# Kannan Prequasi Contraction Maps on Nakano Sequence Spaces

Authors: Awad A. Bakery; O. M. Kalthum S. K. Mohamed
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this article, we explore the concept of the prequasi norm on Nakano special space of sequences (sss) such that its variable exponent in0,1. We evaluate the sufficient setting on it with the definite prequasi norm to configuration prequasi Banach and closed (sss). The Fatou property of different prequasi norms on this (sss) has been investigated. Moreover, the existence of a fixed point of Kannan prequasi norm contraction maps on the prequasi Banach (sss) and the prequasi Banach operator ideal constructed by this (sss) and s− numbers have been examined.


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## Body

## 1. Introduction

Ideal maps and summability theorems [1–6] are extremely significant in mathematical models and have more achievements, such as ideal transformations, normal series, fixed point theory, geometry of Banach spaces, and approximation theory. By RN, we mark the spaces of all sequences of real numbers. We denote the space of all bounded linear maps from a Banach space Z into a Banach space M by LZ,M, and if Z=M, we indicate LZ, the d-th s number by sdW [7], the d-th approximation number by αdW, and ed=0,0,⋯,1,0,0,⋯, where 1 shows at the dth place, for every d∈N=0,1,2,⋯.Notations 1.
The setsSA, SAZ,M, SAapp and SAappZ,M, (cf. [8]) denote
(1)SA≔SAZ,M,whereSAZ,M≔W∈LZ,M:sdWd=0∞∈A.AlsoSAapp≔SAappZ,M,whereSAappZ,M≔W∈LZ,M:αdWd=0∞∈A.Letr=ra∈0,1N, the Nakano sequence space defined and studied in [9–11] is denoted by:
(2)ℓr=v=va∈RN:ϕμv<∞,foranyμ>0,where ϕv=∑a∈Nvara. And ℓr,. is a Banach space, however, ∥v∥=infκ>0:ϕv/κ≤1. Faried and Bakery [8] assumed the hypothesis of prequasi operator ideal that is more established than the quasi operator ideal. Bakery and Abou Elmatty [9] demonstrated the strictly inclusion of the prequasi operator ideal Sℓrapp, for inconsistent powers. It was a small prequasi operator ideal. As the literature of the Banach fixed point theorem [12], many mathematicians created on many actions. Haghi et al. [13, 14] showed that some generalizations in fixed point theory are not real generalizations and investigated some fixed point generalizations to partial metric spaces, which are obtained from the corresponding results in metric spaces. Kannan [15] presented a representation of a class of operators with the same fixed point actions as contractions nevertheless that fails to be continuous. They only try to illustrate Kannan maps [16] in modular vector spaces. The target of this paper is to appraise the concept of prequasi norm on ℓr. The Fatou property of different prequasi norms on this (sss) has been examined. We are delving the sufficient set-up on ℓr equipped with the definite prequasi norm to pattern prequasi Banach and closed (sss). The existence of a fixed point of Kannan prequasi norm contraction mapping on the prequasi Banach (sss) has been given. Finally, the existence of a fixed point of Kannan prequasi norm contraction mapping on the prequasi Banach operator ideal Sℓrϕ has been made current.


## 2. Definitions and Preliminaries

Definition 2 (see [2]).
The linear space of sequencesA is detailed as a special space of sequences (sss), if
(3)eaa∈N⊆A,(1)
A is solid, i.e., let v=va∈RN, t=ta∈A, and va≤ta, for every a∈N, then v∈A(2)
va/2a=0∞∈A, where a/2 marks the integral part of a/2, if vaa=0∞∈ADefinition 3 (see [8]).
A subclassAϕ of A is definite a premodular (sss), if there is ϕ∈0,∞A verifying the set-up:
(i)
Forv∈A, v=θ⇔ϕv=0 with ϕv≥0, where θ is the zero vector of A(ii)
For everyv∈A and η∈R, we have B≥1 for which ϕηv≤Bηϕv,(iii)
ϕv+t≤Jϕv+ϕt, for each v,t∈A, for some J≥1(iv)
Fora∈N and va≤ta, then ϕva≤ϕta(v)
The inequality,ϕva≤ϕva/2≤J0ϕva holds, for some J0≥1(vi)
AssumeF be the space of finite sequences, then F¯=Aϕ(vii)
There isς>0 such that ϕβ,0,0,0,⋯≥ςβϕ1,0,0,0,⋯, for every β∈RDefinition 4 (see [17]).
SupposeA be a (sss). The function ϕ∈0,∞A is called prequasi norm on A, if it provides the conditions (i), (ii), and (iii) of Definition 3.Theorem 5 (see [17]).
Pick upA be a premodular (sss), then it is prequasi normed (sss).Theorem 6 (see [17]).
A is a prequasi normed (sss), if it is quasinormed (sss).Definition 7 (see [3]).
LetL be the class of all bounded linear operators between any two arbitrary Banach spaces. A subclass U of L is named an operator ideal, if every vector UZ,M=U∩LZ,M verifies the next setting:
(i)
IΓ∈U where Γ denotes Banach space of one dimension(ii)
The spaceUZ,M is linear over R(iii)
AssumeW∈LZ0,Z, X∈UZ,M, and Y∈LM,M0, then, YXW∈UZ0,M0, where Z0 and M0 are normed spaces (see [18, 19])The theory of prequasi operator ideal, which is more general than the quasi operator ideal.Definition 8 (see [8]).
A functionϕ∈0,∞U is named a prequasi norm on the ideal U if the following setting includes
(1)
AssumeW∈UZ,M, ϕW≥0, and ϕW=0⇔W=0(2)
There isD≥1 so as to ϕηW≤DηϕW, for every W∈UZ,M and η∈R(3)
There isJ≥1 such that ϕW1+W2≤JϕW1+ϕW2, for each W1,W2∈UZ,M,(4)
There isσ≥1 for to if W∈LZ0,Z, X∈UZ,M and Y∈LM,M0, then ϕYXW≤σYϕXWTheorem 9 (see [20]).
Pick upAϕ be a premodular (sss), then ϕW=ϕsaWa=0∞ be a prequasi norm on SAϕ.Theorem 10 (see [9]).
SupposeZ and M be Banach spaces, and Aϕ be a premodular (sss), then SAϕ,ϕ be a prequasi Banach operator ideal, such that ϕW=ϕsaWa=0∞.Theorem 11 (see [8]).
ϕ is a prequasi norm on the ideal U, if ϕ is a quasinorm on the ideal U.The agreeable inequality [21] will be used in the consequence: Suppose ra∈0,1N and va,ta∈R, for every a∈N, then va+tara≤vara+tara.


