Iterative algorithms for a system of generalized variational inequalities in Hilbert spaces

Abstract

In this paper, a new system of generalized nonlinear variational inequalities involving three operators is introduced. A three-step iterative algorithm is considered for the system of generalized nonlinear variational inequalities. Strong convergence theorems of the three-step iterative algorithm are established.

MSC:47H05, 47H09, 47J25.

1 Introduction

Variational inequalities are among the most interesting and intensively studied classes of mathematical problems and have wide applications in the fields of optimization and control, economics, transportation equilibrium and engineering sciences. There exists a vast amount of literature (see, for instance, [1–26]) on the approximation solvability of nonlinear variational inequalities as well as operator equations.

Iterative algorithms have played a central role in the approximation solvability, especially of nonlinear variational inequalities as well as of nonlinear equations, in several fields such as applied mathematics, mathematical programming, mathematical finance, control theory and optimization, engineering sciences and others. Projection methods have played a significant role in the numerical resolution of variational inequalities based on their convergence analysis. However, the convergence analysis does require some sort of strong monotonicity besides the Lipschitz continuity. There have been some recent developments where convergence analysis for projection methods under somewhat weaker conditions such as cocoercivity [28] and partial relaxed monotonicity [24] is achieved.

Recently, Chang et al. [17] introduced a two-step iterative algorithm for a system of nonlinear variational inequalities and established strong convergence theorems. Huang and Noor [16] introduced the so-called explicit two-step iterative algorithm for a system of nonlinear variational inequalities involving two different nonlinear operators and established strong convergence theorems.

In this paper, we consider, based on the projection method, the approximate solvability of a new system of generalized nonlinear variational inequalities involving three different nonlinear operators in the framework of Hilbert spaces. The results presented in this paper extend and improve the corresponding results announced in Huang and Noor [16], Chang et al. [17], Verma [24–26] and many others.

Let H be a real Hilbert space, whose inner product and norm are denoted by $〈\cdot ,\cdot 〉$ and $\parallel \cdot \parallel$, respectively. Let C be a nonempty closed convex subset of H and $P_C$ be the metric projection from H onto C.

Given nonlinear operators $T,f:C\to H$ and $g:C\to C$, we consider the problem of finding $u\in C$ such that

$$〈g(u)-f(u)+\lambda Tu,v-g(u)〉\ge 0,\mathrm{\forall }v\in C,$$

(1.1)

where $\lambda >0$ is a constant. The variational inequality (1.1) is called the generalized variational inequality involving three operators.

We see that an element $u\in C$ is a solution to the generalized variational inequality (1.1) if and only if $u\in C$ is a fixed point of the mapping

$$I-g+P_C(f-\lambda T),$$

where I is the identity mapping. This equivalence plays an important role in this work.

If $f=g$, then the generalized variational inequality (1.1) is equivalent to the following.

Find $u\in C$ such that

$$〈Tu,v-g(u)〉\ge 0,\mathrm{\forall }v\in C.$$

(1.2)

Further, if $g=I$, then the problem (1.2) is reduced to finding $u\in C$ such that

$$〈Tu,v-u〉\ge 0,\mathrm{\forall }v\in C,$$

(1.3)

which is known as the classical variational inequality originally introduced and studied by Stampacchia [27].

Let $T:C\to H$ be a mapping. Recall the following definitions.

• (1) T is said to be monotone if

  $$〈Tu-Tv,u-v〉\ge 0,\mathrm{\forall }u,v\in C.$$

• (2) T is called 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.

• (3) 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 A is $\frac{1}{\gamma }$-Lipschitz continuous.

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

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

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

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

Let $T_i:C\times C\times C\to H$, $f_i:C\to H$ and $g_i:C\to C$ be nonlinear mappings for each $i=1,2,3$. Consider a system of generalized nonlinear variational inequality (SGNVI) as follows.

