Open Access

A hybrid iterative method for a combination of equilibria problem, a combination of variational inequality problems and a hierarchical fixed point problem

Fixed Point Theory and Applications20142014:163

https://doi.org/10.1186/1687-1812-2014-163

Received: 11 April 2014

Accepted: 27 June 2014

Published: 23 July 2014

Abstract

In this paper, we introduce and analyze a general iterative algorithm for finding a common solution of a combination of variational inequality problems, a combination of equilibria problem, and a hierarchical fixed point problem in the setting of real Hilbert space. Under appropriate conditions we derive the strong convergence results for this method. Several special cases are also discussed. Preliminary numerical experiments are included to verify the theoretical assertions of the proposed method. The results presented in this paper extend and improve some well-known results in the literature.

MSC: 49J30, 47H09, 47J20.

Keywords

combination of equilibria problem variational inequality 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, the equilibrium problem is to find x C such that
F 1 ( x , y ) 0 , y C ,
(1.1)

which was considered and investigated by Blum and Oettli [1]. The solution set of (1.1) is denoted by EP ( F 1 ) . Equilibrium problems theory provides us with a unified, natural, innovative, and general framework to study a wide class of problems arising in finance, economics, network analysis, transportation, elasticity, and optimization. This theory has witnessed an explosive growth in theoretical advances and applications across all disciplines of pure and applied sciences; see [211].

If F 1 ( u , v ) = A u , v u , where A : C H is a nonlinear operator, then problem (1.1) is equivalent to finding a vector u C such that
v u , A u 0 , v C ,
(1.2)
which is known as the classical variational inequality. The solution of (1.2) is denoted by VI ( C , A ) . It is easy to observe that
u VI ( C , A ) u = P C [ u ρ A u ] , where  ρ > 0 .

Variational inequalities are being used as a mathematical programming tool in modeling a large class of problems arising in various branches of pure and applied sciences. In recent years, variational inequalities have been generalized and extended novel and new techniques in several directions. We now have a variety of techniques to suggest and analyze various iterative algorithms for solving variational inequalities and related optimization problems; see [133].

For i = 1 , 2 , , N , let F i : C × C R be bifunctions and a i ( 0 , 1 ) with i = 1 N a i = 1 . Define the mapping i = 1 N a i F i : C × C R . The combination of equilibria problem is to find x C such that
i = 1 N a i F i ( x , y ) 0 , y C ,
(1.3)
which was considered and investigated by Suwannaut and Kangtunyakarn [12]. The set of solutions (1.3) is denoted by
EP ( i = 1 N a i F i ) = i = 1 N EP ( F i ) .

If F i = F 1 , i = 1 , 2 , , N , then the combination of equilibria problem (1.3) reduces to the equilibrium problem (1.1).

For i = 1 , 2 , , N , let A i be a strongly positive linear bounded operator on a Hilbert space H with coefficient ρ i > 0 and b i ( 0 , 1 ) with i = 1 N b i = 1 . The combination of variational inequality problems is to find x C such that
i = 1 N b i A i x , y x 0 , y C .
(1.4)

If A i = A , i = 1 , 2 , , N , then the combination of variational inequality problems (1.4) reduces to the variational inequality problem (1.2).

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)
    strongly positive linear bounded if there exists α > 0 such that
    T x , x α x 2 , x 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
(1.5)
and therefore, we get, for all ( x , y ) H × Fix ( T ) ,
x T ( x ) , y T ( x ) 1 2 T ( x ) x 2 .
(1.6)
The fixed point problem for the mapping T is to find x C such that
T x = x .
(1.7)

We denote by F ( T ) the set of solutions of (1.7). 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 ) .
(1.8)

It is well known that the hierarchical fixed point problem (1.8) links with some monotone variational inequalities and convex programming problems; see [13]. Various methods have been proposed to solve the hierarchical fixed point problem; see [1421]. By combining Korpelevich’s extragradient method and the viscosity approximation method, Ceng et al. [22] introduced and analyzed implicit and explicit iterative schemes for computing a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of the variational inequality for an α-inverse-strongly monotone mapping in a Hilbert space. Under suitable assumptions, they proved the strong convergence of the sequences generated by the proposed schemes.

