#  Originally presented in 2018 at NSSDF (Military Sensing Symposium)
conference Game theory analysis when playing the wrong game

Dan Zwillinger and Paul San Clemente  BAE Systems Information and Electronic
Systems Integration Inc.
600 District Ave
Burlington, MA, 01803
<EMAIL_ADDRESS><EMAIL_ADDRESS>

###### Abstract

In classical game theory, optimal strategies are determined for games with
complete information; this requires knowledge of the opponent’s goals. We
analyze games when a player is mistaken about their opponents goals. For
definitiveness, we study the (common) bimatrix formulation where both player’s
payoffs are matrices. While the payoff matrix weights are arbitrary, we focus
on strict ordinal payoff matrices, which can be enumerated. In this case, a
reasonable error would be for one player to switch two ordinal values in their
opponents payoff matrix. The mathematical formulation of this problem is
stated, and all 78 strict ordinal 2-by-2 bimatrix games are investigated. This
type of incomplete information game has not – to our knowledge – been studied
before.

###### Index Terms:

game theory, optimal strategy, wrong game

##  I Introduction

When playing a game it is important to know both your goals and your opponents
goals. If there is confusion about either of these, then a correct analysis
could result in the wrong strategy.

Brams [Bra93] gives an example from the 1979–1981 Iranian hostage situation.
He analyzed the ”game” that US President Carter was ”playing” and indicates
how he (might have) arrived at his strategy. Then Brams analyzed the different
”game” that Khomeini was ”playing.” Specifically, Bram modeled the Iranian
hostage ”game” using the payoff matrices

$A=\begin{bmatrix}4&2\\\ 3&1\\\ \end{bmatrix}\qquad
B_{\text{Carter}}=\begin{bmatrix}3&4\\\ 2&1\\\ \end{bmatrix}\qquad
B_{\text{Khomeini}}=\begin{bmatrix}2&4\\\ 1&3\\\ \end{bmatrix}$

where $A$ is the USA’s payoff matrix, $B_{\text{Carter}}$ is what Carter
thought Khomeini’s payoff matrix was and $B_{\text{Khomeini}}$ is what
Khomeini’s actual payoff matrix was. Since the games were different, President
Carter’s strategy (derived from analysis of $A$ and $B_{\text{Carter}}$) did
not resolve the game as he thought it would.

In game theory information can be ”complete”, which means that all players
know everything about the game (for example, Go, chess, and checkers) or the
game can be incomplete (for example, poker). There are many types of
incomplete information games. This paper addresses one type of incomplete
information game – the ”cost” of playing a wrong game that is close to the
correct game. We believe this analysis is new. We derive four problem
statements representing different interpretations of ”playing the wrong game”.
These cases represent different understandings of what the other player thinks
you are thinking. We then systematically evaluate these cases for a class of
interesting payoff matrices.

###  I-A Assumptions and Terminology

This paper considers two-player bimatrix games. That is, player 1 has $n$
different strategies available for use, uses strategy ${\mathbf{x}}\in\Delta$,
has payoff matrix $A$ of size $n\times m$, and receives payoff $P_{1}$. Here
$\Delta$ is the simplex representing all possible (mixed and pure)
strategies111As usual,
$\left\|\begin{bmatrix}c_{1}&\dots&c_{k}\end{bmatrix}\right\|_{1}=\sum_{i}^{k}|c_{i}|$.
$\Delta=\left\\{{\mathbf{z}}\mid 0\leq z_{i}\leq
1,\left\|{\mathbf{z}}\right\|_{1}=1\right\\}$. Similarly, player 2 has the
parameters $m$, ${\mathbf{y}}\in\Delta$, $B$, and $P_{2}$. The player payoffs
are $P_{1}={\mathbf{x}}^{\text{T}}A{\mathbf{y}}$ and
$P_{2}={\mathbf{x}}^{\text{T}}B{\mathbf{y}}$.

We assume that the values in $A$ and $B$ are non-negative. This means that the
problem is not zero sum; that is $A$ is not equal to $-B$.

The simultaneous optimal strategies
$\left\\{{\mathbf{x}}^{*},{\mathbf{y}}^{*}\right\\}$ are determined by each
player maximizing their payoff

$\begin{split}{\mathbf{x}}^{*}&=\arg\max_{{\mathbf{x}}\in\Delta}{\mathbf{x}}^{\text{T}}A{\mathbf{y}}^{*}\\\
{\mathbf{y}}^{*}&=\arg\max_{{\mathbf{y}}\in\Delta}{{\mathbf{x}}^{*}}^{\text{T}}B{\mathbf{y}}\\\
\end{split}$ (1)

where a superscript ”T” represents vector transpose. The solution of the
general bimatrix game in (1) can be found using the algorithm by Lemke and
Howson [Lem64].

##  II Problem Statement

This paper makes the following assumptions about the game being played:

  1. 1.

Player 1 uses payoff matrix $A$; both players know this.

  2. 2.

Player 1 believes that player 2 is using payoff matrix $B$.

  3. 3.

Matrix $B$ belongs to a family ${\cal F}(B)$ of related matrices.

  4. 4.

Player 2 is actually using payoff matrix $R$, which is in the family ${\cal
F}(B)$.

  5. 5.

The matrix $R$ may, or may not, be matrix $B$.

  6. 6.

Both payers are playing optimally, based on the information available to them.

That is, player 1 has incomplete information; player 2’s actual payoff matrix
is $R$, which player 1 does not know.

To precisely define the problem, we need to specify exactly what the players
believe to be true. For player 1 we have two assumptions:

E_

player 1 believes that player 2 is using the payoff matrix $B$.

F_

player 1 believes that player 2 is using a payoff matrix in ${\cal F}(B)$.

where we use ”E” to mean ”exact” and ”F” to mean ”family”. For player 2 we
also have two assumptions:

_E

player 2 believes that ”player 1 believes that player 2 is using the payoff
matrix $B$.”

_F

player 2 believes that ”player 1 believes that player 2 is using a payoff
matrix in ${\cal F}(B)$.”

Combining these assumptions, there are four cases to analyze
$\\{EE,EF,FE,FF\\}$. Each case has different solutions, with the payoffs to
player 1 being $\left\\{P_{1}^{EE},P_{1}^{EF},P_{1}^{FE},P_{1}^{FF}\right\\}$.

A last case that could be studied is the ”symmetric case”, when each of the
two players has equal uncertainty about the other player’s payoff matrix. This
game is described in section III-E.

Note that if there were complete information, and player 1 knew that player
2’s payoff matrix was $R$, then the optimal strategies
$\\{{\mathbf{x}}_{R},{\mathbf{y}}_{R}\\}$ would be found from (1)

$\begin{split}{\mathbf{x}}_{R}&=\arg\max_{{\mathbf{x}}\in\Delta}{\mathbf{x}}^{\text{T}}A{\mathbf{y}}_{R}\\\
{\mathbf{y}}_{R}&=\arg\max_{{\mathbf{y}}\in\Delta}{\mathbf{x}}_{R}^{\text{T}}B{\mathbf{y}}\\\
\end{split}$ (2)

Hence, player 1’s payoff would be

$P_{1}^{\text{complete}}={\mathbf{x}}_{R}^{\text{T}}A{\mathbf{y}}_{R}$ (3)

The purpose of this paper is to understand the difference between
$P_{1}^{\text{complete}}$ and
$\left\\{P_{1}^{EE},P_{1}^{EF},P_{1}^{FE},P_{1}^{FF}\right\\}$.

###  II-A Computational Specification

In this paper we evaluate solutions for all strict ordinal 2-by-2 bimatrix
games. An ordinal payoff matrix contains relative integer preferences. That
is, a strategy that returns a ”$k$” is preferred to a strategy that returns a
”$j$”, if $k>j$. A strict ordinal payoff matrix of size $n$-by-$m$ contains
the entries $1,2,\dots,mn$ and each value is used exactly once; there are no
ties among the preferences. Hence, all our payoff matrices contain each of the
values $\\{1,2,3,4\\}$ exactly once. The Carter and Khomeini payoff matrices
shown in the introduction are strict ordinal 2-by-2 payoff matrices.

There are 78 unique strict ordinal 2-by-2 bimatrix games; see Figure 1.
Uniqueness means that 2 matrix pairs cannot be made the same by re-ordering
strategies or switching players. This value is derived in Rapoport and Guyer
[Rap66].

  1. 1.

1 3 2 4 – 1 3 2 4

  2. 2.

4 1 2 3 – 2 4 3 1

  3. 3.

4 1 2 3 – 3 1 2 4

  4. 4.

4 1 2 3 – 4 2 1 3

  5. 5.

4 1 3 2 – 1 3 4 2

  6. 6.

4 1 3 2 – 2 1 3 4

  7. 7.

4 1 3 2 – 2 4 3 1

  8. 8.

4 1 3 2 – 3 1 2 4

  9. 9.

4 1 3 2 – 3 4 2 1

  10. 10.

4 1 3 2 – 4 2 1 3

  11. 11.

4 1 3 2 – 4 3 1 2

  12. 12.

4 2 1 3 – 1 2 4 3

  13. 13.

4 2 1 3 – 1 3 4 2

  14. 14.

4 2 1 3 – 2 1 3 4

  15. 15.

4 2 1 3 – 2 3 4 1

  16. 16.

4 2 1 3 – 2 4 3 1

  17. 17.

4 2 1 3 – 3 1 2 4

  18. 18.

4 2 1 3 – 3 2 1 4

  19. 19.

4 2 1 3 – 3 4 2 1

  20. 20.

4 2 1 3 – 4 1 2 3

  21. 21.

4 2 1 3 – 4 2 1 3

  22. 22.

4 2 1 3 – 4 3 1 2

  23. 23.

4 2 3 1 – 1 2 3 4

  24. 24.

4 2 3 1 – 1 2 4 3

  25. 25.

4 2 3 1 – 1 3 4 2

  26. 26.

4 2 3 1 – 1 4 3 2

  27. 27.

4 2 3 1 – 2 1 3 4

  28. 28.

4 2 3 1 – 2 3 4 1

  29. 29.

4 2 3 1 – 2 4 3 1

  30. 30.

4 2 3 1 – 3 1 2 4

  31. 31.

4 2 3 1 – 3 2 1 4

  32. 32.

4 2 3 1 – 3 4 1 2

  33. 33.

4 2 3 1 – 3 4 2 1

  34. 34.

4 2 3 1 – 4 1 2 3

  35. 35.

4 2 3 1 – 4 2 1 3

  36. 36.

4 2 3 1 – 4 3 1 2

  37. 37.

4 2 3 1 – 4 3 2 1

  38. 38.

4 3 1 2 – 1 2 3 4

  39. 39.

4 3 1 2 – 1 2 4 3

  40. 40.

4 3 1 2 – 1 3 4 2

  41. 41.

4 3 1 2 – 1 4 3 2

  42. 42.

4 3 1 2 – 2 1 3 4

  43. 43.

4 3 1 2 – 2 1 4 3

  44. 44.

4 3 1 2 – 2 3 1 4

  45. 45.

4 3 1 2 – 2 3 4 1

  46. 46.

4 3 1 2 – 2 4 3 1

  47. 47.

4 3 1 2 – 3 1 2 4

  48. 48.

4 3 1 2 – 3 2 1 4

  49. 49.

4 3 1 2 – 3 2 4 1

  50. 50.

4 3 1 2 – 3 4 1 2

  51. 51.

4 3 1 2 – 3 4 2 1

  52. 52.

4 3 1 2 – 4 1 2 3

  53. 53.

4 3 1 2 – 4 1 3 2

  54. 54.

4 3 1 2 – 4 2 1 3

  55. 55.

4 3 1 2 – 4 3 1 2

  56. 56.

4 3 1 2 – 4 3 2 1

  57. 57.

4 3 2 1 – 1 2 3 4

  58. 58.

4 3 2 1 – 1 2 4 3

  59. 59.

4 3 2 1 – 1 3 2 4

  60. 60.

4 3 2 1 – 1 3 4 2

  61. 61.

4 3 2 1 – 1 4 2 3

  62. 62.

4 3 2 1 – 1 4 3 2

  63. 63.

4 3 2 1 – 2 1 3 4

  64. 64.

4 3 2 1 – 2 1 4 3

  65. 65.

4 3 2 1 – 2 3 1 4

  66. 66.

4 3 2 1 – 2 3 4 1

  67. 67.

4 3 2 1 – 2 4 3 1

  68. 68.

4 3 2 1 – 3 1 2 4

  69. 69.

4 3 2 1 – 3 1 4 2

  70. 70.

4 3 2 1 – 3 2 1 4

  71. 71.

4 3 2 1 – 3 2 4 1

  72. 72.

4 3 2 1 – 3 4 1 2

  73. 73.

4 3 2 1 – 3 4 2 1

  74. 74.

4 3 2 1 – 4 1 2 3

  75. 75.

4 3 2 1 – 4 1 3 2

  76. 76.

4 3 2 1 – 4 2 1 3

  77. 77.

4 3 2 1 – 4 2 3 1

  78. 78.

4 3 2 1 – 4 3 1 2

Figure 1: All 78 strict ordinal 2-by-2 bimatrix games; the first 4 elements
are $A$ (column-wise), the next 4 elements are $B$ (column-wise).

For a payoff matrix $B$, we define the family ${\cal F}(B)$ to contain 4
matrices: $B$ and the 3 matrices derivable from $B$ by switching a pair of
adjacent preferences. (These are called ”swaps” by [Rob05].) That is, given
$B$ the matrix $B_{12}$ is obtained by switching the values ”1” and ”2”, the
matrix $B_{23}$ is obtained by switching the values ”2” and ”3”, and the
matrix $B_{34}$ is obtained by switching the values ”3” and ”4”. The
motivation for this type of incomplete information is that we may generally
understand someone else’s preferences, but we may make a single mistake. For
example, suppose you articulated your best friend’s top 4 dessert choices. Are
you confident that you have not switched their 2nd and 3rd preferences?

