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Article

Convergence Criteria of a Three-Step Scheme under the Generalized Lipschitz Condition in Banach Spaces

by
Akanksha Saxena
1,
Jai Prakash Jaiswal
2,*,
Kamal Raj Pardasani
1 and
Ioannis K. Argyros
3,*
1
Department of Mathematics, Maulana Azad National Institute of Technology, Bhopal 462003, India
2
Department of Mathematics, Guru Ghasidas Vishwavidyalaya (A Central University), Bilaspur 495009, India
3
Department of Mathematical Sciences, Cameron University, Lawton, OK 73505, USA
*
Authors to whom correspondence should be addressed.
Mathematics 2022, 10(21), 3946; https://doi.org/10.3390/math10213946
Submission received: 30 September 2022 / Revised: 19 October 2022 / Accepted: 19 October 2022 / Published: 24 October 2022
(This article belongs to the Special Issue Computational Methods in Analysis and Applications, 3rd Edition)

Abstract

:
In the given study, we investigate the three-step NTS’s ball convergence for solving nonlinear operator equations with a convergence order of five in a Banach setting. A nonlinear operator’s first-order derivative is assumed to meet the generalized Lipschitz condition, also known as the κ-average condition. Furthermore, several theorems on the convergence of the same method in Banach spaces are developed with the conditions that the derivative of the operators must satisfy the radius or center-Lipschitz condition with a weak κ-average and that κ is a positive integrable but not necessarily non-decreasing function. This novel approach allows for a more precise convergence analysis even without the requirement for new circumstances. As a result, we broaden the applicability of iterative approaches. The theoretical results are supported further by illuminating examples. The convergence theorem investigates the location of the solution ϵ* and the existence of it. In the end, we achieve weaker sufficient convergence criteria and more specific knowledge on the position of the ϵ* than previous efforts requiring the same computational effort. We obtain the convergence theorems as well as some novel results by applying the results to some specific functions for κ(u). Numerical tests are carried out to corroborate the hypotheses established in this work.

1. Introduction

Let G : Ω χ Y be an operator, i.e., χ and Y are said to be Banach spaces, Ω is a nonempty open convex subset, and G is a nonlinear Fréchet differentiable operator. A popular iterative scheme for resolving the classic equation
G ( x ) = 0 ,
known as Newton’s method, is represented as:
x t + 1 = x t [ G ( x t ) ] 1 G ( x t ) , t 0 .
Newton’s scheme [1] converges quadratically. Kantorovich [2] was the first to investigate this in a Banach space setting. Then, it was re-evaluated in a plethora of papers [1,3,4,5,6,7,8,9,10,11,12].
There have been certain algorithms developed called Newton-like with order three and order four convergence, which do not involve the evaluation of derivatives of second order, given in refs. [13,14,15,16,17]. Despite their speed of convergence, schemes of a larger R-order convergence are frequently not carried out on a regular basis. This is due to the high operational expense. Nevertheless, the technique of increased R-order convergence is employed in stiff applications [2] when fast convergence is necessary. For an application, one can see ref. [18].
The convergence region is critical to the steady behavior of an iterative method from a numerical standpoint. There are two directions of iterative method convergence research: semilocal and local convergence analysis. The semilocal research employs data gathered about a starting point to generate criteria for guaranteeing the convergence. The local one determines the balls relying on the information surrounding ϵ * . A number of authors have investigated the convergence criteria in the local sense for Jarratt-like, Newton-like, Weerakoon-like, and various others in a Banach setup [3,19,20,21,22].
In this paper, we investigate the classical Newton–Traub scheme (NTS) [23]’s ball convergence of order five under the κ -average condition, which is written as:
y t = x t [ G ( x t ) ] 1 G ( x t ) , z t = y t [ G ( x t ) ] 1 G ( y t ) , x t + 1 = z t [ G ( y t ) ] 1 G ( z t ) , t = 0 , 1 , 2 ,
The important feature of scheme (3) is that it is the easiest and most effective five order iterative scheme, requiring only three evaluations of the function G j , per jth iteration, G j , the derivative, with zero evaluations of the G j second derivative, making it mathematically effective. Research on the weakness and/or expansion of the conditions imposed on the underlying operators involved can be found in the literature [3,4,5,6]. Wang [12] created generalized Lipschitz-type conditions to investigate the NTS’s ball convergence, in which a non-decreasing (ND) positive integrable function (P.I.F.) was chosen rather than the normal Lipschitz constant. In the next few years, the results of NTS convergence provided that the operator’s derivative fulfills the center or radius-Lipschitz condition, along with a weak κ -average, as found by Wang and Li [24]. Shakhno [25] explored the convergence criteria of the Secant-type with a two-step approach [2], when the generalized Lipschitz conditions were satisfied by the first-order division differences. Recently, a Newton-type two-step approach to the ball convergence, with a convergence rate of three under generalized Lipschitz conditions was demonstrated by Saxena et al. [10], whose definitions will be used in this article.
Let us revisit the motivational example given by Argyros [3,4]. Consider χ = Y = × × , D = V ¯ ( 0 , 1 ) , and ϵ * = 0 . We describe function H on D with w = ( a , b , c ) t as
H ( w ) = ( e a 1 , e 1 2 b 2 + b , c ) t .
Then, the derivative is found to be
H ( w ) = e a 0 0 0 ( e 1 ) b + 1 0 0 0 1 .
Hence, κ = e 2 , κ 0 = e 1 2 , and κ 0 < κ (see definitions (5) and (6)). So, replacing κ by κ 0 at the denominator provides advantages such as a broader range of initial starters (a larger radius than that in previous research). If κ and κ 0 are not constants, then we may take κ ( u ) = e u 2 , κ 0 ( u ) = ( e 1 ) u 2 , and κ ¯ ( u ) = e 1 ( e 1 ) u 2 (see definitions (7), (8), and (87)).
The fascinating question arises of whether the radius-Lipschitz condition with κ -average and ND of κ is essential for the NTS of the convergence of the fifth-order. In this paper, we derive certain theorems for scheme (3); the abovementioned scientific work in this direction has encouraged and influenced us. Throughout the initial work to explore the ball convergence, generalized Lipschitz conditions were used, where it is significant for enlarging the convergence area without the need for additional assumptions, as well as an estimate of error. The region of uniqueness of solution ϵ * is determined utilizing the center-Lipschitz condition in the latter theorem. A few corollaries are also reported.
The remainder of this section is organized as follows: The definitions for κ -average conditions are found in Section 2. Section 3 and Section 4 discuss its region of uniqueness of the solution and the scheme’s ball convergence, accordingly. The hypothesis that the G’s derivative fulfills the radius and center-Lipschitz continuity condition, i.e., a weak κ -average, notably κ 0 and κ , is improved in Section 5. κ and κ 0 are claimed to be members of a class of P.I.F., which are not strictly ND for the sake of convergence theorems. To demonstrate the significance of the findings, numerical examples are provided.

