Volume 2010, Article ID 687374,14pages doi:10.1155/2010/687374
Research Article
An Iterative Scheme with a Countable Family of
Nonexpansive Mappings for Variational Inequality
Problems in Hilbert Spaces
Yeol Je Cho
1and Shenghua Wang
2, 31Department of Mathematics Education and the RINS, Gyeongsang National University,
Chinju 660-701, South Korea
2School of Applied Mathematics and Physics, North China Electric Power University,
Baoding 071003, China
3Department of Mathematics, Gyeongsang National University, Chinju 660-701, South Korea
Correspondence should be addressed to Shenghua Wang,[email protected]
Received 13 December 2009; Accepted 8 February 2010
Academic Editor: Jong Kim
Copyrightq2010 Y. J. Cho and S. Wang. 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 with a countable family of nonexpansive mappings for the variational inequality problems in Hilbert spaces and prove some strong convergence theorems for the proposed schemes.
1. Introduction
LetHbe a Hilbert space andCbe a nonempty closed convex subset ofH. LetF :H → H be a nonlinear mapping. The classical variational inequality problemfor short, VIF, Cis to find a pointx∈Csuch that
Fx∗, x−x∗ ≥0, ∀x∗∈C. 1.1
This variational inequality was initially studied by Kinderlehrer and Stampacchia
1. Since then, many authors have introduced and studied many kinds of the variational inequality problemsinclusionsand applied them to many fields.
LetT :H → Hbe a mapping. Recall that a mappingT :H → His nonexpansive if Tx−Ty≤x−y, ∀x, y∈H. 1.2
The set of fixed points of T is denoted by FT. Recently, the iterative methods for nonexpansive mappings and some kinds of nonlinear mappings have been applied to solve the convex minimization problemssee3–7.
A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mapping onH:
min
x∈C
1
2Ax, x − x, b, 1.3
whereCis the fixed point set of a nonexpansive mappingTon H,bis a given point inHand
Ais a strongly positive operator, that is, there is a constantγ >0 such that
Ax, x ≥γx2, ∀x∈H. 1.4
Recently, for solving the variational inequality onA, Marino and Xu8introduced the following general iterative scheme:
xn1 I−αnATxnαnγfxn, ∀n≥0, 1.5
whereAis a strongly positive linear bounded operator onH,f is a contraction onHand
{αn} ⊂0,1.
More precisely, they gave the following result.
Theorem MXsee8, Theorem 3.4. Let{xn}be generated by algorithm1.5with the sequence
{αn}satisfying the following conditions:
C1limn→ ∞αn0,
C2∞n0αn∞,
C3either∞n0|αn1−αn|<∞orlimn→ ∞αn1/αn 1.
Then the scheme{xn}defined by1.5converges strongly to an elementx∗∈CFTwhich is the unique solution of the variational inequalityfor short,VIA−γf, C:
A−γfx∗, x−x∗≥0, ∀x∈C. 1.6
Letf : H → H be a contraction with coefficient 0 < α < 1 and letA, B : H → H be two strongly positive linear bounded operators with coefficients γ ∈ 0,1 and β > 0, respectively.
Theorem CGY1see9, Theorem 3.1. Let0 < γα < βandγ ∈0,1. Let{λn}be a sequence in
0,1and{μn}be a sequence in0,min{1,B−1}. Starting with an arbitrary initial guessx 0∈H, generate a sequence{xn}by the following iterative scheme:
xn1 I−λn1ATxnλn1
Txn−μn1
BTxn−γfxn , ∀n≥0. 1.7
Assume that
ilimn→ ∞λn0,
ii∞n1λn∞,
iiieither∞n1|λn1−λn|<∞orlimn→ ∞λn/λn11,
iv 1−γ/β−γα<limn→ ∞μnμ <2−γ/β−γα.
