The convergence analysis of the projection methods for a system of generalized relaxed cocoercive variational inequalities in Hilbert spaces

Abstract

In this paper, the approximate solvability for a system of generalized relaxed cocoercive nonlinear variational inequalities in Hilbert spaces is studied, based on the convergence of the projection methods. The results presented in this paper extend and improve the main results of Refs. (Verma in Comput. Math. Appl. 41:1025-1031, 2001; Verma in Int. J. Differ. Equ. Appl. 6:359-367, 2002; Verma in J. Optim. Theory Appl. 121(1):203-210, 2004; Verma in Appl. Math. Lett. 18(11):1286-1292, 2005; Chang, Lee and Chan in Appl. Math. Lett. 20:329-334, 2007).

MSC:90C33, 65K15, 58E36.

1 Introduction

Variational inequalities are one of the most interesting and intensively studied classes of mathematical problems and there exists a considerable amount of literature [1–15] on the approximate solvability of nonlinear variational inequalities. In this paper, we consider, based on the projection methods, the approximate solvability for a system of generalized relaxed cocoercive nonlinear variational inequalities in Hilbert spaces. The results presented in this paper extend and improve the main results in [1–5].

Throughout this paper, we assume that H is a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the induced norm $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H, and let $P_C$ be the metric projection of H onto C. Let $T_i:C\times C\to H$, $g_i:C\to C$ and $f_i:C\to H$ be relaxed cocoercive mappings for each $i=1,2$. We consider a system of generalized nonlinear variational inequality (SGNVI) problem as follows: find an element $(x^{\ast },y^{\ast })\in C\times C$ such that

$$\{\begin{array}{l}〈\lambda T_1(y^{\ast },x^{\ast })+g_1(x^{\ast })-f_1(y^{\ast }),x-g_1(x^{\ast })〉\ge 0,\mathrm{\forall }x\in C\text{ and }\lambda >0,\\ 〈\mu T_2(x^{\ast },y^{\ast })+g_2(y^{\ast })-f_2(x^{\ast }),x-g_2(y^{\ast })〉\ge 0,\mathrm{\forall }x\in C\text{ and }\mu >0.\end{array}$$

(1.1)

SGNVI problem (1.1) is equivalent to the following projection problem:

$$\{\begin{array}{l}{g}_1(x^{\ast })=P_C(f_1(y^{\ast })-\lambda T_1(y^{\ast },x^{\ast })),\mathrm{\forall }\lambda >0,\\ g_2(y^{\ast })=P_C(f_2(x^{\ast })-\mu T_2(x^{\ast },y^{\ast })),\mathrm{\forall }\mu >0.\end{array}$$

(1.2)

Next we consider some special cases of SGNVI problem (1.1), where I is the identity mapping.

• (1) If $g_1=g_2=I$, then SGNVI problem (1.1) is reduced to the following: find an element $(x^{\ast },y^{\ast })\in C\times C$ such that

  $$\{\begin{array}{l}〈\lambda T_1(y^{\ast },x^{\ast })+x^{\ast }-f_1(y^{\ast }),x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C\text{ and }\lambda >0,\\ 〈\mu T_2(x^{\ast },y^{\ast })+y^{\ast }-f_2(x^{\ast }),x-y^{\ast }〉\ge 0,\mathrm{\forall }x\in C\text{ and }\mu >0.\end{array}$$

  (1.3)

• (2) If $f_1=f_2=I$, then SGNVI problem (1.1) is reduced to the following: find an element $(x^{\ast },y^{\ast })\in C\times C$ such that

  $$\{\begin{array}{l}〈\lambda T_1(y^{\ast },x^{\ast })+g_1(x^{\ast })-y^{\ast },x-g_1(x^{\ast })〉\ge 0,\mathrm{\forall }x\in C\text{ and }\lambda >0,\\ 〈\mu T_2(x^{\ast },y^{\ast })+g_2(y^{\ast })-x^{\ast },x-g_2(y^{\ast })〉\ge 0,\mathrm{\forall }x\in C\text{ and }\mu >0.\end{array}$$

  (1.4)

• (3) If $g_1=g_2=f_1=f_2=I$, then SGNVI problem (1.1) is reduced to the following: find an element $(x^{\ast },y^{\ast })\in C\times C$ such that

  $$\{\begin{array}{l}〈\lambda T_1(y^{\ast },x^{\ast })+x^{\ast }-y^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C\text{ and }\lambda >0,\\ 〈\mu T_2(x^{\ast },y^{\ast })+y^{\ast }-x^{\ast },x-y^{\ast }〉\ge 0,\mathrm{\forall }x\in C\text{ and }\mu >0.\end{array}$$

  (1.5)

• (4) If $T_1$ and $T_2$ are univariate mappings, then SGNVI problem (1.1) is reduced to the following: find an element $(x^{\ast },y^{\ast })\in C\times C$ such that

  $$\{\begin{array}{l}〈\lambda T_1(y^{\ast })+g_1(x^{\ast })-f_1(y^{\ast }),x-g_1(x^{\ast })〉\ge 0,\mathrm{\forall }x\in C\text{ and }\lambda >0,\\ 〈\mu T_2(x^{\ast })+g_2(y^{\ast })-f_2(x^{\ast }),x-g_2(y^{\ast })〉\ge 0,\mathrm{\forall }x\in C\text{ and }\mu >0.\end{array}$$

