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 $D(B)⊂C$ and $D(F)⊂C$ respectively. The purpose of this paper is to introduce a general iterative scheme for finding a point of $F(T)∩ ( A + B ) − 1 0∩ F − 1 0$ 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 and be the sets of all positive integers and real numbers, respectively. Let $φ:C→R$ be a real-valued function, and let $f:C×C→R$ be an equilibrium bifunction, that is, $f(u,u)=0$ for each $u∈C$. The mixed equilibrium problem is to find $x∈C$ such that

(1.1)

Denote the set of solutions of (1.1) by $MEP(f,φ)$. In particular, if $φ=0$, this problem reduces to the equilibrium problem, which is to find $x∈C$ such that

(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 $F(T):={x∈C:Tx=x}$ the set of fixed points of T. A mapping $T:C→C$ is said to be nonexpansive if $∥Tx−Ty∥≤∥x−y∥$ for all $x,y∈C$. The mapping $T:C→C$ is said to be firmly nonexpansive if

(1.3)

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

$2 ∥ T x − T y ∥ 2 ≤ ∥ T x − y ∥ 2 + ∥ x − T y ∥ 2$
(1.4)

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

$∥ T x − T y ∥ 2 ≤ ∥ x − y ∥ 2 +2〈x−Tx,y−Ty〉$
(1.5)

for all $x,y∈C$. 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 $T:C→C$ is said to be generalized hybrid if there are $α,β∈R$ such that

$α ∥ T x − T y ∥ 2 +(1−α) ∥ x − T y ∥ 2 ≤β ∥ T x − y ∥ 2 +(1−β) ∥ x − y ∥ 2$
(1.6)

for all $x,y∈C$. We call such a mapping an $(α,β)$-generalized hybrid mapping. We observe that the mappings above generalize several well-known mappings. For example, an $(α,β)$-generalized hybrid mapping is nonexpansive for $α=1$ and $β=0$, nonspreading for $α=2$ and $β=1$, and hybrid for $α= 3 2$ and $β= 1 2$.

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 $α 1 , α 2 , β 1 , β 2 ∈R$ such that

$α 1 ∥ T 2 x − T y ∥ 2 + α 2 ∥ T x − T y ∥ 2 + ( 1 − α 1 − α 2 ) ∥ x − T y ∥ 2 ≤ β 1 ∥ T 2 x − y ∥ 2 + β 2 ∥ T x − y ∥ 2 + ( 1 − β 1 − β 2 ) ∥ x − y ∥ 2$
(1.7)

for all $x,y∈C$; see [10] for more details. We call such a mapping an $( α 1 , α 2 , β 1 , β 2 )$-generalized hybrid mapping. We can also show that if T is a 2-generalized hybrid mapping and $x=Tx$, then for any $y∈C$,

$α 1 ∥ x − T y ∥ 2 + α 2 ∥ x − T y ∥ 2 + ( 1 − α 1 − α 2 ) ∥ x − T y ∥ 2 ≤ β 1 ∥ x − y ∥ 2 + β 2 ∥ x − y ∥ 2 + ( 1 − β 1 − β 2 ) ∥ x − y ∥ 2 ,$

and hence $∥x−Ty∥≤∥x−y∥$. 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 $(0, α 2 ,0, β 2 )$-generalized hybrid mapping is an $( α 2 , β 2 )$-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 $γ ¯ >0$ with the property

(1.8)

In general, a nonlinear operator $V:H→H$ is called strongly monotone if there exists $γ ¯ >0$ such that

(1.9)

Such V is called $γ ¯$-strongly monotone. A nonlinear operator $V:H→H$ is called Lipschitzian continuous if there exists $L>0$ such that

(1.10)

Such V is called L-Lipschitzian continuous. A mapping $A:C→H$ is said to be α-inverse-strongly monotone if $〈x−y,Ax−Ay〉≥α ∥ A x − A y ∥ 2$ for all $x,y∈C$. It is known that $∥Ax−Ay∥≤( 1 α )∥x−y∥$ for all $x,y∈C$ if A is α-inverse-strongly monotone; see, for example, [1113].

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

$x n + 1 = α n x n +(1− α n )T x n ,$
(1.11)

where the initial guess $x 1 ∈C$ is arbitrary and ${ α n }$ 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

$x n + 1 = α n u+(1− α n )T x n ,$
(1.12)

where $u, x 1 ∈C$ are arbitrary and ${ α n }$ is a real sequence in $[0,1]$ which satisfies $α n →0$, $∑ n = 1 ∞ α n =∞$ and $∑ n = 1 ∞ | α n − α n + 1 |<∞$. 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:

$S n x= 1 n ∑ k = 0 n − 1 T k x$
(1.13)

converges weakly to a fixed point of T for some $x∈C$. 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 $T:C→C$ be a 2-generalized hybrid mapping with $F(T)≠∅$. Suppose that ${ x n }$ is a sequence generated by $x 1 =x∈C$, $u∈C$ and

$x n + 1 = γ n u+(1− γ n ) 1 n ∑ k = 0 n − 1 T k x n ,∀n∈N,$
(1.14)

where $0≤ γ n ≤1$, $lim n → ∞ γ n =0$ and $∑ n = 1 ∞ γ n =∞$. 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, $D(B)={x∈H:Bx≠∅}$. A multi-valued mapping B on H is said to be monotone if $〈x−y,u−v〉≥0$ for all $x,y∈D(B)$, $u∈Bx$, and $v∈By$. 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 $J r = ( I + r B ) − 1 :H→D(B)$, which is called the resolvent of B for r. We denote by $A r = 1 r (I− J r )$ the Yosida approximation of B for $r>0$. We know [21] that

$A r x∈B J r x,∀x∈H,r>0.$
(1.15)

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

$∥ J r x− J r y∥≤〈x−y, J r x− J r y〉,∀x,y∈H.$
(1.16)

Recently, in the case when $T:C→C$ is a nonexpansive mapping, $A:C→H$ is an α-inverse strongly monotone mapping and $B∈H×H$ is a maximal monotone operator, Takahashi et al. [22] proved a strong convergence theorem for finding a point of $F(T)∩ ( A + B ) − 1 0$, 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:

$x n + 1 = β n x n +(1− β n )T ( J λ n ( I − λ n A ) x n ) ,$
(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 $J λ = ( I − λ B ) − 1$; ${ β n }$ and ${ λ n }$ are sequences which satisfy $0 and $0. Then they proved that ${ x n }$ converges weakly to a point $p= lim n → ∞ P F ( T ) ∩ ( A + B ) − 1 ( 0 ) x n$.

