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,somyotp@nu.ac.th 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
F x, 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 EPFis nonempty and proved a strong convergence theorem. LetA:C → Hbe a nonlinear
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
F un, y
1 rn
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−γf
z, z−x
≥0, ∀x∈FS∩EPF, 1.6
which is the optimality condition for the minimization problem
x∈FS∩EPFmin 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
whereθis the zero vector inH. The set of solutions of problem1.8is denoted byIA, M.
IfA 0, then problem1.8becomes the inclusion problem introduced by Rockafellar7.
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
F un, y
1 rn
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
F un, y
1 rn
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 conditions, some strong convergence theorems for approximating to this common elements are proved. Our results extend and improve some corresponding results in3,9and the references therein.
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,
limt→0F
tz 1−tx, y
≤F x, 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 >0 andx∈H. Then, there existsz∈Csuch that
F z, y
1
ry−z, z−x ≥0, ∀y∈C. 2.5 Define a mappingTr :H → Cas follows:
Trx
z∈C:F z, y
1 r
y−z, z−x
≥0, ∀y∈C
, 2.6
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 coefficient 0 < 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−γn
anδn, n≥0, 2.8
where{γn}is a sequence in0,1and{δn}is a sequence inRsuch that 1∞
n1γn∞;
2lim supn→ ∞δn/γn≤0 or∞
n1|δn|<∞.
Then limn→ ∞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:
JM,λu IλM−1u, ∀u∈H, 2.10
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
Then limn→ ∞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∩ IA, M/∅.Letfbe a contraction ofHinto itself with a constantβ∈0,1and letBbe a strongly
bounded linear operator onHwith coefficientγ >0 and 0 < γ < γ/β. Let{xn},{yn}and{un}be sequences generated byx1∈Hand
F un, y
1
rny−un, un−xn ≥0, ∀y∈H xnαnγfxn I−αnBSyn, ynJM,λun−λAun ∀n≥0,
3.3
where λ ∈ 0,2α,{rn} ⊂ 0,∞and {αn} ⊂ 0,1satisfy limn→ ∞αn 0 and lim infn→ ∞rn >
0. Then,{xn}, {yn} and {un} converges strongly to a point z in Ω which solves the variational inequality:
B−γf
z, 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−Bv
1−αnγ
xn−v
≤αnγ
fxn−fv
γfv−Bv
1−αnγ
xn−v
≤αnγβxn−vαnγfv−Bv
1−αnγ
xn−v
1−αn
γ−γβ
xn−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
xn−Synαnγfxn−BSyn −→0 asn−→ ∞. 3.8
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−αnγ2
un−v22αnγfxn−fv, xn−v2αnγfv−Bv, xn−v
≤
1−αnγ2
xn−v2− xn−un2
2αnγβxn−v2 2αnγfv−Bvxn−v
1−2αn γ−γβ
αnγ2
xn−v2−
1−αnγ2
xn−un2 2αnγfv−Bvxn−v
≤ xn−v2αnγ2xn−v2−
1−αnγ2
xn−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 22αnγβxn−v22αnM
≤
1−αnγ2
un−λAun−v−λAv22αnγβxn−v22αnM
≤
1−αnγ2
un−v2λλ−2αAun−Av2
2αnγβxn−v22αnM
≤
1−αnγ2xn−v2
1−αnγ2λλ−2αAun−Av22αnγβxn−v22αnM
1−2αn γ−γβ
αnγ2
xn−v2
1−αnγ2
λλ−2αAun−Av22αnM
≤ xn−v2αnγ2xn−v2
1−αnγ2
λλ−2αAun−Av22αnM,
3.12
which implies that
1−αnγ2
λ2α−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 22λ
un−yn, Aun−Av
−λ2Aun−Av2 . 3.14
Thus, we have
yn−v 2 ≤ un−v2− un−yn 22λ
un−yn, Aun−Av
−λ2Aun−Av2. 3.15
From3.5,3.10, and3.15, we have xn−v2≤
1−αnγ2 yn−v 22αnγβxn−v22αnM
≤
1−αnγ2
un−v2− un−yn 22λ
un−yn, Aun−Av
−λ2Aun−Av2 2αnγβxn−v22αnM
≤
1−αnγ2
xn−v2−
1−αnγ2 un−yn 22
1−αnγ2 λ
un−yn, Aun−Av
−
1−αnγ2
λ2Aun−Av22αnγβxn−v22αnM
1−2αn γ−γβ
αnγ2
xn−v2−
1−αnγ2 un−yn 2 2
1−αnγ2 λ
un−yn, Aun−Av
−
1−αnγ2
λ2Aun−Av22αnM
≤ xn−v2αnγ2xn−v2−
1−αnγ2 un−yn 2 2
1−αnγ2 λ
un−yn, Aun−Av
−
1−αnγ2
λ2Aun−Av22αnM.
