Hybrid iterative scheme for generalized equilibrium problems and fixed point problems of finite family of nonexpansive mappings

Nonlinear Analysis: Hybrid Systems 3 (2009) 296–309

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Nonlinear Analysis: Hybrid Systems journal homepage: www.elsevier.com/locate/nahs

Hybrid iterative scheme for generalized equilibrium problems and fixed point problems of finite family of nonexpansive mappings Atid Kangtunyakarn, Suthep Suantai ∗ Department of Mathematics, Faculty of Science, Chiang Mai University, Chiang Mai, 50200, Thailand

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Article history: Received 14 January 2009 Accepted 21 January 2009 Keywords: Strong convergence Finite families of nonexpansive mapping Fixed point Generalized equilibrium problem Inverse-strongly monotone

In this paper, we introduce a new mapping and a Hybrid iterative scheme for finding a common element of the set of solutions of a generalized equilibrium problem and the set of common fixed points of a finite family of nonexpansive mappings in a Hilbert space. Then, we prove the strong convergence of the proposed iterative algorithm to a common fixed point of a finite family of nonexpansive mappings which is a solution of the generalized equilibrium problem. The results obtained in this paper extend the recent ones of Takahashi and Takahashi [S. Takahashi, W. Takahashi, Strong convergence theorem for a generalized equilibrium problem and a nonexpansive mapping in a Hilbert space, Nonlinear Anal. 69 (2008) 1025–1033]. © 2009 Elsevier Ltd. All rights reserved.

1. Introduction Let H be a real Hilbert space and let C be a nonempty closed convex subset of H and A : C → H be a nonlinear mapping and let PC be the projection of H onto the convex subset C . A mapping T of H into itself is called nonexpansive if kTx−Tyk ≤ kx−yk for all x, y ∈ H. We denote by F (T ) the set of fixed points of T (i.e. F (T ) = {x ∈ H : Tx = x}). Goebel and Kirk [1] showed that F (T ) is always closed convex, and also nonempty provided T has a bounded trajectory. Let {Ti }Ni=1 be a finite family of nonexpansive mappings with i=1 F (Ti ) 6= ∅. Let F : C × C → R be a bifunction. The equilibrium problem for F is to determine its equilibrium points, i.e. the set

TN

EP (F ) = {x ∈ C : F (x, y) ≥ 0, ∀y ∈ C }.

(1.1)

Many problems in physics, optimization, and economics require some elements of EP (F ), see [2–7]. Several iterative methods have been proposed to solve the equilibrium problem, see for instance [3,5–7]. In 2005, Combettes and Hirstoaga [3] introduced an iterative scheme for finding the best approximation to the initial data when EP (F ) is nonempty and proved a strong convergence theorem. The variational inequality problem is to find u ∈ C such that

hAu, v − ui ≥ 0

(1.2)

for all v ∈ C . The set of solutions of the variational inequality is denoted by VI (C , A). For a bifunction F : C × C → R and a nonlinear mapping A : C → H, we consider the following equilibrium problem: Find z ∈ C such that F (z , y) + hAz , y − z i ≥ 0,

∀y ∈ C .

The set of such z ∈ C is denoted by EP, i.e., EP = {z ∈ C : F (z , y) + hAz , y − z i ≥ 0, ∀y ∈ C }.

∗

Corresponding author. E-mail addresses: [email protected] (A. Kangtunyakarn), [email protected] (S. Suantai).

1751-570X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.nahs.2009.01.012

(1.3)

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In the case of A ≡ 0, EP is denoted by EP (F ). In the case of F ≡ 0, EP is also denoted by VI (C , A). Numerous problems in physics, optimization, variational inequalities, minimax problems, the Nash equilibrium problem in noncooperative games, economics reduce to finding a solution of (1.3) see, for instance, [2,4]. A mapping A of C into H is called α -inverse strongly monotone, see [8], if there exists a positive real number α such that

hx − y, Ax − Ayi ≥ αkAx − Ayk2 for all x, y ∈ C . For r > 0, let Tr : H → C be defined by Tr (x) =



z ∈ C : F (z , y) +

1 r

 hy − z , z − xi ≥ 0, ∀y ∈ C .

(1.4)

Combettes and Hirstoaga [9] showed that under some suitable conditions of F , Tr is single-valued and firmly nonexpansive and satisfies F (Tr ) = EP (F ). In 2007, Takahashi and Takahashi [6] introduced a hybrid viscosity approximation method in the framework of a real Hilbert space H. They defined the iterative sequences {xn } and {un } as follows:

 x ∈ H , arbitrarily;   1 1 F (un , y) + hy − un , un − xn i ≥ 0, ∀y ∈ C ,  r n  xn+1 = αn f (xn ) + (1 − αn )Tun , ∀n ∈ N,

(1.5)

where f : H −→ H is a contraction mapping with a constant α ∈ (0, 1) and {αn } ⊂ [0, 1], {rn } ⊂ (0, ∞). They proved, under T some suitable conditions on the sequence {αn }, {rn } and bifunction F , that {xn } and {un } strongly converge to z ∈ F (T ) EP (F ), where z = PF (T ) T EP (F ) f (z ). Recently, in 2008, Takahashi and Takahashi [7] introduced a hybrid iterative method for finding a common element of EP and F (T ). They defined {xn } in the following way:

 u, x ∈ C , arbitrarily;   1 1 F (zn , y) + hAxn , y − zn i + hy − zn , zn − xn i ≥ 0, ∀y ∈ C ,  λ n  xn+1 = βn xn + (1 − βn )T (an u + (1 − an )zn ), ∀n ∈ N,

(1.6)

where A be an α -inverse strongly monotone mapping of C into H with positive real number α , and {an } ∈ [0, 1], {βn } ⊂ T T [0, 1], {λn } ⊂ [0, 2α], and proved strong convergence of the scheme (1.6) to z ∈ Ni=1 F (Ti ) EP, where z = PTN F (T ) T EP u i=1

in the framework of a Hilbert space, under some suitable conditions on {an }, {βn }, {λn } and bifunction F . In 1999, Atsushiba and Takahashi [10] defined the mapping Wn as follows: Un,1 Un,2 Un,3

Un,N −1 Wn

= = = · · · · = =

i

λn,1 T1 + (1 − λn,1 )I , λn,2 T2 Un,1 + (1 − λn,2 )I , λn,3 T3 Un,2 + (1 − λn,3 )I , (1.7)

λn,N −1 TN − 1Un,N −2 + (1 − λn,N −1 )I , Un,N = λn,N TN Un,N −1 + (1 − λn,N )I ,

where {λn,i }Ni ⊆ [0, 1]. This mapping is called the W-mapping generated by T1 , T2 , . . . , TN and λn,1 , λn,2 , . . . , λn,N . In

