Open Access

Regularized hybrid iterative algorithms for triple hierarchical variational inequalities

Journal of Inequalities and Applications20142014:490

https://doi.org/10.1186/1029-242X-2014-490

Received: 31 August 2014

Accepted: 24 November 2014

Published: 12 December 2014

Abstract

In this paper, we introduce and study a triple hierarchical variational inequality (THVI) with constraints of minimization and equilibrium problems. More precisely, let Fix ( T ) be the fixed point set of a nonexpansive mapping, let MEP ( Θ , φ ) be the solution set of a mixed equilibrium problem (MEP), and let Γ be the solution set of a minimization problem (MP) for a convex and continuously Frechet differential functional in Hilbert spaces. We want to find a solution x Fix ( T ) MEP ( Θ , φ ) Γ of a variational inequality with a variational inequality constraint over the intersection of Fix ( T ) , MEP ( Θ , φ ) , and Γ. We propose a hybrid iterative algorithm with regularization to compute approximate solutions of the THVI, and we present the convergence analysis of the proposed iterative algorithm.

MSC:49J40, 47J20, 47H10, 65K05, 47H09.

Keywords

triple hierarchical variational inequality minimization problem mixed equilibrium problem nonexpansive mapping maximal monotone mapping regularization

1 Introduction

Let H be a Hilbert space with inner product , and norm over the real scalar field . Let C be a nonempty closed convex subset of H, and P C be the metric projection of H onto C. Let T : C C be a self-mapping on C. Denote by Fix ( T ) the set of fixed points of T. We say that T is L-Lipschitzian if there exists a constant L 0 such that
T x T y L x y , x , y C .
When L = 1 or 0 L < 1 , we call T a nonexpansive or a contractive mapping, respectively. We say that a mapping A : C H is α-inverse strongly monotone if there exists a constant α > 0 such that
A x A y , x y α A x A y 2 , x , y C ,
and that A is η-strongly monotone (resp. monotone) if there exists a constant η > 0 (resp. η = 0 ) such that
A x A y , x y η x y 2 , x , y C .

It is known that T is nonexpansive if and only if I T is 1 2 -inverse strongly monotone. Moreover, L-Lipschitz continuous mappings are 1 L -inversely strong monotone (see, e.g., [1]).

Let f : C R be a convex and continuously Frechet differentiable functional. Consider the minimization problem (MP):
min x C f ( x )
(1.1)
(assuming the existence of minimizers). We denote by Γ the set of minimizers of problem (1.1). The gradient-projection algorithm (GPA) generates a sequence { x n } determined by the gradient f and the metric projection P C :
x n + 1 : = P C ( x n λ f ( x n ) ) , n 0 .
(1.2)
The convergence of algorithm (1.2) depends on f. It is known that if f is η-strongly monotone and L-Lipschitz continuous, then for 0 < λ < 2 η L 2 , the operator
S : = P C ( I λ f )

is a contraction. Hence, the sequence { x n } defined by the GPA (1.2) converges in norm to the unique solution of (1.1). If the gradient f is only assumed to be Lipschitz continuous, then { x n } can only be weakly convergent when H is infinite-dimensional (a counterexample to the norm convergence of { x n } is given by Xu [[2], Section 5]).

The regularization, in particular the traditional Tikhonov regularization, is usually used to solve ill-posed optimization problems. Consider the regularized minimization problem
min x C f α ( x ) : = f ( x ) + α 2 x 2 ,

where α > 0 is the regularization parameter, and again f is convex with Lipschitz continuous gradient f. While a regularization method provides the possible strong convergence to the minimum-norm solution, its disadvantage is the implicity. Hence explicit iterative methods seem to be attractive. See, e.g., Xu [2, 3].

On the other hand, for a given mapping A : C H , we consider the variational inequality problem (VIP) of finding x C such that
A x , x x 0 , x C .
(1.3)
The solution set of VIP (1.3) is denoted by VI ( C , A ) . It is well known that when A is monotone,
x VI ( C , A ) x = P C ( x λ A x ) , λ > 0 .

Variational inequality theory has been studied quite extensively and has emerged as an important tool in several branches of pure and applied sciences; see, e.g., [1, 48] and the references therein.

When C is the fixed point set Fix ( T ) of a nonexpansive mapping T and A = I S , VIP (1.3) becomes the variational inequality problem of finding x Fix ( T ) such that
( I S ) x , x x 0 , x Fix ( T ) .
(1.4)

This problem, introduced by Moudafi and Maingé [9, 10], is called the hierarchical fixed point problem. It is clear that if S has fixed points, then they are solutions of VIP (1.4). If S is contractive, the solution set of VIP (1.4) is a singleton and it is well known as a viscosity problem. This was previously introduced by Moudafi [11] and also developed by Xu [12]. In this case, solving VIP (1.4) is equivalent to finding a fixed point of the nonexpansive mapping P Fix ( T ) S , where P Fix ( T ) is the metric projection onto the closed and convex set Fix ( T ) . Yao et al. [8] introduced a two-step algorithm to solve VIP (1.4).

Let Θ : C × C R be a bifunction and φ : C R be a function. Consider the mixed equilibrium problem (MEP) of finding x C such that
Θ ( x , y ) + φ ( y ) φ ( x ) 0 , y C ,
(1.5)

which was studied by Ceng and Yao [13]. The solution set of MEP (1.5) is denoted by MEP ( Θ , φ ) . The MEP (1.5) is very general in the sense that it includes, as special cases, fixed point problems, optimization problems, variational inequality problems, minimax problems, Nash equilibrium problems in noncooperative games and others; see, e.g., [1315].

Recently, Iiduka [16, 17] considered a variational inequality with a variational inequality constraint over the set of fixed points of a nonexpansive mapping. Since this problem has a triple structure in contrast with bilevel programming problems or hierarchical constrained optimization problems or hierarchical fixed point problems, it is referred to as a triple hierarchical constrained optimization problem (THCOP). He presented some examples of THCOP and developed iterative algorithms to find the solution of such a problem. Since the original problem is a variational inequality, in this paper, we call it a triple hierarchical variational inequality (THVI). Ceng et al. introduced and considered some THVI in [18]. A nice survey article on THVI is [19]. See also [2022].

