Strong convergence theorems for three-steps iterations for asymptotically nonexpansive mappings in Banach spaces

Abstract

We consider the problem of the convergence of the three-steps iterative sequences for asymptotically nonexpansive mappings in a real Banach space. Under suitable conditions, it has been proved that the iterative sequence converges strongly to a fixed point, which is also a solution of certain variational inequality. The results presented in this paper extend and improve some recent results.

1 Introduction

Let X be a real Banach space with dual $X^{\ast }$, $J:X\to 2^{X^{\ast }}$ denotes the normalized duality mapping from X into $X^{\ast }$ given by

$$J(x)=\{f\in X^{\ast }:〈x,f〉=\parallel x\parallel \parallel f\parallel \wedge \parallel f\parallel =\parallel x\parallel \},\mathrm{\forall }x\in X.$$

Let C be a subset of X. A mapping $T:C\to C$ is called contraction if there exists a constant $\alpha \in (0,1)$ such that $\parallel Tx-Ty\parallel \le \alpha \parallel x-y\parallel$ for any $x,y\in C$. The mapping T is called nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for any $x,y\in C$, and it is called asymptotically nonexpansive if there exists a sequence $\{k_n\}$ in the interval $[1,\mathrm{\infty })$ with ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ and such that

$$\parallel T^{n}x-T^{n}y\parallel \le k_n\parallel x-y\parallel$$

(1)

for all $x,y\in C$ and all $n\in N$, where N is the set of natural numbers.

The class of asymptotically nonexpansive mappings was introduced by Goebel and Kirk [1] as an important generalization of nonexpansive mappings. They proved that if C is a nonempty, bounded, closed and convex subset of a real uniformly convex Banach space and T is an asymptotically nonexpansive self-mapping of C, then T has a fixed point in C. In 2000, Noor [2] introduced a three-steps iterative scheme and studied the approximate solutions of a variational inclusion in Hilbert spaces. In 2002, Xu and Noor [3] introduced and studied a new class of three-steps iterative schemes for solving the nonlinear equation $Tx=x$ for asymptotically nonexpansive mappings T in uniformly convex Banach spaces. In 2006, Nilsrakoo and Saejung [4] defined a three-steps mean value iterative scheme and extended the results of Xu and Noor [3]. In 2007, Yao and Noor [5] made a refinement and improvement of the previously known results.

Now we define a new three-steps iteration scheme for asymptotically nonexpansive mappings as follows.

Definition 1.1 Let X be a Banach space, C be a nonempty convex subset of X, $T:C\to C$ be an asymptotically nonexpansive mapping and $f:C\to C$ be a contraction. For a given $x_1\in C$ and $n\in N$, let us define the sequences $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ by the iterative scheme

$$\{\begin{array}{c}{z}_n=(1-{\gamma }_n)x_n+{\gamma }_n{T}^n{x}_n,\hfill \\ y_n=(1-{\beta }_n)x_n+{\beta }_n{T}^n{z}_n,\hfill \\ x_{n+1}=(1-{\alpha }_n)f(x_n)+{\alpha }_n{T}^n{y}_n,\hfill \end{array}$$

(2)

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are approximate sequences in $[0,1]$.

If ${\gamma }_n\equiv 0$, then the iterations defined in (2) reduces to the two-steps iterations defined as follows.

Definition 1.2 For a given $x_1\in C$ and $n\ge 1$, let us define the sequences $\{x_n\}$ and $\{y_n\}$ by the iterative scheme

$$\{\begin{array}{c}{y}_n=(1-{\beta }_n)x_n+{\beta }_n{T}^n{x}_n,\hfill \\ x_{n+1}=(1-{\alpha }_n)f(x_n)+{\alpha }_n{T}^n{y}_n,\hfill \end{array}$$

(3)

where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are approximate sequences in $[0,1]$.

If ${\beta }_n={\gamma }_n\equiv 0$, then the iterations defined in (2) reduces to the one-step iterations defined as follows.

Definition 1.3 For a given $x_1\in C$ and $n\in \mathbb{N}$, define the sequence $\{x_n\}$ by the iterative scheme

$$x_{n+1}=(1-{\alpha }_n)f(x_n)+{\alpha }_n{T}^n{x}_n,$$

(4)

where $\{{\alpha }_n\}$ is an approximate sequence in $[0,1]$.

The purpose of this paper is to establish a strong convergence theorem of the three-steps iterations for asymptotically nonexpansive mappings in a real Banach space equipped with a uniformly Gâteaux differentiable norm and to present some corollaries. Our results extend and improve the corresponding ones announced by Ceng et al. [6], Chang et al. [7], Lou et al. [8], Shahzad and Udomene [9] and others.

