Skip to content

Advertisement

  • Research
  • Open Access

On solutions of a system of variational inequalities and fixed point problems in Banach spaces

Fixed Point Theory and Applications20132013:176

https://doi.org/10.1186/1687-1812-2013-176

  • Received: 10 April 2013
  • Accepted: 17 June 2013
  • Published:

Abstract

In this paper, considering the problem of solving a system of variational inequalities and a common fixed point problem of an infinite family of nonexpansive mappings in Banach spaces, we propose a two-step relaxed extragradient method which is based on Korpelevich’s extragradient method and viscosity approximation method. Strong convergence results are established.

MSC:49J30, 47H09, 47J20.

Keywords

  • two-step relaxed extragradient method
  • system of variational inequalities
  • nonexpansive mapping
  • sunny nonexpansive retraction
  • fixed point
  • uniform smoothness
  • uniform convexity

1 Introduction

In the last three decades, the theory of variational inequalities has been used as a tool to study the Nash equilibrium problem for a finite or infinite number of players; see, for example, [16] and the references therein. There are two ways to study the Nash equilibrium problem by using variational inequality technique: (1) system of variational inequalities; (2) variational inequalities defined over the product of sets. If the number of players is finite, then the system of variational inequalities is equivalent to the variational inequality defined over the product of sets; see, for example, [7, 8] and the references therein.

Very recently, Cai and Bu [9] considered the following system of two variational inequalities in the setting of Banach spaces.

Let C be a nonempty, closed and convex subset of a real Banach space X, let B 1 , B 2 : C X be two nonlinear mappings and μ 1 and μ 2 be two positive constants. The problem of system of variational inequalities (SVI) [9] is to find ( x , y ) C × C such that
{ μ 1 B 1 y + x y , J ( x x ) 0 , x C , μ 2 B 2 x + y x , J ( x y ) 0 , x C ,
(1.1)

where J is the normalized duality mapping. The set of solutions of GSVI (1.1) is denoted by GSVI ( C , B 1 , B 2 ) . This system could be useful to study the Nash equilibrium problem for two players. They proposed an iterative scheme to compute the approximate solutions of such a system.

In particular, if X = H , a real Hilbert space, then GSVI (1.1) reduces to the following problem of a system of variational inequalities of finding ( x , y ) C × C such that
{ μ 1 B 1 y + x y , x x 0 , x C , μ 2 B 2 x + y x , x y 0 , x C ,
(1.2)

where μ 1 and μ 2 are two positive constants. The set of solutions of problem (1.2) is still denoted by GSVI ( C , B 1 , B 2 ) .

In this paper, we introduce two-step relaxed extragradient method for solving SVI (1.1) and the common fixed point problem of an infinite family { S n } of nonexpansive mappings of C into itself. Here, the two-step relaxed extragradient method is based on Korpelevich’s extragradient method [10] and viscosity approximation method. We first suggest and analyze an implicit iterative algorithm by the two-step relaxed extragradient method in a uniformly convex and 2-uniformly smooth Banach space X, and then another explicit iterative algorithm in a uniformly convex Banach space X with a uniformly Gâteaux differentiable norm. On the other hand, we also propose and analyze a composite explicit iterative algorithm by the two-step relaxed extragradient method for solving SVI (1.1) and the common fixed point problem of { S n } in a uniformly convex and 2-uniformly smooth Banach space. The results presented in this paper improve, extend, supplement and develop the corresponding results that have appeared very recently in the literature.

2 Preliminaries

Let X be the dual of X. The normalized duality mapping J : X 2 X is defined by
J ( x ) = { x X : x , x = x 2 = x 2 } , x X ,

where , denotes the generalized duality pairing.

Let C be a nonempty closed convex subset of a real Banach space X. A mapping A : C X is said to be accretive if for each x , y C there exists j ( x y ) J ( x y ) such that
A x A y , j ( x y ) 0 ,
where J is the normalized duality mapping. A is said to be α-strongly accretive if for each x , y C there exists j ( x y ) J ( x y ) such that
A x A y , j ( x y ) α x y 2
for some α ( 0 , 1 ) . It is said to be β-inverse-strongly-accretive if for each x , y C there exists j ( x y ) J ( x y ) such that
A x A y , j ( x y ) β A x A y 2
for some β > 0 ; and finally A is said to be λ-strictly pseudocontractive if for each x , y C there exists j ( x y ) J ( x y ) such that
A x A y , j ( x y ) x y 2 λ x y ( A x A y ) 2

for some λ ( 0 , 1 ) .

Let D be a subset of C and let Π be a mapping of C into D. Then Π is said to be sunny if
Π [ Π ( x ) + t ( x Π ( x ) ) ] = Π ( x ) ,

whenever Π ( x ) + t ( x Π ( x ) ) C for x C and t 0 . A mapping Π of C into itself is called a retraction if Π 2 = Π . If a mapping Π of C into itself is a retraction, then Π ( z ) = z for every z R ( Π ) , where R ( Π ) is the range of Π. A subset D of C is called a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction from C onto D.

It is well known that if X = H , a Hilbert space, then a sunny nonexpansive retraction Π C is coincident with the metric projection from X onto C; that is, Π C = P C . If C is a nonempty closed convex subset of a strictly convex and uniformly smooth Banach space X and if T : C C is a nonexpansive mapping with the fixed point set Fix ( T ) , then the set Fix ( T ) is a sunny nonexpansive retract of C.

The following lemma concerns the sunny nonexpansive retraction.

Lemma 2.1 (see [11])

Let C be a nonempty closed convex subset of a real smooth Banach space X. Let D be a nonempty subset of C. Let Π be a retraction of C onto D and let J be a normalized duality mapping on X. Then the following are equivalent:
  1. (i)

    Π is sunny and nonexpansive;

     
  2. (ii)

    Π ( x ) Π ( y ) 2 x y , J ( Π ( x ) Π ( y ) ) , x , y C ;

     
  3. (iii)

    x Π ( x ) , J ( y Π ( x ) ) 0 , x C , y D .

     

Next, we present some more lemmas which are crucial for the proofs of our results.

Lemma 2.2 (see [12])

Let { s n } be a sequence of nonnegative real numbers satisfying
s n + 1 ( 1 α n ) s n + α n β n + γ n , n 0 ,
where { α n } , { β n } and { γ n } satisfy the conditions:
  1. (i)

    { α n } [ 0 , 1 ] and n = 0 α n = ;

     
  2. (ii)

    lim sup n β n 0 ;

     
  3. (iii)

    γ n 0 , n 0 , and n = 0 γ n < .

