A viscosity approximation scheme for system of equilibrium problems, nonexpansive mappings and monotone mappings

Nonlinear Analysis 71 (2009) 6001–6010

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A viscosity approximation scheme for system of equilibrium problems, nonexpansive mappings and monotone mappings Jian-Wen Peng a , Jen-Chih Yao b,∗ a

College of Mathematics and Computer Science, Chongqing Normal University, Chongqing 400047, PR China

b

Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 804, Taiwan, ROC

article

info

Article history: Received 26 May 2008 Accepted 11 May 2009 Keywords: System of equilibrium problems Extragradient methods Nonexpansive mapping Monotone mapping Viscosity approximation scheme Strong convergence

abstract In this paper, we introduce a new viscosity approximation scheme based on the extragradient method for finding a common element of the set of solutions to a system of equilibrium problems, the set of fixed points of an infinite family of nonexpansive mappings and the set of solutions to the variational inequality for a monotone, Lipschitz continuous mapping. Several convergence results for the sequences generated by these processes in Hilbert spaces were derived. © 2009 Elsevier Ltd. All rights reserved.

1. Introduction Let H be a real Hilbert space with inner product h·, ·i and norm k · k, respectively. Let C be a nonempty closed convex subset of H and {Fk }k∈Γ be a countable family of bifunctions from C × C to R where R is the set of real numbers. Combettes and Hirstoaga [1] considered the following system of equilibrium problems: Finding x ∈ C

such that Fk (x, y) ≥ 0,

∀k ∈ Γ ∀y ∈ C ,

(1.1)

where Γ is an arbitrary index set. If Γ is a singleton, then problem (1.1) becomes the following equilibrium problem: Finding x ∈ C

such that F (x, y) ≥ 0,

∀y ∈ C .

(1.2)

The set of solutions of (1.2) is denoted by EP(F ). Given a mapping T : C → H, let F (x, y) = hTx, y − xi for all x, y ∈ C . Then problem (1.2) becomes the following variational inequality: Finding x ∈ C

such that hTx, y − xi ≥ 0,

∀y ∈ C .

(1.3)

The set of solutions of (1.3) is denoted by VI(C , A). The problem (1.1) is very general in the sense that it includes, as special cases, optimization problems, variational inequalities, minimax problems, Nash equilibrium problem in noncooperative games and others; see for instance, [2,1,3]. Recall that a mapping S defined on a closed convex subset C of H is nonexpansive [4] if there holds that

kSx − Syk ≤ kx − yk,

∗

for all x, y ∈ C .

Corresponding author. Tel.: +886 7 5253816; fax: +886 7 5253809. E-mail address: [email protected] (J.-C. Yao).

0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.05.028

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J.-W. Peng, J.-C. Yao / Nonlinear Analysis 71 (2009) 6001–6010

We denote the set of fixed points of S by Fix(S ). Combettes and Hirstoaga [5] introduced an iterative scheme for finding a common element of the set of solutions of problem (1.1) in a Hilbert space and obtained a weak convergence theorem. Some methods have been proposed to solve the problem (1.2); see, for instance, [2,6–8,3,9–14] and the references therein. In the present paper, we introduce a new viscosity approximation scheme based on the extragradient method for finding a common element of the set of solutions to a system of equilibrium problems, the set of fixed points of an infinite family of nonexpansive mappings and the set of solutions to the variational inequality for a monotone, Lipschitz continuous mapping. We obtain several strong convergence theorems for the sequences generated by these processes. The results in this paper extend, generalize and improve some well-known results in the literature. 2. Preliminaries Let symbols → and * denote strong and weak convergence, respectively. It is well known that

kλx + (1 − λ)yk2 = λkxk2 + (1 − λ)kyk2 − λ(1 − λ)kx − yk2 for all x, y ∈ H and λ ∈ [0, 1]. For any x ∈ H, there exists a unique nearest point in C denoted by PC (x) such that kx − PC (x)k ≤ kx − yk for all y ∈ C . The mapping PC is called the metric projection of H onto C . We know that PC is a nonexpansive mapping from H onto C . It is also known that PC (x) ∈ C and

hx − PC (x), PC (x) − yi ≥ 0

(2.1)

for all x ∈ H and y ∈ C . It is easy to see that (2.1) is equivalent to the following inequality:

kx − yk2 ≥ kx − PC (x)k2 + ky − PC (x)k2

(2.2)

for all x ∈ H and y ∈ C . For more details, see [15]. A mapping A of C into H is called monotone if hAx − Ay, x − yi ≥ 0 for all x, y ∈ C . The mapping A of C into H is called α inverse-strongly monotone if there exists a positive real number α such that hx − y, Ax − Ayi ≥ αkAx − Ayk2 for all x, y ∈ C . The mapping A : C → H is called k-Lipschitz continuous if there exists a positive real number k such that kAx − Ayk ≤ kkx−yk for all x, y ∈ C . It is easy to see that if A is α -inverse-strongly monotone, then A is monotone and Lipschitz continuous. The converse is not true in general. Let A be a monotone mapping of C into H. It is not difficult to see that the characterization of projection (2.1) implies the following: u ∈ VI(C , A) ⇒ u = PC (u − λAu),

∀λ > 0,

and u = PC (u − λAu)

for some λ > 0 ⇒ u ∈ VI(C , A).

