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Unified Convergence Analysis of Two-Step Iterative Methods for Solving Equations

Ioannis K. Argyros
Department of Mathematical Sciences, Cameron University, Lawton, OK 73505, USA

Correspondence: iargyros@cameron.edu

Abstract. In this paper we consider unified convergence analysis of two-step iterative methods forsolving equations in the Banach space setting. The convergence order four was shown using Taylorexpansions requiring the existence of the fifth derivative not on this method. But these hypotheseslimit the utilization of it to functions which are at least five times differentiable although the methodmay converge. As far as we know no semi-local convergence has been given in this setting. Ourgoal is to extend the applicability of this method in both the local and semi-local convergence caseand in the more general setting of Banach space valued operators. Moreover, we use our idea ofrecurrent functions and conditions only on the first derivative and divided differences which appearon the method. This idea can be used to extend other high convergence multipoint and multistepmethods. Numerical experiments testing the convergence criteria complement this study.

1. Introduction
We consider the problem of approximating a solution x∗ of equation

F (x) = 0, (1.1)
where F : Ω ⊂ B −→ B1 is a continuous operator acting between Banach spaces B and B1 with
Ω 6= ∅. Since a closed form solution is not possible in general, iterative methods are used forsolving (1.1). Many iterative methods are studied for approximating x∗. In this paper, we considerthe iterative methods, defined for n = 0, 1, 2, . . . , by

yn = xn −F ′(xn)−1F (xn)

xn+1 = yn −AnF ′(xn)−1F (yn), (1.2)
An = A(xn,yn), A : Ω×Ω −→ L(B,B1), where A−1 ∈ L(B1,B). Many methods are special casesof (1.2). For example:

Received: 31 Aug 2021.
Key words and phrases. iterative methods; Banach space; convergence criterion; continuous functions.68

https://adac.ee
https://doi.org/10.28924/ada/ma.1.68


Eur. J. Math. Anal. 1 (2021) 69
Traub [35]

yn = xn −F ′(xn)−1F (xn)

xn+1 = yn −F ′(xn)−1F (yn), (1.3)
Newton [6]

yn = xn −F ′(xn)−1F (xn)

xn+1 = yn −F ′(yn)−1F (yn), (1.4)
Ostrowski [25]

yn = xn −F ′(xn)−1F (xn)

xn+1 = yn − (2[xn,yn; F ] −F ′(xn))−1F (yn), (1.5)
Kung-Traub [35–37]

yn = xn −F ′(xn)−1F (xn)

xn+1 = yn − [xn,yn; F ]−1F ′(xn)[xn,yn; F ]−1F (yn), (1.6)
Ostrowski-type [25]

yn = xn −F ′(xn)−1F (xn)

xn+1 = yn − (2[xn,yn; F ]−1 −F ′(xn)−1)F (yn), (1.7)
Sharma type [32]

yn = xn −F ′(xn)−1F (xn)

xn+1 = yn −P (xn,yn)F ′(xn)−1F (yn). (1.8)
To obtain all these special cases choose, An = I,An = F ′(yn)−1F ′(xn), An = (2[xn,yn; F ] −
F ′(xn))F

′(xn), An = [xn,yn; F ]
−1F ′(xn)[xn,yn; F ]

−1F ′(xn), An = (2[xn,yn; F ]
−1−F ′(xn)−1)F ′(xn),

An = P (xn,yn), respectively, where [., .; F ] : Ω × Ω −→ L(B,B1) is a divided difference of orderone and P : Ω×Ω −→ L(B,B1) is weight operator [32] (see also [15,28,40] and reference therein).These special methods were shown to be of order four using Taylor expansion and assumptions onthe fifth order derivative of F, which is not on these methods . So, the assumptions on the fifthderivative reduce the applicability of these methods [1–41].For example: Let B = B1 = R, Ω = [−0.5, 1.5]. Define λ on Ω by
λ(t) =

{
t3 log t2 + t5 − t4 if t 6= 0

0 if t = 0.

Then, we get t∗ = 1, and
λ′′′(t) = 6 log t2 + 60t2 − 24t + 22.