2000, Takahashi and Shimoji [11] proved that if X is a strictly convex Banach space, then F (Wn ) = i=1 F (Ti ), where 0 < λn,i < 1, i = 1, 2, . . . , N . Let X be a real Hilbert space and C a nonempty closed convex subset of X and let {Ti }Ni=1 be a finite family of nonexpansive

TN

(n)

mappings of C into itself. For each n ∈ N, and j = 1, 2, . . . , N, let αj n ,j

n ,j

n ,j

= (α1n,j , α2n,j , α3n,j ) be such that α1n,j , α2n,j , α3n,j ∈ [0, 1]

with α1 + α2 + α3 = 1. We define mapping Sn : C → C as follows: Un,0 Un,1 Un,2 Un,3

Un,N −1 Sn

= = = = · · · = =

I

α1n,1 T1 Un,0 + α2n,1 Un,0 + α3n,1 I α1n,2 T2 Un,1 + α2n,2 Un,1 + α3n,2 I α1n,3 T3 Un,2 + α2n,3 Un,2 + α3n,3 I

α1n,N −1 TN −1 Un,N −2 + α2n,N −1 Un,N −2 + α3n,N −1 I n ,N n ,N n,N Un,N = α1 TN Un,N −1 + α2 Un,N −1 + α3 I .

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(n)

(n)

(n)

The mapping Sn is called the S-mapping generated by T1 , T2 , . . . , TN and α1 , α2 , . . . , αN . For given u ∈ C and x1 ∈ C , let {zn } ⊂ C and {xn } ⊂ C be sequences generated by

(

1

hy − zn , zn − xn i ≥ 0, ∀y ∈ C , (1.8) λn xn+1 = βn xn + (1 − βn )Sn (an u + (1 − an )zn ). ∀n ∈ N. TN n ,j n ,N In this paper, we show that if X is strictly convex, then F (Sn ) = i=1 F (Ti ) if α1 ∈ (0, 1) for all j = 1, 2, . . . , N − 1, α1 ∈ (0, 1] and α2n,j , α3n,j ∈ [0, 1) for all j = 1, 2, . . . , N, and we prove that under some suitable conditions, the sequence {xn } converges strongly to a point z = PTN F (T ) T EP u. i F (zn , y) + hAxn , y − zn i +

i=1

2. Preliminaries In this section, we collect and give some useful lemmas that will be used for our main result in the next section. Let C be the closed convex subset of a real Hilbert space H, let PC be the metric projection of H onto C i.e., for x ∈ H, PC x satisfies the property

kx − PC xk = min kx − yk. y∈C

The following characterizes the projection PC . Lemma 2.1 (See [12]). Given x ∈ H and y ∈ C . Then PC x = y if and only if there holds the inequality hx − y, y − z i ≥ 0 ∀z ∈ C . Lemma 2.2 (See [11]). In a strictly convex Banach space E, if

kxk = kyk = kλx + (1 − λ)yk for all x, y ∈ E and λ ∈ (0, 1), then x = y. Lemma 2.3 (See [13]). Let {sn } be a sequence of nonnegative real numbers satisfying sn+1 = (1 − αn )sn + αn βn , ∀n ≥ 0 where {αn }, {βn } satisfy the conditions

(1) {αn } ⊂ [0, 1],

∞ X

αn = ∞,

n =1

(2) lim sup βn ≤ 0. n→∞

Then limn→∞ sn = 0. Lemma 2.4 (See [14]). Let {xn } and {zn } be bounded sequences in a Banach space X and let {βn } be a sequence in [0, 1] with 0 < lim infn→∞ βn ≤ lim supn→∞ βn < 1. Suppose xn+1 = βn xn + (1 − βn )zn for all integer n ≥ 0 and lim supn→∞ (kzn+1 − zn k − kxn+1 − xn k) ≤ 0. Then limn→∞ kxn − zn k = 0. For solving the equilibrium problem for a bifunction F : C ×C → R, let us assume that F satisfies the following conditions: (A1) F (x, x) = 0 ∀x ∈ C ; (A2) F is monotone, i.e. F (x, y) + F (y, x) ≤ 0, ∀x, y ∈ C ; (A3) ∀x, y, z ∈ C , lim F (tz + (1 − t )x, y) ≤ F (x, y);

t →0+

(A4) ∀x ∈ C , y 7→ F (x, y) is convex and lower semicontinuous. The following lemma appears implicitly in [2]. Lemma 2.5 (See [2]). Let C be a nonempty closed convex subset of H and let F be a bifunction of C × C into R satisfying (A1)–(A4). Let r > 0 and x ∈ H. Then, there exists z ∈ C such that F (z , y) +

1 r

hy − z , z − xi ≥ 0

(2.1)

for all y ∈ C . Lemma 2.6 (See [9]). Assume that F : C × C → R satisfies (A1)–(A4). For r > 0 and x ∈ H, define a mapping Tr : H → C as follows: Tr (x) =



z ∈ C : F (z , y) +

1 r

hy − z , z − xi ≥ 0, ∀y ∈ C

for all z ∈ H. Then, the following hold:

 (2.2)

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299

(1) Tr is single-valued; (2) Tr is firmly nonexpansive i.e. kTr (x) − Tr (y)k2 ≤ hTr (x) − Tr (y), x − yi ∀x, y ∈ H ; (3) F (Tr ) = EP (F ); (4) EP (F ) is closed and convex. Definition 2.7. Let C be a nonempty convex subset of real Banach space. Let {Ti }Ni=1 be a finite family of nonexpansive j

j

j

j

j

j

j

j

j

mappings of C into itself. For each j = 1, 2, . . . , N, let αj = (α1 , α2 , α3 ) where α1 , α2 , α3 ∈ [0, 1] and α1 + α2 + α3 = 1. We define the mapping S : C → C as follows: U0 U1 U2 U3

U N −1 S

= = = = · · · = =

I

α11 T1 U0 + α21 U0 + α31 I α12 T2 U1 + α22 U1 + α32 I α13 T3 U2 + α23 U2 + α33 I (2.3)

α1N −1 TN −1 UN −2 + α2N −1 UN −2 + α3N −1 I UN = α1N TN UN −1 + α2N UN −1 + α3N I .

