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Strong convergence algorithm for approximating the common solutions of a variational inequality, a mixed equilibrium problem and a hierarchical fixed-point problem

Abstract

This paper investigates the common set of solutions of a variational inequality, a mixed equilibrium problem, and a hierarchical fixed-point problem in a Hilbert space. A numerical method is proposed to find the approximate element of this common set. The strong convergence of this method is proved under some conditions. The proposed method is shown to be an improvement and extension of some known results.

MSC: 49J30, 47H09, 47J20.

1 Introduction

Let H be a real Hilbert space, whose inner product and norm are denoted by , and . Let C be a nonempty closed convex subset of H and A be a mapping from C into H. A classical variational inequality problem, denoted by VI(A,C), is to find a vector uC such that

vu,Au0,vC.
(1.1)

The solution of VI(A,C) is denoted by Ω . It is easy to observe that

u Ω u = P C [ u ρ A u ] ,where ρ>0.

We now have a variety of techniques to suggest and analyze various iterative algorithms for solving variational inequalities and the related optimization problems; see [125]. The fixed-point theory has played an important role in the development of various algorithms for solving variational inequalities. Using the projection operator technique, one usually establishes an equivalence between the variational inequalities and the fixed-point problem. This alternative equivalent formulation was used by Lions and Stampacchia [1] to study the existence of a solution of the variational inequalities.

We introduce the following definitions, which are useful in the following analysis.

Definition 1.1 The mapping T:CH is said to be

  1. (a)

    monotone if

    TxTy,xy0,x,yC;
  2. (b)

    strongly monotone if there exists an α>0 such that

    TxTy,xyα x y 2 ,x,yC;
  3. (c)

    α-inverse strongly monotone if there exists an α>0 such that

    TxTy,xyα T x T y 2 ,x,yC;
  4. (d)

    nonexpansive if

    TxTyxy,x,yC;
  5. (e)

    k-Lipschitz continuous if there exists a constant k>0 such that

    TxTykxy,x,yC;
  6. (f)

    contraction on C if there exists a constant 0k<1 such that

    TxTykxy,x,yC.

It is easy to observe that every α-inverse strongly monotone T is monotone and Lipschitz continuous. A mapping T:CH is called k-strict pseudo-contraction if there exists a constant 0k<1 such that

T x T y 2 x y 2 +k ( I T ) x ( I T ) y 2 ,x,yC.
(1.2)

The fixed-point problem for the mapping T is to find xC such that

Tx=x.
(1.3)

We denote by F(T) the set of solutions of (1.3). It is well known that the class of strict pseudo-contractions includes the class of Lipschitzian mappings, then F(T) is closed and convex and P F ( T ) is well defined (see [2]).

The mixed equilibrium problem, denoted by MEP, is to find xC such that

F 1 (x,y)+Dx,yx0,yC,
(1.4)

where F 1 :C×CR is a bifunction, and D:CH is a nonlinear mapping. This problem was introduced and studied by Moudafi and Théra [3] and Moudafi [4]. The set of solutions of (1.4) is denoted by

MEP( F 1 ):= { x C : F 1 ( x , y ) + D x , y x 0 , y C } .
(1.5)

If D=0, then it is reduced to the equilibrium problem, which is to find xC such that

F 1 (x,y)0,yC.
(1.6)

The solution set of (1.6) is denoted by EP( F 1 ). Numerous problems in physics, optimization, and economics reduce to finding a solution of (1.6); see [59]. In 1997, Combettes and Hirstoaga [10] introduced an iterative scheme of finding the best approximation to the initial data when EP(F) is nonempty. Recently Plubtieng and Punpaeng [7] introduced an iterative method for finding the common element of the set F(T) Ω EP( F 1 ).

Let S:CH be a nonexpansive mapping. The following problem is called a hierarchical fixed point problem: Find xF(T) such that

xSx,yx0,yF(T).
(1.7)

It is known that the hierarchical fixed-point problem (1.7) links with some monotone variational inequalities and convex programming problems; see [11, 12, 26]. Various methods have been proposed to solve the hierarchical fixed-point problem; see Moudafi [13], Mainge and Moudafi in [14], Marino and Xu in [15] and Cianciaruso et al. [16]. Very recently, Yao et al. [12] introduced the following strong convergence iterative algorithm to solve problem (1.7):

y n = β n S x n + ( 1 β n ) x n , x n + 1 = P C [ α n f ( x n ) + ( 1 α n ) T y n ] , n 0 ,
(1.8)

where f:CH is a contraction mapping, and { α n } and { β n } are two sequences in (0,1). Under some certain restrictions on the parameters, Yao et al. proved that the sequence { x n } generated by (1.8) converges strongly to zF(T), which is the unique solution of the following variational inequality:

( I f ) z , y z 0,yF(T).
(1.9)

In 2011, Ceng et al. [17] investigated the following iterative method:

x n + 1 = P C [ α n ρ U ( x n ) + ( I α n μ F ) ( T ( y n ) ) ] ,n0,
(1.10)

where U is a Lipschitzian mapping, and F is a Lipschitzian and strongly monotone mapping. They proved that under some approximate assumptions on the operators and parameters, the sequence { x n } generated by (1.10) converges strongly to the unique solution of the variational inequality

ρ U ( z ) μ F ( z ) , x z 0,xFix(T).

In this paper, motivated by the work of Yao et al. [12], Ceng et al. [17], Bnouhachem [18, 19] and by the recent work going in this direction, we give an iterative method for finding the approximate element of the common set of solutions of (1.1), (1.4), and (1.7) in a real Hilbert space. We establish a strong convergence theorem based on this method. We would like to mention that our proposed method is quite general and flexible and includes many known results for solving equilibrium problems, variational inequality problems, and hierarchical fixed-point problems; see, e.g., [11, 12, 1417, 24, 25] and relevant references cited therein.

2 Preliminaries

In this section, we list some fundamental lemmas that are useful in the consequent analysis. The first lemma provides some basic properties of projection onto C.

Lemma 2.1Let P C denote the projection ofHontoC. Then we have the following inequalities:

z P C [ z ] , P C [ z ] v 0,zH,vC;
(2.1)
u v , P C [ u ] P C [ v ] P C [ u ] P C [ v ] 2 ,u,vH;
(2.2)
P C [ u ] P C [ v ] uv,u,vH;
(2.3)
u P C [ z ] 2 z u 2 z P C [ z ] 2 ,zH,uC.
(2.4)

Lemma 2.2 [20]

Let F 1 :C×CRbe a bifunction satisfying the following assumptions:

  1. (i)

    F 1 (x,x)=0, xC;

  2. (ii)

    F 1 is monotone, i.e., F 1 (x,y)+ F 1 (y,x)0, x,yC;

  3. (iii)

    for eachx,y,zC, lim t 0 F 1 (tz+(1t)x,y) F 1 (x,y);

  4. (iv)

    for eachxC, y F 1 (x,y)is convex and lower semicontinuous.

Letr>0andxH. Then there existszCsuch that

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

Lemma 2.3 [10]

Assume that F 1 :C×CRsatisfies assumptions (i)-(iv) of Lemma  2.2, and forr>0andxH, define a mapping T r :HCas follows:

T r (x)= { z C : F 1 ( z , y ) + 1 r y z , z x 0 , y C } .

Then the following hold:

  1. (i)

    T r is single-valued;

  2. (ii)

    T r is firmly nonexpansive, i.e.,

    T r x T r y 2 T r x T r y,xy,x,yH;
  3. (iii)

    F( T r )=EP( F 1 );

  4. (iv)

    EP( F 1 )is closed and convex.

Lemma 2.4 [21]

LetCbe a nonempty closed convex subset of a real Hilbert spaceH. IfT:CCis a nonexpansive mapping withFix(T), then the mappingITis demiclosed at 0, i.e., if{ x n }is a sequence inCweakly converging toxand if{(IT) x n }converges strongly to 0, then(IT)x=0.

Lemma 2.5 [17]

LetU:CHbe aτ-Lipschitzian mapping, and letF:CHbe ak-Lipschitzian andη-strongly monotone mapping, then for0ρτ<μη, μFρUisμηρτ-strongly monotone, i.e.,

( μ F ρ U ) x ( μ F ρ U ) y , x y (μηρτ) x y 2 ,x,yC.

Lemma 2.6 [22]

Suppose thatλ(0,1)andμ>0. LetF:CHbe ak-Lipschitzian andη-strongly monotone operator. In association with a nonexpansive mappingT:CC, define the mapping T λ :CHby

T λ x=TxλμFT(x),xC.

Then T λ is a contraction providedμ< 2 η k 2 , that is,

T λ x T λ y (1λν)xy,x,yC,

whereν=1 1 μ ( 2 η μ k 2 ) .

Lemma 2.7 [23]

Assume that { a n } is a sequence of nonnegative real numbers such that

a n + 1 (1 γ n ) a n + δ n ,

where{ γ n }is a sequence in(0,1), and δ n is a sequence such that

  1. (1)

    n = 1 γ n =;

  2. (2)

    lim sup n δ n / γ n 0or n = 1 | δ n |<.

