A new iterative algorithm for solving common solutions of generalized mixed equilibrium problems, fixed point problems and variational inclusion problems with minimization problems

Abstract

In this article, we introduce a new general iterative method for solving a common element of the set of solutions of fixed point for nonexpansive mappings, the set of solutions of generalized mixed equilibrium problems and the set of solutions of the variational inclusion for a β-inverse-strongly monotone mapping in a real Hilbert space. We prove that the sequence converges strongly to a common element of the above three sets under some mild conditions. Our results improve and extend the corresponding results of Marino and Xu (J. Math. Anal. Appl. 318:43-52, 2006), Su et al. (Nonlinear Anal. 69:2709-2719, 2008), Tan and Chang (Fixed Point Theory Appl. 2011:915629, 2011) and some authors.

MSC:46C05, 47H09, 47H10.

1 Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, respectively. A mapping $S:C\to C$ is said to be nonexpansive if $\parallel Sx-Sy\parallel \le \parallel x-y\parallel$$\mathrm{\forall }x,y\in C$. If C is bounded closed convex and S is a nonexpansive mapping of C into itself, then $F(S):=\{x\in C:Sx=x\}$ is nonempty [1]. A mapping $S:C\to C$ is said to be a k-strictly pseudo-contraction[2] if there exists $0\le k<1$ such that ${\parallel Sx-Sy\parallel }^2\le {\parallel x-y\parallel }^2+k{\parallel (I-S)x-(I-S)y\parallel }^2$$\mathrm{\forall }x,y\in C$, where I denotes the identity operator on C. We denote weak convergence and strong convergence by notations ⇀ and →, respectively. A mapping A of C into H is called monotone if $〈Ax-Ay,x-y〉\ge 0$$\mathrm{\forall }x,y\in C$. A mapping A is called α-inverse-strongly monotone if there exists a positive real number α such that $〈Ax-Ay,x-y〉\ge \alpha {\parallel Ax-Ay\parallel }^2$$\mathrm{\forall }x,y\in C$. A mapping A is called α-strongly monotone if there exists a positive real number α such that $〈Ax-Ay,x-y〉\ge \alpha {\parallel x-y\parallel }^2$$\mathrm{\forall }x,y\in C$. It is obvious that any α-inverse-strongly monotone mappings A is a monotone and $\frac{1}{\alpha }$-Lipschitz continuous mapping. A linear bounded operator A is called strongly positive if there exists a constant $\overline{\gamma }>0$ with the property $〈Ax,x〉\ge \overline{\gamma }{\parallel x\parallel }^2$$\mathrm{\forall }x\in H$. A self mapping $f:C\to C$ is called contraction on C if there exists a constant $\alpha \in (0,1)$ such that $\parallel f(x)-f(y)\parallel \le \alpha \parallel x-y\parallel$$\mathrm{\forall }x,y\in C$.

Let $B:H\to H$ be a single-valued nonlinear mapping and $M:H\to 2^H$ be a set-valued mapping. The variational inclusion problem is to find $x\in H$ such that

$$\theta \in B(x)+M(x),$$

(1.1)

where θ is the zero vector in H. The set of solutions of (1.1) is denoted by $I(B,M)$. The variational inclusion has been extensively studied in the literature. See, e.g.[3–10] and the reference therein.

A set-valued mapping $M:H\to 2^H$ is called monotone if $\mathrm{\forall }x,y\in H$, $f\in M(x)$ and $g\in M(y)$ imply $〈x-y,f-g〉\ge 0$. A monotone mapping M is maximal if its graph $G(M):=\{(f,x)\in H\times H:f\in M(x)\}$ of M is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping M is maximal if and only if for $(x,f)\in H\times H$, $〈x-y,f-g〉\ge 0$ for all $(y,g)\in G(M)$ imply $f\in M(x)$.

Let B be an inverse-strongly monotone mapping of C into H and let $N_{C}v$ be normal cone to C at $v\in C$i.e.$N_{C}v=\{w\in H:〈v-u,w〉\ge 0,\mathrm{\forall }u\in C\}$, and define

$$Mv=\{\begin{array}{cc}Bv+N_{C}v,\hfill & \text{if}v\in C,\hfill \\ \mathrm{\varnothing },\hfill & \text{if}v\notin C.\hfill \end{array}$$

Then M is a maximal monotone and $\theta \in Mv$ if and only if $v\in VI(C,B)$ (see [11]).

Let $M:H\to 2^H$ be a set-valued maximal monotone mapping, then the single-valued mapping $J_{M,\lambda }:H\to H$ defined by

$$J_{M,\lambda }(x)={(I+\lambda M)}^{-1}(x),x\in H$$

(1.2)

is called the resolvent operator associated with M, where λ is any positive number and I is the identity mapping. In the worth mentioning that the resolvent operator is nonexpansive, 1-inverse-strongly monotone and that a solution of problem (1.1) is a fixed point of the operator $J_{M,\lambda }(I-\lambda B)$ for all $\lambda >0$ (see [12]).

Let F be a bifunction of $C\times C$ into $\mathcal{R}$, where $\mathcal{R}$ is the set of real numbers, $\mathrm{\Psi }:C\to H$ be a mapping and $\psi :C\to \mathcal{R}$ be a real-valued function. The generalized mixed equilibrium problem for finding $x\in C$ such that

$$F(x,y)+〈\mathrm{\Psi }x,y-x〉+\psi (y)-\psi (x)\ge 0,\mathrm{\forall }y\in C.$$

(1.3)

The set of solutions of (1.3) is denoted by $GMEP(F,\psi ,\mathrm{\Psi })$, that is

$$GMEP(F,\psi ,\mathrm{\Psi })=\{x\in C:F(x,y)+〈\mathrm{\Psi }x,y-x〉+\psi (y)-\psi (x)\ge 0,\mathrm{\forall }y\in C\}.$$

If $\mathrm{\Psi }\equiv 0$ and $\psi \equiv 0$, the problem (1.3) is reduced into the equilibrium problem (see also [13]) for finding $x\in C$ such that

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

(1.4)

The set of solutions of (1.4) is denoted by $EP(F)$, that is

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

This problem contains fixed point problems, includes as special cases numerous problems in physics, optimization and economics. Some methods have been proposed to solve the equilibrium problem, please consult [14–16].

If $F\equiv 0$ and $\psi \equiv 0$, the problem (1.3) is reduced into the Hartmann-Stampacchia variational inequality[17] for finding $x\in C$ such that

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

(1.5)

The set of solutions of (1.5) is denoted by $VI(C,\mathrm{\Psi })$. The variational inequality has been extensively studied in the literature [18].

If $F\equiv 0$ and $\mathrm{\Psi }\equiv 0$, the problem (1.3) is reduced into the minimize problem for finding $x\in C$ such that

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

(1.6)

The set of solutions of (1.6) is denoted by $Argmin(\psi )$. Iterative methods for nonexpansive mappings have recently been applied to solve convex minimization problems. Convex minimization problems have a great impact and influence in the development of almost all branches of pure and applied sciences. A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbert space H:

$$\theta (x)=\frac{1}{2}〈Ax,x〉-〈x,y〉,\mathrm{\forall }x\in F(S),$$

(1.7)

where A is a linear bounded operator, $F(S)$ is the fixed point set of a nonexpansive mapping S and y is a given point in H[19].

In 2000, Moudafi [20] introduced the viscosity approximation method for nonexpansive mapping and prove that if H is a real Hilbert space, the sequence $\{x_n\}$ defined by the iterative method below, with the initial guess $x_0\in C$ is chosen arbitrarily,

$$x_{n+1}={\alpha }_{n}f(x_n)+(1-{\alpha }_n)S{x}_n,n\ge 0,$$

(1.8)

where $\{{\alpha }_n\}\subset (0,1)$ satisfies certain conditions, converge strongly to a fixed point of S (say $\overline{x}\in C$) which is the unique solution of the following variational inequality:

$$〈(I-f)\overline{x},x-\overline{x}〉\ge 0,\mathrm{\forall }x\in F(S).$$

(1.9)

In 2005, Iiduka and Takahashi [21] introduced following iterative process $x_0\in C$

$$x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)S{P}_C(x_n-{\lambda }_{n}A{x}_n),\mathrm{\forall }n\ge 0,$$

(1.10)

where $u\in C$$\{{\alpha }_n\}\subset (0,1)$ and $\{{\lambda }_n\}\subset [a,b]$ for some a b with $0<a<b<2\beta$. They proved that under certain appropriate conditions imposed on $\{{\alpha }_n\}$ and $\{{\lambda }_n\}$, the sequence $\{x_n\}$ generated by (1.10) converges strongly to a common element of the set of fixed points of nonexpansive mapping and the set of solutions of the variational inequality for an inverse-strongly monotone mapping (say $\overline{x}\in C$) which solve some variational inequality

$$〈\overline{x}-u,x-\overline{x}〉\ge 0,\mathrm{\forall }x\in F(S).$$

(1.11)

In 2006, Marino and Xu [19] introduced a general iterative method for nonexpansive mapping. They defined the sequence $\{x_n\}$ generated by the algorithm $x_0\in C$

$$x_{n+1}={\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)S{x}_n,n\ge 0,$$

(1.12)

where $\{{\alpha }_n\}\subset (0,1)$ and A is a strongly positive linear bounded operator. They proved that if $C=H$ and the sequence $\{{\alpha }_n\}$ satisfies appropriate conditions, then the sequence $\{x_n\}$ generated by (1.12) converge strongly to a fixed point of S (say $\overline{x}\in H$) which is the unique solution of the following variational inequality:

$$〈(A-\gamma f)\overline{x},x-\overline{x}〉\ge 0,\mathrm{\forall }x\in F(S),$$

(1.13)

which is the optimality condition for the minimization problem

$$\underset{x\in F(S)}{min}\frac{1}{2}〈Ax,x〉-h(x),$$

(1.14)

where h is a potential function for γf (i.e.$h^{\prime }(x)=\gamma f(x)$ for $x\in H$).

