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Volume 2011, Article ID 173430,15pages doi:10.1155/2011/173430

Research Article

An Application of Hybrid Steepest Descent Methods for Equilibrium Problems and Strict Pseudocontractions in Hilbert Spaces

Ming Tian

College of Science, Civil Aviation University of China, Tianjin 300300, China

Correspondence should be addressed to Ming Tian,[email protected] Received 9 December 2010; Accepted 13 February 2011

Academic Editor: Shusen Ding

Copyrightq2011 Ming Tian. 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.

We use the hybrid steepest descent methods for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of a strict pseudocontraction mapping in the setting of real Hilbert spaces. We proved strong convergence theorems of the sequence generated by our proposed schemes.

1. Introduction

LetHbe a real Hilbert space andCa closed convex subset ofH, and letφbe a bifunction of C×CintoR, whereRis the set of real numbers. The equilibrium problem forφ:C×CR is to findxCsuch that

EP :φ x, y

≥0 ∀y∈C 1.1

denoted the set of solution by EPφ. Given a mappingT :CH, letφx, y Tx, y−x for allx, yC, thenz ∈ EPφif and only ifTz, y−z ≥ 0 for all yC, that is, z is a solution of the variational inequality. Numerous problems in physics, optimizations, and economics reduce to find a solution of1.1. Some methods have been proposed to solve the equilibrium problem, see, for instance,1,2.

A mappingTofCinto itself is nonexpansive ifTx−Ty ≤ x−y, for allx, yC. The set of fixed points ofTis denoted byFT. In 2007, Plubtieng and Punpaeng3, S. Takahashi and W. Takahashi4, and Tada and W. Takahashi5considered iterative methods for finding an element of EPφ∩FT.

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Recall that an operatorAis strongly positive if there exists a constantγ >0 with the property

Ax, x ≥γx2, ∀x∈H. 1.2

In 2006, Marino and Xu6introduced the general iterative method and proved that for a givenx0H, the sequence{xn}is generated by the algorithm

xn1αnγfxn I−αnATxn, n≥0, 1.3 whereTis a self-nonexpansive mapping onH,fis a contraction ofHinto itself withβ∈0,1 and {αn} ⊂ 0,1 satisfies certain conditions, andA is a strongly positive bounded linear operator onHand converges strongly to a fixed-pointxofT which is the unique solution to the following variational inequality:

γf−Ax, xx ≤ 0, forxFT, and is also the optimality condition for some minimization problem. A mappingS : CH is said to bek-strictly pseudocontractive if there exists a constantk∈0,1such that

SxSy2xy2kI−Sx−I−Sy2, ∀x, y∈C. 1.4 Note that the class ofk-strict pseudo-contraction strictly includes the class of nonex- pansive mapping, that is,Sis nonexpansive if and only ifSis 0-srictly pseudocontractive; it is also said to be pseudocontractive ifk1. Clearly, the class ofk-strict pseudo-contractions falls into the one between classes of nonexpansive mappings and pseudo-contractions.

The set of fixed points ofS is denoted by FS. Very recently, by using the general approximation method, Qin et al.7obtained a strong convergence theorem for finding an element ofFS. On the other hand, Ceng et al.8proposed an iterative scheme for finding an element of EPφ∩FSand then obtained some weak and strong convergence theorems.

Based on the above work, Y. Liu9introduced two iteration schemes by the general iterative method for finding an element of EPφ∩FS.

In 2001, Yamada10introduced the following hybrid iterative method for solving the variational inequality:

xn1TxnμλnFTxn, n≥0, 1.5 whereFisk-Lipschitzian andη-strongly monotone operator withk >0,η >0, 0< μ <2η/k2, then he proved that if {λn} satisfyies appropriate conditions, the {xn} generated by 1.5 converges strongly to the unique solution of variational inequality

Fx, xx ≥ 0, ∀x∈FixT, xFixT. 1.6 Motivated and inspired by these facts, in this paper, we introduced two iteration methods by the hybrid iterative method for finding an element of EPφ∩FS, whereS : CH is ak-strictly pseudocontractive non-self mapping, and then obtained two strong convergence theorems.

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2. Preliminaries

Throughout this paper, we always assume thatCis a nonempty closed convex subset of a Hilbert spaceH. We writexn xto indicate that the sequence{xn}converges weakly to x.xnximplies that{xn}converges strongly tox. For anyxH, there exists a unique nearest point inC, denoted byPCx, such that

x−PCx ≤xy, ∀y∈C. 2.1 Such aPCxis called the metric projection ofHontoC. It is known thatPCis nonexpansive.