Here is a numerical example: the payoff matrix $B=\begin{bmatrix}1&3\\\ 4&2\\\
\end{bmatrix}$ is in the same family as

$B_{12}=\begin{bmatrix}\bf{2}&3\\\ 4&\bf{1}\\\ \end{bmatrix}\quad
B_{23}=\begin{bmatrix}1&\bf{2}\\\ 4&\bf{3}\\\ \end{bmatrix}\quad
B_{34}=\begin{bmatrix}1&\bf{4}\\\ \bf{3}&2\\\ \end{bmatrix}$

##  III Problem Statement – Mathematically

This section contains mathematical descriptions of the different games of
interest.

###  III-A Game EE

In this case

E_

player 1 believes that player 2 is using the payoff matrix $B$.

_E

player 2 believes that ”player 1 believes that player 2 is using the payoff
matrix $B$.”

From assumption ”E_” player 1 solves the system (from (1)) for the strategies
$\left\\{{\mathbf{x}}_{EE},{\mathbf{y}}_{*}\right\\}$:

$\begin{split}{\mathbf{x}}_{EE}&=\arg\max_{{\mathbf{x}}\in\Delta}{\mathbf{x}}^{\text{T}}A{\mathbf{y}}_{*}\\\
{\mathbf{y}}_{*}&=\arg\max_{{\mathbf{y}}\in\Delta}{\mathbf{x}}_{EE}^{\text{T}}B{\mathbf{y}}\\\
\end{split}$ (4)

From assumption ”_E” player 2 knows that player 1 will solve (4), and is using
the strategy ${\mathbf{x}}_{EE}$. However, since player 2 is using payoff
matrix $R$, player 2 determines their strategy from

${\mathbf{y}}_{EE}=\arg\max_{{\mathbf{y}}\in\Delta}\
{\mathbf{x}}_{EE}^{\text{T}}R{\mathbf{y}}$ (5)

This can be readily solved by determining the indices $I$ of the maximum
values of the vector ${\mathbf{x}}_{EE}^{\text{T}}R$; all other indices in
${\mathbf{y}}_{EE}$ have the numerical value zero (that is, $y_{i}=0$ for
$i\not\in I$). Any values can be used for $y_{i}$ when $i\in I$, consistent
with ${\mathbf{y}}\in\Delta$; the payoff to player 2 will be the same.

###  III-B Game FE

In this case

F_

player 1 believes that player 2 is using a payoff matrix in ${\cal F}(B)$.

_E

player 2 believes that ”player 1 believes that player 2 is using the payoff
matrix $B$.”

That is, player 2 is confused about what player 1 knows. By ”_E” player 2
thinks that player 1 will perform the same analysis that player 1 performed in
(4). Then as before, player 2 calculates the quantity in (5), re-written here
is new variables and explicitly showing the dependence on $R$:

${\mathbf{y}}_{FE}(R)=\arg\max_{{\mathbf{y}}\in\Delta}\
{\mathbf{x}}_{EE}^{\text{T}}R{\mathbf{y}}$ (6)

However, by ”F_”, player 1 knows not to use ${\mathbf{x}}_{EE}$ but needs to
incorporate ${\cal F}$ thinking. That is, player 1 needs to solve

${\mathbf{x}}_{FE}=\arg\max_{{\mathbf{x}}}\left(\min_{R\in{\cal
F}(B)}{\mathbf{x}}^{\text{T}}A{\mathbf{y}}_{FE}(R)\right)$ (7)

to ensure a maximal return, regardless of which $R\in{\cal F}(B)(B)$ that
player 2 chooses.

This can be solved as follows. First, define
${\mathbf{a}}_{j}=A{\mathbf{y}}_{EF}(R_{j})$ for each $R_{j}\in{\cal F}(B)$.
Then222The authors thank Peter Kingston for this observation. we note that (7)
can be written as

$\begin{split}{\mathbf{x}}_{FE}&=\arg\max_{{\mathbf{x}}}\left(\min_{R_{j}\in{\cal
F}}{\mathbf{x}}^{\text{T}}{\mathbf{a}}_{j}\right)\\\
&=\arg\max_{{\mathbf{x}}}\left(\max_{z}\left\\{z\mid
z\leq{\mathbf{x}}^{\text{T}}{\mathbf{a}}_{j}\right\\}\right)\\\ \end{split}$
(8)

This, in turn, can be written as a linear programming problem for the values
of ${\mathbf{z}}$ and $z$:

$\begin{split}\max_{\\{{\mathbf{x}},z\\}}\qquad&z\\\
z&\leq{\mathbf{a}}_{j}^{\text{T}}{\mathbf{x}}\qquad\text{for all $j$}\\\
\left\|{\mathbf{x}}\right\|_{1}&=1\\\ 0&\leq x_{i}\leq 1\\\ 0&\leq z\\\
\end{split}$ (9)

The ${\mathbf{x}}$ part of the solution of (9) is now ${\mathbf{x}}_{FE}$.

###  III-C Game FF

In this case

F_

player 1 believes that player 2 is using a payoff matrix in ${\cal F}(B)$.

_F

player 2 believes that ”player 1 believes that player 2 is using a payoff
matrix in ${\cal F}(B)$.”

Here, player 1 does not know player 2’s payoff matrix, but knows the family
(${\cal F}$) that contains it. Hence, player 1 needs to solve the following to
find ${\mathbf{x}}_{FF}$:

$\begin{split}{\mathbf{x}}_{FF}&=\arg\max_{{\mathbf{x}}}\left(\min_{R\in{\cal
F}(B)}{\mathbf{x}}^{\text{T}}A{\mathbf{y}}_{*}(R)\right)\\\
{\mathbf{y}}_{*}(R)&=\arg\max_{{\mathbf{y}}}\
{\mathbf{x}}_{FF}^{\text{T}}R{\mathbf{y}}\\\ \end{split}$ (10)

After player 1 obtains the optimal ${\mathbf{x}}_{FF}$, player 2 will solve
the same problem as in (6), rewritten in FF variables:

${\mathbf{y}}_{FF}=\arg\max_{{\mathbf{y}}\in\Delta}\
{\mathbf{x}}_{FF}^{\text{T}}R{\mathbf{y}}$ (11)

It is possible to determine the optimal strategy (${\mathbf{x}}_{FF}$) for
player 1, as defined by (10), using advanced game theory techniques. For
example, we could represent the problem in sequence form and then solve a
complementary problem, see [Nisan, Theorem 3.14].

Rather than elaborate on that solution technique, let’s refer to a solution of
(10) as ${\mathbf{x}}_{FF}=M(A,{\cal F})$. In section III-F, we show how to
determine $M$ for the payoff matrices that are of interest to us.

###  III-D Game EF

In this case

E_

player 1 believes that player 2 is using the exactly the payoff matrix $B$.

_F

player 2 believes that ”player 1 believes that player 2 is using a payoff
matrix in ${\cal F}(B)$.”

From player 1’s point of view, Game EF is no different from Game EE. Hence,
the solution (${\mathbf{x}}_{EF}$) is the same as the solution in (4);
${\mathbf{x}}_{EF}={\mathbf{x}}_{EE}$. Writing those equations in EF
variables:

$\begin{split}{\mathbf{x}}_{EF}&=\arg\max_{{\mathbf{x}}\in\Delta}{\mathbf{x}}^{\text{T}}A{\mathbf{y}}_{*}\\\
{\mathbf{y}}_{*}&=\arg\max_{{\mathbf{y}}\in\Delta}{\mathbf{x}}_{EF}^{\text{T}}B{\mathbf{y}}\\\
\end{split}$ (12)

From player 2’s point of view, game EF is no different from Game FF. Hence,
the solution (${\mathbf{y}}_{EF}$) is the same as the solution in (10) and
(11). Writing those equations in EF variables:

$\begin{split}{\mathbf{x}}_{EF}&=M(A,{\cal F})\\\
{\mathbf{y}}_{EF}&=\arg\max_{{\mathbf{y}}\in\Delta}\
{\mathbf{x}}_{EF}^{\text{T}}R{\mathbf{y}}\\\ \end{split}$ (13)

###  III-E The symmetric game

In the symmetric game each player has the same knowledge; each knows that the
other player is using a payoff matrix within a known family of payoffs.
Specifically, player 1 knows

  1. 1.

player 1 uses known payoff matrix $R_{A}\in{\cal F}(A)$

  2. 2.

player 2 uses an unknown payoff matrix in ${\cal F}(B)$

while, player 2 knows (symmetrically)

  1. 1.

player 2 uses a known payoff matrix $R_{B}\in{\cal F}(B)$

  2. 2.

player 1 uses an unknown payoff matrix in ${\cal F}(A)$

Assume that players 1 and 2 play the strategies ${\mathbf{x}}_{S}$ and
${\mathbf{y}}_{S}$. Then, player 1 knows that player 2 is going to play one of
the strategies

${\mathbf{y}}_{S}^{i}({\mathbf{x}}_{S})=\arg\max_{{\mathbf{y}}}\
{\mathbf{x}}_{S}^{\text{T}}R_{E}^{i}{\mathbf{y}}\qquad R_{E}^{i}\in{\cal
F}(B)$ (14)

for $i=1,...,4$. Player 1 wants to maximize his payoff, so his strategy is
given by

${\mathbf{x}}_{S}=\arg\max_{{\mathbf{x}}}\left(\min_{i}\
{\mathbf{x}}^{\text{T}}R_{A}{\mathbf{y}}_{S}^{i}({\mathbf{x}})\right)$ (15)

That is, player 1’s strategy only depends on the matrices $A$ and $B$ and is
given by equations (14) and (15). By symmetry, layer 2’s strategy will be

$\begin{split}{\mathbf{x}}_{S}^{j}({\mathbf{y}})&=\arg\max_{{\mathbf{x}}}\
{\mathbf{x}}^{\text{T}}R_{A}^{j}{\mathbf{y}}\qquad R_{A}^{j}\in{\cal F}(A)\\\
{\mathbf{y}}_{S}&=\arg\max_{{\mathbf{y}}}\left(\min_{j}\
{\mathbf{x}}_{S}^{j}({\mathbf{y}})\right)^{\text{T}}R_{E}{\mathbf{y}}\\\
\end{split}$ (16)

Once again, this game can be solved by using the extensive game form and
complementarity. This variation is not considered further in this paper.

###  III-F The function $M(A,{\cal F})$

The examples in this paper have payoff matrices of size 2-by-2. For problems
of this size, we can readily evaluate the $M$ function defined in section
III-C. The following example shows how it can be done.

Consider the payoff matrices $A$ and $B$:

$A=\begin{bmatrix}4&1\\\ 2&3\\\ \end{bmatrix}\qquad B=\begin{bmatrix}1&3\\\
4&2\\\ \end{bmatrix}$ (17)

The family that contains $B$ is ${\cal
F}(B)=\left\\{B,B_{12},B_{23},B_{34}\right\\}$ with

$B_{12}=\begin{bmatrix}2&3\\\ 4&1\\\ \end{bmatrix}\quad
B_{23}=\begin{bmatrix}1&2\\\ 4&3\\\ \end{bmatrix}\quad
B_{34}=\begin{bmatrix}1&4\\\ 3&2\\\ \end{bmatrix}$

Since the payoff matrices are two-dimensional, the strategies (${\mathbf{x}}$
and ${\mathbf{y}}$) are also two-dimensional. Therefore we can, without loss
of generality, write each of the vectors ${\mathbf{x}}$ and ${\mathbf{y}}$ in
terms of one free parameter

$\begin{split}{\mathbf{x}}&=\begin{bmatrix}a&1-a\end{bmatrix}^{\text{T}}\\\
{\mathbf{y}}&=\begin{bmatrix}b&1-b\end{bmatrix}^{\text{T}}\\\ \end{split}$

where $0\leq a\leq 1$ and $0\leq b\leq 1$. Then, by explicit computation, we
can determine the payoffs to each player. Using $P_{N}^{C}$ to denote the
payoff to player $N$ who has payoff matrix $C$:

$\begin{split}P_{1}^{A}&={\mathbf{x}}^{\text{T}}A{\mathbf{y}}=(4b-2)a+(3-b)\\\
P_{2}^{B}&={\mathbf{x}}^{\text{T}}B{\mathbf{y}}=(2-4a)b+(2+a)\\\
P_{2}^{B_{12}}&={\mathbf{x}}^{\text{T}}B_{12}{\mathbf{y}}=(3-4a)b+(1+2a)\\\
P_{2}^{B_{23}}&={\mathbf{x}}^{\text{T}}B_{23}{\mathbf{y}}=(1-2a)b+(3-a)\\\
P_{2}^{B_{34}}&={\mathbf{x}}^{\text{T}}B_{34}{\mathbf{y}}=(1-4a)b+(2+2a)\\\
&\\\ \end{split}$

Consider the different payoffs to player 2; each has the form
$()_{1}b+()_{2}$; where the parenthesis terms do not contain a $b$. Hence, for
player 2 to maximize their payoff, we have:

$\begin{split}&\text{if $()_{1}>0$ then $b=1$}\\\ &\text{if $()_{1}<0$ then
$b=0$}\\\ \end{split}$

and the value of $b$ is undefined otherwise. Using this logic, we find the
following choices for player 2:

$\begin{split}\text{for $B$ player~{}2 plays}&\begin{cases}b=1&\text{if
$a<\leavevmode\kern 1.00006pt\raise
2.15277pt\hbox{\sevenrm1}\kern-1.00006pt/\kern-1.49994pt\lower
1.07639pt\hbox{\sevenrm2}$}\\\ b=0&\text{if $a>\leavevmode\kern
1.00006pt\raise 2.15277pt\hbox{\sevenrm1}\kern-1.00006pt/\kern-1.49994pt\lower
1.07639pt\hbox{\sevenrm2}$}\\\ \end{cases}\\\ \text{for $B_{12}$ player~{}2
plays}&\begin{cases}b=1&\text{if $a<\leavevmode\kern 1.00006pt\raise
2.15277pt\hbox{\sevenrm3}\kern-1.00006pt/\kern-1.49994pt\lower
1.07639pt\hbox{\sevenrm4}$}\\\ b=0&\text{if $a>\leavevmode\kern
1.00006pt\raise 2.15277pt\hbox{\sevenrm3}\kern-1.00006pt/\kern-1.49994pt\lower
1.07639pt\hbox{\sevenrm4}$}\\\ \end{cases}\\\ \text{for $B_{23}$ player~{}2
plays}&\begin{cases}b=1&\text{if $a<\leavevmode\kern 1.00006pt\raise
2.15277pt\hbox{\sevenrm1}\kern-1.00006pt/\kern-1.49994pt\lower
1.07639pt\hbox{\sevenrm2}$}\\\ b=0&\text{if $a>\leavevmode\kern
1.00006pt\raise 2.15277pt\hbox{\sevenrm1}\kern-1.00006pt/\kern-1.49994pt\lower
1.07639pt\hbox{\sevenrm2}$}\\\ \end{cases}\\\ \text{for $B_{34}$ player~{}2
plays}&\begin{cases}b=1&\text{if $a<\leavevmode\kern 1.00006pt\raise
2.15277pt\hbox{\sevenrm1}\kern-1.00006pt/\kern-1.49994pt\lower
1.07639pt\hbox{\sevenrm4}$}\\\ b=0&\text{if $a>\leavevmode\kern
1.00006pt\raise 2.15277pt\hbox{\sevenrm1}\kern-1.00006pt/\kern-1.49994pt\lower
1.07639pt\hbox{\sevenrm4}$}\\\ \end{cases}\\\ &\\\ \end{split}$

Using these values, we can determine the payoff to player 1 (that is,
$P_{1}^{A}$) using the different payoff matrices in ${\cal F}(B)$; see Figure
2. In this case, the maximum value of the payoff to player 1, minimized over
all matrices in ${\cal F}(B)$ is given by $a=\leavevmode\kern 1.00006pt\raise
2.15277pt\hbox{\sevenrm1}\kern-1.00006pt/\kern-1.49994pt\lower
1.07639pt\hbox{\sevenrm4}$ for which $P_{1}(a)=2.5$.

