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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 567948, 14 pages
http://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

Academic Editor: Yonghong Yao

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 𝐻 be a real Hilbert space and 𝐶 a nonempty closed convex subset of 𝐻, and let 𝐹𝐻𝐻 be a nonlinear operator. The variational inequality problem is formulated as finding a point 𝑥𝐶 such that 𝐹𝑥,𝑣𝑥0,𝑣𝐶.(1.1)

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 𝑇 be an operator with domain 𝐷(𝑇) and range 𝑅(𝑇) in 𝐻. A multivalued operator 𝑇𝐻2𝐻 is called monotone if 𝑢𝑣,𝑥𝑦0,(1.2) for any 𝑢𝑇𝑥,𝑣𝑇𝑦, and maximal monotone if it is monotone and its graph 𝐺(𝑇)={(𝑥,𝑦)𝑥𝐷(𝑇),𝑦𝑇𝑥}(1.3) 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 𝑥 so that 𝑥𝑇1(0), where 𝑇1(0) denotes the zero set of the operator 𝑇. 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 𝑥𝑛+1=𝐽𝑇𝑐𝑥𝑛+𝑒𝑛,(1.4)

where, for 𝑐>0, 𝐽𝑇𝑐 denotes the resolvent of 𝑇 given by 𝐽𝑇𝑐=(𝐼+𝑐𝑇)1, with 𝐼 being the identity map on the space 𝐻. If 𝑇1(0), it is known that the sequence generated by (1.4) converges weakly to some point in 𝑇1(0).

Motivated by Lehdili and Moudafi's prox-Tikhonov method [7], Xu [8] considered the following regularization iterative form: for a fixed point 𝑢𝐻, 𝑥𝑛+1=𝐽𝑇𝑐𝑛1𝑡𝑛𝑥𝑛+𝑡𝑛𝑢+𝑒𝑛,𝑛0,(1.5)

where 𝑡𝑛(0,1) and {𝑒𝑛} is a sequence of errors. Then, the iterative sequence converges strongly to 𝑃𝑇1(0)𝑢, provided that(C1)lim𝑛𝑡𝑛=0,(C2)𝑛=0|𝑡𝑛+1𝑡𝑛|<,(C3)0<𝑐𝑐𝑛𝑐,(C4)𝑛=0|𝑐𝑛+1𝑐𝑛|<,(C5)𝑛=0𝑡𝑛=, 𝑛=0𝑒𝑛<.

Recently, Song and Yang [9] removed some strict restrictions in Xu [8]. Under conditions (C1), (C2), (C4) (or 𝑛=0|1(𝑐𝑛/𝑐𝑛+1)|<+), (C5), and (C3′)(C3′)0<liminf𝑛𝑐𝑛, they proved that the sequence generated by (1.5) converges strongly to 𝑃𝑇1(0)𝑢.

Very recently, under conditions (C1), (C3) (or C3′), (C5), and (C4′) (C4′)lim𝑛|1(𝑐𝑛/𝑐𝑛+1)|=0. 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 {𝑥𝑛}, there is at least one parameter sequence converging to zero (i.e., 𝑡𝑛0) 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 {𝑡𝑛} converging to zero?

Question 2. Can we get that the sequence {𝑥𝑛} converges strongly to 𝑥𝑇1(0), 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 𝑥0𝐻, 𝑧𝑛=𝐼𝑡𝑛𝐹𝑥𝑛+𝑡𝑛𝑢+𝑒𝑛,𝑥𝑛+1=𝐽𝑇𝑐𝑛𝑧𝑛,𝑛0,(1.6)

where 𝐹 is a 𝑘-Lipschitzian and 𝜂-strongly monotone operator on 𝐻 and 𝑢 is a fixed point in 𝐻. Without the parameter sequence {𝑡𝑛} converging to zero, we prove that the sequence {𝑥𝑛} generated by the iterative algorithm (1.6) converges strongly to 𝑥𝑇1(0), which solves uniquely the variational inequality 𝐹𝑥𝑢,𝑥𝑝0,  for all 𝑝𝑇1(0). 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 𝐻 be a real Hilbert space with inner product , and norm . For the sequence {𝑥𝑛} in 𝐻, we write 𝑥𝑛𝑥 to indicate that the sequence {𝑥𝑛} converges weakly to 𝑥. 𝑥𝑛𝑥 means that {𝑥𝑛} converges strongly to 𝑥.