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:

(1.18)

where ${ α n }$ 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., [2528] 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:

$θ(x)= min x ∈ C 1 2 〈Vx,x〉−〈x,b〉,$
(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]:

$x n + 1 =(I− α n V)T x n + α n γf( x n ),n≥0,$
(1.20)

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

$〈 ( V − γ f ) x ∗ , x − x ∗ 〉 ≥0,x∈C,$

which is the optimality condition for the minimization problem

$min x ∈ C 1 2 〈Vx,x〉−h(x),$
(1.21)

where h is a potential function for γf (i.e., $h ′ (x)=γf(x)$ for $x∈H$).

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

$x n + 1 = α n γg( x n )+(I−μ α n V)T x n ,$

for all $n∈N$, where $μ,γ∈R$ satisfying $0<μ< 2 γ ¯ L 2$, $0<γ<μ( γ ¯ − L 2 μ 2 )/k$, 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 $p 0 ∈F(T)$ which is a unique solution of the variational inequality

$〈 ( V − γ g ) p 0 , q − p 0 〉 ≥0,∀q∈F(T).$

Very recently, Lin and Takahashi [32] obtained the strong convergence theorem for finding a point $p 0 ∈ ( A + B ) − 1 0∩ F − 1 0$ 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 $D(B)⊂C$ and $D(F)⊂C$, respectively. More precisely, they introduced the following iterative scheme: Let $x 1 =x∈H$ and let ${ x n }⊂H$ be a sequence generated

(1.22)

where ${ α n }⊂(0,1)$, ${ λ n }⊂(0,∞)$ and ${ r n }⊂(0,∞)$ satisfy certain appropriate conditions, $J λ = ( I + λ B ) − 1$ and $T r = ( I + r F ) − 1$ are the resolvents of B for $λ>0$ 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 $D(B)⊂C$ and $D(F)⊂C$ respectively. We introduce a new general iterative scheme for finding a common element of $F(T)∩ ( A + B ) − 1 0∩ F − 1 0$ 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 $〈⋅,⋅〉$ and the norm $∥⋅∥$, 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, $∥x− P C x∥≤∥x−y∥$ for all $x∈H$ and $y∈C$. 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.,

$∥ P C x − P C y ∥ 2 ≤〈 P C x− P C y,x−y〉$
(2.1)

for all $x,y∈H$. Furthermore, $〈 P C x− P C y,x−y〉≤0$ holds for all $x∈H$ and $y∈C$; see [33]. Let $α>0$ 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 $x∈H$, define the resolvent $J r x$. Then the following holds:

$s − t s 〈 J s x− J t x, J s x−x〉≥ ∥ J s x − J t x ∥ 2$
(2.2)

for all $s,t>0$ and $x∈H$.

From Lemma 2.1, we have that

$∥ J λ x− J μ x∥≤ ( | λ − μ | / λ ) ∥x− J λ x∥$
(2.3)

for all $λ,μ>0$ and $x∈H$; 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 $1 α$ -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 $T:C→C$ is a nonexpansive mapping, then $I−T$ is $1 2$-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 $α>0$. 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 $J λ = ( I + λ B ) − 1$ be the resolvent of B for any $λ>0$. Then the following hold:

1. (i)

if $u,v∈ ( A + B ) − 1 (0)$, then $Au=Av$;

2. (ii)

for any $λ>0$, $u∈ ( A + B ) − 1 (0)$ if and only if $u= J λ (I−λA)u$.

Lemma 2.5 [26, 35]

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

$a n + 1 ≤(1− t n ) a n + b n + t n c n ,$

where ${ t n }$, ${ b n }$ and ${ c n }$ satisfy the restrictions:

1. (i)

$∑ n = 0 ∞ t n =∞$;

2. (ii)

$∑ n = 0 ∞ b n <∞$;

3. (iii)

$lim sup n → ∞ c n ≤0$.

Then ${ a n }$ converges to zero as $n→∞$.

Lemma 2.6 [32]

Let H be a Hilbert space, and let $g:H→H$ be a k-contraction with $0. Let V be a $γ ¯$-strongly monotone and L-Lipschitzian continuous operator on H with $γ ¯ >0$ and $L>0$. Let a real number γ satisfy $0<γ< γ ¯ k$. Then $V−γg:H→H$ is a $( γ ¯ −γk)$-strongly monotone and $(L+γk)$-Lipschitzian continuous mapping. Furthermore, let C be a nonempty closed convex subset of H. Then $P C (I−V+γg)$ has a unique fixed point $z 0$ in C. This point $z 0 ∈C$ is also a unique solution of the variational inequality

$〈 ( V − γ f ) z 0 , q − z 0 〉 ≥0,∀q∈C.$

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 $α>0$ and A be an α-inverse-strongly monotone mapping of C into H. Let the set-valued maps $B:D(B)⊂C→ 2 H$ and $F:D(F)⊂C→ 2 H$ be maximal monotone. Let $J λ = ( I + λ B ) − 1$ and $T r = ( I + r F ) − 1$ be the resolvents of B for $λ>0$ and F for $r>0$, respectively. Let $0 and let g be a k-contraction of H into itself. Let V be a $γ ¯$-strongly monotone and L-Lipschitzian continuous operator with $γ ¯ >0$ and $L>0$. Let $T:C→C$ be a 2-generalized hybrid mapping such that $Ω:=F(T)∩ ( A + B ) − 1 0∩ F − 1 0≠∅$. Take $μ,γ∈R$ as follows:

$0<μ< 2 γ ¯ L 2 ,0<γ< γ ¯ − L 2 μ 2 k .$

Let the sequence ${ x n }⊂H$ be generated by

${ x 1 = x ∈ H arbitrarily , z n = J λ n ( I − λ n A ) T r n x n , y n = 1 n ∑ k = 0 n − 1 T k z n , x n + 1 = α n γ g ( x n ) + ( I − α n V ) y n , ∀ n = 1 , 2 , … ,$
(3.1)

where the sequences ${ α n }$, ${ λ n }$ and ${ r n }$ satisfy the following restrictions:

1. (i)

${ α n }⊂[0,1]$, $lim n → ∞ α n =0$ and $∑ n = 1 ∞ α n =∞$;

2. (ii)

there exist constants a and b such that $0 for all $n∈N$;

3. (iii)

$lim inf n → ∞ r n >0$.