Figure 2: The payoffs to player 1 based on different player 2 payoff matrices;
using (17).

With 2-by-2 payoff matrices, the payoffs will always be bilinear in $a$ and
$b$; that is, the payoff has the form $c_{0}+c_{a}a+c_{b}b+c_{ab}ab$ for
constants $\\{c_{0},c_{a},c_{b},c_{ab}\\}$. Hence, the value of $b$
(determined by player 2 to maximize their payoff) will always be zero or one,
on different sides of an $a$ breakpoint.

Then, for each player 2 payoff matrix, player 1’s payoff is composed of two
different linear functions of $a$, the change between the two occurring when
$b$ changes between zero and one; which corresponds to an $a$ breakpoint.
Hence, Figure 2 is representative of many cases (except, sometimes, the linear
functions will have slopes with different signs).

Alternatively, if player 1 knew that all of the matrices in the family ${\cal
F}(B)$ were equally likely (i.e., each has a probability of one-fourth), then
the maximizing solution for player 1 would be to determine the best payoff
using the average value of the payoffs, see Figure 3. In this case, player 1
would select $a=0.5^{-}$ with an average payoff of $P_{1}(a)=2.75$. This
improves the payoff to player 1 by 10%. The notion of average payoff will not
be pursued further in this paper.

Figure 3: The payoffs to player 1, averaging the different player 2 payoff
matrices; using (17).

Another example, which shows that the optimal value of $a$ does not need to
occur at the breakpoint for a single payoff matrix is given by the matrices

$A=\begin{bmatrix}4&1\\\ 2&3\\\ \end{bmatrix}\qquad B=\begin{bmatrix}1&2\\\
3&4\\\ \end{bmatrix}$ (18)

In this case, the results corresponding to graphs (2) and (3) are in graphs
(4) and (5).

Figure 4: The payoffs to player 1 based on different player 2 payoff matrices;
using (18). Figure 5: The payoffs to player 1, averaging the different player
2 payoff matrices; using (18).

##  IV Numerical Results for strict ordinal games

The numerical results in this paper were obtained using Matlab by the
following process:

  1. 1.

Consider every bimatrix pair in the class of strict ordinal 2-by-2 payoffs

  2. 2.

For each, determine the solutions $\\{{\mathbf{x}},{\mathbf{y}}\\}$ for the
four cases $\\{EE,EF,FE,FF\\}$

  3. 3.

For each solution, determine the payoff to player 1 and compare it to the
complete information payoff in (3).

A summary of the numerical results is in the following table

Type | Number of games with no change | average loss
---|---|---
EE | 44 | 1.8676
EF | 56 | 1.7045
FE | 44 | 1.3971
FF | 44 | 0.9632

These values should be compared to the average payoff to player 1 in the
complete information game in (1), which is 3.481. For example, in the “EE”
case, player 1 has (on average) lost 54% of the value that could have been
obtained.

Below are graphs showing the numerical results; both the absolute and relative
change in payoff to Player 1 under the four variations considered. The results
are shown only for those games for which the payoff to player 1 changes.

Figure 6: EE histogram: absolute (left) and relative (right) loss.

Figure 7: EF histogram: absolute (left) and relative (right) loss.

Figure 8: FE histogram: absolute (left) and relative (right) loss.

Figure 9: FF histogram: absolute (left) and relative (right) loss.

##  V Observations and Conclusion

We note the following:

  1. 1.

Under the all variations considered, in the average case, player 1’s payoff is
reduced compared to the perfect information case.

  2. 2.

In every case, adding additional uncertainty to either player 1 or player 2
(by transitioning from ”E” to ”F”) results in a worse payoff for player 1 (on
average over all 78 games). However, in the “FE” case, there are specific
games for which player 1 obtains a small improvement.

  3. 3.

For 22 games the payoffs to player 1 do not change under any of the game
variations considered. Similarly, in 44 games, the payoffs to player 1 are
worse in every game variation. For 12 games the payoff to player 1 is worse,
but only in the “EF” case.

  4. 4.

In the games ”EE” and ”EF” the maximum loss to player 1 can be as large as 3
(which is the maximal loss, going from 4 to 1). In the games ”FE” and ”FF” the
maximum loss to player 1 is only 2.

  5. 5.

There are several ”famous” two person games, here is how they relate to our
game analysis:

    1. (a)

no change in payoff to player 1 under any of the four variations: ”chicken”,
”hero”, ”no conflict”, and ”stag hunt”

    2. (b)

reduced payoff to player 1 under every one of the four variations:
”assurance”, ”battle of the sexes”, ”compromise”, ”coordination game”,
”deadlock”, ”peace”, and ”prisoner’s dilemma”

    3. (c)

reduced payoff only in the “EF” variation: ”harmony”

In all cases, from the perspective of player 1’s average payoff, the games
“EE” and “EF” are worse than the games “FE” and “FF”.

Future work that would extend the analysis in this paper:

  1. 1.

A complete analysis of the symmetric game could be performed.

  2. 2.

An evaluation could be made of how the payoff to player 2 changes under the
variations considered. This would allow questions such as the following to be
answered: “How valuable is it to Player 2 to change what Player 1 is thinking
about Player 2?”

  3. 3.

As we’ve noticed, the payoff to Player 1 does not change under many games;
these are games for which it is not in Player 1’s interest to learn more about
the Player 2’s payoff. Understanding the feaures of these games would be
useful, especially if the information could be extrapolated to larger non-
ordinal games.

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# Convergence Criteria of a Three Step Scheme under generalized Lipschitz
Condition in Banach Spaces

Akanksha Saxena1, J. P. Jaiswal2, K. R. Pardasani1

1 Department of Mathematics, Maulana Azad National Institute of Technology,
Bhopal, M.P., India-462003, Email: AS(akanksha.sai121@gmail.com);
<EMAIL_ADDRESS>
2 Department of Mathematics, Guru Ghasidas Vishwavidyalaya (A Central
University), Bilaspur, C.G., India-495009, Email<EMAIL_ADDRESS>
Abstract. The goal of this study is to investigate the local convergence of a
three-step Newton-Traub technique for solving nonlinear equations in Banach
spaces with a convergence rate of five. The first order derivative of a
nonlinear operator is assumed to satisfy the generalized Lipschitz condition,
i.e. the $\kappa$-average condition. Furthermore, a few results on the
convergence of the same method in Banach spaces are developed under the
assumption that the derivative of the operators satisfies the radius or center
Lipschitz condition with a weak $\kappa$-average, and that $\kappa$ is a
positive integrable function but not necessarily non-decreasing. Our new
notion provides a tighter convergence analysis without the need for new
circumstances. As a result, we broaden the applicability of iterative
approaches. Theoretical results are supported further by illuminating
examples. The existence and uniqueness of the solution $x^{*}$ are examined in
the convergence theorem. In the end, we achieve weaker sufficient convergence
criteria and more specific information on the position of the solution 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 $\kappa(u)$. A numerical test is carried out to
corroborate the hypothesis established in this work.

Mathematics Subject Classification (MSC 2020). 65H10, 65J15, 65G99, 47J25.

Keywords and Phrases. Nonlinear problem, convergence radius, local
convergence, Banach space, Lipschitz condition, $\kappa$-average.

##  1\. Introduction

Let a nonlinear operator $G:\Omega\subseteq X\to Y$ be such that $X$ and $Y$
are two Banach spaces, $\Omega$ is a non-empty open convex subset and $G$ is
Fréchet differentiable nonlinear operator. A popular iterative method for
solving the equation

$\displaystyle G(x)=0,$ (1.1)

is Newton’s scheme, which may be represented as:

$\displaystyle x_{t+1}$ $\displaystyle=$ $\displaystyle
x_{t}-[G^{\prime}(x_{t})]^{-1}G(x_{t}),\ t\geq 0,$ (1.2)

is being preferred though its speed of convergence is low. Newton’s method
[10], is a well known iterative approach that converges quadratically, which
was initially studied by Kantorovich [5] and then re-evaluated by Rall [11].

Some Newton-type methods with third-order, fourth-order convergence that do
not require the computation of second order derivatives have been developed in
the refs.[[4], [7], [9], [15], [18]]. While methods of higher $R$-order
convergence are often not conducted regularly despite having a high speed of
convergence, this is due to the high operational expense. However, the method
of higher $R$-order convergence can be applied in stiff system issues [5]
where quick convergence is necessary.

The convergence domain is critical for the steady behaviour of an iterative
method from a numerical stand point. There are two types of iterative method
convergence research: semilocal and local convergence analysis. The semilocal
convergence study uses information around a starting point to provide criteria
for ensuring the convergence of iterative methods, whereas the local one uses
information around a solution to estimate the radii of the convergence balls.
Numerous authors investigated the local convergence analysis for Newton-type,
Jarratt-type, Weerakoon-type, and other types in Banach space in works [[1],
[3], [6], [8], [14]].

Here, we analyze the local convergence of the classical fifth-order Newton-
Traub method [2] under the $\kappa$-average condition which is written as:

$\displaystyle y_{t}$ $\displaystyle=$ $\displaystyle
x_{t}-[G^{\prime}(x_{t})]^{-1}G(x_{t}),$ $\displaystyle z_{t}$
$\displaystyle=$ $\displaystyle y_{t}-[G^{\prime}(x_{t})]^{-1}G(y_{t}),$
$\displaystyle x_{t+1}$ $\displaystyle=$ $\displaystyle
z_{t}-[G^{\prime}(y_{t})]^{-1}G(z_{t}),\ t=0,1,2,\cdots.$ (1.3)

The important feature of the method $(\ref{eqn:12})$ is that: it is the
easiest and most efficient fifth-order iterative method, requiring only two
evaluations of the function $G_{j}$, per jth iteration, one of the first
derivative $G_{j}^{\prime}$ and no evaluations of the second derivative
$G_{j}^{\prime\prime}$ making it mathematically effective. Several research on
the weakness and/or expansion of hypotheses made on the underlying operators
can be found in the literature. Wang [16] created generalized Lipschitz
conditions to investigate the local convergence of Newton-Traub method, in
which a non-decreasing positive integrable function was chosen rather than the
normal Lipschitz constant. In the coming years, Wang and Li [17] has obtained
some results on Newton-Traub method convergence in Banach spaces when the
derivative of the operators satisfies the radius or center Lipschitz condition
but with a weak $\kappa$-average. Shakhno [13] has explored the local
convergence of the two-step Secant-type approach [5], where the first-order
divided differences satisfy the generalized Lipschitz criteria. Recently, the
local convergence of a two-step Newton type method of convergence rate three
under generalized Lipschitz conditions has been studied by Saxena et. al [12]
whose definitions will be used in this article.

As a motivational example let $X=Y=R^{3},D=\overline{V}(0,1)$ and
$X^{*}=(0,0,0)^{T}$. Define function $t$ on $D$ for $w=(x,y,z)^{T}$ by

$\displaystyle t(w)=(e^{x}-1,\frac{e-1}{2}y^{2}+y,z)^{T}.$

Then, the Fréchet derivative is

$t^{\prime}(w)=\begin{pmatrix}e^{x}&&0&&0\\\ 0&&(e-1)y+1&&0\\\
0&&0&&1\end{pmatrix}.$ (1.4)

Hence, $\kappa=\frac{e}{2},\kappa_{0}=\frac{e-1}{2}$ and $\kappa_{0}<\kappa$
(see definitions $(\ref{eqn:21})$ and $(\ref{eqn:22})$). So, replacing
$\kappa$ by $\kappa_{0}$ at the denominator gives the benefits. If $\kappa$
and $\kappa_{0}$ are not constants then we can take
$\kappa(u)=\frac{eu}{2},\kappa_{0}(u)=\frac{(e-1)u}{2}\ and\
\overline{\kappa}(u)=\frac{e^{\frac{1}{(e-1)}}u}{2}$ (see definitions
$(\ref{eqn:23})$, $(\ref{eqn:24})$ and $(\ref{eqn:556})$).

The fascinating question then arises as to whether the radius Lipschitz
condition with $\kappa$-average and non-decreasing of $\kappa$ are required
for the convergence of fifth-order Newton-Traub method. In this paper, we
derived certain theorems for scheme $(\ref{eqn:12})$, motivated and influenced
by the above-mentioned research efforts in this direction. Throughout the
first result, generalized Lipschitz conditions were utilized to explore local
convergence, which is significant for enlarging the convergence area without
the need for additional assumptions, as well as an error estimate. The domain
of uniqueness of solution has been derived under the center Lipschitz
condition in the second theorem. A few corollaries are also mentioned.