2. Special and Generalized Lipschitz Conditions

Throughout this context, ( ϵ * , δ ) = { x : | | x ϵ * | | < δ } is a ball.
Definition 1.
The constraint placed on G
| | G ( x ) G ( y λ ) | | κ ( 1 λ ) ( | | x ϵ * | | + | | y ϵ * | | ) ; x , y ( ϵ * , δ ) ,
in which y λ = ϵ * + λ ( y ϵ * ) , 0 λ 1 , in the ball ( ϵ * , δ ) , is commonly referred to as the radius-Lipschitz condition with positive constant κ.
Definition 2.
The constraint imposed on G
| | G ( ϵ * ) G ( x ) | | 2 κ 0 | | ϵ * x | | , f o r e a c h x ( ϵ * , δ )
in the ball ( ϵ * , δ ) , is known as the center-Lipschitz condition with positive constant κ 0 , where κ 0 κ .
In this scenario, replacing κ by κ 0 in the case where κ 0 < κ leads to fewer iterations to attain error tolerance, and the solution’s uniqueness ϵ * has been expanded [3,21].
Definition 3.
When κ is not constant but a part of a positive integrable family, criteria (5) is substituted by
| | G ( x ) G ( y λ ) | | λ ( ρ ( x ) + ρ ( y ) ) ρ ( x ) + ρ ( y ) κ ( u ) d u ; x , y ( ϵ * , δ ) , 0 λ 1 .
Definition 4.
When κ 0 is not constant but a part of a positive integrable family, criteria (6) is substituted by
| | G ( x ) G ( ϵ * ) | | 0 2 ρ ( x ) κ 0 ( u ) d u ; x ( ϵ * , δ ) ,
in which ρ ( x ) = | | ϵ * x | | . Then, we have κ 0 ( u ) κ ( u ) . Simultaneously, the κ-average or generalized Lipschitz conditions are the equivalent ’Lipschitz conditions’. Following this, we consider the following auxiliary lemmas, which will be utilized in what follows.
Lemma 1
([10]). Assume that G is continuously differentiable in ( ϵ * , δ ) and [ G ( ϵ * ) ] 1 exists.
(i) 
If [ G ( ϵ * ) ] 1 G satisfies the radius-Lipschitz condition with κ-average
| | [ G ( ϵ * ) ] 1 ( G ( x ) G ( y λ ) ) | | λ ( ρ ( x ) + ρ ( y ) ) ρ ( x ) + ρ ( y ) κ ( u ) d u ; x , y ( ϵ * , δ ) , 0 λ 1 ,
in which ρ ( x ) = | | x ϵ * | | , y λ = ϵ * + λ ( y ϵ * ) , and κ and κ 0 are positive integrable, then
0 1 | | [ G ( ϵ * ) ] 1 ( G ( x ) G ( y λ ) ) | | ρ ( y ) d λ 0 ρ ( x ) + ρ ( y ) κ ( u ) u ρ ( x ) + ρ ( y ) ρ ( y ) d u .
(ii) 
If [ G ( ϵ * ) ] 1 G satisfies the center-Lipschitz condition with κ 0 -average
| | [ G ( ϵ * ) ] 1 ( G ( x λ ) G ( ϵ * ) ) | | 0 2 λ ρ ( x ) κ 0 ( u ) d u ; x , y ( ϵ * , δ ) , 0 λ 1 ,
in which κ 0 is ND, then we obtain
0 1 | | [ G ( ϵ * ) ] 1 ( G ( x λ ) G ( ϵ * ) ) | | ρ ( x ) d λ 0 2 ρ ( x ) κ 0 ( u ) ρ ( x ) u 2 d u .
Proof. 
It is significantly clear from the Lipschitz conditions (9) and (11), respectively,
0 1 | | [ G ( ϵ * ) ] 1 ( G ( x ) G ( y λ ) ) | | ρ ( y ) d λ 0 1 λ ( ρ ( x ) + ρ ( y ) ) ρ ( x ) + ρ ( y ) κ ( u ) d u ρ ( y ) d λ = 0 ρ ( x ) + ρ ( y ) κ ( u ) u ρ ( x ) + ρ ( y ) ρ ( y ) d u , 0 1 | | [ G ( ϵ * ) ] 1 ( G ( x λ ) G ( ϵ * ) ) | | ρ ( x ) d λ 0 1 0 2 λ ρ ( x ) κ 0 ( u ) d u ρ ( x ) d λ = 0 2 ρ ( x ) κ 0 ( u ) ρ ( x ) u 2 d u ,
where we know the definitions y λ = ϵ * + λ ( y ϵ * ) and x λ = ϵ * + λ ( x ϵ * ) . □
The next lemma is required to prove the main convergence theorems under a weak κ -average to show that certainly, in this scenario, the convergence order will decrease.
Lemma 2
([24]). Assume that the function κ α , described as
κ α ( P ) = P 1 α κ ( P ) ,
is ND for some α with 0 α 1 , where κ and κ 0 are positive integrable. Subsequently, f o r e a c h β 0 , the function φ β , α given by
φ β , α ( P ) = 1 P α + β 0 P u β κ ( u ) d u
is also ND.

