A general iterative method for two maximal monotone operators and 2-generalized hybrid mappings in Hilbert spaces

Abstract

Let C be a closed and convex subset of a real Hilbert space H. Let T be a 2-generalized hybrid mapping of C into itself, let A be an α-inverse strongly-monotone mapping of C into H, and let B and F be maximal monotone operators on $D(B)\subset C$ and $D(F)\subset C$ respectively. The purpose of this paper is to introduce a general iterative scheme for finding a point of $F(T)\cap {(A+B)}^{-1}0\cap F^{-1}0$ which is a unique solution of a hierarchical variational inequality, where $F(T)$ is the set of fixed points of T, ${(A+B)}^{-1}0$ and $F^{-1}0$ are the sets of zero points of $A+B$ and F, respectively. A strong convergence theorem is established under appropriate conditions imposed on the parameters. Further, we consider the problem for finding a common element of the set of solutions of a mathematical model related to mixed equilibrium problems and the set of fixed points of a 2-generalized hybrid mapping in a real Hilbert space.

1 Introduction

Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let ℕ and ℝ be the sets of all positive integers and real numbers, respectively. Let $\phi :C\to \mathbb{R}$ be a real-valued function, and let $f:C\times C\to \mathbb{R}$ be an equilibrium bifunction, that is, $f(u,u)=0$ for each $u\in C$. The mixed equilibrium problem is to find $x\in C$ such that

$$f(x,y)+\phi (y)-\phi (x)\ge 0\text{for all }y\in C.$$

(1.1)

Denote the set of solutions of (1.1) by $\mathit{MEP}(f,\phi )$. In particular, if $\phi =0$, this problem reduces to the equilibrium problem, which is to find $x\in C$ such that

$$f(x,y)\ge 0\text{for all }y\in C.$$

(1.2)

The set of solutions of (1.2) is denoted by $EP(f)$. The problem (1.1) is very general in the sense that it includes, as special cases, optimization problems, variational inequalities, min-max problems, the Nash equilibrium problems in noncooperative games and others; see, for example, Blum-Oettli [1] and Moudafi [2]. Numerous problems in physics, optimization and economics reduce to finding a solution of the problem (1.2).

Let T be a mapping of C into C. We denote by $F(T):=\{x\in C:Tx=x\}$ the set of fixed points of T. A mapping $T:C\to C$ is said to be nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all $x,y\in C$. The mapping $T:C\to C$ is said to be firmly nonexpansive if

$${\parallel Tx-Ty\parallel }^2\le 〈x-y,Tx-Ty〉\text{for all }x,y\in C;$$

(1.3)

see, for instance, Browder [3] and Goebel and Kirk [4]. The mapping $T:C\to C$ is said to be firmly nonspreading [5] if

$$2{\parallel Tx-Ty\parallel }^2\le {\parallel Tx-y\parallel }^2+{\parallel x-Ty\parallel }^2$$

(1.4)

for all $x,y\in C$. Iemoto and Takahashi [6] proved that $T:C\to C$ is nonspreading if and only if

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

(1.5)

for all $x,y\in C$. It is not hard to know that a nonspreading mapping is deduced from a firmly nonexpansive mapping; see [7, 8] and a firmly nonexpansive mapping is a nonexpansive mapping.

In 2010, Kocourek et al. [9] introduced a class of nonlinear mappings, say generalized hybrid mappings. A mapping $T:C\to C$ is said to be generalized hybrid if there are $\alpha ,\beta \in \mathbb{R}$ such that

$$\alpha {\parallel Tx-Ty\parallel }^2+(1-\alpha ){\parallel x-Ty\parallel }^2\le \beta {\parallel Tx-y\parallel }^2+(1-\beta ){\parallel x-y\parallel }^2$$

(1.6)

for all $x,y\in C$. We call such a mapping an $(\alpha ,\beta )$-generalized hybrid mapping. We observe that the mappings above generalize several well-known mappings. For example, an $(\alpha ,\beta )$-generalized hybrid mapping is nonexpansive for $\alpha =1$ and $\beta =0$, nonspreading for $\alpha =2$ and $\beta =1$, and hybrid for $\alpha =\frac{3}{2}$ and $\beta =\frac{1}{2}$.

Recently, Maruyama et al. [10] defined a more general class of nonlinear mappings than the class of generalized hybrid mappings. Such a mapping is a 2-generalized hybrid mapping. A mapping T is called 2-generalized hybrid if there exist ${\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2\in \mathbb{R}$ such that

$$\begin{array}{r}{\alpha }_1{\parallel T^{2}x-Ty\parallel }^2+{\alpha }_2{\parallel Tx-Ty\parallel }^2+(1-{\alpha }_1-{\alpha }_2){\parallel x-Ty\parallel }^2\\ \le {\beta }_1{\parallel T^{2}x-y\parallel }^2+{\beta }_2{\parallel Tx-y\parallel }^2+(1-{\beta }_1-{\beta }_2){\parallel x-y\parallel }^2\end{array}$$

(1.7)

for all $x,y\in C$; see [10] for more details. We call such a mapping an $({\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2)$-generalized hybrid mapping. We can also show that if T is a 2-generalized hybrid mapping and $x=Tx$, then for any $y\in C$,

$$\begin{array}{r}{\alpha }_1{\parallel x-Ty\parallel }^2+{\alpha }_2{\parallel x-Ty\parallel }^2+(1-{\alpha }_1-{\alpha }_2){\parallel x-Ty\parallel }^2\\ \le {\beta }_1{\parallel x-y\parallel }^2+{\beta }_2{\parallel x-y\parallel }^2+(1-{\beta }_1-{\beta }_2){\parallel x-y\parallel }^2,\end{array}$$

and hence $\parallel x-Ty\parallel \le \parallel x-y\parallel$. This means that a 2-generalized hybrid mapping with a fixed point is quasi-nonexpansive. We observe that the 2-generalized hybrid mappings above generalize several well-known mappings. For example, a $(0,{\alpha }_2,0,{\beta }_2)$-generalized hybrid mapping is an $({\alpha }_2,{\beta }_2)$-generalized hybrid mapping in the sense of Kocourek et al. [9].

Recall that a linear bounded operator B is strongly positive if there is a constant $\overline{\gamma }>0$ with the property

$$〈Vx,x〉\ge \overline{\gamma }{\parallel x\parallel }^2\text{for all }x\in H.$$

(1.8)

In general, a nonlinear operator $V:H\to H$ is called strongly monotone if there exists $\overline{\gamma }>0$ such that

$$〈x-y,Vx-Vy〉\ge \overline{\gamma }{\parallel x-y\parallel }^2\text{for all }x,y\in H.$$

(1.9)

Such V is called $\overline{\gamma }$-strongly monotone. A nonlinear operator $V:H\to H$ is called Lipschitzian continuous if there exists $L>0$ such that

$$\parallel Vx-Vy\parallel \le L\parallel x-y\parallel \text{for all }x,y\in H.$$

(1.10)

Such V is called L-Lipschitzian continuous. A mapping $A:C\to H$ is said to be α-inverse-strongly monotone if $〈x-y,Ax-Ay〉\ge \alpha {\parallel Ax-Ay\parallel }^2$ for all $x,y\in C$. It is known that $\parallel Ax-Ay\parallel \le (\frac{1}{\alpha })\parallel x-y\parallel$ for all $x,y\in C$ if A is α-inverse-strongly monotone; see, for example, [11–13].

Many studies have been done for structuring the fixed point of a nonexpansive mapping T. In 1953, Mann [14] introduced the iteration as follows: a sequence $\{x_n\}$ defined by

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)T{x}_n,$$

(1.11)

where the initial guess $x_1\in C$ is arbitrary and $\{{\alpha }_n\}$ is a real sequence in $[0,1]$. It is known that under appropriate settings the sequence $\{x_n\}$ converges weakly to a fixed point of T. However, even in a Hilbert space, Mann iteration may fail to converge strongly; for example, see [15]. Some attempts to construct an iteration method guaranteeing the strong convergence have been made. For example, Halpern [16] proposed the so-called Halpern iteration

$$x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)T{x}_n,$$

(1.12)

where $u,x_1\in C$ are arbitrary and $\{{\alpha }_n\}$ is a real sequence in $[0,1]$ which satisfies ${\alpha }_n\to 0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_n-{\alpha }_{n+1}|<\mathrm{\infty }$. Then $\{x_n\}$ converges strongly to a fixed point of T; see [16, 17].

In 1975, Baillon [18] first introduced the nonlinear ergodic theorem in a Hilbert space as follows:

$$S_{n}x=\frac{1}{n}\sum _{k=0}^{n-1}{T}^{k}x$$

(1.13)

converges weakly to a fixed point of T for some $x\in C$. Recently Hojo et al. [19] proved the strong convergence theorem of Halpern type [20] for 2-generalized hybrid mappings in a Hilbert space as follows.

