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# An algorithm for treating asymptotically strict pseudocontractions and monotone operators

Fixed Point Theory and Applications20142014:52

https://doi.org/10.1186/1687-1812-2014-52

• Accepted: 13 February 2014
• Published:

## Abstract

In this paper, an algorithm for treating asymptotically κ-strict pseudocontractions and monotone operators is proposed. Convergence analysis of the algorithm is investigated in the framework of Hilbert spaces.

## Keywords

• asymptotically κ-strict pseudocontraction
• inverse-strongly monotone mapping
• maximal monotone operator
• fixed point

## 1 Introduction-preliminaries

In this paper, we are concerned with the problem of finding a common element in the intersection $F\left(T\right)\cap {\left(A+B\right)}^{-1}\left(0\right)$, where $F\left(T\right)$ denotes the fixed point set of the mapping T and ${\left(A+B\right)}^{-1}\left(0\right)$ denotes the zero point set of the sum of the operator A and the operator B.

The motivation for the common element problem is mainly due to its possible applications to mathematical modeling of concrete complex problems. The common element problems include mini-max problems, complementarity problems, equilibrium problems, common fixed point problems and variational inequalities as special cases; see, for example, [135] and the references therein.

Throughout the article, we always assume that H is a real Hilbert space with the inner product $〈\cdot \phantom{\rule{0.2em}{0ex}},\cdot 〉$ and the norm $\parallel \cdot \parallel$, respectively. Let C be a nonempty closed convex subset of H, and let ${Proj}_{C}$ be the metric projection from H onto C.

Let $A:C\to H$ be a mapping. ${A}^{-1}\left(0\right)$ stands for the zero point set of A; that is, ${A}^{-1}\left(0\right):=\left\{x\in C:Ax=0\right\}$. Recall that A is said to be monotone iff
$〈Ax-Ay,x-y〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
A is said to be α-strongly monotone iff there exists a constant $\alpha >0$ such that
$〈Ax-Ay,x-y〉\ge \alpha {\parallel x-y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
A is said to be α-inverse-strongly monotone iff there exists a constant $\alpha >0$ such that
$〈Ax-Ay,x-y〉\ge \alpha {\parallel Ax-Ay\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
It is not hard to see that α-inverse-strongly monotone mappings are Lipschitz continuous. Indeed, we have
$\alpha {\parallel Ax-Ay\parallel }^{2}\le 〈Ax-Ay,x-y〉\le \parallel Ax-Ay\parallel \parallel x-y\parallel .$

This shows that $\parallel Ax-Ay\parallel \le \frac{1}{\alpha }\parallel x-y\parallel$.

Recall that the classical variational inequality, denoted by $VI\left(C,A\right)$, is to find $u\in C$ such that
$〈Au,v-u〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }v\in C.$
(1.1)

One can see that the variational inequality (1.1) is equivalent to a fixed point problem of the mapping ${Proj}_{C}\left(I-rA\right)$, where I is the identity and r is some positive real number. The element $u\in C$ is a solution of the variational inequality (1.1) iff $u\in C$ satisfies the equation $u={P}_{C}\left(u-rAu\right)$. This alternative equivalent formulation has played a significant role in the studies of variational inequalities and related optimization problems.

A multivalued operator $B:H\to {2}^{H}$ with the domain $D\left(B\right)=\left\{x\in H:Bx\ne \mathrm{\varnothing }\right\}$ and the range $R\left(B\right)=\left\{Bx:x\in D\left(B\right)\right\}$ is said to be monotone if for ${x}_{1}\in D\left(B\right)$, ${x}_{2}\in D\left(B\right)$, ${y}_{1}\in B{x}_{1}$ and ${y}_{2}\in B{x}_{2}$, we have $〈{x}_{1}-{x}_{2},{y}_{1}-{y}_{2}〉\ge 0$. A monotone operator B is said to be maximal if its graph $G\left(B\right)=\left\{\left(x,y\right):y\in Bx\right\}$ is not properly contained in the graph of any other monotone operator. Let I denote the identity operator on H and $B:H\to {2}^{H}$ be a maximal monotone operator. Then we can define, for each $r>0$, a nonexpansive single-valued mapping ${J}_{r}:H\to H$ by ${J}_{r}={\left(I+rB\right)}^{-1}$. It is called the resolvent of B. We know that ${B}^{-1}0=F\left({J}_{r}\right)$ for all $r>0$ and ${J}_{r}$ is firmly nonexpansive.

Let $T:C\to C$ be a mapping. In this paper, we use $F\left(T\right)$ to denote the fixed point set of T; that is, $F\left(T\right):=\left\{x\in C:x=Tx\right\}$. Recall that T is said to be nonexpansive iff
$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
T is said to be asymptotically nonexpansive iff there exists a sequence $\left\{{k}_{n}\right\}\subset \left[1,\mathrm{\infty }\right)$ such that
$\parallel {T}^{n}x-{T}^{n}y\parallel \le {k}_{n}\parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C,n\ge 1.$
T is said to be a κ-strict pseudocontraction iff there exists a constant $\kappa \in \left[0,1\right)$ such that
${\parallel Tx-Ty\parallel }^{2}\le {\parallel x-y\parallel }^{2}+\kappa {\parallel \left(x-Tx\right)-\left(y-Ty\right)\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
Note that the class of κ-strict pseudocontractions strictly includes the class of nonexpansive mappings as a special case. That is, T is nonexpansive iff the coefficient $\kappa =0$. T is said to be an asymptotically κ-strict pseudocontraction iff there exist a constant $\kappa \in \left[0,1\right)$ and a sequence $\left\{{k}_{n}\right\}$ in $\left[1,\mathrm{\infty }\right)$ such that
${\parallel {T}^{n}x-{T}^{n}y\parallel }^{2}\le {k}_{n}{\parallel x-y\parallel }^{2}+\kappa {\parallel \left(x-{T}^{n}x\right)-\left(y-{T}^{n}y\right)\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C,n\ge 1.$

Note that the class of asymptotically κ-strict pseudocontractions strictly includes the class of asymptotically nonexpansive mappings as a special case. That is, T is asymptotically nonexpansive iff the coefficient $\kappa =0$.

