Convergence theorem of κ-strictly pseudo-contractive mapping and a modifıcation of generalized equilibrium problems

Abstract

The purpose of this article, we first introduce strong convergence theorem of κ-strictly pseudo-contractive mapping without assumption of the mapping S = κI + (1 - κ)T. Then, we prove strong convergence of proposed iterative scheme for finding a common element of the set of fixed points of κ-strictly pseudo-contractive mapping and the set of solution of a modification of generalized equilibrium problem. Moreover, by using our main result and a new lemma in the last section we obtain strong convergence theorem for finding a common element of the set of fixed points of κ-strictly pseudo-contractive mapping and two sets of solutions of variational inequalities.

1 Introduction

Throughout this article, we assume that H is a real Hilbert space and C is a nonempty subset of H. A mapping T of C into itself is nonlinear mapping. A point x is called a fixed point of T if Tx = x. We use F(T) to denote the set of fixed point of T. Recalled the following definitions;

Definition 1.1. The mapping T is said to be nonexpansive if

$$\left|\right|Tx-Ty\left|\right|\le \left|\right|x-y\left|\right|,\forall x,y\in H$$

Definition 1.2. The mapping T is said to be strictly pseudo-contractive [1] with the coefficient κ ∈ [0, 1) if

$$\left|\right|Tx-Ty|{|}^2\le ||x-y|{|}^2+\kappa \left|\right|\left(I-T\right)x-\left(I-T\right)y|{|}^2\forall x,y\in H.$$

(1.1)

For such case, T is also said to be a κ-strictly pseudo contractive mapping.

The class of κ-strictly pseudo-contractive mapping strictly includes the class of nonexpansive mapping.

Let A : C → H. The variational inequality problem is to find a point u ∈ C such that

$$〈Au,v-u〉\ge 0$$

(1.2)

for all v ∈ C.

The variational inequality has emerged as a fascinating and interesting branch of mathematical and engineering sciences with a wide range of applications in industry, finance, economics, social, ecology, regional, pure and applied sciences (see, e.g. [2–5]).

A mapping A of C into H is called α-inverse strongly monotone; see [6], if there exists a positive real number α such that

$$〈x-y,Ax-Ay〉\ge \alpha \left|\right|Ax-Ay|{|}^2$$

for all x, y ∈ C.

Let F:C×C→ ℝ be a bifunction. The equilibrium problem for F is to determine its equilibrium points, i.e. the set

$$EP\left(F\right)=\left\{x\in C:F\left(x,y\right)\ge 0,\forall y\in C\right\}.$$

(1.3)

From (1.2) and (1.3), we have the following generalized equilibrium problem, i.e.

$$\text{Find }z\in C\text{such that }F\left(z,y\right)+〈Az,y-z〉\ge 0,\forall y\in C.$$

(1.4)

The set of such z ∈ C is denoted by EP (F, A), i.e.,

$$EP\left(F,A\right)=\left\{z\in C:F\left(z,y\right)+〈Az,y-z〉\ge 0,\forall y\in C\right\}$$

In the case of A ≡ 0, EP (F, A) is denoted by EP(F). In the case of F ≡ 0, EP(F, A) is also denoted by VI(C, A).

Numerous problems in physics, optimization and economics reduce to find a solution of EP(F) (see, for example [7–9]). Recently, many authors considered the iterative scheme for finding a common element of the set of solution of equilibrium problem and the set of solutions of fixed point problem (see, for example [10–14]). In 2005, Combettes and Hirstoaga [8] introduced an iterative scheme for finding the best approximation to the initial data when EP(F) is nonempty and they also proved the strong convergence theorem.

In 2007, Takahashi and Takahashi [11] introduced viscosity approximation method in framework of a real Hilbert space H. They defined the iterative sequence {x _n } and {u _n } as follows:

$$\left\{\begin{array}{c}{x}_1\in H,\text{arbitrarily};\\ F\left(u_n,y\right)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\forall y\in C,\\ x_{n+1}={\alpha }_{n}f\left(x_n\right)+\left(1-{\alpha }_n\right)T{u}_n,\forall n\in \mathbb{N},\end{array}\right.$$

(1.5)

where f : H → H is a contraction mapping with constant α ∈ (0, 1) and {α _n } ⊂ [0,1], {r _n } ⊂ (0, ∞). They proved under some suitable conditions on the sequence {α _n }, {r _n } and bifunction F that {x _n }, {u _n } strongly converge to z ∈ F(T) ∩ EP(F), where z = P_{F(T) ∩ EP(F)}f(z).

Recently, in 2008, Takahashia and Takahashi [14] introduced a general iterative method for finding a common element of EP (F, A) and F(T). They defined {x _n } in the following way:

$$\left\{\begin{array}{c}u,x_1\in C,\text{arbitrarily};\\ F\left(z_n,y\right)+〈A{x}_n,y-z_n〉+\frac{1}{{\lambda }_n}〈y-z_n,z_n-x_n〉\ge 0,\forall y\in C,\\ x_{n+1}={\beta }_n{x}_n+\left(1-{\beta }_n\right)T\left(a_{n}u+\left(1-a_n\right)z_n\right),\forall n\in \mathbb{N},\end{array}\right.$$

(1.6)

where A be an α-inverse strongly monotone mapping of C into H with positive real number α and {a _n } ∈ [0, 1], {β _n } ⊂ [0, 1], {λ _n } ⊂ [0, 2α], and proved strong convergence of the scheme (1.6) to $z\in {\cap }_{i=1}^{N}F\left(T_i\right)\cap EP\left(F,A\right)$, where $z=P_{{\cap }_{i=1}^{N}F\left(T_i\right)\cap EP}u$ in the framework of a Hilbert space, under some suitable conditions on {a _n }, {β _n }, {λ _n } and bifunction F.

In 2009, Inchan [15] proved the following theorem:

Theorem 1.1. Let H be a Hilbert space, C be a nonempty closed convex subset of H such that C ± C ⊂ C, and let T : C → H be a κ-strictly pseudo-contractive mapping with a fixed point for some 0 ≤ κ < 1. Let A be a strongly positive bounded linear operator on C with coefficient $\bar{\gamma }$ and f: C → C be a contraction with the contractive constant (0 < α < 1) such that $0<\gamma <\frac{\bar{\gamma }}{\alpha }$. Let{x _n } be the sequence generated by

$$\left\{\begin{array}{c}{x}_1\in C,\\ x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+{\beta }_n{x}_n+\left(\left(1-{\beta }_n\right)I-{\alpha }_{n}A\right)P_{C}S{x}_n,\end{array}\right.$$

where S : C → H is a mapping defined by

$$Sx=\kappa x+\left(1-\kappa \right)Tx$$

(1.7)

If the control sequence {α _n }, {β _n } ⊂ (0, 1) satisfying

$$\begin{array}{l}\left(i\right)\underset{n\to \infty }{\text{lim}}{\alpha }_n=0and\underset{n\to \infty }{\text{lim}}{\beta }_n=0,\\ \left(ii\right)\sum _{n=1}^{\infty }{\alpha }_n=\infty ,\\ \left(iii\right)\sum _{n=1}^{\infty }|{\alpha }_{n+1}-{\alpha }_n|<\infty ,\sum _{n=1}^{\infty }|{\beta }_{n+1}-{\beta }_n|<\infty .\end{array}$$