In 2010, Yao et al. [13] introduced the following strong convergence iterative algorithm to solve problem (1.8):
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 ,
(1.9)
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 (1.9) 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 ) .
(1.10)
In 2011, Ceng et al. [23] 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 ,
(1.11)
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 (1.11) converges strongly to the unique solution of the variational inequality
ρ U ( z ) μ F ( z ) , x z 0 , x Fix ( T ) .
(1.12)
Very recently, in 2013, Wang and Xu [24] investigated an iterative method for a hierarchical fixed point problem by
y n = β n S x n + ( 1 β n ) x n , x n + 1 = P C [ α n ρ U ( x n ) + ( I α n μ F ) ( T ( y n ) ) ] , n 0 ,
(1.13)

where S : C C is a nonexpansive mapping. They proved that under some approximate assumptions on the operators and parameters, the sequence { x n } generated by (1.13) converges strongly to the unique solution of the variational inequality (1.12).

In this paper, motivated by the work of Ceng et al. [23, 26], Yao et al. [13], Wang and Xu [24], Bnouhachem [15, 25] 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 (1.3), (1.4), and (1.8) 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, equilibrium problems, and hierarchical fixed point problems; see, e.g., [13, 16, 18, 23, 25, 27] 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 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.

Lemma 2.2 [2]

Let C be a nonempty closed convex subset of H. Let F 1 : C × C R satisfy (A1)-(A4). Assume that for r > 0 and x H , define a mapping T r : H C as follows:
T r ( x ) = { z C : F 1 ( z , y ) + 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 ) = EP ( F 1 ) ;

     
  4. (iv)

    EP ( F 1 ) is closed and convex.

     

Lemma 2.3 [12]

Let C be a nonempty closed convex subset of a real Hilbert space H. For i = 1 , 2 , , N , let F i : C × C R be bifunctions satisfying (A1)-(A4) with i = 1 N EP ( F i ) . Then i = 1 N a i F i satisfies (A1)-(A4) and
Fix ( T r ) = EP ( i = 1 N a i F i ) = i = 1 N EP ( F i ) ,

where a i ( 0 , 1 ) for i = 1 , 2 , , N and i = 1 N a i = 1 .

Lemma 2.4 [28]

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 2.5 [23]

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 2.6 [29]

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 2.7 [30]

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 2.8 [31]

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 2.9 [12]

Let C be a nonempty closed convex subset of a real Hilbert space H. For every i = 1 , 2 , , N , let A i be a strongly positive linear bounded operator on a Hilbert space H with coefficient ρ i > 0 , i.e., A i x , x ρ i x 2 , x H , and ρ ¯ = min i = 1 , 2 , , N ρ i . Let { b i } i = 1 N ( 0 , 1 ) with i = 1 N b i = 1 . Then the following properties hold:
  1. (i)

    I λ i = 1 N b i A i 1 λ ρ ¯ and I λ i = 1 N b i A i is a nonexpansive mapping for every 0 < λ < A i 1 ( i = 1 , 2 , , N ).

     
  2. (ii)

    VI ( C , i = 1 N b i A i ) = i = 1 N VI ( C , A i ) .

     

3 The proposed method and some properties

In this section, we suggest and analyze our method for finding common solutions of the combination of equilibria problem (1.3), the combination of variational inequality problems (1.4), and the hierarchical fixed point problem (1.8).

Let C be a nonempty closed convex subset of a real Hilbert space H. For i = 1 , 2 , , N , let F i : C × C R be bifunctions satisfying (A1)-(A4), let A i be a strongly positive linear bounded operator on a Hilbert space H with coefficient ρ i > 0 and ρ ¯ = min i = 1 , 2 , , N ρ i , and let S , T : C C be nonexpansive mappings such that F ( T ) i = 1 N EP ( F i ) i = 1 N VI ( C , A i ) . 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
{ i = 1 N a i F i ( u n , y ) + 1 r n y u n , u n x n 0 , y C ; z n = P C [ u n λ n i = 1 N b i A i u n ] ; y n = β n S x n + ( 1 β n ) u n ; x n + 1 = γ n x n + ( 1 γ n ) P C [ α n ρ U ( x n ) + ( I α n μ F ) ( T ( y n ) ) ] , n 0 .
(3.1)
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 } , { β n } , and { r n } are sequences in ( 0 , 1 ) satisfying the following conditions:
  1. (a)

