Convergence theorems on asymptotically demicontractive and hemicontractivemappings in the intermediate sense

Abstract

In this study, we introduce two classes of nonlinear mappings, the class ofasymptotically demicontractive mappings in the intermediate sense andasymptotically hemicontractive mappings in the intermediate sense and prove theconvergence of Mann-type and Ishikawa-type iterative schemes to their respectivefixed points. Our results are improvements and generalizations of the results ofseveral authors in the literature.

MSC: 47H10, 47H09.

1 Introduction and preliminaries

In the sequel, we give the following definitions of some of the concepts that willfeature prominently in this study.

We define C as a convex subset of a normed space E.

Definition 1.1 Let $T:C\to C$ be a mapping.T is said to be

(1) asymptotically nonexpansive [1] if there exists a sequence $\{k_n\}$ with $k_n\ge 1$ and $lim{k}_n=1$ such that

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

(1.1)

for all integers $n\ge 0$and all $x,y\in C$;

(2) asymptotically strict pseudocontractive [2] if there exist a constant $k\in [0,1)$ and a sequence $\{k_n\}\subset [1,\mathrm{\infty })$ with $k_n\to 1$ as $n\to \mathrm{\infty }$ such that

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

(1.2)

If $k_n=1$and $T^n=T$for all $n\in \mathbb{N}$ in (1.2),then we obtain the class of strict pseudocontractive mappings. The class ofasymptotically strict pseudocontractive mappings was introduced by Qihou in 1987. Weremark that the class of asymptotically strict pseudocontractive mappings is ageneralization of the class of strict pseudocontractive mappings. Observe that if$k=0$ in (1.2),then we obtain (1.1);

(3) asymptotically strict pseudocontractive in the intermediate sense [3] if there exist a constant $k\in [0,1)$ and a sequence $\{k_n\}\subset [1,\mathrm{\infty })$ with $k_n\to 1$ as $n\to \mathrm{\infty }$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{x,y\in C}{sup}({\parallel T^{n}x-T^{n}y\parallel }^2-k_n{\parallel x-y\parallel }^2-k{\parallel (I-T^n)x-(I-T^n)y\parallel }^2)\le 0.$$

(1.3)

Put

$${\zeta }_n=max\{0,\underset{x,y\in C}{sup}({\parallel T^{n}x-T^{n}y\parallel }^2-k_n{\parallel x-y\parallel }^2-k{\parallel (I-T^n)x-(I-T^n)y\parallel }^2)\}.$$

(1.4)

It follows that ${\zeta }_n\to 0$as $n\to \mathrm{\infty }$. Then(1.3) is reduced to the following:

$$\begin{array}{r}{\parallel T^{n}x-T^{n}y\parallel }^2\le k_n{\parallel x-y\parallel }^2+k{\parallel (I-T^n)x-(I-T^n)y\parallel }^2+{\zeta }_n,\\ \mathrm{\forall }n\ge 1,x,y\in C.\end{array}$$

(1.5)

We remark that if ${\zeta }_n=0$$\mathrm{\forall }n\ge 1$in (1.5), then we obtain (1.2), meaning that the class of asymptotically strictpseudocontractive mappings in the intermediate sense contains properly the class ofasymptotically strict pseudocontractive mappings;

(4) asymptotically pseudocontractive [4] if there exists a sequence

$\{k_n\}\subset [1,\mathrm{\infty })$with $k_n\to 1$ as$n\to \mathrm{\infty }$ such that

$$〈T^{n}x-T^{n}y,x-y〉\le k_n{\parallel x-y\parallel }^2,\mathrm{\forall }n\ge 1,x,y\in C.$$

(1.6)

It is easy to show that (1.6) is equivalent to

$${\parallel T^{n}x-T^{n}y\parallel }^2\le (2k_n-1){\parallel x-y\parallel }^2+{\parallel x-y-(T^{n}x-T^{n}y)\parallel }^2,\mathrm{\forall }n\ge 1,x,y\in C.$$

(1.7)

The class of asymptotically pseudocontractive mappings was introduced in 1991 bySchu [5].

Qin et al.[4] in 2010 introduced the following class ofasymptotically pseudocontractive mappings in the intermediate sense. They obtainedsome weak convergence theorems for this class of nonlinear mappings. They alsoestablished a strong convergence theorem without any compact assumption byconsidering the so-called hybrid projection method;

(5) asymptotically pseudocontractive mapping in the intermediate sense [4] if

there exists a sequence $\{k_n\}\subset [1,\mathrm{\infty })$with $k_n\to 1$ as$n\to \mathrm{\infty }$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{x,y\in C}{sup}(〈T^{n}x-T^{n}y,x-y〉-k_n{\parallel x-y\parallel }^2)\le 0.$$

(1.8)

Put

$${\tau }_n=max\{0,\underset{x,y\in C}{sup}(〈T^{n}x-T^{n}y,x-y〉-k_n{\parallel x-y\parallel }^2)\}.$$

(1.9)

It follows that ${\tau }_n\to 0$as $n\to \mathrm{\infty }$. Hence,(1.8) is reduced to the following:

$$〈T^{n}x-T^{n}y,x-y〉\le k_n{\parallel x-y\parallel }^2+{\tau }_n,\mathrm{\forall }n\ge 1,x,y\in C.$$

(1.10)

In real Hilbert spaces, it is easy to check that (1.10) is equivalent to

$$\begin{array}{r}{\parallel T^{n}x-T^{n}y\parallel }^2\le (2k_n-1){\parallel x-y\parallel }^2+{\parallel (I-T^n)x-(I-T^n)y\parallel }^2+2{\tau }_n,\\ \mathrm{\forall }n\ge 1,x,y\in C.\end{array}$$

(1.11)

We remark that if ${\tau }_n=0$$\mathrm{\forall }n\ge 1$,then the class of asymptotically pseudocontractive mappings in the intermediatesense is reduced to the class of asymptotically pseudocontractive mappings;

(6) asymptotically demicontractive mappings [2] if there exists a sequence $\{a_n\}$ such that ${lim}_n{a}_n=1$ and for $0\le k<1$,

$${\parallel T^{n}x-p\parallel }^2\le a_n^2{\parallel x-p\parallel }^2+k{\parallel x-T^{n}x\parallel }^2,\mathrm{\forall }n\in \mathbb{N},x\in C,\mathrm{\forall }p\in F(T).$$

(1.12)

The class of asymptotically demicontractive maps was introduced in 1987 by Liu[6];

(7) asymptotically hemicontractive mappings [2] if there exists a sequence $\{a_n\}$ such that ${lim}_n{a}_n=1$ and

$${\parallel T^{n}x-p\parallel }^2\le a_n{\parallel x-p\parallel }^2+{\parallel x-T^{n}x\parallel }^2,\mathrm{\forall }n\in \mathbb{N},x\in C,\mathrm{\forall }p\in F(T).$$

(1.13)

The class of asymptotically hemicontractive maps was introduced in 1987 by Liu[6], and it properly contains theclass of asymptotically pseudocontractive maps and asymptotically strictpseudocontractive maps in which the fixed point set $F(T):=\{x\in C:Tx=x\}\ne \mathrm{\varnothing }$ is nonempty. Clearly, if$k=1$ in (1.12),then we obtain (1.13).

