Open Access

Viscosity iterative method for a new general system of variational inequalities in Banach spaces

Journal of Inequalities and Applications20132013:249

https://doi.org/10.1186/1029-242X-2013-249

Received: 19 November 2012

Accepted: 2 May 2013

Published: 17 May 2013

Abstract

In this paper, we study a new iterative method for finding a common element of the set of solutions of a new general system of variational inequalities for two different relaxed cocoercive mappings and the set of fixed points of a nonexpansive mapping in real 2-uniformly smooth and uniformly convex Banach spaces. We prove the strong convergence of the proposed iterative method without the condition of weakly sequentially continuous duality mapping. Our result improves and extends the corresponding results announced by many others.

MSC:46B10, 46B20, 47H10, 49J40.

Keywords

a new general system of variational inequalities relaxed cocoercive mapping strong convergence

1 Introduction

Let X be a real Banach space and X be its dual space. Let C be a subset of X and let T be a self-mapping of C. We use F ( T ) to denote the set of fixed points of T. The duality mapping J : X 2 X is defined by J ( x ) = { x X | x , x = x 2 , x = x } , x X . If X is a Hilbert space, then J = I , where I is the identity mapping. It is well-known that if X is smooth, then J is single-valued, which is denoted by j.

Recall that a mapping f : C C is a contraction on C, if there exists a constant α ( 0 , 1 ) such that f ( x ) f ( y ) α x y , x , y C . We use Π C to denote the collection of all contractions on C. This is Π C = { f | f : C C  a contraction } . A mapping T : C C is said to be nonexpansive if T ( x ) T ( y ) x y , x , y C . Let A : C X be a nonlinear mapping. Then A is called
  1. (i)
    L-Lipschitz continuous (or Lipschitzian) if there exists a constant L 0 such that
    A x A y L x y , x , y C ;
     
  2. (ii)
    accretive if there exists j ( x y ) J ( x y ) such that
    A x A y , j ( x y ) 0 , x , y C ;
     
  3. (iii)
    α-inverse strongly accretive if there exist j ( x y ) J ( x y ) and α > 0 such that
    A x A y , j ( x y ) α A x A y 2 , x , y C ;
     
  4. (iv)
    relaxed ( c , d ) -cocoercive if there exist j ( x y ) J ( x y ) and two constants c , d 0 such that
    A x A y , j ( x y ) ( c ) A x A y 2 + d x y 2 , x , y C .
     
Let C be a nonempty closed convex subset of a real Hilbert space H. Recall that the classical variational inequality is to find x C such that
A x , x x 0 , x C ,
(1.1)

where A : C H is a nonlinear mapping. Variational inequality theory has emerged as an important tool in studying a wide class of obstacle, unilateral, free, moving, equilibrium problems arising in several branches of pure and applied sciences in a unified and general framework. The variational inequality problem has been extensively studied in the literature; see [18] and the references cited therein.

In 2006, Aoyama et al. [9] first considered the following generalized variational inequality problem in Banach spaces. Let A : C X be an accretive operator. Find a point x C such that
A x , j ( x x ) 0 , x C .
(1.2)

Problem (1.2) is very interesting as it is connected with the fixed point problem for a nonlinear mapping and the problem of finding a zero point of an accretive operator in Banach spaces; see [1013] and the references cited therein.

In 2010, Yao et al. [14] introduced the following system of general variational inequalities in Banach spaces. For given two operators A , B : C X , they considered the problem of finding ( x , y ) C × C such that
{ A y + x y , j ( x x ) 0 , x C , B x + y x , j ( x y ) 0 , x C ,
(1.3)

which is called the system of general variational inequalities in a real Banach space and the set of solutions of problem (1.3) denoted by Ω 1 . Yao et al. proved the following strong convergence theorem.

Theorem YNNLY Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X which admits a weakly sequentially continuous duality mapping. Let Q C be the sunny nonexpansive retraction from X onto C. Let the mappings A , B : C X be α-inverse-strongly accretive with α K 2 and β-inverse-strongly accretive with β K 2 , respectively, with Ω 1 . For a given x 0 C , let the sequence { x n } be generated iteratively by
{ y n = Q C ( x n B x n ) , x n + 1 = α n u + β n x n + γ n Q C ( y n A y n ) , n 0 ,
where { α n } , { β n } and { γ n } are three sequences in ( 0 , 1 ) . Suppose that the sequences { α n } , { β n } and { γ n } satisfy the following conditions:
  1. (i)

    α n + β n + γ n = 1 , n 0 ;

     
  2. (ii)

    lim n α n = 0 and n = 1 α n = ;

     
  3. (iii)

    0 < lim inf n β n lim sup n β n 1 .

     

Then { x n } converges strongly to Q u where Q is the sunny nonexpansive retraction of C onto  Ω 1 .

In 2011, Katchang and Kumam [15] introduced the following system of general variational inequalities in Banach spaces. For given two operators A , B : C X , they considered the problem of finding ( x , y ) C × C such that
{ λ A y + x y , j ( x x ) 0 , x C , μ B x + y x , j ( x y ) 0 , x C ,
(1.4)

which is called the system of general variational inequalities in a real Banach space and the set of solutions of problem (1.4) denoted by Ω 2 . Katchang and Kumam proved the following strong convergence theorem.

