Strong convergence for solving a general system of variational inequalities and fixed point problems in Banach spaces

Abstract

In this paper, we propose and analyze some iterative algorithms by hybrid viscosity approximation methods for solving a general system of variational inequalities and a common fixed point problem of an infinite family of nonexpansive mappings in a uniformly convex Banach space which has a uniformly Gâteaux differentiable norm, and we prove some strong convergence theorems under appropriate conditions. The results presented in this paper improve, extend, supplement and develop the corresponding results recently obtained in the literature.

MSC:49J30, 47H09, 47J20.

1 Introduction

Let X be a real Banach space whose dual space is denoted by $X^{\ast }$. Let $U=\{x\in X:\parallel x\parallel =1\}$ denote the unit sphere of X. The Banach space X is said to be uniformly convex if for each $ϵ\in (0,2]$ there exists $\delta >0$ such that for all $x,y\in U$,

$$\parallel x-y\parallel \ge ϵ\Rightarrow \parallel x+y\parallel /2\le 1-\delta .$$

It is known that a uniformly convex Banach space is reflexive and strictly convex. The Banach space X is said to be smooth if the limit

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

exists for all $x,y\in U$; in this case, X is also said to have a Gâteaux differentiable norm. X is said to have a uniformly Gâteaux differentiable norm if for each $y\in U$, the limit is attained uniformly for $x\in U$. Moreover, it is said to be uniformly smooth if this limit is attained uniformly for $x,y\in U$. The norm of X is said to be the Fréchet differential if for each $x\in U$, this limit is attained uniformly for $y\in U$. In addition, we define a function $\rho :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, called the modulus of smoothness of X, as follows:

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

It is known that X is uniformly smooth if and only if ${lim}_{\tau \to 0}\rho (\tau )/\tau =0$. Let q be a fixed real number with $1<q\le 2$. Then the Banach space X is said to be q-uniformly smooth if there exists a constant $c>0$ such that $\rho (\tau )\le c{\tau }^q$ for all $\tau >0$. As pointed out in [1], no Banach space is q-uniformly smooth for $q>2$.

Let $X^{\ast }$ be the dual of X. The normalized duality mapping $J:X\to 2^{X^{\ast }}$ is defined by

$$J(x)=\{x^{\ast }\in X^{\ast }:〈x,x^{\ast }〉={\parallel x\parallel }^2={\parallel x^{\ast }\parallel }^2\},\mathrm{\forall }x\in X,$$

where $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing. It is an immediate consequence of the Hahn-Banach theorem that $J(x)$ is nonempty for each $x\in X$. Moreover, it is known that J is single-valued if and only if X is smooth, whereas if X is uniformly smooth, then the mapping J is norm-to-norm uniformly continuous on bounded subsets of X. If X has a uniformly Gateaux differentiable norm, then the duality mapping J is norm-to-weak^∗ uniformly continuous on bounded subsets of X.

Let C be a nonempty closed convex subset of a real Banach space X. A mapping $T:C\to C$ is called nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

The set of fixed points of T is denoted by $Fix(T)$. We use the notation ⇀ to indicate the weak convergence and the one → to indicate the strong convergence.

Definition 1.1 Let $A:C\to X$ be a mapping of C into X. Then A is said to be

• (i) accretive if for each $x,y\in C$ there exists $j(x-y)\in J(x-y)$ such that

  $$〈Ax-Ay,j(x-y)〉\ge 0,$$

where J is the normalized duality mapping;

• (ii) α-strongly accretive if for each $x,y\in C$ there exists $j(x-y)\in J(x-y)$ such that

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

for some $\alpha \in (0,1)$;

• (iii) β-inverse-strongly-accretive if for each $x,y\in C$ there exists $j(x-y)\in J(x-y)$ such that

  $$〈Ax-Ay,j(x-y)〉\ge \beta {\parallel Ax-Ay\parallel }^2,$$

for some $\beta >0$;

• (iv) λ-strictly pseudocontractive [2] (see also [3]) if for each $x,y\in C$ there exists $j(x-y)\in J(x-y)$ such that

  $$〈Ax-Ay,j(x-y)〉\le {\parallel x-y\parallel }^2-\lambda {\parallel x-y-(Ax-Ay)\parallel }^2$$

for some $\lambda \in (0,1)$.

In the literature, much work has been done related to strong convergence for solving variational inequalities and fixed point problems, see [1–48]. It is worth emphasizing that the definition of the inverse strongly accretive mapping is based on that of the inverse strongly monotone mapping, which was studied by so many authors; see, e.g., [15, 30, 31].

Very recently, Cai and Bu [35] considered the following general system of variational inequalities (GSVI) in a real smooth Banach space X, which involves finding $(x^{\ast },y^{\ast })\in C\times C$ such that

$$\{\begin{array}{c}〈{\mu }_1{B}_1{y}^{\ast }+x^{\ast }-y^{\ast },J(x-x^{\ast })〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ 〈{\mu }_2{B}_2{x}^{\ast }+y^{\ast }-x^{\ast },J(x-y^{\ast })〉\ge 0,\mathrm{\forall }x\in C,\hfill \end{array}$$

(1.1)

where C is a nonempty, closed and convex subset of X, $B_1,B_2:C\to X$ are two nonlinear mappings and ${\mu }_1$ and ${\mu }_2$ are two positive constants. Here the set of solutions of GSVI (1.1) is denoted by $GSVI(C,B_1,B_2)$. In particular, if $X=H$, a real Hilbert space, then GSVI (1.1) reduces to the following GSVI of finding $(x^{\ast },y^{\ast })\in C\times C$ such that

$$\{\begin{array}{c}〈{\mu }_1{B}_1{y}^{\ast }+x^{\ast }-y^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ 〈{\mu }_2{B}_2{x}^{\ast }+y^{\ast }-x^{\ast },x-y^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \end{array}$$

(1.2)

where ${\mu }_1$ and ${\mu }_2$ are two positive constants. The set of solutions of problem (1.2) is still denoted by $GSVI(C,B_1,B_2)$. In particular, if $B_1=B_2=A$, then problem (1.2) reduces to the new system of variational inequalities (NSVI), introduced and studied by Verma [5]. Further, if $x^{\ast }=y^{\ast }$ additionally, then the NSVI reduces to the classical variational inequality problem (VIP) of finding $x^{\ast }\in C$ such that

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

(1.3)

The solution set of VIP (1.3) is denoted by $VI(C,A)$. Variational inequality theory has been studied quite extensively and has emerged as an important tool in the study of a wide class of obstacle, unilateral, free, moving, equilibrium problems. It is now well known that the variational inequalities are equivalent to the fixed point problems, the origin of which can be traced back to Lions and Stampacchia [13]. This alternative formulation has been used to suggest and analyze the projection iterative method for solving variational inequalities under the conditions that the involved operator must be strongly monotone and Lipschitz continuous.

Recently, Ceng, Wang and Yao [4] transformed problem (1.2) into a fixed point problem in the following way.

Lemma 1.1 (see [4])

For given$\overline{x},\overline{y}\in C$, $(\overline{x},\overline{y})$is a solution of problem (1.1) if and only if$\overline{x}$is a fixed point of the mapping$G:C\to C$defined by

$$G(x)=P_C[P_C(x-{\mu }_2{B}_{2}x)-{\mu }_1{B}_1{P}_C(x-{\mu }_2{B}_{2}x)],\mathrm{\forall }x\in C,$$

(1.4)

where$\overline{y}=P_C(\overline{x}-{\mu }_2{B}_2\overline{x})$.

In particular, if the mappings$B_i:C\to H$are${\beta }_i$-inverse strongly monotone for$i=1,2$, then the mappingGis nonexpansive provided${\mu }_i\in (0,2{\beta }_i)$for$i=1,2$.

In 1976, Korpelevich [6] proposed an iterative algorithm for solving VIP (1.3) in the Euclidean space ${\mathbf{R}}^n$:

$$\{\begin{array}{c}{y}_n=P_C(x_n-\tau A{x}_n),\hfill \\ x_{n+1}=P_C(x_n-\tau A{y}_n),n\ge 0,\hfill \end{array}$$

with $\tau >0$ a given number, which is known as the extragradient method (see also [7, 8]). The literature on the VIP is vast and Korpelevich’s extragradient method has received great attention given by many authors, who improved it in various ways; see, e.g., [9–27, 35, 40] and references therein, to name but a few.

In particular, whenever X is still a real smooth Banach space, $B_1=B_2=A$ and $x^{\ast }=y^{\ast }$, then GSVI (1.1) reduces to the variational inequality problem (VIP) of finding $x^{\ast }\in C$ such that

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

(1.5)

which was considered by Aoyama, Iiduka and Takahashi [28]. Note that VIP (1.5) is connected with the fixed point problem for a nonlinear mapping (see, e.g., [41]), the problem of finding a zero point of a nonlinear operator (see, e.g., [36]) and so on. It is clear that VIP (1.5) extends VIP (1.3) from Hilbert spaces to Banach spaces.

In order to find a solution of VIP (1.5), Aoyama, Iiduka and Takahashi [28] introduced the following iterative scheme for an accretive operator A:

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n){\Pi }_C(x_n-{\lambda }_{n}A{x}_n),\mathrm{\forall }n\ge 1,$$

where ${\Pi }_C$ is a sunny nonexpansive retraction from X onto C. Then they proved a weak convergence theorem. For related work, see [29] and the references therein.

Recently, Jung [43] introduced and analyzed a composite iterative algorithm by the viscosity approximation method for solving a fixed point problem of a nonexpansive mapping and VIP (1.3) for an inverse-strongly monotone mapping A in a real Hilbert space H.

Theorem 1.1 (see [[43], Theorem 3.1])

LetCbe a nonempty closed convex subset of a real Hilbert spaceH. Let$A:C\to H$be anα-inverse-strongly monotone mapping, let$S:C\to C$be a nonexpansive mapping such that$Fix(S)\cap VI(C,A)\ne \mathrm{\varnothing }$, and let$f:C\to C$be a contraction with coefficient$\rho \in (0,1)$. Let$\{x_n\}$be the sequence generated by

$$\{\begin{array}{c}{x}_0=x\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)S{P}_C(x_n-{\lambda }_{n}A{x}_n),\hfill \\ x_{n+1}=(1-{\beta }_n)y_n+{\beta }_{n}S{P}_C(y_n-{\lambda }_{n}A{y}_n),\mathrm{\forall }n\ge 0,\hfill \end{array}$$

where$\{{\lambda }_n\}\subset [0,2\alpha ]$, $\{{\alpha }_n\}\subset [0,1)$and$\{{\beta }_n\}\subset [0,1]$. If$\{{\alpha }_n\}$, $\{{\lambda }_n\}$and$\{{\beta }_n\}$satisfy the following conditions:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$and${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$,

• (ii) $\{{\beta }_n\}\subset [0,a)$, $\mathrm{\forall }n\ge 0$for some$a\in (0,1)$,

• (iii) $\{{\lambda }_n\}\subset [c,d]$for some$c,d\in (0,2\alpha )$,

• (iv) ${\sum }_{n=0}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$, ${\sum }_{n=0}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|<\mathrm{\infty }$and${\sum }_{n=0}^{\mathrm{\infty }}|{\lambda }_{n+1}-{\lambda }_n|<\mathrm{\infty }$,

then$\{x_n\}$converges strongly to$q\in Fix(S)\cap VI(C,A)$, which solves the VIP

$$〈q-f(q),q-p〉\le 0,\mathrm{\forall }p\in Fix(S)\cap VI(C,A).$$

Beyond doubt, it is an interesting and valuable problem of how to construct some algorithms with strong convergence for solving GSVI (1.1) which contains VIP (1.5) as a special case. Very recently, Cai and Bu [35] constructed an iterative algorithm for solving GSVI (1.1) and a common fixed point problem of an infinite family of nonexpansive mappings in a uniformly convex and 2-uniformly smooth Banach space. They proved strong convergence of the proposed method by virtue of the following inequality in a 2-uniformly smooth Banach space X.

Lemma 1.2 (see [32])

LetXbe a 2-uniformly smooth Banach space. Then

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,J(x)〉+2{\parallel \kappa y\parallel }^2,\mathrm{\forall }x,y\in X,$$

(1.6)

whereκis the 2-uniformly smooth constant ofXandJis the normalized duality mapping fromXinto$X^{\ast }$.

Define the mapping $G:C\to C$ as follows:

$$G(x):={\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)x,\mathrm{\forall }x\in C.$$

(1.7)

The fixed point set of G is denoted by Ω. Then their strong convergence theorem on the proposed method is stated as follows.

Theorem 1.2 (see [[35], Theorem 3.1])

LetCbe a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach spaceX. Let${\Pi }_C$be a sunny nonexpansive retraction fromXontoC. Let the mapping$B_i:C\to X$be${\alpha }_i$-inverse-strongly accretive with$0<{\mu }_i<\frac{{\alpha }_i}{{\kappa }^2}$for$i=1,2$. Letfbe a contraction ofCinto itself with coefficient$\delta \in (0,1)$. Let${\{S_n\}}_{n=1}^{\mathrm{\infty }}$be an infinite family of nonexpansive mappings ofCinto itself such that$F={\bigcap }_{i=1}^{\mathrm{\infty }}Fix(S_i)\cap \Omega \ne \mathrm{\varnothing }$, whereΩis the fixed point set of the mappingGdefined by (1.7). For arbitrarily given$x_1\in C$, let$\{x_n\}$be the sequence generated by

$$\{\begin{array}{c}{x}_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)S_n{y}_n,\hfill \\ y_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)z_n,\hfill \\ z_n={\Pi }_C(u_n-{\mu }_1{B}_1{u}_n),\hfill \\ u_n={\Pi }_C(x_n-{\mu }_2{B}_2{x}_n),\mathrm{\forall }n\ge 1.\hfill \end{array}$$

Suppose that$\{{\alpha }_n\}$and$\{{\beta }_n\}$are two sequences in$(0,1)$satisfying the following conditions:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$and${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$.

Assume that${\sum }_{n=1}^{\mathrm{\infty }}{sup}_{x\in D}\parallel S_{n+1}x-S_{n}x\parallel <\mathrm{\infty }$for any bounded subsetDofC, and letSbe a mapping ofCintoXdefined by$Sx={lim}_{n\to \mathrm{\infty }}{S}_{n}x$for all$x\in C$and suppose that$Fix(S)={\bigcap }_{n=1}^{\mathrm{\infty }}Fix(S_n)$. Then$\{x_n\}$converges strongly to$q\in F$, which solves the following VIP:

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

(1.8)

We remark that in Theorem 1.2, the Banach space X is both uniformly convex and 2-uniformly smooth. According to Lemma 1.2, the 2-uniform smoothness of X guarantees the nonexpansivity of the mapping $I-{\mu }_i{B}_i$ for an ${\alpha }_i$-inverse-strongly accretive mapping $B_i:C\to X$ with $0\le {\mu }_i\le \frac{{\alpha }_i}{{\kappa }^2}$ for $i=1,2$, and hence the composite mapping $G:C\to C$ is nonexpansive where $G={\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)$. In the meantime, for the convenience of implementing the argument techniques in [4], the assumption of real smoothness and uniform convexity on X guarantees that the following inequality holds (see [42]): for any given $r>0$, there exists a strictly increasing, continuous and convex function $g:[0,2r]\to \mathbf{R}$, $g(0)=0$ such that

$$g(\parallel x-y\parallel )\le {\parallel x\parallel }^2-2〈x,J(y)〉+{\parallel y\parallel }^2,\mathrm{\forall }x,y\in B_r,$$

(1.9)

where $B_r=\{x\in X:\parallel x\parallel \le r\}$.

Naturally, we wonder whether the uniform convexity and 2-uniform smoothness of X can be replaced by the weaker geometrical property of X or not. There is no doubt that it is an interesting problem worth investigating.

In this paper, motivated by the above facts, we introduce and study two implicit iterative algorithms and two explicit iterative algorithms by hybrid viscosity approximation methods for finding a common element of the set of solutions of GSVI (1.1) and the set of common fixed points of an infinite family of nonexpansive mappings in a real uniformly convex Banach space which has a uniformly Gâteaux differentiable norm. We prove some strong convergence theorems under appropriate conditions. Our results improve, extend, supplement and develop recent corresponding results in the literature, especially [[35], Theorem 3.1] in the following aspects. First, the assumption of the uniformly convex and 2-uniformly smooth Banach space X in [[35], Theorem 3.1] is weakened to the one of the uniformly convex Banach space X having a uniformly Gâteaux differentiable norm in our results. Second, the proof in [[35], Theorem 3.1] depends on the argument techniques in [4], inequality (1.6) in 2-uniformly smooth Banach spaces and inequality (1.9) in smooth and uniform convex Banach spaces. It is worth emphasizing that the proof in our results depends on no argument techniques in [4] but use the inequality in uniform convex Banach spaces; see Lemma 2.5 in Section 2 of this paper. Third, the four iterative algorithms proposed in this paper are very different from the iterative algorithm in [[35], Theorem 3.1] because two iterative algorithms are implicit ones and the iterative step of computing $x_{n+1}$ in other two explicit iterative algorithms involves the sum of three terms.

