A new mapping for finding a common element of the sets of fixed points of two finite families of nonexpansive and strictly pseudo-contractive mappings and two sets of variational inequalities in uniformly convex and 2-smooth Banach spaces

Abstract

In this paper we introduce a new mapping in a uniformly convex and 2-smooth Banach space to prove a strong convergence theorem for finding a common element of the set of fixed points of a finite family of nonexpansive mappings and the set of fixed points of a finite family of strictly pseudo-contractive mappings and two sets of solutions of variational inequality problems. Moreover, we also obtain a strong convergence theorem for a finite family of the set of solutions of variational inequality problems and the set of fixed points of a finite family of strictly pseudo-contractive mappings by using our main result.

1 Introduction

Throughout this paper, we use E and $E^{\ast }$ to denote a real Banach space and a dual space of E, respectively. For any pair $x\in E$ and $f\in E^{\ast }$, $〈x,f〉$ instead of $f(x)$. The duality mapping $J:E\to 2^{E^{\ast }}$ is defined by $J(x)=\{x^{\ast }\in E^{\ast }:〈x,x^{\ast }〉={\parallel x\parallel }^2,\parallel x\parallel =\parallel x^{\ast }\parallel \}$ for all $x\in E$. It is well known that if E is a Hilbert space, then $J=I$, where I is the identity mapping. Recall the following definitions.

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 A Banach space E is said to be smooth if for each $x\in S_E=\{x\in E:\parallel x\parallel =1\}$, there exists a unique functional $j_x\in E^{\ast }$ such that $〈x,j_x〉=\parallel x\parallel$ and $\parallel j_x\parallel =1$.

It is obvious that if E is smooth, then J is single-valued which is denoted by j.

Definition 1.3 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

$${\rho }_E(t)=sup\{\frac{\parallel x+y\parallel +\parallel x-y\parallel }{2}-1:\parallel x\parallel =1,\parallel y\parallel =t\}.$$

A Banach space E is said to be uniformly smooth if

$$\underset{t\to 0}{lim}\frac{{\rho }_E(t)}{t}=0.$$

It is well known that every uniformly smooth Banach space is smooth.

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

A mapping $T:C\to C$ 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 $\eta \in (0,1)$ such that

$$〈Tx-Ty,j(x-y)〉\le {\parallel x-y\parallel }^2-\eta {\parallel (I-T)x-(I-T)y\parallel }^2$$

(1.1)

for every $x,y\in C$ and for some $j(x-y)\in J(x-y)$. It is clear that (1.1) is equivalent to the following:

$$〈(I-T)x-(I-T)y,j(x-y)〉\ge \eta {\parallel (I-T)x-(I-T)y\parallel }^2$$

(1.2)

for every $x,y\in C$ and for some $j(x-y)\in J(x-y)$. We give some examples for a strictly pseudo-contractive mapping as follows.

Example 1.1 Let ℝ be a real line endowed with the Euclidean norm and let $C=(0,\mathrm{\infty })$. Define the mapping $T:C\to C$ by

$$Tx=\frac{2x^2}{3+2x},\mathrm{\forall }x\in C.$$

Then T is a $\frac{1}{9}$-strictly pseudo-contractive mapping.

Example 1.2 (See [1])

Let ℝ be a real line endowed with the Euclidean norm. Let $C=[-1,1]$ and let $T:C\to C$ be defined by

$$Tx=\{\begin{array}{cc}x\hfill & \text{if }x\in [-1,0];\hfill \\ x-x^2\hfill & \text{if }x\in (0,1].\hfill \end{array}$$

Then T is a λ-strictly pseudo-contractive mapping where $\lambda \le min\{{\lambda }_1,{\lambda }_2\}$ and ${\lambda }_1\le \frac{1}{2}$, ${\lambda }_2<1$.

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 [2] provided $P(x+t(x-P(x)))=P(x)$ for all $x\in C$ and $t\ge 0$, whenever $x+t(x-P(x))\in C$. A 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.

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 $j(x-y)\in J(x-y)$ such that

$$〈Ax-Ay,j(x-y)〉\ge 0,\mathrm{\forall }x,y\in C.$$

A mapping $A:C\to E$ is said to be α-inverse strongly accretive if there exist $j(x-y)\in J(x-y)$ and $\alpha >0$ such that

$$〈Ax-Ay,j(x-y)〉\ge \alpha {\parallel Ax-Ay\parallel }^2,\mathrm{\forall }x,y\in C.$$

Remark 1.3 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 $j(x-x^{\ast })\in J(x-x^{\ast })$,

$$〈A{x}^{\ast },j(x-x^{\ast })〉\ge 0,\mathrm{\forall }x\in C.$$

(1.3)

This problem was considered by Aoyama et al. [3]. The set of solutions of the variational inequality in a Banach space is denoted by $S(C,A)$, that is,

$$S(C,A)=\{u\in C:〈Au,J(v-u)〉\ge 0,\mathrm{\forall }v\in C\}.$$

(1.4)

Several problems in pure and applied science, numerous problems in physics and economics reduce to finding an element in (1.4); see, for instance, [4–6].

Recall that normal Mann’s iterative process was introduced by Mann [7] in 1953. The normal Mann’s iterative process generates a sequence $\{x_n\}$ in the following manner:

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_{n}T{x}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(1.5)

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

In 1967, Halpern has introduced the iteration method guaranteeing the strong convergence as follows:

$$\{\begin{array}{c}{x}_1\in C,\hfill \\ x_{n+1}=(1-{\alpha }_n)x_1+{\alpha }_{n}T{x}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(1.6)

where $\{{\alpha }_n\}\subset (0,1)$. Such an iteration is called Halpern iteration if T is a nonexpansive mapping with a fixed point. He also pointed out that the conditions ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ are necessary for the strong convergence of $\{x_n\}$ to a fixed point of T.

Many authors have modified the iteration (1.6) for a strong convergence theorem; see, for instance, [8–10].

In 2008, Zhou [11] 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 1.4 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 $F(T)\ne \mathrm{\varnothing }$. Given $u,x_0\in C$ and sequences $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$ and $\{{\delta }_n\}$ in $(0,1)$, the following control conditions are satisfied:

$$\begin{array}{r}\text{(i)}a\le {\alpha }_n\le \frac{\lambda }{K^2}\mathit{\text{for some}}a>0\mathit{\text{and for all }}n\ge 0,\\ \text{(ii)}{\beta }_n+{\gamma }_n+{\delta }_n=1\mathit{\text{for all}}n\ge 0,\\ \text{(iii) }\underset{n\to \mathrm{\infty }}{lim}{\beta }_n=0\mathit{\text{and}}\sum _{n=1}^{\mathrm{\infty }}{\beta }_n=\mathrm{\infty },\\ \text{(iv)}{\alpha }_{n+1}-{\alpha }_n\to 0,\mathit{\text{as}}n\to \mathrm{\infty },\\ \text{(v) }0<\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\gamma }_n\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\gamma }_n<1.\end{array}$$

Let a sequence $\{x_n\}$ be generated by

$$\{\begin{array}{c}{y}_n={\alpha }_{n}T{x}_n+(1-{\alpha }_n)x_n,\hfill \\ x_{n+1}={\beta }_{n}u+{\gamma }_n{x}_n+{\delta }_n{y}_n,n\ge 0.\hfill \end{array}$$

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

In 2006, Aoyama et al. introduced a Halpern-type iterative sequence and proved that such a sequence converges strongly to a common fixed point of nonexpansive mappings as follows.

Theorem 1.5 Let E be a uniformly convex Banach space whose norm is uniformly Gâteaux differentiable and let C be a nonempty closed convex subset of E. Let $\{T_n\}$ be a sequence of nonexpansive mappings of C into itself such that ${\bigcap }_{n=1}^{N}F(T_i)$ is nonempty and let $\{{\alpha }_n\}$ be a sequence of $[0,1]$ such that ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$. Let $\{x_n\}$ be a sequence of C defined as follows: $x_1=x\in C$ and

$$x_{n+1}={\alpha }_{n}x+(1-{\alpha }_n)T_n{x}_n$$

for every $n\in \mathbb{N}$. Suppose that ${\sum }_{n=1}^{\mathrm{\infty }}sup\{\parallel T_{n+1}z-T_{n}z\parallel :z\in B\}<\mathrm{\infty }$ for any bounded subset B of C. Let T be a mapping of C into itself defined by $Tz={lim}_{n\to \mathrm{\infty }}{T}_{n}z$ for all $z\in C$ and suppose that $F(T)={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$. If either

$$\begin{array}{r}\text{(i)}\sum _{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }\mathit{\text{or}}\\ \text{(ii)}{\alpha }_n\in (0,1]\mathit{\text{for every}}n\in \mathbb{N}\mathit{\text{and}}\underset{n\to \mathrm{\infty }}{lim}\frac{{\alpha }_n}{{\alpha }_{n+1}},\end{array}$$

then $\{x_n\}$ converges strongly to Qx, where Q is the sunny nonexpansive retraction of E onto $F(T)={\bigcap }_{i=1}^{\mathrm{\infty }}F(T_n)$.

In 2005, Aoyama et al. [3] proved a weak convergence theorem for finding a solution of problem (1.3) as follows.

