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Iterative algorithms with regularization for hierarchical variational inequality problems and convex minimization problems

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Abstract

In this paper, we consider a variational inequality problem which is defined over the set of intersections of the set of fixed points of a ζ-strictly pseudocontractive mapping, the set of fixed points of a nonexpansive mapping and the set of solutions of a minimization problem. We propose an iterative algorithm with regularization to solve such a variational inequality problem and study the strong convergence of the sequence generated by the proposed algorithm. The results of this paper improve and extend several known results in the literature.

1 Introduction

Let H be a real Hilbert space with the inner product , and the norm , let C be a nonempty closed convex subset of H, and let f:CR be a convex and continuously Fréchet differentiable functional. We consider the following minimization problem (MP):

min x C f(x).
(1.1)

We denote by Ξ the set of minimizers of problem (1.1), and we assume that Ξ. The gradient-projection algorithm (GPA) is one of the most elegant methods to solve the minimization problem (1.1). The convergence of the sequence generated by the GPA depends on the behavior of the gradient f. If f is strongly monotone and Lipschitz continuous, then we get the strong convergence of the sequence generated by the GPA to a unique solution of MP (1.1). However, if the gradient f is assumed to be only Lipschitz continuous, then the sequence generated by the GPA converges weakly if H is infinite-dimensional (a counterexample is given in [1]). Since the Lipschitz continuity of the gradient f implies that it is actually inverse strongly monotone (ism) [2], its complement can be an averaged mapping (that is, it can be expressed as a proper convex combination of the identity mapping and a nonexpansive mapping) [1]. Consequently, the GPA can be rewritten as the composite of a projection and an averaged mapping, which is again an averaged mapping. This shows that averaged mappings play an important role in the GPA. Very recently, Xu [1] used averaged mappings to study the convergence analysis of the GPA, which is an operator-oriented approach. He showed that the sequence generated by the GPA converges in norm to a minimizer of MP (1.1), which is also a unique solution of a particular type of variational inequality problem (VIP). It is worth to emphasize that the regularization, in particular the traditional Tikhonov regularization, is usually used to solve ill-posed optimization problems. The advantage of a regularization method is its possible strong convergence to the minimum-norm solution of the optimization problem. In [1], Xu introduced a hybrid gradient-projection algorithm with regularization and proved the strong convergence of the sequence to the minimum-norm solution of MP (1.1). Some iterative algorithms with or without regularization for MP (1.1) are proposed and analyzed in [35] for finding a common solution of MP (1.1) and the set of solutions of a nonexpansive mapping.

On the other hand, the theory of variational inequalities [6, 7] has emerged as an important tool to study a wide class of problems from science, engineering, social sciences. If the underlying set in the formulation of a variational inequality problem is a set of fixed points of a mapping or, more precisely, of a nonexpansive mapping, then the variational inequality problem is called hierarchical variational problem. For further details on hierarchical variational inequalities, we refer to [811] and the references therein.

In this paper, we consider a variational inequality problem which is defined over the set of intersections of the set of fixed points of a ζ-strictly pseudocontractive mapping, the set of fixed points of a nonexpansive mapping and the set of solutions of MP (1.1). We propose an iterative algorithm with regularization to solve such a variational inequality problem and study the strong convergence of the sequence generated by the proposed algorithm. The results of this paper improve and extend several known results in the literature.

2 Preliminaries and formulations

Throughout the paper, unless otherwise specified, we use the following assumptions and notations. Let H be a real Hilbert space whose inner product and norm are denoted by , and , respectively. Let C be a nonempty closed convex subset of H. We write x n x (respectively, x n x) to indicate that the sequence { x n } converges strongly (respectively, weakly) to x. Moreover, we use ω w ( x n ) to denote the weak ω-limit set of the sequence { x n }, that is,

ω w ( x n ):= { x H : x n i x  for some subsequence  { x n i }  of  { x n } } .

The metric (or nearest point) projection from H onto C is the mapping P C :HC which assigns to each point xH the unique point P C xC satisfying

x P C x= inf y C xy=:d(x,C).

Some important properties of projections are gathered in the following proposition.

Proposition 2.1 For given xH and zC, we have

  1. (a)

    z= P C xxz,yz0, yC;

  2. (b)

    z= P C x x z 2 x y 2 y z 2 , yC;

  3. (c)

    P C x P C y,xy P C x P C y 2 , yH, which concludes that P C is nonexpansive and monotone.

Definition 2.1 A mapping T:HH is said to be

  1. (a)

    ζ-strictly pseudocontractive if there exists a constant ζ[0,1) such that

    T x T y 2 x y 2 +ζ ( I T ) x ( I T ) y 2 ,x,yH.

    If ζ=0, then it is called nonexpansive;

  2. (b)

    firmly nonexpansive if 2TI is nonexpansive, or equivalently,

    xy,TxTy T x T y 2 ,x,yH;

    alternatively, T is firmly nonexpansive if and only if T can be expressed as

    T= 1 2 (I+S),

    where S:HH is a nonexpansive mapping.

It can be easily seen that the projection mappings are firmly nonexpansive. It is clear that T:CHC is ζ-strictly pseudocontractive if and only if

TxTy,xy x y 2 1 ζ 2 ( I T ) x ( I T ) y 2 ,x,yC.

Definition 2.2 Let T be a nonlinear operator with domain D(T)H and range R(T)H.

  1. (a)

    T is said to be monotone if

    xy,TxTy0,x,yD(T).
  2. (b)

    Given a number β>0, T is said to be β-strongly monotone if

    xy,TxTyβ x y 2 ,x,yD(T).
  3. (c)

    Given a number ν>0, T is said to be ν-inverse strongly monotone (ν-ism) if

    xy,TxTyν T x T y 2 ,x,yD(T).

