Open Access

Hybrid algorithms of nonexpansive semigroups for mixed equilibrium problems, variational inequalities, and fixed point problems

Journal of Inequalities and Applications20142014:174

https://doi.org/10.1186/1029-242X-2014-174

Received: 22 March 2014

Accepted: 14 April 2014

Published: 12 May 2014

Abstract

The purpose of this paper is to introduce two hybrid algorithms for the variational inequalities and mixed equilibrium problems over the common fixed points set of nonexpansive semigroups in Hilbert space. Under suitable conditions some strong convergence theorems for these two hybrid algorithms are proved. The results presented in the paper extend and improve some recent results.

Keywords

mixed equilibrium problem variational inequalities fixed point nonexpansive semigroups

1 Introduction

Throughout this paper, we always assume that H is a real Hilbert space with inner product , and norm , C is a nonempty closed convex subset of H and P C is the metric projection of H onto C. In the sequel, we denote by → and the strong convergence and weak convergence, respectively. Let ϕ : C be a real-valued function and Θ : C × C be an equilibrium bifunctions, i.e., Θ ( u , u ) = 0 for each u C . We consider the mixed equilibrium problem (MEP) which is to find x C such that
M E P : Θ ( x , y ) + φ ( y ) φ ( x ) 0 , y C .
In particular, if φ 0 , this problem reduces to the equilibrium problem (EP), which is to find x C such that
E P : Θ ( x , y ) 0 .

Denote the set of solutions of MEP by Ω. The MEP includes fixed point problems, optimization problems, variational inequality problems, Nash EPS and the EP as special cases.

Recall that a mapping T : C C is said to be nonexpansive, if T x T y x y , x , y C .

Let C be a closed convex subset of a Hilbert space H. A family of mappings S : = { S ( s ) : 0 s < } : C C is said to be a nonexpansive semigroup, if it satisfies the following conditions:
  1. (i)

    S ( s + t ) = S ( s ) S ( t ) , s , t + and S ( 0 ) = I ;

     
  2. (ii)

    S ( s ) x S ( s ) y x y , x , y C , s 0 ;

     
  3. (iii)

    the mapping t S ( t ) x is continuous for each x C .

     

We denote by F ( S ) the set of common fixed points of , i.e., F ( S ) = s 0 F ( S ( s ) ) . It is well known that F ( S ) is closed and convex.

Now let F : C C be a nonlinear operator. The variational inequality problem is formulated as finding a point x C such that
V I ( F , C ) : F x , v x 0 , v C .
It is well known that the V I ( F , C ) is equivalent to the fixed point equation
x = P C [ x μ F ( x ) ] ,
(1.1)

where μ > 0 is an arbitrarily fixed constant. So, fixed point methods can be implemented to find a solution of the V I ( F , C ) provided F satisfies some conditions and μ > 0 is chosen appropriately. The fixed point formulation (1.1) involves the projection P C , which may not be easy to compute, due to the complexity of the convex set C. In order to reduce the complexity probably caused by the projection P C , Yamada [1] recently introduced a hybrid steepest-descent method for solving the V I ( F , C ) . Assume that F is an η-strongly monotone and κ-Lipschitzian mapping with κ > 0 , η > 0 on C. An equally important problem is how to find an approximate solution of the V I ( F , C ) if any. A great deal of effort has been done in this problem.

In 2007, Ceng and Yao [2] investigate the problem of finding a common element of the set of solutions of a mixed equilibrium problem (MEP) and the set of common fixed points of finitely many nonexpansive mappings in a real Hilbert space. Very recently, Yang et al. [3] introduce two hybrid algorithms for finding a common fixed point of a nonexpansive semigroup in Hilbert space.

Motivated and inspired by Ceng and Yao [2] and Yang et al. [3], the purpose of this paper is to introduce two hybrid algorithms for the variational inequalities and mixed equilibrium problems over the common fixed points set of nonexpansive semigroups in Hilbert space. Under suitable conditions some strong convergence theorem for these two hybrid algorithms are proved. The results presented in the paper extend and improve some recent results.

2 Preliminaries

Let H be a real Hilbert space and C be a nonempty closed convex subset of H. For solving mixed equilibrium problems, let us assume that the function Θ : C × C satisfies the following conditions:

(H1) Θ is monotone, i.e., Θ ( x , y ) + Θ ( y , x ) 0 , x , y C ;

(H2) for each fixed y C , the mapping x Θ ( x , y ) is concave and upper semicontinuous;

(H3) for each fixed x C , the mapping y Θ ( x , y ) is convex.

