(t)dν(t))(

ρ(l(hxn∫−T2xn))

0

φ(t)dν(t)).

Taking the limit asn→ ∞,we get

ρ(c(∫T z−z))

0

φ(t)dν(t) ≤ Ψ(

ρ(l(T z∫ −z))

0

φ(t)dν(t))

<

ρ(l(_∫T z₋z))

0

φ(t)dν(t)<

ρ(c(_∫T z₋z))

0

φ(t)dν(t).

Leading to a contradiction again. Thereforez=T z.Moreover,T(Xρ)⊆h(Xρ)and thus, there exists a pointz1∈Xρsuch

that

z=T z=hz1. (3.11)

From (3.1), we obtain

ρ(c(T2_x n−T z1))

∫

0

φ(t)dν(t)≤Ψ(

ρ(l(hT x∫n−hz1))

0

φ(t)dν(t))

+ϕ(

ρ(l(hT x∫n−T2xn))

0

φ(t)dν(t))(

ρ(l(hz1∫−T2xn))

0

φ(t)dν(t)),

asn→ ∞,and using (3.11), we get

ρ(c(z∫−T z1))

0

φ(t)dν(t)≤Ψ(0) = 0, (3.12)

so that (3.12), we have a contradiction. Therefore by properties ofφ,we get

ρ(c(z−∫T z1))

0

φ(t)dν(t) = 0,from which is follows that

ρ(c(z−T z1)) = 0, or z=T z1=hz1.

Also,hz=hT z1=T hz1=T z=z(see [18]). Thereforezis a common fixed point ofT andh.In addition, if one consider

hto be continuous instead ofT,then by similar argument as above, one can prove thatT z=hz=z.

ρ(c(z_∫−w))

0

φ(t)dν(t) =

ρ(c(T z_∫−T w))

0

φ(t)dν(t)

≤Ψ(

ρ(l(hz∫−hw))

0

φ(t)dν(t)) +ϕ(

ρ(l(hz∫−T z))

0

φ(t)dν(t))(

ρ(l(hw∫−T z))

0

φ(t)dν(t)),

≤Ψ(

ρ(l(_∫z₋w))

0

φ(t)dν(t))≤Ψ(

ρ(c(_∫z₋w))

0

φ(t)dν(t))<(

ρ(c(_∫z₋w))

0

φ(t)dν(t)).

Leading to a contradiction again. Therefore, by the condition onφ,we get

ρ(c(z_∫−w))

0

φ(t)dν(t) = 0,from which it follows that

ρ(c(z−w)) = 0 or z=w.

HenceTandhhave a unique common fixed point.

The following theorem is another version of Theorem 3.1 whenl = c,by adding the restrictions that T, h : B → B,

whereBis aρ−closed andρ−bounded subset ofXρ.

Theorem 3.2

LetXρbe aρ−complete modular space, whereρsatisfies the∆2−condition andBis aρ−closed and

ρ−bounded subset ofXρ.SupposeT, h:B→B,areρ−compatible mappings such thatT(Xρ)⊆h(Xρ)and

ρ(c(T x∫−T y)

0

φ(t)dν(t)≤Ψ(

ρ(c(hx∫−hy))

0

φ(t)dν(t))

+ϕ(

ρ(c(hx_∫−T x))

0

φ(t)dν(t))(

ρ(c(hy_∫−T x))

0

φ(t)dν(t))

for eachx, y∈Xρ,Ψ :R+→R+is a continuous compasion function andν, ϕ:R+ →R+are montone increasing

func-tions, such thatϕ(0) = 0.Letφ:R+ →R+be a Lebesgue−Stieltjes integrable mapping which is summable, nonnegative, and such that for eachϵ >0,

ϵ

∫

0

φ(t)dν(t)>0.

If one ofhorT is continuous, then there exists a unique common fixed point ofhandT.