Clearly,

  • if T is nonexpansive, then IT is monotone;

  • a projection P K is 1-ism;

  • if T is a ζ-strictly pseudocontractive mapping, then IT is 1 ζ 2 -inverse strongly monotone.

Definition 2.3 [1]

A mapping T:HH is said to be an averaged mapping if it can be written as the average of the identity I and a nonexpansive mapping, that is,

T(1α)I+αS,

where α(0,1) and S:HH is a nonexpansive mapping. More precisely, when the last equality holds, we say that T is α-averaged. Thus, firmly nonexpansive mappings (in particular, projections) are 1 2 -averaged maps.

Proposition 2.2 [12]

Let T:HH be a given mapping.

  1. (a)

    T is nonexpansive if and only if the complement IT is 1 2 -ism.

  2. (b)

    If T is ν-ism, then for γ>0, γT is ν γ -ism.

  3. (c)

    T is averaged if and only if the complement IT is ν-ism for some ν>1/2. Indeed, for α(0,1), T is α-averaged if and only if IT is 1 2 α -ism.

Proposition 2.3 [12, 13]

Let S,T,V:HH be given operators.

  1. (a)

    If T=(1α)S+αV for some α(0,1) and if S is averaged and V is nonexpansive, then T is averaged.

  2. (b)

    T is firmly nonexpansive if and only if the complement IT is firmly nonexpansive.

  3. (c)

    If T=(1α)S+αV for some α(0,1) and if S is firmly nonexpansive and V is nonexpansive, then T is averaged.

  4. (d)

    The composite of finitely many averaged mappings is averaged, that is, if each of the mappings { T i } i = 1 N is averaged, then so is the composite T 1 T N . In particular, if T 1 is α 1 -averaged and T 2 is α 2 -averaged, where α 1 , α 2 (0,1), then the composite T 1 T 2 is α-averaged, where α= α 1 + α 2 α 1 α 2 .

Lemma 2.1 [[14], Proposition 2.1]

Let C be a nonempty closed convex subset of a real Hilbert space H, and let T:CC be a mapping.

  1. (a)

    If T is a ζ-strictly pseudocontractive mapping, then T satisfies the Lipschitz condition

    TxTy 1 + ζ 1 ζ xy,x,yC.
  2. (b)

    If T is a ζ-strictly pseudocontractive mapping, then the mapping IT is semiclosed at 0, that is, if { x n } is a sequence in C such that x n x ˜ weakly and (IT) x n 0 strongly, then (IT) x ˜ =0.

  3. (c)

    If T is a ζ-(quasi-)strict pseudocontraction, then the fixed point set Fix(T) of T is closed and convex so that the projection P Fix ( T ) is well defined.

The following lemma is an immediate consequence of an inner product.

Lemma 2.2 In a real Hilbert space H, we have

x + y 2 x 2 +2y,x+y,x,yH.

The following elementary result on real sequences is quite well known.

Lemma 2.3 [15]

Let { a n } be a sequence of nonnegative real numbers such that

a n + 1 (1 s n ) a n + s n t n + ϵ n ,n0,

where { s n }(0,1] and { t n } satisfy the following conditions:

  1. (i)

    n = 0 s n =;

  2. (ii)

    either lim sup n t n 0 or n = 0 s n | t n |<;

  3. (iii)

    n = 0 ϵ n <, where ϵ n 0, n0.

Then lim n a n =0.

Lemma 2.4 [10]

Let C be a nonempty closed convex subset of a real Hilbert space H, and let T:CC be a ζ-strictly pseudocontractive mapping. Let γ and δ be two nonnegative real numbers such that (γ+δ)ζγ. Then

γ ( x y ) + δ ( T x T y ) (γ+δ)xy,x,yC.

The following lemma appeared implicitly in the paper of Reineermann [16].

Lemma 2.5 [16]

Let H be a real Hilbert space. Then, for all x,yH and λ[0,1],

λ x + ( 1 λ ) y 2 =λ x 2 +(1λ) y 2 λ(1λ) x y 2 .

Let C be a nonempty closed convex subset of a real Hilbert space H, and let A:CH be a monotone mapping. The variational inequality problem (VIP) is to find xC such that

Ax,yx0,yC.

The solution set of the VIP is denoted by VI(C,A). It is well known that

xVI(C,A)x= P C (xλAx),λ>0.

A set-valued mapping V:H 2 H is called monotone if for all x,yH, fVx and gVy imply that xy,fg0. A monotone set-valued mapping V:H 2 H is called maximal if its graph Gph(V) is not properly contained in the graph of any other monotone set-valued mapping. It is known that a monotone set-valued mapping V:H 2 H is maximal if and only if for (x,f)H×H, xy,fg0 for every (y,g)Gph(V) implies that fVx. Let A:CH be a monotone and Lipschitz continuous mapping and N C v be the normal cone to C at vC, that is,

N C v= { w H : v u , w 0 , u C } .

Define

Vv={ A v + N C v if  v C , if  v C .

Lemma 2.6 [17]

Let A:CH be a monotone mapping. Then

  1. (i)

    V is maximal monotone;

  2. (ii)

    v V 1 0vVI(C,A).

Throughout the paper, we denote by Fix(T) and Fix(Γ) the set of fixed points of T and Γ, respectively. We also assume that the set Fix(T)Fix(Γ)Ξ is nonempty closed and convex.

Let S,T:CC be nonexpansive mappings and Γ:CC be a ζ-strictly pseudocontractive mapping with ζ[0,1). In this paper, we consider and study the following hierarchical variational inequality problem which is defined on Fix(T)Fix(Γ)Ξ.

