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Trang 1On: 15 September 2014, At: 18:28
Publisher: Taylor & Francis
Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK
Numerical Functional Analysis and Optimization
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Parallel Hybrid Methods for a Finite Family of Relatively Nonexpansive Mappings
Pham Ky Anh a & Cao Van Chung a a
Department of Mathematics , Vietnam National University , Hanoi , Vietnam Accepted author version posted online: 06 Aug 2013.Published online: 01 Apr 2014
To cite this article: Pham Ky Anh & Cao Van Chung (2014) Parallel Hybrid Methods for a Finite Family of Relatively
Nonexpansive Mappings, Numerical Functional Analysis and Optimization, 35:6, 649-664, DOI: 10.1080/01630563.2013.830127
To link to this article: http://dx.doi.org/10.1080/01630563.2013.830127
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Trang 2Copyright © Taylor & Francis Group, LLC
ISSN: 0163-0563 print/1532-2467 online
DOI: 10.1080/01630563.2013.830127
PARALLEL HYBRID METHODS FOR A FINITE FAMILY OF
RELATIVELY NONEXPANSIVE MAPPINGS
Pham Ky Anh and Cao Van Chung
Department of Mathematics, Vietnam National University, Hanoi, Vietnam
In this article, we propose two parallel hybrid methods for finding a common fixed point of
a finite family of relatively nonexpansive mappings The strong convergence of the methods is established and their effectiveness are examined by numerical experiments Thanks to the parallel computation, we can reduce the overall computational effort under widely used assumptions on mappings and spaces.
Keywords Common fixed point; Hybrid method; Parallel computation; Relatively
nonexpansive mapping.
Mathematics Subject Classification 47H09; 47J25; 65J15; 65Y05.
1 INTRODUCTION Various problems of science and engineering, such as the convex feasibility problems with applications in optimization theory, image processing, radiation therapy treatment planning, etc (see [1]), can be reduced to a problem of finding a common fixed point of a family of nonexpansive mappings
In 2005, Matsushita and Takahashi [2] proposed the following hybrid
method, called a CQ algorithm, for finding a fixed point of a relatively
nonexpansive mapping:
x0∈ C chosen arbitrarily,
y n := J−1(n Jx n+ (1 − n )JTx n),
C n := z ∈ C : (z, y n)≤ (z, x n),
Q n := z ∈ C : x n − z, Jx0− Jx n ≥ 0,
x n+1:= C n ∩Q n (x0)
(1.1)
Received 24 March 2012; Revised 21 July 2013; Accepted 25 July 2013.
Address correspondence to Pham Ky Anh, Department of Mathematics, Vietnam National University, 334 Nguyen Trai, Thanh Xuan, Hanoi, Vietnam; E-mail: anhpk@vnu.edu.vn.
649
Trang 3Several attempts to generalize the CQ method (1.1) for finding
a common fixed point of a finite or infinite family of (relatively) nonexpansive mappings have recently made by Takahashi and Zembayashi [3], Plubtieng and Ungchittrakool [4, 5], Reich and Sabach [6, 7], Su, Wang, and Xu [8], Cholamjiak and Suantai [9], etc Very recently, Liu [10]
proposed the following cyclic CQ method for a finite family of N relatively
nonexpansive mappings:
x0 ∈ C chosen arbitrarily,
y n := J−1
n Jx n+ (1 − n )JT [n] x n
,
C n := z ∈ C : (z, y n)≤ n (z, x0)+ (1 − n )(z, x n),
Q n := z ∈ C : x n − z, Jx0− Jx n ≥ 0,
x n+1:= C n ∩Q n (x0), where T [n] := T n(modN )
(1.2)
Clearly, Liu’s algorithm is inherently sequential, hence it will be costly
on a single processor when the number of operators N is large.