This mapping is called S-mapping generated by T1 , . . . , TN and α1 , α2 , . . . , αN . Next, we prove a lemma which is very useful for our consideration. Lemma 2.8. Let C be a nonempty closed convex subset of a strictly convex Banach space. Let {Ti }Ni=1 be a finite family of nonexpansive mappings of C into itself with j 1

j 2

j 3

j

TN

i=1

j 1

j

j

j

j

j

F (Ti ) 6= ∅ and let αj = (α1 , α2 , α3 ), j = 1, 2, 3, . . . , N, where α1 , α2 , α3 ∈ j 2

j 3

[0, 1] , α + α + α = 1, α ∈ (0, 1) for all j = 1, 2, . . . , N − 1, α ∈ (0, 1], α , α ∈ [0, 1) for all j = 1, 2, . . . , N. Let TN S be the mapping generated by T1 , . . . , TN and α1 , α2 , . . . , αN . Then F (S ) = i=1 F (Ti ). TN TN Proof. It is clear that i=1 F (Ti ) ⊆ F (S ). Next, we show that F (S ) ⊆ i=1 F (Ti ). To show this, let x0 ∈ F (S ) and T N x∗ ∈ i=1 F (Ti ). Then we have N 1

kx0 − x∗ k = kSx0 − x∗ k = kα1N (TN UN −1 x0 − x∗ ) + α2N (UN −1 x0 − x∗ ) + α3N (x0 − x∗ )k ≤ α1N kTN UN −1 x0 − x∗ k + α2N kUN −1 x0 − x∗ k + α3N kx0 − x∗ k ≤ (1 − α3N )kUN −1 x0 − x∗ k + (1 − (1 − α3N ))kx0 − x∗ k = (1 − α3N )kα1N −1 (TN −1 UN −2 x0 − x∗ ) + α2N −1 (UN −2 x0 − x∗ ) + α3N −1 (x0 − x∗ )k + (1 − (1 − α3N ))kx0 − x∗ k ≤ (1 − α )(α N 3

(2.4)

kTN −1 UN −2 x0 − x k + α

N −1 1 N 3

∗

N −1 2

kUN −2 x0 − x k + α ∗

N −1 3

∗

kx0 − x k)

+ (1 − (1 − α ))kx0 − x k ≤

N Y

∗

j 3

(1 − α )kUN −2 x0 − x k + 1 − ∗

j =N −1

=

N Y

N Y

! j 3

(1 − α ) kx0 − x∗ k

(2.5)

j =N −1

(1 − α3j )kα1N −2 (TN −2 UN −3 x0 − x∗ ) + α2N −2 (UN −3 x0 − x∗ ) + α3N −2 (x0 − x∗ )k

j =N −1

+

1−

N Y

! j 3

(1 − α ) kx0 − x∗ k

j =N −1

≤

N Y

(1 − α3j )(α1N −2 kTN −2 UN −3 x0 − x∗ k + α2N −2 kUN −3 x0 − x∗ k + α3N −2 kx0 − x∗ k)

j =N −1

+

1−

N Y j =N −1

! j 3

(1 − α ) kx0 − x∗ k

(2.6)

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A. Kangtunyakarn, S. Suantai / Nonlinear Analysis: Hybrid Systems 3 (2009) 296–309 N Y

≤

j 3

(1 − α )kUN −3 x0 − x k + 1 − ∗

j=N −2

N Y

! j 3

(1 − α ) kx0 − x∗ k

j=N −2

≤ · · · ≤

N Y (1 − α3j )kα12 (T2 U1 x0 − x∗ ) + α22 (U1 x0 − x∗ ) + α32 (x0 − x∗ )k j =3

+

1−

N Y

! j 3

( 1 − α ) kx 0 − x ∗ k

(2.7)

j=3

≤

N Y (1 − α3j )(α12 kT2 U1 x0 − x∗ k + α22 kU1 x0 − x∗ k + α32 kx0 − x∗ k) j =3

+

1−

N Y

! j 3

( 1 − α ) kx 0 − x ∗ k

j=3

! N Y Y j j ∗ ≤ (1 − α3 )kU1 x0 − x k + 1 − (1 − α3 ) kx0 − x∗ k N

j =2

(2.8)

j =2

! N N Y Y j j 1 ∗ 1 ∗ (1 − α3 )kα1 (T1 x0 − x ) + (1 − α1 )(x0 − x )k + 1 − (1 − α3 ) kx0 − x∗ k = j =2

≤

(2.9)

j =2

N Y (1 − α3j )(α11 kT1 x0 − x∗ k + (1 − α11 )kx0 − x∗ k) j =2

+

1−

N Y

! j 3

( 1 − α ) kx 0 − x ∗ k

(2.10)

j=2

! N N Y Y j j ∗ ≤ (1 − α3 )kx0 − x k + 1 − (1 − α3 ) kx0 − x∗ k j =2

j =2

= kx0 − x∗ k. This implies by (2.9) that

kx 0 − x ∗ k =

N Y

j 3

(1 − α )kα11 (T1 x0 − x∗ ) + (1 − α11 )(x0 − x∗ )k + 1 −

j =2

N Y

! j 3

(1 − α ) kx0 − x∗ k,

j =2

hence

kx0 − x∗ k = kα11 (T1 x0 − x∗ ) + (1 − α11 )(x0 − x∗ )k.

(2.11)

By (2.10), we obtain ∗

kx 0 − x k =

N Y

j 3

(1 − α )[α kT1 x0 − x k + (1 − α )kx0 − x k] + 1 − 1 1

∗

j =2

1 1

∗

N Y

! j 3

(1 − α ) kx0 − x∗ k,

j =2

which implies

kx0 − x∗ k = α11 kT1 x0 − x∗ k + (1 − α11 )kx0 − x∗ k. It follows that

kx0 − x∗ k = kT1 x0 − x∗ k. From (2.11) and (2.12), we have by Lemma 2.2 that T1 x0 = x0 , that is x0 ∈ F (T1 ). It implies that U1 x0 = λ1 T1 x0 + (1 − λ1 )x0 = x0 .

(2.12)

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301

By (2.7), we have ∗

kx 0 − x k =

N Y

" j 3

(1 − α )kα (T2 U1 x0 − x ) + α (U1 x0 − x ) + α (x0 − x )k + 1 − 2 1

∗

∗

2 2

∗

2 3

j =3

N Y

# j 3

(1 − α ) kx0 − x∗ k.

j =3

It follows that

kx0 − x∗ k = kα12 (T2 U1 x0 − x∗ ) + α22 (U1 x0 − x∗ ) + α32 (x0 − x∗ )k = kα12 (T2 x0 − x∗ ) + (1 − α12 )(x0 − x∗ )k.

(2.13)

By (2.8), we have ∗

kx 0 − x k =

N Y

j 3

(1 − α )(α kT2 U1 x0 − x k + α kU1 x0 − x k + α kx0 − x k) + 1 − 2 1

∗

∗

2 2

∗

2 3

N Y

j =3

! j 3

(1 − α ) kx0 − x∗ k,

j =3

which implies

kx0 − x∗ k = α12 kT2 U1 x0 − x∗ k + α22 kU1 x0 − x∗ k + α32 kx0 − x∗ k = α12 kT2 x0 − x∗ k + (1 − α12 )kx0 − x∗ k. Hence, we obtain

kx0 − x∗ k = kT2 x0 − x∗ k.

