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The hybrid steepest descent method for solving variational inequality over triple hierarchical problems

Abstract

An explicit algorithm is introduced to solve the monotone variational inequality over a triple hierarchical problem. The strong convergence for the proposed algorithm to the solution is guaranteed under some assumptions. Our results extend those of Iiduka (Nonlinear Anal. 71:e1292-e1297, 2009), Marino and Xu (J. Optim. Theory Appl. 149(1):61-78, 2011), Yao et al. (Fixed Point Theory Appl. 2011:794203, 2011) and other authors.

MSC:46C05, 47H06, 47H09, 47H10, 47J20, 47J25, 65J15.

1 Introduction

Let C be a closed convex subset of a real Hilbert space H with the inner product , and the norm . We denote weak convergence and strong convergence by notations and →, respectively. A mapping A is a nonlinear mapping. The Hartmann-Stampacchia variational inequality [1] for finding xC such that

Ax,yx0,yC.
(1.1)

The set of solutions of (1.1) is denoted by VI(C,A). The variational inequality has been extensively studied in the literature [2, 3].

Let f:CC be called a ρ-contraction if there exists a constant ρ[0,1) such that

f ( x ) f ( y ) ρxy,x,yC.

A mapping T:CC is said to be nonexpansive if

TxTyxy,x,yC.

A point xC is a fixed point of T provided Tx=x. Denote by F(T) the set of fixed points of T; that is, F(T)={xC:Tx=x}. If C is bounded closed convex and T is a nonexpansive mapping of C into itself, then F(T) is nonempty [4].

We discuss the following variational inequality problem over the fixed point set of a nonexpansive mapping (see [513]), which is called a hierarchical problem. Consider a monotone, continuous mapping A:HH and a nonexpansive mapping T:HH. Find x VI(F(T),A)={ x F(T):A x ,x x 0,xF(T)}, where F(T). This solution set is denoted by Ξ.

We introduce the following variational inequality problem over the solution set of the variational inequality problem over the fixed point set of a nonexpansive mapping (see [1418]), which is called a triple hierarchical problem. Consider an inverse-strongly monotone A:HH, a strongly monotone and Lipschitz continuous B:HH and a nonexpansive mapping T:HH. Find x VI(Ξ,B)={ x Ξ:B x ,x x 0,xΞ}, where Ξ:=VI(F(T),A).

A mapping A:HH is said to be monotone if

AxAy,xy0,x,yH.

A mapping A:HH is said to be α-strongly monotone if there exists a positive real number α such that

AxAy,xyα x y 2 ,x,yH.

A mapping A:HH is said to be β-inverse-strongly monotone if there exists a positive real number β such that

AxAy,xyβ A x A y 2 ,x,yH.

A mapping A:HH is said to be L-Lipschitz continuous if there exists a positive real number L such that

AxAyLxy,x,yH.

A linear bounded operator A is said to be strongly positive on H if there exists a constant γ ¯ >0 with the property

Ax,x γ ¯ x 2 ,xH.

In 2009, Iiduka [14] introduced an iterative algorithm for the following triple hierarchical constrained optimization problem. The sequence { x n } defined by the iterative method below, with the initial guess x 1 H, is chosen arbitrarily,

{ y n = T ( x n λ n A 1 x n ) , x n + 1 = y n μ α n A 2 y n , n 0 ,
(1.2)

where α n (0,1] and λ n (0,2α] satisfy certain conditions. Let A 1 :HH be an inverse-strongly monotone, A 2 :HH be a strongly monotone and Lipschitz continuous and T:HH be a nonexpansive mapping, then the sequence converges strongly to the set solution of (1.2).

In 2011, Yao et al. [19] studied new algorithms. For x 0 C chosen arbitrarily, let the sequence { x n } be generated iteratively by

x n + 1 = β n x n +(1 β n )T P C [ I α n ( A γ f ) ] x n ,n0,

where the sequences { α n } and { β n } are two sequences in [0,1]. Then { x n } converges strongly to x F(T) which is the unique solution of the variational inequality:

Find  x F(T) such that  ( A γ f ) x , x x 0,xF(T).
(1.3)

Let A:CH be a strongly positive linear bounded operator, f:CH be a ρ-contraction and T:CC be a nonexpansive mapping satisfying some conditions. The solution set of (1.3) is denoted by Ω:=VI(F(T),Aγf).

