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×CR be a bifunction, D:CH be a nonlinear mapping, and φ:CR be a function. Recently, Peng and Yao [1] considered the following generalized mixed equilibrium problem (GMEP), which involves finding xC such that

F 1 (x,y)+φ(y)φ(x)+Dx,yx0,yC.
(1.1)

The set of solutions of (1.1) is denoted by GMEP(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 xC such that

F 1 (x,y)+φ(y)φ(x)0,yC.
(1.2)

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

If φ=0, then the generalized mixed equilibrium problem (1.1) becomes the following generalized equilibrium problem: Find xC such that

F 1 (x,y)+Dx,yx0,yC.
(1.3)

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

If φ=0 and B=0, then the generalized mixed equilibrium problem (1.1) becomes the following equilibrium problem: Find xC such that

F 1 (x,y)0,yC.
(1.4)

The solution set of (1.4) is denoted by EP(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 uC such that

vu,Au0,vC.
(1.5)

The solution set of (1.5) 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.

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:CH is said to be

  1. (a)

    monotone if

    TxTy,xy0,x,yC;
  2. (b)

    strongly monotone if there exists α>0 such that

    TxTy,xyα x y 2 ,x,yC;
  3. (c)

    α-inverse strongly monotone if there exists α>0 such that

    TxTy,xyα T x T y 2 ,x,yC;
  4. (d)

    nonexpansive if

    TxTyxy,x,yC;
  5. (e)

    k-Lipschitz continuous if there exists a constant k>0 such that

    TxTykxy,x,yC;
  6. (f)

    a contraction on C if there exists a constant 0k<1 such that

    TxTykxy,x,yC.

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:HH 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 xC such that

Tx=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:CH be a nonexpansive mapping. The following problem is called a hierarchical fixed point problem: Find xF(T) such that

xSx,yx0,yF(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:CH 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 zF(T), which is the unique solution of the following variational inequality:

( I f ) z , y z 0,yF(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 ) ) ] ,n0,
(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,xFix(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,zH,vC;
(2.1)
u v , P C [ u ] P C [ v ] P C [ u ] P C [ v ] 2 ,u,vH;
(2.2)
P C [ u ] P C [ v ] uv,u,vH;
(2.3)
u P C [ z ] 2 z u 2 z P C [ z ] 2 ,zH,uC.
(2.4)

Assumption 2.1 [1]

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

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

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

(A3) for each x,y,zC, lim t 0 F 1 (tz+(1t)x,y) F 1 (x,y);

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

(B1) for each xH and r>0, there exists a bounded subset K of C and y x Cdom(φ) such that

F 1 (z, y x )+φ( y x )φ(z)+ 1 r y x z,zx<0,zCK;

(B2) C is a bounded set.

Lemma 2.2 [1]

Let C be a nonempty closed convex subset of H. Let F 1 :C×CR satisfy (A1)-(A3), and let φ:CR be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For r>0 and xH, define a mapping T r :HC 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,yH;
  3. (iii)

    F( T r (IrD))=GMEP(F,φ,D);

  4. (iv)

    GMEP(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:CC is a nonexpansive mapping with Fix(T), then the mapping IT is demiclosed at  0, i.e., if { x n } is a sequence in C weakly converging to x and if {(IT) x n } converges strongly to 0, then (IT)x=0.

Lemma 2.4 [21]

Let U:CH be a τ-Lipschitzian mapping, and let F:CH be a k-Lipschitzian and η-strongly monotone mapping, then for 0ρτ<μη, μFρU is μηρτ-strongly monotone, i.e.,

( μ F ρ U ) x ( μ F ρ U ) y , x y (μηρτ) x y 2 ,x,yC.

Lemma 2.5 [27]

Suppose that λ(0,1) and μ>0. Let F:CH be a k-Lipschitzian and η-strongly monotone operator. In association with a nonexpansive mapping T:CC, define the mapping T λ :CH by

T λ x=TxλμFT(x),xC.