Then lim n a n =0.

Lemma 2.8 [27]

LetCbe a closed convex subset ofH. Let{ x n }be a bounded sequence in H. Assume that

  1. (i)

    the weakw-limit set w w ( x n )C, where w w ( x n )={x: x n i x};

  2. (ii)

    for eachzC, lim n x n zexists.

Then{ x n }is weakly convergent to a point inC.

3 The proposed method and some properties

In this section, we suggest and analyze our method for finding the common solutions of the variational inequality (1.1), the mixed equilibrium problem (1.4), and the hierarchical fixed-point problem (1.7).

Let C be a nonempty closed convex subset of a real Hilbert space H. Let D,A:CH be θ, α-inverse strongly monotone mappings, respectively. Let F 1 :C×CR be a bifunction satisfying assumptions (i)-(iv) of Lemma 2.2 and S,T:CC be a nonexpansive mappings such that F(T) Ω MEP( F 1 ). Let F:CC be a k-Lipschitzian mapping and be η-strongly monotone, and let U:CC be a τ-Lipschitzian mapping.

Algorithm 3.1 For an arbitrary given x 0 C, let the iterative sequences { u n }, { x n }, { y n }, and { z n } be generated by

F 1 ( u n , y ) + D x n , y u n + 1 r n y u n , u n x n 0 , y C ; z n = P C [ u n λ n A u n ] ; y n = β n S x n + ( 1 β n ) z n ; x n + 1 = P C [ α n ρ U ( x n ) + ( I α n μ F ) ( T ( y n ) ) ] , n 0 ,
(3.1)

where { λ n }(0,2α), { r n }(0,2θ). Suppose that the parameters satisfy 0<μ< 2 η k 2 , 0ρτ<ν, where ν=1 1 μ ( 2 η μ k 2 ) . Also, { α n } and { β n } are sequences in (0,1) satisfying the following conditions:

  1. (a)

    lim n α n =0 and n = 1 α n =,

  2. (b)

    lim n ( β n / α n )=0,

  3. (c)

    n = 1 | α n 1 α n |< and n = 1 | β n 1 β n |<,

  4. (d)

    lim inf n r n >0 and n = 1 | r n 1 r n |<,

  5. (e)

    lim inf n λ n < lim sup n λ n <2α and n = 1 | λ n 1 λ n |<.

Remark 3.1 Our method can be viewed as an extension and improvement for some well-known results, for example, the following.

  • If A=0, we obtain an extension and improvement of the method of Wang and Xu [24] for finding the approximate element of the common set of solutions of a mixed equilibrium problem and a hierarchical fixed-point problem in a real Hilbert space.

  • If we have the Lipschitzian mapping U=f, F=I, ρ=μ=1, and A=0, we obtain an extension and improvement of the method of Yao et al.[12] for finding the approximate element of the common set of solutions of a mixed equilibrium problem and a hierarchical fixed-point problem in a real Hilbert space.

  • The contractive mapping f with a coefficient α[0,1) in other papers [12, 15, 22, 25] is extended to the cases of the Lipschitzian mapping U with a coefficient constant γ[0,).

This shows that Algorithm 3.1 is quite general and unifying.

Lemma 3.1Let x F(T) Ω MEP( F 1 ). Then{ x n }, { u n }, { z n }, and{ y n }are bounded.

Proof First, we show that the mapping (I r n D) is nonexpansive. For any x,yC,

( I r n D ) x ( I r n D ) y 2 = ( x y ) r n ( D x D y ) 2 = x y 2 2 r n x y , D x D y + r n 2 D x D y 2 x y 2 r n ( 2 θ r n ) D x D y 2 x y 2 .

Similarly, we can show that the mapping (I λ n A) is nonexpansive. It follows from Lemma 2.3 that u n = T r n ( x n r n D x n ). Let x F(T) Ω MEP( F 1 ); we have x = T r n ( x r n D x ).

u n x 2 = T r n ( x n r n D x n ) T r n ( x r n D x ) 2 ( x n r n D x n ) ( x r n D x ) 2 x n x 2 r n ( 2 θ r n ) D x n D x 2 x n x 2 .
(3.2)

Since the mapping A is α-inverse strongly monotone, we have

z n x 2 = P C [ u n λ n A u n ) P C [ x λ n A x ] 2 u n x λ n ( A u n A x ) 2 u n x 2 λ n ( 2 α λ n ) A u n A x 2 u n x 2 x n x 2 .
(3.3)

