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

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.

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

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

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 [1] if there exists a sequence $\{h_n\}\subseteq [1,\mathrm{\infty })$ 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 $j(x-y)\in J(x-y)$ such that $〈Tx-Ty,j(x-y)〉\le {\parallel x-y\parallel }^2$, for all $x,y\in K$. T is called strongly pseudo-contractive, if there exists a constant $\beta \in (0,1)$, $j(x-y)\in J(x-y)$ such that $〈Tx-Ty,j(x-y)〉\le \beta {\parallel x-y\parallel }^2$, for all $x,y\in K$. It is said to be asymptotically pseudo-contractive [4] if there exists a sequence $\{h_n\}\subseteq [1,\mathrm{\infty })$ with ${lim}_{n\to \mathrm{\infty }}{h}_n=1$ and $j(x-y)\in J(x-y)$ such that

$$〈T^{n}x-T^{n}y,j(x-y)〉\le h_n{\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in K,\mathrm{\forall }n\ge 1.$$

(1.1)

It follows from Kato [5] that

$$\parallel x-y\parallel \le \parallel x-y+r[(h_{n}I-T^n)x-(h_{n}I-T^n)y]\parallel ,\mathrm{\forall }x,y\in K,\mathrm{\forall }n\ge 1,r>0.$$

(1.2)

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

It follows from the definition that if T is asymptotically nonexpansive, then for all $j(x-y)\in J(x-y)$,

$$〈T^{n}x-T^{n}y,j(x-y)〉=\parallel x-y\parallel \parallel T^{n}x-T^{n}y\parallel \le h_n{\parallel x-y\parallel }^2.$$

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 $L={sup}_{n\ge 1}\{h_n\}$.

Now consider an example of non-Lipschitzian mapping due to Rhoades [6]. Define a mapping $T:[0,1]\to [0,1]$ by the formula $Tx={\{1-x^{\frac{2}{3}}\}}^{\frac{3}{2}}$, for $x\in [0,1]$. Schu [4] 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

$$|x-y||T^{n}x-T^{n}y|={|x-y|}^2\text{for all }n=2m,m\in \mathbb{N}$$

and

$$\begin{array}{rl}(x-y)(T^{n}x-T^{n}y)& =(x-y)(Tx-Ty)\\ \le 0\\ \le {|x-y|}^2\text{for all }n=2m-1,m\in \mathbb{N}.\end{array}$$

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

The iterative approximation problems for a nonexpansive mapping, an asymptotically nonexpansive mapping, and an asymptotically pseudo-contractive mapping were studied extensively by Browder [7], Kirk [8], Goebel and Kirk [1], Schu [4], Xu [9, 10], Liu [11] in the setting of Hilbert space or uniformly convex Banach space.

In 2001, Xu and Ori [12] introduced the following implicit iteration process for a finite family of nonexpansive self-mappings in Hilbert space:

$$\{\begin{array}{l}{x}_0\in K\text{arbitrary},\\ x_n={\alpha }_n{x}_{n-1}+(1-{\alpha }_n)T_n{x}_n,n\ge 1,\end{array}$$

(1.3)

where $\{{\alpha }_n\}$ be a sequence in $(0,1)$ and $T_n=T_{nmodN}$. They proved in [12] that the sequence $\{x_n\}$ converges weakly to a common fixed point of $T_n$, $n=1,2,\dots ,N$.

Later on Osilike and Akuchu [13], and Chen et al. [14] 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 [15] 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 [16], and Su and Qin [17] introduced the composite implicit iteration process and the general iteration algorithm, respectively, which properly include the implicit iteration process. Recently, Beg and Thakur [18] 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 ${\{T_i\}}_{i=1}^N$ 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 $\{x_n\}$ by

$$\begin{array}{r}{x}_n={\alpha }_n{x}_{n-1}+(1-{\alpha }_n)T_{i(n)}^{k(n)}{y}_n,\\ y_n=r_n{x}_n+s_n{x}_{n-1}+t_n{T}_{i(n)}^{k(n)}{x}_n+w_n{T}_{i(n)}^{k(n)}{x}_{n-1}\end{array}$$

(1.4)

for each $n\ge 1$, which can be written as $n=(k(n)-1)N+i(n)$, where $i(n)=1,2,\dots ,N$ and $k(n)\ge 1$ is a positive integer, with $k(n)\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$. The sequences $\{{\alpha }_n\}$, $\{r_n\}$, $\{s_n\}$, $\{t_n\}$ and $\{w_n\}$ are in $(0,1)$ 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 [19]