Eur. J. Math. Anal. 1 (2021) 70
Obviously λ′′′(t) is not bounded on Ω. So, the convergence of method (1.2) is not guaranteed bythe previous analyses in [1–41].In this paper we introduce a majorant sequence and use our idea of recurrent functions to extendthe applicability of method (1.2). Our analysis includes error bounds and results on uniqueness of
x∗ based on computable Lipschitz constants not given before in [1–41] and in other similar studiesusing Taylor series. Our idea is very general. So, it applies on other methods too.The rest of the paper is set up as follows: In Section 2 we present results on majorizing sequences.Sections 3,4 contain the semi-local and local convergence, respectively, where in Section 4 thenumerical experiments are presented. Concluding remarks are given in the last Section 5.

2. Results on majorizing sequences
We recall the definition followed by convergence results.

Definition 2.1. Let {w̄n} be a sequence in a Banach space. Then, a nondecreasing scalar sequence
{wn} is called majorizing for {w̄n} if

‖w̄n+1 − w̄n‖≤ wn+1 −wn for each n = 0, 1, 2, . . . . (2.1)
Sequence {wn} is used instead to study the convergence of {w̄n} [23–25]. Set M = [0,∞).Let η > 0, P0 : M −→ R, P : M −→ R, a : M × M × M −→ R, ā : M × M × M −→ Rand b : M ×M ×M ×M −→ R be continuous and nondecreasing functions. Set an = a(n) and

ξn = b(n). Define scalar sequences {sn}, {tn} for each n = 0, 1, 2, . . . by t0 = 0,s0 = η,
tn+1 = sn + ᾱn(sn − tn)

sn+1 = tn+1 + βn(tn+1 − sn), (2.2)
where ᾱn = ān ∫ 10 P̄ ((1 −θ)(sn − tn))dθ and

βn =
ξn

1 −P0(tn+1)
, ān =

{
ā, if n = 0

a, if n = 1, 2, . . . ,

P̄ =

{
P0, if n = 0

P, if n = 1, 2, . . .Next, we present results on the convergence of sequence {sn},{tn}.
LEMMA 2.2. Suppose that there exists µ > 0 such that for each n = 0, 1, 2, . . . ,

tn ≤ µ (2.3)
and

P0(µ) < 1. (2.4)



Eur. J. Math. Anal. 1 (2021) 71
Then, sequences {sn},{tn} converge to their unique least upper bound t∗ ∈ [η,µ] and

tn ≤ sn ≤ tn+1.

Proof. It follows from (2.2)-(2.4) that these sequences are nondecreasing, bounded from aboveby µ, and as such they converge to t∗.
�

LEMMA 2.3. If function P0 is increasing then conditions (2.3) and (2.4) can be replaced by

tn ≤ P−10 (1). (2.5)
Proof. Set µ = P−10 (1) in Lemma 2.2.

�

REMARK 2.4. Conditions (2.3)-(2.5) are very general and can be verified only in special cases.
That is why we present stronger conditions that are easier to verify.

Define functions f and g on the interval [0, 1) by
f (t) = a(

η

1 − t
,
η

1 − t
,t2η)

∫ 1
0

P ((1 −θ)t2η)dθ− t

and
g(t) = b(

η

1 − t
,
η

1 − t
,t2η,t3η) + tP0(

η

1 − t
) − t.

Suppose that these functions have minimal zeros λf and λg in (0, 1), respectively. Set λ =
min{λf ,λg} and λ0 = max{α0,β0}. Then, we can show the third result on majorizing sequencefor method (1.2).
LEMMA 2.5. Suppose that

µ0 ≤ λ0 ≤ λ. (2.6)
Then, sequences {sn},{tn} are nondecreasing, bounded from above by t∗∗ = η1−λ, and converge
to t∗ ∈ [0,t∗∗]. Moreover, the following estimates hold for each n = 1, 2, . . .

0 ≤ sn − tn ≤ λ(tn − sn−1) ≤ λ2nη, (2.7)
0 ≤ tn+1 − sn ≤ λ(sn − tn) ≤ λ2n+1η, (2.8)

0 ≤ sn ≤
1 −λ2n+1

1 −λ
η (2.9)

and

0 ≤ tn+1 ≤
1 −λ2n+1

1 −λ
η. (2.10)



Eur. J. Math. Anal. 1 (2021) 72
Proof. Estimates (2.7)-(2.10) hold if

0 ≤ αm ≤ λ, (2.11)
0 ≤ βm ≤ λ, (2.12)

and
tm ≤ sm ≤ tm+1, (2.13)

are true for m = 0, 1, 2, . . . . These estimates hold for m = 0 by (2.6). We suppose that (2.11)-(2.13) are true for m = 1, 2, . . .n. By induction hypotheses, (2.7) and (2.8), we have
sm ≤ tm + λ2mη ≤ sm−1 + λ2m−1η + λ2mη