Furthermore, forxHanduC,upcx,⇔ x−u, uy ≥0, for allyC.

It is widely known thatH satisfies Opial’s condition11, that is, for any sequence {xn}withxn x, the inequality

lim inf

n→ ∞ xnx<lim inf

n→ ∞ xny, 2.2

holds for everyyHwithy /x. In order to solve the equilibrium problem for a bifunction φ:C×CR, let us assume thatφsatisfies the following conditions:

A1φx, x 0, for allxC,

A2φis monotone, that is,φx, y φy, x≤0, for allx, yC, A3For allx, y, zC.

limt↓0φ

tz 1−tx, y

φ x, y

; 2.3

A4For each fixedxC, the functionyφx, yis convex and lower semicontinuous.

Let us recall the following lemmas which will be useful for our paper.

Lemma 2.1see12. Letφbe a bifunction fromC×CintoRsatisfying (A1), (A2),(A3) and (A4) then, for anyr >0 andxH, there existszCsuch that

φ z, y

1

ry−z, zx ≥0, ∀y∈C. 2.4 Further, ifTrx{z∈C:φz, y 1/ry−z, zx ≥0,∀y∈C}, then the following hold:

1Tr is single-valued,

2Tr is firmly nonexpansive, that is, TrxTry2

TrxTry, xy

, ∀x, y∈H; 2.5

3FTr EPφ,

4EPφis nonempty, closed and convex.

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Lemma 2.2see13. IfS:CHis ak-strict pseudo-contraction, then the fixed-point setFS is closed convex, so that the projectionPFSis well difened.

Lemma 2.3see14. LetS : CH be ak-strict pseudo-contraction. DefineT :CHby Tx λx 1−λSx for eachxC, then, asλ ∈ k,1, T is nonexpansive mapping such that FT FS.

Lemma 2.4see15. In a Hilbert spaceH, there holds the inequality xy2 ≤ x22

y, xy

, ∀x, y∈H. 2.6

Lemma 2.5see16. Assume that{an}is a sequence of nonnegative real numbers such that

an1≤ 1−γn

anγnδn, n≥0, 2.7

wheren}is a sequence in (0,1) andn}is a sequence inÊ, such that i

n1γn∞,

iilim supn→ ∞δn0 or

n1nγn|<∞.

Then limn→ ∞an0.

3. Main Results

Throughout the rest of this paper, we always assume thatFis aL-lipschitzian continuous and η-strongly monotone operator withL, η >0 and assume that 0< μ <2η/L2.τ μημL2/2.

Let {Tλn} be mappings defined asLemma 2.1. Define a mappingSn : CH bySnx βnx 1−βnSx, for allxC, whereβn∈k,1, then, byLemma 2.3,Snis nonexpansive. We consider the mappingGnonHdefined by

Gnx

IαnμF

SnTλnx, xH, nN, 3.1

whereαn∈0,1. By Lemmas2.1and 2.3, we have

GnxGny≤1−αnτTλnxTλny

≤1−αnτxy. 3.2 It is easy to see thatGnis a contraction. Therefore, by the Banach contraction principle, Gnhas a unique fixed-pointxFnHsuch that

xFn

IαnμF

SnTλnxFn. 3.3

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For simplicity, we will writexnforxFn provided no confusion occurs. Next, we prove that the sequence{xn}converges strongly to aqFS∩EPφwhich solves the variational inequality

Fq, pq

≥0, ∀p∈FS∩EP φ

. 3.4

Equivalently,qPFS∩EPφI−μFq.

Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H and φ a bifunction from C×C into R satisfying (A1), (A2), (A3), and (A4). Let S : CH be a k- strictly pseudocontractive nonself mapping such thatFS∩EPφ/φ. LetF : HH be an L-Lipschitzian continuous andη-strongly monotone operator onHwithL, η >0 and 0< μ <2η/L2, τ μημL2/2. Let{xn}be asequence generated by

φ un, y

1

λny−un, unxn ≥0, ∀y∈C, ynβnun

1−βn Sun, xn

IαnμF

yn, ∀n∈N,

3.5

whereun Tλnxn,yn Snun, andn} ⊂ 0,∞satisfy lim infn→ ∞λn > 0 ifn} andn} satisfy the following conditions:

i{αn} ⊂0,1, limn→ ∞αn0,

ii0≤kβnλ <1 and limn→ ∞βnλ,

then{xn} converges strongly to a pointqFS∩EPφwhich solves the variational inequality 3.4.

