Volume 2009, Article ID 567147,20pages doi:10.1155/2009/567147
Research Article
A Viscosity Approximation Method for Finding
Common Solutions of Variational Inclusions,
Equilibrium Problems, and Fixed Point Problems in
Hilbert Spaces
Somyot Plubtieng
1, 2and Wanna Sriprad
1, 21Department of Mathematics, Faculty of Science, Naresuan University, Phitsanulok 65000, Thailand 2PERDO National Centre of Excellence in Mathematics, Faculty of Science, Mahidol University, Bangkok
10400, Thailand
Correspondence should be addressed to Somyot Plubtieng,[email protected]
Received 12 February 2009; Accepted 18 May 2009
Recommended by William A. Kirk
We introduce an iterative method for finding a common element of the set of common fixed points of a countable family of nonexpansive mappings, the set of solutions of a variational inclusion with set-valued maximal monotone mapping, and inverse strongly monotone mappings and the set of solutions of an equilibrium problem in Hilbert spaces. Under suitable conditions, some strong convergence theorems for approximating this common elements are proved. The results presented in the paper improve and extend the main results of J. W. Peng et al.2008and many others. Copyrightq2009 S. Plubtieng and W. Sriprad. 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
LetHbe a real Hilbert space whose inner product and norm are denoted by·,·and · , respectively. LetCbe a nonempty closed convex subset ofH, and letF be a bifunction of
C×CintoR, whereRis the set of real numbers. The equilibrium problem forF :C×C → R is to findx∈Csuch that
Fx, y≥0, ∀y∈C. 1.1
The set of solutions of 1.1 is denoted by EPF. Recently, Combettes and Hirstoaga1 introduced an iterative scheme of finding the best approximation to the initial data when
map. The classical variational inequality which is denoted byV IA, Cis to findu∈Csuch that
Au, v−u ≥0, ∀v∈C. 1.2
The variational inequality has been extensively studied in literature. See, for example,2,3 and the references therein. Recall that a mappingTofCinto itself is called nonexpansive if
Su−Sv ≤ u−v, ∀u, v∈C. 1.3
A mappingf:C → Cis called contractive if there exists a constantβ∈0,1such that
fu−fv ≤βu−v, ∀u, v∈C. 1.4
We denote byFSthe set of fixed points ofS.
Some methods have been proposed to solve the equilibrium problem and fixed point problem of nonexpansive mapping; see, for instance,3–6and the references therein. Recently, Plubtieng and Punpaeng6introduced the following iterative scheme. Letx1∈H
and let{xn}, and{un}be sequences generated by
Fun, yrn1 y−un, un−xn≥0, ∀y∈H,
xn1αnγfxn I−αnASun, ∀n∈N.
1.5
They proved that if the sequences{αn}and{rn}of parameters satisfy appropriate conditions,
then the sequences {xn} and {un} both converge strongly to the unique solution of the variational inequality
A−γfz, z−x≥0, ∀x∈FS∩EPF, 1.6
which is the optimality condition for the minimization problem
min
x∈FS∩EPF 1
2Ax, x −hx, 1.7
wherehis a potential function forγf.
LetA : H → Hbe a single-valued nonlinear mapping, and letM : H → 2H be a set-valued mapping. We consider the following variational inclusion, which is to find a point
u∈Hsuch that
θ∈Au Mu, 1.8
IfM ∂δC, whereCis a nonempty closed convex subset ofHandδC :H → 0,∞is the indicator function of C, that is,
δCx ⎧ ⎨ ⎩
0, x∈C,
∞, x /∈C, 1.9
then the variational inclusion problem1.8is equivalent to variational inequality problem
1.2. It is known that1.8provides a convenient framework for the unified study of optimal solutions in many optimization related areas including mathematical programming, com-plementarity, variational inequalities, optimal control, mathematical economics, equilibria, game theory. Also various types of variational inclusions problems have been extended and generalizedsee8and the references therein.
Very recently, Peng et al. 9 introduced the following iterative scheme for finding a common element of the set of solutions to the problem 1.8, the set of solutions of an equilibrium problem, and the set of fixed points of a nonexpansive mapping in Hilbert space. Starting withx1∈H, define sequence,{xn},{yn}, and{un}by
Fun, yr1 n
y−un, un−xn≥0, ∀y∈H,
xn1αnγfxn 1−αnSyn,
ynJM,λun−λAun, ∀n≥0,
1.10
for all n ∈ N, where λ ∈ 0,2α, {αn} ⊂ 0,1 and {rn} ⊂ 0,∞. They proved that
under certain appropriate conditions imposed on{αn}and {rn}, the sequences{xn},{yn},
and {un} generated by 1.10 converge strongly to z ∈ FT ∩IA, M ∩EPF, where
zPFS∩IA,M∩EPFfz.