  (1.6)

• (5) If $g_1=g_2=f_1=f_2=I$, $T_1$ and $T_2$ are univariate mappings, then SGNVI problem (1.1) is reduced to the following: find an element $(x^{\ast },y^{\ast })\in C\times C$ such that

  $$\{\begin{array}{l}〈\lambda T_1(y^{\ast })+x^{\ast }-y^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C\text{ and }\lambda >0,\\ 〈\mu T_2(x^{\ast })+y^{\ast }-x^{\ast },x-y^{\ast }〉\ge 0,\mathrm{\forall }x\in C\text{ and }\mu >0.\end{array}$$

  (1.7)

• (6) If $g_1=g_2=f_1=f_2=I$ and $\mu =0$, then SGNVI problem (1.1) is reduced to the following: find an element $x^{\ast }\in C$ such that

  $$〈T_1(x^{\ast },x^{\ast }),x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C.$$

  (1.8)

2 Preliminaries

In order to prove our main results in the next section, we recall several definitions and lemmas.

Definition 2.1 Let $T:C\to H$ be a mapping.

• (1) T is said to be β-Lipschitz continuous if there exists a constant $\beta >0$ such that

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

• (2) T is said to be monotone if

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

• (3) T is said to be δ-strongly monotone if there exists a constant $\delta >0$ such that

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

This implies that

$$\parallel Tx-Ty\parallel \ge \delta \parallel x-y\parallel ,\mathrm{\forall }x,y\in C,$$

that is, T is δ-expansive.

• (4) T is said to be γ-cocoercive if there exists a constant $\gamma >0$ such that

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

Clearly, every γ-cocoercive mapping T is $\frac{1}{\gamma }$-Lipschitz continuous.

• (5) T is said to be relaxed γ-cocoercive if there exists a constant $\gamma >0$ such that

  $$〈Tx-Ty,x-y〉\ge (-\gamma ){\parallel Tx-Ty\parallel }^2,\mathrm{\forall }x,y\in C.$$

• (6) T is said to be relaxed $(\gamma ,\delta )$-cocoercive if there exist two constants $\gamma ,\delta >0$ such that

  $$〈Tx-Ty,x-y〉\ge (-\gamma ){\parallel Tx-Ty\parallel }^2+\delta {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

Definition 2.2 A mapping $T:C\times C\to H$ is said to be relaxed $(\gamma ,\delta )$-cocoercive if there exist two constants $\gamma ,\delta >0$ such that for all $x,x^{\ast }\in C$,

$$〈T(x,y)-T(x^{\ast },y^{\ast }),x-x^{\ast }〉\ge (-\gamma ){\parallel T(x,y)-T(x^{\ast },y^{\ast })\parallel }^2+\delta {\parallel x-x^{\ast }\parallel }^2,\mathrm{\forall }y,y^{\ast }\in C.$$

Definition 2.3 A mapping $T:C\times C\to H$ is said to be β-Lipschitz continuous in the first variable if there exists a constant $\beta >0$ such that for all $x,x^{\ast }\in C$,

$$\parallel T(x,y)-T(x^{\ast },y^{\ast })\parallel \le \beta \parallel x-x^{\ast }\parallel ,\mathrm{\forall }y,y^{\ast }\in C.$$

Definition 2.4 $P_C:H\to C$ is called a metric projection if for every point $x\in H$, there exists a unique nearest point in C, denoted by $P_{C}x$, such that

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

Lemma 2.1 $P_C:H\to C$ is a metric projection, then $P_C$ is a nonexpansive mapping, i.e.,

$$\parallel P_{C}x-P_{C}y\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in H.$$

Lemma 2.2 [5]

Let $\{a_n\}$, $\{b_n\}$ and $\{c_n\}$ be three nonnegative real sequences such that

$$a_{n+1}\le (1-{\lambda }_n)a_n+b_n+c_n,\mathrm{\forall }n\ge n_0,$$

where $n_0$ is some nonnegative integer, $\{{\lambda }_n\}$ is a sequence in $(0,1)$ with ${\sum }_{n=1}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$, $b_n=o({\lambda }_n)$ and ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$. Then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

3 Main results

In this section, we present the projection methods and give the convergence analysis of SGNVI problem (1.1) involving relaxed $(\gamma ,\delta )$-cocoercive and β-Lipschitz continuous mappings in Hilbert spaces.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_i:C\times C\to H$ be a relaxed $({\gamma }_i,{\delta }_i)$-cocoercive and ${\alpha }_i$-Lipschitz continuous mapping in the first variable, let $g_i:C\to C$ be a relaxed $({\eta }_i,{\rho }_i)$-cocoercive and ${\beta }_i$-Lipschitz continuous mapping, and let $f_i:C\to H$ be a relaxed $({\overline{\eta }}_i,{\overline{\rho }}_i)$-cocoercive and ${\overline{\beta }}_i$-Lipschitz continuous mapping for each $i=1,2$. Suppose that $(x^{\ast },y^{\ast })\in C\times C$ is a solution to SGNVI problem (1.1). For any $(x_0,y_0)\in C\times C$, the sequences $\{x_n\}$ and $\{y_n\}$ are generated by