Very recently, Marino and Xu [20] generated a sequence { x n } through the recursive formula

x n + 1 = λ n f x n +(1 λ n ) ( α n V x n + ( 1 α n ) T x n ) ,n0,

where f is a contraction on C, the initial guess x 0 C is arbitrary and { λ n }, { α n } are two sequences in (0,1) and T,V:CC are two nonexpansive self mappings. Strong convergence of the algorithm is proved under different circumstances of parameter selection.

In this paper, we introduce iterative algorithms and prove a strong convergence theorem for the following variational inequality over the triple hierarchical problem (1.4) below. Let B:CC be a β-strongly monotone and L-Lipschitz continuous. Find x Ω such that

B x , x x 0,xΩ,
(1.4)

where Ω:=VI(F(T),Aγf), T is a nonexpansive mapping, A:CH is a strongly positive linear bounded operator and f:CH is a ρ-contraction. This solution set of (1.4) is denoted by ϒ:=VI(Ω,B):=VI(VI(F(T),Aγf),B). Then the sequence { x n } strongly converges to the unique solution of (3.2) in Section 3.17 and we shall denote the set of such solutions by Θ:=VI(ϒ,Iϕ).

2 Preliminaries

Let H be a real Hilbert space and C be a nonempty closed convex subset of H. Recall that the (nearest point) projection P C from H onto C assigns, to each xH, the unique point in P C xC satisfying the property

x P C x= min y C xy.

The following characterizes the projection P C . We recall some lemmas which will be needed in the rest of this paper.

Lemma 2.1 The function uC is a solution of the variational inequality (1.1) if and only if uC satisfies the relation u= P C (uλAu) for all λ>0.

Lemma 2.2 For a given zH, uC, u= P C zuz,vu0, vC.

It is well known that P C is a firmly nonexpansive mapping of H onto C and satisfies

P C x P C y 2 P C x P C y,xy,x,yH.
(2.1)

Moreover, P C x is characterized by the following properties: P C xC and for all xH, yC,

x P C x,y P C x0.
(2.2)

Lemma 2.3 The following inequality holds in an inner product space H:

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

Lemma 2.4 [21]

Let C be a closed convex subset of a real Hilbert space H and let T:CC be a nonexpansive mapping. Then IT is demiclosed at zero, that is, x n x and x n T x n 0 imply x=Tx.

Lemma 2.5 [22]

Assume A is a strongly positive linear bounded operator on a Hilbert space H with the coefficient γ ¯ >0 and 0<ρ A 1 , then IρA1ρ γ ¯ .

Lemma 2.6 [23]

Each Hilbert space H satisfies Opial’s condition, that is, for any sequence { x n }H with x n x, the inequality

lim inf n x n x< lim inf n x n y

holds for each yH with yx.

Lemma 2.7 [24]

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 integers n0 and lim sup n ( y n + 1 y n x n + 1 x n )0. Then lim n y n x n =0.

Lemma 2.8 [10]

Let B:HH be β-strongly monotone and L-Lipschitz continuous and let μ(0, 2 β L 2 ). For λ[0,1], define T λ :HH by T λ (x):=xλμB(x) for all xH. Then, for all x,yH,

T λ ( x ) T λ ( y ) (1λτ)xy

holds, where τ:=1 1 μ ( 2 β μ L 2 ) (0,1].

Lemma 2.9 [25]

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

a n + 1 (1 γ n ) a n + δ n ,n0,

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

  1. (i)

    n = 1 γ n =,

  2. (ii)

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

Then lim n a n =0.

3 Strong convergence theorem

In this section, we introduce an iterative algorithm for solving the monotone variational inequality over a triple hierarchical problem.

Theorem 3.1 Let H be a real Hilbert space, C be a closed convex subset of H. Let A:CH be a strongly positive linear bounded operator, f:CH be a ρ-contraction, γ be a positive real number such that γ ¯ 1 ρ <γ< γ ¯ ρ . Let T:CC be a nonexpansive mapping, B:CC be a β-strongly monotone and L-Lipschitz continuous. Let ϕ:CC be a k-contraction mapping with k[0,1). Assume that ϒ:=VI(VI(F(T),Aγf),B) is nonempty set. Suppose { x n } is a sequence generated by the following algorithm x 0 C arbitrarily:

{ z n = T P C [ I δ n ( A γ f ) ] x n , y n = ( I μ β n B ) z n , x n + 1 = α n ϕ ( x n ) + ( 1 α n ) y n , n 0 ,
(3.1)

where { α n },{ δ n }[0,1]. If μ(0, 2 β L 2 ) is used and if { β n }(0,1] satisfies the following conditions:

(C1): n = 1 | δ n + 1 δ n |<, n = 1 δ n =;

(C2): n = 1 | β n + 1 β n |<;

(C3): n = 1 | α n + 1 α n |<, lim n α n =0;

(C4): δ n β n and β n α n .