(2.14)

From (2.13) and (2.14), we have by Lemma 2.2 that T2 x0 = x0 , that is x0 ∈ F (T2 ). This implies that U2 x0 = α12 T2 U1 x0 + α22 U1 x0 + α32 x0 = x0 . By continuing in this way, we can show that x0 ∈ F (Ti ) and x0 ∈ F (Ui ) for all i = 1, 2, . . . , N − 1. Finally, we shall show that x0 ∈ F (TN ). Since 0 = Sx0 − x0 = α1N TN UN −1 x0 + α2N UN −1 x0 + α3N x0 − x0

= α1N (TN x0 − x0 ), and α1N ∈ (0, 1], we obtain TN x0 = x0 so that xo ∈ F (TN ). Hence F (S ) ⊆

TN

i=1

F (Ti ).



Lemma 2.9. Let C be a nonempty closed convex subset of Banach space. Let {Ti }Ni=1 be a finite family of nonexpansive mappings

= (α1n,j , α2n,j , α3n,j ), αj = (α1j , α2j , α3j ) where α1n,j , α2n,j , α3n,j ∈ [0, 1], α1j , α2j , α3j ∈ [0, 1], α1n,j + α2n,j + α3n,j = 1 and α1j + α2j + α3j = 1. Suppose αin,j → αij as n → ∞ for i = 1, 3 and j = 1, 2, 3, . . . , N . Let S and Sn be the S-mappings generated by T1 , T2 , . . . , TN and α1 , α2 , . . . , αN and T1 , T2 , . . . , TN (n) (n) (n) and α1 , α2 , . . . , αN , respectively. Then limn→∞ kSn x − Sxk = 0 for every x ∈ C . (n)

of C into itself and for each n ∈ N and j ∈ {1, 2, . . . , N }, let αj

(n)

(n)

(n)

Proof. Let x ∈ C , Uk and Un,k be generated by T1 , T2 , . . . , TN and α1 , α2 , . . . , αN and T1 , T2 , . . . , TN and α1 , α2 , . . . , αN , respectively. For each n ∈ N and for k ∈ {2, 3, . . . , N }, we have

kUn,1 x − U1 xk = kα1n,1 T1 x + (1 − α1n,1 )x − α11 T1 x − (1 − α11 )xk = |α1n,1 − α11 |kT1 x − xk,

(2.15)

and

kUn,k x − Uk xk = kα1n,k Tk Un,k−1 x + α2n,k Un,k−1 x + α3n,k x − α1k Tk Uk−1 x − α2k Uk−1 x − α3k xk = kα1n,k (Tk Un,k−1 x − Tk Uk−1 x) + (α1n,k − α1k )Tk Uk−1 x + (α3n,k − α3k )x + α2n,k (Un,k−1 x − Uk−1 x) + (α2n,k − α2k )Uk−1 xk ≤ α1n,k kTk Un,k−1 x − Tk Uk−1 xk + |α1n,k − α1k |kTk Uk−1 xk + |α3n,k − α3k |kxk + α2n,k kUn,k−1 x − Uk−1 xk + |α2n,k − α2k |kUk−1 xk ≤ α1n,k kUn,k−1 x − Uk−1 xk + |α1n,k − α1k |kTk Uk−1 xk + α2n,k kUn,k−1 x − Uk−1 xk + (|α1k − α1n,k | + |α3n,k − α3k |)kUk−1 xk + |α3n,k − α3k |kxk ≤ kUn,k−1 x − Uk−1 xk + |α1n,k − α1k |(kTk Uk−1 xk + kUk−1 xk) + |α3n,k − α3k |(kUk−1 xk + kxk).

(2.16)

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By (2.15) and (2.16), we have

kSn x − Sxk = kUn,N x − UN xk ≤ |α1n,1 − α11 |kT1 x − xk +

N X

|α1n,j − α1j |(kTj Uj−1 xk + kUN −j xk) +

j=2

N X

|α3n,j − α3j |(kUj−1 xk + kxk).

j=2

This together with our assumption, we can conclude that lim kSn x − Sxk = 0. 

n−→∞

3. Main result

TNIn this section, we prove a strong convergence theorem of the iterative scheme (3.1) to a common element of EP and i=1 F (Ti ) under some control conditions. Theorem 3.1. Let C be a closed convex subset of a real Hilbert space and let F : C × C → R be a bifunction satisfying conditions (A1)–(A4). Let A be an α -inverse strongly monotone mapping of C into H and let {Ti }Ni=1 be a finite family of TN T (n) = (α1n,j , α2n,j , α3n,j ) be nonexpansive mappings of C into itself with i=1 F (Ti ) EP 6= ∅. For j = 1, 2, . . . , N, let αj n ,j

n ,j

n,j

n,j

n ,j

n ,j

n ,j

n,N

such that α1 , α2 , α3 ∈ [0, 1], α1 + α2 + α3 = 1, {α1 }jN=−11 ⊂ [η1 , θ1 ] with 0 < η1 ≤ θ1 < 1, {α1 } ⊂ [ηN , 1] with n ,j n,j 0 < ηN ≤ 1 and {α2 }Nj=1 , {α3 }Nj=1 ⊂ [0, θ3 ] with 0 ≤ θ3 < 1. Let Sn be the S-mappings generated by T1 , T2 , . . . , TN and

α1(n) , α2(n) , . . . , αN(n) . Let u ∈ C and x1 ∈ C and let {zn } ⊂ C and {xn } ⊂ C be sequences generated by ( 1 F (zn , y) + hAxn , y − zn i + hy − zn , zn − xn i ≥ 0, ∀y ∈ C , λn xn+1 = βn xn + (1 − βn )Sn (an u + (1 − an )zn ), ∀n ∈ N,

(3.1)

where {an } ∈ [0, 1], {βn } ⊂ [0, 1] and {λn } ⊂ [0, 2α] satisfy the following conditions:

(i) 0 < a ≤ λn ≤ b < 2α, 0 < c ≤ βn ≤ d < 1; λn = 1; (ii) lim n→∞ λn+1 ∞ X (iii) lim an = 0, an = ∞; n→∞

n=1

(iv) |α1n+1,j − α1n,j | → 0, and |α3n+1,j − α3n,j | → 0 as n → ∞, for all j ∈ {1, 2, 3, . . . , N }. TN T Then {xn } converges strongly to z ∈ i=1 F (Ti ) EP, where z = PTN F (T ) T EP u. i i=1

Proof. First, we show that (I − λn A) is nonexpansive. Let x, y ∈ C . Since A is α -strongly monotone and λn < 2α ∀n ∈ N, we have

k(I − λn A)x − (I − λn A)yk2 = kx − y − λn (Ax − Ay)k2 = kx − yk2 − 2λn hx − y, Ax − Ayi + λ2n kAx − Ayk2 ≤ kx − yk2 − 2αλn kAx − Ayk2 + λ2n kAx − Ayk2 = kx − yk2 + λn (λn − 2α)kAx − Ayk2 ≤ kx − yk2 .

