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Volume 2010, Article ID 165098, 18 pagesdoi:10.1155/2010/165098 Research Article A New Iterative Method for Solving Equilibrium Problems and Fixed Point Problems for Infinite Family of N

Volume 2010, Article ID 165098, 18 pages

doi:10.1155/2010/165098

Research Article

A New Iterative Method for Solving

Equilibrium Problems and Fixed Point Problems for Infinite Family of Nonexpansive Mappings

1 School of Mathematics and Physics, North China Electric Power University, Baoding 071003, China

2 Department of Mathematics Education and the RINS, Gyeongsang National University,

Chinju 660-701, Republic of Korea

3 Department of Mathematics, Hangzhou Normal University, Hangzhou 310036, China

Correspondence should be addressed to Yeol Je Cho,yjcho@gnu.ac.kr

Received 7 January 2010; Revised 21 May 2010; Accepted 11 July 2010

Academic Editor: Simeon Reich

Copyrightq 2010 Shenghua Wang et al 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 introduce a new iterative scheme for finding a common element of the solutions sets of a finite family of equilibrium problems and fixed points sets of an infinite family of nonexpansive mappings in a Hilbert space As an application, we solve a multiobjective optimization problem using the result of this paper

1 Introduction

bifunction ofC × C into R, where R is the set of real numbers The equilibrium problem for

the bifunctionΦ : C × C → R is to find x ∈ C such that

Φx, y≥ 0, ∀y ∈ C. 1.1

The set of solutions of the above inequality is denoted by EPΦ Many problems arising from physics, optimization, and economics can reduce to finding a solution of an equilibrium problem

In 2007, S Takahashi and W Takahashi 1 first introduced an iterative scheme by the viscosity approximation method for finding a common element of the solutions set of equilibrium problem and the set of fixed points of a nonexpansive mapping in a Hilbert space

H and proved a strong convergence theorem which is based on Combettes and Hirstoaga’s

result2 and Wittmann’s result 3 More precisely, they obtained the following theorem

Theorem 1.1 see 1 Let C be a nonempty closed and convex subset of H Let Φ : C × C → R be

a bifunction which satisfies the following conditions:

A1 Φx, x  0 for all x ∈ C;

A2 Φ is monotone, that is, Φx, y  Φy, x ≤ 0 for all x, y ∈ C;

A3 For all x, y, z ∈ C,

lim

A4 For each x ∈ C, y → Φx, y is convex and lower semicontinuous.

Φu n , y 1

rn



x n1  α nfxn   1 − α n Su n , ∀n ≥ 1,

1.3

lim

n1

n1

|α n1 − α n | < ∞,

lim inf

n1

|r n1 − r n | < ∞.

1.4

P is the metric projection of H onto C and PFixS∩EPΦfz denotes nearest point in FixS∩EPΦ

Recently, many results on equilibrium problems and fixed points problems in the context of the Hilbert space and Banach space are introducedsee, e.g., 4 8

Let F : H → H be a nonlinear mapping The variational inequality problem

corresponding to the mappingF is to find a point x∗∈ C such that

Fx∗, x − x∗ ≥ 0, ∀x ∈ C. 1.6

The variational inequality problem is denoted by VIF, C 9

The mapping F is called κ-Lipschitzian and η-strongly monotone if there exist

constantsκ, η > 0 such that

Fx − Fy ≤ κx − y, ∀x,y ∈ H, 1.7



respectively It is well known that if F is strongly monotone and Lipschitzian on C, then

VIF, C has a unique solution An important problem is how to find a solution of VIF, C Recently, there are many results to solve the VIF, C see, e.g., 10–14

be a countable family of nonexpansive mappings, and{Φi}m

satisfying conditionsA1–A4 such that Ω ∞n1FixTn ∩ EPΦ1 ∩ · · · ∩ EPΦm  / ∅ Let

T r i x 

z ∈ C : Φ iz, yr1

i



Lemma 2.5see below shows that, for each 1 ≤ i ≤ m, T r i is firmly nonexpansive and hence nonexpansive and FixTr i  EPΦi  Suppose that F : H → H is a κ-Lipschitzian and