It is also known that H satisfies the Opial condition [16], i.e., for any sequence {xn } ⊂ H with xn * x, the inequality lim inf kxn − xk < lim inf kxn − yk n→∞

n→∞

holds for every y ∈ H with x 6= y. For solving the equilibrium problem, let us assume that the bifunction F satisfies the following conditions which were imposed in [2]: (A1) F (x, x) = 0 for all x ∈ C ; (A2) F is monotone, i.e. F (x, y) + F (y, x) ≤ 0 for any x, y ∈ C ; (A3) for each x, y, z ∈ C , lim F (tz + (1 − t )x, y) ≤ F (x, y); t ↓0

(A4) for each x ∈ C , y 7→ F (x, y) is convex and lower semicontinuous. We recall some lemmas which will be needed in the rest of this paper. Lemma 2.1 ([2]). Let C be a nonempty closed convex subset of H and F be a bifunction from C × C to 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,

for all y ∈ C .

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Lemma 2.2 ([1]). Let C be a nonempty closed convex subset of H and F be a bifunction from C × C to R satisfying (A1)–(A4). For r > 0 and x ∈ H, define a mapping TrF : H → C as follows: TrF (x) =



z ∈ C : F (z , y) +

1 r

hy − z , z − xi ≥ 0, ∀y ∈ C



for all x ∈ H. Then the following statements hold: (1) TrF is single-valued; (2) TrF is firmly nonexpansive, i.e, for any x, y ∈ H ,

kTrF (x) − TrF (y)k2 ≤ hTrF (x) − TrF (y), x − yi; (3) Fix(TrF ) = EP (F ); (4) EP (F ) is closed and convex. Lemma 2.3 ([17,18]). Assume that {αn } is a sequence of nonnegative real numbers such that

αn+1 ≤ (1 − γn )αn + δn , where {γn } is a sequence in (0, 1) and {δn } is a sequence such that (i) n=1 γn = ∞; P∞ (ii) lim supn→∞ γδn ≤ 0 or n=1 |δn | < ∞. n

P∞

Then limn→∞ αn = 0. Lemma 2.4. There holds the following inequality:

kx + yk2 ≤ kxk2 + 2hy, x + yi,

for all x, y ∈ H .

Lemma 2.5 ([19]). Let {xn } and {wn } be bounded sequences in a Banach space and {βn } be a sequence of real numbers such that 0 < lim infn→∞ βn ≤ lim supn→∞ βn < 1 for all n = 0, 1, 2, . . . . Suppose that xn+1 = (1 − βn )wn + βn xn for all n = 0, 1, 2, . . . and lim supn→∞ kwn+1 − wn k + kxn+1 − xn k ≤ 0. Then limn→∞ kwn − xn k = 0. Lemma 2.6 ([5]). Let CP be a nonempty closed convex subset of a Banach space E. Let S1 , S2 , . . . be a sequence of mappings of C ∞ into itself. Suppose that n=1 sup{kSn+1 x − Sn xk : x ∈ C } < ∞. Then for each y ∈ C , {Sn y} converges strongly to some point of C . Moreover, let S be a mapping of C into itself defined by Sy = limn→∞ Sn y for all y ∈ C . Then limn→∞ sup{kSx−Sn xk : x ∈ C } = 0. 3. Strong convergence theorems In this section, we derive a strong convergence result of an iterative algorithm based on both viscosity approximation method and extragradient method which solves the problem of finding a common element of the set of solutions to a system of equilibrium problems, the set of fixed points of an infinite family of nonexpansive mappings and the set of solutions to the variational inequality for a monotone, Lipschitz continuous mapping in a Hilbert space. Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H and f be a contraction of C into itself. Let Fk , k ∈ {1, 2, . . . , M } be bifunctions from C × C to R satisfying (A1)–(A4) and A be a monotone and k-Lipschitz continuous mapping of C into H. Let {Sn } be a sequence of nonexpansive mappings of C into itself such that Ω = ∩∞ i=1 Fix(Si ) ∩ VI (C , A) ∩ (∩M k=1 EP(Fk )) 6= ∅. Let {xn }, {un } and {yn } be sequences generated by

 x1 = x ∈ C ,    F un = TrFMM,n TrMM−−11,n . . . TrF22,n TrF11,n xn ,  y = PC (un − λn Aun ),   n xn+1 = αn f (Sn xn ) + βn xn + γn Sn PC (un − λn Ayn ),

