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Iterative Methods for Variational Inequalities over the Intersection of the Fixed Points Set of a Nonexpansive Semigroup in Banach Spaces

Fixed Point Theory and Applications20102011:620284

https://doi.org/10.1155/2011/620284

Received: 8 November 2010

Accepted: 19 November 2010

Published: 25 November 2010

The Erratum to this article has been published in Fixed Point Theory and Applications 2011 2011:187439

Abstract

This paper presents a framework of iterative methods for finding specific common fixed points of a nonexpansive self-mappings semigroup in a Banach space. We prove, with appropriate conditions, the strong convergence to the solution of some variational inequalities.

1. Introduction

Let be a nonempty closed convex subset of a Hilbert space , and let be a nonlinear map. The classical variational inequality which is denoted by is formulated as finding such that
(1.1)
for all . We recall that is called -strongly monotone, if for each , we have
(1.2)
for a constant , and also -Lipschitzian if for each , we have
(1.3)
for a constant . Existence and uniqueness of solutions are important problems of the . It is known that if is a strongly monotone and Lipschitzian mapping on , then has a unique solution. An important problem is how to find a solution of . It is known that
(1.4)

where is an arbitrarily fixed constant and is the projection of onto . This alternative equivalence has been used to study the existence theory of the solution and to develop several iterative type algorithms for solving variational inequalities. But the fixed point formulation in (1.4) involves the projection , which may not be easy to compute, due to the complexity of the convex set . So, projection methods and their variant forms can be implemented for solving variational inequalities.

In order to reduce the complexity probably caused by the projection , Yamada [1] (see also [2]) introduced a hybrid steepest-descent method for solving . His idea is stated now. Assume that is the fixed point set of a nonexpansive mapping . Recall that is nonexpansive if
(1.5)

Assume that is -strongly monotone and -Lipschitzian on . Take a fixed number and a sequence in satisfying the following conditions:

,

,

.

Starting with an arbitrary initial guess , generate a sequence by the following algorithm:
(1.6)

Yamada [1] proved that the sequence converges strongly to a unique solution of . Xu and Kim [3] further considered and studied the hybrid steepest-descent algorithm (1.6). Their major contribution is that the strong convergence of (1.6) holds with condition being replaced by the following condition:

.

It is clear that condition is strictly weaker than condition , coupled with conditions and . Moreover, includes the important and natural choice for whereas does not. For more related results, see [4, 5].

Let be a Banach space we recall that a nonexpansive semigroup is a family of self-mappings of satisfies the following conditions:

(i) for ,

(ii) for and ,

(iii) for ,
  1. (iv)

    for each is nonexpansive. that is,

     
(1.7)

The problem is to find some fixed point in . For this, so many algorithms have been developed and under some restrictions partial answers have been obtained [611].

Assume that is a strongly monotone and Lipschitzian mapping and is a nonexpansive semigroup of self-mappings on . For an appropriate and starting from an arbitrary initial point , we devise the following implicit, explicit, and modified iterations:
(1.8)
(1.9)
(1.10)

for . With some appropriate assumptions, we prove the strong convergence of (1.8), (1.9), and (1.10) to the unique solution of the variational inequality in , where is the single-valued normalized duality mapping from into .

Our main purpose is to improve some of the conditions and results in the mentioned papers, especially those of Song and Xu [11].

2. Preliminaries

Let be the unit sphere of the Banach space . The space is said to have Gateaux differentiable norm(or is said to be smooth), if the limit
(2.1)

exists for each , and is said to have a uniformly Gateaux differentiable norm if for each , the limit (2.1) converges uniformly for . Further, is said to be uniformly smooth if the limit (2.1) exists uniformly for .

We denote the normalized duality mapping from to defined by
(2.2)

where denotes the generalized duality pairing. It is well known if is smooth then any duality mapping on is single valued, and if has a uniformly Gateaux differentiable norm, then the duality mapping is norm to weak* uniformly continuous on bounded sets.

Recall that a Banach space is said to be strictly convex if and implies . In a strictly convex Banach space , we have that if for and , then .

Now, we recall the concept of uniformly asymptotically regular semigroup. A continuous operator semigroup on is said to be uniformly asymptotically regular on if for all and any bounded subset of , we have
(2.3)

The nonexpansive semigroup is an example of uniformly asymptotically regular operator semigroup [11].

Let be a continuous linear functional on satisfying . Then, we know that is a mean on if and only if
(2.4)
for every . Sometimes, we use instead of . A mean on is called a Banach limit if . We know that if is a Banach limit, then
(2.5)
for every . Thus, if as , then we have
(2.6)

A discussion on these and related concepts can be found in [12].

We make use of the following well-known results throughout the paper.

Lemma 2.1 (see [12, Lemma 4.5.4]).

