A new explicit triple hierarchical problem over the set of fixed points and generalized mixed equilibrium problems

Abstract

In this article, we introduce and consider the triple hierarchical over the fixed point set of a nonexpansive mapping and the generalized mixed equilibrium problem set of an inverse-strongly monotone napping. The strong convergence of the algorithm is proved under some mild conditions. Our results generalize and improve the results of Marino and Xu and some authors.

Mathematics Subject Classification (2000): 47H09; 47H10; 47J20; 49J40; 65J15.

1. Introduction

Let C be a closed convex subset of a real Hilbert space H with the inner product 〈·,·〉 and the norm || · ||. We denote weak convergence and strong convergence by notations ⇀ and →, respectively. Let F be a bifunction of H × H into , where is the set of real numbers. A mapping A be a nonlinear mapping. The generalized mixed equilibrium problem is to find x ∈ C such that

$$F\left(x,y\right)+⟨Ax,y-x⟩+\phi \left(y\right)-\phi \left(x\right)\ge 0,\forall y\in C.$$

(1.1)

The set of solutions of (1.1) is denoted by GMEP (F, φ, A). If φ ≡ 0, the problem (1.1) is reduced into the generalized equilibrium problem is to find x ∈ C such that

$$F\left(x,y\right)+⟨Ax,y-x⟩\ge 0,\forall y\in C.$$

(1.2)

The set of solutions of (1.2) is denoted by GEP(F, A). If A ≡ 0, the problem (1.1) is reduced into the mixed equilibrium problem is to find x ∈ C such that

$$F\left(x,y\right)+\phi \left(y\right)-\phi \left(x\right)\ge 0,\forall y\in C.$$

(1.3)

The set of solutions of (1.3) is denoted by MEP(F, φ). If A ≡ 0 and φ ≡ 0, the problem (1.1) is reduced into the equilibrium problem [1] is to find x ∈ C such that

$$F\left(x,y\right)\ge 0,\forall y\in C.$$

(1.4)

The set of solutions of (1.4) is denoted by EP(F). If F ≡ 0 and φ ≡ 0, the problem (1.1) is reduced into the Hartmann-Stampacchia variational inequality [2] is to find x ∈ C such that

$$⟨Ax,y-x⟩\ge 0,\forall y\in C.$$

(1.5)

The set of solutions of (1.5) is denoted by VI(C, A). The variational inequality has been extensively studied in the literature [3, 4]. A mapping A of C into itself is called an α-inverse-strongly monotone if there exists a positive real number α such that

$$⟨Ax-Ay,x-y⟩\ge \alpha {∥Ax-Ay∥}^2,\forall x,y\in C.$$

A mapping f: C → C is called a ρ-contraction if there exists a constant ρ ∈ [0, 1) such that

$$∥f\left(x\right)-f\left(y\right)∥\le \rho ∥x-y∥,\forall x,y\in C.$$

A mapping S: C → C is called nonexpansive if

$$∥Sx-Sy∥\le ∥x-y∥,\forall x,y\in C.$$

A point x ∈ C is a fixed point of S provided Sx = x. Denote by F(S) the set of fixed points of S; that is, F(S) = {x ∈ C : Sx = x}. If C is bounded closed convex and S is a nonexpansive mapping of C into itself, then F(S) is nonempty [5]. Let A and B are two monotone operators, we consider the hierarchical problem over generalized mixed equilibrium problem: Find a point x* ∈ GMEP(F, φ, B) such that

$$⟨A{x}^{*},y-x^{*}⟩\ge 0,\forall y\in GMEP\left(F,\phi ,B\right).$$

(1.6)

We discuss the hierarchical problem over fixed point: Find a point x* ∈ F(S) such that

$$⟨A{x}^{*},y-x^{*}⟩\ge 0,\forall y\in F\left(S\right).$$

(1.7)

Yao et al. [6] considered the hierarchical problem over generalize equilibrium problem and the set of fixed point, x_s,tbe defined implicitly by

$$x_{s,t}=s\left[tf\left(x_{s,t}\right)+\left(1-t\right)\left(x_{s,t}-\lambda A{x}_{s,t}\right)\right]+\left(1-s\right)T_r\left(x_{s,t}-rB{x}_{s,t}\right),s,t\in \left(0,1\right),$$

(1.8)

for each (s, t) ∈ (0, 1)². The net x_s,thierarchically converges to the unique solution x* of the hierarchical problem: Find a point x* ∈ GEP (F, B) such that

$$⟨A{x}^{*},x-x^{*}⟩\ge 0,\forall x\in GEP\left(F,B\right),$$

(1.9)

where A and B are two monotone operators. The solution set of (1.9) is denoted by Ω.

Marino and Xu [7] studied an explicit algorithm, which generated a sequence {x _n } recursively by the formula: For the initial guess x₀ ∈ C is arbitrary

$$x_{n+1}={\lambda }_{n}f\left(x_n\right)+\left(1-{\lambda }_n\right)\left({\alpha }_{n}V{x}_n+\left(1-{\alpha }_n\right)T{x}_n\right),\forall n\ge 0,$$

(1.10)

where {α _n } and {λ _n } are sequences in (0, 1) satisfy some conditions. Let T, V: C → C are two nonexpansive self mappings and f is a contraction on C. Then {x _n } converges strongly to a solution, which solves another variational inequality. Recently, Jitpeera and Kumam [8] introduced and studied the iterative algorithm for solving a common element of the set of solution of fixed point for a nonexpansive mapping, the set of solution of generalized mixed equilibrium problem, and the set of solution of the variational inclusion. They proved that the sequence converges strongly to a common element of the above three sets under some mild conditions.

In this article, we consider the hierarchical problem over the set of fixed point and generalized mixed equilibrium problem, which contains (1.6) and (1.7): Find a point x* ∈ Ξ: = F(S) ∩ GMEP(F, φ, B) such that

$$⟨A{x}^{*},x-x^{*}⟩\ge 0,\forall x\in \Xi :=F\left(S\right)\cap GMEP\left(F,\phi ,B\right),$$

(1.11)

where A and B are monotone operators. This solution set of (1.11) is denoted by ϒ

We present and construct a new iterative algorithm for solving the problem (1.11). The strong convergence for the proposed algorithm to the solution is derived under some assumptions. Our results generalize and improve the results of Marino and Xu [7] and some authors.

2. Preliminaries

Let H be a real Hilbert space and C be a nonempty closed convex subset of H. Recall that the metric (nearest point) projection P _C from H onto C assigns to each x ∈ H, the unique point in P _C x ∈ C satisfying the property

$$∥x-P_{C}x∥=\underset{y\in C}{\text{min}}∥x-y∥.$$

The following characterizes the projection P _C . We recall some lemmas which will be needed in the rest of this article.

