# Nonexpansive Mappings on New Premodular Special Space of Sequences

Authors: Awad A. Bakery; OM Kalthum S. K. Mohamed
Journal: Journal of Mathematics
 (2022)

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## Abstract

For different premodular, which is a generalization of modular, defined by weighted Orlicz sequence space and its prequasi operator ideal, we have examined the existence of a fixed point for both Kannan contraction and nonexpansive mappings acting on these spaces. Some numerous numerical experiments and practical applications are presented to support our results.


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## Body

## 1. Introduction

The spaces of all, bounded, r-absolutely summable, and null sequences of real numbers will be denoted throughout the article byRZ+, l∞, lr, and c0, respectively, where Z+ is the set of nonnegative integers.Definition 1.
[1, 2] An Orlicz function is a functionM:0,∞⟶0,∞, which is continuous and strictly increasing withM0=0,Mv>0forv>0, andMv⟶∞, asv⟶∞.Definition 2.
An Orlicz functionMis said to satisfyΔ2-condition for every values ofv≥0, if there isk>0, such thatM2v≤kMv. TheΔ2-condition is equivalent toMlv≤klMvfor every values ofl>1andv.Lindentrauss and Tzafriri [3] utilized the idea of a convex Orlicz function to define Orlicz sequence space:(1)lM=v∈RZ+:ρωv<∞for someω>0,whereρv=∑y=0∞Mvy.lM,. is a Banach space with the Luxemburg norm:(2)v=infω>0:ρvω≤1.Every Orlicz sequence space contains a subspace that is isomorphic toc0 or lr, for some 1≤r<∞ ([4], Theorem 4.a.9). The space of all bounded linear operators from a Banach space X into a Banach space Y will be denoted by BX,Y and if X=Y, we write BX. ex=0,0,…,1,0,0,…, while 1 lies in the xth place, with x∈Z+.Definition 3.
[5] Ans-number function is a mapping fromBX,Yinto0,∞Z+which transforms every mapH∈BX,YtosxHx=0∞satisfying the next conditions:(i)
H=s0H≥s1H≥s2H≥⋯≥0, for everyH∈BX,Y,(ii)
sy+x−1H1+H2≤syH1+sxH2, for everyH1,H2∈BX,Y, andy,x∈Z+,(iii)
ideal property:sxUTH≤UsxTH, for everyH∈BX0,X,T∈BX,YandU∈BY,Y0, whereX0andY0are any two Banach spaces,(iv)
forH∈BX,Yandω∈R, we havesxωH=ωsxH,(v)
rank property: IfrankH≤x, thensxH=0, for allH∈BX,Y,(vi)
norming property:sl≥xIx=0orsl<xIx=1, whereIxexplains the unit map on thex-dimensional Hilbert spacel2x.Thexth approximation number, αxW, is defined as(3)αxH=infH−Y:Y∈BX,Yand rankY≤x.Notations 1.
The setsSW, SWX,Y, SWapp, and SWappX,Y (cf. [6]) are defined as follows:(4)SW≔SWX,Y,whereSWX,Y≔H∈BX,Y:sxHx=0∞∈W.AlsoSWapp≔SWappX,Y,whereSWappX,Y≔H∈BX,Y:αxHx=0∞∈W.Fixed point theory, Banach space geometry, normal series theory, ideal transformations, and approximation theory are all examples of ideal operator theorems and summability. Faried and Bakery [6] established the concept of a prequasi operator ideal that encapsulates the quasi operator ideal. Bakery and Abou Elmatty investigated the sufficient (but not necessary) conditions on lγ,r that allowed Slγ,r to build a simple Banach prequasi operator ideal in [7]. For varied weights and powers, the prequasi operator ideal Slγ,rapp was once rigorously contained and small prequasi operator ideal. Several mathematicians were able to investigate many extensions for contraction maps defined on the space or on the space itself thanks to the Banach fixed point theorem [8]. Kannan [9] investigated an example of a class of operators that perform the same fixed point actions as contractions but are not continuous. Kannan operators in modular vector spaces have only been described by Ghoncheh [10]. He demonstrated the existence of a Kannan mapping fixed point in complete modular spaces with Fatou property. For more details on Kannan’s fixed point theorems and modular vector spaces (see [11–14]). Bakery and Mohamed [15] introduced the concept of the prequasi norm on lra with variable exponent in 0,1. They looked at the Fatou property of different prequasi norms on lra, as well as the sufficient requirements on lra with the definite prequasi norm to construct prequasi Banach and closed space. They also demonstrated the existence of a fixed point of Kannan prequasi norm contraction maps on lra and the prequasi Banach operator ideal constructed by lra and s-numbers. Recently, Reich and Zaslavski [16] showed the existence of a unique fixed point for nonlinear contractive self-mappings of a nonbounded closed subset of a Banach space. They extended this conclusion to contractive mappings, which map into a Banach space a closed subset of the space. For nonexpansive mappings defined by an intersection of a finite number of closed bounded and convex nonempty subsets in Banach spaces, Dehici and Redjel [17] obtained certain fixed point results. According to Bendahmane and Bendoukha [18], a p,q-metric space is a generalization of the metric and S-metric spaces. They equipped them a Hausdorff topology and specified several fundamental features. Several well-known findings from fixed point theory are generalized to these new spaces. The paper is structured as follows: we present conditions on the weighted Orlicz sequence space lMλμ, under definite prequasi norm of μ to construct prequasi Banach and closed sequence space in Section 3. The Fatou property of lMλ has been investigated for various prequasi norms. In Section 4, the existence of fixed point for Kannan μ-contraction mapping acting on lMλμ equipped with different prequasi norms are presented. Several numerical experiments are shown to demonstrate our results. In Section 5, the conditions for which the space lMλμ satisfies the property (R) and has the μ-normal structure property are presented. The existence of a fixed point of Kannan prequasi norm nonexpansive mapping on lMλμ has been given. In Section 6, we explain the existence of a fixed point of Kannan prequasi norm contraction mapping in the prequasi Banach operator ideal SlMλμ. In Section 7, we give some applications to the existence of solutions of summable equations.


## 2. Definitions and Preliminaries

Here and after, the space of all functionsμ:Y⟶0,∞ is 0,∞Y, θ is the zero vector of Y, x/2 is the integral part of x/2, F is the space of finite sequences, and B is the class of each bounded linear mapping between any two Banach spaces. Nakano [19] introduced the concept of modular vector spaces.Definition 4.
LetYbe a vector space. A functionμ∈0,∞Yis called modular if the following conditions hold:(i)
Ifβ∈Y,β=θ⇔μβ=0andμβ≥0,(ii)
ifβ∈Yandω=1, thenμωβ=μβ,(iii)
assumeβ,η∈Yandω∈0,1, thenμωβ+1−ωη≤μβ+μη.The concept of premodular vector spaces, which is more general than modular vector spaces.Definition 5.
[6] The linear space of sequences Y is said to be a special space of sequences (sss), if:(1)
exx∈Z+⊆Y,(2)
Yis solid, i.e., forβ=βx∈RZ+,η=ηx∈Yandβx≤ηx, for allx∈Z+, thenβ∈Y,(3)
Ifβxx=0∞∈Y, thenβx/2x=0∞∈Y.Definition 6.
[6] A subclass Yμ of Y is called a premodular (sss), if we have μ∈0,∞Y that satisfies the following conditions:(i)
Whenβ∈Y,β=θ⇔μβ=0,(ii)
For everyβ∈Yandω∈R, then there isB≥1withμωβ≤Bωμβ,(iii)
μβ+η≤Jμβ+μη, for allβ,η∈Y, holds for someJ≥1,(iv)
Ifx∈Z+andβx≤ηx, thenμβx≤μηx,(v)
For someJ0≥1, we haveμβx≤μβx/2≤J0μβx,(vi)
F¯=Yμ,(vii)
There existsς>0such thatμω,0,0,0,…≥ςωμ1,0,0,0,…, for allω∈R.Example 1.
The functionμβ=∑x∈Z+βx55is a premodular (not a modular) on the vector spacel1/5. As for everyβ,η∈l1/5, one has(5)μβ+η2=∑x∈Z+βx+ηx255≤8μβ+μη.Definition 7.
[15] Suppose Y is a (sss). The function μ∈0,∞Y is said to be prequasi norm on Y, if it holds the settings (i), (ii), and (iii) of Definition 6.Theorem 1.
[15] Let Y be a premodular (sss), then it is prequasi normed (sss).Theorem 2.
[15] Y is a prequasi normed (sss), when it is quasi-normed (sss).Definition 8.
[20](i)
The prequasi normμonXμis said to beμ-convex, whenμωβ+1−ωη≤ωμβ+1−ωμη, for allω∈0,1andβ,η∈Xμ.(ii)
βxx∈Z+⊆Xμisμ-convergent toβ∈Xμ, if and only if,limx⟶∞μβx−β=0. If theμ-limit exists, hence it is unique.(iii)
βxx∈Z+⊆Xμisμ-Cauchy, iflimx,h⟶∞μβx−βh=0.(iv)
Φ⊂Xμisμ-closed, if for everyμ-convergingβxx∈Z+⊂Φtoβ, thenβ∈Φ.(v)
Φ⊂Xμisμ-bounded, ifυμΦ=supμβ−η:β,η∈Φ<∞.(vi)
Theμ-ball of radiusr≥0and centerβ, for everyβ∈Xμ, is defined as(6)Bμβ,r=η∈Xμ:μβ−η≤r.(vii)
A prequasi normμonXsatisfies the Fatou property, if for every sequenceηx⊆Xμwithlimx⟶∞μηx−η=0and anyβ∈Xμ, we haveμβ−η≤supminfx≥mμβ−ηx.Recall that theμ-balls are μ-closed under the Fatou property.Definition 9.
[21] A subclass G of B is called an operator ideal, if every vector GX,Y=G∩BX,Y holds the following conditions:(i)
Iγ∈G, whereγindicates Banach space of one dimension.(ii)
The spaceGX,Yis linear overR.(iii)
IfH∈BX0,X,T∈GX,Y, andV∈BY,Y0, thenVTH∈GX0,Y0, whereX0andY0are normed spaces.Recall that the quasi operator ideals are a special case of the prequasi operator ideals.Definition 10.
[6] A function Υ∈0,∞G is said to be a prequasi norm on the ideal G if the following conditions verify:(1)
SupposeH∈GX,Y, ΥH≥0andΥH=0, if and only if,H=0,(2)
there existsD≥1such thatΥωH≤DωΥH, for everyH∈GX,Yandω∈R,(3)
we haveJ≥1so thatΥH1+H2≤JΥH1+ΥH2, for allH1,H2∈GX,Y,(4)
we getω≥1so that ifH∈BX0,X,T∈GX,Y, andV∈BY,Y0, thenΥVTH≤ωVΥTH.Theorem 3.
[15] The function ΥH=μsxHx=0∞ is a prequasi norm on SYμ, when Yμ is a premodular (sss).Theorem 4.
[6] If Υ is a quasi norm on the ideal G, then Υ is a prequasi norm on the ideal G.Lemma 1.
[22, 23] AssumeM:0,∞⟶0,∞is a continuous function and strictly increasing withlimx⟶0Mx=0, and if the functionsMxandlnMexare convex on0,∞, then(7)M−1∑x=0∞λxMβx+ηx≤M−1∑x=0∞λxMβx+M−1∑x=0∞λxMηx.
λx,βx,ηx∈0,∞, for all x∈Z+ and ∑x=0∞λx=1.


