Strong convergence of new iterative algorithms for certain classes of asymptotically pseudocontractions

Abstract

Let C be a nonempty closed convex subset of a real Hilbert space, and let $T:C\to C$ be an asymptotically k-strictly pseudocontractive mapping with $F(T)=\{x\in C:Tx=x\}\ne \mathrm{\varnothing }$. Let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{t_n\}}_{n=1}^{\mathrm{\infty }}$ be real sequences in $(0,1)$. Let ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ be the sequence generated from an arbitrary $x_1\in C$ by

$$\{\begin{array}{l}{\nu }_n=P_C((1-t_n)x_n),n\ge 1,\\ x_{n+1}=(1-{\alpha }_n){\nu }_n+{\alpha }_n{T}^n{\nu }_n,n\ge 1,\end{array}$$

where $P_C:H\to C$ is the metric projection. Under some appropriate mild conditions on ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{t_n\}}_{n=1}^{\mathrm{\infty }}$, we prove that ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to a fixed point of T. Furthermore, if $T:C\to C$ is uniformly L-Lipschitzian and asymptotically pseudocontractive with $F(T)\ne \mathrm{\varnothing }$, we first prove that $(I-T)$ is demiclosed at 0, and then prove that under some suitable conditions on the real sequences ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{t_n\}}_{n=1}^{\mathrm{\infty }}$ in $(0,1)$, the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ generated from an arbitrary $x_1\in C$ by

$$\{\begin{array}{l}{\nu }_n=P_C((1-t_n)x_n),n\ge 1,\\ y_n=(1-{\beta }_n){\nu }_n+{\beta }_n{T}^n{\nu }_n,n\ge 1,\\ x_{n+1}=(1-{\alpha }_n){\nu }_n+{\alpha }_n{T}^n{y}_n,n\ge 1,\end{array}$$

converges strongly to a fixed point of T. No compactness assumption is imposed on T or C and no further requirement is imposed on $F(T)$.

MSC:47H09, 47J25, 65J15.

1 Introduction

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the induced norm $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H. A mapping $T:C\to C$ is said to be L-Lipschitzian if there exists $L\ge 0$ such that

$$\parallel Tx-Ty\parallel \le L\parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

(1.1)

T is said to be a contraction if $L\in [0,1)$, and T is said to be nonexpansive if $L=1$. T is said to be asymptotically nonexpansive (see, for example, [1]) if there exists a sequence ${\{k_n\}}_{n=1}^{\mathrm{\infty }}\subseteq [1,\mathrm{\infty })$ with ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ such that

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

(1.2)

It is well known (see, for example, [1]) that the class of nonexpansive mappings is a proper subclass of the class of asymptotically nonexpansive mappings. T is said to be asymptotically k-strictly pseudocontractive (see, for example, [2]) if there exist $k\in [0,1)$ and a sequence ${\{k_n\}}_{n=1}^{\mathrm{\infty }}\subseteq [1,\mathrm{\infty })$, ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ such that

$${\parallel T^{n}x-T^{n}y\parallel }^2\le k_n{\parallel x-y\parallel }^2+k{\parallel (x-T^{n}x)-(y-T^{n}y)\parallel }^2,\mathrm{\forall }x,y\in C.$$

(1.3)

T is said to be asymptotically pseudocontractive if there exists a sequence ${\{k_n\}}_{n=1}^{\mathrm{\infty }}\subseteq [1,\mathrm{\infty })$, ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ such that

$${\parallel T^{n}x-T^{n}y\parallel }^2\le k_n{\parallel x-y\parallel }^2+{\parallel (x-T^{n}x)-(y-T^{n}y)\parallel }^2,\mathrm{\forall }x,y\in C.$$

(1.4)

It is well known that in real Hilbert spaces, the class of asymptotically nonexpansive maps is a proper subclass of the class of asymptotically k-strictly pseudocontractive maps. Furthermore, the class of asymptotically k-strictly pseudocontractive mappings is a proper subclass of the class of asymptotically pseudocontractive maps. T is said to be uniformly L-Lipschitzian if there exists $L\ge 0$ such that

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

T is said to be demiclosed at p if whenever ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is a sequence in C which converges weakly to $x^{\ast }\in C$ and ${\{T{x}_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to p, then $T{x}^{\ast }=p$. It is well known that if $T:C\to C$ is asymptotically k-strictly pseudocontractive, then T is uniformly L-Lipschitzian (see, for example, [3, 4]), and $(I-T)$ is demiclosed at 0 (see, for example, [5]). The modified Mann iteration scheme ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ generated from an arbitrary $x_1\in C$ by

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

(1.5)

where the control sequence ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ is a real sequence in $(0,1)$ satisfying some appropriate conditions, has been used by several authors for the approximation of fixed points of asymptotically k-strictly pseudocontractive maps (see, for example, [2–10]). The iteration algorithm (1.5) is a modification of the well-known Mann iterative algorithm (see [11]) generated from an arbitrary $x_1\in C$ by

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

(1.6)

where the control sequence ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ is a real sequence in $(0,1)$ satisfying some appropriate conditions.

In real Hilbert spaces, it is known (see, for example, [3–5]) that if C is a nonempty closed convex subset of a real Hilbert space H, and $T:C\to C$ is an asymptotically k-strictly pseudocontractive mapping with a sequence ${\{k_n\}}_{n=1}^{\mathrm{\infty }}\subseteq [1,\mathrm{\infty })$, ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$, and a nonempty fixed point set $F(T)$, then the modified iteration sequence $\{x_n\}$ generated by (1.5) is an approximate fixed point sequence (i.e., $\parallel x_n-T{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$) if ${\alpha }_n\in [a,b]\subseteq (0,1-k)$. This together with the demiclosedness property of $(I-T)$ at 0 yields that $\{x_n\}$ converges weakly to a fixed point of T.

To obtain strong convergence of the modified Mann algorithm (1.5) to a fixed point of an asymptotically k-strictly pseudocontractive mapping, additional conditions are usually required on T and on the subset C (see, for example, [2–10]). Even for nonexpansive maps, additional conditions are required on T or C to obtain strong convergence using the Mann algorithm (1.6). In [12], Genel and Lindenstraus provided an example of a nonexpansive mapping defined on a bounded closed convex subset of a Hilbert space for which the Mann iteration does not converge to a fixed point of T. Recently Yao et al. [13] (see also [14, 15]) studied a modified Mann iteration algorithm $\{x_n\}$ generated from an arbitrary $x_1\in H$ by

$$\{\begin{array}{l}{\nu }_n=(1-t_n)x_n,\\ x_{n+1}=(1-{\alpha }_n){\nu }_n+{\alpha }_{n}T{\nu }_n,\end{array}$$

(1.7)

where $\{t_n\}$ and $\{{\alpha }_n\}$ are real sequences in $(0,1)$ satisfying some appropriate conditions. They proved strong convergence of the modified algorithm to a fixed point of a nonexpansive mapping $T:H\to H$ when $F(T)\ne \mathrm{\varnothing }$. Clearly, the modified Mann iteration algorithm reduces to the normal Mann iteration algorithm when $t_n\equiv 0$.

It is our purpose in this paper to modify algorithm (1.7) and prove that the modified algorithm converges strongly to a fixed point of an asymptotically k-strictly pseudocontractive mapping $T:C\to C$, where C is a nonempty closed convex subset of a real Hilbert space and $F(T)\ne \mathrm{\varnothing }$. Furthermore, we prove that if $T:C\to C$ is a uniformly L-Lipschitzian asymptotically pseudocontractive mapping, then $(I-T)$ is demiclosed at 0. We then introduce an iterative algorithm which converges strongly to a fixed point of a uniformly L-Lipschitzian asymptotically pseudocontractive mapping $T:C\to C$ with $F(T)\ne \mathrm{\varnothing }$. The technique of proof of our convergence theorems follows the one proposed by Maingé [14].

