Open Access

An iterative method for a common solution of generalized mixed equilibrium problems, variational inequalities, and hierarchical fixed point problems

Fixed Point Theory and Applications20142014:155

https://doi.org/10.1186/1687-1812-2014-155

Received: 11 March 2014

Accepted: 6 June 2014

Published: 22 July 2014

Abstract

In this paper, we suggest and analyze an iterative method for finding a common solution of a variational inequality, a generalized mixed equilibrium problem, and a hierarchical fixed point problem in the setting of real Hilbert spaces. We prove the strong convergence of the sequence generated by the proposed method to a common solution of a variational inequality, a generalized mixed equilibrium problem, and a hierarchical fixed point problem. Several special cases are also discussed. The results presented in this paper extend and improve some well-known results in the literature.

MSC:49J30, 47H09, 47J20.

Keywords

variational inequalities generalized mixed problem hierarchical fixed point problem fixed point problem projection method

1 Introduction

Let H be a real Hilbert space whose inner product and norm are denoted by , and . Let C be a nonempty closed convex subset of H. Let F 1 : C × C R be a bifunction, D : C H be a nonlinear mapping, and φ : C R be a function. Recently, Peng and Yao [1] considered the following generalized mixed equilibrium problem (GMEP), which involves finding x C such that
F 1 ( x , y ) + φ ( y ) φ ( x ) + D x , y x 0 , y C .
(1.1)

The set of solutions of (1.1) is denoted by G M E P ( F , φ , D ) . The GMEP is very general in the sense that it includes, as special cases, optimization problems, variational inequalities, minimax problems, and Nash equilibrium problems; see, for example, [25]. For instance, we quote reference [6] for a general system of generalized equilibrium problems.

Very recently, based on Yamada’s hybrid steepest-descent method [7] and Colao, Marino, and Xu’s hybrid viscosity approximation method [8], Ceng et al. [5] introduced a hybrid iterative method for finding a common element of the set of solutions of a generalized mixed equilibrium problem and the set of fixed points of finitely many nonexpansive mappings in a real Hilbert space. Under suitable assumptions, they proved the strong iterative algorithm to a common solution of problem (1.1) and the fixed point problem of finitely many nonexpansive mappings. By combining Korpelevič’s extragradient method [9], the hybrid steepest-descent method in [7], the viscosity approximation method, and the averaged mapping approach to the gradient-projection algorithm in [10], Al-Mazrooei et al. [2] proposed implicit and explicit iterative algorithms for finding a common element of the set of solutions of the convex minimization problem, the set of solutions of a finite family of generalized mixed equilibrium problems, and the set of solutions of a finite family of variational inequality problems for inverse strong monotone mappings in a real Hilbert space. Under very mild control conditions, they proved that the sequences generated by the proposed algorithms converge strongly to a common element of three sets, which is the unique solution of a variational inequality defined over the intersection of three sets.

If B = 0 , then the generalized mixed equilibrium problem (1.1) becomes the following mixed equilibrium problem: Find x C such that
F 1 ( x , y ) + φ ( y ) φ ( x ) 0 , y C .
(1.2)

Problem (1.2) was studied by Ceng and Yao [11]. The set of solutions of (1.2) is denoted by M E P ( F , φ , D ) .

If φ = 0 , then the generalized mixed equilibrium problem (1.1) becomes the following generalized equilibrium problem: Find x C such that
F 1 ( x , y ) + D x , y x 0 , y C .
(1.3)

Problem (1.3) was studied by Takahashi and Takahashi [12]. The set of solutions of (1.3) is denoted by G E P ( F , D ) .

If φ = 0 and B = 0 , then the generalized mixed equilibrium problem (1.1) becomes the following equilibrium problem: Find x C such that
F 1 ( x , y ) 0 , y C .
(1.4)

The solution set of (1.4) is denoted by E P ( F ) . Numerous problems in physics, optimization, and economics reduce to finding a solution of (1.4), see [13, 14].

Let A be a mapping from C into H. A classical variational inequality problem is to find a vector u C such that
v u , A u 0 , v C .
(1.5)
The solution set of (1.5) is denoted by V I ( C , A ) . It is easy to observe that
u V I ( C , A ) u = P C [ u ρ A u ] , where  ρ > 0 .

We now have a variety of techniques to suggest and analyze various iterative algorithms for solving variational inequalities and related optimization problems; see [141]. The fixed point theory has played an important role in the development of various algorithms for solving variational inequalities. Using the projection operator technique, one usually establishes an equivalence between variational inequalities and fixed point problems. We introduce the following definitions, which are useful in the following analysis.

Definition 1.1 The mapping T : C H is said to be
  1. (a)
    monotone if
    T x T y , x y 0 , x , y C ;
     
  2. (b)
    strongly monotone if there exists α > 0 such that
    T x T y , x y α x y 2 , x , y C ;
     
  3. (c)
    α-inverse strongly monotone if there exists α > 0 such that
    T x T y , x y α T x T y 2 , x , y C ;
     
  4. (d)
    nonexpansive if
    T x T y x y , x , y C ;
     
  5. (e)
    k-Lipschitz continuous if there exists a constant k > 0 such that
    T x T y k x y , x , y C ;
     
  6. (f)
    a contraction on C if there exists a constant 0 k < 1 such that
    T x T y k x y , x , y C .
     
