Relaxed and composite viscosity methods for variational inequalities, fixed points of nonexpansive mappings and zeros of accretive operators

Abstract

In this paper, we present relaxed and composite viscosity methods for computing a common solution of a general systems of variational inequalities, common fixed points of infinitely many nonexpansive mappings and zeros of accretive operators in real smooth and uniformly convex Banach spaces. The relaxed and composite viscosity methods are based on Korpelevich’s extragradient method, the viscosity approximation method and the Mann iteration method. Under suitable assumptions, we derive some strong convergence theorems for relaxed and composite viscosity algorithms not only in the setting of a uniformly convex and 2-uniformly smooth Banach space but also in a uniformly convex Banach space having a uniformly Gâteaux differentiable norm. The results presented in this paper improve, extend, supplement, and develop the corresponding results given in the literature.

1 Introduction

The theory of variational inequalities is well established and a tool to solve many problems arising from science, engineering, social sciences, etc., see, for example, [1–4] and the references therein. One of the interesting directions, from the research view point, in the theory of variational inequalities is to develop some new iterative methods for computing the approximate solutions of different kinds of variational inequalities. In 1976, Korpelevich [5] proposed an iterative algorithm for solving variational inequalities (VI) in the finite dimensional space setting, It is now known as the extragradient method. Korpelevich’s extragradient method has received great attention by many authors, who improved it in various ways and in different directions, see, for example [6–16] and the references therein. In the recent past, several iterative methods for solving VI were proposed and analyzed in [17–24] in the setting of Banach spaces. In the last three decades, the system of variational inequalities is used as a tool to study the Nash equilibrium problem for a finite or infinite number of players, see, for example, [2, 3, 25, 26] and the references therein. Cai and Bu [20] considered a system of two variational inequalities (SVI) in the setting of real smooth Banach spaces. They proposed and analyzed an iterative method for computing the approximate solutions of system of variational inequalities. Such a solution is also a common fixed point of a family of nonexpansive mappings.

One of the most interesting problems in nonlinear analysis is to find a zero of an accretive operator. In 2007, Aoyama et al. [27] suggested a Halpern type iterative method for finding a common fixed point of a countable family of nonexpansive mappings and a zero of an accretive operator. They studied the strong convergence of the sequence generated by the proposed method in the setting of a uniformly convex Banach space having a uniformly Gâreaux differentiable norm. Ceng et al. [28] introduced and analyzed the composite iterative scheme to compute a zero of m-accretive operator A defined on a uniformly smooth Banach space or a reflexive Banach space having a weakly sequentially continuous duality mapping. It is shown that the iterative process in each case converges strongly to a zero of A. Subsequently, Jung [29] studied a viscosity approximation method, which generalizes the composite method in [28], to investigate the zero of an accretive operator.

During the last decade, several iterative methods have been proposed and analyzed to find a common solution of two different fixed point problems, a fixed point problem and a variational inequality problem, a fixed point problem for a family of nonexpansive mappings and a variational inequality problem or a fixed point problem and a system of variational inequalities, etc. See, for example, [8, 16, 20, 30, 31] and the references therein.

In the present paper, we mainly propose two different methods, namely, relaxed viscosity method and composite viscosity method, to find a common fixed point of an infinite family of nonexpansive mappings, a system of variational inequalities and zero of an accretive operator in the setting of a uniformly convex and 2-uniformly smooth Banach spaces. These methods are based on Korpelevich’s extragradient method, viscosity approximation method and Mann iteration method. Under suitable assumptions, we derive some strong convergence theorems for relaxed and composite viscosity algorithms not only in the setting of a uniformly convex and 2-uniformly smooth Banach space but also in the setting of uniformly convex Banach spaces having a uniformly Gâteaux differentiable norm. The results presented in this paper improve, extend, supplement, and develop the corresponding results in [10, 20, 24, 29, 30].

2 Preliminaries

Throughout the paper, unless otherwise specified, we adopt the following assumptions and notations.

Let X be a real Banach space whose dual space is denoted by $X^{\ast }$. Let C be a nonempty closed convex subset of X. We denote by ${\Xi }_C$ the set of all contractive mappings from C into itself.

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 $x\in X$.

Let $U=\{x\in X:\parallel x\parallel =1\}$ denote the unite sphere in X. A Banach space X is said to be uniformly convex if for each $ϵ\in (0,2]$, there exists $\delta >0$ such that for all $x,y\in U$,

$$\parallel x-y\parallel \ge ϵ\Rightarrow \frac{\parallel x+y\parallel }{2}\le 1-\delta .$$

It is well known that a uniformly convex Banach space is reflexive and strictly convex. A 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 all $x\in U$. Moreover, it is said to be uniformly smooth if this limit is attained uniformly for all $x,y\in U$. The norm of X is said to be Fréchet differentiable if, for each $x\in U$, this limit is attained uniformly for all $y\in U$. A function $\rho :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ defined by

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

is called the modulus of smoothness of X. It is well 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 a 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 [32], no Banach space is q-uniformly smooth for $q>2$. In addition, it is also 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 Gâteaux differentiable norm then the duality mapping J is norm-to-weak^∗ uniformly continuous on bounded subsets of X. For further details of the geometry of Banach spaces, we refer to [33–35].

Now, we present some lemmas which will be used in the sequel.

Lemma 2.1 [36]

Let X be a 2-uniformly smooth Banach space. Then

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

where κ is the 2-uniformly smooth constant of X.

The following lemma is an immediate consequence of the subdifferential inequality of the function $\frac{1}{2}{\parallel \cdot \parallel }^2$.

Lemma 2.2 [37]

Let X be a real Banach space X. Then, for all $x,y\in X$,

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

• (b) ${\parallel x+y\parallel }^2\ge {\parallel x\parallel }^2+2〈y,j(x)〉$, $\mathrm{\forall }j(x)\in J(x)$.

Lemma 2.3 [36]

Given a number $r>0$. A real Banach space X is 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 2.4 [38]

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

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

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

Proposition 2.1 [22]

Let X be a real smooth and uniform convex Banach space and $r>0$. Then there exists a strictly increasing, continuous, and convex function $g:[0,2r]\to \mathbf{R}$, $g(0)=0$ such that

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

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

Lemma 2.5 [39]

Let C be a nonempty closed convex subset of a strictly convex Banach space X. Let ${\{T_n\}}_{n=0}^{\mathrm{\infty }}$ be a sequence of nonexpansive mappings from C into itself such 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 mapping $S:C\to C$ defined by $Sx={\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n{T}_{n}x$, for all $x\in C$, is well defined and nonexpansive, and $Fix(S)={\bigcap }_{n=0}^{\mathrm{\infty }}Fix(T_n)$.

Lemma 2.6 [40]

Let $\{x_n\}$ and $\{z_n\}$ be bounded sequences in a Banach space X and $\{{\beta }_n\}$ be a sequence of nonnegative numbers in $[0,1]$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose that $x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)z_n$ for all integers $n\ge 0$ 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 ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$.

Lemma 2.7 [41]

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

A mapping $T:C\to C$ is called nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for every $x,y\in C$. The set of fixed points of T is denoted by $Fix(T)$. A mapping $A:C\to X$ is said to be

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

  $$〈Ax-Ay,j(x-y)〉\ge 0;$$

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

  $$〈Ax-Ay,j(x-y)〉\ge \alpha {\parallel x-y\parallel }^2,\text{for some }\alpha \in (0,1);$$

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

  $$〈Ax-Ay,j(x-y)〉\ge \beta {\parallel Ax-Ay\parallel }^2,\text{for some }\beta >0;$$

• (d) λ-strictly pseudocontractive [18, 42] if for each $x,y\in C$, there exists $j(x-y)\in J(x-y)$ such that

  $$〈Ax-Ay,j(x-y)〉\le {\parallel x-y\parallel }^2-\lambda {\parallel x-y-(Ax-Ay)\parallel }^2,\text{for some }\lambda \in (0,1).$$

It is worth to emphasize that the definition of the inverse strongly accretive mapping is based on that of the inverse strongly monotone mapping [43].

Lemma 2.8 [[20], Lemma 2.8]

Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space X and for each $i=1,2$, $B_i:C\to X$ be an ${\alpha }_i$-inverse strongly accretive mapping. Then, for each $i=1,2$,

$${\parallel (I-{\mu }_i{B}_i)x-(I-{\mu }_i{B}_i)y\parallel }^2\le {\parallel x-y\parallel }^2+2{\mu }_i({\mu }_i{\kappa }^2-{\alpha }_i){\parallel B_{i}x-B_{i}y\parallel }^2,\mathrm{\forall }x,y\in C,$$

where ${\mu }_i>0$. In particular, if $0<{\mu }_i\le \frac{{\alpha }_i}{{\kappa }^2}$, then $I-{\mu }_i{B}_i$ is nonexpansive for each $i=1,2$.

Let C be a nonempty closed convex subset of a Banach space X and $T:C\to C$ be a nonexpansive mapping with $Fix(T)\ne \mathrm{\varnothing }$. For all $t\in (0,1)$ and $f\in {\Xi }_C$, let $x_t\in C$ be a 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 2.9 [44, 45]

Let X be an uniformly smooth Banach space, or a reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm. Let C be a nonempty closed convex subset of X, $T:C\to C$ 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

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

Recall that a (possibly set-valued mapping) operator $A\subset X\times X$ with domain $D(A)$ and range $R(A)$ in X is accretive if, for each $x_i\in D(A)$ and $y_i\in A{x}_i$ ($i=1,2$), there exists a $j(x_1-x_2)\in J(x_1-x_2)$ such that $〈y_1-y_2,j(x_1-x_2)〉\ge 0$. An accretive operator A is said to satisfy the range condition if $\overline{D(A)}\subset R(I+rA)$ for all $r>0$. An accretive operator A is m-accretive if $R(I+rA)=X$ for each $r>0$. If A is an accretive operator which satisfies the range condition, then we define a mapping $J_r:R(I+rA)\to D(A)$ by $J_r={(I+rA)}^{-1}$ for each $r>0$, which is called the resolvent of A. It is well known that $J_r$ is nonexpansive and $Fix(J_r)=A^{-1}0$ for all $r>0$. Therefore,

$$Fix(J_r)=A^{-1}0=\{z\in D(A):0\in Az\}.$$

If $A^{-1}0\ne \mathrm{\varnothing }$, then the inclusion $0\in Az$ is solvable.

Proposition 2.2 (Resolvent Identity [46])

For $\lambda >0$, $\mu >0$ and $x\in X$,

$$J_{\lambda }x=J_{\mu }(\frac{\mu }{\lambda }x+(1-\frac{\mu }{\lambda })J_{\lambda }x).$$

Let D be a subset of C. A mapping $\Pi :C\to D$ 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 all $x\in C$ and $t\ge 0$. A mapping $\Pi :C\to C$ is called a retraction if ${\Pi }^2=\Pi$. If a mapping $\Pi :C\to C$ 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.

Lemma 2.10 [23]

Let C be a nonempty closed convex subset of a real smooth Banach space X, D be a nonempty subset of C and Π be a retraction of C onto D. Then the following statements are equivalent:

• (a) Π is sunny and nonexpansive;

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

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

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 2.11 [[20], Lemma 2.9]

Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space X and ${\Pi }_C$ be a sunny nonexpansive retraction from X onto C. For each $i=1,2$, let $B_i:C\to X$ be an ${\alpha }_i$-inverse strongly accretive mapping and $G:C\to C$ be defined by

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

If $0<{\mu }_i\le \frac{{\alpha }_i}{{\kappa }^2}$ for each $i=1,2$, then $G:C\to C$ is nonexpansive.

Let $f\in {\Xi }_C$ with a contractive coefficient $\rho \in (0,1)$, ${\{T_n\}}_{n=0}^{\mathrm{\infty }}$ be a sequence of nonexpansive self-mappings on C and ${\{{\lambda }_n\}}_{n=0}^{\mathrm{\infty }}$ be a sequence of nonnegative numbers in $[0,1]$. For any $n\ge 0$, a self-mapping $W_n$ on C defined by

$$\{\begin{array}{c}{U}_{n,n+1}=I,\hfill \\ U_{n,n}={\lambda }_n{T}_n{U}_{n,n+1}+(1-{\lambda }_n)I,\hfill \\ U_{n,n-1}={\lambda }_{n-1}{T}_{n-1}{U}_{n,n}+(1-{\lambda }_{n-1})I,\hfill \\ \cdots \hfill \\ U_{n,k}={\lambda }_k{T}_k{U}_{n,k+1}+(1-{\lambda }_k)I,\hfill \\ U_{n,k-1}={\lambda }_{k-1}{T}_{k-1}{U}_{n,k}+(1-{\lambda }_{k-1})I,\hfill \\ \cdots \hfill \\ U_{n,1}={\lambda }_1{T}_1{U}_{n,2}+(1-{\lambda }_1)I,\hfill \\ W_n=U_{n,0}={\lambda }_0{T}_0{U}_{n,1}+(1-{\lambda }_0)I\hfill \end{array}$$

(2.1)

is called W-mapping [47] generated by $T_n,T_{n-1},\dots ,T_0$ and ${\lambda }_n,{\lambda }_{n-1},\dots ,{\lambda }_0$.

Lemma 2.12 [[37], Lemma 3.2]

Let C be a nonempty closed convex subset of a strictly convex Banach space X. Let ${\{T_n\}}_{n=0}^{\mathrm{\infty }}$ be a sequence of nonexpansive self-mappings on C such that ${\bigcap }_{n=0}^{\mathrm{\infty }}Fix(T_n)\ne \mathrm{\varnothing }$ and ${\{{\lambda }_n\}}_{n=0}^{\mathrm{\infty }}$ be a sequence of positive numbers in $(0,b]$ for some $b\in (0,1)$. Then, for every $x\in C$ and $k\ge 0$, the limit ${lim}_{n\to \mathrm{\infty }}{U}_{n,k}x$ exists.

B using Lemma 2.12, we define a W-mapping $W:C\to C$ generated by the sequences ${\{T_n\}}_{n=0}^{\mathrm{\infty }}$ and ${\{{\lambda }_n\}}_{n=0}^{\mathrm{\infty }}$ by

$$Wx=\underset{n\to \mathrm{\infty }}{lim}{W}_{n}x=\underset{n\to \mathrm{\infty }}{lim}{U}_{n,0}x,\text{for every }x\in C.$$

Throughout this paper, we assume that ${\{{\lambda }_n\}}_{n=0}^{\mathrm{\infty }}$ is a sequence of positive numbers in $(0,b]$ for some $b\in (0,1)$.

Lemma 2.13 [[37], Lemma 3.3]

Let C be a nonempty closed convex subset of a strictly convex Banach space X. Let ${\{T_n\}}_{n=0}^{\mathrm{\infty }}$ be a sequence of nonexpansive self-mappings on C such that ${\bigcap }_{n=0}^{\mathrm{\infty }}Fix(T_n)\ne \mathrm{\varnothing }$ and let ${\{{\lambda }_n\}}_{n=0}^{\mathrm{\infty }}$ be a sequence of positive numbers in $(0,b]$ for some $b\in (0,1)$. Then $Fix(W)={\bigcap }_{n=0}^{\mathrm{\infty }}Fix(T_n)$.

Let μ be a continuous linear functional on $l^{\mathrm{\infty }}$ and $s=(a_0,a_1,\dots )\in l^{\mathrm{\infty }}$. We write ${\mu }_n(a_n)$ instead of $\mu (s)$. μ is called a Banach limit if μ satisfies $\parallel \mu \parallel ={\mu }_n(1)=1$ and ${\mu }_n(a_{n+1})={\mu }_n(a_n)$ for all $(a_0,a_1,\dots )\in l^{\mathrm{\infty }}$. If μ is a Banach limit, then the following implications hold:

• (a) for all $n\ge 0$, $a_n\le c_n$ implies ${\mu }_n(a_n)\le {\mu }_n(c_n)$;

• (b) ${\mu }_n(a_{n+r})={\mu }_n(a_n)$ for any fixed positive integer r;

• (c) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{a}_n\le {\mu }_n(a_n)\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{a}_n$ for all $(a_0,a_1,\dots )\in l^{\mathrm{\infty }}$.

Lemma 2.14 [48]

Let $a\in \mathbf{R}$ be a real number and a sequence $\{a_n\}\in l^{\mathrm{\infty }}$ satisfy the condition ${\mu }_n(a_n)\le a$ for all Banach limits μ. If ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(a_{n+r}-a_n)\le 0$, then ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{a}_n\le a$.

In particular, if $r=1$ in Lemma 2.14, then we obtain the following corollary.

Corollary 2.1 [49]

Let $a\in \mathbf{R}$ be a real number and a sequence $\{a_n\}\in l^{\mathrm{\infty }}$ satisfy the condition ${\mu }_n(a_n)\le a$ for all Banach limits μ. If ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(a_{n+1}-a_n)\le 0$, then ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{a}_n\le a$.

3 Formulations

Let C be a nonempty closed convex subset of a smooth Banach space X, $B_1,B_2:C\to X$ be nonlinear mappings and ${\mu }_1$ and ${\mu }_2$ be two positive constants. The problem of system of variational inequalities (SVI) in the setting of a real smooth Banach space X is to find $(x^{\ast },y^{\ast })\in C\times C$ such that

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

(3.1)

The set of solutions of SVI (3.1) is denoted by $SVI(C,B_1,B_2)$. Very recently, Cai and Bu [20] constructed an iterative algorithm for solving SVI (3.1) and a common fixed point problem of an infinite family of nonexpansive mappings in a uniformly convex and 2-uniformly smooth Banach space. They studied the strong convergence of the proposed algorithm.

In particular, if $X=H$, a real Hilbert space, then SVI (3.1) reduces to the following problem of SVI of finding $(x^{\ast },y^{\ast })\in C\times C$ such that

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

(3.2)

Further, if $B_1=B_2=A$, where $A:C\to X$ is an operator, and $x^{\ast }=y^{\ast }$, then the SVI (3.2) reduces to the classical variational inequality problem (VIP) of finding $x^{\ast }\in C$ such that

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

(3.3)

The solution set of the VIP (3.3) is denoted by $VI(C,A)$. For details and applications of theory of variational inequalities, we refer to [1–4] and the references therein.

Recently, Ceng et al. [10] transformed problem (3.2) into a fixed point problem in the following way.

Lemma 3.1 [10]

For given $\overline{x},\overline{y}\in C,(\overline{x},\overline{y})$ is a solution of problem (3.2) if and only if $\overline{x}$ is a fixed point of the mapping $G:C\to C$ defined by

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

(3.4)

where $\overline{y}=P_C(\overline{x}-{\mu }_2{B}_2\overline{x})$ and $P_C$ is the projection of H onto C.

In particular, if for each $i=1,2$, $B_i:C\to H$ is a ${\beta }_i$-inverse strongly monotone mapping, then G is a nonexpansive mapping provided ${\mu }_i\in (0,2{\beta }_i)$ for each $i=1,2$.

In particular, whenever X is a real smooth Banach space, $B_1\equiv B_2\equiv A$ and $x^{\ast }=y^{\ast }$, then SVI (3.1) reduces to the variational inequality problem (VIP) of finding $x^{\ast }\in C$ such that

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

(3.5)

which was considered by Aoyama et al. [17]. Note that VIP (3.5) is connected with the fixed point problem for nonlinear mapping [44], the problem of finding a zero point of a nonlinear operator [50] and so on. It is clear that VIP (3.5) extends VIP (3.3) from Hilbert spaces to Banach spaces. For further study on VIP in the setting of Banach spaces, we refer to [17, 21] and the references therein.

Define a mapping $G:C\to C$ by

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

(3.6)

The fixed point set of G is denoted by Ω.

Lemma 3.2 Let C be a nonempty closed convex subset of a smooth Banach space X. Let ${\Pi }_C$ be a sunny nonexpansive retraction from X onto C and $B_1,B_2:C\to X$ be nonlinear mappings. Then $(x^{\ast },y^{\ast })\in C\times C$ is a solution of SVI (3.1) 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 rewrite SVI (3.1) 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 2.10. This completes the proof. □

In terms of Lemma 3.2, 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.

Motivated and inspired by the research going on in this area, we introduce some relaxed and composite viscosity methods for finding a zero of an accretive operator $A\subset X\times X$ such that $\overline{D(A)}\subset C\subset {\bigcap }_{r>0}R(I+rA)$, solving SVI (3.1) and the common fixed point problem of an infinite family $\{T_n\}$ of nonexpansive self-mappings on C. Our methods are based on Korpelevich’s extragradient method, the viscosity approximation method, and Mann’s iteration method. Under suitable assumptions, we derive some strong convergence theorems for relaxed and composite viscosity algorithms not only in the setting of uniformly convex and 2-uniformly smooth Banach space but also in a uniformly convex Banach space having a uniformly Gâteaux differentiable norm. The results presented in this paper improve, extend, supplement, and develop the corresponding results given in [10, 20, 24, 29, 48].

4 Relaxed viscosity algorithms and convergence criteria

In this section, we introduce relaxed viscosity algorithms in the setting of real smooth uniformly convex Banach spaces and study the strong convergence of the sequences generated by the proposed algorithms.

Throughout this paper, we denote by Ω the fixed point set of the mapping $G={\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)$.

Assumption 4.1 Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$, $\{{\delta }_n\}$, $\{{\sigma }_n\}$ be the sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n+{\delta }_n=1$ for all $n\ge 0$. Suppose that the following conditions hold:

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

• (ii) $\{{\gamma }_n\},\{{\delta }_n\}\subset [c,d]$ for some $c,d\in (0,1)$;

• (iii) ${lim}_{n\to \mathrm{\infty }}(|{\sigma }_n-{\sigma }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|)=0$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}|r_n-r_{n-1}|<\mathrm{\infty }$ and $r_n\ge \epsilon >0$ for all $n\ge 0$;

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

Theorem 4.1 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let ${\Pi }_C$ be a sunny nonexpansive retraction from X onto C and $A\subset X\times X$ be an accretive operator such that $\overline{D(A)}\subset C\subset {\bigcap }_{r>0}R(I+rA)$. For each $i=1,2$, let $B_i:C\to X$ be ${\alpha }_i$-inverse strongly accretive mapping and $f:C\to C$ be a contraction with coefficient $\rho \in (0,1)$. Let ${\{T_i\}}_{i=0}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings from C into itself such that $F:={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(T_i)\cap \Omega \cap A^{-1}0\ne \mathrm{\varnothing }$ with $0<{\mu }_i<\frac{{\alpha }_i}{{\kappa }^2}$ for $i=1,2$. Assume that Assumption  4.1 holds. For arbitrarily given $x_0\in C$, let $\{x_n\}$ be a sequence generated by

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

(4.1)

where $W_n$ is the W-mapping generated by (2.1). Then

• (a) ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$;

• (b) the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to some $q\in F$ which is a unique solution of the following variational inequality problem (VIP):

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

provided ${\beta }_n\equiv \beta$ for some fixed $\beta \in (0,1)$.

Proof We first claim that the sequence $\{x_n\}$ is bounded. Indeed, take a fixed $p\in F$ arbitrarily. Then we get $p=Gp$, $p=W_{n}p$, and $p=J_{r_n}p$ for all $n\ge 0$. By Lemma 2.11, G is nonexpansive. Then, from (4.1), we have

$$\begin{array}{rcl}\parallel y_n-p\parallel & \le & {\sigma }_n\parallel x_n-p\parallel +(1-{\sigma }_n)\parallel J_{r_n}G{x}_n-p\parallel \\ \le & {\sigma }_n\parallel x_n-p\parallel +(1-{\sigma }_n)\parallel G{x}_n-p\parallel \\ \le & {\sigma }_n\parallel x_n-p\parallel +(1-{\sigma }_n)\parallel x_n-p\parallel \\ =& \parallel x_n-p\parallel \end{array}$$

(4.2)

and

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & \le & {\alpha }_n\parallel f(y_n)-p\parallel +{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel W_n{y}_n-p\parallel +{\delta }_n\parallel J_{r_n}G{y}_n-p\parallel \\ \le & {\alpha }_n(\parallel f(y_n)-f(p)\parallel +\parallel f(p)-p\parallel )+{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel y_n-p\parallel +{\delta }_n\parallel G{y}_n-p\parallel \\ \le & {\alpha }_n(\rho \parallel y_n-p\parallel +\parallel f(p)-p\parallel )+{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel y_n-p\parallel +{\delta }_n\parallel y_n-p\parallel \\ \le & {\alpha }_n(\rho \parallel x_n-p\parallel +\parallel f(p)-p\parallel )+{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel x_n-p\parallel +{\delta }_n\parallel x_n-p\parallel \\ =& (1-{\alpha }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n(1-\rho )\frac{\parallel f(p)-p\parallel }{1-\rho }\\ \le & max\{\parallel x_n-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-\rho }\}.\end{array}$$

By induction, we obtain

$$\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-\rho }\},\mathrm{\forall }n\ge 0.$$

(4.3)

Hence, $\{x_n\}$ is bounded, and so are the sequences $\{y_n\}$, $\{G{x}_n\}$, $\{G{y}_n\}$, and $\{f(y_n)\}$.

