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General composite implicit iteration process for a finite family of asymptotically pseudo-contractive mappings

Fixed Point Theory and Applications20142014:90

https://doi.org/10.1186/1687-1812-2014-90

• Accepted: 25 March 2014
• Published:

Abstract

In this paper, a modified general composite implicit iteration process is used to study the convergence of a finite family of asymptotically nonexpansive mappings. Weak and strong convergence theorems have been proved, in the framework of a Banach space.

MSC:47H09, 47H10.

Keywords

• implicit iteration process
• asymptotically pseudo-contractive mapping
• common fixed points

1 Introduction

Let K be a nonempty subset of a real Banach space E and let $J:E\to {2}^{{E}^{\ast }}$ is the normalized duality mapping defined by
$J\left(x\right)=\left\{f\in {E}^{\ast }:〈x,f〉=\parallel x\parallel \parallel f\parallel ;\parallel x\parallel =\parallel f\parallel \right\},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in E,$

where ${E}^{\ast }$ denotes the dual space of E and $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing.

It is well known that if ${E}^{\ast }$ is strictly convex, then J is single valued.

In the sequel, we shall denote the single valued normalized duality mapping by j.

Let K be a nonempty subset of E. A mapping $T:K\to K$ is said to be L-Lipschitzian if there exists a constant $L>0$ such that for all $x,y\in K$, we have $\parallel Tx-Ty\parallel \le L\parallel x-y\parallel$. It is said to be nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$, for all $x,y\in K$. T is called asymptotically nonexpansive  if there exists a sequence $\left\{{h}_{n}\right\}\subseteq \left[1,\mathrm{\infty }\right)$ with ${lim}_{n\to \mathrm{\infty }}{h}_{n}=1$ such that $\parallel {T}^{n}x-{T}^{n}y\parallel \le {h}_{n}\parallel x-y\parallel$, for all integers $n\ge 1$ and all $x,y\in K$.

A mapping T is said to be pseudo-contractive [2, 3], if there exists $j\left(x-y\right)\in J\left(x-y\right)$ such that $〈Tx-Ty,j\left(x-y\right)〉\le {\parallel x-y\parallel }^{2}$, for all $x,y\in K$. T is called strongly pseudo-contractive, if there exists a constant $\beta \in \left(0,1\right)$, $j\left(x-y\right)\in J\left(x-y\right)$ such that $〈Tx-Ty,j\left(x-y\right)〉\le \beta {\parallel x-y\parallel }^{2}$, for all $x,y\in K$. It is said to be asymptotically pseudo-contractive  if there exists a sequence $\left\{{h}_{n}\right\}\subseteq \left[1,\mathrm{\infty }\right)$ with ${lim}_{n\to \mathrm{\infty }}{h}_{n}=1$ and $j\left(x-y\right)\in J\left(x-y\right)$ such that
$〈{T}^{n}x-{T}^{n}y,j\left(x-y\right)〉\le {h}_{n}{\parallel x-y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in K,\mathrm{\forall }n\ge 1.$
(1.1)
It follows from Kato  that
$\parallel x-y\parallel \le \parallel x-y+r\left[\left({h}_{n}I-{T}^{n}\right)x-\left({h}_{n}I-{T}^{n}\right)y\right]\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in K,\mathrm{\forall }n\ge 1,r>0.$
(1.2)

We use $F\left(T\right)$ to denote the set of fixed points of T; that is, $F\left(T\right)=\left\{x\in K:x=Tx\right\}$.

It follows from the definition that if T is asymptotically nonexpansive, then for all $j\left(x-y\right)\in J\left(x-y\right)$,
$〈{T}^{n}x-{T}^{n}y,j\left(x-y\right)〉=\parallel x-y\parallel \parallel {T}^{n}x-{T}^{n}y\parallel \le {h}_{n}{\parallel x-y\parallel }^{2}.$

Hence every asymptotically nonexpansive mapping is asymptotically pseudo-contractive.

It can be observed from the definition that an asymptotically nonexpansive mapping is uniformly L-Lipschitzian, where $L={sup}_{n\ge 1}\left\{{h}_{n}\right\}$.

Now consider an example of non-Lipschitzian mapping due to Rhoades . Define a mapping $T:\left[0,1\right]\to \left[0,1\right]$ by the formula $Tx={\left\{1-{x}^{\frac{2}{3}}\right\}}^{\frac{3}{2}}$, for $x\in \left[0,1\right]$. Schu  used this example to show that the class of asymptotically nonexpansive mappings is a subclass of the class of pseudo-contractive mappings. Since T is not Lipschitzian, it cannot be asymptotically nonexpansive. Also ${T}^{2}$ is the identity mapping and T is monotonically decreasing, and it follows that
and

Hence T is asymptotically pseudo-contractive mapping with constant sequence $\left\{1\right\}$.

The iterative approximation problems for a nonexpansive mapping, an asymptotically nonexpansive mapping, and an asymptotically pseudo-contractive mapping were studied extensively by Browder , Kirk , Goebel and Kirk , Schu , Xu [9, 10], Liu  in the setting of Hilbert space or uniformly convex Banach space.

In 2001, Xu and Ori  introduced the following implicit iteration process for a finite family of nonexpansive self-mappings in Hilbert space:
$\left\{\begin{array}{l}{x}_{0}\in K\phantom{\rule{1em}{0ex}}\text{arbitrary},\\ {x}_{n}={\alpha }_{n}{x}_{n-1}+\left(1-{\alpha }_{n}\right){T}_{n}{x}_{n},\phantom{\rule{1em}{0ex}}n\ge 1,\end{array}$
(1.3)

where $\left\{{\alpha }_{n}\right\}$ be a sequence in $\left(0,1\right)$ and ${T}_{n}={T}_{nmodN}$. They proved in  that the sequence $\left\{{x}_{n}\right\}$ converges weakly to a common fixed point of ${T}_{n}$, $n=1,2,\dots ,N$.

Later on Osilike and Akuchu , and Chen et al.  extended the iteration process (1.3) to a finite family of asymptotically pseudo-contractive mapping and a finite family of continuous pseudo-contractive self-mapping, respectively. Zhou and Chang  studied the convergence of a modified implicit iteration process to the common fixed point of a finite family of asymptotically nonexpansive mappings. Then Su and Li , and Su and Qin  introduced the composite implicit iteration process and the general iteration algorithm, respectively, which properly include the implicit iteration process. Recently, Beg and Thakur  introduced a modified general composite implicit iteration process for a finite family of random asymptotically nonexpansive mapping and proved strong convergence theorems.