Extending the works done in [18], we introduce and study in this paper the following triple hierarchical variational inequality with constraints of minimization and equilibrium problems.

The problem to study

Let C be a nonempty closed convex subset of a real Hilbert space H. Let f : C R be convex and continuously Frechet differentiable with Γ being the set of its minimizers. Let T : C C and S : H H be both nonexpansive. Let V : H H be ρ-contractive, and F : C H be κ-Lipschitzian and η-strongly monotone with constants ρ [ 0 , 1 ) and κ , η > 0 . Suppose 0 < μ < 2 η / κ 2 and 0 < γ τ where τ = 1 1 μ ( 2 η μ κ 2 ) .

Let Ξ denote the solution set of the following hierarchical variational inequality (HVI): find z Fix ( T ) MEP ( Θ , φ ) Γ such that
( μ F γ S ) z , z z 0 , z Fix ( T ) MEP ( Θ , φ ) Γ ,

where the solution set Ξ is assumed to be nonempty. Consider the following triple hierarchical variational inequality (THVI).

Find x Ξ such that
( μ F γ V ) x , x x 0 , x Ξ .
(1.6)
Based on the iterative schemes provided by Xu [2] and the two-step iterative scheme provided by Yao et al. [8], by virtue of the viscosity approximation method, hybrid steepest-descent method and the regularization method, we propose the following hybrid iterative algorithm with regularization:
{ Θ ( u n , y ) + φ ( y ) φ ( u n ) + 1 r n y u n , u n x n 0 , y C , y n = θ n γ S x n + ( I θ n μ F ) P C ( u n λ f α n ( u n ) ) , x n + 1 = β n γ V y n + ( I β n μ F ) T P C ( u n λ f α n ( u n ) ) , n 0 .

Here, 0 < λ < 2 / L , { r n } , { α n } ( 0 , + ) , and { β n } , { θ n } ( 0 , 1 ) . It is shown that under appropriate assumptions, the two iterative sequences { x n } and { y n } converge strongly to the unique solution of the THVI (1.6).

2 Preliminaries

Let K be a nonempty closed convex subset of a real Hilbert space H. We write x n x and x n x to indicate that the sequence { x n } converges weakly and strongly to x, respectively. The weak ω-limit set of the sequence { x n } is denoted by
ω w ( x n ) : = { x H : x n i x  for some subsequence  { x n i }  of  { x n } } .
The metric (or nearest point) projection from H onto K is the mapping P K : H K which assigns to each point x H the unique point P K x K satisfying the property
x P K x = inf y K x y = : d ( x , K ) .
Proposition 2.1 For given x H and z K :
  1. (i)

    z = P K x x z , y z 0 , y K ;

     
  2. (ii)

    z = P K x x z 2 x y 2 y z 2 , y K ;

     
  3. (iii)

    P K x P K y , x y P K x P K y 2 , y H .

     

Hence, P K is nonexpansive and monotone.

Definition 2.2 A mapping T : H H is said to be firmly nonexpansive if 2 T I is nonexpansive, or equivalently,
x y , T x T y T x T y 2 , x , y H .
Alternatively, T is firmly nonexpansive if and only if T can be expressed as
T = 1 2 ( I + S ) ,

where S : H H is nonexpansive. Projections are firmly nonexpansive. We call T an averaged mapping if T can be expressed as a proper convex combination of the identity map I and a nonexpansive mapping. In particular, firmly nonexpansive mappings are averaged.

Proposition 2.3 (see [23])

Let T : H H be a given mapping.
  1. (i)

    T is nonexpansive if and only if the complement I T is 1 2 -inverse strongly monotone.

     
  2. (ii)

    If T is ν-inverse strongly monotone, then so is γT for all γ > 0 .

     
  3. (iii)

    T is averaged if and only if the complement I T is ν-inverse strongly monotone for some ν > 1 / 2 . Indeed, for α ( 0 , 1 ) , T is α-averaged if and only if I T is 1 2 α -inverse strongly monotone.

     

Proposition 2.4 (see [23, 24])

Let S , T , V : H H .
  1. (i)

    If T = ( 1 α ) S + α V for some α ( 0 , 1 ) and if S is averaged and V is nonexpansive, then T is averaged.

     
  2. (ii)

    T is firmly nonexpansive if and only if the complement I T is firmly nonexpansive.

     
  3. (iii)

    If T = ( 1 α ) S + α V for some α ( 0 , 1 ) and if S is firmly nonexpansive and V is nonexpansive, then T is averaged.

     
  4. (iv)

    The composition of finitely many averaged mappings is averaged. In particular, if T 1 is α 1 -averaged and T 2 is α 2 -averaged, where α 1 , α 2 ( 0 , 1 ) , then T 1 T 2 is ( α 1 + α 2 α 1 α 2 ) -averaged.

     
  5. (v)
    If the mappings T 1 , , T N are averaged and have a common fixed point, then
    i = 1 N Fix ( T i ) = Fix ( T 1 T N ) .
     
For solving the equilibrium problem for a bifunction Θ : C × C R , let us consider the following conditions:
  • (A1) T ( x , x ) = 0 for all x ? C ;

  • (A2) T is monotone, that is, T ( x , y ) + T ( y , x ) = 0 for all x , y ? C ;

  • (A3) for each x , y , z ? C , lim t ? 0 T ( t z + ( 1 - t ) x , y ) = T ( x , y ) ;

  • (A4) for each x ? C , y ? T ( x , y ) is convex and lower semicontinuous;

  • (A5) for each y ? C , x ? T ( x , y ) is weakly upper semicontinuous;

  • (B1) for each x ? H and r > 0 , there exist a bounded subset D x ? C and y x ? C such that for any z ? C \ D x ,
    T ( z , y x ) + f ( y x ) - f ( z ) + 1 r y x - z , z - x > < 0 ;
  • (B2) C is a bounded set.

Lemma 2.5 (see [14])

Let C be a nonempty closed convex subset of a real Hilbert space H and T : C × C ? R be a bifunction satisfying (A1)-(A4). Let r > 0 and x ? H . Then there exists z ? C such that
T ( z , y + 1 r y - z , z - x > = 0 , ? y ? C .