## 3. Main Results

### 3.1. Prequasi Normed (sss)

We illustrate the adequate set-up onℓr equipped with a prequasi norm ϕ to generate prequasi Banach and closed (sss).Definition 12.
(a)vaa∈N⊆ℓrϕ is ϕ-convergent to v∈ℓrϕ⇔lima⟶∞ϕva−v=0. If the ϕ-limit exists, then it is unique
(b)vaa∈N⊆ℓrϕ is ϕ-Cauchy, if lima,b⟶∞ϕva−vb=0
(c)Λ⊆ℓrϕ is ϕ-closed, if for all ϕ-converging vaa∈N⊂Λ to v, then v∈ΛTheorem 13.
ℓrϕ, where ϕv=∑a∈Nvara, for all v∈ℓr, is a premodular (sss), if raa∈N∈0,1N is an increasing.Proof.
First, we have to proveℓr is a (sss):
(1) Supposev,t∈ℓr. Since ra∈0,1N, we have
(4)ϕv+t=∑a∈Nva+tara≤∑a∈Nvara+∑a∈Ntara=ϕv+ϕt<∞,
sov+t∈ℓr.
(2) Assumeη∈R and v∈ℓr. As ra∈0,1N, one has
(5)ϕηv=∑a∈Nηvara≤supaηra∑a∈Nvara≤D∣η∣ϕv<∞.
Hence,ηv∈ℓr. So, by using Parts (1) and (2), we get ℓr is linear. Also ea∈ℓr, for all a∈N, since ϕea=∑j=0∞eajrj=1.
(3) Letva≤ta, for every a∈N and t∈ℓr. One can see
(6)ϕv=∑a∈Nvara≤∑a∈Ntara=ϕt<∞,
we havev∈ℓr. This implies the sequence space ℓr is solid.
(4) Supposeva∈ℓr and ra be an increasing sequence, one has
(7)ϕva/2=∑a∈Nva/2ra=∑a∈Nvar2a+∑a∈Nvar2a+1≤2∑a∈Nvara=2ϕva,
thenva/2∈ℓr. Secondly, we show that the functional ϕ on ℓr is a premodular:
(i)
Evidently,ϕv≥0 and ϕv=0⇔v=θ(ii)
We haveD=max1,supaηra−1≥1 such that ϕηv≤Dηϕv, for every v∈ℓr and η∈R\0. For η=0, there is D≥1 such that ϕηv≤Dηϕv, for every v∈ℓr(iii)
We haveJ≥1 so that ϕv+t≤Jϕv+ϕt, for every v,t∈ℓr(iv)
Clearly, sinceℓr is solid(v)
From (49), we have J0=2≥1(vi)
Clearly,F¯=ℓr(vii)
There is0<ς≤βr0−1, for β≠0 or ς>0, for β=0 such that ϕβ,0,0,0,⋯≥ςβϕ1,0,0,0,⋯Theorem 14.
Assumera∈0,1N be an increasing, then ℓrϕ be a prequasi Banach (sss), where ϕv=∑a∈Nvara, for every v∈ℓr.Proof.
Let the set-up be verified. From Theorem13, the space ℓrϕ is a premodular (sss). By Theorem 5, the space ℓrϕ is a prequasi normed (sss). To prove that ℓrϕ is a prequasi Banach (sss), assume vp=vapa=0∞ be a Cauchy sequence in ℓrϕ. Hence, for every ε∈0,1, we have p0∈N such that for all p,q≥p0, one has
(8)ϕvp−vq=∑a∈Nvap−vaqra<ε.Therefore, forp,q≥p0 and a∈N, we get vap−vaq<ε. So vaq is a Cauchy sequence in R, for constant a∈N. Which implies limq⟶∞vaq=va0, for fixed a∈N. Hence, ϕvp−v0<ε, for every p≥p0. Then, to show that v0∈ℓr, we have ϕv0=ϕv0−vp+vp≤ϕvp−v0+ϕvp<∞. So v0∈ℓr. This explains that ℓrϕ is a prequasi Banach (sss).Theorem 15.
Pick upra∈0,1N be an increasing, then ℓrϕ be a prequasi closed (sss), where ϕv=∑a∈Nvara, for every v∈ℓr.Proof.
Assume the conditions be verified. From Theorem13, the space ℓrϕ be a premodular (sss). By Theorem 5, the space ℓrϕ is a prequasi normed (sss). To show that ℓrϕ is a prequasi closed (sss), suppose vp=vapa=0∞∈ℓrϕ and limp⟶∞ϕvp−v0=0, then for all ε∈0,1, we have p0∈N so that for all p≥p0, we have ε>ϕvp−v0=∑a∈Nvap−va0ra. Hence, for p≥p0 and a∈N, one has vap−va0<ε. Therefore, vap is a convergent sequence in R, for fixed a∈N. Hence, limp⟶∞vap=va0, for constant a∈N. Finally, to prove that v0∈ℓr, we obtain
(9)ϕv0=ϕv0−vp+vp≤ϕvp−v0+ϕvp<∞,
hence,v0∈ℓr. This gives that ℓrϕ is a prequasi closed (sss).Example 16.
The functionalϕv=∑a∈Nvaa+1/a+2 is a prequasi norm (not a quasinorm) on Nakano special space of sequences ℓa+1/a+2a=0∞.Example 17.
The functionalϕv=∑a∈Nvaa+1/2a+44 is a prequasi norm (not a quasinorm) on Nakano special space of sequences ℓa+1/2a+4a=0∞.Example 18.
The functionalϕv=∑a∈Nvar is a prequasi norm (not a norm) on r-absolutely summable sequences of real numbers ℓr, for all 0<r≤1.Example 19.
Forra∈0,1N, the functional ϕv=infκ>0:∑a∈Nva/κra≤1 is a prequasi norm (a quasinorm and a norm) on Nakano special space of sequences ℓr.