Motivated by the above facts, we now introduce the classes of asymptoticallydemicontractive mappings in the intermediate sense and asymptoticallyhemicontractive mappings in the intermediate sense as generalizations of the classesof asymptotically demicontractive mappings and asymptotically hemicontractivemappings, respectively.

(8) The map $T:C\to C$ is said to be an asymptotically demicontractive mapping in the intermediate sense if there exists a sequence $\{a_n\}$ such that ${lim}_n{a}_n=1$ and for some constant $k\in [0,1)$ if

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{(x,p)\in C\times F(T)}{sup}({\parallel T^{n}x-p\parallel }^2-a_n^2{\parallel x-p\parallel }^2-k{\parallel x-T^{n}x\parallel }^2)\le 0,\\ \mathrm{\forall }(x,p)\in C\times F(T).\end{array}$$

(1.14)

Observe that if we put

$${\nu }_n=max\{0,\underset{(x,p)\in C\times F(T)}{sup}({\parallel T^{n}x-p\parallel }^2-a_n^2{\parallel x-p\parallel }^2-k{\parallel x-T^{n}x\parallel }^2)\},$$

(1.15)

then we get that ${\nu }_n⟶0$as $n\to \mathrm{\infty }$ and (1.14)is reduced to the following:

$${\parallel T^{n}x-p\parallel }^2\le a_n^2{\parallel x-p\parallel }^2+k{\parallel x-T^{n}x\parallel }^2+{\nu }_n.$$

(1.16)

Observe that if ${\nu }_n=0$for all n in (1.16), then we obtain (1.12);

(9) asymptotically hemicontractive mapping in the intermediate sense with sequence $\{a_n\}$ such that ${lim}_n{a}_n=1$ if

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{(x,p)\in C\times F(T)}{sup}({\parallel T^{n}x-p\parallel }^2-a_n^2{\parallel x-p\parallel }^2-{\parallel x-T^{n}x\parallel }^2)\le 0,\\ \mathrm{\forall }(x,p)\in C\times F(T).\end{array}$$

(1.17)

Observe that if we put

$${\nu }_n=max\{0,\underset{(x,p)\in C\times F(T)}{sup}({\parallel T^{n}x-p\parallel }^2-a_n^2{\parallel x-p\parallel }^2-{\parallel x-T^{n}x\parallel }^2)\},$$

(1.18)

then we get that ${\nu }_n⟶0$as $n\to \mathrm{\infty }$ and (1.17)is reduced to the following:

$${\parallel T^{n}x-p\parallel }^2\le a_n^2{\parallel x-p\parallel }^2+{\parallel x-T^{n}x\parallel }^2+{\nu }_n.$$

(1.19)

Observe that if ${\nu }_n=0$for all $n\in \mathbb{N}$ in(1.19), then we obtain (1.13). This means that the class of asymptoticallyhemicontractive maps in the intermediate sense is a generalization of the class ofasymptotically hemicontractive maps. Clearly, if $k=1$ in(1.16), then we obtain (1.19).

The following definition will be useful for our results.

(10) The map $T:C\to C$ is said to be uniformly L-Lipschitzian [2] if

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

(1.20)

for some constant $L>0$for all $n\in \mathbb{N}$ and$x,y\in C$.

Qihou [2] obtained some convergence results ofMann iterative scheme for the class of asymptotically demicontractive mappings.Similarly, Schu [5] proved the convergence ofMann iterative scheme for asymptotically nonexpansive mappings. In this study, weextend the results of Qihou [2] and Schu[5] to the classes of asymptoticallydemicontractive mappings in the intermediate sense and asymptoticallyhemicontractive mappings in the intermediate sense. It is our purpose in this studyto prove strong convergence theorems of Mann and Ishikawa iterative schemes foruniformly L-Lipschitzian asymptotically demicontractive mappings in theintermediate sense and asymptotically hemicontractive maps in the intermediatesense. Our results are extensions and generalizations of the results of Hicks andKubicek [7], Liu [6], Qihou [2] andSchu [5].

Qihou [2] in 1996 proved the followingconvergence theorem for the class of asymptotically demicontractive mappings. Choet al.[8] proved some fixed point theorems for theclass of asymptotically demicontractive mappings in arbitrary real normed linearspaces. Maruster and Maruster [9] introducedthe class of α-demicontractive mappings. They established that thisclass of nonlinear mappings is general than the class of demicontractive mappings.Olaleru and Mogbademu [10, 11] used a three-step iterative scheme to approximate the fixedpoints of strongly successively pseudocontractive maps.

Theorem Q[2]

Let H be a Hilbert space, $C\subset H$be nonempty closed bounded and convex; $T:C\to C$be completely continuous and uniformly L-Lipschitzian and asymptotically demicontractive with sequence$\{a_n\}$,$a_n\in [1,+\mathrm{\infty })$, ${\sum }_{n=0}^{\mathrm{\infty }}(a_n^2-1)<+\mathrm{\infty }$,$ϵ\le {\alpha }_n\le 1-k-ϵ$,for$\mathrm{\forall }n\in \mathbb{N}$and some$ϵ>0$,$\mathrm{\forall }{x}_0\in C$.

$$x_{n+1}={\alpha }_n{T}^n{x}_n+(1-{\alpha }_n)x_n,\mathrm{\forall }n\in \mathbb{N}.$$

(1.21)

Then${\{x_n\}}_{n=0}^{\mathrm{\infty }}$converges strongly to some fixed point of T.

Osilike [12] in 1998 extended the results ofQihou [2] to more generalq-uniformly smooth Banach spaces, $1<q<\mathrm{\infty }$ for theclass of asymptotically demicontractive mappings. Osilike and Aniagbosor[13] in 2001 proved that theboundedness requirement imposed on the subset C in the results of Osilike[12] can be dropped. Moore and Nnoli[14] in 2005 proved the necessaryand sufficient conditions for the strong convergence of the Mann iterative sequenceto a fixed point of an asymptotically demicontractive and uniformlyL-Lipschitzian map. Zegeye et al.[1] in 2011 obtained some strong convergenceresults of the Ishikawa-type iterative scheme for the class of asymptoticallypseudocontractive mappings in the intermediate sense without resorting to the hybridmethod which was the main tool of Qin et al.[4]. Olaleru and Okeke [15] in 2012 established a strong convergence ofNoor-type scheme for uniformly L-Lipschitzian and asymptoticallypseudocontractive mappings in the intermediate sense without assuming any form ofcompactness. It is our purpose in this paper to prove some strong convergenceresults using Ishikawa-type and Mann-type iterative schemes for the classes ofasymptotically demicontractive mappings in the intermediate sense and asymptoticallyhemicontractive mappings in the intermediate sense. Our results generalize andimprove several other results in literature.

The following lemmas will be useful in this study.

Lemma 1.2[2]

Let sequences${\{a_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{b_n\}}_{n=1}^{\mathrm{\infty }}$satisfy$a_{n+1}\le a_n+b_n$,$a_n\ge 0$,$\mathrm{\forall }n\in \mathbb{N}$,${\sum }_{n=1}^{\mathrm{\infty }}{b}_n$is convergent and${\{a_n\}}_{n=1}^{\mathrm{\infty }}$has a subsequence${\{a_{n_k}\}}_{k=1}^{\mathrm{\infty }}$converging to 0. Then we must have

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

(1.22)

Lemma 1.3[1]

Let H be a real Hilbert space. Then the following equality holds:

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

(1.23)

for all$\alpha \in (0,1)$and$x,y\in H$.