(3.1)

for every n = 1, 2, . . . , where {λn }, {rk,n }, k ∈ {1, 2, . . . , M }, {αn }, {βn } and {γn } are sequences of numbers satisfying the following conditions: (C1) (C2) (C3) (C4) (C5)

limn→∞ αn = 0 and n=1 αn = ∞; 1 > lim supn→∞ βn ≥ lim infn→∞ βn > 0; limn→∞ λn = 0; lim infn→∞ rk,n > 0 and limn→∞ |rk,n+1 − rk,n | = 0 for each k ∈ {1, 2, . . . , M }; αn + βn + γn = 1 for all n ≥ 1.

P∞

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J.-W. Peng, J.-C. Yao / Nonlinear Analysis 71 (2009) 6001–6010

Suppose that n=1 sup{kSn+1 x − Sn xk : x ∈ B} < ∞ for any bounded subset B of C . Let S be a mapping of C into itself defined by Sy = limn→∞ Sn y for all y ∈ C and suppose that Fix(S ) = ∩∞ i=1 Fix(Si ). Then the sequences {xn }, {un } and {yn } converge strongly to the same point w ∈ Ω where w = PΩ f (w).

P∞

Proof. We show that PΩ f is a contraction of C into itself. In fact, there exists a ∈ [0, 1) such that kf (x) − f (y)k ≤ akx − yk for all x, y ∈ C from which it follows that

kPΩ f (x) − PΩ f (y)k ≤ kf (x) − f (y)k ≤ akx − yk for all x, y ∈ C . Since H is complete, there exists a unique element v0 ∈ C such that v0 = PΩ f (v0 ). Moreover, by taking Θnk = F

F

F

Trkk,n . . . Tr22,n Tr11,n for k ∈ {1, 2, . . . , M } and Θn0 = I for all n, we have un = ΘnM xn . We divide the proof into several steps. Step 1. The sequence {xn } is bounded. F

F

Let p ∈ Ω . Since for each k ∈ {1, 2, . . . , M }, Trkk,n is nonexpansive and p = Trkk,n p, we have

kun − pk = kΘnM xn − ΘnM pk ≤ kxn − pk

(3.2)

for all n = 1, 2, . . . . From (2.2), the monotonicity of A and the fact p ∈ VI(C , A), we have

ktn − pk2 ≤ = = ≤ ≤ =

kun − λn Ayn − pk2 − kun − λn Ayn − tn k2 kun − pk2 − kun − tn k2 + 2λn hAyn , p − tn i kun − pk2 − kun − tn k2 + 2λn (hAyn − Au, p − yn i + hAp, p − yn i + hAyn , yn − tn i) kun − pk2 − kun − tn k2 + 2λn hAyn , yn − tn i kun − pk2 − kun − yn k2 − 2hun − yn , yn − tn i − kyn − tn k2 + 2λn hAyn , yn − tn i kun − pk2 − kun − yn k2 − kyn − tn k2 + 2hun − λn Ayn − yn , tn − yn i.

Further, since yn = PC (un − λn Aun ) and A is k-Lipschitz continuous, we have

hun − λn Ayn − yn , tn − yn i = hun − λn Aun − yn , tn − yn i + hλn Aun − λn Ayn , tn − yn i ≤ hλn Aun − λn Ayn , tn − yn i ≤ λn kkun − yn kktn − yn k. Thus we have

ktn − uk2 ≤ kun − uk2 − kun − yn k2 − kyn − tn k2 + 2λn kkun − yn kktn − yn k ≤ kun − uk2 − kun − yn k2 − kyn − tn k2 + λn 2 k2 kun − yn k2 + ktn − yn k2 = kun − uk2 + (λn 2 k2 − 1)kun − yn k2 ≤ kun − uk2 .

(3.3)

1 Put M0 = max{kx1 − pk, 1− kf (p) − pk}. It is obvious that kx1 − pk ≤ M0 . Suppose kxn − pk ≤ M0 . Then from (3.2), a (3.3) and xn+1 = αn f (Sn xn ) + βn xn + γn Sn tn , we have p = Sn p and

kxn+1 − pk = kαn f (Sn xn ) + βn xn + γn Sn tn − pk ≤ αn kf (Sn xn ) − f (p)k + αn kf (p) − pk + βn kxn − pk + γn kSn tn − pk ≤ αn akxn − pk + αn kf (p) − pk + βn kxn − pk + (1 − αn − βn )kxn − pk = (1 − a)αn

kf (p) − pk + (1 − (1 − a)αn )kxn − pk ≤ M0 1−a

(3.4)

for every n = 1, 2, . . . . Therefore, by induction the sequence {xn } is bounded. From (3.2) and (3.3), we also obtain that {tn } and {un } are bounded. From yn = PC (un − λn Aun ), the monotonicity and the Lipschitz continuity of A, we have

kyn − uk2 = kPC (un − λn Aun ) − PC (u − λn Au)k2 ≤ kun − λn Aun − (u − λn Au)k2 = kun − uk2 − 2λn hAun − Au, un − ui + λ2n kAun − Auk2 ≤ (1 + λ2n k2 )kun − uk2 .

