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

Fixed Point Theory and Applications20122012:89

https://doi.org/10.1186/1687-1812-2012-89

• Accepted: 23 May 2012
• Published:

## 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.

## Keywords

• nonexpansive mappinga
• strictly pseudo-contractive mapping
• generalized equilibrium problem
• inverse-strongly monotone
• variational inequality problem

## 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
$||Tx-Ty||\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}||x-y||,\phantom{\rule{1em}{0ex}}\forall x,y\in H$
Definition 1.2. The mapping T is said to be strictly pseudo-contractive [1] with the coefficient κ [0, 1) if
$||Tx-Ty|{|}^{2}\le \phantom{\rule{2.77695pt}{0ex}}||x-y|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}\kappa ||\left(I-T\right)x-\left(I-T\right)y|{|}^{2}\phantom{\rule{1em}{0ex}}\forall x,y\in H.$
(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 : CH. 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. [25]).

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

for all x, y C.

Let F:C×C→ be a bifunction. The equilibrium problem for F is to determine its equilibrium points, i.e. the set
$EP\left(F\right)=\left\{x\in C:F\left(x,y\right)\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in C\right\}.$
(1.3)
From (1.2) and (1.3), we have the following generalized equilibrium problem, i.e.
(1.4)
The set of such z C is denoted by EP (F, A), i.e.,
$EP\left(F,A\right)=\left\{z\in C:F\left(z,y\right)+〈Az,\phantom{\rule{2.77695pt}{0ex}}y-z〉\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in C\right\}$

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 [79]). 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 [1014]). 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,\phantom{\rule{2.77695pt}{0ex}}\text{arbitrarily};\\ F\left({u}_{n},y\right)+\frac{1}{{r}_{n}}〈y-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in C,\\ {x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)T{u}_{n},\phantom{\rule{1em}{0ex}}\forall n\in ℕ,\end{array}\right\$
(1.5)

where f : HH 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 = PF(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,\phantom{\rule{2.77695pt}{0ex}}\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,\phantom{\rule{1em}{0ex}}\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),\phantom{\rule{1em}{0ex}}\forall n\in 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 : CH 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 $\stackrel{̄}{\gamma }$ and f: CC be a contraction with the contractive constant (0 < α < 1) such that $0<\gamma <\frac{\stackrel{̄}{\gamma }}{\alpha }$. 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 : CH 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
$\begin{array}{l}\left(i\right)\phantom{\rule{2.77695pt}{0ex}}\underset{n\to \infty }{\text{lim}}{\alpha }_{n}=0\phantom{\rule{2.77695pt}{0ex}}and\phantom{\rule{2.77695pt}{0ex}}\underset{n\to \infty }{\text{lim}}{\beta }_{n}=0,\phantom{\rule{2em}{0ex}}\\ \left(ii\right)\phantom{\rule{2.77695pt}{0ex}}\sum _{n=1}^{\infty }{\alpha }_{n}=\infty ,\phantom{\rule{2em}{0ex}}\\ \left(iii\right)\phantom{\rule{2.77695pt}{0ex}}\sum _{n=1}^{\infty }|{\alpha }_{n+1}-{\alpha }_{n}|\phantom{\rule{2.77695pt}{0ex}}<\infty ,\phantom{\rule{1em}{0ex}}\sum _{n=1}^{\infty }|{\beta }_{n+1}-{\beta }_{n}|\phantom{\rule{2.77695pt}{0ex}}<\infty .\phantom{\rule{2em}{0ex}}\end{array}$
Then {x n } converges strongly to a fixed point q of T, which solves the following solution of variational inequality;
$〈\left(A-\gamma f\right)q,q-x〉\le 0,\phantom{\rule{1em}{0ex}}\forall x\in F\left(T\right).$

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 : CH 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 $\stackrel{̄}{\gamma }$ and f: CC be a contraction with the contractive coefficient 0 < α < 1 such that $0<\gamma <\frac{\stackrel{̄}{\gamma }}{\alpha }$. 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)\phantom{\rule{2.77695pt}{0ex}}0<\underset{n\to \infty }{\text{lim}\text{inf}}{\beta }_{n}\le \underset{n\to \infty }{\text{lim}\text{sup}}{\beta }_{n}
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},\phantom{\rule{1em}{0ex}}n\ge 0,\end{array}\right\$
where S : CH 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;
$〈\left(A-\gamma f\right)q,q-x〉\le 0,\phantom{\rule{1em}{0ex}}\forall x\in F\left(T\right).$

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 : CH be two mappings. By modification of (1.2), we have
(1.9)
From (1.4) and (1.9), we have
$\begin{array}{l}EP\left(F,\left(aA+\left(1-a\right)B\right)\right)=\left\{z\in C:F\left(z,y\right)+〈\left(aA+\left(1-a\right)B\right)z,y-z〉\ge 0,\\ \forall y\in C\text{and}a\in \left(0,1\right)\right\}.\end{array}$

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
$||x-{P}_{C}x||\phantom{\rule{2.77695pt}{0ex}}=\underset{y\in C}{\text{min}}||x-y||.$

