Hindawi Publishing CorporationJournal of Inequalities and Applications Volume 2008, Article ID 598632, 13 pages doi:10.1155/2008/598632 Research Article Finite-Step Relaxed Hybrid Steepe
Trang 1Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2008, Article ID 598632, 13 pages
doi:10.1155/2008/598632
Research Article
Finite-Step Relaxed Hybrid Steepest-Descent
Methods for Variational Inequalities
Yen-Cherng Lin
Department of Occupational Safety and Health, General Education Center,
China Medical University, Taichung 404, Taiwan
Correspondence should be addressed to Yen-Cherng Lin,yclin@mail.cmu.edu.tw
Received 22 August 2007; Revised 2 January 2008; Accepted 13 March 2008
Recommended by Jong Kim
The classical variational inequality problem with a Lipschitzian and strongly monotone operator
on a nonempty closed convex subset in a real Hilbert space was studied A new finite-step relaxed hybrid steepest-descent method for this class of variational inequalities was introduced Strong convergence of this method was established under suitable assumptions imposed on the algorithm parameters
Copyrightq 2008 Yen-Cherng Lin 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
LetH be a real Hilbert space with inner product ·, · and norm · Let C be a nonempty closed
convex subset ofH, and let F : C → H be an operator The classical variational inequality
problem: findu∗∈ C such that
VIF, C Fu∗
, v − u∗
was initially studied by Kinderlehrer and Stampacchia1 It is also known that the VIF, C is
equivalent to the fixed-point equation
u∗ P C
u∗− μFu∗
wherePCis thenearest point projection from H onto C, that is, P Cx argmin y∈C x − y for
eachx ∈ H and where μ > 0 is an arbitrarily fixed constant If F is strongly monotone and
Lipschitzian onC and μ > 0 is small enough, then the mapping determined by the right-hand
side of this equation is a contraction Hence the Banach contraction principle guarantees that the Picard iterates converge in norm to the unique solution of the VIF, C Such a method
is called the projection method However, Zeng and Yao 2 point out that the fixed-point
Trang 22 Journal of Inequalities and Applications equation involves the projectionPC which may not be easy to compute due to the complexity
of the convex setC To reduce the complexity problem probably caused by the projection PC, a class of hybrid steepest-descent methods for solving VIF, C has been introduced and studied recently by many authorssee, e.g., 3,4 Zeng and Yao 2 have established the method of two-step relaxed hybrid steepest-descent for variational inequalities A natural arising problem
is whether there exists a general relaxed hybrid steepest-descent algorithm that is more than two steps for finding approximate solutions of VIF, C or not Motivated and inspired by the recent research work in this direction, we introduce the following finite step relaxed hybrid steepest-descent algorithm for finding approximate solutions of VIF, C and aim to unify the convergence results of this kind of methods
Algorithm 1.1 Let {α k n } ⊂ 0, 1, {λ k n } ⊂ 0, 1, for k 1, 2, , m, and take fixed numbers
t k ∈ 0, 2η/κ2, k 1, 2, , m Starting with arbitrarily chosen initial points u10 ∈ H, compute the
sequences {u k n } such that
u1n1 α1n u1n 1− α1n
Tu2n − λ1n1 t1FTu2n
,
u2n α2n u1n 1− α2n
Tu3n − λ2n1 t2FTu3n
,
u3n α3n u1n 1− α3n
Tu4n − λ3n1 t3FTu4n
,
.
u m n α m n u1n 1− α m n
Tu1n − λ m n1 t m FTu1n
.
1.3
We will prove a strong convergence result for Algorithm 1.1 under suitable restrictions imposed
on the parameters.
2 Preliminaries
The following lemmas will be used for proving the main result of the paper in next section
Lemma 2.1 see 5 Let {s n } be a sequence of nonnegative real numbers satisfying the inequality
sn1 ≤1− α n
where {α n }, {τ n }, and {γ n } satisfy the following conditions:
i {α n } ⊂ 0, 1,∞n0 αn ∞, or equivalently, ∞n0 1 − α n 0;
ii lim supn→∞ τn ≤ 0;
iii {γ n } ⊂ 0, ∞,∞
n0 γn < ∞.
