• Research
• Open Access

# Iterative scheme for a nonexpansive mapping, an η-strictly pseudo-contractive mapping and variational inequality problems in a uniformly convex and 2-uniformly smooth Banach space

Fixed Point Theory and Applications20132013:23

https://doi.org/10.1186/1687-1812-2013-23

• Accepted: 13 January 2013
• Published:

## Abstract

In this paper, we introduce an iterative scheme by the modification of Mann’s iteration process for finding a common element of the set of solutions of a finite family of variational inequality problems and the set of fixed points of an η-strictly pseudo-contractive mapping and a nonexpansive mapping. Moreover, we prove a strong convergence theorem for finding a common element of the set of fixed points of a finite family of ${\eta }_{i}$-strictly pseudo-contractive mappings for every $i=1,2,\dots ,N$ in uniformly convex and 2-uniformly smooth Banach spaces.

## Keywords

• nonexpansive mapping
• strictly pseudo-contractive mapping
• variational inequality problem

## 1 Introduction

Let E be a Banach space with its dual space ${E}^{\ast }$ and let C be a nonempty closed convex subset of E. Throughout this paper, we denote the norm of E and ${E}^{\ast }$ by the same symbol $\parallel \cdot \parallel$. We use the symbol → to denote the strong convergence. Recall the following definition.

Definition 1.1 A Banach space E is said to be uniformly convex iff for any ϵ, $0<ϵ\le 2$, the inequalities $\parallel x\parallel \le 1$, $\parallel y\parallel \le 1$ and $\parallel x-y\parallel \ge ϵ$ imply there exists a $\delta >0$ such that $\parallel \frac{x+y}{2}\parallel \le 1-\delta$.

Definition 1.2 Let E be a Banach space. Then a function ${\rho }_{E}:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is said to be the modulus of smoothness of E if
${\rho }_{E}\left(t\right)=sup\left\{\frac{\parallel x+y\parallel +\parallel x-y\parallel }{2}-1:\parallel x\parallel =1,\parallel y\parallel =t\right\}.$
A Banach space E is said to be uniformly smooth if
$\underset{t\to 0}{lim}\frac{{\rho }_{E}\left(t\right)}{t}=0.$

Let $q>1$. A Banach space E is said to be q-uniformly smooth if there exists a fixed constant $c>0$ such that ${\rho }_{E}\left(t\right)\le c{t}^{q}$. It is easy to see that if E is q-uniformly smooth, then $q\le 2$ and E is uniformly smooth.

Definition 1.3 A mapping J from E onto ${E}^{\ast }$ satisfying the condition

is called the normalized duality mapping of E. The duality pair $〈x,f〉$ represents $f\left(x\right)$ for $f\in {E}^{\ast }$ and $x\in E$.

Definition 1.4 Let C be a nonempty subset of a Banach space E and $T:C\to C$ be a self-mapping. T is called a nonexpansive mapping if
$\parallel Tx-Ty\parallel \le \parallel x-y\parallel$

for all $x,y\in C$.

T is called an η-strictly pseudo-contractive mapping if there exists a constant $\eta \in \left(0,1\right)$ such that
$〈Tx-Ty,j\left(x-y\right)〉\le {\parallel x-y\parallel }^{2}-\eta {\parallel \left(I-T\right)x-\left(I-T\right)y\parallel }^{2}$
(1.1)
for every $x,y\in C$ and for some $j\left(x-y\right)\in J\left(x-y\right)$. It is clear that (1.1) is equivalent to the following:
$〈\left(I-T\right)x-\left(I-T\right)y,j\left(x-y\right)〉\ge \eta {\parallel \left(I-T\right)x-\left(I-T\right)y\parallel }^{2}$
(1.2)

for every $x,y\in C$ and for some $j\left(x-y\right)\in J\left(x-y\right)$.

Let C and D be nonempty subsets of a Banach space E such that C is nonempty closed convex and $D\subset C$, then a mapping $P:C\to D$ is sunny [1] provided $P\left(x+t\left(x-P\left(x\right)\right)\right)=P\left(x\right)$ for all $x\in C$ and $t\ge 0$, whenever $x+t\left(x-P\left(x\right)\right)\in C$. The mapping $P:C\to D$ is called a retraction if $Px=x$ for all $x\in D$. Furthermore, P is a sunny nonexpansive retraction from C onto D if P is a retraction from C onto D which is also sunny and nonexpansive. The subset D of C is called a sunny nonexpansive retraction of C if there exists a sunny nonexpansive retraction from C onto D.

An operator A of C into E is said to be accretive if there exists $j\left(x-y\right)\in J\left(x-y\right)$ such that
$〈Ax-Ay,j\left(x-y\right)〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
A mapping $A:C\to E$ is said to be α-inverse strongly accretive if there exists $j\left(x-y\right)\in J\left(x-y\right)$ and $\alpha >0$ such that
$〈Ax-Ay,j\left(x-y\right)〉\ge \alpha {\parallel Ax-Ay\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

Remark 1.1 From (1.1) and (1.2), if T is an η-strictly pseudo-contractive mapping, then $I-T$ is η-inverse strongly accretive.

The variational inequality problem in a Banach space is to find a point ${x}^{\ast }\in C$ such that for some $j\left(x-{x}^{\ast }\right)\in J\left(x-{x}^{\ast }\right)$,
$〈A{x}^{\ast },j\left(x-{x}^{\ast }\right)〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C.$
(1.3)
This problem was considered by Aoyama et al. [2]. The set of solutions of the variational inequality in a Banach space is denoted by $S\left(C,A\right)$, that is,
$S\left(C,A\right)=\left\{u\in C:〈Au,J\left(v-u\right)〉\ge 0,\mathrm{\forall }v\in C\right\}.$
(1.4)

Numerous problems in physics, optimization, variational inequalities, minimax problems, the Nash equilibrium problem in noncooperative games reduce to find an element of (1.4); see [3, 4].

Recall that the normal Mann’s iterative process was introduced by Mann [5] in 1953. The normal Mann’s iterative process generates a sequence $\left\{{x}_{n}\right\}$ in the following manner:
$\left\{\begin{array}{c}{x}_{1}\in C,\hfill \\ {x}_{n+1}=\left(1-{\alpha }_{n}\right){x}_{n}+{\alpha }_{n}T{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\hfill \end{array}$
(1.5)

where the sequence $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$. If T is a nonexpansive mapping with a fixed point and the control sequence $\left\{{\alpha }_{n}\right\}$ is chosen so that ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}\left(1-{\alpha }_{n}\right)=\mathrm{\infty }$, then the sequence $\left\{{x}_{n}\right\}$ generated by normal Mann’s iterative process (1.5) converges weakly to a fixed point of T.

In 2008, Cho et al. [6] modified the normal Mann’s iterative process and proved strong convergence for a finite family of nonexpansive mappings in the framework of Banach spaces without any commutative assumption as follows.

Theorem 1.2 Let C be a closed convex subset of a uniformly smooth and strictly convex Banach space E. Let $\left\{{T}_{i}\right\}$ be a nonexpansive mapping from C into itself for $i=1,2,\dots ,N$. Assume that $F={\bigcap }_{i=1}^{N}F\left({T}_{i}\right)\ne \mathrm{\varnothing }$. Given a point $u\in C$ and given sequences $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\}\in \left(0,1\right)$, the following conditions are satisfied:
Let $\left\{{x}_{n}\right\}$ be a sequence generated by $u,{x}_{0}=x\in C$ and
$\left\{\begin{array}{c}{y}_{n}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){W}_{n}{x}_{n},\hfill \\ {x}_{n+1}={\alpha }_{n}u+\left(1-{\alpha }_{n}\right){y}_{n},\phantom{\rule{1em}{0ex}}n\ge 0,\hfill \end{array}$
(1.6)

where ${W}_{n}$ is the W-mapping generated by ${T}_{1},{T}_{2},\dots ,{T}_{N}$ and ${\gamma }_{n1},{\gamma }_{n2},\dots ,{\gamma }_{nN}$. Then $\left\{{x}_{n}\right\}$ converges strongly to ${x}^{\ast }\in F$, where ${x}^{\ast }=Q\left(u\right)$ and $Q:C\to F$ is the unique sunny nonexpansive retraction from C onto F.

In 2008, Zhou [7] proved a strong convergence theorem for the modification of normal Mann’s iteration algorithm generated by a strict pseudo-contraction in a real 2-uniformly smooth Banach space as follows.

Theorem 1.3 Let C be a closed convex subset of a real 2-uniformly smooth Banach space E and let $T:C\to C$ be a λ-strict pseudo-contraction such that $F\left(T\right)\ne \mathrm{\varnothing }$. Given $u,{x}_{0}\in C$ and the sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n}\right\}$ and $\left\{{\delta }_{n}\right\}$ in $\left(0,1\right)$, the following control conditions are satisfied:
Let a sequence $\left\{{x}_{n}\right\}$ be generated by
$\left\{\begin{array}{c}{y}_{n}={\alpha }_{n}T{x}_{n}+\left(1-{\alpha }_{n}\right){x}_{n},\hfill \\ {x}_{n+1}={\beta }_{n}u+{\gamma }_{n}{x}_{n}+{\delta }_{n}{y}_{n},\phantom{\rule{1em}{0ex}}n\ge 0.\hfill \end{array}$
(1.7)

Then $\left\{{x}_{n}\right\}$ converges strongly to ${x}^{\ast }\in F\left(T\right)$, where ${x}^{\ast }={Q}_{F\left(T\right)}\left(u\right)$ and ${Q}_{F\left(T\right)}:C\to F\left(T\right)$ is the unique sunny nonexpansive retraction from C onto $F\left(T\right)$.

In 2005, Aoyama et al. [2] proved a weak convergence theorem for finding a solution of problem (1.3) as follows.

Theorem 1.4 Let E be a uniformly convex and 2-uniformly smooth Banach space and let C be a nonempty closed convex subset of E. Let ${Q}_{C}$ be a sunny nonexpansive retraction from E onto C, let $\alpha >0$ and let A be an α-inverse strongly accretive operator of C into E with $S\left(C,A\right)\ne \mathrm{\varnothing }$. Suppose ${x}_{1}=x\in C$ and $\left\{{x}_{n}\right\}$ is given by
${x}_{n+1}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right){Q}_{C}\left({x}_{n}-{\lambda }_{n}A{x}_{n}\right)$

for every $n=1,2,\dots$ , where $\left\{{\lambda }_{n}\right\}$ is a sequence of positive real numbers and $\left\{{\alpha }_{n}\right\}$ is a sequence in $\left[0,1\right]$. If $\left\{{\lambda }_{n}\right\}$ and $\left\{{\alpha }_{n}\right\}$ are chosen so that ${\lambda }_{n}\in \left[a,\frac{\alpha }{{K}^{2}}\right]$ for some $a>0$ and ${\alpha }_{n}\in \left[b,c\right]$ for some b, c with $0, then $\left\{{x}_{n}\right\}$ converges weakly to some element z of $S\left(C,A\right)$, where K is the 2-uniformly smoothness constant of E.