Proof:letx∈B, m, n∈N.Then,

ρ(c(T xn+m∫ −T xm)

0

φ(t)dν(t)≤Ψ(

ρ(c(hxn+m∫ −hxm))

0

φ(t)dν(t))

+ϕ(

ρ(c(hxn+m∫−T xn+m))

0

φ(t)dν(t))(

ρ(c(hxm∫−T xn+m))

0

φ(t)dν(t))

≤ Ψ(

ρ(c(T xn+m∫−1−T xm−1))

0

φ(t)dν(t)) +ϕ(

ρ(c(T xn+m∫−1−T xn+m))

0

φ(t)dν(t))

(

ρ(c(T xm−∫1−T xn+m))

0

Taking the limit asn, m→ ∞and using (3.5), we have

lim

n,m→∞

ρ(c(T xn+m∫ −T xm))

0

φ(t)dν(t)≤ lim

n,m→∞Ψ(

ρ(c(T xn+m∫−1−T xm−1))

0

φ(t)dν(t))

≤ lim

m→∞Ψ m₍

ρ(c(T x∫n−x))

0

φ(t)dν(t))≤ lim

m→∞Ψ m₍

δ∫ρ(B)

0

φ(t)dν(t)).

SinceBisρ−bounded, then,

lim

n,m→∞

ρ(c(T xn+m∫ −T xm))

0

φ(t)dν(t)≤0,

which implies that lim

n,m→∞ρ(c(T xn+m−T xm) = 0.

Therefore by∆2−condition,(T xn)n_∈N isρ−Cauchy. SinceXρisρ−complete, then there existsz∈Xρsuch that ρ(c(T xn−z))→0asn→ ∞.

If T is continuous, then T2_x

n → T z andT hxn → T z. Since ρ(c(hT xn −T hxn)) → 0, then by ρ−compatibility, hT xn→T z.

We now prove thatzis a fixed point ofT,if not, we have from (3.1)

ρ(c(T2x∫n−T xn))

0

φ(t)dν(t)≤Ψ(

ρ(c(hT x∫n−hxn))

0

φ(t)dν(t))

+ϕ(

ρ(c(hT x∫n−T2xn))

0

φ(t)dν(t))(

ρ(c(hxn∫−T2xn))

0

φ(t)dν(t)).

Taking the limit asn→ ∞,we get

ρ(c(∫T z−z))

0

φ(t)dν(t)≤Ψ(

ρ(c(∫T z−z))

0

φ(t)dν(t))<

ρ(c(∫T z−z))

0

φ(t)dν(t).

Leading to a contradiction again. Thereforez=T z.Moreover,T(Xρ)⊆h(Xρ)and thus there exists a pointz1∈Xρsuch

that

z=T z=hz1.

From (3.1), we obtain

ρ(c(T2x∫n−T z1))

0

φ(t)dν(t)≤Ψ(

ρ(c(hT x∫n−hz1))

0

φ(t)dν(t))

+ϕ(

ρ(c(hT x∫n−T2xn))

0

φ(t)dν(t))(

ρ(c(hz∫1−T2xn))

0

φ(t)dν(t)),

asn→ ∞,and using (3.11), we get

ρ(c(z∫−T z1))

0

so that (3.12), we have a contradiction. Therefore by the condition onφ,we get

ρ(c(z∫−T z1))

0

φ(t)dν(t) = 0,from which is follows that

ρ(c(z−T z1)) = 0, orz=T z1=hz1.

Alsohz=hT z1 =T hz1 =T z=z.Thereforezis a common fixed point ofT andh.In addition, if one considershto be

contiuous instead ofT,then by similar argument as above, one can proveT z=hz=z.

For uniqueness, suppose that(z̸=w)are two arbitrary common fixed point ofT andh,then from (3.1), we get

ρ(c(z∫−w))

0

φ(t)dν(t) =

ρ(c(T z∫−T w))

0

φ(t)dν(t)

≤Ψ(

ρ(l(hz∫−hw))

0

φ(t)dν(t)) +ϕ(

ρ(l(hz∫−T z))

0

φ(t)dν(t))(

ρ(l(hw∫−T z))

0

φ(t)dν(t)),

≤Ψ(

ρ(l(∫z−w))

0

φ(t)dν(t))≤Ψ(

ρ(c(∫z−w))

0

φ(t)dν(t))<(

ρ(c(∫z−w))

0

φ(t)dν(t)).

