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# Hybrid methods for common solutions in Hilbert spaces with applications

Journal of Inequalities and Applications20142014:183

https://doi.org/10.1186/1029-242X-2014-183

• Received: 26 February 2014
• Accepted: 29 April 2014
• Published:

## Abstract

In this paper, hybrid methods are investigated for treating common solutions of nonlinear problems. A strong convergence theorem is established in the framework of real Hilbert spaces.

## Keywords

• equilibrium problem
• fixed point
• projection
• variational inequality
• zero point

## 1 Introduction and preliminaries

Common solutions to variational inclusion, equilibrium and fixed point problems have been recently extensively investigated based on iterative methods; see  and the references therein. The motivation for this subject is mainly to its possible applications to mathematical modeling of concrete complex problems, which use more than one constraint. The aim of this paper is to investigate a common solution of variational inclusion, equilibrium and fixed point problems. The organization of this paper is as follows. In Section 1, we provide some necessary preliminaries. In Section 2, a hybrid method is introduced and analyzed. Strong convergence theorems are established in the framework of Hilbert spaces. In Section 3, applications of the main results are discussed.

In what follows, 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, and convex subset of H and let ${P}_{C}$ be the metric projection from H onto C. Let $S:C\to C$ be a mapping. $F\left(S\right)$ stands for the fixed point set of S; that is, $F\left(S\right):=\left\{x\in C:x=Sx\right\}$.

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

If $\alpha =1$, then S is said to be nonexpansive. Let $A:C\to H$ be a mapping. If C is nonempty closed and convex, then the fixed point set of S is nonempty.

Recall that A is said to be monotone iff
$〈Ax-Ay,x-y〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
Recall that 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 also said to be α-strongly monotone. Recall that 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 also said to be α-inverse-strongly monotone.

Recall that a set-valued mapping $M:H⇉H$ is said to be monotone iff, for all $x,y\in H$, $f\in Mx$, and $g\in My$ imply $〈x-y,f-g〉>0$. M is maximal iff the graph $Graph\left(M\right)$ of R is not properly contained in the graph of any other monotone mapping. It is well known that a monotone mapping M is maximal if and only if, for any $\left(x,f\right)\in H×H$, $〈x-y,f-g〉\ge 0$, for all $\left(y,g\right)\in Graph\left(M\right)$ implies $f\in Rx$. For a maximal monotone operator M on H, and $r>0$, we may define the single-valued resolvent ${J}_{r}:H\to D\left(M\right)$, where $D\left(M\right)$ denote the domain of M. It is well known that ${J}_{r}$ is firmly nonexpansive, and ${M}^{-1}\left(0\right)=F\left({J}_{r}\right)$, where $F\left({J}_{r}\right):=\left\{x\in D\left(M\right):x={J}_{r}x\right\}$, and ${M}^{-1}\left(0\right):=\left\{x\in H:0\in Mx\right\}$.

Let $A:C\to H$ be a inverse-strongly monotone mapping, and let F be a bifunction of $C×C$ into , where denotes the set of real numbers. We consider the following generalized equilibrium problem.
(1.1)

In this paper, the set of such an $x\in C$ is denoted by $EP\left(F,A\right)$.

To study the equilibrium problems (1.1), 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 weakly lower semicontinuous.

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

Lemma 1.1 

Assume that $\left\{{\alpha }_{n}\right\}$ is a sequence of nonnegative real numbers such that
${\alpha }_{n+1}\le \left(1-{\gamma }_{n}\right){\alpha }_{n}+{\delta }_{n},$
where $\left\{{\gamma }_{n}\right\}$ is a sequence in $\left(0,1\right)$ and $\left\{{\delta }_{n}\right\}$ is a sequence such that
1. (1)

${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_{n}=\mathrm{\infty }$;

2. (2)

${lim sup}_{n\to \mathrm{\infty }}{\delta }_{n}/{\gamma }_{n}\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n}|<\mathrm{\infty }$.

Then ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$.

Lemma 1.2 

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
$F\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C.$
Define a mapping ${T}_{r}:H\to C$ as follows:
${T}_{r}x=\left\{z\in C:F\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\right\},\phantom{\rule{1em}{0ex}}x\in H,$
then the following conclusions hold:
1. (1)

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

2. (2)
${T}_{r}$ 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. (3)

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

4. (4)

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

Let $\left\{{S}_{i}:C\to C\right\}$ be a family of infinitely nonexpansive mappings and $\left\{{\gamma }_{i}\right\}$ be a nonnegative real sequence with $0\le {\gamma }_{i}<1$, $\mathrm{\forall }i\ge 1$. For $n\ge 1$ define a mapping ${W}_{n}:C\to C$ as follows:
$\begin{array}{r}{U}_{n,n+1}=I,\\ {U}_{n,n}={\gamma }_{n}{S}_{n}{U}_{n,n+1}+\left(1-{\gamma }_{n}\right)I,\\ {U}_{n,n-1}={\gamma }_{n-1}{S}_{n-1}{U}_{n,n}+\left(1-{\gamma }_{n-1}\right)I,\\ ⋮\\ {U}_{n,k}={\gamma }_{k}{S}_{k}{U}_{n,k+1}+\left(1-{\gamma }_{k}\right)I,\\ {U}_{n,k-1}={\gamma }_{k-1}{S}_{k-1}{U}_{n,k}+\left(1-{\gamma }_{k-1}\right)I,\\ ⋮\\ {U}_{n,2}={\gamma }_{2}{S}_{2}{U}_{n,3}+\left(1-{\gamma }_{2}\right)I,\\ {W}_{n}={U}_{n,1}={\gamma }_{1}{S}_{1}{U}_{n,2}+\left(1-{\gamma }_{1}\right)I.\end{array}$
(1.2)

Such a mapping ${W}_{n}$ is nonexpansive from C to C and it is called a W-mapping generated by ${S}_{n},{S}_{n-1},\dots ,{S}_{1}$ and ${\gamma }_{n},{\gamma }_{n-1},\dots ,{\gamma }_{1}$.

Lemma 1.3 

Let $\left\{{S}_{i}:C\to C\right\}$ be a family of infinitely nonexpansive mappings with a nonempty common fixed point set and let $\left\{{\gamma }_{i}\right\}$ be a real sequence such that $0<{\gamma }_{i}\le l<1$, where l is some real number, $\mathrm{\forall }i\ge 1$. Then
1. (1)

${W}_{n}$ is nonexpansive and $F\left({W}_{n}\right)={\bigcap }_{i=1}^{\mathrm{\infty }}F\left({S}_{i}\right)$, for each $n\ge 1$;

2. (2)

for each $x\in C$ and for each positive integer k, the limit ${lim}_{n\to \mathrm{\infty }}{U}_{n,k}$ exists;

3. (3)
the mapping $W:C\to C$ defined by
$Wx:=\underset{n\to \mathrm{\infty }}{lim}{W}_{n}x=\underset{n\to \mathrm{\infty }}{lim}{U}_{n,1}x,\phantom{\rule{1em}{0ex}}x\in C,$
(1.3)

is a nonexpansive mapping satisfying $F\left(W\right)={\bigcap }_{i=1}^{\mathrm{\infty }}F\left({S}_{i}\right)$ and it is called the W-mapping generated by ${S}_{1},{S}_{2},\dots$ and ${\gamma }_{1},{\gamma }_{2},\dots$ .

