A modified projection method for a common solution of a system of variational inequalities, a split equilibrium problem and a hierarchical fixed-point problem

Abstract

In this paper, we suggest and analyze a modified projection method for finding a common solution of a system of variational inequalities, a split equilibrium problem, and a hierarchical fixed-point problem in the setting of real Hilbert spaces. We prove the strong convergence of the sequence generated by the proposed method to a common solution of a system of variational inequalities, a split equilibrium problem, and a hierarchical fixed-point problem. Several special cases are also discussed. The results presented in this paper extend and improve some well-known results in the literature.

MSC: 49J30, 47H09, 47J20.

1 Introduction

Let H be a real Hilbert space, whose inner product and norm are denoted by $〈\cdot ,\cdot 〉$ and $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H. Recently, Ceng et al. [1] considered the general system of variational inequalities, which involves finding $(x^{\ast },y^{\ast })\in C\times C$ such that

$$\{\begin{array}{l}〈{\mu }_1{B}_1{y}^{\ast }+x^{\ast }-y^{\ast },x-x^{\ast }〉\ge 0;\mathrm{\forall }x\in C\text{ and }{\mu }_1>0,\\ 〈{\mu }_2{B}_2{x}^{\ast }+y^{\ast }-x^{\ast },x-y^{\ast }〉\ge 0;\mathrm{\forall }x\in C\text{ and }{\mu }_2>0,\end{array}$$

(1.1)

where $B_i:C\to C$ is a nonlinear mapping for each $i=1,2$. The solution set of (1.1) is denoted by $S^{\ast }$. As pointed out in [2] the system of variational inequalities is used as a tool to study the Nash equilibrium problem, see, for example, [3–6] and the references therein. We believe that the problem (1.1) could be used to study the Nash equilibrium problem for a two players game. The theory of variational inequalities is well established and it has a wide range of applications in science, engineering, management, and social sciences, see, for example, [4–7] and the references therein.

Ceng et al. [1] transformed problem (1.1) into a fixed-point problem (see Lemma 2.2) and introduced an iterative method for finding the common element of the set $Fix(T)\cap S^{\ast }$. Based on the one-step iterative method [8] and the multi-step iterative method [9], Latif et al. [10] proposed a multi-step hybrid viscosity method that generates a sequence via an explicit iterative algorithm to compute the approximate solutions of a system of variational inequalities defined over the intersection of the set of solutions of an equilibrium problem, the set of common fixed points of a finite family of nonexpansive mappings, and the solution set of a nonexpansive mapping. Under very mild conditions, they proved that the sequence converges strongly to a unique solution of system of variational inequalities defined over the set consisting of the set of solutions of an equilibrium problem, the set of common fixed points of nonexpansive mappings, and the set of fixed points of a mapping, and to a unique solution of the triple hierarchical variational inequality problem.

On the other hand, by combining the regularization method, Korpelevich’s extragradient method, the hybrid steepest-descent method, and the viscosity approximation method, Ceng et al. [2] introduced and analyzed implicit and explicit iterative schemes for computing a common element of the solution set of system of variational inequalities and a set of zeros of an accretive operator in Banach space. Under suitable assumptions, they proved the strong convergence of the sequences generated by the proposed schemes.

If $B_1=B_2=B$, then the problem (1.1) reduces to finding $(x^{\ast },y^{\ast })\in C\times C$ such that

$$\{\begin{array}{l}〈{\mu }_{1}B{y}^{\ast }+x^{\ast }-y^{\ast },x-x^{\ast }〉\ge 0;\mathrm{\forall }x\in C\text{ and }{\mu }_1>0,\\ 〈{\mu }_{2}B{x}^{\ast }+y^{\ast }-x^{\ast },x-y^{\ast }〉\ge 0;\mathrm{\forall }x\in C\text{ and }{\mu }_2>0,\end{array}$$

(1.2)

which has been introduced and studied by Verma [11, 12].

If $x^{\ast }=y^{\ast }$ and ${\mu }_1={\mu }_2$, then the problem (1.2) collapses to the classical variational inequality: finding $x^{\ast }\in C$, such that

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

For the recent applications, numerical techniques, and physical formulation, see [1–45].

We introduce the following definitions, which are useful in the following analysis.

Definition 1.1 The mapping $T:C\to H$ is said to be

• (a) monotone, if

  $$〈Tx-Ty,x-y〉\ge 0,\mathrm{\forall }x,y\in C;$$

• (b) strongly monotone, if there exists an $\alpha >0$ such that

  $$〈Tx-Ty,x-y〉\ge \alpha {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C;$$

• (c) α-inverse strongly monotone, if there exists an $\alpha >0$ such that

  $$〈Tx-Ty,x-y〉\ge \alpha {\parallel Tx-Ty\parallel }^2,\mathrm{\forall }x,y\in C;$$

• (d) nonexpansive, if

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

• (e) k-Lipschitz continuous, if there exists a constant $k>0$ such that

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

• (f) a contraction on C, if there exists a constant $0\le k<1$ such that

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

It is easy to observe that every α-inverse strongly monotone T is monotone and Lipschitz continuous. It is well known that every nonexpansive operator $T:H\to H$ satisfies, for all $(x,y)\in H\times H$, the inequality

$$〈(x-T(x))-(y-T(y)),T(y)-T(x)〉\le \frac{1}{2}{\parallel (T(x)-x)-(T(y)-y)\parallel }^2,$$

(1.3)

and therefore, we get, for all $(x,y)\in H\times Fix(T)$,

$$〈x-T(x),y-T(x)〉\le \frac{1}{2}{\parallel T(x)-x\parallel }^2.$$

(1.4)

The fixed-point problem for the mapping T is to find $x\in C$ such that

$$Tx=x.$$

(1.5)

We denote by $F(T)$ the set of solutions of (1.5). It is well known that $F(T)$ is closed and convex, and $P_F(T)$ is well defined.

The equilibrium problem, denoted by EP, is to find $x\in C$ such that

$$F(x,y)\ge 0,\mathrm{\forall }y\in C.$$

(1.6)

The solution set of (1.6) is denoted by $\mathit{EP}(F)$. Numerous problems in physics, optimization, and economics reduce to finding a solution of (1.6), see [25, 38]. In 1994, Censor and Elfving [19] introduced and studied the following split feasibility problem.

Let C and Q be nonempty closed convex subsets of the infinite-dimensional real Hilbert spaces $H_1$ and $H_2$, respectively, and let $A\in B(H_1,H_2)$, where $B(H_1,H_2)$ denotes the collection of all bounded linear operators from $H_1$ to $H_2$. The problem is to find $x^{\ast }$ such that

$$x^{\ast }\in C\text{such that}A{x}^{\ast }\in Q.$$

Very recently, Ceng et al. [22] introduced and analyzed an extragradient method with regularization for finding a common element of the solution set of the split feasibility problem and the set of fixed points of a nonexpansive mapping S in the setting of infinite-dimensional Hilbert spaces. By combining Mann’s iterative method and the extragradient method, Ceng et al. [21] proposed three different kinds of Mann type iterative methods for finding a common element of the solution set of the split feasibility problem and the set of fixed points of a nonexpansive mapping S in the setting of infinite-dimensional Hilbert spaces.

Recently, Censor et al. [23] introduced a new variational inequality problem which we call the split variational inequality problem (SVIP). Let $H_1$ and $H_2$ be two real Hilbert spaces. Given the operators $f:H_1\to H_1$ and $g:H_2\to H_2$, a bounded linear operator $A:H_1\to H_2$, and the nonempty, closed, and convex subsets $C\subseteq H_1$ and $Q\subseteq H_2$, the SVIP is formulated as follows: find a point $x^{\ast }\in C$ such that

$$〈f\left(x^{\ast }\right),x-x^{\ast }〉\ge 0\text{for all }x\in C$$

(1.7)

and such that

$$y^{\ast }=A{x}^{\ast }\in Q\text{solves}〈g\left(y^{\ast }\right),y-y^{\ast }〉\ge 0\text{for all }y\in Q.$$

(1.8)

In [36], Moudafi introduced an iterative method which can be regarded as an extension of the method given by Censor et al. [23] for the following split monotone variational inclusions:

$$\text{Find}{x}^{\ast }\in H_1\text{such that}0\in f\left(x^{\ast }\right)+B_1\left(x^{\ast }\right)$$

and such that

$$y^{\ast }=A{x}^{\ast }\in H_2\text{solves}0\in g\left(y^{\ast }\right)+B_2\left(y^{\ast }\right),$$

where $B_i:H_i\to 2^{H_i}$ is a set-valued mapping for $i=1,2$. Later Byrne et al. [18] generalized and extended the work of Censor et al. [23] and Moudafi [36].

In this paper, we consider the following pair of equilibrium problems, called split equilibrium problems: Let $F_1:C\times C\to R$ and $F_2:Q\times Q\to R$ be nonlinear bifunctions and $A:H_1\to H_2$ be a bounded linear operator, then the split equilibrium problem (SEP) is to find $x^{\ast }\in C$ such that

$$F_1(x^{\ast },x)\ge 0,\mathrm{\forall }x\in C,$$

(1.9)

and such that

$$y^{\ast }=A{x}^{\ast }\in Q\text{solves}{F}_2(y^{\ast },y)\ge 0,\mathrm{\forall }y\in Q.$$

(1.10)

The solution set of SEP (1.9)-(1.10) is denoted by $\mathrm{\Lambda }=\{p\in \mathit{EP}(F_1):Ap\in \mathit{EP}(F_2)\}$.