Lemma 2.6 (see [15])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let Θ : C × C R be a bifunction satisfying (A1)-(A5) and φ : C R be a proper lower semicontinuous and convex function. For r > 0 and x H , define a mapping T r : H C as follows:
T r x : = { z C : Θ ( z , y ) + φ ( y ) φ ( z ) + 1 r y z , z x 0 , y C }

for all x H . Assume that either (B1) or (B2) holds. Then T r is a single-valued firmly nonexpansive map on H, and Fix ( T r ) = MEP ( Θ , φ ) is closed and convex.

Lemma 2.7 (see [25])

Let { a n } be a sequence of nonnegative real numbers such that
a n + 1 ( 1 s n ) a n + s n t n + ϵ n , n 0 .
Here, 0 < s n 1 , 0 ϵ n , and t n R for all n = 0 , 1 , 2 ,  , such that
  1. (i)

    n = 0 s n = + ;

     
  2. (ii)

    either lim sup n t n 0 or n = 0 s n | t n | < + ;

     
  3. (iii)

    n = 0 ϵ n < + .

     

Then lim n a n = 0 .

Lemma 2.8 (Demiclosedness principle; see [1])

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

Lemma 2.9 (see [12])

Let S : H H be a nonexpansive mapping and V : H H be a ρ-contraction with ρ [ 0 , 1 ) , respectively.
  1. (i)
    I S is monotone, i.e.,
    ( I S ) x ( I S ) y , x y 0 , x , y H .
     
  2. (ii)
    I V is ( 1 ρ ) -strongly monotone, i.e.,
    ( I V ) x ( I V ) y , x y ( 1 ρ ) x y 2 , x , y H .
     

Lemma 2.10 ([26])

Let H be a real Hilbert space. Then, for all x , y H and λ [ 0 , 1 ] ,
λ x + ( 1 λ ) y 2 = λ x 2 + ( 1 λ ) y 2 λ ( 1 λ ) x y 2 .
Lemma 2.11 We have the following inequality in an inner product space X:
x + y 2 x 2 + 2 y , x + y , x , y X .
Notations Let λ be a number in ( 0 , 1 ] and let μ , κ , η > 0 . Let F : C H be κ-Lipschitzian and η-strongly monotone. Associated with a nonexpansive mapping T : C C , we define the mapping T λ : C H by
T λ x : = ( I λ μ F ) T x , x C .

Lemma 2.12 (see [[27], Lemma 3.1])

The map T λ is a contraction provided μ < 2 η / κ 2 , that is,
T λ x T λ y ( 1 λ τ ) x y , x , y C ,
where τ = 1 1 μ ( 2 η μ κ 2 ) ( 0 , 1 ] . In particular, if T = I the identity mapping, then
( I λ μ F ) x ( I λ μ F ) y ( 1 λ τ ) x y , x , y C .

A set-valued mapping T ˜ : H 2 H is called monotone if x y , f g 0 for all x , y H , f T ˜ x and g T ˜ y . A monotone set-valued mapping T ˜ : H 2 H is called maximal if its graph Gph ( T ˜ ) is not properly contained in the graph of any other monotone set-valued mapping. It is known that a monotone set-valued mapping T ˜ : H 2 H is maximal if and only if for ( x , f ) H × H , x y , f g 0 for every ( y , g ) Gph ( T ˜ ) implies that f T ˜ x .

Let A : C H be a monotone and Lipschitz continuous mapping and let N C v be the normal cone to C at v C , namely
N C v = { w H : v u , w 0 , u C } .
Define
T ˜ v = { A v + N C v , if  v C , , if  v C .

Lemma 2.13 (see [28])

Let A : C H be a monotone mapping.
  1. (i)

    T ˜ is maximal monotone;

     
  2. (ii)

    v T ˜ 1 0 v VI ( C , A ) .

     

3 Main results

Let us consider the following three-step iterative scheme with regularization:
{ Θ ( u n , y ) + φ ( y ) φ ( u n ) + 1 r n y u n , u n x n 0 , y C , y n = θ n γ S x n + ( I θ n μ F ) P C ( u n λ f α n ( u n ) ) , x n + 1 = β n γ V y n + ( I β n μ F ) T P C ( u n λ f α n ( u n ) ) , n = 0 , 1 , 2 , .
(3.1)

Here,

  • V : H H is a ρ-contraction;

  • S : H H and T : C C are nonexpansive mappings;

  • F : C H is a κ-Lipschitzian and η-strongly monotone mapping;

  • Θ : C × C R and φ : C R are real-valued functions;

  • f : C H is L-Lipschitz continuous with 0 < λ < 2 L ;

  • { r n } and { α n } are sequences in ( 0 , + ) with n = 0 α n < + and lim inf n r n > 0 ;

  • { β n } and { θ n } are sequences in ( 0 , 1 ) ;

  • 0 < μ < 2 η / κ 2 and 0 < γ τ , where τ = 1 1 μ ( 2 η μ κ 2 ) .

Theorem 3.1 Suppose that Θ : C × C R satisfies (A1)-(A5) and that (B1) or (B2) holds. Let { x n } be the bounded sequence generated from any given x 0 C by (3.1). Assume that

(H1) n = 0 β n = + , lim n 1 β n | 1 θ n 1 θ n | = 0 ;

(H2) lim n 1 β n | 1 θ n 1 θ n 1 | = 0 and lim n 1 θ n | 1 β n 1 β n | = 0 ;

(H3) lim n θ n = 0 , lim n α n + β n θ n = 0 , and lim n θ n 2 β n = 0 ;

(H4) lim n | α n α n 1 | β n θ n = 0 and lim n | r n r n 1 | β n θ n = 0 .

Then we have the following:
  1. (i)

    lim n x n + 1 x n θ n = 0 ;

     
  2. (ii)

    ω w ( x n ) Fix ( T ) MEP ( Θ , φ ) Γ ;

     
  3. (iii)

    ω w ( x n ) Ξ if x n y n = o ( θ n ) held in addition, i.e., lim n x n y n θ n = 0 .