In this article we introduce some parallel CQ methods which may
be regarded as a counterpart of the cyclic one Our idea consists of
determining synchronously the vectors y i
n for each operator T i , i =
1, 2, N by N parallel processors Then we find the farthest element y n :=
y i n
n from the previous approximation x n and construct the corresponding
sets C n and Q n The next approximation is defined as the generalized
projection of x0 onto the set C n ∩ Q n
The benifit of our approach is clear Based on the parallel computation we can reduce the overall computational effort under widely used conditions on the space and mappings (cf [2–5, 8, 9]) Moreover, the additional computation cost of our method is essentially negligible Further, for mappings acting in a real Hilbert space, we
modify the algorithm (1.2) by choosing the sets C n , Q n as halfspaces,
hence the generalized projection of x0 onto the set C n ∩ Q n can be computed effectively (see [11]) Finally, we perform both parallel and
sequential CQ methods for finding a common fixed point of two
nonlinear nonexpansive integral operators and compare the obtained results
The remainder of this article is organized as follows In section 2,
we recall some notations and results needed for our further study
Section 3 deals with the convergence analysis of the parallel CQ
algorithm, while in section 4 we provide a modified algorithm for the Hilbert space case A numerical experiment is considered in the final section 5
Trang 42 PRELIMINARIES
Let E be a real Banach space and E∗ be its dual space For each f ∈
E∗ and x ∈ E, we denote x, f := f (x) Let J be the normalized duality
mapping defined by
E = f 2
E∗
We first recall some facts about the geometry of Banach spaces (see [12,
13, 16] for details)
i) E (E∗) is uniformly convex if and only if E∗(E ) is uniformly smooth ii) if E is uniformly convex, then it is reflexive and strictly convex and it
satisfies the Kadec-Klee (or Efimov-Stechkin) property
iii) if E is uniformly smooth and uniformly convex then J and J−1= J∗ are single-valued and uniformly norm-to-norm continuous on bounded
subsets of E and E∗, respectively
Now let E be a smooth, strictly convex and reflexive real Banach space For all x, y ∈ E, we consider the functional : E × E → + defined by
(x, y) = x2− 2x, J (y) + y2 ∀(x, y) ∈ E × E (2.1) Clearly, (x − y)2 ≤ (x, y) ≤ (x + y)2 Let C ⊂ E be a nonempty, convex and closed set The generalized metric projection from E onto C
is defined as follows
C (x):= arg min
z ∈C (z, x) ∀ x ∈ E
It is proved that the minimizer C (x) exists and is unique Besides, in a Hilbert space H , (2.1) reduces to (x, y) = x − y2and C coincides with
the well-known metric projection P C
We have the following properties of the functional and the generalized projection C (see [4, 14–18])
Lemma 2.1 Let E be a reflexive, strictly convex and smooth real Banach space.
Lemma 2.2 Let E be a uniformly convex and smooth real Banach space, x n
Trang 5Lemma 2.3 Let E be a reflexive, strictly convex and smooth real Banach space,
(z, C (x))+ (C (x), x) ≤ (z, x) (2.3)
Let T : C → C be a mapping with a nonempty set of fixed points
A point p in C is said to be an asymptotic fixed point of T if C contains
a sequence x n such that x n p and T (x n)− x n → 0 as n → ∞ We denote the set of all asymptotic fixed point of T by F (T ) The operator T
is said to be relatively nonexpansive if ˆF (T ) ≡ F (T ) and (p, T (x)) ≤
(p, x) for all p ∈ F (T ), x ∈ C.
The structure of the fixed point set of a relatively nonexpansive mapping is described in the following theorem
Lemma 2.4 ([15, 19]) Let E be a strictly convex and smooth real Banach
space, C be a closed convex subset of E , and T be a relatively nonexpansive mapping from C into itself Then F (T ) is closed and convex.
3 STRONG CONVERGENCE OF A PARALLEL HYBRID ALGORITHM
Throughout this section we assume that C is a nonempty, closed and
convex subset of a uniformly convex and uniformly smooth real Banach
space E , and T i : C → C, i = 1, 2, N , is a finite family of relatively
nonexpansive mappings Moreover, suppose that
N
i=1
For finding an element x ∈ F , we propose the following parallel CQ
method
Algorithm 1 Let x0 ∈ C be an arbitrarily chosen initial approximation and
k⊂ (0, 1) be a vanishing numerical sequence For k ≥ 0, assuming x k is known, we
• Calculate
y i := J−1(k J (x k)+ (1 − k )J (T i (x k))), i = 1, 2, , N (3.2)
Trang 6• Find
i k:= arg max
• Define
C k := v ∈ C : (v, y i k
k)≤ (v, x k), and
Q k := u ∈ C : x k − u, J (x0)− J (x k) ≥ 0
• Compute
x k+1:= C k ∩Q k (x0) (3.4)
• If x k+1= x k then stop Else, set k := k + 1 and repeat.
The following lemma shows that the Algorithm 1 is well defined Lemma 3.1 If Algorithm 1 reaches a step k ≥ 0, then F ⊂ C k ∩ Q k and x k+1is well defined.