$$\{\begin{array}{l}{t}_{n+1}=(1-a_n)x_n+a_n(x_n-g_1(x_n)+P_C(f_1(y_n)-\lambda T_1(y_n,x_n))),n\ge 0,\\ x_{n+1}=P_C{t}_{n+1},n\ge 0,\\ z_n=y_n-g_2(y_n)+P_C(f_2(x_n)-\mu T_2(x_n,y_n)),n\ge 1,\\ y_n=(1-b_n)x_n+b_n{P}_C{z}_n,n\ge 1,\end{array}$$

(3.1)

where $\lambda ,\mu >0$ and the following conditions are satisfied:

$$\begin{array}{r}\text{(1)}0<a_n,b_n\le 1;\text{(2)}\sum _{n=0}^{\mathrm{\infty }}{a}_n=\mathrm{\infty };\text{(3)}\sum _{n=1}^{\mathrm{\infty }}(1-b_n)<\mathrm{\infty };\\ \text{(4)}0\le {\theta }_2,{\theta }_5<1;\text{(5)}{\theta }_4+{\theta }_6\ge 1;\\ \text{(6)}0<({\theta }_1+{\theta }_3)({\theta }_4+{\theta }_6)<(1-{\theta }_2)(1-{\theta }_5);\end{array}$$

where

$$\begin{array}{r}{\theta }_1=\sqrt{1-2\lambda {\delta }_1+2\lambda {\gamma }_1{\alpha }_1^2+{\lambda }^2{\alpha }_1^2},{\theta }_2=\sqrt{1-2{\rho }_1+{\beta }_1^2+2{\eta }_1{\beta }_1^2},\\ {\theta }_3=\sqrt{1-2{\overline{\rho }}_1+{\overline{\beta }}_1^2+2{\overline{\eta }}_1{\overline{\beta }}_1^2},{\theta }_4=\sqrt{1-2\mu {\delta }_2+2\mu {\gamma }_2{\alpha }_2^2+{\mu }^2{\alpha }_2^2},\\ {\theta }_5=\sqrt{1-2{\rho }_2+{\beta }_2^2+2{\eta }_2{\beta }_2^2},{\theta }_6=\sqrt{1-2{\overline{\rho }}_2+{\overline{\beta }}_2^2+2{\overline{\eta }}_2{\overline{\beta }}_2^2}.\end{array}$$

Then the sequences $\{x_n\}$ and $\{y_n\}$ converge strongly to $x^{\ast }$ and $y^{\ast }$, respectively.

Proof Since $(x^{\ast },y^{\ast })\in C\times C$ is a solution to SGNVI problem (1.1), it follows that

$$\{\begin{array}{l}{x}^{\ast }=x^{\ast }-g_1(x^{\ast })+P_C(f_1(y^{\ast })-\lambda T_1(y^{\ast },x^{\ast })),\\ y^{\ast }=y^{\ast }-g_2(y^{\ast })+P_C(f_2(x^{\ast })-\mu T_2(x^{\ast },y^{\ast })).\end{array}$$

For $n\ge 1$, we have

$$\begin{array}{rl}\parallel x_{n+1}-x^{\ast }\parallel =& \parallel P_C{t}_{n+1}-P_C{x}^{\ast }\parallel \le \parallel t_{n+1}-x^{\ast }\parallel \\ =& \parallel (1-a_n)x_n+a_n(x_n-g_1(x_n)+P_C(f_1(y_n)-\lambda T_1(y_n,x_n)))-x^{\ast }\parallel \\ \le & (1-a_n)\parallel x_n-x^{\ast }\parallel +a_n\parallel (x_n-g_1(x_n)+P_C(f_1(y_n)-\lambda T_1(y_n,x_n)))\\ -(x^{\ast }-g_1\left(x^{\ast }\right)+P_C(f_1\left(y^{\ast }\right)-\lambda T_1(y^{\ast },x^{\ast })))\parallel \\ \le & (1-a_n)\parallel x_n-x^{\ast }\parallel +a_n\parallel x_n-x^{\ast }-(g_1(x_n)-g_1\left(x^{\ast }\right))\parallel \\ +a_n\parallel P_C(f_1(y_n)-\lambda T_1(y_n,x_n))-P_C(f_1\left(y^{\ast }\right)-\lambda T_1(y^{\ast },x^{\ast }))\parallel \\ \le & (1-a_n)\parallel x_n-x^{\ast }\parallel +a_n\parallel x_n-x^{\ast }-(g_1(x_n)-g_1\left(x^{\ast }\right))\parallel \\ +a_n\parallel f_1(y_n)-f_1\left(y^{\ast }\right)-\lambda (T_1(y_n,x_n)-T_1(y^{\ast },x^{\ast }))\parallel \\ \le & (1-a_n)\parallel x_n-x^{\ast }\parallel +a_n\parallel x_n-x^{\ast }-(g_1(x_n)-g_1\left(x^{\ast }\right))\parallel \\ +a_n\parallel y_n-y^{\ast }-(f_1(y_n)-f_1\left(y^{\ast }\right))\parallel \\ +a_n\parallel y_n-y^{\ast }-\lambda (T_1(y_n,x_n)-T_1(y^{\ast },x^{\ast }))\parallel .\end{array}$$