    0 < a γ n b < 1 ,

     
  2. (b)

    lim n α n = 0 and n = 1 α n = ,

     
  3. (c)

    lim n ( β n / α n ) = 0 ,

     
  4. (d)

    i = 1 N a i = i = 1 N b i = 1 ,

     
  5. (e)

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

     
  6. (f)

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

     
  7. (g)

    lim n λ n = 0 and n = 1 | λ n 1 λ n | < .

     

If for i = 1 , 2 , , N , F i = F and A i = A , then Algorithm 3.1 reduces to Algorithm 3.2 for finding the common solutions of equilibrium problem (1.1), variational inequality problem (1.2) and the hierarchical fixed point problem (1.8).

Algorithm 3.2 For an arbitrarily 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 ) + 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 = γ n x n + ( 1 γ n ) P C [ α n ρ U ( x n ) + ( I α n μ F ) ( T ( y n ) ) ] , n 0 .
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 } , { β n } , and { r n } are sequences in ( 0 , 1 ) satisfying the following conditions:
  1. (a)

    0 < a γ n b < 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 n λ n = 0 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 Wang and Xu [24] and Bnouhachem [25] for finding the approximate element of the common set of solutions of a combination of variational inequality problems, a combination of equilibria problem and a hierarchical fixed point problem in a real Hilbert space.

  • If we have the Lipschitzian mapping U = f , F = I , ρ = μ = 1 , and γ n = 0 , we obtain an extension and improvement of the method of Yao et al.[13] for finding the approximate element of the common set of solutions of a combination of variational inequality problems, a combination of equilibria problem and a hierarchical fixed point problem in a real Hilbert space.

  • The contractive mapping f with a coefficient α [ 0 , 1 ) in other papers [13, 27, 29] 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 ) i = 1 N EP ( F i ) i = 1 N VI ( C , A i ) . Then { x n } , { u n } , { z n } , and { y n } are bounded.