We define V n = α n ρU( x n )+(I α n μF)(T( y n )). Next, we prove that the sequence { x n } is bounded, and without loss of generality we can assume that β n α n for all n1. From (3.1), we have

x n + 1 x = P C [ V n ] P C [ x ] α n ρ U ( x n ) + ( I α n μ F ) ( T ( y n ) ) x α n ρ U ( x n ) μ F ( x ) + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) T ( x ) = α n ρ U ( x n ) ρ U ( x ) + ( ρ U μ F ) x + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) T ( x ) α n ρ τ x n x + α n ( ρ U μ F ) x + ( 1 α n ν ) y n x α n ρ τ x n x + α n ( ρ U μ F ) x + ( 1 α n ν ) β n S x n + ( 1 β n ) z n x α n ρ τ x n x + α n ( ρ U μ F ) x + ( 1 α n ν ) ( β n S x n S x + β n S x x + ( 1 β n ) z n x ) α n ρ τ x n x + α n ( ρ U μ F ) x + ( 1 α n ν ) ( β n S x n S x + β n S x x + ( 1 β n ) x n x ) α n ρ τ x n x + α n ( ρ U μ F ) x + ( 1 α n ν ) ( β n x n x + β n S x x + ( 1 β n ) x n x ) = ( 1 α n ( ν ρ τ ) ) x n x + α n ( ρ U μ F ) x + ( 1 α n ν ) β n S x x ( 1 α n ( ν ρ τ ) ) x n x + α n ( ρ U μ F ) x + β n S x x ( 1 α n ( ν ρ τ ) ) x n x + α n ( ( ρ U μ F ) x + S x x ) = ( 1 α n ( ν ρ τ ) ) x n x + α n ( ν ρ τ ) ν ρ τ ( ( ρ U μ F ) x + S x x ) max { x n x , 1 ν ρ τ ( ( ρ U μ F ) x + S x x ) } ,

where the third inequality follows from Lemma 2.6.

By induction on n, we obtain x n x max{ x 0 x , 1 ν ρ τ ((ρUμF) x +S x x )} for n0 and x 0 C. Hence, { x n } is bounded and, consequently, we deduce that { u n }, { z n }, { v n }, { y n }, {S( x n )}, {T( x n )}, {F(T( y n ))}, and {U( x n )} are bounded. □

Lemma 3.2Let x F(T) Ω MEP( F 1 )and{ x n }be the sequence generated by Algorithm  3.1. Then we have:

  1. (a)

    lim n x n + 1 x n =0.

  2. (b)

    The weakw-limit set w w ( x n )F(T), ( w w ( x n )={x: x n i x}).

Proof From the nonexpansivity of the mapping (I λ n A) and P C , we have

z n z n 1 ( u n λ n A u n ) ( u n 1 λ n 1 A u n 1 ) = ( u n u n 1 ) λ n ( A u n A u n 1 ) ( λ n λ n 1 ) A u n 1 ( u n u n 1 ) λ n ( A u n A u n 1 ) + | λ n λ n 1 | A u n 1 u n u n 1 + | λ n λ n 1 | A u n 1 .
(3.4)

Next, we estimate that

y n y n 1 = β n S x n + ( 1 β n ) z n ( β n 1 S x n 1 + ( 1 β n 1 ) z n 1 ) = β n ( S x n S x n 1 ) + ( β n β n 1 ) S x n 1 + ( 1 β n ) ( z n z n 1 ) + ( β n 1 β n ) z n 1 β n x n x n 1 + ( 1 β n ) z n z n 1 + | β n β n 1 | ( S x n 1 + z n 1 ) .
(3.5)

It follows from (3.4) and (3.5) that

y n y n 1 β n x n x n 1 + ( 1 β n ) { u n u n 1 + | λ n λ n 1 | A u n 1 } + | β n β n 1 | ( S x n 1 + z n 1 ) .
(3.6)

On the other hand, u n = T r n ( x n r n D x n ) and u n 1 = T r n 1 ( x n 1 r n 1 D x n 1 ), we have

F 1 ( u n ,y)+D x n ,y u n + 1 r n y u n , u n x n 0,yC
(3.7)

and

F 1 ( u n 1 ,y)+D x n 1 ,y u n 1 + 1 r n 1 y u n 1 , u n 1 x n 1 0,yC.
(3.8)

Take y= u n 1 in (3.7) and y= u n in (3.8), we get

F 1 ( u n , u n 1 )+D x n , u n 1 u n + 1 r n u n 1 u n , u n x n 0
(3.9)

and

F 1 ( u n 1 , u n )+D x n 1 , u n u n 1 + 1 r n 1 u n u n 1 , u n 1 x n 1 0.
(3.10)