A mapping F : C C is said to be:
  1. (i)
    κ-Lipschitz continuous, if there exists a constant κ > 0 such that
    F x F y κ x y , x , y C ;
     
  2. (ii)
    η-strongly monotone, if there exists a constant η > 0 such that
    F x F y , x y η x y 2 , x , y C .
     
A differentiable function K : C is said to be:
  1. (i)
    ξ-convex [4], if
    K ( y ) K ( x ) K ( x ) , ξ ( y , x ) , x , y C ,
     
where K ( x ) is the Fréchet derivative of K at x;
  1. (ii)
    ξ-strongly convex [5], if there exists a constant σ > 0 such that
    K ( y ) K ( x ) K ( x ) , ξ ( y , x ) ( σ / 2 ) x y 2 , x , y C .
     

The following lemmas will be needed in proving our main results.

Lemma 2.1 [2]

Let C be a nonempty closed convex subset of a real Hilbert space H and φ : C be a lower semicontinuous and convex functional. Let θ : C × C be an equilibrium bifunction satisfying the conditions (H1)-(H3). Assume that
  1. (i)

    ξ : C × C H is λ-Lipschitz continuous such that

     
  2. (a)

    ξ ( x , y ) + ξ ( y , x ) = 0 , x , y C ;

     
  3. (b)

    ξ ( , ) is affine in the first variable;

     
  4. (c)

    for each fixed x C , the mapping y ξ ( x , y ) is sequentially continuous from the weak topology to the weak topology;

     
  1. (ii)

    K : C is ξ-strongly convex with constant σ > 0 , and its derivative K is sequentially continuous from the weak topology to the strong topology;

     
  2. (iii)
    for each x C there exist a bounded subset D x C and a point z x C such that, for any y C D x ,
    Θ ( y , z x ) + φ ( z x ) φ ( y ) + 1 r K ( y ) K ( x ) , ξ ( z x , y ) < 0 .
     
For given r > 0 , let J r : C C be the mapping defined by
J r ( x ) = { y C : Θ ( y , z ) + φ ( z ) φ ( y ) + 1 r K ( y ) K ( x ) , ξ ( z , y ) 0 , z C } , x C .
(2.1)
Then
  1. (i)

    J r is single-valued;

     
  2. (ii)
    J r is nonexpansive if K is Lipschitz continuous with constant ν > 0 and
    K ( x 1 ) K ( x 2 ) , u 1 u 2 K ( u 1 ) K ( u 2 ) , u 1 u 2 , ( x 1 , x 2 ) C × C ,
     
where u i = J r ( x i ) for i = 1 , 2 ;
  1. (iii)
    F ( J r ) = Ω ( Θ , φ ) , where Ω ( Θ , φ ) is the set of solutions of the following mixed equilibrium problem:
    Θ ( x , y ) + φ ( y ) φ ( x ) 0 , y C .
     
  2. (iv)

    Ω ( Θ , φ ) is closed and convex.

     

Lemma 2.2 [6]

Let x n and y n be bounded sequences in a Banach space E and let β n be a sequence in [ 0 , 1 ] with 0 < lim inf n β n lim sup n β n < 1 . Suppose that x n + 1 = ( 1 β n ) y n + β n x n for all integers n 1 and lim sup n ( y n + 1 y n x n + 1 x n ) 0 . Then lim n y n x n = 0 .

Lemma 2.3 [7]

Let { a n } be a sequence of nonnegative real numbers such that
a n + 1 ( 1 λ n ) a n + λ n δ n + γ n , n n 0 ,
where n 0 is some nonnegative integer, { λ n } , { δ n } , and { γ n } are sequences satisfying
  1. (i)

    λ n [ 0 , 1 ] and n = 0 λ n = ,

     
  2. (ii)

    lim sup n δ n 0 or n = 0 λ n δ n < ,

     
  3. (iii)

    γ n 0 ( n 0 ), n = 0 γ n < .

     

Then lim n a n = 0 .

Lemma 2.4 [8]

Let C be a bounded closed convex subset of H and S = { S ( s ) : 0 s < } be a nonexpansive semigroup on C, then for any h > 0
lim s sup x C 1 t 0 t S ( s ) x d s S ( h ) ( 1 t 0 t S ( s ) x d s ) = 0 .

Lemma 2.5 [9]

Let C be a nonempty bounded closed convex subset of H, x n be a sequence in C and S = { S ( s ) : 0 s < } be a nonexpansive semigroup on C. If the following conditions are satisfied:
  1. (i)

    x n z ;

     
  2. (ii)

    lim sup n lim sup n S ( s ) x n x n = 0 ,

     

then z F ( S ) .