Find $(x^{\ast },y^{\ast },z^{\ast })\in C\times C\times C$ such that for all $s,t,r>0$,

$$\{\begin{array}{cc}〈s{T}_1(y^{\ast },z^{\ast },x^{\ast })+g_1(x^{\ast })-f_1(y^{\ast }),x-g_1(x^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈t{T}_2(z^{\ast },x^{\ast },y^{\ast })+g_2(y^{\ast })-f_2(z^{\ast }),x-g_2(x^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈r{T}_3(x^{\ast },y^{\ast },z^{\ast })+g_3(z^{\ast })-f_3(x^{\ast }),x-g_3(x^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C.\hfill \end{array}$$

(1.4)

One can easily see SGNVI (1.4) is equivalent to the following projection problem:

$$\{\begin{array}{cc}{g}_1(x^{\ast })=P_C(f_1(y^{\ast })-s{T}_1(y^{\ast },z^{\ast },x^{\ast })),\hfill & \mathrm{\forall }s>0,\hfill \\ g_2(y^{\ast })=P_C(f_2(z^{\ast })-t{T}_2(z^{\ast },x^{\ast },y^{\ast })),\hfill & \mathrm{\forall }t>0,\hfill \\ g_3(z^{\ast })=P_C(f_3(x^{\ast })-r{T}_3(x^{\ast },y^{\ast },z^{\ast })),\hfill & \mathrm{\forall }r>0.\hfill \end{array}$$

(1.5)

Next, we consider some special classes of SGNVI (1.4) as follows.

• (I) If $g_1=g_2=g_3=I$, then SGNVI (1.4) is reduced to the following.

Find $(x^{\ast },y^{\ast },z^{\ast })\in C\times C\times C$ such that for all $s,t,r>0$,

$$\{\begin{array}{cc}〈s{T}_1(y^{\ast },z^{\ast },x^{\ast })+x^{\ast }-f_1(y^{\ast }),x-x^{\ast }〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈t{T}_2(z^{\ast },x^{\ast },y^{\ast })+y^{\ast }-f_2(z^{\ast }),x-x^{\ast }〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈r{T}_3(x^{\ast },y^{\ast },z^{\ast })+z^{\ast }-f_3(x^{\ast }),x-x^{\ast }〉\ge 0,\hfill & \mathrm{\forall }x\in C.\hfill \end{array}$$

(1.6)

We see that the problem (1.6) is equivalent to the following projection problem:

$$\{\begin{array}{cc}{x}^{\ast }=P_C(f_1(y^{\ast })-s{T}_1(y^{\ast },z^{\ast },x^{\ast })),\hfill & \mathrm{\forall }s>0,\hfill \\ y^{\ast }=P_C(f_2(z^{\ast })-t{T}_2(z^{\ast },x^{\ast },y^{\ast })),\hfill & \mathrm{\forall }t>0,\hfill \\ z^{\ast }=P_C(f_3(x^{\ast })-r{T}_3(x^{\ast },y^{\ast },z^{\ast })),\hfill & \mathrm{\forall }r>0.\hfill \end{array}$$

(1.7)

• (II) If $f_1=f_2=f_3=I$, then SGNVI (1.4) is reduced to the following.

Find $(x^{\ast },y^{\ast },z^{\ast })\in C\times C\times C$ such that for all $s,t,r>0$,

$$\{\begin{array}{cc}〈s{T}_1(y^{\ast },z^{\ast },x^{\ast })+g_1(x^{\ast })-y^{\ast },x-g_1(x^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈t{T}_2(z^{\ast },x^{\ast },y^{\ast })+g_2(y^{\ast })-z^{\ast },x-g_2(x^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈r{T}_3(x^{\ast },y^{\ast },z^{\ast })+g_3(z^{\ast })-x^{\ast },x-g_3(x^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C.\hfill \end{array}$$

(1.8)

We see that the problem (1.8) is equivalent to the following projection problem:

$$\{\begin{array}{cc}{g}_1(x^{\ast })=P_C[y^{\ast }-s{T}_1(y^{\ast },z^{\ast },x^{\ast })],\hfill & \mathrm{\forall }s>0,\hfill \\ g_2(y^{\ast })=P_C[z^{\ast }-t{T}_2(z^{\ast },x^{\ast },y^{\ast })],\hfill & \mathrm{\forall }t>0,\hfill \\ g_3(z^{\ast })=P_C[x^{\ast }-r{T}_3(x^{\ast },y^{\ast },z^{\ast })],\hfill & \mathrm{\forall }r>0.\hfill \end{array}$$

(1.9)

• (III) If $g_1=g_2=g_3=f_1=f_2=f_3=I$, then SGNVI (1.4) is reduced to the following.