3.16
Thus, we get
1−αnγ2 un−yn 2≤αnγ2xn−v22
1−αnγ2 λ
un−yn, Aun−Av
−
1−αnγ2
λ2Aun−Av22αnM. 3.17
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 inf
n→ ∞ yn−Sz ≤lim inf
n→ ∞
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
∈M yn
. 3.19
By the maximal monotonicity ofMA, we have
p−yn, g−Ap−1 λ
un−yn−λAun
≥0, 3.20
and so
p−yn, g
≥
p−yn, Ap 1 λ
un−yn−λAun
p−yn, Ap−AynAyn−Aun1 λ
un−yn
≥0
p−yn, Ayn−Aun
p−yn,1 λ
un−yn .
3.21
It follows fromun−yn → 0,Aun−Ayn → 0 andyn zthat
nlim→ ∞
p−yn, g
p−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 3.4.
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 >0 and 0< γ < γ/β.
Let{xn},{yn}and{un}be sequences generated byx1∈Hand F
un, y 1
rny−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
nlim→ ∞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−Bv
1−αnγ
xn−v
≤αnγ
fxn−fv
γfv−Bv
1−αnγ
xn−v
≤αnγβxn−vαnγfv−Bv
1−αnγ
xn−v
1−αn
γ−γβ
xn−vαnγfv−Bv
1−αn
γ−γα
xn−vαn
γ−γαγfv−Bv γ−γα .
3.25
It follows from3.25and induction that xn−v ≤max
x1−v, 1
γ−γαγf p
−B p
, n≥0. 3.26
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−αnB
Sn1yn−Snyn
αn1−αnγfxnαn1γf
xn1−fxn
≤
1−αn1γ
yn1−yn|αn−αn1| BSn1yn 1−αnγ
Sn1yn−Snyn |αn−αn1| γfxnαn1γβxn1−xn
≤
1−αn1γ yn1−yn αn1γβxn1−xn |αn−αn1|
BSn1ynγfxn
Sn1yn−Snyn
≤
1−αn1γ
un1−unαn1γβxn1−xn|αn−αn1|M sup
Sn1z−Snz:z∈ yn
,
3.28
whereM:sup{max{BSn1yn,γfxn}:n≥0}<∞. On the other hand, we note that
F un, y
1 rn
y−un, un−xn
≥0, ∀y∈H, 3.29 F
un1, y 1
rn1
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− rn
rn1 un1−xn1 ≥0. 3.32
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−xn 1− rn
rn1
un1−xn1
,
3.33
and hence
un1−un ≤ xn1−xn 1
rn1 |rn1−rn| un1−xn1
≤ xn1−xn1
b|rn1−rn|L,
3.34
whereLsup{un−xn:n∈N}. From3.28, we have
xn2−xn1 ≤
1−αn1γ xn1−xn1
b|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−γβ
xn1−xnL
b|rn1−rn||αn−αn1|M sup
Sn1z−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 sup
Sn−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
un−v2≤ xn−v2− xn−un2, 3.37
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−αnγ2
un−v22αnγfxn−fv, xn−v2αnγfv−Bv, xn−v
≤
1−αnγ2
xn−v2− xn−un2
2αnγβxn−v2 2αn γfv−Bv xn−v
1−2αn γ−γβ
αnγ2
xn−v2−
1−αnγ2
xn−un2 2αn γfv−Bv xn−v
≤ xn−v2αnγ2xn−v2−
1−αnγ2
xn−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−αnγ2
un−λAun−v−λAv22αnγβxn−v22αnM
≤
1−αnγ2
un−v2λλ−2αAun−Av2
2αnγβxn−v22αnM
≤
1−αnγ2xn−v2
1−αnγ2λλ−2αAun−Av2 2αnγβxn−v22αnM
1−2αn
γ−γβ
αnγ2
xn−v2
1−αnγ2
λλ−2αAun−Av22αnM
≤ xn−v2αnγ2xn−v2
1−αnγ2
λλ−2αAun−Av22αnM.
3.40