In this paper, motivated by Theorems 1.2, 1.3 and 1.4, we prove a strong convergence theorem for finding a common element of the set of solutions of a finite family of variational inequality problems and the set of fixed points of a nonexpansive mapping and an η-strictly pseudo-contractive mapping in uniformly convex and 2-uniformly smooth spaces. Moreover, by using our main result, we prove a strong convergence theorem for finding a common element of the set of fixed points of a finite family of ${\eta }_{i}$-strictly pseudo-contractive mappings for every $i=1,2,\dots ,N$ in uniformly convex and 2-uniformly smooth Banach spaces.

## 2 Preliminaries

In this section, we collect and prove the following lemmas to use in our main result.

Lemma 2.1 (See [8])

Let E be a real 2-uniformly smooth Banach space with the best smooth constant K. Then the following inequality holds:
${\parallel x+y\parallel }^{2}\le {\parallel x\parallel }^{2}+2〈y,J\left(x\right)〉+2{\parallel Ky\parallel }^{2}$

for any $x,y\in E$.

Definition 2.1 (See [9])

Let C be a nonempty convex subset of a real Banach space. Let ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ be a finite family of nonexpanxive mappings of C into itself and let ${\lambda }_{1},\dots ,{\lambda }_{N}$ be real numbers such that $0\le {\lambda }_{i}\le 1$ for every $i=1,\dots ,N$. Define a mapping $K:C\to C$ as follows:
$\begin{array}{r}{U}_{1}={\lambda }_{1}{T}_{1}+\left(1-{\lambda }_{1}\right)I,\\ {U}_{2}={\lambda }_{2}{T}_{2}{U}_{1}+\left(1-{\lambda }_{2}\right){U}_{1},\\ {U}_{3}={\lambda }_{3}{T}_{3}{U}_{2}+\left(1-{\lambda }_{3}\right){U}_{2},\\ ⋮\\ {U}_{N-1}={\lambda }_{N-1}{T}_{N-1}{U}_{N-2}+\left(1-{\lambda }_{N-1}\right){U}_{N-2},\\ K={U}_{N}={\lambda }_{N}{T}_{N}{U}_{N-1}+\left(1-{\lambda }_{N}\right){U}_{N-1}.\end{array}$
(2.1)

Such a mapping K is called the K-mapping generated by ${T}_{1},\dots ,{T}_{N}$ and ${\lambda }_{1},\dots ,{\lambda }_{N}$.

Lemma 2.2 (See [9])

Let C be a nonempty closed convex subset of a strictly convex Banach space. Let ${\left\{{T}_{i}\right\}}_{i=1}^{N}$ be a finite family of nonexpanxive mappings of C into itself with ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)\ne \mathrm{\varnothing }$ and let ${\lambda }_{1},\dots ,{\lambda }_{N}$ be real numbers such that $0<{\lambda }_{i}<1$ for every $i=1,\dots ,N-1$ and $0<{\lambda }_{N}\le 1$. Let K be the K-mapping generated by ${T}_{1},\dots ,{T}_{N}$ and ${\lambda }_{1},\dots ,{\lambda }_{N}$. Then $F\left(K\right)={\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$.

Remark 2.3 From Lemma 2.2, it is easy to see that the K mapping is a nonexpansive mapping.

Lemma 2.4 (See [10])

Let $\left\{{x}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ be bounded sequences in a Banach space X and let $\left\{{\beta }_{n}\right\}$ be a sequence in $\left[0,1\right]$ with $0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$. Suppose
${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){z}_{n}$
for all integer $n\ge 0$ and
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {z}_{n+1}-{z}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0.$

Then ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{z}_{n}\parallel =0$.

Lemma 2.5 (See [11])

Let X be a uniformly convex Banach space and ${B}_{r}=\left\{x\in X:\parallel x\parallel \le r\right\}$, $r>0$. Then there exists a continuous, strictly increasing and convex function $g:\left[0,\mathrm{\infty }\right]\to \left[0,\mathrm{\infty }\right]$, $g\left(0\right)=0$ such that
${\parallel \alpha x+\beta y+\gamma z\parallel }^{2}\le \alpha {\parallel x\parallel }^{2}+\beta {\parallel y\parallel }^{2}+\gamma {\parallel z\parallel }^{2}-\alpha \beta g\left(\parallel x-y\parallel \right)$

for all $x,y,z\in {B}_{r}$ and all $\alpha ,\beta ,\gamma \in \left[0,1\right]$ with $\alpha +\beta +\gamma =1$.

Lemma 2.6 (See [2])

Let C be a nonempty closed convex subset of a smooth Banach space E. Let ${Q}_{C}$ be a sunny nonexpansive retraction from E onto C and let A be an accretive operator of C into E. Then for all $\lambda >0$,
$S\left(C,A\right)=F\left({Q}_{C}\left(I-\lambda A\right)\right).$

Lemma 2.7 (See [12])

Let C be a closed convex subset of a strictly convex Banach space X. Let $\left\{{T}_{n}:n\in \mathbb{N}\right\}$ be a sequence of nonexpansive mappings on C. Suppose ${\bigcap }_{n=1}^{\mathrm{\infty }}F\left({T}_{n}\right)\ne \mathrm{\varnothing }$ is nonempty. Let $\left\{{\lambda }_{n}\right\}$ be a sequence of positive numbers with ${\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_{n}=1$. Then a mapping S on C defined by $Sx={\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_{n}{T}_{n}x$ for $x\in C$ is well defined, non-expansive and $F\left(S\right)={\bigcap }_{n=1}^{\mathrm{\infty }}F\left({T}_{n}\right)$ holds.

Lemma 2.8 (See [8])

Let $r>0$. If E is uniformly convex, then there exists a continuous, strictly increasing and convex function $g:\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$, $g\left(0\right)=0$ such that for all $x,y\in {B}_{r}\left(0\right)=\left\{x\in E:\parallel x\parallel \le r\right\}$ and for any $\alpha \in \left[0,1\right]$, we have ${\parallel \alpha x+\left(1-\alpha \right)y\parallel }^{2}\le \alpha {\parallel x\parallel }^{2}+\left(1-\alpha \right){\parallel y\parallel }^{2}-\alpha \left(1-\alpha \right)g\left(\parallel x-y\parallel \right)$.

Lemma 2.9 (See [13])

Let X be a uniformly smooth Banach space, C be a closed convex subset of X, $T:C\to C$ be a nonexpansive mapping with $F\left(T\right)\ne \mathrm{\varnothing }$ and let $f\in {\prod }_{C}$ where ${\prod }_{C}$ is to denote the collection of all contractions on C. Then the sequence $\left\{{x}_{t}\right\}$ defined by ${x}_{t}=tf\left({x}_{t}\right)+\left(1-t\right)T{x}_{t}$ converses strongly to a point in $F\left(T\right)$. If we define a mapping $Q:{\prod }_{C}\to F\left(T\right)$ by $Q\left(f\right)={lim}_{t\to 0}{x}_{t}$ for all $f\in {\prod }_{C}$, then $Q\left(f\right)$ solves the following variational inequality:
$〈\left(I-f\right)Q\left(f\right),j\left(Q\left(f\right)-p\right)〉\le 0$

for all $f\in {\prod }_{C}$, $p\in F\left(T\right)$.

Lemma 2.10 (See [14])

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

where $j\left(x+y\right)\in J\left(x+y\right)$.

Lemma 2.11 (See [15])

Let $\left\{{s}_{n}\right\}$ be a sequence of nonnegative real number satisfying
${s}_{n+1}=\left(1-{\alpha }_{n}\right){s}_{n}+{\alpha }_{n}{\beta }_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,$
where $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$ satisfy the conditions

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

Lemma 2.12 Let C be a nonempty closed convex subset of a 2-uniformly smooth Banach space E and let $T:C\to C$ be a nonexpansive mapping and $S:C\to C$ be an η-strictly pseudocontractive mapping with $F\left(S\right)\cap F\left(T\right)\ne \mathrm{\varnothing }$. Define a mapping ${B}_{A}:C\to C$ by ${B}_{A}x=T\left(\left(1-\alpha \right)I+\alpha S\right)x$ for all $x\in C$ and $\alpha \in \left(0,\frac{\eta }{{K}^{2}}\right)$, where K is the 2-uniformly smooth constant of E. Then $F\left({B}_{A}\right)=F\left(S\right)\cap F\left(T\right)$.

Proof It is easy to see that $F\left(T\right)\cap F\left(S\right)\subseteq F\left({B}_{A}\right)$. Let ${x}_{0}\in F\left({B}_{A}\right)$ and ${x}^{\ast }\in F\left(T\right)\cap F\left(S\right)$, we have
$\begin{array}{rcl}{\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}& =& {\parallel T\left(\left(1-\alpha \right){x}_{0}+\alpha S{x}_{0}\right)-{x}^{\ast }\parallel }^{2}\\ \le & {\parallel \left(1-\alpha \right){x}_{0}+\alpha S{x}_{0}-{x}^{\ast }\parallel }^{2}\\ =& {\parallel {x}_{0}-{x}^{\ast }+\alpha \left(S{x}_{0}-{x}_{0}\right)\parallel }^{2}\\ \le & {\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}+2\alpha 〈S{x}_{0}-{x}_{0},j\left({x}_{0}-{x}^{\ast }\right)〉+2{K}^{2}{\alpha }^{2}{\parallel S{x}_{0}-{x}_{0}\parallel }^{2}\\ =& {\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}+2\alpha 〈S{x}_{0}-{x}^{\ast },j\left({x}_{0}-{x}^{\ast }\right)〉+2\alpha 〈{x}^{\ast }-{x}_{0},j\left({x}_{0}-{x}^{\ast }\right)〉\\ +2{K}^{2}{\alpha }^{2}{\parallel S{x}_{0}-{x}_{0}\parallel }^{2}\\ =& {\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}+2\alpha 〈S{x}_{0}-{x}^{\ast },j\left({x}_{0}-{x}^{\ast }\right)〉-2\alpha {\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}+2{K}^{2}{\alpha }^{2}{\parallel S{x}_{0}-{x}_{0}\parallel }^{2}\\ \le & {\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}+2\alpha \left({\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}-\eta {\parallel \left(I-S\right){x}_{0}\parallel }^{2}\right)-2\alpha {\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}\\ +2{K}^{2}{\alpha }^{2}{\parallel S{x}_{0}-{x}_{0}\parallel }^{2}\\ =& {\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}-2\alpha \eta {\parallel {x}_{0}-S{x}_{0}\parallel }^{2}+2{K}^{2}{\alpha }^{2}{\parallel S{x}_{0}-{x}_{0}\parallel }^{2}\\ =& {\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}-2\alpha \left(\eta -{K}^{2}\alpha \right){\parallel {x}_{0}-S{x}_{0}\parallel }^{2}.\end{array}$
(2.2)
(2.2) implies that
$2\alpha \left(\eta -{K}^{2}\alpha \right){\parallel {x}_{0}-S{x}_{0}\parallel }^{2}\le {\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{0}-{x}^{\ast }\parallel }^{2}=0.$

Then we have $S{x}_{0}={x}_{0}$, that is, ${x}_{0}\in F\left(S\right)$.

