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# On asymptotically strict pseudocontractions and equilibrium problems

Journal of Inequalities and Applications20132013:251

https://doi.org/10.1186/1029-242X-2013-251

• Received: 17 March 2013
• Accepted: 30 April 2013
• Published:

## Abstract

In this paper, an equilibrium problem and a fixed point problem of an asymptotically strict pseudocontraction are investigated based on hybrid iterative algorithms. Strong convergence theorems of common solutions to the equilibrium problem and the fixed point problem are obtained in the framework of real Hilbert spaces.

MSC:47H09, 47H10.

## Keywords

• asymptotically strict pseudocontraction
• asymptotically nonexpansive mapping
• nonexpansive mapping
• fixed point
• equilibrium problem

## 1 Introduction

Equilibrium problems have emerged as an effective and powerful tool for studying a wide class of problems which arise in economics, finance, image reconstruction, ecology, transportation, and network; see  and the references therein. Equilibrium problems include fixed point problems, variational inequality problems, variational inclusion problems, saddle point problems, the Nash equilibrium problem, complementarity problems and so on. For the solutions of equilibrium problems, there are several algorithms to solve the problem. The classical algorithm is the Krasnoselskii-Mann iterative algorithm. However, the Krasnoselskii-Mann iterative algorithm is weak convergence for the solutions of equilibrium problems. Haugazeau’s projection method  recently has been considered for the approximation of solutions of equilibrium problems and fixed point problems. The advantage of the projection methods is that strong convergence of iterative sequences can be guaranteed without any compact assumptions. The aim of this paper is to study an equilibrium problem and a fixed point problem of an asymptotically strict pseudocontraction based on hybrid iterative algorithms and establish a strong convergence theorem of common solutions in the framework of Hilbert spaces.

## 2 Preliminaries

Throughout this paper, we always assume that H is a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H, let $A:C\to H$ be a monotone mapping and F be a bifunction of $C×C$ into , where denotes the set of real numbers.

In this paper, we consider the following equilibrium problem.
(2.1)
The set of such an $x\in C$ is denoted by $EP\left(F,A\right)$, i.e.,
$EP\left(F,A\right)=\left\{x\in C:F\left(x,y\right)+〈Ax,y-x〉\ge 0,\mathrm{\forall }y\in C\right\}.$
If $A\equiv 0$, then the problem (2.1) is reduced to the following:
(2.2)
The set of such an $x\in C$ is denoted by $EP\left(F\right)$, i.e.,
$EP\left(F,A\right)=\left\{x\in C:F\left(x,y\right)\ge 0,\mathrm{\forall }y\in C\right\}.$
If $F\equiv 0$, the problem (2.1) is reduced to the classical variational inequality problem.
(2.3)

To study the problems (2.1) and (2.2), we may assume that F satisfies the following conditions:

(A1) $F\left(x,x\right)=0$ for all $x\in C$;

(A2) F is monotone, i.e., $F\left(x,y\right)+F\left(y,x\right)\le 0$ for all $x,y\in C$;

(A3) for each $x,y,z\in C$,
$\underset{t↓0}{lim sup}F\left(tz+\left(1-t\right)x,y\right)\le F\left(x,y\right);$

(A4) for each $x\in C$, $y↦F\left(x,y\right)$ is convex and lower semi-continuous.

Recall that a mapping A is said to be monotone iff
$〈Ax-Ay,x-y〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
A is said to be strongly monotone iff there exists a constant $\alpha >0$ such that
$〈Ax-Ay,x-y〉\ge \alpha {\parallel x-y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
For such a case, A is said to be α-strongly-monotone. A is said to be inverse-strongly monotone iff there exists a constant $\alpha >0$ such that
$〈Ax-Ay,x-y〉\ge \alpha {\parallel Ax-Ay\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

For such a case, A is said to be α-inverse-strongly monotone.

Recall that a set-valued mapping $T:H\to {2}^{H}$ is said to be monotone iff, for all $x,y\in H$, $f\in Tx$ and $g\in Ty$ imply $〈x-y,f-g〉\ge 0$. $T:H\to {2}^{H}$ is maximal if the graph $G\left(T\right)$ of T is not properly contained in the graph of any other monotone mapping.

It is known that a monotone mapping T is maximal if and only if, for any $\left(x,f\right)\in H×H$, $〈x-y,f-g〉\ge 0$ for all $\left(y,g\right)\in G\left(T\right)$ implies $f\in Tx$.

Let $S:C\to C$ be a nonlinear mapping. In this paper, we use $F\left(S\right)$ to denote the fixed point set of S.

Recall that S is said to be nonexpansive iff
$\parallel Sx-Sy\parallel \le \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
Recall that S is said to be asymptotically nonexpansive iff there exists a sequence $\left\{{k}_{n}\right\}\subset \left[1,\mathrm{\infty }\right)$ with ${k}_{n}\to 1$ as $n\to \mathrm{\infty }$ such that
$\parallel {S}^{n}x-{S}^{n}y\parallel \le {k}_{n}\parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C,n\ge 1.$

The class of asymptotically nonexpansive mappings was introduced by Goebel and Kirk  in 1972. Since 1972, a number of authors have studied the convergence problems of the iterative processes for such a class of mappings.

Recall that S is said to be strictly pseudocontractive iff there exists a constant $\kappa \in \left[0,1\right)$ such that
${\parallel Sx-Sy\parallel }^{2}\le {\parallel x-y\parallel }^{2}+\kappa {\parallel \left(I-S\right)x-\left(I-S\right)y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

For such a case, S is also said to be a κ-strict pseudocontraction. The class of strict pseudocontractions is introduced by Browder and Petryshyn  in 1967. It is clear that every nonexpansive mapping is a 0-strict pseudocontraction. We also remark that if $\kappa =1$, then S is said to be pseudocontractive.

Recall that S is said to be an asymptotically strict pseudocontraction iff there exist a sequence $\left\{{k}_{n}\right\}\subset \left[1,\mathrm{\infty }\right)$ with ${k}_{n}\to 1$ as $n\to \mathrm{\infty }$ and a constant $\kappa \in \left[0,1\right)$ such that
${\parallel {S}^{n}x-{S}^{n}y\parallel }^{2}\le {k}_{n}{\parallel x-y\parallel }^{2}+\kappa {\parallel \left(I-{S}^{n}\right)x-\left(I-{S}^{n}\right)y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C,n\ge 1.$

For such a case, S is also said to be an asymptotically κ-strict pseudocontraction. The class of asymptotically strict pseudocontractions is introduced by Qihou  in 1996. It is clear that every asymptotically nonexpansive mapping is an asymptotical 0-strict pseudocontraction. Every nonexpansive mapping is an asymptotically nonexpansive mapping with the sequence $\left\{1\right\}$. We also remark that if $\kappa =1$, then S is said to be an asymptotically pseudocontractive mapping which was introduced by Schu  in 1991.

