Báo cáo hóa học: " Research Article Convergence Theorem for Equilibrium Problems and Fixed Point Problems of Infinite Family of Nonexpansive Mappings" docx

Volume 2007, Article ID 64363, 12 pagesdoi:10.1155/2007/64363 Research Article Convergence Theorem for Equilibrium Problems and Fixed Point Problems of Infinite Family of Nonexpansive Ma

Volume 2007, Article ID 64363, 12 pages

doi:10.1155/2007/64363

Research Article

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

Yonghong Yao, Yeong-Cheng Liou, and Jen-Chih Yao

Received 17 March 2007; Accepted 20 August 2007

Recommended by Billy E Rhoades

We introduce an iterative scheme for finding a common element of the set of solutions

of an equilibrium problem and the set of common fixed points of infinite nonexpan-sive mappings in a Hilbert space We prove a strong-convergence theorem under mild assumptions on parameters

Copyright © 2007 Yonghong Yao 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

1 Introduction

LetH be a real Hilbert space and let C be a nonempty closed convex subset of H Let

h : C × C → R be an equilibrium bifunction, that is, h(u,u) =0 for everyu ∈ C Then one

can define the equilibrium problem that is to find an elementu ∈ C such that

Denote the set of solutions of EP(h) by SEP(h) This problem contains fixed point

problems, optimization problems, variational inequality problems, and Nash equilibrium problems as special cases, see [1] Some methods have been proposed to solve the equi-librium problem, please consult [2–4]

Recently, Combettes and Hirstoaga [2] introduced an iterative scheme of finding the best approximation to the initial data when SEP(h) =∅and proved a strong convergence theorem Motivated by the idea of Combettes and Hirstoaga, very recently, Takahashi and Takahashi [4] introduced a new iterative scheme by the viscosity approximation method for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping in a Hilbert space Their results extend and

improve the corresponding results announced by Combettes and Hirstoaga [2], Moudafi [5], Wittmann [6], and Tada and Takahashi [7]

In this paper, motivated and inspired by Combettes and Hirstoaga [2] and Takahashi and Takahashi [4], we introduce an iterative scheme for finding a common element of the set of solutions of EP(h) and the set of fixed points of infinite nonexpansive mappings in

a Hilbert space We obtain a strong convergence theorem which improves and extends the corresponding results of [2,4]

2 Preliminaries

LetH be a real Hilbert space with inner product ·,· and norm LetC be a nonempty

closed convex subset ofH Then for any x ∈ H, there exists a unique nearest point in C,

denoted byP C(x), such that x − P C(x) x − y for all y ∈ C Such a P Cis called the metric projection ofH onto C We know that P Cis nonexpansive Further, forx ∈ H and

x ∗ ∈ C,

x ∗ = P C(x) ⇐⇒x − x ∗,x ∗ − y≥0 ∀ y ∈ C. (2.1) Recall that a mappingT : C → H is called nonexpansive if Tx − T y x − y for allx, y ∈ C Denote the set of fixed points of T by F(T) It is well known that if C is a

bounded closed convex andT : C → C is nonexpansive, then F(T) =∅; see, for instance, [8] We call a mapping f : H → H contractive if there exists a constant α ∈(0, 1) such that

f (x) − f (y) α x − y for allx, y ∈ H.

For an equilibrium bifunctionh : C × C → R, we call h satisfying condition (A) if h

satisfies the following three conditions:

(i)h is monotone, that is, h(x, y) + h(y,x) ≤0 for allx, y ∈ C;

(ii) for eachx, y, z ∈ C, lim t ↓0h(tz + (1 − t)x, y) ≤ h(x, y);

(iii) for eachx ∈ C, y → h(x, y) is convex and lower semicontinuous.

