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A three-step iterative scheme for solving nonlinear ϕ-strongly accretive operator equations in Banach spaces

Abstract

In this paper, we study a three-step iterative scheme with error terms for solving nonlinear ϕ-strongly accretive operator equations in arbitrary real Banach spaces.

1 Introduction

Let K be a nonempty subset of an arbitrary Banach space X and X be its dual space. The symbols D(T), R(T) and F(T) stand for the domain, the range and the set of fixed points of T respectively (for a single-valued map T:XX, xX is called a fixed point of T iff T(x)=x). We denote by J the normalized duality mapping from E to 2 E defined by

J(x)= { f X : x , f = x 2 = f 2 } .

Let T:D(T)XX be an operator. The following definitions can be found in [115] for example.

Definition 1 T is called Lipshitzian if there exists L>0 such that

TxTyLxy,

for all x,yK. If L=1, then T is called nonexpansive, and if 0<L<1, T is called contraction.

Definition 2

  1. (i)

    T is said to be strongly pseudocontractive if there exists a t>1 such that for each x,yD(T), there exists j(xy)J(xy) satisfying

    Re T x T y , j ( x y ) 1 t x y 2 .
  2. (ii)

    T is said to be strictly hemicontractive if F(T) is nonempty and if there exists a t>1 such that for each xD(T) and qF(T), there exists j(xy)J(xy) satisfying

    Re T x q , j ( x q ) 1 t x q 2 .
  3. (iii)

    T is said to be ϕ-strongly pseudocontractive if there exists a strictly increasing function ϕ:[0,)[0,) with ϕ(0)=0 such that for each x,yD(T), there exists j(xy)J(xy) satisfying

    Re T x T y , j ( x y ) x y 2 ϕ ( x y ) xy.
  4. (iv)

    T is said to be ϕ-hemicontractive if F(T) is nonempty and if there exists a strictly increasing function ϕ:[0,)[0,) with ϕ(0)=0 such that for each xD(T) and qF(T), there exists j(xy)J(xy) satisfying

    Re T x q , j ( x q ) x q 2 ϕ ( x q ) xq.

Clearly, each strictly hemicontractive operator is ϕ-hemicontractive.

Definition 3

  1. (i)

    T is called accretive if the inequality

    xy x y + s ( T x T y )

holds for every x,yD(T) and for all s>0.

  1. (ii)

    T is called strongly accretive if, for all x,yD(T), there exists a constant k>0 and j(xy)J(xy) such that

    T x T y , j ( x y ) k x y 2 .
  2. (iii)

    T is called ϕ-strongly accretive if there exists j(xy)J(xy) and a strictly increasing function ϕ:[0,)[0,) with ϕ(0)=0 such that for each x,yX,

    T x T y , j ( x y ) ϕ ( x y ) xy.

Remark 4 It has been shown in [11, 12] that the class of strongly accretive operators is a proper subclass of the class of ϕ-strongly accretive operators. If I denotes the identity operator, then T is called strongly pseudocontractive (respectively, ϕ-strongly pseudocontractive) if and only if (IT) is strongly accretive (respectively, ϕ-strongly accretive).

Chidume [1] showed that the Mann iterative method can be used to approximate fixed points of Lipschitz strongly pseudocontractive operators in L p (or l p ) spaces for p[2,). Chidume and Osilike [4] proved that each strongly pseudocontractive operator with a fixed point is strictly hemicontractive, but the converse does not hold in general. They also proved that the class of strongly pseudocontractive operators is a proper subclass of the class of ϕ-strongly pseudocontractive operators and pointed out that the class of ϕ-strongly pseudocontractive operators with a fixed point is a proper subclass of the class of ϕ-hemicontractive operators. These classes of nonlinear operators have been studied by various researchers (see, for example, [725]). Liu et al. [26] proved that, under certain conditions, a three-step iteration scheme with error terms converges strongly to the unique fixed point of ϕ-hemicontractive mappings.

In this paper, we study a three-step iterative scheme with error terms for nonlinear ϕ-strongly accretive operator equations in arbitrary real Banach spaces.

2 Preliminaries

We need the following results.