Then ${ x n }$ converges strongly to a point $p 0$ of Ω, where $p 0$ is a unique fixed point of $P Ω (I−V+γg)$. This point $p 0 ∈Ω$ is also a unique solution of the hierarchical variational inequality

$〈 ( V − γ g ) p 0 , q − p 0 〉 ≥0,∀q∈Ω.$
(3.2)

Proof First we prove that ${ x n }$ is bounded and $lim n → ∞ ∥ x n −p∥$ exists for all $p∈Ω$. Let $p∈Ω$, we have that $p= J λ n (I− λ n A)p$ and $p= T r n p$. Putting $u n = T r n x n$, we have that

$∥ z n − p ∥ 2 = ∥ J λ n ( I − λ n A ) T r n x n − J λ n ( I − λ n A ) p ∥ 2 ≤ ∥ ( T r n x n − T r n p ) − λ n ( A T r n x n − A T r n p ) ∥ 2 = ∥ T r n x n − T r n p ∥ 2 − 2 λ n 〈 u n − p , A u n − A p 〉 + λ n 2 ∥ A u n − A p ∥ 2 ≤ ∥ u n − p ∥ 2 − 2 λ n α ∥ A u n − A p ∥ 2 + λ n 2 ∥ A u n − A p ∥ 2 ≤ ∥ x n − p ∥ 2 − λ n ( 2 α − λ n ) ∥ A u n − A p ∥ 2 ≤ ∥ x n − p ∥ 2 .$
(3.3)

This together with quasi-nonexpansiveness of T implies that

$∥ y n − p ∥ = ∥ 1 n ∑ k = 0 n − 1 T k z n − p ∥ ≤ 1 n ∑ k = 0 n − 1 ∥ T k z n − p ∥ ≤ 1 n ∑ k = 0 n − 1 ∥ z n − p ∥ = ∥ z n − p ∥ ≤ ∥ x n − p ∥ .$
(3.4)

Therefore, we have

$∥ x n + 1 − p ∥ = ∥ α n ( γ g ( x n ) − V p ) + ( I − α n V ) y n − ( I − α n V ) p ∥ ≤ α n ∥ γ g ( x n ) − V p ∥ + ∥ ( I − α n V ) y n − ( I − α n V ) p ∥ ≤ α n γ k ∥ x n − p ∥ + α n ∥ γ g ( p ) − V p ∥ + ∥ ( I − α n V ) y n − ( I − α n V ) p ∥ .$
(3.5)

Putting $τ= γ ¯ − L 2 μ 2$, we can calculate the following:

$∥ ( I − α n V ) y n − ( I − α n V ) p ∥ 2 = ∥ ( y n − p ) − α n ( V y n − V p ) ∥ 2 = ∥ y n − p ∥ 2 − 2 α n 〈 y n − p , V y n − V p 〉 + α n 2 ∥ V y n − V p ∥ 2 ≤ ∥ y n − p ∥ 2 − 2 α n γ ¯ ∥ y n − p ∥ 2 + α n 2 L 2 ∥ y n − p ∥ 2 = ( 1 − 2 α n γ ¯ + α n 2 L 2 ) ∥ y n − p ∥ 2 = ( 1 − 2 α n τ − α n L 2 μ + α n 2 L 2 ) ∥ y n − p ∥ 2 ≤ ( 1 − 2 α n τ − α n ( L 2 μ − α n L 2 ) + α n 2 τ 2 ) ∥ y n − p ∥ 2 ≤ ( 1 − 2 α n τ + α n 2 τ 2 ) ∥ y n − p ∥ 2 = ( 1 − α n τ ) 2 ∥ y n − p ∥ 2 .$
(3.6)

Since $1− α n τ>0$, we obtain that

$∥ ( I − α n V ) y n − ( I − α n V ) p ∥ ≤(1− α n τ)∥ y n −p∥.$

Therefore, by (3.5), we have

which yields that the sequence ${∥ x n −p∥}$ 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 $p 0 ∈Ω$ of the hierarchical variational inequality

$〈 ( V − γ g ) p 0 , q − p 0 〉 ≥0,∀q∈Ω.$
(3.7)

We show that $lim sup n → ∞ 〈(V−γg) p 0 , x n − p 0 〉≥0$. We may assume, without loss of generality, that there exists a subsequence ${ x n k }$ of ${ x n }$ converging to $w∈C$, as $k→∞$, such that

$lim sup n → ∞ 〈 ( V − γ g ) p 0 , x n − p 0 〉 = lim k → ∞ 〈 ( V − γ g ) p 0 , x n k − p 0 〉 .$

Since ${∥ x n k −p∥}$ is bounded, there exists a subsequence ${ x n k i }$ of ${ x n k }$ such that $lim i → ∞ ∥ x n k i −p∥$ exists. Now we shall prove that $w∈Ω$.

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

$∥ x n + 1 − y n ∥= ∥ α n γ g ( x n ) + ( I − α n V ) y n − y n ∥ = α n ∥ γ g ( x n ) − V y n ∥ .$

In particular, replacing n by $n k i$ and taking $i→∞$ in the last equality, we have

$lim i → ∞ ∥ x n k i + 1 − y n k i ∥=0,$

so we have $y n k i ⇀w$. Since T is 2-generalized hybrid, there exist $α 1 , α 2 , β 1 , β 2 ∈R$ such that

$α 1 ∥ T 2 x − T y ∥ 2 + α 2 ∥ T x − T y ∥ 2 + ( 1 − α 1 − α 2 ) ∥ x − T y ∥ 2 ≤ β 1 ∥ T 2 x − y ∥ 2 + β 2 ∥ T x − y ∥ 2 + ( 1 − β 1 − β 2 ) ∥ x − y ∥ 2$

for all $x,y∈C$. For any $n∈N$ and $k=0,1,2,…,n−1$, we compute the following:

$0 ≤ β 1 ∥ T 2 T k z n − y ∥ 2 + β 2 ∥ T T k z n − y ∥ 2 + ( 1 − β 1 − β 2 ) ∥ T k z n − y ∥ 2 − α 1 ∥ T 2 T k z n − T y ∥ 2 − α 2 ∥ T T k z n − T y ∥ 2 − ( 1 − α 1 − α 2 ) ∥ T k z n − T y ∥ 2 = β 1 ∥ T k + 2 z n − y ∥ 2 + β 2 ∥ T k + 1 z n − y ∥ 2 + ( 1 − β 1 − β 2 ) ∥ T k z n − y ∥ 2 − α 1 ∥ T k + 2 z n − T y ∥ 2 − α 2 ∥ T k + 1 z n − T y ∥ 2 − ( 1 − α 1 − α 2 ) ∥ T k z n − T y ∥ 2 ≤ β 1 { ∥ T k + 2 z n − T y ∥ 2 + ∥ T y − y ∥ 2 } + β 2 { ∥ T k + 1 z n − T y ∥ 2 + ∥ T y − y ∥ 2 } + ( 1 − β 1 − β 2 ) { ∥ T k z n − T y ∥ 2 + ∥ T y − y ∥ 2 } − α 1 ∥ T k + 2 z n − T y ∥ 2 − α 2 ∥ T k + 1 z n − T y ∥ 2 − ( 1 − α 1 − α 2 ) ∥ T k z n − T y ∥ 2 = β 1 { ∥ T k + 2 z n − T y ∥ 2 + ∥ T y − y ∥ 2 + 2 〈 T k + 2 z n − T y , T y − y 〉 } + β 2 { ∥ T k + 1 z n − T y ∥ 2 + ∥ T y − y ∥ 2 + 2 〈 T k + 1 z n − T y , T y − y 〉 } + ( 1 − β 1 − β 2 ) { ∥ T k z n − T y ∥ 2 + ∥ T y − y ∥ 2 + 2 〈 T k z n − T y , T y − y 〉 } − α 1 ∥ T k + 2 z n − T y ∥ 2 − α 2 ∥ T k + 1 z n − T y ∥ 2 − ( 1 − α 1 − α 2 ) ∥ T k z n − T y ∥ 2 = ( β 1 − α 1 ) ∥ T k + 2 z n − T y ∥ 2 + ( β 2 − α 2 ) ∥ T k + 1 z n − T y ∥ 2 + ( α 1 + α 2 − β 1 − β 2 ) ∥ T k z n − T y ∥ 2 × ( β 1 + β 2 + 1 − β 1 − β 2 ) ∥ T y − y ∥ 2 + 2 〈 β 1 T k + 2 z n − β 1 T y + β 2 T k + 1 z n − β 2 T y + ( 1 − β 1 − β 2 ) T k z n − ( 1 − β 1 − β 2 ) T y , T y − y 〉 = ( β 1 − α 1 ) ∥ T k + 2 z n − T y ∥ 2 + ( β 2 − α 2 ) ∥ T k + 1 z n − T y ∥ 2 − ( ( β 1 − α 1 ) + ( α 2 − β 2 ) ) ∥ T k z n − T y ∥ 2 + ∥ T y − y ∥ 2 + 2 〈 β 1 T k + 2 z n + β 2 T k + 1 z n + ( 1 − β 1 − β 2 ) T k z n − T y , T y − y 〉 = ( β 1 − α 1 ) ( ∥ T k + 2 z n − T y ∥ 2 − ∥ T k z n − T y ∥ 2 ) + ( β 2 − α 2 ) ( ∥ T k + 1 z n − T y ∥ 2 − ∥ T k z n − T y ∥ 2 ) + ∥ T y − y ∥ 2 + 2 〈 β 1 T k + 2 z n + β 2 T k + 1 z n + ( 1 − β 1 − β 2 ) T k z n − T y , T y − y 〉 = ∥ T y − y ∥ 2 + 2 〈 T k z n − T y , T y − y 〉 + 2 〈 β 1 ( T k + 2 z n − T k x n ) + β 2 ( T k + 1 z n − T k z n ) , T y − y 〉 + ( β 1 − α 1 ) ( ∥ T k + 2 z n − T y ∥ 2 − ∥ T k z n − T y ∥ 2 ) + ( β 2 − α 2 ) ( ∥ T k + 1 z n − T y ∥ 2 − ∥ T k z n − T y ∥ 2 ) .$

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

$0 ≤ ∑ k = 0 n − 1 ∥ T y − y ∥ 2 + 2 〈 ∑ k = 0 n − 1 ( T k z n − T y ) , T y − y 〉 + 2 〈 β 1 ∑ k = 0 n − 1 ( T k + 2 z n − T k z n ) + β 2 ∑ k = 0 n − 1 ( T k + 1 z n − T k z n ) , T y − y 〉 + ( β 1 − α 1 ) ∑ k = 0 n − 1 ( ∥ T k + 2 z n − T y ∥ 2 − ∥ T k z n − T y ∥ 2 ) + ( β 2 − α 2 ) ∑ k = 0 n − 1 ( ∥ T k + 1 z n − T y ∥ 2 − ∥ T k z n − T y ∥ 2 ) = n ∥ T y − y ∥ 2 + 2 〈 ∑ k = 0 n − 1 T k z n − n T y , T y − y 〉 + 2 〈 β 1 ( T n + 1 z n − T n z n − z n − T z n ) + β 2 ( T n z n − z n ) , T y − y 〉 + ( β 1 − α 1 ) ( ∥ T n + 1 z n − T y ∥ 2 + ∥ T n z n − T y ∥ 2 − ∥ z n − T y ∥ 2 − ∥ T z n − T y ∥ 2 ) + ( β 2 − α 2 ) ( ∥ T n z n − T y ∥ 2 − ∥ z n − T y ∥ 2 ) .$

Dividing this inequality by n, we get

$0 ≤ ∥ T y − y ∥ 2 + 2 〈 y n − T y , T y − y 〉 + 2 〈 1 n β 1 ( T n + 1 z n − T n z n − z n − T z n ) + 1 n β 2 ( T n z n − z n ) , T y − y 〉 + 1 n ( β 1 − α 1 ) ( ∥ T n + 1 z n − T y ∥ 2 + ∥ T n z n − T y ∥ 2 − ∥ z n − T y ∥ 2 − ∥ T z n − T y ∥ 2 ) + 1 n ( β 2 − α 2 ) ( ∥ T n z n − T y ∥ 2 − ∥ z n − T y ∥ 2 ) .$

Replacing n by $n k i$ and letting $i→∞$ in the last inequality, we have

(3.8)

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

$0≤ ∥ T w − w ∥ 2 +2〈w−Tw,Tw−w〉=− ∥ T w − w ∥ 2 ,$

which ensures that $w∈F(T)$.