Proof. First, takepFS∩EPφ. Sinceun Tλnxnandp Tλnp, fromLemma 2.1, for any nN, we have

unpTλnxnTλnpxnp. 3.6

Then, sinceSnpp, we obtain that

ynpSnunSnpunpxnp. 3.7

Further, we have

xnp−αnμFp

IμαnF yn

IμαnF p

αn−μF

p 1−αnτynp. 3.8 It follows thatxnp ≤ μFp/τ.

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Hence,{xn}is bounded, and we also obtain that{un}and{yn}are bounded. Notice that

unyn≤ unxnxnyn

unxnαn−μFyn. 3.9 ByLemma 2.1, we have

unp2TλnxnTλnp2

xnp, unp 1

2 xnp2unp2− unxn2

. 3.10

It follows that

unp2xnp2− xnun2. 3.11 Thus, fromLemma 2.4,3.7, and3.11, we obtain that

xnp2αn

−μFp

IμαnF yn

IμαnF p2

≤1−αnτ2ynp2n

−μFp, xnp

≤1−αnτ2unp2n

−μFp, xnp

≤1−αnτ2 xnp2− xnun2

n−μFpxnp 1−2αnτ αnτ2xnp2

−1−αnτ2xnun2nμFpxnp

xnp2 αnτ2xnp2−1−αnτ2xnun2n−μFpxnp. 3.12

It follows that

1−αnτ2xnun2 ≤αnτ2xnp2nμFpxnp. 3.13 Sinceαn → 0, therefore

nlim→ ∞xnun0. 3.14

From3.9, we derive that

nlim→ ∞unyn0. 3.15

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DefineT :CHbyTx λx 1−λSx, thenT is nonexpansive withFT FS byLemma 2.3. We note that

TununTunynynunλβnunSunynun. 3.16

So by3.15andβnλ, we obtain that

nlim→ ∞Tunun0. 3.17

Since{un}is bounded, so there exists a subsequence{uni}which converges weakly to q. Next, we show thatqFS∩EPφ. SinceCis closed and convex,Cis weakly closed. So we haveqC. Let us show thatqFS. Assume thatqFT, Sinceuni qandq /Tq, it follows from the Opial’s condition that

lim inf

n→ ∞ uniq<lim inf

n→ ∞ uniTq

≤lim inf

n→ ∞

uniTuniTuniTq

≤lim inf

n→ ∞ uniq.

3.18

This is a contradiction. So, we getqFTandqFS.

Next, we show thatqEPφ. SinceunTλnxn, for anyyC, we obtain φ

un, y 1

λn

yun, unxn

≥0. 3.19

FromA2, we have

1 λn

yun, unxn

φ y, un

. 3.20

Replacingnbyni, we have

yuni,unixni

λni

φ y, uni

. 3.21

Sinceunixnini → 0 anduni q, it follows fromA4that 0 ≥ φy, q, for all yC. Letzt ty 1−tqfor all t ∈ 0,1andyC, then we haveztCand hence φzt, q≤0. Thus, fromA1andA4, we have

0φzt, zt zt, y

1− zt, q

zt, y

, 3.22

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and hence 0 ≤ φzt, y. FromA3, we have 0 ≤ φq, yfor allyCand henceq ∈EPφ.

Therefore,qFS∩EPφ. On the other hand, we note that xnq−αnμFq

IμαnF yn

IμαnF

q. 3.23

Hence, we obtain xnq2

−αnμFq, xnq

IμαnF yn

IμαnF

q, xnq

αn

−μFq, xnq

1−αnτxnq2. 3.24

It follows that

xnq2≤ 1 τ

−μFq, xnq

. 3.25

This implies that

xnq2

−μFq, xnq

τ . 3.26

In particular,

xniq2

−μFq, xniq

τ . 3.27

Sincexni q, it follows from3.27thatxniqasi → ∞. Next, we show thatq solves the variational inequality3.4.

