Iterative algorithms based on hybrid method and Cesàro mean of asymptotically nonexpansive mappings for equilibrium problems

Abstract

Using Cesàro means of a mapping, we modify the progress of Mann’s iteration in hybrid method for asymptotically nonexpansive mappings in Hilbert spaces. Under suitable conditions, we prove that the iterative sequence converges strongly to a fixed point of an asymptotically nonexpansive mapping. We also introduce a new hybrid iterative scheme for finding a common element of the set of common fixed points of asymptotically nonexpansive mappings and the set of solutions of an equilibrium problem in Hilbert spaces.

MSC:47H10, 47J25, 90C33.

1 Introduction

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H. A mapping $T:C\to C$ is said to be asymptotically nonexpansive if for each $n\ge 1$, there exists a nonnegative real number $k_n$ satisfying ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ such that

$$\parallel T^{n}x-T^{n}y\parallel \le k_n\parallel x-y\parallel ,\mathrm{\forall }x,y\in C;$$

when $k_n\equiv 1$, T is called nonexpansive.

The concept of asymptotically nonexpansive mapping was introduced by Goebel and Kirk [1] in 1972. We denote by $F(T):=\{x\in C:Tx=x\}$ the set of fixed points of T. It is well known that if $T:H\to H$ is asymptotically nonexpansive, then $F(T)$ is nonempty convex.

In 1953, Mann [2] introduced the iteration as follows: a sequence $\{x_n\}$ defined by

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)T{x}_n.$$

(1.1)

In an infinite-dimensional Hilbert space, Mann iteration could conclude only weak convergence [3]. Attempts to modify the Mann iteration method (1.1) so that strong convergence is guaranteed have recently been made. Nakajo and Takahashi [4] proposed the following modification of Mann iteration method for a nonexpansive mapping T in a Hilbert space:

$$\{\begin{array}{c}{x}_0\in C\text{is arbitrary},\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)T{x}_n,\hfill \\ C_n=\{z\in C:\parallel y_n-z\parallel \le \parallel x_n-z\parallel \},\hfill \\ Q_n=\{z\in C:〈x_n-z,x_0-x_n〉\ge 0\},\hfill \\ x_{n+1}=P_{C_n\cap Q_n}(x_0),n=0,1,2\dots ,\hfill \end{array}$$

(1.2)

where $P_K$ denotes the metric projection from H onto a closed convex subset K of H. The above method is also called CQ method or hybrid method.

In 2006, Kim and Xu [5] adapted the iteration (1.2) in a Hilbert space. More precisely, they introduced the following iteration process for asymptotically nonexpansive mappings:

$$\{\begin{array}{c}{x}_0\in C\text{is arbitrary,}\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)T^n{x}_n,\hfill \\ C_n=\{z\in C:{\parallel y_n-z\parallel }^2\le {\parallel x_n-z\parallel }^2+{\theta }_n\},\hfill \\ Q_n=\{z\in C:〈x_n-z,x_0-x_n〉\ge 0\},\hfill \\ x_{n+1}=P_{C_n\cap Q_n}(x_0),n=0,1,2\dots ,\hfill \end{array}$$

(1.3)

where

$${\theta }_n=(1-{\alpha }_n)(k_n^2-1){(diamC)}^2\to 0\text{as }n\to \mathrm{\infty }.$$

They proved that $\{x_n\}$ converges in norm to $P_{F(T)}{x}_0$ under some conditions. Several authors (see [6, 7]) have studied the convergence of hybrid method.

Baillon [8] first proved that the following Cesàro mean iterative sequence weakly converges to a fixed point of a nonexpansive mapping in Hilbert spaces:

$$T_{n}x=\frac{1}{n+1}\sum _{i=0}^n{T}^{i}x.$$

Shimizu and Takahashi [9] proved a strong convergence theorem of the above iteration for an asymptotically nonexpansive mapping in Hilbert spaces.

Let C be a nonempty closed convex subset of a real Hilbert space H, let $f:C\times C\to \mathbb{R}$ be a functional, where ℝ is the set of real numbers. The equilibrium problem is to find $\hat{x}\in C$ such that

$$f(x,y)\ge 0,\mathrm{\forall }y\in C.$$

(1.4)

The set of solutions of (1.4) is denoted by $EP(f)$. Given a mapping $T:C\to X^{\ast }$, let $f(x,y)=〈Tx,y-x〉$, $\mathrm{\forall }x,y\in C$, then $z\in EP(f)$ if and only if $〈Tz,y-z〉\ge 0$, $\mathrm{\forall }y\in C$, i.e., z is the solution of the variational inequality.