## 3.1. Prequasi Normed (sss)

We illustrate the adequate set-up onℓr equipped with a prequasi norm ϕ to generate prequasi Banach and closed (sss).Definition 12.
(a)vaa∈N⊆ℓrϕ is ϕ-convergent to v∈ℓrϕ⇔lima⟶∞ϕva−v=0. If the ϕ-limit exists, then it is unique
(b)vaa∈N⊆ℓrϕ is ϕ-Cauchy, if lima,b⟶∞ϕva−vb=0
(c)Λ⊆ℓrϕ is ϕ-closed, if for all ϕ-converging vaa∈N⊂Λ to v, then v∈ΛTheorem 13.
ℓrϕ, where ϕv=∑a∈Nvara, for all v∈ℓr, is a premodular (sss), if raa∈N∈0,1N is an increasing.Proof.
First, we have to proveℓr is a (sss):
(1) Supposev,t∈ℓr. Since ra∈0,1N, we have
(4)ϕv+t=∑a∈Nva+tara≤∑a∈Nvara+∑a∈Ntara=ϕv+ϕt<∞,
sov+t∈ℓr.
(2) Assumeη∈R and v∈ℓr. As ra∈0,1N, one has
(5)ϕηv=∑a∈Nηvara≤supaηra∑a∈Nvara≤D∣η∣ϕv<∞.
Hence,ηv∈ℓr. So, by using Parts (1) and (2), we get ℓr is linear. Also ea∈ℓr, for all a∈N, since ϕea=∑j=0∞eajrj=1.
(3) Letva≤ta, for every a∈N and t∈ℓr. One can see
(6)ϕv=∑a∈Nvara≤∑a∈Ntara=ϕt<∞,
we havev∈ℓr. This implies the sequence space ℓr is solid.
(4) Supposeva∈ℓr and ra be an increasing sequence, one has
(7)ϕva/2=∑a∈Nva/2ra=∑a∈Nvar2a+∑a∈Nvar2a+1≤2∑a∈Nvara=2ϕva,
thenva/2∈ℓr. Secondly, we show that the functional ϕ on ℓr is a premodular:
(i)
Evidently,ϕv≥0 and ϕv=0⇔v=θ(ii)
We haveD=max1,supaηra−1≥1 such that ϕηv≤Dηϕv, for every v∈ℓr and η∈R\0. For η=0, there is D≥1 such that ϕηv≤Dηϕv, for every v∈ℓr(iii)
We haveJ≥1 so that ϕv+t≤Jϕv+ϕt, for every v,t∈ℓr(iv)
Clearly, sinceℓr is solid(v)
From (49), we have J0=2≥1(vi)
Clearly,F¯=ℓr(vii)
There is0<ς≤βr0−1, for β≠0 or ς>0, for β=0 such that ϕβ,0,0,0,⋯≥ςβϕ1,0,0,0,⋯Theorem 14.
Assumera∈0,1N be an increasing, then ℓrϕ be a prequasi Banach (sss), where ϕv=∑a∈Nvara, for every v∈ℓr.Proof.
Let the set-up be verified. From Theorem13, the space ℓrϕ is a premodular (sss). By Theorem 5, the space ℓrϕ is a prequasi normed (sss). To prove that ℓrϕ is a prequasi Banach (sss), assume vp=vapa=0∞ be a Cauchy sequence in ℓrϕ. Hence, for every ε∈0,1, we have p0∈N such that for all p,q≥p0, one has
(8)ϕvp−vq=∑a∈Nvap−vaqra<ε.Therefore, forp,q≥p0 and a∈N, we get vap−vaq<ε. So vaq is a Cauchy sequence in R, for constant a∈N. Which implies limq⟶∞vaq=va0, for fixed a∈N. Hence, ϕvp−v0<ε, for every p≥p0. Then, to show that v0∈ℓr, we have ϕv0=ϕv0−vp+vp≤ϕvp−v0+ϕvp<∞. So v0∈ℓr. This explains that ℓrϕ is a prequasi Banach (sss).Theorem 15.
Pick upra∈0,1N be an increasing, then ℓrϕ be a prequasi closed (sss), where ϕv=∑a∈Nvara, for every v∈ℓr.Proof.
Assume the conditions be verified. From Theorem13, the space ℓrϕ be a premodular (sss). By Theorem 5, the space ℓrϕ is a prequasi normed (sss). To show that ℓrϕ is a prequasi closed (sss), suppose vp=vapa=0∞∈ℓrϕ and limp⟶∞ϕvp−v0=0, then for all ε∈0,1, we have p0∈N so that for all p≥p0, we have ε>ϕvp−v0=∑a∈Nvap−va0ra. Hence, for p≥p0 and a∈N, one has vap−va0<ε. Therefore, vap is a convergent sequence in R, for fixed a∈N. Hence, limp⟶∞vap=va0, for constant a∈N. Finally, to prove that v0∈ℓr, we obtain
(9)ϕv0=ϕv0−vp+vp≤ϕvp−v0+ϕvp<∞,
hence,v0∈ℓr. This gives that ℓrϕ is a prequasi closed (sss).Example 16.
The functionalϕv=∑a∈Nvaa+1/a+2 is a prequasi norm (not a quasinorm) on Nakano special space of sequences ℓa+1/a+2a=0∞.Example 17.
The functionalϕv=∑a∈Nvaa+1/2a+44 is a prequasi norm (not a quasinorm) on Nakano special space of sequences ℓa+1/2a+4a=0∞.Example 18.
The functionalϕv=∑a∈Nvar is a prequasi norm (not a norm) on r-absolutely summable sequences of real numbers ℓr, for all 0<r≤1.Example 19.
Forra∈0,1N, the functional ϕv=infκ>0:∑a∈Nva/κra≤1 is a prequasi norm (a quasinorm and a norm) on Nakano special space of sequences ℓr.


## 4. The Fatou Property

We investigate here the Fatou property of different prequasi normsϕ on ℓr.Definition 20.
A prequasi normϕ on ℓr provides the Fatou property, if for all sequence ta⊆ℓrϕ with lima⟶∞ϕta−t=0 and any v∈ℓrϕ then ϕv−t≤supjinfa≥jϕv−ta.Theorem 21.
The functionϕv=∑a∈Nvara provides the Fatou property, if ra∈0,1N is an increasing, for all v∈ℓr.Proof.
Let the set-up be satisfied andtb⊆ℓrϕ with limb⟶∞ϕtb−t=0. Since the space ℓrϕ is a prequasi closed space, then t∈ℓrϕ. So for every v∈ℓrϕ, one has
(10)ϕv−t=∑a∈Nva−tara≤∑a∈Nva−tabra+∑a∈Ntab−tara≤supjinfb≥jϕv−tb.Theorem 22.
The functionϕv=∑a∈Nvara1/infara does not fulfill the Fatou property, for all v∈ℓr, if ra∈0,1N with infara>0.Proof.
Suppose the set-up be confirmed andtb⊆ℓrϕ with limb⟶∞ϕtb−t=0. Since the space ℓrϕ is a prequasi closed space, then t∈ℓrϕ. Then, for each v∈ℓrϕ, we get
(11)ϕv−t=∑a∈Nva−tara1/infara≤21/infara−1∑a∈Nva−tabra1/infara+∑a∈Ntab−tara1/infara≤21/infara−1supjinfb≥jϕv−tb.So,ϕ does not indulge the Fatou property.