     

Then lim sup n s n = 0 .

Lemma 2.3 (see [12])

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

Lemma 2.4 (see [13])

Let { x n } and { z n } be bounded sequences in a Banach space X and let { α n } be a sequence in [ 0 , 1 ] which satisfies the following condition:
0 < lim inf n α n lim sup n α n < 1 .

Suppose x n + 1 = α n x n + ( 1 α n ) z n , n 0 and lim sup n ( z n + 1 z n x n + 1 x n ) 0 . Then lim n z n x n = 0 .

Lemma 2.5 (see [14])

Given a number r > 0 . A real Banach space X is uniformly convex if and only if there exists a continuous strictly increasing function g : [ 0 , ) [ 0 , ) , g ( 0 ) = 0 , such that
λ x + ( 1 λ ) y 2 λ x 2 + ( 1 λ ) y 2 λ ( 1 λ ) g ( x y )

for all λ [ 0 , 1 ] and x , y X such that x r and y r .

Lemma 2.6 (see [15])

Let C be a nonempty closed convex subset of a Banach space X. Let S 0 , S 1 , be a sequence of mappings of C into itself. Suppose that n = 1 sup { S n x S n 1 x : x C } < . Then, for each y C , { S n y } converges strongly to some point of C. Moreover, let S be a mapping of C into itself defined by S y = lim n S n y for all y C . Then lim n sup { S x S n x : x C } = 0 .

Let C be a nonempty closed convex subset of a Banach space X and let T : C C be a nonexpansive mapping with Fix ( T ) . As previously, let Ξ C be a set of all contractions on C. For t ( 0 , 1 ) and f Ξ C , let x t C be a unique fixed point of the contraction x t f ( x ) + ( 1 t ) T x on C; that is,
x t = t f ( x t ) + ( 1 t ) T x t .

Lemma 2.7 (see [16])

Let X be a uniformly smooth Banach space, or a reflexive and strictly convex Banach space with a uniformly Gateaux differentiable norm. Let C be a nonempty closed convex subset of X, let T : C C be a nonexpansive mapping with Fix ( T ) , and f Ξ C . Then the net { x t } defined by x t = t f ( x t ) + ( 1 t ) T x t converges strongly to a point in Fix ( T ) . If we define a mapping Q : Ξ C Fix ( T ) by Q ( f ) : = s lim t 0 x t , f Ξ C , then Q ( f ) solves the VIP:
( I f ) Q ( f ) , J ( Q ( f ) p ) 0 , f Ξ C , p Fix ( T ) .

Lemma 2.8 (see [17])

Let C be a nonempty closed convex subset of a strictly convex Banach space X. Let { T n } n = 0 be a sequence of nonexpansive mappings on C. Suppose n = 0 Fix ( T n ) is nonempty. Let { λ n } be a sequence of positive numbers with n = 0 λ n = 1 . Then a mapping S on C defined by S x = n = 0 λ n T n x for x C is well defined, nonexpansive and Fix ( S ) = n = 0 Fix ( T n ) holds.

Lemma 2.9 (see [12])

Let C be a nonempty closed convex subset of a smooth Banach space X and let the mapping B i : C X be λ i -strictly pseudocontractive and α i -strongly accretive with α i + λ i 1 for i = 1 , 2 . Then, for μ i ( 0 , 1 ] , we have
( I μ i B i ) x ( I μ i B i ) y { 1 α i λ i + ( 1 μ i ) ( 1 + 1 λ i ) } x y , x , y C ,

for i = 1 , 2 . In particular, if 1 λ i 1 + λ i ( 1 1 α i λ i ) μ i 1 , then I μ i B i is nonexpansive for i = 1 , 2 .

Lemma 2.10 (see [9])

Let C be a nonempty closed convex subset of a smooth Banach space X. Let Π C be a sunny nonexpansive retraction from X onto C and let the mapping B i : C X be λ i -strictly pseudocontractive and α i -strongly accretive with α i + λ i 1 for i = 1 , 2 . Let G : C C be a mapping defined by
G ( x ) = Π C [ Π C ( x μ 2 B 2 x ) μ 1 B 1 Π C ( x μ 2 B 2 x ) ] , x C .

If 1 λ i 1 + λ i ( 1 1 α i λ i ) μ i 1 , then G : C C is nonexpansive.

Lemma 2.11 (see [9])

Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space X. Let the mapping B i : C X be α i -inverse-strongly accretive. Then we have
( I μ i B i ) x ( I μ i B i ) y x y 2 + 2 μ i ( μ i κ 2 α i ) x y , x , y C ,

for i = 1 , 2 , where μ i > 0 . In particular, if 0 < μ i α i κ 2 , then I μ i B i is nonexpansive for i = 1 , 2 .

Lemma 2.12 (see [9])

Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space X. Let Π C be a sunny nonexpansive retraction from X onto C. Let the mapping B i : C X be α i -inverse-strongly accretive for i = 1 , 2 . Let ψ : C C be the mapping defined by
ψ ( x ) = Π C [ Π C ( x μ 2 B 2 x ) μ 1 B 1 Π C ( x μ 2 B 2 x ) ] , x C .

If 0 < μ i α i κ 2 for i = 1 , 2 , then ψ : C C is nonexpansive.

Lemma 2.13 (see [12])

Let C be a nonempty closed convex subset of a smooth Banach space X. Let Π C be a sunny nonexpansive retraction from X onto C and let the mapping B i : C X be λ i -strictly pseudocontractive and α i -strongly accretive for i = 1 , 2 . For given x , y C , ( x , y ) is a solution of GSVI (1.1) if and only if x = Π C ( y μ 1 B 1 y ) , where y = Π C ( x μ 2 B 2 x ) .

By Lemma 2.12, we observe that
x = Π C [ Π C ( x μ 2 B 2 x ) μ 1 B 1 Π C ( x μ 2 B 2 x ) ] ,

which implies that x is a fixed point of the mapping G = Π C ( I μ 1 B 1 ) Π C ( I μ 2 B 2 ) .

Proposition 2.1 (see [18])

Let X be a real smooth and uniform convex Banach space and let r > 0 . Then there exists a strictly increasing, continuous and convex function , g ( 0 ) = 0 such that
g ( x y ) x 2 2 x , J ( y ) + y 2 , x , y B r ,

where B r = { x X : x r } .

3 Two-step relaxed extragradient algorithms

In this section, we first suggest and analyze an implicit iterative algorithm by the two-step relaxed extragradient method in the setting of uniformly convex and 2-uniformly smooth Banach spaces, and then another explicit iterative algorithm in the setting of uniformly convex Banach spaces with a uniformly Gateaux differentiable norm.