Theorem 1.1 Let C be a nonempty, closed and convex subset of a Hilbert space H. Let $T:C\to C$ be a 2-generalized hybrid mapping with $F(T)\ne \mathrm{\varnothing }$. Suppose that $\{x_n\}$ is a sequence generated by $x_1=x\in C$, $u\in C$ and

$$x_{n+1}={\gamma }_{n}u+(1-{\gamma }_n)\frac{1}{n}\sum _{k=0}^{n-1}{T}^k{x}_n,\mathrm{\forall }n\in \mathbb{N},$$

(1.14)

where $0\le {\gamma }_n\le 1$, ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$. Then $\{x_n\}$ converges strongly to $P_{F(T)}u$.

Let B be a mapping of H into $2^H$. The effective domain of B is denoted by $D(B)$, that is, $D(B)=\{x\in H:Bx\ne \mathrm{\varnothing }\}$. A multi-valued mapping B on H is said to be monotone if $〈x-y,u-v〉\ge 0$ for all $x,y\in D(B)$, $u\in Bx$, and $v\in By$. A monotone operator B on H is said to be maximal if its graph is not properly contained in the graph of any other monotone operator on H. For a maximal monotone operator B on H and $r>0$, we may define a single-valued operator $J_r={(I+rB)}^{-1}:H\to D(B)$, which is called the resolvent of B for r. We denote by $A_r=\frac{1}{r}(I-J_r)$ the Yosida approximation of B for $r>0$. We know [21] that

$$A_{r}x\in B{J}_{r}x,\mathrm{\forall }x\in H,r>0.$$

(1.15)

Let B be a maximal monotone operator on H, and let $B^{-1}0=\{x\in H:0\in Bx\}$. It is known that the resolvent $J_r$ is firmly nonexpansive and $B^{-1}0=F(J_r)$ for all $r>0$, i.e.,

$$\parallel J_{r}x-J_{r}y\parallel \le 〈x-y,J_{r}x-J_{r}y〉,\mathrm{\forall }x,y\in H.$$

(1.16)

Recently, in the case when $T:C\to C$ is a nonexpansive mapping, $A:C\to H$ is an α-inverse strongly monotone mapping and $B\in H\times H$ is a maximal monotone operator, Takahashi et al. [22] proved a strong convergence theorem for finding a point of $F(T)\cap {(A+B)}^{-1}0$, where $F(T)$ is the set of fixed points of T and ${(A+B)}^{-1}0$ is the set of zero points of $A+B$. In 2011, for finding a point of the set of fixed points of T and the set of zero points of $A+B$ in a Hilbert space, Manaka and Takahashi [23] introduced an iterative scheme as follows:

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)T(J_{{\lambda }_n}(I-{\lambda }_{n}A)x_n),$$

(1.17)

where T is a nonspreading mapping, A is an α-inverse strongly monotone mapping and B is a maximal monotone operator such that $J_{\lambda }={(I-\lambda B)}^{-1}$; $\{{\beta }_n\}$ and $\{{\lambda }_n\}$ are sequences which satisfy $0<c\le {\beta }_n\le d<1$ and $0<a\le {\lambda }_n\le b<2\alpha$. Then they proved that $\{x_n\}$ converges weakly to a point $p={lim}_{n\to \mathrm{\infty }}{P}_{F(T)\cap {(A+B)}^{-1}(0)}{x}_n$.

Very recently, Liu et al. [24] generalized the iterative algorithm (1.17) for finding a common element of the set of fixed points of a nonspreading mapping T and the set of zero points of a monotone operator $A+B$ (A is an α-inverse strongly monotone mapping and B is a maximal monotone operator). More precisely, they introduced the following iterative scheme:

$$\{\begin{array}{c}{x}_1=x\in H\text{arbitrarily},\hfill \\ z_n=J_{{\lambda }_n}(I-{\lambda }_{n}A)x_n,\hfill \\ y_n=\frac{1}{n}{\sum }_{k=0}^{n-1}{T}^k{z}_n,\hfill \\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)y_n\text{for all }n\in \mathbb{N},\hfill \end{array}$$

(1.18)

where $\{{\alpha }_n\}$ is an appropriate sequence in $[0,1]$. They obtained strong convergence theorems about a common element of the set of fixed points of a nonspreading mapping and the set of zero points of an α-inverse strongly monotone mapping and a maximal monotone operator in a Hilbert space.

On the other hand, iterative methods for nonexpansive mappings have recently been applied to solve convex minimization problems; see, e.g., [25–28] and the references therein. Convex minimization problems have a great impact and influence on the development of almost all branches of pure and applied sciences. A typical problem is to minimize a quadratic function over the set of fixed points a nonexpansive mapping on a real Hilbert space:

$$\theta (x)=\underset{x\in C}{min}\frac{1}{2}〈Vx,x〉-〈x,b〉,$$

(1.19)

where V is a linear bounded operator, C is the fixed point set of a nonexpansive mapping T and b is a given point in H. Let H be a real Hilbert space. In [29], Marino and Xu introduced the following general iterative scheme based on the viscosity approximation method introduced by Moudafi [30]:

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

(1.20)

where V is a strongly positive bounded linear operator on H. They proved that if the sequence $\{{\alpha }_n\}$ of parameters satisfies appropriate conditions, then the sequence $\{x_n\}$ generated by (1.20) converges strongly to the unique solution of the variational inequality

$$〈(V-\gamma f)x^{\ast },x-x^{\ast }〉\ge 0,x\in C,$$

which is the optimality condition for the minimization problem

$$\underset{x\in C}{min}\frac{1}{2}〈Vx,x〉-h(x),$$

(1.21)

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

Recently, Tian [31] introduced the following general iterative scheme based on the viscosity approximation method induced by a $\overline{\gamma }$-strongly monotone and a L-Lipschitzian continuous operator V on H

$$x_{n+1}={\alpha }_n\gamma g(x_n)+(I-\mu {\alpha }_{n}V)T{x}_n,$$

for all $n\in \mathbb{N}$, where $\mu ,\gamma \in \mathbb{R}$ satisfying $0<\mu <\frac{2\overline{\gamma }}{L^2}$, $0<\gamma <\mu (\overline{\gamma }-\frac{L^2\mu }{2})/k$, g is a k-contraction of H into itself and T is a nonexpansive mapping on H. It is proved, under some restrictions on the parameters, in [31] that $\{x_n\}$ converges strongly to a point $p_0\in F(T)$ which is a unique solution of the variational inequality

$$〈(V-\gamma g)p_0,q-p_0〉\ge 0,\mathrm{\forall }q\in F(T).$$

Very recently, Lin and Takahashi [32] obtained the strong convergence theorem for finding a point $p_0\in {(A+B)}^{-1}0\cap F^{-1}0$ which is a unique solution of a hierarchical variational inequality, where A is an α-inverse strongly-monotone mapping of C into H, and B and F are maximal monotone operators on $D(B)\subset C$ and $D(F)\subset C$, respectively. More precisely, they introduced the following iterative scheme: Let $x_1=x\in H$ and let $\{x_n\}\subset H$ be a sequence generated

$$x_{n+1}={\alpha }_n\gamma g(x_n)+(I-{\alpha }_{n}V)J_{{\lambda }_n}(I-{\lambda }_{n}A)T_{r_n}{x}_n\text{for all }n\in \mathbb{N},$$

(1.22)

where $\{{\alpha }_n\}\subset (0,1)$, $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$ and $\{r_n\}\subset (0,\mathrm{\infty })$ satisfy certain appropriate conditions, $J_{\lambda }={(I+\lambda B)}^{-1}$ and $T_r={(I+rF)}^{-1}$ are the resolvents of B for $\lambda >0$ and F for $r>0$, respectively.

In this paper, motivated by the mentioned results, let C be a closed and convex subset of a real Hilbert space H. Let T be a 2-generalized hybrid mapping of C into itself, let A be an α-inverse strongly-monotone mapping of C into H, and let B and F be maximal monotone operators on $D(B)\subset C$ and $D(F)\subset C$ respectively. We introduce a new general iterative scheme for finding a common element of $F(T)\cap {(A+B)}^{-1}0\cap F^{-1}0$ which is a unique solution of a hierarchical variational inequality, where $F(T)$ is the set of fixed points of T, ${(A+B)}^{-1}0$ and $F^{-1}0$ are the sets of zero points of $A+B$ and F, respectively. Then, we prove a strong convergence theorem. Further, we consider the problem for finding a common element of the set of solutions of a mathematical model related to mixed equilibrium problems and the set of fixed points of a 2-generalized hybrid mapping 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. The nearest point projection of H onto C is denoted by $P_C$, that is, $\parallel x-P_{C}x\parallel \le \parallel x-y\parallel$ for all $x\in H$ and $y\in C$. Such $P_C$ is called the metric projection of H onto C. We know that the metric projection $P_C$ is firmly nonexpansive, i.e.,

$${\parallel P_{C}x-P_{C}y\parallel }^2\le 〈P_{C}x-P_{C}y,x-y〉$$

(2.1)

for all $x,y\in H$. Furthermore, $〈P_{C}x-P_{C}y,x-y〉\le 0$ holds for all $x\in H$ and $y\in C$; see [33]. Let $\alpha >0$ be a given constant.