The purpose of this paper is to consider a finite family ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ of asymptotically pseudo-contractive mappings and to establish convergence results in Banach spaces based on the modified general composite implicit iteration:

For ${x}_{0}\in K$, construct a sequence $\left\{{x}_{n}\right\}$ by
$\begin{array}{r}{x}_{n}={\alpha }_{n}{x}_{n-1}+\left(1-{\alpha }_{n}\right){T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n},\\ {y}_{n}={r}_{n}{x}_{n}+{s}_{n}{x}_{n-1}+{t}_{n}{T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}+{w}_{n}{T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n-1}\end{array}$
(1.4)

for each $n\ge 1$, which can be written as $n=\left(k\left(n\right)-1\right)N+i\left(n\right)$, where $i\left(n\right)=1,2,\dots ,N$ and $k\left(n\right)\ge 1$ is a positive integer, with $k\left(n\right)\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$. The sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{r}_{n}\right\}$, $\left\{{s}_{n}\right\}$, $\left\{{t}_{n}\right\}$ and $\left\{{w}_{n}\right\}$ are in $\left(0,1\right)$ such that ${r}_{n}+{s}_{n}+{t}_{n}+{w}_{n}=1$ for all $n\ge 1$.

2 Preliminaries

In what follows we shall use the following results.

Lemma 2.1 

Let E be a Banach space, K be a nonempty closed convex subset of E, and $T:K\to K$ be a continuous and strong pseudo-contraction. Then T has a unique fixed point.

Lemma 2.2 

Let $\left\{{a}_{n}\right\}$, $\left\{{b}_{n}\right\}$, and $\left\{{c}_{n}\right\}$ be three nonnegative sequences satisfying the following condition:

where ${n}_{0}$ is some nonnegative integer, ${\sum }_{n=0}^{\mathrm{\infty }}{b}_{n}<\mathrm{\infty }$ and ${\sum }_{n=0}^{\mathrm{\infty }}{c}_{n}<\mathrm{\infty }$.

Then
1. (i)

${lim}_{n\to \mathrm{\infty }}{a}_{n}$ exists;

2. (ii)

if, in addition, there exists a subsequence $\left\{{a}_{{n}_{i}}\right\}\subset \left\{{a}_{n}\right\}$ such that ${a}_{{n}_{i}}\to 0$, then ${a}_{n}\to 0$ as $n\to \mathrm{\infty }$.

Lemma 2.3 

Let E be a uniformly convex Banach space and let a, b be two constants with $0. Suppose that $\left\{{t}_{n}\right\}\subset \left[a,b\right]$ is a real sequence and $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ are two sequences in E. Then the conditions
$\underset{n\to \mathrm{\infty }}{lim sup}\parallel {x}_{n}\parallel \le d,\phantom{\rule{2em}{0ex}}\underset{n\to \mathrm{\infty }}{lim sup}\parallel {y}_{n}\parallel \le d\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\parallel {t}_{n}{x}_{n}+\left(1-{t}_{n}\right){y}_{n}\parallel =d$

imply that ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{y}_{n}\parallel =0$, where $d\ge 0$ is some constant.

Lemma 2.4 

Let E be a reflexive smooth Banach space with a weakly sequential continuous duality mapping J. Let K be a nonempty bounded and closed convex subset of E and $T:K\to K$ be a uniformly L-Lipschitzian and asymptotical pseudo-contraction. Then $I-T$ is demiclosed at zero, where I is the identical mapping.

We shall denote weak convergence by and strong convergence by →.

A Banach space E is said to satisfy Opial’s condition if for any sequence $\left\{{x}_{n}\right\}\in E$, ${x}_{n}⇀x$ as $n\to \mathrm{\infty }$ implies

We know that a Banach space with a sequentially continuous duality mapping satisfies Opial’s condition (for details, see ).

3 The main results

Throughout this section, E is a uniformly convex Banach space, K a nonempty closed convex subset of E. denotes the set of natural numbers and $I=\left\{1,2,\dots ,N\right\}$, the set of the first N natural numbers. ${T}_{i}$ ($i\in I$) are N uniformly Lipschitzian asymptotically pseudo-contractive self-mappings on K. Let $\mathcal{F}={\bigcap }_{i\in I}F\left({T}_{i}\right)\ne \mathrm{\varnothing }$.

Since ${T}_{i}$ ($i\in I$) are uniformly Lipschitzian, there exist constants ${L}_{i}>0$ such that $\parallel {T}_{i}^{n}x-{T}_{i}^{n}y\parallel \le {L}_{i}\parallel x-y\parallel$, for all $x,y\in K$, $n\in \mathbb{N}$ and $i\in I$. Also, since ${T}_{i}$ ($i\in I$) are asymptotically pseudo-contractive; therefore there exist sequences $\left\{{h}_{n}^{\left(i\right)}\right\}$ such that $〈{T}_{i}^{n}x-{T}_{i}^{n}y,j\left(x-y\right)〉\le {h}_{n}^{\left(i\right)}{\parallel x-y\parallel }^{2}$ for all $x,y\in K$ and $i\in I$.

Take $L={max}_{i\in I}\left({L}_{i}\right)$ and ${h}_{n}={max}_{i\in I}\left({h}_{n}^{\left(i\right)}\right)$.

Before presenting the main results, we first show that the proposed iteration (1.4) is well defined.

Let T be uniformly Lipschitzian asymptotically pseudo-contractive mapping. For every fixed $u\in K$ and $\alpha \in \left(\frac{L+{L}^{2}}{L+{L}^{2}+1},1\right)$, define a mapping ${S}_{n}:K\to K$ by the formula
(3.1)

where $\alpha ,r,s,t,w\in \left(0,1\right)$, with $\left(1-\alpha \right)\left(L+{L}^{2}\right)<1$.

Then, for all $x,y\in K$, $j\left(x-y\right)\in J\left(x-y\right)$, we have
(3.2)
Now
$\begin{array}{rl}〈{T}^{n}a-{T}^{n}b,j\left(x-y\right)〉& =\parallel {T}^{n}a-{T}^{n}b\parallel \parallel x-y\parallel \\ \le L\parallel a-b\parallel \parallel x-y\parallel \\ =L\parallel r\left(x-y\right)+t\left({T}^{n}x-{T}^{n}y\right)\parallel \parallel x-y\parallel \\ \le L\left(r\parallel x-y\parallel +tL\parallel x-y\parallel \right)\parallel x-y\parallel \\ =\left(Lr+t{L}^{2}\right){\parallel x-y\parallel }^{2}\\ \le \left(L+{L}^{2}\right){\parallel x-y\parallel }^{2},\end{array}$
so
$\begin{array}{rl}〈{S}_{n}x-{S}_{n}y,j\left(x-y\right)〉& =\left(1-\alpha \right)〈{T}^{n}a-{T}^{n}b,j\left(x-y\right)〉\\ \le \left(1-\alpha \right)\left(L+{L}^{2}\right){\parallel x-y\parallel }^{2}.\end{array}$
Since $\left(1-\alpha \right)\left(L+{L}^{2}\right)\in \left(0,1\right)$, ${S}_{n}$ is strongly pseudo-contractive, which is also continuous, by Lemma 2.1, ${S}_{n}$ has a unique fixed point ${x}^{\ast }\in K$, i.e.
(3.3)

Thus the implicit iteration (1.4) is defined in K for a finite family $\left\{{T}_{i}\right\}$ of uniformly Lipschitzian asymptotically pseudo-contractive self-mappings on K, provided ${\alpha }_{n}\in \left(\alpha ,1\right)$, where $\alpha =\frac{L+{L}^{2}}{L+{L}^{2}+1}$, for all $n\in \mathbb{N}$, $L={max}_{i\in I}\left({L}_{i}\right)$.