## 3. Main Results

### 3.1. Properties of Different Prequasi Norms

In this section, we have studied some topological structures and the Fatou property of the weighted Orlicz sequence space,lMλ, for various prequasi norms.Lemma 2.
IfMis a concave Orlicz function, thenMx+y≤Mx+My, for allx,y∈0,∞.Proof.
It is easy so omitted.Theorem 5.
lMλμ, whereμβ=∑y=0∞λyMβy, for eachβ∈lMλ, is a premodular (sss), ifMis a concave Orlicz function or convex Orlicz function satisfyingΔ2-condition.Proof.
SupposeM is a convex Orlicz function satisfying Δ2-condition. First, we must demonstrate that lMλ is a (sss):(1)
(i)
Letβ,t∈lMλ. As M is a strictly increasing and convex function satisfying Δ2-condition, we get(8)μβ+η=∑y=0∞λyMβy+ηy≤k2∑y=0∞λyMβy+∑y=0∞λyMηy=k2μβ+μη<∞,this impliesβ+η∈lMλ.(ii)
Supposeω∈R and β∈lMλ. Since M satisfies Δ2-condition, we have(9)μωβ=∑y=0∞λyMωβy≤kω∑y=0∞λyMβy≤Dωμβ<∞.Soωβ∈lMλ. Therefore, from conditions 1 (i) and (ii), one has lMλ is linear. We have ey∈lMλ, for every y∈Z+, as(2)
Letβy≤ηy, for every y∈Z+ and η∈lMλ. Since M is a nondecreasing function, then(10)μβ=∑y=0∞λyMβy≤∑y=0∞λyMηy=μη<∞,one hasβ∈lMλ.(3)
Assumeβy∈lMλ, we get(11)μβy/2=∑y=0∞λyMβy/2≤2∑y=0∞λyMβy=2μβy<∞,thenβy/2∈lMλ. Second, to prove that the functional μ on lMλ is a premodular:(i)
Obviously,μβ≥0 and μβ=0⇔β=θ.(ii)
There areD=max1,k≥1 with μωβ≤Dωμβ, for every β∈lMλ and ω∈R.(iii)
There existsJ=max1,k/2≥1 with μβ+η≤Jμβ+μη, for every β,η∈lMλ.(iv)
Follows the proof part (2).(v)
Follows from the proof part (3) thatJ0=2≥1.(vi)
Obviously,F¯=lMλ.(vii)
There exists0<ς≤M0ω/ωM01, for ω≠0 or ς>0, for ω=0 so that μω,0,0,0,…≥ςωμ1,0,0,0,….
IfM is a concave Orlicz function. By applying Lemma 2 and the parallel proof follows.Theorem 6.
IfMis a concave Orlicz function or convex Orlicz function satisfyingΔ2-condition, thenlMλμis a prequasi Banach (sss), whereμβ=∑y=0∞λyMβy, for eachβ∈lMλ.Proof.
SupposeM is a convex Orlicz function satisfying Δ2-condition. By using Theorem 5, the space lMλμ is a premodular (sss). From Theorem 1, the space lMλμ is a prequasi normed (sss). To prove that lMλμ is a prequasi Banach (sss), let βr=βyry=0∞ be a Cauchy sequence in lMλμ. Therefore, for all ϵ∈0,1, we have that for every r,t≥r0, we get(12)μβr−βt=∑y=0∞λyMβyr−βyt<ϵ.
Hence, forr,t≥r0 and y∈Z+, one has βyr−βyt<ϵ Then βyt is a Cauchy sequence in R, for fixed y∈Z+. This gives limt⟶∞βyt=βy0, for constant y∈Z+. Therefore, μβr−β0<ϵ, for all r≥r0. To investigate that β0∈lMλ, one has μβ0=μβ0−βr+βr≤Jμβr−β0+μβr<∞, so β0∈lMλ. This implies that lMλμ is a prequasi Banach (sss). If M is a concave Orlicz function. By applying Lemma 2 and the parallel proof follows.Theorem 7.
IfM is a concave Orlicz function or convex Orlicz function satisfying Δ2-condition, then lMλμ is a prequasi closed (sss), where μβ=∑y=0∞λyMβy, for every β∈lMλ.Proof.
LetM be a convex Orlicz function satisfying Δ2-condition. According to Theorem 5, the space lMλμ is a premodular (sss). From Theorem 1, the space lMλμ is a prequasi normed (sss). To prove that lMλμ is a prequasi closed (sss), suppose βr=βyry=0∞∈lMλμ and limr⟶∞μβr−β0=0, hence for all ϵ∈0,1, one has r0∈Z+ so that for every r≥r0, we have(13)μβr−β0=∑y=0∞λyMβyr−βy0<ϵ.
Therefore, forr≥r0 and y∈Z+, one has βyr−βy0<ϵ. Hence, βyr is a convergent sequence in R, for constant y∈Z+. So, limr⟶∞βyr=βy0, for constant y∈Z+. Finally to show that β0∈lMλ, one has(14)μβ0=μβ0−βr+βr≤Jμβr−β0+μβr<∞.
Hence,HTML translation failed. This implies that lMλμ is a prequasi closed (sss). If M is a concave Orlicz function, by applying Lemma 2 and the parallel proof follows.Theorem 8.
IfM is a convex Orlicz function satisfying Δ2-condition and lnMex is convex, then the function μβ=M−1∑y=0∞λyMβy verifies the Fatou property, for all β∈lMλ.Proof.
Assume thatηb⊆lMλμ such that limb⟶∞μηb−η=0. As the space lMλμ is a prequasi closed space, one has t∈lMλμ. Hence, for every β∈lMλμ, from Lemma 1, we have(15)μβ−η=M−1∑y=0∞λyMβy−ηy≤M−1∑y=0∞λyMβy−ηyb+M−1∑y=0∞λyMηyb−ηy≤supjinfb≥jμβ−ηb.
Hence,μ satisfies the Fatou property.Theorem 9.
IfMis a concave Orlicz function, then the functionμβ=∑y=0∞λyMβyholds the Fatou property, for allβ∈lMλ.Proof.
Supposeηb⊆lMλμ so that limb⟶∞μηb−η=0. As the space lMλμ is a prequasi closed space; hence, η∈lMλμ. As M is continuous, concave and M0=0. Therefore, for every β∈lMλμ, one has(16)μβ−η=∑y=0∞λyMβy−ηy≤∑y=0∞λyMβy−ηyb+∑y=0∞λyMηyb−ηy≤supjinfb≥jμβ−ηb.
Hence,μ satisfies the Fatou property.Theorem 10.
The functionμβ=∑y=0∞λyMβydoes not satisfy the Fatou property, for allβ∈lMλ, ifMis a strictly convex Orlicz function satisfyingΔ2-condition.Proof.
SinceM is a strictly convex Orlicz function satisfying Δ2-condition, then there exists k>2 such that 2Mu<M2u<kMu, for all u≥0. Let the conditions be fulfilled and ηb⊆lMλμ with limb⟶∞μηb−η=0. As the space lMλμ is a prequasi closed space; hence, η∈lMλμ. Since M is continuous, then for any β∈lMλμ, we have(17)μβ−η=∑y=0∞λyMβy−ηy≤k2∑y=0∞λyMβy−ηyb+∑y=0∞λyMηyb−ηy≤k2supjinfb≥jμβ−ηb.
Therefore,μ does not hold the Fatou property.Example 2.
For everyβ∈lMλ, the functionμβ=ln1+∑y=0∞λyeβy−1is a prequasi norm, not quasi, and not a norm.Example 3.
For allβ∈lMλ, the functionμβ=∑y=0∞λyβy2is a prequasi norm, quasi norm, and not a norm.Example 4.
The functionμβ=infκ>0:∑y=0∞λyMβy/κ≤1is a prequasi norm, a quasi norm, and a norm onlMλ.


