Volume 2010, Article ID 165098,18pages doi:10.1155/2010/165098
Research Article
A New Iterative Method for Solving
Equilibrium Problems and Fixed Point Problems for
Infinite Family of Nonexpansive Mappings
Shenghua Wang,
1Yeol Je Cho,
2and Xiaolong Qin
31School of Mathematics and Physics, North China Electric Power University, Baoding 071003, China 2Department of Mathematics Education and the RINS, Gyeongsang National University,
Chinju 660-701, Republic of Korea
3Department of Mathematics, Hangzhou Normal University, Hangzhou 310036, China
Correspondence should be addressed to Yeol Je Cho,[email protected]
Received 7 January 2010; Revised 21 May 2010; Accepted 11 July 2010
Academic Editor: Simeon Reich
Copyrightq2010 Shenghua Wang 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.
We introduce a new iterative scheme for finding a common element of the solutions sets of a finite family of equilibrium problems and fixed points sets of an infinite family of nonexpansive mappings in a Hilbert space. As an application, we solve a multiobjective optimization problem using the result of this paper.
1. Introduction
LetHbe a Hilbert space andCbe a nonempty, closed, and convex subset ofH. LetΦbe a bifunction ofC×CintoR, whereRis the set of real numbers. The equilibrium problem for the bifunctionΦ:C×C → Ris to findx∈Csuch that
Φx, y≥0, ∀y∈C. 1.1
The set of solutions of the above inequality is denoted by EPΦ. Many problems arising from physics, optimization, and economics can reduce to finding a solution of an equilibrium problem.
Hand proved a strong convergence theorem which is based on Combettes and Hirstoaga’s result2and Wittmann’s result3. More precisely, they obtained the following theorem.
Theorem 1.1see1. LetCbe a nonempty closed and convex subset ofH. LetΦ:C×C → Rbe a bifunction which satisfies the following conditions:
A1 Φx, x 0for allx∈C;
A2 Φis monotone, that is,Φx, y Φy, x≤0for allx, y∈C;
A3For allx, y, z∈C,
lim
t↓0Φ
tz 1−tx, y≤Φx, y; 1.2
A4For eachx∈C,y→Φx, yis convex and lower semicontinuous.
LetS:C → Hbe a nonexpansive mapping withFixS∩EPΦ/∅, whereFixSdenotes
the set of fixed points of the mappingS, and letf :H → Hbe a contraction, if there exists a constant
λ ∈ 0,1such thatfx−fy ≤ λx−yfor allx, y ∈ H. Let{xn}and{un}be the sequences
generated byx1∈Hand
Φun, yr1 n
y−un, un−xn≥0, ∀y∈C,
xn1αnfxn 1−αnSun, ∀n≥1,
1.3
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|<∞.
1.4
Then the sequences{xn}and{un}converge strongly to a pointz∈FixS∩EPΦ, where
zPFixS∩EPΦfz 1.5
Pis the metric projection ofHontoCandPFixS∩EPΦfzdenotes nearest point inFixS∩EPΦ
fromfz.
Let F : H → H be a nonlinear mapping. The variational inequality problem corresponding to the mappingFis to find a pointx∗∈Csuch that
Fx∗, x−x∗ ≥0, ∀x∈C. 1.6
The variational inequality problem is denoted by VIF, C 9.
The mapping F is called κ-Lipschitzian and η-strongly monotone if there exist constantsκ, η >0 such that
Fx−Fy≤κx−y, ∀x, y∈H, 1.7
Fx−Fy, x−y≥ηx−y2, ∀x, y∈H, 1.8
respectively. It is well known that if F is strongly monotone and Lipschitzian on C, then VIF, Chas a unique solution. An important problem is how to find a solution of VIF, C. Recently, there are many results to solve the VIF, C see, e.g.,10–14.