Theorem 1.6 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 $S(C,A)\ne \mathrm{\varnothing }$. Suppose that $x_1=x\in C$ and $\{x_n\}$ is given by

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)Q_C(x_n-{\lambda }_{n}A{x}_n)$$

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

In 2009, Kangtunykarn and Suantai [12] introduced the S-mapping generated by a finite family of mappings and real numbers as follows.

Definition 1.4 Let C be a nonempty convex subset of a real Banach space. Let ${\{T_i\}}_{i=1}^N$ be a finite family of mappings of C into itself. For each $j=1,2,\dots ,N$, let ${\alpha }_j=({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j)\in I\times I\times I$, where $I\in [0,1]$ and ${\alpha }_1^j+{\alpha }_2^j+{\alpha }_3^j=1$. Define the mapping $S:C\to C$ as follows:

$$\begin{array}{r}{U}_0=I,\\ U_1={\alpha }_1^1{T}_1{U}_0+{\alpha }_2^1{U}_0+{\alpha }_3^{1}I,\\ U_2={\alpha }_1^2{T}_2{U}_1+{\alpha }_2^2{U}_1+{\alpha }_3^{2}I,\\ U_3={\alpha }_1^3{T}_3{U}_2+{\alpha }_2^3{U}_2+{\alpha }_3^{3}I,\\ ⋮\\ U_{N-1}={\alpha }_1^{N-1}{T}_{N-1}{U}_{N-2}+{\alpha }_2^{N-1}{U}_{N-2}+{\alpha }_3^{N-1}I,\\ S=U_N={\alpha }_1^N{T}_N{U}_{N-1}+{\alpha }_2^N{U}_{N-1}+{\alpha }_3^{N}I.\end{array}$$

(1.7)

This mapping is called the S-mapping generated by $T_1,T_2,\dots ,T_N$ and ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_N$.

For every $i=1,2,\dots ,N$, put ${\alpha }_3^j=0$ in (1.7), then the S-mapping generated by $T_1,T_2,\dots ,T_N$ and ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_N$ reduces to the K-mapping generated by $T_1,T_2,\dots ,T_N$ and ${\alpha }_1^1,{\alpha }_1^2,\dots ,{\alpha }_1^N$, which is defined by Kangtunyakarn and Suantai [13].

Recently, Kangtunyakarn [14] introduced 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 as follows.

Theorem 1.7 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 $Q_C(I-{\lambda }_i{A}_i)x=G_{i}x$ for all $x\in C$ and $i=1,2,\dots ,N$, where ${\lambda }_i\in (0,\frac{{\alpha }_i}{K^2})$, 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 ${\rho }_i\in (0,1)$, $\mathrm{\forall }i=1,2,\dots ,N-1$ and ${\rho }_N\in (0,1]$. Let $T:C\to C$ be a nonexpansive mapping and $S:C\to C$ be an η-strictly pseudo-contractive mapping with $\mathcal{F}=F(S)\cap F(T)\cap {\bigcap }_{i=1}^{N}S(C,A_i)\ne \mathrm{\varnothing }$. Define a mapping $B_A:C\to C$ by $T((1-\alpha )I+\alpha S)x=B_{A}x$, $\mathrm{\forall }x\in C$ and $\alpha \in (0,\frac{\eta }{K^2})$. Let $\{x_n\}$ be a sequence generated by $x_1\in C$ and

$$x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_{n}B{x}_n+{\delta }_n{B}_A{x}_n,\mathrm{\forall }n\ge 1,$$

(1.8)

where $f:C\to C$ is a contractive mapping and $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\},\{{\delta }_n\}\subseteq [0,1]$, ${\alpha }_n+{\beta }_n+{\gamma }_n+{\delta }_n=1$ and satisfy the following conditions:

$$\begin{array}{r}\text{(i)}\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=0\mathit{\text{and}}\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty },\\ \text{(ii)}\{{\gamma }_n\},\{{\delta }_n\}\subseteq [c,d]\subset (0,1)\mathit{\text{for some}}c,d>0\mathit{\text{and}}\mathrm{\forall }n\ge 1,\\ \text{(iii)}\sum _{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|,\sum _{n=1}^{\mathrm{\infty }}|{\gamma }_{n+1}-{\gamma }_n|,\sum _{n=1}^{\mathrm{\infty }}|{\delta }_{n+1}-{\delta }_n|<\mathrm{\infty },\\ \text{(iv)}0<\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\beta }_n\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\beta }_n<1.\end{array}$$

Then the sequence $\{x_n\}$ converses strongly to $q\in \mathcal{F}$, which solves the following variational inequality:

$$〈q-f(q),j(q-p)〉\le 0,\mathrm{\forall }p\in \mathcal{F}.$$

Question How can we prove a strong convergence theorem for the set of fixed points of a finite family of nonexpansive mappings and the set of fixed points of a finite family of strictly pseudo-contractive mappings and the set of solutions of variational inequality problems in a uniformly convex and 2-uniformly smooth Banach space?

Motivated by the S-mapping, we define a new mapping in the next section to answer the above question, and from Theorems 1.4, 1.5, 1.6 and 1.7 we modify the Halpern iteration for finding a common element of two sets of solutions of (1.3) and the set of fixed points of a finite family of nonexpansive mappings and the set of fixed points of a finite family of strictly pseudo-contractive mappings in a uniformly convex and 2-uniformly smooth Banach space. Moreover, by using our main result, we also obtain a strong convergence theorem for a finite family of the set of solutions of (1.3) and the set of fixed points of a finite family of strictly pseudo-contractive mappings.

2 Preliminaries

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

Lemma 2.1 (See [15])

Let E be a real 2-uniformly smooth Banach space with the best smooth constant K. Then the following inequality holds:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,J(x)〉+2{\parallel Ky\parallel }^2$$

for any $x,y\in E$.

Lemma 2.2 (See [16])

Let X be a uniformly convex Banach space and $B_r=\{x\in X:\parallel x\parallel \le r\}$, $r>0$. Then there exists a continuous, strictly increasing and convex function $g:[0,\mathrm{\infty }]\to [0,\mathrm{\infty }]$, $g(0)=0$ such that

$${\parallel \alpha x+\beta y+\gamma z\parallel }^2\le \alpha {\parallel x\parallel }^2+\beta {\parallel y\parallel }^2+\gamma {\parallel z\parallel }^2-\alpha \beta g(\parallel x-y\parallel )$$

for all $x,y,z\in B_r$ and all $\alpha ,\beta ,\gamma \in [0,1]$ with $\alpha +\beta +\gamma =1$.

Lemma 2.3 (See [3])

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 $\lambda >0$,

$$S(C,A)=F(Q_C(I-\lambda A)).$$

Lemma 2.4 (See [15])

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

Lemma 2.5 (See [17])

Let C be a closed and convex subset of a real uniformly smooth Banach space E and let $T:C\to C$ be a nonexpansive mapping with a nonempty fixed point $F(T)$. If $\{x_n\}\subset C$ is a bounded sequence such that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. Then there exists a unique sunny nonexpansive retraction $Q_{F(T)}:C\to F(T)$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-Q_{F(T)}u,J(x_n-Q_{F(T)}u)〉\le 0$$

for any given $u\in C$.

Lemma 2.6 (See [18])

Let $\{s_n\}$ be a sequence of nonnegative real numbers satisfying

$$s_{n+1}=(1-{\alpha }_n)s_n+{\delta }_n,\mathrm{\forall }n\ge 0,$$

where $\{{\alpha }_n\}$ is a sequence in $(0,1)$ and $\{{\delta }_n\}$ is a sequence such that

$$\begin{array}{r}(1)\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty },\\ (2)\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\delta }_n}{{\alpha }_n}\le 0\mathit{\text{or}}\sum _{n=1}^{\mathrm{\infty }}|{\delta }_n|<\mathrm{\infty }.\end{array}$$

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

From the S-mapping, we define the mapping generated by two sets of finite families of the mappings and real numbers as follows.

Definition 2.1 Let C be a nonempty convex subset of a Banach space. Let ${\{S_i\}}_{i=1}^N$ and ${\{T_i\}}_{i=1}^N$ be two finite families of mappings of C into itself. For each $j=1,2,\dots ,N$, let ${\alpha }_j=({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j)\in I\times I\times I$, where $I\in [0,1]$ and ${\alpha }_1^j+{\alpha }_2^j+{\alpha }_3^j=1$. We define the mapping $S^A:C\to C$ as follows:

$$\begin{array}{r}{U}_0=T_1=I,\\ U_1=T_1({\alpha }_1^1{S}_1{U}_0+{\alpha }_2^1{U}_0+{\alpha }_3^{1}I),\\ U_2=T_2({\alpha }_1^2{S}_2{U}_1+{\alpha }_2^2{U}_1+{\alpha }_3^{2}I),\\ U_3=T_3({\alpha }_1^3{S}_3{U}_2+{\alpha }_2^3{U}_2+{\alpha }_3^{3}I),\\ ⋮\\ U_{N-1}=T_{N-1}({\alpha }_1^{N-1}{S}_{N-1}{U}_{N-2}+{\alpha }_2^{N-1}{U}_{N-2}+{\alpha }_3^{N-1}I),\\ S^A=U_N=T_N({\alpha }_1^N{S}_N{U}_{N-1}+{\alpha }_2^N{U}_{N-1}+{\alpha }_3^{N}I).\end{array}$$

(2.1)

This mapping is called the $S^A$-mapping generated by $S_1,S_2,\dots ,S_N$, $T_1,T_2,\dots ,T_N$ and ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_N$.