(3.2)

Thus (I − λn A) is nonexpansive. Since F (zn , y) + hAxn , y − zn i +

1

λn

hy − zn , zn − xn i ≥ 0,

∀y ∈ C ,

we obtain F (zn , y) +

1

λn

hy − zn , zn − (I − λn A)xn i ≥ 0,

∀y ∈ C .

By Lemma 2.6, we have zn = Tλn (xn − λn Axn ) ∀n ∈ N. Let z ∈

TN

i=1

F (Ti )

T

EP. Then F (z , y) + hy − z , Az i ≥ 0,

∀y ∈ C .

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303

So F (z , y) + λ1 hy − z , z − z + λn Az i ≥ 0, ∀y ∈ C . n Again by Lemma 2.6, we have z = Tλn (z − λn Az ). Since I − λn A and Tλn are nonexpansive, we have

kzn − z k2 = kTλn (xn − λn Axn ) − Tλn (z − λn Az )k2 ≤ kxn − z k2 ,

(3.3)

hence kzn − z k ≤ kxn − z k. Putting yn = an u + (1 − an )zn . Then we have

kyn − z k = kan (u − z ) + (1 − an )(zn − z )k ≤ an ku − z k + (1 − an )kxn − z k.

(3.4)

This implies that

kxn+1 − z k = kβn (xn − z ) + (1 − βn )(Sn yn − z )k ≤ βn kxn − z k + (1 − βn )kyn − z k ≤ βn kxn − z k + (1 − βn )(an ku − z k + (1 − an )kxn − z k).

(3.5)

Putting K = max{kx1 − z k, ku − z k}. By (3.5), we can show by induction that kxn − z k ≤ K , ∀n ∈ N. This implies that {xn } is bounded. Hence {Axn }, {yn }, {Sn yn }, {zn } are bounded. Next we will show that lim kxn+1 − xn k = 0.

(3.6)

n→∞

Putting un = xn − λn Axn . Then, we have zn+1 = Tλn+1 (xn+1 − λn+1 Axn+1 ) = Tλn+1 un+1 , zn = Tλn (xn − λn Axn ) = Tλn un . So we have

kan+1 u + (1 − an+1 )zn+1 − an u − (1 − an )zn k kan+1 u + (1 − an+1 )Tλn+1 un+1 − an u − (1 − an )Tλn un k k(an+1 − an )u + (1 − an+1 )(Tλn+1 un+1 − Tλn+1 un + Tλn+1 un − Tλn un + Tλn un ) − (1 − an )Tλn un k k(an+1 − an )u + (1 − an+1 )(Tλn+1 un+1 − Tλn+1 un ) + (1 − an+1 )(Tλn+1 un − Tλn un ) + (1 − an+1 )Tλn un − (1 − an )Tλn un k ≤ |an+1 − an |kuk + (1 − an+1 )kun+1 − un k

kyn+1 − yn k = = = =

+ (1 − an+1 )kTλn+1 un − Tλn un k + |an+1 − an |kTλn un k.

(3.7)

Since I − λn+1 A is nonexpansive, we have

kun+1 − un k = kxn+1 − λn+1 Axn+1 − xn + λn Axn k = k(I − λn+1 A)xn+1 − (I − λn+1 A)xn + (λn − λn+1 )Axn k ≤ kxn+1 − xn + |λn − λn+1 |Axn k.

(3.8)

By Lemma 2.6, we have F (Tλn un , y) +

1

λn

hy − Tλn un , Tλn un − un i ≥ 0,

∀y ∈ C

and F (Tλn+1 un , y) +

1

λn+1

hy − Tλn+1 un , Tλn+1 un − un i ≥ 0,

∀y ∈ C .

In particular, we have F (Tλn un , Tλn+1 un ) +

1

λn

hTλn+1 un − Tλn un , Tλn un − un i ≥ 0,

(3.9)

and F (Tλn+1 un , Tλn un ) +

1

λn+1

hTλn un − Tλn+1 un , Tλn+1 un − un i ≥ 0.

(3.10)

Summing up (3.9) and (3.10) and using (A2), we obtain 1

λn+1

hTλn un − Tλn+1 un , Tλn+1 un − un i +

1

λn

hTλn+1 un − Tλn un , Tλn un − un i ≥ 0,

∀y ∈ C .

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It then follows that



Tλn un − Tλn+1 un ,

Tλn+1 un − un

λn+1

−

Tλn un − un

λn



≥ 0.

This implies

 λn (Tλ un − un ) 0 ≤ Tλn+1 un − Tλn un , Tλn un − un − λn+1 n+1     λn (Tλn+1 un − un ) . = Tλn+1 un − Tλn un , Tλn un − Tλn+1 un + 1 − λn+1 

It follows that



λn

Tλ un − Tλ un (kTλ un k + kun k). kTλn+1 un − Tλn un k ≤ 1 − n n+1 n+1 λ n +1 2

Hence, we obtain

λn



2

L, u − Tλn un ≤ 1 − n+1 n λn+1

Tλ

(3.11)

where L = sup{kun k + kTλn+1 un k : n ∈ N}. By (3.7), (3.8) and (3.11), we have

kyn+1 − yn k ≤ |an+1 − an |kuk + (1 − an+1 )kun+1 − un k + (1 − an+1 )kTλn+1 un − Tλn un k + |an+1 − an |kTλn un k ≤ |an+1 − an |kuk + (1 − an+1 )(kxn+1 − xn + |λn − λn+1 |Axn k) λn + (1 − an+1 ) 1 − L + |an+1 − an |kTλn un k λn+1



1 − λn Axn x − x + λ ≤ |an+1 − an |kuk + n + 1 n n + 1

λn+1 λn + 1 − L + |an+1 − an |kTλn un k λn+1



λn

Axn ≤ |an+1 − an |kuk + xn+1 − xn + b 1 −

λn+1 λn + 1 − (3.12) L + |an+1 − an |kTλn un k. λ n +1

We can rewrite xn+1 by xn+1 = βn xn + (1 − βn )Sn yn ,

(3.13)

where yn = an u + (1 − an )zn . Next, we show that lim kSn yn − xn k = 0.