2 Preliminaries

Throughout this paper, we will use the following notions. Let X be a real Banach space with the norm $\parallel \cdot \parallel$ and let $X^{\ast }$ be its dual space. When $\{x_n\}$ is a sequence in X, then $x_n\to x$ (respectively $x_n\rightharpoonup x$, $x_n\rightharpoondown x$) will denote the strong (respectively the weak, the weak star) convergence of the sequence $\{x_n\}$ to x. Of course, the weak star convergence is considered in Banach spaces X which are dual spaces. We shall denote the single-valued duality mapping and the set of fixed points for a mapping T by j and $F(T)$, respectively.

Definition 2.1 Let $S_X$ denote the unit sphere of a Banach space X. The space X is said to have a uniformly Gâteaux differentiable norm $\parallel \cdot \parallel$, if for each $y\in C$ the limit

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

exists uniformly with respect to $x\in S_X$.

It is well known [10] that if X is equipped with a uniformly Gâteaux differentiable norm, then any duality mapping on X is single-valued and it is norm-to-weak* uniformly continuous, that is, $x_n\to x$ implies that $j(x_n)\rightharpoondown j(x)$.

Lemma 2.2 [11]

Let X be a real Banach space. Then for each $x,y\in X$, the following inequality holds:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,j(x+y)〉,\mathrm{\forall }j(x+y)\in J(x+y).$$

Lemma 2.3 [12]

Let $\{a_n\}$, $\{b_n\}$, $\{c_n\}$ be three nonnegative real sequences satisfying

$$a_{n+1}\le (1-t_n)a_n+b_n+c_n$$

with $\{t_n\}\subset [0,1]$, ${\sum }_{n=0}^{\mathrm{\infty }}{t}_n=\mathrm{\infty }$, $b_n=o(t_n)$, and ${\sum }_{n=0}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$. Then $a_n\to 0$.

Now, we start with our first result.

Lemma 2.4 Let X be a real Banach space, C be a nonempty convex subset of X and $T:C\to C$ be an asymptotically nonexpansive mapping defined by (1) with $F(T)\ne \mathrm{\varnothing }$ and ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ be the composite process defined by iterative scheme (2). Then the sequence $\{x_n\}$ is bounded.

Proof Let $p\in F(T)$ and ${\gamma }^{\prime }={sup}_n\{3{(k_n)}^3:n\ge 1\}$. We have from (2) that

$$\begin{array}{r}\parallel z_n-p\parallel \le (1-{\gamma }_n)\parallel x_n-p\parallel +{\gamma }_n\parallel T^n{x}_n-p\parallel \le (1+{\gamma }_n(k_n-1))\parallel x_n-p\parallel ,\\ \parallel y_n-p\parallel \le (1-{\beta }_n)\parallel x_n-p\parallel +{\beta }_n\parallel T^n{z}_n-p\parallel \le (1-{\beta }_n)\parallel x_n-p\parallel \\ \phantom{\parallel y_n-p\parallel \le }+{\beta }_n{k}_n\parallel z_n-p\parallel \le (1+{\beta }_n(k_n-1)(1+{\gamma }_n{k}_n))\parallel x_n-p\parallel ,\end{array}$$

and so

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & \le & (1-{\alpha }_n)\parallel f(x_n)-p\parallel +{\alpha }_n\parallel T^n{y}_n-p\parallel \\ \le & \alpha (1-{\alpha }_n)\parallel x_n-p\parallel +(1-{\alpha }_n)\parallel f(p)-p\parallel +{\alpha }_n{k}_n\parallel y_n-p\parallel \\ \le & (\alpha (1-{\alpha }_n)+{\alpha }_n{k}_n(1+{\beta }_n(k_n-1)(1+{\gamma }_n{k}_n)))\parallel x_n-p\parallel +(1-{\alpha }_n)\parallel f(p)-p\parallel \\ \le & (\alpha -\alpha {\alpha }_n+{\alpha }_n{k}_n+{\alpha }_n{\beta }_n{k}_n^2+{\alpha }_n{\beta }_n{k}_n^3-{\alpha }_n{\beta }_n{k}_n-{\alpha }_n{\beta }_n{\gamma }_n{k}_n^2)\parallel x_n-p\parallel \\ +(1-{\alpha }_n)\parallel f(p)-p\parallel \\ \le & (\alpha +{\alpha }_n(3k_n^3-\alpha ))\parallel x_n-p\parallel +(1-{\alpha }_n)\parallel f(p)-p\parallel \\ \le & (\alpha +{\alpha }_n({\gamma }^{\prime }-\alpha ))\parallel x_n-p\parallel +(1-{\alpha }_n)\parallel f(p)-p\parallel \\ =& \frac{(\alpha +{\alpha }_n({\gamma }^{\prime }-\alpha ))}{{\gamma }^{\prime }}{\gamma }^{\prime }\parallel x_n-p\parallel +\frac{(1-{\alpha }_n)({\gamma }^{\prime }-\alpha )}{{\gamma }^{\prime }}\frac{{\gamma }^{\prime }\parallel f(p)-p\parallel }{{\gamma }^{\prime }-\alpha }.\end{array}$$

Hence, it follows by induction that

$$\parallel x_n-p\parallel \le max\{{\gamma }^{\prime }\parallel x_0-p\parallel ,\frac{{\gamma }^{\prime }}{{\gamma }^{\prime }-\alpha }\parallel f(p)-p\parallel \}.$$

Therefore, $\{x_n\}$ is bounded. □

In order to prove our results, we also need the following lemma; see [8].