Motivated and inspired by Plubtieng and Punpaeng6, Peng et al.9and Aoyama et al.10, we introduce an iterative scheme for finding a common element of the set of solutions of the variational inclusion problem 1.8 with multi-valued maximal monotone mapping and inverse-strongly monotone mappings, the set of solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping in Hilbert space. Starting with an arbitrary
x1∈H,define sequences{xn},{yn}and{un}by
Fun, yrn1 y−un, un−xn≥0, ∀y∈H,
xn1 αnγfxn I−αnBSnyn,
ynJM,λun−λAun, ∀n≥0,
1.11
for alln∈N, whereλ∈0,2α,{αn} ⊂0,1, and let{rn} ⊂0,∞;Bbe a strongly bounded linear operator onH, and{Sn}is a sequence of nonexpansive mappings onH. Under suitable
2. Preliminaries
This section collects some lemmas which will be used in the proofs for the main results in the next section.
LetHbe a real Hilbert space with inner product·,·and norm · , respectively. It is wellknown that for allx, y∈Handλ∈0,1, there holds
λx 1−λy 2λx2 1−λ y 2−λ1−λ x−y 2. 2.1
LetCbe a nonempty closed convex subset ofH. Then, for anyx∈H, there exists a unique nearest point ofC, denoted byPCx, such thatx−PCx ≤ x−yfor ally∈C. Such aPCis
called the metric projection fromHintoC. We know thatPCis nonexpansive. It is also known that,PCx∈Cand
x−PCx, PCx−z ≥0, ∀x∈H andz∈C. 2.2
It is easy to see that2.2is equivalent to
x−z2≥ x−PCx2z−PCx2, ∀x∈H, z∈C. 2.3
For solving the equilibrium problem for a bifunctionF :C×C → R, let us assume thatFsatisfies the following conditions:
A1Fx, x 0 for allx∈C;
A2Fis monotone, that is,Fx, y Fy, x≤0 for allx, y∈C;
A3for eachx, y, z∈C,
lim
t→0F
tz 1−tx, y≤Fx, y; 2.4
A4for eachx∈C, y→Fx, yis convex and lower semicontinuous.
The following lemma appears implicitly in11and1.
Lemma 2.1See1,11. LetCbe a nonempty closed convex subset ofHand letFbe a bifunction ofC×Cin toRsatisfying (A1)–(A4). Letr >0andx∈H. Then, there existsz∈Csuch that
Fz, y1ry−z, z−x ≥0, ∀y∈C. 2.5
Define a mappingTr :H → Cas follows:
Trx
z∈C:Fz, y 1
r
y−z, z−x≥0, ∀y∈C
for allz∈H. Then, the following hold:
1Tr is single-valued;
2Tr is firmly nonexpansive, that is, for anyx, y∈H,
Trx−Try2≤Trx−Try, x−y; 2.7
3FTr EPF;
4EPFis closed and convex.
We also need the following lemmas for proving our main result.
Lemma 2.2See12. LetHbe a Hilbert space,Ca nonempty closed convex subset ofH,f :H →
Ha contraction with coefficient0 < α < 1, andBa strongly positive linear bounded operator with coefficientγ >0. Then :
1if 0< γ < γ/α, thenx−y,B−γfx−B−γfy ≥γ−γαx−y2, x, y∈H.
2if 0< ρ <B−1, thenI−ρB ≤1−ργ.
Lemma 2.3See13. Assume{an}is a sequence of nonnegative real numbers such that
an1 ≤
1−γnanδn, n≥0, 2.8
where{γn}is a sequence in0,1and{δn}is a sequence inRsuch that
1∞n1γn∞;
2lim supn→ ∞δn/γn≤0or∞n1|δn|<∞.
Thenlimn→ ∞an0.
Recall that a mapping A : H → H is calledα-inverse-strongly monotone, if there exists a positive numberαsuch that
Au−Av, u−v ≥αAu−Av2, ∀u, v∈H. 2.9
LetIbe the identity mapping onH. It is well known that ifA:H → Hisα -inverse-strongly monotone, thenAis 1/α-Lipschitz continuous and monotone mapping. In addition, if 0< λ≤2α, thenI−λAis a nonexpansive mapping.
A set-valuedM : H → 2H is called monotone if for allx, y ∈ H, f ∈ Mxand g ∈
Myimplyx−y, f −g ≥ 0. A monotone mappingM : H → 2H is maximal if its graph
GM :{x, f ∈H×H |f ∈ Mx}ofMis not properly contained in the graph of any other monotone mapping. It is known that a monotone mappingMis maximal if and only if forx, f∈H×H,x−y, f−g ≥0 for everyy, g∈GMimpliesf∈Mx.