A mapping 𝐹𝐻𝐻 is called 𝑘-Lipschitzian if there exists a positive constant 𝑘 such that 𝐹𝑥𝐹𝑦𝑘𝑥𝑦,𝑥,𝑦𝐻.(2.1)

𝐹 is said to be 𝜂-strongly monotone if there exists a positive constant 𝜂 such that 𝐹𝑥𝐹𝑦,𝑥𝑦𝜂𝑥𝑦2,𝑥,𝑦𝐻.(2.2)

Let 𝐴 be a strongly positive bounded linear operator on 𝐻, that is, there exists a constant ̃𝛾>0 such that 𝐴𝑥,𝑥̃𝛾𝑥2,𝑥𝐻.(2.3)

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 𝐻: min𝑥Fix(𝑊)12,𝐴𝑥,𝑥𝑥,𝑏(2.4)

where 𝑏 is a given point in 𝐻 and Fix(𝑊) is the set of the fixed points of nonexpansive mapping 𝑊.

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

Let 𝑇 be a maximal monotone operator on a real Hilbert space 𝐻 such that 𝑆=𝑇1(0). For 𝑐>0, we use 𝐽𝑇𝑐 to denote the resolvent of 𝑇, that is, 𝐽𝑇𝑐=(𝐼+𝑐𝑇)1.(2.5)

It is well known that 𝐽𝑇𝑐 is firmly nonexpansive and consequently nonexpansive; moreover, 𝑆=Fix(𝐽𝑇𝑐)={𝑥𝐻𝑥=𝐽𝑇𝑐𝑥}.

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

Lemma 2.2 (see [8]). Let 𝑐,𝑡>0. Then, for any 𝑥𝐻, 𝐽𝑇𝑐𝑥=𝐽𝑇𝑡𝑡𝑐𝑡𝑥+1𝑐𝐽𝑇𝑐𝑥.(2.6)

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

Lemma 2.3 (see [11]). Let 𝐹 be a 𝑘-Lipschitzian and 𝜂-strongly monotone operator on a Hilbert space 𝐻 with 0<𝜂𝑘 and 0<𝑡<𝜂/𝑘2. Then, 𝑆=(𝐼𝑡𝐹)𝐻𝐻 is a contraction with contraction coefficient 𝜏𝑡=1𝑡(2𝜂𝑡𝑘2).

Lemma 2.4 (see [12]). 𝑇 is firmly nonexpansive if and only if 2𝑇𝐼 is nonexpansive.

Lemma 2.5 (see [13]). Let 𝐻 be a Hilbert space, 𝐶 a closed convex subset of 𝐻, and 𝑇𝐶𝐶 a nonexpansive mapping with Fix(𝑇); if {𝑥𝑛} is a sequence in 𝐶 weakly converging to 𝑥 and if {(𝐼𝑇)𝑥𝑛} converges strongly to 𝑦, then (𝐼𝑇)𝑥=𝑦.

Lemma 2.6 (see [14]). Let {𝑥𝑛} and {𝑧𝑛} be bounded sequences in Banach space 𝐸 and {𝛾𝑛} a sequence in [0,1] which satisfies the following condition: 0<liminf𝑛𝛾𝑛limsup𝑛𝛾𝑛<1.(2.7) Suppose that 𝑥𝑛+1=𝛾𝑛𝑥𝑛+(1𝛾𝑛)𝑧𝑛, 𝑛0, and limsup𝑛(𝑧𝑛+1𝑧𝑛𝑥𝑛+1𝑥𝑛)0. Then, lim𝑛𝑧𝑛𝑥𝑛=0.

Lemma 2.7 (see [15, 16]). Let {𝑠𝑛} be a sequence of nonnegative real numbers satisfying 𝑠𝑛+11𝜆𝑛𝑠𝑛+𝜆𝑛𝛿𝑛+𝛾𝑛,𝑛0,(2.8) where {𝜆𝑛}, {𝛿𝑛}, and {𝛾𝑛} satisfy the following conditions: (i) {𝜆𝑛}[0,1] and 𝑛=0𝜆𝑛=, (ii) limsup𝑛𝛿𝑛0 or 𝑛=0𝜆𝑛𝛿𝑛<, and (iii) 𝛾𝑛0(𝑛0),𝑛=0𝛾𝑛<. Then, lim𝑛𝑠𝑛=0.