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
$〈x-y,y-z〉\ge 0\phantom{\rule{1em}{0ex}}\forall z\in C.$
Lemma 2.2. [18] Let {s n } be a sequence of nonnegative real number satisfying
${s}_{n+1}=\left(1-{\alpha }_{n}\right){s}_{n}+{\alpha }_{n}{\beta }_{n},\phantom{\rule{1em}{0ex}}\forall n\ge 0$
where {α n }, {β n } satisfy the conditions
$\begin{array}{l}\left(1\right)\phantom{\rule{1em}{0ex}}\left\{{\alpha }_{n}\right\}\subset \left[0,1\right],\phantom{\rule{1em}{0ex}}\sum _{n=1}^{\infty }{\alpha }_{n}=\infty ;\phantom{\rule{2em}{0ex}}\\ \left(2\right)\phantom{\rule{1em}{0ex}}\underset{n\to \infty }{\text{lim}\text{sup}}{\beta }_{n}\le 0\phantom{\rule{2.77695pt}{0ex}}or\phantom{\rule{2.77695pt}{0ex}}\sum _{n=1}^{\infty }|{\alpha }_{n}{\beta }_{n}|<\infty .\phantom{\rule{2em}{0ex}}\end{array}$

Then limn→∞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,
$u={P}_{C}\left(I-\lambda A\right)u⇔u\in VI\left(C,A\right),$

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 infn→∞β n ≤ lim supn→∞β n < 1. Suppose
${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){z}_{n}$
for all n ≥ 0 and
$\underset{n\to \infty }{\text{lim}\text{sup}}\left(||{z}_{n+1}-{z}_{n}||-||{x}_{n+1}-{x}_{n}||\right)\le 0.$

Then limn→∞||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 : CC be a nonexpansive mapping. Then, I - S is demi-closed at zero.

For solving the equilibrium problem for a bifunction F:C×C→ , 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,

limt→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 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→ satisfies (A 1)-(A 4). For r > 0 and x H, define a mapping T r : HC as follows:
${T}_{r}\left(x\right)=\left\{z\in C:F\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in C\right\}.$

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 - yx, 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 : CC 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 : CC is κ-strictly pseudocontractive mapping, we have
$\begin{array}{ll}\hfill ||Tz-T{z}^{*}|{|}^{2}& =||\left(I-A\right)z-\left(I-A\right){z}^{*}|{|}^{2}=\phantom{\rule{2.77695pt}{0ex}}||z-{z}^{*}-\left(Az-A{z}^{*}\right)|{|}^{2}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}||z-{z}^{*}|{|}^{2}-\phantom{\rule{2.77695pt}{0ex}}2〈z-{z}^{*},\phantom{\rule{2.77695pt}{0ex}}Az-A{z}^{*}〉\phantom{\rule{2.77695pt}{0ex}}+||Az-A{z}^{*}|{|}^{2}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}||z-{z}^{*}|{|}^{2}-\phantom{\rule{2.77695pt}{0ex}}2〈z-{z}^{*},\phantom{\rule{2.77695pt}{0ex}}\left(I-T\right)z〉\phantom{\rule{2.77695pt}{0ex}}+||\left(I-T\right)z|{|}^{2}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}||z-{z}^{*}|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}\kappa ||\left(I-T\right)z|{|}^{2}.\phantom{\rule{2em}{0ex}}\end{array}$
It implies that
$\left(1-\kappa \right)||\left(I-T\right)z|{|}^{2}\le 2〈z-{z}^{*},\left(I-T\right)z〉\le 0.$

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
$\begin{array}{ll}\hfill ||Tx-Ty|{|}^{2}& =||\left(I-A\right)x-\left(I-A\right)y|{|}^{2}=\phantom{\rule{2.77695pt}{0ex}}||x-y-\left(Ax-Ay\right)|{|}^{2}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}||x-y|{|}^{2}-\phantom{\rule{2.77695pt}{0ex}}2〈x-y,Ax-Ay〉\phantom{\rule{2.77695pt}{0ex}}+||Ax-Ay|{|}^{2}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}||x-y|{|}^{2}+\kappa ||\left(I-T\right)x-\left(I-T\right)y|{|}^{2}.\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}||x-y|{|}^{2}+\kappa ||Ax-Ay|{|}^{2}.\phantom{\rule{2em}{0ex}}\end{array}$
Then, we have
$〈x-y,Ax-Ay〉\ge \frac{1-\kappa }{2}||Ax-Ay|{|}^{2}.$

## 3 Main result

Theorem 3.1. Let C be a closed convex subset of Hilbert space H and let F:C×C→ be a bifunction satisfying (A1)-(A4), let A, B : CH be α and β-inverse strongly monotone, respectively. Let T : CC 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,\phantom{\rule{1em}{0ex}}\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},\phantom{\rule{1em}{0ex}}\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 and {r n } [0, 2γ], γ = min{α, β} satisfy;
$\begin{array}{l}\left(i\right)\phantom{\rule{2.77695pt}{0ex}}\sum _{n=1}^{\infty }{\alpha }_{n}=\infty ,\phantom{\rule{1em}{0ex}}\underset{n\to \infty }{\text{lim}}{\alpha }_{n}=0;\phantom{\rule{2em}{0ex}}\\ \left(ii\right)\phantom{\rule{2.77695pt}{0ex}}0