≤ η + λη + . . . + λ2mη

=
1 −λ2m+1

1 −λ
η <

η

1 −λ
= t∗∗,

and
tm+1 ≤ sm + λ2m+1η ≤ tm + λ2mη + λ2m+1η

≤ η + λη + . . . + λ2m+1η

=
1 −λ2m+2

1 −λ
η <

η

1 −λ
= t∗∗.

Therefore, by (2.13) and the induction hypotheses, we see that sequences {sm} and {tm} arenondecreasing. Then, (2.11) shall be true if
a(tm,sm,sm − tm)

∫ 1
0

ψ((1 −θ)(sm − tm))dθ ≤ λ

or
a(

1 −λ2m

1 −λ
η,

1 −λ2m+1

1 −λ
η,λ2mη)

∫ 1
0

P ((1 −θ)λ2mη)dθ ≤ λ

or
a(

η

1 −λ
,

η

1 −λ
,λ2η)

∫ 1
0

ψ((1 −θ)λ2η)dθ ≤ λ

or
f (λ) ≤ 0,

which is true by the definition of λf and λ. Similarly, (2.12) shall be true if
b(

1 −λ2m

1 −λ
η,

1 −λ2m

1 −λ
η,λ2mη,λ2m+1η)

+λP0(
1 −λ2m+2

1 −λ
η) ≤ λ,

or
b(

η

1 −λ
,

η

1 −λ
,λ2η,λ3η) + λP0(

η

1 −λ
) ≤ λ



Eur. J. Math. Anal. 1 (2021) 73
or

g(λ) ≤ 0,

which is also true by the definition of λg and λ. Hence, we conclude (2.13) holds and limm−→∞ sm =
limm−→∞ tm = t

∗.

�

3. Semi-local convergence
Let U(x0, r) = {x ∈ B : ‖x − x0‖ < r,r > 0} and U[x0, r] = {x ∈ B : ‖x − x0‖ ≤ r,r > 0}.We use some parameters and functions. Consider M = [0,∞). Suppose that there exists function

P0 : M −→ M which is continuous and nondecreasing such that functions P0(t) − 1 = 0 has aminimal zero s ∈ (0,∞). Set M0 = [0,s). Suppose function P0 : M0 −→ M is continuous andnondecreasing. The following conditions (C) are needed:
(C1) There exists x0 ∈ Ω and η > 0 such that F ′(x0)−1 ∈ L(B1,B) and

‖F ′(x0)−1F (x0)‖≤ η.

(C2) For each x ∈ Ω
‖F ′(x0)−1(F ′(u) −F ′(x0))‖≤ P0(‖u −x0‖).

Set S0 = U(x0,s) ∩ Ω.(C3) For each x,y ∈ S0 the following hold
‖F ′(x0)−1(F ′(y) −F ′(x))‖≤ P (‖y −x‖)

(C4) For each n = 0, 1, 2, . . .
‖AnF ′(xn)−1F ′(x0)‖≤ an

F ′(x0)
−1([y,x; F ] −F ′(x))‖≤ L2‖y −x‖

and
‖F ′(x0)−1Hn‖≤ ξn,

where
Hn = F

′(x0)
−1

∫ 1
0

(F ′(yn + θ(xn+1 −yn)) −F ′(xn)A−1n )dθ.

(C5) Conditions of Lemma 2.2 or Lemma 2.3 or Lemma 2.5 hold.and(C6) U[x0,t∗] ⊂ Ω.Then, we can show the semi-local convergence of method (1.2) using the conditions (C) and thepreceding notation.



Eur. J. Math. Anal. 1 (2021) 74
THEOREM 3.1. Under the conditions (C), sequences {yn},{xn} generated by method (1.2) are
well defined in U[x0,t∗], remain in U[x0,t∗] for each n = 0, 1, 2, . . . and converge to a solution
x∗ ∈ U[x0,t∗] of equation F (x) = 0. Moreover, the following error estimates hold for each n =
0, 1, 2, . . .

‖x∗ −xn‖≤ t∗ − tn.

