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An iterative method for approximating the common solutions of a variational inequality, a mixed equilibrium problem and a hierarchical fixed point problem

Journal of Inequalities and Applications20132013:490

https://doi.org/10.1186/1029-242X-2013-490

Received: 25 July 2013

Accepted: 30 September 2013

Published: 7 November 2013

Abstract

In this paper, we suggest and analyze an iterative scheme for finding the approximate element of the common set of solutions of a generalized equilibrium problem, a variational inequality problem and a hierarchical fixed point problem in a real Hilbert space. We also consider the strong convergence of the proposed method under some conditions. Results proved in this paper may be viewed as an improvement and refinement of the previously known results.

MSC:49J30, 47H09, 47J20.

Keywords

mixed equilibrium problemvariational inequality problemhierarchical fixed point problemprojection methodstrictly pseudo-contractive mapping

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, and A is a mapping from C into H. A classical variational inequality problem, denoted by VI ( A , C ) , is to find a vector u C such that
v u , A u 0 , v C .
(1.1)
The solution of VI ( A , C ) is denoted by Ω . It is easy to observe that
u Ω 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 the related optimization problems, see [122]. 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 the variational inequalities and the fixed-point problem. This alternative equivalent formulation was used by Lions and Stampacchia [8] to study the existence of a solution of the variational inequalities.

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 an 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 an 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)
    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. A mapping T : C H is called k-strict pseudo-contraction if there exists a constant 0 k < 1 such that
T x T y 2 x y 2 + k ( I T ) x ( I T ) y 2 , x , y C .
(1.2)
The fixed point problem for the mapping T is to find x C such that
T x = x .
(1.3)

We denote F ( T ) the set of solutions of (1.3). It is well known that the class of strict pseudo-contractions strictly includes the class of nonexpansive mappings, then F ( T ) is closed and convex, and P F ( T ) is well defined (see [22]).

The mixed equilibrium problem, denoted by MEP, is to find x C such that
F ( x , y ) + D x , y x 0 , y C ,
(1.4)
where F : C × C R is a bifunction, and D : C H is a nonlinear mapping. This problem was introduced and studied by Moudafi and Théra [13] and Moudafi [14]. The set of solutions of (1.4) is denoted by
MEP ( F ) : = { x C : F ( x , y ) + D x , y x 0 , y C } .
(1.5)
If D = 0 , then it is reduced to the equilibrium problem, which is to find x C such that
F ( x , y ) 0 , y C .
(1.6)

The solution set of (1.6) is denoted by EP ( F ) . Numerous problems in physics, optimization, and economics reduce to find a solution of (1.6), see [4, 7, 16, 17]. In 1997, Combettes and Hirstoaga [5] introduced an iterative scheme of finding the best approximation to the initial data when EP ( F ) is nonempty. Recently Plubtieng and Punpaeng [16] introduced an iterative method for finding the common element of the set F ( T ) Ω EP ( F ) .

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 ) .
(1.7)
It is known that the hierarchical fixed point problem (1.7) links with some monotone variational inequalities and convex programming problems; see [6, 20]. Various methods have been proposed to solve the hierarchical fixed point problem; see Moudafi [15], Mainge and Moudafi in [9], Marino and Xu in [11] and Cianciaruso et al. [3]. Very recently, Yao et al. [20] introduced the following strong convergence iterative algorithm to solve problem (1.7):
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 ,
(1.8)
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 parameters, Yao et al. proved that the sequence { x n } generated by (1.8) 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 ) .
(1.9)
By changing the restrictions on parameters, the authors obtained another result on the iterative scheme (1.8), the sequence { x n } generated by (1.8) converges strongly to a point z F ( T ) , which is the unique solution of the following variational inequality:
1 τ ( I f ) z + ( I S ) z , y z 0 , y F ( T ) .
(1.10)
Let S : C H be a nonexpansive mapping, and { T i } i = 1 : C C is a countable family of nonexpansive mappings. Very recently, Gu et al. [6] introduced the following iterative algorithm:
y n = P C [ β n S x n + ( 1 β n ) x n ] , x n + 1 = P C [ α n f ( x n ) + i = 1 n ( α i 1 α i ) T i y n ] , n 1 ,
(1.11)

where α 0 = 1 , { α n } is a strictly decreasing sequence in ( 0 , 1 ) , and { β n } is a sequence in ( 0 , 1 ) . Under some certain conditions on parameters, Gu et al. proved that the sequence { x n } generated by (1.11) converges strongly to z i = 1 F ( T i ) , which is a unique solution of one of the variational inequalities (1.9) and (1.10).