J.-W. Peng, J.-C. Yao / Nonlinear Analysis 71 (2009) 6001–6010

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Hence we obtain that {yn } is bounded. It follows from the Lipschitz continuity of A that {Axn }, {Aun }, {Ayn } are also bounded. Since f and Sn are nonexpansive, we know that {Sn xn }, {f (Sn xn )} and {Sn tn } are also bounded. By the definition of tn , we get

ktn+1 − tn k = ≤ ≤ ≤ ≤

kPC (un+1 − λn+1 Ayn+1 ) − PC (un − λn Ayn )k k(un+1 − λn+1 Ayn+1 ) − (un − λn Ayn )k k(un+1 − λn+1 Aun+1 ) − (un − λn+1 Aun ) + λn+1 (Aun+1 − Ayn+1 − Aun ) + λn Ayn k kun+1 − un k + λn+1 kAun+1 − Aun k + λn+1 kAun+1 − Ayn+1 − Aun k + λn kAyn k kun+1 − un k + kλn+1 kun+1 − un k + λn+1 kAun+1 − Ayn+1 − Aun k + λn kAyn k

≤ kun+1 − un k + (λn+1 + λn )M1 ,

(3.5)

where M1 is a constant such that M1 ≥ sup{kkun+1 − un k + kAun+1 − Ayn+1 − Aun k + kAyn k}. n≥1

Step 2. Let {wn } be a bounded sequence in C . We show that lim kΘnM wn − ΘnM+1 wn k = 0.

(3.6)

n→∞

By Step 2 of the proof of Theorem 3.1 in [8], we have that for k ∈ {1, 2, . . . , M }, lim kTrFkk,n+1 wn − TrFkk,n wn k = 0.

(3.7)

n→∞

F

From the definition of ΘnM and the nonexpansiveness of Trkk,n , we have

kΘnM wn − ΘnM+1 wn k = kTrFMM,n ΘnM −1 wn − TrFMM,n+1 ΘnM+−11 wn k ≤ kTrFMM,n ΘnM −1 wn − TrFMM,n+1 ΘnM −1 wn k + kTrFMM,n+1 ΘnM −1 wn − TrFMM,n+1 ΘnM+−11 wn k ≤ kTrFMM,n ΘnM −1 wn − TrFMM,n+1 ΘnM −1 wn k + kΘnM −1 wn − ΘnM+−11 wn k F

F

F

F

≤ kTrFMM,n ΘnM −1 wn − TrFMM,n+1 ΘnM −1 wn k + kTrMM−−11,n ΘnM −2 wn − TrMM−−11,n+1 ΘnM −2 wn k + kΘnM −2 wn − ΘnM+−12 wn k ≤ kTrFMM,n ΘnM −1 wn − TrFMM,n+1 ΘnM −1 wn k + kTrMM−−11,n ΘnM −2 wn − TrMM−−11,n+1 ΘnM −2 wn k + · · · + kTrF22,n Θn1 wn − TrF22,n+1 Θn1 wn k + kTrF11,n wn − TrF11,n+1 wn k from which (3.6) follows by (3.7). Step 3. limn→∞ kxn+1 − xn k = 0. Since un = ΘnM xn and un+1 = ΘnM+1 xn+1 , we have

kun − un+1 k = kΘnM xn − ΘnM+1 xn+1 k ≤ kΘnM xn − ΘnM+1 xn k + kΘnM+1 xn − ΘnM+1 xn+1 k ≤ kΘnM xn − ΘnM+1 xn k + kxn − xn+1 k.

(3.8)

It follows from (3.5) and (3.8) that

ktn+1 − tn k ≤ kΘnM xn − ΘnM+1 xn k + kxn − xn+1 k + (λn+1 + λn )M1 . Define a sequence {vn } by vn =

xn+1 −βn xn 1−βn

(3.9)

, ∀n ≥ 1. Then we have

xn+2 − βn+1 xn+1 xn+1 − βn xn

kvn+1 − vn k = − 1 − βn+1 1 − βn

αn+1 f (Sn+1 xn+1 ) + (1 − αn+1 − βn+1 )Sn+1 tn+1 αn f (Sn xn ) + (1 − αn − βn )Sn tn