Next we show that

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

(4.4)

We note that $x_{n+1}$ can be rewritten as follows:

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

where $z_n=\frac{{\alpha }_{n}f(y_n)+{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{1-{\beta }_n}$. Observe that

$$\begin{array}{c}\parallel z_n-z_{n-1}\parallel \hfill \\ =\parallel \frac{{\alpha }_{n}f(y_n)+{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{1-{\beta }_n}-\frac{{\alpha }_{n-1}f(y_{n-1})+{\gamma }_{n-1}{W}_{n-1}{y}_{n-1}+{\delta }_{n-1}{J}_{r_{n-1}}G{y}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ =\parallel \frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ =\parallel \frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_n}+\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ \le \parallel \frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_n}\parallel +\parallel \frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}\parallel x_{n+1}-{\beta }_n{x}_n-(x_n-{\beta }_{n-1}{x}_{n-1})\parallel +|\frac{1}{1-{\beta }_n}-\frac{1}{1-{\beta }_{n-1}}|\parallel x_n-{\beta }_{n-1}{x}_{n-1}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}\parallel x_{n+1}-{\beta }_n{x}_n-(x_n-{\beta }_{n-1}{x}_{n-1})\parallel +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{x}_{n-1}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}\hfill \\ \times \parallel {\alpha }_{n}f(y_n)+{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n-{\alpha }_{n-1}f(y_{n-1})-{\gamma }_{n-1}{W}_{n-1}{y}_{n-1}-{\delta }_{n-1}{J}_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{x}_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[{\alpha }_n\parallel f(y_n)-f(y_{n-1})\parallel +{\gamma }_n\parallel W_n{y}_n-W_{n-1}{y}_{n-1}\parallel +{\delta }_n\parallel J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel ]\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{x}_{n-1}\parallel .\hfill \end{array}$$

(4.5)

On the other hand, if $r_{n-1}\le r_n$, using the resolvent identity in Proposition 2.2,

$$J_{r_n}{x}_n=J_{r_{n-1}}(\frac{r_{n-1}}{r_n}{x}_n+(1-\frac{r_{n-1}}{r_n})J_{r_n}{x}_n),$$

we get

$$\begin{array}{rcl}\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel & =& \parallel J_{r_{n-1}}(\frac{r_{n-1}}{r_n}G{x}_n+(1-\frac{r_{n-1}}{r_n})J_{r_n}G{x}_n)-J_{r_{n-1}}G{x}_{n-1}\parallel \\ \le & \frac{r_{n-1}}{r_n}\parallel G{x}_n-G{x}_{n-1}\parallel +(1-\frac{r_{n-1}}{r_n})\parallel J_{r_n}G{x}_n-G{x}_{n-1}\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +\frac{r_n-r_{n-1}}{r_n}\parallel J_{r_n}G{x}_n-G{x}_{n-1}\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +\frac{1}{\epsilon }|r_n-r_{n-1}|\parallel J_{r_n}G{x}_n-G{x}_{n-1}\parallel .\end{array}$$

If $r_n\le r_{n-1}$, then it is easy to see that

$$\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel \le \parallel x_{n-1}-x_n\parallel +\frac{1}{\epsilon }|r_{n-1}-r_n|\parallel J_{r_{n-1}}G{x}_{n-1}-G{x}_n\parallel .$$

By combining the above cases, we obtain

$$\begin{array}{c}\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel \hfill \\ \le \parallel x_{n-1}-x_n\parallel +\frac{|r_{n-1}-r_n|}{\epsilon }\underset{n\ge 1}{sup}\{\parallel J_{r_n}G{x}_n-G{x}_{n-1}\parallel +\parallel J_{r_{n-1}}G{x}_{n-1}-G{x}_n\parallel \},\mathrm{\forall }n\ge 1.\hfill \end{array}$$

Similarly, we have

$$\begin{array}{c}\parallel J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ \le \parallel y_{n-1}-y_n\parallel +\frac{|r_{n-1}-r_n|}{\epsilon }\underset{n\ge 1}{sup}\{\parallel J_{r_n}G{y}_n-G{y}_{n-1}\parallel +\parallel J_{r_{n-1}}G{y}_{n-1}-G{y}_n\parallel \},\mathrm{\forall }n\ge 1.\hfill \end{array}$$

Therefore, we obtain

$$\{\begin{array}{c}\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel \le \parallel x_{n-1}-x_n\parallel +|r_{n-1}-r_n|M_0,\hfill \\ \parallel J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}\parallel \le \parallel y_{n-1}-y_n\parallel +|r_{n-1}-r_n|M_0,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(4.6)

where

$$\underset{n\ge 1}{sup}\left\{\frac{1}{\epsilon }(\parallel J_{r_n}G{x}_n-G{x}_{n-1}\parallel +\parallel J_{r_{n-1}}G{x}_{n-1}-G{x}_n\parallel )\right\}\le M_0,$$

and

$$\underset{n\ge 1}{sup}\left\{\frac{1}{\epsilon }(\parallel J_{r_n}G{y}_n-G{y}_{n-1}\parallel +\parallel J_{r_{n-1}}G{y}_{n-1}-G{y}_n\parallel )\right\}\le M_0,$$

for some $M_0>0$. Since $T_i$ and $U_{n,i}$ are nonexpansive, from (2.1), we deduce that for each $n\ge 1$

$$\begin{array}{c}\parallel W_n{y}_{n-1}-W_{n-1}{y}_{n-1}\parallel =\parallel {\lambda }_0{T}_0{U}_{n,1}{y}_{n-1}-{\lambda }_0{T}_0{U}_{n-1,1}{y}_{n-1}\parallel \hfill \\ \phantom{\parallel W_n{y}_{n-1}-W_{n-1}{y}_{n-1}\parallel }\le {\lambda }_0\parallel U_{n,1}{y}_{n-1}-U_{n-1,1}{y}_{n-1}\parallel \hfill \\ \phantom{\parallel W_n{y}_{n-1}-W_{n-1}{y}_{n-1}\parallel }={\lambda }_0\parallel {\lambda }_1{T}_1{U}_{n,2}{y}_{n-1}-{\lambda }_1{T}_1{U}_{n-1,2}{y}_{n-1}\parallel \hfill \\ \phantom{\parallel W_n{y}_{n-1}-W_{n-1}{y}_{n-1}\parallel }\le {\lambda }_0{\lambda }_1\parallel U_{n,2}{y}_{n-1}-U_{n-1,2}{y}_{n-1}\parallel \hfill \\ \phantom{\parallel W_n{y}_{n-1}-W_{n-1}{y}_{n-1}\parallel }\cdots \hfill \\ \phantom{\parallel W_n{y}_{n-1}-W_{n-1}{y}_{n-1}\parallel }\le \left(\prod _{i=0}^{n-1}{\lambda }_i\right)\parallel U_{n,n}{y}_{n-1}-U_{n-1,n}{y}_{n-1}\parallel \hfill \\ \phantom{\parallel W_n{y}_{n-1}-W_{n-1}{y}_{n-1}\parallel }\le M\prod _{i=0}^{n-1}{\lambda }_i,\text{for some constant }M>0.\hfill \end{array}$$

(4.7)

By simple computations, we obtain

$$\begin{array}{rcl}{y}_n-y_{n-1}& =& {\sigma }_n(x_n-x_{n-1})+({\sigma }_n-{\sigma }_{n-1})(x_{n-1}-J_{r_{n-1}}G{x}_{n-1})\\ +(1-{\sigma }_n)(J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}).\end{array}$$

It follows from (4.6) that

$$\begin{array}{rcl}\parallel y_n-y_{n-1}\parallel & \le & {\sigma }_n\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel x_{n-1}-J_{r_{n-1}}G{x}_{n-1}\parallel \\ +(1-{\sigma }_n)\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel \\ \le & {\sigma }_n\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel x_{n-1}-J_{r_{n-1}}G{x}_{n-1}\parallel \\ +(1-{\sigma }_n)[\parallel x_{n-1}-x_n\parallel +|r_{n-1}-r_n|M_0]\\ \le & \parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel x_{n-1}-J_{r_{n-1}}G{x}_{n-1}\parallel +|r_{n-1}-r_n|M_0.\end{array}$$

(4.8)

Taking into account that $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$, without loss of generality, we may assume that $\{{\beta }_n\}\subset [\hat{c},\hat{d}]$. Utilizing (4.5)-(4.8), we have

$$\begin{array}{c}\parallel z_n-z_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[{\alpha }_n\parallel f(y_n)-f(y_{n-1})\parallel +{\gamma }_n\parallel W_n{y}_n-W_{n-1}{y}_{n-1}\parallel +{\delta }_n\parallel J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel ]\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{x}_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[{\alpha }_n\parallel f(y_n)-f(y_{n-1})\parallel +{\gamma }_n\parallel W_n{y}_n-W_n{y}_{n-1}\parallel +{\delta }_n\parallel J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ +{\gamma }_n\parallel W_n{y}_{n-1}-W_{n-1}{y}_{n-1}\parallel ]+\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{x}_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[{\alpha }_n\rho \parallel y_n-y_{n-1}\parallel +{\gamma }_n\parallel y_n-y_{n-1}\parallel +{\delta }_n(\parallel y_{n-1}-y_n\parallel +|r_{n-1}-r_n|M_0)\hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ +{\gamma }_{n}M\prod _{i=0}^{n-1}{\lambda }_i]+\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{x}_{n-1}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}[(1-{\beta }_n-{\alpha }_n(1-\rho ))\parallel y_n-y_{n-1}\parallel \hfill \\ +\frac{1}{1-{\beta }_n}[{\delta }_n|r_{n-1}-r_n|M_0+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel ]\hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel +{\gamma }_{n}M\prod _{i=0}^{n-1}{\lambda }_i]\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel {\alpha }_{n-1}f(y_{n-1})+{\gamma }_{n-1}{W}_{n-1}{y}_{n-1}+J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ =(1-\frac{{\alpha }_n(1-\rho )}{1-{\beta }_n})\parallel y_n-y_{n-1}\parallel +\frac{1}{1-{\beta }_n}[{\delta }_n|r_{n-1}-r_n|M_0+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel \hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel +{\gamma }_{n}M\prod _{i=0}^{n-1}{\lambda }_i]\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel {\alpha }_{n-1}f(y_{n-1})+{\gamma }_{n-1}{W}_{n-1}{y}_{n-1}+J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ \le \parallel y_n-y_{n-1}\parallel +\frac{1}{1-{\beta }_n}[{\delta }_n|r_{n-1}-r_n|M_0+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel \hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel +{\gamma }_{n}M\prod _{i=0}^{n-1}{\lambda }_i]\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel {\alpha }_{n-1}f(y_{n-1})+{\gamma }_{n-1}{W}_{n-1}{y}_{n-1}+J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ \le \parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel x_{n-1}-J_{r_{n-1}}G{x}_{n-1}\parallel +|r_{n-1}-r_n|M_0\hfill \\ +\frac{1}{1-{\beta }_n}[{\delta }_n|r_{n-1}-r_n|M_0+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel \hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel +{\gamma }_{n}M\prod _{i=0}^{n-1}{\lambda }_i]\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel {\alpha }_{n-1}f(y_{n-1})+{\gamma }_{n-1}{W}_{n-1}{y}_{n-1}+{\delta }_{n-1}{J}_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ \le \parallel x_n-x_{n-1}\parallel +[|{\sigma }_n-{\sigma }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|\hfill \\ +|{\delta }_n-{\delta }_{n-1}|+\prod _{i=0}^{n-1}{\lambda }_i]M_1,\hfill \end{array}$$

(4.9)

where ${sup}_{n\ge 0}\{\frac{1}{{(1-\hat{d})}^2}(\parallel f(y_n)\parallel +\parallel W_n{y}_n\parallel +\parallel J_{r_n}G{y}_n\parallel +\parallel x_n-J_{r_n}G{x}_n\parallel +M+2M_0)\}\le M_1$ for some $M_1>0$. Thus, it follows from (4.9) and conditions (i), (iii), (iv) that

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

Since $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$, by Lemma 2.6, we get

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

Consequently,

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

Now we show that $\parallel x_n-G{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$. Indeed, by Lemma 2.3 and (4.1), we get

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel {\sigma }_n(x_n-p)+(1-{\sigma }_n)(J_{r_n}G{x}_n-p)\parallel }^2\\ \le & {\sigma }_n{\parallel x_n-p\parallel }^2+(1-{\sigma }_n){\parallel J_{r_n}G{x}_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel x_n-J_{r_n}G{x}_n\parallel )\\ \le & {\sigma }_n{\parallel x_n-p\parallel }^2+(1-{\sigma }_n){\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel x_n-J_{r_n}G{x}_n\parallel )\\ =& {\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel x_n-J_{r_n}G{x}_n\parallel ).\end{array}$$

(4.10)

By Lemma 2.2(a), (4.1), and (4.10), we obtain

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ ={\parallel {\alpha }_n(f(y_n)-f(p))+{\beta }_n(x_n-p)+{\gamma }_n(W_n{y}_n-p)+{\delta }_n(J_{r_n}G{y}_n-p)+{\alpha }_n(f(p)-p)\parallel }^2\hfill \\ \le {\parallel {\alpha }_n(f(y_n)-f(p))+{\beta }_n(x_n-p)+{\gamma }_n(W_n{y}_n-p)+{\delta }_n(J_{r_n}G{y}_n-p)\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(x_{n+1}-p)〉\hfill \\ \le {\alpha }_n{\parallel f(y_n)-f(p)\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel W_n{y}_n-p\parallel }^2+{\delta }_n{\parallel J_{r_n}G{y}_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(x_{n+1}-p)〉\hfill \\ \le {\alpha }_n{\rho }^2{\parallel y_n-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2+{\delta }_n{\parallel G{y}_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(x_{n+1}-p)〉\hfill \\ \le {\alpha }_n\rho {\parallel y_n-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2+{\delta }_n{\parallel y_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(x_{n+1}-p)〉\hfill \\ =(1-{\beta }_n-{\alpha }_n(1-\rho )){\parallel y_n-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+2{\alpha }_n〈f(p)-p,J(x_{n+1}-p)〉\hfill \\ \le (1-{\beta }_n-{\alpha }_n(1-\rho ))[{\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel x_n-J_{r_n}G{x}_n\parallel )]+{\beta }_n{\parallel x_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(x_{n+1}-p)〉\hfill \\ =(1-{\alpha }_n(1-\rho )){\parallel x_n-p\parallel }^2-(1-{\beta }_n-{\alpha }_n(1-\rho )){\sigma }_n(1-{\sigma }_n)g(\parallel x_n-J_{r_n}G{x}_n\parallel )\hfill \\ +2{\alpha }_n〈f(p)-p,J(x_{n+1}-p)〉\hfill \\ \le {\parallel x_n-p\parallel }^2-(1-{\beta }_n-{\alpha }_n(1-\rho )){\sigma }_n(1-{\sigma }_n)g(\parallel x_n-J_{r_n}G{x}_n\parallel )\hfill \\ +2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel ,\hfill \end{array}$$

and thus

$$\begin{array}{c}(1-{\beta }_n-{\alpha }_n(1-\rho )){\sigma }_n(1-{\sigma }_n)g(\parallel x_n-J_{r_n}G{x}_n\parallel )\hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel .\hfill \end{array}$$

Since ${\alpha }_n\to 0$ and $\parallel x_{n+1}-x_n\parallel \to 0$, from condition (v) and the boundedness of $\{x_n\}$, it follows that

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

Utilizing the properties of g, we have

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

(4.11)

and thus,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}(1-{\sigma }_n)\parallel J_{r_n}G{x}_n-x_n\parallel =0.$$

(4.12)

For simplicity, we put $q={\Pi }_C(p-{\mu }_2{B}_{2}p)$, $u_n={\Pi }_C(x_n-{\mu }_2{B}_2{x}_n)$ and $v_n={\Pi }_C(u_n-{\mu }_1{B}_1{u}_n)$. Then $v_n=G{x}_n$ for all $n\ge 0$. From Lemma 2.8, we have

$$\begin{array}{rcl}{\parallel u_n-q\parallel }^2& =& {\parallel {\Pi }_C(x_n-{\mu }_2{B}_2{x}_n)-{\Pi }_C(p-{\mu }_2{B}_{2}p)\parallel }^2\\ \le & {\parallel x_n-p-{\mu }_2(B_2{x}_n-B_{2}p)\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-2{\mu }_2({\alpha }_2-{\kappa }^2{\mu }_2){\parallel B_2{x}_n-B_{2}p\parallel }^2,\end{array}$$

(4.13)

and

$$\begin{array}{rcl}{\parallel v_n-p\parallel }^2& =& {\parallel {\Pi }_C(u_n-{\mu }_1{B}_1{u}_n)-{\Pi }_C(q-{\mu }_1{B}_{1}q)\parallel }^2\\ \le & {\parallel u_n-q-{\mu }_1(B_1{u}_n-B_{1}q)\parallel }^2\\ \le & {\parallel u_n-q\parallel }^2-2{\mu }_1({\alpha }_1-{\kappa }^2{\mu }_1){\parallel B_1{u}_n-B_{1}q\parallel }^2.\end{array}$$

(4.14)

By combining (4.13) and (4.14), we obtain

$$\begin{array}{rcl}{\parallel v_n-p\parallel }^2& \le & {\parallel x_n-p\parallel }^2-2{\mu }_2({\alpha }_2-{\kappa }^2{\mu }_2){\parallel B_2{x}_n-B_{2}p\parallel }^2\\ -2{\mu }_1({\alpha }_1-{\kappa }^2{\mu }_1){\parallel B_1{u}_n-B_{1}q\parallel }^2.\end{array}$$

(4.15)

By the convexity of ${\parallel \cdot \parallel }^2$, we have, from (4.1) and (4.15),

$$\begin{array}{c}{\parallel y_n-p\parallel }^2\hfill \\ \le {\sigma }_n{\parallel x_n-p\parallel }^2+(1-{\sigma }_n){\parallel J_{r_n}G{x}_n-p\parallel }^2\hfill \\ \le {\sigma }_n{\parallel x_n-p\parallel }^2+(1-{\sigma }_n){\parallel v_n-p\parallel }^2\hfill \\ \le {\sigma }_n{\parallel x_n-p\parallel }^2+(1-{\sigma }_n)[{\parallel x_n-p\parallel }^2-2{\mu }_2({\alpha }_2-{\kappa }^2{\mu }_2){\parallel B_2{x}_n-B_{2}p\parallel }^2\hfill \\ -2{\mu }_1({\alpha }_1-{\kappa }^2{\mu }_1){\parallel B_1{u}_n-B_{1}q\parallel }^2]\hfill \\ ={\parallel x_n-p\parallel }^2-2(1-{\sigma }_n)[{\mu }_2({\alpha }_2-{\kappa }^2{\mu }_2){\parallel B_2{x}_n-B_{2}p\parallel }^2\hfill \\ +{\mu }_1({\alpha }_1-{\kappa }^2{\mu }_1){\parallel B_1{u}_n-B_{1}q\parallel }^2],\hfill \end{array}$$

and thus

$$\begin{array}{c}2(1-{\sigma }_n)[{\mu }_2({\alpha }_2-{\kappa }^2{\mu }_2){\parallel B_2{x}_n-B_{2}p\parallel }^2+{\mu }_1({\alpha }_1-{\kappa }^2{\mu }_1){\parallel B_1{u}_n-B_{1}q\parallel }^2]\hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel y_n-p\parallel }^2\hfill \\ \le (\parallel x_n-p\parallel +\parallel y_n-p\parallel )\parallel x_n-y_n\parallel .\hfill \end{array}$$

Since $\parallel x_n-y_n\parallel \to 0$ and $0<{\mu }_i<\frac{{\alpha }_i}{{\kappa }^2}$ for $i=1,2$, and $\{x_n\}$ and $\{y_n\}$ are bounded, we obtain from condition (v) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel B_2{x}_n-B_{2}p\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel B_1{u}_n-B_{1}q\parallel =0.$$

(4.16)

Utilizing Proposition 2.2 and Lemma 2.10, we have

$$\begin{array}{rcl}{\parallel u_n-q\parallel }^2& =& {\parallel {\Pi }_C(x_n-{\mu }_2{B}_2{x}_n)-{\Pi }_C(p-{\mu }_2{B}_{2}p)\parallel }^2\\ \le & 〈x_n-{\mu }_2{B}_2{x}_n-(p-{\mu }_2{B}_{2}p),J(u_n-q)〉\\ =& 〈x_n-p,J(u_n-q)〉+{\mu }_2〈B_{2}p-B_2{x}_n,J(u_n-q)〉\\ \le & \frac{1}{2}[{\parallel x_n-p\parallel }^2+{\parallel u_n-q\parallel }^2-g_1\left(\parallel x_n-u_n-(p-q)\parallel \right)]\\ +{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel ,\end{array}$$

which implies that

$${\parallel u_n-q\parallel }^2\le {\parallel x_n-p\parallel }^2-g_1\left(\parallel x_n-u_n-(p-q)\parallel \right)+2{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel .$$

(4.17)

In the same way, we derive

$$\begin{array}{rcl}{\parallel v_n-p\parallel }^2& =& {\parallel {\Pi }_C(u_n-{\mu }_1{B}_1{u}_n)-{\Pi }_C(q-{\mu }_1{B}_{1}q)\parallel }^2\\ \le & 〈u_n-{\mu }_1{B}_1{u}_n-(q-{\mu }_1{B}_{1}q),J(v_n-p)〉\\ =& 〈u_n-q,J(v_n-p)〉+{\mu }_1〈B_{1}q-B_1{u}_n,J(v_n-p)〉\\ \le & \frac{1}{2}[{\parallel u_n-q\parallel }^2+{\parallel v_n-p\parallel }^2-g_2\left(\parallel u_n-v_n+(p-q)\parallel \right)]\\ +{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel ,\end{array}$$

and we get

$${\parallel v_n-p\parallel }^2\le {\parallel u_n-q\parallel }^2-g_2\left(\parallel u_n-v_n+(p-q)\parallel \right)+2{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel .$$

(4.18)

Combining (4.17) and (4.18), we get

$$\begin{array}{rcl}{\parallel v_n-p\parallel }^2& \le & {\parallel x_n-p\parallel }^2-g_1\left(\parallel x_n-u_n-(p-q)\parallel \right)-g_2\left(\parallel u_n-v_n+(p-q)\parallel \right)\\ +2{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel +2{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel .\end{array}$$

(4.19)

By the convexity of ${\parallel \cdot \parallel }^2$, we have, from (4.1) and (4.19),

$$\begin{array}{c}{\parallel y_n-p\parallel }^2\hfill \\ \le {\sigma }_n{\parallel x_n-p\parallel }^2+(1-{\sigma }_n){\parallel J_{r_n}G{x}_n-p\parallel }^2\hfill \\ \le {\sigma }_n{\parallel x_n-p\parallel }^2+(1-{\sigma }_n){\parallel v_n-p\parallel }^2\hfill \\ \le {\sigma }_n{\parallel x_n-p\parallel }^2+(1-{\sigma }_n)[{\parallel x_n-p\parallel }^2-g_1\left(\parallel x_n-u_n-(p-q)\parallel \right)\hfill \\ -g_2\left(\parallel u_n-v_n+(p-q)\parallel \right)+2{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel \hfill \\ +2{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel ]\hfill \\ \le {\parallel x_n-p\parallel }^2-(1-{\sigma }_n)[g_1\left(\parallel x_n-u_n-(p-q)\parallel \right)+g_2\left(\parallel u_n-v_n+(p-q)\parallel \right)]\hfill \\ +2{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel +2{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel ,\hfill \end{array}$$

and hence

$$\begin{array}{c}(1-{\sigma }_n)[g_1\left(\parallel x_n-u_n-(p-q)\parallel \right)+g_2\left(\parallel u_n-v_n+(p-q)\parallel \right)]\hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel y_n-p\parallel }^2+2{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel +2{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel \hfill \\ \le (\parallel x_n-p\parallel +\parallel y_n-p\parallel )\parallel x_n-y_n\parallel +2{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel \hfill \\ +2{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel .\hfill \end{array}$$

From (4.12), (4.16), condition (v), and the boundedness of $\{x_n\}$, $\{y_n\}$, $\{u_n\}$, and $\{v_n\}$, we deduce

$$\underset{n\to \mathrm{\infty }}{lim}{g}_1\left(\parallel x_n-u_n-(p-q)\parallel \right)=0\text{and}\underset{n\to \mathrm{\infty }}{lim}{g}_2\left(\parallel u_n-v_n+(p-q)\parallel \right)=0.$$