The remainder of this paper is organized as follows: The definitions for
$\kappa$-average conditions are found in section $2$. Sections $3$ and $4$
discuss local convergence and its region of uniqueness, accordingly. The
assumption that the derivative of $t$ satisfies the radius and center
Lipschitz condition with weak $\kappa$-average, namely $\kappa$ and
$\kappa_{0}$, is improved in Section $5$. $\kappa$ and $\kappa_{0}$ are
claimed to belong to some family of positive integrable functions that are not
necessarily non-decreasing for convergence theorems. To demonstrate the
significance of the findings, numerical examples are provided.

##  2\. Special and Generalized Lipschitz conditions

Throughout this context, $B(x^{*},\delta)=\\{x:||x-x^{*}||<\delta\\}$ denotes
a ball with radius $\delta$ and center $x^{*}$.
Definition 1. The constraint placed on the function $G$

$||G^{\prime}(x)-G^{\prime}(y^{\tau})||\leq\kappa(1-\tau)(||x-x^{*}||+||y-x^{*}||);\forall\
x,\ y\in B(x^{*},\delta),$ (2.1)

where $y^{\tau}=x^{*}+\tau(y-x^{*}),0\leq\tau\leq 1,$ and $x=y\ if\ \tau=1$,
is usually called radius Lipschitz condition in the ball $B(x^{*},\delta)$
with constant $\kappa$.
Definition 2. The constraint imposed on the function $G$

$\displaystyle||G^{\prime}(x)-G^{\prime}(x^{*})||\leq
2\kappa_{0}||x-x^{*}||;\forall\ x\in B(x^{*},\delta),$ (2.2)

we call it the center Lipschitz condition in the ball $B(x^{*},\delta)$ with
constant $\kappa_{0}$ where $\kappa_{0}\leq\kappa$.

In this scenario, replacing $\kappa$ by $\kappa_{0}$ in case
$\kappa_{0}<\kappa$ leads to a wider range of initial guesses (bigger radius
of convergence than in prior studies) and fewer iterations to attain error
tolerance, and the uniqueness of the solution $x^{*}$ is also expanded ([1],
[8]). When $\kappa$ and $\kappa_{0}$ are not constants but can be a positive
integrable function, in this situation, the criteria $(2.1)-(2.2)$ are
respectively, are substituted by
Definition 3.

$\displaystyle||G^{\prime}(x)-G^{\prime}(y^{\tau})||\leq\int_{\tau(\rho(x)+\rho(y))}^{\rho(x)+\rho(y)}\kappa(u)du;\forall\
x,\ y\in B(x^{*},\delta),0\leq\tau\leq 1,$

Definition 4.

$\displaystyle||G^{\prime}(x)-G^{\prime}(x^{*})||\leq\int_{0}^{2\rho(x)}\kappa_{0}(u)du;\forall\
x\in B(x^{*},\delta),$ (2.4)

where $\rho(x)=||x-x^{*}||$ and we have $\kappa_{0}(u)\leq\kappa(u)$.
Simultaneously, the equivalent’ Lipschitz conditions’ are referred to as
having the $\kappa$-average or generalized Lipschitz conditions. Following
that, we will begin with the below lemmas, that will be used eventually in the
fundamental theorems.

######  Lemma 2.1.

Suppose that $G$ has a continuous derivative in $B(x^{*},\delta)$ and
$[G^{\prime}(x^{*})]^{-1}$ exists.
(i) If $[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies the radius Lipschitz
condition with the $\kappa$-average:

$\displaystyle||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x)-G^{\prime}(y^{\tau}))||\leq\int_{\tau(\rho(x)+\rho(y))}^{\rho(x)+\rho(y)}\kappa(u)du;\forall\
x,\ y\in B(x^{*},\delta),0\leq\tau\leq 1,$

where $y^{\tau}=x^{*}+\tau(y-x^{*})$, $\rho(x)=||x-x^{*}||$ and $\kappa$ is
non-decreasing, then we have

$\int_{0}^{1}||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x)-G^{\prime}(y^{\tau}))||\rho(y)d\tau\leq\int_{0}^{\rho(x)+\rho(y)}\kappa(u)\frac{u}{\rho(x)+\rho(y)}\rho(y)du.$
(2.6)

(ii) If $[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies the center Lipschitz
condition with the $\kappa_{0}$-average:

$||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x^{\tau})-G^{\prime}(x^{*}))||\leq\int_{0}^{2\tau\rho(x)}\kappa_{0}(u)du;\forall\
x,\ y\in B(x^{*},\delta),0\leq\tau\leq 1,$ (2.7)

where $\rho(x)=||x-x^{*}||$ and $\kappa_{0}$ is non-decreasing, then we have

$\int_{0}^{1}||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x^{\tau})-G^{\prime}(x^{*}))||\rho(x)d\tau\leq\int_{0}^{2\rho(x)}\kappa_{0}(u)\left(\rho(x)-\frac{u}{2}\right)du.$
(2.8)

###### Proof.

The Lipschitz conditions $(\ref{eqn:25})$ and $(\ref{eqn:27})$, respectively,
imply that

$\displaystyle\int_{0}^{1}||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x)-G^{\prime}(y^{\tau}))||\rho(y)d\tau$
$\displaystyle\leq$
$\displaystyle\int_{0}^{1}\int_{\tau(\rho(x)+\rho(y))}^{\rho(x)+\rho(y)}\kappa(u)du\rho(y)d\tau$
$\displaystyle=$
$\displaystyle\int_{0}^{\rho(x)+\rho(y)}\kappa(u)\frac{u}{\rho(x)+\rho(y)}\rho(y)du,$
$\displaystyle\int_{0}^{1}||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x^{\tau})-G^{\prime}(x^{*}))||\rho(x)d\tau$
$\displaystyle\leq$
$\displaystyle\int_{0}^{1}\int_{0}^{2\tau\rho(x)}\kappa_{0}(u)du\rho(x)d\tau$
$\displaystyle=$
$\displaystyle\int_{0}^{2\rho(x)}\kappa_{0}(u)\left(\rho(x)-\frac{u}{2}\right)du.$

where $x^{\tau}=x^{*}+\tau(x-x^{*})$ and $y^{\tau}=x^{*}+\tau(y-x^{*})$. ∎

######  Lemma 2.2.

[17] Suppose that $\kappa$ is positive integrable. Assume that the function
$\kappa_{a}$ defined by relation $(\ref{eqn:56})$ is non-decreasing for some
$a$ with $0\leq a\leq 1$. Then, $for\ each\ \beta\geq 0$, the function
$\varphi_{\beta,a}$ defined by

$\varphi_{\beta,a}(P)=\frac{1}{P^{a+\beta}}\int_{0}^{P}u^{\beta}\kappa(u)du,$
(2.9)

is also non-decreasing.

##  3\. Local convergence of Newton-Traub method $(\ref{eqn:12})$

Throughout this section, we prove the existence theorem for the Newton-Traub
method $(\ref{eqn:12})$ under the radius Lipschitz condition.

######  Theorem 3.1.

Suppose that $G(x^{*})=0$, $G$ has a continuous derivative in
$B(x^{*},\delta)$, $[G^{\prime}(x^{*})]^{-1}$ exists and
$[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies $(\ref{eqn:23})$,
$(\ref{eqn:24})$, $\kappa_{0}$ and $\kappa$ are non-decreasing. Let $\delta$
satisfy the relation

$\displaystyle\frac{\int_{0}^{2\delta}\kappa(u)udu}{2\delta(1-\int_{0}^{2\delta}\kappa_{0}(u)du)}\leq
1.$ (3.1)

Then, the three-step Newton-Traub method $(\ref{eqn:12})$ is convergent for
all $x_{0}\in B(x^{*},\delta)$ and

$\displaystyle||y_{t}-x^{*}||\leq\frac{\int_{0}^{2\rho(x_{t})}\kappa(u)udu}{2(1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du)}\leq\frac{C_{1}}{\rho(x_{0})}\rho(x_{t})^{2},$
(3.2)
$\displaystyle||z_{t}-x^{*}||\leq\frac{\int_{0}^{\rho(x_{t})+\rho(y_{t})}\kappa(u)udu}{(\rho(x_{t})+\rho(y_{t}))(1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du)}\rho(y_{t})\leq\frac{C_{1}C_{2}}{\rho(x_{0})\rho(y_{0})}\rho(x_{t})^{3},$
$\displaystyle||x_{t+1}-x^{*}||\leq\frac{\int_{0}^{\rho(y_{t})+\rho(z_{t})}\kappa(u)udu}{(\rho(y_{t})+\rho(z_{t}))(1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du)}\rho(z_{t})\leq\frac{C_{1}^{2}C_{2}C_{3}}{\rho(x_{0})^{2}\rho(y_{0})\rho(z_{0})}\rho(x_{t})^{5},$

where the quantities

$\displaystyle
C_{1}=\frac{\int_{0}^{2\rho(x_{0})}\kappa(u)udu}{2\rho(x_{0})(1-\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du)},\
C_{2}=\frac{\int_{0}^{\rho(x_{0})+\rho(y_{0})}\kappa(u)udu}{(\rho(x_{0})+\rho(y_{0}))(1-\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du)},$
$\displaystyle
C_{3}=\frac{\int_{0}^{\rho(y_{0})+\rho(z_{0})}\kappa(u)udu}{(\rho(y_{0})+\rho(z_{0}))(1-\int_{0}^{2\rho(y_{0})}\kappa_{0}(u)du)},$
(3.5)

are less than $1$. Furthermore,

$\displaystyle||x_{t}-x^{*}||\leq E^{5^{t}-1}||x_{0}-x^{*}||;\
t=1,2,\cdots,E=C_{1}C_{2}\frac{\rho(x_{0})^{2}}{\rho(y_{0})\rho(z_{0})}.$
(3.6)

###### Proof.

By arbitrarily selecting $x_{0}\in B(x^{*},\delta)$, where $\delta$ fulfills
the relation $(\ref{eqn:32})$, $C_{1}$ and $C_{2}$ are determined by the
inequality $(\ref{eqn:36})$ are less than 1. Indeed, because $\kappa$ is
monotone, we obtain

$\displaystyle\left(\frac{1}{f_{2}^{2}}\int_{0}^{f_{2}}-\frac{1}{f_{1}^{2}}\int_{0}^{f_{1}}\right)\kappa(u)udu$
$\displaystyle=\left(\frac{1}{f_{2}^{2}}\int_{f_{1}}^{f_{2}}+\left(\frac{1}{f_{2}^{2}}-\frac{1}{f_{1}^{2}}\right)\int_{0}^{f_{1}}\right)\kappa(u)udu$
$\displaystyle\geq\kappa(f_{1})\left(\frac{1}{f_{2}^{2}}\int_{f_{1}}^{f_{2}}+\left(\frac{1}{f_{2}^{2}}-\frac{1}{f_{1}^{2}}\right)\int_{0}^{f_{1}}\right)udu$
$\displaystyle=\kappa(f_{1})\left(\frac{1}{f_{2}^{2}}\int_{0}^{f_{2}}-\frac{1}{f_{1}^{2}}\int_{0}^{f_{1}}\right)udu=0,$

for $0<f_{1}<f_{2}.$ Thus, $\frac{1}{f^{2}}\int_{0}^{f}\kappa(u)udu$ is non-
decreasing with respect to $f$. Now, we have

$\displaystyle C_{1}$ $\displaystyle=$
$\displaystyle\frac{\int_{0}^{2\rho(x_{0})}\kappa(u)udu}{2\rho(x_{0})^{2}(1-\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du)}\rho(x_{0})$
$\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\delta}\kappa(u)udu}{2\delta^{2}(1-\int_{0}^{2\delta}\kappa_{0}(u)du)}\rho(x_{0})\leq\frac{||x_{0}-x^{*}||}{\delta}<1,$
$\displaystyle C_{2}$ $\displaystyle=$
$\displaystyle\frac{\int_{0}^{\rho(x_{0})+\rho(y_{0})}\kappa(u)udu}{(\rho(x_{0})+\rho(y_{0}))^{2}(1-\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du)}(\rho(x_{0}+\rho(y_{0}))$
$\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\delta}\kappa(u)udu}{2\delta^{2}(1-\int_{0}^{2\delta}\kappa_{0}(u)du)}(\rho(x_{0})+\rho(y_{0}))\leq\frac{||x_{0}-x^{*}||+||y_{0}-x^{*}||}{2\delta}<1,$
$\displaystyle C_{3}$ $\displaystyle=$
$\displaystyle\frac{\int_{0}^{\rho(y_{0})+\rho(z_{0})}\kappa(u)udu}{(\rho(y_{0})+\rho(z_{0}))^{2}(1-\int_{0}^{2\rho(y_{0})}\kappa_{0}(u)du)}(\rho(y_{0}+\rho(z_{0}))$
$\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\delta}\kappa(u)udu}{2\delta^{2}(1-\int_{0}^{2\delta}\kappa_{0}(u)du)}(\rho(y_{0})+\rho(z_{0}))\leq\frac{||x_{0}-x^{*}||+||y_{0}-x^{*}||}{2\delta}<1.$

Clearly, if $x\in B(x^{*},\delta)$, then using center Lipschitz condition with
the $\kappa$-average and the relation $(\ref{eqn:32})$, we can write

$\displaystyle||[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x)-G^{\prime}(x^{*})]||\leq\int_{0}^{2\rho(x)}\kappa_{0}(u)du\leq
1.$ (3.7)

Using the Banach Lemma and the following equation,

$\displaystyle||I-([G^{\prime}(x^{*})]^{-1}G^{\prime}(x)-I)||^{-1}=||[G^{\prime}(x)]^{-1}G^{\prime}(x^{*})||,$

Using the relation $(\ref{eqn:37})$, we arrive to the following inequality.