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
$\begin{array}{ll}\hfill 〈Dx-Dy,x-y〉& =〈aAx+\left(1-a\right)Bx-aAy-\left(1-a\right)By,x-y〉\phantom{\rule{2em}{0ex}}\\ =〈a\left(Ax-Ay\right)+\left(1-a\right)\phantom{\rule{2.77695pt}{0ex}}\left(Bx-By\right),x-y〉\phantom{\rule{2em}{0ex}}\\ =a〈Ax-Ay,x-y〉+\left(1-a\right)\phantom{\rule{2.77695pt}{0ex}}〈Bx-By,x-y〉\phantom{\rule{2em}{0ex}}\\ \ge a\alpha ||Ax-Ay|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}\left(1-a\right)\beta ||Bx-By|{|}^{2}\phantom{\rule{2em}{0ex}}\\ \ge \gamma \left(a||Ax-Ay|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}\left(1-a\right)||Bx-By|{|}^{2}\right)\phantom{\rule{2em}{0ex}}\\ \ge \gamma ||a\left(Ax-Ay\right)+\left(1-a\right)\phantom{\rule{2.77695pt}{0ex}}\left(Bx-By\right)|{|}^{2}\phantom{\rule{2em}{0ex}}\\ =\gamma ||aAx+\left(1-a\right)Bx-aAy-\left(1-a\right)By|{|}^{2}\phantom{\rule{2em}{0ex}}\\ =\gamma ||Dx-Dy|{|}^{2}\phantom{\rule{2em}{0ex}}\end{array}$
(3.2)
Step 2. We show that I - r n D is a nonexpansive mapping for every n and so is P C (I - λ(I - T)). For every n, let x, y C. From step 1, we have
$\begin{array}{ll}\hfill ||\left(I-{r}_{n}D\right)x-\left(I-{r}_{n}D\right)y|{|}^{2}& =||x-y-{r}_{n}\left(Dx-Dy\right)|{|}^{2}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}||x-y|{|}^{2}-2{r}_{n}〈x-y,\phantom{\rule{2.77695pt}{0ex}}Dx-Dy〉+{r}_{n}^{2}||Dx-Dy|{|}^{2}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}||x-y|{|}^{2}-2{r}_{n}\gamma ||Dx-Dy|{|}^{2}+{r}_{n}^{2}||Dx-Dy|{|}^{2}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}||x-y|{|}^{2}+{r}_{n}\left({r}_{n}-2\gamma \right)||Dx-Dy|{|}^{2}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}||x-y|{|}^{2}.\phantom{\rule{2em}{0ex}}\end{array}$
(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. 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
$F\left(z,y\right)+\frac{1}{{r}_{n}}〈y-z,z-z+{r}_{n}Dz〉\ge 0,\phantom{\rule{1em}{0ex}}\forall n\in ℕ\phantom{\rule{2.77695pt}{0ex}}\text{and}\phantom{\rule{2.77695pt}{0ex}}y\in C.$
From Lemma 2.7, we have $z={T}_{{r}_{n}}\left(I-{r}_{n}D\right)z$, n. By nonexpansiveness of ${T}_{{r}_{n}}\left(I-{r}_{n}D\right)$, we have
$\begin{array}{ll}\hfill ||{x}_{n+1}-z||& =||{\alpha }_{n}\left(u-z\right)+{\beta }_{n}\left({x}_{n}-z\right)+{\gamma }_{n}\left({P}_{C}\left(I-\lambda E\right){u}_{n}-z\right)||\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}{\alpha }_{n}||u-z||+{\beta }_{n}||{x}_{n}-z||+{\gamma }_{n}||{P}_{C}\left(I-\lambda E\right){u}_{n}-z||\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}{\alpha }_{n}||u-z||+{\beta }_{n}||{x}_{n}-z||+{\gamma }_{n}||{T}_{{r}_{n}}\left(I-{r}_{n}D\right){x}_{n}-z||\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}{\alpha }_{n}||u-z||+\phantom{\rule{2.77695pt}{0ex}}\left(1-{\alpha }_{n}\right)||{x}_{n}-z||\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}\text{max}\left\{||{x}_{n}-z||,\phantom{\rule{2.77695pt}{0ex}}||u-z||\right\}.\phantom{\rule{2em}{0ex}}\end{array}$