Then limn→∞sn 0.
Lemma 2.2 see 6 Demiclosedness principle: assume that T is a nonexpansive self-mapping on a
nonempty closed convex subset C of a Hilbert space H If T has a fixed point, then I − T is demiclosed; that is, whenever {x n } is a sequence in C weakly converging to some x ∈ C and the sequence {I−Tx n}
strongly converges to some y ∈ H, it follows that I − Tx y Here I is the identity operator of H.
Trang 3Yen-Cherng Lin 3 The following lemma is an immediate consequence of an inner product
Lemma 2.3 In a real Hilbert space H, there holds the inequality
Lemma 2.4 Let C be a nonempty closed convex subset of H For any x, y ∈ H and z ∈ C, the
following statements hold:
i P Cx − x, z − PCx ≥ 0;
ii P Cx − PCy2≤ x − y2− P Cx − x y − PCy2.
3 Convergence theorem
LetH be a real Hilbert space and let C be a nonempty closed convex subset of H Let F :
C → H be an operator such that for some constants κ, η > 0, F is κ-Lipschitzian and η-strongly
monotone onC; that is, F satisfies the conditions
Fx − Fy ≤ κx − y, ∀x, y ∈ C,
respectively Since F is η-strongly monotone, the variational inequality problem VIF, C has a
unique solutionu∗∈ C see, e.g., 7
Assume thatT : H → H is a nonexpansive mapping with the fixed points set FixT C.
Note that obviously FixPC C For any given numbers λ ∈ 0, 1 and μ ∈ 0, 2η/κ2, we define the mappingT λ:H → H by
T λ
μ x : Tx − λμFTx, ∀x ∈ H.
Lemma 3.1 see 3 Let T λ be a contraction provided that 0 < λ < 1 and 0 < μ < 2η/κ2 Indeed,
λ
μ x − T λ
where τ 1 − 1− μ2η − μκ2 ∈ 0, 1.
We now state and prove the main result of this paper
Theorem 3.2 Let H be a real Hilbert space and let C be a nonempty closed convex subset of H.
Let F : C → H be an operator such that for some constants κ, η > 0, F is κ-Lipschitzian and η-strongly monotone on C Assume that T : H → H is a nonexpansive mapping with the fixed points set
FixT C, the real sequences {αk n }, {λ k n }, for k 1, 2, , m, in Algorithm 1.1 satisfy the following conditions:
i∞n1 |α k n − α k n−1 | < ∞, for k 1, 2, , m;
ii limn→∞α1n 0 and lim n→∞α k n 1, for k 2, 3, , m;
Trang 44 Journal of Inequalities and Applications
iii limn→∞λ1n 0, lim n→∞ λ1n /λ1n1 1,∞n1 λ1n ∞;
iv λ1n ≥ max{λ k n :k 2, 3, , m}, for all n ≥ 1.
Then the sequences {u k n } generated by Algorithm 1.1 converge strongly to u∗ which is the unique solution of the VI F, C.