Proof Let x F ( T ) i = 1 N EP ( F i ) i = 1 N VI ( C , A i ) ; we have x = T r n ( x ) . It follows from Lemmas 2.2 and 2.3 that u n = T r n ( x n ) . Since T r n is nonexpansive mapping, we have
u n x x n x .
(3.2)
Since lim n λ n = 0 , without loss of generality, we may assume that 0 < λ n < A i 1 , n 0 and i = 1 , 2 , , N , by Lemma 2.9, the mapping I λ n i = 1 N b i A i is nonexpansive mapping, and we have
z n x = P C [ u n λ n i = 1 N b i A i u n ] P C [ x λ n i = 1 N b i A i x ] ( I λ n i = 1 N b i A i ) u n ( I λ n i = 1 N b i A i ) x u n x x n x .
(3.3)
We define V n = α n ρ U ( x 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 = γ n ( x n x ) + ( 1 γ n ) ( P C [ V n ] P C [ x ] ) γ n x n x + ( 1 γ n ) α n ρ U ( x n ) + ( I α n μ F ) ( T ( y n ) ) x γ n x n x + ( 1 γ n ) ( α n ρ U ( x n ) μ F ( x ) + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) T ( x ) ) = γ n x n x + ( 1 γ n ) ( α n ρ U ( x n ) ρ U ( x ) + ( ρ U μ F ) x + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) T ( x ) ) γ n x n x + ( 1 γ n ) ( α n ρ τ x n x + α n ( ρ U μ F ) x + ( 1 α n ν ) y n x ) = γ n x n x + ( 1 γ n ) ( α n ρ τ x n x + α n ( ρ U μ F ) x + ( 1 α n ν ) β n S x n + ( 1 β n ) z n x ) γ n x n x + α n ρ τ ( 1 γ n ) x n x + α n ( 1 γ n ) ( ρ U μ F ) x + ( 1 α n ν ) ( 1 γ n ) ( β n S x n S x + β n S x x + ( 1 β n ) z n x ) γ n x n x + α n ρ τ ( 1 γ n ) x n x + α n ( 1 γ n ) ( ρ U μ F ) x + ( 1 α n ν ) ( 1 γ n ) ( β n S x n S x + β n S x x + ( 1 β n ) x n x ) γ n x n x + α n ρ τ ( 1 γ n ) x n x + α n ( 1 γ n ) ( ρ U μ F ) x + ( 1 α n ν ) ( 1 γ n ) ( β n x n x + β n S x x + ( 1 β n ) x n x ) = ( 1 α n ( ν ρ τ ) ( 1 γ n ) ) x n x + α n ( 1 γ n ) ( ρ U μ F ) x + ( 1 α n ν ) ( 1 γ n ) β n S x x ( 1 α n ( ν ρ τ ) ( 1 γ n ) ) x n x + α n ( 1 γ n ) ( ρ U μ F ) x + β n ( 1 γ n ) S x x ( 1 α n ( ν ρ τ ) ( 1 γ n ) ) x n x + α n ( 1 γ n ) ( ( ρ U μ F ) x + S x x ) = ( 1 α n ( ν ρ τ ) ( 1 γ n ) ) x n x + α n ( 1 γ 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 2.6 and the fifth inequality follows from (3.3). 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 ) i = 1 N EP ( F i ) i = 1 N VI ( C , A i ) 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 i = 1 N b i A i and P C , we have
z n z n 1 ( u n λ n i = 1 N b i A i u n ) ( u n 1 λ n 1 i = 1 N b i A i u n 1 ) = ( u n λ n i = 1 N b i A i u n ) ( u n 1 λ n i = 1 N b i A i u n 1 ) ( λ n λ n 1 ) i = 1 N b i A i u n 1 ( u n λ n i = 1 N b i A i u n ) ( u n 1 λ n i = 1 N b i A i u n 1 ) + | λ n λ n 1 | i = 1 N b i A i u n 1 u n u n 1 + | λ n λ n 1 | i = 1 N b i A i u n 1 .
(3.4)
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 ) .
(3.5)
It follows from (3.4) and (3.5) 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 | i = 1 N b i A i u n 1 } + | β n β n 1 | ( S x n 1 + z n 1 ) .
(3.6)
On the other hand, u n = T r n ( x n ) and u n 1 = T r n 1 ( x n 1 ) , we obtain
i = 1 N a i F i ( u n , y ) + 1 r n y u n , u n x n 0 , y C
(3.7)
and
i = 1 N a i F i ( u n 1 , y ) + 1 r n 1 y u n 1 , u n 1 x n 1 0 , y C .
(3.8)
Taking y = u n 1 in (3.7) and y = u n in (3.8), we get
i = 1 N a i F i ( u n , u n 1 ) + 1 r n u n 1 u n , u n x n 0
(3.9)
and
i = 1 N a i F i ( u n 1 , u n ) + 1 r n 1 u n u n 1 , u n 1 x n 1 0 .
(3.10)
Adding (3.9) and (3.10) and using the monotonicity of i = 1 N a i F i , we have
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 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 + r n r n 1 x n 1 = u n 1 u n , ( 1 r n r n 1 ) u n 1 x n + 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 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 χ such that r n > χ > 0 for all positive integers n. Then we get
u n 1 u n x n 1 x n + 1 χ | r n 1 r n | u n 1 x n 1 .
(3.11)
It follows from (3.6) and (3.11) that
y n y n 1 β n x n x n 1 + ( 1 β n ) { x n x n 1 + 1 χ | r n r n 1 | u n 1 x n 1 + | λ n λ n 1 | i = 1 N b i A i u n 1 } + | β n β n 1 | ( S x n 1 + z n 1 ) = x n x n 1 + ( 1 β n ) { 1 χ | r n r n 1 | u n 1 x n 1 + | λ n λ n 1 | i = 1 N b i A i u n 1 } + | β n β n 1 | ( S x n 1 + z n 1 ) .
(3.12)
Next, we estimate that
