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

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.

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

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

A Banach space E is said to be uniformly smooth if

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

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.

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

$$J(x)=\{f\in E^{\ast }:〈x,f〉={\parallel x\parallel }^2\text{ and }\parallel f\parallel =\parallel x\parallel \}$$

is called the normalized duality mapping of E. The duality pair $〈x,f〉$ represents $f(x)$ 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 $\eta \in (0,1)$ 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 $j(x-y)\in J(x-y)$.

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 $P(x+t(x-P(x)))=P(x)$ for all $x\in C$ and $t\ge 0$, whenever $x+t(x-P(x))\in C$. 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 $j(x-y)\in J(x-y)$ 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 exists $j(x-y)\in J(x-y)$ and $\alpha >0$ such that

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

Remark 1.1 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 $j(x-x^{\ast })\in J(x-x^{\ast })$,

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

(1.3)

This problem was considered by Aoyama et al. [2]. 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)

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 $\{x_n\}$ 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 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 $\{T_i\}$ be a nonexpansive mapping from C into itself for $i=1,2,\dots ,N$. Assume that $F={\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$. Given a point $u\in C$ and given sequences $\{{\alpha }_n\},\{{\beta }_n\}\in (0,1)$, the following conditions are satisfied:

Let $\{x_n\}$ be a sequence generated by $u,x_0=x\in C$ and

$$\{\begin{array}{c}{y}_n={\beta }_n{x}_n+(1-{\beta }_n)W_n{x}_n,\hfill \\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)y_n,n\ge 0,\hfill \end{array}$$

(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 $\{x_n\}$ converges strongly to $x^{\ast }\in F$, where $x^{\ast }=Q(u)$ 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 $F(T)\ne \mathrm{\varnothing }$. Given $u,x_0\in C$ and the sequences $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$ and $\{{\delta }_n\}$ in $(0,1)$, the following control conditions are satisfied:

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

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

(1.7)

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 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 $S(C,A)\ne \mathrm{\varnothing }$. Suppose $x_1=x\in C$ 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 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:

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

for any $x,y\in E$.

Definition 2.1 (See [9])

Let C be a nonempty convex subset of a real Banach space. Let ${\{T_i\}}_{i=1}^N$ 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:

$$\begin{array}{r}{U}_1={\lambda }_1{T}_1+(1-{\lambda }_1)I,\\ U_2={\lambda }_2{T}_2{U}_1+(1-{\lambda }_2)U_1,\\ U_3={\lambda }_3{T}_3{U}_2+(1-{\lambda }_3)U_2,\\ ⋮\\ U_{N-1}={\lambda }_{N-1}{T}_{N-1}{U}_{N-2}+(1-{\lambda }_{N-1})U_{N-2},\\ K=U_N={\lambda }_N{T}_N{U}_{N-1}+(1-{\lambda }_N)U_{N-1}.\end{array}$$

(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 ${\{T_i\}}_{i=1}^N$ be a finite family of nonexpanxive mappings of C into itself with ${\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$ 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 $F(K)={\bigcap }_{i=1}^{N}F(T_i)$.

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 $\{x_n\}$ and $\{z_n\}$ be bounded sequences in a Banach space X and let $\{{\beta }_n\}$ be a sequence in $[0,1]$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)z_n$$

for all integer $n\ge 0$ and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

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 $B_r=\{x\in X:\parallel x\parallel \le r\}$, $r>0$. 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 2.6 (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 $\lambda >0$,

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

Lemma 2.7 (See [12])

Let C be a closed convex subset of a strictly convex Banach space X. Let $\{T_n:n\in \mathbb{N}\}$ be a sequence of nonexpansive mappings on C. Suppose ${\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)\ne \mathrm{\varnothing }$ is nonempty. Let $\{{\lambda }_n\}$ 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 $F(S)={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$ holds.

Lemma 2.8 (See [8])

Let $r>0$. If E is uniformly convex, then there exists a continuous, strictly increasing and convex function $g:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $g(0)=0$ 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 2.9 (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 $F(T)\ne \mathrm{\varnothing }$ and let $f\in {\prod }_C$ where ${\prod }_C$ is to denote the collection of all contractions on C. Then the sequence $\{x_t\}$ defined by $x_t=tf(x_t)+(1-t)T{x}_t$ converses strongly to a point in $F(T)$. If we define a mapping $Q:{\prod }_C\to F(T)$ by $Q(f)={lim}_{t\to 0}{x}_t$ for all $f\in {\prod }_C$, then $Q(f)$ solves the following variational inequality:

$$〈(I-f)Q(f),j(Q(f)-p)〉\le 0$$

for all $f\in {\prod }_C$, $p\in F(T)$.