3. Main Results

Let 𝐹 be a 𝑘-Lipschitzian and 𝜂-strongly monotone operator on 𝐻 with 0<𝜂𝑘 and 𝐽𝑇𝑐 the resolvent of 𝑇. Let 𝑡(0,𝜂/𝑘2) and 𝜏𝑡=1𝑡(2𝜂𝑡𝑘2)(0,1), and consider a mapping 𝑉𝑡 on 𝐻 defined by 𝑉𝑡𝑥=𝐽𝑇𝑐[](𝐼𝑡𝐹)𝑥+𝑡𝑢,𝑥𝐻,(3.1)

where 𝑐>0 is a fixed constant and 𝑢𝐻 is a fixed point. It is easy to see that 𝑉𝑡 is a contraction. Indeed, from Lemma 2.3, we have 𝑉𝑡𝑥𝑉𝑡𝑦=𝐽𝑇𝑐[](𝐼𝑡𝐹)𝑥+𝑡𝑢𝐽𝑇𝑐[](𝐼𝑡𝐹)𝑦+𝑡𝑢(𝐼𝑡𝐹)𝑥(𝐼𝑡𝐹)𝑦𝜏𝑡𝑥𝑦,(3.2)

for all 𝑥,𝑦𝐻. Hence, it has a unique fixed point, denoted by 𝑣𝑡, which uniquely solves the fixed point equation 𝑣𝑡=𝐽𝑇𝑐(𝐼𝑡𝐹)𝑣𝑡+𝑡𝑢,𝑣𝑡𝐻.(3.3)

Theorem 3.1. For any 𝑐>0 and 𝑢𝐻, let the net {𝑣𝑡} be generated by (3.3). Then, as 𝑡0, the net {𝑣𝑡} converges strongly to 𝑣 of 𝑆, which solves uniquely the variational inequality 𝐹𝑣𝑢,𝑣𝑝0,𝑝𝑆.(3.4)

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 𝐹. Suppose 𝑣𝑆 and ̃𝑣𝑆 both are solutions to (3.4); then, 𝐹𝑣𝑢,𝑣̃̃̃𝑣0,(3.5)𝐹𝑣𝑢,𝑣𝑣0.(3.6)
Adding (3.5) to (3.6), we get 𝐹𝑣̃𝐹𝑣,𝑣̃𝑣0.(3.7)
The strong monotonicity of 𝐹 implies that 𝑣=̃𝑣 and the uniqueness is proved. Below we use 𝑣𝑆 to denote the unique solution of (3.4). Next, we prove that {𝑣𝑡} is bounded. Taking 𝑝𝑆, from (3.3) and using Lemma 2.3, we have 𝑣𝑡=𝐽𝑝𝑇𝑐(𝐼𝑡𝐹)𝑣𝑡+𝑡𝑢𝑝(𝐼𝑡𝐹)𝑣𝑡(𝐼𝑡𝐹)𝑝+𝑡(𝑢𝐹𝑝)𝜏𝑡𝑣𝑡𝑝+𝑡𝑢𝐹𝑝,(3.8)
that is, 𝑣𝑡𝑡𝑝1𝜏𝑡𝑢𝐹𝑝.(3.9)
Observe that lim𝑡0+𝑡1𝜏𝑡=1𝜂.(3.10)
From 𝑡0, we may assume, without loss of generality, that 𝑡𝜂/𝑘2𝜖, where 𝜖 is an arbitrarily small positive number. Thus, we have that 𝑡/(1𝜏𝑡) is continuous, for all 𝑡[0,𝜂/𝑘2𝜖]. Therefore, we obtain 𝑡sup1𝜏𝑡𝜂𝑡0,𝑘2𝜖<+.(3.11) From (3.9) and (3.11), we have {𝑣𝑡} bounded and so is {𝐹𝑣𝑡}. On the other hand, from (3.3), we obtain 𝑣𝑡𝐽𝑇𝑐𝑣𝑡=𝐽𝑇𝑐(𝐼𝑡𝐹)𝑣𝑡+𝑡𝑢𝐽𝑇𝑐𝑣𝑡(𝐼𝑡𝐹)𝑣𝑡+𝑡𝑢𝑣𝑡=𝑡𝑢𝐹𝑣𝑡0(𝑡0).(3.12) To prove that 𝑣𝑡𝑣, for a given  𝑝𝑆, using Lemma 2.3, we have 𝑣𝑡𝑝2=𝐽𝑇𝑐(𝐼𝑡𝐹)𝑣𝑡+𝑡𝑢𝑝2(𝐼𝑡𝐹)𝑣𝑡(𝐼𝑡𝐹)𝑝+𝑡(𝑢𝐹𝑝)2𝜏2𝑡𝑣𝑡𝑝2+𝑡2𝑢𝐹𝑝2+2𝑡(𝐼𝑡𝐹)𝑣𝑡(𝐼𝑡𝐹)𝑝,𝑢𝐹𝑝𝜏𝑡𝑣𝑡𝑝2+𝑡2𝑢𝐹𝑝2+2𝑡𝑣𝑡𝑝,𝑢𝐹𝑝+2𝑡2𝐹𝑝𝐹𝑣𝑡,𝑢𝐹𝑝𝜏𝑡𝑣𝑡𝑝2+𝑡2𝑢𝐹𝑝2+2𝑡𝑣𝑡𝑝,𝑢𝐹𝑝+2𝑡2𝑘𝑝𝑣𝑡𝑢𝐹𝑝𝜏𝑡𝑣𝑡𝑝2+2𝑡2𝑀+2𝑡𝑣𝑡𝑝,𝑢𝐹𝑝,(3.13) where 𝑀=max{𝑢𝐹𝑝2,2𝑘𝑝𝑣𝑡𝑢𝐹𝑝}. Therefore, 𝑣𝑡𝑝22𝑡21𝜏𝑡𝑀+2𝑡1𝜏𝑡𝑣𝑡𝑝,𝑢𝐹𝑝.(3.14) From 𝜏𝑡=1𝑡(2𝜂𝑡𝑘2), we have lim𝑡0(𝑡2/(1𝜏𝑡))=0. Moreover, if 𝑣𝑡𝑝, we have lim𝑡0((2𝑡/(1𝜏𝑡))𝑣𝑡𝑝,𝑢𝐹𝑝)=0.
Since {𝑣𝑡} is bounded, we see that if {𝑡𝑛} is a sequence in (0,𝜂/𝑘2𝜖] such that 𝑡𝑛0 and 𝑣𝑡𝑛̃𝑣, then, by (3.14), we see that 𝑣𝑡𝑛̃𝑣. Moreover, by (3.12) and using Lemma 2.5, we have ̃𝑣𝑆. We next prove that ̃𝑣 solves the variational inequality (3.4). From (3.3) and 𝑝𝑆, we have 𝑣𝑡𝑝2(𝐼𝑡𝐹)𝑣𝑡+𝑡𝑢𝑝2=𝑣𝑡𝑝2+𝑡2𝑢𝐹𝑣𝑡2+2𝑡𝑣𝑡𝑝,𝑢𝐹𝑣𝑡,(3.15) that is, 𝐹𝑣𝑡𝑢,𝑣𝑡𝑡𝑝2𝑢𝐹𝑣𝑡2.(3.16) Now replacing 𝑡 in (3.16) with 𝑡𝑛 and letting 𝑛, we have ̃̃𝐹𝑣𝑢,𝑣𝑝0.(3.17) That is, ̃𝑣𝑆 is a solution of (3.4), and hence ̃𝑣=𝑣 by uniqueness. In a summary, we have shown that each cluster point of {𝑣𝑡} (at 𝑡0) equals 𝑣. Therefore, 𝑣𝑡𝑣 as 𝑡0.