## 5. Kannan Prequasiϕ-Contraction Operator

Now, we explain the definition of Kannanϕ-contraction mapping on the prequasi normed (sss). We study the sufficient setting on ℓrϕ constructed with definite prequasi norm so that there is one and only one fixed point of Kannan prequasi norm contraction mapping.Definition 23.
An operatorW:Aϕ⟶Aϕ is called a Kannan ϕ-contraction, if there is ξ∈0,1/2, so that ϕWv−Wt≤ξϕWv−v+ϕWt−t, for all v,t∈Aϕ.
An elementv∈Aϕ is named a fixed point of W, if Wv=v.Theorem 24.
Assumera∈0,1N be an increasing, and W:ℓrϕ⟶ℓrϕ be Kannan ϕ-contraction mapping, where ϕv=∑a∈Nvara, for all v∈ℓr, then W has one fixed point.Proof.
Let the setting be satisfied. For eachv∈ℓr, then Wpv∈ℓr. As W is a Kannan ϕ-contraction operator, one has
(12)ϕWp+1v−Wpv≤ξϕWp+1v−Wpv+ϕWpv−Wp−1v⇒ϕWp+1v−Wpv≤ξ1−ξϕWpv−Wp−1v≤ξ1−ξ2ϕWp−1v−Wp−2v≤≤ξ1−ξpϕWv−v.So, for allp,q∈N with q>p, one can see
(13)ϕWpv−Wqv≤ξϕWpv−Wp−1v+ϕWqv−Wq−1v≤ξξ1−ξp−1+ξ1−ξq−1ϕWv−v.Therefore,Wpv is a Cauchy sequence in ℓrϕ. As the space ℓrϕ is prequasi Banach space. Hence, there is t∈ℓrϕ so that limp⟶∞Wpv=t. To prove that Wt=t. Since ϕ has the Fatou property, we have
(14)ϕWt−t≤supiinfp≥iϕWp+1v−Wpv≤supiinfp≥iξ1−ξpϕWv−v=0,hence, Wt=t. Then, t is a fixed point of W. To show that the fixed point is unique. Let we have two distinctive fixed points b,t∈ℓrϕ of W. So, we have
(15)ϕb−t≤ϕWb−Wt≤ξϕWb−b+ϕWt−t=0.Therefore,b=t.Corollary 25.
Letra∈0,1N be an increasing, and W:ℓrϕ⟶ℓrϕ be Kannan ϕ-contraction mapping, where ϕv=∑a∈Nvara, for all v∈ℓr, then W has unique fixed point b with
(16)ϕWpv−b≤ξξ1−ξp−1ϕWv−v.Proof.
Pick up the conditions be satisfied. By Theorem24, we have a unique fixed point b of W. Hence, one has
(17)ϕWpv−b=ϕWpv−Wb≤ξϕWpv−Wp−1v+ϕWb−b=ξξ1−ξp−1ϕWv−v.Definition 26.
SupposeAϕ be a prequasi normed (sss), W:Aϕ⟶Aϕ and b∈Aϕ. The operator W is called ϕ-sequentially continuous at b, if and only if, when lima⟶∞ϕva−b=0, then lima⟶∞ϕWva−Wb=0.Theorem 27.
Pick upra∈0,1N with infara>0, and W:ℓrϕ⟶ℓrϕ, where ϕv=∑a∈Nvara1/infara, for all v∈ℓr. The point g∈ℓrϕ is the only fixed point of W, if the following conditions are satisfied:
(a)
W is Kannan ϕ-contraction mapping(b)
W is ϕ-sequentially continuous at g∈ℓrϕ(c)
There isv∈ℓrϕ so that the sequence of iterates Wpv has a subsequence Wpiv converging to gProof.
Let the set-up be verified. Supposeg be not a fixed point of W, then Wg≠g. By the set-up (b) and (c), we have
(18)limpi⟶∞ϕWpiv−g=0andlimpi⟶∞ϕWpi+1v−Wg=0.As the operatorW is Kannan ϕ-contraction, one has
(19)0<ϕWg−g=ϕWg−Wpi+1v+Wpiv−g+Wpi+1v−Wpiv≤22/infara−2ϕWpi+1v−Wg+22/infara−2ϕWpiv−g+21/infara−1ξξ1−ξpi−1ϕWv−v.Aspi⟶∞, we have a contradiction. Therefore, g is a fixed point of W. To prove that the fixed point g is unique. Assume we have two different fixed points g,b∈ℓrϕ of W. So, one can see
