Open Access

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

Fixed Point Theory and Applications20132013:246

https://doi.org/10.1186/1687-1812-2013-246

Received: 18 May 2013

Accepted: 20 August 2013

Published: 7 November 2013

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.

Keywords

2-generalized hybrid mappinginverse strongly monotone mappingmaximal monotone mappinghierarchical variational inequality

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
f ( x , y ) + φ ( y ) φ ( x ) 0 for all  y C .
(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
f ( x , y ) 0 for all  y C .
(1.2)

The set of solutions of (1.2) is denoted by E P ( 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 : T x = x } the set of fixed points of T. A mapping T : C C is said to be nonexpansive if T x T y x y for all x , y C . The mapping T : C C is said to be firmly nonexpansive if
T x T y 2 x y , T x T y for all  x , y C ;
(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 T x , y T y
(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 = T x , 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 T y 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
V x , x γ ¯ x 2 for all  x H .
(1.8)
In general, a nonlinear operator V : H H is called strongly monotone if there exists γ ¯ > 0 such that
x y , V x V y γ ¯ x y 2 for all  x , y H .
(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
V x V y L x y for all  x , y H .
(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 , A x A y α A x A y 2 for all x , y C . It is known that A x A y ( 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 : B x } . 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 B x , and v B y . 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 B x } . 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 < c β n d < 1 and 0 < a λ n b < 2 α . 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:
{ x 1 = x H arbitrarily , z n = J λ n ( I λ n A ) x n , y n = 1 n k = 0 n 1 T k z n , x n + 1 = α n u + ( 1 α n ) y n for all  n N ,
(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 V x , 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 γ f ( 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 V x , x h ( x ) ,
(1.21)

where h is a potential function for γf (i.e., h ( x ) = γ f ( 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 γ g ( 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
x n + 1 = α n γ g ( x n ) + ( I α n V ) J λ n ( I λ n A ) T r n x n for all  n N ,
(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 A u = A v ;

     
  2. (ii)

    for any λ > 0 , u ( A + B ) 1 ( 0 ) if and only if u = J λ ( I λ A ) 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 < k < 1 . 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 γ g : H H is a ( γ ¯ γ k ) -strongly monotone and ( L + γ k ) -Lipschitzian continuous mapping. Furthermore, let C be a nonempty closed convex subset of H. Then P C ( I V + γ g ) 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 < k < 1 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 < a λ n b < 2 α 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 + γ g ) . 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
x n + 1 p α n γ k x n p + α n γ g ( p ) V p + ( 1 α n τ ) y n p α n γ k x n p + α n γ g ( p ) V p + ( 1 α n τ ) x n p = ( 1 α n ( τ γ k ) ) x n p + α n γ g ( p ) V p = ( 1 α n ( τ γ k ) ) x n p + α n ( τ γ k ) γ g ( p ) V p τ γ k max { x n p , γ g ( p ) V p τ γ k } for all  n N ,
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 γ g ) 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
0 T y y 2 + 2 w T y , T y y for all  y C .
(3.8)
In particular, replacing y by w in (3.8), we obtain that
0 T w w 2 + 2 w T w , T w 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 < a λ n b < 2 α and the existence of lim i x n k i p , we have
lim i A u n k i A p = 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 λ A ) 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 , A w v 0 .

Since B is maximal monotone, ( A w ) B w , 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 F w , 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 and f ( 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 and f ( 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 < k < 1 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 ) V I ( 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
{ x 1 = x H arbitrarily , z n = P C ( I λ n A ) P C x 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 for all  n N ,
(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 + γ g ) . 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 V I ( 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 = C and ( A + i C ) 1 = V I ( 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 < k < 1 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 + γ g ) . 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. □

Declarations

Acknowledgements

The first author was partially supported by Naresuan University.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Naresuan University

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© Wangkeeree and Boonkong; licensee Springer. 2013

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