• Research
• Open Access

# Strong convergence of an iterative process for a family of strictly pseudocontractive mappings

Fixed Point Theory and Applications20132013:117

https://doi.org/10.1186/1687-1812-2013-117

• Received: 13 February 2013
• Accepted: 19 April 2013
• Published:

## Abstract

In this article, fixed point problems of a family of strictly pseudocontractive mappings are investigated based on an iterative process. Strong convergence of the iterative process is obtained in a real 2-uniformly Banach space.

MSC:47H09, 47J05, 47J25.

## Keywords

• accretive operator
• iterative process
• fixed point
• nonexpansive mapping
• zero point

## 1 Introduction and preliminaries

Throughout this paper, we always assume that E is a real Banach space. Let ${E}^{\ast }$ be the dual space of E. Let ${J}_{q}$ ($q>1$) denote the generalized duality mapping from E into ${2}^{{E}^{\ast }}$ given by
${J}_{q}\left(x\right)=\left\{{f}^{\ast }\in {E}^{\ast }:〈x,{f}^{\ast }〉={\parallel x\parallel }^{q},\parallel {f}^{\ast }\parallel ={\parallel x\parallel }^{q-1}\right\},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in E,$
where $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing. In particular, ${J}_{2}$ is called the normalized duality mapping, which is usually denoted by J. In this paper, we use j to denote the single-valued normalized duality mapping. It is known that ${J}_{q}\left(x\right)={\parallel x\parallel }^{q-2}J\left(x\right)$ if $x\ne 0$. If E is a Hilbert space, then $J=I$, the identity mapping. Further, we have the following properties of the generalized duality mapping ${J}_{q}$:
1. (1)

${J}_{q}\left(tx\right)={t}^{q-1}{J}_{q}\left(x\right)$ for all $x\in E$ and $t\in \left[0,\mathrm{\infty }\right)$;

2. (2)

${J}_{q}\left(-x\right)=-{J}_{q}\left(x\right)$ for all $x\in E$.

A Banach space E is said to be smooth if the limit
$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$
exists for all $x,y\in {U}_{E}$. It is also said to be uniformly smooth if the limit is attained uniformly for all $x,y\in {U}_{E}$. The norm of E is said to be Fréchet differentiable if, for any $x\in {U}_{E}$, the above limit is attained uniformly for all $y\in {U}_{E}$. The modulus of smoothness of E is the function ${\rho }_{E}:\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ defined by
${\rho }_{E}\left(\tau \right)=sup\left\{\frac{1}{2}\left(\parallel x+y\parallel +\parallel x-y\parallel \right)-1:\parallel x\parallel \le 1,\parallel y\parallel \le \tau \right\},\phantom{\rule{1em}{0ex}}\mathrm{\forall }\tau \ge 0.$

The Banach space E is uniformly smooth if and only if ${lim}_{\tau \to \mathrm{\infty }}\frac{{\rho }_{E}\left(\tau \right)}{\tau }=0$. Let $q>1$. The Banach space E is said to be q-uniformly smooth if there exists a constant $c>0$ such that ${\rho }_{E}\left(\tau \right)\le c{\tau }^{q}$. It is shown in  that there is no Banach space which is q-uniformly smooth with $q>2$. Hilbert spaces, ${L}^{p}$ (or ${l}^{p}$) spaces and Sobolev space ${W}_{m}^{p}$, where $p\ge 2$, are 2-uniformly smooth.

Let C be a nonempty closed convex subset of E and let $T:C\to C$ be a mapping. In this paper, we use $F\left(T\right)$ to denote the fixed point set of T. A mapping T is said to be κ-contractive iff there exists a constant $\kappa \in \left(0,1\right)$ such that
$\parallel Tx-Ty\parallel \le \kappa \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
A mapping T is said to be nonexpansive iff
$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
A mapping T is said to be κ-strictly pseudocontractive iff there exist a constant $\kappa \in \left(0,1\right)$ and $j\left(x-y\right)\in J\left(x-y\right)$ such that
$〈Tx-Ty,j\left(x-y\right)〉\le {\parallel x-y\parallel }^{2}-\kappa {\parallel \left(I-T\right)x-\left(I-T\right)y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
(1.1)
It is clear that (1.1) is equivalent to the following:
$〈\left(I-T\right)x-\left(I-T\right)y,j\left(x-y\right)〉\ge \kappa {\parallel \left(I-T\right)x-\left(I-T\right)y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
(1.2)
The class of κ-strictly pseudocontractive mappings was first introduced by Browder and Petryshyn  in Hilbert spaces. A mapping T is said to be pseudocontractive iff there exists $j\left(x-y\right)\in J\left(x-y\right)$ such that
$〈Tx-Ty,j\left(x-y\right)〉\le {\parallel x-y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
(1.3)
A mapping T is said to be κ-strongly pseudocontractive iff there exist a constant $\kappa \in \left(0,1\right)$ and $j\left(x-y\right)\in J\left(x-y\right)$ such that
$〈Tx-Ty,j\left(x-y\right)〉\le \kappa {\parallel x-y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
(1.4)