In [24], Kamimura and Takahashi investigated the problem of finding zero points of a maximal monotone operator based on the following iterative algorithm:
${x}_{1}\in H,\phantom{\rule{1em}{0ex}}{x}_{n+1}={\alpha }_{n}x+\left(1-{\alpha }_{n}\right){J}_{{r}_{n}}{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
(1.2)

where $\left\{{\alpha }_{n}\right\}$ is a sequence in $\left(0,1\right)$, $\left\{{r}_{n}\right\}$ is a positive real number sequence, $B:H\to {2}^{H}$ is maximal monotone and ${J}_{{r}_{n}}={\left(I+{r}_{n}B\right)}^{-1}$. It is proved that the sequence $\left\{{x}_{n}\right\}$ generated in (1.2) converges strongly to some $z\in {B}^{-1}\left(0\right)$ provided that the control sequence satisfies some restrictions. Further, using this result, they also investigated the case that $B=\partial f$, where $f:H\to \left(-\mathrm{\infty },\mathrm{\infty }\right]$ is a proper lower semicontinuous convex function. Convergence theorems are established in the framework of real Hilbert spaces; for more details, see [24].

Recently, Takahashi et al. studied zero point problems of the sum of two monotone mappings and fixed point problems of a nonexpansive mapping based on the following iterative algorithm:
$\left\{\begin{array}{c}{x}_{1}\in C,\hfill \\ {y}_{n}={\alpha }_{n}x+\left(1-{\alpha }_{n}\right){J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right),\hfill \\ {x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)T{y}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(1.3)

where $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ are real number sequences in $\left(0,1\right)$, $\left\{{r}_{n}\right\}$ is a positive sequence, $T:C\to C$ is a nonexpansive mapping and $A:C\to H$ is an inverse-strongly monotone mapping. It is proved that the sequence $\left\{{x}_{n}\right\}$ generated in (1.3) converges strongly to some $z\in {\left(A+B\right)}^{-1}\left(0\right)\cap F\left(S\right)$ provided that the control sequence satisfies some restrictions; for more details, see [2].

Motivated by the above results, we investigate fixed point problems of asymptotically strict pseudocontractions and zero point problems of the sum of two monotone mappings.

In order to state our main results, we need the following tools.

Recall that a space is said to satisfy Opial’s condition [36] if, for any sequence $\left\{{x}_{n}\right\}\subset H$ with ${x}_{n}⇀x$, where denotes the weak convergence, the inequality
$\underset{n\to \mathrm{\infty }}{lim inf}\parallel {x}_{n}-x\parallel <\underset{n\to \mathrm{\infty }}{lim inf}\parallel {x}_{n}-y\parallel$
holds for every $y\in H$ with $y\ne x$. Indeed, the above inequality is equivalent to the following:
$\underset{n\to \mathrm{\infty }}{lim sup}\parallel {x}_{n}-x\parallel <\underset{n\to \mathrm{\infty }}{lim sup}\parallel {x}_{n}-y\parallel .$

Lemma 1.1 [2]

Let $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ be bounded sequences in a Banach space X, and let ${\beta }_{n}$ 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 integers $n\ge 0$ 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 1.2 [37]

Let C be a nonempty, closed and convex subset of H, let $A:C\to H$ be a mapping, and let $B:H⇉H$ be a maximal monotone operator. Then $F\left({J}_{r}\left(I-rA\right)\right)={\left(A+B\right)}^{-1}\left(0\right)$.

Lemma 1.3 [38]

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. (i)

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

2. (ii)

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

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

Lemma 1.4 [39]

Let C be a nonempty, closed and convex subset of H. Let $T:C\to C$ be an asymptotically strict pseudocontraction. Then T is Lipschitz continuous and $I-T$ is demiclosed at zero.

## 2 Main results

Theorem 2.1 Let C be a nonempty closed convex subset of H. Let $T:C\to C$ be an asymptotically κ-strict pseudocontraction. Let $A:C\to H$ be an α-inverse-strongly monotone mapping, and let B be a maximal monotone operator on H. Assume that $F\left(T\right)\cap {\left(A+B\right)}^{-1}\left(0\right)\ne \mathrm{\varnothing }$. Let $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\gamma }_{n}\right\}$ be real number sequences in $\left(0,1\right)$. Let ${J}_{{r}_{n}}={\left(I+{r}_{n}B\right)}^{-1}$, where $\left\{{r}_{n}\right\}$ is a positive real number sequence. Let $\left\{{x}_{n}\right\}$ be a sequence in C generated by: ${x}_{1}\in C$ is chosen arbitrarily and
$\left\{\begin{array}{c}{z}_{n}={Proj}_{C}\left({\alpha }_{n}u+\left(1-{\alpha }_{n}\right){J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)\right),\hfill \\ {x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)\left({\gamma }_{n}{z}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{z}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.\hfill \end{array}$
Assume that the sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n}\right\}$ and $\left\{{r}_{n}\right\}$ satisfy the following restrictions:
1. (a)

$0, ${lim}_{n\to \mathrm{\infty }}|{r}_{n+1}-{r}_{n}|=0$;

2. (b)

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

3. (c)

$0;

4. (d)

$\kappa \le {\gamma }_{n}\le e<1$, ${lim}_{n\to \mathrm{\infty }}|{\gamma }_{n+1}-{\gamma }_{n}|=0$,

where a, b, c, d and e are some real numbers. If T is asymptotically regular, then the sequence $\left\{{x}_{n}\right\}$ converges strongly to some point $\overline{x}$, where $\overline{x}={P}_{F\left(T\right)\cap {\left(A+B\right)}^{-1}\left(0\right)}u$.