It is easy to observe that every α-inverse strongly monotone T is monotone and Lipschitz continuous. It is well known that every nonexpansive operator T : H H satisfies, for all ( x , y ) H × H , the inequality
( x T ( x ) ) ( y T ( y ) ) , T ( y ) T ( x ) 1 2 ( T ( x ) x ) ( T ( y ) y ) 2
(1.6)
and therefore, we get, for all ( x , y ) H × Fix ( T ) ,
x T ( x ) , y T ( x ) 1 2 T ( x ) x 2 .
(1.7)
The fixed point problem for the mapping T is to find x C such that
T x = x .
(1.8)

We denote by F ( T ) the set of solutions of (1.8). It is well known that F ( T ) is closed and convex, and P F ( T ) is well defined.

Let S : C H be a nonexpansive mapping. The following problem is called a hierarchical fixed point problem: Find x F ( T ) such that
x S x , y x 0 , y F ( T ) .
(1.9)
It is known that the hierarchical fixed point problem (1.9) links with some monotone variational inequalities and convex programming problems; see [15]. Various methods have been proposed to solve the hierarchical fixed point problem; see [1620]. In 2010, Yao et al. [15] introduced the following strong convergence iterative algorithm to solve problem (1.9):
y n = β n S x n + ( 1 β n ) x n , x n + 1 = P C [ α n f ( x n ) + ( 1 α n ) T y n ] , n 0 ,
(1.10)
where f : C H is a contraction mapping and { α n } and { β n } are two sequences in ( 0 , 1 ) . Under some certain restrictions on the parameters, Yao et al. proved that the sequence { x n } generated by (1.10) converges strongly to z F ( T ) , which is the unique solution of the following variational inequality:
( I f ) z , y z 0 , y F ( T ) .
(1.11)
In 2011, Ceng et al. [21] investigated the following iterative method:
x n + 1 = P C [ α n ρ U ( x n ) + ( I α n μ F ) ( T ( y n ) ) ] , n 0 ,
(1.12)
where U is a Lipschitzian mapping, and F is a Lipschitzian and strongly monotone mapping. They proved that under some approximate assumptions on the operators and parameters, the sequence { x n } generated by (1.12) converges strongly to the unique solution of the variational inequality
ρ U ( z ) μ F ( z ) , x z 0 , x Fix ( T ) .

Very recently, Ceng et al. [22] introduced and analyzed hybrid implicit and explicit viscosity iterative algorithms for solving a general system of variational inequalities with hierarchical fixed point problem constraint for a countable family of nonexpansive mappings in a real Banach space, which can be viewed as an extension and improvement of the recent results in the literature.

In this paper, motivated by the work of Ceng et al. [5, 21, 24], Al-Mazrooei et al. [2], Yao et al. [15], Bnouhachem [23] and by the recent work going in this direction, we give an iterative method for finding the approximate element of the common set of solutions of (1.1), (1.5), and (1.9) in a real Hilbert space. We establish a strong convergence theorem based on this method. We would like to mention that our proposed method is quite general and flexible and includes many known results for solving of variational inequality problems, mixed equilibrium problems and hierarchical fixed point problems; see, e.g., [15, 16, 18, 21, 23, 25] and relevant references cited therein.

2 Preliminaries

In this section, we list some fundamental lemmas that are useful in the consequent analysis. The first lemma provides some basic properties of projection onto C.

Lemma 2.1 Let P C denote the projection of H onto C. Then we have the following inequalities:
z P C [ z ] , P C [ z ] v 0 , z H , v C ;
(2.1)
u v , P C [ u ] P C [ v ] P C [ u ] P C [ v ] 2 , u , v H ;
(2.2)
P C [ u ] P C [ v ] u v , u , v H ;
(2.3)
u P C [ z ] 2 z u 2 z P C [ z ] 2 , z H , u C .
(2.4)

Assumption 2.1 [1]

Let F 1 : C × C R be a bifunction and φ : C R be a function satisfying the following assumptions:

(A1) F 1 ( x , x ) = 0 , x C ;

(A2) F 1 is monotone, i.e., F 1 ( x , y ) + F 1 ( y , x ) 0 , x , y C ;

(A3) for each x , y , z C , lim t 0 F 1 ( t z + ( 1 t ) x , y ) F 1 ( x , y ) ;

(A4) for each x C , y F 1 ( x , y ) is convex and lower semicontinuous;

(B1) for each x H and r > 0 , there exists a bounded subset K of C and y x C dom ( φ ) such that
F 1 ( z , y x ) + φ ( y x ) φ ( z ) + 1 r y x z , z x < 0 , z C K ;

(B2) C is a bounded set.

Lemma 2.2 [1]

Let C be a nonempty closed convex subset of H. Let F 1 : C × C R satisfy (A1)-(A3), and let φ : C R be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For r > 0 and x H , define a mapping T r : H C as follows:
T r ( x ) = { z C : F 1 ( z , y ) + φ ( y ) φ ( z ) + 1 r y z , z x 0 , y C } .
Then the following hold:
  1. (i)

    T r is nonempty and single-valued;

     
  2. (ii)
    T r is firmly nonexpansive, i.e.,
    T r ( x ) T r ( y ) 2 T r ( x ) T r ( y ) , x y , x , y H ;
     
  3. (iii)

    F ( T r ( I r D ) ) = G M E P ( F , φ , D ) ;

     
  4. (iv)

    G M E P ( F , φ , D ) is closed and convex.

