Strong convergence theorems for quasi-nonexpansive mappings and maximal monotone operators in Hilbert spaces

Abstract

We present the strong convergence theorem for the iterative scheme for finding a common element of the fixed-point set of a quasi-nonexpansive mapping and the zero set of the sums of maximal monotone operators in Hilbert spaces. Our results extend and improve the recent results of Takahashi et al. (J. Optim. Theory Appl. 147:27-41, 2010) and Takahashi and Takahashi (Nonlinear Anal. 69:1025-1033, 2008).

MSC:47H05, 47H09, 47J25.

1 Introduction

Let H be a real Hilbert space and let K be a nonempty closed convex subset of H. Let $T:K\to K$ be a mapping. We denote by $Fix(T)$ the fixed-point set of T, that is, $Fix(T)=\{x\in K:Tx=x\}$. A mapping $T:K\to K$ is nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all $x,y\in K$. Approximation methods for fixed points of nonexpansive mappings have attracted considerable attention (see [1–5]). A mapping $T:K\to K$ is quasi-nonexpansive if $Fix(T)\ne \mathrm{\varnothing }$ and $\parallel Tx-y\parallel \le \parallel x-y\parallel$ for all $x\in K$ and $y\in Fix(T)$. It is well known that the fixed-point set of a quasi-nonexpansive mapping is closed and convex (see [6, 7]). There are some quasi-nonexpansive mappings which are not nonexpansive (see [8–10]). For example, the level set of a continuous convex function is characterized as the fixed-point set of a nonlinear mapping called the subgradient projection, which is not nonexpansive but quasi-nonexpansive. Quasi-nonexpansive mappings have been discussed in the recent literature (see [9–11]).

We say that a mapping $T:K\to K$ is demiclosed at zero if for any sequence $\{x_n\}\subset K$ which converges weakly to x, the strong convergence of the sequence $\{T{x}_n\}$ to zero implies $Tx=0$. It is well known that $I-T$ is demiclosed whenever T is nonexpansive. In fact, this property is satisfied for more general mappings (see [12, 13]).

Let B be a mapping from H into $2^H$. The effective domain of B is denoted by $dom(B)$, namely, $dom(B)=\{x\in H:Bx\ne \mathrm{\varnothing }\}$. The graph of B is

$$Gra(B)=\{(v,r)\in H\times H:r\in Bv\}.$$

A multi-valued mapping B is said to be monotone if

$$〈x-y,u-v〉\ge 0\text{for all }x,y\in dom(B),u\in Bx,\text{ and }v\in By.$$

A monotone operator B is said to be maximal if its graph is not properly contained in the graph of any other monotone operator. For a maximal monotone operator B on H and $r>0$, we define a single-valued operator $J_{rB}={(I+rB)}^{-1}:H\to dom(B)$, which is called the resolvent of B for r. It is well known that $J_{rB}$ is firmly nonexpansive, that is,

$$〈x-y,J_{rB}x-J_{rB}y〉\ge {\parallel J_{rB}x-J_{rB}y\parallel }^2\text{for any }x,y\in H.$$

A basic problem for maximal monotone operator B is to

$$\text{find }x\in H\text{such that}0\in Bx.$$

(1.1)

The classical method for solving problem (1.1) is the proximal point algorithm which was first introduced by Martinet [14]. Rockafellar [15] obtained the weak convergence of the proximal point algorithm for maximal monotone operators. Güler [16] constructed a proximal point iteration that converges weakly but not strongly. Some researchers have devoted their work to modifications of the proximal point algorithm in order to obtain the strong convergence theorem (see [17, 18]). For a positive constant α, a mapping $A:K\to H$ is said to be α-inverse strongly monotone if

$$〈x-y,Ax-Ay〉\ge \alpha {\parallel Ax-Ay\parallel }^2\text{for all }x,y\in K.$$

We write ${(A+B)}^{-1}0$ for the zero set of $A+B$, that is, ${(A+B)}^{-1}0=\{x\in K:0\in (A+B)x\}$, where the mapping $A:C\to H$ is inverse strongly monotone and B is maximal monotone. It is well known that ${(A+B)}^{-1}0=Fix(J_{\lambda B}(I-\lambda A))$ for all $\lambda >0$ (see [19]). Takahashi et al. [20] presented the following iterative sequence. Let $u\in K$, $x_1=x\in K$ and let $\{x_n\}$ be a sequence generated by

$$x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)J_{{\lambda }_{n}B}(x_n-{\lambda }_{n}A{x}_n).$$