2 Main results

Theorem 2.1 Let H be a Hilbert space, $C\subset H$be a nonempty closed bounded and convex subset of H; $T:C\to C$be a completely continuous and uniformly L-Lipschitzian and asymptotically demicontractive mapping in theintermediate sense with sequence$\{{\nu }_n\}$as defined in (1.16). Assume that$F(T)$is nonempty. Let$\{x_n\}$be a sequence defined by$x_1=x\in C$and

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

(2.1)

where$\{{\alpha }_n\},\{{\beta }_n\}\in [0,1]$. Assume that thefollowing conditions are satisfied:

• (i) the sequence $\{a_n\}$ is such that $a_n\in [1,+\mathrm{\infty })$, $\mathrm{\forall }n\in \mathbb{N}$ and ${\sum }_{n=1}^{\mathrm{\infty }}(a_n^2-1)<+\mathrm{\infty }$,

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{\nu }_n<+\mathrm{\infty }$,

• (iii) $ϵ\le k\le {\alpha }_n\le {\beta }_n\le b$ $\mathrm{\forall }n\in \mathbb{N}$ for some $ϵ>0$, $k\in [0,1)$ and some $b\in (0,L^{-2}[\sqrt{1+L^2}-1])$.

Then$\{x_n\}$converges strongly to a fixed point of T.

Proof Fix $p\in F(T)$. Using (1.16), (2.1) and Lemma 1.3, weobtain

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel {\beta }_n(T^n{x}_n-p)+(1-{\beta }_n)(x_n-p)\parallel }^2\\ =& {\beta }_n{\parallel T^n{x}_n-p\parallel }^2+(1-{\beta }_n){\parallel x_n-p\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ \le & {\beta }_n(a_n^2{\parallel x_n-p\parallel }^2+k{\parallel x_n-T^n{x}_n\parallel }^2+{\nu }_n)+(1-{\beta }_n){\parallel x_n-p\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ =& {\beta }_n{a}_n^2{\parallel x_n-p\parallel }^2+{\beta }_{n}k{\parallel x_n-T^n{x}_n\parallel }^2+{\beta }_n{\nu }_n+(1-{\beta }_n){\parallel x_n-p\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ =& (1+{\beta }_n(a_n^2-1)){\parallel x_n-p\parallel }^2-{\beta }_n(1-{\beta }_n-k){\parallel T^n{x}_n-x_n\parallel }^2\\ +{\beta }_n{\nu }_n.\end{array}$$

(2.2)

Using (1.20), (2.1) and Lemma 1.3, we have

$$\begin{array}{rcl}{\parallel y_n-T^n{y}_n\parallel }^2& =& {\parallel {\beta }_n(T^n{x}_n-T^n{y}_n)+(1-{\beta }_n)(x_n-T^n{y}_n)\parallel }^2\\ =& {\beta }_n{\parallel T^n{x}_n-T^n{y}_n\parallel }^2+(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ \le & {\beta }_n{L}^2{\parallel x_n-y_n\parallel }^2+(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ =& {\beta }_n^3{L}^2{\parallel x_n-T^n{x}_n\parallel }^2+(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2.\end{array}$$

(2.3)

Using (1.16), (2.2) and (2.3), we obtain

$$\begin{array}{rcl}{\parallel T^n{y}_n-p\parallel }^2& \le & a_n^2{\parallel y_n-p\parallel }^2+k{\parallel y_n-T^n{y}_n\parallel }^2+{\nu }_n\\ \le & a_n^2\{(1+{\beta }_n(a_n^2-1)){\parallel x_n-p\parallel }^2-{\beta }_n(1-{\beta }_n-k){\parallel T^n{x}_n-x_n\parallel }^2+{\beta }_n{\nu }_n\}\\ +k\{{\beta }_n^3{L}^2{\parallel x_n-T^n{x}_n\parallel }^2+(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2\}+{\nu }_n\\ =& a_n^2(1+{\beta }_n(a_n^2-1)){\parallel x_n-p\parallel }^2-a_n^2{\beta }_n(1-{\beta }_n-k){\parallel T^n{x}_n-x_n\parallel }^2\\ +a_n^2{\beta }_n{\nu }_n+k{\beta }_n^3{L}^2{\parallel x_n-T^n{x}_n\parallel }^2+k(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ -k{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2+{\nu }_n.\end{array}$$

(2.4)

Using (2.4), Lemma 1.3 and condition (iii), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel {\alpha }_n(T^n{y}_n-p)+(1-{\alpha }_n)(x_n-p)\parallel }^2\\ =& {\alpha }_n{\parallel T^n{y}_n-p\parallel }^2+(1-{\alpha }_n){\parallel x_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel T^n{y}_n-x_n\parallel }^2\\ \le & {\alpha }_n\{a_n^2(1+{\beta }_n(a_n^2-1)){\parallel x_n-p\parallel }^2\\ -a_n^2{\beta }_n(1-{\beta }_n-k){\parallel T^n{x}_n-x_n\parallel }^2+a_n^2{\beta }_n{\nu }_n+k{\beta }_n^3{L}^2{\parallel x_n-T^n{x}_n\parallel }^2\\ +k(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2-k{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2+{\nu }_n\}\\ +(1-{\alpha }_n){\parallel x_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel T^n{y}_n-x_n\parallel }^2\\ \le & [1+{\alpha }_n(a_n^2-1)(1+{\beta }_n{a}_n^2)]{\parallel x_n-p\parallel }^2\\ -{\alpha }_n{\beta }_n[k(1-{\beta }_n-{\beta }_n^2{L}^2)+(1-k-{\beta }_n)a_n^2]{\parallel T^n{x}_n-x_n\parallel }^2\\ +[k(1-{\beta }_n)-{\alpha }_n(1-{\alpha }_n)]{\parallel T^n{y}_n-x_n\parallel }^2+{\alpha }_n(1+a_n^2{\beta }_n){\nu }_n\\ \le & [1+{\alpha }_n(a_n^2-1)(1+{\beta }_n{a}_n^2)]{\parallel x_n-p\parallel }^2\\ -{\alpha }_n{\beta }_n[k(1-{\beta }_n-{\beta }_n^2{L}^2)+(1-k-{\beta }_n)a_n^2]{\parallel T^n{x}_n-x_n\parallel }^2\\ +{\alpha }_n(1+a_n^2{\beta }_n){\nu }_n.\end{array}$$

(2.5)

Observe that by condition (iii), $k(1-{\beta }_n)-{\alpha }_n(1-{\alpha }_n)=-M$,where $M>0$, sothat the term ${\parallel T^n{y}_n-x_n\parallel }^2$can be dropped. Hence, we obtain (2.5).