Theorem KK Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X which admits a weakly sequentially continuous duality mapping. Let S : C C be a nonexpansive mapping and Q C be a sunny nonexpansive retraction from X onto C. Let the mappings A , B : C X be β-inverse-strongly accretive with β λ K 2 and γ-inverse-strongly accretive with γ μ K 2 , respectively, and let K be the 2-uniformly smooth constant of X. Let f be a contraction of C into itself with coefficient α [ 0 , 1 ) . Suppose that F : = Ω 2 F ( S ) . For a given x 0 = x C , let the sequence { x n } be generated iteratively by
{ y n = Q C ( x n μ B x n ) , x n + 1 = α n f ( x n ) + β n x n + γ n S Q C ( y n λ A y n ) , n 0 ,
where { α n } , { β n } and { γ n } are three sequences in ( 0 , 1 ) . Suppose that the sequences { α n } , { β n } and { γ n } satisfy the following conditions:
  1. (i)

    α n + β n + γ n = 1 , n 0 ;

     
  2. (ii)

    lim n α n = 0 and n = 1 α n = ;

     
  3. (iii)

    0 < lim inf n β n lim sup n β n < 1 .

     

Then { x n } converges strongly to x ¯ = Q F f ( x ¯ ) and ( x ¯ , y ¯ ) is a solution of problem (1.4), where y ¯ = Q C ( x ¯ μ B x ¯ ) and Q F is the sunny nonexpansive retraction of C onto F.

The problem of finding solutions of (1.4) by using iterative methods has been studied by many others; see [1619] and the references cited therein.

In this paper, we focus on the problem of finding ( x , y , z ) C × C × C such that
{ λ 1 A 1 y + x y , j ( x x ) 0 , x C , λ 2 A 2 z + y z , j ( x y ) 0 , x C , λ 3 A 3 x + z x , j ( x z ) 0 , x C ,
(1.5)

which is called a new general system of variational inequalities in Banach spaces, where A i : C X are three mappings, λ i > 0 for all i = 1 , 2 , 3 . In particular, if A 3 = 0 and z = x , then problem (1.5) reduces to problem (1.4). If we add up the requirement that λ i = 1 for i = 1 , 2 , then problem (1.5) reduces to problem (1.3).

In this paper, motivated and inspired by the idea of Katchang and Kumam [15] and Yao et al. [14], we introduce a new iterative method for finding a common element of the set of solutions of a new general system of variational inequalities in Banach spaces for two different relaxed cocoercive mappings and the set of fixed points of a nonexpansive mapping in real 2-uniformly smooth and uniformly convex Banach spaces. We prove the strong convergence of the proposed iterative algorithm without the condition of weakly sequentially continuous duality mapping. Our result improves and extends the corresponding results announced by many others.

2 Preliminaries

In this section, we recall the well-known results and give some useful lemmas that are used in the next section.

Let X be a Banach space and let U = { x X : x = 1 } be a unit sphere of X. X is said to be uniformly convex if for each ϵ ( 0 , 2 ] , there exists a constant δ > 0 such that for any x , y U ,
x y ϵ implies x + y 2 1 δ .
The norm on X is said to be Gâteaux differentiable if the limit
lim t 0 x + t y x t
(2.1)
exists for each x , y U and in this case X is said to be smooth. X is said to have a uniformly Frechet differentiable norm if the limit (2.1) is attained uniformly for x , y U and in this case X is said to be uniformly smooth. We define a function ρ : [ 0 , ) [ 0 , ) , called the modulus of smoothness of X, as follows:
ρ ( τ ) = sup { 1 2 ( x + y + x y ) 1 : x , y X , x = 1 , y = τ } .
It is known that X is uniformly smooth if and only if lim τ 0 ρ ( τ ) / τ = 0 . Let q be a fixed real number with 1 < q 2 . Then a Banach space X is said to be q-uniformly smooth if there exists a constant c > 0 such that ρ ( τ ) c τ q for all τ > 0 . For q > 1 , the generalized duality mapping J q : X 2 X is defined by
J q ( x ) = { f X : x , f = x q , f = x q 1 } , x X .
In particular, if q = 2 , the mapping J 2 is called the normalized duality mapping (or duality mapping), and usually we write J 2 = J . If X is a Hilbert space, then J = I . Further, we have the following properties of the generalized duality mapping J q .
  1. (1)

    J q ( x ) = x q 2 J 2 ( x ) for all x X with x 0 .

     
  2. (2)

    J q ( t x ) = t q 1 J q ( x ) for all x X and t [ 0 , ) .

     
  3. (3)

    J q ( x ) = J q ( x ) for all x X .

     

It is known that if X is smooth, then J is a single-valued function, which is denoted by j. Recall that the duality mapping j is said to be weakly sequentially continuous if for each { x n } X with x n x , we have j ( x n ) j ( x ) weakly-. We know that if X admits a weakly sequentially continuous duality mapping, then X is smooth. For details, see [20].

Lemma 2.1 [21]

Let X be a q-uniformly smooth Banach space with 1 q 2 . Then
x + y q x q + q y , J q ( x ) + 2 K y q

for all x , y X , where K is the q-uniformly smooth constant of X.

Lemma 2.2 [22]

In a Banach space X, the following inequality holds:
x + y 2 x 2 + 2 y , j ( x + y ) , x , y X ,

where j ( x + y ) J ( x + y ) .

Lemma 2.3 [23]

Assume that { a n } is a sequence of nonnegative real numbers such that
a n + 1 ( 1 γ n ) a n + δ n , n 1 ,
where { γ n } is a sequence in ( 0 , 1 ) and { δ n } is a sequence such that
  1. (i)

    n = 1 γ n = ;

     
  2. (ii)

    lim sup n δ n / γ n 0 or n = 1 | δ n | < .