Find $(x^{\ast },y^{\ast },z^{\ast })\in C\times C\times C$ such that for all $s,t,r>0$,

$$\{\begin{array}{cc}〈s{T}_1(y^{\ast },z^{\ast },x^{\ast })+x^{\ast }-y^{\ast },x-x^{\ast }〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈t{T}_2(z^{\ast },x^{\ast },y^{\ast })+y^{\ast }-z^{\ast },x-x^{\ast }〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈r{T}_3(x^{\ast },y^{\ast },z^{\ast })+z^{\ast }-x^{\ast },x-x^{\ast }〉\ge 0,\hfill & \mathrm{\forall }x\in C.\hfill \end{array}$$

(1.10)

One can easily get that the problem (1.10) is equivalent to the following projection problem:

$$\{\begin{array}{cc}{x}^{\ast }=P_C(y^{\ast }-s{T}_1(y^{\ast },z^{\ast },x^{\ast })),\hfill & \mathrm{\forall }s>0,\hfill \\ y^{\ast }=P_C(z^{\ast }-t{T}_2(z^{\ast },x^{\ast },y^{\ast })),\hfill & \mathrm{\forall }t>0,\hfill \\ z^{\ast }=P_C(x^{\ast }-r{T}_3(x^{\ast },y^{\ast },z^{\ast })),\hfill & \mathrm{\forall }r>0.\hfill \end{array}$$

(1.11)

• (IV) If $T_1$, $T_2$ and $T_3$ are univariate mappings, then SGNVI (1.4) is reduced to the following.

Find $(x^{\ast },y^{\ast },z^{\ast })\in C\times C\times C$ such that for all $s,t,r>0$,

$$\{\begin{array}{cc}〈s{T}_1{y}^{\ast }+g_1(x^{\ast })-f_1(y^{\ast }),x-g_1(x^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈t{T}_2{z}^{\ast }+g_2(y^{\ast })-f_2(z^{\ast }),x-g_2(x^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈r{T}_3{x}^{\ast }+g_3(z^{\ast })-f_3(x^{\ast }),x-g_3(x^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C.\hfill \end{array}$$

(1.12)

One can easily see that the problem (1.12) is equivalent to the following projection problem:

$$\{\begin{array}{cc}{g}_1(x^{\ast })=P_C(f_1(y^{\ast })-s{T}_1{y}^{\ast }),\hfill & \mathrm{\forall }s>0,\hfill \\ g_2(y^{\ast })=P_C(f_2(z^{\ast })-t{T}_2{z}^{\ast }),\hfill & \mathrm{\forall }t>0,\hfill \\ g_3(z^{\ast })=P_C(f_3(x^{\ast })-r{T}_3{x}^{\ast }),\hfill & \mathrm{\forall }r>0.\hfill \end{array}$$

(1.13)

2 Preliminaries

In this section, to study the approximate solvability of the problems (1.4), (1.6), (1.8), (1.10) and (1.12), we introduce the following three-step algorithms.

Algorithm 2.1 For any $(x_0,y_0,z_0)\in C\times C\times C$, compute the sequences $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ by the following iterative process:

$$\{\begin{array}{cc}{x}_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n(x_n-g_1(x_n)+P_C(f_1(y_n)-s{T}_1(y_n,z_n,x_n))),\hfill & n\ge 0,\hfill \\ g_3(z_{n+1})=P_C(f_3(x_{n+1})-r{T}_3(x_{n+1},y_n,z_n)),\hfill & n\ge 0,\hfill \\ g_2(y_{n+1})=P_C(f_2(z_{n+1})-t{T}_2(z_{n+1},x_n,y_n)),\hfill & n\ge 0,\hfill \end{array}$$

where $r,s,t>0$ are three constants and $\{{\alpha }_n\}$ is a sequence in $[0,1]$.

