Convergence theorems on asymptotically demicontractive and hemicontractivemap**s in the intermediate sense

Abstract

In this study, we introduce two classes of nonlinear map**s, the class ofasymptotically demicontractive map**s in the intermediate sense andasymptotically hemicontractive map**s 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 map**.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 map**s. The class ofasymptotically strict pseudocontractive map**s was introduced by Qihou in 1987. Weremark that the class of asymptotically strict pseudocontractive map**s is ageneralization of the class of strict pseudocontractive map**s. 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 map**s in the intermediate sense contains properly the class ofasymptotically strict pseudocontractive map**s;

(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 map**s was introduced in 1991 bySchu [5].

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

(5) asymptotically pseudocontractive map** 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 map**s in the intermediatesense is reduced to the class of asymptotically pseudocontractive map**s;

(6) asymptotically demicontractive map**s [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 map**s [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 map**s in the intermediate sense and asymptoticallyhemicontractive map**s in the intermediate sense as generalizations of the classesof asymptotically demicontractive map**s and asymptotically hemicontractivemap**s, respectively.

(8) The map $T:C\to C$ is said to be an asymptotically demicontractive map** 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 map** 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 map**s.Similarly, Schu [5] proved the convergence ofMann iterative scheme for asymptotically nonexpansive map**s. In this study, weextend the results of Qihou [2] and Schu[5] to the classes of asymptoticallydemicontractive map**s in the intermediate sense and asymptoticallyhemicontractive map**s 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 map**s 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 map**s. Choet al.[8] proved some fixed point theorems for theclass of asymptotically demicontractive map**s in arbitrary real normed linearspaces. Maruster and Maruster [9] introducedthe class of α-demicontractive map**s. They established that thisclass of nonlinear map**s is general than the class of demicontractive map**s.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 map**s. 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 map**s 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 map**s 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 map**s in the intermediate sense and asymptoticallyhemicontractive map**s 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 map** 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 map** 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-map** 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 map** 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 map** 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 map**s in the intermediatesense is a subclass of the class of asymptotically hemicontractive map**s 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 map** 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.