Perturbed-Steffensen-Aitken projection methods for solving equations with nondifferentiable operators

Abstract

We use perturbed Steffensen-Aitken methods to approximate a locally unique solution of an operator equation in a Banach space. Using projection operators, we reduce the problem to solving a finite linear system of algebraic equations. Since iterates can rarely be computed exactly, we control the residuals to guarantee the convergence of the method. Sufficient convergence conditions as well as an error analysis are given for our method.

Authors

Ioannis K. Argyros
(Cameron University)

Emil Cătinaş
(Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy)

Ion Păvăloiu
(Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy)

Keywords

nonlinear equations in Banach spaces; Steffensen-Aitken methods; projection operator; residuals.

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I. Argyros, E. Cătinaş, I. Păvăloiu, Perturbed-Steffensen-Aitken projection methods for solving equations with nondifferentiable operators, Punjab Univ. J. Math. (Lahore), 33 (2000), pp. 105-113.

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PERTURBED-STEFFENSEN-AITKEN PROJECTION METHODS FOR SOLVING EQUATIONS WITH NONDIFFERENTIABLE OPERATORS

Ioannis K. Argyros
Cameron University
Department of Mathematics
Lawton, OK 73505 U.S.A.
Emil Cătinas and Ion Păaăloiu
Institul de Calcul
Str. Republicii Nr. 37
P.O. Box 68,3400 Cluj-Napoca, Romania

(Received 31 July, 1999)

ABSTRACT In this study we use perturbed-Steffensen- Aitken methods to approximate a locally unique solution of an operator equation in a Banach space. Using projection operators we reduce the problem to solving a linear system of algebraic equations of finite order. Since iterates can rarely be computed exactly we control the residuals to guarantee convergence of the method. Sufficient convergence conditions as well as an error analysis are given for our method.

AMS (MOS) Subject Classification: 65J15, 47H17, 49D15.
Key Words and Phrases Steffensen-Aitken methods, Banach space, projection operator, residuals.
I. INTRODUCTION In this study we are concerned with the problem of
approximating a locally unique fixed point xx^{*} of the nonlinear equation.

T(x)=xT(x)=x (1)

where TT is a continuous operator defined on a convex subset DD of a Banach space EE with values in EE. The differentiability of TT is not assumed. Let T1T_{1} be another nonlinear continuous operator from EE into EE, and let PP be a projection operator (P2=P)\left(P^{2}=P\right) on EE.

We introduce the perturbed-Steffensen-Aitken method

xn+1=T(xn)+PAn(xn+1xn)zn.An=[g1(xn),g2(xn)](n0),x_{n+1}=T\left(x_{n}\right)+PA_{n}\left(x_{n+1}-x_{n}\right)-z_{n}.A_{n}=\left[g_{1}\left(x_{n}\right),g_{2}\left(x_{n}\right)\right]\quad(n\geq 0), (2)

where: [x,y][x,y] denotes a divided difference of order one of T1T_{1} at the points x,yx,y satisfying

[x,y](yx)=T1(y)T1(x) for all x,yD with xy[x,y](y-x)=T_{1}(y)-T_{1}(x)\quad\text{ for all }\quad x,y\in D\quad\text{ with }\quad x\neq y (3)

and

[x,x]=F(x)(xD)[x,x]=F^{\prime}(x)\quad(x\in D) (4)

if T1T_{1} is Frechet-differentiable D;g1,g2:DED;g_{1},g_{2}:D\rightarrow E are continuous operators; the residual points {xn}(n0)\left\{x_{n}\right\}(n\geq 0) are chosen in such a way that iteration {xn}(n0)\left\{x_{n}\right\}(n\geq 0) generated by (2) converges to xx^{*}. The important of studying perturbed SteffensenAitken methods comes from the fact that many commonly used variants can be considered procedures of this type. Indeed the above approximation characterizes any iterative process in which corrections are taken as approximate solutions of the Steffensen-Aitken equations. Moreover we note that if for example an equation on the real line is solved xnT(xn)0(n0)x_{n}-T\left(x_{n}\right)\geq 0(n\geq 0) and IPAnI-PA_{n} overestimates the derivative, xn(IPAn)1(xnT(xn))x_{n}-\left(I-PA_{n}\right)^{-1}\left(x_{n}-T\left(x_{n}\right)\right) is always larger than the corresponding Steffensen-Aitken iterate. In such cases, a positive zn(n0)z_{n}(n\geq 0) correction term is appropriate.