In 1974, Deimling  proved the existence of fixed points of continuous κ-strongly pseudocontractive mappings in Banach spaces; see  for more details. We remark that the class of κ-strongly pseudocontractive mappings is independent of the class of κ-strictly pseudocontractive mappings. This can be seen from Zhou . Lipschitz pseudocontractive mappings may not be κ-strictly pseudocontractive mappings, which can be seen from Chidume and Mutangadura .

One classical way to study nonexpansive mappings is to use contractions to approximate a nonexpansive mapping; for more details, see  and the references therein. More precisely, take $t\in \left(0,1\right)$ and define a contraction ${T}_{t}:C\to C$ by
${T}_{t}x=tu+\left(1-t\right)Tx,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C,$
(1.5)

where $u\in C$ is a fixed point. Banach’s contraction mapping principle guarantees that ${T}_{t}$ has a unique fixed point ${x}_{t}$ in C. In the case of T having a fixed point, Browder  proved that ${x}_{t}$ converges strongly to a fixed point of T in the framework of Hilbert spaces. Reich  extended Browder’s result to the setting of Banach spaces and proved that if E is a uniformly smooth Banach space, then ${x}_{t}$ converges strongly to a fixed point of T and the limit defines the (unique) sunny nonexpansive retraction from C onto $F\left(T\right)$; see  for more details.

Recall that the normal Mann iterative process was introduced by Mann  in 1953. Recently, the construction of fixed points for nonexpansive mappings via the normal Mann iterative process has been extensively investigated by many authors. The normal Mann iterative process generates a sequence $\left\{{x}_{n}\right\}$ in the following manner:
$\left\{\begin{array}{c}{x}_{1}\in C\phantom{\rule{1em}{0ex}}\text{chosen arbitrarily},\hfill \\ {x}_{n+1}=\left(1-{\alpha }_{n}\right){x}_{n}+{\alpha }_{n}T{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(1.6)

where the sequence $\left\{{\alpha }_{n}\right\}$ is in the interval $\left(0,1\right)$.

In an infinite-dimensional Hilbert space, the normal Mann iteration algorithm has only weak convergence; see  for more details. In many disciplines, including economics , image recovery  and control theory , problems arise in infinite dimension spaces. In such problems, strong convergence is often much more desirable than weak convergence for it translates the physically tangible property so that the energy $\parallel {x}_{n}-x\parallel$ of the error between the iterate ${x}_{n}$ and the solution x eventually becomes arbitrarily small.

Recently, many authors have tried to modify the normal Mann iteration process to have strong convergence for nonexpansive mappings and κ-strictly pseudocontractive mappings; see  and the references therein.

Let D be a nonempty subset of C. Let $Q:C\to D$. Q is said to be a contraction iff ${Q}^{2}=Q$; sunny iff for each $x\in C$ and $t\in \left(0,1\right)$, we have $Q\left(tx+\left(1-t\right)Qx\right)=Qx$; sunny nonexpansive retraction iff Q sunny, nonexpansive and contraction. K is said to be a nonexpansive retract of C if there exists a nonexpansive retraction from C onto D. The following result, which was established in  and , describes a characterization of sunny nonexpansive retractions on a smooth Banach space.

Let $Q:E\to C$ be a retraction, and let j be the normalized duality mapping on E. Then the following are equivalent:
1. (1)

Q is sunny and nonexpansive;

2. (2)

${\parallel Qx-Qy\parallel }^{2}\le 〈x-y,j\left(Qx-Qy\right)〉$, $\mathrm{\forall }x,y\in E$;

3. (3)

$〈x-Qx,j\left(y-Qx\right)〉\le 0$, $\mathrm{\forall }x\in E$, $y\in C$.

In this paper, we investigate the problem of modifying the normal Mann iteration process for a family of κ-strictly pseudocontractive mappings. Strong convergence of the purposed iterative process is obtained in a real 2-uniformly Banach space. In order to prove our main results, we need the following tools.