Recently, many authors considered the equilibrium problems (2.1), (2.2) and fixed point problems based on hybrid iterative methods; see, for instance, . In this paper, motivated by these recent results, we consider the shrinking projection algorithm to solve the solutions of the equilibrium problem (2.1) and the fixed point problem of an asymptotically strict pseudocontraction. It is proved that the sequence generated in the purposed iterative process converges strongly to some common element in the solution set of the equilibrium problem (2.1) and in the fixed point set of an asymptotically strict pseudocontraction.

In order to prove our main results, we also need the following lemmas.

Lemma 2.1 

Let C be a nonempty closed convex subset of H and let $F:C×C\to \mathbb{R}$ be a bifunction satisfying (A1)-(A4). Then, for any $r>0$ and $x\in H$, there exists $z\in C$ such that
$F\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C.$
Further, define
${T}_{r}x=\left\{z\in C:F\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\right\}$
for all $r>0$ and $x\in H$. Then the following hold:
1. (a)

${T}_{r}$ is single-valued;

2. (b)
${T}_{r}$ is firmly nonexpansive, i.e., for any $x,y\in H$,
${\parallel {T}_{r}x-{T}_{r}y\parallel }^{2}\le 〈{T}_{r}x-{T}_{r}y,x-y〉;$

3. (c)

$F\left({T}_{r}\right)=EP\left(F\right)$;

4. (d)

$EP\left(F\right)$ is closed and convex.

Lemma 2.2 

In a real Hilbert space, the following inequality holds:
${\parallel ax+\left(1-a\right)y\parallel }^{2}=a{\parallel x\parallel }^{2}+\left(1-a\right){\parallel y\parallel }^{2}-a\left(1-a\right){\parallel x-y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }a\in \left[0,1\right],x,y\in H.$

Lemma 2.3 

Let H be a real Hilbert space, let C be a nonempty closed convex subset of H and let $S:C\to C$ be an asymptotically κ-strict pseudocontraction with the sequence $\left\{{k}_{n}\right\}$. Then
1. (a)

$F\left(S\right)$ is closed and convex;

2. (b)

S is L-Lipschitz continuous.

Lemma 2.4 

Let H be a real Hilbert space, let C be a nonempty closed convex subset of H and let $S:C\to C$ be an asymptotically strict pseudocontraction. Then the mapping $I-S$ is demiclosed at zero, that is, if $\left\{{x}_{n}\right\}$ is a sequence in C such that ${x}_{n}⇀\overline{x}$ and ${x}_{n}-S{x}_{n}\to 0$, then $\overline{x}\in F\left(S\right)$.

## 3 Main results

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${F}_{m}$ be a bifunction from $C×C$ to which satisfies (A1)-(A4) and let ${A}_{m}:C\to H$ be a ${\xi }_{m}$-inverse-strongly monotone mapping for each $1\le m\le N$, where $N\ge 1$ is some positive integer. Let $S:C\to C$ be an asymptotically κ-strict pseudocontraction. Assume that $\mathcal{F}:=F\left(S\right)\cap {\bigcap }_{m=1}^{N}EP\left({F}_{m},{A}_{m}\right)$ is nonempty and bounded. Let $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ be sequences in $\left[0,1\right]$ and let $\left\{{r}_{n,m}\right\}$ be a positive sequence. Let $\left\{{\gamma }_{n,m}\right\}$ be a sequence in $\left[0,1\right]$ for each $1\le m\le N$ such that ${\sum }_{m=1}^{N}{\gamma }_{n,m}=1$. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
$\left\{\begin{array}{c}{x}_{1}\in H,\hfill \\ {C}_{1}=H,\hfill \\ {F}_{m}\left({u}_{n,m},u\right)+〈{A}_{m}{x}_{n},u-{u}_{n,m}〉+\frac{1}{{r}_{n,m}}〈u-{u}_{n,m},{u}_{n,m}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in C,\hfill \\ {y}_{n}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{\sum }_{m=1}^{N}{\gamma }_{n,m}{u}_{n,m}+\left(1-{\beta }_{n}\right){S}^{n}{\sum }_{m=1}^{N}{\gamma }_{n,m}{u}_{n,m}\right),\hfill \\ {C}_{n+1}=\left\{w\in {C}_{n}:{\parallel {y}_{n}-w\parallel }^{2}\le {\parallel {x}_{n}-w\parallel }^{2}+{\theta }_{n}\right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}n\ge 0,\hfill \end{array}$
where ${\theta }_{n}=\left(1-{\alpha }_{n}\right)\left({k}_{n}-1\right){\mathrm{\Theta }}_{n}$ and ${\mathrm{\Theta }}_{n}=sup\left\{{\parallel {x}_{n}-w\parallel }^{2}:w\in \mathcal{F}\right\}$. Assume that the control sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n,1}\right\},\dots ,\left\{{\gamma }_{n,N}\right\}$, $\left\{{r}_{n,1}\right\},\dots$ , and $\left\{{r}_{n,N}\right\}$ satisfy the following restrictions:
1. (a)

$0\le {\alpha }_{n}\le a<1$, $\kappa \le {\beta }_{n}\le b<1$;

2. (b)

$0 and $0 for each $1\le m\le N$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to some point $\overline{x}$, where $\overline{x}={P}_{\mathcal{F}}{x}_{1}$.