3. Ball Convergence

Throughout this section, we prove the existence theorem for the NTS (3) with the radius-Lipschitz continuity condition.
Theorem 1.
Assume that G ( ϵ * ) = 0 , G is continuously differentiable in ( ϵ * , δ ) , [ G ( ϵ * ) ] 1 exists, (7), (8) is satisfied by [ G ( ϵ * ) ] 1 G , and κ and κ 0 are noticed to be ND. Assume that the relation given below is satisfied by δ:
0 2 δ κ 0 ( u ) d u 1 a n d 0 2 δ κ ( u ) u d u 2 δ ( 1 0 2 δ κ 0 ( u ) d u ) 1 .
Subsequently, the NTS (3) converges for all x 0 ( ϵ * , δ ) , and
| | y t ϵ * | | 0 2 ρ ( x t ) κ ( u ) u d u 2 ( 1 0 2 ρ ( x t ) κ 0 ( u ) d u ) C 1 ρ ( x 0 ) ρ ( x t ) 2 ,
| | z t ϵ * | | 0 ρ ( x t ) + ρ ( y t ) κ ( u ) u d u ( ρ ( x t ) + ρ ( y t ) ) ( 1 0 2 ρ ( x t ) κ 0 ( u ) d u ) ρ ( y t ) C 1 2 ρ ( x 0 ) 2 ρ ( x t ) 3 ,
| | x t + 1 ϵ * | | 0 ρ ( y t ) + ρ ( z t ) κ ( u ) u d u ( ρ ( y t ) + ρ ( z t ) ) ( 1 0 2 ρ ( y t ) κ 0 ( u ) d u ) ρ ( z t ) C 1 4 ρ ( x 0 ) 4 ρ ( x t ) 5 ,
where it can be seen that the quantities
C 1 = 0 2 ρ ( x 0 ) κ ( u ) u d u 2 ρ ( x 0 ) ( 1 0 2 ρ ( x 0 ) κ 0 ( u ) d u ) , C 2 = 0 ρ ( x 0 ) + ρ ( y 0 ) κ ( u ) u d u ( ρ ( x 0 ) + ρ ( y 0 ) ) ( 1 0 2 ρ ( x 0 ) κ 0 ( u ) d u ) , C 3 = 0 ρ ( y 0 ) + ρ ( z 0 ) κ ( u ) u d u ( ρ ( y 0 ) + ρ ( z 0 ) ) ( 1 0 2 ρ ( y 0 ) κ 0 ( u ) d u ) ,
are less than 1. Furthermore,
| | x t ϵ * | | C 1 5 t 1 | | x 0 ϵ * | | ; t = 1 , 2 , .
Proof. 
Without loss of generality on selecting x 0 ( ϵ * , δ ) , in which δ fulfills the relation (15), C 1 and C 2 are determined by the inequality (19) to be less than 1. Indeed, because κ is monotone, we obtain
1 f 2 2 0 f 2 1 f 1 2 0 f 1 κ ( u ) u d u = 1 f 2 2 f 1 f 2 + 1 f 2 2 1 f 1 2 0 f 1 κ ( u ) u d u κ ( f 1 ) 1 f 2 2 f 1 f 2 + 1 f 2 2 1 f 1 2 0 f 1 u d u = κ ( f 1 ) 1 f 2 2 0 f 2 1 f 1 2 0 f 1 u d u = 0 ,
for 0 < f 1 < f 2 . Thus, 1 f 2 0 f κ ( u ) u d u is ND with reference to f. Next, we have
C 1 = 0 2 ρ ( x 0 ) κ ( u ) u d u 2 ρ ( x 0 ) 2 ( 1 0 2 ρ ( x 0 ) κ 0 ( u ) d u ) ρ ( x 0 ) 0 2 δ κ ( u ) u d u 2 δ 2 ( 1 0 2 δ κ 0 ( u ) d u ) ρ ( x 0 ) | | x 0 ϵ * | | δ < 1 , C 2 = 0 ρ ( x 0 ) + ρ ( y 0 ) κ ( u ) u d u ( ρ ( x 0 ) + ρ ( y 0 ) ) 2 ( 1 0 2 ρ ( x 0 ) κ 0 ( u ) d u ) ( ρ ( x 0 + ρ ( y 0 ) ) 0 2 δ κ ( u ) u d u 2 δ 2 ( 1 0 2 δ κ 0 ( u ) d u ) ( ρ ( x 0 ) + ρ ( y 0 ) ) | | x 0 ϵ * | | + | | y 0 ϵ * | | 2 δ < 1 , C 3 = 0 ρ ( y 0 ) + ρ ( z 0 ) κ ( u ) u d u ( ρ ( y 0 ) + ρ ( z 0 ) ) 2 ( 1 0 2 ρ ( y 0 ) κ 0 ( u ) d u ) ( ρ ( y 0 + ρ ( z 0 ) ) 0 2 δ κ ( u ) u d u 2 δ 2 ( 1 0 2 δ κ 0 ( u ) d u ) ( ρ ( y 0 ) + ρ ( z 0 ) ) | | x 0 ϵ * | | + | | y 0 ϵ * | | 2 δ < 1 .
Evidently, if x ( ϵ * , δ ) , using the center-Lipschitz condition with the κ -average and the expression (15), we can write
| | [ G ( ϵ * ) ] 1 [ G ( x ) G ( ϵ * ) ] | | 0 2 ρ ( x ) κ 0 ( u ) d u 1 .
Using the Banach Lemma [2] along with the following relation,
| | I ( [ G ( ϵ * ) ] 1 G ( x ) I ) | | 1 = | | [ G ( x ) ] 1 G ( ϵ * ) | | ,
by relation (21), we arrive to the following inequality,
| | [ G ( x ) ] 1 G ( ϵ * ) | | 1 1 0 2 ρ ( x ) κ 0 ( u ) d u .
Next, if x t ( ϵ * , δ ) , we find from expression (3)
| | y t ϵ * | | = | | x t ϵ * [ G ( x t ) ] 1 G ( x t ) | | = | | [ G ( x t ) ] 1 [ G ( x t ) ( x t ϵ * ) G ( x t ) + G ( ϵ * ) ] | | .
With the help of the Taylor’s expansion, expanding G ( x t ) across ϵ * , we obtain
G ( ϵ * ) G ( x t ) + G ( x t ) ( x t ϵ * ) = G ( ϵ * ) 0 1 [ G ( ϵ * ) ] 1 [ G ( x t ) G ( x t λ ) ] d λ ( x t ϵ * ) .
Additionally, from expression (7) and combining Equations (23) and (24), we arrive at
| | y t ϵ * | | | | [ G ( x t ) ] 1 G ( ϵ * ) | | . | | 0 1 [ G ( ϵ * ) ] 1 [ G ( x t ) G ( x t λ ) ] d λ | | . | | ( x t ϵ * ) | | 1 1 0 2 ρ ( x t ) κ 0 ( u ) d u 0 1 2 λ ρ ( x t ) 2 ρ ( x t ) κ ( u ) d u ρ ( x t ) d λ .
As a result of Lemma 1 and the above expression, the first inequality of expression (16) is obtained. Using a parallel analogy and the scheme’s second sub-step (3), we are able to write
| | z t ϵ * | | | | [ G ( x t ) ] 1 G ( ϵ * ) | | . | | 0 1 [ G ( ϵ * ) ] 1 [ G ( x t ) G ( y t λ ) ] d λ | | . | | ( y t ϵ * ) | | 1 1 0 2 ρ ( x t ) κ 0 ( u ) d u 0 1 λ ( ρ ( x t ) + ρ ( y t ) ) ρ ( x t ) + ρ ( y t ) κ ( u ) d u ρ ( y t ) d λ .
In view of Lemma 1 and estimate (26), we can obtain the first inequality of expression (17). Simultaneously, rewriting the scheme’s last sub-step (3), we achieve
| | x t + 1 ϵ * | | | | [ G ( y t ) ] 1 G ( ϵ * ) | | . | | 0 1 [ G ( ϵ * ) ] 1 [ G ( y t ) G ( z t λ ) ] d λ | | . | | ( z t ϵ * ) | | 1 1 0 2 ρ ( y t ) κ 0 ( u ) d u 0 1 λ ( ρ ( y t ) + ρ ( z t ) ) ρ ( y t ) + ρ ( z t ) κ ( u ) d u ρ ( z t ) d λ .
Using Lemma 1 and estimate (27), we can obtain the first inequality of expression (18). Nevertheless, ρ ( y t ) and ρ ( x t ) are monotonically decreasing. Hence, for each t = 0 , 1 , , we find
| | y t ϵ * | | 0 2 ρ ( x t ) κ ( u ) u d u 2 ( 1 0 2 ρ ( x t ) κ 0 ( u ) d u ) 0 2 ρ ( x 0 ) κ ( u ) u d u 2 ρ ( x 0 ) 2 ( 1 0 2 ρ ( x t ) κ 0 ( u ) d u ) 2 ρ ( x t ) 2 C 1 ρ ( x 0 ) ρ ( x t ) 2 .
Taking t = 0 above, we obtain y 0 ϵ * C 1 . ρ ( x 0 ) < ρ ( x 0 ) . Hence, y 0 ( ϵ * , δ ) , which demonstrates that (3) can be repeated indefinitely. All y t belong to ( ϵ * , δ ) by mathematical induction, and ρ ( y t ) = y t ϵ * decreases monotonically. By manipulating the first inequality of expression (17), we also have
| | z t ϵ * | | 0 ρ ( x t ) + ρ ( y t ) κ ( u ) u d u ( ρ ( x t ) + ρ ( y t ) ) 2 ( 1 0 2 ρ ( x t ) κ 0 ( u ) d u ) ρ ( y t ) . [ ρ ( x t ) + ρ ( y t ) ] C 2 ρ ( x 0 ) + ρ ( y 0 ) [ ρ ( x t ) + ρ ( y t ) ] ρ ( y t )
On further simplification,
z t ϵ * C 1 2 ρ ( x 0 ) 2 ρ ( x t ) 3 .
Taking t = 0 in (28), we obtain z 0 ϵ * C 2 . ρ ( y 0 ) < ρ ( x 0 ) . Hence, z 0 ( ϵ * , δ ) , which demonstrates that (3) can be repeated indefinitely. All z t belong to ( ϵ * , δ ) by mathematical induction, and ρ ( z t ) = z t ϵ * decreases monotonically. Lastly, simplifying the first result in the inequality of expression (18), we obtain
| | x t + 1 ϵ * | | 0 ρ ( y t ) + ρ ( z t ) κ ( u ) u d u ( ρ ( y t ) + ρ ( z t ) ) 2 ( 1 0 2 ρ ( y t ) κ 0 ( u ) d u ) ρ ( z t ) . [ ρ ( y t ) + ρ ( z t ) ] C 3 ρ ( y 0 ) + ρ ( z 0 ) [ ρ ( y t ) ρ ( z t ) + ρ ( z t ) 2 ]
On further simplification,
| | x t + 1 ϵ * | | C 1 4 ρ ( x 0 ) 4 ρ ( x t ) 5 .
Thus, we have derived all the expressions of inequalities (16)–(18). Next, to prove inequality (20), we apply mathematical induction. For t = 0 , the inequality (18) gives
| | x 1 ϵ * | | C 1 4 ρ ( x 0 ) 4 ρ ( x 0 ) 5 .
Then, the above inequality may be rewritten as
| | x 1 ϵ * | | C 1 ( 5 1 ) ρ ( x 0 ) .
Thus, expression (20) is true for t = 1 . Then, suppose the expression (20) is proved to hold for some integer t > 1 . The above inequality preserves the form
| | x t + 1 ϵ * | | C 1 5 t + 1 1 ρ ( x 0 ) 4 ρ ( x 0 ) 5 C 1 ( 5 t + 1 1 ) ρ ( x 0 ) .