Lemma 2.6 [10]

Let F be an η-strongly monotone and κ-Lipschitzian operator on a Hilbert space H with 0 < η < κ and 0 < t < η / κ 2 . Then T = ( I t F ) : H H is a contraction with contraction coefficient τ t = 1 t ( 2 η t κ 2 ) .

Lemma 2.7 In a real Hilbert space H, we have the inequality
x y 2 = x 2 2 x , y + y 2 , λ x + ( 1 λ ) y 2 = λ x 2 + ( 1 λ ) y 2 λ ( 1 λ ) x y 2

for all x , y H and λ [ 0 , 1 ] .

Recall that a Banach space E is said to satisfy the Opial condition, if for any sequence x n in E with x n x , then for every y E with y x we have
lim inf n x n x < lim inf n x n y .

It is well known that each Hilbert space satisfies the Opial condition.

3 Main results

Now we will show our main results.

Theorem 3.1 Let H be a real Hilbert space. Let φ : H be a lower semicontinuous and convex functional. Let Θ : H × H be an equilibrium functions satisfying conditions (H1)-(H3). Let S : = { S ( s ) : 0 s < } be a nonexpansive semigroup on H. Let F be an η-strongly monotone and κ-Lipschitzian operator on H. Let { γ t } 0 < t < 1 be a continuous net of positive real numbers such that lim t 0 + γ t = + . Putting τ t = 1 t ( 2 η t κ 2 ) , for each t ( 0 , η / κ 2 ) , let the net { x t } be defined by the following implicit scheme:
x t = 1 γ t 0 γ t S ( s ) [ ( I t F ) J r x t ] d s ,
(3.1)
where J r : H H is the mapping defined by (2.1). Suppose the following conditions are satisfied:
  1. (i)

    ξ : H × H H is λ-Lipschitz continuous such that

     
  2. (a)

    ξ ( x , y ) + ξ ( y , x ) = 0 , x , y H ;

     
  3. (b)

    x ξ ( x , y ) is affine;

     
  4. (c)

    y ξ ( x , y ) is sequentially continuous from the weak topology to the weak topology;

     
  1. (ii)

    K : H is ξ-strongly convex with constant σ > 0 , and its derivative K is not only sequentially continuous from the weak topology to the strong topology but also Lipschitz continuous with a Lipschitz constant ν > 0 and σ λ ν ;

     
  2. (iii)
    for each x H there exist a bounded subset D x H and a point z x H such that, for any y H D x ,
    Θ ( y , z x ) + φ ( z x ) φ ( y ) + 1 r K ( y ) K ( x ) , ξ ( z x , y ) < 0 ,
     
and if Γ : = Ω ( Θ , φ ) F ( S ) . Then, as t 0 + , the net { x t } converges strongly to an element x of Γ provided J r is firmly nonexpansive which is the unique solution of the following variational inequality:
F x , x u 0 , u Γ .
(3.2)

Proof We divide the proof into several steps.

Step 1. First, we note that the net { x t } defined by (3.1) is well defined. In fact, we define a mapping
P t x : = 1 γ t 0 γ t S ( s ) [ ( I t F ) J r x ] d s , t ( 0 , η / κ 2 ) , x H .
(3.3)
Because J r is nonexpansive. It follows from Lemma 2.6 that
P t x P t y 1 γ t 0 γ t S ( s ) [ ( I t F ) J r x ] S ( s ) [ ( I t F ) J r y ] d s ( I t F ) x ( I t F ) y τ t x y .

Hence, the P t is a contraction, and so it has a unique fixed point. Therefore, the net { x t } defined by (3.1) is well defined.

Step 2. We prove that { x t } is bounded. Taking u Γ and using Lemma 2.6, we have
x t u = 1 γ t 0 γ t S ( s ) [ ( I t F ) J r x t ] d s u = 1 γ t 0 γ t S ( s ) [ ( I t F ) J r x t ] d s 1 γ t 0 γ t S ( s ) J r u d s 1 γ t 0 γ t S ( s ) [ ( I t F ) J r x t ] S ( s ) J r u d s ( I t F ) x t u ( I t F ) x t ( I t F ) u t F u τ t x t u + t F u .
It follows that
x t u t 1 τ t F u .
(3.4)
Observe that
lim t 0 + t 1 τ t = 1 η .
(3.5)

Thus, (3.4) and (3.5) imply that the net { x t } is bounded for small enough t. Without loss of generality, we may assume that the net { x t } is bounded for all t ( 0 , η / κ 2 ) . Consequently, we deduce that { F x t } and { J r x t } are also bounded.