Then {x _n } converges strongly to a fixed point q of T, which solves the following solution of variational inequality;

$$〈\left(A-\gamma f\right)q,q-x〉\le 0,\forall x\in F\left(T\right).$$

In 2010, Jung [16] proved the following theorem:

Theorem 1.2. Let H be a Hilbert space, C be a nonempty closed convex subset of H such that C ± C ⊂ C, and let T : C → H be a κ-strictly pseudo-contractive mapping with F(T) ≠ ∅ for some 0 ≤ κ < 1. Let A be a strongly positive bounded linear operator on C with coefficient $\bar{\gamma }$ and f: C → C be a contraction with the contractive coefficient 0 < α < 1 such that $0<\gamma <\frac{\bar{\gamma }}{\alpha }$. Let{α _n } and {β _n } ⊂ (0, 1) be sequences which satisfy the following conditions:

$$\begin{array}{c}\left(C1\right)\underset{n\to \infty }{\text{lim}}{\alpha }_n=0,\\ \left(C2\right)\sum _{n=0}^{\infty }{\alpha }_n=\infty ,\\ \left(B\right)0<\underset{n\to \infty }{\text{lim}\text{inf}}{\beta }_n\le \underset{n\to \infty }{\text{lim}\text{sup}}{\beta }_n<aforsomeaconstanta\in \left(0,1\right).\end{array}$$

Let {x _n } be a sequence in C generated by

$$\left\{\begin{array}{c}{x}_0=x\in C,\\ y_n={\beta }_n{x}_n+\left(1-{\beta }_n\right)P_{C}S{x}_n\\ x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_{n}A\right)y_n,n\ge 0,\end{array}\right.$$

where S : C → H is a^:mapping defined by

$$Sx=\kappa x+\left(1-\kappa \right)Tx$$

(1.8)

Then {x _n } converges strongly to a fixed point q of T, which solves the following solution of variational inequality;

$$〈\left(A-\gamma f\right)q,q-x〉\le 0,\forall x\in F\left(T\right).$$

Question A. How can we prove strong convergence theorem of κ-strictly pseudo-contractive mapping without assumption of the mapping S = κI + (1 - κ)T in Theorems 1.1 and 1.2?

Let A, B : C → H be two mappings. By modification of (1.2), we have

$$\begin{array}{ll}\hfill VI\left(C,aA+\left(1-a\right)B\right)& =\left\{x\in C:〈y-x,\left(aA+\left(1-a\right)B\right)x〉\ge 0,\right.\\ \left.\forall y\in C,a\in \left(0,1\right)\right\}.\end{array}$$

(1.9)

From (1.4) and (1.9), we have

$$\begin{array}{l}EP\left(F,(aA+(1-a)B)\right)=\{z\in C:F(z,y)+〈(aA+(1-a)B)z,y-z〉\ge 0,\\ \forall y\in C\text{and}a\in (0,1)\}.\end{array}$$

In this article, we prove strong convergence theorem to answer question A and to approximate a common element of the set of fixed points of κ-strictly pseudo-contractive mapping and the set of solution of a modification of generalized equilibrium problem. Moreover, by using our main result and a new lemma in the last section we obtain strong convergence theorem for finding a common element of the set of fixed points of κ-strictly pseudo-contractive mapping and two sets of solutions of variational inequalities.

2 Preliminaries

Let H be a real Hilbert space and let C be a nonempty closed convex subset of H, let P _C be the metric projection of H onto C i.e., for x ∈ H, P _C x satisfies the property

$$\left|\right|x-P_{C}x\left|\right|=\underset{y\in C}{\text{min}}\left|\right|x-y\left|\right|.$$

The following characterizes the projection P _C .

Lemma 2.1. [17] Given x ∈ H and y ∈ C. Then P _C x = y if and only if there holds the inequality

$$〈x-y,y-z〉\ge 0\forall z\in C.$$

Lemma 2.2. [18] Let {s _n } be a sequence of nonnegative real number satisfying

$$s_{n+1}=\left(1-{\alpha }_n\right)s_n+{\alpha }_n{\beta }_n,\forall n\ge 0$$

where {α _n }, {β _n } satisfy the conditions

$$\begin{array}{l}\left(1\right)\left\{{\alpha }_n\right\}\subset \left[0,1\right],\sum _{n=1}^{\infty }{\alpha }_n=\infty ;\\ \left(2\right)\underset{n\to \infty }{\text{lim}\text{sup}}{\beta }_n\le 0or\sum _{n=1}^{\infty }\left|{\alpha }_n{\beta }_n\right|<\infty .\end{array}$$

Then lim_n→∞s _n = 0.

Lemma 2.3. [17] Let H be a Hibert space, let C be a nonempty closed convex subset of H and let A be a mapping of C into H. Let u ∈ C. Then for λ > 0,

$$u=P_C\left(I-\lambda A\right)u\iff u\in VI\left(C,A\right),$$

where P _C is the metric projection of H onto C.

Lemma 2.4. [19] Let {x _n } and {z _n } be bounded sequences in a Banach space X and let {β _n } be a sequence in [0,1] with 0 < lim inf_n→∞β _n ≤ lim sup_n→∞β _n < 1. Suppose

$$x_{n+1}={\beta }_n{x}_n+\left(1-{\beta }_n\right)z_n$$

for all n ≥ 0 and

$$\underset{n\to \infty }{\text{lim}\text{sup}}\left(\left|\right|z_{n+1}-z_n\left|\right|-\left|\right|x_{n+1}-x_n\left|\right|\right)\le 0.$$

Then lim_n→∞||x _n - z _n || = 0.

Lemma 2.5. [20] Let E be a uniformly convex Banach space, C be a nonempty closed convex subset of E and S : C → C be a nonexpansive mapping. Then, I - S is demi-closed at zero.

For solving the equilibrium problem for a bifunction F:C×C→ ℝ, let us assume that F satisfies the following conditions:

(A 1) F (x, x) = 0 ∀x ∈ C;

(A 2) F is monotone, i.e. F(x, y) + F(y, x) ≤ 0, ∀x, y ∈ C;

(A 3) ∀x, y, z ∈ C,

lim_t→0+F(tz + (1 - t)x, y) ≤ F(x, y);

(A 4) ∀x ∈ C, y α F(x, y) is convex and lower semicontinuous.

The following lemma appears implicitly in [7].

Lemma 2.6. [7] Let C be a nonempty closed convex subset of H, and let F be a bifunction of C × C into ℝ satisfying (A 1)-(A 4). Let r > 0 and x ∈ H. Then, there exists z ∈ C such that

$$F\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,$$

for all x ∈ C.

Lemma 2.7. [8] Assume that F:C×C→ ℝ satisfies (A 1)-(A 4). For r > 0 and x ∈ H, define a mapping T _r : H → C as follows:

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

for all z ∈ H. Then, the following hold:

(1) T _r is single-valued;

(2) T _r is firmly nonexpansive i.e.

||T _r (x) - T _r (y)||² ≤ 〈T _r (x) - T _r (y), x - y〉 ∀x, y ∈ H;

(3) F(T _r ) = EP(F);

(4) EP(F) is closed and convex.

