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Volume 2011, Article ID 780764,15pages doi:10.1155/2011/780764

Research Article

An Iteration Method for Common Solution of

a System of Equilibrium Problems in Hilbert Spaces

Jong Kyu Kim

1

and Nguyen Buong

2

1Department of Mathematics Education, Kyungnam University, Masan Kyunganm 631-701, Republic of Korea

2Department of Mathematics, Vietnamse Academy of Science and Technology,

Institute of Information Technology, 18, Hoang Quoc Viet, q. Cau Giay, Hanoi 122100, Vietnam

Correspondence should be addressed to Jong Kyu Kim,[email protected]

Received 11 December 2010; Revised 3 March 2011; Accepted 4 March 2011

Academic Editor: Qamrul Hasan Ansari

Copyrightq2011 J. K. Kim and N. Buong. 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.

The strong convergence theorem is proved for finding a common solution for a system of equilibrium problems: findu∗ ∈ S : ∩Ni1EPFi,EPFi : {zC : Fiz, v ≥ 0∀vC}, i 1, . . . , N,whereCis a closed convex subset of a Hilbert spaceHandFiareNbifunctions from C×CintoRgiven exactly or approximatively. As an application, finding a common solution for a system of variational inequality problems is given.

1. Introduction

LetHbe a real Hilbert space with the scalar product and the norm denoted by the symbols

·,· and · , respectively. Let C be a nonempty closed convex subset of H, and let

Fii1, . . . , NbeNbifunctions fromC×CintoR. In this paper, we consider the system of

equilibrium problems:

findu∗∈S:∩iN1EPFi,

EPFi:{zC:Fiz, v≥0∀vC}, i1, . . . , N.

1.1

We assume thatS /∅and the bifunctionsFisatisfy the following conditions.

Condition 1. The bifunctionFsatisfies the following conditions:

A1Fu, u 0 for alluC.

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A3For everyuC,Fu,·:CRis lower semicontinuous and convex.

A4limt→0F1−tutz, vFu, vfor allu, z, vC×C×C.

Definition 1.1. A mappingAofCintoHis called monotone if

AxAy, xy≥0, 1.2

for allx, yC.

Now, we consider the variational inequality problem: findu∗∈Csuch that

Au, xu0, 1.3

for allxC. We denote VIC, Athe set of solutions of the variational inequality problem.

Definition 1.2. A mapping T of C into H is called k-strictly pseudocontractive in the terminology of Browder and Petryshyn1, if there exists a numberk∈0,1such that

TxTy 2xy 2k ITxITy 2, 1.4

whereIis the identity operator inH.

The above inequality is equivalent to

AxAy, xyλ AxAy 2, 1.5

where the operatorA:ITisλ 1−k/2-inverse strongly monotonehence monotone and Lipschitz continuous with the Lipschitz constant 2/1−k. Clearly, whenk 0, T is nonexpansive, that is,

TxTyxy 1.6

for allx, yDT, the domain ofT. It means that the class of k-strictly pseudocontractive mappings strictly includes the class of nonexpansive mappings. Denote byFTthe set of fixed points of the operatorTinC, that is,

FT {xC:xTx}. 1.7

If N 1, then 1.1 is a single equilibrium problem 2, 3 to cover monotone inclusion problems, saddle point problems, variational inequality problems, minimization problems, Nash equilibria in noncooperative games, vector equilibrium problems, as well as certain fixed point problems.

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If N > 1, then 1.1 is a problem of finding a common solution for a system of equilibrium problems which is studied firstly in 5 cf. 16 under the condition that

Fi i 1, . . . , N are bounded, Fr´echet differentiable with respect tov and ∇vFiu, uare

Lipschitz continuous, that is,

vFix, x− ∇vFiy, yL xyx, yC, i1,2, . . . , N, 1.8

whereLis a positive constant. With the case that

Fiu, v ITiu, vu, 1.9

and Ti i 2, . . . , Nare N−1 strictly pseudocontractive mappings, 1.1 is a problem of

finding a solution of an equilibrium problem which is also a common fixed point for a system of a finite family of strictly pseudocontractive mappings17–19.

In addition, whenF1u, v A1u, vu whereA1 is a monotone operator,1.1

is a problem of finding an element which is a solution of a variational inequality problem and a common fixed point for a finite family of strictly pseudocontractive mappings and investigated intensively in 20–32. If all Fi have the form 1.9, then 1.1is a problem of

finding a common fixed point for a finite family of strictly pseudocontractive mappingsTi

fromCintoH14,33–35.

In this paper, we present an iteration method for solving1.1, where the iteration sequence{xn}is defined by

x0xH,

ui

nC:Fi

ui n, v

ui

nxn, vuin ≥0,vC, i1, . . . , N,

xn1xnβn

n

i1

xnuin

αnxn

,

1.10

where{αn},{βn}are two sequences of positive numbers satisfying some conditions.