Step 3. On the other hand, from (3.1) and (3.4), we have
x t S ( s ) x t S ( s ) x t S ( s ) ( 1 γ t 0 γ t S ( s ) x t d s ) + 1 γ t 0 γ t S ( s ) x t d s x t + S ( s ) ( 1 γ t 0 γ t S ( s ) x t d s ) 1 γ t 0 γ t S ( s ) x t d s 2 1 γ t 0 γ t S ( s ) x t d s x t + S ( s ) ( 1 γ t 0 γ t S ( s ) x t d s ) 1 γ t 0 γ t S ( s ) x t d s = 2 1 γ t 0 γ t S ( s ) x t d s 1 γ t 0 γ t S ( s ) [ ( I t F ) J r x t ] d s + S ( s ) ( 1 γ t 0 γ t S ( s ) x t d s ) 1 γ t 0 γ t S ( s ) x t d s 2 1 γ t 0 γ t S ( s ) x t S ( s ) [ ( I t F ) J r x t ] d s + S ( s ) ( 1 γ t 0 γ t S ( s ) x t d s ) 1 γ t 0 γ t S ( s ) x t d s 2 x t ( I t F ) J r x t + S ( s ) ( 1 γ t 0 γ t S ( s ) x t d s ) 1 γ t 0 γ t S ( s ) x t d s 2 x t J r x t + 2 t F J r x t + S ( s ) ( 1 γ t 0 γ t S ( s ) x t d s ) 1 γ t 0 γ t S ( s ) x t d s .
(3.6)
In fact, we have
x t u = 1 γ t 0 γ t S ( s ) [ ( I t F ) J r x t ] d s u = 1 γ t 0 γ t S ( s ) [ ( I t F ) J r x t ] d s 1 γ t 0 γ t S ( s ) u d s 1 γ t 0 γ t S ( s ) [ ( I t F ) J r x t ] S ( s ) u d s ( I t F ) J r x t u ,
so
x t u 2 ( I t F ) J r x t u 2 J r x t u 2 + t 2 F J r x t 2 2 t J r x t u , F J r x t x t u 2 x t J r x t 2 + t 2 F J r x t 2 2 t J r x t u , F J r x t ,
observe that
x t J r x t 2 t 2 F J r x t 2 2 t J r x t u , F J r x t ,
then
lim t 0 + x t J r x t = 0 .
(3.7)
This together with Lemma 2.4 and (3.6) implies that
lim t 0 + x t S ( s ) x t = 0 .
(3.8)
Let { t n } ( 0 , 1 ) be a sequence such that t n 0 as n . Put x n : = x t n and y n = J r x n . Since { y n } is bounded, there exists a subsequence { y n j } of { y n } which converges weakly to w H . Without loss of generality, we can assume that y n j w . Next we prove that
w Γ : = Ω ( Θ , φ ) F ( S ) .
  1. (a)
    In fact, we have
    y n S ( s ) y n = y n x n + x n S ( s ) x n + S ( s ) x n S ( s ) y n 2 y n x n + x n S ( s ) x n .
     
With (3.7) and (3.8), we have
lim n y n S ( s ) y n = 0 .
(3.9)
Indeed, from Lemma 2.5 and (3.9) we know that w F ( S ) , i.e., w = S ( s ) w , s 0 .
  1. (b)
    Now we prove that w Ω = Ω ( Θ , φ ) . In fact, since y n = J r x n , we have
    Θ ( J r x n , x ) + φ ( x ) φ ( J r x n ) + 1 r K ( J r x n ) K ( x n ) , ξ ( x , J r x n ) 0 , x H .
     
From the monotonicity of Θ, we have
1 r K ( J r x n ) K ( x n ) , ξ ( x , J r x n ) + φ ( x ) φ ( J r x n ) Θ ( J r x n , x ) Θ ( x , J r x n )
and hence
K ( J r x n j ) K ( x n j ) r , ξ ( x , J r x n j ) + φ ( x ) φ ( J r x n j ) Θ ( x , J r x n j ) .
(3.10)
Since (3.7), then ( K ( J r x n j ) K ( x n j ) ) / r 0 and y n j w , from the weak lower semicontinuity of φ and Θ ( x , y ) in the second variable y, we have Θ ( x , w ) + φ ( w ) φ ( x ) 0 for all x H . For 0 < t 1 and x H , let x t = t x + ( 1 t ) w . Since x H and w H , we have x t H and hence Θ ( x t , w ) + φ ( w ) φ ( x t ) 0 . From the convexity of equilibrium bifunction Θ ( x , y ) in the second variable y, we have
0 = Θ ( x t , x t ) + φ ( x t ) φ ( x t ) t Θ ( x t , x ) + ( 1 t ) Θ ( x t , w ) + t φ ( x ) + ( 1 t ) φ ( w ) φ ( x t ) t [ Θ ( x t , x ) + φ ( x ) φ ( x t ) ] ,

and hence Θ ( x t , x ) + φ ( x ) φ ( x t ) 0 . Then we have Θ ( w , x ) + φ ( x ) φ ( w ) 0 for all x H . So w Ω .