Under appropriate conditions they proved that the sequence $\{x_n\}$ converges strongly to a point $z_0\in {(A+B)}^{-1}0$. Lin and Takahashi [21] introduced an iterative sequence that converges strongly to an element of ${(A+B)}^{-1}0\cap F^{-1}0$, where F is another maximal monotone operator. Takahashi et al. [22] established an iterative scheme for finding a point of ${(A+B)}^{-1}0\cap Fix(T)$ as follows. Let $x_1=x\in K$ and let $\{x_n\}$ be a sequence generated by

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)T[{\alpha }_{n}x+(1-{\alpha }_n)J_{{\lambda }_{n}B}(x_n-{\lambda }_{n}A{x}_n)],$$

where $T:K\to K$ is a nonexpansive mapping.

Motivated by the above results, especially by Chuang et al. [11] and Takahashi et al. [22], we obtain the strong convergence theorem for the iterative scheme for finding a common element of the fixed-point set of a quasi-nonexpansive mapping and the zero set of the sums of maximal monotone operators in Hilbert spaces. Our results extend and improve the recent results of [22] and [23].

The rest of this paper is organized as follows. Section 2 contains some important facts and tools. In Section 3, we introduce a new iterative scheme for finding a common element of the fixed-point set of a quasi-nonexpansive mapping and the zero set of the sums of maximal monotone operators, and we prove strong convergence theorem in Hilbert spaces.

2 Preliminaries

Throughout this paper, let H be a real Hilbert space with inner product $〈\cdot ,\cdot 〉$ and norm $\parallel \cdot \parallel$, and let K be a nonempty closed convex subset of H. Let ℕ be the set of positive integers. We denote the strong convergence and the weak convergence of $\{x_n\}$ to x by $x_n\to x$ and $x_n\rightharpoonup x$, respectively. For any $x\in H$, there exists a unique point $P_{K}x\in K$ such that

$$\parallel x-P_{K}x\parallel \le \parallel x-y\parallel ,\mathrm{\forall }y\in K.$$

$P_K$ is called the metric projection of H onto K. Note that $P_K$ is a nonexpansive mapping. For $x\in H$ and $z\in K$, we have

$$z=P_{K}x⟺〈x-z,y-z〉\le 0\text{for every }y\in K.$$

(2.1)

Let f be a proper lower semicontinuous convex function of H into $(-\mathrm{\infty },+\mathrm{\infty }]$. The subdifferential ∂f of f is defined as

$$\partial f(x)=\{z\in H:f(y)-f(x)\ge 〈z,y-x〉,\mathrm{\forall }y\in H\}$$

(2.2)

for all $x\in H$. Rockafellar [24] claimed that ∂f is a maximal monotone operator. Let ${\delta }_K$ be the indicator function of K, i.e.,

$${\delta }_K(x)=\{\begin{array}{cc}0,\hfill & x\in K,\hfill \\ +\mathrm{\infty },\hfill & x\notin K.\hfill \end{array}$$

The subdifferential $\partial {\delta }_K$ of ${\delta }_K$ is a maximal monotone operator since ${\delta }_K$ is a proper lower semicontinuous convex function on H. The resolvent $J_{r\partial {\delta }_K}$ of $\partial {\delta }_K$ for r is $P_K$ (see [21]).

Let $A:K\to H$ be a nonlinear mapping. The variational inequality problem is to find $\overline{x}\in K$ such that

$$〈A\overline{x},y-\overline{x}〉\ge 0\text{for every }y\in K.$$

(2.3)

The solution set of (2.3) is denoted by $\mathit{VI}(K,A)$. Some methods have been proposed to study the variational inequality problem (see [25–28] and the references therein). It is easy to see that $\mathit{VI}(K,A)={(A+\partial {\delta }_K)}^{-1}0$, where A is an inverse strongly monotone mapping of K into H (for more details, see [21]).

We collect some useful lemmas.

Lemma 2.1 [29]

Let $A:K\to H$ be an α-inverse strongly monotone mapping. For all $x,y\in K$ and $\lambda >0$, we have

$${\parallel (I-\lambda A)x-(I-\lambda A)y\parallel }^2\le {\parallel x-y\parallel }^2+\lambda (\lambda -2\alpha ){\parallel Ax-Ay\parallel }^2.$$

In particular, if $0<\lambda \le 2\alpha$, then $I-\lambda A$ is a nonexpansive mapping.