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

Corollary 3.2. For any 𝑐>0 and 𝑢𝐻, let 𝐴 be a strongly positive bounded linear operator with coefficient 0<̃𝛾𝐴. For each 𝑡(0,̃𝛾/𝐴2), let the net {𝑣𝑡} be generated by 𝑣𝑡=𝐽𝑇𝑐[(𝐼𝑡𝐴)𝑣𝑡+𝑡𝑢]. Then, as 𝑡0, the net {𝑣𝑡} converges strongly to 𝑣 of 𝑆 which solves uniquely the variational inequality 𝐴𝑣𝑢,𝑣𝑝0,𝑝𝑆.(3.18)

Setting 𝐹=𝐼 and 𝑣=𝑃𝑆𝑢 in Theorem 3.1, we can obtain the following result.

Corollary 3.3 (Xu [8, Theorem 3.1]). For any 𝑐>0 and 𝑢𝐻. For each 𝑡(0,1), let the net {𝑣𝑡} be generated by 𝑣𝑡=𝐽𝑇𝑐[(1𝑡)𝑣𝑡+𝑡𝑢]. Then, as 𝑡0, {𝑣𝑡} converges strongly to the projection of 𝑢 onto 𝑆; that is, lim𝑡0𝑣𝑡=𝑃𝑆𝑢. Moreover, this limit is attained uniformly for 𝑐>0.

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

Theorem 3.4. Let 𝑇 be a maximal monotone operator on a Hilbert space 𝐻 with 𝑆. Let 𝐹 be a 𝑘-Lipschitzian and 𝜂-strongly monotone operator on 𝐻 with 0<𝜂𝑘. Let {𝑡𝑛} be a sequence in (0,1), {𝑐𝑛} a sequence in (0,+), and 𝜖 an arbitrarily small positive number. Assume that the control conditions (C1), (C3), (C4), and (C5) hold for {𝑡𝑛}, {𝑐𝑛}, and {𝑒𝑛}(C1)0<𝑡𝑛𝜂/𝑘2𝜖, forall𝑛𝑛0 for some integer 𝑛00.
For an arbitrary point 𝑥0𝐻, let the sequence {𝑥𝑛} be generated by (1.6). Then, 𝑧𝑛𝑥𝑡𝑛𝑢𝐹𝑥𝑛0(𝑛),(3.19) where 𝑥𝑆 solves the variational inequality 𝐹𝑥𝑢,𝑥𝑝0,𝑝𝑆.(3.20)