Find  x ˜ Fix ( T ) Fix ( Γ ) Ξ such that x ˜ S x ˜ , x ˜ x 0 , x Fix ( T ) Fix ( Γ ) Ξ .
(2.1)

We denote by Ω the solution set of problem (2.1). It is not difficult to verify that solving (2.1) is equivalent to the fixed point problem of finding x ˜ C such that

x ˜ = P Fix ( T ) Fix ( Γ ) Ξ S x ˜ ,

where P Fix ( T ) Fix ( Γ ) Ξ stands for the metric projection onto the closed convex set Fix(T)Fix(Γ)Ξ.

Problem (2.1) contains the hierarchical variational inequality problems considered and studied in [8, 18, 19] and the references therein.

By using the definition of the normal cone to Fix(T)Fix(Γ)Ξ, we have the mapping N Fix ( T ) Fix ( Γ ) Ξ :H 2 H :

x{ { u H : ( y Fix ( T ) Fix ( Γ ) Ξ ) y x , u 0 } , if  x Fix ( T ) Fix ( Γ ) Ξ ; , otherwise ,

and we readily prove that (2.1) is equivalent to the variational inequality

0(IS) x ˜ + N Fix ( T ) Fix ( Γ ) Ξ x ˜ .

By combining the hybrid gradient-projection method of Xu [1] and a two-step method of Yao et al. [11], we introduce the following three-step iterative algorithm:

{ y n = θ n S x n + ( 1 θ n ) x n , z n = β n Q y n + ( 1 β n ) T P C ( y n λ f α n ( y n ) ) , x n + 1 = σ n z n + γ n P C ( z n λ f α n ( z n ) ) + δ n Γ P C ( z n λ f α n ( z n ) ) , n 0 ,
(2.2)

where Q:CC is a ρ-contraction mapping, { α n }(0,), { β n },{ θ n },{ σ n }(0,1) and { γ n },{ δ n }[0,1] with σ n + γ n + δ n =1, n0. It is proven that under appropriate assumptions, the above iterative sequence { x n } converges strongly to an element x ˜ Fix(T)Fix(Γ)Ξ.

3 Main results

Let us consider the following assumptions:

  • the mapping Q:CC is a ρ-contraction;

  • the mapping Γ:CC is a ζ-strict pseudocontraction;

  • S,T:CC are two nonexpansive mappings;

  • f:CH is Lipschitz continuous with 0<λ< 2 L ;

  • { α n } is a sequence in (0,) with n = 0 α n <;

  • { β n }, { θ n }, { σ n } are sequences in (0,1) with 0< lim inf n σ n lim sup n σ n <1;

  • { γ n }, { δ n } are sequences in [0,1] with σ n + γ n + δ n =1, n0;

  • lim inf n δ n >0 and ( γ n + δ n )ζ γ n , n0.

Theorem 3.1 Let { x n } be a bounded sequence generated from any given x 0 C by (2.2). Assume that the following conditions hold:

  1. (H1)

    n = 0 β n =, lim n 1 β n |1 θ n 1 θ n |=0;

  2. (H2)

    lim n 1 β n | 1 θ n 1 θ n 1 |=0, lim n 1 θ n |1 β n 1 β n |=0;

  3. (H3)

    lim n θ n =0 and lim n α n + β n θ n =0;

  4. (H4)

    lim n | α n α n 1 | β n θ n =0, lim n | σ n σ n 1 | β n θ n =0;

  5. (H5)

    lim n 1 β n θ n | γ n 1 σ n γ n 1 1 σ n 1 |=0.

Then the following assertions hold:

  1. (i)

    lim n x n + 1 x n θ n =0;

  2. (ii)

    ω w ( x n )Ω.

Proof First of all, we show that P C (Iλ f α ) is ξ-averaged for each λ(0, 2 α + L ), where

ξ= 2 + λ ( α + L ) 4 (0,1).

Indeed, the Lipschitz continuity of f implies that f is 1 L -ism [2], that is,

f ( x ) f ( y ) , x y 1 L f ( x ) f ( y ) 2 .

Observe that

( α + L ) f α ( x ) f α ( y ) , x y = ( α + L ) [ α x y 2 + f ( x ) f ( y ) , x y ] = α 2 x y 2 + α f ( x ) f ( y ) , x y + α L x y 2 + L f ( x ) f ( y ) , x y α 2 x y 2 + 2 α f ( x ) f ( y ) , x y + f ( x ) f ( y ) 2 = α ( x y ) + f ( x ) f ( y ) 2 = f α ( x ) f α ( y ) 2 .

Therefore, it follows that f α =αI+f is 1 α + L -ism. Thus, by Proposition 2.2(b), λ f α is 1 λ ( α + L ) -ism. From Proposition 2.2(c), the complement Iλ f α is λ ( α + L ) 2 -averaged. Therefore, noting that P C is 1 2 -averaged and utilizing Proposition 2.3(d), we obtain that for each λ(0, 2 α + L ), P C (Iλ f α ) is ξ-averaged with

ξ= 1 2 + λ ( α + L ) 2 1 2 λ ( α + L ) 2 = 2 + λ ( α + L ) 4 (0,1).

This shows that P C (Iλ f α ) is nonexpansive. For λ(0, 2 L ), utilizing the fact that lim n 2 α n + L = 2 L , we may assume that

0<λ< 2 α n + L ,n0.

Consequently, it follows that for each integer n0, P C (Iλ f α n ) is ξ n -averaged with

ξ n = 1 2 + λ ( α n + L ) 2 1 2 λ ( α n + L ) 2 = 2 + λ ( α n + L ) 4 (0,1).

This implies that P C (Iλ f α n ) is nonexpansive for all n0.

The rest of the proof is divided into several steps.

Step 1. lim n x n + 1 x n θ n =0.