(20)ϕg−b≤ϕWg−Wb≤ξϕWg−g+ϕWb−b=0.Therefore,g=b.Example 28.
LetW:ℓa+1/2a+4a=0∞ϕ⟶ℓa+1/2a+4a=0∞ϕ, where ϕv=∑a∈Nvaa+1/2a+4, for all v∈ℓa+1/2a+4a=0∞ and
(21)Wv=v18,ϕv∈0,1,v20,ϕv∈1,∞.Since for allv1,v2∈ℓa+1/2a+4a=0∞ϕ with ϕv1,ϕv2∈0,1, we have
(22)ϕWv1−Wv2=ϕv118−v218≤1174ϕ17v118+ϕ17v218=1174ϕWv1−v1+ϕWv2−v2.For allv1,v2∈ℓa+1/2a+4a=0∞ϕ with ϕv1,ϕv2∈1,∞, we have
(23)ϕWv1−Wv2=ϕv120−v220≤1194ϕ19v120+ϕ19v220=1194ϕWv1−v1+ϕWv2−v2.For allv1,v2∈ℓa+1/2a+4a=0∞ϕ with ϕv1∈0,1 and ϕv2∈1,∞, we have
(24)ϕWv1−Wv2=ϕv118−v220≤1174ϕ17v118+1194ϕ19v220≤1174ϕ17v118+ϕ19v220=1174ϕWv1−v1+ϕWv2−v2.Therefore, the mapW is Kannan ϕ-contraction mapping. Since ϕ satisfies the Fatou property. By Theorem 24, the map W has a unique fixed point θ∈ℓa+1/2a+4a=0∞ϕ.Letvn⊆ℓa+1/2a+4a=0∞ϕ be such that limn⟶∞ϕvn−v0=0, where v0∈ℓa+1/2a+4a=0∞ϕ with ϕv0=1. Since the prequasi norm ϕ is continuous, we have
(25)limn⟶∞ϕWvn−Wv0=limn⟶∞ϕvn18−v020=ϕv0180>0.Hence,W is not ϕ-sequentially continuous at v0. So, the map W is not continuous at v0.Ifϕv=∑a∈Nvaa+1/2a+44, for all v∈ℓa+1/2a+4a=0∞. Since for all v1,v2∈ℓa+1/2a+4a=0∞ϕ with ϕv1,ϕv2∈0,1, we have
(26)ϕWv1−Wv2=ϕv118−v218≤817ϕ17v118+ϕ17v218=817ϕWv1−v1+ϕWv2−v2.For allv1,v2∈ℓa+1/2a+4a=0∞ϕ with ϕv1,ϕv2∈1,∞, we have
(27)ϕWv1−Wv2=ϕv120−v220≤819ϕ19v120+ϕ19v220=819ϕWv1−v1+ϕWv2−v2.For allv1,v2∈ℓa+1/2a+4a=0∞ϕ with ϕv1∈0,1 and ϕv2∈1,∞, we have
(28)ϕWv1−Wv2=ϕv118−v220≤817ϕ17v118+819ϕ19v220≤817ϕ17v118+ϕ19v220=817ϕWv1−v1+ϕWv2−v2.Therefore, the mapW is Kannan ϕ-contraction mapping and Wpv=v/18p,ϕv∈0,1,v/20p,ϕv∈1,∞.It is clear thatW is ϕ-sequentially continuous at θ∈ℓa+1/2a+4a=0∞ϕ and Wpv has a subsequence Wpiv converging to θ. By Theorem 27, the point θ∈ℓa+1/2a+4a=0∞ϕ is the only fixed point of W.Example 29.
LetW:ℓa+1/2a+4a=0∞ϕ⟶ℓa+1/2a+4a=0∞ϕ, where ϕv=∑a∈Nvaa+1/2a+44, for allv∈ℓa+1/2a+4a=0∞ and
(29)Wv=1181+v0,v1,v2,⋯,v0∈−∞,117,1171,0,0,0,⋯,v0=117,1181,0,0,0,⋯,v0∈117,∞.Since for allv,t∈ℓa+1/2a+4a=0∞ϕ with v0,t0∈−∞,1/17, we have
(30)ϕWv−Wt=ϕ118v0−t0,v1−t1,v2−t2,⋯≤817ϕ17v18+ϕ17t18≤817ϕWv−v+ϕWt−t.For allv,t∈ℓa+1/2a+4a=0∞ϕ with v0,t0∈1/17,∞, then for any ε>0, we have
(31)ϕWv−Wt=0≤εϕWv−v+ϕWt−t.For allv,t∈ℓa+1/2a+4a=0∞ϕ with v0∈−∞,1/17 and t0∈1/17,∞, we have
(32)ϕWv−Wt=ϕv18≤117ϕ17v18=117ϕWv−v≤117ϕWv−v+ϕWt−t.Therefore, the mapW is Kannan ϕ-contraction mapping. It is clear that W is ϕ-sequentially continuous at 1/17e0∈ℓa+1/2a+4a=0∞ϕ and there is v∈ℓa+1/2a+4a=0∞ϕ with v0∈−∞,1/17 such that the sequence of iterates Wpv=∑n=1p1/18ne0+1/18pv has a subsequence Wpiv=∑n=1pi1/18ne0+1/18piv converging to 1/17e0. Then, W has one fixed point 1/17e0∈ℓa+1/2a+4a=0∞ϕ. Note that W is not continuous at 1/17e0∈ℓa+1/2a+4a=0∞ϕ.