In this paper, motivated and inspired by the above research results, we introduce the following iterative process for finding an element inΩ: for an arbitrary initial point x1∈ H,

zn  γ1Tr1xn  γ2Tr2xn  · · ·  γ mTr m xn,

i1

α i−1 − α i σ nTixn  1 − α n 1 − σ n T λ n zn, ∀n ≥ 1, 1.10

whereT λ n zn  z n − λ nμFzn , α0  1, {α n}∞n1is a strictly decreasing sequence in0, α with

0< α < 1, {λ n}∞

n1 ⊂ 0, 1, {γ i}m

i1 ⊂ 0, 1 withm

i1 γ i  1, and {σ n}∞

n1 ⊂ a, b with 0 < a, b <

1 Then we prove that the iterative process {x n} defined by 1.10 strongly converge to an elementx∗∈ Ω, which is the unique solution of the variational inequality

Fx∗, x − x∗ ≥ 0, ∀x ∈ Ω. 1.11

As an application of our main result, we solve a multiobjective optimization problem

2 Preliminaries

For all x ∈ FixT and x ∈ H, we have

x − x2≥ Tx − T x2 Tx − x2 Tx − x  x − x2

 Tx − x2 x − x2 2Tx − x, x − x 2.1

and hence

Tx − x2≤ 2x − Tx, x − x, ∀x ∈ FixT, x ∈ H. 2.2

It is well known that, for allx, y ∈ H and t ∈ 0, 1,

tx  1 − ty2

≤ tx2 1 − ty2, 2.3 which implies that







n



i1

t i x i





2

≤n

i1

t i x i2 2.4

for all{x i}n i1 ⊂ H and {t i}n i1 ⊂ 0, 1 withn i1 ti  1

unique nearest point inC, denoted by P C x, such that

Moreover, we have the following:

We need the following lemmas for our main results

Lemma 2.1 see 15 Let C be a nonempty closed and convex subset of a Hilbert space H and T a

Lemma 2.2 see 10, Lemma 3.1b Let H be a Hilbert space and T : H → H be a nonexpansive

Lemma 2.3 see 16 Let {sn }, {c n } be the sequences of nonnegative real numbers and {a n} ⊂

0, 1 Suppose that {b n } is a real number sequence such that

s n1 ≤ 1 − a n s n  b n  c n , ∀n ≥ 0. 2.9

(2) If

∞



n0

a n  ∞, lim sup

n → ∞

bn

a n ≤ 0, 2.10 then lim n → ∞ s n  0.

Lemma 2.4 see 17 Let C be a nonempty closed and convex subset of a Hilbert space H and

Φ : C × C → R be a bifunction which satisfies the conditions (A1)–(A4) Let r > 0 and x ∈ H Then

Lemma 2.5 see 2 Let H be a Hilbert space and C be a nonempty closed and convex subset of H

T r x 

r



Then the following holds:

1 T r is single-valued;

2 T r is firmly nonexpansive, that is, for any x, y ∈ H,

3 FixT r   EPΦ;

4 EPΦ is closed and convex.

The following lemma is an immediate consequence of an inner product.

Lemma 2.6 Let H be a real Hilbert space Then the following identity holds:

x  y2

≤ x2 2y, x  y, ∀x, y ∈ H. 2.14

3 Main Results

First, we prove some lemmas as follows

Lemma 3.1 The sequence {x n } generated by 1.10 is bounded.

firmly-nonexpansive and hence firmly-nonexpansive Hence, for each 1≤ i ≤ m and p ∈ Ω, we have

u in − p  T r i xn − T r i p  ≤ x n − p, ∀n ≥ 1, 3.1

z n − p ≤m

i1

γi u in − p ≤ x n − p, ∀n ≥ 1. 3.2

ByLemma 2.2, we have



whereτ  1 − 1− μ2η − μκ2 ∈ 0, 1 Therefore, by 3.2 and 3.3, we obtain note that {α n } is strictly decreasing and T λ n p − p  −λnμFp