Lemma 2.10 (See [14])

In a Banach space E, the following inequality holds:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,j(x+y)〉,\mathrm{\forall }x,y\in E,$$

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

Lemma 2.11 (See [15])

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

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

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ 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 $F(S)\cap F(T)\ne \mathrm{\varnothing }$. Define a mapping $B_A:C\to C$ by $B_{A}x=T((1-\alpha )I+\alpha S)x$ for all $x\in C$ and $\alpha \in (0,\frac{\eta }{K^2})$, where K is the 2-uniformly smooth constant of E. Then $F(B_A)=F(S)\cap F(T)$.

Proof It is easy to see that $F(T)\cap F(S)\subseteq F(B_A)$. Let $x_0\in F(B_A)$ and $x^{\ast }\in F(T)\cap F(S)$, we have

$$\begin{array}{rcl}{\parallel x_0-x^{\ast }\parallel }^2& =& {\parallel T((1-\alpha )x_0+\alpha S{x}_0)-x^{\ast }\parallel }^2\\ \le & {\parallel (1-\alpha )x_0+\alpha S{x}_0-x^{\ast }\parallel }^2\\ =& {\parallel x_0-x^{\ast }+\alpha (S{x}_0-x_0)\parallel }^2\\ \le & {\parallel x_0-x^{\ast }\parallel }^2+2\alpha 〈S{x}_0-x_0,j(x_0-x^{\ast })〉+2K^2{\alpha }^2{\parallel S{x}_0-x_0\parallel }^2\\ =& {\parallel x_0-x^{\ast }\parallel }^2+2\alpha 〈S{x}_0-x^{\ast },j(x_0-x^{\ast })〉+2\alpha 〈x^{\ast }-x_0,j(x_0-x^{\ast })〉\\ +2K^2{\alpha }^2{\parallel S{x}_0-x_0\parallel }^2\\ =& {\parallel x_0-x^{\ast }\parallel }^2+2\alpha 〈S{x}_0-x^{\ast },j(x_0-x^{\ast })〉-2\alpha {\parallel x_0-x^{\ast }\parallel }^2+2K^2{\alpha }^2{\parallel S{x}_0-x_0\parallel }^2\\ \le & {\parallel x_0-x^{\ast }\parallel }^2+2\alpha ({\parallel x_0-x^{\ast }\parallel }^2-\eta {\parallel (I-S)x_0\parallel }^2)-2\alpha {\parallel x_0-x^{\ast }\parallel }^2\\ +2K^2{\alpha }^2{\parallel S{x}_0-x_0\parallel }^2\\ =& {\parallel x_0-x^{\ast }\parallel }^2-2\alpha \eta {\parallel x_0-S{x}_0\parallel }^2+2K^2{\alpha }^2{\parallel S{x}_0-x_0\parallel }^2\\ =& {\parallel x_0-x^{\ast }\parallel }^2-2\alpha (\eta -K^2\alpha ){\parallel x_0-S{x}_0\parallel }^2.\end{array}$$

(2.2)

(2.2) implies that

$$2\alpha (\eta -K^2\alpha ){\parallel x_0-S{x}_0\parallel }^2\le {\parallel x_0-x^{\ast }\parallel }^2-{\parallel x_0-x^{\ast }\parallel }^2=0.$$

Then we have $S{x}_0=x_0$, that is, $x_0\in F(S)$.

Since $x_0\in F(B_A)$, from the definition of $B_A$, we have

$$x_0=B_A{x}_0=T((1-\alpha )x_0+\alpha S{x}_0)=T{x}_0.$$

Then we have $x_0\in F(T)$. Therefore, $x_0\in F(T)\cap F(S)$. It follows that $F(B_A)\subseteq F(T)\cap F(S)$. Hence, $F(B_A)=F(T)\cap F(S)$. □

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 $Q_C(I-{\lambda }_i{A}_i)x=G_{i}x$ 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 the sequence generated by $x_1\in C$ and

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

(3.1)

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:

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

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

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

$$\begin{array}{rcl}{\parallel G_{i}x-G_{i}y\parallel }^2& =& {\parallel Q_C(I-{\lambda }_i{A}_i)x-Q_C(I-{\lambda }_i{A}_i)y\parallel }^2\\ \le & {\parallel (I-{\lambda }_i{A}_i)x-(I-{\lambda }_i{A}_i)y\parallel }^2\\ =& {\parallel x-y-{\lambda }_i(A_{i}x-A_{i}y)\parallel }^2\\ \le & {\parallel x-y\parallel }^2-2{\lambda }_i〈A_{i}x-A_{i}y,j(x-y)〉+2K^2{\lambda }_i^2{\parallel A_{i}x-A_{i}y\parallel }^2\\ \le & {\parallel x-y\parallel }^2-2{\lambda }_i{\alpha }_i{\parallel A_{i}x-A_{i}y\parallel }^2+2K^2{\lambda }_i^2{\parallel A_{i}x-A_{i}y\parallel }^2\\ =& {\parallel x-y\parallel }^2-2{\lambda }_i({\alpha }_i-K^2{\lambda }_i){\parallel A_{i}x-A_{i}y\parallel }^2\\ \le & {\parallel x-y\parallel }^2.\end{array}$$

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 $F(B)={\bigcap }_{i=1}^{N}F(G_i)$. From Lemma 2.6 and the definition of $G_i$, we have $F(G_i)=S(C,A_i)$ for every $i=1,2,\dots ,N$. Hence, we have

$$F(B)=\bigcap _{i=1}^{N}F(G_i)=\bigcap _{i=1}^{N}S(C,A_i).$$

(3.2)

Next, we will show that the sequence $\{x_n\}$ is bounded.