For simplicity, we put y ˜ n = P C ( y n λ f α n ( y n )) and z ˜ n = P C ( z n λ f α n ( z n )) for every n0. Then z n = β n Q y n +(1 β n )T y ˜ n and x n + 1 = σ n z n + γ n z ˜ n + δ n Γ z ˜ n for every n0.

Taking into account 0< lim inf n σ n lim sup n σ n <1, without loss of generality, we may assume that { σ n }[c,d] for some c,d(0,1). We write x n = σ n 1 z n 1 +(1 σ n 1 ) v n 1 , n1, where v n 1 = x n σ n 1 z n 1 1 σ n 1 . It follows that for all n1,

v n v n 1 = x n + 1 σ n z n 1 σ n x n σ n 1 z n 1 1 σ n 1 = γ n z ˜ n + δ n Γ z ˜ n 1 σ n γ n 1 z ˜ n 1 + δ n 1 Γ z ˜ n 1 1 σ n 1 = γ n ( z ˜ n z ˜ n 1 ) + δ n ( Γ z ˜ n Γ z ˜ n 1 ) 1 σ n + ( γ n 1 σ n γ n 1 1 σ n 1 ) z ˜ n 1 + ( δ n 1 σ n δ n 1 1 σ n 1 ) Γ z ˜ n 1 .
(3.1)

Since ( γ n + δ n )ζ γ n for all n0, by Lemma 2.4, we have

γ n ( z ˜ n z ˜ n 1 ) + δ n ( Γ z ˜ n Γ z ˜ n 1 ) ( γ n + δ n ) z ˜ n z ˜ n 1 .
(3.2)

Now, we estimate z n z n 1 . Observe that for every n1,

y ˜ n y ˜ n 1 P C ( I λ f α n ) y n P C ( I λ f α n ) y n 1 + P C ( I λ f α n ) y n 1 P C ( I λ f α n 1 ) y n 1 y n y n 1 + P C ( I λ f α n ) y n 1 P C ( I λ f α n 1 ) y n 1 y n y n 1 + ( I λ f α n ) y n 1 ( I λ f α n 1 ) y n 1 = y n y n 1 + λ f α n ( y n 1 ) λ f α n 1 ( y n 1 ) = y n y n 1 + λ | α n α n 1 | y n 1 .
(3.3)

Similarly, for all n1, we have

z ˜ n z ˜ n 1 z n z n 1 +λ| α n α n 1 | z n 1 .

From (2.2), we have

{ y n = θ n S x n + ( 1 θ n ) x n , y n 1 = θ n 1 S x n 1 + ( 1 θ n 1 ) x n 1 , n 1 ,

and therefore

y n y n 1 = θ n (S x n S x n 1 )+( θ n θ n 1 )(S x n 1 x n 1 )+(1 θ n )( x n x n 1 ),

which implies that

y n y n 1 θ n S x n S x n 1 + | θ n θ n 1 | S x n 1 x n 1 + ( 1 θ n ) x n x n 1 x n x n 1 + | θ n θ n 1 | S x n 1 x n 1 .
(3.4)

Also, from (2.2) we have

{ z n = β n Q y n + ( 1 β n ) T y ˜ n , z n 1 = β n 1 Q y n 1 + ( 1 β n 1 ) T y ˜ n 1 , n 1 ,

then simple calculations show that

z n z n 1 =(1 β n )(T y ˜ n T y ˜ n 1 )+( β n β n 1 )(Q y n 1 T y ˜ n 1 )+ β n (Q y n Q y n 1 ),

and thus, from (3.3)-(3.4), we have

z n z n 1 ( 1 β n ) T y ˜ n T y ˜ n 1 + | β n β n 1 | Q y n 1 T y ˜ n 1 + β n Q y n Q y n 1 ( 1 β n ) y ˜ n y ˜ n 1 + | β n β n 1 | Q y n 1 T y ˜ n 1 + β n Q y n Q y n 1 ( 1 β n ) ( y n y n 1 + λ | α n α n 1 | y n 1 ) + | β n β n 1 | Q y n 1 T y ˜ n 1 + β n ρ y n y n 1 ( 1 ( 1 ρ ) β n ) y n y n 1 + λ | α n α n 1 | y n 1 + | β n β n 1 | Q y n 1 T y ˜ n 1 ( 1 ( 1 ρ ) β n ) [ x n x n 1 + | θ n θ n 1 | S x n 1 x n 1 ] + λ | α n α n 1 | y n 1 + | β n β n 1 | Q y n 1 T y ˜ n 1 ( 1 ( 1 ρ ) β n ) x n x n 1 + | θ n θ n 1 | S x n 1 x n 1 + λ | α n α n 1 | y n 1 + | β n β n 1 | Q y n 1 T y ˜ n 1 ( 1 ( 1 ρ ) β n ) x n x n 1 + M 1 [ | θ n θ n 1 | + | α n α n 1 | + | β n β n 1 | ] ,
(3.5)

where S x n x n +λ y n +Q y n T y ˜ n M 1 , n0 for some M 1 >0. This together with (3.1)-(3.3) implies that

v n v n 1 γ n ( z ˜ n z ˜ n 1 ) + δ n ( Γ z ˜ n Γ z ˜ n 1 ) 1 σ n + | γ n 1 σ n γ n 1 1 σ n 1 | z ˜ n 1 + | δ n 1 σ n δ n 1 1 σ n 1 | Γ z ˜ n 1 ( γ n + δ n ) z ˜ n z ˜ n 1 1 σ n + | γ n 1 σ n γ n 1 1 σ n 1 | z ˜ n 1 + | γ n 1 σ n γ n 1 1 σ n 1 | Γ z ˜ n 1 = z ˜ n z ˜ n 1 + | γ n 1 σ n γ n 1 1 σ n 1 | ( z ˜ n 1 + Γ y ˜ n 1 ) z n z n 1 + λ | α n α n 1 | z n 1 + | γ n 1 σ n γ n 1 1 σ n 1 | ( z ˜ n 1 + Γ z ˜ n 1 ) ( 1 ( 1 ρ ) β n ) x n x n 1 + M 1 [ | θ n θ n 1 | + | α n α n 1 | + | β n β n 1 | ] + λ | α n α n 1 | z n 1 + | γ n 1 σ n γ n 1 1 σ n 1 | ( z ˜ n 1 + Γ z ˜ n 1 ) ( 1 ( 1 ρ ) β n ) x n x n 1 + M 2 [ | θ n θ n 1 | + 2 | α n α n 1 | + | β n β n 1 | + | γ n 1 σ n γ n 1 1 σ n 1 | ] ,
(3.6)

where M 1 +λ z n + z ˜ n +Γ z ˜ n M 2 , n0 for some M 2 >0.