Since ${x}_{0}\in F\left({B}_{A}\right)$, from the definition of ${B}_{A}$, we have
${x}_{0}={B}_{A}{x}_{0}=T\left(\left(1-\alpha \right){x}_{0}+\alpha S{x}_{0}\right)=T{x}_{0}.$

Then we have ${x}_{0}\in F\left(T\right)$. Therefore, ${x}_{0}\in F\left(T\right)\cap F\left(S\right)$. It follows that $F\left({B}_{A}\right)\subseteq F\left(T\right)\cap F\left(S\right)$. Hence, $F\left({B}_{A}\right)=F\left(T\right)\cap F\left(S\right)$. □

Remark 2.13 Applying (2.2), we have that the mapping ${B}_{A}$ is nonexpansive.

## 3 Main results

Theorem 3.1 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space E. Let ${Q}_{C}$ be the sunny nonexpansive retraction from E onto C. For every $i=1,2,\dots ,N$, let ${A}_{i}:C\to E$ be an ${\alpha }_{i}$-inverse strongly accretive mapping. Define a mapping ${G}_{i}:C\to C$ by ${Q}_{C}\left(I-{\lambda }_{i}{A}_{i}\right)x={G}_{i}x$ for all $x\in C$ and $i=1,2,\dots ,N$, where ${\lambda }_{i}\in \left(0,\frac{{\alpha }_{i}}{{K}^{2}}\right)$, K is the 2-uniformly smooth constant of E. Let $B:C\to C$ be the K-mapping generated by ${G}_{1},{G}_{2},\dots ,{G}_{N}$ and ${\rho }_{1},{\rho }_{2},\dots ,{\rho }_{N}$, where ${\rho }_{i}\in \left(0,1\right)$, $\mathrm{\forall }i=1,2,\dots ,N-1$ and ${\rho }_{N}\in \left(0,1\right]$. Let $T:C\to C$ be a nonexpansive mapping and $S:C\to C$ be an η-strictly pseudo-contractive mapping with $\mathcal{F}=F\left(S\right)\cap F\left(T\right)\cap {\bigcap }_{i=1}^{N}S\left(C,{A}_{i}\right)\ne \mathrm{\varnothing }$. Define a mapping ${B}_{A}:C\to C$ by $T\left(\left(1-\alpha \right)I+\alpha S\right)x={B}_{A}x$, $\mathrm{\forall }x\in C$ and $\alpha \in \left(0,\frac{\eta }{{K}^{2}}\right)$. Let $\left\{{x}_{n}\right\}$ be the sequence generated by ${x}_{1}\in C$ and
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
(3.1)
where $f:C\to C$ is a contractive mapping and $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\},\left\{{\gamma }_{n}\right\},\left\{{\delta }_{n}\right\}\subseteq \left[0,1\right]$, ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}+{\delta }_{n}=1$ and satisfy the following conditions:
Then the sequence $\left\{{x}_{n}\right\}$ converses strongly to $q\in \mathcal{F}$, which solves the following variational inequality:
$〈q-f\left(q\right),j\left(q-p\right)〉\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in \mathcal{F}.$

Proof First, we will show that ${G}_{i}$ is a nonexpansive mapping for every $i=1,2,\dots ,N$.

Let $x,y\in C$. From nonexpansiveness of ${Q}_{C}$, we have
$\begin{array}{rcl}{\parallel {G}_{i}x-{G}_{i}y\parallel }^{2}& =& {\parallel {Q}_{C}\left(I-{\lambda }_{i}{A}_{i}\right)x-{Q}_{C}\left(I-{\lambda }_{i}{A}_{i}\right)y\parallel }^{2}\\ \le & {\parallel \left(I-{\lambda }_{i}{A}_{i}\right)x-\left(I-{\lambda }_{i}{A}_{i}\right)y\parallel }^{2}\\ =& {\parallel x-y-{\lambda }_{i}\left({A}_{i}x-{A}_{i}y\right)\parallel }^{2}\\ \le & {\parallel x-y\parallel }^{2}-2{\lambda }_{i}〈{A}_{i}x-{A}_{i}y,j\left(x-y\right)〉+2{K}^{2}{\lambda }_{i}^{2}{\parallel {A}_{i}x-{A}_{i}y\parallel }^{2}\\ \le & {\parallel x-y\parallel }^{2}-2{\lambda }_{i}{\alpha }_{i}{\parallel {A}_{i}x-{A}_{i}y\parallel }^{2}+2{K}^{2}{\lambda }_{i}^{2}{\parallel {A}_{i}x-{A}_{i}y\parallel }^{2}\\ =& {\parallel x-y\parallel }^{2}-2{\lambda }_{i}\left({\alpha }_{i}-{K}^{2}{\lambda }_{i}\right){\parallel {A}_{i}x-{A}_{i}y\parallel }^{2}\\ \le & {\parallel x-y\parallel }^{2}.\end{array}$
Then we have ${G}_{i}$ is a nonexpansive mapping for every $i=1,2,\dots ,N$. Since $B:C\to C$ is the K-mapping generated by ${G}_{1},{G}_{2},\dots ,{G}_{N}$ and ${\rho }_{1},{\rho }_{2},\dots ,{\rho }_{N}$ and Lemma 2.2, we can conclude that $F\left(B\right)={\bigcap }_{i=1}^{N}F\left({G}_{i}\right)$. From Lemma 2.6 and the definition of ${G}_{i}$, we have $F\left({G}_{i}\right)=S\left(C,{A}_{i}\right)$ for every $i=1,2,\dots ,N$. Hence, we have
$F\left(B\right)=\bigcap _{i=1}^{N}F\left({G}_{i}\right)=\bigcap _{i=1}^{N}S\left(C,{A}_{i}\right).$
(3.2)

Next, we will show that the sequence $\left\{{x}_{n}\right\}$ is bounded.

Let $z\in \mathcal{F}$; from the definition of ${x}_{n}$, we have
$\begin{array}{rcl}\parallel {x}_{n+1}-z\parallel & \le & {\alpha }_{n}\parallel f\left({x}_{n}\right)-z\parallel +{\beta }_{n}\parallel {x}_{n}-z\parallel +{\gamma }_{n}\parallel B{x}_{n}-z\parallel +{\delta }_{n}\parallel {B}_{A}{x}_{n}-z\parallel \\ \le & {\alpha }_{n}\parallel f\left({x}_{n}\right)-z\parallel +\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-z\parallel \\ \le & {\alpha }_{n}\parallel f\left({x}_{n}\right)-f\left(z\right)\parallel +{\alpha }_{n}\parallel f\left(z\right)-z\parallel +\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-z\parallel \\ \le & {\alpha }_{n}a\parallel {x}_{n}-z\parallel +{\alpha }_{n}\parallel f\left(z\right)-z\parallel +\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-z\parallel \\ =& \left(1-{\alpha }_{n}\left(1-a\right)\right)\parallel {x}_{n}-z\parallel +{\alpha }_{n}\parallel f\left(z\right)-z\parallel \\ \le & max\left\{\parallel {x}_{1}-z\parallel ,\frac{\parallel f\left(z\right)-z\parallel }{1-a}\right\}.\end{array}$

By induction, we can conclude that the sequence $\left\{{x}_{n}\right\}$ is bounded and so are $\left\{f\left({x}_{n}\right)\right\}$, $\left\{B{x}_{n}\right\}$, $\left\{{B}_{A}{x}_{n}\right\}$.

Next, we will show that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =0.$
(3.3)
From the definition of ${x}_{n}$, we can rewrite ${x}_{n}$ by
${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){z}_{n},$
(3.4)

where ${z}_{n}=\frac{{\alpha }_{n}f\left({x}_{n}\right)+{\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n}}{1-{\beta }_{n}}$.