(3.2)

Since $g_1$ is a relaxed $({\eta }_1,{\rho }_1)$-cocoercive and ${\beta }_1$-Lipschitz continuous mapping, we have

$$\begin{array}{r}{\parallel x_n-x^{\ast }-(g_1(x_n)-g_1\left(x^{\ast }\right))\parallel }^2\\ ={\parallel x_n-x^{\ast }\parallel }^2-2〈g_1(x_n)-g_1\left(x^{\ast }\right),x_n-x^{\ast }〉+{\parallel g_1(x_n)-g_1\left(x^{\ast }\right)\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2-2((-{\eta }_1){\parallel g_1(x_n)-g_1\left(x^{\ast }\right)\parallel }^2+{\rho }_1{\parallel x_n-x^{\ast }\parallel }^2)+{\beta }_1^2{\parallel x_n-x^{\ast }\parallel }^2\\ =(1-2{\rho }_1+{\beta }_1^2){\parallel x_n-x^{\ast }\parallel }^2+2{\eta }_1{\parallel g_1(x_n)-g_1\left(x^{\ast }\right)\parallel }^2\\ \le (1-2{\rho }_1+{\beta }_1^2){\parallel x_n-x^{\ast }\parallel }^2+2{\eta }_1{\beta }_1^2{\parallel x_n-x^{\ast }\parallel }^2\\ \le {\theta }_2^2{\parallel x_n-x^{\ast }\parallel }^2,\end{array}$$

(3.3)

where ${\theta }_2=\sqrt{1-2{\rho }_1+{\beta }_1^2+2{\eta }_1{\beta }_1^2}$. In a similar way, we can obtain that

$$\parallel y_n-y^{\ast }-(f_1(y_n)-f_1\left(y^{\ast }\right))\parallel \le {\theta }_3\parallel y_n-y^{\ast }\parallel ,$$

(3.4)

where ${\theta }_3=\sqrt{1-2{\overline{\rho }}_1+{\overline{\beta }}_1^2+2{\overline{\eta }}_1{\overline{\beta }}_1^2}$. By the assumption that $T_1$ is a relaxed $({\gamma }_1,{\delta }_1)$-cocoercive and ${\alpha }_1$-Lipschitz continuous mapping in the first variable, we have

$$\begin{array}{r}{\parallel y_n-y^{\ast }-\lambda (T_1(y_n,x_n)-T_1(y^{\ast },x^{\ast }))\parallel }^2\\ ={\parallel y_n-y^{\ast }\parallel }^2-2\lambda 〈T_1(y_n,x_n)-T_1(y^{\ast },x^{\ast }),y_n-y^{\ast }〉+{\lambda }^2{\parallel T_1(y_n,x_n)-T_1(y^{\ast },x^{\ast })\parallel }^2\\ \le {\parallel y_n-y^{\ast }\parallel }^2-2\lambda ((-{\gamma }_1){\parallel T_1(y_n,x_n)-T_1(y^{\ast },x^{\ast })\parallel }^2+{\delta }_1{\parallel y_n-y^{\ast }\parallel }^2)\\ +{\lambda }^2{\parallel T_1(y_n,x_n)-T_1(y^{\ast },x^{\ast })\parallel }^2\\ =(1-2\lambda {\delta }_1){\parallel y_n-y^{\ast }\parallel }^2+(2\lambda {\gamma }_1+{\lambda }^2){\parallel T_1(y_n,x_n)-T_1(y^{\ast },x^{\ast })\parallel }^2\\ \le (1-2\lambda {\delta }_1){\parallel y_n-y^{\ast }\parallel }^2+(2\lambda {\gamma }_1+{\lambda }^2){\alpha }_1^2{\parallel y_n-y^{\ast }\parallel }^2\\ ={\theta }_1^2{\parallel y_n-y^{\ast }\parallel }^2,\end{array}$$

(3.5)

where ${\theta }_1=\sqrt{1-2\lambda {\delta }_1+2\lambda {\gamma }_1{\alpha }_1^2+{\lambda }^2{\alpha }_1^2}$. According to (3.2), (3.3), (3.4) and (3.5), we obtain that

$$\parallel x_{n+1}-x^{\ast }\parallel \le [1-a_n(1-{\theta }_2)]\parallel x_n-x^{\ast }\parallel +a_n({\theta }_1+{\theta }_3)\parallel y_n-y^{\ast }\parallel .$$

(3.6)