Let be a nonempty closed convex subset of a Banach space with a uniformly Gateaux differentiable norm, and let be a bounded sequence in . If , then
(2.7)
if and only if
(2.8)

for all .

Lemma 2.2 (see [13]).

Let , , , and let be sequences such that
(2.9)

for all . Assume also that . Then, the following results hold:

(i)if (where ), then is bounded,

(ii)if we have
(2.10)

then .

Lemma 2.3.

Let be a real normed linear space, and let be the normalized duality mapping on . Then, for any , and , the following inequality holds:
(2.11)
In order to reduce any possible complexity in writing, we set for a nonexpansive semigroup and
(2.12)

where , for all , and is a bounded sequence in .

3. Implicit Iterative Method

Recall that if is the single-valued normalized duality mapping from a Banach space into , a nonlinear operator is called -strongly monotone if for every , the following inequality holds:
(3.1)

for a constant .

The following lemma will be be used to show the convergence of (1.8) and (1.9).

Lemma 3.1.

Let be a Banach space, and let be the single-valued normalized duality mapping from into . Assume also that is -strongly monotone and -Lipschitzian on . Then,
(3.2)

is a contraction on for every .

Proof.

By using Lemma 2.3, we have
(3.3)
Thus, we obtain
(3.4)

and for , we have . That is, is a contraction, and the proof is complete.

In the following theorem, which is the main result in this section, we establish the strong convergence of the sequence defined by (1.8).

Theorem 3.2.

Let be a real Banach space with a uniformly Gateaux differentiable norm, and let be a nonexpansive semigroup from into itself. Let also defined by (1.8) satisfies the following condition:
(3.5)
Assume that is -strongly monotone and -Lipschitzian. Assume also that is a sequence of positive numbers that and . If , then converges strongly to some fixed point , which is the unique solution in to the variational inequality , that is
(3.6)

Proof.

We divide the proof into several steps.

Step 1.

We first prove the uniqueness of the solution to ; for this, we suppose are two solutions of . Thus, we have
(3.7)
By adding up the last two inequalities, we obtain
(3.8)

and so, .

Step 2.

We claim that is bounded. In fact, taking a fixed , we have
(3.9)
Taking and by using induction, we obtain
(3.10)

therefore, is bounded and so is .

Step 3.

The sequence is sequentially compact. To prove this, we assume that the set contains some such that for an arbitrary . So, by using Lemma 2.1, we can obtain
(3.11)
On the other hand, for any , we have
(3.12)
Thus,
(3.13)
Also, we have
(3.14)
It follows that
(3.15)
Combining (3.11) and (3.13) together, we get
(3.16)

This yields . Hence, there exists a subsequence of such as that converges strongly to ; that is, is sequentially compact.

Step 4.

We claim that is the solution of . Since is bounded, for any fixed point , there exist a constant such that . Therefore, we obtain
(3.17)
Hence,
(3.18)
Note that the duality mapping is single valued ( is smooth), and norm topology to weak* uniformly continuous on bounded sets of Banach space with uniformly Gateaux differentiable norm. Therefore,
(3.19)
and by taking limit as in two sides of (3.18), we obtain
(3.20)

Cosequently, is the unique solution of .

Step 5.

in norm. Indeed, we show that each cluster point of the sequence is equal to . Assume that is another strong limit point of in . Thanks to (3.15), we have the following two inequalities:
(3.21)
Therefore,
(3.22)

It yields that , which proves the uniqness of . Thus, itself converges strongly to . This completes the proof.

4. Explicit Iterative Method

In this section, we will present our result of the strong convergence of (1.9), but first, we need to prove, with different approach, the following lemma.

Lemma 4.1.

Let , , , , ,μ, and be as those in Theorem 3.2. If , and there exists a bounded sequence such that
(4.1)
Then,
(4.2)

Proof.

By the uniqueness of and with no loss of generality, we can choose such that
(4.3)
as . Let be the fixed point of the contraction
(4.4)
Then,
(4.5)
Now, by using Lemma 2.3, we have
(4.6)
Therefore,
(4.7)
Because , and are bounded, from (4.3) and (4.7), we conclude that
(4.8)
Moreover, we have
(4.9)
By Theorem 3.2, , as . So, using the boundedness of , we get
(4.10)
On the other hand, noticing that the sequence is bounded and the duality mapping is single-valued and norm to weak* uniformly continuous on bounded subsets of , we conclude that
(4.11)
Therefore, from (4.8) and (4.9), we obtain
(4.12)

This completes the proof.

Next, we prove the strong convergence of explicit iteration scheme (1.9).

Theorem 4.2.

Let be a real Banach space with a uniformly Gateaux differentiable norm, and let be a nonexpansive semigroup from into itself. Let also defined by (1.9) satisfies the following conditions:

(i) ,

(ii) for all .