Theorem 3.1 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let Π C be a sunny nonexpansive retraction from X onto C. Let the mapping B i : C X be α i -inverse-strongly accretive for i = 1 , 2 . Let f : C C be a contraction with coefficient ρ ( 0 , 1 ) . Let { S n } n = 0 be an infinite family of nonexpansive mappings of C into itself such that F = i = 0 Fix ( S i ) Ω , where Ω is a fixed point set of the mapping G. For arbitrarily given x 0 C , let { x n } be a sequence generated by
{ y n = α n f ( y n ) + ( 1 α n ) Π C ( I μ 1 B 1 ) Π C ( I μ 2 B 2 ) x n , x n + 1 = β n x n + ( 1 β n ) S n y n , n 0 ,
(3.1)
where 0 < μ i < α i κ 2 for i = 1 , 2 . Suppose that { α n } and { β n } are sequences in ( 0 , 1 ) satisfying the following conditions:
  1. (i)

    lim n α n = 0 and n = 0 α n = ;

     
  2. (ii)

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

     
Assume that n = 1 sup x D S n x S n 1 x < for any bounded subset D of C and let S be a mapping of C into itself defined by S x = lim n S n x for all x C and suppose that Fix ( S ) = i = 0 Fix ( S i ) . Then { x n } converges strongly to q F , which solves the following VIP:
q f ( q ) , J ( q p ) 0 , p F .
Proof It is easy to see that scheme (3.1) can be rewritten as
{ y n = α n f ( y n ) + ( 1 α n ) G ( x n ) , x n + 1 = β n x n + ( 1 β n ) S n y n , n 0 .
(3.2)
Take a fixed p F arbitrarily. Then by Lemma 2.13 we know that p = G ( p ) . Moreover, by Lemma 2.12 we have
y n p = α n ( f ( y n ) p ) + ( 1 α n ) ( G ( x n ) p ) α n f ( y n ) f ( p ) + α n f ( p ) p + ( 1 α n ) G ( x n ) p α n ρ y n p + α n f ( p ) p + ( 1 α n ) x n p ,
which hence implies that
y n p ( 1 1 ρ 1 α n ρ α n ) x n p + 1 1 α n ρ α n f ( p ) p .
(3.3)
Thus, from (3.2) we have
x n + 1 p = β n ( x n p ) + ( 1 β n ) ( S n y n p ) β n x n p + ( 1 β n ) S n y n p β n x n p + ( 1 β n ) y n p β n x n p + ( 1 β n ) { ( 1 1 ρ 1 α n ρ α n ) x n p + 1 1 α n ρ α n f ( p ) p } = [ 1 ( 1 β n ) ( 1 ρ ) 1 α n ρ α n ] x n p + ( 1 β n ) ( 1 ρ ) 1 α n ρ α n f ( p ) p 1 ρ max { x 0 p , f ( p ) p 1 ρ } .

It immediately follows that { x n } is bounded, and so are the sequences { y n } , { G ( x n ) } due to (3.3) and the nonexpansivity of G.

Let us show that x n + 1 x n 0 as n . As a matter of fact, from (3.2) we have
{ y n = α n f ( y n ) + ( 1 α n ) G ( x n ) , y n 1 = α n 1 f ( y n 1 ) + ( 1 α n 1 ) G ( x n 1 ) , n 1 .
Simple calculations show that
y n y n 1 = α n ( f ( y n ) f ( y n 1 ) ) + ( α n α n 1 ) ( f ( y n 1 ) G ( x n 1 ) ) + ( 1 α n ) ( G ( x n ) G ( x n 1 ) ) .
It follows that
y n y n 1 α n f ( y n ) f ( y n 1 ) + | α n α n 1 | f ( y n 1 ) G ( x n 1 ) + ( 1 α n ) G ( x n ) G ( x n 1 ) α n ρ y n y n 1 + | α n α n 1 | f ( y n 1 ) G ( x n 1 ) + ( 1 α n ) x n x n 1 ,
which hence yields
y n y n 1 ( 1 1 ρ 1 α n ρ α n ) x n x n 1 + | α n α n 1 | 1 α n ρ f ( y n 1 ) G ( x n 1 ) .
(3.4)
Thus we have from (3.4)
S n y n S n 1 y n 1 S n y n S n y n 1 + S n y n 1 S n 1 y n 1 y n y n 1 + S n y n 1 S n 1 y n 1 ( 1 1 ρ 1 α n ρ α n ) x n x n 1 + | α n α n 1 | 1 α n ρ f ( y n 1 ) G ( x n 1 ) + S n y n 1 S n 1 y n 1 x n x n 1 + | α n α n 1 | 1 α n ρ f ( y n 1 ) G ( x n 1 ) + S n y n 1 S n 1 y n 1 ,
which implies that
S n y n S n 1 y n 1 x n x n 1 | α n α n 1 | 1 α n ρ f ( y n 1 ) G ( x n 1 ) + S n y n 1 S n 1 y n 1 .
From condition (i) and the assumption on { S n } , we have
lim sup n ( S n y n S n 1 y n 1 x n x n 1 ) 0 .
It follows from Lemma 2.4 that
lim n S n y n x n = 0 .
(3.5)
Hence we obtain
lim n x n + 1 x n = lim n ( 1 β n ) S n y n x n = 0 .
(3.6)

Next we show that x n G ( x n ) 0 as n .