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 [20]

Let $\{a_n\}$, $\{b_n\}$, and $\{c_n\}$ be three nonnegative sequences satisfying the following condition:

$$a_{n+1}\le (1+b_n)a_n+c_n\mathit{\text{for all }}n\ge n_0,$$

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

• (i) ${lim}_{n\to \mathrm{\infty }}{a}_n$ exists;

• (ii) if, in addition, there exists a subsequence $\{a_{n_i}\}\subset \{a_n\}$ such that $a_{n_i}\to 0$, then $a_n\to 0$ as $n\to \mathrm{\infty }$.

Lemma 2.3 [21]

Let E be a uniformly convex Banach space and let a, b be two constants with $0<a<b<1$. Suppose that $\{t_n\}\subset [a,b]$ is a real sequence and $\{x_n\}$, $\{y_n\}$ are two sequences in E. Then the conditions

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n\parallel \le d,\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel y_n\parallel \le d\mathit{\text{and}}\underset{n\to \mathrm{\infty }}{lim}\parallel t_n{x}_n+(1-t_n)y_n\parallel =d$$

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

Lemma 2.4 [22]

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 $\{x_n\}\in E$, $x_n\rightharpoonup x$ as $n\to \mathrm{\infty }$ implies

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-y\parallel ,\mathrm{\forall }y\in E\text{ with }x\ne y.$$

We know that a Banach space with a sequentially continuous duality mapping satisfies Opial’s condition (for details, see [23]).

3 The main results

Throughout this section, E is a uniformly convex Banach space, K a nonempty closed convex subset of E. ℕ denotes the set of natural numbers and $I=\{1,2,\dots ,N\}$, 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 $\mathcal{F}={\bigcap }_{i\in I}F(T_i)\ne \mathrm{\varnothing }$.

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 $\{h_n^{(i)}\}$ such that $〈T_i^{n}x-T_i^{n}y,j(x-y)〉\le h_n^{(i)}{\parallel x-y\parallel }^2$ for all $x,y\in K$ and $i\in I$.

Take $L={max}_{i\in I}(L_i)$ and $h_n={max}_{i\in I}(h_n^{(i)})$.

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 $\alpha \in (\frac{L+L^2}{L+L^2+1},1)$, define a mapping $S_n:K\to K$ by the formula

$$\begin{array}{c}{S}_{n}x=\alpha u+(1-\alpha )T^{n}a,\hfill \\ a=rx+su+t{T}^{n}x+w{T}^{n}u\text{ for all }x\in K,\hfill \end{array}$$

(3.1)

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

Then, for all $x,y\in K$, $j(x-y)\in J(x-y)$, we have

$$\begin{array}{c}{S}_{n}y=\alpha u+(1-\alpha )T^{n}b,\hfill \\ b=ry+su+t{T}^{n}y+w{T}^{n}u\text{ for all }x\in K.\hfill \end{array}$$

(3.2)

Now

$$\begin{array}{rl}〈T^{n}a-T^{n}b,j(x-y)〉& =\parallel T^{n}a-T^{n}b\parallel \parallel x-y\parallel \\ \le L\parallel a-b\parallel \parallel x-y\parallel \\ =L\parallel r(x-y)+t(T^{n}x-T^{n}y)\parallel \parallel x-y\parallel \\ \le L(r\parallel x-y\parallel +tL\parallel x-y\parallel )\parallel x-y\parallel \\ =(Lr+t{L}^2){\parallel x-y\parallel }^2\\ \le (L+L^2){\parallel x-y\parallel }^2,\end{array}$$

so

$$\begin{array}{rl}〈S_{n}x-S_{n}y,j(x-y)〉& =(1-\alpha )〈T^{n}a-T^{n}b,j(x-y)〉\\ \le (1-\alpha )(L+L^2){\parallel x-y\parallel }^2.\end{array}$$

Since $(1-\alpha )(L+L^2)\in (0,1)$, $S_n$ 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.

$$\begin{array}{c}{S}_n{x}^{\ast }=\alpha u+(1-\alpha )T^{n}a,\hfill \\ a=r{x}^{\ast }+su+t{T}^n{x}^{\ast }+w{T}^{n}u\text{ for all }x\in K.\hfill \end{array}$$

