Open Access

System of variational inequalities and an accretive operator in Banach spaces

Fixed Point Theory and Applications20132013:249

https://doi.org/10.1186/1687-1812-2013-249

Received: 17 June 2013

Accepted: 6 September 2013

Published: 7 November 2013

Abstract

In this paper, we introduce composite Mann iteration methods for a general system of variational inequalities with solutions being also common fixed points of a countable family of nonexpansive mappings and zeros of an accretive operator in real smooth Banach spaces. Here, the composite Mann iteration methods are based on Korpelevich’s extragradient method, viscosity approximation method and the Mann iteration method. We first consider and analyze a composite Mann iterative algorithm in the setting of uniformly convex and 2-uniformly smooth Banach space, and then another composite Mann iterative algorithm in a uniformly convex Banach space having a uniformly Gâteaux differentiable norm. Under suitable assumptions, we derive some strong convergence theorems. The results presented in this paper improve, extend, supplement and develop the corresponding results announced in the earlier and very recent literature.

MSC:49J30, 47H09, 47J20.

Keywords

composite Mann iteration methodsgeneral system of variational inequalitiesaccretive operatornonexpansive mappingsunny nonexpansive retractionfixed pointuniformly Gâteaux differentiable normuniform smoothnessuniform convexity

1 Introduction

Let X be a real Banach space whose dual space is denoted by 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. It is an immediate consequence of the Hahn-Banach theorem that J ( x ) is nonempty for each x X . Let U = { x X : x = 1 } denote the unit sphere of X. A Banach space X is said to be uniformly convex if for each ϵ ( 0 , 2 ] , there exists δ > 0 such that for all x , y U ,
x y ϵ x + y / 2 1 δ .
It is known that a uniformly convex Banach space is reflexive and strict convex. A Banach space X is said to be smooth if the limit
lim t 0 x + t y x t

exists for all x , y U ; in this case, X is also said to have a Gâteaux differentiable norm. X is said to have a uniformly Gâteaux differentiable norm if for each y U , the limit is attained uniformly for x U . Moreover, it is said to be uniformly smooth if this limit is attained uniformly for x , y U . The norm of X is said to be the Frechet differential if for each x U , this limit is attained uniformly for y U . Let C be a nonempty closed convex subset of X. A mapping T : C C is called nonexpansive if T x T y x y for every x , y C . The set of fixed points of T is denoted by Fix ( T ) . 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 .

Recently, Yao et al. [1] combined the viscosity approximation method and the Mann iteration method, and gave the following hybrid viscosity approximation method:

Let C be a nonempty closed convex subset of a real uniformly smooth Banach space X, T : C C a nonexpansive mapping such that Fix ( T ) and f Ξ C with a contractive coefficient ρ ( 0 , 1 ) , where Ξ C is the set of all contractive self-mappings on C. For an arbitrary x 0 C , define { x n } in the following way:
{ y n = α n x n + ( 1 α n ) T x n , x n + 1 = β n f ( x n ) + ( 1 β n ) y n , n 0 ,
(YCY)

where { α n } and { β n } are two sequences in ( 0 , 1 ) . They proved under certain control conditions on the sequences { α n } and { β n } that { x n } converges strongly to a fixed point of T. Subsequently, Ceng and Yao [2] under the convergence of no parameter sequences to zero proved that the sequence { x n } generated by (YCY) converges strongly to a fixed point of T. Such a result includes [[1], Theorem 1] as a special case.

Theorem 1.1 (See [[2], Theorem 3.1])

Let C be a nonempty closed convex subset of a uniformly smooth Banach space X. Let T : C C be a nonexpansive mapping with Fix ( T ) and f Ξ C with contractive coefficient ρ ( 0 , 1 ) . Given sequences { α n } and { β n } in [ 0 , 1 ] , the following control conditions are satisfied:
  1. (i)

    1 β n 1 ρ , n n 0 for some integer n 0 1 ;

     
  2. (ii)

    n = 0 β n = ;

     
  3. (iii)

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

     
  4. (iv)

    lim n ( β n + 1 1 ( 1 β n + 1 ) α n + 1 β n 1 ( 1 β n ) α n ) = 0 .

     
For an arbitrary x 0 C , let { x n } be generated by (YCY). Then
x n q β n ( f ( x n ) x n ) 0 ,
where q Fix ( T ) solves the variational inequality problem (VIP):
q f ( q ) , J ( q p ) 0 , p Fix ( T ) .
On the other hand, Cai and Bu [3] considered the following general system of variational inequalities (GSVI) in a real smooth Banach space X, which involves finding ( 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 C is a nonempty, closed and convex subset of X, B 1 , B 2 : C X are two nonlinear mappings and μ 1 and μ 2 are two positive constants. Here, the set of solutions of GSVI (1.1) is denoted by GSVI ( C , B 1 , B 2 ) . In particular, if X = H in a real Hilbert space, then GSVI (1.1) reduces to the following GSVI 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)
which 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 particular, if B 1 = B 2 = A , then problem (1.2) reduces to the new system of variational inequalities (NSVI), introduced and studied by Verma [4]. Further, if x = y additionally, then the NSVI reduces to the classical variational inequality problem (VIP) of finding x C such that
A x , x x 0 , x C .
(1.3)

The solution set of VIP (1.3) is denoted by VI ( C , A ) . Variational inequality theory has been studied quite extensively and has emerged as an important tool in the study of a wide class of obstacle, unilateral, free, moving, equilibrium problems. It is now well known that the variational inequalities are equivalent to the fixed point problems, the origin of which can be traced back to Lions and Stampacchia [5]. This alternative formulation has been used to suggest and analyze projection iterative method for solving variational inequalities under the conditions that the involved operator must be strongly monotone and Lipschitz continuous.

