On some iterative methods for solving operator equations

9 years ago

Abstract

Let $X,Y$ be two Banach spaces and $P:X\rightarrow Y$ a nonlinear operator. We study the semilocal convergence of the Newton, chord and Steffensen methods for which the derivative $P^{\prime}\left( x\right) $ or the divided differences from each iteration step are approximated by a sequence of operators obtained with the Schultz method:
\begin{equation}
\left\{
\begin{array}
[c]{l}%
x_{n+1}=x_{n}-A_{n}P\left( x_{n}\right) \\
A_{n+1}=A_{n}\left( 2E-\left[ x_{n},x_{n+1};P\right] A_{n}\right)
,\qquad n=0,1,\ldots
\end{array}
\right. \label{f.1.6}%
\end{equation}
and considering the Steffensen method:%
\begin{equation}
\left\{
\begin{array}
[c]{l}%
x_{n+1}=x_{n}-A_{n}P\left( x_{n}\right) \\
A_{n+1}=A_{n}\left( 2E-\left[ x_{n+1},Q\left( x_{n+1}\right) ;P\right]
A_{n}\right) ,\qquad n=0,1,\ldots
\end{array}
\right. \label{f.1.7}%
\end{equation}

Authors

Adrian Diaconu
(Tiberiu Popoviciu Institute of Numerical Analysis)

Ion Păvăloiu
(Tiberiu Popoviciu Institute of Numerical Analysis)

Title

Originnal title (in French)

Sur quelques méthods itératives pour la résolution des equations operationnelles

English translation of the title

On some iterative methods for solving operator equations

Keywords

Newton method; chord method; Steffensen method; Schultz method; semilocal convergence

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Cite this paper as:

A. Diaconu, I. Păvăloiu, Sur quelques méthods itératives pour la résolution des equations operationnelles, Rev. Anal. Numér. Théor. Approx., 1 (1972), pp. 45-61, https://doi.org/10.33993/jnaat11-3 (in French).

About this paper

Journal

Revue d’Analyse Numérique et de Théorie de l’ Approximation

Publisher Name

Academia Republicii S.R.

DOI

https://doi.org/10.33993/jnaat11-3

Print ISBN

Not available yet.

Online ISBN

Not available yet.

References

[1] Collatz, L., Naherungsverfahren hoherer Ordnung fur Gleichungen in Banach-Raumen, Archive for Rational Mechanics and Analysis II (1), 66–75 (1958).

[2] Diaconu, A., Pavaloiu, I., Asupra unor metode iterative pentru rezolvarea ecuatiilor operationale neliniare (I), Revista de analiza numerica si teoria aproximatiei, sous presse 2 (1973), nr. 1, pp. 61–79.

[3] Janko, B., Sur la theorie unitaire des methdes d’iteration pour la resolution des equations operationelles non lineaires. Publications of the Mathematical Institute of the Hungarian Academy of Sciences, II, Ser. A, 302–311 (1961).

[4] Janko, B., Rezolvarea ecuatiilor operationale neliniare in spatii Banach. Bucuresti, Editura Academiei R.S.R. (1969).

[5] Pavaloiu, I., Sur la methode de Steffensen pour la resolution des equations operationnelles non lineaires, Revue Roumaine de mathematiques pures et appliquees, XIII, 6, 857–861 (1968).

[6] Pavaloiu, I., Interpolation dans des espaces lineaires normes et applications. Mathematica (Cluj), 12(35), 1, 149–158 (1970).

[7] Pavaloiu, I., Sur les procedes iteratifs a un ordre eleve de convergence. Mathematica (Cluj), 12 (35), 2, 309–324 (1970).

[8] Pavaloiu, I., Consideratii asupra metodelor iterative obtinute prin interpolare inversa, Studii si cercetari matematice, XXIII, 10, 1545–1549 (1971).

[9] Popoviciu, T., Sur la delimitation de l’erreur dans l’approximation des racines d’une equation par interpolation lineaire ou quadratique. Revue Roumaine de mathematiques pures et appliquees, XIII, 1, 75–78 (1968).

[10] Sergheev, A. S., O metode hord. Sibirski mat. jurnal, XI, 2, 282–289 (1961)

[11] Traub, J. F., Iterative Methods for the Solution of Equations. Prentice-Hall Inc. Englewood Cliffs N.J. (1964).

[12] Ul’m, S., Ob obobscenih razdelenyh raznostjah I. Izv. Akad. Nauk. Estonskoi S.S.R., 16, 1, 13–26 (1967).

[13] Ul’m, S., Ob obobscenih razdelenyh raznostjah II. Izv. Akad. Nauk Estonskoi S.S.R., 16, 2, 146–155 (1967).

[14] Ul’m, S., Ob iteraccionnyh metodah s posledovatel’noi approksimacii obratnovo operatora. Izv. Akad. Nauk Estonskoi S.S.R., 16, 4, 403–411 (1967)

Paper (preprint) in HTML form

On some iterative methods
for solving operator equations

by
A. Diaconu and I. Păvăloiu
(Cluj)

In the application of some iterative methods to the resolution of non-linear operational equations the essential difficulty is the necessity of solving a linear operational equation at each iteration step.

For example, we consider for the resolution of the following equation:

(1)

P\left(x\right)=\theta,

Gold $P:X\rightarrow Y$ is a non-linear operator, $X, Y$ being Banach spaces, the well-known Newton-Kantorovitch method:

(2)

x_{n+1}=x_{n}-\left[P^{\prime}\left(x_{n}\right)\right]^{-1}P\left(x_{n}\right% ),\ \ \ \ n=0,1,\ldots

It is observed that for the application of this method it is necessary to solve at each iteration step the following linear operational equation:

\left(P^{\prime}\left(x_{n}\right)\right)\left(h\right)=-P\left(x_{n}\right),% \ \ \ \ n=0,1,\ldots,

Gold $P^{\prime}\left(x\right):X\rightarrow Y$ For $x$ fixed, is a linear operator, representing the Fréchet derivative of the operator $P$ in the point $x .$

This difficulty can be eliminated if we consider, in addition to the sequence of iterations $\left(x_{n}\right)_{n=0}^{\infty}$ a sequence of linear operators $\left(A_{n}\right)_{n=0}^{\infty};A_{n}:Y\rightarrow X$ tend towards the inverse of the linear operator which intervenes in the iterative method considered.

In the case of Newton's method this problem was studied by S. Ul'm in the work [ 15 ] . For the solution of equation ( 1 ) S. Ul'm considered the following iterative method:

(3)

\left\{\begin{array}[c]{l}x_{n+1}=x_{n}-A_{n}P\left(x_{n}\right)\\ A_{n+1}=A_{n}(2E-P\left(x_{n+1}\right)A_{n}),\qquad n=0,1,\ldots\end{array}\right.

Gold $A_{0}:Y\rightarrow X$ is an arbitrary linear operator. We observe that $A_{n}:Y\rightarrow X$ for each $n=0,1,\ldots$

S. Ul'm's results are based on the assumption that equation ( 1 ) has a solution.

In the work [ 3 ] we studied the convergence of the iterative process ( 3 ), without the hypothesis of the existence of the solution of the equation ( 1 ). We established the following theorem:

Theorem 1 .