x n + 1 x n = ( γ n x n + ( 1 γ n ) P C [ V n ] ) ( γ n 1 x n 1 + ( 1 γ n 1 ) P C [ V n 1 ] ) = γ n ( x n x n 1 ) + ( γ n γ n 1 ) x n 1 + ( 1 γ n ) ( P C [ V n ] P C [ V n 1 ] ) ( γ n γ n 1 ) P C [ V n 1 ] γ n x n x n 1 + | γ n γ n 1 | ( x n 1 + P C [ V n 1 ] ) + ( 1 γ n ) P C [ V n ] P C [ V n 1 ] .
(3.13)
Applying Lemma 2.6 to get
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 ) + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) T ( y n 1 ) + ( I α n μ F ) ( T ( y n 1 ) ) ( I α n 1 μ F ) ( T ( y n 1 ) ) α n ρ τ x n x n 1 + ( 1 α n ν ) y n y n 1 + | α n α n 1 | ( ρ U ( x n 1 ) + μ F ( T ( y n 1 ) ) ) .
(3.14)
From (3.12) and (3.14), we have
P C [ V n ] P C [ V n 1 ] α n ρ τ x n x n 1 + ( 1 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 | i = 1 N b i A i u n 1 ) + | β n β n 1 | ( S x n 1 + z 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 | i = 1 N b i A i u n 1 + | β n β n 1 | ( S x n 1 + z n 1 ) + | α n α n 1 | ( ρ U ( x n 1 ) + μ F ( T ( y n 1 ) ) ) .
(3.15)
Substituting (3.15) into (3.13), we get
x n + 1 x n ( 1 ( ν ρ τ ) ( 1 γ n ) 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 | i = 1 N b i A i u n 1 + | β n β n 1 | ( S x n 1 + z n 1 ) + | α n α n 1 | ( ρ U ( x n 1 ) + μ F ( T ( y n 1 ) ) ) + | γ n γ n 1 | ( x n 1 + P C [ V n 1 ] ) ( 1 ( ν ρ τ ) ( 1 γ n ) 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 | + | γ n γ n 1 | ) .
(3.16)
Here
M = max { sup n 1 u n 1 x n 1 , sup n 1 i = 1 N b i A i u n 1 , sup n 1 ( S x n 1 + z n 1 ) , sup n 1 ( ρ U ( x n 1 ) + μ F ( T ( y n 1 ) ) ) , sup n 1 ( x n 1 + P C [ V n 1 ] ) } .
It follows by conditions (a)-(b), (e)-(g) of Algorithm 3.1 and Lemma 2.7 that
lim n x n + 1 x n = 0 .
Since x n + 1 x n = ( 1 γ n ) ( P C [ V n ] x n ) , we obtain
lim n P C [ V n ] x n = 0 .
(3.17)
Next, we show that lim n u n x n = 0 . Since T r n is firmly nonexpansive, we have
u n x 2 = T r n ( x n ) T r n ( x ) 2 u n x , x n x = 1 2 { u n x 2 + x n x 2 u n x ( x n x ) 2 } .
Hence, we get
u n x 2 x n x 2 u n x n 2 .
From (3.3) and the inequality above, we have
P C [ V n ] x 2 = P C [ V n ] x , P C [ V n ] x = P C [ V n ] V n , P C [ V n ] x + V n x , P C [ V n ] x α n ( ρ U ( x n ) μ F ( x ) ) + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) ( T ( x ) ) , P C [ V n ] x = α n ρ ( U ( x n ) U ( x ) ) , P C [ V n ] x + α n ρ U ( x ) μ F ( x ) , P C [ V n ] x + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) ( T ( x ) ) , P C [ V n ] x α n ρ τ x n x P C [ V n ] x + α n ρ U ( x ) μ F ( x ) , P C [ V n ] x + ( 1 α n ν ) y n x P C [ V n ] x α n ρ τ 2 ( x n x 2 + P C [ V n ] x 2 ) + α n ρ U ( x ) μ F ( x ) , P C [ V n ] x + ( 1 α n ν ) 2 ( y n x 2 + P C [ V n ] x 2 ) ( 1 α n ( ν ρ τ ) ) 2 P C [ V n ] x 2 + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , P C [ V n ] x + ( 1 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 P C [ V n ] x 2 + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , P C [ V n ] x + ( 1 α n ν ) β n 2 S x n x 2 + ( 1 α n ν ) ( 1 β n ) 2 { x n x 2 u n x n 2 } ,
(3.18)
which implies that
P C [ V n ] x 2 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 ) , P C [ V n ] x + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + ( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { x n x 2 u n x n 2 } 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 ) , P C [ V n ] x + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + x n x 2 ( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) u n x n 2 .
Hence,
( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) u n x n 2 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 ) , P C [ V n ] x + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + x n x 2 P C [ V n ] x 2 α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , P C [ V n ] x + ( x n x + P C [ V n ] x ) P C [ V n ] x n .
Since lim n P C [ V n ] x n = 0 , α n 0 , β n 0 , we obtain
lim n u n x n = 0 .
(3.19)
By (2.2) and the nonexpansivity of the mapping I λ n i = 1 N b i A i , we get