Recently, Ceng et al. [6] transformed problem (1.2) into a fixed point problem in the following way.

Lemma 1.1 (See [6])

For given x ¯ , y ¯ C , ( x ¯ , y ¯ ) is a solution of problem (1.2) if and only if x ¯ is a fixed point of the mapping G : C C defined by
G ( x ) = P C [ P C ( x μ 2 B 2 x ) μ 1 B 1 P C ( x μ 2 B 2 x ) ] , x C ,
(1.4)

where y ¯ = P C ( x ¯ μ 2 B 2 x ¯ ) and P C is the projection of H onto C.

In particular, if the mappings B i : C H is β i -inverse strongly monotone for i = 1 , 2 , then the mapping G is nonexpansive provided μ i ( 0 , 2 β i ) for i = 1 , 2 .

In 1976, Korpelevich [7] proposed an iterative algorithm for solving the VIP (1.3) in Euclidean space R n :
{ y n = P C ( x n τ A x n ) , x n + 1 = P C ( x n τ A y n ) , n 0 ,

with τ > 0 a given number, which is known as the extragradient method. The literature on the VIP is vast, and Korpelevich’s extragradient method has received great attention given by many authors, who improved it in various ways; see, e.g., [3, 814] and the references therein, to name but a few.

In particular, whenever X is still a real smooth Banach space, B 1 = B 2 = A and x = y , then GSVI (1.1) reduces to the variational inequality problem (VIP) of finding x C such that
A x , J ( x x ) 0 , x C ,
(1.5)

which was considered by Aoyama et al. [15]. Note that VIP (1.5) is connected with the fixed point problem for nonlinear mapping (see, e.g., [16, 17]), the problem of finding a zero point of a nonlinear operator (see, e.g., [18]) and so on. It is clear that VIP (1.5) extends VIP (1.3) from Hilbert spaces to Banach spaces.

In order to find a solution of VIP (1.5), Aoyama et al. [15] introduced the following Mann iterative scheme for an accretive operator A:
x n + 1 = α n x n + ( 1 α n ) Π C ( x n λ n A x n ) , n 1 ,
(1.6)

where Π C is a sunny nonexpansive retraction from X onto C. Then they proved a weak convergence theorem.

Obviously, it is an interesting and valuable problem of constructing some algorithms with strong convergence for solving GSVI (1.1), which contains VIP (1.5) as a special case. Very recently, Cai and Bu [3] constructed an iterative algorithm for solving GSVI (1.1) and a common fixed point problem of a countable family of nonexpansive mappings in a uniformly convex and 2-uniformly smooth Banach space.

Theorem 1.2 (See [[3], 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 β i -inverse-strongly accretive with 0 < μ i < β i κ 2 for i = 1 , 2 . Let f be a contraction of C into itself with coefficient δ ( 0 , 1 ) . Let { T n } n = 1 be a countable family of nonexpansive mappings of C into itself such that F = i = 1 Fix ( T i ) Ω , where Ω is the fixed point set of the mapping G = Π C ( I μ 1 B 1 ) Π C ( I μ 2 B 2 ) on C. For arbitrarily given x 1 C , let { x n } be the sequence generated by
{ x n + 1 = β n x n + ( 1 β n ) T n y n , y n = α n f ( x n ) + ( 1 α n ) z n , z n = Π C ( u n μ 1 B 1 u n ) , u n = Π C ( x n μ 2 B 2 x n ) , n 1 .
Suppose that { α n } and { β n } are two sequences in ( 0 , 1 ) satisfying the following conditions:
  1. (i)

    lim n α n = 0 and n = 1 α n = ;

     
  2. (ii)

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

     
Assume that n = 1 sup x D T n + 1 x T n x < for any bounded subset D of C, and let T be a mapping of C into X defined by T x = lim n T n x for all x C and suppose that Fix ( T ) = n = 1 Fix ( T n ) . Then { x n } converges strongly to q F , which solves the following VIP:
q f ( q ) , J ( q p ) 0 , p F .
Furthermore, recall that a (possibly multivalued) operator A X × X with domain D ( A ) and range R ( A ) in a real Banach space X is accretive if, for each x i D ( A ) and y i A x i ( i = 1 , 2 ), there exists a j ( x 1 x 2 ) J ( x 1 x 2 ) such that y 1 y 2 , j ( x 1 x 2 ) 0 . (Here J is the duality mapping.) An accretive operator A is said to satisfy the range condition if D ( A ) ¯ R ( I + r A ) for all r > 0 . An accretive operator A is m-accretive if R ( I + r A ) = X for each r > 0 . If A is an accretive operator which satisfies the range condition, then we can define, for each r > 0 a mapping J r : R ( I + r A ) D ( A ) defined by J r = ( I + r A ) 1 , which is called the resolvent of A. We know that J r is nonexpansive and Fix ( J r ) = A 1 0 for all r > 0 . Hence,
Fix ( J r ) = A 1 0 = { z D ( A ) : 0 A z } .