Next, we estimate $\parallel y_n-y^{\ast }\parallel$. From (3.1), we see that

$$\begin{array}{rl}\parallel y_n-y^{\ast }\parallel =& \parallel (1-b_n)x_n+b_n{P}_C{z}_n-y^{\ast }\parallel \\ \le & (1-b_n)\parallel x_n-y^{\ast }\parallel +b_n\parallel P_C{z}_n-P_C{y}^{\ast }\parallel \\ \le & (1-b_n)\parallel x_n-y^{\ast }\parallel +b_n\parallel z_n-y^{\ast }\parallel \\ =& (1-b_n)\parallel x_n-y^{\ast }\parallel +b_n\parallel y_n-g_2(y_n)+P_C(f_2(x_n)-\mu T_2(x_n,y_n))\\ -(y^{\ast }-g_2\left(y^{\ast }\right)+P_C(f_2\left(x^{\ast }\right)-\mu T_2(x^{\ast },y^{\ast })))\parallel \\ \le & (1-b_n)\parallel x_n-y^{\ast }\parallel +b_n\parallel y_n-y^{\ast }-(g_2(y_n)-g_2\left(y^{\ast }\right))\parallel \\ +b_n\parallel P_C(f_2(x_n)-\mu T_2(x_n,y_n))-P_C(f_2\left(x^{\ast }\right)-\mu T_2(x^{\ast },y^{\ast }))\parallel \\ \le & (1-b_n)\parallel x_n-y^{\ast }\parallel +b_n\parallel y_n-y^{\ast }-(g_2(y_n)-g_2\left(y^{\ast }\right))\parallel \\ +b_n\parallel x_n-x^{\ast }-(f_2(x_n)-f_2\left(x^{\ast }\right))\parallel \\ +b_n\parallel x_n-x^{\ast }-\mu (T_2(x_n,y_n)-T_2(x^{\ast },y^{\ast }))\parallel .\end{array}$$

(3.7)

Similarly, we obtain that

$$\parallel y_n-y^{\ast }-(g_2(y_n)-g_2\left(y^{\ast }\right))\parallel \le {\theta }_5\parallel y_n-y^{\ast }\parallel ,$$

(3.8)

where ${\theta }_5=\sqrt{1-2{\rho }_2+{\beta }_2^2+2{\eta }_2{\beta }_2^2}$, and

$$\parallel x_n-x^{\ast }-(f_2(x_n)-f_2\left(x^{\ast }\right))\parallel \le {\theta }_6\parallel x_n-x^{\ast }\parallel ,$$

(3.9)

where ${\theta }_6=\sqrt{1-2{\overline{\rho }}_2+{\overline{\beta }}_2^2+2{\overline{\eta }}_2{\overline{\beta }}_2^2}$, and

$$\parallel x_n-x^{\ast }-\mu (T_2(x_n,y_n)-T_2(x^{\ast },y^{\ast }))\parallel \le {\theta }_4\parallel x_n-x^{\ast }\parallel ,$$

(3.10)

where ${\theta }_4=\sqrt{1-2\mu {\delta }_2+2\mu {\gamma }_2{\alpha }_2^2+{\mu }^2{\alpha }_2^2}$. According to (3.7), (3.8), (3.9) and (3.10), we obtain that

$$\begin{array}{rcl}\parallel y_n-y^{\ast }\parallel & \le & (1-b_n)\parallel x_n-y^{\ast }\parallel +b_n{\theta }_5\parallel y_n-y^{\ast }\parallel \\ +b_n{\theta }_6\parallel x_n-x^{\ast }\parallel +b_n{\theta }_4\parallel x_n-x^{\ast }\parallel \\ \le & (1-b_n)\parallel x_n-x^{\ast }\parallel +(1-b_n)\parallel x^{\ast }-y^{\ast }\parallel \\ +b_n{\theta }_5\parallel y_n-y^{\ast }\parallel +b_n({\theta }_4+{\theta }_6)\parallel x_n-x^{\ast }\parallel ,\end{array}$$

(3.11)

that is,

$$(1-b_n{\theta }_5)\parallel y_n-y^{\ast }\parallel \le (1-b_n)\parallel x^{\ast }-y^{\ast }\parallel +[1-b_n+b_n({\theta }_4+{\theta }_6)]\parallel x_n-x^{\ast }\parallel .$$

(3.12)

By conditions (1), (4) and (5), we have

$$\frac{1}{1-b_n{\theta }_5}\le \frac{1}{1-{\theta }_5}\text{and}1-b_n+b_n({\theta }_4+{\theta }_6)\le {\theta }_4+{\theta }_6.$$

(3.13)

Substituting (3.13) into (3.12), we have

$$\parallel y_n-y^{\ast }\parallel \le \frac{1-b_n}{1-{\theta }_5}\parallel x^{\ast }-y^{\ast }\parallel +\frac{{\theta }_4+{\theta }_6}{1-{\theta }_5}\parallel x_n-x^{\ast }\parallel .$$

(3.14)

According (3.6) and (3.14), for $n\ge 1$, we have

$$\begin{array}{rcl}\parallel x_{n+1}-x^{\ast }\parallel & \le & [1-a_n(1-{\theta }_2-\frac{({\theta }_1+{\theta }_3)({\theta }_4+{\theta }_6)}{1-{\theta }_5})]\parallel x_n-x^{\ast }\parallel \\ +a_n\frac{({\theta }_1+{\theta }_3)}{1-{\theta }_5}(1-b_n)\parallel x^{\ast }-y^{\ast }\parallel \\ \le & [1-a_n(1-{\theta }_2-\frac{({\theta }_1+{\theta }_3)({\theta }_4+{\theta }_6)}{1-{\theta }_5})]\parallel x_n-x^{\ast }\parallel \\ +\frac{({\theta }_1+{\theta }_3)}{1-{\theta }_5}(1-b_n)\parallel x^{\ast }-y^{\ast }\parallel .\end{array}$$

