Volume 2011, Article ID 869458,11pages doi:10.1155/2011/869458
Research Article
Strong Convergence of Modified Halpern Iterations
in CAT(0) Spaces
A. Cuntavepanit
1and B. Panyanak
1, 21Department of Mathematics, Faculty of Science, Chiang Mai University, Chiang Mai 50200, Thailand 2Materials Science Research Center, Faculty of Science, Chiang Mai University,
Chiang Mai 50200, Thailand
Correspondence should be addressed to B. Panyanak,[email protected]
Received 28 November 2010; Accepted 10 January 2011
Academic Editor: Qamrul Hasan Ansari
Copyrightq2011 A. Cuntavepanit and B. Panyanak. 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.
Strong convergence theorems are established for the modified Halpern iterations of nonexpansive mappings in CAT0spaces. Our results extend and improve the recent ones announced by Kim and Xu2005, Hu2008, Song and Chen2008, Saejung2010, and many others.
1. Introduction
LetCbe a nonempty subset of a metric spaceX, d. A mappingT : C → Cis said to be
nonexpansiveif
dTx, Ty≤dx, y, ∀x, y∈C. 1.1
A pointx∈Cis called a fixed point ofT ifx Tx. We will denote byFTthe set of fixed points ofT. In 1967, Halpern1introduced an explicit iterative scheme for a nonexpansive mappingT on a subsetCof a Hilbert space by taking any pointsu, x1 ∈Cand defined the iterative sequence{xn}by
xn1 αnu 1−αnTxn, forn≥1, 1.2
where αn ∈ 0,1. He pointed out that the control conditions:C1 limnαn 0 and C2
∞
n1αn ∞are necessary for the convergence of{xn}to a fixed point ofT. Subsequently,
see, e.g., 2–11 and the references therein. Among other things, Wittmann 7 proved strong convergence of the Halpern iteration under the control conditionsC1,C2, andC4 ∞
n1|αn1−αn|<∞in a Hilbert space. In 2005, Kim and Xu12generalized Wittmann’s result
by introducing a modified Halpern iteration in a Banach space as follows. LetCbe a closed convex subset of a uniformly smooth Banach spaceX, and letT :C → Cbe a nonexpansive mapping. For any pointsu, x1∈C, the sequence{xn}is defined by
xn1βnu
1−βn
Tαnxn 1−αnTxn, forn≥1, 1.3
where {αn} and {βn} are sequences in 0,1. They proved under the following control
conditions:
D1 lim
n αn 0, limn βn0,
D2 ∞
n1
αn∞,
∞
n1
βn∞,
D3 ∞
n1
|αn1−αn|<∞,
∞
n1
βn1−βn<∞,
1.4
that the sequence{xn}converges strongly to a fixed point ofT.
The purpose of this paper is to extend Kim-Xu’s result to a special kind of metric spaces, namely, CAT0spaces. We also prove a strong convergence theorem for another kind of modified Halpern iteration defined by Hu13in this setting.
2. CAT(0) Spaces
A metric space X is a CAT0 space if it is geodesically connected and if every geodesic triangle inXis at least as “thin” as its comparison triangle in the Euclidean plane. The precise definition is given below. It is well known that any complete, simply connected Riemannian manifold having nonpositive sectional curvature is a CAT0space. Other examples include Pre-Hilbert spacessee14,R-treessee15, Euclidean buildingssee16, the complex Hilbert ball with a hyperbolic metricsee17, and many others. For a thorough discussion of these spaces and of the fundamental role they play in geometry, we refer the reader to Bridson and Haefliger14.
Fixed point theory in CAT0 spaces was first studied by Kirk see 18, 19. He showed that every nonexpansive single-valued mapping defined on a bounded closed convex subset of a complete CAT0space always has a fixed point. Since then, the fixed point theory for single-valued and multivalued mappings in CAT0spaces has been rapidly developed, and many papers have appearedsee, e.g.,20–31and the references therein. It is worth mentioning that fixed point theorems in CAT0spacesspecially inR-treescan be applied to graph theory, biology, and computer sciencesee, e.g.,15,32–35.
LetX, dbe a metric space. Ageodesic pathjoiningx∈Xtoy∈Xor, more briefly, a
two points of X are joined by a geodesic, andX is said to be uniquely geodesicif there is exactly one geodesic joiningxandyfor eachx, y∈X. A subsetY ⊆Xis said to beconvexif Y includes every geodesic segment joining any two of its points.