Next, we show that ${lim}_{n\to \mathrm{\infty }}\parallel T^n{x}_n-x_n\parallel =0$. From(2.5), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2-{\parallel x_n-p\parallel }^2& \le & \left[{\alpha }_n(a_n^2-1)(1+{\beta }_n{a}_n^2)\right]{\parallel x_n-p\parallel }^2\\ -{\alpha }_n{\beta }_n[k(1-{\beta }_n-{\beta }_n^2{L}^2)+(1-k-{\beta }_n)a_n^2]\\ \times {\parallel T^n{x}_n-x_n\parallel }^2+{\alpha }_n(1+a_n^2{\beta }_n){\nu }_n.\end{array}$$

(2.6)

Since ${\sum }_{n=1}^{\mathrm{\infty }}(a_n^2-1)<+\mathrm{\infty }$, itfollows that ${lim}_{n\to \mathrm{\infty }}(a_n^2-1)=0$. Hence,${\{a_n^2\}}_{n=0}^{\mathrm{\infty }}$ is bounded. SinceC is bounded and $0\le {\alpha }_n\le {\beta }_n\le 1$,${\{{\alpha }_n(1+{\beta }_n{a}_n^2)\}}_{n=1}^{\mathrm{\infty }}$ and${\{{\alpha }_n(1+{\beta }_n{a}_n^2){\parallel x_n-p\parallel }^2\}}_{n=1}^{\mathrm{\infty }}$ must be bounded. Hence,there exists a constant $M>0$such that

$$0\le {\alpha }_n(1+{\beta }_n{a}_n^2)(1+{\parallel x_n-p\parallel }^2)\le M.$$

(2.7)

Using (2.6) and (2.7), we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2-{\parallel x_n-p\parallel }^2& \le & M(a_n^2-1)-{\alpha }_n{\beta }_n[k(1-{\beta }_n-{\beta }_n^2{L}^2)\\ +(1-k-{\beta }_n)a_n^2]{\parallel T^n{x}_n-x_n\parallel }^2+M{\nu }_n.\end{array}$$

(2.8)

Observe that the condition $b\in (0,L^{-2}[\sqrt{1+L^2}-1])$ implies that $b>0$and $b<L^{-2}[\sqrt{1+L^2}-1]$. This implies that$b{L}^2<\sqrt{1+L^2}-1$,hence $1+b{L}^2<\sqrt{1+L^2}$.On squaring both sides, we obtain ${(1+b{L}^2)}^2<{(\sqrt{1+L^2})}^2$, so that$1+2b{L}^2+b^2{L}^4<1+L^2$,so we obtain $L^2-2b{L}^2-b^2{L}^4>0$,by dividing through by $L^2$,we obtain $1-2b-b^2{L}^2>0$.Hence, $\frac{1-2b-b^2{L}^2}{2}>0$.Since ${lim}_{n\to \mathrm{\infty }}{a}_n=1$,there exists a natural number N such that for $n>N$,

$$k(1-{\beta }_n-{\beta }_n^2{L}^2)+(1-k-{\beta }_n)a_n^2\ge 1-b-a_n^{2}b-L^2{b}^2\ge \frac{1-2b-b^2{L}^2}{2}.$$

(2.9)

Assuming that ${lim}_{n\to \mathrm{\infty }}\parallel T^n{x}_n-x_n\parallel \ne 0$,then there exist ${ϵ}_0>0$and a subsequence ${\{x_{n_r}\}}_{r=1}^{\mathrm{\infty }}$ of ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ such that

$${\parallel x_{n_r}-T^{n_r}{x}_{n_r}\parallel }^2\ge {ϵ}_0.$$

(2.10)

Without loss of generality, we can assume that $n_1>N$.From (2.8), we obtain

$$\begin{array}{r}{\alpha }_n{\beta }_n[k(1-{\beta }_n-{\beta }_n^2{L}^2)+(1-k-{\beta }_n)a_n^2]{\parallel T^n{x}_n-x_n\parallel }^2\\ \le M(a_n^2-1)+{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+M{\nu }_n.\end{array}$$

Hence,

$$\begin{array}{r}\sum _{m=n_1}^{n_r}{\alpha }_m{\beta }_m[k(1-{\beta }_m-{\beta }_m^2{L}^2)+(1-k-{\beta }_m)a_m^2]{\parallel T^m{x}_m-x_m\parallel }^2\\ \le \sum _{m=n_1}^{n_r}M(a_m^2-1)+{\parallel x_{n_1}-p\parallel }^2-{\parallel x_{n_r+1}-p\parallel }^2+\sum _{m=n_1}^{n_r}M{\nu }_m,\\ \sum _{l=1}^r{\alpha }_{n_l}{\beta }_{n_l}[k(1-{\beta }_{n_l}-{\beta }_{n_l}{L}^2)+(1-k-{\beta }_{n_l})a_{n_l}^2]{\parallel T^{n_l}{x}_{n_l}-x_{n_l}\parallel }^2\\ \le \sum _{m=n_1}^{n_r}M(a_m^2-1)+{\parallel x_{n_1}-p\parallel }^2-{\parallel x_{n_r+1}-p\parallel }^2+\sum _{m=n_1}^{n_r}M{\nu }_m.\end{array}$$

(2.11)

From (2.9), (2.10), (2.11) and $0\le ϵ\le {\alpha }_n\le {\beta }_n$,we observe that

$$\begin{array}{r}r{ϵ}^2\left(\frac{1-2b-b^2{L}^2}{2}\right){ϵ}_0\\ \le \sum _{m=n_l}^{n_r}M(a_m^2-1)+{\parallel x_{n_1}-p\parallel }^2-{\parallel x_{n_r+1}-p\parallel }^2+\sum _{m=n_l}^{n_r}M{\nu }_m.\end{array}$$

(2.12)

From ${\sum }_{n=1}^{\mathrm{\infty }}(a_n^2-1)<+\mathrm{\infty }$,${\sum }_{n=1}^{\mathrm{\infty }}{\nu }_n<+\mathrm{\infty }$ and theboundedness of C, we observe that the right-hand side of (2.12) is bounded.However, the left-hand side of (2.12) is positively unbounded when$r\to \mathrm{\infty }$. Hence, acontradiction. Therefore

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

(2.13)

Using (2.1), we have

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & =& \parallel {\alpha }_n{T}^n{y}_n+(1-{\alpha }_n)x_n-x_n\parallel \\ =& \parallel {\alpha }_n(T^n{y}_n-x_n)\parallel \\ \le & {\alpha }_n\parallel T^n{y}_n-x_n\parallel \\ \le & {\alpha }_n(\parallel T^n{y}_n-T^n{x}_n\parallel +\parallel T^n{x}_n-x_n\parallel )\\ \le & {\alpha }_n(L\parallel y_n-x_n\parallel +\parallel T^n{x}_n-x_n\parallel )\\ \le & {\alpha }_n(1+{\beta }_{n}L)\parallel T^n{x}_n-x_n\parallel .\end{array}$$

(2.14)

Using (2.13), we obtain

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

(2.15)

Observe that

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

(2.16)

Using (2.13) and (2.15), we have

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

(2.17)

Since $\{x_n\}$ is bounded, thesequence $\{T{x}_n\}$ has a convergentsubsequence $\{T{x}_{n_r}\}$ say. Let$T{x}_{n_r}\to q$as $r\to \mathrm{\infty }$. Then$x_{n_r}\to q$as $r\to \mathrm{\infty }$ since

$$0=\underset{r\to \mathrm{\infty }}{lim}[x_{n_r}-T{x}_{n_r}]=\underset{r\to \mathrm{\infty }}{lim}{x}_{n_r}-\underset{r\to \mathrm{\infty }}{lim}T{x}_{n_r}=\underset{r\to \mathrm{\infty }}{lim}{x}_{n_r}-q.$$

(2.18)

By the continuity of T, $T{x}_{n_r}\to Tq$ as$r\to \mathrm{\infty }$ but$T{x}_{n_r}\to q$as $r\to \mathrm{\infty }$. Hence,$q=Tq$.