2 Preliminaries

We list some lemmas that will be used in the sequel. Lemma 2.1 can be found in [34]. Lemma 2.2 is an immediate consequence of the subdifferential inequality of the function $\frac{1}{2}{\parallel \cdot \parallel }^2$.

Lemma 2.1 Let $\{s_n\}$ be a sequence of nonnegative real numbers satisfying

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

where$\{{\alpha }_n\}$, $\{{\beta }_n\}$and$\{{\gamma }_n\}$satisfy the conditions:

• (i) $\{{\alpha }_n\}\subset [0,1]$and${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n\le 0$;

• (iii) ${\gamma }_n\ge 0$, $\mathrm{\forall }n\ge 0$, and${\sum }_{n=0}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$.

Then${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{s}_n=0$.

Lemma 2.2In a smooth Banach spaceX, the following inequality holds:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,J(x+y)〉,\mathrm{\forall }x,y\in X.$$

Lemma 2.3 (see [39])

Let$\{x_n\}$and$\{z_n\}$be bounded sequences in a Banach spaceX, and let$\{{\alpha }_n\}$be a sequence in$[0,1]$which satisfies the following condition

$$0<\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\alpha }_n\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\alpha }_n<1.$$

Suppose$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)z_n$, $\mathrm{\forall }n\ge 0$and${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. Then${lim}_{n\to \mathrm{\infty }}\parallel z_n-x_n\parallel =0$.

Let D be a subset of C and let Π be a mapping of C into D. Then Π is said to be sunny if

$$\Pi [\Pi (x)+t(x-\Pi (x))]=\Pi (x),$$

whenever $\Pi (x)+t(x-\Pi (x))\in C$ for $x\in C$ and $t\ge 0$. A mapping Π of C into itself is called a retraction if ${\Pi }^2=\Pi$. If a mapping Π of C into itself is a retraction, then $\Pi (z)=z$ for every $z\in R(\Pi )$, where $R(\Pi )$ is the range of Π. A subset D of C is called a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction from C onto D. The following lemma concerns the sunny nonexpansive retraction.

Lemma 2.4 [33]

LetCbe a nonempty closed convex subset of a real smooth Banach space X. LetDbe a nonempty subset ofC. LetΠbe a retraction ofContoD. Then the following are equivalent:

• (i) Πis sunny and nonexpansive;

• (ii) ${\parallel \Pi (x)-\Pi (y)\parallel }^2\le 〈x-y,J(\Pi (x)-\Pi (y))〉$, $\mathrm{\forall }x,y\in C$;

• (iii) $〈x-\Pi (x),J(y-\Pi (x))〉\le 0$, $\mathrm{\forall }x\in C$, $y\in D$.

It is well known that if $X=H$, a Hilbert space, then a sunny nonexpansive retraction ${\Pi }_C$ is coincident with the metric projection from X onto C; that is, ${\Pi }_C=P_C$. If C is a nonempty closed convex subset of a strictly convex and uniformly smooth Banach space X and if $T:C\to C$ is a nonexpansive mapping with the fixed point set $Fix(T)\ne \mathrm{\varnothing }$, then the set $Fix(T)$ is a sunny nonexpansive retract of C.

Lemma 2.5 (see [32])

Given a number$r>0$, a real Banach spaceXis uniformly convex if and only if there exists a continuous strictly increasing function$g:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $g(0)=0$, such that

$${\parallel \lambda x+(1-\lambda )y\parallel }^2\le \lambda {\parallel x\parallel }^2+(1-\lambda ){\parallel y\parallel }^2-\lambda (1-\lambda )g(\parallel x-y\parallel )$$

for all$\lambda \in [0,1]$and$x,y\in X$such that$\parallel x\parallel \le r$and$\parallel y\parallel \le r$.

Lemma 2.6 (see [37])

LetCbe a nonempty closed convex subset of a Banach spaceX. Let$S_0,S_1,\dots$be a sequence of mappings ofCinto itself. Suppose that${\sum }_{n=1}^{\mathrm{\infty }}sup\{\parallel S_{n}x-S_{n-1}x\parallel :x\in C\}<\mathrm{\infty }$. Then, for each$y\in C$, $\{S_{n}y\}$converges strongly to some point ofC. Moreover, letSbe a mapping ofCinto itself defined by$Sy={lim}_{n\to \mathrm{\infty }}{S}_{n}y$for all$y\in C$. Then${lim}_{n\to \mathrm{\infty }}sup\{\parallel Sx-S_{n}x\parallel :x\in C\}=0$.

Let C be a nonempty closed convex subset of a Banach space X, and let $T:C\to C$ be a nonexpansive mapping with $Fix(T)\ne \mathrm{\varnothing }$. As previously, let ${\Xi }_C$ be the set of all contractions on C. For $t\in (0,1)$ and $f\in {\Xi }_C$, let $x_t\in C$ be the unique fixed point of the contraction $x\mapsto tf(x)+(1-t)Tx$ on C; that is,

$$x_t=tf(x_t)+(1-t)T{x}_t.$$

Lemma 2.7 (see [41])

LetXbe a reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm. LetCbe a nonempty closed convex subset ofX, let$T:C\to C$be a nonexpansive mapping with$Fix(T)\ne \mathrm{\varnothing }$, and$f\in {\Xi }_C$. Then the net$\{x_t\}$defined by$x_t=tf(x_t)+(1-t)T{x}_t$converges strongly to a point in$Fix(T)$. If we define a mapping$Q:{\Xi }_C\to Fix(T)$by$Q(f):=s-{lim}_{t\to 0}{x}_t$, $\mathrm{\forall }f\in {\Xi }_C$, then$Q(f)$solves the VIP

$$〈(I-f)Q(f),J(Q(f)-p)〉\le 0,\mathrm{\forall }f\in {\Xi }_C,p\in Fix(T).$$

Lemma 2.8 (see [38])

LetCbe a nonempty closed convex subset of a strictly convex Banach spaceX. Let${\{T_n\}}_{n=0}^{\mathrm{\infty }}$be a sequence of nonexpansive mappings onC. Suppose that${\bigcap }_{n=0}^{\mathrm{\infty }}Fix(T_n)$is nonempty. Let$\{{\lambda }_n\}$be a sequence of positive numbers with${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n=1$. Then a mappingSonCdefined by$Sx={\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n{T}_{n}x$for$x\in C$is well defined, nonexpansive and$Fix(S)={\bigcap }_{n=0}^{\mathrm{\infty }}Fix(T_n)$holds.

Lemma 2.9LetCbe a nonempty closed convex subset of a smooth Banach spaceXand let the mapping$B_i:C\to X$be${\lambda }_i$-strictly pseudocontractive and${\alpha }_i$-strongly accretive with${\alpha }_i+{\lambda }_i\ge 1$for$i=1,2$. Then, for${\mu }_i\in (0,1]$, we have

$$\parallel (I-{\mu }_i{B}_i)x-(I-{\mu }_i{B}_i)y\parallel \le \{\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}}+(1-{\mu }_i)(1+\frac{1}{{\lambda }_i})\}\parallel x-y\parallel ,\mathrm{\forall }x,y\in C,$$

for$i=1,2$. In particular, if$1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$, then$I-{\mu }_i{B}_i$is nonexpansive for$i=1,2$.

Proof Taking into account the ${\lambda }_i$-strictly pseudocontractivity of $B_i$, we derive for every $x,y\in C$

$$\begin{array}{rl}{\lambda }_i{\parallel (I-B_i)x-(I-B_i)y\parallel }^2& \le 〈(I-B_i)x-(I-B_i)y,J(x-y)〉\\ \le \parallel (I-B_i)x-(I-B_i)y\parallel \parallel x-y\parallel ,\end{array}$$

which implies that

$$\parallel (I-B_i)x-(I-B_i)y\parallel \le \frac{1}{{\lambda }_i}\parallel x-y\parallel .$$

Hence,

$$\parallel B_{i}x-B_{i}y\parallel \le \parallel (I-B_i)x-(I-B_i)y\parallel +\parallel x-y\parallel \le (1+\frac{1}{{\lambda }_i})\parallel x-y\parallel .$$

Utilizing the ${\alpha }_i$-strong accretivity and ${\lambda }_i$-strict pseudocontractivity of $B_i$, we get

$${\lambda }_i{\parallel (I-B_i)x-(I-B_i)y\parallel }^2\le {\parallel x-y\parallel }^2-〈B_{i}x-B_{i}y,J(x-y)〉\le (1-{\alpha }_i){\parallel x-y\parallel }^2.$$

So, we have

$$\parallel (I-B_i)x-(I-B_i)y\parallel \le \sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}}\parallel x-y\parallel .$$

Therefore, for ${\mu }_i\in (0,1]$ we have

$$\begin{array}{rl}\parallel (I-{\mu }_i{B}_i)x-(I-{\mu }_i{B}_i)y\parallel & \le \parallel (I-B_i)x-(I-B_i)y\parallel +(1-{\mu }_i)\parallel B_{i}x-B_{i}y\parallel \\ \le \sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}}\parallel x-y\parallel +(1-{\mu }_i)(1+\frac{1}{{\lambda }_i})\parallel x-y\parallel \\ =\{\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}}+(1-{\mu }_i)(1+\frac{1}{{\lambda }_i})\}\parallel x-y\parallel .\end{array}$$

Since $1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$, it follows immediately that

$$\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}}+(1-{\mu }_i)(1+\frac{1}{{\lambda }_i})\le 1.$$

This implies that $I-{\mu }_i{B}_i$ is nonexpansive for $i=1,2$. □

Lemma 2.10LetCbe a nonempty closed convex subset of a smooth Banach spaceX. Let${\Pi }_C$be a sunny nonexpansive retraction fromXontoC, and let the mapping$B_i:C\to X$be${\lambda }_i$-strictly pseudocontractive and${\alpha }_i$-strongly accretive with${\alpha }_i+{\lambda }_i\ge 1$for$i=1,2$. Let$G:C\to C$be a mapping defined by

$$G(x)={\Pi }_C[{\Pi }_C(x-{\mu }_2{B}_{2}x)-{\mu }_1{B}_1{\Pi }_C(x-{\mu }_2{B}_{2}x)],\mathrm{\forall }x\in C.$$

If$1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$, then$G:C\to C$is nonexpansive.

Proof According to Lemma 2.9, we know that $I-{\mu }_i{B}_i$ is nonexpansive for $i=1,2$. Hence, for all $x,y\in C$, we have

$$\begin{array}{rl}\parallel G(x)-G(y)\parallel =& \parallel {\Pi }_C[{\Pi }_C(x-{\mu }_2{B}_{2}x)-{\mu }_1{B}_1{\Pi }_C(x-{\mu }_2{B}_{2}x)]\\ -{\Pi }_C[{\Pi }_C(y-{\mu }_2{B}_{2}y)-{\mu }_1{B}_1{\Pi }_C(y-{\mu }_2{B}_{2}y)]\parallel \\ =& \parallel {\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)x-{\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)y\parallel \\ \le & \parallel (I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)x-(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)y\parallel \\ \le & \parallel {\Pi }_C(I-{\mu }_2{B}_2)x-{\Pi }_C(I-{\mu }_2{B}_2)y\parallel \\ \le & \parallel (I-{\mu }_2{B}_2)x-(I-{\mu }_2{B}_2)y\parallel \\ \le & \parallel x-y\parallel .\end{array}$$

This shows that $G:C\to C$ is nonexpansive. This completes the proof. □

Lemma 2.11LetCbe a nonempty closed convex subset of a smooth Banach spaceX. Let${\Pi }_C$be a sunny nonexpansive retraction fromXontoC, and let the mapping$B_i:C\to X$be${\lambda }_i$-strictly pseudocontractive and${\alpha }_i$-strongly accretive for$i=1,2$. For given$x^{\ast },y^{\ast }\in C$, $(x^{\ast },y^{\ast })$is a solution of GSVI (1.4) if and only if$x^{\ast }={\Pi }_C(y^{\ast }-{\mu }_1{B}_1{y}^{\ast })$, where$y^{\ast }={\Pi }_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })$.

Proof We can rewrite GSVI (1.4) as

$$\{\begin{array}{c}〈x^{\ast }-(y^{\ast }-{\mu }_1{B}_1{y}^{\ast }),J(x-x^{\ast })〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ 〈y^{\ast }-(x^{\ast }-{\mu }_2{B}_2{x}^{\ast }),J(x-y^{\ast })〉\ge 0,\mathrm{\forall }x\in C,\hfill \end{array}$$

which is obviously equivalent to

$$\{\begin{array}{c}{x}^{\ast }={\Pi }_C(y^{\ast }-{\mu }_1{B}_1{y}^{\ast }),\hfill \\ y^{\ast }={\Pi }_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast }),\hfill \end{array}$$

because of Lemma 2.4. This completes the proof. □

Remark 2.1 By Lemma 2.11, we observe that

$$x^{\ast }={\Pi }_C[{\Pi }_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })-{\mu }_1{B}_1{\Pi }_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })],$$

which implies that $x^{\ast }$ is a fixed point of the mapping $G={\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)$.

3 Two-step implicit iterative algorithm

In this section, we introduce our two-step implicit iterative algorithm and show the strong convergence of the purposed algorithm.

Theorem 3.1LetCbe a nonempty closed convex subset of a uniformly convex Banach spaceXwhich has a uniformly Gâteaux differentiable norm. Let${\Pi }_C$be a sunny nonexpansive retraction fromXontoC. Let the mapping$B_i:C\to X$be${\lambda }_i$-strictly pseudocontractive and${\alpha }_i$-strongly accretive with${\alpha }_i+{\lambda }_i\ge 1$for$i=1,2$. Let$f:C\to C$be a contraction with coefficient$\rho \in (0,1)$. Let${\{S_n\}}_{n=1}^{\mathrm{\infty }}$be an infinite family of nonexpansive mappings ofCinto itself such that$F={\bigcap }_{i=1}^{\mathrm{\infty }}Fix(S_i)\cap \Omega \ne \mathrm{\varnothing }$, whereΩis the fixed point set of the mapping$G={\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)$. For arbitrarily given$x_0\in C$, let$\{x_n\}$be the sequence generated by

$$\{\begin{array}{c}{y}_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)G(x_n),\hfill \\ x_{n+1}={\beta }_n{x}_n+{\gamma }_n{y}_n+{\delta }_n{S}_{n+1}{x}_{n+1},\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(3.1)

where$1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$for$i=1,2$. Suppose that$\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$and$\{{\delta }_n\}$are the sequences in$(0,1)$satisfying the following conditions:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$and${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) ${\beta }_n+{\gamma }_n+{\delta }_n=1$, $\mathrm{\forall }n\ge 0$, and${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}({\beta }_n+{\delta }_n)<1$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}\frac{1}{{\alpha }_n}|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|=0$;

• (v) ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_n-{\alpha }_{n-1}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}{\alpha }_{n-1}/{\alpha }_n=1$;

• (vi) ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|/{\alpha }_n=0$.

Assume that${\sum }_{n=1}^{\mathrm{\infty }}{sup}_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel <\mathrm{\infty }$for any bounded subsetDofC, and letSbe a mapping ofCinto itself defined by$Sx={lim}_{n\to \mathrm{\infty }}{S}_{n}x$for all$x\in C$, and suppose that$Fix(S)={\bigcap }_{i=1}^{\mathrm{\infty }}Fix(S_i)$. Then$\{x_n\}$converges strongly to$q\in F$, which solves the following VIP:

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

(3.2)

Proof Take a fixed $p\in F$ arbitrarily. Then by Lemma 2.11 we know that $p=G(p)$. Moreover, by Lemma 2.10 we have

$$\begin{array}{rl}\parallel y_n-p\parallel & =\parallel {\alpha }_n(f(x_n)-p)+(1-{\alpha }_n)(G(x_n)-p)\parallel \\ \le {\alpha }_n\parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel f(p)-p\parallel +(1-{\alpha }_n)\parallel G(x_n)-p\parallel \\ \le {\alpha }_n\rho \parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel \\ =(1-{\alpha }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel .\end{array}$$

(3.3)

From (3.3) we obtain

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & =\parallel {\beta }_n(x_n-p)+{\gamma }_n(y_n-p)+{\delta }_n(S_{n+1}{x}_{n+1}-p)\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel y_n-p\parallel +{\delta }_n\parallel x_{n+1}-p\parallel ,\end{array}$$

which together with (3.3) implies that

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & \le \frac{1}{{\beta }_n+{\gamma }_n}[{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel y_n-p\parallel ]\\ \le \frac{1}{{\beta }_n+{\gamma }_n}\{{\beta }_n\parallel x_n-p\parallel +{\gamma }_n[(1-{\alpha }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel ]\}\\ =\frac{1}{{\beta }_n+{\gamma }_n}\{({\beta }_n+{\gamma }_n-{\alpha }_n{\gamma }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n{\gamma }_n\parallel f(p)-p\parallel \}\\ =(1-\frac{{\alpha }_n{\gamma }_n(1-\rho )}{{\beta }_n+{\gamma }_n})\parallel x_n-p\parallel +\frac{{\alpha }_n{\gamma }_n}{{\beta }_n+{\gamma }_n}\parallel f(p)-p\parallel \\ =(1-\frac{{\alpha }_n{\gamma }_n(1-\rho )}{{\beta }_n+{\gamma }_n})\parallel x_n-p\parallel +\frac{{\alpha }_n{\gamma }_n(1-\rho )}{{\beta }_n+{\gamma }_n}\cdot \frac{\parallel f(p)-p\parallel }{1-\rho }\\ \le max\{\parallel x_0-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-\rho }\},\end{array}$$

which implies that $\{x_n\}$ is bounded. By Lemma 2.10 we know from (3.3) that $\{G(x_n)\}$ and $\{y_n\}$ both are bounded.

