Superimposed algorithms for the split equilibrium problems and fixed point problems

Abstract

The split common problems for finding the equilibrium points and fixed points have been studied. A parallel superimposed algorithm is introduced to solve this split common problem. Strong convergence theorems are shown with some analysis techniques.

MSC:49J30, 47H09, 65K10.

1 Introduction

In the present article, our main purpose is to study the split problem. First, we recall some relevant background in the literature.

Problem 1: the split feasibility problem

Let C and Q be two nonempty closed convex subsets of Hilbert spaces $H_1$ and $H_2$, respectively, and let $A:H_1\to H_2$ be a bounded linear operator. The problem of finding a point $x^{\ast }$ such that

$$x^{\ast }\in C\text{and}A{x}^{\ast }\in Q$$

(1.1)

is called the split feasibility; it was first introduced by Censor and Elfving [1] in finite dimensional Hilbert spaces. Such problems arise in the field of intensity-modulated radiation therapy when one attempts to describe physical dose constraints and equivalent uniform dose constraints within a single model. When $C\in {\mathbb{R}}^N$ and $Q\in {\mathbb{R}}^M$ are a single pair of sets, Censor and Elfving [1] introduced the simultaneous multi-projections algorithm:

$$x_{n+1}=A^{-1}{({\lambda }_{1}I+{\lambda }_{2}A{A}^{\ast })}^{-1}({\lambda }_1{v}_1^{n+1}+{\lambda }_{2}A{A}^{\ast }{v}_2^{n+1}),n\ge 0,$$

(1.2)

where ${\lambda }_1>0$, ${\lambda }_2>0$, ${\lambda }_1+{\lambda }_2=1$, $v_i^{n+1}=P_{C_i}^{f_i}(b_n)$ ($i=1,2$), and $b_n$ is the solution of the equation

$$\sum _{i=1}^2{\lambda }_i\mathrm{\nabla }{f}_i(b_n)=\sum _{i=1}^2\mathrm{\nabla }{f}_i\left(v_i^n\right).$$

Note that the simultaneous multi-projections algorithm (1.2) involve a matrix inversion $A^{-1}$ at each iterative step. This is very time-consuming, particularly if the dimensions are large. In order to solve this problem, Byrne [2] derived a new algorithm, called the CQ-algorithm:

$$x_{n+1}=P_C(x_n-\tau A^{\ast }(I-P_Q)A{x}_n),n\ge 0,$$

where $\tau \in (0,\frac{2}{L})$ with L being the largest eigenvalue of the matrix $A^{\ast }A$, I is the unit matrix or operator, and $P_C$ and $P_Q$ denote the orthogonal projections onto C and Q, respectively. The CQ-algorithm and its variant forms have now been studied for the split feasibility problem; see, for instance [3–13].

Problem 2: the split common fixed point problem

If every closed convex subset of a Hilbert space is the fixed point set of its associating projection, then the split feasibility problem becomes a special case of the split common fixed point problem of finding a point $x^{\ast }$ with the property:

$$x^{\ast }\in Fix(U)\text{and}A{x}^{\ast }\in Fix(T).$$

(1.3)

This problem was first introduced by Censor and Segal [14] who invented an algorithm which generates a sequence $\{x_n\}$ according to the iterative procedure:

$$x_{n+1}=U(x_n-\gamma A^{\ast }(I-T)A{x}_n),n\in \mathbb{N}.$$

(1.4)

Moudafi [15] extended (1.4) to the following relaxed algorithm:

$$x_{n+1}=U_{{\alpha }_n}(x_n+\gamma A^{\ast }(T_{\beta }-I)A{x}_n),n\in \mathbb{N},$$

where $\beta \in (0,1)$, ${\alpha }_n\in (0,1)$ are relaxation parameters. Consequently, Wang and Xu [16] considered a general cyclic algorithm. Very recently, the split problem has also been extended to solve other problems, such as the split monotone variational inclusions and the split variational inequalities, please refer to [15, 17–22] and [23–25].

Problem 3: the equilibrium problem

Consider the following equilibrium problem: Finding $x^{\ast }\in C$ such that

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

(1.5)

where $F:C\times C\to \mathbb{R}$ is a bifunction. We will denote by $EP(F)$ the set of solutions of (1.5).

The equilibrium problems, in its various forms, found application in optimization problems, fixed point problems, and convex minimization problems; in other words, equilibrium problems are a unified model for problems arising in physics, engineering, economics, and so on (see [26–29]).

Motivated by the split common fixed point problem and the equilibrium problem, He and Du [30] presented the following split equilibrium problem and fixed point problem:

$$\text{Find a point }{x}^{\ast }\in Fix(T)\cap EP(F)\text{ such that }A{x}^{\ast }\in Fix(S)\cap EP(G),$$

(1.6)

where $Fix(S)$ and $Fix(T)$ are the sets of fixed points of two nonlinear mappings S and T, respectively, $EP(F)$ and $EP(G)$ are the solution sets of two equilibrium problems with bifunctions F and G, respectively, and A is a bounded linear mapping. Denote the solution set of (1.6) by

$$\mathrm{\Gamma }=\{x\in Fix(T)\cap EP(F):Ax\in Fix(S)\cap EP(G)\}.$$

Special cases

• 1. If $F=0$ and $G=0$, then (1.6) is reduced to the following split common fixed point problem, which has been considered by many authors, for example, [14, 15, 17] and [21]:

  $$\text{Find a point }{x}^{\ast }\in Fix(T)\text{ such that }A{x}^{\ast }\in Fix(S).$$

  (1.7)

• 2. If $S=P_Q$ and $T=P_C$, then (1.7) is reduced to the split feasibility problem (1.1).