For simplicity, put q = Π C ( p μ 2 B 2 p ) , u n = Π C ( x n μ 2 B 2 x n ) and v n = Π C ( u n μ 1 B 1 u n ) . Then v n = G ( x n ) . From Lemma 2.11 we have
u n q 2 = Π C ( x n μ 2 B 2 x n ) Π C ( p μ 2 B 2 p ) 2 x n p μ 2 ( B 2 x n B 2 p ) 2 x n p 2 2 μ 2 ( α 2 κ 2 μ 2 ) B 2 x n B 2 p 2
(3.7)
and
v n p 2 = Π C ( u n μ 1 B 1 u n ) Π C ( q μ 1 B 1 q ) 2 u n q μ 1 ( B 1 u n B 1 q ) 2 u n q 2 2 μ 1 ( α 1 κ 2 μ 1 ) B 1 u n B 1 q 2 .
(3.8)
Substituting (3.7) into (3.8), we obtain
v n p 2 x n p 2 2 μ 2 ( α 2 κ 2 μ 2 ) B 2 x n B 2 p 2 2 μ 1 ( α 1 κ 2 μ 1 ) B 1 u n B 1 q 2 .
(3.9)
According to Lemma 2.3, we have from (3.2)
y n p 2 = α n ( f ( y n ) f ( p ) ) + ( 1 α n ) ( v n p ) + α n ( f ( p ) p ) 2 α n ( f ( y n ) f ( p ) ) + ( 1 α n ) ( v n p ) 2 + 2 α n f ( p ) p , J ( y n p ) α n f ( y n ) f ( p ) 2 + ( 1 α n ) v n p 2 + 2 α n f ( p ) p , J ( y n p ) α n ρ y n p 2 + ( 1 α n ) v n p 2 + 2 α n f ( p ) p y n p ,
which hence yields
y n p 2 ( 1 1 ρ 1 α n ρ α n ) v n p 2 + 2 α n 1 α n ρ f ( p ) p y n p .
From this together with (3.9) and the convexity of 2 we have
x n + 1 p 2 = β n ( x n p ) + ( 1 β n ) ( S n y n p ) 2 β n x n p 2 + ( 1 β n ) S n y n p 2 β n x n p 2 + ( 1 β n ) y n p 2 β n x n p 2 + ( 1 β n ) { ( 1 1 ρ 1 α n ρ α n ) v n p 2 + 2 α n 1 α n ρ f ( p ) p y n p } β n x n p 2 + ( 1 β n ) ( 1 1 ρ 1 α n ρ α n ) v n p 2 + α n M 1 β n x n p 2 + ( 1 β n ) ( 1 1 ρ 1 α n ρ α n ) [ x n p 2 2 μ 2 ( α 2 κ 2 μ 2 ) B 2 x n B 2 p 2 2 μ 1 ( α 1 κ 2 μ 1 ) B 1 u n B 1 q 2 ] + α n M 1 = ( 1 ( 1 β n ) ( 1 ρ ) 1 α n ρ α n ) x n p 2 2 ( 1 β n ) ( 1 1 ρ 1 α n ρ α n ) [ μ 2 ( α 2 κ 2 μ 2 ) B 2 x n B 2 p 2 + μ 1 ( α 1 κ 2 μ 1 ) B 1 u n B 1 q 2 ] + α n M 1 x n p 2 2 ( 1 β n ) ( 1 1 ρ 1 α n ρ α n ) [ μ 2 ( α 2 κ 2 μ 2 ) B 2 x n B 2 p 2 + μ 1 ( α 1 κ 2 μ 1 ) B 1 u n B 1 q 2 ] + α n M 1 ,
(3.10)
where sup n 0 { 2 ( 1 β n ) 1 α n ρ f ( p ) p y n p } M 1 for some M 1 > 0 . So, it follows that
2 ( 1 β n ) ( 1 1 ρ 1 α n ρ α n ) × [ μ 2 ( α 2 κ 2 μ 2 ) B 2 x n B 2 p 2 + μ 1 ( α 1 κ 2 μ 1 ) B 1 u n B 1 q 2 ] x n p 2 x n + 1 p 2 + α n M 1 ( x n p + x n + 1 p ) x n x n + 1 + α n M 1 .
Since 0 < μ i < α i κ 2 for i = 1 , 2 , from conditions (i), (ii) and (3.6) we obtain
lim n B 2 x n B 2 p = 0 and lim n B 1 u n B 1 q = 0 .
(3.11)
Utilizing Proposition 2.1 and Lemma 2.1, we have
u n q 2 = Π C ( x n μ 2 B 2 x n ) Π C ( p μ 2 B 2 p ) 2 x n μ 2 B 2 x n ( p μ 2 B 2 p ) , J ( u n q ) = x n p , J ( u n q ) + μ 2 B 2 p B 2 x n , J ( u n q ) 1 2 [ x n p 2 + u n q 2 g 1 ( x n u n ( p q ) ) ] + μ 2 B 2 p B 2 x n u n q ,
which implies that
u n q 2 x n p 2 g 1 ( x n u n ( p q ) ) + 2 μ 2 B 2 p B 2 x n u n q .
(3.12)
In the same way, we derive
v n p 2 = Π C ( u n μ 1 B 1 u n ) Π C ( q μ 1 B 1 q ) 2 u n μ 1 B 1 u n ( q μ 1 B 1 q ) , J ( v n p ) = u n q , J ( v n p ) + μ 1 B 1 q B 1 u n , J ( v n p ) 1 2 [ u n q 2 + v n p 2 g 2 ( u n v n + ( p q ) ) ] + μ 1 B 1 q B 1 u n v n p ,
which implies that
v n p 2 u n q 2 g 2 ( u n v n + ( p q ) ) + 2 μ 1 B 1 q B 1 u n v n p .
(3.13)
Substituting (3.12) into (3.13), we get
v n p 2 x n p 2 g 1 ( x n u n ( p q ) ) g 2 ( u n v n + ( p q ) ) + 2 μ 2 B 2 p B 2 x n u n q + 2 μ 1 B 1 q B 1 u n v n p .
(3.14)
From (3.10) and (3.14), we have
x n + 1 p 2 α n M 1 + β n x n p 2 + ( 1 β n ) ( 1 1 ρ 1 α n ρ α n ) [ x n p 2 g 1 ( x n u n ( p q ) ) g 2 ( u n v n + ( p q ) ) + 2 μ 2 B 2 p B 2 x n u n q + 2 μ 1 B 1 q B 1 u n v n p ] α n M 1 + ( 1 ( 1 β n ) ( 1 ρ ) 1 α n ρ α n ) x n p 2 ( 1 β n ) ( 1 1 ρ 1 α n ρ α n ) [ g 1 ( x n u n ( p q ) ) + g 2 ( u n v n + ( p q ) ) ] + 2 μ 2 B 2 p B 2 x n u n q + 2 μ 1 B 1 q B 1 u n v n p α n M 1 + x n p 2 ( 1 β n ) ( 1 1 ρ 1 α n ρ α n ) [ g 1 ( x n u n ( p q ) ) + g 2 ( u n v n + ( p q ) ) ] + 2 μ 2 B 2 p B 2 x n u n q + 2 μ 1 B 1 q B 1 u n v n p ,
which implies that
( 1 β n ) ( 1 1 ρ 1 α n ρ α n ) [ g 1 ( x n u n ( p q ) ) + g 2 ( u n v n + ( p q ) ) ] α n M 1 + x n p 2 x n + 1 p 2 + 2 μ 2 B 2 p B 2 x n u n q + 2 μ 1 B 1 q B 1 u n v n p α n M 1 + ( x n p + x n + 1 p ) x n x n + 1 + 2 μ 2 B 2 p B 2 x n u n q + 2 μ 1 B 1 q B 1 u n v n p .
Utilizing conditions (i), (ii), from (3.6) and (3.11) we have