For: P=I(IP=I(I is the identity operator on E),T(x)=T1(x)(xD),g1(x)=g2(x)(xD)E),T(x)=T_{1}(x)(x\in D),g_{1}(x)=g_{2}(x)(x\in D), and zn=0(n0)z_{n}=0(n\geq 0) we obtain the ordinary Newton method [1], [2]; P=I,T1(x)=T(x)(xD),g1(x)=x(xD)P=I,T_{1}(x)=T(x)(x\in D),g_{1}(x)=x(x\in D), and zn=0(n0)z_{n}=0(n\geq 0) we obtain Steffensen method [4], [5]; P=I,T1(x)=T(x)(xD),g2(x)=g1(xT(x))(xD)P=I,T_{1}(x)=T(x)(x\in D),g_{2}(x)=g_{1}(x-T(x))(x\in D), and zn=0(n0)z_{n}=0(n\geq 0) we obtain Steffensen-Aitken method [4], [5].

It is easy to see that the solution of (2) reduces to solving certain operator equations in the space EpE_{p}. If moreover EpE_{p} is a finite dimensional space of dimension NN, we obtain a system of linear algebraic equations of at most order NN.

We provide sufficient convergence conditions as well as an error analysis for the Steffensen-Aitken method generated by (2).
II. CONVERGENCE ANALYSIS We state the following semilocal convergence theorem.

Theorem Let T,T1,g1,g2:DET,T_{1},g_{1},g_{2}:D\rightarrow E be continuous operators defined on a convex subset DD of a Banach space EE with values in EE, and PP be a projection operator on E. Moreover, assume:
(a) there exists x0Dx_{0}\in D such that B0=IPA0B_{0}=I-PA_{0} is invertible;
(b) there exist nonnegative numbers ai,R,i=0,1,2,,9a_{i},R,\quad i=0,1,2,\cdots,9 such that

B01P([x,y][v,w])a0(xv+yw),\displaystyle\left\|B_{0}^{-1}P([x,y]-[v,w])\right\|\leq a_{0}(\|x-v\|+\|y-w\|), (5)
B01(x0T(x0))a1,\displaystyle\left\|B_{0}^{-1}\left(x_{0}-T\left(x_{0}\right)\right)\right\|\leq a_{1}, (6)
B01P([x,y][g1(x),g2(x)])a2(xg1(x)+yg2(x)),\displaystyle\left\|B_{0}^{-1}P\left([x,y]-\left[g_{1}(x),g_{2}(x)\right]\right)\right\|\leq a_{2}\left(\left\|x-g_{1}(x)\right\|+\left\|y-g_{2}(x)\right\|\right), (7)
B01(QT1(x)QT1(y))a3xy,Q=1P,\displaystyle\left\|B_{0}^{-1}\left(QT_{1}(x)-QT_{1}(y)\right)\right\|\leq a_{3}\|x-y\|,\quad Q=1-P, (8)
B01(F(x)F(y))a4xy,F(x)=T(x)T1(x),\displaystyle\left\|B_{0}^{-1}(F(x)-F(y))\right\|\leq a_{4}\|x-y\|,\quad F(x)=T(x)-T_{1}(x), (9)
xg1(x)a5B1(x)(xT(x)z(x)),B(x)=IPA(x),\displaystyle\left\|x-g_{1}(x)\right\|\leq a_{5}\left\|B^{-1}(x)(x-T(x)-z(x))\right\|,\quad B(x)=I-PA(x),
 for some continuous function z:DE,\displaystyle\text{ for some continuous function }z:D\rightarrow E, (10)
xg2(x)a6B1(x)(xT(x)z(x)),\displaystyle\left\|x-g_{2}(x)\right\|\leq a_{6}\left\|B^{-1}(x)(x-T(x)-z(x))\right\|, (11)
B01(znzn1)a7xnxn1(n1),\displaystyle\left\|B_{0}^{-1}\left(z_{n}-z_{n-1}\right)\right\|\leq a_{7}\left\|x_{n}-x_{n-1}\right\|\quad(n\geq 1), (12)
g1(x)g1(y)a8xy,a8[0,1),\displaystyle\left\|g_{1}(x)-g_{1}(y)\right\|\leq a_{8}\|x-y\|,\quad a_{8}\in[0,1), (13)