There are several other problems, for example, the complementarity problem, fixed point problem and optimization problem, which can also be written in the form of an equilibrium problem. So, equilibrium problems provide us with a systematic framework to study a wide class of problems arising in financial economics, optimization and operation research etc., which motivates the extensive concern. See, for example, [10–14]. In recent years, equilibrium problems have been deeply and thoroughly researched. See, for example, [15–20]. Some methods have been proposed to solve the equilibrium problem in a Hilbert space; see, for instance, [21–23]. In 2011, Jitpeera, Katchang, and Kumam [24] found a common element of the set of solutions for mixed equilibrium problem, the set of solutions of the variational inequality for a β-inverse strongly monotone mapping, and the set of fixed points of a family of finitely nonexpansive mappings in a real Hilbert space by using the viscosity and Cesàro mean approximation method.

Motivated by the above-mentioned results, in this paper we introduce the following iteration process for asymptotically nonexpansive mappings T with C a closed convex bounded subset of a real Hilbert space:

$$\{\begin{array}{c}{x}_0\in C\text{is arbitrary,}\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)\frac{1}{n+1}{\sum }_{j=0}^n{T}^j{x}_n,\hfill \\ C_n=\{z\in C:{\parallel y_n-z\parallel }^2\le {\parallel x_n-z\parallel }^2+{\theta }_n\},\hfill \\ Q_n=\{z\in C:〈x_n-z,x_0-x_n〉\ge 0\},\hfill \\ x_{n+1}=P_{C_n\cap Q_n}(x_0),n=0,1,2\dots ,\hfill \end{array}$$

(1.5)

where

$${\theta }_n=(1-{\alpha }_n)(L_n^2-1){(diamC)}^2\to 0\text{as }n\to \mathrm{\infty }.$$

We shall prove that the above iterative sequence $\{x_n\}$ converges strongly to a fixed point of T under some proper conditions. In addition, we also introduce a new hybrid iterative scheme for finding a common element of the set of common fixed points of asymptotically nonexpansive mappings and the set of solutions of an equilibrium problem in Hilbert spaces.

We will use the notation ⇀ for weak convergence and → for strong convergence.

2 Preliminaries

Let H be a real Hilbert space. Then

$${\parallel x-y\parallel }^2={\parallel x\parallel }^2-{\parallel y\parallel }^2-2〈x-y,y〉$$

(2.1)

and

$${\parallel \lambda x+(1-\lambda )y\parallel }^2=\lambda {\parallel x\parallel }^2+(1-\lambda ){\parallel y\parallel }^2-\lambda (1-\lambda ){\parallel x-y\parallel }^2$$

(2.2)

for all $x,y\in H$ and $\lambda \in [0,1]$. It is also known that H satisfies

• (1) Opial’s condition, that is, for any sequence $\{x_n\}$ with $x_n\rightharpoonup x$, the inequality

  $$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-y\parallel$$

  holds for every $y\in H$ with $y\ne x$.

• (2) The Kadec-Klee property, that is, for any sequence $\{x_n\}$ with $x_n\rightharpoonup x$ and $\parallel x_n\parallel \to \parallel x\parallel$ together implies $\parallel x_n-x\parallel \to 0$.

Let C be a nonempty closed convex subset of H. Then, for any $x\in H$, there exists a unique nearest point in C, denoted by $P_C(x)$, such that

$$\parallel x-P_C(x)\parallel \le \parallel x-y\parallel ,\mathrm{\forall }y\in C.$$

Such a mapping $P_C$ is called the metric projection of H onto C. We know that $P_C$ is nonexpansive. Furthermore, for $x\in H$ and $z\in C$,

$$z=P_C(x)\text{if and only if}〈x-z,z-y〉\ge 0,\mathrm{\forall }y\in C.$$

We also need the following lemmas.