LetCbe a nonempty closed and convex subset of a Hilbert spaceH,{Tn}∞n1:H → H
be a countable family of nonexpansive mappings, and{Φi}mi1:C×C → Rbembifunctions satisfying conditionsA1–A4such thatΩ ∞n1FixTn∩EPΦ1∩ · · · ∩EPΦm/∅. Let
r1, . . . , rm∈0,∞. For eachi1, . . . , m, define the mappingTri :H → Cby
Trix z∈C:Φi
z, yr1 i
y−z, z−x≥0, ∀y∈C
, ∀x∈H. 1.9
Lemma 2.5see belowshows that, for each 1≤i≤m,Tri is firmly nonexpansive and hence nonexpansive and FixTri EPΦi. Suppose thatF:H → His aκ-Lipschitzian and
η-strong monotone operator and letμ∈0,2η/κ2. Assume that VIΦ
iF,Ω/∅.
In this paper, motivated and inspired by the above research results, we introduce the following iterative process for finding an element inΩ: for an arbitrary initial pointx1∈H,
znγ1Tr1xnγ2Tr2xn· · ·γmTrmxn,
xn1αnxn n
i1
αi−1−αiσnTixn 1−αn1−σnTλnzn, ∀n≥1,
1.10
whereTλnzn zn−λnμFzn,α0 1,{αn}∞
n1is a strictly decreasing sequence in0, αwith
0< α <1,{λn}∞n1 ⊂0,1,{γi}mi1⊂0,1withmi1γi1, and{σn}∞n1 ⊂a, bwith 0< a, b < 1. Then we prove that the iterative process {xn} defined by1.10strongly converge to an
elementx∗∈Ω, which is the unique solution of the variational inequality
Fx∗, x−x∗ ≥0, ∀x∈Ω. 1.11
2. Preliminaries
LetHbe a Hilbert space andTa nonexpansive mapping ofHinto itself such that FixT/∅. For allx∈FixTandx∈H, we have
x−x2≥ Tx−Tx2Tx−x2Tx−x x−x2
Tx−x2x−x22Tx−x, x−x 2.1
and hence
Tx−x2≤2x−Tx, x−x, ∀x∈FixT, x∈H. 2.2
It is well known that, for allx, y∈Handt∈0,1,
tx 1−ty2≤tx2 1−ty2, 2.3
which implies that
n
i1
tixi
2
≤n
i1
tixi2 2.4
for all{xi}ni1⊂Hand{ti}ni1⊂0,1with
n
i1ti 1.
LetCbe a nonempty closed and convex subset ofHand, for anyx∈H, there exists unique nearest point inC, denoted byPCx, such that
PCx−x ≤y−x, ∀y∈C. 2.5
Moreover, we have the following:
zPCx⇐⇒x−z, z−y≥0, ∀y∈C. 2.6
LetI denote the identity operator ofHand let{xn}be a sequence in a Hilbert space
Handx∈H. Throughout this paper,xn → xdenotes that{xn}strongly converges toxand
xn xdenotes that{xn}weakly converges tox.
We need the following lemmas for our main results.
Lemma 2.1see15. LetCbe a nonempty closed and convex subset of a Hilbert spaceHandT a nonexpansive mapping fromCinto itself. ThenI−Tis demiclosed at zero, that is,
Lemma 2.2see10, Lemma 3.1b. LetHbe a Hilbert space andT :H → Hbe a nonexpansive
mapping. LetF :H → Hbe a mapping which isκ-Lipschitzian andη-strong monotone onTH.
Assume thatλ∈0,1andμ∈0,2η/κ2. Define a mappingTλ:H → Hby
TλxTx−λμFTx, ∀x∈H. 2.8
ThenTλx−Tλy ≤1−λτx−yfor allx, y∈H, whereτ 1−1−μ2η−μκ2∈0,1.
IfTI,Lemma 2.2still holds.
Lemma 2.3see16. Let{sn},{cn}be the sequences of nonnegative real numbers and {an} ⊂
0,1. Suppose that{bn}is a real number sequence such that
sn1 ≤1−ansnbncn, ∀n≥0. 2.9
Assume that∞n0cn<∞. Then the following results hold.
(1) Ifbn≤βanfor alln≥0, whereβ≥0, then{sn}is a bounded sequence. (2) If
∞
n0
an∞, lim sup n→ ∞
bn
an ≤0, 2.10
thenlimn→ ∞sn0.