= −

1 − βn+1 1 − βn αn+1 αn ≤ (kf (Sn+1 xn+1 )k + kSn+1 tn+1 k) + (kf (Sn xn )k + kSn tn k) + kSn+1 tn+1 − Sn tn k. (3.10) 1 − βn+1 1 − βn

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J.-W. Peng, J.-C. Yao / Nonlinear Analysis 71 (2009) 6001–6010

It follows from (3.9) that

kSn+1 tn+1 − Sn tn k ≤ kSn+1 tn+1 − Sn+1 tn k + kSn+1 tn − Sn tn k ≤ ktn+1 − tn k + kSn+1 tn − Sn tn k ≤ kΘnM xn − ΘnM+1 xn k + kxn − xn+1 k + (λn+1 + λn )M1 + kSn+1 tn − Sn tn k.

(3.11)

Combining (3.10) and (3.11), we have

kvn+1 − vn k − kxn+1 − xn k ≤

αn+1 1 − βn+1

+ ≤

αn

1 − βn

αn+1

1 − βn+1

(kf (Sn+1 xn+1 )k + kSn+1 tn+1 k) (kf (Sn xn )k + kSn tn k) + kSn+1 tn+1 − Sn tn k − kxn+1 − xn k (kf (Sn+1 xn+1 )k + kSn+1 tn+1 k) +

αn 1 − βn

(kf (Sn xn )k + kSn tn k)

+ kΘnM xn − ΘnM+1 xn k + (λn+1 + λn )M1 + sup{kSn+1 t − Sn t k : t ∈ {tn }}.

(3.12)

(C1)–(C3) and (3.6) imply that lim sup(kvn+1 − vn k − kxn+1 − xn k) ≤ 0. n→∞

Hence by Lemma 2.5, we have limn→∞ kvn − xn k = 0. Consequently lim kxn+1 − xn k = lim (1 − βn )kvn − xn k = 0.

n→∞

n→∞

It follows from (3.8), (3.6) and (3.5) that limn→∞ kun+1 − un k = 0 and limn→∞ ktn+1 − tn k = 0. Step 4. kun − tn k → 0. From xn+1 = αn f (Sn xn ) + βn xn + γn Sn tn and (C5), we have

kxn − Sn tn k ≤ kxn+1 − xn k + kxn+1 − Sn tn k ≤ kxn+1 − xn k + αn kf (Sn xn ) − Sn tn k + βn kxn − Sn tn k and thus

kx n − S n t n k ≤

1 1 − βn

(kxn+1 − xn k + αn kf (Sn xn ) − Sn tn k).

It follows from (C1) and (C2) that limn→∞ kxn − Sn tn k = 0. Since xn+1 = αn f (Sn xn ) + βn xn + γn Sn tn for p ∈ Ω , it follows from (3.3) and (3.2) that

kxn+1 − pk2 ≤ αn kf (Sn xn ) − pk2 + βn kxn − pk2 + γn kSn tn − pk2 ≤ αn kf (Sn xn ) − pk2 + βn kxn − pk2 + γn [kun − pk2 + (λn 2 k2 − 1)kun − yn k2 ] ≤ αn kf (Sn xn ) − pk2 + (1 − αn )kxn − pk2 + γn (λn 2 k2 − 1)kun − yn k2

(3.13)

from which it follows that

k un − y n k 2 ≤

αn (kf (Sn xn ) − pk2 − kxn − pk2 ) γn (1 − λn 2 k2 ) +

1

γn (1 − λn 2 k2 )

(kxn − pk − kxn+1 − pk)kxn+1 − xn k.

(C1)–(C3), (C5) and kxn+1 − xn k → 0 imply that kun − yn k → 0. By the same argument as in (3.3), we also have

ktn − uk2 ≤ kun − uk2 − kun − yn k2 − kyn − tn k2 + 2λn kkun − yn kktn − yn k ≤ kun − uk2 − kun − yn k2 − kyn − tn k2 + kun − yn k2 + λn 2 k2 ktn − yn k2 = kun − uk2 + (λn 2 k2 − 1)kyn − tn k2 . Combining the above inequality and (3.13), we have

kxn+1 − pk2 ≤ αn kf (Sn xn ) − pk2 + βn kxn − pk2 + γn ktn − pk2 ≤ αn kf (Sn xn ) − pk2 + βn kxn − pk2 + γn [kun − pk2 + (λ2n k2 − 1)kyn − tn k2 ] ≤ αn kf (Sn xn ) − pk2 + (1 − αn )kxn − pk2 + γn (λn 2 k2 − 1)kyn − tn k2

(3.14)

J.-W. Peng, J.-C. Yao / Nonlinear Analysis 71 (2009) 6001–6010

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and thus

ktn − yn k2 ≤

αn 1 (kf (Sn xn ) − uk2 − kxn − uk2 ) + (kxn − uk − kxn+1 − uk)kxn+1 − xn k 2 2 γn (1 − λn k ) γn (1 − λn 2 k2 )

which implies that ktn − yn k → 0. From kun − tn k ≤ kun − yn k + kyn − tn k, we also have kun − tn k → 0. As A is k-Lipschitz continuous, we have kAyn − Atn k → 0. Step 5. We now show that lim kΘnk xn − Θnk−1 xn k = 0,

n→∞

k = 1, 2, . . . , M .