Proof. We shall show items
(Pm) ‖ym −xm‖≤ sm − tm(Qm) ‖xm+1 −ym‖≤ tm+1 − smusing mathematical induction on integer m. By the first substep of method (1.2) for n = 0 and (C1),we have

‖y0 −x0‖ = ‖F ′(x0)−1F (x0)‖≤ η = s0 − t0 = s0 ≤ t∗,so y0 ∈ U[x0,t∗] and (P0) holds. We can write by the first sustep of method (1.2) that
F (y0) = F (y0) −F (x0) −F ′(x0)(y0 −x0) =

∫ 1
0

(F ′(x0 + θ(y0 −x0)) −F ′(x0))(y0 −x0)dθ,

leading by (C2) and (P0) to
‖F ′(x0)−1F (y0)‖ ≤

∫ 1
0

P0(θ‖y0 −x0‖)dθ‖y0 −x0‖

≤
∫ 1
0

P̄ (θ(s0 − t0))dθ(s0 − t0). (3.1)
Let z ∈ U(x0,t∗). In view of (C2), we get

‖F ′(x0)−1(F ′(z) −F ′(x0))‖ ≤ P0(‖z −x0‖)

≤ P0(t∗) < 1, (3.2)
so

‖F ′(z)−1F ′(x0)‖≤
1

1 −P0(‖z −x0‖)
(3.3)

holds by a lemma on invertible linear operators due to Banach [24] and (3.2). Therefore, iterate x1is well defined and we can write in turn by (C3) and (3.3) (for z = x0,y0)
‖x1 −y0‖ = ‖A0F ′(x0)−1F (y0)‖

≤ ‖A0F ′(x0)−1F ′(x0)‖‖
∫ 1
0

F ′(x0)
−1(F ′(x0 + θ(y0 −x0)) −F ′(x0))dθ(y0 −x0)‖

≤
a0

∫ 1
0
P̄ ((1 −θ)‖y0 −x0‖)dθ‖y0 −x0‖

1 −P0(‖x0 −x0‖)

≤
a0

∫ 1
0
P̄ ((1 −θ)(s0 − t0))dθ

1 −P0(0)
(s0 − t0) = t1 − s0, (3.4)



Eur. J. Math. Anal. 1 (2021) 75
showing (Q0). Then, we have

‖x1 −x0‖≤‖x0 −y0‖ + ‖y0 −x0‖≤ t1 − s0 + s0 − t0 = t1 ≤ t∗,

so x1 ∈ U[x0,t∗]. Moreover, we can write
F (x1) = F (x1) −F (y0) + F (y0)

= F (x1) −F (y0) −F ′(x0)A−10 (x1 −y0)

=

∫ 1
0

(F ′(y0 + θ(x1 −x0)) −F ′(x0)A−10 )dθ(x1 −y0)

= H0(x1 −y0), (3.5)
since by the second substep of method (1.2), we have F (y0) = −F ′(x0)A−10 (x1−y0). By (C3), (3.4)and (3.5), we obtain

‖F ′(x0)−1F (x1)‖ ≤ ‖F ′(x0)−1H0‖‖x1 −y0‖

≤ ξ0(t1 − s0), (3.6)
so

‖y1 −x1‖ ≤ ‖F ′(x1)−1F ′(x0)‖‖F ′(x0)−1F (x1)‖

≤
ξ0(t1 − s0)
1 −P0(t1)

= s1 − t1, (3.7)
showing (P1) for m = 1. Suppose (Pm), (Qm) hold ym and xm+1 ∈ U[x0,t∗]. Then, by repeatingthese computations with xm,ym,xm+1 replacing x0,y0,x1, respectively, we complete the induction.Moreover, sequence {xm} is complete in a Banach space, so it converges to some x∗ ∈ U[x0,t∗].Finally, by letting m −→∞ in the estimation

‖F ′(x0)−1F (xm+1)‖≤ ξm(tm+1 − sm) (3.8)
and using the continuity of F, we conclude F (x∗) = 0.

�Next, we present a result for uniqueness of the solution x∗.
PROPOSITION 3.2. Suppose
(a) x∗ is a solution of F (x) = 0
(b) there exists s̃ ≥ t∗ such that ∫ 1

0

P0((1 −θ)s̃ + θt∗)dθ < 1. (3.9)
Set S1 = U[x0, s̃] ∩ Ω. Then, the only solution of equation F (x) = 0 in the region S1 is x∗.