As an application, we find a common solution for a system ofNvariational inequality problems with monotone mappings.

2. Main Results

The strong and weak convergence of any sequence are denoted by → and, respectively. We formulate the following facts which are necessary in the proof of our main results.

Lemma 2.1see5. LetCbe a nonempty closed convex subset of a Hilbert spaceH, and letFbe a bifunction ofC×CintoRsatisfying the Condition1. Letr >0andxH. Then, there existszC such that

Fz, v 1

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Lemma 2.2see5. Assume thatF :C×CRsatisfies the Condition1. Forr >0andxH, define a mappingTr :HCas follows:

Trx

zC:Fz, v 1

rzx, vz ≥0,vC

. 2.2

Then, the following hold:

iTr is single-valued;

iiTr is firmly nonexpansive, that is, for anyx, yH,

TrxTry 2 ≤TrxTry, xy; 2.3

iiiFTr EPF;

ivEPFis closed and convex.

Lemma 2.3. LetFhu, vbe a bifunction approximating the bifunctionFu, vin the sense

Fhu, vFu, vhg u uvu, vC, h >0, 2.4

wheregtis a real positive function. Then, for eachr >0andxH, we have

Th

rxTrxrhg Trx , 2.5

where

Th rx

zC:Fhz, v 1

rzx, vz ≥0∀vC

. 2.6

Proof. Letxbe an arbitrary element ofH. By replacingvbyzin2.2and byzin2.6, we obtain

Fz,z Fhz, z 1

rxz,zzzx,zz. 2.7

Therefore, by virtue ofA2in Condition1, we can write

Fz,zFhz,z 1

r zz 2. 2.8

Consequently,

zzrhg Trx . 2.9

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Lemma 2.4 see36. Let {an},{bn}, and{cn} be sequences of positive numbers satisfying the conditions:

ian1≤1−bnancn, bn<1, ii∞n0bn, limn→∞cn/bn 0. Then,limn→∞an0.

Lemma 2.5see37. Assume thatT is a nonexpansive mapping of a closed convex subsetCof a Hilbert spaceH. ThenIT is demiclosed at zero; that is whenever{xn}is a sequence inCweakly converging to some xCand the sequence {ITxn} strongly converges to zero, it follows ITx 0.

Lemma 2.6see17. LetAbe aλ-inverse strongly monotone mapping fromCintoHsuch that

SA/, whereSA{xC:Ax 0}. Then,SA VIC, A.

Now, consider the firmly nonexpansive mappingsTidefined by

Tix {zC:Fiz, v zx, vz ≥0,vC}, i1, . . . , N. 2.10

By virtue ofLemma 2.2, we can seeTi is nonexpansive. Consequently,Ai : ITiis1/2

-inverse strongly monotone and Lipschitz continuous with the Lipschitz constantLi 2,i

1, . . . , N.

We construct a Tikhonov regularization solutionynfor1.1by solving the following

operator equation: findynHsuch that

N

i1

Aiynαnyn0, 2.11

where the positive regularization parameterαn → 0 asn → ∞. We have the following

result.

Theorem 2.7. iFor eachαn >0, problem2.11has a unique solutionyn. iilimn→∞ynu,u∗∈S, u∗ ≤ y , for allyS.

iii ynym ≤|αnαm|/αn u.

Proof. i Since the mapping ni1Ai is a monotone and Lipschitz continuous mapping

defined onH, it is maximal monotone. Therefore,2.11has a unique solution for eachαn>0 38.

iiFor eachyS, on the base ofLemma 2.2, we have thatAiy 0,i 1, . . . , N.

Thus, from2.11it follows that

N

i1

AiynAiy, ynyαnyn, yny0. 2.12

Since everyAiis monotone, from the last equality, we obtain

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Hence,

yny ,yS. 2.14

It means that the sequence{yn}is bounded. Let{ynk}be a subsequence of the sequence{yn}

such thatynk yask → ∞.

Again, letybe an arbitrary element ofS. From the1/2-inverse strongly monotone property ofAl, andAly 0,l1, . . . , N, it implies that

1

2 ynkTl

ynk 2≤Alynk, ynky

N

i1

Aiynk, ynky

≤ −αnkynk, ynky

αnkynky, ynkyαnky, ynky

≤ −αnky, ynky

αnk2 y 2,

2.15

that is,

ynkTlynk ≤2 y αnk. 2.16

Therefore,

lim

k→ ∞ Al

ynk 0. 2.17

ByLemma 2.5,Aly 0, that is,yFTl,l 1, . . . , N. It means thatyS. BecauseSis

a closed convex subset in Hilbert space, it has a unique minimal elementu∗in norm. From

2.14and the weak convergence of{ynk}to y u∗, it also follows that ynku∗ , as

k → ∞. Moreover, the sequence{yn}converges strongly tou∗asn → ∞. iiiFrom2.11,2.14, and the monotone property ofAi, it follows

αnyn, ynymαmym, ynym≤0 2.18

or

ynym ≤ |αnααm|

n ym

|αnαm|

αn u

, 2.19

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Theorem 2.8. Suppose thatαn, βnsatisfy the following conditions:

αn, βn>0 αn≤1, nlim

→ ∞αnn→ ∞lim

|αnαn1|

α2

nβn

0,

n0

αnβn, lim n→ ∞βn

2Nαn2

αn <1.