Proof Since F is η-strongly monotone, by 7, the VIF, C has the unique solution u∗∈ C Next
we divide the rest of the proof into several steps
Step 1 Let {u k n } is bounded for each k 1, 2, , m Indeed, let us denote that T λ
t u∗ Tu∗−
λtFTu∗, then we have
1
n1 − u∗ 1
n u1n 1− α1n
T λ1n1
t1 u2n − u∗
≤ α1n 1n − u∗ 1
1
n1
t1 u2n − u∗
≤ α1n 1n − u∗ 1
1
n1
t1 u2n − T λ1n1
t1 u∗ λ1n1
t1 u∗− u∗
≤ α1n 1n − u∗ 1
n
1− λ1n1 τ1 2
n − u∗ 1
n1 t1 ∗
3.3
whereτ1 1 − 1− t12η − t1κ2 ∈ 0, 1 Moreover, we also have
2
n − u∗ 2
n u1n 1− α2n
T λ2n1
t2 u3n − u∗
≤ α2n 1n − u∗ 2
2
n1
t2 u3n − T λ2n1
t2 u∗ λ2n1
t2 u∗− u∗
≤ α2n 1n − u∗ 2
n
1− λ2n1 τ2 3
n − u∗ 2
n1 t2 ∗
≤ α2n 1n − u∗ 2
n 3n − u∗ 2
n
λ2n1 t2 ∗
3.4
whereτ2 1 − 1− t22η − t2κ2 ∈ 0, 1, and for k 2, 3, , m − 1,
k
n − u∗ k
n u1n 1− α k n
T λ k n1
t k u k1 n − u∗
≤ α k n 1n − u∗ k
k
n1
t k u k1 n − T λ k n1
t k u∗ λ k n1
t k u∗− u∗
≤ α k n 1n − u∗ k
n
1− λ k n1 τ k k1
n − u∗ k
n1 t k ∗
≤ α k n 1n − u∗ k
n
λ k n1 t k ∗
3.5
Trang 5Yen-Cherng Lin 5 whereτ k 1 − 1− t k 2η − t k κ2 ∈ 0, 1, and
m
n − u∗ m
n u1n 1− α m n
T t λ m m u1n − u∗
≤ α m n 1n − u∗ m
m
t m u1n − T t λ m m u∗ λ m
t m u∗− u∗
≤ α m n 1n − u∗ m
n
1− λ m n1 τ m 1
n − u∗ m
n1 t m ∗
≤ 1n − u∗ m
n1 t m ∗
3.6 whereτ m 1 − 1− t m 2η − t m κ2 ∈ 0, 1.
Thus we obtain
m−1
n − u∗
n 1n − u∗ m
max
λ m n1 , λ m−1 n1 t m t m−1 ∗
k
n − u∗ 1
n max
k≤j≤m
λ j n1
m
jk
t j
∗
3.7 fork 2, 3, , m − 1 In particular,
2
n − u∗ 1
n max
2≤j≤m
λ j n1
m
j2
t j
Hence, substituting3.8 in 3.3 and by condition iv, we obtain
1
n1 − u∗
≤ α1n 1n − u∗ 1
n
1− λ1n1 τ1 2
n1 t1 ∗
≤ α1n 1n − u∗ 1
n
1− λ1n1 τ1 1
n − u∗
max
2≤j≤m{λ j n1}m
j2
t j ∗ 1
n1 t1 ∗
≤ α1n 1n − u∗ 1
n
1− λ1n1 τ1 1
n − u∗
1≤j≤m
λ j n1 m j1
t j ∗
.
3.9
By induction, it is easy to see that
1
Trang 66 Journal of Inequalities and Applications where M max{u10 − u∗, m j1 t j /τ1Fu∗} Indeed, for n 0, from 3.9 we obtain
1
0
1
0
1− λ11 τ1 1
0 − u∗
1≤j≤mλ j1
m
j1
t j
∗
≤ α10 M 1− α10 1− λ11 τ1 M λ1
1 τ1M M.
3.11 Suppose thatu1n − u∗ ≤ M, for n ≥ 1 We want to claim that u1n1 − u∗ ≤ M Indeed,
1
n1 − u∗ 1
n 1n − u∗ 1
n
1− λ1n1 τ1 1
n − u∗ 1
n1 m
j1
t j ∗
≤ α1n M 1− α1
n
1− λ1n1 τ1 M λ1
n1 τ1M M.
