# A Modified Regularization Method for the Proximal Point Algorithm

Authors: Shuang Wang
Journal: Journal of Applied Mathematics
 (2012)

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## 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.


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## Body

## 1. Introduction

LetH be a real Hilbert space and C a nonempty closed convex subset of H, and let F:H→H be a nonlinear operator. The variational inequality problem is formulated as finding a point x*∈C such that
(1.1)〈Fx*,v-x*〉≥0,∀v∈C.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 [1–5].LetT be an operator with domain D(T) and range R(T) in H. A multivalued operator T:H→2H is called monotone if
(1.2)〈u-v,x-y〉≥0,
for any u∈Tx,v∈Ty, and maximal monotone if it is monotone and its graph
(1.3)G(T)={(x,y):x∈D(T),y∈Tx}
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 pointx so that x∈T-1(0), where T-1(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
(1.4)xn+1=JcT(xn+en),where, forc>0, JcT denotes the resolvent of T given by JcT=(I+cT)-1, with I being the identity map on the space H. If T-1(0)≠∅, it is known that the sequence generated by (1.4) converges weakly to some point in T-1(0).Motivated by Lehdili and Moudafi's prox-Tikhonov method [7], Xu [8] considered the following regularization iterative form: for a fixed point u∈H,
(1.5)xn+1=JcnT((1-tn)xn+tnu+en),n≥0,wheretn∈(0,1) and {en} is a sequence of errors. Then, the iterative sequence converges strongly to PT-1(0)u, provided that(C1)
limn→∞tn=0,(C2)
∑n=0∞‍|tn+1-tn|<∞,(C3)
0<c_≤cn≤c¯,(C4)
∑n=0∞‍|cn+1-cn|<∞,(C5)
∑n=0∞‍tn=∞, ∑n=0∞‍∥en∥<∞.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<liminfn→∞cn,they proved that the sequence generated by (1.5) converges strongly to PT-1(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., tn→0) 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 x*∈T-1(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 arbitraryx0∈H,
(1.6)zn=(I-tnF)xn+tnu+en,xn+1=JcnTzn,n≥0,whereF 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 x*∈T-1(0), which solves uniquely the variational inequality 〈Fx*-u,x*-p〉≤0,  for all p∈T-1(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 [6–10].


## 2. Preliminaries

LetH be a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥. For the sequence {xn} in H, we write xn⇀x to indicate that the sequence {xn} converges weakly to x. xn→x means that {xn} converges strongly to x.A mappingF:H→H is called k-Lipschitzian if there exists a positive constant k such that
(2.1)∥Fx-Fy∥≤k∥x-y∥,∀x,y∈H.F is said to be η-strongly monotone if there exists a positive constant η such that
(2.2)〈Fx-Fy,x-y〉≥η∥x-y∥2,∀x,y∈H.LetA be a strongly positive bounded linear operator on H, that is, there exists a constant γ~>0 such that
(2.3)〈Ax,x〉≥γ~∥x∥2,∀x∈H.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 spaceH:
(2.4)minx∈Fix(W){12〈Ax,x〉-〈x,b〉},whereb 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 ofA, we note that a strongly positive bounded linear operator A is a ∥A∥-Lipschitzian and γ~-strongly monotone operator.LetT be a maximal monotone operator on a real Hilbert space H such that S:=T-1(0)≠∅. For c>0, we use JcT to denote the resolvent of T, that is,
(2.5)JcT=(I+cT)-1.It is well known thatJcT is firmly nonexpansive and consequently nonexpansive; moreover, S=Fix(JcT)={x∈H:x=JcTx}.The following lemma is known as the resolvent identity of maximal monotone operators.Lemma 2.2 (see [8]).
Letc,t>0. Then, for any x∈H,
(2.6)JcTx=JtT(tcx+(1-tc)JcTx).In order to prove our main results, we need the following lemmas.Lemma 2.3 (see [11]).
LetF be a k-Lipschitzian and η-strongly monotone operator on a Hilbert space H with 0<η≤k and 0<t<η/k2. Then, S=(I-tF):H→H is a contraction with contraction coefficient τt=1-t(2η-tk2).Lemma 2.4 (see [12]).
T is firmly nonexpansive if and only if 2T-I is nonexpansive.Lemma 2.5 (see [13]).
LetH be a Hilbert space, C a closed convex subset of H, and T:C→C a nonexpansive mapping with Fix(T)≠∅; if {xn} is a sequence in C weakly converging to x and if {(I-T)xn} converges strongly to y, then (I-T)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:
(2.7)0<liminfn→∞γn≤limsupn→∞γn<1.
Suppose that xn+1=γnxn+(1-γn)zn, n≥0, and limsupn→∞(∥zn+1-zn∥-∥xn+1-xn∥)≤0. Then, limn→∞∥zn-xn∥=0.Lemma 2.7 (see [15, 16]).
Let{sn} be a sequence of nonnegative real numbers satisfying
(2.8)sn+1≤(1-λn)sn+λnδn+γn,n≥0,
where {λn}, {δn}, and {γn} satisfy the following conditions: (i) {λn}⊂[0,1] and ∑n=0∞‍λn=∞, (ii) limsupn→∞δn≤0 or ∑n=0∞‍λnδn<∞, and (iii) γn≥0(n≥0),∑n=0∞‍γn<∞. Then, limn→∞sn=0.