Lemma 2.5 Let X be a real Banach space equipped with a uniformly Gâteaux differentiable norm, C a bounded, closed and convex subset of X, $T:C\to C$ an asymptotically nonexpansive mapping defined by (1) with $F(T)\ne \mathrm{\varnothing }$, $f:C\to C$ a contraction with the contraction constant α. For any $\{{\alpha }_n\}\subset (0,1)$ define the sequence of contractions $T_n^f:C\to C$ by $T_n^f(\tilde{z})=(1-{\alpha }_n)f(\tilde{z})+{\alpha }_n{T}^n\tilde{z}$, where ${\alpha }_n<\frac{1-\alpha }{k_n-\alpha }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=1$ and ${lim}_{n\to \mathrm{\infty }}\frac{k_n-1}{1-{\alpha }_n}=0$. Let ${\tilde{z}}_n\in C$ be the unique fixed points of $T_n^f$, that is,

$${\tilde{z}}_n=(1-{\alpha }_n)f({\tilde{z}}_n)+{\alpha }_n{T}^n{\tilde{z}}_n,\mathrm{\forall }n\in N.$$

(5)

Then the sequence $\{{\tilde{z}}_n\}$ converges strongly to the unique solution of the following variational solution p:

$$p\in F(T)\mathit{\text{and}}〈(I-f)p,J(p-x^{\ast })〉\le 0,\mathrm{\forall }{x}^{\ast }\in F(T).$$

(6)

Lemma 2.6 Let C be a closed convex subset of a real Banach space X, $T:C\to C$ be an asymptotically nonexpansive mapping and $f:C\to C$ be a contraction with the contraction constant α. Let us assume that there are given three sequences $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ in $[0,1]$ satisfying the following conditions:

• (i) ${\alpha }_n<\frac{1-\alpha }{k_n-\alpha }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=1$, ${\sum }_{n=0}^{\mathrm{\infty }}(1-{\alpha }_n)=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}\frac{k_n-1}{1-{\alpha }_n}=0$;

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|)<\mathrm{\infty }$

and that ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is the composite process defined by the iterative scheme (2). Then we have the following assertions:

• (a) ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$;

• (b) if ${lim\hspace{0.17em}sup}_n{\beta }_n(1+{\gamma }_n)<1$, then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$.

Proof (a) By Lemma 2.4, we know that the sequence $\{x_n\}$ is bounded. Hence, it follows that the sequences $\{f(x_n)\}$, $\{y_n\}$, $\{T^n{x}_n\}$ and $\{T^n{z}_n\}$ are also bounded. Therefore, we have from (2) that

$$\begin{array}{rcl}\parallel z_n-z_{n-1}\parallel & =& \parallel (1-{\gamma }_n)(x_n-x_{n-1})+{\gamma }_n(T^n{x}_n-T^n{x}_{n-1})\\ +{\gamma }_n(T^n{x}_{n-1}-T^{n-1}{x}_{n-1})+({\gamma }_{n-1}-{\gamma }_n)(x_{n-1}-T^{n-1}{x}_{n-1})\parallel \\ \le & (1-{\gamma }_n)\parallel x_n-x_{n-1}\parallel +{\gamma }_n{k}_n\parallel x_n-x_{n-1}\parallel \\ +{\gamma }_n\parallel T^n{x}_{n-1}-T^{n-1}{x}_{n-1}\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel x_{n-1}-T^{n-1}{x}_{n-1}\parallel \\ =& (1+{\gamma }_n(k_n-1))\parallel x_n-x_{n-1}\parallel +{\gamma }_n\parallel T^n{x}_{n-1}-T^{n-1}{x}_{n-1}\parallel \\ +|{\gamma }_n-{\gamma }_{n-1}|\parallel x_{n-1}-T^{n-1}{x}_{n-1}\parallel \end{array}$$