Let us show that $\parallel x_{n+1}-x_n\parallel \to 0$ and $\parallel x_n-y_n\parallel \to 0$ as $n\to \mathrm{\infty }$. As a matter of fact, from (3.1) we have

$$y_n-y_{n-1}={\alpha }_n(f(x_n)-f(x_{n-1}))+({\alpha }_n-{\alpha }_{n-1})(f(x_{n-1})-G(x_{n-1}))+(1-{\alpha }_n)(G(x_n)-G(x_{n-1})).$$

It follows that

$$\begin{array}{rl}\parallel y_n-y_{n-1}\parallel \le & {\alpha }_n\parallel f(x_n)-f(x_{n-1})\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-G(x_{n-1})\parallel \\ +(1-{\alpha }_n)\parallel G(x_n)-G(x_{n-1})\parallel \\ \le {\alpha }_n\rho \parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-G(x_{n-1})\parallel \\ +(1-{\alpha }_n)\parallel x_n-x_{n-1}\parallel \\ =& (1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-G(x_{n-1})\parallel .\end{array}$$

(3.4)

Taking into consideration that $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}({\beta }_n+{\delta }_n)<1$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n=1-{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}({\beta }_n+{\delta }_n)>0$, we may assume that $\{{\beta }_n\}\subset [c,d]$ and ${\gamma }_n\subset [c,1)$ for some $c,d\in (0,1)$. First, we write $x_n={\beta }_{n-1}{x}_{n-1}+(1-{\beta }_{n-1})v_{n-1}$, $\mathrm{\forall }n\ge 1$, where $v_{n-1}=\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}$. It follows that for all $n\ge 1$,

$$\begin{array}{rl}{v}_n-v_{n-1}=& \frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}\\ =& \frac{{\gamma }_n{y}_n+{\delta }_n{S}_{n+1}{x}_{n+1}}{1-{\beta }_n}-\frac{{\gamma }_{n-1}{y}_{n-1}+{\delta }_{n-1}{S}_n{x}_n}{1-{\beta }_{n-1}}\\ =& \frac{{\gamma }_n(y_n-y_{n-1})+{\delta }_n(S_{n+1}{x}_{n+1}-S_n{x}_n)}{1-{\beta }_n}\\ +(\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}})y_{n-1}+(\frac{{\delta }_n}{1-{\beta }_n}-\frac{{\delta }_{n-1}}{1-{\beta }_{n-1}})S_n{x}_n.\end{array}$$

(3.5)

This together with (3.4) implies that

$$\begin{array}{r}\parallel v_n-v_{n-1}\parallel \\ \le \frac{\parallel {\gamma }_n(y_n-y_{n-1})+{\delta }_n(S_{n+1}{x}_{n+1}-S_n{x}_n)\parallel }{1-{\beta }_n}+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\parallel y_{n-1}\parallel \\ +|\frac{{\delta }_n}{1-{\beta }_n}-\frac{{\delta }_{n-1}}{1-{\beta }_{n-1}}|\parallel S_n{x}_n\parallel \\ \le \frac{{\gamma }_n\parallel y_n-y_{n-1}\parallel +{\delta }_n(\parallel S_{n+1}{x}_{n+1}-S_{n+1}{x}_n\parallel +\parallel S_{n+1}{x}_n-S_n{x}_n\parallel )}{1-{\beta }_n}\\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\parallel y_{n-1}\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\parallel S_n{x}_n\parallel \\ \le \frac{{\gamma }_n\parallel y_n-y_{n-1}\parallel +{\delta }_n\parallel x_{n+1}-x_n\parallel }{1-{\beta }_n}+\frac{{\delta }_n}{1-{\beta }_n}\parallel S_{n+1}{x}_n-S_n{x}_n\parallel \\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|(\parallel y_{n-1}\parallel +\parallel S_n{x}_n\parallel )\\ \le \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}[(1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-G(x_{n-1})\parallel ]\\ +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel x_{n+1}-x_n\parallel +\frac{{\delta }_n}{1-{\beta }_n}\parallel S_{n+1}{x}_n-S_n{x}_n\parallel \\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|(\parallel y_{n-1}\parallel +\parallel S_n{x}_n\parallel )\\ \le \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel +\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}|{\alpha }_n-{\alpha }_{n-1}|M\\ +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel x_{n+1}-x_n\parallel +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel S_{n+1}{x}_n-S_n{x}_n\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|M,\end{array}$$

(3.6)

where ${sup}_{n\ge 0}\{\parallel f(x_n)-G(x_n)\parallel \}\le M$ and ${sup}_{n\ge 1}\{\parallel y_{n-1}\parallel +\parallel S_n{x}_n\parallel \}\le M$ for some $M>0$.

Now we observe that

$$\begin{array}{c}{x}_{n+1}-x_n=({\beta }_n-{\beta }_{n-1})(x_n-v_n)+{\beta }_{n-1}(x_n-x_{n-1})+(1-{\beta }_{n-1})(v_n-v_{n-1}),\hfill \\ 1-(1-{\beta }_{n-1})\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}=\frac{{\gamma }_n+{\beta }_{n-1}{\delta }_n}{{\gamma }_n+{\delta }_n},\hfill \end{array}$$

and

$$\begin{array}{r}{\beta }_{n-1}+(1-{\beta }_{n-1})\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(1-(1-\rho ){\alpha }_n)\\ ={\beta }_{n-1}(\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n})+(1-{\beta }_{n-1})\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(1-(1-\rho ){\alpha }_n)\\ =\frac{{\beta }_{n-1}{\delta }_n}{{\gamma }_n+{\delta }_n}+\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}[{\beta }_{n-1}+(1-{\beta }_{n-1})(1-(1-\rho ){\alpha }_n)]\\ =\frac{{\beta }_{n-1}{\delta }_n}{{\gamma }_n+{\delta }_n}+\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}[1-(1-{\beta }_{n-1})(1-\rho ){\alpha }_n]\\ =\frac{{\gamma }_n+{\beta }_{n-1}{\delta }_n}{{\gamma }_n+{\delta }_n}-(1-{\beta }_{n-1})(1-\rho )\frac{{\alpha }_n{\gamma }_n}{{\gamma }_n+{\delta }_n}.\end{array}$$

Hence from (3.6) it follows that

$$\begin{array}{r}\parallel x_{n+1}-x_n\parallel \\ \le |{\beta }_n-{\beta }_{n-1}|\parallel x_n-v_n\parallel +{\beta }_{n-1}\parallel x_n-x_{n-1}\parallel +(1-{\beta }_{n-1})\parallel v_n-v_{n-1}\parallel \\ \le |{\beta }_n-{\beta }_{n-1}|\parallel x_n-v_n\parallel +{\beta }_{n-1}\parallel x_n-x_{n-1}\parallel \\ +(1-{\beta }_{n-1})\{\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel +\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}|{\alpha }_n-{\alpha }_{n-1}|M\\ +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel x_{n+1}-x_n\parallel +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel S_{n+1}{x}_n-S_n{x}_n\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|M\}\\ \le |{\beta }_n-{\beta }_{n-1}|M_1+[{\beta }_{n-1}+(1-{\beta }_{n-1})\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(1-(1-\rho ){\alpha }_n)]\parallel x_n-x_{n-1}\parallel \\ +|{\alpha }_n-{\alpha }_{n-1}|M_1+(1-{\beta }_{n-1})\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel x_{n+1}-x_n\parallel +\parallel S_{n+1}{x}_n-S_n{x}_n\parallel \\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|M_1\\ =[\frac{{\gamma }_n+{\beta }_{n-1}{\delta }_n}{{\gamma }_n+{\delta }_n}-(1-{\beta }_{n-1})(1-\rho )\frac{{\alpha }_n{\gamma }_n}{{\gamma }_n+{\delta }_n}]\parallel x_n-x_{n-1}\parallel \\ +(1-{\beta }_{n-1})\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel x_{n+1}-x_n\parallel \\ +M_1(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|)+\parallel S_{n+1}{x}_n-S_n{x}_n\parallel ,\end{array}$$

where ${sup}_{n\ge 0}\{M+\parallel x_n-v_n\parallel \}\le M_1$ for some $M_1>0$. Therefore, we get

$$\begin{array}{r}\parallel x_{n+1}-x_n\parallel \\ \le [1-(1-{\beta }_{n-1})(1-\rho )\frac{{\alpha }_n{\gamma }_n}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}]\parallel x_n-x_{n-1}\parallel \\ +\frac{{\gamma }_n+{\delta }_n}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}[M_1(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|)\\ +\parallel S_{n+1}{x}_n-S_n{x}_n\parallel ]\\ \le [1-c(1-d)(1-\rho )\frac{{\alpha }_n}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}]\parallel x_n-x_{n-1}\parallel \\ +\frac{1}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}[M_1(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|)\\ +\parallel S_{n+1}{x}_n-S_n{x}_n\parallel ]\\ =[1-c(1-d)(1-\rho )\frac{{\alpha }_n}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}]\parallel x_n-x_{n-1}\parallel \\ +c(1-d)(1-\rho )\frac{1}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}\cdot \frac{1}{c(1-d)(1-\rho )}\left[M_1\right(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|\\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|)+\parallel S_{n+1}{x}_n-S_n{x}_n\parallel ].\end{array}$$

Utilizing Lemma 2.1, from conditions (i), (iv)-(vi) and the assumption on $\{S_n\}$, we deduce that

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

(3.7)

Also, we note that for $p\in F$,

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2& ={\parallel {\alpha }_n(f(x_n)-p)+(1-{\alpha }_n)(G(x_n)-G(p))\parallel }^2\\ \le {\parallel (1-{\alpha }_n)(G(x_n)-G(p))\parallel }^2+2{\alpha }_n〈f(x_n)-p,J(y_n-p)〉\\ \le {\parallel x_n-p\parallel }^2+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel .\end{array}$$

(3.8)

Since $\{x_n\}$ and $\{y_n\}$ both are bounded, by Lemma 2.5 there exists a continuous strictly increasing function $g:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $g(0)=0$ such that

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ ={\parallel ({\beta }_n+{\delta }_n)(\frac{{\beta }_n}{{\beta }_n+{\delta }_n}(x_n-p)+\frac{{\delta }_n}{{\beta }_n+{\delta }_n}(S_{n+1}{x}_{n+1}-p))+{\gamma }_n(y_n-p)\parallel }^2\\ \le ({\beta }_n+{\delta }_n){\parallel \frac{{\beta }_n}{{\beta }_n+{\delta }_n}(x_n-p)+\frac{{\delta }_n}{{\beta }_n+{\delta }_n}(S_{n+1}{x}_{n+1}-p)\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2\\ \le ({\beta }_n+{\delta }_n)[\frac{{\beta }_n}{{\beta }_n+{\delta }_n}{\parallel x_n-p\parallel }^2+\frac{{\delta }_n}{{\beta }_n+{\delta }_n}\parallel S_{n+1}{x}_{n+1}-p\parallel \\ -\frac{{\beta }_n{\delta }_n}{{({\beta }_n+{\delta }_n)}^2}g(\parallel x_n-S_{n+1}{x}_{n+1}\parallel )]+{\gamma }_n{\parallel y_n-p\parallel }^2\\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\delta }_n{\parallel x_{n+1}-p\parallel }^2-\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}g(\parallel x_n-S_{n+1}{x}_{n+1}\parallel )+{\gamma }_n{\parallel y_n-p\parallel }^2\\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\delta }_n{\parallel x_{n+1}-p\parallel }^2-\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}g(\parallel x_n-S_{n+1}{x}_{n+1}\parallel )\\ +{\gamma }_n[{\parallel x_n-p\parallel }^2+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel ]\\ =({\beta }_n+{\gamma }_n){\parallel x_n-p\parallel }^2+{\delta }_n{\parallel x_{n+1}-p\parallel }^2-\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}g(\parallel x_n-S_{n+1}{x}_{n+1}\parallel )\\ +2{\alpha }_n{\gamma }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel ,\end{array}$$

which immediately yields

$$\begin{array}{r}{\beta }_n{\delta }_{n}g(\parallel x_n-S_{n+1}{x}_{n+1}\parallel )\\ \le \frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}g(\parallel x_n-S_{n+1}{x}_{n+1}\parallel )\\ \le ({\beta }_n+{\gamma }_n)({\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2)+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel \\ \le (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel .\end{array}$$

Since ${\alpha }_n\to 0$, $\parallel x_{n+1}-x_n\parallel \to 0$, ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$, we get ${lim}_{n\to \mathrm{\infty }}g(\parallel x_n-S_{n+1}{x}_{n+1}\parallel )=0$ and hence

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

(3.9)

In the meantime, according to condition (iii), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\gamma }_n=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(1-{\beta }_n-{\delta }_n)=1-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}({\beta }_n+{\delta }_n)>0.$$

Thus, from (3.7) and (3.9) it follows that

$$\begin{array}{rl}{\gamma }_n\parallel y_n-x_n\parallel & =\parallel (x_{n+1}-x_n)-{\delta }_n(S_{n+1}{x}_{n+1}-x_n)\parallel \\ \le \parallel x_{n+1}-x_n\parallel +{\delta }_n\parallel S_{n+1}{x}_{n+1}-x_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

That is,

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

(3.10)

This together with (3.1) leads to

$$\begin{array}{rl}(1-{\alpha }_n)\parallel G(x_n)-x_n\parallel & =\parallel (y_n-x_n)-{\alpha }_n(f(x_n)-x_n)\parallel \\ \le \parallel y_n-x_n\parallel +{\alpha }_n\parallel f(x_n)-x_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

That is,

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

(3.11)

On the other hand, we observe that

$$y_n-G(x_n)={\alpha }_n(f(x_n)-G(x_n)),$$

which together with ${\alpha }_n\to 0$ implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-G(x_n)\parallel =0.$$

(3.12)

We note that

$$\begin{array}{r}\parallel S_{n}G(x_n)-G(x_n)\parallel \\ \le \parallel S_{n}G(x_n)-S_{n+1}{x}_{n+1}\parallel +\parallel S_{n+1}{x}_{n+1}-x_n\parallel +\parallel x_n-G(x_n)\parallel \\ \le \parallel S_{n}G(x_n)-S_n{x}_{n+1}\parallel +\parallel S_n{x}_{n+1}-S_{n+1}{x}_{n+1}\parallel +\parallel S_{n+1}{x}_{n+1}-x_n\parallel +\parallel x_n-G(x_n)\parallel \\ \le \parallel G(x_n)-x_{n+1}\parallel +\parallel S_n{x}_{n+1}-S_{n+1}{x}_{n+1}\parallel +\parallel S_{n+1}{x}_{n+1}-x_n\parallel +\parallel x_n-G(x_n)\parallel \\ \le \parallel G(x_n)-x_n\parallel +\parallel x_n-x_{n+1}\parallel +\parallel S_n{x}_{n+1}-S_{n+1}{x}_{n+1}\parallel +\parallel S_{n+1}{x}_{n+1}-x_n\parallel \\ +\parallel x_n-G(x_n)\parallel \\ \le 2\parallel G(x_n)-x_n\parallel +\parallel x_n-x_{n+1}\parallel +\parallel S_n{x}_{n+1}-S_{n+1}{x}_{n+1}\parallel +\parallel S_{n+1}{x}_{n+1}-x_n\parallel .\end{array}$$

From (3.7), (3.9), (3.11) and the assumption on $\{S_n\}$, we obtain that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel S_{n}G(x_n)-G(x_n)\parallel =0.$$

(3.13)