(3.3)

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

Lemma 3.1 Let E, K, and $T_i$ ($i\in I$) be as defined above and let $\{x_n\}$ be the sequence defined by (1.4), where $\{{\alpha }_n\}$ 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 ${\sum }_{n=1}^{\mathrm{\infty }}(1-{\alpha }_n)<\mathrm{\infty }$ 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

• (i) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists, for all $p\in \mathcal{F}$,

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

• (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

$$\begin{array}{rcl}{\parallel x_n-p\parallel }^2& =& 〈x_n-p,j(x_n-p)〉\\ \le & {\alpha }_n〈x_{n-1}-p,j(x_n-p)〉+(1-{\alpha }_n)〈T_{i(n)}^{k(n)}{y}_n-T_{i(n)}^{k(n)}{x}_n,j(x_n-p)〉\\ +(1-{\alpha }_n)h_{k(n)}{\parallel x_n-p\parallel }^2\\ =& {\alpha }_n\parallel x_{n-1}-p\parallel \parallel x_n-p\parallel +(1-{\alpha }_n)L\parallel y_n-x_n\parallel \parallel x_n-p\parallel \\ +(1-{\alpha }_n)h_{k(n)}{\parallel x_n-p\parallel }^2.\end{array}$$

(3.4)

Using (1.4), we obtain

$$\begin{array}{rcl}\parallel y_n-x_n\parallel & =& \parallel s_n(x_{n-1}-x_n)+t_n(T_{i(n)}^{k(n)}{x}_n-x_n)+w_n(T_{i(n)}^{k(n)}{x}_{n-1}-x_n)\parallel \\ \le & s_n\parallel x_{n-1}-p\parallel +s_n\parallel x_n-p\parallel +t_{n}L\parallel x_n-p\parallel +t_n\parallel x_n-p\parallel \\ +w_{n}L\parallel x_{n-1}-p\parallel +w_n\parallel x_n-p\parallel .\end{array}$$

(3.5)

Substituting (3.5) in (3.4), we get

$$\begin{array}{rcl}{\parallel x_n-p\parallel }^2& \le & ({\alpha }_n+(1-{\alpha }_n)L(s_n+w_{n}L))\parallel x_{n-1}-p\parallel \parallel x_n-p\parallel \\ +(1-{\alpha }_n)[(s_n+t_n+w_n+t_{n}L)L+h_{k(n)}]{\parallel x_n-p\parallel }^2\\ \le & ({\alpha }_n+(1-{\alpha }_n)(1+L)L)\parallel x_{n-1}-p\parallel \parallel x_n-p\parallel \\ +(1-{\alpha }_n)[(1+L)L+h_{k(n)}]{\parallel x_n-p\parallel }^2\\ \le & ({\alpha }_n+(1-{\alpha }_n)(1+L)L)\parallel x_{n-1}-p\parallel \parallel x_n-p\parallel \\ +[(1-{\alpha }_n)(1+L)L+(1-{\alpha }_n+{\mu }_{k(n)})]{\parallel x_n-p\parallel }^2,\end{array}$$

(3.6)

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

Therefore, we have

$$\begin{array}{rl}\parallel x_n-p\parallel & \le \frac{({\alpha }_n+(1-{\alpha }_n)(1+L)L)}{{\alpha }_n-{\mu }_{k(n)}-(1-{\alpha }_n)(1+L)L}\parallel x_{n-1}-p\parallel \\ \le [1+\frac{{\mu }_{k(n)}+2(1-{\alpha }_n)(1+L)L}{{\alpha }_n-{\mu }_{k(n)}-(1-{\alpha }_n)(1+L)L}]\parallel x_{n-1}-p\parallel \\ \le [1+\frac{{\mu }_{k(n)}+2(1-{\alpha }_n)(1+L)L}{1-(1-{\alpha }_n+{\mu }_{k(n)}+(1-{\alpha }_n)(1+L)L)}]\parallel x_{n-1}-p\parallel .\end{array}$$

(3.7)

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

Now, we consider the second term on the right side of (3.7). We have

$$(1-{\alpha }_n+{\mu }_{k(n)}+(1-{\alpha }_n)(1+L)L)\le (1-{\alpha }_n)[1+M+(1+L)L].$$