Then { x n } converges strongly to x ϒ, which is the unique solution of the variational inequality:

Find x ϒ such that ( I ϕ ) x , x x 0,xϒ.
(3.2)

Proof We will divide the proof into four steps.

Step 1. We will show { x n } is bounded. For any qϒ, we have

z n q = T P C [ I δ n ( A γ f ) ] x n T P C q [ I δ n ( A γ f ) ] x n q δ n γ f ( x n ) γ f ( q ) + δ n γ f ( q ) A q + I δ n A x n q δ n γ ρ x n q + δ n γ f ( q ) A q + ( 1 δ n γ ¯ ) x n q = [ 1 ( γ ¯ γ ρ ) δ n ] x n q + δ n γ f ( q ) A q .
(3.3)

By Lemma 2.8, it is found that

y n q = ( I μ β n B ) z n ( I μ β n B ) q ( 1 β n τ ) z n q ( 1 β n τ ) { [ 1 ( γ ¯ γ ρ ) δ n ] x n q + δ n γ f ( q ) A q } .
(3.4)

From (3.1), we get

x n + 1 q α n ϕ ( x n ) ϕ ( q ) + α n ϕ ( q ) q + ( 1 α n ) y n q α n k x n q + α n ϕ ( q ) ϕ ( q ) + ( 1 α n ) y n q α n k x n q + ( 1 α n ) ( 1 β n τ ) × { [ 1 ( γ ¯ γ ρ ) δ n ] x n q + δ n γ f ( q ) A q } α n x n q + ( 1 α n ) ( 1 β n τ ) [ 1 ( γ ¯ γ ρ ) δ n ] x n q + ( 1 α n ) ( 1 β n τ ) δ n γ f ( q ) A q = α n x n q + ( 1 α n ) [ 1 ( γ ¯ γ ρ ) δ n β n τ + β n τ ( γ ¯ γ ρ ) δ n ] x n q + ( 1 α n ) ( 1 β n τ ) δ n γ f ( q ) A q = α n x n q + ( 1 α n ) [ 1 { ( γ ¯ γ ρ ) δ n + β n τ β n τ ( γ ¯ γ ρ ) δ n } ] x n q + ( 1 α n ) ( 1 β n τ ) δ n γ f ( q ) A q = α n x n q + [ 1 α n { ( γ ¯ γ ρ ) δ n + β n τ β n τ ( γ ¯ γ ρ ) δ n } ( 1 α n ) ] x n q + ( 1 α n ) ( 1 β n τ ) δ n γ f ( q ) A q = [ 1 ( 1 α n ) { ( γ ¯ γ ρ ) δ n + β n τ β n τ ( γ ¯ γ ρ ) δ n } ] x n q + ( 1 α n ) ( 1 β n τ ) δ n γ f ( q ) A q = [ 1 ( 1 α n ) { ( γ ¯ γ ρ ) δ n ( 1 β n τ ) + β n τ } ] x n q + ( 1 α n ) ( 1 β n τ ) δ n γ f ( q ) A q = [ 1 ( 1 α n ) ( γ ¯ γ ρ ) δ n ( 1 β n τ ) ( 1 α n ) β n τ ] x n q + ( 1 α n ) ( 1 β n τ ) δ n γ f ( q ) A q = [ 1 ( 1 α n ) ( γ ¯ γ ρ ) δ n ( 1 β n τ ) ] x n q ( 1 α n ) β n τ x n q + ( 1 α n ) ( 1 β n τ ) δ n γ f ( q ) A q [ 1 ( γ ¯ γ ρ ) ( 1 α n ) ( 1 β n τ ) δ n ] x n q + ( γ ¯ γ ρ ) ( 1 α n ) ( 1 β n τ ) δ n γ f ( q ) A q γ ¯ γ ρ = ( 1 σ n ) x n q + σ n γ f ( q ) A q γ ¯ γ ρ ,

where σ n :=( γ ¯ γρ)(1 α n )(1 β n τ) δ n . Then the mathematical induction implies that

x n qmax { x 0 q , γ f ( q ) A q γ ¯ γ ρ } ,n0.

Therefore, { x n } is bounded and so are { y n }, { z n }, {A x n }, {B x n }, {ϕ( x n )} and {f( x n )}.