Hence, ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ has a subsequence whichconverges to the fixed point q of T. Using (2.9), there existssome natural number N, when $n>N$,$k(1-{\beta }_n-{\beta }_n^2{L}^2)+(1-k-{\beta }_n)a_n^2\ge \frac{1-2b-b^2{L}^2}{2}>0$$\mathrm{\forall }n>N$.Using (2.7), $0\le {\alpha }_n(1+{\beta }_n{a}_n^2){\nu }_n+{\alpha }_n(1+{\beta }_n{a}_n^2)(a_n^2-1){\parallel x_n-q\parallel }^2\le M(a_n^2-1)+M{\nu }_n$.From (2.6),

$${\parallel x_{n+1}-q\parallel }^2\le {\parallel x_n-q\parallel }^2+M(a_n^2-1+{\nu }_n).$$

(2.19)

But ${\sum }_{n=1}^{\mathrm{\infty }}(a_n^2-1)<+\mathrm{\infty }$ and${\sum }_{n=1}^{\mathrm{\infty }}{\nu }_n<+\mathrm{\infty }$ imply that${\sum }_{n=1}^{\mathrm{\infty }}M(a_n^2-1+{\nu }_n)<+\mathrm{\infty }$. From(2.18), it follows that there exists a subsequence ${\{{\parallel x_{n_r}-q\parallel }^2\}}_{r=1}^{\mathrm{\infty }}$ of ${\{{\parallel x_n-q\parallel }^2\}}_{n=1}^{\mathrm{\infty }}$, which converges to 0.Hence, using (2.19) and Lemma 1.2, ${lim}_{n\to \mathrm{\infty }}{\parallel x_n-q\parallel }^2=0$.This means that ${lim}_{n\to \mathrm{\infty }}{x}_n=q$.The proof of the theorem is complete. □

Remark 2.2 Theorem 2.1 extends the results of Osilike [12], Osilike and Aniagbosor [13], Igbokwe [16]in the framework of Hilbert spaces since the class of asymptotically demicontractivemaps considered by these authors is a subclass of the class of asymptoticallydemicontractive maps in the intermediate sense introduced in this article.

Theorem 2.3 Let H be a Hilbert space, $C\subset H$be a nonempty closed bounded and convex subset of H; $T:C\to C$be a completely continuous and asymptotically demicontractive mapping in theintermediate sense with sequence$\{{\nu }_n\}$as defined in (1.16). Assume that$F(T)$is nonempty. Let$\{x_n\}$be a sequence defined by$x_0=x\in C$and

$$x_{n+1}={\alpha }_n{T}^n{x}_n+(1-{\alpha }_n)x_n,\mathrm{\forall }n\in \mathbb{N},$$

(2.20)

where${\alpha }_n\in [0,1]$. Assume that thefollowing conditions are satisfied:

• (i) the sequence $\{a_n\}$ is such that $a_n\in [1,+\mathrm{\infty })$ and ${\sum }_{n=0}^{\mathrm{\infty }}(a_n^2-1)<+\mathrm{\infty }$,

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{\nu }_n<+\mathrm{\infty }$ and

• (iii) $ϵ\le {\alpha }_n\le 1-k-ϵ$ $\mathrm{\forall }n\in \mathbb{N}$ for some $ϵ>0$ and $k\in [0,1)$.

Then${\{x_n\}}_{n=0}^{\mathrm{\infty }}$converges strongly to a fixed point of T.

Proof Using (1.16), we obtain

$${\parallel T^n{x}_n-p\parallel }^2\le a_n^2{\parallel x_n-p\parallel }^2+k{\parallel x_n-T^n{x}_n\parallel }^2+{\nu }_n.$$

(2.21)

From (2.21) and Lemma 1.3, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel {\alpha }_n(T^n{x}_n-p)+(1-{\alpha }_n)(x_n-p)\parallel }^2\\ =& {\alpha }_n{\parallel T^n{x}_n-p\parallel }^2+(1-{\alpha }_n){\parallel x_n-p\parallel }^2\\ -{\alpha }_n(1-{\alpha }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ \le & {\alpha }_n(a_n^2{\parallel x_n-p\parallel }^2+k{\parallel x_n-T^n{x}_n\parallel }^2+{\nu }_n)+(1-{\alpha }_n){\parallel x_n-p\parallel }^{2}x\\ -{\alpha }_n(1-{\alpha }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ =& {\alpha }_n{a}_n^2{\parallel x_n-p\parallel }^2+{\alpha }_{n}k{\parallel x_n-T^n{x}_n\parallel }^2+{\alpha }_n{\nu }_n+(1-{\alpha }_n){\parallel x_n-p\parallel }^2\\ -{\alpha }_n(1-{\alpha }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ =& {\parallel x_n-p\parallel }^2+{\alpha }_n(a_n^2-1){\parallel x_n-p\parallel }^2\\ -{\alpha }_n(1-{\alpha }_n-k){\parallel T^n{x}_n-x_n\parallel }^2+{\alpha }_n{\nu }_n.\end{array}$$

(2.22)

Now, we show that ${lim}_{n\to \mathrm{\infty }}\parallel T^n{x}_n-x_n\parallel =0$. But$0<ϵ\le {\alpha }_n\le 1-k-ϵ$,$1-k-{\alpha }_n\ge ϵ$.Hence ${\alpha }_n(1-k-{\alpha }_n)\ge {ϵ}^2$and ${\nu }_n⟶0$as $n\to \mathrm{\infty }$. From(2.22), we have

$${\parallel x_{n+1}-p\parallel }^2\le {\parallel x_n-p\parallel }^2+{\alpha }_n(a_n^2-1){\parallel x_n-p\parallel }^2-{ϵ}^2{\parallel T^n{x}_n-x_n\parallel }^2+{\alpha }_n{\nu }_n.$$

(2.23)

Since C is bounded and T is a self-mapping on C, itfollows that there exists some $M>0$such that ${\parallel x_n-p\parallel }^2\le M$,$\mathrm{\forall }n\in \mathbb{N}$. But${\alpha }_n\in [0,1]$, from (2.23) weobtain

$${\parallel x_{n+1}-p\parallel }^2\le {\parallel x_n-p\parallel }^2+M(a_n^2-1)+{\nu }_n-{ϵ}^2{\parallel T^n{x}_n-x_n\parallel }^2.$$

(2.24)

Hence,

$${ϵ}^2{\parallel T^n{x}_n-x_n\parallel }^2\le {\parallel x_n-p\parallel }^2+M(a_n^2-1)+{\nu }_n-{\parallel x_{n+1}-p\parallel }^2,$$

(2.25)

$$\begin{array}{rl}\sum _{n=1}^m{ϵ}^2{\parallel T^n{x}_n-x_n\parallel }^2& \le {\parallel x_1-p\parallel }^2-{\parallel x_{m+1}-p\parallel }^2+\sum _{n=1}^m[M(a_n^2-1)+{\nu }_n]\\ \le 2M+\sum _{n=1}^{\mathrm{\infty }}[M(a_n^2-1)+{\nu }_n].\end{array}$$

