## Journal of Inequalities and Applications

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# Finite-Step Relaxed Hybrid Steepest-Descent Methods for Variational Inequalities

Journal of Inequalities and Applications20082008:598632

DOI: 10.1155/2008/598632

Received: 22 August 2007

Accepted: 13 March 2008

Published: 8 April 2008

## Abstract

The classical variational inequality problem with a Lipschitzian and strongly monotone operator on a nonempty closed convex subset in a real Hilbert space was studied. A new finite-step relaxed hybrid steepest-descent method for this class of variational inequalities was introduced. Strong convergence of this method was established under suitable assumptions imposed on the algorithm parameters.

## 1. Introduction

Let be a real Hilbert space with inner product and norm . Let be a nonempty closed convex subset of and let be an operator. The classical variational inequality problem: find such that
(1.1)
was initially studied by Kinderlehrer and Stampacchia [1]. It is also known that the is equivalent to the fixed-point equation
(1.2)

where is the (nearest point) projection from onto , that is, for each and where is an arbitrarily fixed constant. If is strongly monotone and Lipschitzian on and is small enough, then the mapping determined by the right-hand side of this equation is a contraction. Hence the Banach contraction principle guarantees that the Picard iterates converge in norm to the unique solution of the . Such a method is called the projection method. However, Zeng and Yao [2] point out that the fixed-point equation involves the projection which may not be easy to compute due to the complexity of the convex set . To reduce the complexity problem probably caused by the projection , a class of hybrid steepest-descent methods for solving VI( has been introduced and studied recently by many authors (see, e.g., [3, 4]). Zeng and Yao [2] have established the method of two-step relaxed hybrid steepest-descent for variational inequalities. A natural arising problem is whether there exists a general relaxed hybrid steepest-descent algorithm that is more than two steps for finding approximate solutions of VI( or not. Motivated and inspired by the recent research work in this direction, we introduce the following finite step relaxed hybrid steepest-descent algorithm for finding approximate solutions of VI( and aim to unify the convergence results of this kind of methods.

Algorithm 1.1

Let , , for , and take fixed numbers , . Starting with arbitrarily chosen initial points , compute the sequences such that
(1.3)

We will prove a strong convergence result for Algorithm 1.1 under suitable restrictions imposed on the parameters.

## 2. Preliminaries

The following lemmas will be used for proving the main result of the paper in next section.

Lemma 2.1 (See [5]).

Let be a sequence of nonnegative real numbers satisfying the inequality
(2.1)

where and satisfy the following conditions:

(i) , or equivalently,

(ii) ;

(iii) .

Then

Lemma 2.2 (See [6]).

Demiclosedness principle: assume that is a nonexpansive self-mapping on a nonempty closed convex subset of a Hilbert space If has a fixed point, then is demiclosed; that is, whenever is a sequence in weakly converging to some and the sequence strongly converges to some , it follows that Here is the identity operator of

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

Lemma 2.3.

In a real Hilbert space there holds the inequality
(2.2)

Lemma 2.4.

Let be a nonempty closed convex subset of . For any and , the following statements hold:

(i) ;

(ii) .

## 3. Convergence Theorem

Let be a real Hilbert space and let be a nonempty closed convex subset of Let be an operator such that for some constants is -Lipschitzian and -strongly monotone on that is, satisfies the conditions
(3.1)

respectively. Since F is -strongly monotone, the variational inequality problem has a unique solution (see, e.g., [7]).

Assume that is a nonexpansive mapping with the fixed points set Note that obviously For any given numbers and , we define the mapping by

Lemma 3.1 (See [3]).

Let be a contraction provided that and Indeed,
(3.2)

where

We now state and prove the main result of this paper.

Theorem 3.2.

Let be a real Hilbert space and let be a nonempty closed convex subset of Let be an operator such that for some constants is -Lipschitzian and -strongly monotone on . Assume that is a nonexpansive mapping with the fixed points set the real sequences , for , in Algorithm 1.1 satisfy the following conditions:

(i) , for ;

(ii) and for ;

(iii) ;

(iv) , for all .

Then the sequences generated by Algorithm 1.1 converge strongly to which is the unique solution of the .

Proof.

Since is -strongly monotone, by [7], the has the unique solution . Next we divide the rest of the proof into several steps.

Step 1.

Let is bounded for each . Indeed, let us denote that , then we have
(3.3)
where . Moreover, we also have
(3.4)
where , and for ,
(3.5)
where , and
(3.6)

where .

Thus we obtain
(3.7)
for In particular,
(3.8)
Hence, substituting (3.8) in (3.3) and by condition (iv), we obtain
(3.9)
By induction, it is easy to see that
(3.10)
where . Indeed, for , from (3.9) we obtain
(3.11)
Suppose that for We want to claim that Indeed,
(3.12)
Therefore, we have , for all and , for all . In this case, from (3.8), it follows that
(3.13)

Step 2.

Let Indeed by Step 1, is bounded for and so are and for . Thus from the conditions that , for and , we have, for ,
(3.14)
and so
(3.15)

Step 3.

Let Indeed, we observe that
(3.16)
and, for ,
(3.17)
(3.18)
(3.19)
Hence it follows from the above inequalities (3.17)–(3.19) that
(3.20)
Let us substitute (3.19) into (3.20), then we have
(3.21)
where
(3.22)
We put
(3.23)
Then and
(3.24)

From (ii)–(iv), we obtain as . Furthermore, from (i), By Lemma 2.1, we deduce that as .

Step 4.

Let as . From Steps 2 and 3, we have
(3.25)

as .

Step 5.

Let , for . Let be a subsequence of such that
(3.26)
Without loss of generality, we assume that weakly for some . By Step 4, we derive weakly. But by Lemma 2.2 and Step 4, we have Since is the unique solution of the we obtain
(3.27)
From the proof of Step 2,
(3.28)
for Then
(3.29)

for

Step 6.

Let in norm and so does for Indeed using Lemma 2.3 and (3.7) we get
(3.30)
From (ii), (iii), and Step 5, we have for and for , and is bounded; by Lemma 2.4, we conclude that
(3.31)

Consequently from Lemma 2.1, we obtain and hence it follows from for that for .

## Declarations

### Acknowledgment

This research was partially supported by Grant no. NSC95-2115-M-039-001- from the National Science Council of Taiwan.

## Authors’ Affiliations

(1)
Department of Occupational Safety and Health, General Education Center, China Medical University

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