     

Lemma 2.3 [26]

Let C be a nonempty closed convex subset of a real Hilbert space H. If T : C C is a nonexpansive mapping with Fix ( T ) , then the mapping I T is demiclosed at  0, i.e., if { x n } is a sequence in C weakly converging to x and if { ( I T ) x n } converges strongly to 0, then ( I T ) x = 0 .

Lemma 2.4 [21]

Let U : C H be a τ-Lipschitzian mapping, and let F : C H be a k-Lipschitzian and η-strongly monotone mapping, then for 0 ρ τ < μ η , μ F ρ U is μ η ρ τ -strongly monotone, i.e.,
( μ F ρ U ) x ( μ F ρ U ) y , x y ( μ η ρ τ ) x y 2 , x , y C .

Lemma 2.5 [27]

Suppose that λ ( 0 , 1 ) and μ > 0 . Let F : C H be a k-Lipschitzian and η-strongly monotone operator. In association with a nonexpansive mapping T : C C , define the mapping T λ : C H by
T λ x = T x λ μ F T ( x ) , x C .
Then T λ is a contraction provided μ < 2 η k 2 , that is,
T λ x T λ y ( 1 λ ν ) x y , x , y C ,

where ν = 1 1 μ ( 2 η μ k 2 ) .

Lemma 2.6 [28]

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

    n = 1 υ n = ;

     
  2. (2)

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

     

Then lim n a n = 0 .

Lemma 2.7 [29]

Let C be a closed convex subset of H. Let { x n } be a bounded sequence in H. Assume that
  1. (i)

    the weak w-limit set w w ( x n ) C , where w w ( x n ) = { x : x n i x } ;

     
  2. (ii)

    for each z C , lim n x n z exists.

     

Then { x n } is weakly convergent to a point in C.

3 The proposed method and some properties

In this section, we suggest and analyze our method for finding common solutions of the generalized mixed equilibrium problem (1.1), the variational inequality (1.5), and the hierarchical fixed point problem (1.9).

Let C be a nonempty closed convex subset of a real Hilbert space H. Let D , A : C H be θ , α -inverse strongly monotone mappings, respectively. Let F 1 : C × C R satisfy (A1)-(A4), and let φ : C R be a proper lower semicontinuous and convex function. Let S , T : C C be nonexpansive mappings such that F ( T ) V I ( C , A ) G M E P ( F , φ , D ) . Let F : C C be a k-Lipschitzian mapping and be η-strongly monotone, and let U : C C be a τ-Lipschitzian mapping.

Algorithm 3.1 For an arbitrarily given x 0 C , let the iterative sequences { u n } , { x n } , { y n } , and { z n } be generated by
{ F 1 ( u n , y ) + D x n , y u n + φ ( y ) φ ( u n ) + 1 r n y u n , u n x n 0 , y C ; z n = P C [ u n λ n A u n ] ; y n = β n S x n + ( 1 β n ) z n ; x n + 1 = P C [ α n ρ U ( x n ) + γ n x n + ( ( 1 γ n ) I α n μ F ) ( T ( y n ) ) ] , n 0 ,
where λ n ( 0 , 2 α ) , { r n } ( 0 , 2 θ ) . Suppose that the parameters satisfy 0 < μ < 2 η k 2 , 0 ρ τ < ν , where ν = 1 1 μ ( 2 η μ k 2 ) . Also { γ n } , { α n } , and { β n } are sequences in ( 0 , 1 ) satisfying the following conditions:
  1. (a)

    lim n γ n = 0 , γ n + α n < 1 ,

     
  2. (b)

    lim n α n = 0 and n = 1 α n = ,

     
  3. (c)

    lim n ( β n / α n ) = 0 ,

     
  4. (d)

    n = 1 | α n 1 α n | < , n = 1 | γ n 1 γ n | < , and n = 1 | β n 1 β n | < ,

     
  5. (e)

    lim inf n r n > 0 and n = 1 | r n 1 r n | < ,

     
  6. (f)

    lim inf n λ n < lim sup n λ n < 2 α and n = 1 | λ n 1 λ n | < .

     

Remark 3.1 Our method can be viewed as an extension and improvement for some well-known results, for example, the following.

  • If γ n = 0 , the proposed method is an extension and improvement of the method of Bnouhachem [23] and Wang and Xu [30] for finding the approximate element of the common set of solutions of generalized mixed equilibrium and hierarchical fixed point problems in a real Hilbert space.

  • If we have the Lipschitzian mapping U = f , F = I , ρ = μ = 1 , γ n = 0 , and A = 0 , we obtain an extension and improvement of the method of Yao et al.[15] for finding the approximate element of the common set of solutions of generalized mixed equilibrium and hierarchical fixed point problems in a real Hilbert space.

  • The contractive mapping f with a coefficient α [ 0 , 1 ) in other papers [15, 25, 27] is extended to the cases of the Lipschitzian mapping U with a coefficient constant γ [ 0 , ) .

This shows that Algorithm 3.1 is quite general and unifying.

Lemma 3.1 Let x F ( T ) V I ( C , D ) G M E P ( F , φ , D ) . Then { x n } , { u n } , { z n } , and { y n } are bounded.