4. Uniqueness Ball

We show the theorem with uniqueness in the NTS (3), using the center-Lipschitz condition.
Theorem 2.
Assume that G ( ϵ * ) = 0 , G is continuously differentiable in ( ϵ * , δ ) , [ G ( ϵ * ) ] 1 exists, and expression (8) is satisfied by [ G ( ϵ * ) ] 1 G . Assume that the relation given below is satisfied by δ:
0 2 δ κ 0 ( u ) ( 2 δ u ) d u 2 δ < 1 .
Consequently, in ( ϵ * , δ ) , the relation G ( x ) = 0 always has a unique solution ϵ * .
Proof. 
By randomly selecting y * ( ϵ * , δ ) , where G ( y * ) = 0 , ϵ * y * , and evaluating the step, we obtain
| | y * ϵ * | | = | | y * ϵ * [ G ( ϵ * ) ] 1 G ( y * ) | | . = | | [ G ( ϵ * ) ] 1 [ G ( ϵ * ) ( y * ϵ * ) + G ( ϵ * ) G ( y * ) ] | | .
From the help of Taylor’s Expansion, extending G ( y * ) , we are able to write
G ( ϵ * ) G ( y * ) + G ( ϵ * ) ( y * ϵ * ) = 0 1 [ G ( ϵ * ) ] 1 [ G ( y * λ ) G ( ϵ * ) ] d λ ( y * ϵ * ) .
Using the expression (8) as well as merging inequalities (33) and (34), we arrive at
| | y * ϵ * | | | | [ G ( ϵ * ) ] 1 G ( ϵ * ) | | . | | 0 1 [ G ( ϵ * ) ] 1 [ G ( y * λ ) G ( ϵ * ) ] d λ | | . | | ( y * ϵ * ) | | 0 1 0 2 λ ρ ( y * ) κ 0 ( u ) d u ρ ( y * ) d λ .
As a result of Lemma (1) and expression (35), we obtain
| | y * ϵ * | | 1 2 ρ ( y * ) 0 2 ρ ( y * ) κ 0 ( u ) [ 2 ρ ( y * ) u ] d u ( y * ϵ * ) 0 2 δ κ 0 ( u ) ( 2 δ u ) d u 2 δ ρ ( y * ) < | | y * ϵ * | | .
However, it contradicts our hypotheses. Hence, we conclude that y * = ϵ * . □
Specifically, claiming κ and κ 0 are constants, we have Corollaries (1) and (2) derived from Theorems 1 and 2, respectively.
Corollary 1.
Assume that G ( ϵ * ) = 0 is satisfied by ϵ * , G is continuously differentiable in ( ϵ * , δ ) , [ G ( ϵ * ) ] 1 exists, and (5), (6) is satisfied by [ G ( ϵ * ) ] 1 G . Assume that the relation given below is satisfied by δ:
δ 1 2 κ 0 + κ .
Then, the three-step NTS (3) converges for all x 0 ( ϵ * , δ ) , and
| | y t ϵ * | | C 1 ρ ( x 0 ) ρ ( x t ) 2 ,
| | z t ϵ * | | C 1 2 ρ ( x 0 ) 2 ρ ( x t ) 3 ,
| | x t + 1 ϵ * | | C 1 4 ρ ( x 0 ) 4 ρ ( x t ) 5 ,
where the following quantity
C 1 = κ ρ ( x 0 ) 1 2 κ 0 ρ ( x 0 ) ,
is proved to be less than 1.
Corollary 2.
Assume that G ( ϵ * ) = 0 is satisfied by ϵ * , G is continuously differentiable in ( ϵ * , δ ) , [ G ( ϵ * ) ] 1 exists, and (6) is satisfied by [ G ( ϵ * ) ] 1 G . Assume that the relation given below is satisfied by δ:
δ < 1 κ 0 .
Consequently, in ( ϵ * , δ ) , the relation G ( x ) = 0 always has a unique solution ϵ * . Additionally, the radius of the ball δ is solely determined by κ 0 .
Following this, we will use our fundamental theorems for some specific values of the function κ and discover the resulting corollaries:
Corollary 3.
Assume that G ( ϵ * ) = 0 is satisfied by ϵ * , G is continuously differentiable in ( ϵ * , δ ) , [ G ( ϵ * ) ] 1 exists, and (7), (8) is satisfied by [ G ( ϵ * ) ] 1 G , in which γ, κ > 0 , and κ 0 > 0 are given fixed positive constants. We take κ 0 ( u ) = γ + κ 0 u , κ ( u ) = γ + κ u , and we attain
| | [ G ( ϵ * ) ] 1 ( G ( x ) G ( y λ ) ) | | γ ( 1 λ ) ( | | x ϵ * | | + | | y ϵ * | | ) + κ 2 ( 1 λ 2 ) ( | | x ϵ * | | + | | y ϵ * | | ) 2 ,
and
| | [ G ( ϵ * ) ] 1 ( G ( x ) G ( ϵ * ) ) | | 2 | | x ϵ * | | ( γ + κ 0 | | x ϵ * | | ) ,
x , y G ( ϵ * , δ ) , 0 λ 1 , in which y λ = ϵ * + λ ( y ϵ * ) , ρ ( x ) = | | x ϵ * | | . Assume that the relation given below is satisfied by δ:
δ = 3 γ + 9 γ 2 + ( 16 / 3 ) κ + 8 κ 0 8 / 3 κ + 4 κ 0 a n d 9 γ 2 + ( 16 / 3 ) κ + 8 κ 0 0 .
Then, the NTS (3) with the three-step always converges for all x 0 ( ϵ * , δ ) , and further,
| | y t ϵ * | | C 1 ρ ( x 0 ) ρ ( x t ) 2 ,
| | z t ϵ * | | C 1 2 ρ ( x 0 ) 2 ρ ( x t ) 3 ,
| | x t + 1 ϵ * | | C 1 4 ρ ( x 0 ) 4 ρ ( x t ) 5 ,
where the following quantity
C 1 = ρ ( x 0 ) [ γ + 4 / 3 κ ρ ( x 0 ) ] [ 1 2 γ ρ ( x 0 ) 2 κ 0 ρ ( x 0 ) 2 ] ,
is found to be less than 1.
Corollary 4
([10]).Assume that G ( ϵ * ) = 0 is satisfied by ϵ * , G is continuously differentiable in ( ϵ * , δ ) , [ G ( ϵ * ) ] 1 exists, and (8) is satisfied by [ G ( ϵ * ) ] 1 G , in which γ, κ > 0 , and κ 0 > 0 are given fixed positive constants. We take κ 0 ( u ) = γ + κ 0 u , and we attain
| | [ G ( ϵ * ) ] 1 ( G ( x ) G ( ϵ * ) ) | | 2 | | x ϵ * | | ( γ + κ 0 | | x ϵ * | | ) ; x ( ϵ * , δ ) ,
in which ρ ( x ) = | | x ϵ * | | . Assume that the relation given below is satisfied by δ:
δ = 2 γ 4 γ 2 ( 16 / 3 ) κ 0 ( 8 / 3 ) κ 0 a n d 4 γ 2 ( 16 / 3 ) κ 0 0 .
Consequently, in ( ϵ * , δ ) , the relation G ( x ) = 0 always has a unique solution ϵ * . Additionally, the radius of the ball δ is solely determined by κ 0 and γ.

