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Affine algorithms for the split variational inequality and equilibrium problems

Fixed Point Theory and Applications20132013:140

https://doi.org/10.1186/1687-1812-2013-140

Received: 11 October 2012

Accepted: 14 May 2013

Published: 29 May 2013

Abstract

An affine algorithm for the split variational inequality and equilibrium problems is presented. Strong convergence result is given.

Keywords

affine algorithm split method variational inequality equilibrium problem

1 Introduction

In the present manuscript, we focus on the following split variational inequality and equilibrium problem: Finding a point x such that
x GVI ( B , ψ , C ) and ψ ( x ) EP ( F , A ) ,
(1.1)
where GVI ( B , ψ , C ) is the solution set of the generalized variational inequality of finding u C , ψ ( u ) C such that
B u , ψ ( v ) ψ ( u ) 0 , ψ ( v ) C ,
(1.2)
and EP ( F , A ) is the solution set of the equilibrium problem, which is to find x C such that
F ( x , y ) + A x , y x 0 , y C .
(1.3)

Our main motivations are inspired by the following reasons.

Reason 1 Recently, the split problems have been considered by some authors. Especially, the split feasibility problem which can mathematically be formulated as the problem of finding a point x ˜ with the property
x ˜ C and g ( x ˜ ) Q

has received much attention due to its applications in signal processing and image reconstruction with particular progress in intensity modulated radiation therapy [113]. Note that the involved operator g is a bounded linear operator. However, in the present paper, the involved mapping ψ in (1.1) is a nonlinear mapping.

Reason 2 The variational inequality problem [1424] and equilibrium problem [2327], which include the fixed point problems and optimization problems [2830], have been studied by many authors. It is an interesting topic associated with the analytical and algorithmic approach to the variational inequality and equilibrium problems.

Motivated and inspired by the results in the literature, we present an affine algorithm for solving the split problem (1.1). Strong convergence theorem is given under some mild assumptions.

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.

2.1 Monotonicity and convexity

An operator A : C H is said to be monotone if x y , A x A y 0 for all x , y C . A : C H is said to be strongly monotone if there exists a constant γ > 0 such that x y , A x A y γ x y 2 for all x , y C . A : C H is called an inverse-strongly-monotone operator if there exists α > 0 such that x y , A x A y α A x A y 2 for all x , y C . Let g : C C be a nonlinear operator. A : C H is said to be α-inverse strongly g-monotone iff g ( x ) g ( y ) , A x A y α A x A y 2 for all x , y C and for some α > 0 . Let B be a mapping of H into 2 H . The effective domain of B is denoted by dom ( B ) , that is, dom ( B ) = { x H : B x } . A multi-valued mapping B is said to be a monotone operator on H iff x y , u v 0 for all x , y dom ( B ) , u B x , and v B y . A monotone operator B on H is said to be maximal iff its graph is not strictly contained in the graph of any other monotone operator on H.

A function F : H R is said to be convex if for any x , y H and for any λ [ 0 , 1 ] , F ( λ x + ( 1 λ ) y ) λ F ( x ) + ( 1 λ ) F ( y ) .

2.2 Nonexpansivity and continuity

A mapping T : C C is said to be nonexpansive [3138] if T x T y x y for all x , y C . We use Fix ( T ) to denote the set of fixed points of T. T : C C is called a firmly nonexpansive mapping if, for all x , y C , T x T y 2 x y , T x T y . It is known that T is firmly nonexpansive if and only if a mapping 2 T I is nonexpansive, where I is the identity mapping on H. T : C H is said to be L-Lipschitz continuous if there exists a constant L > 0 such that T x T y L x y for all x , y C . In such a case, T is said to be L-Lipschitz continuous. Given a nonempty, closed convex subset C of H, the mapping that assigns every point x H to its unique nearest point in C is called a metric projection onto C and denoted by P C , that is, P C x C and x P C x = inf { x y : y C } . The metric projection P C is a typical firmly nonexpansive mapping. The characteristic inequality of the projection is x P C x , y P C x 0 for all x H , y C .

2.3 Equilibrium problem

In this paper, we consider the split problem (1.1). In the sequel, we assume that the solution set S of (1.1) is nonempty.

Problem 2.1 Assume that
  1. (A1)

    B : C H is an α-inverse strongly ψ-monotone mapping;

     
  2. (A2)

    ψ : C C is a weakly continuous and γ-strongly monotone mapping such that R ( ψ ) = C ;

     
  3. (A3)

    F : C × C R is a bifunction;

     
  4. (A4)

    A : C H is a β-inverse-strongly monotone mapping.