Lemma 1.4 

Let $\left\{{S}_{i}:C\to C\right\}$ be a family of infinitely nonexpansive mappings with a nonempty common fixed point set and let $\left\{{\gamma }_{i}\right\}$ be a real sequence such that $0<{\gamma }_{i}\le l<1$, $\mathrm{\forall }i\ge 1$. If K is any bounded subset of C, then
$\underset{n\to \mathrm{\infty }}{lim}\underset{x\in K}{sup}\parallel Wx-{W}_{n}x\parallel =0.$

Throughout this paper, we always assume that $0<{\gamma }_{i}\le l<1$, $\mathrm{\forall }i\ge 1$.

Lemma 1.5 

Let $B:C\to H$ be a mapping and let $M:H⇉H$ be a maximal monotone operator. Then $F\left({J}_{r}\left(I-sB\right)\right)={\left(B+M\right)}^{-1}\left(0\right)$.

Lemma 1.6 

Let $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ be bounded sequences in H and let $\left\{{\beta }_{n}\right\}$ be a sequence in $\left(0,1\right)$ with $0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$. Suppose that ${x}_{n+1}=\left(1-{\beta }_{n}\right){y}_{n}+{\beta }_{n}{x}_{n}$ for all $n\ge 0$ and
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {y}_{n+1}-{y}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0.$

Then ${lim}_{n\to \mathrm{\infty }}\parallel {y}_{n}-{x}_{n}\parallel =0$.

Lemma 1.7 

Let $A:C\to H$ a Lipschitz monotone mapping and let ${N}_{C}x$ be the normal cone to C at $x\in C$; that is, ${N}_{C}x=\left\{y\in H:〈x-u,y〉,\mathrm{\forall }u\in C\right\}$. Define
$Dx=\left\{\begin{array}{cc}Ax+{N}_{C}x,\hfill & x\in C,\hfill \\ \mathrm{\varnothing }\hfill & x\notin C.\hfill \end{array}$

Then D is maximal monotone and $0\in Dx$ if and only if $x\in VI\left(C,A\right)$.

## 2 Main results

Theorem 2.1 Let C be a nonempty closed convex subset of a Hilbert space H and F a bifunction from $C×C$ to which satisfies (A1)-(A4). Let ${A}_{1}:C\to H$ be a ${\delta }_{1}$-inverse-strongly monotone mapping, ${A}_{2}:C\to H$ be a ${\delta }_{2}$-inverse-strongly monotone mapping, ${A}_{3}:C\to H$ be a ${\delta }_{3}$-inverse-strongly monotone mapping, ${M}_{1}:H⇉H$ a maximal monotone operator such that $Dom\left({M}_{1}\right)\subset C$ and ${M}_{2}:H⇉H$ a maximal monotone operator such that $Dom\left({M}_{2}\right)\subset C$. Let $\left\{{S}_{i}:C\to C\right\}$ be a family of infinitely nonexpansive mappings. Assume that $\mathrm{\Omega }:={\bigcap }_{i=1}^{\mathrm{\infty }}F\left({S}_{i}\right)\cap EP\left(F,{A}_{3}\right)\cap {\left({A}_{1}+{M}_{1}\right)}^{-1}\left(0\right)\cap {\left({A}_{2}+{M}_{2}\right)}^{-1}\left(0\right)\ne \mathrm{\varnothing }$. Let ${x}_{1}\in C$ and $\left\{{x}_{n}\right\}$ be a sequence generated by
$\left\{\begin{array}{c}{z}_{n}={J}_{{s}_{n}}\left({u}_{n}-{s}_{n}{A}_{2}{u}_{n}\right),\hfill \\ {x}_{n+1}={\alpha }_{n}u+{\beta }_{n}{x}_{n}+{\gamma }_{n}{W}_{n}{J}_{{r}_{n}}\left({z}_{n}-{r}_{n}{A}_{1}{z}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
where u is a fixed element in C, ${u}_{n}$ is such that
$F\left({u}_{n},y\right)+〈{A}_{3}{x}_{n},y-{u}_{n}〉+\frac{1}{{\lambda }_{n}}〈y-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C,$
$\left\{{W}_{n}:C\to C\right\}$ is the sequence generated in (1.2), $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, and $\left\{{\gamma }_{n}\right\}$ are sequences in $\left(0,1\right)$ such that ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$ for each $n\ge 1$ and $\left\{{r}_{n}\right\}$, $\left\{{s}_{n}\right\}$, and $\left\{{\lambda }_{n}\right\}$ are positive number sequences. Assume that the above control sequences satisfy the following restrictions:
1. (a)

$0, $0<{a}^{\prime }\le {r}_{n}\le {b}^{\prime }<2{\delta }_{1}$, $0<\overline{a}\le {s}_{n}\le \overline{b}<2{\delta }_{2}$;

2. (b)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

3. (c)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

4. (d)

${lim}_{n\to \mathrm{\infty }}|{\lambda }_{n}-{\lambda }_{n+1}|={lim}_{n\to \mathrm{\infty }}|{s}_{n}-{s}_{n+1}|={lim}_{n\to \mathrm{\infty }}|{r}_{n}-{r}_{n+1}|=0$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to $\overline{x}\in \mathrm{\Omega }$, where $\overline{x}={P}_{\mathrm{\Omega }}u$.