     

Then lim n a n = 0 .

Let C be a nonempty closed convex subset of a smooth Banach space X and let D be a nonempty subset of C. A mapping Q : C D is said to be sunny if
Q ( Q x + t ( x Q x ) ) = Q x ,

whenever Q x + t ( x Q x ) C for x C and t 0 . A mapping Q : C D is called a retraction if Q x = x for all x D . Furthermore, Q is a sunny nonexpansive retraction from C onto D if Q is a retraction from C onto D, which is also sunny and nonexpansive. A subset D of C is called a sunny nonexpansive retraction of C if there exists a sunny nonexpansive retraction from C onto D.

It is well known that if X is a Hilbert space, then a sunny nonexpansive retraction Q C is coincident with the metric projection from X onto C.

Lemma 2.4 [24]

Let C be a closed convex subset of a smooth Banach space X. Let D be a nonempty subset of C and Q : C D be a retraction. Then the following are equivalent:
  1. (a)

    Q is sunny and nonexpansive.

     
  2. (b)

    Q x Q y 2 x y , j ( Q x Q y ) , x , y C .

     
  3. (c)

    x Q x , j ( y Q x ) 0 , x C , y D .

     

Lemma 2.5 [25]

If X is strictly convex and uniformly smooth and if T : C C is a nonexpansive mapping having a nonempty fixed point set F ( T ) , then the set F ( T ) is a sunny nonexpansive retraction of C.

Lemma 2.6 [26]

Let { x n } and { y n } be bounded sequences in a Banach space X and let { b n } be a sequence in [ 0 , 1 ] with 0 < lim inf n b n lim sup n b n < 1 . Suppose that x n + 1 = ( 1 b n ) y n + b n x n for all integers n 1 and lim sup n ( y n + 1 y n x n + 1 x n ) 0 . Then lim n y n x n = 0 .

Lemma 2.7 [27]

Let C be a closed convex subset of a strictly convex Banach space X. Let T 1 and T 2 be two nonexpansive mappings from C into itself with F ( T 1 ) F ( T 2 ) . Define a mapping S by
S x = λ T 1 x + ( 1 λ ) T 2 x , x C ,

where λ is a constant in ( 0 , 1 ) . Then S is nonexpansive and F ( S ) = F ( T 1 ) F ( T 2 ) .

Lemma 2.8 [28]

Let X be a real smooth and uniformly convex Banach space and let r > 0 . Then there exists a strictly increasing, continuous and convex function g : [ 0 , 2 r ] R such that g ( 0 ) = 0 and g ( x y ) x 2 2 x , j ( y ) + y 2 for all x , y B r .

Lemma 2.9 [23]

Let X be a uniformly smooth Banach space, let C be a closed convex subset of X, let T : C C be a nonexpansive mapping with F ( T ) and let f Π C . Then the sequence { x t } defined by x t = t f ( x t ) + ( 1 t ) T x t converges strongly to a point in F(T) as t 0 . If we define a mapping Q : Π C F ( T ) by Q ( f ) : = lim t 0 x t , f Π C , then Q ( f ) solves the following variational inequality:
( I f ) Q ( f ) , j ( Q ( f ) p ) 0 , f Π C , p F ( T ) .

Lemma 2.10 [17]

Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space X. Let the mapping A : C X be relaxed ( c , d ) -cocoercive and L A -Lipschitzian. Then we have
( I λ A ) x ( I λ A ) y 2 x y 2 + 2 ( λ c L A 2 λ d + K 2 λ 2 L A 2 ) x y 2 ,

where λ > 0 and K is the 2-uniformly smooth constant of X. In particular, if 0 < λ d c L A 2 K 2 L A 2 , then I λ A is a nonexpansive mapping.

In order to prove our main result, the next lemma is crucial for proving the main theorem.

Lemma 2.11 Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space X with the 2-uniformly smooth constant K. Let Q C be the sunny nonexpansive retraction from X onto C and let A i : C X be a relaxed ( c i , d i ) -cocoercive and L i -Lipschitzian mapping for i = 1 , 2 , 3 . Let G : C C be a mapping defined by
G ( x ) = Q C [ Q C ( Q C ( x λ 3 A 3 x ) λ 2 A 2 Q C ( x λ 3 A 3 x ) ) λ 1 A 1 Q C ( Q C ( x λ 3 A 3 x ) λ 2 A 2 Q C ( x λ 3 A 3 x ) ) ] , x C .

If 0 < λ i d i c i L i 2 K 2 L i 2 for all i = 1 , 2 , 3 , then G : C C is nonexpansive.

