# Quasi-Contractive Mappings in Modular Metric Spaces

Authors: Yeol J. E. Cho; Reza Saadati; Ghadir Sadeghi
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We prove the existence of fixed point and uniqueness of quasi-contractive mappings in modular metric spaces which was introduced by Ćirić


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## Body

## 1. Introduction and Preliminaries

In this paper, we prove the existence and uniqueness of fixed points of quasi-contractive mappings in modular metric spaces which develop the theory of metric spaces generated by modulars. Throughout the paper𝔛 is a nonempty set and λ>0. The notion of a metric modular was introduced by Chistyakov [1] as follows.Definition 1.1.
A functionω:(0,∞)×𝔛×𝔛→[0,∞] is said to be a metric modular on 𝔛 (or, simply, a modular if no ambiguity arises) if it satisfies three axioms:(i)
for anyx,y∈𝔛,  ωλ(x,y)=0 for all λ>0 if and only if x=y;(ii)
ωλ(x,y)=ωλ(y,x) for all λ>0, and x,y∈𝔛;(iii)
ωλ+μ(x,y)≤ωλ(x,z)+ωμ(z,y) for all λ,μ>0 and x,y∈𝔛.Definition 1.2.
Let(𝔛,ω) be a metric modular space.(1)
A sequence{xn} in 𝔛ω is said to be ω-convergent to a point x∈𝔛 if, for all λ>0,
(1.1)ωλ(xn,x)⟶0
as n→∞.(2)
A subsetℭ of 𝔛ω is said to be ω-closed if the ω-limit of a ω-convergent sequence of ℭ always belongs to ℭ.(3)
A subsetℭ of 𝔛ω is said to be ω-complete if every ω-Cauchy sequence in ℭ is ω-convergent and its ω-limit is in ℭ.Definition 1.3.
The metric modularω is said to have the Fatou property if
(1.2)ωλ(x,y)≤liminf⁡n→∞ω(xn,y)
for all y∈𝔛ω and λ∈(0,∞), where {xn}ω-converges to x.