Leading to a contradiction again. Therefore, by the conditions onφ,we get

ρ(c(z∫−w))

0

φ(t)dν(t) = 0,from which it follows that

ρ(c(z−w)) = 0orz=w.

HenceTandhhave a unique common fixed point.

Now, we study the existence of a common fixed point for ρ−compatible mappings in modular spaces involving altering distances of integral type in the following Theorem.

Theorem 3.3

Let Xρ be aρ−complete modular space, whereρ satisfies the ∆2−condition. Suppose c, l ∈ R+,

c > landT, h:Xρ→Xρare twoρ−compatible mappings such thatT(Xρ)⊆h(Xρ)and

ψ(

ρ(c(T x∫−T y))

0

u(t)dt)≤ψ(θ(x, y))−Φ(θ(x, y)), (3.13)

for eachx, y∈Xρwith non−negative real numbersζ, β, γsuch that

2ζ+β+ 2γ <1,whereψ,Φare altering distances, and

θ(x, y) =ζ

ρ(l(hx−T x)+∫ l(hy−T y))

0

u(t)dt

+β

ρ(l(hx_∫−hy))

0

u(t)dt+γ

max_{ρ(l(hx₋T y_∫)),ρ(l(hy₋T x))_}

0

u(t)dt, (3.14)

whereu(t) :R+ →R+be a Lebesgue−integrable mapping which is summable, subadditive on each subset ofR+, nonneg-ative such that for each

ϵ >0, ϵ

∫

0

u(t)dt >0. (3.15)

If one ofhorT is continuous, then there exists a unique fixed point ofhandT

Proof: Letα ∈ R+ _{be the conjugate of} c

l such that l c +

1

α = 1.Letxbe an arbitrary point of Xρ and generate

θ(xn+1, xn) =ζ

ρ(l(hxn+1−T xn+1∫ )+l(hxn−T xn))

0

u(t)dt

+β

ρ(l(hxn+1∫ −hxn))

0

u(t)dt+γ

max{ρ(l(hxn+1−T x∫n)),ρ(l(hxn−T xn+1))}

0

u(t)dt

=ζ

ρ(l(T xn−T xn+1∫)+l(T xn−1−T xn))

0

u(t)dt

+β

ρ(l(T xn∫−T xn−1))

0

u(t)dt+γ

max_{ρ(l(T xn−T xn))∫,ρ(l(T xn−1−T xn+1))}

0

u(t)dt, (3.16)

by subadditive ofu,we have

θ(xn+1, xn) =ζ

ρ(l(T xn∫−T xn+1))

0

u(t)dt+ζ

ρ(l(T xn∫−1−T xn)

0

u(t)dt

+β

ρ(l(T xn∫−T xn−1))

0

u(t)dt+γ

ρl(T xn−∫1−T xn+1)

0

u(t)dt.

Moreover,(c > l)and

ρ(l(T xn−1−T xn+1))≤ρ(

αl

α(T xn−1−T xn)) +ρ( cl

c(T xn−T xn+1))

≤ρ(αl(T xn−1−T xn)) +ρ(c(T xn−T xn+1)),

which implies

θ(xn+1, xn)≤ζ

ρ(c(T xn∫−T xn+1))

0

u(t)dt

+ζ

ρ(c(T xn∫−1−T xn))

0

u(t)dt+β

ρ(c(T xn∫−T xn−1))

0

u(t)dt

+γ

ρ(c(T xn∫−1−T xn))

0

u(t)dt+γ

ρ(c(T xn∫−T xn+1))

0

u(t)dt. (3.17)

From (3.13) and (3.17), we have

ψ(

ρ(c(T xn+1∫ −T xn))

0

u(t)dt) ≤ ψ(θ(xn+1, xn))−Φ(θ(xn+1, xn))≤ψ(θ(xn+1, xn))

= ψ(ζ

ρ(c(T xn∫−T xn+1))