Let the set-valued mapping M : H → 2H be maximal monotone. We define the resolvent operatorJM,λassociated withMandλas follows:
where λ is a positive number. It is worth mentioning that the resolvent operator JM,λ is single-valued, nonexpansive and 1-inverse-strongly monotone, see for example14and that a solution of problem1.8is a fixed point of the operatorJM,λI−λAfor allλ >0, see for instance15.
Lemma 2.4See14. LetM:H → 2Hbe a maximal monotone mapping andA:H → Hbe a Lipschitz-continuous mapping. Then the mappingSMA:H → 2His a maximal monotone mapping.
Remark 2.5See9. Lemma 2.4implies thatIA, Mis closed and convex ifM:H → 2His a maximal monotone mapping andA:H → Hbe an inverse strongly monotone mapping.
Lemma 2.6See10. LetCbe a nonempty closed subset of a Banach space and let{Sn}a sequence of mappings ofCinto itself. Suppose that∞n1sup{Sn1z−Snz :z ∈ C} < ∞. Then, for each x∈C,{Snx}converges strongly to some point ofC. Moreover, letSbe a mapping fromCinto itself defined by
Sx lim
n→ ∞Snx, ∀x∈C. 2.11
Thenlimn→ ∞sup{Sz−Snz:z∈C}0.
3. Main Results
We begin this section by proving a strong convergence theorem of an implicit iterative sequence {xn} obtained by the viscosity approximation method for finding a common
element of the set of solutions of the variational inclusion, the set of solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping.
Throughout the rest of this paper, we always assume thatfis a contraction ofHinto itself with coefficientβ ∈ 0,1, andB is a strongly positive bounded linear operator with coefficientγand 0< γ < γ/β. LetSbe a nonexpansive mapping ofHintoH. LetA:H → H be anα-inverse-strongly monotone mapping,M:H → 2Hbe a maximal monotone mapping and letJM,λbe defined as in2.10. Let{Trn}be a sequence of mappings defined asLemma 2.1. Consider a sequence of mappings{Sn}onHdefined by
Snxαnγfx I−αnBSJM,λI−λATrnx, x∈H, n≥1, 3.1
where{αn} ⊂ 0,B−1.By Lemma 2.2, we note thatSn is a contraction. Therefore, by the
Banach contraction principle,Snhas a unique fixed pointxn∈Hsuch that
xnαnγfxn I−αnBSJM,λI−λATrnxn. 3.2
Theorem 3.1. LetH be a real Hilbert space, let F be a bifunction from H×H → Rsatisfying (A1)–(A4) and letSbe a nonexpansive mapping onH. LetA :H → H be anα-inverse-strongly monotone mapping,M:H → 2Hbe a maximal monotone mapping such thatΩ:FS∩EPF∩
bounded linear operator onHwith coefficientγ >0and0 < γ < γ/β. Let{xn},{yn}and{un}be sequences generated byx1∈Hand
Fun, yr1
ny−un, un−xn ≥0, ∀y∈H xnαnγfxn I−αnBSyn,
ynJM,λun−λAun ∀n≥0,
3.3
where λ ∈ 0,2α,{rn} ⊂ 0,∞and {αn} ⊂ 0,1satisfylimn→ ∞αn 0 andlim infn→ ∞rn >
0. Then, {xn}, {yn} and {un} converges strongly to a point z in Ω which solves the variational inequality:
B−γfz, z−x ≤0, x∈Ω. 3.4
Equivalently, we havezPΩI−Bγfz.
Proof. First, we assume thatαn∈0,B−1. ByLemma 2.2, we obtainI−αnB ≤1−αnγ. Let v∈Ω.SinceunTrnxn,we have
un−vTrnxn−Trnv ≤ xn−v ∀n∈N. 3.5
We note fromv∈ΩthatvJM,λv−λAv. AsI−λAis nonexpansive, we have
yn−vJM,λun−λAun−JM,λv−λAv
≤ un−λAun−v−λAv ≤ un−v ≤ xn−v
3.6
for alln∈N.Thus, we have
xn−vαnγfxn I−αnBSyn−v ≤αnγfxn−BvI−αnByn−v ≤αnγfxn−Bv1−αnγxn−v
≤αnγfxn−fvγfv−Bv1−αnγxn−v
≤αnγβxn−vαnγfv−Bv1−αnγxn−v
1−αnγ−γβxn−vαnγfv−Bv.