Let $S:C\to H$ be a nonexpansive mapping. The following problem is called a hierarchical fixed-point problem: find $x\in F(T)$ such that

$$〈x-Sx,y-x〉\ge 0,\mathrm{\forall }y\in F(T).$$

(1.11)

It is known that the hierarchical fixed-point problem (1.11) links with some monotone variational inequalities and convex programming problems; see [45]. Various methods have been proposed to solve the hierarchical fixed-point problem; see Mainge and Moudafi [34] and Cianciaruso et al. [26]. In 2010, Yao et al. [45] introduced the following strong convergence iterative algorithm to solve the problem (1.11):

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

(1.12)

where $f:C\to H$ is a contraction mapping and $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two sequences in $(0,1)$. Under some certain restrictions on the parameters, Yao et al. proved that the sequence $\{x_n\}$ generated by (1.12) converges strongly to $z\in F(T)$, which is the unique solution of the following variational inequality:

$$〈(I-f)z,y-z〉\ge 0,\mathrm{\forall }y\in F(T).$$

(1.13)

In 2011, Ceng et al. [20] investigated the following iterative method:

$$x_{n+1}=P_C[{\alpha }_n\rho U(x_n)+(I-{\alpha }_n\mu F)(T(y_n))],\mathrm{\forall }n\ge 0,$$

(1.14)

where U is a Lipschitzian mapping, and F is a Lipschitzian and strongly monotone mapping. They proved that under some approximate assumptions on the operators and parameters, the sequence $\{x_n\}$ generated by (1.14) converges strongly to the unique solution of the variational inequality

$$〈\rho U(z)-\mu F(z),x-z〉\ge 0,\mathrm{\forall }x\in Fix(T).$$

In this paper, motivated by the work of Censor et al. [23], Moudafi [36], Byrne et al. [18], Yao et al. [45], Ceng et al. [20], Bnouhachem [15–17] and by the recent work going in this direction, we give an iterative method for finding the approximate element of the common set of solutions of (1.1), (1.9)-(1.10), and (1.11) in real Hilbert space. We establish a strong convergence theorem based on this method. We would like to mention that our proposed method is quite general and flexible and includes many known results for solving a system of variational inequality problems, split equilibrium problems, and hierarchical fixed-point problems, see, e.g. [20, 26, 34, 40, 45] and relevant references cited therein.

2 Preliminaries

In this section, we list some fundamental lemmas that are useful in the consequent analysis. The first lemma provides some basic properties of projection onto C.

Lemma 2.1 Let $P_C$ denote the projection of H onto C. Then we have the following inequalities:

$$〈z-P_C[z],P_C[z]-v〉\ge 0,\mathrm{\forall }z\in H,v\in C;$$

(2.1)

$$〈u-v,P_C[u]-P_C[v]〉\ge {\parallel P_C[u]-P_C[v]\parallel }^2,\mathrm{\forall }u,v\in H;$$

(2.2)

$$\parallel P_C[u]-P_C[v]\parallel \le \parallel u-v\parallel ,\mathrm{\forall }u,v\in H;$$

(2.3)

$${\parallel u-P_C[z]\parallel }^2\le {\parallel z-u\parallel }^2-{\parallel z-P_C[z]\parallel }^2,\mathrm{\forall }z\in H,u\in C.$$

(2.4)

Lemma 2.2 [1]

For any $(x^{\ast },y^{\ast })\in C\times C$, $(x^{\ast },y^{\ast })$ is a solution of (1.1) if and only if $x^{\ast }$ is a fixed point of the mapping $Q:C\to C$ defined by

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

(2.5)

where $y^{\ast }=P_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }]$, ${\mu }_i\in (0,2{\theta }_i)$ and $B_i:C\to C$ is for the ${\theta }_i$-inverse strongly monotone mappings for each $i=1,2$.

Assumption 2.1 [14]

Let $F:C\times C\to \mathbb{R}$ be a bifunction satisfying the following assumptions:

• (i) $F(x,x)=0$, $\mathrm{\forall }x\in C$;

• (ii) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$, $\mathrm{\forall }x,y\in C$;

• (iii) for each $x,y,z\in C$, ${lim}_{t\to 0}F(tz+(1-t)x,y)\le F(x,y)$;

• (iv) for each $x\in C$, $y\to F(x,y)$ is convex and lower semicontinuous;

• (v) for fixed $r>0$ and $z\in C$, there exists a bounded subset K of $H_1$ and $x\in C\cap K$ such that

  $$F(y,x)+\frac{1}{r}〈y-x,x-z〉\ge 0,\mathrm{\forall }y\in C\mathrm{\setminus }K.$$

Lemma 2.3 [28]

Assume that $F_1:C\times C\to \mathbb{R}$ satisfies Assumption  2.1. For $r>0$ and $\mathrm{\forall }x\in H_1$, define a mapping $T_r^{F_1}:H_1\to C$ as follows:

$$T_r^{F_1}(x)=\{z\in C:F_1(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\}.$$

Then the following hold:

• (i) $T_r^{F_1}$ is nonempty and single-valued;

• (ii) $T_r^{F_1}$ is firmly nonexpansive, i.e.,

  $${\parallel T_r^{F_1}(x)-T_r^{F_1}(y)\parallel }^2\le 〈T_r^{F_1}(x)-T_r^{F_1}(y),x-y〉,\mathrm{\forall }x,y\in H_1;$$

• (iii) $F(T_r^{F_1})=\mathit{EP}(F_1)$;

• (iv) $\mathit{EP}(F_1)$ is closed and convex.

Assume that $F_2:Q\times Q\to \mathbb{R}$ satisfies Assumption 2.1, and for $s>0$ and $\mathrm{\forall }u\in H_2$, define a mapping $T_s^{F_2}:H_2\to Q$ as follows:

$$T_s^{F_2}(u)=\{v\in Q:F_2(v,w)+\frac{1}{s}〈w-v,v-u〉\ge 0,\mathrm{\forall }w\in Q\}.$$

Then $T_s^{F_2}$ satisfies conditions (i)-(iv) of Lemma 2.3. $F(T_s^{F_2})=\mathit{EP}(F_2,Q)$, where $\mathit{EP}(F_2,Q)$ is the solution set of the following equilibrium problem:

$$\text{Find}{y}^{\ast }\in Q\text{such that}{F}_2(y^{\ast },y)\ge 0,\mathrm{\forall }y\in Q.$$

Lemma 2.4 [27]

Assume that $F_1:C\times C\to \mathbb{R}$ satisfies Assumption  2.1, and let $T_r^{F_1}$ be defined as in Lemma  2.3. Let $x,y\in H_1$ and $r_1,r_2>0$. Then

$$\parallel T_{r_2}^{F_1}(y)-T_{r_1}^{F_1}(x)\parallel \le \parallel y-x\parallel +\left|\frac{r_2-r_1}{r_2}\right|\parallel T_{r_2}^{F_1}(y)-y\parallel .$$

Lemma 2.5 [31]

Let C be a nonempty closed convex subset of a real Hilbert space H. If $T:C\to C$ is a nonexpansive mapping with $Fix(T)\ne \mathrm{\varnothing }$, then the mapping $I-T$ is demiclosed at 0, i.e., if $\{x_n\}$ is a sequence in C weakly converges to x and if $\{(I-T)x_n\}$ converges strongly to 0, then $(I-T)x=0$.

Lemma 2.6 [20]

Let $U:C\to H$ be a τ-Lipschitzian mapping and let $F:C\to H$ be a k-Lipschitzian and η-strongly monotone mapping, then for $0\le \rho \tau <\mu \eta$, $\mu F-\rho U$ is $\mu \eta -\rho \tau$-strongly monotone i.e.,

$$〈(\mu F-\rho U)x-(\mu F-\rho U)y,x-y〉\ge (\mu \eta -\rho \tau ){\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

Lemma 2.7 [39]

Suppose that $\lambda \in (0,1)$ and $\mu >0$. Let $F:C\to H$ be a k-Lipschitzian and η-strongly monotone operator. In association with a nonexpansive mapping $T:C\to C$, define the mapping $T^{\lambda }:C\to H$ by

$$T^{\lambda }x=Tx-\lambda \mu FT(x),\mathrm{\forall }x\in C.$$

Then $T^{\lambda }$ is a contraction provided $\mu <\frac{2\eta }{k^2}$, that is,

$$\parallel T^{\lambda }x-T^{\lambda }y\parallel \le (1-\lambda \nu )\parallel x-y\parallel ,\mathrm{\forall }x,y\in C,$$

where $\nu =1-\sqrt{1-\mu (2\eta -\mu k^2)}$.

Lemma 2.8 [43]

Assume $\{a_n\}$ is a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\delta }_n,$$

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

• (1) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$;

• (2) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n/{\gamma }_n\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_n|<\mathrm{\infty }$.

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

Lemma 2.9 [13]

Let C be a closed convex subset of H. Let $\{x_n\}$ be a bounded sequence in H. Assume that

• (i) the weak w-limit set $w_w(x_n)\subset C$ where $w_w(x_n)=\{x:x_{n_i}\rightharpoonup x\}$;

• (ii) for each $z\in C$, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z\parallel$ exists.

Then $\{x_n\}$ is weakly convergent to a point in C.

3 The proposed method and some properties

In this section, we suggest and analyze our method for finding the common solutions of the system of the variational inequality problem (1.1), the split equilibrium problem (1.9)-(1.10), and the hierarchical fixed-point problem (1.11).

Let $H_1$ and $H_2$ be two real Hilbert spaces and $C\subseteq H_1$ and $Q\subseteq H_2$ be nonempty closed convex subsets of Hilbert spaces $H_1$ and $H_2$, respectively. Let $A:H_1\to H_2$ be a bounded linear operator. Assume that $F_1:C\times C\to \mathbb{R}$ and $F_2:Q\times Q\to \mathbb{R}$ are the bifunctions satisfying Assumption 2.1 and $F_2$ is upper semicontinuous in the first argument. Let $B_i:C\to H$ be a ${\theta }_i$-inverse strongly monotone mapping for each $i=1,2$ and $S,T:C\to C$ a nonexpansive mappings such that $S^{\ast }\cap \mathrm{\Lambda }\cap F(T)\ne \mathrm{\varnothing }$. Let $F:C\to C$ be a k-Lipschitzian mapping and be η-strongly monotone, and let $U:C\to C$ be a τ-Lipschitzian mapping.