Proof For all x , y C , by Lemma 2.10, we have
G ( x ) G ( y ) = Q C [ Q C ( Q C ( I λ 3 A 3 ) x λ 2 A 2 Q C ( I λ 3 A 3 ) x ) λ 1 A 1 Q C ( Q C ( I λ 3 A 3 ) x λ 2 A 2 Q C ( I λ 3 A 3 ) x ) ] Q C [ Q C ( Q C ( I λ 3 A 3 ) y λ 2 A 2 Q C ( I λ 3 A 3 ) y ) λ 1 A 1 Q C ( Q C ( I λ 3 A 3 ) y λ 2 A 2 Q C ( I λ 3 A 3 ) y ) ] Q C ( Q C ( I λ 3 A 3 ) x λ 2 A 2 Q C ( I λ 3 A 3 ) x ) λ 1 A 1 Q C ( Q C ( I λ 3 A 3 ) x λ 2 A 2 Q C ( I λ 3 A 3 ) x ) [ Q C ( Q C ( I λ 3 A 3 ) y λ 2 A 2 Q C ( I λ 3 A 3 ) y ) λ 1 A 1 Q C ( Q C ( I λ 3 A 3 ) y λ 2 A 2 Q C ( I λ 3 A 3 ) y ) ] = ( I λ 1 A 1 ) Q C ( I λ 2 A 2 ) Q C ( I λ 3 A 3 ) x ( I λ 1 A 1 ) Q C ( I λ 2 A 2 ) Q C ( I λ 3 A 3 ) y x y ,

which implies that G is nonexpansive. □

Lemma 2.12 [29]

Let C be a nonempty closed convex subset of a real smooth Banach space X. Let Q C be the sunny nonexpansive retraction from X onto C. Let A i : C X be three possibly nonlinear mappings. For given x , y , z C , ( x , y , z ) is a solution of problem (1.5) if and only if x F ( G ) , y = Q C ( z λ 2 A 2 z ) and z = Q C ( x λ 3 A 3 x ) , where G is the mapping defined as in Lemma  2.11.

3 Main results

We are now in a position to state and prove our main result.

Theorem 3.1 Let X be a uniformly convex and 2-uniformly smooth Banach space with the 2-uniformly smooth constant K, let C be a nonempty closed convex subset of X and Q C be a sunny nonexpansive retraction from X onto C. Let the mappings A i : C X be relaxed ( c i , d i ) -cocoercive and L i -Lipschitzian with 0 < λ i < d i c i L i 2 K 2 L i 2 for all i = 1 , 2 , 3 . Let f be a contractive mapping with the constant α ( 0 , 1 ) and let S : C C be a nonexpansive mapping such that Ω = F ( S ) F ( G ) , where G is the mapping defined as in Lemma  2.11. For a given x 1 C , let { x n } , { y n } and { z n } be the sequences generated by
{ z n = Q C ( x n λ 3 A 3 x n ) , y n = Q C ( z n λ 2 A 2 z n ) , x n + 1 = a n f ( x n ) + b n x n + ( 1 a n b n ) S Q C ( y n λ 1 A 1 y n ) , n 1 ,
(3.1)

where { a n } and { b n } are two sequences in ( 0 , 1 ) such that

(C1) lim n a n = 0 and n = 1 a n = ;

(C2) 0 < lim inf n b n lim sup n b n < 1 .

Then { x n } converges strongly to q Ω , which solves the following variational inequality:
q f ( q ) , j ( q p ) 0 , f Π C , p Ω .

Proof Step 1. We show that { x n } is bounded.

Let x Ω and t n = Q C ( y n λ 1 A 1 y n ) . It follows from Lemma 2.12 that
x = Q C [ Q C ( Q C ( x λ 3 A 3 x ) λ 2 A 2 Q C ( x λ 3 A 3 x ) ) λ 1 A 1 Q C ( Q C ( x λ 3 A 3 x ) λ 2 A 2 Q C ( x λ 3 A 3 x ) ) ] .
Put y = Q C ( z λ 2 A 2 z ) and z = Q C ( x λ 3 A 3 x ) . Then x = Q C ( y λ 1 A 1 y ) and
x n + 1 = a n f ( x n ) + b n x n + ( 1 a n b n ) S t n .
From Lemma 2.10, we have I λ i A i ( i = 1 , 2 , 3 ) is nonexpansive. Therefore
t n x = Q C ( y n λ 1 A 1 y n ) Q C ( y λ 1 A 1 y ) y n y = Q C ( z n λ 2 A 2 z n ) Q C ( z λ 2 A 2 z ) z n z = Q C ( x n λ 3 A 3 x n ) Q C ( x λ 3 A 3 x ) x n x
(3.2)
and S t n x t n x . It follows from (3.2) that
x n + 1 x = a n f ( x n ) + b n x n + ( 1 a n b n ) S t n x a n f ( x n ) x + b n x n x + ( 1 a n b n ) t n x a n f ( x n ) x + ( 1 a n ) x n x a n α x n x + a n f ( x ) x + ( 1 a n ) x n x = a n f ( x ) x + ( 1 a n ( 1 α ) ) x n x .
By induction, we have
x n + 1 x max { f ( x ) x 1 α , x 1 x } .

Therefore, { x n } is bounded. Hence { y n } , { z n } , { t n } , { A 1 y n } , { A 2 z n } , { S t n } , { f ( x n ) } and { A 3 x n } are also bounded.

Step 2. We show that lim n x n + 1 x n = 0 .