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-u_n-(p-q)\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-v_n+(p-q)\parallel =0.$$

(4.20)

Hence,

$$\parallel x_n-v_n\parallel \le \parallel x_n-u_n-(p-q)\parallel +\parallel u_n-v_n+(p-q)\parallel \to 0\text{as }n\to \mathrm{\infty },$$

that is,

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

(4.21)

Next, we show that

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

Indeed, observe that $x_{n+1}$ can be rewritten as

$$\begin{array}{rcl}{x}_{n+1}& =& {\alpha }_{n}f(y_n)+{\beta }_n{x}_n+{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n\\ =& {\alpha }_{n}f(y_n)+{\beta }_n{x}_n+({\gamma }_n+{\delta }_n)\frac{{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{{\gamma }_n+{\delta }_n}\\ =& {\alpha }_{n}f(y_n)+{\beta }_n{x}_n+e_n{\hat{z}}_n,\end{array}$$

(4.22)

where $e_n={\gamma }_n+{\delta }_n$ and ${\hat{z}}_n=\frac{{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{{\gamma }_n+{\delta }_n}$. Utilizing Lemma 2.4 and (4.22), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel {\alpha }_n(f(y_n)-p)+{\beta }_n(x_n-p)+e_n({\hat{z}}_n-p)\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+e_n{\parallel {\hat{z}}_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\\ =& {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\\ +e_n{\parallel \frac{{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{{\gamma }_n+{\delta }_n}-p\parallel }^2\\ =& {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\\ +e_n{\parallel \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(W_n{y}_n-p)+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}(J_{r_n}G{y}_n-p)\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\\ +e_n[\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}{\parallel W_n{y}_n-p\parallel }^2+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel J_{r_n}G{y}_n-p\parallel }^2]\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\\ +e_n[\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}\parallel y_n-p\parallel +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel y_n-p\parallel }^2]\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\\ +e_n[\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}\parallel x_n-p\parallel +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel x_n-p\parallel }^2]\\ =& {\alpha }_n{\parallel f(y_n)-p\parallel }^2+(1-{\alpha }_n){\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel ),\end{array}$$

which implies that

$$\begin{array}{rcl}{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )& \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel .\end{array}$$

Utilizing (4.4), conditions (i), (ii), (v), and the boundedness of $\{x_n\}$ and $\{f(y_n)\}$, we obtain

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

From the properties of $g_3$, we have

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

Utilizing Lemma 2.3 and the definition of ${\hat{z}}_n$, we have

$$\begin{array}{rcl}{\parallel {\hat{z}}_n-p\parallel }^2& =& {\parallel \frac{{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{{\gamma }_n+{\delta }_n}-p\parallel }^2\\ =& {\parallel \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(W_n{y}_n-p)+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}(J_{r_n}G{y}_n-p)\parallel }^2\\ \le & \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}{\parallel W_n{y}_n-p\parallel }^2+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel J_{r_n}G{y}_n-p\parallel }^2\\ -\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_4(\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel )\\ \le & {\parallel y_n-p\parallel }^2-\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_4(\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel )\\ \le & {\parallel x_n-p\parallel }^2-\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_4(\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel ),\end{array}$$

and thus

$$\begin{array}{rcl}\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_4(\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel )& \le & {\parallel x_n-p\parallel }^2-{\parallel {\hat{z}}_n-p\parallel }^2\\ \le & (\parallel x_n-p\parallel +\parallel {\hat{z}}_n-p\parallel )\parallel x_n-{\hat{z}}_n\parallel .\end{array}$$

Since $\{x_n\}$ and $\{{\hat{z}}_n\}$ are bounded and $\parallel {\hat{z}}_n-x_n\parallel \to 0$ as $n\to \mathrm{\infty }$, we deduce from condition (ii) that

$$\underset{n\to \mathrm{\infty }}{lim}{g}_4(\parallel W_n{y}_n-J_{r_n}G{y}_n\parallel )=0.$$

From the properties of $g_4$, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel W_n{y}_n-J_{r_n}G{y}_n\parallel =0.$$

(4.23)

On the other hand, $x_{n+1}$ can also be rewritten as

$$\begin{array}{rcl}{x}_{n+1}& =& {\alpha }_{n}f(y_n)+{\beta }_n{x}_n+{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n\\ =& {\beta }_n{x}_n+{\gamma }_n{W}_n{y}_n+({\alpha }_n+{\delta }_n)\frac{{\alpha }_{n}f(y_n)+{\delta }_n{J}_{r_n}G{y}_n}{{\alpha }_n+{\delta }_n}\\ =& {\beta }_n{x}_n+{\gamma }_n{W}_n{y}_n+d_n{\tilde{z}}_n,\end{array}$$

where $d_n={\alpha }_n+{\delta }_n$ and ${\tilde{z}}_n=\frac{{\alpha }_{n}f(y_n)+{\delta }_n{J}_{r_n}G{y}_n}{{\alpha }_n+{\delta }_n}$. Utilizing Lemma 2.4 and the convexity of ${\parallel \cdot \parallel }^2$, we have

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ ={\parallel {\beta }_n(x_n-p)+{\gamma }_n(W_n{y}_n-p)+d_n({\tilde{z}}_n-p)\parallel }^2\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel W_n{y}_n-p\parallel }^2+d_n{\parallel {\tilde{z}}_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_5(\parallel x_n-W_n{y}_n\parallel )\hfill \\ ={\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel W_n{y}_n-p\parallel }^2+d_n{\parallel \frac{{\alpha }_{n}f(y_n)+{\delta }_n{J}_{r_n}G{y}_n}{{\alpha }_n+{\delta }_n}-p\parallel }^2\hfill \\ -{\beta }_n{\gamma }_n{g}_5(\parallel x_n-W_n{y}_n\parallel )\hfill \\ ={\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel W_n{y}_n-p\parallel }^2+d_n{\parallel \frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}(f(y_n)-p)+\frac{{\delta }_n}{{\alpha }_n+{\delta }_n}(J_{r_n}G{y}_n-p)\parallel }^2\hfill \\ -{\beta }_n{\gamma }_n{g}_5(\parallel x_n-W_n{y}_n\parallel )\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2+d_n[\frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}{\parallel f(y_n)-p\parallel }^2+\frac{{\delta }_n}{{\alpha }_n+{\delta }_n}{\parallel J_{r_n}G{y}_n-p\parallel }^2]\hfill \\ -{\beta }_n{\gamma }_n{g}_5(\parallel x_n-W_n{y}_n\parallel )\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2+d_n[\frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}{\parallel f(y_n)-p\parallel }^2+\frac{{\delta }_n}{{\alpha }_n+{\delta }_n}{\parallel y_n-p\parallel }^2]\hfill \\ -{\beta }_n{\gamma }_n{g}_5(\parallel x_n-W_n{y}_n\parallel )\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+({\beta }_n+{\gamma }_n){\parallel x_n-p\parallel }^2+{\delta }_n{\parallel x_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_5(\parallel x_n-W_n{y}_n\parallel )\hfill \\ ={\alpha }_n{\parallel f(y_n)-p\parallel }^2+(1-{\alpha }_n){\parallel x_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_5(\parallel x_n-W_n{y}_n\parallel )\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_5(\parallel x_n-W_n{y}_n\parallel ),\hfill \end{array}$$

which implies that

$$\begin{array}{rcl}{\beta }_n{\gamma }_n{g}_5(\parallel x_n-W_n{y}_n\parallel )& \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel .\end{array}$$

From (4.4), conditions (i), (ii), (v), and the boundedness of $\{x_n\}$ and $\{f(y_n)\}$, we have

$$\underset{n\to \mathrm{\infty }}{lim}{g}_5(\parallel x_n-W_n{y}_n\parallel )=0.$$

Utilizing the properties of $g_5$, we have

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

(4.24)

which together with (4.12) and (4.24), implies that

$$\begin{array}{rcl}\parallel x_n-W_n{x}_n\parallel & \le & \parallel x_n-W_n{y}_n\parallel +\parallel W_n{y}_n-W_n{x}_n\parallel \\ \le & \parallel x_n-W_n{y}_n\parallel +\parallel y_n-x_n\parallel \to 0\text{as }n\to \mathrm{\infty },\end{array}$$

that is,

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

(4.25)

We note that

$$\begin{array}{c}\parallel x_n-J_{r_n}{x}_n\parallel \hfill \\ \le \parallel x_n-W_n{y}_n\parallel +\parallel W_n{y}_n-J_{r_n}G{y}_n\parallel +\parallel J_{r_n}G{y}_n-J_{r_n}G{x}_n\parallel +\parallel J_{r_n}G{x}_n-J_{r_n}{x}_n\parallel \hfill \\ \le \parallel x_n-W_n{y}_n\parallel +\parallel W_n{y}_n-J_{r_n}G{y}_n\parallel +\parallel y_n-x_n\parallel +\parallel G{x}_n-x_n\parallel .\hfill \end{array}$$

Thus, from (4.12), (4.21), (4.23), and (4.24), it follows that

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

(4.26)

Now, we claim that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-J_r{x}_n\parallel =0$ for a fixed number r such that $\epsilon >r>0$. In fact, using the resolvent identity in Proposition 2.2, we have

$$\begin{array}{rcl}\parallel J_{r_n}{x}_n-J_r{x}_n\parallel & =& \parallel J_r(\frac{r}{r_n}{x}_n+(1-\frac{r}{r_n})J_{r_n}{x}_n)-J_r{x}_n\parallel \\ \le & (1-\frac{r}{r_n})\parallel x_n-J_{r_n}{x}_n\parallel \\ \le & \parallel x_n-J_{r_n}{x}_n\parallel .\end{array}$$

(4.27)

Thus, from (4.26) and (4.27), we get

$$\begin{array}{rcl}\parallel x_n-J_r{x}_n\parallel & \le & \parallel x_n-J_{r_n}{x}_n\parallel +\parallel J_{r_n}{x}_n-J_r{x}_n\parallel \\ \le & \parallel x_n-J_{r_n}{x}_n\parallel +\parallel x_n-J_{r_n}{x}_n\parallel \\ =& 2\parallel x_n-J_{r_n}{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty },\end{array}$$

that is,

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

(4.28)

Suppose that ${\beta }_n\equiv \beta$ for some fixed $\beta ,\gamma \in (0,1)$ such that ${\alpha }_n+\beta +{\gamma }_n+{\delta }_n=1$ for all $n\ge 0$. Define a mapping $Vx=(1-{\theta }_1-{\theta }_2)J_{r}x+{\theta }_{1}Wx+{\theta }_{2}Gx$, where ${\theta }_1,{\theta }_2\in (0,1)$ are two constants with ${\theta }_1+{\theta }_2<1$. Then, by Lemmas 2.5 and 2.13, we have $Fix(V)=Fix(J_r)\cap Fix(W)\cap Fix(G)=F$. For each $k\ge 1$, let $\{p_k\}$ be a unique element of C such that

$$p_k=\frac{1}{k}f(p_k)+(1-\frac{1}{k})V{p}_k.$$

From Lemma 2.9, we conclude that $p_k\to q\in Fix(V)=F$ as $k\to \mathrm{\infty }$. Observe that for every n, k

$$\begin{array}{c}\parallel x_{n+1}-W{p}_k\parallel \hfill \\ =\parallel {\alpha }_n(f(y_n)-W{p}_k)+\beta (x_n-W{p}_k)+{\gamma }_n(W_n{y}_n-W{p}_k)+{\delta }_n(J_{r_n}G{y}_n-W{p}_k)\parallel \hfill \\ \le {\alpha }_n\parallel f(y_n)-W{p}_k\parallel +\beta \parallel x_n-W{p}_k\parallel +{\gamma }_n\parallel W_n{y}_n-W{p}_k\parallel \hfill \\ +{\delta }_n(\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel +\parallel W_n{y}_n-W{p}_k\parallel )\hfill \\ ={\alpha }_n\parallel f(y_n)-W{p}_k\parallel +\beta \parallel x_n-W{p}_k\parallel +({\gamma }_n+{\delta }_n)\parallel W_n{y}_n-W{p}_k\parallel +{\delta }_n\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel \hfill \\ ={\alpha }_n\parallel f(y_n)-W{p}_k\parallel +\beta \parallel x_n-W{p}_k\parallel +(1-{\alpha }_n-\beta )\parallel W_n{y}_n-W{p}_k\parallel \hfill \\ +{\delta }_n\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel \hfill \\ \le {\alpha }_n\parallel f(y_n)-W{p}_k\parallel +\beta \parallel x_n-W{p}_k\parallel \hfill \\ +(1-{\alpha }_n-\beta )[\parallel W_n{y}_n-W_n{p}_k\parallel +\parallel W_n{p}_k-W{p}_k\parallel ]+{\delta }_n\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel \hfill \\ \le {\alpha }_n\parallel f(y_n)-W{p}_k\parallel +\beta \parallel x_n-W{p}_k\parallel +(1-{\alpha }_n-\beta )[\parallel y_n-p_k\parallel +\parallel W_n{p}_k-W{p}_k\parallel ]\hfill \\ +{\delta }_n\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel \hfill \\ \le {\alpha }_n\parallel f(y_n)-W{p}_k\parallel +\beta \parallel x_n-W{p}_k\parallel \hfill \\ +(1-\beta )[\parallel x_n-p_k\parallel +\parallel y_n-x_n\parallel +\parallel W_n{p}_k-W{p}_k\parallel ]+{\delta }_n\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel \hfill \\ ={\Delta }_n+\beta \parallel x_n-W{p}_k\parallel +(1-\beta )\parallel x_n-p_k\parallel ,\hfill \end{array}$$

(4.29)

where ${\Delta }_n={\alpha }_n\parallel f(y_n)-W{p}_k\parallel +(1-\beta )[\parallel y_n-x_n\parallel +\parallel W_n{p}_k-W{p}_k\parallel ]+{\delta }_n\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel$. Since ${lim}_{n\to \mathrm{\infty }}{\alpha }_n={lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel W_n{p}_k-W{p}_k\parallel ={lim}_{n\to \mathrm{\infty }}\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel =0$, we know that ${\Delta }_n\to 0$ as $n\to \mathrm{\infty }$.

From (4.29), we obtain

$$\begin{array}{c}{\parallel x_{n+1}-W{p}_k\parallel }^2\hfill \\ \le {(\beta \parallel x_n-W{p}_k\parallel +(1-\beta )\parallel x_n-p_k\parallel )}^2\hfill \\ +{\Delta }_n[2(\beta \parallel x_n-W{p}_k\parallel +(1-\beta )\parallel x_n-p_k\parallel )+{\Delta }_n]\hfill \\ ={\beta }^2{\parallel x_n-W{p}_k\parallel }^2+{(1-\beta )}^2{\parallel x_n-p_k\parallel }^2+2\beta (1-\beta )\parallel x_n-W{p}_k\parallel \parallel x_n-p_k\parallel +{\tau }_n\hfill \\ \le {\beta }^2{\parallel x_n-W{p}_k\parallel }^2+{(1-\beta )}^2{\parallel x_n-p_k\parallel }^2\hfill \\ +\beta (1-\beta )({\parallel x_n-W{p}_k\parallel }^2+{\parallel x_n-p_k\parallel }^2)+{\tau }_n\hfill \\ =\beta {\parallel x_n-W{p}_k\parallel }^2+(1-\beta ){\parallel x_n-p_k\parallel }^2+{\tau }_n,\hfill \end{array}$$

(4.30)

where ${\tau }_n={\Delta }_n[2(\beta \parallel x_n-W{p}_k\parallel +(1-\beta )\parallel x_n-p_k\parallel )+{\Delta }_n]\to 0$ as $n\to \mathrm{\infty }$.

For any Banach limit μ, from (4.30), we have

$${\mu }_n{\parallel x_n-W{p}_k\parallel }^2={\mu }_n{\parallel x_{n+1}-W{p}_k\parallel }^2\le {\mu }_n{\parallel x_n-p_k\parallel }^2.$$

(4.31)

In addition, note that

$$\begin{array}{rcl}{\parallel x_n-G{p}_k\parallel }^2& \le & {\parallel x_n-G{x}_n+G{x}_n-G{p}_k\parallel }^2\\ \le & {(\parallel x_n-G{x}_n\parallel +\parallel x_n-p_k\parallel )}^2\\ =& {\parallel x_n-p_k\parallel }^2+\parallel x_n-G{x}_n\parallel (2\parallel x_n-p_k\parallel +\parallel x_n-G{x}_n\parallel ),\end{array}$$

and

$$\begin{array}{rcl}{\parallel x_n-J_r{p}_k\parallel }^2& \le & {\parallel x_n-J_r{x}_n+J_r{x}_n-J_r{p}_k\parallel }^2\\ \le & {(\parallel x_n-J_r{x}_n\parallel +\parallel x_n-p_k\parallel )}^2\\ =& {\parallel x_n-p_k\parallel }^2+\parallel x_n-J_r{x}_n\parallel (2\parallel x_n-p_k\parallel +\parallel x_n-J_r{x}_n\parallel ).\end{array}$$

It is easy to see from (4.21) and (4.28) that

$${\mu }_n{\parallel x_n-G{p}_k\parallel }^2\le {\mu }_n{\parallel x_n-p_k\parallel }^2\text{and}{\mu }_n{\parallel x_n-J_r{p}_k\parallel }^2\le {\mu }_n{\parallel x_n-p_k\parallel }^2.$$

(4.32)

Utilizing (4.31) and (4.32), we have

$$\begin{array}{rcl}{\mu }_n{\parallel x_n-V{p}_k\parallel }^2& =& {\mu }_n{\parallel (1-{\theta }_1-{\theta }_2)(x_n-J_r{p}_k)+{\theta }_1(x_n-W{p}_k)+{\theta }_2(x_n-G{p}_k)\parallel }^2\\ \le & (1-{\theta }_1-{\theta }_2){\mu }_n{\parallel x_n-J_r{p}_k\parallel }^2+{\theta }_1{\mu }_n{\parallel x_n-W{p}_k\parallel }^2+{\theta }_2{\mu }_n{\parallel x_n-G{p}_k\parallel }^2\\ \le & {\mu }_n{\parallel x_n-p_k\parallel }^2.\end{array}$$

(4.33)

Also, observe that

$$x_n-p_k=\frac{1}{k}(x_n-f(p_k))+(1-\frac{1}{k})(x_n-V{p}_k),$$

that is,

$$(1-\frac{1}{k})(x_n-V{p}_k)=x_n-p_k-\frac{1}{k}(x_n-f(p_k)).$$

(4.34)

It follows from Lemma 2.2(ii) and (4.34) that

$$\begin{array}{rcl}{(1-\frac{1}{k})}^2{\parallel x_n-V{p}_k\parallel }^2& \ge & {\parallel x_n-p_k\parallel }^2-\frac{2}{k}〈x_n-p_k+p_k-f(p_k),J(x_n-p_k)〉\\ =& (1-\frac{2}{k}){\parallel x_n-p_k\parallel }^2+\frac{2}{k}〈f(p_k)-p_k,J(x_n-p_k)〉.\end{array}$$

(4.35)

So by (4.33) and (4.35), we have

$${(1-\frac{1}{k})}^2{\mu }_n{\parallel x_n-p_k\parallel }^2\ge (1-\frac{2}{k}){\mu }_n{\parallel x_n-p_k\parallel }^2+\frac{2}{k}{\mu }_n〈f(p_k)-p_k,J(x_n-p_k)〉,$$

and hence,

$$\frac{1}{k^2}{\mu }_n{\parallel x_n-p_k\parallel }^2\ge \frac{2}{k}{\mu }_n〈f(p_k)-p_k,J(x_n-p_k)〉.$$

This implies that

$$\frac{1}{2k}{\mu }_n{\parallel x_n-p_k\parallel }^2\ge {\mu }_n〈f(p_k)-p_k,J(x_n-p_k)〉.$$

(4.36)

Since $p_k\to q\in Fix(V)=F$ as $k\to \mathrm{\infty }$, by the uniform Fréchet differentiability of the norm of X, we have

$${\mu }_n〈f(q)-q,J(x_n-q)〉\le 0.$$

On the other hand, from (4.4) and the norm-to-norm uniform continuity of J on bounded subsets of X, we have

$$\underset{n\to \mathrm{\infty }}{lim}|〈f(q)-q,J(x_{n+1}-q)〉-〈f(q)-q,J(x_n-q)〉|=0.$$

(4.37)

Utilizing Lemma 2.14, we deduce from (4.36) and (4.37) that

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

Finally, we show that $x_n\to q$ as $n\to \mathrm{\infty }$. It is easy to see from (4.1) that

$${\parallel y_n-q\parallel }^2\le {\sigma }_n{\parallel x_n-q\parallel }^2+(1-{\sigma }_n){\parallel J_{r_n}G{x}_n-q\parallel }^2\le {\parallel x_n-q\parallel }^2.$$

Utilizing Lemma 2.2(a), from (4.1) and the convexity of ${\parallel \cdot \parallel }^2$ we get

$$\begin{array}{c}{\parallel x_{n+1}-q\parallel }^2\hfill \\ ={\parallel {\alpha }_n(f(y_n)-f(q))+{\beta }_n(x_n-q)+{\gamma }_n(W_n{y}_n-q)+{\delta }_n(J_{r_n}G{y}_n-q)+{\alpha }_n(f(q)-q)\parallel }^2\hfill \\ \le {\parallel {\alpha }_n(f(y_n)-f(q))+{\beta }_n(x_n-q)+{\gamma }_n(W_n{y}_n-q)+{\delta }_n(J_{r_n}G{y}_n-q)\parallel }^2\hfill \\ +2{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\hfill \\ \le {\alpha }_n{\parallel f(y_n)-f(q)\parallel }^2+{\beta }_n{\parallel x_n-q\parallel }^2+{\gamma }_n{\parallel W_n{y}_n-q\parallel }^2+{\delta }_n{\parallel J_{r_n}G{y}_n-q\parallel }^2\hfill \\ +2{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\hfill \\ \le {\alpha }_n\rho {\parallel y_n-q\parallel }^2+{\beta }_n{\parallel x_n-q\parallel }^2+{\gamma }_n{\parallel y_n-q\parallel }^2+{\delta }_n{\parallel y_n-q\parallel }^2\hfill \\ +2{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\hfill \\ \le {\alpha }_n\rho {\parallel x_n-q\parallel }^2+{\beta }_n{\parallel x_n-q\parallel }^2+{\gamma }_n{\parallel x_n-q\parallel }^2+{\delta }_n{\parallel x_n-q\parallel }^2\hfill \\ +2{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\hfill \\ =(1-{\alpha }_n(1-\rho )){\parallel x_n-q\parallel }^2+2{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\hfill \\ =(1-{\alpha }_n(1-\rho )){\parallel x_n-q\parallel }^2+{\alpha }_n(1-\rho )\frac{2〈f(q)-q,J(x_{n+1}-q)〉}{1-\rho }.\hfill \end{array}$$

(4.38)

Applying Lemma 2.7 to (4.38), we obtain $x_n\to q$ as $n\to \mathrm{\infty }$. This completes the proof. □

Corollary 4.1 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X and ${\Pi }_C$ be a sunny nonexpansive retraction from X onto C. Let $A\subset X\times X$ be an accretive operator such that $\overline{D(A)}\subset C\subset {\bigcap }_{r>0}R(I+rA)$. Let $V:C\to C$ be α-strictly pseudocontractive mapping and $f:C\to C$ be a contraction with coefficient $\rho \in (0,1)$. Let ${\{T_i\}}_{i=0}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings from C into itself such that $F={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(T_i)\cap Fix(V)\cap A^{-1}0\ne \mathrm{\varnothing }$. Suppose that Assumption  4.1 holds. For arbitrarily given $x_0\in C$, let $\{x_n\}$ be the sequence generated by

$$\{\begin{array}{c}{y}_n={\sigma }_n{x}_n+(1-{\sigma }_n)J_{r_n}((1-l)I+lV)x_n,\hfill \\ x_{n+1}={\alpha }_{n}f(y_n)+{\beta }_n{x}_n+{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}((1-l)I+lV)y_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(4.39)

where $0<l<\frac{\alpha }{{\kappa }^2}$, $W_n$ is the W-mapping generated by (2.1). Then

• (a) ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$;

• (b) the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to some $q\in F$ which is a unique solution of the following variational inequality problem (VIP):

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

provided ${\beta }_n\equiv \beta$ for some fixed $\beta \in (0,1)$.