Algorithm 3.1 For an arbitrary given $x_0\in C$, let the iterative sequences $\{u_n\}$, $\{x_n\}$, $\{y_n\}$, and $\{z_n\}$ be generated by

$$\{\begin{array}{l}{u}_n=T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n);\\ z_n=P_C[P_C[u_n-{\mu }_2{B}_2{u}_n]-{\mu }_1{B}_1{P}_C[u_n-{\mu }_2{B}_2{u}_n]];\\ y_n={\beta }_{n}S{x}_n+(1-{\beta }_n)z_n;\\ x_{n+1}=P_C[{\alpha }_n\rho U(x_n)+(I-{\alpha }_n\mu F)(T(y_n))],\mathrm{\forall }n\ge 0,\end{array}$$

(3.1)

where ${\mu }_i\in (0,2{\theta }_i)$ for each $i=1,2$, $\{r_n\}\subset (0,2\varsigma )$ and $\gamma \in (0,1/L)$, L is the spectral radius of the operator $A^{\ast }A$, and $A^{\ast }$ is the adjoint of A. Suppose the parameters satisfy $0<\mu <\frac{2\eta }{k^2}$, $0\le \rho \tau <\nu$, where $\nu =1-\sqrt{1-\mu (2\eta -\mu k^2)}$. Also, $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are sequences in $(0,1)$ satisfying the following conditions:

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

• (b) ${lim}_{n\to \mathrm{\infty }}({\beta }_n/{\alpha }_n)=0$,

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

• (d) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n<{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{r}_n<2\varsigma$ and ${\sum }_{n=1}^{\mathrm{\infty }}|r_{n-1}-r_n|<\mathrm{\infty }$.

Remark 3.1 Our method can be viewed as an extension and improvement for some well-known results, for example the following.

• The proposed method is an extension and improvement of the method of Wang and Xu [42] for finding the approximate element of the common set of solutions of a split equilibrium problem and a hierarchical fixed-point problem in a real Hilbert space.

• If we have the Lipschitzian mapping $U=f$, $F=I$, $\rho =\mu =1$, and $B_1=B_2=0$, we obtain an extension and improvement of the method of Yao et al.[45] for finding the approximate element of the common set of solutions of a split equilibrium problem and a hierarchical fixed-point problem in a real Hilbert space.

• The contractive mapping f with a coefficient $\alpha \in [0,1)$ in other papers [39, 40, 45] is extended to the cases of the Lipschitzian mapping U with a coefficient constant $\gamma \in [0,\mathrm{\infty })$.

This shows that Algorithm 3.1 is quite general and unifying.

Lemma 3.1 Let $x^{\ast }\in S^{\ast }\cap \mathrm{\Lambda }\cap F(T)$. Then $\{x_n\}$, $\{u_n\}$, $\{z_n\}$, and $\{y_n\}$ are bounded.

Proof Let $x^{\ast }\in S^{\ast }\cap \mathrm{\Lambda }\cap F(T)$; we have $x^{\ast }=T_{r_n}^{F_1}(x^{\ast })$ and $A{x}^{\ast }=T_{r_n}^{F_2}(A{x}^{\ast })$. Then

$$\begin{array}{rcl}{\parallel u_n-x^{\ast }\parallel }^2& =& {\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-x^{\ast }\parallel }^2\\ =& {\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-T_{r_n}^{F_1}\left(x^{\ast }\right)\parallel }^2\\ \le & {\parallel x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n-x^{\ast }\parallel }^2\\ =& {\parallel x_n-x^{\ast }\parallel }^2+{\gamma }^2{\parallel A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n\parallel }^2+2\gamma 〈x_n-x^{\ast },A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉\\ =& {\parallel x_n-x^{\ast }\parallel }^2+{\gamma }^2〈(T_{r_n}^{F_2}-I)A{x}_n,A{A}^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉\\ +2\gamma 〈x_n-x^{\ast },A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉.\end{array}$$

(3.2)

From the definition of L it follows that

$$\begin{array}{rcl}{\gamma }^2〈(T_{r_n}^{F_2}-I)A{x}_n,A{A}^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉& \le & L{\gamma }^2〈(T_{r_n}^{F_2}-I)A{x}_n,(T_{r_n}^{F_2}-I)A{x}_n〉\\ =& L{\gamma }^2{\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2.\end{array}$$

(3.3)

It follows from (1.4) that

$$\begin{array}{c}2\gamma 〈x_n-x^{\ast },A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉\hfill \\ =2\gamma 〈A(x_n-x^{\ast }),(T_{r_n}^{F_2}-I)A{x}_n〉\hfill \\ =2\gamma 〈A(x_n-x^{\ast })+(T_{r_n}^{F_2}-I)A{x}_n-(T_{r_n}^{F_2}-I)A{x}_n,(T_{r_n}^{F_2}-I)A{x}_n〉\hfill \\ =2\gamma (〈T_{r_n}^{F_2}A{x}_n-A{x}^{\ast },(T_{r_n}^{F_2}-I)A{x}_n〉-{\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2)\hfill \\ \le 2\gamma (\frac{1}{2}{\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2-{\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2)\hfill \\ =-\gamma {\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2.\hfill \end{array}$$

(3.4)

Applying (3.4) and (3.3) to (3.2) and from the definition of γ, we get

$$\begin{array}{rcl}{\parallel u_n-x^{\ast }\parallel }^2& \le & {\parallel x_n-x^{\ast }\parallel }^2+\gamma (L\gamma -1){\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2.\end{array}$$

(3.5)

Let $x^{\ast }\in S^{\ast }\cap \mathrm{\Lambda }\cap F(T)$; we have

$$x^{\ast }=P_C[y^{\ast }-{\mu }_1{B}_1{y}^{\ast }],$$

where

$$y^{\ast }=P_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }].$$

We set $v_n=P_C[u_n-{\mu }_2{B}_2{u}_n]$. Since $B_2$ is a ${\theta }_2$-inverse strongly monotone mapping, it follows that

$$\begin{array}{rcl}{\parallel v_n-y^{\ast }\parallel }^2& =& {\parallel P_C[u_n-{\mu }_2{B}_2{u}_n]-P_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }]\parallel }^2\\ \le & {\parallel u_n-x^{\ast }-{\mu }_2(B_2{u}_n-B_2{x}^{\ast })\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2-{\mu }_2(2{\theta }_2-{\mu }_2){\parallel B_2{u}_n-B_2{x}^{\ast }\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2.\end{array}$$

(3.6)

Since $B_i$ is ${\theta }_i$-inverse strongly monotone mappings, for each $i=1,2$, we get

$$\begin{array}{rcl}{\parallel z_n-x^{\ast }\parallel }^2& =& \parallel P_C[P_C[u_n-{\mu }_2{B}_2{u}_n]-{\mu }_1{B}_1{P}_C[u_n-{\mu }_2{B}_2{u}_n]]\\ {-P_C[P_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }]-{\mu }_1{B}_1{P}_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }]]\parallel }^2\\ \le & \parallel P_C[u_n-{\mu }_2{B}_2{u}_n]-{\mu }_1{B}_1{P}_C[u_n-{\mu }_2{B}_2{u}_n]\\ {-(P_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }]-{\mu }_1{B}_1{P}_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }])\parallel }^2\\ =& \parallel P_C[u_n-{\mu }_2{B}_2{u}_n]-P_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }]\\ {-{\mu }_1(B_1{P}_C[u_n-{\mu }_2{B}_2{u}_n]-B_1{P}_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }])\parallel }^2\\ \le & {\parallel P_C[u_n-{\mu }_2{B}_2{u}_n]-P_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }]\parallel }^2\\ {-{\mu }_1(2{\theta }_1-{\mu }_1)\parallel B_1{P}_C[u_n-{\mu }_2{B}_2{u}_n]-B_1{P}_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }]\parallel }^2\\ \le & {\parallel (u_n-{\mu }_2{B}_2{u}_n)-(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })\parallel }^2\\ {-{\mu }_1(2{\theta }_1-{\mu }_1)\parallel B_1{P}_C[u_n-{\mu }_2{B}_2{u}_n]-B_1{P}_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }]\parallel }^2\\ \le & {\parallel u_n-x^{\ast }\parallel }^2-{\mu }_2(2{\theta }_2-{\mu }_2){\parallel B_2{u}_n-B_2{x}^{\ast }\parallel }^2-{\mu }_1(2{\theta }_1-{\mu }_1){\parallel B_1{v}_n-B_1{y}^{\ast }\parallel }^2\\ \le & {\parallel u_n-x^{\ast }\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2.\end{array}$$

(3.7)

We denote $V_n={\alpha }_n\rho U(x_n)+(I-{\alpha }_n\mu F)(T(y_n))$. Next, we prove that the sequence $\{x_n\}$ is bounded, and without loss of generality we can assume that ${\beta }_n\le {\alpha }_n$ for all $n\ge 1$. From (3.1), we have