     
Our objective is to
find  x GVI ( B , ψ , C )  such that  ψ ( x ) EP ( F , A ) ,
where F satisfies the following conditions:
  1. (F1)

    F ( x , x ) = 0 for all x C ;

     
  2. (F2)

    F is monotone, i.e., F ( x , y ) + F ( y , x ) 0 for all x , y C ;

     
  3. (F3)

    for each x , y , z C , lim t 0 F ( t z + ( 1 t ) x , y ) F ( x , y ) ;

     
  4. (F4)

    for each x C , y F ( x , y ) is convex and lower semicontinuous.

     

In order to solve Problem 2.1, we need the following useful lemmas.

2.4 Useful lemmas

The following three lemmas are important tools for our main results in the next section. Note that these lemmas are used extensively in the literature.

Lemma 2.2 (Combettes and Hirstoaga’s lemma [26])

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 (F1)-(F4). Let λ > 0 and x C . Then there exists z C such that
F ( z , y ) + 1 λ y z , z x 0 , y C .
Further, if T λ ( x ) = { z C : F ( z , y ) + 1 λ y z , z x 0 for all y C } , then the following hold:
  1. (a)

    T λ is single-valued and T λ is firmly nonexpansive;

     
  2. (b)

    EP ( F ) is closed and convex and EP ( F ) = Fix ( T λ ) .

     

Lemma 2.3 (Suzuki’s lemma [39])

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

Lemma 2.4 (Xu’s lemma [40])

Assume that { a n } is a sequence of nonnegative real numbers such that a n + 1 ( 1 γ n ) a n + δ n γ n , where { γ n } is a sequence in ( 0 , 1 ) and { δ n } is a sequence such that n = 1 γ n = and lim sup n δ n 0 (or n = 1 | δ n γ n | < ). Then lim n a n = 0 .

3 Algorithms and convergence analysis

In this section, we first present our algorithm for solving Problem 2.1. Assume that the conditions in Problem 2.1 are all satisfied.

Algorithm 3.1 Let C be a nonempty closed and convex subset of a real Hilbert space H.

Step 0. (Initialization)
x 0 C .
Step 1. (Projection step) For { x n } , let the sequence { u n } be generated iteratively by
u n = P C [ α n δ φ ( x n ) + ( 1 α n ) ( ψ ( x n ) μ n B x n ) ] , n 0 ,

where P C is the metric projection, { α n } [ 0.1 ] is a real number sequence, φ : C H is an L-Lipschitz continuous mapping and δ > 0 is a constant.

Step 2. (Proximal step) Find { z n } such that
F ( z n , y ) + A u n , y z n + 1 λ n y z n , z n u n 0 , y C ,

where { λ n } ( 0 , ) is a real number sequence.

Step 3. (Affine step) For the above sequences { x n } and { z n } , let the ( n + 1 ) th sequence { x n + 1 } be generated by
ψ ( x n + 1 ) = β n ψ ( x n ) + ( 1 β n ) z n , n 0 ,

where { β n } [ 0 , 1 ] is a real number sequence.

Theorem 3.2 Suppose S . Assume that the following restrictions are satisfied:
  1. (C1)

    λ n ( a , b ) ( 0 , 2 β ) , μ n ( c , d ) ( 0 , 2 α ) and γ ( L δ , 2 α ) ;

     
  2. (C2)

    lim n ( μ n + 1 μ n ) = 0 and lim n ( λ n + 1 λ n ) = 0 ;

     
  3. (C3)

    lim n α n = 0 and n α n = ;

     
  4. (C4)

    β n [ ξ 1 , ξ 2 ] ( 0 , 1 ) .