2 Preliminaries

In what follows, we shall need the following results.

Lemma 2.1 [16]

Let ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-{\lambda }_n)a_n+{\lambda }_n{\gamma }_n+{\sigma }_n,n\ge 1,$$

where $\{{\lambda }_n\}\subseteq (0,1)$, $\{{\gamma }_n\}\subseteq \mathrm{\Re }$, $\{{\sigma }_n\}$ is a sequence of nonnegative real numbers and

• (i) ${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$, or equivalently, ${\prod }_{n=0}^{\mathrm{\infty }}(1-{\lambda }_n)=0$,

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

• (iii) ${\sum }_{n=0}^{\mathrm{\infty }}{\sigma }_n<\mathrm{\infty }$.

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

Let C be a closed convex subset of a real Hilbert space H. Let $P_C:H\to C$ denote the metric projection (the proximity map) which assigns to each point $x\in H$ the unique nearest point in C, denoted by $P_C(x)$. It is well known that

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

(2.1)

and that $P_C$ is nonexpansive.

It is also well known that in real Hilbert spaces H, we have the following (see, for example, [17]):

$$\text{(i)}{\parallel x+y\parallel }^2\le {\parallel y\parallel }^2+2〈x,x+y〉,\mathrm{\forall }x,y\in H;$$

(2.2)

$$\begin{array}{rc}\text{(ii)}& {\parallel \alpha x+(1-\alpha )y\parallel }^2=\alpha {\parallel x\parallel }^2+(1-\alpha ){\parallel y\parallel }^2-\alpha (1-\alpha ){\parallel x-y\parallel }^2,\\ \mathrm{\forall }x,y\in H\text{ and }\alpha \in [0,1];\end{array}$$

(2.3)

$$\begin{array}{rc}\text{(iii)}& \text{if }{\{x_n\}}_{n=1}^{\mathrm{\infty }}\text{ is a sequence in }H\text{ which converges weakly to }z,\text{ then}\\ \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-y\parallel }^2=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-z\parallel }^2+{\parallel z-y\parallel }^2,\mathrm{\forall }y\in H.\end{array}$$

(2.4)

3 Main results

3.1 Strong convergence of an iterative algorithm for asymptotically k-strictly pseudocontractive maps

We now introduce the following iterative algorithm analogous to one studied in [13].

Modified averaging Mann algorithm Let C be a nonempty closed convex subset of a real Hilbert space H, and let $T:C\to C$ be a given mapping. For arbitrary $x_1\in C$, our iteration sequence $\{x_n\}$ is given by

$$\{\begin{array}{l}{\nu }_n=P_C((1-t_n)x_n),\\ x_{n+1}=(1-{\alpha }_n){\nu }_n+{\alpha }_n{T}^n{\nu }_n,\end{array}$$

(3.1)

where $\{t_n\}$ and $\{{\alpha }_n\}$ are suitable real sequences in $(0,1)$ satisfying some appropriate conditions that will be made precise in our strong convergence theorem.

We now prove the following convergence theorem.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space, and let $T:C\to C$ be an asymptotically k-strictly pseudocontractive mapping with a sequence ${\{k_n\}}_{n=1}^{\mathrm{\infty }}\subseteq [1,\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$. Let $F(T)\ne \mathrm{\varnothing }$, and let ${\{t_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ satisfying the conditions:

• (c1) ${lim}_{n\to \mathrm{\infty }}{t}_n=0$;

• (c2) ${\sum }_{n=1}^{\mathrm{\infty }}{t}_n=\mathrm{\infty }$;

• (c3) ${lim}_{n\to \mathrm{\infty }}\frac{1}{t_n}(k_n-1)=0$;

• (c4) $0<ϵ\le {\alpha }_n<\frac{1}{2}(1-t_n)(1-k)$, $\mathrm{\forall }n\ge 1$ and for some ϵ.

Then the modified averaging iteration sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ generated from $x_1\in C$ by (3.1) converges strongly to a fixed point of T.

Proof Observe that (1.3) is equivalent to each of the following inequalities:

$$\begin{array}{rl}2〈(I-T^n)x-(I-T^n)y,x-y〉\ge & (1-k){\parallel (I-T^n)x-(I-T^n)y\parallel }^2\\ -(k_n-1){\parallel x-y\parallel }^2,\end{array}$$

(3.2)

$$2〈T^{n}x-T^{n}y,x-y〉\le (k_n+1){\parallel x-y\parallel }^2-(1-k){\parallel (I-T^n)x-(I-T^n)y\parallel }^2.$$

(3.3)

Let $p\in F(T)$ be arbitrary. Then, using (1.3), (2.3), (3.1) and (3.3), we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel (1-{\alpha }_n)({\nu }_n-p)+{\alpha }_n(T^n{\nu }_n-p)\parallel }^2\\ =& (1-{\alpha }_n){\parallel {\nu }_n-p\parallel }^2+{\alpha }_n{\parallel T^n{\nu }_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel {\nu }_n-T^n{\nu }_n\parallel }^2\\ \le & [1+{\alpha }_n(k_n-1)]{\parallel {\nu }_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n-k){\parallel {\nu }_n-T^n{\nu }_n\parallel }^2.\end{array}$$

(3.4)

Hence

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & \le & [1+{\alpha }_n(k_n-1)]\parallel {\nu }_n-p\parallel \\ =& [1+{\alpha }_n(k_n-1)]\parallel P_C((1-t_n)x_n)-p\parallel \\ \le & [1+{\alpha }_n(k_n-1)]\parallel (1-t_n)x_n-p\parallel \\ =& [1+{\alpha }_n(k_n-1)]\parallel (1-t_n)(x_n-p)-t_{n}p\parallel \\ \le & [1+{\alpha }_n(k_n-1)][(1-t_n)\parallel x_n-p\parallel +t_n\parallel p\parallel ]\\ \le & [1+{\alpha }_n(k_n-1)]max\{\parallel x_n-p\parallel ,\parallel p\parallel \}\\ ⋮\\ \le & \prod _{j=1}^n[1+{\alpha }_j(k_j-1)]max\{\parallel x_1-p\parallel ,\parallel p\parallel \}.\end{array}$$

Since ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$, it follows from (3.4) that ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded. Hence ${\{{\nu }_n\}}_{n=1}^{\mathrm{\infty }}$ is also bounded. Furthermore, it follows from (2.2) that

$$\begin{array}{rcl}{\parallel x_n-x_{n+1}\parallel }^2& =& {\parallel x_n-{\nu }_n+{\nu }_n-x_{n+1}\parallel }^2\\ \le & {\parallel {\nu }_n-x_{n+1}\parallel }^2+2〈x_n-{\nu }_n,x_n-x_{n+1}〉\\ \le & {\parallel {\nu }_n-x_{n+1}\parallel }^2+2\parallel x_n-{\nu }_n\parallel \parallel x_n-x_{n+1}\parallel \\ \le & {\parallel {\nu }_n-x_{n+1}\parallel }^2+2t_n\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel .\end{array}$$

(3.5)

From (3.1) and (3.5) we obtain

$$\begin{array}{rcl}{\parallel {\nu }_n-T^n{\nu }_n\parallel }^2& =& \frac{1}{{\alpha }_n^2}{\parallel {\nu }_n-x_{n+1}\parallel }^2\\ \ge & \frac{1}{{\alpha }_n^2}[{\parallel x_n-x_{n+1}\parallel }^2-2t_n\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel ].\end{array}$$

(3.6)

