A note on the hybrid steepest descent methods

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by  Koji Aoyama

A NOTE ON THE HYBRID STEEPEST DESCENT METHODS
KOJIAOYAMAANDYASUNORIKIMURA
1
1
Abstract. Theaimofthispaperistoprovethat, inanappropriatesetting,
0
every iterative sequence generated by the hybrid steepest descent method is
2
convergent whenever so is every iterative sequence generated by the Halpern

n typeiterativemethod.
a
J

5

1. Introduction
]
A
In this paper, we consider the following variational inequality problem in a real
F
Hilbert space H: Find z ∈F such that
.
h
hy−z,Azi≥0 for all y ∈F,
t
a
m where F is the set of common fixed points of a sequence {Tn} of nonexpansive
mappings on H and A is a strongly monotone and lipschitzian mapping on H.
[
Then we study convergence of the iterative sequence {xn} defined by x1 ∈H and
1
v (1.1) xn+1 =(I −λnA)Tnxn
1 for n∈N in orderto approximatethe solution, where I is the identity mapping on
8
H. If A=I −u for some u∈H, then it is clear that A is strongly monotone and
8
0 lipschitzian, and (1.1) is reduced to
.
1 (1.2) xn+1 =λnu+(1−λn)Tnxn.
0
We dealwith these two types ofiterations,and especially we focus onthe relation-
1
1 ship between them; see §3.
: The iterative method defined by (1.1) is called the hybrid steepest descent
v
i method, which was introduced by Yamada [32]. He considered the variational
X
inequality problem over the set of common fixed points of a finite family of non-
r expansive mappings and proved strong convergence of the sequence generated by
a
the method. We know many results by using the hybrid steepest descent method;
see [2,10–12,14,15,17–20,25,30,31,33–38].
Theiterativemethoddefinedby(1.2)iscalledtheHalperntypeiterativemethod;
see Halpern [13], Wittmann [27], and Shioji and Takahashi [21]; see also [1,4,5].
2. Preliminaries
Throughout the present paper, H denotes a real Hilbert space with the inner
product h·, ·i and the norm k·k, I the identity mapping on H, and N the set of
positive integers.
Date:January6,2011.
2010 Mathematics Subject Classification. 47J20,47H09,47H10.
Key words and phrases. hybrid steepest descent method, variational ineqality, nonexpansive
mapping,fixedpoint.
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2 KOJIAOYAMAANDYASUNORIKIMURA
Let C be a nonempty closed convex subset of H. A mapping S: C →C is said
to be lipschitzian if there exists a constant η >0 such that kSx−Syk≤ηkx−yk
for all x,y ∈ C. In this case, S is called an η-lipschitzian mapping. In particular,
an η-lipschitzian mapping is said to be nonexpansive if η = 1; an η-lipschitzian
mapping is said to be a contraction if 0≤η <1. It is known that Fix(S) is closed
and convex if S is nonexpansive, where Fix(S) denotes the set of fixed points of
S. The metric projection of H onto C is denoted by P and we know that P is
C C
nonexpansive. We also know the following; see [23].
Lemma2.1. Letx∈H andz ∈C. Then z =P (x)if andonlyif hy−z,x−zi≤
C
0 for all y ∈C.
Let C be a nonempty closed convex subset of H. Let {S } be a sequence of
n
self-mappings of C. We say that {S } satisfies the condition (Z) if every weak
n
cluster point of{x } is a commonfixed point of {S } whenever {x } is a bounded
n n n
sequence in C and x −S x →0; see [1,3,6–9].
n n n
A mapping A: H → H is said to be strongly monotone if there is a constant
2
κ > 0 such that hx−y,Ax−Ayi ≥ κkx−yk for all x,y ∈ H. In this case, A is
calledaκ-stronglymonotonemapping. In§3,wedealwiththefollowingvariational
inequality problem:
Problem 2.2. Let κ and η be positive real numbers such that η2 < 2κ. Let F a
nonempty closed convex subset of H and A: H → H a κ-strongly monotone and
η-lipschitzian mapping. Then find z ∈F such that
hy−z,Azi≥0 for all y ∈F.
The set of solution of Problem 2.2 is denoted by VI(F,A). It is known that the
solution set is a singleton; see Lemma 2.4 below.
