Abstract.

We introduce an iterative technique with an inertial term that converges strongly to a fixed point of mappings satisfying Condition (E). Our results extend existing work by providing a robust numerical method for solving fixed point problems in Banach spaces. To demonstrate the effectiveness of our approach, we present numerical examples of a mapping that is not nonexpansive but satisfies Condition (E). Furthermore, we illustrate the convergence behavior of our algorithm for different choices of initial guesses and coefficients, using MATLAB to validate the theoretical results. This work contributes to the broader framework of fixed point theory and offers practical insights for solving nonlinear problems in applied mathematics.

keywords:
condition (E); uniformly convex Banach space; strong convergence.
MSC:
47H09; 47H10; 54A20.

1. Introduction

Fixed point theory is a fundamental area of nonlinear analysis, offering powerful tools to address a wide range of problems in mathematics and its applications. It plays a pivotal role in optimization, variational inequalities, differential equations, and many other fields. In this study, we explore the fixed point problem within the framework of Banach spaces, a setting that provides a rich structure for investigating operator properties and convergence behaviors. Let be a nonempty, closed, and convex subset of a real Banach space Σ, equipped with its dual space Σ. Central to our analysis is the normalized duality mapping J:Σ2Σ, defined such that for any vΣ, it satisfies

v,χ=v2=χ2,

where , denotes the duality pairing [8]. This mapping becomes single-valued when Σ is strictly convex, a property we leverage throughout this work.

Banach spaces exhibit diverse geometric characteristics, such as uniform convexity and smoothness, which significantly influence the behavior of iterative methods. A Banach space Σ is uniformly convex if sequences on the unit sphere satisfying a specific limit condition collapse to the same point [1, 24]. Similarly, uniform smoothness is characterized via the modulus of smoothness ρΣ(τ), with Σ being uniformly smooth if ρΣ(τ)/τ0 as τ0. A refined notion, q-uniform smoothness, imposes a bound ρΣ(τ)Kqτq for some q>1 and constant Kq>0, implying uniform smoothness. These properties underpin our investigation of mappings Φ:, particularly those that are L-Lipschitzian, where

Φ(v)Φ(ϑ)Lvϑfor all v,ϑ.

When L=1, the mapping is nonexpansive, and various methods have been explored to numerically solve the fixed point equation v=Φ(v), assuming that the fixed point set (Φ)={vv=Φ(v)} is nonempty.

The simplest approach to finding fixed points, known as the Banach-Picard iteration, involves the recursive formula vm+1=Φ(vm). For contraction mappings (L<1), this method guarantees strong convergence to a unique fixed point, as established by the Banach-Picard theorem. However, when Φ is merely nonexpansive, convergence is not assured, as demonstrated by the counterexample Φ=Id with a nonzero starting point. To address this limitation, Krasnoselskii introduced an averaged iteration, applying the Banach-Picard scheme to the operator (1/2)Id+(1/2)Φ [14]. This evolved into the Krasnoselskii-Mann iteration, given by

vm+1=(1ηm)vm+ηmΦ(vm),

where {ηm}(0,1) is a control sequence. Under mild conditions on {ηm} and the existence of fixed points, this iteration yields weak convergence to a point in (Φ) [13, 7]. Despite its success, weak convergence remains a limitation, especially in Hilbert spaces, prompting further refinements [12].

A key step in establishing the convergence of the iterates in (1.3) involves demonstrating that vmΦ(vm)0 as m+, as shown by Browder and Petryshyn in [18] for the constant case where ηmη(0,1). Subsequently, the weak convergence of these iterates was investigated across various contexts in [18, 17].

It is worth noting that, even in real Hilbert spaces, all prior adaptations of the Krasnoselskii-Mann method for nonexpansive mappings yield only weak convergence; additional details can be found in [19].