Since ${\{{\nu }_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded, then

$${\parallel {\nu }_n-p\parallel }^2\le D,\mathrm{\forall }n\ge 1\text{ and for some }D>0,$$

and hence using condition (c4) and (3.6) in (3.4), we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & [1+{\alpha }_n(k_n-1)]{\parallel {\nu }_n-p\parallel }^2\\ -{\alpha }_n(1-{\alpha }_n-k){\parallel {\nu }_n-T^n{\nu }_n\parallel }^2\\ \le & {\parallel {\nu }_n-p\parallel }^2-\frac{(1-{\alpha }_n-k)}{{\alpha }_n}[{\parallel x_n-x_{n+1}\parallel }^2\\ -2t_n\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel ]+{\alpha }_n(k_n-1)D\\ \le & {\parallel (1-t_n)x_n-p\parallel }^2-\frac{(1-{\alpha }_n-k)}{{\alpha }_n}[{\parallel x_n-x_{n+1}\parallel }^2\\ -2t_n\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel ]+{\alpha }_n(k_n-1)D\\ \le & {\parallel x_n-p\parallel }^2-2t_n〈x_n,x_n-p〉+t_n^2{\parallel x_n\parallel }^2\\ -{\sigma }_1{\parallel x_n-x_{n+1}\parallel }^2+2{\sigma }_2{t}_n\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel \\ +{\alpha }_n(k_n-1)D(\text{where }{\sigma }_1:=\frac{1}{2}(1-k)>0;{\sigma }_2=\frac{1}{ϵ})\\ =& {\parallel x_n-p\parallel }^2-{\sigma }_1{\parallel x_n-x_{n+1}\parallel }^2+t_n[t_n{\parallel x_n\parallel }^2\\ +2{\sigma }_2\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel -2〈x_n,x_n-p〉]\\ +{\alpha }_n(k_n-1)D.\end{array}$$

(3.7)

Since ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded, we have that there exists $M>0$ such that

$$t_n{\parallel x_n\parallel }^2+2{\sigma }_2\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel -2〈x_n,x_n-p〉\le M,\mathrm{\forall }n\ge 1.$$

(3.8)

From (3.7) and (3.8) we obtain

$${\parallel x_{n+1}-p\parallel }^2-{\parallel x_n-p\parallel }^2+{\sigma }_1{\parallel x_n-x_{n+1}\parallel }^2\le M{t}_n+{\alpha }_n(k_n-1)D.$$

(3.9)

To complete the proof, we now consider the following two cases.

Case 1. Suppose that ${\{\parallel x_n-p\parallel \}}_{n=1}^{\mathrm{\infty }}$ is a monotone sequence, then we may assume that $\{\parallel x_n-p\parallel \}$ is monotone decreasing. Then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists and it follows from (3.9), conditions (c1) and ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ that

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

(3.10)

Furthermore,

$$\begin{array}{c}\parallel {\nu }_n-x_n\parallel \le t_n\parallel x_n\parallel \to 0\text{as }n\to \mathrm{\infty },\text{ and}\hfill \\ \parallel {\nu }_n-x_{n+1}\parallel \le \parallel {\nu }_n-x_n\parallel +\parallel x_n-x_{n+1}\parallel \to 0\text{as }n\to \mathrm{\infty }.\hfill \end{array}$$

Hence

$$\parallel {\nu }_n-T^n{\nu }_n\parallel \le \frac{1}{{\alpha }_n}\parallel {\nu }_n-x_{n+1}\parallel \le \frac{1}{ϵ}\parallel {\nu }_n-x_{n+1}\parallel \to 0\text{as }n\to \mathrm{\infty },$$

and

$$\begin{array}{rcl}\parallel x_n-T^n{x}_n\parallel & \le & \parallel x_n-{\nu }_n\parallel +\parallel {\nu }_n-T^n{\nu }_n\parallel +\parallel T^n{\nu }_n-T^n{x}_n\parallel \\ \le & (1+k_n)\parallel x_n-{\nu }_n\parallel +\parallel {\nu }_n-T^n{\nu }_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

Observe also that since T is uniformly L-Lipschitzian, we obtain

$$\begin{array}{rcl}\parallel {\nu }_n-T{\nu }_n\parallel & \le & \parallel {\nu }_n-T^n{\nu }_n\parallel +\parallel T^n{\nu }_n-T{\nu }_n\parallel \\ \le & \parallel {\nu }_n-T^n{\nu }_n\parallel +L\parallel T^{n-1}{\nu }_n-{\nu }_n\parallel \\ \le & \parallel {\nu }_n-T^n{\nu }_n\parallel +L\parallel T^{n-1}{\nu }_n-T^{n-1}{\nu }_{n-1}\parallel \\ +L\parallel T^{n-1}{\nu }_{n-1}-{\nu }_{n-1}\parallel +L\parallel {\nu }_{n-1}-{\nu }_n\parallel \\ \le & \parallel {\nu }_n-T^n{\nu }_n\parallel +L\parallel T^{n-1}{\nu }_{n-1}-{\nu }_{n-1}\parallel \\ +L(1+L)\parallel {\nu }_n-{\nu }_{n-1}\parallel \\ \le & \parallel {\nu }_n-T^n{\nu }_n\parallel +L\parallel T^{n-1}{\nu }_{n-1}-{\nu }_{n-1}\parallel \\ +L(1+L)[\parallel {\nu }_n-x_n\parallel +\parallel x_n-x_{n-1}\parallel \\ +\parallel x_{n-1}-{\nu }_{n-1}\parallel ]\to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.11)

Furthermore,

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

(3.12)

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel {\nu }_n-T{\nu }_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel {\nu }_n-x_n\parallel =0$, then the demiclosedness property of $(I-T)$, (2.4) and the usual standard argument yield that ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\nu }_n\}}_{n=1}^{\mathrm{\infty }}$ converge weakly to some $x^{\ast }\in F(T)$.

Since ${\alpha }_n(1-{\alpha }_n-k)\ge \frac{1}{ϵ}(1-k)>0$, and since ${\parallel {\nu }_n-x^{\ast }\parallel }^2\le D_2$, $\mathrm{\forall }n\ge 1$, and for some $D_2>0$, then using (3.4) we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\parallel {\nu }_n-x^{\ast }\parallel }^2+{\alpha }_n(k_n-1)D_2\\ \le & {\parallel (1-t_n)(x_n-x^{\ast })-t_n{x}^{\ast }\parallel }^2+{\alpha }_n(k_n-1)D_2\\ =& {(1-t_n)}^2{\parallel x_n-x^{\ast }\parallel }^2-2t_n(1-t_n)〈x_n-x^{\ast },x^{\ast }〉\\ +t_n^2{\parallel x^{\ast }\parallel }^2+{\alpha }_n(k_n-1)D_2\\ \le & (1-t_n){\parallel x_n-x^{\ast }\parallel }^2-2t_n(1-t_n)〈x_n-x^{\ast },x^{\ast }〉\\ +t_n^2{\parallel x^{\ast }\parallel }^2+{\alpha }_n(k_n-1)D_2.\end{array}$$

(3.13)

Thus

$${\parallel x_{n+1}-x^{\ast }\parallel }^2\le (1-t_n){\parallel x_n-x^{\ast }\parallel }^2+t_n{\gamma }_n+{\sigma }_n,\mathrm{\forall }n\ge 1,$$

where ${\gamma }_n:=-2(1-t_n)〈x_n-x^{\ast },x^{\ast }〉+t_n{\parallel x^{\ast }\parallel }^2$, and ${\sigma }_n={\alpha }_n(k_n-1)D_2$, with ${\sum }_{n=1}^{\mathrm{\infty }}{\sigma }_n<\mathrm{\infty }$. Since ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges weakly to $x^{\ast }$, then ${lim}_{n\to \mathrm{\infty }}〈x_n-x^{\ast },x^{\ast }〉=0$, and this together with condition (c1) (i.e., ${lim}_{n\to \mathrm{\infty }}{t}_n=0$) implies that ${\gamma }_n:=-2(1-t_n)〈x_n-x^{\ast },x^{\ast }〉+t_n{\parallel x^{\ast }\parallel }^2\to 0$ as $n\to \mathrm{\infty }$. It now follows from Lemma 2.1 that ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to $x^{\ast }$. Consequently, ${\{{\nu }_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to $x^{\ast }$.