Remark 2.3. Theassumptionthatη2 <2κinProblem2.2isnotrestrictive. Indeed,
suppose that a κ-strongly monotone and η-lipschitzian mapping A is given. Let
us choose a positive constant µ such that µ < 2κ/η2, and define κ′ = µκ and
η′ = µη. Then it is easy to verify that (η′)2 < 2κ′, µA is κ′-strongly monotone
′
andη -lipschitzian,andmoreover,VI(F,A)=VI(F,µA)foreverynonemptyclosed
convex subset F of H.
Lemma 2.4. Under the assumptions of Problem 2.2, the following hold:
(1) κ ≤ η, 0 ≤ 1 − 2κ +η2 < 1 and I − A is a θ-contraction, where θ =
p1−2κ+η2.
(2) Problem 2.2 has a unique solution and VI(F,A)=Fix P (I −A) .
(cid:0) F (cid:1)
Proof. We first prove (1). Since A is κ-strongly monotone and η-lipschitzian, it
follows that
2 2
κkx−yk ≤hx−y,Ax−Ayi≤kx−ykkAx−Ayk≤ηkx−yk
and
2 2 2
k(I−A)x−(I −A)yk =kx−yk −2hx−y,Ax−Ayi+kAx−Ayk
≤kx−yk2−2κkx−yk2+η2kx−yk2
= 1−2κ+η2 kx−yk2
(cid:0) (cid:1)
A NOTE ON THE HYBRID STEEPEST DESCENT METHODS 3
for all x,y ∈ H. Therefore, κ ≤ η. By assumption, it is easy to check that
0≤1−2κ+η2 <1, and thus I −A is a θ-contraction.
We nextprove(2). Since I−Aisa contractionby(1)andthe metricprojection
P is nonexpansive, P (I −A) is a contraction on H. The Banach contraction
F F
principle guarantees that P (I −A) has a unique fixed point. On the other hand,
F
Lemma 2.1 shows that VI(F,A) = Fix P (I −A) and thus Problem 2.2 has a
(cid:0) F (cid:1)
unique solution. (cid:3)
We know the following result; see [2,5] and see also [3,9].
Theorem 2.5. Let H be a Hilbert space, C a nonempty closed convex subset of H,
{T } a sequence of nonexpansive self-mappings of C with a common fixed point, F
n
the set of common fixed points of {T }, and {λ } a sequence in [0,1] such that
n n
∞ ∞
(2.1) λn →0, Xλn =∞, and X|λn+1−λn|<∞.
n=1 n=1
Suppose that {T } satisfies the condition (Z) and
n
∞
(2.2) Xsup{kTn+1y−Tnyk:y ∈D}<∞
n=1
for every nonempty bounded subset D of C. Let x,u be points in C and {x } a
n
sequence defined by x1 = x and (1.2) for n ∈ N. Then {xn} converges strongly to
P (u).
F
We also know the following result; see [1,4].
Theorem 2.6. Let H be a Hilbert space, C a nonempty closed convex subset of H,
{S } a sequence of nonexpansive self-mappings of C with a common fixed point, F
n
the set of common fixed points of {S }. Let {λ } and {β } be a sequences in [0,1]
n n n
such that
∞
λn →0, Xλn =∞, and 0<liminfβn ≤limsupβn <1.
n→∞ n→∞
n=1
Suppose that {S } satisfies the condition (Z) and
n
(2.3) sup{kSn+1y−Snyk:y ∈D}→0
for every nonempty bounded subset D of C. Let x,u be points in C and {x } a
n
sequence defined by x1 =x and
xn+1 =λnu+(1−λn)(cid:0)βnxn+(1−βn)Snxn(cid:1)
for n∈N. Then {x } converges strongly to P (u).
n F
The following lemma is well known; see [5,16,26,28,29].
Lemma 2.7. Let {ǫ } be a sequence of nonnegative real numbers, {γ } a sequence
n n
ofrealnumbers,and{λn}asequencein[0,1]. Supposethatǫn+1 ≤(1−λn)ǫn+λnγn
for every n∈N, limsupn→∞γn ≤0, and P∞n=1λn =∞. Then ǫn →0.
4 KOJIAOYAMAANDYASUNORIKIMURA
3. Convergence theorems by the hybrid steepest descent method
In this section, we deal with the variational inequality problem over the set of
common fixed points of a sequence of nonexpansive mappings; see Problem 3.1
below. We first investigate the relationship between the hybrid steepest descent
method and the Halpern type iterative method (Theorem 3.2). Then, by using
Theorems 2.5 and 2.6, we show two convergence theorems by the hybrid steepest
descent method for this problem.
Problem 3.1. Let H be a Hilbert space, {T } a sequence of nonexpansive self-
n
mappings of H with a common fixed point, F the set of common fixed points of
{T }, and A: H →H a κ-strongly monotone and η-lipschitzian mapping, where κ
n
and η are positive real numbers such that η2 <2κ. Then find z ∈F such that
hy−z,Azi≥0 for all y ∈F.
Usingthetechniquein[22],wecanprovethefollowingtheorem,whichshowsthat