(b) We prove that $w∈ ( A + B ) − 1 0$. From (3.3), (3.4) and (3.6), we have

$∥ x n + 1 − p ∥ 2 ≤ ∥ ( I − α n V ) y n − ( I − α n V ) p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ( 1 − α n τ ) 2 ∥ y n − p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ( 1 − α n τ ) 2 ∥ z n − p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ( 1 − α n τ ) 2 { ∥ x n − p ∥ 2 − λ n ( 2 α − λ n ) ∥ A u n − A p ∥ 2 } + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 = ( 1 − 2 α n τ + α n 2 τ 2 ) ∥ x n − p ∥ 2 − ( 1 − α n τ ) 2 λ n ( 2 α − λ n ) ∥ A u n − A p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ∥ x n − p ∥ 2 + α n 2 τ 2 ∥ x n − p ∥ 2 − ( 1 − α n τ ) 2 λ n ( 2 α − λ n ) ∥ A u n − A p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ,$
(3.9)

and hence

$( 1 − α n τ ) 2 λ n ( 2 α − λ n ) ∥ A u n − A p ∥ 2 ≤ ∥ x n − p ∥ 2 − ∥ x n + 1 − p ∥ 2 + α n 2 τ 2 ∥ x n − p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 .$
(3.10)

Replacing n by $n k i$ in (3.10), we have

$( 1 − α n k i τ ) 2 λ n k i ( 2 α − λ n k i ) ∥ A u n k i − A p ∥ 2 ≤ ∥ x n k i − p ∥ 2 − ∥ x n k i + 1 − p ∥ 2 + α n k i 2 τ 2 ∥ x n k i − p ∥ 2 + 2 α n k i 〈 γ g ( x n ) − V p , x n k i + 1 − p 〉 .$

Since $lim n → ∞ α n =0$, $0 and the existence of $lim i → ∞ ∥ x n k i −p∥$, we have

$lim i → ∞ ∥A u n k i −Ap∥=0.$
(3.11)

We also have from (1.16) that

$2 ∥ u n − p ∥ 2 = 2 ∥ T r n x n − T r n p ∥ 2 ≤ 2 〈 x n − p , u n − p 〉 = ∥ x n − p ∥ 2 + ∥ u n − p ∥ 2 − ∥ u n − x n ∥ 2 ,$

and hence

$∥ u n − p ∥ 2 ≤ ∥ x n − p ∥ 2 − ∥ u n − x n ∥ 2 .$
(3.12)

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

$∥ x n + 1 − p ∥ 2 ≤ ∥ ( I − α n V ) y n − ( I − α n V ) p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ( 1 − α n τ ) 2 ∥ y n − p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ( 1 − α n τ ) 2 ∥ z n − p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ( 1 − α n τ ) 2 { ∥ u n − p ∥ 2 − λ n ( 2 α − λ n ) ∥ A u n − A p ∥ 2 } + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ( 1 − α n τ ) 2 { ∥ x n − p ∥ 2 − ∥ u n − x n ∥ 2 } − ( 1 − α n τ ) 2 λ n ( 2 α − λ n ) ∥ A u n − A p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ( 1 − 2 α n τ + α n 2 τ 2 ) ∥ x n − p ∥ 2 − ( 1 − α n τ ) 2 ∥ u n − x n ∥ 2 − ( 1 − α n τ ) 2 λ n ( 2 α − λ n ) ∥ A u n − A p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ∥ x n − p ∥ 2 + α n 2 τ 2 ∥ x n − p ∥ 2 − ( 1 − α n τ ) 2 ∥ u n − x n ∥ 2 − ( 1 − α n τ ) 2 λ n ( 2 α − λ n ) ∥ A u n − A p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ,$

and hence

$( 1 − α n τ ) 2 ∥ u n − x n ∥ 2 ≤ ∥ x n − p ∥ 2 − ∥ x n + 1 − p ∥ 2 + α n 2 τ 2 ∥ x n − p ∥ 2 − ( 1 − α n τ ) 2 λ n ( 2 α − λ n ) ∥ A u n − A p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 .$
(3.13)

Replacing n by $n k i$ in (3.13), we have

$( 1 − α n k i τ ) 2 ∥ u n k i − x n k i ∥ 2 ≤ ∥ x n k i − p ∥ 2 − ∥ x n k i + 1 − p ∥ 2 + α n k i 2 τ 2 ∥ x n k i − p ∥ 2 − ( 1 − α n k i τ ) 2 λ n k i ( 2 α − λ n k i ) ∥ A u n k i − A p ∥ 2 + 2 α n k i 〈 γ g ( x n k i ) − V p , x n k i + 1 − p 〉 .$

From (3.11), $lim n → ∞ α n =0$ and the existence of $lim i → ∞ ∥ x n k i −p∥$, we have

$lim i → ∞ ∥ u n k i − x n k i ∥=0.$
(3.14)

On the other hand, since $J λ n$ is firmly nonexpansive and $u n = T r n x n$, we have that

$∥ z n − p ∥ 2 = ∥ J λ n ( I − λ n A ) u n − J λ n ( I − λ n A ) p ∥ 2 ≤ 〈 z n − p , ( I − λ n A ) u n − ( I − λ n A ) p 〉 = 1 2 ( ∥ z n − p ∥ 2 + ∥ ( I − λ n A ) u n − ( I − λ n A ) p ∥ 2 − ∥ z n − p − ( I − λ n A ) u n + ( I − λ n A ) p ∥ 2 ) ≤ 1 2 { ∥ z n − p ∥ 2 + ∥ u n − p ∥ 2 − ∥ z n − p − ( I − λ n A ) u n + ( I − λ n A ) p ∥ 2 } ≤ 1 2 ( ∥ z n − p ∥ 2 + ∥ x n − p ∥ 2 − ∥ z n − u n ∥ 2 − 2 λ n 〈 z n − u n , A u n − A p 〉 − λ n 2 ∥ A u n − A p ∥ 2 ) ,$

and hence

$∥ z n − p ∥ 2 ≤ ∥ x n − p ∥ 2 − ∥ z n − u n ∥ 2 − 2 λ n 〈 z n − u n , A u n − A p 〉 − λ n 2 ∥ A u n − A p ∥ 2 .$
(3.15)