2.20

Then,

lim

n→ ∞xnu

S, 2.21

wherexnis defined by1.10.

Proof. Letynbe a solution of2.11. SetΔn xnyn . Then,

Δn1 xn1−yn1 ≤ xn1−yn yn1−yn ,

xn1−yn xnynβn

N

i0

AixnAiynαnxnyn.

2.22

From the monotone and Lipschitz continuous properties ofAi,i1, . . . , N,2.11, anduin

Tixn, we can write

xnynβn

N

i1

AixnAiynαnxnyn

2

xnyn 2β2n

N

i1

AixnAiynαnxnyn

2

−2βn

N

i1

AixnAiynαnxnyn, xnyn

xnyn 2

1−2βnαnβ2n2Nαn2

.

2.23

Hence,

xn1−yn ≤Δn

1−2βnαnβ2n2Nαn2 1/2

. 2.24

Therefore,

Δn1≤Δn

1−2βnαnβ2n2Nαn2 1/2

|αnααn1|

n u

≤Δn1−αnβn1/2|αnααn1|

n u

.

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Thus, applying the inequality

ab21ε

a2b2 ε

ε >0, ε αnβn

2 , 2.26

we obtain

0≤Δ2n1≤Δ2n1−αnβn1 1

2αnβn

α

nαn1 αn u

∗ 2 2

αnβn

11 2αnβn

a2

n

1−1 2αnβn

1 2

αnβn2

αnαn1 α2

nβn u

2

2αnβn

1 1 2αnβn

.

2.27

Set

bnαnβn

1 2

1 2αnβn

,

cn

αnαn1 α2

nβn

u

2

2αnβn

11 2αnβn

.

2.28

It is not difficult to check thatbnandcn satisfy the conditions inLemma 2.4for sufficiently

largen. Hence, limn→∞Δ2n0. Since limn→ ∞ynu∗, we have

lim

n→ ∞xnu

S.

2.29

Now, letFniu, v:Fihnu, vbe bifunctions approximating the bifunctionsFiu, vin

the sense2.4wherehn → 0, asn → ∞, andgtis a real positive and boundedthe image

of any bounded set is boundedfunction. Then, the sequence of iterations{xn}is defined by

x0xH,

ui

nC:Fin

ui n, v

ui

nxn, vuni ≥0 ∀vC, i1, . . . , N,

xn1xnβn

n

i1

xnuin

αnxn

,

2.30

where{αn},{βn}are two sequences of positive numbers satisfying some conditions.

We have the following result.

Theorem 2.9. Suppose thatαn, βn, andhnsatisfy the conditions inTheorem 2.8and

lim

n→ ∞

hnhn1 α2

nβn

(9)

Then, we have

lim

n→ ∞xnu

S,

2.32

wherexnis defined by2.30.

Proof. Letynbe a solution of the following equation:

N

i1 An

i

ynαnyn0, Ani ITin, 2.33

where eachTinis defined by

Tn ix

zC:Finz, v zx, vz ≥0,vC, i1, . . . , N. 2.34

Since

xnu∗ ≤ xnyn ynyn ynu, 2.35

and limn→ ∞ynu∗, in order to prove that limn→ ∞xnu∗, it is necessary to prove that

lim

n→ ∞ xnyn nlim→ ∞ ynyn 0. 2.36

For this purpose, first we estimate the value ynyn . On the basis ofLemma 2.3, we have

AixAnix TixTinxhng Tix . 2.37

Therefore, from2.11,2.33, and the monotone property ofAni it implies that

ynyn 2 α1 n

N

i1

An i

ynAiyn, ynyn

≤ 1 αn

N

i1

An i

ynAiyn, ynyn.

2.38

Consequently, we have

ynynα1 n

N

i1 An

i

ynAiyn

Nhn αng

Tiyn .

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On the other hand,

Tiyn TiynTiuu

ynuu

yn 2 u

≤3 u.

2.40

Therefore,

ynynC0N hn

αn, 2.41

whereC0sup{gt: 0< t≤3 u∗ }. It means that limn→ ∞ynu∗because limn→ ∞hn/αn

0.