Then the scheme{xn}defined by1.7converges strongly to an elementx∗∈CFTwhich is the unique solution of the variational inequalityfor short,VIA−IμB−γf, C:
A−IμB−γf x∗, x−x∗≥0, ∀x∈C. 1.8
Theorem CGY2see9, Theorem 3.2. Let0< γα < βandγ ∈0,1. Let{λn}be a sequence in
0,1and{μn}be a sequence in0,min{1,B−1}. Starting with an arbitrary initial guessx0∈H, generate a sequence{xn}by the following iterative scheme:
xn1 I−λn1ATn1xnλn1
Tn1xn−μn1
BTn1xn−γfxn , ∀n≥0. 1.9
Assume that
ilimn→ ∞λn0,
ii∞n1λn∞,
iiieither∞n1|λnN−λn|<∞orlimn→ ∞λn/λnN 1,
iv 1−γ/β−γα<limn→ ∞μnμ <2−γ/β−γα.
In addition, assume that
CN
i1
FTi FT1T2· · ·TN FTNT1· · ·T2
· · ·FT2T3· · ·TNT1.
1.10
Then the scheme{xn}defined by1.9converges strongly to an elementx∗∈CFTwhich is the unique solution of the variational inequalityfor short,VIA−IμB−γf, C:
A−IμB−γf x∗, x−x∗≥0, ∀x∈C. 1.11
More precisely, let H be a Hilbert space and {Ti}∞i1 be a countable family of nonexpansive mappings fromH toH such thatC ∞i1FTi/∅. Let f : H → Hbe a contraction with coefficient 0< α <1 andA, Bbe strongly positive linear bounded operators with coefficientsη ∈0,1andβ > 0, respectively. Let{λn}∞n1 ⊂ 0,1and{αn}∞n0 ⊂ 0,1 with α0 1. Take three fixed numbersγ,μ1 andμ2 such that 0 < γα < β,μ1 ∈ 0,1and
μ2 ∈ 1−ημ1/β−γα,min{1,B−1,2−ημ1/β−γα}. For anyx1 ∈ H, generate the iterative scheme{xn}by
xn1αn
I−λnμ1A
xnλnxn−μ2
Bxn−γfxn
n
i1
αi−1−αiTixn, ∀n≥1.
1.12
We prove that the iterative scheme {xn} defined by 1.12strongly converges to an elementx∗∈Cwhich is the unique solution of the variational inequalityfor short, VIμ1A−
Iμ2B−γf, C:
μ1A−Iμ2
B−γf x∗, x−x∗≥0, ∀x∈C. 1.13
2. Preliminaries
LetHbe a Hilbert space andTbe a nonexpansive mapping ofHinto itself such thatFT/∅. For allx∈FTandx∈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∈FT, x∈H. 2.2
Let{xn}be a sequence in a Hilbert spaceH and letx ∈ H. Throughout this paper, xn → xand xn x denote that{xn} strongly converges to x ∈ H and {xn} converges
weakly to a pointx∈H, respectively.
Lemma 2.1see10. LetCbe a closed convex subset of a Hilbert spaceHandTbe a nonexpansive mapping fromCinto itself. ThenI−Tis demiclosed at zero, that is,
xn x, xn−Txn−→0 implies xTx. 2.3
The following lemma is an immediate consequence of the equality:
Lemma 2.2. LetHbe a real Hilbert space. Then the following identity holds:
xy2
≤ x22y, xy, ∀x, y∈H. 2.5
Lemma 2.3see4,11. Let{sn},{cn}be the sequences of nonnegative real numbers and let{an} ⊂
0,1. Suppose that{bn}is a sequence of real numbers such that
sn1 ≤1−ansnbncn, ∀n≥0. 2.6
Assume that∞n0cn<∞. Then the following results hold.
1Ifbn≤βan, whereβ≥0, then{sn}is a bounded sequence.
2If one has
∞
n0
an∞, lim sup
n→ ∞
bn
an ≤0, 2.7
thenlimn→ ∞sn0.