Lemma 2.7 Let C be a nonempty closed convex subset of a 2-uniformly smooth and uniformly convex Banach space. Let ${\{S_i\}}_{i=1}^N$ be a finite family of ${\kappa }_i$-strict pseudo-contractions of C into itself and let ${\{T_i\}}_{i=1}^N$ be a finite family of nonexpansive mappings of C into itself with ${\bigcap }_{i=1}^{N}F(S_i)\cap {\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$ and $\kappa =min\{{\kappa }_i:i=1,2,\dots ,N\}$ with $K^2\le \kappa$, where K is the 2-uniformly smooth constant of E. Let ${\alpha }_j=({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j)\in I\times I\times I$, where $I=[0,1]$, ${\alpha }_1^j+{\alpha }_2^j+{\alpha }_3^j=1$, ${\alpha }_1^j\in (0,1]$, ${\alpha }_2^j\in [0,1]$ and ${\alpha }_3^j\in (0,1)$ for all $j=1,2,\dots ,N$. Let $S^A$ be the $S^A$-mapping generated by $S_1,S_2,\dots ,S_N$, $T_1,T_2,\dots ,T_N$ and ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_N$. Then $F(S^A)={\bigcap }_{i=1}^{N}F(S_i)\cap {\bigcap }_{i=1}^{N}F(T_i)$ and $S^A$ is a nonexpansive mapping.

Proof Let $x_0\in F(S^A)$ and $x^{\ast }\in {\bigcap }_{i=1}^{N}F(S_i)\cap {\bigcap }_{i=1}^{N}F(T_i)$, we have

$$\begin{array}{rcl}{\parallel x_0-x^{\ast }\parallel }^2& =& {\parallel T_N({\alpha }_1^N{S}_N{U}_{N-1}+{\alpha }_2^N{U}_{N-1}+{\alpha }_3^{N}I)x_0-x^{\ast }\parallel }^2\\ \le & {\parallel {\alpha }_1^N(S_N{U}_{N-1}{x}_0-x^{\ast })+{\alpha }_2^N(U_{N-1}{x}_0-x^{\ast })+{\alpha }_3^N(x_0-x^{\ast })\parallel }^2\\ =& \parallel (1-{\alpha }_3^N)(\frac{{\alpha }_1^N}{1-{\alpha }_3^N}(S_N{U}_{N-1}{x}_0-x^{\ast })+\frac{{\alpha }_2^N}{1-{\alpha }_3^N}(U_{N-1}{x}_0-x^{\ast }))\\ +{\alpha }_3^N(x_0-x^{\ast }){\parallel }^2\\ \le & (1-{\alpha }_3^N){\parallel \frac{{\alpha }_1^N}{1-{\alpha }_3^N}(S_N{U}_{N-1}{x}_0-x^{\ast })+\frac{{\alpha }_2^N}{1-{\alpha }_3^N}(U_{N-1}{x}_0-x^{\ast })\parallel }^2\\ +{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ =& (1-{\alpha }_3^N){\parallel \frac{{\alpha }_1^N}{1-{\alpha }_3^N}(S_N{U}_{N-1}{x}_0-x^{\ast })+(1-\frac{{\alpha }_1^N}{1-{\alpha }_3^N})(U_{N-1}{x}_0-x^{\ast })\parallel }^2\\ +{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ =& (1-{\alpha }_3^N){\parallel \frac{{\alpha }_1^N}{1-{\alpha }_3^N}(S_N{U}_{N-1}{x}_0-U_{N-1}{x}_0)+U_{N-1}{x}_0-x^{\ast }\parallel }^2+{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ \le & (1-{\alpha }_3^N)({\parallel U_{N-1}{x}_0-x^{\ast }\parallel }^2\\ +2\frac{{\alpha }_1^N}{1-{\alpha }_3^N}〈S_N{U}_{N-1}{x}_0-U_{N-1}{x}_0,j(U_{N-1}{x}_0-x^{\ast })〉\\ +2K^2{\left(\frac{{\alpha }_1^N}{1-{\alpha }_3^N}\right)}^2{\parallel S_N{U}_{N-1}{x}_0-U_{N-1}{x}_0\parallel }^2)+{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ =& (1-{\alpha }_3^N)({\parallel U_{N-1}{x}_0-x^{\ast }\parallel }^2+2\frac{{\alpha }_1^N}{1-{\alpha }_3^N}〈S_N{U}_{N-1}{x}_0-x^{\ast },j(U_{N-1}{x}_0-x^{\ast })〉\\ +2\frac{{\alpha }_1^N}{1-{\alpha }_3^N}〈x^{\ast }-U_{N-1}{x}_0,j(U_{N-1}{x}_0-x^{\ast })〉\\ +2K^2{\left(\frac{{\alpha }_1^N}{1-{\alpha }_3^N}\right)}^2{\parallel S_N{U}_{N-1}{x}_0-U_{N-1}{x}_0\parallel }^2)+{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ \le & (1-{\alpha }_3^N)({\parallel U_{N-1}{x}_0-x^{\ast }\parallel }^2\\ +2\frac{{\alpha }_1^N}{1-{\alpha }_3^N}({\parallel U_{N-1}{x}_0-x^{\ast }\parallel }^2-\kappa {\parallel (I-S_N)U_{N-1}{x}_0\parallel }^2)\\ -2\frac{{\alpha }_1^N}{1-{\alpha }_3^N}{\parallel x^{\ast }-U_{N-1}{x}_0\parallel }^2+2K^2{\left(\frac{{\alpha }_1^N}{1-{\alpha }_3^N}\right)}^2{\parallel S_N{U}_{N-1}{x}_0-U_{N-1}{x}_0\parallel }^2)\\ +{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ =& (1-{\alpha }_3^N)({\parallel U_{N-1}{x}_0-x^{\ast }\parallel }^2-2\frac{{\alpha }_1^N}{1-{\alpha }_3^N}\kappa {\parallel (I-S_N)U_{N-1}{x}_0\parallel }^2\\ +2K^2{\left(\frac{{\alpha }_1^N}{1-{\alpha }_3^N}\right)}^2{\parallel S_N{U}_{N-1}{x}_0-U_{N-1}{x}_0\parallel }^2)+{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ =& (1-{\alpha }_3^N)({\parallel U_{N-1}{x}_0-x^{\ast }\parallel }^2\\ -2\frac{{\alpha }_1^N}{1-{\alpha }_3^N}(\kappa -K^2\left(\frac{{\alpha }_1^N}{1-{\alpha }_3^N}\right)){\parallel (I-S_N)U_{N-1}{x}_0\parallel }^2)+{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ \le & (1-{\alpha }_3^N){\parallel U_{N-1}{x}_0-x^{\ast }\parallel }^2+{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ \le & (1-{\alpha }_3^N)((1-{\alpha }_3^{N-1}){\parallel U_{N-2}{x}_0-x^{\ast }\parallel }^2+{\alpha }_3^{N-1}{\parallel x_0-x^{\ast }\parallel }^2)+{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=N-1}^N(1-{\alpha }_3^j){\parallel U_{N-2}{x}_0-x^{\ast }\parallel }^2+(1-\prod _{j=N-1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ ⋮\\ \le & \prod _{j=3}^N(1-{\alpha }_3^j){\parallel U_2{x}_0-x^{\ast }\parallel }^2+(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=3}^N(1-{\alpha }_3^j){\parallel T_2({\alpha }_1^2{S}_2{U}_1+{\alpha }_2^2{U}_1+{\alpha }_3^{2}I)x_0-x^{\ast }\parallel }^2\\ +(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=3}^N(1-{\alpha }_3^j){\parallel {\alpha }_1^2(S_2{U}_1{x}_0-x^{\ast })+{\alpha }_2^2(U_1{x}_0-x^{\ast })+{\alpha }_3^2(x_0-x^{\ast })\parallel }^2\\ +(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=3}^N(1-{\alpha }_3^j)\parallel (1-{\alpha }_3^2)(\frac{{\alpha }_1^2}{1-{\alpha }_3^2}(S_2{U}_1{x}_0-x^{\ast })+\frac{{\alpha }_2^2}{1-{\alpha }_3^2}(U_1{x}_0-x^{\ast }))\\ +{\alpha }_3^2(x_0-x^{\ast }){\parallel }^2+(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=3}^N(1-{\alpha }_3^j)((1-{\alpha }_3^2)\parallel \frac{{\alpha }_1^2}{1-{\alpha }_3^2}(S_2{U}_1{x}_0-x^{\ast })\\ +\frac{{\alpha }_2^2}{1-{\alpha }_3^2}(U_1{x}_0-x^{\ast }){\parallel }^2+{\alpha }_3^2{\parallel x_0-x^{\ast }\parallel }^2)+(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=3}^N(1-{\alpha }_3^j)((1-{\alpha }_3^2){\parallel \frac{{\alpha }_1^2}{1-{\alpha }_3^2}(S_2{U}_1{x}_0-x^{\ast })+(1-\frac{{\alpha }_1^2}{1-{\alpha }_3^2})(U_1{x}_0-x^{\ast })\parallel }^2\\ +{\alpha }_3^2{\parallel x_0-x^{\ast }\parallel }^2)+(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=3}^N(1-{\alpha }_3^j)((1-{\alpha }_3^2){\parallel \frac{{\alpha }_1^2}{1-{\alpha }_3^2}(S_2{U}_1{x}_0-U_1{x}_0)+U_1{x}_0-x^{\ast }\parallel }^2\\ +{\alpha }_3^2{\parallel x_0-x^{\ast }\parallel }^2)+(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=3}^N(1-{\alpha }_3^j)\left((1-{\alpha }_3^2)\right({\parallel U_1{x}_0-x^{\ast }\parallel }^2\\ +2\frac{{\alpha }_1^2}{1-{\alpha }_3^2}〈S_2{U}_1{x}_0-U_1{x}_0,j(U_1{x}_0-x^{\ast })〉\\ +2K^2\left(\frac{{\alpha }_1^2}{1-{\alpha }_3^2}\right){\parallel S_2{U}_1{x}_0-U_1{x}_0\parallel }^2)+{\alpha }_3^2{\parallel x_0-x^{\ast }\parallel }^2)\\ +(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=3}^N(1-{\alpha }_3^j)\left((1-{\alpha }_3^2)\right({\parallel U_1{x}_0-x^{\ast }\parallel }^2\\ -2\frac{{\alpha }_1^2}{1-{\alpha }_3^2}(\kappa -K^2\left(\frac{{\alpha }_1^2}{1-{\alpha }_3^2}\right)){\parallel (I-S_2)U_1{x}_0\parallel }^2)\\ +{\alpha }_3^2{\parallel x_0-x^{\ast }\parallel }^2)+(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=3}^N(1-{\alpha }_3^j)(1-{\alpha }_3^2)({\parallel U_1{x}_0-x^{\ast }\parallel }^2+{\alpha }_3^2{\parallel x_0-x^{\ast }\parallel }^2)\\ +(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=2}^N(1-{\alpha }_3^j){\parallel U_1{x}_0-x^{\ast }\parallel }^2+(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=2}^N(1-{\alpha }_3^j){\parallel {\alpha }_1^1(S_1{U}_0{x}_0-x^{\ast })+{\alpha }_2^1(U_0{x}_0-x^{\ast })+{\alpha }_3^1(x_0-x^{\ast })\parallel }^2\\ +(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=2}^N(1-{\alpha }_3^j){\parallel {\alpha }_1^1(S_1{x}_0-x^{\ast })+(1-{\alpha }_1^1)(x_0-x^{\ast })\parallel }^2\\ +(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=2}^N(1-{\alpha }_3^j){\parallel {\alpha }_1^1(S_1{x}_0-x_0)+x_0-x^{\ast }\parallel }^2+(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=2}^N(1-{\alpha }_3^j)({\parallel x_0-x^{\ast }\parallel }^2+2{\alpha }_1^1〈S_1{x}_0-x_0,j(x_0-x^{\ast })〉\\ +2K^2{\left({\alpha }_1^1\right)}^2{\parallel S_1{x}_0-x_0\parallel }^2)+(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=2}^N(1-{\alpha }_3^j)({\parallel x_0-x^{\ast }\parallel }^2+2{\alpha }_1^1〈S_1{x}_0-x^{\ast },j(x_0-x^{\ast })〉\\ +2{\alpha }_1^1〈x^{\ast }-x_0,j(x_0-x^{\ast })〉\\ +2K^2{\left({\alpha }_1^1\right)}^2{\parallel S_1{x}_0-x_0\parallel }^2)+(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=2}^N(1-{\alpha }_3^j)({\parallel x_0-x^{\ast }\parallel }^2+2{\alpha }_1^1(\parallel x_0-x^{\ast }\parallel -\kappa {\parallel S_1{x}_0-x_0\parallel }^2)\\ -2{\alpha }_1^1{\parallel x^{\ast }-x_0\parallel }^2\\ +2K^2{\left({\alpha }_1^1\right)}^2{\parallel S_1{x}_0-x_0\parallel }^2)+(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=2}^N(1-{\alpha }_3^j)({\parallel x_0-x^{\ast }\parallel }^2-2{\alpha }_1^1(\kappa -K^2{\alpha }_1^1){\parallel S_1{x}_0-x_0\parallel }^2)\\ +(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& {\parallel x_0-x^{\ast }\parallel }^2-\prod _{j=2}^N(1-{\alpha }_3^j)2{\alpha }_1^1(\kappa -K^2{\alpha }_1^1){\parallel S_1{x}_0-x_0\parallel }^2\\ \le & {\parallel x_0-x^{\ast }\parallel }^2.\end{array}$$