$$\begin{array}{rcl}\parallel x_{n+1}-x^{\ast }\parallel & =& \parallel P_C[V_n]-P_C\left[x^{\ast }\right]\parallel \\ \le & \parallel {\alpha }_n\rho U(x_n)+(I-{\alpha }_n\mu F)(T(y_n))-x^{\ast }\parallel \\ \le & {\alpha }_n\parallel \rho U(x_n)-\mu F\left(x^{\ast }\right)\parallel +\parallel (I-{\alpha }_n\mu F)(T(y_n))-(I-{\alpha }_n\mu F)T\left(x^{\ast }\right)\parallel \\ =& {\alpha }_n\parallel \rho U(x_n)-\rho U\left(x^{\ast }\right)+(\rho U-\mu F)x^{\ast }\parallel \\ +\parallel (I-{\alpha }_n\mu F)(T(y_n))-(I-{\alpha }_n\mu F)T\left(x^{\ast }\right)\parallel \\ \le & {\alpha }_n\rho \tau \parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel (\rho U-\mu F)x^{\ast }\parallel +(1-{\alpha }_n\nu )\parallel y_n-x^{\ast }\parallel \\ \le & {\alpha }_n\rho \tau \parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel (\rho U-\mu F)x^{\ast }\parallel \\ +(1-{\alpha }_n\nu )\parallel {\beta }_{n}S{x}_n+(1-{\beta }_n)z_n-x^{\ast }\parallel \\ \le & {\alpha }_n\rho \tau \parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel (\rho U-\mu F)x^{\ast }\parallel +(1-{\alpha }_n\nu )({\beta }_n\parallel S{x}_n-S{x}^{\ast }\parallel \\ +{\beta }_n\parallel S{x}^{\ast }-x^{\ast }\parallel +(1-{\beta }_n)\parallel z_n-x^{\ast }\parallel )\\ \le & {\alpha }_n\rho \tau \parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel (\rho U-\mu F)x^{\ast }\parallel +(1-{\alpha }_n\nu )({\beta }_n\parallel S{x}_n-S{x}^{\ast }\parallel \\ +{\beta }_n\parallel S{x}^{\ast }-x^{\ast }\parallel +(1-{\beta }_n)\parallel x_n-x^{\ast }\parallel )\\ \le & {\alpha }_n\rho \tau \parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel (\rho U-\mu F)x^{\ast }\parallel +(1-{\alpha }_n\nu )({\beta }_n\parallel x_n-x^{\ast }\parallel \\ +{\beta }_n\parallel S{x}^{\ast }-x^{\ast }\parallel +(1-{\beta }_n)\parallel x_n-x^{\ast }\parallel )\\ =& (1-{\alpha }_n(\nu -\rho \tau ))\parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel (\rho U-\mu F)x^{\ast }\parallel \\ +(1-{\alpha }_n\nu ){\beta }_n\parallel S{x}^{\ast }-x^{\ast }\parallel \\ \le & (1-{\alpha }_n(\nu -\rho \tau ))\parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel (\rho U-\mu F)x^{\ast }\parallel +{\beta }_n\parallel S{x}^{\ast }-x^{\ast }\parallel \\ \le & (1-{\alpha }_n(\nu -\rho \tau ))\parallel x_n-x^{\ast }\parallel +{\alpha }_n(\parallel (\rho U-\mu F)x^{\ast }\parallel +\parallel S{x}^{\ast }-x^{\ast }\parallel )\\ =& (1-{\alpha }_n(\nu -\rho \tau ))\parallel x_n-x^{\ast }\parallel \\ +\frac{{\alpha }_n(\nu -\rho \tau )}{\nu -\rho \tau }(\parallel (\rho U-\mu F)x^{\ast }\parallel +\parallel S{x}^{\ast }-x^{\ast }\parallel )\\ \le & max\{\parallel x_n-x^{\ast }\parallel ,\frac{1}{\nu -\rho \tau }(\parallel (\rho U-\mu F)x^{\ast }\parallel +\parallel S{x}^{\ast }-x^{\ast }\parallel )\},\end{array}$$

where the third inequality follows from Lemma 2.7.

By induction on n, we obtain $\parallel x_n-x^{\ast }\parallel \le max\{\parallel x_0-x^{\ast }\parallel ,\frac{1}{\nu -\rho \tau }(\parallel (\rho U-\mu F)x^{\ast }\parallel +\parallel S{x}^{\ast }-x^{\ast }\parallel )\}$, for $n\ge 0$ and $x_0\in C$. Hence $\{x_n\}$ is bounded and, consequently, we deduce that $\{u_n\}$, $\{z_n\}$, $\{v_n\}$, $\{y_n\}$, $\{S(x_n)\}$, $\{T(x_n)\}$, $\{F(T(y_n))\}$, and $\{U(x_n)\}$ are bounded. □

Lemma 3.2 Let $x^{\ast }\in S^{\ast }\cap \mathrm{\Lambda }\cap F(T)$ and $\{x_n\}$ be the sequence generated by Algorithm  3.1. Then we have:

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

• (b) The weak w-limit set $w_w(x_n)\subset F(T)$ ($w_w(x_n)=\{x:x_{n_i}\rightharpoonup x\}$).

Proof Since $u_n=T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)$ and $u_{n-1}=T_{r_{n-1}}^{F_1}(x_{n-1}+\gamma A^{\ast }(T_{r_{n-1}}^{F_2}-I)A{x}_{n-1})$. It follows from Lemma 2.4 that

$$\begin{array}{rcl}\parallel u_n-u_{n-1}\parallel & \le & \parallel x_n-x_{n-1}+\gamma (A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n-A^{\ast }(T_{r_{n-1}}^{F_2}-I)A{x}_{n-1})\parallel \\ +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel \\ \le & \parallel x_n-x_{n-1}-\gamma A^{\ast }A(x_n-x_{n-1})\parallel +\gamma \parallel A\parallel \parallel T_{r_n}^{F_2}A{x}_n-T_{r_{n-1}}^{F_2}A{x}_{n-1}\parallel \\ +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel \\ \le & {({\parallel x_n-x_{n-1}\parallel }^2-2\gamma {\parallel A(x_n-x_{n-1})\parallel }^2+{\gamma }^2{\parallel A\parallel }^4{\parallel x_n-x_{n-1}\parallel }^2)}^{\frac{1}{2}}\\ +\gamma \parallel A\parallel (\parallel A(x_n-x_{n-1})\parallel +|1-\frac{r_{n-1}}{r_n}\left|\parallel T_{r_n}^{F_2}A{x}_n-A{x}_n\parallel \right)\\ +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel \\ \le & {(1-2\gamma {\parallel A\parallel }^2+{\gamma }^2{\parallel A\parallel }^4)}^{\frac{1}{2}}\parallel x_n-x_{n-1}\parallel +\gamma {\parallel A\parallel }^2\parallel x_n-x_{n-1}\parallel \\ +\gamma \parallel A\parallel |1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_2}A{x}_n-A{x}_n\parallel \\ +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel \\ =& (1-\gamma {\parallel A\parallel }^2)\parallel x_n-x_{n-1}\parallel +\gamma {\parallel A\parallel }^2\parallel x_n-x_{n-1}\parallel \\ +\gamma \parallel A\parallel |1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_2}A{x}_n-A{x}_n\parallel \\ +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel \\ =& \parallel x_n-x_{n-1}\parallel +\gamma \parallel A\parallel |1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_2}A{x}_n-A{x}_n\parallel \\ +|1-\frac{r_{n-1}}{r_n}|\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel \\ =& \parallel x_n-x_{n-1}\parallel +\left|\frac{r_n-r_{n-1}}{r_n}\right|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n),\end{array}$$

where ${\sigma }_n:=\parallel T_{r_n}^{F_2}A{x}_n-A{x}_n\parallel$ and ${\chi }_n:=\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)\parallel$. Without loss of generality, let us assume that there exists a real number μ such that $r_n>\mu >0$, for all positive integers n. Then we get

$$\parallel u_{n-1}-u_n\parallel \le \parallel x_{n-1}-x_n\parallel +\frac{1}{\mu }|r_{n-1}-r_n|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n).$$

(3.8)

Next, we estimate

$$\begin{array}{rcl}{\parallel z_n-z_{n-1}\parallel }^2& =& \parallel P_C[P_C[u_n-{\mu }_2{B}_2{u}_n]-{\mu }_1{B}_1{P}_C[u_n-{\mu }_2{B}_2{u}_n]]\\ {-P_C[P_C[u_{n-1}-{\mu }_2{B}_2{u}_{n-1}]-{\mu }_1{B}_1{P}_C[u_{n-1}-{\mu }_2{B}_2{u}_{n-1}]]\parallel }^2\\ \le & \parallel P_C[u_n-{\mu }_2{B}_2{u}_n]-{\mu }_1{B}_1{P}_C[u_n-{\mu }_2{B}_2{u}_n]\\ {-(P_C[u_{n-1}-{\mu }_2{B}_2{u}_{n-1}]-{\mu }_1{B}_1{P}_C[u_{n-1}-{\mu }_2{B}_2{u}_{n-1}])\parallel }^2\\ =& \parallel P_C[u_n-{\mu }_2{B}_2{u}_n]-P_C[u_{n-1}-{\mu }_2{B}_2{u}_{n-1}]\\ {-{\mu }_1(B_1{P}_C[u_n-{\mu }_2{B}_2{u}_n]-B_1{P}_C[u_{n-1}-{\mu }_2{B}_2{u}_{n-1}])\parallel }^2\\ \le & {\parallel P_C[u_n-{\mu }_2{B}_2{u}_n]-P_C[u_{n-1}-{\mu }_2{B}_2{u}_{n-1}]\parallel }^2\\ -{\mu }_1(2{\theta }_1-{\mu }_1){\parallel B_1{P}_C[u_n-{\mu }_2{B}_2{u}_n]-B_1{P}_C[u_{n-1}-{\mu }_2{B}_2{u}_{n-1}]\parallel }^2\\ \le & {\parallel P_C[u_n-{\mu }_2{B}_2{u}_n]-P_C[u_{n-1}-{\mu }_2{B}_2{u}_{n-1}]\parallel }^2\\ \le & {\parallel (u_n-u_{n-1})-{\mu }_2(B_2{u}_n-B_2{u}_{n-1})\parallel }^2\\ \le & {\parallel u_n-u_{n-1}\parallel }^2-{\mu }_2(2{\theta }_2-{\mu }_2){\parallel B_2{u}_n-B_2{u}_{n-1}\parallel }^2\\ \le & {\parallel u_n-u_{n-1}\parallel }^2.\end{array}$$

(3.9)

It follows from (3.8) and (3.9) that

$$\parallel z_n-z_{n-1}\parallel \le \parallel x_{n-1}-x_n\parallel +\frac{1}{\mu }|r_{n-1}-r_n|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n).$$

From (3.1) and the above inequality, we get

$$\begin{array}{rcl}\parallel y_n-y_{n-1}\parallel & =& \parallel {\beta }_{n}S{x}_n+(1-{\beta }_n)z_n-({\beta }_{n-1}S{x}_{n-1}+(1-{\beta }_{n-1})z_{n-1})\parallel \\ =& \parallel {\beta }_n(S{x}_n-S{x}_{n-1})+({\beta }_n-{\beta }_{n-1})S{x}_{n-1}\\ +(1-{\beta }_n)(z_n-z_{n-1})+({\beta }_{n-1}-{\beta }_n)z_{n-1}\parallel \\ \le & {\beta }_n\parallel x_n-x_{n-1}\parallel +(1-{\beta }_n)\parallel z_n-z_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel )\\ \le & {\beta }_n\parallel x_n-x_{n-1}\parallel +(1-{\beta }_n)\{\parallel x_{n-1}-x_n\parallel +\frac{1}{\mu }|r_{n-1}-r_n|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n)\}\\ +|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel )\\ \le & \parallel x_n-x_{n-1}\parallel +\frac{1}{\mu }|r_{n-1}-r_n|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n)\\ +|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel ).\end{array}$$

(3.10)

