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Strong convergence theorems for a generalized mixed equilibrium problem and variational inequality problems

Abstract

In this paper, a new iterative scheme based on the extragradient-like method for finding a common element of the set of common fixed points of a finite family of nonexpansive mappings, the set of solutions of variational inequalities for a strongly positive linear bounded operator and the set of solutions of a mixed equilibrium problem is proposed. A strong convergence theorem for this iterative scheme in Hilbert spaces is established. Our results extend recent results announced by many others.

MSC:49J30, 49J40, 47J25, 47H09.

1 Introduction

Let H be a real Hilbert space with the inner product , and the norm . Let C be a nonempty closed convex subset of H. Recall that a mapping T:CC is said to be nonexpansive if

TxTyxy,x,yC.

We denote by F(T) the set of fixed points of T. Let P C be the projection of H onto the convex subset C. Moreover, we also denote by the set of all real numbers.

Peng and Yao [1] considered the generalized mixed equilibrium problem of finding xC such that

Θ(x,y)+φ(y)φ(x)+Fx,yx0,yC,
(1.1)

where F:CH is a nonlinear mapping and φ:CR is a function and Θ:C×CR is a bifunction. The set of solutions of problem (1.1) is denoted by GMEP.

In the case of F=0, problem (1.1) reduces to the mixed equilibrium problem of finding xC such that

Θ(x,y)+φ(y)φ(x)0,yC,

which was considered by Ceng and Yao [2]. GMEP is denoted by MEP.

In the case of φ=0, problem (1.1) reduces to the generalized equilibrium problem of finding xC such that

Θ(x,y)+Fx,yx0,yC,

which was studied by Takahashi and Takahashi [3] and many others, for example, [410].

In the case of φ=0 and F=0, problem (1.1) reduces to the equilibrium problem of finding xC such that

Θ(x,y)0,yC.
(1.2)

The set of solutions of (1.2) is denoted by EP(Θ).

In the case Θ=0, φ=0 and F=A, problem (1.1) reduces to the classical variational inequality problem of finding xC such that

Ax,yx0,yC.
(1.3)

The set of solutions of problem (1.3) is denoted by VI(A,C).

Problem (1.1) is very general in the sense that it includes, as special cases, optimization problems, variational inequalities, minimax problems, the Nash equilibrium problem in noncooperative games and others; see, for instance, [2, 3, 11]. Peng and Yao [1] considered iterative methods for finding a common element of the set of solutions of problem (1.1), the set of solutions of problem (1.2) and the set of fixed points of a nonexpansive mapping.

Let G 1 , G 2 :C×CR be two bifunctions and let B 1 , B 2 :CH be two nonlinear mappings. We consider the generalized equilibrium problem ( x ¯ , y ¯ )C×C such that

{ G 1 ( x ¯ , x ) + B 1 y ¯ , x x ¯ + 1 μ 1 x ¯ y ¯ , x x ¯ 0 , x C , G 2 ( y ¯ , y ) + B 2 x ¯ , y y ¯ + 1 μ 2 y ¯ x ¯ , y y ¯ 0 , y C ,
(1.4)

where μ 1 >0 and μ 2 >0 are two constants.

In the case G 1 = G 2 =0, problem (1.4) reduces to the general system of variational inequalities of finding ( x ¯ , y ¯ )C×C such that

{ μ 1 B 1 y ¯ + x ¯ y ¯ , x x ¯ 0 , x C , μ 2 B 2 x ¯ + y ¯ x ¯ , y y ¯ 0 , y C ,
(1.5)

where μ 1 >0 and μ 2 >0 are two constants, which was considered by Ceng, Wang and Yao [12]. In particular, if B 1 = B 2 =A, then problem (1.5) reduces to the system of variational inequalities of finding ( x ¯ , y ¯ )C×C such that

{ μ 1 A y ¯ + x ¯ y ¯ , x x ¯ 0 , x C , μ 2 A x ¯ + y ¯ x ¯ , y y ¯ 0 , y C ,
(1.6)

which was studied by Verma [13].

If x ¯ = y ¯ in (1.6), then (1.6) reduces to the classical variational inequality (1.3). Further, problem (1.6) is equivalent to the following projection formulas:

{ x ¯ = P C ( I μ 1 A ) y ¯ , y ¯ = P C ( I μ 2 A ) x ¯ .

Recently, Ceng et al. [12] introduced and studied a relaxed extragradient method for finding solutions of problem (1.5).

Let { T i } be an infinite family of nonexpansive mappings of C into itself and { λ n 1 },{ λ n 2 },,{ λ n N } be real sequences such that λ n 1 , λ n 2 ,, λ n N (0,1] for every nN. For any n1, we define a mapping W n of C into itself as follows:

U n 0 = I , U n 1 = λ n 1 T 1 U n 0 + ( 1 λ n 1 ) I , U n 2 = λ n 2 T 2 U n 1 + ( 1 λ n 2 ) I , U n , N 1 = λ n , N 1 T N 1 U n , N 2 + ( 1 λ n , N 1 ) I , W n = U n , N = λ n , N T N U n , N 1 + ( 1 λ n , N ) I .

Such a mapping W n is called the W-mapping generated by T 1 , T 2 ,, T N and { λ n 1 },{ λ n 2 },,{ λ n N }. Nonexpansivity of each T i ensures the nonexpansivity of W n . Moreover, in [1], it is shown that F( W n )= i = 1 N F( T i ).

Throughout this article, let us assume that a bifunction Θ:C×CR and a convex function φ:CR satisfy the following conditions:

  1. (H1)

    Θ(x,x)=0 for all xC;

  2. (H2)

    Θ is monotone, i.e., Θ(x,y)+Θ(y,x)0 for all x,yC;

  3. (H3)

    for each yC, xΘ(x,y) is weakly upper semicontinuous;

  4. (H4)

    for each xC, yΘ(x,y) is convex and lower semicontinuous;

  5. (A1)

    for each xH and r>0, there exists a bounded subset D x C and y x C such that for any zC D x ,

    Θ(z, y x )+φ( y x )φ(z)+ 1 r y x z,zx<0;
  6. (A2)

    C is a bounded set.

Recently, Qin et al. [8] studied the problem of finding a common element of the set of common fixed points of a finite family of nonexpansive mappings, the set of solutions of variational inequalities for a relaxed cocoercive mapping and the set of solutions of an equilibrium problem. More precisely, they proved the following theorem.