5. Convergence under the Weak κ -Average

This section presents the findings of a re-examination of the requirements and the convergence’s radius of the given method, which were previously stated in Theorem 1, although κ is not regarded a ND function. The convergence order has been observed to be decreasing. This section’s second theorem yields a result identical to Theorem 3. However, the center-Lipschitz condition is our new assumption.
Theorem 3.
Assume that G ( ϵ * ) = 0 , G is continuously differentiable in ( ϵ * , δ ) , [ G ( ϵ * ) ] 1 exists, (7), (8) is satisfied by [ G ( ϵ * ) ] 1 G , and κ and κ 0 are noticed to be positive integrable. Assume that the relation given below is satisfied by δ:
0 2 δ κ 0 ( u ) d u 1 a n d 0 2 δ ( κ ( u ) + κ 0 ( u ) ) d u 1 .
Then, the three-step NTS (3) is said to converge for all x 0 ( ϵ * , δ ) , and
| | y t ϵ * | | 0 2 ρ ( x t ) κ ( u ) u d u 2 ( 1 0 2 ρ ( x t ) κ 0 ( u ) d u ) q 1 ρ ( x t ) ,
| | z t ϵ * | | 0 ρ ( x t ) + ρ ( y t ) κ ( u ) u d u ( ρ ( x t ) + ρ ( y t ) ) ( 1 0 2 ρ ( x t ) κ 0 ( u ) d u ) ρ ( y t ) q 2 q 1 ρ ( x t ) ,
| | x t + 1 ϵ * | | 0 ρ ( y t ) + ρ ( z t ) κ ( u ) u d u ( ρ ( y t ) + ρ ( z t ) ) ( 1 0 2 ρ ( y t ) κ 0 ( u ) d u ) ρ ( z t ) q 3 q 2 q 1 ρ ( x t ) ,
where the quantities
q 1 = 0 2 ρ ( x 0 ) κ ( u ) d u 1 0 2 ρ ( x 0 ) κ 0 ( u ) d u , q 2 = 0 ρ ( x 0 ) + ρ ( y 0 ) κ ( u ) d u 1 0 2 ρ ( x 0 ) κ 0 ( u ) d u , q 3 = 0 ρ ( y 0 ) + ρ ( z 0 ) κ ( u ) d u 1 0 2 ρ ( y 0 ) κ 0 ( u ) d u ,
are less than 1. Moreover,
| | x t ϵ * | | ( q 1 q 2 q 3 ) t | | x 0 ϵ * | | , t = 1 , 2 ,
Nevertheless, assuming the function κ a described by the expression (13) to be ND for some α with 0 < α 1 , and δ is satisfied by
1 2 δ 0 2 δ ( 2 δ κ 0 ( u ) + u κ ( u ) ) d u 1 ,
then the three-step NTS (3) is said to converge for all x 0 ( ϵ * , δ ) , and
| | x t ϵ * | | C 1 ( 1 + 3 α + α 2 ) t ( α 2 + 2 α 2 ) | | x 0 ϵ * | | , t = 1 , 2 , ,
where the said quantities C 1 are defined in equation (19) and are less than 1.
Proof. 
We can show that the quantities q 1 , q 2 , and q 3 described by Equation (49) are less than 1, without loss of generality, by picking x 0 ( ϵ * , δ ) , where δ fulfills the relation (45). Indeed, because κ is a positive integrable function, we obtain
q 1 = 0 2 ρ ( x 0 ) κ ( u ) d u 1 0 2 ρ ( x 0 ) κ 0 ( u ) d u 0 2 δ κ ( u ) d u 1 0 2 δ κ 0 ( u ) d u < 1 , q 2 = 0 ρ ( x 0 ) + ρ ( y 0 ) κ ( u ) d u 1 0 2 ρ ( x 0 ) κ 0 ( u ) d u 0 2 δ κ ( u ) d u 1 0 2 δ κ 0 ( u ) d u < 1 , q 3 = 0 ρ ( y 0 ) + ρ ( z 0 ) κ ( u ) d u 1 0 2 ρ ( y 0 ) κ 0 ( u ) d u 0 2 δ κ ( u ) d u 1 0 2 δ κ 0 ( u ) d u < 1 .
Evidently, if x ( ϵ * , δ ) , we can write with the help of the center-Lipschitz condition with the κ -average and expression (45),
| | [ G ( ϵ * ) ] 1 [ G ( x ) G ( ϵ * ) ] | | 0 2 ρ ( x ) κ 0 ( u ) d u ; x ( ϵ * , δ ) 1 .
Using the Banach lemma along with the following relation
| | I ( [ G ( ϵ * ) ] 1 G ( x ) I ) | | 1 = | | [ G ( x ) ] 1 G ( ϵ * ) | | ,
and using Equation (53), we arrive to the following expression,
| | [ G ( x ) ] 1 G ( ϵ * ) | | 1 1 0 2 ρ ( x ) κ 0 ( u ) d u .
Hence, if x t ( ϵ * , δ ) , we can begin writing from the scheme’s first sub-step (3)
| | y t ϵ * | | = | | x t ϵ * [ G ( x t ) ] 1 G ( x t ) | | = | | [ G ( x t ) ] 1 [ G ( x t ) ( x t ϵ * ) G ( x t ) + G ( ϵ * ) ] | | .
Expanding G ( x t ) around ϵ * from Taylor’s expansion, it can be written as
G ( ϵ * ) G ( x t ) + G ( x t ) ( x t ϵ * ) = G ( ϵ * ) 0 1 [ G ( ϵ * ) ] 1 [ G ( x t ) G ( x t λ ) ] d λ ( x t ϵ * ) .
Following definition (7) and compacting the expressions (55) and (56), we may obtain
| | y t ϵ * | | | | [ G ( x t ) ] 1 G ( ϵ * ) | | . | | 0 1 [ G ( ϵ * ) ] 1 [ G ( x t ) G ( x t λ ) ] d λ | | . | | ( x t ϵ * ) | | 1 0 2 ρ ( x t ) κ 0 ( u ) d u 0 1 2 λ ρ ( x t ) 2 ρ ( x t ) κ ( u ) d u ρ ( x t ) d λ .
Using the results of Lemma 1 and the inequality (54) in the above expression, we obtain the intermediate expression of (46). We use the parallel analogy for the method’s second sub-step (3),
| | z t ϵ * | | | | [ G ( x t ) ] 1 G ( ϵ * ) | | . | | 0 1 [ G ( ϵ * ) ] 1 [ G ( x t ) G ( y t λ ) ] d λ | | . | | ( y t ϵ * ) | | 1 0 2 ρ ( x t ) κ 0 ( u ) d u 0 1 λ ( ρ ( x t ) + ρ ( y t ) ) ρ ( x t ) + ρ ( y t ) κ ( u ) d u ρ ( y t ) d λ .
This way, we obtain the first inequality of (47) using the Lemma (1) in the preceding expression. Similarly, repeating the scheme’s final sub-step (3), we achieve
| | x t + 1 ϵ * | | | | [ G ( y t ) ] 1 G ( ϵ * ) | | . | | 0 1 [ G ( ϵ * ) ] 1 [ G ( y t ) G ( z t λ ) ] d λ | | . | | ( z t ϵ * ) | | 1 0 2 ρ ( y t ) κ 0 ( u ) d u 0 1 λ ( ρ ( y t ) + ρ ( z t ) ) ρ ( y t ) + ρ ( z t ) κ ( u ) d u ρ ( z t ) d λ .
Using Lemma 1 in the above expression, we can obtain the first inequality of (48). Moreover, ρ ( x t ) , ρ ( y t ) , and ρ ( z t ) are monotonically decreasing; thus, for each t = 0 , 1 , . . . , we find
| | y t ϵ * | | 0 2 ρ ( x t ) κ ( u ) u d u 2 ( 1 0 2 ρ ( x t ) κ 0 ( u ) d u ) 0 2 ρ ( x 0 ) κ ( u ) d u 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( x t ) q 1 ρ ( x t ) .
The second inequality of relation (46) gives
| | z t ϵ * | | 0 ρ ( x t ) + ρ ( y t ) κ ( u ) u d u ( ρ ( x t ) + ρ ( y t ) ) ( 1 0 2 ρ ( x t ) κ 0 ( u ) d u ) ρ ( y t ) 0 ρ ( x 0 ) + ρ ( y 0 ) κ ( u ) d u ( 1 0 2 ρ ( x 0 ) κ 0 ( u ) d u ) ρ ( y t ) q 2 q 1 ρ ( x t ) .
Next, with the help of the inequality of expression (47), we are able to reach
| | x t + 1 ϵ * | | 0 ρ ( y t ) + ρ ( z t ) κ ( u ) u d u ( ρ ( y t ) + ρ ( z t ) ) ( 1 0 2 ρ ( y t ) κ 0 ( u ) d u ) ρ ( z t ) 0 ρ ( y 0 ) + ρ ( z 0 ) κ ( u ) d u 1 0 2 ρ ( y 0 ) κ 0 ( u ) d u ρ ( z t ) q 3 q 2 q 1 ρ ( x t ) .
Therefore, relation (50) can be simply deduced with inequality (61). Henceforth, if relation (13) describes the function κ α , which is ND for some α with 0 α 1 , and δ is determined by the expression (51), Lemma 2 and expression (46)’s first relation implies
| | y t ϵ * | | φ 1 , α ( 2 ρ ( x t ) ) 2 α 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( x t ) α + 1 φ 1 , α ( 2 ρ ( x 0 ) ) 2 α 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( x t ) α + 1 = C 1 ρ ( x 0 ) α ρ ( x t ) α + 1 .
Moreover, from the first part of inequality (47) and Lemma 2, we may write