## 6. Kannan Contraction Maps on Prequasi Ideal

We account the being present of a fixed point of Kannan prequasi norm contraction operator on the prequasi Banach operator ideal investigated byℓrϕ and s− numbers.Theorem 30.
LetZ and M be Banach spaces, and ra∈0,1N be an increasing, then Sℓrϕ,Φ, where ΦW=ϕsaWa=0∞ be a prequasi Banach operator ideal.Proof.
Pick up the conditions be verified. By Theorem13, the space ℓrϕ is a premodular (sss). Therefore, from Theorem 9, one has ΦW=ϕsaWa=0∞ is a prequasi norm on Sℓrϕ. So, from Theorem 10, we obtain the space Sℓrϕ,Φ is a prequasi Banach operator ideal.Theorem 31.
Pick upZ and M be Banach spaces, and ra∈0,1N be an increasing, then Sℓrϕ,Φ, where ΦW=ϕsaWa=0∞ be a prequasi closed operator ideal.Proof.
By Theorem13, the space ℓrϕ is a premodular (sss). Therefore, from Theorem 9, we have ΦW=ϕsaWa=0∞ is a prequasi norm on Sℓrϕ. Assume Wq∈SℓrϕZ,M, for each q∈N and limq⟶∞ΦWq−W=0. Hence, we have ς>0 and since LZ,M⊇SℓrϕZ,M, we have
(33)ΦWq−W=ϕsaWq−Wa=0∞≥ϕs0Wq−W,0,0,0,⋯=ϕWq−W,0,0,0,⋯≥ςWq−W.SoWqq∈N is convergent in LZ,M, i.e., limq⟶∞Wq−W=0 and as saWqa=0∞∈ℓrϕ, for every q∈N and ℓrϕ is a premodular (sss). Therefore, we get
(34)ΦW=ϕsaWa=0∞=ϕsaW−Wq+Wqa=0∞≤ϕsa/2W−Wqa=0∞+ϕsa/2Wqa=0∞≤ϕWq−Wa=0∞+2ϕsaWqa=0∞<ε,we havesaWa=0∞∈ℓrϕ, so W∈SℓrϕZ,M.Definition 32.
A prequasi normΦon the idealSAϕ, whereΦW=ϕsaWa=0∞, provides the Fatou property if for every sequenceWaa∈N⊆SAϕZ,Mwithlima⟶∞ΦWa−W=0and allV∈SAϕZ,M, then(35)ΦV−W≤supainfi≥aΦV−Wi.Theorem 33.
The prequasi normΦW=∑a∈NsaWra, for allW∈SℓrϕZ,Mdoes not satisfy the Fatou property, ifra∈0,1Nis increasing.Proof.
Let the setting be provided andWpp∈N⊆SℓrϕZ,M with limp⟶∞ΦWp−W=0. Since the space Sℓrϕ is a prequasi closed ideal, so W∈SℓrϕZ,M. Therefore, for every V∈SℓrϕZ,M, we have
(36)ΦV−W=∑a∈NsaV−Wra≤∑a∈Nsa/2V−Wira+∑a∈Nsa/2Wi−Wra≤2suppinfi≥p∑a∈NsaV−Wira.Hence,Φ does not support the Fatou property.Now, we introduce the definition of KannanΦ-contraction operator on the prequasi operator ideal.Definition 34.
For the prequasi normΦon the idealSAϕ, whereΦW=ϕsaWa=0∞. An operatorG:SAϕZ,M⟶SAϕZ,Mis called a KannanΦ-contraction, if we haveξ∈0,1/2so thatΦGW−GA≤ξΦGW−W+ΦGA−A,for allW,A∈SAϕZ,M.Definition 35.