If $g_1=g_2=g_3=I$, then Algorithm 2.1 is reduced to the following.

Algorithm 2.2 For any $(x_0,y_0,z_0)\in C\times C\times C$, compute the sequences $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ by the following iterative process:

$$\{\begin{array}{cc}{x}_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n{P}_C(f_1(y_n)-s{T}_1(y_n,z_n,x_n)),\hfill & n\ge 0,\hfill \\ z_{n+1}=P_C(f_3(x_{n+1})-r{T}_3(x_{n+1},y_n,z_n)),\hfill & n\ge 0,\hfill \\ y_{n+1}=P_C(f_2(z_{n+1})-t{T}_2(z_{n+1},x_n,y_n)),\hfill & n\ge 0,\hfill \end{array}$$

where $r,s,t>0$ are three constants and $\{{\alpha }_n\}$ is a sequence in $[0,1]$.

If $f_1=f_2=f_3=I$, the identity mapping, then Algorithm 2.1 is reduced to the following.

Algorithm 2.3 For any $(x_0,y_0,z_0)\in C\times C\times C$, compute the sequences $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ by the following iterative process:

$$\{\begin{array}{cc}{x}_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n(x_n-g_1(x_n)+P_C(y_n-s{T}_1(y_n,z_n,x_n))),\hfill & n\ge 0,\hfill \\ g_3(z_{n+1})=P_C(x_{n+1}-r{T}_3(x_{n+1},y_n,z_n)),\hfill & n\ge 0,\hfill \\ g_2(y_{n+1})=P_C(z_{n+1}-t{T}_2(z_{n+1},x_n,y_n)),\hfill & n\ge 0,\hfill \end{array}$$

where $r,s,t>0$ are three constants and $\{{\alpha }_n\}$ is a sequence in $[0,1]$.

If $f_1=f_2=f_3=g_1=g_2=g_3=I$, the identity mapping, then Algorithm 2.1 is reduced to the following.

Algorithm 2.4 For any $(x_0,y_0,z_0)\in C\times C\times C$, compute the sequences $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ by the following iterative process:

$$\{\begin{array}{cc}{x}_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n{P}_C(y_n-s{T}_1(y_n,z_n,x_n)),\hfill & n\ge 0,\hfill \\ z_{n+1}=P_C(x_{n+1}-r{T}_3(x_{n+1},y_n,z_n)),\hfill & n\ge 0,\hfill \\ y_{n+1}=P_C(z_{n+1}-t{T}_2(z_{n+1},x_n,y_n)),\hfill & n\ge 0,\hfill \end{array}$$

where $r,s,t>0$ are three constants and $\{{\alpha }_n\}$ is a sequence in $[0,1]$.

• (IV) If $T_1$, $T_2$ and $T_3$ are univariate mappings, then Algorithm 2.1 is reduced to the following.

Algorithm 2.5 For any $(x_0,y_0,z_0)\in C\times C\times C$, compute the sequences $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ by the following iterative process:

$$\{\begin{array}{cc}{x}_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n(x_n-g_1(x_n)+P_C(f_1(y_n)-s{T}_1{y}_n)),\hfill & n\ge 0,\hfill \\ g_3(z_{n+1})=P_C(f_3(x_{n+1})-r{T}_3{x}_{n+1}),\hfill & n\ge 0,\hfill \\ g_2(y_{n+1})=P_C(f_2(z_{n+1})-t{T}_2{z}_{n+1}),\hfill & n\ge 0,\hfill \end{array}$$

where $r,s,t>0$ are three constants and $\{{\alpha }_n\}$ is a sequence in $[0,1]$.

In order to prove our main results, we also need the following lemma and definitions.