3.12
Therefore, we haveu1n −u∗ ≤ M, for all n ≥ 0, and u m n −u∗ ≤ Mλ m n1 τ1M ≤ 1τ 1 M,
for alln ≥ 0 In this case, from 3.8, it follows that
k
n − u∗ M max
k≤j≤m
λ j n1 τ1M ≤ 1 τ1 M, ∀n ≥ 0, ∀k 2, 3, , m − 1. 3.13
Step 2 Let u1n1 − Tu1n → 0, n → ∞ Indeed byStep 1,{u k n } is bounded for 1 ≤ k ≤ m and
so are{Tu k n } and {FTu k n } for 1 ≤ k ≤ m Thus from the conditions that lim n→∞α1n 0, limn→∞ α k n 1, for k 2, 3, , m and lim n→∞λ1n 0, we have, for k 2, , m,
k
n − u1n
k n u1n 1− α k n
Tu k1 n − λ k n1 t k FTu k1 n
− u1n
u1n 1− α k n
Tu k1 n − λ k n1 t k FTu k1 n
n
3.14
and so
1
n1 − Tu1n
1n u1n 1− α1n
Tu2n − λ1n1 t1FTu2n
− Tu1n
1n
u1n − Tu1n
1− α1n
Tu2n − Tu11 − λ1n1 t1FTu2n
≤ α1n 1n − Tu1n 1n 2n − Tu1n 1n
λ1n1 t1 2
n
≤ α1n 1n − Tu1n 2n − u1n 1n1 t1 2
n
3.15
Trang 7Yen-Cherng Lin 7
Step 3 Let u1n1 − u1n → 0, as n → ∞ Indeed, we observe that
m
n − u m n−1
m n u1n − α m n−1 u1n−11− α m n
T t λ m m u1n −1− α m n−1T λ m n
t m u1n−1
≤ α m n 1n − u1n−1 m n − α m n−1 1n−1 m n λ
m
t m u1n − T t λ m m u1n−1
1− α m n
T t λ m m u1n−1−1− α m n−1T λ m n
t m u1n−1
≤ α m n 1n − u1n−1 m n − α m n−1 1n−1 m n
1− λ m n1 τ m 1
n − u1n−1
α m
n − α m n−1 1n−1 m n
λ m n1−1− α m n−1λ m n · t m 1
n−1
1−1− α m n
λ m n1 τ m 1
n − u1n−1
1− α m
n
λ m n1−1− α m n−1λ m n · t m 1
n−1 m n − α m n−1 1n−1 1n−1
≤ 1n − u1n−1 m n
λ m n1−1− α m n−1λ m n · t m 1
n−1
α m
n − α m n−1 1n−1 1n−1
3.16 and, for 2≤ k ≤ m − 1,
k
n − u k n−1
k n u1n − α k n−1 u1n−11− α k n
T λ k n1
t k u k1 n −1− α k n−1T λ k n
t k u k1 n−1
≤ α k n 1n − u1n−1 k n − α k n−1 1n−1 k n λ
k
n1
t k u k1 n − T λ k n1
t k u k1 n−1
1− α k n
T λ k n1
t k u k1 n−1 −1− α k n−1T λ k n
t k u k1 n−1
≤ α k n 1n − u1n−1 k n − α k n−1 1n−1 k n
1− λ k n1 τ k k1
n − u k1 n−1
αk
λ k n1−1− α k n−1λ k n · t k k1
n−1
α k n 1n − u1n−1 k n − α k n−1 1n−1 k1 n−1
1− α k n
1− λ k n1 τ k k1
λ k n1−1− α k n−1λ k n · t k k1
n−1
3.17
Trang 88 Journal of Inequalities and Applications
m
n − u m n−1 1n − u1n−1 m n − α m n−1 1n−1
λ m n1−1− α m n−1λ m n t m 1
n−1
m−1
λ m n1−1− α m n−1λ m n t m 1
n−1
1− α m−1
λ m−1 n1 −1− α m−1 n−1 λ m−1 n t m−1 m
n−1
α m−1
α m
n − α m n−1 1n−1 1n−1
.