Next, we estimate

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & =& \parallel P_C[V_n]-P_C[V_{n-1}]\parallel \\ \le & \parallel {\alpha }_n\rho (U(x_n)-U(x_{n-1}))+({\alpha }_n-{\alpha }_{n-1})\rho U(x_{n-1})+(I-{\alpha }_n\mu F)(T(y_n))\\ -(I-{\alpha }_n\mu F)T(y_{n-1})+(I-{\alpha }_n\mu F)(T(y_{n-1}))-(I-{\alpha }_{n-1}\mu F)(T(y_{n-1}))\parallel \\ \le & {\alpha }_n\rho \tau \parallel x_n-x_{n-1}\parallel +(1-{\alpha }_n\nu )\parallel y_n-y_{n-1}\parallel \\ +|{\alpha }_n-{\alpha }_{n-1}|(\parallel \rho U(x_{n-1})\parallel +\parallel \mu F(T(y_{n-1}))\parallel ),\end{array}$$

(3.11)

where the second inequality follows from Lemma 2.7. From (3.10) and (3.11), we have

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & \le & {\alpha }_n\rho \tau \parallel x_n-x_{n-1}\parallel +(1-{\alpha }_n\nu )\{\parallel x_n-x_{n-1}\parallel +\frac{1}{\mu }|r_{n-1}-r_n|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n)\\ +|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel )\}\\ +|{\alpha }_n-{\alpha }_{n-1}|(\parallel \rho U(x_{n-1})\parallel +\parallel \mu F(T(y_{n-1}))\parallel )\\ \le & (1-(\nu -\rho \tau ){\alpha }_n)\parallel x_n-x_{n-1}\parallel \\ +\frac{1}{\mu }|r_{n-1}-r_n|(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n)+|{\beta }_n-{\beta }_{n-1}|(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel )\\ +|{\alpha }_n-{\alpha }_{n-1}|(\parallel \rho U(x_{n-1})\parallel +\parallel \mu F(T(y_{n-1}))\parallel )\\ \le & (1-(\nu -\rho \tau ){\alpha }_n)\parallel x_n-x_{n-1}\parallel \\ +M(\frac{1}{\mu }|r_{n-1}-r_n|+|{\beta }_n-{\beta }_{n-1}|+|{\alpha }_n-{\alpha }_{n-1}|).\end{array}$$

(3.12)

Here

$$\begin{array}{rcl}M& =& max\{\underset{n\ge 1}{sup}(\gamma \parallel A\parallel {\sigma }_n+{\chi }_n),\underset{n\ge 1}{sup}(\parallel S{x}_{n-1}\parallel +\parallel z_{n-1}\parallel ),\\ \underset{n\ge 1}{sup}(\parallel \rho U(x_{n-1})\parallel +\parallel \mu F(T(y_{n-1}))\parallel )\}.\end{array}$$

It follows by conditions (a)-(d) of Algorithm 3.1 and Lemma 2.8 that

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

Next, we show that ${lim}_{n\to \mathrm{\infty }}\parallel u_n-x_n\parallel =0$. Since $x^{\ast }\in S^{\ast }\cap \mathrm{\Lambda }\cap F(T)$ by using (3.2), (3.5), and (3.7), we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& =& 〈P_C[V_n]-x^{\ast },x_{n+1}-x^{\ast }〉\\ =& 〈P_C[V_n]-V_n,P_C[V_n]-x^{\ast }〉+〈V_n-x^{\ast },x_{n+1}-x^{\ast }〉\\ \le & 〈{\alpha }_n(\rho U(x_n)-\mu F\left(x^{\ast }\right))+(I-{\alpha }_n\mu F)(T(y_n))\\ -(I-{\alpha }_n\mu F)\left(T\left(x^{\ast }\right)\right),x_{n+1}-x^{\ast }〉\\ =& 〈{\alpha }_n\rho (U(x_n)-U\left(x^{\ast }\right)),x_{n+1}-x^{\ast }〉+{\alpha }_n〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +〈(I-{\alpha }_n\mu F)(T(y_n))-(I-{\alpha }_n\mu F)\left(T\left(x^{\ast }\right)\right),x_{n+1}-x^{\ast }〉\\ \le & {\alpha }_n\rho \tau \parallel x_n-x^{\ast }\parallel \parallel x_{n+1}-x^{\ast }\parallel +{\alpha }_n〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +(1-{\alpha }_n\nu )\parallel y_n-x^{\ast }\parallel \parallel x_{n+1}-x^{\ast }\parallel \\ \le & \frac{{\alpha }_n\rho \tau }{2}({\parallel x_n-x^{\ast }\parallel }^2+{\parallel x_{n+1}-x^{\ast }\parallel }^2)+{\alpha }_n〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu )}{2}({\parallel y_n-x^{\ast }\parallel }^2+{\parallel x_{n+1}-x^{\ast }\parallel }^2)\\ \le & \frac{(1-{\alpha }_n(\nu -\rho \tau ))}{2}{\parallel x_{n+1}-x^{\ast }\parallel }^2+\frac{{\alpha }_n\rho \tau }{2}{\parallel x_n-x^{\ast }\parallel }^2\\ +{\alpha }_n〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu )}{2}({\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(1-{\beta }_n){\parallel z_n-x^{\ast }\parallel }^2)\\ \le & \frac{(1-{\alpha }_n(\nu -\rho \tau ))}{2}{\parallel x_{n+1}-x^{\ast }\parallel }^2+\frac{{\alpha }_n\rho \tau }{2}{\parallel x_n-x^{\ast }\parallel }^2\\ +{\alpha }_n〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉+\frac{(1-{\alpha }_n\nu ){\beta }_n}{2}{\parallel S{x}_n-x^{\ast }\parallel }^2\\ +\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{2}\{{\parallel u_n-x^{\ast }\parallel }^2-{\mu }_2(2{\theta }_2-{\mu }_2){\parallel B_2{u}_n-B_2{x}^{\ast }\parallel }^2\\ -{\mu }_1(2{\theta }_1-{\mu }_1){\parallel B_1{v}_n-B_1{y}^{\ast }\parallel }^2\}\\ \le & \frac{(1-{\alpha }_n(\nu -\rho \tau ))}{2}{\parallel x_{n+1}-x^{\ast }\parallel }^2+\frac{{\alpha }_n\rho \tau }{2}{\parallel x_n-x^{\ast }\parallel }^2\\ +{\alpha }_n〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉+\frac{(1-{\alpha }_n\nu ){\beta }_n}{2}{\parallel S{x}_n-x^{\ast }\parallel }^2\\ +\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{2}\{{\parallel x_n-x^{\ast }\parallel }^2+\gamma (L\gamma -1){\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2\\ -{\mu }_2(2{\theta }_2-{\mu }_2){\parallel B_2{u}_n-B_2{x}^{\ast }\parallel }^2-{\mu }_1(2{\theta }_1-{\mu }_1){\parallel B_1{v}_n-B_1{y}^{\ast }\parallel }^2\},\end{array}$$

(3.13)

which implies that

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & \frac{{\alpha }_n\rho \tau }{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-x^{\ast }\parallel }^2\\ +\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu ){\beta }_n}{1+{\alpha }_n(\nu -\rho \tau )}{\parallel S{x}_n-x^{\ast }\parallel }^2\\ +\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{1+{\alpha }_n(\nu -\rho \tau )}\{{\parallel x_n-x^{\ast }\parallel }^2+\gamma (L\gamma -1){\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2\\ -{\mu }_2(2{\theta }_2-{\mu }_2){\parallel B_2{u}_n-B_2{x}^{\ast }\parallel }^2-{\mu }_1(2{\theta }_1-{\mu }_1){\parallel B_1{v}_n-B_1{y}^{\ast }\parallel }^2\}\\ \le & \frac{{\alpha }_n\rho \tau }{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-x^{\ast }\parallel }^2\\ +\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +{\parallel x_n-x^{\ast }\parallel }^2+\frac{(1-{\alpha }_n\nu ){\beta }_n}{1+{\alpha }_n(\nu -\rho \tau )}{\parallel S{x}_n-x^{\ast }\parallel }^2\\ -\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{1+{\alpha }_n(\nu -\rho \tau )}\{\gamma (1-L\gamma ){\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2\\ +{\mu }_2(2{\theta }_2-{\mu }_2){\parallel B_2{u}_n-B_2{x}^{\ast }\parallel }^2+{\mu }_1(2{\theta }_1-{\mu }_1){\parallel B_1{v}_n-B_1{y}^{\ast }\parallel }^2\}.\end{array}$$

Then from the above inequality, we get

$$\begin{array}{c}\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{1+{\alpha }_n(\nu -\rho \tau )}\{\gamma (1-L\gamma ){\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel }^2+{\mu }_2(2{\theta }_2-{\mu }_2){\parallel B_2{u}_n-B_2{x}^{\ast }\parallel }^2\hfill \\ +{\mu }_1(2{\theta }_1-{\mu }_1){\parallel B_1{v}_n-B_1{y}^{\ast }\parallel }^2\}\hfill \\ \le \frac{{\alpha }_n\rho \tau }{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-x^{\ast }\parallel }^2+\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\hfill \\ +{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2\hfill \\ \le \frac{{\alpha }_n\rho \tau }{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-x^{\ast }\parallel }^2+\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\hfill \\ +{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel .\hfill \end{array}$$

Since $\gamma (1-L\gamma )>0$, $2{\theta }_1-{\mu }_1>0$, $2{\theta }_2-{\mu }_2>0$, ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${\alpha }_n\to 0$, and ${\beta }_n\to 0$, we obtain

$$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}\parallel B_2{u}_n-B_2{x}^{\ast }\parallel =0,\hfill \\ \underset{n\to \mathrm{\infty }}{lim}\parallel B_1{v}_n-B_1{y}^{\ast }\parallel =0\hfill \end{array}$$

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel =0.$$

(3.14)