We also know the following lemma from [22].

Lemma 2.1 Let H be a real Hilbert space, and let B be a maximal monotone operator on H. For $r>0$ and $x\in H$, define the resolvent $J_{r}x$. Then the following holds:

$$\frac{s-t}{s}〈J_{s}x-J_{t}x,J_{s}x-x〉\ge {\parallel J_{s}x-J_{t}x\parallel }^2$$

(2.2)

for all $s,t>0$ and $x\in H$.

From Lemma 2.1, we have that

$$\parallel J_{\lambda }x-J_{\mu }x\parallel \le (|\lambda -\mu |/\lambda )\parallel x-J_{\lambda }x\parallel$$

(2.3)

for all $\lambda ,\mu >0$ and $x\in H$; see also [33, 34]. To prove our main result, we need the following lemmas.

Remark 2.2 It is not hard to know that if A is an α-inverse strongly monotone mapping, then it is $\frac{1}{\alpha }$ -Lipschitzian and hence uniformly continuous. Clearly, the class of monotone mappings includes the class of α-inverse strongly monotone mappings.

Remark 2.3 It is well known that if $T:C\to C$ is a nonexpansive mapping, then $I-T$ is $\frac{1}{2}$-inverse strongly monotone, where I is the identity mapping on H; see, for instance, [21]. It is known that the resolvent $J_r$ is firmly nonexpansive and $B^{-1}0=F(J_r)$ for all $r>0$.

Lemma 2.4 [23]

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let $\alpha >0$. Let A be an α-inverse strongly monotone mapping of C into H, and let B be a maximal monotone operator on H such that the domain of B is included in C. Let $J_{\lambda }={(I+\lambda B)}^{-1}$ be the resolvent of B for any $\lambda >0$. Then the following hold:

• (i) if $u,v\in {(A+B)}^{-1}(0)$, then $Au=Av$;

• (ii) for any $\lambda >0$, $u\in {(A+B)}^{-1}(0)$ if and only if $u=J_{\lambda }(I-\lambda A)u$.

Lemma 2.5 [26, 35]

Let $\{a_n\}$ be a sequence of nonnegative real numbers satisfying the property

$$a_{n+1}\le (1-t_n)a_n+b_n+t_n{c}_n,$$

where $\{t_n\}$, $\{b_n\}$ and $\{c_n\}$ satisfy the restrictions:

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

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{b}_n<\mathrm{\infty }$;

• (iii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{c}_n\le 0$.

Then $\{a_n\}$ converges to zero as $n\to \mathrm{\infty }$.

Lemma 2.6 [32]

Let H be a Hilbert space, and let $g:H\to H$ be a k-contraction with $0<k<1$. Let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator on H with $\overline{\gamma }>0$ and $L>0$. Let a real number γ satisfy $0<\gamma <\frac{\overline{\gamma }}{k}$. Then $V-\gamma g:H\to H$ is a $(\overline{\gamma }-\gamma k)$-strongly monotone and $(L+\gamma k)$-Lipschitzian continuous mapping. Furthermore, let C be a nonempty closed convex subset of H. Then $P_C(I-V+\gamma g)$ has a unique fixed point $z_0$ in C. This point $z_0\in C$ is also a unique solution of the variational inequality

$$〈(V-\gamma f)z_0,q-z_0〉\ge 0,\mathrm{\forall }q\in C.$$

3 Main results

In this section, we are in a position to propose a new general iterative sequence for 2-generalized hybrid mappings and establish a strong convergence theorem for the proposed sequence.

Theorem 3.1 Let H be a real Hilbert space, and let C be a nonempty, closed and convex subset of H. Let $\alpha >0$ and A be an α-inverse-strongly monotone mapping of C into H. Let the set-valued maps $B:D(B)\subset C\to 2^H$ and $F:D(F)\subset C\to 2^H$ be maximal monotone. Let $J_{\lambda }={(I+\lambda B)}^{-1}$ and $T_r={(I+rF)}^{-1}$ be the resolvents of B for $\lambda >0$ and F for $r>0$, respectively. Let $0<k<1$ and let g be a k-contraction of H into itself. Let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$. Let $T:C\to C$ be a 2-generalized hybrid mapping such that $\mathrm{\Omega }:=F(T)\cap {(A+B)}^{-1}0\cap F^{-1}0\ne \mathrm{\varnothing }$. Take $\mu ,\gamma \in \mathbb{R}$ as follows:

$$0<\mu <\frac{2\overline{\gamma }}{L^2},0<\gamma <\frac{\overline{\gamma }-\frac{L^2\mu }{2}}{k}.$$

Let the sequence $\{x_n\}\subset H$ be generated by

$$\{\begin{array}{c}{x}_1=x\in H\mathit{\text{arbitrarily}},\hfill \\ z_n=J_{{\lambda }_n}(I-{\lambda }_{n}A)T_{r_n}{x}_n,\hfill \\ y_n=\frac{1}{n}{\sum }_{k=0}^{n-1}{T}^k{z}_n,\hfill \\ x_{n+1}={\alpha }_n\gamma g(x_n)+(I-{\alpha }_{n}V)y_n,\mathrm{\forall }n=1,2,\dots ,\hfill \end{array}$$

(3.1)

where the sequences $\{{\alpha }_n\}$, $\{{\lambda }_n\}$ and $\{r_n\}$ satisfy the following restrictions:

• (i) $\{{\alpha }_n\}\subset [0,1]$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) there exist constants a and b such that $0<a\le {\lambda }_n\le b<2\alpha$ for all $n\in \mathbb{N}$;

• (iii) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$.

Then $\{x_n\}$ converges strongly to a point $p_0$ of Ω, where $p_0$ is a unique fixed point of $P_{\mathrm{\Omega }}(I-V+\gamma g)$. This point $p_0\in \mathrm{\Omega }$ is also a unique solution of the hierarchical variational inequality

$$〈(V-\gamma g)p_0,q-p_0〉\ge 0,\mathrm{\forall }q\in \mathrm{\Omega }.$$

(3.2)

Proof First we prove that $\{x_n\}$ is bounded and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists for all $p\in \mathrm{\Omega }$. Let $p\in \mathrm{\Omega }$, we have that $p=J_{{\lambda }_n}(I-{\lambda }_{n}A)p$ and $p=T_{r_n}p$. Putting $u_n=T_{r_n}{x}_n$, we have that

$$\begin{array}{rcl}{\parallel z_n-p\parallel }^2& =& {\parallel J_{{\lambda }_n}(I-{\lambda }_{n}A)T_{r_n}{x}_n-J_{{\lambda }_n}(I-{\lambda }_{n}A)p\parallel }^2\\ \le & {\parallel (T_{r_n}{x}_n-T_{r_n}p)-{\lambda }_n(A{T}_{r_n}{x}_n-A{T}_{r_n}p)\parallel }^2\\ =& {\parallel T_{r_n}{x}_n-T_{r_n}p\parallel }^2-2{\lambda }_n〈u_n-p,A{u}_n-Ap〉+{\lambda }_n^2{\parallel A{u}_n-Ap\parallel }^2\\ \le & {\parallel u_n-p\parallel }^2-2{\lambda }_n\alpha {\parallel A{u}_n-Ap\parallel }^2+{\lambda }_n^2{\parallel A{u}_n-Ap\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{u}_n-Ap\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2.\end{array}$$

(3.3)

This together with quasi-nonexpansiveness of T implies that

$$\begin{array}{rcl}\parallel y_n-p\parallel & =& \parallel \frac{1}{n}\sum _{k=0}^{n-1}{T}^k{z}_n-p\parallel \\ \le & \frac{1}{n}\sum _{k=0}^{n-1}\parallel T^k{z}_n-p\parallel \\ \le & \frac{1}{n}\sum _{k=0}^{n-1}\parallel z_n-p\parallel \\ =& \parallel z_n-p\parallel \le \parallel x_n-p\parallel .\end{array}$$

(3.4)

Therefore, we have

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & =& \parallel {\alpha }_n(\gamma g(x_n)-Vp)+(I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel \\ \le & {\alpha }_n\parallel \gamma g(x_n)-Vp\parallel +\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel \\ \le & {\alpha }_n\gamma k\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma g(p)-Vp\parallel +\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel .\end{array}$$

(3.5)