(2.2)

For every $j=1,2,\dots ,N$ and (2.2), we have

$${\parallel U_j{x}_0-x^{\ast }\parallel }^2\le {\parallel x_0-x^{\ast }\parallel }^2.$$

(2.3)

For every $k=1,2,\dots ,N-1$ and (2.2) we have

$$\begin{array}{rcl}{\parallel x_0-x^{\ast }\parallel }^2& \le & \prod _{j=k+1}^N(1-{\alpha }_3^j){\parallel U_k{x}_0-x^{\ast }\parallel }^2+(1-\prod _{j=k+1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=k+1}^N(1-{\alpha }_3^j){\parallel T_k({\alpha }_1^k{S}_k{U}_{k-1}+{\alpha }_2^k{U}_{k-1}+{\alpha }_3^{k}I)x_0-x^{\ast }\parallel }^2\\ +(1-\prod _{j=k+1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=k+1}^N(1-{\alpha }_3^j){\parallel {\alpha }_1^k(S_k{U}_{k-1}{x}_0-x^{\ast })+{\alpha }_2^k(U_{k-1}{x}_0-x^{\ast })+{\alpha }_3^k(x_0-x^{\ast })\parallel }^2\\ +(1-\prod _{j=k+1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=k+1}^N(1-{\alpha }_3^j)\parallel (1-{\alpha }_3^k)(\frac{{\alpha }_1^k}{1-{\alpha }_3^k}(S_k{U}_{k-1}{x}_0-x^{\ast })+\frac{{\alpha }_2^k}{1-{\alpha }_3^k}(U_{k-1}{x}_0-x^{\ast }))\\ +{\alpha }_3^k(x_0-x^{\ast }){\parallel }^2+(1-\prod _{j=k+1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=k+1}^N(1-{\alpha }_3^j)((1-{\alpha }_3^k){\parallel \frac{{\alpha }_1^k}{1-{\alpha }_3^k}(S_k{U}_{k-1}{x}_0-x^{\ast })+\frac{{\alpha }_2^k}{1-{\alpha }_3^k}(U_{k-1}{x}_0-x^{\ast })\parallel }^2\\ +{\alpha }_3^k{\parallel x_0-x^{\ast }\parallel }^2)+(1-\prod _{j=k+1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=k+1}^N(1-{\alpha }_3^j)((1-{\alpha }_3^k)\parallel \frac{{\alpha }_1^k}{1-{\alpha }_3^k}(S_k{U}_{k-1}{x}_0-x^{\ast })\\ +(1-\frac{{\alpha }_1^k}{1-{\alpha }_3^k})(U_{k-1}{x}_0-x^{\ast }){\parallel }^2\\ +{\alpha }_3^k{\parallel x_0-x^{\ast }\parallel }^2)+(1-\prod _{j=k+1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=k+1}^N(1-{\alpha }_3^j)((1-{\alpha }_3^k){\parallel \frac{{\alpha }_1^k}{1-{\alpha }_3^k}(S_k{U}_{k-1}{x}_0-U_{k-1}{x}_0)+U_{k-1}{x}_0-x^{\ast }\parallel }^2\\ +{\alpha }_3^k{\parallel x_0-x^{\ast }\parallel }^2)+(1-\prod _{j=k+1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=k+1}^N(1-{\alpha }_3^j)\left((1-{\alpha }_3^k)\right({\parallel U_{k-1}{x}_0-x^{\ast }\parallel }^2\\ +2\frac{{\alpha }_1^k}{1-{\alpha }_3^k}〈S_k{U}_{k-1}{x}_0-U_{k-1}{x}_0,j(U_{k-1}{x}_0-x^{\ast })〉\\ +2K^2{\left(\frac{{\alpha }_1^k}{1-{\alpha }_3^k}\right)}^2{\parallel S_k{U}_{k-1}{x}_0-U_{k-1}{x}_0\parallel }^2)+{\alpha }_3^k{\parallel x_0-x^{\ast }\parallel }^2)\\ +(1-\prod _{j=k+1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=k+1}^N(1-{\alpha }_3^j)\left((1-{\alpha }_3^k)\right({\parallel U_{k-1}{x}_0-x^{\ast }\parallel }^2\\ +2\frac{{\alpha }_1^k}{1-{\alpha }_3^k}〈S_k{U}_{k-1}{x}_0-x^{\ast },j(U_{k-1}{x}_0-x^{\ast })〉\\ +2\frac{{\alpha }_1^k}{1-{\alpha }_3^k}〈x^{\ast }-U_{k-1}{x}_0,j(U_{k-1}{x}_0-x^{\ast })〉\\ +2K^2{\left(\frac{{\alpha }_1^k}{1-{\alpha }_3^k}\right)}^2{\parallel S_k{U}_{k-1}{x}_0-U_{k-1}{x}_0\parallel }^2)+{\alpha }_3^k{\parallel x_0-x^{\ast }\parallel }^2)\\ +(1-\prod _{j=k+1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=k+1}^N(1-{\alpha }_3^j)\left((1-{\alpha }_3^k)\right({\parallel U_{k-1}{x}_0-x^{\ast }\parallel }^2\\ +2\frac{{\alpha }_1^k}{1-{\alpha }_3^k}({\parallel U_{k-1}{x}_0-x^{\ast }\parallel }^2-\kappa \parallel (I-S_k)U_{k-1}{x}_0\parallel )\\ -2\frac{{\alpha }_1^k}{1-{\alpha }_3^k}{\parallel x^{\ast }-U_{k-1}{x}_0\parallel }^2\\ +2K^2{\left(\frac{{\alpha }_1^k}{1-{\alpha }_3^k}\right)}^2{\parallel S_k{U}_{k-1}{x}_0-U_{k-1}{x}_0\parallel }^2)+{\alpha }_3^k{\parallel x_0-x^{\ast }\parallel }^2)\\ +(1-\prod _{j=k+1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=k+1}^N(1-{\alpha }_3^j)\left((1-{\alpha }_3^k)\right({\parallel U_{k-1}{x}_0-x^{\ast }\parallel }^2\\ -2\frac{{\alpha }_1^k}{1-{\alpha }_3^k}(\kappa -K^2\left(\frac{{\alpha }_1^k}{1-{\alpha }_3^k}\right)){\parallel (I-S_k)U_{k-1}{x}_0\parallel }^2)+{\alpha }_3^k{\parallel x_0-x^{\ast }\parallel }^2)\\ +(1-\prod _{j=k+1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=k+1}^N(1-{\alpha }_3^j)\left((1-{\alpha }_3^k)\right({\parallel x_0-x^{\ast }\parallel }^2\\ -2\frac{{\alpha }_1^k}{1-{\alpha }_3^k}(\kappa -K^2\left(\frac{{\alpha }_1^k}{1-{\alpha }_3^k}\right)){\parallel (I-S_k)U_{k-1}{x}_0\parallel }^2)+{\alpha }_3^k{\parallel x_0-x^{\ast }\parallel }^2)\\ +(1-\prod _{j=k+1}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2,\end{array}$$