     
Proof First, let us show that P C ( I λ f α ) is ξ-averaged for each λ ( 0 , 2 α + L ) , where
ξ = 2 + λ ( α + L ) 4 ( 0 , 1 ) .
The Lipschitz condition implies that the gradient f is 1 L -inverse strongly monotone [1], that is,
f ( x ) f ( y ) , x y 1 L f ( x ) f ( y ) 2 .
Observe that
( α + L ) f α ( x ) f α ( y ) , x y = ( α + L ) [ α x y 2 + f ( x ) f ( y ) , x y ] = α 2 x y 2 + α f ( x ) f ( y ) , x y + α L x y 2 + L f ( x ) f ( y ) , x y α 2 x y 2 + 2 α f ( x ) f ( y ) , x y + f ( x ) f ( y ) 2 = α ( x y ) + f ( x ) f ( y ) 2 = f α ( x ) f α ( y ) 2 .
Hence, f α = α I + f is 1 α + L -inverse strongly monotone. Thus, λ f α is 1 λ ( α + L ) -inverse strongly monotone by Proposition 2.3(ii). By Proposition 2.3(iii) the complement I λ f α is λ ( α + L ) 2 -averaged. Noting that P C is 1 2 -averaged and utilizing Proposition 2.4(iv), we know that for each λ ( 0 , 2 α + L ) , the map P C ( I λ f α ) is ξ-averaged with
ξ = 1 2 + λ ( α + L ) 2 1 2 λ ( α + L ) 2 = 2 + λ ( α + L ) 4 ( 0 , 1 ) .
In particular, P C ( I λ f α ) is nonexpansive. Furthermore, for λ ( 0 , 2 L ) , utilizing the fact that lim n 2 α n + L = 2 L , we may assume
0 < λ < 2 α n + L , n 0 .
Consequently, for each integer n 0 , P C ( I λ f α n ) is ξ n -averaged with
ξ n = 1 2 + λ ( α n + L ) 2 1 2 λ ( α n + L ) 2 = 2 + λ ( α n + L ) 4 ( 0 , 1 ) .

This immediately implies that P C ( I λ f α n ) is nonexpansive for all n 0 .

We divide the proof into several steps.

Step 1. lim n x n + 1 x n θ n = 0 .

For simplicity, put u ˜ n = P C ( u n λ f α n ( u n ) ) . Then y n = θ n γ S x n + ( I θ n μ F ) u ˜ n and x n + 1 = β n γ V y n + ( I β n μ F ) T u ˜ n for every n 0 . We observe that
u ˜ n u ˜ n 1 P C ( I λ f α n ) u n P C ( I λ f α n ) u n 1 + P C ( I λ f α n ) u n 1 P C ( I λ f α n 1 ) u n 1 u n u n 1 + P C ( I λ f α n ) u n 1 P C ( I λ f α n 1 ) u n 1 u n u n 1 + ( I λ f α n ) u n 1 ( I λ f α n 1 ) u n 1 = u n u n 1 + λ f α n ( u n 1 ) λ f α n 1 ( u n 1 ) = u n u n 1 + λ | α n α n 1 | u n 1 , n 1 .
(3.2)
Moreover, from (3.1) we have
{ y n = θ n γ S x n + ( I θ n μ F ) u ˜ n , y n 1 = θ n 1 γ S x n 1 + ( I θ n 1 μ F ) u ˜ n 1 , n 1 .
Thus
y n y n 1 = θ n ( γ S x n γ S x n 1 ) + ( θ n θ n 1 ) ( γ S x n 1 μ F u ˜ n 1 ) + ( I θ n μ F ) u ˜ n ( I θ n μ F ) u ˜ n 1 .
Utilizing Lemma 2.12 from (3.2) we deduce that
y n y n 1 θ n γ S x n γ S x n 1 + | θ n θ n 1 | γ S x n 1 μ F u ˜ n 1 + ( I θ n μ F ) u ˜ n ( I θ n μ F ) u ˜ n 1 θ n γ x n x n 1 + | θ n θ n 1 | γ S x n 1 μ F u ˜ n 1 + ( 1 θ n τ ) u ˜ n u ˜ n 1 θ n γ x n x n 1 + | θ n θ n 1 | γ S x n 1 μ F u ˜ n 1 + ( 1 θ n τ ) ( u n u n 1 + λ | α n α n 1 | u n 1 ) ,
(3.3)
where τ = 1 1 μ ( 2 η μ κ 2 ) . Taking into consideration that u n = T r n x n and u n 1 = T r n 1 x n 1 , we have
Θ ( u n , y ) + φ ( y ) φ ( u n ) + 1 r n y u n , u n x n 0 , y C
(3.4)
and
Θ ( u n 1 , y ) + φ ( y ) φ ( u n 1 ) + 1 r n 1 y u n 1 , u n 1 x n 1 0 , y C .
(3.5)
Putting y = u n 1 in (3.4) and y = u n in (3.5), we obtain
Θ ( u n , u n 1 ) + φ ( u n 1 ) φ ( u n ) + 1 r n u n 1 u n , u n x n 0 , y C
and
Θ ( u n 1 , u n ) + φ ( u n ) φ ( u n 1 ) + 1 r n 1 u n u n 1 , u n 1 x n 1 0 , y C .
Adding the last two inequalities, by (A2) we get
u n u n 1 , u n 1 x n 1 r n 1 u n x n r n 0 ,
and hence
u n u n 1 , u n 1 u n + u n x n 1 r n 1 r n ( u n x n ) 0 .
Since lim inf n r n > 0 , we may assume, without loss of generality, that there exists a positive number c such that r n c > 0 for all n 0 . Thus we have
u n u n 1 2 u n u n 1 , x n x n 1 + ( 1 r n 1 r n ) ( u n x n ) u n u n 1 [ x n x n 1 + | 1 r n 1 r n | u n x n ] ,
and hence
u n u n 1 x n x n 1 + | 1 r n 1 r n | u n x n x n x n 1 + M 0 c | r n r n 1 | .
(3.6)

Here, M 0 = sup { u n x n : n 0 } < + .

Substituting (3.6) into (3.3) we derive
y n y n 1 θ n γ x n x n 1 + | θ n θ n 1 | γ S x n 1 μ F u ˜ n 1 + ( 1 θ n τ ) ( x n x n 1 + M 0 c | r n r n 1 | + λ | α n α n 1 | u n 1 ) ( 1 θ n ( τ γ ) ) x n x n 1 + | θ n θ n 1 | γ S x n 1 μ F u ˜ n 1 + M 0 c | r n r n 1 | + λ | α n α n 1 | u n 1 x n x n 1 + M 1 ( | θ n θ n 1 | + | r n r n 1 | + | α n α n 1 | ) .
(3.7)

Here, γ S x n μ F u ˜ n + M 0 c + λ u n M 1 for some M 1 0 .