Theorem 1.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let Θ be a bifunction from C×C to which satisfies (H 1)-(H 4). Let T 1 , T 2 ,, T N be a finite family of nonexpansive mappings of C into H and let B be a μ-Lipschitz, relaxed (u,v)-cocoercive mapping of C into H such that F= i = 1 N F( T i )EP(Θ)VI(A,C)ϕ. Let f be a contraction of H into itself with a coefficient α (0<α<1) and let A be a strongly positive linear bounded operator with a coefficient γ ¯ >0 such that A1. Assume that 0<γ< γ ¯ α . Let { x n } and { y n } be sequences generated by x 1 H and

{ Θ ( y n , η ) + 1 r n η y n , y n x n 0 , η C , x n + 1 = α n γ f ( W n x n ) + ( 1 α n A ) W n P C ( I s n B ) y n , n 1 ,

where α n (0,1] and { r n },{ s n }[0,) satisfy

  1. (i)

    lim n α n =0 and n = 1 α n =;

  2. (ii)

    n = 1 | α n + 1 α n |<, n = 1 | r n + 1 r n |< and n = 1 | s n + 1 s n |<;

  3. (iii)

    lim inf n r n >0;

  4. (iv)

    { s n }[a,b] for some a, b with 0ab 2 ( v u μ 2 ) μ 2 , vu μ 2 ;

  5. (v)

    n = 0 | λ n , i λ n 1 , i |< for all i=1,2,,N.

Then, both { x n } and { y n } converge strongly to qF, where q= P F (γf+(IA))(q), which solves the following variational inequality:

γ f ( q ) A q , p q 0,pF.

In this paper, motivated by Takahashi and Takahashi [3], Ceng, Wang and Yao [12], Peng and Yao [1] and Qin, Shang and Su [8], we introduce the general iterative scheme for finding a common element of the set of common fixed points of a finite family of nonexpansive mappings, the set of solutions of the generalized mixed equilibrium problem (1.1) and the set of solutions of the generalized equilibrium problem (1.4), which solves the variational inequality

( A γ f ) x , x x 0,xF,

where F= i = 1 N F( T i )GMEPΩ and Ω is the set of solutions of the generalized equilibrium problem (1.4). The results obtained in this paper improve and extend the recent results announced by Qin et al. [8], Chen et al. [14], Combetters and Hirstoaga [15], Iiduka and Takahashi [16], Marino and Xu [17], Takahashi and Takahashi [18], Wittmann [19] and many others.

2 Preliminaries

Let C be a nonempty closed convex subset of a real Hilbert space H. For every point xH, there exists a unique nearest point of C, denoted by P C x, such that x P C xxy for all yC. Such a P C is called the metric projection of H onto C. We know that P C is a firmly nonexpansive mapping of H onto C, i.e.,

xy, P C x P C y P C x P C y 2 ,x,yH.

Further, for any xH and zC, z= P C x if and only if

xz,zy0,yC.

It is also known that H satisfies Opial’s condition [20] if for each sequence { x n } n = 1 in H which converges weakly to a point xH, we have

lim inf n x n x< lim inf n x n y,yH,yx.

Moreover, we assume that A is a bounded strongly positive operator on H with a constant γ ¯ , that is, there exists γ ¯ >0 such that

Ax,x γ ¯ x 2 ,xH.

A mapping B:CH is called β-inverse strongly monotone if there exists β>0 such that

xy,BxByβ B x B y 2 ,x,yC.

It is obvious that any inverse strongly monotone mapping is Lipschitz continuous.

In order to prove our main results in the next section, we need the following lemmas and proposition.

Lemma 2.1 [2]

Let C be a nonempty closed convex subset of H. Let Θ:C×CR be a bifunction satisfying conditions (H1)-(H4) and let φ:CR be a lower semicontinuous and convex function. For r>0 and xH, define a mapping

T r ( Θ , φ ) (x)= { z C : Θ ( z , y ) + φ ( y ) φ ( z ) + 1 r y z , z x 0 , y C }

for all xH. Assume that either (A1) or (A2) holds. Then the following results hold:

  1. (i)

    T r ( Θ , φ ) (x)ϕ for each xH and T r ( Θ , φ ) is single-valued;

  2. (ii)

    T r ( Θ , φ ) is firmly nonexpansive, i.e., for any x,yH,

    T r ( Θ , φ ) x T r ( Θ , φ ) y 2 T r ( Θ , φ ) x T r ( Θ , φ ) y , x y ;
  3. (iii)

    F( T r ( Θ , φ ) )=MEP(Θ,φ);

  4. (iv)

    MEP(Θ,φ) is closed and convex.

Remark 2.1 If φ=0, then T r ( Θ , φ ) is rewritten as T r Θ .

By a similar argument as that in the proof of Lemma 2.1 in [12], we have the following result.

Lemma 2.2 Let C be a nonempty closed convex subset of H. Let G 1 , G 2 :C×CR be two bifunctions satisfying conditions (H1)-(H4) and let the mappings B 1 , B 2 :CH be β 1 -inverse strongly monotone and β 2 -inverse strongly monotone, respectively. Then, for given x ¯ , y ¯ C, ( x ¯ , y ¯ ) is a solution of (1.4) if and only if x ¯ is a fixed point of the mapping Γ:CC defined by

Γ(x)= T μ 1 G 1 [ T μ 2 G 2 ( x μ 2 B 2 x ) μ 1 B 1 T μ 2 G 2 ( x μ 2 B 2 x ) ] ,xC,

where y ¯ = T μ 2 G 2 ( x ¯ μ 2 B 2 x ¯ ).

The set of fixed points of the mapping Γ is denoted by Ω.

Proposition 2.1 [3]

Let C, H, Θ, φ and T r ( Θ , φ ) be as in Lemma  2.1. Then the following holds:

T s ( Θ , φ ) x T t ( Θ , φ ) x 2 s t s T s ( Θ , φ ) x T t ( Θ , ρ ) x , T s ( Θ , φ ) x x

for all s,t>0 and xH.

Lemma 2.3 [21]

Assume that T is a nonexpansive self-mapping of a nonempty closed convex subset C of H. If T has a fixed point, then IT is demi-closed; that is, when { x n } is a sequence in C converging weakly to some xC and the sequence {(IT) x n } converges strongly to some y, it follows that (IT)x=y.