Further, we observe that

{ x n + 1 = σ n z n + ( 1 σ n ) v n , x n = σ n 1 z n 1 + ( 1 β n 1 ) v n 1 , n 1 ,

and then by simple calculations, we have

x n + 1 x n =(1 σ n )( v n v n 1 )+( σ n σ n 1 )( z n 1 v n 1 )+ σ n ( z n z n 1 ).

By taking norm and using (3.5)-(3.6), we get

x n + 1 x n ( 1 σ n ) v n v n 1 + | σ n σ n 1 | z n 1 v n 1 + σ n z n z n 1 ( 1 σ n ) { ( 1 ( 1 ρ ) β n ) x n x n 1 + M 2 [ | θ n θ n 1 | + 2 | α n α n 1 | + | β n β n 1 | + | γ n 1 σ n γ n 1 1 σ n 1 | ] } + | σ n σ n 1 | z n 1 v n 1 + σ n { ( 1 ( 1 ρ ) β n ) x n x n 1 + M 1 [ | θ n θ n 1 | + | α n α n 1 | + | β n β n 1 | ] } ( 1 ( 1 ρ ) β n ) x n x n 1 + M 2 [ | θ n θ n 1 | + 2 | α n α n 1 | + | β n β n 1 | + | γ n 1 σ n γ n 1 1 σ n 1 | ] + | σ n σ n 1 | z n 1 v n 1 ( 1 ( 1 ρ ) β n ) x n x n 1 + M 3 [ | θ n θ n 1 | + 2 | α n α n 1 | + | β n β n 1 | + | γ n 1 σ n γ n 1 1 σ n 1 | + | σ n σ n 1 | ] ,

where M 2 + z n v n M 3 , n0 for some M 3 0. Therefore,

x n + 1 x n θ n ( 1 ( 1 ρ ) β n ) x n x n 1 θ n + M 3 [ | θ n θ n 1 | θ n + 2 | α n α n 1 | θ n + | β n β n 1 | θ n + 1 θ n | γ n 1 σ n γ n 1 1 σ n 1 | + | σ n σ n 1 | θ n ] = ( 1 ( 1 ρ ) β n ) x n x n 1 θ n 1 + ( 1 ( 1 ρ ) β n ) ( x n x n 1 θ n x n x n 1 θ n 1 ) + M 3 [ | θ n θ n 1 | θ n + 2 | α n α n 1 | θ n + | β n β n 1 | θ n + 1 θ n | γ n 1 σ n γ n 1 1 σ n 1 | + | σ n σ n 1 | θ n ] ( 1 ( 1 ρ ) β n ) x n x n 1 θ n 1 + M [ | 1 θ n 1 θ n 1 | + | θ n θ n 1 | θ n + 2 | α n α n 1 | θ n + | β n β n 1 | θ n + 1 θ n | γ n 1 σ n γ n 1 1 σ n 1 | + | σ n σ n 1 | θ n ] = ( 1 ( 1 ρ ) β n ) x n x n 1 θ n 1 + ( 1 ρ ) β n M 1 ρ { 1 β n | 1 θ n 1 θ n 1 | + 1 β n | 1 θ n 1 θ n | + 2 | α n α n 1 | β n θ n + 1 θ n | 1 β n 1 β n | + 1 β n θ n | γ n 1 σ n γ n 1 1 σ n 1 | + | σ n σ n 1 | β n θ n } ,
(3.7)

where M 3 + x n x n 1 M, n1 for some M0. From (H1)-(H5), it follows that n = 0 (1ρ) β n = and

lim n M 1 ρ { 1 β n | 1 θ n 1 θ n 1 | + 1 β n | 1 θ n 1 θ n | + 2 | α n α n 1 | β n θ n + 1 θ n | 1 β n 1 β n | + 1 β n θ n | γ n 1 σ n γ n 1 1 σ n 1 | + | σ n σ n 1 | β n θ n } = 0 .

Thus, by applying Lemma 2.3 to (3.7), we conclude that

lim n x n + 1 x n θ n =0,

which implies that

lim n x n + 1 x n =0.
(3.8)

Step 2. lim n x n z n =0.

Indeed, let pFix(T)Fix(Γ)Ξ. Then we have

y ˜ n p = P C ( I λ f α n ) y n P C ( I λ f ) p P C ( I λ f α n ) y n P C ( I λ f α n ) p + P C ( I λ f α n ) p P C ( I λ f ) p y n p + P C ( I λ f α n ) p P C ( I λ f ) p y n p + λ α n p .
(3.9)

Similarly, we get

z ˜ n p z n p+λ α n p.