On the other hand, from (3.1) we have
{ x n + 1 = β n γ V y n + ( I β n μ F ) T u ˜ n , x n = β n 1 γ V y n 1 + ( I β n 1 μ F ) T u ˜ n 1 , n 1 .
Simple calculations show that
x n + 1 x n = ( I β n μ F ) T u ˜ n ( I β n μ F ) T u ˜ n 1 + ( β n β n 1 ) ( γ V y n 1 μ F T u ˜ n 1 ) + β n ( γ V y n γ V y n 1 ) .
Utilizing Lemma 2.12 from (3.2), (3.6), and (3.7) we deduce that
x n + 1 x n ( I β n μ F ) T u ˜ n ( I β n μ F ) T u ˜ n 1 + | β n β n 1 | γ V y n 1 μ F T u ˜ n 1 + β n γ V y n γ V y n 1 ( 1 β n τ ) u ˜ n u ˜ n 1 + | β n β n 1 | γ V y n 1 μ F T u ˜ n 1 + β n γ ρ y n y n 1 ( 1 β n τ ) ( u n u n 1 + λ | α n α n 1 | u n 1 ) + | β n β n 1 | γ V y n 1 μ F T u ˜ n 1 + β n γ ρ y n y n 1 ( 1 β n τ ) ( x n x n 1 + M 0 c | r n r n 1 | + λ | α n α n 1 | u n 1 ) + | β n β n 1 | γ V y n 1 μ F T u ˜ n 1 + β n γ ρ [ x n x n 1 + M 1 ( | θ n θ n 1 | + | r n r n 1 | + | α n α n 1 | ) ] ( 1 β n τ ) [ x n x n 1 + M 1 ( | r n r n 1 | + | α n α n 1 | ) ] + | β n β n 1 | γ V y n 1 μ F T u ˜ n 1 + β n γ ρ [ x n x n 1 + M 1 ( | θ n θ n 1 | + | r n r n 1 | + | α n α n 1 | ) ] ( 1 β n τ ) x n x n 1 + ( 1 β n τ ) M 1 ( | r n r n 1 | + | α n α n 1 | ) + | β n β n 1 | γ V y n 1 μ F T u ˜ n 1 + β n γ ρ x n x n 1 + β n τ M 1 ( | θ n θ n 1 | + | r n r n 1 | + | α n α n 1 | ) ( 1 β n ( τ γ ρ ) ) x n x n 1 + | β n β n 1 | γ V y n 1 μ F T u ˜ n 1 + M 1 ( | θ n θ n 1 | + | r n r n 1 | + | α n α n 1 | ) ( 1 β n ( τ γ ρ ) ) x n x n 1 + M ( | α n α n 1 | + | β n β n 1 | + | θ n θ n 1 | + | r n r n 1 | ) ,
where M 1 + γ V y n μ F T u ˜ n M , n 0 for some M 0 . Therefore,
x n + 1 x n θ n ( 1 β n ( τ γ ρ ) ) x n x n 1 θ n + M ( | α n α n 1 | θ n + | β n β n 1 | θ n + | θ n θ n 1 | θ n + | r n r n 1 | θ n ) = ( 1 ( τ γ ρ ) β n ) x n x n 1 θ n 1 + ( 1 ( τ γ ρ ) β n ) x n x n 1 ( 1 θ n 1 θ n 1 ) + M ( | α n α n 1 | θ n + | β n β n 1 | θ n + | θ n θ n 1 | θ n + | r n r n 1 | θ n ) ( 1 ( τ γ ρ ) β n ) x n x n 1 θ n 1 + ( τ γ ρ ) β n 1 τ γ ρ { x n x n 1 1 β n | 1 θ n 1 θ n 1 | + M ( | α n α n 1 | β n θ n + | β n β n 1 | β n θ n + | θ n θ n 1 | β n θ n + | r n r n 1 | β n θ n ) } ( 1 ( τ γ ρ ) β n ) x n x n 1 θ n 1 + ( τ γ ρ ) β n M ˜ τ γ ρ { 1 β n | 1 θ n 1 θ n 1 | + | α n α n 1 | β n θ n + 1 θ n | 1 β n 1 β n | + 1 β n | 1 θ n 1 θ n | + | r n r n 1 | β n θ n } ,
(3.8)
where M + x n x n 1 M ˜ , n 1 for some M ˜ 0 . From (H1), (H2), and (H4), it follows that n = 0 ( τ γ ρ ) β n = + and
lim n M ˜ τ γ ρ { 1 β n | 1 θ n 1 θ n 1 | + | α n α n 1 | β n θ n + 1 θ n | 1 β n 1 β n | + 1 β n | 1 θ n 1 θ n | + | r n r n 1 | β n θ n } = 0 .
Applying Lemma 2.7 to (3.8), we immediately conclude that
lim n x n + 1 x n θ n = 0 .
In particular, from (H3) it follows that
lim n x n + 1 x n = 0 .

Step 2. lim n x n u n = 0 and lim n y n u ˜ n = 0 .