Proof First, we show that the mapping $I-{r}_{n}{A}_{1}$, $I-{s}_{n}{A}_{2}$, and $I-{\lambda }_{n}{A}_{3}$ are nonexpansive. Indeed, we find from the restriction (a) that
$\begin{array}{r}{\parallel \left(I-{r}_{n}{A}_{1}\right)x-\left(I-{r}_{n}{A}_{1}\right)y\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel x-y\parallel }^{2}-2{r}_{n}〈x-y,{A}_{1}x-{A}_{1}y〉+{r}_{n}^{2}{\parallel {A}_{1}x-{A}_{1}y\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel x-y\parallel }^{2}-2{r}_{n}{\delta }_{1}{\parallel {A}_{1}x-{A}_{1}y\parallel }^{2}+{r}_{n}^{2}{\parallel {A}_{1}x-{A}_{1}y\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel x-y\parallel }^{2}+{r}_{n}\left({r}_{n}-2{\delta }_{1}\right){\parallel {A}_{1}x-{A}_{1}y\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel x-y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C,\end{array}$
which implies that the mapping $I-{r}_{n}{A}_{1}$ is nonexpansive. In the same way, we find $I-{s}_{n}{A}_{2}$ and $I-{\lambda }_{n}{A}_{3}$ are also nonexpansive. Put ${y}_{n}={J}_{{r}_{n}}\left({z}_{n}-{r}_{n}{A}_{1}{z}_{n}\right)$. Fixing ${x}^{\ast }\in \mathrm{\Omega }$, we find
$\begin{array}{rl}\parallel {y}_{n}-{x}^{\ast }\parallel & =\parallel {J}_{{r}_{n}}\left({z}_{n}-{r}_{n}{A}_{1}{z}_{n}\right)-{J}_{{r}_{n}}\left({x}^{\ast }-{r}_{n}{A}_{1}{x}^{\ast }\right)\parallel \\ \le \parallel {z}_{n}-{x}^{\ast }\parallel \\ =\parallel {J}_{{s}_{n}}\left({u}_{n}-{s}_{n}{A}_{2}{u}_{n}\right)-{J}_{{s}_{n}}\left({x}^{\ast }-{s}_{n}{A}_{2}{x}^{\ast }\right)\parallel \\ \le \parallel {T}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{A}_{3}\right){x}_{n}-{T}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{A}_{3}\right){x}^{\ast }\parallel \\ \le \parallel {x}_{n}-{x}^{\ast }\parallel .\end{array}$
It follows that
$\begin{array}{rl}\parallel {x}_{n+1}-{x}^{\ast }\parallel & =\parallel {\alpha }_{n}u+{\beta }_{n}{x}_{n}+{\gamma }_{n}{W}_{n}{y}_{n}-{x}^{\ast }\parallel \\ \le {\alpha }_{n}\parallel u-{x}^{\ast }\parallel +{\beta }_{n}\parallel {x}_{n}-{x}^{\ast }\parallel +{\gamma }_{n}\parallel {W}_{n}{y}_{n}-{x}^{\ast }\parallel \\ \le {\alpha }_{n}\parallel u-{x}^{\ast }\parallel +\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-{x}^{\ast }\parallel .\end{array}$
This implies that $\left\{{x}_{n}\right\}$ is bounded, and so are $\left\{{y}_{n}\right\}$, $\left\{{z}_{n}\right\}$, and $\left\{{u}_{n}\right\}$. Without loss of generality, we can assume that there exists a bounded set $K\subset C$ such that ${x}_{n},{y}_{n},{z}_{n},{u}_{n}\in K$. Notice that
$F\left({u}_{n+1},y\right)+\frac{1}{{\lambda }_{n+1}}〈y-{u}_{n+1},{u}_{n+1}-\left(I-{r}_{n+1}{A}_{3}\right){x}_{n+1}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C,$
(2.1)
and
$F\left({u}_{n},y\right)+\frac{1}{{\lambda }_{n}}〈y-{u}_{n},{u}_{n}-\left(I-{r}_{n}{A}_{3}\right){x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C.$
(2.2)
Let $y={u}_{n}$ in (2.1) and $y={u}_{n+1}$ in (2.2). By adding these two inequalities, we obtain
$〈{u}_{n+1}-{u}_{n},\frac{{u}_{n}-\left(I-{\lambda }_{n}{A}_{3}\right){x}_{n}}{{\lambda }_{n}}-\frac{{u}_{n+1}-\left(I-{\lambda }_{n+1}{A}_{3}\right){x}_{n+1}}{{\lambda }_{n+1}}〉\ge 0.$
It follows that
$\begin{array}{rl}{\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}\le & 〈{u}_{n+1}-{u}_{n},\left(I-{\lambda }_{n+1}{A}_{3}\right){x}_{n+1}-\left(I-{\lambda }_{n}{A}_{3}\right){x}_{n}\\ +\left(1-\frac{{\lambda }_{n}}{{\lambda }_{n+1}}\right)\left({u}_{n+1}-\left(I-{\lambda }_{n+1}{A}_{3}\right){x}_{n+1}\right)〉\\ \le & \parallel {u}_{n+1}-{u}_{n}\parallel \left(\parallel \left(I-{\lambda }_{n+1}{A}_{3}\right){x}_{n+1}-\left(I-{\lambda }_{n}{A}_{3}\right){x}_{n}\parallel \\ +|1-\frac{{\lambda }_{n}}{{\lambda }_{n+1}}|\parallel {u}_{n+1}-\left(I-{\lambda }_{n+1}{A}_{3}\right){x}_{n+1}\right)\parallel \right).\end{array}$
It follows that
$\begin{array}{rl}\parallel {u}_{n+1}-{u}_{n}\parallel \le & \parallel \left(I-{\lambda }_{n+1}{A}_{3}\right){x}_{n+1}-\left(I-{\lambda }_{n}{A}_{3}\right){x}_{n}\parallel \\ +\frac{|{\lambda }_{n+1}-{\lambda }_{n}|}{{\lambda }_{n+1}}\parallel {u}_{n+1}-\left(I-{\lambda }_{n+1}{A}_{3}\right){x}_{n+1}\parallel \\ =& \parallel \left(I-{\lambda }_{n+1}{A}_{3}\right){x}_{n+1}-\left(I-{\lambda }_{n+1}{A}_{3}\right){x}_{n}+\left(I-{\lambda }_{n+1}{A}_{3}\right){x}_{n}-\left(I-{\lambda }_{n}{A}_{3}\right){x}_{n}\parallel \\ +\frac{|{\lambda }_{n+1}-{\lambda }_{n}|}{{\lambda }_{n+1}}\parallel {u}_{n+1}-\left(I-{\lambda }_{n+1}{A}_{3}\right){x}_{n+1}\parallel \\ \le & \parallel {x}_{n+1}-{x}_{n}\parallel +|{\lambda }_{n+1}-{\lambda }_{n}|{M}_{1},\end{array}$
(2.3)
where ${M}_{1}$ is an appropriate constant such that
${M}_{1}=\underset{n\ge 1}{sup}\left\{\parallel {A}_{3}{x}_{n}\parallel +\frac{\parallel {u}_{n+1}-\left(I-{\lambda }_{n+1}{A}_{3}\right){x}_{n+1}\parallel }{a}\right\}.$
Since ${J}_{{s}_{n}}$ is firmly nonexpansive, we find that
$\begin{array}{r}\parallel {z}_{n+1}-{z}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {u}_{n+1}-{s}_{n+1}{A}_{2}{u}_{n+1}-\left({u}_{n}-{s}_{n}{A}_{2}{u}_{n}\right)\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel \left(I-{s}_{n+1}{A}_{2}\right){u}_{n+1}-\left(I-{s}_{n+1}{A}_{2}\right){u}_{n}+\left({s}_{n}-{s}_{n+1}\right){A}_{2}{u}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {u}_{n+1}-{u}_{n}\parallel +|{s}_{n}-{s}_{n+1}|\parallel {A}_{2}{u}_{n}\parallel .\end{array}$
(2.4)
Combining (2.3) with (2.4) yields
$\parallel {z}_{n+1}-{z}_{n}\parallel \le \parallel {x}_{n+1}-{x}_{n}\parallel +|{\lambda }_{n+1}-{\lambda }_{n}|{M}_{1}+|{s}_{n}-{s}_{n+1}|\parallel {A}_{2}{u}_{n}\parallel .$
(2.5)
Since ${J}_{{r}_{n}}$ is also firmly nonexpansive, we find that
$\parallel {y}_{n+1}-{y}_{n}\parallel \le \parallel {z}_{n+1}-{z}_{n}\parallel +|{r}_{n}-{r}_{n+1}|\parallel {A}_{1}{z}_{n}\parallel .$
(2.6)
Substituting (2.5) into (2.6), we see that
$\parallel {y}_{n+1}-{y}_{n}\parallel \le \parallel {x}_{n+1}-{x}_{n}\parallel +\left(|{r}_{n+1}-{r}_{n}|+|{\lambda }_{n}-{\lambda }_{n+1}|+|{s}_{n}-{s}_{n+1}|\right){M}_{2},$
(2.7)
where ${M}_{2}$ is an appropriate constant such that
${M}_{2}=max\left\{\underset{n\ge 1}{sup}\left\{\parallel {A}_{1}{z}_{n}\parallel \right\},\underset{n\ge 1}{sup}\left\{\parallel {A}_{2}{u}_{n}\parallel \right\},{M}_{1}\right\}.$
Since ${W}_{n}$ is nonexpansive, we find that
$\begin{array}{r}\parallel {W}_{n+1}{y}_{n+1}-{W}_{n}{y}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel {W}_{n+1}{y}_{n+1}-W{y}_{n+1}+W{y}_{n+1}-W{y}_{n}+W{y}_{n}-{W}_{n}{y}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {W}_{n+1}{y}_{n+1}-W{y}_{n+1}\parallel +\parallel W{y}_{n+1}-W{y}_{n}\parallel +\parallel W{y}_{n}-{W}_{n}{y}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le \underset{x\in K}{sup}\left\{\parallel {W}_{n+1}x-Wx\parallel +\parallel Wx-{W}_{n}x\parallel \right\}+\parallel {y}_{n+1}-{y}_{n}\parallel ,\end{array}$
(2.8)
where K is the bounded subset of C defined as above. Substituting (2.7) into (2.8), we find that
$\begin{array}{rl}\parallel {W}_{n+1}{y}_{n+1}-{W}_{n}{y}_{n}\parallel \le & \underset{x\in K}{sup}\left\{\parallel {W}_{n+1}x-Wx\parallel +\parallel Wx-{W}_{n}x\parallel \right\}+\parallel {x}_{n+1}-{x}_{n}\parallel \\ +\left(|{r}_{n+1}-{r}_{n}|+|{\lambda }_{n}-{\lambda }_{n+1}|+|{s}_{n}-{s}_{n+1}|\right){M}_{2}.\end{array}$
(2.9)
Letting
${x}_{n+1}=\left(1-{\beta }_{n}\right){v}_{n}+{\beta }_{n}{x}_{n},$
we see that
$\begin{array}{rl}{v}_{n+1}-{v}_{n}=& \frac{{\alpha }_{n+1}u+{\gamma }_{n+1}{W}_{n+1}{y}_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_{n}u+{\gamma }_{n}{W}_{n}{y}_{n}}{1-{\beta }_{n}}\\ =& \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}u+\frac{1-{\alpha }_{n+1}-{\beta }_{n+1}}{1-{\beta }_{n+1}}{W}_{n+1}{y}_{n+1}\\ -\left(\frac{{\alpha }_{n}}{1-{\beta }_{n}}u+\frac{1-{\alpha }_{n}-{\beta }_{n}}{1-{\beta }_{n}}{W}_{n}{y}_{n}\right)\\ =& \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\left(u-{W}_{n+1}{y}_{n+1}\right)-\frac{{\alpha }_{n}}{1-{\beta }_{n}}\left(u-{W}_{n}{y}_{n}\right)\\ +{W}_{n+1}{y}_{n+1}-{W}_{n}{y}_{n}.\end{array}$
Hence, we have
$\begin{array}{rl}\parallel {v}_{n+1}-{v}_{n}\parallel \le & \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel u-{W}_{n+1}{y}_{n+1}\parallel +\frac{{\alpha }_{n}}{1-{\beta }_{n}}\parallel u-{W}_{n}{y}_{n}\parallel \\ +\parallel {W}_{n+1}{y}_{n+1}-{W}_{n}{y}_{n}\parallel .\end{array}$
(2.10)
Substituting (2.9) into (2.10), we find that
$\begin{array}{rl}\parallel {v}_{n+1}-{v}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \le & \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel u-{W}_{n+1}{y}_{n+1}\parallel +\frac{{\alpha }_{n}}{1-{\beta }_{n}}\parallel u-{W}_{n}{y}_{n}\parallel \\ +\underset{x\in K}{sup}\left\{\parallel {W}_{n+1}x-Wx\parallel +\parallel Wx-{W}_{n}x\parallel \right\}\\ +\left(|{r}_{n+1}-{r}_{n}|+|{\lambda }_{n}-{\lambda }_{n+1}|+|{s}_{n}-{s}_{n+1}|\right){M}_{2}.\end{array}$
It follows from Lemma 1.4 that
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {v}_{n+1}-{v}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0.$
In view of Lemma 1.6, we find that ${lim}_{n\to \mathrm{\infty }}\parallel {v}_{n}-{x}_{n}\parallel =0$. It follows that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =0.$
(2.11)