Lemma 2.4 [22]

Assume that { a n } is a sequence of nonnegative real numbers such that

a n + 1 (1 γ n ) a n + δ n ,n1,

where { γ n } is a sequence in (0,1) and { δ n } is a sequence such that

  1. (i)

    n = 1 γ n =;

  2. (ii)

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

Then lim n a n =0.

Lemma 2.5 [17]

Assume A is a strong positive linear bounded operator on a Hilbert space H with a coefficient γ ¯ >0 and 0<ρ A 1 . Then IρA1ρ γ ¯ .

The following lemma is an immediate consequence of an inner product.

Lemma 2.6 In a real Hilbert space H, the following inequality holds:

x + y 2 x 2 +2y,x+y

for all x,yH.

3 Main results

Now we state and prove our main results.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let Θ, G 1 , G 2 :C×CR be three bifunctions which satisfy assumptions (H1)-(H4) and φ:CR be a lower semicontinuous and convex function with assumption (A1) or (A2). Let the mappings F, B 1 , B 2 :CH be ζ-inverse strongly monotone, β 1 -inverse strongly monotone and β 2 -inverse strongly monotone, respectively. Let T 1 , T 2 ,, T N be a finite family of nonexpansive mappings of C into H such that F= i = 1 N F( T i )GMEPΩϕ. Let f be a contraction of C into itself with a constant α (0<α<1) and let A be a strongly positive linear bounded operator with a coefficient γ ¯ >0 such that A1. Assume that 0<γ< γ ¯ α . Let x 1 C and let { x n } be a sequence defined by

{ z n = T δ n ( Θ , φ ) ( x n δ n F x n ) , y n = T μ 1 G 1 [ T μ 2 G 2 ( z n μ 2 B 2 z n ) μ 1 B 1 T μ 2 G 2 ( z n μ 2 B 2 z n ) ] , x n + 1 = α n γ f ( W n x n ) + ( 1 α n A ) W n y n , n 1 ,
(3.1)

where α n [0,1], μ 1 (0,2 β 1 ), μ 2 (0,2 β 2 ) and { δ n }[0,2ζ] satisfy the following conditions:

  1. (i)

    lim n α n =0, n = 1 α n =0 and n = 1 | α n + 1 α n |<;

  2. (ii)

    0< lim inf n δ n lim sup n δ n <2ζ and n = 1 | δ n + 1 δ n |<;

  3. (iii)

    lim n λ n , i =0 and n = 1 | λ n , i λ n 1 , i |< for all i=1,2,,N.

Then { x n } converges strongly to x = P F (γf+(IA))( x ), which solves the following variational inequality:

( A γ f ) x , x x 0,xF,

and ( x , y ) is a solution of problem (1.4), where y = T μ 2 G 2 ( x μ 2 B 2 x ).

Proof We divide the proof into several steps.

Step 1. { x n } is bounded.

Indeed, take pF= i = 1 N F( T i )GMEPΩϕ arbitrarily. Since p= T δ n ( Θ , φ ) (p δ n Fp), F is ζ-inverse strongly monotone and 0 δ n 2ζ, we obtain that for any n1,

z n p 2 = T δ n ( Θ , φ ) ( x n δ n F x n ) T δ n ( Θ , φ ) ( p δ n F p ) 2 ( x n p ) δ n ( F x n F p ) 2 = x n p 2 2 δ n x n p , F x n F p + δ n 2 F x n F p 2 x n p 2 2 δ n ζ F x n F p 2 + δ n 2 F x n F p 2 = x n p 2 + δ n ( δ n 2 ζ ) F x n F p 2 x n p 2 .
(3.2)

Putting u n = T μ 2 G 2 ( z n μ 2 B 2 z n ) and u= T μ 2 G 2 (p μ 2 B 2 p), we have

u n u 2 = T μ 2 G 2 ( z n μ 2 B 2 z n ) T μ 2 G 2 ( p μ 2 B 2 p ) 2 ( z n p ) μ 2 ( B 2 z n B 2 p ) 2 = z n p 2 2 μ 2 z n p , B 2 z n B 2 p + μ 2 2 B 2 z n B 2 p 2 z n p 2 2 μ 2 β 2 B 2 z n B 2 p 2 + μ 2 2 B 2 z n B 2 p 2 = z n p 2 + μ 2 ( μ 2 2 β 2 ) B 2 z n B 2 p 2 z n p 2 .
(3.3)

And since p= T μ 1 G 1 (u μ 1 B 1 u), we know that for any n1,

y n p 2 = T μ 1 G 1 ( u n μ 1 B 1 u n ) T μ 1 G 1 ( u μ 1 B 1 u ) 2 ( u n u ) μ 1 ( B 1 u n B 1 u ) 2 u n u 2 2 μ 1 u n u , B 1 u n B 1 u + μ 1 2 B 1 u n B 1 u 2 u n u 2 2 μ 1 β 1 B 1 u n B 1 u + μ 1 B 1 u n b 1 u 2 u n u 2 + μ 1 ( μ 1 2 β 1 ) B 1 u n B 1 u u n u 2 z n p 2 .
(3.4)

Furthermore, from (3.1), we have

x n + 1 p = α n γ f ( W n x n ) + ( 1 α n A ) W n y n p = α n ( γ f ( W n x n ) A p ) + ( 1 α n A ) ( W n y n p ) α n γ f ( W n x n ) A p + I α n A W n y n p α n ( γ f ( W n x n ) f ( p ) + γ f ( p ) A p ) + ( 1 α n γ ¯ ) y n p α n ( γ α x n p + γ f ( p ) A p ) + ( 1 α n γ ¯ ) x n p [ 1 ( γ ¯ γ α ) α n ] x n p + α n γ f ( p ) A p .

By induction, we obtain that for all n1,

x n pmax { x 1 p , 1 γ ¯ γ α γ f ( p ) A p } .

Hence { x n } is bounded. Consequently, we deduce immediately that { z n }, { y n }, {f( W n x n )} and { W n ( y n )} are bounded.

Step 2. lim n W n + 1 y n W n y n =0.