and

$$\begin{array}{rcl}\parallel y_n-y_{n-1}\parallel & =& \parallel (1-{\beta }_n)x_n+{\beta }_n{T}^n{z}_n-(1-{\beta }_{n-1})x_{n-1}-{\beta }_{n-1}{T}^{n-1}{z}_{n-1}\parallel \\ =& \parallel (1-{\beta }_n)(x_n-x_{n-1})+{\beta }_n(T^n{z}_n-T^n{z}_{n-1})\\ +{\beta }_n(T^n{z}_{n-1}-T^{n-1}{z}_{n-1})+({\beta }_{n-1}-{\beta }_n)(x_{n-1}-T^{n-1}{z}_{n-1})\parallel \\ \le & (1-{\beta }_n)\parallel x_n-x_{n-1}\parallel +{\beta }_n{k}_n\parallel z_n-z_{n-1}\parallel \\ +{\beta }_n\parallel T^n{z}_{n-1}-T^{n-1}{z}_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}-T^{n-1}{z}_{n-1}\parallel \\ \le & (1-{\beta }_n)\parallel x_n-x_{n-1}\parallel +{\beta }_n{k}_n(1+{\gamma }_n(k_n-1))\parallel x_n-x_{n-1}\parallel \\ +k_n{\beta }_n{\gamma }_n\parallel T^n{x}_{n-1}-T^{n-1}{x}_{n-1}\parallel +{\beta }_n\parallel T^n{z}_{n-1}-T^{n-1}{z}_{n-1}\parallel \\ +k_n{\beta }_n|{\gamma }_n-{\gamma }_{n-1}|\parallel x_{n-1}-T^{n-1}{x}_{n-1}\parallel \\ +|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}-T^{n-1}{z}_{n-1}\parallel ,\end{array}$$

where it follows that

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & =& \parallel (1-{\alpha }_n)(f(x_n)-f(x_{n-1}))+{\alpha }_n(T^n{y}_{n-1}-T^{n-1}{y}_{n-1})\\ +({\alpha }_{n-1}-{\alpha }_n)(f(x_{n-1})-T^{n-1}{y}_{n-1})+{\alpha }_n(T^n{y}_n-T^n{y}_{n-1})\parallel \\ \le & \alpha (1-{\alpha }_n)\parallel x_n-x_{n-1}\parallel +{\alpha }_n\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-T^{n-1}{y}_{n-1}\parallel +{\alpha }_n{k}_n\parallel y_n-y_{n-1}\parallel \\ \le & (\alpha +{\alpha }_n(1-\alpha ))\parallel x_n-x_{n-1}\parallel \\ +{\alpha }_n(k_n-1)(1+{\beta }_n{k}_n(1+{\gamma }_n{k}_n))\parallel x_n-x_{n-1}\parallel \\ +{\alpha }_n\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel +{\alpha }_n{\beta }_n{k}_n\parallel T^n{z}_{n-1}-T^{n-1}{z}_{n-1}\parallel \\ +{\alpha }_n{\beta }_n{\gamma }_n{k}_n^2\parallel T^n{x}_{n-1}-T^{n-1}{x}_{n-1}\parallel \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-T^{n-1}{y}_{n-1}\parallel \\ +{\alpha }_n{k}_n|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}-T^{n-1}{z}_{n-1}\parallel \\ +{\alpha }_n{\beta }_n{k}_n^2|{\gamma }_n-{\gamma }_{n-1}|\parallel x_{n-1}-T^{n-1}{x}_{n-1}\parallel \\ \le & (1-(1-\alpha )(1-{\alpha }_n))\parallel x_n-x_{n-1}\parallel \\ +{\alpha }_n(k_n-1)(1+{\beta }_n{k}_n(1+{\gamma }_n{k}_n))\parallel x_n-x_{n-1}\parallel \\ +{\alpha }_n\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel +{\alpha }_n{k}_n\parallel T^n{z}_{n-1}-T^{n-1}{z}_{n-1}\parallel \\ +{\alpha }_n{k}_n^2\parallel T^n{x}_{n-1}-T^{n-1}{x}_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-T^{n-1}{y}_{n-1}\parallel \\ +{\alpha }_n{k}_n|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}-T^{n-1}{z}_{n-1}\parallel \\ +{\alpha }_n{\beta }_n{k}_n^2|{\gamma }_n-{\gamma }_{n-1}|\parallel x_{n-1}-T^{n-1}{x}_{n-1}\parallel \\ \le & (1-(1-\alpha )(1-{\alpha }_n))\parallel x_n-x_{n-1}\parallel \\ +{\alpha }_n(k_n-1)(1+{\beta }_n{k}_n(1+{\gamma }_n{k}_n))\parallel x_n-x_{n-1}\parallel \\ +{\alpha }_n(k_n^{12}-1)M+M(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|)\\ \le & (1-(1-\alpha )(1-{\alpha }_n))\parallel x_n-x_{n-1}\parallel +(1-\alpha )(1-{\alpha }_n)\frac{{\alpha }_n(k_n-1)}{(1-\alpha )(1-{\alpha }_n)}\\ \times [(1+{\beta }_n{k}_n(1+{\gamma }_n{k}_n))+(k_n^6+1)(k_n^3+1)(k_n^2+k_n+1)]M\\ +M(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|),\end{array}$$

where

$$\begin{array}{rcl}M& =& max\{\parallel x_n-x_{n-1}\parallel ,\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel ,\parallel f(x_{n-1})-T^{n-1}{y}_{n-1}\parallel ,\\ k_n(\parallel x_{n-1}-T^{n-1}{z}_{n-1}\parallel ,\parallel T^n{z}_{n-1}-T^{n-1}{z}_{n-1}\parallel ),\\ k_n^2(\parallel x_{n-1}-T^{n-1}{x}_{n-1}\parallel ,\parallel T^n{x}_{n-1}-T^{n-1}{x}_{n-1}\parallel )\}.\end{array}$$