| | z t ϵ * | | φ 1 , α ( ρ ( x t ) + ρ ( y t ) ) ( ρ ( x t ) + ρ ( y t ) ) α 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( y t ) , φ 1 , α ( 2 ρ ( x 0 ) ) ( ρ ( x t ) + ρ ( y t ) ) α 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( y t ) , C 1 ( 2 ρ ( x 0 ) ) α ( ρ ( x t ) + ρ ( y t ) ) α ρ ( y t ) , C 1 2 ρ ( x 0 ) 2 α ρ ( x t ) 2 α + 1 .
Similarly, with Lemma 2 and the first relation of (48), we arrive at
| | x t + 1 ϵ * | | φ 1 , α ( ρ ( y t ) + ρ ( z t ) ) ( ρ ( y t ) + ρ ( z t ) ) α 1 0 2 ρ ( y t ) κ 0 ( u ) d u ρ ( z t ) φ 1 , α ( 2 ρ ( x 0 ) ) ( ρ ( y t ) + ρ ( z t ) ) α 1 0 2 ρ ( y t ) κ 0 ( u ) d u ρ ( z t ) = C 1 α + 3 ρ ( x 0 ) 3 α + α 2 ρ ( x t ) ( α + 1 ) 2 + α ,
where the quantity C 1 < 1 is determined by the expression (19). Next, to prove inequality (52), we apply mathematical induction. With t = 0 , the said inequality becomes
| | x 1 ϵ * | | C 1 3 + α ρ ( x 0 ) .
Then, carrying out the calculations, we obtain
| | x 1 ϵ * | | C 1 ( ( 1 + 3 α + α 2 ) ( 2 + 2 α + α 2 ) ) ρ ( x 0 ) .
Thus, the expression (52) is true for t = 1 . Next, suppose the inequality (52) is true for some integer t > 1 . Then, using the inequalities (52) for t = t , (63), (62) after re-arranging the terms, the above inequality preserves the form
| | x t + 1 ϵ * | | C 1 α + 3 ρ ( x k ) ( 1 + 3 α + α 2 ) ρ ( x 0 ) 3 α + α 2 C 1 ( 1 + 3 α + α 2 ) t + 1 ( 2 + 2 α + α 2 ) ρ ( x 0 ) .
which shows that the result is true for t = t + 1 . Hence, the inequality (52) is true for all natural numbers, and consequently, x t converges to ϵ * . □
Theorem 4.
Assume that G ( ϵ * ) = 0 , G is continuously differentiable in ( ϵ * , δ ) , [ G ( ϵ * ) ] 1 exists, (8) is satisfied by [ G ( ϵ * ) ] 1 G , and κ 0 is noticed to be positive integrable. Assume that the relation given below is satisfied by δ:
0 2 δ κ 0 ( u ) d u 1 3 .
Then, the three-step NTS (3) always converges for all x 0 ( ϵ * , δ ) , and
| | y t ϵ * | | 2 0 2 ρ ( x t ) κ 0 ( u ) d u 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( x t ) q 1 ρ ( x t ) , | | z t ϵ * | | 0 2 ρ ( x t ) κ 0 ( u ) d u + 0 2 ρ ( y t ) κ 0 ( u ) d u 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( y t ) q 2 q 1 ρ ( x t ) , | | x t + 1 ϵ * | | 0 2 ρ ( y t ) κ 0 ( u ) d u + 0 2 ρ ( z t ) κ 0 ( u ) d u 1 0 2 ρ ( y t ) κ 0 ( u ) d u ρ ( z t ) q 3 q 2 q 1 ρ ( x t ) ,
where the quantities
q 1 = 2 0 2 ρ ( x 0 ) κ 0 ( u ) d u 1 0 2 ρ ( x 0 ) κ 0 ( u ) d u , q 2 = 0 2 ρ ( x 0 ) κ 0 ( u ) d u + 0 2 ρ ( y 0 ) κ 0 ( u ) d u 1 0 2 ρ ( x 0 ) κ 0 ( u ) d u , q 3 = 0 2 ρ ( y 0 ) κ 0 ( u ) d u + 0 2 ρ ( z 0 ) κ 0 ( u ) d u 1 0 2 ρ ( y 0 ) κ 0 ( u ) d u ,
are less than 1. Moreover,
| | x t ϵ * | | ( q 1 q 2 q 3 ) t | | x 0 ϵ * | | , t = 1 , 2 , . . . .
Additionally, assume the function κ α described with the expression (13) is ND for some α with 0 < α 1 , then
| | x t ϵ * | | C 1 ( 1 + 3 α + α 2 ) t ( α 2 + 2 α 2 ) | | x 0 ϵ * | | , t = 1 , 2 , ,
and the first expression of Equation (68) gives q 1 , q 2 .
Proof. 
Assume x 0 ( ϵ * , δ ) , and x t denotes the sequence generated by NTS three-step (3). Let δ , q 1 , and q 2 be determined by the expressions (66) and (68), respectively. Assume that x t ( ϵ * , δ ) . Then,
| | y t ϵ * | | = | | y t ϵ * [ G ( x t ) ] 1 G ( x t ) | | = | | [ G ( x t ) ] 1 [ G ( x t ) ( x t ϵ * ) G ( x t ) + G ( ϵ * ) ] | | .
with the help of the Taylor series expansion, expanding G ( x t ) across ϵ * , we obtain
G ( ϵ * ) G ( x t ) + G ( x t ) ( x t ϵ * ) = G ( ϵ * ) 0 1 [ G ( ϵ * ) ] 1 [ G ( x t ) G ( x t λ ) ] d λ ( x t ϵ * ) .
Following the theorem’s assumptions (8) along with utilizing expressions (71) and (72), we may write
| | y t ϵ * | | | | [ G ( x t ) ] 1 G ( ϵ * ) | | . | | 0 1 [ G ( ϵ * ) ] 1 [ G ( x t ) G ( ϵ * ) + G ( ϵ * ) G ( x t λ ) ] d λ | | . | | ( x t ϵ * ) | | 1 1 0 2 ρ ( x t ) κ 0 ( u ) d u . 0 1 0 2 λ ρ ( x t ) κ 0 ( u ) d u ρ ( x t ) d λ + 0 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( x t ) d λ .
By virtue of Lemma (1), the above inequality becomes
| | y t ϵ * | | 2 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( x t ) 1 2 0 2 ρ ( x t ) κ 0 ( u ) u d u 1 0 2 ρ ( x t ) κ 0 ( u ) d u 2 0 2 ρ ( x t ) κ 0 ( u ) d u 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( x t ) = q 1 ρ ( x t ) ,
which is equivalent to the first inequality of (67). We use a parallel analogy for the method’s second sub-step (3),
| | z t ϵ * | | | | [ G ( x t ) ] 1 G ( ϵ * ) | | . | | 0 1 [ G ( ϵ * ) ] 1 [ G ( x t ) G ( ϵ * ) ] d λ | | . | | ( y t ϵ * ) | | + | | 0 1 [ G ( ϵ * ) ] 1 [ G ( ϵ * ) G ( y t λ ) ] d λ | | . | | ( y t ϵ * ) | | 1 1 0 2 ρ ( x t ) κ 0 ( u ) d u . 0 1 0 2 λ ρ ( y t ) κ 0 ( u ) d u ρ ( y t ) d λ + 0 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( y t ) d λ .
As a result of Lemma (1), the above expression takes the form
| | z t ϵ * | | 0 2 ρ ( x n ) κ 0 ( u ) d u ρ ( y t ) + 0 2 ρ ( y t ) κ 0 ( u ) d u ρ ( y t ) 1 2 0 2 ρ ( y t ) κ 0 ( u ) u d u 1 0 2 ρ ( x t ) κ 0 ( u ) d u 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( y t ) + 0 2 ρ ( y t ) κ 0 ( u ) d u ρ ( y t ) 1 0 2 ρ ( x t ) κ 0 ( u ) d u = q 2 q 1 ρ ( x t ) ,
Simultaneously, the final step of the method (3) gives
| | x t + 1 ϵ * | | | | [ G ( y t ) ] 1 G ( ϵ * ) | | . | | 0 1 [ G ( ϵ * ) ] 1 [ G ( y t ) G ( ϵ * ) ] d λ | | . | | ( z t ϵ * ) | | + | | 0 1 [ G ( ϵ * ) ] 1 [ G ( ϵ * ) G ( z t λ ) ] d λ | | . | | ( z t ϵ * ) | | 1 1 0 2 ρ ( y t ) κ 0 ( u ) d u . 0 1 0 2 λ ρ ( y t ) κ 0 ( u ) d u ρ ( z t ) d λ + 0 1 0 2 ρ ( z t ) κ 0 ( u ) d u ρ ( z t ) d λ .
Because of Lemma 1, the above expression leads to
| | x t + 1 ϵ * | | 0 2 ρ ( y t ) κ 0 ( u ) d u ρ ( z t ) + 0 2 ρ ( z t ) κ 0 ( u ) d u ρ ( z t ) 1 2 0 2 ρ ( z t ) κ 0 ( u ) u d u 1 0 2 ρ ( y t ) κ 0 ( u ) d u 0 2 ρ ( y t ) κ 0 ( u ) d u ρ ( z t ) + 0 2 ρ ( z t ) κ 0 ( u ) d u ρ ( z t ) 1 0 2 ρ ( y t ) κ 0 ( u ) d u = q 3 q 2 q 1 ρ ( x n ) ,
where the relation (68) gives q 1 < 1 , q 2 < 1 , and q 3 < 1 . Moreover, inequality (69) can be readily deduced from the second Equation (67), implying that x t is converging to ϵ * .
Additionally, if relation (13) describes the function κ α , which is ND for some α with 0 α 1 , and δ is determined by the expression (66), then, based on Lemma 2 and the inequality’s first statement (67), it follows that