Lemma 2.4 see17. Let C be a nonempty closed and convex subset of a Hilbert space H and
Φ:C×C → Rbe a bifunction which satisfies the conditions (A1)–(A4). Letr >0andx∈H. Then there existsz∈Csuch that
Φz, y1ry−z, z−x≥0, ∀y∈C. 2.11
Lemma 2.5see2. LetHbe a Hilbert space andCbe a nonempty closed and convex subset ofH. Assume thatΦ:C×C → Rsatisfies the conditions (A1)–(A4). For allr >0andx∈H, define a
mappingTr :H → Cas follows:
Trx z∈C:Φz, y r1y−z, z−x≥0, ∀y∈C
, ∀x∈H. 2.12
Then the following holds:
1Tr is single-valued;
2Tr is firmly nonexpansive, that is, for anyx, y∈H,
Trx−Try2≤T
rx−Try, x−y; 2.13
3FixTr EPΦ;
The following lemma is an immediate consequence of an inner product.
Lemma 2.6. Let H be a real Hilbert space. Then the following identity holds:
xy2
≤ x22y, xy, ∀x, y∈H. 2.14
3. Main Results
First, we prove some lemmas as follows.
Lemma 3.1. The sequence{xn}generated by1.10is bounded.
Proof. Let uin Trixn for each i 1,2, . . . , m. Lemma 2.5 shows that each Tri is
firmly-nonexpansive and hence firmly-nonexpansive. Hence, for each 1≤i≤mandp∈Ω, we have
uin−pTrixn−Trip≤xn−p, ∀n≥1, 3.1
zn−p≤m i1
γiuin−p≤xn−p, ∀n≥1. 3.2
ByLemma 2.2, we have
Tλnx−Tλny≤1−λ
nτx−y, ∀x, y∈H, 3.3
whereτ 1−1−μ2η−μκ2 ∈ 0,1. Therefore, by3.2and3.3, we obtainnote that
{αn}is strictly decreasing andTλnp−p−λnμFp
xn1−p
αnxn−p n
i1
αi−1−αiσnTixn−p 1−αn1−σn
Tλnz n−p
≤αnxn−p n
i1
αi−1−αiσnTixn−p 1−αn1−σnTλnzn−p
≤αnxn−p n
i1
αi−1−αiσnxn−p
1−αn1−σnTλnzn−TλnpTλnp−p
≤αnxn−p n
i1
αi−1−αiσnxn−p
≤αnxn−p n
i1
αi−1−αiσnxn−p
1−αn1−σn1−λnτxn−pλnμFp
1−1−αn1−σnλnτxn−p 1−αn1−σnλnμFp.
3.4
By induction, we obtainxn1 ≤max{x1−p,μ/τFp}. Hence{xn}is bounded and so
are{zn}and{uin}for eachi1,2, . . . , m. SinceFisκ-Lipschitzian, we have
Fzn ≤Fzn−FpFp
≤κzn−pFp≤κznκpFp,
3.5
which shows that{Fzn}is bounded. This completes the proof.
Lemma 3.2. If the following conditions hold:
∞
n1
λn∞, ∞
n1
|λn−λn1|<∞, ∞
n1
|σn−σn1|<∞, 3.6
thenlimn→ ∞xn1−xn0.
Proof. For eachi1,2, . . . , m, since eachTri is nonexpansive, we have
uin−1−uinTrixn−1−Trixn ≤ xn−1−xn, ∀n≥1. 3.7
By3.7, we have
zn−zn−1γ1u1n−u1n−1 γ2u2n−u2n−1 · · ·γmumn−umn−1
≤m
i1
γiuin−uin−1 ≤ m
i1
γixn−xn−1
xn−1−xn, ∀n≥1.