Proof. On the one hand, suppose that 𝑡𝑛(𝑢𝐹𝑥𝑛)0(𝑛). We proceed with the following steps.
Step 1. We claim that {𝑥𝑛} is bounded. In fact, taking 𝑝𝑆, from (1.6) and (C1) and using Lemma 2.3, we have 𝑥𝑛+1=𝐽𝑝𝑇𝑐𝑛𝑧𝑛𝑝𝐼𝑡𝑛𝐹𝑥𝑛+𝑡𝑛𝑢+𝑒𝑛𝑝𝐼𝑡𝑛𝐹𝑥𝑛𝐼𝑡𝑛𝐹𝑝+𝑡𝑛(𝑢𝐹𝑝)+𝑒𝑛𝜏𝑡𝑛𝑥𝑛𝑝+𝑡𝑛𝑒𝑢𝐹𝑝+𝑛11𝜏𝑡𝑛𝑥𝑛+𝑝1𝜏𝑡𝑛𝑡𝑛1𝜏𝑡𝑛𝑒𝑢𝐹𝑝+𝑛𝑥max𝑛,𝑡𝑝𝑛1𝜏𝑡𝑛+𝑒𝑢𝐹𝑝𝑛,(3.21) for all𝑛𝑛0 for some integer 𝑛00, where 𝜏𝑡𝑛=1𝑡𝑛(2𝜂𝑡𝑛𝑘2)(0,1). By induction, we have 𝑥𝑛𝑥𝑝max0𝑝,𝑢𝐹𝑝𝑀1+𝑛1𝑗=0𝑒𝑗,(3.22) for all𝑛𝑛0 for some integer 𝑛00, where 𝑀1=sup{𝑡𝑛/(1𝜏𝑡𝑛)0<𝑡𝑛𝜂/𝑘2𝜖}<+. Therefore, {𝑥𝑛} is bounded. We also obtain that {𝑧𝑛} and {𝐹𝑥𝑛} are bounded.
Step 2. We claim that lim𝑛𝑥𝑛+1𝑥𝑛=0. In fact, write 𝐽𝑛=𝐽𝑇𝑐𝑛 and 𝑇𝑛=2𝐽𝑛𝐼. Then, 𝐽𝑛 is firmly nonexpansive and 𝑇𝑛 is nonexpansive (see Lemma 2.4).
Observe that 𝑥𝑛+1=𝐽𝑛𝑧𝑛=𝐼+𝑇𝑛2𝑧𝑛=12𝑧𝑛+12𝑇𝑛𝑧𝑛=12𝑥𝑛+12𝑡𝑛𝑢𝐹𝑥𝑛+𝑒𝑛+𝑇𝑛𝑧𝑛=12𝑥𝑛+12𝑦𝑛,(3.23) where 𝑦𝑛=𝑡𝑛(𝑢𝐹𝑥𝑛)+𝑒𝑛+𝑇𝑛𝑧𝑛. Therefore, 𝑦𝑛+1𝑦𝑛=𝑡𝑛+1𝑢𝐹𝑥𝑛+1+𝑒𝑛+1+𝑇𝑛+1𝑧𝑛+1𝑡𝑛𝑢𝐹𝑥𝑛𝑒𝑛𝑇𝑛𝑧𝑛𝑡𝑛+1𝑢𝐹𝑥𝑛+1+𝑡𝑛𝑢𝐹𝑥𝑛+𝑒𝑛+1+𝑒𝑛+𝑇𝑛+1𝑧𝑛+1𝑇𝑛𝑧𝑛.(3.24)
It follows from the resolvent identity that 𝑇𝑛+1𝑥𝑇𝑛𝑥𝐽=2𝑛+1𝑥𝐽𝑛𝑥𝐽=2𝑛𝑐𝑛𝑐𝑛+1𝑐𝑥+1𝑛𝑐𝑛+1𝐽𝑛+1𝑥𝐽𝑛𝑥||||𝑐21𝑛𝑐𝑛+1||||𝐽𝑛+1||||𝑐𝑥𝑥1𝑛𝑐𝑛+1||||𝑇𝑛+1𝑥𝑥(3.25) for any 𝑥𝐻. From (1.6), we get 𝑧𝑛+1𝑧𝑛=𝐼𝑡𝑛+1𝐹𝑥𝑛+1+𝑡𝑛+1𝑢+𝑒𝑛+1𝐼𝑡𝑛𝐹𝑥𝑛𝑡𝑛𝑢𝑒𝑛𝑥𝑛+1𝑥𝑛+𝑡𝑛+1𝑢𝐹𝑥𝑛+1+𝑡𝑛𝑢𝐹𝑥𝑛+𝑒𝑛+1+𝑒𝑛.