In 2008, Su et al.[22] introduced the following iterative scheme by the viscosity approximation method in a real Hilbert space: $x_1,u_n\in H$

$$\{\begin{array}{c}F(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\alpha }_{n}f(x_n)+(1-{\alpha }_n)S{P}_C(u_n-{\lambda }_{n}A{u}_n),\hfill \end{array}$$

(1.15)

for all $n\in \mathbb{N}$, where $\{{\alpha }_n\}\subset [0,1)$ and $\{r_n\}\subset (0,\mathrm{\infty })$ satisfy some appropriate conditions. Furthermore, they proved $\{x_n\}$ and $\{u_n\}$ converge strongly to the same point $z\in F(S)\cap VI(C,A)\cap EP(F)$ where $z=P_{F(S)\cap VI(C,A)\cap EP(F)}f(z)$.

In 2011, Tan and Chang [10] introduced following iterative process for $\{T_n:C\to C\}$ is a sequence of nonexpansive mappings. Let $\{x_n\}$ be the sequence defined by

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)\left(S{P}_C((1-t_n)J_{M,\lambda }(I-\lambda A)T_n(I-\mu B))x_n\right),\mathrm{\forall }n\ge 0,$$

(1.16)

where $\{{\alpha }_n\}\subset (0,1)$$\lambda \in (0,2\alpha ]$ and $\mu \in (0,2\beta ]$. The sequence $\{x_n\}$ defined by (1.16) converges strongly to a common element of the set of fixed points of nonexpansive mappings, the set of solutions of the variational inequality and the generalized equilibrium problem.

In this article, we mixed and modified the iterative methods (1.12), (1.15) and (1.16) by purposing the following new general viscosity iterative method: $x_0,u_n\in C$ and

$$\{\begin{array}{c}{u}_n=T_{r_n}^{(F_1,{\psi }_1)}(x_n-r_n{B}_1{x}_n),\hfill \\ v_n=T_{s_n}^{(F_2,{\psi }_2)}(x_n-s_n{B}_2{x}_n),\hfill \\ x_{n+1}={\xi }_n{P}_C[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)S{J}_{M,\lambda }(I-\lambda B)u_n]+(1-{\xi }_n)v_n,n\ge 0,\hfill \end{array}$$

where $\{{\alpha }_n\},\{{\xi }_n\}\subset (0,1)$, $\lambda \in (0,2\beta )$ such that $0<a\le \lambda \le b<2\beta$, $\{r_n\}\in (0,2\eta )$ with $0<c\le d\le 1-\eta$ and $\{s_n\}\in (0,2\rho )$ with $0<e\le f\le 1-\rho$ satisfy some appropriate conditions. The purpose of this article, we show that under some control conditions the sequence $\{x_n\}$ converges strongly to a common element of the set of fixed points of nonexpansive mappings, the common solutions of the generalized mixed equilibrium problem and the set of solutions of the variational inclusion in a real Hilbert space.

2 Preliminaries

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, respectively. Let C be a nonempty closed convex subset of H. Recall that the metric (nearest point) projection $P_C$ from H onto C assigns to each $x\in H$, the unique point in $P_{C}x\in C$ satisfying the property

$$\parallel x-P_{C}x\parallel =\underset{y\in C}{min}\parallel x-y\parallel .$$

The following characterizes the projection $P_C$. We recall some lemmas which will be needed in the rest of this article.

Lemma 2.1 The function$u\in C$is a solution of the variational inequality (1.5) if and only if$u\in C$satisfies the relation$u=P_C(u-\lambda \mathrm{\Psi }u)$for all$\lambda >0$.

Lemma 2.2 For a given$z\in H$, $u\in C$, $u=P_{C}z\iff 〈u-z,v-u〉\ge 0$, $\mathrm{\forall }v\in C$.

It is well known that $P_C$ is a firmly nonexpansive mapping of H onto C and satisfies

$${\parallel P_{C}x-P_{C}y\parallel }^2\le 〈P_{C}x-P_{C}y,x-y〉,\mathrm{\forall }x,y\in H.$$

(2.1)

Moreover, $P_{C}x$is characterized by the following properties: $P_{C}x\in C$and for all$x\in H$, $y\in C$,

$$〈x-P_{C}x,y-P_{C}x〉\le 0.$$

(2.2)

Lemma 2.3 ([23])

Let$M:H\to 2^H$be a maximal monotone mapping and let$B:H\to H$be a monotone and Lipshitz continuous mapping. Then the mapping$L=M+B:H\to 2^H$is a maximal monotone mapping.

Lemma 2.4 ([24])

Each Hilbert space H satisfies Opial’s condition, that is, for any sequence$\{x_n\}\subset H$with$x_n\rightharpoonup x$, the inequality${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel x_n-x\parallel <{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel x_n-y\parallel$, hold for each$y\in H$with$y\ne x$.

Lemma 2.5 ([25])

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

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\delta }_n,\mathrm{\forall }n\ge 0,$$

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

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

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

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

Lemma 2.6 ([26])

Let C be a closed convex subset of a real Hilbert space H and let$T:C\to C$be a nonexpansive mapping. Then$I-T$is demiclosed at zero, that is,

$$x_n\rightharpoonup x,x_n-T{x}_n\to 0$$

implies$x=Tx$.

For solving the generalized mixed equilibrium problem, let us assume that the bifunction $F:C\times C\to \mathcal{R}$, the nonlinear mapping $\mathrm{\Psi }:C\to H$ is continuous monotone and $\psi :C\to \mathcal{R}$ satisfies the following conditions:

(A1) $F(x,x)=0$ for all $x\in C$;

(A2) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$ for any $x,y\in C$;

(A3) for each fixed $y\in C$, $x\mapsto F(x,y)$ is weakly upper semicontinuous;

(A4) for each fixed $x\in C$, $y\mapsto F(x,y)$ is convex and lower semicontinuous;

(B1) for each $x\in C$ and $r>0$, there exist a bounded subset $D_x\subseteq C$ and $y_x\in C$ such that for any $z\in C\setminus D_x$,

$$F(z,y_x)+\psi (y_x)-\psi (z)+\frac{1}{r}〈y_x-z,z-x〉<0,$$

(2.3)

(B2) C is a bounded set.

Lemma 2.7 ([27])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let$F:C\times C\to \mathcal{R}$be a bifunction mapping satisfies (A 1)-(A 4) and let$\psi :C\to \mathcal{R}$is convex and lower semicontinuous such that$C\cap dom\psi \ne \mathrm{\varnothing }$. Assume that either (B 1) or (B 2) holds. For$r>0$and$x\in H$, then there exists$u\in C$such that

$$F(u,y)+\psi (y)-\psi (u)+\frac{1}{r}〈y-u,u-x〉\ge 0.$$

Define a mapping$T_r^{(F,\psi )}:H\to C$as follows:

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

(2.4)

for all$x\in H$. Then, the following hold:

• (i) $T_r^{(F,\psi )}$ is single-valued;

• (ii) $T_r^{(F,\psi )}$ is firmly nonexpansive, i.e., for any $x,y\in H$,

  $${\parallel T_r^{(F,\psi )}x-T_r^{(F,\psi )}y\parallel }^2\le 〈T_r^{(F,\psi )}x-T_r^{(F,\psi )}y,x-y〉;$$

• (iii) $F(T_r^{(F,\psi )})=MEP(F,\psi )$;

• (iv) $MEP(F,\psi )$ is closed and convex.

Lemma 2.8 ([19])

Assume A is a strongly positive linear bounded operator on a Hilbert space H with coefficient$\overline{\gamma }>0$and$0<\rho \le {\parallel A\parallel }^{-1}$, then$\parallel I-\rho A\parallel \le 1-\rho \overline{\gamma }$.

Lemma 2.9 ([28])

Let H be a real Hilbert space and$A:H\to H$a mapping.

• (i) If A is δ-strongly monotone and μ-strictly pseudo-contraction with $\delta +\mu >1$, then $I-A$ is contraction with constant $\sqrt{(1-\delta )/\mu }$.

• (ii) If A is δ-strongly monotone and μ-strictly pseudo-contraction with $\delta +\mu >1$, then for any fixed number $\tau \in (0,1)$, $I-\tau A$ is contraction with constant $1-\tau (1-\sqrt{(1-\delta )/\mu })$.

3 Strong convergence theorems

In this section, we show a strong convergence theorem which solves the problem of finding a common element of $F(S)$, $GMEP(F_1,{\psi }_1,B_1)$, $GMEP(F_2,{\psi }_2,B_2)$ and $I(B,M)$ of an inverse-strongly monotone mapping in a real Hilbert space.

Theorem 3.1 Let H be a real Hilbert space, C be a closed convex subset of H. Let$F_1$, $F_2$be two bifunctions of$C\times C$into$\mathcal{R}$satisfying (A 1)-(A 4) and$B,B_1,B_2:C\to H$be$\beta ,\eta ,\rho$-inverse-strongly monotone mappings, ${\psi }_1,{\psi }_2:C\to \mathcal{R}$be convex and lower semicontinuous function, $f:C\to C$be a contraction with coefficient α ($0<\alpha <1$), $M:H\to 2^H$be a maximal monotone mapping and A be a δ-strongly monotone and μ-strictly pseudo-contraction mapping with$\delta +\mu >1$, γ is a positive real number such that$\gamma <\frac{1}{\alpha }(1-\sqrt{\frac{1-\delta }{\mu }})$. Assume that either (B 1) or (B 2) holds. Let S be a nonexpansive mapping of H into itself such that

$$\mathrm{\Theta }:=F(S)\cap GMEP(F_1,{\psi }_1,B_1)\cap GMEP(F_2,{\psi }_2,B_2)\cap I(B,M)\ne \mathrm{\varnothing }.$$

Suppose$\{x_n\}$is a sequence generated by the following algorithm$x_0\in C$arbitrarily:

$$\{\begin{array}{c}{u}_n=T_{r_n}^{(F_1,{\psi }_1)}(x_n-r_n{B}_1{x}_n),\hfill \\ v_n=T_{s_n}^{(F_2,{\psi }_2)}(x_n-s_n{B}_2{x}_n),\hfill \\ x_{n+1}={\xi }_n{P}_C[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)S{J}_{M,\lambda }(I-\lambda B)u_n]+(1-{\xi }_n)v_n,n\ge 0,\hfill \end{array}$$

(3.1)

where$\{{\alpha }_n\},\{{\xi }_n\}\subset (0,1)$, $\lambda \in (0,2\beta )$such that$0<a\le \lambda \le b<2\beta$, $\{r_n\}\in (0,2\eta )$with$0<c\le d\le 1-\eta$and$\{s_n\}\in (0,2\rho )$with$0<e\le f\le 1-\rho$satisfy the following conditions:

(C1): ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\mathrm{\Sigma }}_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$, ${\mathrm{\Sigma }}_{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$,

(C2): $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\xi }_n<{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\xi }_n<1$, ${\mathrm{\Sigma }}_{n=1}^{\mathrm{\infty }}|{\xi }_{n+1}-{\xi }_n|<\mathrm{\infty }$,

(C3): ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0\mathit{\text{and}}{lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$,

(C4): ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{s}_n>0\mathit{\text{and}}{lim}_{n\to \mathrm{\infty }}|s_{n+1}-s_n|=0$.