Case 2. Suppose that ${\{\parallel x_n-p\parallel \}}_{n=1}^{\mathrm{\infty }}$ is not a monotone decreasing sequence, then set ${\mathrm{\Gamma }}_n:={\parallel x_n-p\parallel }^2$, and let $\tau :\mathbb{N}\to \mathbb{N}$ be a mapping defined for all $n\ge N_0$ for some sufficiently large $N_0$ by

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

Then τ is a non-decreasing sequence such that $\tau (n)\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$ and ${\mathrm{\Gamma }}_{\tau (n)}\le {\mathrm{\Gamma }}_{\tau (n)+1}$ for $n\ge N_0$. Using (c1) and (c2) in (3.9), we obtain

$${\parallel x_{\tau (n)+1}-x_{\tau (n)}\parallel }^2\le \frac{1}{{\sigma }_1}[M{t}_{\tau (n)}+{\alpha }_{\tau (n)}(k_{\tau (n)}-1)D]\to 0\text{as }n\to \mathrm{\infty }.$$

(3.14)

Following the same argument as in Case 1, we obtain

$$\parallel {\nu }_{\tau (n)}-T{\nu }_{\tau (n)}\parallel \to 0\text{as }n\to \mathrm{\infty }\text{and}\parallel x_{\tau (n)}-T{x}_{\tau (n)}\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

As in Case 1, we also obtain that $\{x_{\tau (n)}\}$ and $\{{\nu }_{\tau (n)}\}$ converge weakly to some $x^{\ast }$ in $F(T)$. Furthermore, for all $n\ge N_0$, we obtain from (3.13) that

$$\begin{array}{rcl}0& \le & {\parallel x_{\tau (n)+1}-x^{\ast }\parallel }^2-{\parallel x_{\tau (n)}-x^{\ast }\parallel }^2\\ \le & t_{\tau (n)}[-2(1-t_{\tau (n)})〈x_{\tau (n)}-x^{\ast },x^{\ast }〉+t_{\tau (n)}{\parallel x^{\ast }\parallel }^2\\ +D_2{\alpha }_{\tau (n)}\frac{(k_{\tau (n)}-1)}{t_{\tau (n)}}-{\parallel x_{\tau (n)}-x^{\ast }\parallel }^2].\end{array}$$

(3.15)

It follows from (3.15) that

$$\begin{array}{rcl}{\parallel x_{\tau (n)}-x^{\ast }\parallel }^2& \le & 2(1-t_{\tau (n)})〈x^{\ast }-x_{\tau (n)},x^{\ast }〉+t_{\tau (n)}{\parallel x^{\ast }\parallel }^2\\ +D_2{\alpha }_{\tau (n)}\frac{(k_{\tau (n)}-1)}{t_{\tau (n)}}\to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

Thus

$$\underset{n\to \mathrm{\infty }}{lim}{\mathrm{\Gamma }}_{\tau (n)}=\underset{n\to \mathrm{\infty }}{lim}{\mathrm{\Gamma }}_{\tau (n)+1}.$$

Furthermore, for $n\ge N_0$, we have ${\mathrm{\Gamma }}_n\le {\mathrm{\Gamma }}_{\tau (n)+1}$ if $n\ne \tau (n)$ (i.e., $\tau (n)<n$), because ${\mathrm{\Gamma }}_j>{\mathrm{\Gamma }}_{j+1}$ for $\tau (n)+1\le j\le n$. It then follows that for all $n\ge N_0$ we have

$$0\le {\mathrm{\Gamma }}_n\le max\{{\mathrm{\Gamma }}_{\tau (n)},{\mathrm{\Gamma }}_{\tau (n)+1}\}={\mathrm{\Gamma }}_{\tau (n)+1}.$$

This implies ${lim}_{n\to \mathrm{\infty }}{\mathrm{\Gamma }}_n=0$, and hence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to $x^{\ast }\in F(T)$. □

Corollary 3.1 Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H with $0\in C$. Let $T:C\to C$ be an asymptotically k-strictly pseudocontractive mapping with a sequence ${\{k_n\}}_{n=1}^{\mathrm{\infty }}\subseteq [1,\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$. Let $F(T)\ne \mathrm{\varnothing }$, and let ${\{t_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ satisfying the conditions:

• (c1) ${lim}_{n\to \mathrm{\infty }}{t}_n=0$;

• (c2) ${\sum }_{n=1}^{\mathrm{\infty }}{t}_n=\mathrm{\infty }$;

• (c3) ${lim}_{n\to \mathrm{\infty }}\frac{1}{t_n}(k_n-1)=0$;

• (c4) $0<ϵ\le {\alpha }_n<\frac{1}{2}(1-t_n)(1-k)$, $\mathrm{\forall }n\ge 1$ and for some ϵ.

Then the modified averaging iteration sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$, generated from $x_1\in C$ by

$$\{\begin{array}{l}{\nu }_n:=(1-t_n)x_n,\\ x_{n+1}:=(1-{\alpha }_n){\nu }_n+{\alpha }_n{T}^n{\nu }_n,\end{array}$$

converges strongly to a fixed point of T.

Remark 3.1 Prototypes for our real sequences ${\{t_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ are:

$$t_n:=\sqrt{k_n-1+\frac{1}{n+1}},n\ge 1;{\alpha }_n:=\frac{n}{2(n+1)}(1-k)(1-t_n),n\ge 1.$$

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space, and let $T:C\to C$ be an asymptotically nonexpansive mapping with a sequence ${\{k_n\}}_{n=1}^{\mathrm{\infty }}\subseteq [1,\mathrm{\infty })$. Let $F(T)\ne \mathrm{\varnothing }$, and let ${\{t_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ satisfying the conditions:

• (c1) ${lim}_{n\to \mathrm{\infty }}{t}_n=0$;

• (c2) ${\sum }_{n=1}^{\mathrm{\infty }}{t}_n=\mathrm{\infty }$;

• (c3) $0<ϵ\le {\alpha }_n<\frac{1}{2}(1-t_n)$, $\mathrm{\forall }n\ge 1$ and for some ϵ.

Then the modified averaging iteration sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$, generated from $x_1\in C$ by (3.1), converges strongly to a fixed point of T.

3.2 Demiclosedness principle and strong convergence results for uniformly Lipschitzian asymptotically pseudocontractive maps

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. For uniformly L-Lipschitzian asymptotically pseudocontractive maps $T:C\to C$, we first prove that $(I-T)$ is demiclosed at 0 and then introduce a modified averaging Ishikawa iteration algorithm and prove that it converges strongly to a fixed point of $T:C\to C$ without any compactness assumption on T or C and without further requirement on $F(T)$. Our demiclosedness principle does not require the boundedness of C imposed in the result of [18].

Theorem 3.2 Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let $T:C\to C$ be a uniformly L-Lipschitzian asymptotically pseudocontractive mapping. Then $(I-T)$ is demiclosed at 0.