(3.14)

n→∞

For k ∈ {2, 3, . . . , N }, we have

kUn+1,k yn − Un,k yn k = kα1n+1,k Tk Un+1,k−1 yn + α2n+1,k Un+1,k−1 yn + α3n+1,k yn − α1n,k Tk Un,k−1 yn − α2n,k Un,k−1 yn − α3n,k yn k = kα1n+1,k (Tk Un+1,k−1 yn − Tk Un,k−1 yn ) + (α1n+1,k − α1n,k )Tk Un,k−1 yn + (α3n+1,k − α3n,k )yn + α2n+1,k (Un+1,k−1 yn − Un,k−1 yn ) + (α2n+1,k − α2n,k )Un,k−1 yn k ≤ α1n+1,k kUn+1,k−1 yn − Un,k−1 yn k + |α1n+1,k − α1n,k |kTk Un,k−1 yn k + |α3n+1,k − α3n,k |kyn k + α2n+1,k kUn+1,k−1 yn − Un,k−1 yn k + |α2n+1,k − α2n,k |kUn,k−1 yn k = (α1n+1,k + α2n+1,k )kUn+1,k−1 yn − Un,k−1 yn k + |α1n+1,k − α1n,k |kTk Un,k−1 yn k + |α3n+1,k − α3n,k |kyn k + |α2n+1,k − α2n,k |kUn,k−1 yn k ≤ kUn+1,k−1 yn − Un,k−1 yn k + |α1n+1,k − α1n,k |kTk Un,k−1 yn k

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305

+ |α3n+1,k − α3n,k |kyn k + |(α1n,k − α1n+1,k ) + (α3n,k − α3n+1,k )|kUn,k−1 yn k ≤ kUn+1,k−1 yn − Un,k−1 yn k + |α1n+1,k − α1n,k |kTk Un,k−1 yn k + |α3n+1,k − α3n,k |kyn k + |α1n,k − α1n+1,k |kUn,k−1 yn k + |α3n,k − α3n+1,k |kUn,k−1 yn k = kUn+1,k−1 yn − Un,k−1 yn k + |α1n+1,k − α1n,k |(kTk Un,k−1 yn k + kUn,k−1 yn k) + |α3n+1,k − α3n,k |(kyn k + kUn,k−1 yn k).

(3.15)

By (3.15), we obtain that for each n ∈ N,

kSn+1 yn − Sn yn k = kUn+1,N yn − Un,N yn k ≤ kUn+1,1 yn − Un,1 yn k +

N X

|α1n+1,j − α1n,j |(kTj Un,j−1 yn k + kUn,j−1 yn k)

j =2

+

N X

|α3n+1,j − α3n,j |(kyn k + kUn,j−1 yn k)

j =2

= |α1n+1,1 − α1n,1 |kT1 yn − yn k +

N X

|α1n+1,j − α1n,j |(kTj Un,j−1 yn k + kUn,j−1 yn k)

j =2

+

N X

|α3n+1,j − α3n,j |(kyn k + kUn,j−1 yn k).

j =2

This together with condition (iv), we obtain lim kSn+1 yn − Sn yn k = 0.

(3.16)

n→∞

By (3.12), we have

kSn+1 yn+1 − Sn yn k ≤ kyn+1 − yn k + kSn+1 yn − Sn yn k λn ≤ |an+1 − an |kuk + kxn+1 − xn k + b 1 − kAxn k λ n +1 λn L + |an+1 − an |kTλn un k + kSn+1 yn − Sn yn k. + 1 − λ

(3.17)

n+1

This together with (3.16) and conditions (ii) and (iii), we obtain lim sup(kSn+1 yn+1 − Sn yn k − kxn+1 − xn k) ≤ 0.

(3.18)

n→∞

It follows from (3.13) and (3.17) and Lemma 2.4, limn→∞ kSn yn − xn k = 0. This implies that lim kxn+1 − xn k = lim (1 − βn )kSn yn − xn k = 0.

n→∞

n→∞

(3.19)

Next, we show that lim kxn − zn k = 0.

n→∞

(3.20)

By monotonicity of A and nonexpansiveness of Tλn , we have

kxn+1 − z k2 = kβn (xn − z ) + (1 − βn )(Sn yn − z )k2 ≤ βn kxn − z k2 + (1 − βn )kyn − z k2 = βn kxn − z k2 + (1 − βn )kan (u − z ) + (1 − an )(zn − z )k2 ≤ βn kxn − z k2 + (1 − βn )(an ku − z k2 + (1 − an )kzn − z k2 ) = βn kxn − z k2 + (1 − βn )(an ku − z k2 + (1 − an )kTλn (xn − λn Axn ) − Tλn (z − λn Az )k2 ) ≤ βn kxn − z k2 + (1 − βn )(an ku − z k2 + (1 − an )k(xn − λn Axn ) − (z − λn Az )k2 ) = βn kxn − z k2 + (1 − βn )(an ku − z k2 + (1 − an )k(xn − z ) − λn (Axn − Az )k2 ) = βn kxn − z k2 + (1 − βn )(an ku − z k2 + (1 − an )(kxn − z k2 − 2λn hxn − z , Axn − Az i + λ2n kAxn − Az k2 ))

(3.21)

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≤ βn kxn − z k2 + (1 − βn )(an ku − z k2 + (1 − an )(kxn − z k2 − 2λn αkAxn − Az k2 + λ2n kAxn − Az k2 )) = βn kxn − z k2 + (1 − βn )(an ku − z k2 + (1 − an )(kxn − z k2 + λn (λn − 2α)kAxn − Az k2 )) ≤ kxn − z k2 + (1 − βn )an ku − z k2 + (1 − an )(1 − βn )λn (λn − 2α)kAxn − Az k2 .

(3.22)

(1 − an )(1 − βn )λn (2α − λn )kAxn − Az k2 ≤ kxn − z k2 − kxn+1 − z k2 + (1 − βn )an ku − z k2 .

(3.23)

By (3.22), we have

Since 0 < a ≤ λn ≤ b < 2α and 0 < c ≤ βn ≤ d < 1, we have

(1 − an )(1 − d)a(2α − λn )kAxn − Az k2 ≤ kxn − z k2 − kxn+1 − z k2 + (1 − βn )an ku − z k2 ≤ kxn+1 − xn k(kxn − z k + kxn+1 − z k) + (1 − βn )an ku − z k2 .

(3.24)

This implies, by (3.19) and condition (iii), that lim kAxn − Az k = 0.

(3.25)

n→∞

Since Tλn is a firmly nonexpansive, we have

kzn − z k2 = kTλn (xn − λn Axn ) − Tλn (z − λn Az )k2 ≤ h(xn − λn Axn ) − (z − λn Az ), zn − z i = ≤ =

1 2 1 2 1 2

(k(xn − λn Axn ) − (z − λn Az )k2 + kzn − z k2 − k(xn − λn Axn ) − (z − λn Az ) − (zn − z )k2 ) (kxn − z k2 + kzn − z k2 − k(xn − zn ) − λn (Axn − Az )k2 ) (kxn − z k2 + kzn − z k2 − kxn − zn k2 + 2λn hxn − zn , Axn − Az i − λ2N kAxn − Az k2 ).