Let $z\in \mathcal{F}$; from the definition of $x_n$, we have

$$\begin{array}{rcl}\parallel x_{n+1}-z\parallel & \le & {\alpha }_n\parallel f(x_n)-z\parallel +{\beta }_n\parallel x_n-z\parallel +{\gamma }_n\parallel B{x}_n-z\parallel +{\delta }_n\parallel B_A{x}_n-z\parallel \\ \le & {\alpha }_n\parallel f(x_n)-z\parallel +(1-{\alpha }_n)\parallel x_n-z\parallel \\ \le & {\alpha }_n\parallel f(x_n)-f(z)\parallel +{\alpha }_n\parallel f(z)-z\parallel +(1-{\alpha }_n)\parallel x_n-z\parallel \\ \le & {\alpha }_{n}a\parallel x_n-z\parallel +{\alpha }_n\parallel f(z)-z\parallel +(1-{\alpha }_n)\parallel x_n-z\parallel \\ =& (1-{\alpha }_n(1-a))\parallel x_n-z\parallel +{\alpha }_n\parallel f(z)-z\parallel \\ \le & max\{\parallel x_1-z\parallel ,\frac{\parallel f(z)-z\parallel }{1-a}\}.\end{array}$$

By induction, we can conclude that the sequence $\{x_n\}$ is bounded and so are $\{f(x_n)\}$, $\{B{x}_n\}$, $\{B_A{x}_n\}$.

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

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)z_n,$$

(3.4)

where $z_n=\frac{{\alpha }_{n}f(x_n)+{\gamma }_{n}B{x}_n+{\delta }_n{B}_A{x}_n}{1-{\beta }_n}$.

Since

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & =& \parallel \frac{{\alpha }_{n+1}f(x_{n+1})+{\gamma }_{n+1}B{x}_{n+1}+{\delta }_{n+1}{B}_A{x}_{n+1}}{1-{\beta }_{n+1}}\\ -\left(\frac{{\alpha }_{n}f(x_n)+{\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}-(x_{n+1}-{\beta }_n{x}_n)\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}-(x_{n+1}-{\beta }_n{x}_n)\parallel \\ +\frac{|{\beta }_{n+1}-{\beta }_n|}{(1-{\beta }_n)(1-{\beta }_{n+1})}\parallel x_{n+1}-{\beta }_n{x}_n\parallel \\ =& \frac{1}{1-{\beta }_{n+1}}\parallel {\alpha }_{n+1}f(x_{n+1})+{\gamma }_{n+1}B{x}_{n+1}+{\delta }_{n+1}{B}_A{x}_{n+1}\\ -({\alpha }_{n}f(x_n)+{\gamma }_{n}B{x}_n+{\delta }_n{B}_A{x}_n)\parallel +\frac{|{\beta }_{n+1}-{\beta }_n|}{(1-{\beta }_n)(1-{\beta }_{n+1})}\parallel x_{n+1}-{\beta }_n{x}_n\parallel \\ =& \frac{1}{1-{\beta }_{n+1}}(\parallel {\alpha }_{n+1}f(x_{n+1})-{\alpha }_{n}f(x_n)\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 )\\ +\frac{|{\beta }_{n+1}-{\beta }_n|}{(1-{\beta }_n)(1-{\beta }_{n+1})}\parallel x_{n+1}-{\beta }_n{x}_n\parallel \\ \le & \frac{1}{1-{\beta }_{n+1}}({\alpha }_{n+1}\parallel f(x_{n+1})\parallel +{\alpha }_n\parallel f(x_n)\parallel +({\gamma }_{n+1}+{\delta }_{n+1})\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 )\\ +\frac{|{\beta }_{n+1}-{\beta }_n|}{(1-{\beta }_n)(1-{\beta }_{n+1})}\parallel x_{n+1}-{\beta }_n{x}_n\parallel \\ =& \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel f(x_{n+1})\parallel +\frac{{\alpha }_n}{1-{\beta }_{n+1}}\parallel f(x_n)\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|}{(1-{\beta }_n)(1-{\beta }_{n+1})}\parallel x_{n+1}-{\beta }_n{x}_n\parallel \\ \le & \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel f(x_{n+1})\parallel +\frac{{\alpha }_n}{1-{\beta }_{n+1}}\parallel f(x_n)\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|}{(1-{\beta }_n)(1-{\beta }_{n+1})}\parallel x_{n+1}-{\beta }_n{x}_n\parallel .\end{array}$$