• 3. If S and T are all identity operators, then (1.6) is reduced to the split equilibrium problem which has been considered in [18]:

  $$\text{Find a point }{x}^{\ast }\in EP(F)\text{ such that }A{x}^{\ast }\in EP(G).$$

Based on the work in this direction, in this paper we will develop new algorithms to solve the split equilibrium problem and the fixed point problem (1.6). We first introduce a parallel superimposed algorithm. Consequently, strong convergence theorems are shown with some analysis techniques.

2 Preliminaries

Let H be a real Hilbert space with inner product $〈\cdot ,\cdot 〉$ and norm $\parallel \cdot \parallel$, respectively. Let C be a nonempty closed convex subset of H.

Definition 2.1 A mapping $T:C\to C$ is called nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel$$

for all $x,y\in C$.

We will use $Fix(T)$ to denote the set of fixed points of T, that is, $Fix(T)=\{x\in C:x=Tx\}$.

Definition 2.2 A mapping $f:C\to C$ is called contractive if

$$\parallel f(x)-f(y)\parallel \le \rho \parallel x-y\parallel$$

for all $x,y\in C$ and for some constant $\rho \in (0,1)$. In this case, we call f is a ρ-contraction.

Definition 2.3 A linear bounded operator $B:H\to H$ is called strongly positive if there exists a constant $\gamma >0$ such that

$$〈Bx,x〉\ge \gamma {\parallel x\parallel }^2$$

for all $x,y\in H$.

Definition 2.4 We call $P_C:H\to C$ the metric projection if for each $x\in H$

$$\parallel x-P_C(x)\parallel =inf\{\parallel x-y\parallel :y\in C\}.$$

It is well known that the metric projection $P_C:H\to C$ is characterized by

$$〈x-P_C(x),y-P_C(x)〉\le 0$$

for all $x\in H$, $y\in C$. From this, we can deduce that $P_C$ is firmly nonexpansive, that is,

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

(2.1)

for all $x,y\in H$. Hence $P_C$ is also nonexpansive.

It is well known that in a real Hilbert space H, the following two equalities hold:

$${\parallel tx+(1-t)y\parallel }^2=t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2-t(1-t){\parallel x-y\parallel }^2$$

(2.2)

for all $x,y\in H$ and $t\in [0,1]$, and

$${\parallel x+y\parallel }^2={\parallel x\parallel }^2+2〈x,y〉+{\parallel y\parallel }^2$$

(2.3)

for all $x,y\in H$. It follows that

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,x+y〉$$

(2.4)

for all $x,y\in H$.

Throughout this paper, we assume that a bifunction $F:C\times C\to \mathbb{R}$ satisfies the following conditions:

• (H1) $F(x,x)=0$ for all $x\in C$;

• (H2) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$ for all $x,y\in C$;

• (H3) for each $x,y,z\in C$, ${lim}_{t\downarrow 0}F(tz+(1-t)x,y)\le F(x,y)$;

• (H4) for each $x\in C$, $y\mapsto F(x,y)$ is convex and lower semicontinuous.

Lemma 2.5 ([31])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $F:C\times C\to \mathbb{R}$ be a bifunction which satisfies conditions (H1)-(H4). Let $r>0$ and $x\in C$. Then there exists $z\in C$ such that

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

Further, if $U_r^F(x)=\{z\in C:F(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\}$, then the following hold:

• (i) $U_r^F$ is single-valued and $U_r^F$ is firmly nonexpansive, i.e., for any $x,y\in H$, ${\parallel U_r^{F}x-U_r^{F}y\parallel }^2\le 〈U_r^{F}x-U_r^{F}y,x-y〉$;

• (ii) $EP(F)$ is closed and convex and $EP(F)=Fix(U_r^F)$.

Lemma 2.6 ([32])

Let the mapping $U_r^F$ be defined as in Lemma  2.5. Then, for $r,s>0$ and $x,y\in H$,

$$\parallel U_r^F(x)-U_s^F(y)\parallel \le \parallel x-y\parallel +\frac{|s-r|}{s}\parallel U_s^F(y)-y\parallel .$$

Lemma 2.7 ([33])

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

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

for all $n\ge 0$ and

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

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

Lemma 2.8 ([34])

Let C be a closed convex subset of a real Hilbert space H and let $S:C\to C$ be a nonexpansive mapping. Then the mapping $I-S$ is demiclosed. That is, if $\{x_n\}$ is a sequence in C such that $x_n\to x^{\ast }$ weakly and $(I-S)x_n\to y$ strongly, then $(I-S)x^{\ast }=y$.

Lemma 2.9 ([35])

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

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

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

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

3 Main results

In this section, we introduce our algorithm and prove our main results.

Let $H_1$ and $H_2$ be two real Hilbert spaces and let C and D be two nonempty closed convex subsets of $H_1$ and $H_2$, respectively. Let $A:H_1\to H_2$ be a bounded linear operator with its adjoint $A^{\ast }$, B be a strongly positive bounded linear operator on $H_1$ with coefficient $\gamma >0$. Let $f:C\to C$ be a ρ-contraction and $F:C\times C\to \mathbb{R}$ and $G:D\times D\to \mathbb{R}$ be two bifunctions satisfying the conditions (H1)-(H4). Let $S:D\to D$ and $T:C\to C$ be two nonexpansive mappings.