(2.26)

But ${\sum }_{n=1}^{\mathrm{\infty }}[M(a_n^2-1)+{\nu }_n]<+\mathrm{\infty }$,${\sum }_{n=1}^{\mathrm{\infty }}{ϵ}^2{\parallel T^n{x}_n-x_n\parallel }^2<+\mathrm{\infty }$. Hence, weobtain ${lim}_{n\to \mathrm{\infty }}{\parallel T^n{x}_n-x_n\parallel }^2=0$.So that

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

(2.27)

Since ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is a bounded sequence andT is completely continuous, hence there is a subsequence${\{T{x}_{n_k}\}}_{k=0}^{\mathrm{\infty }}$ of ${\{T{x}_n\}}_{n=0}^{\mathrm{\infty }}$. Using (2.27),${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ must have a convergentsubsequence ${\{x_{n_k}\}}_{k=0}^{\mathrm{\infty }}$. Assume${lim}_{k\to \mathrm{\infty }}{x}_{n_k}=x^{\ast }$.From the continuity of T and using (2.27), we obtain$x^{\ast }=T{x}^{\ast }$,meaning that $x^{\ast }$is a fixed point of T. Hence, ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ has a subsequence whichconverges to a fixed point $x^{\ast }$of T.

Since ${\sum }_{n=1}^{\mathrm{\infty }}{ϵ}^2{\parallel T^n{x}_n-x_n\parallel }^2<+\mathrm{\infty }$ and${\sum }_{n=1}^{\mathrm{\infty }}[M(a_n^2-1)+{\nu }_n]<+\mathrm{\infty }$ and usingLemma 1.2, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel x_n-x^{\ast }\parallel }^2=0.$$

(2.28)

Hence, ${lim}_{n\to \mathrm{\infty }}{x}_n=x^{\ast }$.The proof of Theorem 2.3 is completed. □

Corollary 2.4 Let H be a Hilbert space, $C\subset H$be a nonempty closed bounded and convex subset of H; $T:C\to C$be a completely continuous and uniformly L-Lipschitzian and asymptotically demicontractive mapping withsequence$\{a_n\}$as defined in (1.12). Assume that$F(T)$is nonempty. Let$\{x_n\}$be a sequence defined by$x_0=x\in C$and

$$x_{n+1}={\alpha }_n{T}^n{x}_n+(1-{\alpha }_n)x_n,\mathrm{\forall }n\in \mathbb{N},$$

(2.29)

where${\alpha }_n\in [0,1]$. Assume that thefollowing conditions are satisfied:

• (i) the sequence $\{a_n\}$ is such that $a_n\in [1,+\mathrm{\infty })$ and ${\sum }_{n=0}^{\mathrm{\infty }}(a_n^2-1)<+\mathrm{\infty }$ and

• (ii) $ϵ\le {\alpha }_n\le 1-k-ϵ$, $\mathrm{\forall }n\in \mathbb{N}$, $k\in [0,1)$ and some $ϵ>0$.

Then${\{x_n\}}_{n=0}^{\mathrm{\infty }}$converges strongly to a fixed point of T.

Remark 2.5 Corollary 2.4 is Theorem 1 of Qihou [2] when ${\nu }_n=0$for all $n\in \mathbb{N}$ inTheorem 2.3.

Theorem 2.6 Let H be a Hilbert space, $C\subset H$be a nonempty closed bounded and convex subset of H; $T:C\to C$be a completely continuous and uniformly L-Lipschitzian and asymptotically hemicontractive mapping in theintermediate sense with sequence$\{{\nu }_n\}$as defined in (1.19). Assume that$F(T)$is nonempty. Let$\{x_n\}$be a sequence defined by$x_1=x\in C$and

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

(2.30)

where${\alpha }_n,{\beta }_n\in [0,1]$. Assume that thefollowing conditions are satisfied:

• (i) the sequence $\{a_n\}$ is such that $a_n\in [1,+\mathrm{\infty })$ $\mathrm{\forall }n\in \mathbb{N}$ and ${\sum }_{n=1}^{\mathrm{\infty }}(a_n^2-1)<+\mathrm{\infty }$,

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{\nu }_n<+\mathrm{\infty }$,

• (iii) $ϵ\le {\alpha }_n\le {\beta }_n\le b$ $\mathrm{\forall }n\in \mathbb{N}$ for some $ϵ>0$ and some $b\in (0,L^{-2}[\sqrt{1+L^2}-1])$.

Then$\{x_n\}$converges strongly to a fixed point of T.

Proof Fix $p\in F(T)$. Using (1.19), (2.30) and Lemma 1.3, weobtain

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel {\beta }_n(T^n{x}_n-p)+(1-{\beta }_n)(x_n-p)\parallel }^2\\ =& {\beta }_n{\parallel T^n{x}_n-p\parallel }^2+(1-{\beta }_n){\parallel x_n-p\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ \le & {\beta }_n(a_n^2{\parallel x_n-p\parallel }^2+{\parallel x_n-T^n{x}_n\parallel }^2+{\nu }_n)+(1-{\beta }_n){\parallel x_n-p\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ =& {\beta }_n{a}_n^2{\parallel x_n-p\parallel }^2+{\beta }_n{\parallel x_n-T^n{x}_n\parallel }^2+{\beta }_n{\nu }_n+(1-{\beta }_n){\parallel x_n-p\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ =& (1+{\beta }_n(a_n^2-1)){\parallel x_n-p\parallel }^2+{\beta }_n^2{\parallel T^n{x}_n-x_n\parallel }^2\\ +{\beta }_n{\nu }_n.\end{array}$$

(2.31)

Using (1.20), (2.30) and Lemma 1.3, we have

$$\begin{array}{rcl}{\parallel y_n-T^n{y}_n\parallel }^2& =& {\parallel {\beta }_n(T^n{x}_n-T^n{y}_n)+(1-{\beta }_n)(x_n-T^n{y}_n)\parallel }^2\\ =& {\beta }_n{\parallel T^n{x}_n-T^n{y}_n\parallel }^2+(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ \le & {\beta }_n{L}^2{\parallel x_n-y_n\parallel }^2+(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2\\ =& {\beta }_n^3{L}^2{\parallel x_n-T^n{x}_n\parallel }^2+(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2.\end{array}$$

(2.32)

Using (1.19), (2.31) and (2.32), we obtain

$$\begin{array}{rcl}{\parallel T^n{y}_n-p\parallel }^2& \le & a_n^2{\parallel y_n-p\parallel }^2+{\parallel y_n-T^n{y}_n\parallel }^2+{\nu }_n\\ \le & a_n^2\{(1+{\beta }_n(a_n^2-1)){\parallel x_n-p\parallel }^2+{\beta }_n^2{\parallel T^n{x}_n-x_n\parallel }^2+{\beta }_n{\nu }_n\}\\ +{\beta }_n^3{L}^2{\parallel x_n-T^n{x}_n\parallel }^2+(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2+{\nu }_n\\ =& a_n^2(1+{\beta }_n(a_n^2-1)){\parallel x_n-p\parallel }^2+a_n^2{\beta }_n^2{\parallel T^n{x}_n-x_n\parallel }^2\\ +a_n^2{\beta }_n{\nu }_n+{\beta }_n^3{L}^2{\parallel x_n-T^n{x}_n\parallel }^2+(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2+{\nu }_n.\end{array}$$

