Volume 2008, Article ID 720371,15pages doi:10.1155/2008/720371
Research Article
Common Solutions of an Iterative Scheme for
Variational Inclusions, Equilibrium Problems, and
Fixed Point Problems
Jian-Wen Peng,1Yan Wang,1David S. Shyu,2and Jen-Chih Yao3
1College of Mathematics and Computer Science, Chongqing Normal University, Chongqing 400047, China 2Department of Finance, National Sun Yat-Sen University, Kaohsiung 80424, Taiwan
3Department of Applied Mathematics, National Sun Yat-Sen University, Kaohsiung 80424, Taiwan
Correspondence should be addressed to David S. Shyu,[email protected]
Received 24 October 2008; Accepted 6 December 2008
Recommended by Ram U. Verma
We introduce an iterative scheme by the viscosity approximate method for finding a common element of the set of solutions of a variational inclusion with set-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. We obtain a strong convergence theorem for the sequences generated by these processes in Hilbert spaces. The results in this paper unify, extend, and improve some well-known results in the literature.
Copyrightq2008 Jian-Wen Peng et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Throughout this paper, we always assume thatHis a real Hilbert space with norm and inner product denoted by · and ·,·, respectively. 2H denotes the family of all the nonempty subsets ofH.
LetA :H → Hbe a single-valued nonlinear mapping andM :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, 1.1
whereθis the zero vector inH. The set of solutions of problem1.1is denoted byIA, M. If
HRm, then problem1.1becomes the generalized equation introduced by Robinson1. If
in many optimization-related areas including mathematical programming, complementarity, variational inequalities, optimal control, mathematical economics, equilibria, game theory, and so forth. Also various types of variational inclusions problems have been extended and generalized, for more details, please see3–20and the references therein.
There are many algorithms for solving variational inclusionssee3,5,7, 8, 10–13,
16–20 and the references therein. We note that recently Zhang et al.20 introduced the following new iterative scheme for finding a common element of the set of solutions to the problem1.1and the set of fixed points of nonexpansive mappings in Hilbert spaces. Starting with an arbitrary pointx1x∈H, define sequences{xn}by
xn1αnx
1−αnSyn,
ynJM,λxn−λAxn, ∀n≥0,
1.2
whereJM,λ IλM−1is the resolvent operator associated withMand a positive number
λ,{αn}is a sequence in the interval0,1.
Let F be a bifunction from C × C to R, where R is the set of real numbers. The equilibrium problem forF:C×C → Ris to findx∈Csuch that
Fx, y≥0, ∀y∈C. 1.3
The set of solutions of1.3is denoted byEPF. The problem1.3is 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; for more detailssee21.
Recall that a mappingSof a closed convex subsetCinto itself is nonexpansive if there holds that
Sx−Sy ≤ x−y, ∀x, y∈C. 1.4
A mappingf:C → Cis called contractive if there exists a constantα∈0,1such that
fx−fy ≤αx−y, ∀x, y∈C. 1.5
We denote the set of fixed points of S by FixS. It is known that FixS is closed convex, but possibly empty. There are some methods for approximation of fixed points of a nonexpansive mapping. In 2000, Moudafi 22 introduced the viscosity approximation method for nonexpansive mappings.
Some methods have been proposed to solve the equilibrium problem; see, for instance,
23–30and the references therein. Recently, S. Takahashi and W. Takahashi23introduced the following Mann iterative scheme by the viscosity approximation method for finding a common element of the set of solutions of problem 1.3 and the set of fixed points of a nonexpansive mapping in a Hilbert space. Starting with an arbitrary pointx1 ∈ H, define sequences{xn}and{un}by
Fun, y r1 n
y−un, un−xn≥0, ∀y∈C,
xn1αnfxn1−αnSun, ∀n∈N.
They proved that under certain appropriate conditions imposed on {αn} and {rn}, the
sequences{xn}and{un}generated by1.6converge strongly toz ∈ FS∩EPF, where
zPFixS∩EPFfz.