Remark 2.8. If C is nonempty closed convex subset of H and T : C → C is κ-strictly pseudocontractive mapping with F(T) ≠ ∅. Then F(T) = VI(C, (I - T)). To show this, put A = I - T. Let z ∈ VI(C, (I - T)) and z* ∈ F(T). Since z ∈ VI(C, (I - T)), 〈y - z, (I - T)z〉 ≥ 0, ∀y ∈ C. Since T : C → C is κ-strictly pseudocontractive mapping, we have

$$\begin{array}{ll}\hfill \left|\right|Tz-T{z}^{*}|{|}^2& =\left|\right|\left(I-A\right)z-\left(I-A\right)z^{*}|{|}^2=||z-z^{*}-\left(Az-A{z}^{*}\right)|{|}^2\\ =\left|\right|z-z^{*}|{|}^2-2〈z-z^{*},Az-A{z}^{*}〉+||Az-A{z}^{*}|{|}^2\\ =\left|\right|z-z^{*}|{|}^2-2〈z-z^{*},\left(I-T\right)z〉+|\left|\left(I-T\right)z\right|{|}^2\\ \le \left|\right|z-z^{*}|{|}^2+\kappa |\left|\left(I-T\right)z\right|{|}^2.\end{array}$$

It implies that

$$\left(1-\kappa \right)\left|\right|\left(I-T\right)z|{|}^2\le 2〈z-z^{*},\left(I-T\right)z〉\le 0.$$

Then, we have z = Tz, therefore z ∈ F(T). Hence VI(C, (I - T)) ⊆ F(T). It is easy to see that F(T) ⊆ VI(C, (I - T)).

Remark 2.9. A = I - T is $\frac{1-\kappa }{2}$- inverse strongly monotone mapping. To show this, let x, y ∈ C, we have

$$\begin{array}{ll}\hfill \left|\right|Tx-Ty|{|}^2& =\left|\right|\left(I-A\right)x-\left(I-A\right)y|{|}^2=||x-y-\left(Ax-Ay\right)|{|}^2\\ =\left|\right|x-y|{|}^2-2〈x-y,Ax-Ay〉+||Ax-Ay|{|}^2\\ \le \left|\right|x-y|{|}^2+\kappa ||\left(I-T\right)x-\left(I-T\right)y|{|}^2.\\ =\left|\right|x-y|{|}^2+\kappa ||Ax-Ay|{|}^2.\end{array}$$

Then, we have

$$〈x-y,Ax-Ay〉\ge \frac{1-\kappa }{2}\left|\right|Ax-Ay|{|}^2.$$

3 Main result

Theorem 3.1. Let C be a closed convex subset of Hilbert space H and let F:C×C→ ℝ be a bifunction satisfying (A₁)-(A₄), let A, B : C → H be α and β-inverse strongly monotone, respectively. Let T : C → C be κ-strictly pseudo contractive mapping with $F=F\left(T\right)\cap EP\left(F,aA+\left(1-a\right)B\right)\ne \varnothing$ for all a ∈ (0, 1). Let {x _n } and {u _n } be the sequences generated by x₁, u ∈ C and

$$\left\{\begin{array}{c}F\left(u_n,y\right)+〈\left(aA+\left(1-a\right)B\right)x_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\forall y\in C,\\ x_{n+1}={\alpha }_{n}u+{\beta }_n{x}_n+{\gamma }_n{P}_C\left(I-\lambda \left(I-T\right)\right)u_n,\forall n\ge 1,\end{array}\right.$$

(3.1)

where {α _n }, {β _n }, {γ _n } ⊂ [0, 1], λ ∈ (0, 1 - κ), α _n + β _n + γ _n = 1, ∀n∈ℕ and {r _n } ⊂ [0, 2γ], γ = min{α, β} satisfy;

$$\begin{array}{l}\left(i\right)\sum _{n=1}^{\infty }{\alpha }_n=\infty ,\underset{n\to \infty }{\text{lim}}{\alpha }_n=0;\\ \left(ii\right)0<c\le {\beta }_n\le d<1,0<e\le r_n\le f<2\gamma ;\\ \left(iii\right)\underset{n\to \infty }{\text{lim}}|r_{n+1}-r_n|=0.\end{array}$$

Then {x _n } converges strongly to $z_0=P_{F}u$.

Proof. We divide the proof into seven steps.

Step 1. For every a ∈ (0, 1), we prove that aA + (1 - a)B is γ-inverse strongly monotone mapping. Put D = aA + (1 - a)B. For x, y ∈ C, we have

$$\begin{array}{ll}\hfill 〈Dx-Dy,x-y〉& =〈aAx+\left(1-a\right)Bx-aAy-\left(1-a\right)By,x-y〉\\ =〈a\left(Ax-Ay\right)+\left(1-a\right)\left(Bx-By\right),x-y〉\\ =a〈Ax-Ay,x-y〉+\left(1-a\right)〈Bx-By,x-y〉\\ \ge a\alpha \left|\right|Ax-Ay|{|}^2+\left(1-a\right)\beta ||Bx-By|{|}^2\\ \ge \gamma \left(a\left|\right|Ax-Ay|{|}^2+\left(1-a\right)||Bx-By|{|}^2\right)\\ \ge \gamma \left|\right|a\left(Ax-Ay\right)+\left(1-a\right)\left(Bx-By\right)|{|}^2\\ =\gamma \left|\right|aAx+\left(1-a\right)Bx-aAy-\left(1-a\right)By|{|}^2\\ =\gamma \left|\right|Dx-Dy|{|}^2\end{array}$$

(3.2)

Step 2. We show that I - r _n D is a nonexpansive mapping for every n∈ℕ and so is P _C (I - λ(I - T)). For every n∈ℕ, let x, y ∈ C. From step 1, we have

$$\begin{array}{ll}\hfill \left|\right|\left(I-r_{n}D\right)x-\left(I-r_{n}D\right)y|{|}^2& =\left|\right|x-y-r_n\left(Dx-Dy\right)|{|}^2\\ =\left|\right|x-y|{|}^2-2r_n〈x-y,Dx-Dy〉+r_n^2||Dx-Dy|{|}^2\\ \le \left|\right|x-y|{|}^2-2r_n\gamma ||Dx-Dy|{|}^2+r_n^2\left|\right|Dx-Dy|{|}^2\\ =\left|\right|x-y|{|}^2+r_n\left(r_n-2\gamma \right)||Dx-Dy|{|}^2\\ \le \left|\right|x-y|{|}^2.\end{array}$$

(3.3)

Then I - r _n D is a nonexpansive mapping.

Putting E = I - T, from Remark 2.9, we have E is η-inverse strong monotone mapping, where $\eta =\frac{1-\kappa }{2}$. By using the same method as (3.3), we have I - λE is nonexpansive mapping. Then, we have P _C (I - λ(I - T)) is a nonexpansive mapping.