For any ${x}^{\ast }\in \mathrm{\Omega }$, we see that
${\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}\le {\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}+{\beta }_{n}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\gamma }_{n}{\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}.$
(2.12)
Since
$\begin{array}{rl}{\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}& \le {\parallel \left(I-{r}_{n}{A}_{1}\right){z}_{n}-\left(I-{r}_{n}{A}_{1}\right){x}^{\ast }\parallel }^{2}\\ ={\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}-2{r}_{n}〈{z}_{n}-{x}^{\ast },{A}_{1}{z}_{n}-{A}_{1}{x}^{\ast }〉+{r}_{n}^{2}{\parallel {A}_{1}{z}_{n}-{A}_{1}{x}^{\ast }\parallel }^{2}\\ \le {\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}-2{r}_{n}{\delta }_{1}{\parallel {A}_{1}{z}_{n}-{A}_{1}{x}^{\ast }\parallel }^{2}+{r}_{n}^{2}{\parallel {A}_{1}{z}_{n}-{A}_{1}{x}^{\ast }\parallel }^{2}\\ ={\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{r}_{n}\left({r}_{n}-2{\delta }_{1}\right){\parallel {A}_{1}{z}_{n}-{A}_{1}{x}^{\ast }\parallel }^{2},\end{array}$
we find that
$\begin{array}{r}{\gamma }_{n}{r}_{n}\left(2{\delta }_{1}-{r}_{n}\right){\parallel {A}_{1}{z}_{n}-{A}_{1}{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}+\left(\parallel {x}_{n}-{x}^{\ast }\parallel +\parallel {x}_{n+1}-{x}^{\ast }\parallel \right)\parallel {x}_{n}-{x}_{n+1}\parallel .\end{array}$
Using the restrictions (a) and (b), we obtain
$\underset{n\to \mathrm{\infty }}{lim}\parallel {A}_{1}{z}_{n}-{A}_{1}{x}^{\ast }\parallel =0.$
(2.13)
It follows from (2.12) that
${\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}\le {\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}+{\beta }_{n}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\gamma }_{n}{\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}.$
Since
$\begin{array}{rl}{\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}& ={\parallel {J}_{{r}_{n}}\left({u}_{n}-{s}_{n}{A}_{2}{u}_{n}\right)-{x}^{\ast }\parallel }^{2}\\ \le {\parallel \left(I-{s}_{n}{A}_{2}\right){u}_{n}-\left(I-{s}_{n}{A}_{2}\right){x}^{\ast }\parallel }^{2}\\ ={\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}-2{s}_{n}〈{u}_{n}-{x}^{\ast },{A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }〉+{s}_{n}^{2}{\parallel {A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }\parallel }^{2}\\ \le {\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}-2{s}_{n}{\delta }_{2}{\parallel {A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }\parallel }^{2}+{s}_{n}^{2}{\parallel {A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }\parallel }^{2}\\ ={\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}+{s}_{n}\left({s}_{n}-2{\delta }_{2}\right){\parallel {A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }\parallel }^{2},\end{array}$
we have
${\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}\le {\alpha }_{n}{\parallel f\left({x}_{n}\right)-{x}^{\ast }\parallel }^{2}+{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\gamma }_{n}{s}_{n}\left({s}_{n}-2{\delta }_{2}\right){\parallel {A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }\parallel }^{2},$
which implies that
$\begin{array}{r}{\gamma }_{n}{s}_{n}\left(2{\delta }_{2}-{s}_{n}\right){\parallel {A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}{\parallel f\left({x}_{n}\right)-{x}^{\ast }\parallel }^{2}+\left(\parallel {x}_{n}-{x}^{\ast }\parallel +\parallel {x}_{n+1}-{x}^{\ast }\parallel \right)\parallel {x}_{n}-{x}_{n+1}\parallel .\end{array}$
Using the restrictions (a) and (b), we obtain
$\underset{n\to \mathrm{\infty }}{lim}\parallel {A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }\parallel =0.$
(2.14)
Note that
$\begin{array}{rl}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}\le & {\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}+{\beta }_{n}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\gamma }_{n}{\parallel {x}_{n}-{x}^{\ast }-{r}_{n}\left({A}_{3}{x}_{n}-{A}_{3}{x}^{\ast }\right)\parallel }^{2}\\ \le & {\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}+{\beta }_{n}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\gamma }_{n}\left({\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}\\ +{\lambda }_{n}^{2}{\parallel {A}_{3}{x}_{n}-{A}_{3}{x}^{\ast }\parallel }^{2}-2{\lambda }_{n}〈{A}_{3}{x}_{n}-{A}_{3}{x}^{\ast },{x}_{n}-{x}^{\ast }〉\right)\\ \le & {\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}+{\beta }_{n}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\gamma }_{n}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}\\ -{\lambda }_{n}{\gamma }_{n}\left(2{\delta }_{3}-{\lambda }_{n}\right){\parallel {A}_{3}{x}_{n}-{A}_{3}{x}^{\ast }\parallel }^{2}.\end{array}$
This implies that
$\begin{array}{r}{\lambda }_{n}{\gamma }_{n}\left(2{\delta }_{3}-{\lambda }_{n}\right){\parallel {A}_{3}{x}_{n}-{A}_{3}{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}{\parallel u-{x}^{\ast }\parallel }^{2}+\left(\parallel {x}_{n}-{x}^{\ast }\parallel +\parallel {x}_{n+1}-{x}^{\ast }\parallel \right)\parallel {x}_{n}-{x}_{n+1}\parallel .\end{array}$
Using the restrictions (a) and (b), we see that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {A}_{3}{x}_{n}-{A}_{3}{x}^{\ast }\parallel =0.$
(2.15)
Since ${T}_{{\lambda }_{n}}$ is firmly nonexpansive, we find that
$\begin{array}{rl}{\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}& \le 〈\left(I-{\lambda }_{n}{A}_{3}\right){x}_{n}-\left(I-{\lambda }_{n}{A}_{3}\right){x}^{\ast },{u}_{n}-{x}^{\ast }〉\\ \le \frac{1}{2}\left({\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n}-{u}_{n}\parallel }^{2}-{\lambda }_{n}^{2}{\parallel {A}_{3}{x}_{n}-{A}_{3}{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}+2{\lambda }_{n}〈{A}_{3}{x}_{n}-{A}_{3}{x}^{\ast },{x}_{n}-{u}_{n}〉\right).\end{array}$
This in turn implies that
$\begin{array}{rl}{\gamma }_{n}{\parallel {x}_{n}-{u}_{n}\parallel }^{2}\le & {\alpha }_{n}{\parallel f\left({x}_{n}\right)-{x}^{\ast }\parallel }^{2}+\left(\parallel {x}_{n}-{x}^{\ast }\parallel +\parallel {x}_{n+1}-{x}^{\ast }\parallel \right)\parallel {x}_{n}-{x}_{n+1}\parallel \\ +2{\lambda }_{n}\parallel {A}_{3}{x}_{n}-{A}_{3}{x}^{\ast }\parallel \parallel {x}_{n}-{u}_{n}\parallel .\end{array}$
Using the restrictions (a) and (b), we see that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{u}_{n}\parallel =0.$
(2.16)
Since ${J}_{{s}_{n}}$ is also firmly nonexpansive mapping, we see that
$\begin{array}{rl}{\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}\le & 〈\left(I-{s}_{n}{A}_{2}\right){u}_{n}-\left(I-{s}_{n}{A}_{2}\right){x}^{\ast },{z}_{n}-{x}^{\ast }〉\\ \le & \frac{1}{2}\left({\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}+{\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {u}_{n}-{z}_{n}-{s}_{n}\left({A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }\right)\parallel }^{2}\right)\\ \le & \frac{1}{2}\left({\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}+{\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {u}_{n}-{z}_{n}\parallel }^{2}\\ +2{s}_{n}\parallel {u}_{n}-{z}_{n}\parallel \parallel {A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }\parallel -{s}_{n}^{2}{\parallel {A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }\parallel }^{2}\right),\end{array}$
which implies that
${\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}\le {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {u}_{n}-{z}_{n}\parallel }^{2}+2{s}_{n}\parallel {u}_{n}-{z}_{n}\parallel \parallel {A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }\parallel .$
It follows that
$\begin{array}{rl}{\gamma }_{n}{\parallel {u}_{n}-{z}_{n}\parallel }^{2}\le & {\alpha }_{n}{\parallel f\left({x}_{n}\right)-{x}^{\ast }\parallel }^{2}+\left(\parallel {x}_{n}-{x}^{\ast }\parallel +\parallel {x}_{n+1}-{x}^{\ast }\parallel \right)\parallel {x}_{n}-{x}_{n+1}\parallel \\ +2{s}_{n}\parallel {u}_{n}-{z}_{n}\parallel \parallel {A}_{2}{u}_{n}-{A}_{2}{x}^{\ast }\parallel .\end{array}$
Using the restrictions (a) and (b), we obtain
$\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n}-{z}_{n}\parallel =0$
(2.17)
and
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{z}_{n}\parallel =0.$
(2.18)
Note that
$\left(1-{\beta }_{n}\right)\parallel {W}_{n}{y}_{n}-{x}_{n}\parallel \le \parallel {x}_{n}-{x}_{n+1}\parallel +{\alpha }_{n}\parallel u-{W}_{n}{y}_{n}\parallel .$
Using the restrictions (b) and (c), we obtain
$\underset{n\to \mathrm{\infty }}{lim}\parallel {W}_{n}{y}_{n}-{x}_{n}\parallel =0.$
(2.19)
On the other hand, one has
$\parallel {W}_{n}{y}_{n}-{y}_{n}\parallel \le \parallel {y}_{n}-{z}_{n}\parallel +\parallel {z}_{n}-{u}_{n}\parallel +\parallel {u}_{n}-{x}_{n}\parallel +\parallel {x}_{n}-{W}_{n}{y}_{n}\parallel .$
Using (2.6), (2.7), (2.8), and (2.9), we find that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {W}_{n}{y}_{n}-{y}_{n}\parallel =0.$
(2.20)
Next, we prove that
$\underset{n\to \mathrm{\infty }}{lim sup}〈u-\overline{x},{x}_{n}-\overline{x}〉\le 0.$
To see this, we choose a subsequence $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that
$\underset{n\to \mathrm{\infty }}{lim sup}〈u-\overline{x},{x}_{n}-\overline{x}〉=\underset{i\to \mathrm{\infty }}{lim}〈u-\overline{x},{x}_{{n}_{i}}-\overline{x}〉.$
(2.21)