Since
$\begin{array}{rcl}\parallel {z}_{n+1}-{z}_{n}\parallel & =& \parallel \frac{{\alpha }_{n+1}f\left({x}_{n+1}\right)+{\gamma }_{n+1}B{x}_{n+1}+{\delta }_{n+1}{B}_{A}{x}_{n+1}}{1-{\beta }_{n+1}}\\ -\left(\frac{{\alpha }_{n}f\left({x}_{n}\right)+{\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n}}{1-{\beta }_{n}}\right)\parallel \\ =& \parallel \frac{{x}_{n+2}-{\beta }_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}-\frac{{x}_{n+1}-{\beta }_{n}{x}_{n}}{1-{\beta }_{n}}\parallel \\ =& \parallel \frac{{x}_{n+2}-{\beta }_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}-\frac{{x}_{n+1}-{\beta }_{n}{x}_{n}}{1-{\beta }_{n+1}}+\frac{{x}_{n+1}-{\beta }_{n}{x}_{n}}{1-{\beta }_{n+1}}-\frac{{x}_{n+1}-{\beta }_{n}{x}_{n}}{1-{\beta }_{n}}\parallel \\ \le & \parallel \frac{{x}_{n+2}-{\beta }_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}-\frac{{x}_{n+1}-{\beta }_{n}{x}_{n}}{1-{\beta }_{n+1}}\parallel +\parallel \frac{{x}_{n+1}-{\beta }_{n}{x}_{n}}{1-{\beta }_{n+1}}-\frac{{x}_{n+1}-{\beta }_{n}{x}_{n}}{1-{\beta }_{n}}\parallel \\ =& \frac{1}{1-{\beta }_{n+1}}\parallel {x}_{n+2}-{\beta }_{n+1}{x}_{n+1}-\left({x}_{n+1}-{\beta }_{n}{x}_{n}\right)\parallel \\ +|\frac{1}{1-{\beta }_{n+1}}-\frac{1}{1-{\beta }_{n}}|\parallel {x}_{n+1}-{\beta }_{n}{x}_{n}\parallel \\ =& \frac{1}{1-{\beta }_{n+1}}\parallel {x}_{n+2}-{\beta }_{n+1}{x}_{n+1}-\left({x}_{n+1}-{\beta }_{n}{x}_{n}\right)\parallel \\ +\frac{|{\beta }_{n+1}-{\beta }_{n}|}{\left(1-{\beta }_{n}\right)\left(1-{\beta }_{n+1}\right)}\parallel {x}_{n+1}-{\beta }_{n}{x}_{n}\parallel \\ =& \frac{1}{1-{\beta }_{n+1}}\parallel {\alpha }_{n+1}f\left({x}_{n+1}\right)+{\gamma }_{n+1}B{x}_{n+1}+{\delta }_{n+1}{B}_{A}{x}_{n+1}\\ -\left({\alpha }_{n}f\left({x}_{n}\right)+{\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n}\right)\parallel +\frac{|{\beta }_{n+1}-{\beta }_{n}|}{\left(1-{\beta }_{n}\right)\left(1-{\beta }_{n+1}\right)}\parallel {x}_{n+1}-{\beta }_{n}{x}_{n}\parallel \\ =& \frac{1}{1-{\beta }_{n+1}}\left(\parallel {\alpha }_{n+1}f\left({x}_{n+1}\right)-{\alpha }_{n}f\left({x}_{n}\right)\parallel +{\gamma }_{n+1}\parallel B{x}_{n+1}-B{x}_{n}\parallel \\ +{\delta }_{n+1}\parallel {B}_{A}{x}_{n+1}-{B}_{A}{x}_{n}\parallel +|{\gamma }_{n+1}-{\gamma }_{n}|\parallel B{x}_{n}\parallel +|{\delta }_{n+1}-{\delta }_{n}|\parallel {B}_{A}{x}_{n}\parallel \right)\\ +\frac{|{\beta }_{n+1}-{\beta }_{n}|}{\left(1-{\beta }_{n}\right)\left(1-{\beta }_{n+1}\right)}\parallel {x}_{n+1}-{\beta }_{n}{x}_{n}\parallel \\ \le & \frac{1}{1-{\beta }_{n+1}}\left({\alpha }_{n+1}\parallel f\left({x}_{n+1}\right)\parallel +{\alpha }_{n}\parallel f\left({x}_{n}\right)\parallel +\left({\gamma }_{n+1}+{\delta }_{n+1}\right)\parallel {x}_{n+1}-{x}_{n}\parallel \\ +|{\gamma }_{n+1}-{\gamma }_{n}|\parallel B{x}_{n}\parallel +|{\delta }_{n+1}-{\delta }_{n}|\parallel {B}_{A}{x}_{n}\parallel \right)\\ +\frac{|{\beta }_{n+1}-{\beta }_{n}|}{\left(1-{\beta }_{n}\right)\left(1-{\beta }_{n+1}\right)}\parallel {x}_{n+1}-{\beta }_{n}{x}_{n}\parallel \\ =& \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel f\left({x}_{n+1}\right)\parallel +\frac{{\alpha }_{n}}{1-{\beta }_{n+1}}\parallel f\left({x}_{n}\right)\parallel +\frac{{\gamma }_{n+1}+{\delta }_{n+1}}{1-{\beta }_{n+1}}\parallel {x}_{n+1}-{x}_{n}\parallel \\ +\frac{|{\gamma }_{n+1}-{\gamma }_{n}|}{1-{\beta }_{n+1}}\parallel B{x}_{n}\parallel +\frac{|{\delta }_{n+1}-{\delta }_{n}|}{1-{\beta }_{n+1}}\parallel {B}_{A}{x}_{n}\parallel \\ +\frac{|{\beta }_{n+1}-{\beta }_{n}|}{\left(1-{\beta }_{n}\right)\left(1-{\beta }_{n+1}\right)}\parallel {x}_{n+1}-{\beta }_{n}{x}_{n}\parallel \\ \le & \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel f\left({x}_{n+1}\right)\parallel +\frac{{\alpha }_{n}}{1-{\beta }_{n+1}}\parallel f\left({x}_{n}\right)\parallel +\parallel {x}_{n+1}-{x}_{n}\parallel +\frac{|{\gamma }_{n+1}-{\gamma }_{n}|}{1-{\beta }_{n+1}}\parallel B{x}_{n}\parallel \\ +\frac{|{\delta }_{n+1}-{\delta }_{n}|}{1-{\beta }_{n+1}}\parallel {B}_{A}{x}_{n}\parallel +\frac{|{\beta }_{n+1}-{\beta }_{n}|}{\left(1-{\beta }_{n}\right)\left(1-{\beta }_{n+1}\right)}\parallel {x}_{n+1}-{\beta }_{n}{x}_{n}\parallel .\end{array}$
(3.5)
From (3.5) and the conditions (i)-(iv), we have
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {z}_{n+1}-{z}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0.$
(3.6)
From Lemma 2.4 and (3.4), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{n}-{x}_{n}\parallel =0.$
(3.7)
From (3.4), we have
$\parallel {x}_{n+1}-{x}_{n}\parallel =\left(1-{\beta }_{n}\right)\parallel {z}_{n}-{x}_{n}\parallel ,$
and from the condition (iv) and (3.7), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =0.$
Next, we will show that
$\underset{n\to \mathrm{\infty }}{lim}\parallel B{x}_{n}-{x}_{n}\parallel =0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\parallel {B}_{A}{x}_{n}-{x}_{n}\parallel =0.$
From the definition of ${x}_{n}$, we can rewrite ${x}_{n+1}$ by
$\begin{array}{rcl}{x}_{n+1}& =& {\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n}\\ =& {\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+\left({\gamma }_{n}+{\delta }_{n}\right)\frac{\left({\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n}\right)}{{\gamma }_{n}+{\delta }_{n}}\\ =& {\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{e}_{n}{z}_{n}^{\mathrm{\prime }},\end{array}$
(3.8)

where ${e}_{n}={\gamma }_{n}+{\delta }_{n}$ and ${z}_{n}^{\mathrm{\prime }}=\frac{\left({\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n}\right)}{{\gamma }_{n}+{\delta }_{n}}$.

From Lemma 2.5 and (3.8), we have
$\begin{array}{rcl}{\parallel {x}_{n+1}-z\parallel }^{2}& =& {\parallel {\alpha }_{n}\left(f\left({x}_{n}\right)-z\right)+{\beta }_{n}\left({x}_{n}-z\right)+{e}_{n}\left({z}_{n}^{\mathrm{\prime }}-z\right)\parallel }^{2}\\ \le & {\alpha }_{n}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}+{\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{e}_{n}{\parallel {z}_{n}^{\mathrm{\prime }}-z\parallel }^{2}-{\beta }_{n}{e}_{n}{g}_{1}\left(\parallel {z}_{n}^{\mathrm{\prime }}-{x}_{n}\parallel \right)\\ =& {\alpha }_{n}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}+{\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}-{\beta }_{n}{e}_{n}{g}_{1}\left(\parallel {z}_{n}^{\mathrm{\prime }}-{x}_{n}\parallel \right)\\ +{e}_{n}{\parallel \frac{\left({\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n}\right)}{{\gamma }_{n}+{\delta }_{n}}-z\parallel }^{2}\\ =& {\alpha }_{n}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}+{\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}-{\beta }_{n}{e}_{n}{g}_{1}\left(\parallel {z}_{n}^{\mathrm{\prime }}-{x}_{n}\parallel \right)\\ +{e}_{n}{\parallel \left(1-\frac{{\delta }_{n}}{{\gamma }_{n}+{\delta }_{n}}\right)\left(B{x}_{n}-z\right)+\frac{{\delta }_{n}}{{\gamma }_{n}+{\delta }_{n}}\left({B}_{A}{x}_{n}-z\right)\parallel }^{2}\\ \le & {\alpha }_{n}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}+{\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}-{\beta }_{n}{e}_{n}{g}_{1}\left(\parallel {z}_{n}^{\mathrm{\prime }}-{x}_{n}\parallel \right)\\ +{e}_{n}{\left(\left(1-\frac{{\delta }_{n}}{{\gamma }_{n}+{\delta }_{n}}\right)\parallel B{x}_{n}-z\parallel +\frac{{\delta }_{n}}{{\gamma }_{n}+{\delta }_{n}}\parallel {B}_{A}{x}_{n}-z\parallel \right)}^{2}\\ \le & {\alpha }_{n}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}+{\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}-{\beta }_{n}{e}_{n}{g}_{1}\left(\parallel {z}_{n}^{\mathrm{\prime }}-{x}_{n}\parallel \right)+{e}_{n}{\parallel {x}_{n}-z\parallel }^{2}\\ \le & {\alpha }_{n}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}+{\parallel {x}_{n}-z\parallel }^{2}-{\beta }_{n}{e}_{n}{g}_{1}\left(\parallel {z}_{n}^{\mathrm{\prime }}-{x}_{n}\parallel \right),\end{array}$
which implies that
$\begin{array}{rcl}{\beta }_{n}{e}_{n}{g}_{1}\left(\parallel {z}_{n}^{\mathrm{\prime }}-{x}_{n}\parallel \right)& \le & {\alpha }_{n}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}+{\parallel {x}_{n}-z\parallel }^{2}-{\parallel {x}_{n+1}-z\parallel }^{2}\\ \le & {\alpha }_{n}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}+\left(\parallel {x}_{n}-z\parallel +\parallel {x}_{n+1}-z\parallel \right)\parallel {x}_{n+1}-{x}_{n}\parallel .\end{array}$
(3.9)
From the conditions (i), (ii), (iv) and (3.3), we have
$\underset{n\to \mathrm{\infty }}{lim}{g}_{1}\left(\parallel {z}_{n}^{\mathrm{\prime }}-{x}_{n}\parallel \right)=0.$
From the properties of ${g}_{1}$, we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{n}^{\mathrm{\prime }}-{x}_{n}\parallel =0.$
(3.10)
From Lemma 2.8 and the definition of ${z}_{n}^{\mathrm{\prime }}$, we have
$\begin{array}{rcl}{\parallel {z}_{n}^{\mathrm{\prime }}-z\parallel }^{2}& =& {\parallel \frac{\left({\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n}\right)}{{\gamma }_{n}+{\delta }_{n}}-z\parallel }^{2}\\ =& {\parallel \left(1-\frac{{\delta }_{n}}{{\delta }_{n}+{\gamma }_{n}}\right)\left(B{x}_{n}-z\right)+\frac{{\delta }_{n}}{{\delta }_{n}+{\gamma }_{n}}\left({B}_{A}{x}_{n}-z\right)\parallel }^{2}\\ \le & \left(1-\frac{{\delta }_{n}}{{\delta }_{n}+{\gamma }_{n}}\right){\parallel B{x}_{n}-z\parallel }^{2}+\frac{{\delta }_{n}}{{\delta }_{n}+{\gamma }_{n}}{\parallel {B}_{A}{x}_{n}-z\parallel }^{2}\\ -\left(1-\frac{{\delta }_{n}}{{\delta }_{n}+{\gamma }_{n}}\right)\frac{{\delta }_{n}}{{\delta }_{n}+{\gamma }_{n}}{g}_{2}\left(\parallel B{x}_{n}-{B}_{A}{x}_{n}\parallel \right)\\ \le & {\parallel {x}_{n}-z\parallel }^{2}-\left(1-\frac{{\delta }_{n}}{{\delta }_{n}+{\gamma }_{n}}\right)\frac{{\delta }_{n}}{{\delta }_{n}+{\gamma }_{n}}{g}_{2}\left(\parallel B{x}_{n}-{B}_{A}{x}_{n}\parallel \right),\end{array}$
which implies that
$\begin{array}{rcl}\left(1-\frac{{\delta }_{n}}{{\delta }_{n}+{\gamma }_{n}}\right)\frac{{\delta }_{n}}{{\delta }_{n}+{\gamma }_{n}}{g}_{2}\left(\parallel B{x}_{n}-{B}_{A}{x}_{n}\parallel \right)& \le & {\parallel {x}_{n}-z\parallel }^{2}-{\parallel {z}_{n}^{\mathrm{\prime }}-z\parallel }^{2}\\ \le & \left(\parallel {x}_{n}-z\parallel +\parallel {z}_{n}^{\mathrm{\prime }}-z\parallel \right)\parallel {z}_{n}^{\mathrm{\prime }}-{x}_{n}\parallel .\end{array}$
From the condition (iii) and (3.10), we have
$\underset{n\to \mathrm{\infty }}{lim}{g}_{2}\left(\parallel B{x}_{n}-{B}_{A}{x}_{n}\parallel \right)=0.$
From the properties of ${g}_{2}$, we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel B{x}_{n}-{B}_{A}{x}_{n}\parallel =0.$
(3.11)
From the definition of ${x}_{n}$, we can rewrite ${x}_{n+1}$ by
$\begin{array}{rcl}{x}_{n+1}& =& {\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n}\\ =& {\beta }_{n}{x}_{n}+{\gamma }_{n}B{x}_{n}+\left({\alpha }_{n}+{\delta }_{n}\right)\frac{{\alpha }_{n}f\left({x}_{n}\right)+{\delta }_{n}{B}_{A}{x}_{n}}{{\alpha }_{n}+{\delta }_{n}}\\ =& {\beta }_{n}{x}_{n}+{\gamma }_{n}B{x}_{n}+{d}_{n}{z}_{n}^{\mathrm{\prime }\mathrm{\prime }},\end{array}$
(3.12)