Lemma 5 [27]

Let { a n }, { b n } and { c n } be three sequences of nonnegative real numbers with n = 1 b n < and n = 1 c n <. If

a n + 1 (1+ b n ) a n + c n ,n1,

then the limit lim n a n exists.

Lemma 6 [28]

Let x,yX. Then xx+ry for every r>0 if and only if there is fJ(x) such that Rey,f0.

Lemma 7 [9]

Suppose that X is an arbitrary Banach space and A:EE is a continuous ϕ-strongly accretive operator. Then the equation Ax=f has a unique solution for any fE.

3 Strong convergence of a three-step iterative scheme to a solution of the system of nonlinear operator equations

For the rest of this section, L denotes the Lipschitz constant of T 1 , T 2 , T 3 :XX, L =(1+L) and R( T 1 ), R( T 2 ) and R( T 3 ) denote the ranges of T 1 , T 2 and T 3 respectively. We now prove our main results.

Theorem 8 Let X be an arbitrary real Banach space and T 1 , T 2 , T 3 :XX Lipschitz ϕ-strongly accretive operators. Let fR( T 1 )R( T 2 )R( T 3 ) and generate { x n } from an arbitrary x 0 X by

x n + 1 = a n x n + b n ( f + ( I T 1 ) y n ) + c n v n , y n = a n x n + b n ( f + ( I T 2 ) z n ) + c n u n , z n = a n x n + b n ( f + ( I T 3 ) x n ) + c n w n , n 0 ,
(3.1)

where { v n } n = 0 , { u n } n = 0 and { w n } n = 0 are bounded sequences in X and { a n }, { c n }, { a n }, { b n }, { c n }, { a n }, { b n }, { c n } are sequences in [0,1] and { b n } in (0,1) satisfying the following conditions: (i) a n + b n + c n =1= a n + b n + c n = a n + b n + c n , (ii) n = 0 b n =, (iii) n = 0 b n 2 <, n = 0 b n <, (iv) n = 0 c n <, n = 0 c n < and n = 0 c n <. Then the sequence { x n } converges strongly to the solution of the system T i x=f; i=1,2,3.

Proof By Lemma 7, the system T i x=f; i=1,2,3 has the unique solution x X. Following the techniques of [5, 812, 26, 29], define S i :XX by S i x=f+(I T i )x; i=1,2,3; then each S i is demicontinuous and x is the unique fixed point of S i ; i=1,2,3, and for all x,yX, we have

( I S i ) x ( I S i ) y , j ( x y ) ϕ i ( x y ) x y ϕ i ( x y ) ( 1 + ϕ i ( x y ) + x y ) x y 2 = θ i ( x , y ) x y 2 ,

where θ i (x,y)= ϕ i ( x y ) ( 1 + ϕ i ( x y ) + x y ) [0,1) for all x,yX; i=1,2,3. Let x i = 1 3 F( S i ) be the fixed point set of S i , and let θ(x,y)= inf min i { θ i (x,y)}[0,1]. Thus

( I S i ) x ( I S i ) y , j ( x y ) θ(x,y) x y 2 ;i=1,2,3.
(3.2)

It follows from Lemma 6 and inequality (3.2) that

xy x y + λ [ ( I S i ) x θ ( x , y ) x ( ( I S i ) y θ ( x , y ) y ) ] ,
(3.3)

for all x,yX and for all λ>0; i=1,2,3.

Set α n = b n + c n , β n = b n + c n and γ n = b n + c n , then (3.1) becomes

x n + 1 = ( 1 α n ) x n + α n S 1 y n + c n ( v n S 1 y n ) , y n = ( 1 β n ) x n + β n S 2 z n + c n ( u n S 2 z n ) , z n = ( 1 γ n ) x n + γ n S 3 x n + c n ( w n S 3 x n ) , n 0 .
(3.4)

We have

x n = ( 1 + α n ) x n + 1 + α n [ ( I S 1 ) x n + 1 θ ( x n + 1 , x ) x n + 1 ] ( 1 θ ( x n + 1 , x ) ) α n x n + ( 2 θ ( x n + 1 , x ) ) α n 2 ( x n S 1 y n ) + α n ( S 1 x n + 1 S 1 y n ) [ 1 + ( 2 θ ( x n + 1 , x ) ) α n ] c n ( v n S 1 y n ) .