(3.5)

From (3.5) and the conditions (i)-(iv), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

(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

$$\parallel x_{n+1}-x_n\parallel =(1-{\beta }_n)\parallel z_n-x_n\parallel ,$$

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel B{x}_n-x_n\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel B_A{x}_n-x_n\parallel =0.$$

From the definition of $x_n$, we can rewrite $x_{n+1}$ by

$$\begin{array}{rcl}{x}_{n+1}& =& {\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_{n}B{x}_n+{\delta }_n{B}_A{x}_n\\ =& {\alpha }_{n}f(x_n)+{\beta }_n{x}_n+({\gamma }_n+{\delta }_n)\frac{({\gamma }_{n}B{x}_n+{\delta }_n{B}_A{x}_n)}{{\gamma }_n+{\delta }_n}\\ =& {\alpha }_{n}f(x_n)+{\beta }_n{x}_n+e_n{z}_n^{\mathrm{\prime }},\end{array}$$

(3.8)

where $e_n={\gamma }_n+{\delta }_n$ and $z_n^{\mathrm{\prime }}=\frac{({\gamma }_{n}B{x}_n+{\delta }_n{B}_A{x}_n)}{{\gamma }_n+{\delta }_n}$.

From Lemma 2.5 and (3.8), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& =& {\parallel {\alpha }_n(f(x_n)-z)+{\beta }_n(x_n-z)+e_n(z_n^{\mathrm{\prime }}-z)\parallel }^2\\ \le & {\alpha }_n{\parallel f(x_n)-z\parallel }^2+{\beta }_n{\parallel x_n-z\parallel }^2+e_n{\parallel z_n^{\mathrm{\prime }}-z\parallel }^2-{\beta }_n{e}_n{g}_1\left(\parallel z_n^{\mathrm{\prime }}-x_n\parallel \right)\\ =& {\alpha }_n{\parallel f(x_n)-z\parallel }^2+{\beta }_n{\parallel x_n-z\parallel }^2-{\beta }_n{e}_n{g}_1\left(\parallel z_n^{\mathrm{\prime }}-x_n\parallel \right)\\ +e_n{\parallel \frac{({\gamma }_{n}B{x}_n+{\delta }_n{B}_A{x}_n)}{{\gamma }_n+{\delta }_n}-z\parallel }^2\\ =& {\alpha }_n{\parallel f(x_n)-z\parallel }^2+{\beta }_n{\parallel x_n-z\parallel }^2-{\beta }_n{e}_n{g}_1\left(\parallel z_n^{\mathrm{\prime }}-x_n\parallel \right)\\ +e_n{\parallel (1-\frac{{\delta }_n}{{\gamma }_n+{\delta }_n})(B{x}_n-z)+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}(B_A{x}_n-z)\parallel }^2\\ \le & {\alpha }_n{\parallel f(x_n)-z\parallel }^2+{\beta }_n{\parallel x_n-z\parallel }^2-{\beta }_n{e}_n{g}_1\left(\parallel z_n^{\mathrm{\prime }}-x_n\parallel \right)\\ +e_n{((1-\frac{{\delta }_n}{{\gamma }_n+{\delta }_n})\parallel B{x}_n-z\parallel +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel B_A{x}_n-z\parallel )}^2\\ \le & {\alpha }_n{\parallel f(x_n)-z\parallel }^2+{\beta }_n{\parallel x_n-z\parallel }^2-{\beta }_n{e}_n{g}_1\left(\parallel z_n^{\mathrm{\prime }}-x_n\parallel \right)+e_n{\parallel x_n-z\parallel }^2\\ \le & {\alpha }_n{\parallel f(x_n)-z\parallel }^2+{\parallel x_n-z\parallel }^2-{\beta }_n{e}_n{g}_1\left(\parallel z_n^{\mathrm{\prime }}-x_n\parallel \right),\end{array}$$

which implies that

$$\begin{array}{rcl}{\beta }_n{e}_n{g}_1\left(\parallel z_n^{\mathrm{\prime }}-x_n\parallel \right)& \le & {\alpha }_n{\parallel f(x_n)-z\parallel }^2+{\parallel x_n-z\parallel }^2-{\parallel x_{n+1}-z\parallel }^2\\ \le & {\alpha }_n{\parallel f(x_n)-z\parallel }^2+(\parallel x_n-z\parallel +\parallel x_{n+1}-z\parallel )\parallel x_{n+1}-x_n\parallel .\end{array}$$