Recently, Bot et al. [5] introduced a novel formulation of Mann’s method to tackle these challenges. Given an arbitrary v0 in a real Hilbert space , for all m0, the iteration is defined as:

vm+1=ηmvm+ζm(Φ(ηmvm)ηmvm).

They demonstrated that the sequence of iterates in (1.4) exhibits strong convergence.

Recent advancements, such as those by [5] and Baewnoi et al. [4], have sought to achieve strong convergence. Baewnoi et al. demonstrated that an inertial-enhanced algorithm in uniformly convex Banach spaces with uniformly Gâteaux differentiable norms ensures strong convergence to fixed points of nonexpansive mappings. Building on this, Suzuki [23] introduced a broader class of mappings satisfying condition (C), later generalized by Garcia-Falset as condition (E), offering new avenues for analysis.

A mapping Φ: is said to satisfy condition (C) (also known as Suzuki-generalized nonexpansiveness) if for all v,ϑ,

12vΦ(v)vϑΦ(v)Φ(ϑ)vϑ.

This condition generalizes the concept of nonexpansiveness and has been widely studied due to its applicability in fixed point theory [23]. García-Falset [11] considered the class of mapping satisfying condition (E). A mapping Φ: satisfies condition (E) if there exists μ1 such that for all v,ϑ,

vΦ(ϑ)μvΦ(v)+vϑ.

This condition extends the class of mappings that can be analyzed using fixed point methods, providing a broader framework for solving nonlinear problems in Banach spaces. For further insights, we refer the reader to some interesting results derived from condition (E) and the related literature cited therein [22, 21]. Recently, Paimsang and Thianwan [15] introduced modified Noor iterative scheme and proved strong and weak convergence results for the mappings satisfying condition (E).

In this paper, we extend the results of Baewnoi et al. [4] to the more general class of Suzuki-generalized nonexpansive mappings and those satisfying condition (E). We propose a numerical method for solving the fixed point problem and establish a strong convergence theorem under suitable conditions. Additionally, we provide an illustrative example and use MATLAB to analyze the convergence behavior, contributing to the ongoing development of robust iterative techniques in Banach spaces.

2. Preliminaries

Throughout this paper, the space Σ is a real uniformly convex Banach space (RUCBS, in short), for the mapping Φ: we denote the set of fixed points of the mapping Φ by F(Φ).

Definition 2.1.

A mapping Φ:ΣΣ is said to be

  1. (a)

    monotone, if

    Φ(ζ)Φ(η),ζη0,ζ,ηΣ;
  2. (b)

    pseudomonotone, if

    Φ(ζ),ηζ0Φ(η),ηζ0,ζ,ηΣ;
  3. (c)

    contraction, if a constant k(0,1) such that

    Φ(ζ)Φ(η)kζη,ζ,ηΣ;
  4. (d)

    L-Lipschitz continuous, if

    Φ(ζ)Φ(η)Lζη,ζ,ηΣ;
  5. (e)

    nonexpansive mapping if

    Φ(ζ)Φ(η)ζη,ζ,ηΣ;
  6. (f)

    mapping satisfying Condition (C) if

    12ζΦ(ζ)ζηΦ(ζ)Φ(η)ζη,ζ,ηΣ;
Lemma 2.2 ([25]).

Suppose that Σ is a RUCBS. For any ν>0, Ξν(0)={μΣ:μν} and β[0,1] then there exists a continuous, convex and strictly increasing function r:[0,2ν],r(0)=0 such that

βμ+(1β)ϑ2βμ2+(1β)ϑ2β(1β)r(μϑ).
Lemma 2.3 ([9]).

Suppose Σ is a real normed linear space, then for any μ,ϑΣ, j(μ+ϑ)J(μ+ϑ), then the following holds

μ+ϑ2μ2+2ϑ,j(μ+ϑ).
Definition 2.4.