     
Then the sequence { x n } generated by Algorithm 3.1 converges strongly to x S , which solves the following variational inequality:
δ φ ( x ) ψ ( x ) , ψ ( x ) ψ ( x ) 0 , x Ω .
(3.1)
Remark 3.3 The solution of variational inequality (3.1) is unique. As a matter of fact, if x ˜ S also solves (3.1), we have
δ φ ( x ) ψ ( x ) , ψ ( x ˜ ) ψ ( x ) 0 and δ φ ( x ˜ ) ψ ( x ˜ ) , ψ ( x ) ψ ( x ˜ ) 0 .
Adding up the above two inequalities, we deduce
δ φ ( x ˜ ) ψ ( x ˜ ) δ φ ( x ) + ψ ( x ) , ψ ( x ) ψ ( x ˜ ) 0 .
It follows that
ψ ( x ) ψ ( x ˜ ) 2 δ φ ( x ) φ ( x ˜ ) , ψ ( x ) ψ ( x ˜ ) δ φ ( x ) φ ( x ˜ ) ψ ( x ) ψ ( x ˜ ) ,
which implies that
ψ ( x ) ψ ( x ˜ ) δ φ ( x ) φ ( x ˜ ) .
Since ψ is γ-strongly monotone, we have
γ x x ˜ 2 ψ ( x ) ψ ( x ˜ ) , x x ˜ ψ ( x ) ψ ( x ˜ ) x x ˜ .
Hence,
γ x x ˜ ψ ( x ) ψ ( x ˜ ) δ φ ( x ) φ ( x ˜ ) δ L x x ˜ .

This deduces the contraction because of δ L < γ by the assumption. Therefore, x = x ˜ . So, the solution of variational inequality (3.1) is unique.

Remark 3.4 Using the characteristic inequality of the projection, we have
x ˘ GVI ( B , ψ , C ) ψ ( x ˘ ) = P C ( ψ ( x ˘ ) ν B x ˘ ) , ν > 0 .
Remark 3.5
( ψ ( x ) μ B x ) ( ψ ( y ) μ B y ) 2 ψ ( x ) ψ ( y ) 2 + μ ( μ 2 α ) B x B y 2 .
In fact,
( ψ ( x ) μ B x ) ( ψ ( y ) μ B y ) 2 = ψ ( x ) ψ ( y ) 2 2 μ B x B y , ψ ( x ) ψ ( y ) + μ 2 B x B y 2 ψ ( x ) ψ ( y ) 2 2 μ α B x B y 2 + μ 2 B x B y 2 ψ ( x ) ψ ( y ) 2 + μ ( μ 2 α ) B x B y 2 .

Next, we prove Theorem 3.2.