Step 3. We prove that the sequence {x _n } is bounded. From $F\ne \varnothing$ and (3.1), we have $u_n=T_{r_n}\left(I-r_{n}D\right)x_n$, ∀n∈ℕ. Let $z\in F$. From Remark 2.8 and Lemma 2.3, we have z = P _C (I - λE)z, where E = I - T. Since z ∈ EP(F, D), we have F(z, y) + 〈y - z, Dz〉 ≥ 0 ∀y ∈ C, so we have

$$F\left(z,y\right)+\frac{1}{r_n}〈y-z,z-z+r_{n}Dz〉\ge 0,\forall n\in \mathbb{N}\text{and}y\in C.$$

From Lemma 2.7, we have $z=T_{r_n}\left(I-r_{n}D\right)z$, ∀n∈ℕ. By nonexpansiveness of $T_{r_n}\left(I-r_{n}D\right)$, we have

$$\begin{array}{ll}\hfill \left|\right|x_{n+1}-z\left|\right|& =\left|\right|{\alpha }_n\left(u-z\right)+{\beta }_n\left(x_n-z\right)+{\gamma }_n\left(P_C\left(I-\lambda E\right)u_n-z\right)\left|\right|\\ \le {\alpha }_n\left|\right|u-z\left|\right|+{\beta }_n\left|\right|x_n-z\left|\right|+{\gamma }_n\left|\right|P_C\left(I-\lambda E\right)u_n-z\left|\right|\\ \le {\alpha }_n\left|\right|u-z\left|\right|+{\beta }_n\left|\right|x_n-z\left|\right|+{\gamma }_n\left|\right|T_{r_n}\left(I-r_{n}D\right)x_n-z\left|\right|\\ \le {\alpha }_n\left|\right|u-z\left|\right|+\left(1-{\alpha }_n\right)\left|\right|x_n-z\left|\right|\\ \le \text{max}\left\{\left|\right|x_n-z\left|\right|,\left|\right|u-z\left|\right|\right\}.\end{array}$$

By induction we can prove that {x _n } is bounded and so are {u _n }, {P _C (I - λE)u _n }.

Step 4. We will show that

$$\underset{n\to \infty }{\text{lim}}\left|\right|x_{n+1}-x_n\left|\right|=0.$$

(3.4)

Let $p_n=\frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}$, we have

$$x_{n+1}=\left(1-{\beta }_n\right)p_n+{\beta }_n{x}_n.$$

(3.5)

From (3.5), we have

$$\begin{array}{ll}\hfill \left|\right|p_{n+1}-p_n\left|\right|& =∥\frac{x_{n+2}-{\beta }_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}-\frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}∥\\ =∥\frac{{\alpha }_{n+1}u+{\gamma }_{n+1}{P}_C\left(I-\lambda E\right)u_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_{n}u+{\gamma }_n{P}_C\left(I-\lambda E\right)u_n}{1-{\beta }_n}∥\\ =∥\left(\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}\right)u+\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}\left(P_C\left(I-\lambda E\right)u_{n+1}-P_C\left(I-\lambda E\right)u_n\right)\\ +\left(\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n}\right)P_C\left(I-\lambda E\right)u_n∥\\ \le \left|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}\right|\left|\right|u\left|\right|+\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}\left|\right|u_{n+1}-u_n\left|\right|\\ +\left|\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n}\right|\left|\right|P_C\left(I-\lambda E\right)u_n\left|\right|\\ =\left|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}\right|\left|\right|u\left|\right|+\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}\left|\right|u_{n+1}-u_n\left|\right|\\ +\left|\frac{1-{\beta }_{n+1}-{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{1-{\beta }_n-{\alpha }_n}{1-{\beta }_n}\right|\left|\right|P_C\left(I-\lambda E\right)u_n\left|\right|\\ =\left|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}\right|\left|\right|u\left|\right|+\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}\left|\right|u_{n+1}-u_n\left|\right|\\ +\left|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}\right|\left|\right|P_C\left(I-\lambda E\right)u_n\left|\right|\\ =\left|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}\right|\left(\left|\right|u\left|\right|+\left|\right|P_C\left(I-\lambda E\right)u_n\left|\right|\right)\\ +\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}\left|\right|u_{n+1}-u_n\left|\right|.\end{array}$$

(3.6)

Putting v _n = x _n - r _n Dx _n , we have $u_n=T_{r_n}\left(x_n-r_{n}D{x}_n\right)=T_{r_n}{v}_n$. From definition of u _n , we have

$$F\left(u_n,y\right)+\frac{1}{r_n}〈y-u_n,u_n-v_n〉\ge 0,\forall y\in C,$$

(3.7)

and

$$F\left(u_{n+1},y\right)+\frac{1}{r_{n+1}}〈y-u_{n+1},u_{n+1}-v_{n+1}〉\ge 0,\forall y\in C.$$

(3.8)

Putting y = u_n+1in (3.7) and y = u _n in (3.8), we have

$$F\left(u_n,u_{n+1}\right)+\frac{1}{r_n}〈u_{n+1}-u_n,u_n-v_n〉\ge 0,$$

(3.9)

and

$$F\left(u_{n+1},u_n\right)+\frac{1}{r_{n+1}}〈u_n-u_{n+1},u_{n+1}-v_{n+1}〉\ge 0.$$

(3.10)

Summing up (3.9) and (3.10) and using (A 2), we have

$$\begin{array}{ll}\hfill 0& \le \frac{1}{r_n}〈u_{n+1}-u_n,u_n-v_n〉+\frac{1}{r_{n+1}}〈u_n-u_{n+1},u_{n+1}-v_{n+1}〉\\ =〈u_{n+1}-u_n,\frac{u_n-v_n}{r_n}〉+〈u_n-u_{n+1},\frac{u_{n+1}-v_{n+1}}{r_{n+1}}〉\\ =〈u_{n+1}-u_n,\frac{u_n-v_n}{r_n}-\frac{u_{n+1}-v_{n+1}}{r_{n+1}}〉.\end{array}$$

It implies that

$$\begin{array}{ll}\hfill 0& \le 〈u_{n+1}-u_n,u_n-v_n-\frac{r_n}{r_{n+1}}\left(u_{n+1}-v_{n+1}\right)〉\\ =〈u_{n+1}-u_n,u_n-u_{n+1}+u_{n+1}-v_n-\frac{r_n}{r_{n+1}}\left(u_{n+1}-v_{n+1}\right)〉.\end{array}$$

It implies that

$$\begin{array}{ll}\hfill \left|\right|u_{n+1}-u_n|{|}^2& \le 〈u_{n+1}-u_n,u_{n+1}-v_n-\frac{r_n}{r_{n+1}}\left(u_{n+1}-v_{n+1}\right)〉\\ =〈u_{n+1}-u_n,u_{n+1}-v_{n+1}+v_{n+1}-v_n-\frac{r_n}{r_{n+1}}\left(u_{n+1}-v_{n+1}\right)〉\\ =〈u_{n+1}-u_n,v_{n+1}-v_n+\left(1-\frac{r_n}{r_{n+1}}\right)\left(u_{n+1}-v_{n+1}\right)〉\\ \le \left|\right|u_{n+1}-u_n\left|\right|\left(\left|\right|v_{n+1}-v_n\left|\right|+\frac{1}{r_{n+1}}|r_{n+1}-r_n|\left|\right|u_{n+1}-v_{n+1}\left|\right|\right).\end{array}$$