The purpose of this paper is twofold. On one hand, it is easy to see that the authors in28–32introduced some algorithms for finding a common element of the set of solutions of an equilibrium problem, the set of fixed points of a nonexpansive mapping and the set of solutions of a variational inequality. It is natural to raise and to give an answer to the following question: can one construct algorithms for finding a common element of the set of solutions of an equilibrium problem, the set of fixed points of a nonexpansive mapping, and the set of solutions of a variational inclusion? In other words, can we construct algorithms for finding a solution of a mathematical model which consists of an equilibrium problem and a variational inclusion if this mathematical model has a solution? In this paper, we will give a positive answer to this question. By combining the algorithm1.2for a variational inclusion problem and the algorithm 1.6 for an equilibrium problem, we introduce the following iterative scheme by the viscosity approximation method for finding a common element of the set of solutions of the variational inclusion with a set-valued maximal monotone mapping and an inverse strongly monotone mapping, the set of solutions of equilibrium problem and the set of fixed points of a nonexpansive mapping in a Hilbert space. Starting with an arbitrary pointx1∈H, define sequences{xn}, {yn}, and{un}by
Fun, y r1 n
y−un, un−xn≥0, ∀y∈C,
xn1 αnfxn1−αnSyn,
ynJM,λun−λAun, ∀n≥0,
1.7
where{αn} ⊂0,1and{rn} ⊂0,∞.
On the other hand, we will show that the sequences{xn},{yn}, and{un}generated
by1.7converge strongly toz ∈ FS∩EPF∩IA, Mif the parameter sequences{αn}
and{rn}satisfy appropriate conditions. The results in this paper unify, extend, and improve
some corresponding results in20,23and the references therein.
2. Preliminaries
In a real Hilbert spaceH, it is well known that
λx 1−λy2λx2 1−λy2−λ1−λx−y2 2.1
for allx, y∈Handλ∈0,1.
For anyx∈H, there exists a unique nearest point inC, denoted byPCx, such that
x−PCx ≤ x−yfor ally∈C. The mappingPCis called the metric projection ofHonto
C. We know thatPCis a nonexpansive mapping fromHontoC. It is also known thatPCx∈C
and
x−PCx, PCx−y≥0 2.2
for allx∈Handy∈C.
It is easy to see that2.2is equivalent to
x−y2≥x−P
Cx2y−PCx2 2.3
Recall that a mapping A : H → H is calledα-inverse strongly monotone, if there exists anα >0 such that
Ax−Ay, x−y ≥αAx−Ay2, ∀x, y∈H. 2.4
LetI be the identity mapping onH. It is well known that ifA : H → H is anα -inverse strongly monotone, thenAis1/α-Lipschitz continuous and monotone mapping. In addition, if 0< λ≤2α, thenI−λAis a nonexpansive mapping.
Let symbols → anddenote strong and weak convergence, respectively. It is known thatHsatisfies the Opial condition33, that is, for any sequence{xn} ⊂Hwithxn x, the
inequality
lim inf
n→ ∞ xn−x<lim infn→ ∞ xn−y 2.5
holds for everyy ∈H withx /y. A set-valued mappingM :H → 2H is called monotone if for allx, y ∈ H,f ∈ Mx, andg ∈ Myimplyx−y, f−g ≥ 0. A monotone mapping
M : H → 2H is maximal if its graph GM : {f, x ∈ H×H | f ∈ Mx}of M is 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, 2.6
whereλis a positive number. It is worth mentioning that the resolvent operatorJM,λis
single-valued, nonexpansive, and 1-inverse strongly monotone,see, e.g.,34, and that a solution of problem1.1is a fixed point of the operatorJM,λI−λAfor allλ >0see, e.g.,35.
For solving the equilibrium problem, let us assume that the bifunctionF satisfies the following conditions:
A1Fx, x 0 for allx∈C;
A2Fis monotone, that is,Fx, y Fy, x≤0 for anyx, y∈C;
A3for eachx, y, z∈C,
lim
t↓0 F
tz 1−tx, y≤Fx, y, 2.7
A4for eachx∈C, y→Fx, yis convex and lower semicontinuous.
We recall some lemmas which will be needed in the rest of this paper.
Lemma 2.1see21. LetCbe a nonempty closed convex subset ofH, letF be a bifunction from
C×CtoRsatisfying (A1)–(A4). Letr >0andx∈H. Then, there existsz∈Csuch that
Fz, y 1
Lemma 2.2see23,24. LetCbe a nonempty closed convex subset ofH, letFbe a bifunction from
C×CtoRsatisfying (A1)–(A4). Forr >0andx∈H,define a mappingTr :H → Cas follows:
Trx
z∈C:Fz, y 1
ry−z, z−x ≥0,∀y∈C
2.9
for allx∈H. Then, the following statements hold
1Tr is single-valued;
2Tr is firmly nonexpansive, that is, for anyx, y∈H Trx−Try2≤T
rx−Try, x−y, 2.10
3FTr EPF;
4EPFis closed and convex.