We can obtain w Γ : = Ω F ( S ) and x n w .

Step 4. Finally, from (3.1), we have
x t u 2 = 1 γ t 0 γ t S ( s ) [ ( I t F ) J r x t ] d s u 2 = 1 γ t 0 γ t [ S ( s ) ( I t F ) J r x t S ( s ) J r u ] d s 2 ( I t F ) x t ( I t F ) u t F u 2 τ t 2 x t u 2 + t 2 F u 2 + 2 t ( I t F ) u ( I t F ) x t , F u τ t x t u 2 + t 2 F u 2 + 2 t u x t , F u + 2 t 2 F x t F u , F u τ t x t u 2 + t 2 F u 2 + 2 t u x t , F u + 2 κ t 2 x t u F u .
Therefore,
x t u 2 t 2 1 τ t F u 2 + 2 t 1 τ t u x t , F u + 2 κ t 2 1 τ t x t u F u .
(3.11)
It follows that
x n w 2 t n 2 1 τ t n F w 2 + 2 t n 1 τ t n w x n , F w + 2 κ t n 2 1 τ t n x n w F w .

Thus, x n w implies that x n w .

Again, from (3.11), we obtain
x n u 2 t n 2 1 τ t n F u 2 + 2 t n 1 τ t n u x n , F u + 2 κ t n 2 1 τ t n x n u F u .
(3.12)

It is clear that lim n ( t n 2 / 1 τ t n ) = 0 , lim n ( 2 t n / 1 τ t n ) = 2 / η , and lim n ( 2 κ t n 2 / 1 τ t n ) = 0 . We deduce immediately from (3.12) that F u , w u 0 , which is equivalent to its dual variational inequality F w , w u 0 . That is, w Γ is a solution of the variational inequality (3.2).

Suppose that x Γ and w Γ both are solutions to the variational inequality (3.2); then
F x , x w 0 , F w , w x 0 .
(3.13)

Adding up (3.13) and the last inequality yields F x F w , x w 0 . The strong monotonicity of F implies that x = w and the uniqueness is proved. Later, we will use x Γ to denote the unique solution of (3.2). This completes the proof. □

Next we introduce an explicit algorithm for finding an element of Γ.

Theorem 3.2 Let H be a real Hilbert space. Let φ : H be a lower semicontinuous and convex functional. Let Θ : H × H be an equilibrium functions satisfying conditions (H1)-(H3). Let S : = { S ( s ) : 0 s < } be a nonexpansive semigroup on H such that F ( S ) . Let F be an η-strongly monotone and κ-Lipschitzian operator on H with 0 < η < κ . For given x 0 H arbitrarily, define a sequence { x n } iteratively by
y n = x n λ n F ( x n ) , x n + 1 = ( 1 α n ) y n + α n 1 t n 0 t n S ( s ) J r y n d s , n 0 ,
(3.14)
where { λ n } , { t n } are sequences in ( 0 , ) , { α n } is a sequence in [ 0 , 1 ] , and J r : H H is the mapping defined by (2.1). Suppose the following conditions are satisfied:
  1. (i)

    ξ : H × H H is λ-Lipschitz continuous such that

     
  2. (a)

    ξ ( x , y ) + ξ ( y , x ) = 0 , x , y H ;

     
  3. (b)

    x ξ ( x , y ) is affine;

     
  4. (c)

    y ξ ( x , y ) is sequentially continuous from the weak topology to the weak topology;

     
  1. (ii)

    K : H is ξ-strongly convex with constant σ > 0 , and its derivative K is not only sequentially continuous from the weak topology to the strong topology but also Lipschitz continuous with a Lipschitz constant ν > 0 and σ λ ν ;

     
  2. (iii)
    For each x H there exist a bounded subset D x H and a point z x H such that, for any y H D x ,
    Θ ( y , z x ) + φ ( z x ) φ ( y ) + 1 r K ( y ) K ( x ) , ξ ( z x , y ) < 0 ;
     
  3. (iv)

    lim sup n λ n < η / κ 2 and n = 1 λ n = ;

     
  4. (vi)

    lim n t n = and lim n ( t n + 1 / t n ) = 1 ;

     
  5. (vii)

    0 < γ lim inf n α n lim sup n α n < 1 , for some γ ( 0 , 1 ) .