Ageodesic triangleΔ x1, x2, x3in a geodesic metric spaceX, dconsists of three points x1, x2, andx3inXtheverticesofΔand a geodesic segment between each pair of vertices
the edges of Δ. A comparison triangle for the geodesic triangle Δx1, x2, x3 in X, d is a triangleΔx1, x2, x3: Δx1, x2, x3in the Euclidean planeE2such thatdE2xi, xj dxi, xj
fori, j∈ {1,2,3}.
A geodesic space is said to be a CAT0 space if all geodesic triangles satisfy the following comparison axiom.
CAT0: letΔbe a geodesic triangle inX, and letΔbe a comparison triangle forΔ. Then,Δis said to satisfy the CAT0inequalityif for allx, y ∈ Δand all comparison points x, y∈Δ,
dx, y≤dE2x, y. 2.1
Letx, y ∈X, and by Lemma 2.1ivof23for eacht∈ 0,1, there exists a unique pointz∈x, ysuch that
dx, z tdx, y, dy, z 1−tdx, y. 2.2
From now on, we will use the notation1−tx⊕tyfor the unique pointzsatisfying2.2. We now collect some elementary facts about CAT0spaces which will be used in the proofs of our main results.
Lemma 2.1. LetXbe aCAT0space. Then,
i(see [23, Lemma 2.4]) for eachx, y, z∈Xandt∈0,1, one has
d1−tx⊕ty, z≤1−tdx, z tdy, z, 2.3
ii(see [21]) for eachx, y∈Xandt, s∈0,1, one has
d1−tx⊕ty,1−sx⊕sy|t−s|dx, y, 2.4
iii(see [19, Lemma 3]) for eachx, y, z∈Xandt∈0,1, one has
d1−tz⊕tx,1−tz⊕ty≤tdx, y, 2.5
iv(see [23, Lemma 2.5]) for eachx, y, z∈Xandt∈0,1, one has
d1−tx⊕ty, z2≤1−tdx, z2tdy, z2−t1−tdx, y2. 2.6
Recall that a continuous linear functionalμon∞, the Banach space of bounded real sequences, is called aBanach limitifμμ1,1, . . . 1 andμnan μnan1for all{an} ∈
Lemma 2.2see8, Proposition 2. Let{a1, a2, . . .} ∈∞be such thatμnan≤0for all Banach
limitsμandlim supnan1−an≤0. Then,lim supnan≤0.
Lemma 2.3see28, Lemma 2.1. LetCbe a closed convex subset of a completeCAT0spaceX, and letT :C → Cbe a nonexpansive mapping. Letu∈Cbe fixed. For eacht∈0,1, the mapping St:C → Cdefined by
Stztu⊕1−tTz, forz∈C 2.7
has a unique fixed pointzt∈C, that is,
ztStzt tu⊕1−tTzt. 2.8
Lemma 2.4see28, Lemma 2.2. LetCandTbe as the preceding lemma. Then,FT/∅if and only if{zt}given by2.8remains bounded ast → 0. In this case, the following statements hold:
1{zt}converges to the unique fixed pointzofT which is nearestu,
2d2u, z ≤ μ
nd2u, xn for all Banach limits μ and all bounded sequences {xn} with
limndxn, Txn 0.
Lemma 2.5 see 10, Lemma 2.1. Let {αn}∞n1 be a sequence of nonnegative real numbers
satisfying the condition
αn1≤
1−γn
αnγnσn, n≥1, 2.9
where{γn}and{σn}are sequences of real numbers such that
1{γn} ⊂0,1and
∞
n1γn∞,
2eitherlim supn→ ∞σn≤0or∞n1|γnσn|<∞.
Then,limn→ ∞αn0.
Lemma 2.6see27,36. Let{xn}and{yn}be bounded sequences in aCAT0spaceX, and let {αn}be a sequence in0,1with0<lim infnαn≤lim supnαn <1. Suppose thatxn1αnyn⊕1−
αnxnfor alln∈Nand
lim sup
n→ ∞
dyn1, yn
−dxn1, xn
≤0. 2.10
Then,limndxn, yn 0.
3. Main Results
Theorem 3.1. Let Cbe a nonempty closed convex subset of a complete CAT0 spaceX, and let
T :C → Cbe a nonexpansive mapping such thatFT/∅. Given a pointu∈Cand sequences{αn}
and{βn}in0,1, the following conditions are satisfied:
(A1)limnαn0and
∞
n1|αn1−αn|<∞,
(A2)limnβn0,∞n1βn∞and∞n1|βn1−βn|<∞.
Define a sequence{xn}inCbyx1x∈Carbitrarily, and
xn1βnu⊕
1−βn
αnxn⊕1−αnTxn, ∀n≥1. 3.1
Then,{xn}converges to a fixed pointz∈FTwhich is nearestu.