3.7
It follows thatxn −v ≤ γfv−Bv/γ−γβ,∀n ≥ 1.Hence{xn} is bounded and we also obtain that{un},{yn},{fxn},{Syn}and{Aun}are bounded. Next, we show thatyn− Syn → 0. Sinceαn → 0, we note that
Moreover, it follows fromLemma 2.1that
un−v2Trnxn−Trnv2≤ Trnxn−Trnv, xn−vun−v, xn−v
1
2
un−v2xn−v2− xn−un2
, 3.9
and henceun−v2≤ xn−v2− xn−un2.Therefore, we have
xn−v2 αnγfxn I−αnBSyn−v 2
I−αnBSyn−v αnγfxn−Bv 2
≤1−αnγ2 Syn−v 22αnγfxn−Bv, xn−v
≤1−αnγ2 yn−v 22αnγfxn−fv, xn−v2αnγfv−Bv, xn−v
≤1−αnγ2un−v22αnγfxn−fv, xn−v2αnγfv−Bv, xn−v
≤1−αnγ2
xn−v2− xn−un2
2αnγβxn−v2
2αnγfv−Bvxn−v
1−2αnγ−γβαnγ2xn−v2−1−αnγ2xn−un2
2αnγfv−Bvxn−v
≤ xn−v2αnγ2xn−v2−1−αnγ2xn−un22αnγfv−Bvxn−v,
3.10
and hence
1−αnγ2xn−un2≤αnγ2xn−v22αnγfv−Bvxn−v. 3.11
Since{xn}is bounded andαn → 0,it follows thatxn−un → 0 asn → ∞.
PutMsupn≥1{γfv−Bvxn−v}. From3.10, it follows by the nonexpansive of JM,λand the inverse strongly monotonicity ofAthat
xn−v2≤1−αnγ2 yn−v 2
2αnγβxn−v22αnM
≤1−αnγ2un−λAun−v−λAv22αnγβxn−v22αnM
≤1−αnγ2
un−v2λλ−2αAun−Av2
2αnγβxn−v22αnM
≤1−αnγ2xn−v21−αnγ2λλ−2αAun−Av22αnγβxn−v22αnM
1−2αnγ−γβαnγ2xn−v21−αnγ2λλ−2αAun−Av22αnM
≤ xn−v2αnγ2xn−v21−αnγ2λλ−
2αAun−Av22αnM,
which implies that
1−αnγ2λ2α−λAun−Av2 ≤αnγ2xn−v22αnM. 3.13
Sinceαn → 0, we haveAun−Av → 0 asn → ∞. SinceJM,λis 1–inverse-strongly monotone andI−λAis nonexpansive, we have
yn−v 2JM,λun−λAun−JM,λv−λAv2≤un−λAun−v−λAv, yn−v
1
2
un−λAun−v−λAv2 yn−v 2− un−λAun−v−λAv−yn−v 2
≤ 1
2
un−v2 yn−v 2− un−yn−λAun−Av 2
1
2
un−v2 yn−v 2− un−yn 22λun−yn, Aun−Av−λ2Aun−Av2
.
3.14
Thus, we have
yn−v 2 ≤ un−v2− un−yn 22λun−yn, Aun−Av−λ2Aun−Av2. 3.15
From3.5,3.10, and3.15, we have
xn−v2≤1−αnγ2 yn−v 22αnγβxn−v22αnM
≤1−αnγ2un−v2− un−yn 22λun−yn, Aun−Av−λ2Aun−Av2
2αnγβxn−v22αnM
≤1−αnγ2xn−v2−1−αnγ2 un−yn 221−αnγ2λun−yn, Aun−Av
−1−αnγ2λ2Aun−Av22αnγβxn−v22αnM
1−2αnγ−γβαnγ2xn−v2−1−αnγ2 un−yn 2
21−αnγ2λun−yn, Aun−Av−1−αnγ2λ2Aun−Av22αnM
≤ xn−v2αnγ2xn−v2−1−αnγ2 un−yn 2
21−αnγ2λun−yn, Aun−Av−1−αnγ2λ2Aun−Av22αnM.
3.16
Thus, we get
1−αnγ2 un−yn 2≤αnγ2xn−v221−αnγ2λun−yn, Aun−Av
−1−αnγ2λ2Aun−Av22αnM.
Sinceαn → 0,Aun−Av → 0 asn → ∞, we haveun−yn → 0 asn → ∞. It follows from the inequalityyn−Syn ≤ yn−unun−xnxn−Synthatyn−Syn → 0 as n → ∞. Moreover, we havexn−yn ≤ xn−unun−yn → 0 asn → ∞.
PutU ≡SJM,λI−λA. Since bothSandJM,λI−λAare nonexpansive, we haveU is a nonexpansive mapping onHand then we havexnαnγfxn I−αnBUTrnxnfor all n∈N. It follows byTheorem 3.1of Plubtieng and Punpaeng6that{xn}converges strongly
toz∈FU∩EPF, wherezPFU∩EPFγf I−BzandB−γfz, u−z ≥0, for all
u∈FU∩EPF. We will show thatz ∈FS∩IA, M. Since{xn}converges strongly to z, we also havexn z. Let us showz ∈FS.Assumez /∈FS.Sincexn−yn → 0 and xn z, we haveyn zSincez /Sz,it follows by the Opial’s condition that
lim inf
n→ ∞ yn−z<lim infn→ ∞ yn−Sz ≤lim infn→ ∞
yn−SynSyn−Sz
≤lim inf
n→ ∞ yn−z.