Proof Let ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence in C which converges weakly to p and ${\{x_n-T{x}_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to 0. We prove that $p\in F(T)$. Since ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges weakly, it is bounded. For each $x\in H$, define $f:H\to [0,\mathrm{\infty })$ by

$$f(x):=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-x\parallel }^2.$$

(3.16)

Observe that for arbitrary but fixed integer $m\ge 1$, we have

$$\begin{array}{rcl}\parallel x_n-T^m{x}_n\parallel & \le & \parallel x_n-T{x}_n\parallel +\parallel T{x}_n-T^2{x}_n\parallel +\cdots +\parallel T^{m-1}{x}_n-T^m{x}_n\parallel \\ \le & mL\parallel x_n-T{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

Set

$$G_{m}x:=T^m((1-\beta )x+\beta T^{m}x),$$

where $\beta \in (0,\frac{2}{(1+\lambda )+\sqrt{{(1+\lambda )}^2+4L^2}})$, and $\lambda :={sup}_{n\ge 1}{k}_n$. Then

$$\parallel (1-\beta )x_n+\beta T^m{x}_n-T^m{x}_n\parallel =(1-\beta )\parallel x_n-T^m{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty },$$

and

$$\begin{array}{rcl}\parallel T^m{x}_n-G_m{x}_n\parallel & =& \parallel T^m{x}_n-T^m((1-\beta )x_n+\beta T^m{x}_n)\parallel \\ \le & L\beta \parallel x_n-T^m{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

Hence

$$\begin{array}{rcl}\parallel (1-\beta )x_n+\beta T^m{x}_n-G_m{x}_n\parallel & \le & \parallel (1-\beta )x_n+\beta T^m{x}_n-T^m{x}_n\parallel \\ +\parallel T^m{x}_n-G_m{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

Also

$$\parallel x_n-G_m{x}_n\parallel \le \parallel x_n-T^m{x}_n\parallel +\parallel T^m{x}_n-G_m{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

From (2.4) we obtain

$$f(x)=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-p\parallel }^2+{\parallel p-x\parallel }^2,\mathrm{\forall }x\in H.$$

Thus

$$f(x)=f(p)+{\parallel p-x\parallel }^2,\mathrm{\forall }x\in H,$$

and hence

$$f(G_{m}p)=f(p)+{\parallel p-G_{m}p\parallel }^2.$$

(3.17)

Observe that

$$\begin{array}{rcl}f(G_{m}p)& =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-G_{m}p\parallel }^2\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-G_m{x}_n+G_m{x}_n-G_{m}p\parallel }^2\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel G_m{x}_n-G_{m}p\parallel }^2\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel T^m((1-\beta )x_n+\beta T^m{x}_n)-T^m((1-\beta )p+\beta T^{m}p)\parallel }^2\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}[k_m{\parallel (1-\beta )x_n+\beta T^m{x}_n-((1-\beta )p+\beta T^{m}p)\parallel }^2\\ +{\parallel (1-\beta )x_n+\beta T^m{x}_n-G_m{x}_n-((1-\beta )p+\beta T^{m}p-G_{m}p)\parallel }^2]\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}[k_m{\parallel (1-\beta )(x_n-p)+\beta (T^m{x}_n-T^{m}p)\parallel }^2\\ +{\parallel (1-\beta )(p-G_{m}p)+\beta (T^{m}p-G_{m}p)\parallel }^2]\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\left[k_m\right((1-\beta ){\parallel x_n-p\parallel }^2+\beta {\parallel T^m{x}_n-T^{m}p\parallel }^2\\ -\beta (1-\beta ){\parallel x_n-T^m{x}_n-(p-T^{m}p)\parallel }^2)+(1-\beta ){\parallel p-G_{m}p\parallel }^2\\ +\beta {\parallel T^{m}p-G_{m}p\parallel }^2-\beta (1-\beta ){\parallel p-T^{m}p\parallel }^2]\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}[k_m(1-\beta +k_m\beta ){\parallel x_n-p\parallel }^2+k_m\beta {\parallel x_n-T^m{x}_n-(p-T^{m}p)\parallel }^2\\ -k_m\beta (1-\beta ){\parallel x_n-T^m{x}_n-(p-T^{m}p)\parallel }^2+(1-\beta ){\parallel p-G_{m}p\parallel }^2\\ +{\beta }^3{L}^2{\parallel p-T^{m}p\parallel }^2-\beta (1-\beta ){\parallel p-T^{m}p\parallel }^2]\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}[k_m(1+\beta (k_m-1)){\parallel x_n-p\parallel }^2+(1-\beta ){\parallel p-G_{m}p\parallel }^2\\ -\beta [1-\beta (1+k_m)-{\beta }^2{L}^2]{\parallel p-T^{m}p\parallel }^2]\\ \le & k_m(1+\beta (k_m-1))f(p)+(1-\beta ){\parallel p-G_{m}p\parallel }^2.\end{array}$$

(3.18)

Equations (3.17) and (3.18) imply that

$$f(p)+{\parallel p-G_{m}p\parallel }^2\le k_m(1+\beta (k_m-1))f(p)+(1-\beta ){\parallel p-G_{m}p\parallel }^2,$$

from which it follows that

$$\beta {\parallel p-G_{m}p\parallel }^2\le [k_m(1+\beta (k_m-1))-1]f(p)\to 0\text{as }m\to \mathrm{\infty }.$$

Thus

$$\begin{array}{c}\parallel p-G_{m}p\parallel \to 0\text{as }m\to \mathrm{\infty },\hfill \\ \begin{array}{rl}\parallel p-T^{m}p\parallel & \le \parallel p-G_{m}p\parallel +\parallel G_{m}p-T^{m}p\parallel \\ \le \parallel p-G_{m}p\parallel +L\parallel (1-\beta )p+\beta T^{m}p-p\parallel \\ =\parallel p-G_{m}p\parallel +L\beta \parallel p-T^{m}p\parallel .\end{array}\hfill \end{array}$$

Hence

$$(1-L\beta )\parallel p-T^{m}p\parallel \le \parallel p-G_{m}p\parallel \to 0\text{as }m\to \mathrm{\infty }.$$

It now follows that $T^{m}p\to p$ as $m\to \mathrm{\infty }$. Since T is continuous, we have that $T^{m+1}p\to Tp$ as $m\to \mathrm{\infty }$, and hence $Tp=p$. □

We now introduce the following iterative algorithm for uniformly L-Lipschitzian asymptotically pseudocontractive maps.

Modified averaging Ishikawa algorithm For arbitrary $x_1\in C$, the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is given by

$$\{\begin{array}{l}{\nu }_n=P_C((1-t_n)x_n),n\ge 1,\\ y_n=(1-{\beta }_n){\nu }_n+{\beta }_n{T}^n{\nu }_n,n\ge 1,\\ x_{n+1}=(1-{\alpha }_n){\nu }_n+{\alpha }_n{T}^n{y}_n,n\ge 1.\end{array}$$

(3.19)

Theorem 3.3 Let C be a nonempty closed convex subset of a real Hilbert space H, and let $T:C\to C$ be a uniformly L-Lipschitzian asymptotically pseudocontractive mapping with a sequence ${\{k_n\}}_{n=1}^{\mathrm{\infty }}\subseteq [1,\mathrm{\infty })$, ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$ and $F(T)\ne \mathrm{\varnothing }$. Let ${\{t_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ be real sequences in $(0,1)$ satisfying the conditions:

• (c1) ${lim}_{n\to \mathrm{\infty }}{t}_n=0$;

• (c2) ${\sum }_{n=1}^{\mathrm{\infty }}{t}_n=\mathrm{\infty }$;

• (c3) $0<ϵ\le {\alpha }_n\le (1-t_n){\beta }_n\le {\beta }_n\le b<\frac{2}{(1+\lambda )+\sqrt{{(1+\lambda )}^2+4L^2}}$, where $\lambda ={sup}_n{k}_n$;

• (c4) ${lim}_{n\to \mathrm{\infty }}\frac{(k_n-1)}{t_n}=0$.