Lemma 2.3see34. LetM:H → 2H be a maximal monotone mapping andA :H → Hbe a Lipschitz continuous mapping. Then the mappingSMA:H → 2H is a maximal monotone mapping.
Remark 2.4. Lemma 2.3implies that IA, M is closed and convex if M : H → 2H is a maximal monotone mapping andA:H → Hbe an inverse strongly monotone mapping.
Lemma 2.5see36,37. Assume that{αn}is a sequence of nonnegative real numbers such that
αn1≤
1−γnαnδn, 2.11
whereγnis a sequence in0,1and{δn}is a sequence such that
i ∞
n1 γn∞,
iilim sup
n→ ∞
δn
γn ≤0 or
∞
n1
δn<∞.
2.12
Then,limn→ ∞αn0.
3. Strong convergence theorem
In this section, we establish a strong convergence theorem which solves the problem of finding a common element of the set of solutions of variational inclusion, the set of solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping in Hilbert space.
Theorem 3.1. LetCbe a nonempty closed convex subset ofH. LetFbe a bifunction fromC×Cto
Rsatisfying (A1)–(A4) and letSbe a nonexpansive mapping ofCintoH. LetA :H → Hbe an
ξ ∈ 0,1. Let x1 be an arbitrary point in H, {xn}, {un}, and{yn} be sequences generated by algorithm1.7. If{αn} ⊂0,1and{rn} ⊂0,∞satisfy the following conditions:
lim
n→ ∞αn0, ∞
n1
αn∞,
∞
n1
αn1−αn<∞,
lim inf
n→ ∞ rn>0, ∞
n1
rn1−rn<∞. 3.1
Then,{xn},{yn}, and{un}converge strongly toz∈Ω,wherezPΩfz.
Proof. We show thatPΩfis a contraction mapping. In fact, we have
PΩfx−PΩfy≤fx−fy≤ξx−y 3.2
for anyx, y ∈ H. Since H is complete, there exists a unique elementz ∈ Hsuch thatz
PΩfz. Letv∈Ω. Then, fromunTrnxn,we have
un−vTrnxn−Trnv≤xn−v, 3.3
for alln∈N.It is easy to see thatvJM,λv−λAv. AsI−λAis nonexpansive, we have
yn−vJM,λun−λAun−JM,λv−λAv
≤un−λAun−v−λAv
≤un−v
≤xn−v,
3.4
for alln∈N.It follows from1.7and3.4that
xn1−vαnfxn1−αnSyn−v
αnfxn−v1−αnSyn−v
≤αnfxn−v1−αnSyn−v
≤αnfxn−fvfv−v1−αnyn−v
≤αnξxn−vαnfv−v1−αnxn−v
1−αn1−ξxn−vαn1−ξ 1
1−ξfv−v
≤maxxn−v; 1
1−ξfv−v
≤ · · ·
≤maxx1−v; 1
1−ξfv−v
for alln≥1. This implies that{xn}is bounded. From which it follows that{un},{yn},{fxn},
{Syn}, and{Aun}are also bounded. Next, we show that limn→ ∞xn1−xn0.In fact, since
I−λAis nonexpansive, we have
yn1−ynJM,λ
un1−λAun1
−JM,λun−λAun
≤un1−λAun1−
un−λAun
≤un1−un.
3.6
It follows from1.7and3.6that
xn1−xnαnf
xn1−αnSyn−αn−1fxn−1
−1−αn−1
Syn−1
αnfxn−αnfxn−1
αnfxn−1
−αn−1f
xn−1
1−αnSyn−1−αnSyn−1
1−αnSyn−1−
1−αn−1
Syn−1
≤αnξxn−xn−1Kαn−αn−1
1−αnSyn−Syn−1
≤αnξxn−xn−1Kαn−αn−1
1−αnyn−yn−1
≤αnξxn−xn−1Kαn−αn−1
1−αnun−un−1,
3.7
whereKsup{fxnSyn, n≥1}.