(3.15) F

Indeed, let p ∈ Ω . It follows from the firmly nonexpansiveness of Trkk,n that we have for each k ∈ {1, 2, . . . , M },

kΘnk xn − pk2 = kTrFkk,n Θnk−1 xn − TrFkk,n pk2 ≤ hΘnk xn − p, Θnk−1 xn − pi 1

=

2

(kΘnk xn − pk2 + kΘnk−1 xn − pk2 − kΘnk xn − Θnk−1 xn k2 ).

Thus we get

kΘnk xn − pk2 ≤ kΘnk−1 xn − pk2 − kΘnk xn − Θnk−1 xn k2 ,

k = 1, 2, . . . , M

which implies that for each k ∈ {1, 2, . . . , M },

kΘnk xn − pk2 ≤ kΘn0 xn − pk2 − kΘnk xn − Θnk−1 xn k2 − kΘnk−1 xn − Θnk−2 xn k2 − · · · − kΘn2 xn − Θn1 xn k2 − kΘn1 xn − Θn0 xn k2 ≤ kxn − pk2 − kΘnk xn − Θnk−1 xn k2 .

(3.16) F

By (3.13), un = ΘnM xn , (3.16) and the nonexpansiveness of Trkk,n , we have for each k ∈ {1, 2, . . . , M },

kxn+1 − pk2 ≤ αn kf (Sn xn ) − pk2 + βn kxn − pk2 + γn kun − pk2 ≤ αn kf (Sn xn ) − pk2 + βn kxn − pk2 + γn kΘnk xn − pk2 ≤ αn [kf (Sn xn ) − pk2 + kxn − pk2 ] − γn kΘnk xn − Θnk−1 xn k2 . Consequently, we have

γn kΘnk xn − Θnk−1 xn k2 ≤ αn kf (Sn xn ) − pk2 + kxn − pk2 − kxn+1 − pk2 . Now by condition (C1) and using kxn+1 − xn k → 0 we obtain (3.15) and Step 5 is proved. Step 6. limn→∞ kSyn − yn k = 0. We observe that

kSn yn − yn k ≤ kSn yn − Sn tn k + kSn tn − xn k + kxn − Θn1 xn k + kΘn1 xn − Θn2 xn k + · · · + kΘnm−1 xn − Θnm xn k + kun − yn k ≤ kyn − tn k + kSn tn − xn k + kxn − Θn1 xn k + kΘn1 xn − Θn2 xn k + · · · + kΘnm−1 xn − Θnm xn k + kun − yn k. It follows that limn→∞ kSn yn − yn k = 0. At the same time, observe that

kSyn − yn k ≤ kSyn − Sn yn k + kSn yn − yn k.

(3.17)

Thus by (3.17) and Lemma 2.6, we have limn→∞ kSyn − yn k = 0. Step 7. We show that lim suphf (v0 ) − v0 , xn − v0 i ≤ 0, n→∞

where v0 = PΩ f (v0 ). To show this inequality, we can choose a subsequence {xni } of {xn } such that lim hf (v0 ) − v0 , xnj − v0 i = lim suphf (v0 ) − v0 , xn − v0 i.

n→∞

n→∞

Since {xni } is bounded, there exists a subsequence {xni } of {xni } which converges weakly to w . Without loss of generality, j

we can assume that {xni } * w . Since kΘnk xn − Θnk−1 xn k → 0 for each k = 1, 2, . . . , m, we obtain that Θnki xni * w for

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k = 1, 2, . . . , M. It follows from kun − yn k → 0 and kun − tn k → 0 that yni * w and tni * w . Since {uni } ⊂ C and C is closed and convex, we obtain w ∈ C . In order to show that w ∈ Ω , we first show w ∈ ∩M k=1 EP(Fk ). Indeed by Lemma 2.2, we have that for each k = 1, 2, . . . , M, Fk (Θnk xn , y) +

1 rn

hy − Θnk xn , Θnk xn − Θnk−1 xn i ≥ 0,

∀y ∈ C .

It follows from (A2) that 1 rn

hy − Θnk xn , Θnk xn − Θnk−1 xn i ≥ Fk (y, Θnk xn ),

∀y ∈ C .

Hence,

* y−

Θnki xni

,

Θnki xni − Θnki−1 xni

Θnk xni −Θnki−1 xni i

By (A4),

rni

Fk (y, w) ≤ 0,

rni

+ ≥ Fk (y, Θnki xni ),

∀y ∈ C .