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
$\underset{n\to \infty }{\text{lim}}||{x}_{n+1}-{x}_{n}||\phantom{\rule{2.77695pt}{0ex}}=0.$
(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
(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
$F\left({u}_{n},y\right)+\frac{1}{{r}_{n}}〈y-{u}_{n},{u}_{n}-{v}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in C,$
(3.7)
and
$F\left({u}_{n+1},y\right)+\frac{1}{{r}_{n+1}}〈y-{u}_{n+1},{u}_{n+1}-{v}_{n+1}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in C.$
(3.8)
Putting y = un+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
$\begin{array}{ll}\hfill 0& \le \phantom{\rule{2.77695pt}{0ex}}\frac{1}{{r}_{n}}〈{u}_{n+1}-{u}_{n},{u}_{n}-{v}_{n}〉+\frac{1}{{r}_{n+1}}〈{u}_{n}-{u}_{n+1},\phantom{\rule{2.77695pt}{0ex}}{u}_{n+1}-{v}_{n+1}〉\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}〈{u}_{n+1}-{u}_{n},\frac{{u}_{n}-{v}_{n}}{{r}_{n}}〉+〈{u}_{n}-{u}_{n+1},\frac{{u}_{n+1}-{v}_{n+1}}{{r}_{n+1}}〉\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}〈{u}_{n+1}-{u}_{n},\phantom{\rule{2.77695pt}{0ex}}\frac{{u}_{n}-{v}_{n}}{{r}_{n}}-\frac{{u}_{n+1}-{v}_{n+1}}{{r}_{n+1}}〉.\phantom{\rule{2em}{0ex}}\end{array}$
It implies that
$\begin{array}{ll}\hfill 0& \le 〈{u}_{n+1}-{u}_{n},\phantom{\rule{2.77695pt}{0ex}}{u}_{n}-{v}_{n}-\frac{{r}_{n}}{{r}_{n+1}}\left({u}_{n+1}-{v}_{n+1}\right)〉\phantom{\rule{2em}{0ex}}\\ =〈{u}_{n+1}-{u}_{n},\phantom{\rule{2.77695pt}{0ex}}{u}_{n}-{u}_{n+1}+{u}_{n+1}-{v}_{n}-\frac{{r}_{n}}{{r}_{n+1}}\left({u}_{n+1}-{v}_{n+1}\right)〉.\phantom{\rule{2em}{0ex}}\end{array}$
It implies that
$\begin{array}{ll}\hfill ||{u}_{n+1}-{u}_{n}|{|}^{2}& \le 〈{u}_{n+1}-{u}_{n},\phantom{\rule{2.77695pt}{0ex}}{u}_{n+1}-{v}_{n}-\frac{{r}_{n}}{{r}_{n+1}}\left({u}_{n+1}-{v}_{n+1}\right)〉\phantom{\rule{2em}{0ex}}\\ =〈{u}_{n+1}-{u}_{n},\phantom{\rule{2.77695pt}{0ex}}{u}_{n+1}-{v}_{n+1}+{v}_{n+1}-{v}_{n}-\frac{{r}_{n}}{{r}_{n+1}}\left({u}_{n+1}-{v}_{n+1}\right)〉\phantom{\rule{2em}{0ex}}\\ =〈{u}_{n+1}-{u}_{n},\phantom{\rule{2.77695pt}{0ex}}{v}_{n+1}-{v}_{n}+\left(1-\frac{{r}_{n}}{{r}_{n+1}}\right)\left({u}_{n+1}-{v}_{n+1}\right)〉\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}||{u}_{n+1}-{u}_{n}||\left(||{v}_{n+1}-{v}_{n}||+\frac{1}{{r}_{n+1}}|{r}_{n+1}-{r}_{n}|\phantom{\rule{2.77695pt}{0ex}}||{u}_{n+1}-{v}_{n+1}||\right).\phantom{\rule{2em}{0ex}}\end{array}$
It follows that
$||{u}_{n+1}-{u}_{n}||\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}||{v}_{n+1}-{v}_{n}||+\frac{1}{e}|{r}_{n+1}-{r}_{n}|\phantom{\rule{2.77695pt}{0ex}}||{u}_{n+1}-{v}_{n+1}||.$
(3.11)
Since v n = x n - r n Dx n , we have
$\begin{array}{ll}\hfill ||{v}_{n+1}-{v}_{n}||& =||{x}_{n+1}-{r}_{n+1}D{x}_{n+1}-{x}_{n}+{r}_{n}D{x}_{n}||\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}||\left(I-{r}_{n+1}D\right){x}_{n+1}-\left(I-{r}_{n+1}D\right){x}_{n}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left(I-{r}_{n+1}D\right){x}_{n}-\left(I-{r}_{n}D\right){x}_{n}||\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}||\left(I-{r}_{n+1}D\right){x}_{n+1}-\left(I-{r}_{n+1}D\right){x}_{n}||\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+||\left({r}_{n}-{r}_{n+1}\right)D{x}_{n}||\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}||{x}_{n+1}-{x}_{n}||+|{r}_{n}-{r}_{n+1}|\phantom{\rule{2.77695pt}{0ex}}||D{x}_{n}||.\phantom{\rule{2em}{0ex}}\end{array}$
(3.12)
Substitute (3.12) into (3.11), we have
$\begin{array}{ll}\hfill ||{u}_{n+1}-{u}_{n}||& \le ||{v}_{n+1}-{v}_{n}||+\frac{1}{e}|{r}_{n+1}-{r}_{n}|\phantom{\rule{2.77695pt}{0ex}}||{u}_{n+1}-{v}_{n+1}||\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}||{x}_{n+1}-{x}_{n}||+|{r}_{n}-{r}_{n+1}|\phantom{\rule{2.77695pt}{0ex}}||D{x}_{n}||\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{1}{e}|{r}_{n+1}-{r}_{n}|\phantom{\rule{2.77695pt}{0ex}}||{u}_{n+1}-{v}_{n+1}||\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}||{x}_{n+1}-{x}_{n}||+|{r}_{n}-{r}_{n+1}|L+\frac{1}{e}|{r}_{n+1}-{r}_{n}|L,\phantom{\rule{2em}{0ex}}\end{array}$
(3.13)
where L=maxn{||Dx n ||, ||u n n ||}. Substitute (3.13) into (3.6), we have
$\begin{array}{ll}\hfill ||{p}_{n+1}-{p}_{n}||& \le \left|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_{n}}{1-{\beta }_{n}}\right|\left(||u||+||{P}_{C}\left(I-\lambda E\right){u}_{n}||\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}||{u}_{n+1}-{u}_{n}||\phantom{\rule{2em}{0ex}}\\ \le \left|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_{n}}{1-{\beta }_{n}}\right|\left(||u||+||{P}_{C}\left(I-\lambda E\right){u}_{n}||\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+||{x}_{n+1}-{x}_{n}||+|{r}_{n}-{r}_{n+1}|L+\frac{1}{e}|{r}_{n+1}-{r}_{n}|L,\phantom{\rule{2em}{0ex}}\end{array}$
(3.14)
From conditions (i), (iii) and (3.14), we have
$\underset{n\to \infty }{\text{lim}\text{sup}}\phantom{\rule{2.77695pt}{0ex}}\left(||{p}_{n+1}-{p}_{n}||-||{x}_{n+1}-{x}_{n}||\right)\le 0.$
(3.15)
From Lemma 2.4, (3.15) and (3.5), we have
$\underset{n\to \infty }{\text{lim}}||{p}_{n}-{x}_{n}||\phantom{\rule{2.77695pt}{0ex}}=0.$
(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
$\underset{n\to \infty }{\text{lim}}||{x}_{n+1}-{x}_{n}||\phantom{\rule{2.77695pt}{0ex}}=0.$
(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
$\underset{n\to \infty }{\text{lim}}||{P}_{C}\left(I-\lambda E\right){u}_{n}-{x}_{n}||\phantom{\rule{2.77695pt}{0ex}}=0,$
(3.19)

where E = I - T .