Putting $\tau =\overline{\gamma }-\frac{L^2\mu }{2}$, we can calculate the following:

$$\begin{array}{rcl}{\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel }^2& =& {\parallel (y_n-p)-{\alpha }_n(V{y}_n-Vp)\parallel }^2\\ =& {\parallel y_n-p\parallel }^2-2{\alpha }_n〈y_n-p,V{y}_n-Vp〉+{\alpha }_n^2{\parallel V{y}_n-Vp\parallel }^2\\ \le & {\parallel y_n-p\parallel }^2-2{\alpha }_n\overline{\gamma }{\parallel y_n-p\parallel }^2+{\alpha }_n^2{L}^2{\parallel y_n-p\parallel }^2\\ =& (1-2{\alpha }_n\overline{\gamma }+{\alpha }_n^2{L}^2){\parallel y_n-p\parallel }^2\\ =& (1-2{\alpha }_n\tau -{\alpha }_n{L}^2\mu +{\alpha }_n^2{L}^2){\parallel y_n-p\parallel }^2\\ \le & (1-2{\alpha }_n\tau -{\alpha }_n(L^2\mu -{\alpha }_n{L}^2)+{\alpha }_n^2{\tau }^2){\parallel y_n-p\parallel }^2\\ \le & (1-2{\alpha }_n\tau +{\alpha }_n^2{\tau }^2){\parallel y_n-p\parallel }^2\\ =& {(1-{\alpha }_n\tau )}^2{\parallel y_n-p\parallel }^2.\end{array}$$

(3.6)

Since $1-{\alpha }_n\tau >0$, we obtain that

$$\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel \le (1-{\alpha }_n\tau )\parallel y_n-p\parallel .$$

Therefore, by (3.5), we have

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & \le & {\alpha }_n\gamma k\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma g(p)-Vp\parallel +(1-{\alpha }_n\tau )\parallel y_n-p\parallel \\ \le & {\alpha }_n\gamma k\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma g(p)-Vp\parallel +(1-{\alpha }_n\tau )\parallel x_n-p\parallel \\ =& (1-{\alpha }_n(\tau -\gamma k))\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma g(p)-Vp\parallel \\ =& (1-{\alpha }_n(\tau -\gamma k))\parallel x_n-p\parallel +{\alpha }_n(\tau -\gamma k)\frac{\parallel \gamma g(p)-Vp\parallel }{\tau -\gamma k}\\ \le & max\{\parallel x_n-p\parallel ,\frac{\parallel \gamma g(p)-Vp\parallel }{\tau -\gamma k}\}\text{for all }n\in \mathbb{N},\end{array}$$

which yields that the sequence $\{\parallel x_n-p\parallel \}$ is bounded, so are $\{x_n\}$, $\{y_n\}$, $\{V{y}_n\}$, $\{g(x_n)\}$ and $\{T^n{z}_n\}$. Using Lemma 2.6, we can take a unique $p_0\in \mathrm{\Omega }$ of the hierarchical variational inequality

$$〈(V-\gamma g)p_0,q-p_0〉\ge 0,\mathrm{\forall }q\in \mathrm{\Omega }.$$

(3.7)

We show that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈(V-\gamma g)p_0,x_n-p_0〉\ge 0$. We may assume, without loss of generality, that there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ converging to $w\in C$, as $k\to \mathrm{\infty }$, such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(V-\gamma g)p_0,x_n-p_0〉=\underset{k\to \mathrm{\infty }}{lim}〈(V-\gamma g)p_0,x_{n_k}-p_0〉.$$

Since $\{\parallel x_{n_k}-p\parallel \}$ is bounded, there exists a subsequence $\{x_{n_{k_i}}\}$ of $\{x_{n_k}\}$ such that ${lim}_{i\to \mathrm{\infty }}\parallel x_{n_{k_i}}-p\parallel$ exists. Now we shall prove that $w\in \mathrm{\Omega }$.

(a) We first prove $w\in F(T)$. We notice that

$$\parallel x_{n+1}-y_n\parallel =\parallel {\alpha }_n\gamma g(x_n)+(I-{\alpha }_{n}V)y_n-y_n\parallel ={\alpha }_n\parallel \gamma g(x_n)-V{y}_n\parallel .$$

In particular, replacing n by $n_{k_i}$ and taking $i\to \mathrm{\infty }$ in the last equality, we have

$$\underset{i\to \mathrm{\infty }}{lim}\parallel x_{n_{k_i}+1}-y_{n_{k_i}}\parallel =0,$$

so we have $y_{n_{k_i}}\rightharpoonup w$. Since T is 2-generalized hybrid, there exist ${\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2\in \mathbb{R}$ such that

$$\begin{array}{r}{\alpha }_1{\parallel T^{2}x-Ty\parallel }^2+{\alpha }_2{\parallel Tx-Ty\parallel }^2+(1-{\alpha }_1-{\alpha }_2){\parallel x-Ty\parallel }^2\\ \le {\beta }_1{\parallel T^{2}x-y\parallel }^2+{\beta }_2{\parallel Tx-y\parallel }^2+(1-{\beta }_1-{\beta }_2){\parallel x-y\parallel }^2\end{array}$$

for all $x,y\in C$. For any $n\in \mathbb{N}$ and $k=0,1,2,\dots ,n-1$, we compute the following:

$$\begin{array}{rcl}0& \le & {\beta }_1{\parallel T^2{T}^k{z}_n-y\parallel }^2+{\beta }_2{\parallel T{T}^k{z}_n-y\parallel }^2+(1-{\beta }_1-{\beta }_2){\parallel T^k{z}_n-y\parallel }^2\\ -{\alpha }_1{\parallel T^2{T}^k{z}_n-Ty\parallel }^2-{\alpha }_2{\parallel T{T}^k{z}_n-Ty\parallel }^2-(1-{\alpha }_1-{\alpha }_2){\parallel T^k{z}_n-Ty\parallel }^2\\ =& {\beta }_1{\parallel T^{k+2}{z}_n-y\parallel }^2+{\beta }_2{\parallel T^{k+1}{z}_n-y\parallel }^2+(1-{\beta }_1-{\beta }_2){\parallel T^k{z}_n-y\parallel }^2\\ -{\alpha }_1{\parallel T^{k+2}{z}_n-Ty\parallel }^2-{\alpha }_2{\parallel T^{k+1}{z}_n-Ty\parallel }^2-(1-{\alpha }_1-{\alpha }_2){\parallel T^k{z}_n-Ty\parallel }^2\\ \le & {\beta }_1\{{\parallel T^{k+2}{z}_n-Ty\parallel }^2+{\parallel Ty-y\parallel }^2\}+{\beta }_2\{{\parallel T^{k+1}{z}_n-Ty\parallel }^2+{\parallel Ty-y\parallel }^2\}\\ +(1-{\beta }_1-{\beta }_2)\{{\parallel T^k{z}_n-Ty\parallel }^2+{\parallel Ty-y\parallel }^2\}-{\alpha }_1{\parallel T^{k+2}{z}_n-Ty\parallel }^2\\ -{\alpha }_2{\parallel T^{k+1}{z}_n-Ty\parallel }^2-(1-{\alpha }_1-{\alpha }_2){\parallel T^k{z}_n-Ty\parallel }^2\\ =& {\beta }_1\{{\parallel T^{k+2}{z}_n-Ty\parallel }^2+{\parallel Ty-y\parallel }^2+2〈T^{k+2}{z}_n-Ty,Ty-y〉\}\\ +{\beta }_2\{{\parallel T^{k+1}{z}_n-Ty\parallel }^2+{\parallel Ty-y\parallel }^2+2〈T^{k+1}{z}_n-Ty,Ty-y〉\}\\ +(1-{\beta }_1-{\beta }_2)\{{\parallel T^k{z}_n-Ty\parallel }^2+{\parallel Ty-y\parallel }^2+2〈T^k{z}_n-Ty,Ty-y〉\}\\ -{\alpha }_1{\parallel T^{k+2}{z}_n-Ty\parallel }^2-{\alpha }_2{\parallel T^{k+1}{z}_n-Ty\parallel }^2-(1-{\alpha }_1-{\alpha }_2){\parallel T^k{z}_n-Ty\parallel }^2\\ =& ({\beta }_1-{\alpha }_1){\parallel T^{k+2}{z}_n-Ty\parallel }^2+({\beta }_2-{\alpha }_2){\parallel T^{k+1}{z}_n-Ty\parallel }^2\\ +({\alpha }_1+{\alpha }_2-{\beta }_1-{\beta }_2){\parallel T^k{z}_n-Ty\parallel }^2\\ \times ({\beta }_1+{\beta }_2+1-{\beta }_1-{\beta }_2){\parallel Ty-y\parallel }^2+2〈{\beta }_1{T}^{k+2}{z}_n-{\beta }_{1}Ty+{\beta }_2{T}^{k+1}{z}_n-{\beta }_{2}Ty\\ +(1-{\beta }_1-{\beta }_2)T^k{z}_n-(1-{\beta }_1-{\beta }_2)Ty,Ty-y〉\\ =& ({\beta }_1-{\alpha }_1){\parallel T^{k+2}{z}_n-Ty\parallel }^2+({\beta }_2-{\alpha }_2){\parallel T^{k+1}{z}_n-Ty\parallel }^2\\ -(({\beta }_1-{\alpha }_1)+({\alpha }_2-{\beta }_2)){\parallel T^k{z}_n-Ty\parallel }^2+{\parallel Ty-y\parallel }^2\\ +2〈{\beta }_1{T}^{k+2}{z}_n+{\beta }_2{T}^{k+1}{z}_n+(1-{\beta }_1-{\beta }_2)T^k{z}_n-Ty,Ty-y〉\\ =& ({\beta }_1-{\alpha }_1)({\parallel T^{k+2}{z}_n-Ty\parallel }^2-{\parallel T^k{z}_n-Ty\parallel }^2)\\ +({\beta }_2-{\alpha }_2)({\parallel T^{k+1}{z}_n-Ty\parallel }^2-{\parallel T^k{z}_n-Ty\parallel }^2)\\ +{\parallel Ty-y\parallel }^2+2〈{\beta }_1{T}^{k+2}{z}_n+{\beta }_2{T}^{k+1}{z}_n+(1-{\beta }_1-{\beta }_2)T^k{z}_n-Ty,Ty-y〉\\ =& {\parallel Ty-y\parallel }^2+2〈T^k{z}_n-Ty,Ty-y〉\\ +2〈{\beta }_1(T^{k+2}{z}_n-T^k{x}_n)+{\beta }_2(T^{k+1}{z}_n-T^k{z}_n),Ty-y〉\\ +({\beta }_1-{\alpha }_1)({\parallel T^{k+2}{z}_n-Ty\parallel }^2-{\parallel T^k{z}_n-Ty\parallel }^2)\\ +({\beta }_2-{\alpha }_2)({\parallel T^{k+1}{z}_n-Ty\parallel }^2-{\parallel T^k{z}_n-Ty\parallel }^2).\end{array}$$