If A 1 0 , then the inclusion 0 A z is solvable. The following resolvent identity is well known to us; see [19], where more details on accretive operators can be found.

Proposition 1.1 (Resolvent identity)

For λ > 0 , μ > 0 and x X ,
J λ x = J μ ( μ λ x + ( 1 μ λ ) J λ x ) .
Recently, Aoyama et al. [20] studied the following iterative scheme in a uniformly convex Banach space having a uniformly Gâteaux differentiable norm: for resolvents J r n of an accretive operator A such that A 1 0 and D ( A ) ¯ C r > 0 R ( I + r A ) and { α n } ( 0 , 1 )
{ x 0 = x C , x n + 1 = α n x + ( 1 α n ) J r n x n .
(1.7)
They proved that the sequence { x n } generated by (1.7) converges strongly to a zero of A under appropriate assumptions on { α n } and { r n } . Subsequently, Ceng et al. [21] introduced and analyzed the following composite iterative scheme in either a uniformly smooth Banach space or a reflexive Banach space having a weakly sequentially continuous duality mapping
{ x 0 = x X , y n = α n u + ( 1 α n ) J r n x n , x n + 1 = ( 1 β n ) y n + β n J r n y n ,
(1.8)
where u D ( A ) ¯ is an arbitrary (but fixed) element, under the following control conditions:
  1. (H1)

    lim n α n = 0 ;

     
  2. (H2)

    n = 0 α n = , or, equivalently, n = 0 ( 1 α n ) = 0 ;

     
  3. (H3)

    n = 1 | α n α n 1 | < ;

     
  4. (H4)

    r n ε , n 0 , for some ε > 0 and n = 1 | r n r n 1 | < ;

     
  5. (H5)

    β n [ 0 , a ) for some a ( 0 , 1 ) and n = 1 | β n β n 1 | < .

     
Further, as the viscosity approximation method, Jung [22] purposed and analyzed the following composite iterative scheme for finding a zero of an accretive operator A: for resolvent J r n of an accretive operator A such that A 1 0 and D ( A ) ¯ C r > 0 R ( I + r A ) , f Ξ C ( Ξ C denotes the set of all contractions on C) and { α n } , { β n } ( 0 , 1 ) ,
{ x 0 = x C , y n = α n f ( x n ) + ( 1 α n ) J r n x n , x n + 1 = ( 1 β n ) y n + β n J r n y n .
(JS)

Theorem 1.3 (See [[22], Theorem 3.1])

Let X be a strictly convex and reflexive Banach space having a uniformly Gâteaux differentiable norm. Let C be a nonempty closed convex subset of X and A X × X an accretive operator in X such that A 1 0 and D ( A ) ¯ C r > 0 R ( I + r A ) . Let { α n } and { β n } be sequences in ( 0 , 1 ) which satisfy the conditions:
  1. (i)

    lim n α n = 0 and n = 0 α n = ;

     
  2. (ii)

    β n [ 0 , a ) for some 0 < a < 1 for all n 0 .

     
Let f Ξ C and x 0 C be chosen arbitrarily. Let { x n } be a sequence generated by (JS) for r n > 0 . If { x n } is asymptotically regular, i.e., lim n x n + 1 x n = 0 , then { x n } converges strongly to q A 1 0 , which is the unique solution of the variational inequality problem (VIP)
( I f ) q , J ( q p ) 0 , f Ξ C , p A 1 0 .

Let C be a nonempty closed convex subset of a real smooth Banach space X. Let Π C be a sunny nonexpansive retraction from X onto C, and let f : C C be a contraction with coefficient ρ ( 0 , 1 ) . Motivated and inspired by the research going on in this area, we introduce the composite Mann iteration methods for finding solutions of GSVI (1.1), which are also common fixed points of a countable family of nonexpansive mappings and zeros of an accretive operator A X × X such that D ( A ) ¯ C r > 0 R ( I + r A ) . Here, the composite Mann iteration methods are based on Korpelevich’s extragradient method, viscosity approximation method and the Mann iteration method. We first consider and analyze a composite Mann iterative algorithm in the setting of uniformly convex and 2-uniformly smooth Banach space, and then another composite Mann iterative algorithm in a uniformly convex Banach space having a uniformly Gâteaux differentiable norm. Under suitable assumptions, we derive some strong convergence theorems. The results presented in this paper improve, extend, supplement and develop the corresponding results announced in the earlier and very recent literature; see, e.g., [2, 3, 6, 8, 22].

2 Preliminaries

Let X be a real Banach space. 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 . As pointed out in [23], no Banach space is q-uniformly smooth for q > 2 . In addition, it is also known that J is single-valued if and only if X is smooth, whereas if X is uniformly smooth, then J is norm-to-norm uniformly continuous on bounded subsets of X. If X has a uniformly Gâteaux differentiable norm, then the duality mapping J is norm-to-weak uniformly continuous on bounded subsets of X. We use the notation to indicate the weak convergence and the one → to indicate the strong convergence.