z n x 2 = P C [ u n λ n i = 1 N b i A i u n ] P C [ x λ n i = 1 N b i A i x ] 2 z n x , ( u n λ n i = 1 N b i A i u n ) ( x λ n i = 1 N b i A i x ) = 1 2 { z n x 2 + ( I λ n i = 1 N b i A i ) u n ( I λ n i = 1 N b i A i ) x 2 u n x λ n ( i = 1 N b i A i u n i = 1 N b i A i x ) ( z n x ) 2 } 1 2 { z n x 2 + u n x 2 u n z n λ n ( i = 1 N b i A i u n i = 1 N b i A i x ) 2 } 1 2 { z n x 2 + u n x 2 u n z n 2 + 2 λ n u n z n , i = 1 N b i A i u n i = 1 N b i A i x } 1 2 { z n x 2 + u n x 2 u n z n 2 + 2 λ n u n z n i = 1 N b i A i u n i = 1 N b i A i x } .
Hence
z n x 2 u n x 2 u n z n 2 + 2 λ n u n z n i = 1 N b i A i u n i = 1 N b i A i x x n x 2 u n z n 2 + 2 λ n u n z n i = 1 N b i A i u n i = 1 N b i A i x ,
where the second inequality follows from (3.2). From (3.18), and the inequality above, we have
P C [ V n ] x 2 ( 1 α n ( ν ρ τ ) ) 2 P C [ V n ] x 2 + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , P C [ V n ] x + ( 1 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 P C [ V n ] x 2 + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , P C [ V n ] x + ( 1 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 i = 1 N b i A i u n i = 1 N b i A i x ) } ,
which implies that
P C [ V n ] x 2 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 ) , P C [ V n ] x + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + ( 1 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 i = 1 N b i A i u n i = 1 N b i A i x } α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , P C [ V n ] x + x n x 2 + ( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { u n z n 2 + 2 λ n u n z n i = 1 N b i A i u n i = 1 N b i A i x } .
Hence,
( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) u n z n 2 α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , P C [ V n ] x + x n x 2 P C [ V n ] x 2 + ( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { 2 λ n u n z n i = 1 N b i A i u n i = 1 N b i A i x } α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , P C [ V n ] x + ( x n x + P C [ V n ] x ) P C [ V n ] x n + 2 λ n u n z n i = 1 N b i A i u n i = 1 N b i A i x .
Since lim n P C [ V n ] x n = 0 , α n 0 , β n 0 , and lim n λ n = 0 , we obtain
lim n u n z n = 0 .
(3.20)
It follows from (3.19) and (3.20) that
lim n x n z n = 0 .
(3.21)
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 + γ n ( x n T ( x n ) ) + ( 1 γ n ) ( P C [ V n ] T ( x n ) ) x n x n + 1 + γ n x n T ( x n ) + ( 1 γ n ) α n ( ρ U ( x n ) μ F ( T ( y n ) ) ) + T ( y n ) T ( x n ) x n x n + 1 + γ n x n T ( x n ) + α n ( 1 γ n ) ρ U ( x n ) μ F ( T ( y n ) ) + ( 1 γ n ) y n x n = x n x n + 1 + γ n x n T ( x n ) + α n ( 1 γ n ) ρ U ( x n ) μ F ( T ( y n ) ) + ( 1 γ n ) β n S x n + ( 1 β n ) z n x n x n x n + 1 + γ n x n T ( x n ) + α n ( 1 γ n ) ρ U ( x n ) μ F ( T ( y n ) ) + β n ( 1 γ n ) S x n x n + ( 1 β n ) ( 1 γ n ) z n x n ,
which implies that
x n T ( x n ) 1 1 γ n x n x n + 1 + α n ρ U ( x n ) μ F ( 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 , α n 0 , β n 0 , and ρ U ( x n ) μ F ( T ( y n ) ) and S x n x n are bounded, and lim n z n x 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 2.4 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 F ( T ) i = 1 N EP ( F i ) i = 1 N VI ( C , A i ) .
(3.22)
Proof Since { x n } is bounded x n w and from Lemma 3.2, we have w F ( T ) . Next, we show that w i = 1 N EP ( F i ) . Since u n = T r n ( x n ) , we have
i = 1 N a i F i ( u n , y ) + 1 r n y u n , u n x n 0 , y C .
It follows from the monotonicity of i = 1 N a i F i that
1 r n y u n , u n x n i = 1 N a i F i ( y , u n ) , y C
and
y u n k , u n k x n k r n k i = 1 N a i F i ( y , u n k ) , y C .
(3.23)
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 let us have y t C . Then from (3.23), we obtain
0 y t u n k , u n k x n k r n k + i = 1 N a i F i ( y t , u n k ) .
(3.24)
Since u n k w , it follows from (3.24) that
0 i = 1 N a i F i ( y t , w ) .
(3.25)
Since i = 1 N a i F i satisfies (A1)-(A4), it follows from (3.25) that
0 = i = 1 N a i F i ( y t , y t ) t i = 1 N a i F i ( y t , y ) + ( 1 t ) i = 1 N a i F i ( y t , w ) t i = 1 N a i F i ( y t , y ) ,
(3.26)
which implies that i = 1 N a i F i ( y t , y ) 0 . Letting t 0 + , we have
i = 1 N a i F i ( w , y ) 0 , y C ,