Lemma 1.1 

Let E be a real 2-uniformly smooth Banach space with the best smooth constant K. Then the following inequality holds:
${\parallel x+y\parallel }^{2}\le {\parallel x\parallel }^{2}+2〈y,Jx〉+2{\parallel Ky\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in E.$

Lemma 1.2 

Let C be a nonempty subset of a real 2-uniformly smooth Banach space E and let $T:C\to C$ be a κ-strict pseudocontraction. For $\alpha \in \left(0,1\right)$, we define ${T}_{\alpha }x=\left(1-\alpha \right)x+\alpha Tx$ for every $x\in C$ . Then, as $\alpha \in \left(0,\frac{\kappa }{{K}^{2}}\right]$, ${T}_{\alpha }$ is nonexpansive such that $F\left({T}_{\alpha }\right)=F\left(T\right)$.

Lemma 1.3 

Assume that $\left\{{\alpha }_{n}\right\}$ is a sequence of nonnegative real numbers such that
${\alpha }_{n+1}\le \left(1-{\gamma }_{n}\right){\alpha }_{n}+{\delta }_{n},$
where $\left\{{\gamma }_{n}\right\}$ is a sequence in $\left(0,1\right)$ and $\left\{{\delta }_{n}\right\}$ is a sequence such that
1. (i)

${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_{n}=\mathrm{\infty }$;

2. (ii)

${lim sup}_{n\to \mathrm{\infty }}{\delta }_{n}/{\gamma }_{n}\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n}|<\mathrm{\infty }$.

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

Lemma 1.4 

Let E be a real smooth Banach space. Then the following inequality holds:
${\parallel x+y\parallel }^{2}\le {\parallel x\parallel }^{2}+2〈y,J\left(x+y\right)〉,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in E.$

Lemma 1.5 

Let E be a smooth Banach space and let C be a nonempty convex subset of E. Given an integer $N\ge 1$, assume that ${\left\{{T}_{i}\right\}}_{i=1}^{N}:C\to C$ is a finite family of ${\kappa }_{i}$-strict pseudocontractions such that ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)\ne \mathrm{\varnothing }$. Assume that ${\left\{{\lambda }_{i}\right\}}_{i=1}^{r}$ is a positive sequence such that ${\sum }_{i=1}^{N}{\lambda }_{i}=1$. Then $F\left({\sum }_{i=1}^{N}F\left({T}_{i}\right)\right)={\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$.

Lemma 1.6 

Let E be a real uniformly smooth Banach space and let C be a nonempty closed convex subset of E. Let $T:C\to C$ be a nonexpansive mapping with a fixed point and let $f:C\to C$ be a contraction. For each $t\in \left(0,1\right)$, let ${z}_{t}$ be the unique solution of the equation $x=tf\left(x\right)+\left(1-t\right)Tx$. Then $\left\{{z}_{t}\right\}$ converges to a fixed point of T as $t\to 0$ and $Q\left(f\right)=s-{lim}_{t\to 0}{z}_{t}$ defines the unique sunny nonexpansive retraction from C onto $F\left(T\right)$.

## 2 Main results

Theorem 2.1 Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space E with the best smooth constant K and let N be some positive integer. Let ${T}_{i}:C\to C$ be a ${\kappa }_{i}$-strictly pseudocontractive mapping for each $1\le i\le N$. Assume that ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)\ne \mathrm{\varnothing }$. Let f be an α-contractive mapping. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following process:
$\left\{\begin{array}{c}{x}_{0}\in C\phantom{\rule{1em}{0ex}}\mathit{\text{arbitrarily chosen}},\hfill \\ {y}_{n}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){\sum }_{i=1}^{N}{\lambda }_{i}{T}_{i}{x}_{n},\hfill \\ {x}_{n+1}={\alpha }_{n}f\left({y}_{n}\right)+\left(1-{\alpha }_{n}\right){y}_{n},\phantom{\rule{1em}{0ex}}n\ge 0,\hfill \end{array}$
where $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\lambda }_{i}\right\}$ are real number sequences in $\left[0,1\right]$ satisfying the following restrictions:
1. (a)

${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n}-{\alpha }_{n-1}|<\mathrm{\infty }$,

2. (b)

$1-\frac{\kappa }{{K}^{2}}\le {\beta }_{n}\le \beta <1$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n}-{\beta }_{n-1}|<\mathrm{\infty }$,

3. (c)

${\sum }_{n=1}^{N}{\lambda }_{i}=1$,

where β is some real number, and $\kappa :=min\left\{{\kappa }_{i}:1\le i\le N\right\}$. Then $\left\{{x}_{n}\right\}$ converges strongly as $n\to \mathrm{\infty }$ to some point in ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$, which is the unique solution in ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$ to the following variational inequality:
$〈f\left({x}^{\ast }\right)-{x}^{\ast },j\left({x}^{\ast }-p\right)〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in \bigcap _{i=1}^{N}F\left({T}_{i}\right).$

Proof The proof is split into four steps.