Step 5. We will show that
$\underset{n\to \infty }{\text{lim}}||{u}_{n}-{x}_{n}||\phantom{\rule{2.77695pt}{0ex}}=0.$
(3.20)
Since ${u}_{n}={T}_{{r}_{n}}\left({x}_{n}-{r}_{n}D{x}_{n}\right)$, we have
it implies that
$||{u}_{n}-z|{|}^{2}\le \phantom{\rule{2.77695pt}{0ex}}||{x}_{n}-z|{|}^{2}-||{x}_{n}-{u}_{n}|{|}^{2}-{r}_{n}^{2}||D{x}_{n}-Dz|{|}^{2}+2{r}_{n}〈{x}_{n}-{u}_{n},D{x}_{n}-Dz〉.$
(3.21)
By nonexpansiveness of ${T}_{{r}_{n}}$ and using the same method as (3.3), we have
$\begin{array}{ll}\hfill ||{u}_{n}-z|{|}^{2}& =||{T}_{{r}_{n}}\left(I-{r}_{n}D\right){x}_{n}-{T}_{{r}_{n}}\left(I-{r}_{n}D\right)z|{|}^{2}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}||\left(I-{r}_{n}D\right){x}_{n}-\left(I-{r}_{n}D\right)z|{|}^{2}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}||{x}_{n}-z|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}{r}_{n}\left({r}_{n}-2\gamma \right)||D{x}_{n}-Dz|{|}^{2}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}||{x}_{n}-z|{|}^{2}-{r}_{n}\left(2\gamma -{r}_{n}\right)||D{x}_{n}-Dz|{|}^{2}.\phantom{\rule{2em}{0ex}}\end{array}$
(3.22)
By nonexpansiveness of P C (I - λE) and (3.22), we have
$\begin{array}{l}||{x}_{n+1}-z|{|}^{2}=||{\alpha }_{n}\left(u-z\right)+{\beta }_{n}\left({x}_{n}-z\right)+{\gamma }_{n}\left({P}_{C}\left(I-\lambda E\right){u}_{n}-z\right)|{|}^{2}\\ \le {\alpha }_{n}||u-z|{|}^{2}+{\beta }_{n}||{x}_{n}-z|{|}^{2}+{\gamma }_{n}||{u}_{n}-z|{|}^{2}\\ \le {\alpha }_{n}||u-z|{|}^{2}+{\beta }_{n}||{x}_{n}-z|{|}^{2}+{\gamma }_{n}\left(||{x}_{n}-z|{|}^{2}\\ -{r}_{n}\left(2\gamma -{r}_{n}\right)||D{x}_{n}-Dz|{|}^{2}\right)\\ \le {\alpha }_{n}||u-z|{|}^{2}+||{x}_{n}-z|{|}^{2}-{r}_{n}{\gamma }_{n}\left(2\gamma -{r}_{n}\right)||D{x}_{n}-Dz|{|}^{2},\end{array}$
(3.23)
it implies that
$\begin{array}{l}{r}_{n}{\gamma }_{n}\left(2\gamma -{r}_{n}\right)||D{x}_{n}-Dz|{|}^{2}\le {\alpha }_{n}||u-z|{|}^{2}+||{x}_{n}-z|{|}^{2}-||{x}_{n+1}-z|{|}^{2}\\ \le {\alpha }_{n}||u-z|{|}^{2}+\left(||{x}_{n}-z||\\ +||{x}_{n+1}-z||\right)||{x}_{n+1}-{x}_{n}||.\end{array}$
(3.24)
From (3.18), (3.24), conditions (i) and (ii), we have
$\underset{n\to \infty }{\text{lim}}||D{x}_{n}-Dz||\phantom{\rule{2.77695pt}{0ex}}=0$
(3.25)
From (3.23) and (3.21). we have
which implies that
$\begin{array}{ll}\hfill {\gamma }_{n}||{x}_{n}-{u}_{n}|{|}^{2}& \le {\alpha }_{n}||u-z|{|}^{2}+||{x}_{n}-z|{|}^{2}-||{x}_{n+1}-z|{|}^{2}+2{r}_{n}{\gamma }_{n}||{x}_{n}-{u}_{n}||\phantom{\rule{2.77695pt}{0ex}}||D{x}_{n}-Dz||\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}{\alpha }_{n}||u-z|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}\left(||{x}_{n}-z||+||{x}_{n+1}-z||\right)||{x}_{n+1}-{x}_{n}||\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+2{r}_{n}{\gamma }_{n}||{x}_{n}-{u}_{n}||\phantom{\rule{2.77695pt}{0ex}}||D{x}_{n}-Dz||,\phantom{\rule{2em}{0ex}}\end{array}$
from condition (i), (3.25) and (3.18), we have
$\underset{n\to \infty }{\text{lim}}||{x}_{n}-{u}_{n}||\phantom{\rule{2.77695pt}{0ex}}=0.$
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 ${x}_{{n}_{k}}⇀\omega$ 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 ${u}_{{n}_{k}}⇀\omega$ as k → ∞. Since${u}_{n}={T}_{{r}_{n}}\left({x}_{n}-{r}_{n}D{x}_{n}\right)$, we obtain
$F\left({u}_{n},y\right)+〈D{x}_{n},\phantom{\rule{2.77695pt}{0ex}}y-{u}_{n}〉+\frac{1}{{r}_{n}}〈y-{u}_{n},\phantom{\rule{2.77695pt}{0ex}}{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in C.$
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
$〈D{x}_{{n}_{k}},y-{u}_{{n}_{k}}〉+\frac{1}{{r}_{{n}_{k}}}〈y-{u}_{{n}_{k}},{u}_{{n}_{k}}-{x}_{{n}_{k}}〉\ge F\left(y,\phantom{\rule{2.77695pt}{0ex}}{u}_{{n}_{k}}\right),\phantom{\rule{1em}{0ex}}\forall y\in C.$
(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
