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Finite-Step Relaxed Hybrid Steepest-Descent Methods for Variational Inequalities
Journal of Inequalities and Applications volume 2008, Article number: 598632 (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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ1_HTML.gif)
was initially studied by Kinderlehrer and Stampacchia [1]. It is also known that the is equivalent to the fixed-point equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ2_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ3_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ4_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ5_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ6_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equa_HTML.gif)
Lemma 3.1 (See [3]).
Let be a contraction provided that
and
Indeed,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ7_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ8_HTML.gif)
where . Moreover, we also have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ9_HTML.gif)
where , and for
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ10_HTML.gif)
where , and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ11_HTML.gif)
where .
Thus we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ12_HTML.gif)
for In particular,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ13_HTML.gif)
Hence, substituting (3.8) in (3.3) and by condition (iv), we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ14_HTML.gif)
By induction, it is easy to see that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ15_HTML.gif)
where . Indeed, for
, from (3.9) we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ16_HTML.gif)
Suppose that for
We want to claim that
Indeed,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ17_HTML.gif)
Therefore, we have , for all
and
, for all
. In this case, from (3.8), it follows that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ18_HTML.gif)
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
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ19_HTML.gif)
and so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ20_HTML.gif)
Step 3.
Let Indeed, we observe that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ21_HTML.gif)
and, for ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ22_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ23_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ24_HTML.gif)
Hence it follows from the above inequalities (3.17)–(3.19) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ25_HTML.gif)
Let us substitute (3.19) into (3.20), then we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ26_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ27_HTML.gif)
We put
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ28_HTML.gif)
Then and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ29_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ30_HTML.gif)
as .
Step 5.
Let , for
. Let
be a subsequence of
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ31_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ32_HTML.gif)
From the proof of Step 2,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ33_HTML.gif)
for Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ34_HTML.gif)
for
Step 6.
Let in norm and so does
for
Indeed using Lemma 2.3 and (3.7) we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ35_HTML.gif)
From (ii), (iii), and Step 5, we have for
and
for
,
and
is bounded; by Lemma 2.4, we conclude that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F598632/MediaObjects/13660_2007_Article_1834_Equ36_HTML.gif)
Consequently from Lemma 2.1, we obtain and hence it follows from
for
that
for
.
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Acknowledgment
This research was partially supported by Grant no. NSC95-2115-M-039-001- from the National Science Council of Taiwan.
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Lin, YC. Finite-Step Relaxed Hybrid Steepest-Descent Methods for Variational Inequalities. J Inequal Appl 2008, 598632 (2008). https://doi.org/10.1155/2008/598632
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DOI: https://doi.org/10.1155/2008/598632