and

g2(x)g2(y)a9xy,a9[0,1)\left\|g_{2}(x)-g_{2}(y)\right\|\leq a_{9}\|x-y\|,\quad a_{9}\in[0,1) (14)

for all x,y,v,wU(x0,R)={xExx0R}D;x,y,v,w\in U\left(x_{0},R\right)=\left\{x\in E\left\|x-x_{0}\right\|\leq R\right\}\subseteq D;
(c) the sequence {zn}(n0)\left\{z_{n}\right\}(n\geq 0) is null;
(d) there exists a minumum nonnegative number rr^{*} satisfying

G(r)r and rRG\left(r^{*}\right)\leq r^{*}\quad\text{ and }\quad r^{*}\leq R (15)

where

G(r)=a1+a2(1+a8+a9)r+(a3+a4+a7)[aa0(a8+a9)r][1a2(a5+a6)rβ(r)]rG(r)=a_{1}+\frac{a_{2}\left(1+a_{8}+a_{9}\right)r+\left(a_{3}+a_{4}+a_{7}\right)}{\left[a-a_{0}\left(a_{8}+a_{9}\right)r\right]\left[1-a_{2}\left(a_{5}+a_{6}\right)r\beta(r)\right]}r (16)

and

β(r)=[1a0(a8+a9)r]1\beta(r)=\left[1-a_{0}\left(a_{8}+a_{9}\right)r\right]^{-1} (17)

(e) the numbers r,Rr^{*},R also satisfy

r<1a2(a5+a6)+a0(a8+a9)\displaystyle r^{*}<\frac{1}{a_{2}\left(a_{5}+a_{6}\right)+a_{0}\left(a_{8}+a_{9}\right)} (18)
rg1(x0)x01a8\displaystyle r^{*}\geq\frac{\left\|g_{1}\left(x_{0}\right)-x_{0}\right\|}{1-a_{8}} (19)
rg2(x0)x01a9\displaystyle r^{*}\geq\frac{\left\|g_{2}\left(x_{0}\right)-x_{0}\right\|}{1-a_{9}} (19)
b=α(r,R)<1.\displaystyle b=\alpha(r,R)<1. (21)

where

α(s,t)=a2(1+a8+a9)(s+t)+a3+a4[1a0(a8+a9)s][1a2(a5a6)(s+t)β(s)],s,t[0,R]\alpha(s,t)=\frac{a_{2}\left(1+a_{8}+a_{9}\right)(s+t)+a_{3}+a_{4}}{\left[1-a_{0}\left(a_{8}+a_{9}\right)s\right]\left[1-a_{2}\left(a_{5}-a_{6}\right)(s+t)\beta(s)\right]},\quad s,t\in[0,R] (22)

and

limnqn=0\lim_{n\rightarrow\infty}q_{n}=0 (23)

where

qn=m=0nbnmcm,cm=zn,Bn=IPAn(n0)q_{n}=\sum_{m=0}^{n}b^{n-m}c_{m},\quad c_{m}=\left\|z_{n}\right\|,\quad B_{n}=I-PA_{n}\quad(n\geq 0) (24)