Step 2. We claim that lim n x n + 1 x n =0 and lim n x n T x n =0. From (3.1), we have

z n + 1 z n = T P C [ I δ n + 1 ( A γ f ) ] x n + 1 T P C [ I δ n ( A γ f ) ] x n P C [ I δ n + 1 ( A γ f ) ] x n + 1 P C [ I δ n ( A γ f ) ] x n [ I δ n + 1 ( A γ f ) ] x n + 1 [ I δ n ( A γ f ) ] x n = δ n + 1 ( γ f ( x n + 1 ) γ f ( x n ) ) + ( δ n + 1 δ n ) γ f ( x n ) + ( I δ n + 1 A ) ( x n + 1 x n ) + ( δ n δ n + 1 ) A x n δ n + 1 γ f ( x n + 1 ) f ( x n ) + ( 1 δ n + 1 γ ¯ ) x n + 1 x n + | δ n + 1 δ n | ( γ f ( x n ) + A x n ) δ n + 1 γ ρ x n + 1 x n + ( 1 δ n + 1 γ ¯ ) x n + 1 x n + | δ n + 1 δ n | ( γ f ( x n ) + A x n ) = [ 1 ( γ ¯ γ ρ ) δ n + 1 ] x n + 1 x n + | δ n + 1 δ n | ( γ f ( x n ) + A x n )
(3.5)

and

y n + 1 y n = ( I μ β n + 1 B ) z n + 1 ( I μ β n B ) z n ( I μ β n + 1 B ) z n + 1 ( I μ β n + 1 B ) z n + ( I μ β n + 1 B ) z n ( I μ β n B ) z n ( 1 β n τ ) z n + 1 z n + μ | β n + 1 β n | B z n .
(3.6)

Using (3.5) and (3.6), we get

x n + 2 x n + 1 = α n + 1 ϕ ( x n + 1 ) + ( 1 α n + 1 ) y n + 1 α n ϕ ( x n ) ( 1 α n ) y n α n + 1 ϕ ( x n + 1 ) ϕ ( x n ) + | α n + 1 α n | ϕ ( x n + 1 ) + ( 1 α n + 1 ) y n + 1 y n + | α n + 1 α n | y n α n + 1 k x n + 1 x n + | α n + 1 α n | ( ϕ ( x n + 1 ) + y n ) + ( 1 α n + 1 ) y n + 1 y n α n + 1 k x n + 1 x n + | α n + 1 α n | ( ϕ ( x n + 1 ) + y n ) + ( 1 α n + 1 ) { ( 1 β n τ ) z n + 1 z n + μ | β n + 1 β n | B z n } = α n + 1 k x n + 1 x n + | α n + 1 α n | ( ϕ ( x n + 1 ) + y n ) + ( 1 α n + 1 ) ( 1 β n τ ) z n + 1 z n + ( 1 α n + 1 ) μ | β n + 1 β n | B z n α n + 1 x n + 1 x n + | α n + 1 α n | ( ϕ ( x n + 1 ) + y n ) + ( 1 α n + 1 ) μ | β n + 1 β n | B z n + ( 1 α n + 1 ) { [ 1 ( γ ¯ γ ρ ) δ n + 1 ] x n + 1 x n + | δ n + 1 δ n | ( γ f ( x n ) + A x n ) } α n + 1 x n + 1 x n + | α n + 1 α n | ( ϕ ( x n + 1 ) + y n ) + ( 1 α n + 1 ) μ | β n + 1 β n | B z n + ( 1 α n + 1 ) [ 1 ( γ ¯ γ ρ ) δ n + 1 ] x n + 1 x n + ( 1 α n + 1 ) | δ n + 1 δ n | ( γ f ( x n ) + A x n ) [ 1 ( γ ¯ γ ρ ) δ n + 1 ( 1 α n + 1 ) ] x n + 1 x n + | α n + 1 α n | ( ϕ ( x n + 1 ) + y n ) + μ | β n + 1 β n | B z n + | δ n + 1 δ n | ( γ f ( x n ) + A x n ) [ 1 ( γ ¯ γ ρ ) δ n + 1 ( 1 α n + 1 ) ] x n + 1 x n + { | α n + 1 α n | + | β n + 1 β n | + | δ n + 1 δ n | } M ,

where M is some constant such that

sup n 0 { ϕ ( x n ) + y n , μ B z n , γ f ( x n ) + A x n } M.