By Lemma 2.5 and (3.9), we have

z n p 2 = β n ( Q y n p ) + ( 1 β n ) ( T y ˜ n p ) 2 β n Q y n p 2 + ( 1 β n ) y ˜ n p 2 β n Q y n p 2 + y ˜ n p 2 β n Q y n p 2 + ( y n p + λ α n p ) 2 = β n Q y n p 2 + y n p 2 + λ α n p ( 2 y n p + λ α n p ) β n Q y n p 2 + θ n S x n p 2 + ( 1 θ n ) x n p 2 + λ α n p ( 2 y n p + λ α n p ) β n Q y n p 2 + θ n S x n p 2 + x n p 2 + λ α n p ( 2 y n p + λ α n p ) .

Since ( γ n + δ n )ζ γ n for all n0, utilizing Lemma 2.4, we obtain

x n + 1 p 2 = σ n ( z n p ) + γ n ( z ˜ n p ) + δ n ( Γ z ˜ n p ) 2 = σ n ( z n p ) + ( γ n + δ n ) 1 γ n + δ n [ γ n ( z ˜ n p ) + δ n ( Γ z ˜ n p ) ] 2 = σ n z n p 2 + ( γ n + δ n ) 1 γ n + δ n [ γ n ( z ˜ n p ) + δ n ( Γ z ˜ n p ) ] 2 σ n ( γ n + δ n ) ( z n p ) 1 γ n + δ n [ γ n ( z ˜ n p ) + δ n ( Γ z ˜ n p ) ] 2 = σ n z n p 2 + ( γ n + δ n ) 1 γ n + δ n [ γ n ( z ˜ n p ) + δ n ( Γ z ˜ n p ) ] 2 σ n ( γ n + δ n ) 1 γ n + δ n [ γ n ( z ˜ n z n ) + δ n ( Γ z ˜ n z n ) ] 2 = σ n z n p 2 + ( γ n + δ n ) 1 γ n + δ n [ γ n ( z ˜ n p ) + δ n ( Γ z ˜ n p ) ] 2 σ n γ n + δ n x n + 1 z n 2 σ n z n p 2 + ( γ n + δ n ) z ˜ n p 2 σ n γ n + δ n x n + 1 z n 2 = σ n z n p 2 + ( 1 σ n ) z ˜ n p 2 σ n 1 σ n x n + 1 z n 2 σ n z n p 2 + ( 1 σ n ) [ z n p 2 + λ α n p ( 2 z n p + λ α n p ) ] σ n 1 σ n x n + 1 z n 2 z n p 2 + λ α n p ( 2 z n p + λ α n p ) σ n 1 σ n x n + 1 z n 2 β n Q y n p 2 + θ n S x n p 2 + x n p 2 + λ α n p ( 2 y n p + λ α n p ) + λ α n p ( 2 z n p + λ α n p ) σ n 1 σ n x n + 1 z n 2 = x n p 2 + β n Q y n p 2 + θ n S x n p 2 + 2 λ α n p ( y n p + z n p + λ α n p ) σ n 1 σ n x n + 1 z n 2 .

Since 0< lim inf n σ n lim sup n σ n <1, we may assume that { σ n }[c,d] for some c,d(0,1). Therefore, we deduce

c 1 c x n + 1 z n 2 σ n 1 σ n x n + 1 z n 2 x n p 2 x n + 1 p 2 + β n Q y n p 2 + θ n S x n p 2 + 2 λ α n p ( y n p + z n p + λ α n p ) ( x n p + x n + 1 p ) x n x n + 1 + β n Q y n p 2 + θ n S x n p 2 + 2 λ α n p ( y n p + z n p + λ α n p ) .

Since α n 0, β n 0, θ n 0 and x n x n + 1 0 as n, we conclude from the boundedness of { x n }, { y n } and { z n } that x n + 1 z n 0 as n. This together with x n x n + 1 0 implies that

lim n x n z n =0.
(3.10)

Step 3. lim n y n y ˜ n =0 and lim n z n z ˜ n =0.

Let pFix(T)Fix(Γ)Ξ. Then, by Lemmas 2.2 and 2.5, we have

z n p 2 = β n ( Q y n p ) + ( 1 β n ) ( T y ˜ n p ) 2 β n Q y n p 2 + ( 1 β n ) y ˜ n p 2 = β n Q y n p 2 + ( 1 β n ) P C ( I λ f α n ) y n P C ( I λ f ) p 2 β n Q y n p 2 + ( 1 β n ) ( I λ f ) y n ( I λ f ) p λ α n y n 2 β n Q y n p 2 + ( 1 β n ) [ ( I λ f ) y n ( I λ f ) p 2 2 λ α n y n , ( I λ f α n ) y n ( I λ f ) p ] β n Q y n p 2 + ( 1 β n ) [ y n p 2 + λ ( λ 2 L ) f ( y n ) f ( p ) 2 + 2 λ α n y n ( I λ f α n ) y n ( I λ f ) p ] β n Q y n p 2 + ( 1 β n ) [ θ n S x n p 2 + ( 1 θ n ) x n p 2 + λ ( λ 2 L ) f ( y n ) f ( p ) 2 + 2 λ α n y n ( I λ f α n ) y n ( I λ f ) p ] β n Q y n p 2 + θ n S x n p 2 + x n p 2 + ( 1 β n ) λ ( λ 2 L ) f ( y n ) f ( p ) 2 + 2 λ α n y n ( I λ f α n ) y n ( I λ f ) p .

Therefore, we obtain

( 1 β n ) λ ( 2 L λ ) f ( y n ) f ( p ) 2 β n Q y n p 2 + θ n S x n p 2 + x n p 2 z n p 2 + 2 λ α n y n ( I λ f α n ) y n ( I λ f ) p β n Q y n p 2 + θ n S x n p 2 + ( x n p + z n p ) ( x n p z n p ) + 2 λ α n y n ( I λ f α n ) y n ( I λ f ) p β n Q y n p 2 + θ n S x n p 2 + ( x n p + z n p ) x n z n + 2 λ α n y n ( I λ f α n ) y n ( I λ f ) p .