Lemma 2.1 (See [25])

Let T be an asymptotically nonexpansive mapping defined on a bounded closed convex subset C of a Hilbert space H. Assume that $\{x_n\}$ is a sequence in C with the properties: (i) $x_n\rightharpoonup z$; (ii) $T{x}_n-x_n\to 0$. Then $z\in F(T)$.

Lemma 2.2 (See [9])

Let C be a nonempty bounded subset of a Hilbert space. Let T be an asymptotically nonexpansive mapping from C into itself such that $F(T)$ is nonempty. Then, for any $\epsilon >0$, there exists a positive integer $l_{\epsilon }$ such that for any integer $l\ge l_{\epsilon }$, there is a positive integer $n_l$ satisfying

$$\parallel \frac{1}{n+1}\sum _{j=0}^n{T}^{j}x-T^l\left(\frac{1}{n+1}\sum _{j=0}^n{T}^{j}x\right)\parallel <\epsilon ,\mathrm{\forall }x\in C,\mathrm{\forall }n\ge n_l.$$

The equilibrium problem is to find $\hat{x}\in C$ such that

$$f(\hat{x},y)\ge 0,\mathrm{\forall }y\in C.$$

The set of solutions of the above inequality is denoted by $EP(f)$. For solving the equilibrium problem, let us assume that a bifunction f satisfies the following conditions:

(A1) $f(x,x)=0$ for all $x\in C$;

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

(A3) for each $x,y,z\in C$,

$$\underset{t\to 0^{+}}{lim\hspace{0.17em}sup}f(tz+(1-t)x,y)\le f(x,y);$$

(A4) for each $x\in C$, $f(x,\cdot )$ is convex and lower semi-continuous.

The following lemma appears implicitly in [21].

Lemma 2.3 ([21])

Let C be a nonempty closed convex subset of H, let f be a bifunction of $C\times C$ into ℝ satisfying (A1)-(A4), and let $r>0$ and $x\in H$. Then there exists $z\in C$ such that

$$f(z,y)+\frac{1}{r}〈y-z,Jz-Jx〉\ge 0,\mathrm{\forall }y\in C.$$

The following lemma was also given in [26].

Lemma 2.4 ([26])

Assume that $f:C\times C\to \mathbb{R}$ satisfies (A1)-(A4). For $r>0$ and $x\in H$, define a mapping $T_r:X\to C$ as follows:

$$T_r(x)=\{z\in C:f(z,y)+\frac{1}{r}〈y-z,Jz-Jx〉\ge 0,\mathrm{\forall }y\in C\}$$

for all $z\in H$. Then the following hold:

• (1) $T_r$ is single-valued;

• (2) $T_r$ is firmly nonexpansive, i.e.,

  $${\parallel T_{r}x-T_{r}y\parallel }^2\le 〈T_{r}x-T_{r}y,x-y〉,\mathrm{\forall }x,y\in H;$$

• (3) $F(T_r)=EP(f)$;

• (4) $EP(f)$ is closed and convex.

Lemma 2.5 ([26])

Let C be a nonempty closed convex subset of H, let $f:C\times C\to \mathbb{R}$ be a functional, satisfying (A1)-(A4), and let $r>0$. Then, for $x\in H$ and $q\in F(T_r)$,

$${\parallel q-T_{r}x\parallel }^2+{\parallel T_{r}x-x\parallel }^2\le {\parallel q-x\parallel }^2.$$

3 Strong convergence theorem of modified Mann iteration based on hybrid method

Inspired by Kim and Xu’s results (see [5]), Mann-type iteration (1.3) is modified to obtain the strong convergence theorem as follows.

Theorem 3.1 Let C be a nonempty bounded closed convex subset of a real Hilbert space H and let $T:C\to C$ be an asymptotically nonexpansive mapping with $k_n$, denote $L_n=\frac{1}{n+1}{\sum }_{j=0}^n{k}_j$. Assume that ${\{{\alpha }_n\}}_{n=0}^{\mathrm{\infty }}$ is a sequence in $(0,1)$ such that ${\alpha }_n\le a$ for all n and for some $0<a<1$. Define a sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ in C by the following algorithm:

$$\{\begin{array}{c}{x}_0\in C\mathit{\text{is arbitrary}},\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)\frac{1}{n+1}{\sum }_{j=0}^n{T}^j{x}_n,\hfill \\ C_n=\{z\in C:{\parallel y_n-z\parallel }^2\le {\parallel x_n-z\parallel }^2+{\theta }_n\},\hfill \\ Q_n=\{z\in C:〈x_n-z,x_0-x_n〉\ge 0\},\hfill \\ x_{n+1}=P_{C_n\cap Q_n}(x_0),n=0,1,2\dots ,\hfill \end{array}$$

(3.1)

where

$${\theta }_n=(1-{\alpha }_n)(L_n^2-1){(diamC)}^2\to 0\mathit{\text{as }}n\to \mathrm{\infty }.$$

Then $\{x_n\}$ converges strongly to $P_{F(T)}(x_0)$.