Assume that is -strongly monotone and -Lipschitzian, and that is a sequence of positive numbers that . Assume also that the sequence in satisfies the control condition
(4.13)
If , then converges strongly to some fixed point , which is the unique solution in for the following variational inequality:
(4.14)

Proof.

Existence and uniqueness of the solution of is attained from Theorem 3.2. Now, we claim that is bounded. Indeed, taking a fixed , we have
(4.15)

Taking , , , and using Lemma 2.2, we conclude that is bounded and so is .

Next, we prove that converges strongly to the unique solution of . By definition of the algorithm and taking , we have
(4.16)

Taking , , and and using Lemma 4.1 together with Lemma 2.2 lead to , that is, in norm. This completes the proof.

Corollary 4.3.

Let be a real reflexive strictly convex Banach space with a uniformly Gateaux differentiable norm. Let also be a nonexpansive semigroup from into itself such that . Assume that defined by (1.9) satisfies condition in Theorem 4.2, then condition holds.

Proof.

Clearly, is a convex and continuous function. Because is a reflexive Banach space, according to [12, Theorem ], is nonempty. Also, by convexity and continuity of , the set is a closed convex subset of . Since for every and , we have
(4.17)
So, , and therefore . Suppose that . Because every nonempty closed convex subset of a strictly convex and reflexive Banach space is a Chebyshev set, according to [14, Corollary ], there exists a unique such that
(4.18)
On the other hand, for all , and is nonexpansive, so we get
(4.19)

that is, , by uniqueness of . Thus, . This completes the proof.

Corollary 4.4.

Let be a real Banach space, and let be a nonexpansive uniformly asymptotically regular semigroup from into itself. If is defined by (1.9), where satisfies , then condition in Theorem 4.2 holds.

Proof.

From (1.9), , and the boundedness of , we conclude that
(4.20)
as . On the other hand, the semigroup is uniformly asymptotically regular, , and is a bounded subset in , so for all , we have
(4.21)
Hence,
(4.22)
So, from (4.20), (4.21), and (4.22), we get
(4.23)

and it completes the proof.

Remark 4.5.

According to Corollaries 4.3 and 4.4, our assumptions are weaker than those of Song and Xu [11]. Also, noticing that for a contraction , the mapping is strongly monotone and Lipschitzian. So, by replacing by in (1.8) and (1.9), the following schemes are, respectively, obtained:
(4.24)

Remark 4.6.

In the same way and with the same conditions mentioned in Theorem 4.2, it's easy to see that the sequence defined by
(4.25)

converges strongly to the variational inequality .

5. Modified Iterative Method

In this section, we show that the modified sequence defined by (1.10) also converges strongly to the solution of variational equality , but first, we need to prove the following lemma.

Lemma 5.1.

Let be a Banach space. Assume that is -strongly monotone and -Lipschitzian nonlinear operator and a nonexpansive mapping. If , where , then
(5.1)

is a contraction on .

Proof.

Considering the inequality
(5.2)
from Lemma 2.3 in a Banach space , where is the normalized single-valued duality, we have
(5.3)
Noticing that
(5.4)
we obtain
(5.5)
Thus, we have
(5.6)

Note that for , we conclude . That is, is a contraction and the proof is complete.

Theorem 5.2.

Let be a real Banach space with a uniformly Gateaux differentiable norm and a nonexpansive semigroup from into itself. Let also defined by (1.10) satisfies the following conditions:

(i) ,

(ii) for all .

Assume that is -strongly monotone and -Lipschitzian and a sequence of positive numbers that . Assume also that the sequences , where , and in satisfy the following control conditions:

,

does not take 0 as it's limit point.

Then, converges strongly to some fixed point , which is the unique solution in for the variational inequality .

Proof.

Existence and uniqueness of the solution of is obtained from Theorem 3.2. We claim that is bounded. Indeed, taking a fixed , we have
(5.7)
Noticing that
(5.8)
and by assumption that there exists so that , for all . Thus, we get
(5.9)

and from Lemma 2.2, we conclude that is bounded.

By Theorem 3.2, there exists a unique solution to . We prove that converges strongly to
(5.10)
Taking , , and and using Lemma 4.1 together with Lemma 2.2 imply , that is,
(5.11)

This completes the proof.

Remark 5.3.

In Theorem 5.2, if , then , and therefore we can remove , also turns to .

Remark 5.4.

We can easily see that under some restrictions all the strongly monotone and Lipschitzian nonlinear operators used in this paper are replaceable by strongly accretive and strictly pseudocontractive ones (see [15]).

Notes

Authors’ Affiliations

(1)
Department of Mathematics, Islamic Azad University

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© Issa Mohamadi. 2011

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