lim n g 1 ( x n u n ( p q ) ) = 0 , lim n g 2 ( u n v n + ( p q ) ) = 0 .
(3.15)
Utilizing the properties of g 1 and g 2 , we deduce that
lim n x n u n ( p q ) = 0 , lim n u n v n + ( p q ) = 0 .
(3.16)
From (3.16) we obtain
x n v n x n u n ( p q ) + u n v n + ( p q ) 0 as  n .
That is,
lim n x n G ( x n ) = 0 .
(3.17)
On the other hand, we observe that
y n G ( x n ) = α n ( f ( y n ) G ( x n ) ) .
Since α n 0 as n , we have
lim n y n G ( x n ) = 0 .
(3.18)
We note that
S n G ( x n ) G ( x n ) S n G ( x n ) S n y n + S n y n x n + x n G ( x n ) G ( x n ) y n + S n y n x n + x n G ( x n ) .
From (3.5), (3.17) and (3.18), we obtain that
lim n S n G ( x n ) G ( x n ) = 0 .
(3.19)
By (3.19) and Lemma 2.6, we have
S G ( x n ) G ( x n ) S G ( x n ) S n G ( x n ) + S n G ( x n ) G ( x n ) 0 as  n .
(3.20)
In terms of (3.17) and (3.20), we have
x n S x n x n G ( x n ) + G ( x n ) S G ( x n ) + S G ( x n ) S x n 2 x n G ( x n ) + G ( x n ) S G ( x n ) 0 as  n .
(3.21)
Define a mapping W x = ( 1 θ ) S x + θ G ( x ) , θ ( 0 , 1 ) is a constant. Then by Lemma 2.8 we have that Fix ( W ) = Fix ( S ) Fix ( G ) = F . We observe that
x n W x n = ( 1 θ ) ( x n S x n ) + θ ( x n G ( x n ) ) ( 1 θ ) x n S x n + θ x n G ( x n ) .
From (3.17) and (3.21), we obtain
lim n x n W x n = 0 .
(3.22)
Now, we claim that
lim sup n f ( q ) q , J ( x n q ) 0 ,
(3.23)
where q = s lim t 0 x t with x t being a fixed point of the contraction
x t f ( x ) + ( 1 t ) W x .
Then x t solves the fixed point equation x t = t f ( x t ) + ( 1 t ) W x t . Thus we have
x t x n = ( 1 t ) ( W x t x n ) + t ( f ( x t ) x n ) .
By Lemma 2.3 we conclude 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 ( x t x n + W x n x n ) 2 + 2 t f ( x t ) x n , J ( x t x n ) = ( 1 t ) 2 [ x t x n 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 , J ( x t x n ) = ( 1 2 t + t 2 ) x t x n 2 + f n ( t ) + 2 t f ( x t ) x t , J ( x t x n ) + 2 t x t x n 2 ,
(3.24)
where
f n ( t ) = ( 1 t ) 2 ( 2 x t x n + x n W x n ) x n W x n 0 as  n .
(3.25)
It follows from (3.24) that
x t f ( x t ) , J ( x t x n ) t 2 x t x n 2 + 1 2 t f n ( t ) .
(3.26)
Letting n in (3.26) and noticing (3.25), we derive
lim sup n x t f ( x t ) , J ( x t x n ) t 2 M 2 ,
(3.27)
where M 2 > 0 is a constant such that x t x n 2 M 2 for all t ( 0 , 1 ) and n 0 . Taking t 0 in (3.27), we have
lim sup t 0 lim sup n x t f ( x t ) , J ( x t x n ) 0 .
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 + ρ q x t lim sup n x n x t + lim sup n f ( x t ) x t , J ( x n x t ) .
Taking into account that x t q as t 0 , we have
lim sup n f ( q ) q , J ( x n q ) = lim sup t 0 lim sup n f ( q ) q , J ( x n q ) lim sup t 0 lim sup n f ( q ) q , J ( x n q ) J ( x n x t ) .
(3.28)
Since X has a uniformly Fréchet differentiable norm, the duality mapping J is norm-to-norm uniformly continuous on bounded subsets of X. Consequently, the two limits are interchangeable and hence (3.23) holds. From (3.17) and (3.18) we get ( y n q ) ( x n q ) 0 . Noticing that J is norm-to-norm uniformly continuous on bounded subsets of X, we deduce from (3.23) that
lim sup n f ( q ) q , J ( y n q ) = lim sup n ( f ( q ) q , J ( x n q ) + f ( q ) q , J ( y n q ) J ( x n q ) ) = lim sup n f ( q ) q , J ( x n q ) 0 .
(3.29)
Finally, let us show that x n q as n . We observe that
y n q 2 = α n ( f ( y n ) f ( q ) ) + ( 1 α n ) ( G ( x n ) q ) + α n ( f ( q ) q ) 2 α n ( f ( y n ) f ( q ) ) + ( 1 α n ) ( G ( x n ) q ) 2 + 2 α n f ( q ) q , J ( y n q ) α n f ( y n ) f ( q ) 2 + ( 1 α n ) G ( x n ) q 2 + 2 α n f ( q ) q , J ( y n q ) α n ρ y n q 2 + ( 1 α n ) x n q 2 + 2 α n f ( q ) q , J ( y n q ) ,
which implies that
y n q 2 ( 1 1 ρ 1 α n ρ α n ) x n q 2 + α n ( 1 ρ ) 1 α n ρ 2 f ( q ) q , J ( y n q ) 1 ρ .
(3.30)
By the convexity of 2 and (3.2), we get
x n + 1 q 2 β n x n q 2 + ( 1 β n ) y n q 2 ,
which together with (3.30) leads to
x n + 1 q 2 β n x n q 2 + ( 1 β n ) { ( 1 1 ρ 1 α n ρ α n ) x n q 2 + α n ( 1 ρ ) 1 α n ρ 2 f ( q ) q , J ( y n q ) 1 ρ } = [ 1 ( 1 β n ) ( 1 ρ ) 1 α n ρ α n ] x n q 2 + ( 1 β n ) ( 1 ρ ) 1 α n ρ α n 2 f ( q ) q , J ( y n q ) 1 ρ .
(3.31)