It follows from the definition of W n that

W n + 1 y n W n y n = λ n + 1 , N T N U n + 1 , N 1 y n ( 1 λ n + 1 , N ) y n λ n , N T N U n , N 1 y n ( 1 λ n , N ) y n | λ n + 1 , N λ n , N | y n + λ n + 1 , N T N U n + 1 , N 1 y n λ n , N T N U n , N 1 y n | λ n + 1 , N λ n , N | y n + λ n + 1 , N ( T N U n + 1 , N 1 y n T N U n , N 1 y n ) + | λ n + 1 , N λ n , N | T N U n , N 1 y n | λ n + 1 , N λ n , N | y n + λ n + 1 , N U n + 1 , N 1 y n U n , N 1 y n + | λ n + 1 , N λ n , N | T N U n , N 1 y n .

Since { y n } is bounded and T k , U n , k are nonexpansive, lim n W n + 1 y n W n y n =0.

Step 3. lim n x n + 1 x n =0.

We estimate y n + 1 y n , W n + 1 x n + 1 W n x n and W n + 1 y n + 1 W n y n . From (3.1) we have

(3.5)

and

z n + 1 z n = T δ n + 1 ( Θ , φ ) ( x n + 1 δ n + 1 F x n + 1 ) T δ n ( Θ , φ ) ( x n δ n F x n ) T δ n + 1 ( Θ , φ ) ( x n + 1 δ n + 1 F x n + 1 ) T δ n + 1 ( Θ , φ ) ( x n δ n F x n ) + T δ n + 1 ( Θ , φ ) ( x n δ n F x n ) T δ n ( Θ , φ ) ( x n δ n F x n ) ( x n + 1 δ n + 1 F x n + 1 ) ( x n δ n F x n ) + T δ n + 1 ( Θ , φ ) ( x n δ n F x n ) T δ n ( Θ , φ ) ( x n δ n F x n ) x n + 1 x n + | δ n + 1 δ n | F x n + T δ n + 1 ( Θ , φ ) ( x n δ n F x n ) T δ n ( Θ , φ ) ( x n δ n F x n ) .
(3.6)

It follows from (3.5) and (3.6) that

y n + 1 y n z n + 1 z n x n + 1 x n + | δ n + 1 δ n | F x n + T δ n + 1 ( Θ , φ ) ( x n δ n F x n ) T δ n ( Θ , φ ) ( x n δ n F x n ) .
(3.7)

Without loss of generality, let us assume that there exists a real number a such that δ n >a>0 for all n. Utilizing Proposition 2.1, we have

T δ n + 1 ( Θ , φ ) ( x n δ n F x n ) T δ n ( Θ , φ ) ( x n δ n F x n ) | δ n + 1 δ n | δ n + 1 T δ n + 1 ( Θ , φ ) ( I δ n F ) x n | δ n + 1 δ n | a T δ n + 1 ( Θ , φ ) ( I δ n F ) x n .
(3.8)

It follows from the definition of W n that

W n + 1 y n W n y n = λ n + 1 , N T N U n + 1 , N 1 y n + ( 1 λ n + 1 , N ) y n λ n , N T N U n , N 1 y n ( 1 λ n , N ) y n | λ n + 1 , N λ n , N | y n + λ n + 1 , N T N U n + 1 , N 1 y n λ n , N T N U n , N 1 y n | λ n + 1 , N λ n , N | y n + λ n + 1 , N ( T N U n + 1 , N 1 y n T N U n , N 1 y n ) + | λ n + 1 , N λ n , N | T N U n , N 1 y n | λ n + 1 , N λ n , N | y n + λ n + 1 , N U n + 1 , N 1 y n U n , N 1 y n + | λ n + 1 , N λ n , N | T N U n , N 1 y n M 1 | λ n + 1 , N λ n , N | + λ n + 1 , N U n + 1 , N 1 y n U n , N 1 y n ,
(3.9)

where M 1 is a constant such that M 1 2max{ sup n 1 y n , sup n 1 T N U n , N 1 y n }. Next, we consider

U n + 1 , N 1 y n U n , N 1 y n = λ n + 1 , N 1 T N 1 U n + 1 , N 2 y n + ( 1 λ n + 1 , N 1 ) y n λ n , N 1 T N 1 U n , N 2 y n ( 1 λ n , N 1 ) y n | λ n + 1 , N 1 λ n , N 1 | y n + λ n + 1 , N 1 T N 1 U n + 1 , N 2 y n λ n , N 1 T N 1 U n , N 2 y n | λ n + 1 , N 1 λ n , N 1 | y n + λ n + 1 , N 1 T N 1 U n + 1 , N 2 y n T N 1 U n , N 2 y n + | λ n + 1 , N 1 λ n , N 1 | T N 1 U n , N 2 y n M 2 | λ n + 1 , N 1 λ n , N 1 | + U n + 1 , N 2 y n U n , N 2 y n ,

where M 2 is a constant that M 2 2max{ sup n 1 y n , sup n 1 T N 1 U n , N 2 y n }. In a similar way, we obtain

U n + 1 , N 1 y n U n , N 1 y n M 3 i = 1 N 1 | λ n + 1 , i λ n , i |,
(3.10)

where M 3 is an appropriate constant. Substituting (3.10) into (3.9), we have that

W n + 1 y n W n y n M 1 | λ n + 1 , N λ n , N | + λ n + 1 , N M 3 i = 1 N 1 | λ n + 1 , i λ n , i | M 4 i = 1 N | λ n + 1 , i λ n , i | ,
(3.11)

where M 4 is a constant such that M 4 max{ M 1 , M 3 }. Similarly, we have

W n + 1 x n W n x n M 5 i = 1 N | λ n + 1 , i λ n , i |,
(3.12)

where M 5 is an appropriate constant. Hence it follows from (3.1), (3.7), (3.8), (3.11) and (3.12) that

