Abstract
Motivated by Halpern's result, we prove strong convergence theorem of an iterative sequence in CAT(0) spaces. We apply our result to find a common fixed point of a family of nonexpansive mappings. A convergence theorem for nonself mappings is also discussed.
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1. Introduction
Let 
be a metric space and 
with 
. A geodesic path from 
to 
is an isometry 
such that 
and 
. The image of a geodesic path is called a geodesic segment. A metric space 
is a (uniquely) geodesic space if every two points of 
are joined by only one geodesic segment. A geodesic triangle
in a geodesic space 
consists of three points 
of 
and three geodesic segments joining each pair of vertices. A comparison triangle of a geodesic triangle 
is the triangle 
in the Euclidean space 
such that 
for all 
.
A geodesic space 
is a CAT(0) space if for each geodesic triangle 
in 
and its comparison triangle 
in 
, the CAT(0) inequality
is satisfied by all 
and 
. The meaning of the CAT(0) inequality is that a geodesic triangle in 
is at least thin as its comparison triangle in the Euclidean plane. A thorough discussion of these spaces and their important role in various branches of mathematics are given in [1, 2]. The complex Hilbert ball with the hyperbolic metric is an example of a CAT(0) space (see [3]).
The concept of 
-convergence introduced by Lim in 1976 was shown by Kirk and Panyanak [4] in CAT(0) spaces to be very similar to the weak convergence in Banach space setting. Several convergence theorems for finding a fixed point of a nonexpansive mapping have been established with respect to this type of convergence (e.g., see [5–7]). The purpose of this paper is to prove strong convergence of iterative schemes introduced by Halpern [8] in CAT(0) spaces. Our results are proved under weaker assumptions as were the case in previous papers and we do not use 
-convergence. We apply our result to find a common fixed point of a countable family of nonexpansive mappings. A convergence theorem for nonself mappings is also discussed.
In this paper, we write 
for the the unique point 
in the geodesic segment joining from 
to 
such that
We also denote by 
the geodesic segment joining from 
to 
, that is, 
. A subset 
of a CAT(0) space is convex if 
for all 
. For elementary facts about CAT(0) spaces, we refer the readers to [1] (or, briefly in [5]).
The following lemma plays an important role in our paper.
Lemma 1.1.
A geodesic space 
is a CAT(0) space if and only if the following inequality
is satisfied by all 
and all 
. In particular, if 
are points in a CAT(0) space and 
, then
Recall that a continuous linear functional 
on 
, the Banach space of bounded real sequences, is called a Banach limit if 
and 
for all 
.
Lemma 1.2 (see [9, Proposition 
]).
Let 
be such that 
for all Banach limits 
and 
. Then 
.
Lemma 1.3 (see [10, Lemma 
]).
Let 
be a sequence of nonnegative real numbers, 
a sequence of real numbers in 
with 
, 
a sequence of nonnegative real numbers with 
, and 
a sequence of real numbers with 
. Suppose that
Then 
.
2. Halpern's Iteration for a Single Mapping
Lemma 2.1.
Let 
be a closed convex subset of a complete CAT(0) space 
and let 
be a nonexpansive mapping. Let 
be fixed. For each 
, the mapping 
defined by
has a unique fixed point 
, that is,
Proof.
For 
, we consider the triangle 
and its comparison triangle and we have the following:
This implies that 
is a contraction mapping and hence the conclusion follows.
The following result is proved by Kirk in [11, Theorem 
] under the boundedness assumption on 
. We present here a new proof which is modified from Kirk's proof.
Lemma 2.2.
Let 
, 
be as the preceding lemma. Then 
if and only if 
given by the formula (2.2) remains bounded as 
. In this case, the following statements hold:
(1)
converges to the unique fixed point 
of 
which is nearest 
;
(2)
for all Banach limits 
and all bounded sequences 
with 
.
Proof.