→ 0 and Θnki xni * w, we have for each k = 1, 2, . . . , M, ∀y ∈ C .

For t with 0 < t ≤ 1 and y ∈ C , let yt = ty + (1 − t )w. Since y ∈ C and w ∈ C , we obtain yt ∈ C and hence Fk (yt , w) ≤ 0. So by (A4), we have 0 = Fk (yt , yt ) ≤ tFk (yt , y) + (1 − t )Fk (yt , w) ≤ tFk (yt , y). Dividing both sides of the above inequality by t, we get that for each k = 1, 2, . . . , m Fk (yt , y) ≥ 0. Letting t → 0, it follows from (A3) that for each k = 1, 2, . . . , m Fk (w, y) ≥ 0 for all y ∈ C and hence w ∈ EP(Fk ) for k = 1, 2, . . . , M, i.e., w ∈ ∩M k=1 EP(Fk ). By exactly the same argument as that in the proof of Theorem 3.1 in [9], we can show that w ∈ VI(C , A). We next show that w ∈ Fix(S ). Assume w 6∈ Fix(S ). Since yni * w and w 6= S w , from the Opial condition we have lim inf kyni − wk < lim inf kyni − S wk i→∞

i→∞

≤ lim inf{kyni − Syni k + kSyni − S wk} i→∞

≤ lim inf kyni − wk i→∞

which is a contradiction. Thus we get w ∈ Fix(S ) = ∩∞ i=1 Fix(Si ) which implies that w ∈ Ω . Therefore, we have lim suphf (v0 ) − v0 , xn − v0 i = lim hf (v0 ) − v0 , xnj − v0 i = hf (v0 ) − v0 , w − v0 i ≤ 0. n→∞

n→∞

(3.18)

Finally, we show that xn → v0 where v0 = PΩ f (v0 ). From Lemma 2.4, we have

kxn+1 − v0 k2 = ≤ ≤ ≤ ≤

kαn (f (Sn xn ) − v0 ) + βn (xn − v0 ) + γn (Sn tn − v0 )k2 kβn (xn − v0 ) + γn (Sn tn − v0 )k2 + 2αn hf (Sn xn ) − v0 , xn+1 − v0 i [γn ktn − v0 k + βn kxn − v0 k]2 + 2αn hf (Sn xn ) − v0 , xn+1 − v0 i (1 − αn )2 kxn − v0 k2 + 2αn akxn − v0 kkxn+1 − v0 k + 2αn hf (v0 ) − v0 , xn+1 − v0 i (1 − αn )2 kxn − v0 k2 + αn a(kxn − v0 k2 + kxn+1 − v0 k2 ) + 2αn hf (v0 ) − v0 , xn+1 − v0 i

and thus

  2αn (1 − a) kxn+1 − v0 k2 ≤ 1 − kx n − v 0 k2 1 − aαn   2αn (1 − a) αn 1 + kxn − v0 k2 + h2f (v0 ) − 2v0 , xn+1 − v0 i . 1 − aαn 2(1 − a) 1−a

(3.19)

It follows from Lemma 2.3, (3.18) and (3.19) that limn→∞ kxn − v0 k = 0. (3.15) implies that un → v0 . From kyn − un k → 0, we have yn → v0 . The proof is now complete. 

J.-W. Peng, J.-C. Yao / Nonlinear Analysis 71 (2009) 6001–6010

6009

Theorem 3.2. Let C be a nonempty closed convex subset of a real Hilbert space H. Let Fk , k ∈ {1, 2, . . . , M } be bifunctions from C × C to R satisfying (A1)–(A4)and A be a monotone and k-Lipschitz continuous mapping of C into H. Let {Sn } be a sequence M of nonexpansive mappings of C into itself such that Ω = ∩∞ i=1 Fix(Si ) ∩ VI (C , A) ∩ (∩k=1 EP(Fk )) 6= ∅. Assume that v is an arbitrary element in C . Let {xn }, {un } and {yn } be sequences generated by

 x1 = x ∈ C ,    F un = TrFMM,n TrMM−−11,n . . . TrF22,n TrF11,n xn ,  y = PC (un − λn Aun ),   n xn+1 = αn v + βn xn + γn Sn PC (un − λn Ayn ),

(3.20)

for every n = 1, 2, . . . , where {λnP }, {rk,n }, k ∈ {1, 2, . . . , M }, {αn }, {βn } and {γn } are sequences of numbers satisfying the ∞ conditions (C1)–(C5). Suppose that n=1 sup{kSn+1 x − Sn xk : x ∈ B} < ∞ for any bounded subset B of C . Let S be a mapping of C into itself defined by Sy = limn→∞ Sn y for all y ∈ C and suppose that Fix(S ) = ∩∞ i=1 Fix(Si ). Then the sequences {xn }, {un } and {yn } converge strongly to the same point w ∈ Ω where w = PΩ (v). Proof. Let f (x) = v for all x ∈ C . By Theorem 3.1 we obtain the desired result.