Let C be a nonempty closed convex subset of X. Recall that a mapping A : C X is said to be
  1. (i)
    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 ) ;

     
  2. (ii)
    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 ;

     
  3. (iii)
    λ-strictly pseudocontractive [24] 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 ) .

     

It is worth emphasizing that the definition of the inverse strongly accretive mapping is based on that of the inverse strongly monotone mapping, which was studied by so many authors; see, e.g., [9, 25, 26].

Proposition 2.1 (See [27])

Let X be a 2-uniformly smooth Banach space. Then
x + y 2 x 2 + 2 y , J ( x ) + 2 κ y 2 , x , y X ,

where κ is the 2-uniformly smooth constant of X, and J is the normalized duality mapping from X into X .

Proposition 2.2 (See [28])

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 , 2 r ] R , 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 } .

Next, we list some lemmas that will be used in the sequel. Lemma 2.1 can be found in [29]. Lemma 2.2 is an immediate consequence of the subdifferential inequality of the function 1 2 2 .

Lemma 2.1 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 2.2 In a real smooth Banach space X, the following inequality holds:
x + y 2 x 2 + 2 y , J ( x + y ) , x , y X .
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. The following lemma concerns the sunny nonexpansive retraction.

Lemma 2.3 (See [30])

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. 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 .

     

It is well known that if X = H in 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.

Lemma 2.4 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 B 1 , B 2 : C X be nonlinear mappings. 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 ) .

Proof We can rewrite GSVI (1.1) as
{ x ( y μ 1 B 1 y ) , J ( x x ) 0 , x C , y ( x μ 2 B 2 x ) , J ( x y ) 0 , x C ,
which is obviously equivalent to
{ x = Π C ( y μ 1 B 1 y ) , y = Π C ( x μ 2 B 2 x ) ,

because of Lemma 2.3. This completes the proof. □

In terms of Lemma 2.4, 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. Throughout this paper, the set of fixed points of the mapping G is denoted by Ω.

Lemma 2.5 (See [27])

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 [31])

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 the set of all contractions on C. For t ( 0 , 1 ) and f Ξ C , let x t C be the 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 [17, 32])

Let X be a uniformly smooth Banach space, or a reflexive and strictly convex Banach space with a uniformly Gâteaux 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 [33])

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 that 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 defined well, nonexpansive and Fix ( S ) = n = 0 Fix ( T n ) holds.

Lemma 2.9 (See [15])

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 A : C X be an accretive mapping. Then for all λ > 0 ,
VI ( C , A ) = Fix ( Π C ( I λ A ) ) .

Lemma 2.10 (See [34])

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

Lemma 2.11 (See [35])

Let X be a uniformly convex Banach space and B r = { x X : x r } , r > 0 . Then there exists a continuous, strictly increasing and convex function g : [ 0 , ] [ 0 , ] , g ( 0 ) = 0 such that
α x + β y + γ z 2 α x 2 + β y 2 + γ z 2 α β g ( x y )

for all x , y , z B r and all α , β , γ [ 0 , 1 ] with α + β + γ = 1 .

3 Composite Mann iterative algorithms in uniformly convex and 2-uniformly smooth Banach spaces

In this section, we introduce our composite Mann iterative algorithms in uniformly convex and 2-uniformly smooth Banach spaces and show the strong convergence theorems. We will use some useful lemmas in the sequel.

Lemma 3.1 (See [[3], Lemma 2.8])

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 2 x y 2 + 2 μ i ( μ i κ 2 α i ) B i x B i y 2 , 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 3.2 (See [[3], Lemma 2.9])

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 G : C C be the 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 0 < μ i α i κ 2 for i = 1 , 2 , then G : C C is nonexpansive.

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 A X × X be an accretive operator in X such that D ( A ) ¯ C r > 0 R ( I + r A ) . Let 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 i } i = 0 be a countable family of nonexpansive mappings of C into itself such that F = i = 0 Fix ( S i ) Ω A 1 0 , where Ω is the fixed point set of the mapping G = Π C ( I μ 1 B 1 ) Π C ( I μ 2 B 2 ) with 0 < μ i < α i κ 2 for i = 1 , 2 . For arbitrarily given x 0 C , let { x n } be the sequence generated by
{ y n = β n x n + γ n S n x n + δ n J r n G x n , x n + 1 = α n f ( x n ) + ( 1 α n ) y n , n 0 ,
(3.1)
where { α n } , { β n } , { γ n } and { δ n } are the sequences in [ 0 , 1 ] such that β n + γ n + δ n = 1 for all n 0 . Suppose that the following conditions hold:
  1. (i)

    n = 0 α n = and 0 α n 1 ρ , n n 0 for some integer n 0 0 ;

     
  2. (ii)

    lim inf n γ n > 0 and lim inf n δ n > 0 ;

     
  3. (iii)

    lim n ( | α n + 1 1 ( 1 α n + 1 ) β n + 1 α n 1 ( 1 α n ) β n | + | δ n + 1 1 β n + 1 δ n 1 β n | ) = 0 ;

     
  4. (iv)

    lim n | r n + 1 r n | = 0 and r n ε > 0 for all n 0 ;

     
  5. (v)

