`Journal of Applied MathematicsVolume 2012 (2012), Article ID 567948, 14 pageshttp://dx.doi.org/10.1155/2012/567948`
Research Article

## A Modified Regularization Method for the Proximal Point Algorithm

School of Mathematical Sciences, Yancheng Teachers University, Yancheng 224051, China

Received 15 April 2012; Accepted 28 May 2012

Copyright © 2012 Shuang Wang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Under some weaker conditions, we prove the strong convergence of the sequence generated by a modified regularization method of finding a zero for a maximal monotone operator in a Hilbert space. In addition, an example is also given in order to illustrate the effectiveness of our generalizations. The results presented in this paper can be viewed as the improvement, supplement, and extension of the corresponding results.

#### 1. Introduction

Let be a real Hilbert space and a nonempty closed convex subset of , and let be a nonlinear operator. The variational inequality problem is formulated as finding a point such that

In 1964, Stampacchia [1] introduced and studied variational inequality initially. It is now well known that variational inequalities cover as diverse disciplines as partial differential equations, optimal control, optimization, mathematical programming, mechanics, and finance, see [15].

Let be an operator with domain and range in . A multivalued operator is called monotone if for any , and maximal monotone if it is monotone and its graph is not properly contained in the graph of any other monotone operator.

One of the major problems in the theory of monotone operators is to find a point in the zero set, which can be formulated as finding a point so that , where denotes the zero set of the operator . A variety of problems, including convex programming and variational inequalities, can be formulated as finding a zero of maximal monotone operators. A classical way to solve such problem is Rockafellar's proximal point algorithm [6], which generates an iterative sequence as

where, for , denotes the resolvent of given by , with being the identity map on the space . If , it is known that the sequence generated by (1.4) converges weakly to some point in .

Motivated by Lehdili and Moudafi's prox-Tikhonov method [7], Xu [8] considered the following regularization iterative form: for a fixed point ,

where and is a sequence of errors. Then, the iterative sequence converges strongly to , provided that(C1),(C2),(C3),(C4),(C5), .

Recently, Song and Yang [9] removed some strict restrictions in Xu [8]. Under conditions (C1), (C2), (C4) (or ), (C5), and (C3′)(C3′), they proved that the sequence generated by (1.5) converges strongly to .

Very recently, under conditions (C1), (C3) (or C3′), (C5), and (C4′) (C4′). Wang [10] proved the strong convergence of the sequence generated by (1.5). It is easy to see that conditions (C3′) and (C4′) are strictly weaker than conditions (C3) and (C4), respectively.

We remind the reader of the following fact: in order to guarantee the strong convergence of the iterative sequence , there is at least one parameter sequence converging to zero (i.e., ) in the result of Xu [8], Song and Yang [9], and Wang [10]. So the above results bring us to the following natural questions.

Question 1. Can we obtain the strong convergence theorem without the parameter sequence converging to zero?

Question 2. Can we get that the sequence converges strongly to , which solves uniquely some variational inequalities?

In this work, motivated by the above results, we consider the following modified regularization method for the proximal point algorithm: for an arbitrary ,

where is a -Lipschitzian and -strongly monotone operator on and is a fixed point in . Without the parameter sequence converging to zero, we prove that the sequence generated by the iterative algorithm (1.6) converges strongly to , which solves uniquely the variational inequality ,  for all . In addition, an example is also given in order to illustrate the effectiveness of our generalizations. The results presented in this paper can be viewed as the improvement, supplement, and extension of the results obtained in [610].

#### 2. Preliminaries

Let be a real Hilbert space with inner product and norm . For the sequence in , we write to indicate that the sequence converges weakly to . means that converges strongly to .

A mapping is called -Lipschitzian if there exists a positive constant such that

is said to be -strongly monotone if there exists a positive constant such that

Let be a strongly positive bounded linear operator on , that is, there exists a constant such that

A typical problem is that of minimizing a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbert space :

where is a given point in and is the set of the fixed points of nonexpansive mapping .

Remark 2.1 (see [11]). From the definition of , we note that a strongly positive bounded linear operator is a -Lipschitzian and -strongly monotone operator.

Let be a maximal monotone operator on a real Hilbert space such that . For , we use to denote the resolvent of , that is,

It is well known that is firmly nonexpansive and consequently nonexpansive; moreover, .

The following lemma is known as the resolvent identity of maximal monotone operators.

Lemma 2.2 (see [8]). Let . Then, for any ,

In order to prove our main results, we need the following lemmas.

Lemma 2.3 (see [11]). Let be a -Lipschitzian and -strongly monotone operator on a Hilbert space with and . Then, is a contraction with contraction coefficient .

Lemma 2.4 (see [12]). is firmly nonexpansive if and only if is nonexpansive.

Lemma 2.5 (see [13]). Let be a Hilbert space, a closed convex subset of , and a nonexpansive mapping with ; if is a sequence in weakly converging to and if converges strongly to , then .

Lemma 2.6 (see [14]). Let and be bounded sequences in Banach space and a sequence in which satisfies the following condition: Suppose that , , and . Then, .

Lemma 2.7 (see [15, 16]). Let be a sequence of nonnegative real numbers satisfying where , , and satisfy the following conditions: (i) and , (ii) or , and (iii) . Then, .