## 3. Main Results

LetF be a k-Lipschitzian and η-strongly monotone operator on H with 0<η≤k and JcT the resolvent of T. Let t∈(0,η/k2) and τt=1-t(2η-tk2)∈(0,1), and consider a mapping Vt on H defined by
(3.1)Vtx=JcT[(I-tF)x+tu],x∈H,wherec>0 is a fixed constant and u∈H is a fixed point. It is easy to see that Vt is a contraction. Indeed, from Lemma 2.3, we have
(3.2)∥Vtx-Vty∥=∥JcT[(I-tF)x+tu]-JcT[(I-tF)y+tu]∥≤∥(I-tF)x-(I-tF)y∥≤τt∥x-y∥,for allx,y∈H. Hence, it has a unique fixed point, denoted by vt, which uniquely solves the fixed point equation
(3.3)vt=JcT[(I-tF)vt+tu],vt∈H.Theorem 3.1.
For anyc>0 and u∈H, let the net {vt} be generated by (3.3). Then, as t→0, the net {vt} converges strongly to v* of S, which solves uniquely the variational inequality
(3.4)〈Fv*-u,v*-p〉≤0,∀p∈S.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 v*∈S and v~∈S both are solutions to (3.4); then,
(3.5)〈Fv*-u,v*-v~〉≤0,(3.6)〈Fv~-u,v~-v*〉≤0.
Adding (3.5) to (3.6), we get
(3.7)〈Fv*-Fv~,v*-v~〉≤0.
The strong monotonicity ofF implies that v*=v~ and the uniqueness is proved. Below we use v*∈S to denote the unique solution of (3.4). Next, we prove that {vt} is bounded. Taking p∈S, from (3.3) and using Lemma 2.3, we have
(3.8)∥vt-p∥=∥JcT[(I-tF)vt+tu]-p∥≤∥(I-tF)vt-(I-tF)p+t(u-Fp)∥≤τt∥vt-p∥+t∥u-Fp∥,
that is,(3.9)∥vt-p∥≤t1-τt∥u-Fp∥.
Observe that(3.10)limt→0+t1-τt=1η.
Fromt→0, we may assume, without loss of generality, that t≤η/k2-ϵ, where ϵ is an arbitrarily small positive number. Thus, we have that t/(1-τt) is continuous, for all t∈[0,η/k2-ϵ]. Therefore, we obtain
(3.11)sup{t1-τt:t∈(0,ηk2-ϵ]}<+∞.
From (3.9) and (3.11), we have {vt} bounded and so is {Fvt}. On the other hand, from (3.3), we obtain
(3.12)∥vt-JcTvt∥=∥JcT[(I-tF)vt+tu]-JcTvt∥≤∥(I-tF)vt+tu-vt∥=t∥u-Fvt∥→0(t→0).
To prove that vt→v*, for a given  p∈S, using Lemma 2.3, we have
(3.13)∥vt-p∥2=∥JcT[(I-tF)vt+tu]-p∥2≤∥(I-tF)vt-(I-tF)p+t(u-Fp)∥2≤τt2∥vt-p∥2+t2∥u-Fp∥2+2t〈(I-tF)vt-(I-tF)p,u-Fp〉≤τt∥vt-p∥2+t2∥u-Fp∥2+2t〈vt-p,u-Fp〉+2t2〈Fp-Fvt,u-Fp〉≤τt∥vt-p∥2+t2∥u-Fp∥2+2t〈vt-p,u-Fp〉+2t2k∥p-vt∥∥u-Fp∥≤τt∥vt-p∥2+2t2M+2t〈vt-p,u-Fp〉,
where M=max{∥u-Fp∥2,2k∥p-vt∥∥u-Fp∥}. Therefore,
(3.14)∥vt-p∥2≤2t21-τtM+2t1-τt〈vt-p,u-Fp〉.
From τt=1-t(2η-tk2), we have limt→0(t2/(1-τt))=0. Moreover, if vt⇀p, we have limt→0((2t/(1-τt))〈vt-p,u-Fp〉)=0.
Since{vt} is bounded, we see that if {tn} is a sequence in (0,η/k2-ϵ] such that tn→0 and vtn⇀v~, then, by (3.14), we see that vtn→v~. Moreover, by (3.12) and using Lemma 2.5, we have v~∈S. We next prove that v~ solves the variational inequality (3.4). From (3.3) and p∈S, we have