x n1 − p α n

x n − pn

i1

α i−1 − α i σ n

T i x n − p 1 − α n 1 − σ nT λ n z n − p



≤ α n x n − p n

i1

α i−1 − α i σ n T ixn − p  1 − α n 1 − σ nT λ n zn − p

≤ α n x n − p n

i1

α i−1 − α i σ n x n − p

 1 − α n 1 − σ nT λ n zn − T λ n p T λ n p − p

≤ α n x n − p n

i1

α i−1 − α i σ n x n − p

 1 − α n 1 − σ n1 − λ nτ z n − p  λ n μ Fp

≤ α n x n − p n

i1

α i−1 − α i σ n x n − p

 1 − α n 1 − σ n1 − λ nτ x n − p  λ nμ Fp

 1 − 1 − α n 1 − σ n λ nτ x n − p  1 − α n 1 − σ n λ nμ Fp.

3.4

By induction, we obtainx n1 ≤ max{x1− p, μ/τFp} Hence {x n} is bounded and so are{z n } and {u in } for each i  1, 2, , m Since F is κ-Lipschitzian, we have

Fz n ≤Fz n  − Fp   Fp

≤ κz n − p  Fp ≤ κz n  κp  Fp, 3.5

which shows that{Fz n} is bounded This completes the proof

Lemma 3.2 If the following conditions hold:

∞



n1

n1

|λ n − λ n1 | < ∞, ∞

n1

|σ n − σ n1 | < ∞, 3.6

then lim n → ∞ x n1 − x n  0.

u in−1 − u in  T r i x n−1 − T r i x n ≤ x n−1 − x n , ∀n ≥ 1. 3.7

By3.7, we have

z n − z n−1 γ1u1n − u1n−1   γ2u2n − u2n−1   · · ·  γ m u mn − u mn−1

≤m

i1

γ i u in − u in−1 ≤m

i1

γ i x n − x n−1

 x n−1 − x n , ∀n ≥ 1.