If an equilibrium bifunctionh : C × C → R satisfies condition (A), then we have the

fol-lowing two important results You can find the first lemma in [1] and the second one in [2]

Lemma 2.1 Let C be a nonempty closed convex subset of H and let h be an equilibrium bifunction of C × C into R, satisfying condition (A) Let r > 0 and x ∈ H Then there exists

y ∈ C such that

Lemma 2.2 Assume that h satisfies the same assumptions as Lemma 2.1 For r > 0 and

x ∈ H, define a mapping S r:H → C as follows:

S r(x) =



y ∈ C : h(y,z) +1r  z − y, y − x ≥0,∀ z ∈ C (2.3)

for all y ∈ H Then the following holds:

(1)S r is single-valued and S r is firmly nonexpansive, that is, for any x, y ∈ H,

S r x − S r y 2

≤S r x − S r y,x − y; (2.4)

(2)F(S r)=SEP(h) and SEP(h) is closed and convex.

We also need the following lemmas for proving our main results.

Lemma 2.3 (see [9]) Let { x n } and { y n } be bounded sequences in a Banach space X and let { β n } be a sequence in [0, 1] with 0 < liminf n →∞ β n ≤lim supn →∞ β n < 1 Suppose x n+1 =

(1− β n)y n+β n x n for all integers n ≥ 0 and lim sup n →∞( y n+1 − y n x n+1 − x n )≤ 0 Then lim n →∞ y n − x n 0.

Lemma 2.4 (see [10]) Assume { a n } is a sequence of nonnegative real numbers such that

a n+1 ≤(1− γ n)a n+δ n , where { γ n } is a sequence in (0, 1) and { δ n } is a sequence such that

∞

n =1γ n = ∞ and lim sup n →∞ δ n /γ n ≤ 0 Then lim n →∞ a n = 0.

3 Iterative scheme and strong convergence theorems

In this section, we first introduce our iterative scheme Consequently, we will establish strong convergence theorems for this iteration scheme To be more specific, letT1,T2,

be infinite mappings ofC into C and let λ1,λ2, be real numbers such that 0 ≤ λ i ≤1 for everyi ∈ N For any n ∈ N, define a mapping W nofC into C as follows:

U n,n+1 = I,

U n,n = λ n T n U n,n+1+

1− λ n

I,

U n,n −1= λ n −1T n −1U n,n+

1− λ n −1

I,

U n,k = λ k T k U n,k+1+

1− λ k

I,

U n,2 = λ2T2U n,3+

1− λ2

I,

W n = U n,1 = λ1T1U n,2+

1− λ1

I.

(3.1)

Such a mappingW n is called theW-mapping generated by T n,T n −1, ,T1 andλ n,

λ n −1, ,λ1; see [11]

Now we introduce the following iteration scheme: Let f be a contraction of H into

it-self with coefficient α∈(0, 1) and givenx0∈ H arbitrarily Suppose the sequences { x n } ∞ n =1

and{ y n } ∞ n =1are generated iteratively by

hy n,x + 1

r n



x − y n,y n − x n

≥0, ∀ x ∈ C,

x n+1 = α n fx n

+β n x n+γ n W n y n,

(3.2)

where{ α n },{ β n }, and{ γ n }are three sequences in (0, 1) such thatα n+β n+γ n =1,{ r n }

is a real sequence in (0,∞),h is an equilibrium bifunction, and W nis theW-mapping

defined by (3.1)

We have the following crucial conclusions concerningW n You can find them in [12,

13] Now we only need the following similar version in Hilbert spaces

Lemma 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H Let T1,

T2, be nonexpansive mappings of C into C such that ∞

i =1F(T i ) is nonempty, and let

λ1,λ2, be real numbers such that 0 < λ i ≤ b < 1 for any i ∈ N Then for every x ∈ C and

k ∈ N, the limit lim n →∞ U n,k x exists.

Remark 3.2 From Lemma 3.1, we have that ifC is bounded, then for all ε > 0, there

exists a common positive integer numberN0such that forn > N0, U n,k x − U k(x) < ε

for allx ∈ C Indeed, by the similar argument to Lemma 3.2 in [13], letw ∈ ∞ n =1F(T n) SinceC is bounded, there exists a constant M > 0 such that x − w M for all x ∈ C.