#### 3. Main Results

Let be a -Lipschitzian and -strongly monotone operator on with and the resolvent of . Let and , and consider a mapping on defined by

where is a fixed constant and is a fixed point. It is easy to see that is a contraction. Indeed, from Lemma 2.3, we have

for all . Hence, it has a unique fixed point, denoted by , which uniquely solves the fixed point equation

Theorem 3.1. For any and , let the net be generated by (3.3). Then, as , the net converges strongly to of , which solves uniquely the variational inequality

Proof. We first show the uniqueness of a solution of the variational inequality (3.4), which is indeed a consequence of the strong monotonicity of . Suppose and both are solutions to (3.4); then,
Adding (3.5) to (3.6), we get
The strong monotonicity of implies that and the uniqueness is proved. Below we use to denote the unique solution of (3.4). Next, we prove that is bounded. Taking , from (3.3) and using Lemma 2.3, we have
that is,
Observe that
From , we may assume, without loss of generality, that , where is an arbitrarily small positive number. Thus, we have that is continuous, for all . Therefore, we obtain From (3.9) and (3.11), we have bounded and so is . On the other hand, from (3.3), we obtain To prove that , for a given  , using Lemma 2.3, we have where . Therefore, From , we have . Moreover, if , we have .
Since is bounded, we see that if is a sequence in such that and , then, by (3.14), we see that . Moreover, by (3.12) and using Lemma 2.5, we have . We next prove that solves the variational inequality (3.4). From (3.3) and , we have that is, Now replacing in (3.16) with and letting , we have That is, is a solution of (3.4), and hence by uniqueness. In a summary, we have shown that each cluster point of (at ) equals . Therefore, as .

Setting in Theorem 3.1, we can obtain the following result.

Corollary 3.2. For any and , let be a strongly positive bounded linear operator with coefficient . For each , let the net be generated by . Then, as , the net converges strongly to of which solves uniquely the variational inequality

Setting and in Theorem 3.1, we can obtain the following result.

Corollary 3.3 (Xu [8, Theorem 3.1]). For any and . For each , let the net be generated by . Then, as , converges strongly to the projection of onto ; that is, . Moreover, this limit is attained uniformly for .

The next result gives a strong convergence theorem on algorithm (1.6) with a weaker restriction on the sequence .

Theorem 3.4. Let be a maximal monotone operator on a Hilbert space with . Let be a -Lipschitzian and -strongly monotone operator on with . Let be a sequence in , a sequence in , and an arbitrarily small positive number. Assume that the control conditions , , , and hold for , , and , for some integer .
For an arbitrary point , let the sequence be generated by (1.6). Then, where solves the variational inequality

Proof. On the one hand, suppose that . We proceed with the following steps.
Step 1. We claim that is bounded. In fact, taking , from (1.6) and and using Lemma 2.3, we have for all for some integer , where . By induction, we have for all for some integer , where . Therefore, is bounded. We also obtain that and are bounded.
Step 2. We claim that . In fact, write and . Then, is firmly nonexpansive and is nonexpansive (see Lemma 2.4).
Observe that where . Therefore,
It follows from the resolvent identity that for any . From (1.6), we get By (3.25) and (3.26), we have Substituting (3.27) into (3.24) at once gives that is, where . Observing , , and , it follows that From (3.23) and using Lemma 2.6, we have . Therefore, Step 3. We claim that . Since , there exist and a positive integer such that for all , . From Lemma 2.2, for each , we have Observe that Thus, it follows from (3.32), (3.33), and that Since , then Step 4. We claim that , where and is defined by (3.3). Since is bounded, there exists a subsequence of which converges weakly to . From , we obtain . From Lemma 2.5, we have . Hence, by Theorem 3.1, we have Step 5. We claim that converges strongly to . From (1.6), for an appropriate constant , we have for all for some integer . For every , put and . It follows that It is easy to see that and . Hence, by Lemma 2.7, the sequence converges strongly to . Observe that Thus, it follows that the sequence converges strongly to .
On the other hand, suppose that as . From (1.6), we have Therefore,

Setting and in Theorem 3.4, we can obtain the following result.

Corollary 3.5. Let be a maximal monotone operator on a Hilbert space with . Let be a sequence in , a sequence in , and an arbitrarily small positive number. Assume that the control conditions , , , and hold for , , and .(), for all for some integer .
For an arbitrary point , let the sequence be generated by Then,

Corollary 3.6 ([Wang [10], Theorem  4]). Let , , and satisfy , ,or , and . In addition, if , then the sequence generated by (1.5) converges strongly to .

Proof. Since , it is easy to see that , for all for some integer . Without loss of generality, we assume that , for all for some integer . Repeating the same argument as in the proof of Theorem  4 in Wang [10], we know that is bounded. Thus, we have that . Therefore, all conditions of Corollary 3.5 are satisfied. Using Corollary 3.5, we have that converges strongly to , with . Therefore,

Remark 3.7. Corollary 3.5 is more general than Theorem 4 of Wang [10]. The following example is given in order to illustrate the effectiveness of our generalizations.

Example 3.8. Let   be the set of real numbers, and for all . Define a maximal monotone operator as follows: , for all . It is easy to see that and . Given sequences and , and , for all . For an arbitrary , let be defined by (3.42), that is, Observe that Hence, we have for all . This implies that converges strongly to . Thus, Furthermore, it is easy to see that there hold the following:(B1), for some integer ,(B2),(B3) and ,(B4).

Hence there is no doubt that all conditions of Corollary 3.5 are satisfied. Since , the condition of Wang [10, Theorem   4] is not satisfied. So, by Corollary 3.5, we obtain that the sequence and converges strongly to zero but Theorem  4 of Wang [10] cannot be applied to and in this example.

#### Acknowledgment

This paper is supported by the Natural Science Foundation of Yancheng Teachers University under Grant 11YCKL009.

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