Proof First, we show that ${C}_{n}$ is closed and convex for each $n\ge 1$. It is easy to see that ${C}_{n}$ is closed for each $n\ge 1$. We only show that ${C}_{n}$ is convex for each $n\ge 1$. Note that ${C}_{1}=H$ is convex. Suppose that ${C}_{i}$ is convex for some positive integer i. Next, we show that ${C}_{i+1}$ is convex for the same i. Note that
${\parallel {y}_{i}-w\parallel }^{2}\le {\parallel {x}_{i}-w\parallel }^{2}+{\theta }_{i}$
is equivalent to
$2〈{x}_{i}-{y}_{i},w〉\le {\parallel {x}_{i}\parallel }^{2}-{\parallel {y}_{i}\parallel }^{2}+{\theta }_{i}.$
(3.1)
Take ${w}_{1}$ and ${w}_{2}$ in ${C}_{i+1}$ and put $\overline{w}=t{w}_{1}+\left(1-t\right){w}_{2}$. It follows that ${w}_{1}\in {C}_{i}$, ${w}_{2}\in {C}_{i}$,
$2〈{x}_{i}-{y}_{i},{w}_{1}〉\le {\parallel {x}_{i}\parallel }^{2}-{\parallel {y}_{i}\parallel }^{2}+{\theta }_{i}$
(3.2)
and
$2〈{x}_{i}-{y}_{i},{w}_{2}〉\le {\parallel {x}_{i}\parallel }^{2}-{\parallel {y}_{i}\parallel }^{2}+{\theta }_{i}.$
(3.3)
Combining (3.2) with (3.3), we can obtain that $2〈{x}_{i}-{y}_{i},\overline{w}〉\le {\parallel {x}_{i}\parallel }^{2}-{\parallel {y}_{i}\parallel }^{2}+{\theta }_{i}$, that is, ${\parallel {y}_{i}-w\parallel }^{2}\le {\parallel {x}_{i}-\overline{x}\parallel }^{2}+{\theta }_{i}$. In view of the convexity of ${C}_{i}$, we see that $\overline{w}\in {C}_{i}$. This shows that $\overline{w}\in {C}_{i+1}$. This concludes that ${C}_{n}$ is closed and convex for each $n\ge 1$. Notice that $\left(I-{r}_{n,m}{A}_{m}\right)$ is nonexpansive. Indeed, for any $x,y\in C$, we find from the restriction (b) that
$\begin{array}{c}{\parallel \left(I-{r}_{n,m}{A}_{m}\right)x-\left(I-{r}_{n,m}{A}_{m}\right)y\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel \left(x-y\right)-{r}_{n,m}\left({A}_{m}x-{A}_{m}y\right)\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel x-y\parallel }^{2}-2{r}_{n,m}〈x-y,{A}_{m}x-{A}_{m}y〉+{r}_{n,m}^{2}{\parallel {A}_{m}x-{A}_{m}y\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel x-y\parallel }^{2}-2{r}_{n,m}{\xi }_{m}{\parallel {A}_{m}x-{A}_{m}y\parallel }^{2}+{r}_{n,m}^{2}{\parallel {A}_{m}x-{A}_{m}y\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel x-y\parallel }^{2}+{r}_{n,m}\left({r}_{n,m}-2{\xi }_{m}\right){\parallel {A}_{m}x-{A}_{m}y\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel x-y\parallel }^{2}.\hfill \end{array}$
It follows that
$\begin{array}{rcl}{\parallel {u}_{n,m}-p\parallel }^{2}& =& {\parallel {T}_{{r}_{n,m}}\left({x}_{n}-{r}_{n,m}{A}_{m}{x}_{n}\right)-{T}_{{r}_{n,m}}\left(p-{r}_{n,m}{A}_{m}p\right)\parallel }^{2}\\ \le & {\parallel \left({x}_{n}-{r}_{n,m}{A}_{m}{x}_{n}\right)-\left(p-{r}_{n,m}{A}_{m}p\right)\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}.\end{array}$
(3.4)
Put ${z}_{n}={\sum }_{m=1}^{N}{\gamma }_{n,m}{u}_{n,m}$ for each $n\ge 1$. Next, we show that $\mathcal{F}\subset {C}_{n}$ for all $n\ge 1$. It is easy to see that $\mathcal{F}\subset {C}_{1}=H$. Suppose that $\mathcal{F}\subset {C}_{h}$ for some integer $h\ge 1$. We intend to claim that $\mathcal{F}\subset {C}_{h+1}$ for the same h. For any $p\in \mathcal{F}\subset {C}_{h}$, we have from Lemma 2.2 and the restriction (a) that
$\begin{array}{c}{\parallel {y}_{h}-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel {\alpha }_{h}{x}_{h}+\left(1-{\alpha }_{h}\right)\left({\beta }_{h}{z}_{h}+\left(1-{\beta }_{h}\right){S}^{h}{z}_{h}\right)-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{h}{\parallel {x}_{h}-p\parallel }^{2}+\left(1-{\alpha }_{h}\right){\parallel {\beta }_{h}\left({z}_{h}-p\right)+\left(1-{\beta }_{h}\right)\left({S}^{h}{z}_{h}-{S}^{h}p\right)\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\alpha }_{h}{\parallel {x}_{h}-p\parallel }^{2}+\left(1-{\alpha }_{h}\right)\left({\beta }_{h}{\parallel {z}_{h}-p\parallel }^{2}+\left(1-{\beta }_{h}\right){\parallel {S}^{h}{z}_{h}-{S}^{h}p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-{\beta }_{h}\left(1-{\beta }_{h}\right){\parallel {z}_{h}-p-\left({S}^{h}{z}_{h}-{S}^{h}p\right)\parallel }^{2}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{h}{\parallel {x}_{h}-p\parallel }^{2}+\left(1-{\alpha }_{h}\right){\beta }_{h}{\parallel {z}_{h}-p\parallel }^{2}+\left(1-{\alpha }_{h}\right)\left(1-{\beta }_{h}\right)\left({k}_{h}{\parallel {z}_{h}-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\kappa {\parallel {z}_{h}-p-\left({S}^{h}{z}_{h}-{S}^{h}p\right)\parallel }^{2}\right)-\left(1-{\alpha }_{h}\right){\beta }_{h}\left(1-{\beta }_{h}\right){\parallel {z}_{h}-p-\left({S}^{h}{z}_{h}-{S}^{h}p\right)\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{h}{\parallel {x}_{h}-p\parallel }^{2}+\left(1-{\alpha }_{h}\right){k}_{h}{\parallel {z}_{h}-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\left(1-{\alpha }_{h}\right)\left(1-{\beta }_{h}\right)\left({\beta }_{h}-\kappa \right){\parallel {z}_{h}-p-\left({S}^{h}{z}_{h}-{S}^{h}p\right)\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{h}{\parallel {x}_{h}-p\parallel }^{2}+\left(1-{\alpha }_{h}\right){k}_{h}{\parallel {z}_{h}-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{h}{\parallel {x}_{h}-p\parallel }^{2}+\left(1-{\alpha }_{h}\right){k}_{h}\sum _{m=1}^{N}{\gamma }_{h,m}{\parallel {T}_{{r}_{h,m}}\left({x}_{n}-{r}_{h,m}{A}_{m}{x}_{h}\right)-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{h}{\parallel {x}_{h}-p\parallel }^{2}+\left(1-{\alpha }_{h}\right){k}_{h}{\parallel {x}_{h}-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{h}-p\parallel }^{2}+{\theta }_{h}.\hfill \end{array}$
This shows that $p\in {C}_{h+1}$. This proves that $\mathcal{F}\subset {C}_{n}$ for all $n\ge 1$. Since ${x}_{n}={P}_{{C}_{n}}{x}_{1}$ and ${x}_{n+1}={P}_{{C}_{n+1}}{x}_{1}\in {C}_{n+1}\subset {C}_{n}$, we have that
$\begin{array}{rcl}0& \le & 〈{x}_{1}-{x}_{n},{x}_{n}-{x}_{n+1}〉\\ =& 〈{x}_{1}-{x}_{n},{x}_{n}-{x}_{1}+{x}_{1}-{x}_{n+1}〉\\ \le & -{\parallel {x}_{1}-{x}_{n}\parallel }^{2}+\parallel {x}_{1}-{x}_{n}\parallel \parallel {x}_{1}-{x}_{n+1}\parallel .\end{array}$
(3.5)
It follows that
$\parallel {x}_{1}-{x}_{n}\parallel \le \parallel {x}_{1}-{x}_{n+1}\parallel .$
(3.6)
On the other hand, for any $p\in \mathcal{F}\subset {C}_{n}$, we see that $\parallel {x}_{1}-{x}_{n}\parallel \le \parallel {x}_{1}-p\parallel$. In particular, we have
$\parallel {x}_{1}-{x}_{n}\parallel \le \parallel {x}_{1}-{P}_{\mathcal{F}}{x}_{1}\parallel .$
This shows that the sequence $\left\{{x}_{n}\right\}$ is bounded. In view of (3.6), we see that ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{x}_{1}\parallel$ exists. It follows from (3.5) that
$\begin{array}{c}{\parallel {x}_{n}-{x}_{n+1}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel {x}_{n}-{x}_{1}+{x}_{1}-{x}_{n+1}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel {x}_{n}-{x}_{1}\parallel }^{2}+2〈{x}_{n}-{x}_{1},{x}_{1}-{x}_{n+1}〉+{\parallel {x}_{1}-{x}_{n+1}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel {x}_{n}-{x}_{1}\parallel }^{2}+2〈{x}_{n}-{x}_{1},{x}_{1}-{x}_{n}+{x}_{n}-{x}_{n+1}〉+{\parallel {x}_{1}-{x}_{n+1}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel {x}_{n}-{x}_{1}\parallel }^{2}-2{\parallel {x}_{n}-{x}_{1}\parallel }^{2}+2〈{x}_{n}-{x}_{1},{x}_{n}-{x}_{n+1}〉+{\parallel {x}_{1}-{x}_{n+1}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{1}-{x}_{n+1}\parallel }^{2}-{\parallel {x}_{n}-{x}_{1}\parallel }^{2},\hfill \end{array}$
which implies that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{x}_{n+1}\parallel =0.$