x n + 2 x n + 1 = ( I α n + 1 A ) ( W n + 1 y n + 1 W n y n ) ( α n + 1 α n ) A W n y n + γ [ α n + 1 ( f ( W n + 1 x n + 1 ) f ( W n x n ) ) + f ( W n x n ) ( α n + 1 α n ) ] ( 1 α n + 1 γ ¯ ) ( W n + 1 y n + 1 W n + 1 y n + W n + 1 y n W n y n ) + | α n + 1 α n | A W n y n + γ [ α n + 1 f ( W n + 1 x n + 1 ) f ( W n x n ) + f ( W n x n ) ( α n + 1 α n ) ] ( 1 α n + 1 γ ¯ ) ( y n + 1 y n + W n + 1 y n W n y n ) + | α n + 1 α n | A W n y n + γ [ α n + 1 α ( x n + 1 x n + W n + 1 x n W n x n ) + | α n + 1 α n | f ( W n x n ) ] ( 1 α n + 1 γ ¯ ) ( x n + 1 x n + | δ n + 1 δ n | F x n + T δ n + 1 ( Θ , φ ) ( x n δ n F x n ) T δ n ( Θ , φ ) ( x n δ n F x n ) + M 4 i = 1 N | λ n + 1 , i λ n , i | ) + | α n + 1 α n | A W n y n + γ α α n + 1 x n + 1 x n + γ α α n + 1 W n + 1 x n W n x n + γ | α n + 1 α n | f ( W n x n ) [ 1 α n + 1 ( γ ¯ γ α ) ] x n + 1 x n + | δ n + 1 δ n | F x n + | δ n + 1 δ n | a T δ n + 1 ( Θ , φ ) ( I δ n F ) x n + M 4 i = 1 N | λ n + 1 , i λ n , i | + | α n + 1 α n | A W n y n + γ α α n + 1 M 5 i = 1 N | λ n + 1 , i λ n , i | + γ | α n + 1 α n | f ( W n x n ) [ 1 α n + 1 ( γ ¯ γ α ) ] x n + 1 x n + M 6 [ 2 | δ n + 1 δ n | + i = 1 N | λ n + 1 , i λ n , i | + ( 1 + γ ) | α n + 1 α n | ] ,

where M 6 is a constant such that M 6 max{ sup n 1 F x n , 1 a sup n 1 T δ n + 1 ( Θ , φ ) (I δ n F) x n , M 4 +γ M 5 , sup n 1 A W n y n , sup n 1 f( W n x n )}. By Lemma 2.4, we get lim n x n + 1 x n =0.

Step 4. lim n F x n Fp=0, lim n B 1 u n B 1 u=0 and lim n B 2 z n B 2 p=0.

Indeed, from (3.1)-(3.4) we get

x n + 1 p 2 = α n ( γ f ( W n x n ) A p ) + ( I α n A ) ( W n y n p ) 2 ( α n γ f ( W n x n ) A p + ( 1 α n γ ¯ ) y n p ) 2 α n γ f ( W n x n ) A p 2 + ( 1 α n γ ¯ ) y n p 2 + 2 α n γ f ( W n x n ) A p y n p α n γ f ( W n x n ) A p 2 + ( 1 α n γ ¯ ) [ u n u 2 + μ 1 ( μ 1 2 β 1 ) B 1 u n B 1 u 2 ] + 2 α n γ f ( W n x n ) A p y n p α n γ f ( W n x n ) A p 2 + ( 1 α n γ ¯ ) [ z n p 2 + μ 2 ( μ 2 2 β 2 ) B 2 z n B 2 p 2 + μ 1 ( μ 1 2 β 1 ) B 1 u n B 1 u 2 ] + 2 α n γ f ( W n x n ) A p y n p α n γ f ( W n x n ) A p 2 + x n p 2 + δ n ( δ n 2 ζ ) F n F p 2 + μ 2 ( μ 2 2 β 2 ) B 2 z n B 2 p 2 + μ 1 ( μ 1 2 β 1 ) B 1 u n B 1 u 2 + 2 α n γ f ( W n x n ) A p y n p .

Therefore

δ n ( 2 ζ δ n ) F x n F p 2 + μ 2 ( 2 β 2 μ 2 ) B 2 z n B 2 p 2 + μ 1 ( 2 β 1 μ 1 ) B 1 u n B 1 u 2 α n γ f ( W n x n ) A p 2 + x n p x n + 1 p 2 + 2 α n γ f ( W n x n ) A p y n p α n γ f ( W n x n ) A p 2 + ( x n p + x n + 1 p ) x n x n + 1 + 2 α n γ f ( W n x n ) A p y n p .

Since α n 0 and x n x n + 1 0 as n, we have lim n F x n Fp=0, lim n B 1 u n B 1 u=0 and lim n B 2 z n B 2 p=0.

Step 5. lim n x n z n =0, lim n z n y n =0 and lim n x n y n =0.

Indeed, from (3.2), (3.3) and Lemma 2.1, we have

u n u 2 = T μ 2 G 2 ( z n μ 2 B 2 z n ) T μ 2 G 2 ( p μ 2 B 2 p ) 2 ( z n μ 2 B 2 z n ) ( p μ 2 B 2 p ) , u n u = 1 2 [ ( z n μ 2 B 2 z n ) ( p μ 2 B 2 p ) 2 + u n u 2 ( z n μ 2 B 2 z n ) ( p μ 2 B 2 p ) ( u n u ) 2 ] 1 2 [ z n p 2 + u n u 2 ( z n u n ) μ 2 ( B 2 z n B 2 p ) ( p u ) 2 ] 1 2 [ x n p 2 + u n u 2 ( z n u n ) ( p u ) 2 + 2 μ 2 ( z n u n ) ( p u ) , B 2 z n B 2 p μ 2 2 B 2 z n B 2 p 2 ]

and

y n p 2 = T μ 1 G 1 ( u n μ 1 B 1 u n ) T μ 1 G 1 ( u μ 1 B 1 u ) 2 ( u n μ 1 B 1 u n ) ( u μ 1 B 1 u ) , y n p = 1 2 [ ( u n μ 1 B 1 u n ) ( u μ 1 B 1 u ) 2 + y n p 2 ( u n μ 1 B 1 u n ) ( u μ 1 B 1 u ) ( y n p ) 2 ] 1 2 [ u n u 2 + y n p 2 ( u n y n ) + ( p u ) 2 + 2 μ 1 B 1 u n B 1 u , ( u n y n ) + ( p u ) μ 1 2 B 1 u n B 1 u 2 ] 1 2 [ x n p 2 + y n p 2 ( u n y n ) + ( p u ) 2 + 2 μ 1 B 1 u n B 1 u , ( u n y n ) + ( p u ) ] ,

which imply that

u n u 2 x n p 2 ( z n u n ) ( p u ) 2 + 2 μ 2 ( z n u n ) ( p u ) , B 2 z n B 2 p μ 2 2 B 2 z n B 2 p 2
(3.13)

and

y n p 2 x n p 2 ( u n y n ) + ( p u ) 2 + 2 μ 1 B 1 u n B 1 u ( u n y n ) + ( p u ) .
(3.14)