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

     
Assume that n = 0 sup x D S n + 1 x S n 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 q α n ( f ( x n ) x n ) 0 ,
where q F solves the following VIP:
q f ( q ) , J ( q p ) 0 , p F .
Proof First of all, let us show that the sequence { x n } is bounded. Indeed, take a fixed p F arbitrarily. Then we get p = G p , p = S n p and p = J r n p for all n 0 . By Lemma 3.2, we know that G is nonexpansive. Then from (3.1), we have
y n p β n x n p + γ n S n x n p + δ n J r n G x n p β n x n p + γ n x n p + δ n G x n p β n x n p + γ n x n p + δ n x n p = x n p ,
(3.2)
and hence
x n + 1 p α n f ( x n ) p + ( 1 α n ) y n p α n ( f ( x n ) f ( p ) + f ( p ) p ) + ( 1 α n ) y n p α n ( ρ x n p + f ( p ) p ) + ( 1 α n ) x n p = ( 1 α n ( 1 ρ ) ) x n p + α n ( 1 ρ ) f ( p ) p 1 ρ max { x n p , f ( p ) p 1 ρ } .
By induction, we obtain
x n p max { x 0 p , f ( p ) p 1 ρ } , n 0 .
(3.3)

Thus, { x n } is bounded, and so are the sequences { y n } , { G x n } and { f ( x n ) } .