Lemma 3.1 Let E, K, and ${T}_{i}$ ($i\in I$) be as defined above and let $\left\{{x}_{n}\right\}$ be the sequence defined by (1.4), where $\left\{{\alpha }_{n}\right\}$ is a sequence of real numbers such that $0<\alpha <{\alpha }_{n}\le \beta <1$ for $\alpha =\frac{L+{L}^{2}}{L+{L}^{2}+1}$ and β is some constant and satisfying the conditions ${\sum }_{n=1}^{\mathrm{\infty }}\left(1-{\alpha }_{n}\right)<\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}\frac{{h}_{n}-1}{1-{\alpha }_{n}}=0$. Let $b>0$ be a real number such that ${t}_{n}+{w}_{n}\le b/L<1$. Then
1. (i)

${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-p\parallel$ exists, for all $p\in \mathcal{F}$,

2. (ii)

${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},\mathcal{F}\right)$ exists, where $d\left({x}_{n},\mathcal{F}\right)={inf}_{p\in \mathcal{F}}\parallel {x}_{n}-p\parallel$,

3. (iii)

${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{T}_{l}{x}_{n}\parallel =0$, $\mathrm{\forall }l\in I$.

Proof Let $p\in \mathcal{F}$. Using (1.4), we have
$\begin{array}{rcl}{\parallel {x}_{n}-p\parallel }^{2}& =& 〈{x}_{n}-p,j\left({x}_{n}-p\right)〉\\ \le & {\alpha }_{n}〈{x}_{n-1}-p,j\left({x}_{n}-p\right)〉+\left(1-{\alpha }_{n}\right)〈{T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}-{T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n},j\left({x}_{n}-p\right)〉\\ +\left(1-{\alpha }_{n}\right){h}_{k\left(n\right)}{\parallel {x}_{n}-p\parallel }^{2}\\ =& {\alpha }_{n}\parallel {x}_{n-1}-p\parallel \parallel {x}_{n}-p\parallel +\left(1-{\alpha }_{n}\right)L\parallel {y}_{n}-{x}_{n}\parallel \parallel {x}_{n}-p\parallel \\ +\left(1-{\alpha }_{n}\right){h}_{k\left(n\right)}{\parallel {x}_{n}-p\parallel }^{2}.\end{array}$
(3.4)
Using (1.4), we obtain
$\begin{array}{rcl}\parallel {y}_{n}-{x}_{n}\parallel & =& \parallel {s}_{n}\left({x}_{n-1}-{x}_{n}\right)+{t}_{n}\left({T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-{x}_{n}\right)+{w}_{n}\left({T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n-1}-{x}_{n}\right)\parallel \\ \le & {s}_{n}\parallel {x}_{n-1}-p\parallel +{s}_{n}\parallel {x}_{n}-p\parallel +{t}_{n}L\parallel {x}_{n}-p\parallel +{t}_{n}\parallel {x}_{n}-p\parallel \\ +{w}_{n}L\parallel {x}_{n-1}-p\parallel +{w}_{n}\parallel {x}_{n}-p\parallel .\end{array}$
(3.5)
Substituting (3.5) in (3.4), we get
$\begin{array}{rcl}{\parallel {x}_{n}-p\parallel }^{2}& \le & \left({\alpha }_{n}+\left(1-{\alpha }_{n}\right)L\left({s}_{n}+{w}_{n}L\right)\right)\parallel {x}_{n-1}-p\parallel \parallel {x}_{n}-p\parallel \\ +\left(1-{\alpha }_{n}\right)\left[\left({s}_{n}+{t}_{n}+{w}_{n}+{t}_{n}L\right)L+{h}_{k\left(n\right)}\right]{\parallel {x}_{n}-p\parallel }^{2}\\ \le & \left({\alpha }_{n}+\left(1-{\alpha }_{n}\right)\left(1+L\right)L\right)\parallel {x}_{n-1}-p\parallel \parallel {x}_{n}-p\parallel \\ +\left(1-{\alpha }_{n}\right)\left[\left(1+L\right)L+{h}_{k\left(n\right)}\right]{\parallel {x}_{n}-p\parallel }^{2}\\ \le & \left({\alpha }_{n}+\left(1-{\alpha }_{n}\right)\left(1+L\right)L\right)\parallel {x}_{n-1}-p\parallel \parallel {x}_{n}-p\parallel \\ +\left[\left(1-{\alpha }_{n}\right)\left(1+L\right)L+\left(1-{\alpha }_{n}+{\mu }_{k\left(n\right)}\right)\right]{\parallel {x}_{n}-p\parallel }^{2},\end{array}$
(3.6)

where ${\mu }_{k\left(n\right)}={h}_{k\left(n\right)}-1$ for all $n\ge 1$, by condition ${\sum }_{n=1}^{\mathrm{\infty }}\left({h}_{k\left(n\right)}-1\right)<\mathrm{\infty }$, we have ${\sum }_{n=1}^{\mathrm{\infty }}{\mu }_{k\left(n\right)}<\mathrm{\infty }$.

Therefore, we have
$\begin{array}{rl}\parallel {x}_{n}-p\parallel & \le \frac{\left({\alpha }_{n}+\left(1-{\alpha }_{n}\right)\left(1+L\right)L\right)}{{\alpha }_{n}-{\mu }_{k\left(n\right)}-\left(1-{\alpha }_{n}\right)\left(1+L\right)L}\parallel {x}_{n-1}-p\parallel \\ \le \left[1+\frac{{\mu }_{k\left(n\right)}+2\left(1-{\alpha }_{n}\right)\left(1+L\right)L}{{\alpha }_{n}-{\mu }_{k\left(n\right)}-\left(1-{\alpha }_{n}\right)\left(1+L\right)L}\right]\parallel {x}_{n-1}-p\parallel \\ \le \left[1+\frac{{\mu }_{k\left(n\right)}+2\left(1-{\alpha }_{n}\right)\left(1+L\right)L}{1-\left(1-{\alpha }_{n}+{\mu }_{k\left(n\right)}+\left(1-{\alpha }_{n}\right)\left(1+L\right)L\right)}\right]\parallel {x}_{n-1}-p\parallel .\end{array}$
(3.7)

Since ${lim}_{n\to \mathrm{\infty }}\frac{{h}_{k\left(n\right)}-1}{1-{\alpha }_{n}}={lim}_{n\to \mathrm{\infty }}\frac{{\mu }_{k\left(n\right)}}{1-{\alpha }_{n}}=0$, there exists a M such that $\frac{{\mu }_{k\left(n\right)}}{1-{\alpha }_{n}}.