Lemma 2.1. The function x ∈ C is a solution of the variational inequality (1.5) if and only if x ∈ C satisfies the relation x = P _C (x - λAx) for all λ > 0.

Lemma 2.2. For a given z ∈ H, u ∈ C, u = P _C z ⇔ 〈u - z, v - u〉 ≥ 0, ∀v ∈ C. It is well known that P _C is a firmly nonexpansive mapping of H onto C and satisfies

$${∥P_{C}x-P_{C}y∥}^2\le ⟨P_{C}x-P_{C}y,x-y⟩,\forall x,y\in H.$$

(2.1)

Moreover, P _C x is characterized by the following properties: P _C x ∈ C and for all x ∈ H, y ∈ C,

$$⟨x-P_{C}x,y-P_{C}x⟩\le 0.$$

(2.2)

Lemma 2.3. There holds the following inequality in an inner product space H

$${∥x+y∥}^2\le {∥x∥}^2+2⟨y,x+y⟩,\forall x,y\in H.$$

Lemma 2.4. [9] Let C be a closed convex subset of a real Hilbert space H and let S: C → C be a nonexpansive mapping. Then I - S is demiclosed at zero, that is,

$$x_n\rightharpoonup xand{x}_n-S{x}_n\to 0$$

imply x = Sx.

For solving the generalized mixed equilibrium problem and the mixed equilibrium problem, let us give the following assumptions for the bifunction F, φ and the set C:

(A1) F(x, x) = 0 for all x ∈ C;

(A2) F is monotone, i.e., F(x, y) + F(y, x) ≤ 0 for all x, y ∈ C;

(A3) for each y ∈ C, x ↦ F(x, y) is weakly upper semicontinuous;

(A4) for each x ∈ C, y ↦ F(x, y) is convex;

(A5) for each x ∈ C, y ↦ F(x, y) is lower 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 ,

$$F\left(z,y_x\right)+\phi \left(y_x\right)-\phi \left(z\right)+\frac{1}{r}⟨y_x-z,z-x⟩<0;$$

(2.3)

(B2) C is a bounded set;

(B3) 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 ,

$$\phi \left(y_x\right)-\phi \left(z\right)+\frac{1}{r}⟨y_x-z,z-x⟩<0;$$

(B4) 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 ,

$$F\left(z,y_x\right)+\frac{1}{r}⟨y_x-z,z-x⟩<0.$$

Lemma 2.5. [10] Let C be a nonempty closed convex subset of a real Hilbert space H. Let F be a bifunction from C × C to satisfying (A 1) - (A 5) and let $\phi :C\to ℛ$ be a proper lower semicontinuous and convex function. For r > 0 and x ∈ H, define a mapping T _r : H → C as follows.

$$T_r\left(x\right)=\left\{z\in C:F\left(z,y\right)+\phi \left(y\right)-\phi \left(z\right)+\frac{1}{r}⟨y-z,z-x⟩\ge 0,\forall y\in C\right\}$$

(2.4)

for all x ∈ H. Assume that either (B 1) or (B 2) holds. Then, the following results hold:

• (1) For each x ∈ H, T _r (x) ≠ ∅;

• (2) T _r is single-valued;

• (3) T _r is firmly nonexpansive, i.e., for any x, y ∈ H, ||T _r x - T _r y||² ≤ 〈T _r x - T _r y, x - y〉;

• (4) F(T _r ) = MEP(F, φ);

• (5) MEP(F, φ) is closed and convex.

Lemma 2.6. [11] Assume {a _n } is a sequence of nonnegative real numbers such that

$$a_{n+1}\le \left(1-{\gamma }_n\right)a_n+{\gamma }_n{\delta }_n+{\beta }_n,\forall n\ge 0,$$

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

(i) ${\sum }_{n=1}^{\infty }{\gamma }_n=\infty$;

(ii) either lim sup_n→∞δ _n ≤ 0 or ${\sum }_{n=1}^{\infty }{\gamma }_n\left|{\delta }_n\right|<\infty$;

(iii) ${\sum }_{n=1}^{\infty }{\beta }_n<\infty$.

Then lim_n→∞a _n = 0.

3. Strong convergence theorems

In this section, we introduce an iterative algorithm for solving some the hierarchical problem over the set of fixed point and generalized mixed equilibrium problem.

Theorem 3.1. Let H be a real Hilbert space, A: C → C be an α-inverse-strongly monotone, f : C → C be a ρ-contraction with coefficient ρ ∈ [0, 1) and S, V: C → C be two nonexpansive mappings. Let B: C → C be a β-inverse-strongly monotone and F be a bifunction from $C\times C\to ℛ$ satisfying (A1)-(A5) and let $\phi :C\to ℛ$ is convex and lower semicontinuous with either (B1) or (B2). Assume that Ξ: = F(S) ∩ GMEP(F, φ, B) is nonempty. Suppose {x _n } is a sequence generated by the following algorithm with x₀ ∈ C arbitrarily:

$$x_{n+1}={\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)\left[{\alpha }_{n}V\left(I-{\lambda }_{n}A\right)x_n+\left(1-{\alpha }_n\right)S{T}_{r_n}\left(x_n-r_{n}B{x}_n\right)\right],$$

(3.1)

where {α _n } and {β _n } ⊂ (0, 1) and λ _n ∈ (0, 2α), r _n ∈ (0, 2β) satisfy the following conditions:

(C1): α _n λ _n < α _n < γβ _n for all n and some constant γ;

(C2): lim_n→∞β _n = 0, ${\sum }_{n=1}^{\infty }{\beta }_n=\infty$, $\underset{n\to \infty }{\text{lim}}\frac{{\beta }_{n-1}}{{\beta }_n}=1$;

(C3): $\underset{n\to \infty }{\text{lim}}\frac{{\alpha }_{n-1}}{{\alpha }_n}=1$;

(C4): ${\sum }_{n=1}^{\infty }\left|{\lambda }_n-{\lambda }_{n-1}\right|<\infty$;

(C5): ${\sum }_{n=1}^{\infty }\left|r_n-r_{n-1}\right|<\infty$, lim inf_n→∞r _n > 0.

Then {x _n } converges strongly to x* ∈ ϒ, which is the unique solution of the variational inequality:

$$⟨\left(I-f\right)x^{*},x-x^{*}⟩\ge 0,\forall x\in \Upsilon .$$

(3.2)

Proof. We will divide the proof into five steps.