By condition ${\sum }_{n=1}^{\mathrm{\infty }}(1-{\alpha }_n)<\mathrm{\infty }$, we have ${lim}_{n\to \mathrm{\infty }}(1-{\alpha }_n)=0$, then there exists a natural number $N_1$ such that if $n>N_1$, then

$$1-(1-{\alpha }_n+{\mu }_{k(n)}+(1-{\alpha }_n)(1+L)L)\ge \frac{1}{2}.$$

Therefore, it follows from (3.7) that

$$\begin{array}{rl}\parallel x_n-p\parallel & \le [1+2\{{\mu }_{k(n)}+2(1-{\alpha }_n)(1+L)L\}]\parallel x_{n-1}-p\parallel \\ =(1+{\sigma }_n)\parallel x_{n-1}-p\parallel ,\end{array}$$

(3.8)

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

Taking the infimum over $p\in \mathcal{F}$, we have

$$d(x_n,\mathcal{F})\le (1+{\sigma }_n)d(x_{n-1},\mathcal{F}).$$

(3.9)

Since ${\sum }_{n=1}^{\mathrm{\infty }}{\mu }_{k(n)}<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}(1-{\alpha }_n)<\mathrm{\infty }$, 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 ${lim}_{n\to \mathrm{\infty }}d(x_n,\mathcal{F})$ exist.

Without loss of generality, we assume

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-p\parallel =d^1.$$

(3.10)

Set $v_{k(n)}=\frac{h_{k(n)}-1}{h_{k(n)}}$, and from (1.2), we have

$$\begin{array}{rcl}\parallel x_n-p\parallel & \le & \parallel x_n-p+\frac{1-{\alpha }_n}{2{\alpha }_n{h}_{k(n)}}[(h_{k(n)}I-T_{i(n)}^{k(n)})x_n-(h_{k(n)}I-T_{i(n)}^{k(n)})p]\parallel \\ \le & \parallel x_n-p+\frac{1-{\alpha }_n}{2{\alpha }_n}[{\alpha }_n(x_{n-1}-T_{i(n)}^{k(n)}{x}_n)+(1-{\alpha }_n)(T_{i(n)}^{k(n)}{y}_n-T_{i(n)}^{k(n)}{x}_n)]\parallel \\ +\left(\frac{1-{\alpha }_n}{2{\alpha }_n}\right)\left(\frac{h_{k(n)}-1}{h_{k(n)}}\right)\parallel T_{i(n)}^{k(n)}{x}_n-p\parallel \\ =& \parallel x_n-p+\frac{1-{\alpha }_n}{2}(x_{n-1}-T_{i(n)}^{k(n)}{x}_n)+\frac{{(1-{\alpha }_n)}^2}{2{\alpha }_n}(T_{i(n)}^{k(n)}{y}_n-T_{i(n)}^{k(n)}{x}_n)\parallel \\ +\left(\frac{1-{\alpha }_n}{2{\alpha }_n}\right)v_{k(n)}\parallel T_{i(n)}^{k(n)}{x}_n-p\parallel \\ \le & \parallel x_n-p+\frac{1}{2}(x_{n-1}-x_n)\parallel +\left(\frac{1-{\alpha }_n}{2{\alpha }_n}\right)v_{k(n)}\parallel T_{i(n)}^{k(n)}{x}_n-p\parallel \\ +\frac{{(1-{\alpha }_n)}^2}{2{\alpha }_n}L\parallel y_n-x_n\parallel \\ \le & \parallel \frac{1}{2}(x_n-p)+\frac{1}{2}(x_{n-1}-p)\parallel +\left(\frac{1-{\alpha }_n}{2{\alpha }_n}\right)v_{k(n)}\parallel T_{i(n)}^{k(n)}{x}_n-p\parallel \\ +\frac{{(1-{\alpha }_n)}^2}{2{\alpha }_n}L\parallel y_n-x_n\parallel .\end{array}$$

Thus

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-p\parallel & \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel \frac{1}{2}(x_n-p)+\frac{1}{2}(x_{n-1}-p)\parallel \\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\left(\frac{1-{\alpha }_n}{2{\alpha }_n}\right)v_{k(n)}\parallel T_{i(n)}^{k(n)}{x}_n-p\parallel \\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{{(1-{\alpha }_n)}^2}{2{\alpha }_n}L\parallel y_n-x_n\parallel .\end{array}$$