Proof First, we show that the mapping ( I r n D ) is nonexpansive. For any x , y C ,
( I r n D ) x ( I r n D ) y 2 = ( x y ) r n ( D x D y ) 2 = x y 2 2 r n x y , D x D y + r n 2 D x D y 2 x y 2 r n ( 2 θ r n ) D x D y 2 x y 2 .
Similarly, we can show that the mapping ( I λ n A ) is nonexpansive. It follows from Lemma 2.2 that u n = T r n ( x n r n D x n ) . Let x F ( T ) V I ( C , D ) G M E P ( F , φ , D ) , we have x = T r n ( x r n D x ) .
u n x 2 = T r n ( x n r n D x n ) T r n ( x r n D x ) 2 ( x n r n D x n ) ( x r n D x ) 2 x n x 2 r n ( 2 θ r n ) D x n D x 2 x n x 2 .
(3.1)
Since the mapping A is α-inverse strongly monotone, we have
z n x 2 = P C [ u n λ n A u n ] P C [ x λ n A x ] 2 u n x λ n ( A u n A x ) 2 u n x 2 λ n ( 2 α λ n ) A u n A x 2 u n x 2 x n x 2 .
(3.2)
We define V n = α n ρ U ( x n ) + γ n x n + ( ( 1 γ n ) I α n μ F ) ( T ( y n ) ) . Next, we prove that the sequence { x n } is bounded, and without loss of generality we can assume that β n α n for all n 1 . From (3.1), we have
x n + 1 x = P C [ V n ] P C [ x ] α n ρ U ( x n ) + γ n x n + ( ( 1 γ n ) I α n μ F ) ( T ( y n ) ) x = α n ( ρ U ( x n ) μ F ( x ) ) + γ n ( x n x ) + ( ( 1 γ n ) I α n μ F ) ( T ( y n ) ) ( ( 1 γ n ) I α n μ F ) ( T ( x ) ) α n ρ U ( x n ) μ F ( x ) + γ n x n x + ( 1 γ n ) ( I α n μ 1 γ n F ) ( T ( y n ) ) ( I α n μ 1 γ n F ) T ( x ) = α n ρ U ( x n ) ρ U ( x ) + ( ρ U μ F ) x + γ n x n x + ( 1 γ n ) ( I α n μ 1 γ n F ) ( T ( y n ) ) ( I α n μ 1 γ n F ) T ( x ) α n ρ τ x n x + α n ( ρ U μ F ) x + γ n x n x + ( 1 γ n ) ( 1 α n ν 1 γ n ) y n x α n ρ τ x n x + α n ( ρ U μ F ) x + γ n x n x + ( 1 γ n α n ν ) β n S x n + ( 1 β n ) z n x α n ρ τ x n x + α n ( ρ U μ F ) x + γ n x n x + ( 1 γ n α n ν ) ( β n S x n S x + β n S x x + ( 1 β n ) z n x ) α n ρ τ x n x + α n ( ρ U μ F ) x + γ n x n x + ( 1 γ n α n ν ) ( β n S x n S x + β n S x x + ( 1 β n ) x n x ) α n ρ τ x n x + α n ( ρ U μ F ) x + γ n x n x + ( 1 γ n α n ν ) ( β n x n x + β n S x x + ( 1 β n ) x n x ) = ( 1 α n ( ν ρ τ ) ) x n x + α n ( ρ U μ F ) x + ( 1 γ n α n ν ) β n S x x ( 1 α n ( ν ρ τ ) ) x n x + α n ( ρ U μ F ) x + β n S x x ( 1 α n ( ν ρ τ ) ) x n x + α n ( ( ρ U μ F ) x + S x x ) = ( 1 α n ( ν ρ τ ) ) x n x + α n ( ν ρ τ ) ν ρ τ ( ( ρ U μ F ) x + S x x ) max { x n x , 1 ν ρ τ ( ( ρ U μ F ) x + S x x ) } ,

where the third inequality follows from Lemma 2.5. By induction on n, we obtain x n x max { x 0 x , 1 ν ρ τ ( ( ρ U μ F ) x + S x x ) } for n 0 and x 0 C . Hence { x n } is bounded, and consequently, we deduce that { u n } , { z n } , { v n } , { y n } , { S ( x n ) } , { T ( x n ) } , { F ( T ( y n ) ) } , and { U ( x n ) } are bounded. □

Lemma 3.2 Let x F ( T ) V I ( C , D ) G M E P ( F , φ , D ) and { x n } be the sequence generated by Algorithm 3.1. Then we have
  1. (a)

    lim n x n + 1 x n = 0 .

     
  2. (b)

    The weak w-limit set w w ( x n ) F ( T ) ( w w ( x n ) = { x : x n i x } ).