Lemma 2.4see8. LetHbe a real Hilbert space,f :H → Hbe a contraction with coefficient 0< α <1andBbe a strongly positive linear bounded operator with coefficientβ >0. Then, for anyγ with0< γ < β/α,
x−y,B−γfx−B−γfy≥β−γαx−y2, ∀x, y∈H, 2.8
that is,B−γfis strongly monotone with coefficientβ−γα.
Lemma 2.5. AssumeAis a strongly monotone linear bounded operator on a Hilbert spaceHwith coefficientα >0. Take a fixed numberρsuch that0< ρ≤ A−1. ThenI−ρA ≤1−ρα.
Proof. The proof method is mainly from the idea of Marino and Xu8, Lemma 2.5. It is known that the norm of a linear bounded self-adjoint operatorV onHis as follows:
Now, for allx∈Hwithx1, we see thathere 0 denotes zero point inH
I−ρAsupI−ρAx, x:x∈H, x1
supI−ρAx−I−ρA0, x−0:x∈H, x1
supx−0−ρAx−A0, x−0:x∈H, x1
supx2−ρAx−A0, x−0:x∈H, x1
≤sup1−ραx−02:x∈H, x1
1−ρα.
2.10
This completes the proof.
Remark 2.6. Lemma 2.5still holds if Ais a strongly positive linear bounded operator see
8, Lemma 2.5. That is,Lemma 2.5in this section and Lemma 2.5 in8both hold whenA is a strongly monotone linear bounded operator or a strongly positive linear bounded one because an operator on a Hilbert space is strongly monotone linear if and only if it is strongly positive linear.
In fact, ifAis a strongly monotone linear operator with coefficientα >0 on a Hilbert spaceH, then, for allx∈H,
Ax, xAx−A0, x−0 ≥αx−02αx2, 2.11
which shows thatAis strongly positive linear. Assume thatAis a strongly positive linear operator with coefficientα >0 onH. Then, for allx, y∈H,
Ax−Ay, x−yAx−y, x−y≥αx−y2, 2.12
which shows thatAis strongly monotone and linear.
3. Main Results
coefficientβ−γα > 0. For any fixed numbersσ1 ∈ 0,1andσ2 ∈ 1−ησ1/β−γα,2−
ησ1/β−γα, we haveθ ησ1−1σ2β−γα∈0,1, which can be seen easily from the following:
σ2<
2−ησ1
β−γα ⇐⇒σ2
β−γα<2−ησ1
⇐⇒θησ1−1σ2
β−γα<1,
3.1
1−ησ1
β−γα < σ2⇐⇒σ2
β−γαησ1>1
⇐⇒θησ1−1σ2
β−γα>0.
3.2
Moreover, observe that
σ1A−Iσ2
B−γfx−σ1A−Iσ2
B−γfy
σ1A−I
x−yσ2
B−γfx−y
≤ σ1A−Ix−yσ2B
x−yγfx−fy
≤σ1A−Iσ2
Bγα x−y,
3.3
which implies thatσ1A−Iσ2B−γfis Lipschitzian with coefficientσ1A−Iσ2Bγα> 0.
On the other hand, fromLemma 2.4, it follows that
σ1A−Iσ2
B−γfx−σ1A−Iσ2
B−γfy, x−y
σ1
Ax−Ay, x−yσ2
B−γfx−B−γfy, x−y−x−y2
≥σ1ηx−y2σ2
β−γαx−y2−x−y2
θx−y2,
3.4
which implies thatσ1A−Iσ2B−γfis strongly monotone with coefficientθ >0. Hence the variational inequalityfor short, VIσ1A−Iσ2B−γf, C
σ1A−Iσ2
B−γfx∗, x−x∗≥0, ∀x∈C 3.5
has the unique solution.