Proof First, we show that the mapping $I-{r}_{n}A$ is nonexpansive. Indeed, we have
$\begin{array}{r}{\parallel \left(I-{r}_{n}A\right)x-\left(I-{r}_{n}A\right)y\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel x-y\parallel }^{2}-2{r}_{n}〈x-y,Ax-Ay〉+{{r}_{n}}^{2}{\parallel Ax-Ay\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel x-y\parallel }^{2}-{r}_{n}\left(2\alpha -{r}_{n}\right){\parallel Ax-Ay\parallel }^{2}.\end{array}$
It follows from Restriction (a) that $I-{r}_{n}A$ is nonexpansive. Put ${y}_{n}={\gamma }_{n}{z}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{z}_{n}$ and fix $p\in F\left(T\right)\cap {\left(A+B\right)}^{-1}\left(0\right)$. It follows from Lemma 1.2 that
$\begin{array}{rl}\parallel {z}_{n}-p\parallel & =\parallel {Proj}_{C}\left({\alpha }_{n}u+\left(1-{\alpha }_{n}\right){J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)\right)-p\parallel \\ \le {\alpha }_{n}\parallel u-p\parallel +\left(1-{\alpha }_{n}\right)\parallel {J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)-p\parallel \\ \le {\alpha }_{n}\parallel u-p\parallel +\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-p\parallel .\end{array}$
(2.1)
In view of Restriction (d), we find that
$\begin{array}{rl}{\parallel {y}_{n}-p\parallel }^{2}\le & {\parallel {\gamma }_{n}\left({z}_{n}-p\right)+\left(1-{\gamma }_{n}\right)\left({T}^{n}{z}_{n}-{T}^{n}p\right)\parallel }^{2}\\ =& {\gamma }_{n}{\parallel {z}_{n}-p\parallel }^{2}+\left(1-{\gamma }_{n}\right){\parallel {T}^{n}{z}_{n}-{T}^{n}p\parallel }^{2}\\ -{\gamma }_{n}\left(1-{\gamma }_{n}\right){\parallel \left({z}_{n}-p\right)-\left({T}^{n}{z}_{n}-{T}^{n}p\right)\parallel }^{2}\\ \le & {\gamma }_{n}{\parallel {z}_{n}-p\parallel }^{2}+\left(1-{\gamma }_{n}\right)\left({\parallel {z}_{n}-p\parallel }^{2}+\kappa {\parallel \left({z}_{n}-p\right)-\left({T}^{n}{z}_{n}-{T}^{n}p\right)\parallel }^{2}\right)\\ -{\gamma }_{n}\left(1-{\gamma }_{n}\right){\parallel \left({z}_{n}-p\right)-\left({T}^{n}{z}_{n}-{T}^{n}p\right)\parallel }^{2}\\ =& {\parallel {z}_{n}-p\parallel }^{2}-\left(1-{\gamma }_{n}\right)\left({\gamma }_{n}-\kappa \right){\parallel \left({z}_{n}-p\right)-\left({T}^{n}{z}_{n}-{T}^{n}p\right)\parallel }^{2}\\ \le & {\parallel {z}_{n}-p\parallel }^{2}.\end{array}$
(2.2)
Substituting (2.1) into (2.2), we obtain that
$\begin{array}{rl}\parallel {x}_{n+1}-p\parallel & \le {\beta }_{n}\parallel {x}_{n}-p\parallel +\left(1-{\beta }_{n}\right)\parallel {y}_{n}-p\parallel \\ \le {\beta }_{n}\parallel {x}_{n}-p\parallel +\left(1-{\beta }_{n}\right)\left({\alpha }_{n}\parallel u-p\parallel +\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-p\parallel \right)\\ \le \left(1-{\alpha }_{n}\left(1-{\beta }_{n}\right)\right)\parallel {x}_{n}-p\parallel +{\alpha }_{n}\left(1-{\beta }_{n}\right)\parallel u-p\parallel .\end{array}$
Putting $M=max\left\{\parallel {x}_{1}-p\parallel ,\parallel u-p\parallel \right\}$, we find that $\parallel {x}_{n}-p\parallel \le M$ for all $n\ge 1$. Indeed, it is clear that $\parallel {x}_{1}-p\parallel \le M$. Suppose that $\parallel {x}_{m}-p\parallel \le M$ for some positive integer m. It follows that
$\begin{array}{rl}\parallel {x}_{m+1}-p\parallel & \le \left(1-{\alpha }_{m}\left(1-{\beta }_{m}\right)\right)\parallel {x}_{m}-p\parallel +{\alpha }_{m}\left(1-{\beta }_{m}\right)\parallel u-p\parallel \\ \le \left(1-{\alpha }_{m}\left(1-{\beta }_{m}\right)\right)M+{\alpha }_{m}\left(1-{\beta }_{n}\right)M\\ =M.\end{array}$
This finds that $\left\{{x}_{n}\right\}$ is bounded. Putting ${\rho }_{n}={J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)$, we find that
$\begin{array}{rl}\parallel {\rho }_{n+1}-{\rho }_{n}\parallel \le & \parallel {J}_{{r}_{n+1}}\left({x}_{n+1}-{r}_{n+1}A{x}_{n+1}\right)-{J}_{{r}_{n+1}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)\parallel \\ +\parallel {J}_{{r}_{n+1}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)-{J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)\parallel \\ \le & \parallel \left({x}_{n+1}-{r}_{n+1}A{x}_{n+1}\right)-\left({x}_{n}-{r}_{n}A{x}_{n}\right)\parallel \\ +\parallel {J}_{{r}_{n+1}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)-{J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)\parallel \\ \le & \parallel {x}_{n+1}-{x}_{n}\parallel +|{r}_{n+1}-{r}_{n}|\parallel A{x}_{n}\parallel \\ +\parallel {J}_{{r}_{n+1}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)-{J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)\parallel .\end{array}$
(2.3)
On the other hand, we have
$\begin{array}{rl}\parallel {z}_{n+1}-{z}_{n}\parallel & \le \parallel \left({\alpha }_{n+1}u+\left(1-{\alpha }_{n+1}\right){\rho }_{n+1}\right)-\left({\alpha }_{n}u+\left(1-{\alpha }_{n}\right){\rho }_{n}\right)\parallel \\ \le \left(1-{\alpha }_{n+1}\right)\parallel {\rho }_{n+1}-{\rho }_{n}\parallel +|{\alpha }_{n+1}-{\alpha }_{n}|\parallel {\rho }_{n}-u\parallel .\end{array}$
(2.4)
Substituting (2.3) into (2.4), we find that
$\begin{array}{rl}\parallel {z}_{n+1}-{z}_{n}\parallel \le & \parallel \left({\alpha }_{n+1}u+\left(1-{\alpha }_{n+1}\right){\rho }_{n+1}\right)-\left({\alpha }_{n}u+\left(1-{\alpha }_{n}\right){\rho }_{n}\right)\parallel \\ \le & \left(1-{\alpha }_{n+1}\right)\parallel {x}_{n+1}-{x}_{n}\parallel +|{r}_{n+1}-{r}_{n}|\parallel A{x}_{n}\parallel +|{\alpha }_{n+1}-{\alpha }_{n}|\parallel {\rho }_{n}-u\parallel \\ +\left(1-{\alpha }_{n+1}\right)\parallel {J}_{{r}_{n+1}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)-{J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)\parallel .\end{array}$
(2.5)
Put ${\xi }_{n}={x}_{n}-{r}_{n}A{x}_{n}$. Since B is monotone, we find that
$〈{J}_{{r}_{n+1}}{\xi }_{n}-{J}_{{r}_{n}}{\xi }_{n},\frac{{\xi }_{n}-{J}_{{r}_{n+1}}{\xi }_{n}}{{r}_{n+1}}-\frac{{\xi }_{n}-{J}_{{r}_{n}}{\xi }_{n}}{{r}_{n}}〉\ge 0.$
It follows that $〈{J}_{{r}_{n+1}}{\xi }_{n}-{J}_{{r}_{n}}{\xi }_{n},\left(1-\frac{{r}_{n+1}}{{r}_{n}}\right)\left({\xi }_{n}-{J}_{{r}_{n}}{\xi }_{n}\right)〉\ge {\parallel {J}_{{r}_{n+1}}{\xi }_{n}-{J}_{{r}_{n}}{\xi }_{n}\parallel }^{2}$. This yields that $|{r}_{n+1}-{r}_{n}|\parallel {\xi }_{n}-{J}_{{r}_{n}}{\xi }_{n}\parallel \ge {r}_{n}\parallel {J}_{{r}_{n+1}}{\xi }_{n}-{J}_{{r}_{n}}{\xi }_{n}\parallel$. This combines with (2.5) to yield that