therefore, w EP ( i = 1 N a i F i ) = i = 1 N EP ( F i ) .

Furthermore, we show that w i = 1 N VI ( C , A i ) . Let
T v = { i = 1 N b i A i 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 VI ( C , i = 1 N b i A i ) (see [33]). Let G ( T ) denote the graph of T, and let ( v , u ) G ( T ) ; since u i = 1 N b i A i v N C v and z n C , we have
v z n , u i = 1 N b i A i v 0 .
(3.27)
It follows from z n = P C [ u n λ n i = 1 N b i A i u n ] and v C that
v z n , z n ( u n λ n i = 1 N b i A i u n ) 0
and
v z n , z n u n λ n + i = 1 N b i A i u n 0 .
Therefore, from (3.27) and strongly positivity of i = 1 N b i A i , we have
v z n k , u v z n k , i = 1 N b i A i v v z n k , i = 1 N b i A i v v z n k , z n k u n k λ n k + i = 1 N b i A i u n k = v z n k , i = 1 N b i A i v i = 1 N b i A i z n k + v z n k , i = 1 N b i A i z n k i = 1 N b i A i u n k v z n k , z n k u n k λ n k = v z n k , i = 1 N b i A i ( v z n k ) + v z n k , i = 1 N b i A i z n k i = 1 N b i A i u n k v z n k , z n k u n k λ n k v z n k , i = 1 N b i A i z n k i = 1 N b i A i 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 VI ( C , i = 1 N b i A i ) = i = 1 N VI ( C , A i ) . Thus we have
w F ( T ) i = 1 N EP ( F i ) i = 1 N VI ( C , A i ) .
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 2.5, 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 (3.22) and denote it by z F ( T ) i = 1 N EP ( F i ) i = 1 N VI ( C , A i ) .

