# Some Fixed Point Results in Premodular Special Space of Sequences and Their Associated Pre-Quasi-Operator Ideal

Authors: Awad A. Bakery; Elsayed A. E. Mohamed; O. M. Kalthum S. K. Mohamed
Journal: Journal of Function Spaces
 (2021)

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## Abstract

A weighted Nakano sequence space and thes-numbers it contains are the subject of this article, which explains the concept of the pre-quasi-norm and its operator ideal. We show that both Kannan contraction and nonexpansive mappings acting on these spaces have a fixed point. A slew of numerical experiments back up our findings. The presence of summable equations’ solutions is shown to be useful in a number of ways. Weight and power of the weighted Nakano sequence space are used to define the parameters for this technique, resulting in customizable solutions.


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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 by ℝℕ0, ℓ∞, ℓr, and c0, respectively, where ℕ0 is the set of nonnegative integers. ey=0,0,⋯,1,0,0,⋯, while 1 lies in the yth place, with y∈ℕ0.Definition 1 (see [1]).
A functions:LE,H⟶0,∞ℕ0, where LE,H is the space of all bounded linear operators from a Banach space E into a Banach space H and if E=H, we write LE, which transforms every mapping J∈LE,H to syJy=0∞ is said to be s-number, if it is satisfying the following conditions:
(a)
J=s0J≥s1J≥s2J≥⋯≥0, for every J∈LE,H(b)
sx+y−1J1+J2≤sxJ1+syJ2, for every J1,J2∈LE,H and x, y∈ℕ0(c)
Ideal property:syWHJ≤WsyHJ, for every J∈LE0,E, H∈LE,H, and W∈LH,H0, where E0 and H0 are any two Banach spaces(d)
IfJ∈LE,H and δ∈ℝ, we have syδJ=∣δ∣syJ(e)
Rank property: ifrankJ≤y, then syJ=0, for all J∈LE,H(f)
Norming property:sk≥yIy=0 or sk<yIy=1Theyth approximation number, αyJ, is defined as
(1)αyJ=infJ−K:K∈LE,HandrankK≤y.Notations 2 (see [2]).
SC≔SCE,H, where SCE,H≔J∈LE,H:syJy=0∞∈C. Also, SCappE,H,whereSCappE,H≔J∈LE,H:αyJy=0∞∈C.Fixed point theory, Banach space geometry, normal series theory, ideal transformations, and approximation theory are all examples of ideal operator theorems and summability. The concept of a pre-quasi-operator ideal is introduced and studied by Faried and Bakery [2]. Bakery and Abou Elmatty investigated some topological and geometric structures of ℓγ,λ in [3]. They proved that the space Sℓγ,λapp is a small pre-quasi-operator ideal and gave a strictly inclusion relation for different weights and powers. Several mathematicians were able to investigate many extensions for contraction mappings defined on the space or on the space itself thanks to the Banach fixed point theorem [4]. Kannan [5] investigated an example of a class of operators that perform the same fixed point actions as contractions but are not continuous. Ghoncheh [6] demonstrated the existence of a Kannan mapping fixed point in complete modular spaces with Fatou property (also see [7–10]). Bakery and Mohamed [11] examined the sufficient requirements on ℓλz, variable exponent in 0,1 under definite pre-quasi-norm so that there is a fixed point of Kannan pre-quasi-norm contraction mappings on this space. For the construction of pre-quasi-Banach and closed spaces, we use a weighted Nakano sequence space, ℓγ,λP, with various pre-quasi-norms in this study. Weighted Nakano sequence space’s pre-quasi-normal structural features, including the fixed point idea of Kannan pre-quasi-norm contraction and the Kannan pre-quasi-norm nonexpansive mapping in weighted Nakano sequence space, are improved. The existence of a fixed point for the Kannan pre-quasi-norm contraction mapping has been demonstrated using weighted Nakano sequence space and s-numbers. Our talk concluded with various instances of how the information gathered could be put to good use in resolving a problem.


## 2. Preliminaries and Definitions

We indicate the space of all mappingsP:A⟶0,∞, by 0,∞A.Definition 3 (see [12]).
IfA is a vector space and θ=0,0,0,, a mapping P∈0,∞A is said to be modular:
(a)
Ifg∈A, g=θ⟺Pg=0 with Pg≥0(b)
Pδg=Pg holds, for each g∈A and ∣δ∣=1(c)
The inequalityPδg+1−δf≤Pg+Pf verifies, for every g,f∈A and δ∈0,1.Definition 4 (see [2]).
If the following conditions hold:(2)eyy∈ℕ0⊆A.(1)
A is solid. This means if g=gy∈ℝℕ0, f=fy∈A, and ∣gy∣≤∣fy∣, for every y∈ℕ0, then g∈A(2)
gy/2y=0∞∈A, where y/2 denotes the integral part of y/2, when gyy=0∞∈A
Then,A is said to be a special space of sequences (sss).Definition 5 (see [2]).
If we haveP∈0,∞A with the following:
(i)
ifg∈A, g=θ⟺Pg=0(ii)
supposeg∈A and δ∈ℝ, then there is B≥1 for which Pδg≤B∣δ∣Pg(iii)
the inequality,Pg+f≤JPg+Pf, for each g,f∈A, verifies for some J≥1(iv)
ifz∈ℕ0 and ∣gz∣≤∣fz∣, then Pgz≤Pfz(v)
the inequality,Pgz≤Pgz/2≤J0Pgz, satisfies, for some J0≥1(vi)
assumeF is the space of finite sequences, one has F¯=AP(vii)
we haveσ>0 so that Pη,0,0,0,⋯≥σ∣η∣P1,0,0,0,⋯, for every η∈ℝ
Then,AP⊆A is said to be a premodular sss.Example 1.
Since for allv,t∈ℓa+1/2a+5a=0∞, we have
(3)Pv+t2=∑a∈ℕ0va+ta2a+1/2a+55≤8Pv+Pt.
Hence,Pv=∑a∈ℕ0vaa+1/2a+55 is a premodular (not a modular) on ℓa+1/2a+5a=0∞.Definition 6 (see [11]).
AssumeA is a sss. The function P∈0,∞A is called a pre-quasi-norm on A, if it satisfies the conditions (i), (ii), and (iii) of Definition 5.Theorem 7 (see [11].
SupposeA is a premodular sss; then, A is a pre-quasi-normed sss.Theorem 8 (see [11]).
Quasinormed sss is contained in pre-quasi-normed sss.Definition 9 (see [13]).
(a)
The pre-quasi-normP on ℓγ,Δ is called P-convex, if Pεg+1−εf≤εPg+1−εPf, for each ε∈0,1 and g,f∈ℓγ,Δ(b)
gzz∈ℕ0⊆ℓγ,ΔP is P-convergent to g∈ℓγ,ΔP, if and only if limz⟶∞Pgz−g=0. If the P-limit exists, then it is unique(c)
gzz∈ℕ0⊆ℓγ,ΔP is P-Cauchy, if limz,y⟶∞Pgz−gy=0(d)
Δ⊂ℓγ,ΔP is P-closed, if for every P-converging gzz∈ℕ0⊂Δ to g, then g∈Δ(e)
Δ⊂ℓγ,λP is P-bounded, if δPΔ=supPg−f:g,f∈Δ<∞(f)
TheP-ball of radius l≥0 and center g, for every g∈ℓγ,λP, is defined as(4)BPg,l=f∈ℓγ,λP:Pg−f≤l(g)
A pre-quasi-normP on ℓγ,λ satisfies the Fatou property, when for every sequence fz⊆ℓγ,λP with limz⟶∞Pfz−f=0 and every g∈ℓγ,λP then Pg−f≤supyinfz≥yPg−fz
Recall that the Fatou property implies theP-closed of the P-balls.Definition 10 (see [14]).
LetL be the class of each bounded linear operators between any two Banach spaces. A subclass U of L is known as an operator ideal, if all element UE,H=U∩LE,H fulfills the following conditions:
(i)
IΓ∈U, where Γ indicates Banach space of one dimension(ii)
The spaceUE,H is linear over ℝ(iii)
IfG1∈LE0,E, G2∈UE,H, and G3∈LH,H0, then G3G2G1∈UE0,H0, where E0 and H0 are normed spacesPre-quasi-operator ideals are more general than quasioperator ideals.Definition 11 (see [2]).
A mappingℙ∈0,∞U is called a pre-quasi-norm when
(a)
letJ∈UE,H, ℙJ≥0, and ℙJ=0⇔J=0(b)
we haveD≥1 with ℙγJ≤D∣γ∣ℙJ, when J∈UE,H and γ∈ℝ(c)
we haveJ≥1 so that ℙJ1+J2≤JℙJ1+ℙJ2, for all J1,J2∈UE,H(d)
we haveτ≥1 such that J∈LE0,E, J1∈UE,H and J2∈LH,H0 then ℙJ2J1J≤τ∥J2∥ℙJ1∥J∥Theorem 12 (see [11]).
SupposeAP is a premodular sss, then ℙJ=PszJz=0∞ is a pre-quasi-norm on SAP.Theorem 13 (see [2]).
Quasi-normed ideal is contained in pre-quasi-normed ideal.Lemma 14 (see [15]).
Ifλ≥2 and for all g,f∈ℝ, then
(5)g+f2λ+g−f2λ≤12gλ+fλ.Lemma 15 (see [16]).
Let1<λ≤2 and g,f∈ℝ with ∣g∣+∣f∣≠0; then,
(6)g+f2λ+λλ−12g−f∣g∣+∣f∣2−λg−f2λ≤12gλ+fλ.Lemma 16 (see [17]).
Supposeλz>0 and gz,fz∈ℝ, for all z∈ℕ0; then,
(7)gz+fzλz≤2K−1gzλz+fzλz,where K=max1,supzλz.