Adding (3.9) and (3.10) and using the monotonicity of F 1 , we have

D x n 1 D x n , u n u n 1 + u n u n 1 , u n 1 x n 1 r n 1 u n x n r n 0,

which implies that

0 u n u n 1 , r n ( D x n 1 D x n ) + r n r n 1 ( u n 1 x n 1 ) ( u n x n ) = u n 1 u n , u n u n 1 + ( 1 r n r n 1 ) u n 1 + ( x n 1 r n D x n 1 ) ( x n r n D x n ) x n 1 + r n r n 1 x n 1 = u n 1 u n , ( 1 r n r n 1 ) u n 1 + ( x n 1 r n D x n 1 ) ( x n r n D x n ) x n 1 + r n r n 1 x n 1 u n u n 1 2 = u n 1 u n , ( 1 r n r n 1 ) ( u n 1 x n 1 ) + ( x n 1 r n D x n 1 ) ( x n r n D x n ) u n u n 1 2 u n 1 u n { | 1 r n r n 1 | u n 1 x n 1 + ( x n 1 r n D x n 1 ) ( x n r n D x n } u n u n 1 2 u n 1 u n { | 1 r n r n 1 | u n 1 x n 1 + x n 1 x n } u n u n 1 2 ,

and then

u n 1 u n |1 r n r n 1 | u n 1 x n 1 + x n 1 x n .

Without loss of generality, let us assume that there exists a real number μ such that r n >μ>0 for all positive integers n. Then we get

u n 1 u n x n 1 x n + 1 μ | r n 1 r n | u n 1 x n 1 .
(3.11)

It follows from (3.6) and (3.11) that

y n y n 1 β n x n x n 1 + ( 1 β n ) { x n x n 1 + 1 μ | r n r n 1 | u n 1 x n 1 + | λ n λ n 1 | A u n 1 } + | β n β n 1 | ( S x n 1 + z n 1 ) = x n x n 1 + ( 1 β n ) { 1 μ | r n r n 1 | u n 1 x n 1 + | λ n λ n 1 | A u n 1 } + | β n β n 1 | ( S x n 1 + z n 1 ) .
(3.12)

Next, we estimate that

x n + 1 x n = P C [ V n ] P C [ V n 1 α n ρ ( U ( x n ) U ( x n 1 ) ) + ( α n α n 1 ) ρ U ( x n 1 ) + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) T ( y n 1 ) + ( I α n μ F ) ( T ( y n 1 ) ) ( I α n 1 μ F ) ( T ( y n 1 ) ) α n ρ τ x n x n 1 + ( 1 α n ν ) y n y n 1 + | α n α n 1 | ( ρ U ( x n 1 ) + μ F ( T ( y n 1 ) ) ) ,
(3.13)

where the second inequality follows from Lemma 2.6. From (3.12) and (3.13), we have

x n + 1 x n α n ρ τ x n x n 1 + ( 1 α n ν ) × ( x n x n 1 + 1 μ | r n r n 1 | u n 1 x n 1 + | λ n λ n 1 | A u n 1 ) + | β n β n 1 | ( S x n 1 + z n 1 ) + | α n α n 1 | ( ρ U ( x n 1 ) + μ F ( T ( y n 1 ) ) ) ( 1 ( ν ρ τ ) α n ) x n x n 1 + 1 μ | r n r n 1 | u n 1 x n 1 + | λ n λ n 1 | A u n 1 + | β n β n 1 | ( S x n 1 + z n 1 ) + | α n α n 1 | ( ρ U ( x n 1 ) + μ F ( T ( y n 1 ) ) ) ( 1 ( ν ρ τ ) α n ) x n x n 1 + M ( 1 μ | r n r n 1 | + | λ n λ n 1 | + | β n β n 1 | + | α n α n 1 | ) .
(3.14)

Here

M = max { sup n 1 u n 1 x n 1 , sup n 1 A u n 1 , sup n 1 ( S x n 1 + z n 1 ) , sup n 1 ( ρ U ( x n 1 ) + μ F ( T ( y n 1 ) ) ) } .

It follows by conditions (a)-(e) of Algorithm 3.1 and Lemma 2.7 that

lim n x n + 1 x n =0.