On the other hand, fromunTrnxnandun1 Trn1xn1, we have
Fun, yr1 n
y−un, un−xn≥0 ∀y∈C, 3.8
Fun1, y
1
rn1
y−un1, un1−xn1
≥0 ∀y∈C. 3.9
Puttingyun1in3.8andyunin3.9, we have
Fun, un1
r1
n
un1−un, un−xn≥0,
Fun1, unr1 n1
un−un1, un1−xn1
≥0.
3.10
It follows from the monotonicity ofFthat
un1−un, unr−xn n −
un1−xn1 rn1
≥0. 3.11
Thus,
un1−un, un−un1un1−xn−rrn n1
un1−xn1
≥0. 3.12
Without loss of generality, let us assume that there exists a real numberbsuch thatrn> b >0
for alln∈N.Then, we have
un1−un2≤
un1−un, xn1−xn
1− rn
rn1
un1−xn1
≤un1−unxn1−xn1− rrn
n1
un1−xn1
.
It follows that
un1−un≤xn1−xnr1 n1
rn1−rnun1−xn1
≤xn1−xnLbrn1−rn,
3.14
whereLsup{un−xn:n∈N}.From3.7and3.14, we have xn1−xn≤αnξxn−xn−1Kαn−αn−1
1−αnxn−xn−1
1−αnLbrn−rn−1
≤1−αn1−ξxn−xn−1Kαn−αn−1
L
brn−rn−1.
3.15
It follows fromLemma 2.5that
lim
n→ ∞xn1−xn0. 3.16 From3.14and|rn1−rn| → 0,we have
lim
n→ ∞un1−un0. 3.17 It follows from3.6that
lim
n→ ∞yn1−yn0. 3.18 Sincexnαn−1fxn−1 1−αn−1Syn−1,we have
xn−Syn≤xn−Syn−1Syn−1−Syn
≤αn−1f
xn−1
−Syn−1yn−1−yn.
3.19
It follows fromαn → 0 thatxn−Syn → 0.
Now we prove that for any givenv∈Ω,
Aun−Av−→0. 3.20
In fact, forv∈Ω,we have
un−v2T
rnxn−Trnv
2
≤Trnxn−Trnv, xn−v
un−v, xn−v
1
2un−v
2x
n−v2−xn−un2,
3.21
and hence,
un−v2≤x
It follows from1.7,3.4, and3.22that
xn1−v2α
nfxn1−αnSyn−v2
≤αnfxn−v21−αnSyn−v2
≤αnfxn−v21−αnyn−v2
≤αnfxn−v21−αnxn−v2−xn−un2
≤αnfxn−v2xn−v2−1−αnxn−un2.
3.23
Thus, we have
1−αnxn−un2 ≤αnfxn−v2xn−v2−xn1−v2
≤αnfxn−v2xn−xn1xn−vxn1−v.
3.24
Sinceαn → 0,xn−xn1 → 0, and{xn}is bounded, we havexn−un → 0. Fromun−
xn1 ≤ un−xnxn−xn1, we getun−xn1 → 0.
Equation3.23, the nonexpansiveness ofJM,λ, and the inverse strongly monotonicity
ofAimply that
xn1−v2≤αnfxn−v21−αnyn−v2
≤αnfxn−v21−αnun−λAun−v−λAv2
≤αnfxn−v21−αnun−v2λλ−2αAun−Av2
≤αnfxn−v2un−v21−αnλλ−2αAun−Av2.
3.25
Thus, we have
1−αnλλ−2αAun−Av2 ≤αnfxn−v2un−v2−xn1−v2
≤αnfxn−v2un−xn1un−vxn1−v.
3.26
Sinceαn → 0, un−xn1 → 0, {un}and{xn}are bounded, we have
Aun−Av−→0. 3.27
Next, we show thatSyn−yn → 0. Indeed, for anyv∈Ω, yn−v2JM,λun−λAun−JM,λv−λAv2
≤un−λAun−v−λAv, yn−v
1
2un−λAun−
v−λAv2y
n−v2−un−λAun−v−λAv−yn−v2
≤ 1
2un−v
2y
n−v2−un−yn−λAun−Av2
1
2un−v
2y
Thus, we have
yn−v2≤u
n−v2−un−yn22λun−yn, Aun−Av−λ2Aun−Av2. 3.29
As the function·2is convex, by3.29and3.23, we have
xn1−v2≤αnf
xn−v21−αnyn−v2
≤αnfxn−v21−αn
×un−v2−un−yn22λun−yn, Aun−Av−λ2Aun−Av2.