Suppose that J is a normalized duality mapping on a real Banach space 𝒳, J:𝒳2𝒳 defined as J(x)={h𝒳:x,h=xh,h=x}, for all x𝒳, where 𝒳 is dual space of 𝒳 and the pair , denotes the generalized duality pairing. If the dual space 𝒳 is strictly convex, then the normalized duality J is single valued, and the single valued duality mapping is denoted by j.

Definition 2.5 ([2]).

Let Σ be a Banach space and SΣ={ζΣ:ζ=1} the unit sphere of Σ. Then

  1. (1)

    the norm of Σ is said to be Gâteaux differentiable at point ζSΣ if for ηSΣ

    limt0ζ+tηζt, (1)

    exists. The norm of Σ is said to be Gâteaux differentiable if it is Gâteaux differentiable at each point of SΣ.

  2. (2)

    the Banach space Σ is said to be smooth if the limit (1) exists for all ζ,ηSΣ.

  3. (3)

    the norm of the Banach space Σ is said to be Frećhet differentiable if for each ζ,ηSΣ, the limit (1) exists uniformly.

Lemma 2.6 ([6]).

Suppose Σ is a uniformly convex Banach space and is a closed, and convex subset of Σ. Suppose that Φ: is a mapping with F(Φ) and {ζn} is a sequence in such that {ζn} converges weakly to ζ and limnζnΦ(ζn)=d then limnζΦ(ζ)=d.

Lemma 2.7 ([26]).

Suppose (μ0,μ1,μ2,), is so that Λnμn0 for all Banach limits Λ. If lim supn(μn+1μn)0, then lim supnμn0.

Lemma 2.8 ([20]).

Suppose {τn} is a sequence of nonnegative real numbers such that

τn+1(1ρn)τn+ρnσn+δn,n1.

If

  1. (1)

    {ρn}[0,1],ρn=,lim supσn0,

  2. (2)

    for any n0,δn0,δn<,

then limnτn=0.

Lemma 2.9.

Let Φ be a mapping on a subset of a Banach space Σ. Assume that Φ satisfies the condition (C). Then

χΦ(ϑ)3Φ(χ)χ+χϑ

holds for all χ,ϑ.

Definition 2.10 ([16]).

Let be a nonempty subset of a Banach space Σ. A mapping Φ: is called a generalized α-Reich-Suzuki nonexpansive mapping if, for some α[0,1), the following condition holds:

12χΦ(χ)χϑ implies Φ(χ)TΦ(ϑ)max{P(χ,ϑ),Q(χ,ϑ)}

for all χ,ϑ, where

P(χ,ϑ)=αΦ(χ)χ+αΦ(ϑ)ϑ+(12α)χϑ
Q(χ,ϑ)=αΦ(χ)ϑ+αΦ(ϑ)χ+(12α)χϑ.
Definition 2.11 ([3]).

Let be a nonempty subset of a Banach space Σ. A mapping T: is called a generalized contraction of Suzuki type if there exist λ(0,1) and α1,α2,α3[0,1] where α1+2α2+2α3=1 such that for all x,yX

12χΦ(χ)χϑΦ(χ)Φ(ϑ) α1χϑ+α2[Φ(χ)χ+Φ(ϑ)ϑ]
+α3[Φ(ϑ)χ+Φ(χ)ϑ].
Proposition 2.12 ([11]).

Let be a nonempty subset of a Banach space Σ. If Φ: is a mapping satisfying condition (E) with F(Φ) then Φ is quasi-nonexpansive.

3. Main Results

Theorem 3.1.

Suppose is a convex and closed subset of a RUCBS Σ having uniformly G\a^ateaux differentiable norm, and Φ: is a mapping satisfying condition (E) with F(Φ). Assume the following conditions satisfied:

  1. (a)

    limnωn=0,limnρn=0,n=1ρn=,

  2. (b)

    υn0,n and n=1υn<,

  3. (c)

    ωn,ρn(0,1),γn[a,b](0,1).