Furthermore,

x =(1+ α n ) x + α n [ ( I S 1 ) x θ ( x n + 1 , x ) x ] ( 1 θ ( x n + 1 , x ) ) α n x ,

so that

x n x = ( 1 + α n ) ( x n + 1 x ) + α n [ ( I S 1 ) x n + 1 θ ( x n + 1 , x ) x n + 1 ( ( I S 1 ) x θ ( x n + 1 , x ) x ) ] ( 1 θ ( x n + 1 , x ) ) α n ( x n x ) + ( 2 θ ( x n + 1 , x ) ) α n 2 ( x n S 1 y n ) + α n ( S 1 x n + 1 S 1 y n ) [ 1 + ( 2 θ ( x n + 1 , x ) ) α n ] c n ( v n S 1 y n ) .

Hence,

x n x ( 1 + α n ) x n + 1 x + α n ( 1 + α n ) [ ( I S 1 ) x n + 1 θ ( x n + 1 , x ) x n + 1 ( ( I S 1 ) x θ ( x n + 1 , x ) x ) ] ( 1 θ ( x n + 1 , x ) ) α n x n x ( 2 θ ( x n + 1 , x ) ) α n 2 x n S 1 y n α n S 1 x n + 1 S 1 y n [ 1 + ( 2 θ ( x n + 1 , x ) ) α n ] c n v n S 1 y n ( 1 + α n ) x n + 1 x ( 1 θ ( x n + 1 , x ) ) α n x n x ( 2 θ ( x n + 1 , x ) ) α n 2 x n S 1 y n α n S 1 x n + 1 S 1 y n [ 1 + ( 2 θ ( x n + 1 , x ) ) α n ] c n v n S 1 y n .

Hence,

x n + 1 x [ 1 + ( 1 θ ( x n + 1 , x ) ) α n ] ( 1 + α n ) x n x + 2 α n 2 x n S 1 y n + α n S 1 x n + 1 S 1 y n + [ 1 + ( 2 θ ( x n + 1 , x ) ) α n ] c n v n S 1 y n [ 1 + ( 1 θ ( x n + 1 , x ) ) α n ] [ 1 α n + α n 2 ] x n x + 2 α n 2 x n S 1 y n + α n S 1 x n + 1 S 1 y n + 3 c n v n S 1 y n [ 1 θ ( x n + 1 , x ) α n + α n 2 ] x n x + 2 α n 2 x n S 1 y n + α n S 1 x n + 1 S 1 y n + 3 c n v n S 1 y n .
(3.5)

Furthermore, we have the following estimates:

(3.6)
(3.7)
(3.8)
(3.9)

Using (3.4) and (3.6),

x n y n = β n ( x n S 2 z n ) c n ( u n S 2 z n ) β n x n S 2 z n + c n u n S 2 z n [ [ 1 + L ( 3 L 1 ) ] β n + L ( 3 L 1 ) c n ] x n x + L ( β n + c n ) c n w n x + c n u n x [ [ 1 + L ( 3 L 1 ) ] β n + L ( 3 L 1 ) c n ] x n x + 3 L c n w n x + c n u n x .
(3.10)

Using (3.7),

S 1 y n y n S 1 y n x + y n x ( 1 + L ) y n x ( 1 + L ) [ 3 L ( 3 L 1 ) 1 ] x n x + 3 L ( 1 + L ) c n w n x + ( 1 + L ) c n u n x .
(3.11)

Again, using (3.7),

v n S 1 y n v n x + L y n x L [ 3 L ( 3 L 1 ) 1 ] x n x + v n x + 3 L 2 c n w n x + L c n u n x .
(3.12)

Substituting (3.10)-(3.12) in (3.9), we obtain

S 1 x n + 1 S 1 y n L [ 1 + L ( 3 L 1 ) ] β n + L ( 3 L 1 ) c n + [ 3 L ( 3 L 1 ) 1 ] [ ( 1 + L ) α n + L c n ] x n x + 3 L [ L c n + [ ( 1 + L ) α n + L c n ] c n ] w n x + L [ c n + [ ( 1 + L ) α n + L c n ] c n ] u n x + L c n v n x .
(3.13)