Fixk ∈ N Then for all x ∈ C and any n ∈ N with n ≥ k, we have U n+1,k x − U n,k x

2(n+1

i = k λ i) x − w 2M(n+1 i = k λ i)

Let ε > 0 Then there exists n0∈ N with n0≥ k such that for all x ∈ C, b n0 − k+2 < ε(1 − b)/2M So for all x ∈ C and every m, n with m > n > n0, we have

U m,k x − U n,k x

≤

m−1

j = n

U j+1,k x − U j,k x  ≤ m−1

j = n

2

j+1

i = k

λ i



x − w



≤2M m −

1

j = n b j − k+2 ≤2Mb n − k+2

1− b < ε.

(3.3)

Remark 3.3 Using Lemma 3.1, one can define a mapping W of C into C as Wx =

limn →∞ W n x =limn →∞ U n,1 x for every x ∈ C Such a W is called the W-mapping

gen-erated byT1,T2, and λ1,λ2, We observe that if { x n }is a bounded sequence inC,

then we have

lim

n →∞Wx n − W n x n  =0. (3.4) Indeed, fromRemark 3.1, we have: for anyε > 0, there is n0such that Wx − W n x

ε for all x ∈ { x n } and for alln ≥ n0 In particular, Wx n − W n x n ε for all n ≥ n0 Consequently, limn →∞ Wx n − W n x n 0, as claimed

Throughout this paper, we will assume that 0< λ i ≤ b < 1 for every i ∈ N.

Lemma 3.4 Let C be a nonempty closed convex subset of a real Hilbert space H Let T1,

T2, be nonexpansive mappings of C into C such that ∞

i =1F(T i ) is nonempty, and let

λ1,λ2, be real numbers such that 0 < λ i ≤ b < 1 for any i ∈ N Then F(W) = ∞ i =1F(T i ).

Now we state and prove our main results

Theorem 3.5 Let C be a nonempty closed convex subset of a real Hilbert space H Let

h : C × C → R be an equilibrium bifunction satisfying condition (A) and let { T i } ∞ i =1 be an infinite family of nonexpansive mappings of C into C such that ∞

i =1F(T i)∩SEP(h) =∅ Suppose { α n } , { β n } , and { γ n } are three sequences in (0, 1) such that α n+β n+γ n = 1 and { r n } ⊂(0,∞ ) Suppose the following conditions are satisfied:

(i) limn →∞ α n = 0 and∞

n =0α n = ∞ ;

(ii) 0< liminf n →∞ β n ≤lim supn →∞ β n < 1;

(iii) lim infn →∞ r n > 0 and lim n →∞(r n+1 − r n)= 0.

Let f be a contraction of H into itself and given x0∈ H arbitrarily Then the sequences { x n } and { y n } generated iteratively by ( 3.2 ) converge strongly to x ∗ ∈ ∞ i =1F(T i)∩SEP(h), where

x ∗ = P ∞

i =1F(Ti)∩SEP(h) f (x ∗ ).

Proof Let Q = P ∞

i =1F(T i)∩SEP(h) Note that f is a contraction mapping with coefficient α ∈

(0, 1) Then Q f (x) − Q f (y) f (x) − f (y) α x − y for allx, y ∈ H Therefore,

Q f is a contraction of H into itself, which implies that there exists a unique element

x ∗ ∈ H such that x ∗ = Q f (x ∗) At the same time, we note thatx ∗ ∈ C.

Letp ∈ ∞ i =1F(T i)∩SEP(h) From the definition of S r, we note thaty n = S r n x n It fol-lows that y n − p S r n x n − S r n p x n − p Next, we prove that{ x n }and{ y n }are bounded From (3.1) and (3.2), we obtain

x n+1 − p  ≤ α nf

x n

− p+β nx n − p+γ nW n y n − p

≤ α nf

x n

− f (p)+f (p) − p +β nx n − p+γ ny n − p

≤ α n

αx n − p+f (p) − p +

1− α n x n − p

≤max

x0− p, 1

1− αf (p) − p.