Then
(i) the scalar sequence {tn}(n0)\left\{t_{n}\right\}\quad(n\geq 0) generated by

t0=0,t1=a1x1x0\displaystyle t_{0}=0,\quad t_{1}=a_{1}\geq\left\|x_{1}-x_{0}\right\| (25)
tn+1=tn+a2(1+a8+a9)(tntn1)+a3+a4+a7[1a0(a8+a9)tn][1a2(a5+a6)(tntn1βn](tntn1)(n1)\displaystyle t_{n+1}=t_{n}+\frac{a_{2}\left(1+a_{8}+a_{9}\right)\left(t_{n}-t_{n-1}\right)+a_{3}+a_{4}+a_{7}}{\left[1-a_{0}\left(a_{8}+a_{9}\right)t_{n}\right]\left[1-a_{2}\left(a_{5}+a_{6}\right)\left(t_{n}-t_{n-1}\beta_{n}\right]\right.}\left(t_{n}-t_{n-1}\right)\quad(n\geq 1) (26)

is monotonically increasing, bounded above by rr^{*} and limntn=r\lim_{n\rightarrow\infty}t_{n}=r^{*}, with βn=[1a0(a8+a9)tn]1(n0)\beta_{n}=\left[1-a_{0}\left(a_{8}+a_{9}\right)t_{n}\right]^{-1}\quad(n\geq 0).
(ii) The perturbed-Steffensen-Aitken method generated by (2) is well defined, remains in U(x0,r)U\left(x_{0},r^{*}\right) for all n0n\geq 0, converges to a unique fixed point xx^{*} of TT in U(x0,R)U\left(x_{0},R\right).

Moreover the following error bounds hold:

xn+1xna2(1+a8+a9)xnxn1+a3+a4+a7[1a0(a8+a9)xnx0][1a2(a5+a6)xnxn1βn]\displaystyle\left\|x_{n+1}-x_{n}\right\|\leq\frac{a_{2}\left(1+a_{8}+a_{9}\right)\left\|x_{n}-x_{n-1}\right\|+a_{3}+a_{4}+a_{7}}{\left[1-a_{0}\left(a_{8}+a_{9}\right)\left\|x_{n}-x_{0}\right\|\right]\left[1-a_{2}\left(a_{5}+a_{6}\right)\left\|x_{n}-x_{n-1}\right\|\beta_{n}\right]}
xnxn1(n1)\displaystyle\left\|x_{n}-x_{n-1}\right\|(n\geq 1) (27)
xn+1xntn+1tn(n0)\displaystyle\left\|x_{n+1}-x_{n}\right\|\leq t_{n+1}-t_{n}\quad(n\geq 0) (28)

and

xnxrtn(n0)\left\|x_{n}-x^{*}\right\|\leq r^{*}-t_{n}\quad(n\geq 0) (29)

where β¯n=[1a0(a8a9)xnx0]1(n0)\bar{\beta}_{n}=\left[1-a_{0}\left(a_{8}-a_{9}\right)\left\|x_{n}-x_{0}\right\|\right]^{-1}\quad(n\geq 0)

Proof (i). By (15) and (25) we get 0t0t1r0\leq t_{0}\leq t_{1}\leq r^{*}. Let us assume 0tk1tkr0\leq t_{k-1}\leq t_{k}\leq r^{*} for k=1,2,,nk=1,2,\cdots,n. It follows from (18) and (26) that 0tktk+10\leq t_{k}\leq t_{k+1}. Hence, the sequence {tn}(n0)\left\{t_{n}\right\}(n\geq 0) is monotonically increasing. Moreover using (26) we get in turn