Let us show that
lim n x n + 1 x n = 0 .
(3.4)
As a matter of fact, put σ n = ( 1 α n ) β n , n 0 . Then it follows from (i) and (v) that
β n σ n = ( 1 α n ) β n ( 1 ( 1 ρ ) ) β n = ρ β n , n n 0 ,
and hence
0 < lim inf n σ n lim inf n σ n < 1 .
(3.5)
Define
x n + 1 = σ n x n + ( 1 σ n ) z n .
(3.6)
Observe that
z n + 1 z n = x n + 2 σ n + 1 x n + 1 1 σ n + 1 x n + 1 σ n x n 1 σ n = α n + 1 f ( x n + 1 ) + ( 1 α n + 1 ) y n + 1 σ n + 1 x n + 1 1 σ n + 1 α n f ( x n ) + ( 1 α n ) y n σ n x n 1 σ n = ( α n + 1 f ( x n + 1 ) 1 σ n + 1 α n f ( x n ) 1 σ n ) ( 1 α n ) [ β n x n + γ n S n x n + δ n J r n G x n ] σ n x n 1 σ n + ( 1 α n + 1 ) [ β n + 1 x n + 1 + γ n + 1 S n + 1 x n + 1 + δ n + 1 J r n + 1 G x n + 1 ] σ n + 1 x n + 1 1 σ n + 1 = ( α n + 1 f ( x n + 1 ) 1 σ n + 1 α n f ( x n ) 1 σ n ) + 1 α n + 1 1 σ n + 1 ( γ n + 1 S n + 1 x n + 1 + δ n + 1 J r n + 1 G x n + 1 ) 1 α n 1 σ n ( γ n S n x n + δ n J r n G x n ) = ( α n + 1 f ( x n + 1 ) 1 σ n + 1 α n f ( x n ) 1 σ n ) + ( 1 α n + 1 ) ( 1 β n + 1 ) 1 σ n + 1 [ γ n + 1 S n + 1 x n + 1 + δ n + 1 J r n + 1 G x n + 1 1 β n + 1 γ n S n x n + δ n J r n G x n 1 β n ] + [ ( 1 α n + 1 ) ( 1 β n + 1 ) 1 σ n + 1 ( 1 α n ) ( 1 β n ) 1 σ n ] γ n S n x n + δ n J r n G x n 1 β n = α n + 1 1 σ n + 1 ( f ( x n + 1 ) f ( x n ) ) + ( α n + 1 1 σ n + 1 α n 1 σ n ) f ( x n ) + ( 1 α n + 1 ) ( 1 β n + 1 ) 1 σ n + 1 [ γ n + 1 γ n + 1 + δ n + 1 ( S n + 1 x n + 1 S n x n ) + ( γ n + 1 γ n + 1 + δ n + 1 γ n γ n + δ n ) S n x n + δ n + 1 γ n + 1 + δ n + 1 ( J r n + 1 G x n + 1 J r n G x n ) + ( δ n + 1 γ n + 1 + δ n + 1 δ n γ n + δ n ) J r n G x n ] ( α n + 1 1 σ n + 1 α n 1 σ n ) γ n S n x n + δ n J r n G x n γ n + δ n = α n + 1 1 σ n + 1 ( f ( x n + 1 ) f ( x n ) ) + ( α n + 1 1 σ n + 1 α n 1 σ n ) ( f ( x n ) γ n S n x n + δ n J r n G x n γ n + δ n ) + 1 σ n + 1 α n + 1 1 σ n + 1 [ γ n + 1 γ n + 1 + δ n + 1 ( S n + 1 x n + 1 S n x n ) + ( γ n + 1 γ n + 1 + δ n + 1 γ n γ n + δ n ) S n x n + δ n + 1 γ n + 1 + δ n + 1 ( J r n + 1 G x n + 1 J r n G x n ) + ( δ n + 1 γ n + 1 + δ n + 1 δ n γ n + δ n ) J r n G x n ] ,
and hence
z n + 1 z n α n + 1 1 σ n + 1 f ( x n + 1 ) f ( x n ) + | α n + 1 1 σ n + 1 α n 1 σ n | f ( x n ) γ n S n x n + δ n J r n G x n γ n + δ n + 1 σ n + 1 α n + 1 1 σ n + 1 γ n + 1 γ n + 1 + δ n + 1 ( S n + 1 x n + 1 S n x n ) + ( γ n + 1 γ n + 1 + δ n + 1 γ n γ n + δ n ) S n x n + δ n + 1 γ n + 1 + δ n + 1 ( J r n + 1 G x n + 1 J r n G x n ) + ( δ n + 1 γ n + 1 + δ n + 1 δ n γ n + δ n ) J r n G x n ρ α n + 1 1 σ n + 1 x n + 1 x n + | α n + 1 1 σ n + 1 α n 1 σ n | ( f ( x n ) + S n x n + J r n G x n ) + 1 σ n + 1 α n + 1 1 σ n + 1 [ γ n + 1 γ n + 1 + δ n + 1 S n + 1 x n + 1 S n x n + | γ n + 1 γ n + 1 + δ n + 1 γ n γ n + δ n | S n x n + δ n + 1 γ n + 1 + δ n + 1 J r n + 1 G x n + 1 J r n G x n + | δ n + 1 γ n + 1 + δ n + 1 δ n γ n + δ n | J r n G x n ] .
(3.7)
On the other hand, if r n r n + 1 , using the resolvent identity in Proposition 1.1,