By the firm nonexpansivity of T r n , if v MEP ( Θ , φ ) = Fix ( T r n ) , we have
u n v 2 = T r n x n T r n v 2 T r n x n T r n v , x n v = 1 2 { u n v 2 + x n v 2 T r n x n T r n v ( x n v ) 2 } .
This immediately yields
u n v 2 x n v 2 x n u n 2 .
(3.9)
Let p Fix ( T ) MEP ( Θ , φ ) Γ . We have
u ˜ n p = P C ( I λ f α n ) u n P C ( I λ f ) p P C ( I λ f α n ) u n P C ( I λ f α n ) p + P C ( I λ f α n ) p P C ( I λ f ) p u n p + P C ( I λ f α n ) p P C ( I λ f ) p u n p + λ α n p .
(3.10)
Note that
y n p = θ n γ S x n θ n μ F p + ( I θ n μ F ) u ˜ n ( I θ n μ F ) p = θ n ( γ S x n μ F p ) + ( 1 θ n ) ( u ˜ n p ) + θ n [ ( I μ F ) u ˜ n ( I μ F ) p ] = θ n ( γ S x n + ( I μ F ) u ˜ n p ) + ( 1 θ n ) ( u ˜ n p ) .
Hence we have
y n u ˜ n = θ n ( γ S x n + ( I μ F ) u ˜ n u ˜ n ) .
By Lemmas 2.10 and 2.12, we have from (3.9) and (3.10) that
y n p 2 = θ n ( γ S x n + ( I μ F ) u ˜ n p ) + ( 1 θ n ) ( u ˜ n p ) 2 = θ n γ S x n + ( I μ F ) u ˜ n p 2 + ( 1 θ n ) u ˜ n p 2 θ n ( 1 θ n ) γ S x n + ( I μ F ) u ˜ n u ˜ n 2 = θ n γ S x n + ( I μ F ) u ˜ n p 2 + ( 1 θ n ) u ˜ n p 2 1 θ n θ n y n u ˜ n 2 θ n γ S x n + ( I μ F ) u ˜ n p 2 + u ˜ n p 2 1 θ n θ n y n u ˜ n 2 .
(3.11)
Furthermore, utilizing Lemmas 2.11 and 2.12 we have from (3.9) and (3.10) that
x n + 1 p 2 = β n γ V y n β n μ F T p + ( I β n μ F ) T u ˜ n ( I β n μ F ) T p 2 ( β n γ V y n β n μ F T p + ( I β n μ F ) T u ˜ n ( I β n μ F ) T p ) 2 ( β n γ V y n μ F p + ( 1 β n τ ) u ˜ n p ) 2 β n 1 τ γ V y n μ F p 2 + ( 1 β n τ ) u ˜ n p 2 β n 1 τ [ γ V y n γ V p 2 + 2 γ V p μ F p , γ V y n μ F p ] + ( 1 β n τ ) u ˜ n p 2 β n γ 2 ρ 2 τ y n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p + ( 1 β n τ ) u ˜ n p 2 β n γ 2 ρ 2 τ [ θ n γ S x n + ( I μ F ) u ˜ n p 2 + u ˜ n p 2 1 θ n θ n y n u ˜ n 2 ] + β n 2 τ γ V p μ F p γ V y n μ F p + ( 1 β n τ ) u ˜ n p 2 = ( 1 β n ( τ γ 2 ρ 2 τ ) ) u ˜ n p 2 + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 β n γ 2 ρ 2 ( 1 θ n ) τ θ n y n u ˜ n 2 + β n 2 τ γ V p μ F p γ V y n μ F p u ˜ n p 2 + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 β n γ 2 ρ 2 ( 1 θ n ) τ θ n y n u ˜ n 2 + β n 2 τ γ V p μ F p γ V y n μ F p ( u n p + λ α n p ) 2 + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 β n γ 2 ρ 2 ( 1 θ n ) τ θ n y n u ˜ n 2 + β n 2 τ γ V p μ F p γ V y n μ F p u n p 2 + λ α n p ( 2 u n p + λ α n p ) + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 β n γ 2 ρ 2 ( 1 θ n ) τ θ n y n u ˜ n 2 + β n 2 τ γ V p μ F p γ V y n μ F p x n p 2 x n u n 2 + λ α n p ( 2 u n p + λ α n p ) + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 β n γ 2 ρ 2 ( 1 θ n ) τ θ n y n u ˜ n 2 + β n 2 τ γ V p μ F p γ V y n μ F p .
(3.12)
It turns out therefore that
x n u n 2 + β n γ 2 ρ 2 ( 1 θ n ) τ θ n y n u ˜ n 2 x n p 2 x n + 1 p 2 + λ α n p ( 2 u n p + λ α n p ) + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p ( x n p + x n + 1 p ) x n x n + 1 + λ α n p ( 2 u n p + λ α n p ) + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p .
Then it is clear that
x n u n 2 ( x n p + x n + 1 p ) x n x n + 1 + λ α n p ( 2 u n p + λ α n p ) + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p .
Since α n 0 , β n 0 , θ n 0 , and x n + 1 x n 0 , we conclude that
lim n x n u n = 0 .
Furthermore,
β n γ 2 ρ 2 ( 1 θ n ) τ θ n y n u ˜ n 2 ( x n p + x n + 1 p ) x n x n + 1 + λ α n p ( 2 u n p + λ α n p ) + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p .
This yields
γ 2 ρ 2 ( 1 θ n ) τ y n u ˜ n 2 ( x n p + x n + 1 p ) θ n β n x n x n + 1 + α n θ n β n λ p ( 2 u n p + λ α n p ) + θ n 2 γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + θ n 2 τ γ V p μ F p γ V y n μ F p .
Since x n + 1 x n θ n 0 , α n + β n θ n 0 , and θ n 2 β n 0 as n , we have
lim n θ n β n x n x n + 1 = lim n x n x n + 1 θ n θ n 2 β n = 0
and
lim n α n θ n β n = lim n α n θ n θ n 2 β n = 0 .
Therefore, from the last inequality we have
lim n y n u ˜ n = 0 .

Step 3. lim n u n u ˜ n = 0 and lim n x n y n = 0 .