$\displaystyle||[G^{\prime}(x)]^{-1}G^{\prime}(x^{*})||$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\int_{0}^{2\rho(x)}\kappa_{0}(u)du}.$ (3.8)

Now, if $x_{t}\in B(x^{*},\delta)$ then we may write from expression
$(\ref{eqn:12})$

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle=$
$\displaystyle||x_{t}-x^{*}-[G^{\prime}(x_{t})]^{-1}G(x_{t})||$ (3.9)
$\displaystyle=$
$\displaystyle||[G^{\prime}(x_{t})]^{-1}[G^{\prime}(x_{t})(x_{t}-x^{*})-G(x_{t})+G(x^{*})]||.$

Expanding $G(x_{t})$ along $x^{*}$ from the Taylor series expansion, we may
get

$\displaystyle
G(x^{*})-G(x_{t})+t^{\prime}(x_{t})(x_{t}-x^{*})=G^{\prime}(x^{*})\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x_{t})-G^{\prime}(x^{\tau})]d\tau(x_{t}-x^{*}).$

Additionally, from the expression $(\ref{eqn:23})$ and combining the equations
$(\ref{eqn:39})$ and $(\ref{eqn:310})$, we reach to

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle||[G^{\prime}(x_{t})]^{-1}G^{\prime}(x^{*})||.||\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x_{t})-G^{\prime}(x^{\tau})]d\tau||.||(x_{t}-x^{*})||$
(3.11) $\displaystyle\leq$
$\displaystyle\frac{1}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\int_{0}^{1}\int_{2\tau\rho(x_{n})}^{2\rho(x_{n})}\kappa(u)du\rho(x_{t})d\tau.$

As a result of Lemma $(\ref{lm:21})$ and the above expression, The first
inequality of expression $(\ref{eqn:34})$ can be obtained. By similar analogy
and using the second sub-step of the scheme $(\ref{eqn:12})$, we can write

$\displaystyle||z_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle||[G^{\prime}(x_{t})]^{-1}G^{\prime}(x^{*})||.||\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x_{t})-G^{\prime}(y^{\tau})]d\tau||.||(y_{t}-x^{*})||$
(3.12) $\displaystyle\leq$
$\displaystyle\frac{1}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\int_{0}^{1}\int_{\tau(\rho(x_{t})+\rho(y_{t}))}^{\rho(x_{t})+\rho(y_{t})}\kappa(u)du\rho(y_{t})d\tau.$

Using Lemma $(\ref{lm:21})$ and above expression, we can get the first
inequality of expression $(\ref{eqn:34b})$. Simultaneously, re-writing the
last sub-step of the scheme $(\ref{eqn:12})$, we achieve

$\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle||[G^{\prime}(y_{t})]^{-1}G^{\prime}(x^{*})||.||\int_{0}^{1}[t^{\prime}(x^{*})]^{-1}[G^{\prime}(y_{t})-G^{\prime}(z^{\tau})]d\tau||.||(z_{t}-x^{*})||$
(3.13) $\displaystyle\leq$
$\displaystyle\frac{1}{1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}\int_{0}^{1}\int_{\tau(\rho(y_{t})+\rho(z_{t}))}^{\rho(y_{t})+\rho(z_{t})}\kappa(u)du\rho(z_{n})d\tau.$

Using Lemma $(\ref{lm:21})$ and above expression, we can get the first
inequality of expression $(\ref{eqn:34a})$. Furthermore, $\rho(x_{t})$ and
$\rho(y_{t})$ are decreasing monotonically, therefore for all $t=0,1,...$, we
have

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(x_{t})}\kappa(u)udu}{2(1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du)}$
$\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(x_{0})}\kappa(u)udu}{2\rho(x_{0})^{2}(1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du)}2\rho(x_{t})^{2}\leq\frac{C_{1}}{\rho(x_{0})}\rho(x_{t})^{2}.$

Also, by manipulating first inequality of expression $(\ref{eqn:34b})$, we
have

$\displaystyle||z_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{\rho(x_{t})+\rho(y_{t})}\kappa(u)udu}{(\rho(x_{t})+\rho(y_{t}))^{2}(1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du)}\rho(y_{t}).[\rho(x_{t})+\rho(y_{t})]$
$\displaystyle\leq$
$\displaystyle\frac{C_{2}}{\rho(x_{0})+\rho(y_{0})}[\rho(x_{t})+\rho(y_{t})]\rho(y_{t})$
$\displaystyle\leq$
$\displaystyle\frac{C_{1}C_{2}}{\rho(x_{0})\rho(y_{0})}\rho(x_{t})^{3}.$

Lastly, simplifying the first result in the inequality of expression
$(\ref{eqn:34a})$, we have

$\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{\rho(y_{t})+\rho(z_{t})}\kappa(u)udu}{(\rho(y_{t})+\rho(z_{t}))^{2}(1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du)}\rho(z_{t}).[\rho(y_{t})+\rho(z_{t})]$
$\displaystyle\leq$
$\displaystyle\frac{C_{3}}{\rho(y_{0})+\rho(z_{0})}[\rho(y_{t})\rho(z_{t})+\rho(z_{t})^{2}]\leq\frac{C_{1}^{2}C_{2}C_{3}}{\rho(x_{0}^{2})\rho(y_{0})\rho(z_{0})}\rho(x_{t})^{5}.$

Thus, we have derived all the expressions of inequalities $(\ref{eqn:34})$,
$(\ref{eqn:34b})$ and $(\ref{eqn:34a})$. Now, we use mathematical induction to
prove the inequality $(\ref{eqn:33})$. For $t=0$, the inequality
$(\ref{eqn:34a})$ gives

$\displaystyle||x_{1}-x^{*}||\leq\frac{C_{1}^{2}C_{2}C_{3}}{\rho(x_{0})^{2}\rho(y_{0})\rho(z_{0})}\rho(x_{0})^{5}.$

Multiplying the numerator and denominator by $\rho(y_{0})^{3}\rho(z_{0})^{3}$
and using the inequalities $(\ref{eqn:34b})$ and $(\ref{eqn:313})$ for $t=0$
and then doing some mathematical manipulations, the above inequality may be
re-written as

$\displaystyle||x_{1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\left[C_{1}^{2}C_{2}\frac{\rho(x_{0})^{2}}{\rho(y_{0})\rho(z_{0})}\right]^{4}\rho(x_{0})$
$\displaystyle\leq$ $\displaystyle E^{(5-1)}\rho(x_{0}).$

Thus the expression $(\ref{eqn:33})$ is true for $t=1$. Now, suppose the
inequality $(\ref{eqn:33})$ is true for some integer $t>1$. Again multiplying
the numerator and denominator by $\rho(y_{0})^{3}\rho(z_{0})^{3}$ and using
the inequalities $(\ref{eqn:33})$ for $t=t$, $(\ref{eqn:34b})$ and
$(\ref{eqn:313})$ for $t=0$, the above inequality preserves the form

$\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\left[C_{1}^{2}C_{2}\frac{\rho(x_{0})^{2}}{\rho(y_{0})\rho(z_{0})}\right]^{5^{t+1}-1}\rho(x_{0})$
$\displaystyle\leq$ $\displaystyle E^{(5^{t+1}-1)}\rho(x_{0}).$

Thus we reach at completion of the proof. ∎

##  4\. The uniqueness ball for the optimal solution

We will show the uniqueness theorem for the Newton-Traub approach
$(\ref{eqn:12})$, using the center Lipschitz condition.

######  Theorem 4.1.

Suppose that $G(x^{*})=0$, $G$ has a continuous derivative in
$B(x^{*},\delta)$, $[G^{\prime}(x^{*})]^{-1}$ exists and
$[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies $(\ref{eqn:24})$. Let $\delta$
satisfy the relation

$\displaystyle\frac{\int_{0}^{2\delta}\kappa_{0}(u)(2\delta-u)du}{2\delta}\leq
1.$ (4.1)

Then, the equation $G(x)=0$ has a unique solution $x^{*}$ in
$B(x^{*},\delta)$.

###### Proof.

By arbitrarily choosing $y^{*}\in B(x^{*},r)$, $y^{*}\neq x^{*}$ and
evaluating the iteration, we get

$\displaystyle||y^{*}-x^{*}||$ $\displaystyle=$
$\displaystyle||y^{*}-x^{*}-[G^{\prime}(x^{*})]^{-1}G(y^{*})||.$
$\displaystyle=$
$\displaystyle||[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x^{*})(y^{*}-x^{*})-G(y^{*})+G(x^{*})]||$

Expanding $G(y^{*})$ along $\ x^{*}$ from the Taylor’s expansion, we have

$\displaystyle
G(x^{*})-G(y^{*})+G^{\prime}(x^{*})(y^{*}-x^{*})=\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(y^{*\tau})-G^{\prime}(x^{*})]d\tau(y^{*}-x^{*}),$

Using the expression $(\ref{eqn:24})$ and combining the inequalities
$(\ref{eqn:43})$ and $(\ref{eqn:44})$, we may write

$\displaystyle||y^{*}-x^{*}||$ $\displaystyle\leq$
$\displaystyle||[G^{\prime}(x^{*})]^{-1}G^{\prime}(x^{*})||.||\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(y^{*\tau})-G^{\prime}(x^{*})]d\tau||.||(y^{*}-x^{*})||$
(4.4) $\displaystyle\leq$
$\displaystyle\int_{0}^{1}\int_{0}^{2\tau\rho(y^{*})}L_{0}(u)du\rho(y^{*})d\tau.$

As a result of Lemma $(\ref{lm:21})$ and expression $(\ref{eqn:45})$, we
obtain

$\displaystyle||y^{*}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{1}{2\rho(y^{*})}\int_{0}^{2\rho(y^{*})}\kappa_{0}(u)[2\rho(y^{*})-u]du(y^{*}-x^{*})$
(4.5) $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\delta}\kappa_{0}(u)(2\delta-u)du}{2\delta}\rho(y^{*})\leq||y^{*}-x^{*}||.$

But this contradicts our hypotheses. Thus, we see that $y^{*}=x^{*}$. This
concludes the theorem’s proof. ∎

Specifically, assuming that $\kappa$ and $\kappa_{0}$ are constants, we have
the Corollaries $(\ref{cr:51})$ and $(\ref{cr:52})$ derived from Theorems
$(\ref{th:31})$ and $(\ref{th:41})$, respectively.

######  Corollary 4.1.

Suppose that $x^{*}$ satisfies $G(x^{*})=0$, $G$ has a continuous derivative
in $B(x^{*},\delta)$, $[G^{\prime}(x^{*})]^{-1}$ exists and
$[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies $(\ref{eqn:21})$ and
$(\ref{eqn:22})$. Let $\delta$ satisfy the relation

$\displaystyle\delta$ $\displaystyle=$
$\displaystyle\frac{1}{2\kappa_{0}+\kappa}.$ (4.6)

Then, the three-step Newton-Traub method $(\ref{eqn:12})$ is convergent for
all $x_{0}\in B(x^{*},\delta)$ and

$\displaystyle||y_{t}-x^{*}||\leq\frac{C_{1}}{\rho(x_{0})}\rho(x_{t})^{2},$
$\displaystyle||z_{t}-x^{*}||\leq\frac{C_{1}C_{2}}{\rho(x_{0})\rho(y_{0})}\rho(x_{t})^{3},$
$\displaystyle||x_{t+1}-x^{*}||\leq\frac{C_{1}^{2}C_{2}C_{3}}{\rho(x_{0})^{2}\rho(y_{0})\rho(z_{0})}\rho(x_{t})^{5},$

where the quantities

$\displaystyle
C_{1}=\frac{\kappa\rho(x_{0})}{1-2\kappa_{0}\rho(x_{0})},C_{2}=\frac{\kappa(\rho(x_{0})+\rho(y_{0}))}{2(1-2\kappa_{0}\rho(x_{0}))},C_{3}=\frac{\kappa(\rho(y_{0})+\rho(z_{0}))}{2(1-2\kappa_{0}\rho(y_{0}))},$

are less than $1$.

######  Corollary 4.2.

Suppose that $x^{*}$ satisfies $G(x^{*})=0$, $G$ has a continuous derivative
in $B(x^{*},\delta)$, $[G^{\prime}(x^{*})]^{-1}$ exists and
$[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies the assumption
$(\ref{eqn:22})$. Let $\delta$ fulfill the condition

$\displaystyle\delta$ $\displaystyle=$ $\displaystyle\frac{1}{\kappa_{0}}.$
(4.8)

Then, the equation $G(x)=0$ has a unique solution $x^{*}$ in
$B(x^{*},\delta)$. Moreover, the ball radius $\delta$ depends only on
$\kappa_{0}$.

Following that, we will apply our fundamental theorems to some specific
function $\kappa$ and find the following corollaries:

######  Corollary 4.3.

Suppose that $x^{*}$ satisfies $G(x^{*})=0$, $G$ has a continuous derivative
in $B(x^{*},\delta)$, $[G^{\prime}(x^{*})]^{-1}$ exists and
$[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies $(\ref{eqn:23})$,
$(\ref{eqn:24})$ where given fixed positive constants $\gamma$, $\kappa>0$ and
$\kappa_{0}>0$ with $\kappa(u)=\gamma+\kappa u$ and
$\kappa_{0}(u)=\gamma+\kappa_{0}u$ i.e.:

$\displaystyle||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x)-G^{\prime}(y^{\tau}))||$
$\displaystyle\leq$ $\displaystyle\gamma(1-\tau)(||x-x^{*}||+||y-x^{*}||)$
$\displaystyle+\frac{\kappa}{2}(1-\tau^{2})(||x-x^{*}||+||y-x^{*}||)^{2},$

and

$\displaystyle||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x)-G^{\prime}(x^{*}))||\leq
2||x-x^{*}||(\gamma+\kappa_{0}||x-x^{*}||),$ (4.10)

$\forall\ x,\ y\in G(x^{*},\delta),0\leq\tau\leq 1$, where
$y^{\tau}=x^{*}+\tau(y-x^{*})$, $\rho(x)=||x-x^{*}||$. Let $\delta$ satisfy
the relation

$\displaystyle\delta$ $\displaystyle=$
$\displaystyle\frac{-3\gamma+\sqrt{9\gamma^{2}+(16/3)\kappa+8\kappa_{0}}}{8/3\kappa+4\kappa_{0}}\
and\ 9\gamma^{2}+(16/3)\kappa+8\kappa_{0}\geq 0.$

Then, the three-step Newton-Traub method $(\ref{eqn:12})$ is convergent for
all $x_{0}\in B(x^{*},\delta)$ and

$\displaystyle||y_{t}-x^{*}||\leq\frac{C_{1}}{\rho(x_{0})}\rho(x_{t})^{2},$
$\displaystyle||z_{t}-x^{*}||\leq\frac{C_{1}C_{2}}{\rho(x_{0})\rho(y_{0})}\rho(x_{t})^{3},$
$\displaystyle||x_{t+1}-x^{*}||\leq\frac{C_{1}^{2}C_{2}C_{3}}{\rho(x_{0})^{2}\rho(y_{0})\rho(z_{0})}\rho(x_{t})^{5},$

where the quantities

$\displaystyle C_{1}$ $\displaystyle=$
$\displaystyle\frac{\rho(x_{0})[\gamma+4/3\kappa\rho(x_{0})]}{[1-2\gamma\rho(x_{0})-2\kappa_{0}\rho(x_{0})^{2}]},$
$\displaystyle C_{2}$ $\displaystyle=$
$\displaystyle\frac{\rho(x_{0})+\rho(y_{0})[\gamma/2+\kappa/3(\rho(x_{0})+\rho(y_{0})]}{[1-2\gamma\rho(x_{0})-2\kappa_{0}\rho(x_{0})^{2}]},$
$\displaystyle C_{3}$ $\displaystyle=$
$\displaystyle\frac{\rho(y_{0})+\rho(z_{0})[\gamma/2+\kappa/3(\rho(y_{0})+\rho(z_{0})]}{[1-2\gamma\rho(y_{0})-2\kappa_{0}\rho(y_{0})^{2}]},$
(4.12)

are less than $1$.