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 .
By (3.17), we deduce
lim sup n ρ U ( z ) μ F ( z ) , P C [ V n ] z lim sup n ρ U ( z ) μ F ( z ) , P C [ V n ] x n + lim sup n ρ U ( z ) μ F ( z ) , x n z lim sup n ρ U ( z ) μ F ( z ) , x n z 0 .
Next, we show that x n z . Note that
P C [ V n ] z 2 = P C [ V n ] z , P C [ V n ] z = P C [ V n ] V n , P C [ V n ] z + V n z , P C [ V n ] z α n ( ρ U ( x n ) μ F ( z ) ) + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) ( T ( z ) ) , P C [ V n ] z = α n ρ ( U ( x n ) U ( z ) ) , P C [ V n ] z + α n ρ U ( z ) μ F ( z ) , P C [ V n ] z + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) ( T ( z ) ) , P C [ V n ] z α n ρ τ x n z P C [ V n ] z + α n ρ U ( z ) μ F ( z ) , P C [ V n ] z + ( 1 α n ν ) y n z P C [ V n ] z α n ρ τ x n z P C [ V n ] z + α n ρ U ( z ) μ F ( z ) , P C [ V n ] z + ( 1 α n ν ) { β n S x n S z + β n S z z + ( 1 β n ) z n z } P C [ V n ] z α n ρ τ x n z P C [ V n ] z + α n ρ U ( z ) μ F ( z ) , P C [ V n ] z + ( 1 α n ν ) { β n x n z + β n S z z + ( 1 β n ) x n z } P C [ V n ] z = ( 1 α n ( ν ρ τ ) ) x n z P C [ V n ] z + α n ρ U ( z ) μ F ( z ) , P C [ V n ] z + ( 1 α n ν ) β n S z z P C [ V n ] z 1 α n ( ν ρ τ ) 2 ( x n z 2 + P C [ V n ] z 2 ) + α n ρ U ( z ) μ F ( z ) , P C [ V n ] z + ( 1 α n ν ) β n S z z P C [ V n ] z ,
which implies that
P C [ V n ] z 2 1 α n ( ν ρ τ ) 1 + α n ( ν ρ τ ) x n z 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( z ) μ F ( z ) , P C [ V n ] z + 2 ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S z z P C [ V n ] z ( 1 α n ( ν ρ τ ) ) x n z 2 + 2 α n ( ν ρ τ ) 1 + α n ( ν ρ τ ) { 1 ν ρ τ ρ U ( z ) μ F ( z ) , P C [ V n ] z + ( 1 α n ν ) β n α n ( ν ρ τ ) S z z P C [ V n ] z } .
From (3.1) and the inequality above, we get
x n + 1 z 2 γ n x n z 2 + ( 1 γ n ) P C ( V n ) z 2 γ n x n z 2 + ( 1 α n ( ν ρ τ ) ) ( 1 γ n ) x n z 2 + 2 α n ( 1 γ n ) ( ν ρ τ ) 1 + α n ( ν ρ τ ) { 1 ν ρ τ ρ U ( z ) μ F ( z ) , P C [ V n ] z + ( 1 α n ν ) β n α n ( ν ρ τ ) S z z P C [ V n ] z } = ( 1 α n ( ν ρ τ ) ( 1 γ n ) ) x n z 2 + 2 α n ( 1 γ n ) ( ν ρ τ ) 1 + α n ( ν ρ τ ) { 1 ν ρ τ ρ U ( z ) μ F ( z ) , P C [ V n ] z + ( 1 α n ν ) β n α n ( ν ρ τ ) S z z P C [ V n ] z } .
Let
υ n = α n ( 1 γ n ) ( ν ρ τ )
and
δ n = 2 α n ( 1 γ n ) ( ν ρ τ ) 1 + α n ( ν ρ τ ) { 1 ν ρ τ ρ U ( z ) μ F ( z ) , P C [ V n ] z + ( 1 α n ν ) β n α n ( ν ρ τ ) S z z P C [ V n ] z } .
We have
n = 1 α n =
and
lim sup n { 1 ν ρ τ ρ U ( z ) μ F ( z ) ,</