0

u(t)dt+ζ

ρ(c(T xn∫−1−T xn))

0

u(t)dt

+β

ρ(c(T xn∫−T xn−1))

0

u(t)dt+γ

ρ(c(T xn∫−1−T xn))

0

u(t)dt

+γ

ρc(T xn∫−T xn+1))

0

By the factψis non-decreasing, we get

ρ(c(T xn+1∫ −T xn))

0

u(t)dt≤(θ(xn+1, xn))≤(ζ+γ)

ρ(c(T xn+1∫ −T xn))

0

u(t)dt

+(ζ+β+γ)

ρ(c(T xn∫−T xn−1))

0

u(t)dt,

which implies that

ρ(c(T xn+1∫ −T xn))

0

u(t)dt≤ζ+β+γ

1−ζ−γ

ρ(c(T xn∫−T xn−1))

0

u(t)dt.

Lettingh= ζ₁+₋β_ζ−+_γγ.By hypotheses onζ, βandγ,we geth∈[0,1).By induction, we have

ρ(c(T xn+1∫ −T xn))

0

u(t)dt≤h

ρ(c(T xn∫−T xn−1))

0

u(t)dt≤hn

ρ(c(∫T x−x))

0

u(t)dt.

Taking the limit asn→ ∞yields,

lim

n

ρ(c(T xn+1∫ −T xn))

0

u(t)dt≤0,

thus inequality (3.15) implies that

lim

n→∞ρ(c(T xn+1−T xn))→0. (3.18)

Now, we show that(T xn)n∈N isρ−Cauchy. If not, then, there exists

anϵ >0and two sequences of integers{n(s)},{m(s)},withn(s)> m(s)≥s,such that

ρ(l(T xn(s)−T xm(s)))≥ϵfors= 1,2, ... (3.19)

We can assume that

ρ(l(T xn(s)−1−T xm(s)))< ϵ. (3.20)

Again from (3.14), we have

θ(xm(s), xn(s)) =ζ

ρ(l(hxm(s)−T xm(s)∫)+l(hxn(s)−T xn(s)))

0

u(t)dt

+β

ρ(l(hxm(s)∫−hxn(s)))

0

u(t)dt+γ

max{ρ(l(hxm(s)−T xn(s)∫)),ρ(l(hxn(s)−T xm(s)))}

0

u(t)dt

=ζ

ρ(l(T xm(s)−1−T xm(s)∫)+l(T xn(s)−1−T xn(s)))

0

u(t)dt+β

ρ(l(T xm(s)−∫1−T xn(s)−1))

0

u(t)dt

+γ

max_{ρ(l(T xm(s)−1−T xn(s)∫)),ρ(l(T xn(s)−1−T xm(s)))}

0

moreover,

ρ(l(T xm(s)−1−T xn(s)−1)) ≤ ρ(

αl

α(T xm(s)−1−T xm(s))) +ρ( cl

c(T xm(s)−T xn(s)−1))

≤ ρ(αl(T xm(s)−1−T xm(s))) +ρ(c(T xm(s)−T xn(s)−1)),

using the∆2−condition and (18), we get

lim

s→∞ρ(αl(T xm(s)−1−T xm(s))) = 0. (3.22)

Therefore

lim

s→∞

ρ(l(T xm(s)−∫1−T xn(s)−1))

0

u(t)dt≤ ϵ

∫

0

u(t)dt, (3.23)

also

ρ(l(T xm(s)₋1−T xn(s))) ≤ ρ(

αl

α(T xm(s)−1−T xm(s))) +ρ( cl

c(T xm(s)−T xn(s)))

≤ ρ(αl(T xm(s)−1−T xm(s))) +ρ(c(T xm(s)−T xn(s))),

using the∆2−condition and (3.19), (3.20), (3.22), we have

lim

s→∞

max_{ρ(l(T xm(s)−1−T xn(s)∫)),ρ(l(T xn(s)−1−T xm(s)))}

0

u(t)dt≤ ϵ

∫

0

u(t)dt. (3.24)