Since $v_{k(n)}=\frac{h_{k(n)}-1}{h_{k(n)}}\in (0,1)$, we have ${lim}_{n\to \mathrm{\infty }}{v}_{k(n)}=0$ and from ${\sum }_{n=1}^{\mathrm{\infty }}(1-{\alpha }_n)<\mathrm{\infty }$, we have ${lim}_{n\to \mathrm{\infty }}(1-{\alpha }_n)=0$ and using (3.10), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel \frac{1}{2}(x_n-p)+\frac{1}{2}(x_{n-1}-p)\parallel \ge d^1.$$

(3.11)

On the other hand, we obtain

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel \frac{1}{2}(x_n-p)+\frac{1}{2}(x_{n-1}-p)\parallel \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}[\frac{1}{2}\parallel x_n-p\parallel +\frac{1}{2}\parallel x_{n-1}-p\parallel ]=d^1,$$

(3.12)

from (3.11) and (3.12), we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel \frac{1}{2}(x_n-p)+\frac{1}{2}(x_{n-1}-p)\parallel =d^1.$$

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

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_{i(n)}^{k(n)}{y}_n\parallel & =& \underset{n\to \mathrm{\infty }}{lim}\frac{{\alpha }_n}{1-{\alpha }_n}\parallel x_n-x_{n-1}\parallel \\ \le & \frac{1}{1-\beta }\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x_{n-1}\parallel =0.\end{array}$$

(3.15)

On the other hand, from (3.13) and (3.15)

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n-1}-T_{i(n)}^{k(n)}{y}_n\parallel \le \underset{n\to \mathrm{\infty }}{lim}\parallel x_{n-1}-x_n\parallel +\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_{i(n)}^{k(n)}{y}_n\parallel =0.$$

(3.16)

Now,

$$\begin{array}{rl}\parallel T_{i(n)}^{k(n)}{x}_n-x_n\parallel & \le \parallel x_n-x_{n-1}\parallel +\parallel T_{i(n)}^{k(n)}{y}_n-x_{n-1}\parallel +\parallel T_{i(n)}^{k(n)}{y}_n-T_{i(n)}^{k(n)}{x}_n\parallel \\ \le (1+L)\parallel x_n-x_{n-1}\parallel +\parallel T_{i(n)}^{k(n)}{y}_n-x_{n-1}\parallel +L\parallel y_n-x_{n-1}\parallel .\end{array}$$

(3.17)

Again, by using (1.4), we obtain

$$\begin{array}{rl}\parallel y_n-x_{n-1}\parallel & \le \parallel r_n{x}_n+s_n{x}_{n-1}+t_n{T}_{i(n)}^{k(n)}{x}_n+w_n{T}_{i(n)}^{k(n)}{x}_{n-1}-x_{n-1}\parallel \\ \le t_n\parallel T_{i(n)}^{k(n)}{x}_n-x_n\parallel +w_n\parallel T_{i(n)}^{k(n)}{x}_{n-1}-x_n\parallel +(r_n+t_n+w_n)\parallel x_n-x_{n-1}\parallel \\ \le (t_n+w_n)\parallel T_{i(n)}^{k(n)}{x}_n-x_n\parallel +(r_n+t_n+w_n+w_{n}L)\parallel x_n-x_{n-1}\parallel .\end{array}$$

(3.18)

Substituting (3.18) into (3.17), we get

$$\begin{array}{rcl}\parallel T_{i(n)}^{k(n)}{x}_n-x_n\parallel & \le & (1+L)\parallel x_n-x_{n-1}\parallel +\parallel T_{i(n)}^{k(n)}{y}_n-x_{n-1}\parallel +L(t_n+w_n)\parallel T_{i(n)}^{k(n)}{x}_n-x_n\parallel \\ +L(r_n+t_n+w_n+w_{n}L)\parallel x_n-x_{n-1}\parallel .\end{array}$$

Since $t_n+w_n\le b/L<1$, the above inequality gives

$$(1-b)\parallel T_{i(n)}^{k(n)}{x}_n-x_n\parallel \le [1+L(1+r_n+t_n+w_n+w_{n}L)]\parallel x_n-x_{n-1}\parallel +\parallel T_{i(n)}^{k(n)}{y}_n-x_{n-1}\parallel .$$

Then from (3.13), (3.16), and the above inequality, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_{i(n)}^{k(n)}{x}_n-x_n\parallel =0.$$

(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 $n=(k(n)-1)N+i(n)$, $i(n)\in I$.

