Volume 2007, Article ID 32870, 8 pagesdoi:10.1155/2007/32870 Research Article Iterative Algorithm for Approximating Solutions of Maximal Monotone Operators in Hilbert Spaces Yonghong Yao
Trang 1Volume 2007, Article ID 32870, 8 pages
doi:10.1155/2007/32870
Research Article
Iterative Algorithm for Approximating Solutions of
Maximal Monotone Operators in Hilbert Spaces
Yonghong Yao and Rudong Chen
Received 11 October 2006; Revised 8 December 2006; Accepted 11 December 2006 Recommended by Nan-Jing Huang
We first introduce and analyze an algorithm of approximating solutions of maximal monotone operators in Hilbert spaces Using this result, we consider the convex mini-mization problem of finding a minimizer of a proper lower-semicontinuous convex func-tion and the variafunc-tional problem of finding a solufunc-tion of a variafunc-tional inequality Copyright © 2007 Y Yao and R Chen 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 assume thatH is a real Hilbert space and T : H →2H is a maximal monotone operator A well-known method for solving the equation 0∈ Tv in a
Hilbert spaceH is the proximal point algorithm: x1= x ∈ H and
where{ r n } ⊂(0,∞) andJ r =(I + rT) −1for allr > 0 This algorithm was first introduced
by Martinet [1] Rockafellar [2] proved that if lim infn→∞ r n > 0 and T −10= ∅, then the sequence{ x n }defined by (1.1) converges weakly to an element ofT −10 Later, many re-searchers have studied the convergence of the sequence defined by (1.1) in a Hilbert space; see, for instance, [3–6] and the references mentioned therein In particular, Kamimura and Takahashi [7] proved the following result
Theorem 1.1 Let T : H →2H be a maximal monotone operator Let { x n } be a sequence defined as follows: x1= u ∈ H and
x n+1 = α n u +
1− α n
Trang 2where { α n } ⊂ [0, 1] and { r n } ⊂(0,∞ ) satisfy lim n→∞ α n = 0,∞
n=1α n = ∞ , and lim n→∞ r n =
∞ If T −10= ∅ , then the sequence { x n } defined by ( 1.2 ) converges strongly to Pu, where P
is the metric projection of H onto T −10.
Motivated and inspired by the above result, in this paper, we suggest and analyze an it-erative algorithm which has strong convergence Further, using this result, we consider the convex minimization problem of finding a minimizer of a proper lower-semicontinuous convex function and the variational problem of finding a solution of a variational in-equality
2 Preliminaries
Recall that a mappingU : H → H is said to be nonexpansive if Ux − U y ≤ x − y for allx, y ∈ H We denote the set of all fixed points of U by F(U) A multivalued operator
T : H →2H with domainD(T) and range R(T) is said to be monotone if for each x i ∈
D(T) and y i ∈ Tx i, =1, 2, we have x1− x2,y1− y2 ≥0
A monotone operatorT is said to be maximal if its graph G(T) = {(x, y) : y ∈ Tx }is not properly contained in the graph of any other monotone operator LetI denote the
identity operator onH and let T : H →2H be a maximal monotone operator Then we can define, for eachr > 0, a nonexpansive single-valued mapping J r:H → H by J r =(I + rT) −1 It is called the resolvent (or the proximal mapping) ofT We also define the Yosida
approximationA r byA r =(I − J r)/r We know that A r x ∈ TJ r x and A r x ≤inf{ y :
y ∈ Tx }for allx ∈ H.
Before starting the main result of this paper, we include some lemmas
Lemma 2.1 (see [8]) Let { x n } and { z n } be bounded sequences in a Banach space X and let { α n } be a sequence in [0, 1] with 0 < lim inf n→∞ α n ≤lim supn→∞ α n < 1 Suppose x n+1 =
α n x n+ (1− α n) n for all integers n ≥ 0 and lim sup n→∞( z n+1 − z n − x n+1 − x n )≤ 0 Then, lim n→∞ z n − x n = 0.
Lemma 2.2 (the resolvent identity) For λ, μ > 0, there holds the identity
J λ x = J μ
μ
λ x +
1− μ
λ
J λ x
Lemma 2.3 (see [9]) Let E be a real Banach space Then for all x, y ∈ E
x + y 2≤ x 2+ 2
y, j(x + y)
∀ j(x + y) ∈ J(x + y). (2.2)
Lemma 2.4 (see[10]) Let { a n } be a sequence of nonnegative real numbers satisfying the property a n+1 ≤(1− s n)a n+s n t n , n ≥ 0, where { s n } ⊂ (0, 1) and { t n } are such that
(i)∞
n=0s n = ∞ ,
(ii) either lim sup n→∞ t n ≤ 0 or∞
n=0| s n t n | < ∞ Then { a n } converges to zero.