Since $\left\{{x}_{{n}_{i}}\right\}$ is bounded, there exists a subsequence $\left\{{x}_{{n}_{{i}_{j}}}\right\}$ of $\left\{{x}_{{n}_{i}}\right\}$ which converges weakly to w. Without loss of generality, we may assume that ${x}_{{n}_{i}}⇀q$. Therefore, we see that ${y}_{{n}_{i}}⇀q$. We also have ${z}_{{n}_{i}}⇀q$.

Next, we show that $q\in {\bigcap }_{i=1}^{\mathrm{\infty }}F\left({S}_{i}\right)$. Suppose the contrary, $q\notin CFPS$, i.e., $Wq\ne q$. Since ${y}_{{n}_{i}}⇀q$, we see from Opial’s condition that
$\begin{array}{rl}\underset{i\to \mathrm{\infty }}{lim inf}\parallel {y}_{{n}_{i}}-q\parallel & <\underset{i\to \mathrm{\infty }}{lim inf}\parallel {y}_{{n}_{i}}-Wq\parallel \\ \le \underset{i\to \mathrm{\infty }}{lim inf}\left\{\parallel {y}_{{n}_{i}}-W{y}_{{n}_{i}}\parallel +\parallel W{y}_{{n}_{i}}-Wq\parallel \right\}\\ \le \underset{i\to \mathrm{\infty }}{lim inf}\left\{\parallel {y}_{{n}_{i}}-W{y}_{{n}_{i}}\parallel +\parallel {y}_{{n}_{i}}-q\parallel \right\}.\end{array}$
(2.22)
On the other hand, we have
$\begin{array}{rl}\parallel W{y}_{n}-{y}_{n}\parallel & \le \parallel W{y}_{n}-{W}_{n}{y}_{n}\parallel +\parallel {W}_{n}{y}_{n}-{y}_{n}\parallel \\ \le \underset{x\in K}{sup}\parallel Wx-{W}_{n}x\parallel +\parallel {W}_{n}{y}_{n}-{y}_{n}\parallel .\end{array}$