Step 1. Show that $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ are bounded.

Putting $T:={\sum }_{i=1}^{N}{\lambda }_{i}{T}_{i}$, we see that T is a κ-strictly pseudocontractive mapping. Indeed, we have the following:
$\begin{array}{r}〈Tx-Ty,j\left(x-y\right)〉\\ \phantom{\rule{1em}{0ex}}={\lambda }_{1}〈{T}_{1}x-{T}_{1}y,j\left(x-y\right)〉+{\lambda }_{2}〈{T}_{2}x-{T}_{2}y,j\left(x-y\right)〉+\cdots \\ \phantom{\rule{2em}{0ex}}+{\lambda }_{N}〈{T}_{N}x-{T}_{N}y,j\left(x-y\right)〉\\ \phantom{\rule{1em}{0ex}}\le {\lambda }_{1}\left({\parallel x-y\parallel }^{2}-{\kappa }_{1}{\parallel \left(I-{T}_{1}\right)x-\left(I-{T}_{1}\right)y\parallel }^{2}\right)\\ \phantom{\rule{2em}{0ex}}+{\lambda }_{2}\left({\parallel x-y\parallel }^{2}-{\kappa }_{2}{\parallel \left(I-{T}_{2}\right)x-\left(I-{T}_{2}\right)y\parallel }^{2}\right)+\cdots \\ \phantom{\rule{2em}{0ex}}+{\lambda }_{N}\left({\parallel x-y\parallel }^{2}-{\kappa }_{N}{\parallel \left(I-{T}_{N}\right)x-\left(I-{T}_{N}\right)y\parallel }^{2}\right)\\ \phantom{\rule{1em}{0ex}}\le {\parallel x-y\parallel }^{2}-\kappa \left({\lambda }_{1}{\parallel \left(I-{T}_{1}\right)x-\left(I-{T}_{1}\right)y\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+{\lambda }_{2}{\parallel \left(I-{T}_{2}\right)x-\left(I-{T}_{2}\right)y\parallel }^{2}+\cdots +{\lambda }_{N}{\parallel \left(I-{T}_{N}\right)x-\left(I-{T}_{N}\right)y\parallel }^{2}\right)\\ \phantom{\rule{1em}{0ex}}\le {\parallel x-y\parallel }^{2}-\kappa {\parallel \left(I-T\right)x-\left(I-T\right)y\parallel }^{2}.\end{array}$
This proves that T is a κ-strictly pseudocontractive mapping. Fix $p\in {\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$. It follows from Lemma 1.1 that
$\begin{array}{rcl}{\parallel {y}_{n}-p\parallel }^{2}& =& {\parallel \left({x}_{n}-p\right)+\left(1-{\beta }_{n}\right)\left(T{x}_{n}-{x}_{n}\right)\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}+2\left(1-{\beta }_{n}\right)〈T{x}_{n}-{x}_{n},j\left({x}_{n}-p\right)〉\\ +2{K}^{2}{\left(1-{\beta }_{n}\right)}^{2}{\parallel \left(T{x}_{n}-{x}_{n}\right)\parallel }^{2}\\ =& {\parallel {x}_{n}-p\parallel }^{2}+2\left(1-{\beta }_{n}\right)〈T{x}_{n}-p,j\left({x}_{n}-p\right)〉-2\left(1-{\beta }_{n}\right){\parallel {x}_{n}-p\parallel }^{2}\\ +2{K}^{2}{\left(1-{\beta }_{n}\right)}^{2}{\parallel \left(T{x}_{n}-{x}_{n}\right)\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}+2\left(1-{\beta }_{n}\right)\left({\parallel {x}_{n}-p\parallel }^{2}-\kappa {\parallel T{x}_{n}-{x}_{n}\parallel }^{2}\right)\\ -2\left(1-{\beta }_{n}\right){\parallel {x}_{n}-p\parallel }^{2}+2{K}^{2}{\left(1-{\beta }_{n}\right)}^{2}{\parallel \left(T{x}_{n}-{x}_{n}\right)\parallel }^{2}\\ =& {\parallel {x}_{n}-p\parallel }^{2}-2\left(1-{\beta }_{n}\right)\left(\kappa -{K}^{2}\left(1-{\beta }_{n}\right)\right){\parallel T{x}_{n}-{x}_{n}\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}.\end{array}$
(2.1)
This implies that
$\begin{array}{rl}\parallel {x}_{n+1}-p\parallel & =\parallel {\alpha }_{n}\left(f\left({y}_{n}\right)-p\right)+\left(1-{\alpha }_{n}\right)\left({y}_{n}-p\right)\parallel \\ \le {\alpha }_{n}\parallel f\left({y}_{n}\right)-p\parallel +\left(1-{\alpha }_{n}\right)\parallel {y}_{n}-p\parallel \\ \le \left(1-{\alpha }_{n}\left(1-\alpha \right)\right)\parallel {y}_{n}-p\parallel +{\alpha }_{n}\parallel f\left(p\right)-p\parallel \\ \le \left(1-{\alpha }_{n}\left(1-\alpha \right)\right)\parallel {x}_{n}-p\parallel +{\alpha }_{n}\parallel f\left(p\right)-p\parallel \\ \le max\left\{\parallel {x}_{n}-p\parallel ,\frac{\parallel f\left(p\right)-p\parallel }{1-\alpha }\right\}.\end{array}$
This in turn implies that
$\parallel {x}_{n}-p\parallel \le max\left\{\parallel {x}_{0}-p\parallel ,\frac{\parallel p-f\left(p\right)\parallel }{1-\alpha }\right\},$