Proof Obviously, Q k is closed and convex for all k ≥ 0 Further, the relation
(v, y i k
k)≤ (v, x k)
is equivalent to
v, J (x k)− J (y i k
k) ≤ 1
2xk2− y i k
k2
,
which shows that C k is also convex and closed Thus, both sets C k and Q k
are closed and convex
Noting that the function : C → , (x) := x2 is convex, for all p ∈
F ⊂ F (T i k), we have
(p, y i k
k)= (p, J−1(k J (x k)+ (1 − k )J (T i k (x k))))
= p2− 2p, k J (x k)+ (1 − k )J (T i k (x k))
+ k J (x k)+ (1 − k )J (T i k (x k))2
≤ p2− 2k p, J (x k) − 2(1 − k)p, J (Ti k (x k))
+ k x k2+ (1 − k)Ti k (x k)2
≤ k (p, x k)+ (1 − k )(p, T i k (x k))
≤ (p, x )+ (1 − )(p, x )= (p, x )
Trang 7Hence, p ∈ C k Thus, F ⊂ C k for all k ≥ 0.
Clearly, F ⊂ Q0 ≡ C Assume that F ⊂ Q k−1 for some k ≥ 1, we will
show that F ⊂ Q k Indeed, since x k= C k−1∩Q k−1(x0), property (2.2) implies that x k − z, J (x0)− J (x k) ≥ 0 for all z ∈ C k−1∩ Q k−1 Therefore, from F ⊂
C k−1∩ Q k−1 we have x k − z, J (x0)− J (x k) ≥ 0 for all z ∈ F Hence, the definition of Q k ensures that F ⊂ Q k
Thus,
F ⊂ C k ∩ Q k for all k ≥ 0
Since F = ∅, the next approximation x k+1 is well defined
We have the following convergent results of Algorithm 1
Lemma 3.2 If Algorithm 1 finishes at a finite iteration k < ∞, then x k is a
Proof For all x ∈ E, C (x) is uniquely defined From stopping rule of Algorithm 1, we see that if it finishes at step k < ∞, then x k = x k+1:=
C k ∩Q k (x0)∈ C k Using the definition of C k , we have (x k , y i k
k)≤ (x k , x k)=
0 Hence, (x k , y i k
k)= 0 From Lemma 2.1, it follows x k = y i k
k By the
definition of i k , we have x k = y i
k for i = 1, N Taking into account y i
k = x k and (3.2), we have
J (x k)= k J (x k)+ (1 − k )J (T i (x k))
Hence, J (x k)= J (T i (x k )), i = 1, N
Since E is reflexive, uniformly convex and uniformly smooth, J and J−1
is uniformly norm-to-norm continuous Therefore, x k = T i (x k ), i = 1, N
Otherwise, we are able to prove the main convergence theorem Theorem 3.1 Let x k be the (infinite) sequence generated by Algorithm 1 and
Proof Since T i is relatively nonexpansive, F (T i) is convex and closed for
i = 1, N Therefore, F is a convex, closed set and there exists a unique element x† = F (x0) Using (2.2) and the definition of Q k , we get x k=
Q k (x0) From x k+1= C k ∩Q k (x0)∈ Q k and Lemma 2.3, it follows (x k , x0)≤
(x k+1, x0) Therefore, the sequence (x k , x0) is nondecreasing Taking
into account the relations x†∈ F ⊂ Q k and x k= Q k (x0) and using (2.3),
we find
(x , x )≤ (x†
, x )− (x†
, x )≤ (x†
, x ), ∀ k ≥ 0 (3.5)
Trang 8The last inequalities ensure the boundedness of x k and (x k , x0) Hence, there exists a finite limit limk→∞(x k , x0) From x k+1= C k ∩Q k (x0)∈ Q k and
x k = Q k (x0), we have
(x k+1, x k)≤ (x k+1, x0)− (x k , x0)
The last relation implies that limk→∞(x k+1, x k)= 0, hence, by Lemma 2.2, limk→∞x k+1− x k = 0 Using the definition of C k and the inclusion x k+1∈
C k, we also have
(x k+1, y i k
k)≤ (x k+1, x k)
Due to limk→∞(x k+1, x k)= 0, from the last inequality we have
(x k+1, y i k
k)→ 0 as k → ∞
Again, Lemma 2.2 ensures that limk→∞x k+1− y i k
k = 0
Using this relation andx k − y i k
k ≤ x k+1− y i k
k + x k − x k+1, we find
x k − y i k
k → 0 as k → ∞
By the definition of i k, we also have x k − y i
k → 0 as k → ∞ for i =
1, 2, , N Hence, y i
k is also bounded for i = 1, 2, , N
From (3.2) it follows that
J (T i (x k))− J (y i
k) = J (Ti (x k))− n J (x k)+ (1 − n )J (T i (x k))