Suppose χ0,χ1 then for all n1 the sequence {χn} generated by

{ϑn=χn+υn(χnχn1),ξn=(1ωn)(1ρn)ϑn,χn+1=(1γn)ξn+γnΦ(ξn), (2)

converges strongly to a point χF(Φ).

Proof 3.2.

Suppose χF(Φ), using (2), we have

χn+1χ =(1γn)ξn+γnΦ(ξn)χ
=γn(Φ(ξn)χ)+(1γn)(ξnχ)
γnΦ(ξn)χ+(1γn)ξnχ.

Using Proposition 2.12, we get

χn+1χ γnξnχ+(1γn)ξnχ
=ξnχ. (3)

Let (1ρn)ϑn=ϖn, we have

ξnχ =(1ωn)(1ρn)ϑnχ=(1ωn)ϖnχ
=(1ωn)(ϖnχ)ωnχ
(1ωn)ϖnχ+ωnχ
(1ωn)(1ρn)ϑnχ+ωnχ
=(1ωn)(1ρn)(ϑnχ)ρnχ+ωnχ
(1ωn)[(1ρn)ϑnχ+ρnχ]+ωnχ
=(1ωn)(1ρn)ϑnχ+(1ωn)ρnχ+ωnχ
(1ωn)(1ρn)ϑnχ+(1ωn)χ+ωnχ
(1ρn)ϑnχ+χ
=(1ρn)χn+υn(χnχn1)χ+χ
=(1ρn)(χnχ)+υn(χnχn1)+χ
(1ρn)χnχ+(1ρn)υnχnχn1+χ
max{χnχ,χnχn1,χ}.

From here we can say that the sequence {χn} is bounded and hence {ϑn}, {ξn} are also bounded.

Since (1ρn)ϑn=ϖn then ξn=(1ωn)ϖn, using Lemma 2.2 and Lemma 2.3, we get

χn+1χ2 =(1γn)ξn+γnΦ(ξn)χ2
(1γn)ξnχ2+γnΦ(ξn)χ2γn(1γn)r(Φ(ξn)ξn)
(1γn)ξnχ2+γnξnχ2γn(1γn)r(Φ(ξn)ξn)
=ξnχ2γn(1γn)r(Φ(ξn)ξn)
=(1ωn)ϖnχ2γn(1γn)r(Φ(ξn)ξn)
ϖnχ2+2ωnϖnχ,j(ξnχ)γn(1γn)r(Φ(ξn)ξn)
ϑnχ2+2ρnϑnχ,j(ϖnχ)+2ωnϖnχ,j(ξnχ)
γn(1γn)r(Φ(ξn)ξn)
χnχ2+2υnχnχ,j(ϑnχ)+2ρnϑnχ,j(ϖnχ)
+2ωnϖnχ,j(ξnχ)γn(1γn)r(Φ(ξn)ξn).

Simplifying the above equation we get

γn(1γn)r(Φ(ξn)ξn) χnχ2χn+1χ2+2υnχnχ,j(ϑnχ)
+2ρnϑnχ,j(ϖnχ)+2ωnϖnχ,j(ξnχ). (4)

Since the sequences {χn}, {ϑn}, {ϖn} and {ξn} are bounded and if we take Γ1,Γ2,Γ3>0 then for all n1

{χnχ,j(ϑnχ)Γ1,ϑnχ,j(ϖnχ)Γ2,ϖnχ,j(ξnχ)Γ3. (5)

Applying (5) in (3.2), we get

γn(1γn)r(Φ(ξn)ξn) χnχ2χn+1χ2+2υnΓ1+2ρnΓ2+2ωnΓ3. (6)

It gives that the sequence {χn} converges to χ. To prove strong convergence we consider the following cases.

  • Case(1):

    If the sequence {χnχ} is monotonically decreasing. We have

    χn+1χ2χnχ20, as n.

    From (6), we get

    limnγn(1γn)r(Φ(ξn)ξn)=0.