Obviously, by condition (i), we have $t_n=(1-\alpha )(1-{\alpha }_n)\to 0$ and

$$\begin{array}{rcl}\frac{b_n}{t_n}& =& \frac{{\alpha }_n(k_n-1)}{(1-\alpha )(1-{\alpha }_n)}[(1+{\beta }_n{k}_n(1+{\gamma }_n{k}_n))+(k_n^6+1)(k_n^3+1)(k_n^2+k_n+1)]M\\ \to & 0(n\to \mathrm{\infty }).\end{array}$$

It follows from Lemma 2.3 and condition (ii) that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$.

(b) By (a), ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, so it follows from (2) and (i) that $\parallel x_{n+1}-T^n{y}_n\parallel =(1-{\alpha }_n)\parallel f(x_n)-T^n{y}_n\parallel \to 0$ and

$$\parallel x_n-T^n{y}_n\parallel \le \parallel x_{n+1}-x_n\parallel +\parallel x_{n+1}-T^n{y}_n\parallel \to 0.$$

(7)

Now, we will prove that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T^n{x}_n\parallel =0$. We have from (2) that

$$\begin{array}{rcl}\parallel x_n-T^n{x}_n\parallel & \le & \parallel x_n-T^n{y}_n\parallel +\parallel T^n{y}_n-T^n{x}_n\parallel \\ \le & \parallel x_n-T^n{y}_n\parallel +k_n\parallel x_n-y_n\parallel \\ =& \parallel x_n-T^n{y}_n\parallel +{\beta }_n{k}_n\parallel x_n-T^n{z}_n\parallel \\ \le & \parallel x_n-T^n{y}_n\parallel +{\beta }_n{k}_n\parallel x_n-T^n{x}_n\parallel +{\beta }_n{k}_n\parallel T^n{x}_n-T^n{z}_n\parallel \\ \le & \parallel x_n-T^n{y}_n\parallel +{\beta }_n{k}_n\parallel x_n-T^n{x}_n\parallel +{\beta }_n{k}_n^2\parallel x_n-z_n\parallel \\ =& \parallel x_n-T^n{y}_n\parallel +{\beta }_n{k}_n(1+{\gamma }_n{k}_n)\parallel x_n-T^n{x}_n\parallel .\end{array}$$

It follows from (7) that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T^n{x}_n\parallel \le {\beta }_n(1+{\gamma }_n){lim}_{n\to \mathrm{\infty }}\parallel x_n-T^n{x}_n\parallel$, where, by the condition ${lim\hspace{0.17em}sup}_n{\beta }_n(1+{\gamma }_n)<1$, we have that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T^n{x}_n\parallel =0.$$

(8)

Finally, we will show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. In fact, according to (8), we have

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

Hence,

$$\begin{array}{rcl}\parallel x_{n+1}-T{x}_{n+1}\parallel & \le & \parallel x_{n+1}-T^{n+1}{x}_{n+1}\parallel +\parallel T^{n+1}{x}_{n+1}-T{x}_{n+1}\parallel \\ \le & \parallel x_{n+1}-T^{n+1}{x}_{n+1}\parallel +k_1\parallel T^n{x}_{n+1}-x_{n+1}\parallel \to 0,\end{array}$$

which implies that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. □

3 Main results

Theorem 3.1 Let X be a real Banach space equipped with a uniformly Gâteaux differentiable norm, C be a bounded, closed and convex subset of X, $T:C\to C$ be an asymptotically nonexpansive mapping defined by (1) with $F(T)\ne \mathrm{\varnothing }$ and $f:C\to C$ be a contraction with the contraction constant α. Let $\{x_n\}$ be the sequence defined by the iterative scheme (2) with $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ satisfying the following conditions:

• (C₁) ${\alpha }_n<\frac{1-\alpha }{k_n-\alpha }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=1$, ${\sum }_{n=0}^{\mathrm{\infty }}(1-{\alpha }_n)=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}\frac{k_n-1}{1-{\alpha }_n}=0$;

• (C₂) ${lim\hspace{0.17em}sup}_n{\beta }_n(1+{\gamma }_n)<1$;

• (C₃) ${\sum }_{n=1}^{\mathrm{\infty }}(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|)<\mathrm{\infty }$.