Then$\{x_n\}$converges strongly to$q\in \mathrm{\Theta }$, where$q=P_{\mathrm{\Theta }}(\gamma f+I-A)(q)$which solves the following variational inequality:

$$〈(\gamma f-A)q,p-q〉\le 0,\mathrm{\forall }p\in \mathrm{\Theta }$$

which is the optimality condition for the minimization problem

$$\underset{q\in \mathrm{\Theta }}{min}\frac{1}{2}〈Aq,q〉-h(q),$$

(3.2)

where h is a potential function for γf (i.e., $h^{\prime }(q)=\gamma f(q)$for$q\in H$).

Proof Since B is β-inverse-strongly monotone mappings, we have

$$\begin{array}{rcl}{\parallel (I-\lambda B)x-(I-\lambda B)y\parallel }^2& =& {\parallel (x-y)-\lambda (Bx-By)\parallel }^2\\ =& {\parallel x-y\parallel }^2-2\lambda 〈x-y,Bx-By〉+{\lambda }^2{\parallel Bx-By\parallel }^2\\ \le & {\parallel x-y\parallel }^2+\lambda (\lambda -2\beta ){\parallel Bx-By\parallel }^2\\ \le & {\parallel x-y\parallel }^2.\end{array}$$

(3.3)

And $B_1$, $B_2$ are $\eta ,\rho$-inverse-strongly monotone mappings, we have

$$\begin{array}{rcl}{\parallel (I-r_n{B}_1)x-(I-r_n{B}_1)y\parallel }^2& =& {\parallel (x-y)-r_n(B_{1}x-B_{1}y)\parallel }^2\\ =& {\parallel x-y\parallel }^2-2r_n〈x-y,B_{1}x-B_{1}y〉+r_n^2{\parallel B_{1}x-B_{1}y\parallel }^2\\ \le & {\parallel x-y\parallel }^2+r_n(r_n-2\eta ){\parallel B_{1}x-B_{1}y\parallel }^2\\ \le & {\parallel x-y\parallel }^2.\end{array}$$

(3.4)

In similar way, we can obtain

$${\parallel (I-s_n{B}_2)x-(I-s_n{B}_2)y\parallel }^2\le {\parallel x-y\parallel }^2.$$

(3.5)

It is clear that if $0<\lambda <2\beta$, $0<r_n<2\eta$, $0<s_n\le 2\rho$ then $I-\lambda B$, $I-r_n{B}_1$, $I-s_n{B}_2$ are all nonexpansive. We will divide the proof into six steps.

Step 1. We will show $\{x_n\}$ is bounded. Put $y_n=J_{M,\lambda }(u_n-\lambda B{u}_n)$, $n\ge 0$. It follows that

$$\begin{array}{rcl}\parallel y_n-q\parallel & =& \parallel J_{M,\lambda }(u_n-\lambda B{u}_n)-J_{M,\lambda }(q-\lambda Bq)\parallel \\ \le & \parallel u_n-q\parallel .\end{array}$$

(3.6)

By Lemma 2.7, we have $u_n=T_{r_n}^{(F_1,{\psi }_1)}(x_n-r_n{B}_1{x}_n)$ for all $n\ge 0$. Then, we note that

$$\begin{array}{rcl}{\parallel u_n-q\parallel }^2& =& {\parallel T_{r_n}^{(F_1,{\psi }_1)}(x_n-r_n{B}_1{x}_n)-T_{r_n}^{(F_1,{\psi }_1)}(q-r_n{B}_{1}q)\parallel }^2\\ \le & {\parallel (x_n-r_n{B}_1{x}_n)-(q-r_n{B}_{1}q)\parallel }^2\\ \le & {\parallel x_n-q\parallel }^2+r_n(r_n-2\eta ){\parallel B_1{x}_n-B_{1}q\parallel }^2\\ \le & {\parallel x_n-q\parallel }^2.\end{array}$$

(3.7)

In similar way, we can obtain

$$\begin{array}{rcl}{\parallel v_n-q\parallel }^2& =& {\parallel T_{s_n}^{(F_2,{\psi }_2)}(x_n-s_n{B}_2{x}_n)-T_{s_n}^{(F_2,{\psi }_2)}(q-s_n{B}_{2}q)\parallel }^2\\ \le & {\parallel (x_n-s_n{B}_2{x}_n)-(q-s_n{B}_{2}q)\parallel }^2\\ \le & {\parallel x_n-q\parallel }^2+s_n(s_n-2\rho ){\parallel B_2{x}_n-B_{2}q\parallel }^2\\ \le & {\parallel x_n-q\parallel }^2.\end{array}$$

(3.8)

Put $z_n=P_C[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)S{y}_n]$ for all $n\ge 0$. From (3.1) and Lemma 2.9(ii), we deduce that

(3.9)

It follows from induction that

$$\parallel x_n-q\parallel \le max\{\parallel x_0-q\parallel ,\frac{\parallel \gamma f(q)-Aq\parallel }{1-\sqrt{\frac{1-\delta }{\mu }}-\gamma \alpha }\},n\ge 0.$$

Therefore $\{x_n\}$ is bounded, so are $\{v_n\}$, $\{y_n\}$, $\{z_n\}$, $\{S{y}_n\}$, $\{f(x_n)\}$ and $\{AS{y}_n\}$.

Step 2. We claim that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+2}-x_{n+1}\parallel =0$. From (3.1), we have

$$\begin{array}{rcl}\parallel x_{n+2}-x_{n+1}\parallel & =& \parallel {\xi }_{n+1}{z}_{n+1}+(1-{\xi }_{n+1})v_{n+1}-{\xi }_n{z}_n-(1-{\xi }_n)v_n\parallel \\ =& \parallel {\xi }_{n+1}(z_{n+1}-z_n)+({\xi }_{n+1}-{\xi }_n)z_n\\ +(1-{\xi }_{n+1})(v_{n+1}-v_n)+({\xi }_{n+1}-{\xi }_n)v_n\parallel \\ \le & {\xi }_{n+1}\parallel z_{n+1}-z_n\parallel +(1-{\xi }_{n+1})\parallel v_{n+1}-v_n\parallel \\ +|{\xi }_{n+1}-{\xi }_n|(\parallel z_n\parallel +\parallel v_n\parallel ).\end{array}$$

(3.10)

We will estimate $\parallel v_{n+1}-v_n\parallel$. On the other hand, from $v_{n-1}=T_{s_{n-1}}^{(F_2,{\psi }_2)}(x_{n-1}-s_{n-1}{B}_2{x}_{n-1})$ and $v_n=T_{s_n}^{(F_2,{\psi }_2)}(x_n-s_n{B}_2{x}_n)$, it follows that

(3.11)

and

$$F_2(v_n,y)+〈B_2{x}_n,y-v_n〉+{\psi }_2(y)-{\psi }_2(v_n)+\frac{1}{s_n}〈y-v_n,v_n-x_n〉\ge 0,\mathrm{\forall }y\in C.$$

(3.12)

Substituting $y=v_n$ in (3.11) and $y=v_{n-1}$ in (3.12), we get

$$F_2(v_{n-1},v_n)+〈B_2{x}_{n-1},v_n-v_{n-1}〉+{\psi }_2(v_n)-{\psi }_2(v_{n-1})+\frac{1}{s_{n-1}}〈v_n-v_{n-1},v_{n-1}-x_{n-1}〉\ge 0$$

and

$$F_2(v_n,v_{n-1})+〈B_2{x}_n,v_{n-1}-v_n〉+{\psi }_2(v_{n-1})-{\psi }_2(v_n)+\frac{1}{s_n}〈v_{n-1}-v_n,v_n-x_n〉\ge 0.$$

From (A2), we obtain

$$〈v_n-v_{n-1},B_2{x}_{n-1}-B_2{x}_n+\frac{v_{n-1}-x_{n-1}}{s_{n-1}}-\frac{v_n-x_n}{s_n}〉\ge 0,$$

and then

$$〈v_n-v_{n-1},s_{n-1}(B_2{x}_{n-1}-B_2{x}_n)+v_{n-1}-x_{n-1}-\frac{s_{n-1}}{s_n}(v_n-x_n)〉\ge 0,$$

so

$$〈v_n-v_{n-1},s_{n-1}{B}_2{x}_{n-1}-s_{n-1}{B}_2{x}_n+v_{n-1}-v_n+v_n-x_{n-1}-\frac{s_{n-1}}{s_n}(v_n-x_n)〉\ge 0.$$