Algorithm 3.1 Taking $x_0\in H_1$ arbitrarily, we define a sequence $\{x_n\}$ by the following:

$$x_{n+1}={\alpha }_n\sigma f(x_n)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_{n}B)T{U}_{{\lambda }_n}^F(x_n+\delta A^{\ast }(S{U}_{{\gamma }_n}^G-I)A{x}_n)$$

(3.1)

for all $n\in \mathbb{N}$, where $\{{\lambda }_n\}$ and $\{{\gamma }_n\}$ are two real number sequences in $(0,\mathrm{\infty })$, $\delta \in (0,\frac{1}{{\parallel A\parallel }^2})$ and $\sigma >0$ are two constants and $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two real number sequences in $(0,1)$.

Theorem 3.2 Suppose $\mathrm{\Gamma }\ne \mathrm{\varnothing }$ and suppose the following conditions hold:

• (C1): ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (C2): $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

• (C3): ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\lambda }_n>0$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\lambda }_{n+1}}{{\lambda }_n}=1$;

• (C4): ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n>0$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\gamma }_{n+1}}{{\gamma }_n}=1$;

• (C5): $\sigma \rho <\gamma$.

Then the sequence $\{x_n\}$ generated by algorithm (3.1) converges strongly to $p={Proj}_{\mathrm{\Gamma }}(\sigma f+I-B)p$, which solves the following VI:

$$〈(\sigma f-B)x,y-x〉\le 0,\mathrm{\forall }y\in \mathrm{\Gamma }.$$

(3.2)

Proof First, we know that the solution of (3.2) is unique. We denote the unique solution by p. That is, $p={Proj}_{\mathrm{\Gamma }}(\sigma f+I-B)p$. Then we have $p\in Fix(T)\cap EP(F)$ and $Ap\in Fix(S)\cap EP(G)$. Set $z_n=U_{{\gamma }_n}^{G}A{x}_n$, $y_n=x_n+\delta A^{\ast }(S{U}_{{\gamma }_n}^G-I)A{x}_n$ and $u_n=U_{{\lambda }_n}^F(x_n+\delta A^{\ast }(S{U}_{{\gamma }_n}^G-I)A{x}_n)$ for all $n\in \mathbb{N}$. Then $u_n=U_{{\lambda }_n}^F{y}_n$. From Lemma 2.5, we know that $U_{{\lambda }_n}^F$ and $U_{{\gamma }_n}^G$ are firmly nonexpansive. By these facts, we have the following conclusions:

$$\parallel z_n-Ap\parallel =\parallel U_{{\gamma }_n}^{G}A{x}_n-Ap\parallel \le \parallel A{x}_n-Ap\parallel ,$$

(3.3)

$$\parallel u_n-p\parallel =\parallel U_{{\lambda }_n}^F{y}_n-p\parallel \le \parallel y_n-p\parallel$$

(3.4)

and

$$\begin{array}{rcl}{\parallel S{U}_{{\gamma }_n}^{G}A{x}_n-Ap\parallel }^2& \le & {\parallel U_{{\gamma }_n}^{G}A{x}_n-Ap\parallel }^2\\ \le & {\parallel A{x}_n-Ap\parallel }^2-{\parallel U_{{\gamma }_n}^{G}A{x}_n-A{x}_n\parallel }^2.\end{array}$$

(3.5)

Applying Lemma 2.6, we deduce

$$\begin{array}{rcl}\parallel u_{n+1}-u_n\parallel & =& \parallel U_{{\lambda }_{n+1}}^F{y}_{n+1}-U_{{\lambda }_n}^F{y}_n\parallel \\ \le & \parallel y_{n+1}-y_n\parallel +\left|\frac{{\lambda }_{n+1}-{\lambda }_n}{{\lambda }_{n+1}}\right|\parallel u_{n+1}-y_{n+1}\parallel \end{array}$$

(3.6)

and

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & =& \parallel U_{{\gamma }_{n+1}}^{G}A{x}_{n+1}-U_{{\gamma }_n}^{G}A{x}_n\parallel \\ \le & \parallel A{x}_{n+1}-A{x}_n\parallel +\left|\frac{{\gamma }_{n+1}-{\gamma }_n}{{\gamma }_{n+1}}\right|\parallel z_{n+1}-A{x}_n\parallel .\end{array}$$

(3.7)

From (3.1), we have

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & =& \parallel {\alpha }_n(\sigma f(x_n)-Bp)+{\beta }_n(x_n-p)+((1-{\beta }_n)I-{\alpha }_{n}B)(T{u}_n-p)\parallel \\ \le & {\alpha }_n\sigma \parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel \sigma f(p)-Bp\parallel +{\beta }_n\parallel x_n-p\parallel \\ +(1-{\beta }_n-{\alpha }_n\gamma )\parallel u_n-p\parallel .\end{array}$$

(3.8)

Using (2.3), we get

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel x_n-p+\delta A^{\ast }(S{z}_n-A{x}_n)\parallel }^2\\ =& {\parallel x_n-p\parallel }^2+{\delta }^2{\parallel A^{\ast }(S{z}_n-A{x}_n)\parallel }^2\\ +2\delta 〈x_n-p,A^{\ast }(S{z}_n-A{x}_n)〉.\end{array}$$

(3.9)

Since A is a linear operator with its adjoint $A^{\ast }$, we have

$$\begin{array}{rcl}〈x_n-p,A^{\ast }(S{z}_n-A{x}_n)〉& =& 〈A(x_n-p),S{z}_n-A{x}_n〉\\ =& 〈A{x}_n-Ap+S{z}_n-A{x}_n-(S{z}_n-A{x}_n),S{z}_n-A{x}_n〉\\ =& 〈S{z}_n-Ap,S{z}_n-A{x}_n〉-{\parallel z_n-A{x}_n\parallel }^2.\end{array}$$