Step 1. We will show {x _n } is bounded. Since A, B are α, β-inverse-strongly monotone mappings, we have

$$\begin{array}{ll}\hfill {∥\left(I-{\lambda }_{n}A\right)x-\left(I-{\lambda }_{n}A\right)y∥}^2& ={∥\left(x-y\right)-{\lambda }_n\left(Ax-Ay\right)∥}^2\\ ={∥x-y∥}^2-2{\lambda }_n⟨x-y,Ax-Ay⟩+{\lambda }_n^2{∥Ax-Ay∥}^2\\ \le {∥x-y∥}^2+{\lambda }_n\left({\lambda }_n-2\alpha \right){∥Ax-Ay∥}^2\\ \le {∥x-y∥}^2.\end{array}$$

(3.3)

By Lemma 2.5, we have $u_n=T_{r_n}\left(x_n-r_{n}B{x}_n\right)$ for all n ≥ 0. Then, we have

$$\begin{array}{ll}\hfill {∥u_n-q∥}^2& \le {∥T_{r_n}\left(x_n-r_{n}B{x}_n\right)-T_{r_n}\left(q-r_{n}Bq\right)∥}^2\\ \le {∥\left(x_n-r_{n}B{x}_n\right)-\left(q-r_{n}Bq\right)∥}^2\\ \le {∥x_n-q∥}^2+r_n\left(r_n-2\beta \right){∥B{x}_n-Bq∥}^2\\ \le {∥x_n-q∥}^2.\end{array}$$

For any q ∈ Ξ. Since V, I - λ _n A and $T_{r_n}$ are nonexpansive mappings, we have

$$\begin{array}{ll}\hfill ∥x_{n+1}-q∥& =∥{\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)\left[{\alpha }_{n}V\left(I-{\lambda }_{n}A\right)x_n+\left(1-{\alpha }_n\right)S{T}_{r_n}\left(x_n-r_{n}B{x}_n\right)\right]-q∥\\ \le {\beta }_n∥f\left(x_n\right)-q∥+\left(1-{\beta }_n\right)∥{\alpha }_{n}V\left(I-{\lambda }_{n}A\right)x_n+\left(1-{\alpha }_n\right)S{T}_{r_n}\left(x_n-r_{n}B{x}_n\right)-q∥\\ \le {\beta }_n∥f\left(x_n\right)-f\left(q\right)∥+{\beta }_n∥f\left(q\right)-q∥+\left(1-{\beta }_n\right){\alpha }_n∥V\left(I-{\lambda }_{n}A\right)x_n-q∥\\ +\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)∥S{T}_{r_n}\left(x_n-r_{n}B{x}_n\right)-q∥\\ \le {\beta }_n\rho ∥x_n-q∥+{\beta }_n∥f\left(q\right)-q∥+\left(1-{\beta }_n\right){\alpha }_n\left(∥V\left(I-{\lambda }_{n}A\right)x_n-V\left(I-{\lambda }_{n}A\right)q∥\right.\\ \left.+∥V\left(I-{\lambda }_{n}A\right)q-q∥\right)+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)∥T_{r_n}\left(x_n-r_{n}B{x}_n\right)-q∥\\ \le {\beta }_n\rho ∥x_n-q∥+{\beta }_n∥f\left(q\right)-q∥+\left(1-{\beta }_n\right){\alpha }_n\left(∥x_n-q∥+∥Vq-q∥+{\lambda }_n∥VAq∥\right)\\ +\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)∥x_n-q∥\\ ={\beta }_n\rho ∥x_n-q∥+{\beta }_n∥f\left(q\right)-q∥+\left(1-{\beta }_n\right){\alpha }_n∥x_n-q∥+\left(1-{\beta }_n\right){\alpha }_n∥Vq-q∥\\ +\left(1-{\beta }_n\right){\alpha }_n{\lambda }_n∥VAq∥+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)∥x_n-q∥\\ \le {\beta }_n\rho ∥x_n-q∥+{\beta }_n∥f\left(q\right)-q∥+\left(1-{\beta }_n\right)∥x_n-q∥+{\alpha }_n∥Vq-q∥+{\alpha }_n{\lambda }_n∥VAq∥\\ \le \left[1-\left(1-\rho \right){\beta }_n\right]∥x_n-q∥+{\beta }_n\left(∥f\left(q\right)-q∥+\gamma ∥Vq-q∥+\gamma ∥VAq∥\right).\end{array}$$

By induction, it follows that

$$∥x_n-q∥\le \text{max}\left\{∥x_0-q∥,\frac{1}{1-\rho }\left(∥f\left(q\right)-q∥+\gamma ∥Vq-q∥+\gamma ∥VAq∥\right)\right\},\forall n\ge 0.$$

Therefore {x _n } is bounded and so are {u _n }, {Ax _n }, {V x _n }, and {f(x _n )}.

Step 2. We claim that lim_n→∞||x_n+1- x _n || = 0. Setting y _n = (I - λ _n A)x _n , since I - λ _n A be nonexpansive, we have

$$\begin{array}{ll}\hfill ∥y_n-y_{n-1}∥& =∥\left(I-{\lambda }_{n}A\right)x_n-\left(I-{\lambda }_{n-1}A\right)x_{n-1}∥\\ \le ∥\left(I-{\lambda }_{n}A\right)x_n-\left(I-{\lambda }_{n}A\right)x_{n-1}∥\\ +∥\left(I-{\lambda }_{n}A\right)x_{n-1}-\left(I-{\lambda }_{n-1}A\right)x_{n-1}∥\\ \le ∥x_n-x_{n-1}∥+\left|{\lambda }_n-{\lambda }_{n-1}\right|∥A{x}_{n-1}∥\\ \le ∥x_n-x_{n-1}∥+M_1\left|{\lambda }_n-{\lambda }_{n-1}\right|,\end{array}$$

where $M_1=\text{sup}\left\{∥A{x}_n∥:n\in \mathbb{N}\right\}.$ On the other hand, from $u_{n-1}=T_{r_{n-1}}\left(x_{n-1}-r_{n-1}B{x}_{n-1}\right)$ and $u_n=T_{r_n}\left(x_n-r_{n}B{x}_n\right)$, it follows that

$$F\left(u_{n-1},y\right)+⟨B{x}_{n-1},y-u_{n-1}⟩+\phi \left(y\right)-\phi \left(u_{n-1}\right)+\frac{1}{r_{n-1}}⟨y-u_{n-1},u_{n-1}-x_{n-1}⟩\ge 0,\forall y\in C$$

(3.4)

and

$$F\left(u_n,y\right)+⟨B{x}_n,y-u_n⟩+\phi \left(y\right)-\phi \left(u_n\right)+\frac{1}{r_n}⟨y-u_n,u_n-x_n⟩\ge 0,\forall y\in C.$$

(3.5)