3.8
By the definition of the iterative sequence1.10, we have
xn1−xnαnxn−xn−1 αnxn−1 n
i1
αi−1−αiσnTixn−Tixn−1
n
i1
αi−1−αiσnTixn−1 1−αn1−σn
Tλnz
n−Tλnzn−1
1−αn1−σnTλnzn−1−αn−1xn−1− n−1
i1
αi−1−αiσn−1Tixn−1
αnxn−xn−1 αn−αn−1xn−1 n
i1
αi−1−αiσnTixn−Tixn−1
1−αn1−σn
Tλnz
n−Tλnzn−1
n
i1
αi−1−αiσnTixn−1
−n−1
i1
αi−1−αiσn−1Tixn−1 1−αn1−σnTλnzn−1
−1−αn−11−σn−1Tλn−1z n−1
αnxn−xn−1 αn−αn−1xn−1 n
i1
αi−1−αiσnTixn−Tixn−1
1−αn1−σnTλnz
n−Tλnzn−1
n−1 i1
αi−1−αiσn−σn−1Tixn−1
αn−1−αnσnTnxn−1 αn−1−αn1−σn σn−1−σn1−αn−1zn−1 {1−αn−11−σn−1λn−1−λn
−αn−1−αn1−σn σn−1−σn1−αn−1λn}μFzn−1,
3.9
and hence
xn1−xn ≤αnxn−xn−1 αn−1−αnxn−1 n
i1
αi−1−αiσnxn−xn−1
1−αn1−σn1−λnτzn−zn−1
n−1
i1
αi−1−αi|σn−σn−1|Tixn−1
αn−1−αnTnxn−1 αn−1−αn |σn−1−σn|zn−1
|λn−1−λn| αn−1−αn |σn−1−σn|μFzn−1
αnxn−xn−1 n
i1
αi−1−αiσnxn−xn−1
1−αn1−σn1−λnτzn−zn−1
αn−1−αnxn−1Tnxn−1zn−1μFzn−1
n−1 i1
αi−1−αi|σn−σn−1|Tixn−1|σn−1−σn|zn−1μFzn−1
|λn−1−λn|μFzn−1.
It follows from3.8and3.10that
xn1−xn ≤αnxn−xn−1 n
i1
αi−1−αiσnxn−xn−1
1−αn1−σn1−λnτxn−1−xn
αn−1−αnxn−1Tnxn−1zn−1μFzn−1
n−1 i1
αi−1−αi|σn−σn−1|Tixn−1|σn−1−σn|zn−1μFzn−1
|λn−1−λn|μFzn−1
≤1−1−αn1−σnλnτxn−xn−1 αn−1−αn3μM |σn−σn−1|2μM|λn−1−λn|μM
≤1−1−α1−bλnτxn−xn−1 αn−1−αn3μM |σn−σn−1|2μM|λn−1−λn|μM,
3.11
where M max{supn≥1xn,supn≥1zn,supi≥1,n≥1Tixn,supn≥1Fzn}. Since {αn} is
strictly decreasing, we have ∞n2αn−1−αn α1 < ∞. Further, from the assumptions, it
follows that
∞
n2
αn−1−αn3μM|σn−σn−1|2μM|λn−1−λn|μM<∞. 3.12
Therefore, byLemma 2.3, we have limn→ ∞xn1−xn0. This completes the proof.
Lemma 3.3. If the following conditions hold:
lim
n→ ∞λn 0, ∞
n1
λn∞, ∞
n1
|λn−λn1|<∞, ∞
n1
|σn−σn1|<∞, 3.13
then lim
n→ ∞xn−uin0for eachi1,2, . . . , m.
Proof. For anyp∈Ωandi1,2, . . . , m, it follows fromLemma 2.52that
uin−p2 T
rixn−Trip 2≤ T
rixn−Trip, xn−puin−p, xn−p
1
2
uin−p2xn−p2− uin−xn2
and henceuin−p2 ≤ xn−p2− uin−xn2. Further, we have
zn−p2
m
i1
γiuin−p
2
≤m
i1
γiuin−p2
≤m
i1
γixn−p2− uin−xn2
xn−p2− m
i1
γiuin−xn2, ∀n≥1.
3.15
Therefore, from2.4and3.3, we have
xn1−p2
αnxn−p
n
i1
αi−1−αiσnTixn−p 1−αn1−σnTλnzn−p
2
≤αnxn−p2 n
i1
αi−1−αiσnTixn−p2 1−αn1−σnTλnzn−p 2
≤αnxn−p2 1−αnσnxn−p2
1−αn1−σnTλnz
n−TλnpTλnp−p
2
≤αnxn−p2 1−αnσnxn−p2
1−αn1−σn1−λnτzn−pλnμFp2
≤αnxn−p2 1−αnσnxn−p2 1−αn1−σn
×1−λnτzn−p22λn1−λnτμzn−pFpλnμ2Fp2
≤αnxn−p2 1−αnσnxn−p2 1−αn1−σn
×
1−λnτ
xn−p2−m i1
γiuin−xn2
2λn1−λnτμzn−pFpλnμ2Fp2
1−1−αn1−σnλnτxn−p2−1−αn1−σn1−λnτ m
i1
γiuin−xn2
≤xn−p2−1−αn1−σn1−λnτ m
i1
γiuin−xn2
1−αn1−σnλnμ2Fp2
2λnμ1−αn1−σn1−λnτzn−pFp.