tk+1\displaystyle t_{k+1} tk+a2(1+a8+a9)r+a3+a4+a7[1a0(a8+a9)r][1a2(a5+a6)rβ(r)](tktk1)\displaystyle\leq t_{k}+\frac{a_{2}\left(1+a_{8}+a_{9}\right)r^{*}+a_{3}+a_{4}+a_{7}}{\left[1-a_{0}\left(a_{8}+a_{9}\right)r^{*}\right]\left[1-a_{2}\left(a_{5}+a_{6}\right)r^{*}\beta\left(r^{*}\right)\right]}\left(t_{k}-t_{k-1}\right)
a1+a2(1+a8+a9)ra3+a4+a7[1a0(a8+a9)r][aa2(a5+a6)rβ(r)](tkt0)\displaystyle\leq\cdots\leq a_{1}+\frac{a_{2}\left(1+a_{8}+a_{9}\right)r^{*}a_{3}+a_{4}+a_{7}}{\left[1-a_{0}\left(a_{8}+a_{9}\right)r^{*}\right]\left[a-a_{2}\left(a_{5}+a_{6}\right)r^{*}\beta\left(r^{*}\right)\right]}\left(t_{k}-t_{0}\right)
G(r)r( by (15))\displaystyle\leq G\left(r^{*}\right)\leq r^{*}\quad(\text{ by }(5))

That is the sequence {tn}(n0)\left\{t_{n}\right\}(n\geq 0) is also bounded above by rr^{*}. Since rr^{*} is the minimum nonnegative number satisfying G(r)rG\left(r^{*}\right)\leq r^{*}, it follows that limntn=r\lim_{n\rightarrow\infty}t_{n}=r^{*}.
(ii) By hypothesis (15) and the choice of a1a_{1} it follows that x1U(x0,r)x_{1}\in U\left(x_{0},r^{*}\right). From (19) and (20) we get g1(x0),g2(x0)U(x0,r)g_{1}\left(x_{0}\right),g_{2}\left(x_{0}\right)\in U\left(x_{0},r^{*}\right). Let us assume xk+1,g1(xk),g2(xk)U(x0,r)x_{k+1},g_{1}\left(x_{k}\right),g_{2}\left(x_{k}\right)\in U\left(x_{0},r^{*}\right) for k=0,1,,n1k=0,1,\cdots,n-1. Then from (13), (14), (19) and (20) we get

g1(xk)x0g1(xk)g1(x0)+g1(x0)x0a8xkx0+g1(x0)x0a8r+g1(x0)x0r\begin{gathered}\left\|g_{1}\left(x_{k}\right)-x_{0}\right\|\leq\left\|g_{1}\left(x_{k}\right)-g_{1}\left(x_{0}\right)\right\|+\left\|g_{1}\left(x_{0}\right)-x_{0}\right\|\leq a_{8}\left\|x_{k}-x_{0}\right\|+\left\|g_{1}\left(x_{0}\right)-x_{0}\right\|\\ \leq a_{8}r^{*}+\left\|g_{1}\left(x_{0}\right)-x_{0}\right\|\leq r^{*}\end{gathered}

and

g2(xk)x0g2(xk)g2(x0)+g2(x0)x0a9xkx0+g2(x0)x0a9r+g2(x0)x0r\begin{gathered}\left\|g_{2}\left(x_{k}\right)-x_{0}\right\|\leq\left\|g_{2}\left(x_{k}\right)-g_{2}\left(x_{0}\right)\right\|+\left\|g_{2}\left(x_{0}\right)-x_{0}\right\|\leq a_{9}\left\|x_{k}-x_{0}\right\|+\left\|g_{2}\left(x_{0}\right)-x_{0}\right\|\\ \leq a_{9}r^{*}+\left\|g_{2}\left(x_{0}\right)-x_{0}\right\|\leq r^{*}\end{gathered}

Hence g1(xn),g2(xn)U(x0,r)g_{1}\left(x_{n}\right),g_{2}\left(x_{n}\right)\in U\left(x_{0},r^{*}\right). Using (5), (13), (14) and (17) we obtain