Eur. J. Math. Anal. 1 (2021) 76
Proof. Set T = ∫ 1

0
F ′(x̃ + θ(x∗ − x̃))dθ for some x̃ ∈ S1 with F (x̃) = 0. Using (C2) and (3.9),we get

‖F ′(x0)−1(T −F ′(x0))‖ ≤
∫ 1
0

P0(‖x̃ + θ(x∗ − x̃) −x0‖dθ

≤
∫ 1
0

P0((1 −θ)‖x̃ −x0‖ + θ‖x∗ −x0‖)dθ

≤
∫ 1
0

P0((1 −θ)s̃ + θt∗)dθ < 1,

leading to x̃ = x∗, where we used the identity T (x∗ − x̃) = F (x∗) −F (x̃) = 0 − 0 = 0 and theinvertability of T.
�

REMARK 3.3. Let us specialize operators An to see how sequences {sn},{tn},{an},{ξn},{αn}
and {βn} are defined. Choose the case of Newton’s method (1.4). Then, we have

‖AnF ′(xn)−1F ′(x0)‖ = ‖F ′(yn)−1F ′(x0)‖

≤
1

1 −P0(‖yn −x0‖)
,

and

‖F ′(x0)−1Hn‖ = ‖
∫ 1
0

F ′(x0)
−1(F ′(yn + θ(xn+1 −yn)) −F ′(xn)A−1n )dθ‖

= ‖
∫ 1
0

F ′(x0)
−1(F ′(yn + θ(xn+1 −yn)) −F ′(yn))dθ‖

≤
∫ 1
0

P̄ (θ‖xn+1 −yn‖)dθ

≤
∫ 1
0

P̄ (θ(tn+1 − sn))dθ,

so we can choose

an =
1

1 −P0(sn)
(3.10)

and

ξn =

∫ 1
0

P̄ (θ(tn+1 − sn))dθ. (3.11)
In this case we can show another result on majorizing sequences which is weaker than Lemma 2.5
for the interesting case P0(t) = L0t and P (t) = Lt. We get in this special case that

αn =
L(sn − tn)

2(1 −L0sn)
(3.12)

and

βn =
L(tn+1 − sn)

2(1 −L0tn+1)
. (3.13)



Eur. J. Math. Anal. 1 (2021) 77
Define sequences of function {f (1)n },{f (2)n } on the interval [0, 1) by

f
(1)
n (t) =

L

2
t2n−1η + L0(1 + t + . . . + t

2n)η − 1,

f
(2)
n (t) =

L

2
t2nη + L0(1 + t + . . . + t

2n+1)η − 1,

and polynomial ϕ by
ϕ(t) = L0t

3 + (L0 +
L

2
)t2 −

L

2
.

Notice that ϕ(0) = −L
2

and ϕ(1) = 2L0. Denote by ρ the smallest zero of polynomial ϕ in (0, 1)assured to exist by the intermediate value theorem.
LEMMA 3.4. Suppose that

λ0 ≤ ρ < 1 −L0η. (3.14)
Then, the conclusions of Lemma 2.5 hold for sequences {sn},{tn} with ρ replacing λ.

Proof. We must show this time
0 ≤

L(sm − tm)
2(1 −L0sm)

≤ ρ, (3.15)
0 ≤

L(tm+1 − sm)
2(1 −L0tm+1)

≤ ρ (3.16)
and

tm ≤ sm ≤ tm+1. (3.17)These estimates hold for m = 0 by (3.14) and the definition of these sequences. Then, as in Lemma2.5 we can show instead for (3.15) that
L

2
ρ2mη + ρL0(1 + ρ + . . . + ρ

2m)η − 1 ≤ 0. (3.18)
This estimate motivates us to define recurrent functions f (1)m by

f
(1)
m (t) =

L

2
t2m−1η + L0(1 + t + . . . + t

2m)η − 1. (3.19)
We shall find a relationship between recurrent functions f (1)m+1 and f (1)m . By definition (3.9), we havein turn that

f
(1)
m+1(t) =

L

2
t2m+1η + L0(1 + t + . . . + t

2m+2)η − 1

−
L

2
t2m−1η −L0(1 + t + . . . + t2m)η + 1 + f

(1)
m (t)