Summing up these inequalities from $k=0$ to $n-1$, we get

$$\begin{array}{rcl}0& \le & \sum _{k=0}^{n-1}{\parallel Ty-y\parallel }^2+2〈\sum _{k=0}^{n-1}(T^k{z}_n-Ty),Ty-y〉\\ +2〈{\beta }_1\sum _{k=0}^{n-1}(T^{k+2}{z}_n-T^k{z}_n)+{\beta }_2\sum _{k=0}^{n-1}(T^{k+1}{z}_n-T^k{z}_n),Ty-y〉\\ +({\beta }_1-{\alpha }_1)\sum _{k=0}^{n-1}({\parallel T^{k+2}{z}_n-Ty\parallel }^2-{\parallel T^k{z}_n-Ty\parallel }^2)\\ +({\beta }_2-{\alpha }_2)\sum _{k=0}^{n-1}({\parallel T^{k+1}{z}_n-Ty\parallel }^2-{\parallel T^k{z}_n-Ty\parallel }^2)\\ =& n{\parallel Ty-y\parallel }^2+2〈\sum _{k=0}^{n-1}{T}^k{z}_n-nTy,Ty-y〉\\ +2〈{\beta }_1(T^{n+1}{z}_n-T^n{z}_n-z_n-T{z}_n)+{\beta }_2(T^n{z}_n-z_n),Ty-y〉\\ +({\beta }_1-{\alpha }_1)({\parallel T^{n+1}{z}_n-Ty\parallel }^2+{\parallel T^n{z}_n-Ty\parallel }^2-{\parallel z_n-Ty\parallel }^2-{\parallel T{z}_n-Ty\parallel }^2)\\ +({\beta }_2-{\alpha }_2)({\parallel T^n{z}_n-Ty\parallel }^2-{\parallel z_n-Ty\parallel }^2).\end{array}$$

Dividing this inequality by n, we get

$$\begin{array}{rcl}0& \le & {\parallel Ty-y\parallel }^2+2〈y_n-Ty,Ty-y〉\\ +2〈\frac{1}{n}{\beta }_1(T^{n+1}{z}_n-T^n{z}_n-z_n-T{z}_n)+\frac{1}{n}{\beta }_2(T^n{z}_n-z_n),Ty-y〉\\ +\frac{1}{n}({\beta }_1-{\alpha }_1)({\parallel T^{n+1}{z}_n-Ty\parallel }^2+{\parallel T^n{z}_n-Ty\parallel }^2-{\parallel z_n-Ty\parallel }^2-{\parallel T{z}_n-Ty\parallel }^2)\\ +\frac{1}{n}({\beta }_2-{\alpha }_2)({\parallel T^n{z}_n-Ty\parallel }^2-{\parallel z_n-Ty\parallel }^2).\end{array}$$

Replacing n by $n_{k_i}$ and letting $i\to \mathrm{\infty }$ in the last inequality, we have

$$0\le {\parallel Ty-y\parallel }^2+2〈w-Ty,Ty-y〉\text{for all }y\in C.$$

(3.8)

In particular, replacing y by w in (3.8), we obtain that

$$0\le {\parallel Tw-w\parallel }^2+2〈w-Tw,Tw-w〉=-{\parallel Tw-w\parallel }^2,$$

which ensures that $w\in F(T)$.

(b) We prove that $w\in {(A+B)}^{-1}0$. From (3.3), (3.4) and (3.6), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & {\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel y_n-p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel z_n-p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {(1-{\alpha }_n\tau )}^2\{{\parallel x_n-p\parallel }^2-{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{u}_n-Ap\parallel }^2\}\\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ =& (1-2{\alpha }_n\tau +{\alpha }_n^2{\tau }^2){\parallel x_n-p\parallel }^2-{(1-{\alpha }_n\tau )}^2{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{u}_n-Ap\parallel }^2\\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {\parallel x_n-p\parallel }^2+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2-{(1-{\alpha }_n\tau )}^2{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{u}_n-Ap\parallel }^2\\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉,\end{array}$$

(3.9)

and hence

$$\begin{array}{rcl}{(1-{\alpha }_n\tau )}^2{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{u}_n-Ap\parallel }^2& \le & {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2\\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉.\end{array}$$

(3.10)

Replacing n by $n_{k_i}$ in (3.10), we have

$$\begin{array}{r}{(1-{\alpha }_{n_{k_i}}\tau )}^2{\lambda }_{n_{k_i}}(2\alpha -{\lambda }_{n_{k_i}}){\parallel A{u}_{n_{k_i}}-Ap\parallel }^2\\ \le {\parallel x_{n_{k_i}}-p\parallel }^2-{\parallel x_{n_{k_i}+1}-p\parallel }^2+{\alpha }_{n_{k_i}}^2{\tau }^2{\parallel x_{n_{k_i}}-p\parallel }^2\\ +2{\alpha }_{n_{k_i}}〈\gamma g(x_n)-Vp,x_{n_{k_i}+1}-p〉.\end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, $0<a\le {\lambda }_n\le b<2\alpha$ and the existence of ${lim}_{i\to \mathrm{\infty }}\parallel x_{n_{k_i}}-p\parallel$, we have

$$\underset{i\to \mathrm{\infty }}{lim}\parallel A{u}_{n_{k_i}}-Ap\parallel =0.$$

(3.11)

We also have from (1.16) that

$$\begin{array}{rcl}2{\parallel u_n-p\parallel }^2& =& 2{\parallel T_{r_n}{x}_n-T_{r_n}p\parallel }^2\\ \le & 2〈x_n-p,u_n-p〉\\ =& {\parallel x_n-p\parallel }^2+{\parallel u_n-p\parallel }^2-{\parallel u_n-x_n\parallel }^2,\end{array}$$

and hence

$${\parallel u_n-p\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel u_n-x_n\parallel }^2.$$

(3.12)

From (3.3), (3.4), (3.6) and (3.12), we obtain the following:

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & {\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel y_n-p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel z_n-p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {(1-{\alpha }_n\tau )}^2\{{\parallel u_n-p\parallel }^2-{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{u}_n-Ap\parallel }^2\}\\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {(1-{\alpha }_n\tau )}^2\{{\parallel x_n-p\parallel }^2-{\parallel u_n-x_n\parallel }^2\}\\ -{(1-{\alpha }_n\tau )}^2{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{u}_n-Ap\parallel }^2\\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & (1-2{\alpha }_n\tau +{\alpha }_n^2{\tau }^2){\parallel x_n-p\parallel }^2-{(1-{\alpha }_n\tau )}^2{\parallel u_n-x_n\parallel }^2\\ -{(1-{\alpha }_n\tau )}^2{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{u}_n-Ap\parallel }^2\\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {\parallel x_n-p\parallel }^2+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2-{(1-{\alpha }_n\tau )}^2{\parallel u_n-x_n\parallel }^2\\ -{(1-{\alpha }_n\tau )}^2{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{u}_n-Ap\parallel }^2\\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉,\end{array}$$

and hence

$$\begin{array}{rcl}{(1-{\alpha }_n\tau )}^2{\parallel u_n-x_n\parallel }^2& \le & {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2\\ -{(1-{\alpha }_n\tau )}^2{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{u}_n-Ap\parallel }^2\\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉.\end{array}$$

(3.13)