If in the sphere $S=\left\{x:\left|\left|x-x_{0}\right|\right|\leq r\right\}$ the following conditions are met:

1)

The operator $P$ admits derivatives of Fréchet type up to and including order 2, the first-order derivative of the operator $P,$ admits a bounded inverse, 'that is, for each $x\in S,$ $\|\left[P^{\prime}\left(x\right)\right]^{-1}\|\leq B<+\infty$ And $\left\|P^{\prime\prime}\left(x\right)\right\|\leq M<+\infty.$
2)

The initial element $x_{0},$ the initial operator $A_{0}$ and the constants $B$ And $M$ satisfy the condition:

$\max\left\{2MB^{2}\left\|P\left(x_{0}\right)\right\|,\ \left\|E-P^{\prime}% \left(x_{0}\right)A_{0}\right\|\right\}\leq\tfrac{1}{9}d,$

Gold $d<1$ And $r=\frac{d}{9MB\left(1-d^{p-1}\right)},p=2-\varepsilon,\varepsilon>0$ being arbitrarily small, then the following properties hold:
1)

The sequels $\left(x_{n}\right)_{n=0}^{\infty},\left(A_{n}\right)_{n=0}^{\infty}$ are convergent
2)

Equation ( 1 ) admits the solution $x^{\ast}\in S$ which can be obtained as the limit of the sequence $\left(x_{n}\right)_{n=0}^{\infty}\$ generated by ( 3 ) and:

$A^{\ast}=\lim_{n\rightarrow\infty}A_{n}=\left[P^{\prime}\left(x^{\ast}\right)% \right]^{-1}.$

We have the following inequalities:

(4) $\left\|x_{n}-x^{\ast}\right\|\leq\tfrac{d^{p^{n}}}{9MB\left(1-d^{p^{n}\left(p-% 1\right)}\right)},\ \;\;n=0,1,\ldots$

(5) $\left\|A_{n}-A^{\ast}\right\|\leq\tfrac{Bd^{p^{n}}}{9\left(1-d^{p^{n}\left(p-1% \right)}\right)},\ \;\;n=0,1,\ldots$

The result presented above, establishes besides the convergence of the iterative process ( 3 ) the existence of the solution of the equation ( 1 ). The inequalities ( 4 ) and ( 5 ) give the speed of convergence, an evaluation of the error and show that the iterative process ( 3 ) has the order of convergence $p=2-\varepsilon,$ $\varepsilon>0$ arbitrarily small. Finally, the iterative process ( 3 ) has an order of convergence arbitrarily close to the order of convergence of Newton's method without, however, reaching it.

In the work [ 3 ] we studied at the same time other iterative processes obtained from well-known iterative methods. Starting from the rope method we considered the following iterative process

(6)

\left\{\begin{array}[c]{l}x_{n+1}=x_{n}-A_{n}P\left(x_{n}\right)\\ A_{n+1}=A_{n}\left(2E-\left[x_{n},x_{n+1};P\right]A_{n}\right),\qquad n=0,1,% \ldots\end{array}\right.

and starting from Steffensen's method the process:

(7)

\left\{\begin{array}[c]{l}x_{n+1}=x_{n}-A_{n}P\left(x_{n}\right)\\ A_{n+1}=A_{n}\left(2E-\left[x_{n+1},Q\left(x_{n+1}\right);P\right]A_{n}\right)% ,\qquad n=0,1,\ldots\end{array}\right.

Gold $x_{0}$ is an arbitrary element of space $X,$ the operator $A_{0}\$ is an arbitrary linear operator that transforms the space $Y$ In $X,\left[x,y;P\right]:X\rightarrow Y$ is the divided difference, [ 7 ] of the operator $P$ on the knots $x,y\in X$ and the operator $Q$ is an iterative operator attached to equation ( 1 ), [ 8 ] .

With respect to the processes ( 6 ) and ( 7 ) we have established the following theorems:

Theorem 2 .

If in the sphere $S=\left\{x:\left\|x-x_{0}\right\|\leq r\right\}$ the following conditions are met.

1)

The operator $P$ admits divided differences up to and including order 2, the divided difference of the first order admits an inverse and: $\|\left[x,y;P\right]^{-1}\|\leq B<+\infty$ for each $x,y\in S$ And $\left\|\left[x,y,z;P\right]\right\|\leq M<+\infty$ for each $x,y,z\in S,$ (by $\left[x,y,z;P\right]:X\rightarrow\left(X\rightarrow Y\right)$ we have designated the divided difference of the order $2$ of the operator $P$ ).
2)

The operator $A_{0}$ is limited and $\left\|A_{0}\right\|\leq 2B.$
3)

We have the inequalities:

$\max\left\{4MB^{2}\left\|P\left(x_{0}\right)\right\|,\tfrac{1}{9}\sqrt{4MB^{2}% \left\|P\left(x_{1}\right)\right\|},\left\|E-\left[x_{0},x_{1};P\right]A_{0}% \right\|^{2}\right\}\leq\tfrac{1}{9}d,$

Or $d<1$ And $r=\frac{d}{18MB\left(1-d^{p-1}\right)},p=\frac{1+\sqrt{5}}{2}-\varepsilon,\ % \varepsilon>0,$ arbitrarily small,

then the following propositions take place:
1)

The sequels $\left(x_{n}\right)_{n=0}^{\infty}$ And $\left(A_{n}\right)_{n=0}^{\infty},$ generated by ( 6 ), are convergent.
2)

Equation ( 1 ) has the solution $x^{\ast}\in S$ And $\lim\limits_{n\rightarrow\infty}x_{n}=x^{\ast}.$
3)

If $A^{\ast}=\lim_{n\rightarrow\infty}A_{n},$ SO $A^{\ast}=\lim_{n\rightarrow\infty}\left[x_{n},x_{n+1};P\right]^{-1}$ (the operator $\left[x_{n},x_{n+1};P\right]^{-1}$ exists for each $n=0,1,\ldots,$ because $x_{n}\in S,$ $n=0,1,\ldots).$
4)

We have the following inequalities:

(8) $\left\|x^{\ast}-x_{n}\right\|\leq\tfrac{d^{p^{n}}}{18MB\left(1-d^{p^{n}\left(p% -1\right)}\right)},\ \;\;n=0,1,\ldots,$

(9) $\left\|A^{\ast}-A_{n}\right\|\leq\tfrac{2B}{9}\left[2d^{p^{n-1}}+\tfrac{3d^{p^% {n}}}{1-d^{p^{n}\left(p-1\right)}}\right],\ \;\;n=0,1,\ldots$

Theorem 3 .

If in the sphere $S=\{x:\left\|x-x_{0}\right\|\leq r\}$ the following conditions are met:

1)

The operator $P$ admits divided differences up to order $2$ inclusively, the divided difference of the first order admits an inverse, $\|\left[x,y;P\right]^{-1}\|\leq B<+\infty$ for each $x,y,\in S$ And $\left\|\left[x,y,z;P\right]\right\|\leq M<+\infty$ for each $x,y,z\in S.$
2)

The operator $Q$ meets the following conditions:

$\displaystyle\left\|Q\left(x\right)-Q\left(y\right)\right\|$ $\displaystyle\leq L\left\|x-y\right\|,\ \ \text{pour chaque\ \ \ }x,y\in S,\ % \ L\in\mathbb{R}_{+},$

$\displaystyle\left\|x-Q\left(x\right)\right\|$ $\displaystyle\leq\alpha\left\|P\left(x\right)\right\|,\ \ \text{pour chaque \ % \ }x\in S,\ \ \alpha\in\mathbb{R}_{+},$

The following conditions are met:

		$\displaystyle\max\left\{2MB\left(2B+\alpha\right)\left\\|P\left(x_{0}\right)% \right\\|,\left\\|E-\left[x_{0},Q\left(x_{0}\right);P\right]A_{0}\right\\|\right% \}\leq cd$
	$\displaystyle\text{o\`{u} }d$	$\displaystyle<1,\ r=\tfrac{\alpha cd}{2MB\left(2B+\alpha\right)}+\tfrac{cd}{M% \left(2B+\alpha\right)\left(1-d^{p-1}\right)},$
	$\displaystyle c$	$\displaystyle=\min\left\{\tfrac{1}{2},\tfrac{1}{1+\alpha^{\prime}}\right\},\ % \alpha^{\prime}=\tfrac{1+L}{2B+\alpha},\ p=2-\varepsilon,\ \varepsilon>0\$

arbitrarily small, then:

1)

The sequels $\left(x_{n}\right)_{n=0}^{\infty}$ And $\left(A_{n}\right)_{n=0}^{\infty},$ generated by ( 7 ), are convergent.
2)

Equation ( 1 ) has the solution $x^{\ast}\in S$ which can be obtained as the limit of the sequence $\left(x_{n}\right)_{n=0}^{\infty}$ given by ( 7 ) and:

$A^{\ast}=\lim_{n\rightarrow\infty}A_{n}=\left[x^{\ast},Q\left(x^{\ast}\right);% P\right]^{-1}=\left[P^{\prime}\left(x^{\ast}\right)\right]^{-1}$

(so we can conclude the existence of the Fréchet type derivative of the operator $P$ in the point $x^{\ast}$ )
3)

The following inequalities are verified:

(10) $\left\|x^{\ast}-x_{n}\right\|\leq\tfrac{cd^{p^{n}}}{M\left(2B+\alpha\right)% \left(1-d^{p^{n}\left(p-1\right)}\right)},\ \;\;n=0,1,\ldots$

(11) $\left\|A^{\ast}-A_{n}\right\|\leq Bcd^{p^{n}}\left[1+\tfrac{2B\left(1+L\right)% }{\left(2B+\alpha\right)\left(1-d^{p^{n}\left(p-1\right)}\right)}\right],\;\;% \ n=0,1,\ldots,$

Theorems ( 2 ) and ( 3 ) show that the order of convergence of processes ( 6 ) and ( 7 ) is $\frac{1+\sqrt{5}}{2}-\varepsilon$ respectively $2-\varepsilon,\ \varepsilon>0$ arbitrarily approached the order of convergence of the string method, respectively of the Steffensen method, without however reaching the order of these methods.

To prove the stated theorems, we used the method of systems of recurring inequalities.

For processes ( 3 ) and ( 7 ) the following system is used:

(12)

\left\{\begin{array}[c]{l}\rho_{n+1}\leq\rho_{n}^{2}+\rho_{n}\delta_{n}\\ \delta_{n+1}\leq\left(\delta_{n}+a\rho_{n}\right)^{2},\qquad n=0,1,\ldots\end{% array}\right.

Or $a=2$ in the case of method ( 3 ) and $a=\alpha^{\prime}=\frac{1+L}{2B+\alpha}$ in the case of process ( 7 ). We choose $\rho_{n}=2MB^{2}\left\|P\left(x_{n}\right)\right\|$ in the case of method ( 3 ) and $\rho_{n}=2MB\left(2B+\alpha\right)\left\|P\left(x_{n}\right)\right\|$ in the case of method ( 7 ), $\delta_{n}=\left\|E-P^{\prime}\left(x_{n}\right)A_{n}\right\|$ in the case of method ( 3 ) and $\delta_{n}=\left\|E-\left[x_{n},Q\left(x_{n}\right);P\right]A_{n}\right\|$ in the case of the process ( 7 ). With the above notations $\rho_{n}$ And $\delta_{n}$ check the system ( 12 ).

Relative to the process ( 6 ) we choose $\rho_{n}=4MB^{2}\left\|P\left(x_{n}\right)\right\|$ And $\delta_{n}=\left\|E-\left[x_{n-1},x_{n};P\right]A_{n}\right\|$ and we demonstrate that we have the following inequalities:

(13)		$\displaystyle\rho_{n+1}$	$\displaystyle\leq\rho_{n}^{2}+\rho_{n}\rho_{n-1}+\rho_{n},\delta_{n}$
	$\displaystyle\delta_{n+1}$	$\displaystyle\leq\left(\delta_{n}+\rho_{n}+\rho_{n-1}\right)^{2}$

From systems ( 12 ) and ( 13 ) we easily deduce that: $\rho_{n}\leq cd^{p^{n}}$ And $\delta_{n}\leq cd^{p^{n}}$ for processes ( 3 ) and ( 7 ) and $\rho_{n}\leq cd^{p^{n}},$ $\delta_{n}\leq cd^{p^{n-1}}$ for the process ( 6 ). The constant $c=\frac{1}{9}$ for processes ( 3 ) and ( 6 ) and $c=\min\left\{\frac{1}{2},\frac{1}{1+\alpha^{\prime}}\right\}$ for the process ( 7 ), $p=2-\varepsilon$ for processes ( 3 ) and ( 7 ) and $p=\frac{1+\sqrt{5}}{2}-\varepsilon$ for the process ( 6 ), $\varepsilon>0$ arbitrarily small. In

using these inequalities we easily deduce in each case that the sequences $\left(x_{n}\right)_{n=0}^{\infty}$ are convergent.

The other conclusions are directly verified by calculation.

We now ask the following question: if instead of the Schultz-type method used for the approximation of linear operators $(\left[P^{\prime}\left(x_{n}\right)\right]^{-1})_{n=0}^{\infty},(\left[x_{n},x% _{n+1};P\right]^{-1})_{n=0}^{\infty}$ And $\left(\left[x_{n+1},Q\left(x_{n+1}\right);P\right]\right)_{n=0}^{\infty}$ If we used a more rapidly converging method, could the order of convergence of the obtained methods become equal to the order of convergence of the Newton and Steffensen methods, respectively, or the string method?

For the approximation of the inverse of a linear operator $A:Y\rightarrow X$ The following method is well known.

A_{n+1}=A_{n}\big{(}3E-3AA_{n}+\left(AA_{n}\right)^{2}\big{)},\ \ \;\;n=0,1,\ldots,

which is the Tchébischeff method for calculating the inverse of linear operators. This method has convergence order 3.

This method can be combined with Newton's method, the string method and Steffensen's method. We then obtain the following three processes:

(14)

\left\{\begin{array}[c]{l}x_{n+1}=x_{n}-A_{n}P\left(x_{n}\right)\\ A_{n+1}=A_{n}\left[3E-3P\left(x_{n+1}\right)A_{n}+\left(P^{\prime}\left(x_{n+1% }\right)A_{n}\right)^{2}\right],\quad n=0,1,\ldots\end{array}\right.

(15)

\left\{\begin{array}[c]{l}x_{n+1}=x_{n}-A_{n}P\left(x_{n}\right)\\ A_{n+1}=A_{n}\left[3E-3\left[x_{n},x_{n+1};P\right]A_{n}+\left(\left[x_{n},x_{% n+1};P\right]A_{n}\right)^{2}\right],\ \;\;n=0,1,\ldots\end{array}\right.

(18)		$\displaystyle\left\{\begin{array}[c]{l}x_{n+1}=x_{n}-A_{n}P\left(x_{n}\right)% \\ A_{n+1}=A_{n}\left[3E-3\left[x_{n+1},Q\left(x_{n+1}\right);P\right]A_{n}+\left% (\left[x_{n+1},Q\left(x_{n+1}\right);P\right]A_{n}\right)^{2}\right],\end{% array}\right.$
	$\displaystyle n=0,1,\ldots$

To study the convergences of the iterative processes ( 14 ) and ( 18 ) we use the following lemma.