Then the sequence $\{x_n\}$ converges strongly to the unique solution p of the variational inequality:

$$p\in F(T)\mathit{\text{and}}〈(I-f)p,J(p-x^{\ast })〉\le 0,\mathrm{\forall }{x}^{\ast }\in F(T).$$

(9)

Proof Since C is closed, by Lemma 2.4, $\{x_n\}$ is bounded, so $\{f(x_n)\}$, $\{y_n\}$, $\{T^n{x}_n\}$ and $\{T^n{z}_n\}$ are also bounded. Let $\{{\tilde{z}}_n\}$ be the sequence defined by

$${\tilde{z}}_n=(1-{\alpha }_n)f({\tilde{z}}_n)+{\alpha }_n{T}^n{\tilde{z}}_n,\mathrm{\forall }n\in N.$$

(10)

It follows from Lemma 2.5 that the sequence $\{{\tilde{z}}_n\}$ converges strongly to a fixed point p of T and p is also the unique solution of the variational inequality (6). We will next prove that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(p)-p,j(x_n-p)〉\le 0.$$

(11)

By Lemma 2.6(b), ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. It is easy to show that

$$\parallel T^m{x}_n-x_n\parallel \to 0(n\to \mathrm{\infty }),\mathrm{\forall }m\in N,$$

where if we put $P_n(m)=\parallel T^m{x}_n-x_n\parallel (2k_m\parallel {\tilde{z}}_m-x_n\parallel +\parallel T^m{\tilde{z}}_n-x_n\parallel )$, then $P_n(m)\to 0$ as $n\to \mathrm{\infty }$. On the other hand, we have from (10) that

$${\tilde{z}}_m-x_n=(1-{\alpha }_m)(f({\tilde{z}}_m)-x_n)+{\alpha }_m(T^m{\tilde{z}}_m-x_n).$$

It follows by Lemma 2.2 that

$$\begin{array}{rcl}{\parallel {\tilde{z}}_m-x_n\parallel }^2& \le & {\alpha }_m^2{\parallel T^m{\tilde{z}}_m-x_n\parallel }^2+2(1-{\alpha }_m)〈f({\tilde{z}}_m)-x_n,j({\tilde{z}}_m-x_n)〉\\ \le & {\alpha }_m^2{(\parallel T^m{\tilde{z}}_m-T^m{x}_n\parallel +\parallel T^m{x}_n-x_n\parallel )}^2\\ +2(1-{\alpha }_m)(〈f({\tilde{z}}_m)-{\tilde{z}}_m,j({\tilde{z}}_m-x_n)〉+{\parallel {\tilde{z}}_m-x_n\parallel }^2)\\ \le & {\alpha }_m^2{(k_m\parallel {\tilde{z}}_m-x_n\parallel +\parallel T^m{x}_n-x_n\parallel )}^2\\ +2(1-{\alpha }_m)(〈f({\tilde{z}}_m)-{\tilde{z}}_m,j({\tilde{z}}_m-x_n)〉+k_m^2{\parallel {\tilde{z}}_m-x_n\parallel }^2)\\ =& {\alpha }_m^2{k}_m^2{\parallel {\tilde{z}}_m-x_n\parallel }^2+{\alpha }_m^2{P}_n(m)+2(1-{\alpha }_m)k_m^2{\parallel {\tilde{z}}_m-x_n\parallel }^2\\ +2(1-{\alpha }_m)〈f({\tilde{z}}_m)-{\tilde{z}}_m,j({\tilde{z}}_m-x_n)〉\\ =& k_m^2(1+{(1-{\alpha }_m)}^2){\parallel {\tilde{z}}_m-x_n\parallel }^2+{\alpha }_m^2{P}_n(m)\\ +2(1-{\alpha }_m)〈f({\tilde{z}}_m)-{\tilde{z}}_m,j({\tilde{z}}_m-x_n)〉.\end{array}$$

Hence,

$$〈{\tilde{z}}_m-f({\tilde{z}}_m),j({\tilde{z}}_m-x_n)〉\le \frac{k_m^2-1+k_m^2{(1-{\alpha }_m)}^2}{2(1-{\alpha }_m)}{\parallel {\tilde{z}}_m-x_n\parallel }^2+\frac{{\alpha }_m^2{P}_n(m)}{2(1-{\alpha }_m)}.$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$ and the sequences $\{{\tilde{z}}_n\}$ and $\{x_n\}$ are bounded it follows that for some constant $M>{sup}_{m,n}\parallel {\tilde{z}}_m-x_n\parallel$, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈{\tilde{z}}_m-f({\tilde{z}}_m),j({\tilde{z}}_m-x_n)〉\le \frac{M}{2}((k_m+1)\frac{k_m-1}{1-{\alpha }_m}+2k_m^2(1-{\alpha }_m)).$$