## 3. Main Results

### 3.1. The Sequence Spaceℓγ,λ

Assumeλ=λz∈ℝ+ℕ0 and γ=γz∈ℝ+ℕ0, where ℝ+ denotes the set of positive reals. In [3], the weighted Nakano sequence space was defined as
(8)ℓγ,λ=v=va∈ℝℕ0:Pμv<∞,forsomeμ>0,while Pv=∑a=0∞γavaλa.Theorem 17.
Ifλa∈ℓ∞, then
(9)ℓγ,λ=v=va∈ℝℕ0:Pμv<∞,foranyμ>0.Proof.
(10)ℓγ,λ=v=va∈ℝℕ0:Pμv<∞,forsomeμ>0=v=va∈ℝℕ0:infaμλa∑a=0∞γavaλa≤∑a=0∞γaμvaλa<∞,forsomeμ>0=v=va∈ℝℕ0:∑a=0∞γavaλa<∞=v=va∈ℝℕ0:Pμv<∞,foranyμ>0.(1)
Ifγa=1, with a∈ℕ0, then ℓγ,λ=ℓλa defined and considered in [18, 19](2)
Ifγa=1/λa, with a∈ℕ0, then ℓγ,λ=ℓλ. examined by many authors [16, 20, 21]Theorem 18.
The spaceℓγ,λ,∥.∥ is a Banach space, where ∥v∥=infκ>0:Pv/κ≤1.Proof.
Since we have the following:(i)
∥v∥≥0, for each v∈ℓγ,λ and ∥v∥=0, if and only if, v=θ(ii)
supposeν∈ℝ, v∈ℓγ,λ without loss of generality, let ν≠0 then(11)∥νv∥=infζ>0:Pνvζ≤1=infνμ>0:Pvμ≤1=νv(iii)
assumev,t∈ℓγ,λ, then there are ζ1>0 and ζ2>0 be such that Pv/ζ1≤1andPt/ζ2≤1. Let ζ=ζ1+ζ2; since P is nondecreasing and convex, one has(12)Pv+tζ=Pv+tζ1+ζ2≤ζ1ζ1+ζ2Pvζ1+ζ2ζ1+ζ2Ptζ2≤1
As theζ’s are nonnegative, one can see
(13)∥v+t∥=infζ>0:Pv+tζ≤1≤infζ1>0:Pvζ1≤1+infζ2>0:Ptζ2≤1=∥v∥+∥t∥.
Then, the spaceℓγ,λ,∥.∥ is a normed space. Next, let gx=gzxz=0∞ be a Cauchy sequence in ℓγ,λ. Therefore, for every ε∈0,1, we have x0∈ℕ0 such that for all x,y≥x0, we obtain
(14)∥gx−gy∥=infζ>0:Pgx−gyζ≤1<ε.
So, forx,y≥x0 and z∈ℕ0, one can see ∣gzx−gzy∣<ε. Hence, gzy is a Cauchy sequence in ℝ, for fixed z∈ℕ0. This implies limy⟶∞gzy=gz0, for fixed z∈ℕ0. Hence, ∥gx−g0∥<ε, for every x≥x0. Since ∥g0∥≤∥gx−g0∥+∥gx∥<∞, therefore, g0∈ℓγ,λ. This implies that ℓγ,λ,∥.∥ is a Banach space.


## 3.1. The Sequence Spaceℓγ,λ

Assumeλ=λz∈ℝ+ℕ0 and γ=γz∈ℝ+ℕ0, where ℝ+ denotes the set of positive reals. In [3], the weighted Nakano sequence space was defined as
(8)ℓγ,λ=v=va∈ℝℕ0:Pμv<∞,forsomeμ>0,while Pv=∑a=0∞γavaλa.Theorem 17.
Ifλa∈ℓ∞, then
(9)ℓγ,λ=v=va∈ℝℕ0:Pμv<∞,foranyμ>0.Proof.
(10)ℓγ,λ=v=va∈ℝℕ0:Pμv<∞,forsomeμ>0=v=va∈ℝℕ0:infaμλa∑a=0∞γavaλa≤∑a=0∞γaμvaλa<∞,forsomeμ>0=v=va∈ℝℕ0:∑a=0∞γavaλa<∞=v=va∈ℝℕ0:Pμv<∞,foranyμ>0.(1)
Ifγa=1, with a∈ℕ0, then ℓγ,λ=ℓλa defined and considered in [18, 19](2)
Ifγa=1/λa, with a∈ℕ0, then ℓγ,λ=ℓλ. examined by many authors [16, 20, 21]Theorem 18.
The spaceℓγ,λ,∥.∥ is a Banach space, where ∥v∥=infκ>0:Pv/κ≤1.Proof.
Since we have the following:(i)
∥v∥≥0, for each v∈ℓγ,λ and ∥v∥=0, if and only if, v=θ(ii)
supposeν∈ℝ, v∈ℓγ,λ without loss of generality, let ν≠0 then(11)∥νv∥=infζ>0:Pνvζ≤1=infνμ>0:Pvμ≤1=νv(iii)
assumev,t∈ℓγ,λ, then there are ζ1>0 and ζ2>0 be such that Pv/ζ1≤1andPt/ζ2≤1. Let ζ=ζ1+ζ2; since P is nondecreasing and convex, one has(12)Pv+tζ=Pv+tζ1+ζ2≤ζ1ζ1+ζ2Pvζ1+ζ2ζ1+ζ2Ptζ2≤1
As theζ’s are nonnegative, one can see
(13)∥v+t∥=infζ>0:Pv+tζ≤1≤infζ1>0:Pvζ1≤1+infζ2>0:Ptζ2≤1=∥v∥+∥t∥.
Then, the spaceℓγ,λ,∥.∥ is a normed space. Next, let gx=gzxz=0∞ be a Cauchy sequence in ℓγ,λ. Therefore, for every ε∈0,1, we have x0∈ℕ0 such that for all x,y≥x0, we obtain
(14)∥gx−gy∥=infζ>0:Pgx−gyζ≤1<ε.
So, forx,y≥x0 and z∈ℕ0, one can see ∣gzx−gzy∣<ε. Hence, gzy is a Cauchy sequence in ℝ, for fixed z∈ℕ0. This implies limy⟶∞gzy=gz0, for fixed z∈ℕ0. Hence, ∥gx−g0∥<ε, for every x≥x0. Since ∥g0∥≤∥gx−g0∥+∥gx∥<∞, therefore, g0∈ℓγ,λ. This implies that ℓγ,λ,∥.∥ is a Banach space.