It follows that

$$\left|\right|u_{n+1}-u_n\left|\right|\le \left|\right|v_{n+1}-v_n\left|\right|+\frac{1}{e}|r_{n+1}-r_n|\left|\right|u_{n+1}-v_{n+1}\left|\right|.$$

(3.11)

Since v _n = x _n - r _n Dx _n , we have

$$\begin{array}{ll}\hfill \left|\right|v_{n+1}-v_n\left|\right|& =\left|\right|x_{n+1}-r_{n+1}D{x}_{n+1}-x_n+r_{n}D{x}_n\left|\right|\\ =\left|\right|\left(I-r_{n+1}D\right)x_{n+1}-\left(I-r_{n+1}D\right)x_n\\ +\left(I-r_{n+1}D\right)x_n-\left(I-r_{n}D\right)x_n\left|\right|\\ \le \left|\right|\left(I-r_{n+1}D\right)x_{n+1}-\left(I-r_{n+1}D\right)x_n\left|\right|\\ +\left|\right|\left(r_n-r_{n+1}\right)D{x}_n\left|\right|\\ \le \left|\right|x_{n+1}-x_n\left|\right|+|r_n-r_{n+1}|\left|\right|D{x}_n\left|\right|.\end{array}$$

(3.12)

Substitute (3.12) into (3.11), we have

$$\begin{array}{ll}\hfill \left|\right|u_{n+1}-u_n\left|\right|& \le \left|\right|v_{n+1}-v_n\left|\right|+\frac{1}{e}|r_{n+1}-r_n|\left|\right|u_{n+1}-v_{n+1}\left|\right|\\ \le \left|\right|x_{n+1}-x_n\left|\right|+|r_n-r_{n+1}|\left|\right|D{x}_n\left|\right|\\ +\frac{1}{e}|r_{n+1}-r_n|\left|\right|u_{n+1}-v_{n+1}\left|\right|\\ \le \left|\right|x_{n+1}-x_n\left|\right|+|r_n-r_{n+1}|L+\frac{1}{e}|r_{n+1}-r_n|L,\end{array}$$

(3.13)

where L=max_n∈ℕ{||Dx _n ||, ||u _n -ν _n ||}. Substitute (3.13) into (3.6), we have

$$\begin{array}{ll}\hfill \left|\right|p_{n+1}-p_n\left|\right|& \le \left|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}\right|\left(\left|\right|u\left|\right|+\left|\right|P_C\left(I-\lambda E\right)u_n\left|\right|\right)\\ +\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}\left|\right|u_{n+1}-u_n\left|\right|\\ \le \left|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}\right|\left(\left|\right|u\left|\right|+\left|\right|P_C\left(I-\lambda E\right)u_n\left|\right|\right)\\ +\left|\right|x_{n+1}-x_n\left|\right|+|r_n-r_{n+1}|L+\frac{1}{e}|r_{n+1}-r_n|L,\end{array}$$

(3.14)

From conditions (i), (iii) and (3.14), we have

$$\underset{n\to \infty }{\text{lim}\text{sup}}\left(\left|\right|p_{n+1}-p_n\left|\right|-\left|\right|x_{n+1}-x_n\left|\right|\right)\le 0.$$

(3.15)

From Lemma 2.4, (3.15) and (3.5), we have

$$\underset{n\to \infty }{\text{lim}}\left|\right|p_n-x_n\left|\right|=0.$$

(3.16)

From (3.5), we have

$$x_{n+1}-x_n=\left(1-{\beta }_n\right)\left(p_n-x_n\right).$$

(3.17)

From (3.16), (3.17) and condition (ii), we have

$$\underset{n\to \infty }{\text{lim}}\left|\right|x_{n+1}-x_n\left|\right|=0.$$

(3.18)

Since

$$x_{n+1}-x_n={\alpha }_n\left(u-x_n\right)+{\gamma }_n\left(P_C\left(I-\lambda \left(I-T\right)\right)u_n-x_n\right),$$

from conditions (i), (ii) and (3.18), we have

$$\underset{n\to \infty }{\text{lim}}\left|\right|P_C\left(I-\lambda E\right)u_n-x_n\left|\right|=0,$$

(3.19)

where E = I - T .

Step 5. We will show that

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

(3.20)

Since $u_n=T_{r_n}\left(x_n-r_{n}D{x}_n\right)$, we have

$$\begin{array}{ll}\hfill \parallel u_n-z{\parallel }^2& =\parallel T_{r_n}\left(x_n-r_{n}D{x}_n\right)-T_{r_n}\left(I-r_{n}D\right)z{\parallel }^2\\ \le 〈\left(I-r_{n}D\right)x_n-\left(I-r_{n}D\right)z,u_n-z〉\\ =\frac{1}{2}\left(\parallel \left(I-r_{n}D\right)x_n-\left(I-r_{n}D\right)z{\parallel }^2+\parallel u_n-z{\parallel }^2\right.\\ \left.-\parallel \left(I-r_{n}D\right)x_n-\left(I-r_{n}D\right)z-u_n+z{\parallel }^2\right)\\ \le \frac{1}{2}\left(\parallel x_n-z{\parallel }^2+\parallel u_n-z{\parallel }^2-\parallel \left(x_n-u_n\right)-r_n\left(D{x}_n-Dz\right){\parallel }^2\right)\\ \le \frac{1}{2}\left(\parallel x_n-z{\parallel }^2+\parallel u_n-z{\parallel }^2-\parallel x_n-u_n{\parallel }^2-r_n^2\parallel D{x}_n-Dz{\parallel }^2\right.\\ \left.+2r_n〈x_n-u_n,D{x}_n-Dz〉\right),\end{array}$$

it implies that

$$\left|\right|u_n-z|{|}^2\le ||x_n-z|{|}^2-\left|\right|x_n-u_n|{|}^2-r_n^2||D{x}_n-Dz|{|}^2+2r_n〈x_n-u_n,D{x}_n-Dz〉.$$

(3.21)

By nonexpansiveness of $T_{r_n}$ and using the same method as (3.3), we have

$$\begin{array}{ll}\hfill \left|\right|u_n-z|{|}^2& =\left|\right|T_{r_n}\left(I-r_{n}D\right)x_n-T_{r_n}\left(I-r_{n}D\right)z|{|}^2\\ \le \left|\right|\left(I-r_{n}D\right)x_n-\left(I-r_{n}D\right)z|{|}^2\\ \le \left|\right|x_n-z|{|}^2+r_n\left(r_n-2\gamma \right)||D{x}_n-Dz|{|}^2\\ =\left|\right|x_n-z|{|}^2-r_n\left(2\gamma -r_n\right)||D{x}_n-Dz|{|}^2.\end{array}$$

(3.22)