Lemma 1 .

Let the consequences be given $\left(\rho_{n}\right)_{n=0}^{\infty},$ $\rho_{n}\geq 0,$ $\left(\delta_{n}\right)_{n=0}^{\infty},$ $\delta_{n}\geq 0$ which satisfy the following recurring inequalities:

(19)

\left\{\begin{array}[c]{l}\rho_{n+1}\leq\rho_{n}^{2}+\rho_{n}\delta_{n}\\ \delta_{n+1}\leq\left(\delta_{n}+a\rho_{n}\right)^{3},\ \ \ \ \;\;a>1,\ n=0,1,% \ldots\end{array}\right.

If $\rho_{0}\leq\alpha d,$ $\delta_{0}\leq\beta d,$ Or $\beta<1,$ $\alpha<1-\beta$ And $d<\frac{\beta}{\left[a+\left(1-a\right)\beta\right]^{3}},$ then the elements of the given sequences satisfy the inequalities:

(20)

\rho_{n}\leq\alpha d^{2^{n}},\ \ \delta_{n}\leq\beta d^{2^{n}}

And

\lim_{n\rightarrow\infty}\rho_{n}=0,\ \ \lim_{n\rightarrow\infty}\delta_{n}=0.

Demonstration..

We will demonstrate using the method of mathematical induction that for $n=0,1,\ldots$ the following inequalities are true.

(21)

\rho_{n}\leq\theta_{n}d^{2^{n}}\leq\alpha d^{2^{n}},\ \ \delta_{n}\leq\mu_{n}d% ^{2^{n}}\leq\beta d^{2^{n}}

Or:

(22)

\left\{\begin{array}[c]{c}\theta_{1}=\theta_{0}^{2}+\theta_{0}\mu_{0}\\ \mu_{1}=\left(\mu_{0}+a\theta_{0}\right)^{3}d\end{array}\right.

And

(23)

\left\{\begin{array}[c]{l}\theta_{n+1}=\theta_{n}^{2}+\mu_{n}\theta_{n}\\ \mu_{n+1}=\left(\mu_{n}+a\theta_{n}\right)^{3}d^{2^{n}},\ \ n=1,2,\ldots\end{% array}\right.

For $n=0$ the inequalities ( 21 ) result from the hypotheses of the lemma. For $n=0$ from ( 19 ) we have:

	$\displaystyle\rho_{1}$	$\displaystyle\leq\rho_{0}^{2}+\rho_{0}\delta_{0}$
	$\displaystyle\delta_{1}$	$\displaystyle\leq\left(\delta_{0}+a\rho_{0}\right)^{3}$

from where

	$\displaystyle\rho_{1}$	$\displaystyle\leq\left(\theta_{n}^{2}+\theta_{0}\mu_{0}\right)d^{2}\leq\left(% \alpha^{2}+\alpha\beta\right)d^{2}<\alpha d^{2}$
	$\displaystyle\delta_{1}$	$\displaystyle\leq\left(\mu_{0}+a\theta_{0}\right)^{3}dd^{2}\leq\left(\beta+a% \alpha\right)^{3}dd^{2}<\beta d^{2}$

because if $\beta<1$ And $d<\frac{\beta}{\left[a+\left(1-a\right)\beta\right]^{3}},$ $a>1,$ SO:

\left\{\begin{array}[c]{c}\alpha^{2}+\alpha\beta<\alpha\\ \left(\beta+a\alpha\right)^{3}d<\beta.\end{array}\right.

We now assume that inequalities ( 21 ) are true for $n=k$ and we will show that they take place for $n=k+1$

Indeed from ( 19 ) we deduce:

\left\{\begin{array}[c]{c}\rho_{k+1}\leq\rho_{k}^{2}+\rho_{k}\delta_{k}\\ \delta_{k+1}\leq\left(\delta_{k}^{-}+a\rho_{k}\right)^{3}\end{array}\right.

from which, taking into account ( 21 ) and the induction hypothesis we have:

	$\displaystyle\rho_{k+1}$	$\displaystyle\leq\left(\theta_{k}^{2}+\theta_{k}\mu_{k}\right)d^{2^{k+1}}\leq% \left(\alpha^{2}+\alpha\beta\right)d^{2^{k+1}}<\alpha d^{2^{k+1}},$
	$\displaystyle\delta_{k+1}$	$\displaystyle\leq\left(\mu_{k}+a\theta_{k}\right)^{3}d^{3.2^{k}}\leq\left(% \beta+a\alpha\right)^{3}d^{2^{k}}d^{2^{k-1}}+\beta d^{2^{k-1}},$

because from the hypotheses of the theorem it results $d<1.$

It follows that the inequalities ( 21 ) are true for each $n$ and taking into account the fact that $d<1$ We have $\lim\limits_{n\rightarrow\infty}\rho_{n}=0,$ $\lim\limits_{n\rightarrow\infty}\delta_{n}=0.$ The lemma is proven. ∎

For method ( 15 ) we will now establish the:

Lemma 2 .

(24)

\left\{\begin{array}[c]{c}\rho_{n+1}\leq\rho_{n}^{2}+\rho_{n}\rho_{n-1}+\rho_{% n}\delta_{n}\\ \delta_{n+1}\leq\left(\delta_{n}+\rho_{n}+\rho_{n-1}\right)^{3},\ \ \ n=0,1,% \ldots\end{array}\right.

If $\rho_{0}<\alpha d,$ $\rho_{1}<\alpha d^{p},$ $\delta_{1}<\beta d^{p},$ Or $p=\frac{1+\sqrt{5}}{2}$ And $\alpha,\beta$ And $d$ meet the conditions: $d<1,$ $\left(\beta+2\alpha\right)^{3}<\beta<1,$ SO:

\rho_{n}\leq\alpha d^{p^{n}},\ \delta_{n}\leq\beta d^{p^{n}}

And

\lim_{n\rightarrow\infty}\rho_{n}=0,\ \ \lim_{n\rightarrow\infty}\delta_{n}=0.

Demonstration..

We will show using the method of mathematical induction that for $n=0,1,\ldots$ the following inequalities are true:

(25)

\rho_{n}\leq\theta_{n}d^{p^{n}}\leq\alpha d^{p^{n}},\ \ \ \;\;n=0,1,\ldots

(26)

\delta_{n}\leq\mu_{n}d^{p^{n}}\leq\beta d^{p^{n}},\ \ \;\;n=1,2,\ldots

	$\displaystyle\theta_{n+1}$	$\displaystyle=\theta_{n}^{2}d^{2p^{n}-p^{n+1}}+\theta_{n}\theta_{n-1}d^{p^{n}+% p^{n-1}-p^{n+1}}+\theta_{n}\mu_{n}d^{2p^{n}-p^{n+1}},\ \;\;n=0,1,\ldots$
	$\displaystyle\mu_{n+1}$	$\displaystyle=\left(\mu_{n}d^{p^{n}-p^{n-1}}+\theta_{n}d^{p^{n}-p^{n-1}}+% \theta_{n-1}\right)^{3}d^{3p^{n-1}-p^{n+1}},\ \;\;n=1,2,\ldots$

Under the assumptions made it follows that the inequalities ( 25 ) are true for $n=0$ And $n=1$ , and inequalities ( 26 ) are true for $n=1,$ $\theta_{0}=\theta_{1}=\alpha,$ $\mu_{1}=\beta.$

We assume that inequalities ( 25 )-( 26 ) are true for each $n\leq k.$ We have:

	$\displaystyle\rho_{k+1}$	$\displaystyle\leq\theta_{k}^{2}d^{2p^{k}}+\theta_{k}\theta_{k-1}d^{p^{k}+p^{k-% 1}}+\theta_{k}\mu_{k}d^{2p^{k}}$
		$\displaystyle\leq\big{(}\theta_{k}^{2}d^{2p^{k}-p^{k+1}}+\theta_{k}\theta_{k+1% }d^{p^{k}+p^{k-1}-p^{k+1}}+\theta_{k}\mu_{k}d^{2p^{k}-p^{k+1}}\big{)}d^{p^{k+1}}$
		$\displaystyle=\theta_{k+1}d^{p^{k+1}}.$

We will show that: $\theta_{k+1}\leq\alpha.$ In fact, in the hypotheses of the lemma and the induction hypothesis it results:

\theta_{k+1}\leq\left(2\alpha^{2}+\alpha\beta\right)=\alpha\left(\beta+2\alpha% \right)\leq\alpha,

because:

2p^{k}-p^{k+1}>0,\ p^{k}+p^{k-1}-p^{k+1}=0\ \text{et\ }d<1

Likewise

	$\displaystyle\delta_{k+1}$	$\displaystyle\leq\left(\mu_{k}d^{p^{k}}+\theta_{k}d^{p^{k}}+\theta_{k+1}d^{p^{% k-1}}\right)^{3}$
		$\displaystyle=\left(\mu_{k}d^{p^{k}-p^{k-1}}+\theta_{k}d^{p^{k}-p^{k-1}}+% \theta_{k-1}\right)d^{3p^{k-1}-p^{k+1}}d^{p^{k+1}}$
		$\displaystyle=\mu_{k+1}d^{p^{k+1}}$
	$\displaystyle\mu_{k+1}$	$\displaystyle\leq\left(\beta+2\alpha\right)^{2}\leq\beta,\qquad\text{par ce % que }3p^{k-1}-p^{k+1}>0.$

So it follows that for each $n=1,2,\ldots$ we have the inequalities ( 25 ), ( 26 ) from which we have:

\lim_{n\rightarrow\infty}\rho_{n}=0,\ \ \ \lim_{n\rightarrow\infty}\delta_{n}=0.

The lemma is proven. ∎

Based on Lemma 1 , we will establish the following two theorems:

Theorem 4 .

If in the sphere $S=\left\{x\in X\left|\left\|x-x_{0}\right\|\leq r\right.\right\}$ the following conditions are met:

1)

The operator $P$ admits Fréchet-type derivatives up to order $2$ inclusively, the derivative of order $1$ is reversible and we have $\|[P^{\prime}\left(x\right)]^{-1}\|\leq B<+\infty,\ x\in S,$ And $\left\|P^{\prime\prime}\left(x\right)\right\|\leq M<+\infty,$ $x\in S.$
2)

$2MB^{2}\left\|P\left(x_{0}\right)\right\|\leq\alpha d,$ $\left\|E-P^{\prime}\left(x_{0}\right)A_{0}\right\|\leq\beta d,$ Or

$\beta<1,\alpha<1-\beta,\ d<\tfrac{\beta}{\left(2-\beta\right)^{3}},\ r=\tfrac{% \alpha d}{MB\left(1-d\right)},$

so we have the following properties:
1)

The iterative method ( 14 ) is convergent.
2)

The operational equation admits the solution $x^{\ast}\in S,$ which can be obtained as the limit of the sequence given by method ( 14 ) .
3)

We have the following delimitations:

$\left\|x^{\ast}-x_{n}\right\|\leq\tfrac{\alpha d^{2^{n}}}{MB\left(1-d^{2^{n}}% \right)},\ \ \;\;n=0,1,\ldots$

if $A^{\ast}=\left[P^{\prime}\left(x^{\ast}\right)\right]^{-1},$ the sequel $\left(A_{n}\right)_{n=1}^{\infty}$ tends towards $A^{\ast}$ and we have:

$\left\|A^{\ast}-A_{n}\right\|\leq\tfrac{2Bd^{2^{n}}\left(2\alpha+\beta-\beta d% ^{2^{n}}\right)}{1-d^{2^{n}}}.$

Demonstration..

We will demonstrate using mathematical induction that for each $n=0,1,\ldots$ the following properties are true:

(has)		$\displaystyle x_{n}$	$\displaystyle\in S,$
(b)		$\displaystyle\rho_{n}$	$\displaystyle=2MB^{2}\left\\|P\left(x_{n}\right)\right\\|\leq\alpha d^{2^{n}},$
	$\displaystyle\delta_{n}$	$\displaystyle=\left\\|E-P^{\prime}\left(x_{n}\right)A_{n}\right\\|\leq\beta d^{2% ^{n}},$
(c)		$\displaystyle\left\\|A_{n}\right\\|$	$\displaystyle\leq 2B.$

For $n=0$ properties a) and b) are verified in the hypotheses of the theorem. For c) we have:

	$\displaystyle\big{\\|}A_{0}\big{\\|}$	$\displaystyle\leq\big{\\|}\left[P^{\prime}\left(x_{0}\right)\right]^{-1}+A_{0}-% \left[P^{\prime}\left(x_{0}\right)\right]^{-1}\big{\\|}$
		$\displaystyle\leq\big{\\|}\left[P^{\prime}\left(x_{0}\right)\right]^{-1}\big{\\|% }\left(1+\left\\|E-P^{\prime}\left(x_{0}\right)A_{0}\right\\|\right)$
		$\displaystyle\leq B\left(1+\alpha d\right)<2B.$

We assume that a), b), c) are true for $n=0,1,\ldots,k.$ For $n=k+1$ we have:

	$\displaystyle\left\\|x_{k+1}-x_{0}\right\\|$	$\displaystyle\leq\sum_{i=0}^{k}\left\\|x_{i+1}-x_{i}\right\\|\leq\sum_{i=0}^{k}% \left\\|A_{i}\right\\|\cdot\left\\|P\left(x_{i}\right)\right\\|\leq 2B\tfrac{% \alpha}{2MB^{2}}\sum_{i=0}^{k}d^{2^{i}}$
		$\displaystyle=\tfrac{\alpha}{MB}\left(d+d^{2}+d^{2^{2}}+\ldots+d^{2^{k}}\right% )<\tfrac{\alpha d}{MB\left(1-d\right)},$

from which it results $x_{n+1}\in S.$

For b) proceeding as in [ 3 ] we deduce the relations:

\left\|P\left(x_{k+1}\right)\right\|\leq\tfrac{M}{2}\left\|A_{k}\right\|^{2}% \left\|P\left(x_{k}\right)\right\|^{2}+\left\|P\left(x_{k}\right)\right\|\left% \|E-P^{\prime}\left(x_{k}\right)A_{k}\right\|

And

\left\|E-P^{\prime}\left(x_{k+1}\right)A_{k+1}\right\|\leq\left(\left\|E-P^{% \prime}\left(x_{k}\right)A_{k}\right\|+M\left\|A_{k}\right\|^{2}\left\|P\left(% x_{k}\right)\right\|\right)^{3},

from which with the introduced notations we have:

\left\{\begin{array}[c]{c}\rho_{k+1}\leq\rho_{k}^{2}+\rho_{k}\delta_{k}\\ \delta_{k+1}\leq\left(\rho_{k}+2\rho_{k}\right)^{3}\end{array}\right.