Lemma 2.2 [30]

Let $\{{\mathrm{\Gamma }}_n\}$ be a sequence of real numbers that does not decrease at infinity in the sense that there exists a subsequence $\{{\mathrm{\Gamma }}_{n_j}\}$ of $\{{\mathrm{\Gamma }}_n\}$ such that

$${\mathrm{\Gamma }}_{n_j}<{\mathrm{\Gamma }}_{n_j+1}\mathit{\text{for all}}j\in \mathbb{N}.$$

Define the sequence ${\{\tau (n)\}}_{n\ge n_0}$ of integers as follows:

$$\tau (n)=\underset{k}{max}\{k\le n:{\mathrm{\Gamma }}_k<{\mathrm{\Gamma }}_{k+1}\},$$

where $n_0\in \mathbb{N}$ such that $\{k\le n_0:{\mathrm{\Gamma }}_k<{\mathrm{\Gamma }}_{k+1}\}\ne \mathrm{\varnothing }$. Then, for all $n\ge n_0$, the following hold:

• (1) $\tau (n)\le \tau (n+1)\le \cdots$ and $\tau (n)\to \mathrm{\infty }$;

• (2) ${\mathrm{\Gamma }}_{\tau (n)}\le {\mathrm{\Gamma }}_{\tau (n)+1}$ and ${\mathrm{\Gamma }}_n\le {\mathrm{\Gamma }}_{\tau (n)+1}$.

Lemma 2.3 [22]

Let B be a maximal monotone operator on H. Then the following holds:

$$\frac{s-t}{s}〈J_{sB}x-J_{tB}x,J_{sB}x-x〉\ge {\parallel J_{sB}x-J_{tB}x\parallel }^2$$

for all $s,t>0$ and $x\in H$.

The following lemma is an immediate consequence of the inner product on H.

Lemma 2.4 For all $x,y\in H$, the inequality ${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,x+y〉$ holds.

Lemma 2.5 [31]

Let $\{a_n\}$ be a sequence of nonnegative real numbers satisfying $a_{n+1}\le (1-{\alpha }_n)a_n+{\alpha }_n{\beta }_n$, where

• (i) $\{{\alpha }_n\}\subset (0,1)$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n\le 0$.

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

3 Strong convergence theorems

In this section, a new iterative scheme for finding a common element of the fixed-point set of a quasi-nonexpansive mapping and the zero set of the sums of maximal monotone operators is presented.

Theorem 3.1 Let K be a nonempty closed convex subset of a real Hilbert space H. Let $A:K\to H$ and $C:K\to H$ be α-inverse strongly monotone and γ-inverse strongly monotone, respectively. Suppose that B and D are maximal monotone operators on H such that the domains of B and D are contained in K and that $T:K\to K$ is a quasi-nonexpansive mapping such that $I-T$ is demiclosed at zero. Assume that $\mathrm{\Omega }:=Fix(T)\cap {(A+B)}^{-1}0\cap {(C+D)}^{-1}0\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ be sequences in $(0,1)$ and let $\{u_n\}$ be a sequence in K. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{x}_1\in K\mathit{\text{chosen arbitrarily}},\hfill \\ y_n={\alpha }_n{u}_n+(1-{\alpha }_n)J_{{\lambda }_{n}B}(x_n-{\lambda }_{n}A{x}_n),\hfill \\ z_n=J_{{\psi }_{n}D}(y_n-{\psi }_{n}C{y}_n),\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)T{z}_n.\hfill \end{array}$$

(3.1)

Suppose the following conditions are satisfied:

(c1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

(c2) ${lim}_{n\to \mathrm{\infty }}{u}_n=u$;

(c3) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

(c4) $0<a\le {\lambda }_n\le b<2\alpha$;

(c5) $0<c\le {\psi }_n\le d<2\gamma$.

Then the sequence $\{x_n\}$ converges strongly to $P_{\mathrm{\Omega }}u$.

Proof Observe that the set Ω is closed and convex since $Fix(T)$, ${(A+B)}^{-1}0$ and ${(C+D)}^{-1}0$ are closed and convex.