(3.15)

From conditions (1), (2), (3) and (6), we get

$$a_n(1-{\theta }_2-\frac{({\theta }_1+{\theta }_3)({\theta }_4+{\theta }_6)}{1-{\theta }_5})\in (0,1),\sum _{n=1}^{\mathrm{\infty }}{a}_n(1-{\theta }_2-\frac{({\theta }_1+{\theta }_3)({\theta }_4+{\theta }_6)}{1-{\theta }_5})=\mathrm{\infty }$$

and

$$\sum _{n=1}^{\mathrm{\infty }}\frac{({\theta }_1+{\theta }_3)}{1-{\theta }_5}(1-b_n)\parallel x^{\ast }-y^{\ast }\parallel <\mathrm{\infty }.$$

The conditions in Lemma 2.2 are satisfied, then $\parallel x_n-x^{\ast }\parallel \to 0$ (as $n\to \mathrm{\infty }$), i.e., $x_n\to x^{\ast }$ (as $n\to \mathrm{\infty }$).

On the one hand, from condition (3) we know that $1-b_n\to 0$ (as $n\to \mathrm{\infty }$). On the other hand, from (3.14) and the result that $x_n\to x^{\ast }$ (as $n\to \mathrm{\infty }$), we can get $\parallel y_n-y^{\ast }\parallel \to 0$ (as $n\to \mathrm{\infty }$), i.e., $y_n\to y^{\ast }$ (as $n\to \mathrm{\infty }$).

This completes the proof of Theorem 3.1. □

When $g_1=g_2=I$, $f_1=f_2=I$, then we have ${\theta }_2={\theta }_5=0$, ${\theta }_3={\theta }_6=0$, respectively. And from Theorem 3.1 we can get the following results immediately.

Corollary 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_i:C\times C\to H$ be a relaxed $({\gamma }_i,{\delta }_i)$-cocoercive and ${\alpha }_i$-Lipschitz continuous mapping in the first variable, and let $f_i:C\to H$ be a relaxed $({\overline{\eta }}_i,{\overline{\rho }}_i)$-cocoercive and ${\overline{\beta }}_i$-Lipschitz continuous mapping for each $i=1,2$. Suppose that $(x^{\ast },y^{\ast })\in C\times C$ is a solution to problem (1.3). For any $(x_0,y_0)\in C\times C$, the sequences $\{x_n\}$ and $\{y_n\}$ are generated by

$$\{\begin{array}{l}{x}_{n+1}=(1-a_n)x_n+a_n{P}_C(f_1(y_n)-\lambda T_1(y_n,x_n)),n\ge 0,\\ y_n=(1-b_n)x_n+b_n{P}_C(f_2(x_n)-\mu T_2(x_n,y_n)),n\ge 1,\end{array}$$

(3.16)

where $\lambda ,\mu >0$ and the following conditions are satisfied:

$$\begin{array}{r}\text{(1)}0<a_n,b_n\le 1;\text{(2)}\sum _{n=0}^{\mathrm{\infty }}{a}_n=\mathrm{\infty };\text{(3)}\sum _{n=1}^{\mathrm{\infty }}(1-b_n)<\mathrm{\infty };\\ \text{(4)}{\theta }_4+{\theta }_6\ge 1;\text{(5)}0<({\theta }_1+{\theta }_3)({\theta }_4+{\theta }_6)<1;\end{array}$$

where

$$\begin{array}{r}{\theta }_1=\sqrt{1-2\lambda {\delta }_1+2\lambda {\gamma }_1{\alpha }_1^2+{\lambda }^2{\alpha }_1^2},{\theta }_3=\sqrt{1-2{\overline{\rho }}_1+{\overline{\beta }}_1^2+2{\overline{\eta }}_1{\overline{\beta }}_1^2},\\ {\theta }_4=\sqrt{1-2\mu {\delta }_2+2\mu {\gamma }_2{\alpha }_2^2+{\mu }^2{\alpha }_2^2},{\theta }_6=\sqrt{1-2{\overline{\rho }}_2+{\overline{\beta }}_2^2+2{\overline{\eta }}_2{\overline{\beta }}_2^2}.\end{array}$$

Then the sequences $\{x_n\}$ and $\{y_n\}$ converge strongly to $x^{\ast }$ and $y^{\ast }$, respectively.