Proof. For eachn ≥ 1, we letyn : αnxn ⊕1−αnTxn. We divide the proof into 3 steps.
i We will show that {xn}, {yn}, and {Txn} are bounded sequences. ii We show that
limndxn, Txn 0. Finally, we show thatiii{xn}converges to a fixed pointz∈FTwhich
is nearestu.
i As in the first part of the proof of 12, Theorem 1, we can show that {xn} is
bounded and so is{yn}and{Txn}. Notice also that
dyn, p
≤dxn, p
, ∀p∈FT. 3.2
iiIt suffices to show that
lim
n→ ∞dxn, xn1 0. 3.3
Indeed, if3.3holds, we obtain
dxn, Txn≤dxn, xn1 d
xn1, yn
dyn, Txn
dxn, xn1 d
βnu⊕
1−βn
yn, yn
dαnxn⊕1−αnTxn, Txn ≤dxn, xn1 βnd
u, yn
αndxn, Txn−→0, asn−→ ∞.
3.4
By usingLemma 2.1, we get
dxn1, xn d
βnu⊕
1−βn
yn, βn−1u⊕1−βn−1
yn−1
≤dβnu⊕
1−βn
yn, βnu⊕
1−βn
yn−1
dβnu⊕
1−βn
yn−1, βn−1u⊕1−βn−1
≤1−βn
dyn, yn−1
βn−βn−1d
u, yn−1
1−βn
dαnxn⊕1−αnTxn, αn−1xn−1⊕1−αn−1Txn−1
βn−βn−1du, αn−1xn−1⊕1−αn−1Txn−1
≤1−βn
dαnxn⊕1−αnTxn, αnxn−1⊕1−αnTxn
dαnxn−1⊕1−αnTxn, αnxn−1⊕1−αnTxn−1
dαnxn−1⊕1−αnTxn−1, αn−1xn−1⊕1−αn−1Txn−1
βn−βn−1αn−1du, xn−1 1−αn−1du, Txn−1
≤1−βn
αndxn, xn−1 1−αndTxn, Txn−1 |αn−αn−1|dxn−1, Txn−1
βn−βn−1αn−1du, xn−1 1−αn−1du, Txn−1
1−βn
dxn, xn−1
1−βn
|αn−αn−1|dxn−1, Txn−1
βn−βn−1αn−1du, xn−1 βn−βn−11−αn−1du, Txn−1
≤1−βn
dxn, xn−1
1−βn
|αn−αn−1|dxn−1, Txn−1
βn−βn−1αn−1du, Txn−1 dTxn−1, xn−1
βn−βn−1du, Txn−1−βn−βn−1αn−1du, Txn−1
1−βn
dxn, xn−1
1−βn
|αn−αn−1|dxn−1, Txn−1
βn−βn−1αn−1dxn−1, Txn−1 βn−βn−1du, Txn−1.
3.5
Hence,
dxn1, xn≤
1−βn
dxn, xn−1 γ
|αn−αn−1|2βn−βn−1, 3.6
whereγ > 0 is a constant such thatγ ≥ max{du, Txn−1, dxn−1, Txn−1} for alln ∈ N. By assumptions, we have
lim
n→ ∞βn0, ∞
n1
βn∞, ∞
n1
|αn−αn−1|2βn−βn−1<∞. 3.7
iiiFromLemma 2.3, letzlimt→0zt, whereztis given by2.8. Then,zis the point
ofFTwhich is nearestu. We observe that
d2xn1, z d2βnu⊕
1−βn
yn, z
≤βnd2u, z
1−βn
d2yn, z
−βn
1−βn
d2u, yn
≤βnd2u, z
1−βn
d2x
n, z−βn
1−βn
d2u, y
n
1−βn
d2xn, z βn
d2u, z−1−βn
d2u, yn
.
3.8
ByLemma 2.4, we haveμnd2u, z−d2u, xn ≤ 0 for all Banach limitμ. Moreover, since
limndxn1, xn 0,
lim sup
n→ ∞ d
2u, z−d2u, x
n1
− d2u, z−d2u, xn
0. 3.9
It follows from limndyn, xn 0 andLemma 2.2that
lim sup
n→ ∞
d2u, z−1−βn
d2u, yn
lim sup
n→ ∞
d2u, z−d2u, xn
≤0. 3.10
Hence, the conclusion follows fromLemma 2.5.