It follows that

Without loss of generality, let us assume that there exists a real number e such that $s_{n-1}>e>0$, for all $n\in \mathbb{N}$. Then, we have

$$\begin{array}{rcl}{\parallel v_n-v_{n-1}\parallel }^2& \le & 〈v_n-v_{n-1},x_n-x_{n-1}+(1-\frac{s_{n-1}}{s_n})(v_n-x_n)〉\\ \le & \parallel v_n-v_{n-1}\parallel \{\parallel x_n-x_{n-1}\parallel +|1-\frac{s_{n-1}}{s_n}|\parallel v_n-x_n\parallel \}\end{array}$$

and hence

$$\begin{array}{rcl}\parallel v_n-v_{n-1}\parallel & \le & \parallel x_n-x_{n-1}\parallel +\frac{1}{s_n}|s_n-s_{n-1}|\parallel v_n-x_n\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +\frac{M_1}{e}|s_n-s_{n-1}|,\end{array}$$

(3.13)

where $M_1=sup\{\parallel v_n-x_n\parallel :n\in \mathbb{N}\}$. Substituting (3.13) into (3.10) that

$$\begin{array}{rcl}\parallel x_{n+2}-x_{n+1}\parallel & \le & {\xi }_{n+1}\parallel z_{n+1}-z_n\parallel +(1-{\xi }_{n+1})\{\parallel x_{n+1}-x_n\parallel +\frac{M_1}{e}|s_n-s_{n-1}|\}\\ +|{\xi }_{n+1}-{\xi }_n|(\parallel z_n\parallel +\parallel v_n\parallel ).\end{array}$$

(3.14)

We note that

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & =& \parallel P_C[{\alpha }_{n+1}\gamma f(x_{n+1})+(I-{\alpha }_{n+1}A)S{y}_{n+1}]-P_C[{\alpha }_n\gamma f(x_n)-(I-{\alpha }_{n}A)S{y}_n]\parallel \\ \le & \parallel {\alpha }_{n+1}\gamma f(x_{n+1})+(I-{\alpha }_{n+1}A)S{y}_{n+1}-({\alpha }_n\gamma f(x_n)-(I-{\alpha }_{n}A)S{y}_n)\parallel \\ \le & \parallel {\alpha }_{n+1}\gamma (f(x_{n+1})-f(x_n))+({\alpha }_{n+1}-{\alpha }_n)\gamma f(x_n)+(I-{\alpha }_{n+1}A)(S{y}_{n+1}-S{y}_n)\\ +({\alpha }_n-{\alpha }_{n+1})AS{y}_n\parallel \\ \le & {\alpha }_{n+1}\gamma \alpha \parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|\parallel \gamma f(x_n)\parallel \\ +(1-{\alpha }_{n+1}(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel y_{n+1}-y_n\parallel \\ +|{\alpha }_{n+1}-{\alpha }_n|\parallel AS{y}_n\parallel \\ \le & {\alpha }_{n+1}\gamma \alpha \parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|(\parallel \gamma f(x_n)\parallel +\parallel AS{y}_n\parallel )\\ +(1-{\alpha }_{n+1}(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel y_{n+1}-y_n\parallel .\end{array}$$

(3.15)

Since $I-\lambda B$ be nonexpansive, we have

$$\begin{array}{rcl}\parallel y_{n+1}-y_n\parallel & =& \parallel J_{M,\lambda }(u_{n+1}-\lambda B{u}_{n+1})-J_{M,\lambda }(u_n-\lambda B{u}_n)\parallel \\ \le & \parallel (u_{n+1}-\lambda B{u}_{n+1})-(u_n-\lambda B{u}_n)\parallel \\ \le & \parallel u_{n+1}-u_n\parallel .\end{array}$$

(3.16)

On the other hand, from $u_{n-1}=T_{r_{n-1}}^{(F_1,{\psi }_1)}(x_{n-1}-r_{n-1}{B}_1{x}_{n-1})$ and $u_n=T_{r_n}^{(F_1,{\psi }_1)}(x_n-r_n{B}_1{x}_n)$, it follows that

(3.17)

and

$$F_1(u_n,y)+〈B_1{x}_n,y-u_n〉+{\psi }_1(y)-{\psi }_1(u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C.$$

(3.18)

Substituting $y=u_n$ in (3.17) and $y=u_{n-1}$ in (3.18), we get

$$F_1(u_{n-1},u_n)+〈B_1{x}_{n-1},u_n-u_{n-1}〉+{\psi }_1(u_n)-{\psi }_1(u_{n-1})+\frac{1}{r_{n-1}}〈u_n-u_{n-1},u_{n-1}-x_{n-1}〉\ge 0$$

and

$$F_1(u_n,u_{n-1})+〈B_1{x}_n,u_{n-1}-u_n〉+{\psi }_1(u_{n-1})-{\psi }_1(u_n)+\frac{1}{r_n}〈u_{n-1}-u_n,u_n-x_n〉\ge 0.$$

From (A2), we obtain

$$〈u_n-u_{n-1},B_1{x}_{n-1}-B_1{x}_n+\frac{u_{n-1}-x_{n-1}}{r_{n-1}}-\frac{u_n-x_n}{r_n}〉\ge 0,$$

and then

$$〈u_n-u_{n-1},r_{n-1}(B_1{x}_{n-1}-B_1{x}_n)+u_{n-1}-x_{n-1}-\frac{r_{n-1}}{r_n}(u_n-x_n)〉\ge 0,$$

so

$$〈u_n-u_{n-1},r_{n-1}{B}_1{x}_{n-1}-r_{n-1}{B}_1{x}_n+u_{n-1}-u_n+u_n-x_{n-1}-\frac{r_{n-1}}{r_n}(u_n-x_n)〉\ge 0.$$

It follows that

Without loss of generality, let us assume that there exists a real number c such that $r_{n-1}>c>0$, for all $n\in \mathbb{N}$. Then, we have

$$\begin{array}{rcl}{\parallel u_n-u_{n-1}\parallel }^2& \le & 〈u_n-u_{n-1},x_n-x_{n-1}+(1-\frac{r_{n-1}}{r_n})(u_n-x_n)〉\\ \le & \parallel u_n-u_{n-1}\parallel \{\parallel x_n-x_{n-1}\parallel +|1-\frac{r_{n-1}}{r_n}|\parallel u_n-x_n\parallel \}\end{array}$$

and hence

$$\begin{array}{rcl}\parallel u_n-u_{n-1}\parallel & \le & \parallel x_n-x_{n-1}\parallel +\frac{1}{r_n}|r_n-r_{n-1}|\parallel u_n-x_n\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +\frac{M_2}{c}|r_n-r_{n-1}|,\end{array}$$

(3.19)

where $M_2=sup\{\parallel u_n-x_n\parallel :n\in \mathbb{N}\}$. Substituting (3.19) into (3.16), we have

$$\parallel y_n-y_{n-1}\parallel \le \parallel x_n-x_{n-1}\parallel +\frac{M_2}{c}|r_n-r_{n-1}|.$$

(3.20)

Substituting (3.20) into (3.15), we obtain that

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & \le & {\alpha }_{n+1}\gamma \alpha \parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|(\parallel \gamma f(x_n)\parallel +\parallel AS{y}_n\parallel )\\ +(1-{\alpha }_{n+1}(1-\sqrt{\frac{1-\delta }{\mu }}))\{\parallel x_n-x_{n-1}\parallel +\frac{M_2}{c}|r_n-r_{n-1}|\}.\end{array}$$

(3.21)

And substituting (3.13), (3.21) into (3.10), we get

$$\begin{array}{rcl}\parallel x_{n+2}-x_{n+1}\parallel & \le & {\xi }_{n+1}\{{\alpha }_{n+1}\gamma \alpha \parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|(\parallel \gamma f(x_n)\parallel +\parallel AS{y}_n\parallel )\\ +(1-{\alpha }_{n+1}(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel x_n-x_{n-1}\parallel +\frac{M_2}{c}|r_n-r_{n-1}|\}\\ +(1-{\xi }_{n+1})\{\parallel x_n-x_{n-1}\parallel +\frac{M_1}{e}|s_n-s_{n-1}|\}+|{\xi }_{n+1}-{\xi }_n|(\parallel z_n\parallel +\parallel v_n\parallel )\\ \le & (1-((1-\sqrt{\frac{1-\delta }{\mu }})-\gamma \alpha ){\xi }_{n+1}{\alpha }_{n+1})\parallel x_{n+1}-x_n\parallel +(|{\alpha }_{n+1}-{\alpha }_n|\\ +|{\xi }_{n+1}-{\xi }_n|)M_3+\frac{M_1}{e}|s_n-s_{n-1}|+\frac{M_2}{c}|r_n-r_{n-1}|,\end{array}$$

(3.22)

where $M_3>0$ is a constant satisfying

$$\underset{n}{sup}\{\parallel \gamma f(x_n)\parallel +\parallel AS{y}_n\parallel ,\parallel z_n\parallel +\parallel v_n\parallel \}\le M_3.$$

This together with (C1)-(C4) and Lemma 2.5, imply that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+2}-x_{n+1}\parallel =0.$$

(3.23)

From (3.20), we also have $\parallel y_{n+1}-y_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Step 3. We show the followings:

• (i) ${lim}_{n\to \mathrm{\infty }}\parallel B{u}_n-Bq\parallel =0$;

• (ii) ${lim}_{n\to \mathrm{\infty }}\parallel B_1{x}_n-B_{1}q\parallel =0$;

• (iii) ${lim}_{n\to \mathrm{\infty }}\parallel B_2{x}_n-B_{2}q\parallel =0$.