For the prequasi normΦon the idealSAϕ, whereΦW=ϕsaWa=0∞, G:SAϕZ,M⟶SAϕZ,MandB∈SAϕZ,M.The operatorGis calledΦ-sequentially continuous atB, if and only if, whenlimp⟶∞ΦWp−B=0,thenlimp⟶∞ΦGWp−GB=0.Theorem 36.
Set upra∈0,1Nbe an increasing andG:SℓrϕZ,M⟶SℓrϕZ,M, whereΦW=∑a∈NsaWra, for everyW∈SℓrϕZ,M. The pointA∈SℓrϕZ,Mis the unique fixed point ofG, if the following set up are satisfied:(a)
G is Kannan Φ-contraction mapping(b)
G is Φ-sequentially continuous at a point A∈SℓrϕZ,M(c)
There isB∈SℓrϕZ,M such that the sequence of iterates GpB has a subsequence GpiB converging to AProof.
Let the conditions be verified. IfA is not a fixed point of G, then GA≠A. From the setting (b) and (c), we have
(37)limpi⟶∞ΦGpiB−A=0andlimpi⟶∞ΦGpi+1B−GA=0.SinceG is Kannan Φ-contraction mapping, one can see
(38)0<ΦGA−A=ΦGA−Gpi+1B+GpiB−A+Gpi+1B−GpiB≤2ΦGpi+1B−GA+4ΦGpiB−A+4ξξ1−ξpi−1ΦGB−B.Aspi⟶∞, this implies a contradiction. Therefore, A is a fixed point of G. To show that the fixed point A is unique. Let we have two different fixed points A,D∈SℓrϕZ,M of G. Hence, one has
(39)ΦA−D≤ΦGA−GD≤ξΦGA−A+ΦGD−D=0.Therefore,A=D.Example 37.
LetZandMbe Banach spaces, G:Sℓa+1/a+2a=0∞ϕZ,M⟶Sℓa+1/a+2a=0∞ϕZ,M, whereΦW=∑a∈NsaWa+1/a+2, for everyW∈Sℓa+1/a+2a=0∞ϕZ,Mand(40)GW=W26,ΦW∈0,1,W37,ΦW∈1,∞.Since for allW1,W2∈Sℓa+1/a+2a=0∞ϕ with ΦW1,ΦW2∈0,1, we have
(41)ΦGW1−GW2=ΦW126−W226≤25Φ25W126+Φ25W226=25ΦGW1−W1+ΦGW2−W2.For allW1,W2∈Sℓa+1/a+2a=0∞ϕ with ΦW1,ΦW2∈1,∞, we have
(42)ΦGW1−GW2=ΦW137−W237≤13Φ36W137+Φ36W237=13ΦGW1−W1+ΦGW2−W2.For allW1,W2∈Sℓa+1/a+2a=0∞ϕ with ΦW1∈0,1 and ΦW2∈1,∞, we have
(43)ΦGW1−GW2=ΦW126−W237≤25Φ25W126+13Φ36W237≤25Φ25W126+Φ36W237=25ΦGW1−W1+ΦGW2−W2.Therefore, the mapW is Kannan Φ-contraction mapping and GpW=W/26p,ΦW∈0,1,W/37p,ΦW∈1,∞.It is clear thatG is Φ-sequentially continuous at the zero operator Θ∈Sℓa+1/a+2a=0∞ϕ and GpW has a subsequence GpiW converging to Θ. By Theorem 36, the zero operator Θ∈Sℓa+1/a+2a=0∞ϕ is the only fixed point of G. Let Wn⊆Sℓa+1/a+2a=0∞ϕ be such that limn⟶∞ΦWn−W0=0, where W0∈Sℓa+1/a+2a=0∞ϕ with ΦW0=1. Since the prequasi norm Φ is continuous, we have
(44)limn⟶∞ΦGWn−GW0=limn⟶∞ΦWn26−W037=Φ11W0962>0.HenceG is not Φ-sequentially continuous at W0. So, the map G is not continuous at W0.