(3.5)

Therefore,{ x n }is bounded We also obtain that{ y n },{ W n x n }, and{ f (x n)}are all bounded We shall useM to denote the possible different constants appearing in the

fol-lowing reasoning

Settingx n+1 = β n x n+ (1− β n)z nfor alln ≥0, we have that

z n+1 − z n = x n+2 − β n+1 x n+1

1− β n+1 − x n+1 − β n x n

1− β n

= α n+1

1− β n+1



fx n+1

− fx n

+

 α n+1

1− β n+1 − α n

1− β n



fx n

+ γ n+1

1− β n+1



W n+1 y n+1 − W n y n

+

 γ n+1

1− β n+1 − γ n

1− β n



W n y n

(3.6)

So we have

z n+1 − z n  ≤ αα n+1

1− β n+1

x n+1 − x n

+

 α n+1

1− β n+1 − α n

1− β n



fx n +W n y n

+ γ n+1

1− β n+1

W n+1 y n+1 − W n y n.

(3.7)

SinceT iandU n,iare nonexpansive from (3.1), we have

W n+1 y n − W n y n  =  λ1T1U n+1,2 y n − λ1T1U n,2 y n

≤ λ1 U n+1,2 y n − U n,2 y n

≤ λ1λ2 U n+1,3 y n − U n,3 y n

≤ ···

≤ λ1λ2··· λ nU n+1,n+1 y n − U n,n+1 y n

≤ Mn

i =1

λ i,

(3.8)

and hence

W n+1 y n+1 − W n y n  ≤  W n+1 y n+1 − W n+1 y n+W n+1 y n − W n y n

≤y n+1 − y n+Mn

i =1

Substituting (3.9) into (3.7), we have

z n+1 − z n  ≤ αα n+1

1− β n+1

x n+1 − x n+ α n+1

1− β n+1 − α n

1− β n



fx n +W n y n

+ γ n+1

1− β n+1

y n+1 − y n+ Mγ n+1

1− β n+1

n



i =1

λ i

(3.10)

On the other hand, fromy n = S r n x nandy n+1 = S r n+1 x n+1, we have

hy n,x + 1

r n



x − y n,y n − x n

≥0 ∀ x ∈ C, (3.11)

hy n+1,x + 1

r n+1



x − y n+1,y n+1 − x n+1

≥0 ∀ x ∈ C. (3.12)

Puttingx = y n+1in (3.11) andx = y nin (3.12), we have

hy n,y n+1

+ 1

r n



y n+1 − y n,y n − x n

hy n+1,y n

+ 1

r n+1



y n − y n+1,y n+1 − x n+1

≥0. (3.14) From the monotonicity ofh, we have

hy n,y n+1

+hy n+1,y n

So from (3.13), we can conclude that



y n+1 − y n,y n − x n

r n − y n+1 − x n+1

r n+1



and hence



y n+1 − y n,y n − y n+1+y n+1 − x n − r n

r n+1



y n+1 − x n+1 

≥0. (3.17) Since lim infn →∞ r n > 0, without loss of generality, we may assume that there exists a real

numberτ such that r n > τ > 0 for all n ∈ N Then we have

y n+1 − y n 2

≤



y n+1 − y n,x n+1 − x n+



1− r n

r n+1



y n+1 − x n+1 

≤y n+1 − y nx n+1 − x n+1− r n

r n+1



 y n+1 − x n+1, (3.18)

and hence

y n+1 − y n  ≤  x n+1 − x n+M

τr n+1 − r n. (3.19)

Substituting (3.19) into (3.10), we have

z n+1 − z n  ≤ αα n+1

1− β n+1

x n+1 − x n+ α n+1

1− β n+1 − α n

1− β n



fx n +W n y n

+ γ n+1

1− β n+1

x n+1 − x n+ γ n+1

1− β n+1 × M

τr n+1 − r n+ Mγ n+1

1− β n+1

n



i =1

λ i

≤x n+1 − x n+ α n+1

1− β n+1 − α n

1− β n



fx n

+W n y n

+ γ n+1

1− β n+1 × M

τr n+1 − r n+ Mγ n+1

1− β n+1

n



i =1

λ i

(3.20) This together withα n →0 andr n+1 − r n →0 imply that lim supn →∞( z n+1 − z n x n+1 −

x n )≤0 Hence byLemma 2.3, we obtain z n − x n 0 asn →∞ Consequently,

lim

n →∞x n+1 − x n  =lim

n →∞



1− β n z n − x n  =0. (3.21)