## 4. Pre-Quasi-Normed Sequence Space

To create pre-quasi-Banach and closed sequence space, we study the conditions onℓγ,λP, where Pg=∑z=0∞γzgzλz1/K, for each g∈ℓγ,λ. The Fatou property of ℓγ,λP has been investigated for various P.Theorem 19.
(a1)
Letλz∈ℝ+ℕ0∩ℓ∞ be an increase.(a2)
Eitherγz is a monotonic decrease or monotonic increase so that there is E≥1, where γ2a+1≤Eγz.
Then,ℓγ,λP is a premodular sss.Proof.
(i)
Evidently,Pg≥0 and Pg=0⟺g=θ.
(1-i) and (iii). Letg,f∈ℓγ,λ. As λz∈ℝ+ℕ0∩ℓ∞, one gets
(15)Pg+f=∑z=0∞γzgz+fzλz1/K≤∑z=0∞γzgzλz1/K+∑z=0∞γzfzλz1/K=Pg+Pf<∞.
Hence,g+f∈ℓγ,λ.
(1-ii) and (ii). Letη∈ℝ and g∈ℓγ,λ. Since λz∈ℝ+ℕ0∩ℓ∞, one has
(16)Pηg=∑z=0∞γzηgzλz1/K≤supzηλz/K∑z=0∞γzgzλz1/K≤D∣η∣Pg<∞,where D=max1,supzηλz/K−1≥1. Therefore, ηg∈ℓγ,λ. From conditions (1-i) and (1-ii), one hasℓγ,λ which is linear. And ez∈ℓγ,λ, for all z∈ℕ0, as
(17)Pez=∑r=0∞γrezrλr1/K=γzλz/K.
(2) and (iv). Let∣gz∣≤∣fz∣, for all z∈ℕ0 and f∈ℓγ,λ. Since γz>0, for all z∈ℕ0, then
(18)Pg=∑z=0∞γzgzλz1/K≤∑z=0∞γzfzλz1/K=Pf<∞,we get g∈ℓγ,λ.
(3) and (v). Supposegz∈ℓγ,λ and γz is increasing. There is E≥1 so that γ2z+1≤Eγz and λz∈ℝ+ℕ0∩ℓ∞ is increasing; one can see
(19)Pgz2=∑z=0∞γzgz/2λz1/K=∑z=0∞γ2zgzλ2z+∑z=0∞γ2z+1gzλ2z+11/K≤∑z=0∞γ2zgzλz+∑z=0∞γ2z+1gzλz1/K≤2E1/K∑z=0∞γzgzλz1/K=2E1/KPgz,
Then,gz/2∈ℓγ,λ.
(vi) Obviously,F¯=ℓγ,λ
(vii) We have0<κ≤ξλ0/K−1, for ξ≠0 or κ>0, for ξ=0 such that
(20)Pξ,0,0,0,⋯≥κξP1,0,0,0,⋯Theorem 20.
Let the conditions (a1) and (a2) of Theorem19 be satisfied, then ℓγ,λP be a pre-quasi-Banach sss.Proof.
From Theorems19 and 7, we have ℓγ,λP which is a pre-quasi-normed sss. Suppose gx=gzxz=0∞ is a Cauchy sequence in ℓγ,λP. Therefore, for all ε∈0,1, we have x0∈ℕ0 such that for all x,y≥x0, we get
(21)Pgx−gy=∑z=0∞γzgzx−gzyλz1/K<ε.
Hence, forx,y≥x0 and z∈ℕ0, one obtains ∣gzx−gzy∣<ε. This implies gzy is a Cauchy sequence in ℝ, for fixed z∈ℕ0. This explains limy⟶∞gzy=gz0, with fixed z∈ℕ0. Therefore, Pgx−g0<ε, for all x≥x0. Also, one has Pg0=Pg0−gx+gx≤Pgx−g0+Pgx<∞; hence, g0∈ℓγ,λ.Theorem 21.
The spaceℓγ,λP is a pre-quasi-closed sss, whenever the conditions (a1) and (a2) of Theorem 19 are satisfied.Proof.
Letgx=gzxz=0∞∈ℓγ,λP and limx⟶∞Pgx−g0=0; hence, for all ε∈0,1, one has x0∈ℕ0 such that for every x≥x0, we obtain
(22)ε>Pgx−g0=∑z=0∞γzgzx−gz0λz1/K.
This gives∣gzx−gz0∣<ε. Therefore, gzx is a convergent sequence in ℝ, for constant z∈ℕ0. Hence, limx⟶∞gzx=gz0, with constant z∈ℕ0. Also, one gets
(23)Pg0=Pg0−gx+gx≤Pgx−g0+Pgx<∞.
Hence,g0∈ℓγ,λ.Theorem 22.
The functionPg=∑z=0∞γzgzλz1/K, for every g∈ℓγ,λ, has the Fatou property, if the conditions (a1) and (a2) of Theorem 19conditions (a1) and (a2) of Theorem 19 are satisfied.Proof.
Assumefy⊆ℓγ,λP and limy⟶∞Pfy−f=0. As ℓγ,λP is a pre-quasi-closed space, this implies f∈ℓγ,λP. Hence, for all g∈ℓγ,λP, we have
(24)Pg−f=∑z=0∞γzgz−fzλz1/K≤∑z=0∞γzgz−fzyλz1/K+∑z=0∞γzfzy−fzλz1/K≤suplinfy≥lPg−fy.Theorem 23.
The functionPg=∑z=0∞γzgzλz does not hold the Fatou property, if the setups (a1) and (a2) of Theorem 19 are satisfied with λ0>1.Proof.
Assumefy⊆ℓγ,λP and limy⟶∞Pfy−f=0. As ℓγ,λP is a pre-quasi-closed space, this implies f∈ℓγ,λP. Hence, for all g∈ℓγ,λP, we have
(25)Pg−f=∑z=0∞γzgz−fzλz≤2supzλz−1∑z=0∞γzgz−fzyλz+∑z=0∞γzfzy−fzλz≤2supzλz−1suplinfy≥lPg−fy.Example 2.
The functionPg=∑z=0∞γzgzλz1/K is a pre-quasi-norm and not quasi-norm, for all g∈ℓγ,λ.Example 3.
The functionPg=∑z=0∞γzgzλ1/λ is a pre-quasi-norm, quasi-norm, and not a norm on ℓγ,λ, for 0<λ<1.Example 4.
The functionPg=infκ>0:∑z=0∞γz∣gz∣/κλz≤1 is a norm on ℓγ,λ.