In this paper, motivated by the work of Yao et al. [20] and Gu et al. [6] 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), (1.4) and (1.7) for a strictly pseudo-contraction mapping in a real Hilbert space. We establish a strong convergence theorem based on this method. The presented method improves and generalizes many known results for solving equilibrium problems, variational inequality problems and hierarchical fixed point problems, see, e.g., [3, 6, 9, 20] 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)

Lemma 2.2 [2]

Let F : C × C R be a bifunction satisfying the following assumptions:
  1. (i)

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

     
  2. (ii)

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

     
  3. (iii)

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

     
  4. (iv)

    For each x C , y F ( x , y ) is convex and lower semicontinuous.

     
Let r > 0 and x H . Then there exists z C such that
F ( z , y ) + 1 r y z , z x 0 , y C .

Lemma 2.3 [5]

Assume that F : C × C R satisfies assumptions (i)-(iv) of Lemma 2.2. For r > 0 and x H , define a mapping T r : H C as follows:
T r ( x ) = { z C : F ( z , y ) + 1 r y z , z x 0 , y C } .
Then the following hold:
  1. (i)

    T r is 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 ) = EP ( F ) ;

     
  4. (iv)

    EP ( F ) is closed and convex.

     

Lemma 2.4 [21]

Let C be a nonempty closed convex subset of a real Hilbert space H. If T : C C is a k-strict pseudo-contraction, then
  1. (i)

    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 ;

     
  2. (ii)

    The set F ( T ) of T is closed and convex, so that the projection P F ( T ) is well defined.

     

Lemma 2.5 [10]

Let H be a real Hilbert space. Then the following inequality holds:
x + y 2 x 2 + 2 y , x + y , x , y H .

Lemma 2.6 [19]

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

    n = 1 γ n = ;

     
  2. (2)

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

     

Then lim n a n = 0 .

Lemma 2.7 [1]

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.

Lemma 2.8 [22]

Let H be a Hilbert space, C be a closed and convex subset of H, and T : C C be a k-strict pseudo-contraction mapping. Define a mapping V : C H by V x = λ x + ( 1 λ ) T x , x C . Then, as k λ < 1 , V is a nonexpansive mapping such that F ( V ) = F ( T ) .

Lemma 2.9 [6]

Let H be a Hilbert space, C be a closed and convex subset of H, and T : C C be a nonexpansive mapping such that F ( T ) . Then
T x x 2 2 x T x , x x , x F ( T ) , x C .

3 The proposed method and some properties

In this section, we suggest and analyze our method for finding the common solutions of the variational inequality (1.1), the mixed equilibrium problem (1.4) and the hierarchical fixed point problem (1.7).

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 : C × C R be a bifunction satisfying assumptions (i)-(iv) of Lemma 2.2, S : C H be a nonexpansive mapping and { T i } i = 1 : C C is a countable family of k i -strict pseudo-contraction mappings such that F ( T ) Ω MEP ( F ) , where F ( T ) = i = 1 F ( T i ) . Let f be a ρ-contraction mapping.