By (3.13) and Lemma 2.6, we have

$$\begin{array}{rl}\parallel SG(x_n)-G(x_n)\parallel & \le \parallel SG(x_n)-S_{n}G(x_n)\parallel +\parallel S_{n}G(x_n)-G(x_n)\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.14)

In terms of (3.11) and (3.14), we have

$$\begin{array}{rl}\parallel x_n-S{x}_n\parallel & \le \parallel x_n-G(x_n)\parallel +\parallel G(x_n)-SG(x_n)\parallel +\parallel SG(x_n)-S{x}_n\parallel \\ \le 2\parallel x_n-G(x_n)\parallel +\parallel G(x_n)-SG(x_n)\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.15)

Define a mapping $Wx=(1-\theta )Sx+\theta G(x)$, where G is defined by (1.7), $\theta \in (0,1)$ is a constant. Then by Lemma 2.8 we have that $Fix(W)=Fix(S)\cap Fix(G)=F$. We observe that

$$\begin{array}{rl}\parallel x_n-W{x}_n\parallel & =\parallel (1-\theta )(x_n-S{x}_n)+\theta (x_n-G(x_n))\parallel \\ \le (1-\theta )\parallel x_n-S{x}_n\parallel +\theta \parallel x_n-G(x_n)\parallel .\end{array}$$

From (3.11) and (3.15), we obtain

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

(3.16)

Now, we claim that

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

(3.17)

where $q=s-{lim}_{t\to 0}{x}_t$ with $x_t$ being the fixed point of the contraction

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

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

$$\parallel x_t-x_n\parallel =\parallel (1-t)(W{x}_t-x_n)+t(f(x_t)-x_n)\parallel .$$

By Lemma 2.2 we conclude that

$$\begin{array}{r}{\parallel x_t-x_n\parallel }^2\\ ={\parallel (1-t)(W{x}_t-x_n)+t(f(x_t)-x_n)\parallel }^2\\ \le {(1-t)}^2{\parallel W{x}_t-x_n\parallel }^2+2t〈f(x_t)-x_n,J(x_t-x_n)〉\\ \le {(1-t)}^2{(\parallel W{x}_t-W{x}_n\parallel +\parallel W{x}_n-x_n\parallel )}^2+2t〈f(x_t)-x_n,J(x_t-x_n)〉\\ \le {(1-t)}^2{(\parallel x_t-x_n\parallel +\parallel W{x}_n-x_n\parallel )}^2+2t〈f(x_t)-x_n,J(x_t-x_n)〉\\ ={(1-t)}^2[{\parallel x_t-x_n\parallel }^2+2\parallel x_t-x_n\parallel \parallel W{x}_n-x_n\parallel +{\parallel W{x}_n-x_n\parallel }^2]\\ +2t〈f(x_t)-x_t,J(x_t-x_n)〉+2t〈x_t-x_n,J(x_t-x_n)〉\\ =(1-2t+t^2){\parallel x_t-x_n\parallel }^2+f_n(t)+2t〈f(x_t)-x_t,J(x_t-x_n)〉+2t{\parallel x_t-x_n\parallel }^2,\end{array}$$

(3.18)

where

$$f_n(t)={(1-t)}^2(2\parallel x_t-x_n\parallel +\parallel x_n-W{x}_n\parallel )\parallel x_n-W{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

(3.19)

It follows from (3.18) that

$$〈x_t-f(x_t),J(x_t-x_n)〉\le \frac{t}{2}{\parallel x_t-x_n\parallel }^2+\frac{1}{2t}{f}_n(t).$$

(3.20)

Letting $n\to \mathrm{\infty }$ in (3.20) and noticing (3.19), we derive

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x_t-f(x_t),J(x_t-x_n)〉\le \frac{t}{2}{M}_2,$$

(3.21)

where $M_2>0$ is a constant such that ${\parallel x_t-x_n\parallel }^2\le M_2$ for all $t\in (0,1)$ and $n\ge 0$. Taking $t\to 0$ in (3.21), we have

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

(3.22)

On the other hand, we have

$$\begin{array}{r}〈f(q)-q,J(x_n-q)〉\\ =〈f(q)-q,J(x_n-q)〉-〈f(q)-q,J(x_n-x_t)〉+〈f(q)-q,J(x_n-x_t)〉\\ -〈f(q)-x_t,J(x_n-x_t)〉+〈f(q)-x_t,J(x_n-x_t)〉-〈f(x_t)-x_t,J(x_n-x_t)〉\\ +〈f(x_t)-x_t,J(x_n-x_t)〉\\ =〈f(q)-q,J(x_n-q)-J(x_n-x_t)〉+〈x_t-q,J(x_n-x_t)〉\\ +〈f(q)-f(x_t),J(x_n-x_t)〉+〈f(x_t)-x_t,J(x_n-x_t)〉.\end{array}$$

It follows that

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

Taking into account that $x_t\to q$ as $t\to 0$, we have from (3.22)

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,J(x_n-q)〉& =\underset{t\to 0}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,J(x_n-q)〉\\ \le \underset{t\to 0}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,J(x_n-q)-J(x_n-x_t)〉.\end{array}$$

Since X has a uniformly Gâteaux differentiable norm, the duality mapping J is norm-to-weak^∗ uniformly continuous on bounded subsets of X. Consequently, the two limits are interchangeable and hence (3.17) holds. From (3.10) we get $(y_n-q)-(x_n-q)\to 0$. Noticing that J is norm-to-weak^∗ uniformly continuous on bounded subsets of X, we deduce from (3.17) that

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,J(y_n-q)〉& =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,J(x_n-q)〉\\ \le 0.\end{array}$$

(3.23)

Finally, let us show that $x_n\to q$ as $n\to \mathrm{\infty }$. We observe that

$$\begin{array}{r}{\parallel y_n-q\parallel }^2\\ ={\alpha }_n〈f(x_n)-q,J(y_n-q)〉+(1-{\alpha }_n)〈G(x_n)-q,J(y_n-q)〉\\ ={\alpha }_n〈f(x_n)-f(q),J(y_n-q)〉+{\alpha }_n〈f(q)-q,J(y_n-q)〉+(1-{\alpha }_n)〈G(x_n)-q,J(y_n-q)〉\\ \le {\alpha }_n\rho \parallel x_n-q\parallel \parallel y_n-q\parallel +(1-{\alpha }_n)\parallel G(x_n)-q\parallel \parallel y_n-q\parallel +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ \le {\alpha }_n\rho \parallel x_n-q\parallel \parallel y_n-q\parallel +(1-{\alpha }_n)\parallel x_n-q\parallel \parallel y_n-q\parallel +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ =[1-{\alpha }_n(1-\rho )]\parallel x_n-q\parallel \parallel y_n-q\parallel +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ \le \frac{1-{\alpha }_n(1-\rho )}{2}({\parallel x_n-q\parallel }^2+{\parallel y_n-q\parallel }^2)+{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ \le \frac{1-{\alpha }_n(1-\rho )}{2}{\parallel x_n-q\parallel }^2+\frac{1}{2}{\parallel y_n-q\parallel }^2+{\alpha }_n〈f(q)-q,J(y_n-q)〉,\end{array}$$

which implies that

$${\parallel y_n-q\parallel }^2\le [1-{\alpha }_n(1-\rho )]{\parallel x_n-q\parallel }^2+{\alpha }_n(1-\rho )\frac{2〈f(q)-q,J(y_n-q)〉}{1-\rho }.$$

(3.24)

Also, by the convexity of ${\parallel \cdot \parallel }^2$ and (3.1), we get

$${\parallel x_{n+1}-q\parallel }^2\le {\beta }_n{\parallel x_n-q\parallel }^2+{\gamma }_n{\parallel y_n-q\parallel }^2+{\delta }_n{\parallel x_{n+1}-q\parallel }^2,$$

which together with (3.24) leads to

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2\le & \frac{{\beta }_n}{{\beta }_n+{\gamma }_n}{\parallel x_n-q\parallel }^2+\frac{{\gamma }_n}{{\beta }_n+{\gamma }_n}{\parallel y_n-q\parallel }^2\\ \le & \frac{{\beta }_n}{{\beta }_n+{\gamma }_n}{\parallel x_n-q\parallel }^2+\frac{{\gamma }_n}{{\beta }_n+{\gamma }_n}[(1-{\alpha }_n(1-\rho )){\parallel x_n-q\parallel }^2\\ +{\alpha }_n(1-\rho )\frac{2〈f(q)-q,J(y_n-q)〉}{1-\rho }]\\ =& [1-\frac{{\alpha }_n{\gamma }_n(1-\rho )}{{\beta }_n+{\gamma }_n}]{\parallel x_n-q\parallel }^2\\ +\frac{{\alpha }_n{\gamma }_n(1-\rho )}{{\beta }_n+{\gamma }_n}\cdot \frac{2〈f(q)-q,J(y_n-q)〉}{1-\rho }.\end{array}$$

(3.25)

Applying Lemma 2.1 to (3.25), we obtain that $x_n\to q$ as $n\to \mathrm{\infty }$. This completes the proof. □

Corollary 3.1LetCbe a nonempty closed convex subset of a uniformly convex Banach spaceXwhich has a uniformly Gâteaux differentiable norm. Let${\Pi }_C$be a sunny nonexpansive retraction fromXontoC. Let the mapping$B_i:C\to X$be${\lambda }_i$-strictly pseudocontractive and${\alpha }_i$-strongly accretive with${\alpha }_i+{\lambda }_i\ge 1$for$i=1,2$. Let$f:C\to C$be a contraction with coefficient$\rho \in (0,1)$. LetSbe a nonexpansive mapping ofCinto itself such that$F=Fix(S)\cap \Omega \ne \mathrm{\varnothing }$, whereΩis the fixed point set of the mapping$G={\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)$. For arbitrarily given$x_0\in C$, let$\{x_n\}$be the sequence generated by

$$\{\begin{array}{c}{y}_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)G(x_n),\hfill \\ x_{n+1}={\beta }_n{x}_n+{\gamma }_n{y}_n+{\delta }_{n}S{x}_{n+1},\mathrm{\forall }n\ge 0,\hfill \end{array}$$

where$1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$for$i=1,2$. Suppose that$\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$and$\{{\delta }_n\}$are the sequences in$(0,1)$satisfying the following conditions:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$and${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) ${\beta }_n+{\gamma }_n+{\delta }_n=1$, $\mathrm{\forall }n\ge 0$, and${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}({\beta }_n+{\delta }_n)<1$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}\frac{1}{{\alpha }_n}|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|=0$;

• (v) ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_n-{\alpha }_{n-1}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}{\alpha }_{n-1}/{\alpha }_n=1$;

• (vi) ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|/{\alpha }_n=0$.

Then$\{x_n\}$converges strongly to$q\in F$, which solves the following VIP:

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

4 Three-step implicit iterative algorithm

In this section, we introduce our three-step implicit iterative algorithm and show strong convergence of the purposed algorithm.

Theorem 4.1LetCbe a nonempty closed convex subset of a uniformly convex Banach spaceXwhich has a uniformly Gâteaux differentiable norm. Let${\Pi }_C$be a sunny nonexpansive retraction fromXontoC. Let the mapping$B_i:C\to X$be${\lambda }_i$-strictly pseudocontractive and${\alpha }_i$-strongly accretive with${\alpha }_i+{\lambda }_i\ge 1$for$i=1,2$. Let$f:C\to C$be a contraction with coefficient$\rho \in (0,1)$. Let${\{S_n\}}_{n=1}^{\mathrm{\infty }}$be an infinite family of nonexpansive mappings ofCinto itself such that$F={\bigcap }_{i=1}^{\mathrm{\infty }}Fix(S_i)\cap \Omega \ne \mathrm{\varnothing }$, whereΩis the fixed point set of the mapping$G={\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)$. For arbitrarily given$x_0\in C$, let$\{x_n\}$be the sequence generated by

$$\{\begin{array}{c}{z}_n={\sigma }_{n}G(x_n)+(1-{\sigma }_n)S_{n}G(x_n),\hfill \\ y_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)z_n,\hfill \\ x_{n+1}={\beta }_n{x}_n+{\gamma }_n{y}_n+{\delta }_n{S}_{n+1}G(x_{n+1}),\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(4.1)

where$1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$for$i=1,2$. Suppose that$\{{\sigma }_n\}$, $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$and$\{{\delta }_n\}$are the sequences in$(0,1)$satisfying the following conditions:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\sigma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\sigma }_n<1$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$and${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (iii) ${\beta }_n+{\gamma }_n+{\delta }_n=1$, $\mathrm{\forall }n\ge 0$, and${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$;

• (iv) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}({\beta }_n+{\delta }_n)<1$;

• (v) ${\sum }_{n=1}^{\mathrm{\infty }}|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}\frac{1}{{\alpha }_n}|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|=0$;

• (vi) ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_n-{\alpha }_{n-1}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}{\alpha }_{n-1}/{\alpha }_n=1$;

• (vii) ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|/{\alpha }_n=0$;

• (viii) ${\sum }_{n=1}^{\mathrm{\infty }}|{\sigma }_n-{\sigma }_{n-1}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}|{\sigma }_n-{\sigma }_{n-1}|/{\alpha }_n=0$.

Assume that${\sum }_{n=1}^{\mathrm{\infty }}{sup}_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel <\mathrm{\infty }$for any bounded subsetDofC, and letSbe a mapping ofCinto itself defined by$Sx={lim}_{n\to \mathrm{\infty }}{S}_{n}x$for all$x\in C$and suppose that$Fix(S)={\bigcap }_{i=1}^{\mathrm{\infty }}Fix(S_i)$. Then$\{x_n\}$converges strongly to$q\in F$, which solves the following VIP:

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

(4.2)

Proof Take a fixed $p\in F$ arbitrarily. Then by Lemma 2.11 we know that $p=G(p)$ and $p=S_{n}p$ for all $n\ge 1$. Moreover, by Lemma 2.10 we have

$$\begin{array}{rl}\parallel z_n-p\parallel & =\parallel {\sigma }_n(G(x_n)-p)+(1-{\sigma }_n)(S_{n}G(x_n)-p)\parallel \\ \le {\sigma }_n\parallel G(x_n)-p\parallel +(1-{\sigma }_n)\parallel S_{n}G(x_n)-p\parallel \\ \le {\sigma }_n\parallel G(x_n)-p\parallel +(1-{\sigma }_n)\parallel G(x_n)-p\parallel \\ =\parallel G(x_n)-p\parallel \\ \le \parallel x_n-p\parallel ,\end{array}$$

(4.3)

and

$$\begin{array}{rl}\parallel y_n-p\parallel & =\parallel {\alpha }_n(f(x_n)-p)+(1-{\alpha }_n)(z_n-p)\parallel \\ \le {\alpha }_n\parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel f(p)-p\parallel +(1-{\alpha }_n)\parallel z_n-p\parallel \\ \le {\alpha }_n\rho \parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel \\ =(1-{\alpha }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel .\end{array}$$

(4.4)

From (4.1) we obtain

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & =\parallel {\beta }_n(x_n-p)+{\gamma }_n(y_n-p)+{\delta }_n(S_{n+1}G(x_{n+1})-p)\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel y_n-p\parallel +{\delta }_n\parallel x_{n+1}-p\parallel ,\end{array}$$

which together with (4.4) implies that

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & \le \frac{1}{{\beta }_n+{\gamma }_n}[{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel y_n-p\parallel ]\\ \le \frac{1}{{\beta }_n+{\gamma }_n}\{{\beta }_n\parallel x_n-p\parallel +{\gamma }_n[(1-{\alpha }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel ]\}\\ =\frac{1}{{\beta }_n+{\gamma }_n}\{({\beta }_n+{\gamma }_n-{\alpha }_n{\gamma }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n{\gamma }_n\parallel f(p)-p\parallel \}\\ =(1-\frac{{\alpha }_n{\gamma }_n(1-\rho )}{{\beta }_n+{\gamma }_n})\parallel x_n-p\parallel +\frac{{\alpha }_n{\gamma }_n}{{\beta }_n+{\gamma }_n}\parallel f(p)-p\parallel \\ =(1-\frac{{\alpha }_n{\gamma }_n(1-\rho )}{{\beta }_n+{\gamma }_n})\parallel x_n-p\parallel +\frac{{\alpha }_n{\gamma }_n(1-\rho )}{{\beta }_n+{\gamma }_n}\cdot \frac{\parallel f(p)-p\parallel }{1-\rho }\\ \le max\{\parallel x_0-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-\rho }\},\end{array}$$

which implies that $\{x_n\}$ is bounded. By Lemma 2.10 we know from (4.3) and (4.4) that $\{y_n\}$, $\{z_n\}$, $\{G(x_n)\}$ and $\{G(y_n)\}$ are bounded.