If 
, then it is clear that 
is bounded. Conversely, suppose that 
is bounded. Let 
be any sequence in 
such that 
and define 
by
for all 
. By the boundedness of 
, we have 
. We choose a sequence 
in 
such that 
. It follows from Lemma 1.1 that
Then, by the convexity of 
,
This implies that 
is a Cauchy sequence in 
and hence it converges to a point 
. Suppose that 
is a point in 
satisfying 
. It follows then that
and hence 
. Moreover, 
is a fixed point of 
. To see this, we consider
and
This implies that 
and hence 
.
-
(1)
is proved in [12, Theorem
]. In fact, it is shown that 
is the nearest point of 
to 
. Finally, we prove (2). Suppose that 
is a sequence given by the formula (2.2), where 
is a sequence in 
such that 
. We also assume that 
is the nearest point of 
to 
. By the first inequality in Lemma 1.1, we have
]. In fact, it is shown that 
is the nearest point of 
to 
. Finally, we prove (2). Suppose that 
is a sequence given by the formula (2.2), where 
is a sequence in 
such that 
. We also assume that 
is the nearest point of 
to 
. By the first inequality in Lemma 1.1, we have
Let 
be a Banach limit. Then
This implies that
Letting 
gives
In particular,
Inspired by the results of Wittmann [13] and of Shioji and Takahashi [9], we use the iterative scheme introduced by Halpern to obtain a strong convergence theorem for a nonexpansive mapping in CAT(0) space setting. A part of the following theorem is proved in [14].
Theorem 2.3.
Let 
be a closed convex subset of a complete CAT(0) space 
and let 
be a nonexpansive mapping with a nonempty fixed point set 
. Suppose that 
are arbitrarily chosen and 
is iteratively generated by
where 
is a sequence in 
satisfying
(C1)
;
(C2)
;
(C3)
or 
.
Then 
converges to 
which is the nearest point of 
to 
.
Proof.
We first show that the sequence 
is bounded. Let 
. Then
By induction, we have
for all 
. This implies that 
is bounded and so is the sequence 
.
Next, we show that 
. To see this, we consider the following:
By the conditions (C2) and (C3), we have
Consequently, by the condition (C1),
From Lemma 2.2, let 
where 
is given by the formula (2.2). Then 
is the nearest point of 
to 
. We next consider the following:
By Lemma 2.2, we have 
for all Banach limits 
. Moreover, since 
,
It follows from 
and Lemma 1.2 that
Hence the conclusion follows by Lemma 1.3.
3. Halpern's Iteration for a Family of Mappings
3.1. Finitely Many Mappings
We use the "cyclic method" [15] and Bauschke's condition [16] to obtain the following strong convergence theorem for a finite family of nonexpansive mappings.
Theorem 3.1.
Let 
be a complete CAT(0) space and 
a closed convex subset of 
. Let 
be nonexpansive mappings with 
and let 
be arbitrarily chosen. Define an iterative sequence 
by
where 
is a sequence in 
satisfying
(C1)
;
(C2)
;
(C3)
or 
.
Suppose, in addition, that
Then 
converges to 
which is nearest 
.
Here the 
function takes values in 
.
Proof.
By [16, Theorem 
], we have
The proof line now follows from the proofs of Theorem 2.3 and [15, Theorem 
].
3.2. Countable Mappings
The following concept is introduced by Aoyama et al. [10]. Let 
be a complete CAT(0) space and 
a subset of 
. Let 
be a countable family of mappings from 
into itself. We say that a family 
satisfies AKTT-condition if
for each bounded subset of 
of 
.
If 
is a closed subset and 
satisfies AKTT-condition, then we can define 
such that
In this case, we also say that 
satisfies AKTT-condition.
Theorem 3.2.
Let 
be a complete CAT(0) space and 
a closed convex subset of 
. Let 
be a countable family of nonexpansive mappings with 
. Suppose that 
are arbitrarily chosen and 
is defined by
where 
is a sequence in 
satisfying
(C1)
;
(C2)
;
(C3)
or 
.