(3.15)∥vt-p∥2≤∥(I-tF)vt+tu-p∥2=∥vt-p∥2+t2∥u-Fvt∥2+2t〈vt-p,u-Fvt〉,
that is,
(3.16)〈Fvt-u,vt-p〉≤t2∥u-Fvt∥2.
Now replacing t in (3.16) with tn and letting n→∞, we have
(3.17)〈Fv~-u,v~-p〉≤0.
That is, v~∈S 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 t→0) equals v*. Therefore, vt→v* as t→0.SettingF=A in Theorem 3.1, we can obtain the following result.Corollary 3.2.
For anyc>0 and u∈H, let A be a strongly positive bounded linear operator with coefficient 0<γ~≤∥A∥. For each t∈(0,γ~/∥A∥2), let the net {vt} be generated by vt=JcT[(I-tA)vt+tu]. Then, as t→0, the net {vt} converges strongly to v* of S which solves uniquely the variational inequality
(3.18)〈Av*-u,v*-p〉≤0,∀p∈S.SettingF=I and v*=PSu in Theorem 3.1, we can obtain the following result.Corollary 3.3 (Xu [8, Theorem 3.1]).
For anyc>0 and u∈H. For each t∈(0,1), let the net {vt} be generated by vt=JcT[(1-t)vt+tu]. Then, as t→0, {vt} converges strongly to the projection of u onto S; that is, limt→0vt=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.
LetT 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<η≤k. 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<tn≤η/k2-ϵ, foralln≥n0 for some integer n0≥0.
For an arbitrary pointx0∈H, let the sequence {xn} be generated by (1.6). Then,
(3.19)zn→x*⟺tn(u-Fxn)→0(n→∞),
where x*∈S solves the variational inequality
(3.20)〈Fx*-u,x*-p〉≤0,∀p∈S.Proof.
On the one hand, suppose thattn(u-Fxn)→0(n→∞). We proceed with the following steps.
Step
1. We claim that {xn} is bounded. In fact, taking p∈S, from (1.6) and  (C1′) and using Lemma 2.3, we have
(3.21)∥xn+1-p∥=∥JcnTzn-p∥≤∥(I-tnF)xn+tnu+en-p∥≤∥(I-tnF)xn-(I-tnF)p+tn(u-Fp)+en∥≤τtn∥xn-p∥+tn∥u-Fp∥+∥en∥≤[1-(1-τtn)]∥xn-p∥+(1-τtn)tn1-τtn∥u-Fp∥+∥en∥≤max{∥xn-p∥,tn1-τtn∥u-Fp∥}+∥en∥,
for alln≥n0 for some integer n0≥0, where τtn=1-tn(2η-tnk2)∈(0,1). By induction, we have
(3.22)∥xn-p∥≤max{∥x0-p∥,∥u-Fp∥M1}+∑j=0n-1‍∥ej∥,
for alln≥n0 for some integer n0≥0, where M1=sup{tn/(1-τtn):0<tn≤η/k2-ϵ}<+∞. Therefore, {xn} is bounded. We also obtain that {zn} and {Fxn} are bounded.
Step
2. We claim that limn→∞∥xn+1-xn∥=0. In fact, write Jn=JcnT and Tn=2Jn-I. Then, Jn is firmly nonexpansive and Tn is nonexpansive (see Lemma 2.4).
Observe that(3.23)xn+1=Jnzn=(I+Tn2)zn=12zn+12Tnzn=12xn+12[tn(u-Fxn)+en+Tnzn]=12xn+12yn,
where yn=tn(u-Fxn)+en+Tnzn. Therefore,