(2.33)

Using (2.33), Lemma 1.3 and condition (iii), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel {\alpha }_n(T^n{y}_n-p)+(1-{\alpha }_n)(x_n-p)\parallel }^2\\ =& {\alpha }_n{\parallel T^n{y}_n-p\parallel }^2+(1-{\alpha }_n){\parallel x_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel T^n{y}_n-x_n\parallel }^2\\ \le & {\alpha }_n\{a_n^2(1+{\beta }_n(a_n^2-1)){\parallel x_n-p\parallel }^2\\ +a_n^2{\beta }_n^2{\parallel T^n{x}_n-x_n\parallel }^2+a_n^2{\beta }_n{\nu }_n+{\beta }_n^3{L}^2{\parallel x_n-T^n{x}_n\parallel }^2\\ +(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel T^n{x}_n-x_n\parallel }^2+{\nu }_n\}\\ +(1-{\alpha }_n){\parallel x_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel T^n{y}_n-x_n\parallel }^2\\ \le & [1+{\alpha }_n(a_n^2-1)(1+{\beta }_n{a}_n^2)]{\parallel x_n-p\parallel }^2\\ -{\alpha }_n{\beta }_n[1-{\beta }_n-{\beta }_n^2{L}^2-{\beta }_n{a}_n^2]{\parallel T^n{x}_n-x_n\parallel }^2\\ +[(1-{\beta }_n)-{\alpha }_n(1-{\alpha }_n)]{\parallel T^n{y}_n-x_n\parallel }^2+{\alpha }_n(1+a_n^2{\beta }_n){\nu }_n\\ \le & [1+{\alpha }_n(a_n^2-1)(1+{\beta }_n{a}_n^2)]{\parallel x_n-p\parallel }^2\\ -{\alpha }_n{\beta }_n[1-{\beta }_n-{\beta }_n^2{L}^2-{\beta }_n{a}_n^2]{\parallel T^n{x}_n-x_n\parallel }^2\\ +{\alpha }_n(1+a_n^2{\beta }_n){\nu }_n.\end{array}$$

(2.34)

Next, we show that ${lim}_{n\to \mathrm{\infty }}\parallel T^n{x}_n-x_n\parallel =0$. From(2.34), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2-{\parallel x_n-p\parallel }^2& \le & \left[{\alpha }_n(a_n^2-1)(1+{\beta }_n{a}_n^2)\right]{\parallel x_n-p\parallel }^2\\ -{\alpha }_n{\beta }_n[1-{\beta }_n-{\beta }_n^2{L}^2-{\beta }_n{a}_n^2]\\ \times {\parallel T^n{x}_n-x_n\parallel }^2+{\alpha }_n(1+a_n^2{\beta }_n){\nu }_n.\end{array}$$

(2.35)

Since ${\sum }_{n=1}^{\mathrm{\infty }}(a_n^2-1)<+\mathrm{\infty }$, itfollows that ${lim}_{n\to \mathrm{\infty }}(a_n^2-1)=0$. Hence,${\{a_n^2\}}_{n=0}^{\mathrm{\infty }}$ is bounded. SinceC is bounded and $0\le {\alpha }_n\le {\beta }_n\le 1$,${\{{\alpha }_n(1+{\beta }_n{a}_n^2)\}}_{n=1}^{\mathrm{\infty }}$ and${\{{\alpha }_n(1+{\beta }_n{a}_n^2){\parallel x_n-p\parallel }^2\}}_{n=1}^{\mathrm{\infty }}$ must be bounded. Hence,there exists a constant $M>0$such that

$$0\le {\alpha }_n(1+{\beta }_n{a}_n^2)(1+{\parallel x_n-p\parallel }^2)\le M.$$

(2.36)

Using (2.35) and (2.36), we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2-{\parallel x_n-p\parallel }^2& \le & M(a_n^2-1)-{\alpha }_n{\beta }_n[1-{\beta }_n-{\beta }_n^2{L}^2-{\beta }_n{a}_n^2]\\ \times {\parallel T^n{x}_n-x_n\parallel }^2+M{\nu }_n.\end{array}$$

(2.37)

Observe that the condition $b\in (0,L^{-2}[\sqrt{1+L^2}-1])$ implies that $b>0$and $b<L^{-2}[\sqrt{1+L^2}-1]$. This implies that$b{L}^2<\sqrt{1+L^2}-1$,hence $1+b{L}^2<\sqrt{1+L^2}$.On squaring both sides, we obtain ${(1+b{L}^2)}^2<{(\sqrt{1+L^2})}^2$, so that$1+2b{L}^2+b^2{L}^4<1+L^2$,so we obtain $L^2-2b{L}^2-b^2{L}^4>0$,by dividing through by $L^2$,we obtain $1-2b-b^2{L}^2>0$.Hence, $\frac{1-2b-b^2{L}^2}{2}>0$.Since ${lim}_{n\to \mathrm{\infty }}{a}_n=1$,there exists a natural number N such that for $n>N$,

$$1-{\beta }_n-{\beta }_n^2{L}^2-{\beta }_n{a}_n^2\ge 1-b-a_n^{2}b-L^2{b}^2\ge \frac{1-2b-b^2{L}^2}{2}.$$

(2.38)

Assuming that ${lim}_{n\to \mathrm{\infty }}\parallel T^n{x}_n-x_n\parallel \ne 0$,then there exist ${ϵ}_0>0$and a subsequence ${\{x_{n_r}\}}_{r=1}^{\mathrm{\infty }}$ of ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ such that

$${\parallel x_{n_r}-T^{n_r}{x}_{n_r}\parallel }^2\ge {ϵ}_0.$$

(2.39)

Without loss of generality, we can assume that $n_1>N$.From (2.37), we obtain

$$\begin{array}{r}{\alpha }_n{\beta }_n[1-{\beta }_n-{\beta }_n^2{L}^2-{\beta }_n{a}_n^2]{\parallel T^n{x}_n-x_n\parallel }^2\\ \le M(a_n^2-1)+{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+M{\nu }_n.\end{array}$$

Hence,

$$\begin{array}{r}\sum _{m=n_1}^{n_r}{\alpha }_m{\beta }_m[1-{\beta }_m-{\beta }_m^2{L}^2-{\beta }_m{a}_m^2]{\parallel T^m{x}_m-x_m\parallel }^2\\ \le \sum _{m=n_1}^{n_r}M(a_m^2-1)+{\parallel x_{n_1}-p\parallel }^2-{\parallel x_{n_r+1}-p\parallel }^2+\sum _{m=n_1}^{n_r}M{\nu }_m,\\ \sum _{l=1}^r{\alpha }_{n_l}{\beta }_{n_l}[1-{\beta }_{n_l}-{\beta }_{n_l}{L}^2-{\beta }_{n_l}{a}_{n_l}^2]{\parallel T^{n_l}{x}_{n_l}-x_{n_l}\parallel }^2\\ \le \sum _{m=n_1}^{n_r}M(a_m^2-1)+{\parallel x_{n_1}-p\parallel }^2-{\parallel x_{n_r+1}-p\parallel }^2+\sum _{m=n_1}^{n_r}M{\nu }_m.\end{array}$$

(2.40)