By Lemma 2.1, for any $p\in \mathrm{\Omega }$, we have

$$\parallel z_n-p\parallel =\parallel J_{{\psi }_{n}D}(y_n-{\psi }_{n}C{y}_n)-J_{{\psi }_{n}D}(p-{\psi }_{n}Cp)\parallel \le \parallel y_n-p\parallel$$

and

$$\begin{array}{rl}\parallel y_n-p\parallel & \le {\alpha }_n\parallel u_n-p\parallel +(1-{\alpha }_n)\parallel J_{{\lambda }_{n}B}(I-{\lambda }_{n}A)x_n-p\parallel \\ \le {\alpha }_n\parallel u_n-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel .\end{array}$$

It follows that

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)\parallel z_n-p\parallel \\ \le [1-{\alpha }_n(1-{\beta }_n)]\parallel x_n-p\parallel +{\alpha }_n(1-{\beta }_n)\parallel u_n-p\parallel \\ \le max\{\parallel x_n-p\parallel ,\parallel u_n-p\parallel \}.\end{array}$$

The sequence $\{u_n\}$ is bounded due to condition (c2). Hence there exists a positive number L such that ${sup}_n\{\parallel u_n-p\parallel \}\le L$. By a simple inductive process, we have

$$\parallel x_{n+1}-p\parallel \le max\{\parallel x_1-p\parallel ,L\},$$

which shows that $\{x_n\}$ is bounded. So are $\{y_n\}$ and $\{z_n\}$.

Note that

$$2〈x_{n+1}-x_n,x_n-p〉={\parallel x_{n+1}-p\parallel }^2-{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-x_n\parallel }^2$$

and

$$x_{n+1}-x_n=(1-{\beta }_n)(T{z}_n-x_n).$$

Thus we get

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2-{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-x_n\parallel }^2\\ =2〈x_{n+1}-x_n,x_n-p〉\\ =(1-{\beta }_n)[{\parallel T{z}_n-p\parallel }^2-{\parallel T{z}_n-x_n\parallel }^2-{\parallel x_n-p\parallel }^2]\\ \le (1-{\beta }_n)[{\parallel z_n-p\parallel }^2-{\parallel T{z}_n-x_n\parallel }^2-{\parallel x_n-p\parallel }^2]\\ \le {\alpha }_n(1-{\beta }_n){\parallel u_n-p\parallel }^2-(1-{\beta }_n){\parallel T{z}_n-x_n\parallel }^2\end{array}$$

and

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2-{\parallel x_n-p\parallel }^2-{(1-{\beta }_n)}^2{\parallel T{z}_n-x_n\parallel }^2\\ \le {\alpha }_n(1-{\beta }_n){\parallel u_n-p\parallel }^2-(1-{\beta }_n){\parallel T{z}_n-x_n\parallel }^2.\end{array}$$

This implies that

$$(1-{\beta }_n){\beta }_n{\parallel T{z}_n-x_n\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n(1-{\beta }_n){\parallel u_n-p\parallel }^2.$$

(3.2)

Set ${\mathrm{\Gamma }}_n={\parallel x_n-p_0\parallel }^2$, where $p_0=P_{\mathrm{\Omega }}u$. We divide the rest proof into two cases.

Case 1. Suppose that ${\mathrm{\Gamma }}_{n+1}\le {\mathrm{\Gamma }}_n$ for all $n\in \mathbb{N}$. In this case, the limit ${lim}_{n\to \mathrm{\infty }}{\mathrm{\Gamma }}_n$ exists and then ${lim}_{n\to \mathrm{\infty }}({\mathrm{\Gamma }}_{n+1}-{\mathrm{\Gamma }}_n)=0$. We obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T{z}_n-x_n\parallel =0,$$

(3.3)

which implies

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}(1-{\beta }_n)\parallel T{z}_n-x_n\parallel =0.$$

(3.4)

Note that

$$\begin{array}{r}{\parallel J_{{\lambda }_{n}B}(x_n-{\lambda }_{n}A{x}_n)-p_0\parallel }^2\\ \le {\parallel (I-{\lambda }_{n}A)x_n-(I-{\lambda }_{n}A)p_0\parallel }^2\\ \le {\parallel x_n-p_0\parallel }^2+{\lambda }_n({\lambda }_n-2\alpha ){\parallel A{x}_n-A{p}_0\parallel }^2.\end{array}$$