(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

$$\begin{array}{rcl}{\parallel z_n^{\mathrm{\prime }}-z\parallel }^2& =& {\parallel \frac{({\gamma }_{n}B{x}_n+{\delta }_n{B}_A{x}_n)}{{\gamma }_n+{\delta }_n}-z\parallel }^2\\ =& {\parallel (1-\frac{{\delta }_n}{{\delta }_n+{\gamma }_n})(B{x}_n-z)+\frac{{\delta }_n}{{\delta }_n+{\gamma }_n}(B_A{x}_n-z)\parallel }^2\\ \le & (1-\frac{{\delta }_n}{{\delta }_n+{\gamma }_n}){\parallel B{x}_n-z\parallel }^2+\frac{{\delta }_n}{{\delta }_n+{\gamma }_n}{\parallel B_A{x}_n-z\parallel }^2\\ -(1-\frac{{\delta }_n}{{\delta }_n+{\gamma }_n})\frac{{\delta }_n}{{\delta }_n+{\gamma }_n}{g}_2(\parallel B{x}_n-B_A{x}_n\parallel )\\ \le & {\parallel x_n-z\parallel }^2-(1-\frac{{\delta }_n}{{\delta }_n+{\gamma }_n})\frac{{\delta }_n}{{\delta }_n+{\gamma }_n}{g}_2(\parallel B{x}_n-B_A{x}_n\parallel ),\end{array}$$

which implies that

$$\begin{array}{rcl}(1-\frac{{\delta }_n}{{\delta }_n+{\gamma }_n})\frac{{\delta }_n}{{\delta }_n+{\gamma }_n}{g}_2(\parallel B{x}_n-B_A{x}_n\parallel )& \le & {\parallel x_n-z\parallel }^2-{\parallel z_n^{\mathrm{\prime }}-z\parallel }^2\\ \le & (\parallel x_n-z\parallel +\parallel z_n^{\mathrm{\prime }}-z\parallel )\parallel z_n^{\mathrm{\prime }}-x_n\parallel .\end{array}$$

From the condition (iii) and (3.10), we have

$$\underset{n\to \mathrm{\infty }}{lim}{g}_2(\parallel B{x}_n-B_A{x}_n\parallel )=0.$$

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

$$\begin{array}{rcl}{x}_{n+1}& =& {\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_{n}B{x}_n+{\delta }_n{B}_A{x}_n\\ =& {\beta }_n{x}_n+{\gamma }_{n}B{x}_n+({\alpha }_n+{\delta }_n)\frac{{\alpha }_{n}f(x_n)+{\delta }_n{B}_A{x}_n}{{\alpha }_n+{\delta }_n}\\ =& {\beta }_n{x}_n+{\gamma }_{n}B{x}_n+d_n{z}_n^{\mathrm{\prime }\mathrm{\prime }},\end{array}$$

(3.12)

where $d_n={\alpha }_n+{\delta }_n$ and $z_n^{\mathrm{\prime }\mathrm{\prime }}=\frac{{\alpha }_{n}f(x_n)+{\delta }_n{B}_A{x}_n}{{\alpha }_n+{\delta }_n}$.

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(x_n-z)+{\gamma }_n(B{x}_n-z)+d_n(z_n^{\mathrm{\prime }\mathrm{\prime }}-z)\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(\parallel x_n-B{x}_n\parallel )\\ =& {\beta }_n{\parallel x_n-z\parallel }^2+{\gamma }_n{\parallel B{x}_n-z\parallel }^2+d_n{\parallel \frac{{\alpha }_{n}f(x_n)+{\delta }_n{B}_A{x}_n}{{\alpha }_n+{\delta }_n}-z\parallel }^2\\ -{\beta }_n{\gamma }_n{g}_3(\parallel x_n-B{x}_n\parallel )\\ =& {\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}(f(x_n)-z)\\ +(1-\frac{{\alpha }_n}{{\alpha }_n+{\delta }_n})(B_A{x}_n-z){\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel x_n-B{x}_n\parallel )\\ \le & {{\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(x_n)-z\parallel }^2\\ +(1-\frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}){\parallel B_A{x}_n-z\parallel }^2)-{\beta }_n{\gamma }_n{g}_3(\parallel x_n-B{x}_n\parallel )\\ =& {\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(x_n)-z\parallel }^2\\ +d_n(1-\frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}){\parallel B_A{x}_n-z\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel x_n-B{x}_n\parallel )\\ \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(x_n)-z\parallel }^2\\ +d_n{\parallel x_n-z\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel x_n-B{x}_n\parallel )\\ \le & {\parallel x_n-z\parallel }^2+d_n\frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}{\parallel f(x_n)-z\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel x_n-B{x}_n\parallel ),\end{array}$$

(3.13)

which implies that

$$\begin{array}{rcl}{\beta }_n{\gamma }_n{g}_3(\parallel x_n-B{x}_n\parallel )& \le & {\parallel x_n-z\parallel }^2-{\parallel x_{n+1}-z\parallel }^2+d_n\frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}{\parallel f(x_n)-z\parallel }^2\\ \le & (\parallel x_n-z\parallel +\parallel x_{n+1}-z\parallel )\parallel x_{n+1}-x_n\parallel \\ +d_n\frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}{\parallel f(x_n)-z\parallel }^2.\end{array}$$