     
If Γ : = Ω ( Θ , φ ) F ( S ) , then the sequences { x n } and { y n } converge strongly to an element x of Γ provided J r is firmly nonexpansive if and only if λ n F ( x n ) 0 , where x is the unique solution of the following variational inequality:
F x , x u 0 , u Γ .

Proof The necessity is obvious. We only need to prove the sufficiency. Suppose that λ n F ( x n ) 0 .

Step 1. First, we show that { x n } , { y n } , { S ( s ) y n } , { F x n } , and { J r x n } are bounded. In fact, letting u Γ , we have u = S ( s ) J r u .

Then
x n + 1 u = ( 1 α n ) y n + α n 1 t n 0 t n S ( s ) J r y n d s u = ( 1 α n ) ( y n u ) + α n ( 1 t n 0 t n S ( s ) J r y n d s u ) ( 1 α n ) y n u + α n 1 t n 0 t n S ( s ) J r y n S ( s ) J r u d s ( 1 α n ) y n u + α n y n u = y n u .
(3.15)
From condition (iv), without loss of generality, we can assume that λ n a < η / κ 2 , n 0 . By (3.14) and Lemma 2.6, we have
y n u = x n λ n F ( x n ) u = ( I λ n F ) x n ( I λ n F ) u λ n F u ( I λ n F ) x n ( I λ n F ) u + λ n F u τ λ n x n u + λ n F u ,
(3.16)

where τ λ n = 1 λ n ( 2 η λ n κ 2 ) ( 0 , 1 ) .

Then, from (3.15) and (3.16), we obtain
x n + 1 u τ λ n x n u + λ n F u = [ 1 ( 1 τ λ n ) ] x n u + ( 1 τ λ n ) λ n 1 τ λ n F u max { x n u , λ n 1 τ λ n F u } .
Observe that lim n ( λ n / 1 τ λ n ) = 1 / η , we have by induction
x n + 1 u max { x 0 u , M 1 F u } ,
(3.17)

where M 1 = sup n { λ n / 1 τ λ n } < . Hence { x n } is bounded. Consequently, we deduce that { y n } , { S ( s ) y n } , { F x n } and { J r x n } are also bounded.

Step 2. Define x n + 1 = ( 1 α n ) x n + α n u n , n 0 , then u n = 1 α n [ x n + 1 ( 1 α n ) x n ] .