where ${d}_{n}={\alpha }_{n}+{\delta }_{n}$ and ${z}_{n}^{\mathrm{\prime }\mathrm{\prime }}=\frac{{\alpha }_{n}f\left({x}_{n}\right)+{\delta }_{n}{B}_{A}{x}_{n}}{{\alpha }_{n}+{\delta }_{n}}$.

From Lemma 2.5 and the convexity of ${\parallel \cdot \parallel }^{2}$, we have
$\begin{array}{rcl}{\parallel {x}_{n+1}-z\parallel }^{2}& =& {\parallel {\beta }_{n}\left({x}_{n}-z\right)+{\gamma }_{n}\left(B{x}_{n}-z\right)+{d}_{n}\left({z}_{n}^{\mathrm{\prime }\mathrm{\prime }}-z\right)\parallel }^{2}\\ \le & {\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{\gamma }_{n}{\parallel B{x}_{n}-z\parallel }^{2}+{d}_{n}{\parallel {z}_{n}^{\mathrm{\prime }\mathrm{\prime }}-z\parallel }^{2}-{\beta }_{n}{\gamma }_{n}{g}_{3}\left(\parallel {x}_{n}-B{x}_{n}\parallel \right)\\ =& {\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{\gamma }_{n}{\parallel B{x}_{n}-z\parallel }^{2}+{d}_{n}{\parallel \frac{{\alpha }_{n}f\left({x}_{n}\right)+{\delta }_{n}{B}_{A}{x}_{n}}{{\alpha }_{n}+{\delta }_{n}}-z\parallel }^{2}\\ -{\beta }_{n}{\gamma }_{n}{g}_{3}\left(\parallel {x}_{n}-B{x}_{n}\parallel \right)\\ =& {\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{\gamma }_{n}{\parallel B{x}_{n}-z\parallel }^{2}+{d}_{n}\parallel \frac{{\alpha }_{n}}{{\alpha }_{n}+{\delta }_{n}}\left(f\left({x}_{n}\right)-z\right)\\ +\left(1-\frac{{\alpha }_{n}}{{\alpha }_{n}+{\delta }_{n}}\right)\left({B}_{A}{x}_{n}-z\right){\parallel }^{2}-{\beta }_{n}{\gamma }_{n}{g}_{3}\left(\parallel {x}_{n}-B{x}_{n}\parallel \right)\\ \le & {{\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{\gamma }_{n}{\parallel B{x}_{n}-z\parallel }^{2}+{d}_{n}\left(\frac{{\alpha }_{n}}{{\alpha }_{n}+{\delta }_{n}}\parallel f\left({x}_{n}\right)-z\parallel }^{2}\\ +\left(1-\frac{{\alpha }_{n}}{{\alpha }_{n}+{\delta }_{n}}\right){\parallel {B}_{A}{x}_{n}-z\parallel }^{2}\right)-{\beta }_{n}{\gamma }_{n}{g}_{3}\left(\parallel {x}_{n}-B{x}_{n}\parallel \right)\\ =& {\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{\gamma }_{n}{\parallel B{x}_{n}-z\parallel }^{2}+{d}_{n}\frac{{\alpha }_{n}}{{\alpha }_{n}+{\delta }_{n}}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}\\ +{d}_{n}\left(1-\frac{{\alpha }_{n}}{{\alpha }_{n}+{\delta }_{n}}\right){\parallel {B}_{A}{x}_{n}-z\parallel }^{2}-{\beta }_{n}{\gamma }_{n}{g}_{3}\left(\parallel {x}_{n}-B{x}_{n}\parallel \right)\\ \le & {\beta }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{\gamma }_{n}{\parallel {x}_{n}-z\parallel }^{2}+{d}_{n}\frac{{\alpha }_{n}}{{\alpha }_{n}+{\delta }_{n}}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}\\ +{d}_{n}{\parallel {x}_{n}-z\parallel }^{2}-{\beta }_{n}{\gamma }_{n}{g}_{3}\left(\parallel {x}_{n}-B{x}_{n}\parallel \right)\\ \le & {\parallel {x}_{n}-z\parallel }^{2}+{d}_{n}\frac{{\alpha }_{n}}{{\alpha }_{n}+{\delta }_{n}}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}-{\beta }_{n}{\gamma }_{n}{g}_{3}\left(\parallel {x}_{n}-B{x}_{n}\parallel \right),\end{array}$
(3.13)
which implies that
$\begin{array}{rcl}{\beta }_{n}{\gamma }_{n}{g}_{3}\left(\parallel {x}_{n}-B{x}_{n}\parallel \right)& \le & {\parallel {x}_{n}-z\parallel }^{2}-{\parallel {x}_{n+1}-z\parallel }^{2}+{d}_{n}\frac{{\alpha }_{n}}{{\alpha }_{n}+{\delta }_{n}}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}\\ \le & \left(\parallel {x}_{n}-z\parallel +\parallel {x}_{n+1}-z\parallel \right)\parallel {x}_{n+1}-{x}_{n}\parallel \\ +{d}_{n}\frac{{\alpha }_{n}}{{\alpha }_{n}+{\delta }_{n}}{\parallel f\left({x}_{n}\right)-z\parallel }^{2}.\end{array}$
(3.14)
From the conditions (i), (ii), (iv) (3.14) and (3.3), we have
$\underset{n\to \mathrm{\infty }}{lim}{g}_{3}\left(\parallel {x}_{n}-B{x}_{n}\parallel \right)=0.$
From the properties of ${g}_{3}$, we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-B{x}_{n}\parallel =0.$
(3.15)
From (3.11), (3.15) and
$\parallel {x}_{n}-{B}_{A}{x}_{n}\parallel \le \parallel {x}_{n}-B{x}_{n}\parallel +\parallel B{x}_{n}-{B}_{A}{x}_{n}\parallel ,$
we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{B}_{A}{x}_{n}\parallel =0.$
(3.16)

Define a mapping $L:C\to C$ by $Lx=\left(1-ϵ\right)Bx+ϵ{B}_{A}x$ for all $x\in C$ and $ϵ\in \left(0,1\right)$. From Lemma 2.7, 2.12 and (3.2), we have $F\left(L\right)=F\left(B\right)\cap F\left({B}_{A}\right)={\bigcap }_{i=1}^{N}S\left(C,{A}_{i}\right)\cap F\left(S\right)\cap F\left(T\right)=\mathcal{F}$.

From (3.15) and (3.16) and
$\begin{array}{rcl}\parallel {x}_{n}-L{x}_{n}\parallel & =& \parallel \left(1-ϵ\right)\left({x}_{n}-B{x}_{n}\right)+ϵ\left({x}_{n}-{B}_{A}{x}_{n}\right)\parallel \\ \le & \left(1-ϵ\right)\parallel {x}_{n}-B{x}_{n}\parallel +ϵ\parallel {x}_{n}-{B}_{A}{x}_{n}\parallel ,\end{array}$
we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-L{x}_{n}\parallel =0.$
(3.17)
Next, we will show that
$\underset{n\to \mathrm{\infty }}{lim sup}〈f\left(q\right)-q,j\left({x}_{n}-q\right)〉\le 0,$
(3.18)
where ${lim}_{t\to 0}{x}_{t}=q\in \mathcal{F}$ and ${x}_{t}$ begins the fixed point of the contraction
$x↦tf\left(x\right)+\left(1-t\right)Lx.$

Then ${x}_{t}$ solves the fixed point equation ${x}_{t}=tf\left({x}_{t}\right)+\left(1-t\right)L{x}_{t}$.