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_i:C\times C\to H$ be a relaxed $({\gamma }_i,{\delta }_i)$-cocoercive and ${\alpha }_i$-Lipschitz continuous mapping in the first variable, and let $g_i:C\to C$ be a relaxed $({\eta }_i,{\rho }_i)$-cocoercive and ${\beta }_i$-Lipschitz continuous mapping for each $i=1,2$. Suppose that $(x^{\ast },y^{\ast })\in C\times C$ is a solution to problem (1.4). For any $(x_0,y_0)\in C\times C$, the sequences $\{x_n\}$ and $\{y_n\}$ are generated by

$$\{\begin{array}{l}{t}_{n+1}=(1-a_n)x_n+a_n(x_n-g_1(x_n)+P_C(y_n-\lambda T_1(y_n,x_n))),n\ge 0,\\ x_{n+1}=P_C{t}_{n+1},n\ge 0,\\ z_n=y_n-g_2(y_n)+P_C(x_n-\mu T_2(x_n,y_n)),n\ge 1,\\ y_n=(1-b_n)x_n+b_n{P}_C{z}_n,n\ge 1,\end{array}$$

(3.17)

where $\lambda ,\mu >0$ and the following conditions are satisfied:

$$\begin{array}{r}\text{(1)}0<a_n,b_n\le 1;\text{(2)}\sum _{n=0}^{\mathrm{\infty }}{a}_n=\mathrm{\infty };\text{(3)}\sum _{n=1}^{\mathrm{\infty }}(1-b_n)<\mathrm{\infty };\\ \text{(4)}0\le {\theta }_2,{\theta }_5<1;\text{(5)}{\theta }_4\ge 1;\text{(6)}0<{\theta }_1{\theta }_4<(1-{\theta }_2)(1-{\theta }_5);\end{array}$$

where

$$\begin{array}{r}{\theta }_1=\sqrt{1-2\lambda {\delta }_1+2\lambda {\gamma }_1{\alpha }_1^2+{\lambda }^2{\alpha }_1^2},{\theta }_2=\sqrt{1-2{\rho }_1+{\beta }_1^2+2{\eta }_1{\beta }_1^2},\\ {\theta }_4=\sqrt{1-2\mu {\delta }_2+2\mu {\gamma }_2{\alpha }_2^2+{\mu }^2{\alpha }_2^2},{\theta }_5=\sqrt{1-2{\rho }_2+{\beta }_2^2+2{\eta }_2{\beta }_2^2}.\end{array}$$

Then the sequences $\{x_n\}$ and $\{y_n\}$ converge strongly to $x^{\ast }$ and $y^{\ast }$, respectively.

Corollary 3.3 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_i:C\times C\to H$ be a relaxed $({\gamma }_i,{\delta }_i)$-cocoercive and ${\alpha }_i$-Lipschitz continuous mapping in the first variable for each $i=1,2$. Suppose that $(x^{\ast },y^{\ast })\in C\times C$ is a solution to problem (1.5). For any $(x_0,y_0)\in C\times C$, the sequences $\{x_n\}$ and $\{y_n\}$ are generated by

$$\{\begin{array}{l}{x}_{n+1}=(1-a_n)x_n+a_n{P}_C(y_n-\lambda T_1(y_n,x_n)),n\ge 0,\\ y_n=(1-b_n)x_n+b_n{P}_C(x_n-\mu T_2(x_n,y_n)),n\ge 1,\end{array}$$

(3.18)

where $\lambda ,\mu >0$ and the following conditions are satisfied:

$$\begin{array}{r}\text{(1)}0<a_n,b_n\le 1;\text{(2)}\sum _{n=0}^{\mathrm{\infty }}{a}_n=\mathrm{\infty };\text{(3)}\sum _{n=1}^{\mathrm{\infty }}(1-b_n)<\mathrm{\infty };\\ \text{(4)}{\theta }_4\ge 1;\text{(5)}0<{\theta }_1{\theta }_4<1;\end{array}$$

where

$${\theta }_1=\sqrt{1-2\lambda {\delta }_1+2\lambda {\gamma }_1{\alpha }_1^2+{\lambda }^2{\alpha }_1^2},{\theta }_4=\sqrt{1-2\mu {\delta }_2+2\mu {\gamma }_2{\alpha }_2^2+{\mu }^2{\alpha }_2^2}.$$

Then the sequences $\{x_n\}$ and $\{y_n\}$ converge strongly to $x^{\ast }$ and $y^{\ast }$, respectively.

Corollary 3.4 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_i:C\to H$ be a relaxed $({\gamma }_i,{\delta }_i)$-cocoercive and ${\alpha }_i$-Lipschitz continuous mapping, let $g_i:C\to C$ be a relaxed $({\eta }_i,{\rho }_i)$-cocoercive and ${\beta }_i$-Lipschitz continuous mapping, and let $f_i:C\to H$ be a relaxed $({\overline{\eta }}_i,{\overline{\rho }}_i)$-cocoercive and ${\overline{\beta }}_i$-Lipschitz continuous mapping for each $i=1,2$. Suppose that $(x^{\ast },y^{\ast })\in C\times C$ is a solution to problem (1.6). For any $(x_0,y_0)\in C\times C$, the sequences $\{x_n\}$ and $\{y_n\}$ are generated by

$$\{\begin{array}{l}{t}_{n+1}=(1-a_n)x_n+a_n(x_n-g_1(x_n)+P_C(f_1(y_n)-\lambda T_1(y_n))),n\ge 0,\\ x_{n+1}=P_C{t}_{n+1},n\ge 0,\\ z_n=y_n-g_2(y_n)+P_C(f_2(x_n)-\mu T_2(x_n)),n\ge 1,\\ y_n=(1-b_n)x_n+b_n{P}_C{z}_n,n\ge 1,\end{array}$$