From (3.19) and limn →∞(r n+1 − r n)=0, we have limn →∞ y n+1 − y n 0 Sincex n+1 =

α n f (x n) +β n x n+γ n W n y n, we have

x n − W n y n  ≤  x n − x n+1+x n+1 − W n y n

≤x n − x n+1+α nf

x n

− W n y n+β nx n − W n y n, (3.22) that is,

x n − W n y n  ≤ 1

1− β n

x n − x n+1+ α n

1− β n

f

x n

− W n y n. (3.23)

Hence we have limn →∞ x n − W n y n 0 For p ∈ ∞ i =1F(T i)∩SEP(h), note that S r is firmly nonexpansive Then we have

y n − p 2

=S r

n x n − S r n p 2

≤S r n x n − S r n p,x n − p

=y n − p,x n − p

=1

2



y n − p 2

+x n − p 2

−x n − y n 2 

,

(3.24)

and hence

y n − p 2

≤x n − p 2

−x n − y n 2

Therefore, we have

x n+1 − p 2

≤ α nf

x n

− p 2 +β nx n − p 2

+γ nW n y n − p 2

≤ α nf

x n

− p 2 +β nx n − p 2

+γ ny n − p 2

≤ α nf

x n

− p 2 +β nx n − p 2 +γ n

x n − p 2

−x n − y n 2 

≤ α nf

x n

− p 2 +x n − p 2

− γ nx n − y n 2

.

(3.26)

Then we have

γ nx n − y n 2

≤ α nf

x n

− p 2 +x n − p+x n+1 − p

×x n − p − x n+1 − p

≤ α nf

x n

− p 2 +x n − x n+1x n − p+x n+1 − p . (3.27)

It is easily seen that lim infn →∞ γ n > 0 So we have

lim

n →∞x n − y n  =0. (3.28) From W n y n − y n W n y n − x n + x n − y n , we also have W n y n − y n 0 At that same time, we note that

W y n − y n  ≤  W y n − W n y n+W n y n − y n. (3.29)

It follows from (3.29) andRemark 3.2that limn →∞ W y n − y n 0 Next, we show that

lim sup

n →∞



fx ∗

− x ∗,x n − x ∗

wherex ∗ = P F(W) ∩SEP(h) f (x ∗) First, we can choose a subsequence{ y n j } of{ y n }such that

lim

j →∞



fx ∗

− x ∗,y n j − x ∗

=lim sup

n →∞



fx ∗

− x ∗,y n − x ∗

. (3.31)

Since{ y n j }is bounded, there exists a subsequence{ y n ji }of{ y n j }, which converges weakly

tow Without loss of generality, we can assume that y n j → w weakly From W y n − y n 0,

we obtainW y n j → w weakly Now we show w ∈SEP(h).

By y n = S r n x n, we haveh(y n,x) + (1/r n) x − y n,y n − x n ≥0 for allx ∈ C From the

monotonicity ofh, we have (1/r n) x − y n,y n − x n ≥ − h(y n,x) ≥ h(x, y n), and hence



x − y n j,y n j − x n j

r n j



≥ hx, y n j

Since (y n j − x n j)/r n j →0 andy n j → w weakly, from the lower semicontinuity of h(x, y) on

the second variabley, we have h(x,w) ≤0 for allx ∈ C For t with 0 < t ≤1 andx ∈ C,

letx t = tx + (1 − t)w Since x ∈ C and w ∈ C, we have x t ∈ C, and hence h(x t,w) ≤0 So from the convexity of equilibrium bifunctionh(x, y) on the second variable y, we have

0= hx t,x t

≤ thx t,x + (1− t)hx t,w ≤ thx t,x . (3.33)

Henceh(x t,x) ≥0 Therefore, we haveh(w,x) ≥0 for allx ∈ C, and hence w ∈SEP(h).