For $q\in \mathrm{\Theta }$ and $q=J_{M,\lambda }(q-\lambda Bq)$, then we get

$$\begin{array}{rcl}{\parallel y_n-q\parallel }^2& =& {\parallel J_{M,\lambda }(u_n-\lambda B{u}_n)-J_{M,\lambda }(q-\lambda Bq)\parallel }^2\\ \le & {\parallel (u_n-\lambda B{u}_n)-(q-\lambda Bq)\parallel }^2\\ \le & {\parallel u_n-q\parallel }^2+\lambda (\lambda -2\beta ){\parallel B{u}_n-Bq\parallel }^2\\ \le & {\parallel x_n-q\parallel }^2+\lambda (\lambda -2\beta ){\parallel B{u}_n-Bq\parallel }^2.\end{array}$$

(3.24)

It follows that

$$\begin{array}{rcl}{\parallel z_n-q\parallel }^2& =& {\parallel P_C({\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)S{y}_n)-P_C(q)\parallel }^2\\ \le & {\parallel {\alpha }_n(\gamma f(x_n)-Aq)+(I-{\alpha }_{n}A)(S{y}_n-q)\parallel }^2\\ \le & {\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2+(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }})){\parallel y_n-q\parallel }^2\\ +2{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel \\ \le & {\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2+2{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel \\ +(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\{{\parallel x_n-q\parallel }^2+\lambda (\lambda -2\beta ){\parallel B{u}_n-Bq\parallel }^2\}\\ \le & {\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2+2{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel \\ +{\parallel x_n-q\parallel }^2+(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\lambda (\lambda -2\beta ){\parallel B{u}_n-Bq\parallel }^2.\end{array}$$

(3.25)

By the convexity of the norm $\parallel \cdot \parallel$, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& =& {\parallel {\xi }_n{z}_n+(1-{\xi }_n)v_n-q\parallel }^2\\ \le & {\parallel {\xi }_n(z_n-q)+(1-{\xi }_n)(v_n-q)\parallel }^2\\ \le & {\xi }_n{\parallel z_n-q\parallel }^2+(1-{\xi }_n){\parallel v_n-q\parallel }^2.\end{array}$$

(3.26)

Substituting (3.8), (3.25) into (3.26), we obtain

So, we obtain

where ${ϵ}_n=2{\xi }_n{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel$. Since conditions (C1)-(C3) and ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, then we obtain that $\parallel B{u}_n-Bq\parallel \to 0$ as $n\to \mathrm{\infty }$. We consider this inequality in (3.25) that

$$\begin{array}{rcl}{\parallel z_n-q\parallel }^2& \le & {\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2+(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }})){\parallel y_n-q\parallel }^2\\ +2{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel .\end{array}$$

(3.27)

Substituting (3.6) and (3.8) into (3.27), we have

$$\begin{array}{rcl}{\parallel z_n-q\parallel }^2& \le & {\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2+(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\\ \times \{{\parallel x_n-q\parallel }^2+r_n(r_n-2\eta ){\parallel B_1{x}_n-B_{1}q\parallel }^2\}\\ +2{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel \\ =& {\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2+(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }})){\parallel x_n-q\parallel }^2\\ +(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))r_n(r_n-2\eta ){\parallel B_1{x}_n-B_{1}q\parallel }^2\\ +2{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel \\ \le & {\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2+{\parallel x_n-q\parallel }^2\\ +(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))r_n(r_n-2\eta ){\parallel B_1{x}_n-B_{1}q\parallel }^2\\ +2{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel .\end{array}$$

(3.28)

Substituting (3.7) and (3.28) into (3.26), we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& \le & {\xi }_n\{{\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2+{\parallel x_n-q\parallel }^2\\ +(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))r_n(r_n-2\eta ){\parallel B_1{x}_n-B_{1}q\parallel }^2\\ +2{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel \}\\ +(1-{\xi }_n){\parallel x_n-q\parallel }^2\\ =& {\xi }_n{\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2\\ +{\xi }_n{\parallel x_n-q\parallel }^2+{\xi }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))r_n(r_n-2\eta ){\parallel B_1{x}_n-B_{1}q\parallel }^2\\ +2{\xi }_n{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel \\ +(1-{\xi }_n){\parallel x_n-q\parallel }^2.\end{array}$$

(3.29)

So, we also have

where ${ϵ}_n=2{\xi }_n{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel$. Since conditions (C1)-(C3), ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, then we obtain that $\parallel B_1{x}_n-B_{1}q\parallel \to 0$ as $n\to \mathrm{\infty }$. Substituting (3.24) into (3.27), we have

(3.30)

Substituting (3.8) and (3.30) into (3.26), we obtain

(3.31)

So, we also have

where ${ϵ}_n=2{\xi }_n{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel$. Since conditions (C1), (C2), (C4), ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel B{u}_n-Bq\parallel =0$, then we obtain that $\parallel B_2{x}_n-B_{2}q\parallel \to 0$ as $n\to \mathrm{\infty }$.

Step 4. We show the followings:

• (i) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-u_n\parallel =0$;

• (ii) ${lim}_{n\to \mathrm{\infty }}\parallel u_n-y_n\parallel =0$;

• (iii) ${lim}_{n\to \mathrm{\infty }}\parallel y_n-S{y}_n\parallel =0$.

Since $T_{r_n}^{(F_1,{\psi }_1)}$ is firmly nonexpansive, we observe that

$$\begin{array}{rcl}{\parallel u_n-q\parallel }^2& =& {\parallel T_{r_n}^{(F_1,{\psi }_1)}(x_n-r_n{B}_1{x}_n)-T_{r_n}^{(F_1,{\psi }_1)}(q-r_n{B}_{1}q)\parallel }^2\\ \le & 〈(x_n-r_n{B}_1{x}_n)-(q-r_n{B}_{1}q),u_n-q〉\\ =& \frac{1}{2}({\parallel (x_n-r_n{B}_1{x}_n)-(q-r_n{B}_{1}q)\parallel }^2+{\parallel u_n-q\parallel }^2\\ -{\parallel (x_n-r_n{B}_1{x}_n)-(q-r_n{B}_{1}q)-(u_n-q)\parallel }^2)\\ \le & \frac{1}{2}({\parallel x_n-q\parallel }^2+{\parallel u_n-q\parallel }^2-{\parallel (x_n-u_n)-r_n(B_1{x}_n-B_{1}q)\parallel }^2)\\ =& \frac{1}{2}({\parallel x_n-q\parallel }^2+{\parallel u_n-q\parallel }^2-{\parallel x_n-u_n\parallel }^2\\ +2r_n〈B_1{x}_n-B_{1}q,x_n-u_n〉-r_n^2{\parallel B_1{x}_n-B_{1}q\parallel }^2).\end{array}$$

Hence, we have

$${\parallel u_n-q\parallel }^2\le {\parallel x_n-q\parallel }^2-{\parallel x_n-u_n\parallel }^2+2r_n\parallel B_1{x}_n-B_{1}q\parallel \parallel x_n-u_n\parallel .$$

(3.32)

Since $J_{M,\lambda }$ is 1-inverse-strongly monotone, we compute

$$\begin{array}{rcl}{\parallel y_n-q\parallel }^2& =& {\parallel J_{M,\lambda }(u_n-\lambda B{u}_n)-J_{M,\lambda }(q-\lambda Bq)\parallel }^2\\ \le & 〈(u_n-\lambda B{u}_n)-(q-\lambda Bq),y_n-q〉\\ =& \frac{1}{2}({\parallel (u_n-\lambda B{u}_n)-(q-\lambda Bq)\parallel }^2+{\parallel y_n-q\parallel }^2\\ -{\parallel (u_n-\lambda B{u}_n)-(q-\lambda Bq)-(y_n-q)\parallel }^2)\\ =& \frac{1}{2}({\parallel u_n-q\parallel }^2+{\parallel y_n-q\parallel }^2-{\parallel (u_n-y_n)-\lambda (B{u}_n-Bq)\parallel }^2)\\ \le & \frac{1}{2}({\parallel u_n-q\parallel }^2+{\parallel y_n-q\parallel }^2-{\parallel u_n-y_n\parallel }^2\\ +2\lambda 〈u_n-y_n,B{u}_n-Bq〉-{\lambda }^2{\parallel B{u}_n-Bq\parallel }^2),\end{array}$$

(3.33)

which implies that

$${\parallel y_n-q\parallel }^2\le {\parallel u_n-q\parallel }^2-{\parallel u_n-y_n\parallel }^2+2\lambda \parallel u_n-y_n\parallel \parallel B{u}_n-Bq\parallel .$$

(3.34)

Substitute (3.32) into (3.34), we have

$$\begin{array}{rcl}{\parallel y_n-q\parallel }^2& \le & \{{\parallel x_n-q\parallel }^2-{\parallel x_n-u_n\parallel }^2+2r_n\parallel B_1{x}_n-B_{1}q\parallel \parallel x_n-u_n\parallel \}\\ -{\parallel u_n-y_n\parallel }^2+2\lambda \parallel u_n-y_n\parallel \parallel B{u}_n-Bq\parallel .\end{array}$$

(3.35)

Substitute (3.35) into (3.27), we have

$$\begin{array}{rcl}{\parallel z_n-q\parallel }^2& \le & {\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2+(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\{{\parallel x_n-q\parallel }^2-{\parallel x_n-u_n\parallel }^2\\ +2r_n\parallel B_1{x}_n-B_{1}q\parallel \parallel x_n-u_n\parallel -{\parallel u_n-y_n\parallel }^2+2\lambda \parallel u_n-y_n\parallel \parallel B{u}_n-Bq\parallel \}\\ +2{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel \\ \le & {\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2+{\parallel x_n-q\parallel }^2-{\parallel x_n-u_n\parallel }^2\\ +2(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))r_n\parallel B_1{x}_n-B_{1}q\parallel \parallel x_n-u_n\parallel -{\parallel u_n-y_n\parallel }^2\\ +2(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\lambda \parallel u_n-y_n\parallel \parallel B{u}_n-Bq\parallel \\ +2{\alpha }_n(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel \gamma f(x_n)-Aq\parallel \parallel y_n-q\parallel .\end{array}$$

(3.36)