Since α n 0, β n 0, θ n 0, x n z n 0 and 0<λ< 2 L , from the boundedness of { x n }, { y n } and { z n }, we obtain lim n f( y n )f(p)=0, and hence

lim n f α n ( y n ) f ( p ) =0.

Also, since

y n z n y n x n + x n z n = θ n S x n x n + x n z n ,

from θ n 0 and x n z n 0, it follows that

lim n y n z n =0and lim n f α n ( z n ) f ( p ) =0.
(3.11)

Furthermore, from the firm nonexpansiveness of P C , we obtain

y ˜ n p 2 = P C ( I λ f α n ) y n P C ( I λ f ) p 2 ( I λ f α n ) y n ( I λ f ) p , y ˜ n p = 1 2 { ( I λ f α n ) y n ( I λ f ) p 2 + y ˜ n p 2 ( I λ f α n ) y n ( I λ f ) p ( y ˜ n p ) 2 } 1 2 { y n p 2 + 2 λ f α n ( y n ) f ( p ) ( I λ f α n ) y n ( I λ f ) p + y ˜ n p 2 y n y ˜ n 2 + 2 λ y n y ˜ n , f α n ( y n ) f ( p ) λ 2 f α n ( y n ) f ( p ) 2 } ,

and so,

y ˜ n p 2 y n p 2 y n y ˜ n 2 + 2 λ f α n ( y n ) f ( p ) ( I λ f α n ) y n ( I λ f ) p + 2 λ y n y ˜ n , f α n ( y n ) f ( p ) λ 2 f α n ( y n ) f ( p ) 2 .

Similarly, we have

z ˜ n p 2 z n p 2 z n z ˜ n 2 + 2 λ f α n ( z n ) f ( p ) ( I λ f α n ) z n ( I λ f ) p + 2 λ z n z ˜ n , f α n ( z n ) f ( p ) λ 2 f α n ( z n ) f ( p ) 2 .
(3.12)

Thus, we have

z n p 2 β n Q y n p 2 + ( 1 β n ) y ˜ n p 2 β n Q y n p 2 + y ˜ n p 2 β n Q y n p 2 + y n p 2 y n y ˜ n 2 + 2 λ f α n ( y n ) f ( p ) ( I λ f α n ) y n ( I λ f ) p + 2 λ y n y ˜ n , f α n ( y n ) f ( p ) λ 2 f α n ( y n ) f ( p ) 2 β n Q y n p 2 + y n p 2 y n y ˜ n 2 + 2 λ f α n ( y n ) f ( p ) ( I λ f α n ) y n ( I λ f ) p + 2 λ y n y ˜ n , f α n ( y n ) f ( p ) β n Q y n p 2 + y n p 2 y n y ˜ n 2 + 2 λ f α n ( y n ) f ( p ) ( ( I λ f α n ) y n ( I λ f ) p + y n y ˜ n ) ,

which implies that

y n y ˜ n 2 β n Q y n p 2 + y n p 2 z n p 2 + 2 λ f α n ( y n ) f ( p ) ( ( I λ f α n ) y n ( I λ f ) p + y n y ˜ n ) β n Q y n p 2 + ( y n p + z n p ) y n z n + 2 λ f α n ( y n ) f ( p ) ( ( I λ f α n ) y n ( I λ f ) p + y n y ˜ n ) .

Since β n 0, y n z n 0 and f α n ( y n )f(p)0, from the boundedness of { x n }, { y n }, { z n } and { y ˜ n }, it follows that

lim n y n y ˜ n =0.

In addition, since ( γ n + δ n )ζ γ n for all n0, utilizing Lemma 2.4, we get from (3.12)

x n + 1 p 2 σ n z n p 2 + ( γ n + δ n ) z ˜ n p 2 = σ n z n p 2 + ( 1 σ n ) z ˜ n p 2 σ n z n p 2 + ( 1 σ n ) { z n p 2 z n z ˜ n 2 + 2 λ f α n ( z n ) f ( p ) ( I λ f α n ) z n ( I λ f ) p + 2 λ z n z ˜ n , f α n ( z n ) f ( p ) λ 2 f α n ( z n ) f ( p ) 2 } σ n z n p 2 + ( 1 σ n ) { z n p 2 z n z ˜ n 2 + 2 λ f α n ( z n ) f ( p ) ( I λ f α n ) z n ( I λ f ) p + 2 λ z n z ˜ n f α n ( z n ) f ( p ) } z n p 2 ( 1 σ n ) z n z ˜ n 2 + 2 λ f α n ( z n ) f ( p ) ( ( I λ f α n ) z n ( I λ f ) p + z n z ˜ n ) ,

which implies that

( 1 σ n ) z n z ˜ n 2 z n p 2 x n + 1 p 2 + 2 λ f α n ( z n ) f ( p ) ( ( I λ f α n ) z n ( I λ f ) p + z n z ˜ n ) ( z n p + x n + 1 p ) z n x n + 1 + 2 λ f α n ( z n ) f ( p ) ( ( I λ f α n ) z n ( I λ f ) p + z n z ˜ n ) .

Since { σ n }[c,d], z n x n + 1 0 and f α n ( z n )f(p)0, from the boundedness of { x n }, { z n } and { z ˜ n }, it follows that

lim n z n z ˜ n =0.

Step 4. ω w ( x n )Ω.

Let p ω w ( x n ). Then there exists a subsequence { x n i } of { x n } such that x n i p . Since

z n y n = β n ( Q y n y n ) + ( 1 β n ) ( T y ˜ n y n ) = β n ( Q y n y n ) + ( 1 β n ) ( T y ˜ n y ˜ n ) + ( 1 β n ) ( y ˜ n y n ) ,

we have

( 1 β n ) T y ˜ n y ˜ n = z n y n β n ( Q y n y n ) ( 1 β n ) ( y ˜ n y n ) z n y n + β n Q y n y n + ( 1 β n ) y ˜ n y n z n y n + β n Q y n y n + y ˜ n y n .