Let us show that $\parallel x_{n+1}-x_n\parallel \to 0$ and $\parallel x_n-y_n\parallel \to 0$ as $n\to \mathrm{\infty }$. As a matter of fact, from (4.1) we have

$$\begin{array}{rl}{z}_n-z_{n-1}=& {\sigma }_n(G(x_n)-G(x_{n-1}))+({\sigma }_n-{\sigma }_{n-1})(G(x_{n-1})-S_{n-1}G(x_{n-1}))\\ +(1-{\sigma }_n)(S_{n}G(x_n)-S_{n-1}G(x_{n-1})),\end{array}$$

and

$$\begin{array}{rl}{y}_n-y_{n-1}=& {\alpha }_n(f(x_n)-f(x_{n-1}))+({\alpha }_n-{\alpha }_{n-1})(f(x_{n-1})-z_{n-1})\\ +(1-{\alpha }_n)(z_n-z_{n-1}).\end{array}$$

It follows that

$$\begin{array}{rl}\parallel z_n-z_{n-1}\parallel \le & {\sigma }_n\parallel G(x_n)-G(x_{n-1})\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +(1-{\sigma }_n)\parallel S_{n}G(x_n)-S_{n-1}G(x_{n-1})\parallel \\ \le & {\sigma }_n\parallel G(x_n)-G(x_{n-1})\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +(1-{\sigma }_n)(\parallel S_{n}G(x_n)-S_{n}G(x_{n-1})\parallel +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel )\\ \le & {\sigma }_n\parallel G(x_n)-G(x_{n-1})\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +(1-{\sigma }_n)(\parallel G(x_n)-G(x_{n-1})\parallel +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel )\\ \le & \parallel G(x_n)-G(x_{n-1})\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel ,\end{array}$$

and

$$\begin{array}{rl}\parallel y_n-y_{n-1}\parallel \le & {\alpha }_n\parallel f(x_n)-f(x_{n-1})\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-z_{n-1}\parallel \\ +(1-{\alpha }_n)\parallel z_n-z_{n-1}\parallel \\ \le & {\alpha }_n\rho \parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-z_{n-1}\parallel \\ +(1-{\alpha }_n)\parallel z_n-z_{n-1}\parallel \\ \le & {\alpha }_n\rho \parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-z_{n-1}\parallel \\ +(1-{\alpha }_n)[\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel ]\\ \le & (1-{\alpha }_n(1-\rho ))\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-z_{n-1}\parallel +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel .\end{array}$$

(4.5)

Taking into consideration that $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}({\beta }_n+{\delta }_n)<1$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n=1-{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}({\beta }_n+{\delta }_n)>0$, we may assume that $\{{\beta }_n\}\subset [c,d]$ and $\{{\gamma }_n\}\subset [c,1)$ for some $c,d\in (0,1)$. Now, we write $x_n={\beta }_{n-1}{x}_{n-1}+(1-{\beta }_{n-1})v_{n-1}$, $\mathrm{\forall }n\ge 1$, where $v_{n-1}=\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}$. It follows that for all $n\ge 1$,

$$\begin{array}{rl}{v}_n-v_{n-1}=& \frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}\\ =& \frac{{\gamma }_n{y}_n+{\delta }_n{S}_{n+1}G(x_{n+1})}{1-{\beta }_n}-\frac{{\gamma }_{n-1}{y}_{n-1}+{\delta }_{n-1}{S}_{n}G(x_n)}{1-{\beta }_{n-1}}\\ =& \frac{{\gamma }_n(y_n-y_{n-1})+{\delta }_n(S_{n+1}G(x_{n+1})-S_{n}G(x_n))}{1-{\beta }_n}\\ +(\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}})y_{n-1}+(\frac{{\delta }_n}{1-{\beta }_n}-\frac{{\delta }_{n-1}}{1-{\beta }_{n-1}})S_{n}G(x_n).\end{array}$$

(4.6)

This together with (4.5) implies that

$$\begin{array}{r}\parallel v_n-v_{n-1}\parallel \\ \le \frac{\parallel {\gamma }_n(y_n-y_{n-1})+{\delta }_n(S_{n+1}G(x_{n+1})-S_{n}G(x_n))\parallel }{1-{\beta }_n}+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\parallel y_{n-1}\parallel \\ +|\frac{{\delta }_n}{1-{\beta }_n}-\frac{{\delta }_{n-1}}{1-{\beta }_{n-1}}|\parallel S_{n}G(x_n)\parallel \\ \le \frac{{\gamma }_n\parallel y_n-y_{n-1}\parallel +{\delta }_n(\parallel S_{n+1}G(x_{n+1})-S_{n+1}G(x_n)\parallel +\parallel S_{n+1}G(x_n)-S_{n}G(x_n)\parallel )}{1-{\beta }_n}\\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\parallel y_{n-1}\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\parallel S_{n}G(x_n)\parallel \\ \le \frac{{\gamma }_n\parallel y_n-y_{n-1}\parallel +{\delta }_n\parallel x_{n+1}-x_n\parallel }{1-{\beta }_n}+\frac{{\delta }_n}{1-{\beta }_n}\parallel S_{n+1}G(x_n)-S_{n}G(x_n)\parallel \\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|(\parallel y_{n-1}\parallel +\parallel S_{n}G(x_n)\parallel )\\ \le \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}[(1-{\alpha }_n(1-\rho ))\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-z_{n-1}\parallel +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel ]\\ +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel x_{n+1}-x_n\parallel +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel S_{n+1}G(x_n)-S_{n}G(x_n)\parallel \\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|(\parallel y_{n-1}\parallel +\parallel S_{n}G(x_n)\parallel )\\ \le \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}[(1-{\alpha }_n(1-\rho ))\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|M\\ +|{\alpha }_n-{\alpha }_{n-1}|M+\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel ]+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel x_{n+1}-x_n\parallel \\ +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel S_{n+1}G(x_n)-S_{n}G(x_n)\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|M,\end{array}$$

(4.7)

where ${sup}_{n\ge 0}\{\parallel f(x_n)-z_n\parallel +\parallel G(x_n)-S_{n}G(x_n)\parallel \}\le M$ and ${sup}_{n\ge 1}\{\parallel y_{n-1}\parallel +\parallel S_n{x}_n\parallel \}\le M$ for some $M>0$.

Now we observe that

$$\begin{array}{c}{x}_{n+1}-x_n=({\beta }_n-{\beta }_{n-1})(x_n-v_n)+{\beta }_{n-1}(x_n-x_{n-1})+(1-{\beta }_{n-1})(v_n-v_{n-1}),\hfill \\ 1-(1-{\beta }_{n-1})\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}=\frac{{\gamma }_n+{\beta }_{n-1}{\delta }_n}{{\gamma }_n+{\delta }_n},\hfill \end{array}$$

and

$$\begin{array}{r}{\beta }_{n-1}+(1-{\beta }_{n-1})\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(1-(1-\rho ){\alpha }_n)\\ ={\beta }_{n-1}(\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n})+(1-{\beta }_{n-1})\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(1-(1-\rho ){\alpha }_n)\\ =\frac{{\beta }_{n-1}{\delta }_n}{{\gamma }_n+{\delta }_n}+\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}[{\beta }_{n-1}+(1-{\beta }_{n-1})(1-(1-\rho ){\alpha }_n)]\\ =\frac{{\beta }_{n-1}{\delta }_n}{{\gamma }_n+{\delta }_n}+\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}[1-(1-{\beta }_{n-1})(1-\rho ){\alpha }_n]\\ =\frac{{\gamma }_n+{\beta }_{n-1}{\delta }_n}{{\gamma }_n+{\delta }_n}-(1-{\beta }_{n-1})(1-\rho )\frac{{\alpha }_n{\gamma }_n}{{\gamma }_n+{\delta }_n}.\end{array}$$

Hence from (4.7) it follows that

$$\begin{array}{r}\parallel x_{n+1}-x_n\parallel \\ \le |{\beta }_n-{\beta }_{n-1}|\parallel x_n-v_n\parallel +{\beta }_{n-1}\parallel x_n-x_{n-1}\parallel +(1-{\beta }_{n-1})\parallel v_n-v_{n-1}\parallel \\ \le |{\beta }_n-{\beta }_{n-1}|\parallel x_n-v_n\parallel +{\beta }_{n-1}\parallel x_n-x_{n-1}\parallel \\ +(1-{\beta }_{n-1})\left\{\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}\right[(1-{\alpha }_n(1-\rho ))\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|M\\ +|{\alpha }_n-{\alpha }_{n-1}|M+\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel ]+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel x_{n+1}-x_n\parallel \\ +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel S_{n+1}G(x_n)-S_{n}G(x_n)\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|M\}\\ \le |{\beta }_n-{\beta }_{n-1}|M_1+[{\beta }_{n-1}+(1-{\beta }_{n-1})\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(1-(1-\rho ){\alpha }_n)]\parallel x_n-x_{n-1}\parallel \\ +|{\sigma }_n-{\sigma }_{n-1}|M_1+|{\alpha }_n-{\alpha }_{n-1}|M_1+\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +(1-{\beta }_{n-1})\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel x_{n+1}-x_n\parallel +\parallel S_{n+1}G(x_n)-S_{n}G(x_n)\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|M_1\\ =[\frac{{\gamma }_n+{\beta }_{n-1}{\delta }_n}{{\gamma }_n+{\delta }_n}-(1-{\beta }_{n-1})(1-\rho )\frac{{\alpha }_n{\gamma }_n}{{\gamma }_n+{\delta }_n}]\parallel x_n-x_{n-1}\parallel \\ +(1-{\beta }_{n-1})\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel x_{n+1}-x_n\parallel \\ +M_1(|{\sigma }_n-{\sigma }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|)\\ +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel +\parallel S_{n+1}G(x_n)-S_{n}G(x_n)\parallel ,\end{array}$$

where ${sup}_{n\ge 0}\{M+\parallel x_n-v_n\parallel \}\le M_1$ for some $M_1>0$. Therefore, we get

$$\begin{array}{r}\parallel x_{n+1}-x_n\parallel \\ \le [1-(1-{\beta }_{n-1})(1-\rho )\frac{{\alpha }_n{\gamma }_n}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}]\parallel x_n-x_{n-1}\parallel +\frac{{\gamma }_n+{\delta }_n}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}\left[M_1\right(|{\sigma }_n-{\sigma }_{n-1}|\\ +|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|)\\ +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel +\parallel S_{n+1}G(x_n)-S_{n}G(x_n)\parallel ]\\ \le [1-c(1-d)(1-\rho )\frac{{\alpha }_n}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}]\parallel x_n-x_{n-1}\parallel \\ +\frac{1}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}[M_1(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|)\\ +\parallel S_{n+1}{x}_n-S_n{x}_n\parallel ]\\ =[1-c(1-d)(1-\rho )\frac{{\alpha }_n}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}]\parallel x_n-x_{n-1}\parallel \\ +c(1-d)(1-\rho )\frac{1}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}\cdot \frac{1}{c(1-d)(1-\rho )}\left[M_1\right(|{\sigma }_n-{\sigma }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|\\ +|{\beta }_n-{\beta }_{n-1}|+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|)+\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +\parallel S_{n+1}G(x_n)-S_{n}G(x_n)\parallel ]\\ \le [1-c(1-d)(1-\rho )\frac{{\alpha }_n}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}]\parallel x_n-x_{n-1}\parallel \\ +\frac{1}{{\gamma }_n+{\beta }_{n-1}{\delta }_n}[M_1(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|)\\ +\parallel S_{n+1}{x}_n-S_n{x}_n\parallel ].\end{array}$$

Utilizing Lemma 2.1, from conditions (ii), (v)-(viii) and the assumption on $\{S_n\}$, we deduce that

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

(4.8)

Also, taking into account the boundedness of $\{G(x_n)\}$ and $\{S_{n}G(x_n)\}$, by Lemma 2.5 there exists a continuous strictly increasing function $g_1:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $g_1(0)=0$ such that for $p\in F$,

$$\begin{array}{r}{\parallel z_n-p\parallel }^2\\ ={\parallel {\sigma }_n(G(x_n)-p)+(1-{\sigma }_n)(S_{n}G(x_n)-p)\parallel }^2\\ \le {\sigma }_n{\parallel G(x_n)-p\parallel }^2+(1-{\sigma }_n){\parallel S_{n}G(x_n)-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)\\ \le {\sigma }_n{\parallel x_n-p\parallel }^2+(1-{\sigma }_n){\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)\\ ={\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right),\end{array}$$

and hence

$$\begin{array}{r}{\parallel y_n-p\parallel }^2\\ ={\parallel {\alpha }_n(f(x_n)-p)+(1-{\alpha }_n)(z_n-p)\parallel }^2\\ \le {\parallel (1-{\alpha }_n)(z_n-p)\parallel }^2+2{\alpha }_n〈f(x_n)-p,J(y_n-p)〉\\ \le {\parallel z_n-p\parallel }^2+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel \\ \le {\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel .\end{array}$$

(4.9)

Since $\{x_n\}$ and $\{S_{n}G(x_n)\}$ both are bounded, by Lemma 2.5 there exists a continuous strictly increasing function $g_2:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $g(0)=0$ such that

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ ={\parallel ({\beta }_n+{\delta }_n)(\frac{{\beta }_n}{{\beta }_n+{\delta }_n}(x_n-p)+\frac{{\delta }_n}{{\beta }_n+{\delta }_n}(S_{n+1}G(x_{n+1})-p))+{\gamma }_n(y_n-p)\parallel }^2\\ \le ({\beta }_n+{\delta }_n){\parallel \frac{{\beta }_n}{{\beta }_n+{\delta }_n}(x_n-p)+\frac{{\delta }_n}{{\beta }_n+{\delta }_n}(S_{n+1}G(x_{n+1})-p)\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2\\ \le ({\beta }_n+{\delta }_n)[\frac{{\beta }_n}{{\beta }_n+{\delta }_n}{\parallel x_n-p\parallel }^2+\frac{{\delta }_n}{{\beta }_n+{\delta }_n}\parallel S_{n+1}G(x_{n+1})-p\parallel \\ -\frac{{\beta }_n{\delta }_n}{{({\beta }_n+{\delta }_n)}^2}{g}_2\left(\parallel x_n-S_{n+1}G(x_{n+1})\parallel \right)]+{\gamma }_n{\parallel y_n-p\parallel }^2\\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\delta }_n{\parallel x_{n+1}-p\parallel }^2-\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}{g}_2\left(\parallel x_n-S_{n+1}G(x_{n+1})\parallel \right)+{\gamma }_n{\parallel y_n-p\parallel }^2\\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\delta }_n{\parallel x_{n+1}-p\parallel }^2-\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}{g}_2\left(\parallel x_n-S_{n+1}G(x_{n+1})\parallel \right)\\ +{\gamma }_n[{\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel ]\\ =({\beta }_n+{\gamma }_n){\parallel x_n-p\parallel }^2+{\delta }_n{\parallel x_{n+1}-p\parallel }^2-{\gamma }_n{\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)\\ -\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}{g}_2(\parallel x_n-S_{n+1}{x}_{n+1}\parallel )+2{\alpha }_n{\gamma }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel ,\end{array}$$

which immediately yields

$$\begin{array}{r}{\gamma }_n{\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)+{\beta }_n{\delta }_n{g}_2\left(\parallel x_n-S_{n+1}G(x_{n+1})\parallel \right)\\ \le {\gamma }_n{\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)+\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}g(\parallel x_n-S_{n+1}{x}_{n+1}\parallel )\\ \le ({\beta }_n+{\gamma }_n)({\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2)+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel \\ \le (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel .\end{array}$$

According to condition (iv), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\gamma }_n=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(1-{\beta }_n-{\delta }_n)=1-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}({\beta }_n+{\delta }_n)>0.$$

Since ${\alpha }_n\to 0$, $\parallel x_{n+1}-x_n\parallel \to 0$, ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$, we obtain from condition (i) that

$$\underset{n\to \mathrm{\infty }}{lim}{g}_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)=0\text{and}\underset{n\to \mathrm{\infty }}{lim}{g}_2\left(\parallel x_n-S_{n+1}G(x_{n+1})\parallel \right)=0.$$

Utilizing the properties of $g_1$ and $g_2$, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel G(x_n)-S_{n}G(x_n)\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-S_{n+1}G(x_{n+1})\parallel =0.$$

(4.10)

Thus, from (4.8) and (4.10) it follows that

$$\begin{array}{rl}{\gamma }_n\parallel y_n-x_n\parallel & =\parallel (x_{n+1}-x_n)-{\delta }_n(S_{n+1}G(x_{n+1})-x_n)\parallel \\ \le \parallel x_{n+1}-x_n\parallel +{\delta }_n\parallel S_{n+1}G(x_{n+1})-x_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

That is,

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

(4.11)