We will showw ∈ F(W) Assume w ∈ F(W) Since y n j → w weakly and w = Ww, from

Opial’s condition, we have

lim inf

j →∞

y n

j − w< liminf

j →∞

y n

j − Ww

≤lim inf

j →∞ y n

j − W y n j+W y n

j − Ww

≤lim inf

j →∞ y n

This is a contradiction So we getw ∈ F(W) = ∞ i =1F(T i) Therefore,w ∈ ∞ i =1F(T i)∩

SEP(h) Since x ∗ = P ∞

i =1F(Ti)∩SEP(h) f (x ∗), we have lim sup

n →∞



fx ∗

− x ∗,x n − x ∗

=lim

j →∞



fx ∗

− x ∗,x n j − x ∗

=lim

j →∞



fx ∗

− x ∗,y n j − x ∗

=fx ∗

− x ∗,w − x ∗

≤0.

(3.35)

First, we prove that{ x n }converges strongly tox ∗ ∈ ∞ i =1F(T i)∩SEP(h) From (3.2),

we have

x n+1 − x ∗ 2

≤β n

x n − x ∗

+γ n

W n y n − x ∗  2

+ 2α n

fx n

− x ∗,x n+1 − x ∗

≤β nx n − x ∗+γ nW n y n − x ∗  2

+ 2α n

fx ∗

− x ∗,x n+1 − x ∗

+ 2α n

fx n

− fx ∗ ,x n+1 − x ∗

≤β nx n − x ∗+γ ny n − x ∗ 2

+ 2αα nx n − x ∗x n+1 − x ∗

+ 2α n

fx ∗

− x ∗,x n+1 − x ∗

≤1− α n 2 x n − x ∗ 2

+αα nx n+1 − x ∗ 2

+x n − x ∗ 2

+ 2α n

fx ∗

− x ∗,x n+1 − x ∗

,

(3.36) which implies that

x n+1 − x ∗ 2

≤



1− α n 2

+αα n

1− αα n

x n − x ∗ 2

+ 2α n

1− αα n



fx ∗

− x ∗,x n+1 − x ∗

=1−2α n+αα n

1− αα n

x n − x ∗ 2

+ α2

n

1− αα n

x n − x ∗ 2

+ 2α n

1− αα n



fx ∗

− x ∗,x n+1 − x ∗

≤



1−2(1− α)α n

1− αα n



x n − x ∗ 2

+2(1− α)α n

1− αα n

× Mα n

2(1− α)+

1

1− α



fx ∗

− x ∗,x n+1 − x ∗

=1− ϕ n x n − x ∗ 2

+φ n ϕ n,

(3.37)

whereϕ n =2(1− α)α n /(1 − αα n) andφ n = Mα n /2(1 − α) + 1/(1 − α)  f (x ∗)− x ∗,x n+1 −

x ∗ It is easily seen that∞

n =0ϕ n = ∞and lim supn →∞ φ n ≤0 Now applyingLemma 2.4

and (3.35) to (3.37), we conclude thatx n → x ∗ (n →∞) Consequently, from (3.28), we havey n → x ∗(n →∞) This completes the proof 

Corollary 3.6 Let C be a nonempty closed convex subset of a real Hilbert space H Let

h : C × C → R be an equilibrium bifunction satisfying condition (A) such that SEP(h) =∅ Let { α n } , { β n } , and { γ n } be three sequences in (0, 1) such that α n+β n+γ n = 1 and { r n } ⊂

(0,∞ ) is a real sequence Suppose the following conditions are satisfied:

(i) limn →∞ α n = 0 and∞

n =0α n = ∞ ;

(ii) 0< liminf n →∞ β n ≤lim supn →∞ β n < 1;

(iii) lim infn →∞ r n > 0 and lim n →∞(r n+1 − r n)= 0.

Lemma 3.4 Let C be a nonempty... (3.12)

Puttingx = y n+1in (3.11) and< i>x =... =0. (3.21)

From (3.19) and limn →∞(r n+1