Algorithm 3.1 For a given x 0 C arbitrarily, let the iterative sequences { u n } , { x n } , { y n } and { z n } be generated by
F ( u n , y ) + D x n , y 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 = P C [ β n S x n + ( 1 β n ) z n ] ; x n + 1 = P C [ α n f ( x n ) + i = 1 n ( α i 1 α i ) V i y n ] , n 0 ,
(3.1)
where V i = k i I + ( 1 k i ) T i , 0 k i < 1 , { λ n } ( 0 , 2 α ) , { r n } ( 0 , 2 θ ] , α 0 = 1 , { α n } is a strictly decreasing sequence in ( 0 , 1 ) , and { β n } is a sequence in ( 0 , 1 ) satisfying the following conditions:
  1. (a)

    lim n α n = 0 and n = 1 α n = ,

     
  2. (b)

    lim n ( β n / α n ) = 0 ,

     
  3. (c)

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

     
  4. (d)

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

     
  5. (e)

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

     

Remark 3.1 It is easy to verify that Algorithm 3.1 includes some existing methods (e.g., [3, 6, 9, 20]) as special cases. Therefore, the new algorithm is expected to be widely applicable.

Lemma 3.1 Let x F ( T ) Ω MEP ( F ) . 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 2.3 that u n = T r n ( x n r n D x n ) . Let x F ( T ) Ω MEP ( F ) , 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 .
(3.2)
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 .
(3.3)
Next, we prove that the sequence { x n } is bounded, 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 α n f ( x n ) + i = 1 n ( α i 1 α i ) V i y n α n x i = 1 n ( α i 1 α i ) V i x α n f ( x n ) f ( x ) + α n f ( x ) x + i = 1 n ( α i 1 α i ) V i y n V i x α n f ( x n ) f ( x ) + α n f ( x ) x + i = 1 n ( α i 1 α i ) y n x = α n f ( x n ) f ( x ) + α n f ( x ) x + ( 1 α n ) β n S x n + ( 1 β n ) z n x α n f ( x n ) f ( x ) + α n f ( x ) x + ( 1 α n ) ( β n S x n S x + β n S x x + ( 1 β n ) z n x ) α n ρ x n x + α n f ( x ) x + ( 1 α n ) ( β n x n x + β n S x x + ( 1 β n ) x n x ) = ( 1 α n ( 1 ρ ) ) x n x + α n f ( x ) x + ( 1 α n ) β n S x x ( 1 α n ( 1 ρ ) ) x n x + α n f ( x ) x + β n S x x ( 1 α n ( 1 ρ ) ) x n x + α n ( f ( x ) x + S x x ) = ( 1 α n ( 1 ρ ) ) x n x + α n ( 1 ρ ) 1 ρ ( f ( x ) x + S x x ) max { x n x , 1 1 ρ ( f ( x ) x + S x x ) } .
(3.4)

By induction on n, we obtain x n x max { x 0 x , 1 1 ρ ( f ( x ) 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 } and { y n } are bounded. □