(3.10)

Again from (2.3), we obtain

$$〈S{z}_n-Ap,S{z}_n-A{x}_n〉=\frac{1}{2}({\parallel S{z}_n-Ap\parallel }^2+{\parallel S{z}_n-A{x}_n\parallel }^2-{\parallel A{x}_n-Ap\parallel }^2).$$

(3.11)

From (3.5), (3.10), and (3.11), we have

$$\begin{array}{rcl}〈x_n-p,A^{\ast }(S{z}_n-A{x}_n)〉& =& \frac{1}{2}({\parallel S{z}_n-Ap\parallel }^2+{\parallel S{z}_n-A{x}_n\parallel }^2-{\parallel A{x}_n-Ap\parallel }^2)\\ -{\parallel S{z}_n-A{x}_n\parallel }^2\\ \le & \frac{1}{2}({\parallel A{x}_n-Ap\parallel }^2-{\parallel z_n-A{x}_n\parallel }^2+{\parallel S{z}_n-A{x}_n\parallel }^2\\ -{\parallel A{x}_n-Ap\parallel }^2)-{\parallel S{z}_n-A{x}_n\parallel }^2\\ =& -\frac{1}{2}{\parallel z_n-A{x}_n\parallel }^2-\frac{1}{2}{\parallel S{z}_n-A{x}_n\parallel }^2.\end{array}$$

(3.12)

Substituting (3.12) into (3.9) to deduce

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& \le & {\parallel x_n-p\parallel }^2+{\delta }^2{\parallel A\parallel }^2{\parallel S{z}_n-A{x}_n\parallel }^2+2\delta (-\frac{1}{2}{\parallel z_n-A{x}_n\parallel }^2-\frac{1}{2}{\parallel S{z}_n-A{x}_n\parallel }^2)\\ =& {\parallel x_n-p\parallel }^2+({\delta }^2{\parallel A\parallel }^2-\delta ){\parallel S{z}_n-A{x}_n\parallel }^2-\delta {\parallel z_n-A{x}_n\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2.\end{array}$$

It follows that

$$\parallel y_n-p\parallel \le \parallel x_n-p\parallel .$$

Thus, from (3.8), we get

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & \le & {\alpha }_n\sigma \rho \parallel x_n-p\parallel +{\alpha }_n\parallel \sigma f(p)-Bp\parallel +{\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n-{\alpha }_n\gamma )\parallel x_n-p\parallel \\ =& [1-(\gamma -\sigma \rho ){\alpha }_n]\parallel x_n-p\parallel +{\alpha }_n\parallel \sigma f(p)-Bp\parallel \\ \le & max\{\parallel x_n-p\parallel ,\frac{\parallel \sigma f(p)-Bp\parallel }{\gamma -\sigma \rho }\}.\end{array}$$

The boundedness of the sequence $\{x_n\}$ follows.

Next, we estimate $\parallel u_{n+1}-u_n\parallel$. Observe that

$$\begin{array}{c}{\parallel y_{n+1}-y_n\parallel }^2\hfill \\ ={\parallel x_{n+1}-x_n+\delta [A^{\ast }(S{z}_{n+1}-A{x}_{n+1})-A^{\ast }(S{z}_n-A{x}_n)]\parallel }^2\hfill \\ ={\parallel x_{n+1}-x_n\parallel }^2+{\delta }^2{\parallel A^{\ast }(S{z}_{n+1}-A{x}_{n+1})-A^{\ast }(S{z}_n-A{x}_n)\parallel }^2\hfill \\ +2\delta 〈x_{n+1}-x_n,A^{\ast }[(S{z}_{n+1}-A{x}_{n+1})-(S{z}_n-A{x}_n)]〉\hfill \\ \le {\parallel x_{n+1}-x_n\parallel }^2+{\delta }^2{\parallel A\parallel }^2{\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\hfill \\ +2\delta 〈A{x}_{n+1}-A{x}_n,S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)〉\hfill \\ ={\parallel x_{n+1}-x_n\parallel }^2+{\delta }^2{\parallel A\parallel }^2{\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\hfill \\ +2\delta 〈S{z}_{n+1}-S{z}_n,S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)〉\hfill \\ -2\delta {\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\hfill \\ ={\parallel x_{n+1}-x_n\parallel }^2+{\delta }^2{\parallel A\parallel }^2{\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\hfill \\ +\delta ({\parallel S{z}_{n+1}-S{z}_n\parallel }^2+{\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\hfill \\ -{\parallel A{x}_{n+1}-A{x}_n\parallel }^2)-2\delta {\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\hfill \\ ={\parallel x_{n+1}-x_n\parallel }^2+({\delta }^2{\parallel A\parallel }^2-\delta ){\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\hfill \\ +\delta ({\parallel S{z}_{n+1}-S{z}_n\parallel }^2-{\parallel A{x}_{n+1}-A{x}_n\parallel }^2)\hfill \\ \le {\parallel x_{n+1}-x_n\parallel }^2+({\delta }^2{\parallel A\parallel }^2-\delta ){\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\hfill \\ +\delta ({\parallel z_{n+1}-z_n\parallel }^2-{\parallel A{x}_{n+1}-A{x}_n\parallel }^2).\hfill \end{array}$$

(3.13)