(3.26) By (3.25) and (3.26), we have 𝑇𝑛+1𝑧𝑛+1𝑇𝑛𝑧𝑛𝑇𝑛+1𝑧𝑛+1𝑇𝑛𝑧𝑛+1+𝑇𝑛𝑧𝑛+1𝑇𝑛𝑧𝑛||||𝑐1𝑛𝑐𝑛+1||||𝑇𝑛+1𝑧𝑛+1𝑧𝑛+1+𝑧𝑛+1𝑧𝑛||||𝑐1𝑛𝑐𝑛+1||||𝑇𝑛+1𝑧𝑛+1𝑧𝑛+1+𝑥𝑛+1𝑥𝑛+𝑡𝑛+1𝑢𝐹𝑥𝑛+1+𝑡𝑛𝑢𝐹𝑥𝑛+𝑒𝑛+1+𝑒𝑛.(3.27) Substituting (3.27) into (3.24) at once gives 𝑦𝑛+1𝑦𝑛𝑡2𝑛+1𝑢𝐹𝑥𝑛+1𝑡+2𝑛𝑢𝐹𝑥𝑛𝑒+2𝑛+1𝑒+2𝑛+||||𝑐1𝑛𝑐𝑛+1||||𝑀2+𝑥𝑛+1𝑥𝑛,(3.28) that is, 𝑦𝑛+1𝑦𝑛𝑥𝑛+1𝑥𝑛𝑡2𝑛+1𝑢𝐹𝑥𝑛+1𝑡+2𝑛𝑢𝐹𝑥𝑛𝑒+2𝑛+1𝑒+2𝑛+||||𝑐1𝑛𝑐𝑛+1||||𝑀2,(3.29) where 𝑀2=sup{𝑇𝑛+1𝑧𝑛+1𝑧𝑛+1,𝑛0}. Observing 𝑡𝑛(𝑢𝐹𝑥𝑛)0, 𝑒𝑛0, and |1(𝑐𝑛/𝑐𝑛+1)|0(𝑛), it follows that limsup𝑛𝑦𝑛+1𝑦𝑛𝑥𝑛+1𝑥𝑛0.(3.30) From (3.23) and using Lemma 2.6, we have lim𝑛𝑦𝑛𝑥𝑛=0. Therefore, lim𝑛𝑥𝑛+1𝑥𝑛=lim𝑛12𝑦𝑛𝑥𝑛=0.(3.31)Step 3. We claim that lim𝑛𝑥𝑛𝐽𝑇𝑐𝑥𝑛=0. Since liminf𝑛𝑐𝑛>0, there exist 𝛼>0 and a positive integer 𝑁 such that for all 𝑛𝑁, 𝑐𝑛𝛼. From Lemma 2.2, for each 𝑐(0,𝛼), we have 𝐽𝑛𝑥𝑛𝐽𝑇𝑐𝑥𝑛=𝐽𝑇𝑐𝑐𝑐𝑛𝑥𝑛+𝑐1𝑐𝑛𝐽𝑛𝑥𝑛𝐽𝑇𝑐𝑥𝑛𝑐𝑐𝑛𝑥𝑛+𝑐1𝑐𝑛𝐽𝑛𝑥𝑛𝑥𝑛=||||𝑐1𝑐𝑛||||𝐽𝑛𝑥𝑛𝑥𝑛𝐽𝑛𝑥𝑛𝑥𝑛+1+𝑥𝑛+1𝑥𝑛.(3.32) Observe that 𝐽𝑛𝑥𝑛𝑥𝑛+1𝑥𝑛𝑧𝑛𝑡𝑛𝑢𝐹𝑥𝑛+𝑒𝑛0.(3.33) Thus, it follows from (3.32), (3.33), and Step2 that lim𝑛𝐽𝑛𝑥𝑛𝐽𝑇𝑐𝑥𝑛=0.(3.34) Since 𝑥𝑛𝐽𝑇𝑐𝑥𝑛𝑥𝑛𝑥𝑛+1+𝑥𝑛+1𝐽𝑛𝑥𝑛+𝐽𝑛𝑥𝑛𝐽𝑇𝑐𝑥𝑛, then lim𝑛𝑥𝑛𝐽𝑇𝑐𝑥𝑛=0.