######  Corollary 4.4.

Suppose that $x^{*}$ satisfies $G(x^{*})=0$, $G$ has a continuous derivative
in $B(x^{*},\delta)$, $[G^{\prime}(x^{*})]^{-1}$ exists and
$[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies $(\ref{eqn:24})$ where given
fixed positive constants $\gamma$ and $\kappa_{0}>0$ with
$\kappa_{0}(u)=\gamma+\kappa_{0}u$ i.e.:

$\displaystyle||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x)-G^{\prime}(x^{*}))||\leq
2||x-x^{*}||(\gamma+\kappa_{0}||x-x^{*}||);\forall\ x\in B(x^{*},\delta),$

where $\rho(x)=||x-x^{*}||$. Let $\delta$ satisfy the relation

$\displaystyle\delta$ $\displaystyle=$
$\displaystyle\frac{2\gamma-\sqrt{4\gamma^{2}-(16/3)\kappa_{0}}}{(8/3)\kappa_{0}}\
and\ 4\gamma^{2}-(16/3)\kappa_{0}\geq 0.$ (4.14)

Then, the equation $G(x)=0$ has a unique solution $x^{*}$ in
$B(x^{*},\delta)$. Moreover, the ball radius $\delta$ depends only on
$\kappa_{0}$ and $\gamma$.

##  5\. Local Convergence under weak $\kappa$-average

This section presents the findings of a re-examination of the requirements and
radius of convergence of the considered scheme, which were previously stated
in the first theorem, although $\kappa$ is not regarded a non-decreasing
function. The convergence order has been observed to be decreasing. The second
theorem in this section yields a result identical to theorem $(\ref{th:51})$,
but with the center Lipschitz condition assumed.

######  Theorem 5.1.

Suppose that $G(x^{*})=0$, $G$ has a continuous derivative in
$B(x^{*},\delta)$, $[G^{\prime}(x^{*})]^{-1}$ exists and
$[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies the assumptions
$(\ref{eqn:23})$, $(\ref{eqn:24})$, $\kappa_{0}$ and $\kappa$ are positive
integrable. Let $\delta$ satisfy

$\displaystyle\int_{0}^{2\delta}(\kappa(u)+\kappa_{0}(u))du\leq 1.$ (5.1)

Then, the three-step Newton-Traub method $(\ref{eqn:12})$ is convergent for
all $x_{0}\in B(x^{*},\delta)$ and

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(x_{t})}\kappa(u)udu}{2(1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du)}\leq
q_{1}\rho(x_{t}),$ (5.2) $\displaystyle||z_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{\rho(x_{t})+\rho(y_{t})}\kappa(u)udu}{(\rho(x_{t})+\rho(y_{t}))(1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du)}\rho(y_{t})\leq
q_{2}q_{1}\rho(x_{t}),$ $\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{\rho(y_{t})+\rho(z_{t})}\kappa(u)udu}{(\rho(y_{t})+\rho(z_{t}))(1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du)}\rho(z_{t})\leq
q_{3}q_{2}q_{1}\rho(x_{t}),$

where the quantities

$\displaystyle
q_{1}=\frac{\int_{0}^{2\rho(x_{0})}\kappa(u)du}{1-\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du},\
q_{2}=\frac{\int_{0}^{\rho(x_{0})+\rho(y_{0})}\kappa(u)du}{1-\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du},$
$\displaystyle
q_{3}=\frac{\int_{0}^{\rho(y_{0})+\rho(z_{0})}\kappa(u)du}{1-\int_{0}^{2\rho(y_{0})}\kappa_{0}(u)du},$
(5.5)

are less than $1$. Moreover,

$\displaystyle||x_{t}-x^{*}||\leq(q_{1}q_{2}q_{3})^{t}||x_{0}-x^{*}||,\
t=1,2,....$ (5.6)

Furthermore, suppose that the function $\kappa_{a}$ is defined by

$\displaystyle\kappa_{a}(P)=P^{1-a}\kappa(P),$ (5.7)

is non-decreasing for some $a$ with $0\leq a\leq 1$ and $\delta$ satisfies

$\frac{1}{2\delta}\int_{0}^{2\delta}(2\delta\kappa_{0}(u)+u\kappa(u))du\leq
1.$ (5.8)

Then, the three-step Newton-Traub method $(\ref{eqn:12})$ is convergent for
all $x_{0}\in B(x^{*},\delta)$ and

$\displaystyle||x_{t}-x^{*}||\leq E^{(1+3a+a^{2})^{t}-1}||x_{0}-x^{*}||,\
t=1,2,\cdots,E=C_{1}^{2}C_{2}\frac{\rho(x_{0})^{2}}{\rho(y_{0})\rho(z_{0})},$

where the quantities $C_{1}$ and $C_{2}$ are defined in the equation
$(\ref{eqn:36})$ and are less than $1$.

###### Proof.

We can show that the quantities $q_{1}$, $q_{2}$ and $q_{3}$ described by the
equation $(\ref{eqn:55})$ as less than 1 by arbitrarily picking $x_{0}\in
B(x^{*},\delta)$, where $\delta$ fulfills the relation $(\ref{eqn:52})$.
Indeed, because $\kappa$ is positive integrable function, we may obtain

$\displaystyle q_{1}$ $\displaystyle=$
$\displaystyle\frac{\int_{0}^{2\rho(x_{0})}\kappa(u)du}{1-\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du}\leq\frac{\int_{0}^{2\delta}\kappa(u)du}{1-\int_{0}^{2\delta}\kappa_{0}(u)du}<1,$
$\displaystyle q_{2}$ $\displaystyle=$
$\displaystyle\frac{\int_{0}^{\rho(x_{0})+\rho(y_{0})}\kappa(u)du}{1-\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du}\leq\frac{\int_{0}^{2\delta}\kappa(u)du}{1-\int_{0}^{2\delta}\kappa_{0}(u)du}<1,$
$\displaystyle q_{3}$ $\displaystyle=$
$\displaystyle\frac{\int_{0}^{\rho(y_{0})+\rho(z_{0})}\kappa(u)du}{1-\int_{0}^{2\rho(y_{0})}\kappa_{0}(u)du}\leq\frac{\int_{0}^{2\delta}\kappa(u)du}{1-\int_{0}^{2\delta}\kappa_{0}(u)du}<1.$

Obviously, if $x\in B(x^{*},\delta)$, then using center Lipschitz condition
with the $\kappa$-average and the relation $(\ref{eqn:52})$, we have

$\displaystyle||[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x)-G^{\prime}(x^{*})]||\leq\int_{0}^{2\rho(x)}\kappa_{0}(u)du;\forall\
x\in B(x^{*},\delta)\leq 1.$

Using the Banach Lemma and the following equation

$\displaystyle||I-([G^{\prime}(x^{*})]^{-1}G^{\prime}(x)-I)||^{-1}=||[G^{\prime}(x)]^{-1}G^{\prime}(x^{*})||,$

we come to following inequality using the relation $(\ref{eqn:510})$

$\displaystyle||[G^{\prime}(x)]^{-1}G^{\prime}(x^{*})||$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\int_{0}^{2\rho(x)}\kappa_{0}(u)du}.$ (5.11)

Hence, if $x_{t}\in B(x^{*},\delta)$, then we may write from first sub-step of
scheme $(\ref{eqn:12})$

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle=$
$\displaystyle||x_{t}-x^{*}-[G^{\prime}(x_{t})]^{-1}G(x_{t})||$
$\displaystyle=$
$\displaystyle||[G^{\prime}(x_{t})]^{-1}[G^{\prime}(x_{t})(x_{t}-x^{*})-G(x_{t})+G(x^{*})]||.$

Expanding $G(x_{t})$ around $x^{*}$ from the Taylor’s expansion, it can
written as

$\displaystyle
G(x^{*})-G(x_{t})+G^{\prime}(x_{t})(x_{t}-x^{*})=G^{\prime}(x^{*})\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x_{t})-G^{\prime}(x_{t}^{\tau})]d\tau(x_{t}-x^{*}).$

Following the assumptions $(\ref{eqn:23})$ and combining the inequalities
$(\ref{eqn:512})$ and $(\ref{eqn:513})$, we may write

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle||[G^{\prime}(x_{t})]^{-1}G^{\prime}(x^{*})||.||\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x_{t})-G^{\prime}(x_{t}^{\tau})]d\tau||.||(x_{t}-x^{*})||$
(5.14) $\displaystyle\leq$
$\displaystyle\frac{1}{\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\int_{0}^{1}\int_{2\tau\rho(x_{t})}^{2\rho(x_{t})}\kappa(u)du\rho(x_{t})d\tau.$

Using the results of Lemma $(\ref{lm:21})$ and the inequality
$(\ref{eqn:511})$ in the above expression we can obtain the first inequality
of $(\ref{eqn:54})$. By similar analogy for the second sub-step of the scheme
$(\ref{eqn:12})$, we can write

$\displaystyle||z_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle||[G^{\prime}(x_{t})]^{-1}G^{\prime}(x^{*})||.||\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x_{t})-G^{\prime}(y_{t}^{\tau})]d\tau||.||(y_{t}-x^{*})||$
(5.15) $\displaystyle\leq$
$\displaystyle\frac{1}{\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\int_{0}^{1}\int_{\tau(\rho(x_{t})+\rho(y_{t}))}^{\rho(x_{t})+\rho(y_{t})}\kappa(u)du\rho(y_{t})d\tau.$

This way, we get the first inequality of $(\ref{eqn:54b})$ using the Lemma
$(\ref{lm:21})$ in the above expression. Similarly, rewriting the last sub-
step of the scheme $(\ref{eqn:12})$, we achieve

$\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle||[G^{\prime}(y_{t})]^{-1}G^{\prime}(x^{*})||.||\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(y_{t})-G^{\prime}(z_{t}^{\tau})]d\tau||.||(z_{t}-x^{*})||$
(5.16) $\displaystyle\leq$
$\displaystyle\frac{1}{\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}\int_{0}^{1}\int_{\tau(\rho(y_{t})+\rho(z_{t}))}^{\rho(y_{t})+\rho(z_{t})}\kappa(u)du\rho(z_{t})d\tau.$

Using Lemma $(\ref{lm:21})$ in the above expression, we can get the first
inequality of $(\ref{eqn:54a})$. Moreover, $\rho(x_{t})$, $\rho(y_{t})$ and
$\rho(z_{t})$ are decreasing monotonically, therefore for all $t=0,1,...,$ we
have

$\displaystyle||y_{t}-x^{*}||\leq\frac{\int_{0}^{2\rho(x_{t})}\kappa(u)udu}{2(1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du)}\leq\frac{\int_{0}^{2\rho(x_{0})}\kappa(u)du}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(x_{t})\leq
q_{1}\rho(x_{t}).$

Using the second inequality of expression $(\ref{eqn:54})$, we arrive at

$\displaystyle||z_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{\rho(x_{t})+\rho(y_{t})}\kappa(u)udu}{(\rho(x_{t})+\rho(y_{t}))(1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du)}\rho(y_{t})$
(5.17) $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{\rho(x_{0})+\rho(y_{0})}\kappa(u)du}{(1-\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du)}\rho(y_{t})\leq
q_{2}q_{1}\rho(x_{t}).$

Next, with the help of inequality of expression $(\ref{eqn:54b})$, we are able
to reach at

$\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{\rho(y_{t})+\rho(z_{t})}\kappa(u)udu}{(\rho(y_{t})+\rho(z_{t}))(1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du)}\rho(z_{t})$
(5.18) $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{\rho(y_{0})+\rho(z_{0})}\kappa(u)du}{1-\int_{0}^{2\rho(y_{0})}\kappa_{0}(u)du}\rho(z_{t})\leq
q_{3}q_{2}q_{1}\rho(x_{t}).$

Therefore, the inequality $(\ref{eqn:53})$ may be easily deduced from the
inequality $(\ref{eqn:517a})$. Furthermore, if the function $\kappa_{a}$
defined by the relation $(\ref{eqn:56})$ is non-decreasing for some $a$ with
$0\leq a\leq 1$ and $\delta$ is determined by the expression $(\ref{eqn:57})$,
the first inequality of expression $(\ref{eqn:54})$ and Lemma $(\ref{lm:22})$
imply that

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\varphi_{1,a}(2\rho(x_{t}))2^{a}}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(x_{t})^{a+1}$
(5.19) $\displaystyle\leq$
$\displaystyle\frac{\varphi_{1,a}(2\rho(x_{0}))2^{a}}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(x_{t})^{a+1}=\frac{C_{1}}{\rho(x_{0})^{a}}\rho(x_{t})^{a+1}\
$