## 3.1. Properties of Different Prequasi Norms

In this section, we have studied some topological structures and the Fatou property of the weighted Orlicz sequence space,lMλ, for various prequasi norms.Lemma 2.
IfMis a concave Orlicz function, thenMx+y≤Mx+My, for allx,y∈0,∞.Proof.
It is easy so omitted.Theorem 5.
lMλμ, whereμβ=∑y=0∞λyMβy, for eachβ∈lMλ, is a premodular (sss), ifMis a concave Orlicz function or convex Orlicz function satisfyingΔ2-condition.Proof.
SupposeM is a convex Orlicz function satisfying Δ2-condition. First, we must demonstrate that lMλ is a (sss):(1)
(i)
Letβ,t∈lMλ. As M is a strictly increasing and convex function satisfying Δ2-condition, we get(8)μβ+η=∑y=0∞λyMβy+ηy≤k2∑y=0∞λyMβy+∑y=0∞λyMηy=k2μβ+μη<∞,this impliesβ+η∈lMλ.(ii)
Supposeω∈R and β∈lMλ. Since M satisfies Δ2-condition, we have(9)μωβ=∑y=0∞λyMωβy≤kω∑y=0∞λyMβy≤Dωμβ<∞.Soωβ∈lMλ. Therefore, from conditions 1 (i) and (ii), one has lMλ is linear. We have ey∈lMλ, for every y∈Z+, as(2)
Letβy≤ηy, for every y∈Z+ and η∈lMλ. Since M is a nondecreasing function, then(10)μβ=∑y=0∞λyMβy≤∑y=0∞λyMηy=μη<∞,one hasβ∈lMλ.(3)
Assumeβy∈lMλ, we get(11)μβy/2=∑y=0∞λyMβy/2≤2∑y=0∞λyMβy=2μβy<∞,thenβy/2∈lMλ. Second, to prove that the functional μ on lMλ is a premodular:(i)
Obviously,μβ≥0 and μβ=0⇔β=θ.(ii)
There areD=max1,k≥1 with μωβ≤Dωμβ, for every β∈lMλ and ω∈R.(iii)
There existsJ=max1,k/2≥1 with μβ+η≤Jμβ+μη, for every β,η∈lMλ.(iv)
Follows the proof part (2).(v)
Follows from the proof part (3) thatJ0=2≥1.(vi)
Obviously,F¯=lMλ.(vii)
There exists0<ς≤M0ω/ωM01, for ω≠0 or ς>0, for ω=0 so that μω,0,0,0,…≥ςωμ1,0,0,0,….
IfM is a concave Orlicz function. By applying Lemma 2 and the parallel proof follows.Theorem 6.
IfMis a concave Orlicz function or convex Orlicz function satisfyingΔ2-condition, thenlMλμis a prequasi Banach (sss), whereμβ=∑y=0∞λyMβy, for eachβ∈lMλ.Proof.
SupposeM is a convex Orlicz function satisfying Δ2-condition. By using Theorem 5, the space lMλμ is a premodular (sss). From Theorem 1, the space lMλμ is a prequasi normed (sss). To prove that lMλμ is a prequasi Banach (sss), let βr=βyry=0∞ be a Cauchy sequence in lMλμ. Therefore, for all ϵ∈0,1, we have that for every r,t≥r0, we get(12)μβr−βt=∑y=0∞λyMβyr−βyt<ϵ.
Hence, forr,t≥r0 and y∈Z+, one has βyr−βyt<ϵ Then βyt is a Cauchy sequence in R, for fixed y∈Z+. This gives limt⟶∞βyt=βy0, for constant y∈Z+. Therefore, μβr−β0<ϵ, for all r≥r0. To investigate that β0∈lMλ, one has μβ0=μβ0−βr+βr≤Jμβr−β0+μβr<∞, so β0∈lMλ. This implies that lMλμ is a prequasi Banach (sss). If M is a concave Orlicz function. By applying Lemma 2 and the parallel proof follows.Theorem 7.
IfM is a concave Orlicz function or convex Orlicz function satisfying Δ2-condition, then lMλμ is a prequasi closed (sss), where μβ=∑y=0∞λyMβy, for every β∈lMλ.Proof.
LetM be a convex Orlicz function satisfying Δ2-condition. According to Theorem 5, the space lMλμ is a premodular (sss). From Theorem 1, the space lMλμ is a prequasi normed (sss). To prove that lMλμ is a prequasi closed (sss), suppose βr=βyry=0∞∈lMλμ and limr⟶∞μβr−β0=0, hence for all ϵ∈0,1, one has r0∈Z+ so that for every r≥r0, we have(13)μβr−β0=∑y=0∞λyMβyr−βy0<ϵ.
Therefore, forr≥r0 and y∈Z+, one has βyr−βy0<ϵ. Hence, βyr is a convergent sequence in R, for constant y∈Z+. So, limr⟶∞βyr=βy0, for constant y∈Z+. Finally to show that β0∈lMλ, one has(14)μβ0=μβ0−βr+βr≤Jμβr−β0+μβr<∞.
Hence,HTML translation failed. This implies that lMλμ is a prequasi closed (sss). If M is a concave Orlicz function, by applying Lemma 2 and the parallel proof follows.Theorem 8.
IfM is a convex Orlicz function satisfying Δ2-condition and lnMex is convex, then the function μβ=M−1∑y=0∞λyMβy verifies the Fatou property, for all β∈lMλ.Proof.
Assume thatηb⊆lMλμ such that limb⟶∞μηb−η=0. As the space lMλμ is a prequasi closed space, one has t∈lMλμ. Hence, for every β∈lMλμ, from Lemma 1, we have(15)μβ−η=M−1∑y=0∞λyMβy−ηy≤M−1∑y=0∞λyMβy−ηyb+M−1∑y=0∞λyMηyb−ηy≤supjinfb≥jμβ−ηb.
Hence,μ satisfies the Fatou property.Theorem 9.
IfMis a concave Orlicz function, then the functionμβ=∑y=0∞λyMβyholds the Fatou property, for allβ∈lMλ.Proof.
Supposeηb⊆lMλμ so that limb⟶∞μηb−η=0. As the space lMλμ is a prequasi closed space; hence, η∈lMλμ. As M is continuous, concave and M0=0. Therefore, for every β∈lMλμ, one has(16)μβ−η=∑y=0∞λyMβy−ηy≤∑y=0∞λyMβy−ηyb+∑y=0∞λyMηyb−ηy≤supjinfb≥jμβ−ηb.
Hence,μ satisfies the Fatou property.Theorem 10.
The functionμβ=∑y=0∞λyMβydoes not satisfy the Fatou property, for allβ∈lMλ, ifMis a strictly convex Orlicz function satisfyingΔ2-condition.Proof.
SinceM is a strictly convex Orlicz function satisfying Δ2-condition, then there exists k>2 such that 2Mu<M2u<kMu, for all u≥0. Let the conditions be fulfilled and ηb⊆lMλμ with limb⟶∞μηb−η=0. As the space lMλμ is a prequasi closed space; hence, η∈lMλμ. Since M is continuous, then for any β∈lMλμ, we have(17)μβ−η=∑y=0∞λyMβy−ηy≤k2∑y=0∞λyMβy−ηyb+∑y=0∞λyMηyb−ηy≤k2supjinfb≥jμβ−ηb.
Therefore,μ does not hold the Fatou property.Example 2.
For everyβ∈lMλ, the functionμβ=ln1+∑y=0∞λyeβy−1is a prequasi norm, not quasi, and not a norm.Example 3.
For allβ∈lMλ, the functionμβ=∑y=0∞λyβy2is a prequasi norm, quasi norm, and not a norm.Example 4.
The functionμβ=infκ>0:∑y=0∞λyMβy/κ≤1is a prequasi norm, a quasi norm, and a norm onlMλ.