3.18
This is a contradiction. Hencez ∈ FS.We now show thatz ∈ IA, M. In fact, since A isα–inverse-strongly monotone,Ais an 1/α-Lipschitz continuous monotone mapping and
DA H. It follows fromLemma 2.4thatMAis maximal monotone. Letp, g∈GMA, that is,g−Ap∈Mp. Again sinceynJM,λun−λAun, we haveun−λAun∈IλMyn, that is,
1
λ
un−yn−λAun∈Myn. 3.19
By the maximal monotonicity ofMA, we have
p−yn, g−Ap−1λun−yn−λAun≥0, 3.20
and so
p−yn, g≥
p−yn, Ap 1λun−yn−λAun
p−yn, Ap−AynAyn−Aun1λun−yn
≥0p−yn, Ayn−Aun
p−yn,1λun−yn.
3.21
It follows fromun−yn → 0,Aun−Ayn → 0 andyn zthat
lim
n→ ∞
p−yn, gp−z, g≥0. 3.22
SinceAMis maximal monotone, this implies thatθ ∈ MAz,that is,z ∈ IA, M. Hence,z∈Ω:FS∩EPF∩IA, M. SinceFS∩IA, M FS∩FJM,λI−λA⊂FU, we haveΩ⊂FU∩EPF. It implies thatzis the unique solution of the variational inequality
Now we prove the following theorem which is the main result of this paper.
Theorem 3.2. LetH be a real Hilbert space, let F be a bifunction from H×H → Rsatisfying (A1)–(A4) and let{Sn}be a sequence of nonexpansive mappings on H. LetA : H → H be an
α-inverse-strongly monotone mapping,M : H → 2H be a maximal monotone mapping such that
Ω : ∞n1FSn∩EPF∩IA, M/∅.Let f be a contraction ofH into itself with a constant β∈0,1and letBbe a strongly bounded linear operator onHwith coefficientγ >0and0< γ < γ/β. Let{xn},{yn}and{un}be sequences generated byx1∈Hand
Fun, yr1
ny−un, un−xn ≥0, ∀y∈H xn1 αnγfxn I−αnBSnyn,
ynJM,λun−λAun ∀n≥0,
3.23
for alln∈N, whereλ∈0,2α,{αn} ⊂0,1and{rn} ⊂0,∞satisfy
lim
n→ ∞αn0, ∞
n1
αn∞, ∞ n1
|αn1−αn|<∞,
lim inf
n→ ∞ rn>0, ∞
n1
|rn1−rn|<∞.
3.24
Suppose that∞n1sup{Sn1z−Snz :z ∈ K} < ∞for any bounded subsetKofH. LetSbe a
mapping ofHinto itself defined by Sx limn→ ∞Snx, for all x ∈ H and suppose thatFS ∞
n1FSn. Then,{xn},{yn}and{un}converges strongly toz, wherez PΩI−Bγfzis a
unique solution of the variational inequalities3.4.
Proof. Sinceαn → 0, we may assume thatαn≤ B−1for alln. First we will prove that{xn}is
bonded. Letv∈Ω.Then, we have
xn1−vαnγfxn I−αnBSnyn−v
≤αnγfxn−BvI−αnByn−v
≤αnγfxn−Bv1−αnγxn−v
≤αnγfxn−fvγfv−Bv1−αnγxn−v
≤αnγβxn−vαnγfv−Bv1−αnγxn−v
1−αnγ−γβxn−vαnγfv−Bv
1−αnγ−γαxn−vαnγ−γαγfv−Bv
γ−γα .
3.25
It follows from3.25and induction that
xn−v ≤max
x1−v,γ−1γαγf
p−Bp
Hence {xn} is bounded and therefore {un}, {yn},{fxn},{Snyn} and {Aun} are also bounded. Next, we show that xn1 −xn → 0. Since I−λA is nonexpansive, it follows
that
yn1−ynJM,λun1−λAun1−JM,λun−λAun
≤ un1−λAun1−un−λAun ≤ un1−un.