     
Proof From the nonexpansivity of the mapping ( I λ n A ) and P C , we have
z n z n 1 ( u n λ n A u n ) ( u n 1 λ n 1 A u n 1 ) = ( u n u n 1 ) λ n ( A u n A u n 1 ) ( λ n λ n 1 ) A u n 1 ( u n u n 1 ) λ n ( A u n A u n 1 ) + | λ n λ n 1 | A u n 1 u n u n 1 + | λ n λ n 1 | A u n 1 .
(3.3)
Next, we estimate that
y n y n 1 β n S x n + ( 1 β n ) z n ( β n 1 S x n 1 + ( 1 β n 1 ) z n 1 ) = β n ( S x n S x n 1 ) + ( β n β n 1 ) S x n 1 + ( 1 β n ) ( z n z n 1 ) + ( β n 1 β n ) z n 1 β n x n x n 1 + ( 1 β n ) z n z n 1 + | β n β n 1 | ( S x n 1 + z n 1 ) .
(3.4)
It follows from (3.3) and (3.4) that
y n y n 1 β n x n x n 1 + ( 1 β n ) { u n u n 1 + | λ n λ n 1 | A u n 1 } + | β n β n 1 | ( S x n 1 + z n 1 ) .
(3.5)
On the other hand, u n = T r n ( x n r n D x n ) and u n 1 = T r n 1 ( x n 1 r n 1 D x n 1 ) , we have
F 1 ( u n , y ) + φ ( y ) φ ( u n ) + D x n , y u n + 1 r n y u n , u n x n 0 , y C
(3.6)
and
F 1 ( u n 1 , y ) + φ ( y ) φ ( u n 1 ) + D x n 1 , y u n 1 + 1 r n 1 y u n 1 , u n 1 x n 1 0 , y C .
(3.7)
Taking y = u n 1 in (3.6) and y = u n in (3.7), we get
F 1 ( u n , u n 1 ) + φ ( u n 1 ) φ ( u n ) + D x n , u n 1 u n + 1 r n u n 1 u n , u n x n 0
(3.8)
and
F 1 ( u n 1 , u n ) + φ ( u n ) φ ( u n 1 ) + D x n 1 , u n u n 1 + 1 r n 1 u n u n 1 , u n 1 x n 1 0 .
(3.9)
Adding (3.8) and (3.9) and using the monotonicity of F 1 , we have
D x n 1 D x n , u n u n 1 + u n u n 1 , u n 1 x n 1 r n 1 u n x n r n 0 ,
which implies that
0 u n u n 1 , r n ( D x n 1 D x n ) + r n r n 1 ( u n 1 x n 1 ) ( u n x n ) = u n 1 u n , u n u n 1 + ( 1 r n r n 1 ) u n 1 + ( x n 1 r n D x n 1 ) ( x n r n D x n ) x n 1 + r n r n 1 x n 1 = u n 1 u n , ( 1 r n r n 1 ) u n 1 + ( x n 1 r n D x n 1 ) ( x n r n D x n ) x n 1 + r n r n 1 x n 1 u n u n 1 2 = u n 1 u n , ( 1 r n r n 1 ) ( u n 1 x n 1 ) + ( x n 1 r n D x n 1 ) ( x n r n D x n ) u n u n 1 2 u n 1 u n { | 1 r n r n 1 | u n 1 x n 1 + ( x n 1 r n D x n 1 ) ( x n r n D x n ) } u n u n 1 2 u n 1 u n { | 1 r n r n 1 | u n 1 x n 1 + x n 1 x n } u n u n 1 2 ,
and then
u n 1 u n | 1 r n r n 1 | u n 1 x n 1 + x n 1 x n .
Without loss of generality, let us assume that there exists a real number μ such that r n > μ > 0 for all positive integers n. Then we get
u n 1 u n x n 1 x n + 1 μ | r n 1 r n | u n 1 x n 1 .
(3.10)
It follows from (3.5) and (3.10) that
y n y n 1 β n x n x n 1 + ( 1 β n ) { x n x n 1 + 1 μ | r n r n 1 | u n 1 x n 1 + | λ n λ n 1 | A u n 1 } + | β n β n 1 | ( S x n 1 + z n 1 ) = x n x n 1 + ( 1 β n ) { 1 μ | r n r n 1 | u n 1 x n 1 + | λ n λ n 1 | A u n 1 } + | β n β n 1 | ( S x n 1 + z n 1 ) .
(3.11)
Next, we estimate
x n + 1 x n = P C [ V n ] P C [ V n 1 ] α n ρ ( U ( x n ) U ( x n 1 ) ) + ( α n α n 1 ) ρ U ( x n 1 ) + γ n ( x n x n 1 ) + ( γ n γ n 1 ) x n 1 + ( 1 γ n ) [ ( I α n μ 1 γ n F ) ( T ( y n ) ) ( I α n μ 1 γ n F ) T ( y n 1 ) ] + ( ( 1 γ n ) I α n μ F ) ( T ( y n 1 ) ) ( ( 1 γ n 1 ) I α n 1 μ F ) ( T ( y n 1 ) ) α n ρ τ x n x n 1 + γ n x n x n 1 + ( 1 γ n ) ( 1 α n ν 1 γ n ) y n y n 1 + | γ n γ n 1 | ( x n 1 + T ( y n 1 ) ) + | α n α n 1 | ( ρ U ( x n 1 ) + μ F ( T ( y n 1 ) ) ) ,
(3.12)
where the second inequality follows from Lemma 2.5. From (3.11) and (3.12), we have