Now, we consider the second term on the right side of (3.7). We have
$\left(1-{\alpha }_{n}+{\mu }_{k\left(n\right)}+\left(1-{\alpha }_{n}\right)\left(1+L\right)L\right)\le \left(1-{\alpha }_{n}\right)\left[1+M+\left(1+L\right)L\right].$
By condition ${\sum }_{n=1}^{\mathrm{\infty }}\left(1-{\alpha }_{n}\right)<\mathrm{\infty }$, we have ${lim}_{n\to \mathrm{\infty }}\left(1-{\alpha }_{n}\right)=0$, then there exists a natural number ${N}_{1}$ such that if $n>{N}_{1}$, then
$1-\left(1-{\alpha }_{n}+{\mu }_{k\left(n\right)}+\left(1-{\alpha }_{n}\right)\left(1+L\right)L\right)\ge \frac{1}{2}.$
Therefore, it follows from (3.7) that
$\begin{array}{rl}\parallel {x}_{n}-p\parallel & \le \left[1+2\left\{{\mu }_{k\left(n\right)}+2\left(1-{\alpha }_{n}\right)\left(1+L\right)L\right\}\right]\parallel {x}_{n-1}-p\parallel \\ =\left(1+{\sigma }_{n}\right)\parallel {x}_{n-1}-p\parallel ,\end{array}$
(3.8)

where ${\sigma }_{n}=2\left\{{\mu }_{k\left(n\right)}+2\left(1-{\alpha }_{n}\right)\left(1+L\right)L\right\}$.

Taking the infimum over $p\in \mathcal{F}$, we have
$d\left({x}_{n},\mathcal{F}\right)\le \left(1+{\sigma }_{n}\right)d\left({x}_{n-1},\mathcal{F}\right).$
(3.9)
Since ${\sum }_{n=1}^{\mathrm{\infty }}{\mu }_{k\left(n\right)}<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}\left(1-{\alpha }_{n}\right)<\mathrm{\infty }$, we have
$\sum _{n=1}^{\mathrm{\infty }}{\sigma }_{n}<\mathrm{\infty }.$

Thus, by Lemma 2.2, ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-p\parallel$ and ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},\mathcal{F}\right)$ exist.