Since $\delta \in (0,\frac{1}{{\parallel A\parallel }^2})$, we derive by virtue of (3.7) and (3.13) that

$$\begin{array}{rcl}{\parallel y_{n+1}-y_n\parallel }^2& \le & {\parallel x_{n+1}-x_n\parallel }^2\\ +\delta \left|\frac{{\gamma }_{n+1}-{\gamma }_n}{{\gamma }_{n+1}}\right|(\parallel z_{n+1}-z_n\parallel +\parallel A{x}_{n+1}-A{x}_n\parallel ).\end{array}$$

(3.14)

According to (3.6) and (3.14), we have

$$\begin{array}{rcl}{\parallel u_{n+1}-u_n\parallel }^2& =& {\parallel U_{{\lambda }_{n+1}}^F{y}_{n+1}-U_{{\lambda }_n}^F{y}_n\parallel }^2\\ \le & {(\parallel y_{n+1}-y_n\parallel +|\frac{{\lambda }_{n+1}-{\lambda }_n}{{\lambda }_{n+1}}|\parallel u_{n+1}-y_{n+1}\parallel )}^2\\ \le & {\parallel y_{n+1}-y_n\parallel }^2+\left|\frac{{\lambda }_{n+1}-{\lambda }_n}{{\lambda }_{n+1}}\right|(2\parallel y_{n+1}-y_n\parallel \parallel u_{n+1}-y_{n+1}\parallel \\ +\left|\frac{{\lambda }_{n+1}-{\lambda }_n}{{\lambda }_{n+1}}\right|{\parallel u_{n+1}-y_{n+1}\parallel }^2)\\ \le & {\parallel x_{n+1}-x_n\parallel }^2+\left|\frac{{\lambda }_{n+1}-{\lambda }_n}{{\lambda }_{n+1}}\right|(2\parallel y_{n+1}-y_n\parallel \parallel u_{n+1}-y_{n+1}\parallel \\ +\left|\frac{{\lambda }_{n+1}-{\lambda }_n}{{\lambda }_{n+1}}\right|{\parallel u_{n+1}-y_{n+1}\parallel }^2)\\ +\delta \left|\frac{{\gamma }_{n+1}-{\gamma }_n}{{\gamma }_{n+1}}\right|(\parallel z_{n+1}-z_n\parallel +\parallel A{x}_{n+1}-A{x}_n\parallel )\\ \le & {\parallel x_{n+1}-x_n\parallel }^2+\left(\right|\frac{{\lambda }_{n+1}-{\lambda }_n}{{\lambda }_{n+1}}|+\delta |\frac{{\gamma }_{n+1}-{\gamma }_n}{{\gamma }_{n+1}}\left|\right)M,\end{array}$$

(3.15)

where $M>0$ is a constant such that

$$\begin{array}{c}\underset{n}{sup}\{2\parallel y_{n+1}-y_n\parallel \parallel u_{n+1}-y_{n+1}\parallel +|\frac{{\lambda }_{n+1}-{\lambda }_n}{{\lambda }_{n+1}}|{\parallel u_{n+1}-y_{n+1}\parallel }^2\hfill \\ +\delta (\parallel z_{n+1}-z_n\parallel +\parallel A{x}_{n+1}-A{x}_n\parallel )\}\le M.\hfill \end{array}$$

Therefore,

$$\parallel u_{n+1}-u_n\parallel \le \parallel x_{n+1}-x_n\parallel +\sqrt{\left(\right|\frac{{\lambda }_{n+1}-{\lambda }_n}{{\lambda }_{n+1}}|+\delta |\frac{{\gamma }_{n+1}-{\gamma }_n}{{\gamma }_{n+1}}\left|\right)M}.$$

(3.16)

From (3.1), we write $x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)w_n$ where $w_n=T{u}_n+\frac{{\alpha }_n}{1-{\beta }_n}(\sigma f(x_n)-BT{u}_n)$ for all $n\in \mathbb{N}$. Then we have

$$\begin{array}{c}\parallel w_{n+1}-w_n\parallel \hfill \\ =\parallel T{u}_{n+1}-T{u}_n+\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}(\sigma f(x_{n+1})-BT{u}_{n+1})-\frac{{\alpha }_n}{1-{\beta }_n}(\sigma f(x_n)-BT{u}_n)\parallel \hfill \\ \le \parallel T{u}_{n+1}-T{u}_n\parallel +\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel \sigma f(x_{n+1})-BT{u}_{n+1}\parallel +\frac{{\alpha }_n}{1-{\beta }_n}\parallel \sigma f(x_n)-BT{u}_n\parallel \hfill \\ \le \parallel u_{n+1}-u_n\parallel +\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel \sigma f(x_{n+1})-BT{u}_{n+1}\parallel +\frac{{\alpha }_n}{1-{\beta }_n}\parallel \sigma f(x_n)-BT{u}_n\parallel \hfill \\ \le \parallel x_{n+1}-x_n\parallel +\sqrt{\left(\right|\frac{{\lambda }_{n+1}-{\lambda }_n}{{\lambda }_{n+1}}|+\delta |\frac{{\gamma }_{n+1}-{\gamma }_n}{{\gamma }_{n+1}}\left|\right)M}\hfill \\ +\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel \sigma f(x_{n+1})-BT{u}_{n+1}\parallel +\frac{{\alpha }_n}{1-{\beta }_n}\parallel \sigma f(x_n)-BT{u}_n\parallel .\hfill \end{array}$$

Noting the condition (C1) and the boundedness of the sequences $\{u_{n+1}\}$, $\{y_{n+1}\}$, $\{z_{n+1}\}$, $\{A{x}_n\}$, $\{f(x_n)\}$, and $\{BT{u}_n\}$, we have