Since $T_{r_n}^{F_1}$ is firmly nonexpansive, we have

$$\begin{array}{rcl}{\parallel u_n-x^{\ast }\parallel }^2& =& {\parallel T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)-T_{r_n}^{F_1}\left(x^{\ast }\right)\parallel }^2\\ \le & 〈u_n-x^{\ast },x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n-x^{\ast }〉\\ =& \frac{1}{2}\{{\parallel u_n-x^{\ast }\parallel }^2+{\parallel x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n-x^{\ast }\parallel }^2\\ -{\parallel u_n-x^{\ast }-[x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n-x^{\ast }]\parallel }^2\}\\ =& \frac{1}{2}\{{\parallel u_n-x^{\ast }\parallel }^2+{\parallel x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n-x^{\ast }\parallel }^2\\ -{\parallel u_n-x_n-\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel u_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2\\ -{\parallel u_n-x_n-\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n\parallel }^2\}\\ =& \frac{1}{2}\{{\parallel u_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2\\ -[{\parallel u_n-x_n\parallel }^2+{\gamma }^2{\parallel A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n\parallel }^2-2\gamma 〈u_n-x_n,A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉]\}.\end{array}$$

Hence, we get

$${\parallel u_n-x^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-x_n\parallel }^2+2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel .$$

(3.15)

From (3.13), (3.7), and the above inequality, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & \frac{(1-{\alpha }_n(\nu -\rho \tau ))}{2}{\parallel x_{n+1}-x^{\ast }\parallel }^2+\frac{{\alpha }_n\rho \tau }{2}{\parallel x_n-x^{\ast }\parallel }^2\\ +{\alpha }_n〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu )}{2}({\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(1-{\beta }_n){\parallel z_n-x^{\ast }\parallel }^2)\\ \le & \frac{(1-{\alpha }_n(\nu -\rho \tau ))}{2}{\parallel x_{n+1}-x^{\ast }\parallel }^2+\frac{{\alpha }_n\rho \tau }{2}{\parallel x_n-x^{\ast }\parallel }^2\\ +{\alpha }_n〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu )}{2}({\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(1-{\beta }_n){\parallel u_n-x^{\ast }\parallel }^2)\\ \le & \frac{(1-{\alpha }_n(\nu -\rho \tau ))}{2}{\parallel x_{n+1}-x^{\ast }\parallel }^2+\frac{{\alpha }_n\rho \tau }{2}{\parallel x_n-x^{\ast }\parallel }^2\\ +{\alpha }_n〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu )}{2}\{{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(1-{\beta }_n)({\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-x_n\parallel }^2\\ +2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel \left)\right\},\end{array}$$

which implies that

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & \frac{{\alpha }_n\rho \tau }{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-x^{\ast }\parallel }^2\\ +\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu ){\beta }_n}{1+{\alpha }_n(\nu -\rho \tau )}{\parallel S{x}_n-x^{\ast }\parallel }^2\\ +\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{1+{\alpha }_n(\nu -\rho \tau )}\{{\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-x_n\parallel }^2\\ +2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel \}\\ \le & \frac{{\alpha }_n\rho \tau }{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-x^{\ast }\parallel }^2\\ +\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu ){\beta }_n}{1+{\alpha }_n(\nu -\rho \tau )}{\parallel S{x}_n-x^{\ast }\parallel }^2\\ +{\parallel x_n-x^{\ast }\parallel }^2+\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{1+{\alpha }_n(\nu -\rho \tau )}\{-{\parallel u_n-x_n\parallel }^2\\ +2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel \}.\end{array}$$

Hence

$$\begin{array}{rcl}\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{1+{\alpha }_n(\nu -\rho \tau )}{\parallel u_n-x_n\parallel }^2& \le & \frac{{\alpha }_n\rho \tau }{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-x^{\ast }\parallel }^2\\ +\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu ){\beta }_n}{1+{\alpha }_n(\nu -\rho \tau )}{\parallel S{x}_n-x^{\ast }\parallel }^2\\ +\frac{2(1-{\alpha }_n\nu )(1-{\beta }_n)\gamma }{1+{\alpha }_n(\nu -\rho \tau )}\parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel \\ +{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2\\ \le & \frac{{\alpha }_n\rho \tau }{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-x^{\ast }\parallel }^2\\ +\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu ){\beta }_n}{1+{\alpha }_n(\nu -\rho \tau )}{\parallel S{x}_n-x^{\ast }\parallel }^2\\ +\frac{2(1-{\alpha }_n\nu )(1-{\beta }_n)\gamma }{1+{\alpha }_n(\nu -\rho \tau )}\parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel \\ +(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel .\end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${\alpha }_n\to 0$, ${\beta }_n\to 0$, and ${lim}_{n\to \mathrm{\infty }}\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel =0$, we obtain

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

(3.16)

From (2.2), we get

$$\begin{array}{rcl}{\parallel v_n-y^{\ast }\parallel }^2& =& {\parallel P_C[u_n-{\mu }_2{B}_2{u}_n]-P_C[x^{\ast }-{\mu }_2{B}_2{x}^{\ast }]\parallel }^2\\ \le & 〈v_n-y^{\ast },(u_n-{\mu }_2{B}_2{u}_n)-(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })〉\\ =& \frac{1}{2}\{{\parallel v_n-y^{\ast }\parallel }^2+{\parallel u_n-x^{\ast }-{\mu }_2(B_2{u}_n-B_2{x}^{\ast })\parallel }^2\\ -{\parallel u_n-x^{\ast }-{\mu }_2(B_2{u}_n-B_2{x}^{\ast })-(v_n-y^{\ast })\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel v_n-y^{\ast }\parallel }^2+{\parallel u_n-x^{\ast }\parallel }^2-{\mu }_2(2{\theta }_2-{\mu }_2){\parallel B_2{u}_n-B_2{x}^{\ast }\parallel }^2\\ -{\parallel u_n-x^{\ast }-{\mu }_2(B_2{u}_n-B_2{x}^{\ast })-(v_n-y^{\ast })\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel v_n-y^{\ast }\parallel }^2+{\parallel u_n-x^{\ast }\parallel }^2-{\parallel u_n-v_n-{\mu }_2(B_2{u}_n-B_2{x}^{\ast })-(x^{\ast }-y^{\ast })\parallel }^2\}\\ =& \frac{1}{2}\{{\parallel v_n-y^{\ast }\parallel }^2+{\parallel u_n-x^{\ast }\parallel }^2-{\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel }^2\\ +2{\mu }_2〈u_n-v_n-(x^{\ast }-y^{\ast }),B_2{u}_n-B_2{x}^{\ast }〉-{\mu }_2^2{\parallel B_2{u}_n-B_2{x}^{\ast }\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel v_n-y^{\ast }\parallel }^2+{\parallel u_n-x^{\ast }\parallel }^2-{\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel }^2\\ +2{\mu }_2\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{u}_n-B_2{x}^{\ast }\parallel \}.\end{array}$$

Hence

$$\begin{array}{rcl}{\parallel v_n-y^{\ast }\parallel }^2& \le & {\parallel u_n-x^{\ast }\parallel }^2-{\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel }^2\\ +2{\mu }_2\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{u}_n-B_2{x}^{\ast }\parallel \\ \le & {\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-x_n\parallel }^2+2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel \\ -{\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel }^2\\ +2{\mu }_2\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{u}_n-B_2{x}^{\ast }\parallel ,\end{array}$$

(3.17)

where the last inequality follows from (3.15). On the other hand, from (3.1) and (2.2), we obtain

$$\begin{array}{rcl}{\parallel z_n-x^{\ast }\parallel }^2& =& {\parallel P_C[v_n-{\mu }_1{B}_1{v}_n]-P_C[y^{\ast }-{\mu }_1{B}_1{y}^{\ast }]\parallel }^2\\ \le & 〈z_n-x^{\ast },(v_n-{\mu }_1{B}_1{v}_n)-(y^{\ast }-{\mu }_1{B}_1{y}^{\ast })〉\\ =& \frac{1}{2}\{{\parallel z_n-x^{\ast }\parallel }^2+{\parallel v_n-y^{\ast }-{\mu }_1(B_1{v}_n-B_1{y}^{\ast })\parallel }^2\\ -{\parallel v_n-y^{\ast }-{\mu }_1(B_1{v}_n-B_1{y}^{\ast })-(z_n-x^{\ast })\parallel }^2\}\\ =& \frac{1}{2}\{{\parallel z_n-x^{\ast }\parallel }^2+{\parallel v_n-y^{\ast }\parallel }^2-2{\mu }_1〈v_n-y^{\ast },B_1{v}_n-B_1{y}^{\ast }〉\\ +{\mu }_1^2{\parallel B_1{v}_n-B_1{y}^{\ast }\parallel }^2-{\parallel v_n-y^{\ast }-{\mu }_1(B_1{v}_n-B_1{y}^{\ast })-(z_n-x^{\ast })\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel z_n-x^{\ast }\parallel }^2+{\parallel v_n-y^{\ast }\parallel }^2-{\mu }_1(2{\theta }_1-{\mu }_1){\parallel B_1{v}_n-B_1{y}^{\ast }\parallel }^2\\ -{\parallel v_n-y^{\ast }-{\mu }_1(B_1{v}_n-B_1{y}^{\ast })-(z_n-x^{\ast })\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel z_n-x^{\ast }\parallel }^2+{\parallel v_n-y^{\ast }\parallel }^2\\ -{\parallel v_n-z_n-{\mu }_1(B_1{v}_n-B_1{y}^{\ast })+(x^{\ast }-y^{\ast })\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel z_n-x^{\ast }\parallel }^2+{\parallel v_n-y^{\ast }\parallel }^2-{\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel }^2\\ +2{\mu }_1〈v_n-z_n+(x^{\ast }-y^{\ast }),B_1{v}_n-B_1{y}^{\ast }〉\}\\ \le & \frac{1}{2}\{{\parallel z_n-x^{\ast }\parallel }^2+{\parallel v_n-y^{\ast }\parallel }^2-{\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel }^2\\ +2{\mu }_1\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{v}_n-B_1{y}^{\ast }\parallel \}.\end{array}$$

