Volume 2008, Article ID 745010, 7 pagesdoi:10.1155/2008/745010 Research Article Strong Convergence Theorems for Nonexpansive Semigroups without Bochner Integrals Satit Saejung Department
Trang 1Volume 2008, Article ID 745010, 7 pages
doi:10.1155/2008/745010
Research Article
Strong Convergence Theorems for Nonexpansive Semigroups without Bochner Integrals
Satit Saejung
Department of Mathematics, Faculty of Science, Khon Kaen University, Khon Kaen 40002, Thailand
Correspondence should be addressed to Satit Saejung, saejung@kku.ac.th
Received 28 November 2007; Revised 15 January 2008; Accepted 30 January 2008
Recommended by William A Kirk
We prove a convergence theorem by the new iterative method introduced by Takahashi et al 2007 Our result does not use Bochner integrals so it is different from that by Takahashi et al We also cor-rect the strong convergence theorem recently proved by He and Chen 2007.
Copyright q 2008 Satit Saejung 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 the inner product ·, · and the norm · Let {Tt : t ≥ 0}
be a family of mappings from a subsetC of H into itself We call it a nonexpansive semigroup
onC if the following conditions are satisfied:
1 T0x x for all x ∈ C;
2 Ts t TsTt for all s, t ≥ 0;
3 for each x ∈ C the mapping t → Ttx is continuous;
4 Ttx − Tty ≤ x − y for all x, y ∈ C and t ≥ 0.
Motivated by Suzuki’s result1 and Nakajo-Takahashi’s results 2, He and Chen 3 recently proved a strong convergence theorem for nonexpansive semigroups in Hilbert spaces by hy-brid method in the mathematical programming However, their proof of the main result3, Theorem 2.3 is very questionable Indeed, the existence of the subsequence {sj} such that
2.16 of 3 are satisfied, that is,
s j −→ 0, x j − Ts j
x j
s j −→ 0, 1.1 needs to be proved precisely So, the aim of this short paper is to correct He-Chen’s result and also to give a new result by using the method recently introduced by Takahashi et al
Trang 2We need the following lemma proved by Suzuki4, Lemma 1.
Lemma 1.1 Let {tn } be a real sequence and let τ be a real number such that lim inf n t n ≤ τ ≤
lim supn t n Suppose that either of the following holds:
i lim supn t n1 − t n ≤ 0, or
ii lim infn t n1 − t n ≥ 0.
Then τ is a cluster point of {t n } Moreover, for ε > 0, k, m ∈ N, there exists m0≥ m such that |t j −τ| < ε
for every integer j with m0≤ j ≤ m0 k.
2 Results
2.1 The shrinking projection method
The following method is introduced by Takahashi et al in5 We use this method to approx-imate a common fixed point of a nonexpansive semigroup without Bochner integrals as was the case in5, Theorem 4.4
Theorem 2.1 Let C be a closed convex subset of a real Hilbert space H Let {Tt : t ≥ 0} be a
nonexpansive semigroup on C with a nonempty common fixed point F, that is, F ∩ t≥0 FTt / ∅ Suppose that {x n } is a sequence iteratively generated by the following scheme:
x0∈ H taken arbitrary,
C1 C,
x1 P C1
x0
,
y n α n x n1− α n Tt n
x n ,
C n1z ∈ C n:y n − z ≤ x n − z,
x n1 P C n1
x0
.
2.1
where {α n } ⊂ 0, a ⊂ 0, 1, lim inf n t n 0, lim sup n t n > 0, and lim n t n1 − t n 0 Then x n →
P F x0.
Proof It is well known that F is closed and convex We first show that the iterative scheme is
well defined To see that eachC nis nonempty, it suffices to show that F ⊂ Cn The proof is by induction Clearly,F ⊂ C1 Suppose thatF ⊂ C k Then, forz ∈ F ⊂ C k,
y k − z ≤ α k x k − z 1 − α k Tt k
x k − z
≤ α k x k − z 1 − α k x k − z
x k − z.