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

By Lemma 2.7, we deduce

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

Hence,

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

(3.17)

Since $x_{n+1}-x_n={\alpha }_n(\sigma f(x_n)-BT{u}_n)+(1-{\beta }_n)(T{u}_n-x_n)$, we obtain

$$\parallel T{u}_n-x_n\parallel \le \frac{1}{{\beta }_n}\{{\alpha }_n\parallel \sigma f(x_n)-BT{u}_n\parallel +\parallel x_{n+1}-x_n\parallel \}.$$

Thus,

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

(3.18)

Using the firmly nonexpansiveness of $U_{{\lambda }_n}^F$, we have

$$\begin{array}{rcl}{\parallel u_n-p\parallel }^2& =& {\parallel U_{{\lambda }_n}^F{y}_n-p\parallel }^2\\ \le & {\parallel y_n-p\parallel }^2-{\parallel U_{{\lambda }_n}^F{y}_n-y_n\parallel }^2\\ =& {\parallel y_n-p\parallel }^2-{\parallel u_n-y_n\parallel }^2\\ =& {\parallel y_n-p\parallel }^2-{\parallel u_n-x_n-\delta A^{\ast }(S{U}_{{\gamma }_n}^G-I)A{x}_n\parallel }^2\\ =& {\parallel y_n-p\parallel }^2-{\parallel u_n-x_n\parallel }^2-{\delta }^2{\parallel A^{\ast }(S{U}_{{\gamma }_n}^G-I)A{x}_n\parallel }^2\\ +2\delta 〈u_n-x_n,A^{\ast }(S{U}_{{\gamma }_n}^G-I)A{x}_n〉.\end{array}$$

(3.19)

Applying (2.4) to (3.1) to deduce

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ ={\parallel {\alpha }_n(\sigma f(x_n)-Bp)+{\beta }_n(x_n-T{u}_n)+(I-{\alpha }_{n}B)(T{u}_n-p)\parallel }^2\hfill \\ \le {\parallel (I-{\alpha }_{n}B)(T{u}_n-p)+{\beta }_n(x_n-T{u}_n)\parallel }^2+2{\alpha }_n〈\sigma f(x_n)-Bp,x_{n+1}-p〉\hfill \\ \le {[\parallel I-{\alpha }_{n}B\parallel \parallel T{u}_n-p\parallel +{\beta }_n\parallel x_n-T{u}_n\parallel ]}^2+2{\alpha }_n\parallel \sigma f(x_n)-Bp\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le {[(1-{\alpha }_n\gamma )\parallel u_n-p\parallel +{\beta }_n\parallel x_n-T{u}_n\parallel ]}^2+2{\alpha }_n\parallel \sigma f(x_n)-Bp\parallel \parallel x_{n+1}-p\parallel \hfill \\ ={(1-{\alpha }_n\gamma )}^2{\parallel u_n-p\parallel }^2+{\beta }_n^2{\parallel x_n-T{u}_n\parallel }^2+2(1-{\alpha }_n\gamma ){\beta }_n\parallel u_n-p\parallel \parallel x_n-T{u}_n\parallel \hfill \\ +2{\alpha }_n\parallel \sigma f(x_n)-Bp\parallel \parallel x_{n+1}-p\parallel .\hfill \end{array}$$

(3.20)

It follows from (3.19) that

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & {\parallel x_n-p\parallel }^2-{\parallel u_n-y_n\parallel }^2+{\beta }_n^2{\parallel x_n-T{u}_n\parallel }^2\\ +2(1-{\alpha }_n\gamma ){\beta }_n\parallel u_n-p\parallel \parallel x_n-T{u}_n\parallel \\ +2{\alpha }_n\parallel \sigma f(x_n)-Bp\parallel \parallel x_{n+1}-p\parallel .\end{array}$$

(3.21)

Then

$$\begin{array}{rcl}{\parallel u_n-y_n\parallel }^2& \le & {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\beta }_n^2{\parallel x_n-T{u}_n\parallel }^2\\ +2(1-{\alpha }_n\gamma ){\beta }_n\parallel u_n-p\parallel \parallel x_n-T{u}_n\parallel +2{\alpha }_n\parallel \sigma f(x_n)-Bp\parallel \parallel x_{n+1}-p\parallel \\ \le & (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_{n+1}-x_n\parallel +{\beta }_n^2{\parallel x_n-T{u}_n\parallel }^2\\ +2(1-{\alpha }_n\gamma ){\beta }_n\parallel u_n-p\parallel \parallel x_n-T{u}_n\parallel +2{\alpha }_n\parallel \sigma f(x_n)-Bp\parallel \parallel x_{n+1}-p\parallel .\end{array}$$

This together with (C1), (3.17), and (3.18) implies that

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

(3.22)

From (3.20), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & {(1-{\alpha }_n\gamma )}^2{\parallel u_n-p\parallel }^2+{\beta }_n^2{\parallel x_n-T{u}_n\parallel }^2\\ +2(1-{\alpha }_n\gamma ){\beta }_n\parallel u_n-p\parallel \parallel x_n-T{u}_n\parallel +2{\alpha }_n\parallel \sigma f(x_n)-Bp\parallel \parallel x_{n+1}-p\parallel \\ \le & {\parallel y_n-p\parallel }^2+{\beta }_n^2{\parallel x_n-T{u}_n\parallel }^2+2(1-{\alpha }_n\gamma ){\beta }_n\parallel u_n-p\parallel \parallel x_n-T{u}_n\parallel \\ +2{\alpha }_n\parallel \sigma f(x_n)-Bp\parallel \parallel x_{n+1}-p\parallel \\ \le & {\parallel x_n-p\parallel }^2+({\delta }^2{\parallel A\parallel }^2-\delta ){\parallel S{z}_n-A{x}_n\parallel }^2-\delta {\parallel z_n-A{x}_n\parallel }^2\\ +{\beta }_n^2{\parallel x_n-T{u}_n\parallel }^2+2(1-{\alpha }_n\gamma ){\beta }_n\parallel u_n-p\parallel \parallel x_n-T{u}_n\parallel \\ +2{\alpha }_n\parallel \sigma f(x_n)-Bp\parallel \parallel x_{n+1}-p\parallel .\end{array}$$