From the definition of ${x}_{t}$, we have
$\begin{array}{rcl}{\parallel {x}_{t}-{x}_{n}\parallel }^{2}& =& {\parallel t\left(f\left({x}_{t}\right)-{x}_{n}\right)+\left(1-t\right)\left(L{x}_{t}-{x}_{n}\right)\parallel }^{2}\\ \le & {\left(1-t\right)}^{2}{\parallel L{x}_{t}-{x}_{n}\parallel }^{2}+2t〈f\left({x}_{t}\right)-{x}_{n},j\left({x}_{t}-{x}_{n}\right)〉\\ \le & {\left(1-t\right)}^{2}{\left(\parallel L{x}_{t}-L{x}_{n}\parallel +\parallel L{x}_{n}-{x}_{n}\parallel \right)}^{2}+2t〈f\left({x}_{t}\right)-{x}_{n},j\left({x}_{t}-{x}_{n}\right)〉\\ \le & {\left(1-t\right)}^{2}{\left(\parallel {x}_{t}-{x}_{n}\parallel +\parallel L{x}_{n}-{x}_{n}\parallel \right)}^{2}+2t〈f\left({x}_{t}\right)-{x}_{n},j\left({x}_{t}-{x}_{n}\right)〉\\ =& {\left(1-t\right)}^{2}\left({\parallel {x}_{t}-{x}_{n}\parallel }^{2}+2\parallel {x}_{t}-{x}_{n}\parallel \parallel L{x}_{n}-{x}_{n}\parallel +{\parallel L{x}_{n}-{x}_{n}\parallel }^{2}\right)\\ +2t〈f\left({x}_{t}\right)-{x}_{n},j\left({x}_{t}-{x}_{n}\right)〉\\ =& {\left(1-t\right)}^{2}\left({\parallel {x}_{t}-{x}_{n}\parallel }^{2}+2\parallel {x}_{t}-{x}_{n}\parallel \parallel L{x}_{n}-{x}_{n}\parallel +{\parallel L{x}_{n}-{x}_{n}\parallel }^{2}\right)\\ +2t〈f\left({x}_{t}\right)-{x}_{t},j\left({x}_{t}-{x}_{n}\right)〉+2t〈{x}_{t}-{x}_{n},j\left({x}_{t}-{x}_{n}\right)〉\\ =& \left(1-2t+{t}^{2}\right){\parallel {x}_{t}-{x}_{n}\parallel }^{2}+{\left(1-t\right)}^{2}\left(2\parallel {x}_{t}-{x}_{n}\parallel \parallel L{x}_{n}-{x}_{n}\parallel +{\parallel L{x}_{n}-{x}_{n}\parallel }^{2}\right)\\ +2t〈f\left({x}_{t}\right)-{x}_{t},j\left({x}_{t}-{x}_{n}\right)〉+2t{\parallel {x}_{t}-{x}_{n}\parallel }^{2}\\ =& \left(1+{t}^{2}\right){\parallel {x}_{t}-{x}_{n}\parallel }^{2}+{f}_{n}\left(t\right)+2t〈f\left({x}_{t}\right)-{x}_{t},j\left({x}_{t}-{x}_{n}\right)〉,\end{array}$
(3.19)
where ${f}_{n}\left(t\right)={\left(1-t\right)}^{2}\left(2\parallel {x}_{t}-{x}_{n}\parallel \parallel L{x}_{n}-{x}_{n}\parallel +{\parallel L{x}_{n}-{x}_{n}\parallel }^{2}\right)$. From (3.17), we have
$\underset{n\to \mathrm{\infty }}{lim}{f}_{n}\left(t\right)=0.$
(3.20)
(3.19) implies that
$\begin{array}{rcl}〈{x}_{t}-f\left({x}_{t}\right),j\left({x}_{t}-{x}_{n}\right)〉& \le & \frac{t}{2}{\parallel {x}_{t}-{x}_{n}\parallel }^{2}+\frac{1}{2t}{f}_{n}\left(t\right)\\ \le & \frac{t}{2}D+\frac{1}{2t}{f}_{n}\left(t\right),\end{array}$
(3.21)
where $D>0$ such that ${\parallel {x}_{t}-{x}_{n}\parallel }^{2}\le D$ for all $t\in \left(0,1\right)$ and $n\ge 1$. From (3.20) and (3.21), we have
$\underset{n\to \mathrm{\infty }}{lim sup}〈{x}_{t}-f\left({x}_{t}\right),j\left({x}_{t}-{x}_{n}\right)〉\le \frac{t}{2}D.$
(3.22)
From (3.22) taking $t\to 0$, we have
$\underset{t\to 0}{lim sup}\phantom{\rule{0.2em}{0ex}}\underset{n\to \mathrm{\infty }}{lim sup}〈{x}_{t}-f\left({x}_{t}\right),j\left({x}_{t}-{x}_{n}\right)〉\le 0.$
(3.23)
Since
$\begin{array}{rcl}〈f\left(q\right)-q,j\left({x}_{n}-q\right)〉& =& 〈f\left(q\right)-q,j\left({x}_{n}-q\right)〉-〈f\left(q\right)-q,j\left({x}_{n}-{x}_{t}\right)〉+〈f\left(q\right)-q,j\left({x}_{n}-{x}_{t}\right)〉\\ -〈f\left(q\right)-{x}_{t},j\left({x}_{n}-{x}_{t}\right)〉+〈f\left(q\right)-{x}_{t},j\left({x}_{n}-{x}_{t}\right)〉\\ -〈f\left({x}_{t}\right)-{x}_{t},j\left({x}_{n}-{x}_{t}\right)〉+〈f\left({x}_{t}\right)-{x}_{t},j\left({x}_{n}-{x}_{t}\right)〉\\ =& 〈f\left(q\right)-q,j\left({x}_{n}-q\right)-j\left({x}_{n}-{x}_{t}\right)〉+〈{x}_{t}-q,j\left({x}_{n}-{x}_{t}\right)〉\\ +〈f\left(q\right)-f\left({x}_{t}\right),j\left({x}_{n}-{x}_{t}\right)〉+〈f\left({x}_{t}\right)-{x}_{t},j\left({x}_{n}-{x}_{t}\right)〉\\ \le & 〈f\left(q\right)-q,j\left({x}_{n}-q\right)-j\left({x}_{n}-{x}_{t}\right)〉+\parallel {x}_{t}-q\parallel \parallel {x}_{n}-{x}_{t}\parallel \\ +a\parallel q-{x}_{t}\parallel \parallel {x}_{n}-{x}_{t}\parallel +〈f\left({x}_{t}\right)-{x}_{t},j\left({x}_{n}-{x}_{t}\right)〉,\end{array}$
it follows that
$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim sup}〈f\left(q\right)-q,j\left({x}_{n}-q\right)〉& \le & \underset{n\to \mathrm{\infty }}{lim sup}〈f\left(q\right)-q,j\left({x}_{n}-q\right)-j\left({x}_{n}-{x}_{t}\right)〉\\ +\parallel {x}_{t}-q\parallel \underset{n\to \mathrm{\infty }}{lim sup}\parallel {x}_{n}-{x}_{t}\parallel +a\parallel q-{x}_{t}\parallel \underset{n\to \mathrm{\infty }}{lim sup}\parallel {x}_{n}-{x}_{t}\parallel \\ +\underset{n\to \mathrm{\infty }}{lim sup}〈f\left({x}_{t}\right)-{x}_{t},j\left({x}_{n}-{x}_{t}\right)〉.\end{array}$
(3.24)
Since j is norm-to-norm uniformly continuous on a bounded subset of C and (3.24), then we have
$\underset{n\to \mathrm{\infty }}{lim sup}〈f\left(q\right)-q,j\left({x}_{n}-q\right)〉=\underset{t\to 0}{lim sup}\phantom{\rule{0.2em}{0ex}}\underset{n\to \mathrm{\infty }}{lim sup}〈f\left(q\right)-q,j\left({x}_{n}-q\right)〉\le 0.$
Finally, we will show the sequence $\left\{{x}_{n}\right\}$ converses strongly to $q\in \mathcal{F}$. From the definition of ${x}_{n}$, we have
$\begin{array}{rcl}{\parallel {x}_{n+1}-q\parallel }^{2}& =& {\parallel {\alpha }_{n}\left(f\left({x}_{n}\right)-q\right)+{\beta }_{n}\left({x}_{n}-q\right)+{\gamma }_{n}\left(B{x}_{n}-q\right)+{\delta }_{n}\left({B}_{A}{x}_{n}-q\right)\parallel }^{2}\\ \le & {\parallel {\beta }_{n}\left({x}_{n}-q\right)+{\gamma }_{n}\left(B{x}_{n}-q\right)+{\delta }_{n}\left({B}_{A}{x}_{n}-q\right)\parallel }^{2}\\ +2{\alpha }_{n}〈f\left({x}_{n}\right)-q,j\left({x}_{n+1}-q\right)〉\\ \le & {\left({\beta }_{n}\parallel {x}_{n}-q\parallel +{\gamma }_{n}\parallel B{x}_{n}-q\parallel +{\delta }_{n}\parallel {B}_{A}{x}_{n}-q\parallel \right)}^{2}\\ +2{\alpha }_{n}〈f\left({x}_{n}\right)-f\left(q\right),j\left({x}_{n+1}-q\right)〉+2{\alpha }_{n}〈f\left(q\right)-q,j\left({x}_{n+1}-q\right)〉\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{\parallel {x}_{n}-q\parallel }^{2}+2{\alpha }_{n}〈f\left({x}_{n}\right)-f\left(q\right),j\left({x}_{n+1}-q\right)〉\\ +2{\alpha }_{n}〈f\left(q\right)-q,j\left({x}_{n+1}-q\right)〉\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{\parallel {x}_{n}-q\parallel }^{2}+2a{\alpha }_{n}\parallel {x}_{n}-q\parallel \parallel {x}_{n+1}-q\parallel \\ +2{\alpha }_{n}〈f\left(q\right)-q,j\left({x}_{n+1}-q\right)〉\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{\parallel {x}_{n}-q\parallel }^{2}+a{\alpha }_{n}{\parallel {x}_{n}-q\parallel }^{2}+a{\alpha }_{n}{\parallel {x}_{n+1}-q\parallel }^{2}\\ +2{\alpha }_{n}〈f\left(q\right)-q,j\left({x}_{n+1}-q\right)〉\\ =& \left(1-2{\alpha }_{n}+{\alpha }_{n}^{2}\right){\parallel {x}_{n}-q\parallel }^{2}+a{\alpha }_{n}{\parallel {x}_{n}-q\parallel }^{2}+a{\alpha }_{n}{\parallel {x}_{n+1}-q\parallel }^{2}\\ +2{\alpha }_{n}〈f\left(q\right)-q,j\left({x}_{n+1}-q\right)〉\\ =& \left(1-2{\alpha }_{n}+a{\alpha }_{n}\right){\parallel {x}_{n}-q\parallel }^{2}+{\alpha }_{n}^{2}{\parallel {x}_{n}-q\parallel }^{2}+a{\alpha }_{n}{\parallel {x}_{n+1}-q\parallel }^{2}\\ +2{\alpha }_{n}〈f\left(q\right)-q,j\left({x}_{n+1}-q\right)〉\\ =& \left(1-a{\alpha }_{n}-2{\alpha }_{n}+2a{\alpha }_{n}\right){\parallel {x}_{n}-q\parallel }^{2}+{\alpha }_{n}^{2}{\parallel {x}_{n}-q\parallel }^{2}+a{\alpha }_{n}{\parallel {x}_{n+1}-q\parallel }^{2}\\ +2{\alpha }_{n}〈f\left(q\right)-q,j\left({x}_{n+1}-q\right)〉\\ =& \left(1-a{\alpha }_{n}-2{\alpha }_{n}\left(1-a\right)\right){\parallel {x}_{n}-q\parallel }^{2}+{\alpha }_{n}^{2}{\parallel {x}_{n}-q\parallel }^{2}+a{\alpha }_{n}{\parallel {x}_{n+1}-q\parallel }^{2}\\ +2{\alpha }_{n}〈f\left(q\right)-q,j\left({x}_{n+1}-q\right)〉,\end{array}$
which implies that
$\begin{array}{rcl}{\parallel {x}_{n+1}-q\parallel }^{2}& \le & \left(1-\frac{2{\alpha }_{n}\left(1-a\right)}{1-a{\alpha }_{n}}\right){\parallel {x}_{n}-q\parallel }^{2}\\ +\frac{{\alpha }_{n}}{1-a{\alpha }_{n}}\left({\alpha }_{n}{\parallel {x}_{n}-q\parallel }^{2}+2〈f\left(q\right)-q,j\left({x}_{n+1}-q\right)〉\right)\\ \le & \left(1-\frac{2{\alpha }_{n}\left(1-a\right)}{1-a{\alpha }_{n}}\right){\parallel {x}_{n}-q\parallel }^{2}\\ +\frac{2{\alpha }_{n}\left(1-a\right)}{1-a{\alpha }_{n}}\cdot \frac{1}{2\left(1-a\right)}\left({\alpha }_{n}{\parallel {x}_{n}-q\parallel }^{2}+2〈f\left(q\right)-q,j\left({x}_{n+1}-q\right)〉\right).\end{array}$