which implies that

$$U_{k-1}{x}_0=S_k{U}_{k-1}{x}_0$$

(2.4)

for every $k=1,2,\dots ,N-1$.

From (2.2), it implies that $x_0=S_1{x}_0$, that is, $x_0\in F(S)$. From the definition of $S^A$, we have

$$U_1{x}_0=T_1({\alpha }_1^1{S}_1{U}_0{x}_0+{\alpha }_2^1{U}_0{x}_0+{\alpha }_3^1{x}_0)=T_1{x}_0=x_0.$$

(2.5)

From (2.2) and $U_1{x}_0=x_0$, we have

$$\begin{array}{rcl}{\parallel x_0-x^{\ast }\parallel }^2& \le & \prod _{j=3}^N(1-{\alpha }_3^j)\left((1-{\alpha }_3^2)\right({\parallel U_1{x}_0-x^{\ast }\parallel }^2\\ -2\frac{{\alpha }_1^2}{1-{\alpha }_3^2}(\kappa -K^2\left(\frac{{\alpha }_1^2}{1-{\alpha }_3^2}\right)){\parallel (I-S_2)U_1{x}_0\parallel }^2)\\ +{\alpha }_3^2{\parallel x_0-x^{\ast }\parallel }^2)+(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=3}^N(1-{\alpha }_3^j)\left((1-{\alpha }_3^2)\right({\parallel x_0-x^{\ast }\parallel }^2\\ -2\frac{{\alpha }_1^2}{1-{\alpha }_3^2}(\kappa -K^2\left(\frac{{\alpha }_1^2}{1-{\alpha }_3^2}\right)){\parallel (I-S_2)x_0\parallel }^2)\\ +{\alpha }_3^2{\parallel x_0-x^{\ast }\parallel }^2)+(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=3}^N(1-{\alpha }_3^j)(1-{\alpha }_3^2)({\parallel x_0-x^{\ast }\parallel }^2\\ -2\frac{{\alpha }_1^2}{1-{\alpha }_3^2}(\kappa -K^2\left(\frac{{\alpha }_1^2}{1-{\alpha }_3^2}\right)){\parallel (I-S_2)x_0\parallel }^2)\\ +\prod _{j=3}^N(1-{\alpha }_3^j){\alpha }_3^2{\parallel x_0-x^{\ast }\parallel }^2+(1-\prod _{j=3}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=2}^N(1-{\alpha }_3^j)({\parallel x_0-x^{\ast }\parallel }^2-2\frac{{\alpha }_1^2}{1-{\alpha }_3^2}(\kappa -K^2\left(\frac{{\alpha }_1^2}{1-{\alpha }_3^2}\right)){\parallel (I-S_2)x_0\parallel }^2)\\ +(1-\prod _{j=2}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2.\end{array}$$

It implies that $x_0=S_2{x}_0$.

From the definition of $S^A$ and $x_0=S_2{x}_0$, we have

$$U_2{x}_0=T_2({\alpha }_1^2{S}_2{U}_1+{\alpha }_2^2{U}_1+{\alpha }_3^{2}I)x_0=T_2{x}_0.$$

(2.6)

From the definition of $U_3$ and (2.4), we have

$$U_3{x}_0=T_3({\alpha }_1^3{S}_3{U}_2+{\alpha }_2^3{U}_2+{\alpha }_3^{3}I)x_0=T_3((1-{\alpha }_3^3)U_2{x}_0+{\alpha }_3^3{x}_0).$$

(2.7)

From (2.2), (2.6), (2.7) and E is uniformly convex, we have

$$\begin{array}{rcl}{\parallel x_0-x^{\ast }\parallel }^2& \le & \prod _{j=4}^N(1-{\alpha }_3^j){\parallel U_3{x}_0-x^{\ast }\parallel }^2+(1-\prod _{j=4}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=4}^N(1-{\alpha }_3^j){\parallel T_3((1-{\alpha }_3^3)U_2{x}_0+{\alpha }_3^3{x}_0)-x^{\ast }\parallel }^2\\ +(1-\prod _{j=4}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=4}^N(1-{\alpha }_3^j){\parallel (1-{\alpha }_3^3)(U_2{x}_0-x^{\ast })+{\alpha }_3^3(x_0-x^{\ast })\parallel }^2\\ +(1-\prod _{j=4}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ =& \prod _{j=4}^N(1-{\alpha }_3^j){\parallel (1-{\alpha }_3^3)(T_2{x}_0-x^{\ast })+{\alpha }_3^3(x_0-x^{\ast })\parallel }^2\\ +(1-\prod _{j=4}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=4}^N(1-{\alpha }_3^j)((1-{\alpha }_3^3){\parallel T_2{x}_0-x^{\ast }\parallel }^2+{\alpha }_3^3{\parallel x_0-x^{\ast }\parallel }^2\\ -{\alpha }_3^3(1-{\alpha }_3^3)g_2(\parallel T_2{x}_0-x_0\parallel ))\\ +(1-\prod _{j=4}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2\\ \le & \prod _{j=4}^N(1-{\alpha }_3^j)({\parallel x_0-x^{\ast }\parallel }^2-{\alpha }_3^3(1-{\alpha }_3^3)g_2(\parallel T_2{x}_0-x_0\parallel ))\\ +(1-\prod _{j=4}^N(1-{\alpha }_3^j)){\parallel x_0-x^{\ast }\parallel }^2.\end{array}$$

It implies that

$$g_2(\parallel T_2{x}_0-x_0\parallel )=0.$$

(2.8)

Assume that $T_2{x}_0\ne x_0$, then we have $\parallel T_2{x}_0-x_0\parallel >0$. From the properties of $g_2$, we have

$$0=g(0)<g(\parallel T_2{x}_0-x_0\parallel )=0.$$

(2.9)

This is a contradiction. Then we have $T_2{x}_0=x_0$. From (2.6), we have $x_0=T_2{x}_0=U_2{x}_0$.