$\begin{array}{ll}\hfill 〈{z}_{t}-{u}_{{n}_{k}},D{z}_{t}〉& \ge 〈{z}_{t}-{u}_{{n}_{k}},D{z}_{t}〉-〈{z}_{t}-{u}_{{n}_{k}},D{x}_{{n}_{k}}〉-〈{z}_{t}-{u}_{{n}_{k}},\frac{{u}_{{n}_{k}}-{x}_{{n}_{k}}}{{r}_{{n}_{k}}}〉+F\left({z}_{t},\phantom{\rule{2.77695pt}{0ex}}{u}_{{n}_{k}}\right)\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}〈{z}_{t}-{u}_{{n}_{k}},\phantom{\rule{2.77695pt}{0ex}}D{z}_{t}-D{u}_{{n}_{k}}〉+〈{z}_{t}-{u}_{{n}_{k}},\phantom{\rule{2.77695pt}{0ex}}D{u}_{{n}_{k}}-D{x}_{{n}_{k}}〉\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-〈{z}_{t}-{u}_{{n}_{k}},\phantom{\rule{2.77695pt}{0ex}}\frac{{u}_{{n}_{k}}-{x}_{{n}_{k}}}{{r}_{{n}_{k}}}〉+F\left({z}_{t},\phantom{\rule{2.77695pt}{0ex}}{u}_{{n}_{k}}\right).\phantom{\rule{2em}{0ex}}\end{array}$
Since $||{u}_{{n}_{k}}-{x}_{{n}_{k}}||\phantom{\rule{2.77695pt}{0ex}}\to 0$, we have $||D{u}_{{n}_{k}}-D{x}_{{n}_{k}}||\phantom{\rule{2.77695pt}{0ex}}\to 0$. 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
(3.29)
From (A 1), (A 4) and (3.29), we also have
$\begin{array}{ll}\hfill 0& =\phantom{\rule{2.77695pt}{0ex}}F\left({z}_{t},\phantom{\rule{2.77695pt}{0ex}}{z}_{t}\right)\le tF\left({z}_{t},\phantom{\rule{2.77695pt}{0ex}}y\right)+\left(1-t\right)F\left({z}_{t},\phantom{\rule{2.77695pt}{0ex}}\omega \right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}tF\left({z}_{t},\phantom{\rule{2.77695pt}{0ex}}y\right)+\left(1-t\right)〈{z}_{t}-\omega ,\phantom{\rule{2.77695pt}{0ex}}D{z}_{t}〉\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}tF\left({z}_{t},\phantom{\rule{2.77695pt}{0ex}}y\right)+\left(1-t\right)t〈y-\omega ,\phantom{\rule{2.77695pt}{0ex}}D{z}_{t}〉,\phantom{\rule{2em}{0ex}}\end{array}$
hence
$0\le F\left({z}_{t},\phantom{\rule{2.77695pt}{0ex}}y\right)+\left(1-t\right)〈y-\omega ,\phantom{\rule{2.77695pt}{0ex}}D{z}_{t}〉.$
Letting t → 0, we have
$0\le F\left(\omega ,y\right)+〈y-\omega ,D\omega 〉\phantom{\rule{1em}{0ex}}\forall y\in C.$
(3.30)
Therefore ω EP(F, D), where D = aA + (1 - a)B, a [0,1]. Since
$||{P}_{C}\left(I-\lambda E\right){u}_{n}-{u}_{n}||\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}||\phantom{\rule{2.77695pt}{0ex}}||{P}_{C}\left(I-\lambda E\right){u}_{n}-{x}_{n}||\phantom{\rule{2.77695pt}{0ex}}||+||{x}_{n}-{u}_{n}||,$
where E = I - T from (3.19) and (3.20), we have
$\underset{n\to 0}{\text{lim}}||{P}_{C}\left(I-\lambda E\right){u}_{n}-{u}_{n}||\phantom{\rule{2.77695pt}{0ex}}=0.$
(3.31)
Since ${u}_{{n}_{k}}\to \omega$ as kω, (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
$\begin{array}{ll}\hfill ||{x}_{n+1}-{z}_{0}|{|}^{2}& =||{\alpha }_{n}\left(u-{z}_{0}\right)+{\beta }_{n}\left({x}_{n}-{z}_{0}\right)+{\gamma }_{n}\left({P}_{C}\left(I-\lambda \left(I-T\right)\right){u}_{n}-{z}_{0}\right)|{|}^{2}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}||{\beta }_{n}\left({x}_{n}-{z}_{0}\right)+{\gamma }_{n}\left({P}_{C}\left(I-\lambda \left(I-T\right)\right){u}_{n}-{z}_{0}\right)|{|}^{2}+2{\alpha }_{n}〈u-{z}_{0},\phantom{\rule{2.77695pt}{0ex}}{x}_{n+1}-{z}_{0}〉\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}{\beta }_{n}||{x}_{n}-{z}_{0}|{|}^{2}+{\gamma }_{n}||{P}_{C}\left(I-\lambda \left(I-T\right)\right){u}_{n}-{z}_{0}|{|}^{2}+2{\alpha }_{n}〈u-{z}_{0},\phantom{\rule{2.77695pt}{0ex}}{x}_{n+1}-{z}_{0}〉\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}{\beta }_{n}||{x}_{n}-{z}_{0}|{|}^{2}+{\gamma }_{n}||{T}_{{r}_{n}}\left(I-{r}_{n}D\right){x}_{n}-{z}_{0}|{|}^{2}+2{\alpha }_{n}〈u-{z}_{0},\phantom{\rule{2.77695pt}{0ex}}{x}_{n+1}-{z}_{0}〉\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}\left(1-{\alpha }_{n}\right)||{x}_{n}-{z}_{0}|{|}^{2}+2{\alpha }_{n}〈u-{z}_{0},\phantom{\rule{2.77695pt}{0ex}}{x}_{n+1}-{z}_{0}〉\phantom{\rule{2em}{0ex}}\end{array}$

