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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:CR be a real-valued function, and let f:C×CR be an equilibrium bifunction, that is, f(u,u)=0 for each uC. The mixed equilibrium problem is to find xC such that

f(x,y)+φ(y)φ(x)0for all yC.
(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 xC such that

f(x,y)0for all yC.
(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):={xC:Tx=x} the set of fixed points of T. A mapping T:CC is said to be nonexpansive if TxTyxy for all x,yC. The mapping T:CC is said to be firmly nonexpansive if

T x T y 2 xy,TxTyfor all x,yC;
(1.3)

see, for instance, Browder [3] and Goebel and Kirk [4]. The mapping T:CC 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,yC. Iemoto and Takahashi [6] proved that T:CC is nonspreading if and only if

T x T y 2 x y 2 +2xTx,yTy
(1.5)

for all x,yC. 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:CC is said to be generalized hybrid if there are α,βR such that

α T x T y 2 +(1α) x T y 2 β T x y 2 +(1β) x y 2
(1.6)

for all x,yC. 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,yC; 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 yC,

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 xTyxy. 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

Vx,x γ ¯ x 2 for all xH.
(1.8)

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

xy,VxVy γ ¯ x y 2 for all x,yH.
(1.9)

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

VxVyLxyfor all x,yH.
(1.10)

Such V is called L-Lipschitzian continuous. A mapping A:CH is said to be α-inverse-strongly monotone if xy,AxAyα A x A y 2 for all x,yC. It is known that AxAy( 1 α )xy for all x,yC 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 xC. 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:CC be a 2-generalized hybrid mapping with F(T). Suppose that { x n } is a sequence generated by x 1 =xC, uC and

x n + 1 = γ n u+(1 γ n ) 1 n k = 0 n 1 T k x n ,nN,
(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)={xH:Bx}. A multi-valued mapping B on H is said to be monotone if xy,uv0 for all x,yD(B), uBx, and vBy. 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 :HD(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 xB J r x,xH,r>0.
(1.15)

Let B be a maximal monotone operator on H, and let B 1 0={xH:0Bx}. 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 yxy, J r x J r y,x,yH.
(1.16)

Recently, in the case when T:CC is a nonexpansive mapping, A:CH is an α-inverse strongly monotone mapping and BH×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<2α. 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 Vx,xx,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 ),n0,
(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,xC,

which is the optimality condition for the minimization problem

min x C 1 2 Vx,xh(x),
(1.21)

where h is a potential function for γf (i.e., h (x)=γf(x) for xH).

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 )+(Iμ α n V)T x n ,

for all nN, where μ,γR 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,qF(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 =xH and let { x n }H be a sequence generated

x n + 1 = α n γg( x n )+(I α n V) J λ n (I λ n A) T r n x n for all nN,
(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 xxy for all xH and yC. 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,xy
(2.1)

for all x,yH. Furthermore, P C x P C y,xy0 holds for all xH and yC; 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 xH, define the resolvent J r x. Then the following holds:

s t s J s x J t x, J s xx J s x J t x 2
(2.2)

for all s,t>0 and xH.

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 xH; 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:CC is a nonexpansive mapping, then IT 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 (IλA)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:HH 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 Vγg:HH 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 (IV+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,qC.

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:CC be a 2-generalized hybrid mapping such that Ω:=F(T) ( A + B ) 1 0 F 1 0. Take μ,γR 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<2α for all nN;

  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 (IV+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,qΩ.
(3.2)

Proof First we prove that { x n } is bounded and lim n x n p exists for all pΩ. Let pΩ, 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,qΩ.
(3.7)

We show that lim sup n (Vγg) 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 wC, 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 wΩ.

(a) We first prove wF(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,yC. For any nN and k=0,1,2,,n1, 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 n1, 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 +2wTy,Tyyfor all yC.
(3.8)

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

0 T w w 2 +2wTw,Tww= T w w 2 ,

which ensures that wF(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<2α 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 (IλA) 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

wu,Awv0.

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

uw,v0.

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

wF(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 xH; 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 xH. We have that for any xH and uC,

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×CR be a bifunction. For solving the equilibrium problem, let us assume that the bifunction f:C×CR 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 xC;

  2. (A2)

    f is monotone, i.e., f(x,y)+f(y,x)0 for any x,yC;

  3. (A3)

    for all x,y,zC,

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

    for all xC, f(x,) is convex and lower semicontinuous;

  5. (B1)

    for each xH and r>0, there exist a bounded subset D x C and y x C such that for any zC D x ,

    f(z, y x )+φ( y x )+ 1 r y x z,zx<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 xH. Then there exists a unique zC such that

f(z,y)+ 1 r yz,zx0,yC.

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×CR be a bifunction which satisfies conditions (A1)-(A4), and let φ:CR{+} be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For r>0 and xH, define a mapping T r :HC as follows:

T r (x)= { z C : f ( z , y ) + φ ( y ) + 1 r y z , z x φ ( z ) , y C }

for all xH. Then following conclusions hold:

  1. (1)

    For each xH, T r (x);

  2. (2)

    T r is single-valued;

  3. (3)

    T r is firmly nonexpansive, i.e., for any x,yH,

    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×CR 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 xH 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×CR satisfy (A1)-(A4), and let φ:CR{+} 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 xH 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 yC,

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

0f( 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 xH and r>0. Then, from Lemma 4.2, there exists zC such that

f(z,y)+φ(y)φ(z)+ 1 r yz,zx0,yC.

So, we have that

f(z,y)+φ(y)φ(z) y z , 1 r ( x z ) ,yC.

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

A ( f , φ ) z 1 r (xz),

and hence xz+r A ( f , φ ) z. Therefore, HR(I+r A ( f , φ ) ) and R(I+r A ( f , φ ) )=H. Also, xz+r A ( f , φ ) z implies that T r x= ( I + r A ( f , φ ) ) 1 x for all xH 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:CC be a 2-generalized hybrid mapping such that Γ:=F(T)VI(C,A). Take μ,γR 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 (IV+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,qVI(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<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×CR be a bifunction satisfying conditions (A1)-(A4), and let φ:CR{+} be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. Let T:CC be a 2-generalized hybrid mapping with Θ:=F(T) ( A + B ) 1 0MEP(f,φ). Take μ,γR 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 (IV+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,qΘ.
(4.5)

Proof Since f is a bifunction of C×C into satisfying conditions (A1)-(A4) and φ:CR{+} 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