Lemma 2.6 [29]

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

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

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

Definition 2.7 A mapping $T:C\times C\times C\to H$ is said to be relaxed $(\gamma ,\delta )$-cocoercive if there exist constants $\gamma ,\delta >0$ such that for all $x,x^{\prime }\in C$,

Definition 2.8 A mapping $T:C\times C\times C\to H$ is said to be β-Lipschitz continuous in the first variable if there exists a constant $\beta >0$ such that for all $x,x^{\prime }\in C$,

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

3 Main results

Theorem 3.1 Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let $T_i:C\times C\times C\to H$ be a relaxed $({\gamma }_i,{\delta }_i)$-cocoercive and ${\beta }_i$-Lipschitz continuous mapping in the first variable, $f_i:C\to H$ be a relaxed $({\eta }_i,{\rho }_i)$-cocoercive and ${\lambda }_i$-Lipschitz continuous mapping and $g_i:C\to C$ be a relaxed $({\overline{\eta }}_i,{\overline{\rho }}_i)$-cocoercive and ${\overline{\lambda }}_i$-Lipschitz continuous mapping for each $i=1,2,3$. Suppose that $(x^{\ast },y^{\ast },z^{\ast })\in C\times C\times C$ is a solution to the problem (1.4). Let $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ be the sequences generated by Algorithm 2.1. Assume that the following conditions are satisfied:

• (a) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (b) $0\le {\theta }_6$, ${\theta }_9<1$;

• (c) $({\theta }_1+{\theta }_2)({\theta }_4+{\theta }_5)({\theta }_7+{\theta }_8)\le (1-{\theta }_3)(1-{\theta }_6)(1-{\theta }_9)$,

where

and

$${\theta }_9=\sqrt{1-2{\overline{\rho }}_3+{\overline{\lambda }}_3^2+2{\overline{\eta }}_3{\overline{\lambda }}_3^2}.$$

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

Proof In view of $(x^{\ast },y^{\ast },z^{\ast })$ being a solution to the problem (1.4), we see that

$$\{\begin{array}{cc}{x}^{\ast }=(1-{\alpha }_n)x^{\ast }+{\alpha }_n(x^{\ast }-g_1(x^{\ast })+P_C(f_1(y^{\ast })-s{T}_1(y^{\ast },z^{\ast },x^{\ast }))),\hfill & n\ge 0,\hfill \\ g_3(z^{\ast })=P_C(f_3(x^{\ast })-r{T}_3(x^{\ast },y^{\ast },z^{\ast })),\hfill & n\ge 0,\hfill \\ g_2(y^{\ast })=P_C(f_2(z^{\ast })-t{T}_2(z^{\ast },x^{\ast },y^{\ast })),\hfill & n\ge 0.\hfill \end{array}$$

It follows from Algorithm (2.1) that

(3.1)

By the assumption that $T_1$ is relaxed $({\gamma }_1,r_1)$-cocoercive and ${\beta }_1$-Lipschitz continuous in the first variable, we obtain that

(3.2)

where ${\theta }_1=\sqrt{1-2s{\delta }_1+2s{\gamma }_1{\beta }_1^2+s^2{\beta }_1^2}$. On the other hand, it follows from the assumption that $f_1$ is relaxed $({\eta }_1,{\rho }_1)$-cocoercive and ${\lambda }_1$-Lipschitz continuous that

(3.3)

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

$$\parallel x_n-x^{\ast }-(g_1(x_n)-g_1\left(x^{\ast }\right))\parallel \le {\theta }_3\parallel x_n-x^{\ast }\parallel ,$$

(3.4)

where ${\theta }_3=\sqrt{1-2{\overline{\rho }}_1+{\overline{\lambda }}_1^2+2{\overline{\eta }}_1{\overline{\lambda }}_1^2}$. Substituting (3.2), (3.3) and (3.4) into (3.1), we arrive at

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

(3.5)