Next, we show that lim n u n x n =0. Since x F(T) Ω MEP( F 1 ), by using (3.2) and (3.3), we obtain

x n + 1 x 2 = P C ( V n ) x , x n + 1 x = P C ( V n ) V n , P C ( V n ) x + V n x , x n + 1 x α n ( ρ U ( x n ) μ F ( x ) ) + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) ( T ( x ) ) , x n + 1 x = α n ρ ( U ( x n ) U ( x ) ) , x n + 1 x + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) ( T ( x ) ) , x n + 1 x α n ρ τ x n x x n + 1 x + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) y n x x n + 1 x α n ρ τ 2 ( x n x 2 + x n + 1 x 2 ) + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) 2 ( y n x 2 + x n + 1 x 2 ) ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) 2 ( β n S x n x 2 + ( 1 β n ) z n x 2 ) ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) β n 2 S x n x 2 + ( 1 α n ν ) ( 1 β n ) 2 { x n x 2 r n ( 2 θ r n ) D x n D x 2 λ n ( 2 α λ n ) A u n A x 2 } ,
(3.15)

which implies that

x n + 1 x 2 α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + ( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { x n x 2 r n ( 2 θ r n ) D x n D x 2 λ n ( 2 α λ n ) A u n A x 2 } α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + x n x 2 + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 ( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { r n ( 2 θ r n ) D x n D x 2 + λ n ( 2 α λ n ) A u n A x 2 } .

Then, from the inequality above, we get

( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { r n ( 2 θ r n ) D x n D x 2 + λ n ( 2 α λ n ) A u n A x 2 } α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + β n S x n x 2 + x n x 2 x n + 1 x 2 α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + β n S x n x 2 + ( x n x + x n + 1 x ) x n + 1 x n .

Since lim inf n λ n lim sup n λ n <2α, { r n }(0,2θ), lim n x n + 1 x n =0, α n 0, and β n 0, we obtain lim n D x n D x =0 and lim n A u n A x =0.

Since T r n is firmly nonexpansive, we have

u n x 2 = T r n ( x n r n D x n ) T r n ( x r n D x ) 2 u n x , ( x n r n D x n ) ( x r n D x ) = 1 2 { u n x 2 + ( x n r n D x n ) ( x r n D x ) 2 u n x [ ( x n r n D x n ) ( x r n D x ) ] 2 } .

Hence,

u n x 2 ( x n r n D x n ) ( x r n D x ) 2 u n x n + r n ( D x n D x ) 2 x n x 2 u n x n + r n ( D x n D x ) 2 x n x 2 u n x n 2 + 2 r n u n x n D x n D x .

From (3.15), (3.3), and the inequality above, we have

x n + 1 x 2 ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) 2 ( β n S x n x 2 + ( 1 β n ) z n x 2 ) ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) 2 ( β n S x n x 2 + ( 1 β n ) u n x 2 ) ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) 2 { β n S x n x 2 + ( 1 β n ) ( x n x 2 u n x n 2 + 2 r n u n x n D x n D x ) } ,

which implies that

x n + 1 x 2 α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + ( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { x n x 2 u n x n 2 + 2 r n u n x n D x n D x } α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + x n x 2 + ( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { u n x n 2 + 2 r n u n x n D x n D x } .

Hence,

( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) u n x n 2 α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + 2 ( 1 α n ν ) ( 1 β n ) r n 1 + α n ( ν ρ τ ) u n x n D x n D x + x n x 2 x n + 1 x 2 = α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + 2 ( 1 α n ν ) ( 1 β n ) r n 1 + α n ( ν ρ τ ) u n x n D x n D x + ( x n x + x n + 1 x ) x n + 1 x n .

Since lim n x n + 1 x n =0, α n 0, β n 0, and lim n D x n D x =0, we obtain

lim n u n x n =0.
(3.16)

From (2.2), we get

z n x 2 = P C [ u n λ n A u n ] P C [ x λ n A x ] 2 z n x , ( u n λ n A u n ) ( x λ n A x ) = 1 2 { z n x 2 + u n x λ n ( A u n A x ) 2 u n x λ n ( A u n A x ) ( z n x ) 2 } 1 2 { z n x 2 + u n x 2 u n z n λ n ( A u n A x ) 2 } 1 2 { z n x 2 + u n x 2 u n z n 2 + 2 λ n u n z n , A u n A x } 1 2 { z n x 2 + u n x 2 u n z n 2 + 2 λ n u n z n A u n A x } .

Hence,

z n x 2 u n x 2 u n z n 2 + 2 λ n u n z n A u n A x x n x 2 u n z n 2 + 2 λ n u n z n A u n A x .