As a matter of fact, we have xn

IαnμF yn

IαnμF

SnTλnxn, 3.28

and we have

μFxn− 1 αn

I−SnTλnxnμαnFxnFSnTλnxn

. 3.29

Hence, forpFS∩EPφ, μF

xn, xnp − 1

αn

I−SnTλnxnμαnFxnFSnTλnxn

, xnp

− 1 αn

I−SnTλnxn−I−SnTλnp, xnp μ

FxnFSnTλnxn, xnp . 3.30

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SinceISnTλn is monotonei.e.,x−y,I−SnTλnx−I−SnTλny ≥0, for allx, yH. This is due to the nonexpansivity ofSnTλn.

Now replacingnin3.30withniand lettingi → ∞, we obtain μF

q, qp lim

i→ ∞

μFxni, xnip

≤ lim

i→ ∞μ

FxniFSnTλnxni, xnip

0. 3.31

That is,qFS∩EPφis a solution of3.4. To show that the sequence{xn}converges strongly toq, we assume thatxnkx. Similiary to the proof above, we derive xFS∩ EPφ. Moreover, it follows from the inequality3.31that

μF

q, qx

≤0. 3.32

Interchangeqandxto obtain μF

x,xq

≤0. 3.33

Adding up3.32and3.33yields μηqx2

qx, μF

qμF

x

≤0. 3.34

Hence,qx, and therefore xnqasn → ∞, IμF

qq, qp

≥0,∀p∈FS∩EP φ

. 3.35

This is equivalent to the fixed-point equation PFS∩EPφ

IμF

qq. 3.36

Theorem 3.2. LetCbe a nonempty closed convex subset of a real Hilbert spaceHandφa bifunction from C × C into R satisfying (A1), (A2), (A3) and (A4). Let S : CH be a k-strictly pseudocontractive nonself mapping such thatFS∩EPφ/φ. LetF:HHbe anL-Lipschitzian continuous andη-strongly monotone operator onH withL, η > 0. Suppose that 0 < μ < 2η/L2, τ μημL2/2. Let{xn}and{un}be sequences generated byx1Hand

φ un, y

1 λn

yun, unxn

≥0, ∀y∈C, ynβnun

1−βn Sun, xn1

IαnμF

yn, ∀n∈N,

3.37

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whereunTλnxn,ynSnunifn},{βn}, and{λn}satisfy the following conditions:

i{αn} ⊂0,1, limn→ ∞αn0,

n1αn∞,

n1n1αn|<∞, ii0≤kβnλ <1 and limn→ ∞βnλ,

n1n1βn|<∞, iii{λn} ∈0,∞, limn→ ∞λn>0 and

n1n1λn|<∞,

then{xn}and{un}converge strongly to a pointqFS∩EPφwhich solves the variational inequality3.4.

Proof. We first show that{xn}is bounded. Indeed, pick anypFS∩EPφto derive that xn1p−αnμFp

IμαnF yn

IμαnF p

αn−μF

p 1−αnτxnp

≤1−αnτxnn−μF p.

3.38

By induction, we have

xnp ≤max

x1p,1

τμF p

, ∀n∈N, 3.39

and hence {xn} is bounded. From3.6 and 3.7, we also derive that {un} and {yn} are bounded. Next, we show thatxn1xn → 0. We have

xn1xnIαnμF yn

Iαn−1μF yn−1 IαnμF

yn

IαnμF

yn−1

IαnμF

yn−1

Iαn−1μF yn−1

≤1−αnτynyn−1nαn−1|μFyn−1

≤1−αnτynyn−1K|αnαn−1|,

3.40

where

KsupμFyn:nN

<∞. 3.41

On the other hand, we have

ynyn−1SnunSn−1un−1

≤ SnunSnun−1Snun−1Sn−1un−1

≤ unun−1Snun−1Sn−1un−1.

3.42

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Fromun1Tλn1xn1andun Tλnxn, we note that φ

un1, y 1

λn1

yun1, un1xn1

≥0, ∀y∈C, 3.43 φ

un, y 1

λn

yun, unxn

≥0, ∀y∈C. 3.44

Puttingyunin3.43andyun1in3.44, we have φun1, un 1

λn1unun1, un1xn1 ≥0, φun, un1 1

λnun1un, unxn ≥0. 3.45 So, fromA2, we have

un1un,unxn

λnun1xn1

λn1

≥0 , 3.46

and hence

un1un, unun1un1xnλn

λn1un1xn1

≥0. 3.47

Since limn→ ∞λn > 0, without loss of generality, let us assume that there exists a real number a such thatλn> a >0 for allnN. Thus, we have

un1un2

un1un, xn1xn

1− λn λn1

un1xn1

≤ un1un

xn1xn 1− λn

λn1

un1xn1

un1un ≤ xn1xn 1

an1λn|M0,

3.48

whereM0sup{unxn:nN}. Next, we estimateSnun−1Sn−1un−1. Notice that Snun−1Sn−1un−1βnun−1