It follows from (3.14) that

x n + 1 p 2 = α n ( γ f ( W n x n ) A p ) + ( I α n A ) ( W n y n p ) 2 α n γ f ( W n x n ) A p 2 + ( 1 α n γ ¯ ) y n p 2 + 2 α n γ f ( W n x n ) A p y n p α n γ f ( W n x n ) A p 2 + ( 1 α n γ ¯ ) [ x n p 2 ( u n y n ) + ( p u ) 2 + 2 μ 1 B 1 u n B 1 u ( u n y n ) + ( p u ) ] + 2 α n γ f ( W n x n ) A p y n p ,

which gives that

( 1 α n γ ¯ ) ( u n y n ) + ( p u ) 2 α n γ f ( W n x n ) A p 2 + x n p 2 x n + 1 p 2 + 2 μ 1 ( 1 α n γ ¯ ) B 1 u n B 1 u ( u n y n ) + ( p u ) + 2 α n γ f ( W n x n ) A p y n p α n γ f ( W n x n ) A p 2 + ( x n p + x n + 1 p ) x n x n + 1 + 2 μ 1 ( 1 α n γ ¯ ) B 1 u n B 1 u ( u n y n ) + ( p u ) + 2 α n γ f ( W n x n ) A p y n p .

Since α n 0, x n + 1 x n 0 and B 1 u n B 1 u0 as n, we have

lim n ( u n y n ) + ( p u ) 2 =0.
(3.15)

Also, from (3.4) and (3.13), we have

x n + 1 p 2 = α n ( γ f ( W n x n ) A p ) + ( I α n A ) ( W n y n p ) 2 α n γ f ( W n x n ) A p 2 + ( 1 α n γ ¯ ) y n p 2 + 2 α n γ f ( W n x n ) A p y n p α n γ f ( W n x n ) A p 2 + ( 1 α n γ ¯ ) u n u 2 + 2 α n γ f ( W n x n ) A p y n p α n γ f ( W n x n ) A p 2 + ( 1 α n γ ¯ ) [ x n p 2 ( z n u n ) ( p u ) 2 + 2 μ 2 ( z n u n ) ( p u ) , B 2 z n B 2 p ] + 2 α n γ f ( W n x n ) A p y n p .

So, we have

( 1 α n γ ¯ ) ( z n u n ) ( p u ) 2 α n γ f ( W n x n ) A p 2 + x n p 2 x n + 1 p 2 + 2 μ 2 ( z n u n ) ( p u ) B 2 z n B 2 p + 2 α n γ f ( W n x n ) A p ) y n p α n γ f ( W n x n ) A p 2 + ( x n p + x n + 1 p ) x n x n + 1 + 2 μ 2 ( z n u n ) ( p u ) B 2 z n B 2 p + 2 α n γ f ( W n x n ) A p y n p .

Note that B 2 z n B 2 p0 as n. Then we have

lim n ( z n u n ) ( p u ) =0.
(3.16)

In addition, from the firm nonexpansivity of T δ n ( Θ , φ ) , we have

z n p 2 = T δ n ( Θ , φ ) ( x n δ n F x n ) T δ n ( Θ , φ ) ( p δ n F p ) 2 ( x n δ n F x n ) ( p δ n F p ) , z n p = 1 2 [ ( x n δ n F x n ) ( p δ n F p ) 2 + z n p 2 ( x n δ n F x n ) ( p δ n F p ) ( z n p ) 2 ] 1 2 [ x n p 2 + z n p 2 x n z n δ n ( F x n F p ) 2 ] = 1 2 [ x n p 2 + z n p 2 x n z n 2 + 2 δ n F x n F p , x n z n δ n 2 F x n F p 2 ] ,

which implies that

z n p 2 x n p 2 x n z n 2 +2 δ n F x n Fp x n z n .
(3.17)

From (3.1), (3.4) and (3.17), we have

x n + 1 p 2 = α n ( γ f ( W n x n ) A p ) + ( I α n A ) ( W n y n p ) 2 α n γ f ( W n x n ) A p 2 + ( 1 α n γ ¯ ) y n p 2 + 2 α n γ f ( W n x n ) A p y n p α n γ f ( W n x n ) A p 2 + ( 1 α n γ ¯ ) z n p 2 + 2 α n γ f ( W n x n ) A p y n p α n γ f ( W n x n ) A p 2 + ( 1 α n γ ¯ ) [ x n p 2 x n z n 2 + 2 δ n F x n F p x n z n ] + 2 α n γ f ( W n x n ) A p y n p .

It follows that

( 1 α n γ ¯ ) x n z n 2 α n γ f ( W n x n ) A p 2 + x n p 2 x n + 1 p 2 + 2 ( 1 α n γ ¯ ) δ n F x n F p x n z n + 2 α n γ f ( W n x n ) A p y n p α n γ f ( W n x n ) A p 2 + ( x n p + x n + 1 p ) x n x n + 1 + 2 ( 1 α n γ ¯ ) δ n F x n F p x n z n + 2 α n γ f ( W n x n ) A p y n p .

Since F x n Fp0 as n, we obtain

lim n x n z n =0.
(3.18)

Thus, from (3.15), (3.16) and (3.18), we obtain that

lim n z n y n = lim n ( z n u n ) ( p u ) + ( u n y n ) + ( p u ) lim n z n u n ( p u ) + lim n u n y n + ( p u ) = 0

and

lim n x n y n lim n x n z n + lim n z n y n = 0 .

Step 6. lim n y n W n y n =0.

Indeed, observe that

x n W n y n x n x n + 1 + x n + 1 W n y n = x n x n + 1 + α n γ f ( W n x n ) + ( I α n A ) W n y n W n y n x n x n + 1 + α n [ γ f ( W n x n ) + A W n y n ] .

From Step 3 and α n 0 as n, we have lim n x n W n y n =0. Consequently,

lim n y n W n y n lim n ( y n x n + x n W n y n ) = 0 .

Step 7. lim sup n γf( x )A x , x n x 0, where x = P F (γf+(IA))( x ).

Indeed, take a subsequence { x n i } of { x n } such that

lim sup n γ f ( x ) A x , x n x = lim i γ f ( x ) A x , x n i x .