Proof In Theorem 4.1, we put $B_1=I-V$, $B_2=0$ and ${\mu }_1=l$ where $0<l<\frac{\alpha }{{\kappa }^2}$. Then SVI (3.1) is equivalent to the VIP of finding $x^{\ast }\in C$ such that

$$〈B_1{x}^{\ast },J(x-x^{\ast })〉\ge 0,\mathrm{\forall }x\in C.$$

In this case, $B_1:C\to X$ is α-inverse strongly accretive. It is not hard to see that $Fix(V)=VI(C,B_1)$. Indeed, for $l>0$, we have

$$\begin{array}{rcl}u\in VI(C,B_1)& \iff & 〈B_{1}u,J(y-u)〉\ge 0\mathrm{\forall }y\in C\\ \iff & 〈u-l{B}_{1}u-u,J(u-y)〉\ge 0\mathrm{\forall }y\in C\\ \iff & u={\Pi }_C(u-l{B}_{1}u)\\ \iff & u={\Pi }_C(u-lu+lVu)\\ \iff & 〈u-lu+lVu-u,J(u-y)〉\ge 0\mathrm{\forall }y\in C\\ \iff & 〈u-Vu,J(u-y)〉\le 0\mathrm{\forall }y\in C\\ \iff & u=Vu\\ \iff & u\in Fix(V).\end{array}$$

Accordingly, we have $F={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(T_i)\cap \Omega \cap A^{-1}0={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(T_i)\cap Fix(V)\cap A^{-1}0$, and

$${\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)x_n={\Pi }_C(I-{\mu }_1{B}_1)x_n={\Pi }_C((1-l)x_n+lV{x}_n)=((1-l)I+lV)x_n.$$

Similarly, we get

$${\Pi }_C(I-{\mu }_1{B}_1){\Pi }_C(I-{\mu }_2{B}_2)y_n=((1-l)I+lV)y_n.$$

So, the scheme (4.1) reduces to (4.39), and therefore, the desired result follows from Theorem 4.1. □

We give the following important lemmas which will be used in our next result.

Lemma 4.1 Let C be a nonempty closed convex subset of a smooth Banach space X and $B_i:C\to X$ be ${\lambda }_i$-strictly pseudocontractive mappings and ${\alpha }_i$-strongly accretive with ${\alpha }_i+{\lambda }_i\ge 1$ for $i=1,2$. Then, for ${\mu }_i\in (0,1]$,

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

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 Using the ${\lambda }_i$-strict pseudocontractivity of $B_i$, we derive for every $x,y\in C$

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

which implies that

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

Hence,

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

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

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

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

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

Since $1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$, it follows 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 4.2 Let C be a nonempty closed convex subset of a smooth Banach space X and ${\Pi }_C$ be a sunny nonexpansive retraction from X onto C. For each $i=1,2$, let $B_i:C\to X$ be ${\lambda }_i$-strictly pseudocontractive and ${\alpha }_i$-strongly accretive with ${\alpha }_i+{\lambda }_i\ge 1$. Let $G:C\to C$ be the 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 By Lemma 4.1, $I-{\mu }_i{B}_i$ is nonexpansive for $i=1,2$. Therefore, for all $x,y\in C$, we have

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

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

Theorem 4.2 Let C be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gâteaux differentiable norm. Let ${\Pi }_C$ be a sunny nonexpansive retraction from X onto C and $A\subset X\times X$ be an accretive operator in X such that $\overline{D(A)}\subset C\subset {\bigcap }_{r>0}R(I+rA)$. For each $i=1,2$, let $B_i:C\to X$ be ${\lambda }_i$-strictly pseudocontractive and ${\alpha }_i$-strongly accretive with ${\alpha }_i+{\lambda }_i\ge 1$ and $f:C\to C$ be a contraction with coefficient $\rho \in (0,1)$. Let ${\{T_i\}}_{i=0}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings from C into itself such that $F={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(T_i)\cap \Omega \cap A^{-1}0\ne \mathrm{\varnothing }$ with $1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$ for $i=1,2$. For arbitrarily given $x_0\in C$, let $\{x_n\}$ be the sequence generated by

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

(4.40)

where $W_n$ is the W-mapping generated by (2.1). Assume that Assumption  4.1 holds except condition (iii), which is replaced by the following condition:

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}(|{\sigma }_n-{\sigma }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|)<\mathrm{\infty }$.

Then

• (a) ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$;

• (b) the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to some $q\in F$ which is the unique solution of the following variational inequality problem (VIP):

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

provided ${\beta }_n\equiv \beta$ for some fixed $\beta \in (0,1)$.

Proof Take a fixed $p\in F$ arbitrarily. Then we obtain $p=Gp$, $p=W_{n}p$ and $J_{r_n}p=p$ for all $n\ge 0$. By using Lemma 4.2 and the same argument as in the proof beginning of the proof of Theorem 4.1, we have $\{x_n\}$, $\{y_n\}$, $\{G{x}_n\}$, $\{G{y}_n\}$, $\{f(y_n)\}$ are bounded sequences. Let us show that $\parallel x_{n+1}-x_n\parallel \to 0$ as $n\to \mathrm{\infty }$. In fact, repeating the same argument as those in the proof of Theorem 4.1, we obtain

$$\{\begin{array}{c}\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel \le \parallel x_{n-1}-x_n\parallel +|r_{n-1}-r_n|M_0,\hfill \\ \parallel J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}\parallel \le \parallel y_{n-1}-y_n\parallel +|r_{n-1}-r_n|M_0,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(4.41)

where

$$\underset{n\ge 1}{sup}\left\{\frac{1}{\epsilon }(\parallel J_{r_n}G{x}_n-G{x}_{n-1}\parallel +\parallel J_{r_{n-1}}G{x}_{n-1}-G{x}_n\parallel )\right\}\le M_0$$

and

$$\underset{n\ge 1}{sup}\left\{\frac{1}{\epsilon }(\parallel J_{r_n}G{y}_n-G{y}_{n-1}\parallel +\parallel J_{r_{n-1}}G{y}_{n-1}-G{y}_n\parallel )\right\}\le M_0,$$

for some $M_0>0$. By (4.40) and simple calculations, we have

$$\begin{array}{rcl}{y}_n-y_{n-1}& =& {\sigma }_n(G{x}_n-G{x}_{n-1})+({\sigma }_n-{\sigma }_{n-1})(G{x}_{n-1}-J_{r_{n-1}}G{x}_{n-1})\\ +(1-{\alpha }_n)(J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}).\end{array}$$

It follows that

$$\begin{array}{rcl}\parallel y_n-y_{n-1}\parallel & \le & {\sigma }_n\parallel G{x}_n-G{x}_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G{x}_{n-1}-J_{r_{n-1}}G{x}_{n-1}\parallel \\ +(1-{\alpha }_n)\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel \\ \le & {\sigma }_n\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G{x}_{n-1}-J_{r_{n-1}}G{x}_{n-1}\parallel \\ +(1-{\sigma }_n)(\parallel x_{n-1}-x_n\parallel +|r_{n-1}-r_n|M_0)\\ \le & \parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G{x}_{n-1}-J_{r_{n-1}}G{x}_{n-1}\parallel +|r_n-r_{n-1}|M_0.\end{array}$$

(4.42)

Repeating the same argument as in (4.7) in the proof of Theorem 4.1, we get

$$\parallel W_n{y}_{n-1}-W_{n-1}{y}_{n-1}\parallel \le M\prod _{i=0}^{n-1}{\lambda }_i,\text{for some constant }M>0.$$

(4.43)

Considering condition (v), without loss of generality, we may assume that $\{{\beta }_n\}\subset [\hat{c},\hat{d}]$ for some $\hat{c},\hat{d}\in (0,1)$. From (4.40), it follows that $x_{n+1}$ can be rewritten as

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

(4.44)

where $z_n=\frac{{\alpha }_{n}f(y_n)+{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{1-{\beta }_n}$. Utilizing (4.3) and (4.42) we have

$$\begin{array}{c}\parallel z_n-z_{n-1}\parallel \hfill \\ =\parallel \frac{{\alpha }_{n}f(y_n)+{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{1-{\beta }_n}-\frac{{\alpha }_{n-1}f(y_{n-1})+{\gamma }_{n-1}{W}_{n-1}{y}_{n-1}+{\delta }_{n-1}{J}_{r_{n-1}}G{y}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ =\parallel \frac{x_{n+1}-{\beta }_n{y}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ =\parallel \frac{x_{n+1}-{\beta }_n{y}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_n}+\frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ \le \parallel \frac{x_{n+1}-{\beta }_n{y}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_n}\parallel +\parallel \frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}\parallel x_{n+1}-{\beta }_n{y}_n-(x_n-{\beta }_{n-1}{y}_{n-1})\parallel +|\frac{1}{1-{\beta }_n}-\frac{1}{1-{\beta }_{n-1}}|\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}\parallel x_{n+1}-{\beta }_n{y}_n-(x_n-{\beta }_{n-1}{y}_{n-1})\parallel +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}\hfill \\ \times \parallel {\alpha }_{n}f(y_n)+{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n-{\alpha }_{n-1}f(y_{n-1})-{\gamma }_{n-1}{W}_{n-1}{y}_{n-1}-{\delta }_{n-1}{J}_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[{\alpha }_n\parallel f(y_n)-f(y_{n-1})\parallel +{\gamma }_n\parallel W_n{y}_n-W_{n-1}{y}_{n-1}\parallel +{\delta }_n\parallel J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel ]\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[{\alpha }_n\rho \parallel y_n-y_{n-1}\parallel +{\gamma }_n\parallel W_n{y}_n-W_n{y}_{n-1}\parallel +{\delta }_n[\parallel y_{n-1}-y_n\parallel +|r_{n-1}-r_n|M_0]\hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +{\gamma }_n\parallel W_n{y}_{n-1}-W_{n-1}{y}_{n-1}\parallel \hfill \\ +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel ]+\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[({\alpha }_n\rho +{\gamma }_n+{\delta }_n)\parallel y_{n-1}-y_n\parallel +|r_{n-1}-r_n|M_0+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel \hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +{\gamma }_{n}M\prod _{i=0}^{n-1}{\lambda }_i+|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel ]\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ \le (1-\frac{(1-\rho ){\alpha }_n}{1-{\beta }_n})\parallel y_n-y_{n-1}\parallel +\frac{1}{1-{\beta }_n}[|r_{n-1}-r_n|M_0+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel \hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel ]+M\prod _{i=0}^{n-1}{\lambda }_i\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel .\hfill \end{array}$$

(4.45)

By simple calculations, (4.44) implies that

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

This together with (4.42) and (4.45) we have

$$\begin{array}{c}\parallel x_{n+1}-x_n\parallel \hfill \\ \le {\beta }_n\parallel y_n-y_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel y_{n-1}-z_{n-1}\parallel +(1-{\beta }_n)\parallel z_n-z_{n-1}\parallel \hfill \\ \le {\beta }_n\parallel y_n-y_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel y_{n-1}-z_{n-1}\parallel +(1-{\beta }_n)\{(1-\frac{(1-\rho ){\alpha }_n}{1-{\beta }_n})\parallel y_n-y_{n-1}\parallel \hfill \\ +\frac{1}{1-{\beta }_n}[|r_{n-1}-r_n|M_0+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel \hfill \\ +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel ]+M\prod _{i=0}^{n-1}{\lambda }_i+\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \}\hfill \\ \le (1-(1-\rho ){\alpha }_n)\parallel y_n-y_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel y_{n-1}-z_{n-1}\parallel +|r_{n-1}-r_n|M_0\hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ +M\prod _{i=0}^{n-1}{\lambda }_i+\frac{|{\beta }_n-{\beta }_{n-1}|}{1-{\beta }_{n-1}}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ \le (1-(1-\rho ){\alpha }_n)[\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G{x}_{n-1}-J_{r_{n-1}}G{x}_{n-1}\parallel +|r_n-r_{n-1}|M_0]\hfill \\ +|{\beta }_n-{\beta }_{n-1}|\parallel y_{n-1}-z_{n-1}\parallel +|r_{n-1}-r_n|M_0+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel \hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|\parallel W_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel +M\prod _{i=0}^{n-1}{\lambda }_i\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{1-{\beta }_{n-1}}\parallel {\alpha }_{n-1}f(y_{n-1})+{\gamma }_{n-1}{W}_{n-1}{y}_{n-1}+{\delta }_{n-1}{J}_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ \le (1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel +(|{\sigma }_n-{\sigma }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|\hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|+|r_{n-1}-r_n|)\hat{M}+M\prod _{i=0}^{n-1}{\lambda }_i,\hfill \end{array}$$

where $\frac{1}{1-\hat{d}}{sup}_{n\ge 0}\{\parallel f(y_n)\parallel +\parallel W_n{y}_n\parallel +\parallel J_{r_n}G{y}_n\parallel +\parallel G{x}_n-J_{r_n}G{x}_n\parallel +\parallel y_n-z_n\parallel +2M_0\}\le \hat{M}$ for some $\hat{M}>0$. By Lemma 2.7 and conditions (i), (iii), and (iv), we conclude that (noting that $0<{\lambda }_i\le b<1$, $\mathrm{\forall }i\ge 0$)

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

(4.46)

Next we show that $\parallel x_n-G{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$. Indeed, utilizing Lemma 2.3, we get from (4.40)

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel {\sigma }_n(G{x}_n-p)+(1-{\sigma }_n)(J_{r_n}G{x}_n-p)\parallel }^2\\ \le & {\sigma }_n{\parallel G{x}_n-p\parallel }^2+(1-{\sigma }_n){\parallel J_{r_n}G{x}_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel G{x}_n-J_{r_n}G{x}_n\parallel )\\ \le & {\sigma }_n{\parallel x_n-p\parallel }^2+(1-{\sigma }_n){\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel G{x}_n-J_{r_n}G{x}_n\parallel )\\ =& {\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel G{x}_n-J_{r_n}G{x}_n\parallel ).\end{array}$$

(4.47)

By Lemma 2.2 (a), (4.40), and (4.47), we have

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ ={\parallel {\alpha }_n(f(y_n)-f(p))+{\beta }_n(y_n-p)+{\gamma }_n(W_n{y}_n-p)+{\delta }_n(J_{r_n}G{y}_n-p)+{\alpha }_n(f(p)-p)\parallel }^2\hfill \\ \le {\parallel {\alpha }_n(f(y_n)-f(p))+{\beta }_n(y_n-p)+{\gamma }_n(W_n{y}_n-p)+{\delta }_n(J_{r_n}G{y}_n-p)\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(x_{n+1}-p)〉\hfill \\ \le {\alpha }_n{\parallel f(y_n)-f(p)\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel W_n{y}_n-p\parallel }^2+{\delta }_n{\parallel J_{r_n}G{y}_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(x_{n+1}-p)〉\hfill \\ \le {\alpha }_n\rho {\parallel y_n-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2+{\delta }_n{\parallel y_n-p\parallel }^2\hfill \\ +2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel \hfill \\ =(1-{\alpha }_n(1-\rho )){\parallel y_n-p\parallel }^2+2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le {\parallel y_n-p\parallel }^2+2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le {\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel G{x}_n-J_{r_n}G{x}_n\parallel )+2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel ,\hfill \end{array}$$

which yields

$$\begin{array}{c}{\sigma }_n(1-{\sigma }_n)g(\parallel G{x}_n-J_{r_n}G{x}_n\parallel )\hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel .\hfill \end{array}$$

Since ${\alpha }_n\to 0$ and $\parallel x_{n+1}-x_n\parallel \to 0$, from condition (v) and the boundedness of $\{x_n\}$, it follows that

$$\underset{n\to \mathrm{\infty }}{lim}g(\parallel G{x}_n-J_{r_n}G{x}_n\parallel )=0.$$

Utilizing the properties of g, we have

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

(4.48)

On the other hand, observe that $x_{n+1}$ can be rewritten as

$$\begin{array}{rcl}{x}_{n+1}& =& {\alpha }_{n}f(y_n)+{\beta }_n{y}_n+{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n\\ =& {\alpha }_{n}f(y_n)+{\beta }_n{y}_n+({\gamma }_n+{\delta }_n)\frac{{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{{\gamma }_n+{\delta }_n}\\ =& {\alpha }_{n}f(y_n)+{\beta }_n{y}_n+e_n{\hat{z}}_n,\end{array}$$

(4.49)

where $e_n={\gamma }_n+{\delta }_n$ and ${\hat{z}}_n=\frac{{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{{\gamma }_n+{\delta }_n}$. By Lemma 2.4, (4.3), and (4.49), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel {\alpha }_n(f(y_n)-p)+{\beta }_n(y_n-p)+e_n({\hat{z}}_n-p)\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2+e_n{\parallel {\hat{z}}_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ =& {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ +e_n{\parallel \frac{{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{{\gamma }_n+{\delta }_n}-p\parallel }^2\\ =& {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ +e_n{\parallel \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(W_n{y}_n-p)+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}(J_{r_n}G{y}_n-p)\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ +e_n[\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}{\parallel W_n{y}_n-p\parallel }^2+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel J_{r_n}G{y}_n-p\parallel }^2]\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ +e_n[\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}{\parallel y_n-p\parallel }^2+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel y_n-p\parallel }^2]\\ =& {\alpha }_n{\parallel f(y_n)-p\parallel }^2+(1-{\alpha }_n){\parallel y_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel y_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel ),\end{array}$$

which implies that

$$\begin{array}{rcl}{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )& \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel .\end{array}$$

Utilizing (4.46), conditions (i), (ii), (v), and the boundedness of $\{x_n\}$ and $\{f(y_n)\}$, we get

$$\underset{n\to \mathrm{\infty }}{lim}{g}_1(\parallel {\hat{z}}_n-y_n\parallel )=0.$$

From the properties of $g_1$, we have

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

(4.50)

Utilizing Lemma 2.3 and the definition of ${\hat{z}}_n$, we have

$$\begin{array}{rcl}{\parallel {\hat{z}}_n-p\parallel }^2& =& {\parallel \frac{{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{{\gamma }_n+{\delta }_n}-p\parallel }^2\\ =& {\parallel \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(W_n{y}_n-p)+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}(J_{r_n}G{y}_n-p)\parallel }^2\\ \le & \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}{\parallel W_n{y}_n-p\parallel }^2+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel J_{r_n}G{y}_n-p\parallel }^2\\ -\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_2(\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel )\\ \le & {\parallel y_n-p\parallel }^2-\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_2(\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel ),\end{array}$$

which leads to

$$\begin{array}{rcl}\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_2(\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel )& \le & {\parallel y_n-p\parallel }^2-{\parallel {\hat{z}}_n-p\parallel }^2\\ \le & (\parallel y_n-p\parallel +\parallel {\hat{z}}_n-p\parallel )\parallel y_n-{\hat{z}}_n\parallel .\end{array}$$

Since $\{y_n\}$ and $\{{\hat{z}}_n\}$ are bounded, from (4.50) and condition (ii), we deduce

$$\underset{n\to \mathrm{\infty }}{lim}{g}_2(\parallel W_n{y}_n-J_{r_n}G{y}_n\parallel )=0.$$

From the properties of $g_2$, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel W_n{y}_n-J_{r_n}G{y}_n\parallel =0.$$

(4.51)

Furthermore, $x_{n+1}$ can also be rewritten as

$$\begin{array}{rcl}{x}_{n+1}& =& {\alpha }_{n}f(y_n)+{\beta }_n{y}_n+{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n\\ =& {\beta }_n{y}_n+{\gamma }_n{W}_n{y}_n+({\alpha }_n+{\delta }_n)\frac{{\alpha }_{n}f(y_n)+{\delta }_n{J}_{r_n}G{y}_n}{{\alpha }_n+{\delta }_n}\\ =& {\beta }_n{y}_n+{\gamma }_n{W}_n{y}_n+d_n{\tilde{z}}_n,\end{array}$$

where $d_n={\alpha }_n+{\delta }_n$ and ${\tilde{z}}_n=\frac{{\alpha }_{n}f(y_n)+{\delta }_n{J}_{r_n}G{y}_n}{{\alpha }_n+{\delta }_n}$. Utilizing Lemma 2.4, the convexity of ${\parallel \cdot \parallel }^2$, and (4.3), we have

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ ={\parallel {\beta }_n(y_n-p)+{\gamma }_n(W_n{y}_n-p)+d_n({\tilde{z}}_n-p)\parallel }^2\hfill \\ \le {\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel W_n{y}_n-p\parallel }^2+d_n{\parallel {\tilde{z}}_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel y_n-W_n{y}_n\parallel )\hfill \\ ={\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel W_n{y}_n-p\parallel }^2+d_n{\parallel \frac{{\alpha }_{n}f(y_n)+{\delta }_n{J}_{r_n}G{y}_n}{{\alpha }_n+{\delta }_n}-p\parallel }^2\hfill \\ -{\beta }_n{\gamma }_n{g}_3(\parallel y_n-W_n{y}_n\parallel )\hfill \\ ={\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel W_n{y}_n-p\parallel }^2+d_n{\parallel \frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}(f(y_n)-p)+\frac{{\delta }_n}{{\alpha }_n+{\delta }_n}(J_{r_n}G{y}_n-p)\parallel }^2\hfill \\ -{\beta }_n{\gamma }_n{g}_3(\parallel y_n-W_n{y}_n\parallel )\hfill \\ \le {\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2+d_n[\frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}{\parallel f(y_n)-p\parallel }^2+\frac{{\delta }_n}{{\alpha }_n+{\delta }_n}{\parallel J_{r_n}G{y}_n-p\parallel }^2]\hfill \\ -{\beta }_n{\gamma }_n{g}_3(\parallel y_n-W_n{y}_n\parallel )\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+({\beta }_n+{\gamma }_n){\parallel y_n-p\parallel }^2+{\delta }_n{\parallel y_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel y_n-W_n{y}_n\parallel )\hfill \\ ={\alpha }_n{\parallel f(y_n)-p\parallel }^2+(1-{\alpha }_n){\parallel y_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel y_n-W_n{y}_n\parallel )\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel y_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel y_n-W_n{y}_n\parallel )\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel y_n-W_n{y}_n\parallel ),\hfill \end{array}$$

which implies that

$$\begin{array}{rcl}{\beta }_n{\gamma }_n{g}_3(\parallel y_n-W_n{y}_n\parallel )& \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel .\end{array}$$

From (4.46), conditions (i), (ii), (v), and the boundedness of $\{x_n\}$ and $\{f(y_n)\}$, we have

$$\underset{n\to \mathrm{\infty }}{lim}{g}_3(\parallel y_n-W_n{y}_n\parallel )=0.$$

Utilizing the properties of $g_3$, we have

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

(4.52)

Thus, from (4.51) and (4.52), we get

$$\parallel y_n-J{r}_{n}G{y}_n\parallel \le \parallel y_n-W_n{y}_n\parallel +\parallel W_n{y}_n-J{r}_{n}G{y}_n\parallel \to 0\text{as }n\to \mathrm{\infty },$$

that is,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-J{r}_{n}G{y}_n\parallel =0.$$

(4.53)

Therefore, from (4.40), (4.46), (4.52), (4.53), and ${\alpha }_n\to 0$, we have

$$\begin{array}{c}\parallel x_n-y_n\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(y_n)-y_n\parallel +{\gamma }_n\parallel W_n{y}_n-y_n\parallel +{\delta }_n\parallel J{r}_{n}G{y}_n-y_n\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(y_n)-y_n\parallel +\parallel W_n{y}_n-y_n\parallel +\parallel J{r}_{n}G{y}_n-y_n\parallel \to 0\text{as }n\to \mathrm{\infty },\hfill \end{array}$$

that is,

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

(4.54)

Utilizing (4.40), (4.48), and (4.54), we obtain

$$\begin{array}{rcl}\parallel x_n-G{x}_n\parallel & \le & \parallel x_n-y_n\parallel +\parallel y_n-G{x}_n\parallel \\ =& \parallel x_n-y_n\parallel +(1-{\sigma }_n)\parallel J_{r_n}G{x}_n-G{x}_n\parallel \\ \le & \parallel x_n-y_n\parallel +\parallel J_{r_n}G{x}_n-G{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty },\end{array}$$

that is,

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

(4.55)

In addition, from (4.52) and (4.54), we have

$$\begin{array}{rcl}\parallel x_n-W_n{x}_n\parallel & \le & \parallel x_n-y_n\parallel +\parallel y_n-W_n{y}_n\parallel +\parallel W_n{y}_n-W_n{x}_n\parallel \\ \le & 2\parallel x_n-y_n\parallel +\parallel y_n-W_n{y}_n\parallel \to 0\text{as }n\to \mathrm{\infty },\end{array}$$

that is,

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

(4.56)

Note that

$$\begin{array}{rcl}\parallel x_n-J_{r_n}{x}_n\parallel & \le & \parallel x_n-G{x}_n\parallel +\parallel G{x}_n-J_{r_n}G{x}_n\parallel +\parallel J_{r_n}G{x}_n-J_{r_n}{x}_n\parallel \\ \le & 2\parallel x_n-G{x}_n\parallel +\parallel G{x}_n-J_{r_n}G{x}_n\parallel .\end{array}$$

So, from (4.48) and (4.55), it follows that

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

(4.57)

Repeating the same argument as in (4.28) in the proof of Theorem 4.1, we get

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

(4.58)

for a fixed number r such that $\epsilon >r>0$.