everysequencegeneratedbythe hybridsteepestdescentmethod forProblem3.1is
convergent whenever so is every sequence generated by the Halpern type iterative
method for the sequence of nonexpansive mappings.
Theorem 3.2. Let H, {T }, F, κ, η, and A be the same as in Problem 3.1.
n
∞
Let {λ } be a sequence in [0,1] such that λ = ∞. Suppose that for any
n Pn=1 n
(x,u)∈H ×H, the sequence {xn} defined by x1 =x and
(3.1) xn+1 =λnu+(1−λn)Tnxn
for n∈N converges strongly to P (u). Let y be a point in H and {y } a sequence
F n
defined by y1 =y and
(3.2) yn+1 =(I −λnA)Tnyn
for n∈N. Then {y } converges strongly to the unique solution of Problem 3.1.
n
Proof. Setf =(I−A)T forn∈N. SinceT isnonexpansive,f isaθ-contraction
n n n n
on H by Lemma 2.4, where θ =p1−2κ+η2. Let w be the fixed point of PF ◦f1
and {xn} a sequence defined by x1 =y and
xn+1 =λnf1(w)+(1−λn)Tnxn
for n∈N. Then
(3.3) xn →PF(cid:0)f1(w)(cid:1)=w
by assumption. Since T is nonexpansive and f is a θ-contraction, it follows from
n n
f1(w)=fn(w) that
kxn+1−yn+1k=(cid:13)(1−λn)(Tnxn−Tnyn)+λn(cid:0)f1(w)−fn(yn)(cid:1)(cid:13)
(cid:13) (cid:13)
≤(1−λ )kT x −T y k+λ kf (w)−f (y )k
n n n n n n n n n
≤(1−λ )kx −y k+λ θkw−y k
n n n n n
≤(1−λ )kx −y k+λ θ(kw−x k+kx −y k)
n n n n n n n
θ
≤ 1−(1−θ)λ kx −y k+(1−θ)λ kx −wk
(cid:0) n(cid:1) n n n n
1−θ
for every n ∈ N. Since ∞ (1 − θ)λ = ∞, (3.3) and Lemma 2.7 show that
Pn=1 n
x −y →0. Therefore,weconclude that {y } convergesstronglyto w =P (I−
n n n F(cid:0)
A NOTE ON THE HYBRID STEEPEST DESCENT METHODS 5
A)T1w(cid:1)=PF(I−A)w, whichisthe unique solutionofProblem3.1byLemma2.4.
This completes the proof. (cid:3)
Using Theorems 2.5 and 3.2, we obtain the following:
Theorem 3.3. Let H, {T }, F, κ, η, and A be the same as in Problem 3.1. Let
n
{λ } be a sequence in [0,1] such that (2.1) holds. Suppose that {T } satisfies the
n n
condition (Z) and (2.2) holds for every nonempty bounded subset D of C. Let y be
a point in H and {yn} a sequence defined by y1 = y and (3.2) for n ∈ N. Then
{y } converges strongly to the unique solution of Problem 3.1.
n
Proof. Let (x,u) ∈ H ×H be fixed. Then it follows from Theorem 2.5 that the
sequence {xn} defined by x1 =x and (3.1) for n∈N converges strongly to PF(u).
Therefore, Theorem 3.2 implies the conclusion. (cid:3)
Using Theorem 3.2 and other known results, we also obtain the following:
Theorem 3.4 (Iemoto and Takahashi [14, Theorem 3.1]). Let H, {T }, F, κ, η,
n
and A be the same as in Problem 3.1. Let {λ } be a sequence in [0,1] such that
n
∞
λn →0 and Xλn =∞
n=1
and {γ } a sequence in [a,b], where 0 < a ≤ b < 1. For each n ∈ N and k ∈
n
{1,2,...,n+1}, define a mapping Un,k by Un,n+1 =I and
Un,k =γkTkUn,k+1+(1−γk)I.
Let y be a point in H and {yn} a sequence defined by y1 =y and
(3.4) yn+1 =(I−λnA)Un,1yn
for n∈N. Then {y } converges strongly to the unique solution of Problem 3.1.
n
Proof. Set Sn = T1Un,2 for n ∈ N. Then it is clear that each Sn is nonexpansive.
It is known that
n
Fix(Sn)=Fix(Un,1)= \ Fix(Tk)
k=1
by [24, Lemma 3.2]; see also [7, Lemma 4.2]. Hence we have
∞ ∞ ∞ n
\ Fix(Sn)= \ Fix(Un,1)= \ \ Fix(Tk)=F.
n=1 n=1 n=1k=1
It is also known that {S } satisfies the condition (Z) and (2.3) holds for every
n
nonempty bounded subset D of H; see [7], [9], [3], and [4]. Thus, for any (x,u) ∈
H×H, it follows from Theorem2.6 that the sequence {xn} defined by x1 =x and
xn+1 =λnu+(1−λn)Un,1xn =λnu+(1−λn)(cid:0)(1−γ1)xn+γ1Snxn(cid:1)
for n∈N convergesstronglyto P (u). Therefore,Theorem3.2 implies the conclu-
F
sion. (cid:3)
6 KOJIAOYAMAANDYASUNORIKIMURA
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(KojiAoyama)DepartmentofEconomics,ChibaUniversity,Yayoi-cho,Inage-ku,Chi-
ba-shi,Chiba 263-8522,Japan
E-mail address: [email protected]
(Yasunori Kimura) Department of Mathematical and Computing Sciences, Tokyo In-
stitute of Technology,Ookayama,Meguro-ku,Tokyo 152-8552,Japan
E-mail address: [email protected]