Proof First note that T has a fixed point in C (see [1]); that is, $F(T)$ is nonempty. We divide the proof of this theorem into four steps as below.

Step 1. We show that $C_n$ and $Q_n$ are closed and convex for each $n\in \mathbb{N}\cup \{0\}$.

From the definition of $C_n$ and $Q_n$, it is obvious that $C_n$ is closed and $Q_n$ is closed and convex for each $n\in \mathbb{N}\cup \{0\}$. We prove that $C_n$ is convex. Since ${\parallel y_n-z\parallel }^2\le {\parallel x_n-z\parallel }^2+{\theta }_n$ is equivalent to

$$2〈x_n-y_n,z〉\le {\parallel x_n\parallel }^2-{\parallel y_n\parallel }^2+{\theta }_n,$$

it follows that $C_n$ is convex.

Step 2. We show that $F(T)\subset C_n$, $\mathrm{\forall }n\in \mathbb{N}\cup \{0\}$.

Let $p\in F(T)$ and $n\in \mathbb{N}\cup \{0\}$. Then from

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel {\alpha }_n{x}_n+(1-{\alpha }_n)\frac{1}{n+1}\sum _{j=0}^n{T}^j{x}_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\left(\frac{1}{n+1}\sum _{j=0}^n\parallel T^j{x}_n-p\parallel \right)}^2\\ \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){(\frac{1}{n+1}\sum _{j=0}^n{k}_j\parallel x_n-p\parallel )}^2\\ \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n)L_n^2{\parallel x_n-p\parallel }^2\\ =& {\parallel x_n-p\parallel }^2+(1-{\alpha }_n)(L_n^2-1){\parallel x_n-p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2+{\theta }_n,\end{array}$$

we have $p\in C_n$. Next, we show that $F(T)\subset Q_n$, $\mathrm{\forall }n\in \mathbb{N}\cup \{0\}$. We prove this by induction. For $n=0$, we have $F(T)\subset C=Q_0$. Suppose that $F(T)\subset Q_n$, then $\mathrm{\varnothing }\ne F(T)\subset C_n\cap Q_n$ and there exists a unique element $x_{n+1}\in C_n\cap Q_n$ such that $x_{n+1}=P_{C_n\cap Q_n}(x_0)$. Then

$$〈x_{n+1}-z,x_0-x_{n+1}〉\ge 0,\mathrm{\forall }z\in C_n\cap Q_n.$$

In particular,

$$〈x_{n+1}-p,x_0-x_{n+1}〉\ge 0,\mathrm{\forall }p\in F(T).$$

It follows that $F(T)\subset Q_{n+1}$ and hence $F(T)\subset Q_n$ for each n. Thus we obtain $F(T)\subset C_n\cap Q_n$, $\mathrm{\forall }n\in \mathbb{N}\cup \{0\}$. This means that $\{x_n\}$ is well defined.

Step 3. We show that $\{x_n\}$ is bounded and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$.

It follows from the definition of $Q_n$ that $x_n=P_{Q_n}(x_0)$. Therefore

$$\parallel x_n-x_0\parallel \le \parallel z-x_0\parallel \text{for all }z\in Q_n\text{ and all }n\in \mathbb{N}\cup \{0\}.$$

Let $z\in F(T)\subset Q_n$ for all $n\in \mathbb{N}\cup \{0\}$. Then

$$\parallel x_n-x_0\parallel \le \parallel z-x_0\parallel \text{for all }n\in \mathbb{N}\cup \{0\}.$$