From the definition of $U_3$, we have

$$U_3{x}_0=T_3((1-{\alpha }_3^3)U_2{x}_0+{\alpha }_3^3{x}_0)=T_3{x}_0.$$

By using the same method as above, we have

$$x_0=U_3{x}_0=T_3{x}_0.$$

Continuing on this way, we can conclude that

$$x_0=U_i{x}_0=T_i{x}_0$$

(2.10)

for every $i=1,2,\dots ,N-1$. From (2.2) and (2.10), we have

$$\begin{array}{rcl}{\parallel x_0-x^{\ast }\parallel }^2& \le & (1-{\alpha }_3^N)({\parallel U_{N-1}{x}_0-x^{\ast }\parallel }^2\\ -2\frac{{\alpha }_1^N}{1-{\alpha }_3^N}(\kappa -K^2\left(\frac{{\alpha }_1^N}{1-{\alpha }_3^N}\right)){\parallel (I-S_N)U_{N-1}{x}_0\parallel }^2)+{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2\\ =& (1-{\alpha }_3^N)({\parallel x_0-x^{\ast }\parallel }^2-2\frac{{\alpha }_1^N}{1-{\alpha }_3^N}(\kappa -K^2\left(\frac{{\alpha }_1^N}{1-{\alpha }_3^N}\right)){\parallel (I-S_N)x_0\parallel }^2)\\ +{\alpha }_3^N{\parallel x_0-x^{\ast }\parallel }^2.\end{array}$$

It implies that

$$x_0=S_N{x}_0.$$

(2.11)

From the definition of $S^A$ and (2.10), we have

$$x_0=S^A{x}_0=U_N{x}_0=T_N({\alpha }_1^N{S}_N{U}_{N-1}+{\alpha }_2^N{U}_{N-1}+{\alpha }_3^{N}I)x_0=T_N{x}_0.$$

Then we have

$$x_0\in \bigcap _{i=1}^{N}F(T_i)\text{and}{x}_0\in \bigcap _{i=1}^{N}F(U_i).$$

(2.12)

Since $S_k{U}_{k-1}{x}_0=U_{k-1}{x}_0$ for every $k=1,2,\dots ,N-1$ and $x_0\in {\bigcap }_{i=1}^{N}F(U_i)$, then we have

$$S_k{x}_0=x_0$$

for every $k=1,2,\dots ,N-1$. From (2.11), it implies that

$$x_0\in \bigcap _{i=1}^{N}F(S_i).$$

(2.13)

From (2.12) and (2.13), we have

$$x_0\in \bigcap _{i=1}^{N}F(T_i)\cap \bigcap _{i=1}^{N}F(S_i).$$

(2.14)

Hence, $F(S^A)\subseteq {\bigcap }_{i=1}^{N}F(T_i)\cap {\bigcap }_{i=1}^{N}F(S_i)$. It is easy to see that ${\bigcap }_{i=1}^{N}F(T_i)\cap {\bigcap }_{i=1}^{N}F(S_i)\subseteq F(S^A)$.

Applying (2.2), we have that the mapping $S^A$ is nonexpansive. □

Lemma 2.8 [19]

Let C be a closed convex subset of a strictly convex Banach space E. Let $T_1$ and $T_2$ be two nonexpansive mappings from C into itself with $F(T_1)\cap F(T_2)\ne \mathrm{\varnothing }$. Define a mapping S by

$$Sx=\lambda T_{1}x+(1-\lambda )T_{2}x,\mathrm{\forall }x\in C,$$

where λ is a constant in $(0,1)$. Then S is nonexpansive and $F(S)=F(T_1)\cap F(T_2)$.

Applying Lemma 2.8, we have the following lemma.

Lemma 2.9 Let C be a closed convex subset of a strictly convex Banach space E. Let $T_1$, $T_2$ and $T_3$ be three nonexpansive mappings from C into itself with $F(T_1)\cap F(T_2)\cap F(T_3)\ne \mathrm{\varnothing }$. Define a mapping S by

$$Sx=\alpha T_{1}x+\beta T_{2}x+\gamma T_{3}x,\mathrm{\forall }x\in C,$$

where α, β, γ is a constant in $(0,1)$ and $\alpha +\beta +\gamma =1$. Then S is nonexpansive and $F(S)=F(T_1)\cap F(T_2)\cap F(T_3)$.

Proof For every $x\in C$ and the definition of the mapping S, we have

$$\begin{array}{rcl}Sx& =& \alpha T_{1}x+\beta T_{2}x+\gamma T_{3}x\\ =& \alpha T_{1}x+(1-\alpha )(\frac{\beta }{1-\alpha }{T}_{2}x+\frac{\gamma }{1-\alpha }{T}_{3}x)\\ =& \alpha T_{1}x+(1-\alpha )(\frac{\beta }{1-\alpha }{T}_{2}x+(1-\frac{\beta }{1-\alpha })T_{3}x)\\ =& \alpha T_{1}x+(1-\alpha )S_{1}x,\end{array}$$

(2.15)

where $S_1=\frac{\beta }{1-\alpha }{T}_2+(1-\frac{\beta }{1-\alpha })T_3$. From Lemma 2.8, we have $F(S_1)=F(T_2)\cap F(T_3)$ and $S_1$ is a nonexpansive mapping. From Lemma 2.8 and (2.15), we have $F(S)=F(T_1)\cap F(S_1)$ and S is a nonexpansive mapping. Hence we have $F(S)=F(T_1)\cap F(T_2)\cap F(T_3)$. □

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 a sunny nonexpansive retraction from E onto C and let A, B be α- and β-inverse strongly accretive mappings of C into E, respectively. Let ${\{S_i\}}_{i=1}^N$ be a finite family of ${\kappa }_i$-strict pseudo-contractions of C into itself and let ${\{T_i\}}_{i=1}^N$ be a finite family of nonexpansive mappings of C into itself with $\mathcal{F}={\bigcap }_{i=1}^{N}F(S_i)\cap {\bigcap }_{i=1}^{N}F(T_i)\cap S(C,A)\cap S(C,B)\ne \mathrm{\varnothing }$ and $\kappa =min\{{\kappa }_i:i=1,2,\dots ,N\}$ with $K^2\le \kappa$, where K is the 2-uniformly smooth constant of E. Let ${\alpha }_j=({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j)\in I\times I\times I$, where $I=[0,1]$, ${\alpha }_1^j+{\alpha }_2^j+{\alpha }_3^j=1$, ${\alpha }_1^j\in (0,1]$, ${\alpha }_2^j\in [0,1]$ and ${\alpha }_3^j\in (0,1)$ for all $j=1,2,\dots ,N$. Let $S^A$ be the $S^A$-mapping generated by $S_1,S_2,\dots ,S_N$, $T_1,T_2,\dots ,T_N$ and ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_N$. Let $\{x_n\}$ be the sequence generated by $x_1,u\in C$ and

$$x_{n+1}={\alpha }_{n}u+{\beta }_n{x}_n+{\gamma }_n{Q}_C(I-aA)x_n+{\delta }_n{Q}_C(I-bB)x_n+{\eta }_n{S}^A{x}_n,\mathrm{\forall }n\ge 1,$$

(3.1)

where $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\},\{{\delta }_n\},\{{\eta }_n\}\in [0,1]$ and ${\alpha }_n+{\beta }_n+{\gamma }_n+{\delta }_n+{\eta }_n=1$ and satisfy the following conditions:

$$\begin{array}{r}\text{(i)}\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=0,\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty },\\ \text{(ii)}\{{\gamma }_n\},\{{\delta }_n\},\{{\eta }_n\}\subseteq [c,d]\subset (0,1),\mathit{\text{for some}}c,d>0,\mathrm{\forall }n\ge 1,\\ \text{(iii)}\sum _{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|,\sum _{n=1}^{\mathrm{\infty }}|{\gamma }_{n+1}-{\gamma }_n|,\sum _{n=1}^{\mathrm{\infty }}|{\delta }_{n+1}-{\delta }_n|,\\ \phantom{\text{(iii)}}\sum _{n=1}^{\mathrm{\infty }}|{\eta }_{n+1}-{\eta }_n|,\sum _{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty },\\ \text{(iv)}0<\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\beta }_n\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\beta }_n<1,\\ \text{(v)}a\in (0,\frac{\alpha }{K^2})\mathit{\text{and}}b\in (0,\frac{\beta }{K^2}).\end{array}$$

Then $\{x_n\}$ converges strongly to $z_0=Q_{\mathcal{F}}u$, where $Q_{\mathcal{F}}$ is the sunny nonexpansive retraction of C onto ℱ.

Proof First we show that $Q_C(I-aA)$ and $Q_C(I-bB)$ are nonexpansive mappings. Let $x,y\in C$, we have

$$\begin{array}{rcl}{\parallel Q_C(I-aA)x-Q_C(I-aA)y\parallel }^2& \le & {\parallel x-y-a(Ax-Ay)\parallel }^2\\ \le & {\parallel x-y\parallel }^2-2a〈Ax-Ay,j(x-y)〉+2K^2{a}^2{\parallel Ax-Ay\parallel }^2\\ \le & {\parallel x-y\parallel }^2-2a\alpha {\parallel Ax-Ay\parallel }^2+2K^2{a}^2{\parallel Ax-Ay\parallel }^2\\ =& {\parallel x-y\parallel }^2-2a(\alpha -K^{2}a){\parallel Ax-Ay\parallel }^2\\ \le & {\parallel x-y\parallel }^2.\end{array}$$

(3.2)

Then we have $Q_C(I-aA)$ is a nonexpansive mapping. By using the same methods as (3.2), we have $Q_C(I-bB)$ is a nonexpansive mapping.