## 5. Kannan Contraction’s Fixed Points

Here,P-Lipschitzian mapping acting on ℓγ,λP as Kannan P-Lipschitzian mapping has been defined. We investigate the adequate requirements for a fixed point of Kannan contraction mapping on ℓγ,λP equipped with various pre-quasi-norms.Definition 24.
A mappingJ:AP⟶AP is said to be a Kannan P-Lipschitzian, if there exists ι≥0, such that
(26)PJg−Jf≤ιPJg−g+PJf−f,for every g,f∈AP.
(1)
Letι∈0,1/2; then, the operator J is called Kannan P-contraction(2)
Forι=1/2, then the operator J is said to be Kannan P-non-expansive
A vectorv∈AP is said to be a fixed point of J, if Jg=g.Theorem 25.
Assume the conditions (a1) and (a2) of Theorem19 are satisfied, and J:ℓγ,λP⟶ℓγ,λP is Kannan P-contraction mapping, where Pg=∑z=0∞γzgzλz1/K, for all g∈ℓγ,λ; hence, J has a unique fixed point.Proof.
Let the setups be satisfied. Assumeg∈ℓγ,λ; hence, Jxg∈ℓγ,λ. Since J is a Kannan P-contraction mapping, we obtain
(27)PJx+1g−Jxg≤ιPJx+1g−Jxg+PJxg−Jx−1g⟹PJx+1g−Jxg≤ι1−ιPJxg−Jx−1g≤ι1−ι2PJx−1g−Jx−2g≤≤ι1−ιxPJg−g.
Therefore, fory>x with x,y∈ℕ0, one has
(28)PJxg−Jyg≤ιPJxg−Jx−1g+PJyg−Jy−1g≤ιι1−ιx−1+ι1−ιy−1PJg−g.
Hence,Jxg is a Cauchy sequence inℓγ,λP, since ℓγ,λP is pre-quasi-Banach space. We have f∈ℓγ,λP withlimx⟶∞Jxg=f, to show that Jf=f. As P verifies the Fatou property, we get
(29)PJf−f≤suplinfx≥lPJx+1g−Jxg≤suplinfx≥lι1−ιpPJg−g=0.
Hence,Jf=f. So f is a fixed point of J. To show the uniqueness of f, let us have two different fixed points f,g∈ℓγ,λP of J. So, one has
(30)Pf−g≤PJf−Jg≤ιPJf−f+PJg−g=0.
This impliesf=g.Example 5.
AssumeJ:ℓa+2/2a+3a=0∞,2a+3/a+2a=0∞P⟶ℓa+2/2a+3a=0∞,2a+3/a+2a=0∞P, wherePg=∑a∈ℕ0a+2/2a+3ga2a+3/a+2, for every g∈ℓa+2/2a+3a=0∞,2a+3/a+2a=0∞ and
(31)Jg=g4,Pg∈0,1,g5,Pg∈1,∞.
As for eachg1,g2∈ℓa+2/2a+3a=0∞,2a+3/a+2a=0∞P with Pg1,Pg2∈0,1, one has
(32)PJg1−Jg2=Pg14−g24≤1274P3g14+P3g24=1274PJg1−g1+PJg2−g2.
For allg1,g2∈ℓa+2/2a+3a=0∞,2a+3/a+2a=0∞P with Pg1,Pg2∈1,∞, one has
(33)PJg1−Jg2=Pg15−g25≤1644P4g15+P4g25=1644PJg1−g1+PJg2−g2.
For allg1,g2∈ℓa+2/2a+3a=0∞,2a+3/a+2a=0∞P with Pg1∈0,1 and Pg2∈1,∞, we get
(34)PJg1−Jg2=Pg14−g25≤1274P3g14+1644P4g25≤1274P3g14+P4g25=1274PJg1−g1+PJg2−g2.Hence,J is Kannan P-contraction and holds one element θ∈ℓa+2/2a+3a=0∞,2a+3/a+2a=0∞P, so that Jθ=θ, by Theorem 25.Corollary 26.
Let conditions (a1) and (a2) of Theorem19 be satisfied, and J:ℓγ,λP⟶ℓγ,λP is Kannan P-contraction mapping, where Pg=∑z=0∞γzgzλz1/K, for all g∈ℓγ,λ; then, J has a unique fixed point f with PJxg−f≤ιι/1−ιx−1PJg−g.Proof.
In view of Theorem25, we have a unique fixed point f of J. Therefore, one gets
(35)PJxg−f=PJxg−Jf≤ιPJxg−Jx−1g+PJf−f=ιι1−ιx−1PJg−g.Definition 27.
AssumeAP is a pre-quasi-normed sss, J:AP⟶AP and f∈AP. The operator J is said to be P-sequentially continuous at f, if and only if, when limx⟶∞Pgx−f=0, then limx⟶∞PJgx−Jf=0.Example 6.
SupposeJ:ℓ2z+4/z+1z=0∞,z+1/2z+4z=0∞P⟶ℓ2z+4/z+1z=0∞,z+1/2z+4z=0∞P, where Pg=∑z∈N2z+4/z+1gzz+1/2z+44, for every g∈ℓ2z+4/z+1z=0∞,z+1/2z+4z=0∞ and
(36)Jg=118e0+g,g0∈−∞,117,117e0,g0=117,118e0,g0∈117,∞.J is clearly both P-sequentially continuous and discontinuous at 1/17e0∈ℓ2z+4/z+1z=0∞,z+1/2z+4z=0∞P.Example 7.
AssumeJ is defined as in Example 5. Suppose gn⊆ℓz+2/2z+3z=0∞,2z+3/z+2z=0∞P is such that limn⟶∞Pgn−g0=0, where g0∈ℓz+2/2z+3z=0∞,2z+3/z+2z=0∞P with Pg0=1.
As the pre-quasi-normP is continuous, we have
(37)limn⟶∞PJgn−Jg0=limn⟶∞Pgn4−g05=Pg020>0.Therefore,J is not P-sequentially continuous at g0.Theorem 28.
If the conditions (a1) and (a2) of Theorem19 are satisfied with λ0>1 and J:ℓγ,λP⟶ℓγ,λP, where Pg=∑z=0∞γzgzλz, for all g∈ℓγ,λ,
(1)
supposeJ is Kannan P-contraction mapping(2)
assumeJ is P-sequentially continuous at a point g∈ℓγ,λP(3)
we havet∈ℓγ,λP such that Jxt has a subsequence Jxjt converging to f; then, f∈ℓγ,λP is the only fixed point of JProof.
Suppose the settings are verified. Assumef is not a fixed point of J, then Jf≠f. From parts (54) and (55), one gets
(38)limxj⟶∞PJxjt−f=0andlimxj⟶∞PJxj+1t−Jf=0.
SinceJ is Kannan P-contraction, we obtain
(39)0<PJf−f=PJf−Jxj+1t+Jxjt−f+Jxj+1t−Jxjt≤22supjλj−2PJxj+1t−Jf+22supjλj−2PJxjt−f+2supjλj−1ιι1−ιxj−1PJt−t.
We get a contradiction whenxj⟶∞. To show the uniqueness of f, suppose we have two different fixed points f,g∈ℓγ,λP of J. Therefore, one obtains
(40)Pf−g≤PJf−Jg≤ιPJf−f+PJg−g=0.
Hence,f=g.Example 8.
AssumeJ is defined as in Example 5. Let Pg=∑a∈ℕ0a+2/2a+3ga2a+3/a+2, for every g∈ℓa+2/2a+3a=0∞,2a+3/a+2a=0∞.
As for eachg1,g2∈ℓa+2/2a+3a=0∞,2a+3/a+2a=0∞P with Pg1,Pg2∈0,1, one has
(41)PJg1−Jg2=Pg14−g24≤227P3g14+P3g24=227PJg1−g1+PJg2−g2.
For allg1,g2∈ℓa+2/2a+3a=0∞,2a+3/a+2a=0∞P with Pg1,Pg2∈1,∞, one has
(42)PJg1−Jg2=Pg15−g25≤14P4g15+P4g25=14PJg1−g1+PJg2−g2.
For allg1,g2∈ℓa+2/2a+3a=0∞,2a+3/a+2a=0∞P with Pg1∈0,1 and Pg2∈1,∞, we obtain
(43)PJg1−Jg2=Pg14−g25≤227P3g14+14P4g25≤227P3g14+P4g25=227PJg1−g1+PJg2−g2.So, the mappingJ is Kannan P-contraction and
(44)Jpg=g4p,Pg∈0,1,g5p,Pg∈1,∞.Obviously,J is P-sequentially continuous at θ∈(ℓ(a+2/2a+3a=0∞,2a+3/a+2a=0∞P and Jpg contains a subsequence Jpig converging to θ. From Theorem 28, the vector θ∈ℓa+2/2a+3a=0∞,2a+3/a+2a=0∞P is the unique fixed point of J.