Substituting y = u _n into (3.4) and y = u_n-1into (3.5), we have

$$F\left(u_{n-1},u_n\right)+⟨B{x}_{n-1},u_n-u_{n-1}⟩+\phi \left(u_n\right)-\phi \left(u_{n-1}\right)+\frac{1}{r_{n-1}}⟨u_n-u_{n-1},u_{n-1}-x_{n-1}⟩\ge 0$$

and

$$F\left(u_n,u_{n-1}\right)+⟨B{x}_n,u_{n-1}-u_n⟩+\phi \left(u_{n-1}\right)-\phi \left(u_n\right)+\frac{1}{r_n}⟨u_{n-1}-u_n,u_n-x_n⟩\ge 0.$$

From (A2), we have

$$⟨u_n-u_{n-1},B{x}_{n-1}-B{x}_n+\frac{u_{n-1}-x_{n-1}}{r_{n-1}}-\frac{u_n-x_n}{r_n}⟩\ge 0,$$

and then

$$⟨u_n-u_{n-1},r_{n-1}\left(B{x}_{n-1}-B{x}_n\right)+u_{n-1}-x_{n-1}-\frac{r_{n-1}}{r_n}\left(u_n-x_n\right)⟩\ge 0,$$

so

$$⟨u_n-u_{n-1},r_{n-1}B{x}_{n-1}-r_{n-1}B{x}_n+u_{n-1}-u_n+u_n-x_{n-1}-\frac{r_{n-1}}{r_n}\left(u_n-x_n\right)⟩\ge 0.$$

It follows that

$$⟨u_n-u_{n-1},\left(I-r_{n-1}B\right)x_n-\left(I-r_{n-1}B\right)x_{n-1}+u_{n-1}-u_n+u_n-x_n-\frac{r_{n-1}}{r_n}\left(u_n-x_n\right)⟩\ge 0,$$

$$⟨u_n-u_{n-1},u_{n-1}-u_n⟩+⟨u_n-u_{n-1},x_n-x_{n-1}+\left(1-\frac{r_{n-1}}{r_n}\right)\left(u_n-x_n\right)⟩\ge 0.$$

Without loss of generality, let us assume that there exists a real number c such that r_n-1> c > 0, for all $n\in \mathbb{N}$. Then, we have

$$\begin{array}{c}\hfill {∥u_n-u_{n-1}∥}^2\le ⟨u_n-u_{n-1},x_n-x_{n-1}+\left(1-\frac{r_{n-1}}{r_n}\right)\left(u_n-x_n\right)⟩\hfill \\ \hfill \le ∥u_n-u_{n-1}∥\left\{∥x_n-x_{n-1}∥+\left|1-\frac{r_{n-1}}{r_n}\right|∥u_n-x_n∥\right\}\hfill \end{array}$$

and hence

$$\begin{array}{c}\hfill ∥u_n-u_{n-1}∥\le ∥x_n-x_{n-1}∥+\frac{1}{r_n}\left|r_n-r_{n-1}\right|∥u_n-x_n∥\hfill \\ \hfill \le ∥x_n-x_{n-1}∥+\frac{M_2}{c}\left|r_n-r_{n-1}\right|,\hfill \end{array}$$

(3.6)

where $M_2=\text{sup}\left\{∥u_n-x_n∥:n\in \mathbb{N}\right\}$. From (3.1), we have

$$\begin{array}{ll}\hfill ∥x_{n+1}-x_n∥& =∥{\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)\left[{\alpha }_{n}V{y}_n+\left(1-{\alpha }_n\right)S{u}_n\right]\\ -{\beta }_{n-1}f\left(x_{n-1}\right)-\left(1-{\beta }_{n-1}\right)\left[{\alpha }_{n-1}V{y}_{n-1}+\left(1-{\alpha }_{n-1}\right)S{u}_{n-1}\right]∥\\ \le {\beta }_n\rho ∥x_n-x_{n-1}∥+\left|{\beta }_n-{\beta }_{n-1}\right|∥f\left(x_{n-1}\right)∥+∥\left(1-{\beta }_n\right)\left[{\alpha }_{n}V{y}_n+\left(1-{\alpha }_n\right)S{u}_n\right]\\ -\left(1-{\beta }_{n-1}\right)\left[{\alpha }_{n-1}V{y}_{n-1}+\left(1-{\alpha }_{n-1}\right)S{u}_{n-1}\right]∥\\ ={\beta }_n\rho ∥x_n-x_{n-1}∥+\left|{\beta }_n-{\beta }_{n-1}\right|∥f\left(x_n-1\right)∥\\ +∥\left(1-{\beta }_n\right)\left[{\alpha }_{n}V{y}_n-{\alpha }_{n}V{y}_{n-1}+\left(1-{\alpha }_n\right)S{u}_n-\left(1-{\alpha }_n\right)S{u}_{n-1}\right]\\ +\left(1-{\beta }_n\right){\alpha }_{n}V{y}_{n-1}-\left(1-{\beta }_{n-1}\right){\alpha }_{n-1}V{y}_{n-1}\\ +\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)S{u}_{n-1}-\left(1-{\beta }_{n-1}\right)\left(1-{\alpha }_{n-1}\right)S{u}_{n-1}∥\\ \le {\beta }_n\rho ∥x_n-x_{n-1}∥+\left|{\beta }_n-{\beta }_{n-1}\right|∥f\left(x_{n-1}\right)∥\\ +\left(1-{\beta }_n\right)∥{\alpha }_n\left[V{y}_n-V{y}_{n-1}\right]+\left(1-{\alpha }_n\right)\left[s{u}_n-S{u}_{n-1}\right]∥\\ +∥\left({\alpha }_n-{\beta }_n{\alpha }_n-{\alpha }_{n-1}+{\beta }_{n-1}{\alpha }_{n-1}\right)V{y}_{n-1}\\ +\left(1-{\beta }_n-{\alpha }_n+{\beta }_n{\alpha }_n-1+{\beta }_{n-1}+{\alpha }_{n-1}-{\beta }_{n-1}{\alpha }_{n-1}\right)S{u}_{n-1}∥\\ \le {\beta }_n\rho ∥x_n-x_{n-1}∥+\left|{\beta }_n-{\beta }_{n-1}\right|∥f\left(x_{n-1}\right)∥\end{array}$$