By nonexpansiveness of Q C and I λ i A i ( i = 1 , 2 , 3 ), we have
t n + 1 t n = Q C ( y n + 1 λ 1 A 1 y n + 1 ) Q C ( y n λ 1 A 1 y n ) y n + 1 y n = Q C ( z n + 1 λ 2 A 2 z n + 1 ) Q C ( z n λ 2 A 2 z n ) z n + 1 z n = Q C ( x n + 1 λ 3 A 3 x n + 1 ) Q C ( x n λ 3 A 3 x n ) x n + 1 x n .
(3.3)
Let w n = x n + 1 b n x n 1 b n , n N . Then x n + 1 = b n x n + ( 1 b n ) w n for all n N and
w n + 1 w n = x n + 2 b n + 1 x n + 1 1 b n + 1 x n + 1 b n x n 1 b n = a n + 1 f ( x n + 1 ) + ( 1 a n + 1 b n + 1 ) S t n + 1 1 b n + 1 a n f ( x n ) + ( 1 a n b n ) S t n 1 b n = a n + 1 1 b n + 1 ( f ( x n + 1 ) S t n + 1 ) + a n 1 b n ( S t n f ( x n ) ) + S t n + 1 S t n .
(3.4)
By (3.3), (3.4) and nonexpansiveness of S, we have
w n + 1 w n x n + 1 x n a n + 1 1 b n + 1 f ( x n + 1 ) S t n + 1 + a n 1 b n S t n f ( x n ) .
By this together with (C1) and (C2), we obtain that
lim sup n w n + 1 w n x n + 1 x n 0 .
Hence, by Lemma 2.6, we get x n w n 0 as n . Consequently,
lim n x n + 1 x n = lim n ( 1 b n ) w n x n = 0 .
(3.5)

Step 3. We show that lim n S x n x n = 0 .