3.8

By the definition of the iterative sequence1.10, we have

xn1 − x n  α n x n − x n−1   α nxn−1n

i1

α i−1 − α i σ n T ixn − T ixn−1

n

i1

α i−1 − α i σ nTixn−1  1 − α n 1 − σ nT λ n zn − T λ n zn−1

 1 − α n 1 − σ n T λ n z n−1 − α n−1 x n−1−n−1

i1

α i−1 − α i σ n−1 T i x n−1

− 1 − α n−1 1 − σ n−1 T λ n−1 z n−1

 α n x n − x n−1   α n − α n−1 x n−1n

i1

α i−1 − α i σ n T ixn − T ixn−1

 1 − α n 1 − σ nT λ n zn − T λ n zn−1n

i1

α i−1 − α i σ nTixn−1

−n−1

i1

α i−1 − α i σ n−1Tixn−1  1 − α n 1 − σ n T λ n zn−1

− 1 − α n−1 1 − σ n−1 T λ n−1 z n−1

 α n x n − x n−1   α n − α n−1 x n−1n

i1

α i−1 − α i σ n T ixn − T ixn−1

 1 − α n 1 − σ nT λ n z n − T λ n z n−1n−1

i1

α i−1 − α i σ n − σ n−1 T i x n−1

 α n−1 − α n σ n T n x n−1  α n−1 − α n 1 − σ n   σ n−1 − σ n 1 − α n−1 z n−1

 {1 − α n−1 1 − σ n−1 λ n−1 − λ n

−α n−1 − α n 1 − σ n   σ n−1 − σ n 1 − α n−1 λ n }μFz n−1 ,

3.9

and hence

x n1 − x n ≤ α n x n − x n−1  α n−1 − α n x n−1 n

i1

α i−1 − α i σ n x n − x n−1

 1 − α n 1 − σ n 1 − λ nτzn − z n−1 n−1

i1

α i−1 − α i |σ n − σ n−1 |T i x n−1

 α n−1 − α n T n x n−1  α n−1 − α n   |σ n−1 − σ n |z n−1

 |λ n−1 − λ n |  α n−1 − α n   |σ n−1 − σ n |μFz n−1

 α n x n − x n−1 n

i1

α i−1 − α i σ n x n − x n−1

 1 − α n 1 − σ n 1 − λ nτzn − z n−1

 α n−1 − α nx n−1  T nxn−1  z n−1  μFz n−1

n−1

i1

α i−1 − α i |σ n − σ n−1 |T i x n−1  |σ n−1 − σ n|z n−1  μFz n−1

 |λ n−1 − λ n |μFz n−1 

3.10

It follows from3.8 and 3.10 that

x n1 − x n ≤ α n x n − x n−1 n

i1

α i−1 − α i σ n x n − x n−1

 1 − α n 1 − σ n 1 − λ nτxn−1 − x n

 α n−1 − α nx n−1  T n x n−1  z n−1  μFz n−1

n−1

i1

α i−1 − α i |σ n − σ n−1 |T ixn−1  |σ n−1 − σ n|z n−1  μFz n−1

 |λ n−1 − λ n |μFz n−1

≤ 1 − 1 − α n 1 − σ n λ nτxn − x n−1  α n−1 − α n3 μM

 |σ n − σ n−1|2 μM  |λn−1 − λ n |μM

≤ 1 − 1 − α1 − bλ nτxn − x n−1  α n−1 − α n3 μM

 |σ n − σ n−1|2 μM  |λn−1 − λ n |μM,

3.11

where M  max{sup n≥1 x n , sup n≥1 z n , sup i≥1,n≥1 T ixn , sup n≥1 Fz n } Since {α n} is strictly decreasing, we have ∞

n2 α n−1 − α n   α1 < ∞ Further, from the assumptions, it

follows that

∞



n2



Therefore, byLemma 2.3, we have limn → ∞ x n1 − x n  0 This completes the proof

Lemma 3.3 If the following conditions hold:

lim

n1

n1

|λ n − λ n1 | < ∞, ∞

n1

|σ n − σ n1 | < ∞, 3.13

then lim

n → ∞ x n − u in  0 for each i  1, 2, , m.

u in − p2 T r i x n − T r i p2≤ T r i x n − T r i p, x n − p  u in − p, x n − p

 1 2

u

in − p2x n − p2

− u in − x n2

and henceu in − p2 ≤ x n − p2− u in − x n2 Further, we have

z n − p2





m



i1

γi u in − p





2

≤m

i1

γi u in − p2

≤m

i1

γ ix

n − p2

− u in − x n2

x n − p2−m

i1

γ i u in − x n2, ∀n ≥ 1.

3.15

Therefore, from2.4 and 3.3, we have

x n1 − p2



α n x n − p 

n



i1

α i−1 − α i σ n T ixn − p  1 − α n 1 − σ n T λ n zn − p





2

≤ α n x n − p2n

i1

α i−1 − α i σ n T ixn − p2

 1 − α n 1 − σ nT λ n zn − p2

≤ α n x n − p2 1 − α n σ n x n − p2

 1 − α n 1 − σ nT λ n z n − T λ n p T λ n p − p2

≤ α n x n − p2

 1 − α n σ n x n − p2

 1 − α n 1 − σ n1 − λ n τ z n − p  λ n μ Fp2

≤ α n x n − p2

 1 − α n σ n x n − p2

 1 − α n 1 − σ n

×1 − λ nτ z n − p2 2λ n 1 − λ nτμ z n − pFp  λ n μ2Fp2

≤ α n x n − p2 1 − α n σ n x n − p2 1 − α n 1 − σ n

×



1 − λ nτ



x n − p2−m

i1

γ i u in − x n2



2λ n 1 − λ nτμ z n − pFp  λ nμ2Fp2



 1 − 1 − α n 1 − σ n λ nτ x n − p2

− 1 − α n 1 − σ n 1 − λ nτm

i1

γi u in − x n2

 2λ nμ1 − αn 1 − σ n 1 − λ nτ z n − pFp  1 − α n 1 − σ n λ n μ2Fp2

≤x n − p2

− 1 − α n 1 − σ n 1 − λ nτm

i1

γi u in − x n2

 1 − α n 1 − σ n λ n μ2Fp2

 2λ nμ1 − αn 1 − σ n 1 − λ nτ z n − pFp.