| | y t ϵ * | | 2 φ 0 , α ( 2 ρ ( x t ) ) 2 α 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( x t ) α + 1 , 2 φ 0 , α ( 2 ρ ( x 0 ) ) 2 α 1 0 2 ρ ( x 0 ) κ 0 ( u ) d u ρ ( x t ) α + 1 = q 1 ρ ( x 0 ) α ρ ( x t ) α + 1 .
Nevertheless, with Lemma 2 and the inequality’s second relation (67),
| | z t ϵ * | | φ 0 , α ( 2 ρ ( x t ) ) + φ 0 , α ( 2 ρ ( y t ) ) ( 2 ρ ( x t ) ) α ρ ( y t ) 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( y t ) , φ 0 , α ( 2 ρ ( x 0 ) ) + φ 0 , α ( 2 ρ ( x 0 ) ) ( 2 ρ ( x t ) ) α ρ ( y n ) 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( y t ) , q 1 2 ρ ( x 0 ) 2 α ρ ( x t ) 2 α + 1 .
Finally, the second inequality of expression (67) of the last sub-step and Lemma 2 gives
| | x t + 1 ϵ * | | φ 0 , α ( 2 ρ ( y t ) ) + φ 0 , α ( 2 ρ ( z t ) ) ( 2 ρ ( y t ) ) α ρ ( z t ) 1 0 2 ρ ( y t ) κ 0 ( u ) d u ρ ( z t ) , φ 0 , α ( 2 ρ ( x 0 ) ) + φ 0 , α ( 2 ρ ( x 0 ) ) ( 2 ρ ( y t ) ) α ρ ( z t ) 1 0 2 ρ ( y t ) κ 0 ( u ) d u ρ ( z t ) , = q 1 α + 3 ρ ( x 0 ) 3 α + α 2 ρ ( x t ) ( α + 1 ) 2 + α ,
Next, we can prove that the expression (70) is true for all integers t 1 through mathematical induction along the same lines as the previous result. Accordingly, sequence { x t } converges to ϵ * . □
Next, the outcomes of Theorems 3 and 4 are recaptured using our new and improved theorems on a variety of special functions κ .
Corollary 5.
Assume that G ( ϵ * ) = 0 , G is continuously differentiable in ( ϵ * , δ ) , [ G ( ϵ * ) ] 1 exists, and (7), (8) is satisfied by [ G ( ϵ * ) ] 1 G , where κ 0 ( u ) = c 0 a u a 1 , and κ ( u ) = c a u a 1 :
| | [ G ( ϵ * ) ] 1 ( G ( x ) G ( y λ ) ) | | c ( 1 λ a ) ( | | x ϵ * | | + | | y ϵ * | | ) a ,
and
| | [ G ( ϵ * ) ] 1 ( G ( x ) G ( ϵ * ) ) | | c 0 2 a | | x ϵ * | | a ,
x , y ( ϵ * , δ ) , 0 λ 1 , in which ρ ( x ) = | | x ϵ * | | , y λ = ϵ * + λ ( y ϵ * ) , 0 < a < 1 , c > 0 , and c 0 > 0 . Assume that the relation given below is satisfied by δ:
δ = a + 1 2 a ( c 0 ( a + 1 ) + c a ) 1 a .
Thus, the NTS three-step (3) always converges for all x 0 ( ϵ * , δ ) with
| | y t ϵ * | | 0 2 ρ ( x t ) κ ( u ) u d u 2 ( 1 0 2 ρ ( x t ) κ 0 ( u ) d u ) q 1 ρ ( x t ) ,
| | z t ϵ * | | 0 ρ ( x t ) + ρ ( y t ) κ ( u ) u d u ( ρ ( x t ) + ρ ( y t ) ) ( 1 0 2 ρ ( x t ) κ 0 ( u ) d u ) ρ ( y t ) q 2 q 1 ρ ( x t ) ,
| | x t + 1 ϵ * | | 0 ρ ( y t ) + ρ ( z t ) κ ( u ) u d u ( ρ ( y t ) + ρ ( z t ) ) ( 1 0 2 ρ ( y t ) κ 0 ( u ) d u ) ρ ( z t ) q 3 q 2 q 1 ρ ( x t ) ,
where the following quantities
q 1 = c a 2 a ρ ( x 0 ) a ( 1 + a ) [ 1 2 a c 0 ρ ( x 0 ) a ] , q 2 = c a ( ρ ( x 0 ) + ρ ( y 0 ) ) a ( a + 1 ) ( 1 2 a c 0 ρ ( x 0 ) a ) , q 3 = c a ( ρ ( y 0 ) + ρ ( z 0 ) ) a ( a + 1 ) ( 1 2 a c 0 ρ ( y 0 ) a ) ,
are found to be less than 1.
Corollary 6.
Assume that G ( ϵ * ) = 0 , G is continuously differentiable in ( ϵ * , δ ) , [ G ( ϵ * ) ] 1 exists, and (8) is satisfied by [ G ( ϵ * ) ] 1 G , where κ 0 ( u ) = c 0 a u a 1 :
| | [ G ( ϵ * ) ] 1 ( G ( x ) G ( ϵ * ) ) | | c 0 2 a | | x ϵ * | | a , x ( ϵ * , δ ) ,
in which ρ ( x ) = | | x ϵ * | | , 0 < a < 1 , and c 0 > 0 . Assume that the relation given below is satisfied by δ:
δ = 1 3 c 0 2 a 1 a .
Thus, the NTS three-step scheme (3) always converges for all x 0 ( ϵ * , δ ) with
| | y t ϵ * | | 2 0 2 ρ ( x t ) κ 0 ( u ) d u 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( x t ) q 1 ρ ( x t ) , | | z t ϵ * | | 0 2 ρ ( x t ) κ 0 ( u ) d u + 0 2 ρ ( y t ) κ 0 ( u ) d u 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( y t ) q 2 q 1 ρ ( x t ) , | | x t + 1 ϵ * | | 0 2 ρ ( y t ) κ 0 ( u ) d u + 0 2 ρ ( z t ) κ 0 ( u ) d u 1 0 2 ρ ( y t ) κ 0 ( u ) d u ρ ( z t ) q 3 q 2 q 1 ρ ( x t ) ,
where the following quantities
q 1 = c 0 2 a + 1 ρ ( x 0 ) a 1 2 a c 0 ρ ( x 0 ) a , q 2 = c 0 2 a ( ρ ( x 0 ) a + ρ ( y 0 ) a ) 1 2 a c 0 ρ ( x 0 ) a , q 3 = c 0 2 a ( ρ ( y 0 ) a + ρ ( z 0 ) a ) 1 2 a c 0 ρ ( y 0 ) a ,
are found to be less than 1.
Corollary 7.
Assume that G ( ϵ * ) = 0 , G is continuously differentiable in ( ϵ * , δ ) , [ G ( ϵ * ) ] 1 exists, and (8) is satisfied by [ G ( ϵ * ) ] 1 G , where κ 0 ( u ) = 2 γ c 0 ( 1 γ u ) 3 :
| | [ G ( ϵ * ) ] 1 ( G ( x ) G ( ϵ * ) ) | | c 0 ( 1 2 γ ρ ( x ) ) 2 c 0 , x ( ϵ * , δ ) ,
in which ρ ( x ) = | | x ϵ * | | , γ > 0 , and c 0 > 0 . Assume that the relation given below is satisfied by δ:
δ = 3 c 0 + 1 3 c 0 ( 3 c 0 + 1 ) 2 γ ( 3 c 0 + 1 ) .
Then, three-step NTS (3) always converges for all x 0 ( ϵ * , δ ) with
| | y t ϵ * | | 2 0 2 ρ ( x t ) κ 0 ( u ) d u 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( x t ) q 1 ρ ( x t ) , | | z t ϵ * | | 0 2 ρ ( x t ) κ 0 ( u ) d u + 0 2 ρ ( y t ) κ 0 ( u ) d u 1 0 2 ρ ( x t ) κ 0 ( u ) d u ρ ( y t ) q 2 q 1 ρ ( x t ) , | | x t + 1 ϵ * | | 0 2 ρ ( y t ) κ 0 ( u ) d u + 0 2 ρ ( z t ) κ 0 ( u ) d u 1 0 2 ρ ( y t ) κ 0 ( u ) d u ρ ( z t ) q 3 q 2 q 1 ρ ( x t ) ,
where the following quantities
q 1 = 2 c 0 2 c 0 ( 1 2 γ ρ ( x 0 ) ) 2 [ 1 2 γ ρ ( x 0 ) ] 2 ( 1 + c 0 ) c 0 , q 2 = [ c 0 c 0 ( 1 2 γ ρ ( x 0 ) ) 2 ] ( 1 2 γ ρ ( y 0 ) ) 2 ) + [ c 0 c 0 ( 1 2 γ ρ ( y 0 ) ) 2 ] ( 1 2 γ ρ ( x 0 ) ) 2 ) ( [ 1 2 γ ρ ( x 0 ) ] 2 ( 1 + c 0 ) c 0 ) ( 1 2 γ ρ ( y 0 ) ) 2 ) , q 3 = [ c 0 c 0 ( 1 2 γ ρ ( y 0 ) ) 2 ] ( 1 2 γ ρ ( z 0 ) ) 2 ) + [ c 0 c 0 ( 1 2 γ ρ ( z 0 ) ) 2 ] ( 1 2 γ ρ ( y 0 ) ) 2 ) ( [ 1 2 γ ρ ( y 0 ) ] 2 ( 1 + c 0 ) c 0 ) ( 1 2 γ ρ ( z 0 ) ) 2 ) ,
are found to be less than 1.
Remark 1.
( a ) If κ 0 = κ , then our results specialize to earlier ones [3,5,6,12,14,19,23,24]. However, if κ 0 < κ , the advantages described in the abstract are thus attained (see also Examples 1 and 2). ( b ) A further expansion is possible as follows. Assume (6) is valid, as is the equation 2 κ 0 ( u ) u 1 = 0 having a minimum positive value of zero δ ¯ . We describe
~ = ( ϵ * , δ ) ( ϵ * , δ ¯ ) . Furthermore, we consider
| | G ( x ) G ( y λ ) | | λ ( ρ ( x ) + ρ ( y ) ) ρ ( x ) + ρ ( y ) κ ¯ ( u ) d u ,
when κ ¯ is as κ and x , y , 0 λ 1 . We arrive at
κ ¯ ( u ) κ ( u ) f o r a l l u [ 0 , m i n { δ , δ ¯ } ] .
Therefore, given the proofs, κ ¯ may be used to substitute κ in all results. However, if
κ ¯ ( u ) < κ ( u ) ,
the advantages mentioned in the introduction are expanded yet further. We have, in the case of the motivational example,
κ 0 < κ ¯ = e 1 ( e 1 ) 2 < κ .