Since it follows from (4.1) that

$$y_n-G(x_n)={\alpha }_n(f(x_n)-G(x_n))+(1-{\alpha }_n)(1-{\sigma }_n)(S_{n}G(x_n)-G(x_n)),$$

we conclude from (4.10) and ${\alpha }_n\to 0$ that

$$\begin{array}{rl}\parallel y_n-G(x_n)\parallel & =\parallel {\alpha }_n(f(x_n)-G(x_n))+(1-{\alpha }_n)(1-{\sigma }_n)(S_{n}G(x_n)-G(x_n))\parallel \\ \le {\alpha }_n\parallel f(x_n)-G(x_n)\parallel +(1-{\alpha }_n)(1-{\sigma }_n)\parallel S_{n}G(x_n)-G(x_n)\parallel \\ \le {\alpha }_n\parallel f(x_n)-G(x_n)\parallel +\parallel S_{n}G(x_n)-G(x_n)\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

That is,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-G(x_n)\parallel =0.$$

(4.12)

This together with (4.11) leads to

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

(4.13)

By (4.10) and Lemma 2.6, we have

$$\begin{array}{rl}\parallel SG(x_n)-G(x_n)\parallel & \le \parallel SG(x_n)-S_{n}G(x_n)\parallel +\parallel S_{n}G(x_n)-G(x_n)\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(4.14)

In terms of (4.13) and (4.14), we have

$$\begin{array}{rl}\parallel x_n-S{x}_n\parallel & \le \parallel x_n-G(x_n)\parallel +\parallel G(x_n)-SG(x_n)\parallel +\parallel SG(x_n)-S{x}_n\parallel \\ \le 2\parallel x_n-G(x_n)\parallel +\parallel G(x_n)-SG(x_n)\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(4.15)

Define a mapping $Wx=(1-\theta )Sx+\theta G(x)$, where G is defined by (1.7), $\theta \in (0,1)$ is a constant. Then by Lemma 2.8 we have that $Fix(W)=Fix(S)\cap Fix(G)=F$. We observe that

$$\begin{array}{rl}\parallel x_n-W{x}_n\parallel & =\parallel (1-\theta )(x_n-S{x}_n)+\theta (x_n-G(x_n))\parallel \\ \le (1-\theta )\parallel x_n-S{x}_n\parallel +\theta \parallel x_n-G(x_n)\parallel .\end{array}$$

From (4.13) and (4.15), we obtain

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

(4.16)

Utilizing the arguments similar to those of (3.17) in the proof of Theorem 3.1, we can obtain

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

(4.17)

where $q=s-{lim}_{t\to 0}{x}_t$ with $x_t$ being the fixed point of the contraction $x\mapsto tf(x)+(1-t)Wx$; that is, $x_t$ solves the fixed point equation $x_t=tf(x_t)+(1-t)W{x}_t$. Noticing that J is norm-to-weak^∗ uniformly continuous on bounded subsets of X, we deduce from (4.11) and (4.17) that

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

(4.18)

Finally, let us show that $x_n\to q$ as $n\to \mathrm{\infty }$. We observe that

$$\begin{array}{r}{\parallel y_n-q\parallel }^2\\ ={\alpha }_n〈f(x_n)-q,J(y_n-q)〉+(1-{\alpha }_n)〈{\sigma }_n(G(x_n)-q)+(1-{\sigma }_n)(S_{n}G(x_n)-q),J(y_n-q)〉\\ ={\alpha }_n〈f(x_n)-f(q),J(y_n-q)〉+{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ +(1-{\alpha }_n)〈{\sigma }_n(G(x_n)-q)+(1-{\sigma }_n)(S_{n}G(x_n)-q),J(y_n-q)〉\\ \le {\alpha }_n\rho \parallel x_n-q\parallel \parallel y_n-q\parallel +(1-{\alpha }_n)\parallel {\sigma }_n(G(x_n)-q)+(1-{\sigma }_n)(S_{n}G(x_n)-q)\parallel \parallel y_n-q\parallel \\ +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ \le {\alpha }_n\rho \parallel x_n-q\parallel \parallel y_n-q\parallel +(1-{\alpha }_n)({\sigma }_n\parallel G(x_n)-q\parallel +(1-{\sigma }_n)\parallel S_{n}G(x_n)-q\parallel )\parallel y_n-q\parallel \\ +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ \le {\alpha }_n\rho \parallel x_n-q\parallel \parallel y_n-q\parallel +(1-{\alpha }_n)({\sigma }_n\parallel x_n-q\parallel +(1-{\sigma }_n)\parallel x_n-q\parallel )\parallel y_n-q\parallel \\ +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ ={\alpha }_n\rho \parallel x_n-q\parallel \parallel y_n-q\parallel +(1-{\alpha }_n)\parallel x_n-q\parallel \parallel y_n-q\parallel +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ =(1-{\alpha }_n(1-\rho ))\parallel x_n-q\parallel \parallel y_n-q\parallel +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ \le \frac{1-{\alpha }_n(1-\rho )}{2}({\parallel x_n-q\parallel }^2+{\parallel y_n-q\parallel }^2)+{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ \le \frac{1-{\alpha }_n(1-\rho )}{2}{\parallel x_n-q\parallel }^2+\frac{1}{2}{\parallel y_n-q\parallel }^2+{\alpha }_n〈f(q)-q,J(y_n-q)〉,\end{array}$$

which implies that

$${\parallel y_n-q\parallel }^2\le (1-{\alpha }_n(1-\rho )){\parallel x_n-q\parallel }^2+{\alpha }_n(1-\rho )\frac{2〈f(q)-q,J(y_n-q)〉}{1-\rho }.$$

(4.19)

Also, by (4.1) and the convexity of ${\parallel \cdot \parallel }^2$, we get

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2& ={\parallel {\beta }_n(x_n-q)+{\gamma }_n(y_n-q)+{\delta }_n(S_{n+1}G(x_{n+1})-q)\parallel }^2\\ \le {\beta }_n{\parallel x_n-q\parallel }^2+{\gamma }_n{\parallel y_n-q\parallel }^2+{\delta }_n{\parallel x_{n+1}-q\parallel }^2,\end{array}$$

which together with (4.19) implies that

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2\le & \frac{{\beta }_n}{{\beta }_n+{\gamma }_n}{\parallel x_n-q\parallel }^2+\frac{{\gamma }_n}{{\beta }_n+{\gamma }_n}{\parallel y_n-q\parallel }^2\\ \le & \frac{{\beta }_n}{{\beta }_n+{\gamma }_n}{\parallel x_n-q\parallel }^2+\frac{{\gamma }_n}{{\beta }_n+{\gamma }_n}[(1-{\alpha }_n(1-\rho )){\parallel x_n-q\parallel }^2\\ +{\alpha }_n(1-\rho )\frac{2〈f(q)-q,J(y_n-q)〉}{1-\rho }]\\ =& (1-\frac{{\alpha }_n{\gamma }_n(1-\rho )}{{\beta }_n+{\gamma }_n}){\parallel x_n-q\parallel }^2\\ +\frac{{\alpha }_n{\gamma }_n(1-\rho )}{{\beta }_n+{\gamma }_n}\cdot \frac{2〈f(q)-q,J(y_n-q)〉}{1-\rho }.\end{array}$$

(4.20)

Applying Lemma 2.1 to (4.20), we conclude that $x_n\to q$ as $n\to \mathrm{\infty }$. This completes the proof. □

Corollary 4.1LetCbe a nonempty closed convex subset of a uniformly convex Banach spaceXwhich has a uniformly Gâteaux differentiable norm. Let${\Pi }_C$be a sunny nonexpansive retraction fromXontoC. Let the mapping$B_i:C\to X$be${\lambda }_i$-strictly pseudocontractive and${\alpha }_i$-strongly accretive with${\alpha }_i+{\lambda }_i\ge 1$for$i=1,2$. Let$f:C\to C$be a contraction with coefficient$\rho \in (0,1)$. LetSbe a nonexpansive mapping ofCinto itself such that$F=Fix(S)\cap \Omega \ne \mathrm{\varnothing }$, whereΩis the fixed point set of the mapping$G={\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)$. For arbitrarily given$x_0\in C$, let$\{x_n\}$be the sequence generated by

$$\{\begin{array}{c}{z}_n={\sigma }_{n}G(x_n)+(1-{\sigma }_n)SG(x_n),\hfill \\ y_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)z_n,\hfill \\ x_{n+1}={\beta }_n{x}_n+{\gamma }_n{y}_n+{\delta }_{n}SG(x_{n+1}),\mathrm{\forall }n\ge 0,\hfill \end{array}$$

where$1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$for$i=1,2$. Suppose that$\{{\sigma }_n\}$, $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$and$\{{\delta }_n\}$are the sequences in$(0,1)$satisfying the following conditions:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\sigma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\sigma }_n<1$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$and${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (iii) ${\beta }_n+{\gamma }_n+{\delta }_n=1$, $\mathrm{\forall }n\ge 0$, and${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$;

• (iv) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}({\beta }_n+{\delta }_n)<1$;

• (v) ${\sum }_{n=1}^{\mathrm{\infty }}|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}\frac{1}{{\alpha }_n}|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|=0$;

• (vi) ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_n-{\alpha }_{n-1}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}{\alpha }_{n-1}/{\alpha }_n=1$;

• (vii) ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|/{\alpha }_n=0$;

• (viii) ${\sum }_{n=1}^{\mathrm{\infty }}|{\sigma }_n-{\sigma }_{n-1}|<\mathrm{\infty }$or${lim}_{n\to \mathrm{\infty }}|{\sigma }_n-{\sigma }_{n-1}|/{\alpha }_n=0$.

Then$\{x_n\}$converges strongly to$q\in F$, which solves the following VIP:

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

5 Two-step explicit iterative algorithm

In this section, we introduce our two-step explicit iterative algorithm and show strong convergence of the purposed algorithm.

Theorem 5.1LetCbe a nonempty closed convex subset of a uniformly convex Banach spaceXwhich has a uniformly Gâteaux differentiable norm. Let${\Pi }_C$be a sunny nonexpansive retraction fromXontoC. Let the mapping$B_i:C\to X$be${\lambda }_i$-strictly pseudocontractive and${\alpha }_i$-strongly accretive with${\alpha }_i+{\lambda }_i\ge 1$for$i=1,2$. Let$f:C\to C$be a contraction with coefficient$\rho \in (0,1)$. Let${\{S_n\}}_{n=0}^{\mathrm{\infty }}$be an infinite family of nonexpansive mappings ofCinto itself such that$F={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(S_i)\cap \Omega \ne \mathrm{\varnothing }$, whereΩis the fixed point set of the mapping$G={\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)$. For arbitrarily given$x_0\in C$, let$\{x_n\}$be the sequence generated by

$$\{\begin{array}{c}{y}_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)G(x_n),\hfill \\ x_{n+1}={\beta }_n{x}_n+{\gamma }_n{y}_n+{\delta }_n{S}_n{y}_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(5.1)

where$1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$for$i=1,2$. Suppose that$\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$and$\{{\delta }_n\}$are the sequences in$(0,1)$satisfying the following conditions:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$and${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) ${\beta }_n+{\gamma }_n+{\delta }_n=1$, $\mathrm{\forall }n\ge 0$, and${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}({\beta }_n+{\delta }_n)<1$;

• (iv) ${lim}_{n\to \mathrm{\infty }}|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|=0$.

Assume that${\sum }_{n=1}^{\mathrm{\infty }}{sup}_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel <\mathrm{\infty }$for any bounded subsetDofCand letSbe a mapping ofCinto itself defined by$Sx={lim}_{n\to \mathrm{\infty }}{S}_{n}x$for all$x\in C$and suppose that$Fix(S)={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(S_i)$. Then$\{x_n\}$converges strongly to$q\in F$, which solves the following VIP:

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

(5.2)

Proof Take a fixed $p\in F$ arbitrarily. Then by Lemma 2.11 we know that $p=G(p)$. Moreover, by Lemma 2.10 we have

$$\begin{array}{rl}\parallel y_n-p\parallel & =\parallel {\alpha }_n(f(x_n)-p)+(1-{\alpha }_n)(G(x_n)-p)\parallel \\ \le {\alpha }_n\parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel f(p)-p\parallel +(1-{\alpha }_n)\parallel G(x_n)-p\parallel \\ \le {\alpha }_n\rho \parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel \\ =(1-{\alpha }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel .\end{array}$$

(5.3)

From (5.3) we obtain

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & =\parallel {\beta }_n(x_n-p)+{\gamma }_n(y_n-p)+{\delta }_n(S_n{y}_n-p)\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel y_n-p\parallel +{\delta }_n\parallel y_n-p\parallel \\ ={\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)\parallel y_n-p\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)[(1-{\alpha }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel ]\\ =[1-(1-{\beta }_n)(1-\rho ){\alpha }_n]\parallel x_n-p\parallel +(1-{\beta }_n)(1-\rho ){\alpha }_n\cdot \frac{\parallel f(p)-p\parallel }{1-\rho }\\ \le max\{\parallel x_0-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-\rho }\},\end{array}$$

which implies that $\{x_n\}$ is bounded. By Lemma 2.10 we know from (5.3) that $\{G(x_n)\}$ and $\{y_n\}$ both are bounded.

Let us show that $\parallel x_{n+1}-x_n\parallel \to 0$ and $\parallel x_n-y_n\parallel \to 0$ as $n\to \mathrm{\infty }$. As a matter of fact, from (5.1) we have

$$y_n-y_{n-1}={\alpha }_n(f(x_n)-f(x_{n-1}))+({\alpha }_n-{\alpha }_{n-1})(f(x_{n-1})-G(x_{n-1}))+(1-{\alpha }_n)(G(x_n)-G(x_{n-1})).$$

It follows that

$$\begin{array}{rl}\parallel y_n-y_{n-1}\parallel \le & {\alpha }_n\parallel f(x_n)-f(x_{n-1})\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-G(x_{n-1})\parallel \\ +(1-{\alpha }_n)\parallel G(x_n)-G(x_{n-1})\parallel \\ \le & {\alpha }_n\rho \parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-G(x_{n-1})\parallel \\ +(1-{\alpha }_n)\parallel x_n-x_{n-1}\parallel \\ =& (1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-G(x_{n-1})\parallel .\end{array}$$

(5.4)

Now, we write $x_n={\beta }_{n-1}{x}_{n-1}+(1-{\beta }_{n-1})v_{n-1}$, $\mathrm{\forall }n\ge 1$, where $v_{n-1}=\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}$. It follows that for all $n\ge 1$,

$$\begin{array}{rl}{v}_n-v_{n-1}=& \frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}\\ =& \frac{{\gamma }_n{y}_n+{\delta }_n{S}_n{y}_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}{y}_{n-1}+{\delta }_{n-1}{S}_{n-1}{y}_{n-1}}{1-{\beta }_{n-1}}\\ =& \frac{{\gamma }_n(y_n-y_{n-1})+{\delta }_n(S_n{y}_n-S_{n-1}{y}_{n-1})}{1-{\beta }_n}\\ +(\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}})y_{n-1}+(\frac{{\delta }_n}{1-{\beta }_n}-\frac{{\delta }_{n-1}}{1-{\beta }_{n-1}})S_{n-1}{y}_{n-1}.\end{array}$$

(5.5)

This together with (5.4) implies that

$$\begin{array}{r}\parallel v_n-v_{n-1}\parallel \\ \le \frac{\parallel {\gamma }_n(y_n-y_{n-1})+{\delta }_n(S_n{y}_n-S_{n-1}{y}_{n-1})\parallel }{1-{\beta }_n}+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\parallel y_{n-1}\parallel \\ +|\frac{{\delta }_n}{1-{\beta }_n}-\frac{{\delta }_{n-1}}{1-{\beta }_{n-1}}|\parallel S_{n-1}{y}_{n-1}\parallel \\ \le \frac{{\gamma }_n\parallel y_n-y_{n-1}\parallel +{\delta }_n(\parallel S_n{y}_n-S_n{y}_{n-1}\parallel +\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel )}{1-{\beta }_n}\\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\parallel y_{n-1}\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\parallel S_{n-1}{y}_{n-1}\parallel \\ \le \frac{{\gamma }_n\parallel y_n-y_{n-1}\parallel +{\delta }_n\parallel y_n-y_{n-1}\parallel }{{\gamma }_n+{\delta }_n}+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel \\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|(\parallel y_{n-1}\parallel +\parallel S_{n-1}{y}_{n-1}\parallel )\\ \le \parallel y_n-y_{n-1}\parallel +\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|(\parallel y_{n-1}\parallel +\parallel S_{n-1}{y}_{n-1}\parallel )\\ \le (1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-G(x_{n-1})\parallel \\ +\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|(\parallel y_{n-1}\parallel +\parallel S_{n-1}{y}_{n-1}\parallel )\\ \le (1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|(\parallel f(x_{n-1})\parallel +\parallel G(x_{n-1})\parallel )\\ +\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|(\parallel y_{n-1}\parallel +\parallel S_{n-1}{y}_{n-1}\parallel )\\ \le \parallel x_n-x_{n-1}\parallel +M(|{\alpha }_n-{\alpha }_{n-1}|+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|)+\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel ,\end{array}$$