## 6. Kannan Nonexpansive Fixed Points

The uniform convexity (UUC 2) defined in [22] of the spaceℓγ,λP has been investigated, where Pg=∑d=0∞γdgdλd1/K, for all g∈ℓγ,λ. The property (R) and the P-normal structure property of this space have been discussed. Finally, we present the sufficient conditions on this space such that the Kannan pre-quasi-norm nonexpansive mapping on it has a fixed point.Definition 29 (see [23, 24]).
(1)
[25] Suppose r>0 and h>0. Let(45)ℍ1r,h=g,f:g,h∈AP,Pg≤r,Pf≤h,Pg−h≥rh
Ifℍ1r,h≠∅, we put
(46)H1r,h=inf1−1rPg+f2:g,f∈ℍ1r,h.
Ifℍ1r,h=∅, we put H1r,h=1. The function P holds the uniform convexity (UC), if for all r>0 and h>0, one has H1r,h>0. Note that for every r>0, then ℍ1r,h≠∅, for very small h>0.
(2)
[22] The function P holds (UUC1), if for every x≥0 and h>0, we have β1x,h with(47)H1r,h>β1x,h>0,andr>x(3)
[22] Let r>0 and h>0. Suppose(48)ℍ2r,h=g,f:g,f∈AP,Pg≤r,Pf≤r,Pg−f2≥rh
Whenℍ2r,h≠∅, we put
(49)H2r,h=inf1−1rPg+f2:g,f∈ℍ2r,h.
Ifℍ2r,h=∅, we place H2r,h=1. The function P supports (UC 2), if for every r>0 and h>0, one has H2r,h>0. Note that for all r>0, ℍ2r,h≠∅, with very small h>0.
(4)
[22] P verifies (UUC 2); when x≥0 and h>0, one has β2x,h with(50)H2r,h>β2x,h>0,forr>x(5)
[25] The function P is strictly convex, (SC); when g,f∈AP with Pg=Pf and Pg+f/2=Pg+Pf/2, one obtains g=f
We will need the following comment here and later:PVg=∑m∈Vγmgmλm1/K, for all V⊂ℕ0 and g∈ℓγ,λP. If V=∅, we set PVg=0.Theorem 30.
If the conditions (a1) and (a2) of Theorem19 are satisfied with λ0>1, then the pre-quasi-norm P on ℓγ,λ is (UUC2).Proof.
Assume the settings are satisfied,r>x≥0 and h>0. Let g,t∈ℓγ,λ so that
(51)Pg≤r,Pf≤r,Pg−f2≥rh.
From the definition ofP, we have
(52)rh≤Pg−f2=∑y=0∞γygy−fy2λy1/K≤2−λ0∑y=0∞γygy−fyλy1/K≤2−λ0/K∑y=0∞γygyλy1/K+∑y=0∞γyfyλy1/K=2−λ0/KPg+Pf≤2r,which implies h≤2. Consequent, put A=y∈ℕ0:λy≥2 and B=y∈ℕ0:1<λy<2=ℕ0\A. Let κ∈ℓγ,λ; we get PKκ=PAKκ+PBKκ. By using the conditions, we get PAg−f/2≥rh/2 or PBg−f/2≥rh/2. Assume first PAg−f/2≥rh/2. Using Lemma 14, one gets
(53)PAKg+f2+PAKg−f2≤PAKg+PAKf2.
This explains(54)PAKg+f2≤PAKg+PAKf2−rh2K.
Since(55)PBKg+f2≤PBKg+PBKf2,
by adding inequalities 2 and 3, and from inequality 1, we have(56)PKg+f2≤PKg+PKf2−rh2K≤rK1−h2K.
This gives(57)Pg+f2≤r1−h2K1/K.
Next, supposePBg−f/2≥rh/2. Set D=h/4K,(58)B1=y∈Q:gy−fy≤Dgy+fy,B2=B\B1.
AsD≤1 and the power function is convex, so
(59)PB1Kg−f2≤∑y∈B1Dλyγy∣gy∣+∣fy∣2λy≤D2λ0PB1Kg+PB1Kf≤D2PBKg+PBKf≤DrK.
SincePBg−f/2≥rh/2, we get
(60)PB2Kg−f2=PBKg−f2−PB1Kg−f2≥rKh2K−h4K.
For anyd∈B2, we have
(61)λ0−1<λ0λ0−1≤≤λy−1λy−1−1≤λyλy−1,D≤D2−λy≤gy−fygy+fy2−λy.
By Lemma15, one gets
(62)γygy+fy2λy+λ0−1D2γygy−fy2λy≤12γygyλy+γyfyλy.
Hence,(63)PB2Kg+f2+λ0−1D2PB2Kg−f2≤PB2Kg+PB2Kf2.
This investigates(64)PB2Kg+f2≤PB2Kg+PB2Kf2−λ0−1D2rKh2K−h4K.
Since(65)PB1Kg+f2≤PB1Kg+PB1Kf2,by adding inequalities 5 and 6, one has
(66)PBKg+f2≤PBKg+PBKf2−λ0−1D2rKh2K−h4K≤PBKg+PBKf2−λ0−12h42KrK2K−1≤PBKg+PBKf2−λ0−12K−1h42KrK.
Since(67)PAKg+f2≤PAKg+PAKf2,by adding inequalities 7 and 8, and from inequality 1, we obtain
(68)PKg+f2≤PKg+PKf2−λ0−12K−1h42KrK≤rK1−λ0−12K−1h42K.
This implies(69)Pg+f2≤r1−λ0−12K−1h42K1/K.
It is clear that(70)1<λ0≤K<2K⇒0<λ0−12K−1<1.
By using inequalities 4 and 9 and Definition29, if we put
(71)β2x,h=min1−1−h2K1/K,1−1−λ0−12K−1h42K1/K,therefore, we have H2r,h>β2x,h>0; this implies P is (UUC2).Definition 31.
SpaceAP is said to satisfy the property (R), if for all decreasing sequence Δjj∈ℕ0 of P-closed and P-convex nonempty subsets of AP with supj∈ℕ0dPg,Δj<∞, for some g∈AP, then one has ⋂j∈ℕ0Δj≠∅.Theorem 32.
If conditions (a1) and (a2) of Theorem19 are satisfied with λ0>1, then
(1)
AssumeΔ is a nonempty P-closed and P-convex subset of ℓγ,λP. Let g∈ℓγ,λP be with(72)dPg,Δ=infPg−f:f∈Δ<∞
Hence, one has a uniqueη∈Δ so that dPg,Δ=Pg−η.(2)
The property (R) holds onℓγ,λPProof.
Suppose the conditions are satisfied. To show (51), assume g∉Δ as Δ is P-closed. So, one has D≔dPg,Δ>0. Hence, for every x∈ℕ0, one has fx∈Δ with Pg−fx<D1+1/x. Assume fx/2 is not P-Cauchy. Therefore, we get a subsequence fmx/2 and y0>0 with Pfmx−fma/2≥y0, for every x>a≥0. Furthermore, we have H2D1+1/x,y0/2D>ι≔β2D1+1/x,y0/2D>0, for all x∈ℕ0. As
(73)maxPg−fmx,Pg−fma≤D1+1maand
(74)Pfmx−fma2≥y0≥D1+1may02D,for all x>a≥0, we obtain
(75)Pg−fmx+fma2≤D1+1ma1−ι.
So(76)D=dPg,Δ≤D1+1ma1−ι,for any a∈ℕ0. If we let a⟶∞, we get