Since $T_{s_n}^{(F_2,{\psi }_2)}$ is firmly nonexpansive, we observe that

$$\begin{array}{rcl}{\parallel v_n-q\parallel }^2& =& {\parallel T_{s_n}^{(F_2,{\psi }_2)}(x_n-s_n{B}_2{x}_n)-T_{s_n}^{(F_2,{\psi }_2)}(q-s_n{B}_{2}q)\parallel }^2\\ \le & 〈(x_n-s_n{B}_2{x}_n)-(q-s_n{B}_{2}q),v_n-q〉\\ =& \frac{1}{2}({\parallel (x_n-s_n{B}_2{x}_n)-(q-s_n{B}_{2}q)\parallel }^2+{\parallel v_n-q\parallel }^2\\ -{\parallel (x_n-s_n{B}_2{x}_n)-(q-s_n{B}_{2}q)-(v_n-q)\parallel }^2)\\ \le & \frac{1}{2}({\parallel x_n-q\parallel }^2+{\parallel v_n-q\parallel }^2-{\parallel (x_n-v_n)-s_n(B_2{x}_n-B_{2}q)\parallel }^2)\\ =& \frac{1}{2}({\parallel x_n-q\parallel }^2+{\parallel v_n-q\parallel }^2-{\parallel x_n-v_n\parallel }^2\\ +2s_n〈B_2{x}_n-B_{2}q,x_n-v_n〉-s_n^2{\parallel B_2{x}_n-B_{2}q\parallel }^2).\end{array}$$

Hence, we have

$${\parallel v_n-q\parallel }^2\le {\parallel x_n-q\parallel }^2-{\parallel x_n-v_n\parallel }^2+2s_n\parallel B_2{x}_n-B_{2}q\parallel \parallel x_n-v_n\parallel .$$

(3.37)

Substitute (3.36) and (3.37) into (3.26), we obtain

(3.38)

Then, we derive

(3.39)

By conditions (C1)-(C4), ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel B{u}_n-Bq\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel B_1{x}_n-B_{1}q\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel B_2{x}_n-B_{2}q\parallel =0$. So, we have $\parallel x_n-u_n\parallel \to 0$, $\parallel u_n-y_n\parallel \to 0$, $\parallel x_n-v_n\parallel \to 0$ as $n\to \mathrm{\infty }$. We note that $x_{n+1}-x_n={\xi }_n(z_n-x_n)+(1-{\xi }_n)(v_n-x_n)$. From ${lim}_{n\to \mathrm{\infty }}\parallel x_n-v_n\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, and hence

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

(3.40)

It follows that

$$\parallel x_n-y_n\parallel \le \parallel x_n-u_n\parallel +\parallel u_n-y_n\parallel \to 0,\text{as}n\to \mathrm{\infty }.$$

(3.41)

Since

$$\parallel z_n-y_n\parallel \le \parallel z_n-x_n\parallel +\parallel x_n-y_n\parallel .$$

So, by (3.40) and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$, we obtain

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

(3.42)

Therefore, we observe that

$$\begin{array}{rcl}\parallel S{y}_n-z_n\parallel & =& \parallel P_{C}S{y}_n-P_C({\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)S{y}_n)\parallel \\ \le & \parallel S{y}_n-{\alpha }_n\gamma f(x_n)-(I-{\alpha }_{n}A)S{y}_n\parallel \\ =& {\alpha }_n\parallel \gamma f(x_n)-AS{y}_n\parallel .\end{array}$$

(3.43)

By condition (C1), we have $\parallel S{y}_n-z_n\parallel \to 0$ as $n\to \mathrm{\infty }$. Next, we observe that

$$\parallel S{y}_n-y_n\parallel \le \parallel S{y}_n-z_n\parallel +\parallel z_n-y_n\parallel .$$

By (3.42) and (3.43), we have $\parallel S{y}_n-y_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Step 5. We show that $q\in \mathrm{\Theta }:=F(S)\cap GMEP(F_1,{\psi }_1,B_1)\cap GMEP(F_2,{\psi }_2,B_2)\cap I(B,M)$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈(\gamma f-A)q,S{y}_n-q〉\le 0$. It is easy to see that $P_{\mathrm{\Theta }}(\gamma f+(I-A))$ is a contraction of H into itself. In fact, from Lemma 2.9, we have

$$\begin{array}{c}\parallel P_{\mathrm{\Theta }}(\gamma f+(I-A))x-P_{\mathrm{\Theta }}(\gamma f+(I-A))y\parallel \hfill \\ \le \parallel (\gamma f+(I-A))x-(\gamma f+(I-A))y\parallel \hfill \\ \le \gamma \parallel f(x)-f(y)\parallel +(I-A)\parallel x-y\parallel \hfill \\ \le \gamma \alpha \parallel x-y\parallel +(1-(1-\sqrt{\frac{1-\delta }{\mu }}))\parallel x-y\parallel \hfill \\ =(\sqrt{\frac{1-\delta }{\mu }}+\gamma \alpha )\parallel x-y\parallel .\hfill \end{array}$$

Hence H is complete, there exists a unique fixed point $q\in H$ such that $q=P_{\mathrm{\Theta }}(\gamma f+(I-A))(q)$. By Lemma 2.2 we obtain that $〈(\gamma f-A)q,w-q〉\le 0$ for all $w\in \mathrm{\Theta }$.

Next, we show that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈(\gamma f-A)q,S{y}_n-q〉\le 0$, where $q=P_{\mathrm{\Theta }}(\gamma f+I-A)(q)$ is the unique solution of the variational inequality $〈(\gamma f-A)q,p-qr〉\ge 0$, $\mathrm{\forall }p\in \mathrm{\Theta }$. We can choose a subsequence $\{y_{n_i}\}$ of $\{y_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma f-A)q,S{y}_n-q〉=\underset{i\to \mathrm{\infty }}{lim}〈(\gamma f-A)q,S{y}_{n_i}-q〉.$$

As $\{y_{n_i}\}$ is bounded, there exists a subsequence $\{y_{n_{i_j}}\}$ of $\{y_{n_i}\}$ which converges weakly to w. We may assume without loss of generality that $y_{n_i}\rightharpoonup w$. We claim that $w\in \mathrm{\Theta }$. Since $\parallel y_n-S{y}_n\parallel \to 0$ and by Lemma 2.6, we have $w\in F(S)$.

Next, we show that $w\in GMEP(F_1,{\psi }_1,B_1)$. Since $u_n=T_{r_n}^{(F_1,{\psi }_1)}(x_n-r_n{B}_1{x}_n)$, we know that

$$F_1(u_n,y)+{\psi }_1(y)-{\psi }_1(u_n)+〈B_1{x}_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C.$$

It follows by (A2) that

$${\psi }_1(y)-{\psi }_1(u_n)+〈B_1{x}_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge F_1(y,u_n),\mathrm{\forall }y\in C.$$

Hence,

$${\psi }_1(y)-{\psi }_1(u_{n_i})+〈B_1{x}_{n_i},y-u_{n_i}〉+\frac{1}{r_{n_i}}〈y-u_{n_i},u_{n_i}-x_{n_i}〉\ge F_1(y,u_{n_i}),\mathrm{\forall }y\in C.$$

(3.44)

For $t\in (0,1]$ and $y\in H$, let $y_t=ty+(1-t)w$. From (3.44), we have

$$\begin{array}{c}〈y_t-u_{n_i},B_1{y}_t〉\hfill \\ \ge 〈y_t-u_{n_i},B_1{y}_t〉-{\psi }_1(y_t)+{\psi }_1(u_{n_i})-〈B_1{x}_{n_i},y_t-u_{n_i}〉\hfill \\ -\frac{1}{r_{n_i}}〈y_t-u_{n_i},u_{n_i}-x_{n_i}〉+F_1(y_t,u_{n_i})\hfill \\ =〈y_t-u_{n_i},B_1{y}_t-B_1{u}_{n_i}〉+〈y_t-u_{n_i},B_1{u}_{n_i}-B_1{x}_{n_i}〉-{\psi }_1(y_t)+{\psi }_1(u_{n_i})\hfill \\ -\frac{1}{r_{n_i}}〈y_t-u_{n_i},u_{n_i}-x_{n_i}〉+F_1(y_t,u_{n_i}).\hfill \end{array}$$

From $\parallel u_{n_i}-x_{n_i}\parallel \to 0$, we have $\parallel B_1{u}_{n_i}-B_1{x}_{n_i}\parallel \to 0$. Further, from (A4) and the weakly lower semicontinuity of ${\psi }_1$, $\frac{u_{n_i}-x_{n_i}}{r_{n_i}}\to 0$ and $u_{n_i}\rightharpoonup w$, we have

$$〈y_t-w,B_1{y}_t〉\ge -{\psi }_1(y_t)+{\psi }_1(w)+F_1(y_t,w).$$

(3.45)

From (A1), (A4) and (3.45), we have

$$\begin{array}{rcl}0& =& F_1(y_t,y_t)-{\psi }_1(y_t)+{\psi }_1(y_t)\\ \le & t{F}_1(y_t,y)+(1-t)F_1(y_t,w)+t{\psi }_1(y)+(1-t){\psi }_1(w)-{\psi }_1(y_t)\\ =& t[F_1(y_t,y)+{\psi }_1(y)-{\psi }_1(y_t)]+(1-t)[F_1(y_t,w)+{\psi }_1(w)-{\psi }_1(y_t)]\\ \le & t[F_1(y_t,y)+{\psi }_1(y)-{\psi }_1(y_t)]+(1-t)〈y_t-w,B_1{y}_t〉\\ =& t[F_1(y_t,y)+{\psi }_1(y)-{\psi }_1(y_t)]+(1-t)t〈y-w,B_1{y}_t〉,\end{array}$$

and hence

$$0\le F_1(y_t,y)+{\psi }_1(y)-{\psi }_1(y_t)+(1-t)〈y-w,B_1{y}_t〉.$$

Letting $t\to 0$, we have, for each $y\in C$,

$$F_1(w,y)+{\psi }_1(y)-{\psi }_1(w)+〈y-w,B_{1}w〉\ge 0.$$

This implies that $w\in GMEP(F_1,{\psi }_1,B_1)$. By following the same arguments, we can show that $w\in GMEP(F_2,{\psi }_2,B_2)$.