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 : CH be α and β-inverse strongly monotone mappings, respectively, with α, β > 0 and VI(C, A) ∩ VI(C, B) ≠ . Then
$VI\left(C,\phantom{\rule{2.77695pt}{0ex}}aA+\left(1-a\right)B\right)=VI\left(C,\phantom{\rule{2.77695pt}{0ex}}A\right)\bigcap VI\left(C,\phantom{\rule{2.77695pt}{0ex}}B\right),\phantom{\rule{2.77695pt}{0ex}}\forall a\in \left(0,1\right).$
(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
$〈y-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}A{x}^{*}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in C,$
and
$〈y-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}B{x}^{*}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in C.$
For every a (0, 1), we have
$〈y-{x}^{*},aA{x}^{*}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in C,$
(4.2)
and
$〈y-{x}^{*},\left(1-a\right)B{x}^{*}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in C.$
(4.3)
By monotonicity of A, B and x*, x0 C, we have
$\begin{array}{ll}\hfill 〈{x}^{*}-{x}_{0},aA{x}_{0}〉& =〈{x}^{*}-{x}_{0},aA{x}_{0}+\left(1-a\right)B{x}_{0}-\left(1-a\right)B{x}_{0}〉\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}aA{x}_{0}+\left(1-a\right)B{x}_{0}〉-〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}\left(1-a\right)B{x}_{0}〉\phantom{\rule{2em}{0ex}}\\ \ge \phantom{\rule{2.77695pt}{0ex}}\left(1-a\right)〈{x}_{0}-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}B{x}_{0}〉\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}\left(1-a\right)\left(〈{x}_{0}-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}B{x}_{0}-B{x}^{*}〉+〈{x}_{0}-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}B{x}^{*}〉\right)\phantom{\rule{2em}{0ex}}\\ \ge 0.\phantom{\rule{2em}{0ex}}\end{array}$
(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
$\begin{array}{ll}\hfill 0& \le 〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}A{x}_{0}〉\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}A{x}_{0}-A{x}^{*}+A{x}^{*}〉\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}A{x}_{0}-A{x}^{*}〉+〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}A{x}^{*}〉\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}-\alpha ||A{x}^{*}-A{x}_{0}|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}A{x}^{*}〉\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}-\alpha ||A{x}^{*}-A{x}_{0}|{|}^{2},\phantom{\rule{2em}{0ex}}\end{array}$
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
$\begin{array}{ll}\hfill 〈y-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}A{x}_{0}〉& =〈y-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}A{x}_{0}〉+〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}A{x}_{0}〉\phantom{\rule{2em}{0ex}}\\ \ge \phantom{\rule{2.77695pt}{0ex}}〈y-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}A{x}^{*}〉\ge 0.\phantom{\rule{2em}{0ex}}\end{array}$
Then, we have
${x}_{0}\in VI\left(C,\phantom{\rule{2.77695pt}{0ex}}A\right).$
(4.7)
From (4.4), we have
$\begin{array}{ll}\hfill \left(1-a\right)〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}B{x}_{0}〉& \ge \phantom{\rule{2.77695pt}{0ex}}a〈{x}_{0}-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}A{x}_{0}〉\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}a\left(〈{x}_{0}-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}A{x}_{0}-A{x}^{*}〉+〈{x}_{0}-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}A{x}^{*}〉\right)\phantom{\rule{2em}{0ex}}\\ \ge 0.\phantom{\rule{2em}{0ex}}\end{array}$
(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
$\begin{array}{ll}\hfill 0& \le \phantom{\rule{2.77695pt}{0ex}}〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}B{x}_{0}〉\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}B{x}_{0}-B{x}^{*}+B{x}^{*}〉\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{2.77695pt}{0ex}}〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}B{x}_{0}-B{x}^{*}〉+〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}B{x}^{*}〉\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}-\beta ||B{x}^{*}-B{x}_{0}|{|}^{2}+\phantom{\rule{2.77695pt}{0ex}}〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}B{x}^{*}〉\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{2.77695pt}{0ex}}-\beta ||B{x}^{*}-B{x}_{0}|{|}^{2},\phantom{\rule{2em}{0ex}}\end{array}$
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
$\begin{array}{ll}\hfill 〈y-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}B{x}_{0}〉& =〈y-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}B{x}_{0}〉+〈{x}^{*}-{x}_{0},\phantom{\rule{2.77695pt}{0ex}}B{x}_{0}〉\phantom{\rule{2em}{0ex}}\\ \ge \phantom{\rule{2.77695pt}{0ex}}〈y-{x}^{*},\phantom{\rule{2.77695pt}{0ex}}B{x}^{*}〉\ge 0.\phantom{\rule{2em}{0ex}}\end{array}$
Then, we have
${x}_{0}\in VI\left(C,\phantom{\rule{2.77695pt}{0ex}}B\right).$
(4.11)
By (4.7) and (4.11), we have x0 VI(C, A) ∩ VI(C, B). Hence, we have
$VI\phantom{\rule{2.77695pt}{0ex}}\left(C,\phantom{\rule{2.77695pt}{0ex}}aA+\left(1-a\right)B\right)\subseteq VI\left(C,\phantom{\rule{2.77695pt}{0ex}}A\right)\bigcap VI\left(C,\phantom{\rule{2.77695pt}{0ex}}B\right).$
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
$\begin{array}{l}{‖\left(I-\gamma \left(aA+\left(1-a\right)B\right)\right)x-\left(I-\gamma \left(aA+\left(1-a\right)B\right)\right)y‖}^{2}\\ ={‖x-y-\gamma \left(\left(aA+\left(1-a\right)B\right)x-\left(aA+\left(1-a\right)B\right)y\right)‖}^{2}\\ ={‖x-y-\gamma \left(a\left(Ax-Ay\right)+\left(1-a\right)\left(Bx-By\right)\right)‖}^{2}\\ =||x-y|{|}^{2}-2\gamma 〈a\left(Ax-Ay\right)+\left(1-a\right)\left(Bx-By\right),x-y〉\\ +{\gamma }^{2}||a\left(Ax-Ay\right)+\left(1-a\right)\left(Bx-By\right)|{|}^{2}\\ \le ||x-y|{|}^{2}-2\gamma a〈Ax-Ay,x-y〉-2\gamma \left(1-a\right)〈Bx-By,x-y〉\\ +a{\gamma }^{2}||Ax-Ay|{|}^{2}+\left(1-a\right){\gamma }^{2}||Bx-By|{|}^{2}\\ \le ||x-y|{|}^{2}-2\gamma a\alpha ||Ax-Ay|{|}^{2}-2\gamma \left(1-a\right)\beta ||Bx-By|{|}^{2}\\ +a{\gamma }^{2}||Ax-Ay|{|}^{2}+\left(1-a\right){\gamma }^{2}||Bx-By|{|}^{2}\\ =||x-y|{|}^{2}+a\gamma \left(\gamma -2\alpha \right)||Ax-Ay|{|}^{2}+\left(1-a\right)\gamma \left(\gamma -2\beta \right)||Bx-By|{|}^{2}\\ \le ||x-y|{|}^{2}.\end{array}$
(4.12)