From (C1)-(C3) and the boundedness of { x n }, { y n }, {A x n }, {B z n }, {ϕ( x n )} and {f( x n )}, by Lemma 2.9, we have

lim n x n + 1 x n =0.
(3.7)

On the other hand, we note that

z n T x n = T P C [ I δ n ( A γ f ) ] x n T x n = T P C [ I δ n ( A γ f ) ] x n T P C x n [ I δ n ( A γ f ) ] x n x n δ n ( A γ f ) ] x n ,

by (C3)-(C4) and it follows that

lim n z n T x n =0.
(3.8)

From (3.1), we compute

x n + 1 z n = α n ϕ ( x n ) + ( 1 α n ) y n z n = α n ϕ ( x n ) + ( 1 α n ) ( I μ β n B ) z n z n α n ϕ ( x n ) z n + ( 1 α n ) ( I μ β n B ) z n z n α n k x n z n + α n ϕ ( z n ) z n + ( 1 α n ) μ β n B z n .

By (C3) and (C4), it follows that

lim n x n + 1 z n =0.
(3.9)

Since

x n T x n x n x n + 1 + x n + 1 z n + z n T x n .

By (3.7), (3.8) and (3.9), we obtain

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

From (3.1), we compute

x n + 1 y n = α n ϕ ( x n ) + ( 1 α n ) y n y n = α n ϕ ( x n ) + y n α n y n y n α n ϕ ( x n ) y n .
(3.11)

By (C3), it follows that

lim n x n + 1 y n =0.
(3.12)

Since

x n y n x n x n + 1 + x n + 1 y n .

From (3.7) and (3.12), we obtain

lim n x n y n =0.
(3.13)

Step 3. First, lim sup n u n x ,γf( x )A x 0 is proven. Choose a subsequence { x n i } of { x n } such that

lim sup n x n x , γ f ( x ) A x = lim i x n i x , γ f ( x ) A x .

The boundedness of { x n i } implies the existences of a subsequence { x n i j } of { x n i } and a point x ˆ H such that { x n i j } converges weakly to x ˆ . We may assume, without loss of generality, that lim i x n i ,w= x ˆ ,w, wH. Assume x ˆ T( x ˆ ). lim n x n T x n =0 with F(T) guarantees that

which is a contradiction. Therefore, x ˆ F(T). From x VI(F(T),Aγf), we find

lim sup n x n x , γ f ( x ) A x = lim i x n i x , γ f ( x ) A x = x ˆ x , γ f ( x ) A x 0 .

Setting u n =[I δ n (Aγf)] x n , by (C3)-(C4), we notice that

u n x n δ n ( A γ f ) 0.

Hence, we get

lim sup n u n x , γ f ( x ) A x 0.
(3.14)

Second, lim sup n x x n + 1 ,B x 0 is proven. lim n x n + 1 x n =0 guarantees the existences of a subsequence { x n k + 1 } of { x n k } and a point x ¯ H such that lim sup n x x n + 1 ,B x = lim k x x n k + 1 ,B x and lim k x n k ,w= lim k x n k + 1 ,w= x ¯ ,w, wH. By the same discussion as in the proof of x ˆ F(T), we have x ¯ F(T). Let yF(T) be fixed arbitrarily. Then it follows that T:CC is a nonexpansive mapping with F(T), A:CH is a strongly positive linear bounded operator and f:CH is a contraction for all nN. From (3.1),

z n y = T P C u n T P C y u n y .
(3.15)

By (C3)-(C4), we observe that

u n y = [ I δ n ( A γ f ) ] x n y x n y + δ n ( A γ f ) x n x n y .
(3.16)

Using (3.15) and (3.16),

u n y 2 = [ I δ n ( A γ f ) ] x n y 2 = δ n ( γ f ( x n ) A y ) + ( I δ n A ) ( x n y ) 2 ( 1 δ n γ ¯ ) 2 x n y 2 + 2 δ n γ f ( x n ) A y , u n y ( 1 2 δ n γ ¯ + δ n 2 γ ¯ 2 ) x n y 2 + 2 δ n γ ρ x n y u n y + 2 δ n γ f ( y ) A y , u n y ( 1 2 δ n γ ¯ + δ n 2 γ ¯ 2 ) x n y 2 + 2 δ n γ ρ x n y 2 + 2 δ n γ f ( y ) A y , u n y = [ 1 2 δ n ( γ ¯ γ ρ ) ] x n y 2 + δ n 2 γ ¯ 2 x n y 2 + 2 δ n γ f ( y ) A y , u n y ,