$$\begin{array}{c}\hfill +\left(1-{\beta }_n\right){\alpha }_n∥y_n-y_{n-1}∥+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)∥u_n-u_{n-1}∥\hfill \\ \hfill +∥\left({\alpha }_n-{\alpha }_{n-1}-{\beta }_n{\alpha }_n+{\beta }_n{\alpha }_{n-1}-{\beta }_n{\alpha }_{n-1}+{\beta }_{n-1}{\alpha }_{n-1}\right)V{y}_{n-1}\hfill \\ \hfill +\left(-{\beta }_n+{\beta }_{n-1}-{\alpha }_n+{\alpha }_{n-1}+{\beta }_n{\alpha }_n-{\beta }_n{\alpha }_{n-1}+{\beta }_n{\alpha }_{n-1}-{\beta }_{n-1}{\alpha }_{n-1}\right)S{u}_{n-1}∥\hfill \\ \hfill \le {\beta }_n\rho ∥x_n-x_{n-1}∥+\left|{\beta }_n-{\beta }_{n-1}\right|∥f\left(x_{n-1}\right)∥\hfill \\ \hfill +\left(1-{\beta }_n\right){\alpha }_n\left\{∥x_n-x_{n-1}∥+M_1\left|{\lambda }_n-{\lambda }_{n-1}\right|\right\}\hfill \\ \hfill +\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)\left\{∥x_n-x_{n-1}∥+\frac{M_2}{c}\left|r_n-r_{n-1}\right|\right\}\hfill \\ \hfill +∥\left[\left({\alpha }_n-{\alpha }_{n-1}\right)-{\beta }_n\left({\alpha }_n-{\alpha }_{n-1}\right)-\left({\beta }_n-{\beta }_{n-1}\right){\alpha }_{n-1}\right]V{y}_{n-1}\hfill \\ \hfill +\left[-\left({\beta }_n-{\beta }_{n-1}\right)-\left({\alpha }_n-{\alpha }_{n-1}\right)+{\beta }_n\left({\alpha }_n-{\alpha }_{n-1}\right)+\left({\beta }_n-{\beta }_{n-1}\right){\alpha }_{n-1}\right]S{u}_{n-1}∥\hfill \\ \hfill ={\beta }_n\rho ∥x_n-x_{n-1}∥+\left|{\beta }_n-{\beta }_{n-1}\right|∥f\left(x_{n-1}\right)∥\hfill \\ \hfill +\left(1-{\beta }_n\right){\alpha }_n∥x_n-x_{n-1}∥+\left(1-{\beta }_n\right){\alpha }_n{M}_1\left|{\lambda }_n-{\lambda }_{n-1}\right|\hfill \\ \hfill +\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)∥x_n-x_{n-1}∥+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)\frac{M_2}{c}\left|r_n-r_{n-1}\right|\hfill \\ \hfill +∥\left[\left(1-{\beta }_n\right)\left({\alpha }_n-{\alpha }_{n-1}\right)-\left({\beta }_n-{\beta }_{n-1}\right){\alpha }_{n-1}\right]V{y}_{n-1}\hfill \\ \hfill +\left[\left({\beta }_n-1\right)\left({\alpha }_n-{\alpha }_{n-1}\right)+\left({\beta }_n-{\beta }_{n-1}\right)\left({\alpha }_{n-1}-1\right)\right]S{u}_{n-1}∥\hfill \\ \hfill \le {\beta }_n\rho ∥x_n-x_{n-1}∥+\left(1-{\beta }_n\right)∥x_n-x_{n-1}∥\hfill \\ \hfill +\left|{\beta }_n-{\beta }_{n-1}\right|∥f\left(x_{n-1}\right)∥+\left(1-{\beta }_n\right){\alpha }_n{M}_1\left|{\lambda }_n-{\lambda }_{n-1}\right|\hfill \\ \hfill +\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)\frac{M_2}{c}\left|r_n-r_{n-1}\right|\hfill \\ \hfill +\left(1-{\beta }_n\right)\left({\alpha }_n-{\alpha }_{n-1}\right)∥V{y}_{n-1}-S{u}_{n-1}∥\hfill \\ \hfill +∥\left({\beta }_n-{\beta }_{n-1}\right)\left({\alpha }_{n-1}-1\right)S{u}_{n-1}-\left({\beta }_n-{\beta }_{n-1}\right){\alpha }_{n-1}V{y}_{n-1}∥\hfill \end{array}$$

$$\begin{array}{c}\hfill ={\beta }_n\rho ∥x_n-x_{n-1}∥+\left(1-{\beta }_n\right)∥x_n-x_{n-1}∥\hfill \\ \hfill +\left(1-{\beta }_n\right){\alpha }_n{M}_1\left|{\lambda }_n-{\lambda }_{n-1}\right|+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)\frac{M_2}{c}\left|r_n-r_{n-1}\right|\hfill \\ \hfill +\left(1-{\beta }_n\right)\left({\alpha }_n-{\alpha }_{n-1}\right)∥V{y}_{n-1}-S{u}_{n-1}∥\hfill \\ \hfill +\left|{\beta }_n-{\beta }_{n-1}\right|\left(∥f\left(x_{n-1}\right)∥+{\alpha }_{n-1}∥V{y}_{n-1}∥+\left|1-{\alpha }_{n-1}\right|∥S{u}_{n-1}∥\right)\hfill \\ \hfill \le {\beta }_n\rho ∥x_n-x_{n-1}∥+\left(1-{\beta }_n\right)∥x_n-x_{n-1}∥+{\alpha }_n{M}_1\left|{\lambda }_n-{\lambda }_{n-1}\right|\hfill \\ \hfill +\frac{M_2}{c}\left|r_n-r_{n-1}\right|+\left({\alpha }_n-{\alpha }_{n-1}\right)∥V{y}_{n-1}-S{u}_{n-1}∥\hfill \\ \hfill +\left|{\beta }_n-{\beta }_{n-1}\right|\left(∥f\left(x_{n-1}\right)∥+{\alpha }_{n-1}∥V{y}_{n-1}∥+\left|1-{\alpha }_{n-1}\right|∥S{u}_{n-1}∥\right)\hfill \\ \hfill \le \left[1-\left(1-\rho \right){\beta }_n\right]∥x_n-x_{n-1}∥+{\alpha }_n{M}_1\left|{\lambda }_n-{\lambda }_{n-1}\right|+\frac{M_2}{c}\left|r_n-r_{n-1}\right|\hfill \\ \hfill +\left(\left|{\alpha }_n-{\alpha }_{n-1}\right|+\left|{\beta }_n-{\beta }_{n-1}\right|\right)M_3\hfill \\ \hfill =\left[1-\left(1-\rho \right){\beta }_n\right]∥x_n-x_{n-1}∥+{\alpha }_n{M}_1\left|{\lambda }_n-{\lambda }_{n-1}\right|+\frac{M_2}{c}\left|r_n-r_{n-1}\right|\hfill \\ \hfill +\left(\frac{\left|{\alpha }_n-{\alpha }_{n-1}\right|}{{\beta }_n}+\frac{\left|{\beta }_n-{\beta }_{n-1}\right|}{{\beta }_n}\right){\beta }_n{M}_3\hfill \\ \hfill \le \left[1-\left(1-\rho \right){\beta }_n\right]∥x_n-x_{n-1}∥+{\alpha }_n{M}_1\left|{\lambda }_n-{\lambda }_{n-1}\right|+\frac{M_2}{c}\left|r_n-r_{n-1}\right|\hfill \\ \hfill +\left(\frac{\gamma \left|{\alpha }_n-{\alpha }_{n-1}\right|}{{\alpha }_n}+\frac{\left|{\beta }_n-{\beta }_{n-1}\right|}{{\beta }_n}\right){\beta }_n{M}_3,\hfill \end{array}$$

where M₃ = sup{max{||Vy_n-1||, ||Su_n-1||, ||f (x_n-1)||}}. Since conditions (C1)-(C5) by Lemma 2.6, we have ||x_n+1- x _n || → 0 as n → ∞.