Proof Let x Ω . Hence x GVI ( B , ψ , C ) and ψ ( x ) EP ( F , A ) . Since μ n > 0 , from Remark 3.4 we have ψ ( x ) = P C [ ψ ( x ) μ n B x ] for all n 0 . Thus,
u n ψ ( x ) = P C [ α n δ φ ( x n ) + ( 1 α n ) ( ψ ( x n ) μ n B x n ) ] P C [ ψ ( x ) μ n B x ] α n ( δ φ ( x n ) ψ ( x ) + μ n B x ) + ( 1 α n ) ( ( ψ ( x n ) μ n B x n ) ( ψ ( x ) μ n B x ) ) α n δ φ ( x n ) δ φ ( x ) + α n δ φ ( x ) ψ ( x ) + μ n B x + ( 1 α n ) ( ψ ( x n ) μ n B x n ) ( ψ ( x ) μ n B x ) α n δ L x n x + α n δ φ ( x ) ψ ( x ) + μ n B x + ( 1 α n ) ψ ( x n ) ψ ( x ) α n δ L / γ ψ ( x n ) ψ ( x ) + α n δ φ ( x ) ψ ( x ) + μ n B x + ( 1 α n ) ψ ( x n ) ψ ( x ) = [ 1 ( 1 δ L / γ ) α n ] ψ ( x n ) ψ ( x ) + α n δ φ ( x ) ψ ( x ) + μ n B x [ 1 ( 1 δ L / γ ) α n ] ψ ( x n ) ψ ( x ) + α n ( δ φ ( x ) ψ ( x ) + 2 α B x ) .
(3.2)
By Algorithm 3.1, we have z n = T λ n ( I λ n A ) u n for all n 0 . Noting that ψ ( x ) EP ( F , A ) , we deduce ψ ( x ) = T λ n ( I λ n A ) ψ ( x ) for all n 0 . It follows that
ψ ( x n + 1 ) ψ ( x ) β n ψ ( x n ) ψ ( x ) + ( 1 β n ) T λ n ( I λ n A ) u n T λ n ( I λ n A ) ψ ( x ) β n ψ ( x n ) ψ ( x ) + ( 1 β n ) u n ψ ( x ) β n ψ ( x n ) ψ ( x ) + ( 1 β n ) [ 1 ( 1 δ L / γ ) α n ] ψ ( x n ) ψ ( x ) + ( 1 β n ) α n ( δ φ ( x ) ψ ( x ) + 2 α B x ) = [ 1 ( 1 δ L / γ ) ( 1 β n ) α n ] ψ ( x n ) ψ ( x ) + ( 1 δ L / γ ) ( 1 β n ) α n δ φ ( x ) ψ ( x ) + 2 α B x 1 δ L / γ .
By induction
ψ ( x n ) ψ ( x ) max { ψ ( x 0 ) ψ ( x ) , δ φ ( x ) ψ ( x ) + 2 α B x 1 δ L / γ } .
Hence, { ψ ( x n ) } is bounded. Since ψ is γ-strongly monotone, we can get (by a similar technique as that in Remark 3.3) γ x n x ψ ( x n ) ψ ( x ) . So, x n x 1 γ ψ ( x n ) ψ ( x ) 1 γ max { ψ ( x 0 ) ψ ( x ) , δ φ ( x ) ψ ( x ) + 2 α B x 1 δ L / γ } . This implies that { x n } is bounded. Next, we show x n + 1 x n 0 . From Step 2 in Algorithm 3.1, we have
F ( z n , y ) + 1 λ n y z n , z n ( u n λ n A u n ) 0 , y C .
Taking y = z n + 1 , we get
F ( z n , z n + 1 ) + 1 λ n z n + 1 z n , z n ( u n λ n A u n ) 0 .
Similarly, we also have
F ( z n + 1 , z n ) + 1 λ n + 1 z n z n + 1 , z n + 1 ( u n + 1 λ n + 1 A u n + 1 ) 0 .
Adding up the above two inequalities, we get
F ( z n , z n + 1 ) + F ( z n + 1 , z n ) + A u n A u n + 1 , z n + 1 z n + z n + 1 z n , z n u n λ n z n + 1 u n + 1 λ n + 1 0 .
By the monotonicity of F, we have
F ( z n , z n + 1 ) + F ( z n + 1 , z n ) 0 .
So,
A u n A u n + 1 , z n + 1 z n + z n + 1 z n , z n u n λ n z n + 1 u n + 1 λ n + 1 0 .
Thus,
λ n A u n A u n + 1 , z n + 1 z n + z n + 1 z n , z n z n + 1 + z n + 1 u n λ n λ n + 1 ( z n + 1 u n + 1 ) 0 .
It follows that