From (3.15) and the inequality above, we have

x n + 1 x 2 ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) 2 ( β n S x n x 2 + ( 1 β n ) z n x 2 ) ( 1 α n ( ν ρ τ ) ) 2 x n + 1 x 2 + α n ρ τ 2 x n x 2 + α n ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) 2 { β n S x n x 2 + ( 1 β n ) ( x n x 2 u n z n 2 + 2 λ n u n z n A u n A x ) } ,

which implies that

x n + 1 x 2 α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + ( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) { x n x 2 u n z n 2 + 2 λ n u n z n A u n A x } .

Hence,

( 1 α n ν ) ( 1 β n ) 1 + α n ( ν ρ τ ) u n z n 2 α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + x n x 2 x n + 1 x 2 + 2 λ n u n z n A u n A x = α n ρ τ 1 + α n ( ν ρ τ ) x n x 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( x ) μ F ( x ) , x n + 1 x + ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S x n x 2 + ( x n x + x n + 1 x ) x n + 1 x n + 2 λ n u n z n A u n A x .

Since lim n x n + 1 x n =0, α n 0, β n 0, and lim n A u n A x =0, we obtain

lim n u n z n =0.
(3.17)

It follows from (3.16) and (3.17) that

lim n x n z n =0.
(3.18)

Since T( x n )C, we have

x n T ( x n ) x n x n + 1 + x n + 1 T ( x n ) = x n x n + 1 + P C [ V n ] P C [ T ( x n ) ] x n x n + 1 + α n ( ρ U ( x n ) μ F ( T ( y n ) ) ) + T ( y n ) T ( x n ) x n x n + 1 + α n ρ U ( x n ) μ F ( T ( y n ) ) + y n x n x n x n + 1 + α n ρ U ( x n ) μ F ( T ( y n ) ) + β n S x n + ( 1 β n ) z n x n x n x n + 1 + α n ρ U ( x n ) μ F ( T ( y n ) ) + β n S x n x n + ( 1 β n ) z n x n .

Since lim n x n + 1 x n =0, α n 0, β n 0, and ρU( x n )μF(T( y n )) and S x n x n are bounded, and lim n x n z n =0, we obtain

lim n x n T ( x n ) =0.

Since { x n } is bounded, without loss of generality we can assume that x n x C. It follows from Lemma 2.4 that x F(T). Therefore, w w ( x n )F(T). □

Theorem 3.1The sequence{ x n }generated by Algorithm  3.1 converges strongly toz, which is the unique solution of the variational inequality

ρ U ( z ) μ F ( z ) , x z 0,x Ω MEP( F 1 )F(T).
(3.19)

Proof Since { x n } is bounded x n w and from Lemma 3.2, we have wF(T). Next, we show that wMEP( F 1 ). Since u n = T r n ( x n r n D x n ), we have

F 1 ( u n ,y)+D x n ,y u n + 1 r n y u n , u n x n 0,yC.

It follows from the monotonicity of F 1 that

D x n ,y u n + 1 r n y u n , u n x n F 1 (y, u n ),yC

and

D x n k ,y u n k + y u n k , u n k x n k r n k F 1 (y, u n k ),yC.
(3.20)

Since lim n u n x n =0, and x n w, it is easy to observe that u n k w. For any 0<t1 and yC, let y t =ty+(1t)w, and we have y t C. Then from (3.20), we obtain

D y t , y t u n k D y t , y t u n k D x n k , y t u n k y t u n k , u n k x n k r n k + F 1 ( y t , u n k ) = D y t D u n k , y t u n k + D u n k D x n k , y t u n k y t u n k , u n k x n k r n k + F 1 ( y t , u n k ) .
(3.21)

Since D is Lipschitz continuous and lim n u n x n =0, we obtain lim k D u n k D x n k =0. From the monotonicity of D and u n k w, it follows from (3.21) that

D y t , y t w F 1 ( y t ,w).
(3.22)

Hence, from assumptions (i)-(iv) of Lemma 2.2 and (3.22), we have

0 = F 1 ( y t , y t ) t F 1 ( y t , y ) + ( 1 t ) F 1 ( y t , w ) t F 1 ( y t , y ) + ( 1 t ) D y t , y t w t F 1 ( y t , y ) + ( 1 t ) t D y t , y w ,
(3.23)

which implies that F 1 ( y t ,y)+(1t)D y t ,yw0. Letting t 0 + , we have

F 1 (w,y)+Dw,yw0,yC,

which implies that wMEP( F 1 ).

Furthermore, we show that w Ω . Let

Tv= { A v + N C v , v C , , otherwise,

where N C v:={wH:w,vu0,uC} is the normal cone to C at vC. Then T is maximal monotone and 0Tv if and only if v Ω (see [28]). Let G(T) denote the graph of T, and let (v,u)G(T); since uAv N C v and z n C, we have

v z n ,uAv0.
(3.24)

On the other hand, it follows from z n = P C [ u n λ n A u n ] and vC that

v z n , z n ( u n λ n A u n ) 0

and

v z n , z n u n λ n + A u n 0.