2.2
That is,z ∈ C k1as required
Notice that
Trang 3
is convex since
y n − z ≤ x n − z ⇐⇒ 2x n − y n , z ≤x n2−y n2. 2.4
This implies that each subsetC n C ∩ C1∩ · · · ∩ C n−1is convex It is also clear thatC nis closed Hence the first claim is proved
Next, we prove that{x n } is bounded As x n P C n x0,
x n − x0 ≤ z − x0 ∀z ∈ C n 2.5
In particular, forz ∈ F ⊂ C n for alln ∈ N, the sequence {x n − x0} is bounded and hence so is
{x n}
Next, we show that{x n } is a Cauchy sequence As x n1 ∈ C n1 ⊂ C nandx n P C n x0,
x n − x0 ≤ x n1 − x0 ∀n. 2.6 Moreover, since the sequence{x n} is bounded,
lim
n→∞ x n − x0 exists. 2.7 Note that
x0− x n , x n − v ≥ 0 ∀v ∈ C n 2.8
In particular, sincex nk ∈ C nk ⊂ C nfor allk ∈ N,
x nk − x n2x nk − x02−x n − x02− 2x nk − x n , x n − x0
≤x nk − x02−x n − x02. 2.9
It then follows from the existence of limn x n − x02that{x n} is a Cauchy sequence In fact, for
ε > 0, there exists a natural number N such that, for all n ≥ N,
n − x02− a
wherea lim n x n − x02 In particular, ifn ≥ N and k ∈ N, then
x nk − x n2≤x nk − x02−x n − x02
≤ a ε
2 −
a − ε
2
Moreover,
x n1 − x n −→ 0. 2.12
Trang 4We now assume thatx n → p for some p ∈ C Now since α n ≤ a < 1 for all n ∈ N and
x n1 ∈ C n,
x n − Tt n
x n 1
1− α n y n − x n
≤ 1
1− a y n − x n1 x n1 − x n
≤ 2
1− a x n1 − x n −→ 0.
2.13
The last convergence follows from2.12 We choose a sequence {t n k} of positive real number such that
t n k −→ 0, t1
n k
x n k − Tt n k
x n k −→ 0. 2.14
We now show that how such a special subsequence can be constructed First we fixδ > 0 such
that
lim inf
n t n 0 < δ < lim sup
n t n 2.15 From2.13, there exists m1 ∈ N such that Tt n x n − x n < 1/32for alln ≥ m1 ByLemma 1.1,
δ/2 is a cluster point of {t n } In particular, there exists n1 > m1such thatδ/3 < t n1 < δ Next,
we choose m2 > n1such that Tt n x n − x n < 1/42 for alln ≥ m2 Again, byLemma 1.1,δ/3
is a cluster point of{t n } and this implies that there exists n2 > m2 such thatδ/4 < t n2 < δ/2.
Continuing in this way, we obtain a subsequence{n k } of {n} satisfying
Tt n k
x n k − x n k < 1
k 22, k 2 δ < t n k < δ k ∀k ∈ N. 2.16 Consequently,2.14 is satisfied
We next show thatp ∈ F To see this, we fix t > 0,
x n k − Ttp
≤
t/t nk−1
j0
Tjt n k
x n k − Tj 1t n k
x n k
T t n t
k
t n k
x n k − T
t
t n k
t n k
p
T t n t
k
t n k
p − Ttp
≤
t
t n k
x n k − Tt n k x n k x n k − p
T t −
t
t n k
t n k
p − p
≤ t t
n k
x n k − Tt n k
x n k x n k − p supTsp − p : 0 ≤ s ≤ t n k
.
2.17
Asx n k → p and 2.14, we have x n k → Ttp and so Ttp p.