J r n + 1 G x n + 1 = J r n ( r n r n + 1 G x n + 1 + ( 1 r n r n + 1 ) J r n + 1 G x n + 1 ) ,
we get
J r n + 1 G x n + 1 J r n G x n = J r n ( r n r n + 1 G x n + 1 + ( 1 r n r n + 1 ) J r n + 1 G x n + 1 ) J r n G x n r n r n + 1 G x n + 1 G x n + ( 1 r n r n + 1 ) J r n + 1 G x n + 1 G x n x n + 1 x n + r n + 1 r n r n + 1 J r n + 1 G x n + 1 G x n x n + 1 x n + 1 ε | r n + 1 r n | J r n + 1 G x n + 1 G x n .
If r n + 1 r n , we derive in the similar way
J r n + 1 G x n + 1 J r n G x n x n x n + 1 + 1 ε | r n r n + 1 | J r n G x n G x n + 1 .
Thus, combining the above cases, we obtain
J r n + 1 G x n + 1 J r n G x n x n x n + 1 + M 0 | r n r n + 1 | , n 0 ,
(3.8)
where sup n 0 { 1 ε ( J r n + 1 G x n + 1 G x n + J r n G x n G x n + 1 ) } M 0 for some M 0 > 0 . Substituting (3.8) for (3.7), we have
z n + 1 z n ρ α n + 1 1 σ n + 1 x n + 1 x n + | α n + 1 1 σ n + 1 α n 1 σ n | ( f ( x n ) + S n x n + J r n G x n ) + 1 σ n + 1 α n + 1 1 σ n + 1 [ γ n + 1 γ n + 1 + δ n + 1 ( S n + 1 x n + 1 S n + 1 x n + S n + 1 x n S n x n ) + | γ n + 1 γ n + 1 + δ n + 1 γ n γ n + δ n | S n x n + δ n + 1 γ n + 1 + δ n + 1 ( x n x n + 1 + M 0 | r n r n + 1 | ) + | δ n + 1 γ n + 1 + δ n + 1 δ n γ n + δ n | J r n G x n ] ρ α n + 1 1 σ n + 1 x n + 1 x n + | α n + 1 1 σ n + 1 α n 1 σ n | ( f ( x n ) + S n x n + J r n G x n ) + 1 σ n + 1 α n + 1 1 σ n + 1 [ γ n + 1 γ n + 1 + δ n + 1 ( x n + 1 x n + S n + 1 x n S n x n ) + δ n + 1 γ n + 1 + δ n + 1 ( x n x n + 1 + M 0 | r n r n + 1 | ) + | δ n + 1 γ n + 1 + δ n + 1 δ n γ n + δ n | ( S n x n + J r n G x n ) ] = 1 σ n + 1 α n + 1 ( 1 ρ ) 1 σ n + 1 x n + 1 x n + | α n + 1 1 σ n + 1 α n 1 σ n | ( f ( x n ) + S n x n + J r n G x n ) + 1 σ n + 1 α n + 1 1 σ n + 1 [ γ n + 1 γ n + 1 + δ n + 1 S n + 1 x n S n x n + δ n + 1 γ n + 1 + δ n + 1 M 0 | r n r n + 1 | + | δ n + 1 γ n + 1 + δ n + 1 δ n γ n + δ n | ( S n x n + J r n G x n ) ] x n + 1 x n + | α n + 1 1 σ n + 1 α n 1 σ n | M + S n + 1 x n S n x n + M | r n r n + 1 | + | δ n + 1 γ n + 1 + δ n + 1 δ n γ n + δ n | M = x n + 1 x n + M ( | α n + 1 1 σ n + 1 α n 1 σ n | + | δ n + 1 γ n + 1 + δ n + 1 δ n γ n + δ n | + | r n + 1 r n | ) + S n + 1 x n S n x n ,
which hence yields
z n + 1 z n x n + 1 x n M ( | α n + 1 1 σ n + 1 α n 1 σ n | + | δ n + 1 γ n + 1 + δ n + 1 δ n γ n + δ n | + | r n + 1 r n | ) + S n + 1 x n S n x n ,
(3.9)
where sup n 0 { f ( x n ) + S n x n + J r n G x n + M 0 } M for some M > 0 . So, from (3.9), conditions (iii), (iv) and the assumption on { S n } , it follows that
lim sup n ( z n + 1 z n x n + 1 x n ) 0 .
Consequently, by Lemma 2.10, we have
lim n z n x n = 0 .
(3.10)
It follows from (3.5) and (3.6) that
lim n x n + 1 x n = lim n ( 1 σ n ) z n x n = 0 .
(3.11)
From (3.1), we have
x n + 1 x n = α n ( f ( x n ) x n ) + ( 1 α n ) ( y n x n ) ,
which hence implies that
ρ y n x n = ( 1 ( 1 ρ ) ) y n x n ( 1 α n ) y n x n = x n + 1 x n α n ( f ( x n ) x n ) x n + 1 x n + α n ( f ( x n ) x n ) .
Since x n + 1 x n 0 and α n ( f ( x n ) x n ) 0 , we get
lim n y n x n = 0 .
(3.12)