Without loss of generality, we assume
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-p\parallel ={d}^{1}.$
(3.10)
Set ${v}_{k\left(n\right)}=\frac{{h}_{k\left(n\right)}-1}{{h}_{k\left(n\right)}}$, and from (1.2), we have
$\begin{array}{rcl}\parallel {x}_{n}-p\parallel & \le & \parallel {x}_{n}-p+\frac{1-{\alpha }_{n}}{2{\alpha }_{n}{h}_{k\left(n\right)}}\left[\left({h}_{k\left(n\right)}I-{T}_{i\left(n\right)}^{k\left(n\right)}\right){x}_{n}-\left({h}_{k\left(n\right)}I-{T}_{i\left(n\right)}^{k\left(n\right)}\right)p\right]\parallel \\ \le & \parallel {x}_{n}-p+\frac{1-{\alpha }_{n}}{2{\alpha }_{n}}\left[{\alpha }_{n}\left({x}_{n-1}-{T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}\right)+\left(1-{\alpha }_{n}\right)\left({T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}-{T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}\right)\right]\parallel \\ +\left(\frac{1-{\alpha }_{n}}{2{\alpha }_{n}}\right)\left(\frac{{h}_{k\left(n\right)}-1}{{h}_{k\left(n\right)}}\right)\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-p\parallel \\ =& \parallel {x}_{n}-p+\frac{1-{\alpha }_{n}}{2}\left({x}_{n-1}-{T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}\right)+\frac{{\left(1-{\alpha }_{n}\right)}^{2}}{2{\alpha }_{n}}\left({T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}-{T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}\right)\parallel \\ +\left(\frac{1-{\alpha }_{n}}{2{\alpha }_{n}}\right){v}_{k\left(n\right)}\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-p\parallel \\ \le & \parallel {x}_{n}-p+\frac{1}{2}\left({x}_{n-1}-{x}_{n}\right)\parallel +\left(\frac{1-{\alpha }_{n}}{2{\alpha }_{n}}\right){v}_{k\left(n\right)}\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-p\parallel \\ +\frac{{\left(1-{\alpha }_{n}\right)}^{2}}{2{\alpha }_{n}}L\parallel {y}_{n}-{x}_{n}\parallel \\ \le & \parallel \frac{1}{2}\left({x}_{n}-p\right)+\frac{1}{2}\left({x}_{n-1}-p\right)\parallel +\left(\frac{1-{\alpha }_{n}}{2{\alpha }_{n}}\right){v}_{k\left(n\right)}\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-p\parallel \\ +\frac{{\left(1-{\alpha }_{n}\right)}^{2}}{2{\alpha }_{n}}L\parallel {y}_{n}-{x}_{n}\parallel .\end{array}$
Thus
$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim inf}\parallel {x}_{n}-p\parallel & \le & \underset{n\to \mathrm{\infty }}{lim inf}\parallel \frac{1}{2}\left({x}_{n}-p\right)+\frac{1}{2}\left({x}_{n-1}-p\right)\parallel \\ +\underset{n\to \mathrm{\infty }}{lim inf}\left(\frac{1-{\alpha }_{n}}{2{\alpha }_{n}}\right){v}_{k\left(n\right)}\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-p\parallel \\ +\underset{n\to \mathrm{\infty }}{lim inf}\frac{{\left(1-{\alpha }_{n}\right)}^{2}}{2{\alpha }_{n}}L\parallel {y}_{n}-{x}_{n}\parallel .\end{array}$
Since ${v}_{k\left(n\right)}=\frac{{h}_{k\left(n\right)}-1}{{h}_{k\left(n\right)}}\in \left(0,1\right)$, we have ${lim}_{n\to \mathrm{\infty }}{v}_{k\left(n\right)}=0$ and from ${\sum }_{n=1}^{\mathrm{\infty }}\left(1-{\alpha }_{n}\right)<\mathrm{\infty }$, we have ${lim}_{n\to \mathrm{\infty }}\left(1-{\alpha }_{n}\right)=0$ and using (3.10), we have
$\underset{n\to \mathrm{\infty }}{lim inf}\parallel \frac{1}{2}\left({x}_{n}-p\right)+\frac{1}{2}\left({x}_{n-1}-p\right)\parallel \ge {d}^{1}.$
(3.11)
On the other hand, we obtain
$\underset{n\to \mathrm{\infty }}{lim sup}\parallel \frac{1}{2}\left({x}_{n}-p\right)+\frac{1}{2}\left({x}_{n-1}-p\right)\parallel \le \underset{n\to \mathrm{\infty }}{lim sup}\left[\frac{1}{2}\parallel {x}_{n}-p\parallel +\frac{1}{2}\parallel {x}_{n-1}-p\parallel \right]={d}^{1},$
(3.12)
from (3.11) and (3.12), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel \frac{1}{2}\left({x}_{n}-p\right)+\frac{1}{2}\left({x}_{n-1}-p\right)\parallel ={d}^{1}.$
It follows from Lemma 2.3 that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{x}_{n-1}\parallel =0.$
(3.13)
Thus, for any $i\in I$, we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{x}_{n+i}\parallel =0.$
(3.14)
Since $0<\alpha <{\alpha }_{n}\le \beta <1$ and from (1.4) and (3.13), we get
$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}\parallel & =& \underset{n\to \mathrm{\infty }}{lim}\frac{{\alpha }_{n}}{1-{\alpha }_{n}}\parallel {x}_{n}-{x}_{n-1}\parallel \\ \le & \frac{1}{1-\beta }\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{x}_{n-1}\parallel =0.\end{array}$
(3.15)
On the other hand, from (3.13) and (3.15)
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n-1}-{T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}\parallel \le \underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n-1}-{x}_{n}\parallel +\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}\parallel =0.$
(3.16)
Now,
$\begin{array}{rl}\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-{x}_{n}\parallel & \le \parallel {x}_{n}-{x}_{n-1}\parallel +\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}-{x}_{n-1}\parallel +\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}-{T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}\parallel \\ \le \left(1+L\right)\parallel {x}_{n}-{x}_{n-1}\parallel +\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}-{x}_{n-1}\parallel +L\parallel {y}_{n}-{x}_{n-1}\parallel .\end{array}$
(3.17)
Again, by using (1.4), we obtain
$\begin{array}{rl}\parallel {y}_{n}-{x}_{n-1}\parallel & \le \parallel {r}_{n}{x}_{n}+{s}_{n}{x}_{n-1}+{t}_{n}{T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}+{w}_{n}{T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n-1}-{x}_{n-1}\parallel \\ \le {t}_{n}\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-{x}_{n}\parallel +{w}_{n}\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n-1}-{x}_{n}\parallel +\left({r}_{n}+{t}_{n}+{w}_{n}\right)\parallel {x}_{n}-{x}_{n-1}\parallel \\ \le \left({t}_{n}+{w}_{n}\right)\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-{x}_{n}\parallel +\left({r}_{n}+{t}_{n}+{w}_{n}+{w}_{n}L\right)\parallel {x}_{n}-{x}_{n-1}\parallel .\end{array}$
(3.18)
Substituting (3.18) into (3.17), we get
$\begin{array}{rcl}\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-{x}_{n}\parallel & \le & \left(1+L\right)\parallel {x}_{n}-{x}_{n-1}\parallel +\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}-{x}_{n-1}\parallel +L\left({t}_{n}+{w}_{n}\right)\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-{x}_{n}\parallel \\ +L\left({r}_{n}+{t}_{n}+{w}_{n}+{w}_{n}L\right)\parallel {x}_{n}-{x}_{n-1}\parallel .\end{array}$
Since ${t}_{n}+{w}_{n}\le b/L<1$, the above inequality gives
$\left(1-b\right)\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-{x}_{n}\parallel \le \left[1+L\left(1+{r}_{n}+{t}_{n}+{w}_{n}+{w}_{n}L\right)\right]\parallel {x}_{n}-{x}_{n-1}\parallel +\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}-{x}_{n-1}\parallel .$
Then from (3.13), (3.16), and the above inequality, we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{x}_{n}-{x}_{n}\parallel =0.$
(3.19)
From (3.13), (3.18), and (3.19), we get
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{x}_{n-1}\parallel =0.$
(3.20)
On the other hand, from (3.13) and (3.20) we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{x}_{n}\parallel \le \underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{x}_{n-1}\parallel +\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n-1}-{x}_{n}\parallel =0.$
(3.21)

Since for any positive integer $n>N$, we can write $n=\left(k\left(n\right)-1\right)N+i\left(n\right)$, $i\left(n\right)\in I$.