Substituting (3.8), (3.12) and (3.13) in (3.5), we obtain

x n + 1 x [ 1 + [ 3 + L ( 3 + L ) 3 L ( 3 L 1 ) 1 ] ] α n 2 + L [ 3 L ( 3 L 1 ) 1 ] α n β n + L 2 ( 3 L 1 ) α n c n + L [ 3 L ( 3 L 1 ) 1 ] α n c n + 3 L [ 3 L ( 3 L 1 ) 1 ] c n x n x θ ( x n + 1 , x ) α n x n x + [ 3 L ( 1 + 3 L ) α n 2 c n + 3 L 2 α n c n + 3 L 2 α n c n c n + 9 L 2 c n c n ] w n x + [ L ( 3 + L ) α n 2 c n + L α n c n + L 2 α n c n c n + 3 L c n c n ] u n x + ( 2 L + 3 ) c n v n x .
(3.14)

Since { v n }, { u n } and { w n } are bounded, we set

M= sup n 0 v n x + sup n 0 u n x + sup n 0 w n x <.

Then it follows from (3.14) that

x n + 1 x [ 1 + [ 3 + L ( 3 + L ) [ 3 L ( 3 L 1 ) 1 ] ] α n 2 + L [ 3 L ( 3 L 1 ) 1 ] α n β n + L 2 ( 3 L 1 ) α n c n + L [ 3 L ( 3 L 1 ) 1 ] α n c n + 3 L [ 3 L ( 3 L 1 ) 1 ] c n ] x n x θ ( x n + 1 , x ) α n x n x + [ 3 L ( 1 + 3 L ) α n 2 c n + 3 L 2 α n c n + 3 L 2 α n c n c n + 9 L 2 c n c n ] M + [ L ( 3 + L ) α n 2 c n + L α n c n + L 2 α n c n c n + 3 L c n c n ] M + ( 2 L + 3 ) c n M = ( 1 + δ n ) x n x θ ( x n + 1 , x ) α n x n x + σ n ( 1 + δ n ) x n x + σ n ,
(3.15)

where

Since b n (0,1), the conditions (iii) and (iv) imply that n = 0 δ n < and n = 0 σ n <. It then follows from Lemma 5 that lim n x n x exists. Let lim n x n x =δ0. We now prove that δ=0. Assume that δ>0. Then there exists a positive integer N 0 such that x n x δ 2 for all n N 0 . Since

θ ( x n + 1 , x ) x n x = ϕ ( x n + 1 x ) 1 + ϕ ( x n + 1 x ) + x n + 1 x x n x ϕ ( δ 2 ) δ 2 ( 1 + ϕ ( D ) + D ) ,

for all n N 0 , it follows from (3.15) that

x n + 1 x x n x ϕ ( δ 2 ) δ 2 ( 1 + ϕ ( D ) + D ) α n + λ n for all n N 0 .

Hence,

ϕ ( δ 2 ) δ 2 ( 1 + ϕ ( D ) + D ) α n x n x x n + 1 x + λ n for all n N 0 .

This implies that

ϕ ( δ 2 ) δ 2 ( 1 + ϕ ( D ) + D ) j = N 0 n α j x N 0 x + j = N 0 n λ j .

Since b n α n ,

ϕ ( δ 2 ) δ 2 ( 1 + ϕ ( D ) + D ) j = N 0 n b j x N 0 x + j = N 0 n λ j

yields n = 0 b n <, contradicting the fact that n = 0 b n =. Hence, lim n x n x =0. □

Corollary 9 Let X be an arbitrary real Banach space and T 1 , T 2 , T 3 :XX be three Lipschitz ϕ-strongly accretive operators, where ϕ is in addition continuous. Suppose lim inf r ϕ(r)>0 or T i x as x; i=1,2,3. Let { a n }, { b n }, { c n }, { a n }, { b n }, { c n }, { a n }, { b n }, { c n }, { w n }, { u n }, { v n }, { y n } and { x n } be as in Theorem  8. Then, for any given fX, the sequence { x n } converges strongly to the solution of the system T i x=f; i=1,2,3.