Taking the limit ass→ ∞in (3.21), using (3.18), (3.23) and (3.24), we have

lim

s_→∞θ(xm(s), xn(s))≤(β+γ) ϵ

∫

0

u(t)dt. (3.25)

On the other hand, by (3.13)

ψ(

ρ(c(T xm(s)∫ −T xn(s))

0

u(t)dt)≤ψ(θ(xm(s), xn(s)))−Φ(θ(xm(s), xn(s)))

Takings→ ∞and using the continuity ofψandΦ,we have from (3.19), (3.25)

ψ(

ϵ

∫

0

u(t)dt)≤ψ(

ρ(c(T xm(s)∫−T xn(s))

0

u(t)dt)

≤ψ((β+γ)

ϵ

∫

0

u(t)dt)−Φ((β+γ)

ϵ

∫

0

u(t)dt)

≤ψ(

ϵ

∫

0

u(t)dt)−Φ((β+γ)

ϵ

∫

0

u(t)dt),

which implies thatΦ((β+γ)

ϵ

∫

0

u(t)dt) = 0,so by a property of Φ, we get

ϵ

∫

0

(T xn)n_∈N isρ−Cauchy.

SinceXρisρ−complete, then there existsz∈Xρsuch that

ρ(c(T xn−z))→0asn→ ∞.

If T is continuous, then T2_x

n → T z and T hxn → T z. Since ρ(c(hT xn −T hxn)) → 0, then by ρ−compatibility, hT xn→T z.

We now prove thatzis a fixed point ofT,we have from (3.14)

θ(T xn, xn) =ζ

ρ(l(hT xn−T2xn∫)+l(hxn−T xn))

0

u(t)dt

+β

ρ(l(hT x∫n−hxn))

0

u(t)dt+γ

max_{ρ(l(hT xn−T x∫n)),ρ(l(hxn−T2xn))}

0

u(t)dt.

Taking the limit asn→ ∞,yields

lim

n_→∞θ(T xn, xn) = (β+γ)

ρ(l(∫T z−z))

0

u(t)dt. (3.26)

Again, from (3.13)

ψ(

ρ(c(T2_x n−T xn))

∫

0

u(t)dt)≤ψ(θ(T xn, xn))−Φ(θ(T xn, xn)),

asn→ ∞and using (3.26), we get

ψ(

ρ(c(∫T z−z))

0

u(t)dt)≤ψ((β+γ)

ρ(l(∫T z−z))

0

u(t)dt)−Φ((β+γ)

ρ(l(∫T z−z))

0

u(t)dt),

using the continuity ofψandΦ,we obtain

ψ(

ρ(c(_∫T z₋z))

0

u(t)dt) ≤ ψ((β+γ)

ρ(c(_∫T z₋z))

0

u(t)dt)

−Φ((β+γ)

ρ(c(∫T z−z))

0

u(t)dt)

≤ ψ(

ρ(c(_∫T z₋z))

0

u(t)dt)−Φ((β+γ)

ρ(c(_∫T z₋z))

0

u(t)dt),

which implies thatΦ((β+γ)

ρ(c(T z∫−z))

0

u(t)dt) = 0,so by properties ofΦandu, we getρ(c(T z−z)) = 0orz =T z.

Moreover,T(Xρ)⊆h(Xρ)and thus there exists a pointz1∈Xρsuch that

z=T z=hz1. (3.27)

θ(T xn, z1) =ζ

ρ(l(hT xn−T2x∫n)+l(hz1−T z1))

0

u(t)dt

+β

ρ(l(hT x∫n−hz1))

0

u(t)dt+γ

max{ρ(l(hT xn−T z∫1)),ρ(l(hz1−T2xn))}

0

u(t)dt,

asn→ ∞,yields

lim

n_→∞θ(T xn, z1) =ζ

ρ(l(T z−T z)+∫l(hz1−T z1))

0

u(t)dt

+β

ρ(l(T z∫−hz1))

0

u(t)dt+γ

max{ρ(l(T z−T z∫1)),ρ(l(hz1−T z))}

0

u(t)dt,

using (3.27), we get

lim

n_→∞θ(T xn, z1) = (ζ+γ)

ρ(l(z∫−T z1))

0

u(t)dt. (3.28)

Again from (3.13), we have

ψ(

ρ(c(T2_x n−T z1))

∫

0

u(t)dt)≤ψ(θ(T xn, z1))−Φ(θ(T xn, z1)).