z n + 1 z n 2 λ n A u n A u n + 1 , z n + 1 z n + z n + 1 z n , u n + 1 u n + ( 1 λ n λ n + 1 ) ( z n + 1 u n + 1 ) = ( I λ n A ) u n + 1 ( I λ n A ) u n , z n + 1 z n + z n + 1 z n , ( 1 λ n λ n + 1 ) ( z n + 1 u n + 1 ) ( I λ n A ) u n + 1 ( I λ n A ) u n z n + 1 z n + | 1 λ n λ n + 1 | z n + 1 z n z n + 1 u n + 1 z n + 1 z n ( u n + 1 u n + | 1 λ n λ n + 1 | z n + 1 u n + 1 )
and hence
z n + 1 z n u n + 1 u n + | 1 λ n λ n + 1 | z n + 1 u n + 1 u n + 1 u n + 1 a | λ n + 1 λ n | z n + 1 u n + 1 .
By Algorithm 3.1, we have
u n + 1 u n = P C [ α n + 1 δ φ ( x n + 1 ) + ( 1 α n + 1 ) ( ψ ( x n + 1 ) μ n + 1 B x n + 1 ) ] P C [ α n δ φ ( x n ) + ( 1 α n ) ( ψ ( x n ) μ n B x n ) ] [ α n + 1 δ φ ( x n + 1 ) + ( 1 α n + 1 ) ( ψ ( x n + 1 ) μ n + 1 B x n + 1 ) ] [ α n δ φ ( x n ) + ( 1 α n ) ( ψ ( x n ) μ n B x n ) ] α n + 1 δ φ ( x n + 1 ) φ ( x n ) + δ | α n + 1 α n | φ ( x n ) + ( 1 α n + 1 ) ψ ( x n + 1 ) μ n + 1 B x n + 1 ( ψ ( x n ) μ n + 1 B x n ) + | α n + 1 α n | ψ ( x n ) + | μ n + 1 μ n | B ( x n ) + | α n + 1 μ n + 1 α n μ n | B ( x n ) α n + 1 δ L x n + 1 x n + ( 1 α n + 1 ) ψ ( x n + 1 ) ψ ( x n ) + | α n + 1 α n | ( δ φ ( x n ) + ψ ( x n ) ) + | μ n + 1 μ n | B ( x n ) + | α n + 1 μ n + 1 α n μ n | B ( x n ) α n + 1 ( δ L / γ ) ψ ( x n + 1 ) ψ ( x n ) + ( 1 α n + 1 ) ψ ( x n + 1 ) ψ ( x n ) + | α n + 1 α n | ( δ φ ( x n ) + ψ ( x n ) ) + | μ n + 1 μ n | B ( x n ) + | α n + 1 μ n + 1 α n μ n | B ( x n ) = [ 1 ( 1 δ L / γ ) α n + 1 ] ψ ( x n + 1 ) ψ ( x n ) + | α n + 1 α n | ( δ φ ( x n ) + ψ ( x n ) ) + | μ n + 1 μ n | B ( x n ) + | α n + 1 μ n + 1 α n μ n | B ( x n ) .
Therefore,
z n + 1 z n [ 1 ( 1 δ L / γ ) α n + 1 ] ψ ( x n + 1 ) ψ ( x n ) + | α n + 1 α n | ( δ φ ( x n ) + ψ ( x n ) ) + | μ n + 1 μ n | B ( x n ) + | α n + 1 μ n + 1 α n μ n | B ( x n ) + 1 a | λ n + 1 λ n | z n + 1 u n + 1 .
It follows that
z n + 1 z n ψ ( x n + 1 ) ψ ( x n ) | α n + 1 α n | ( δ φ ( x n ) + ψ ( x n ) ) + | μ n + 1 μ n | B ( x n ) + | α n + 1 μ n + 1 α n μ n | B ( x n ) + 1 a | λ n + 1 λ n | z n + 1 u n + 1 .
Since lim n α n = 0 , lim n ( μ n + 1 μ n ) = 0 , lim n ( λ n + 1 λ n ) = 0 and the sequences { φ ( x n ) } , { ψ ( x n ) } , { z n } , { u n } and { B x n } are bounded, we have
lim sup n ( z n + 1 z n ψ ( x n + 1 ) ψ ( x n ) ) 0 .
By Lemma 2.3, we obtain
lim n z n ψ ( x n ) = 0 .
Hence,
lim n ψ ( x n + 1 ) ψ ( x n ) = lim n ( 1 β n ) z n ψ ( x n ) = 0 .
This together with the γ-strong monotonicity of ψ implies that
lim n x n + 1 x n = 0 .
By the convexity of the norm, we have
ψ ( x n + 1 ) ψ ( x ) 2 = β n ( ψ ( x n ) ψ ( x ) ) + ( 1 β n ) ( z n ψ ( x ) ) 2 β n ψ ( x n ) ψ ( x ) 2 + ( 1 β n ) z n ψ ( x ) 2 β n ψ ( x n ) ψ ( x ) 2 + ( 1 β n ) [ α n ( δ φ ( x n ) ψ ( x ) + μ n B x ) + ( 1 α n ) ( ( ψ ( x n ) μ n B x n ) ( ψ ( x ) μ n B x ) ) ] 2 β n ψ ( x n ) ψ ( x ) 2 + ( 1 β n ) [ α n δ φ ( x n ) ψ ( x ) + μ n B x 2 + ( 1 α n ) ( ψ ( x n ) μ n B x n ) ( ψ ( x ) μ n B x ) 2 + 2 α n ( 1 α n ) δ φ ( x n ) ψ ( x ) + μ n B x ( ψ ( x n ) μ n B x n ) ( ψ ( x ) μ n B x ) ] β n ψ ( x n ) ψ ( x ) 2 + ( 1 β n ) ( 1 α n ) ( ψ ( x n ) μ n B x n ) ( ψ ( x ) μ n B x ) 2 + α n M ,