Then the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ generated from an arbitrary $x_1\in C$ by (3.19) converges strongly to a fixed point of T.

Proof Since T is asymptotically pseudocontractive, it follows that

$$2〈(I-T^n)x-(I-T^n)y,x-y〉\ge -(k_n-1){\parallel x-y\parallel }^2$$

(3.20)

and

$$2〈T^{n}x-T^{n}y,x-y〉\le (k_n+1){\parallel x-y\parallel }^2.$$

(3.21)

Set

$$G_n{\nu }_n=T^n((1-{\beta }_n){\nu }_n+{\beta }_n{T}^n{\nu }_n),n\ge 1.$$

Then, for arbitrary $p\in F(T)$, we obtain

$$\begin{array}{rcl}{\parallel G_n{\nu }_n-p\parallel }^2& =& {\parallel T^n((1-{\beta }_n){\nu }_n+{\beta }_n{T}^n{\nu }_n)-T^{n}p\parallel }^2\\ \le & k_n{\parallel (1-{\beta }_n)({\nu }_n-p)+{\beta }_n(T^n{\nu }_n-p)\parallel }^2\\ +{\parallel (1-{\beta }_n){\nu }_n+{\beta }_n{T}^n{\nu }_n-G_n{\nu }_n\parallel }^2\\ =& k_n(1-{\beta }_n){\parallel {\nu }_n-p\parallel }^2+k_n{\beta }_n{\parallel T^n{\nu }_n-p\parallel }^2\\ -k_n{\beta }_n(1-{\beta }_n){\parallel {\nu }_n-T^n{\nu }_n\parallel }^2\\ +{\parallel (1-{\beta }_n)({\nu }_n-G_n{\nu }_n)+{\beta }_n(T^n{\nu }_n-G_n{\nu }_n)\parallel }^2\\ =& [k_n(1-{\beta }_n)+k_n^2{\beta }_n]{\parallel {\nu }_n-p\parallel }^2+k_n{\beta }_n{\parallel {\nu }_n-T^n{\nu }_n\parallel }^2\\ -k_n(1-{\beta }_n){\beta }_n{\parallel {\nu }_n-T^n{\nu }_n\parallel }^2+(1-{\beta }_n){\parallel {\nu }_n-G_n{\nu }_n\parallel }^2\\ +{\beta }_n{\parallel T^n{\nu }_n-G_n{\nu }_n\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel {\nu }_n-T^n{\nu }_n\parallel }^2\\ \le & [1+(k_n^2-1)]{\parallel {\nu }_n-p\parallel }^2+k_n{\beta }_n{\parallel {\nu }_n-T^n{\nu }_n\parallel }^2\\ -k_n(1-{\beta }_n){\beta }_n{\parallel {\nu }_n-T^n{\nu }_n\parallel }^2+(1-{\beta }_n){\parallel {\nu }_n-G_n{\nu }_n\parallel }^2\\ +{\beta }_n^3{L}^2{\parallel {\nu }_n-T^n{\nu }_n\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel {\nu }_n-T^n{\nu }_n\parallel }^2\\ =& [1+(k_n^2-1)]{\parallel {\nu }_n-p\parallel }^2+(1-{\beta }_n){\parallel {\nu }_n-G_n{\nu }_n\parallel }^2\\ -{\beta }_n[1-(1+k_n){\beta }_n-{\beta }_n^2{L}^2]{\parallel {\nu }_n-T^n{\nu }_n\parallel }^2.\end{array}$$

Thus

$$\begin{array}{rcl}{\parallel G_n{\nu }_n-p\parallel }^2& \le & [1+(k_n^2-1)]{\parallel {\nu }_n-p\parallel }^2+(1-{\beta }_n){\parallel {\nu }_n-G_n{\nu }_n\parallel }^2\\ -{\beta }_n[1-(1+k_n){\beta }_n-{\beta }_n^2{L}^2]{\parallel {\nu }_n-T^n{\nu }_n\parallel }^2.\end{array}$$

(3.22)

From (3.22) we obtain

$$\begin{array}{rcl}2〈{\nu }_n-G_n{\nu }_n,{\nu }_n-p〉& \ge & {\beta }_n{\parallel {\nu }_n-G_n{\nu }_n\parallel }^2+{\beta }_n[1-(1+k_n){\beta }_n-{\beta }_n^2{L}^2]{\parallel {\nu }_n-T{\nu }_n\parallel }^2\\ -(k_n^2-1){\parallel {\nu }_n-p\parallel }^2\end{array}$$

(3.23)

and

$$\begin{array}{rcl}2〈G_n{\nu }_n-p,{\nu }_n-p〉& \le & (1+k_n^2){\parallel {\nu }_n-p\parallel }^2-{\beta }_n{\parallel {\nu }_n-G_n{\nu }_n\parallel }^2\\ -{\beta }_n[1-(1+k_n){\beta }_n-{\beta }_n^2{L}^2]{\parallel {\nu }_n-T^n{\nu }_n\parallel }^2.\end{array}$$

(3.24)

Observe that

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel (1-{\alpha }_n){\nu }_n+{\alpha }_n{G}_n{\nu }_n-p\parallel }^2\\ =& (1-{\alpha }_n){\parallel {\nu }_n-p\parallel }^2+{\alpha }_n{\parallel G_n{\nu }_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel {\nu }_n-G_n{\nu }_n\parallel }^2\\ \le & (1-{\alpha }_n){\parallel {\nu }_n-p\parallel }^2+{\alpha }_n[[1+(k_n^2-1)]{\parallel {\nu }_n-p\parallel }^2+(1-{\beta }_n){\parallel {\nu }_n-G_n{\nu }_n\parallel }^2\\ -{\beta }_n[1-(1+k_n){\beta }_n-{\beta }_n^2{L}^2]{\parallel {\nu }_n-T^n{\nu }_n\parallel }^2]\\ -{\alpha }_n(1-{\alpha }_n){\parallel {\nu }_n-G_n{\nu }_n\parallel }^2\\ \le & [1+(k_n^2-1)]{\parallel {\nu }_n-p\parallel }^2-{\alpha }_n({\beta }_n-{\alpha }_n){\parallel {\nu }_n-G_n{\nu }_n\parallel }^2\\ -{\alpha }_n{\beta }_n[1-(1+k_n){\beta }_n-{\beta }_n^2{L}^2]{\parallel {\nu }_n-T^n{\nu }_n\parallel }^2.\end{array}$$

(3.25)

Hence

$${\parallel x_{n+1}-p\parallel }^2\le [1+(k_n^2-1)]{\parallel {\nu }_n-p\parallel }^2$$

and it follows, as in the proof of Theorem 3.1, that ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded. Observe that

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

Hence

$${\parallel {\nu }_n-x_{n+1}\parallel }^2\ge {\parallel x_n-x_{n+1}\parallel }^2-2t_n\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel .$$

(3.26)

Furthermore,

$$\begin{array}{rcl}\parallel {\nu }_n-G_n{\nu }_n\parallel & \le & \parallel {\nu }_n-T^n{\nu }_n\parallel +\parallel T^n{\nu }_n-G_n{\nu }_n\parallel \\ \le & \parallel {\nu }_n-T^n{\nu }_n\parallel +L{\beta }_n\parallel {\nu }_n-T^n{\nu }_n\parallel \\ =& (1+L{\beta }_n)\parallel {\nu }_n-T^n{\nu }_n\parallel .\end{array}$$