Lemma 3.2 Let x F ( T ) Ω MEP ( F ) 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 .
(3.5)
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 ) .
(3.6)
It follows from (3.5) and (3.6) 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 ) .
(3.7)
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 ( u n , y ) + D x n , y u n + 1 r n y u n , u n x n 0 , y C
(3.8)
and
F ( u n 1 , y ) + 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 .
(3.9)
Take y = u n 1 in (3.8) and y = u n in (3.9), we get
F ( u n , u n 1 ) + D x n , u n 1 u n + 1 r n u n 1 u n , u n x n 0
(3.10)
and
F ( u n 1 , u n ) + 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 .
(3.11)
Adding (3.10) and (3.11) and using the monotonicity of F, 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 .
(3.12)
It follows from (3.7) and (3.12) 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 ) .
(3.13)
Next, we estimate that
x n + 1 x n α n f ( x n ) + i = 1 n ( α i 1 α i ) V i y n ( α n 1 f ( x n 1 ) + i = 1 n 1 ( α i 1 α i ) V i y n 1 ) = α n ( f ( x n ) f ( x n 1 ) ) + ( α n α n 1 ) f ( x n 1 ) + i = 1 n ( α i 1 α i ) ( V i y n V i y n 1 ) + ( α n 1 α n ) V n y n 1 α n f ( x n ) f ( x n 1 ) + i = 1 n ( α i 1 α i ) V i y n V i y n 1 + | α n α n 1 | ( f ( x n 1 ) + V n y n 1 ) α n ρ x n x n 1 + i = 1 n ( α i 1 α i ) y n y n 1 + | α n α n 1 | ( f ( x n 1 ) + V n y n 1 ) = α n ρ x n x n 1 + ( 1 α n ) y n y n 1 + | α n α n 1 | ( f ( x n 1 ) + V n y n 1 ) .
(3.14)
From (3.13) and (3.14), we have
x n + 1 x n α n ρ x n x n 1 + ( 1 α n ) { 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 ) } + | α n α n 1 | ( f ( x n 1 ) + V n y n 1 ) ( 1 ( 1 ρ ) 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 ) + | α n α n 1 | ( f ( x n 1 ) + V n y n 1 ) ( 1 ( 1 ρ ) α n ) x n x n 1 + M ( 1 μ | r n r n 1 | + | λ n λ n 1 | + | β n β n 1 | + | α n α n 1 | ) .
(3.15)
Where
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 ( f ( x n 1 ) + V n 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 ) Ω MEP ( F ) and α n + i = 1 n ( α i 1 α i ) = 1 , by using (3.2) and (3.3), we obtain
x n + 1 x 2 α n f ( x n ) + i = 1 n ( α i 1 α i ) V i y n α n x i = 1 n ( α i 1 α i ) V i x 2 α n f ( x n ) x 2 + i = 1 n ( α i 1 α i ) V i y n V i x 2 α n f ( x n ) x 2 + i = 1 n ( α i 1 α i ) y n x 2 α n f ( x n ) x 2 + ( 1 α n ) ( β n S x n x 2 + ( 1 β n ) z n x 2 ) α n f ( x n ) x 2 + ( 1 α n ) β n S x n x 2 + ( 1 α n ) ( 1 β n ) { 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 } α n f ( x n ) x 2 + β n S x n x 2 + x n x 2 ( 1 α n ) ( 1 β n ) { r n ( 2 θ r n ) D x n D x 2 + λ n ( 2 α λ n ) A u n A x 2 } .
(3.16)
Then from the inequality above, we get
( 1 α n ) ( 1 β n ) { r n ( 2 θ r n ) D x n D x 2 + λ n ( 2 α λ n ) A u n A x 2 } α n f ( x n ) x 2 + β n S x n x 2 + x n x 2 x n + 1 x 2 α n f ( x n ) x 2 + 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 , lim n x n + 1 x 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,
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 (3.16), (3.3) and the inequality above, we have
x n + 1 x 2 α n f ( x n ) x 2 + ( 1 α n ) ( β n S x n x 2 + ( 1 β n ) z n x 2 ) α n f ( x n ) x 2 + ( 1 α n ) ( β n S x n x 2 + ( 1 β n ) u n x 2 ) α n f ( x n ) x 2 + ( 1 α n ) { 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 ) } α n f ( x n ) x 2 + β n S x n x 2 + x n x 2 ( 1 α n ) ( 1 β n ) u n x n 2 + 2 r n u n x n D x n D x .
Hence,
( 1 α n ) ( 1 β n ) u n x n 2 α n f ( x n ) x 2 + β n S x n x 2 + x n x 2 x n + 1 x 2 + 2 r n u n x n D x n D x α n f ( x n ) x 2 + β n S x n x 2 + ( x n x + x n + 1 x ) x n + 1 x n + 2 r n u n x n D x n D x .
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 .
(3.17)
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 (3.16) and the inequality above, we have
x n + 1 x 2 α n f ( x n ) x 2 + ( 1 α n ) ( β n S x n x 2 + ( 1 β n ) z n x 2 ) α n f ( x n ) x 2 + ( 1 α n ) { 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 ) } α n f ( x n ) x 2 + β n S x n x 2 + x n x 2 ( 1 α n ) ( 1 β n ) u n z n 2 + 2 λ n u n z n A u n A x .
Hence,
( 1 α n ) ( 1 β n ) u n z n 2 α n f ( x n ) x 2 + β n 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 α n f ( x n ) x 2 + β n 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 , α n 0 , β n 0 and lim n A u n A x = 0 , we obtain
lim n u n z n = 0 .
(3.18)
It follows from (3.17) and (3.18) that
lim n x n z n = 0 .
(3.19)
Now, let z F ( T ) Ω MEP ( F ) , since for each i 1 , V i x n C and α n + i = 1 n ( α i 1 α i ) = 1 , we have i = 1 n ( α i 1 α i ) V i x n + α n z C . And
i = 1 n ( α i 1 α i ) ( x n V i x n ) = P C [ α n f ( x n ) + i = 1 n ( α i 1 α i ) V i y n ] + ( 1 α n ) x n ( i = 1 n ( α i 1 α i ) V i x n + α n z ) + α n z x n + 1 = P C [ α n f ( x n ) + i = 1 n ( α i 1 α i ) V i y n ] + α n ( z x n + 1 ) P C [ i = 1 n ( α i 1 α i ) V i x n + α n z ] + ( 1 α n ) ( x n x n + 1 ) .
It follows that
i = 1 n ( α i 1 α i ) x n V i x n , x n x = P C [ α n f ( x n ) + i = 1 n ( α i 1 α i ) V i y n ] P C [ i = 1 n ( α i 1 α i ) V i x n + α n z ] , x n x + α n z x n + 1 , x n x + ( 1 α n ) x n x n + 1 , x n x α n ( f ( x n ) z ) + i = 1 n ( α i 1 α i ) ( V i y n V i x n ) x n x + α n z x n + 1 x n x + ( 1 α n ) x n x n + 1 x n x α n f ( x n ) z x n x + i = 1 n ( α i 1 α i ) y n x n x n x + α n z x n + 1 x n x + ( 1 α n ) x n x n + 1 x n x = α n f ( x n ) z x n x + ( 1 α n ) y n x n x n x + α n z x n + 1 x n x + ( 1 α n ) x n x n + 1 x n x α n f ( x n ) z x n x + ( 1 α n ) β n S x n + ( 1 β n ) z n x n x n x + α n z x n + 1 x n x + ( 1 α n ) x n x n + 1 x n x α n f ( x n ) z x n x + ( 1 α n ) β n S x n x n x n x + ( 1 α n ) ( 1 β n ) z n x n x n x + α n z x n + 1 x n x + ( 1 α n ) x n x n + 1 x n x .
From Lemma 2.9 and the inequality above, we get
1 2 i = 1 n ( α i 1 α i ) x n V i x n 2 i = 1 n ( α i 1 α i ) x n V i x n , x n x α n f ( x n ) z x n x + ( 1 α n ) β n S x n x n x n x + ( 1 α n ) ( 1 β n ) z n x n x n x + α n z x n + 1 x n x + ( 1 α n ) x n x n + 1 x n x .
Since lim n x n + 1 x n = 0 , α n 0 , β n 0 and lim n x n z n = 0 , we obtain
lim n i = 1 n ( α i 1 α i ) x n V i x n 2 = 0 .
Since ( α i 1 α i ) x n V i x n 2 i = 1 n ( α i 1 α i ) x n V i x n 2 and { α n } is strictly decreasing, we have
lim n x n V i x n = 0 .
Hence, we obtain
lim n x n T i x n = lim n x n V i x n ( 1 k i ) = 0 , i 1 .