Then T λ is a contraction provided μ< 2 η k 2 , that is,

T λ x T λ y (1λν)xy,x,yC,

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 zC, 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:CH be θ,α-inverse strongly monotone mappings, respectively. Let F 1 :C×CR satisfy (A1)-(A4), and let φ:CR be a proper lower semicontinuous and convex function. Let S,T:CC be nonexpansive mappings such that F(T)VI(C,A)GMEP(F,φ,D). Let F:CC be a k-Lipschitzian mapping and be η-strongly monotone, and let U:CC 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 (0,2α), { r n }(0,2θ). Suppose that the parameters satisfy 0<μ< 2 η k 2 , 0ρτ<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 <2α 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)VI(C,D)GMEP(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,yC,

( 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)VI(C,D)GMEP(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 ρU( x n )+ γ n x n +((1 γ n )I α n μF)(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 n1. 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 ((ρUμF) x +S x x )} for n0 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)VI(C,D)GMEP(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,yC
(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)VI(C,D)GMEP(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 <2α, { r n }(0,2θ), 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, ρU( x n )μF(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,xVI(C,A)GMEP(F,φ,D)F(T).
(3.19)

Proof Since { x n } is bounded x n w and from Lemma 3.2, we have wF(T). Next, we show that wGMEP(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,yC.

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 ),yC

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 ),yC.
(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<t1 and yC, let y t =ty+(1t)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 wφ(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 )+(1t)D y t ,yw0. Letting t 0 + , we have

F 1 ( y t ,y)+φ(y)φ(w)+Dw,yw0,yC,

which implies that wGMEP(F,φ,D).

Furthermore, we show that w Ω . Let

Tv= { A v + N C v , v C , , otherwise ,

where N C v:={wH:w,vu0,uC} is the normal cone to C at vC. Then T is maximal monotone and 0Tv if and only if v Ω (see [31]). Let G(T) denote the graph of T, and let (v,u)G(T); since uAv N C v and z n C, we have

v z n ,uAv0.
(3.24)

On the other hand, it follows from z n = P C [ u n λ n A u n ] and vC 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 vw,u0. Since T is maximal monotone, we have w T 1 0, and hence wVI(C,A). Thus we have

wVI(C,A)GMEP(F,φ,D)F(T).

Observe that the constants satisfy 0ρτ<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 μFρU is μηρτ-strongly monotone, and we get the uniqueness of the solution of the variational inequality (3.19) and denote it by zVI(C,A)GMEP(F,φ,D)F(T).

Next, we claim that lim sup n ρU(z)μF(z), x n z0. 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 ( x n z 2 + x n + 1 z 2 ) + 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 ,

which implies that

x n + 1 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 ) , x n + 1 z + 2 ( 1 γ n α n ν ) β n 1 + α n ( ν ρ τ ) S z z x n + 1 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 ) , x n + 1 z + ( 1 γ n α n ν ) β n α n ( ν ρ τ ) S z z x n + 1 z } .

Let υ n = α n (νρτ) and δ n = 2 α n ( ν ρ τ ) 1 + α n ( ν ρ τ ) { 1 ν ρ τ ρU(z)μF(z), x n + 1 z+ ( 1 γ n α n ν ) β n α n ( ν ρ τ ) Szz x n + 1 z}.

We have

n = 1 α n =

and

lim sup n { 1 ν ρ τ ρ U ( z ) μ F ( z ) , x n + 1 z + ( 1 γ n α n ν ) β n α n ( ν ρ τ ) S z z x n + 1 z } 0.

It follows that

n = 1 υ n =and lim sup n δ n υ n 0.

Thus all the conditions of Lemma 2.6 are satisfied. Hence we deduce that x n z. This completes the proof. □

4 Applications

In this section, we obtain the following results by using a special case of the proposed method for example.

Putting γ n =0 and A=0 in Algorithm 3.1, we obtain the following result which can be viewed as an extension and improvement of the method of Wang and Xu [30] for finding the approximate element of the common set of solutions of a generalized mixed equilibrium problem and a hierarchical fixed point problem in a real Hilbert space.