From (2.38), (2.39), (2.40) and $0\le ϵ\le {\alpha }_n\le {\beta }_n$,we observe that

$$\begin{array}{r}r{ϵ}^2\left(\frac{1-2b-b^2{L}^2}{2}\right){ϵ}_0\\ \le \sum _{m=n_l}^{n_r}M(a_m^2-1)+{\parallel x_{n_1}-p\parallel }^2-{\parallel x_{n_r+1}-p\parallel }^2+\sum _{m=n_l}^{n_r}M{\nu }_m.\end{array}$$

(2.41)

From ${\sum }_{n=1}^{\mathrm{\infty }}(a_n^2-1)<+\mathrm{\infty }$,${\sum }_{n=1}^{\mathrm{\infty }}{\nu }_n<+\mathrm{\infty }$ and theboundedness of C, we observe that the right-hand side of (2.41) is bounded.However, the left-hand side of (2.41) is positively unbounded when$r\to \mathrm{\infty }$. Hence, acontradiction. Therefore

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

(2.42)

Using (2.30), we have

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & =& \parallel {\alpha }_n{T}^n{y}_n+(1-{\alpha }_n)x_n-x_n\parallel \\ =& \parallel {\alpha }_n(T^n{y}_n-x_n)\parallel \\ \le & {\alpha }_n\parallel T^n{y}_n-x_n\parallel \\ \le & {\alpha }_n(\parallel T^n{y}_n-T^n{x}_n\parallel +\parallel T^n{x}_n-x_n\parallel )\\ \le & {\alpha }_n(L\parallel y_n-x_n\parallel +\parallel T^n{x}_n-x_n\parallel )\\ \le & {\alpha }_n(1+{\beta }_{n}L)\parallel T^n{x}_n-x_n\parallel .\end{array}$$

(2.43)

Using (2.42), we obtain

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

(2.44)

Observe that

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

(2.45)

Using (2.42) and (2.44), we have

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

(2.46)

Since $\{x_n\}$ is bounded, thesequence $\{T{x}_n\}$ has a convergentsubsequence $\{T{x}_{n_r}\}$ say. Let$T{x}_{n_r}\to q$as $r\to \mathrm{\infty }$. Then$x_{n_r}\to q$as $r\to \mathrm{\infty }$ since

$$0=\underset{r\to \mathrm{\infty }}{lim}[x_{n_r}-T{x}_{n_r}]=\underset{r\to \mathrm{\infty }}{lim}{x}_{n_r}-\underset{r\to \mathrm{\infty }}{lim}T{x}_{n_r}=\underset{r\to \mathrm{\infty }}{lim}{x}_{n_r}-q.$$

(2.47)

By the continuity of T, $T{x}_{n_r}\to Tq$ as$r\to \mathrm{\infty }$ but$T{x}_{n_r}\to q$as $r\to \mathrm{\infty }$. Hence,$q=Tq$.

Hence, ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ has a subsequence whichconverges to the fixed point q of T. Using (2.38), there existssome natural number N, when $n>N$,$(1-{\beta }_n-{\beta }_n^2{L}^2)-{\beta }_n{a}_n^2\ge \frac{1-2b-b^2{L}^2}{2}>0$$\mathrm{\forall }n>N$.Using (2.36), $0\le {\alpha }_n(1+{\beta }_n{a}_n^2){\nu }_n+{\alpha }_n(1+{\beta }_n{a}_n^2)(a_n^2-1){\parallel x_n-q\parallel }^2\le M(a_n^2-1)+M{\nu }_n$.From (2.35),

$${\parallel x_{n+1}-q\parallel }^2\le {\parallel x_n-q\parallel }^2+M(a_n^2-1+{\nu }_n).$$

(2.48)

But ${\sum }_{n=1}^{\mathrm{\infty }}(a_n^2-1)<+\mathrm{\infty }$ and${\sum }_{n=1}^{\mathrm{\infty }}{\nu }_n<+\mathrm{\infty }$ imply that${\sum }_{n=1}^{\mathrm{\infty }}M(a_n^2-1+{\nu }_n)<+\mathrm{\infty }$. From(2.47), it follows that there exists a subsequence ${\{{\parallel x_{n_r}-q\parallel }^2\}}_{r=1}^{\mathrm{\infty }}$ of ${\{{\parallel x_n-q\parallel }^2\}}_{n=1}^{\mathrm{\infty }}$, which converges to 0.Hence, using (2.48) and Lemma 1.2, ${lim}_{n\to \mathrm{\infty }}{\parallel x_n-q\parallel }^2=0$.This means that ${lim}_{n\to \mathrm{\infty }}{x}_n=q$.The proof of the theorem is complete. □

Observe that if ${\nu }_n=0$for all $n\in \mathbb{N}$ inTheorem 2.6, then we obtain Theorem 3 of Qihou [2].

Corollary 2.7 [[2], Theorem 3]

Let H be a Hilbert space, $C\subset H$be nonempty closed bounded and convex; $T:C\to C$be completely continuous and uniformly L-Lipschitzian and asymptotically hemicontractive with sequence$\{a_n\}$,$a_n\in [1,+\mathrm{\infty })$; $\mathrm{\forall }n\in \mathbb{N}$,${\sum }_{n=1}^{\mathrm{\infty }}(a_n-1)<+\mathrm{\infty }$;$\{{\alpha }_n\},\{{\beta }_n\}\in [0,1]$;$ϵ\le {\alpha }_n\le {\beta }_n\le b$for$\mathrm{\forall }n\in \mathbb{N}$,some$ϵ>0$,and some$b\in (0,L^{-2}[{(1+L^2)}^{\frac{1}{2}}-1])$; $x_1\in C$for$\mathrm{\forall }n\in \mathbb{N}$define

$$\{\begin{array}{l}{z}_n={\beta }_n{T}^n{x}_n+(1-{\beta }_n)x_n,\\ x_{n+1}={\alpha }_n{T}^n{z}_n+(1-{\alpha }_n)x_n,n\ge 1.\end{array}$$

Then$\{x_n\}$converges strongly to some fixed point of T.

Since the class of asymptotically pseudocontractive mappings in the intermediatesense is a subclass of the class of asymptotically hemicontractive mappings in theintermediate sense, we obtain the following corollary.

Corollary 2.8 [[1], Theorem 2.1]

Let C be a nonempty, closed and convex subset of a real Hilbert space H and$T:C\to C$be a uniformly L-Lipschitzian and asymptotically pseudocontractive mapping in theintermediate sense with sequences$\{k_n\}\subset [1,\mathrm{\infty })$and$\{{\tau }_n\}\subset [0,\mathrm{\infty })$as defined in (1.11). Assume that the interior of$F(T)$is nonempty. Let$\{x_n\}$be a sequence defined by$x_1=x\in C$and

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

(2.49)

where$\{{\alpha }_n\}$and$\{{\beta }_n\}$are sequences in$(0,1)$. Assume that thefollowing conditions are satisfied:

• (i) ${\sum }_{n=1}^{\mathrm{\infty }}{\tau }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}(q_n^2-1)<\mathrm{\infty }$, where $q_n=2k_n-1$;

• (ii) $a\le {\alpha }_n\le {\beta }_n\le b$ for some $a>0$ and $b\in (0,L^{-2}[\sqrt{1+L^2}-1])$.

Then the sequence$\{x_n\}$generated by (2.49) converges strongly to a fixed point of T.