Hence

$$\begin{array}{rcl}{\parallel z_n-x^{\ast }\parallel }^2& \le & {\parallel v_n-y^{\ast }\parallel }^2-{\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel }^2\\ +2{\mu }_1\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{v}_n-B_1{y}^{\ast }\parallel \\ \le & {\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-x_n\parallel }^2+2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel \\ -{\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel }^2+2{\mu }_2\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{u}_n-B_2{x}^{\ast }\parallel \\ -{\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel }^2+2{\mu }_1\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{v}_n-B_1{y}^{\ast }\parallel ,\end{array}$$

where the last inequality follows from (3.17). From (3.13) and the above inequality, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & \frac{(1-{\alpha }_n(\nu -\rho \tau ))}{2}{\parallel x_{n+1}-x^{\ast }\parallel }^2+\frac{{\alpha }_n\rho \tau }{2}{\parallel x_n-x^{\ast }\parallel }^2\\ +{\alpha }_n〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu )}{2}({\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(1-{\beta }_n){\parallel z_n-x^{\ast }\parallel }^2)\\ \le & \frac{(1-{\alpha }_n(\nu -\rho \tau ))}{2}{\parallel x_{n+1}-x^{\ast }\parallel }^2+\frac{{\alpha }_n\rho \tau }{2}{\parallel x_n-x^{\ast }\parallel }^2\\ +{\alpha }_n〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu )}{2}\{{\beta }_n{\parallel S{x}_n-x^{\ast }\parallel }^2+(1-{\beta }_n)({\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-x_n\parallel }^2\\ +2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel )\\ +(1-{\beta }_n)(-{\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel }^2\\ +2{\mu }_2\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{u}_n-B_2{x}^{\ast }\parallel )\\ +(1-{\beta }_n)(-{\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel }^2\\ +2{\mu }_1\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{v}_n-B_1{y}^{\ast }\parallel \left)\right\},\end{array}$$

which implies that

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & \frac{{\alpha }_n\rho \tau }{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-x^{\ast }\parallel }^2\\ +\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu ){\beta }_n}{1+{\alpha }_n(\nu -\rho \tau )}{\parallel S{x}_n-x^{\ast }\parallel }^2\\ +\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{1+{\alpha }_n(\nu -\rho \tau )}\{{\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-x_n\parallel }^2\\ +2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel \}\\ +\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{1+{\alpha }_n(\nu -\rho \tau )}(-{\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel }^2\\ +2{\mu }_2\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{u}_n-B_2{x}^{\ast }\parallel )\\ +\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{1+{\alpha }_n(\nu -\rho \tau )}(-{\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel }^2\\ +2{\mu }_1\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{v}_n-B_1{y}^{\ast }\parallel )\\ \le & \frac{{\alpha }_n\rho \tau }{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-x^{\ast }\parallel }^2\\ +\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +\frac{(1-{\alpha }_n\nu ){\beta }_n}{1+{\alpha }_n(\nu -\rho \tau )}{\parallel S{x}_n-x^{\ast }\parallel }^2\\ +{\parallel x_n-x^{\ast }\parallel }^2+2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel \\ +2{\mu }_2\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{u}_n-B_2{x}^{\ast }\parallel \\ +2{\mu }_1\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{v}_n-B_1{y}^{\ast }\parallel \\ -\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{1+{\alpha }_n(\nu -\rho \tau )}({\parallel u_n-x_n\parallel }^2+{\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel }^2\\ +{\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel }^2).\end{array}$$

Hence

$$\begin{array}{c}\frac{(1-{\alpha }_n\nu )(1-{\beta }_n)}{1+{\alpha }_n(\nu -\rho \tau )}({\parallel u_n-x_n\parallel }^2+{\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel }^2+{\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel }^2)\hfill \\ \le \frac{{\alpha }_n\rho \tau }{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-x^{\ast }\parallel }^2\hfill \\ +\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉+\frac{(1-{\alpha }_n\nu ){\beta }_n}{1+{\alpha }_n(\nu -\rho \tau )}{\parallel S{x}_n-x^{\ast }\parallel }^2\hfill \\ +{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2+2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel \hfill \\ +2{\mu }_2\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{u}_n-B_2{x}^{\ast }\parallel \hfill \\ +2{\mu }_1\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{v}_n-B_1{y}^{\ast }\parallel \hfill \\ =\frac{{\alpha }_n\rho \tau }{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-x^{\ast }\parallel }^2\hfill \\ +\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U\left(x^{\ast }\right)-\mu F\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉+\frac{(1-{\alpha }_n\nu ){\beta }_n}{1+{\alpha }_n(\nu -\rho \tau )}{\parallel S{x}_n-x^{\ast }\parallel }^2\hfill \\ +(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel +2\gamma \parallel A{u}_n-A{x}_n\parallel \parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel \hfill \\ +2{\mu }_2\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{u}_n-B_2{x}^{\ast }\parallel \hfill \\ +2{\mu }_1\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{v}_n-B_1{y}^{\ast }\parallel .\hfill \end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${\alpha }_n\to 0$, ${\beta }_n\to 0$, and ${lim}_{n\to \mathrm{\infty }}\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel B_2{u}_n-B_2{x}^{\ast }\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel B_1{v}_n-B_1{y}^{\ast }\parallel =0$, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel =0.$$

Since

$$\parallel u_n-z_n\parallel \le \parallel u_n-v_n-(x^{\ast }-y^{\ast })\parallel +\parallel v_n-z_n+(x^{\ast }-y^{\ast })\parallel$$

we get

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

(3.18)

It follows from (3.16) and (3.18) that

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

(3.19)

Since $T(x_n)\in C$, we have

$$\begin{array}{rcl}\parallel x_n-T(x_n)\parallel & \le & \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-T(x_n)\parallel \\ =& \parallel x_n-x_{n+1}\parallel +\parallel P_C[V_n]-P_C[T(x_n)]\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +\parallel {\alpha }_n(\rho U(x_n)-\mu F(T(y_n)))+T(y_n)-T(x_n)\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel \rho U(x_n)-\mu F(T(y_n))\parallel +\parallel y_n-x_n\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel \rho U(x_n)-\mu F(T(y_n))\parallel +\parallel {\beta }_{n}S{x}_n+(1-{\beta }_n)z_n-x_n\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel \rho U(x_n)-\mu F(T(y_n))\parallel \\ +{\beta }_n\parallel S{x}_n-x_n\parallel +(1-{\beta }_n)\parallel z_n-x_n\parallel .\end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${\alpha }_n\to 0$, ${\beta }_n\to 0$, and $\parallel \rho U(x_n)-\mu F(T(y_n))\parallel$ and $\parallel S{x}_n-x_n\parallel$ are bounded and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$, we obtain

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

Since $\{x_n\}$ is bounded, without loss of generality we can assume that $x_n\rightharpoonup x^{\ast }\in C$. It follows from Lemma 2.5 that $x^{\ast }\in F(T)$. Therefore $w_w(x_n)\subset F(T)$. □

Theorem 3.1 The sequence $\{x_n\}$ generated by Algorithm  3.1 converges strongly to z, which is the unique solution of the variational inequality

$$〈\rho U(z)-\mu F(z),x-z〉\le 0,\mathrm{\forall }x\in S^{\ast }\cap \mathrm{\Lambda }\cap F(T).$$

(3.20)

Proof Since $\{x_n\}$ is bounded $x_n\rightharpoonup w$ and from Lemma 3.2, we have $w\in F(T)$. Next, we show that $w\in \mathit{EP}(F_1)$. Since $u_n=T_{r_n}^{F_1}(x_n+\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n)$, we have

$$F_1(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉-\frac{1}{r_n}〈y-u_n,\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉\ge 0,\mathrm{\forall }y\in C.$$

It follows from monotonicity of $F_1$ that

$$-\frac{1}{r_n}〈y-u_n,\gamma A^{\ast }(T_{r_n}^{F_2}-I)A{x}_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge F_1(y,u_n),\mathrm{\forall }y\in C$$

and

$$-\frac{1}{r_{n_k}}〈y-u_{n_k},\gamma A^{\ast }(T_{r_{n_k}}^{F_2}-I)A{x}_{n_k}〉+〈y-u_{n_k},\frac{u_{n_k}-x_{n_k}}{r_{n_k}}〉\ge F_1(y,u_{n_k}),\mathrm{\forall }y\in C.$$

(3.21)

Since ${lim}_{n\to \mathrm{\infty }}\parallel u_n-x_n\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel (T_{r_n}^{F_2}-I)A{x}_n\parallel =0$, and $x_n\rightharpoonup w$, it is easy to observe that $u_{n_k}\to w$. It follows by Assumption 2.1(iv) that $F_1(y,w)\le 0$, $\mathrm{\forall }y\in C$.

For any $0<t\le 1$ and $y\in C$, let $y_t=ty+(1-t)w$, and we have $y_t\in C$. Then, from Assumption 2.1(i) and (iv), we have

$$\begin{array}{rcl}0=F_1(y_t,y_t)& \le & t{F}_1(y_t,y)+(1-t)F_1(y_t,w)\\ \le & t{F}_1(y_t,y).\end{array}$$

Therefore $F_1(y_t,y)\ge 0$. From Assumption 2.1(iii), we have $F_1(w,y)\ge 0$, which implies that $w\in \mathit{EP}(F_1)$.

Next, we show that $Aw\in \mathit{EP}(F_2)$. Since $\{x_n\}$ is bounded and $x_n\rightharpoonup w$, there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $x_{n_k}\to w$, and since A is a bounded linear operator $A{x}_{n_k}\to Aw$. Now we set $v_{n_k}=A{x}_{n_k}-T_{r_{n_k}}^{F_2}A{x}_{n_k}$. It follows from (3.14) that ${lim}_{k\to \mathrm{\infty }}{v}_{n_k}=0$ and $A{x}_{n_k}-v_{n_k}=T_{r_{n_k}}^{F_2}A{x}_{n_k}$. Therefore from the definition of $T_{r_{n_k}}^{F_2}$, we have

$$F_2(A{x}_{n_k}-v_{n_k},y)+\frac{1}{r_{n_k}}〈y-(A{x}_{n_k}-v_{n_k}),(A{x}_{n_k}-v_{n_k})-A{x}_{n_k}〉\ge 0,\mathrm{\forall }y\in C.$$

Since $F_2$ is upper semicontinuous in the first argument, taking lim sup in the above inequality as $k\to \mathrm{\infty }$ and using Assumption 2.1(iv), we obtain

$$F_2(Aw,y)\ge 0,\mathrm{\forall }y\in C,$$

which implies that $Aw\in \mathit{EP}(F_2)$ and hence $w\in \mathrm{\Lambda }$.