3.16
It follows that
γi1−αn1−σn1−λnτuin−xn2
≤xn−pxn1−pxn−xn1λn
μ2Fp22μz
n−pFp
3.17
for each i 1,2, . . . , m. Note that 0 < γi < 1 for i 1,2, . . . , m. From the assumptions,
Lemma 3.2, and the previous inequality, we conclude thatuin −xn → 0 asn → ∞for
eachi1,2, . . . , m. Further, we have
zn−xn ≤ m
i1
γiuin−xn −→0 n−→ ∞. 3.18
This completes the proof.
Lemma 3.4. If the following conditions hold:
lim
n→ ∞λn 0, ∞
n1
λn∞, ∞
n1
|λn−λn1|<∞, ∞
n1
|σn−σn1|<∞, 3.19
thenlimn→ ∞xn−Tixn0for alli≥1.
Proof. By the definition of the iterative sequence1.10, we have
xn1 n
i1
αi−1−αiσnxn−Tixn−1−αnσnxnαnxn 1−αn1−σnTλnzn, 3.20
that is,
n
i1
αi−1−αiσnxn−Tixn xn−xn1−xnαnxn 1−αnσnxn 1−αn1−σnTλnzn
xn−xn1 1−αnσn−1xn 1−αn1−σnTλnzn
xn−xn1 1−αn1−σn
Tλnz n−xn
.
Hence, for anyp∈Ω, we get
n
i1
αi−1−αiσnxn−Tixn, xn−p 1−αn1−σnTλnzn−xn, xn−pxn−xn1, xn−p.
3.22
Since eachTiis nonexpansive, by2.2, we have
Tixn−xn2≤2xn−Tixn, xn−p. 3.23
Hence, combining this inequality with3.22, we get
1 2
n
i1
αi−1−αiσnTixn−xn2≤1−αn1−σnTλnzn−xn, xn−pxn−xn1, xn−p,
3.24
which implies thatnote that{αn}is a strictly decreasing sequence
Tixn−xn2≤ 21α−αn1−σn i−1−αiσn T
λnz
n−xn, xn−p α 2
i−1−αiσnxn−xn1, xn−p
≤ 21−αn1−σn
αi−1−αiσn
Tλnz
n−xnxn−pα 2
i−1−αiσnxn−xn1
xn−p.
3.25
FromLemma 3.3, limn→ ∞λn0, and the inequality
Tλnz
n−xn≤ zn−xnλnμFzn, 3.26
we obtain
lim
n→ ∞
Tλnz
n−xn0. 3.27
Therefore, fromLemma 3.2,3.25, and3.27, it follows that
lim
n→ ∞Tixn−xn0, ∀i≥1. 3.28
Next we prove the main results of this paper.
Theorem 3.5. Assume that the following conditions hold:
lim
n→ ∞λn0, ∞
n1
λn∞, ∞
n1
|λn−λn1|<∞,
∞
n1
γn−γn1<∞, ∞ n1
|σn−σn1|<∞.
3.29
Then the sequence{xn}generated by1.10converges strongly to an element inΩ, which is the unique solution of the variational inequalityVIF,Ω.
Proof. Since VIF,Ω/∅, we can select an elementx∗∈VIF,Ω, which implies that
Fx∗, x∗−x ≥0, ∀x∈Ω. 3.30
First, we prove that
lim sup
n→ ∞ −Fx ∗, x
n1−x∗ ≤0. 3.31
Since{xn}is bounded, there exists a subsequence{xnj}of{xn}such that
lim sup
n→ ∞ −Fx ∗, x
n−x∗jlim→ ∞−Fx∗, xnj−x∗. 3.32
Without loss of generality, we may further assume that xnj x for some x ∈ H. From
Lemmas 3.4 and 2.1, we get x ∈ FixTn for all n ≥ 1. Hence we have x ∈ ∞n1FixTn.