= f
(1)
m (t) + (

L

2
t2 −

L

2
+ L0(t

2 + t3))t2m−1η

= f
(1)
m (t) + p(t)t

2m−1η. (3.20)
In particular, we have

fm+1(ρ) = fm(ρ), (3.21)



Eur. J. Math. Anal. 1 (2021) 78
so evidently (3.8) holds if

f
(1)
m (ρ) ≤ 0. (3.22)Define f (1)∞ (t) = limm−→∞ f (1)m (t). Then, we have

f∞(t) =
L0η

1 − t
− 1. (3.23)

Then, (3.22) holds if
f∞(ρ) ≤ 0, (3.24)which is true by (3.14). Similarly, (3.16) holds if

L

2
ρ2m+1η + ρL0(1 + ρ + . . . + ρ

2m+1)η −ρ ≤ 0 (3.25)
or

f
(2)
m (ρ) ≤ 0. (3.26)As in (3.20), we get in turn that

f
(2)
m+1(t) =

L

2
t2m+2η + L0(1 + t + . . . + t

2m+3)η − 1

−
L

2
t2mη −L0(1 + t + . . . + t2m+1)η + 1 + f

(2)
m (t)

= f
(2)
m (t) + ϕ(t)t

2mη. (3.27)
Define f (2)∞ (t) = lim(2)m−→∞(t). Then, we get again

f (2)∞ (t) = f
(1)
∞ (t),

so
f (2)∞ (ρ) ≤ 0,can be shown instead of (3.26). But this is true by (3.14). The induction for items (3.15)-(3.17) iscompleted. The rest of the proof follows as in Lemma 2.2.

4. Local Convergence
We shall introduce real parameters and functions to be used in the convergence analysis. Set

M = [0,∞).Suppose function(i) ψ0(t) − 1 = 0 has a smallest zero R0 ∈ M −{0}, where function ψ0 : M −→ M is continuousand nondecreasing. Set M0 = [0,R0).(ii) ψ1(t)−1 = 0, has a smallest zero R1 ∈ M0−{0}, where function ψ : M0 −→ M is continuousand nondecreasing and ψ1 : M0 −→ M is defined by
ψ1(t) =

∫ 1
0
ψ((1 −θ)t)dθ
1 −ψ0(t)

.



Eur. J. Math. Anal. 1 (2021) 79
(iii) ψ0(ψ1(t)t)−1 has a smallest zero R̄1 ∈ M0−{0}. Ser R̄2 = min{R0, R̄1} and M1 = [0, R̄2).(iv) ψ2(t) − 1 = 0 has a smallest zero R2 ∈ M1 −{0}, where

ψ2(t) = [ψ1(ψ1(t)t) +
(ψ0(t) + h(t,ψ1(t)t))

∫ 1
0
ω(θψ1(t)t)dθ

(1 −ψ0(t))(1 −ψ0(ψ1(t)t))
]ψ1(t),

where ω : M1 −→ M and h : M ×M1 −→ M are continuous and nondecreasing. We shall showthat
R = min{R1,R2}, (4.1)is a convergence radius for method (1.2). Set M2 = [0,R). These definitions, imply that for each

t ∈ M2
0 ≤ ψ0(t) < 1, (4.2)

0 ≤ ψ0(ψ1(t)t) < 1, (4.3)and
0 ≤ ψi (t) < 1, i = 1, 2. (4.4)The conditions (H) shall be used provided that x∗ is a simple solution of equation F (x) = 0.Suppose:

(H1) For each x ∈ Ω
‖F ′(x∗)−1(F ′(x) −F ′(x∗))‖≤ ψ0(‖x −x0‖).

Set Ω0 = U(x∗,R0) ∩ Ω.(H2) For each x,y ∈ Ω0
‖F ′(x∗)−1(F ′(y) −F ′(x))‖≤ ψ(‖y −x‖),

‖F ′(x∗)−1F ′(x)‖≤ ω(‖x −x∗‖),

and
‖F ′(x∗)−1(F ′(x∗) −A(x,y))‖≤ h(‖x −x∗‖,‖y −x∗‖).

(H3) U[x∗,R] ⊂ Ω.
Next, we show the local convergence of method (1.2) based on the preceding notation and conditions(H)..
THEOREM 4.1. Under conditions (H) further suppose that x0 ∈ U(x∗,R) − {x∗}. Then, we
conclude limn−→∞xn = x∗.