3.30
Thus, we get
1−αnun−yn2≤αnfxn−v2un−v2−xn1−v2
21−αnλun−yn, Aun−Av−1−αnλ2Aun−Av2
≤αnfxn−v2un−xn1un−v−xn1−v
21−αnλun−yn, Aun−Av−1−αnλ2Aun−Av2.
3.31
Sinceαn → 0, Aun−Av → 0, andun−xn1 → 0, we haveun−yn → 0. From the
triangle inequalityxn−ynxn−unun−yn, we deduce thatxn−yn → 0.It then
follows from the inequalitySyn−yn ≤ Syn−xnxn−unun−ynthatSyn−yn → 0.
Next, we show that
lim sup
n→ ∞
fz−z, xn−z≤0, 3.32
wherezPΩfz. To show this inequality, we choose a subsequence{xni}of{xn}such that
lim
ni→ ∞
fz−z, xni−z
lim sup
n→ ∞
fz−z, xn−z. 3.33
Since{uni} is bounded, there exists a subsequence {unij} of {uni} which converges weakly tow. Without loss of generality, we can assume that{uni} w.Fromun−yn → 0, we also obtain thatyni w. Since{uni} ⊂CandCis closed and convex, we obtainw∈C.
Let us showw∈EPF.ByunTrnxn,we have
Fun, yr1 n
y−un, un−xn≥0, ∀y∈C. 3.34
FromA2, we also have
1
rn
y−un, un−xn≥Fy, un, 3.35
and hence,
y−uni,
uni−xni
rni
≥Fy, uni
Sinceuni−xni/rni → 0 anduni w, fromA4, we have
Fy, w≤0, ∀y∈C. 3.37
Fortwith 0< t≤1 andy∈C, letytty 1−tw.Since,y∈Candw∈C, we obtainyt∈C
and hence,Fyt, w≤0. So we have
0Fyt, yt≤tFyt, y1−tFyt, w≤tFyt, y. 3.38
Dividing byt, we get
Fyt, y≥0. 3.39
Lettingt → 0 and fromA3, we get
Fw, y≥0 3.40
for ally∈Cand hencew∈EPF.
We next show thatw ∈FixS.Assumew /∈FixS.Sinceyni wandw /Sw, from the Opial theorem33we have
lim inf
i→ ∞ yni−w<lim infi→ ∞ yni−Sw
≤lim inf
i→ ∞ yni−SyniSyni−Sw
≤lim inf
i→ ∞ yni−w.
3.41
This is a contradiction. So, we getw∈FixS.
We now show thatw ∈ IA, M. In fact, sinceAis anα-inverse strongly monotone,
A is an 1/α-Lipschitz continuous monotone mapping and DA H. It follows from
Lemma 2.3thatMAis maximal monotone. Letp, g∈GMA,that is,g−Ap∈Mp. Again sinceyni JM,λuni−λAuni,we haveuni−λAuni ∈IλMyni, that is,
1
λ
uni−yni−λAuni
∈Myni
. 3.42
By virtue of the maximal monotonicity ofMA, we have
p−yni, g−Ap− 1
λ
uni−yni−λAuni
≥0, 3.43
and so
p−yni, g
≥
p−yni, Ap 1
λ
uni−yni−λAuni
p−yni, Ap−AyniAyni−Auni 1
λ
uni −yni
≥0p−yni, Ayni−Auni
p−yni, 1
λ
uni−yni
.
It follows fromun−yn → 0, Aun−Ayn → 0 andyni wthat
lim
ni→ ∞
p−yni, g
p−w, g≥0. 3.45
It follows from the maximal monotonicity ofAMthatθ∈MAw,that is,w∈IA, M. This implies thatw∈Ω.
SincezPΩfz, we have
lim sup
n→ ∞
fz−z, xn−zilim
→ ∞
fz−z, xni−z
lim
i→ ∞
fz−z, uni−z
fz−z, w−z≤0.
3.46
Fromxn1−zαnfxn−z 1−αnSyn−zand3.4, we have xn1−z2
αnfxn−z1−αnSyn−z, xn1−z
≤αnfxn−fz, xn1−z
αnfz−z, xn1−z
1−αnSyn−zxn1−z
≤αnξxn−zxn1−zαnfz−z, xn1−z
1−αnyn−zxn1−z
≤ 1
2αnξxn−z
2x
n1−z2
αnfz−z, xn1−z
1
2
1−αnyn−z2xn1−z2
≤ 1
2αnξxn−z
2x
n1−z2
αnfz−z, xn1−z
1
2
1−αnxn−z2xn1−z2
≤ 1
2
1−αn1−ξxn−z21
2xn1−z
2α
nfz−z, xn1−z
.