(3.3)
where M > 0 is some constant. From Remark 3.5, we derive
( ψ ( x n ) μ n B x n ) ( ψ ( x ) μ n B x ) 2 ψ ( x n ) ψ ( x ) 2 + μ n ( μ n 2 α ) B x n B x 2 .
Thus,
ψ ( x n + 1 ) ψ ( x ) 2 β n ψ ( x n ) ψ ( x ) 2 + ( 1 β n ) ( 1 α n ) ( ψ ( x n ) ψ ( x ) 2 + μ n ( μ n 2 α ) B x n B x 2 ) + α n M ψ ( x n ) ψ ( x ) 2 + ( 1 β n ) ( 1 α n ) μ n ( μ n 2 α ) B x n B x 2 + α n M .
So,
( 1 β n ) ( 1 α n ) μ n ( 2 α μ n ) B x n B x 2 ψ ( x n ) ψ ( x ) 2 ψ ( x n + 1 ) ψ ( x ) 2 + α n M ( ψ ( x n ) ψ ( x ) + ψ ( x n + 1 ) ψ ( x ) ) ψ ( x n + 1 ) ψ ( x n ) + α n M .
Since α n 0 , ψ ( x n + 1 ) ψ ( x n ) 0 and lim inf n ( 1 β n ) ( 1 α n ) μ n ( 2 α μ n ) > 0 , we obtain
lim n B x n B x = 0 .
Set y n = ψ ( x n ) μ n B x n ( ψ ( x ) μ n B x ) for all n. By using the property of projection, we get
u n ψ ( x ) 2 = P C [ α n δ φ ( x n ) + ( 1 α n ) ( ψ ( x n ) μ n B x n ) ] P C [ ψ ( x ) μ n B x ] 2 α n ( δ φ ( x n ) ψ ( x ) + μ n B x ) + ( 1 α n ) y n , u n ψ ( x ) = 1 2 { α n ( δ φ ( x n ) ψ ( x ) + μ n B x ) + ( 1 α n ) y n 2 + u n ψ ( x ) 2 α n ( δ φ ( x n ) ψ ( x ) + μ n B x ) + ( 1 α n ) y n u n + ψ ( x ) 2 } 1 2 { α n δ φ ( x n ) ψ ( x ) + μ n B x 2 + ( 1 α n ) ψ ( x n ) ψ ( x ) 2 + u n ψ ( x ) 2 α n ( δ φ ( x n ) ψ ( x ) + μ n B x y n ) + ψ ( x n ) u n μ n ( B x n B x ) 2 } = 1 2 { α n δ φ ( x n ) ψ ( x ) + μ n B x 2 + ( 1 α n ) ψ ( x n ) ψ ( x ) 2 + u n ψ ( x ) 2 ψ ( x n ) u n 2 μ n 2 B x n B x α n 2 δ φ ( x n ) ψ ( x ) + μ n B x y n 2 + 2 μ n α n B x n B x , δ φ ( x n ) ψ ( x ) + μ n B x y n + 2 μ n ψ ( x n ) u n , B x n B x 2 α n ψ ( x n ) u n , δ φ ( x n ) ψ ( x ) + μ n B x y n } .
(3.4)
It follows that
u n ψ ( x ) 2 α n δ φ ( x n ) ψ ( x ) + μ n B x 2 + ( 1 α n ) ψ ( x n ) ψ ( x ) 2 ψ ( x n ) u n 2 + 2 μ n α n B x n B x δ φ ( x n ) ψ ( x ) + μ n B x y n + 2 μ n ψ ( x n ) u n B x n B x + 2 α n ψ ( x n ) u n δ φ ( x n ) ψ ( x ) + μ n B x y n .
(3.5)
From (3.3) and (3.5), we have
ψ ( x n + 1 ) ψ ( x ) 2 β n ψ ( x n ) ψ ( x ) 2 + ( 1 β n ) u n ψ ( x ) 2 β n ψ ( x n ) ψ ( x ) 2 + ( 1 β n ) α n δ φ ( x n ) ψ ( x ) + μ n B x 2 + ( 1 α n ) ( 1 β n ) ψ ( x n ) ψ ( x ) 2 ( 1 β n ) ψ ( x n ) u n 2 + 2 μ n ( 1 β n ) α n B x n B x δ φ ( x n ) ψ ( x ) + μ n B x y n + 2 μ n ( 1 β n ) ψ ( x n ) u n B x n B x + 2 ( 1 β n ) α n ψ ( x n ) u n δ φ ( x n ) ψ ( x ) + μ n B x y n ψ ( x n ) ψ ( x ) 2 + α n δ φ ( x n ) ψ ( x ) + μ n B x 2 ( 1 β n ) ψ ( x n ) u n 2 + 2 μ n α n B x n B x δ φ ( x n ) ψ ( x ) + μ n B x y n + 2 μ n ψ ( x n ) u n B x n B x + 2 α n ψ ( x n ) u n δ φ ( x n ) ψ ( x ) + μ n B x y n .
Then we obtain