Let $x^{\ast }\in \mathcal{F}$. From Lemma 2.3, we have $x^{\ast }\in F(Q_C(I-aA))$ and $x^{\ast }\in F(Q_C(I-bB))$. By the definition of $x_n$, we have

$$\begin{array}{rcl}\parallel x_{n+1}-x^{\ast }\parallel & \le & {\alpha }_n\parallel u-x^{\ast }\parallel +{\beta }_n\parallel x_n-x^{\ast }\parallel +{\gamma }_n\parallel Q_C(I-aA)x_n-x^{\ast }\parallel \\ +{\delta }_n\parallel Q_C(I-bB)x_n-x^{\ast }\parallel +{\eta }_n\parallel S^A{x}_n-x^{\ast }\parallel \\ \le & {\alpha }_n\parallel u-x^{\ast }\parallel +(1-{\alpha }_n)\parallel x_n-x^{\ast }\parallel \\ \le & max\{\parallel u-x^{\ast }\parallel ,\parallel x_1-x^{\ast }\parallel \}.\end{array}$$

By induction, we have $\parallel x_n-x^{\ast }\parallel \le max\{\parallel u-x^{\ast }\parallel ,\parallel x_1-x^{\ast }\parallel \}$. We can imply that the sequence $\{x_n\}$ is bounded and so are $\{S^A{x}_n\}$, $\{Q_C(I-aA)x_n\}$ and $\{Q_C(I-bB)x_n\}$.

Next, we show that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$. From the definition of $x_n$, we have

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & =& \parallel {\alpha }_{n}u+{\beta }_n{x}_n+{\gamma }_n{Q}_C(I-aA)x_n+{\delta }_n{Q}_C(I-bB)x_n+{\eta }_n{S}^A{x}_n\\ -{\alpha }_{n-1}u-{\beta }_{n-1}{x}_{n-1}-{\gamma }_{n-1}{Q}_C(I-aA)x_{n-1}-{\delta }_{n-1}{Q}_C(I-bB)x_{n-1}\\ -{\eta }_{n-1}{S}^A{x}_{n-1}\parallel \\ \le & |{\alpha }_n-{\alpha }_{n-1}|\parallel u\parallel +{\beta }_n\parallel x_n-x_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}\parallel \\ +{\gamma }_n\parallel Q_C(I-aA)x_n-Q_C(I-aA)x_{n-1}\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel Q_C(I-aA)x_{n-1}\parallel \\ +{\delta }_n\parallel Q_C(I-bB)x_n-Q_C(I-bB)x_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel Q_C(I-bB)x_{n-1}\parallel \\ +{\eta }_n\parallel S^A{x}_n-S^A{x}_{n-1}\parallel +|{\eta }_{n-1}-{\eta }_n|\parallel S^A{x}_n\parallel \\ \le & (1-{\alpha }_n)\parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel u\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}\parallel \\ +|{\gamma }_n-{\gamma }_{n-1}|\parallel Q_C(I-aA)x_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel Q_C(I-bB)x_{n-1}\parallel \\ +|{\eta }_{n-1}-{\eta }_n|\parallel S^A{x}_n\parallel .\end{array}$$

Applying Lemma 2.6, we have

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

(3.3)

Next, we show that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel Q_C(I-aA)x_n-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel Q_C(I-bB)x_n-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel S^A{x}_n-x_n\parallel =0.$$

(3.4)

From the definition of $x_n$, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& =& \parallel {\alpha }_n(u-x^{\ast })+{\beta }_n(x_n-x^{\ast })+{\gamma }_n(Q_C(I-aA)x_n-x^{\ast })\\ +{\delta }_n(Q_C(I-bB)x_n-x^{\ast })+{\eta }_n(S^A{x}_n-x^{\ast }){\parallel }^2\\ =& \parallel {\beta }_n(x_n-x^{\ast })+{\gamma }_n(Q_C(I-aA)x_n-x^{\ast })+({\alpha }_n+{\delta }_n+{\eta }_n)(\frac{{\alpha }_n(u-x^{\ast })}{{\alpha }_n+{\delta }_n+{\eta }_n}\\ +\frac{{\delta }_n(Q_C(I-bB)x_n-x^{\ast })}{{\alpha }_n+{\delta }_n+{\eta }_n}+\frac{{\eta }_n(S^A{x}_n-x^{\ast })}{{\alpha }_n+{\delta }_n+{\eta }_n}){\parallel }^2\\ =& {\parallel {\beta }_n(x_n-x^{\ast })+{\gamma }_n(Q_C(I-aA)x_n-x^{\ast })+c_n{z}_n\parallel }^2,\end{array}$$

where $c_n={\alpha }_n+{\delta }_n+{\eta }_n$ and $z_n=\frac{{\alpha }_n(u-x^{\ast })}{{\alpha }_n+{\delta }_n+{\eta }_n}+\frac{{\delta }_n(Q_C(I-bB)x_n-x^{\ast })}{{\alpha }_n+{\delta }_n+{\eta }_n}+\frac{{\eta }_n(S^A{x}_n-x^{\ast })}{{\alpha }_n+{\delta }_n+{\eta }_n}$.

From Lemma 2.2, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\beta }_n{\parallel x_n-x^{\ast }\parallel }^2+{\gamma }_n\parallel Q_C(I-aA)x_n-x^{\ast }\parallel +c_n{\parallel z_n\parallel }^2\\ -{\beta }_n{\gamma }_n{g}_1\left(\parallel x_n-Q_C(I-aA)x_n\parallel \right)\\ \le & ({\beta }_n+{\gamma }_n){\parallel x_n-x^{\ast }\parallel }^2-{\beta }_n{\gamma }_n{g}_1\left(\parallel x_n-Q_C(I-aA)x_n\parallel \right)\\ +c_n(\frac{{\alpha }_n{\parallel u-x^{\ast }\parallel }^2}{{\alpha }_n+{\delta }_n+{\eta }_n}+\frac{{\delta }_n{\parallel Q_C(I-bB)x_n-x^{\ast }\parallel }^2}{{\alpha }_n+{\delta }_n+{\eta }_n}+\frac{{\eta }_n{\parallel S^A{x}_n-x^{\ast }\parallel }^2}{{\alpha }_n+{\delta }_n+{\eta }_n})\\ \le & ({\beta }_n+{\gamma }_n){\parallel x_n-x^{\ast }\parallel }^2-{\beta }_n{\gamma }_n{g}_1\left(\parallel x_n-Q_C(I-aA)x_n\parallel \right)\\ +{\alpha }_n{\parallel u-x^{\ast }\parallel }^2+({\delta }_n+{\eta }_n){\parallel x_n-x^{\ast }\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2-{\beta }_n{\gamma }_n{g}_1\left(\parallel x_n-Q_C(I-aA)x_n\parallel \right)+{\alpha }_n{\parallel u-x^{\ast }\parallel }^2,\end{array}$$

which implies that

$$\begin{array}{rcl}{\beta }_n{\gamma }_n{g}_1\left(\parallel x_n-Q_C(I-aA)x_n\parallel \right)& \le & {\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2+{\alpha }_n{\parallel u-x^{\ast }\parallel }^2\\ \le & (\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel \\ +{\alpha }_n{\parallel u-x^{\ast }\parallel }^2.\end{array}$$

(3.5)

From (3.3) and condition (i), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}{g}_1\left(\parallel x_n-Q_C(I-aA)x_n\parallel \right)=0.$$

(3.6)

From the property of $g_1$, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-Q_C(I-aA)x_n\parallel =0.$$

(3.7)

By using the same method as (3.7), we can imply that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-Q_C(I-bB)x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-S^A{x}_n\parallel =0.$$

Define $Gx=\alpha S^{A}x+\beta Q_C(I-aA)x+\gamma Q_C(I-bB)x$ for all $x\in C$ and $\alpha +\beta +\gamma =1$. From Lemma 2.9, we have $F(G)=F(Q_C(I-aA))\cap F(Q_C(I-bB))\cap F(S^A)$. From Lemmas 2.3 and 2.7, we have $\mathcal{F}=F(G)={\bigcap }_{i=1}^{N}F(T_i)\cap {\bigcap }_{i=1}^{N}F(S_i)\cap S(C,A)\cap S(C,B)$. By the definition of G, we obtain

$$\parallel G{x}_n-x_n\parallel \le \alpha \parallel S^A{x}_n-x_n\parallel +\beta \parallel Q_C(I-aA)x_n-x_n\parallel +\gamma \parallel Q_C(I-bB)x_n-x_n\parallel .$$

From (3.4), we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel G{x}_n-x_n\parallel =0.$$

(3.8)

From Lemma 2.5 and (3.8), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-z_0,j(x_n-z_0)〉\le 0,$$

(3.9)

where $z_0=Q_{\mathcal{F}}u$. Finally, we prove strong convergence of the sequence $\{x_n\}$ to $z_0=Q_{\mathcal{F}}u$. From the definition of $x_n$, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-z_0\parallel }^2& =& \parallel {\alpha }_n(u-z_0)+{\beta }_n(x_n-z_0)+{\gamma }_n(Q_C(I-aA)x_n-z_0)\\ +{\delta }_n(Q_C(I-bB)x_n-z_0)+{\eta }_n(S^A{x}_n-z_0){\parallel }^2\\ =& \parallel {\alpha }_n(u-z_0)+(1-{\alpha }_n)(\frac{{\beta }_n(x_n-z_0)}{1-{\alpha }_n}+\frac{{\gamma }_n(Q_C(I-aA)x_n-z_0)}{1-{\alpha }_n}\\ +\frac{{\delta }_n(Q_C(I-bB)x_n-z_0)}{1-{\alpha }_n}+\frac{{\eta }_n(S^A{x}_n-z_0)}{1-{\alpha }_n}){\parallel }^2\\ \le & \parallel (1-{\alpha }_n)(\frac{{\beta }_n(x_n-z_0)}{1-{\alpha }_n}+\frac{{\gamma }_n(Q_C(I-aA)x_n-z_0)}{1-{\alpha }_n}\\ +\frac{{\delta }_n(Q_C(I-bB)x_n-z_0)}{1-{\alpha }_n}+\frac{{\eta }_n(S^A{x}_n-z_0)}{1-{\alpha }_n}){\parallel }^2+2{\alpha }_n〈u-x_0,j(x_{n+1}-z_0)〉\\ \le & (1-{\alpha }_n){\parallel x_n-z_0\parallel }^2+2{\alpha }_n〈u-x_0,j(x_{n+1}-z_0)〉.\end{array}$$

Applying Lemma 2.6 and condition (i), we have ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_0\parallel =0$. This completes the proof. □

4 Applications

From our main results, we obtain strong convergence theorems in a Banach space. Before proving these theorem, we need the following lemma which is the result from Lemma 2.7 and Definition 1.4. Therefore, we omit the proof.