Since { x n } is bounded, without loss of generality, we can assume that x n w C . It follows from Lemma 2.4 that w 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 = P Ω MEP ( F ) F ( T ) f ( z ) , which is the unique solution of the variational inequality
( I f ) z , x z 0 , x Ω MEP ( F ) F ( T ) .
(3.20)
Proof Since { x n } is bounded x n w and from Lemma 3.2, we have w F ( T ) . Next, we show that w MEP ( F ) . Since u n = T r n ( x n r n D x n ) , we have
F ( u n , y ) + D x n , y u n + 1 r n y u n , u n x n 0 , y C .
It follows from monotonicity of F that
D x n , y u n + 1 r n y u n , u n x n F ( y , u n ) , y C
and
D x n k , y u n k + y u n k , u n k x n k r n k F ( y , u n k ) , y C .
(3.21)
Since lim n u n x n = 0 and x n w , it easy to observe that u n k w . For any 0 < t 1 and y C , let y t = t y + ( 1 t ) w , we have y t C . Then from (3.21), we obtain
D y t , y t u n k 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 ( y t , u n k ) = 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 ( y t , u n k ) .
(3.22)
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 and u n k w , it follows from (3.22) that
D y t , y t w F ( y t , w ) .
(3.23)
Hence, from assumptions (i)-(iv) of Lemma 2.2 and (3.23), we have
0 = F ( y t , y t ) t F ( y t , y ) + ( 1 t ) F ( y t , w ) t F ( y t , y ) + ( 1 t ) D y t , y t w t F ( y t , y ) + ( 1 t ) t D y t , y w ,
(3.24)
which implies that F ( y t , y ) + ( 1 t ) D y t , y w 0 . Letting t 0 + , we have
F ( w , y ) + D w , y w 0 , y C ,