Replacing n by $n_{k_i}$ in (3.13), we have

$$\begin{array}{rcl}{(1-{\alpha }_{n_{k_i}}\tau )}^2{\parallel u_{n_{k_i}}-x_{n_{k_i}}\parallel }^2& \le & {\parallel x_{n_{k_i}}-p\parallel }^2-{\parallel x_{n_{k_i}+1}-p\parallel }^2+{\alpha }_{n_{k_i}}^2{\tau }^2{\parallel x_{n_{k_i}}-p\parallel }^2\\ -{(1-{\alpha }_{n_{k_i}}\tau )}^2{\lambda }_{n_{k_i}}(2\alpha -{\lambda }_{n_{k_i}}){\parallel A{u}_{n_{k_i}}-Ap\parallel }^2\\ +2{\alpha }_{n_{k_i}}〈\gamma g(x_{n_{k_i}})-Vp,x_{n_{k_i}+1}-p〉.\end{array}$$

From (3.11), ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and the existence of ${lim}_{i\to \mathrm{\infty }}\parallel x_{n_{k_i}}-p\parallel$, we have

$$\underset{i\to \mathrm{\infty }}{lim}\parallel u_{n_{k_i}}-x_{n_{k_i}}\parallel =0.$$

(3.14)

On the other hand, since $J_{{\lambda }_n}$ is firmly nonexpansive and $u_n=T_{r_n}{x}_n$, we have that

$$\begin{array}{rcl}{\parallel z_n-p\parallel }^2& =& {\parallel J_{{\lambda }_n}(I-{\lambda }_{n}A)u_n-J_{{\lambda }_n}(I-{\lambda }_{n}A)p\parallel }^2\\ \le & 〈z_n-p,(I-{\lambda }_{n}A)u_n-(I-{\lambda }_{n}A)p〉\\ =& \frac{1}{2}({\parallel z_n-p\parallel }^2+{\parallel (I-{\lambda }_{n}A)u_n-(I-{\lambda }_{n}A)p\parallel }^2\\ -{\parallel z_n-p-(I-{\lambda }_{n}A)u_n+(I-{\lambda }_{n}A)p\parallel }^2)\\ \le & \frac{1}{2}\{{\parallel z_n-p\parallel }^2+{\parallel u_n-p\parallel }^2-{\parallel z_n-p-(I-{\lambda }_{n}A)u_n+(I-{\lambda }_{n}A)p\parallel }^2\}\\ \le & \frac{1}{2}({\parallel z_n-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel z_n-u_n\parallel }^2-2{\lambda }_n〈z_n-u_n,A{u}_n-Ap〉\\ -{\lambda }_n^2{\parallel A{u}_n-Ap\parallel }^2),\end{array}$$

and hence

$$\begin{array}{r}{\parallel z_n-p\parallel }^2\\ \le {\parallel x_n-p\parallel }^2-{\parallel z_n-u_n\parallel }^2-2{\lambda }_n〈z_n-u_n,A{u}_n-Ap〉-{\lambda }_n^2{\parallel A{u}_n-Ap\parallel }^2.\end{array}$$

(3.15)

From (3.3), (3.4), (3.6) and (3.15), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & {\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel y_n-p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel z_n-p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {(1-{\alpha }_n\tau )}^2({\parallel x_n-z\parallel }^2-{\parallel z_n-u_n\parallel }^2-2{\lambda }_n〈z_n-u_n,A{u}_n-Ap〉\\ -{\lambda }_n^2{\parallel A{u}_n-Ap\parallel }^2)+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\\ \le & {\parallel x_n-p\parallel }^2+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2-{(1-{\alpha }_n\tau )}^2\parallel z_n-u_n\parallel \\ -2{(1-{\alpha }_n\tau )}^2{\lambda }_n({\lambda }_n-2\alpha )\parallel z_n-u_n\parallel \parallel A{u}_n-Ap\parallel \\ -{(1-{\alpha }_n\tau )}^2{\lambda }_n^2{\parallel A{u}_n-Ap\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉,\end{array}$$

and hence

$$\begin{array}{r}{(1-{\alpha }_n\tau )}^2\parallel z_n-u_n\parallel \\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2\\ -2{(1-{\alpha }_n\tau )}^2{\lambda }_n({\lambda }_n-2\alpha )\parallel z_n-u_n\parallel \parallel A{u}_n-Ap\parallel \\ -{(1-{\alpha }_n\tau )}^2{\lambda }_n^2{\parallel A{u}_n-Ap\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉.\end{array}$$

(3.16)

Replacing n by $n_{k_i}$ in (3.16), we have

$$\begin{array}{r}{(1-{\alpha }_{n_{k_i}}\tau )}^2{\parallel z_{n_{k_i}}-u_{n_{k_i}}\parallel }^2\\ \le {\parallel x_{n_{k_i}}-p\parallel }^2-{\parallel x_{n_{k_i}+1}-p\parallel }^2+{\alpha }_{n_{k_i}}^2{\tau }^2{\parallel x_{n_{k_i}}-p\parallel }^2\\ -2{(1-{\alpha }_{n_{k_i}}\tau )}^2{\lambda }_{n_{k_i}}({\lambda }_{n_{k_i}}-2\alpha )\parallel z_{n_{k_i}}-u_{n_{k_i}}\parallel \parallel A{u}_{n_{k_i}}-Ap\parallel \\ -{(1-{\alpha }_{n_{k_i}}\tau )}^2{\lambda }_{n_{k_i}}^2{\parallel A{u}_{n_{k_i}}-Ap\parallel }^2+2{\alpha }_{n_{k_i}}〈\gamma g(x_{n_{k_i}})-Vp,x_{n_{k_i}+1}-p〉.\end{array}$$

From (3.11), ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and the existence of ${lim}_{i\to \mathrm{\infty }}\parallel x_{n_{k_i}}-p\parallel$, we obtain that

$$\underset{i\to \mathrm{\infty }}{lim}\parallel z_{n_{k_i}}-u_{n_{k_i}}\parallel =0.$$

(3.17)

Since $\parallel z_{n_{k_i}}-x_{n_{k_i}}\parallel \le \parallel z_{n_{k_i}}-u_{n_{k_i}}\parallel +\parallel u_{n_{k_i}}-x_{n_{k_i}}\parallel$, by (3.14) and (3.17), we obtain that

$$\underset{i\to \mathrm{\infty }}{lim}\parallel z_{n_{k_i}}-x_{n_{k_i}}\parallel =0.$$

(3.18)

Since A is Lipschitz continuous, we also obtain

$$\underset{i\to \mathrm{\infty }}{lim}\parallel A{z}_{n_{k_i}}-A{x}_{n_{k_i}}\parallel =0.$$

(3.19)

Since $z_n=J_{\lambda }(I-\lambda A)u_n$, we have that

$$\begin{array}{rcl}{z}_n& =& {(I+{\lambda }_{n}B)}^{-1}(I-{\lambda }_{n}A)u_n\\ \iff & (I-{\lambda }_{n}A)u_n\in (I+{\lambda }_{n}B)z_n=z_n+{\lambda }_{n}B{z}_n\\ \iff & u_n-z_n-{\lambda }_{n}A{u}_n\in {\lambda }_{n}B{z}_n\\ \iff & \frac{1}{{\lambda }_n}(u_n-z_n-{\lambda }_{n}A{u}_n)\in B{z}_n.\end{array}$$

Since B is monotone, we have that for $(u,v)\in B$,

$$〈z_n-u,\frac{1}{{\lambda }_n}(u_n-z_n-{\lambda }_{n}A{u}_n)-v〉\ge 0,$$

and hence

$$〈z_n-u,u_n-z_n-{\lambda }_n(A{u}_n+v)〉\ge 0.$$

(3.20)

Replacing n by $n_{k_i}$ in (3.20), we have that

$$〈z_{n_{k_i}}-u,u_{n_{k_i}}-z_{n_{k_i}}-{\lambda }_{n_{k_i}}(A{u}_{n_{k_i}}+v)〉\ge 0.$$

(3.21)

Since $x_{n_{k_i}}\rightharpoonup w$ and $x_{n_{k_i}}-u_{n_{k_i}}\to 0$, so $u_{n_{k_i}}\rightharpoonup w$. From (3.17), we get that $z_{n_{k_i}}\rightharpoonup w$, together with (3.21), we have that

$$〈w-u,-Aw-v〉\ge 0.$$

Since B is maximal monotone, $(-Aw)\in Bw$, that is, $w\in {(A+B)}^{-1}0$.