Applying Lemma 2.2 to (3.31), we obtain that x n q as n . This completes the proof. □

Corollary 3.1 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let Π C be a sunny nonexpansive retraction from X onto C. Let the mapping B i : C X be α i -inverse-strongly accretive for i = 1 , 2 . Let f : C C be a contraction with coefficient ρ ( 0 , 1 ) . Let S be a nonexpansive mapping of C into itself such that F = Fix ( S ) Ω , where Ω is the fixed point set of the mapping G. For arbitrarily given x 0 C , let { x n } be a sequence generated by
{ y n = α n f ( y n ) + ( 1 α n ) Π C ( I μ 1 B 1 ) Π C ( I μ 2 B 2 ) x n , x n + 1 = β n x n + ( 1 β n ) S y n , n 0 ,
where 0 < μ i < α i κ 2 for i = 1 , 2 . Suppose that { α n } and { β n } are sequences in ( 0 , 1 ) satisfying the following conditions:
  1. (i)

    lim n α n = 0 and n = 0 α n = ;

     
  2. (ii)

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

     
Then { x n } converges strongly to q F , which solves the following VIP:
q f ( q ) , J ( q p ) 0 , p F .
Theorem 3.2 Let C be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gâteaux differentiable norm. Let Π C be a sunny nonexpansive retraction from X onto C. Let the mapping B i : C X be λ i -strictly pseudocontractive and α i -strongly accretive with α i + λ i 1 for i = 1 , 2 . Let f : C C be a contraction with coefficient ρ ( 0 , 1 ) . Let { S n } n = 0 be an infinite family of nonexpansive mappings of C into itself such that F = i = 0 Fix ( S i ) Ω , where Ω is a fixed point set of the mapping G. For arbitrarily given x 0 C , let { x n } be a sequence generated by
{ y n = α n x n + ( 1 α n ) Π C ( I μ 1 B 1 ) Π C ( I μ 2 B 2 ) x n , x n + 1 = β n f ( x n ) + ( 1 β n ) S n y n , n 0 ,
(3.32)
where 1 λ i 1 + λ i ( 1 1 α i λ i ) μ i 1 for i = 1 , 2 . Suppose that { α n } and { β n } are sequences in ( 0 , 1 ) satisfying the following conditions:
  1. (i)

    0 < lim inf n α n lim sup n α n < 1 ;

     
  2. (ii)

    lim n β n = 0 and n = 0 β n = ;

     
  3. (iii)

    n = 1 | α n α n 1 | < or lim n | α n α n 1 | / β n = 0 ;

     
  4. (iv)

    n = 1 | β n β n 1 | < or lim n β n 1 / β n = 1 .

     
Assume that n = 1 sup x D S n x S n 1 x < for any bounded subset D of C and let S be a mapping of C into itself defined by S x = lim n S n x for all x C and suppose that Fix ( S ) = i = 0 Fix ( S i ) . Then { x n } converges strongly to q F , which solves the following VIP:
q f ( q ) , J ( q p ) 0 , p F .
Proof It is easy to see that scheme (3.32) can be rewritten as
{ y n = α n x n + ( 1 α n ) G ( x n ) , x n + 1 = β n f ( x n ) + ( 1 β n ) S n y n , n 0 .
(3.33)
Take a fixed p F arbitrarily. Then by Lemma 2.13 we know that p = G ( p ) . Moreover, by Lemma 2.10 we have
y n p = α n ( x n p ) + ( 1 α n ) ( G ( x n ) p ) α n x n p + ( 1 α n ) G ( x n ) p α n x n p + ( 1 α n ) x n p = x n p .
(3.34)
From (3.34) we obtain
x n + 1 p = β n ( f ( x n ) p ) + ( 1 β n ) ( S n y n p ) β n ( f ( x n ) f ( p ) + f ( p ) p ) + ( 1 β n ) S n y n p β n ρ x n p + β n f ( p ) p + ( 1 β n ) y n p β n ρ x n p + β n f ( p ) p + ( 1 β n ) x n p = ( 1 β n ( 1 ρ ) ) x n p + β n f ( p ) p = ( 1 β n ( 1 ρ ) ) x n p + β n ( 1 ρ ) f ( p ) p 1 ρ max { x 0 p , f ( p ) p 1 ρ } ,

which implies that { x n } is bounded. By Lemma 2.10 we know from (3.34) that { G ( x n ) } and { y n } both are bounded.