Let ${\mathcal{A}}_n=\parallel T_{i(n)}^{k(n)}{y}_n-x_{n-1}\parallel$, then from (3.16), we have ${\mathcal{A}}_n\to 0$. Also,

$$\begin{array}{rcl}\parallel x_{n-1}-T_n{x}_n\parallel & \le & \parallel x_{n-1}-T_{i(n)}^{k(n)}{y}_n\parallel +\parallel T_{i(n)}^{k(n)}{y}_n-T_n{x}_n\parallel \\ =& {\mathcal{A}}_n+\parallel T_{i(n)}^{k(n)}{y}_n-T_{i(n)}{x}_n\parallel \le {\mathcal{A}}_n+L\parallel T_{i(n)}^{k(n)-1}{y}_n-x_n\parallel \\ \le & {\mathcal{A}}_n+L\{\parallel T_{i(n)}^{k(n)-1}{y}_n-T_{i(n-N)}^{k(n)-1}{x}_{n-N}\parallel \\ +\parallel T_{i(n-N)}^{k(n)-1}{x}_{n-N}-T_{i(n-N)}^{k(n)-1}{y}_{n-N}\parallel \\ +\parallel T_{i(n-N)}^{k(n)-1}{y}_{n-N}-x_{(n-N)-1}\parallel +\parallel x_{(n-N)-1}-x_n\parallel \}.\end{array}$$

(3.22)

Since for each $n>N$, $n=(n-N)(modN)$ and $n=(k(n)-1)N+i(n)$, $n-N=((k(n)-1)-1)N+i(n)=(k(n-N)-1)N+i(n-N)$, i.e.

$$k(n-N)=k(n)-1\text{and}i(n-N)=i(n).$$

Therefore from (3.22), we have

$$\begin{array}{rcl}\parallel x_{n-1}-T_n{x}_n\parallel & \le & {\mathcal{A}}_n+L\{\parallel T_{i(n)}^{k(n)-1}{y}_n-T_{i(n)}^{k(n)-1}{x}_{n-N}\parallel \\ +\parallel T_{i(n-N)}^{k(n-N)}{x}_{n-N}-T_{i(n-N)}^{k(n-N)}{y}_{n-N}\parallel \\ +\parallel T_{i(n-N)}^{k(n-N)}{y}_{n-N}-x_{(n-N)-1}\parallel +\parallel x_{(n-N)-1}-x_n\parallel \}\\ \le & {\mathcal{A}}_n+L\{L\parallel y_n-x_{n-N}\parallel +L\parallel x_{n-N}-y_{n-N}\parallel \\ +{\mathcal{A}}_{n-N}+\parallel x_{(n-N)-1}-x_n\parallel \}\\ \le & {\mathcal{A}}_n+L^2(\parallel y_n-x_n\parallel +\parallel x_n-x_{n-N}\parallel +\parallel x_{n-N}-y_{n-N}\parallel )\\ +L({\mathcal{A}}_{n-N}+\parallel x_{(n-N)-1}-x_n\parallel ).\end{array}$$

(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

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

(3.25)

Consequently, for any $i\in I$, from (3.14), (3.25), we obtain

$$\begin{array}{rl}\parallel x_n-T_{n+i}{x}_n\parallel & \le \parallel x_n-x_{n+i}\parallel +\parallel x_{n+i}-T_{n+i}{x}_{n+i}\parallel +\parallel T_{n+i}{x}_{n+i}-T_{n+i}{x}_n\parallel \\ \le (1+L)\parallel x_n-x_{n+i}\parallel +\parallel x_{n+i}-T_{n+i}{x}_{n+i}\parallel \to 0,\end{array}$$

as $n\to \mathrm{\infty }$. This implies that the sequence

$$\bigcup _{i=1}^N{\{\parallel x_n-T_{n+i}{x}_n\parallel \}}_{n=1}^{\mathrm{\infty }}\to 0,\text{as }n\to \mathrm{\infty }.$$

Since for each $l=1,2,\dots ,N$, $\{\parallel x_n-T_l{x}_n\parallel \}$ is a subsequence of ${\bigcup }_{i=1}^N\{\parallel x_n-T_{n+i}{x}_n\parallel \}$, therefore, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_l{x}_n\parallel =0,\mathrm{\forall }l\in I.$$

(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 $\{{\alpha }_n\}$ be a sequence of real numbers as in Lemma  3.1. Then the sequence $\{x_n\}$ generated by (1.4) converges strongly to a member of ℱ if and only if ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}d(x_n,\mathcal{F})=0$.