By using the similar technique as in the proof of Theorem 3.1, we can obtain a strong convergence theorem which is an analog of 13, Theorem 3.1 see also37,38for subsequence comments.
Theorem 3.2. Let C be a nonempty closed and convex subset of a completeCAT0spaceX, and let
T :C → Cbe a nonexpansive mapping such thatFT/∅. Given a pointu∈Cand an initial value
x1∈C. The sequence{xn}is defined iteratively by
xn1βnxn⊕
1−βn
αnu⊕1−αnTxn, n≥1. 3.11
Suppose that both{αn}and{βn}are sequences in0,1satisfying
(B1)limn→ ∞αn0,
(B2)∞n1αn∞,
(B3)0<lim infn→ ∞βn≤lim supn→ ∞βn<1.
Proof. Letyn:αnu⊕1−αnTxn. We divide the proof into 3 steps.
Step 1. We show that{xn},{yn}, and{Txn}are bounded sequences. Letp ∈ FT, then we
have
dxn1, pdβnxn⊕
1−βn
αnu⊕1−αnTxn, p
≤βnd
xn, p
1−βn
dαnu⊕1−αnTxn, p
≤βnd
xn, p
1−βn
αnd
u, p1−βn
1−αnd
Txn, p
≤βn
1−βn
1−αn
dxn, p
1−βn
αnd
u, p
1−1−βn
αn
dxn, p
1−βn
αnd
u, p
≤maxdxn, p
, du, p.
3.12
Now, an induction yields
dxn1, p≤maxdx1, p, du, p, n≥1. 3.13
Hence,{xn}is bounded and so are{yn}and{Txn}.
Step 2. We show that limndxn, Txn 0. By usingLemma 2.1, we get
dyn1, yn
dαn1u⊕1−αn1Txn1, αnu⊕1−αnTxn ≤αndαn1u⊕1−αn1Txn1, u
1−αndαn1u⊕1−αn1Txn1, Txn ≤αn1−αn1dTxn1, u 1−αnαn1du, Txn
1−αn1−αn1dTxn1, Txn
≤αn1−αn1dTxn1, u 1−αnαn1du, Txn
1−αn1−αn1dxn1, xn.
3.14
This implies that
dyn1, yn
−dxn1, xn≤αn1−αn1dTxn1, u 1−αnαn1du, Txn
αnαn1−αn−αn1dxn1, xn.
3.15
Since{xn}and{Txn}are bounded and limn→ ∞αn0, it follows that
lim sup
n→ ∞
dyn1, yn
−dxn1, xn
Hence, byLemma 2.6, we get
lim
n→ ∞d
xn, yn
0. 3.17
On the other hand,
dyn, Txn
dαnu⊕1−αnTxn, Txn≤αndu, Txn−→0, asn−→ ∞. 3.18
Using3.17and3.18, we get
dxn, Txn≤d
xn, yn
dyn, Txn
−→0, asn−→ ∞. 3.19
Step 3. We show that{xn}converges to a fixed point ofT. Letzlimt→0zt, whereztis given
by2.8, thenz∈FT. Finally, we show that limnxn z
d2xn1, z d2βnxn⊕
1−βn
yn, z
≤βnd2xn, z
1−βn
d2yn, z
−βn
1−βn
d2xn, yn
≤βnd2xn, z
1−βn
d2α
nu⊕1−αnTxn, z−βn
1−βn
d2x
n, yn
≤1−βn
αnd2u, z 1−αnd2Txn, z−αn1−αnd2u, Txn
−βn
1−βn
d2xn, yn
βnd2xn, z
≤βn
1−βn
1−αn
d2x
n, z
1−βn
αn
d2u, z−1−α
nd2u, Txn
1−1−βn
αn
d2xn, z
1−βn
αn
d2u, z−1−αnd2u, Txn
.
3.20
ByLemma 2.4, we haveμnd2u, z−d2u, xn≤0 for all Banach limitμ. Moreover, since
dxn1, xn d
βnxn⊕
1−βn
yn, xn
≤1−βn
dyn, xn
−→0, asn−→ ∞,
lim sup
n→ ∞ d
2u, z d2u, x
n1−d2u, z−d2u, xn
0,
3.21
it follows from conditionB1, limndxn, Txn 0 andLemma 2.2that
lim sup
n→ ∞ d
2u, z−1−α
nd2u, Txn
lim sup
n→ ∞ d
2u, z−d2u, x
n
≤0. 3.22
Acknowledgments
The authors are grateful to Professor Sompong Dhompongsa for his suggestions and advices during the preparation of the paper. This research was supported by the National Research University Project under Thailand’s Office of the Higher Education Commission.
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