x n + 1 x n α n ρ τ x n x n 1 + γ n x n x n 1 + ( 1 γ n α n ν ) { x n x n 1 + 1 μ | r n r n 1 | u n 1 x n 1 + | λ n λ n 1 | A u n 1 } + | β n β n 1 | ( S x n 1 + z n 1 ) + | γ n γ n 1 | ( x n 1 + T ( y n 1 ) ) + | α n α n 1 | ( ρ U ( x n 1 ) + μ F ( T ( y n 1 ) ) ) ( 1 ( ν ρ τ ) α n ) x n x n 1 + 1 μ | r n r n 1 | u n 1 x n 1 + | λ n λ n 1 | A u n 1 + | β n β n 1 | ( S x n 1 + z n 1 ) + | γ n γ n 1 | ( x n 1 + T ( y n 1 ) ) + | α n α n 1 | ( ρ U ( x n 1 ) + μ F ( T ( y n 1 ) ) ) ( 1 ( ν ρ τ ) α n ) x n x n 1 + M ( 1 μ | r n 1 r n | + | λ n λ n 1 | + | β n β n 1 | + | γ n γ n 1 | + | α n α n 1 | ) .
(3.13)
Here
M = max { sup n 1 u n 1 x n 1 , sup n 1 A u n 1 , sup n 1 ( S x n 1 + z n 1 ) , sup n 1 ( x n 1 + T ( y n 1 ) ) , sup n 1 ( ρ U ( x n 1 ) + μ F ( T ( y n 1 ) ) ) } .
It follows by conditions (a)-(e) of Algorithm 3.1 and Lemma 2.6 that
lim n x n + 1 x n = 0 .
Next, we show that lim n u n x n = 0 . Since x F ( T ) V I ( C , D ) G M E P ( F , φ , D ) , by using (3.1) and (3.2), we obtain
x n + 1 x 2 = P C [ V n ] x , x n + 1 x = P C [ V n ] V n , P C [ V n ] x + V n x , x n + 1 x α n ( ρ U ( x n ) μ F ( x ) ) + γ n ( x n x ) + ( ( 1 γ n ) I α n μ F ) ( T ( y n ) ) ( ( 1 γ n ) I α n μ F ) ( T ( x ) ) , x n + 1 x = α n ρ ( U ( x n ) U ( x ) ) , x n + 1 x + α n ρ U ( x ) μ F ( x ) , x n + 1 x + γ n ( x n x ) , x n + 1 x + ( 1 γ n ) ( I α n μ 1 γ n F ) ( T ( y n ) ) ( I α n μ 1 γ n F ) ( T ( x ) ) , x n + 1 x ( α n ρ τ + γ n ) x n x x n + 1 x + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) y n x x n + 1 x γ n + α n ρ τ 2 ( x n x 2 + x n + 1 x 2 ) + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) 2 ( y n x 2 + x n + 1 x 2 ) ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + γ n + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) 2 ( β n S x n x 2 + ( 1 β n ) z n x 2 ) ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + γ n + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) β n 2 S x n x 2 + ( 1 γ n α n ν ) ( 1 β n ) 2 { u n x 2 λ n ( 2 α λ n ) A u n A x 2 } ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + γ n + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) β n 2 S x n x 2 + ( 1 γ n α n ν ) ( 1 β n ) 2 { x n x 2 r n ( 2 θ r n ) D x n D x 2 λ n ( 2 α λ n ) A u n A x 2 } ,
(3.14)
which implies that
x n + 1 x 2 γ n + α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + ( 1 γ n α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { x n x 2 r n ( 2 θ r n ) D x n D x 2 λ n ( 2 α λ n ) A u n A x 2 } γ n + α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + x n x 2 + ( 1 γ n α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 ( 1 γ n α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { r n ( 2 θ r n ) D x n D x 2 + λ n ( 2 α λ n ) A u n A x 2 } .
Then from the above inequality, we get
( 1 γ n α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { r n ( 2 θ r n ) D x n D x 2 + λ n ( 2 α λ n ) A u n A x 2 } γ n + α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + β n S x n x 2 + x n x 2 x n + 1 x 2 γ n + α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + β n S x n x 2 + ( x n x + x n + 1 x ) x n + 1 x n .