which implies that w MEP ( F ) .

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 [18]). 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 .
(3.25)
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 (3.25) and inverse strongly 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 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 Ω . Thus, we have
w Ω MEP ( F ) F ( T ) .

Next, we claim that lim sup n f ( z ) z , x n z 0 , where z = P Ω MEP ( F ) F ( T ) f ( z ) .

Since { x n } is bounded, there exists a subsequence { x n k } of { x n } such that
lim sup n f ( z ) z , x n z = lim sup k f ( z ) z , x n k z = f ( z ) z , w z 0 .
Next, we show that x n z .
x n + 1 z 2 = x n + 1 α n f ( x n ) i = 1 n ( α i 1 α i ) V i y n , x n + 1 z + α n f ( x n ) + i = 1 n ( α i 1 α i ) V i y n z , x n + 1 z α n f ( x n ) + i = 1 n ( α i 1 α i ) V i y n z , x n + 1 z α n f ( x n ) f ( z ) , x n + 1 z + α n f ( z ) z , x n + 1 z + i = 1 n ( α i 1 α i ) V i y n z , x n + 1 z α n f ( x n ) f ( z ) x n + 1 z + α n f ( z ) z , x n + 1 z + i = 1 n ( α i 1 α i ) V i y n z x n + 1 z α n ρ x n z x n + 1 z + α n f ( z ) z , x n + 1 z + i = 1 n ( α i 1 α i ) y n z x n + 1 z α n ρ x n z x n + 1 z + α n f ( z ) z , x n + 1 z + ( 1 α n ) { β n S x n S z + β n S z z + ( 1 β n ) z n z } x n + 1 z α n ρ x n z x n + 1 z + α n f ( z ) z , x n + 1 z + ( 1 α n ) { β n x n z + β n S z z + ( 1 β n ) x n z } x n + 1 z ( 1 α n ( 1 ρ ) ) x n z x n + 1 z + α n f ( z ) z , x n + 1 z + ( 1 α n ) β n S z z x n + 1 z 1 α n ( 1 ρ ) 2 ( x n z 2 + x n + 1 z 2 ) + α n f ( z ) z , x n + 1 z + ( 1 α n ) β n S z z x n + 1 z ,
which implies that
x n + 1 z 2 ( 1 2 α n ( 1 ρ ) 1 + α n ( 1 ρ ) ) x n z 2 + 2 α n 1 + α n ( 1 ρ ) f ( z ) z , x n + 1 z + 2 ( 1 α n ) β n 1 + α n ( 1 ρ ) S z z x n + 1 z ( 1 2 α n ( 1 ρ ) 1 + α n ( 1 ρ ) ) x n z 2 + 2 α n ( 1 ρ ) 1 + α n ( 1 ρ ) { 1 1 ρ f ( z ) z , x n + 1 z + ( 1 α n ) β n α n ( 1 ρ ) S z z x n + 1 z } .