( 1 β n ) ψ ( x n ) u n 2 ( ψ ( x n ) ψ ( x ) + ψ ( x n + 1 ) ψ ( x ) ) ψ ( x n + 1 ) ψ ( x n ) + α n δ φ ( x n ) ψ ( x ) + μ n B x 2 + 2 μ n α n B x n B x δ φ ( x n ) ψ ( x ) + μ n B x y n + 2 μ n ψ ( x n ) u n B x n B x + 2 α n ψ ( x n ) u n δ φ ( x n ) ψ ( x ) + μ n B x y n .
Since lim n α n = 0 , lim n ψ ( x n + 1 ) ψ ( x n ) = 0 and lim n B x n B x = 0 , we have
lim n ψ ( x n ) u n = 0 .
(3.6)
Next, we prove lim sup n δ φ ( x ) ψ ( x ) , u n ψ ( x ) 0 , where x is the unique solution of (3.1). We take a subsequence { u n i } of { u n } such that
lim sup n δ φ ( x ) ψ ( x ) , u n ψ ( x ) = lim i δ φ ( x ) ψ ( x ) , u n i ψ ( x ) = lim i δ φ ( x ) ψ ( x ) , ψ ( x n i ) ψ ( x ) .
(3.7)
Since { x n i } is bounded, there exists a subsequence { x n i j } of { x n i } which converges weakly to some point z C . Without loss of generality, we may assume that x n i z . This implies that ψ ( x n i ) ψ ( z ) due to the weak continuity of ψ. Now, we show z S . We firstly show z EP ( F , A ) . Since z n = T λ n ( u n λ n A u n ) , for any y C , we have
F ( z n , y ) + 1 λ n y z n , z n ( u n λ n A u n ) 0 .
From the monotonicity of F, we have
1 λ n y z n , z n ( u n λ n A u n ) F ( y , z n ) , y C .
Hence,
y z n i , z n i u n i λ n i + A u n i F ( y , z n i ) , y C .
(3.8)
Put v t = t y + ( 1 t ) z for all t ( 0 , 1 ] and y C . Then we have v t C . So, from (3.8) we have
v t z n i , A v t v t z n i , A v t v t z n i , z n i u n i λ n i + A u n i + F ( v t , z n i ) = v t z n i , A v t A z n i + v t z n i , A z n i A u n i v t z n i , z n i u n i λ n i + F ( v t , z n i ) .
(3.9)
Note that A z n i A u n i 1 β z n i u n i 0 . Further, from the monotonicity of A, we have v t z n i , A v t A z n i 0 . Letting i in (3.9), we have v t z , A v t F ( v t , z ) . This together with (F1), (F4) implies that
0 = F ( v t , v t ) t F ( v t , y ) + ( 1 t ) F ( v t , z ) t F ( v t , y ) + ( 1 t ) v t z , A v t = t F ( v t , y ) + ( 1 t ) t y z , A v t ,
and hence 0 F ( v t , y ) + ( 1 t ) A v t , y z . Letting t 0 , we have 0 F ( z , y ) + y z , A z . This implies that z EP ( F , A ) . Next, we only need to prove z GVI ( B , ψ , C ) . Set
R v = { B v + N C ( v ) , v C , , v C .
By [41], we know that R is maximal ψ-monotone. Let ( v , w ) G ( R ) . Since w B v N C ( v ) and x n C , we have ψ ( v ) ψ ( x n ) , w B v 0 . Noting that u n = P C [ α n δ φ ( x n ) + ( 1 α n ) ( ψ ( x n ) μ n B x n ) ] , we get
ψ ( v ) u n , u n [ α n δ φ ( x n ) + ( 1 α n ) ( ψ ( x n ) μ n B x n ) ] 0 .
It follows that
ψ ( v ) u n , u n ψ ( x n ) μ n + B x n α n μ n ( δ φ ( x n ) ψ ( x n ) + μ n B x n ) 0 .
Then