Let ${\mathcal{A}}_{n}=\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}-{x}_{n-1}\parallel$, then from (3.16), we have ${\mathcal{A}}_{n}\to 0$. Also,
$\begin{array}{rcl}\parallel {x}_{n-1}-{T}_{n}{x}_{n}\parallel & \le & \parallel {x}_{n-1}-{T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}\parallel +\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}-{T}_{n}{x}_{n}\parallel \\ =& {\mathcal{A}}_{n}+\parallel {T}_{i\left(n\right)}^{k\left(n\right)}{y}_{n}-{T}_{i\left(n\right)}{x}_{n}\parallel \le {\mathcal{A}}_{n}+L\parallel {T}_{i\left(n\right)}^{k\left(n\right)-1}{y}_{n}-{x}_{n}\parallel \\ \le & {\mathcal{A}}_{n}+L\left\{\parallel {T}_{i\left(n\right)}^{k\left(n\right)-1}{y}_{n}-{T}_{i\left(n-N\right)}^{k\left(n\right)-1}{x}_{n-N}\parallel \\ +\parallel {T}_{i\left(n-N\right)}^{k\left(n\right)-1}{x}_{n-N}-{T}_{i\left(n-N\right)}^{k\left(n\right)-1}{y}_{n-N}\parallel \\ +\parallel {T}_{i\left(n-N\right)}^{k\left(n\right)-1}{y}_{n-N}-{x}_{\left(n-N\right)-1}\parallel +\parallel {x}_{\left(n-N\right)-1}-{x}_{n}\parallel \right\}.\end{array}$
(3.22)
Since for each $n>N$, $n=\left(n-N\right)\left(modN\right)$ and $n=\left(k\left(n\right)-1\right)N+i\left(n\right)$, $n-N=\left(\left(k\left(n\right)-1\right)-1\right)N+i\left(n\right)=\left(k\left(n-N\right)-1\right)N+i\left(n-N\right)$, i.e.
$k\left(n-N\right)=k\left(n\right)-1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}i\left(n-N\right)=i\left(n\right).$
Therefore from (3.22), we have
$\begin{array}{rcl}\parallel {x}_{n-1}-{T}_{n}{x}_{n}\parallel & \le & {\mathcal{A}}_{n}+L\left\{\parallel {T}_{i\left(n\right)}^{k\left(n\right)-1}{y}_{n}-{T}_{i\left(n\right)}^{k\left(n\right)-1}{x}_{n-N}\parallel \\ +\parallel {T}_{i\left(n-N\right)}^{k\left(n-N\right)}{x}_{n-N}-{T}_{i\left(n-N\right)}^{k\left(n-N\right)}{y}_{n-N}\parallel \\ +\parallel {T}_{i\left(n-N\right)}^{k\left(n-N\right)}{y}_{n-N}-{x}_{\left(n-N\right)-1}\parallel +\parallel {x}_{\left(n-N\right)-1}-{x}_{n}\parallel \right\}\\ \le & {\mathcal{A}}_{n}+L\left\{L\parallel {y}_{n}-{x}_{n-N}\parallel +L\parallel {x}_{n-N}-{y}_{n-N}\parallel \\ +{\mathcal{A}}_{n-N}+\parallel {x}_{\left(n-N\right)-1}-{x}_{n}\parallel \right\}\\ \le & {\mathcal{A}}_{n}+{L}^{2}\left(\parallel {y}_{n}-{x}_{n}\parallel +\parallel {x}_{n}-{x}_{n-N}\parallel +\parallel {x}_{n-N}-{y}_{n-N}\parallel \right)\\ +L\left({\mathcal{A}}_{n-N}+\parallel {x}_{\left(n-N\right)-1}-{x}_{n}\parallel \right).\end{array}$
(3.23)
From (3.14), (3.21), and ${\mathcal{A}}_{n}\to 0$, we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n-1}-{T}_{n}{x}_{n}\parallel =0.$
(3.24)
It follows from (3.13) and (3.24) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{T}_{n}{x}_{n}\parallel \le \underset{n\to \mathrm{\infty }}{lim}\left\{\parallel {x}_{n}-{x}_{n-1}\parallel +\parallel {x}_{n-1}-{T}_{n}{x}_{n}\parallel \right\}=0.$
(3.25)
Consequently, for any $i\in I$, from (3.14), (3.25), we obtain
$\begin{array}{rl}\parallel {x}_{n}-{T}_{n+i}{x}_{n}\parallel & \le \parallel {x}_{n}-{x}_{n+i}\parallel +\parallel {x}_{n+i}-{T}_{n+i}{x}_{n+i}\parallel +\parallel {T}_{n+i}{x}_{n+i}-{T}_{n+i}{x}_{n}\parallel \\ \le \left(1+L\right)\parallel {x}_{n}-{x}_{n+i}\parallel +\parallel {x}_{n+i}-{T}_{n+i}{x}_{n+i}\parallel \to 0,\end{array}$
as $n\to \mathrm{\infty }$. This implies that the sequence
Since for each $l=1,2,\dots ,N$, $\left\{\parallel {x}_{n}-{T}_{l}{x}_{n}\parallel \right\}$ is a subsequence of ${\bigcup }_{i=1}^{N}\left\{\parallel {x}_{n}-{T}_{n+i}{x}_{n}\parallel \right\}$, therefore, we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{T}_{l}{x}_{n}\parallel =0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }l\in I.$
(3.26)

This completes the proof. □

3.1 Strong convergence theorems

First, we prove necessary and sufficient conditions for the strong convergence of the modified general composite implicit iteration process to a common fixed point of a finite family of asymptotically pseudo-contractive mappings.

Theorem 3.1 Let E, K, and ${T}_{i}$ ($i\in I$) be as defined above and $\left\{{\alpha }_{n}\right\}$ be a sequence of real numbers as in Lemma  3.1. Then the sequence $\left\{{x}_{n}\right\}$ generated by (1.4) converges strongly to a member of if and only if ${lim inf}_{n\to \mathrm{\infty }}d\left({x}_{n},\mathcal{F}\right)=0$.

Proof The necessity of the condition is obvious. Thus, we will only prove the sufficiency.

Let ${lim inf}_{n\to \mathrm{\infty }}d\left({x}_{n},\mathcal{F}\right)=0$. Then from (ii) in Lemma 3.1, we have ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},\mathcal{F}\right)=0$.

Next, we show that $\left\{{x}_{n}\right\}$ is a Cauchy sequence in K. For any given $\epsilon >0$, since ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},\mathcal{F}\right)=0$, there exists a natural number ${n}_{1}$ such that $d\left({x}_{n},\mathcal{F}\right)<\epsilon /4$ when $n\ge {n}_{1}$.

Since ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-p\parallel$ exists for all $p\in \mathcal{F}$, we have $\parallel {x}_{n}-p\parallel <{M}^{\prime }$, for all $n\ge 1$ and some positive number ${M}^{\prime }$.

Furthermore ${\sum }_{n=1}^{\mathrm{\infty }}{\sigma }_{n}<\mathrm{\infty }$ implies that there exists a positive integer ${n}_{2}$ such that ${\sum }_{j=n}^{\mathrm{\infty }}{\sigma }_{j}<\epsilon /4{M}^{\prime }$ for all $n\ge {n}_{2}$. Let ${N}^{\prime }=max\left\{{n}_{1},{n}_{2}\right\}$. It follows from (3.8) that
$\parallel {x}_{n}-p\parallel \le \parallel {x}_{n-1}-p\parallel +{M}^{\prime }{\sigma }_{n}.$
Now, for all $n,m\ge {N}^{\prime }$ and for all $p\in \mathcal{F}$, we have
$\begin{array}{rl}\parallel {x}_{n}-{x}_{m}\parallel & \le \parallel {x}_{n}-p\parallel +\parallel {x}_{m}-p\parallel \\ \le \parallel {x}_{{N}^{\prime }}-p\parallel +{M}^{\prime }\sum _{j={N}^{\prime }+1}^{n}{\sigma }_{j}+\parallel {x}_{{N}^{\prime }}-p\parallel +{M}^{\prime }\sum _{j={N}^{\prime }+1}^{m}{\sigma }_{j}\\ \le 2\parallel {x}_{{N}^{\prime }}-p\parallel +2{M}^{\prime }\sum _{j={N}^{\prime }}^{\mathrm{\infty }}{\sigma }_{j}.\end{array}$
Taking the infimum over all $p\in \mathcal{F}$, we obtain
$\parallel {x}_{n}-{x}_{m}\parallel \le 2d\left({x}_{{N}^{\prime }},\mathcal{F}\right)+2{M}^{\prime }\sum _{j={N}^{\prime }}^{\mathrm{\infty }}{\sigma }_{j}<\epsilon .$

This implies that $\left\{{x}_{n}\right\}$ is a Cauchy sequence. Since E is complete, therefore $\left\{{x}_{n}\right\}$ is convergent.

Suppose ${lim}_{n\to \mathrm{\infty }}{x}_{n}=q$.

Since K is closed, we get $q\in K$, then $\left\{{x}_{n}\right\}$ converges strongly to q.