Proof. Let v ∈ U(x∗,R) −{x∗}. Using (4.1), (4.2), and (H1) we obtain in turn that
‖F ′(x∗)−1(F ′(v) −F ′(x∗))‖≤ ψ0(‖v −x∗‖) ≤ ψ0(R) < 1,

so
‖F ′(v)−1F ′(x∗)‖≤

1

1 −ψ0(‖v −x∗‖)
. (4.5)



Eur. J. Math. Anal. 1 (2021) 80
In particular, iterate is well defined for v = x0 and the first substep of method (1.2), from which wecan also write

y0 −x∗ = x0 −x∗ −F ′(x0)−1F (x0)

= (F ′(x0)
−1F ′(x∗))

×(
∫ 1
0

F ′(x∗)−1(F ′(x∗ + θ(x0 −x∗)) −F ′(x0))dθ(x0 −x∗).

(4.6)
By (4.1), (4.4) (for i = 1), (4.5) (for v = x0), (4.6) and (H2), we get in turn that

‖y0 −x∗‖ ≤
∫ 1
0
ψ̄((1 −θ)‖x0 −x∗‖)dθ‖x0 −x∗‖

1 −ψ0(‖x0 −x∗‖)

≤ ‖x0 −x∗‖ < R, (4.7)
so y0 ∈ U(x∗,R). We also have that (4.5) holds for v = y0, and iterate x1 is well defined fromwhich we can write in turn that

x1 −x∗ = y0 −x∗ −F ′(y0)−1F (x0)

+(F ′(y0)
−1 −A0F ′(x0)−1)F (y0)

= y0 −x∗ −F ′(y0)−1F (y0) + F ′(y0)−1(F ′(x0) −A0)F ′(x0)1F (y0).

(4.8)
In view of (4.1), (4.4) (for i = 2), (4.5)(for v = x0,y0), (4.7), (4.8) and (H2), we obtain in turn
‖x1 −x∗‖ ≤ [ψ1(ψ1(‖x0 −x∗‖))

+
(ψ0(‖x0 −x∗‖) + h(‖x0 −x∗‖,‖y0 −x∗‖))

∫ 1
0
ω(θ‖y0 −x∗‖)dθ

(1 −ψ0(‖y0 −x∗‖))(1 −ψ0(‖x0 −x∗‖))
]‖y0 −x∗‖

≤ ψ2(‖x0 −x∗‖)‖x0 −x∗‖≤‖x0 −x∗‖ < R, (4.9)
so x1 ∈ U(x∗,R). Simply, switch x0,y0,x1 by xm,ym,xm+1, respectively in the preceding calcula-tions to get

‖ym −x∗‖≤ ψ1(‖xm −x∗‖)‖xm −x∗‖≤‖xm −x∗‖ < R (4.10)and
‖xm+1 −x∗‖≤ ψ2(‖xm −x∗‖)‖xm −x∗‖≤‖xm −x∗‖. (4.11)Then, by the estimation

‖xm+1 −x∗‖≤ d‖xm −x∗‖ < R, (4.12)where d = ψ2(‖x0 −x∗‖) ∈ [0, 1), we get limm−→∞xm = x∗ and xm+1 ∈ U(x∗,R).
�Next, we present a uniqueness result.



Eur. J. Math. Anal. 1 (2021) 81
PROPOSITION 4.2. Suppose:
(i) There exists a simple solution x∗ of equation F (x) = 0
(ii) There exists R∗ ≥ R such that ∫ 1

0

ψ0(θR
∗)dθ < 1. (4.13)

Set Ω2 = Ω ∩U[x∗,R∗]. Then, the only solution of equation F (x) = 0 in the region Ω2 is x∗.

Proof. Consider x̃ ∈ Ω1 with F (x̃) = 0. Set T = ∫ 10 F ′(x∗ + θ(x̃−x∗))dθ. Then, using (H1) and(4.13), we get in turn that
‖F ′(x∗)−1(T −F ′(x∗))‖ ≤

∫ 1
0

ψ0(θ‖x̃ −x∗‖dθ

≤
∫ 1
0

ψ0(θR
∗)dθ < 1,

so x̃ = x∗, follows by T−1 ∈ L(B1,B) and T (x̃ −x∗) = F (x̃) −F (x∗) = 0 − 0 = 0.
�

5. Numerical Experiments
We provide some examples in this section.

EXAMPLE 5.1. Define function

q(t) = ξ0t + ξ1 + ξ2 sin ξ3t, x0 = 0,

where ξj, j = 0, 1, 2, 3 are parameters. Choose P0(t) = L0t and P (t) = Lt. Notice that L0 and L
are the center Lipschitz and Lipschitz constants, respectively. Then, from the graph of q(t) clearly
for ξ3 large and ξ2 small, L0L can be small (arbitrarily). Notice that

L0
L
−→ 0.