On the other hand, from $x_{n+1}=P_{C_n\cap Q_n}(x_0)\in Q_n$, we have

$$\parallel x_n-x_0\parallel \le \parallel x_{n+1}-x_0\parallel \text{for all }n\in \mathbb{N}\cup \{0\}.$$

Therefore, the sequence $\{\parallel x_n-x_0\parallel \}$ is nondecreasing. Since C is bounded, we obtain that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_0\parallel$ exists. This implies that $\{x_n\}$ is bounded. Noticing again that $x_{n+1}=P_{C_n\cap Q_n}(x_0)\in Q_n$ and $x_n=P_{Q_n}(x_0)$, we have $〈x_{n+1}-x_n,x_n-x_0〉\ge 0$. It follows from (2.1) that

$$\begin{array}{rcl}{\parallel x_{n+1}-x_n\parallel }^2& =& {\parallel (x_{n+1}-x_0)-(x_n-x_0)\parallel }^2\\ =& {\parallel x_{n+1}-x_0\parallel }^2-{\parallel x_n-x_0\parallel }^2-2〈x_{n+1}-x_n,x_n-x_0〉\\ \le & {\parallel x_{n+1}-x_0\parallel }^2-{\parallel x_n-x_0\parallel }^2\end{array}$$

for all $n\in \mathbb{N}\cup \{0\}$. This implies that

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

Since $x_{n+1}=P_{C_n\cap Q_n}\in C_n$, we have ${\parallel y_n-x_{n+1}\parallel }^2\le {\parallel x_n-x_{n+1}\parallel }^2+{\theta }_n$. Noticing that ${lim}_{n\to \mathrm{\infty }}{k}_n=1$, then $L_n=\frac{1}{n+1}{\sum }_{j=0}^n{k}_j\to 1$ as $n\to \mathrm{\infty }$. Hence

$$\parallel y_n-x_{n+1}\parallel \le \parallel x_n-x_{n+1}\parallel +\sqrt{{\theta }_n}\to 0.$$

Let $A_n=\frac{1}{n+1}{\sum }_{j=0}^n{T}^j$. Also since ${\alpha }_n\le a<1$ for all n, then

$$\begin{array}{rcl}\parallel x_n-A_n{x}_n\parallel & =& \frac{1}{1-{\alpha }_n}\parallel y_n-x_n\parallel \\ \le & \frac{1}{1-a}(\parallel y_n-x_{n+1}\parallel +\parallel x_n-x_{n+1}\parallel )\to 0.\end{array}$$

From Lemma 2.2, we have

$$\underset{l\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel A_n{x}_n-T^l(A_n{x}_n)\parallel \le \underset{l\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{x\in C}{sup}\parallel A_{n}x-T^l(A_{n}x)\parallel =0.$$

Therefore,

$$\underset{l\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel A_n{x}_n-T^l(A_n{x}_n)\parallel =0.$$

Put $k_{\mathrm{\infty }}=sup\{k_n:n\ge 1\}<\mathrm{\infty }$, then

$$\begin{array}{rcl}\parallel x_n-T^l{x}_n\parallel & \le & \parallel x_n-A_n{x}_n\parallel +\parallel A_n{x}_n-T^l(A_n{x}_n)\parallel +\parallel T^l(A_n{x}_n)-T^l{x}_n\parallel \\ \le & \parallel x_n-A_n{x}_n\parallel +\parallel A_n{x}_n-T^l(A_n{x}_n)\parallel +k_l\parallel x_n-A_n{x}_n\parallel \\ \le & (1+k_{\mathrm{\infty }})\parallel x_n-A_n{x}_n\parallel +\parallel A_n{x}_n-T^l(A_n{x}_n)\parallel \to 0\end{array}$$

as $n,l\to \mathrm{\infty }$. Thus, we have

$$\begin{array}{rcl}\parallel T{x}_n-x_n\parallel & \le & \parallel T{x}_n-T^{l+1}{x}_n\parallel +\parallel T^{l+1}{x}_n-T^{l+1}{x}_{n+1}\parallel \\ +\parallel T^{l+1}{x}_{n+1}-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel \\ \le & k_{\mathrm{\infty }}\parallel x_n-T^l{x}_n\parallel +\parallel x_{n+1}+T^l{x}_{n+1}\parallel +(k_{\mathrm{\infty }}+1)\parallel x_{n+1}-x_n\parallel \to 0.\end{array}$$

Step 4. We show that $\{x_n\}$ converges strongly to $P_{F(T)}(x_0)$.