Let γ n = 2 α n ( 1 ρ ) 1 + α n ( 1 ρ ) and δ n = 2 α n ( 1 ρ ) 1 + α n ( 1 ρ ) { 1 1 ρ f ( z ) z , x n + 1 z + ( 1 α n ) β n α n ( 1 ρ ) S z z x n + 1 z } .

Since
n = 1 α n = , 1 + α n ( 1 ρ ) 2 and lim sup n { 1 1 ρ f ( z ) z , x n + 1 z + ( 1 α n ) β n α n ( 1 ρ ) S z z x n + 1 z } 0 .
It follows that
n = 1 γ n = and lim sup n δ n γ n 0 .

Thus, all the conditions of Lemma 2.6 are satisfied. Hence, we deduce that x n z .

Since P Ω MEP ( F ) F ( T ) f is a contraction, there exists a unique z C such that z = P Ω MEP ( F ) F ( T ) f ( z ) . From (2.1), it follows that z is the unique solution of problem (3.20). This completes the proof. □

Theorem 3.2 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 : C × C R be a bifunction satisfying the assumptions (i)-(iv) of Lemma 2.2, S : C H be a nonexpansive mapping, and { T i } i = 1 : C C is a countable family of k i -strict pseudo-contraction mappings such that F ( T ) Ω MEP ( F ) , where F ( T ) = i = 1 F ( T i ) . Let f be a ρ-contraction mapping. For a given x 0 C arbitrarily, let the iterative sequences { u n } , { x n } , { y n } and { z n } be generated by
F ( u n , y ) + D x n , y 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 f ( x n ) + i = 1 n ( α i 1 α i ) V i y n ] , n 0 ,
(3.26)
where V i = k i I + ( 1 k i ) T i , 0 k i < 1 , α 0 = 1 , { α n } is a strictly decreasing sequence in ( 0 , 1 ) , and { β n } is a sequence in ( 0 , 1 ) satisfying the following conditions:
  1. (a)

    lim n α n = 0 and n = 1 α n = ,

     
  2. (b)

    lim n β n α n = τ ( 0 , ) ,

     
  3. (c)

    n = 1 ( α n 1 α n ) < and n = 1 | β n 1 β n | < ,

     
  4. (d)

    lim n 1 μ | r n r n 1 | + | λ n λ n 1 | + | α n 1 α n | + | β n 1 β n | α n β n = 0 ,

     
  5. (e)

    there exists a constant K > 0 such that 1 α n | 1 β n 1 β n 1 | K ,

     
  6. (f)

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

     
  7. (g)