## 7. Application to the Existence of Solutions of Summable Equations

Summable equations like (45) were studied by Salimi et al. [22], Agarwal et al. [23], and Hussain et al. [24]. In this section, we search for a solution to (45) in ℓrϕ, where ra∈0,1N be an increasing and ϕv=∑a∈Nvara, for all v∈ℓr. Consider the summable equations
(45)va=pa+∑m=0∞Aa,mfm,vm,and let W:ℓrϕ⟶ℓrϕ defined by
(46)Wvaa∈N=pa+∑m=0∞Aa,mfm,vma∈N.Theorem 38.
The summable equations ((45)) has a solution inℓrϕ,ifA:N2⟶R,f:N×R⟶R,p:N⟶R,and for alla∈N, there isξ∈0,1/2, so that(47)∑m∈NAa,mfm,vm−fm,tmra≤ξpa−va+∑m=0∞Aa,mfm,vmra+pa−ta+∑m=0∞Aa,mfm,tmra.Proof.
Let the conditions be verified. Consider the mappingW:ℓrϕ⟶ℓrϕ defined by (46). We have
(48)ϕWv−Wt=∑a∈NWva−Wtara=∑a∈N∑m∈NAa,mfm,vm−fm,tmra≤ξ∑a∈Npa−va+∑m=0∞Aa,mfm,vmra+∑a∈Npa−ta+∑m=0∞Aa,mfm,tmra=ξϕWv−v+ϕWt−t.Then, from Theorem24, we have a solution of equation (45) in ℓrϕ.Example 39.
Given the sequence spaceℓa+1/a+2a=0∞ϕ, whereϕv=∑a∈Nvaa+1/a+2, for allv∈ℓa+1/a+2a=0∞. Consider the summable equations(49)va=e−3a+6+∑m=0∞−1a+mvaa2+m!+1q,
whereq>2 and let W:ℓa+1/a+2a=0∞ϕ⟶ℓa+1/a+2a=0∞ϕ defined by
(50)Wvaa∈N=e−3a+6+∑m=0∞−1a+mvaa2+m!+1qa∈N.It is easy to see that(51)∑m=0∞−1avaa2+m!+1q−1m−−1ma+1/a+2≤13e−3a+6−va+∑m=0∞−1a+mvaa2+m!+1qa+1/a+2+e−3a+6−ta+∑m=0∞−1a+mvaa2+m!+1qa+1/a+2.By Theorem38, the summable equations (49) has a solution in ℓa+1/a+2a=0∞ϕ.



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