Taking the limit asn→ ∞and using (3.27) and (3.28), we get

ψ(

ρ(c(z∫−T z1))

0

u(t)dt)≤ψ((ζ+γ)

ρ(l(z∫−T z1))

0

u(t)dt)−Φ((ζ+γ)

ρ(l(z∫−T z1))

0

u(t)dt)

≤ψ((ζ+γ)

ρ(c(z∫−T z1))

0

u(t)dt)−Φ((ζ+γ)

ρ(c(z∫−T z1))

0

u(t)dt)

≤ψ(

ρ(l(z∫₋T z1))

0

u(t)dt)−Φ((ζ+γ)

ρ(l(z∫₋T z1))

0

u(t)dt),

which implies thatΦ((ζ+γ)

ρ(l(z−∫T z1))

0

u(t)dt) = 0,so by properties ofΦandu,we getρ(c(z−T z1)) = 0, therefore

z=T z1=hz1and alsohz=hT z1=T hz1=T z=z.

In addition, if one considerhto be a continuous in stead ofT,

then by similar argument (as above), one can prove

hz=T z=z.

Finally, suppose thatzandware two arbitrary common fixed point ofTandh,(w̸=z),then from (3.14), we get

θ(z, w) =ζ

ρ(l(hz−T z)+∫l(hw−T w))

0

u(t)dt

+β

ρ(l(hz∫−hw))

0

u(t)dt+γ

max{ρ(l(hz−T w∫)),ρ(l(hw−T z))}

0

= (β+γ)

ρ(l(_∫z₋w))

0

u(t)dt.

From (3.13), we have

ψ(

ρ(c(T z∫−T w))

0

u(t)dt)≤ψ(θ(z, w))−Φ(θ(z, w))

ψ(

ρ(c(∫z−w))

0

u(t)dt) = ψ(

ρ(c(T z∫−T w))

0

u(t)dt)

≤ ψ((β+γ)

ρ(l(∫z−w))

0

u(t)dt)−Φ((β+γ)

ρ(l(∫z−w))

0

u(t)dt)

≤ ψ((β+γ)

ρ(c(_∫z₋w))

0

u(t)dt)−Φ((β+γ)

ρ(c(_∫z₋w))

0

u(t)dt)

≤ ψ(

ρ(c(∫z−w))

0

u(t)dt)−Φ((β+γ)

ρ(c(∫z−w))

0

u(t)dt),

also by properties ofΦandu,we get

Φ((β+γ)

ρ(c(∫z−w))

0

u(t)dt) = 0 =⇒ρ(c(z−w)) = 0orz=w.

This complete the prove of this Theorem.

If we takeψ(t) =₂t andΦ(t) =₄t in Theorem 3.3 we get the following Corollary

Corollary 3.1

Let Xρ be a ρ−complete modular space, where ρsatisfies the ∆2−condition. Suppose c, l ∈ R+,

c > landT, h:Xρ→Xρare twoρ−compatible mappings such thatT(Xρ)⊆h(Xρ)and

ρ(c(T x_∫−T y))

0

u(t)dt≤ζ

ρ(l(hx₋T x_∫)+l(hy₋T y))

0

u(t)dt

+β

ρ(l(hx∫−hy))

0

u(t)dt+γ

max{ρ(l(hx−T y∫)),ρ(l(hy−T x))}

0

u(t)dt

for eachx, y∈Xρwith non−negative real numbersζ, β, γsuch that2ζ+β+2γ <1, whereψ,Φare altering distences, where

u(t) :R+ _→R+ _{be a Lebesgue−integrable mapping which is summable, subadditive on each subset ofR}+_{,nonnegative,}

and such that for each

ϵ >0, ϵ

∫

0

u(t)dt >0.

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