Trang 33 Main result
LetT : H →2Hbe a maximal monotone operator and letJ r:H → H be the resolvent of
T for each r > 0 Then we consider the following algorithm: for fixed u ∈ H and given
x0∈ H arbitrarily, let the sequence { x n }is generated by
y n≈J r n x n,
x n+1 = α n u + β n x n+δ n y n, (3.1)
where{ α n },{ β n },{ δ n }are three real numbers in [0, 1] and{ r n } ⊂(0,∞) Here the crite-rion for the approximate computation ofy nin (3.1) will be
y n − J r
where∞
n=0σ n < ∞
Theorem 3.1 Let T : H →2H be a maximal monotone operator Assume { α n } , { β n } , { δ n } , and { r n } satisfy the following control conditions:
(i)α n+β n+δ n = 1;
(ii) limn→∞ α n = 0 and∞
n=0α n = ∞ ;
(iii) 0< lim inf n→∞ β n ≤lim supn→∞ β n < 1;
(iv)r n ≥ > 0 for all n and r n+1 − r n →0(n → ∞ ).
If T −10= ∅ , then { x n } defined by ( 3.1 ) under criterion ( 3.2 ) converges strongly to Pu, where
P is the metric projection of H onto T −10.
Proof From T −10= ∅, there existsp ∈ T −10 such thatJ s p = p for all s > 0 Then we have
x n+1 − p ≤ α n u − p +β n x n − p +δ n y n − p
≤ α n u − p +β n x n − p +δ n
σ n+ J r
n x n − p
≤ α n u − p +β n x n − p +δ n x n − p +δ n σ n
= α n u − p +
1− α n x n − p +δ n σ n .
(3.3)
An induction gives that
x n − p ≤max u − p , x0− p + n
k=0
for alln ≥0 This implies that{ x n }is bounded Hence{ J r n x n }and{ y n }are also bounded Define a sequence{ z n }by
x n+1 = β n x n+
1− β n
Trang 4Then we observe that
z n+1 − z n = x n+2 − β n+1 x n+1
1− β n+1 − x n+1 − β n x n
1− β n
=
α
n+1
1− β n+1 − α n
1− β n
u + δ n+1
1− β n+1
y n+1 − y n
+
δ n+1
1− β n+1 − δ n
1− β n
y n
(3.6)
Ifr n−1≤ r n, fromLemma 2.2, using the resolvent identity
J r n x n = J r n −1
r
n−1
r n x n+
1− r n−1
r n
J r n x n
we obtain
J r
n x n − J r n −1x n−1 ≤ r n−1
r n x n − x n−1 +r n − r n−1
r n
J r n x n − x n−1
≤ x n − x n−1 +1
r n−1− r n J r n x n − x n−1 . (3.8)
Similarly, we can prove that the last inequality holds ifr n−1≥ r n
On the other hand, from (3.2), we have
y n+1 − y n ≤ y n+1 − J r n+1 x n+1 + y n − J r
n x n + J r
n+1 x n+1 − J r n x n
≤ σ n+1+σ n+ J r
Thus it follows from (3.5) that
z n+1 − z n − x n+1 − x n
≤ α n+1
1− β n+1 − α n
1− β n
u + y n + δ n+1
1− β n+1 x n+1 − x n
+ δ n+1
1− β n+1
1
r n+1 − r n J r n+1 x n+1 − x n +σ n+1+σ n − x n+1 − x n
≤ α n+1
1− β n+1 − α n
1− β n
u + y n +σ n+1+σ n
+ δ n+1
1− β n+1
1
r n+1 − r n J r n+1 x n+1 − x n ,
(3.10)
which implies that lim supn→∞( z n+1 − z n − x n+1 − x n )≤0 Hence, byLemma 2.1, we have limn→∞ z n − x n =0 Consequently, it follows from (3.5) that
lim
n→∞ x n+1 − x n =lim
n→∞
1− β n z n − x n =0. (3.11)
On the other hand,
x n − y n ≤ x n+1 − x n + x n+1 − y n
≤ x n+1 − x n +α n u − y n +β n x n − y n , (3.12)
Trang 5and so, by (ii), (iii), (3.11), and (3.12), we have limn→∞ x n − y n =0 It follows that
A r
n x n ≤ 1
r n x n − y n + y n − J r
n x n ≤1
x n − y n +σ n
We next prove that
lim sup
n→∞
u − Pu, x n+1 − Pu
whereP is the metric projection of H onto T −10 To prove this, it is sufficient to show lim supn→∞ u − Pu, J r n x n − Pu ≤0, becausex n+1 − J r n x n →0 Now there exists a subse-quence{ x n i } ⊂ { x n }such that
lim
i→∞
u − Pu, J r ni x n i − Pu
=lim sup
n→∞
u − Pu, J r n x n − Pu
Since { J r n x n }is bounded, we may assume that{ J r ni x n i } converges weakly to somev ∈
H Then it follows that v ∈ T −10 Indeed, sinceA r n x n ∈ TJ r n x nandT is monotone, we
have s − J r ni x n i,s − A r ni x n i ≥0, wheres ∈ Ts From A r n x n →0, we obtain s − v, s ≥0 whenevers ∈ Ts Hence, from the maximality of T, we have v ∈ T −10 SinceP is the
metric projection ofH onto T −10, we obtain
lim sup
n→∞
u − Pu, J r n x n − Pu
=lim
i→∞
u − Pu, J r ni x n i − Pu
u − Pu, v − Pu ≤0 (3.16)
That is, (3.14) holds
Finally, to prove thatx n → p, we applyLemma 2.3to get
x n+1 − Pu 2
≤ β n
x n − Pu
+δ n
y n − Pu 2
+ 2α n
u − Pu, x n+1 − Pu