Put $w=P_{F(T)}(x_0)$. Since $\{x_n\}$ is bounded, let $\{x_{n_k}\}$ be a subsequence of $\{x_n\}$ such that $x_{n_k}\rightharpoonup w^{\prime }$. By Lemma 2.1, we get $w^{\prime }\in F(T)$. Since $x_n=P_{Q_n}(x_0)$ and $w\in F(T)\subset Q_n$, we have $\parallel x_n-x_0\parallel \le \parallel w-x_0\parallel$. It follows from $w=P_{F(T)}(x_0)$ and the weak lower semicontinuity of the norm that

$$\parallel w-x_0\parallel \le \parallel w^{\prime }-x_0\parallel \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_k}-x_0\parallel \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-x_0\parallel \le \parallel w-x_0\parallel .$$

Thus, we obtain that ${lim}_{k\to \mathrm{\infty }}\parallel x_{n_k}-x_0\parallel =\parallel w^{\prime }-x_0\parallel =\parallel w-x_0\parallel$. Using the Kadec-Klee property of H, we get ${lim}_{k\to \mathrm{\infty }}{x}_{n_k}=w^{\prime }=w$. Since $\{x_{n_k}\}$ is an arbitrary subsequence of $\{x_n\}$, we can conclude that $\{x_n\}$ converges strongly to $P_{F(T)}(x_0)$. □

Remark 3.2 It is not difficult to see from the proof above that the boundedness of C can be discarded if T is a nonexpansive mapping.

4 Strong convergence theorem for equilibrium problems

In this section, we prove a strong convergence theorem for finding a common element of the set of zero points of an asymptotically nonexpansive mapping T and the set of solutions of an equilibrium problem in a Hilbert space.

Theorem 4.1 Let H be a real Hilbert space, and let C be a nonempty bounded closed convex subset of H, let $f:C\times C\to \mathbb{R}$ be a functional, satisfying (A1)-(A4). Let $T:C\to C$ be an asymptotically nonexpansive mapping with $k_n$, denote $L_n=\frac{1}{n+1}{\sum }_{j=0}^n{k}_j$ such that $F(T)\cap EP(f)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{x}_0\in C\mathit{\text{is arbitrary}},\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)\frac{1}{n+1}{\sum }_{j=0}^n{T}^j{x}_n,\hfill \\ u_n\in C,\mathit{\text{such that }}f(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-y_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ C_n=\{z\in C:{\parallel u_n-z\parallel }^2\le {\parallel x_n-z\parallel }^2+{\theta }_n\},\hfill \\ Q_n=\{z\in C:〈x_n-z,x_0-x_n〉\ge 0\},\hfill \\ x_{n+1}=P_{C_n\cap Q_n}(x_0),n=0,1,2,\dots ,\hfill \end{array}$$

(4.1)

where

$${\theta }_n=(1-{\alpha }_n)(L_n^2-1){(diamC)}^2\to 0,n\to \mathrm{\infty }.$$

Suppose that $\{{\alpha }_n\}\subset [0,1]$ and there exists $a\in (0,1)$ such that ${\alpha }_n\le a$, $\mathrm{\forall }n\in \mathbb{N}$, and $\{r_n\}\subset (0,\mathrm{\infty })$ such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$. Then $\{x_n\}$ converges strongly to $P_{F(T)\cap EP(f)}(x_0)$.

Proof We divide the proof of this theorem into four steps as below.

Step 1. Similar to the proof of Step 1 in Theorem 3.1, it is easy to see that $C_n$ and $Q_n$ are closed convex sets for each $n\in \mathbb{N}\cup \{0\}$.

Step 2. We show that $F(T)\cap EP(f)\subset C_n\cap Q_n$, $\mathrm{\forall }n\in \mathbb{N}\cup \{0\}$.