(c) Next, we show that $w\in F^{-1}0$. Since F is a maximal monotone operator, we have from (1.15) that $A_{r_{n_{k_i}}}{x}_{n_{k_i}}\in F{T}_{r_{n_{k_i}}}{x}_{n_{k_i}}$, where $A_r$ is the Yosida approximation of F for $r>0$. Furthermore, we have that for any $(u,v)\in F$,

$$〈u-u_{n_{k_i}},v-\frac{x_{n_{k_i}}-u_{n_{k_i}}}{r_{n_{k_i}}}〉\ge 0.$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$, $u_{n_{k_i}}\rightharpoonup w$ and $x_{n_{k_i}}-u_{n_{k_i}}\to 0$, we have

$$〈u-w,v〉\ge 0.$$

Since F is a maximal monotone operator, we have $0\in Fw$, that is, $w\in F^{-1}0$. By (a), (b) and (c), we conclude that

$$w\in F(T)\cap {(A+B)}^{-1}0\cap F^{-1}0.$$

Using (3.7), we obtain

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(V-\gamma g)p_0,x_n-p_0〉& =& \underset{k\to \mathrm{\infty }}{lim}〈(V-\gamma g)p_0,x_{n_k}-p_0)〉\\ =& 〈(V-\gamma g)p_0,w-p_0)〉\ge 0.\end{array}$$

Finally, we prove that $x_n\to p_0$. Notice that

$$x_{n+1}-p_0={\alpha }_n(\gamma g(x_n)-p_0)+(I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p_0,$$

we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p_0\parallel }^2& \le & {(1-{\alpha }_n\tau )}^2{\parallel y_n-p_0\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-V{p}_0,x_{n+1}-p_0〉\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel x_n-p_0\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-V{p}_0,x_{n+1}-p_0〉\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel x_n-p_0\parallel }^2+2{\alpha }_n\gamma k\parallel x_n-p_0\parallel \parallel x_{n+1}-p_0\parallel \\ +2{\alpha }_n〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel x_n-p_0\parallel }^2+{\alpha }_n\gamma k({\parallel x_n-p_0\parallel }^2+{\parallel x_{n+1}-p_0\parallel }^2)\\ +2{\alpha }_n〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉\\ \le & \{{(1-{\alpha }_n\tau )}^2+{\alpha }_n\gamma k\}{\parallel x_n-p_0\parallel }^2+{\alpha }_n\gamma k{\parallel x_{n+1}-p_0\parallel }^2\\ +2{\alpha }_n〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉,\end{array}$$

and hence

$$\begin{array}{rcl}{\parallel x_{n+1}-p_0\parallel }^2& \le & \frac{1-2{\alpha }_n\tau +{({\alpha }_n\tau )}^2+{\alpha }_n\gamma k}{1-{\alpha }_n\gamma k}{\parallel x_n-p_0\parallel }^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma k}〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉\\ =& \{1-\frac{2(\tau -\gamma k){\alpha }_n}{1-{\alpha }_n\gamma k}\}{\parallel x_n-p_0\parallel }^2+\frac{{({\alpha }_n\tau )}^2}{1-{\alpha }_n\gamma k}{\parallel x_n-p_0\parallel }^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma k}〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉\\ =& \{1-\frac{2(\tau -\gamma k){\alpha }_n}{1-{\alpha }_n\gamma k}\}{\parallel x_n-p_0\parallel }^2+\frac{{\alpha }_n\cdot {\alpha }_n{\tau }^2}{1-{\alpha }_n\gamma k}{\parallel x_n-p_0\parallel }^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma k}〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉\\ =& (1-{\beta }_n){\parallel x_n-p_0\parallel }^2\\ +{\beta }_n\{\frac{{\alpha }_n{\tau }^2{\parallel x_n-p_0\parallel }^2}{2(\tau -\gamma k)}+\frac{1}{\tau -\gamma k}〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉\},\end{array}$$

(3.22)

where ${\beta }_n=\frac{2(\tau -\gamma k){\alpha }_n}{1-{\alpha }_n\gamma k}$. Since ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_n=\mathrm{\infty }$, we have from Lemma 2.5 and (3.22) that $x_n\to p_0$. This completes the proof. □

4 Applications

Let H be a Hilbert space, and let f be a proper lower semicontinuous convex function of H into $(-\mathrm{\infty },\mathrm{\infty }]$. Then the subdifferential ∂f of f is defined as follows:

$$\partial f(x)=\{z\in H:f(x)+〈z,y-x〉\le f(y),y\in H\}$$

for all $x\in H$; see, for instance, [36]. From Rockafellar [37], we know that ∂f is maximal monotone. Let C be a nonempty closed convex subset of H, and let $i_C$ be the indicator function of C, i.e.,

$$i_C(x)=\{\begin{array}{cc}0,\hfill & x\in C,\hfill \\ \mathrm{\infty },\hfill & x\notin C.\hfill \end{array}$$

Then $i_C$ is a proper lower semicontinuous convex function of H into $(-\mathrm{\infty },\mathrm{\infty }]$, and then the subdifferential ${\partial }_{i_C}$ of $i_C$ is a maximal monotone operator. So, we can define the resolvent $J_{\lambda }$ of ${\partial }_{i_C}$ for $\lambda >0$, i.e.,

$$J_{\lambda }x={(I+\lambda {\partial }_{i_C})}^{-1}x$$

for all $x\in H$. We have that for any $x\in H$ and $u\in C$,

$$\begin{array}{rcl}u=J_{\lambda }x& \iff & x\in u+\lambda {\partial }_{i_C}u\\ \iff & x\in u+\lambda N_{C}u\\ \iff & x-u\in \lambda N_{C}u\\ \iff & \frac{1}{\lambda }〈x-u,v-u〉\le 0,\mathrm{\forall }v\in C\\ \iff & 〈x-u,v-u〉\le 0,\mathrm{\forall }v\in C\\ \iff & u=P_{C}x,\end{array}$$

where $N_{C}u$ is the normal cone to C at u, i.e.,

$$N_{C}u=\{x\in H:〈z,v-u〉\le 0,\mathrm{\forall }v\in C\}.$$

Let C be a nonempty, closed and convex subset of H, and let $f:C\times C\to \mathbb{R}$ be a bifunction. For solving the equilibrium problem, let us assume that the bifunction $f:C\times C\to \mathbb{R}$ satisfies the following conditions.

For solving the mixed equilibrium problem, let us give the following assumptions for the bifunction F, φ and the set C:

• (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 all $x,y,z\in C$,

  $$\underset{t\downarrow 0}{lim\hspace{0.17em}sup}f(tz+(1-t)x,y)\le f(x,y);$$

• (A4) for all $x\in C$, $f(x,\cdot )$ is convex and lower semicontinuous;

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

  $$f(z,y_x)+\phi (y_x)+\frac{1}{r}〈y_x-z,z-x〉<\phi (z);$$

• (B2) C is a bounded set.

We know the following lemma which appears implicitly in Blum and Oettli [1].

Lemma 4.1 [1]

Let C be a nonempty closed convex subset of H, and let f be a bifunction of $C\times C$ into ℝ satisfying (A1)-(A5). Let $r>0$ and $x\in H$. Then there exists a unique $z\in C$ such that

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

By a similar argument as that in [[38], Lemma 2.3], we have the following result.

Lemma 4.2 [38]

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $f:C\times C\to \mathbb{R}$ be a bifunction which satisfies conditions (A1)-(A4), and let $\phi :C\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For $r>0$ and $x\in H$, define a mapping $T_r:H\to C$ as follows:

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

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

• (1) For each $x\in H$, $T_r(x)\ne \mathrm{\varnothing }$;

• (2) $T_r$ is single-valued;

• (3) $T_r$ is firmly nonexpansive, i.e., for any $x,y\in H$,

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

• (4) $Fix(T_r)=\mathit{MEP}(f,\phi )$;

• (5) $\mathit{MEP}(f,\phi )$ is closed and convex.

We call such $T_r$ the resolvent of f for $r>0$. Using Lemmas 4.1 and 4.2, Takahashi et al. [22] obtained the following lemma. See [39] for a more general result.

Lemma 4.3 [22]

Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let $f:C\times C\to \mathbb{R}$ satisfy (A1)-(A5). Let $A_f$ be a set-valued mapping of H into itself defined by

$$A_{f}x=\{\begin{array}{c}\{z\in H:f(x,y)\ge 〈y-x,z〉,\mathrm{\forall }y\in C\},\mathrm{\forall }x\in C,\hfill \\ \mathrm{\varnothing },\mathrm{\forall }x\notin C.\hfill \end{array}$$

Then $\mathit{MEP}(f)=A_f^{-1}0$ and $A_f$ is a maximal monotone operator with $dom{A}_f\subset C$. Furthermore, for any $x\in H$ and $r>0$, the resolvent $T_r$ of f coincides with the resolvent of $A_f$, i.e.,

$$T_{r}x={(I+r{A}_f)}^{-1}x.$$

Applying the idea of the proof in Lemma 4.3, we have the following results.