(3.19)

where $\lambda ,\mu >0$ and the following conditions are satisfied:

$$\begin{array}{r}\text{(1)}0<a_n,b_n\le 1;\text{(2)}\sum _{n=0}^{\mathrm{\infty }}{a}_n=\mathrm{\infty };\text{(3)}\sum _{n=1}^{\mathrm{\infty }}(1-b_n)<\mathrm{\infty };\\ \text{(4)}0\le {\theta }_2,{\theta }_5<1;\text{(5)}{\theta }_4+{\theta }_6\ge 1;\\ \text{(6)}0<({\theta }_1+{\theta }_3)({\theta }_4+{\theta }_6)<(1-{\theta }_2)(1-{\theta }_5);\end{array}$$

where

$$\begin{array}{r}{\theta }_1=\sqrt{1-2\lambda {\delta }_1+2\lambda {\gamma }_1{\alpha }_1^2+{\lambda }^2{\alpha }_1^2},{\theta }_2=\sqrt{1-2{\rho }_1+{\beta }_1^2+2{\eta }_1{\beta }_1^2},\\ {\theta }_3=\sqrt{1-2{\overline{\rho }}_1+{\overline{\beta }}_1^2+2{\overline{\eta }}_1{\overline{\beta }}_1^2},{\theta }_4=\sqrt{1-2\mu {\delta }_2+2\mu {\gamma }_2{\alpha }_2^2+{\mu }^2{\alpha }_2^2},\\ {\theta }_5=\sqrt{1-2{\rho }_2+{\beta }_2^2+2{\eta }_2{\beta }_2^2},{\theta }_6=\sqrt{1-2{\overline{\rho }}_2+{\overline{\beta }}_2^2+2{\overline{\eta }}_2{\overline{\beta }}_2^2}.\end{array}$$

Then the sequences $\{x_n\}$ and $\{y_n\}$ converge strongly to $x^{\ast }$ and $y^{\ast }$, respectively.

Corollary 3.5 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_i:C\to H$ be a relaxed $({\gamma }_i,{\delta }_i)$-cocoercive and ${\alpha }_i$-Lipschitz continuous mapping for each $i=1,2$. Suppose that $(x^{\ast },y^{\ast })\in C\times C$ is a solution to problem (1.7). For any $(x_0,y_0)\in C\times C$, the sequences $\{x_n\}$ and $\{y_n\}$ are generated by

$$\{\begin{array}{l}{x}_{n+1}=(1-a_n)x_n+a_n{P}_C(y_n-\lambda T_1(y_n)),n\ge 0,\\ y_n=(1-b_n)x_n+b_n{P}_C(x_n-\mu T_2(x_n)),n\ge 1,\end{array}$$

(3.20)

where $\lambda ,\mu >0$ and the following conditions are satisfied:

$$\begin{array}{r}\text{(1)}0<a_n,b_n\le 1;\text{(2)}\sum _{n=0}^{\mathrm{\infty }}{a}_n=\mathrm{\infty };\text{(3)}\sum _{n=1}^{\mathrm{\infty }}(1-b_n)<\mathrm{\infty };\\ \text{(4)}{\theta }_4\ge 1;\text{(5)}0<{\theta }_1{\theta }_4<1;\end{array}$$

where

$${\theta }_1=\sqrt{1-2\lambda {\delta }_1+2\lambda {\gamma }_1{\alpha }_1^2+{\lambda }^2{\alpha }_1^2},{\theta }_4=\sqrt{1-2\mu {\delta }_2+2\mu {\gamma }_2{\alpha }_2^2+{\mu }^2{\alpha }_2^2}.$$

Then the sequences $\{x_n\}$ and $\{y_n\}$ converge strongly to $x^{\ast }$ and $y^{\ast }$, respectively.

For $\mu =0$, $b_n=1$ in Corollary 3.3, we arrive at the following result.

Corollary 3.6 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_1:C\times C\to H$ be a relaxed $({\gamma }_1,{\delta }_1)$-cocoercive and ${\alpha }_1$-Lipschitz continuous mapping in the first variable. Suppose that $x^{\ast }\in C$ is a solution to problem (1.8). For any $x_0\in C$, the sequence $\{x_n\}$ is generated by

$$x_{n+1}=(1-a_n)x_n+a_n{P}_C(x_n-\lambda T_1(x_n,x_n)),n\ge 0,$$

(3.21)

where $\lambda >0$ and the following conditions are satisfied:

$$\text{(1)}0<a_n\le 1;\text{(2)}\sum _{n=0}^{\mathrm{\infty }}{a}_n=\mathrm{\infty };\text{(3)}0<{\theta }_1<1;$$

where

$${\theta }_1=\sqrt{1-2\lambda {\delta }_1+2\lambda {\gamma }_1{\alpha }_1^2+{\lambda }^2{\alpha }_1^2}.$$

Then the sequence $\{x_n\}$ converges strongly to $x^{\ast }$.