Thus

$${\parallel {\nu }_n-T^n{\nu }_n\parallel }^2\ge \frac{1}{{(1+L{\beta }_n)}^2}{\parallel {\nu }_n-G_n{\nu }_n\parallel }^2.$$

(3.27)

Observe also that

$$\parallel x_{n+1}-{\nu }_n\parallel =\parallel (1-{\alpha }_n){\nu }_n+{\alpha }_n{G}_n{\nu }_n-{\nu }_n\parallel ={\alpha }_n\parallel {\nu }_n-G_n{\nu }_n\parallel .$$

(3.28)

Using (3.26) and (3.28), we obtain

$$\begin{array}{rcl}{\parallel {\nu }_n-G_n{\nu }_n\parallel }^2& =& \frac{1}{{\alpha }_n^2}{\parallel x_{n+1}-{\nu }_n\parallel }^2\\ \ge & \frac{1}{{\alpha }_n^2}[{\parallel x_n-x_{n+1}\parallel }^2-2t_n\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel ].\end{array}$$

It now follows from (3.27) that

$$\begin{array}{rcl}{\parallel {\nu }_n-T^n{\nu }_n\parallel }^2& \ge & \frac{1}{{(1+L{\beta }_n)}^2}{\parallel {\nu }_n-G_n{\nu }_n\parallel }^2\\ \ge & \frac{1}{{\alpha }_n^2{(1+L{\beta }_n)}^2}[{\parallel x_n-x_{n+1}\parallel }^2-2t_n\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel ].\end{array}$$

(3.29)

Using (3.29) in (3.25), we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & [1+(k_n^2-1)]{\parallel {\nu }_n-p\parallel }^2\\ -{\alpha }_n{\beta }_n[1-(1+k_n){\beta }_n-{\beta }_n^2{L}^2]\left\{\frac{1}{{\alpha }_n^2{(1+L{\beta }_n)}^2}\right[{\parallel x_n-x_{n+1}\parallel }^2\\ -2t_n\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel \left]\right\}\\ \le & {\parallel {\nu }_n-p\parallel }^2+(k_n^2-1)D\\ -\frac{{\beta }_n[1-(1+k_n){\beta }_n-{\beta }_n^2{L}^2]}{{\alpha }_n{(1+L{\beta }_n)}^2}[{\parallel x_n-x_{n+1}\parallel }^2\\ -2t_n\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel ]\\ \le & {\parallel x_n-p\parallel }^2-2t_n〈x_n,x_n-p〉+t_n^2{\parallel x_n\parallel }^2-{\sigma }_3{\parallel x_n-x_{n+1}\parallel }^2\\ +2{\sigma }_4{t}_n\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel \\ (\text{where }{\sigma }_3=\frac{ϵ[1-(1+\lambda )b-b^2{L}^2]}{b{(1+Lϵ)}^2},{\sigma }_4=\frac{1}{ϵ{(1+Lϵ)}^2})\\ =& {\parallel x_n-p\parallel }^2-{\sigma }_3\parallel x_n-x_{n+1}\parallel +t_n[-〈x_n,x_n-p〉\\ +t_n{\parallel x_n\parallel }^2+2{\sigma }_4\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel ]+(k_n^2-1)D.\end{array}$$

(3.30)

Since $\{x_n\}$ is bounded, we have that there exists $M>0$ such that

$$-〈x_n,x_n-p〉+t_n{\parallel x_n\parallel }^2+2{\sigma }_4\parallel x_n\parallel \parallel x_n-x_{n+1}\parallel \le M,\mathrm{\forall }n\ge 1.$$

(3.31)

From (3.30) and (3.31) we obtain

$${\parallel x_{n+1}-p\parallel }^2-{\parallel x_n-p\parallel }^2+{\sigma }_3\parallel x_n-x_{n+1}\parallel \le M{t}_n+(k_n^2-1)D.$$

(3.32)

To complete the proof, we now consider the following two cases.

Case 1. Suppose that ${\{\parallel x_n-p\parallel \}}_{n=1}^{\mathrm{\infty }}$ is a monotone sequence, then we may assume that $\{\parallel x_n-p\parallel \}$ is monotone decreasing. Then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists and it follows from (3.32), conditions (c1) and ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ that

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

(3.33)

Furthermore,

$$\begin{array}{c}\parallel {\nu }_n-x_n\parallel \le t_n\parallel x_n\parallel \to 0\text{as }n\to \mathrm{\infty },\text{ and }\hfill \\ \parallel {\nu }_n-x_{n+1}\parallel \le \parallel {\nu }_n-x_n\parallel +\parallel x_n-x_{n+1}\parallel \to 0\text{as }n\to \mathrm{\infty }.\hfill \end{array}$$

Hence

$$\parallel {\nu }_n-G_n{\nu }_n\parallel =\frac{1}{{\alpha }_n}\parallel {\nu }_n-G_n{\nu }_n\parallel \le \frac{1}{ϵ}\parallel {\nu }_n-x_{n+1}\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

Furthermore,

$$\begin{array}{rcl}\parallel {\nu }_n-T^n{\nu }_n\parallel & \le & \parallel {\nu }_n-G_n{\nu }_n\parallel +\parallel G_n{\nu }_n-T^n{\nu }_n\parallel \\ \le & \parallel {\nu }_n-G_n{\nu }_n\parallel +L{\beta }_n\parallel {\nu }_n-T^n{\nu }_n\parallel .\end{array}$$

Thus

$${\parallel {\nu }_n-T^n{\nu }_n\parallel }^2\le \frac{1}{1-L{\beta }_n}\parallel {\nu }_n-G_n{\nu }_n\parallel \le \frac{1}{1-Lb}\parallel {\nu }_n-G_n{\nu }_n\parallel \to 0\text{as }n\to \mathrm{\infty },$$

and

$$\begin{array}{rcl}\parallel x_n-T^n{x}_n\parallel & \le & \parallel x_n-{\nu }_n\parallel +\parallel {\nu }_n-T^n{\nu }_n\parallel \\ +\parallel T^n{\nu }_n-T^n{x}_n\parallel \\ \le & (1+k_n)\parallel x_n-{\nu }_n\parallel \\ +\parallel {\nu }_n-T^n{\nu }_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

Observe also that since T is uniformly L-Lipschitzian, we obtain

$$\begin{array}{rcl}\parallel {\nu }_n-T{\nu }_n\parallel & \le & \parallel {\nu }_n-T^n{\nu }_n\parallel +\parallel T^n{\nu }_n-T{\nu }_n\parallel \\ \le & \parallel {\nu }_n-T^n{\nu }_n\parallel +L\parallel T^{n-1}{\nu }_n-{\nu }_n\parallel \\ \le & \parallel {\nu }_n-T^n{\nu }_n\parallel +L\parallel T^{n-1}{\nu }_n-T^{n-1}{\nu }_{n-1}\parallel \\ +L\parallel T^{n-1}{\nu }_{n-1}-{\nu }_{n-1}\parallel +L\parallel {\nu }_{n-1}-{\nu }_n\parallel \\ \le & \parallel {\nu }_n-T^n{\nu }_n\parallel +L\parallel T^{n-1}{\nu }_{n-1}-{\nu }_{n-1}\parallel +L(1+L)\parallel {\nu }_n-{\nu }_{n-1}\parallel \\ \le & \parallel {\nu }_n-T^n{\nu }_n\parallel +L\parallel T^{n-1}{\nu }_{n-1}-{\nu }_{n-1}\parallel \\ +L(1+L)[\parallel {\nu }_n-x_n\parallel +\parallel x_n-x_{n-1}\parallel \\ +\parallel x_{n-1}-{\nu }_{n-1}\parallel ]\to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.34)

Furthermore,

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

(3.35)