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

$∥ x n + 1 − p ∥ 2 ≤ ∥ ( I − α n V ) y n − ( I − α n V ) p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ( 1 − α n τ ) 2 ∥ y n − p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ( 1 − α n τ ) 2 ∥ z n − p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ( 1 − α n τ ) 2 ( ∥ x n − z ∥ 2 − ∥ z n − u n ∥ 2 − 2 λ n 〈 z n − u n , A u n − A p 〉 − λ n 2 ∥ A u n − A p ∥ 2 ) + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ≤ ∥ x n − p ∥ 2 + α n 2 τ 2 ∥ x n − p ∥ 2 − ( 1 − α n τ ) 2 ∥ z n − u n ∥ − 2 ( 1 − α n τ ) 2 λ n ( λ n − 2 α ) ∥ z n − u n ∥ ∥ A u n − A p ∥ − ( 1 − α n τ ) 2 λ n 2 ∥ A u n − A p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 ,$

and hence

$( 1 − α n τ ) 2 ∥ z n − u n ∥ ≤ ∥ x n − p ∥ 2 − ∥ x n + 1 − p ∥ 2 + α n 2 τ 2 ∥ x n − p ∥ 2 − 2 ( 1 − α n τ ) 2 λ n ( λ n − 2 α ) ∥ z n − u n ∥ ∥ A u n − A p ∥ − ( 1 − α n τ ) 2 λ n 2 ∥ A u n − A p ∥ 2 + 2 α n 〈 γ g ( x n ) − V p , x n + 1 − p 〉 .$
(3.16)

Replacing n by $n k i$ in (3.16), we have

$( 1 − α n k i τ ) 2 ∥ z n k i − u n k i ∥ 2 ≤ ∥ x n k i − p ∥ 2 − ∥ x n k i + 1 − p ∥ 2 + α n k i 2 τ 2 ∥ x n k i − p ∥ 2 − 2 ( 1 − α n k i τ ) 2 λ n k i ( λ n k i − 2 α ) ∥ z n k i − u n k i ∥ ∥ A u n k i − A p ∥ − ( 1 − α n k i τ ) 2 λ n k i 2 ∥ A u n k i − A p ∥ 2 + 2 α n k i 〈 γ g ( x n k i ) − V p , x n k i + 1 − p 〉 .$

From (3.11), $lim n → ∞ α n =0$ and the existence of $lim i → ∞ ∥ x n k i −p∥$, we obtain that

$lim i → ∞ ∥ z n k i − u n k i ∥=0.$
(3.17)

Since $∥ z n k i − x n k i ∥≤∥ z n k i − u n k i ∥+∥ u n k i − x n k i ∥$, by (3.14) and (3.17), we obtain that

$lim i → ∞ ∥ z n k i − x n k i ∥=0.$
(3.18)

Since A is Lipschitz continuous, we also obtain

$lim i → ∞ ∥A z n k i −A x n k i ∥=0.$
(3.19)

Since $z n = J λ (I−λA) u n$, we have that

$z n = ( I + λ n B ) − 1 ( I − λ n A ) u n ⇔ ( I − λ n A ) u n ∈ ( I + λ n B ) z n = z n + λ n B z n ⇔ u n − z n − λ n A u n ∈ λ n B z n ⇔ 1 λ n ( u n − z n − λ n A u n ) ∈ B z n .$

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

$〈 z n − u , 1 λ n ( u n − z n − λ n A u n ) − v 〉 ≥0,$

and hence

$〈 z n − u , u n − z n − λ n ( A u n + v ) 〉 ≥0.$
(3.20)

Replacing n by $n k i$ in (3.20), we have that

$〈 z n k i − u , u n k i − z n k i − λ n k i ( A u n k i + v ) 〉 ≥0.$
(3.21)

Since $x n k i ⇀w$ and $x n k i − u n k i →0$, so $u n k i ⇀w$. From (3.17), we get that $z n k i ⇀w$, together with (3.21), we have that

$〈w−u,−Aw−v〉≥0.$

Since B is maximal monotone, $(−Aw)∈Bw$, that is, $w∈ ( A + B ) − 1 0$.

(c) Next, we show that $w∈ F − 1 0$. Since F is a maximal monotone operator, we have from (1.15) that $A r n k i x n k i ∈F T r n k i x n k i$, where $A r$ is the Yosida approximation of F for $r>0$. Furthermore, we have that for any $(u,v)∈F$,

$〈 u − u n k i , v − x n k i − u n k i r n k i 〉 ≥0.$

Since $lim inf n → ∞ r n >0$, $u n k i ⇀w$ and $x n k i − u n k i →0$, we have

$〈u−w,v〉≥0.$

Since F is a maximal monotone operator, we have $0∈Fw$, that is, $w∈ F − 1 0$. By (a), (b) and (c), we conclude that

$w∈F(T)∩ ( A + B ) − 1 0∩ F − 1 0.$

Using (3.7), we obtain

$lim sup n → ∞ 〈 ( V − γ g ) p 0 , x n − p 0 〉 = lim k → ∞ 〈 ( V − γ g ) p 0 , x n k − p 0 ) 〉 = 〈 ( V − γ g ) p 0 , w − p 0 ) 〉 ≥ 0 .$

Finally, we prove that $x n → p 0$. Notice that

$x n + 1 − p 0 = α n ( γ g ( x n ) − p 0 ) +(I− α n V) y n −(I− α n V) p 0 ,$

we have

$∥ x n + 1 − p 0 ∥ 2 ≤ ( 1 − α n τ ) 2 ∥ y n − p 0 ∥ 2 + 2 α n 〈 γ g ( x n ) − V p 0 , x n + 1 − p 0 〉 ≤ ( 1 − α n τ ) 2 ∥ x n − p 0 ∥ 2 + 2 α n 〈 γ g ( x n ) − V p 0 , x n + 1 − p 0 〉 ≤ ( 1 − α n τ ) 2 ∥ x n − p 0 ∥ 2 + 2 α n γ k ∥ x n − p 0 ∥ ∥ x n + 1 − p 0 ∥ + 2 α n 〈 γ g ( p 0 ) − V p 0 , x n + 1 − p 0 〉 ≤ ( 1 − α n τ ) 2 ∥ x n − p 0 ∥ 2 + α n γ k ( ∥ x n − p 0 ∥ 2 + ∥ x n + 1 − p 0 ∥ 2 ) + 2 α n 〈 γ g ( p 0 ) − V p 0 , x n + 1 − p 0 〉 ≤ { ( 1 − α n τ ) 2 + α n γ k } ∥ x n − p 0 ∥ 2 + α n γ k ∥ x n + 1 − p 0 ∥ 2 + 2 α n 〈 γ g ( p 0 ) − V p 0 , x n + 1 − p 0 〉 ,$