## 4. Kannanμ-Contraction Operator

We now define Kannanμ-Lipschitzian mapping acting on lMλμ. The sufficient conditions for a fixed point of Kannan contraction mapping on lMλμ under various prequasi norms are investigated.Definition 11.
An operatorH:lMλμ⟶lMλμis called a Kannanμ-Lipschitzian, if there existsν≥0, so that(18)μHβ−Hη≤νμHβ−β+μHη−η,for every β,η∈lMλμ.(1)
The operatorH is said to be Kannan μ-contraction, when ν∈0,1/2.(2)
The operatorH is said to be Kannan μ-nonexpansive, whenever ν=1/2.
A vectorβ∈lMλμ is called a fixed point of H, when Hβ=β.Theorem 11.
IfMis a convex Orlicz function satisfyingΔ2-condition andlnMexis convex, andH:lMλμ⟶lMλμis Kannanμ-contraction mapping, whereμβ=M−1∑y=0∞λyMβy, for allβ∈lMλ; hence,Hhas a unique fixed point.Proof.
Assume thatβ∈lMλ, one has Htβ∈lMλ. Since H is a Kannan μ-contraction mapping, we have(19)μHt+1β−Htβ≤νμHt+1β−Htβ+μHtβ−Ht−1β⇒μHt+1β−Htβ≤ν1−νμHtβ−Ht−1β≤ν1−ν2μHt−1β−Ht−2β≤⋯≤ν1−νtμHβ−β.
Therefore, for everyt,v∈Z+ with v>t, then we get(20)μHtβ−Hvβ≤νμHtβ−Ht−1β+μHvβ−Hv−1β≤νν1−νt−1+ν1−νv−1μHβ−β.
So,Htβ is a Cauchy sequence in lMλμ. As the space lMλμ is prequasi Banach space. Therefore, there is η∈lMλμ such that limt⟶∞Htβ=η. To prove that Hη=η. As μ holds the Fatou property, we obtain(21)μHη−η≤suppinft≥pμHt+1β−Htβ≤suppinft≥pν1−νtμHβ−β=0,hence Hη=η. Hence, η is a fixed point of H. To prove the uniqueness of the fixed point. For different fixed points ζ,η∈lMλμ of H. We have that(22)μζ−η≤μHζ−Hη≤νμHζ−ζ+μHη−η=0.
Therefore,ζ=η.Corollary 1.
LetMbe a convex Orlicz function satisfyingΔ2-condition andlnMexbe convex, andH:lMλμ⟶lMλμbe Kannanμ-contraction mapping, withμβ=M−1∑y=0∞λyMβy, for everyβ∈lMλ, thenHhas a unique fixed pointζsuch thatμHtβ−ζ≤νν/1−νt−1μHβ−β.Proof.
From Theorem11, there is a unique fixed point ζ of H. Hence, one has(23)μHtβ−ζ=μHtβ−Hζ≤νμHtβ−Ht−1β+μHζ−ζ=νν1−νt−1μHβ−β.Theorem 12.
SupposeMis a concave Orlicz function, andH:lMλμ⟶lMλμis Kannanμ-contraction mapping, whereμβ=∑y=0∞λyMβy, for allβ∈lMλ; hence,Hhas a unique fixed point.Proof.
It is easy so omitted.Definition 13.
AssumelMλμis a pr-quasi normed (sss),H:lMλμ⟶lMλμandζ∈lMλμ. The operatorHis calledμ-sequentially continuous atζ, if and only if, whenlimy⟶∞μβy−ζ=0, thenlimy⟶∞μHβy−Hζ=0.Theorem 14.
LetMbe a strictly convex Orlicz function satisfyingΔ2-condition, andH:lMλμ⟶lMλμ, whereμβ=∑y=0∞λyMβy, for everyβ∈lMλ. The elementη∈lMλμis the unique fixed point ofH, if the next conditions are satisfied:(i)
H is Kannan μ-contraction mapping,(ii)
H is μ-sequentially continuous at a point η∈lMλμ,(iii)
There existsβ∈lMλμ such that the sequence of iterates Htβ has a subsequence Htpβ converging to η.Proof.
SinceM is a strictly convex Orlicz function satisfying Δ2-condition, then there exists k>2 such that 2Mu<M2u<kMu, for all u≥0. Let the conditions be verified. If η is not a fixed point of H, then Hη≠η. By the conditions (ii) and (iii), we have(24)limtp⟶∞μHtpβ−η=0,limtp⟶∞μHtp+1β−Hη=0.
As the operatorH is Kannan μ-contraction, one can see(25)0<μHη−η=μHη−Htp+1β+Htpβ−η+Htp+1β−Htpβ≤k24μHtp+1β−Hη+k24μHtpβ−η+k2νν1−νtp−1μHβ−β.
Sincetp⟶∞, this gives a contradiction. Hence, η is a fixed point of H. To prove that the uniqueness of the fixed point η. For different fixed points η,ζ∈lMλμ of H. Therefore, one has(26)μη−ζ≤μHη−Hζ≤νμHη−η+μHζ−ζ=0.
So,η=ζ.Example 15.
AssumeH:lMλμ⟶lMλμ, whereMt=t3+t4andμβ=∑y=0∞λyMβy, for allβ∈lMλand(27)Hβ=β18,μβ∈0,1,β20,μβ∈1,∞.
As for eachβ1,β2∈lMλμ with μβ1,μβ2∈0,1, one has(28)μHβ1−Hβ2=μβ118−β218≤1174μ17β118+μ17β218=1174μHβ1−β1+μHβ2−β2.
For allβ1,β2∈lMλμ with μβ1,μβ2∈1,∞, one has(29)μHβ1−Hβ2=μβ120−β220≤1194μ19β120+μ19β220=1194μHβ1−β1+μHβ2−β2.
For allβ1,β2∈lMλμ with μβ1∈0,1 and μβ2∈1,∞, we obtain(30)μHβ1−Hβ2=μβ118−β220≤1174μ17β118+1194μ19β220≤1174μ17β118+μ19β220=1174μHβ1−β1+μHβ2−β2.
Hence, the operatorH is Kannan μ-contraction. As μ verifies the Fatou property. From Theorem 11, the operator H has a unique fixed point θ∈lMλμ.
Assumeβy⊆lMλμ is such that limy⟶∞μβy−β0=0, where
β0∈lMλμ with μβ0=1.
As the prequasi normμ is continuous, one can see(31)limtp⟶∞μHβy−Hβ0=limtp⟶∞μβy18−β020=μβ0180>0.
Therefore,H is not μ-sequentially continuous at β0. Hence, the operator H is not continuous at β0.
Letμβ=∑y=0∞λyMβy4, for all β∈lMλ.
As for allβ1,β2∈lMλμ with μβ1,μβ2∈0,1, one has(32)μHβ1−Hβ2=μβ118−β218≤817μ17β118+μ17β218=817μHβ1−β1+μHβ2−β2.
For allβ1,β2∈lMλμ with μβ1,μβ2∈1,∞, one has(33)μHβ1−Hβ2=μβ120−β220≤819μ19β120+μ19β220=819μHβ1−β1+μHβ2−β2.