Since
x n + 1 x n = a n ( f ( x n ) x n ) + ( 1 a n b n ) ( S t n x n ) ,
therefore
S t n x n 0 as  n .
(3.6)
Next, we prove that lim n x n t n = 0 . From Lemma 2.1 and nonexpansiveness of Q C , we have
z n z 2 = Q C ( x n λ 3 A 3 x n ) Q C ( x λ 3 A 3 x ) 2 x n x λ 3 ( A 3 x n A 3 x ) 2 x n x 2 2 λ 3 A 3 x n A 3 x , j ( x n x ) + 2 K 2 λ 3 2 A 3 x n A 3 x 2 x n x 2 2 λ 3 ( c 3 A 3 x n A 3 x 2 + d 3 x n x 2 ) + 2 K 2 λ 3 2 A 3 x n A 3 x 2 x n x 2 + 2 λ 3 c 3 A 3 x n A 3 x 2 2 λ 3 d 3 L 3 2 A 3 x n A 3 x 2 + 2 K 2 λ 3 2 A 3 x n A 3 x 2 = x n x 2 2 λ 3 ( d 3 L 3 2 c 3 K 2 λ 3 ) A 3 x n A 3 x 2
(3.7)
and
y n y 2 = Q C ( z n λ 2 A 2 z n ) Q C ( z λ 2 A 2 z ) 2 z n z λ 2 ( A 2 z n A 2 z ) 2 z n z 2 2 λ 2 A 2 z n A 2 z , j ( z n z ) + 2 K 2 λ 2 2 A 2 z n A 2 z 2 z n z 2 2 λ 2 ( c 2 A 2 z n A 2 z 2 + d 2 z n z 2 ) + 2 K 2 λ 2 2 A 2 z n A 2 z 2 z n z 2 + 2 λ 2 c 2 A 2 z n A 2 z 2 2 λ 2 d 2 L 2 2 A 2 z n A 2 z 2 + 2 K 2 λ 2 2 A 2 z n A 2 z 2 = z n z 2 2 λ 2 ( d 2 L 2 2 c 2 K 2 λ 2 ) A 2 z n A 2 z 2 .
(3.8)
Similarly, we have
t n x 2 = Q C ( y n λ 1 A 1 y n ) Q C ( y λ 1 A 1 y ) 2 y n y λ 1 ( A 1 y n A 1 y ) 2 y n y 2 2 λ 1 A 1 y n A 1 y , j ( y n y ) + 2 K 2 λ 1 2 A 1 y n A 1 y 2 y n y 2 2 λ 1 ( c 1 A 1 y n A 1 y 2 + d 1 y n y 2 ) + 2 K 2 λ 1 2 A 1 y n A 1 y 2 y n y 2 + 2 λ 1 c 1 A 1 y n A 1 y 2 2 λ 1 d 1 L 1 2 A 1 y n A 1 y 2 + 2 K 2 λ 1 2 A 1 y n A 1 y 2 = y n y 2 2 λ 1 ( d 1 L 1 2 c 1 K 2 λ 1 ) A 1 y n A 1 y 2 .
(3.9)
Substituting (3.7) and (3.8) into (3.9), we have
t n x 2 x n x 2 2 λ 3 ( d 3 L 3 2 c 3 K 2 λ 3 ) A 3 x n A 3 x 2 2 λ 2 ( d 2 L 2 2 c 2 K 2 λ 2 ) A 2 z n A 2 z 2 2 λ 1 ( d 1 L 1 2 c 1 K 2 λ 1 ) A 1 y n A 1 y 2 .
(3.10)
By the convexity of 2 , we obtain
x n + 1 x 2 = a n f ( x n ) + b n x n + ( 1 a n b n ) S t n x 2 a n f ( x n ) x 2 + b n x n x 2 + ( 1 a n b n ) S t n x 2 a n f ( x n ) x 2 + b n x n x 2 + ( 1 a n b n ) t n x 2 .
(3.11)
Substituting (3.10) into (3.11), we have
x n + 1 x 2 a n f ( x n ) x 2 + b n x n x 2 + ( 1 a n b n ) ( x n x 2 2 λ 3 ( d 3 L 3 2 c 3 K 2 λ 3 ) A 3 x n A 3 x 2 2 λ 2 ( d 2 L 2 2 c 2 K 2 λ 2 ) A 2 z n A 2 z 2 2 λ 1 ( d 1 L 1 2 c 1 K 2 λ 1 ) A 1 y n A 1 y 2 ) = a n f ( x n ) x 2 + ( 1 a n ) x n x 2 ( 1 a n b n ) 2 λ 3 ( d 3 L 3 2 c 3 K 2 λ 3 ) A 3 x n A 3 x 2 ( 1 a n b n ) 2 λ 2 ( d 2 L 2 2 c 2 K 2 λ 2 ) A 2 z n A 2 z 2 ( 1 a n b n ) 2 λ 1 ( d 1 L 1 2 c 1 K 2 λ 1 ) A 1 y n A 1 y 2 ,
which implies
( 1 a n b n ) 2 λ 3 ( d 3 L 3 2 c 3 K 2 λ 3 ) A 3 x n A 3 x 2 + ( 1 a n b n ) 2 λ 2 ( d 2 L 2 2 c 2 K 2 λ 2 ) A 2 z n A 2 z 2 + ( 1 a n b n ) 2 λ 1 ( d 1 L 1 2 c 1 K 2 λ 1 ) A 1 y n A 1 y 2 a n f ( x n ) x 2 + x n x 2 x n + 1 x 2 a n f ( x n ) x 2 + x n x n + 1 ( x n x + x n + 1 x ) .
By the conditions (C1), (C2), (3.5) and 0 < λ i < d i c i L i 2 K 2 L i 2 for each i = 1 , 2 , 3 , we obtain
lim n A 3 x n A 3 x = 0 , lim n A 2 z n A 2 z = 0 and lim n A 1 y n A 1 y = 0 .
(3.12)
Let r = sup n 1 { x n x , z n z , y n y , t n x } . By Lemma 2.4(b) and Lemma 2.8, we obtain
t n x 2 = Q C ( y n λ 1 A 1 y n ) Q C ( y λ 1 A 1 y ) 2 y n λ 1 A 1 y n ( y λ 1 A 1 y ) , j ( t n x ) = y n y , j ( t n x ) λ 1 A 1 y n A 1 y , j ( t n x ) 1 2 [ y n y 2 + t n x 2 g ( y n y ( t n x ) ) ] + λ 1 A 1 y A 1 y n , j ( t n x ) ,
which implies
t n x 2 y n y 2 g ( y n y ( t n x ) ) + 2 λ 1 A 1 y A 1 y n , j ( t n x ) y n y 2 g ( y n y ( t n x ) ) + 2 λ 1 A 1 y A 1 y n t n x .
(3.13)
And
y n y 2 = Q C ( z n λ 2 A 2 z n ) Q C ( z λ 2 A 2 z ) 2 z n λ 2 A 2 z n ( z λ 2 A 2 z ) , j ( y n y ) = z n z , j ( y n y ) λ 2 A 2 z n A 2 z , j ( y n y ) 1 2 [ z n z 2 + y n y 2 g ( z n z ( y n y ) ) ] + λ 2 A 2 z A 2 z n , j ( y n y ) ,
which implies
y n y 2 z n z 2 g ( z n z ( y n y ) ) + 2 λ 2 A 2 z A 2 z n , j ( y n y ) z n z 2 g ( z n z ( y n y ) ) + 2 λ 2 A 2 z A 2 z n y n y .
(3.14)
Similarly, we have
z n z 2 = Q C ( x n λ 3 A 3 x n ) Q C ( x λ 3 A 3 x ) 2 x n λ 3 A 3 x n ( x λ 3 A 3 x ) , j ( z n z ) = x n x , j ( z n z ) λ 3 A 3 x n A 3 x , j ( z n z ) 1 2 [ x n x 2 + z n z 2 g ( x n x ( z n z ) ) ] + λ 3 A 3 x A 3 x n , j ( z n z ) ,
which implies
z n z 2 x n x 2 g ( x n x ( z n z ) ) + 2 λ 3 A 3 x A 3 x n , j ( z n z ) x n x 2 g ( x n x ( z n z ) ) + 2 λ 3 A 3 x A 3 x n z n z .
(3.15)
From (3.11), (3.13), (3.14) and (3.15), we have