$\begin{array}{rl}\parallel {z}_{n+1}-{z}_{n}\parallel \le & \left(1-{\alpha }_{n+1}\right)\parallel {x}_{n+1}-{x}_{n}\parallel +|{r}_{n+1}-{r}_{n}|\parallel A{x}_{n}\parallel \\ +|{\alpha }_{n+1}-{\alpha }_{n}|\parallel {\rho }_{n}-u\parallel +\frac{|{r}_{n+1}-{r}_{n}|}{{r}_{n}}\parallel {\xi }_{n}-{J}_{{r}_{n}}{\xi }_{n}\parallel .\end{array}$
(2.6)
On the other hand, we have
$\begin{array}{rl}\parallel {y}_{n+1}-{y}_{n}\parallel \le & {\gamma }_{n+1}\parallel {z}_{n+1}-{z}_{n}\parallel +|{\gamma }_{n+1}-{\gamma }_{n}|\parallel {z}_{n}-{T}^{n}{z}_{n}\parallel \\ +\left(1-{\gamma }_{n+1}\right)\parallel {T}^{n+1}{z}_{n+1}-{T}^{n}{z}_{n}\parallel .\end{array}$
(2.7)
Substituting (2.6) into (2.7), we find that
$\begin{array}{r}\parallel {y}_{n+1}-{y}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le |{r}_{n+1}-{r}_{n}|\parallel A{x}_{n}\parallel +|{\alpha }_{n+1}-{\alpha }_{n}|\parallel {\rho }_{n}-u\parallel \\ \phantom{\rule{2em}{0ex}}+\frac{|{r}_{n+1}-{r}_{n}|}{{r}_{n}}\parallel {\xi }_{n}-{J}_{{r}_{n}}{\xi }_{n}\parallel +|{\gamma }_{n+1}-{\gamma }_{n}|\parallel {z}_{n}-{T}^{n}{z}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+\left(1-{\gamma }_{n+1}\right)\parallel {T}^{n+1}{z}_{n+1}-{T}^{n}{z}_{n}\parallel .\end{array}$
It follows from Restrictions (a), (c) and (d) that
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {y}_{n+1}-{y}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0.$
It follows from Lemma 1.1 that ${lim}_{n\to \mathrm{\infty }}\parallel {y}_{n}-{x}_{n}\parallel =0$. Since ${x}_{n+1}-{x}_{n}=\left(1-{\beta }_{n}\right)\left({y}_{n}-{x}_{n}\right)$, we find that ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n+1}-{x}_{n}\parallel =0$. Notice that
$\begin{array}{r}{\parallel {J}_{{r}_{n}}\left(I-{r}_{n}A\right){x}_{n}-{J}_{{r}_{n}}\left(I-{r}_{n}A\right)p\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel \left({x}_{n}-p\right)-{r}_{n}\left(A{x}_{n}-Ap\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel {x}_{n}-p\parallel }^{2}-2{r}_{n}〈{x}_{n}-p,A{x}_{n}-Ap〉+{{r}_{n}}^{2}{\parallel A{x}_{n}-Ap\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-p\parallel }^{2}-{r}_{n}\left(2\alpha -{r}_{n}\right){\parallel A{x}_{n}-Ap\parallel }^{2}.\end{array}$
(2.8)
Since the norm is convex, we see from (2.2) and (2.8) that
$\begin{array}{rl}{\parallel {x}_{n+1}-p\parallel }^{2}\le & {\beta }_{n}{\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {y}_{n}-p\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {z}_{n}-p\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {\alpha }_{n}\left(u-p\right)+\left(1-{\alpha }_{n}\right)\left({J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)-p\right)\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-p\parallel }^{2}+{\alpha }_{n}\left(1-{\beta }_{n}\right){\parallel u-p\parallel }^{2}\\ +\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel {J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)-p\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}+{\alpha }_{n}{\parallel u-p\parallel }^{2}-{r}_{n}\left(2\alpha -{r}_{n}\right)\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel A{x}_{n}-Ap\parallel }^{2}.\end{array}$
(2.9)
This yields that
$\begin{array}{r}{r}_{n}\left(2\alpha -{r}_{n}\right)\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel A{x}_{n}-Ap\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-p\parallel }^{2}+{\alpha }_{n}{\parallel u-p\parallel }^{2}-{\parallel {x}_{n+1}-p\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(\parallel {x}_{n}-p\parallel +\parallel {x}_{n+1}-p\parallel \right)\parallel {x}_{n+1}-{x}_{n}\parallel +{\alpha }_{n}{\parallel u-p\parallel }^{2}.\end{array}$
In view of Restrictions (a), (b) and (c), we obtain that
$\underset{n\to \mathrm{\infty }}{lim}\parallel A{x}_{n}-Ap\parallel =0.$
(2.10)
Notice that
$\begin{array}{rl}{\parallel {\rho }_{n}-p\parallel }^{2}=& {\parallel {J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)-{J}_{{r}_{n}}\left(p-{r}_{n}Ap\right)\parallel }^{2}\\ \le & 〈\left({x}_{n}-{r}_{n}A{x}_{n}\right)-\left(p-{r}_{n}Ap\right),{\rho }_{n}-p〉\\ =& \frac{1}{2}\left({\parallel \left({x}_{n}-{r}_{n}A{x}_{n}\right)-\left(p-{r}_{n}Ap\right)\parallel }^{2}+{\parallel {\rho }_{n}-p\parallel }^{2}\\ -{\parallel \left({x}_{n}-{r}_{n}A{x}_{n}\right)-\left(p-{r}_{n}Ap\right)-\left({\rho }_{n}-p\right)\parallel }^{2}\right)\\ \le & \frac{1}{2}\left({\parallel {x}_{n}-p\parallel }^{2}+{\parallel {\rho }_{n}-p\parallel }^{2}-{\parallel {x}_{n}-{\rho }_{n}-{r}_{n}\left(A{x}_{n}-Ap\right)\parallel }^{2}\right)\\ \le & \frac{1}{2}\left({\parallel {x}_{n}-p\parallel }^{2}+{\parallel {\rho }_{n}-p\parallel }^{2}-{\parallel {x}_{n}-{\rho }_{n}\parallel }^{2}-{r}_{n}^{2}{\parallel A{x}_{n}-Ap\parallel }^{2}\\ +2{r}_{n}\parallel {x}_{n}-{\rho }_{n}\parallel \parallel A{x}_{n}-Ap\parallel \right)\\ \le & \frac{1}{2}\left({\parallel {x}_{n}-p\parallel }^{2}+{\parallel {\rho }_{n}-p\parallel }^{2}-{\parallel {x}_{n}-{\rho }_{n}\parallel }^{2}+2{r}_{n}\parallel {x}_{n}-{\rho }_{n}\parallel \parallel A{x}_{n}-Ap\parallel \right).\end{array}$
It follows that
${\parallel {\rho }_{n}-p\parallel }^{2}\le {\parallel {x}_{n}-p\parallel }^{2}-{\parallel {x}_{n}-{\rho }_{n}\parallel }^{2}+2{r}_{n}\parallel {x}_{n}-{\rho }_{n}\parallel \parallel A{x}_{n}-Ap\parallel .$
(2.11)
This yields that
$\begin{array}{rl}{\parallel {z}_{n}-p\parallel }^{2}& \le {\alpha }_{n}{\parallel u-p\parallel }^{2}+\left(1-{\alpha }_{n}\right){\parallel {\rho }_{n}-p\parallel }^{2}\\ \le {\alpha }_{n}{\parallel u-p\parallel }^{2}+{\parallel {x}_{n}-p\parallel }^{2}-\left(1-{\alpha }_{n}\right){\parallel {x}_{n}-{\rho }_{n}\parallel }^{2}+2{r}_{n}\parallel {x}_{n}-{\rho }_{n}\parallel \parallel A{x}_{n}-Ap\parallel .\end{array}$
It follows from (2.2) that
$\begin{array}{rl}{\parallel {x}_{n+1}-p\parallel }^{2}\le & {\beta }_{n}{\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {y}_{n}-p\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-p\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {z}_{n}-p\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}+{\alpha }_{n}{\parallel u-p\parallel }^{2}-\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel {x}_{n}-{\rho }_{n}\parallel }^{2}\\ +2{r}_{n}\left(1-{\beta }_{n}\right)\parallel {x}_{n}-{\rho }_{n}\parallel \parallel A{x}_{n}-Ap\parallel .\end{array}$