Lemma 4.4 Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let $f:C\times C\to \mathbb{R}$ satisfy (A1)-(A4), and let $\phi :C\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) hold. Let $A_{(f,\phi )}$ be a set-valued mapping of H into itself defined by

$$A_{(f,\phi )}x=\{\begin{array}{c}\{z\in H:f(x,y)+\phi (y)-\phi (x)\ge 〈y-x,z〉,\mathrm{\forall }y\in C\},\mathrm{\forall }x\in C,\hfill \\ \mathrm{\varnothing },\mathrm{\forall }x\notin C.\hfill \end{array}$$

(4.1)

Then $\mathit{MEP}(f,\phi )=A_{(f,\phi )}^{-1}0$ and $A_{(f,\phi )}$ is a maximal monotone operator with $dom{A}_{(f,\phi )}\subset C$. Furthermore, for any $x\in H$ and $r>0$, the resolvent $T_r$ of f coincides with the resolvent of $A_{(f,\phi )}$, i.e.,

$$T_{r}x={(I+r{A}_{(f,\phi )})}^{-1}x.$$

Proof It is obvious that $\mathit{MEP}(f,\phi )=A_{(f,\phi )}^{-1}0$. In fact, we have that

$$\begin{array}{rcl}z\in \mathit{MEP}(f,\phi )& \iff & f(z,y)+\phi (y)-\phi (z)\ge 0,\mathrm{\forall }y\in C\\ \iff & f(z,y)+\phi (y)-\phi (z)\ge 〈y-z,0〉,\mathrm{\forall }y\in C\\ \iff & 0\in A_{(f,\phi )}z\\ \iff & z\in A_{(f,\phi )}^{-1}0.\end{array}$$

We show that $A_{(f,\phi )}$ is monotone. Let $(x_1,z_1),(x_2,z_2)\in A_{(f,\phi )}$ be given. Then we have, for all $y\in C$,

$$f(x_1,y)+\phi (y)-\phi (x_1)\ge 〈y-x_1,z_1〉\text{and}f(x_2,y)+\phi (y)-\phi (x_2)\ge 〈y-x_2,z_2〉,$$

and hence

$$f(x_1,x_2)+\phi (x_2)-\phi (x_1)\ge 〈x_2-x_1,z_1〉\text{and}f(x_2,x_1)+\phi (x_1)-\phi (x_2)\ge 〈x_1-x_2,z_2〉.$$

It follows from (A2) that

$$0\ge f(x_1,x_2)+f(x_2,x_1)\ge 〈x_2-x_1,z_1〉+〈x_1-x_2,z_2〉=-〈x_1-x_2,z_1-z_2〉.$$

This implies that $A_{(f,\phi )}$ is monotone. We next prove that $A_{(f,\phi )}$ is maximal monotone. To show that $A_{(f,\phi )}$ is maximal monotone, it is sufficient to show from [33] that $R(I+r{A}_{(f,\phi )})=H$ for all $r>0$, where $R(I+r{A}_{(f,\phi )})$ is the range of $I+r{A}_{(f,\phi )}$. Let $x\in H$ and $r>0$. Then, from Lemma 4.2, there exists $z\in C$ such that

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

So, we have that

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

By the definition of $A_{(f,\phi )}$, we get

$$A_{(f,\phi )}z\ni \frac{1}{r}(x-z),$$

and hence $x\in z+r{A}_{(f,\phi )}z$. Therefore, $H\subset R(I+r{A}_{(f,\phi )})$ and $R(I+r{A}_{(f,\phi )})=H$. Also, $x\in z+r{A}_{(f,\phi )}z$ implies that $T_{r}x={(I+r{A}_{(f,\phi )})}^{-1}x$ for all $x\in H$ and $r>0$. □

Using Theorem 3.1, we obtain the following results for an inverse-strongly monotone mapping.

Theorem 4.5 Let H be a real Hilbert space, and let C be a nonempty, closed and convex subset of H. Let $\alpha >0$ and let A be an α-inverse-strongly monotone mapping of C into H. Let $0<k<1$ and let g be a k-contraction of H into itself. Let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$. Let $T:C\to C$ be a 2-generalized hybrid mapping such that $\mathrm{\Gamma }:=F(T)\cap VI(C,A)\ne \mathrm{\varnothing }$. Take $\mu ,\gamma \in \mathbb{R}$ as follows:

$$0<\mu <\frac{2\overline{\gamma }}{L^2},0<\gamma <\frac{\overline{\gamma }-\frac{L^2\mu }{2}}{k}.$$

Let $\{x_n\}\subset H$ be a sequence generated by

$$\{\begin{array}{c}{x}_1=x\in H\mathit{\text{arbitrarily}},\hfill \\ z_n=P_C(I-{\lambda }_{n}A)P_C{x}_n,\hfill \\ y_n=\frac{1}{n}{\sum }_{k=0}^{n-1}{T}^k{z}_n,\mathrm{\forall }n=1,2,\dots ,\hfill \\ x_{n+1}={\alpha }_n\gamma g(x_n)+(I-{\alpha }_{n}V)y_n\mathit{\text{for all }}n\in \mathbb{N},\hfill \end{array}$$

(4.2)

where $\{{\alpha }_n\}\subset (0,1)$ and $\{r_n\}\subset (0,\mathrm{\infty })$ satisfy

$$\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=0,\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }\mathit{\text{and}}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{r}_n>0.$$

Then $\{x_n\}$ converges strongly to a point $p_0$ of Γ, where $p_0$ is a unique fixed point of $P_{\mathrm{\Gamma }}(I-V+\gamma g)$. This point $p_0\in \mathrm{\Gamma }$ is also a unique solution of the hierarchical variational inequality

$$〈(V-\gamma g)p_0,q-p_0〉\ge 0,\mathrm{\forall }q\in VI(C,A).$$

(4.3)

Proof Put $B=F=\partial i_C$ in Theorem 3.1. Then, for ${\lambda }_n>0$ and $r_n>0$, we have that

$$J_{{\lambda }_n}=T_{r_n}=P_C.$$

Furthermore we have, from the proof of [[32], Theorem 12], that

$${(\partial i_C)}^{-1}0=C\text{and}{(A+\partial i_C)}^{-1}=VI(C,A).$$

Thus we obtained the desired results by Theorem 3.1. □

Using Theorem 3.1, we finally prove a strong convergence theorem for inverse-strongly monotone operators and equilibrium problems in a Hilbert space.

Theorem 4.6 Let H be a real Hilbert space, and let C be a nonempty, closed and convex subset of H. Let $\alpha >0$ and let A be an α-inverse-strongly monotone mapping of C into H. Let $B:D(B)\subset C\to 2^H$ be maximal monotone. Let $J_{\lambda }={(I+\lambda B)}^{-1}$ be the resolvent of B for $\lambda >0$. Let $0<k<1$ and let g be a k-contraction of H into itself. Let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$. Let $f:C\times C\to \mathbb{R}$ be a bifunction satisfying conditions (A1)-(A4), and let $\phi :C\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. Let $T:C\to C$ be a 2-generalized hybrid mapping with $\mathrm{\Theta }:=F(T)\cap {(A+B)}^{-1}0\cap \mathit{MEP}(f,\phi )\ne \mathrm{\varnothing }$. Take $\mu ,\gamma \in \mathbb{R}$ as follows:

$$0<\mu <\frac{2\overline{\gamma }}{L^2},0<\gamma <\frac{\overline{\gamma }-\frac{L^2\mu }{2}}{k}.$$

Let $\{x_n\}\subset H$ be a sequence generated by

$$\{\begin{array}{c}{x}_1=x\in H\mathit{\text{arbitrarily}},\hfill \\ f(u_n,y)+\phi (y)-\phi (u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ z_n=J_{{\lambda }_n}(I-{\lambda }_{n}A)u_n,\hfill \\ y_n=\frac{1}{n}{\sum }_{k=0}^{n-1}{T}^k{z}_n,\mathrm{\forall }n=1,2,\dots ,\hfill \\ x_{n+1}={\alpha }_n\gamma g(x_n)+(I-{\alpha }_{n}V)y_n,\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

(4.4)

where $\{{\alpha }_n\}\subset (0,1)$ and $\{r_n\}\subset (0,\mathrm{\infty })$ satisfy

$$\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=0,\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }\mathit{\text{and}}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{r}_n>0.$$

Then $\{x_n\}$ converges strongly to a point $p_0$ of Θ, where $p_0$ is a unique fixed point of $P_{\mathrm{\Theta }}(I-V+\gamma g)$. This point $p_0\in \mathrm{\Theta }$ is also a unique solution of the hierarchical variational inequality

$$〈(V-\gamma g)p_0,q-p_0〉\ge 0,\mathrm{\forall }q\in \mathrm{\Theta }.$$

(4.5)

Proof Since f is a bifunction of $C\times C$ into ℝ satisfying conditions (A1)-(A4) and $\phi :C\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ is a proper lower semicontinuous and convex function, we have that the mapping $A_f^{\phi }$ defined by (4.1) is a maximal monotone operator with $dom{A}_f^{\phi }\subset C$. Put $F=A_f^{\phi }$ in Theorem 3.1. Then we obtain that $u_n=T_{r_n}{x}_n$. Therefore, we arrive at the desired results. □