Let p Fix ( T ) Fix ( Γ ) Ξ . Utilizing Lemmas 2.6 and 2.11 we have from (3.12) that
x n + 1 p 2 u ˜ n p 2 + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 β n γ 2 ρ 2 ( 1 θ n ) τ θ n y n u ˜ n 2 + β n 2 τ γ V p μ F p γ V y n μ F p P C ( I λ f α n ) u n P C ( I λ f ) p 2 + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p ( I λ f ) u n ( I λ f ) p λ α n u n 2 + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p ( I λ f ) u n ( I λ f ) p 2 2 λ α n u n , ( I λ f α n ) u n ( I λ f ) p + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p u n p 2 + λ ( λ 2 L ) f ( u n ) f ( p ) 2 + 2 λ α n u n ( I λ f α n ) u n ( I λ f ) p + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p x n p 2 + λ ( λ 2 L ) f ( u n ) f ( p ) 2 + 2 λ α n u n ( I λ f α n ) u n ( I λ f ) p + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p .
Hence,
λ ( 2 L λ ) f ( u n ) f ( p ) 2 x n p 2 x n + 1 p 2 + 2 λ α n u n ( I λ f α n ) u n ( I λ f ) p + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p ( x n p + x n + 1 p ) x n x n + 1 + 2 λ α n u n ( I λ f α n ) u n ( I λ f ) p + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p .
Since α n 0 , β n 0 , θ n 0 , and x n + 1 x n 0 , it follows from 0 < λ < 2 L that lim n f ( u n ) f ( p ) = 0 , and hence
lim n f α n ( u n ) f ( p ) = 0 .
Furthermore, from the firm nonexpansiveness of P C we obtain
u ˜ n p 2 = P C ( I λ f α n ) u n P C ( I λ f ) p 2 ( I λ f α n ) u n ( I λ f ) p , u ˜ n p = 1 2 { ( I λ f α n ) u n ( I λ f ) p 2 + u ˜ n p 2 ( I λ f α n ) u n ( I λ f ) p ( u ˜ n p ) 2 } 1 2 { u n p 2 + 2 λ f α n ( u n ) f ( p ) ( I λ f α n ) u n ( I λ f ) p + u ˜ n p 2 u n u ˜ n 2 + 2 λ u n u ˜ n , f α n ( u n ) f ( p ) λ 2 f α n ( u n ) f ( p ) 2 } .
Consequently,
u ˜ n p 2 u n p 2 u n u ˜ n 2 + 2 λ f α n ( u n ) f ( p ) ( I λ f α n ) u n ( I λ f ) p + 2 λ u n u ˜ n , f α n ( u n ) f ( p ) λ 2 f α n ( u n ) f ( p ) 2 .
Thus, from (3.12) we have
x n + 1 p 2 u ˜ n p 2 + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 β n γ 2 ρ 2 ( 1 θ n ) τ θ n y n u ˜ n 2 + β n 2 τ γ V p μ F p γ V y n μ F p u n p 2 u n u ˜ n 2 + 2 λ f α n ( u n ) f ( p ) ( I λ f α n ) u n ( I λ f ) p + 2 λ u n u ˜ n , f α n ( u n ) f ( p ) λ 2 f α n ( u n ) f ( p ) 2 + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 β n γ 2 ρ 2 ( 1 θ n ) τ θ n y n u ˜ n 2 + β n 2 τ γ V p μ F p γ V y n μ F p x n p 2 u n u ˜ n 2 + 2 λ f α n ( u n ) f ( p ) ( I λ f α n ) u n ( I λ f ) p + 2 λ u n u ˜ n f α n ( u n ) f ( p ) + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p .
This implies that
u n u ˜ n 2 x n p 2 x n + 1 p 2 + 2 λ f α n ( u n ) f ( p ) ( I λ f α n ) u n ( I λ f ) p + 2 λ u n u ˜ n f α n ( u n ) f ( p ) + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p ( x n p + x n + 1 p ) x n x n + 1 + 2 λ f α n ( u n ) f ( p ) ( I λ f α n ) u n ( I λ f ) p + 2 λ u n u ˜ n f α n ( u n ) f ( p ) + β n θ n γ 2 ρ 2 τ γ S x n + ( I μ F ) u ˜ n p 2 + β n 2 τ γ V p μ F p γ V y n μ F p .
Since θ n 0 , β n 0 , x n x n + 1 0 , and f α n ( u n ) f ( p ) 0 , it follows that
lim n u n u ˜ n = 0 .
This, together with x n u n 0 and y n u ˜ n 0 (due to Step 2), implies that
x n y n x n u n + u n u ˜ n + u ˜ n y n 0 as  n ,
and thus
lim n x n y n = 0 .

Step 4. ω w ( x n ) Fix ( T ) MEP ( Θ , φ ) Γ ; moreover, if x n y n = o ( θ n ) in addition, then ω w ( x n ) Ξ .

Let p ω w ( x n ) . Then there exists a subsequence { x n i } of { x n } such that x n i p . Since
x n + 1 u ˜ n = β n γ S y n + ( I β n μ F ) T u ˜ n u ˜ n = β n ( γ S y n μ F T u ˜ n ) + T u ˜ n u ˜ n ,
we have
T u ˜ n u ˜ n = x n + 1 u ˜ n β n ( γ S y n μ F T u ˜ n ) x n + 1 u ˜ n + β n γ S y n μ F T u ˜ n x n + 1 x n + x n u n + u n u ˜ n + β n γ S y n μ F T u ˜ n .
Hence from x n + 1 x n 0 , x n u n 0 , u n u ˜ n 0 , and β n 0 , we get
lim n T u ˜ n u ˜ n = 0 .

Since x n u n 0 and u n u ˜ n 0 , we have u ˜ n i p . Utilizing Lemma 2.8 we derive p Fix ( T ) .

Let us show that p MEP ( Θ , φ ) . As a matter of fact, since u n = T r n x n , for any y C we have
Θ ( u n , y ) + φ ( y ) φ ( u n ) + 1 r n y u n , u n x n 0 .
It follows from (A2) that
φ ( y ) φ ( u n ) + 1 r n y u n , u n x n Θ ( y , u n ) .
Replacing n by n i , we have
φ ( y ) φ ( u n i ) + 1 r n i y u n i , u n i x n i Θ ( y , u n i ) .
Since u n i x n i r n i 0 and u n i p , it follows from (A4) that
0 φ ( y ) + φ ( p ) + Θ ( y , p ) , y C .
Put z t = t y + ( 1 t ) p for all t ( 0 , 1 ] and y C . We have z t C and
0 φ ( z t ) + φ ( p ) + Θ ( z t , p ) .
(3.13)
Utilizing (A1), (A4), and (3.13), we have
0 = Θ ( z t , z t ) + φ ( z t ) φ ( z t ) t Θ ( z t , y ) + ( 1 t ) Θ ( z t , p ) + t φ ( y ) + ( 1 t ) φ ( p ) φ ( z t ) t ( Θ ( z t , y ) + φ ( y ) φ ( z t ) ) + ( 1 t ) ( Θ ( z t , p ) + φ ( p ) φ ( z t ) ) t ( Θ ( z t , y ) + φ ( y ) φ ( z t ) ) ,
and hence
0 Θ ( z t , y ) + φ ( y ) φ ( z t ) .
(3.14)
Letting t 0 in (3.14) and utilizing (A3), we get, for each y C ,
0 Θ ( p , y ) + φ ( y ) φ ( p ) .