(5.6)

where ${sup}_{n\ge 0}\{\parallel f(x_n)\parallel +\parallel G(x_n)\parallel +\parallel y_n\parallel +\parallel S_n{y}_n\parallel \}\le M$ for some $M>0$. Since ${\alpha }_n\to 0$, $|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\to 0$ and the assumption on $\{S_n\}$, we have

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

Utilizing Lemma 2.3, from condition (iii) we obtain that

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

So, we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}(1-{\beta }_n)\parallel v_n-x_n\parallel =0.$$

(5.7)

Also, we note that for $p\in F$,

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2& ={\parallel {\alpha }_n(f(x_n)-p)+(1-{\alpha }_n)(G(x_n)-G(p))\parallel }^2\\ \le {\parallel (1-{\alpha }_n)(G(x_n)-G(p))\parallel }^2+2{\alpha }_n〈f(x_n)-p,J(y_n-p)〉\\ \le {\parallel x_n-p\parallel }^2+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel .\end{array}$$

(5.8)

Since $\{x_n\}$ and $\{y_n\}$ both are bounded, by Lemma 2.5 there exists a continuous strictly increasing function $g:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $g(0)=0$ such that

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ ={\parallel ({\beta }_n+{\delta }_n)(\frac{{\beta }_n}{{\beta }_n+{\delta }_n}(x_n-p)+\frac{{\delta }_n}{{\beta }_n+{\delta }_n}(S_n{y}_n-p))+{\gamma }_n(y_n-p)\parallel }^2\\ \le ({\beta }_n+{\delta }_n){\parallel \frac{{\beta }_n}{{\beta }_n+{\delta }_n}(x_n-p)+\frac{{\delta }_n}{{\beta }_n+{\delta }_n}(S_n{y}_n-p)\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2\\ \le ({\beta }_n+{\delta }_n)[\frac{{\beta }_n}{{\beta }_n+{\delta }_n}{\parallel x_n-p\parallel }^2+\frac{{\delta }_n}{{\beta }_n+{\delta }_n}\parallel S_n{y}_n-p\parallel -\frac{{\beta }_n{\delta }_n}{{({\beta }_n+{\delta }_n)}^2}g(\parallel x_n-S_n{y}_n\parallel )]\\ +{\gamma }_n{\parallel y_n-p\parallel }^2\\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\delta }_n{\parallel S_n{y}_n-p\parallel }^2-\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}g(\parallel x_n-S_n{y}_n\parallel )+{\gamma }_n{\parallel y_n-p\parallel }^2\\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\delta }_n{\parallel y_n-p\parallel }^2-\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}g(\parallel x_n-S_n{y}_n\parallel )+{\gamma }_n{\parallel y_n-p\parallel }^2\\ ={\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel y_n-p\parallel }^2-\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}g(\parallel x_n-S_n{y}_n\parallel )\\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)[{\parallel x_n-p\parallel }^2+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel ]\\ -\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}g(\parallel x_n-S_n{y}_n\parallel )\\ \le {\parallel x_n-p\parallel }^2+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel -\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}g(\parallel x_n-S_n{y}_n\parallel ),\end{array}$$

which immediately yields

$$\begin{array}{rl}{\beta }_n{\delta }_{n}g(\parallel x_n-S_n{y}_n\parallel )& \le \frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}g(\parallel x_n-S_n{y}_n\parallel )\\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel \\ \le (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel .\end{array}$$

Since ${\alpha }_n\to 0$, $\parallel x_{n+1}-x_n\parallel \to 0$, ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$, we get ${lim}_{n\to \mathrm{\infty }}g(\parallel x_n-S_n{y}_n\parallel )=0$ and hence

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

(5.9)

In the meantime, according to condition (iii), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\gamma }_n=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(1-{\beta }_n-{\delta }_n)=1-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}({\beta }_n+{\delta }_n)>0.$$

Thus, from (5.7) and (5.9) it follows that

$$\begin{array}{rl}{\gamma }_n\parallel y_n-x_n\parallel & =\parallel (x_{n+1}-x_n)-{\delta }_n(S_n{y}_n-x_n)\parallel \\ \le \parallel x_{n+1}-x_n\parallel +{\delta }_n\parallel S_n{y}_n-x_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

That is,

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

(5.10)

This together with (5.1) leads to

$$\begin{array}{rl}(1-{\alpha }_n)\parallel G(x_n)-x_n\parallel & =\parallel (y_n-x_n)-{\alpha }_n(f(x_n)-x_n)\parallel \\ \le \parallel y_n-x_n\parallel +{\alpha }_n\parallel f(x_n)-x_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

That is,

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

(5.11)

On the other hand, we observe that

$$y_n-G(x_n)={\alpha }_n(f(x_n)-G(x_n)),$$

which together with ${\alpha }_n\to 0$ implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-G(x_n)\parallel =0.$$

(5.12)

We note that

$$\begin{array}{rl}\parallel S_{n}G(x_n)-G(x_n)\parallel & \le \parallel S_{n}G(x_n)-S_n{y}_n\parallel +\parallel S_n{y}_n-x_n\parallel +\parallel x_n-G(x_n)\parallel \\ \le \parallel G(x_n)-y_n\parallel +\parallel S_n{y}_n-x_n\parallel +\parallel x_n-G(x_n)\parallel .\end{array}$$

From (5.9), (5.11) and (5.12), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel S_{n}G(x_n)-G(x_n)\parallel =0.$$

(5.13)

By (5.13) and Lemma 2.6, we have

$$\begin{array}{rl}\parallel SG(x_n)-G(x_n)\parallel & \le \parallel SG(x_n)-S_{n}G(x_n)\parallel +\parallel S_{n}G(x_n)-G(x_n)\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(5.14)

In terms of (5.11) and (5.14), we have

$$\begin{array}{rl}\parallel x_n-S{x}_n\parallel & \le \parallel x_n-G(x_n)\parallel +\parallel G(x_n)-SG(x_n)\parallel +\parallel SG(x_n)-S{x}_n\parallel \\ \le 2\parallel x_n-G(x_n)\parallel +\parallel G(x_n)-SG(x_n)\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(5.15)

Define a mapping $Wx=(1-\theta )Sx+\theta G(x)$, where G is defined by (1.7), $\theta \in (0,1)$ is a constant. Then by Lemma 2.8 we have that $Fix(W)=Fix(S)\cap Fix(G)=F$. We observe that

$$\begin{array}{rl}\parallel x_n-W{x}_n\parallel & =\parallel (1-\theta )(x_n-S{x}_n)+\theta (x_n-G(x_n))\parallel \\ \le (1-\theta )\parallel x_n-S{x}_n\parallel +\theta \parallel x_n-G(x_n)\parallel .\end{array}$$

From (5.11) and (5.15), we obtain

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

(5.16)

Utilizing the arguments similar to those of (3.17) in the proof of Theorem 3.1, we can obtain

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

(5.17)

where $q=s-{lim}_{t\to 0}{x}_t$ with $x_t$ being the fixed point of the contraction $x\mapsto tf(x)+(1-t)Wx$; that is, $x_t$ solves the fixed point equation $x_t=tf(x_t)+(1-t)W{x}_t$. Noticing that J is norm-to-weak^∗ uniformly continuous on bounded subsets of X, we deduce from (5.10) and (5.17) that

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

(5.18)

Finally, let us show that $x_n\to q$ as $n\to \mathrm{\infty }$. We observe that

$$\begin{array}{r}{\parallel y_n-q\parallel }^2\\ ={\alpha }_n〈f(x_n)-q,J(y_n-q)〉+(1-{\alpha }_n)〈G(x_n)-q,J(y_n-q)〉\\ ={\alpha }_n〈f(x_n)-f(q),J(y_n-q)〉+{\alpha }_n〈f(q)-q,J(y_n-q)〉+(1-{\alpha }_n)〈G(x_n)-q,J(y_n-q)〉\\ \le {\alpha }_n\rho \parallel x_n-q\parallel \parallel y_n-q\parallel +(1-{\alpha }_n)\parallel G(x_n)-q\parallel \parallel y_n-q\parallel +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ \le {\alpha }_n\rho \parallel x_n-q\parallel \parallel y_n-q\parallel +(1-{\alpha }_n)\parallel x_n-q\parallel \parallel y_n-q\parallel +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ =[1-{\alpha }_n(1-\rho )]\parallel x_n-q\parallel \parallel y_n-q\parallel +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ \le \frac{1-{\alpha }_n(1-\rho )}{2}({\parallel x_n-q\parallel }^2+{\parallel y_n-q\parallel }^2)+{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ \le \frac{1-{\alpha }_n(1-\rho )}{2}{\parallel x_n-q\parallel }^2+\frac{1}{2}{\parallel y_n-q\parallel }^2+{\alpha }_n〈f(q)-q,J(y_n-q)〉,\end{array}$$

which implies that

$${\parallel y_n-q\parallel }^2\le [1-{\alpha }_n(1-\rho )]{\parallel x_n-q\parallel }^2+{\alpha }_n(1-\rho )\frac{2〈f(q)-q,J(y_n-q)〉}{1-\rho }.$$

(5.19)

Also, by the convexity of ${\parallel \cdot \parallel }^2$ and (5.1), we get

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2& \le {\beta }_n{\parallel x_n-q\parallel }^2+{\gamma }_n{\parallel y_n-q\parallel }^2+{\delta }_n{\parallel y_n-q\parallel }^2\\ ={\beta }_n{\parallel x_n-q\parallel }^2+(1-{\beta }_n){\parallel y_n-q\parallel }^2.\end{array}$$

It follows from (5.19) that

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2\le & [1-{\alpha }_n(1-{\beta }_n)(1-\rho )]{\parallel x_n-q\parallel }^2\\ +{\alpha }_n(1-{\beta }_n)(1-\rho )\frac{2〈f(q)-q,J(y_n-q)〉}{1-\rho }.\end{array}$$

(5.20)

Applying Lemma 2.1 to (5.20), we obtain that $x_n\to q$ as $n\to \mathrm{\infty }$. This completes the proof. □

Corollary 5.1LetCbe a nonempty closed convex subset of a uniformly convex Banach spaceXwhich has a uniformly Gâteaux differentiable norm. Let${\Pi }_C$be a sunny nonexpansive retraction fromXontoC. Let the mapping$B_i:C\to X$be${\lambda }_i$-strictly pseudocontractive and${\alpha }_i$-strongly accretive with${\alpha }_i+{\lambda }_i\ge 1$for$i=1,2$. Let$f:C\to C$be a contraction with coefficient$\rho \in (0,1)$. LetSbe a nonexpansive mapping ofCinto itself such that$F=Fix(S)\cap \Omega \ne \mathrm{\varnothing }$, whereΩis the fixed point set of the mapping$G={\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)$. For arbitrarily given$x_0\in C$, let$\{x_n\}$be the sequence generated by

$$\{\begin{array}{c}{y}_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)G(x_n),\hfill \\ x_{n+1}={\beta }_n{x}_n+{\gamma }_n{y}_n+{\delta }_{n}S{y}_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

where$1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$for$i=1,2$. Suppose that$\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$and$\{{\delta }_n\}$are the sequences in$(0,1)$satisfying the following conditions:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$and${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) ${\beta }_n+{\gamma }_n+{\delta }_n=1$, $\mathrm{\forall }n\ge 0$, and${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}({\beta }_n+{\delta }_n)<1$;

• (iv) ${lim}_{n\to \mathrm{\infty }}|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|=0$.

Then$\{x_n\}$converges strongly to$q\in F$, which solves the following VIP:

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

6 Three-step explicit iterative algorithm

In this section, we introduce our three-step explicit iterative algorithm and show strong convergence of the proposed algorithm.

Theorem 6.1LetCbe a nonempty closed convex subset of a uniformly convex Banach spaceXwhich has a uniformly Gâteaux differentiable norm. Let${\Pi }_C$be a sunny nonexpansive retraction fromXontoC. Let the mapping$B_i:C\to X$be${\lambda }_i$-strictly pseudocontractive and${\alpha }_i$-strongly accretive with${\alpha }_i+{\lambda }_i\ge 1$for$i=1,2$. Let$f:C\to C$be a contraction with coefficient$\rho \in (0,1)$. Let${\{S_n\}}_{n=0}^{\mathrm{\infty }}$be an infinite family of nonexpansive mappings ofCinto itself such that$F={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(S_i)\cap \Omega \ne \mathrm{\varnothing }$, whereΩis the fixed point set of the mapping$G={\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)$. For arbitrarily given$x_0\in C$, let$\{x_n\}$be the sequence generated by

$$\{\begin{array}{c}{z}_n={\sigma }_{n}G(x_n)+(1-{\sigma }_n)S_{n}G(x_n)\hfill \\ y_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)z_n,\hfill \\ x_{n+1}={\beta }_n{x}_n+{\gamma }_n{y}_n+{\delta }_n{S}_{n}G(y_n),\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(6.1)

where$1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$for$i=1,2$. Suppose that$\{{\sigma }_n\}$, $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$and$\{{\delta }_n\}$are the sequences in$(0,1)$satisfying the following conditions:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\sigma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\sigma }_n<1$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$and${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (iii) ${\beta }_n+{\gamma }_n+{\delta }_n=1$, $\mathrm{\forall }n\ge 0$, and${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$;

• (iv) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}({\beta }_n+{\delta }_n)<1$;

• (v) ${lim}_{n\to \mathrm{\infty }}|{\sigma }_n-{\sigma }_{n-1}|=0$and${lim}_{n\to \mathrm{\infty }}|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|=0$.

Assume that${\sum }_{n=1}^{\mathrm{\infty }}{sup}_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel <\mathrm{\infty }$for any bounded subsetDofC, and letSbe a mapping ofCinto itself defined by$Sx={lim}_{n\to \mathrm{\infty }}{S}_{n}x$for all$x\in C$and suppose that$Fix(S)={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(S_i)$. Then$\{x_n\}$converges strongly to$q\in F$, which solves the following VIP:

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

(6.2)

Proof Take a fixed $p\in F$ arbitrarily. Then by Lemma 2.11 we know that $p=G(p)$ and $S_{n}p=p$ for all $n\ge 0$. Moreover, by Lemma 2.10 we have

$$\begin{array}{rl}\parallel z_n-p\parallel & \le {\sigma }_n\parallel G(x_n)-p\parallel +(1-{\sigma }_n)\parallel S_{n}G(x_n)-p\parallel \\ \le \parallel x_n-p\parallel ,\end{array}$$

(6.3)

and

$$\begin{array}{rl}\parallel y_n-p\parallel & \le {\alpha }_n\parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel f(p)-p\parallel +(1-{\alpha }_n)\parallel z_n-p\parallel \\ \le {\alpha }_n\rho \parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel \\ =(1-{\alpha }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel .\end{array}$$

(6.4)

From (6.1) and (6.4) we obtain

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & \le {\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel y_n-p\parallel +{\delta }_n\parallel S_{n}G(y_n)-p\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)\parallel y_n-p\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)[(1-{\alpha }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel ]\\ =[1-(1-{\beta }_n)(1-\rho ){\alpha }_n]\parallel x_n-p\parallel +(1-{\beta }_n)(1-\rho ){\alpha }_n\cdot \frac{\parallel f(p)-p\parallel }{1-\rho }\\ \le max\{\parallel x_0-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-\rho }\},\end{array}$$

which implies that $\{x_n\}$ is bounded. By Lemma 2.10 we know from (6.3) and (6.4) that $\{y_n\}$, $\{z_n\}$, $\{G(x_n)\}$ and $\{G(y_n)\}$ are bounded.