Suppose, in addition, that
(M1)
satisfies AKTT-condition;
(M2)
.
Then 
converges to 
which is nearest 
.
Proof.
Since the proof of this theorem is very similar to that of Theorem 2.3, we present here only the sketch proof. First, we notice that both sequences 
and 
are bounded and
By conditions (C2), (C3), AKTT-condition, and Lemma 1.3, we have
Consequently, 
and hence
Let 
be the nearest point of 
to 
. As in the proof of Theorem 2.3, we have 
for all Banach limits 
and 
. We observe that
and this implies that
Therefore, 
and hence 
converges to 
.
We next show how to generate a family of mappings from a given family of mappings to satisfy conditions (M1) and (M2) of the preceding theorem. The following is an analogue of Bruck's result [17] in CAT(0) space setting. The idea using here is from [10].
Theorem 3.3.
Let 
be a complete CAT(0) space and 
a closed convex subset of 
. Suppose that 
is a countable family of nonexpansive mappings with 
. Then there exist a family of nonexpansive mappings 
and a nonexpansive mapping 
such that
(M1)
satisfies AKTT-condition;
(M2)
.
Lemma 3.4.
Let 
and 
be as above. Suppose that 
are nonexpansive mappings and 
. Then, for any 
, the mapping 
is nonexpansive and 
.
Proof.
To see that 
is nonexpansive, we only apply the triangle inequality and two applications of the second inequality in Lemma 1.1. We next prove the latter. It is clear that 
. To see the reverse inclusion, let 
and 
. Then, by the first inequality of Lemma 1.1,
This implies 
. As 
, we have 
, as desired.
Proof of Theorem 3.3.
We first define a family of mappings 
by
By Lemma 3.4, each 
is a nonexpansive mapping satisfying 
. Notice that, for fixed 
,
From the estimation above, we have
for each bounded subset 
of 
. In particular, 
is a Cauchy sequence for each 
. We now define the nonexpansive mapping 
by
Finally, we prove that
The latter equality is clearly verified and 
holds. On the other hand, let 
and 
. We consider the following:
Then
Letting 
yields
Because 
, we have 
. Continuing this procedure we obtain that 
and hence 
. This completes the proof.
4. Nonself Mappings
From Bridson and Haefliger's book (page 176), the following result is proved.
Theorem 4.1.
Let 
be a complete CAT(0) space and 
a closed convex subset of 
. Then the followings hold true.
(i)For each 
, there exists an element 
such that
(ii)
for all 
.
(iii)The mapping 
is nonexpansive.
The mapping 
in the preceding theorem is called the metric projection from
onto
. From this, we have the following result.
Theorem 4.2.
Let 
be a complete CAT(0) space and 
a closed convex subset of 
. Let 
be a nonself nonexpansive mapping with 
and 
the metric projection from 
onto 
. Then the mapping 
is nonexpansive and 
.
Proof.
It follows from Theorem 4.1 that 
is nonexpansive. To see the latter, it suffices to show that 
. Let 
and 
. Since
we have 
and this finishes the proof.
By the preceding theorem and Theorem 2.3, we obtain the following result.
Theorem 4.3.
Let 
, 
, 
, and 
be as the same as Theorem 4.2. Suppose that 
are arbitrarily chosen and the sequence 
is defined by
where 
is a sequence in 
satisfying
(C1)
;
(C2)
;
(C3)
or 
.
Then 
converges to 
which is nearest 
.
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Acknowledgments
The author would like to thank the referee for the information that a part of Theorem 2.3 was proved in [14]. This work was supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.
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Saejung, S. Halpern's Iteration in CAT(0) Spaces. Fixed Point Theory Appl 2010, 471781 (2009). https://doi.org/10.1155/2010/471781
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DOI: https://doi.org/10.1155/2010/471781