We therefore obtain that
$\begin{array}{r}\left(1-{\alpha }_{n}\right)\left(1-{\beta }_{n}\right){\parallel {x}_{n}-{\rho }_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-p\parallel }^{2}+{\alpha }_{n}{\parallel u-p\parallel }^{2}-{\parallel {x}_{n+1}-p\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2{r}_{n}\left(1-{\beta }_{n}\right)\parallel {x}_{n}-{\rho }_{n}\parallel \parallel A{x}_{n}-Ap\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(\parallel {x}_{n}-p\parallel +\parallel {x}_{n+1}-p\parallel \right)\parallel {x}_{n}-{x}_{n+1}\parallel +{\alpha }_{n}{\parallel u-p\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2{r}_{n}\left(1-{\beta }_{n}\right)\parallel {x}_{n}-{\rho }_{n}\parallel \parallel A{x}_{n}-Ap\parallel .\end{array}$
In view of Restrictions (a), (b) and (c), we find from (2.10) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{\rho }_{n}\parallel =0.$
(2.12)
Next, we show that ${lim sup}_{n\to \mathrm{\infty }}〈u-\overline{x},{\rho }_{n}-\overline{x}〉\le 0$, where $\overline{x}={P}_{F\left(T\right)\cap {\left(A+B\right)}^{-1}\left(0\right)}u$. To show it, we can choose a subsequence $\left\{{\rho }_{{n}_{i}}\right\}$ of $\left\{{\rho }_{n}\right\}$ such that
$\underset{n\to \mathrm{\infty }}{lim sup}〈u-\overline{x},{\rho }_{n}-\overline{x}〉=\underset{i\to \mathrm{\infty }}{lim}〈u-\overline{x},{\rho }_{{n}_{i}}-\overline{x}〉.$
Since ${\rho }_{{n}_{i}}$ is bounded, we can choose a subsequence $\left\{{\rho }_{{n}_{{i}_{j}}}\right\}$ of $\left\{{\rho }_{{n}_{i}}\right\}$ which converges weakly to some point x. We may assume, without loss of generality, that ${\rho }_{{n}_{i}}$ converges weakly to x. Since ${\rho }_{n}={J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)$, we find that $\frac{{x}_{n}-{\rho }_{n}}{{r}_{n}}-A{x}_{n}\in B{\rho }_{n}$. Since B is monotone, we get, for any $\left(\mu ,\nu \right)\in B$, that $〈{\rho }_{n}-\mu ,\frac{{x}_{n}-{\rho }_{n}}{{r}_{n}}-A{x}_{n}-\nu 〉\ge 0$. Replacing n by ${n}_{i}$ and letting $i\to \mathrm{\infty }$, we obtain from (2.12) that $〈x-\mu ,-Ax-\nu 〉\ge 0$. This means $-Ax\in Bx$, that is, $0\in \left(A+B\right)\left(x\right)$. Hence we get $x\in {\left(A+B\right)}^{-1}\left(0\right)$. Next, we show that $x\in F\left(T\right)$. Notice that
$\begin{array}{rl}\parallel {z}_{n}-{x}_{n}\parallel & \le \parallel {Proj}_{C}\left({\alpha }_{n}u+\left(1-{\alpha }_{n}\right){J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)\right)-{x}_{n}\parallel \\ \le {\alpha }_{n}\parallel u-{x}_{n}\parallel +\left(1-{\alpha }_{n}\right)\parallel {J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)-{x}_{n}\parallel .\end{array}$
In view of Restriction (a), we find from (2.12) that ${lim}_{n\to \mathrm{\infty }}\parallel {z}_{n}-{x}_{n}\parallel =0$. Note that
$\begin{array}{r}\parallel \left({\gamma }_{n}{x}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{x}_{n}\right)-{x}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel \left({\gamma }_{n}{x}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{x}_{n}\right)-\left({\gamma }_{n}{z}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{z}_{n}\right)\parallel \\ \phantom{\rule{2em}{0ex}}+\parallel \left({\gamma }_{n}{z}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{z}_{n}\right)-{x}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le {\gamma }_{n}\parallel {x}_{n}-{z}_{n}\parallel +\left(1-{\gamma }_{n}\right)\parallel {T}^{n}{x}_{n}-{T}^{n}{z}_{n}\parallel +\parallel \left({\gamma }_{n}{z}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{z}_{n}\right)-{x}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le L\parallel {x}_{n}-{z}_{n}\parallel +\parallel {y}_{n}-{x}_{n}\parallel .\end{array}$
It follows from (2.8) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel \left({\gamma }_{n}{x}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{x}_{n}\right)-{x}_{n}\parallel =0.$
(2.13)
Note that
$\begin{array}{rl}\parallel {T}^{n}{x}_{n}-{x}_{n}\parallel \le & \parallel {T}^{n}{x}_{n}-\left({\gamma }_{n}{x}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{x}_{n}\right)\parallel \\ +\parallel \left({\gamma }_{n}{x}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{x}_{n}\right)-{x}_{n}\parallel \\ \le & {\gamma }_{n}\parallel {T}^{n}{x}_{n}-{x}_{n}\parallel +\parallel \left({\gamma }_{n}{x}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{x}_{n}\right)-{x}_{n}\parallel .\end{array}$
It follows that $\left(1-{\gamma }_{n}\right)\parallel {T}^{n}{x}_{n}-{x}_{n}\parallel \le \parallel \left({\gamma }_{n}{x}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{x}_{n}\right)-{x}_{n}\parallel$. This implies from Restriction (d) and (2.13) that ${lim}_{n\to \mathrm{\infty }}\parallel {T}^{n}{x}_{n}-{x}_{n}\parallel =0$. Since T is uniformly L-Lipschitz continuous, we can obtain that ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_{n}-{x}_{n}\parallel =0$. In view of Lemma 1.4, we find that $x\in F\left(T\right)$. This implies that
$\underset{n\to \mathrm{\infty }}{lim sup}〈u-\overline{x},{\rho }_{n}-\overline{x}〉=〈u-\overline{x},x-\overline{x}〉\le 0.$
On the other hand, we have
$\begin{array}{rl}{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}\le & {\beta }_{n}{\parallel {x}_{n}-\overline{x}\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {y}_{n}-\overline{x}\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-x\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {z}_{n}-\overline{x}\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-x\parallel }^{2}+\left(1-{\beta }_{n}\right){\parallel {\alpha }_{n}\left({u}_{n}-\overline{x}\right)+\left(1-{\alpha }_{n}\right)\left({\rho }_{n}-\overline{x}\right)\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-x\parallel }^{2}+{\left(1-{\alpha }_{n}\right)}^{2}\left(1-{\beta }_{n}\right){\parallel {\rho }_{n}-x\parallel }^{2}\\ +2{\alpha }_{n}\left(1-{\beta }_{n}\right)〈u-\overline{x},{\rho }_{n}-x〉\\ \le & \left(1-{\alpha }_{n}\left(1-{\beta }_{n}\right)\right){\parallel {x}_{n}-\overline{x}\parallel }^{2}+2{\alpha }_{n}\left(1-{\beta }_{n}\right)〈u-\overline{x},{\rho }_{n}-\overline{x}〉.\end{array}$