By nonexpansiveness of P _C (I - λE) and (3.22), we have

$$\begin{array}{l}\left|\right|x_{n+1}-z|{|}^2=||{\alpha }_n(u-z)+{\beta }_n(x_n-z)+{\gamma }_n(P_C(I-\lambda E)u_n-z)|{|}^2\\ \le {\alpha }_n\left|\right|u-z|{|}^2+{\beta }_n||x_n-z|{|}^2+{\gamma }_n\left|\right|u_n-z|{|}^2\\ \le {\alpha }_n\left|\right|u-z|{|}^2+{\beta }_n||x_n-z|{|}^2+{\gamma }_n(||x_n-z|{|}^2\\ -r_n(2\gamma -r_n)\left|\right|D{x}_n-Dz|{|}^2)\\ \le {\alpha }_n\left|\right|u-z|{|}^2+||x_n-z|{|}^2-r_n{\gamma }_n(2\gamma -r_n)\left|\right|D{x}_n-Dz|{|}^2,\end{array}$$

(3.23)

it implies that

$$\begin{array}{l}{r}_n{\gamma }_n(2\gamma -r_n)\left|\right|D{x}_n-Dz|{|}^2\le {\alpha }_n||u-z|{|}^2+\left|\right|x_n-z|{|}^2-||x_{n+1}-z|{|}^2\\ \le {\alpha }_n\left|\right|u-z|{|}^2+(\left|\right|x_n-z\left|\right|\\ +\left|\right|x_{n+1}-z\left|\right|)||x_{n+1}-x_n||.\end{array}$$

(3.24)

From (3.18), (3.24), conditions (i) and (ii), we have

$$\underset{n\to \infty }{\text{lim}}\left|\right|D{x}_n-Dz\left|\right|=0$$

(3.25)

From (3.23) and (3.21). we have

$$\begin{array}{ll}\hfill \left|\right|x_{n+1}-z|{|}^2& \le \left|\right|{\alpha }_n\left(u-z\right)+{\beta }_n\left(x_n-z\right)+{\gamma }_n\left(P_C\left(I-\lambda E\right)u_n-z\right)|{|}^2\\ \le {\alpha }_n\left|\right|u-z|{|}^2+{\beta }_n||x_n-z|{|}^2+{\gamma }_n\left|\right|u_n-z|{|}^2\\ \le {\alpha }_n\left|\right|u-z|{|}^2+{\beta }_n||x_n-z|{|}^2+{\gamma }_n\left(\left|\right|x_n-z|{|}^2-||x_n-u_n|{|}^2\right.\\ \left.-r_n^2\left|\right|D{x}_n-Dz|{|}^2+2r_n〈x_n-u_n,D{x}_n-Dz〉\right)\\ \le {\alpha }_n\left|\right|u-z|{|}^2+{\beta }_n||x_n-z|{|}^2+{\gamma }_n\left|\right|x_n-z|{|}^2-{\gamma }_n||x_n-u_n|{|}^2\\ +2r_n{\gamma }_n\left|\right|x_n-u_n\left|\right|\left|\right|D{x}_n-Dz\left|\right|\\ \le {\alpha }_n\left|\right|u-z|{|}^2+||x_n-z|{|}^2-{\gamma }_n\left|\right|x_n-u_n|{|}^2+2r_n{\gamma }_n||x_n-u_n|\left|\right||D{x}_n-Dz||,\end{array}$$

which implies that

$$\begin{array}{ll}\hfill {\gamma }_n\left|\right|x_n-u_n|{|}^2& \le {\alpha }_n\left|\right|u-z|{|}^2+||x_n-z|{|}^2-\left|\right|x_{n+1}-z|{|}^2+2r_n{\gamma }_n||x_n-u_n|\left|\right||D{x}_n-Dz||\\ \le {\alpha }_n\left|\right|u-z|{|}^2+\left(\left|\right|x_n-z\left|\right|+\left|\right|x_{n+1}-z\left|\right|\right)||x_{n+1}-x_n||\\ +2r_n{\gamma }_n\left|\right|x_n-u_n\left|\right|\left|\right|D{x}_n-Dz\left|\right|,\end{array}$$

from condition (i), (3.25) and (3.18), we have

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

Step 6. We prove that

$$\underset{n\to \infty }{\text{lim}\text{sup}}〈u-z_0,x_n-z_0〉\le 0,$$

(3.26)

where $z_0=P_{F}u$. To show this equality, take a subsequence $\left\{x_{n_k}\right\}$ of {x _n } such that

$$\underset{n\to \infty }{\text{lim}\text{sup}}〈u-z_0,x_n-z_0〉=\underset{k\to \infty }{\text{lim}}〈u-z_0,x_{n_k}-z_0〉,$$

(3.27)

Without loss of generality, we may assume that $x_{n_k}\rightharpoonup \omega$ as k → ∞ where ω ∈ C. We first show ω ∈ EP(F, D), where D = aA + (1 - a)B, ∀a ∈ [0,1]. From (3.20), we have $u_{n_k}\rightharpoonup \omega$ as k → ∞. Since$u_n=T_{r_n}\left(x_n-r_{n}D{x}_n\right)$, we obtain

$$F\left(u_n,y\right)+〈D{x}_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\forall y\in C.$$

From (A 2), we have $〈D{x}_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge F\left(y,u_n\right)$. Then

$$〈D{x}_{n_k},y-u_{n_k}〉+\frac{1}{r_{n_k}}〈y-u_{n_k},u_{n_k}-x_{n_k}〉\ge F\left(y,u_{n_k}\right),\forall y\in C.$$

(3.28)

Put z _t = ty + (1 - t)ω for all t ∈ (0, 1] and y ∈ C. Then, we have z _t ∈ C. So, from (3.28) we have

$$\begin{array}{ll}\hfill 〈z_t-u_{n_k},D{z}_t〉& \ge 〈z_t-u_{n_k},D{z}_t〉-〈z_t-u_{n_k},D{x}_{n_k}〉-〈z_t-u_{n_k},\frac{u_{n_k}-x_{n_k}}{r_{n_k}}〉+F\left(z_t,u_{n_k}\right)\\ =〈z_t-u_{n_k},D{z}_t-D{u}_{n_k}〉+〈z_t-u_{n_k},D{u}_{n_k}-D{x}_{n_k}〉\\ -〈z_t-u_{n_k},\frac{u_{n_k}-x_{n_k}}{r_{n_k}}〉+F\left(z_t,u_{n_k}\right).\end{array}$$

Since $\left|\right|u_{n_k}-x_{n_k}\left|\right|\to 0$, we have $\left|\right|D{u}_{n_k}-D{x}_{n_k}\left|\right|\to 0$. Further, from monotonicity of D, we have $〈z_t-u_{n_k},D{z}_t-D{u}_{n_k}〉\ge 0$. So, from (A4) we have

$$〈z_t-\omega ,D{z}_t〉\ge F\left(z_t,\omega \right)\text{as }k\to \infty .$$

(3.29)

From (A 1), (A 4) and (3.29), we also have

$$\begin{array}{ll}\hfill 0& =F\left(z_t,z_t\right)\le tF\left(z_t,y\right)+\left(1-t\right)F\left(z_t,\omega \right)\\ \le tF\left(z_t,y\right)+\left(1-t\right)〈z_t-\omega ,D{z}_t〉\\ =tF\left(z_t,y\right)+\left(1-t\right)t〈y-\omega ,D{z}_t〉,\end{array}$$

hence

$$0\le F\left(z_t,y\right)+\left(1-t\right)〈y-\omega ,D{z}_t〉.$$

Letting t → 0, we have

$$0\le F\left(\omega ,y\right)+〈y-\omega ,D\omega 〉\forall y\in C.$$

(3.30)