which implies that

0 ( x n y 2 u n y 2 ) 2 δ n ( γ ¯ γ ρ ) x n y 2 + δ n 2 γ ¯ 2 x n y 2 + 2 δ n γ f ( y ) A y , u n y = ( x n y + u n y ) ( x n y u n y ) 2 δ n ( γ ¯ γ ρ ) x n y 2 + δ n 2 γ ¯ 2 x n y 2 + 2 δ n γ f ( y ) A y , u n y M 2 x n u n 2 δ n ( γ ¯ γ ρ ) x n y 2 + δ n 2 γ ¯ 2 x n y 2 + 2 δ n γ f ( y ) A y , u n y ,

where M 2 :=sup{ x n y+ u n y:nN}< for every nN. By the weak convergence of { u n i } to x ¯ F(T), lim n u n x n =0 and (C3)-(C4), we get (γfA)y, x ¯ y0 for all yF(T). A mapping A being a strongly positive linear bounded operator and f being a contraction ensure (γfA)y, x ¯ y0 for all yF(T), that is, x ¯ VI(F(T),Aγf). Thus, x VI(VI(F(T),Aγf),B), we have

lim sup n x x n , B x = lim sup i x x n i , B x = x x ¯ , B x 0 .

From (3.13), we notice that

lim sup n x y n , B x 0.
(3.17)

Third, lim sup n x n x ,ϕ( x ) x 0 is proven. Choose a subsequence { x n g } of { x n } such that

lim sup n x n x , ϕ ( x ) x = lim g x n g x , ϕ ( x ) x .

The boundedness of { x n g } implies the existence of a subsequence { x n g h } of { x n g } and a point x ˜ H such that { x n g h } converges weakly to x ˜ . By lim n x n + 1 x n =0, we have lim h x n g h + 1 ,w= x ˜ ,w, wH. We may assume, without loss of generality, that lim i x n g ,w= x ˜ ,w, wH. Assume x ˜ T( x ˜ ). lim n x n T x n =0 with F(T) guarantees that

lim inf g x n g x ˜ < lim inf g x n g T ( x ˜ ) = lim inf g x n g T ( x n g ) + T ( x n g ) T ( x ˜ ) = lim inf g T ( x n g ) T ( x ˜ ) lim inf g x n g x ˜ .

This is a contradiction, that is, x ˜ F(T). From x VI(VI(VI(F(T),Aγf),B),Iϕ), we find

(3.18)