    Since γn[a,b](0,1), using the property of r as given in Lemma 2.2, we get

    limnΦ(ξn)ξn=0. (7)

    Now we get

    limnχn+1ξn=limnγn(Φ(ξn)ξn)=0, (8)

    and using (a), we get

    limnξnϖn=limnωnϖn=0, (9)

    and

    limnϖnϑn=limnρnϑn=0. (10)

    Now, using (9) and (10), we have

    limnξnϑnlimnξnϖn+limnϖnϑn=0. (11)

    Since n=1υnχnχn1<, we have

    limnϑnχn=limnυnχnχn1=0. (12)

    We can also get

    limnξnχnlimnξnϑn+limnϑnχn=0, (13)

    and

    limnχn+1χnlimnχn+1ξn+limnξnχn=0,

    and

    Φ(ϖn)ϖn Φ(ϖn)ξn+ξnϖn
    μξnΦ(ξn)+ξnϖn+ξnϖn.

    Applying limn and using (7), (9), we get

    limnΦ(ϖn)ϖn=0. (14)

    Since the sequence {χn} is bounded so there is a subsequence {χnk} of {χn} such that it converges weakly to χΣ. Using Lemma 2.6, we get χF(Φ).
    We also have

    ξn =(1ωn)ϖn
    =(1ωn)(1ρn)ϑn
    (1ρn)ϑn.

    Thus

    ξnχ2 (1ρn)ϑnχ2
    (1ρn)(ϑnχ)ρnχ2. (15)

    By applying inequality (Case(1):), Lemma 2.3, and n=1υnχnχn1<, we get

    χn+1χ2 =(1γn)(ξnχ)+γn(Φ(ξn)ζ)2
    (1γn)ξnχ2+γnΦ(ξn)χ2
    ξnχ2
    (1ρn)(ϑnχ)ρnχ2
    =(1ρn)ϑnχ2+2ρnχ,j(ϖnχ)
    (1ρn)(χnχ)+υn(χnχn1)2+2ρnχ,j(ϖnχ)
    =(1ρn)χnχ2+2ρnχ,j(ϖnχ). (16)

    Now, we prove that χ,j(ϖnχ)0.

    Define a continuous convex function f from into [0,) by

    f(χ)=lim supnϖnχ for all χ.

    Since is closed and bounded, Σ is uniformly convex Banach space, and f is weakly lower semicontinuous, there exists z such that

    f(z)=min{f(χ):χ}.

    Define

    ^={v:f(v)=minaf(a)}.

    Let z^, we have

    ϖnΦ(z)μΦ(ϖn)ϖn+ϖnz.

    From (14), we have f(Φz)f(z). Since f(z) is the minimum, f(Φz)=f(z) holds. If Φzz, then since f is strictly quasiconvex (see [23, Lemma 2]), we have

    f(z)f(z+Φz2)<max{f(z),f(Φz)}=f(z).

    This is a contradiction. Hence Φ(z)=z. Hence, Φ has a fixed point in ^, so ^(Φ). Without losing the general case, as a particular instance, suppose that z=χ^(Φ). Let δ(0,1) then we can easily get f(χ)f(χδχ). Since f(χ)[0,), we can have [f(χ)]2[f(χδχ)]2. Applying Lemma 2.3 to the expression ϖnχ+δχ2, we get

    ϖnχ+δχ2ϖnχ2+2δχ,j(ϖnχ+δχ).

    Applying lim sup both the sides, we have

    lim supnϖnχ+δχ2 lim supnϖnχ2
    +2lim supnδχ,j(ϖnχ+δχ).

    Using the property of f that f(χ)=lim supnϖnχ we get

    [f(χδχ)]2[f(χ)]2+2lim supnδχ,j(ϖnχ+δχ).

    And

    2lim supnδχ,j(ϖnχ+δχ)[f(χ)]2[f(χδχ)]20.