which gives that the sequence $\left\{{x}_{n}\right\}$ is bounded, so is $\left\{{y}_{n}\right\}$. This completes step 1.

Step 2. Show that $\parallel {T}_{\frac{\kappa }{{K}^{2}}}{x}_{n}-{x}_{n}\parallel \to 0$ as $n\to \mathrm{\infty }$.

Put ${T}_{{\beta }_{n}}x={\beta }_{n}x+\left(1-{\beta }_{n}\right)Tx$, $\mathrm{\forall }x\in C$. It follows from Lemma 1.2 that
$\begin{array}{rl}\parallel {y}_{n}-{y}_{n-1}\parallel & =\parallel {T}_{{\beta }_{n}}{x}_{n}-{T}_{{\beta }_{n-1}}{x}_{n-1}\parallel \\ \le \parallel {T}_{{\beta }_{n}}{x}_{n}-{T}_{{\beta }_{n}}{x}_{n-1}\parallel +\parallel {T}_{{\beta }_{n}}{x}_{n-1}-{T}_{{\beta }_{n-1}}{x}_{n-1}\parallel \\ \le \parallel {x}_{n}-{x}_{n-1}\parallel +\parallel {\beta }_{n}{x}_{n-1}+\left(1-{\beta }_{n}\right)T{x}_{n-1}-{\beta }_{n-1}{x}_{n-1}-\left(1-{\beta }_{n-1}\right)T{x}_{n-1}\parallel \\ \le \parallel {x}_{n}-{x}_{n-1}\parallel +|{\beta }_{n}-{\beta }_{n-1}|\parallel {x}_{n-1}-T{x}_{n-1}\parallel .\end{array}$
(2.2)
Notice that
${x}_{n+1}-{x}_{n}={\alpha }_{n}\left(f\left({y}_{n}\right)-f\left({y}_{n-1}\right)\right)+\left(1-{\alpha }_{n}\right)\left({y}_{n}-{y}_{n-1}\right)+\left({\alpha }_{n}-{\alpha }_{n-1}\right)\left(f\left({y}_{n-1}\right)-{y}_{n-1}\right).$
It follows from (2.2) that
$\begin{array}{r}\parallel {x}_{n+1}-{x}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}\parallel f\left({y}_{n}\right)-f\left({y}_{n-1}\right)\parallel +\left(1-{\alpha }_{n}\right)\parallel {y}_{n}-{y}_{n-1}\parallel +|{\alpha }_{n}-{\alpha }_{n-1}|\parallel f\left({y}_{n-1}\right)-{y}_{n-1}\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(1-{\alpha }_{n}\left(1-\alpha \right)\right)\parallel {y}_{n}-{y}_{n-1}\parallel +|{\alpha }_{n}-{\alpha }_{n-1}|\parallel f\left({y}_{n-1}\right)-{y}_{n-1}\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(1-{\alpha }_{n}\left(1-\alpha \right)\right)\parallel {x}_{n}-{x}_{n-1}\parallel +|{\beta }_{n}-{\beta }_{n-1}|\parallel {x}_{n-1}-T{x}_{n-1}\parallel \\ \phantom{\rule{2em}{0ex}}+|{\alpha }_{n}-{\alpha }_{n-1}|\parallel f\left({y}_{n-1}\right)-{y}_{n-1}\parallel .\end{array}$
In view of Lemma 1.3, we obtain from the restrictions (a) and (b) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =0.$
(2.3)
Notice that
${x}_{n+1}-{x}_{n}={\alpha }_{n}\left(f\left({x}_{n}\right)-{x}_{n}\right)+\left(1-{\alpha }_{n}\right)\left({y}_{n}-{x}_{n}\right).$
In view of the restriction (a), we obtain that ${lim}_{n\to \mathrm{\infty }}\parallel {y}_{n}-{x}_{n}\parallel =0$. On the other hand, we have ${y}_{n}-{x}_{n}=\left(1-{\beta }_{n}\right)\left(T{x}_{n}-{x}_{n}\right)$. This in turn implies that ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_{n}-{x}_{n}\parallel =0$. It follows from the restriction (b) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {T}_{\frac{\kappa }{{K}^{2}}}{x}_{n}-{x}_{n}\parallel =0.$
(2.4)