Step 4. Finally, we prove lim n x n x =0. By Lemma 2.8, we compute

x n + 1 x 2 = α n ϕ ( x n ) + ( 1 α n ) y n x 2 = α n ( ϕ ( x n ) ϕ ( x ) ) + α n ( ϕ ( x ) x ) + ( 1 α n ) ( y n x ) 2 α n ϕ ( x n ) ϕ ( x ) 2 + ( 1 α n ) y n x 2 + 2 α n ϕ ( x ) x , x n + 1 x α n k 2 x n x 2 + ( 1 α n ) ( I μ β n B ) z n x 2 + 2 α n ϕ ( x ) x , x n + 1 x = α n k 2 x n x 2 + ( 1 α n ) ( z n μ β n B z n ) ( x μ β n B x ) μ β n B x 2 + 2 α n ϕ ( x ) x , x n + 1 x α n k 2 x n x 2 + ( 1 α n ) × { ( z n μ β n B z n ) ( x μ β n B x ) 2 + 2 μ β n x y n , B x } + 2 α n ϕ ( x ) x , x n + 1 x α n k 2 x n x 2 + ( 1 α n ) ( 1 τ β n ) 2 z n x 2 + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x α n k 2 x n x 2 + ( 1 α n ) ( 1 τ β n ) u n x 2 + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x = α n k 2 x n x 2 + ( 1 α n ) ( 1 τ β n ) [ I δ n ( A γ f ) ] x n x 2 + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x = α n k 2 x n x 2 + ( 1 α n ) ( 1 τ β n ) × ( I δ n A ) ( x n x ) + δ n ( γ f ( x n ) A x ) 2 + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x α n k 2 x n x 2 + ( 1 α n ) ( 1 τ β n ) × { ( 1 δ n γ ¯ ) 2 x n x 2 + 2 δ n γ f ( x n ) A x , u n x } + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x α n k x n x 2 + ( 1 α n ) ( 1 τ β n ) { ( 1 2 δ n γ ¯ + δ n 2 γ ¯ 2 ) x n x 2 + 2 δ n γ f ( x n ) γ f ( x ) , u n x + 2 δ n γ f ( x ) A x , u n x } + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x α n k x n x 2 + ( 1 α n ) ( 1 τ β n ) { ( 1 2 δ n γ ¯ ) x n x 2 + δ n 2 γ ¯ 2 x n x 2 + 2 δ n γ ρ x n x u n x + 2 δ n γ f ( x ) A x , u n x } + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x α n x n x 2 + ( 1 α n ) ( 1 τ β n ) [ 1 2 δ n ( γ ¯ γ ρ ) ] x n x 2 + ( 1 α n ) ( 1 τ β n ) δ n 2 γ ¯ 2 x n x 2 + 2 δ n γ f ( x ) A x , u n x + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x = α n x n x 2 + ( 1 α n ) [ 1 2 δ n ( γ ¯ γ ρ ) τ β n + τ β n 2 δ n ( γ ¯ γ ρ ) ] x n x 2 + ( 1 α n ) ( 1 τ β n ) δ n 2 γ ¯ 2 x n x 2 + 2 δ n γ f ( x ) A x , u n x + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x = α n x n x 2 + ( 1 α n ) [ 1 { 2 δ n ( γ ¯ γ ρ ) + τ β n τ β n 2 δ n ( γ ¯ γ ρ ) } ] x n x 2 + ( 1 α n ) ( 1 τ β n ) δ n 2 γ ¯ 2 x n x 2 + 2 δ n γ f ( x ) A x , u n x + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x = [ 1 ( 1 α n ) { 2 δ n ( γ ¯ γ ρ ) + τ β n τ β n 2 δ n ( γ ¯ γ ρ ) } ] x n x 2 + ( 1 α n ) ( 1 τ β n ) δ n 2 γ ¯ 2 x n x 2 + 2 δ n γ f ( x ) A x , u n x + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x = [ 1 ( 1 α n ) { 2 δ n ( γ ¯ γ ρ ) ( 1 τ β n ) + τ β n } ] x n x 2 + ( 1 α n ) ( 1 τ β n ) δ n 2 γ ¯ 2 x n x 2 + 2 δ n γ f ( x ) A x , u n x + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x = [ 1 ( 1 α n ) 2 δ n ( γ ¯ γ ρ ) ( 1 τ β n ) ] x n x 2 ( 1 α n ) τ β n x n x 2 + ( 1 α n ) ( 1 τ β n ) δ n 2 γ ¯ 2 x n x 2 + 2 δ n γ f ( x ) A x , u n x + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x [ 1 2 ( γ ¯ γ ρ ) ( 1 α n ) ( 1 τ β n ) δ n ] x n x 2 + ( 1 α n ) ( 1 τ β n ) δ n 2 γ ¯ 2 x n x 2 + 2 δ n γ f ( x ) A x , u n x + 2 μ β n x y n , B x + 2 α n ϕ ( x ) x , x n + 1 x .
(3.19)

Since { x n }, {A x n }, {B x n }, {ϕ( x n )} and {f( x n )} are all bounded, we can choose a constant M 1 >0 such that

sup n 1 γ ¯ γ ρ { δ n γ ¯ 2 2 x n x 2 } M 1 .

It follows that

x n + 1 x 2 [ 1 2 ( γ ¯ γ ρ ) ( 1 α n ) ( 1 τ β n ) δ n ] x n x 2 + 2 ( γ ¯ γ ρ ) ( 1 α n ) ( 1 τ β n ) δ n ς n ,
(3.20)

where

ς n = δ n M 1 + 1 ( γ ¯ γ ρ ) ( 1 α n ) ( 1 τ β n ) γ f ( x ) A x , u n x + μ β n ( γ ¯ γ ρ ) ( 1 α n ) ( 1 τ β n ) δ n x y n , B x + α n ( γ ¯ γ ρ ) ( 1 α n ) ( 1 τ β n ) δ n ϕ ( x ) x , x n + 1 x .

By (3.14), (3.17), (3.18) and (C3)-(C4), we get lim sup n ς n 0. Applying Lemma 2.9, we can conclude that x n x . This completes the proof. □

4 An example

Next, the following example shows that all the conditions of Theorem 3.1 are satisfied.

Example 4.1 For instance, let α n = 1 n , β n = 1 2 n and δ n = 1 3 n . We will show that the condition (C1) is achieved. Then, clearly, the sequence { δ n }

n = 1 δ n = n = 1 1 3 n =

and

n = 1 | δ n + 1 δ n | = n = 1 | 1 3 ( n + 1 ) 1 3 n | | 1 3 1 1 3 2 | + | 1 3 2 1 3 3 | + | 1 3 3 1 3 4 | + = 1 3 .