LetT :H → Hbe a nonexpansive mapping. Take two fixed numbersμ1andμ2such thatμ1 ∈ 0,1andμ2 ∈ 0,min{1,B−1}and, for allλ ∈ 0,min{1,A−1/μ1}, define a mappingTλ:H → Hby
TλxI−λμ
1A
TxλTx−μ2
BTx−γfx , ∀x∈H. 3.6
Lemma 3.1. Ifμ2∈1−ημ1/β−γλ,2−ημ1/β−γλ, thenTλis a contraction with coefficient 1−λτ, whereτημ1−1μ2β−γα∈0,1, that is,
Tλx−Tλy≤1−λτx−y, ∀x, y∈H. 3.7
Proof. FromLemma 2.5andRemark 2.6, it follows that, for allx, y∈H,
Tλx−TλyI−λμ
1A
TxλTx−μ2
BTx−γfx
−I−λμ1A
Ty−λTy−μ2
BTy−γfy
≤I−λμ1A
Tx−I−λμ1A
Ty
λTx−μ2
BTx−γfx−Ty−μ2
BTy−γfy
≤I−λμ1ATx−Ty
λI−μ2B
Tx−I−μ2B
Tyμ2γfx−f
y
≤1−λμ1ηx−yλI−μ2BTx−Tyμ2γαx−y
≤1−λμ1ηλ
1−μ2
β−γα x−y
1−λμ1η−1μ2
β−γα x−y
1−λτx−y.
3.8
This completes the proof.
Let{Ti}∞n1be a countable family of nonexpansive mappings fromHinto itself such thatC∞n1FTi/∅. Since eachFTiis closed and convex, thenCis closed and convex.
Throughout this paper, letf : H → Hbe a contraction with coefficient 0 < α < 1. LetA, B :H → Hbe strongly positive linear bounded mapping with coefficientη ∈ 0,1 andβ >0, respectively. Take a fixed numberγsuch that 0< γα < β. Suppose thatμ1∈0,1,
μ2∈1−ημ1/β−γα,min{1,B−1,2−ημ1/β−γα} assuming that1−ημ1/β−γα< min{1,B−1}such that1−ημ
1/β−γλ,min{1,B−1,2−ημ1/β−γλ}is nonempty,
{λn}∞n1⊂0,min{1,A−1/μ1}with lim infn→ ∞λn>0 and{αn}∞n0⊂0,1withα01. Now, we can rewrite the iterative scheme1.12as follows:
xn1αnTλnxn n
i1
αi−1−αiTixn, ∀n≥1, 3.9
whereTλnx
n I−λnμ1Axnλnxn−μ2Bxn−γfxn. Then, byLemma 3.1, for allx, y∈H,
we have
Tλn
n x−Tnλny≤1−λnτx−y, ∀n≥1, 3.10
Lemma 3.2. If{αn}is strictly decreasing, then the scheme{xn}defined by3.9is bounded.
Proof. SinceTλnp−pλ
nμ1A−Iμ2B−γfp, it follows from3.10that, for allp∈C,
xn1−p αn
Tλnxn−p
n
i1
αi−1−αi
Tixn−p
≤αnTλnxn−p n
i1
αi−1−αiTixn−p
≤αnTλnxn−TλnpαnTλnp−p 1−αnxn−p
≤αn1−λnτxn−pαnλnμ1A−Iμ2
B−γfp 1−αnxn−p
1−αnλnτxn−pαnλnμ1A−Iμ2
B−γfp.
3.11
By induction, we obtain
xn1−p≤maxx1−p, 1
τμ1A−Iμ2
B−γfp
. 3.12
Hence {xn} is bounded and so are{Tλnxn}and {Tixn} for eachi ≥ 1. This completes the
proof.
Lemma 3.3. If{αn}is strictly decreasing and the following conditions hold:
∞
n1
αn∞, ∞ n1
|λn−λn1|<∞, 3.13
thenlimn→ ∞xn1−xn0.