It follows from Lemma 2.5that each Tri is firmly nonexpansive and hence nonexpansive. Lemma 3.3shows thatTrixn−xn → 0 asn → ∞. Therefore, fromLemma 2.1, it follows
thatx ∈FixTrifor eachi1, . . . , m, which shows thatx ∈
m
i1FixTri.Lemma 2.5shows
that FixTri EPΦi for each i 1, . . . , m. Hence x ∈ mi1EPΦi. By using the above argument, we conclude that
x∈Ω ∞ n1
FixTn∩EPΦ1∩ · · · ∩EPΦm. 3.33
Noting thatx∗is a solution of the VIF,Ω, we obtain
lim sup
n→ ∞ −Fx ∗, x
It follows fromLemma 2.6that
xn1−x∗2
αnxn−x∗ n
i1
αi−1−αnσnTixn−x∗ 1−αn1−σn
Tλnz
n−Tλnx∗
1−αn1−σn
Tλnx∗−x∗
2
≤αnxn−x∗ n
i1
αi−1−αnσnTixn−x∗ 1−αn1−σnTλnzn−Tλnx∗
2
21−αn1−σnTλnx∗−x∗, xn1−x∗
≤αnxn−x∗2 n
i1
αi−1−αnσnTixn−x∗2 1−αn1−σnTλnzn−Tλnx∗ 2
21−αn1−σnλnμ−Fx∗, xn1−x∗
≤αnxn−x∗2 n
i1
αi−1−αnσnxn−x∗2 1−αn1−σn1−λnτzn−x∗2
21−αn1−σnλnμ−Fx∗, xn1−x∗
αnxn−x∗2 1−αnσnxn−x∗2 1−αn1−σn1−λnτzn−x∗2 21−αn1−σnλnμ−Fx∗, xn1−x∗
≤αnxn−x∗2 1−αnσnxn−x∗2 1−αn1−σn1−λnτxn−x∗2 21−αn1−σnλnμ−Fx∗, xn1−x∗
1−1−αn1−σnλnτxn−x∗21−αn1−σnλnμ−Fx∗, xn1−x∗.
3.35
Letan 1−αn1−σnλnτ andbn 21−αn1−σnλnμ−Fx∗, xn1−x∗for alln ≥1.
Then, from the assumptions and3.31, we have
0< an<1,
∞
n1
an∞, lim sup n→ ∞
bn
an 0. 3.36
Therefore, by applyingLemma 2.3 to3.35, we conclude that the sequence {xn} strongly
converges to a pointx∗.
In order to prove the uniqueness of solution of the VIF,Ω, we assume that u∗ is another solution of VIF,Ω. Similarly, we can conclude that {xn} converges strongly to a
As direct consequences ofTheorem 3.5, we obtain the following corollaries.
Corollary 3.6. Let Cbe a nonempty closed and convex subset of a Hilbert space H. For each i
1,2, . . . , m let Φi : C×C → R bem bifunctions which satisfy conditions (A1)–(A4) such that
m
i1EPΦi/∅. Letμ∈0,2, and let{αn}∞n1⊂0, αbe a strictly decreasing sequence with0< α <
1,{λn}∞n1 ⊂ 0,1,{γi}mi1 ⊂ 0,1withmi1γi 1,r1, r2, . . . , rm ∈0,∞, and{σn}nn1 ⊂a, b
with0< a, b <1. For an arbitrary initialx1∈H, define the iterative sequence{xn}by
znγ1Tr1xnγ2Tr2xn· · ·γmTrmxn,
xn1 αn 1−αnσnxn 1−αn1−σn1−λnμzn, ∀n≥1.
3.37
If the following conditions hold:
lim
n→ ∞λn 0, ∞
n1
λn∞, ∞
n1
|λn−λn1|<∞, ∞
n1
|σn−σn1|<∞, 3.38
then the sequence{xn}converges strongly to an elementx∗∈mi1EPΦi.