From Lemma 1.3, we find that ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-\overline{x}\parallel =0$. This completes the proof. □

If T is asymptotically nonexpansive, then we find the following result.

Corollary 2.2 Let C be a nonempty closed convex subset of H. Let $T:C\to C$ be an asymptotically nonexpansive mapping. Let $A:C\to H$ be an α-inverse-strongly monotone mapping, and let B be a maximal monotone operator on H. Assume that $F\left(T\right)\cap {\left(A+B\right)}^{-1}\left(0\right)\ne \mathrm{\varnothing }$. Let $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ be real number sequences in $\left(0,1\right)$. Let ${J}_{{r}_{n}}={\left(I+{r}_{n}B\right)}^{-1}$, where $\left\{{r}_{n}\right\}$ is a positive real number sequence. Let $\left\{{x}_{n}\right\}$ be a sequence in C generated by: ${x}_{1}\in C$ is chosen arbitrarily and
$\left\{\begin{array}{c}{z}_{n}={P}_{C}\left({\alpha }_{n}u+\left(1-{\alpha }_{n}\right){J}_{{r}_{n}}\left({x}_{n}-{r}_{n}A{x}_{n}\right)\right),\hfill \\ {x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){T}^{n}{z}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.\hfill \end{array}$
Assume that the sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\gamma }_{n}\right\}$ and $\left\{{r}_{n}\right\}$ satisfy the following restrictions:
1. (a)