    It gives us

    χ,j(ϖnχ+δχ)0.

    We can also have

    χ,j(ϖnχ) χ,j(ϖnχ)j(ϖnχ+δχ)
    +χ,j(ϖnχ+δχ)
    χ,j(ϖnχ)j(ϖnχ+δχ). (17)

    Since the normalized duality mapping is norm to weak* uniformly continuous on bounded subsets of Σ. We also have, for fixed n δ0, thus

    χ,j(ϖnχ)j(ϖnχ+δχ)
    χ,j(ϖnχ)χ,j(ϖnχ+δχ)0.

    Thus for each ε>0 there exists ςε>0 such that for all δ(0,ςε)

    χ,j(ϖnχ)χ,j(ϖnχ+δχ)<ε.

    Since ε is arbitrary, using (10) we get

    χ,j(ϖnχ)0.

    Now

    ϖn+1ϖnϖn+1ϑn+1+ϑn+1χn+1+χn+1ξn+ξnϖn.

    Applying above conditions and n, we get

    limnϖn+1ϖn=0.

    Again since the normalized duality mapping is norm to weak* uniformly continuous on bounded subsets of Σ, we get

    limn(χ,j(ϖnχ)χ,j(ϖn+1χ))=0.

    Applying Lemma 2.7, we get

    lim supn(χ,j(ϖnχ)0.

    Thus applying Lemma 2.8 at (Case(1):), we get the sequence {χn} converges to χ.

  • Case(2):

    If the sequence {χnχ} is not monotonically decreasing, let n=χnχ2. Let : defined as

    (n)max{ϑ:ϑn:ϑϑ+1}.

    Here, is a nonincreasing sequence so limn(n)= and (n)(n)+1 nn0 for some sufficient large n0. Now using (6), we get

    γ(n)(1γ(n))r(Φ(ξ(n))ξ(n)) χ(n)χ2χ(n)+1χ2
    +2υ(n)Γ1+2ρ(n)Γ2+2ω(n)Γ3
    =(n)(n)+1
    +2υ(n)Γ1+2ρ(n)Γ2+2ω(n)Γ3
    2υ(n)Γ1+2ρ(n)Γ2+2ω(n)Γ3.

    Since

    limn2υ(n)Γ1+limn2ρ(n)Γ2+limn2ω(n)Γ3=0.

    We can also get

    limnΦ(ξ(n))ξ(n)=0.

    Following the same proof as in Case(a), we can get χ(n)χ as (n) and lim sup(n)χ,j(ϖ(n)χ)0. For all nn0, from (Case(1):), we get

    0χ(n)+1χ2χ(n)χ2ρ(n)[2χ,j(ϖ(n)χ)χ(n)χ2].

    It gives us

    χ(n)χ22χ,j(ϖ(n)χ).

    Since lim sup(n)χ,j(ϖ(n)χ)0, we get

    limnχ(n)χ2=0.

    Thus

    limn(n)=limn(n)+1=0.

    Moreover, for each nn0, it can be easily seen that n(n)+1 if n(n) i.e. (n)<n. Since i>i+1 for (n)+1in. For all nn0, we get

    0nmax{(n),(n)+1}=(n)+1.

    Hence, limnn=0, it gives us that the sequence {χn} converges strongly to χ.

Corollary 3.3.

Suppose is a convex and closed subset of a RUCBS Σ having uniformly G\a^ateaux differentiable norm, and Φ: is a mapping satisfying condition (C) with F(Φ). Assume the following conditions satisfied:

  1. (a)

    limnωn=0,limnρn=0,n=1ρn=,

  2. (b)

    υn0,n and n=1υn<,

  3. (c)

    ωn,ρn(0,1),γn[a,b](0,1).

Suppose χ0,χ1 then n1 the sequence {χn} generated by

{ϑn=χn+υn(χnχn1),ξn=(1ωn)(1ρn)ϑn,χn+1=(1γn)ξn+γnΦ(ξn), (18)

converges strongly to a point χF(Φ).