Next, we estimate $\parallel y_n-y^{\ast }\parallel$. From Algorithm 2.1, we see that

$$\begin{array}{rcl}\parallel g_2(y_{n+1})-g_2\left(y^{\ast }\right)\parallel & =& \parallel P_C(f_2(z_{n+1})-t{T}_2(z_{n+1},x_n,y_n))\\ -P_C(f_2\left(z^{\ast }\right)-t{T}_2(z^{\ast },x^{\ast },y^{\ast }))\parallel \\ \le & \parallel (f_2(z_{n+1})-t{T}_2(z_{n+1},x_n,y_n))-(f_2\left(z^{\ast }\right)-t{T}_2(z^{\ast },x^{\ast },y^{\ast }))\parallel \\ \le & \parallel (z_{n+1}-z^{\ast })-(f_2(z_{n+1})-f_2\left(z^{\ast }\right))\parallel \\ +\parallel (z_{n+1}-z^{\ast })-t(T_2(z_{n+1},x_n,y_n)-T_2(z^{\ast },x^{\ast },y^{\ast }))\parallel .\end{array}$$

(3.6)

By the assumption that $T_2$ is relaxed $({\gamma }_2,r_2)$-cocoercive and ${\beta }_2$-Lipschitz continuous in the first variable, we obtain that

(3.7)

where ${\theta }_4=\sqrt{1-2t{\delta }_2+2t{\gamma }_2{\beta }_2^2+t^2{\beta }_2^2}$. It follows from the assumption that $f_2$ is relaxed $({\eta }_2,{\rho }_2)$-cocoercive and ${\lambda }_2$-Lipschitz continuous that

(3.8)

where ${\theta }_5=\sqrt{1-2{\rho }_2+{\lambda }_2^2+2{\eta }_2{\lambda }_2^2}$. Substituting (3.7) and (3.8) into (3.6), we see that

$$\parallel g_2(y_{n+1})-g_2\left(y^{\ast }\right)\parallel \le ({\theta }_4+{\theta }_5)\parallel z_{n+1}-z^{\ast }\parallel .$$

(3.9)

On the other hand, we have

$$\parallel y_{n+1}-y^{\ast }\parallel \le \parallel y_{n+1}-y^{\ast }-(g_2(y_{n+1})-g_2\left(y^{\ast }\right))\parallel +\parallel g_2(y_{n+1})-g_2\left(y^{\ast }\right)\parallel .$$

(3.10)

From the proof of (3.8), we arrive at

$$\parallel (y_{n+1}-y^{\ast })-(g_2(y_{n+1})-g_2\left(y^{\ast }\right))\parallel \le {\theta }_6\parallel y_{n+1}-y^{\ast }\parallel ,$$

(3.11)

where ${\theta }_6=\sqrt{1-2{\overline{\rho }}_2+{\overline{\lambda }}_2^2+2{\overline{\eta }}_2{\overline{\lambda }}_2^2}$. Substituting (3.9) and (3.11) into (3.10), we see that

$$\parallel y_{n+1}-y^{\ast }\parallel \le {\theta }_6\parallel y_{n+1}-y^{\ast }\parallel +({\theta }_4+{\theta }_5)\parallel z_{n+1}-z^{\ast }\parallel .$$

It follows from the condition (b) that

$$\parallel y_{n+1}-y^{\ast }\parallel \le \frac{{\theta }_4+{\theta }_5}{1-{\theta }_6}\parallel z_{n+1}-z^{\ast }\parallel .$$

That is,

$$\parallel y_n-y^{\ast }\parallel \le \frac{{\theta }_4+{\theta }_5}{1-{\theta }_6}\parallel z_n-z^{\ast }\parallel .$$

(3.12)

Finally, we estimate $\parallel z_n-z^{\ast }\parallel$. It follows from Algorithm 2.1 that

(3.13)

In a similar way, we can show that

$$\parallel (x_{n+1}-x^{\ast })-r(T_3(x_{n+1},y_n,z_n)-T_3(x^{\ast },y^{\ast },z^{\ast }))\parallel \le {\theta }_7\parallel x_{n+1}-x^{\ast }\parallel$$