B01(BkB0)a0(g1(x0)g1(xk)+g2(x0)g2(xk))a0(a8+a9)x0xka0(a8+a9)r<1\begin{gathered}\left\|B_{0}^{-1}\left(B_{k}-B_{0}\right)\right\|\leq a_{0}\left(\left\|g_{1}\left(x_{0}\right)-g_{1}\left(x_{k}\right)\right\|+\left\|g_{2}\left(x_{0}\right)-g_{2}\left(x_{k}\right)\right\|\right)\\ \leq a_{0}\left(a_{8}+a_{9}\right)\left\|x_{0}-x_{k}\right\|\leq a_{0}\left(a_{8}+a_{9}\right)r^{*}<1\end{gathered}

It follows from the Banach lemma on invertible operators [3] that BkB_{k} is invertible and

Bk1B011a0(a8+a9)xkx0=β¯k\left\|B_{k}^{-1}B_{0}\right\|\leq\frac{1}{1-a_{0}\left(a_{8}+a_{9}\right)\left\|x_{k}-x_{0}\right\|}=\bar{\beta}_{k} (30)

Using (2) we obtain the approximation

xk+1xk=Bk1(T(xk)xkzk)=(Bk1B0)B01\displaystyle x_{k+1}-x_{k}=B_{k}^{-1}\left(T\left(x_{k}\right)-x_{k}-z_{k}\right)=\left(B_{k}^{-1}B_{0}\right)B_{0}^{-1}
{(PT1(xk)PT1(xk1)P[g1(xk1),g2(xk1)](xkxk1)\displaystyle\left\{\left(PT_{1}\left(x_{k}\right)-PT_{1}\left(x_{k-1}\right)-P\left[g_{1}\left(x_{k-1}\right),g_{2}\left(x_{k-1}\right)\right]\left(x_{k}-x_{k-1}\right)\right.\right.
+(QT1(xk)QT1(xk1)+(F(xk)F(xk1))+(zk1zk)}\displaystyle+\left(QT_{1}\left(x_{k}\right)-QT_{1}\left(x_{k-1}\right)+\left(F\left(x_{k}\right)-F\left(x_{k-1}\right)\right)+\left(z_{k-1}-z_{k}\right)\right\} (31)

From (7), we get

B01[PT1(xk)PT1(xk1)PAk1(xkxk1)]B01P([xk1,xk]Ak1)(xkxk1)\displaystyle\left\|B_{0}^{-1}\left[PT_{1}\left(x_{k}\right)-PT_{1}\left(x_{k-1}\right)-PA_{k-1}\left(x_{k}-x_{k-1}\right)\right]\right\|\leq\left\|B_{0}^{-1}P\left(\left[x_{k-1},x_{k}\right]-A_{k-1}\right)\left(x_{k}-x_{k-1}\right)\right\|
a2(xk1g1(xk1)+xkg2(xk1))xkxk1\displaystyle\leq a_{2}\left(\left\|x_{k-1}-g_{1}\left(x_{k-1}\right)\right\|+\left\|x_{k}-g_{2}\left(x_{k-1}\right)\right\|\right)\left\|x_{k}-x_{k-1}\right\| (32)

and since by (10), (11), (13), (14)

xk1g1(xk1)\displaystyle\left\|x_{k-1}-g_{1}\left(x_{k-1}\right)\right\| xk1xk+g1(xk)g1(xk1)+xkg1(xk)\displaystyle\leq\left\|x_{k-1}-x_{k}\right\|+\left\|g_{1}\left(x_{k}\right)-g_{1}\left(x_{k-1}\right)\right\|+\left\|x_{k}-g_{1}\left(x_{k}\right)\right\|
xkxk1+a8xkxk1+a5Bk1(xkT(xk)zk\displaystyle\leq\left\|x_{k}-x_{k-1}\right\|+a_{8}\left\|x_{k}-x_{k-1}\right\|+a_{5}\|B_{k}^{-1}\left(x_{k}-T\left(x_{k}\right)-z_{k}\|\right.
xkg2(xk1)\displaystyle\left\|x_{k}-g_{2}\left(x_{k-1}\right)\right\| xkg2(xk)+g2(xk)g2(xk1)\displaystyle\leq x_{k}-g_{2}\left(x_{k}\right)\|+\|g_{2}\left(x_{k}\right)-g_{2}\left(x_{k-1}\right)\|
a6Bk1(xkT(xk)zk)+a9xkxk1\displaystyle\leq a_{6}\left\|B_{k}^{-1}\left(x_{k}-T\left(x_{k}\right)-z_{k}\right)\right\|+a_{9}\left\|x_{k}-x_{k-1}\right\|