= n J (x k)− J (T i (x k)), i = 1, 2, , N The relative nonexpansiveness of T i and the boundedness of x k yield the
boundedness of T i (x k ) for all i = 1, N Since E is uniformly smooth, J
is uniformly continuous on every bounded subsets Therefore, J (x k)−
J (T i (x k)) is bounded Using limk→∞k = 0, we find
J (T i (x k))− J (y i
k) → 0 as k → ∞, (i = 1, N )
Since E is reflexive, uniformly convex, J−1 is uniformly norm-to-norm continuous on every bounded subsets Hence,
T i (x k)− y i
k → 0 as k → ∞, (i = 1, N )
Using this relation together with limk→∞x k − y i
k → 0 and x k − T i (x k) ≤
x k − y i
k + T i (x k)− y i
k, we have lim
k→∞x k − T i (x k) = 0, i = 1, N
Trang 9From the boundedness of x k , there exists a subsequence x k j such that
x k j ˜x and lim j→∞x k j − T i (x k j) = 0 Since Ti is relatively nonexpansive,
we have ˜x ∈ F (T i)= F (T i ), i = 1, N Hence, ˜x ∈ F
Using x k+1= C k ∩Q k (x0) and x†∈ F ⊂ C k ∩ Q k , we have (x k+1, x0)≤
(x†, x0) On the other hand, from the weak lower semicontinuity of the norm, we obtain
(˜x, x0)= ˜x2− 2˜x, J (x0) + x02
≤ lim inf
j→∞ xk j2− 2x k j , J (x0) + x02
≤ lim inf
j→∞ (x k j , x0)≤ lim sup
j→∞ (x k j , x0)
≤ (x†
, x0)
From definition of x† = F (x0), it follows ˜x = x† Hence, limj→∞(x k j , x0)=
(x†, x0) and limj→∞x k j = x† Using the Kadec-Klee property of E, we obtain that x k j converges strongly to F (x0) Since x k j is an arbitrary
weakly convergent sequence of x k , we have x k→ F (x0) as k → ∞ Remark 3.1 At each step k ≥ 0, y i
k can be computed simultaneously by
N parallel processors.
Remark 3.2 It is easy to find the optimal index i k by (3.3) Therefore, the additional computation cost in Algorithm 1 is essentially negligible Remark 3.3 In general, it is not easy to determine the generalized
projection x k+1 by (3.4) However in a Hilbert space we can modify
Algorithm 1, so that the sets C k and Q k are half spaces, hence x k+1 can be effectively computed
4 A PARALLEL HYBRID ALGORITHM IN HILBERT SPACES
Now suppose C is a nonempty, closed and convex subset of a real Hilbert space H In this case, for all x, y ∈ H , we have (x, y) ≡ x − y2,
J ≡ I (identity mapping in H ) and C (x) ≡ P C (x) Obviously, the relations
(2.2) and (2.3) still hold Moreover, we have ([22])
P C (x) − P C (y)2≤ x − y2− (P C (x) − x) − (P C (y) − y)2 ≤ x − y2
(4.1)
Let T i (i = 1, N ) be a family of relatively nonexpansive mappings from C into itself and assume that the set F := N
i=1F (T i) is not empty
We consider the following modified algorithm
Trang 10Algorithm 2 Let x0 ∈ C be an arbitrarily chosen element and k⊂ (0, 1) be
• Perform
z k := P C (x k)
• Calculate
y k i := k z k+ (1 − k )T i (z k), i = 1, 2, , N (4.2)
• Find
i k:= arg max
k − x k
• If y i k
k − x k = 0 then stop Else:
• Define
C k := v ∈ H : v − y i k
k ≤ v − x k , and
(4.3)
Q k := u ∈ H : x0− x k , x k − u ≥ 0
• Compute
x k+1:= P C k ∩Q k (x0) (4.4)
• If x k+1= x k then stop Else, set k := k + 1 and repeat.
For each k ≥ 0, it is easy to see that Q k is a halfspace or Q k = H
Further, the relation v − y i k
k ≤ v − x k is equivalent to v, x k − y i k
k ≤ 1
k2 or
2(x k + y i k
k ), x k − y i k
k
Hence, for all k ≥ 0, C k is a halfspace in H or C k = H An explicit formula for P C k ∩Q k (x0) can be obtained similarly as in [11] Therefore, if C k ∩ Q k=
∅ then x k+1is easily computed by (4.4) We have the following convergence results for Algorithm 2
Lemma 4.1 If Algorithm 2 finishes at a step k < ∞, then x k is a common fixed