Step 3. We claim that lim_n→∞||x _n - Sx _n || = 0. For each q ∈ Ξ, we note that since $T_{r_n}$ is firmly nonexpansive, then we have

$$\begin{array}{ll}\hfill {∥u_n-q∥}^2& ={∥T_{r_n}\left(x_n-r_{n}B{x}_n\right)-T_{r_n}\left(q-r_{n}Bq\right)∥}^2\\ \le ⟨T_{r_n}\left(x_n-r_{n}B{x}_n\right)-T_{r_n}\left(q-r_{n}Bq\right),u_n-q⟩\\ =⟨\left(x_n-r_{n}B{x}_n\right)-\left(q-r_{n}Bq\right),u_n-q⟩\\ =\frac{1}{2}\left\{{∥\left(x_n-r_{n}B{x}_n\right)-\left(q-r_{n}Bq\right)∥}^2+{∥u_n-q∥}^2\right.\\ \left.-{∥\left(x_n-r_{n}B{x}_n\right)-\left(q-r_{n}Bq\right)-\left(u_n-q\right)∥}^2\right\}\\ \le \frac{1}{2}\left\{{∥x_n-q∥}^2+{∥u_n-q∥}^2-{∥x_n-u_n-r_n\left(B{x}_n-Bq\right)∥}^2\right\}\\ \le \frac{1}{2}\left\{{∥x_n-q∥}^2+{∥u_n-q∥}^2-{∥x_n-u_n∥}^2\right.\\ \left.+2r_n⟨x_n-u_n,B{x}_n-Bq⟩-r_n^2{∥B{x}_n-Bq∥}^2\right\},\end{array}$$

which imply that

$${∥u_n-q∥}^2\le {∥x_n-q∥}^2-{∥x_n-u_n∥}^2+2r_n∥x_n-u_n∥∥B{x}_n-Bq∥.$$

(3.7)

From (3.1) and set w _n : = α _n V(I - λ _n A)x _n + (1 - α _n )Su _n , when y _n = (I - λ _n A)x _n , then we have

$$\begin{array}{ll}\hfill {∥w_n-q∥}^2& ={∥{\alpha }_{n}V{y}_n+\left(1-{\alpha }_n\right)S{u}_n-q∥}^2\\ ={∥{\alpha }_{n}V{y}_n+\left(1-{\alpha }_n\right)S{u}_n-\left(1-{\alpha }_n\right)Sq+\left(1-{\alpha }_n\right)Sq-q∥}^2\\ ={∥{\alpha }_n\left(V{y}_n-Sq\right)+\left(1-{\alpha }_n\right)\left(S{u}_n-Sq\right)+Sq-q∥}^2\\ \le {\alpha }_n{∥V{y}_n-Sq∥}^2+\left(1-{\alpha }_n\right){∥u_n-q∥}^2.\end{array}$$

(3.8)

On the other hand, we note that

$$\begin{array}{ll}\hfill {∥u_n-q∥}^2& ={∥T_{r_n}\left(x_n-r_{n}B{x}_n\right)-T_{r_n}\left(q-r_{n}Bq\right)∥}^2\\ \le {∥\left(x_n-r_{n}B{x}_n\right)-\left(q-r_{n}Bq\right)∥}^2\\ ={∥\left(x_n-q\right)-r_n\left(B{x}_n-Bq\right)∥}^2\\ \le {∥x_n-q∥}^2-2r_n⟨x_n-q,B{x}_n-Bq⟩+r_n^2{∥B{x}_n-Bq∥}^2\\ \le {∥x_n-q∥}^2-2r_n\beta {∥B{x}_n-Bq∥}^2+r_n^2{∥B{x}_n-Bq∥}^2.\end{array}$$

(3.9)

Using (3.8) and (3.9), we note that

$$\begin{array}{ll}\hfill {∥x_{n+1}-q∥}^2& ={∥{\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)w_n-q∥}^2\\ \le {\beta }_n{∥f\left(x_n\right)-q∥}^2+\left(1-{\beta }_n\right){∥w_n-q∥}^2\\ \le {\beta }_n{\rho }^2{∥x_n-q∥}^2+\left(1-{\beta }_n\right)\left\{{\alpha }_n{∥V{y}_n-Sq∥}^2+\left(1-{\alpha }_n\right){∥u_n-q∥}^2\right\}\\ \le {\beta }_n\rho {∥x_n-q∥}^2+\left(1-{\beta }_n\right){\alpha }_n{∥V{y}_n-Sq∥}^2+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right){∥u_n-q∥}^2\\ \le {\beta }_n\rho {∥x_n-q∥}^2+{\alpha }_n{∥V{y}_n-Sq∥}^2+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)\\ \times \left\{{∥x_n-q∥}^2-2r_n\beta {∥B{x}_n-Bq∥}^2+r_n^2{∥B{x}_n-Bq∥}^2\right\}\\ ={\beta }_n\rho {∥x_n-q∥}^2+{\alpha }_n{∥V{y}_n-Sq∥}^2+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)\\ \times \left\{{∥x_n-q∥}^2-r_n\left(r_n-2\beta \right){∥B{x}_n-Bq∥}^2\right\}\\ ={\beta }_n\rho {∥x_n-q∥}^2+{\alpha }_n{∥V{y}_n-Sq∥}^2+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right){∥x_n-q∥}^2\\ +\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)r_n\left(r_n-2\beta \right){∥B{x}_n-Bq∥}^2\\ \le {\beta }_n\rho {∥x_n-q∥}^2+{\alpha }_n{∥V{y}_n-Sq∥}^2+\left(1-{\beta }_n\right){∥x_n-q∥}^2\\ +\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)r_n\left(r_n-2\beta \right){∥B{x}_n-Bq∥}^2\\ \le \left[1-\left(1-\rho \right){\beta }_n\right]{∥x_n-q∥}^2+\gamma {\beta }_n{∥V{y}_n-Sq∥}^2\\ +\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)r_n\left(r_n-2\beta \right){∥B{x}_n-Bq∥}^2.\end{array}$$

(3.10)