In the hypotheses of the theorem it results that for $a=2$ the hypotheses of lemma 1 are verified, in fact we have:

\rho_{k+1}\leq\alpha d^{2^{k+1}},\ \ \delta_{k+1}\leq\beta d^{2^{k+1}}

For c) we deduce in a manner analogous to that of the case

\left\|A_{k+1}\right\|\leq 2B.

Properties a), b), c) being true for $n=k+1$ also it results their validity for each $n=0,1,\ldots$

We now have:

	$\displaystyle\left\\|x_{n+m}+x_{n}\right\\|$	$\displaystyle\leq\sum_{i=n}^{n+m-1}\left\\|x_{i+1}-x_{i}\right\\|$
		$\displaystyle\leq\tfrac{\alpha d^{2^{n}}}{MB}\big{(}1+d^{2^{n+1}-2^{n}}+d^{2^{% n+2}-2^{n}}+\ldots+d^{2^{n+m-1}-2^{n}}\big{)}$
		$\displaystyle<\tfrac{\alpha d^{2^{n}}}{MB\left(1-d^{2^{n}}\right)},$

from which it follows that the following $\left(x_{n}\right)_{n=0}^{\infty}$ being fundamental, it will be convergent towards $x^{\ast}.$ From b) it follows $\lim\limits_{n\rightarrow\infty}\left\|P\left(x_{n}\right)\right\|=0,$ from where $P\left(x^{\ast}\right)=0.$ In the above inequality we easily see that $x^{\ast}\in S$ and that we have the following inequality

\left\|x^{\ast}-x_{n}\right\|\leq\tfrac{\alpha d^{2^{n}}}{MB\left(1-d^{2^{n}}% \right)},\ \;\;n=0,1,\ldots

if $A^{\ast}=\left[P^{\prime}\left(x^{\ast}\right)\right]^{-1}$ we have:

	$\displaystyle\left\\|A^{\ast}-A_{n}\right\\|$	$\displaystyle=\big{\\|}\left[P^{\prime}\left(x^{\ast}\right)\right]^{-1}-A_{n}% \big{\\|}$
		$\displaystyle\leq\big{\\|}\left[P^{\prime}\left(x^{\ast}\right)\right]^{-1}\big% {\\|}\left\\|E-P^{\prime}\left(x^{\ast}\right)A_{n}\right\\|$
		$\displaystyle\leq 2B\left(\left\\|E-P^{\prime}\left(x_{n}\right)A_{n}\right\\|+M% \left\\|x^{\ast}-x_{n}\right\\|\left\\|A_{n}\right\\|\right)$
		$\displaystyle\leq\tfrac{2Bd^{2^{n}}\left(2\alpha+\beta-\beta d^{2n}\right)}{1-% d^{2^{n}}},$

from which it follows that $\left(A_{n}\right)_{n=0}^{\infty}$ tends towards $A^{\ast}et$ that the given delimitation takes place. The theorem is proven. ∎

Theorem 5 .

If in the sphere $S=\left\{x\in X:\left\|x-x_{0}\right\|\leq r\right\}$ the following conditions are met:

1)

The operator $P$ admits the divided differences up to the order $2$ inclusively, the divided difference of the first order admits a bounded inverse, that is to say $\big{\|}\left[x,y;P\right]^{-1}\big{\|}\leq B<+\infty$ for each $x,y\in S$ And $\left\|\left[x,y,z;P\right]\right\|\leq M<+\infty$ for each $x,y,z\in S.$
2)

The iterative operator $Q$ meets the following conditions:

$\displaystyle\left\|Q\left(x\right)-Q\left(y\right)\right\|$ $\displaystyle\leq L\left\|x-y\right\|,\;\;\ \text{pour\ chaque\ }x,y\in S,$

$\displaystyle\left\|x-Q\left(x\right)\right\|$ $\displaystyle\leq C\left\|P\left(x\right)\right\|,\;\;\ \text{pour chaque }x% \in S.$
3)

The initial element $x_{0}$ can be chosen in such a way that the following conditions are met:

$\displaystyle 2MB\left(2B+C\right)\left\|P\left(x_{0}\right)\right\|$ $\displaystyle\leq\alpha d,$

$\displaystyle\left\|E-\left[x_{0},Q\left(x_{0}\right);P\right]A_{0}\right\|$ $\displaystyle\leq\beta d,$

Or

$\beta<1,\ \alpha<1-\beta,\ d<\tfrac{\beta}{\left[a+\left(1-a\right)\beta\right% ]^{3}},$

with

$a=\tfrac{1+L}{2B+C}>1\ \text{et\ }r=\tfrac{\alpha d\left[2B+C\left(1-d\right)% \right]}{2MB\left(2B+C\right)\left(1-d\right)},$

SO:
1)

The sequels $\left(x_{n}\right)_{n=0}^{\infty}$ And $\left(A_{n}\right)_{n=0}^{\infty}$ data by the iterative method ( 18 ) are convergent.
2)

Equation ( 1 ) has a solution $x^{\ast}\in S$ which can be obtained as the limit of the sequence $\left(x_{n}\right)_{n=0}^{\infty}$ And $A^{\ast}=\lim\limits_{n\rightarrow\infty}A_{n}=\left[x^{\ast},Q\left(x^{\ast}% \right);P\right]^{-1}.$
3)

We have:

$\left\|x^{\ast}-x_{n}\right\|\leq\tfrac{\alpha d^{2^{n}}}{M\left(2B+C\right)% \left(1-d^{2^{n}}\right)},\;\;\ n=0,1,\ldots$

And

$\left\|A^{\ast}-A_{n}\right\|\leq Bd^{2^{n}}\left[\beta+\tfrac{2\alpha B\left(% 1+L\right)}{\left(2B+C\right)\left(1-d^{2^{n}}\right)}\right],\;\;\ n=0,1,\ldots$

The proof of this theorem is absolutely analogous to the proof of Theorem 4 , taking into account Lemma 1 and the relations established during the proof of Theorem 3 , [ 3 ] .

Theorem 6 .

If in the sphere $S=\left\{\alpha\in X\left|\ \left\|x-x_{0}\right\|\leq r\right.\right\}$ the following conditions are met:

1)

The operator $P$ admits divided differences up to and including order 2, the divided difference of the first order admits a bounded inverse, that is to say $\left\|\left[x,y;P\right]^{-1}\right\|\leq B<+\infty$ for each $x,y\in S$ And $\left\|\left[x,y,z;P\right]\right\|\leq M<+\infty$ For $x,y,z\in S$
2)

The operator $A_{0}\$ is limited and $\left\|A_{0}\right\|\leq B.$
3)

We have the inequalities:

$\displaystyle 4MB^{2}\left\|P\left(x_{0}\right)\right\|$ $\displaystyle<\alpha d,$

$\displaystyle 4MB^{2}\left\|P\left(x_{1}\right)\right\|$ $\displaystyle\leq\alpha d^{p},$

$\left\|E-\left[x_{0},x_{1};P\right]A_{0}\right\|\leq\sqrt[3]{\beta}d^{\frac{p}% {3}},$

Or

$p=\tfrac{1+\sqrt{5}}{2},\ d<1,\ \left(\beta+2\alpha\right)^{3}<\beta<1,$

$r=\tfrac{\alpha d}{2MB\left(1-d^{\frac{\sqrt{5-1}}{2}}\right)}$

SO:
1)

The sequels $\left(x_{n}\right)_{n=0}^{\infty}$ And $\left(A_{n}\right)_{n=0}^{\infty}$ given by ( 15 ) are convergent
2)