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

     
Then sequence { x n } generated by (3.26) converges strongly to x Ω MEP ( F ) F ( T ) , which is the unique solution of the variational inequality
1 τ ( I f ) x + ( I S ) x , x x 0 , x Ω MEP ( F ) F ( T ) .
(3.27)
Proof From lim n ( β n / α n ) = τ ( 0 , ) , without loss of generality, we can assume that β n ( 1 + τ ) α n for all n 1 . Hence, β n 0 . By similar argument as that in Lemmas 3.1 and 3.2, we can deduce that { x n } is bounded, lim n x n + 1 x n = 0 , lim n x n z n = 0 (see (3.19)) and ( I V i ) x n 0 . Then we have
y n x n β n x n S x n + ( 1 β n ) x n z n 0 as  n .
(3.28)
It follows that for all i 1 ,
y n V i x n y n x n + x n V i x n 0 as  n .
(3.29)
From (3.28) and (3.29), we have
y n V i y n y n V i x n + V i x n V i y n y n V i x n + y n x n 0 as  n .
Set w n = α n f ( x n ) + i = 1 n ( α i 1 α i ) V i y n . From (3.14) and (3.15), we obtain
x n + 1 x n β n w n w n 1 β n ( 1 ( 1 ρ ) α n ) x n x n 1 β n + M ( 1 μ | r n r n 1 | β n + | λ n λ n 1 | β n + | β n β n 1 | β n + | α n α n 1 | β n ) = ( 1 ( 1 ρ ) α n ) x n x n 1 β n 1 + ( 1 ( 1 ρ ) α n ) x n x n 1 ( 1 β n 1 β n 1 ) + M ( 1 μ | r n r n 1 | β n + | λ n λ n 1 | β n + | β n β n 1 | β n + | α n α n 1 | β n ) ( 1 ( 1 ρ ) α n ) x n x n 1 β n 1 + x n x n 1 | 1 β n 1 β n 1 | + M ( 1 μ | r n r n 1 | β n + | λ n λ n 1 | β n + | β n β n 1 | β n + | α n α n 1 | β n ) ( 1 ( 1 ρ ) α n ) x n x n 1 β n 1 + α n K x n x n 1 + M ( 1 μ | r n r n 1 | β n + | λ n λ n 1 | β n + | β n β n 1 | β n + | α n α n 1 | β n ) ( 1 ( 1 ρ ) α n ) w n 1 w n 2 β n 1 + α n K x n x n 1 + M ( 1 μ | r n r n 1 | β n + | λ n λ n 1 | β n + | β n β n 1 | β n + | α n α n 1 | β n ) .
Let γ n = ( 1 ρ ) α n and δ n = α n K x n x n 1 + M ( 1 μ | r n r n 1 | β n + | λ n λ n 1 | β n + | β n β n 1 | β n + | α n α n 1 | β n ) . From conditions (a) and (d) of Theorem 3.2, we have
n = 1 γ n = and lim n δ n γ n = 0 .
By Lemma 2.6, we obtain
lim n x n + 1 x n β n = 0 , lim n w n + 1 w n β n = lim n w n + 1 w n α n = 0 .
From (3.26), we have
x n + 1 = P C [ w n ] w n + α n f ( x n ) + i = 1 n ( α i 1 α i ) ( V i y n y n ) + ( 1 α n ) y n .
Hence, it follows that
x n x n + 1 = ( 1 α n ) x n + α n x n ( P C [ w n ] w n + α n f ( x n ) + i = 1 n ( α i 1 α i ) ( V i y n y n ) + ( 1 α n ) y n ) = ( 1 α n ) [ β n ( x n S x n ) + ( 1 β n ) ( x n z n ) ] + ( w n P C [ w n ] ) + i = 1 n ( α i 1 α i ) ( y n V i y n ) + α n ( x n f ( x n ) ) ,
and hence,
x n x n + 1 ( 1 α n ) β n = x n S x n + ( 1 β n ) β n ( x n z n ) + 1 ( 1 α n ) β n ( w n P C [ w n ] ) + 1 ( 1 α n ) β n i = 1 n ( α i 1 α i ) ( y n V i y n ) + α n ( 1 α n ) β n ( x n f ( x n ) ) .
Let v n = x n x n + 1 ( 1 α n ) β n . For any z Ω MEP ( F ) F ( T ) , we have
v n , x n z = 1 ( 1 α n ) β n w n P C [ w n ] , P C [ w n 1 ] z + α n ( 1 α n ) β n ( I f ) x n , x n z + x n S x n , x n z + ( 1 β n ) β n x n z n , x n z + 1 ( 1 α n ) β n i = 1 n ( α i 1 α i ) y n V i y n , x n z .
(3.30)
Since S is a nonexpansive mapping, f is a ρ-contraction mapping, and V i is a k i -strict pseudo-contraction mapping. Then ( I S ) and ( I V i ) are monotones, and f is strongly monotone with coefficient ( 1 ρ ) . We can deduce
x n S x n , x n z = ( I S ) x n ( I S ) z , x n z + ( I S ) z , x n z ( I S ) z , x n z ,
(3.31)
( I f ) x n , x n z = ( I f ) x n ( I f ) z , x n z + ( I f ) z , x n z ( 1 ρ ) x n z 2 + ( I f ) z , x n z ,
(3.32)