For allβ1,β2∈lMλμ with μβ1∈0,1 and μβ2∈1,∞, we get(34)μHβ1−Hβ2=μβ118−β220≤817μ17β118+819μ19β220≤817μ17β118+μ19β220=817μHβ1−β1+μHβ2−β2.
So, the operatorH is Kannan μ-contraction and Htβ=β/18tμβ∈0,1β/20tμβ∈1,∞
Clearly,H is μ-sequentially continuous at θ∈lMλμ and Htβ contains a subsequence Htpβ converging to θ. From Theorem 14, then θ∈lMλμ is the unique fixed point of H.Example 5.
AssumeH:lMλμ⟶lMλμ, whereMt=t2andμβ=∑y=0∞λyMβy, for allβ∈lMλand(35)Hβ=β4,μβ∈0,1,β5,μβ∈1,∞.
As for eachβ1,β2∈lMλμ with μβ1,μβ2∈0,1, one has(36)μHβ1−Hβ2=μβ14−β24≤13μ3β14+μ3β24=13μHβ1−β1+μHβ2−β2.
For allβ1,β2∈lMλμ with μβ1,μβ2∈1,∞, one has(37)μHβ1−Hβ2=μβ15−β25≤14μ4β15+μ4β25=14μHβ1−β1+μHβ2−β2.
For allβ1,β2∈lMλμ with μβ1∈0,1 and μβ2∈1,∞, we get(38)μHβ1−Hβ2=μβ14−β25≤13μ3β14+14μ4β25≤13μ3β14+μ4β25=13μHβ1−β1+μHβ2−β2.
Hence, the operatorH is Kannan μ-contraction. As μ satisfies the Fatou property. From Theorem 11, the operator H has one fixed point θ∈lMλμ.
Supposeβy⊆lMλμ is so that limy⟶∞μβy−β0=0, where.
β0∈lMλμ with μβ0=1. As the prequasi norm μ is continuous, one can see(39)limy⟶∞μHβy−Hβ0=limy⟶∞μβy4−β05=μβ020>0.
Therefore,H is not μ-sequentially continuous at β0. Hence, the map H is not continuous at β0.
Letμβ=∑y=0∞λyMβy, for every β∈lMλ.
As for eachβ1,β2∈lMλμ with μβ1,μβ2∈0,1, one has(40)μHβ1−Hβ2=μβ14−β24≤29μ3β14+μ3β24=29μHβ1−β1+μHβ2−β2.
For allβ1,β2∈lMλμ with μβ1,μβ2∈1,∞, one has(41)μHβ1−Hβ2=μβ15−β25≤18μ4β15+μ4β25=18μHβ1−β1+μHβ2−β2.
For allβ1,β2∈lMλμ with μβ1∈0,1 and μβ2∈1,∞, we obtain(42)μHβ1−Hβ2=μβ14−β25≤29μ3β14+18μ4β25≤29μ3β14+μ4β25=29μHβ1−β1+μHβ2−β2.
So, the operatorH is Kannan μ-contraction and. Htβ=β/4tμβ∈0,1β/5tμβ∈1,∞
Obviously,H is μ-sequentially continuous at θ∈lMλμ and Htβ has a subsequence Htpβ converging to θ. From Theorem 14, then θ∈lMλμ is the unique fixed point of H.Example 16.
SupposeH:lMλμ⟶lMλμ, whereMt=t3+t4andμβ=∑y=0∞λyMβy4, for everyβ∈lMλand(43)Hβ=118e0+β,β0∈−∞,117,117e0,β0=117,118e0,β0∈117,∞.
As for eachβ,η∈lMλμ with β0,η0∈−∞,1/17, one has(44)μHβ−Hη=μ118β0−η0,β1−η1,β2−η2,…≤817μ17β18+μ17t18≤817μHβ−β+μHη−η.
For everyβ,η∈lMλμ with β0,η0∈1/17,∞, then for all ϵ>0 one has(45)μHβ−Hη=0≤ϵμHβ−β+μHη−η.
For everyβ,η∈lMλμ with β0∈−∞,1/17 and η0∈1/17,∞, we get(46)μHβ−Hη=μβ18≤117μ17β18=117μHβ−β≤117μHβ−β+μHη−η.
Hence, the operatorH is Kannan μ-contraction. Evidently, H is μ-sequentially continuous at 1/17e0∈lMλμ, and we have β∈lMλμ with β0∈−∞,1/17 under Htβ=∑n=1t1/18ne0+1/18tβ contains a subsequence Htpβ=∑n=1tp1/18ne0+1/18tpβ converging to 1/17e0. From Theorem 14, the map H has a unique fixed point 1/17e0∈lMλμ. Observe that H is not continuous at 1/17e0∈lMλμ.
Ifμβ=∑y∈Z+λyMβy, for every β∈lMλ. As for all β,η∈lMλμ with β0,η0∈−∞,1/17, one has(47)μHβ−Hη=μ118β0−η0,β1−η1,β2−η2,…≤1174μ17β18+μ17t18≤1174μHβ−β+μHη−η.
For eachβ,η∈lMλμ with β0,η0∈1/17,∞, then for all ϵ>0 we get(48)μHβ−Hη=0≤ϵμHβ−β+μHη−η.
For everyβ,η∈lMλμ with β0∈−∞,1/17 and η0∈1/17,∞, this gives(49)μHβ−Hη=μβ18≤1174μ17β18=1174μHβ−β≤1174μHβ−β+μHη−η.
So, the operatorH is Kannan μ-contraction. As μ satisfies the Fatou property. From Theorem 11, the operator H holds one fixed point 1/17e0∈lMλμ.Example 6.
AssumeH:lMλμ⟶lMλμ, where Mt=t2+2t and μβ=∑y=0∞λyMβy, for every β∈lMλ and(50)Hβ=16e1+β,β0∈−∞,15,15e1,β0=15,16e1,β0∈15,∞.
As for eachβ,η∈lMλμ with β0,η0∈−∞,1/5, one has(51)μHβ−Hη=μ16β0−η0,β1−η1,β2−η2,…≤25μ5β6+μ5t6≤25μHβ−β+μHv−η.
Supposeβ,η∈lMλμ with β0,η0∈1/5,∞, then for any ϵ>0 we obtain(52)μHβ−Hη=0≤ϵμHβ−β+μHη−η.
Assumeβ,η∈lMλμ with β0∈−∞,1/5 and η0∈1/5,∞, one can see(53)μHβ−Hη=μβ6≤15μ5β6=15μHβ−β≤15μHβ−β+μHη−η.
Hence, the operatorH is Kannan μ-contraction. Clearly, H is μ-sequentially continuous at 1/5e1∈lMλμ and there exists β∈lMλμ with β0∈−∞,1/5 under Htβ=∑n=1n=11/6ne1+1/6tβ contains a subsequence Htpβ=∑n=1tp1/6ne1+1/6tpβ converging to 1/5e1. From Theorem 14, the operator H holds a unique fixed point 1/5e1∈lMλμ. Observe that H is not continuous at 1/5e1∈lMλμ.
IfMt=t2 and μβ=∑y=0∞λyMβy, for every β∈lMλ.
Since for allβ,η∈lMλμ with β0,η0∈−∞,1/5, one has(54)μHβ−Hη=μ16β0−η0,β1−η1,β2−η2,…≤15μ5β6+μ5η6≤15μHβ−β+μHη−η.
Ifβ,η∈lMλμ with β0,η0∈1/5,∞, then for all ϵ>0 one has(55)μHβ−Hη=0≤ϵμHβ−β+μHη−η.
Assumeβ,η∈lMλμ with β0∈−∞,1/5 and η0∈1/5,∞, we get(56)μHβ−Hη=μβ6≤15μ5β6=15μHβ−β≤15μHβ−β+μHη−η.
So, the operatorH is Kannan μ-contraction. As μ satisfies the Fatou property. From Theorem 11, the operator H contains one fixed point 1/5e1∈lMλμ.