Therefore ω ∈ EP(F, D), where D = aA + (1 - a)B, ∀a ∈ [0,1]. Since

$$\left|\right|P_C\left(I-\lambda E\right)u_n-u_n\left|\right|\le \left|\right|\left|\right|P_C\left(I-\lambda E\right)u_n-x_n\left|\right|\left|\right|+\left|\right|x_n-u_n\left|\right|,$$

where E = I - T from (3.19) and (3.20), we have

$$\underset{n\to 0}{\text{lim}}\left|\right|P_C\left(I-\lambda E\right)u_n-u_n\left|\right|=0.$$

(3.31)

Since $u_{n_k}\to \omega$ as k → ω, (3.31) and Lemma 2.5, we have ω ∈ F(P _C (I - λE)). From Lemma 2.3 and Remark 2.8, we have ω ∈ F(T). Therefore $\omega \in F$. Since $x_{n_k}\to \omega$ as k → ∞ and $\omega \in F$, we have

$$\underset{n\to \infty }{\text{lim}\text{sup}}〈u-z_0,x_n-z_0〉=\underset{n\to \infty }{\text{lim}}〈u-z_0,x_{n_k}-z_0〉=〈u-z_0,\omega -z_0〉\le 0.$$

Step 7. Finally, we show that {x _n } converses strongly to $z_0=P_{F}u$. From definition of x _n , we have

$$\begin{array}{ll}\hfill \left|\right|x_{n+1}-z_0|{|}^2& =\left|\right|{\alpha }_n\left(u-z_0\right)+{\beta }_n\left(x_n-z_0\right)+{\gamma }_n\left(P_C\left(I-\lambda \left(I-T\right)\right)u_n-z_0\right)|{|}^2\\ \le \left|\right|{\beta }_n\left(x_n-z_0\right)+{\gamma }_n\left(P_C\left(I-\lambda \left(I-T\right)\right)u_n-z_0\right)|{|}^2+2{\alpha }_n〈u-z_0,x_{n+1}-z_0〉\\ \le {\beta }_n\left|\right|x_n-z_0|{|}^2+{\gamma }_n||P_C\left(I-\lambda \left(I-T\right)\right)u_n-z_0|{|}^2+2{\alpha }_n〈u-z_0,x_{n+1}-z_0〉\\ \le {\beta }_n\left|\right|x_n-z_0|{|}^2+{\gamma }_n||T_{r_n}\left(I-r_{n}D\right)x_n-z_0|{|}^2+2{\alpha }_n〈u-z_0,x_{n+1}-z_0〉\\ \le \left(1-{\alpha }_n\right)\left|\right|x_n-z_0|{|}^2+2{\alpha }_n〈u-z_0,x_{n+1}-z_0〉\end{array}$$

From (3.26) and Lemma 2.2, we have {x _n } converses strongly to $z_0=P_{F}u$. This completes the prove. □

4 Applications

To prove strong convergence theorem in this section, we needed the following lemma.

Lemma 4.1. Let C be a nonempty closed convex subset of a real Hilbert space H and let A, B : C → H be α and β-inverse strongly monotone mappings, respectively, with α, β > 0 and VI(C, A) ∩ VI(C, B) ≠ ∅. Then

$$VI\left(C,aA+\left(1-a\right)B\right)=VI\left(C,A\right)\bigcap VI\left(C,B\right),\forall a\in \left(0,1\right).$$

(4.1)

Furthermore if 0 < γ < 2η, where η = min{α, β}, we have I - γ(aA + (1 - a)B) is a nonexpansive mapping.

Proof. It is easy to see that VI(C, A) ∩ VI(C, B) ⊆ VI(C, aA + (1 - a)B). Next, we will show that VI(C, aA + (1 - a)B) ⊆ VI(C, A) ∩ VI(C, B). Let x₀ ∈ VI(C, aA + (1 - a)B) and x* ∈ VI(C, A) ∩ VI(C, B). Then, we have

$$〈y-x^{*},A{x}^{*}〉\ge 0,\forall y\in C,$$

and

$$〈y-x^{*},B{x}^{*}〉\ge 0,\forall y\in C.$$

For every a ∈ (0, 1), we have

$$〈y-x^{*},aA{x}^{*}〉\ge 0,\forall y\in C,$$

(4.2)

and

$$〈y-x^{*},\left(1-a\right)B{x}^{*}〉\ge 0,\forall y\in C.$$

(4.3)

By monotonicity of A, B and x*, x₀ ∈ C, we have

$$\begin{array}{ll}\hfill 〈x^{*}-x_0,aA{x}_0〉& =〈x^{*}-x_0,aA{x}_0+\left(1-a\right)B{x}_0-\left(1-a\right)B{x}_0〉\\ =〈x^{*}-x_0,aA{x}_0+\left(1-a\right)B{x}_0〉-〈x^{*}-x_0,\left(1-a\right)B{x}_0〉\\ \ge \left(1-a\right)〈x_0-x^{*},B{x}_0〉\\ =\left(1-a\right)\left(〈x_0-x^{*},B{x}_0-B{x}^{*}〉+〈x_0-x^{*},B{x}^{*}〉\right)\\ \ge 0.\end{array}$$

(4.4)

It implies that

$$〈x^{*}-x_0,A{x}_0〉\ge 0.$$

(4.5)

By monotonicity of A, x* ∈ VI(C, A) and (4.5), we have

$$\begin{array}{ll}\hfill 0& \le 〈x^{*}-x_0,A{x}_0〉\\ =〈x^{*}-x_0,A{x}_0-A{x}^{*}+A{x}^{*}〉\\ =〈x^{*}-x_0,A{x}_0-A{x}^{*}〉+〈x^{*}-x_0,A{x}^{*}〉\\ \le -\alpha \left|\right|A{x}^{*}-A{x}_0|{|}^2+〈x^{*}-x_0,A{x}^{*}〉\\ \le -\alpha \left|\right|A{x}^{*}-A{x}_0|{|}^2,\end{array}$$

it implies that

$$A{x}^{*}=A{x}_0.$$

(4.6)

For every y ∈ C, from (4.5), (4.6) and x* ∈ VI(C, A), we have

$$\begin{array}{ll}\hfill 〈y-x_0,A{x}_0〉& =〈y-x^{*},A{x}_0〉+〈x^{*}-x_0,A{x}_0〉\\ \ge 〈y-x^{*},A{x}^{*}〉\ge 0.\end{array}$$

Then, we have

$$x_0\in VI\left(C,A\right).$$

(4.7)

From (4.4), we have

$$\begin{array}{ll}\hfill \left(1-a\right)〈x^{*}-x_0,B{x}_0〉& \ge a〈x_0-x^{*},A{x}_0〉\\ =a\left(〈x_0-x^{*},A{x}_0-A{x}^{*}〉+〈x_0-x^{*},A{x}^{*}〉\right)\\ \ge 0.\end{array}$$