(77)0<D≤D1+1ma1−ι<D,which is a contradiction. So, fx/2 is P-Cauchy. Since ℓγ,λP is P-complete, then fx/2P-converges to some f. For every a∈ℕ0, one has the sequence fx+fa/2P converges to f+fa/2. As Δ is P-closed and P-convex, one gets f+fa/2∈Δ. Surely f+fa/2P converges to 2f, which implies 2f∈Δ. By setting η=2f and using Theorem 22, as P satisfies the Fatou property, we obtain
(78)dPg,Δ≤Pg−η≤supiinfa≥iPg−f+fa2≤supiinfa≥isupiinfx≥iPg−fx+fa2≤12supiinfa≥isupiinfx≥iPg−fx+Pg−fa=dPg,Δ.
Hence,Pg−λ=dPg,Δ. As P is (UUC2), then P is (SC); this implies η is only one. To show (2), let g∉Δx0, for some x0∈ℕ0. Since dPg,Δxx∈ℕ0∈ℓ∞ is increasing. Set limx⟶∞dPg,Δx=D. If D>0. Else g∈Δx, for all x∈ℕ0. From (51), we get one point fx∈Δx with dPg,Δx=Pg−fx, for every x∈ℕ0. A consistent proof will show that fx/2P converges to some f∈ℓγ,λP. Since Δx are P-convex, decreasing, and P-closed, we get 2f∈∩x∈ℕ0Δx.Definition 33.
ℓγ,λP satisfies the P-normal structure property, if for all nonempty P-bounded, P-convex, and P-closed subset Δ of ℓγ,λP not decreased to one point, we have g∈Δ with
(79)supf∈ΔPg−f<δPΔ≔supPg−f:g,f∈Δ<∞.Theorem 34.
If the conditions (a1) and (a2) of Theorem19 are satisfied with λ0>1, then ℓγ,λP has the P-normal structure property.Proof.
Suppose the setups are satisfied. Theorem30 implies that P is (UUC2). Let Δ be a P-bounded, P-convex, and P-closed subset of ℓγ,λP not decreased to one point. Hence, δPΔ>0. Put D=δPΔ. Suppose g,f∈Δ with g≠f. Hence, Pg−f/2=y>0. For every η∈Δ, we have Pg−η≤D and Pf−η≤D. Since Δ is P-convex, one obtains g+f/2∈Δ. Hence,
(80)Pg+f2−η=Pg−η+f−η2≤D1−H2D,yD,for every η∈Δ. Hence,
(81)supη∈ΔPg+f2−η≤D1−H2D,yD<D=δPΔ.Lemma 35.
LetΔ be a nonempty P-bounded, P-convex, and 𝑃-closed subset of ℓγ,λP, where ℓγ,λP verifies the (R) property and the P-quasi-normal property, and J:Δ⟶Δ be a Kannan P-nonexpansive mapping. For z>0, suppose Gz=g∈Δ:Pg−Jg≤z≠∅. Take
(82)Δz=⋂BPx,y:JKz⊂BPx,y∩Δ.
Then,Δz is a nonempty, P-convex, and P-closed subset of Δ and
(83)JΔz⊂Δz⊂KzandδPΔz≤z.Proof.
AsJKz⊂Δz, which implies Δz≠∅, Δz is a P-closed and P-convex subset of Δ, as the P-balls are P-convex and P-closed. Assume g∈Δz. When Pg−Jg=0, we get g∈Kz. Else, suppose Pg−Jg>0. Take
(84)x=supPJκ−Jg:κ∈Kz.
From the definition ofx, so JKz⊂BPJg,x. Hence, Δz⊂BPJg,x, which implies Pg−Jg≤x. Assume r>0. Hence, there is κ∈Kz with x−r≤PJκ−Jg. Then,
(85)Pg−Jg−r≤x−r≤PJκ−Jg≤12Pg−Jg+Pκ−Jκ≤12Pg−Jg+z.
Sincer is randomly positive, one has Pg−Jg≤z; then, we have g∈Kz. This implies Δz⊂Kz. As JKz⊂Δz, we get JΔz⊂JKz⊂Δz; this indicates Δz is J-invariant, consequent to prove that δPΔz≤z. As
(86)PJg−Jf≤12Pg−Jg+Pf−Jf,for every g,f∈Kz, let g∈Kz. Then, JKz⊂BPJg,z. The definition of Δz implies Δz⊂BPJg,z. So, Jg∈⋂f∈ΔzBPf,z. Therefore, we obtain Pf−κ≤z, for every f,κ∈Δz; this gives δPΔz≤z.Theorem 36.
LetΔ be a nonempty, P-convex, P-closed, and P-bounded subset of ℓγ,λP, where ℓγ,λP holds the P-quasi-normal property and the (R) property, and J:Δ⟶Δ be a Kannan P-non-expansive mapping. Then, J has a fixed point.Proof.
Takez0=infPg−Jg:g∈Δ and zx=z0+1/x, for every x≥1. In view of the definition of z0, we get Kzx=g∈Δ:Pg−Jg≤zx≠∅, for all x≥1. Let Δzx as defined in Lemma 35. Clearly, Δzx is a decreasing sequence of nonempty P-bounded, P-closed, and P-convex subsets of Δ. One has Δ∞=⋂x≥1Δzx≠∅, from the property (R). Let g∈Δ∞; we have Pg−Jg≤zx, for every x≥1. Suppose x⟶∞; one has Pg−Jg≤z0, which implies Pg−Jg=z0. Then, Kz0≠∅. One gets z0=0. Otherwise, z0>0; this implies that J has no a fixed point. Assume Δz0 is as defined in Lemma 35. Since J has not a fixed point and Δz0 is J-invariant, then δPΔz0>0. From the P-quasi-normal property, we have g∈Δz0 with
(87)Pg−f<δPΔz0≤z0,for every f∈Δz0. According to Lemma 35, one gets Δz0⊂Kz0. From definition of Δz0, hence, Jg∈Kz0⊂Δz0. Obviously, this explains
(88)Pg−Jg<δPΔz0≤z0,which contradicts the definition of z0. Therefore, z0=0, which means that J has a fixed point in Δ.We have the next corollary according to Theorems32, 34, and 36.Corollary 37.
Pick up the conditions (a1) and (a2) of Theorem19 to be satisfied with λ0>1. Assume Δ is a nonempty, P-convex, P-closed, and P-bounded subset of ℓγ,λP, and J:Δ⟶Δ is a Kannan P-nonexpansive operator. Hence, J has a fixed point.Example 9.
SupposeJ:Δ⟶Δ, where
(89)Jg=g4,Pg∈0,1,g5,Pg∈1,∞,where Δ=g∈ℓz+2/2z+3z=0∞,2z+3/z+2z=0∞P:g0=g1=0 and Pg=∑z∈ℕ0z+2/2z+3gz2z+3/z+2, for every g∈ℓz+2/2z+3z=0∞,2z+3/z+2z=0∞P. According to Example 5, the operator J is Kannan P-contraction. Hence, it is a Kannan P-nonexpansive mapping. Evidently, Δ is a nonempty, P-convex, P-closed, and P-bounded subset of ℓz+2/2z+3z=0∞,2z+3/z+2z=0∞P. In view of Corollary 37, the operator J has a fixed point θ∈Δ.