(3.14)

and

$$\parallel (x_{n+1}-x^{\ast })-(f_3(x_{n+1})-f_3\left(x^{\ast }\right))\parallel \le {\theta }_8\parallel x_{n+1}-x^{\ast }\parallel ,$$

(3.15)

where ${\theta }_7=\sqrt{1-2r{\delta }_3+2r{\gamma }_3{\beta }_3^2+r^2{\beta }_3^2}$ and ${\theta }_8=\sqrt{1-2{\rho }_3+{\lambda }_3^2+2{\eta }_3{\lambda }_3^2}$. Substituting (3.14) and (3.15) into (3.13), we arrive at

$$\parallel g_3(z_{n+1})-g_3\left(z^{\ast }\right)\parallel \le ({\theta }_7+{\theta }_8)\parallel x_{n+1}-x^{\ast }\parallel .$$

(3.16)

Note that

$$\parallel z_{n+1}-z^{\ast }\parallel \le \parallel z_{n+1}-z^{\ast }-(g_3(z_{n+1})-g_3\left(z^{\ast }\right))\parallel +\parallel g_3(z_{n+1})-g_3\left(z^{\ast }\right)\parallel .$$

(3.17)

On the other hand, we have

$$\parallel z_{n+1}-z^{\ast }-(g_3(z_{n+1})-g_3\left(z^{\ast }\right))\parallel \le {\theta }_9\parallel z_{n+1}-z^{\ast }\parallel ,$$

(3.18)

${\theta }_9=\sqrt{1-2{\overline{\rho }}_3+{\overline{\lambda }}_3^2+2{\overline{\eta }}_3{\overline{\lambda }}_3^2}$. Substituting (3.16) and (3.18) into (3.17), we arrive at

$$\parallel z_{n+1}-z^{\ast }\parallel \le {\theta }_9\parallel z_{n+1}-z^{\ast }\parallel +({\theta }_7+{\theta }_8)\parallel x_{n+1}-x^{\ast }\parallel .$$

It follows from the condition (b) that

$$\parallel z_{n+1}-z^{\ast }\parallel \le \frac{{\theta }_7+{\theta }_8}{1-{\theta }_9}\parallel x_{n+1}-x^{\ast }\parallel .$$

That is,

$$\parallel z_n-z^{\ast }\parallel \le \frac{{\theta }_7+{\theta }_8}{1-{\theta }_9}\parallel x_n-x^{\ast }\parallel .$$

(3.19)

Combining (3.5), (3.12) with (3.19), we obtain that

$$\parallel x_{n+1}-x^{\ast }\parallel \le (1-{\alpha }_n(1-{\theta }_3-({\theta }_1+{\theta }_2)\frac{{\theta }_4+{\theta }_5}{1-{\theta }_6}\frac{{\theta }_7+{\theta }_8}{1-{\theta }_9}))\parallel x_n-x^{\ast }\parallel .$$

Since ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ and the condition (c), we can conclude the desired conclusion easily from Lemma 2.6. This completes the proof. □

Remark 3.2 Theorem 3.1 includes the corresponding results in Huang and Noor [16] Chang et al. [17], and Verma [24–26] as special cases.

From Theorem 3.1, we can get the following results immediately.

Corollary 3.3 Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let $T_i:C\times C\times C\to H$ be a relaxed $({\gamma }_i,{\delta }_i)$-cocoercive and ${\beta }_i$-Lipschitz continuous mapping in the first variable and $f_i:C\to H$ be a relaxed $({\eta }_i,{\rho }_i)$-cocoercive and ${\lambda }_i$-Lipschitz continuous mapping for each $i=1,2,3$. Suppose that $(x^{\ast },y^{\ast },z^{\ast })\in C\times C\times C$ is a solution to the problem (1.6). Let $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ be the sequences generated by Algorithm 2.2. Assume that the following conditions are satisfied:

• (a) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (b) $({\theta }_1+{\theta }_2)({\theta }_4+{\theta }_5)({\theta }_7+{\theta }_8)\le 1$,

where

and

$${\theta }_7=\sqrt{1-2r{\delta }_3+2r{\gamma }_3{\beta }_3^2+r^2{\beta }_3^2},{\theta }_8=\sqrt{1-2{\rho }_3+{\lambda }_3^2+2{\eta }_3{\lambda }_3^2}.$$

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

Corollary 3.4 Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let $T_i:C\times C\times C\to H$ be a relaxed $({\gamma }_i,{\delta }_i)$-cocoercive and ${\beta }_i$-Lipschitz continuous mapping in the first variable and $g_i:C\to C$ be a relaxed $({\overline{\eta }}_i,{\overline{\rho }}_i)$-cocoercive and ${\overline{\lambda }}_i$-Lipschitz continuous mapping for each $i=1,2,3$. Suppose that $(x^{\ast },y^{\ast },z^{\ast })\in C\times C\times C$ is a solution to the problem (1.8). Let $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ be the sequences generated by Algorithm 2.3. Assume that the following conditions are satisfied:

• (a) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (b) $0\le {\theta }_6$, ${\theta }_9<1$;

• (c) ${\theta }_1{\theta }_4{\theta }_7\le (1-{\theta }_3)(1-{\theta }_6)(1-{\theta }_9)$,

where

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

and

$${\theta }_7=\sqrt{1-2r{\delta }_3+2r{\gamma }_3{\beta }_3^2+r^2{\beta }_3^2},{\theta }_9=\sqrt{1-2{\overline{\rho }}_3+{\overline{\lambda }}_3^2+2{\overline{\eta }}_3{\overline{\lambda }}_3^2}.$$

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

Corollary 3.5 Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let $T_i:C\times C\times C\to H$ be a relaxed $({\gamma }_i,{\delta }_i)$-cocoercive and ${\beta }_i$-Lipschitz continuous mapping in the first variable for each $i=1,2,3$. Suppose that $(x^{\ast },y^{\ast },z^{\ast })\in C\times C\times C$ is a solution to the problem (1.10). Let $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ be the sequences generated by Algorithm 2.4. Assume that the following conditions are satisfied:

• (a) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (b) ${\theta }_1{\theta }_4{\theta }_7\le 1$,

where

$${\theta }_1=\sqrt{1-2s{\delta }_1+2s{\gamma }_1{\beta }_1^2+s^2{\beta }_1^2},{\theta }_4=\sqrt{1-2t{\delta }_2+2t{\gamma }_2{\beta }_2^2+t^2{\beta }_2^2},$$

and

$${\theta }_7=\sqrt{1-2r{\delta }_3+2r{\gamma }_3{\beta }_3^2+r^2{\beta }_3^2}.$$

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

Corollary 3.6 Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let $T_i:C\to H$ be a relaxed $({\gamma }_i,{\delta }_i)$-cocoercive and ${\beta }_i$-Lipschitz continuous mapping, $f_i:C\to H$ be a relaxed $({\eta }_i,{\rho }_i)$-cocoercive and ${\lambda }_i$-Lipschitz continuous mapping and $g_i:C\to C$ be a relaxed $({\overline{\eta }}_i,{\overline{\rho }}_i)$-cocoercive and ${\overline{\lambda }}_i$-Lipschitz continuous mapping for each $i=1,2,3$. Suppose that $(x^{\ast },y^{\ast },z^{\ast })\in C\times C\times C$ is a solution to the problem (1.12). Let $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ be the sequences generated by Algorithm 2.5. Assume that the following conditions are satisfied:

• (a) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (b) $0\le {\theta }_6$, ${\theta }_9<1$;

• (c) $({\theta }_1+{\theta }_2)({\theta }_4+{\theta }_5)({\theta }_7+{\theta }_8)\le (1-{\theta }_3)(1-{\theta }_6)(1-{\theta }_9)$,

where

and

$${\theta }_9=\sqrt{1-2{\overline{\rho }}_3+{\overline{\lambda }}_3^2+2{\overline{\eta }}_3{\overline{\lambda }}_3^2}.$$

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