Let us show that $\parallel x_{n+1}-x_n\parallel \to 0$ and $\parallel x_n-y_n\parallel \to 0$ as $n\to \mathrm{\infty }$. As a matter of fact, from (6.1), we have

$$\begin{array}{rl}{z}_n-z_{n-1}=& {\sigma }_n(G(x_n)-G(x_{n-1}))+({\sigma }_n-{\sigma }_{n-1})(G(x_{n-1})-S_{n-1}G(x_{n-1}))\\ +(1-{\sigma }_n)(S_{n}G(x_n)-S_{n-1}G(x_{n-1})),\end{array}$$

and

$$y_n-y_{n-1}={\alpha }_n(f(x_n)-f(x_{n-1}))+({\alpha }_n-{\alpha }_{n-1})(f(x_{n-1})-z_{n-1})+(1-{\alpha }_n)(z_n-z_{n-1}).$$

It follows that

$$\begin{array}{rl}\parallel z_n-z_{n-1}\parallel \le & {\sigma }_n\parallel G(x_n)-G(x_{n-1})\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +(1-{\sigma }_n)(\parallel S_{n}G(x_n)-S_{n}G(x_{n-1})\parallel +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel )\\ \le & \parallel G(x_n)-G(x_{n-1})\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel ,\end{array}$$

and

$$\begin{array}{rl}\parallel y_n-y_{n-1}\parallel \le & {\alpha }_n\rho \parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-z_{n-1}\parallel \\ +(1-{\alpha }_n)[\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel ]\\ \le & (1-{\alpha }_n(1-\rho ))\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-z_{n-1}\parallel +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel .\end{array}$$

(6.5)

Now, we write $x_n={\beta }_{n-1}{x}_{n-1}+(1-{\beta }_{n-1})v_{n-1}$, $\mathrm{\forall }n\ge 1$, where $v_{n-1}=\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}$. It follows that for all $n\ge 1$,

$$\begin{array}{rl}{v}_n-v_{n-1}=& \frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}\\ =& \frac{{\gamma }_n{y}_n+{\delta }_n{S}_{n}G(y_n)}{1-{\beta }_n}-\frac{{\gamma }_{n-1}{y}_{n-1}+{\delta }_{n-1}{S}_{n-1}G(y_{n-1})}{1-{\beta }_{n-1}}\\ =& \frac{{\gamma }_n(y_n-y_{n-1})+{\delta }_n(S_{n}G(y_n)-S_{n-1}G(y_{n-1}))}{1-{\beta }_n}\\ +(\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}})y_{n-1}+(\frac{{\delta }_n}{1-{\beta }_n}-\frac{{\delta }_{n-1}}{1-{\beta }_{n-1}})S_{n-1}G(y_{n-1}).\end{array}$$

(6.6)

This together with (6.5) implies that

$$\begin{array}{r}\parallel v_n-v_{n-1}\parallel \\ \le \frac{\parallel {\gamma }_n(y_n-y_{n-1})+{\delta }_n(S_{n}G(y_n)-S_{n-1}G(y_{n-1}))\parallel }{1-{\beta }_n}+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\parallel y_{n-1}\parallel \\ +|\frac{{\delta }_n}{1-{\beta }_n}-\frac{{\delta }_{n-1}}{1-{\beta }_{n-1}}|\parallel S_{n-1}G(y_{n-1})\parallel \\ \le \frac{{\gamma }_n\parallel y_n-y_{n-1}\parallel +{\delta }_n(\parallel S_{n}G(y_n)-S_{n}G(y_{n-1})\parallel +\parallel S_{n}G(y_{n-1})-S_{n-1}G(y_{n-1})\parallel )}{1-{\beta }_n}\\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\parallel y_{n-1}\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\parallel S_{n-1}G(y_{n-1})\parallel \\ \le \frac{{\gamma }_n\parallel y_n-y_{n-1}\parallel +{\delta }_n\parallel y_n-y_{n-1}\parallel }{{\gamma }_n+{\delta }_n}+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}\parallel S_{n}G(y_{n-1})-S_{n-1}G(y_{n-1})\parallel \\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|(\parallel y_{n-1}\parallel +\parallel S_{n-1}G(y_{n-1})\parallel )\\ \le \parallel y_n-y_{n-1}\parallel +\parallel S_{n}G(y_{n-1})-S_{n-1}G(y_{n-1})\parallel \\ +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|(\parallel y_{n-1}\parallel +\parallel S_{n-1}G(y_{n-1})\parallel )\\ \le (1-{\alpha }_n(1-\rho ))\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})-z_{n-1}\parallel +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel \\ +\parallel S_{n}G(y_{n-1})-S_{n-1}G(y_{n-1})\parallel +|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|(\parallel y_{n-1}\parallel +\parallel S_{n-1}G(y_{n-1})\parallel )\\ \le \parallel x_n-x_{n-1}\parallel +M(|{\sigma }_n-{\sigma }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|)\\ +\parallel S_{n}G(x_{n-1})-S_{n-1}G(x_{n-1})\parallel +\parallel S_{n}G(y_{n-1})-S_{n-1}G(y_{n-1})\parallel ,\end{array}$$

(6.7)

where ${sup}_{n\ge 0}\{\parallel G(x_n)\parallel +\parallel S_{n}G(x_n)\parallel +\parallel f(x_n)\parallel +\parallel z_n\parallel +\parallel y_n\parallel +\parallel S_n{y}_n\parallel \}\le M$ for some $M>0$. Since $|{\sigma }_n-{\sigma }_{n-1}|\to 0$, ${\alpha }_n\to 0$, $|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|\to 0$ and the assumption on $\{S_n\}$, we have

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

Utilizing Lemma 2.3, from condition (iv) we obtain that

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

So, we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}(1-{\beta }_n)\parallel v_n-x_n\parallel =0.$$

(6.8)

Also, taking into account the boundedness of $\{G(x_n)\}$ and $\{S_{n}G(x_n)\}$, by Lemma 2.5 there exists a continuous strictly increasing function $g_1:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $g_1(0)=0$ such that for $p\in F$,

$$\begin{array}{r}{\parallel z_n-p\parallel }^2\\ \le {\sigma }_n{\parallel G(x_n)-p\parallel }^2+(1-{\sigma }_n){\parallel S_{n}G(x_n)-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)\\ \le {\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right),\end{array}$$

and hence

$$\begin{array}{r}{\parallel y_n-p\parallel }^2\\ \le {\parallel (1-{\alpha }_n)(z_n-p)\parallel }^2+2{\alpha }_n〈f(x_n)-p,J(y_n-p)〉\\ \le {\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel .\end{array}$$

(6.9)

Since $\{x_n\}$ and $\{S_{n}G(y_n)\}$ both are bounded, by Lemma 2.5 there exists a continuous strictly increasing function $g_2:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $g(0)=0$ such that

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ ={\parallel ({\beta }_n+{\delta }_n)(\frac{{\beta }_n}{{\beta }_n+{\delta }_n}(x_n-p)+\frac{{\delta }_n}{{\beta }_n+{\delta }_n}(S_{n}G(y_n)-p))+{\gamma }_n(y_n-p)\parallel }^2\\ \le ({\beta }_n+{\delta }_n){\parallel \frac{{\beta }_n}{{\beta }_n+{\delta }_n}(x_n-p)+\frac{{\delta }_n}{{\beta }_n+{\delta }_n}(S_{n}G(y_n)-p)\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2\\ \le ({\beta }_n+{\delta }_n)[\frac{{\beta }_n}{{\beta }_n+{\delta }_n}{\parallel x_n-p\parallel }^2+\frac{{\delta }_n}{{\beta }_n+{\delta }_n}\parallel S_{n}G(y_n)-p\parallel \\ -\frac{{\beta }_n{\delta }_n}{{({\beta }_n+{\delta }_n)}^2}{g}_2\left(\parallel x_n-S_{n}G(y_n)\parallel \right)]+{\gamma }_n{\parallel y_n-p\parallel }^2\\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\delta }_n{\parallel S_{n}G(y_n)-p\parallel }^2-\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}{g}_2\left(\parallel x_n-S_{n}G(y_n)\parallel \right)+{\gamma }_n{\parallel y_n-p\parallel }^2\\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\delta }_n{\parallel y_n-p\parallel }^2-\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}{g}_2\left(\parallel x_n-S_{n}G(y_n)\parallel \right)+{\gamma }_n{\parallel y_n-p\parallel }^2\\ ={\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel y_n-p\parallel }^2-\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}{g}_2\left(\parallel x_n-S_{n}G(y_n)\parallel \right)\\ \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)[{\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)\\ +2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel ]-\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}{g}_2\left(\parallel x_n-S_{n}G(y_n)\parallel \right)\\ \le {\parallel x_n-p\parallel }^2+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel -(1-{\beta }_n){\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)\\ -\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}{g}_2\left(\parallel x_n-S_{n}G(y_n)\parallel \right),\end{array}$$

which immediately yields

$$\begin{array}{r}(1-{\beta }_n){\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)+{\beta }_n{\delta }_n{g}_2\left(\parallel x_n-S_{n}G(y_n)\parallel \right)\\ \le (1-{\beta }_n){\sigma }_n(1-{\sigma }_n)g_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)+\frac{{\beta }_n{\delta }_n}{{\beta }_n+{\delta }_n}{g}_2\left(\parallel x_n-S_{n}G(y_n)\parallel \right)\\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel \\ \le (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +2{\alpha }_n\parallel f(x_n)-p\parallel \parallel y_n-p\parallel .\end{array}$$

Since ${\alpha }_n\to 0$, $\parallel x_{n+1}-x_n\parallel \to 0$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$, we deduce from conditions (i) and (iv) that

$$\underset{n\to \mathrm{\infty }}{lim}{g}_1\left(\parallel G(x_n)-S_{n}G(x_n)\parallel \right)=0\text{and}\underset{n\to \mathrm{\infty }}{lim}{g}_2\left(\parallel x_n-S_{n}G(y_n)\parallel \right)=0.$$

Utilizing the properties of $g_1$ and $g_2$, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel G(x_n)-S_{n}G(x_n)\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-S_{n}G(y_n)\parallel =0.$$

(6.10)

In the meantime, according to condition (iv), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\gamma }_n=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(1-{\beta }_n-{\delta }_n)=1-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}({\beta }_n+{\delta }_n)>0.$$

Thus, from (6.8) and (6.10) it follows that

$$\begin{array}{rl}{\gamma }_n\parallel y_n-x_n\parallel & =\parallel (x_{n+1}-x_n)-{\delta }_n(S_{n}G(y_n)-x_n)\parallel \\ \le \parallel x_{n+1}-x_n\parallel +{\delta }_n\parallel S_{n}G(y_n)-x_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

That is,

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

(6.11)

Since it follows from (6.1) that

$$\begin{array}{rl}\parallel y_n-G(x_n)\parallel & =\parallel {\alpha }_n(f(x_n)-G(x_n))+(1-{\alpha }_n)(1-{\sigma }_n)(S_{n}G(x_n)-G(x_n))\parallel \\ \le {\alpha }_n\parallel f(x_n)-G(x_n)\parallel +(1-{\alpha }_n)(1-{\sigma }_n)\parallel S_{n}G(x_n)-G(x_n)\parallel \\ \le {\alpha }_n\parallel f(x_n)-G(x_n)\parallel +\parallel S_{n}G(x_n)-G(x_n)\parallel ,\end{array}$$

we obtain from (6.10) and ${\alpha }_n\to 0$ that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-G(x_n)\parallel =0,$$

(6.12)

which together with (6.11) leads to

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

(6.13)

By (6.10) and Lemma 2.6, we have

$$\begin{array}{rl}\parallel SG(x_n)-G(x_n)\parallel & \le \parallel SG(x_n)-S_{n}G(x_n)\parallel +\parallel S_{n}G(x_n)-G(x_n)\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(6.14)

In terms of (6.13) and (6.14), we have

$$\begin{array}{rl}\parallel x_n-S{x}_n\parallel & \le \parallel x_n-G(x_n)\parallel +\parallel G(x_n)-SG(x_n)\parallel +\parallel SG(x_n)-S{x}_n\parallel \\ \le 2\parallel x_n-G(x_n)\parallel +\parallel G(x_n)-SG(x_n)\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(6.15)

Define a mapping $Wx=(1-\theta )Sx+\theta G(x)$, where G is defined by (1.7) and $\theta \in (0,1)$ is a constant. Then by Lemma 2.8 we have that $Fix(W)=Fix(S)\cap Fix(G)=F$. We observe that

$$\begin{array}{rl}\parallel x_n-W{x}_n\parallel & =\parallel (1-\theta )(x_n-S{x}_n)+\theta (x_n-G(x_n))\parallel \\ \le (1-\theta )\parallel x_n-S{x}_n\parallel +\theta \parallel x_n-G(x_n)\parallel .\end{array}$$

From (6.13) and (6.15), we obtain

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

(6.16)

Utilizing the arguments similar to those of (3.17) in the proof of Theorem 3.1, we can obtain

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

(6.17)

where $q=s-{lim}_{t\to 0}{x}_t$ with $x_t$ being the fixed point of the contraction $x\mapsto tf(x)+(1-t)Wx$; that is, $x_t$ solves the fixed point equation $x_t=tf(x_t)+(1-t)W{x}_t$. Noticing that J is norm-to-weak^∗ uniformly continuous on bounded subsets of X, we deduce from (6.11) and (6.17) that

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

(6.18)

Finally, let us show that $x_n\to q$ as $n\to \mathrm{\infty }$. We observe that

$$\begin{array}{r}{\parallel y_n-q\parallel }^2\\ ={\alpha }_n〈f(x_n)-f(q),J(y_n-q)〉+{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ +(1-{\alpha }_n)〈{\sigma }_n(G(x_n)-q)+(1-{\sigma }_n)(S_{n}G(x_n)-q),J(y_n-q)〉\\ \le {\alpha }_n\rho \parallel x_n-q\parallel \parallel y_n-q\parallel +(1-{\alpha }_n)\parallel x_n-q\parallel \parallel y_n-q\parallel +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ =(1-{\alpha }_n(1-\rho ))\parallel x_n-q\parallel \parallel y_n-q\parallel +{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ \le \frac{1-{\alpha }_n(1-\rho )}{2}({\parallel x_n-q\parallel }^2+{\parallel y_n-q\parallel }^2)+{\alpha }_n〈f(q)-q,J(y_n-q)〉\\ \le \frac{1-{\alpha }_n(1-\rho )}{2}{\parallel x_n-q\parallel }^2+\frac{1}{2}{\parallel y_n-q\parallel }^2+{\alpha }_n〈f(q)-q,J(y_n-q)〉,\end{array}$$

which implies that

$${\parallel y_n-q\parallel }^2\le (1-{\alpha }_n(1-\rho )){\parallel x_n-q\parallel }^2+{\alpha }_n(1-\rho )\frac{2〈f(q)-q,J(y_n-q)〉}{1-\rho }.$$

(6.19)

Also, by (6.1) and the convexity of ${\parallel \cdot \parallel }^2$, we get

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2& \le {\beta }_n{\parallel x_n-q\parallel }^2+{\gamma }_n{\parallel y_n-q\parallel }^2+{\delta }_n{\parallel y_n-q\parallel }^2\\ ={\beta }_n{\parallel x_n-q\parallel }^2+(1-{\beta }_n){\parallel y_n-q\parallel }^2.\end{array}$$

It follows from (6.19) that

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2\le & (1-{\alpha }_n(1-{\beta }_n)(1-\rho )){\parallel x_n-q\parallel }^2\\ +{\alpha }_n(1-{\beta }_n)(1-\rho )\frac{2〈f(q)-q,J(y_n-q)〉}{1-\rho }.\end{array}$$

(6.20)

Applying Lemma 2.1 to (6.20), we conclude that $x_n\to q$ as $n\to \mathrm{\infty }$. This completes the proof. □

Corollary 6.1LetCbe a nonempty closed convex subset of a uniformly convex Banach spaceXwhich has a uniformly Gâteaux differentiable norm. Let${\Pi }_C$be a sunny nonexpansive retraction fromXontoC. Let the mapping$B_i:C\to X$be${\lambda }_i$-strictly pseudocontractive and${\alpha }_i$-strongly accretive with${\alpha }_i+{\lambda }_i\ge 1$for$i=1,2$. Let$f:C\to C$be a contraction with coefficient$\rho \in (0,1)$. LetSbe a nonexpansive mapping ofCinto itself such that$F=Fix(S)\cap \Omega \ne \mathrm{\varnothing }$, whereΩis the fixed point set of the mapping$G={\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)$. For arbitrarily given$x_0\in C$, let$\{x_n\}$be the sequence generated by

$$\{\begin{array}{c}{z}_n={\sigma }_{n}G(x_n)+(1-{\sigma }_n)SG(x_n),\hfill \\ y_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)z_n,\hfill \\ x_{n+1}={\beta }_n{x}_n+{\gamma }_n{y}_n+{\delta }_{n}SG(y_n),\mathrm{\forall }n\ge 0,\hfill \end{array}$$

where$1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$for$i=1,2$. Suppose that$\{{\sigma }_n\}$, $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$and$\{{\delta }_n\}$are the sequences in$(0,1)$satisfying the following conditions:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\sigma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\sigma }_n<1$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$and${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (iii) ${\beta }_n+{\gamma }_n+{\delta }_n=1$, $\mathrm{\forall }n\ge 0$, and${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$;

• (iv) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}({\beta }_n+{\delta }_n)<1$;

• (v) ${lim}_{n\to \mathrm{\infty }}|{\sigma }_n-{\sigma }_{n-1}|=0$and${lim}_{n\to \mathrm{\infty }}|\frac{{\gamma }_n}{1-{\beta }_n}-\frac{{\gamma }_{n-1}}{1-{\beta }_{n-1}}|=0$.

Then$\{x_n\}$converges strongly to$q\in F$, which solves the following VIP:

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