## 7. Kannan Contraction’s Fixed Points on Pre-Quasi-Ideal

In this part, we supposeE and H are Banach spaces. The Kannan contraction’s fixed points on Sℓγ,λP,ℙ, where ℙC=PsaCa=0∞, has been examined.Theorem 38 (see [3]).
If the conditions (a1) and (a2) of Theorem19 are fulfilled, then Sℓγ,λP,ℙ is a pre-quasi-Banach operator ideal.Theorem 39.
If the conditions (a1) and (a2) of Theorem19 are satisfied, then Sℓγ,λP,ℙ is a pre-quasi-closed operator ideal.Proof.
The classSℓγ,λP is a pre-quasi-operator ideal and follows from Theorems 19 and 12. Let Cy∈Sℓγ,λPE,H, for every y∈ℕ0 and limy⟶∞ℙCy−C=0. Hence, there is ς>0 and since LE,H⊇Sℓγ,λPE,H, one obtains
(90)ℙCy−C=PsxCy−Cx=0∞≥Ps0Cy−C,0,0,0,⋯=P∥Cy−C∥,0,0,0,⋯≥ςγ01/K∥Cy−C∥.
Hence,Cyy∈ℕ0 is convergent in LE,H. This implies limy⟶∞∥Cy−C∥=0 and while sxCyx=0∞∈ℓγ,λP, for every y∈ℕ0 and ℓγ,λP is a premodular sss. Hence, there exists E≥1 with
(91)ℙC=PsxCx=0∞=PsxC−Cy+Cyx=0∞≤Psx2C−Cyx=0∞+Psx2Cyx=0∞≤P∥Cy−C∥x=0∞+2E1/KPsxCyx=0∞<ε.
We havesxCx=0∞∈ℓγ,λP, then C∈Sℓγ,λPE,H.Definition 40.
The functionℙ on SAP satisfies the Fatou property, if for any sequence Cxx∈ℕ0⊆SAPE,H with limx⟶∞ℙCx−C=0 and any V∈SAPE,H, then
(92)ℙV−C≤supxinfj≥xPV−Cj.Theorem 41.
Suppose the conditions (a1) and (a2) of Theorem19 are satisfied, then the function ℙC=∑x=0∞γxsxCλx1/K, for all C∈Sℓγ,λPE,H, does not hold the Fatou property.Proof.
AssumeCpp∈ℕ0⊆Sℓγ,λPE,H with limp⟶∞ℙCp−C=0. Since the space Sℓγ,λP is a pre-quasi-closed ideal, then C∈Sℓγ,λPE,H. Hence, for any V∈Sℓγ,λPE,H, we have
(93)ℙV−C=∑x=0∞γxsxV−Cλx1/K≤∑x=0∞γxsx/2V−Ciλx1/K+∑x=0∞γxsx/2Ci−Cλx1/K≤2E1/Ksuppinfi≥p∑x=0∞γxsxV−Ciλx1/K.Definition 42.
A mappingG:SAPE,H⟶SAPE,H is said to be a Kannan ℙ-Lipschitzian, if we have ι≥0, such that
(94)ℙGC−GA≤ιℙGC−C+ℙGA−A,for every C,A∈SAPE,H.
(1)
The mappingG is said to be Kannan ℙ-contraction, whenever ι∈0,1/2(2)
The mappingG is said to be Kannan ℙ-non-expansive, whenever ι=1/2Definition 43.
SupposeG:SAPE,H⟶SAPE,H and B∈SAPE,H.G is said to be ℙ-sequentially continuous at B, if and only if, if limp⟶∞ℙCp−B=0, then limp⟶∞ℙGCp−GB=0.Example 10.
AssumeG:Sℓa+3/2a+1a=0∞,2a+1/a+3a=0∞PE,H⟶Sℓa+3/2a+1a=0∞,2a+1/a+3a=0∞PE,H, where
ℙC=∑a=0∞a+3/2a+1saC2a+1/a+3, for every C∈Sℓa+3/2a+1a=0∞,2a+1/a+3a=0∞PE,H and
(95)GC=C263170,ℙC∈0,1,C263171,ℙC∈1,∞.
Evidently,G is ℙ-sequentially continuous at the zero operator Θ∈Sℓa+3/2a+1a=0∞,2a+1/a+3a=0∞P.
LetCn⊆Sℓa+3/2a+1a=0∞,2a+1/a+3a=0∞P with limm⟶∞ℙCm−C0=0, where C0∈Sℓa+3/2a+1a=0∞,2a+1/a+3a=0∞P with ℙC0=1. As ℙ is continuous, one has
(96)limn⟶∞ℙGCn−GC0=limn⟶∞ℙCn263170−C0263171=ℙC069258712070>0.
SoG is not ℙ-sequentially continuous at C0.Theorem 44.
Assume the conditions (a1) and (a2) of Theorem19 are satisfied and G:Sℓγ,λPE,H⟶Sℓγ,λPE,H, where ℙC=∑y=0∞γysyCλy1/K, for every C∈Sℓγ,λPE,H.
(i)
LetG is Kannan ℙ-contraction mapping(ii)
G is ℙ-sequentially continuous at a point A∈Sℓγ,λPE,H(iii)
We haveY∈Sℓγ,λPE,H such that the sequence GpY has a subsequence GpiY converging to X
Then,X∈Sℓγ,λPE,H is the only fixed point of G.Proof.
AssumeX is not a fixed point of G; we have GX≠X. By parts (ii) and (iii), one can see
(97)limpi⟶∞ℙGpiY−X=0,limpi⟶∞ℙGpi+1Y−GX=0.
AsG is Kannan ℙ-contraction mapping, we get
(98)0<ℙGX−X=ℙGX−Gpi+1Y+GpiY−X+Gpi+1Y−GpiY≤2E1/KℙGpi+1Y−GX+2E2/KℙGpiY−X+2E2/Kιι1−ιpi−1ℙGY−Y.This gives a contradiction aspi⟶∞. Hence, X is a fixed point of G. For the uniqueness of the fixed point X, assume we have two different fixed points X,Y∈Sℓγ,λPE,H of G. Therefore, one gets ℙX−Y≤ℙGX−GY≤ιℙGX−X+ℙGY−Y=0. This implies X=Y.Example 11.
AssumeG:Sℓy+2/y+1y=0∞,y+1/y+2y=0∞PE,H⟶Sℓy+2/y+1y=0∞,y+1/y+2y=0∞PE,H, where ℙC=∑y=0∞y+2/y+1syCy+1/y+2, for every C∈Sℓy+2/y+1y=0∞,y+1/y+2y=0∞PE,H and
(99)GC=C26,ℙC∈0,1,C37,ℙC∈1,∞.
IfC1,C2∈Sℓa+2/a+1a=0∞,a+1/a+2a=0∞P with ℙC1,ℙC2∈0,1, one has
(100)ℙGC1−GC2=ℙC126−C226≤25ℙ25C126+ℙ25C226=25ℙGC1−C1+ℙGC2−C2.
For eachC1,C2∈Sℓa+2/a+1a=0∞,a+1/a+2a=0∞P with ℙC1,ℙC2∈1,∞, we get
(101)ℙGC1−GC2=ℙC137−C237≤13ℙ36C137+ℙ36C237=13ℙGC1−C1+ℙGC2−C2.
For eachC1,C2∈Sℓa+2/a+1a=0∞,a+1/a+2a=0∞P with ℙC1∈0,1 and ℙC2∈1,∞, one can see
(102)ℙGC1−GC2=ℙC126−C237≤25ℙ25C126+13ℙ36C237≤25ℙ25C126+ℙ36C237=25ℙGC1−C1+ℙGC2−C2.
So, the mappingC is Kannan ℙ-contraction and
(103)GpC=C26p,ℙC∈0,1,C37p,ℙC∈1,∞.
From Theorem44, Θ∈Sℓa+2/a+1a=0∞,a+1/a+2a=0∞P is the unique fixed point of G, since G is ℙ-sequentially continuous and GpiC converging to Θ.