This completes step 2.

Step 3. Show that
$\underset{n\to \mathrm{\infty }}{lim sup}〈z-f\left(z\right),j\left(z-{x}_{n}\right)〉\le 0,$
(2.5)
where $z=Qf\left(z\right)$, where Q is a sunny nonexpansive retraction from C onto ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$, is the strong limit of the sequence ${z}_{t}$ defined by
${z}_{t}=tf\left({z}_{t}\right)+\left(1-t\right){T}_{\frac{\kappa }{{K}^{2}}}{z}_{t},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left(0,1\right).$
It follows that
${z}_{t}-{x}_{n}=\left(1-t\right)\left({T}_{\frac{\kappa }{{K}^{2}}}{z}_{t}-{x}_{n}\right)+t\left(f\left({z}_{t}\right)-{x}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left(0,1\right).$
For any $t\in \left(0,1\right)$, we see from Lemma 1.4 that
$\begin{array}{rcl}{\parallel {z}_{t}-{x}_{n}\parallel }^{2}& \le & {\left(1-t\right)}^{2}{\parallel {T}_{\frac{\kappa }{{K}^{2}}}{z}_{t}-{x}_{n}\parallel }^{2}+2t〈f\left({z}_{t}\right)-{x}_{n},j\left({z}_{t}-{x}_{n}\right)〉\\ \le & {\left(1-t\right)}^{2}\left({\parallel {T}_{\frac{\kappa }{{K}^{2}}}{z}_{t}-{T}_{\frac{\kappa }{{K}^{2}}}{x}_{n}\parallel }^{2}+{\parallel {T}_{\frac{\kappa }{{K}^{2}}}{x}_{n}-{x}_{n}\parallel }^{2}\\ +2\parallel {T}_{\frac{\kappa }{{K}^{2}}}{z}_{t}-{T}_{\frac{\kappa }{{K}^{2}}}{x}_{n}\parallel \parallel {T}_{\frac{\kappa }{{K}^{2}}}{x}_{n}-{x}_{n}\parallel \right)+2t〈f\left({z}_{t}\right)-{z}_{t},j\left({z}_{t}-{x}_{n}\right)〉\\ +2t〈{z}_{t}-{x}_{n},j\left({z}_{t}-{x}_{n}\right)〉\\ \le & {\left(1-t\right)}^{2}{\parallel {z}_{t}-{x}_{n}\parallel }^{2}+{\lambda }_{n}\left(t\right)+2t〈f\left({z}_{t}\right)-{z}_{t},j\left({z}_{t}-{x}_{n}\right)〉+2t{\parallel {z}_{t}-{x}_{n}\parallel }^{2},\end{array}$
(2.6)
where
${\lambda }_{n}\left(t\right)={\parallel {T}_{\frac{\kappa }{{K}^{2}}}{x}_{n}-{x}_{n}\parallel }^{2}+2\parallel {z}_{t}-{x}_{n}\parallel \parallel {T}_{\frac{\kappa }{{K}^{2}}}{x}_{n}-{x}_{n}\parallel .$
It follows from (2.6) that
$〈{z}_{t}-f\left({z}_{t}\right),j\left({z}_{t}-{x}_{n}\right)〉\le \frac{t}{2}{\parallel {z}_{t}-{x}_{n}\parallel }^{2}+\frac{1}{2t}{\lambda }_{n}\left(t\right).$
This implies that
$\underset{n\to \mathrm{\infty }}{lim sup}〈{z}_{t}-f\left({z}_{t}\right),j\left({z}_{t}-{x}_{n}\right)〉\le \frac{t}{2}{\parallel {z}_{t}-{x}_{n}\parallel }^{2}.$
Since E is 2-uniformly smooth, $J:E\to {E}^{\ast }$ is uniformly continuous on any bounded sets of E, which ensures that the ${lim sup}_{n\to \mathrm{\infty }}$ and ${lim sup}_{t\to 0}$ are interchangeable, and hence
$\underset{n\to \mathrm{\infty }}{lim sup}〈z-f\left(z\right),j\left(z-{x}_{n}\right)〉\le 0.$