1−βn Sun−1

βn−1un−1

1−βn−1

Sun−1

βnβn−1un−1Sun−1. 3.49 From3.48,3.49, and3.42, we obtain that

ynyn−1≤ xnxn−1M0

anλn−1|βnβn−1un−1Sun−1

≤ xnxn−1nλn−1|M1βnβn−1M1,

3.50

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whereM1is an appropriate constant such that M1M0

a un−1Sun−1, ∀n∈N. 3.51 From3.41and3.50, we obtain

xn1xnK|αnαn−1| 1−αnτ

xnxn−1nλn−1|M1βnβn−1M1

≤1−αnτxnxn−1M

nαn−1||λnλn−1|βnβn−1, 3.52

whereMmaxK, M1. Hence, few byLemma 2.5, we have

nlim→ ∞xn1xn0. 3.53

From3.48and3.50,|λnλn−1| → 0 and|βnβn−1| → 0, we have

nlim→ ∞un1un0, lim

n→ ∞yn1yn0. 3.54

Since

xn1

IαnμF

yn, 3.55

it follows that

xnyn≤ xnxn1xn1yn

xnxn1αn−μFyn. 3.56

Fromαn → 0 and3.53, we have

nlim→ ∞xnyn0. 3.57

ForpFS∩EPφ, we have

unp2TλnxnTλnp2

xnp, unp 1

2 xnp2unp2− unxn2

. 3.58

This implies that

unp2xnp2− unxn2. 3.59

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Then, from3.7and3.59, we derive that xn1p2−μαnFp

IμαnF yn

IμαnF p2

≤1−αnτ2ynp2α2n−μFp2n−μFpynp

unp2α2n−μFp2n−μFpynp

xnp2− xnun2α2n−μFp2n−μFpynp.

3.60

Sinceαn → 0,xnxn1 → 0, we have

nlim→ ∞xnun0. 3.61

From3.57and3.61, we obtain that

unyn ≤ unxnxnyn → 0, asn → ∞. 3.62

DefineT :CHbyTx λx 1−λSx, thenT is nonexpansive withFT FS byLemma 2.3. Notice that

TununTunynynun

λβnunSunynun. 3.63

By3.62andβnλ, we obtain that

nlim→ ∞Tunun0. 3.64

Next, we show that lim supn→ ∞μFq, q−xn ≤0, whereq PFS∩EPφI−μFqis a unique solution of the variational inequality3.4. Indeed, take a subsequence{xni}of{xn} such that

ilim→ ∞

μFq, qxni

lim sup

n→ ∞

μFq, qxn

. 3.65

Since{xni} is bounded, there exists a subsequence {xnij} of {uni} which converges weakly tow.

Without loss of generality, we can assume thatuni w. From3.61and3.64, we obtainxni wandTuni w. By the same argument as in the proof ofTheorem 3.1, we have wFS∩EPφ. SinceqPFS∩EPφI−μFq, it follows that

lim sup

n→ ∞

μFq, qxn

μFq, qw

≤0. 3.66

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Fromxn1q−αnμFq I−μαnFyn−I−μαnFq, we have xn1q2IμαnF

yn

IμαnF

q2n

−μFq, xn1q

≤1−αnτ2xnq2n

−μFq, xn1q .

3.67

This implies that xn1q2

1−2αnτ αnτ2xnq2n

−μFq, xn1q 1−2αnτxnq2 αnτ2xnq2n

−μFq, xn1q

1−2αnτxnq2nτ αnτ2

M1 τ

−μFq, xn1q

1−γnxnq2γnδn,

3.68

whereMsup{xn−q2:nN},γnnτ, andδn αnτ2/2τM1/τ−μFq, xn1−q.

It is easy to see thatγn → 0,

n1γn∞, and lim supn→ ∞δn ≤0 by3.66. Hence by Lemma 2.5, the sequence{xn}converges strongly toq.

Acknowledgments

M. Tian was supported in part by the Science Research Foundation of Civil Aviation University of Chinano. 2010kys02. He was also Supported in part by The Fundamental Research Funds for the Central UniversitiesGrant no. ZXH2009D021.

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