(3.7)
In view of ${x}_{n+1}={P}_{{C}_{n+1}}{x}_{1}\in {C}_{n+1}$, we find that
${\parallel {y}_{n}-{x}_{n+1}\parallel }^{2}\le {\parallel {x}_{n}-{x}_{n+1}\parallel }^{2}+{\theta }_{n}.$
This combines with (3.7) yielding that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{x}_{n+1}\parallel =0.$
(3.8)
Notice that $\parallel {y}_{n}-{x}_{n}\parallel \le \parallel {y}_{n}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-{x}_{n}\parallel$. Combining (3.7) with (3.8), we find that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{y}_{n}\parallel =0.$
(3.9)
Note that
$\begin{array}{rcl}\parallel {x}_{n}-{y}_{n}\parallel & =& \parallel {x}_{n}-{\alpha }_{n}{x}_{n}-\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{z}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{z}_{n}\right)\parallel \\ =& \left(1-{\alpha }_{n}\right)\parallel {x}_{n}-\left({\beta }_{n}{z}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{z}_{n}\right)\parallel .\end{array}$
In view of the restriction (a), we obtain from (3.9) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-\left({\beta }_{n}{z}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{z}_{n}\right)\parallel =0.$
(3.10)
Notice that
$\begin{array}{rcl}{\parallel {u}_{n,m}-p\parallel }^{2}& =& {\parallel {T}_{{r}_{n,m}}\left(I-{r}_{n,m}{A}_{m}\right){x}_{n}-{T}_{{r}_{n,m}}\left(I-{r}_{n,m}{A}_{m}\right)p\parallel }^{2}\\ \le & 〈\left(I-{r}_{n,m}{A}_{m}\right){x}_{n}-\left(I-{r}_{n,m}{A}_{m}\right)p,{u}_{n,m}-p〉\\ =& \frac{1}{2}\left({\parallel \left(I-{r}_{n,m}{A}_{m}\right){x}_{n}-\left(I-{r}_{n,m}{A}_{m}\right)p\parallel }^{2}+{\parallel {u}_{n,m}-p\parallel }^{2}\\ -{\parallel \left(I-{r}_{n,m}{A}_{m}\right){x}_{n}-\left(I-{r}_{n,m}{A}_{m}\right)p-\left({u}_{n,m}-p\right)\parallel }^{2}\right)\\ \le & \frac{1}{2}\left({\parallel {x}_{n}-p\parallel }^{2}+{\parallel {u}_{n,m}-p\parallel }^{2}-{\parallel \left({x}_{n}-{u}_{n,m}\right)-{r}_{n,m}\left({A}_{m}{x}_{n}-{A}_{m}p\right)\parallel }^{2}\right)\\ =& \frac{1}{2}\left({\parallel {x}_{n}-p\parallel }^{2}+{\parallel {u}_{n,m}-p\parallel }^{2}-{\parallel {x}_{n}-{u}_{n,m}\parallel }^{2}\\ +2{r}_{n,m}〈{x}_{n}-{u}_{n,m},{A}_{m}{x}_{n}-{A}_{m}p〉-{r}_{n,m}^{2}{\parallel {A}_{m}{x}_{n}-{A}_{m}p\parallel }^{2}\right)\end{array}$
and hence
$\begin{array}{c}{\parallel {u}_{n,m}-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-p\parallel }^{2}-{\parallel {x}_{n}-{u}_{n,m}\parallel }^{2}+2{r}_{n,m}〈{x}_{n}-{u}_{n,m},{A}_{m}{x}_{n}-{A}_{m}p〉-{r}_{n,m}^{2}{\parallel {A}_{m}{x}_{n}-{A}_{m}p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-p\parallel }^{2}-{\parallel {x}_{n}-{u}_{n,m}\parallel }^{2}+2{r}_{n,m}\parallel {x}_{n}-{u}_{n,m}\parallel \parallel {A}_{m}{x}_{n}-{A}_{m}\parallel .\hfill \end{array}$
(3.11)
It follows from Lemma 2.2 and the restriction (a) that
$\begin{array}{c}{\parallel {y}_{n}-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel {\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{z}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{z}_{n}\right)-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}{\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\alpha }_{n}\right){\parallel {\beta }_{n}\left({z}_{n}-p\right)+\left(1-{\beta }_{n}\right)\left({S}^{n}{z}_{n}-{S}^{n}p\right)\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\alpha }_{h}{\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{\parallel {z}_{n}-p\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {S}^{n}{z}_{n}-{S}^{n}p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-{\beta }_{n}\left(1-{\beta }_{n}\right){\parallel {z}_{n}-p-\left({S}^{n}{z}_{n}-{S}^{n}p\right)\parallel }^{2}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}{\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\alpha }_{n}\right){\beta }_{n}{\parallel {z}_{n}-p\parallel }^{2}+\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right)\left({k}_{n}{\parallel {z}_{n}-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\kappa {\parallel {z}_{n}-p-\left({S}^{n}{z}_{n}-{S}^{n}p\right)\parallel }^{2}\right)-\left(1-{\alpha }_{n}\right){\beta }_{n}\left(1-{\beta }_{n}\right){\parallel {z}_{n}-p-\left({S}^{n}{z}_{n}-{S}^{n}p\right)\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}{\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\alpha }_{n}\right){k}_{n}{\parallel {z}_{n}-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right)\left({\beta }_{n}-\kappa \right){\parallel {z}_{n}-p-\left({S}^{n}{z}_{n}-{S}^{n}p\right)\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}{\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\alpha }_{n}\right){k}_{n}{\parallel {z}_{n}-p\parallel }^{2}.\hfill \end{array}$
(3.12)
It follows that
$\begin{array}{rcl}{\parallel {y}_{n}-p\parallel }^{2}& \le & \left(1-{\alpha }_{n}\right){k}_{n}\sum _{m=1}^{N}{\gamma }_{n,m}{\parallel {T}_{{r}_{n,m}}\left({x}_{n}-{r}_{n,m}{A}_{m}{x}_{n}\right)-{T}_{{r}_{n,m}}\left(p-{r}_{n,m}{A}_{m}p\right)\parallel }^{2}\\ +{\alpha }_{n}{\parallel {x}_{n}-p\parallel }^{2}\\ \le & {\alpha }_{n}{\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\alpha }_{n}\right){k}_{n}\sum _{m=1}^{N}{\gamma }_{n,m}{\parallel \left({x}_{n}-{r}_{n,m}{A}_{m}{x}_{n}\right)-\left(p-{r}_{n,m}{A}_{m}p\right)\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\alpha }_{n}\right)\left({k}_{n}-1\right){\parallel {x}_{n}-p\parallel }^{2}\\ -{r}_{n,m}\left(2{\xi }_{m}-{r}_{n,m}\right)\left(1-{\alpha }_{n}\right){k}_{n}\sum _{m=1}^{N}{\gamma }_{n,m}{\parallel {A}_{m}{x}_{n}-{A}_{m}p\parallel }^{2}.\end{array}$
This implies that
$\begin{array}{c}{r}_{n,m}\left(2{\xi }_{m}-{r}_{n,m}\right)\left(1-{\alpha }_{n}\right){k}_{n}{\gamma }_{n,m}{\parallel {A}_{m}{x}_{n}-{A}_{m}p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-p\parallel }^{2}-{\parallel {y}_{n}-p\parallel }^{2}+\left(1-{\alpha }_{n}\right)\left({k}_{n}-1\right){\parallel {x}_{n}-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(\parallel {x}_{n}-p\parallel +\parallel {y}_{n}-p\parallel \right)\parallel {x}_{n}-{y}_{n}\parallel +\left(1-{\alpha }_{n}\right)\left({k}_{n}-1\right){\parallel {x}_{n}-p\parallel }^{2}.\hfill \end{array}$
In view of the restrictions (a) and (b), we find from (3.9) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {A}_{m}{x}_{n}-{A}_{m}p\parallel =0.$
(3.13)
On the other hand, we find from (3.11) and (3.12) that
$\begin{array}{rcl}{\parallel {y}_{n}-p\parallel }^{2}& \le & {\alpha }_{n}{\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\alpha }_{n}\right){k}_{n}\sum _{m=1}^{N}{\gamma }_{n,m}{\parallel {u}_{n,m}-p\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\alpha }_{n}\right)\left({k}_{n}-1\right){\parallel {x}_{n}-p\parallel }^{2}-\left(1-{\alpha }_{n}\right){k}_{n}\sum _{m=1}^{N}{\gamma }_{n,m}{\parallel {x}_{n}-{u}_{n,m}\parallel }^{2}\\ +2{r}_{n,m}\left(1-{\alpha }_{n}\right){k}_{n}\sum _{m=1}^{N}{\gamma }_{n,m}\parallel {x}_{n}-{u}_{n,m}\parallel \parallel {A}_{m}{x}_{n}-{A}_{m}p\parallel .\end{array}$
This implies that
