- Research Article
- Open Access

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

- Yen-Cherng Lin
^{1}Email author

**2008**:598632

https://doi.org/10.1155/2008/598632

© Yen-Cherng Lin. 2008

**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.

## Keywords

- Unique Solution
- Variational Inequality
- Nonexpansive Mapping
- Strong Convergence
- Convergence Result

## 1. Introduction

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

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]).

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.

Lemma 2.4.

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

(i) ;

(ii) .

## 3. Convergence Theorem

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

Lemma 3.1 (See [3]).

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.

where .

Step 2.

Step 3.

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

Step 4.

as .

Step 5.

for

Step 6.

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

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## Copyright

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