Let us show that x n + 1 x n 0 and x n y n 0 as n . As a matter of fact, from (3.3) we have
{ y n = α n x n + ( 1 α n ) G ( x n ) , y n 1 = α n 1 x n 1 + ( 1 α n 1 ) G ( x n 1 ) , n 1 .
Simple calculations show that
y n y n 1 = α n ( x n x n 1 ) + ( α n α n 1 ) ( x n 1 G ( x n 1 ) ) + ( 1 α n ) ( G ( x n ) G ( x n 1 ) ) .
It follows that
y n y n 1 α n x n x n 1 + | α n α n 1 | x n 1 G ( x n 1 ) + ( 1 α n ) G ( x n ) G ( x n 1 ) α n x n x n 1 + | α n α n 1 | x n 1 G ( x n 1 ) + ( 1 α n ) x n x n 1 = x n x n 1 + | α n α n 1 | x n 1 G ( x n 1 ) .
(3.35)
Furthermore, from (3.33) we have
{ x n + 1 = β n f ( x n ) + ( 1 β n ) S n y n , x n = β n 1 f ( x n 1 ) + ( 1 β n 1 ) S n 1 y n 1 , n 1 .
Also, simple calculations show that
x n + 1 x n = β n ( f ( x n ) f ( x n 1 ) ) + ( β n β n 1 ) ( f ( x n 1 ) S n 1 y n 1 ) + ( 1 β n ) ( S n y n S n 1 y n 1 ) .
(3.36)
It follows from (3.35) and (3.36) that
x n + 1 x n β n f ( x n ) f ( x n 1 ) + | β n β n 1 | f ( x n 1 ) S n 1 y n 1 + ( 1 β n ) S n y n S n 1 y n 1 β n ρ x n x n 1 + | β n β n 1 | f ( x n 1 ) S n 1 y n 1 + ( 1 β n ) ( S n y n S n y n 1 + S n y n 1 S n 1 y n 1 ) β n ρ x n x n 1 + | β n β n 1 | f ( x n 1 ) S n 1 y n 1 + ( 1 β n ) ( y n y n 1 + S n y n 1 S n 1 y n 1 ) β n ρ x n x n 1 + | β n β n 1 | f ( x n 1 ) S n 1 y n 1 + ( 1 β n ) [ x n x n 1 + | α n α n 1 | x n 1 G ( x n 1 ) ] + S n y n 1 S n 1 y n 1 = ( 1 β n ( 1 ρ ) ) x n x n 1 + | β n β n 1 | f ( x n 1 ) S n 1 y n 1 + | α n α n 1 | x n 1 G ( x n 1 ) + S n y n 1 S n 1 y n 1 ( 1 β n ( 1 ρ ) ) x n x n 1 + M ( | α n α n 1 | + | β n β n 1 | ) + S n y n 1 S n 1 y n 1 ,
(3.37)
where sup n 0 { f ( x n ) S n y n + x n G ( x n ) } M for some M > 0 . Utilizing Lemma 2.2, from conditions (ii)-(iv) and the assumption on { S n } , we deduce that
lim n x n + 1 x n = 0 .
(3.38)
Since { x n } and { G ( x n ) } both are bounded, by Lemma 2.5 there exists a continuous strictly increasing function g : [ 0 , ) [ 0 , ) , g ( 0 ) = 0 such that for p F
y n p 2 α n x n p 2 + ( 1 α n ) G ( x n ) p 2 α n ( 1 α n ) g ( x n G ( x n ) ) α n x n p 2 + ( 1 α n ) x n p 2 α n ( 1 α n ) g ( x n G ( x n ) ) = x n p 2 α n ( 1 α n ) g ( x n G ( x n ) ) ,
which together with (3.33) implies that
x n + 1 p 2 = β n ( f ( x n ) f ( p ) ) + ( 1 β n ) ( S n y n p ) + β n ( f ( p ) p ) 2 β n ( f ( x n ) f ( p ) ) + ( 1 β n ) ( S n y n p ) 2 + 2 β n f ( p ) p , J ( x n + 1 p ) β n f ( x n ) f ( p ) 2 + ( 1 β n ) S n y n p 2 + 2 β n f ( p ) p x n + 1 p β n ρ 2 x n p 2 + ( 1 β n ) y n p 2 + 2 β n f ( p ) p x n + 1 p β n ρ x n p 2 + ( 1 β n ) [ x n p 2 α n ( 1 α n ) g ( x n G ( x n ) ) ] + 2 β n f ( p ) p x n + 1 p = ( 1 β n ( 1 ρ ) ) x n p 2 ( 1 β n ) α n ( 1 α n ) g ( x n G ( x n ) ) + 2 β n f ( p ) p x n + 1 p x n p 2 ( 1 β n ) α n ( 1 α n ) g ( x n G ( x n ) ) + 2 β n f ( p ) p x n + 1 p .
(3.39)
It immediately follows that
( 1 β n ) α n ( 1 α n ) g ( x n G ( x n ) ) x n p 2 x n + 1 p 2 + 2 β n f ( p ) p x n + 1 p ( x n p + x n + 1 p ) x n x n + 1 + 2 β n f ( p ) p x n + 1 p .
Since β n 0 , x n + 1 x n 0 and 0 < lim inf n α n lim sup n α n < 1 , we get lim n g ( x n G ( x n ) ) = 0 and hence
lim n x n G ( x n ) = 0 .
(3.40)
Thus, from (3.33) and (3.40) it follows that
lim n y n x n = lim n ( 1 α n ) G ( x n ) x n = 0 .
(3.41)
On the other hand, we observe that
x n + 1 x n = β n ( f ( x n ) x n ) + ( 1 β n ) ( S n y n x n ) = β n ( f ( x n ) x n ) + ( 1 β n ) ( S n y n y n ) + ( 1 β n ) ( y n x n ) .
Then we have
( 1 β n ) S n y n y n = x n + 1 x n β n ( f ( x n ) x n ) ( 1 β n ) ( y n x n ) x n + 1 x n + β n f ( x n ) x n + ( 1 β n ) y n x n x n + 1 x n + β n f ( x n ) x n + y n x n .
Since β n 0 , x n + 1 x n 0 and x n y n 0 as n , we get
lim n S n y n y n = 0 and lim n S n y n x n = 0 .
(3.42)
In the meantime, since x n G ( x n ) 0 and x n y n 0 as n , we also get
lim n y n G ( x n ) = 0 .
(3.43)
We note that
S n G ( x n ) G ( x n ) S n G ( x n ) S n y n + S n y n x n + x n G ( x n ) G ( x n ) y n + S n y n x n + x n G ( x n ) .
From (3.40), (3.42) and (3.43), we obtain
lim n S n G ( x n ) G ( x n ) = 0 .
(3.44)
By (3.44) and Lemma 2.6, we have
S G ( x n ) G ( x n ) S G ( x n ) S n G ( x n ) + S n G ( x n ) G ( x n ) 0 as  n .
(3.45)
In terms of (3.40) and (3.45), we have
x n S x n x n G ( x n ) + G ( x n ) S G ( x n ) + S G ( x n ) S x n 2 x n G ( x n ) + G ( x n ) S G ( x n ) 0 as  n .
(3.46)
Define a mapping W x = ( 1 θ ) S x + θ G ( x ) , θ ( 0 , 1 ) is a constant. Then by Lemma 2.8 we have that Fix ( W ) = Fix ( S ) Fix ( G ) = F . We observe that
x n W x n = ( 1 θ ) ( x n S x n ) + θ ( x n G ( x n ) ) ( 1 θ ) x n S x n + θ x n G ( x n ) .
From (3.40) and (3.46), we obtain
lim n x n W x n = 0 .
(3.47)
Now, we claim that
lim sup n f ( q ) q , J ( x n q ) 0 ,
(3.48)
where q = s lim t 0 x t with x t being the fixed point of the contraction x t f ( x ) + ( 1 t ) W x . Then x t solves the fixed point equation x t = t f ( x t ) + ( 1 t ) W x t . Utilizing the arguments similar to those of (3.28) in the proof of Theorem 3.1, we can deduce 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 ) lim sup t 0 lim sup n f ( q ) q , J ( x n q ) J ( x n x t ) .
Since X has a uniformly Gâteaux differentiable norm, the duality mapping J is norm-to-weak uniformly continuous on bounded subsets of X. Consequently, the two limits are interchangeable and hence the following holds:
lim sup n f ( q ) q , J ( x n q ) 0 .
(3.49)
From (3.38) we get ( x n + 1 q ) ( x n q ) 0 . Noticing the norm-to-weak uniform continuity of J on bounded subsets of X, we deduce from (3.48) that
lim sup n f ( q ) q , J ( x n + 1 q ) = lim sup n ( f ( q ) q , J ( x n + 1 q ) J ( x n q ) + f ( q ) q , J ( x n + 1 q ) ) = lim sup n f ( q ) q , J ( x n q ) 0 .
(3.50)
Finally, let us show that x n q as n . We observe that
y n q = α n ( x n q ) + ( 1 α n ) ( G ( x n ) q ) α n x n q + ( 1 α n ) x n q = x n q ,
and
x n + 1 q 2 = β n ( f ( x n ) f ( q ) ) + ( 1 β n ) ( S n y n q ) + β n ( f ( q ) q ) 2 β n ( f ( x n ) f ( q ) ) + ( 1 β n ) ( S n y n q ) 2 + 2 β n f ( q ) q , J ( x n + 1 q ) β n f ( x n ) f ( q ) 2 + ( 1 β n ) S n y n q 2 + 2 β n f ( q ) q , J ( x n + 1 q ) β n ρ x n q 2 + ( 1 β n ) y n q 2 + 2 β n f ( q ) q , J ( x n + 1 q ) β n ρ x n q 2 + ( 1 β n ) x n q 2 + 2 β n f ( q ) q , J ( x n + 1 q ) = ( 1 β n ( 1 ρ ) ) x n q 2 + 2 β n f ( q ) q , J ( x n + 1 q ) .
(3.51)