Next, we show that x n G x n 0 as n .

Indeed, 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 for all n 0 . From Lemma 3.1, 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 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 ,
(3.13)
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 .
(3.14)
Substituting (3.13) for (3.14), 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 .
(3.15)
From (3.1) and (3.15), we have
y n p 2 β n x n p 2 + γ n S n x n p 2 + δ n J r n G x n p 2 β n x n p 2 + γ n x n p 2 + δ n v n p 2 β n x n p 2 + γ n x n p 2 + δ 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 ] = x n p 2 2 δ n [ 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 ] ,
(3.16)
which hence implies that
2 δ 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 y n p 2 ( x n p + y n p ) x n y n .
(3.17)
Since 0 < μ i < α i κ 2 for i = 1 , 2 , and { x n } is bounded, we obtain from (3.12), (3.17) and condition (ii) that
lim n B 2 x n B 2 p = 0 and lim n B 1 u n B 1 q = 0 .
(3.18)
Utilizing Proposition 2.2 and Lemma 2.3, 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 .
(3.19)
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 .
(3.20)
Substituting (3.19) for (3.20), 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 .
(3.21)
By Lemma 2.2, we have from (3.16) and (3.21)
y n p 2 β n x n p 2 + γ n x n p 2 + δ n v n p 2 β n x n p 2 + γ n x n p 2 + δ 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 ] x n p 2 δ 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 hence leads to
δ n [ g 1 ( x n u n ( p q ) ) + g 2 ( u n v n + ( p q ) ) ] x n p 2 y n 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 ( x n p + y n p ) x n y n + 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 .
(3.22)
From (3.18), (3.22), condition (ii) and the boundedness of { x n } , { y n } , { u n } and { v n } , we deduce that
lim n g 1 ( x n u n ( p q ) ) = 0 and lim n g 2 ( u n v n + ( p q ) ) = 0 .
Utilizing the properties of g 1 and g 2 , we deduce that
lim n x n u n ( p q ) = 0 and lim n u n v n + ( p q ) = 0 .
(3.23)
From (3.23), we get
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 .
(3.24)
Next, let us show that
lim n S n x n x n = 0 and lim n J r n x n x n = 0 .
Indeed, utilizing Lemma 2.5 and (3.1), we have
y n p 2 = δ n ( J r n G x n p ) + ( β n + γ n ) ( β n x n + γ n S n x n β n + γ n p ) 2 δ n J r n G x n p 2 + ( β n + γ n ) β n x n + γ n S n x n β n + γ n p 2 = δ n J r n G x n p 2 + ( β n + γ n ) β n β n + γ n ( x n p ) + γ n β n + γ n ( S n x n p ) 2 δ n G x n p 2 + ( β n + γ n ) [ β n β n + γ n x n p 2 + γ n β n + γ n S n x n p 2 β n γ n ( β n + γ n ) 2 g 3 ( x n S n x n ) ] δ n x n p 2 + β n x n p 2 + γ n x n p 2 β n γ n β n + γ n g 3 ( x n S n x n ) = x n p 2 β n γ n β n + γ n g 3 ( x n S n x n ) ,
which immediately implies that
β n γ n g 3 ( x n S n x n ) β n γ n β n + γ n g 3 ( x n S n x n ) x n p 2 y n p 2 ( x n p + y n p ) x n y n .
So, from (3.12), the boundedness of { x n } , { y n } and conditions (ii), (v), it follows that
lim n g 3 ( x n S n x n ) = 0 .
From the properties of g 3 , we have
lim n x n S n x n = 0 .
(3.25)
Taking into account that
y n x n = γ n ( S n x n x n ) + δ n ( J r n G x n x n ) ,
we have
δ n J r n G x n x n = y n x n γ n ( S n x n x n ) y n x n + γ n S n x n x n y n x n + S n x n x n .
From (3.12), (3.25) and condition (ii), it follows that
lim n J r n G x n x n = 0 .
(3.26)
Note that
x n S x n x n S n x n + S n x n S x n .
So, in terms of (3.25) and Lemma 2.6, we have
lim n x n S x n = 0 .
(3.27)
Also, note that
x n J r n x n x n J r n G x n + J r n G x n J r n x n x n J r n G x n + G x n x n .
From (3.24) and (3.26), we have
lim n x n J r n x n = 0 .
(3.28)
Furthermore, we claim that lim n x n J r x n = 0 for a fixed number r such that ε > r > 0 . In fact, taking into account the resolvent identity in Proposition 1.1, we have
J r n x n J r x n = J r ( r r n x n + ( 1 r r n ) J r n x n ) J r x n ( 1 r r n ) x n J r n x n x n J r n x n .
(3.29)
Thus, we get from (3.28) and (3.29)
x n J r x n x n J r n x n + J r n x n J r x n x n J r n x n + x n J r n x n = 2 x n J r n x n 0 as  n .
That is,
lim n x n J r x n = 0 .
(3.30)
Define a mapping W x = ( 1 θ 1 θ 2 ) J r x + θ 1 S x + θ 2 G x , where θ 1 , θ 2 ( 0 , 1 ) are two constants with θ 1 + θ 2 < 1 . Then by Lemma 2.8, we have that Fix ( W ) = Fix ( J r ) Fix ( S ) Fix ( G ) = F . We observe that
x n W x n = ( 1 θ 1 θ 2 ) ( x n J r x n ) + θ 1 ( x n S x n ) + θ 2 ( x n G x n ) ( 1 θ 1 θ 2 ) x n J r x n + θ 1 x n S x n + θ 2 x n G x n .
From (3.24), (3.27) and (3.30), we obtain
lim n x n W x n = 0 .
(3.31)
Now, we claim that
lim sup n f ( q ) q , J ( x n q ) 0 ,
(3.32)
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 . Thus, we have
x t x n = ( 1 t ) ( W x t x n ) + t ( f ( x t ) x n ) .
By Lemma 2.2, 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 ,
(3.33)
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 .
(3.34)
It follows from (3.33) that
x t f ( x t ) , J ( x t x n ) t 2 x t x n 2 + 1 2 t f n ( t ) .
(3.35)
Letting n in (3.35) and noticing (3.34), we derive
lim sup n x t f ( x t ) , J ( x t x n ) t 2 M 2 ,
(3.36)
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 (3.36), 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 ) .
(3.37)
Since X has a uniformly Frechet 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 (3.32) holds. From (3.4), we get ( x n + 1 q ) ( x n q ) 0 . Noticing the norm-to-norm uniform continuity of J on bounded subsets of X, we deduce from (3.32) that
lim sup n f ( q ) q , J ( x n + 1 q ) = lim sup n ( f ( q ) q , J ( x n q ) + f ( q ) q , J ( x n + 1 q ) J ( x n q ) ) = lim sup n f ( q ) q , J ( x n q ) 0 .
(3.38)
Finally, let us show that x n q as n . Utilizing Lemma 2.2, from (3.1) and the convexity of 2 , we get
y n q 2 β n x n q 2 + γ n S n x n q 2 + δ n J r n G x n q 2 β n x n q 2 + γ n x n q 2 + δ n x n q 2 = x n q 2 ,
and
x n + 1 q 2 = α n ( f ( x n ) f ( q ) ) + ( 1 α n ) ( y n q ) + α n ( f ( q ) q ) 2 α n ( f ( x n ) f ( q ) ) + ( 1 α n ) ( y n q ) 2 + 2 α n f ( q ) q , J ( x n + 1 q ) α n f ( x n ) f ( q ) 2 + ( 1 α n ) y n q 2 + 2 α n f ( q ) q , J ( x n + 1 q ) α n ρ x n q 2 + ( 1 α n ) x n q 2 + 2 α n f ( q ) q , J ( x n + 1 q ) = ( 1 α n ( 1 ρ ) ) x n q 2 + 2 α n f ( q ) q , J ( x n + 1 q ) = ( 1 α n ( 1 ρ ) ) x n q 2 + α n ( 1 ρ ) 2 f ( q ) q , J ( x n + 1 q ) 1 ρ .
(3.39)

Applying Lemma 2.1 to (3.39), 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 A X × X be an accretive operator in X such that D ( A ) ¯ C r > 0 R ( I + r A ) . Let V : C C be an α-strictly pseudocontractive mapping. Let f : C C be a contraction with coefficient ρ ( 0 , 1 ) . Let { S i } i = 0 be a countable family of nonexpansive mappings of C into itself such that F = i = 0 Fix ( S i ) Fix ( V ) A 1 0 . For arbitrarily given x 0 C , let { x n } be the sequence generated by
{ y n = β n x n + γ n S n x n + δ n J r n ( ( 1 l ) I + l V ) x n , x n + 1 = α n f ( x n ) + ( 1 α n ) y n , n 0 ,
(3.40)
where 0 < l < α κ 2 , and { α n } , { β n } , { γ n } and { δ n } are the sequences in [ 0 , 1 ] such that β n + γ n + δ n = 1 for all n 0 . Suppose that the following conditions hold:
  1. (i)

    n = 0 α n = and 0 α n 1 ρ , n n 0 for some integer n 0