The sequence { δ n } satisfies the condition (C1).

Next, we will show that the condition (C2) is achieved. We compute

n = 1 | β n + 1 β n | = n = 1 | 1 2 ( n + 1 ) 1 2 n | | 1 2 1 1 2 2 | + | 1 2 2 1 2 3 | + | 1 2 3 1 2 4 | + = 1 2 .

The sequence { β n } satisfies the condition (C2).

Next, we will show that the condition (C3) is achieved. We compute

n = 1 | α n + 1 α n | = n = 1 | 1 n + 1 1 n | | 1 1 1 2 | + | 1 2 1 3 | + | 1 3 1 4 | + = 1

and

lim n α n = lim n 1 n =0.

The sequence { α n } satisfies the condition (C3).

Finally, we will show that the condition (C4) is achieved.

1 3 n < 1 2 n and 1 2 n < 1 n .

Corollary 4.2 Let H be a real Hilbert space, C be a closed convex subset of H. Let A:CH be inverse-strongly monotone. Let T:CC be a nonexpansive mapping. Let B:CC be β-strongly monotone and L-Lipschitz continuous. Assume that VI(F(T),A) is nonempty set. Suppose { x n } is a sequence generated by the following algorithm x 0 C arbitrarily:

{ z n = T ( I δ n A ) x n , y n = ( I μ β n B ) z n , x n + 1 = ( 1 α n ) y n , n 0 ,
(4.1)

{ α n },{ δ n }[0,1]. If μ(0, 2 β L 2 ) is used and if { β n }(0,1] satisfies the following conditions:

(C1): n = 1 | δ n + 1 δ n |<, n = 1 δ n =;

(C2): n = 1 | β n + 1 β n |<;

(C3): n = 1 | α n + 1 α n |<, lim n α n =0;

(C4): δ n β n and β n α n .

Then { x n } converges strongly to x VI(F(T),A), which is the unique solution of the variational inequality:

Find x VI ( F ( T ) , A ) such that B x , x x 0,xVI ( F ( T ) , A ) .
(4.2)

Proof Putting P C is the identity and f,ϕ0 in Theorem 3.1, we can obtain the desired conclusion immediately. □

Remark 4.3 Corollary 4.2 generalizes and improves the results of Iiduka [14].

Corollary 4.4 Let H be a real Hilbert space, C be a closed convex subset of H. Let A:CH be a strongly positive linear bounded operator, f:CH be a ρ-contraction, γ be a positive real number such that γ ¯ 1 ρ <γ< γ ¯ ρ . Let T:CC be a nonexpansive mapping. Assume that Ω is nonempty set. Suppose { x n } is a sequence generated by the following algorithm x 0 C arbitrarily:

{ z n = T P C [ I δ n ( A γ f ) ] x n , y n = ( I μ β n B ) z n , x n + 1 = α n ( x n ) + ( 1 α n ) y n , n 0 ,
(4.3)

where { α n },{ δ n }[0,1]. If μ(0, 2 β L 2 ) is used and if { β n }(0,1] satisfies the following conditions:

(C1): n = 1 | δ n + 1 δ n |<, n = 1 δ n =;

(C2): n = 1 | β n + 1 β n |<;

(C3): n = 1 | α n + 1 α n |<, lim n α n =0;

(C4): δ n β n and β n α n .

Then { x n } converges strongly to x Ω, which is the unique solution of the variational inequality:

Find x Ω such that B x , x x 0,xΩ.
(4.4)

Proof Putting ϕ is the identity in Theorem 3.1, we can obtain the desired conclusion immediately. □

Remark 4.5 Corollary 4.4 generalizes and improves the results of Marino and Xu [20].

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Acknowledgements

The authors thank to the Higher Education Research Promotion and National Research University Project of Thailand, the Office of the Higher Education Commission for financial support during the preparation of this paper.

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Wairojjana, N., Jitpeera, T. & Kumam, P. The hybrid steepest descent method for solving variational inequality over triple hierarchical problems. J Inequal Appl 2012, 280 (2012). https://doi.org/10.1186/1029-242X-2012-280

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Keywords

  • nonexpansive mapping
  • strongly monotone mapping
  • strongly positive linear bounded operator
  • Lipchitz continuous
  • variational inequality
  • hierarchical fixed point