6. Numerical Examples

Example 1.
Returning to the motivational example presented in the study’s introduction, using (15) and G ( ϵ * ) = ( 1 , 1 , 1 ) t , we obtain the following.
Case κ 0 ( u ) = κ ( u ) = e 2 provides
δ 0 = 0.245253 .
Case κ 0 ( u ) = e 1 2 a n d κ ( u ) = e 2 gives
δ 1 = 0.324947 .
Case κ 0 ( u ) = e 1 2 a n d κ ¯ ( u ) = e 1 ( e 1 ) 2 gives
δ 2 = 0.382692 .
Hence, we conclude
δ 0 < δ 1 < δ 2 .
Figure 1 also explains the advantages of different cases of κ on the convergence radius. The radius δ 0 was given independently by Rheinboldt [1] and Traub [16], whereas radii δ 1 and δ 2 were reported by Argyros [3,4] for Newton’s method.
Example 2.
Assuming = V ¯ ( 0 , 1 ) and ϵ * = 0 , χ = Y = C [ 0 , 1 ] , we describe M on ℜ such as
M ( v ) ( x ) = v ( x ) 0 1 x λ v ( λ ) 3 d λ .
So,
M ( v ( p ) ) ( x ) = p ( x ) 3 0 1 x λ v ( λ ) 2 p ( λ ) d λ f o r a l l p .
Then, we obtain
κ 0 ( u ) = 1.5 u < κ ( u ) = κ ¯ ( u ) = 3 u .
As a result, by solving (15), we obtain the same advantages as in Example 1.
Old case κ ( u ) = κ 0 ( u ) = 3 u gives
δ 0 = 0.316228 .
Case κ ( u ) = 3 u a n d κ 0 ( u ) = 1.5 u gives
δ 1 = 0.377964 .
Case κ ¯ ( u ) = 3 u a n d κ 0 ( u ) = 1.5 u gives
δ 2 = 0.377964 .
Hence, we conclude
δ 0 < δ 1 δ 2 .
This can also be explained by Figure 2, which extends the applicability of the method through different cases of κ ( u ) .
Example 3
([10]). Assume χ = Y = . We describe
G ( u ) = 0 u 1 + 2 u sin π u d u , u ( , ) .
Therefore,
G ( u ) = 1 + 2 u sin π u , u 0 , 1 , u = 0 , .
Evidently, ϵ * = 0 , G ( u ) = 0 , and with the help of the assumption on G
| | [ G ( ϵ * ) ] 1 ( G ( u ) G ( ϵ * ) ) | | = 2 u sin π u 2 | u ϵ * | , u R .
Theorem 4 demonstrates this for any u 0 ( ϵ * , 1 / 6 ) ,
| | u t ϵ * | | E 5 t 1 | | u 0 ϵ * | | , t = 1 , 2 , , E = 16 | u 0 | 4 [ 2 | u 0 | + 2 | y 0 | ] [ 1 2 | u 0 | ] 3 | y 0 | | z 0 | .
Moreover, no other P.I.F. κ satisfies the expression (7). Observe, for example,
| | [ G ( ϵ * ) ] 1 ( G ( u ) G ( y λ ) ) | | = 2 u sin π u 2 y λ sin π y λ = 4 2 i + 1 ,
where u = 1 / i , y = 1 / i , λ = 2 i 2 i + 1 , and i = 1 , 2 , Thus, if there exists a P.I.F. κ, i.e., relation (7) belongs in ( ϵ * , δ ) with some δ > 0 , as a result, there exists some n 0 > 1 , i.e.,
0 2 r δ κ ( u ) d u i = n 0 + 4 2 i + 1 2 i κ ( u ) d u i = n 0 + 4 2 i + 1 = + ,
which is counterintuitive. This example demonstrates that Theorem 4 is a major extension of Theorem 3, when the radius is neglected.

7. Conclusions

A novel technique was developed in order to provide a finer ball convergence analysis without making additional assumptions than in earlier studies. The method is quite generic. It turns out that while the criteria are more generic, they are also more flexible, which results in some benefits with no more computational cost. As a result, we are able to expand the usage of classified NTS to situations not previously covered. The methodology opens the door for future studies to be more effective in NTS’s convergence and other iterative procedures [9,10,11,13,14,15,19,20,21,23,25,26,27,28,29] along the same lines.

Author Contributions

A.S. and J.P.J. wrote the framework and the original draft of this paper. K.R.P. and I.K.A. reviewed and validated the paper. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

The authors would like to offer their sincere thanks to the reviewer for their constructive suggestions.

Conflicts of Interest

The authors declare no conflict of interest.

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Figure 1. Convergence radius for different cases of κ ( u ) .
Figure 1. Convergence radius for different cases of κ ( u ) .
Mathematics 10 03946 g001
Figure 2. Convergence radius for different cases of κ .
Figure 2. Convergence radius for different cases of κ .
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Saxena, A.; Jaiswal, J.P.; Pardasani, K.R.; Argyros, I.K. Convergence Criteria of a Three-Step Scheme under the Generalized Lipschitz Condition in Banach Spaces. Mathematics 2022, 10, 3946. https://doi.org/10.3390/math10213946

AMA Style

Saxena A, Jaiswal JP, Pardasani KR, Argyros IK. Convergence Criteria of a Three-Step Scheme under the Generalized Lipschitz Condition in Banach Spaces. Mathematics. 2022; 10(21):3946. https://doi.org/10.3390/math10213946

Chicago/Turabian Style

Saxena, Akanksha, Jai Prakash Jaiswal, Kamal Raj Pardasani, and Ioannis K. Argyros. 2022. "Convergence Criteria of a Three-Step Scheme under the Generalized Lipschitz Condition in Banach Spaces" Mathematics 10, no. 21: 3946. https://doi.org/10.3390/math10213946

APA Style

Saxena, A., Jaiswal, J. P., Pardasani, K. R., & Argyros, I. K. (2022). Convergence Criteria of a Three-Step Scheme under the Generalized Lipschitz Condition in Banach Spaces. Mathematics, 10(21), 3946. https://doi.org/10.3390/math10213946

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