From the condition (i) and Lemma 2.11, we can imply that $\left\{{x}_{n}\right\}$ converses strongly to $q\in \mathcal{F}$. This completes the proof. □

The following results can be obtained from Theorem 3.1. We, therefore, omit the proof.

Corollary 3.2 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space E. Let ${Q}_{C}$ be the sunny nonexpansive retraction from E onto C. For every $i=1,2,\dots ,N$, let $A:C\to E$ be a ν-inverse strongly accretive mapping. Let $T:C\to C$ be a nonexpansive mapping and $S:C\to C$ be an η-strictly pseudo-contractive mapping with $\mathcal{F}=F\left(S\right)\cap F\left(T\right)\cap S\left(C,A\right)\ne \mathrm{\varnothing }$. Define a mapping ${B}_{A}:C\to C$ by $T\left(\left(1-\alpha \right)I+\alpha S\right)x={B}_{A}x$, $\mathrm{\forall }x\in C$ and $\alpha \in \left(0,\frac{\eta }{{K}^{2}}\right)$, where K is the 2-uniformly smooth constant of E. Let $\left\{{x}_{n}\right\}$ be the sequence generated by ${x}_{1}\in C$ and
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}{Q}_{C}\left(I-\lambda A\right){x}_{n}+{\delta }_{n}{B}_{A}{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
where $f:C\to C$ is a contractive mapping and $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\},\left\{{\gamma }_{n}\right\},\left\{{\delta }_{n}\right\}\subseteq \left[0,1\right]$, ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}+{\delta }_{n}=1$, $\lambda \in \left(0,\frac{\nu }{{K}^{2}}\right)$ and satisfy the following conditions:
Then the sequence $\left\{{x}_{n}\right\}$ converses strongly to $q\in \mathcal{F}$, which solves the following variational inequality:
$〈q-f\left(q\right),j\left(q-p\right)〉\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in \mathcal{F}.$
Corollary 3.3 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space E. Let ${Q}_{C}$ be the sunny nonexpansive retraction from E onto C. For every $i=1,2,\dots ,N$, let ${A}_{i}:C\to E$ be an ${\alpha }_{i}$-inverse strongly accretive mapping. Define a mapping ${G}_{i}:C\to C$ by ${Q}_{C}\left(I-{\lambda }_{i}{A}_{i}\right)x={G}_{i}x$ for all $x\in C$ and $i=1,2,\dots ,N$, where ${\lambda }_{i}\in \left(0,\frac{{\alpha }_{i}}{{K}^{2}}\right)$, K is the 2-uniformly smooth constant of E. Let $B:C\to C$ be the K-mapping generated by ${G}_{1},{G}_{2},\dots ,{G}_{N}$ and ${\rho }_{1},{\rho }_{2},\dots ,{\rho }_{N}$, where ${\rho }_{i}\in \left(0,1\right)$, $\mathrm{\forall }i=1,2,\dots ,N-1$ and ${\rho }_{N}\in \left(0,1\right]$. Let $T:C\to C$ be a nonexpansive mapping with $\mathcal{F}=F\left(T\right)\cap {\bigcap }_{i=1}^{N}S\left(C,{A}_{i}\right)\ne \mathrm{\varnothing }$. Let $\left\{{x}_{n}\right\}$ be the sequence generated by ${x}_{1}\in C$ and
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}B{x}_{n}+{\delta }_{n}T{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
where $f:C\to C$ is a contractive mapping and $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\},\left\{{\gamma }_{n}\right\},\left\{{\delta }_{n}\right\}\subseteq \left[0,1\right]$, ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}+{\delta }_{n}=1$ and satisfy the following conditions:
Then the sequence $\left\{{x}_{n}\right\}$ converses strongly to $q\in \mathcal{F}$, which solves the following variational inequality:
$〈q-f\left(q\right),j\left(q-p\right)〉\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in \mathcal{F}.$
Corollary 3.4 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space E. Let ${Q}_{C}$ be the sunny nonexpansive retraction from E onto C. For every $i=1,2,\dots ,N$, let ${A}_{i}:C\to E$ be an ${\alpha }_{i}$-inverse strongly accretive mapping. Define a mapping ${G}_{i}:C\to C$ by ${Q}_{C}\left(I-{\lambda }_{i}{A}_{i}\right)x={G}_{i}x$ for all $x\in C$ and $i=1,2,\dots ,N$, where ${\lambda }_{i}\in \left(0,\frac{{\alpha }_{i}}{{K}^{2}}\right)$, K is the 2-uniformly smooth constant of E. Let $B:C\to C$ be the K-mapping generated by ${G}_{1},{G}_{2},\dots ,{G}_{N}$ and ${\rho }_{1},{\rho }_{2},\dots ,{\rho }_{N}$, where ${\rho }_{i}\in \left(0,1\right)$, $\mathrm{\forall }i=1,2,\dots ,N-1$ and ${\rho }_{N}\in \left(0,1\right]$. Let $S:C\to C$ be an η-strictly pseudo-contractive mapping with $\mathcal{F}=F\left(S\right)\cap {\bigcap }_{i=1}^{N}S\left(C,{A}_{i}\right)\ne \mathrm{\varnothing }$. Define a mapping ${B}_{A}:C\to C$ by $\left(1-\alpha \right)x+\alpha Sx={B}_{A}x$, $\mathrm{\forall }x\in C$ and $\alpha \in \left(0,\frac{\eta }{{K}^{2}}\right)$. Let $\left\{{x}_{n}\right\}$ be the sequence generated by ${x}_{1}\in C$ and
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
where $f:C\to C$ is a contractive mapping and $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\},\left\{{\gamma }_{n}\right\},\left\{{\delta }_{n}\right\}\subseteq \left[0,1\right]$, ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}+{\delta }_{n}=1$ and satisfy the following conditions:
Then the sequence $\left\{{x}_{n}\right\}$ converses strongly to $q\in \mathcal{F}$, which solves the following variational inequality:
$〈q-f\left(q\right),j\left(q-p\right)〉\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in \mathcal{F}.$

## 4 Applications

To prove the next theorem, we needed the following lemma.

Lemma 4.1 Let C be a nonempty closed convex subset of a Banach space E and let $P:C\to C$ be an η-strictly pseudo-contractive mapping with $F\left(P\right)\ne \mathrm{\varnothing }$. Then $F\left(P\right)=S\left(C,I-P\right)$.

Proof It is easy to see that $F\left(P\right)\subseteq S\left(C,I-P\right)$. Put $A=I-P$ and ${z}^{\ast }\in F\left(P\right)$. Let ${z}_{0}\in S\left(C,I-P\right)$, then there exists $j\left(x-{z}_{0}\right)\in J\left(x-{z}_{0}\right)$ such that
$〈\left(I-P\right){z}_{0},j\left(x-{z}_{0}\right)〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C.$
(4.1)
Since P is an η-strictly pseudo-contractive mapping, then there exists $j\left({z}_{0}-{z}^{\ast }\right)$ such that
$\begin{array}{rcl}〈P{z}_{0}-P{z}^{\ast },j\left({z}_{0}-{z}^{\ast }\right)〉& =& 〈\left(I-A\right){z}_{0}-\left(I-A\right){z}^{\ast },j\left({z}_{0}-{z}^{\ast }\right)〉\\ =& 〈{z}_{0}-{z}^{\ast }-\left(A{z}_{0}-A{z}^{\ast }\right),j\left({z}_{0}-{z}^{\ast }\right)〉\\ =& 〈{z}_{0}-{z}^{\ast },j\left({z}_{0}-{z}^{\ast }\right)〉-〈A{z}_{0}-A{z}^{\ast },j\left({z}_{0}-{z}^{\ast }\right)〉\\ =& {\parallel {z}_{0}-{z}^{\ast }\parallel }^{2}-〈A{z}_{0},j\left({z}_{0}-{z}^{\ast }\right)〉\\ \le & {\parallel {z}_{0}-z\parallel }^{2}-\eta {\parallel \left(I-P\right){z}_{0}\parallel }^{2}.\end{array}$
(4.2)
From (4.1), (4.2), we have
$\eta {\parallel {z}_{0}-P{z}_{0}\parallel }^{2}\le 〈A{z}_{0},j\left({z}_{0}-{z}^{\ast }\right)〉=-〈A{z}_{0},j\left({z}^{\ast }-{z}_{0}\right)〉\le 0.$

It implies that ${z}_{0}=P{z}_{0}$, that is, ${z}_{0}\in F\left(P\right)$. Then we have $S\left(C,I-P\right)\subseteq F\left(P\right)$. Hence, we have $S\left(C,I-P\right)=F\left(P\right)$. □

Remark 4.2 If C is a closed convex subset of a smooth Banach space E and ${Q}_{C}$ is a sunny nonexpansive retraction from E onto C, from Remark 1.1, Lemma 2.6 and 4.1, we have
$F\left(P\right)=S\left(C,I-P\right)=F\left({Q}_{C}\left(I-\lambda \left(I-P\right)\right)\right)$
(4.3)

for all $\lambda >0$.