Corollary 4.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let D:CH be a θ-inverse strongly monotone mapping. Let F 1 :C×CR be a bifunction satisfying assumptions (i)-(iv) of Lemma  2.2 and S,T:CC be nonexpansive mappings such that F(T)MEP( F 1 ). Let F:CC be a k-Lipschitzian mapping and η-strongly monotone, and let U:CC be a τ-Lipschitzian mapping. 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 ; y n = β n S x n + ( 1 β n ) u n ; x n + 1 = P C [ α n ρ U ( x n ) + ( I α n μ F ) ( T ( y n ) ) ] , n 0 ,

where { r n }(0,2θ), { α n }(0,1), { β n }(0,1). Suppose that the parameters satisfy 0<μ< 2 η k 2 , 0ρτ<ν, where ν=1 1 μ ( 2 η μ k 2 ) . Also { α n }, { β n }, and { r n } are sequences satisfying conditions (b)-(e) of Algorithm 3.1. The sequence { x n } converges strongly to z, which is the unique solution of the variational inequality

ρ U ( z ) μ F ( z ) , x z 0,xMEP( F 1 )F(T).

Putting 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 a generalized mixed equilibrium problem and a hierarchical fixed point problem in a real Hilbert space.

Corollary 4.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let D:CH be a θ-inverse strongly monotone mapping. Let F 1 :C×CR be a bifunction satisfying assumptions (i)-(iv) of Lemma  2.2 and S,T:CC be nonexpansive mappings such that F(T)MEP( F 1 ). Let f:CC be a τ-Lipschitzian mapping. 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 ; y n = β n S x n + ( 1 β n ) u n ; x n + 1 = P C [ α n f ( x n ) + ( 1 α n ) T ( y n ) ] , n 0 ,

where { r n }(0,2θ), { α n }, { β n } are sequences in (0,1) satisfying conditions (b)-(e) of Algorithm 3.1. The sequence { x n } converges strongly to z, which is the unique solution of the variational inequality

f ( z ) z , x z 0,xMEP( F 1 )F(T).

Next, the following example shows that conditions (a)-(f) of Algorithm 3.1 are satisfied.

Example 4.1 Let α n = 1 2 n t , γ n = 1 2 n t , β n = n s (with 0<t<s1), λ n = 1 2 ( n + 1 ) , and r n = n n + 1 .

We have

α n + γ n = 1 n t < 1 , lim n α n = lim n γ n = lim n 1 2 n t = 0

and

n = 1 α n = n = 1 1 2 n t =.

Conditions (a) and (b) are satisfied.

lim n β n α n = lim n 1 2 n s t =0.

Condition (c) is satisfied. We compute

α n 1 α n = 1 2 ( 1 ( n 1 ) t 1 n t ) = 1 2 ( n 1 ) t ( 1 ( 1 1 n ) t ) t 2 n t + 1 .

It is easy to show n = 1 | α n 1 α n |<. Similarly, we can show n = 1 | β n 1 β n |< and n = 1 | γ n 1 γ n |<. The sequences { α n }, { γ n }, and { β n } satisfy condition (d). We have

lim inf n r n = lim inf n n n + 1 =1

and

n = 1 | r n 1 r n | = n = 1 | n 1 n n n + 1 | = n = 1 1 n ( n + 1 ) n = 1 1 n 2 < .

Then the sequence { r n } satisfies condition (e). We compute

n = 1 | λ n 1 λ n | < = n = 1 | 1 2 n 1 2 ( n + 1 ) | = 1 2 .

Then the sequence { λ n } satisfies condition (f).

Remark 4.1 In the hierarchical fixed point problem (1.9), if S=I(ρUμF), then we can get the variational inequality (3.19). In (3.19), if U=0 then we get the variational inequality F(z),xz0, xVI(C,A)GMEP(F,φ,D)F(T), which just is the variational inequality studied by Suzuki [27] extending the common set of solutions of a system of variational inequalities, a generalized mixed equilibrium problem and a hierarchical fixed point problem.

5 Conclusions

In this paper, we suggest and analyze 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, which can be viewed as a refinement and improvement of some existing methods for solving a variational inequality problem, a generalized mixed equilibrium problem, and a hierarchical fixed point problem. Some existing methods (e.g., [15, 16, 18, 21, 23, 25]) can be viewed as special cases of Algorithm 3.1. Therefore, the new algorithm is expected to be widely applicable.