Correspondingly, there exists a subsequence { y n i } of { y n }. Since { y n i } is bounded, there exists a subsequence of y n i which converges weakly to w. Without loss of generality, we can assume that y n i w. Next we show wF. First, we prove that wΩ. Utilizing Lemma 2.1, we have for all x,yC

Γ ( x ) Γ ( y ) 2 = T μ 1 G 1 [ T μ 2 G 2 ( x μ 2 B 2 x ) μ 1 B 1 T μ 2 G 2 ( x μ 2 B 2 x ) ] T μ 1 G 1 [ T μ 2 G 2 ( y μ 2 B 2 y ) μ 1 B 1 T μ 2 G 2 ( y μ 2 B 2 y ) ] 2 T μ 2 G 2 ( x μ 2 B 2 x ) T μ 2 G 2 ( y μ 2 B 2 y ) μ 1 [ B 1 T μ 2 G 2 ( x μ 2 B 2 x ) B 1 T μ 2 G 2 ( y μ 2 B 2 y ) ] 2 T μ 2 G 2 ( x μ 2 B 2 x ) T μ 2 G 2 ( y μ 2 B 2 y ) 2 + μ 1 ( μ 1 2 β 1 ) B 1 T μ 2 G 2 ( x μ 2 B 2 x ) B 1 T μ 2 G 2 ( y μ 2 B 2 y ) 2 T μ 2 G 2 ( x μ 2 B 2 x ) T μ 2 G 2 ( y μ 2 B 2 y ) 2 x y μ 2 ( B 2 x B 2 y ) 2 x y 2 + μ 2 ( μ 2 2 β 2 ) B 2 x B 2 y 2 x y 2 .

This shows that Γ:CC is nonexpansive. Note that

y n Γ ( y n ) = Γ ( z n ) Γ ( y n ) z n y n 0 as  n .

According to Lemma 2.2 and Lemma 2.3, we obtain wΩ.

Next, let us show that wGMEP. From z n = T δ n ( Θ , φ ) ( x n δ n F x n ), we obtain

Θ( z n ,y)+φ(y)φ( z n )+ 1 δ n y z n , z n ( x n δ n F x n ) 0,yC.

It follows from (H2) that

φ(y)φ( z n )+y z n ,F x n + 1 δ n y z n , z n x n Θ(y, z n ),yC.
(3.19)

Replacing n by n i , we have

φ(y)φ( z n i )+y z n i ,F x n i + y z n i , z n i x n i δ n i Θ(y, z n i ),yC.

Let z t =ty+(1t)w for all t[0,1] and yC. Then we have z t C. It follows from (3.19) that

z t z n i , F z t z t z n i , F z t φ ( z t ) + φ ( z n i ) z t z n i , F x n i z t z n i , z n i x n i δ n i + Θ ( z t , z n i ) = z t z n i , F z t F z n i + z t z n i , F z n i F x n i φ ( z t ) + φ ( z n i ) z t z n i , z n i x n i δ n i + Θ ( z t , z n i ) .

Since z n i x n i 0, we have F z n i F x n i 0. From the monotonicity of F, we have

F z t F z n i , z t z n i 0.

From (H4), the weakly lower semicontinuity of φ, z n i x n i δ n i 0 and z n i w, we have

z t w,F z t φ( z t )+φ(w)+Θ( z t ,w)
(3.20)

as i. By (H1), (H4) and (3.20), we obtain

0 = Θ ( z t , z t ) + φ ( z t ) φ ( z t ) = Θ ( z t , t y + ( 1 t ) w ) + φ ( t y + ( 1 t ) w ) φ ( z t ) t Θ ( z t , y ) + ( 1 t ) Θ ( z t , w ) + t φ ( y ) + ( 1 t ) φ ( w ) φ ( z t ) = t [ Θ ( z t , y ) + φ ( y ) φ ( z t ) ] + ( 1 t ) [ Θ ( z t , w ) + φ ( w ) φ ( z t ) ] t [ Θ ( z t , y ) + φ ( y ) φ ( z t ) ] + ( 1 t ) z t w , F z t = t [ Θ ( z t , y ) + φ ( y ) φ ( z t ) ] + ( 1 t ) t y w , F z t .

Hence we obtain

0Θ( z t ,y)+φ(y)φ( z t )+(1t)yw,F z t .

Putting t0, we have

0Θ(w,y)+φ(y)φ(w)+yw,Fw,yC.

This implies that wGMEP.

Since Hilbert spaces satisfy Opial’s condition, it follows from Step 5 that

lim inf i y n i w < lim inf i y n i W n w lim inf i [ y n i W n y n i + W n y n i W n w ] lim sup i y n i W n y n i + lim inf i W n y n i W n w = lim inf i W n y n i W n w lim inf i y n i w ,

which derives a contraction. This implies that wF( W n ). It follows from F( W n )= i = 1 N F( T i ) that w i = 1 N F( T i ).

Since x = P F (γf+(IA))( x ), we have

lim sup n γ f ( x ) A x , x n x = lim n γ f ( x ) A x , x n i x = γ f ( x ) A x , w x 0 .

Step 8. x n x as n.

Indeed, from Lemma 2.6 and (3.4), we have

x n + 1 x 2 = α n γ f ( W n x n ) + ( I α n A ) W n y n x 2 = ( I α n A ) ( W n y n x ) + α n ( γ f ( W n x n ) A x ) 2 ( I α n A ) ( W n y n x ) 2 + 2 α n γ f ( W n x n ) A x , x n + 1 x ( 1 α n γ ¯ ) 2 y n x 2 + 2 α n γ f ( W n x n ) A x , x n + 1 x ( 1 α n γ ¯ ) 2 x n x 2 + 2 α n γ f ( W n x n ) f ( x ) , x n + 1 x + 2 α n γ f ( x ) A x , x n + 1 x ( 1 α n γ ¯ ) 2 x n x 2 + 2 α n γ α x n x x n + 1 x + 2 α n γ f ( x ) A x , x n + 1 x ( 1 α n γ ¯ ) 2 x n x 2 + α n γ α ( x n x 2 + x n + 1 x 2 ) + 2 α n γ f ( x ) A x , x n + 1 x ,

which implies that

x n + 1 x 2 ( 1 α n γ ¯ ) 2 + α n γ α 1 α n γ α x n x 2 + 2 α n 1 α n γ α γ f ( x ) A x , x n + 1 x = ( 1 2 α n ( γ ¯ α γ ) 1 α n γ α ) x n x 2 + 2 α n ( γ ¯ α γ ) 1 α n γ α [ α n γ ¯ 2 2 ( γ ¯ α γ ) x n x 2 + 1 γ ¯ α γ γ f ( x ) A x , x n + 1 x ] .