Moreover, from the first part of inequality $(\ref{eqn:54b})$ and Lemma
$(\ref{lm:22})$, we may write

$\displaystyle||z_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\varphi_{1,a}(\rho(x_{t})+\rho(y_{t}))(\rho(x_{t})+\rho(y_{t}))^{a}}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(y_{t}),$
$\displaystyle\leq$
$\displaystyle\frac{\varphi_{1,a}(\rho(x_{0})+\rho(y_{0}))(\rho(x_{t})+\rho(y_{t}))^{a}}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(y_{t}),$
$\displaystyle\leq$
$\displaystyle\frac{C_{2}}{(\rho(x_{0})+\rho(y_{0}))^{a}}(\rho(x_{t})+\rho(y_{t}))^{a}\rho(y_{t}),$
$\displaystyle\leq$
$\displaystyle\frac{C_{1}C_{2}}{\rho(x_{0})^{a}\rho(y_{0})^{a}}\rho(x_{t})^{2a+1}.$

Similarly, from the first inequality of $(\ref{eqn:54a})$ and Lemma
$(\ref{lm:22})$, we arrive at

$\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\varphi_{1,a}(\rho(y_{t})+\rho(z_{t}))(\rho(y_{t})+\rho(z_{t}))^{a}}{1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}\rho(z_{t})$
(5.21) $\displaystyle\leq$
$\displaystyle\frac{\varphi_{1,a}(\rho(y_{0})+\rho(z_{0}))(\rho(y_{t})+\rho(z_{t}))^{a}}{1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}\rho(z_{t})$
$\displaystyle=$
$\displaystyle\frac{C_{1}^{a+1}C_{2}C_{3}}{\rho(x_{0})^{a+a^{2}}\rho(y_{0})^{a}\rho(z_{0})^{a}}\rho(x_{t})^{(a+1)^{2}+a},$

where $C_{1}<1$, $C_{2}<1$ and $C_{3}<1$ are determined by the expression
$(\ref{eqn:36})$. Now, we use mathematical induction to prove the inequality
$(\ref{eqn:58})$. For $t=0$, the above inequality becomes

$\displaystyle||x_{1}-x^{*}||\leq\frac{C_{1}^{1+a}C_{2}C_{3}}{\rho(x_{0})^{a^{2}+a}\rho(y_{0})^{a}\rho(z_{0})^{a}}\rho(x_{0})^{(a+1)^{2}+a}.$

Multiplying the numerator and denominator by
$\rho(y_{0})^{a^{2}+2a}\rho(z_{0})^{a^{2}+2a}$ and using the inequality
$(\ref{eqn:4444b})$ for $t=0$ and then doing the calculations, we can get

$\displaystyle||x_{1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\left[C_{1}^{2}C_{2}\frac{\rho(x_{0})^{2}}{\rho(y_{0})\rho(z_{0})}\right]^{a^{2}+3a}\rho(x_{0})$
(5.22) $\displaystyle\leq$ $\displaystyle E^{((1+3a+a^{2})-1)}\rho(x_{0}).$

Thus the expression $(\ref{eqn:58})$ is true for $t=1$. Now, suppose the
inequality $(\ref{eqn:58})$ is true for some integer $t>1$. Again multiplying
in the numerator and denominator of the inequality $(\ref{eqn:4444a})$ by
$\rho(y_{0})^{a^{2}+2a}\rho(z_{0})^{a^{2}+2a}$ and then using the inequalities
$(\ref{eqn:58})$ for $t=t$, $(\ref{eqn:4444b})$, $(\ref{eqn:4444c})$ after re-
arranging the terms, the above inequality preserves the form

$\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\left[C_{1}^{2}C_{2}\frac{\rho(x_{0})^{2}}{\rho(y_{0})\rho(z_{0})}\right]^{(1+3a+a^{2})^{t+1}-1}\rho(x_{0})$
$\displaystyle\leq$ $\displaystyle E^{(1+3a+a^{2})^{t+1}-1)}\rho(x_{0}).$

which shows that the result is true for $t=t+1$, hence the inequality
$(\ref{eqn:58})$ is true for all natural number and consequently $x_{t}$
converges to $x^{*}$. Hence, the proof is finished. ∎

######  Theorem 5.2.

Suppose that $G(x^{*})=0$, $G$ has a continuous derivative in
$B(x^{*},\delta)$, $[G^{\prime}(x^{*})]^{-1}$ exists and
$[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies the assumption $(\ref{eqn:24})$
and $\kappa_{0}$ is positive integrable function. Let $\delta$ satisfy

$\displaystyle\int_{0}^{2\delta}\kappa_{0}(u)du\leq\frac{1}{3}.$ (5.23)

Then, the three-step Newton-Traub method $(\ref{eqn:12})$ is convergent for
all
$x_{0}\in B(x^{*},\delta)$ and

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{2\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(x_{t})\leq
q_{1}\rho(x_{t}),$ $\displaystyle||z_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du+\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(y_{t})\leq
q_{2}q_{1}\rho(x_{t}),$ $\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du+\int_{0}^{2\rho(z_{t})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}\rho(z_{t})\leq
q_{3}q_{2}q_{1}\rho(x_{t}),$

where the quantities

$\displaystyle
q_{1}=\frac{2\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du},\
q_{2}=\frac{\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du+\int_{0}^{2\rho(y_{0})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du},$
$\displaystyle
q_{3}=\frac{\int_{0}^{2\rho(y_{0})}\kappa_{0}(u)du+\int_{0}^{2\rho(z_{0})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(y_{0})}\kappa_{0}(u)du},$
(5.25)

are less than $1$. Moreover,

$\displaystyle||x_{t}-x^{*}||\leq(q_{1}q_{2}q_{3})^{t}||x_{0}-x^{*}||,\
t=1,2,....$ (5.26)

Furthermore, suppose that the function $\kappa_{a}$ defined by the relation
$(\ref{eqn:56})$ is non-decreasing for some $a$ with $0\leq a\leq 1$, then

$\displaystyle||x_{t}-x^{*}||\leq F^{(1+3a+a^{2})^{t}-1}||x_{0}-x^{*}||,\
t=1,2,\cdots,F=q_{1}^{2}q_{2}\frac{\rho(x_{0})^{2}}{\rho(y_{0})\rho(z_{0})},$

and $q_{1}$, $q_{2}$ are given by the first expression of the equation
$(\ref{eqn:524})$.

###### Proof.

Let $x_{0}\in B(x^{*},\delta)$ and ${x_{t}}$ be the sequence generated by
three-step Newton-Traub method given in $(\ref{eqn:12})$. Let $\delta$,
$q_{1}$ and $q_{2}$ be determined by the expressions $(\ref{eqn:522})$ and
$(\ref{eqn:524})$, respectively. Assume that $x_{t}\in B(x^{*},\delta)$. Then

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle=$
$\displaystyle||y_{t}-x^{*}-[G^{\prime}(x_{t})]^{-1}G(x_{t})||$
$\displaystyle=$
$\displaystyle||[G^{\prime}(x_{t})]^{-1}[G^{\prime}(x_{t})(x_{t}-x^{*})-G(x_{t})+G(x^{*})]||.$

Expanding $G(x_{t})$ along $\ x^{*}$ from Taylor’s expansion, we have

$\displaystyle
G(x^{*})-G(x_{t})+G^{\prime}(x_{t})(x_{t}-x^{*})=G^{\prime}(x^{*})\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x_{t})-G^{\prime}(x_{t}^{\tau})]d\tau(x_{t}-x^{*}).$

Following the assumptions $(\ref{eqn:24})$ of the theorem and using the
equations $(\ref{eqn:526})$ and $(\ref{eqn:527})$, it can be written as

$\displaystyle||y_{t}-x^{*}||\leq||[G^{\prime}(x_{t})]^{-1}G^{\prime}(x^{*})||.||\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x_{t})-G^{\prime}(x^{*})+G^{\prime}(x^{*})-G^{\prime}(x_{t}^{\tau})]d\tau||$
$\displaystyle.||(x_{t}-x^{*})||$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\left(\int_{0}^{1}\int_{0}^{2\tau\rho(x_{t})}\kappa_{0}(u)du\rho(x_{t})d\tau+\int_{0}^{1}\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du\rho(x_{t})d\tau\right).$

In virtue of Lemma $(\ref{lm:21})$, the above inequality takes the form

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{2\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du\rho(x_{t})-\frac{1}{2}\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)udu}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}$
$\displaystyle\leq$
$\displaystyle\frac{2\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(x_{t})=q_{1}\rho(x_{t}),$

which is equivalent to the first inequality of $(\ref{eqn:520a})$. Using a
similar analogy and the second sub-step of the technique $(\ref{eqn:12})$, we
may write

$\displaystyle||z_{t}-x^{*}||\leq||[G^{\prime}(x_{t})]^{-1}G^{\prime}(x^{*})||\left(||\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x_{t})-G^{\prime}(x^{*})]d\tau||.||(y_{t}-x^{*})||\right)$
$\displaystyle+\left(||\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x^{*})-G^{\prime}(y_{t}^{\tau})]d\tau||.||(y_{t}-x^{*})||\right)$
$\displaystyle\leq$
$\displaystyle\frac{1}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\left(\int_{0}^{1}\int_{0}^{2\tau\rho(y_{t})}\kappa_{0}(u)du\rho(y_{t})d\tau+\int_{0}^{1}\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du\rho(y_{t})d\tau\right).$

As a result of Lemma $(\ref{lm:21})$, the above expression takes the form

$\displaystyle||z_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(x_{n})}\kappa_{0}(u)du\rho(y_{t})+\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du\rho(y_{t})-\frac{1}{2}\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)udu}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}$
$\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du\rho(y_{t})+\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du\rho(y_{t})}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}$
$\displaystyle=$ $\displaystyle q_{2}q_{1}\rho(x_{t}),$

Simultaneously, from the final sub-step of the scheme $(\ref{eqn:12})$, we can
write

$\displaystyle||x_{t+1}-x^{*}||\leq||[G^{\prime}(y_{t})]^{-1}G^{\prime}(x^{*})||\left(||\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(y_{t})-G^{\prime}(x^{*})]d\tau||.||(z_{t}-x^{*})||\right)$
$\displaystyle+\left(||\int_{0}^{1}[G^{\prime}(x^{*})]^{-1}[G^{\prime}(x^{*})-G^{\prime}(z_{t}^{\tau})]d\tau||.||(z_{t}-x^{*})||\right)$
$\displaystyle\leq$
$\displaystyle\frac{1}{1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}\left(\int_{0}^{1}\int_{0}^{2\tau\rho(y_{t})}\kappa_{0}(u)du\rho(z_{t})d\tau+\int_{0}^{1}\int_{0}^{2\rho(z_{t})}\kappa_{0}(u)du\rho(z_{t})d\tau\right).$

Because of Lemma $(\ref{lm:21})$, the above expression leads to

$\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du\rho(z_{t})+\int_{0}^{2\rho(z_{t})}\kappa_{0}(u)du\rho(z_{t})-\frac{1}{2}\int_{0}^{2\rho(z_{t})}\kappa_{0}(u)udu}{1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}$
$\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du\rho(z_{t})+\int_{0}^{2\rho(z_{t})}\kappa_{0}(u)du\rho(z_{t})}{1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}$
$\displaystyle=$ $\displaystyle q_{3}q_{2}q_{1}\rho(x_{n}),$

where $q_{1}<1$, $q_{2}<1$ and $q_{3}<1$ are determined by the relation
$(\ref{eqn:524})$. Also, it can be seen that inequality $(\ref{eqn:523})$ may
be easily derived from the second expression $(\ref{eqn:520a})$ and hence
$x_{t}$ converges to $x^{*}$.

Furthermore, if the function $\kappa_{a}$ defined by the relation
$(\ref{eqn:56})$ is non-decreasing for some $a$ with $0\leq a\leq 1$ and
$\delta$ is determined by the inequality $(\ref{eqn:522})$, it follows from
the first inequality of the expression $(\ref{eqn:520a})$ and Lemma
$(\ref{lm:22})$ that

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{2\varphi_{0,a}(2\rho(x_{t}))2^{a}}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(x_{t})^{a+1},$
$\displaystyle\leq$
$\displaystyle\frac{2\varphi_{0,a}(2\rho(x_{0}))2^{a}}{1-\int_{0}^{2\rho(x_{0})}\kappa_{0}(u)du}\rho(x_{t})^{a+1}=\frac{q_{1}}{\rho(x_{0})^{a}}\rho(x_{t})^{a+1}.$

Moreover, from the second inequality of expression $(\ref{eqn:520a})$ and
Lemma $(\ref{lm:22})$, we get

$\displaystyle||z_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\varphi_{0,a}(2\rho(x_{t}))+\varphi_{0,a}(2\rho(y_{t}))(2\rho(x_{t}))^{a}\rho(y_{t})}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(y_{t}),$
$\displaystyle\leq$
$\displaystyle\frac{\varphi_{0,a}(2\rho(x_{0}))+\varphi_{0,a}(2\rho(y_{0}))(2\rho(x_{t}))^{a}\rho(y_{n})}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(y_{t}),$
$\displaystyle=$
$\displaystyle\frac{q_{1}q_{2}}{\rho(x_{0})^{a}\rho(y_{0})^{a}}\rho(x_{t})^{2a+1}.$

Finally, the second inequality of expression $(\ref{eqn:520a})$ of the last
sub-step and Lemma $(\ref{lm:22})$ gives

$\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\varphi_{0,a}(2\rho(y_{t}))+\varphi_{0,a}(2\rho(z_{t}))(2\rho(y_{t}))^{a}\rho(z_{t})}{1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}\rho(z_{t}),$
(5.33) $\displaystyle\leq$
$\displaystyle\frac{\varphi_{0,a}(2\rho(y_{0}))+\varphi_{0,a}(2\rho(z_{0}))(2\rho(y_{t}))^{a}\rho(z_{t})}{1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}\rho(z_{t}),$
$\displaystyle=$
$\displaystyle\frac{q_{1}^{a+1}q_{2}q_{3}}{\rho(x_{0})^{a+a^{2}}\rho(y_{0})^{a}\rho(z_{0})^{a}}\rho(x_{t})^{(a+1)^{2}+a}.$

Now, we can prove that the expression $(\ref{eqn:525})$ is true for all
integers $t\geq 1$ through mathematical induction in the same line as previous
result. Accordingly, $x_{t}$ eventually converges to $x^{*}$. ∎

Next, the outcomes of Theorems $(\ref{th:51})$ and $(\ref{th:52})$ will then
be re-captured using our newly improved theorems on some special functions
$\kappa$.