Hence,

$$\begin{array}{c}(\delta -{\delta }^2{\parallel A\parallel }^2){\parallel S{z}_n-A{x}_n\parallel }^2+\delta {\parallel z_n-A{x}_n\parallel }^2\hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\beta }_n^2{\parallel x_n-T{u}_n\parallel }^2\hfill \\ +2(1-{\alpha }_n\gamma ){\beta }_n\parallel u_n-p\parallel \parallel x_n-T{u}_n\parallel +2{\alpha }_n\parallel \sigma f(x_n)-Bp\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_{n+1}-x_n\parallel +{\beta }_n^2{\parallel x_n-T{u}_n\parallel }^2\hfill \\ +2(1-{\alpha }_n\gamma ){\beta }_n\parallel u_n-p\parallel \parallel x_n-T{u}_n\parallel +2{\alpha }_n\parallel \sigma f(x_n)-Bp\parallel \parallel x_{n+1}-p\parallel ,\hfill \end{array}$$

which implies that

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

So,

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

(3.23)

Note that

$$\begin{array}{rl}\parallel y_n-x_n\parallel & =\parallel \delta A^{\ast }(S{U}_{{\gamma }_n}^G-I)A{x}_n\parallel \\ \le \delta \parallel A\parallel \parallel S{z}_n-A{x}_n\parallel .\end{array}$$

Therefore,

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

(3.24)

From (3.18), (3.22), and (3.24), we get

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

(3.25)

Now, we show that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈(\sigma f-B)p,x_n-p〉\le 0$. Choose a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\sigma f-B)p,x_n-p〉=\underset{i\to \mathrm{\infty }}{lim}〈(\sigma f-B)p,x_{n_i}-p〉.$$

(3.26)

Since the sequence $\{x_{n_i}\}$ is bounded, we can choose a subsequence $\{x_{n_{i_j}}\}$ of $\{x_{n_i}\}$ such that $x_{n_{i_j}}\rightharpoonup z$. For the sake of convenience, we assume (without loss of generality) that $x_{n_i}\rightharpoonup z$. Consequently, we derive from the above conclusions that

$$y_{n_i}\rightharpoonup z,u_{n_i}\rightharpoonup z,A{x}_{n_i}\rightharpoonup z\text{and}{z}_{n_i}\rightharpoonup Az.$$

(3.27)

By the demi-closed principle of the nonexpansive mappings S and T (see Lemma 2.8), we deduce $z\in Fix(T)$ and $Az\in Fix(S)$ (according to (3.25) and (3.23), respectively).

Next, we show that $z\in EP(F)$. Since $u_n=U_{{\lambda }_n}^F{y}_n$, we have

$$F(u_n,y)+\frac{1}{{\lambda }_n}〈y-u_n,u_n-y_n〉\ge 0,\mathrm{\forall }y\in C.$$

(3.28)

It follows from the monotonicity of F that

$$\frac{1}{{\lambda }_n}〈y-u_n,u_n-y_n〉\ge F(y,u_n),$$

(3.29)

and hence

$$〈y-u_{n_i},\frac{u_{n_i}-y_{n_i}}{{\lambda }_{n_i}}〉\ge F(y,u_{n_i}).$$

(3.30)

Since $\parallel u_n-y_n\parallel \to 0$, $u_{n_i}\rightharpoonup z$, and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\lambda }_n>0$, we obtain $\frac{u_{n_i}-y_{n_i}}{{\lambda }_{n_i}}\to 0$. It follows that $0\ge F(y,z)$. For t with $0<t\le 1$ and $y\in C$, let $y_t=ty+(1-t)z\in C$. It follows that $F(y_t,z)\le 0$. So,

$$0=F(y_t,y_t)\le tF(y_t,y)+(1-t)F(y_t,z)\le tF(y_t,y).$$

(3.31)

Therefore, $0\le F(y_t,y)$. Thus $0\le F(z,y)$. This implies that $z\in EP(F)$. Similarly, we can prove that $Az\in EP(G)$. To this end, we deduce $z\in Fix(T)\cap EP(F)$ and $Az\in Fix(S)\cap EP(G)$. That is to say, $z\in \mathrm{\Gamma }$. Therefore,

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\sigma f-B)p,x_n-p〉& =\underset{i\to \mathrm{\infty }}{lim}〈(\sigma f-B)p,x_{n_i}-p〉\\ =\underset{i\to \mathrm{\infty }}{lim}〈(\sigma f-B)p,z-p〉\\ \le 0.\end{array}$$

(3.32)