(32) gives

B01[PT1(xk)\displaystyle\|B_{0}^{-1}\left[PT_{1}\left(x_{k}\right)\right. PT1(xk1)PAk1(xkxk1)a2(1+a8+a9)xkxk12\displaystyle-PT_{1}\left(x_{k-1}\right)-PA_{k-1}\left(x_{k}-x_{k-1}\right)\left\|\leq a_{2}\left(1+a_{8}+a_{9}\right)\right\|x_{k}-x_{k-1}\|^{2}
+a2(a5+a6)Bk1(xkT(xk)zk)xkxk1\displaystyle+a_{2}\left(a_{5}+a_{6}\right)\left\|B_{k}^{-1}\left(x_{k}-T\left(x_{k}\right)-z_{k}\right)\right\|\left\|x_{k}-x_{k-1}\right\| (33)

Moreover from (8), (9) and (12) we obtain respectively

B01(QT1(xk)QT1(xk1)a3xkxk1(k1)\displaystyle\|B_{0}^{-1}\left(QT_{1}\left(x_{k}\right)-QT_{1}\left(x_{k-1}\right)\left\|\leq a_{3}\right\|x_{k}-x_{k-1}\|\quad(k\geq 1)\right. (34)
B01(F(xk)F(xk1))a4xkxk1(k1)\displaystyle\left\|B_{0}^{-1}\left(F\left(x_{k}\right)-F\left(x_{k-1}\right)\right)\right\|\leq a_{4}\left\|x_{k}-x_{k-1}\right\|\quad(k\geq 1) (35)

and

B01(zkzk1)a7xkxk1(k1)\left\|B_{0}^{-1}\left(z_{k}-z_{k-1}\right)\right\|\leq a_{7}\left\|x_{k}-x_{k-1}\right\|\quad(k\geq 1) (36)

Furthermore (31) because of (30), (33)-(36) finally gives (27) for n=kn=k.

Estimate (28) is true for n=0n=0 by (25). Assume (28) is true for k=0,1,2,,nk=0,1,2,\cdots,n- 1. Then from (26), (27) and the induction hypothesis it follows that (28) is true for k=nk=n. By (28) and part (i) it follows that iteration {xn}(n0)\left\{x_{n}\right\}(n\geq 0) is Cauchy in a Banach space EE and as such it converges to some xU(x0,r)x^{*}\in U\left(x_{0},r^{*}\right) (since U(x0,r)U\left(x_{0},r^{*}\right) is a closed set). Using hypothesis ( c ) and letting nn\rightarrow\infty in (2) we get x=T(x)x^{*}=T\left(x^{*}\right). That is xx^{*} is a fixed point of TT. Estimate (29) follows immediately from (28) using standard majorization techniques [2], [3].

Finally to show uniqueness let us assume yU(x0,R)y^{*}\in U\left(x_{0},R\right) is a fixed point of equation (1). As in (31) we start from the approximation.