Proof. By the iterative scheme3.9, we have
xn1−xnαnTλnxn n
i1
αi−1−αiTixn−
αn−1Tλn−1xn−1
n−1
i1
αi−1−αiTixn−1
αnTλnxn−Tλnxn −1
n
i1
αi−1−αiTixn−Tixn−1
n
i1
αi−1−αiTixn−1−
n−1
i1
αi−1−αiTixn−1αnTλnxn−1−αn−1Tλn−1xn−1
αn
Tλnx
n−Tλnxn−1
n
i1
αi−1−αiTixn−Tixn−1
αn−1−αnTnxn−1 αn−1λn−1−αnλnμ1A−Iμ2
B−γfxn−1
αnTλnxn−Tλnxn −1
n
i1
αi−1−αiTixn−Tixn−1
αn−1−αnTnxn−1 αn−αn−1xn−1
αn−1−αnλn λn−1−λnαn−1
μ1A−Iμ2
B−γfxn−1
3.14
and hence
xn1−xn ≤αnTλnxn−Tλnxn−1
n
i1
αi−1−αnTixn−Tixn−1
αn−1−αnTnxn−1 αn−1−αnxn−1
αn−1−αnλn|λn−1−λn|αn−1μ1A−Iμ2
B−γfxn−1
≤αn1−λnτxn−xn−1 1−αnxn−xn−1
αn−1−αnTnxn−1 αn−1−αnxn−1
αn−1−αn |λn−1−λn|μ1A−Iμ2
B−γfxn−1
≤1−αnλnτxn−xn−1 αn−1−αnM|λn−1−λn|M,
3.15
whereMis a constant. Since{λn} ⊂0,min{1,A−1/μ
1}, there exists a constantλ>0 such thatλn≥λfor alln≥1. Therefore, we have
xn1−xn ≤1−αnλτxn−xn−1 αn−1−αn |λn−1−λn|M. 3.16
Putcn αn−1−αn|λn−1−λn|M. Since{αn}is a strictly decreasing sequence and∞n2|λn−1−
λn|<∞, we have∞n2cn<∞. ByLemma 2.3, it follows thatxn1−xn → 0 asn → ∞. This
completes the proof.
Lemma 3.4. If{αn}is strictly decreasing and the following conditions hold:
lim
n→ ∞αn0, ∞
n1
αn∞, ∞ n1
|λn−λn1|<∞ 3.17
thenlimn→ ∞xn−Tixn0, for alli≥1.
Proof. By the iterative scheme3.9, we have
xn1
n
i1
that is,
n
i1
αi−1−αixn−Tixn αn
Tλnx
n−xn
xn−xn1. 3.19
Hence, for anyp∈C, we get
n
i1
αi−1−αixn−Tixn, xn−pαn
Tλnx
n−xn, xn−p
xn−xn1, xn−p. 3.20
Since eachTiis nonexpansive, it follows from2.2that
Tixn−xn2≤2xn−Tixn, xn−p. 3.21
Hence, combining3.21with3.20, it follows that
n
i1
αi−1−αiTixn−xn2≤2
n
i1
αi−1−αi
xn−Tixn, xn−p
2αnTλnxn−xn, xn−p2xn−x
n1, xn−p
,
3.22
which implies that
1 2
n
i1
αi−1−αiTixn−xn2≤αn
Tλnx
n−xn, xn−p
xn−xn1, xn−p. 3.23
Since eachαi−1−αiTixn−xn2≥0 andαi−1−αi>0, then we have
1
2αi−1−αiTixn−xn 2≤α
n
Tλnx
n−xn, xn−p
xn−xn1, xn−p, 3.24
that is,
Tixn−xn2≤ α 2αn i−1−αi
Tλnx
n−xn, xn−p
α 2
i−1−αi
xn−xn1, xn−p. 3.25
Since{xn}and{Tλnxn}are both bounded, there exists a constantM>0 such that
Tixn−xn2≤M
αn
αi−1−αi
1
αi−1−αixn−xn1
. 3.26
ByLemma 3.3and the assumption condition limn→ ∞αn0, it follows that
lim
n→ ∞Tixn−xn0, ∀i≥1. 3.27
Finally, we give the main result in this paper.