Next, we show that $w\in S^{\ast }$. Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$ and there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $x_{n_k}\to w$, it is easy to observe that $z_{n_k}\to w$. For any $x,y\in C$, using (2.5), we have

$$\begin{array}{rcl}{\parallel Q(x)-Q(y)\parallel }^2& =& \parallel P_C[P_C[x-{\mu }_2{B}_{2}x]-{\mu }_1{B}_1{P}_C[x-{\mu }_2{B}_{2}x]]\\ {-P_C[P_C[y-{\mu }_2{B}_{2}y]-{\mu }_1{B}_1{P}_C[y-{\mu }_2{B}_{2}y]]\parallel }^2\\ \le & \parallel (P_C[x-{\mu }_2{B}_{2}x]-P_C[y-{\mu }_2{B}_{2}y])\\ {-{\mu }_1(B_1{P}_C[x-{\mu }_2{B}_{2}x]-B_1{P}_C[y-{\mu }_2{B}_{2}y])\parallel }^2\\ \le & {\parallel P_C[x-{\mu }_2{B}_{2}x]-P_C[y-{\mu }_2{B}_{2}y]\parallel }^2\\ {-{\mu }_1(2{\theta }_1-{\mu }_1)\parallel B_1{P}_C[x-{\mu }_2{B}_{2}x]-B_1{P}_C[y-{\mu }_2{B}_{2}y]\parallel }^2\\ \le & {\parallel P_C[x-{\mu }_2{B}_{2}x]-P_C[y-{\mu }_2{B}_{2}y]\parallel }^2\\ \le & {\parallel (x-{\mu }_2{B}_{2}x)-(y-{\mu }_2{B}_{2}y)\parallel }^2\\ \le & {\parallel x-y\parallel }^2-{\mu }_2(2{\theta }_2-{\mu }_2){\parallel B_{2}x-B_{2}y\parallel }^2\\ \le & {\parallel x-y\parallel }^2.\end{array}$$

This implies that $Q:C\to C$ is nonexpansive. On the other hand

$$\begin{array}{rcl}{\parallel z_n-Q(z_n)\parallel }^2& =& {\parallel P_C[P_C[u_n-{\mu }_2{B}_2{u}_n]-{\mu }_1{B}_1{P}_C[u_n-{\mu }_2{B}_2{u}_n]]-Q(z_n)\parallel }^2\\ =& {\parallel Q(u_n)-Q(z_n)\parallel }^2\\ \le & {\parallel u_n-z_n\parallel }^2.\end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel u_n-z_n\parallel =0$ (see (3.18)), we have ${lim}_{n\to \mathrm{\infty }}\parallel z_n-Q(z_n)\parallel =0$. It follows from Lemma 2.5 that $w=Q(w)$, which implies from Lemma 2.2 that $w\in S^{\ast }$.

Thus we have

$$w\in S^{\ast }\cap \mathrm{\Lambda }\cap F(T).$$

Observe that the constants satisfy $0\le \rho \tau <\nu$ and

$$\begin{array}{rcl}k\ge \eta & ⟺& k^2\ge {\eta }^2\\ ⟺& 1-2\mu \eta +{\mu }^2{k}^2\ge 1-2\mu \eta +{\mu }^2{\eta }^2\\ ⟺& \sqrt{1-\mu (2\eta -\mu k^2)}\ge 1-\mu \eta \\ ⟺& \mu \eta \ge 1-\sqrt{1-\mu (2\eta -\mu k^2)}\\ ⟺& \mu \eta \ge \nu ;\end{array}$$

therefore, from Lemma 2.6, the operator $\mu F-\rho U$ is $\mu \eta -\rho \tau$ strongly monotone, and we get the uniqueness of the solution of the variational inequality (3.20) and denote it by $z\in S^{\ast }\cap \mathrm{\Lambda }\cap F(T)$.

Next, we claim that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈\rho U(z)-\mu F(z),x_n-z〉\le 0$. Since $\{x_n\}$ is bounded, there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\rho U(z)-\mu F(z),x_n-z〉& =& \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\rho U(z)-\mu F(z),x_{n_k}-z〉\\ =& 〈\rho U(z)-\mu F(z),w-z〉\le 0.\end{array}$$

Next, we show that $x_n\to z$. We have

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& =& 〈P_C[V_n]-z,x_{n+1}-z〉\\ =& 〈P_C[V_n]-V_n,P_C[V_n]-z〉+〈V_n-z,x_{n+1}-z〉\\ \le & 〈{\alpha }_n(\rho U(x_n)-\mu F(z))+(I-{\alpha }_n\mu F)(T(y_n))-(I-{\alpha }_n\mu F)(T(z)),x_{n+1}-z〉\\ \le & 〈{\alpha }_n\rho (U(x_n)-U(z)),x_{n+1}-z〉+{\alpha }_n〈\rho U(z)-\mu F(z),x_{n+1}-z〉\\ +〈(I-{\alpha }_n\mu F)(T(y_n))-(I-{\alpha }_n\mu F)(T(z)),x_{n+1}-z〉\\ \le & {\alpha }_n\rho \tau \parallel x_n-z\parallel \parallel x_{n+1}-z\parallel +{\alpha }_n〈\rho U(z)-\mu F(z),x_{n+1}-z〉\\ +(1-{\alpha }_n\nu )\parallel y_n-z\parallel \parallel x_{n+1}-z\parallel \\ \le & {\alpha }_n\rho \tau \parallel x_n-z\parallel \parallel x_{n+1}-z\parallel +{\alpha }_n〈\rho U(z)-\mu F(z),x_{n+1}-z〉\\ +(1-{\alpha }_n\nu )\{{\beta }_n\parallel S{x}_n-Sz\parallel +{\beta }_n\parallel Sz-z\parallel +(1-{\beta }_n)\parallel z_n-z\parallel \}\parallel x_{n+1}-z\parallel \\ \le & {\alpha }_n\rho \tau \parallel x_n-z\parallel \parallel x_{n+1}-z\parallel +{\alpha }_n〈\rho U(z)-\mu F(z),x_{n+1}-z〉\\ +(1-{\alpha }_n\nu )\{{\beta }_n\parallel x_n-z\parallel +{\beta }_n\parallel Sz-z\parallel +(1-{\beta }_n)\parallel x_n-z\parallel \}\parallel x_{n+1}-z\parallel \\ =& (1-{\alpha }_n(\nu -\rho \tau ))\parallel x_n-z\parallel \parallel x_{n+1}-z\parallel +{\alpha }_n〈\rho U(z)-\mu F(z),x_{n+1}-z〉\\ +(1-{\alpha }_n\nu ){\beta }_n\parallel Sz-z\parallel \parallel x_{n+1}-z\parallel \\ \le & \frac{1-{\alpha }_n(\nu -\rho \tau )}{2}({\parallel x_n-z\parallel }^2+{\parallel x_{n+1}-z\parallel }^2)+{\alpha }_n〈\rho U(z)-\mu F(z),x_{n+1}-z〉\\ +(1-{\alpha }_n\nu ){\beta }_n\parallel Sz-z\parallel \parallel x_{n+1}-z\parallel ,\end{array}$$

which implies that

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& \le & \frac{1-{\alpha }_n(\nu -\rho \tau )}{1+{\alpha }_n(\nu -\rho \tau )}{\parallel x_n-z\parallel }^2+\frac{2{\alpha }_n}{1+{\alpha }_n(\nu -\rho \tau )}〈\rho U(z)-\mu F(z),x_{n+1}-z〉\\ +\frac{2(1-{\alpha }_n\nu ){\beta }_n}{1+{\alpha }_n(\nu -\rho \tau )}\parallel Sz-z\parallel \parallel x_{n+1}-z\parallel \\ \le & (1-{\alpha }_n(\nu -\rho \tau )){\parallel x_n-z\parallel }^2\\ +\frac{2{\alpha }_n(\nu -\rho \tau )}{1+{\alpha }_n(\nu -\rho \tau )}\{\frac{1}{\nu -\rho \tau }〈\rho U(z)-\mu F(z),x_{n+1}-z〉\\ +\frac{(1-{\alpha }_n\nu ){\beta }_n}{{\alpha }_n(\nu -\rho \tau )}\parallel Sz-z\parallel \parallel x_{n+1}-z\parallel \}.\end{array}$$

Let ${\gamma }_n={\alpha }_n(\nu -\rho \tau )$ and ${\delta }_n=\frac{2{\alpha }_n(\nu -\rho \tau )}{1+{\alpha }_n(\nu -\rho \tau )}\{\frac{1}{\nu -\rho \tau }〈\rho U(z)-\mu F(z),x_{n+1}-z〉+\frac{(1-{\alpha }_n\nu ){\beta }_n}{{\alpha }_n(\nu -\rho \tau )}\parallel Sz-z\parallel \parallel x_{n+1}-z\parallel \}$.

We have

$$\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$$

and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{\frac{1}{\nu -\rho \tau }〈\rho U(z)-\mu F(z),x_{n+1}-z〉+\frac{(1-{\alpha }_n\nu ){\beta }_n}{{\alpha }_n(\nu -\rho \tau )}\parallel Sz-z\parallel \parallel x_{n+1}-z\parallel \}\le 0.$$

It follows that

$$\sum _{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }\text{and}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\delta }_n}{{\gamma }_n}\le 0.$$

Thus all the conditions of Lemma 2.8 are satisfied. Hence we deduce that $x_n\to z$. This completes the proof. □

Remark 3.2 In hierarchical fixed-point problem (1.11), if $S=I-(\rho U-\mu F)$, then we can get the variational inequality (3.20). In (3.20), if $U=0$ then we get the variational inequality $〈F(z),x-z〉\ge 0$, $\mathrm{\forall }x\in S^{\ast }\cap \mathrm{\Lambda }\cap F(T)$, which just is the variational inequality studied by Suzuki [39] extending the common set of solutions of a system of variational inequalities, a split equilibrium problem, and a hierarchical fixed-point problem.

4 Conclusions

In this paper, we suggest and analyze an iterative method for finding the approximate element of the common set of solutions of (1.1), (1.9)-(1.10), and (1.11) in real Hilbert space, which can be viewed as a refinement and improvement of some existing methods for solving a system of variational inequality problem, a split equilibrium problem, and a hierarchical fixed-point problem. Some existing methods (e.g. [20, 26, 34, 40, 45]) can be viewed as special cases of Algorithm 3.1. Therefore, the new algorithm is expected to be widely applicable.