Suppose that ${\beta }_n\equiv \beta$ for some fixed $\beta ,\gamma \in (0,1)$ satisfying ${\alpha }_n+\beta +{\gamma }_n+{\delta }_n=1$ for all $n\ge 0$. Define a mapping $Vx=(1-{\theta }_1-{\theta }_2)J_{r}x+{\theta }_{1}Wx+{\theta }_{2}Gx$, where ${\theta }_1,{\theta }_2\in (0,1)$ are two constants with ${\theta }_1+{\theta }_2<1$. Then, by Lemmas 2.5 and 2.13, we have $Fix(V)=Fix(J_r)\cap Fix(W)\cap Fix(G)=F$. For each $k\ge 1$, let $\{p_k\}$ be a unique element of C such that

$$p_k=\frac{1}{k}f(p_k)+(1-\frac{1}{k})V{p}_k.$$

From Lemma 2.9, we conclude that $p_k\to q\in Fix(V)=F$ as $k\to \mathrm{\infty }$. Observe that for every n, k

$$\begin{array}{c}\parallel x_{n+1}-W{p}_k\parallel \hfill \\ \le {\alpha }_n\parallel f(y_n)-W{p}_k\parallel +\beta \parallel y_n-W{p}_k\parallel +{\gamma }_n\parallel W_n{y}_n-W{p}_k\parallel \hfill \\ +{\delta }_n(\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel +\parallel W_n{y}_n-W{p}_k\parallel )\hfill \\ ={\alpha }_n\parallel f(y_n)-W{p}_k\parallel +\beta \parallel y_n-W{p}_k\parallel +({\gamma }_n+{\delta }_n)\parallel W_n{y}_n-W{p}_k\parallel +{\delta }_n\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel \hfill \\ ={\alpha }_n\parallel f(y_n)-W{p}_k\parallel +\beta \parallel y_n-W{p}_k\parallel +(1-{\alpha }_n-\beta )\parallel W_n{y}_n-W{p}_k\parallel \hfill \\ +{\delta }_n\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel \hfill \\ \le {\alpha }_n\parallel f(y_n)-W{p}_k\parallel +\beta \parallel y_n-W{p}_k\parallel \hfill \\ +(1-{\alpha }_n-\beta )[\parallel W_n{y}_n-W_n{p}_k\parallel +\parallel W_n{p}_k-W{p}_k\parallel ]+{\delta }_n\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel \hfill \\ \le {\alpha }_n\parallel f(y_n)-W{p}_k\parallel +\beta \parallel y_n-W{p}_k\parallel +(1-{\alpha }_n-\beta )(\parallel y_n-p_k\parallel +\parallel W_n{p}_k-W{p}_k\parallel )\hfill \\ +{\delta }_n\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel \hfill \\ \le {\alpha }_n\parallel f(y_n)-W{p}_k\parallel +\beta (\parallel x_n-W{p}_k\parallel +\parallel y_n-x_n\parallel )+(1-\beta )[\parallel x_n-p_k\parallel +\parallel y_n-x_n\parallel \hfill \\ +\parallel W_n{p}_k-W{p}_k\parallel ]+{\delta }_n\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel \hfill \\ ={\Delta }_n+\beta \parallel x_n-W{p}_k\parallel +(1-\beta )\parallel x_n-p_k\parallel ,\hfill \end{array}$$

(4.59)

where ${\Delta }_n={\alpha }_n\parallel f(y_n)-W{p}_k\parallel +(1-\beta )(\frac{1}{1-\beta }\parallel y_n-x_n\parallel +\parallel W_n{p}_k-W{p}_k\parallel )+{\delta }_n\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel$. Since ${lim}_{n\to \mathrm{\infty }}{\alpha }_n={lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel W_n{p}_k-W{p}_k\parallel ={lim}_{n\to \mathrm{\infty }}\parallel J_{r_n}G{y}_n-W_n{y}_n\parallel =0$, we know that ${\Delta }_n\to 0$ as $n\to \mathrm{\infty }$.

Repeating the same argument as in (4.31) and (4.32) in the proof of Theorem 4.1, we conclude that for any Banach limit μ,

$${\mu }_n{\parallel x_n-W{p}_k\parallel }^2={\mu }_n{\parallel x_{n+1}-W{p}_k\parallel }^2\le {\mu }_n{\parallel x_n-p_k\parallel }^2,$$

(4.60)

and

$${\mu }_n{\parallel x_n-G{p}_k\parallel }^2\le {\mu }_n{\parallel x_n-p_k\parallel }^2\text{and}{\mu }_n{\parallel x_n-J_r{p}_k\parallel }^2\le {\mu }_n{\parallel x_n-p_k\parallel }^2.$$

(4.61)

Utilizing (4.60) and (4.61), we obtain

$$\begin{array}{rcl}{\mu }_n{\parallel x_n-V{p}_k\parallel }^2& =& {\mu }_n{\parallel (1-{\theta }_1-{\theta }_2)(x_n-J_r{p}_k)+{\theta }_1(x_n-W{p}_k)+{\theta }_2(x_n-G{p}_k)\parallel }^2\\ \le & (1-{\theta }_1-{\theta }_2){\mu }_n{\parallel x_n-J_r{p}_k\parallel }^2+{\theta }_1{\mu }_n{\parallel x_n-W{p}_k\parallel }^2+{\theta }_2{\mu }_n{\parallel x_n-G{p}_k\parallel }^2\\ \le & {\mu }_n{\parallel x_n-p_k\parallel }^2.\end{array}$$

(4.62)

Repeating the same argument as in (4.36) in the proof of Theorem 4.1, we get

$$\frac{1}{2k}{\mu }_n{\parallel x_n-p_k\parallel }^2\ge {\mu }_n〈f(p_k)-p_k,J(x_n-p_k)〉.$$

(4.63)

Since $p_k\to q\in Fix(V)=F$ as $k\to \mathrm{\infty }$, by the uniform Gâteaux differentiability of the norm of X, we have

$${\mu }_n〈f(q)-q,J(x_n-q)〉\le 0.$$

On the other hand, from (4.4) and the norm-to-weak^∗ uniform continuity of J on bounded subsets of X, it follows that

$$\underset{n\to \mathrm{\infty }}{lim}|〈f(q)-q,J(x_{n+1}-q)〉-〈f(q)-q,J(x_n-q)〉|=0.$$

(4.64)

Using Lemma 2.14, we deduce from (4.63) and (4.64) that

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

Finally, we show that $x_n\to q$ as $n\to \mathrm{\infty }$. It is easy to see from (4.1) that

$${\parallel y_n-q\parallel }^2\le {\sigma }_n{\parallel G{x}_n-q\parallel }^2+(1-{\sigma }_n){\parallel J_{r_n}G{x}_n-q\parallel }^2\le {\parallel x_n-q\parallel }^2.$$

Utilizing Lemma 2.2(a), from (4.1) and the convexity of ${\parallel \cdot \parallel }^2$ we get

$$\begin{array}{c}{\parallel x_{n+1}-q\parallel }^2\hfill \\ ={\parallel {\alpha }_n(f(y_n)-f(q))+{\beta }_n(y_n-q)+{\gamma }_n(W_n{y}_n-q)+{\delta }_n(J_{r_n}G{y}_n-q)+{\alpha }_n(f(q)-q)\parallel }^2\hfill \\ \le {\alpha }_n{\parallel f(y_n)-f(q)\parallel }^2+{\beta }_n{\parallel y_n-q\parallel }^2+{\gamma }_n{\parallel W_n{y}_n-q\parallel }^2+{\delta }_n{\parallel J_{r_n}G{y}_n-q\parallel }^2\hfill \\ +2{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\hfill \\ \le {\alpha }_n\rho {\parallel y_n-q\parallel }^2+{\beta }_n{\parallel y_n-q\parallel }^2+{\gamma }_n{\parallel y_n-q\parallel }^2+{\delta }_n{\parallel y_n-q\parallel }^2\hfill \\ +2{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\hfill \\ =(1-{\alpha }_n(1-\rho )){\parallel y_n-q\parallel }^2+2{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\hfill \\ \le (1-{\alpha }_n(1-\rho )){\parallel x_n-q\parallel }^2+{\alpha }_n(1-\rho )\frac{2〈f(q)-q,J(x_{n+1}-q)〉}{1-\rho }.\hfill \end{array}$$

(4.65)

Applying Lemma 2.7 to (4.65), we obtain $x_n\to q$ as $n\to \mathrm{\infty }$. This completes the proof. □

Corollary 4.2 Let C be a nonempty closed convex subset of a uniformly convex Banach space X which has an uniformly Gâteaux differentiable norm. Let ${\Pi }_C$ be a sunny nonexpansive retraction from X onto C and $A\subset X\times X$ be an accretive operator in X such that $\overline{D(A)}\subset C\subset {\bigcap }_{r>0}R(I+rA)$. Let $V:C\to C$ be a mapping such that $I-V:C\to X$ is ζ-strictly pseudocontractive and θ-strongly accretive with $\theta +\zeta \ge 1$. Let $f:C\to C$ be a contraction with coefficient $\rho \in (0,1)$ and ${\{T_i\}}_{i=0}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings of C into itself such that $F={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(T_i)\cap Fix(V)\cap A^{-1}0\ne \mathrm{\varnothing }$. For arbitrarily given $x_0\in C$, let $\{x_n\}$ be the sequence generated by

$$\{\begin{array}{c}{y}_n={\sigma }_n((1-l)I+lV)x_n+(1-{\sigma }_n)J_{r_n}((1-l)I+lV)x_n,\hfill \\ x_{n+1}={\alpha }_{n}f(y_n)+{\beta }_n{y}_n+{\gamma }_n{W}_n{y}_n+{\delta }_n{J}_{r_n}((1-l)I+lV)y_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(4.66)

where $1-\frac{\zeta }{1+\zeta }(1-\sqrt{\frac{1-\theta }{\zeta }})\le l\le 1$, $W_n$ is the W-mapping generated by (2.1). Assume that Assumption  4.1 holds except condition (iii), which is replaced by the following condition:

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}(|{\sigma }_n-{\sigma }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|)<\mathrm{\infty }$.

Then

• (a) ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$;

• (b) the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to some $q\in F$ which is a unique solution of the following variational inequality problem (VIP):

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

provided ${\beta }_n\equiv \beta$ for some fixed $\beta \in (0,1)$.

Proof In Theorem 4.2, we put $B_1=I-V$, $B_2=0$ and ${\mu }_1=l$ where $1-\frac{\zeta }{1+\zeta }(1-\sqrt{\frac{1-\theta }{\zeta }})\le l\le 1$. Then SVI (3.1) is equivalent to the VIP of finding $x^{\ast }\in C$ such that

$$〈B_1{x}^{\ast },J(x-x^{\ast })〉\ge 0,\mathrm{\forall }x\in C.$$

In this case, $B_1:C\to X$ is ζ-strictly pseudocontractive and θ-strongly accretive. Repeating the same arguments as in the proof of Corollary 4.1, we can infer that $Fix(V)=VI(C,B_1)$. Accordingly, $F={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(T_i)\cap \Omega \cap A^{-1}0={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(T_i)\cap Fix(V)\cap A^{-1}0$,

$$G{x}_n=((1-l)I+lV)x_n\text{and}G{y}_n=((1-l)I+lV)y_n.$$

So, the scheme (4.40) reduces to (4.66). Therefore, the desired result follows from Theorem 4.2. □

Remark 4.1 Theorems 4.1 and 4.2 improve and extend [[30], Theorem 3.2], [[20], Theorem 3.1] and [[29], Theorem 3.1] in the following aspects.

• (a) The problem of finding a point $q\in {\bigcap }_{n}Fix(T_n)\cap \Omega \cap A^{-1}0$ in Theorems 4.1 and 4.2 is more general and more subtle than the problem of finding a point $q\in {\bigcap }_{n}Fix(T_n)$ in [[30], Theorem 3.2], the problem of finding a point $q\in {\bigcap }_{n}Fix(T_n)\cap \Omega$ in [[20], Theorem 3.1] and the problem of finding a point $q\in A^{-1}0$ in [[29], Theorem 3.1].

• (b) Theorems 4.1 and 4.2 are proved without the assumption of the asymptotical regularity of $\{x_n\}$ in [[29], Theorem 3.1] (that is, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_{n+1}\parallel =0$).

• (c) The iterative scheme in [[20], Theorem 3.1] is extended to develop the iterative schemes (4.1) and (4.40) in Theorems 4.1 and 4.2 by virtue of the iterative schemes of [[30], Theorem 3.2] and [[29], Theorem 3.1]. The iterative schemes (4.1) and (4.40) in Theorems 4.1 and 4.2 are more advantageous and more flexible than the iterative scheme in [[20], Theorem 3.1] because they involve several parameter sequences.

• (d) The iterative schemes (4.1) and (4.40) in Theorems 4.1 and 4.2 are different from the iterative schemes in [[30], Theorem 3.2], [[20], Theorem 3.1] and [[29], Theorem 3.1] because the mapping G in [[20], Theorem 3.1] and the mapping $J_{r_n}$ in [[29], Theorem 3.1] are replaced by the composite mapping $J_{r_n}G$ in Theorems 4.1 and 4.2.

• (e) The proof of [[20], Theorem 3.1] depends on the argument techniques in [10], the inequality in 2-uniformly smooth Banach spaces, and the inequality in smooth and uniform convex Banach spaces. Because the composite mapping $J_{r_n}G$ appears in the iterative scheme (4.1) of Theorem 4.1, the proof of Theorem 4.1 depends on the argument techniques in [10], the inequality in 2-uniformly smooth Banach spaces, the inequality in smooth and uniform convex Banach spaces, the inequalities in uniform convex Banach spaces, and the properties of the W-mapping and the Banach limit. However, the proof of our Theorem 4.2 does not depend on the argument techniques in [10], the inequality in 2-uniformly smooth Banach spaces, and the inequality in smooth and uniform convex Banach spaces. It depends on only the inequalities in uniform convex Banach spaces and the properties of the W-mapping and the Banach limit.

• (f) The assumption of the uniformly convex and 2-uniformly smooth Banach space X in [[20], Theorem 3.1] is weakened to the uniformly convex Banach space X having a uniformly Gâteaux differentiable norm in Theorem 4.2.

5 Composite viscosity algorithms and convergence criteria

In this section, we introduce composite viscosity algorithms in real smooth and uniformly convex Banach spaces and study the strong convergence theorems. We first state the following important and useful lemma which will be used in the sequel.

Lemma 5.1 [27]

Let C be a nonempty closed convex subset of a Banach space X and $S_0,S_1,\dots$ be a sequence of mappings of C into 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 in C. Moreover, let $S:C\to C$ be a mapping defined by $Sy={lim}_{n\to \mathrm{\infty }}{S}_{n}y$ for all $y\in C$. Then ${lim}_{n\to \mathrm{\infty }}sup\{\parallel Sx-S_{n}x\parallel :x\in C\}=0$.

Assumption 5.1 Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$, $\{{\delta }_n\}$, $\{{\sigma }_n\}$ be the sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n+{\delta }_n=1$ for all $n\ge 0$. Suppose that the following conditions hold:

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

• (ii) $\{{\gamma }_n\},\{{\delta }_n\}\subset [c,d]$ for some $c,d\in (0,1)$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}(|{\sigma }_n-{\sigma }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|)<\mathrm{\infty }$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}|r_n-r_{n-1}|<\mathrm{\infty }$ and $r_n\ge \epsilon >0$ for all $n\ge 0$;

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

We now state and prove our first result on the composite implicit viscosity algorithm.

Theorem 5.1 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let ${\Pi }_C$ be a sunny nonexpansive retraction from X onto C and $A\subset X\times X$ be an accretive operator on X such that $\overline{D(A)}\subset C\subset {\bigcap }_{r>0}R(I+rA)$. Let the mapping $B_i:C\to X$ be ${\alpha }_i$-inverse strongly accretive for $i=1,2$, and $f:C\to C$ be a contraction with coefficient $\rho \in (0,1)$. Let ${\{S_i\}}_{i=0}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings of C into itself such that $F={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(S_i)\cap \Omega \cap A^{-1}0\ne \mathrm{\varnothing }$ with $0<{\mu }_i<\frac{{\alpha }_i}{{\kappa }^2}$ for $i=1,2$. Suppose that Assumption  5.1 holds. For arbitrarily given $x_0\in C$, let $\{x_n\}$ be the sequence generated by

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

(5.1)

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 subset D of C, $S:C\to C$ be a mapping defined by $Sx={lim}_{n\to \mathrm{\infty }}{S}_{n}x$ for all $x\in C$, and $Fix(S)={\bigcap }_{n=0}^{\mathrm{\infty }}Fix(S_n)$. Then the sequence $\{x_n\}$ converges strongly to $q\in F$, which solves the following VIP:

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

Proof First of all, let us show that the sequence $\{x_n\}$ is bounded. Indeed, take a fixed $p\in F$ arbitrarily. Then we get $p=Gp$, $p=S_{n}p$ and $p=J_{r_n}p$ for all $n\ge 0$. By Lemma 2.11, G is nonexpansive. Then, from (5.1), we have

$$\begin{array}{rcl}\parallel y_n-p\parallel & \le & {\alpha }_n\parallel f(y_n)-p\parallel +{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel S_n{x}_n-p\parallel +{\delta }_n\parallel J_{r_n}G{x}_n-p\parallel \\ \le & {\alpha }_n(\parallel f(y_n)-f(p)\parallel +\parallel f(p)-p\parallel )+{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel x_n-p\parallel +{\delta }_n\parallel G{x}_n-p\parallel \\ \le & {\alpha }_n(\rho \parallel y_n-p\parallel +\parallel f(p)-p\parallel )+{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel x_n-p\parallel +{\delta }_n\parallel x_n-p\parallel \\ =& (1-{\alpha }_n)\parallel x_n-p\parallel +{\alpha }_n\rho \parallel y_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel ,\end{array}$$

which implies that

$$\parallel y_n-p\parallel \le (1-\frac{(1-\rho ){\alpha }_n}{1-{\alpha }_n\rho })\parallel x_n-p\parallel +\frac{{\alpha }_n}{1-{\alpha }_n\rho }\parallel f(p)-p\parallel .$$

(5.2)

So, we have

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & \le & {\sigma }_n\parallel y_n-p\parallel +(1-{\sigma }_n)\parallel J_{r_n}G{y}_n-p\parallel \\ \le & {\sigma }_n\parallel y_n-p\parallel +(1-{\sigma }_n)\parallel G{y}_n-p\parallel \\ \le & {\sigma }_n\parallel y_n-p\parallel +(1-{\sigma }_n)\parallel y_n-p\parallel \\ =& \parallel y_n-p\parallel \\ \le & (1-\frac{(1-\rho ){\alpha }_n}{1-{\alpha }_n\rho })\parallel x_n-p\parallel +\frac{{\alpha }_n}{1-{\alpha }_n\rho }\parallel f(p)-p\parallel \\ =& (1-\frac{(1-\rho ){\alpha }_n}{1-{\alpha }_n\rho })\parallel x_n-p\parallel +\frac{(1-\rho ){\alpha }_n}{1-{\alpha }_n\rho }\frac{\parallel f(p)-p\parallel }{1-\rho }\\ \le & max\{\parallel x_n-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-\rho }\}.\end{array}$$

By induction, we obtain

$$\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-\rho }\},\mathrm{\forall }n\ge 0.$$

(5.3)

Hence, $\{x_n\}$ is bounded, and so are the sequences $\{y_n\}$, $\{G{x}_n\}$, $\{G{y}_n\}$, and $\{f(y_n)\}$.

Let us show that

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

(5.4)

Observe that $y_n$ can be rewritten as

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

where $z_n=\frac{{\alpha }_{n}f(y_n)+{\gamma }_n{S}_n{x}_n+{\delta }_n{J}_{r_n}G{x}_n}{1-{\beta }_n}$. Note that

$$\begin{array}{c}\parallel z_n-z_{n-1}\parallel \hfill \\ =\parallel \frac{{\alpha }_{n}f(y_n)+{\gamma }_n{S}_n{x}_n+{\delta }_n{J}_{r_n}G{x}_n}{1-{\beta }_n}-\frac{{\alpha }_{n-1}f(y_{n-1})+{\gamma }_{n-1}{S}_{n-1}{x}_{n-1}+{\delta }_{n-1}{J}_{r_{n-1}}G{x}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ =\parallel \frac{y_n-{\beta }_n{x}_n}{1-{\beta }_n}-\frac{y_{n-1}-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ =\parallel \frac{y_n-{\beta }_n{x}_n}{1-{\beta }_n}-\frac{y_{n-1}-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_n}+\frac{y_{n-1}-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_n}-\frac{y_{n-1}-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ \le \parallel \frac{y_n-{\beta }_n{x}_n}{1-{\beta }_n}-\frac{y_{n-1}-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_n}\parallel +\parallel \frac{y_{n-1}-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_n}-\frac{y_{n-1}-{\beta }_{n-1}{x}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}\parallel y_n-{\beta }_n{x}_n-(y_{n-1}-{\beta }_{n-1}{x}_{n-1})\parallel +|\frac{1}{1-{\beta }_n}-\frac{1}{1-{\beta }_{n-1}}|\parallel y_{n-1}-{\beta }_{n-1}{x}_{n-1}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}\parallel y_n-{\beta }_n{x}_n-(y_{n-1}-{\beta }_{n-1}{x}_{n-1})\parallel +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel y_{n-1}-{\beta }_{n-1}{x}_{n-1}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}\parallel {\alpha }_{n}f(y_n)+{\gamma }_n{S}_n{x}_n+{\delta }_n{J}_{r_n}G{x}_n-{\alpha }_{n-1}f(y_{n-1})-{\gamma }_{n-1}{S}_{n-1}{x}_{n-1}-{\delta }_{n-1}{J}_{r_{n-1}}G{x}_{n-1}\parallel \hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel y_{n-1}-{\beta }_{n-1}{x}_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[{\alpha }_n\parallel f(y_n)-f(y_{n-1})\parallel +{\gamma }_n\parallel S_n{x}_n-S_{n-1}{x}_{n-1}\parallel +{\delta }_n\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel \hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel S_{n-1}{x}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{x}_{n-1}\parallel ]\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel y_{n-1}-{\beta }_{n-1}{x}_{n-1}\parallel .\hfill \end{array}$$

(5.5)

On the other hand, if $r_{n-1}\le r_n$, using the resolvent identity in Proposition 2.2,

$$J_{r_n}{x}_n=J_{r_{n-1}}(\frac{r_{n-1}}{r_n}{x}_n+(1-\frac{r_{n-1}}{r_n})J_{r_n}{x}_n),$$

we get

$$\begin{array}{rcl}\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel & =& \parallel J_{r_{n-1}}(\frac{r_{n-1}}{r_n}G{x}_n+(1-\frac{r_{n-1}}{r_n})J_{r_n}G{x}_n)-J_{r_{n-1}}G{x}_{n-1}\parallel \\ \le & \frac{r_{n-1}}{r_n}\parallel G{x}_n-G{x}_{n-1}\parallel +(1-\frac{r_{n-1}}{r_n})\parallel J_{r_n}G{x}_n-G{x}_{n-1}\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +\frac{r_n-r_{n-1}}{r_n}\parallel J_{r_n}G{x}_n-G{x}_{n-1}\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +\frac{1}{\epsilon }|r_n-r_{n-1}|\parallel J_{r_n}G{x}_n-G{x}_{n-1}\parallel .\end{array}$$

If $r_n\le r_{n-1}$, then it is easy to see that

$$\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel \le \parallel x_{n-1}-x_n\parallel +\frac{1}{\epsilon }|r_{n-1}-r_n|\parallel J_{r_{n-1}}G{x}_{n-1}-G{x}_n\parallel .$$

Thus, combining the above cases, we obtain

$$\begin{array}{c}\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel \hfill \\ \le \parallel x_{n-1}-x_n\parallel +\frac{|r_{n-1}-r_n|}{\epsilon }\underset{n\ge 1}{sup}\{\parallel J_{r_n}G{x}_n-G{x}_{n-1}\parallel +\parallel J_{r_{n-1}}G{x}_{n-1}-G{x}_n\parallel \},\mathrm{\forall }n\ge 1.\hfill \end{array}$$