Corollary 3.4.

Suppose is a convex and closed subset of a RUCBS Σ having uniformly G\a^ateaux differentiable norm, and Φ: is a generalized contraction of Suzuki type mapping with F(Φ). Assume the following conditions satisfied:

  1. (a)

    limnωn=0,limnρn=0,n=1ρn=,

  2. (b)

    υn0,n and n=1υn<,

  3. (c)

    ωn,ρn(0,1),γn[a,b](0,1).

Suppose χ0,χ1 then n1 the sequence {χn} generated by

{ϑn=χn+υn(χnχn1),ξn=(1ωn)(1ρn)ϑn,χn+1=(1γn)ξn+γnΦ(ξn), (19)

converges strongly to a point χF(Φ).

Corollary 3.5.

Suppose is a convex and closed subset of a RUCBS Σ having uniformly G\a^ateaux differentiable norm, and Φ: is a generalized α-Reich-Suzuki nonexpansive mapping with F(Φ). Assume the following conditions satisfied:

  1. (a)

    limnωn=0,limnρn=0,n=1ρn=,

  2. (b)

    υn0,n and n=1υn<,

  3. (c)

    ωn,ρn(0,1),γn[a,b](0,1).

Suppose χ0,χ1 then for all n1 the sequence {χn} generated by

{ϑn=χn+υn(χnχn1),ξn=(1ωn)(1ρn)ϑn,χn+1=(1γn)ξn+γnΦ(ξn), (20)

converges strongly to a point χF(Φ).

4. Examples

Example 4.1.

Define the mapping Φ:[0,1][0,1] by

Φ(ζ)={1ζ, if ζ[0,16),ζ+56, if ζ[16,1].

First, we verify that the mapping Φ satisfies condition (E). We follow these cases

  1. Case (1):

    Let ζ[0,16), then

    12ζΦ(ζ)=12ζ2(26,12],

    for 12ζΦ(ζ)ζη, we must have

    12ζ2ηζ, i.e. 12η, hence η[12,1].

    We have

    Φ(ζ)Φ(η) =(1ζ)(η+56)=|η+6ζ16|<16,

    and

    ζη=|ζη|>13.

    Hence

    12ζΦ(ζ)ζηΦ(ζ)Φ(η)ζη.
  2. Case (2):

    Let ζ[16,1], then

    12ζΦ(ζ)=12ζ+56ζ=1ζ2[0,512],

    for 12ζΦ(ζ)ζη we must have 1ζ2ηζ which gives two possibilities:

    1. Case (a):

      ζ<η, then 1ζ2ηζη1+ζ2η[712,1][16,1]. So

      Φ(ζ)Φ(η)=ζ+56η+56=16ζηζη.

      Hence

      12ζΦ(ζ)ζηΦ(ζ)Φ(η)ζη.
    2. Case (b):

      Let ζ>η, then

      1ζ2ζηη3ζ12η[324,1].

      Since η[0,1], so

      η3ζ12ζ2η+13ζ[13,1].

      So the case is ζ[13,1] and η[0,1]. Since ζ[13,1] and η[16,1] is already included in the Case (a). So let ζ[13,1] and η[0,16), then

      Φ(ζ)Φ(η)=ζ+56(1η)=|ζ+6η16|.

      First, we consider ζ[13,12] and η[0,16], then

      Φ(ζ)Φ(η)=|ζ+6η16|112,

      and

      ζη|1316|=16.

      Hence

      Φ(ζ)Φ(η)ζη.

      Next consider ζ[12,1], η[0,16). Then

      Φ(ζ)Φ(η)16 and ζη13Φ(ζ)Φ(η)ζη.

      Hence

      12ζΦ(ζ)ζηΦ(ζ)Φ(η)ζη.

      Hence the mapping Φ is satisfying Condition (C).