Therefore, from (3.24) and the inverse strong monotonicity of A, we have

v z n k , u v z n k , A v v z n k , A v v z n k , z n k u n k λ n k + A u n k v z n k , A v A z n k + v z n k , A z n k A u n k v z n k , z n k u n k λ n k v z n k , A z n k A u n k v z n k , z n k u n k λ n k .

Since lim n u n z n =0 and u n k w, it is easy to observe that z n k w. Hence, we obtain vw,u0. Since T is maximal monotone, we have w T 1 0, and hence w Ω . Thus we have

w Ω MEP( F 1 )F(T).

Observe that the constants satisfy 0ρτ<ν and

k η k 2 η 2 1 2 μ η + μ 2 k 2 1 2 μ η + μ 2 η 2 1 μ ( 2 η μ k 2 ) 1 μ η μ η 1 1 μ ( 2 η μ k 2 ) μ η ν ,

therefore, from Lemma 2.5, the operator μFρU is μηρτ strongly monotone, and we get the uniqueness of the solution of the variational inequality (3.19) and denote it by z Ω MEP( F 1 )F(T).

Next, we claim that lim sup n ρU(z)μF(z), x n z0. Since { x n } is bounded, there exists a subsequence { x n k } of { x n } such that

lim sup n ρ U ( z ) μ F ( z ) , x n z = lim sup k ρ U ( z ) μ F ( z ) , x n k z = ρ U ( z ) μ F ( z ) , w z 0 .

Next, we show that x n z. We have

x n + 1 z 2 = P C ( V n ) z , x n + 1 z = P C ( V n ) V n , P C ( V n ) z + V n z , x n + 1 z α n ( ρ U ( x n ) μ F ( z ) ) + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) ( T ( z ) ) , x n + 1 z = α n ρ ( U ( x n ) U ( z ) ) , x n + 1 z + α n ρ U ( z ) μ F ( z ) , x n + 1 z + ( I α n μ F ) ( T ( y n ) ) ( I α n μ F ) ( T ( z ) ) , x n + 1 z α n ρ τ x n z x n + 1 z + α n ρ U ( z ) μ F ( z ) , x n + 1 z + ( 1 α n ν ) y n z x n + 1 z α n ρ τ x n z x n + 1 z + α n ρ U ( z ) μ F ( z ) , x n + 1 z + ( 1 α n ν ) { β n S x n S z + β n S z z + ( 1 β n ) z n z } x n + 1 z α n ρ τ x n z x n + 1 z + α n ρ U ( z ) μ F ( z ) , x n + 1 z + ( 1 α n ν ) { β n x n z + β n S z z + ( 1 β n ) x n z } x n + 1 z = ( 1 α n ( ν ρ τ ) ) x n z x n + 1 z + α n ρ U ( z ) μ F ( z ) , x n + 1 z + ( 1 α n ν ) β n S z z x n + 1 z 1 α n ( ν ρ τ ) 2 ( x n z 2 + x n + 1 z 2 ) + α n ρ U ( z ) μ F ( z ) , x n + 1 z + ( 1 α n ν ) β n S z z x n + 1 z ,

which implies that

x n + 1 z 2 1 α n ( ν ρ τ ) 1 + α n ( ν ρ τ ) x n z 2 + 2 α n 1 + α n ( ν ρ τ ) ρ U ( z ) μ F ( z ) , x n + 1 z + 2 ( 1 α n ν ) β n 1 + α n ( ν ρ τ ) S z z x n + 1 z ( 1 α n ( ν ρ τ ) ) x n z 2 + 2 α n ( ν ρ τ ) 1 + α n ( ν ρ τ ) × { 1 ν ρ τ ρ U ( z ) μ F ( z ) , x n + 1 z + ( 1 α n ν ) β n α n ( ν ρ τ ) S z z x n + 1 z } .

Let γ n = α n (νρτ) and δ n = 2 α n ( ν ρ τ ) 1 + α n ( ν ρ τ ) { 1 ν ρ τ ρU(z)μF(z), x n + 1 z+ ( 1 α n ν ) β n α n ( ν ρ τ ) Szz x n + 1 z}.

We have

n = 1 α n =

and

lim sup n { 1 ν ρ τ ρ U ( z ) μ F ( z ) , x n + 1 z + ( 1 α n ν ) β n α n (