and hence

$∥ x n + 1 − p 0 ∥ 2 ≤ 1 − 2 α n τ + ( α n τ ) 2 + α n γ k 1 − α n γ k ∥ x n − p 0 ∥ 2 + 2 α n 1 − α n γ k 〈 γ g ( p 0 ) − V p 0 , x n + 1 − p 0 〉 = { 1 − 2 ( τ − γ k ) α n 1 − α n γ k } ∥ x n − p 0 ∥ 2 + ( α n τ ) 2 1 − α n γ k ∥ x n − p 0 ∥ 2 + 2 α n 1 − α n γ k 〈 γ g ( p 0 ) − V p 0 , x n + 1 − p 0 〉 = { 1 − 2 ( τ − γ k ) α n 1 − α n γ k } ∥ x n − p 0 ∥ 2 + α n ⋅ α n τ 2 1 − α n γ k ∥ x n − p 0 ∥ 2 + 2 α n 1 − α n γ k 〈 γ g ( p 0 ) − V p 0 , x n + 1 − p 0 〉 = ( 1 − β n ) ∥ x n − p 0 ∥ 2 + β n { α n τ 2 ∥ x n − p 0 ∥ 2 2 ( τ − γ k ) + 1 τ − γ k 〈 γ g ( p 0 ) − V p 0 , x n + 1 − p 0 〉 } ,$
(3.22)

where $β n = 2 ( τ − γ k ) α n 1 − α n γ k$. Since $∑ n = 1 ∞ β n =∞$, we have from Lemma 2.5 and (3.22) that $x n → p 0$. 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 $(−∞,∞]$. Then the subdifferential ∂f of f is defined as follows:

$∂f(x)= { z ∈ H : f ( x ) + 〈 z , y − x 〉 ≤ f ( y ) , y ∈ H }$

for all $x∈H$; 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.,

$i C (x)={ 0 , x ∈ C , ∞ , x ∉ C .$

Then $i C$ is a proper lower semicontinuous convex function of H into $(−∞,∞]$, and then the subdifferential $∂ i C$ of $i C$ is a maximal monotone operator. So, we can define the resolvent $J λ$ of $∂ i C$ for $λ>0$, i.e.,

$J λ x= ( I + λ ∂ i C ) − 1 x$

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

$u = J λ x ⇔ x ∈ u + λ ∂ i C u ⇔ x ∈ u + λ N C u ⇔ x − u ∈ λ N C u ⇔ 1 λ 〈 x − u , v − u 〉 ≤ 0 , ∀ v ∈ C ⇔ 〈 x − u , v − u 〉 ≤ 0 , ∀ v ∈ C ⇔ u = P C x ,$

where $N C u$ is the normal cone to C at u, i.e.,

$N C u= { x ∈ H : 〈 z , v − u 〉 ≤ 0 , ∀ v ∈ C } .$

Let C be a nonempty, closed and convex subset of H, and let $f:C×C→R$ be a bifunction. For solving the equilibrium problem, let us assume that the bifunction $f:C×C→R$ 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:

1. (A1)

$f(x,x)=0$ for all $x∈C$;

2. (A2)

f is monotone, i.e., $f(x,y)+f(y,x)≤0$ for any $x,y∈C$;

3. (A3)

for all $x,y,z∈C$,

$lim sup t ↓ 0 f ( t z + ( 1 − t ) x , y ) ≤f(x,y);$
4. (A4)

for all $x∈C$, $f(x,⋅)$ is convex and lower semicontinuous;

5. (B1)

for each $x∈H$ and $r>0$, there exist a bounded subset $D x ⊆C$ and $y x ∈C$ such that for any $z∈C∖ D x$,

$f(z, y x )+φ( y x )+ 1 r 〈 y x −z,z−x〉<φ(z);$
6. (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 $C×C$ into satisfying (A1)-(A5). Let $r>0$ and $x∈H$. Then there exists a unique $z∈C$ such that

$f(z,y)+ 1 r 〈y−z,z−x〉≥0,∀y∈C.$

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 $f:C×C→R$ be a bifunction which satisfies conditions (A1)-(A4), and let $φ:C→R∪{+∞}$ be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For $r>0$ and $x∈H$, define a mapping $T r :H→C$ as follows:

$T r (x)= { z ∈ C : f ( z , y ) + φ ( y ) + 1 r 〈 y − z , z − x 〉 ≥ φ ( z ) , ∀ y ∈ C }$

for all $x∈H$. Then following conclusions hold:

1. (1)

For each $x∈H$, $T r (x)≠∅$;

2. (2)

$T r$ is single-valued;

3. (3)

$T r$ is firmly nonexpansive, i.e., for any $x,y∈H$,

$∥ T r ( x ) − T r ( y ) ∥ 2 ≤ 〈 T r ( x ) − T r ( y ) , x − y 〉 ;$
4. (4)

$Fix( T r )=MEP(f,φ)$;

5. (5)

$MEP(f,φ)$ 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 $f:C×C→R$ satisfy (A1)-(A5). Let $A f$ be a set-valued mapping of H into itself defined by

$A f x={ { z ∈ H : f ( x , y ) ≥ 〈 y − x , z 〉 , ∀ y ∈ C } , ∀ x ∈ C , ∅ , ∀ x ∉ C .$

Then $MEP(f)= A f − 1 0$ and $A f$ is a maximal monotone operator with $dom A f ⊂C$. Furthermore, for any $x∈H$ 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 $f:C×C→R$ satisfy (A1)-(A4), and let $φ:C→R∪{+∞}$ be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) hold. Let $A ( f , φ )$ be a set-valued mapping of H into itself defined by

$A ( f , φ ) x={ { z ∈ H : f ( x , y ) + φ ( y ) − φ ( x ) ≥ 〈 y − x , z 〉 , ∀ y ∈ C } , ∀ x ∈ C , ∅ , ∀ x ∉ C .$
(4.1)

Then $MEP(f,φ)= A ( f , φ ) − 1 0$ and $A ( f , φ )$ is a maximal monotone operator with $dom A ( f , φ ) ⊂C$. Furthermore, for any $x∈H$ and $r>0$, the resolvent $T r$ of f coincides with the resolvent of $A ( f , φ )$, i.e.,

$T r x= ( I + r A ( f , φ ) ) − 1 x.$

Proof It is obvious that $MEP(f,φ)= A ( f , φ ) − 1 0$. In fact, we have that

$z ∈ MEP ( f , φ ) ⇔ f ( z , y ) + φ ( y ) − φ ( z ) ≥ 0 , ∀ y ∈ C ⇔ f ( z , y ) + φ ( y ) − φ ( z ) ≥ 〈 y − z , 0 〉 , ∀ y ∈ C ⇔ 0 ∈ A ( f , φ ) z ⇔ z ∈ A ( f , φ ) − 1 0 .$