Theorem 4.3 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space E. Let ${Q}_{C}$ be the sunny nonexpansive retraction from E onto C. For every $i=1,2,\dots ,N$, let ${S}_{i}:C\to E$ be an ${\eta }_{i}$-strictly pseudo-contractive mapping. Define a mapping ${G}_{i}:C\to C$ by ${Q}_{C}\left(I-{\lambda }_{i}\left(I-{S}_{i}\right)\right)x={G}_{i}x$ for all $x\in C$ and $i=1,2,\dots ,N$, where ${\lambda }_{i}\in \left(0,\frac{{\eta }_{i}}{{K}^{2}}\right)$, K is the 2-uniformly smooth constant of E. Let $B:C\to C$ be the K-mapping generated by ${G}_{1},{G}_{2},\dots ,{G}_{N}$ and ${\rho }_{1},{\rho }_{2},\dots ,{\rho }_{N}$, where ${\rho }_{i}\in \left(0,1\right)$, $\mathrm{\forall }i=1,2,\dots ,N-1$ and ${\rho }_{N}\in \left(0,1\right]$. Let $T:C\to C$ be a nonexpansive mapping and $S:C\to C$ be an η-strictly pseudo-contractive mapping with $\mathcal{F}=F\left(S\right)\cap F\left(T\right)\cap {\bigcap }_{i=1}^{N}F\left({S}_{i}\right)\ne \mathrm{\varnothing }$. Define a mapping ${B}_{A}:C\to C$ by $T\left(\left(1-\alpha \right)I+\alpha S\right)x={B}_{A}x$, $\mathrm{\forall }x\in C$ and $\alpha \in \left(0,\frac{\eta }{{K}^{2}}\right)$. Let $\left\{{x}_{n}\right\}$ be the sequence generated by ${x}_{1}\in C$ and
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
where $f:C\to C$ is a contractive mapping and $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\},\left\{{\gamma }_{n}\right\},\left\{{\delta }_{n}\right\}\subseteq \left[0,1\right]$, ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}+{\delta }_{n}=1$ and satisfy the following conditions:
Then the sequence $\left\{{x}_{n}\right\}$ converses strongly to $q\in \mathcal{F}$, which solves the following variational inequality:
$〈q-f\left(q\right),j\left(q-p\right)〉\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in \mathcal{F}.$

Proof Since ${S}_{i}$ is an ${\eta }_{i}$-strictly pseudo-contractive mapping, then we have $\left(I-{S}_{i}\right)$ is an ${\eta }_{i}$-inverse strongly accretive mapping for every $i=1,2,\dots ,N$. For every $i=1,2,\dots ,N$, putting ${A}_{i}=I-{S}_{i}$ in Theorem 3.1, from Remark 4.2 and Theorem 3.1, we can conclude the desired results. □

Next corollaries are derived from Theorem 4.3. We, therefore, omit the proof.

Corollary 4.4 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space E. Let ${Q}_{C}$ be the sunny nonexpansive retraction from E onto C. For every $i=1,2,\dots ,N$, let ${S}_{i}:C\to E$ be an ${\eta }_{i}$-strictly pseudo contractive mapping. Define a mapping ${G}_{i}:C\to C$ by ${Q}_{C}\left(I-{\lambda }_{i}\left(I-{S}_{i}\right)\right)x={G}_{i}x$ for all $x\in C$ and $i=1,2,\dots ,N$, where ${\lambda }_{i}\in \left(0,\frac{{\eta }_{i}}{{K}^{2}}\right)$, K is the 2-uniformly smooth constant of E. Let $B:C\to C$ be the K-mapping generated by ${G}_{1},{G}_{2},\dots ,{G}_{N}$ and ${\rho }_{1},{\rho }_{2},\dots ,{\rho }_{N}$, where ${\rho }_{i}\in \left(0,1\right)$, $\mathrm{\forall }i=1,2,\dots ,N-1$ and ${\rho }_{N}\in \left(0,1\right]$. Let $T:C\to C$ be a nonexpansive mapping with $\mathcal{F}=F\left(T\right)\cap {\bigcap }_{i=1}^{N}F\left({S}_{i}\right)\ne \mathrm{\varnothing }$. Let $\left\{{x}_{n}\right\}$ be the sequence generated by ${x}_{1}\in C$ and
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}B{x}_{n}+{\delta }_{n}T{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
where $f:C\to C$ is a contractive mapping and $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\},\left\{{\gamma }_{n}\right\},\left\{{\delta }_{n}\right\}\subseteq \left[0,1\right]$, ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}+{\delta }_{n}=1$ and satisfy the following conditions:
Then the sequence $\left\{{x}_{n}\right\}$ converses strongly to $q\in \mathcal{F}$, which solves the following variational inequality:
$〈q-f\left(q\right),j\left(q-p\right)〉\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in \mathcal{F}.$
Corollary 4.5 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space E. Let ${Q}_{C}$ be the sunny nonexpansive retraction from E onto C. For every $i=1,2,\dots ,N$, let ${S}_{i}:C\to E$ be an ${\eta }_{i}$-strictly pseudo contractive mapping. Define a mapping ${G}_{i}:C\to C$ by ${Q}_{C}\left(I-{\lambda }_{i}\left(I-{S}_{i}\right)\right)x={G}_{i}x$ for all $x\in C$ and $i=1,2,\dots ,N$, where ${\lambda }_{i}\in \left(0,\frac{{\eta }_{i}}{{K}^{2}}\right)$, K is the 2-uniformly smooth constant of E. Let $B:C\to C$ be the K-mapping generated by ${G}_{1},{G}_{2},\dots ,{G}_{N}$ and ${\rho }_{1},{\rho }_{2},\dots ,{\rho }_{N}$, where ${\rho }_{i}\in \left(0,1\right)$, $\mathrm{\forall }i=1,2,\dots ,N-1$ and ${\rho }_{N}\in \left(0,1\right]$. $S:C\to C$ be an η-strictly pseudo contractive mapping with $\mathcal{F}=F\left(S\right)\cap {\bigcap }_{i=1}^{N}F\left({S}_{i}\right)\ne \mathrm{\varnothing }$. Define a mapping ${B}_{A}:C\to C$ by $\left(1-\alpha \right)x+\alpha Sx={B}_{A}x$, $\mathrm{\forall }x\in C$ and $\alpha \in \left(0,\frac{\eta }{{K}^{2}}\right)$. Let $\left\{{x}_{n}\right\}$ be a sequence generated by ${x}_{1}\in C$ and
${x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}B{x}_{n}+{\delta }_{n}{B}_{A}{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
where $f:C\to C$ is a contractive mapping and $\left\{{\alpha }_{n}\right\},\left\{{\beta }_{n}\right\},\left\{{\gamma }_{n}\right\},\left\{{\delta }_{n}\right\}\subseteq \left[0,1\right]$, ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}+{\delta }_{n}=1$ and satisfy the following conditions:
Then the sequence $\left\{{x}_{n}\right\}$ converses strongly to $q\in \mathcal{F}$, which solves the following variational inequality:
$〈q-f\left(q\right),j\left(q-p\right)〉\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }p\in \mathcal{F}.$

## Declarations

### Acknowledgements

This research was supported by the Research Administration Division of King Mongkut’s Institute of Technology Ladkrabang.

## Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, King Mongkut’s Institute of Technology Ladkrabang, Bangkok, 10520, Thailand

## References

1. Reich S: Asymptotic behavior of contractions in Banach spaces. J. Math. Anal. Appl. 1973, 44: 57–70. 10.1016/0022-247X(73)90024-3
2. Aoyama K, Iiduka H, Takahashi W: Weak convergence of an iterative sequence for accretive operators in Banach spaces. Fixed Point Theory Appl. 2006., 2006: Article ID 35390. doi:10.1155/FPTA/2006/35390Google Scholar
3. Chang SS, Lee HWJ, Chan CK: A new method for solving equilibrium problem fixed point problem and variational inequality problem with application to optimization. Nonlinear Anal. 2009, 70: 3307–3319. 10.1016/j.na.2008.04.035
4. Kangtunyakarn A: A new iterative algorithm for the set of fixed-point problems of nonexpansive mappings and the set of equilibrium problem and variational inequality problem. Abstr. Appl. Anal. 2011., 2011: Article ID 562689. doi:10.1155/2011/562689Google Scholar
5. Mann WR: Mean value methods in iteration. Proc. Am. Math. Soc. 1953, 4(3):506–510. 10.1090/S0002-9939-1953-0054846-3
6. Cho YJ, Kang SM, Qin X: Convergence theorems of fixed points for a finite family of nonexpansive mappings in Banach spaces. Fixed Point Theory Appl. 2008., 2008: Article ID 856145. doi:10.1155/2008/856145Google Scholar
7. Zhou H: Convergence theorems for λ -strict pseudo-contractions in 2-uniformly smooth Banach spaces. Nonlinear Anal. 2008, 69: 3160–3173. 10.1016/j.na.2007.09.009
8. Xu HK: Inequalities in Banach spaces with applications. Nonlinear Anal. 1991, 16: 1127–1138. 10.1016/0362-546X(91)90200-K
9. Kangtunyakarn A, Suantai S: A new mapping for finding common solutions of equilibrium problems and fixed point problems of finite family of nonexpansive mappings. Nonlinear Anal. 2009, 71: 4448–4460. 10.1016/j.na.2009.03.003
10. Suzuki T: Strong convergence of Krasnoselskii and Mann’s type sequences for one-parameter nonexpansive semigroups without Bochner integrals. J. Math. Anal. Appl. 2005, 305(1):227–239. 10.1016/j.jmaa.2004.11.017
11. Cho YJ, Zhou HY, Guo G: Weak and strong convergence theorems for three-step iterations with errors for asymptotically nonexpansive mappings. Comput. Math. Appl. 2004, 47: 707–717. 10.1016/S0898-1221(04)90058-2
12. Bruck RE: Properties of fixed point sets of nonexpansive mappings in Banach spaces. Trans. Am. Math. Soc. 1973, 179: 251–262.
13. Xu HK: Viscosity approximation methods for nonexpansive mappings. J. Math. Anal. Appl. 2004, 298: 279–291. 10.1016/j.jmaa.2004.04.059
14. Chang SS: On Chidumes open questions and approximate solutions of multivalued strongly accretive mapping equations in Banach spaces. J. Math. Anal. Appl. 1997, 216: 94–111. 10.1006/jmaa.1997.5661
15. Xu HK: An iterative approach to quadratic optimization. J. Optim. Theory Appl. 2003, 116(3):659–678. 10.1023/A:1023073621589