(3.35)Step 4. We claim that limsup𝑛𝑥𝑛𝑥,𝑢𝐹𝑥0, where 𝑥=lim𝑡0𝑣𝑡 and 𝑣𝑡 is defined by (3.3). Since 𝑥𝑛 is bounded, there exists a subsequence {𝑥𝑛𝑘} of {𝑥𝑛} which converges weakly to 𝜔. From Step3, we obtain 𝐽𝑇𝑐𝑥𝑛𝜔. From Lemma 2.5, we have 𝜔𝑆. Hence, by Theorem 3.1, we have limsup𝑛𝑥𝑛𝑥,𝑢𝐹𝑥=lim𝑘𝑥𝑛𝑘𝑥,𝑢𝐹𝑥=𝜔𝑥,𝑢𝐹𝑥0.(3.36)Step 5. We claim that {𝑧𝑛} converges strongly to 𝑥𝑆. From (1.6), for an appropriate constant 𝛾>0, we have𝑥𝑛+1𝑥2=𝐽𝑛𝑧𝑛𝑥2𝑧𝑛𝑥2=(𝐼𝑡𝑛𝐹)𝑥𝑛+𝑡𝑛𝑢+𝑒𝑛𝑥2(𝐼𝑡𝑛𝐹)𝑥𝑛+𝑡𝑛𝑢𝑥2𝑒+𝛾𝑛𝐼𝑡𝑛𝐹𝑥𝑛𝐼𝑡𝑛𝐹𝑥+𝑡𝑛𝑢𝐹𝑥2𝑒+𝛾𝑛𝜏2𝑡𝑛𝑥𝑛𝑥2+𝑡2𝑛𝑢𝐹𝑥2+2𝑡𝑛𝐼𝑡𝑛𝐹𝑥𝑛𝐼𝑡𝑛𝐹𝑥,𝑢𝐹𝑥𝑒+𝛾𝑛𝜏𝑡𝑛𝑥𝑛𝑥2+𝑡2𝑛𝑢𝐹𝑥2+2𝑡𝑛𝑥𝑛𝑥𝑡𝑛𝐹𝑥𝑛+𝑡𝑛𝑢,𝑢𝐹𝑥+2𝑡2𝑛𝐹𝑥𝑢,𝑢𝐹𝑥𝑒+𝛾𝑛𝜏𝑡𝑛𝑥𝑛𝑥2+2𝑡𝑛𝑥𝑛𝑥,𝑢𝐹𝑥+2𝑡𝑛𝑡𝑛𝑢𝐹𝑥𝑛𝑢𝐹𝑥𝑒+𝛾𝑛11𝜏𝑡𝑛𝑥𝑛𝑥2+1𝜏𝑡𝑛2𝑀1𝑥𝑛𝑥,𝑢𝐹𝑥+2𝑀1𝑡𝑛𝑢𝐹𝑥𝑛𝑢𝐹𝑥𝑒+𝛾𝑛,(3.37) for all 𝑛𝑛0 for some integer 𝑛00. For every 𝑛𝑛0, put 𝜇𝑛=1𝜏𝑡𝑛 and 𝛿𝑛=2𝑀1𝑥𝑛𝑥,𝑢𝐹𝑥+2𝑀1𝑡𝑛(𝑢𝐹𝑥𝑛)𝑢𝐹𝑥. It follows that 𝑥𝑛+1𝑥21𝜇𝑛𝑥𝑛𝑥2+𝜇𝑛𝛿𝑛𝑒+𝛾𝑛,𝑛𝑛0.(3.38) It is easy to see that 𝑛=1𝜇𝑛= and limsup𝑛𝛿𝑛0. Hence, by Lemma 2.7, the sequence {𝑥𝑛} converges strongly to 𝑥𝑆. Observe that 𝑧𝑛𝑥=𝐼𝑡𝑛𝐹𝑥𝑛+𝑡𝑛𝑢+𝑒𝑛𝑥𝑥𝑛𝑥+𝑡𝑛𝑢𝐹𝑥𝑛+𝑒𝑛.(3.39) Thus, it follows that the sequence {𝑧𝑛} converges strongly to 𝑥𝑆.
On the other hand, suppose that 𝑧𝑛𝑥𝑆 as 𝑛. From (1.6), we have 𝑥𝑛+1𝑥=𝐽𝑛𝑧𝑛𝑥𝑧𝑛𝑥0.(3.40) Therefore, 𝑡𝑛𝑢𝐹𝑥𝑛=𝑧𝑛𝑥𝑛𝑒𝑛𝑧𝑛𝑥𝑛+𝑒𝑛𝑧𝑛𝑥+𝑥𝑛𝑥+𝑒𝑛0.(3.41)