Let $p\in F(T)\cap EP(f)$. Putting $u_n=T_{r_n}{y}_n$, $\mathrm{\forall }n\in \mathbb{N}\cup \{0\}$, by (2) of Lemma 2.4, we have $T_{r_n}$ is relatively nonexpansive. Noticing that relatively nonexpansive mappings are nonexpansive in Hilbert spaces, then for any $n\in \mathbb{N}\cup \{0\}$,

$${\parallel u_n-p\parallel }^2={\parallel T_{r_n}{y}_n-p\parallel }^2={\parallel T_{r_n}{y}_n-T_{r_n}p\parallel }^2\le {\parallel y_n-p\parallel }^2.$$

From the proof of Step 2 in Theorem 3.1, we have ${\parallel u_n-p\parallel }^2\le {\parallel x_n-p\parallel }^2+{\theta }_n$. Thus $p\in C_n$, hence $F(T)\cap EP(f)\subset C_n$. Similar to the proof of Step 2 in Theorem 3.1, it is easy to see that $F(T)\cap EP(f)\subset Q_n$. Therefore we have $F(T)\cap EP(f)\subset C_n\cap Q_n$, $\mathrm{\forall }n\in \mathbb{N}\cup \{0\}$. This means that $\{x_n\}$ is well defined.

Step 3. We show that $\{x_n\}$ is bounded and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$.

Similar to the proof of Step 3 in Theorem 3.1, we may obtain that $\{x_n\}$ is bounded and

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

(4.2)

Since $x_{n+1}=P_{C_n\cap Q_n}\in C_n$, then ${\parallel u_n-x_{n+1}\parallel }^2\le {\parallel x_n-x_{n+1}\parallel }^2+{\theta }_n$. Noticing that ${\theta }_n\to 0$, we have

$$\parallel u_n-x_{n+1}\parallel \le \parallel x_n-x_{n+1}\parallel +\sqrt{{\theta }_n}\to 0.$$

Hence

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-u_n\parallel \le \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x_{n+1}\parallel +\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-u_n\parallel \to 0.$$

(4.3)

For $p\in F(T)\cap EP(f)\subset C_n$, we have

$${\parallel u_n-p\parallel }^2\le {\parallel x_n-p\parallel }^2+{\theta }_n.$$

(4.4)

Since $u_n=T_{r_n}{y}_n$, by (4.4) and Lemma 2.5, we get that

$$\begin{array}{rcl}{\parallel u_n-y_n\parallel }^2& =& {\parallel T_{r_n}{y}_n-y_n\parallel }^2\\ \le & {\parallel y_n-p\parallel }^2-{\parallel T_{r_n}{y}_n-p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2+{\theta }_n-{\parallel u_n-p\parallel }^2\to 0.\end{array}$$

(4.5)

Let $A_n=\frac{1}{n+1}{\sum }_{j=0}^n{T}^j$. Since ${\alpha }_n\le a<1$, $\mathrm{\forall }n\in \mathbb{N}$, then by (4.3) and (4.5), we have

$$\begin{array}{rcl}\parallel x_n-A_n{x}_n\parallel & =& \frac{1}{1-{\alpha }_n}\parallel y_n-x_n\parallel \\ \le & \frac{1}{1-a}(\parallel y_n-u_n\parallel +\parallel u_n-x_n\parallel )\to 0.\end{array}$$

(4.6)

From Lemma 2.2, it follows that

$$\underset{l\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel A_n{x}_n-T^l(A_n{x}_n)\parallel \le \underset{l\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{x\in C}{sup}\parallel A_{n}x-T^l(A_{n}x)\parallel =0.$$

Therefore

$$\underset{l\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel A_n{x}_n-T^l(A_n{x}_n)\parallel =0.$$

(4.7)

Put $k_{\mathrm{\infty }}=sup\{k_n:n\ge 1\}<\mathrm{\infty }$, then

$$\begin{array}{rcl}\parallel y_n-T^l{x}_n\parallel & =& \parallel {\alpha }_n{x}_n+(1-{\alpha }_n)A_n{x}_n-T^l{x}_n\parallel \\ \le & {\alpha }_n\parallel x_n-A_n{x}_n\parallel +\parallel A_n{x}_n-T^l(A_n{x}_n)\parallel +\parallel T^l(A_n{x}_n)-T^l{x}_n\parallel \\ \le & {\alpha }_n\parallel x_n-A_n{x}_n\parallel +\parallel A_n{x}_n-T^l(A_n{x}_n)\parallel +k_l\parallel x_n-A_n{x}_n\parallel \\ \le & (1+k_{\mathrm{\infty }})\parallel x_n-A_n{x}_n\parallel +\parallel A_n{x}_n-T^l(A_n{x}_n)\parallel .\end{array}$$