In view of Lemma 1.4, we obtain that ${lim}_{n\to \mathrm{\infty }}\parallel W{y}_{n}-{y}_{n}\parallel =0$. This implies from (2.22) that ${lim inf}_{i\to \mathrm{\infty }}\parallel {y}_{{n}_{i}}-q\parallel <{lim inf}_{i\to \mathrm{\infty }}\parallel {y}_{{n}_{i}}-q\parallel$. This is a contradiction. Thus, we have $q\in {\bigcap }_{i=1}^{\mathrm{\infty }}F\left({S}_{i}\right)$.

Now, we are in a position to prove that $q\in {\left({A}_{1}+{M}_{1}\right)}^{-1}\left(0\right)$. Notice that $\frac{{z}_{n}-{y}_{n}}{{r}_{n}}-{A}_{1}{z}_{n}\in {M}_{1}{y}_{n}$. Let $\mu \in {M}_{1}\nu$. Since ${M}_{1}$ is monotone, we find that
$〈\frac{{z}_{n}-{y}_{n}}{{r}_{n}}-{A}_{1}{z}_{n}-\mu ,{y}_{n}-\nu 〉\ge 0.$

This implies that $〈-{A}_{1}q-\mu ,q-\nu 〉\ge 0$. This implies that $-{A}_{1}q\in {M}_{1}q$, that is, $q\in {\left({A}_{1}+{M}_{1}\right)}^{-1}\left(0\right)$.