Proof. Put F I and Ti I for each i ≥ 1 in Theorem 3.5. Then we know that F is
1-Lipschitzian and 1-strongly monotone,ni1αi−1−αiTixn 1−αnxnandTλnz
n 1−λnμzn.
Therefore, byTheorem 3.5, we conclude the desired result.
Corollary 3.7. Let C be a nonempty closed and convex subset of a Hilbert space H. Let{Ti}∞i1 be
a countable family of nonexpansive mappings of H such that C ∞i1FixTi and F : H → H
an operator which isκ-Lipschitzian andη-strong monotone onH. Letμ∈0,2η/κ2. Assume that
VIF,C/∅. Let{αn}∞n1 ⊂0, αwith0< α <1be a strictly decreasing sequence,{λn}∞n1 ⊂0,1
and{σn}∞n1 ⊂a, bwith0 < a, b <1. For an arbitrary initialx1 ∈H, define the iterative sequence
{xn}by
xn1αnxn n
i1
αi−1−αiσnTixn 1−αn1−σnPCxn−λnμFPCxn
, ∀n≥1, 3.39
whereα01. If the following conditions hold:
lim
n→ ∞λn 0, ∞
n1
λn∞, ∞
n1
|λn−λn1|<∞, ∞
n1
|σn−σn1|<∞, 3.40
then the sequence{xn}strongly converges to an elementx∗ ∈C, which is the unique solution of the variational inequality
Fx∗, x−x∗ ≥0, ∀x∈C. 3.41
Proof. PutΦix, y 0 for each i 1,2, . . . , m and x, y ∈ C. Set r1 r2 · · · rm 1
inTheorem 3.5. Then, by2.6, we haveTr1xn Tr2xn · · · Trmxn PCxn. Therefore, by
Remark 3.8. 1 Recently, many authors have studied the iteration sequences for infinite family of nonexpansive mappings. But our iterative sequence1.10 is very different from others because we do not useW-mapping generated by the infinite family of nonexpansive mappings and we have no any restriction with the infinite family of nonlinear mappings.
2 We do not use Suzuki’s lemma18for obtaining the result that limn→ ∞xn1 −
xn0. However, many authors have used Suzuki’s lemma18for obtaining the result that
limn→ ∞xn1 −xn 0 in the process of studying the similar algorithms. For example, see 5,19,20and so on.
4. Application
In this section, we study a kind of multiobjective optimization problem based on the result of this paper. That is, we give an iterative sequence which solves the following multiobjective optimization problem with nonempty set of solutions:
minh1x, minh2x,
x∈C,
4.1
where h1xand h2xare both convex and lower semicontinuous functions defined on a nonempty closed and convex subset ofCof a Hilbert spaceH. We denote byAthe set of solutions of4.1and assume thatA /∅.
We denote the sets of solutions of the following two optimization problems byA1and
A2, respectively,
minh1x
x∈C,
minh2x
x∈C.
4.2
Obviously, if we find a solutionx∈A1∩A2, then one must havex∈A.
Now, let Φ1 and Φ2 be two bifunctions from C ×C to R defined by Φ1x, y
h1y−h1xandΦ2x, y h2y−h2x, respectively. It is easy to see that EPΦ1 A1
and EPΦ2 A2, where EPΦidenotes the set of solutions of the equilibrium problem:
Φix, y≥0, ∀y∈C, i1,2, 4.3
respectively. In addition, it is easy to see thatΦ1 and Φ2 satisfy the conditionsA1–A4. Therefore, by settingm2 inCorollary 3.6, we know that, for any initial guessx1∈H,
h1
y−h1u1n r1
1ny−u1n, u1n−xn ≥0, ∀y∈C,
h2
y−h2u2n r1
2ny−u2n, u2n−xn ≥0, ∀y∈C,
zn γ1u1n1−γ1
u2n,
xn1 αn 1−αnσnxn 1−αn1−σn1−λnμzn, ∀n≥1.
By Corollary 3.6, we know that the sequence {xn} converges strongly to a solutionx∗ ∈
EPΦ1∩EPΦ2 A1∩A2, which is a solution of the multiobjective optimization problem 4.1.
Acknowledgment
This work was supported by the Korea Research Foundation Grant funded by the Korean GovernmentKRF-2008-313-C00050.
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