x n + 1 x 2 a n f ( x n ) x 2 + b n x n x 2 + ( 1 a n b n ) t n x 2 a n f ( x n ) x 2 + b n x n x 2 + ( 1 a n b n ) [ y n y 2 g ( y n y ( t n x ) ) + 2 λ 1 A 1 y A 1 y n t n x ] a n f ( x n ) x 2 + b n x n x 2 + ( 1 a n b n ) [ z n z 2 g ( z n z ( y n y ) ) + 2 λ 2 A 2 z A 2 z n y n y g ( y n y ( t n x ) ) + 2 λ 1 A 1 y A 1 y n t n x ] a n f ( x n ) x 2 + b n x n x 2 + ( 1 a n b n ) [ x n x 2 g ( x n x ( z n z ) ) + 2 λ 3 A 3 x A 3 x n z n z g ( z n z ( y n y ) ) + 2 λ 2 A 2 z A 2 z n y n y g ( y n y ( t n x ) ) + 2 λ 1 A 1 y A 1 y n t n x ] = a n f ( x n ) x 2 + ( 1 a n ) x n x 2 + ( 1 a n b n ) ( 2 λ 1 A 1 y A 1 y n t n x ) + ( 1 a n b n ) ( 2 λ 2 A 2 z A 2 z n y n y ) + ( 1 a n b n ) ( 2 λ 3 A 3 x A 3 x n z n z ) ( 1 a n b n ) g ( y n y ( t n x ) ) ( 1 a n b n ) g ( z n z ( y n y ) ) ( 1 a n b n ) g ( x n x ( z n z ) ) ,
which implies
( 1 a n b n ) g ( y n y ( t n x ) ) + ( 1 a n b n ) g ( z n z ( y n y ) ) + ( 1 a n b n ) g ( x n x ( z n z ) ) a n f ( x n ) x 2 + x n x 2 x n + 1 x 2 + ( 1 a n b n ) ( 2 λ 1 A 1 y A 1 y n t n x ) + ( 1 a n b n ) ( 2 λ 2 A 2 z A 2 z n y n y ) + ( 1 a n b n ) ( 2 λ 3 A 3 x A 3 x n z n z ) a n f ( x n ) x 2 + x n x n + 1 ( x n x + x n + 1 x ) + ( 1 a n b n ) ( 2 λ 1 A 1 y A 1 y n t n x ) + ( 1 a n b n ) ( 2 λ 2 A 2 z A 2 z n y n y ) + ( 1 a n b n ) ( 2 λ 3 A 3 x A 3 x n z n z ) .
By the conditions (C1), (C2), (3.5) and (3.12), we obtain
lim n g ( y n y ( t n x ) ) = 0 , lim n g ( z n z ( y n y ) ) = 0 and lim n g ( x n x ( z n z ) ) = 0 .
It follows from the properties of g that
lim n y n y ( t n x ) = 0 , lim n z n z ( y n y ) = 0 and lim n x n x ( z n z ) = 0 .
Therefore
x n t n x n z n ( x z ) + z n y n ( z y ) + y n t n ( y x ) 0 as  n .
(3.16)
By (3.6) and (3.16), we have
S x n x n S x n S t n + S t n x n x n t n + S t n x n 0 as  n .
(3.17)
Define a mapping W : C C as
W x = η S x + ( 1 η ) G x , x C ,
where η is a constant in ( 0 , 1 ) . Then it follows from Lemma 2.7 that F ( W ) = F ( G ) F ( S ) and W is nonexpansive. From (3.16) and (3.17), we have
x n W x n = x n ( η S x n + ( 1 η ) G x n ) = η ( x n S x n ) + ( 1 η ) ( x n G x n ) η x n S x n + ( 1 η ) x n G x n = η x n S x n + ( 1 η ) x n t n 0 as  n .
(3.18)
Step 4. We claim that
lim sup n f ( q ) q , j ( x n q ) 0 ,
(3.19)
where q = lim t 0 x t with x t being the fixed point of the contraction
x t f ( x ) + ( 1 t ) W x .
From Lemma 2.9, we have q F ( W ) = F ( G ) F ( S ) = Ω and
( I f ) q , j ( q p ) 0 , f Π C , p Ω .
Since x t = t f ( x t ) + ( 1 t ) W x t , we have
x t x n = t f ( x t ) + ( 1 t ) W x t x n = ( 1 t ) ( W x t x n ) + t ( f ( x t ) x n ) .
It follows from (3.18) and Lemma 2.2 that
x t x n 2 = ( 1 t ) ( W x t x n ) + t ( f ( x t ) x n ) 2 ( 1 t ) 2 W x t x n 2 + 2 t f ( x t ) x n , j ( x t x n ) ( 1 t ) 2 ( W x t W x n + W x n x n ) 2 + 2 t f ( x t ) x n , j ( x t x n ) = ( 1 t ) 2 ( W x t W x n 2 + 2 W x t W x n W x n x n + W x n x n 2 ) + 2 t f ( x t ) x t , j ( x t x n ) + 2 t x t x n , j ( x t x n ) ( 1 2 t + t 2 ) x t x n 2 + ( 1 t ) 2 ( 2 x t x n W x n x n + W x n x n 2 ) + 2 t f ( x t ) x t , j ( x t x n ) + 2 t x t x n 2 = ( 1 + t 2 ) x t x n 2 + f n ( t ) + 2 t f ( x t ) x t , j ( x t x n ) ,
(3.20)
where f n ( t ) = ( 1 t ) 2 ( 2 x t x n + W x n x n ) W x n x n 0 as n . It follows from (3.20) that
x t f ( x t ) , j ( x t x n ) t 2 x t x n 2 + f n ( t ) 2 t .
(3.21)
Let n in (3.21), we obtain that
lim sup n x t f ( x t ) , j ( x t x n ) t 2 M ,
(3.22)
where M > 0 is a constant such that M x t x n 2 for all t ( 0 , 1 ) and n 1 . Let t 0 in (3.22), we obtain
lim sup t 0 lim sup n x t f ( x t ) , j ( x t x n ) 0 .
(3.23)
On the other hand, we have
f ( q ) q , j ( x n q ) = f ( q ) q , j ( x n q ) f ( q ) q , j ( x n x t ) + f ( q ) q , j ( x n x t ) f ( q ) x t , j ( x n x t ) + f ( q ) x t , j ( x n x t ) f ( x t ) x t , j ( x n x t ) + f ( x t ) x t , j ( x n x t ) = f ( q ) q , j ( x n q ) j ( x n x t ) + x t q , j ( x n x t ) + f ( q ) f ( x t ) , j ( x n x t ) + f ( x t ) x t , j ( x n x t ) .
It follows that
lim sup n f ( q ) q , j ( x n q ) lim sup n f ( q ) q , j ( x n q ) j ( x n x t ) + x t q lim sup n x n x t + α x t q lim sup n x n x t + lim sup n f ( x t ) x t , j ( x n x t ) .
Noticing that j is norm-to-norm uniformly continuous on a bounded subset of C, it follows from (3.23) and lim t 0 x t = q that
lim sup n f ( q ) q , j ( x n q ) = lim sup t 0 lim sup n f ( q ) q , j ( x n q ) 0 .