Now, we prove that $q\in {\left({A}_{2}+{M}_{2}\right)}^{-1}\left(0\right)$. Notice that $\frac{{u}_{n}-{z}_{n}}{{s}_{n}}-{A}_{2}{u}_{n}\in {M}_{2}{z}_{n}$. Let ${\mu }^{\prime }\in {M}_{2}{\nu }^{\prime }$. Since ${M}_{2}$ is monotone, we find that
$〈\frac{{u}_{n}-{z}_{n}}{{s}_{n}}-{A}_{2}{u}_{n}-{\mu }^{\prime },{z}_{n}-{\nu }^{\prime }〉\ge 0.$

This implies that $〈-{A}_{2}q-{\mu }^{\prime },q-{\nu }^{\prime }〉\ge 0$. This implies that $-{A}_{2}q\in {M}_{2}q$, that is, $q\in {\left({A}_{2}+{M}_{2}\right)}^{-1}\left(0\right)$.

Next, we show that $q\in EP\left(F,{A}_{3}\right)$. Since ${u}_{n}={T}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{A}_{3}\right){x}_{n}$, for any $y\in C$, we have
$F\left({u}_{n},y\right)+〈{A}_{3}{x}_{n},y-{u}_{n}〉+\frac{1}{{\lambda }_{n}}〈y-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0.$
Replacing n by ${n}_{i}$, we find from (A2) that
$〈{A}_{3}{x}_{{n}_{i}},y-{u}_{{n}_{i}}〉+〈y-{u}_{{n}_{i}},\frac{{u}_{{n}_{i}}-{x}_{{n}_{i}}}{{\lambda }_{{n}_{i}}}〉\ge F\left(y,{u}_{{n}_{i}}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C.$
Putting ${y}_{t}=ty+\left(1-t\right)q$ for any $t\in \left(0,1\right]$ and $y\in C$, we see that ${y}_{t}\in C$. It follows that
$\begin{array}{r}〈{y}_{t}-{u}_{{n}_{i}},{A}_{3}{y}_{t}〉\\ \phantom{\rule{1em}{0ex}}\ge 〈{y}_{t}-{u}_{{n}_{i}},{A}_{3}{y}_{t}〉-〈{A}_{3}{x}_{{n}_{i}},{y}_{t}-{u}_{{n}_{i}}〉-〈{y}_{t}-{u}_{{n}_{i}},\frac{{u}_{{n}_{i}}-{x}_{{n}_{i}}}{{\lambda }_{{n}_{i}}}〉+F\left({y}_{t},{u}_{{n}_{i}}\right)\\ \phantom{\rule{1em}{0ex}}=〈{y}_{t}-{u}_{{n}_{i}},{A}_{3}{y}_{t}-{A}_{3}{u}_{{n}_{i}}〉+〈{y}_{t}-{u}_{{n}_{i}},{A}_{3}{u}_{{n}_{i}}-{A}_{3}{x}_{{n}_{i}}〉-〈{y}_{t}-{u}_{{n}_{i}},\frac{{u}_{{n}_{i}}-{x}_{{n}_{i}}}{{\lambda }_{{n}_{i}}}〉\\ \phantom{\rule{2em}{0ex}}+F\left({y}_{t},{u}_{{n}_{i}}\right).\end{array}$
In view of the monotonicity of ${A}_{3}$, and the restriction (a), we obtain from (A4) that
$〈{y}_{t}-q,{A}_{3}{y}_{t}〉\ge F\left({y}_{t},q\right).$
From (A1) and (A4), we see that
$\begin{array}{rl}0& =F\left({y}_{t},{y}_{t}\right)\le tF\left({y}_{t},y\right)+\left(1-t\right)F\left({y}_{t},q\right)\\ \le tF\left({y}_{t},y\right)+\left(1-t\right)〈{y}_{t}-q,{A}_{3}{y}_{t}〉\\ =tF\left({y}_{t},y\right)+\left(1-t\right)t〈y-q,{A}_{3}{y}_{t}〉.\end{array}$
It follows that
$0\le F\left({y}_{t},y\right)+\left(1-t\right)〈y-w,{A}_{3}{y}_{t}〉,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C.$
It follows from (A3) that $q\in EP\left(F,{A}_{3}\right)$. Hence,
$\underset{n\to \mathrm{\infty }}{lim sup}〈u-\overline{x},{x}_{n}-\overline{x}〉\le 0.$
Finally, we show that ${x}_{n}\to \overline{x}$. Note that
$\begin{array}{r}{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}〈u-\overline{x},{x}_{n+1}-\overline{x}〉+{\beta }_{n}\parallel {x}_{n}-\overline{x}\parallel \parallel {x}_{n+1}-\overline{x}\parallel +{\gamma }_{n}\parallel {y}_{n}-\overline{x}\parallel \parallel {x}_{n+1}-\overline{x}\parallel \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}〈u-\overline{x},{x}_{n+1}-\overline{x}〉+\frac{1-{\alpha }_{n}}{2}\left({\parallel {x}_{n}-\overline{x}\parallel }^{2}+{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}\right).\end{array}$
This implies that
${\parallel {x}_{n+1}-\overline{x}\parallel }^{2}\le 2{\alpha }_{n}〈u-\overline{x},{x}_{n+1}-\overline{x}〉+\left(1-{\alpha }_{n}\right){\parallel {x}_{n}-\overline{x}\parallel }^{2}.$

Using Lemma 1.1, we find that ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-\overline{x}\parallel =0$. This completes the proof. □

## 3 Applications

In this section, we consider some applications of the main results.

Recall that the classical variational inequality is to find an $x\in C$ such that
$〈Ax,y-x〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C.$

In this paper, we use $VI\left(C,A\right)$ to denote the solution set of the inequality. It is well known that $x\in C$ is a solution of the inequality iff x is a fixed point of the mapping ${P}_{C}\left(I-rA\right)$, where $r>0$ is a constant, I stands for the identity mapping. If A is α-inverse-strongly monotone and $r\in \left(0,2\alpha \right]$, then the mapping $I-rA$ is nonexpansive. It follows that $VI\left(C,A\right)$ is closed and convex.

Let $g:H\to \left(-\mathrm{\infty },+\mathrm{\infty }\right]$ be a proper convex lower semicontinuous function. Then the subdifferential ∂g of g is defined as follows:
$\partial fg\left(x\right)=\left\{y\in H:g\left(z\right)\ge g\left(x\right)+〈z-x,y〉,z\in H\right\},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in H.$

From Rockafellar , we know that ∂g is maximal monotone. It is not hard to verify that $0\in \partial g\left(x\right)$ if and only if $g\left(x\right)={min}_{y\in H}g\left(y\right)$.

Let ${I}_{C}$ be the indicator function of C, i.e.,
${I}_{C}\left(x\right)=\left\{\begin{array}{cc}0,\hfill & x\in C,\hfill \\ +\mathrm{\infty },\hfill & x\notin C.\hfill \end{array}$

Since ${I}_{C}$ is a proper lower semicontinuous convex function on H, we see that the subdifferential $\partial {I}_{C}$ of ${I}_{C}$ is a maximal monotone operator. It is clearly that ${J}_{r}x={P}_{C}x$, $\mathrm{\forall }x\in H$, ${\left({A}_{1}+\partial {I}_{C}\right)}^{-1}\left(0\right)=VI\left(C,{A}_{1}\right)$ and ${\left({A}_{2}+\partial {I}_{C}\right)}^{-1}\left(0\right)=VI\left(C,{A}_{2}\right)$.