(4.8)

It implies that

$$〈x^{*}-x_0,B{x}_0〉\ge 0.$$

(4.9)

By monotonicity of B, x* ∈ VI(C, B) and (4.9), we have

$$\begin{array}{ll}\hfill 0& \le 〈x^{*}-x_0,B{x}_0〉\\ =〈x^{*}-x_0,B{x}_0-B{x}^{*}+B{x}^{*}〉\\ =〈x^{*}-x_0,B{x}_0-B{x}^{*}〉+〈x^{*}-x_0,B{x}^{*}〉\\ \le -\beta \left|\right|B{x}^{*}-B{x}_0|{|}^2+〈x^{*}-x_0,B{x}^{*}〉\\ \le -\beta \left|\right|B{x}^{*}-B{x}_0|{|}^2,\end{array}$$

it implies that

$$B{x}^{*}=B{x}_0.$$

(4.10)

For every y ∈ C, from (4.9), (4.10) and x* ∈ VI(C, B), we have

$$\begin{array}{ll}\hfill 〈y-x_0,B{x}_0〉& =〈y-x^{*},B{x}_0〉+〈x^{*}-x_0,B{x}_0〉\\ \ge 〈y-x^{*},B{x}^{*}〉\ge 0.\end{array}$$

Then, we have

$$x_0\in VI\left(C,B\right).$$

(4.11)

By (4.7) and (4.11), we have x₀ ∈ VI(C, A) ∩ VI(C, B). Hence, we have

$$VI\left(C,aA+\left(1-a\right)B\right)\subseteq VI\left(C,A\right)\bigcap VI\left(C,B\right).$$

Next, we will show that I - γ(aA + (1 - a)B) is a nonexpansive mapping. To show this let x, y ∈ C, then we have

$$\begin{array}{l}{\Vert \left(I-\gamma (aA+(1-a)B)\right)x-\left(I-\gamma (aA+(1-a)B)\right)y\Vert }^2\\ ={\Vert x-y-\gamma \left((aA+(1-a)B\right)x-\left(aA+(1-a)B)y\right)\Vert }^2\\ ={\Vert x-y-\gamma \left(a(Ax-Ay)+(1-a)(Bx-By)\right)\Vert }^2\\ =\left|\right|x-y|{|}^2-2\gamma 〈a(Ax-Ay)+(1-a)(Bx-By),x-y〉\\ +{\gamma }^2\left|\right|a(Ax-Ay)+(1-a)(Bx-By)|{|}^2\\ \le \left|\right|x-y|{|}^2-2\gamma a〈Ax-Ay,x-y〉-2\gamma (1-a)〈Bx-By,x-y〉\\ +a{\gamma }^2\left|\right|Ax-Ay|{|}^2+(1-a){\gamma }^2||Bx-By|{|}^2\\ \le \left|\right|x-y|{|}^2-2\gamma a\alpha ||Ax-Ay|{|}^2-2\gamma (1-a)\beta \left|\right|Bx-By|{|}^2\\ +a{\gamma }^2\left|\right|Ax-Ay|{|}^2+(1-a){\gamma }^2||Bx-By|{|}^2\\ =\left|\right|x-y|{|}^2+a\gamma (\gamma -2\alpha )||Ax-Ay|{|}^2+(1-a)\gamma (\gamma -2\beta )\left|\right|Bx-By|{|}^2\\ \le \left|\right|x-y|{|}^2.\end{array}$$

(4.12)

□

Theorem 4.2. Let C be a closed convex subset of Hilbert space H and let A, B : C → H be α and β-inverse strongly monotone, respectively. Let T be κ-strictly pseudo contractive mapping with $F=F\left(T\right)\cap VI\left(C,A\right)\cap VI\left(C,B\right)\ne \varnothing$. Let {x _n } be the sequence generated by x₁, u ∈ C and

$$x_{n+1}={\alpha }_{n}u+{\beta }_n{x}_n+{\gamma }_n{P}_C\left(I-\lambda \left(I-T\right)\right)P_C\left(I-r_n\left(aA+\left(1-a\right)B\right)\right)x_n,\forall n\ge 1,$$

(4.13)

where {α _n }, {β _n }, {γ _n } ⊂ [0, 1], a ∈ (0, 1), λ ∈ (0, 1 - κ), α _n + β _n + γ _n = 1, ∀n∈ℕ and {r _n } ⊂ [0, 2γ], γ = min{α, β} satisfy;

$$\begin{array}{l}\left(i\right)\sum _{n=1}^{\infty }{\alpha }_n=\infty ,\underset{n\to \infty }{\text{lim}}{\alpha }_n=0;\\ \left(ii\right)0<c\le {\beta }_n\le d<1,0<e\le r_n\le f<2\gamma ;\\ \left(iii\right)\underset{n\to \infty }{\text{lim}}|r_{n+1}-r_n|=0.\end{array}$$

Then {x _n } converges strongly to $z_0=P_{F}u$.

Proof. From 3.1 putting F ≡ 0 in Theorem 3.1, we have

$$〈y-u_n,u_n-\left(I-r_{n}D\right)x_n〉\ge 0,\forall y\in C,$$

where D = aA + (1 - a)B, ∀a ∈ [0,1] It implies that

$$u_n=P_C\left(I-r_{n}D\right)x_n.$$

Then, we have (4.13). From Theorem 3.1 and Lemma 4.1, we can conclude the desired conclusion. □

Theorem 4.3. Let C be a closed convex subset of Hilbert space H and let F:C×C→ℝ be a bifunction satisfying (A₁)-(A₄), let A : C → H be α-inverse strongly monotone. Let T : C → C be κ-strictly pseudo contractive mapping with $F=F\left(T\right)\cap EP\left(F,A\right)\ne \varnothing$. Let {x _n } and {u _n } be the sequences generated by x₁, u ∈ C and

$$\left\{\begin{array}{c}F\left(u_n,y\right)+〈A{x}_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\forall y\in C,\\ x_{n+1}={\alpha }_{n}u+{\beta }_n{x}_n+{\gamma }_n{P}_C\left(I-\lambda \left(I-T\right)\right)u_n,\forall n\ge 1,\end{array}\right.$$

(4.14)

where {α _n }, {β _n }, {γ _n } ⊂ [0, 1], λ ∈ (0, 1 - κ), α _n + β _n + γ _n = 1, ∀n∈ℕ and {r _n } ⊂ [0, 2γ], γ = min{α, β} satisfy;

$$\begin{array}{l}\left(i\right)\sum _{n=1}^{\infty }{\alpha }_n=\infty ,\underset{n\to \infty }{\text{lim}}{\alpha }_n=0;\\ \left(ii\right)0<c\le {\beta }_n\le d<1,0<e\le r_n\le f<2\gamma ;\\ \left(iii\right)\underset{n\to \infty }{\text{lim}}|r_{n+1}-r_n|=0.\end{array}$$

Then {x _n } converges strongly to $z_0=P_{F}u$.

Proof. From Theorem 3.1, putting A ≡ B, we can conclude the desired conclusion. □