Remark 3.1. (i) There are some examples of sequences of nonexpansive mappings satisfying the conditions in Theorem 3.1. The first example can be found in [5, Section 4]. The other example is given as follows: Let S1 , S2 , . . . be an infinite family of nonexpansive mappings of C into itself and let ξ1 , ξ2 , . . . be real numbers such that 0 ≤ ξi ≤ 1 for every i ∈ N. For any n ∈ N, define a mapping Wn of C into C as follows: Un,n+1 = I , Un,n = ξn Sn Un,n+1 + (1 − ξn )I , Un,n−1 = ξn−1 Sn−1 Un,n + (1 − ξn−1 )I ,

... Un,k = ξk Sk Un,k+1 + (1 − ξk )I , Un,k−1 = ξk−1 Sk−1 Un,k + (1 − ξk−1 )I ,

... Un,2 = ξ2 S2 Un,3 + (1 − ξ2 )I , Wn = Un,1 = ξ1 S1 Un,2 + (1 − ξ1 )I . Such a mapping Wn is called the W -mapping generated by Sn , Sn−1 , . . . , S1 and ξn , ξn−1 , . . . , ξ1 ; see [20]. Using Lemma 3.2 in [20], one can define mappings U∞,k and W of C into itself as follows: U∞,k x = lim Un,k x n→∞

and Wx = limn→∞ Wn x = limn→∞ Un,1 x for every x ∈ C . Such a mapping W is called the W -mapping generated by S1 , S2 , . . . and ξ1 , ξ2 , . . . . Let Wn be the W -mapping of C into itself generated by Sn , Sn−1 , . . . , S1 and ξn , ξn−1 , . . . , ξ1 satisfying P∞ the conditions of Theorem 3.1 in [9]. Then {Wn } is a sequence of nonexpansive mappings satisfying condition n=1 sup{kWn+1 x − Wn xk : x ∈ B} < ∞ for any bounded subset B of C . Indeed, for x ∈ B, we have

kWn+1 x − Wn xk = = ≤ = ≤

kUn+1,1 x − Un,1 xk kξ1 S1 Un+1,2 x + (1 − ξ1 )x − {ξ1 S1 Un,2 x + (1 − ξ1 )x}k ξ1 kUn+1,2 x − Un,2 xk ξ1 kξ2 S2 Un+1,3 x + (1 − ξ2 )x − {ξ2 S2 Un,3 x + (1 − ξ2 )x}k ξ1 ξ2 kUn+1,3 x − Un,3 xk ... n Y ≤ ξi kUn+1,n+1 x − Un,n+1 xk i =1

=

n Y

ξi kξn+1 Sn+1 x + (1 − ξn+1 )x − xk

i =1

=

n +1 Y

ξi kSn+1 x − xk ≤ dn+1 M2 ,

i =1

P∞

where M2 is an approximate constant such that M2 ≥ supn≥1 {kSn+1 tn − xk : x ∈ B}. Thus n=1 sup{kWn+1 x − Wn xk : x ∈ B} < ∞. Moreover, let W be the W -mapping generated by S1 , S2 , . . . and ξ1 , ξ2 , . . . . Then from [20], we know that Wy = limn→∞ Wn y for all y ∈ C and that Fix(W ) = ∩∞ i=1 Fix(Wi ).

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(ii) Theorems 3.1 and 3.2 extend and generalize main results in [2,6–8,3,9–14] from the case of single equilibrium problem to that of the simultaneous equilibrium problems. (iii) Since α -inverse-strong monotonicity of A has been weakened by the monotonicity and Lipschitz continuity of A, Theorems 3.1 and 3.2 improve main results in [7,8,10]. Theorems 3.1 and 3.2 also improve main results in [9,10,12] by removing the Cn and Dn conditions in the corresponding algorithms. (iv) We observed that Tn of Procedure 3.1 in [1] is not easy to be found even if Γ = {1, 2, . . . , M } for any n ≥ 1. Thus the Algorithm 2.5 for problem (1.1) in [1] is more complicated than those in Theorems 3.1 and 3.2. Acknowledgements The authors are grateful to the referees for the detailed comments and helpful suggestions which improved the original manuscript greatly. The first author’s research was supported by the National Natural Science Foundation of China (Grant Nos. 10771228 and 10831009), the Science and Technology Research Project of Chinese Ministry of Education (Grant No.206123), the Research Project of Chongqing Normal University(Grant No. 08XLZ05). The second author’s research was partially supported by the grant NSC 96-2119-M-110-001. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]

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