(3.14)

From the conditions (i), (ii), (iv) (3.14) and (3.3), we have

$$\underset{n\to \mathrm{\infty }}{lim}{g}_3(\parallel x_n-B{x}_n\parallel )=0.$$

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 $Lx=(1-ϵ)Bx+ϵB_{A}x$ for all $x\in C$ and $ϵ\in (0,1)$. From Lemma 2.7, 2.12 and (3.2), we have $F(L)=F(B)\cap F(B_A)={\bigcap }_{i=1}^{N}S(C,A_i)\cap F(S)\cap F(T)=\mathcal{F}$.

From (3.15) and (3.16) and

$$\begin{array}{rcl}\parallel x_n-L{x}_n\parallel & =& \parallel (1-ϵ)(x_n-B{x}_n)+ϵ(x_n-B_A{x}_n)\parallel \\ \le & (1-ϵ)\parallel x_n-B{x}_n\parallel +ϵ\parallel x_n-B_A{x}_n\parallel ,\end{array}$$

we have

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

(3.17)

Next, we will show that

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

(3.18)

where ${lim}_{t\to 0}{x}_t=q\in \mathcal{F}$ and $x_t$ begins the fixed point of the contraction

$$x\mapsto tf(x)+(1-t)Lx.$$

Then $x_t$ solves the fixed point equation $x_t=tf(x_t)+(1-t)L{x}_t$.

From the definition of $x_t$, we have

$$\begin{array}{rcl}{\parallel x_t-x_n\parallel }^2& =& {\parallel t(f(x_t)-x_n)+(1-t)(L{x}_t-x_n)\parallel }^2\\ \le & {(1-t)}^2{\parallel L{x}_t-x_n\parallel }^2+2t〈f(x_t)-x_n,j(x_t-x_n)〉\\ \le & {(1-t)}^2{(\parallel L{x}_t-L{x}_n\parallel +\parallel L{x}_n-x_n\parallel )}^2+2t〈f(x_t)-x_n,j(x_t-x_n)〉\\ \le & {(1-t)}^2{(\parallel x_t-x_n\parallel +\parallel L{x}_n-x_n\parallel )}^2+2t〈f(x_t)-x_n,j(x_t-x_n)〉\\ =& {(1-t)}^2({\parallel x_t-x_n\parallel }^2+2\parallel x_t-x_n\parallel \parallel L{x}_n-x_n\parallel +{\parallel L{x}_n-x_n\parallel }^2)\\ +2t〈f(x_t)-x_n,j(x_t-x_n)〉\\ =& {(1-t)}^2({\parallel x_t-x_n\parallel }^2+2\parallel x_t-x_n\parallel \parallel L{x}_n-x_n\parallel +{\parallel L{x}_n-x_n\parallel }^2)\\ +2t〈f(x_t)-x_t,j(x_t-x_n)〉+2t〈x_t-x_n,j(x_t-x_n)〉\\ =& (1-2t+t^2){\parallel x_t-x_n\parallel }^2+{(1-t)}^2(2\parallel x_t-x_n\parallel \parallel L{x}_n-x_n\parallel +{\parallel L{x}_n-x_n\parallel }^2)\\ +2t〈f(x_t)-x_t,j(x_t-x_n)〉+2t{\parallel x_t-x_n\parallel }^2\\ =& (1+t^2){\parallel x_t-x_n\parallel }^2+f_n(t)+2t〈f(x_t)-x_t,j(x_t-x_n)〉,\end{array}$$

(3.19)

where $f_n(t)={(1-t)}^2(2\parallel x_t-x_n\parallel \parallel L{x}_n-x_n\parallel +{\parallel L{x}_n-x_n\parallel }^2)$. From (3.17), we have

$$\underset{n\to \mathrm{\infty }}{lim}{f}_n(t)=0.$$

(3.20)

(3.19) implies that

$$\begin{array}{rcl}〈x_t-f(x_t),j(x_t-x_n)〉& \le & \frac{t}{2}{\parallel x_t-x_n\parallel }^2+\frac{1}{2t}{f}_n(t)\\ \le & \frac{t}{2}D+\frac{1}{2t}{f}_n(t),\end{array}$$

(3.21)

where $D>0$ such that ${\parallel x_t-x_n\parallel }^2\le D$ for all $t\in (0,1)$ and $n\ge 1$. From (3.20) and (3.21), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x_t-f(x_t),j(x_t-x_n)〉\le \frac{t}{2}D.$$

(3.22)

From (3.22) taking $t\to 0$, we have

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

(3.23)