3.47
It follows that
xn1−z2 ≤
1−αn1−ξxn−z22αnfz−z, xn1−z
. 3.48
Lemma 2.5implies thatxn → z PΩfz.Sincexn−un → 0 andyn−un → 0,
we also obtain thatun → zandyn → z. The proof is now complete.
By Theorem 3.1, we can obtain some new and interesting results as follows: let
V IC, Adenote the solution set of the following variational inequality: findingu∈ Csuch that
Au, v−u≥0, ∀v∈C, 3.49
Theorem 3.2. LetCbe a nonempty closed convex subset ofH. LetFbe a bifunction fromC×Cto
Rsatisfying (A1)–(A4) and letSbe a nonexpansive mapping ofCintoH. LetA:C → Hbe anα -inverse strongly monotone mapping such thatΓ FixS∩EFF∩V IC, A/∅, whereV IC, A is the set of solutions of problem3.49. Letf :H → Hbe a contraction mapping with a constant
ξ∈0,1andx1be an arbitrary point inH. Let{xn},{un}, and{yn}be sequences generated by
Fun, yr1 n
y−un, un−xn≥0, ∀y∈C,
xn1 αnfxn1−αnSyn,
ynPCun−λAun
3.50
for alln∈N. Ifλ∈0,2α,{αn} ⊂0,1, and{rn} ⊂0,∞satisfy the following conditions:
lim
n→ ∞αn0, ∞
n1
αnref∞,
∞
n1
αn1−αn<∞,
lim inf
n→ ∞ rn>0, ∞
n1
rn1−rn<∞.
3.51
Then,{xn},{yn}, and{un}converge strongly toz∈ΓwherezPΓfz.
Proof. InTheorem 3.1takeM ∂δC : H → 2H,where δC : H → 0,∞is the indicator
function ofC, that is,
δCx
⎧ ⎨ ⎩
0, x∈C,
∞, v /∈C. 3.52
Then, the problem1.1is equivalent to problem3.49. Again, sinceM∂δC, thenJM,λPC,
so we have
ynPCun−λAun. 3.53
The conclusion ofTheorem 3.2can be obtained fromTheorem 3.1immediately.
Theorem 3.3. LetCbe a nonempty closed convex subset ofH. LetFbe a bifunction fromC×Cto
Rsatisfying (A1)–(A4) and letSbe a nonexpansive mapping ofCintoH, LetA :H → Hbe an
α-inverse strongly monotone mapping,M : H → 2H be a maximal monotone mapping such that Ω FixS∩EPF∩IA, M/∅.Letx1xbe an arbitrary point inHand let{xn}, {un}, and
{yn}be sequences generated by
Fun, yr1 n
y−un, un−xn≥0, ∀y∈C,
xn1αnx
1−αnSyn,
ynJM,λun−λAun
3.54
for alln≥1, where{αn} ⊂0,1and{rn} ⊂0,∞satisfy the following conditions:
lim
n→ ∞αn0, ∞
n1
αn∞,
∞
n1
αn1−αn<∞,
lim inf
n→ ∞ rn>0, ∞
n1
rn1−rn<∞.
3.55
Proof. Puttingfy x,for ally ∈ H inTheorem 3.1. Then,fxn xfor all n ≥ 1. By
Theorem 3.1, we obtained the desired result.
Remark 3.4. Putting A 0 inTheorem 3.2, thenyn un. ByTheorem 3.2, we recover 23,
Theorem 3.1. Putting C H and Fx, y 0 for allx, y ∈ H inTheorem 3.3, thenun
PHxn xn. ByTheorem 3.3, we recover20, Theorem 2.1. Hence, Theorem 3.1unifies and
extends the main results in20,23and the references therein.
Acknowledgments
The authors would like to thank the referees for helpful suggestions. The first author was supported by the National Natural Science Foundation of ChinaGrant no. 10771228 and Grant no. 10831009, the Science and Technology Research Project of Chinese Ministry of Education Grant no. 206123, the Education Committee project Research Foundation of Chongqing Grant no. KJ070816. The Third author was partially supported by the grant NSC96-2628-E-110-014-MY3.
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