$\begin{array}{c}\left(1-{\alpha }_{n}\right){k}_{n}{\gamma }_{n,m}{\parallel {u}_{n,m}-{x}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-p\parallel }^{2}-{\parallel {y}_{n}-p\parallel }^{2}+\left(1-{\alpha }_{n}\right)\left({k}_{n}-1\right){\parallel {x}_{n}-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+2{r}_{n,m}\left(1-{\alpha }_{n}\right){k}_{n}\sum _{m=1}^{N}{\gamma }_{n,m}\parallel {x}_{n}-{u}_{n,m}\parallel \parallel {A}_{m}{x}_{n}-{A}_{m}p\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \left(\parallel {x}_{n}-p\parallel +\parallel {y}_{n}-p\parallel \right)\parallel {x}_{n}-{y}_{n}\parallel +\left(1-{\alpha }_{n}\right)\left({k}_{n}-1\right){\parallel {x}_{n}-p\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+2{r}_{n,m}\left(1-{\alpha }_{n}\right){k}_{n}\sum _{m=1}^{N}{\gamma }_{n,m}\parallel {x}_{n}-{u}_{n,m}\parallel \parallel {A}_{m}{x}_{n}-{A}_{m}p\parallel .\hfill \end{array}$
In view of the restrictions (a) and (b), we obtain from (3.9) and (3.13) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n,m}-{x}_{n}\parallel =0.$
(3.14)
Since $\left\{{x}_{n}\right\}$ is bounded, there exists a subsequence $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that ${x}_{{n}_{i}}⇀q$. Next, we show that $q\in F\left(S\right)$. Note that
$\parallel {z}_{n}-{x}_{n}\parallel =\parallel \sum _{m=1}^{N}{\gamma }_{n,m}{u}_{n,m}-{x}_{n}\parallel \le \sum _{m=1}^{N}{\gamma }_{n,m}\parallel {u}_{n,m}-{x}_{n}\parallel .$
It follows from (3.14) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{n}-{x}_{n}\parallel =0.$
(3.15)
On the other hand, we have
$\begin{array}{c}\parallel {x}_{n}-\left({\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{x}_{n}\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel \left({\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{x}_{n}\right)-\left({\beta }_{n}{z}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{z}_{n}\right)\parallel +\parallel \left({\beta }_{n}{z}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{z}_{n}\right)-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le {\beta }_{n}\parallel {x}_{n}-{z}_{n}\parallel +\left(1-{\beta }_{n}\right)\parallel {S}^{n}{x}_{n}-{S}^{n}{z}_{n}\parallel +\parallel \left({\beta }_{n}{z}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{z}_{n}\right)-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le {\beta }_{n}\parallel {x}_{n}-{z}_{n}\parallel +\left(1-{\beta }_{n}\right)L\parallel {x}_{n}-{z}_{n}\parallel +\parallel \left({\beta }_{n}{z}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{z}_{n}\right)-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le L\parallel {x}_{n}-{z}_{n}\parallel +\parallel \left({\beta }_{n}{z}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{z}_{n}\right)-{x}_{n}\parallel .\hfill \end{array}$
From (3.10) and (3.15), we find that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-\left({\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{x}_{n}\right)\parallel =0.$
(3.16)
Note that
$\begin{array}{rcl}\parallel {S}^{n}{x}_{n}-{x}_{n}\parallel & \le & \parallel {S}^{n}{x}_{n}-\left({\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{x}_{n}\right)\parallel +\parallel \left({\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{x}_{n}\right)-{x}_{n}\parallel \\ \le & {\beta }_{n}\parallel {S}^{n}{x}_{n}-{x}_{n}\parallel +\parallel \left({\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{x}_{n}\right)-{x}_{n}\parallel ,\end{array}$
which yields that
$\left(1-{\beta }_{n}\right)\parallel {S}^{n}{x}_{n}-{x}_{n}\parallel \le \parallel \left({\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{x}_{n}\right)-{x}_{n}\parallel .$
In view of the restriction (a), we find from (3.16) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {S}^{n}{x}_{n}-{x}_{n}\parallel =0.$
(3.17)
Note that
$\begin{array}{rcl}\parallel {x}_{n}-S{x}_{n}\parallel & \le & \parallel {x}_{n}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-{S}^{n+1}{x}_{n+1}\parallel \\ +\parallel {S}^{n+1}{x}_{n+1}-{S}^{n+1}{x}_{n}\parallel +\parallel {S}^{n+1}{x}_{n}-S{x}_{n}\parallel \\ \le & \left(1+L\right)\parallel {x}_{n}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-{S}^{n+1}{x}_{n+1}\parallel +L\parallel {S}^{n}{x}_{n}-{x}_{n}\parallel .\end{array}$
It follows from (3.7) and (3.17) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel S{x}_{n}-{x}_{n}\parallel =0.$
With the aid of Lemma 2.4, we find that $q\in F\left(S\right)$. Next, we prove $q\in {\bigcap }_{m=1}^{N}EP\left({F}_{m}\right)$. In view of (3.14), we find that ${u}_{{n}_{i},m}⇀q$ for each $1\le m\le N$. From (3.14) and the restriction (b), we see that
$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel {u}_{n,m}-{x}_{n}\parallel }{{r}_{n,m}}=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }1\le m\le N.$
(3.18)
Notice that
${F}_{m}\left({u}_{n,m},u\right)+〈{A}_{m}{x}_{n},u-{u}_{n,m}〉+\frac{1}{{r}_{n,m}}〈u-{u}_{n,m},{u}_{n,m}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in C.$
From (A2), we see that
$〈{A}_{m}{x}_{n},u-{u}_{n,m}〉+\frac{1}{{r}_{n,m}}〈u-{u}_{n,m},{u}_{n,m}-{x}_{n}〉\ge {F}_{m}\left(u,{u}_{n,m}\right).$
Replacing n by ${n}_{i}$, we arrive at
$〈{A}_{m}{x}_{{n}_{i}},u-{u}_{{n}_{i},m}〉+〈u-{u}_{{n}_{i},m},\frac{{u}_{{n}_{i},m}-{x}_{{n}_{i}}}{{r}_{{n}_{i},m}}〉\ge {F}_{m}\left(u,{u}_{{n}_{i},m}\right).$
(3.19)
For any t with $0 and $u\in C$, let ${\rho }_{t}=tu+\left(1-t\right)q$. Since $u\in C$ and $q\in C$, we have ${\rho }_{t}\in C$. It follows from (3.19) that
$\begin{array}{c}〈{\rho }_{t}-{u}_{{n}_{i},m},{A}_{m}{\rho }_{t}〉\hfill \\ \phantom{\rule{1em}{0ex}}\ge 〈{\rho }_{t}-{u}_{{n}_{i},m},{A}_{m}{\rho }_{t}〉-〈{A}_{m}{x}_{{n}_{i}},{\rho }_{t}-{u}_{{n}_{i},m}〉-〈{\rho }_{t}-{u}_{{n}_{i},m},\frac{{u}_{{n}_{i},m}-{x}_{{n}_{i}}}{{r}_{{n}_{i},m}}〉+{F}_{m}\left({\rho }_{t},{u}_{{n}_{i},m}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=〈{\rho }_{t}-{u}_{{n}_{i},m},{A}_{m}{\rho }_{t}-{A}_{m}{u}_{n,i,m}〉+〈{\rho }_{t}-{u}_{{n}_{i},m},{A}_{m}{u}_{n,i,m}-{A}_{m}{x}_{{n}_{i}}〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-〈{\rho }_{t}-{u}_{{n}_{i},m},\frac{{u}_{{n}_{i},m}-{x}_{{n}_{i}}}{{r}_{{n}_{i},m}}〉+F\left({\rho }_{t},{u}_{{n}_{i},m}\right).\hfill \end{array}$
(3.20)
Since ${A}_{m}$ is Lipschitz continuous, we find from (3.14) that ${A}_{m}{u}_{n,i,m}-{A}_{m}{x}_{{n}_{i}}\to 0$ as $i\to \mathrm{\infty }$. From the monotonicity of ${A}_{m}$, we get that
$〈{\rho }_{t}-{u}_{{n}_{i},m},{A}_{m}{\rho }_{t}-{A}_{m}{u}_{n,i,m}〉\ge 0.$
In view of (A4), we find from (3.20) that
$〈{\rho }_{t}-q,{A}_{m}{\rho }_{t}〉\ge F\left({\rho }_{t},q\right).$
(3.21)
With the aid of (A1), (A4), we obtain from (3.21) that
$\begin{array}{rcl}0& =& F\left({\rho }_{t},{\rho }_{t}\right)\le tF\left({\rho }_{t},u\right)+\left(1-t\right)F\left({\rho }_{t},q\right)\\ \le & tF\left({\rho }_{t},u\right)+\left(1-t\right)〈{\rho }_{t}-q,{A}_{m}{\rho }_{t}〉\\ =& tF\left({\rho }_{t},u\right)+\left(1-t\right)t〈u-q,{A}_{m}{\rho }_{t}〉,\end{array}$
which implies that
$F\left({\rho }_{t},u\right)+\left(1-t\right)〈u-q,{A}_{m}{\rho }_{t}〉\ge 0.$
Letting $t\to 0$ in the above inequality, we arrive at
$F\left(q,u\right)+〈u-q,{A}_{m}q〉\ge 0.$
This shows that $q\in EP\left({F}_{m},{A}_{m}\right)$, $\mathrm{\forall }1\le m\le N$. This completes the proof that $q\in \mathcal{F}$. Put $\overline{x}={P}_{\mathcal{F}}{x}_{1}$, we obtain that
$\begin{array}{rcl}\parallel {x}_{1}-\overline{x}\parallel & \le & \parallel {x}_{1}-q\parallel \le \underset{i\to \mathrm{\infty }}{lim inf}\parallel {x}_{1}-{x}_{{n}_{i}}\parallel \\ \le & \underset{i\to \mathrm{\infty }}{lim sup}\parallel {x}_{1}-{x}_{{n}_{i}}\parallel \le \parallel {x}_{1}-\overline{x}\parallel ,\end{array}$
which yields that
$\underset{i\to \mathrm{\infty }}{lim}\parallel {x}_{1}-{x}_{{n}_{i}}\parallel =\parallel {x}_{1}-q\parallel =\parallel {x}_{1}-\overline{x}\parallel .$