## 5. Kannan Nonexpansive Operator

We have presented in this section the uniform convexity of the spacelMλμ, where(57)lMλ=u∈RZ+:ϱωu<∞,for someω>0,and ϱu=∑y=0∞λyMuy, under the Luxemburg norm(58)μu=infω>0:ϱuω≤1.Definition 12.
(1)
The continuous functionMis called strictly convex (SC), if(59)Mv+t2<Mv+Mt2,for allv,t∈0,∞ and v≠t.(2)
[24] The following statements are equivalent:(i)
M is a uniformly convex function on 0,∞.(ii)
For anyϵ>0 and u0>0, there exists a number δ∈0,1 such that for all u, v, and u−v≥ϵmaxu,v≥ϵu0 imply(60)Mu+v2≤1−δ2Mu+Mv,ifu≥u0.(iii)
For anyu0>0 and a∈0,1, there exists a number δ∈0,1 such thatifu≥u0.(3)
[25] A normed space X,μ is said to be strictly convex if for any u,v∈X and b>0 satisfying μu≤b, μv≤b, and μu−v>0 imply μu+v/2<b.(4)
[26] A normed space X,μ is said to be uniformly convex if for any b>0 and ϵ>0, there exists δ>0 such that for all u,v∈X satisfying μu≤b, μv≤b and μu−v≥ϵ imply μu+v/2≤b−δ.Theorem 17.
Iflimn⟶∞μxn=b,limn⟶∞μyn=bandlimn⟶∞μxn+yn/2=bimplylimn⟶∞μxn−yn=0, for allxn,yn⊂lMλandb>0, thenlMλis uniformly convex, whereMis a convex Orlicz function satisfyingΔ2-condition.Proof.
Let the conditions be satisfied andlMλ is not uniformly convex, then there exists ϵ0>0 and xn,yn⊂lMλ such that μxn≤b, μyn≤b, μxn−yn≥ϵ0 we get μxn+yn/2>b−1/n, for some b>0. To prove that limn⟶∞μxn=b, let limn⟶∞μxn=b1<b and limn⟶∞μyn=b. Since M is satisfying Δ2-condition, we have limn⟶∞ϱxn/b1=1 and limn⟶∞ϱxn/b=1. Hence,(61)limn⟶∞ϱxn+yn2b≤b12blimn⟶∞ϱxnb1+12limn⟶∞ϱynb<1.
This is equivalent tolim¯n⟶∞μxn+yn/2<b. This contradicts limn⟶∞μxn+yn/2>b, so limn⟶∞ϱxn=b. Similarly, we can prove that limn⟶∞ϱynb. Also since(62)1<lim¯n⟶∞ϱxn+yn2b≤limn⟶∞ϱxn+yn2b≤lim¯n⟶∞ϱxn+yn2b≤12lim¯n⟶∞ϱxnb+lim¯n⟶∞ϱynb=1.
Thenlimn⟶∞ϱxn+yn/2b=1. This implies limn⟶∞μxn+yn/2=b. But limn⟶∞μxn−yn≥ϵ0>0, this gives a contradiction.Theorem 18.
The spacelMλis uniformly convex, ifMis a uniformly convex Orlicz function satisfyingΔ2-condition.Proof.
Assume the settings are satisfied,limn⟶∞μxn=b, limn⟶∞μyn=b, and limn⟶∞μxn+yn/2=b, we will prove that limn⟶∞μxn−yn=0. For any ϵ∈0,1/2, let us choose u0>0 such that M2u0<ϵ. Since M is uniformly convex, then there exists δ∈0,1 such that u−v≥ϵmaxu,v≥ϵu0 imply(63)Mu+v2≤1−δ2Mu+Mv.
For eachn∈Z+, put(64)Gn=i∈Z+:xnib,ynib<u0,En=i∈Z+:xni−ynib<ϵmaxxnib,ynib<ϵu0,Fn=i∈Z+:xni−ynib≥ϵmaxxnib,ynib≥ϵu0.
Then we deduce(65)∑i∈GnMxni−ynib≤M2u0,and thus(66)∑i∈EnMxni−ynib≤2ϵ∑i∈Z+Mxni+yni2b≤ϵ∑i∈Z+Mxnib+∑i∈Z+Mynib≤2ϵ.
Hence, we get(67)0←μxn+μyn2−μxn+yn2=∑i∈Z+Mxni/b+∑i∈Z+Myni/b2−∑i∈Z+Mxni+yni2b≥∑i∈FnMxni/b+∑i∈FnMyni/b2−∑i∈FnMxni+yni2b≥∑i∈FnMxni/b+∑i∈FnMyni/b2−1−δ2∑i∈FnMxni/b+∑i∈FnMyni/b=δ2∑i∈FnMxni/b+∑i∈FnMyni/b.
Sinceu0 and ϵ are arbitrary, then limn⟶∞ϱxn−yn/2b=0. As M verifies Δ2-condition. Therefore, limn⟶∞μxn−yn=0. From Theorem 17, the proof follows.
Here, we discuss the property (R) and the μ-normal structure property of the space lMλμ.Definition 13.
The spaceYμholds the property (R), if for all decreasing sequenceΦxx∈Z+ofμ-closed andμ-convex nonempty subsets ofYμso thatsupx∈Z+dμβ,Φx<∞, for someβ∈Yμ; hence, we have∩x∈Z+Φx≠∅.Definition 14.
The spaceYμholds theμ-normal structure property if for all nonemptyμ-bounded,μ-convex, andμ-closed subsetΦofYμnot decreased to one point, there existsβ∈Φwith(68)supη∈Φμβ−η<υμΦ≔supμβ−η:β,η∈Φ<∞.Theorem 19.
IfMis a uniformly convex Orlicz function satisfyingΔ2-condition, then(1)
AssumeΦ is a nonempty μ-closed and μ-convex subset of lMλμ. For β∈lMλμ with(69)dμβ,Φ=infμβ−η:η∈Φ<∞.Therefore, we have oneϕ∈Φ with dμβ,Φ=μβ−ϕ.(2)
lMλμ satisfies the property (R).Proof.
For (1), assumeβ∉Φ as Φ is μ-closed. So, one has D≔dμβ,Φ>0. Therefore, there is ηt∈Φ so that limt⟶∞μβ−ηt=D. To prove that ηt is a μ-Cauchy. For any two subsequences ηta and ηtb⊂ηt, we have μηta+ηtb/2−β≥D, as lima⟶∞μβ−ηta=D and limb⟶∞μβ−ηtb=D. Moreover,(70)μηta+ηtb2−β=μηta−β2+ηtb−β2<12D+D=D.
Therefore, we havelima,b⟶∞μηta+ηtb/2−β=D. Since the space lMλμ is uniformly convex, we get(71)lima,b⟶∞μηta−β−ηtb−β=lima,b⟶∞μηta−ηtb=0.
Thus,ηt is a μ-Cauchy in Φ. Since Φ is closed and the space lMλμ is complete, then there exists ϕ∈Φ with μβ−ϕ=dμβ,Φ. Since the space lMλμ is uniformly convex, then it is (SC), which implies the uniqueness of ϕ. To show (2), for some t0∈Z+, suppose β∉Φt0. Since dμβ,Φtt∈Z+∈l∞ is increasing. Set limt⟶∞dμβ,Φt=D, when D>0. Otherwise, β∈Φt, for each t∈Z+. From (1), we have a unique ηt∈Φt with dμβ,Φt=μβ−ηt, for all t∈Z+. A consistent proof will show that ηt/2μ-converges to some η∈lMλμ. Since Φt are μ-convex, decreasing and μ-closed, we get 2η∈∩t∈Z+Φt.Theorem 20.
IfMis a uniformly convex Orlicz function satisfyingΔ2-condition, thenlMλμhas theμ-normal structure property.Proof.
Let the conditions are satisfied. Theorem18 gives that lMλμ is uniformly convex. Assume Φ is a μ-bounded, μ-convex, and μ-closed subset of lMλμ not decreased to one point. Hence, υμΦ>0. Set D=υμΦ. Let β,η∈Φ with β≠η. Hence, μβ−η/2>0. For every ϕ∈Φ, one has μβ−ϕ≤D and μη−ϕ≤D. As Φ is μ-convex, then β+η/2∈Φ. Hence,(72)μβ+η2−ϕ=μβ−ϕ+η−ϕ2<D,for every ϕ∈Φ. So(73)supϕ∈Φμβ+η2−ϕ<D=υμΦ.Lemma 3.
Let the spacelMλμ verify the (R) property and the μ-quasi-normal property. Assume Φ is a nonempty μ-bounded, μ-convex, and μ-closed subset of lMλμ. Suppose H:Φ⟶Φ is a Kannan μ-nonexpansive mapping. For x>0. If Wx=β∈Φ:μβ−Hβ≤x≠∅. Set(74)Φx=∩Bμt,v:HWx⊂Bμt,v∩Φ.ThenΦx is a nonempty, μ-convex, μ-closed subset of Φ with HΦx⊂Φx⊂WxandυμΦx≤x.Proof.
AsHWx⊂Φx, this gives Φx≠∅. Since the μ-balls are μ-convex, and μ-closed, then Φx is a μ-closed and μ-convex subset of Φ. To prove that Φx⊂Wx. Assume β∈Φx. If μβ−Hβ=0, we have β∈Wx. Otherwise, suppose μβ−Hβ>0. Set(75)t=supμHζ−Hβ:ζ∈Wx.
From the definition oft, then HWx⊂BμHβ,t. Hence, Φx⊂BμHβ,t, which implies μβ−Hβ≤t. Assume d>0. Hence, there is ζ∈Wx so that t−d≤μHζ−Hβ. Then(76)μβ−Hβ−d≤t−d≤μHζ−Hβ≤12μβ−Hβ+μζ−Hζ≤12μβ−Hβ+x.
Sinced is arbitrarily positive, we have μβ−Hβ≤x, then we have β∈Wx. For HWx⊂Φx, we get HΦx⊂HWx⊂Φx, this indicates Φx is H-invariant. Consequent to prove that υμΦx≤x. As(77)μHβ−Hη≤12μβ−Hβ+μη−Hη,
For everyβ,η∈Wx. Let β∈Wx. So HWx⊂BμHβ,x. From the definition of Φx, one has Φx⊂BμHβ,x. Hence, Hβ∈∩η∈ΦxBμη,x. Therefore, we have μη−ζ≤x, for every η,ζ∈Φx, which implies υμΦx≤x. This finishes the proof.
In this part, we give enough settings onlMλμ so that the Kannan μ− nonexpansive mapping defined on it contains a fixed point.Theorem 21.
LetlMλμhold theμ-quasinormal property and the (R) property. AssumeΦis a nonempty,μ-convex,μ-closed, andμ-bounded subset oflMλμ. IfH:Φ⟶Φis a Kannanμ-nonexpansive mapping, thenHhas a fixed point.Proof.
Letxt=x0+1/t, for all t≥1, where x0=infμβ−Hβ:β∈Φ. We have for each t≥1 that Wxt=β∈Φ:μβ−Hβ≤xt≠∅. Suppose Φxt explained as in Lemma 3. Clearly, Φxt is a decreasing sequence of nonempty μ-bounded, μ-closed, and μ-convex subsets of Φ. The property (R) gives that Φ∞=∩t≥1Φxt≠∅. Let β∈Φ∞, we have μβ−Hβ≤xt, for every t≥1. If t⟶∞, one has μβ−Hβ≤x0, which implies μβ−Hβ=x0. Hence, Wx0≠∅. So x0=0. Otherwise, x0>0 which investigates that H has no fixed point. Assume Φx0 as defined in Lemma 3. Since H has no fixed point and Φx0 is H-invariant, hence Φx0 holds more than one point, which gives, υμΦx0>0. By the μ-quasinormal property, one has β∈Φx0 with(78)μβ−η<υμΦx0≤x0,for every η∈Φx0. By Lemma 3, we have Φx0⊂Wx0. By definition of Φx0, then Hβ∈Wx0⊂Φx0. Obviously, one has(79)μβ−Hβ<υμΦx0≤x0,which contradicts the definition of x0. So x0=0 this implies that any point in Wx0 is a fixed point of H, i.e., H has a fixed point in Φ.
According to Theorem19, Theorem 20, and Theorem 21, we obtain the next corollary:Corollary 2.
IfMis a uniformly convex Orlicz function satisfyingΔ2-condition. AssumeΦis a nonempty,μ-convex,μ-closed, andμ-bounded subset oflMλμ. SupposeH:Φ⟶Φis a Kannanμ-nonexpansive operator. ThenHholds a fixed point.Example 7.
LetH:Φ⟶Φ with Hβ=β/4,μβ∈0,1,β/5,μβ∈1,∞,
whereΦ=β∈lMλμ:β0=β1=0, where ϱβ=∑x=0∞λxβx2, for every β∈lMλμ. As Example 5, the operator H is Kannan μ-contraction mapping. So it is Kannan μ-nonexpansive operator. Clearly, Φ is a nonempty, μ-convex, μ-closed and μ-bounded subset of lMλμ. By Corollary 2, the operator H has a fixed point in Φ.