(3.26)

It follows that

kzn − z k2 ≤ kxn − z k2 − kxn − zn k2 + 2λn kxn − zn kkAxn − Az k.

(3.27)

By (3.21) and (3.27), we have

kxn+1 − z k2 ≤ βn kxn − z k2 + (1 − βn )[an ku − z k2 + (1 − an )kzn − z k2 ] ≤ βn kxn − z k2 + an ku − z k2 + (1 − βn )kzn − z k2 ≤ βn kxn − z k2 + an ku − z k2 + (1 − βn )(kxn − z k2 − kxn − zn k2 + 2λn kxn − zn kkAxn − Az k) ≤ kxn − z k2 + an ku − z k2 − (1 − βn )kxn − zn k2 + 2λn kxn − zn kkAxn − Az k.

(3.28)

This implies

(1 − βn )kxn − zn k2 ≤ kxn − z k2 − kxn+1 − z k2 + an ku − z k2 + 2λn kxn − zn kkAxn − Az k. Hence

(1 − d)kxn − zn k2 ≤ kxn+1 − xn k(kxn − z k + kxn+1 − z k) + an ku − z k2 + 2λn kxn − zn kkAxn − Az k. By (3.19) and (3.25), we obtain lim kxn − zn k = 0.

(3.29)

n→∞

Since yn = an u + (1 − an )zn , we have kyn − zn k = an ku − zn k. This implies limn→∞ kyn − zn k = 0. By (3.14) and (3.29), we have

kSn yn − yn k ≤ kSn yn − xn k + kxn − zn k + kzn − yn k → 0 as n → ∞. Next, putting z0 =

PTN

i=1 F (Ti )

T

EP

(3.30)

u, we shall show that

lim suphu − z0 , yn − z0 i ≤ 0.

(3.31)

n→∞

To show this inequality, take a subsequence {ynk } of {yn } such that lim suphu − z0 , yn − z0 i = lim suphu − z0 , ynk − z0 i. n→∞

k→∞

(3.32)

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307

Without loss of generality, we may assume that ynk * ω as k → ∞ where ω ∈ C . We first show ω ∈ EP. We have znk * ω as k → ∞. Since zn = Tλn (xn − λn Axn ), we obtain F (zn , y) + hAxn , y − zn i +

1

λn

hy − zn , zn − xn i ≥ 0,

∀y ∈ C .

From (A2), we have hAxn , y − zn i + λ1 hy − zn , zn − xn i ≥ F (y, zn ). Then n

hAxnk , y − znk i +

1

λnk

hy − znk , zni − xnk i ≥ F (y, znk ),

∀y ∈ C .

(3.33)

Put zt = ty + (1 − t )ω for all t ∈ (0, 1] and y ∈ C . Then, we have zt ∈ C . So, from (3.33) we have

  zn − xnk + F (zt , znk ) hzt − znk , Azt i ≥ hzt − znk , Azt i − hzt − znk , Axni i − zt − znk , i λnk   z n − xn k + F (zt , znk ). = hzt − znk , Azt − Aznk i + hzt − znk , Aznk − Axnk i − zt − znk , k λ nk Since kznk − xnk k → 0, we have kAznk − Axnk k → 0. Further, from the monotonicity of A, we have hzt − znk , Azt − Aznk i ≥ 0. So, from (A4) we have

hzt − ω, Azt i ≥ F (zt , ω) as k → ∞.

(3.34)

From (A1), (A4) and (3.34), we also have 0 = F (zt , zt ) ≤ tF (zt , y) + (1 − t )F (zt , ω)

≤ tF (zt , y) + (1 − t )hzt − ω, Azt i = tF (zt , y) + (1 − t )t hy − ω, Azt i, hence 0 ≤ F (zt , y) + (1 − t )hy − ω, Azt i. Letting t → 0, we have 0 ≤ F (ω, y) + hy − ω, Aωi

∀y ∈ C .

(3.35)

Therefore ω ∈ EP. TN Next, we show that ω ∈ i=1 F (Ti ). We can assume that n ,j

n ,N

α1 k → α1j ∈ (0, 1) and α1 k

→ α1N ∈ (0, 1] as k → ∞ for j = 1, 2, . . . , N − 1

(3.36)

and n ,j

α3 k → α3j ∈ [0, 1) as k → ∞ for j = 1, 2, . . . , N .

(3.37) j

j

j

Let S be the S-mappings generated by T1 , T2 , . . . , TN and β1 , β2 , . . . , βN where βj = (α1 , α2 , α3 ), for j = 1, 2, . . . , N . By Lemma 2.9, we have lim kSnk x − Sxk = 0

(3.38)

k→∞

for all x ∈ C . TN By Lemma 2.8, we have i=1 F (Ti ) = F (S ). Assume that S ω 6= ω. By using the Opial property and (3.30) and (3.38), we have lim inf kynk − ωk < lim inf kynk − S ωk k→∞

k→∞

≤ lim inf(kynk − Snk ynk k + kSnk ynk − Snk ωk + kSnk ω − S ωk) k→∞

≤ lim inf kynk − ωk, k→∞

which is a contradiction. Thus S ω = ω, so ω ∈ F (S ) =

T Hence ω ∈ i=1 F (Ti ) EP. TN T Since ynk * ω and ω ∈ i=1 F (Ti ) EP, we have TN

TN

i =1

F (Ti ).

lim suphu − z0 , yn − z0 i = lim suphu − z0 , ynk − z0 i = hu − z0 , ω − z0 i ≤ 0. n→∞

k→∞

(3.39)

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By using (3.3), we have

kxn+1 − z0 k2 = kβn (xn − z0 ) + (1 − βn )(Sn yn − z0 )k2 ≤ βn kxn − z0 k2 + (1 − βn )kyn − z0 k2 = βn kxn − z0 k2 + (1 − βn )kan u + (1 − an )zn − z0 k2 = βn kxn − z0 k2 + (1 − βn )kan (u − z0 ) + (1 − an )(zn − z0 )k2 ≤ βn kxn − z0 k2 + (1 − βn )((1 − an )2 kzn − z0 k2 + 2an hu − z0 , yn − z0 i) ≤ βn kxn − z0 k2 + (1 − βn )(1 − an )kzn − z0 k2 + 2(1 − βn )an hu − z0 , yn − z0 i ≤ βn kxn − z0 k2 + (1 − βn )(1 − an )kxn − z0 k2 + 2(1 − βn )an hu − z0 , yn − z0 i = (1 − (1 − βn )an )kxn − z0 k2 + 2(1 − βn )an hu − z0 , yn − z0 i. P∞ Since i=1 (1 − βn )an = ∞ and lim supn→∞ 2hu − z0 , yn − z0 i ≤ 0, we can conclude from Lemma 2.3 that lim kxn − z0 k = 0. 