Since

$$\begin{array}{rcl}〈f(q)-q,j(x_n-q)〉& =& 〈f(q)-q,j(x_n-q)〉-〈f(q)-q,j(x_n-x_t)〉+〈f(q)-q,j(x_n-x_t)〉\\ -〈f(q)-x_t,j(x_n-x_t)〉+〈f(q)-x_t,j(x_n-x_t)〉\\ -〈f(x_t)-x_t,j(x_n-x_t)〉+〈f(x_t)-x_t,j(x_n-x_t)〉\\ =& 〈f(q)-q,j(x_n-q)-j(x_n-x_t)〉+〈x_t-q,j(x_n-x_t)〉\\ +〈f(q)-f(x_t),j(x_n-x_t)〉+〈f(x_t)-x_t,j(x_n-x_t)〉\\ \le & 〈f(q)-q,j(x_n-q)-j(x_n-x_t)〉+\parallel x_t-q\parallel \parallel x_n-x_t\parallel \\ +a\parallel q-x_t\parallel \parallel x_n-x_t\parallel +〈f(x_t)-x_t,j(x_n-x_t)〉,\end{array}$$

it follows that

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,j(x_n-q)〉& \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,j(x_n-q)-j(x_n-x_t)〉\\ +\parallel x_t-q\parallel \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-x_t\parallel +a\parallel q-x_t\parallel \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-x_t\parallel \\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(x_t)-x_t,j(x_n-x_t)〉.\end{array}$$

(3.24)

Since j is norm-to-norm uniformly continuous on a bounded subset of C and (3.24), then we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,j(x_n-q)〉=\underset{t\to 0}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,j(x_n-q)〉\le 0.$$

Finally, we will show the sequence $\{x_n\}$ converses strongly to $q\in \mathcal{F}$. From the definition of $x_n$, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& =& {\parallel {\alpha }_n(f(x_n)-q)+{\beta }_n(x_n-q)+{\gamma }_n(B{x}_n-q)+{\delta }_n(B_A{x}_n-q)\parallel }^2\\ \le & {\parallel {\beta }_n(x_n-q)+{\gamma }_n(B{x}_n-q)+{\delta }_n(B_A{x}_n-q)\parallel }^2\\ +2{\alpha }_n〈f(x_n)-q,j(x_{n+1}-q)〉\\ \le & {({\beta }_n\parallel x_n-q\parallel +{\gamma }_n\parallel B{x}_n-q\parallel +{\delta }_n\parallel B_A{x}_n-q\parallel )}^2\\ +2{\alpha }_n〈f(x_n)-f(q),j(x_{n+1}-q)〉+2{\alpha }_n〈f(q)-q,j(x_{n+1}-q)〉\\ \le & {(1-{\alpha }_n)}^2{\parallel x_n-q\parallel }^2+2{\alpha }_n〈f(x_n)-f(q),j(x_{n+1}-q)〉\\ +2{\alpha }_n〈f(q)-q,j(x_{n+1}-q)〉\\ \le & {(1-{\alpha }_n)}^2{\parallel x_n-q\parallel }^2+2a{\alpha }_n\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel \\ +2{\alpha }_n〈f(q)-q,j(x_{n+1}-q)〉\\ \le & {(1-{\alpha }_n)}^2{\parallel x_n-q\parallel }^2+a{\alpha }_n{\parallel x_n-q\parallel }^2+a{\alpha }_n{\parallel x_{n+1}-q\parallel }^2\\ +2{\alpha }_n〈f(q)-q,j(x_{n+1}-q)〉\\ =& (1-2{\alpha }_n+{\alpha }_n^2){\parallel x_n-q\parallel }^2+a{\alpha }_n{\parallel x_n-q\parallel }^2+a{\alpha }_n{\parallel x_{n+1}-q\parallel }^2\\ +2{\alpha }_n〈f(q)-q,j(x_{n+1}-q)〉\\ =& (1-2{\alpha }_n+a{\alpha }_n){\parallel x_n-q\parallel }^2+{\alpha }_n^2{\parallel x_n-q\parallel }^2+a{\alpha }_n{\parallel x_{n+1}-q\parallel }^2\\ +2{\alpha }_n〈f(q)-q,j(x_{n+1}-q)〉\\ =& (1-a{\alpha }_n-2{\alpha }_n+2a{\alpha }_n){\parallel x_n-q\parallel }^2+{\alpha }_n^2{\parallel x_n-q\parallel }^2+a{\alpha }_n{\parallel x_{n+1}-q\parallel }^2\\ +2{\alpha }_n〈f(q)-q,j(x_{n+1}-q)〉\\ =& (1-a{\alpha }_n-2{\alpha }_n(1-a)){\parallel x_n-q\parallel }^2+{\alpha }_n^2{\parallel x_n-q\parallel }^2+a{\alpha }_n{\parallel x_{n+1}-q\parallel }^2\\ +2{\alpha }_n〈f(q)-q,j(x_{n+1}-q)〉,\end{array}$$