Observe that
u n + 1 u n = x n + 2 ( 1 α n + 1 ) x n + 1 α n + 1 x n + 1 ( 1 α n ) x n α n = ( 1 α n + 1 ) y n + 1 + α n + 1 1 t n + 1 0 t n + 1 S ( s ) J r y n + 1 d s ( 1 α n + 1 ) x n + 1 α n + 1 ( 1 α n ) y n + α n 1 t n 0 t n S ( s ) J r y n d s ( 1 α n ) x n α n = α n + 1 1 t n + 1 0 t n + 1 S ( s ) J r y n + 1 d s ( 1 α n + 1 ) λ n + 1 F ( x n + 1 ) α n + 1 α n 1 t n 0 t n S ( s ) J r y n d s ( 1 α n ) λ n F ( x n ) α n 1 t n + 1 0 t n + 1 S ( s ) J r y n + 1 d s 1 t n 0 t n S ( s ) J r y n d s + 1 α n + 1 α n + 1 λ n + 1 F ( x n + 1 ) + 1 α n α n λ n F ( x n ) .
(3.18)
Next, we estimate
1 t n + 1 0 t n + 1 S ( s ) J r y n + 1 d s 1 t n 0 t n S ( s ) J r y n d s 1 t n + 1 0 t n + 1 S ( s ) J r y n + 1 d s 1 t n + 1 0 t n + 1 S ( s ) J r y n d s + 1 t n + 1 0 t n + 1 S ( s ) J r y n d s 1 t n 0 t n S ( s ) J r y n d s y n + 1 y n + 1 t n + 1 0 t n + 1 S ( s ) J r y n d s 1 t n 0 t n S ( s ) J r y n d s y n + 1 y n + | 1 t n + 1 1 t n | 0 t n S ( s ) J r y n d s + 1 t n + 1 t n t n + 1 S ( s ) J r y n d s x n + 1 λ n + 1 F ( x n + 1 ) x n + λ n F ( x n ) + | t n t n + 1 1 | M 2 x n + 1 x n + λ n + 1 F ( x n + 1 ) + λ n F ( x n ) + | t n t n + 1 1 | M 2 ,
(3.19)
where M 2 = sup n { 2 S ( s ) J r y n } < . From (3.18) and (3.19), we have
u n + 1 u n 1 α n + 1 α n + 1 λ n + 1 F ( x n + 1 ) + 1 α n α n λ n F ( x n ) + x n + 1 x n + λ n + 1 F ( x n + 1 ) + λ n F ( x n ) + | t n t n + 1 1 | M 2 .
This together with condition (vii) implies that
u n + 1 u n 1 γ ( λ n + 1 F ( x n + 1 ) + λ n F ( x n ) ) + x n + 1 x n + | t n t n + 1 1 | M 2 .
(3.20)
Namely,
u n + 1 u n x n + 1 x n 1 γ ( λ n + 1 F ( x n + 1 ) + λ n F ( x n ) ) + | t n t n + 1 1 | M 2 .
Since λ n F ( x n ) 0 and condition (vi), we get
lim sup n ( u n + 1 u n x n + 1 x n ) 0 .
Consequently, by Lemma 2.2, we deduce lim n u n x n = 0 . Therefore,
x n + 1 x n = α n u n x n 0 ( n ) .
(3.21)
Step 3. Next, we claim that lim n x n S ( s ) x n = 0 . Observe that
x n S ( s ) x n S ( s ) x n S ( s ) ( 1 t n 0 t n S ( s ) x n d s ) + 1 t n 0 t n S ( s ) x n d s x n + S ( s ) ( 1 t n 0 t n S ( s ) x n d s ) 1 t n 0 t n S ( s ) x n d s 2 1 t n 0 t n S ( s ) x n d s x n + S ( s ) ( 1 t n 0 t n S ( s ) x n d s ) 1 t n 0 t n S ( s ) x n d s .
(3.22)
Note that
1 t n 0 t n S ( s ) x n d s x n x n x n + 1 + x n + 1 1 t n 0 t n S ( s ) x n d s = x n x n + 1 + ( 1 α n ) y n + α n 1 t n 0 t n S ( s ) J r y n d s 1 t n 0 t n S ( s ) x n d s x n x n + 1 + ( 1 α n ) y n 1 t n 0 t n S ( s ) x n d s + α n 1 t n 0 t n S ( s ) x n d s 1 t n 0 t n S ( s ) J r y n d s x n x n + 1 + ( 1 α n ) y n x n + ( 1 α n ) x n 1 t n 0 t n S ( s ) x n d s + α n x n J r y n .
It follows that
1 t n 0 t n S ( s ) x n d s x n 1 α n ( x n x n + 1 + ( 1 α n ) λ n F ( x n ) + α n x n J r y n ) .
(3.23)
From (3.15), we have
x n + 1 u 2 = ( 1 α n ) y n + α n 1 t n 0 t n S ( s ) J r y n d s u 2 ( 1 α n ) y n u 2 + α n 1 t n 0 t n S ( s ) J r y n d s u 2 ( 1 α n ) x n λ n F ( x n ) u 2 + α n J r y n u 2 ( 1 α n ) x n u 2 + ( 1 α n ) λ n F ( x n ) 2 + α n ( x n u 2 x n J r y n 2 ) ( 1 α n ) λ n F ( x n ) 2 + x n u 2 α n x n J r y n 2 ,
then
α n x n J r y n 2 ( 1 α n ) λ n F ( x n ) 2 + x n u 2 x n + 1 u 2 ( 1 α n ) λ n F ( x n ) 2 + x n x n + 1 ( x n u + x n + 1 u ) .
This together with condition (vii), λ n F ( x n ) 0 , and (3.21), we have
lim n x n J r y n = 0 .
(3.24)
By Lemma 2.4, (3.21)-(3.24), and λ n F ( x n ) 0 , we derive
lim n x n S ( s ) x n = 0 .
(3.25)
Step 4. Next, we show that lim sup n F x , x x n 0 , where x = lim n x t n and x t n is defined by x t n = ( 1 / t n ) 0 t n S ( s ) [ ( I t n F ) J r x t n ] d s . Since { x n } is bounded, there exists a subsequence { x n k } of { x n } that converges weakly to w. Similarly be able to prove w Γ like Theorem 3.1. Hence, by Theorem 3.1, we have
lim sup n F x , x x n = lim k F x , x x n k = F x , x w 0 .
(3.26)
Step 5. Finally, we prove that { x n } converges strongly to x Γ . From (3.14), we have
x n + 1 x 2 = ( 1 α n ) y n + α n 1 t n 0 t n S ( s ) J r y n d s x 2 ( 1 α n ) y n x 2 + α n 1 t n 0 t n S ( s ) J r y n d s x 2 ( 1 α n ) y n x 2 + α n y n x 2 y n x 2 = x n λ n F ( x n ) x 2 = ( I λ n F ) x n ( I λ n F ) x λ n F x 2 τ λ n 2 x n x 2 + λ n 2 F ( x ) 2 + 2 λ n ( I λ n F ) x ( I λ n F ) x n , F ( x ) τ λ n x n x 2 + λ n 2 F ( x ) 2 + 2 λ n x x n , F x + 2 λ n λ n F x n , F x 2 λ n 2 F x 2 [ 1 ( 1 τ λ n ) ] x n x 2 + 2 λ n x x n , F x + 2 λ n λ n F ( x n ) F x λ n 2 F x 2 [ 1 ( 1 τ λ n ) ] x n x 2 + ( 1 τ λ n ) [ 2 λ n 1 τ λ n x x n , F x + 2 λ n F x 1 τ λ n λ n F ( x n ) ] = ( 1 δ n ) x n x 2 + δ n ρ n ,
(3.27)