Let $n,l\to \mathrm{\infty }$, we get that $\parallel y_n-T^l{x}_n\parallel \to 0$. Thus, by (4.3) and (4.5), we have

$$\parallel x_n-T^l{x}_n\parallel \le \parallel x_n-u_n\parallel +\parallel u_n-y_n\parallel +\parallel y_n-T^l{x}_n\parallel \to 0.$$

(4.8)

Therefore, by (4.8) and (4.2), we obtain that

$$\begin{array}{rcl}\parallel T{x}_n-x_n\parallel & \le & \parallel T{x}_n-T^{l+1}{x}_n\parallel +\parallel T^{l+1}{x}_n-T^{l+1}{x}_{n+1}\parallel +\parallel T^{l+1}{x}_{n+1}-x_{n+1}\parallel \\ +\parallel x_{n+1}-x_n\parallel \\ \le & k_{\mathrm{\infty }}\parallel x_n-T^l{x}_n\parallel +(k_{\mathrm{\infty }}+1)\parallel x_{n+1}-x_n\parallel +\parallel T^{l+1}{x}_{n+1}-x_{n+1}\parallel \\ \to & 0.\end{array}$$

Step 4. We show that $\{x_n\}$ converges strongly to $P_{F(T)\cap EP(f)}(x_0)$.

Since $\{x_n\}$ is bounded, there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $x_{n_k}\rightharpoonup w^{\prime }$. By Lemma 2.1, we have $w^{\prime }\in F(T)$. Next we show $w^{\prime }\in EP(f)$.

From (4.3) and (4.5), we get that $u_{n_k}\rightharpoonup w^{\prime }$, $y_{n_k}\rightharpoonup w^{\prime }$. Since $u_n=T_{r_n}{y}_n$, then

$$f(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-y_n〉\ge 0,\mathrm{\forall }y\in C.$$

Replacing n by $n_k$, we have from Condition (A2) that

$$\frac{1}{r_{n_k}}〈y-u_{n_k},u_{n_k}-y_{n_k}〉\ge -f(u_{n_k},y)\ge f(y,u_{n_k}),\mathrm{\forall }y\in C.$$

Let $k\to \mathrm{\infty }$, since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$, by (4.5) and Condition (A4), we get that

$$f(y,w^{\prime })\le 0,\mathrm{\forall }y\in C.$$

For $t\in (0,1)$, $y\in C$, let $y_t=ty+(1-t)w^{\prime }$, then $y_t\in C$, thus $f(y_t,w^{\prime })\le 0$. By Condition (A1), we get that

$$0=f(y_t,y_t)\le tf(y_t,y)+(1-t)f(y_t,w^{\prime })\le tf(y_t,y).$$

Dividing by t, we have

$$f(y_t,y)\ge 0,\mathrm{\forall }y\in C.$$

Let $t\to 0$. From Condition (A3), we obtain that $f(w^{\prime },y)\ge 0$, $\mathrm{\forall }y\in C$. Therefore, $w^{\prime }\in EP(f)$.

Denote $w=P_{F(T)\cap EP(f)}(x_0)$. Since $x_{n+1}=P_{C_n\cap Q_n}(x_0)$, $w\in F(T)\cap EP(f)\subset C_n\cap Q_n$, then $\parallel x_{n+1}-x_0\parallel \le \parallel w-x_0\parallel$. Since the norm is weakly lower semicontinuous, we have

$$\parallel w-x_0\parallel \le \parallel w^{\prime }-x_0\parallel \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_k}-x_0\parallel \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-x_0\parallel \le \parallel w-x_0\parallel .$$

Hence ${lim}_{k\to \mathrm{\infty }}\parallel x_{n_k}-x_0\parallel =\parallel w^{\prime }-x_0\parallel =\parallel w-x_0\parallel$. Using the Kadec-Klee property of H, we get ${lim}_{k\to \mathrm{\infty }}{x}_{n_k}=w^{\prime }=w$. Since $\{x_{n_k}\}$ is an arbitrary subsequence of $\{x_n\}$, we can conclude that $\{x_n\}$ converges strongly to $P_{F(T)\cap EP(f)}(x_0)$. □