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

Let ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}d(x_n,\mathcal{F})=0$. Then from (ii) in Lemma 3.1, we have ${lim}_{n\to \mathrm{\infty }}d(x_n,\mathcal{F})=0$.

Next, we show that $\{x_n\}$ is a Cauchy sequence in K. For any given $\epsilon >0$, since ${lim}_{n\to \mathrm{\infty }}d(x_n,\mathcal{F})=0$, there exists a natural number $n_1$ such that $d(x_n,\mathcal{F})<\epsilon /4$ 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 ${\sum }_{j=n}^{\mathrm{\infty }}{\sigma }_j<\epsilon /4M^{\prime }$ for all $n\ge n_2$. Let $N^{\prime }=max\{n_1,n_2\}$. 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

$$\begin{array}{rl}\parallel x_n-x_m\parallel & \le \parallel x_n-p\parallel +\parallel x_m-p\parallel \\ \le \parallel x_{N^{\prime }}-p\parallel +M^{\prime }\sum _{j=N^{\prime }+1}^n{\sigma }_j+\parallel x_{N^{\prime }}-p\parallel +M^{\prime }\sum _{j=N^{\prime }+1}^m{\sigma }_j\\ \le 2\parallel x_{N^{\prime }}-p\parallel +2M^{\prime }\sum _{j=N^{\prime }}^{\mathrm{\infty }}{\sigma }_j.\end{array}$$

Taking the infimum over all $p\in \mathcal{F}$, we obtain

$$\parallel x_n-x_m\parallel \le 2d(x_{N^{\prime }},\mathcal{F})+2M^{\prime }\sum _{j=N^{\prime }}^{\mathrm{\infty }}{\sigma }_j<\epsilon .$$

This implies that $\{x_n\}$ is a Cauchy sequence. Since E is complete, therefore $\{x_n\}$ is convergent.

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

Since K is closed, we get $q\in K$, then $\{x_n\}$ converges strongly to q.

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

Notice that

$$|d(q,\mathcal{F})-d(x_n,\mathcal{F})|\le \parallel q-x_n\parallel ,\mathrm{\forall }n\in \mathbb{N},$$

since ${lim}_{n\to \mathrm{\infty }}{x}_n=q$ and ${lim}_{n\to \mathrm{\infty }}d(x_n,\mathcal{F})=0$, 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 $\{x_n\}$ generated by (1.4) converges strongly to $u\in \mathcal{F}$ if and only if $\{x_n\}$ has a subsequence $\{x_{n_j}\}$ which converges strongly to $u\in \mathcal{F}$.

A mapping $T:K\to K$ with $F(T)\ne \mathrm{\varnothing }$ is said to satisfy condition (A) [24] on K if there exists a nondecreasing function $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, with $f(0)=0$ and $f(r)>r$, for all $r\in (0,\mathrm{\infty })$, such that for all $x\in K$,

$$\parallel x-Tx\parallel \ge f\left(d(x,F(T))\right).$$

A family ${\{T_i\}}_{i=1}^N$ of N self-mappings of K with $\mathcal{F}={\bigcap }_{i\in I}F(T_i)\ne \mathrm{\varnothing }$ is said to satisfy

• (1) condition (B) on K [25] if there is a nondecreasing function $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $f(0)=0$ and $f(r)>r$ for all $r\in (0,\mathrm{\infty })$ such that for all $x\in K$ such that

  $$\underset{1\le l\le N}{max}\{\parallel x-T_{l}x\parallel \}\ge f(d(x,\mathcal{F}));$$

• (2) condition ($\overline{\mathrm{C}}$) on K [26] if there is a nondecreasing function $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $f(0)=0$ and $f(r)>r$ for all $r\in (0,\mathrm{\infty })$ such that for all $x\in K$ such that

  $$\{\parallel x-T_{l}x\parallel \}\ge f(d(x,\mathcal{F}))$$

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 [26]).