It remains to show that $q\in \mathcal{F}$.

Notice that
$|d\left(q,\mathcal{F}\right)-d\left({x}_{n},\mathcal{F}\right)|\le \parallel q-{x}_{n}\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N},$

since ${lim}_{n\to \mathrm{\infty }}{x}_{n}=q$ and ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},\mathcal{F}\right)=0$, we obtain $q\in \mathcal{F}$.

This completes the proof. □

Corollary 3.1 Suppose that the conditions are the same as in Theorem  3.1. Then the sequence $\left\{{x}_{n}\right\}$ generated by (1.4) converges strongly to $u\in \mathcal{F}$ if and only if $\left\{{x}_{n}\right\}$ has a subsequence $\left\{{x}_{{n}_{j}}\right\}$ which converges strongly to $u\in \mathcal{F}$.

A mapping $T:K\to K$ with $F\left(T\right)\ne \mathrm{\varnothing }$ is said to satisfy condition (A)  on K if there exists a nondecreasing function $f:\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$, with $f\left(0\right)=0$ and $f\left(r\right)>r$, for all $r\in \left(0,\mathrm{\infty }\right)$, such that for all $x\in K$,
$\parallel x-Tx\parallel \ge f\left(d\left(x,F\left(T\right)\right)\right).$
A family ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ of N self-mappings of K with $\mathcal{F}={\bigcap }_{i\in I}F\left({T}_{i}\right)\ne \mathrm{\varnothing }$ is said to satisfy
1. (1)
condition (B) on K  if there is a nondecreasing function $f:\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ with $f\left(0\right)=0$ and $f\left(r\right)>r$ for all $r\in \left(0,\mathrm{\infty }\right)$ such that for all $x\in K$ such that
$\underset{1\le l\le N}{max}\left\{\parallel x-{T}_{l}x\parallel \right\}\ge f\left(d\left(x,\mathcal{F}\right)\right);$

2. (2)
condition ($\overline{\mathrm{C}}$) on K  if there is a nondecreasing function $f:\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ with $f\left(0\right)=0$ and $f\left(r\right)>r$ for all $r\in \left(0,\mathrm{\infty }\right)$ such that for all $x\in K$ such that
$\left\{\parallel x-{T}_{l}x\parallel \right\}\ge f\left(d\left(x,\mathcal{F}\right)\right)$

for at least one ${T}_{l}$, $l=1,2,\dots ,N$ or, in other words, at least one of the ${T}_{l}$’s satisfies condition (A).

Condition (B) reduces to condition (A) when all but one of the ${T}_{l}$’s are identities. Also condition (B) and condition ($\overline{\mathrm{C}}$) are equivalent (see ).

Senter and Dotson  established a relation between condition (A) and demicompactness that the condition (A) is weaker than demicompactness for a nonexpansive mapping T defined on a bounded set. Every compact operator is demicompact. Since every completely continuous mapping $T:K\to K$ is continuous and demicompact, it satisfies condition (A).

Therefore in the next result, instead of complete continuity of mappings ${T}_{1},{T}_{2},\dots ,{T}_{N}$, we use condition ($\overline{\mathrm{C}}$).

Theorem 3.2 Let E and K be as defined above, ${T}_{i}$ ($i\in I$) be N asymptotically pseudo-contractive mappings as defined above and satisfying condition ($\overline{\mathrm{C}}$) and $\left\{{\alpha }_{n}\right\}$ be a sequence of real numbers as in Lemma  3.1. Then the sequence $\left\{{x}_{n}\right\}$ generated by (1.4) converges strongly to a member of .

Proof By Lemma 3.1, we see that ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-p\parallel$ and ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},\mathcal{F}\right)$ exist.

Let one of the ${T}_{i}$’s, say ${T}_{l}$, $l\in I$, satisfy condition (A).

By Lemma 3.1, we have ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{T}_{l}{x}_{n}\parallel =0$. Therefore we have ${lim}_{n\to \mathrm{\infty }}f\left(d\left({x}_{n},\mathcal{F}\right)\right)=0$. By the nature of f and the fact that ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},\mathcal{F}\right)$ exists, we have ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},\mathcal{F}\right)=0$. By Theorem 3.1, we find that $\left\{{x}_{n}\right\}$ converges strongly to a common fixed point in .

This completes the proof. □

A mapping $T:K\to K$ is said to be semicompact, if the sequence $\left\{{x}_{n}\right\}$ in K such that $\parallel {x}_{n}-T{x}_{n}\parallel \to 0$, as $n\to \mathrm{\infty }$, has a convergent subsequence.

Theorem 3.3 Let E and K be as defined above, and let ${T}_{i}$ ($i\in I$) be N asymptotically pseudo-contractive mappings as defined above such that one of the mappings in ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ is semicompact, and let $\left\{{\alpha }_{n}\right\}$ be a sequence of real numbers as in Lemma  3.1. Then the sequence $\left\{{x}_{n}\right\}$ generated by (1.4) converges strongly to a member of .

Proof Without loss of generality, we may assume that ${T}_{s}$ is semicompact for some fixed $s\in \left\{1,2,\dots ,N\right\}$. Then by Lemma 3.1, we have ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{T}_{s}{x}_{n}\parallel =0$. So by definition of semicompactness, there exists a subsequence $\left\{{x}_{{n}_{j}}\right\}$ of $\left\{{x}_{n}\right\}$ such that $\left\{{x}_{{n}_{j}}\right\}$ converges strongly to ${x}^{\ast }\in K$. Now again by Lemma 3.1, we have
$\underset{{n}_{j}\to \mathrm{\infty }}{lim}\parallel {x}_{{n}_{j}}-{T}_{l}{x}_{{n}_{j}}\parallel =0$

for all $l\in I$. By continuity of ${T}_{l}$, we have ${T}_{l}{x}_{{n}_{j}}\to {T}_{l}{x}^{\ast }$ for all $l\in I$.

Thus ${lim}_{j\to \mathrm{\infty }}\parallel {x}_{{n}_{j}}-{T}_{l}{x}_{{n}_{j}}\parallel =\parallel {x}^{\ast }-{T}_{l}{x}^{\ast }\parallel =0$ for all $l\in I$. This implies that ${x}^{\ast }\in \mathcal{F}$. Also, ${lim inf}_{n\to \mathrm{\infty }}d\left({x}_{n},\mathcal{F}\right)=0$. By Theorem 3.1, we find that $\left\{{x}_{n}\right\}$ converges strongly to a common fixed point in . □

3.2 Weak convergence theorem

Theorem 3.4 Let E be a uniformly convex and smooth Banach space which admits a weakly sequentially continuous duality mapping, and let K and ${T}_{i}$ ($i\in I$) be as defined above and $\left\{{\alpha }_{n}\right\}$ be a sequence of real numbers as in Lemma  3.1. Then the sequence $\left\{{x}_{n}\right\}$ generated by (1.4) converges weakly to a member of .