Since ${\tilde{z}}_m\to p\in F(T)$ as $m\to \mathrm{\infty }$ and the duality mapping is norm-to-weakly* uniformly continuous, we obtain that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(p)-p,j(x_n-p)〉\le 0.$$

(12)

Finally, we will show that $x_n\to p$. We have

$$\begin{array}{r}\parallel z_n-p\parallel \le (1-{\gamma }_n)\parallel x_n-p\parallel +{\gamma }_n\parallel T^n{x}_n-p\parallel \le k_n\parallel x_n-p\parallel ,\\ \parallel y_n-p\parallel \le (1-{\beta }_n)\parallel x_n-p\parallel +{\beta }_n\parallel T^n{z}_n-p\parallel \\ \phantom{\parallel y_n-p\parallel }\le (1-{\beta }_n)\parallel x_n-p\parallel +{\beta }_n{k}_n\parallel z_n-p\parallel \\ \phantom{\parallel y_n-p\parallel }\le (1-{\beta }_n+{\beta }_n{k}_n^2)\parallel x_n-p\parallel \le k_n^2\parallel x_n-p\parallel \end{array}$$

and so

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel (1-{\alpha }_n)(f(x_n)-p)+{\alpha }_n(T^n{y}_n-p)\parallel }^2\\ \le & {\alpha }_n^2{k}_n^2{\parallel y_n-p\parallel }^2+2(1-{\alpha }_n)〈f(x_n)-p,J(x_{n+1}-p)〉\\ \le & {\alpha }_n^2{k}_n^6{\parallel x_n-p\parallel }^2+\alpha (1-{\alpha }_n){(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )}^2\\ +2(1-{\alpha }_n)〈f(p)-p,J(x_{n+1}-p)〉\\ \le & (\alpha (1-{\alpha }_n)+{\alpha }_n^2){\parallel x_n-p\parallel }^2+\alpha (1-{\alpha }_n){\parallel x_{n+1}-p\parallel }^2\\ {\alpha }_n^2(k_n^6-1){\parallel x_n-p\parallel }^2+2(1-{\alpha }_n)〈f(p)-p,J(x_{n+1}-p)〉\\ \le & (1-(2-\alpha )(1-{\alpha }_n)){\parallel x_n-p\parallel }^2+\alpha (1-{\alpha }_n){\parallel x_{n+1}-p\parallel }^2\\ +({(1-{\alpha }_n)}^2+(k_n-1))M+2(1-{\alpha }_n){\tilde{\gamma }}_{n+1},\end{array}$$

where ${\tilde{\gamma }}_{n+1}=max\{〈f(p)-p,J(x_{n+1}-p)〉,0\}$. By (11), we have ${\tilde{\gamma }}_{n+1}\to 0$, and $M>{sup}_n(k_n^5+k_n^4+k_n^3+k_n^2+k_n+1)\parallel x_n-p\parallel$. Then it follows that

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & \frac{1-(2-\alpha )(1-{\alpha }_n)}{1-\alpha (1-{\alpha }_n)}{\parallel x_n-p\parallel }^2+\frac{2(1-{\alpha }_n)}{1-\alpha (1-{\alpha }_n)}{\tilde{\gamma }}_{n+1}\\ +(\frac{{(1-{\alpha }_n)}^2}{1-\alpha (1-{\alpha }_n)}+\frac{(k_n-1)}{1-\alpha (1-{\alpha }_n)})M.\end{array}$$

If we define $a_n={\parallel x_n-p\parallel }^2$, $t_n=\frac{2(1-\alpha )(1-{\alpha }_n)}{1-\alpha (1-{\alpha }_n)}$ and $b_n=\frac{{(1-{\alpha }_n)}^2+(k_n-1)}{1-\alpha (1-{\alpha }_n)}M$, then applying Lemma 2.3 we conclude that $x_n\to p$. Moreover, it follows from (12) that p satisfies condition (9). In order to show that p is unique, let $p^{\ast }\in F$ be another solution of (9) in F. Then adding the inequalities $〈f(p)-p,j(p^{\ast }-p)〉\le 0$ and $〈f(p^{\ast })-p^{\ast },j(p-p^{\ast })〉\le 0$, we get that $(1-\alpha ){\parallel p-p^{\ast }\parallel }^2\le 0$, which implies the equality $p=p^{\ast }$. □

The following example gives a mapping T, which is not nonexpansive but satisfying all the assumptions of Theorem 3.1.

Example 3.2 Let B denote the unit ball in the Hilbert space $l^2$ and let $T:B\to B$ be defined as follows:

$$T:(x_1,x_2,x_3,\dots )\to (0,2x_1,A_2{x}_2,A_3{x}_3,\dots ),$$

where $A_n={(1-\frac{1}{n})}^2(\frac{2}{n+1}-\frac{1}{n(n+2)})$, $n=2,3,\dots$ . Then it is easy to verify that T is an asymptotically nonexpansive mapping with $k_n=2{\prod }_{i=2}^n{A}_i=1+\frac{1}{n^2}$, but it is not nonexpansive. If we set ${\alpha }_n=1-\frac{1}{n}$, $f(x)=\frac{1}{2}x$ and ${\beta }_n={\gamma }_n=\frac{1}{2}$, then the real sequences $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ satisfy conditions (C₁), (C₂) and (C₃) from Theorem 3.1, and it is easy to prove that 0 is the unique fixed point of T in B.