3.16

It follows that

γi 1 − α n 1 − σ n 1 − λ nτuin − x n2

≤x n − p  x n1 − px n − x n1  λ nμ2Fp2 2μz n − pFp 3.17

for each i  1, 2, , m Note that 0 < γi < 1 for i  1, 2, , m From the assumptions,

Lemma 3.2, and the previous inequality, we conclude thatu in − x n → 0 as n → ∞ for

z n − x n ≤m

i1

γi u in − x n −→ 0 n −→ ∞. 3.18

This completes the proof

Lemma 3.4 If the following conditions hold:

lim

n1

n1

|λ n − λ n1 | < ∞, ∞

n1

|σ n − σ n1 | < ∞, 3.19

then lim n → ∞ x n − T ixn  0 for all i ≥ 1.

i1

α i−1 − α i σ n x n − T ixn  − 1 − α n σ nxn  α nxn  1 − α n 1 − σ n T λ n zn, 3.20

that is,

n



i1

α i−1 − α i σ n x n − T ixn   x n − x n1 − x n  α nxn  1 − α n σ nxn  1 − α n 1 − σ n T λ n zn

 x n − x n1  1 − α n σ n − 1x n  1 − α n 1 − σ n T λ n z n

 x n − x n1  1 − α n 1 − σ nT λ n z n − x n.

3.21

Hence, for anyp ∈ Ω, we get

n



i1

α i−1 − α i σ n

xn − T ixn, xn − p 1 − α n 1 − σ n T λ n zn − x n, xn − p  x n − x n1, xn − p.

3.22 Since eachTiis nonexpansive, by2.2, we have

T ixn − x n2≤ 2x n − T ixn, xn − p. 3.23 Hence, combining this inequality with3.22, we get

1

2

n



i1

α i−1 − α i σ n T ixn − x n2≤ 1 − α n 1 − σ n T λ n zn − x n, xn − p  x n − x n1, xn − p,

3.24 which implies thatnote that {α n} is a strictly decreasing sequence

T i x n − x n2≤ 21 − αn 1 − σ n

α i−1 − α i σ n T λ n z n − x n , x n − p  2

α i−1 − α i σ n x n − x n1 , x n − p

≤ 21 − αn 1 − σ n

α i−1 − α i σ n



T λ n z n − x nxn − p  2

α i−1 − α i σ n x n − x n1xn − p.

3.25

FromLemma 3.3, limn → ∞ λ n 0, and the inequality



T λ n zn − x n ≤ z

n − x n  λ nμFzn , 3.26

we obtain

lim

n → ∞



T λ n zn − x n  0. 3.27 Therefore, fromLemma 3.2,3.25, and 3.27, it follows that

lim

This completes the proof

Next we prove the main results of this paper.

Theorem 3.5 Assume that the following conditions hold:

lim

n1

n1

|λ n − λ n1 | < ∞,

∞



n1 γ n − γ n1  < ∞, ∞

n1

|σ n − σ n1 | < ∞.

3.29

Fx∗, x∗− x ≥ 0, ∀x ∈ Ω. 3.30 First, we prove that

lim sup

Since{x n } is bounded, there exists a subsequence {x n j } of {x n} such that

lim sup

n → ∞ −Fx∗, x n − x∗  lim

Without loss of generality, we may further assume that xn j  x for some x ∈ H From

Lemmas 3.4 and 2.1, we get x ∈ FixT n  for all n ≥ 1 Hence we have x ∈ ∞n1FixTn

It follows from Lemma 2.5that each T r i is firmly nonexpansive and hence nonexpansive Lemma 3.3shows thatT r i xn − x n → 0 as n → ∞ Therefore, fromLemma 2.1, it follows that x ∈ FixT r i  for each i  1, , m, which shows that x ∈ m i1FixTr i.Lemma 2.5shows that FixTr i  EPΦi  for each i  1, , m Hence x ∈ m

i1EPΦi By using the above argument, we conclude that

x ∈ Ω ∞

n1

FixTn ∩ EPΦ1 ∩ · · · ∩ EPΦm . 3.33

Noting thatx∗is a solution of the VIF, Ω, we obtain

lim sup

Lemma 2.2 see... class="text_page_counter">Trang 10

and henceu in − p2 ≤ x n − p2− u in...