Finally, we prove $x_n\to p$. From (3.1), we have

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ =〈{\alpha }_n(\sigma f(x_n)-Bp)+{\beta }_n(x_n-p)+((1-{\beta }_n)I-{\alpha }_{n}B)(T{u}_n-p),x_{n+1}-p〉\hfill \\ ={\alpha }_n〈\sigma f(x_n)-Bp,x_{n+1}-p〉+{\beta }_n〈x_n-p,x_{n+1}-p〉\hfill \\ +〈((1-{\beta }_n)I-{\alpha }_{n}B)(T{u}_n-p),x_{n+1}-p〉\hfill \\ \le {\alpha }_n\sigma 〈f(x_n)-f(p),x_{n+1}-p〉+{\alpha }_n〈\sigma f(p)-Bp,x_{n+1}-p〉\hfill \\ +{\beta }_n\parallel x_n-p\parallel \parallel x_{n+1}-p\parallel +(1-{\beta }_n-{\alpha }_n\gamma )\parallel T{u}_n-p\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le [1-(\gamma -\sigma \rho ){\alpha }_n]\parallel x_n-p\parallel \parallel x_{n+1}-p\parallel +{\alpha }_n〈\sigma f(p)-Bp,x_{n+1}-p〉\hfill \\ \le \frac{1-(\gamma -\sigma \rho ){\alpha }_n}{2}{\parallel x_n-p\parallel }^2+\frac{1}{2}{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n〈\sigma f(p)-Bp,x_{n+1}-p〉.\hfill \end{array}$$

(3.33)

It follows that

$${\parallel x_{n+1}-p\parallel }^2\le [1-(\gamma -\sigma \rho ){\alpha }_n]{\parallel x_n-p\parallel }^2+2{\alpha }_n〈\sigma f(p)-Bp,x_{n+1}-p〉.$$

(3.34)

Applying Lemma 2.9 and (3.32) to (3.34), we deduce $x_n\to p$. The proof is completed. □

Algorithm 3.3 Taking $x_0\in H_1$ arbitrarily, we define a sequence $\{x_n\}$ by the following:

$$x_{n+1}={\alpha }_n\sigma f(x_n)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_{n}B)T(x_n+\delta A^{\ast }(S-I)A{x}_n)$$

(3.35)

for all $n\in \mathbb{N}$, where $\delta \in (0,\frac{1}{{\parallel A\parallel }^2})$ and $\sigma >0$ are two constants and $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two real number sequences in $(0,1)$.

Corollary 3.4 Suppose ${\mathrm{\Gamma }}_1=\{x\in Fix(T):Ax\in Fix(S)\}\ne \mathrm{\varnothing }$ and suppose the following conditions hold:

• (C1): ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (C2): $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

• (C3): $\sigma \rho <\gamma$.

Then the sequence $\{x_n\}$ generated by algorithm (3.35) converges strongly to $p={Proj}_{{\mathrm{\Gamma }}_1}(\sigma f+I-B)p$, which solves the following VI:

$$〈(\sigma f-B)x,y-x〉\le 0,\mathrm{\forall }y\in {\mathrm{\Gamma }}_1.$$

Algorithm 3.5 Taking $x_0\in H_1$ arbitrarily, we define a sequence $\{x_n\}$ by the following:

$$x_{n+1}={\alpha }_n\sigma f(x_n)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_{n}B)U_{{\lambda }_n}^F(x_n+\delta A^{\ast }(U_{{\gamma }_n}^G-I)A{x}_n)$$

(3.36)

for all $n\in \mathbb{N}$, where $\{{\lambda }_n\}$ and $\{{\gamma }_n\}$ are two real number sequences in $(0,\mathrm{\infty })$, $\delta \in (0,\frac{1}{{\parallel A\parallel }^2})$ and $\sigma >0$ are two constants and $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two real number sequences in $(0,1)$.

Corollary 3.6 Suppose ${\mathrm{\Gamma }}_2=\{x\in EP(F):Ax\in EP(G)\}\ne \mathrm{\varnothing }$ and suppose the following conditions hold:

• (C1): ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (C2): $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

• (C3): ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\lambda }_n>0$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\lambda }_{n+1}}{{\lambda }_n}=1$;

• (C4): ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n>0$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\gamma }_{n+1}}{{\gamma }_n}=1$;

• (C5): $\sigma \rho <\gamma$.

Then the sequence $\{x_n\}$ generated by algorithm (3.36) converges strongly to $p={Proj}_{{\mathrm{\Gamma }}_2}(\sigma f+I-B)p$, which solves the following VI:

$$〈(\sigma f-B)x,y-x〉\le 0,\mathrm{\forall }y\in {\mathrm{\Gamma }}_2.$$

Algorithm 3.7 Taking $x_0\in H_1$ arbitrarily, we define a sequence $\{x_n\}$ by the following:

$$x_{n+1}={\alpha }_n\sigma f(x_n)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_{n}B)P_C(x_n+\delta A^{\ast }(P_Q-I)A{x}_n)$$

(3.37)

for all $n\in \mathbb{N}$, where $\delta \in (0,\frac{1}{{\parallel A\parallel }^2})$ and $\sigma >0$ are two constants and $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two real number sequences in $(0,1)$.

Corollary 3.8 Suppose ${\mathrm{\Gamma }}_3=\{x\in C:Ax\in Q\}\ne \mathrm{\varnothing }$ and suppose the following conditions hold:

• (C1): ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (C2): $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

• (C3): $\sigma \rho <\gamma$.

Then the sequence $\{x_n\}$ generated by algorithm (3.37) converges strongly to $p={Proj}_{{\mathrm{\Gamma }}_3}(\sigma f+I-B)p$, which solves the following VI:

$$〈(\sigma f-B)x,y-x〉\le 0,\mathrm{\forall }y\in {\mathrm{\Gamma }}_3.$$