This shows that (2.5) holds. This completes the proof of step 3.

Step 4. Show that ${x}_{n}\to z$ as $n\to \mathrm{\infty }$.

It follows from (2.1) that $\parallel {y}_{n}-z\parallel \le \parallel {x}_{n}-z\parallel$. In view of Lemma 1.4, we see that
$\begin{array}{rcl}{\parallel {x}_{n+1}-z\parallel }^{2}& =& {\parallel \left(1-{\alpha }_{n}\right)\left({y}_{n}-z\right)+{\alpha }_{n}\left(f\left({y}_{n}\right)-z\right)\parallel }^{2}\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{\parallel {y}_{n}-z\parallel }^{2}+2{\alpha }_{n}〈f\left({y}_{n}\right)-z,J\left({x}_{n+1}-z\right)〉\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{\parallel {x}_{n}-z\parallel }^{2}+2{\alpha }_{n}〈f\left({y}_{n}\right)-f\left(z\right),J\left({x}_{n+1}-z\right)〉\\ +2{\alpha }_{n}〈f\left(z\right)-z,J\left({x}_{n+1}-z\right)〉\\ \le & \left({\left(1-{\alpha }_{n}\right)}^{2}+{\alpha }_{n}\alpha \right){\parallel {x}_{n}-z\parallel }^{2}+{\alpha }_{n}\alpha {\parallel {x}_{n+1}-z\parallel }^{2}+2{\alpha }_{n}〈f\left(z\right)-z,J\left({x}_{n+1}-z\right)〉.\end{array}$
It then follows that
$\begin{array}{rcl}{\parallel {x}_{n+1}-z\parallel }^{2}& \le & \frac{1-\left(2-\alpha \right){\alpha }_{n}+{\alpha }^{2}}{1-{\alpha }_{n}\alpha }{\parallel {x}_{n}-z\parallel }^{2}+\frac{2{\alpha }_{n}}{1-{\alpha }_{n}\alpha }〈f\left(z\right)-z,J\left({x}_{n+1}-z\right)〉\\ \le & \frac{1-\left(2-\alpha \right){\alpha }_{n}}{1-{\alpha }_{n}\alpha }{\parallel {x}_{n}-z\parallel }^{2}+\frac{2{\alpha }_{n}}{1-{\alpha }_{n}\alpha }〈f\left(z\right)-z,J\left({x}_{n+1}-z\right)〉\\ +\frac{{\alpha }_{n}^{2}}{1-{\alpha }_{n}\alpha }{\parallel {x}_{n}-z\parallel }^{2}\\ \le & \left(1-\frac{2\left(1-\alpha \right){\alpha }_{n}}{1-{\alpha }_{n}\alpha }\right){\parallel {x}_{n}-z\parallel }^{2}+\frac{2\left(1-\alpha \right){\alpha }_{n}}{1-{\alpha }_{n}\alpha }\left(\frac{1}{1-\alpha }〈f\left(z\right)-z,J\left({x}_{n+1}-z\right)〉\\ +\frac{{\alpha }_{n}}{2\left(1-\alpha \right)}{\parallel {x}_{n}-z\parallel }^{2}\right).\end{array}$
It follows from the restrictions (a) and (b) that
$\underset{n\to \mathrm{\infty }}{lim}\frac{2\left(1-\alpha \right){\alpha }_{n}}{1-{\alpha }_{n}\alpha }=0,\phantom{\rule{2em}{0ex}}\sum _{n=1}^{\mathrm{\infty }}\frac{2\left(1-\alpha \right){\alpha }_{n}}{1-{\alpha }_{n}\alpha }=\mathrm{\infty }$
and
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\frac{1}{1-\alpha }〈f\left(z\right)-z,J\left({x}_{n+1}-z\right)〉+\frac{{\alpha }_{n}}{2\left(1-\alpha \right)}{\parallel {x}_{n}-z\parallel }^{2}\right)\le 0.$

This implies from Lemma 1.3 that ${x}_{n}\to z$ as $n\to \mathrm{\infty }$. This completes the proof. □

For a single mapping, we have the following.