######  Corollary 5.1.

Suppose that $x^{*}$ satisfies $G(x^{*})=0$, $G$ has a continuous derivative
in $B(x^{*},\delta)$, $[G^{\prime}(x^{*})]^{-1}$ exists and
$[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies $(\ref{eqn:23})$,
$(\ref{eqn:24})$ with $\kappa(u)=cau^{a-1}$ and $\kappa_{0}(u)=c_{0}au^{a-1}$
i.e.:

$||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x)-G^{\prime}(y^{\tau}))||\leq
c(1-\tau^{a})(||x-x^{*}||+||y-x^{*}||)^{a},$ (5.34)

and

$\displaystyle||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x)-G^{\prime}(x^{*}))||\leq
c_{0}2^{a}||x-x^{*}||^{a},$ (5.35)

$\forall\ x,\ y\in B(x^{*},\delta),0\leq\tau\leq 1$, where
$y^{\tau}=x^{*}+\tau(y-x^{*})$, $\rho(x)=||x-x^{*}||$,$0<a<1$, $c>0$ and
$c_{0}>0$. Let $\delta$ satisfy

$\displaystyle\delta$ $\displaystyle=$
$\displaystyle\left(\frac{a+1}{2^{a}(c_{0}(a+1)+ca)}\right)^{\frac{1}{a}}.$
(5.36)

Then, the three-step Newton-Traub method $(\ref{eqn:12})$ is convergent for
all $x_{0}\in B(x^{*},\delta)$ and

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(x_{t})}\kappa(u)udu}{2(1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du)}\leq
q_{1}\rho(x_{t}),$ $\displaystyle||z_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{\rho(x_{t})+\rho(y_{t})}\kappa(u)udu}{(\rho(x_{t})+\rho(y_{t}))(1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du)}\rho(y_{t})\leq
q_{2}q_{1}\rho(x_{t}),$ $\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{\rho(y_{t})+\rho(z_{t})}\kappa(u)udu}{(\rho(y_{t})+\rho(z_{t}))(1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du)}\rho(z_{t})\leq
q_{3}q_{2}q_{1}\rho(x_{t}),$

where the quantities

$\displaystyle
q_{1}=\frac{ca2^{a}\rho(x_{0})^{a}}{(1+a)[1-2^{a}c_{0}\rho(x_{0})^{a}]},\
q_{2}=\frac{ca(\rho(x_{0})+\rho(y_{0}))^{a}}{(a+1)(1-2^{a}c_{0}\rho(x_{0})^{a})},$
$\displaystyle
q_{3}=\frac{ca(\rho(y_{0})+\rho(z_{0}))^{a}}{(a+1)(1-2^{a}c_{0}\rho(y_{0})^{a})},$
(5.37)

are less than $1$.

######  Corollary 5.2.

Suppose that $x^{*}$ satisfies $G(x^{*})=0$, $G$ has a continuous derivative
in $B(x^{*},\delta)$, $[G^{\prime}(x^{*})]^{-1}$ exists and
$[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies $(\ref{eqn:24})$ with
$\kappa_{0}(u)=c_{0}au^{a-1}$ i.e.:

$\displaystyle||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x)-G^{\prime}(x^{*}))||\leq
c_{0}2^{a}||x-x^{*}||^{a},\forall\ x\in B(x^{*},\delta),$ (5.38)

where $\rho(x)=||x-x^{*}||$, $0<a<1$ and $c_{0}>0$. Let $\delta$ satisfy

$\displaystyle\delta$ $\displaystyle=$
$\displaystyle\left(\frac{1}{3c_{0}2^{a}}\right)^{\frac{1}{a}}.$ (5.39)

Then, the three-step Newton-Traub method $(\ref{eqn:12})$ is convergent for
all $x_{0}\in B(x^{*},\delta)$ and

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{2\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(x_{t})\leq
q_{1}\rho(x_{t}),$ $\displaystyle||z_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du+\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(y_{t})\leq
q_{2}q_{1}\rho(x_{t}),$ $\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du+\int_{0}^{2\rho(z_{t})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}\rho(z_{t})\leq
q_{3}q_{2}q_{1}\rho(x_{t}),$

where the quantities

$\displaystyle
q_{1}=\frac{c_{0}2^{a+1}\rho(x_{0})^{a}}{1-2^{a}c_{0}\rho(x_{0})^{a}},\
q_{2}=\frac{c_{0}2^{a}(\rho(x_{0})^{a}+\rho(y_{0})^{a})}{1-2^{a}c_{0}\rho(x_{0})^{a}},$
$\displaystyle
q_{3}=\frac{c_{0}2^{a}(\rho(y_{0})^{a}+\rho(z_{0})^{a})}{1-2^{a}c_{0}\rho(y_{0})^{a}},$
(5.40)

are less than 1.

######  Corollary 5.3.

Suppose that $x^{*}$ satisfies $G(x^{*})=0$, $G$ has a continuous derivative
in $B(x^{*},\delta)$, $[G^{\prime}(x^{*})]^{-1}$ exists and
$[G^{\prime}(x^{*})]^{-1}G^{\prime}$ satisfies $(\ref{eqn:24})$ with
$\kappa_{0}(u)=\frac{2\gamma c_{0}}{(1-\gamma u)^{3}}$ i.e.:

$\displaystyle||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x)-G^{\prime}(x^{*}))||\leq\frac{c_{0}}{(1-2\gamma\rho(x))^{2}}-c_{0},\forall\
x\in B(x^{*},\delta),$

where $\rho(x)=||x-x^{*}||$, $\gamma>0$ and $c_{0}>0$. Let $\delta$ satisfy

$\displaystyle\delta$ $\displaystyle=$
$\displaystyle\frac{3c_{0}+1-\sqrt{3c_{0}(3c_{0}+1)}}{2\gamma(3c_{0}+1)}.$
(5.42)

Then, three-step Newton-Traub method $(\ref{eqn:12})$ is convergent for all
$x_{0}\in B(x^{*},\delta)$ and

$\displaystyle||y_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{2\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(x_{t})\leq
q_{1}\rho(x_{t}),$ $\displaystyle||z_{t}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du+\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(x_{t})}\kappa_{0}(u)du}\rho(y_{t})\leq
q_{2}q_{1}\rho(x_{t}),$ $\displaystyle||x_{t+1}-x^{*}||$ $\displaystyle\leq$
$\displaystyle\frac{\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du+\int_{0}^{2\rho(z_{t})}\kappa_{0}(u)du}{1-\int_{0}^{2\rho(y_{t})}\kappa_{0}(u)du}\rho(z_{t})\leq
q_{3}q_{2}q_{1}\rho(x_{t}),$

where the quantities

$\displaystyle q_{1}$ $\displaystyle=$
$\displaystyle\frac{2c_{0}-2c_{0}(1-2\gamma\rho(x_{0}))^{2}}{[1-2\gamma\rho(x_{0})]^{2}(1+c_{0})-c_{0}},$
$\displaystyle q_{2}$ $\displaystyle=$
$\displaystyle\frac{[c_{0}-c_{0}(1-2\gamma\rho(x_{0}))^{2}](1-2\gamma\rho(y_{0}))^{2})+[c_{0}-c_{0}(1-2\gamma\rho(y_{0}))^{2}](1-2\gamma\rho(x_{0}))^{2})}{([1-2\gamma\rho(x_{0})]^{2}(1+c_{0})-c_{0})(1-2\gamma\rho(y_{0}))^{2})},$
$\displaystyle q_{3}$ $\displaystyle=$
$\displaystyle\frac{[c_{0}-c_{0}(1-2\gamma\rho(y_{0}))^{2}](1-2\gamma\rho(z_{0}))^{2})+[c_{0}-c_{0}(1-2\gamma\rho(z_{0}))^{2}](1-2\gamma\rho(y_{0}))^{2})}{([1-2\gamma\rho(y_{0})]^{2}(1+c_{0})-c_{0})(1-2\gamma\rho(z_{0}))^{2})},$

are less than 1.

Remark 5.4 $(a)$ If $\kappa_{0}=\kappa$, then our results specialize to
earlier ones [[3], [7], [2], [16], [17]]. But if $\kappa_{0}<\kappa$, then the
benefits stated in the abstract are obtained (see also Example $6.1$ and
Example $6.2$).
$(b)$ A further extension can be achieved as follows. Suppose $(\ref{eqn:22})$
holds and equation $2\kappa_{0}(u)u-1=0$ has a minimal positive zero
$\overline{\delta}$. Define
B̃$=B(x^{*},\delta)\cap B(x^{*},\overline{\delta})$. Moreover, suppose

$\displaystyle||G(x)-G(y^{\tau})||\leq\int_{\tau(\rho(x)+\rho(y))}^{\rho(x)+\rho(y)}\overline{\kappa}(u)du,$
(5.44)

where $\forall\ x,\ y\in$B̃, $0\leq\tau\leq 1,\ and\ \overline{\kappa}$ is as
$\kappa$. Then, we have

$\displaystyle\overline{\kappa}(u)$ $\displaystyle\leq$
$\displaystyle\kappa(u)\ for\ all\ u\in[0,min\\{\delta,\overline{\delta}\\}].$

Then, in view of the proofs $\overline{\kappa}$ can replace $\kappa$ in all
results with $\kappa$. But, then if

$\displaystyle\overline{\kappa}(u)<\kappa(u),$

the benefits stated in the introduction are extended even further. In the case
of the motivational example, we have

$\displaystyle\kappa_{0}<\overline{\kappa}=\frac{e^{\frac{1}{(e-1)}}}{2}<\kappa.$

##  6\. Numerical Examples

######  Example 6.1.

Returning to the motivational example presented in the study’s introduction,
using $(\ref{eqn:32})$ and $G^{\prime}(x^{*})=(1,1,1)^{T}$, we have:
Old case $\kappa_{0}(u)=\kappa(u)=\frac{e}{2}$ gives

$\displaystyle\delta_{0}=0.245253.$

Case $\kappa_{0}(u)=\frac{e-1}{2}\ and\ \kappa(u)=\frac{e}{2}$ gives

$\displaystyle\delta_{1}=0.324947.$

Case $\kappa_{0}(u)=\frac{e-1}{2}\ and\
\overline{\kappa}(u)=\frac{e^{\frac{1}{(e-1)}}}{2}$ gives

$\displaystyle\delta_{2}=0.382692.$

Notice that

$\displaystyle\delta_{0}<\delta_{1}<\delta_{2}.$

######  Example 6.2.

Choose $X=Y=C[0,1],\Omega=\overline{V}(0,1)$ and $x^{*}=0$. Then, define $t$
on $\Omega$ as

$\displaystyle t(h)(x)=h(x)-\int_{0}^{1}x\tau h(\tau)^{3}d\tau.$

So,

$t^{\prime}(h(p))(x)=p(x)-3\int_{0}^{1}x\tau h(\tau)^{2}p(\tau)d\tau\ for\
all\ p\in\Omega.$

Then, we get

$\displaystyle\kappa_{0}(u)=1.5u<\kappa(u)=\overline{\kappa}(u)=3u.$

As a result, we get the same advantages as in Example $6.1$ by solving
$(3.1)$.

######  Example 6.3.

Let $X=Y=R$, the reals. Define

$\displaystyle G(x)=\int_{0}^{x}\left(1+2x\sin\frac{\pi}{x}\right)dx,\ \forall
x\in R.$

Then

$G^{\prime}(x)=\begin{cases}1+2x\sin\frac{\pi}{x},&x\neq 0,\\\
1,&x=0,\end{cases}$

Obviously, $x^{*}=0$ is a zero of $G$ and $G^{\prime}$ satisfies that

$\displaystyle||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x)-G^{\prime}(x^{*}))||=\left|2x\sin\frac{\pi}{x}\right|\leq
2|x-x^{*}|,\forall\ x\in R.$

It follows from Theorem $(\ref{th:52})$ that for any $x_{0}\in B(x^{*},1/6)$

$\displaystyle||x_{t}-x^{*}||\leq F^{5^{t}-1}||x_{0}-x^{*}||,\
t=1,2,\cdots,F=\left(\frac{16|x_{0}|^{4}[2|x_{0}|+2|y_{0}|]}{[1-2|x_{0}|]^{3}|y_{0}||z_{0}|}\right).$

However, there is no positive integrable function $L$ such that the
inequality $(\ref{eqn:23})$ is satisfied. In fact, notice that

$||[G^{\prime}(x^{*})]^{-1}(G^{\prime}(x)-G^{\prime}(y^{\tau}))||=\left|2x\sin\frac{\pi}{x}-2y\tau\sin\frac{\pi}{y\tau}\right|=\frac{4}{2k+1},$

for $x=1/k,y=1/k,\tau=\frac{2k}{2k+1}$ and $k=1,2,\cdots$ Thus, if there was a
positive integrable function $L$ such that the inequality $(\ref{eqn:23})$
holds on $V(x^{*},r)$ for some $r>0$, it follows that there exists some
$n_{0}>1$ such that

$\displaystyle\int_{0}^{2r\delta}\kappa(u)du\geq\sum_{k=n_{0}}^{+\infty}\int_{\frac{4}{2k+1}}^{\frac{2}{k}}L(u)du\geq\sum_{k=n_{0}}^{+\infty}\frac{4}{2k+1}=+\infty,$

which is a contradiction. This example indicates that Theorem $(\ref{th:52})$
is a significant improvement over Theorem $(\ref{th:51})$ if the radius of the
convergence ball is neglected.

Conclusion

A novel technique is developed in order to provide a finer local 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. Hence, we have extended the applicability of modified
Newton’s method in cases not covered before. Our approach paves the way for
future research to improve local results for Newton-type methods and other
iterative procedures.

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