where δ n = 1 τ λ n and ρ n = 2 λ n 1 τ λ n x x n , F x + 2 λ n F x 1 τ λ n λ n F ( x n ) . Obviously, n = 1 δ n = and lim sup n ρ n 0 . Hence, all conditions of Lemma 2.3 are satisfied. Therefore, we immediately deduce that the sequence { x n } converges strongly to x Γ .

Observe that
y n x y n x n + x n x λ n F ( x n ) + x n x 0 , n .
(3.28)

It is clear that { y n } converges strongly to x Γ . From x = lim t 0 x t and Theorem 3.1, we see that x is the unique solution of the variational inequality (3.2). This completes the proof. □

Declarations

Acknowledgements

The research was supported by Fujian Nature Science Foundation.

Authors’ Affiliations

(1)
Institute of Mathematics and Computer, Fuzhou University

References

  1. Yamada I: The hybrid steepest descent method for the variational inequality problem over the intersection of fixed point sets of nonexpansive mappings. 8. In Inherently Parallel Algorithms in Feasibility and Optimization and Their Applications. Edited by: Butnariu D, Censor Y, Reich S. North-Holland, Amsterdam; 2001:473–504.View ArticleGoogle Scholar
  2. Ceng LC, Yao JC: A hybrid iterative scheme for mixed equilibrium problems and fixed point problems. J. Comput. Appl. Math. 2008, 214: 186–201. 10.1016/j.cam.2007.02.022MathSciNetView ArticleMATHGoogle Scholar
  3. Yang PX, Yao YH, Liou Y-C, Chen R: Hybrid algorithms of nonexpansive semigroups for variational inequalities. J. Appl. Math. 2012., 2012: Article ID 634927Google Scholar
  4. Hanson MA: On sufficiency of the Kuhn-Tucker conditions. J. Math. Anal. Appl. 1981, 80: 545–550. 10.1016/0022-247X(81)90123-2MathSciNetView ArticleMATHGoogle Scholar
  5. Ansari QH, Yao JC: Iterative schemes for solving mixed variational-like inequalities. J. Optim. Theory Appl. 2001,108(3):527–541. 10.1023/A:1017531323904MathSciNetView ArticleMATHGoogle Scholar
  6. Suzuki T: Strong convergence of Krasnoselskii and Mann’s type sequences for one-parameter nonexpansive semigroups without Bochner integrals. J. Math. Anal. Appl. 2005, 305: 227–239. 10.1016/j.jmaa.2004.11.017MathSciNetView ArticleMATHGoogle Scholar
  7. Xu HK: A regularization method for the proximal point algorithm. J. Glob. Optim. 2006,36(1):115–125. 10.1007/s10898-006-9002-7View ArticleMathSciNetMATHGoogle Scholar
  8. Shimizu T, Takahashi W: Strong convergence to common fixed points of families of nonexpansive mappings. J. Math. Anal. Appl. 1997, 211: 71–83. 10.1006/jmaa.1997.5398MathSciNetView ArticleMATHGoogle Scholar
  9. Tan KK, Xu HK: The nonlinear ergodic theorem for asymptotically nonexpansive mappings in Banach spaces. Proc. Am. Math. Soc. 1992, 114: 399–404. 10.1090/S0002-9939-1992-1068133-2MathSciNetView ArticleMATHGoogle Scholar
  10. Wang S, Hu C: Two new iterative methods for a countable family of nonexpansive mappings in Hilbert spaces. Fixed Point Theory Appl. 2010., 2010: Article ID 852030Google Scholar

Copyright

© Jiang and Wang; licensee Springer. 2014

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.