(3.24)∥yn+1-yn∥=∥tn+1(u-Fxn+1)+en+1+Tn+1zn+1-tn(u-Fxn)-en-Tnzn∥≤∥tn+1(u-Fxn+1)∥+∥tn(u-Fxn)∥+∥en+1∥+∥en∥+∥Tn+1zn+1-Tnzn∥.
It follows from the resolvent identity that(3.25)∥Tn+1x-Tnx∥=2∥Jn+1x-Jnx∥=2∥Jn(cncn+1x+(1-cncn+1)Jn+1x)-Jnx∥≤2|1-cncn+1|∥Jn+1x-x∥≤|1-cncn+1|∥Tn+1x-x∥
for any x∈H. From (1.6), we get
(3.26)∥zn+1-zn∥=∥(I-tn+1F)xn+1+tn+1u+en+1-(I-tnF)xn-tnu-en∥≤∥xn+1-xn∥+∥tn+1(u-Fxn+1)∥+∥tn(u-Fxn)∥+∥en+1∥+∥en∥.
By (3.25) and (3.26), we have
(3.27)∥Tn+1zn+1-Tnzn∥≤∥Tn+1zn+1-Tnzn+1∥+∥Tnzn+1-Tnzn∥≤|1-cncn+1|∥Tn+1zn+1-zn+1∥+∥zn+1-zn∥≤|1-cncn+1|∥Tn+1zn+1-zn+1∥+∥xn+1-xn∥+∥tn+1(u-Fxn+1)∥+∥tn(u-Fxn)∥+∥en+1∥+∥en∥.
Substituting (3.27) into (3.24) at once gives
(3.28)∥yn+1-yn∥≤2∥tn+1(u-Fxn+1)∥+2∥tn(u-Fxn)∥+2∥en+1∥+2∥en∥+|1-cncn+1|M2+∥xn+1-xn∥,
that is,
(3.29)∥yn+1-yn∥-∥xn+1-xn∥≤2∥tn+1(u-Fxn+1)∥+2∥tn(u-Fxn)∥+2∥en+1∥+2∥en∥+|1-cncn+1|M2,
where M2=sup{∥Tn+1zn+1-zn+1∥,n≥0}. Observing tn(u-Fxn)→0, en→0, and |1-(cn/cn+1)|→0(n→∞), it follows that
(3.30)limsupn→∞(∥yn+1-yn∥-∥xn+1-xn∥)≤0.
From (3.23) and using Lemma 2.6, we have limn→∞∥yn-xn∥=0. Therefore,
(3.31)limn→∞∥xn+1-xn∥=limn→∞12∥yn-xn∥=0.Step3. We claim that limn→∞∥xn-JcTxn∥=0. Since liminfn→∞cn>0, there exist α>0 and a positive integer N such that for all n≥N, cn≥α. From Lemma 2.2, for each c∈(0,α), we have
(3.32)∥Jnxn-JcTxn∥=∥JcT(ccnxn+(1-ccn)Jnxn)-JcTxn∥≤∥ccnxn+(1-ccn)Jnxn-xn∥=|1-ccn|∥Jnxn-xn∥≤∥Jnxn-xn+1∥+∥xn+1-xn∥.
Observe that
(3.33)∥Jnxn-xn+1∥≤∥xn-zn∥≤∥tn(u-Fxn)∥+∥en∥→0.
Thus, it follows from (3.32), (3.33), and Step 2 that
(3.34)limn→∞∥Jnxn-JcTxn∥=0.
Since ∥xn-JcTxn∥≤∥xn-xn+1∥+∥xn+1-Jnxn∥+∥Jnxn-JcTxn∥, then
(3.35)limn→∞∥xn-JcTxn∥=0.Step4. We claim that limsupn→∞〈xn-x*,u-Fx*〉≤0, where x*=limt→0vt and vt is defined by (3.3). Since xn is bounded, there exists a subsequence {xnk} of {xn} which converges weakly to ω. From Step 3, we obtain JcTxn⇀ω. From Lemma 2.5, we have ω∈S. Hence, by Theorem 3.1, we have
(3.36)limsupn→∞〈xn-x*,u-Fx*〉=limk→∞〈xnk-x*,u-Fx*〉=〈ω-x*,u-Fx*〉≤0.Step5. We claim that {zn} converges strongly to x*∈S. From (1.6), for an appropriate constant γ>0, we have(3.37)∥xn+1-x*∥2=∥Jnzn-x*∥2≤∥zn-x*∥2=∥(I-tnF)xn+tnu+en-x*∥2≤∥(I-tnF)xn+tnu-x*∥2+γ∥en∥≤∥(I-tnF)xn-(I-tnF)x*+tn(u-Fx*)∥2+γ∥en∥≤τtn2∥xn-x*∥2+tn2∥u-Fx*∥2+2tn〈(I-tnF)xn-(I-tnF)x*,u-Fx*〉+γ∥en∥≤τtn∥xn-x*∥2+tn2∥u-Fx*∥2+2tn〈xn-x*-tnFxn+tnu,u-Fx*〉+2tn2〈Fx*-u,u-Fx*〉+γ∥en∥≤τtn∥xn-x*∥2+2tn〈xn-x*,u-Fx*〉+2tn∥tn(u-Fxn)∥∥u-Fx*∥+γ∥en∥≤[1-(1-τtn)]∥xn-x*∥2+(1-τtn)[2M1〈xn-x*,u-Fx*〉+2M1∥tn(u-Fxn)∥∥u-Fx*∥]+γ∥en∥,