ψ ( v ) ψ ( x n i ) , w ψ ( v ) ψ ( x n i ) , B v ψ ( v ) ψ ( x n i ) , B v ψ ( v ) u n i , u n i ψ ( x n i ) μ n i ψ ( v ) u n i , B x n i + α n i μ n i ψ ( v ) u n i , δ φ ( x n i ) ψ ( x n i ) + μ n i B x n i = ψ ( v ) ψ ( x n i ) , B v B x n i + ψ ( v ) ψ ( x n i ) , B x n i ψ ( v ) u n i , u n i ψ ( x n i ) μ n i ψ ( v ) u n i , B x n i + α n i μ n i ψ ( v ) u n i , δ φ ( x n i ) ψ ( x n i ) + μ n i B x n i ψ ( v ) u n i , u n i ψ ( x n i ) μ n i ψ ( x n i ) u n i , B x n i + α n i μ n i ψ ( v ) u n i , δ φ ( x n i ) ψ ( x n i ) + μ n i B x n i .
(3.10)
Since ψ ( x n i ) u n i 0 and ψ ( x n i ) ψ ( z ) , we deduce that ψ ( v ) ψ ( z ) , w 0 by taking i in (3.10). Thus, z R 1 0 by the maximal ψ-monotonicity of R. Hence, z GVI ( B , ψ , C ) . Therefore, z S . From (3.7), we obtain
lim sup n δ φ ( x ) ψ ( x ) , u n ψ ( x ) = lim i δ φ ( x ) ψ ( x ) , ψ ( x n i ) ψ ( x ) = δ φ ( x ) ψ ( x ) , ψ ( z ) ψ ( x ) 0 .
Note that
u n ψ ( x ) 2 α n ( δ φ ( x n ) ψ ( x ) ) + ( 1 α n ) y n , u n ψ ( x ) α n δ φ ( x n ) φ ( x ) , u n ψ ( x ) + α n δ φ ( x ) ψ ( x ) , u n ψ ( x ) + ( 1 α n ) ψ ( x n ) μ n B x n ( ψ ( x ) μ n B x ) u n ψ ( x ) α n L δ x n x u n ψ ( x ) + α n δ φ ( x ) ψ ( x ) , u n ψ ( x ) + ( 1 α n ) ψ ( x n ) ψ ( x ) u n ψ ( x ) α n ( δ L / γ ) ψ ( x n ) ψ ( x ) u n ψ ( x ) + α n δ φ ( x ) ψ ( x ) , u n ψ ( x ) + ( 1 α n ) ψ ( x n ) ψ ( x ) u n ψ ( x ) = [ 1 ( 1 L δ / γ ) α n ] ψ ( x n ) ψ ( x ) u n ψ ( x ) + α n δ φ ( x ) ψ ( x ) , u n ψ ( x ) = 1 ( 1 L δ / γ ) α n 2 ψ ( x n ) ψ ( x ) 2 + 1 2 u n ψ ( x ) 2 + α n δ φ ( x ) ψ ( x ) , u n ψ ( x ) .
It follows that
u n ψ ( x ) 2 [ 1 ( 1 L δ / γ ) α n ] ψ ( x n ) ψ ( x ) 2 + 2 α n δ φ ( x ) ψ ( x ) , u n ψ ( x ) .
Therefore,
ψ ( x n + 1 ) ψ ( x ) 2 β n ψ ( x n ) ψ ( x ) 2 + ( 1 β n ) u n ψ ( x ) 2 β n ψ ( x n ) ψ ( x ) 2 + ( 1 β n ) [ 1 ( 1 δ L / γ ) α n ] ψ ( x n ) ψ ( x ) 2 + 2 ( 1 β n ) α n δ φ ( x ) ψ ( x ) , u n ψ ( x ) = [ 1 ( 1 δ L / γ ) ( 1 β n ) α n ] ψ ( x n ) ψ ( x ) 2 + 2 ( 1 β n ) α n δ φ ( x ) ψ ( x ) , u n ψ ( x ) = [ 1 ( 1 δ L / γ ) ( 1 β n ) α n ] ψ ( x n ) ψ ( x ) 2 + ( 1 δ L / γ ) ( 1 β n ) × α n ( 2 1 δ L / γ δ φ ( x ) ψ ( x ) , u n ψ ( x ) ) = ( 1 γ n ) ψ ( x n ) ψ ( x ) 2 + δ n γ n ,

where γ n = ( 1 δ L / γ ) ( 1 β n ) α n and δ n = 2 1 δ L / γ δ φ ( x ) ψ ( x ) , u n ψ ( x ) . It is easily seen that n γ n = and lim sup n δ n 0 . We can therefore apply Lemma 2.4 to conclude that ψ ( x n ) ψ ( x ) and x n x . This completes the proof. □

Declarations

Acknowledgements

Yonghong Yao was supported in part by NSFC 11071279 and NSFC 71161001-G0105. Rudong Chen was supported in part by NSFC 11071279. Yeong-Cheng Liou was supported in part by NSC 101-2628-E-230-001-MY3 and NSC 101-2622-E-230-005-CC3.

Authors’ Affiliations

(1)
Department of Mathematics, Tianjin Polytechnic University
(2)
Department of Information Management, Cheng Shiu University

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