Since lim inf n λ n lim sup n λ n < 2 α , { r n } ( 0 , 2 θ ) , lim n x n + 1 x n = 0 , γ n 0 , α n 0 , and β n 0 , we obtain lim n D x n D x = 0 and lim n A u n A x = 0 .

Since T r n is firmly nonexpansive, we have
u n x 2 = T r n ( x n r n D x n ) T r n ( x r n D x ) 2 u n x , ( x n r n D x n ) ( x r n D x ) = 1 2 { u n x 2 + ( x n r n D x n ) ( x r n D x ) 2 u n x [ ( x n r n D x n ) ( x r n D x ) ] 2 } .
Hence, we get
u n x 2 ( x n r n D x n ) ( x r n D x ) 2 u n x n + r n ( D x n D x ) 2 x n x 2 u n x n + r n ( D x n D x ) 2 x n x 2 u n x n 2 + 2 r n u n x n D x n D x .
From (3.14), (3.2), and the above inequality, we have
x n + 1 x 2 ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + γ n + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) 2 ( β n S x n x 2 + ( 1 β n ) z n x 2 ) ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + γ n + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) 2 ( β n S x n x 2 + ( 1 β n ) u n x 2 ) ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + γ n + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) 2 { β n S x n x 2 + ( 1 β n ) ( x n x 2 u n x n 2 + 2 r n u n x n D x n D x ) } ,
which implies that
x n + 1 x 2 γ n + α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + ( 1 γ n α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { x n x 2 u n x n 2 + 2 r n u n x n D x n D x } γ n + α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + x n x 2 + ( 1 γ n α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) × { u n x n 2 + 2 r n u n x n D x n D x } .
Hence
( 1 γ n α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) u n x n 2 γ n + α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + 2 ( 1 γ n α n ν ) ( 1 β n ) r n 1 + α n ( ν ρ τ ) u n x n D x n D x + x n x 2 x n + 1 x 2 γ n + α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + 2 ( 1 γ n α n ν ) ( 1 β n ) r n 1 + α n ( ν ρ τ ) u n x n D x n D x + ( x n x + x n + 1 x ) x n + 1 x n .
Since lim n x n + 1 x n = 0 , α n 0 , β n 0 , and lim n D x n D x = 0 , we obtain
lim n u n x n = 0 .
(3.15)
From (2.2), we get
z n x 2 = P C [ u n λ n A u n ] P C [ x λ n A x ] 2 z n x , ( u n λ n A u n ) ( x λ n A x ) = 1 2 { z n x 2 + u n x λ n ( A u n A x ) 2 u n x λ n ( A u n A x ) ( z n x ) 2 } 1 2 { z n x 2 + u n x 2 u n z n λ n ( A u n A x ) 2 } 1 2 { z n x 2 + u n x 2 u n z n 2 + 2 λ n u n z n , A u n A x } 1 2 { z n x 2 + u n x 2 u n z n 2 + 2 λ n u n z n A u n A x } .
Hence
z n x 2 u n x 2 u n z n 2 + 2 λ n u n z n A u n A x x n x 2 u n z n 2 + 2 λ n u n z n A u n A x .
From (3.14) and the inequality above, we have
x n + 1 x 2 ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + γ n + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) 2 ( β n S x n x 2 + ( 1 β n ) z n x 2 ) ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + γ n + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) 2 { β n S x n x 2 + ( 1 β n ) ( x n x 2 u n z n 2 + 2 λ n u n z n A u n A x ) } ,
(3.16)
which implies that
x n + 1 x 2 γ n + α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + ( 1 γ n α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { x n x 2 u n z n 2 + 2 λ n u n z n A u n A x } .
Hence
( 1 γ n α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) u n x n 2 γ n + α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + x n x 2 x n + 1 x 2 + 2 λ n u n z n A u n A x = γ n + α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 γ n α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + ( x n x + x n + 1 x ) x n + 1 x n + 2 λ n u n z n A u n A x .
Since lim n x n + 1 x n = 0 , γ n 0 , α n 0 , β n 0 , and lim n A u n A x = 0 , we get
lim n u n z n = 0 .
(3.17)
It follows from (3.15) and (3.17) that
lim n x n z n = 0 , x n T ( y n ) x n x n + 1 + x n + 1 T ( y n ) = x n x n + 1 + P C [ V n ] P C [ T ( y n ) ] x n x n + 1 + α n ( ρ U ( x n ) μ F ( T ( y n ) ) ) + γ n ( x n T ( y n ) ) x n x n + 1 + α n ρ U ( x n ) μ F ( T ( y n ) ) + γ n x n T ( y n ) ,
(3.18)
which implies that
x n T ( y n ) 1 1 γ n x n x n + 1 + α n 1 γ n ρ U ( x n ) μ F ( T ( y n ) ) .
Since lim n x n + 1 x n = 0 , α n 0 , we obtain
lim n x n T ( y n ) = 0 .
Since T ( x n ) C , we have
x n T ( x n ) x n x n + 1 + x n + 1 T ( x n ) = x n x n + 1 + P C [ V n ] P C [ T ( x n ) ] x n x n + 1 + α n ( ρ U ( x n ) μ F ( T ( y n ) ) ) + γ n ( x n T ( y n ) ) + T ( y n ) T ( x n ) x n x n + 1 + α n ρ U ( x n ) μ F ( T ( y n ) ) + γ n x n T ( y n ) + y n x n x n x n + 1 + α n ρ U ( x n ) μ F ( T ( y n ) ) + γ n x n T ( y n ) + β n S x n + ( 1 β n ) z n x n x n x n + 1 + α n ρ U ( x n ) μ F ( T ( y n ) ) + γ n x n T ( y n ) + β n S x n x n + ( 1 β n ) z n x n .
Since lim n x n + 1 x n = 0 , γ n 0 , α n 0 , β n 0 , lim n x n T ( y n ) = 0 , ρ U ( x n ) μ F ( T ( y n ) ) and S x n x n are bounded and lim n x n z n = 0 , we obtain
lim n x n T ( x n ) = 0 .

Since { x n } is bounded, without loss of generality we can assume that x n x C . It follows from Lemma 2.3 that x F ( T ) . Therefore w w ( x n ) F ( T ) . □

Theorem 3.1 The sequence { x n } generated by Algorithm 3.1 converges strongly to z, which is the unique solution of the variational inequality
ρ U ( z ) μ F ( z ) , x z 0 , x V I ( C , A ) G M E P ( F , φ , D ) F ( T ) .
(3.19)
Proof Since { x n } is bounded x n w and from Lemma 3.2, we have w F ( T ) . Next, we show that w G M E P ( F , φ , D ) . Since u n = T r n ( x n r n D x n ) , we have
F 1 ( u n , y ) + φ ( y ) φ ( u n ) + D x n , y u n + 1 r n y u n , u n x n 0 , y C .
It follows from the monotonicity of F 1 that
φ ( y ) φ ( u n ) + D x n , y u n + 1 r n y u n , u n x n F 1 ( y , u n ) , y C
and
φ ( y ) φ ( u n k ) + D x n k , y u n k + y u n k , u n k x n k r n k F 1 ( y , u n k ) , y C .
(3.20)
Since lim n u n x n = 0 and x n w , it is easy to observe that u n k w . For any 0 < t 1 and y C , let y t = t y + ( 1 t ) w , and we have y t C . Then from (3.20), we obtain
D y t , y t u n k φ ( u n k ) φ ( y t ) + D y t , y t u n k D x n k , y t u n k y t u n k , u n k x n k r n k + F 1 ( y t , u n k ) = φ ( u n k ) φ ( y t ) + D y t D u n k , y t u n k + D u n k D x n k , y t u n k y t u n k , u n k x n k r n k + F 1 ( y t , u n k ) .
(3.21)
Since D is Lipschitz continuous and lim n u n x n = 0 , we obtain lim k D u n k D x n k = 0 . From the monotonicity of D, the weakly lower semicontinuity of φ and u n k w , it follows from (3.21) that
D y t , y t w φ ( w ) φ ( y t ) + F 1 ( y t , w ) .
(3.22)
Hence, from assumptions (A1)-(A4) and (3.22), we have
0 = F 1 ( y t , y t ) + φ ( y t ) φ ( y t ) t F 1 ( y t , y ) + ( 1 t ) F 1 ( y t , w ) + t φ ( y ) + ( 1 t ) φ ( w ) φ ( y t ) = t [ F 1 ( y t , y ) + φ ( y ) φ ( y t ) ] + ( 1 t ) [ F 1 ( y t , w ) + φ ( w ) φ ( y t ) ] t [ F 1 ( y t , y ) + φ ( y ) φ ( y t ) ] + ( 1 t ) t D y t , y w ,
(3.23)
which implies that F 1 ( y t , y ) + φ ( y ) φ ( y t ) + ( 1 t ) D y t , y w 0 . Letting t 0 + , we have
F 1 ( y t , y ) + φ ( y ) φ ( w ) + D w , y w 0 , y C ,