Finally, we show thatp P F x0 Since F ⊂ C n1andx n1 P C n1 x0,
x n1 − x0 ≤ q − x0 ∀n ∈ N, q ∈ F. 2.18
Trang 5Butx n → p; we have
p − x0 ≤ q − x0 ∀q ∈ F. 2.19 Hencep P F x0 as required This completes the proof
2.2 The hybrid method
We consider the iterative scheme computing by the hybrid methodsome authors call the CQ-method The following result is proved by He and Chen 3 However, the important part
of the proof seems to be overlooked Here we present the correction under some additional restriction on the parameter{t n}
Theorem 2.2 Let C be a closed convex subset of a real Hilbert space H Let {Tt : t ≥ 0} be a
nonexpansive semigroup on C with a nonempty common fixed point F, that is, F ∩ t≥0 FTt / ∅ Suppose that {x n } is a sequence iteratively generated by the following scheme:
x0∈ C taken arbitrary,
y n α n x n1− α n
Tt n
x n ,
C nz ∈ C : y n − z ≤ x n − z,
Q nz ∈ C :x n − x0, z − x n ≥ 0,
x n1 P C n ∩Q n
x0
,
2.20
where {α n } ⊂ 0, a ⊂ 0, 1, lim inf n t n 0, lim sup n t n > 0, and lim n t n1 − t n 0 Then x n →
P F x0.
Proof For the sake of clarity, we give the whole sketch proof even though some parts of the
proof are the same as3 To see that the scheme is well defined, it suffices to show that both
C n and Q n are closed and convex, and C n ∩ Q n / ∅ for all n ∈ N It follows easily from the
definition thatC nandQ nare just the intersection ofC and the half-spaces, respectively,
C n :z ∈ H : 2x n − y n , z ≤x n2−y n2
,
Q n :z ∈ H :x n − x0, z − x n
≥ 0.
2.21
As in the proof of the preceding theorem, we haveF ⊂ C n for alln ∈ N Clearly, F ⊂ C Q1 Suppose thatF ⊂ Q kfor somek ∈ N, we have p ∈ C k ∩Q k In particular,x k1 −x0, p−x k1 ≥ 0, that is,p ∈ Q k1 It follows from the induction thatF ⊂ Q nfor alln ∈ N This proves the claim.
We next show thatx n − Tt n x n→ 0 To see this, we first prove that
x n1 − x n −→ 0. 2.22
Asx n1 ∈ Q nandx n P Q n x0,
x n − x0 ≤ x n1 − x0 ∀n ∈ N. 2.23
Trang 6For fixedz ∈ F It follows from F ⊂ Q nfor alln ∈ N that
x n − x0 ≤ z − x0 ∀n ∈ N. 2.24 This implies that sequence{x n} is bounded and
lim
n→∞ x n − x0 exists. 2.25 Notice that
x n1 − x n , x n − x0
This implies that
x n1 − x n2x n1 − x02−x n − x02− 2x n1 − x n , x n − x0
≤x n1 − x02−x n − x02−→ 0. 2.27
It then follows fromx n1 ∈ C nthaty n − x n1 ≤ x n − x n1 and hence
Tt n
x n − x n 1α
n y n − x n
≤ α1
n y n − x n1 x n1 − x n −→ 0. 2.28
As inTheorem 2.1, we can choose a subsequence{n k } of {n} such that
x n k
w
−−−→ p ∈ C, t n k −→ 0, t1
n k
x n k − Tt n k
x n k −→ 0. 2.29 Consequently, for anyt > 0,
x n k − Ttp ≤ t t
n k
x n k − Tt n k
x n k x n k − p supTsp − p : 0 ≤ s ≤ t n k
.
2.30 This implies that
lim sup
k→∞ x n k − Ttp ≤ limsup
k→∞ x n k − p. 2.31
In virtue of Opial’s condition ofH, we have p Ttp for all t > 0, that is, p ∈ F Next, we
observe that
x0− P F
x0 ≤ x0− p ≤ liminf
k→∞ x0− x n k ≤ limsup
k→∞ x0− x n k ≤ x0− P F
x0.
2.32 This implies that
lim
k→∞ x0− x n k x0− P F
x0 x0− p. 2.33 Consequently,
x n k −→ P F
x0
Hence the whole sequence must converge toP F x0 p, as required.
Trang 7The author would like to thank the referees for his comments and suggestions on the manuscript This work is supported by the Commission on Higher Education and the Thai-land Research FundGrant MRG4980022
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