Senter and Dotson [24] 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 $\{{\alpha }_n\}$ be a sequence of real numbers as in Lemma  3.1. Then the sequence $\{x_n\}$ generated by (1.4) converges strongly to a member of ℱ.

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

This completes the proof. □

A mapping $T:K\to K$ is said to be semicompact, if the sequence $\{x_n\}$ 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 ${\{T_i\}}_{i=1}^N$ is semicompact, and let $\{{\alpha }_n\}$ be a sequence of real numbers as in Lemma  3.1. Then the sequence $\{x_n\}$ generated by (1.4) converges strongly to a member of ℱ.

Proof Without loss of generality, we may assume that $T_s$ is semicompact for some fixed $s\in \{1,2,\dots ,N\}$. 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 $\{x_{n_j}\}$ of $\{x_n\}$ such that $\{x_{n_j}\}$ 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, ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}d(x_n,\mathcal{F})=0$. By Theorem 3.1, we find that $\{x_n\}$ converges strongly to a common fixed point in ℱ. □

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 $\{{\alpha }_n\}$ be a sequence of real numbers as in Lemma  3.1. Then the sequence $\{x_n\}$ generated by (1.4) converges weakly to a member of ℱ.

Proof Since $\{x_n\}$ is a bounded sequence in K, there exists a subsequence $\{x_{n_k}\}\subset \{x_n\}$ such that $\{x_{n_k}\}$ converges weakly to $q\in K$. Hence from Lemma 3.1, we have

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

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

Next we prove that $\{x_n\}$ converges weakly to q.

If $\{x_n\}$ has another subsequence $\{x_{n_j}\}$ 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 $\{x_n\}$ converges weakly to $q\in \mathcal{F}$. □

Remark 3.1 Our results improve and generalize the corresponding results of Browder [7], Kirk [8], Goebel and Kirk [1], Schu [4], Xu [9, 10], Liu [11], Zhou and Chang [15], Osilike [27], Osilike and Akuchu [13], Su and Li [16], Su and Qin [17], 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, $\gamma (D)$, is defined as

$$\gamma (D)=inf\{d>0:D\text{ can be covered by a finite number of sets of diameter}\le d\}.$$

The ball-measure of compactness of D, $\chi (D)$, is defined as

$$\begin{array}{rcl}\chi (D)& =& inf\{r>0:D\text{ can be covered by a finite family of balls with centers in }E\\ \text{and radius }r\}.\end{array}$$

A bounded continuous mapping $T:K\to E$ is called

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

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

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

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

A mapping $T:K\to E$ is called

• (5) compact if $cl(T(A))$ is compact whenever $A\subset K$ is bounded;

• (6) completely continuous if it maps weakly convergence sequences into strongly convergent sequences;

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

• (8) a mapping $T:E\to E$ is called uniformly strictly contractive (relative to E) if for each $x\in E$ there exists $k(x)<1$ such that $\parallel Tx-Ty\parallel \le k(x)\parallel x-y\parallel$ 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

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

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

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

For details refer to [28–30].

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 [[25], Proposition 3.4]:

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

• (ii) T is a generalized contraction;

• (iii) T is uniformly strictly contractive;

• (iv) T is strictly semicontractive;

• (v) T is of strictly semicontractive type;

• (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 [31]

Let $E=\mathbb{R}=(-\mathrm{\infty },\mathrm{\infty })$ with usual norm and $K=[0,1]$ and define $T:K\to K$ by

$$Tx=\{\begin{array}{ll}0& \text{if }x=0,\\ \frac{1}{9}& \text{if }x=1,\\ x-\frac{1}{3^{n+1}}& \text{if }\frac{1}{3^{n+1}}\le x<\frac{1}{3}(\frac{1}{3^{n+1}}+\frac{1}{3^n}),\\ \frac{1}{3^n}-x& \text{if }\frac{1}{3}(\frac{1}{3^{n+1}}+\frac{1}{3^n})\le x<\frac{1}{3^n}\end{array}$$

for all $n\ge 0$. Then $F(T)=\{0\}$ and for any $x\in K$, there exists $j(x-0)\in J(x-0)$ satisfying

$$〈T^{n}x-T^{n}0,j(x-0)〉=T^{n}x\cdot x\le \frac{1}{3}{\parallel x\parallel }^2<{\parallel x\parallel }^2$$

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