In a similar way, we derive

$$\begin{array}{c}\parallel J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ \le \parallel y_{n-1}-y_n\parallel +\frac{|r_{n-1}-r_n|}{\epsilon }\underset{n\ge 1}{sup}\{\parallel J_{r_n}G{y}_n-G{y}_{n-1}\parallel +\parallel J_{r_{n-1}}G{y}_{n-1}-G{y}_n\parallel \},\mathrm{\forall }n\ge 1.\hfill \end{array}$$

Therefore, we have

$$\{\begin{array}{c}\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel \le \parallel x_{n-1}-x_n\parallel +|r_{n-1}-r_n|M_0,\hfill \\ \parallel J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}\parallel \le \parallel y_{n-1}-y_n\parallel +|r_{n-1}-r_n|M_0,\hfill \end{array}$$

(5.6)

for all $n\ge 1$, where

$$\underset{n\ge 1}{sup}\left\{\frac{1}{\epsilon }(\parallel J_{r_n}G{x}_n-G{x}_{n-1}\parallel +\parallel J_{r_{n-1}}G{x}_{n-1}-G{x}_n\parallel )\right\}\le M_0,$$

and

$$\underset{n\ge 1}{sup}\left\{\frac{1}{\epsilon }(\parallel J_{r_n}G{y}_n-G{y}_{n-1}\parallel +\parallel J_{r_{n-1}}G{y}_{n-1}-G{y}_n\parallel )\right\}\le M_0,$$

for some $M_0>0$. Combining (5.6) and (5.5), we have

$$\begin{array}{c}\parallel z_n-z_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[{\alpha }_n\parallel f(y_n)-f(y_{n-1})\parallel +{\gamma }_n\parallel S_n{x}_n-S_{n-1}{x}_{n-1}\parallel \hfill \\ +{\delta }_n(\parallel x_{n-1}-x_n\parallel +M_0|r_{n-1}-r_n|)+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel S_{n-1}{x}_{n-1}\parallel \hfill \\ +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{x}_{n-1}\parallel ]+\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel y_{n-1}-{\beta }_{n-1}{x}_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[{\alpha }_n\rho \parallel y_n-y_{n-1}\parallel +{\gamma }_n\parallel S_n{x}_n-S_n{x}_{n-1}\parallel +{\delta }_n(\parallel x_{n-1}-x_n\parallel +M_0|r_{n-1}-r_n|)\hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel S_{n-1}{x}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{x}_{n-1}\parallel \hfill \\ +{\gamma }_n\parallel S_n{x}_{n-1}-S_{n-1}{x}_{n-1}\parallel ]\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel {\alpha }_{n-1}f(y_{n-1})+{\gamma }_{n-1}{S}_{n-1}{x}_{n-1}+{\delta }_{n-1}{J}_{r_{n-1}}G{x}_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[{\alpha }_n\rho \parallel y_n-y_{n-1}\parallel +({\gamma }_n+{\delta }_n)\parallel x_{n-1}-x_n\parallel +M(|{\alpha }_n-{\alpha }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|\hfill \\ +|{\delta }_n-{\delta }_{n-1}|+|r_{n-1}-r_n|)+{\gamma }_n\parallel S_n{x}_{n-1}-S_{n-1}{x}_{n-1}\parallel ]\hfill \\ +\frac{1}{(1-{\beta }_{n-1})(1-{\beta }_n)}|{\beta }_n-{\beta }_{n-1}|M,\hfill \end{array}$$

(5.7)

where ${sup}_{n\ge 0}\{M_0+\parallel f(y_n)\parallel +\parallel S_n{x}_n\parallel +\parallel J_{r_n}G{x}_n\parallel \}\le M$ for some $M>0$. By simple calculations, we have

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

(5.8)

Taking into account condition (v), without loss of generality, we may assume that $\{{\beta }_n\}\subset [a,b]$ for some $a,b\in (0,1)$. Hence, from (5.7) and (5.8), we deduce

$$\begin{array}{c}\parallel y_n-y_{n-1}\parallel \hfill \\ \le {\beta }_n\parallel x_n-x_{n-1}\parallel +(1-{\beta }_n)\parallel z_n-z_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}-z_{n-1}\parallel \hfill \\ \le {\beta }_n\parallel x_n-x_{n-1}\parallel +(1-{\beta }_n)\left\{\frac{1}{1-{\beta }_n}\right[{\alpha }_n\rho \parallel y_n-y_{n-1}\parallel +({\gamma }_n+{\delta }_n)\parallel x_{n-1}-x_n\parallel \hfill \\ +M(|{\alpha }_n-{\alpha }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|+|r_{n-1}-r_n|)+{\gamma }_n\parallel S_n{x}_{n-1}-S_{n-1}{x}_{n-1}\parallel ]\hfill \\ +\frac{1}{(1-{\beta }_{n-1})(1-{\beta }_n)}|{\beta }_n-{\beta }_{n-1}|M\}+|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}-z_{n-1}\parallel \hfill \\ =(1-{\alpha }_n)\parallel x_{n-1}-x_n\parallel +{\alpha }_n\rho \parallel y_n-y_{n-1}\parallel +M(|{\alpha }_n-{\alpha }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|\hfill \\ +|r_{n-1}-r_n|)+{\gamma }_n\parallel S_n{x}_{n-1}-S_{n-1}{x}_{n-1}\parallel \hfill \\ +\frac{1}{1-{\beta }_{n-1}}|{\beta }_n-{\beta }_{n-1}|M+|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}-z_{n-1}\parallel \hfill \\ \le (1-{\alpha }_n)\parallel x_{n-1}-x_n\parallel +{\alpha }_n\rho \parallel y_n-y_{n-1}\parallel +M_1(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|\hfill \\ +|{\delta }_n-{\delta }_{n-1}|+|r_{n-1}-r_n|)+\parallel S_n{x}_{n-1}-S_{n-1}{x}_{n-1}\parallel ,\hfill \end{array}$$

where ${sup}_{n\ge 0}\{\frac{M}{1-b}+\parallel x_n-z_n\parallel \}\le M_1$ for some $M_1>0$. This leads to

$$\begin{array}{c}\parallel y_n-y_{n-1}\parallel \hfill \\ \le (1-\frac{(1-\rho ){\alpha }_n}{1-{\alpha }_n\rho })\parallel x_{n-1}-x_n\parallel +\frac{M_1}{1-{\alpha }_n\rho }(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|\hfill \\ +|{\delta }_n-{\delta }_{n-1}|+|r_{n-1}-r_n|)+\frac{1}{1-{\alpha }_n\rho }\parallel S_n{x}_{n-1}-S_{n-1}{x}_{n-1}\parallel .\hfill \end{array}$$

(5.9)

Again by simple calculations, we have

$$\begin{array}{rcl}{x}_{n+1}-x_n& =& {\sigma }_n(y_n-y_{n-1})+({\sigma }_n-{\sigma }_{n-1})(y_{n-1}-J_{r_{n-1}}G{y}_{n-1})\\ +(1-{\sigma }_n)(J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}).\end{array}$$

This together with (5.6) and (5.9) implies that

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & \le & {\sigma }_n\parallel y_n-y_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel y_{n-1}-J_{r_{n-1}}G{y}_{n-1}\parallel \\ +(1-{\sigma }_n)\parallel J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}\parallel \\ \le & {\sigma }_n\parallel y_n-y_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel y_{n-1}-J_{r_{n-1}}G{y}_{n-1}\parallel \\ +(1-{\sigma }_n)(\parallel y_{n-1}-y_n\parallel +|r_{n-1}-r_n|M_0)\\ \le & \parallel y_n-y_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel y_{n-1}-J_{r_{n-1}}G{y}_{n-1}\parallel +|r_{n-1}-r_n|M_0\\ \le & (1-\frac{(1-\rho ){\alpha }_n}{1-{\alpha }_n\rho })\parallel x_{n-1}-x_n\parallel +\frac{M_1}{1-{\alpha }_n\rho }(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|\\ +|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|+|r_{n-1}-r_n|)+\frac{1}{1-{\alpha }_n\rho }\parallel S_n{x}_{n-1}-S_{n-1}{x}_{n-1}\parallel \\ +|{\sigma }_n-{\sigma }_{n-1}|\parallel y_{n-1}-J_{r_{n-1}}G{y}_{n-1}\parallel +|r_{n-1}-r_n|M_0\\ \le & (1-\frac{(1-\rho ){\alpha }_n}{1-{\alpha }_n\rho })\parallel x_{n-1}-x_n\parallel +\tilde{M}(|{\sigma }_n-{\sigma }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|\\ +|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|+|r_{n-1}-r_n|+\parallel S_n{x}_{n-1}-S_{n-1}{x}_{n-1}\parallel ),\end{array}$$

where ${sup}_{n\ge 0}\{\frac{M_1+1}{1-{\alpha }_n\rho }+M_0+\parallel y_n-J_{r_n}G{y}_n\parallel \}\le \tilde{M}$ for some $\tilde{M}>0$. Noting that $\frac{(1-\rho ){\alpha }_n}{1-{\alpha }_n\rho }\ge (1-\rho ){\alpha }_n$ for all $n\ge 0$, from condition (i), we know that ${\sum }_{n=0}^{\mathrm{\infty }}\frac{(1-\rho ){\alpha }_n}{1-{\alpha }_n\rho }=\mathrm{\infty }$. Utilizing Lemma 2.7, we conclude from conditions (iii), (iv), and the assumption on $\{S_n\}$ that

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

Next we show that $\parallel x_n-G{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Indeed, according to Lemma 2.2(a), we have from (5.1)

$$\begin{array}{c}{\parallel y_n-p\parallel }^2\hfill \\ ={\parallel {\alpha }_n(f(y_n)-f(p))+{\beta }_n(x_n-p)+{\gamma }_n(S_n{x}_n-p)+{\delta }_n(J_{r_n}G{x}_n-p)+{\alpha }_n(f(p)-p)\parallel }^2\hfill \\ \le {\parallel {\alpha }_n(f(y_n)-f(p))+{\beta }_n(x_n-p)+{\gamma }_n(S_n{x}_n-p)+{\delta }_n(J_{r_n}G{x}_n-p)\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(y_n-p)〉\hfill \\ \le {\alpha }_n{\parallel f(y_n)-f(p)\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel S_n{x}_n-p\parallel }^2+{\delta }_n{\parallel J_{r_n}G{x}_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(y_n-p)〉\hfill \\ \le {\alpha }_n{\rho }^2{\parallel y_n-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel x_n-p\parallel }^2+{\delta }_n{\parallel G{x}_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(y_n-p)〉\hfill \\ \le {\alpha }_n\rho {\parallel y_n-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel x_n-p\parallel }^2+{\delta }_n{\parallel x_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(y_n-p)〉\hfill \\ ={\alpha }_n\rho {\parallel y_n-p\parallel }^2+(1-{\alpha }_n){\parallel x_n-p\parallel }^2+2{\alpha }_n〈f(p)-p,J(y_n-p)〉,\hfill \end{array}$$

(5.10)

which implies that

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

Utilizing Lemma 2.3, we get from (5.1) and (5.10)

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel {\sigma }_n(y_n-p)+(1-{\sigma }_n)(J_{r_n}G{y}_n-p)\parallel }^2\\ \le & {\sigma }_n{\parallel y_n-p\parallel }^2+(1-{\sigma }_n){\parallel J_{r_n}G{y}_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel y_n-J_{r_n}G{y}_n\parallel )\\ \le & {\sigma }_n{\parallel y_n-p\parallel }^2+(1-{\sigma }_n){\parallel y_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel y_n-J_{r_n}G{y}_n\parallel )\\ =& {\parallel y_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel y_n-J_{r_n}G{y}_n\parallel )\\ \le & (1-\frac{(1-\rho ){\alpha }_n}{1-{\alpha }_n\rho }){\parallel x_n-p\parallel }^2+\frac{2{\alpha }_n}{1-{\alpha }_n\rho }〈f(p)-p,J(y_n-p)〉\\ -{\sigma }_n(1-{\sigma }_n)g(\parallel y_n-J_{r_n}G{y}_n\parallel )\\ \le & {\parallel x_n-p\parallel }^2+\frac{2{\alpha }_n}{1-{\alpha }_n\rho }\parallel f(p)-p\parallel \parallel y_n-p\parallel -{\sigma }_n(1-{\sigma }_n)g(\parallel y_n-J_{r_n}G{y}_n\parallel ),\end{array}$$

and hence

$$\begin{array}{c}{\sigma }_n(1-{\sigma }_n)g(\parallel y_n-J_{r_n}G{y}_n\parallel )\hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+\frac{2{\alpha }_n}{1-{\alpha }_n\rho }\parallel f(p)-p\parallel \parallel y_n-p\parallel \hfill \\ \le (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +\frac{2{\alpha }_n}{1-{\alpha }_n\rho }\parallel f(p)-p\parallel \parallel y_n-p\parallel .\hfill \end{array}$$

Since ${\alpha }_n\to 0$ and $\parallel x_{n+1}-x_n\parallel \to 0$, from condition (v) and the boundedness of $\{x_n\}$ and $\{y_n\}$, it follows that

$$\underset{n\to \mathrm{\infty }}{lim}g(\parallel y_n-J_{r_n}G{y}_n\parallel )=0.$$

Utilizing the properties of g, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-J_{r_n}G{y}_n\parallel =0.$$

(5.11)

Observe that

$$\begin{array}{rcl}\parallel x_n-y_n\parallel & \le & \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel \\ =& \parallel x_n-x_{n+1}\parallel +(1-{\sigma }_n)\parallel J_{r_n}G{y}_n-y_n\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +\parallel J_{r_n}G{y}_n-y_n\parallel .\end{array}$$

From (5.4) and (5.11), we have

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

(5.12)

For simplicity, put $q={\Pi }_C(p-{\mu }_2{B}_{2}p)$, $u_n={\Pi }_C(x_n-{\mu }_2{B}_2{x}_n)$ and $v_n={\Pi }_C(u_n-{\mu }_1{B}_1{u}_n)$. Then $v_n=G{x}_n$ for all $n\ge 0$. From Lemma 2.8, we have

$$\begin{array}{rcl}{\parallel u_n-q\parallel }^2& =& {\parallel {\Pi }_C(x_n-{\mu }_2{B}_2{x}_n)-{\Pi }_C(p-{\mu }_2{B}_{2}p)\parallel }^2\\ \le & {\parallel x_n-p-{\mu }_2(B_2{x}_n-B_{2}p)\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-2{\mu }_2({\alpha }_2-{\kappa }^2{\mu }_2){\parallel B_2{x}_n-B_{2}p\parallel }^2,\end{array}$$

(5.13)

and

$$\begin{array}{rcl}{\parallel v_n-p\parallel }^2& =& {\parallel {\Pi }_C(u_n-{\mu }_1{B}_1{u}_n)-{\Pi }_C(q-{\mu }_1{B}_{1}q)\parallel }^2\\ \le & {\parallel u_n-q-{\mu }_1(B_1{u}_n-B_{1}q)\parallel }^2\\ \le & {\parallel u_n-q\parallel }^2-2{\mu }_1({\alpha }_1-{\kappa }^2{\mu }_1){\parallel B_1{u}_n-B_{1}q\parallel }^2.\end{array}$$

(5.14)

Combining (5.13) and (5.14), we obtain

$$\begin{array}{rcl}{\parallel v_n-p\parallel }^2& \le & {\parallel x_n-p\parallel }^2-2{\mu }_2({\alpha }_2-{\kappa }^2{\mu }_2){\parallel B_2{x}_n-B_{2}p\parallel }^2\\ -2{\mu }_1({\alpha }_1-{\kappa }^2{\mu }_1){\parallel B_1{u}_n-B_{1}q\parallel }^2.\end{array}$$

(5.15)

By Lemma 2.2(a), (5.1), and (5.15), we have

$$\begin{array}{c}{\parallel y_n-p\parallel }^2\hfill \\ ={\parallel {\alpha }_n(f(y_n)-f(p))+{\beta }_n(x_n-p)+{\gamma }_n(S_n{x}_n-p)+{\delta }_n(J_{r_n}G{x}_n-p)+{\alpha }_n(f(p)-p)\parallel }^2\hfill \\ \le {\alpha }_n{\parallel f(y_n)-f(p)\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel S_n{x}_n-p\parallel }^2+{\delta }_n{\parallel J_{r_n}G{x}_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(y_n-p)〉\hfill \\ \le {\alpha }_n{\rho }^2{\parallel y_n-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel x_n-p\parallel }^2+{\delta }_n{\parallel v_n-p\parallel }^2\hfill \\ +2{\alpha }_n\parallel f(p)-p\parallel \parallel y_n-p\parallel \hfill \\ \le {\alpha }_n\rho {\parallel y_n-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel x_n-p\parallel }^2+{\delta }_n[{\parallel x_n-p\parallel }^2\hfill \\ -2{\mu }_2({\alpha }_2-{\kappa }^2{\mu }_2){\parallel B_2{x}_n-B_{2}p\parallel }^2-2{\mu }_1({\alpha }_1-{\kappa }^2{\mu }_1){\parallel B_1{u}_n-B_{1}q\parallel }^2]\hfill \\ +2{\alpha }_n\parallel f(p)-p\parallel \parallel y_n-p\parallel \hfill \\ \le {\alpha }_n{\parallel y_n-p\parallel }^2+(1-{\alpha }_n){\parallel x_n-p\parallel }^2-2{\delta }_n[{\mu }_2({\alpha }_2-{\kappa }^2{\mu }_2){\parallel B_2{x}_n-B_{2}p\parallel }^2\hfill \\ +{\mu }_1({\alpha }_1-{\kappa }^2{\mu }_1){\parallel B_1{u}_n-B_{1}q\parallel }^2]+2{\alpha }_n\parallel f(p)-p\parallel \parallel y_n-p\parallel .\hfill \end{array}$$

Thus, we have

$$\begin{array}{c}2{\delta }_n[{\mu }_2({\alpha }_2-{\kappa }^2{\mu }_2){\parallel B_2{x}_n-B_{2}p\parallel }^2+{\mu }_1({\alpha }_1-{\kappa }^2{\mu }_1){\parallel B_1{u}_n-B_{1}q\parallel }^2]\hfill \\ \le (1-{\alpha }_n){\parallel x_n-p\parallel }^2-(1-{\alpha }_n){\parallel y_n-p\parallel }^2+2{\alpha }_n\parallel f(p)-p\parallel \parallel y_n-p\parallel \hfill \\ \le (1-{\alpha }_n)(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\parallel x_n-y_n\parallel +2{\alpha }_n\parallel f(p)-p\parallel \parallel y_n-p\parallel .\hfill \end{array}$$

Since $0<{\mu }_i<\frac{{\alpha }_i}{{\kappa }^2}$ for $i=1,2$, from (5.12) and conditions (i), (ii), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel B_2{x}_n-B_{2}p\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel B_1{u}_n-B_{1}q\parallel =0.$$

(5.16)

Utilizing Proposition 2.2 and Lemma 2.10, we have

$$\begin{array}{rcl}{\parallel u_n-q\parallel }^2& =& {\parallel {\Pi }_C(x_n-{\mu }_2{B}_2{x}_n)-{\Pi }_C(p-{\mu }_2{B}_{2}p)\parallel }^2\\ \le & 〈x_n-{\mu }_2{B}_2{x}_n-(p-{\mu }_2{B}_{2}p),J(u_n-q)〉\\ =& 〈x_n-p,J(u_n-q)〉+{\mu }_2〈B_{2}p-B_2{x}_n,J(u_n-q)〉\\ \le & \frac{1}{2}[{\parallel x_n-p\parallel }^2+{\parallel u_n-q\parallel }^2-g_1\left(\parallel x_n-u_n-(p-q)\parallel \right)]\\ +{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel ,\end{array}$$

which implies that

$${\parallel u_n-q\parallel }^2\le {\parallel x_n-p\parallel }^2-g_1\left(\parallel x_n-u_n-(p-q)\parallel \right)+2{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel .$$

(5.17)

In the same way, we derive

$$\begin{array}{rcl}{\parallel v_n-p\parallel }^2& =& {\parallel {\Pi }_C(u_n-{\mu }_1{B}_1{u}_n)-{\Pi }_C(q-{\mu }_1{B}_{1}q)\parallel }^2\\ \le & 〈u_n-{\mu }_1{B}_1{u}_n-(q-{\mu }_1{B}_{1}q),J(v_n-p)〉\\ =& 〈u_n-q,J(v_n-p)〉+{\mu }_1〈B_{1}q-B_1{u}_n,J(v_n-p)〉\\ \le & \frac{1}{2}[{\parallel u_n-q\parallel }^2+{\parallel v_n-p\parallel }^2-g_2\left(\parallel u_n-v_n+(p-q)\parallel \right)]\\ +{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel ,\end{array}$$

which implies that

$${\parallel v_n-p\parallel }^2\le {\parallel u_n-q\parallel }^2-g_2\left(\parallel u_n-v_n+(p-q)\parallel \right)+2{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel .$$

(5.18)

Combining (5.17) and (5.18), we get

$$\begin{array}{rcl}{\parallel v_n-p\parallel }^2& \le & {\parallel x_n-p\parallel }^2-g_1\left(\parallel x_n-u_n-(p-q)\parallel \right)-g_2\left(\parallel u_n-v_n+(p-q)\parallel \right)\\ +2{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel +2{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel .\end{array}$$

(5.19)

By Lemma 2.2(a), (5.1), and (5.19), we have

$$\begin{array}{c}{\parallel y_n-p\parallel }^2\hfill \\ ={\parallel {\alpha }_n(f(y_n)-f(p))+{\beta }_n(x_n-p)+{\gamma }_n(S_n{x}_n-p)+{\delta }_n(J_{r_n}G{x}_n-p)+{\alpha }_n(f(p)-p)\parallel }^2\hfill \\ \le {\alpha }_n{\parallel f(y_n)-f(p)\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel S_n{x}_n-p\parallel }^2+{\delta }_n{\parallel J_{r_n}G{x}_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(y_n-p)〉\hfill \\ \le {\alpha }_n\rho {\parallel y_n-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel x_n-p\parallel }^2+{\delta }_n{\parallel v_n-p\parallel }^2+2{\alpha }_n\parallel f(p)-p\parallel \parallel y_n-p\parallel \hfill \\ \le {\alpha }_n\rho {\parallel y_n-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel x_n-p\parallel }^2+{\delta }_n[{\parallel x_n-p\parallel }^2\hfill \\ -g_1\left(\parallel x_n-u_n-(p-q)\parallel \right)-g_2\left(\parallel u_n-v_n+(p-q)\parallel \right)+2{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel \hfill \\ +2{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel ]+2{\alpha }_n\parallel f(p)-p\parallel \parallel y_n-p\parallel \hfill \\ \le {\alpha }_n{\parallel y_n-p\parallel }^2+(1-{\alpha }_n){\parallel x_n-p\parallel }^2-{\delta }_n[g_1\left(\parallel x_n-u_n-(p-q)\parallel \right)\hfill \\ +g_2\left(\parallel u_n-v_n+(p-q)\parallel \right)]+2{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel \hfill \\ +2{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel +2{\alpha }_n\parallel f(p)-p\parallel \parallel y_n-p\parallel ,\hfill \end{array}$$

and hence

$$\begin{array}{c}{\delta }_n[g_1\left(\parallel x_n-u_n-(p-q)\parallel \right)+g_2\left(\parallel u_n-v_n+(p-q)\parallel \right)]\hfill \\ \le (1-{\alpha }_n){\parallel x_n-p\parallel }^2-(1-{\alpha }_n){\parallel y_n-p\parallel }^2+2{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel \hfill \\ +2{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel +2{\alpha }_n\parallel f(p)-p\parallel \parallel y_n-p\parallel \hfill \\ \le (1-{\alpha }_n)(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\parallel x_n-y_n\parallel +2{\mu }_2\parallel B_{2}p-B_2{x}_n\parallel \parallel u_n-q\parallel \hfill \\ +2{\mu }_1\parallel B_{1}q-B_1{u}_n\parallel \parallel v_n-p\parallel +2{\alpha }_n\parallel f(p)-p\parallel \parallel y_n-p\parallel .\hfill \end{array}$$

Utilizing conditions (i), (ii), from (5.12) and (5.16), we have

$$\underset{n\to \mathrm{\infty }}{lim}{g}_1\left(\parallel x_n-u_n-(p-q)\parallel \right)=0\text{and}\underset{n\to \mathrm{\infty }}{lim}{g}_2\left(\parallel u_n-v_n+(p-q)\parallel \right)=0.$$

(5.20)

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-u_n-(p-q)\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-v_n+(p-q)\parallel =0.$$

(5.21)

From (5.21), we get

$$\parallel x_n-v_n\parallel \le \parallel x_n-u_n-(p-q)\parallel +\parallel u_n-v_n+(p-q)\parallel \to 0\text{as }n\to \mathrm{\infty },$$

that is,

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

(5.22)