Theorem 3.1 Let C be a nonempty closed convex subset of a Hilbert space H and F a bifunction from $C×C$ to which satisfies (A1)-(A4). Let ${A}_{1}:C\to H$ be a ${\delta }_{1}$-inverse-strongly monotone mapping, ${A}_{2}:C\to H$ be a ${\delta }_{2}$-inverse-strongly monotone mapping, ${A}_{3}:C\to H$ be a ${\delta }_{3}$-inverse-strongly monotone mapping, and $\left\{{S}_{i}:C\to C\right\}$ be a family of infinitely nonexpansive mappings. Assume that $\mathrm{\Omega }:={\bigcap }_{i=1}^{\mathrm{\infty }}F\left({S}_{i}\right)\cap EP\left(F,{A}_{3}\right)\cap VI\left(C,{A}_{1}\right)\cap VI\left(C,{A}_{2}\right)\ne \mathrm{\varnothing }$. Let ${x}_{1}\in C$ and $\left\{{x}_{n}\right\}$ be a sequence generated by
$\left\{\begin{array}{c}{z}_{n}={P}_{C}\left({u}_{n}-{s}_{n}{A}_{2}{u}_{n}\right),\hfill \\ {x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}{W}_{n}{P}_{C}\left({z}_{n}-{r}_{n}{A}_{1}{z}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
where ${u}_{n}$ is such that
$F\left({u}_{n},y\right)+〈{A}_{3}{x}_{n},y-{u}_{n}〉+\frac{1}{{\lambda }_{n}}〈y-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C,$
$\left\{{W}_{n}:C\to C\right\}$ is the sequence generated in (1.2), $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, and $\left\{{\gamma }_{n}\right\}$ are sequences in $\left(0,1\right)$ such that ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$ for each $n\ge 1$ and $\left\{{r}_{n}\right\}$, $\left\{{s}_{n}\right\}$, and $\left\{{\lambda }_{n}\right\}$ are positive number sequences. Assume that the above control sequences satisfy the following restrictions:
1. (a)

$0, $0<{a}^{\prime }\le {r}_{n}\le {b}^{\prime }<2{\delta }_{1}$, $0<\overline{a}\le {s}_{n}\le \overline{b}<2{\delta }_{2}$;

2. (b)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

3. (c)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

4. (d)

${lim}_{n\to \mathrm{\infty }}|{\lambda }_{n}-{\lambda }_{n+1}|={lim}_{n\to \mathrm{\infty }}|{s}_{n}-{s}_{n+1}|={lim}_{n\to \mathrm{\infty }}|{r}_{n}-{r}_{n+1}|=0$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to $\overline{x}={P}_{\mathrm{\Omega }}u$.

Recall that a mapping $T:C\to C$ is said to be a k-strict pseudo-contraction if there exists a constant $k\in \left[0,1\right)$ such that
${\parallel Tx-Ty\parallel }^{2}\le {\parallel x-y\parallel }^{2}+k{\parallel \left(I-T\right)x-\left(I-T\right)y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

Putting $A=I-T$, where $T:C\to C$ is a k-strict pseudo-contraction, we find that A is $\frac{1-k}{2}$-inverse-strongly monotone.

Next, we consider fixed points of strict pseudo-contractions.

Theorem 3.2 Let C be a nonempty closed convex subset of a Hilbert space H and F a bifunction from $C×C$ to which satisfies (A1)-(A4). Let ${T}_{1}:C\to H$ be a ${k}_{1}$-strict pseudo-contraction, ${T}_{2}:C\to H$ be a ${k}_{2}$-strict pseudo-contraction, ${A}_{3}:C\to H$ be a δ-inverse-strongly monotone mapping, and $\left\{{S}_{i}:C\to C\right\}$ be a family of infinitely nonexpansive mappings. Assume that $\mathrm{\Omega }:={\bigcap }_{i=1}^{\mathrm{\infty }}F\left({S}_{i}\right)\cap EP\left(F,{A}_{3}\right)\cap F\left({T}_{1}\right)\cap F\left({T}_{2}\right)\ne \mathrm{\varnothing }$. Let ${x}_{1}\in C$ and $\left\{{x}_{n}\right\}$ be a sequence generated by
$\left\{\begin{array}{c}{z}_{n}=\left(1-{s}_{n}\right){u}_{n}+{s}_{n}{T}_{2}{u}_{n},\hfill \\ {y}_{n}=\left(1-{r}_{n}\right){u}_{n}+{r}_{n}{T}_{1}{u}_{n},\hfill \\ {x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}{W}_{n}{y}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
where ${u}_{n}$ is such that
$F\left({u}_{n},y\right)+〈{A}_{3}{x}_{n},y-{u}_{n}〉+\frac{1}{{\lambda }_{n}}〈y-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C,$
$\left\{{W}_{n}:C\to C\right\}$ is the sequence generated in (1.2), $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, and $\left\{{\gamma }_{n}\right\}$ are sequences in $\left(0,1\right)$ such that ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$ for each $n\ge 1$ and $\left\{{r}_{n}\right\}$, $\left\{{s}_{n}\right\}$, and $\left\{{\lambda }_{n}\right\}$ are positive number sequences. Assume that the above control sequences satisfy the following restrictions:
1. (a)

$0, $0<{a}^{\prime }\le {r}_{n}\le {b}^{\prime }<1-{k}_{1}$, $0<\overline{a}\le {s}_{n}\le \overline{b}<1-{k}_{2}$;

2. (b)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

3. (c)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

4. (d)

${lim}_{n\to \mathrm{\infty }}|{\lambda }_{n}-{\lambda }_{n+1}|={lim}_{n\to \mathrm{\infty }}|{s}_{n}-{s}_{n+1}|={lim}_{n\to \mathrm{\infty }}|{r}_{n}-{r}_{n+1}|=0$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to $\overline{x}={P}_{\mathrm{\Omega }}u$.

Proof Taking ${A}_{i}=I-{T}_{i}$, wee see that ${A}_{i}:C\to H$ is a ${\delta }_{i}$-strict pseudo-contraction with ${\delta }_{i}=\frac{1-{k}_{i}}{2}$ and $F\left({T}_{i}\right)=VI\left(C,{A}_{i}\right)$ for $i=1,2$. In view of Theorem 3.1, we find the desired conclusion immediately. □

## Declarations

### Acknowledgements

The author is very grateful to the reviewers for useful suggestions which improved the contents of this paper.

## Authors’ Affiliations

(1)
Kaifeng Vocational College of Culture and Arts, Kaifeng, Henan, China

## References 