Then, we have

$$\begin{array}{ll}\hfill \left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)c\left(2\beta -d\right){∥B{x}_n-Bq∥}^2& \le \gamma {\beta }_n{∥V{y}_n-Sq∥}^2+{∥x_n-q∥}^2-{∥x_{n+1}-q∥}^2\\ \le \gamma {\beta }_n{∥V{y}_n-Sq∥}^2\\ +∥x_n-x_{n+1}∥\left(∥x_n-q∥+∥x_{n+1}-q∥\right).\end{array}$$

From (C2), {r _n } ⊂ [c, d] ⊂ (0, 2β) and lim_n→∞||x_n+1- x _n || = 0, we obtain

$$\underset{n\to \infty }{\text{lim}}∥B{x}_n-Bq∥=0.$$

(3.11)

Using (3.7), (3.8) and (3.10), it follows that

$$\begin{array}{ll}\hfill {∥x_{n+1}-q∥}^2& \le {\beta }_n\rho {∥x_n-q∥}^2+\left(1-{\beta }_n\right){\alpha }_n{∥V{y}_n-Sq∥}^2+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right){∥u_n-q∥}^2\\ \le {\beta }_n\rho {∥x_n-q∥}^2+{\alpha }_n{∥V{y}_n-Sq∥}^2+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)\\ \times \left\{{∥x_n-q∥}^2-{∥x_n-u_n∥}^2+2r_n∥x_n-u_n∥∥B{x}_n-Bq∥\right\}\\ ={\beta }_n\rho {∥x_n-q∥}^2+{\alpha }_n{∥V{y}_n-Sq∥}^2+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right){∥x_n-q∥}^2\\ -\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right){∥x_n-u_n∥}^2+2\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)r_n∥x_n-u_n∥∥B{x}_n-Bq∥\\ \le {\beta }_n\rho {∥x_n-q∥}^2+{\alpha }_n{∥V{y}_n-Sq∥}^2+\left(1-{\beta }_n\right){∥x_n-q∥}^2\\ -\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right){∥x_n-u_n∥}^2+2r_n∥x_n-u_n∥∥B{x}_n-Bq∥\\ \le \left[1-\left(1-\rho \right){\beta }_n\right]{∥x_n-q∥}^2+\gamma {\beta }_n{∥V{y}_n-Sq∥}^2\\ -\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right){∥x_n-u_n∥}^2+2r_n∥x_n-u_n∥∥B{x}_n-Bq∥.\end{array}$$

(3.12)

Then, we have

$$\begin{array}{ll}\hfill \left(1-{\beta }_n\right)\left(1-{\alpha }_n\right){∥x_n-u_n∥}^2& \le {∥x_n-q∥}^2-{∥x_{n+1}-q∥}^2+\gamma {\beta }_n{∥V{y}_n-Sq∥}^2\\ +2r_n∥x_n-u_n∥∥B{x}_n-B_q∥\\ \le ∥x_n-x_{n+1}∥\left(∥x_n-q∥+∥x_{n+1}-q∥\right)+\gamma {\beta }_n{∥V{y}_n-Sq∥}^2\\ +2r_n∥x_n-u_n∥∥B{x}_n-Bq∥.\end{array}$$

From (C1), (C2), (3.13) and lim_n→∞||x_n+1- x _n || = 0, we obtain

$$\underset{n\to \infty }{\text{lim}}∥x_n-u_n∥=0.$$

(3.13)

By (C5), we obtain

$$\underset{n\to \infty }{\text{lim}}∥\frac{x_n-u_n}{r_n}∥=\underset{n\to \infty }{\text{lim}}\frac{1}{r_n}∥x_n-u_n∥=0.$$

(3.14)

From (3.1), it follows that

$$\begin{array}{ll}\hfill ∥x_{n+1}-S{u}_n∥& =∥{\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)\left[{\alpha }_{n}V{y}_n+\left(1-{\alpha }_n\right)S{u}_n\right]-S{u}_n∥\\ =∥{\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right){\alpha }_{n}V{y}_n+\left(1-{\beta }_n\right)\left(1-{\alpha }_n\right)S{u}_n-S{u}_n∥\\ =∥{\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right){\alpha }_{n}V{y}_n+\left(1-{\beta }_n\right)S{u}_n+\left(1-{\beta }_n\right){\alpha }_{n}S{u}_n-S{u}_n∥\\ \le {\beta }_n∥f\left(x_n\right)-S{u}_n∥+\left(1-{\beta }_n\right){\alpha }_n∥V{y}_n-S{u}_n∥.\end{array}$$

(3.15)

By (C1) and (C2), then we get

$$\underset{n\to \infty }{\text{lim}}∥x_{n+1}-S{u}_n∥=0.$$

(3.16)

Since

$$\begin{array}{ll}\hfill ∥x_n-S{x}_n∥& \le ∥x_n-x_{n+1}∥+∥x_{n+1}-S{u}_n∥+∥S{u}_n-S{x}_n∥\\ \le ∥x_n-x_{n+1}∥+∥x_{n+1}-S{u}_n∥+∥u_n-x_n∥.\end{array}$$

By lim_n→∞||x_n+1- x _n || = 0, (3.13) and (3.16), so we obtain

$$\underset{n\to \infty }{\text{lim}}∥x_n-S{x}_n∥=0.$$

(3.17)

Step 4. Next, we will show that

$$\underset{n\to \infty }{\lim \sup }〈\left(I-f\right)x^{\ast },x_n-x^{\ast }〉\le 0.$$

Indeed, we choose a subsequence $\left\{x_{n_i}\right\}$ of {x _n } such that

$$\underset{n\to \infty }{\lim \sup }〈\left(I-f\right)x^{\ast },x_n-x^{\ast }〉=\underset{n\to \infty }{\lim \sup }〈\left(I-f\right)x^{\ast },x_{n_i}-x^{\ast }〉.$$

(3.18)

Since ?82? is bounded, there exists a subsequence $\left\{x_{n_{i_j}}\right\}$ of ?82? which converge weakly to z ∈ C. Without loss of generality, we can assume that $x_{n_i}\rightharpoonup z$. From ||x _n -- Sx _n || → 0, we obtain $S{x}_{n_i}\rightharpoonup z$. Now, we will show that z ∈ Ξ: = F (S) ∩ GMEP (F, φ, B). Let us show z ∈ F(S). Assume that z ∉ F(S). Since ?85? and Sz ≠ z. By the Opial's condition, we obtain

$$\begin{array}{c}\underset{i\to \infty }{\lim \inf }\Vert x_{n_i}-z\Vert <\underset{i\to \infty }{\lim \inf }\Vert x_{n_i}-Sz\Vert \\ =\underset{i\to \infty }{\lim \inf }\Vert x_{n_i}-S{x}_{n_i}+S{x}_{n_i}-Sz\Vert \\ \le \underset{i\to \infty }{\lim \inf }\left(\Vert x_{n_i}-S{x}_{n_i}\Vert +\Vert S{x}_{n_i}-Sz\Vert \right)\\ =\underset{i\to \infty }{\lim \inf }\Vert S{x}_{n_i}-Sz\Vert \\ \le \underset{i\to \infty }{\lim \inf }\Vert x_{n_i}-z\Vert .\end{array}$$

This is a contradiction. Thus, we have z ∈ F(S).