Proof Since $\left\{{x}_{n}\right\}$ is a bounded sequence in K, there exists a subsequence $\left\{{x}_{{n}_{k}}\right\}\subset \left\{{x}_{n}\right\}$ such that $\left\{{x}_{{n}_{k}}\right\}$ converges weakly to $q\in K$. Hence from Lemma 3.1, we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{{n}_{k}}-{T}_{l}{x}_{{n}_{k}}\parallel =0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }l\in I.$

By Lemma 2.4, we find that $\left(I-{T}_{l}\right)$ is demiclosed at zero, i.e. $\left(I-{T}_{l}\right)q=0$, so that $q\in F\left({T}_{l}\right)$. By the arbitrariness of $l\in I$, we know that $q\in \mathcal{F}={\bigcap }_{l\in I}F\left({T}_{l}\right)$.

Next we prove that $\left\{{x}_{n}\right\}$ converges weakly to q.

If $\left\{{x}_{n}\right\}$ has another subsequence $\left\{{x}_{{n}_{j}}\right\}$ which converges weakly to ${q}_{1}\ne q$, then we must have ${q}_{1}\in \mathcal{F}$, and since ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{q}_{1}\parallel$ exists and since the Banach space E has a weakly sequentially duality mapping, it satisfies Opial’s condition, and it follows from a standard argument that ${q}_{1}=q$. Thus $\left\{{x}_{n}\right\}$ converges weakly to $q\in \mathcal{F}$. □

Remark 3.1 Our results improve and generalize the corresponding results of Browder , Kirk , Goebel and Kirk , Schu , Xu [9, 10], Liu , Zhou and Chang , Osilike , Osilike and Akuchu , Su and Li , Su and Qin , and many others.

Let K be a nonempty subset of a real Banach space E. Let D be a nonempty bounded subset of K. The set-measure of noncompactness of D, $\gamma \left(D\right)$, is defined as
The ball-measure of compactness of D, $\chi \left(D\right)$, is defined as
A bounded continuous mapping $T:K\to E$ is called
1. (1)

k-set-contractive if $\gamma \left(T\left(D\right)\right)\le k\gamma \left(D\right)$, for each bounded subset D of K and some constant $k\ge 0$;

2. (2)

k-set-condensing if $\gamma \left(T\left(D\right)\right)<\gamma \left(D\right)$, for each bounded subset D of K with $\gamma \left(D\right)>0$;

3. (3)

k-ball-contractive if $\chi \left(T\left(D\right)\right)\le k\chi \left(D\right)$, for each bounded subset D of K and some constant $k\ge 0$;

4. (4)

k-ball-condensing if $\chi \left(T\left(D\right)\right)<\chi \left(D\right)$, for each bounded subset D of K with $\chi \left(D\right)>0$.

A mapping $T:K\to E$ is called
1. (5)

compact if $cl\left(T\left(A\right)\right)$ is compact whenever $A\subset K$ is bounded;

2. (6)

completely continuous if it maps weakly convergence sequences into strongly convergent sequences;

3. (7)

a generalized contraction if for each $x\in K$ there exists $k\left(x\right)<1$ such that $\parallel Tx-Ty\parallel \le k\left(x\right)\parallel x-y\parallel$ for all $y\in K$;

4. (8)

a mapping $T:E\to E$ is called uniformly strictly contractive (relative to E) if for each $x\in E$ there exists $k\left(x\right)<1$ such that $\parallel Tx-Ty\parallel \le k\left(x\right)\parallel x-y\parallel$ for all $y\in K$. Every k-set-contractive mapping with $k<1$ is set-condensing and also every compact mapping is set-condensing.

Let K be a nonempty closed bounded subset of E and $T:K\to E$ a continuous mapping. Then
1. (a)

T is strictly semicontractive if there exists a continuous mapping $V:E×E\to E$ with $T\left(x\right)=V\left(x,x\right)$ for $x\in E$ such that for each $x\in E$, $V\left(\cdot ,x\right)$ is a k-contraction with $k<1$ and $V\left(x,\cdot \right)$ is compact;

2. (b)

T is of strictly semicontractive type if there exists a continuous mapping $V:K×K\to E$ with $T\left(x\right)=V\left(x,x\right)$, for $x\in K$ such that for each $x\in K$, $V\left(\cdot ,x\right)$ is a k-contraction with some $k<1$ independent of x and $x↦V\left(\cdot ,x\right)$ is compact from K into the space of continuous mapping of K into E with the uniform metric;

3. (c)

T is of strongly semicontractive type relative to X if there exists a mapping $V:E×K\to E$ with $T\left(x\right)=V\left(x,x\right)$, for $x\in K$ such that $x\in K$, $V\left(\cdot ,x\right)$ is uniformly strictly contractive on K relative to E and $V\left(x,\cdot \right)$ is a completely continuous from K to E, uniformly for $x\in K$.

For details refer to .

Let K be a nonempty closed convex bounded subset of a uniformly convex Banach space E. Suppose $T:K\to K$. Then T is semicompact if T satisfies any one of the following conditions [, Proposition 3.4]:
1. (i)

T is either set-condensing or ball-condensing (or compact);

2. (ii)

T is a generalized contraction;

3. (iii)

T is uniformly strictly contractive;

4. (iv)

T is strictly semicontractive;

5. (v)

T is of strictly semicontractive type;

6. (vi)

T is of strongly semicontractive type.

Remark 3.2 In view of the above, it is possible to replace the semicompactness assumption in Theorem 3.3 with any of the contractive assumptions (i)-(vi).

We now give an example of asymptotically pseudo-contractive mapping with nonempty fixed point set.

Example 3.1 

Let $E=\mathbb{R}=\left(-\mathrm{\infty },\mathrm{\infty }\right)$ with usual norm and $K=\left[0,1\right]$ and define $T:K\to K$ by
for all $n\ge 0$. Then $F\left(T\right)=\left\{0\right\}$ and for any $x\in K$, there exists $j\left(x-0\right)\in J\left(x-0\right)$ satisfying
$〈{T}^{n}x-{T}^{n}0,j\left(x-0\right)〉={T}^{n}x\cdot x\le \frac{1}{3}{\parallel x\parallel }^{2}<{\parallel x\parallel }^{2}$

for all $n\ge 1$. That is, T is an asymptotically pseudo-contractive mapping with sequence $\left\{{k}_{n}\right\}=1$.

Authors’ Affiliations

(1)
School of Studies in Mathematics, Pt. Ravishankar Shukla University, Raipur, CG, 492010, India
(2)
Department of Mathematics and Informatics, University Politehnica of Bucharest, Bucharest, 060042, Romania

References 