Setting 𝐹=𝐼 and 𝑥=𝑃𝑆𝑢 in Theorem 3.4, we can obtain the following result.

Corollary 3.5. Let 𝑇 be a maximal monotone operator on a Hilbert space 𝐻 with 𝑆. Let {𝑡𝑛} be a sequence in (0,1), {𝑐𝑛} a sequence in (0,+), and 𝜖 an arbitrarily small positive number. Assume that the control conditions (C1), (C3), (C4), and (C5) hold for {𝑡𝑛}, {𝑐𝑛}, and {𝑒𝑛}.(C1)0<𝑡𝑛1𝜖, for all 𝑛𝑛0 for some integer 𝑛00.
For an arbitrary point 𝑥0𝐻, let the sequence {𝑥𝑛} be generated by 𝑧𝑛=1𝑡𝑛𝑥𝑛+𝑡𝑛𝑢+𝑒𝑛,𝑥𝑛+1=𝐽𝑇𝑐𝑛𝑧𝑛,𝑛0.(3.42) Then, 𝑧𝑛𝑃𝑆𝑢𝑡𝑛𝑢𝑥𝑛0(𝑛).(3.43)

Corollary 3.6 ([Wang [10], Theorem  4]). Let {𝑐𝑛}, {𝑡𝑛}, and {𝑒𝑛} satisfy (C1), (C3),(or (C3)), (C4) and (C5). In addition, if 𝑆, then the sequence generated by (1.5) converges strongly to 𝑃𝑆𝑢.

Proof. Since lim𝑛𝑡𝑛=0, it is easy to see that 𝑡𝑛𝜂/𝑘2𝜖, for all 𝑛𝑛0 for some integer 𝑛00. Without loss of generality, we assume that 0<𝑡𝑛𝜂/𝑘2𝜖, for all 𝑛𝑛0 for some integer 𝑛00. Repeating the same argument as in the proof of Theorem  4 in Wang [10], we know that {𝑥𝑛} is bounded. Thus, we have that 𝑡𝑛(𝑢𝑥𝑛)0. Therefore, all conditions of Corollary 3.5 are satisfied. Using Corollary 3.5, we have that {𝑧𝑛} converges strongly to 𝑃𝑆𝑢𝑆, with 𝑧𝑛=(1𝑡𝑛)𝑥𝑛+𝑡𝑛𝑢+𝑒𝑛. Therefore, 𝑥𝑛+1𝑃𝑆𝑢𝑧𝑛𝑃𝑆𝑢0.(3.44)

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 𝐻=𝑅  be the set of real numbers, 𝑢=0 and 𝑐𝑛=1/2 for all 𝑛0. Define a maximal monotone operator 𝑇 as follows: 𝑇𝑥=2𝑥, for all 𝑥𝑅. It is easy to see that 𝐽𝑇𝑐𝑛=(1/2)𝐼 and 𝑆={0}. Given sequences {𝑡𝑛} and {𝑒𝑛}, 𝑡𝑛=1/2 and 𝑒𝑛=0, for all 𝑛0. For an arbitrary 𝑥0𝑅, let {𝑥𝑛} be defined by (3.42), that is, 𝑧𝑛=12𝑥𝑛,𝑥𝑛+1=12𝑧𝑛=14𝑥𝑛,𝑛0.(3.45) Observe that 𝑥𝑛+1=104𝑥𝑛=104𝑥𝑛.0(3.46) Hence, we have 𝑥𝑛+10=(1/4)𝑛+1𝑥00 for all 𝑛0. This implies that {𝑥𝑛} converges strongly to 0=𝑃𝑆0. Thus, 𝑡𝑛𝑢𝑥𝑛=12𝑥𝑛0(𝑛).(3.47) Furthermore, it is easy to see that there hold the following:(B1)0<𝑡𝑛=1/21𝜖, forall𝑛𝑛0 for some integer 𝑛00,(B2)𝑛=0𝑡𝑛=𝑛=0(1/2)=,(B3)liminf𝑛𝑐𝑛=1/2>0 and lim𝑛|1(𝑐𝑛/𝑐𝑛+1)|=0,(B4)𝑛=0𝑒𝑛=𝑛=00=0<.

Hence there is no doubt that all conditions of Corollary 3.5 are satisfied. Since 𝑡𝑛=1/20, the condition 𝑡𝑛0 of Wang [10, Theorem   4] is not satisfied. So, by Corollary 3.5, we obtain that the sequence {𝑥𝑛} and {𝑧𝑛} converges strongly to zero but Theorem  4 of Wang [10] cannot be applied to {𝑥𝑛} and {𝑧𝑛} in this example.

Acknowledgment

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

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