for all n≥n0 for some integer n0≥0. For every n≥n0, put μn=1-τtn and δn=2M1〈xn-x*,u-Fx*〉+2M1∥tn(u-Fxn)∥∥u-Fx*∥. It follows that
(3.38)∥xn+1-x*∥2≤(1-μn)∥xn-x*∥2+μnδn+γ∥en∥,∀n≥n0.
It is easy to see that ∑n=1∞‍μn=∞ and limsupn→∞δn≤0. Hence, by Lemma 2.7, the sequence {xn} converges strongly to x*∈S. Observe that
(3.39)∥zn-x*∥=∥(I-tnF)xn+tnu+en-x*∥≤∥xn-x*∥+∥tn(u-Fxn)∥+∥en∥.
Thus, it follows that the sequence {zn} converges strongly to x*∈S.
On the other hand, suppose thatzn→x*∈S as n→∞. From (1.6), we have
(3.40)∥xn+1-x*∥=∥Jnzn-x*∥≤∥zn-x*∥→0.
Therefore,
(3.41)∥tn(u-Fxn)∥=∥zn-xn-en∥≤∥zn-xn∥+∥en∥≤∥zn-x*∥+∥xn-x*∥+∥en∥→0.SettingF=I and x*=PSu in Theorem 3.4, we can obtain the following result.Corollary 3.5.
LetT 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<tn≤1-ϵ, for all n≥n0 for some integer n0≥0.
For an arbitrary pointx0∈H, let the sequence {xn} be generated by
(3.42)zn=(1-tn)xn+tnu+en,xn+1=JcnTzn,n≥0.
Then,
(3.43)zn→PSu⟺tn(u-xn)→0(n→∞).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.
Sincelimn→∞tn=0, it is easy to see that tn≤η/k2-ϵ, for all n≥n0 for some integer n0≥0. Without loss of generality, we assume that 0<tn≤η/k2-ϵ, for all n≥n0 for some integer n0≥0. 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(u-xn)→0. Therefore, all conditions of Corollary 3.5 are satisfied. Using Corollary 3.5, we have that {zn} converges strongly to PSu∈S, with zn=(1-tn)xn+tnu+en. Therefore,
(3.44)∥xn+1-PSu∥≤∥zn-PSu∥→0.Remark 3.7.
Corollary3.5 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.
LetH=R  be the set of real numbers, u=0 and cn=1/2 for all n≥0. Define a maximal monotone operator T as follows: Tx=2x, for all x∈R. It is easy to see that JcnT=(1/2)I and S={0}. Given sequences {tn} and {en}, tn=1/2 and en=0, for all n≥0. For an arbitrary x0∈R, let {xn} be defined by (3.42), that is,
(3.45)zn=12xn,xn+1=12zn=14xn,n≥0.
Observe that
(3.46)∥xn+1-0∥=∥14xn-0∥=14∥xn-0∥.
Hence, we have ∥xn+1-0∥=(1/4)n+1∥x0-0∥ for all n≥0. This implies that {xn} converges strongly to 0=PS0. Thus,
(3.47)∥tn(u-xn)∥=12∥xn∥→0(n→∞).
Furthermore, it is easy to see that there hold the following:(B1)
0<tn=1/2≤1-ϵ, foralln≥n0 for some integer n0≥0,(B2)
∑n=0∞‍tn=∑n=0∞‍(1/2)=∞,(B3)
liminfn→∞cn=1/2>0 and limn→∞|1-(cn/cn+1)|=0,(B4)
∑n=0∞‍∥en∥=∑n=0∞‍0=0<∞.Hence there is no doubt that all conditions of Corollary3.5 are satisfied. Since tn=1/2↛0, the condition tn→0 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.



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*Source: 567948-2012-08-30.xml*