Since n = 0 β n = and lim sup n f ( q ) q , J ( x n + 1 q ) 0 , by Lemma 2.2 we conclude from (3.51) that x n q as n . This completes the proof. □

Corollary 3.2 Let C be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gâteaux differentiable norm. Let Π C be a sunny nonexpansive retraction from X onto C. Let the mapping B i : C X be λ i -strictly pseudocontractive and α i -strongly accretive with α i + λ i 1 for i = 1 , 2 . Let f : C C be a contraction with coefficient ρ ( 0 , 1 ) . Let S be a nonexpansive mapping of C into itself such that F = Fix ( S ) Ω , where Ω is a fixed point set of the mapping G. For arbitrarily given x 0 C , let { x n } be a sequence generated by
{ y n = α n x n + ( 1 α n ) Π C ( I μ 1 B 1 ) Π C ( I μ 2 B 2 ) x n , x n + 1 = β n f ( x n ) + ( 1 β n ) S y n , n 0 ,
where 1 λ i 1 + λ i ( 1 1 α i λ i ) μ i 1 for i = 1 , 2 . Suppose that { α n } and { β n } are sequences in ( 0 , 1 ) satisfying the following conditions:
  1. (i)

    0 < lim inf n α n lim sup n α n < 1 ;

     
  2. (ii)

    lim n β n = 0 and n = 0 β n = ;

     
  3. (iii)

    n = 1 | α n α n 1 | < or lim n | α n α n 1 | / β n = 0 ;

     
  4. (iv)

    n = 1 | β n β n 1 | < or lim n β n 1 / β n = 1 .

     
Then { x n } converges strongly to q F , which solves the following VIP:
q f ( q ) , J ( q p ) 0 , p F .

Remark 3.1 Theorems 3.1 and 3.2 improve, extend, supplement and develop [[14], Theorem 3.1] in the following aspects. Although the iterative algorithm in Theorem 3.1 is an implicit algorithm, we can derive the strong convergence of the proposed algorithm under the same conditions on the parameter sequences { α n } , { β n } as in [[14], Theorem 3.1]. The assumption of the uniformly convex and 2-uniformly smooth Banach space X in [[14], Theorem 3.1] is weakened to the one of the uniformly convex Banach space X having a uniformly Gâteaux differentiable norm in Theorem 3.2.

4 Relaxed extragradient composite algorithms

In this section, we propose and analyze a composite explicit iterative algorithm by the two-step relaxed extragradient method for solving GSVI (1.1) and the common fixed point problem of an infinite family of nonexpansive self-mappings { S n } in a 2-uniformly smooth and uniformly convex Banach space.

Theorem 4.1 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let Π C be a sunny nonexpansive retraction from X onto C. Let the mapping B i : C X be α i -inverse-strongly accretive for i = 1 , 2 . Let f : C C be a contraction with coefficient ρ ( 0 , 1 ) . Let { S n } n = 0 be an infinite family of nonexpansive mappings of C into itself such that F = i = 0 Fix ( S i ) Ω , where Ω is a fixed point set of the mapping G. For arbitrarily given x 0 C , let { x n } be a sequence generated by
{ y n = α n f ( x n ) + ( 1 α n ) S n Π C ( I μ 1 B 1 ) Π C ( I μ 2 B 2 ) x n , x n + 1 = β n y n + ( 1 β n ) S n Π C ( I μ 1 B 1 ) Π C ( I μ 2 B 2 ) y n , n 0 ,
(4.1)
where 0 < μ i < α i κ 2 for i = 1 , 2 . Suppose that { α n } and { β n } are sequences in ( 0 , 1 ] satisfying the following conditions:
  1. (i)

    lim n α n = 0 and n = 0 α n = ;

     
  2. (ii)

    { β n } [ a , 1 ] for some a ( 0 , 1 ) ;

     
  3. (iii)

    n = 1 | α n α n 1 | < or lim n α n 1 / α n = 1 ;

     
  4. (iv)

    n = 1 | β n β n 1 | < or lim n | β n β n 1 | / α n = 0 .

     
Assume that n = 1 sup x D S n x S n 1 x < for any bounded subset D of C and let S be a mapping of C into itself defined by S x = lim n S n x for all x C and suppose that Fix ( S ) = i = 0 Fix ( S i ) . Then { x n } converges strongly to q F , which solves the following VIP:
q f ( q ) , J ( q p ) 0 , p F .
Proof It is easy to see that scheme (4.1) can be rewritten as
{ y n = α n f ( x n ) + ( 1 α n ) S n G ( x n ) , x n + 1 = β n y n + ( 1 β n ) S n G ( y n ) , n 0 .
(4.2)
Take a fixed p F arbitrarily. Then by Lemma 2.13 we know that p = G ( p ) . Moreover, by Lemma 2.12 we have
y n p = α n ( f ( x n ) p ) + ( 1 α n ) ( S n G ( x n ) p ) α n f ( x n ) f ( p ) + α n f ( p ) p + ( 1 α n ) S n G ( x n ) p α n ρ x n p + α n f ( p ) p + ( 1 α n ) G ( x n ) p α n ρ x n p + α n f ( p ) p + ( 1 α n ) x n p = ( 1 α n ( 1 ρ ) ) x n p + α n f ( p ) p .
(4.3)
From (4.3) we have
x n + 1 p = β n ( y n p ) + ( 1 β n ) ( S n G ( y n ) p ) β n y n p + ( 1 β n ) S n G ( y n ) p β n y n p + ( 1 β n ) G ( y n ) p β n y n p +