We show that $A ( f , φ )$ is monotone. Let $( x 1 , z 1 ),( x 2 , z 2 )∈ A ( f , φ )$ be given. Then we have, for all $y∈C$,

$f( x 1 ,y)+φ(y)−φ( x 1 )≥〈y− x 1 , z 1 〉andf( x 2 ,y)+φ(y)−φ( x 2 )≥〈y− x 2 , z 2 〉,$

and hence

$f( x 1 , x 2 )+φ( x 2 )−φ( x 1 )≥〈 x 2 − x 1 , z 1 〉andf( x 2 , x 1 )+φ( x 1 )−φ( x 2 )≥〈 x 1 − x 2 , z 2 〉.$

It follows from (A2) that

$0≥f( x 1 , x 2 )+f( x 2 , x 1 )≥〈 x 2 − x 1 , z 1 〉+〈 x 1 − x 2 , z 2 〉=−〈 x 1 − x 2 , z 1 − z 2 〉.$

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

$f(z,y)+φ(y)−φ(z)+ 1 r 〈y−z,z−x〉≥0,∀y∈C.$

So, we have that

$f(z,y)+φ(y)−φ(z)≥ 〈 y − z , 1 r ( x − z ) 〉 ,∀y∈C.$

By the definition of $A ( f , φ )$, we get

$A ( f , φ ) z∋ 1 r (x−z),$

and hence $x∈z+r A ( f , φ ) z$. Therefore, $H⊂R(I+r A ( f , φ ) )$ and $R(I+r A ( f , φ ) )=H$. Also, $x∈z+r A ( f , φ ) z$ implies that $T r x= ( I + r A ( f , φ ) ) − 1 x$ for all $x∈H$ 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 $α>0$ and let A be an α-inverse-strongly monotone mapping of C into H. Let $0 and let g be a k-contraction of H into itself. Let V be a $γ ¯$-strongly monotone and L-Lipschitzian continuous operator with $γ ¯ >0$ and $L>0$. Let $T:C→C$ be a 2-generalized hybrid mapping such that $Γ:=F(T)∩VI(C,A)≠∅$. Take $μ,γ∈R$ as follows:

$0<μ< 2 γ ¯ L 2 ,0<γ< γ ¯ − L 2 μ 2 k .$

Let ${ x n }⊂H$ be a sequence generated by

(4.2)

where ${ α n }⊂(0,1)$ and ${ r n }⊂(0,∞)$ satisfy

$lim n → ∞ α n =0, ∑ n = 1 ∞ α n =∞and lim inf n → ∞ r n >0.$

Then ${ x n }$ converges strongly to a point $p 0$ of Γ, where $p 0$ is a unique fixed point of $P Γ (I−V+γg)$. This point $p 0 ∈Γ$ is also a unique solution of the hierarchical variational inequality

$〈 ( V − γ g ) p 0 , q − p 0 〉 ≥0,∀q∈VI(C,A).$
(4.3)

Proof Put $B=F=∂ i C$ in Theorem 3.1. Then, for $λ n >0$ and $r n >0$, we have that

$J λ n = T r n = P C .$

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

$( ∂ i C ) − 1 0=Cand ( A + ∂ i C ) − 1 =VI(C,A).$

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 $α>0$ and let A be an α-inverse-strongly monotone mapping of C into H. Let $B:D(B)⊂C→ 2 H$ be maximal monotone. Let $J λ = ( I + λ B ) − 1$ be the resolvent of B for $λ>0$. Let $0 and let g be a k-contraction of H into itself. Let V be a $γ ¯$-strongly monotone and L-Lipschitzian continuous operator with $γ ¯ >0$ and $L>0$. Let $f:C×C→R$ be a bifunction satisfying conditions (A1)-(A4), and let $φ:C→R∪{+∞}$ be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. Let $T:C→C$ be a 2-generalized hybrid mapping with $Θ:=F(T)∩ ( A + B ) − 1 0∩MEP(f,φ)≠∅$. Take $μ,γ∈R$ as follows:

$0<μ< 2 γ ¯ L 2 ,0<γ< γ ¯ − L 2 μ 2 k .$

Let ${ x n }⊂H$ be a sequence generated by

${ x 1 = x ∈ H arbitrarily , f ( u n , y ) + φ ( y ) − φ ( u n ) + 1 r n 〈 y − u n , u n − x n 〉 ≥ 0 , ∀ y ∈ C , z n = J λ n ( I − λ n A ) u n , y n = 1 n ∑ k = 0 n − 1 T k z n , ∀ n = 1 , 2 , … , x n + 1 = α n γ g ( x n ) + ( I − α n V ) y n , ∀ n ∈ N ,$
(4.4)

where ${ α n }⊂(0,1)$ and ${ r n }⊂(0,∞)$ satisfy

$lim n → ∞ α n =0, ∑ n = 1 ∞ α n =∞and lim inf n → ∞ r n >0.$

Then ${ x n }$ converges strongly to a point $p 0$ of Θ, where $p 0$ is a unique fixed point of $P Θ (I−V+γg)$. This point $p 0 ∈Θ$ is also a unique solution of the hierarchical variational inequality

$〈 ( V − γ g ) p 0 , q − p 0 〉 ≥0,∀q∈Θ.$
(4.5)

Proof Since f is a bifunction of $C×C$ into satisfying conditions (A1)-(A4) and $φ:C→R∪{+∞}$ is a proper lower semicontinuous and convex function, we have that the mapping $A f φ$ defined by (4.1) is a maximal monotone operator with $dom A f φ ⊂C$. Put $F= A f φ$ in Theorem 3.1. Then we obtain that $u n = T r n x n$. Therefore, we arrive at the desired results. □

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Acknowledgements

The first author was partially supported by Naresuan University.

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Correspondence to Rabian Wangkeeree.

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Wangkeeree, R., Boonkong, U. A general iterative method for two maximal monotone operators and 2-generalized hybrid mappings in Hilbert spaces. Fixed Point Theory Appl 2013, 246 (2013). https://doi.org/10.1186/1687-1812-2013-246

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Keywords

• 2-generalized hybrid mapping
• inverse strongly monotone mapping
• maximal monotone mapping
• hierarchical variational inequality