Corollary 2.2 Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space E with the best smooth constant K. Let $T:C\to C$ be a κ-strictly pseudocontractive mapping such that $F\left(T\right)\ne \mathrm{\varnothing }$. Let f be an α-contractive mapping. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following process:
$\left\{\begin{array}{c}{x}_{0}\in C\phantom{\rule{1em}{0ex}}\mathit{\text{arbitrarily chosen}},\hfill \\ {y}_{n}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)T{x}_{n},\hfill \\ {x}_{n+1}={\alpha }_{n}f\left({y}_{n}\right)+\left(1-{\alpha }_{n}\right){y}_{n},\phantom{\rule{1em}{0ex}}n\ge 0,\hfill \end{array}$
where $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\lambda }_{i}\right\}$ are real number sequences in $\left[0,1\right]$ satisfying the following restrictions:
1. (a)

${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n}-{\alpha }_{n-1}|<\mathrm{\infty }$;

2. (b)

$1-\frac{\kappa }{{K}^{2}}\le {\beta }_{n}\le \beta <1$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n}-{\beta }_{n-1}|<\mathrm{\infty }$,

where β is some real number, and $\kappa :=min\left\{{\kappa }_{i}:1\le i\le N\right\}$. Then $\left\{{x}_{n}\right\}$ converges strongly as $n\to \mathrm{\infty }$ to some point in $F\left(T\right)$, which is the unique solution in $F\left(T\right)$, to the following variational inequality:
$〈f\left({x}^{\ast }\right)-{x}^{\ast },j\left({x}^{\ast }-p\right)〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in F\left(T\right).$

If E is a Hilbert space, then the best smooth constant $K=\frac{\sqrt{2}}{2}$. The following result can be deduced from Theorem 2.1 immediately.

Corollary 2.3 Let C be a nonempty closed convex subset of a real Hilbert space E and let N be some positive integer. Let ${T}_{i}:C\to C$ be a ${\kappa }_{i}$-strictly pseudocontractive mapping for each $1\le i\le N$. Assume that ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)\ne \mathrm{\varnothing }$. Let f be an α-contractive mapping. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following process:
$\left\{\begin{array}{c}{x}_{0}\in C\phantom{\rule{1em}{0ex}}\mathit{\text{arbitrarily chosen}},\hfill \\ {y}_{n}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){\sum }_{i=1}^{N}{\lambda }_{i}{T}_{i}{x}_{n},\hfill \\ {x}_{n+1}={\alpha }_{n}f\left({y}_{n}\right)+\left(1-{\alpha }_{n}\right){y}_{n},\phantom{\rule{1em}{0ex}}n\ge 0,\hfill \end{array}$
where $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ and $\left\{{\lambda }_{i}\right\}$ are real number sequences in $\left[0,1\right]$ satisfying the following restrictions:
1. (a)

${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n}-{\alpha }_{n-1}|<\mathrm{\infty }$;

2. (b)

$1-2\kappa \le {\beta }_{n}\le \beta <1$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n}-{\beta }_{n-1}|<\mathrm{\infty }$;

3. (c)

${\sum }_{n=1}^{N}{\lambda }_{i}=1$,

where β is some real number, and $\kappa :=min\left\{{\kappa }_{i}:1\le i\le N\right\}$. Then $\left\{{x}_{n}\right\}$ converges strongly as $n\to \mathrm{\infty }$ to some point in ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$, which is the unique solution in ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$, to the following variational inequality:
$〈f\left({x}^{\ast }\right)-{x}^{\ast },j\left({x}^{\ast }-p\right)〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in \bigcap _{i=1}^{N}F\left({T}_{i}\right).$

## Declarations

### Acknowledgements

This research was supported by the Natural Science Foundation of Hebei Province (A2010001943), the Science Foundation of Shijiazhuang Science and Technology Bureau (121130971) and the Science Foundation of Beijing Jiaotong University (2011YJS075). The authors are grateful to the referees for their valuable comments and suggestions which improved the contents of the article.

## Authors’ Affiliations

(1)
Department of Mathematics, Hangzhou Normal University, Hangzhou, 310036, China
(2)
Department of Mathematics, Gyeongsang National University, Jinju, 660-701, Korea
(3)
Department of Mathematics, School of Science, Beijing Jiaotong University, Beijing, 100044, China
(4)
Department of Mathematics, Shijiazhuang University, Shijiazhuang, 050035, China

## References 