## 6. KannanΥ− Contraction Mapping on SlMλμ

For any two Banach spacesX and Y, we examine in this section the existence of a fixed point of Kannan Υ− contraction mapping on SlMλμ, where ΥQ=M−1∑y=0∞λyMsyQ, for all Q∈SlMλμX,Y.Theorem 22.
IfM is a convex Orlicz function satisfying Δ2-condition and lnMex is convex, then SlMλμ,Υ is a prequasi Banach operator ideal, where ΥQ=μsyQy=0∞.Proof.
As Theorem5, the space lMλμ is a premodular (sss). Therefore, from Theorem 3, one has ΥQ=μsyQy=0∞ is a prequasi norm on SlMλμ. Suppose Qr∈SlMλμX,Y is a Cauchy sequence. As BX,Y⊇SlMλμX,Y, one obtains(80)ΥQr−Qt=μsyQr−Qty=0∞≥μs0Qr−Qt,0,0,0,…,=M−1λ0Qr−Qt.
HenceQrr∈Z+ is a Cauchy sequence in BX,Y. Since BX,Y is a Banach space, so there is Q∈BX,Y with limr⟶∞Qr−Q=0. Since syQry=0∞∈lMλμ, for every r∈Z+. We have(81)ΥQ=μsyQy=0∞=μsyQ−Qr+Qry=0∞≤μsy/2Q−Qry=0∞+μsy/2Qry=0∞≤μQr−Qy=0∞+2μsyQry=0∞<ϵ.
Therefore,syQy=0∞∈lMλμ, this implies Q∈SlMλμX,Y.Theorem 23.
IfMis a convex Orlicz function satisfyingΔ2-condition andlnMexis convex, thenSlMλμ,Υis a prequasi closed operator ideal, whereΥQ=μsyQy=0∞.Proof.
As Theorem5, the space lMλμ is a premodular (sss). Therefore, from Theorem 3, one has ΥQ=μsyQy=0∞ is a prequasi norm on SlMλμ. Assume Qr∈SlMλμX,Y, for every r∈Z+ and limr⟶∞ΥQr−Q=0. Hence, there is ς>0 and since BX,Y⊇SlMλμX,Y, we get(82)ΥQr−Q=μsyQr−Qy=0∞≥μs0Qr−Q,0,0,0,…=M−1λ0Qr−Q.
HenceQrr∈Z+ is convergent in BX,Y. i.e., limr⟶∞Qr−Q=0 and while syQry=0∞∈lMλμ, for all r∈Z+ and lMλμ is a premodular (sss). Hence, we have(83)ΥQ=μsyQy=0∞=μsyQ−Qr+Qry=0∞≤μsy/2Q−Qry=0∞+μsy/2Qry=0∞≤μQr−Qy=0∞+2μsyQry=0∞<ϵ,we have syQy=0∞∈lMλμ, then Q∈SlMλμX,Y.Definition 15.
A prequasi normΥon the idealSlMλμ, whereΥQ=μsyQy=0∞, satisfies the Fatou property if for any sequenceQyy∈Z+⊆SlMλΥX,Ywithlimy⟶∞ΥQy−Q=0and anyV∈SlMλμX,Y, thenΥV−Q≤supyinfj≥yμV−Qj.Theorem 25.
The prequasinormΥQ=M−1∑y=0∞λyMsyQ, for all Q∈SlMλμX,Y does not satisfy the Fatou property, ifMis a convex Orlicz function satisfyingΔ2-condition andlnMexis convex.Proof.
Assume the settings are satisfied andQtt∈Z+⊆SlMλμX,Y with limt⟶∞ΥQt−Q=0. Since the space SlMλμ is a prequasi closed ideal, then, Q∈SlMλμX,Y. Hence, for any V∈SlMλμX,Y, we have(84)ΥV−Q=M−1∑y=0∞λyMsyV−Q≤M−1∑y=0∞λyMsy/2V−Qj+M−1∑y=0∞λyMsy/2Qj−Q≤2suptinfj≥tM−1∑y=0∞λyMsyV−Qj.
Hence,Υ does not satisfy the Fatou property.Definition 16.
An operatorP:SlMλμX,Y⟶SlMλμX,Yis called a KannanΥ-Lipschitzian, if there existsν≥0, so that for everyQ,T∈SlMλμX,Y, we have(85)ΥPQ−PT≤νΥPQ−Q+ΥPT−T.(1)
Ifν∈0,1/2, the operator P is said to be Kannan Υ-contraction.(2)
Ifν=1/2, the operator P is said to be Kannan Υ-nonexpansive.Definition 17.
An operatorP:SlMλμX,Y⟶SlMλμX,Y is said to be Υ-sequentially continuous at V, if and only if, when limt⟶∞ΥQt−V=0, then limt⟶∞ΥPQt−PV=0.Theorem 26.
IfMis a convex Orlicz function satisfyingΔ2-condition andlnMexis conve, andP:SlMλμX,Y⟶SlMλμX,Y. The pointT∈SlMλΥX,Yis the unique fixed point ofP, when the following conditions are satisfied:(i)
P is Kannan Υ-contraction mapping,(ii)
P is Υ-sequentially continuous at a point T∈SlMλμX,Y,(iii)
There existsV∈SlMλμX,Y so that the sequence of iterates PtV has a subsequence PtjV converging to T.Proof.
Let the conditions be verified. IfT is not a fixed point of P, then PT≠T. From the conditions (ii) and (iii), we have(86)limtj⟶∞ΥPtjV−T=0,limtj⟶∞ΥPtj+1V−PT=0.
SinceP is Kannan Υ-contraction mapping, one can see(87)0<ΥPT−T=ΥPT−Ptj+1V+PtjV−T+Ptj+1V−PtjV≤2ΥPtj+1V−PT+4ΥPtjV−T+4νν1−νtj−1ΥPV−V.
Sincetj⟶∞, we have a contradiction. Hence, T is a fixed point of P. To prove the uniqueness of the fixed point T. Let we have two different fixed points T,U∈SlMλμX,Y of P. Therefore, one has(88)ΥT−U≤ΥPT−PU≤νΥPT−T+ΥPU−U=0.
Hence,T=U.Example 8.
AssumeP:SlMλμX,Y⟶SlMλΥX,Y, whereΥQ=∑y=0∞λysyQ, for everyQ∈SlMλμX,Yand(89)PQ=Q26,ΥQ∈0,1,Q37,ΥQ∈1,∞.
As for everyQ1,Q2∈SlMλμ with ΥQ1,ΥQ2∈0,1, one has(90)ΥPQ1−PQ2=ΥQ126−Q226≤25Υ25Q126+Υ25Q226=25ΥPQ1−Q1+ΥPQ2−Q2.
For eachQ1,Q2∈SlMλμ with ΥQ1,ΥQ2∈1,∞, we get(91)ΥPQ1−PQ2=ΥQ137−Q237≤13Υ36Q137+Υ36Q237=13ΥPQ1−Q1+ΥPQ2−Q2.
For eachQ1,Q2∈SlMλμ with ΥQ1∈0,1 and ΥQ2∈1,∞, one can see(92)ΥPQ1−PQ2=ΥQ126−Q237≤25Υ25Q126+13Υ36Q237≤25Υ25Q126+Υ36Q237=25ΥPQ1−Q1+ΥPQ2−Q2.
So, the operatorQ is Kannan Υ-contraction and PtQ=Q/26t,ΥQ∈0,1,Q/37t,ΥQ∈1,∞.
Clearly,P is Υ-sequentially continuous at the zero operator Θ∈SlMλμ and PtQ has a subsequence PtjQ converging to Θ. From Theorem 27, the zero operator Θ∈SlMλμ is the unique fixed point of P. Suppose with limt⟶∞ΥQt−Q0=0−b±b2−4ac/2a, where Q0∈SlMλμ with ΥQ0=1. From the continuously of the prequasi norm Υ, one has(93)limt⟶∞ΥPQt−PQ0=limt⟶∞ΥQt26−Q037=Υ11Q0962>0.
SoP is not Υ-sequentially continuous at Q0. This implies the operator P is not continuous at Q0.Example 9.
SupposeP:SlMλμX,Y⟶SlMλΥX,Y, whereΥQ=∑y=0∞λysyQ2, for everyQ∈SlMλΥX,Yand(94)PQ=Q5,ΥQ∈0,1,Q6,ΥQ∈1,∞.
As for eachQ1,Q2∈SlMλμ with ΥQ1,ΥQ2∈0,1, one can see(95)ΥPQ1−PQ2=ΥQ15−Q25≤24Υ4Q15+Υ4Q25=24ΥPQ1−Q1+ΥPQ2−Q2.
For everyQ1,Q2∈SlMλμ with ΥQ1,ΥQ2∈1,∞, this implies(96)ΥPQ1−PQ2=ΥQ16−Q26≤25Υ5Q16+Υ5Q26=25ΥPQ1−Q1+ΥPQ2−Q2.
For eachQ1,Q2∈SlMλμ with ΥQ1∈0,1 and ΥQ2∈1,∞, one obtains(97)ΥPQ1−PQ2=ΥQ15−Q26≤24Υ4Q15+25Υ5Q26≤24Υ4Q15+Υ5Q26=24ΥPQ1−Q1+ΥPQ2−Q2.
So, the operatorQ is Kannan Υ-contraction and. PtQ=Q/5t,ΥQ∈0,1,Q/6t,ΥQ∈1,∞.
Evidently,P is Υ-sequentially continuous at the zero operator Θ∈SlMλμ and PtQ has a subsequence PtjQ converging to Θ. From Theorem 27, the zero operator Θ∈SlMλμ is the unique fixed point of P. Suppose Qt⊆SlMλμ with limt⟶∞ΥQt−Q0=0, where Q0∈SlMλμ with ΥQ0=1. From the continuously of the prequasi norm Υ, one has(98)limt⟶∞ΥPQt−PQ0=limt⟶∞ΥQt5−Q06=ΥQ030>0.
SoP is not Υ-sequentially continuous at Q0. Hence, the operator P is not continuous at Q0.