3.18
2
n − u2n−1 1n − u1n−1 m−1
k2
1− α k
n
λ k n1−1− α k n−1λ k n t k k1
n−1
m−1
k2
α k
3.19 Hence it follows from the above inequalities3.17–3.19 that
1
n1 − u1n
1n u1n − α1n−1 u1n−11− α1n
T λ1n1
t1 u2n −1− α1n−1T λ1n
t1u2n−1
≤ α1n 1n − u1n−1 1n − α1n−1 1n−1 1n λ
1
n1
t1 u2n − T λ1n1
t1 u2n−1
1− α1n
T λ1n1
t1 u2n−1−1− α1n−1T λ1n
t1u2n−1
≤ α1n 1n − u1n−1 1n − α1n−1 1n−1 1n
1− λ1n1 τ1 2
n − u2n−1
α1
n − α1n−1 2n−1 1n
λ1n1−1− α1n−1λ1n · t1 2
n−1
α1n 1n − u1n−1 1n − α k n−1 1n−1 k1 n−1
1− α1n
1− λ1n1 τ1 2
n − u2n−1
1− α1
n
λ1n1−1− α1n−1λ1n · t1 2
n−1
3.20
Trang 9Yen-Cherng Lin 9 Let us substitute3.19 into 3.20, then we have
1
n1 − u1n
≤α1n 1− α1n
1− λ1n1 τ1 1
n − u1n−1 m−1
k1
α k
m−1
k1
1− α k
n
λ k n1−1− α k n−1λ k n t k· k1 n−1
1− α m n
λ m n1−1− α m n−1λ m n t m 1
n−1 m n − α m n−1 1n−1 1n−1
1−1− α1n
λ1n1 τ1 1
n − u1n−1 1n
λ1n1 τ1νn δ n,
3.21 where
δnm−1
k1
α k
1 − α1n λ1n1 τ1
1− α m
n
λ m n1−1− α m n−1λ m n t m 1
n−1
m−1
k1
1− α k
n
λ k n1−1− α k n−1λ k n t k k1
n−1
.
3.22
We put
ξ sup 1n :n ≥ 0 sup k n :n ≥ 0, k 1, 2, , m
sup k n :n ≥ 0, k 1, 2, , m ,
M ∗
k1
t k
∗
3.23
Thenδn ≤ 2Mm k1 |α k n − α k n−1 | → 0, as n → ∞, and
νn≤
m
k1
t k
1− α1n
λ1n1 τ1
×
m
k2
1− α k
n
λ k n1−1− α k n−1λ k n 1 − α1
n
λ1n1−1− α1n−1λ1n . 3.24
Trang 1010 Journal of Inequalities and Applications Fromii–iv, we obtain ν n → 0 as n → ∞ Furthermore, from i,∞n1 δn < ∞ ByLemma 2.1,
we deduce thatu1n1 − u1n → 0 as n → ∞.
Step 4 Let u1n − Tu1n → 0 as n → ∞ From Steps2and3, we have
1
n − Tu1n 1n1 − u1n 1n1 − Tu1n 3.25
asn → ∞.
Step 5 Let lim sup n→∞ −Fu∗, Tu k n − u∗ ≤ 0, for k 2, 3, , m Let {Tu1n i } be a subsequence
of{Tu1n } such that
lim sup
n→∞
− Fu∗
, Tu1n − u∗
lim
i→∞
− Fu∗
, Tu1n i − u∗
Without loss of generality, we assume thatTu1n i → u1 weakly for some u1 ∈ H ByStep 4,
we deriveu1n i → u1weakly But byLemma 2.2andStep 4, we have u1 ∈ FixT C Since
u∗is the unique solution of the VIF, C, we obtain
lim sup
n→∞
− Fu∗
, Tu1n − u∗
− Fu∗
, u1− u∗
From the proof ofStep 2,
Tu k n − Tu1n ≤ u2n − u1n −→ 0, as n −→ ∞, 3.28 fork 2, 3, , m Then
lim sup
n→∞
− Fu∗
, Tu k n − u∗
lim sup
n→∞
− Fu∗
, Tu k n − Tu1n
− Fu∗
, Tu1n − u∗
≤ lim sup
n→∞
− Fu∗
, Tu k n − Tu1n
lim sup
n→∞
− Fu∗
, Tu1n − u∗
lim sup
n→∞
− Fu∗
, Tu1n − u∗
≤ 0,
3.29 fork 2, 3, , m.
...Step Let u1n1 − Tu1n → 0, n → ∞ Indeed byStep 1,{u k n } is bounded for ≤ k ≤... weakly for some u1 ∈ H ByStep 4,
we deriveu1n i → u1weakly But byLemma 2.2andStep 4,... n } for ≤ k ≤ m Thus from the conditions that lim n→∞α1n 0, limn→∞ α k n 1, for k 2,