Hence from z n y n 0, β n 0 and y ˜ n y n 0, we get lim n T y ˜ n y ˜ n =0. Since x n y n 0 and y n y ˜ n 0, we have y ˜ n i p . By Lemma 2.1(b) (demiclosedness principle), we obtain p Fix(T).

Meanwhile, observe that

x n + 1 z n = γ n ( z ˜ n z n ) + δ n ( Γ z ˜ n z ˜ n ) + δ n ( z ˜ n z n ) = ( γ n + δ n ) ( z ˜ n z n ) + δ n ( Γ z ˜ n z ˜ n ) = ( 1 σ n ) ( z ˜ n z n ) + δ n ( Γ z ˜ n z ˜ n ) .

Thus,

δ n Γ z ˜ n z ˜ n = x n + 1 z n ( 1 σ n ) ( z ˜ n z n ) x n + 1 z n + ( 1 σ n ) z ˜ n z n x n + 1 z n + z ˜ n z n 0 as  n .

This together with lim inf n δ n >0 yields lim n Γ z ˜ n z ˜ n =0. Since x n z n 0 and z n z ˜ n 0, we have z ˜ n i p . By Lemma 2.1(b) (demiclosedness principle), we have p Fix(Γ).

Further, let us show p Ξ. Indeed, from x n y n 0 and y ˜ n y n 0, we have y n i p and y ˜ n i p . Define

Vv={ f ( v ) + N C v if  v C , if  v C ,

where N C v={wH:vu,w0,uC}. Then V is maximal monotone and 0Vv if and only if vVI(C,f) (see [17]). Let (v,w)graph(V). Then we have

wVv=f(v)+ N C v,

and hence

wf(v) N C v.

Therefore, we have

v u , w f ( v ) 0,uC.

On the other hand, from

y ˜ n = P C ( y n λ f α n ( y n ) ) andvC,

we have

y n λ f α n ( y n ) y ˜ n , y ˜ n v 0,

and hence

v y ˜ n , y ˜ n y n λ + f α n ( y n ) 0.

Therefore, from

wf(v) N C (v)and y ˜ n i C,

we have

v y ˜ n i , w v y ˜ n i , f ( v ) v y ˜ n i , f ( v ) v y ˜ n i , y ˜ n i y n i λ + f α n i ( y n i ) = v y ˜ n i , f ( v ) v y ˜ n i , y ˜ n i y n i λ + f ( y n i ) α n i v y ˜ n i , y n i = v y ˜ n i , f ( v ) f ( y ˜ n i ) + v y ˜ n i , f ( y ˜ n i ) f ( y n i ) v y ˜ n i , y ˜ n i y n i λ α n i v y ˜ n i , y n i v y ˜ n i , f ( y ˜ n i ) f ( y n i ) v y ˜ n i , y ˜ n i y n i λ α n i v y ˜ n i , y n i .

Hence, we obtain

v p , w 0as i.

Since V is maximal monotone, we have p V 1 0, and hence p VI(C,f), which leads to pΞ. Consequently, p Fix(T)Fix(Γ)Ξ. This shows that ω w ( x n )Fix(T)Fix(Γ)Ξ.

Finally, let us show p Ω. Indeed, it follows from (2.2) that for every pFix(T)Fix(Γ)Ξ,

y n p 2 = ( 1 θ n ) ( x n p ) + θ n ( S x n S p ) + θ n ( S p p ) 2 ( 1 θ n ) ( x n p ) + θ n ( S x n S p ) 2 + 2 θ n S p p , y n p ( 1 θ n ) x n p 2 + θ n S x n S p 2 + 2 θ n S p p , y n p x n p 2 + 2 θ n S p p , y n p ,

and hence

z n p 2 β n Q y n p 2 + ( 1 β n ) y ˜ n p 2 β n Q y n p 2 + y ˜ n p 2 β n Q y n p 2 + ( y n p + λ α n p ) 2 β n Q y n p 2 + y n p 2 + λ α n p ( 2 y n p + λ α n p ) β n Q y n p 2 + x n p 2 + 2 θ n S p p , y n p + λ α n p ( 2 y n p + λ α n p ) .

Since ( γ n + δ n )ζ γ n for all n0, by Lemma 2.4, we have

x n + 1 p 2 σ n z n p 2 + ( γ n + δ n ) z ˜ n p 2 σ n z n p 2 + ( 1 σ n ) ( z n p + λ α n p ) 2 z n p 2 + λ α n p ( 2 z n p + λ α n p ) β n Q y n p 2 + x n p 2 + 2 θ n S p p , y n p + λ α n p ( 2 y n p + λ α n p ) + λ α n p ( 2 z n p + λ α n p ) = x n p 2 + β n Q y n p 2 + 2 θ n S p p , y n p + 2 λ α n p ( y n p + z n p + λ α n p ) ,

which implies that

2 p S p , y n p 1 θ n ( x n p 2 x n + 1 p 2 ) + β n θ n Q y n p 2 + α n θ n 2 λ p ( y n p + z n p + λ α n p ) x n x n + 1 θ n ( x n p + x n + 1 p ) + β n θ n Q y n p 2 + α n θ n 2 λ p ( y n p + z n p + λ α n p ) .

Since α n + β n θ n 0 and x n x n + 1 θ n 0 as n, from the boundedness of { x n }, { y n } and { z n }, we deduce that

lim sup n pSp, y n p0,pFix(T)Fix(Γ)Ξ.

So, from y n i p , we get

p S p , p p 0,pFix(T)Fix(Γ)Ξ.

Taking into con