Next, we will show that z ∈ GMEP (F, φ, B). Since $u_n=T_{r_n}\left(x_n-r_{n}B{x}_n\right)$, we have

$$F\left(u_n,y\right)+⟨B{x}_n,y-u_n⟩+\phi \left(y\right)-\phi \left(u_n\right)+\frac{1}{r_n}⟨y-u_n,u_n-x_n⟩\ge 0,\forall y\in C.$$

From (A2), we also have

$$⟨B{x}_n,y-u_n⟩+\phi \left(y\right)-\phi \left(u_n\right)+\frac{1}{r_n}⟨y-u_n,u_n-x_n⟩\ge F\left(y,u_n\right),\forall y\in C.$$

and hence

$$⟨B{x}_{n_i},y-u_{n_i}⟩+\phi \left(y\right)-\phi \left(u_{n_i}\right)+⟨y-u_{n_i},\frac{u_{n_i}-x_{n_i}}{r_{n_i}}⟩\ge F\left(y,u_{n_i}\right),\forall y\in C.$$

(3.19)

For t with 0 < t ≤ 1 and y ∈ C, let y _t = ty + (1 - t)z. Since y ∈ C and z ∈ C, we have y _t ∈ C. So, from (3.19), we have

$$\begin{array}{ll}\hfill ⟨y_t-u_{n_i},B{y}_t⟩& \ge ⟨y_t-u_{n_i},B{y}_t⟩-\phi \left(y_t\right)+\phi \left(u_{n_i}\right)-⟨y_t-u_{n_i},B{x}_{n_i}⟩-⟨y_t-u_{n_i},\frac{u_{n_i}-x_{n_i}}{r_{n_i}}⟩\\ +F\left(y_t,u_{n_i}\right)\\ =⟨y_t-u_{n_i},B{y}_t-B{u}_{n_i}⟩+⟨y_t-u_{n_i},B{u}_{n_i}-B{x}_{n_i}⟩-\phi \left(y_t\right)+\phi \left(u_{n_i}\right)\\ -⟨y_t-u_{n_i},\frac{u_{n_i}-x_{n_i}}{r_{n_i}}⟩+F\left(y_t,u_{n_i}\right).\end{array}$$

Since $∥u_{n_i}-x_{n_i}∥\to 0$, we have $∥B{u}_{n_i}-B{x}_{n_i}∥\to 0$. Further, from the inverse strongly monotonicity of B, we have $⟨y_t-u_{n_i},B{y}_t-B{u}_{n_i}⟩\ge 0$. So, from (A4), (A5), and the weak lower semicontinuity of $\phi ,\frac{u_{n_i}-x_{n_i}}{r_{n_i}}\to 0$ and $u_{n_i}\rightharpoonup z$, we have at the limit

$$⟨y_t-z,B{y}_t⟩\ge -\phi \left(y_t\right)+\phi \left(z\right)+F\left(y_t,z\right)$$

(3.20)

as i → ∞. From (A1), (A4), and (3.20), we also get

$$\begin{array}{c}\hfill 0=F\left(y_t,y_t\right)+\phi \left(y_t\right)-\phi \left(y_t\right)\hfill \\ \hfill \le tF\left(y_t,y\right)+\left(1-t\right)F\left(y_t,z\right)+t\phi \left(y\right)-\left(1-t\right)\phi \left(z\right)-\phi \left(y_t\right)\hfill \\ \hfill =t\left[F\left(y_t,y\right)+\phi \left(y\right)-\phi \left(y_t\right)\right]+\left(1-t\right)\left[F\left(y_t,z\right)+\phi \left(z\right)-\phi \left(y_t\right)\right]\hfill \\ \hfill \le t\left[F\left(y_t,y\right)+\phi \left(y\right)-\phi \left(y_t\right)\right]+\left(1-t\right)⟨y_t-z,B{y}_t⟩\hfill \\ \hfill =t\left[F\left(y_t,y\right)+\phi \left(y\right)-\phi \left(y_t\right)\right]+\left(1-t\right)t⟨y-z,B{y}_t⟩,\hfill \\ \hfill 0\le F\left(y_t,y\right)+\phi \left(y\right)-\phi \left(y_t\right)+\left(1-t\right)⟨y-z,B{y}_t⟩.\hfill \end{array}$$

Letting t → 0, we have, for each y ∈ C,

$$F\left(z,y\right)+\phi \left(y\right)-\phi \left(z\right)+⟨y-z,Bz⟩\ge 0.$$

This implies that z ∈ GMEP (F, φ, B). Therefore x* ∈ Ξ. It is easy to see that P_ϒ(I - f)(x*) is a contraction of H into itself. Hence H is complete, there exists a unique fixed point x* ∈ H, such that x* = P_ϒ(I - f)(x*). Since x* = P_ϒ(I - f)(x*), we have

$$\begin{array}{l}\underset{n\to \infty }{\lim \sup }〈\left(I-f\right)x^{\ast },x_n-x^{\ast }〉=\underset{n\to \infty }{\lim \sup }〈\left(I-f\right)x^{\ast },S{x}_n-x^{\ast }〉\hfill \\ =\underset{n\to \infty }{\lim \sup }〈\left(I-f\right)x^{\ast },S{x}_{n_i}-x^{\ast }〉\hfill \\ =〈\left(I-f\right)x^{*},z-x^{*}〉\le 0.\hfill \end{array}$$

(3.21)

Step 5. Last, we will prove x _n → x* ∈ ϒ. It follows from (3.1) that, we compute

$$\begin{array}{ll}\hfill {∥x_{n+1}-x^{*}∥}^2& ={∥{\beta }_{n}f\left(x_n\right)+\left(1-{\beta }_n\right)\left[{\alpha }_{n}V\left(I-{\lambda }_{n}A\right)x_n+\left(1-{\alpha }_n\right)S{T}_{r_n}\left(x_n-r_{n}B{x}_n\right)\right]-x^{*}∥}^2\\ =∥{\beta }_n\left[f\left(x_n\right)-f\left(x^{*}\right)\right]+\left(1-{\beta }_n\right)\left\{{\alpha }_n\left[V\left(I-{\lambda }_{n}A\right)x_n-V\left(I-{\lambda }_{n}A\right)x^{*}\right]\right.\\ \left.+\left(1-{\alpha }_n\right)\right\}\end{array}$$