3.27
Then, we have
xn2−xn1αn1γfxn1 I−αn1BSn1yn1−αnγfxn−I−αnBSnyn
αn1γfxn1 I−αn1BSn1yn1−αnγfxn−I−αnBSnyn
− I−αn1BSn1yn I−αn1BSn1yn−I−αnBSn1yn I−αnBSn1yn
I−αn1B
Sn1yn1−Sn1yn
αn−αn1BSn1yn
I−αnBSn1yn−Snyn αn1−αnγfxnαn1γf
xn1−fxn
≤1−αn1γ
yn1−yn|αn−αn1| BSn1yn
1−αnγSn1yn−Snyn
|αn−αn1| γfxnαn1γβxn1−xn
≤1−αn1γ yn1−yn αn1γβxn1−xn
|αn−αn1|
BSn1ynγfxn
Sn1yn−Snyn
≤1−αn1γ
un1−unαn1γβxn1−xn|αn−αn1|M
supSn1z−Snz:z∈yn,
3.28
whereM:sup{max{BSn1yn,γfxn}:n≥0}<∞. On the other hand, we note that
Fun, yr1
n
y−un, un−xn≥0, ∀y∈H, 3.29
Fun1, y
r1
n1
y−un1, un1−xn1
≥0, ∀y∈H. 3.30
Puttingy un1 in3.29andy un in3.30,Fun, un1 1/rnun1−un, un−xn ≥0
andFun1, un 1/rn1un−un1, un1−xn1 ≥0.ByA2, we have
un1−un,
un−xn rn −
un1−xn1
rn1
≥0 3.31
and hence
un1−un, un−un1un1−xn−rnrn
1
Since lim infn→ ∞rn >0,we assume that there exists a real numberbsuch thatrn > b >0 for
alln∈N.Thus, we have
un1−un2 ≤
un1−un, xn1−xn
1− rn
rn1
un1−xn1
≤ un1−un
xn1−xn1−rrn n1
un1−xn1
,
3.33
and hence
un1−un ≤ xn1−xnrn1
1 |
rn1−rn| un1−xn1
≤ xn1−xn
1
b|rn1−rn|L,
3.34
whereLsup{un−xn:n∈N}. From3.28, we have
xn2−xn1 ≤
1−αn1γ xn1−xn1b|rn1−rn|L
αn1γβxn1−xn
|αn−αn1|Msup
Sn1z−Snz:z∈yn
1−αn1γαn1γβ
xn1−xn
1−αn1γ
b |rn1−rn|L
|αn−αn1|Msup
Sn1z−Snz:z∈
yn
≤1−αn1
γ−γβxn1−xn
L
b|rn1−rn||αn−αn1|M
supSn1z−Snz:z∈
yn.
3.35
Since{yn}is bounded, it follows that ∞n1sup{Sn1z−Snz : z ∈ {yn}} < ∞. Hence, by Lemma 2.3, we havexn1−xn → ∞asn → ∞. From3.34and|rn1−rn| → 0, we have
limn→ ∞un1−un0. Moreover, we have from3.27that limn→ ∞yn1−yn0.
We note from3.23thatxnαn−1γfxn−1 1−αn−1BSn−1yn−1. Then, we have
xn−Snyn ≤ xn−Sn−1yn−1Sn−1yn−1−Sn−1ynSn−1yn−Snyn
≤αn−1γfxn−1−BSn−1yn−1yn−1−yn
supSn−1z−Snz:z∈
yn.
3.36
Sinceαn → 0,yn−1−yn → 0 and sup{Sn−1z−Snz:z∈ {yn}} → 0, we getxn−Snyn →
0. From the proof ofTheorem 3.1, we have
for allv∈Ω. Therefore, we have
xn1−v2 αnγfxn I−αnBSnyn−v 2
I−αnBSnyn−v αnγfxn−Bv 2
≤1−αnγ2 Snyn−v 22αnγfxn−Bv, xn−v
≤1−αnγ2 yn−v 22αnγfxn−fv, xn−v2αnγfv−Bv, xn−v
≤1−αnγ2un−v22αnγfxn−fv, xn−v2αnγfv−Bv, xn−v
≤1−αnγ2
xn−v2− xn−un2
2αnγβxn−v2
2αn γfv−Bv xn−v
1−2αnγ−γβαnγ2
xn−v2−1−αnγ2xn−un2
2αn γfv−Bv xn−v
≤ xn−v2αnγ2xn−v2−1−αnγ2xn−un22αnγfv−Bvxn−v
3.38
and hence
1−αnγ2xn−un2≤αnγ2xn−v22αnγfv−Bvxn−vxn−v2− xn1−v2
≤αnγ2xn−v22αnγfv−Bvxn−v
xn−xn1xn−vxn1−v.
3.39
Since{xn}is bounded,αn → 0 andxn−xn1 → 0, it follows thatxn−un → 0 asn → ∞.
PutM supn≥1{γfv−Bvxn−v}. It follows from3.38, the nonexpansive of JM,λand the inverse strongly monotonicity ofAthat
xn1−v2≤
1−αnγ2 yn−v 22αnγβxn−v22αnM
≤1−αnγ2un−λAun−v−λAv22αnγβxn−v22αnM
≤1−αnγ2
un−v2λλ−2αAun−Av2
2αnγβxn−v22αnM
≤1−αnγ2xn−v21−αnγ2λλ−2αAun−Av2
2αnγβxn−v22αnM
1−2αnγ−γβαnγ2xn−v21−αnγ2λλ−2αAun−Av22αnM
≤ xn−v2αnγ2xn−v21−αnγ2λλ−
2αAun−Av22αnM.