Theorem 3.5. If{αn}is strictly decreasing and the following conditions hold:
lim
n→ ∞αn 0, ∞
n1
αn∞,
∞
n1
|λn−λn1|<∞, 3.28
then the scheme{xn}defined by3.9converges strongly to an elementx∗ ∈Cwhich is the unique solution of the variational inequalityVIμ1A−Iμ2B−γf, C:
μ1A−Iμ2
B−γf x∗, x−x∗≥0, ∀x∈C. 3.29
Proof. First, we prove that lim supn→ ∞−μ1A−Iμ2B−γfx∗, xn1−x∗ ≤0. To prove this, we pick a subsequence{xni}of{xn}such that
lim sup
n→ ∞
−μ1A−Iμ2
B−γf x∗, xn−x∗ lim
i→ ∞
−μ1A−Iμ2
B−γf x∗, xn i−x∗
.
3.30
Without loss of generality, we may, further, assume thatxni xfor somex∈H. From
Lemmas2.1and3.3, it follows thatx∈FTifor eachi≥1 and sox∈C∩∞i1FTi. Sincex∗ is the unique solution of the problem VIμ1A−Iμ2B−γf, C, we obtain
lim sup
n→ ∞
−μ1A−Iμ2
B−γf x∗, x
n−x∗ −μ1A−Iμ2
B−γf x∗,x−x∗≤0.
3.31
It follows fromLemma 2.2and3.10that
xn1−x∗2
αnTλnxn−Tλnx∗n
i1
αi−1−αnTixn−x∗
αnTλnx∗−x∗
2
≤αn
Tλnx
n−Tλnx∗
n
i1
αi−1−αnTixn−x∗
2
2αnTλnx∗−x∗, xn
1−x∗
≤αn1−λnτxn−x∗ 1−αnxn−x∗2
2αn−μ1A−Iμ2
B−γf x∗, x
n1−x∗
≤1−αnλnτxn−x∗22αn−μ1A−Iμ2
B−γf x∗, x
n1−x∗
≤1−αnλτxn−x∗22αn−μ1A−Iμ2
B−γf x∗, xn
1−x∗
,
where λ > 0 is a constant such that λn ≥ λ for all n ≥ 1. Since ∞n0αn ∞ and lim supn→ ∞−μ1A−Iμ2B−γfx∗, xn1−x∗ ≤ 0, byLemma 2.3, we conclude that the scheme{xn}converges strongly tox∗. This completes the proof.
Remark 3.6. 1For eachn ≥ 1, a simple example on control parameters is αn 1/nand
λnλ, whereλis a constant in0,min{1,A−1/μ1}.
2We obtain the desired results without any assumptions on the family{Ti}∞i1. Foe example, in Theorem CGY2, the authors gave the strong condition1.10.
Remark 3.7. 1IfT1 T2 · · · Tn · · · T in3.9, then we have the following iterative
scheme:
xn1αnI−λnμ1A
xnλnxn−μ2
Bxn−γfxn 1−αnTxn, ∀n≥1, 3.33
and the scheme{xn}defined by3.33converges strongly to an elementx∗ ∈Cwhich is the unique solution of the variational inequalityVIμ1A−Iμ2B−γf, C:
μ1A−Iμ2
B−γf x∗, x−x∗≥0, ∀x∈C. 3.34
2IfAIandμ11 in3.33, then we have the following iterative scheme:
xn1αnI−λnμ2
B−γfxn n
i1
αi−1−αiTixn, ∀n≥1, 3.35
and the scheme{xn}defined by3.35converges strongly to an elementx∗ ∈Cwhich is the
unique solution of the variational inequalityVIB−γf, C:
μ2
B−γfx∗, x−x∗≥0, ∀x∈C. 3.36
3Furthermore, if μ2 1 andγ 0 in3.35, then we have the following iterative scheme:
xn1 αnI−λnBxn n
i1
αi−1−αiTixn, ∀n≥1, 3.37
and the scheme{xn}defined by3.37converges strongly to an elementx∗ ∈Cwhich is the unique solution of the variational inequalityVIB, C, which is Stampacchia’s variational inequality:
Bx∗, x−x∗ ≥0, ∀x∈C. 3.38
Acknowledgment
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