Hence, p MEP ( Θ , φ ) .

Let us show that p Γ . From x n u n 0 and u n u ˜ n 0 , we know that u n i p and u ˜ n i p . Define
T ˜ v = { f ( v ) + N C v , if  v C , , if  v C ,
where
N C v = { w H : v u , w 0 , u C } .
Then T ˜ is maximal monotone and 0 T ˜ v if and only if v VI ( C , f ) ; see [28] for more details. Let ( v , w ) Gph ( T ˜ ) . Then we have
w T ˜ v = f ( v ) + N C v
and hence,
w f ( v ) N C v .
Therefore,
v u , w f ( v ) 0 , u C .
On the other hand, from
u ˜ n = P C ( u n λ f α n ( u n ) ) and v C ,
we have
u n λ f α n ( u n ) u ˜ n , u ˜ n v 0 ,
and hence
v u ˜ n , u ˜ n u n λ + f α n ( u n ) 0 .
Therefore, from
w f ( v ) N C ( v ) and u ˜ n i C ,
we have
v u ˜ n i , w v u ˜ n i , f ( v ) v u ˜ n i , f ( v ) v u ˜ n i , u ˜ n i u n i λ + f α n i ( u n i ) = v u ˜ n i , f ( v ) v u ˜ n i , u ˜ n i u n i λ + f ( u n i ) α n i v u ˜ n i , u n i = v u ˜ n i , f ( v ) f ( u ˜ n i ) + v u ˜ n i , f ( u ˜ n i ) f ( u n i ) v u ˜ n i , u ˜ n i u n i λ α n i v u ˜ n i , u n i v u ˜ n i , f ( u ˜ n i ) f ( u n i ) v u ˜ n i , u ˜ n i u n i λ α n i v u ˜ n i , u n i .
Hence, we obtain
v p , w 0 as  i .

Since T ˜ is maximal monotone, we have p T ˜ 1 0 , and hence, p VI ( C , f ) , which leads to p Γ . Consequently, p Fix ( T ) MEP ( Θ , φ ) Γ . This shows that ω w ( x n ) Fix ( T ) MEP ( Θ , φ ) Γ .

Utilizing Lemmas 2.11 and 2.12, we have for every p Fix ( T ) MEP ( Θ , φ ) Γ ,
y n p 2 = θ n γ S x n + ( I μ F ) u ˜ n p 2 = θ n ( γ S p μ F p ) + θ n ( γ S x n γ S p ) + ( I θ n μ F ) u ˜ n ( I θ n μ F ) p 2 θ n ( γ S x n γ S p ) + ( I θ n μ F ) u ˜ n ( I θ n μ F ) p 2 + 2 θ n γ S p μ F p , y n p [ θ n ( γ S x n γ S p ) + ( I θ n μ F ) u ˜ n ( I θ n μ F ) p ] 2 + 2 θ n γ S p μ F p , y n p [ θ n γ x n p + ( 1 θ n τ ) u ˜ n p ] 2 + 2 θ n γ S p μ F p , y n p [ θ n γ x n p + ( 1 θ n τ ) ( u n p + λ α n p ) ] 2 + 2 θ n γ S p μ F p , y n p [ θ n γ x n p + ( 1 θ n τ ) ( x n p + λ α n p ) ] 2 + 2 θ n γ S p μ F p , y n p [ ( 1 θ n ( τ γ ) ) x n p + λ α n p ] 2 + 2 θ n γ S p μ F p , y n p ( x n p + λ α n p ) 2 + 2 θ n γ S p μ F p , y n p .
(3.15)
Suppose now that x n y n = o ( θ n ) in addition. It follows from (3.15) that
2 γ S p μ F p , y n p 1 θ n [ ( x n p + λ α n p ) 2 y n p 2 ] 1 θ n ( x n p + λ α n p + y n p ) ( x n p + λ α n p y n p ) 1 θ n ( x n p + λ α n p + y n p ) ( x n y n + λ α n p ) .
This, together with α n θ n 0 and x n y n θ n 0 , leads to
lim sup n γ S p μ F p , y n p 0 .
Observe that
lim sup n γ S p μ F p , x n p = lim sup n ( γ S p μ F p , x n y n + γ S p μ F p , y n p ) = lim sup n γ S p μ F p , y n p 0 .
So, it follows from x n i p that
γ S p μ F p , p p 0 , p Fix ( T ) MEP ( Θ , φ ) Γ .
Note also that 0 < γ τ and
μ η τ μ η 1 1 μ ( 2 η μ κ 2 ) 1 μ ( 2 η μ κ 2 ) 1 μ η 1 2 μ η + μ 2 κ 2 1 2 μ η + μ 2 η 2 κ 2 η 2 κ η .
It is clear that
( μ F γ S ) x ( μ F γ S ) y , x y ( μ η γ ) x y 2 , x , y H .
Hence, it follows from 0 < γ τ μ η that μ F γ S is monotone. Since p ω w ( x n ) Fix ( T ) MEP ( Θ , φ ) Γ , by Minty’s lemma [1] we have
γ S p μ F p , p p 0 , p Fix ( T ) MEP ( Θ , φ ) Γ ;

that is, p Ξ . This shows that ω w ( x n ) Ξ . □

Theorem 3.2 Assuming the conditions in Theorem  3.1. We have:
  1. (i)
    { x n } and { y n } both converge strongly to an element x Fix ( T ) MEP ( Θ , φ ) Γ , which is a unique solution of the variational inequality
    γ V x μ F x , x x 0 , x Fix ( T ) MEP ( Θ , φ ) Γ .
     
  2. (ii)

    { x n } and { y n } both converge strongly to a unique solution of THVI (1.6) if x n y n = o ( θ n ) in addition.

     
Proof Utilizing Lemmas 2.11 and 2.12 we get from (3.15)
x n + 1 p 2 = β n γ V y n + ( I μ F ) T u ˜ n p 2 = β n ( γ V p μ F p ) + β n ( γ V y n γ V p ) + ( I β n μ F ) T u ˜ n ( I β n μ F ) T p 2 β n ( γ V y n γ V p ) + ( I θ n μ F ) T