Equation ( 1 ) has the solution $x^{\ast}\in S,$ which can be obtained as the limit of the sequence $\left(x_{n}\right)_{n=0}^{\infty}$ given by ( 15 ).
3)

If $A^{\ast}$ is the limit of the sequence of operators $\left(A_{n}\right)_{n=0}^{\infty}$ it is at the same time the limit of the sequence of operators $(\left[x_{n},x_{n+1};P\right]^{-1})_{n=0}^{\infty},$ $(\left[x_{n},x_{n+1};P\right]^{-1}$ exists for each $n=0,1,\ldots$ because $x_{n}\in S$ what results from the demonstration);
4)

We have:

$\left\|x^{\ast}-x_{n}\right\|\leq\frac{\alpha d^{p^{n}}}{2MB\left(1-d^{p^{n% \frac{\sqrt{5-1}}{2}}}\right)},$

$\left\|A^{\ast}-A\right\|\leq d^{p^{n-1}}\left[2B\left(\alpha+\beta\right)% \tfrac{d^{p^{n-1\frac{\sqrt{5-1}}{2}}}}{1-d^{p^{n-1\frac{\sqrt{5-1}}{2}}}}+2% \alpha B\tfrac{1}{1-d^{p^{n-1\frac{\sqrt{5-1}}{2}}}}\right].$

Demonstration..

Using the method of mathematical induction we will demonstrate the following properties.

has)

$x_{n}\in S,\qquad n=0,1,\ldots,$
b)

$\rho_{n}=4MB^{2}\left|\left|P\left(x_{n}\right)\right|\right|\leq\alpha d^{p^{% n}},\qquad n=0,1,\ldots,$
c)

$\delta_{n}=\left|\left|E-\left[x_{n-1},x_{n};P\right]A_{n}\right|\right|\leq% \beta d^{p^{n}},\qquad n=1,2,\ldots,$
d)

$\left|\left|A_{n}\right|\right|\leq 2B,\qquad n=0,1,\ldots,$

In the hypotheses of the theorem it follows that properties a), b), d) are verified for $n=0$ and c) for $n=1.$

We assume that properties a)–d) hold for $n=k$ and we will show their validity for $n=k+1.$

For property a) we have:

	$\displaystyle\left\\|x_{k+1}-x_{0}\right\\|$	$\displaystyle\leq\sum_{i=0}^{k}\left\\|x_{i+1}-x_{i}\right\\|\leq\sum_{i=0}^{k}% \left\\|A_{i}\right\\|\cdot\left\\|P\left(x_{i}\right)\right\\|$
		$\displaystyle<2B\sum_{i=0}^{k}\left\\|P\left(x_{i}\right)\right\\|\leq\tfrac{% \alpha d}{2MB}\left(1+d^{p-1}+d^{p^{2}-1}+\ldots\right)$
		$\displaystyle<\tfrac{\alpha d}{2MB\left(1-d^{p-1}\right)}=\tfrac{\alpha d}{2MB% \left(1-d^{\frac{\sqrt{5-1}}{2}}\right)}=r,$

SO $x_{k+1}\in S.$

For properties b), c) we establish in the same way as in [ 3 ] the inequalities:

	$\displaystyle\left\\|P\left(x_{k+1}\right)\right\\|\leq$	$\displaystyle M\left\\|A_{k}\right\\|\cdot\left\\|P\left(x_{k}\right)\right\\|% \left(\left\\|A_{k-1}\right\\|\cdot\left\\|P\left(x_{k}\right)\right\\|+\left\\|A_{% k-1}\right\\|\cdot\left\\|P\left(x_{k-1}\right)\right\\|\right)$
		$\displaystyle+\left\\|P\left(x_{k}\right)\right\\|\cdot\left\\|E-\left[x_{k-1},x_% {k};P\right]A_{k}\right\\|,$

	$\displaystyle\left\\|E-\left[x_{k},x_{k+1};P\right]A_{k+1}\right\\|\leq$
	$\displaystyle\leq\!\Big{\{}\!\!\left\\|E\!-\!\left[x_{k-1},x_{k};P\right]A_{k}% \right\\|\!+\!M\left\\|A_{k}^{-1}\right\\|\!\Big{(}\!\left\\|A_{k}\right\\|\left\\|P% \left(x_{k}\right)\right\\|\!+\!\left\\|A_{k-1}\right\\|\left\\|P\left(x_{k-1}% \right)\right\\|\!\Big{)}\!\!\Big{\}}^{3},$

hence seen that $\left\|A_{k}\right\|\leq 2B,\left\|A_{k-1}\right\|\leq 2B$ with the introduced notations, it results:

\left\{\begin{array}[c]{l}\rho_{k+1}\leq\rho_{k}^{2}+\rho_{k}\rho_{k-1}+\rho_{% k}\delta_{k},\\ \delta_{k+1}\leq\left(\delta_{k}+\rho_{k}+\rho_{k-1}\right)^{3}.\end{array}\right.

Since the assumptions of Lemma 2) are verified

\rho_{k+1}\leq\alpha d^{p^{k+1}}\ \ \text{et\ \ \ }\delta_{k+1}\leq\beta d^{p^% {k+1}}.

For property d) we have:

	$\displaystyle\big{\\|}A_{k+1}\big{\\|}$	$\displaystyle\leq\big{\\|}\left[x_{k},x_{k+1};P\right]^{-1}\big{\\|}\left(1+% \left\\|E-\left[x_{k},x_{k+1};P\right]A_{k+1}\right\\|\right)$
		$\displaystyle\leq B\left(1+\delta_{k+1}\right)<2B.$

In accordance with the principle of mathematical induction it follows that properties a), b), d) are true for each $n=0,1,\ldots,$ and property c) for each $n=1,2,\ldots$

In the same way as in the previous theorem it results:

\left\|x_{m+n}-x_{n}\right\|\leq\tfrac{\alpha d^{p^{n}}}{2MB\left(1-d^{p^{n% \frac{\sqrt{5-1}}{2}}}\right)},

from which it follows that the following $\left(x_{n}\right)_{n=0}^{\infty}$ being fundamental it will converge towards $x^{\ast}$ given by the inequality:

\left\|x^{\ast}-x_{n}\right\|\leq\tfrac{\alpha d^{p^{n}}}{2MB\left(1-d^{p^{n% \frac{\sqrt{5-1}}{2}}}\right)},

hence, by doing $n=0$ it results $x^{\ast}\in S.$ Because $\lim\limits_{n\rightarrow\infty}\rho_{n=0,}il$ results that $P\left(x^{\ast}\right)=0.$

For the demonstration of the convergence of the sequence $\left(A_{n}\right)_{n=0}^{\infty}$ and from the last inequality we evaluate:

	$\displaystyle\left\\|A_{i+1}-A_{i}\right\\|\leq$
	$\displaystyle\leq\left\\|A_{i}\right\\|\left\\|E-\left[x_{i},x_{i+1};P\right]A_{i% }\right\\|$
	$\displaystyle\leq 2B\{\left\\|E-\left[x_{i-1},x_{i};P\right]A_{i}\right\\|+\left% \\|\left[x_{i-1,}x_{i},x_{i+1};P\right]\right\\|\cdot\left\\|x_{i+1}-x_{i}\right% \\|\cdot\left\\|A_{i}\right\\|\}$
	$\displaystyle\leq 2B\left(\alpha+\beta\right)d^{p^{i}}+2\alpha Bd^{p^{i-1}}$

This results in

	$\displaystyle\left\\|A_{n+m}-A_{n}\right\\|$	$\displaystyle\leq\sum_{i=n}^{n+m-1}\left\\|A_{i+1}-A_{i}\right\\|$
		$\displaystyle<d^{p^{n-1}}\left[2B\left(\alpha+\beta\right)\tfrac{d^{p^{n-1% \frac{\sqrt{5-1}}{2}}}}{1-d^{p^{n\frac{\sqrt{5-1}}{2}}}}+2\alpha B\tfrac{1}{1-% d^{p^{n-1}\frac{\sqrt{5-1}}{2}}}\right],$

from which it follows that the following $\left(A_{n}\right)_{n=0}^{\infty}$ being fundamental it will be convergent. By designating by $A^{\ast}$ its limit, it results in the delimitation expressed by the inequality of 4).