Hence (3.19) holds.

Step 5. Finally, we show that x n q as n .

From (3.2), we have
x n + 1 q 2 = x n + 1 q , j ( x n + 1 q ) = a n ( f ( x n ) q ) + b n ( x n q ) + ( 1 a n b n ) ( S t n q ) , j ( x n + 1 q ) = a n f ( x n ) f ( q ) , j ( x n + 1 q ) + b n x n q , j ( x n + 1 q ) + ( 1 a n b n ) S t n q , j ( x n + 1 q ) + a n f ( q ) q , j ( x n + 1 q ) a n α x n q x n + 1 q + b n x n q x n + 1 q + ( 1 a n b n ) S t n q x n + 1 q + a n f ( q ) q , j ( x n + 1 q ) a n α x n q x n + 1 q + b n x n q x n + 1 q + ( 1 a n b n ) x n q x n + 1 q + a n f ( q ) q , j ( x n + 1 q ) = ( 1 a n ( 1 α ) ) x n q x n + 1 q + a n f ( q ) q , j ( x n + 1 q ) 1 a n ( 1 α ) 2 ( x n q 2 + x n + 1 q 2 ) + a n f ( q ) q , j ( x n + 1 q ) 1 a n ( 1 α ) 2 x n q 2 + 1 2 x n + 1 q 2 + a n f ( q ) q , j ( x n + 1 q ) ,
which implies
x n + 1 q 2 ( 1 a n ( 1 α ) ) x n q 2 + a n ( 1 α ) 2 f ( q ) q , j ( x n + 1 q ) 1 α .

It follows from Lemma 2.3, (3.19) and condition (C1) that { x n } converges strongly to q. This completes the proof. □

Example 3.2 Let X = R and C = [ 0 , 1 ] . Define the mappings S , f : C C and A 1 , A 2 , A 3 : C X as follows:
S ( x ) = x 3 , f ( x ) = x 2 + 3 , A 1 ( x ) = x , A 2 ( x ) = 2 x and A 3 ( x ) = 3 x .
Then it is obvious that S is nonexpansive, f is contractive with a constant α = 1 2 , A 1 is relaxed ( 1 2 , 1 ) -cocoercive and 1-Lipschitzian, A 2 is relaxed ( 1 4 , 2 ) -cocoercive and 2-Lipschitzian and A 3 is relaxed ( 1 9 , 3 ) -cocoercive and 3-Lipschitzian. In this case, we have Ω = F ( S ) F ( G ) = { 0 } . In the terms of Theorem 3.1, we choose the parameters λ 1 , λ 2 , λ 3 . Then the sequence { x n } generated by (3.1) converges to q = 0 Ω , which solves the following variational inequality:
q f ( q ) , j ( q p ) 0 , p Ω .

Let A 3 = 0 in Theorem 3.1, we obtain the following result.

Corollary 3.3 Let X be a uniformly convex and 2-uniformly smooth Banach space with the 2-uniformly smooth constant K, let C be a nonempty closed convex subset of X and Q C a sunny nonexpansive retraction from X onto C. Let the mappings A i : C X be relaxed ( c i , d i ) -cocoercive and L i -Lipschitzian with 0 < λ i < d i c i L i 2 K 2 L i 2 , for all i = 1 , 2 . Let f be a contractive mapping with the constant α ( 0 , 1 ) and let S : C C be a nonexpansive mapping such that F = F ( S ) Ω 2 , where Ω 2 is the set of solutions of problem (1.4). For a given x 1 C , let { x n } and { y n } be the sequences generated by
{ y n = Q C ( x n λ 2 A 2 x n ) , x n + 1 = a n f ( x n ) + b n x n + ( 1 a n b n ) S Q C ( y n λ 1 A 1 y n ) , n 1 ,

where { a n } and { b n } are two sequences in ( 0 , 1 ) such that

(C1) lim n a n = 0 and n = 1 a n = ;

(C2) 0 < lim inf n b n lim sup n b n < 1 .

Then { x n } converges strongly to q F , which solves the following variational inequality:
q f ( q ) , j ( q p ) 0 , f Π C , p F .
Remark 3.4 (i) Since L p for all p 2 is uniformly convex and 2-uniformly smooth, we see that Theorem 3.1 is applicable to L p for all p 2 .
  1. (ii)

    The problem of finding solutions for a finite number of variational inequalities can use the same idea of a new general system of variational inequalities in Banach spaces.

     

Declarations

Acknowledgements

The author would like to thank Professor Dr. Suthep Suantai and the reviewer for careful reading, valuable comment and suggestions on this paper. This research is partially supported by the Center of Excellence in Mathematics, the Commission on Higher Education, Thailand. The author also thanks the Thailand Research Fund and Thaksin university for their financial support.

Authors’ Affiliations

(1)
Department of Mathematics and Statistics, Faculty of Science, Thaksin University
(2)
Centre of Excellence in Mathematics, CHE

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© Imnang; licensee Springer. 2013

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