Secondly, to prove

lim

n→ ∞ xnyn 0, 2.42

as in the proof ofTheorem 2.8, first we need to estimate the value yn1−yn . By the argument

as in the proof ofTheorem 2.7, we have

N

i1

An i

ynAni1

yn1

,ynyn1 αnyn,ynyn1

αn1

yn1,ynyn1

0. 2.43

Thus,

ynyn1 2 αnααn1 n

yn1,ynyn1

α1

n N

i1

An1

i

yn1

An i

yn,ynyn1

αnαn1 αn

yn,ynyn1

1

αn N

i1

An1

i

yn1

An i

yn,ynyn1

αnαn1

αn yn ynyn1

1

αn N

i1

An1

i

ynAni

yn,ynyn1 .

2.44

Therefore,

ynyn1 ≤ αnααn1

n yn

1

αn N

i1 An1

i

ynAiyn AiynAni

yn

αnαn1

αn yn N

hnhn1 αn g

yn .

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Using2.14and2.41, we have

ynuC0N hn

αn. 2.46

Consequently, there exists a positive constantCsuch that ynCforn≥0. Finally, we have

ynyn1 ≤ αnααn1

n CNC1

hnhn1

αn , 2.47

whereC1sup{gt: 0< t < C}. Now, setΔn xnyn . It is not difficult to verify that

xn1−yn ≤Δn

1−2βnαnβ2n2Nαn2 1/2

,

Δn1≤Δn1−αnβn1/2|αnααn1|

n CNC1

hnhn1 αn .

2.48

Therefore, limn→ ∞Δn0. The proof is completed.

Remark. The sequencesαn 1n−p, 0< p <1/2, andβnγ0αnwith

0< γ0<

1

202

, 2.49

satisfy all the necessary conditions inTheorem 2.8.

3. Applications

Consider the following problem: find an elementu∗∈Csuch that

Aiu, vu∗ ≥0,vC, i1, . . . , N, 3.1

whereAi areN monotone hemicontinuous mappings from a closed convex subsetC of a

Hilbert spaceHintoH.

Theorem 3.1. Letx0xbe an arbitrary element inH. If{αn},{βn}are chosen as inTheorem 2.8, and the iteration sequence{xn}is defined as follows:

ui nC,

Ai

ui n

, vui

n

ui

nxn, vuin ≥0,vC, i1, . . . , N,

xn1xnβn

N

i1

xnuin

αnxn

,

3.2

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IfCH, then we have a problem of finding a common zero for a system of monotone hemicontinous mappingsAi,i1, . . . , N. In this case, variational inequality in3.2has the

formAiuin uinxn. Therefore, we have the following result.

Theorem 3.2. LetAi, i1, . . . , NbeNhemicontinuous monotone mappings defined onH. Letx0 xbe an arbitrary element inH, let{αn}and{βn}be the sequences that are chosen as inTheorem 2.8, and, the iteration sequence{xn}be defined as follows:

ui n:Ai

ui n

ui

nxn,

xn1xnβn

N

i1

xnuin

αnxn

.

3.3

Then the sequence{xn}converges strongly to an elementusuch that

Aiu∗ 0, i1, . . . , N. 3.4

Without the strongly or uniformly monotone property forAi, each problem of3.1,

in general, is ill-posed. Some methods for finding a solution of each variational inequality in

3.1are presented in39.

Here we show an iterative regularization method for finding a common solution of these problems. Suppose that instead ofAi, we have their monotone approximationsAni such

thatDAni Cand

An

ixAixhng x , i1, . . . , N, 3.5

where the positive parameterhn → 0 as n → ∞, andgtis a real positive and bounded

function. Obviously, the bifunctions

Fnu, v:An

iu, vu

, i1, . . . , N, 3.6

satisfy the Condition1and2.4. Therefore, we have the following theorem.

Theorem 3.3. Letx0xbe an arbitrary element inH. If{αn},{βn}are chosen as inTheorem 2.9, and the iteration sequence{xn}is defined as follows:

ui nC:

Ai

ui n

, vui

n

ui

nxn, vuni ≥0 ∀vC, i1, . . . , N,

xn1xnβn

N

i1

xnuin

αnxn

,

3.7

then the sequence{xn}converges strongly to a common solution for3.1.

IfCH, then a common zero for a system of monotone hemicontinuous mappings

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Theorem 3.4. LetAi,i1, . . . , NbeNhemicontinuous monotone mappings defined onH. Letx0 xbe an arbitrary element inH, let{αn}and{βn}be the sequences that are chosen as inTheorem 2.9, and the iteration sequence{xn}be defined as follows:

ui n:Ai

ui n

ui

nxn,

xn1xnβn

N

i1

xnuin

αnxn

.

3.8

Then the sequence{xn}converges strongly to an elementusuch that

Aiu∗ 0, i1, . . . , N. 3.9

Acknowledgment

This work was supported by the Kyungnam University Research Fund, 2010.

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References

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