## 2. Main Results

Definition 2.1.
Let(𝔛,ω) be a metric modular space, and let ℭ be a nonempty subset of 𝔛ω. The self-mapping T:ℭ→ℭ is said to be quasi-contraction if there exists 0<k<1 such that
(2.1)ωλ(T(x),T(y))≤kmax⁡{ωλ(x,y),ωλ(x,T(x)),ωλ(y,T(y)),ωλ(x,T(y)),ωλ(T(x),y)}
for any x,y∈𝔛 and λ∈(0,∞).LetT:ℭ→ℭ be a mapping, and let ℭ be a nonempty subset of 𝔛ω. For any x∈ℭ, define the orbit(2.2)O(x)={x,T(x),T2(x),…}
and its ω-diameter by(2.3)δω(x)=diam⁡(O(x))=sup⁡{ωλ(Tn(x),Tm(x)):n,m∈N)}.Lemma 2.2.
Let(𝔛,ω) be a metric modular space, and let ℭ be a nonempty subset of 𝔛ω. Let T:ℭ→ℭ be a quasi-contractive mapping, and let x∈ℭ be such that δω(x)<∞. Then, for any n≥1, one has
(2.4)δω(T(x))≤knδω(x),
where k is the constant associated with the mapping of T. Moreover, one has
(2.5)ωλ(Tn(x),Tn+m(x))≤knδω(x)
for any n,m≥1 and λ∈(0,∞).Proof.
For eachn,m≥1, we have
(2.6)ωλ(Tn(x),Tm(y))≤kmax⁡{ωλ(Tn-1(x),Tm-1(y)),ωλ(Tn-1(x),Tn(x)),ωλ(Tm-1(y),Tm(y)),ωλ(Tn-1(x),Tm(y)),ωλ(Tn(x),Tm-1(y))}
for any x,y∈ℭ and λ∈(0,∞). This obviously implies that
(2.7)δω(Tn(x))≤kδω(Tn-1(x))
for any n≥1. Hence, for any n≥1, we have
(2.8)δω(Tn(x))≤knδω(x).
Moreover, for any n,m≥1, we have
(2.9)ωλ(Tn(x),Tn+m(x)≤δω(Tn(x))≤knδω(x).
This completes the proof.The next lemma is helpful to prove the main result in this paper.Lemma 2.3.
Let(𝔛,ω) be a modular metric space, and let ℭ be a ω-complete nonempty subset of 𝔛ω. Let T:ℭ→ℭ be quasi-contractive mapping, and let x∈ℭ be such that δω(x)<∞. Then {Tn(x)}ω-converges to a point ν∈ℭ. Moreover, one has
(2.10)ωλ(Tn(X)-ν)≤knδω(x)
for all n≥1 and λ∈(0,∞).Proof.
From Lemma2.2, we know that {Tn(x)} is a ω-Cauchy sequence in ℭ. Since ℭ is ω-complete, then there exists ν∈ℭ such that {Tn(x)}ω-converges to ν. Since
(2.11)ωλ(Tn(x),Tn+m(x))≤knδω(x)
for any n,m≥1 and ω satisfies the Fatou property, and letting m→∞, we have
(2.12)ωλ(Tn(x),ν)≤liminf⁡m→∞ωλ(Tn(x),Tn+m(x))≤knδω(x).
This completes the proof.Next, we prove thatν is, in fact, a fixed point of T and it is unique provided some extra assumptions.Theorem 2.4.
LetT,ℭ, and x be as in Lemma 2.3. Suppose that ωλ(ν,T(ν))<∞ and ωλ(x,T(x))<∞ for all λ∈(0,∞). Then the ω-limit of {Tn(x)} is a fixed point of T, that is, T(ν)=ν. Moreover, if ν* is any fixed point of T in ℭ such that ωλ(ν,ν*)<∞ for all λ∈(0,∞), then one has ν=ν*.Proof.
We have(2.13)ωλ(T(x),T(ν))≤kmax⁡{ωλ(x,ν),ωλ(x,T(x)),ωλ(ν,T(ν)),ωλ(x,T(ν)),ωλ(T(x),ν)}.
From Lemma 2.3, it follows that
(2.14)ωλ(T(x),T(ν))≤kmax⁡{δω(x),ωλ(ν,T(ν)),ωλ(x,T(ν))}.
Suppose that, for each n≥1,
(2.15)ωλ(Tn(x),T(ν))≤max⁡{knδω(x),kωλ(ν,T(ν)),knωλ(x,T(ν))}.
Then we have
(2.16)ωλ(Tn+1(x),T(ν))≤kmax⁡{ωλ(Tn(x),ν),ωλ(Tn(x),Tn+1(x)),ωλ(ν,T(ν)),ωλ(Tn(x),T(ν)),ωλ(Tn+1(x),ν)}.
Hence we have
(2.17)ωλ(Tn+1(x),T(ν))≤kmax⁡{knδω(x),kωλ(ν,T(ν)),ωλ(Tn(x),T(ν))}.
Using our previous assumption, we get
(2.18)ωλ(Tn+1(x),T(ν))≤max⁡{kn+1δω(x),kωλ(ν,T(ν)),kn+1ωλ(x,T(ν))}.
Thus, by induction, we have
(2.19)ωλ(Tn(x),T(ν))≤max⁡{knδω(x),kωλ(ν,T(ν)),knωλ(x,T(ν))}
for any n≥1 and λ∈(0,∞). Therefore, we have
(2.20)limsup⁡n→∞ωλ(Tn(x),T(x))≤ω(ν,T(ν))
for all λ∈(0,∞). Using the Fatou property for the metric modular ω, we get
(2.21)ωλ(ν,T(ν))liminf⁡n→∞ωλ(Tn(x),T(ν))≤kω(ν,T(ν))
for all λ∈(0,∞). Since k<1, we get ωλ(ν,T(ν))=0 for all λ∈(0,∞), and so T(ν)=ν.
Letν* be another fixed point of T such that ωλ(ν,ν*)<∞ for all λ∈(0,∞). Then we have
(2.22)ωλ(ν,ν*)=ωλ(T(ν),T(ν*))≤kωλ(ν,ν*),
which implies that
(2.23)ωλ(ν,ν*)=0
for all λ∈(0,∞). Hence ν=ν*. This complete the proof.



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*Source: 907951-2012-02-26.xml*