Theorem 4.2. Let C be a closed convex subset of Hilbert space H and let A, B : CH 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
${x}_{n+1}={\alpha }_{n}u+{\beta }_{n}{x}_{n}+{\gamma }_{n}{P}_{C}\left(I-\lambda \left(I-T\right)\right){P}_{C}\left(I-{r}_{n}\left(aA+\left(1-a\right)B\right)\right){x}_{n},\phantom{\rule{1em}{0ex}}\forall n\ge 1,$
(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 and {r n } [0, 2γ], γ = min{α, β} satisfy;
$\begin{array}{l}\left(i\right)\phantom{\rule{2.77695pt}{0ex}}\sum _{n=1}^{\infty }{\alpha }_{n}=\infty ,\phantom{\rule{1em}{0ex}}\underset{n\to \infty }{\text{lim}}{\alpha }_{n}=0;\phantom{\rule{2em}{0ex}}\\ \left(ii\right)\phantom{\rule{2.77695pt}{0ex}}0

Then {x n } converges strongly to ${z}_{0}={P}_{F}u$.

Proof. From 3.1 putting F ≡ 0 in Theorem 3.1, we have
$〈y-{u}_{n},\phantom{\rule{2.77695pt}{0ex}}{u}_{n}-\left(I-{r}_{n}D\right){x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall y\in C,$
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 be a bifunction satisfying (A1)-(A4), let A : CH be α-inverse strongly monotone. Let T : CC 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},\phantom{\rule{2.77695pt}{0ex}}y\right)+〈A{x}_{n},\phantom{\rule{2.77695pt}{0ex}}y-{u}_{n}〉+\frac{1}{{r}_{n}}〈y-{u}_{n},\phantom{\rule{2.77695pt}{0ex}}{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\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},\phantom{\rule{1em}{0ex}}\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 and {r n } [0, 2γ], γ = min{α, β} satisfy;
$\begin{array}{l}\left(i\right)\phantom{\rule{2.77695pt}{0ex}}\sum _{n=1}^{\infty }{\alpha }_{n}=\infty ,\phantom{\rule{1em}{0ex}}\underset{n\to \infty }{\text{lim}}{\alpha }_{n}=0;\phantom{\rule{2em}{0ex}}\\ \left(ii\right)\phantom{\rule{2.77695pt}{0ex}}0

Then {x n } converges strongly to ${z}_{0}={P}_{F}u$.

Proof. From Theorem 3.1, putting AB, we can conclude the desired conclusion. □

## Declarations

### Acknowledgements

This research was supported by the Research Administration Division of King Mongkut's Institute of Technology Ladkrabang.

## Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, King Mongkut's Institute of Technology Ladkrabang, Bangkok, 10520, Thailand

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