which implies that

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& \le & (1-\frac{2{\alpha }_n(1-a)}{1-a{\alpha }_n}){\parallel x_n-q\parallel }^2\\ +\frac{{\alpha }_n}{1-a{\alpha }_n}({\alpha }_n{\parallel x_n-q\parallel }^2+2〈f(q)-q,j(x_{n+1}-q)〉)\\ \le & (1-\frac{2{\alpha }_n(1-a)}{1-a{\alpha }_n}){\parallel x_n-q\parallel }^2\\ +\frac{2{\alpha }_n(1-a)}{1-a{\alpha }_n}\cdot \frac{1}{2(1-a)}({\alpha }_n{\parallel x_n-q\parallel }^2+2〈f(q)-q,j(x_{n+1}-q)〉).\end{array}$$

From the condition (i) and Lemma 2.11, we can imply that $\{x_n\}$ 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 $\mathcal{F}=F(S)\cap F(T)\cap S(C,A)\ne \mathrm{\varnothing }$. 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 $\alpha \in (0,\frac{\eta }{K^2})$, where K is the 2-uniformly smooth constant of E. Let $\{x_n\}$ be the sequence generated by $x_1\in C$ and

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

where $f:C\to C$ is a contractive mapping and $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\},\{{\delta }_n\}\subseteq [0,1]$, ${\alpha }_n+{\beta }_n+{\gamma }_n+{\delta }_n=1$, $\lambda \in (0,\frac{\nu }{K^2})$ and satisfy the following conditions:

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

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

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 $Q_C(I-{\lambda }_i{A}_i)x=G_{i}x$ 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 with $\mathcal{F}=F(T)\cap {\bigcap }_{i=1}^{N}S(C,A_i)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be the sequence generated by $x_1\in C$ and

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

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:

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

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

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

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

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:

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

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

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 $F(P)\ne \mathrm{\varnothing }$. Then $F(P)=S(C,I-P)$.

Proof It is easy to see that $F(P)\subseteq S(C,I-P)$. Put $A=I-P$ and $z^{\ast }\in F(P)$. Let $z_0\in S(C,I-P)$, then there exists $j(x-z_0)\in J(x-z_0)$ such that

$$〈(I-P)z_0,j(x-z_0)〉\ge 0,\mathrm{\forall }x\in C.$$

(4.1)

Since P is an η-strictly pseudo-contractive mapping, then there exists $j(z_0-z^{\ast })$ such that

$$\begin{array}{rcl}〈P{z}_0-P{z}^{\ast },j(z_0-z^{\ast })〉& =& 〈(I-A)z_0-(I-A)z^{\ast },j(z_0-z^{\ast })〉\\ =& 〈z_0-z^{\ast }-(A{z}_0-A{z}^{\ast }),j(z_0-z^{\ast })〉\\ =& 〈z_0-z^{\ast },j(z_0-z^{\ast })〉-〈A{z}_0-A{z}^{\ast },j(z_0-z^{\ast })〉\\ =& {\parallel z_0-z^{\ast }\parallel }^2-〈A{z}_0,j(z_0-z^{\ast })〉\\ \le & {\parallel z_0-z\parallel }^2-\eta {\parallel (I-P)z_0\parallel }^2.\end{array}$$

(4.2)

From (4.1), (4.2), we have

$$\eta {\parallel z_0-P{z}_0\parallel }^2\le 〈A{z}_0,j(z_0-z^{\ast })〉=-〈A{z}_0,j(z^{\ast }-z_0)〉\le 0.$$

It implies that $z_0=P{z}_0$, that is, $z_0\in F(P)$. Then we have $S(C,I-P)\subseteq F(P)$. Hence, we have $S(C,I-P)=F(P)$. □

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

$$F(P)=S(C,I-P)=F\left(Q_C(I-\lambda (I-P))\right)$$

(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 $Q_C(I-{\lambda }_i(I-S_i))x=G_{i}x$ for all $x\in C$ and $i=1,2,\dots ,N$, where ${\lambda }_i\in (0,\frac{{\eta }_i}{K^2})$, 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 $\mathcal{F}=F(S)\cap F(T)\cap {\bigcap }_{i=1}^{N}F(S_i)\ne \mathrm{\varnothing }$. 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 the sequence generated by $x_1\in C$ and

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

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:

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

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

Proof Since $S_i$ is an ${\eta }_i$-strictly pseudo-contractive mapping, then we have $(I-S_i)$ 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 $Q_C(I-{\lambda }_i(I-S_i))x=G_{i}x$ for all $x\in C$ and $i=1,2,\dots ,N$, where ${\lambda }_i\in (0,\frac{{\eta }_i}{K^2})$, 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 with $\mathcal{F}=F(T)\cap {\bigcap }_{i=1}^{N}F(S_i)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be the sequence generated by $x_1\in C$ and

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

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:

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

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

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

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

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:

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

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