It follows that $\left\{{x}_{{n}_{i}}\right\}$ converges strongly to $\overline{x}$. Therefore, we can conclude that the sequence $\left\{{x}_{n}\right\}$ converges strongly to $\overline{x}={P}_{\mathcal{F}}{x}_{1}$. This completes the proof. □

Based on Theorem 3.1, we have the following results.

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let F be a bifunction from $C×C$ to which satisfies (A1)-(A4) and let $A:C\to H$ be a ξ-inverse-strongly monotone mapping. Let $S:C\to C$ be an asymptotically κ-strict pseudocontraction. Assume that $\mathcal{F}:=F\left(S\right)\cap EP\left(F,A\right)$ is nonempty and bounded. Let $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ be sequences in $\left[0,1\right]$ and $\left\{{r}_{n}\right\}$ be a positive sequence. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
$\left\{\begin{array}{c}{x}_{1}\in H,\hfill \\ {C}_{1}=H,\hfill \\ F\left({u}_{n},u\right)+〈A{x}_{n},u-{u}_{n}〉+\frac{1}{{r}_{n}}〈u-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in C,\hfill \\ {y}_{n}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{u}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{u}_{n}\right),\hfill \\ {C}_{n+1}=\left\{w\in {C}_{n}:{\parallel {y}_{n}-w\parallel }^{2}\le {\parallel {x}_{n}-w\parallel }^{2}+{\theta }_{n}\right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}n\ge 0,\hfill \end{array}$
where ${\theta }_{n}=\left(1-{\alpha }_{n}\right)\left({k}_{n}-1\right){\mathrm{\Theta }}_{n}$ and ${\mathrm{\Theta }}_{n}=sup\left\{{\parallel {x}_{n}-w\parallel }^{2}:w\in \mathcal{F}\right\}$. Assume that the control sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{r}_{n}\right\}$ satisfy the following restrictions:
1. (a)