If ${\gamma }_n\equiv 0$ in Theorem 3.1, then we have by (2) that $z_n=x_n$. In fact, we have the following corollary.

Corollary 3.3 Let X be a real Banach space equipped with a uniformly Gâteaux differentiable norm, C be a bounded closed convex subset of X, $T:C\to C$ be an asymptotically nonexpansive mapping defined by (1) with $F(T)\ne \mathrm{\varnothing }$ and $f:C\to C$ be a contraction with the contraction constant α. Let $\{x_n\}$ be the sequence defined by the iterative scheme (3) with $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ satisfying the following conditions:

• (C₁) ${\alpha }_n<\frac{1-\alpha }{k_n-\alpha }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=1$, ${\sum }_{n=0}^{\mathrm{\infty }}(1-{\alpha }_n)=\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}\frac{k_n-1}{1-{\alpha }_n}=0$;

• (C₂) ${lim\hspace{0.17em}sup}_n{\beta }_n<1$;

• (C₃) ${\sum }_{n=1}^{\mathrm{\infty }}(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|)<\mathrm{\infty }$.

Then the sequence $\{x_n\}$ converges strongly to the unique solution p of the variational inequality (9).

If ${\beta }_n={\gamma }_n\equiv 0$ in Theorem 3.1, then we have by (2) that $z_n=y_n=x_n$. Hence, it follows that the following result is satisfied.

Corollary 3.4 Let X be a real Banach space equipped with a uniformly Gâteaux differentiable norm, C be a bounded closed convex subset of X, $T:C\to C$ be an asymptotically nonexpansive mapping defined by (1) with $F(T)\ne \mathrm{\varnothing }$ and $f:C\to C$ be a contraction with the contraction constant α. Let $\{x_n\}$ be the sequence defined by the iterative scheme (4) with $\{{\alpha }_n\}$ satisfying the following conditions:

• (C₁) ${\alpha }_n<\frac{1-\alpha }{k_n-\alpha }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=1$ and ${lim}_{n\to \mathrm{\infty }}\frac{k_n-1}{1-{\alpha }_n}=0$;

• (C₂) ${\sum }_{n=0}^{\mathrm{\infty }}(1-{\alpha }_n)=\mathrm{\infty }$;

• (C₃) either ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_n-{\alpha }_{n-1}|<\mathrm{\infty }$ or ${lim}_{n\to \mathrm{\infty }}({\alpha }_{n+1}/{\alpha }_n)=1$.

Then the sequence $\{x_n\}$ converges strongly to the unique solution of the variational inequality (9).

Remarks 3.5

• 1. If ${\gamma }_n\equiv 0$ and $f=u$ is a constant function in Theorem 3.1, then the iterative scheme (2) reduces to the following iterative scheme:

  $$\{\begin{array}{c}{y}_n=(1-{\beta }_n)x_n+{\beta }_n{T}^n{x}_n,\hfill \\ x_{n+1}=(1-{\alpha }_n)u+{\alpha }_n{T}^n{y}_n.\hfill \end{array}$$

  In consequence, Corollary 3.3 improves Theorem 1 of Chang et al. from [7].

• 2. Let in Theorem 3.1 ${\gamma }_n\equiv 0$ and the iterative scheme (3) be replaced by the following scheme:

  $$\{\begin{array}{c}{y}_n=(1-{\beta }_n)x_n+{\beta }_n{T}^n{x}_n,\hfill \\ x_{n+1}=(1-{\alpha }_n)f(x_n)+{\alpha }_n{y}_n.\hfill \end{array}$$

  Then by Theorem 3.1, we have the more general result than the result of Lou et al. from [8] and Corollary 3.6 of Ceng et al. in [6]. If T and $\{{\alpha }_n\}$ are as in Corollary 3.4, assume that $\{{\beta }_n\},\{{\gamma }_n\}\in [0,1]$, ${\alpha }_n+{\beta }_n+{\gamma }_n=1$ and $0<{lim\hspace{0.17em}inf}_n{\beta }_n\le {lim\hspace{0.17em}sup}_n{\beta }_n<1$. Then the sequence $\{x_n\}$ defined by $x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n{T}^n{x}_n$ converges strongly to the unique solution of the variational inequality (9).

• 3. Theorem 3.1 and Corollary 3.4 extend Theorem 3.3 of Shahzad and Udomene in [9] to a more wide class of spaces.