This implies that
1−αnγ2λ2α−λAun−Av2≤αnγ2xn−v22αnMxn−v2− xn1−v2
≤αnγ2xn−v22αnMxn−xn
1xn−vxn1−v.
3.41
Sinceαn → 0 andxn−xn1 → 0, we haveAun−Av → 0 asn → ∞.From3.5,3.15
and3.38, we have
xn1−v2≤
1−αnγ2 yn−v 22αnγβxn−v22αnM
≤1−αnγ2un−v2− un−yn 22λun−yn, Aun−Av−λ2Aun−Av2
2αnγβxn−v22αnM
≤1−αnγ2xn−v2−1−αnγ2 un−yn 221−αnγ2λun−yn, Aun−Av
−1−αnγ2λ2Aun−Av22αnγβxn−v22αnM
1−2αnγ−γβαnγ2xn−v2−1−αnγ2 un−yn 2
21−αnγ2λun−yn, Aun−Av−1−αnγ2λ2Aun−Av22αnM
≤ xn−v2αnγ2xn−v2−1−αnγ2 un−yn 2
21−αnγ2λun−yn, Aun−Av−1−αnγ2λ2Aun−Av22αnM.
3.42
Thus, we obtain
1−αnγ2 un−yn 2≤αnγ2xn−v221−αnγ2λun−yn, Aun−Av
−1−αnγ2λ2Aun−Av22αnMxn−v2− xn1−v2
≤αnγ2xn−v221−αnγ2λun−yn, Aun−Av
−1−αnγ2λ2Aun−Av22αnMxn−xn1xn−vxn1−v
3.43
Sinceαn → 0,Aun−Av → 0 andxn−xn1 → 0, we haveun−yn → 0 asn → ∞. It
follows from the inequalityyn−Snyn ≤ yn−unun−xnxn−Snynthatyn−Snyn → 0
asn → ∞. Moreover, we note thatxn−yn ≤ xn−unun−yn → 0 asn → ∞. Since
Syn−yn ≤ Syn−SnynSnyn−yn
for alln∈N, it follows that limn→ ∞Syn−yn0. Next, we show that
lim sup
n→ ∞
B−γfz, z−xn≤0, 3.45
wherezPΩI−Bγfzis a unique solution of the variational inequality3.4. To show this inequality, we choose a subsequence{xni}of{xn}such that
lim
i→ ∞
B−γfz, z−xni
lim sup
n→ ∞
B−γfz, z−xn. 3.46
Since{uni}is bounded, there exists a subsequence{unij}of{uni}which converges weakly to w. Without loss of generality, we can assume thatuni w.Fromun−yn → 0,we obtain yniw. Let us showw∈EPF. It follows by3.23andA2that
1
rn
y−un, un−xn≥Fy, un 3.47
and hence
y−uni,
uni−xni rni
≥Fy, uni
. 3.48
Sinceuni−xni/rni → 0 anduni w,it follows byA4that 0≥Fy, wfor ally∈H.Fort with 0< t≤1 andy∈H,letytty 1−tw.Sincey∈Handw∈H,we haveyt∈Hand henceFyt, w≤0.So, fromA1andA4we have
0Fyt, yt≤tFyt, y 1−tFyt, w≤tFyt, y, 3.49
and hence 0≤Fyt, y. FromA3, we have 0≤Fw, yfor ally∈Hand hencew∈EPF. By the same argument as in proof ofTheorem 3.1, we havew ∈ FS∩IA, Mand hence
w∈Ω. This implies that
lim sup
n→ ∞
B−γfz, z−xn lim i→ ∞
B−γfz, z−xni
Finally we prove thatxn → z. From3.23, we have
xn1−z2 αnγfxn I−αnBSnyn−z 2
αnγfxn−Bz I−αnBSnyn−z 2
≤ I−αnBSnyn−z 22αnγfxn−Bz, xn1−z
≤1−αnγ2 yn−z 22αnγfxn−fz, xn1−z2αnγfz−Bz, xn1−z
≤1−αnγ2xn−z22αnγβxn−zxn1−z2αnγfz−Bz, xn1−z
≤1−αnγ2xn−z2αnγβxn−z2xn1−z2
2αnγfz−Bz, xn1−z.