## 7. Applications on Summable Equations

We investigate here a solution to (101), which studied by many authors (see [27–29]), in lMλμ.(99)βx=rx+∑y=0∞Dx,yhy,βy.SupposeH:lMλμ⟶lMλμ constructed by(100)Hβxx∈Z+=rx+∑y=0∞Dx,yhy,βyx∈Z+.Theorem 27.
IfMis a convex Orlicz function satisfyingΔ2-condition andlnMexis convex,D:Z+2⟶R,h:Z+×R⟶R,r:Z+⟶R, and for allx∈Z+, there existsν∈0,1/2, with(101)M∑y∈Z+Dx,yhy,βy−hy,ηy≤MνMrx−βx+∑y=0∞Dx,yfy,βy+Mrx−ηx+∑y=0∞Dx,yfy,ηy,then equation (101) hold a solution in lMλμ, where μβ=M−1∑x=0∞Mβx, for every β∈lMλ.Proof.
Suppose the setups are verified. We have(102)μHβ−Hη=M−1∑x∈Z+λxMHβx−Hηx=M−1∑x∈Z+λxM∑m∈Z+Dx,yhy,βy−hy,ηy≤νM−1∑x∈Z+λxMrx−βx+∑m=0∞Dx,yhy,βy+νM−1∑x∈Z+λxMrx−ηx+∑m=0∞Dx,yhy,ηy=νμHβ−β+μHη−η.
In view of Theorem11, there exists a unique solution of equation (101) in lMλμ.Example 10.
For the spacelMλμ, where μβ=∑x∈Z+βx44, for all β∈lMλ. Assume the summable equations are defined as(103)βx=e−3x+6+∑y=0∞−1xβxx2+y2+1vsiny,where v>2 and let H:lMλμ⟶lMλμ is defined by(104)Hβxx∈Z+=e−3x+6+∑y=0∞−1xβxx2+y2+1vsinyx∈Z+.
We have(105)∑y=0∞−1xβxx2+y2+1vsiny−siny4≤181e−3x+6−βx+∑y=0∞−1xβxx2+y2+1vsiny4+e−3x+6−ηx+∑y=0∞−1xηxx2+y2+1vsiny4.
By Theorem27, the summable equations (105) have one solution in lMλμ.Theorem 30.
IfMis a concave Orlicz function,D:Z+2⟶R,h:Z+×R⟶R,r:Z+⟶R, and for everyx∈Z+, there existsν∈0,1/2, with(106)M∑y∈Z+Dx,yhy,βy−hy,ηy≤νMrx−βx+∑y=0∞Dx,yfy,βy+Mrx−ηx+∑y=0∞Dx,yfy,ηy,then equation (101) contains one solution in lMλμ, where μβ=∑x=0∞Mβx, for each β∈lMλ.Proof.
Suppose the setups are verified. One has(107)μHβ−Hη=∑x∈Z+λxMHβx−Hηx=∑x∈Z+λxM∑y∈Z+Dx,yhy,βy−hy,ηy≤ν∑x∈Z+λxMrx−βx+∑y=0∞Dx,yhy,βy+∑x∈Z+λxMrx−ηx+∑y=0∞Dx,yhy,ηy=νμHβ−β+μHη−η.
In view of Theorem12, there exists a unique solution of equation (101) in lMλμ.Example 11.
For the spacelMλμ, whereμβ=∑x∈Z+βx3, for everyβ∈lMλ. Assume the summable equations(108)βx=e−3x+6+∑y=0∞−1x+yeβxx2+y2+1v,where v>2 and let H:lMλμ⟶lMλμ is defined by(109)Hβxx∈Z+=e−3x+6+∑y=0∞−1x+yeβxx2+y2+1vx∈Z+.
It is easy to see that(110)∑y=0∞−1xeβxx2+y2+1v−1y−−1y1/3≤13e−3x+6−βx+∑y=0∞−1x+yeβxx2+y2+1v1/3+e−3x+6−ηx+∑y=0∞−1x+yeβxx2+y2+1v1/3.
By Theorem30, the summable equation (105) has an unique solution in lMλμ.Example 12.
Given the sequence spacelMλμ, whereμβ=∑x∈Z+βx2, for allβ∈lMλ. Consider the summable equations (110), withx≥2andv>2and letH:Φ⟶Φ, whereΦ=β∈lMλμ:β0=β1=0, defined by(111)Hβxx≥2=e−3x+6+∑y=0∞−1x+yeβxx2+y2+1vx≥2.
Obviously,Φ is a nonempty, μ-convex, μ-closed, and μ-bounded subset of lMλμ. It is easy to see that(112)∑y=0∞−1xeβxx2+y2+1v−1y−−1y2≤19e−3x+6−βx+∑y=0∞−1x+yeβxx2+y2+1v2+e−3x+6−ηx+∑y=0∞−1x+yeηxx2+y2+1v2.
By Theorem27 and Corollary 2, the summable equation (110) have a solution in Φ.


## 8. Conclusion

We explored the presence of a fixed point for both Kannan contraction and nonexpansive mappings working on various premodular, which is a generalization of modular, defined by weighted Orlicz sequence space and its prequasi operator ideal. Numerous numerical experiments and practical applications are used to substantiate our findings.



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