Lastly, we show that $w\in I(B,M)$. In fact, since B is a β-inverse-strongly monotone, B is monotone and Lipschitz continuous mapping. It follows from Lemma 2.3 that $M+B$ is a maximal monotone. Let $(v,g)\in G(M+B)$, since $g-Bv\in M(v)$. Again since $y_{n_i}=J_{M,\lambda }(u_{n_i}-\lambda B{u}_{n_i})$, we have $u_{n_i}-\lambda B{u}_{n_i}\in (I+\lambda M)(y_{n_i})$, that is, $\frac{1}{\lambda }(u_{n_i}-y_{n_i}-\lambda B{u}_{n_i})\in M(y_{n_i})$. By virtue of the maximal monotonicity of $M+B$, we have

$$〈v-y_{n_i},g-Bv-\frac{1}{\lambda }(u_{n_i}-y_{n_i}-\lambda B{u}_{n_i})〉\ge 0,$$

and hence

$$\begin{array}{rcl}〈v-y_{n_i},g〉& \ge & 〈v-y_{n_i},Bv+\frac{1}{\lambda }(u_{n_i}-y_{n_i}-\lambda B{u}_{n_i})〉\\ =& 〈v-y_{n_i},Bv-B{y}_{n_i}〉+〈v-y_{n_i},B{y}_{n_i}-B{u}_{n_i}〉\\ +〈v-y_{n_i},\frac{1}{\lambda }(u_{n_i}-y_{n_i})〉.\end{array}$$

It follows from ${lim}_{n\to \mathrm{\infty }}\parallel u_n-y_n\parallel =0$, we have ${lim}_{n\to \mathrm{\infty }}\parallel B{u}_n-B{y}_n\parallel =0$ and $y_{n_i}\rightharpoonup w$ that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈v-y_n,g〉=〈v-w,g〉\ge 0.$$

It follows from the maximal monotonicity of $B+M$ that $\theta \in (M+B)(w)$, that is, $w\in I(B,M)$. Therefore, $w\in \mathrm{\Theta }$. It follows that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma f-A)q,S{y}_n-q〉=\underset{i\to \mathrm{\infty }}{lim}〈(\gamma f-A)q,S{y}_{n_i}-q〉=〈(\gamma f-A)q,w-q〉\le 0.$$

Step 6. We prove $x_n\to q$. By using (3.1) and together with Schwarz inequality, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& =& {\parallel {\xi }_n{P}_C[({\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)S{y}_n)-q]+(1-{\xi }_n)(v_n-q)\parallel }^2\\ \le & {\xi }_n{\parallel P_C[({\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)S{y}_n)-P_C(q)]\parallel }^2+(1-{\xi }_n){\parallel v_n-q\parallel }^2\\ \le & {\xi }_n{\parallel {\alpha }_n(\gamma f(x_n)-Aq)+(I-{\alpha }_{n}A)(S{y}_n-q)\parallel }^2+(1-{\xi }_n){\parallel x_n-q\parallel }^2\\ \le & {\xi }_n{(I-{\alpha }_{n}A)}^2{\parallel S{y}_n-q\parallel }^2+{\xi }_n{\alpha }_n^2{\parallel \gamma f(x_n)-Aq\parallel }^2\\ +2{\xi }_n{\alpha }_n〈(I-{\alpha }_{n}A)(S{y}_n-q),\gamma f(x_n)-Aq〉+(1-{\xi }_n){\parallel x_n-q\parallel }^2\\ \le & {\xi }_n{(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))}^2{\parallel y_n-q\parallel }^2+{\xi }_n{\alpha }_n^2{\parallel \gamma f(x_n)-Aq\parallel }^2\\ +2{\xi }_n{\alpha }_n〈S{y}_n-q,\gamma f(x_n)-Aq〉-2{\xi }_n{\alpha }_n^2〈A(S{y}_n-q),\gamma f(x_n)-Aq〉\\ +(1-{\xi }_n){\parallel x_n-q\parallel }^2\\ \le & {\xi }_n{(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))}^2{\parallel x_n-q\parallel }^2+{\xi }_n{\alpha }_n^2{\parallel \gamma f(x_n)-Aq\parallel }^2\\ +2{\xi }_n{\alpha }_n〈S{y}_n-q,\gamma f(x_n)-\gamma f(q)〉+2{\xi }_n{\alpha }_n〈S{y}_n-q,\gamma f(q)-Aq〉\\ -2{\xi }_n{\alpha }_n^2〈A(S{y}_n-q),\gamma f(x_n)-Aq〉+(1-{\xi }_n){\parallel x_n-q\parallel }^2\\ \le & {\xi }_n{(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))}^2{\parallel x_n-q\parallel }^2+{\xi }_n{\alpha }_n^2{\parallel \gamma f(x_n)-Aq\parallel }^2\\ +2{\xi }_n{\alpha }_n\parallel S{y}_n-q\parallel \parallel \gamma f(x_n)-\gamma f(q)\parallel +2{\xi }_n{\alpha }_n〈S{y}_n-q,\gamma f(q)-Aq〉\\ -2{\xi }_n{\alpha }_n^2〈A(S{y}_n-q),\gamma f(x_n)-Aq〉+(1-{\xi }_n){\parallel x_n-q\parallel }^2\\ \le & {\xi }_n{(1-{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }}))}^2{\parallel x_n-q\parallel }^2+{\xi }_n{\alpha }_n^2{\parallel \gamma f(x_n)-Aq\parallel }^2\\ +2{\xi }_n\gamma \alpha {\alpha }_n\parallel y_n-q\parallel \parallel x_n-q\parallel +2{\xi }_n{\alpha }_n〈S{y}_n-q,\gamma f(q)-Aq〉\\ -2{\xi }_n{\alpha }_n^2〈A(S{y}_n-q),\gamma f(x_n)-Aq〉+(1-{\xi }_n){\parallel x_n-q\parallel }^2\\ \le & ({\xi }_n-2{\xi }_n{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }})+{\xi }_n{\alpha }_n^2{(1-\sqrt{\frac{1-\delta }{\mu }})}^2){\parallel x_n-q\parallel }^2\\ +{\xi }_n{\alpha }_n^2{\parallel \gamma f(x_n)-Aq\parallel }^2+2{\xi }_n\gamma \alpha {\alpha }_n{\parallel x_n-q\parallel }^2\\ +2{\xi }_n{\alpha }_n〈S{y}_n-q,\gamma f(q)-Aq〉-2{\xi }_n{\alpha }_n^2〈A(S{y}_n-q),\gamma f(x_n)-Aq〉\\ +(1-{\xi }_n){\parallel x_n-q\parallel }^2\\ \le & (1-2{\xi }_n{\alpha }_n(1-\sqrt{\frac{1-\delta }{\mu }})+2{\xi }_n\gamma \alpha {\alpha }_n){\parallel x_n-q\parallel }^2\\ +{\alpha }_n\{{\xi }_n{\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2\\ +2{\xi }_n〈S{y}_n-q,\gamma f(q)-Aq〉-2{\xi }_n{\alpha }_n\parallel A(S{y}_n-q)\parallel \parallel \gamma f(x_n)-Aq\parallel \\ +{\xi }_n{\alpha }_n{(1-\sqrt{\frac{1-\delta }{\mu }})}^2{\parallel x_n-q\parallel }^2\}\\ =& (1-2((1-\sqrt{\frac{1-\delta }{\mu }})-\gamma \alpha ){\xi }_n{\alpha }_n){\parallel x_n-q\parallel }^2+{\alpha }_n\{{\xi }_n{\alpha }_n{\parallel \gamma f(x_n)-Aq\parallel }^2\\ +2{\xi }_n〈S{y}_n-q,\gamma f(q)-Aq〉-2{\xi }_n{\alpha }_n\parallel A(S{y}_n-q)\parallel \parallel \gamma f(x_n)-Aq\parallel \\ +{\xi }_n{\alpha }_n{(1-\sqrt{\frac{1-\delta }{\mu }})}^2{\parallel x_n-q\parallel }^2\}.\end{array}$$

Since $\{x_n\}$ is bounded, where $\eta \ge {\xi }_n{\parallel \gamma f(x_n)-Aq\parallel }^2-2{\xi }_n\parallel A(S{y}_n-q)\parallel \parallel \gamma f(x_n)-Aq\parallel +{\xi }_n{(1-\sqrt{\frac{1-\delta }{\mu }})}^2{\parallel x_n-q\parallel }^2$ for all $n\ge 0$. It follows that

$${\parallel x_{n+1}-q\parallel }^2\le (1-2((1-\sqrt{\frac{1-\delta }{\mu }})-\gamma \alpha ){\xi }_n{\alpha }_n){\parallel x_n-q\parallel }^2+{\alpha }_n{\varsigma }_n,$$

(3.46)

where ${\varsigma }_n=2{\xi }_n〈S{y}_n-q,\gamma f(q)-Aq〉+\eta {\alpha }_n$. By ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈(\gamma f-A)q,S{y}_n-q〉\le 0$, we get ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\varsigma }_n\le 0$. Applying Lemma 2.5, we can conclude that $x_n\to q$. This completes the proof. □

Corollary 3.2 Let H be a real Hilbert space, C be a closed convex subset of H. Let$F_1$, $F_2$be two bifunctions of$C\times C$into$\mathcal{R}$satisfying (A 1)-(A 4) and$B,B_1,B_2:C\to H$be$\beta ,\eta ,\rho$-inverse-strongly monotone mappings, ${\psi }_1,{\psi }_2:C\to \mathcal{R}$be convex and lower semicontinuous function, $f:C\to C$be a contraction with coefficient α ($0<\alpha <1$), $M:H\to 2^H$be a maximal monotone mapping. Assume that either (B 1) or (B 2) holds. Let S be a nonexpansive mapping of H into itself such that

$$\mathrm{\Theta }:=F(S)\cap GMEP(F_1,{\psi }_1,B_1)\cap GMEP(F_2,{\psi }_2,B_2)\cap I(B,M)\ne \mathrm{\varnothing }.$$

Suppose$\{x_n\}$is a sequence generated by the following algorithm$x_0\in C$arbitrarily:

$$\{\begin{array}{c}{u}_n=T_{r_n}^{(F_1,{\psi }_1)}(x_n-r_n{B}_1{x}_n),\hfill \\ v_n=T_{s_n}^{(F_2,{\psi }_2)}(x_n-s_n{B}_2{x}_n),\hfill \\ x_{n+1}={\xi }_n{P}_C[{\alpha }_{n}f(x_n)+(I-{\alpha }_n)S{J}_{M,\lambda }(I-\lambda B)u_n]+(1-{\xi }_n)v_n,\hfill \end{array}$$

where$\{{\alpha }_n\},\{{\xi }_n\}\subset (0,1)$, $\lambda \in (0,2\beta )$such that$0<a\le \lambda \le b<2\beta$, $\{r_n\}\in (0,2\eta )$with$0<c\le d\le 1-\eta$and$\{s_n\}\in (0,2\rho )$with$0<e\le f\le 1-\rho$satisfy the conditions (C 1)-(C 4).