$0\le {\alpha }_{n}\le a<1$, $0<\kappa \le {\beta }_{n}\le b<1$;

2. (b)

$0.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to some point $\overline{x}$, where $\overline{x}={P}_{\mathcal{F}}{x}_{1}$.

Proof Putting ${F}_{m}\equiv F$, ${A}_{m}\equiv A$ and ${r}_{n,m}\equiv {r}_{n}$ in Theorem 3.1, we see that ${\sum }_{m=1}^{N}{\gamma }_{n,m}{u}_{n,m}\equiv {u}_{n}$. With the aid of Theorem 3.1, we can easily conclude the desired conclusion. □

If S is asymptotically nonexpansive, then Corollary 3.2 is reduced to the following.

Corollary 3.3 Let C be a nonempty closed convex subset of a real Hilbert space H. Let F be a bifunction from $C×C$ to which satisfies (A1)-(A4) and let $A:C\to H$ be a ξ-inverse-strongly monotone mapping. Let $S:C\to C$ be an asymptotically nonexpansive mapping. Assume that $\mathcal{F}:=F\left(S\right)\cap EP\left(F,A\right)$ is nonempty and bounded. Let $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ be sequences in $\left[0,1\right]$ and let $\left\{{r}_{n}\right\}$ be a positive sequence. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
$\left\{\begin{array}{c}{x}_{1}\in H,\hfill \\ {C}_{1}=H,\hfill \\ F\left({u}_{n},u\right)+〈A{x}_{n},u-{u}_{n}〉+\frac{1}{{r}_{n}}〈u-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in C,\hfill \\ {y}_{n}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{u}_{n}+\left(1-{\beta }_{n}\right){S}^{n}{u}_{n}\right),\hfill \\ {C}_{n+1}=\left\{w\in {C}_{n}:{\parallel {y}_{n}-w\parallel }^{2}\le {\parallel {x}_{n}-w\parallel }^{2}+{\theta }_{n}\right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}n\ge 0,\hfill \end{array}$
where ${\theta }_{n}=\left(1-{\alpha }_{n}\right)\left({k}_{n}-1\right){\mathrm{\Theta }}_{n}$ and ${\mathrm{\Theta }}_{n}=sup\left\{{\parallel {x}_{n}-w\parallel }^{2}:w\in \mathcal{F}\right\}$. Assume that the control sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{r}_{n}\right\}$ satisfy the following restrictions:
1. (a)

$0\le {\alpha }_{n}\le a<1$, $0\le {\beta }_{n}\le b<1$;

2. (b)

$0.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to some point $\overline{x}$, where $\overline{x}={P}_{\mathcal{F}}{x}_{1}$.

Putting ${\beta }_{n}\equiv 0$ in Corollary 3.3, we have the following.

Corollary 3.4 Let C be a nonempty closed convex subset of a real Hilbert space H. Let F be a bifunction from $C×C$ to which satisfies (A1)-(A4) and let $A:C\to H$ be a ξ-inverse-strongly monotone mapping. Let $S:C\to C$ be an asymptotically nonexpansive mapping. Assume that $\mathcal{F}:=F\left(S\right)\cap EP\left(F,A\right)$ is nonempty and bounded. Let $\left\{{\alpha }_{n}\right\}$ be a sequence in $\left[0,1\right]$ and $\left\{{r}_{n}\right\}$ be a positive sequence. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
$\left\{\begin{array}{c}{x}_{1}\in H,\hfill \\ {C}_{1}=H,\hfill \\ F\left({u}_{n},u\right)+〈A{x}_{n},u-{u}_{n}〉+\frac{1}{{r}_{n}}〈u-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in C,\hfill \\ {y}_{n}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right){S}^{n}{u}_{n},\hfill \\ {C}_{n+1}=\left\{w\in {C}_{n}:{\parallel {y}_{n}-w\parallel }^{2}\le {\parallel {x}_{n}-w\parallel }^{2}+{\theta }_{n}\right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}n\ge 0,\hfill \end{array}$
where ${\theta }_{n}=\left(1-{\alpha }_{n}\right)\left({k}_{n}-1\right){\mathrm{\Theta }}_{n}$ and ${\mathrm{\Theta }}_{n}=sup\left\{{\parallel {x}_{n}-w\parallel }^{2}:w\in \mathcal{F}\right\}$. Assume that the control sequences $\left\{{\alpha }_{n}\right\}$ and $\left\{{r}_{n}\right\}$ satisfy the following restrictions:
1. (a)

$0\le {\alpha }_{n}\le a<1$;

2. (b)

$0.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to some point $\overline{x}$, where $\overline{x}={P}_{\mathcal{F}}{x}_{1}$.