Using Lemma 2.9, we can say that the mapping Φ is satisfying condition (E) for μ=3.

On the other hand if we take ζ=319, η=16

Φ(ζ)Φ(η)=(1319)(16+56)=13684>1114=ζη.

Therefore Φ is not a nonexpansive mapping.

Example 4.2.

Let Σ=2 and ={ζ=(ζ1,ζ2)[0,1]×[0,1]} be a subset of Σ with norm ζ=(ζ1,ζ2)=(|ζ1|2+|ζ2|2)1/2. The mapping Φ: is defined by

Φ(ζ1,ζ2)={(1ζ1,1ζ2),(ζ1,ζ2)[0,12]×[0,1],13(1+ζ1,1+ζ2),(ζ1,ζ2)(12,1]×[0,1].

For this, we consider the following cases:

  1. Case(a):

    ζ,η[0,12]×[0,1]. Then

    ζΦ(η) ζΦ(ζ)+Φ(ζ)Φ(η)
    =ζΦ(ζ)+(|ζ1η1|2+|ζ2η2|2)1/2
    =ζΦ(ζ)+ζη.
  2. Case(b):

    ζ,η(12,1]×[0,1]. Then

    ζΦ(η) ζΦ(ζ)+Φ(ζ)Φ(η)
    =ζΦ(ζ)+13(|ζ1η1|2+|ζ2η2|2)1/2
    ζΦ(ζ)+(|ζ1η1|2+|ζ2η2|2)1/2
    =ζΦ(ζ)+ζη.
  3. Case(c):

    ζ[0,12]×[0,1],η(12,1]×[0,1]. Then

    ζΦ(η) =(ζ1,ζ2)(1+η13,1+η23)
    =(ζ1(1+η13)),(ζ2(1+η23))
    =(3ζ1η113,3ζ2η213)
    =(|3ζ1η113|2+|3ζ2η213|2)1/2,

    and

    ζΦ(ζ)=(|2ζ11|2+|2ζ21|2)1/2.

    Since

    |3ζ1η113||2ζ11+ζ1η1||2ζ11|+|ζ1η1|,
    |3ζ2η213|<|2ζ21+ζ2η2||2ζ21|+|ζ2η2|,

    we have,

    ζΦ(η)ζΦ(ζ)+ζη.

Therefore in all the cases Φ satisfies condition (E).

On the other hand, for ζ=(0,0) and η=(51100,25100), we have

Φ(ζ)Φ(η)=0.766>0.567=ζη.

Therefore Φ is not a nonexpansive mapping.

Now, we present the convergence behaviour of the Algorithm (2) for the Example (4.1). We present the convergence behaviour for the two cases, first we present for the different choices of initial guesses and then for the different choices of υn, ωn, ρn, γn.

[Uncaptioned image]

Figure 1: Convergence behaviour for the different choices of initial guesses.

[Uncaptioned image]

Figure 2: Convergence behaviour for different choices of υn, ωn, ρn, γn.

Now, we present the comparison of Algorithm (2) with the algorithms presented in Baewnoi et al. [4] and Dong et al. [10]. We have taken Example (4.1) for Algorithm (2) and mapping Φ(ζ)=ζ3 for the algorithms presented in Baewnoi et al. [4] and Dong et al. [10].

[Uncaptioned image]

Figure 3: Comparison.

Open Problem

We conclude this paper by proposing the following open question:

Question 4.3.

Can Algorithm (2) be extended to approximate fixed points of quasi-nonexpansive mappings in the setting of Banach spaces?

Acknowledgements.
We would like to say thank to the editor and both the reviewers for reading the manuscript carefully and giving their suggestions which have been useful for the improvement of this paper.
Funding.
This research has not received external funding.
Author Contributions.
Formal analysis, writing - original draft, writing - review & editing, P. P.; supervision, validation, writing - review & editing, R. S. All authors have read and agreed to the published version of the manuscript.

References