Next, let us show that

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

Indeed, observe that $y_n$ can be rewritten as

$$\begin{array}{rcl}{y}_n& =& {\alpha }_{n}f(y_n)+{\beta }_n{x}_n+{\gamma }_n{S}_n{x}_n+{\delta }_n{J}_{r_n}G{x}_n\\ =& {\alpha }_{n}f(y_n)+{\beta }_n{x}_n+({\gamma }_n+{\delta }_n)\frac{{\gamma }_n{S}_n{x}_n+{\delta }_n{J}_{r_n}G{x}_n}{{\gamma }_n+{\delta }_n}\\ =& {\alpha }_{n}f(y_n)+{\beta }_n{x}_n+e_n{\hat{z}}_n,\end{array}$$

(5.23)

where $e_n={\gamma }_n+{\delta }_n$ and ${\hat{z}}_n=\frac{{\gamma }_n{S}_n{x}_n+{\delta }_n{J}_{r_n}G{x}_n}{{\gamma }_n+{\delta }_n}$. Utilizing Lemma 2.4 and (5.23), we have

$$\begin{array}{c}{\parallel y_n-p\parallel }^2\hfill \\ ={\parallel {\alpha }_n(f(y_n)-p)+{\beta }_n(x_n-p)+e_n({\hat{z}}_n-p)\parallel }^2\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+e_n{\parallel {\hat{z}}_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\hfill \\ ={\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\hfill \\ +e_n{\parallel \frac{{\gamma }_n{S}_n{x}_n+{\delta }_n{J}_{r_n}G{x}_n}{{\gamma }_n+{\delta }_n}-p\parallel }^2\hfill \\ ={\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\hfill \\ +e_n{\parallel \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(S_n{x}_n-p)+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}(J_{r_n}G{x}_n-p)\parallel }^2\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\hfill \\ +e_n[\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}{\parallel S_n{x}_n-p\parallel }^2+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel J_{r_n}G{x}_n-p\parallel }^2]\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\hfill \\ +e_n[\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}\parallel x_n-p\parallel +\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel x_n-p\parallel }^2]\hfill \\ ={\alpha }_n{\parallel f(y_n)-p\parallel }^2+(1-{\alpha }_n){\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel ),\hfill \end{array}$$

which implies that

$$\begin{array}{rcl}{\beta }_n{e}_n{g}_3(\parallel {\hat{z}}_n-x_n\parallel )& \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel y_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\parallel x_n-y_n\parallel .\end{array}$$

Utilizing (5.12), conditions (i), (ii), (v), and the boundedness of $\{x_n\}$, $\{y_n\}$, and $\{f(y_n)\}$, we get

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

From the properties of $g_3$, we have

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

Utilizing Lemma 2.3 and the definition of ${\hat{z}}_n$, we have

$$\begin{array}{rcl}{\parallel {\hat{z}}_n-p\parallel }^2& =& {\parallel \frac{{\gamma }_n{S}_n{x}_n+{\delta }_n{J}_{r_n}G{x}_n}{{\gamma }_n+{\delta }_n}-p\parallel }^2\\ =& {\parallel \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(S_n{x}_n-p)+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}(J_{r_n}G{x}_n-p)\parallel }^2\\ \le & \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}{\parallel S_n{x}_n-p\parallel }^2+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel J_{r_n}G{x}_n-p\parallel }^2\\ -\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_4(\parallel J_{r_n}G{x}_n-S_n{x}_n\parallel )\\ \le & {\parallel x_n-p\parallel }^2-\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_4(\parallel J_{r_n}G{x}_n-S_n{x}_n\parallel ),\end{array}$$

which leads to

$$\begin{array}{rcl}\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_4(\parallel J_{r_n}G{x}_n-S_n{x}_n\parallel )& \le & {\parallel x_n-p\parallel }^2-{\parallel {\hat{z}}_n-p\parallel }^2\\ \le & (\parallel x_n-p\parallel +\parallel {\hat{z}}_n-p\parallel )\parallel x_n-{\hat{z}}_n\parallel .\end{array}$$

Since $\{x_n\}$ and $\{{\hat{z}}_n\}$ are bounded and $\parallel {\hat{z}}_n-x_n\parallel \to 0$ as $n\to \mathrm{\infty }$, we deduce from condition (ii) that

$$\underset{n\to \mathrm{\infty }}{lim}{g}_4(\parallel S_n{x}_n-J_{r_n}G{x}_n\parallel )=0.$$

From the properties of $g_4$, we have

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

(5.24)

On the other hand, $y_n$ can also be rewritten as

$$\begin{array}{rcl}{y}_n& =& {\alpha }_{n}f(y_n)+{\beta }_n{x}_n+{\gamma }_n{S}_n{x}_n+{\delta }_n{J}_{r_n}G{x}_n\\ =& {\beta }_n{x}_n+{\gamma }_n{S}_n{x}_n+({\alpha }_n+{\delta }_n)\frac{{\alpha }_{n}f(y_n)+{\delta }_n{J}_{r_n}G{x}_n}{{\alpha }_n+{\delta }_n}={\beta }_n{x}_n+{\gamma }_n{S}_n{x}_n+d_n{\tilde{z}}_n,\end{array}$$

where $d_n={\alpha }_n+{\delta }_n$ and ${\tilde{z}}_n=\frac{{\alpha }_{n}f(y_n)+{\delta }_n{J}_{r_n}G{x}_n}{{\alpha }_n+{\delta }_n}$. Utilizing Lemma 2.4 and the convexity of ${\parallel \cdot \parallel }^2$, we have

$$\begin{array}{c}{\parallel y_n-p\parallel }^2\hfill \\ ={\parallel {\beta }_n(x_n-p)+{\gamma }_n(S_n{x}_n-p)+d_n({\tilde{z}}_n-p)\parallel }^2\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel S_n{x}_n-p\parallel }^2+d_n{\parallel {\tilde{z}}_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_5(\parallel x_n-S_n{x}_n\parallel )\hfill \\ ={\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel S_n{x}_n-p\parallel }^2+d_n{\parallel \frac{{\alpha }_{n}f(y_n)+{\delta }_n{J}_{r_n}G{x}_n}{{\alpha }_n+{\delta }_n}-p\parallel }^2\hfill \\ -{\beta }_n{\gamma }_n{g}_5(\parallel x_n-S_n{x}_n\parallel )\hfill \\ ={\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel S_n{x}_n-p\parallel }^2+d_n{\parallel \frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}(f(y_n)-p)+\frac{{\delta }_n}{{\alpha }_n+{\delta }_n}(J_{r_n}G{x}_n-p)\parallel }^2\hfill \\ -{\beta }_n{\gamma }_n{g}_5(\parallel x_n-S_n{x}_n\parallel )\hfill \\ \le {\beta }_n{\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel x_n-p\parallel }^2+d_n[\frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}{\parallel f(y_n)-p\parallel }^2+\frac{{\delta }_n}{{\alpha }_n+{\delta }_n}{\parallel J_{r_n}G{x}_n-p\parallel }^2]\hfill \\ -{\beta }_n{\gamma }_n{g}_5(\parallel x_n-S_n{x}_n\parallel )\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+({\beta }_n+{\gamma }_n){\parallel x_n-p\parallel }^2+{\delta }_n{\parallel x_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_5(\parallel x_n-S_n{x}_n\parallel )\hfill \\ ={\alpha }_n{\parallel f(y_n)-p\parallel }^2+(1-{\alpha }_n){\parallel x_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_5(\parallel x_n-S_n{x}_n\parallel )\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_5(\parallel x_n-S_n{x}_n\parallel ),\hfill \end{array}$$

which implies that

$$\begin{array}{rcl}{\beta }_n{\gamma }_n{g}_5(\parallel x_n-S_n{x}_n\parallel )& \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel y_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+(\parallel x_n-p\parallel +\parallel y_n-p\parallel )\parallel x_n-y_n\parallel .\end{array}$$

From (5.12), conditions (i), (ii), (v), and the boundedness of $\{x_n\}$, $\{y_n\}$, and $\{f(y_n)\}$, we have

$$\underset{n\to \mathrm{\infty }}{lim}{g}_5(\parallel x_n-S_n{x}_n\parallel )=0.$$

Utilizing the properties of $g_5$, we have

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

(5.25)

By Lemma 5.1, we get

$$\parallel x_n-S{x}_n\parallel \le \parallel x_n-S_n{x}_n\parallel +\parallel S_n{x}_n-S{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty },$$

that is,

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

(5.26)

We note that

$$\begin{array}{rcl}\parallel x_n-J_{r_n}{x}_n\parallel & \le & \parallel x_n-S_n{x}_n\parallel +\parallel S_n{x}_n-J_{r_n}G{x}_n\parallel +\parallel J_{r_n}G{x}_n-J_{r_n}{x}_n\parallel \\ \le & \parallel x_n-S_n{x}_n\parallel +\parallel S_n{x}_n-J_{r_n}G{x}_n\parallel +\parallel G{x}_n-x_n\parallel .\end{array}$$

So, from (5.22), (5.24), and (5.25), it follows that

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

(5.27)

Furthermore, we claim that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-J_r{x}_n\parallel =0$ for a fixed number r such that $\epsilon >r>0$. In fact, taking into account the resolvent identity in Proposition 2.2, we have

$$\begin{array}{rcl}\parallel J_{r_n}{x}_n-J_r{x}_n\parallel & =& \parallel J_r(\frac{r}{r_n}{x}_n+(1-\frac{r}{r_n})J_{r_n}{x}_n)-J_r{x}_n\parallel \\ \le & (1-\frac{r}{r_n})\parallel x_n-J_{r_n}{x}_n\parallel \le \parallel x_n-J_{r_n}{x}_n\parallel .\end{array}$$

(5.28)

From (5.27) and (5.8), we get

$$\begin{array}{rcl}\parallel x_n-J_r{x}_n\parallel & \le & \parallel x_n-J_{r_n}{x}_n\parallel +\parallel J_{r_n}{x}_n-J_r{x}_n\parallel \le \parallel x_n-J_{r_n}{x}_n\parallel +\parallel x_n-J_{r_n}{x}_n\parallel \\ =& 2\parallel x_n-J_{r_n}{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty },\end{array}$$

that is,

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

(5.29)

Define a mapping $Wx=(1-{\theta }_1-{\theta }_2)J_{r}x+{\theta }_{1}Sx+{\theta }_{2}Gx$, where ${\theta }_1,{\theta }_2\in (0,1)$ are two constants with ${\theta }_1+{\theta }_2<1$. Then by Lemma 2.5, we have $Fix(W)=Fix(J_r)\cap Fix(S)\cap Fix(G)=F$. We observe that

$$\begin{array}{rcl}\parallel x_n-W{x}_n\parallel & =& \parallel (1-{\theta }_1-{\theta }_2)(x_n-J_r{x}_n)+{\theta }_1(x_n-S{x}_n)+{\theta }_2(x_n-G{x}_n)\parallel \\ \le & (1-{\theta }_1-{\theta }_2)\parallel x_n-J_r{x}_n\parallel +{\theta }_1\parallel x_n-S{x}_n\parallel +{\theta }_2\parallel x_n-G{x}_n\parallel .\end{array}$$

From (5.22), (5.26), and (5.29), we obtain

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

(5.30)

Now, we claim that

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

(5.31)

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(a), we obtain

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

(5.32)

where

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

(5.33)

It follows from (5.32) that

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

(5.34)

Letting $n\to \mathrm{\infty }$ in (5.34) and noticing (5.33), we derive

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

(5.35)

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 (5.35), we have

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

(5.36)

On the other hand, we have

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

It follows that

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

Taking into account that $x_t\to q$ as $t\to 0$, we have from (5.36)

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

(5.37)

Since X has a uniformly Fréchet differentiable norm, the duality mapping J is norm-to-norm uniformly continuous on bounded subsets of X. Consequently, the two limits are interchangeable and hence (5.31) holds. From (5.12) we get $(y_n-q)-(x_n-q)\to 0$. Noticing that J is norm-to-norm uniformly continuous on bounded subsets of X, we deduce from (5.31) that

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

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

$$\begin{array}{c}{\parallel y_n-q\parallel }^2\hfill \\ ={\parallel {\alpha }_n(f(y_n)-f(q))+{\beta }_n(x_n-q)+{\gamma }_n(S_n{x}_n-q)+{\delta }_n(J_{r_n}G{x}_n-q)+{\alpha }_n(f(q)-q)\parallel }^2\hfill \\ \le {\parallel {\alpha }_n(f(y_n)-f(q))+{\beta }_n(x_n-q)+{\gamma }_n(S_n{x}_n-q)+{\delta }_n(J_{r_n}G{x}_n-q)\parallel }^2\hfill \\ +2{\alpha }_n〈f(q)-q,J(y_n-q)〉\hfill \\ \le {\alpha }_n{\parallel f(y_n)-f(q)\parallel }^2+{\beta }_n{\parallel x_n-q\parallel }^2+{\gamma }_n{\parallel S_n{x}_n-q\parallel }^2+{\delta }_n{\parallel J_{r_n}G{x}_n-q\parallel }^2\hfill \\ +2{\alpha }_n〈f(q)-q,J(y_n-q)〉\hfill \\ \le {\alpha }_n\rho {\parallel y_n-q\parallel }^2+(1-{\alpha }_n){\parallel x_n-q\parallel }^2+2{\alpha }_n〈f(q)-q,J(y_n-q)〉,\hfill \end{array}$$

which implies that

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

(5.38)

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

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& \le & {\sigma }_n{\parallel y_n-q\parallel }^2+(1-{\sigma }_n){\parallel J_{r_n}G{y}_n-q\parallel }^2\\ \le & {\parallel y_n-q\parallel }^2\\ \le & (1-\frac{{\alpha }_n(1-\rho )}{1-{\alpha }_n\rho }){\parallel x_n-q\parallel }^2+\frac{{\alpha }_n(1-\rho )}{1-{\alpha }_n\rho }\cdot \frac{2〈f(q)-q,J(y_n-q)〉}{1-\rho }.\end{array}$$

(5.39)

Applying Lemma 2.7 to (5.39), we obtain $x_n\to q$ as $n\to \mathrm{\infty }$. This completes the proof. □

Corollary 5.1 Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X. Let ${\Pi }_C$ be a sunny nonexpansive retraction from X onto C and $A\subset X\times X$ be an accretive operator on X such that $\overline{D(A)}\subset C\subset {\bigcap }_{r>0}R(I+rA)$. Let the mapping $B_i:C\to X$ be ${\alpha }_i$-inverse strongly accretive for $i=1,2$, and $f:C\to C$ be a contraction with coefficient $\rho \in (0,1)$. Let $S:C\to C$ be a nonexpansive mapping such that $F=Fix(S)\cap \Omega \cap A^{-1}0\ne \mathrm{\varnothing }$ with $0<{\mu }_i<\frac{{\alpha }_i}{{\kappa }^2}$ for $i=1,2$. For arbitrarily given $x_0\in C$, let $\{x_n\}$ be the sequence generated by

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

Suppose that Assumption  5.1 holds. 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 subset D of C, $S:C\to C$ is a mapping defined by $Sx={lim}_{n\to \mathrm{\infty }}{S}_{n}x$ for all $x\in C$, and $Fix(S)={\bigcap }_{n=0}^{\mathrm{\infty }}Fix(S_n)$. Then the sequence $\{x_n\}$ converges strongly to $q\in F$, which solves the following VIP:

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

We now establish the following strong convergence result on the composite explicit viscosity algorithm.

Theorem 5.2 Let C be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gâteaux differentiable norm. Let ${\Pi }_C$ be a sunny nonexpansive retraction from X onto C and $A\subset X\times X$ be an accretive operator on X such that $\overline{D(A)}\subset C\subset {\bigcap }_{r>0}R(I+rA)$. For each $i=1,2$, let $B_i:C\to X$ be a ${\lambda }_i$-strictly pseudocontractive and ${\alpha }_i$-strongly accretive mapping with ${\alpha }_i+{\lambda }_i\ge 1$. Let $f:C\to C$ be a contraction with coefficient $\rho \in (0,1)$ and ${\{S_i\}}_{i=0}^{\mathrm{\infty }}$ be an infinite family of nonexpansive mappings $S_i:C\to C$ such that $F={\bigcap }_{i=0}^{\mathrm{\infty }}Fix(S_i)\cap \Omega \cap A^{-1}0\ne \mathrm{\varnothing }$ with $1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$ for $i=1,2$. Suppose that Assumption  5.1 holds. For arbitrarily given $x_0\in C$, let $\{x_n\}$ be the sequence generated by

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

(5.40)

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 subset D of C, $S:C\to C$ is a mapping defined by $Sx={lim}_{n\to \mathrm{\infty }}{S}_{n}x$ for all $x\in C$, and $Fix(S)={\bigcap }_{n=0}^{\mathrm{\infty }}Fix(S_n)$. Then $\{x_n\}$ converges strongly to $q\in F$, which solves the following VIP:

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

Proof Take a fixed $p\in F$ arbitrarily. Then we obtain $p=Gp$, $p=S_{n}p$ and $J_{r_n}p=p$ for all $n\ge 0$. Moreover, by Lemma 4.2, we have

$$\begin{array}{rcl}\parallel y_n-p\parallel & \le & {\sigma }_n\parallel G{x}_n-p\parallel +(1-{\sigma }_n)\parallel J_{r_n}G{x}_n-p\parallel \\ \le & {\sigma }_n\parallel x_n-p\parallel +(1-{\sigma }_n)\parallel x_n-p\parallel \\ =& \parallel x_n-p\parallel ,\end{array}$$

(5.41)

and therefore

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & \le & {\alpha }_n\parallel f(y_n)-p\parallel +{\beta }_n\parallel y_n-p\parallel +{\gamma }_n\parallel S_n{y}_n-p\parallel +{\delta }_n\parallel J_{r_n}G{y}_n-p\parallel \\ \le & {\alpha }_n(\parallel f(y_n)-f(p)\parallel +\parallel f(p)-p\parallel )+{\beta }_n\parallel y_n-p\parallel \\ +{\gamma }_n\parallel y_n-p\parallel +{\delta }_n\parallel y_n-p\parallel \\ \le & {\alpha }_n\rho \parallel y_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel +({\beta }_n+{\gamma }_n+{\delta }_n)\parallel y_n-p\parallel \\ =& (1-{\alpha }_n(1-\rho ))\parallel y_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel \\ \le & (1-{\alpha }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel \\ =& (1-{\alpha }_n(1-\rho ))\parallel x_n-p\parallel +{\alpha }_n(1-\rho )\cdot \frac{\parallel f(p)-p\parallel }{1-\rho }\\ \le & max\{\parallel x_n-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-\rho }\}.\end{array}$$

By induction, we get

$$\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-\rho }\},\mathrm{\forall }n\ge 0,$$

which implies that $\{x_n\}$ is bounded and so are the sequences $\{y_n\}$, $\{G{x}_n\}$, $\{G{y}_n\}$, $\{f(y_n)\}$.

Let us show that $\parallel x_{n+1}-x_n\parallel \to 0$ as $n\to \mathrm{\infty }$. As a matter of fact, repeating the same arguments as those in the proof of Theorem 4.1, we obtain

$$\{\begin{array}{c}\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel \le \parallel x_{n-1}-x_n\parallel +|r_{n-1}-r_n|M_0,\hfill \\ \parallel J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}\parallel \le \parallel y_{n-1}-y_n\parallel +|r_{n-1}-r_n|M_0,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(5.42)

where

$$\underset{n\ge 1}{sup}\left\{\frac{1}{\epsilon }(\parallel J_{r_n}G{x}_n-G{x}_{n-1}\parallel +\parallel J_{r_{n-1}}G{x}_{n-1}-G{x}_n\parallel )\right\}\le M_0,$$

and

$$\underset{n\ge 1}{sup}\left\{\frac{1}{\epsilon }(\parallel J_{r_n}G{y}_n-G{y}_{n-1}\parallel +\parallel J_{r_{n-1}}G{y}_{n-1}-G{y}_n\parallel )\right\}\le M_0,$$

for some $M_0>0$. By (5.40) and simple calculations, we have

$$\begin{array}{rcl}{y}_n-y_{n-1}& =& {\sigma }_n(G{x}_n-G{x}_{n-1})+({\sigma }_n-{\sigma }_{n-1})(G{x}_{n-1}-J_{r_{n-1}}G{x}_{n-1})\\ +(1-{\alpha }_n)(J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}).\end{array}$$

It follows that

$$\begin{array}{rcl}\parallel y_n-y_{n-1}\parallel & \le & {\sigma }_n\parallel G{x}_n-G{x}_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G{x}_{n-1}-J_{r_{n-1}}G{x}_{n-1}\parallel \\ +(1-{\alpha }_n)\parallel J_{r_n}G{x}_n-J_{r_{n-1}}G{x}_{n-1}\parallel \\ \le & {\sigma }_n\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G{x}_{n-1}-J_{r_{n-1}}G{x}_{n-1}\parallel \\ +(1-{\sigma }_n)(\parallel x_{n-1}-x_n\parallel +|r_{n-1}-r_n|M_0)\\ \le & \parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G{x}_{n-1}-J_{r_{n-1}}G{x}_{n-1}\parallel +|r_n-r_{n-1}|M_0.\end{array}$$

(5.43)

Taking into account condition (v), without loss of generality we may assume that $\{{\beta }_n\}\subset [a,b]$ for some $a,b\in (0,1)$. From (5.40), $x_{n+1}$ can be rewritten as

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

(5.44)

where $z_n=\frac{{\alpha }_{n}f(y_n)+{\gamma }_n{S}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{1-{\beta }_n}$. Utilizing (5.42) and (5.43), we have

$$\begin{array}{c}\parallel z_n-z_{n-1}\parallel \hfill \\ =\parallel \frac{{\alpha }_{n}f(y_n)+{\gamma }_n{S}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{1-{\beta }_n}-\frac{{\alpha }_{n-1}f(y_{n-1})+{\gamma }_{n-1}{S}_{n-1}{y}_{n-1}+{\delta }_{n-1}{J}_{r_{n-1}}G{y}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ =\parallel \frac{x_{n+1}-{\beta }_n{y}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ =\parallel \frac{x_{n+1}-{\beta }_n{y}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_n}+\frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ \le \parallel \frac{x_{n+1}-{\beta }_n{y}_n}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_n}\parallel +\parallel \frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_n}-\frac{x_n-{\beta }_{n-1}{y}_{n-1}}{1-{\beta }_{n-1}}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}\parallel x_{n+1}-{\beta }_n{y}_n-(x_n-{\beta }_{n-1}{y}_{n-1})\parallel +|\frac{1}{1-{\beta }_n}-\frac{1}{1-{\beta }_{n-1}}|\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}\parallel x_{n+1}-{\beta }_n{y}_n-(x_n-{\beta }_{n-1}{y}_{n-1})\parallel +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ =\frac{1}{1-{\beta }_n}\parallel {\alpha }_{n}f(y_n)+{\gamma }_n{S}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n-{\alpha }_{n-1}f(y_{n-1})-{\gamma }_{n-1}{S}_{n-1}{y}_{n-1}-{\delta }_{n-1}{J}_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[{\alpha }_n\parallel f(y_n)-f(y_{n-1})\parallel +{\gamma }_n\parallel S_n{y}_n-S_{n-1}{y}_{n-1}\parallel +{\delta }_n\parallel J_{r_n}G{y}_n-J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel S_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel ]\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[{\alpha }_n\rho \parallel y_n-y_{n-1}\parallel +{\gamma }_n\parallel S_n{y}_n-S_n{y}_{n-1}\parallel +{\delta }_n[\parallel y_{n-1}-y_n\parallel +|r_{n-1}-r_n|M_0]\hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel S_{n-1}{y}_{n-1}\parallel +{\gamma }_n\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel \hfill \\ +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel ]+\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ \le \frac{1}{1-{\beta }_n}[({\alpha }_n\rho +{\gamma }_n+{\delta }_n)\parallel y_{n-1}-y_n\parallel +|r_{n-1}-r_n|M_0+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel \hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|\parallel S_{n-1}{y}_{n-1}\parallel +{\gamma }_n\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel ]\hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ \le (1-\frac{(1-\rho ){\alpha }_n}{1-{\beta }_n})\parallel y_n-y_{n-1}\parallel +\frac{1}{1-{\beta }_n}[|r_{n-1}-r_n|M_0+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel \hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|\parallel S_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel ]+\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel \hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel .\hfill \end{array}$$

(5.45)