Corollary 3.5 Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${A}_{m}:C\to H$ be a ${\xi }_{m}$-inverse-strongly monotone mapping, for each $1\le m\le N$, where $N\ge 1$ is some positive integer. Let $S:C\to C$ be an asymptotically κ-strict pseudocontraction. Assume that $\mathcal{F}:=F\left(S\right)\cap {\bigcap }_{m=1}^{N}VI\left(C,{A}_{m}\right)$ is nonempty and bounded. Let $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ be sequences in $\left[0,1\right]$ and $\left\{{r}_{n,m}\right\}$ be a positive sequence. Let $\left\{{\gamma }_{n,m}\right\}$ be a sequence in $\left[0,1\right]$ for each $1\le m\le N$ such that ${\sum }_{m=1}^{N}{\gamma }_{n,m}=1$. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
$\left\{\begin{array}{c}{x}_{1}\in H,\hfill \\ {C}_{1}=H,\hfill \\ {u}_{n,m}={P}_{C}\left({x}_{n}-{r}_{n,m}{A}_{m}{x}_{n}\right),\hfill \\ {y}_{n}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{\sum }_{m=1}^{N}{\gamma }_{n,m}{u}_{n,m}+\left(1-{\beta }_{n}\right){S}^{n}{\sum }_{m=1}^{N}{\gamma }_{n,m}{u}_{n,m}\right),\hfill \\ {C}_{n+1}=\left\{w\in {C}_{n}:{\parallel {y}_{n}-w\parallel }^{2}\le {\parallel {x}_{n}-w\parallel }^{2}+{\theta }_{n}\right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}n\ge 0,\hfill \end{array}$
where ${\theta }_{n}=\left(1-{\alpha }_{n}\right)\left({k}_{n}-1\right){\mathrm{\Theta }}_{n}$ and ${\mathrm{\Theta }}_{n}=sup\left\{{\parallel {x}_{n}-w\parallel }^{2}:w\in \mathcal{F}\right\}$. Assume that the control sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n,1}\right\},\dots ,\left\{{\gamma }_{n,N}\right\}$, $\left\{{r}_{n,1}\right\},\dots$ and $\left\{{r}_{n,N}\right\}$ satisfy the following restrictions:
1. (a)

$0\le {\alpha }_{n}\le a<1$, $\kappa \le {\beta }_{n}\le b<1$;

2. (b)

$0 and $0 for each $1\le m\le N$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to some point $\overline{x}$, where $\overline{x}={P}_{\mathcal{F}}{x}_{1}$.

Proof In Theorem 3.1, put ${F}_{m}\left(x,y\right)=0$ for all $x,y\in C$. From
$〈{A}_{m}{x}_{n},u-{u}_{n,m}〉+\frac{1}{{r}_{n,m}}〈u-{u}_{n,m},{u}_{n,m}-{x}_{n}〉,$
we have
$〈u-{u}_{n,m},{x}_{n}-{u}_{n,m}-{r}_{n,m}{A}_{m}{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in C.$
This implies that
${u}_{n,m}={P}_{C}\left({x}_{n}-{r}_{n,m}{A}_{m}{x}_{n}\right).$

In view of Theorem 3.1, we can immediately obtain the desired conclusion. This completes the proof. □

Corollary 3.6 Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${F}_{m}$ be a bifunction from $C×C$ to which satisfies (A1)-(A4) for each $1\le m\le N$, where $N\ge 1$ is some positive integer. Let $S:C\to C$ be an asymptotically κ-strict pseudocontraction. Assume that $\mathcal{F}:=F\left(S\right)\cap {\bigcap }_{m=1}^{N}EP\left({F}_{m}\right)$ is nonempty and bounded. Let $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ be sequences in $\left[0,1\right]$ and $\left\{{r}_{n,m}\right\}$ be a positive sequence. Let $\left\{{\gamma }_{n,m}\right\}$ be a sequence in $\left[0,1\right]$ for each $1\le m\le N$ such that ${\sum }_{m=1}^{N}{\gamma }_{n,m}=1$. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
$\left\{\begin{array}{c}{x}_{1}\in H,\hfill \\ {C}_{1}=H,\hfill \\ {F}_{m}\left({u}_{n,m},u\right)+\frac{1}{{r}_{n,m}}〈u-{u}_{n,m},{u}_{n,m}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in C,\hfill \\ {y}_{n}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{\sum }_{m=1}^{N}{\gamma }_{n,m}{u}_{n,m}+\left(1-{\beta }_{n}\right){S}^{n}{\sum }_{m=1}^{N}{\gamma }_{n,m}{u}_{n,m}\right),\hfill \\ {C}_{n+1}=\left\{w\in {C}_{n}:{\parallel {y}_{n}-w\parallel }^{2}\le {\parallel {x}_{n}-w\parallel }^{2}+{\theta }_{n}\right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}{x}_{1},\phantom{\rule{1em}{0ex}}n\ge 0,\hfill \end{array}$
where ${\theta }_{n}=\left(1-{\alpha }_{n}\right)\left({k}_{n}-1\right){\mathrm{\Theta }}_{n}$ and ${\mathrm{\Theta }}_{n}=sup\left\{{\parallel {x}_{n}-w\parallel }^{2}:w\in \mathcal{F}\right\}$. Assume that the control sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n,1}\right\},\dots ,\left\{{\gamma }_{n,N}\right\}$, $\left\{{r}_{n,1}\right\},\dots$ and $\left\{{r}_{n,N}\right\}$ satisfy the following restrictions:
1. (a)

$0\le {\alpha }_{n}\le a<1$, $\kappa \le {\beta }_{n}\le b<1$;

2. (b)

$0 and $0 for each $1\le m\le N$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to some point $\overline{x}$, where $\overline{x}={P}_{\mathcal{F}}{x}_{1}$.

Proof In Theorem 3.1, put ${A}_{m}=0$. Then, for any ${\xi }_{m}>0$, we see that
$〈x-y,{A}_{m}x-{A}_{m}y〉\ge {\xi }_{m}{\parallel {A}_{m}x-{A}_{m}y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

Let $\left\{{r}_{n,m}\right\}$ be a sequence satisfying the restriction $d\le {r}_{n}\le e$, where $d,e\in \left(0,\mathrm{\infty }\right)$. Then we can obtain the desired conclusion easily from Theorem 3.1. This completes the proof. □

## Declarations

### Acknowledgements

The authors are grateful to the referees for their valuable comments and suggestions which improved the contents of the article.

## Authors’ Affiliations

(1)
School of Mathematics and Information Science, North China University of Water Resources and Electric Power, Zhengzhou, China
(2)
College of Science, Hebei University of Engineering, Handan, 056038, China

## References

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