xn+1y=\displaystyle x_{n+1}-y^{*}= (Bn1B0)B01{[PT1(xn)PT1(y)PAn(xny)]\displaystyle\left(B_{n}^{-1}B_{0}\right)B_{0}^{-1}\left\{\left[PT_{1}\left(x_{n}\right)-PT_{1}\left(y^{*}\right)-PA_{n}\left(x_{n}-y^{*}\right)\right]\right.
+[QT1(xn)QT1(y)]+[F(xn)F(y)]zn}\displaystyle\left.+\left[QT_{1}\left(x_{n}\right)-QT_{1}\left(y^{*}\right)\right]+\left[F\left(x_{n}\right)-F\left(y^{*}\right)\right]-z_{n}\right\}

and using (5), (7)-(11), (13), (14), (21), (22) and (24) we get

xn+1ybxny+cnbn+1x0y+qn(n0)\left\|x_{n+1}-y^{*}\right\|\leq b\left\|x_{n}-y^{*}\right\|+c_{n}\leq\cdots\leq b^{n+1}\left\|x_{0}-y^{*}\right\|+q_{n}\quad(n\geq 0) (37)

By letting nn\rightarrow\infty as using (21) and (23) we get limnxn=y\lim_{n\rightarrow\infty}x_{n}=y^{*}. It follows from the uniqueness of the limit that x=yx^{*}=y^{*}.

That completes the proof of the Theorem.

Remarks

(1) Conditions (19) and (20) guarantee g1(x),g2(x)U(x0,r)g_{1}(x),g_{2}(x)\in U\left(x_{0},r^{*}\right) for xU(x0,r)x\in U\left(x_{0},r^{*}\right). Hence condition (7) can be dropped and we can set a2=a0a_{2}=a_{0}. However it is hoped that a2a0a_{2}\leq a_{0}.
(2) It can easily be seen that the first inequality in (15) can be replaced by the system of inequalities (17), (18) and

f(r)0f\left(r^{*}\right)\leq 0

where

f(r)=d2r2+d1r+d0f(r)=d_{2}r^{2}+d_{1}r+d_{0}

with

b1=a2(1+a8+a9),b2=a3+a4+a7,b3=a0(a8+a9)+a2(a5+a6)d2=b1+b3d1=b11b3a1\begin{gathered}b_{1}=a_{2}\left(1+a_{8}+a_{9}\right),\quad b_{2}=a_{3}+a_{4}+a_{7},\quad b_{3}=a_{0}\left(a_{8}+a_{9}\right)+a_{2}\left(a_{5}+a_{6}\right)\\ d_{2}=b_{1}+b_{3}\\ d_{1}=b_{1}-1-b_{3}a_{1}\end{gathered}

and d0=a1d_{0}=a_{1}.
(3) Condition (23) is satisfied if and only if zn=0(n0)z_{n}=0(n\geq 0)
(4) It can easily be seen from (10) and (11) that conditions (19) and (20) will be satisfied if a5+a81a_{5}+a_{8}\leq 1 and a6+a91a_{6}+a_{9}\leq 1 for r0r^{*}\neq 0. Indeed from (10) we have x0g1(x0)a5x1x0a5r\left\|x_{0}-g_{1}\left(x_{0}\right)\right\|\leq a_{5}\left\|x_{1}-x_{0}\right\|\leq a_{5}r^{*}. Hence (19) will be certainly satisfied if a5r(1a8)ra_{5}r^{*}\leq\left(1-a_{8}\right)r^{*}. That is if a5+a81a_{5}+a_{8}\leq 1. We argue similarly for (20).

REFERENCES

[1] Argyros, I.K. On some projection methods for the solution of nonlinear operator equations with non-differentiable operators, Tamkang J. Math. 24, 1, (1993), 1-8.
[2] Argyros, I. K. and Szidarovszky, F. The theory and application of iteration methods, C.R.C. Press, Inc. Boca Raton, Florida, 1993.
[3] Kantorovich, L.V. and Akilov, G.P. Functional analysis in normed spaces, Academic Press, New York, 1978.
[4] Păvăloiu, I. Sur une generalisation de la methode de Steffensen, Revue d’analyse Numerique et de theorie de I’approximation, 21, 1, (1992), 59-65.
[5] Pǎvăloiu, I. Bilateral approximations for the solutions of scalar equations, Revue d’analyse numerique et de theorie de l’approximation, 23, 1, (1994), 95100 .

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