which implies that w G M E P ( F , φ , D ) .

Furthermore, we show that w Ω . Let
T v = { A v + N C v , v C , , otherwise ,
where N C v : = { w H : w , v u 0 , u C } is the normal cone to C at v C . Then T is maximal monotone and 0 T v if and only if v Ω (see [31]). Let G ( T ) denote the graph of T, and let ( v , u ) G ( T ) ; since u A v N C v and z n C , we have
v z n , u A v 0 .
(3.24)
On the other hand, it follows from z n = P C [ u n λ n A u n ] and v C that
v z n , z n ( u n λ n A u n ) 0
and
v z n , z n u n λ n + A u n 0 .
Therefore, from (3.24) and the inverse strong monotonicity of A, we have
v z n k , u v z n k , A v v z n k , A v v z n k , z n k u n k λ n k + A u n k v z n k , A v A z n k + v z n k , A z n k A u n k v z n k , z n k u n k λ n k v z n k , A z n k A u n k v z n k , z n k u n k λ n k .
Since lim n u n z n = 0 and u n k w , it is easy to observe that z n k w . Hence, we obtain v w , u 0 . Since T is maximal monotone, we have w T 1 0 , and hence w V I ( C , A ) . Thus we have
w V I ( C , A ) G M E P ( F , φ , D ) F ( T ) .
Observe that the constants satisfy 0 ρ τ < ν and
k η k 2 η 2 1 2 μ η + μ 2 k 2 1 2 μ η + μ 2 η 2 1 μ ( 2 η μ k 2 ) 1 μ η μ η 1 1 μ ( 2 η μ k 2 ) μ η ν ;

therefore, from Lemma 2.4, the operator μ F ρ U is μ η ρ τ -strongly monotone, and we get the uniqueness of the solution of the variational inequality (3.19) and denote it by z V I ( C , A ) G M E P ( F , φ , D ) F ( T ) .

Next, we claim that lim sup n ρ U ( z ) μ F ( z ) , x n z 0 . Since { x n } is bounded, there exists a subsequence { x n k } of { x n } such that
lim sup n ρ U ( z ) μ F ( z ) , x n z = lim sup k ρ U ( z ) μ F ( z ) , x n k z = ρ U ( z ) μ F ( z ) , w z 0 .
Next, we show that x n z . We have
x n + 1 z 2 = P C [ V n ] z , x n + 1 z = P C [ V n ] V n , P C [ V n ] z + V n z , x n + 1 z α n ( ρ U ( x n ) μ F ( z ) ) + γ n ( x n z ) + ( 1 γ n ) [ ( I α n μ 1 γ n F ) ( T ( y n ) ) ( I α n μ 1 γ n F ) ( T ( z ) ) ] , x n + 1 z = α n ρ ( U ( x n ) U ( z ) ) , x n + 1 z + α n ρ U ( z ) μ F ( z ) , x n + 1 z + γ n x n z , x n + 1 z + ( 1 γ n ) ( I α n μ 1 γ n F ) ( T ( y n ) ) ( I α n μ 1 γ n F ) ( T ( z ) ) , x n + 1 z ( γ n + α n ρ τ ) x n z x n + 1 z + α n ρ U ( z ) μ F ( z ) , x n + 1 z + ( 1 γ n α n ν ) y n z x n + 1 z ( γ n + α n ρ τ ) x n z x n + 1 z + α n ρ U ( z ) μ F ( z ) , x n + 1 z + ( 1 γ n α n ν ) { β n S x n S z + β n S z z + ( 1 β n ) z n z } × x n + 1 z ( γ n + α n ρ τ ) x n z x n + 1 z + α n ρ U ( z ) μ F ( z ) , x n + 1 z + ( 1 γ n α n ν ) { β n x n z + β n S z z + ( 1 β n ) x n z } x n + 1 z = ( 1 α n ( ν ρ τ ) ) x n z x n + 1 z + α n ρ U ( z ) μ F ( z ) , x n + 1 z + ( 1 γ n α n ν ) β n S z z x n + 1 z 1 α n ( ν ρ τ ) 2 (