Lemma 4.1 Let C be a nonempty closed convex subset of a 2-uniformly smooth and uniformly convex Banach space. Let ${\{S_i\}}_{i=1}^N$ be a finite family of ${\kappa }_i$-strict pseudo-contractions of C into itself with ${\bigcap }_{i=1}^{N}F(S_i)\ne \mathrm{\varnothing }$ and $\kappa =min\{{\kappa }_i:i=1,2,\dots ,N\}$ with $K^2\le \kappa$, where K is the 2-uniformly smooth constant of E. Let ${\alpha }_j=({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j)\in I\times I\times I$, where $I=[0,1]$, ${\alpha }_1^j+{\alpha }_2^j+{\alpha }_3^j=1$, ${\alpha }_1^j\in (0,1]$, ${\alpha }_2^j\in [0,1]$ and ${\alpha }_3^j\in (0,1)$ for all $j=1,2,\dots ,N$. Let S be the S-mapping generated by $S_1,S_2,\dots ,S_N$ and ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_N$. Then $F(S)={\bigcap }_{i=1}^{N}F(S_i)$ and S is a nonexpansive mapping.

Theorem 4.2 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space E. Let $Q_C$ be a sunny nonexpansive retraction from E onto C and let A, B be α- and β-inverse strongly accretive mappings of C into E, respectively. Let ${\{S_i\}}_{i=1}^N$ be a finite family of ${\kappa }_i$-strict pseudo-contractions of C into itself with $\mathcal{F}={\bigcap }_{i=1}^{N}F(S_i)\cap S(C,A)\cap S(C,B)\ne \mathrm{\varnothing }$ and $\kappa =min\{{\kappa }_i:i=1,2,\dots ,N\}$ with $K^2\le \kappa$, where K is the 2-uniformly smooth constant of E. Let ${\alpha }_j=({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j)\in I\times I\times I$, where $I=[0,1]$, ${\alpha }_1^j+{\alpha }_2^j+{\alpha }_3^j=1$, ${\alpha }_1^j\in (0,1]$, ${\alpha }_2^j\in [0,1]$ and ${\alpha }_3^j\in (0,1)$ for all $j=1,2,\dots ,N$. Let S be the S-mapping generated by $S_1,S_2,\dots ,S_N$ and ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_N$. Let $\{x_n\}$ be the sequence generated by $x_1,u\in C$ and

$$x_{n+1}={\alpha }_{n}u+{\beta }_n{x}_n+{\gamma }_n{Q}_C(I-aA)x_n+{\delta }_n{Q}_C(I-bB)x_n+{\eta }_{n}S{x}_n,\mathrm{\forall }n\ge 1,$$

where $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\},\{{\delta }_n\},\{{\eta }_n\}\in [0,1]$ and ${\alpha }_n+{\beta }_n+{\gamma }_n+{\delta }_n+{\eta }_n=1$ and satisfy the following conditions:

$$\begin{array}{r}\text{(i)}\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=0,\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty },\\ \text{(ii)}\{{\gamma }_n\},\{{\delta }_n\},\{{\eta }_n\}\subseteq [c,d]\subset (0,1)\mathit{\text{for some}}c,d>0,\mathrm{\forall }n\ge 1,\\ \text{(iii)}\sum _{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|,\sum _{n=1}^{\mathrm{\infty }}|{\gamma }_{n+1}-{\gamma }_n|,\sum _{n=1}^{\mathrm{\infty }}|{\delta }_{n+1}-{\delta }_n|,\\ \phantom{\text{(iii)}}\sum _{n=1}^{\mathrm{\infty }}|{\eta }_{n+1}-{\eta }_n|,\sum _{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty },\\ \text{(iv)}0<\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\beta }_n\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\beta }_n<1,\\ \text{(v)}a\in (0,\frac{\alpha }{K^2})\mathit{\text{and}}b\in (0,\frac{\beta }{K^2}).\end{array}$$

Then $\{x_n\}$ converges strongly to $z_0=Q_{\mathcal{F}}u$, where $Q_{\mathcal{F}}$ is the sunny nonexpansive retraction of C onto ℱ.

Proof Put $I=T_1=T_2=\cdots =T_N$ in Theorem 3.1. From Lemma 4.1 and Theorem 3.1 we can conclude the desired result. □

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 a sunny nonexpansive retraction from E onto C. For every $i=1,2,\dots ,N$, let $A_i$, A, B be ${\alpha }_i$-, α- and β-inverse strongly accretive mappings of C into E, respectively. Define a mapping $G_i:C\to C$ by $Q_C(I-{\lambda }_i{A}_i)x=G_{i}x$, where ${\lambda }_i\in (0,\frac{{\alpha }_i}{K^2})$, K is the 2-uniformly smooth constant of E, for all $x\in C$ and $i=1,2,\dots ,N$. Let ${\{S_i\}}_{i=1}^N$ be a finite family of ${\kappa }_i$-strict pseudo-contractions of C into itself and with $\mathcal{F}={\bigcap }_{i=1}^{N}F(S_i)\cap {\bigcap }_{i=1}^{N}S(C,A_i)\cap S(C,A)\cap S(C,B)\ne \mathrm{\varnothing }$ and $\kappa =min\{{\kappa }_i:i=1,2,\dots ,N\}$ with $K^2\le \kappa$. Let ${\alpha }_j=({\alpha }_1^j,{\alpha }_2^j,{\alpha }_3^j)\in I\times I\times I$, where $I=[0,1]$, ${\alpha }_1^j+{\alpha }_2^j+{\alpha }_3^j=1$, ${\alpha }_1^j\in (0,1]$, ${\alpha }_2^j\in [0,1]$ and ${\alpha }_3^j\in (0,1)$ for all $j=1,2,\dots ,N$. Let $S^A$ be the $S^A$-mapping generated by $S_1,S_2,\dots ,S_N$, $G_1,G_2,\dots ,G_N$ and ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_N$. Let $\{x_n\}$ be the sequence generated by $x_1,u\in C$ and

$$x_{n+1}={\alpha }_{n}u+{\beta }_n{x}_n+{\gamma }_n{Q}_C(I-aA)x_n+{\delta }_n{Q}_C(I-bB)x_n+{\eta }_n{S}^A{x}_n,\mathrm{\forall }n\ge 1,$$

where $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\},\{{\delta }_n\},\{{\eta }_n\}\in [0,1]$ and ${\alpha }_n+{\beta }_n+{\gamma }_n+{\delta }_n+{\eta }_n=1$ and satisfy the following conditions:

$$\begin{array}{r}\text{(i)}\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=0,\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty },\\ \text{(ii)}\{{\gamma }_n\},\{{\delta }_n\},\{{\eta }_n\}\subseteq [c,d]\subset (0,1)\mathit{\text{for some}}c,d>0,\mathrm{\forall }n\ge 1,\\ \text{(iii)}\sum _{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|,\sum _{n=1}^{\mathrm{\infty }}|{\gamma }_{n+1}-{\gamma }_n|,\sum _{n=1}^{\mathrm{\infty }}|{\delta }_{n+1}-{\delta }_n|,\\ \phantom{\text{(iii)}}\sum _{n=1}^{\mathrm{\infty }}|{\eta }_{n+1}-{\eta }_n|,\sum _{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty },\\ \text{(iv)}0<\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\beta }_n\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\beta }_n<1,\\ \text{(v)}a\in (0,\frac{\alpha }{K^2})\mathit{\text{and}}b\in (0,\frac{\beta }{K^2}).\end{array}$$

Then $\{x_n\}$ converges strongly to $z_0=Q_{\mathcal{F}}u$, where $Q_{\mathcal{F}}$ is the sunny nonexpansive retraction of C onto ℱ.

Proof By using the same method as (3.2), we can conclude that ${\{G_i\}}_{i=1}^N$ is a nonexpansive mapping. From Lemma 2.3, we have $F(G_i)=S(C,A_i)$ for all $i=1,2,\dots ,N$. From Theorem 3.1 we can conclude the desired conclusion. □