## 8. The Presence of Solutions to Summable Equations

The solution to (104), which is studied by some authors (see [26–28]), in ℓγ,λP, where λl∈ℝ+ℕ0 is an increase and Pg=∑z=0∞γzgzλz1/K, for all g∈ℓγ,λ, has been examined.Consider the summable equations(104)gz=xz+∑r=0∞Gz,rWr,gr,and suppose J:ℓγ,λP⟶ℓγ,λP is defined by
(105)Jgzz∈ℕ0=xz+∑r=0∞Gz,rWr,grz∈ℕ0.Theorem 45.
AssumeG:ℕ02⟶ℝ,W:ℕ0×ℝ⟶ℝ,x:ℕ0⟶ℝ, and for all z∈ℕ0, there exists ι∈0,1/2 with
(106)∑r∈ℕ0Gz,rWr,gr−Wr,frλz≤ιKxz−gz+∑r=0∞Gz,rWr,grλz+xz−fz+∑r=0∞Gz,rWr,frλz.
Then, the summable equation (104) has a solution in ℓγ,λP.Proof.
SupposeJ:ℓγ,λP⟶ℓγ,λP is defined by (105). We have
(107)PJg−Jf=∑z∈ℕ0γzJgz−Jfzλz1/K=∑z∈ℕ0γz∑r∈ℕ0Gz,rWr,gr−Wr,frλz1/K≤ι∑z∈ℕ0γzxz−gz+∑r=0∞Gz,rWr,grλz1/K+∑z∈ℕ0γzxz−fz+∑r=0∞Gz,rWr,frλz1/K=ιPJg−g+PJf−f.Hence, there is a unique solution of equation (104) in ℓγ,λP according to Theorem 25.Example 12.
Suppose we haveℓz+2/z+1z=0∞,z+1/z+2z=0∞P, where Pg=∑z∈ℕ0z+2/z+1gzz+1/z+2, for all g∈ℓz+2/z+1z=0∞,z+1/z+2z=0∞. Consider the summable equations
(108)gz=e−3z+6+∑r=0∞−1z+re∣gz∣z2+r2+1l,where l>2, and let J:ℓz+2/z+1z=0∞,z+1/z+2z=0∞P⟶ℓz+2/z+1z=0∞,z+1/z+2z=0∞P be defined by
(109)Jgzz∈ℕ0=e−3z+6+∑r=0∞−1z+re∣gz∣z2+r2+1lz∈ℕ0.
Obviously,(110)∑r=0∞−1ze∣gz∣z2+r2+1l−1r−−1rz+1/z+2≤13e−3z+6−gz+∑r=0∞−1z+regzz2+r2+1lz+1/z+2+e−3z+6−fz+∑r=0∞−1z+refzz2+r2+1lz+1/z+2.By Theorem45, the summable equation (108) has a unique solution in ℓz+2/z+1z=0∞,z+1/z+2z=0∞P.Example 13.
Assume we haveℓz+3/2z+1z=0∞,2z+1/z+3z=0∞P, where Pg=∑z∈ℕ0z+3/2z+1gz2z+1/z+3, for all g∈ℓz+3/2z+1z=0∞,2z+1/z+3z=0∞. Consider the summable equations
(111)gz=e−3z+6+∑r=0∞−1z+rgzz2+r2+1l,where l>2, and let J:ℓz+3/2z+1z=0∞,2z+1/z+3z=0∞P⟶ℓz+3/2z+1z=0∞,2z+1/z+3z=0∞P be defined by
(112)Jgzz∈ℕ0=e−3z+6+∑r=0∞−1z+rgzz2+r2+1lz∈ℕ0.
It is easy to see that(113)∑r=0∞−1zgzz2+r2+1l−1r−−1r2z+1/z+3≤19e−3z+6−gz+∑r=0∞−1z+rgzz2+r2+1l2z+1/z+3+e−3z+6−fz+∑r=0∞−1z+rfzz2+r2+1l2z+1/z+3.By Theorem45, the summable equation (111) has one solution in ℓz+3/2z+1z=0∞,2z+1/z+3z=0∞P.Example 14.
If we haveℓz+2/2z+3z=0∞,2z+3/z+2z=0∞P, where Pg=∑z∈ℕ0z+2/2z+3gz2z+3/z+2, for all g∈ℓz+2/2z+3z=0∞,2z+3/z+2z=0∞. Consider the summable equations
(114)gz=e−3z+6+∑r=0∞−1z+rgzz2+r2+1l,such that z≥2 and l>2, and assume J:Δ⟶Δ, where Δ=v∈ℓz+2/2z+3z=0∞,2z+3/z+2z=0∞P:g0=g1=0 is defined by
(115)Jgzz≥2=e−3z+6+∑r=0∞−1z+rgzz2+r2+1lz≥2,∑r=0∞−1zgzz2+r2+1l−1r−−1r2z+3/z+2≤19e−3z+6−gz+∑r=0∞−1z+rgzz2+r2+1l2z+3/z+2+e−3z+6−fz+∑r=0∞−1z+rfzz2+r2+1l2z+3/z+2.According to Theorem45, the summable equation (114) has a solution in Δ.


## 9. Conclusion

There is a pre-quasi-normed space theorem that is more general than quasi-normed space. In weighted Nakano sequence space with the well-known pre-quasi-norm, we investigate the appropriate conditions for the generation of pre-quasi-Banach and closed spaces. Pre-quasi-normal structural properties of weighted Nakano sequence space, including the fixed point idea of Kannan pre-quasi-norm contraction and Kannan pre-quasi-norm nonexpansive mapping in weighted Nakano sequence space, are improved. Using weighted Nakano sequence space ands-numbers, the presence of a fixed point for Kannan pre-quasi-norm contraction mapping has been proved. Toward the end of our discussion, we provided several examples of how the collected data could be used to solve an issue. The weight and power of the weighted Nakano sequence space can be used to define a wide range of circumstances under which existence findings can be found. Banach lattices are introduced in this article, and a new space of solutions for many difference equations is introduced, the spectrum of any bounded linear operator between any two Banach spaces with s-numbers in this sequence space.



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