3.51
This implies that
xn1−z2≤
1−2αnγαnγ2αnγβ
1−αnγβ xn−z
2 2αn
1−αnγβγfz−Az, xn1−z
1− 2
γ−γβαn
1−αnγβ
xn−z2
αnγ2
1−αnγβxn−z
2
2αn
1−αnγβγfz−Az, xn1−z
≤1−γnxn−z2δn,
3.52
whereγn : 2αn γ−γβ/1−αn γβandδn :αn/1−αn γβ{αnγ2xn−z22γfz−
Az, xn1−z}. It easily verified thatγn → 0,n∞1γn ∞and lim supn→ ∞δn/γn ≤0. Hence,
byLemma 2.1, the sequence{xn}converges strongly toz.
As in10, Theorem 4.1, we can generate a sequence{Sn}of nonexpansive mappings satisfying condition∞n1sup{Sn1z−Snz:z∈K}<∞. for any boundedKofHby using
convex combination of a general sequence{Tk}of nonexpansive mappings with a common
fixed point.
Corollary 3.3. LetH be a real Hilbert space, let F be a bifunction fromH×H → Rsatisfying (A1)–(A4) and letA :H → H be anα-inverse-strongly monotone mapping,M : H → 2H be a maximal monotone mapping. Letf be a contraction ofHinto itself with a constantβ∈0,1and let
Bbe a strongly bounded linear operator onHwith coefficientγ >0and0< γ < γ/β. Let{βkn}be a
family of nonnegative numbers with indicesn, k∈Nwithk≤nsuch that
ik1βnk1,for alln∈N;
iilimn→ ∞βnk>0,for everyk∈N;
Let {Tk} be a sequence of nonexpansive mappings on H with Ω : F∞k1FTk∩ EPF ∩
IA, M/∅and let{xn},{yn}and{un}be sequences generated byx1∈Hand
Fun, yr1
ny−un, un−xn ≥0, ∀y∈H
xn1αnγfxn I−αnB n
k1
βk nTkyn,
ynJM,λun−λAun ∀n≥0,
3.53
for alln∈N, whereλ∈0,2α,{αn} ⊂0,1and{rn} ⊂0,∞satisfy
lim
n→ ∞αn0, ∞
n1
αn∞, ∞ n1
|αn1−αn|<∞,
lim inf
n→ ∞ rn>0, ∞
n1
|rn1−rn|<∞.
3.54
Then,{xn},{yn}and{un}converges strongly tozinΩwhich solves the variational inequality:
B−γfz, z−x ≥0, x∈Ω. 3.55
IfSnS, B≡Iandγ1inTheorem 3.2, we obtain the following corollary.
Corollary 3.4see Peng et al.9. LetHbe a real Hilbert space, letFbe a bifunction fromC×C →
Rsatisfying (A1)–(A4) and letS be a nonexpansive mapping onH. Let A : H → H be an α -inverse-strongly monotone mapping,M : H → 2H be a maximal monotone mapping such that
Ω:FS∩EPF∩IA, M/∅.Letfbe a contraction ofHinto itself with a constantβ∈0,1. Let{xn},{yn}and{un}be sequences generated byx1∈Hand
Fun, yrn1y−un, un−xn ≥0, ∀y∈H
xn1αnfxn I−αnSyn,
ynJM,λun−λAun ∀n≥0,
3.56
for alln∈N, whereλ∈0,2α,{αn} ⊂0,1and{rn} ⊂0,∞satisfy
lim
n→ ∞αn0, ∞
n1
αn∞, ∞
n1
|αn1−αn|<∞,
lim inf
n→ ∞ rn>0, ∞
n1
|rn1−rn|<∞.
3.57
IfSnS, A≡0 andM≡0 inTheorem 3.2, we obtain the following corollary.
Corollary 3.5see S. Plubtieng and R. Punpaeng6. LetHbe a real Hilbert space, letF be a bifunction fromH×H → Rsatisfying (A1)–(A4) and letSbe a nonexpansive mapping onHsuch thatΩ:FS∩EPF/∅.Letfbe a contraction ofHinto itself with a constantβ∈0,1and let
Bbe a strongly bounded linear operator onHwith coefficientγ >0and0< γ < γ/β. Let{xn},{un}
and be sequences generated byx1∈Hand
Fun, yr1
ny−un, un−xn ≥0, ∀y∈H
xn1αnγfxn I−αnBSun,
3.58
for alln∈N, where{αn} ⊂0,1and{rn} ⊂0,∞satisfy
lim
n→ ∞αn0, ∞
n1
αn∞, ∞
n1
|αn1−αn|<∞,
lim inf
n→ ∞ rn>0,
∞
n1
|rn1−rn|<∞.
3.59
Then,{xn}and{un}converges strongly to a pointzinΩwhich solves the variational inequality:
B−γfz, z−x ≥0, x∈Ω. 3.60
Acknowledgments
The first author thank the National Research Council of Thailand to Naresuan University, 2009 for the financial support. Moreover, the second author would like to thank the “National Centre of Excellence in Mathematics”, PERDO, under the Commission on Higher Education, Ministry of Education, Thailand.
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