EXAMPLE 5.2. Let B = B1 = C[0, 1] and Ω = U[0, 1]. It is well known that the boundary value
problem [16].

ς(0) = 0, (1) = 1,

ς′′ = −ς −σς2

can be given as a Hammerstein-like nonlinear integral equation

ς(s) = s +

∫ 1
0

Q(s,t)(ς3(t) + σς2(t))dt

where σ is a parameter. Then, define F : Ω −→ B1 by

[F (x)](s) = x(s) − s −
∫ 1
0

Q(s,t)(x3(t) + σx2(t))dt.



Eur. J. Math. Anal. 1 (2021) 82
Choose ς0(s) = s and Ω = U(ς0,ρ0). Then, clearly U(ς0,ρ0) ⊂ U(0,ρ0 + 1), since ‖ς0‖ = 1.
Suppose 2σ < 5. Then, conditions (A) are satisfied for

L0 =
2σ + 3ρ0 + 6

8
, L =

σ + 6ρ0 + 3

4
,

and η = 1+σ
5−2σ. Notice that L0 < L.

In the last two examples we consider Traub’s method (1.3). So, we take A(x,y) = I and
h(s,t) = 0.

EXAMPLE 5.3. Consider the motion system

G′1(v1) = e
v1, G′2(y) = (e − 1)v2 + 1, G

′
3(v3) = 1

with G1(0) = G2(0) = G3(0) = 0. Let G = (G1,G2,G3). Let B = B1 = R3, Ω = Ū(0, 1),x∗ =
(0, 0, 0)T . Define function G on Ω for v = (v1,v2,v3)T by

G(v) = (ev1 − 1,
e − 1

2
v22 + v2,v3)

T .

Then, we get

G′(v) =


ex 0 0

0 (e − 1)v2 + 1 0
0 0 1

 ,
so ψ0(t) = (e−1)t, ψ(t) = e

1
e−1t, ω(t) = e

1
e−1 and K = e is the Lipschitz constant on Ω and ρT

is given in [29, 35]. Then, the radii:

R1 = 0.3827 = ρA =
2

2(e − 1) + e
1
e−1
, R2 = 0.3061 = R, ρT =

2

3K
= 0.2453.

EXAMPLE 5.4. Consider B = B1 = C[0, 1], Ω = U(0, 1) and Q : Ω −→ B1 defined by

Q(ς)(x) = %(x) − 5
∫ 1
0

xθς(θ)3dθ. (5.1)
We obtain

Q′(ς(ξ))(x) = ξ(x) − 15
∫ 1
0

xθς(θ)2ξ(θ)dθ, for each ξ ∈ D.
Then, since x∗ = 0, we set ψ0(t) = 7.5t, ψ(t) = 15t, ω(t) = 15 and K = 15. Then, the radii:

R1 = 0.0667 = ρA =
2

2(7.5) + 15
, R2 = 0.0290 = R, ρT =

2

3K
= 0.0444.

Notice that in the last two examples ρA is the radius given by us in [1–7] and is the largest.



Eur. J. Math. Anal. 1 (2021) 83
6. Conclusion

We have provided sufficient convergence criterion for the semi-local and local convergence oftwo-step methods. Upon specializing the parameters involved we show that although our majorizingsequence is more general than earlier ones: Convergence criteria are weaker (i.e., the utility of themethods is extended); the upper error estimates are more accurate (i.e. at least as few iterates arerequired to achieve a predecided error tolerance) and we have an at least as large ball containingthe solution. These benefits are obtained without additional hypotheses. According to our newtechnique we locate a more accurate domain than before containing the iterates resulting to moreaccurate (at least as small) Lipschitz condition.Our theoretical results are further justified using numerical experiments.
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	1. Introduction
	2. Results on majorizing sequences
	3. Semi-local convergence
	4. Local Convergence
	5. Numerical Experiments
	6. Conclusion
	References