Put

γ n = 2 α n ( γ ¯ α γ ) 1 α n γ α

and

ξ n = 2 α n ( γ ¯ α γ ) 1 α n γ α [ α n γ ¯ 2 2 ( γ ¯ α γ ) x n x 2 + 1 γ ¯ α γ γ f ( x ) A x , x n + 1 x ] .

Then we can write the last inequality as

a n + 1 (1 γ n ) a n + ξ n .

It follows from condition (i) and Step 6 that

n = 1 γ n =+

and

lim sup n ξ n γ n = lim sup n { α n γ ¯ 2 2 ( γ ¯ α γ ) x n x 2 + 1 γ ¯ α γ γ f ( x ) A x , x n + 1 x } 0 .

Hence, applying Lemma 2.4, we immediately obtain that x n x as n. This completes the proof. □

As corollaries of Theorem 3.1, we have the following results.

Corollary 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let Θ, G 1 , G 2 :C×CR be three bifunctions which satisfy assumptions (H1)-(H4) and φ:CR be a lower semicontinuous and convex function satisfying (A1) or (A2). Let the mappings B 1 , B 2 :CH be β 1 -inverse strongly monotone and β 2 -inverse strongly monotone, respectively. Let T 1 , T 2 ,, T N be a finite family of nonexpansive mappings of C into H such that F= i = 1 N F( T i )MEPΩϕ. Let f be a contraction of H into itself with a constant α (0<α<1) and let A be a strongly positive linear bounded operator with a coefficient γ ¯ >0 such that A1. Assume that 0<γ< γ ¯ α . Let x 1 C and let { x n } be a sequence defined by

{ Θ ( z n , y ) + φ ( y ) φ ( z n ) + 1 δ n y z n , z n x n 0 , y C , y n = T μ 1 G 1 [ T μ 2 G 2 ( z n μ 2 B 2 z n ) μ 1 B 1 T μ 2 G 2 ( z n μ 2 B 2 z n ) ] , x n + 1 = α n γ f ( W n x n ) + ( 1 α n A ) W n y n , n 1 ,

where α n [0,1], μ 1 (0,2 β 1 ), μ 2 (0,2 β 2 ) and { δ n }(0,) satisfy the following conditions:

  1. (i)

    lim n α n =0, n = 1 α n = and n = 1 | α n + 1 α n |<;

  2. (ii)

    lim n λ n , i =0, n = 1 | λ n , i λ n 1 , i |< for all i=1,2,,N,

  3. (iii)

    0< lim inf n δ n lim sup n δ n < and n = 1 N | δ n + 1 δ n |<.

Then { x n } converges strongly to x = P i = 1 N F ( T i ) MEP Ω ( γ ¯ f+(IA))( x ) and ( x , y ) is a solution of problem (1.4), where y = T μ 2 G 2 ( x μ 2 B 2 x ), which solves the following variational inequality:

( A γ f ) x , x x 0,x i = 1 N F( T i )MEPΩ.

Proof In Theorem 3.1, for all n0, z n = T δ n ( Θ , φ ) ( x n δ n F x n ) is equivalent to

Θ( z n ,y)+φ(y)φ( z n )+F x n ,y z n + 1 δ n y z n , z n x n 0,yC.
(3.21)

Putting F0, we obtain

Θ( z n ,y)+φ(y)φ( z n )+ 1 δ n y z n , z n x n 0,yC.

 □

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H and let G 1 , G 2 :C×CR be two bifunctions with satisfy assumptions (H1)-(H4). Let the mappings F, B 1 , B 2 :CH be ζ-inverse strongly monotone, β 1 -inverse strongly monotone and β 2 -inverse strongly monotone, respectively. Let T 1 , T 2 ,, T N be a finite family of nonexpansive mappings of C into H such that i = 1 N F( T i )VI(A,C)Ωϕ. Let f be a contraction of H into itself with a constant α (0<α<1) and let A be a strongly positive linear bounded operator with a coefficient γ ¯ >0 such that A1. Assume that 0<γ< γ ¯ α . Let x 1 C and let { x n } be a sequence defined by

{ z n = P C ( x n δ F x n ) , y n = T μ 1 G 1 [ T μ 2 G 2 ( z n μ 2 B 2 z n ) μ 1 B 1 T μ 2 G 2 ( z n μ 2 B 2 z n ) ] , x n + 1 = α n γ f ( W n x n ) + ( 1 α n A ) W n y n , n 1 ,

where α n [0,1], μ 1 (0,2 β 1 ), μ 2 (0,2 β 2 ) and { δ n }[0,2ζ] satisfy the following conditions:

  1. (i)

    lim n α n =0, n = 1 α n =0 and n = 1 | α n + 1 α n |<;

  2. (ii)

    lim n λ n , i =0 and n = 1 | λ n , i λ n 1 , i |< for all i=1,2,,N;

  3. (iii)

    lim inf n δ n >0 and n = 1 | δ n + 1 δ n |<.

Then { x n } converges strongly to x = P i = 1 N F ( T i ) VI ( A , C ) Ω (γf+(IA))( x ) and ( x , y ) is a solution of problem (1.4), where y = T μ 2 G 2 ( x μ 2 B 2 x ), which solves the following variational inequality:

( A γ f ) x , x x 0,x i = 1 N F( T i )VI(A,C)Ω.

Proof Put Θ=0 and φ=0 in Theorem 3.1. Then we have from (3.21) that

F x n ,y z n + 1 δ n y z n , z n x n 0,yC,n1.

That is,

y z n , x n δ n F x n z n 0,yC.

It follows that P C ( x n δ n F x n )= z n for all n1. We can obtain the desired conclusion easily. □

Remark 3.1 We can see easily that Takahashi and Takahashi [18], Peng and Yao’s [1] results are special cases of Theorem 3.1.

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Jeong, J.U. Strong convergence theorems for a generalized mixed equilibrium problem and variational inequality problems. Fixed Point Theory Appl 2013, 65 (2013). https://doi.org/10.1186/1687-1812-2013-65

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  • DOI: https://doi.org/10.1186/1687-1812-2013-65

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