Proof The existence of a unique solution to the system T i x=f; i=1,2,3 follows from [9] and the result follows from Theorem 8. □

Theorem 10 Let X be a real Banach space and K be a nonempty closed convex subset of X. Let T 1 , T 2 , T 3 :KK be three Lipschitz ϕ-strong pseudocontractions with a nonempty fixed point set. Let { a n }, { b n }, { c n }, { a n }, { b n }, { c n }, { a n }, { b n }, { c n }, { w n }, { u n } and { v n } be as in Theorem  8. Let { x n } be the sequence generated iteratively from an arbitrary x 0 K by

x n + 1 = a n x n + b n T 1 y n + c n v n , y n = a n x n + b n T 2 z n + c n u n , z n = a n x n + b n T 3 x n + c n w n , n 0 .

Then { x n } converges strongly to the common fixed point of T 1 , T 2 , T 3 .

Proof As in the proof of Theorem 8, set α n = b n + c n , β n = b n + c n , γ n = b n + c n to obtain

x n + 1 = ( 1 α n ) x n + α n T 1 y n + c n ( v n T 1 y n ) , y n = ( 1 β n ) x n + β n T 2 z n + c n ( u n T 2 z n ) , z n = ( 1 γ n ) x n + γ n T 3 x n + c n ( w n T 3 x n ) , n 0 .

Since each T i ; i=1,2,3 is a ϕ-strong pseudocontraction, (I T i ) is ϕ-strongly accretive so that for all x,yX, there exist j(xy)J(xy) and a strictly increasing function ϕ:(0,)(0,) with ϕ(0)=0 such that

( I T i ) x ( I T i ) y , j ( x y ) ϕ ( x y ) xyθ(x,y) x y 2 ;i=1,2,3.

The rest of the argument now follows as in the proof of Theorem 8. □

Remark 11 The example in [4] shows that the class of ϕ-strongly pseudocontractive operators with nonempty fixed point sets is a proper subclass of the class of ϕ-hemicontractive operators. It is easy to see that Theorem 8 easily extends to the class of ϕ-hemicontractive operators.

Remark 12

  1. (i)

    If we set b n =0= c n for all n0 in our results, we obtain the corresponding results for the Ishikawa iteration scheme with error terms in the sense of Xu [15].

  2. (ii)

    If we set b n =0= c n = b n =0= c n for all n0 in our results, we obtain the corresponding results for the Mann iteration scheme with error terms in the sense of Xu [15].

Remark 13 Let { α n } and { β n } be real sequences satisfying the following conditions:

  1. (i)

    0 α n , β n 1, n0,

  2. (ii)

    lim n α n = lim n β n =0,

  3. (iii)

    n = 0 α n =,

  4. (iv)

    n = 0 β n <, and

  5. (v)

    n = 0 α n 2 <.

If we set a n =(1 β n ), b n = β n , c n =0, a n =(1 α n ), b n = α n , c n =0, b n =0= c n for all n0 in Theorems 8 and 10 respectively, we obtain the corresponding convergence theorems for the original Ishikawa [18] and Mann [30] iteration schemes.

Remark 14

  1. (i)

    Gurudwan and Sharma [29] studied a strong convergence of multi-step iterative scheme to a common solution for a finite family of ϕ-strongly accretive operator equations in a reflexive Banach space with weakly continuous duality mapping. Some remarks on their work can be seen in [31].

  2. (ii)

    All the above results can be extended to a finite family of ϕ-strongly accretive operators.

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Acknowledgements

The last author gratefully acknowledges the support from the Deanship of Scientific Research (DSR) at King Abdulaziz University (KAU) during this research.

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Khan, S.H., Rafiq, A. & Hussain, N. A three-step iterative scheme for solving nonlinear ϕ-strongly accretive operator equations in Banach spaces. Fixed Point Theory Appl 2012, 149 (2012). https://doi.org/10.1186/1687-1812-2012-149

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