Then$\{x_n\}$converges strongly to$q\in \mathrm{\Theta }$, where$q=P_{\mathrm{\Theta }}(f+I)(q)$which solves the following variational inequality:

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

Proof Putting $A\equiv I$ and $\gamma \equiv 1$ in Theorem 3.1, we can obtain desired conclusion immediately. □

Corollary 3.3 Let H be a real Hilbert space, C be a closed convex subset of H. Let$F_1$, $F_2$be two bifunctions of$C\times C$into$\mathcal{R}$satisfying (A 1)-(A 4) and$B,B_1,B_2:C\to H$be$\beta ,\eta ,\rho$-inverse-strongly monotone mappings, ${\psi }_1,{\psi }_2:C\to \mathcal{R}$be convex and lower semicontinuous function. Assume that either (B 1) or (B 2) holds. Let S be a nonexpansive mapping of H into itself such that

$$\mathrm{\Theta }:=F(S)\cap GMEP(F_1,{\psi }_1,B_1)\cap GMEP(F_2,{\psi }_2,B_2)\cap I(B,M)\ne \mathrm{\varnothing }.$$

Suppose$\{x_n\}$is a sequence generated by the following algorithm$x_0\in C$arbitrarily:

$$\{\begin{array}{c}{u}_n=T_{r_n}^{(F_1,{\psi }_1)}(x_n-r_n{B}_1{x}_n),\hfill \\ v_n=T_{s_n}^{(F_2,{\psi }_2)}(x_n-s_n{B}_2{x}_n),\hfill \\ x_{n+1}={\xi }_n{P}_C[{\alpha }_{n}u+(I-{\alpha }_n)S{J}_{M,\lambda }(I-\lambda B)u_n]+(1-{\xi }_n)v_n,\hfill \end{array}$$

where$\{{\alpha }_n\},\{{\xi }_n\}\subset (0,1)$, $\lambda \in (0,2\beta )$such that$0<a\le \lambda \le b<2\beta$, $\{r_n\}\in (0,2\eta )$with$0<c\le d\le 1-\eta$and$\{s_n\}\in (0,2\rho )$with$0<e\le f\le 1-\rho$satisfy the conditions (C 1)-(C 4).

Then$\{x_n\}$converges strongly to$q\in \mathrm{\Theta }$, where$q=P_{\mathrm{\Theta }}(q)$which solves the following variational inequality:

$$〈u-q,p-q〉\le 0,\mathrm{\forall }p\in \mathrm{\Theta }.$$

Proof Putting $f\equiv u\in C$ is a constant in Corollary 3.2, we can obtain desired conclusion immediately. □

Corollary 3.4 Let H be a real Hilbert space, C be a closed convex subset of H. Let$F_1$, $F_2$be two bifunctions of$C\times C$into$\mathcal{R}$satisfying (A 1)-(A 4) and$B,B_1,B_2:C\to H$be$\beta ,\eta ,\rho$-inverse-strongly monotone mappings, ${\psi }_1,{\psi }_2:C\to \mathcal{R}$be convex and lower semicontinuous function, $f:C\to C$be a contraction with coefficient α ($0<\alpha <1$) and A is δ-strongly monotone and μ-strictly pseudo-contraction with$\delta +\mu >1$, γ is a positive real number such that$\gamma <\frac{1}{\alpha }(1-\sqrt{\frac{1-\delta }{\mu }})$. Assume that either (B 1) or (B 2) holds. Let S be a nonexpansive mapping of C into itself such that

$$\mathrm{\Theta }:=F(S)\cap GMEP(F_1,{\psi }_1,B_1)\cap GMEP(F_2,{\psi }_2,B_2)\cap VI(C,B)\ne \mathrm{\varnothing }.$$

Suppose$\{x_n\}$is a sequence generated by the following algorithm$x_0\in C$arbitrarily:

$$\{\begin{array}{c}{u}_n=T_{r_n}^{(F_1,{\psi }_1)}(x_n-r_n{B}_1{x}_n),\hfill \\ v_n=T_{s_n}^{(F_2,{\psi }_2)}(x_n-s_n{B}_2{x}_n),\hfill \\ x_{n+1}={\xi }_n{P}_C[{\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)S{P}_C(I-\lambda B)u_n]+(1-{\xi }_n)v_n,\hfill \end{array}$$

where$\{{\alpha }_n\},\{{\xi }_n\}\subset (0,1)$, $\lambda \in (0,2\beta )$such that$0<a\le \lambda \le b<2\beta$, $\{r_n\}\in (0,2\eta )$with$0<c\le d\le 1-\eta$and$\{s_n\}\in (0,2\rho )$with$0<e\le f\le 1-\rho$satisfy the conditions (C 1)-(C 4).

Then$\{x_n\}$converges strongly to$q\in \mathrm{\Theta }$, where$q=P_{\mathrm{\Theta }}(\gamma f+I-A)(q)$which solves the following variational inequality:

$$〈(\gamma f-A)q,p-q〉\le 0,\mathrm{\forall }p\in \mathrm{\Theta }.$$

Proof Taking $J_{M,\lambda }=P_C$ in Theorem 3.1, we can obtain desired conclusion immediately. □

Corollary 3.5 Let H be a real Hilbert space, C be a closed convex subset of H. Let$f:C\to C$be a contraction with coefficient α ($0<\alpha <1$), A is δ-strongly monotone and μ-strictly pseudo-contraction with$\delta +\mu >1$, γ is a positive real number such that$\gamma <\frac{1}{\alpha }(1-\sqrt{\frac{1-\delta }{\mu }})$. Let S be a nonexpansive mapping of C into itself such that

$$\mathrm{\Theta }:=F(S)\ne \mathrm{\varnothing }.$$

Suppose$\{x_n\}$is a sequence generated by the following algorithm$x_0\in C$arbitrarily:

$$x_{n+1}={\alpha }_n\gamma f(x_n)+(I-{\alpha }_{n}A)S{x}_n,$$

where$\{{\alpha }_n\}\subset (0,1)$and satisfy the condition${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$. Then$\{x_n\}$converges strongly to$q\in \mathrm{\Theta }$, where$q=P_{\mathrm{\Theta }}(\gamma f+I-A)(q)$which solves the following variational inequality:

$$〈(\gamma f-A)q,p-q〉\le 0,\mathrm{\forall }p\in \mathrm{\Theta }.$$

Proof Taking ${\xi }_n\equiv 1$, $P_C\equiv I$ and $B,B_1,B_2\equiv 0$ in Corollary 3.4, we can obtain desired conclusion immediately. □

Remark 3.6 Corollary 3.5 generalizes and improves the result of Marino and Xu [19].

Corollary 3.7 Let H be a real Hilbert space, C be a closed convex subset of H. Let$F_1$, $F_2$be two bifunctions of$C\times C$into$\mathcal{R}$satisfying (A 1)-(A 4) and$B_1,B_2:C\to H$be$\eta ,\rho$-inverse-strongly monotone mappings, ${\psi }_1,{\psi }_2:C\to \mathcal{R}$be convex and lower semicontinuous function, $f:C\to C$be a contraction with coefficient α ($0<\alpha <1$). Assume that either (B 1) or (B 2) holds. Let S be a nonexpansive mapping of C into itself such that

$$\mathrm{\Theta }:=F(S)\cap GMEP(F_1,{\psi }_1,B_1)\cap GMEP(F_2,{\psi }_2,B_2)\ne \mathrm{\varnothing }.$$

Suppose$\{x_n\}$is a sequence generated by the following algorithm$x_0\in C$arbitrarily:

$$\{\begin{array}{c}{u}_n=T_{r_n}^{(F_1,{\psi }_1)}(x_n-r_n{B}_1{x}_n),\hfill \\ v_n=T_{s_n}^{(F_2,{\psi }_2)}(x_n-s_n{B}_2{x}_n),\hfill \\ x_{n+1}={\xi }_n{P}_C[{\alpha }_{n}f(x_n)+(I-{\alpha }_n)S{u}_n]+(1-{\xi }_n)v_n,\hfill \end{array}$$

where$\{{\alpha }_n\},\{{\xi }_n\}\subset (0,1)$, $\{r_n\}\in (0,2\eta )$with$0<c\le d\le 1-\eta$and$\{s_n\}\in (0,2\rho )$with$0<e\le f\le 1-\rho$satisfy the conditions (C 1)-(C 4).

Then$\{x_n\}$converges strongly to$q\in \mathrm{\Theta }$, where$q=P_{\mathrm{\Theta }}(f+I)(q)$which solves the following variational inequality:

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

Proof Taking $\gamma \equiv 1$, $A\equiv I$, $J_{M,\lambda }\equiv I$ and $B\equiv 0$ in Theorem 3.1, we can obtain desired conclusion immediately. □

Remark 3.8 Corollary 3.7 generalizes and improves the result of Yao and Liou [29].