It follows that

$$\begin{array}{r}{\parallel x_{n+1}-p_0\parallel }^2\\ \le {\beta }_n{\parallel x_n-p_0\parallel }^2+(1-{\beta }_n)[{\alpha }_n{\parallel u_n-p_0\parallel }^2+(1-{\alpha }_n){\parallel J_{{\lambda }_{n}B}(x_n-{\lambda }_{n}A{x}_n)-p_0\parallel }^2]\\ \le {\beta }_n{\parallel x_n-p_0\parallel }^2+{\alpha }_n(1-{\beta }_n){\parallel u_n-p_0\parallel }^2\\ +(1-{\alpha }_n)(1-{\beta }_n)[{\parallel x_n-p_0\parallel }^2+{\lambda }_n({\lambda }_n-2\alpha ){\parallel A{x}_n-A{p}_0\parallel }^2]\\ \le {\parallel x_n-p_0\parallel }^2+{\alpha }_n(1-{\beta }_n){\parallel u_n-p_0\parallel }^2\\ +(1-{\alpha }_n)(1-{\beta }_n){\lambda }_n({\lambda }_n-2\alpha ){\parallel A{x}_n-A{p}_0\parallel }^2,\end{array}$$

which yields

$$\begin{array}{r}(1-{\alpha }_n)(1-{\beta }_n){\lambda }_n(2\alpha -{\lambda }_n){\parallel A{x}_n-A{p}_0\parallel }^2\\ \le {\parallel x_n-p_0\parallel }^2-{\parallel x_{n+1}-p_0\parallel }^2+{\alpha }_n(1-{\beta }_n){\parallel u_n-p_0\parallel }^2.\end{array}$$

Therefore we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel A{x}_n-A{p}_0\parallel =0.$$

(3.5)

In a similar way, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel C{y}_n-C{p}_0\parallel =0.$$

(3.6)

Letting $h_n=J_{{\lambda }_{n}B}(I-{\lambda }_{n}A)x_n$, we have

$$\begin{array}{rl}{\parallel h_n-p_0\parallel }^2& ={\parallel J_{{\lambda }_{n}B}(I-{\lambda }_{n}A)x_n-J_{{\lambda }_{n}B}(I-{\lambda }_{n}A)p_0\parallel }^2\\ \le 〈(I-{\lambda }_{n}A)x_n-(I-{\lambda }_{n}A)p_0,h_n-p_0〉\\ \le \frac{1}{2}[{\parallel x_n-p_0\parallel }^2+{\parallel h_n-p_0\parallel }^2-{\parallel (x_n-h_n)-{\lambda }_n(A{x}_n-A{p}_0)\parallel }^2],\end{array}$$

from which one deduces that

$${\parallel h_n-p_0\parallel }^2\le {\parallel x_n-p_0\parallel }^2-{\parallel (x_n-h_n)-{\lambda }_n(A{x}_n-A{p}_0)\parallel }^2.$$

Using (3.1), we see that

$$\begin{array}{r}{\parallel x_{n+1}-p_0\parallel }^2\\ \le {\beta }_n{\parallel x_n-p_0\parallel }^2+(1-{\beta }_n)[{\alpha }_n{\parallel u_n-p_0\parallel }^2+(1-{\alpha }_n){\parallel h_n-p_0\parallel }^2]\\ \le {\beta }_n{\parallel x_n-p_0\parallel }^2+{\alpha }_n(1-{\beta }_n){\parallel u_n-p_0\parallel }^2\\ +(1-{\alpha }_n)(1-{\beta }_n)({\parallel x_n-p_0\parallel }^2-{\parallel x_n-h_n\parallel }^2+2{\lambda }_n〈A{x}_n-A{p}_0,x_n-h_n〉)\\ \le {\parallel x_n-p_0\parallel }^2+{\alpha }_n(1-{\beta }_n){\parallel u_n-p_0\parallel }^2\\ -(1-{\alpha }_n)(1-{\beta }_n){\parallel x_n-h_n\parallel }^2+2(1-{\alpha }_n)(1-{\beta }_n){\lambda }_n〈A{x}_n-A{p}_0,x_n-h_n〉.\end{array}$$