By simple calculations and (5.44), we get

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

This together with (5.43) and (5.45) implies that

$$\begin{array}{c}\parallel x_{n+1}-x_n\parallel \hfill \\ \le {\beta }_n\parallel y_n-y_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel y_{n-1}-z_{n-1}\parallel +(1-{\beta }_n)\parallel z_n-z_{n-1}\parallel \hfill \\ \le {\beta }_n\parallel y_n-y_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel y_{n-1}-z_{n-1}\parallel +(1-{\beta }_n)\{(1-\frac{(1-\rho ){\alpha }_n}{1-{\beta }_n})\parallel y_n-y_{n-1}\parallel \hfill \\ +\frac{1}{1-{\beta }_n}[|r_{n-1}-r_n|M_0+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel S_{n-1}{y}_{n-1}\parallel \hfill \\ +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel ]+\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel \hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{(1-{\beta }_{n-1})(1-{\beta }_n)}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \}\hfill \\ \le (1-(1-\rho ){\alpha }_n)\parallel y_n-y_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel y_{n-1}-z_{n-1}\parallel +|r_{n-1}-r_n|M_0\hfill \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel S_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ +\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel +\frac{|{\beta }_n-{\beta }_{n-1}|}{1-{\beta }_{n-1}}\parallel x_n-{\beta }_{n-1}{y}_{n-1}\parallel \hfill \\ \le (1-(1-\rho ){\alpha }_n)[\parallel x_n-x_{n-1}\parallel +|{\sigma }_n-{\sigma }_{n-1}|\parallel G{x}_{n-1}-J_{r_{n-1}}G{x}_{n-1}\parallel +|r_n-r_{n-1}|M_0]\hfill \\ +|{\beta }_n-{\beta }_{n-1}|\parallel y_{n-1}-z_{n-1}\parallel +|r_{n-1}-r_n|M_0+|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})\parallel \hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|\parallel S_{n-1}{y}_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel J_{r_{n-1}}G{y}_{n-1}\parallel +\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel \hfill \\ +\frac{|{\beta }_n-{\beta }_{n-1}|}{1-{\beta }_{n-1}}\parallel {\alpha }_{n-1}f(y_{n-1})+{\gamma }_{n-1}{S}_{n-1}{y}_{n-1}+{\delta }_{n-1}{J}_{r_{n-1}}G{y}_{n-1}\parallel \hfill \\ \le (1-(1-\rho ){\alpha }_n)\parallel x_n-x_{n-1}\parallel +(|{\sigma }_n-{\sigma }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|\hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|+|{\delta }_n-{\delta }_{n-1}|+|r_{n-1}-r_n|)M+\parallel S_n{y}_{n-1}-S_{n-1}{y}_{n-1}\parallel ,\hfill \end{array}$$

where $\frac{1}{1-b}{sup}_{n\ge 0}\{\parallel f(y_n)\parallel +\parallel S_n{y}_n\parallel +\parallel J_{r_n}G{y}_n\parallel +\parallel G{x}_n-J_{r_n}G{x}_n\parallel +\parallel y_n-z_n\parallel +2M_0\}\le M$ for some $M>0$. So, in terms of Lemma 2.7 and conditions (i), (iii), and (iv), we conclude that

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

(5.46)

Next we show that $\parallel x_n-G{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Indeed, utilizing Lemma 2.3 and (5.40), we get

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel {\sigma }_n(G{x}_n-p)+(1-{\sigma }_n)(J_{r_n}G{x}_n-p)\parallel }^2\\ \le & {\sigma }_n{\parallel G{x}_n-p\parallel }^2+(1-{\sigma }_n){\parallel J_{r_n}G{x}_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel G{x}_n-J_{r_n}G{x}_n\parallel )\\ \le & {\sigma }_n{\parallel x_n-p\parallel }^2+(1-{\sigma }_n){\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel G{x}_n-J_{r_n}G{x}_n\parallel )\\ =& {\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel G{x}_n-J_{r_n}G{x}_n\parallel ).\end{array}$$

(5.47)

According to Lemma 2.2, we have from (5.40) and (5.47)

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ ={\parallel {\alpha }_n(f(y_n)-f(p))+{\beta }_n(y_n-p)+{\gamma }_n(S_n{y}_n-p)+{\delta }_n(J_{r_n}G{y}_n-p)+{\alpha }_n(f(p)-p)\parallel }^2\hfill \\ \le {\parallel {\alpha }_n(f(y_n)-f(p))+{\beta }_n(y_n-p)+{\gamma }_n(S_n{y}_n-p)+{\delta }_n(J_{r_n}G{y}_n-p)\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(x_{n+1}-p)〉\hfill \\ \le {\alpha }_n{\parallel f(y_n)-f(p)\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel S_n{y}_n-p\parallel }^2+{\delta }_n{\parallel J_{r_n}G{y}_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(x_{n+1}-p)〉\hfill \\ \le {\alpha }_n{\rho }^2{\parallel y_n-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2+{\delta }_n{\parallel G{y}_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈f(p)-p,J(x_{n+1}-p)〉\hfill \\ \le {\alpha }_n\rho {\parallel y_n-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2+{\delta }_n{\parallel y_n-p\parallel }^2\hfill \\ +2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel \hfill \\ =(1-{\alpha }_n(1-\rho )){\parallel y_n-p\parallel }^2+2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le {\parallel y_n-p\parallel }^2+2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le {\parallel x_n-p\parallel }^2-{\sigma }_n(1-{\sigma }_n)g(\parallel G{x}_n-J_{r_n}G{x}_n\parallel )+2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel ,\hfill \end{array}$$

which hence yields

$$\begin{array}{c}{\sigma }_n(1-{\sigma }_n)g(\parallel G{x}_n-J_{r_n}G{x}_n\parallel )\hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +2{\alpha }_n\parallel f(p)-p\parallel \parallel x_{n+1}-p\parallel .\hfill \end{array}$$

Since ${\alpha }_n\to 0$ and $\parallel x_{n+1}-x_n\parallel \to 0$, from condition (v) and the boundedness of $\{x_n\}$, it follows that

$$\underset{n\to \mathrm{\infty }}{lim}g(\parallel G{x}_n-J_{r_n}G{x}_n\parallel )=0.$$

Utilizing the properties of g, we have

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

(5.48)

On the other hand, $x_{n+1}$ can be rewritten as

$$\begin{array}{rcl}{x}_{n+1}& =& {\alpha }_{n}f(y_n)+{\beta }_n{y}_n+{\gamma }_n{S}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n\\ =& {\alpha }_{n}f(y_n)+{\beta }_n{y}_n+({\gamma }_n+{\delta }_n)\frac{{\gamma }_n{S}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{{\gamma }_n+{\delta }_n}\\ =& {\alpha }_{n}f(y_n)+{\beta }_n{y}_n+e_n{\hat{z}}_n,\end{array}$$

(5.49)

where $e_n={\gamma }_n+{\delta }_n$ and ${\hat{z}}_n=\frac{{\gamma }_n{S}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{{\gamma }_n+{\delta }_n}$. Utilizing Lemma 2.4, from (5.41) and (5.49), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel {\alpha }_n(f(y_n)-p)+{\beta }_n(y_n-p)+e_n({\hat{z}}_n-p)\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2+e_n{\parallel {\hat{z}}_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ =& {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ +e_n{\parallel \frac{{\gamma }_n{S}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{{\gamma }_n+{\delta }_n}-p\parallel }^2\\ =& {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ +e_n{\parallel \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(S_n{y}_n-p)+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}(J_{r_n}G{y}_n-p)\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ +e_n[\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}{\parallel S_n{y}_n-p\parallel }^2+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel J_{r_n}G{y}_n-p\parallel }^2]\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\beta }_n{\parallel y_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ +e_n[\frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}{\parallel y_n-p\parallel }^2+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel y_n-p\parallel }^2]\\ =& {\alpha }_n{\parallel f(y_n)-p\parallel }^2+(1-{\alpha }_n){\parallel y_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel y_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel ),\end{array}$$

which hence implies that

$$\begin{array}{rcl}{\beta }_n{e}_n{g}_1(\parallel {\hat{z}}_n-y_n\parallel )& \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel .\end{array}$$

Utilizing (5.46), conditions (i), (ii), (v), and the boundedness of $\{x_n\}$ and $\{f(y_n)\}$, we get

$$\underset{n\to \mathrm{\infty }}{lim}{g}_1(\parallel {\hat{z}}_n-y_n\parallel )=0.$$

From the properties of $g_1$, we have

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

(5.50)

Utilizing Lemma 2.3 and the definition of ${\hat{z}}_n$, we have

$$\begin{array}{rcl}{\parallel {\hat{z}}_n-p\parallel }^2& =& {\parallel \frac{{\gamma }_n{S}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n}{{\gamma }_n+{\delta }_n}-p\parallel }^2\\ =& {\parallel \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}(S_n{y}_n-p)+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}(J_{r_n}G{y}_n-p)\parallel }^2\\ \le & \frac{{\gamma }_n}{{\gamma }_n+{\delta }_n}{\parallel S_n{y}_n-p\parallel }^2+\frac{{\delta }_n}{{\gamma }_n+{\delta }_n}{\parallel J_{r_n}G{y}_n-p\parallel }^2\\ -\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_2(\parallel J_{r_n}G{y}_n-S_n{y}_n\parallel )\\ \le & {\parallel y_n-p\parallel }^2-\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_2(\parallel J_{r_n}G{y}_n-S_n{y}_n\parallel ),\end{array}$$

which leads to

$$\begin{array}{rcl}\frac{{\gamma }_n{\delta }_n}{{({\gamma }_n+{\delta }_n)}^2}{g}_2(\parallel J_{r_n}G{y}_n-S_n{y}_n\parallel )& \le & {\parallel y_n-p\parallel }^2-{\parallel {\hat{z}}_n-p\parallel }^2\\ \le & (\parallel y_n-p\parallel +\parallel {\hat{z}}_n-p\parallel )\parallel y_n-{\hat{z}}_n\parallel .\end{array}$$

Since $\{y_n\}$ and $\{{\hat{z}}_n\}$ are bounded, we deduce from (5.50) and condition (ii) that

$$\underset{n\to \mathrm{\infty }}{lim}{g}_2(\parallel S_n{y}_n-J_{r_n}G{y}_n\parallel )=0.$$

From the properties of $g_2$, we have

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

(5.51)

Furthermore, $x_{n+1}$ can also be rewritten as

$$\begin{array}{rcl}{x}_{n+1}& =& {\alpha }_{n}f(y_n)+{\beta }_n{y}_n+{\gamma }_n{S}_n{y}_n+{\delta }_n{J}_{r_n}G{y}_n\\ =& {\beta }_n{y}_n+{\gamma }_n{S}_n{y}_n+({\alpha }_n+{\delta }_n)\frac{{\alpha }_{n}f(y_n)+{\delta }_n{J}_{r_n}G{y}_n}{{\alpha }_n+{\delta }_n}\\ =& {\beta }_n{y}_n+{\gamma }_n{S}_n{y}_n+d_n{\tilde{z}}_n,\end{array}$$

where $d_n={\alpha }_n+{\delta }_n$ and ${\tilde{z}}_n=\frac{{\alpha }_{n}f(y_n)+{\delta }_n{J}_{r_n}G{y}_n}{{\alpha }_n+{\delta }_n}$. Utilizing Lemma 2.4 and the convexity of ${\parallel \cdot \parallel }^2$, we have from (5.41)

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ ={\parallel {\beta }_n(y_n-p)+{\gamma }_n(S_n{y}_n-p)+d_n({\tilde{z}}_n-p)\parallel }^2\hfill \\ \le {\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel S_n{y}_n-p\parallel }^2+d_n{\parallel {\tilde{z}}_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel y_n-S_n{y}_n\parallel )\hfill \\ ={\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel S_n{y}_n-p\parallel }^2+d_n{\parallel \frac{{\alpha }_{n}f(y_n)+{\delta }_n{J}_{r_n}G{y}_n}{{\alpha }_n+{\delta }_n}-p\parallel }^2\hfill \\ -{\beta }_n{\gamma }_n{g}_3(\parallel y_n-S_n{y}_n\parallel )\hfill \\ ={\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel S_n{y}_n-p\parallel }^2+d_n{\parallel \frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}(f(y_n)-p)+\frac{{\delta }_n}{{\alpha }_n+{\delta }_n}(J_{r_n}G{y}_n-p)\parallel }^2\hfill \\ -{\beta }_n{\gamma }_n{g}_3(\parallel y_n-S_n{y}_n\parallel )\hfill \\ \le {\beta }_n{\parallel y_n-p\parallel }^2+{\gamma }_n{\parallel y_n-p\parallel }^2+d_n[\frac{{\alpha }_n}{{\alpha }_n+{\delta }_n}{\parallel f(y_n)-p\parallel }^2+\frac{{\delta }_n}{{\alpha }_n+{\delta }_n}{\parallel J_{r_n}G{y}_n-p\parallel }^2]\hfill \\ -{\beta }_n{\gamma }_n{g}_3(\parallel y_n-S_n{y}_n\parallel )\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+({\beta }_n+{\gamma }_n){\parallel y_n-p\parallel }^2+{\delta }_n{\parallel y_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel y_n-S_n{y}_n\parallel )\hfill \\ ={\alpha }_n{\parallel f(y_n)-p\parallel }^2+(1-{\alpha }_n){\parallel y_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel y_n-S_n{y}_n\parallel )\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel y_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel y_n-S_n{y}_n\parallel )\hfill \\ \le {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel y_n-S_n{y}_n\parallel ),\hfill \end{array}$$

which implies that

$$\begin{array}{rcl}{\beta }_n{\gamma }_n{g}_3(\parallel y_n-S_n{y}_n\parallel )& \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-p\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel .\end{array}$$

From (5.46), conditions (i), (ii), (v), and the boundedness of $\{x_n\}$ and $\{f(y_n)\}$, we have

$$\underset{n\to \mathrm{\infty }}{lim}{g}_3(\parallel y_n-S_n{y}_n\parallel )=0.$$

Utilizing the properties of $g_3$, we have

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

(5.52)

Thus, from (5.51) and (5.52), we get

$$\parallel y_n-J{r}_{n}G{y}_n\parallel \le \parallel y_n-S_n{y}_n\parallel +\parallel S_n{y}_n-J{r}_{n}G{y}_n\parallel \to 0\text{as }n\to \mathrm{\infty },$$

that is,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-J{r}_{n}G{y}_n\parallel =0.$$

(5.53)

Therefore, from (5.40), (5.46), (5.52), (5.53), and ${\alpha }_n\to 0$, it follows that

$$\begin{array}{c}\parallel x_n-y_n\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(y_n)-y_n\parallel +{\gamma }_n\parallel S_n{y}_n-y_n\parallel +{\delta }_n\parallel J{r}_{n}G{y}_n-y_n\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(y_n)-y_n\parallel +\parallel S_n{y}_n-y_n\parallel +\parallel J{r}_{n}G{y}_n-y_n\parallel \to 0\text{as }n\to \mathrm{\infty },\hfill \end{array}$$

that is,

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

(5.54)

Utilizing (5.40), (5.48), and (5.54), we obtain

$$\begin{array}{rcl}\parallel x_n-G{x}_n\parallel & \le & \parallel x_n-y_n\parallel +\parallel y_n-G{x}_n\parallel =\parallel x_n-y_n\parallel +(1-{\sigma }_n)\parallel J_{r_n}G{x}_n-G{x}_n\parallel \\ \le & \parallel x_n-y_n\parallel +\parallel J_{r_n}G{x}_n-G{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty },\end{array}$$

that is,

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

(5.55)

In addition, from (5.52) and (5.54), we have

$$\begin{array}{rcl}\parallel x_n-S_n{x}_n\parallel & \le & \parallel x_n-y_n\parallel +\parallel y_n-S_n{y}_n\parallel +\parallel S_n{y}_n-S_n{x}_n\parallel \\ \le & 2\parallel x_n-y_n\parallel +\parallel y_n-S_n{y}_n\parallel \to 0\text{as }n\to \mathrm{\infty },\end{array}$$

that is,

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

(5.56)

In terms of (5.56) and Lemma 2.6, we have

$$\parallel x_n-S{x}_n\parallel \le \parallel x_n-S_n{x}_n\parallel +\parallel S_n{x}_n-S{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty },$$

that is,

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

(5.57)

We note that

$$\begin{array}{rcl}\parallel x_n-J_{r_n}{x}_n\parallel & \le & \parallel x_n-y_n\parallel +\parallel y_n-J_{r_n}G{y}_n\parallel +\parallel J_{r_n}G{y}_n-J_{r_n}G{x}_n\parallel +\parallel J_{r_n}G{x}_n-J_{r_n}{x}_n\parallel \\ \le & 2\parallel x_n-y_n\parallel +\parallel y_n-J_{r_n}G{y}_n\parallel +\parallel G{x}_n-x_n\parallel .\end{array}$$

So, from (5.53), (5.54), and (5.55), we obtain

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

(5.58)

Furthermore, repeating the same arguments as those of (5.29) in the proof of Theorem 4.1, we can derive

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

(5.59)

for a fixed number $r\in (0,\epsilon )$. Define a mapping $Wx=(1-{\theta }_1-{\theta }_2)J_{r}x+{\theta }_{1}Sx+{\theta }_{2}Gx$, where ${\theta }_1,{\theta }_2\in (0,1)$ are two constants with ${\theta }_1+{\theta }_2<1$. Then by Lemma 2.5, we have $Fix(W)=Fix(J_r)\cap Fix(S)\cap Fix(G)=F$. We observe that

$$\begin{array}{rcl}\parallel x_n-W{x}_n\parallel & =& \parallel (1-{\theta }_1-{\theta }_2)(x_n-J_r{x}_n)+{\theta }_1(x_n-S{x}_n)+{\theta }_2(x_n-G{x}_n)\parallel \\ \le & (1-{\theta }_1-{\theta }_2)\parallel x_n-J_r{x}_n\parallel +{\theta }_1\parallel x_n-S{x}_n\parallel +{\theta }_2\parallel x_n-G{x}_n\parallel .\end{array}$$

From (5.55), (5.57), and (5.59), we obtain

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

(5.60)

Now, we claim that

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

(5.61)

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 $x_t=tf(x_t)+(1-t)W{x}_t$. Repeating the same arguments as those of (5.36) in the proof of Theorem 4.1, we derive

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

(5.62)

Repeating the same arguments as those of (5.37) in the proof of Theorem 4.1, we obtain

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

(5.63)

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 (5.61) holds. From (5.46), we get $(x_{n+1}-q)-(x_n-q)\to 0$. Noticing the norm-to-weak^∗ uniform continuity of J on bounded subsets of X, we deduce from (5.61) that

$$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,J(x_{n+1}-q)〉\hfill \\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(〈f(q)-q,J(x_{n+1}-q)-J(x_n-q)〉+〈f(q)-q,J(x_n-q)〉)\hfill \\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,J(x_n-q)〉\le 0.\hfill \end{array}$$

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

$$\begin{array}{rcl}\parallel y_n-q\parallel & =& \parallel {\alpha }_n(G(x_n)-q)+(1-{\alpha }_n)(J_{r_n}G(x_n)-q)\parallel \\ \le & {\alpha }_n\parallel x_n-q\parallel +(1-{\alpha }_n)\parallel x_n-q\parallel =\parallel x_n-q\parallel ,\end{array}$$

and hence

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& =& {\alpha }_n〈f(y_n)-f(q)+f(q)-q,J(x_{n+1}-q)〉\\ +〈{\beta }_n(y_n-q)+{\gamma }_n(S_n{y}_n-q)+{\delta }_n(J_{r_n}G(y_n)-q),J(x_{n+1}-q)〉\\ \le & {\alpha }_n\parallel f(y_n)-f(q)\parallel \parallel x_{n+1}-q\parallel +{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\\ +\parallel {\beta }_n(y_n-q)+{\gamma }_n(S_n{y}_n-q)+{\delta }_n(J_{r_n}G(y_n)-q)\parallel \parallel x_{n+1}-q\parallel \\ \le & {\alpha }_n\rho \parallel y_n-q\parallel \parallel x_{n+1}-q\parallel +{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\\ +({\beta }_n\parallel y_n-q\parallel +{\gamma }_n\parallel y_n-q\parallel +{\delta }_n\parallel y_n-q\parallel )\parallel x_{n+1}-q\parallel \\ =& {\alpha }_n\rho \parallel y_n-q\parallel \parallel x_{n+1}-q\parallel +{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\\ +(1-{\alpha }_n)\parallel y_n-q\parallel \parallel x_{n+1}-q\parallel \\ \le & (1-{\alpha }_n(1-\rho ))\parallel y_n-q\parallel \parallel x_{n+1}-q\parallel +{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\\ \le & (1-{\alpha }_n(1-\rho ))\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel +{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\\ =& \frac{1-{\alpha }_n(1-\rho )}{2}({\parallel x_n-q\parallel }^2+{\parallel x_{n+1}-q\parallel }^2)+{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\\ \le & \frac{1-{\alpha }_n(1-\rho )}{2}{\parallel x_n-q\parallel }^2+\frac{1}{2}{\parallel x_{n+1}-q\parallel }^2+{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉.\end{array}$$

Thus, we have

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

(5.64)

Since ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈f(q)-q,J(x_{n+1}-q)〉\le 0$, by Lemma 2.7, we conclude from (5.64) that $x_n\to q$ as $n\to \mathrm{\infty }$. This completes the proof. □

Corollary 5.2 Let C be a nonempty closed convex subset of a uniformly convex Banach space X which has a uniformly Gâteaux differentiable norm. Let ${\Pi }_C$ be a sunny nonexpansive retraction from X onto C and $A\subset X\times X$ be an accretive operator on X such that $\overline{D(A)}\subset C\subset {\bigcap }_{r>0}R(I+rA)$. 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 $\rho \in (0,1)$ and $S:C\to C$ be a nonexpansive mapping such that $F=Fix(S)\cap \Omega \cap A^{-1}0\ne \mathrm{\varnothing }$ with $1-\frac{{\lambda }_i}{1+{\lambda }_i}(1-\sqrt{\frac{1-{\alpha }_i}{{\lambda }_i}})\le {\mu }_i\le 1$ for $i=1,2$. Suppose that Assumption  5.1 holds. For arbitrarily given $x_0\in C$, let $\{x_n\}$ be the sequence generated by

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

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

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

Remark 5.1 Our Theorems 5.1 and 5.2 improve and extend [[30], Theorem 3.2], [[20], Theorem 3.1] and [[29], Theorem 3.1] in the following aspects.

• (a) The problem of finding a point $q\in {\bigcap }_{n}Fix(S_n)\cap \Omega \cap A^{-1}0$ in Theorems 5.1 and 5.2 is more general and more subtle than the problem of finding $q\in {\bigcap }_{n}Fix(T_n)$ in [[30], Theorem 3.2], the problem of finding $q\in {\bigcap }_{n}Fix(T_n)\cap \Omega$ in [[20], Theorem 3.1] and the problem of finding $q\in A^{-1}0$ in [[29], Theorem 3.1].

• (b) Theorems 5.1 and 5.2 are proved without the asymptotical regularity assumption of $\{x_n\}$ in [[29], Theorem 3.1] (that is, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_{n+1}\parallel =0$).

• (c) The iterative scheme in [[20], Theorem 3.1] is extended to develop the iterative schemes (5.1) and (5.40) in Theorems 5.1 and 5.2 by virtue of the iterative schemes of [[30], Theorem 3.2] and [[29], Theorem 3.1]. The iterative schemes (5.1) and (5.40) in Theorems 5.1 and 5.2 are more advantageous and more flexible than the iterative scheme in [[20], Theorem 3.1] because they involves several parameter sequences.

• (d) The iterative schemes (5.1) and (5.40) in Theorems 5.1 and 5.2 are different from the one given in [[30], Theorem 3.2], [[20], Theorem 3.1] and [[29], Theorem 3.1] because the first iteration step in (5.1) is implicit and because the mapping G in [[20], Theorem 3.1] and the mapping $J_{r_n}$ in [[29], Theorem 3.1] are replaced by the same composite mapping $J_{r_n}G$ in Theorems 5.1 and 5.2.

• (e) The proof of [[20], Theorem 3.1] depends on the argument techniques in [10], the inequality in 2-uniformly smooth Banach spaces and the inequality in smooth and uniform convex Banach spaces. Because the composite mapping $J_{r_n}G$ appears in the iterative scheme (5.1) in Theorem 5.1, the proof of Theorem 5.1 depends on the argument techniques in [10], the inequality in 2-uniformly smooth Banach spaces, the inequality in smooth and uniform convex Banach spaces, and the inequalities in uniform convex Banach spaces. However, the proof of our Theorem 5.1 does not depend on the argument techniques in [10], the inequality in 2-uniformly smooth Banach spaces, and the inequality in smooth and uniform convex Banach spaces. It depends on only the inequalities in uniform convex Banach spaces.

• (f) The assumption of the uniformly convex and 2-uniformly smooth Banach space X in [[20], Theorem 3.1] is weakened to the one of the uniformly convex Banach space X having a uniformly Gâteaux differentiable norm in Theorem 5.2.