We will now establish that the sequence of operators $\big{(}\left[x_{n},x_{n+1};P\right]^{-1}\big{)}_{n=0}^{\infty}$ also tends towards $A^{\ast}.$ Indeed:

	$\displaystyle\big{\\|}A^{\ast}-\left[x_{n},x_{n+1};P\right]^{-1}\big{\\|}\leq$
	$\displaystyle\leq\big{\\|}A^{\ast}-A_{n}\big{\\|}+\big{\\|}A_{n}-\left[x_{n},x_{n% +1};P\right]^{-1}\big{\\|}$
	$\displaystyle=\big{\\|}A^{\ast}-A_{n}\big{\\|}+\big{\\|}\left[x_{n},x_{n+1};P% \right]^{-1}\big{\\|}\cdot\left\\|E-\left[x_{n},x_{n+1};P\right]A_{n}\right\\|$
	$\displaystyle\leq\left\\|A^{\ast}-A_{n}\right\\|+B\left(\alpha+\beta\right)d^{p^% {n}}+\alpha B^{p^{n-1}},$

from which it follows that: $\lim\limits_{n\rightarrow\infty}\left[x_{n},x_{n+1};P\right]^{-1}=A^{\ast}.$

The theorem is proven. ∎

Bibliography

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[5] Jankó, B., Rezolvarea ecuaţiilor operaţionale neliniare în spaţii Banach . Bucuresti, Editura Academiei RSR (1969).
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Received on 18. IX.1971.

1972

On the chromatic number K(p,G) of graphs

August 19, 2025

Differential and integral equations

April 25, 2025

Course of Mathematical Analysis. Part II, Functions. Limits

26 Feb at 8:24 am

	$\displaystyle\big{\\|}A_{0}\big{\\|}$	$\displaystyle\leq\big{\\|}\left[P^{\prime}\left(x_{0}\right)\right]^{-1}+A_{0}-% \left[P^{\prime}\left(x_{0}\right)\right]^{-1}\big{\\|}$
		$\displaystyle\leq\big{\\|}\left[P^{\prime}\left(x_{0}\right)\right]^{-1}\big{\\|% }\left(1+\left\\|E-P^{\prime}\left(x_{0}\right)A_{0}\right\\|\right)$
		$\displaystyle\leq B\left(1+\alpha d\right)<2B.$

	$\displaystyle\left\\|A^{\ast}-A_{n}\right\\|$	$\displaystyle=\big{\\|}\left[P^{\prime}\left(x^{\ast}\right)\right]^{-1}-A_{n}% \big{\\|}$
		$\displaystyle\leq\big{\\|}\left[P^{\prime}\left(x^{\ast}\right)\right]^{-1}\big% {\\|}\left\\|E-P^{\prime}\left(x^{\ast}\right)A_{n}\right\\|$
		$\displaystyle\leq 2B\left(\left\\|E-P^{\prime}\left(x_{n}\right)A_{n}\right\\|+M% \left\\|x^{\ast}-x_{n}\right\\|\left\\|A_{n}\right\\|\right)$
		$\displaystyle\leq\tfrac{2Bd^{2^{n}}\left(2\alpha+\beta-\beta d^{2n}\right)}{1-% d^{2^{n}}},$

	$\displaystyle\left\\|x_{k+1}-x_{0}\right\\|$	$\displaystyle\leq\sum_{i=0}^{k}\left\\|x_{i+1}-x_{i}\right\\|\leq\sum_{i=0}^{k}% \left\\|A_{i}\right\\|\cdot\left\\|P\left(x_{i}\right)\right\\|$
		$\displaystyle<2B\sum_{i=0}^{k}\left\\|P\left(x_{i}\right)\right\\|\leq\tfrac{% \alpha d}{2MB}\left(1+d^{p-1}+d^{p^{2}-1}+\ldots\right)$
		$\displaystyle<\tfrac{\alpha d}{2MB\left(1-d^{p-1}\right)}=\tfrac{\alpha d}{2MB% \left(1-d^{\frac{\sqrt{5-1}}{2}}\right)}=r,$

	$\displaystyle\left\\|A_{i+1}-A_{i}\right\\|\leq$
	$\displaystyle\leq\left\\|A_{i}\right\\|\left\\|E-\left[x_{i},x_{i+1};P\right]A_{i% }\right\\|$
	$\displaystyle\leq 2B\{\left\\|E-\left[x_{i-1},x_{i};P\right]A_{i}\right\\|+\left% \\|\left[x_{i-1,}x_{i},x_{i+1};P\right]\right\\|\cdot\left\\|x_{i+1}-x_{i}\right% \\|\cdot\left\\|A_{i}\right\\|\}$
	$\displaystyle\leq 2B\left(\alpha+\beta\right)d^{p^{i}}+2\alpha Bd^{p^{i-1}}$

	$\displaystyle\big{\\|}A^{\ast}-\left[x_{n},x_{n+1};P\right]^{-1}\big{\\|}\leq$
	$\displaystyle\leq\big{\\|}A^{\ast}-A_{n}\big{\\|}+\big{\\|}A_{n}-\left[x_{n},x_{n% +1};P\right]^{-1}\big{\\|}$
	$\displaystyle=\big{\\|}A^{\ast}-A_{n}\big{\\|}+\big{\\|}\left[x_{n},x_{n+1};P% \right]^{-1}\big{\\|}\cdot\left\\|E-\left[x_{n},x_{n+1};P\right]A_{n}\right\\|$
	$\displaystyle\leq\left\\|A^{\ast}-A_{n}\right\\|+B\left(\alpha+\beta\right)d^{p^% {n}}+\alpha B^{p^{n-1}},$

	$\displaystyle\left\\|Q\left(x\right)-Q\left(y\right)\right\\|$	$\displaystyle\leq L\left\\|x-y\right\\|,\ \ \text{pour chaque\ \ \ }x,y\in S,\ % \ L\in\mathbb{R}_{+},$
	$\displaystyle\left\\|x-Q\left(x\right)\right\\|$	$\displaystyle\leq\alpha\left\\|P\left(x\right)\right\\|,\ \ \text{pour chaque \ % \ }x\in S,\ \ \alpha\in\mathbb{R}_{+},$

	$\displaystyle 2MB\left(2B+C\right)\left\\|P\left(x_{0}\right)\right\\|$	$\displaystyle\leq\alpha d,$
	$\displaystyle\left\\|E-\left[x_{0},Q\left(x_{0}\right);P\right]A_{0}\right\\|$	$\displaystyle\leq\beta d,$

	$\displaystyle 4MB^{2}\left\\|P\left(x_{0}\right)\right\\|$	$\displaystyle<\alpha d,$
	$\displaystyle 4MB^{2}\left\\|P\left(x_{1}\right)\right\\|$	$\displaystyle\leq\alpha d^{p},$

On some iterative methods for solving operator equations

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On some iterative methods for solving operator equations

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