Thus,

$$\begin{array}{r}(1-{\alpha }_n)(1-{\beta }_n){\parallel x_n-h_n\parallel }^2\\ \le {\parallel x_n-p_0\parallel }^2-{\parallel x_{n+1}-p_0\parallel }^2+{\alpha }_n(1-{\beta }_n){\parallel u_n-p_0\parallel }^2\\ +2(1-{\alpha }_n)(1-{\beta }_n){\lambda }_n〈A{x}_n-A{p}_0,x_n-h_n〉.\end{array}$$

It follows from (3.5) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-h_n\parallel =0.$$

(3.7)

This implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-y_n\parallel =0.$$

(3.8)

By (3.6), a similar argument shows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-z_n\parallel =0.$$

(3.9)

As

$$\parallel T{z}_n-z_n\parallel \le \parallel T{z}_n-x_n\parallel +\parallel x_n-y_n\parallel +\parallel y_n-z_n\parallel ,$$

combining (3.3), (3.8) and (3.9) gives

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T{z}_n-z_n\parallel =0.$$

(3.10)

Next we prove that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-p_0,y_n-p_0〉\le 0.$$

(3.11)

To show this inequality, we choose a subsequence $\{y_{n_j}\}$ of $\{y_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-p_0,y_n-p_0〉=\underset{j\to \mathrm{\infty }}{lim}〈u-p_0,y_{n_j}-p_0〉.$$

(3.12)

In view of the boundedness of $\{y_{n_j}\}$, without loss of generality, we assume that $y_{n_j}\rightharpoonup \omega$. Now we show that $\omega \in \mathrm{\Omega }$. According to the fact that $\{y_n\}$ is contained in K and K is a closed convex set, one has $\omega \in \mathrm{\Omega }$.

Note that the expressions (3.8) and (3.9) yield $x_{n_j}\rightharpoonup \omega$ and $z_{n_j}\rightharpoonup \omega$. By the fact that $I-T$ is demiclosed at zero, the expression (3.10) implies $\omega \in Fix(T)$.

We prove that $\omega \in {(A+B)}^{-1}0$. Due to (c4), there is a subsequence $\{{\lambda }_{n_{j_i}}\}$ of $\{{\lambda }_{n_j}\}$ such that ${\lambda }_{n_{j_i}}\to {\lambda }_0\in [a,b]$. Without loss of generality, we assume that ${\lambda }_{n_j}\to {\lambda }_0$. Observe that

$$\begin{array}{r}\parallel x_{n_j}-J_{{\lambda }_{0}B}(I-{\lambda }_{0}A)x_{n_j}\parallel \\ \le \parallel x_{n_j}-y_{n_j}\parallel +\parallel y_{n_j}-[{\alpha }_{n_j}{u}_{n_j}+(1-{\alpha }_{n_j})J_{{\lambda }_{0}B}(I-{\lambda }_{0}A)x_{n_j}]\parallel \\ +\parallel {\alpha }_{n_j}{u}_{n_j}+(1-{\alpha }_{n_j})J_{{\lambda }_{0}B}(I-{\lambda }_{0}A)x_{n_j}-J_{{\lambda }_{0}B}(I-{\lambda }_{0}A)x_{n_j}\parallel \\ \le \parallel x_{n_j}-y_{n_j}\parallel +(1-{\alpha }_{n_j})\parallel J_{{\lambda }_{n_j}B}(I-{\lambda }_{n_j}A)x_{n_j}-J_{{\lambda }_{0}B}(I-{\lambda }_{0}A)x_{n_j}\parallel \\ +{\alpha }_{n_j}\parallel u_{n_j}-J_{{\lambda }_{0}B}(I-{\lambda }_{0}A)x_{n_j}\parallel \\ \le \parallel x_{n_j}-y_{n_j}\parallel +(1-{\alpha }_{n_j})[\parallel J_{{\lambda }_{n_j}B}(I-{\lambda }_{n_j}A)x_{n_j}-J_{{\lambda }_{0}B}(I-{\lambda }_{n_j}A)x_{n_j}\parallel \\ +\parallel J_{{\lambda }_{0}B}(I-{\lambda }_{n_j}A)x_{n_j}-J_{{\lambda }_{0}B}(I-{\lambda }_{0}A)x_{n_j}\parallel ]+{\alpha }_{n_j}\parallel u_{n_j}-J_{{\lambda }_{0}B}(I-{\lambda }_{0}A)x_{n_j}\parallel \\ \le \parallel x_{n_j}-y_{n_j}\parallel +(1-{\alpha }_{n_j})[\frac{|{\lambda }_0-{\lambda }_{n_j}|}{{\lambda }_0}\parallel J_{{\lambda }_{0}B}(I-{\lambda }_{n_j}A)x_{n_j}-(I-{\lambda }_{n_j}A)x_{n_j}\parallel \\ +|{\lambda }_{n_j}-{\lambda }_0|\parallel A{x}_{n_j}\parallel ]+{\alpha }_{n_j}\parallel u_{n_j}-J_{{\lambda }_{0}B}(I-{\lambda }_{0}A)x_{n_j}\parallel .\end{array}$$