$0, ${lim}_{n\to \mathrm{\infty }}|{r}_{n+1}-{r}_{n}|=0$;

2. (b)

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

3. (c)

$0,

where a, b, c and d are some real numbers. If T is asymptotically regular, then the sequence $\left\{{x}_{n}\right\}$ converges strongly to some point $\overline{x}$, where $\overline{x}={P}_{F\left(T\right)\cap {\left(A+B\right)}^{-1}\left(0\right)}u$.

## 3 Applications

In this section, we shall consider equilibrium problems and variational inequalities.

Let F be a bifunction of $C×C$ into , where denotes the set of real numbers. Recall the following equilibrium problem:

In this work, we use $EP\left(F\right)$ to denote the solution set of the equilibrium problem.

To study the equilibrium problems, 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 semi-continuous.

Putting $F\left(x,y\right)=〈Ax,y-x〉$ for every $x,y\in C$, we see that the equilibrium problem is reduced to the variational inequality (1.1).

The following lemma can be found in [40].

Lemma 3.1 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
${T}_{r}x=\left\{z\in C:F\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\right\}$
(3.1)
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 3.2 [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 ${A}_{F}$ be a multivalued mapping of H into itself defined by
${A}_{F}x=\left\{\begin{array}{cc}\left\{z\in H:F\left(x,y\right)\ge 〈y-x,z〉,\mathrm{\forall }y\in C\right\},\hfill & x\in C,\hfill \\ \mathrm{\varnothing },\hfill & x\notin C.\hfill \end{array}$
(3.2)
Then ${A}_{F}$ is a maximal monotone operator with the domain $D\left({A}_{F}\right)\subset C$, $EP\left(F\right)={A}_{F}^{-1}\left(0\right)$ and
${T}_{r}x={\left(I+r{A}_{F}\right)}^{-1}x,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in H,r>0,$

where ${T}_{r}$ is defined as in (3.1).

The following result is not derived based on Theorem 2.1 and Lemma 3.2.

Theorem 3.3 Let C be a nonempty closed convex subset of H. Let $T:C\to C$ be an asymptotically κ-strict pseudocontraction. Let F be a bifunction from $C×C$ to which satisfies (A1)-(A4). Assume that $F\left(T\right)\cap EP\left(F\right)\ne \mathrm{\varnothing }$. Let $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\gamma }_{n}\right\}$ be real number sequences in $\left(0,1\right)$. Let $\left\{{x}_{n}\right\}$ be a sequence in C generated by: ${x}_{1}\in C$ is chosen arbitrarily and
Assume that the sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n}\right\}$ and $\left\{{r}_{n}\right\}$ satisfy the following restrictions:
1. (a)

$0, ${lim}_{n\to \mathrm{\infty }}|{r}_{n+1}-{r}_{n}|=0$;

2. (b)

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

3. (c)

$0;

4. (d)

$\kappa \le {\gamma }_{n}\le e<1$, ${lim}_{n\to \mathrm{\infty }}|{\gamma }_{n+1}-{\gamma }_{n}|=0$,

where a, b, c, d and e are some real numbers. If T is asymptotically regular, then the sequence $\left\{{x}_{n}\right\}$ converges strongly to some point $\overline{x}$, where $\overline{x}={P}_{F\left(T\right)\cap EP\left(F\right)}u$.