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel {\nu }_n-T{\nu }_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel {\nu }_n-x_n\parallel =0$, then the demiclosedness property of $(I-T)$, (2.4) and the usual standard argument yield that ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\nu }_n\}}_{n=1}^{\mathrm{\infty }}$ converge weakly to some $x^{\ast }\in F(T)$. Since ${\parallel {\nu }_n-x^{\ast }\parallel }^2\le D_2$, $\mathrm{\forall }n\ge 1$, and for some $D_2>0$, then using (3.25) we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\parallel {\nu }_n-x^{\ast }\parallel }^2+{\alpha }_n(k_n^2-1)D_2\\ \le & {\parallel (1-t_n)(x_n-x^{\ast })-t_n{x}^{\ast }\parallel }^2+{\alpha }_n(k_n^2-1)D_2\\ =& {(1-t_n)}^2{\parallel x_n-x^{\ast }\parallel }^2-2t_n(1-t_n)〈x_n-x^{\ast },x^{\ast }〉\\ +t_n^2{\parallel x^{\ast }\parallel }^2+{\alpha }_n(k_n^2-1)D_2\\ \le & (1-t_n){\parallel x_n-x^{\ast }\parallel }^2-2t_n(1-t_n)〈x_n-x^{\ast },x^{\ast }〉\\ +t_n^2{\parallel x^{\ast }\parallel }^2+{\alpha }_n(k_n^2-1)D_2.\end{array}$$

(3.36)

Thus

$${\parallel x_{n+1}-x^{\ast }\parallel }^2\le (1-t_n){\parallel x_n-x^{\ast }\parallel }^2+t_n{\gamma }_n+{\sigma }_n,\mathrm{\forall }n\ge 1,$$

where ${\gamma }_n:=-2(1-t_n)〈x_n-x^{\ast },x^{\ast }〉+t_n{\parallel x^{\ast }\parallel }^2\to 0$ as $n\to \mathrm{\infty }$, and ${\sigma }_n={\alpha }_n(k_n^2-1)D_2$ with ${\sum }_{n=1}^{\mathrm{\infty }}{\sigma }_n<\mathrm{\infty }$. It now follows from Lemma 2.1 that ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to $x^{\ast }$. Consequently, ${\{{\nu }_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to $x^{\ast }$.

Case 2. Suppose that ${\{\parallel x_n-p\parallel \}}_{n=1}^{\mathrm{\infty }}$ is not a monotone decreasing sequence, then set ${\mathrm{\Gamma }}_n:={\parallel x_n-p\parallel }^2$, and let $\tau :\mathbb{N}\to \mathbb{N}$ be a mapping defined for all $n\ge N_0$ for some sufficiently large $N_0$ by

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

Then τ is a non-decreasing sequence such that $\tau (n)\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$ and ${\mathrm{\Gamma }}_{\tau (n)}\le {\mathrm{\Gamma }}_{\tau (n)+1}$ for $n\ge N_0$. Using (c1) and (c2) in (3.32), we obtain

$${\parallel x_{\tau (n)+1}-x_{\tau (n)}\parallel }^2\le \frac{1}{{\sigma }_4}[M{t}_{\tau (n)}+(k_{\tau (n)}^2-1)D]\to 0\text{as }n\to \mathrm{\infty }.$$

(3.37)

Following the same argument as in Case 1, we obtain

$$\parallel {\nu }_{\tau (n)}-T{\nu }_{\tau (n)}\parallel \to 0\text{as }n\to \mathrm{\infty }\text{and}\parallel x_{\tau (n)}-T{x}_{\tau (n)}\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

As in Case 1 we also obtain that $\{x_{\tau (n)}\}$ and $\{{\nu }_{\tau (n)}\}$ converge weakly to some $x^{\ast }$ in $F(T)$.

Furthermore, for all $n\ge N_0$, we obtain from (3.36) that

$$\begin{array}{rcl}0& \le & {\parallel x_{\tau (n)+1}-x^{\ast }\parallel }^2-{\parallel x_{\tau (n)}-x^{\ast }\parallel }^2\\ \le & t_{\tau (n)}[-2(1-t_{\tau (n)})〈x_{\tau (n)}-x^{\ast },x^{\ast }〉+t_{\tau (n)}{\parallel x^{\ast }\parallel }^2\\ +D_2{\alpha }_{\tau (n)}\frac{(k_{\tau (n)}^2-1)}{t_{\tau (n)}}-{\parallel x_{\tau (n)}-x^{\ast }\parallel }^2].\end{array}$$

(3.38)

It follows from (3.38) that

$$\begin{array}{rcl}{\parallel x_{\tau (n)}-x^{\ast }\parallel }^2& \le & 2(1-t_{\tau (n)})〈x^{\ast }-x_{\tau (n)},x^{\ast }〉+t_{\tau (n)}{\parallel x^{\ast }\parallel }^2\\ +D_2{\alpha }_{\tau (n)}\frac{(k_{\tau (n)}^2-1)}{t_{\tau (n)}}\to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

Thus

$$\underset{n\to \mathrm{\infty }}{lim}{\mathrm{\Gamma }}_{\tau (n)}=\underset{n\to \mathrm{\infty }}{lim}{\mathrm{\Gamma }}_{\tau (n)+1}.$$

Furthermore, for $n\ge N_0$, we have ${\mathrm{\Gamma }}_n\le {\mathrm{\Gamma }}_{\tau (n)+1}$ if $n\ne \tau (n)$ (i.e., $\tau (n)<n$), because ${\mathrm{\Gamma }}_j>{\mathrm{\Gamma }}_{j+1}$ for $\tau (n)+1\le j\le n$. It then follows that for all $n\ge N_0$ we have

$$0\le {\mathrm{\Gamma }}_n\le max\{{\mathrm{\Gamma }}_{\tau (n)},{\mathrm{\Gamma }}_{\tau (n)+1}\}={\mathrm{\Gamma }}_{\tau (n)+1}.$$

This implies ${lim}_{n\to \mathrm{\infty }}{\mathrm{\Gamma }}_n=0$, and hence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to $x^{\ast }\in F(T)$. □

Corollary 3.3 Let C be a nonempty closed convex subset of a real Hilbert space H with $0\in C$, and let $T:C\to C$ be a uniformly L-Lipschitzian asymptotically pseudocontractive mapping with a sequence ${\{k_n\}}_{n=1}^{\mathrm{\infty }}\subseteq [1,\mathrm{\infty })$, ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$ and $F(T)\ne \mathrm{\varnothing }$. Let ${\{t_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ be real sequences in $(0,1)$ satisfying the conditions:

• (c1) ${lim}_{n\to \mathrm{\infty }}{t}_n=0$;

• (c2) ${\sum }_{n=1}^{\mathrm{\infty }}{t}_n=\mathrm{\infty }$;

• (c3) $0<ϵ\le {\alpha }_n\le (1-t_n){\beta }_n\le {\beta }_n\le b<\frac{2}{(1+\lambda )+\sqrt{{(1+\lambda )}^2+4L^2}}$, where $\lambda ={sup}_n{k}_n$;

• (c4) ${lim}_{n\to \mathrm{\infty }}\frac{(k_n-1)}{t_n}=0$.

Then the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ generated from an arbitrary $x_1\in C$ by

$$\{\begin{array}{l}{\nu }_n=(1-t_n)x_n,n\ge 1,\\ y_n=(1-{\beta }_n){\nu }_n+{\beta }_n{T}^n{\nu }_n,n\ge 1,\\ x_{n+1}=(1-{\alpha }_n){\nu }_n+{\alpha }_n{T}^n{y}_n,n\ge 1,\end{array}$$

converges strongly to a fixed point of T.