Hence,

$$\underset{j\to \mathrm{\infty }}{lim}\parallel x_{n_j}-J_{{\lambda }_{0}B}(I-{\lambda }_{0}A)x_{n_j}\parallel =0.$$

(3.13)

Since $J_{{\lambda }_{0}B}(I-{\lambda }_{0}A)$ is nonexpansive, the demiclosedness for a nonexpansive mapping implies that $\omega \in Fix(J_{{\lambda }_{0}B}(I-{\lambda }_{0}A))$, that is, $\omega \in {(A+B)}^{-1}0$.

Note that

$$\begin{array}{r}\parallel y_{n_i}-J_{{\psi }_{0}D}(I-{\psi }_{0}C)y_{n_i}\parallel \\ \le \parallel y_{n_i}-z_{n_i}\parallel +\parallel J_{{\psi }_{n_i}D}(I-{\psi }_{n_i}C)y_{n_i}-J_{{\psi }_{0}D}(I-{\psi }_{0}C)y_{n_i}\parallel \\ \le \parallel y_{n_i}-z_{n_i}\parallel +\parallel J_{{\psi }_{n_i}D}(I-{\psi }_{n_i}C)y_{n_i}-J_{{\psi }_{n_i}D}(I-{\psi }_{0}C)y_{n_i}\parallel \\ +\parallel J_{{\psi }_{n_i}D}(I-{\psi }_{0}C)y_{n_i}-J_{{\psi }_{0}D}(I-{\psi }_{0}C)y_{n_i}\parallel .\end{array}$$

Using a similar argument, we get $\omega \in {(C+D)}^{-1}0$. In fact, we have obtained $\omega \in \mathrm{\Omega }$.

By (3.12) and (2.1), we have

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-p_0,y_n-p_0〉& =\underset{j\to \mathrm{\infty }}{lim}〈u-p_0,y_{n_j}-p_0〉\\ =〈u-p_0,\omega -p_0〉\\ \le 0.\end{array}$$

The inequality (3.11) is obtained.

Finally, we prove that $x_n\to p_0$. With the help of Lemma 2.4, we obtain

$$\begin{array}{r}{\parallel x_{n+1}-p_0\parallel }^2\\ \le {\beta }_n{\parallel x_n-p_0\parallel }^2+(1-{\beta }_n){\parallel y_n-p_0\parallel }^2\\ \le {\beta }_n{\parallel x_n-p_0\parallel }^2+(1-{\beta }_n)[(1-{\alpha }_n){\parallel x_n-p_0\parallel }^2+2{\alpha }_n〈u_n-p_0,y_n-p_0〉]\\ \le [1-{\alpha }_n(1-{\beta }_n)]{\parallel x_n-p_0\parallel }^2+2{\alpha }_n(1-{\beta }_n)(〈u_n-u,y_n-p_0〉+〈u-p_0,y_n-p_0〉).\end{array}$$

It follows from (3.11) and Lemma 2.5 that $\{x_n\}$ converges strongly to $p_0$.

Case 2. Suppose that there exists a subsequence $\{{\mathrm{\Gamma }}_{n_i}\}$ of $\{{\mathrm{\Gamma }}_n\}$ such that

$${\mathrm{\Gamma }}_{n_i}<{\mathrm{\Gamma }}_{n_i+1}\text{for all }i\in \mathbb{N}.$$

We define $\tau :\mathbb{N}\to \mathbb{N}$ by

$$\tau (n)=max\{k\le n:{\mathrm{\Gamma }}_k<{\mathrm{\Gamma }}_{k+1}\}.$$

Lemma 2.2 shows that ${\mathrm{\Gamma }}_{\tau (n)}\le {\mathrm{\Gamma }}_{\tau (n)+1}$. Therefore we have from (3.2)