If T is asymptotically nonexpansive, then Theorem 3.3 is reduced to the following.

Corollary 3.4 Let C be a nonempty closed convex subset of H. Let $T:C\to C$ be an asymptotically nonexpansive mapping. Let F be a bifunction from $C×C$ to which satisfies (A1)-(A4). Assume that $F\left(T\right)\cap EP\left(F\right)\ne \mathrm{\varnothing }$. Let $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ be real number sequences in $\left(0,1\right)$. Let $\left\{{x}_{n}\right\}$ be a sequence in C generated by: ${x}_{1}\in C$ is chosen arbitrarily and
Assume that the sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n}\right\}$ and $\left\{{r}_{n}\right\}$ satisfy the following restrictions:
1. (a)

$0, ${lim}_{n\to \mathrm{\infty }}|{r}_{n+1}-{r}_{n}|=0$;

2. (b)

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

3. (c)

$0,

where a, b, c and d are some real numbers. If T is asymptotically regular, then the sequence $\left\{{x}_{n}\right\}$ converges strongly to some point $\overline{x}$, where $\overline{x}={P}_{F\left(T\right)\cap EP\left(F\right)}u$.

Let H be a Hilbert space and $f:H\to \left(-\mathrm{\infty },+\mathrm{\infty }\right]$ be a proper convex lower semicontinuous function. Then the subdifferential ∂f of f is defined as follows:
$\partial f\left(x\right)=\left\{y\in H:f\left(z\right)\ge f\left(x\right)+〈z-x,y〉,z\in H\right\},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in H.$
From Rockafellar [41], we find that ∂f is maximal monotone. It is easy to verify that $0\in \partial f\left(x\right)$ if and only if $f\left(x\right)={min}_{y\in H}f\left(y\right)$. 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.

Lemma 3.5 [2]

Let C be a nonempty closed convex subset of a real Hilbert space H, ${Proj}_{C}$ the metric projection from H onto C, $\partial {I}_{C}$ the subdifferential of ${I}_{C}$, where ${I}_{C}$ is defined above and ${J}_{\lambda }={\left(I+\lambda \partial {I}_{C}\right)}^{-1}$. Then
$y={J}_{\lambda }x\phantom{\rule{1em}{0ex}}⟺\phantom{\rule{1em}{0ex}}y={Proj}_{C}x,\phantom{\rule{1em}{0ex}}x\in H,y\in C.$

Now, we consider a variation inequality problem.

Theorem 3.6 Let C be a nonempty closed convex subset of H. Let $T:C\to C$ be an asymptotically κ-strict pseudocontraction. Let $A:C\to H$ be an α-inverse-strongly monotone mapping. Assume that $F\left(T\right)\cap VI\left(C,A\right)\ne \mathrm{\varnothing }$. Let $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\gamma }_{n}\right\}$ be real number sequences in $\left(0,1\right)$. Let $\left\{{x}_{n}\right\}$ be a sequence in C generated by: ${x}_{1}\in C$ is chosen arbitrarily and
$\left\{\begin{array}{c}{z}_{n}={P}_{C}\left({\alpha }_{n}u+\left(1-{\alpha }_{n}\right){P}_{C}\left({x}_{n}-{r}_{n}A{x}_{n}\right)\right),\hfill \\ {y}_{n}={\gamma }_{n}{z}_{n}+\left(1-{\gamma }_{n}\right){T}^{n}{z}_{n},\hfill \\ {x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){y}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.\hfill \end{array}$
Assume that the sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n}\right\}$ and $\left\{{r}_{n}\right\}$ satisfy the following restrictions:
1. (a)

$0, ${lim}_{n\to \mathrm{\infty }}|{r}_{n+1}-{r}_{n}|=0$;

2. (b)

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

3. (c)

$0;

4. (d)

$\kappa \le {\gamma }_{n}\le e<1$, ${lim}_{n\to \mathrm{\infty }}|{\gamma }_{n+1}-{\gamma }_{n}|=0$,

where a, b, c, d and e are some real numbers. If T is asymptotically regular, then the sequence $\left\{{x}_{n}\right\}$ converges strongly to some point $\overline{x}$, where $\overline{x}={P}_{F\left(T\right)\cap VI\left(C,A\right)}u$.

Proof Put $Bx=\partial {I}_{C}$. Next, we show that $VI\left(C,A\right)={\left(A+\partial {I}_{C}\right)}^{-1}\left(0\right)$. Notice that
$\begin{array}{rl}x\in {\left(A+\partial {I}_{C}\right)}^{-1}\left(0\right)& \phantom{\rule{1em}{0ex}}⟺\phantom{\rule{1em}{0ex}}0\in Ax+\partial {I}_{C}x\\ \phantom{\rule{1em}{0ex}}⟺\phantom{\rule{1em}{0ex}}-Ax\in \partial {I}_{C}x\\ \phantom{\rule{1em}{0ex}}⟺\phantom{\rule{1em}{0ex}}〈Ax,y-x〉\ge 0\\ \phantom{\rule{1em}{0ex}}⟺\phantom{\rule{1em}{0ex}}x\in VI\left(C,A\right).\end{array}$

From Lemma 3.5, we can conclude the desired conclusion immediately. □

## Declarations

### Acknowledgements

The author is grateful to the referees for useful suggestions which improved the contents of the article.

## Authors’ Affiliations

(1)
School of Mathematics and Information Science, Henan University, Kaifeng, 475000, China

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