≤β n x n − Pu +δ n x n − Pu +δ n σ n 2
+ 2α n
u − Pu, x n+1 − Pu
=1− α n x n − Pu +δ n σ n 2
+ 2α n
u − Pu, x n+1 − Pu
≤1− α n x n − Pu 2
+ 2α n
u − Pu, x n+1 − Pu
+Mσ n,
(3.17)
whereM > 0 is some constant such that 2(1 − α n)δ n x n − Pu +δ2
n σ n ≤ M An
applica-tion ofLemma 2.4yields that x n − Pu →0 This completes the proof
Remark 3.2 It is clear that the algorithm (3.1) includes the algorithm (1.2) as a special case Our result can be considered as a complement of Kamimura and Takahashi [7] and others
4 Applications
Let f : H →(−∞,∞] be a proper lower semicontinuous convex function Then we can define the subdifferential ∂ f of f by
∂ f (x) = z ∈ H : f (y) ≥ f (x) + y − x, z ∀ y ∈ H
(4.1)
Trang 6for allx ∈ H It is well known that ∂ f is a maximal monotone operator of H into itself;
see Minty [11] and Rockafellar [12,13]
In this section, we investigate our algorithm in the case ofT = ∂ f Our discussion
fol-lows Rockafellar [14, Section 4] IfT = ∂ f , the algorithm (3.1) is reduced to the following algorithm:
y n≈arg min
z∈H
f (z) + 1
2r n
z − x n 2
,
x n+1 = α n u + β n x n+δ n y n, n ∈ N,
(4.2)
with the following criterion:
d
0,S n
y n
≤ σ n
where∞
n=0σ n < ∞,S n(z) = ∂ f (z) + (z − x n)/r n, andd(0, A) =inf{ x :x ∈ A } About (4.3), the following lemma was proved in Rockafellar [2, Proposition 3]
Lemma 4.1 If y n is chosen according to criterion ( 4.3 ), then y n − J r n x n ≤ σ n holds, where
J r n =(I + r n ∂ f ) −1.
Theorem 4.2 Let f : H →(−∞,∞ ] be a proper lower semicontinuous convex function Assume { α n } , { β n } , { δ n } , and { r n } satisfy the same conditions (i)–(iv) as in Theorem 3.1
If (∂ f ) −10= ∅ , then { x n } defined by ( 4.2 ) with criterion ( 4.3 ) converges strongly to v ∈ H, which is the minimizer of f nearest to u.
Proof Putting g n(z) = f (z) + z − x n 2/2r n, we obtain
∂g n(z) = ∂ f (z) + 1
r n
z − x n
for allz ∈ H and J r n x n =(I + r n ∂ f ) −1x n =arg minz∈H g n(z) It follows fromTheorem 3.1 andLemma 4.1that{ x n }converges strongly tov ∈ H and f (v) =minz∈H f (z) This
Next we consider a variational inequality LetC be a nonempty closed convex subset
ofH and let T be a single-valued operator of C into H We denote by VI(C, T) the set of
solutions of the variational inequality, that is,
VI(C, T) = w ∈ X : s − w, Tw ≥0,∀ s ∈ C
A single-valued operatorT is called semicontinuous if T is continuous from each line
segment ofC to H with the weak topology Let F be a single-valued monotone and
semi-continuous operator ofC into H and let N C z be the normal cone to C at z ∈ C, that is,
N C z = { w ∈ H : z − s, w ≥0,∀ s ∈ C } Letting
Az =
⎧
⎨
⎩
Fz + N C z, z ∈ C,
Trang 7we have thatA is a maximal monotone operator (see Rockafellar [14, Theorem 3]) We can also check that 0∈ Av if and only if v ∈VI(C, F) and that J r x =VI(C, F r,x) for all
r > 0 and x ∈ H, where F r,x z = Fz + (z − x)/r for all z ∈ C Then we have the following
result
Corollary 4.3 Let F be a single-valued monotone and semicontinuous operator of C into
H For fixed u ∈ H, let the sequence { x n } be generated by
y n≈VI
C, F r n,x n
,
x n+1 = α n u + β n x n+δ n y n (4.7) Here the criterion for the approximate computation of y n in ( 4.7 ) will be
y n −VI
C, F r n,x n ≤ σ n, (4.8)
where∞
n=0σ n < ∞ Assume { α n } , { β n } , { δ n } , and { r n } satisfy the same conditions (i)–(iv)
as in Theorem 3.1 If VI(C, F) = ∅ , then { x n } defined by ( 4.7 ) with criterion ( 4.8 ) converges strongly to the point of VI(C, F) nearest to u.
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Yonghong Yao: Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, China
Email address:yuyanrong@tjpu.edu.cn
Rudong Chen: Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, China
Email address:chenrd@tjpu.edu.cn
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Hilbert space,” Mathematical Programming, vol 87, no 1, pp 189–202, 2000.