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T{z}_{\tau (n)}-x_{\tau (n)}\parallel =0$$

(3.14)

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{\tau (n)+1}-x_{\tau (n)}\parallel =0.$$

(3.15)

As in the proof of Case 1, we obtain

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-p_0,y_{\tau (n)}-p_0〉\le 0.$$

(3.16)

Observe that

$$\begin{array}{r}{\parallel x_{\tau (n)+1}-p_0\parallel }^2\\ \le [1-{\alpha }_{\tau (n)}(1-{\beta }_{\tau (n)})]{\parallel x_{\tau (n)}-p_0\parallel }^2\\ +2{\alpha }_{\tau (n)}(1-{\beta }_{\tau (n)})(〈u_{\tau (n)}-u,y_{\tau (n)}-p_0〉+〈u-p_0,y_{\tau (n)}-p_0〉).\end{array}$$

It follows that

$${\parallel x_{\tau (n)}-p_0\parallel }^2\le 2(〈u_{\tau (n)}-u,y_{\tau (n)}-p_0〉+〈u-p_0,y_{\tau (n)}-p_0〉),$$

(3.17)

which implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{\tau (n)}-p_0\parallel }^2\le 0.$$

Thus we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{\tau (n)}-p_0\parallel =0.$$

(3.18)

It follows from (3.15) and (3.18) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{\tau (n)+1}-p_0\parallel =0.$$

(3.19)

Lemma 2.2 implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-p_0\parallel =0.$$

The proof is completed. □

The following result is a direct consequence of Theorem 3.1.

Corollary 3.2 Let K be a nonempty closed convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone operator of K into H and let B be a maximal monotone operator on H such that the domain of B is contained in K. Let $T:K\to K$ be a quasi-nonexpansive mapping such that $I-T$ is demiclosed at zero. Assume that $Fix(T)\cap {(A+B)}^{-1}0\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ be sequences in $(0,1)$ and let $\{u_n\}$ be a sequence in K. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{x}_1\in K\mathit{\text{chosen arbitrarily}},\hfill \\ y_n={\alpha }_n{u}_n+(1-{\alpha }_n)J_{{\lambda }_{n}B}(x_n-{\lambda }_{n}A{x}_n),\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)T{y}_n.\hfill \end{array}$$

(3.20)

If conditions (c1)-(c4) are satisfied, then the sequence $\{x_n\}$ converges strongly to the element $P_{Fix(T)\cap {(A+B)}^{-1}0}u$.

Proof Letting $C=0$ and $D=\partial {\delta }_K$ in Theorem 3.1, the desired result follows. □

Let us consider the variational inequality problem. Recall that the subdifferential $\partial {\delta }_K$ of ${\delta }_K$ is a maximal monotone operator and $\mathit{VI}(K,A)={(A+\partial {\delta }_K)}^{-1}0$, where A is an inverse strongly monotone mapping. We obtain the following result.

Corollary 3.3 Let K be a nonempty closed convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone operator of K into H and let $T:K\to K$ be a quasi-nonexpansive mapping such that $I-T$ is demiclosed at zero. Assume that $Fix(T)\cap \mathit{VI}(K,A)\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ be sequences in $(0,1)$ and let $\{u_n\}$ be a sequence in K. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{x}_1\in K\mathit{\text{chosen arbitrarily}},\hfill \\ y_n={\alpha }_n{u}_n+(1-{\alpha }_n)P_K(x_n-{\lambda }_{n}A{x}_n),\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)T{y}_n.\hfill \end{array}$$

(3.21)

If conditions (c1)-(c4) are satisfied, then the sequence $\{x_n\}$ converges strongly to the element $P_{Fix(T)\cap \mathit{VI}(K,A)}u$.

Proof Corollary 3.2 easily yields the desired result. □

Remark Corollaries 3.2 and 3.3 improve and extend Theorem 3.1 of Takahashi et al. [22] and Theorem 4.2 of Takahashi and Takahashi [23] in the following aspects, respectively.

• (1) The nonexpansive mapping is extended to the quasi-nonexpansive mapping.

• (2) The constant vector u is replaced by the variables $u_n$ with ${lim}_{n\to \mathrm{\infty }}{u}_n=u$.

• (3) The condition ${lim}_{n\to \mathrm{\infty }}({\lambda }_{n+1}-{\lambda }_n)=0$ is removed.