n→∞

4. Applications Using our main theorem (Theorem 3.1), we obtain the following strong convergence theorems in a real Hilbert space. Theorem 4.1. Let C be a closed convex subset of a real Hilbert space and let F : C × C → R be a bifunction satisfying TN T conditions (A1)–(A4). Let {Ti }Ni=1 be a finite family of nonexpansive mappings of C into itself with i=1 F (Ti ) EP (F ) 6= ∅. For

= (α1n,j , α2n,j , α3n,j ) be such that α1n,j , α2n,j , α3n,j ∈ [0, 1], α1n,j + α2n,j + α3n,j = 1, {α1n,j }Nj=−11 ⊂ [η1 , θ1 ] n ,j n ,j n ,N with 0 < η1 ≤ θ1 < 1, {α1 } ⊂ [ηN , 1] with 0 < ηN ≤ 1 and {α2 }Nj=1 , {α3 }Nj=1 ⊂ [0, θ3 ] with 0 ≤ θ3 < 1. Let Sn be (n) (n) (n) the S-mappings generated by T1 , T2 , . . . , TN and α1 , α2 , . . . , αN . Let u ∈ C and x1 ∈ C and let {zn } ⊂ C and {xn } ⊂ C be (n)

j = 1, 2, . . . , N, let αj

sequences generated by

(

x n +1

1

hy − zn , zn − xn i ≥ 0, ∀y ∈ C , λn = βn xn + (1 − βn )Sn (an u + (1 − an )zn ), ∀n ∈ N,

F (zn , y) +

(4.1)

where {an } ∈ [0, 1], {βn } ⊂ [0, 1] and {λn } ⊂ [0, 2α] satisfy the following conditions:

(i) 0 < a ≤ λn ≤ b < 2α, 0 < c ≤ βn ≤ d < 1; λn (ii) lim = 1; n→∞ λn+1 ∞ X (iii) lim an = 0, ; an = ∞ n→∞

n=1

(iv) |α1n+1,j − α1n,j | → 0, and |α3n+1,j − α3n,j | → 0 as n → ∞, for all j ∈ {1, 2, 3, . . . , N }. TN T Then {xn } converges strongly to z ∈ i=1 F (Ti ) EP (F ), where z = PTN F (T ) T EP (F ) u. i i=1

Proof. Put A ≡ 0 in Theorem 3.1. Then, from Theorem 3.1, we can get the desired conclusion.



Theorem 4.2. Let C be a closed convex subset of a real Hilbert space and let F : C × C → R be a bifunction satisfying conditions (A1)–(A4). Let A be an α -inverse strongly monotone mapping of C into H and let {Ti }Ni=1 be a finite family of TN T n,j n,j nonexpansive mappings of C into itself with i=1 F (Ti ) EP 6= ∅. For j = 1, 2, . . . , N, let {α1 }Nj=1 ∈ [0, 1], {α1 }jN=−11 ⊂

[η1 , θ1 ] with 0 < η1 ≤ θ1 < 1, {α1n,N } ⊂ [ηN , 1] with 0 < ηN ≤ 1, ∀n ∈ N. Let Wn be the W-mappings generated by n,1 n ,2 n ,N T1 , T2 , . . . , TN and α1 , α1 , . . . , α1 . Let u ∈ C and x1 ∈ C and let {zn } ⊂ C and {xn } ⊂ C be sequences generated by ( 1 F (zn , y) + hAxn , y − zn i + hy − zn , zn − xn i ≥ 0, ∀y ∈ C , (4.2) λn xn+1 = βn xn + (1 − βn )Wn (an u + (1 − an )zn ), ∀n ∈ N, where {an } ∈ [0, 1], {βn } ⊂ [0, 1] and {λn } ⊂ [0, 2α] satisfy the following conditions:

(i) 0 < a ≤ λn ≤ b < 2α, 0 < c ≤ βn ≤ d < 1; λn (ii) lim = 1; n→∞ λn+1

A. Kangtunyakarn, S. Suantai / Nonlinear Analysis: Hybrid Systems 3 (2009) 296–309

(iii) lim an = 0, n→∞

∞ X

309

an = ∞;

n =1

(iv) |α1n+1,j − α1n,j | → 0, as n → ∞, for all j ∈ {1, 2, 3, . . . , N }. TN T Then {xn } converges strongly to z ∈ i=1 F (Ti ) EP, where z = PTN F (T ) T EP u. i i=1

n.j 2

Proof. Put α follows. 

= 0 for all j ∈ {1, 2, 3, . . . , N }, and all n ∈ N in Theorem 3.1. Then, by Theorem 3.1 the conclusion

Corollary 4.3 ([7], Theorem 3.1). Let C be a closed convex subset of a real Hilbert space and let F : C × C → R be a bifunction satisfying conditions (A1)–(A4). Let TA be an α -inverse strongly monotone mapping of C into H and let T be nonexpansive mappings of C into itself with F (T ) EP 6= ∅. Let u, x1 ∈ C and let {zn }, {xn } ⊂ C be sequences generated by

(

x n +1

1

hy − zn , zn − xn i ≥ 0, ∀y ∈ C , λn = βn xn + (1 − βn )T1 (an u + (1 − an )zn ), ∀n ∈ N,

F (zn , y) + hAxn , y − zn i +

(4.3)

where {an } ∈ [0, 1], {βn } ⊂ [0, 1] and {λn } ⊂ [0, 2α] satisfy the following conditions:

(i) 0 < a ≤ λn ≤ b < 2α, 0 < c ≤ βn ≤ d < 1; λn (ii) lim = 1; n→∞ λn+1 ∞ X an = ∞. (iii) lim an = 0, n→∞

n =1

Then {xn } converges strongly to z ∈

TN

i =1

n,1

F (Ti ) n ,1

Proof. Put N = 1 and T1 = T and α2 , α3 from Theorem 3.1. 

T

EP, where z = PTN

i=1 F (Ti )

T

EP

u.

= 0 ∀n ∈ N in Theorem 3.1. Then Sn = T . Hence, we obtain the desired result n,1

n ,1

Remark. In Theorem 3.1, by taking N = 1 and α2 , α3 = 0 for all n ∈ N, one can easily see that Theorems 4.1, 4.2, 4.3 of Takahashi and Takahashi [7] are special cases of Theorem 3.1.

Acknowledgments The authors would like to thank the Thailand Research Fund and the commission on Higher Education for their financial support during the preparation of this paper. The first author was supported by the graduate school Chiang Mai University. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14]

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