Open Access

Halpern's Iteration in CAT(0) Spaces

Fixed Point Theory and Applications20092010:471781

https://doi.org/10.1155/2010/471781

Received: 26 September 2009

Accepted: 24 November 2009

Published: 7 December 2009

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.

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

(1.1)

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 [57]). 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
(1.2)

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
(1.3)
is satisfied by all and all . In particular, if are points in a CAT(0) space and , then
(1.4)

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
(1.5)

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
(2.1)
has a unique fixed point , that is,
(2.2)

Proof.

For , we consider the triangle and its comparison triangle and we have the following:
(2.3)

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
(2.4)
for all . By the boundedness of , we have . We choose a sequence in such that . It follows from Lemma 1.1 that
(2.5)
Then, by the convexity of ,
(2.6)
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
(2.7)
and hence . Moreover, is a fixed point of . To see this, we consider
(2.8)
and
(2.9)
This implies that and hence .
  1. (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

     
(2.10)
Let be a Banach limit. Then
(2.11)
This implies that
(2.12)
Letting gives
(2.13)
In particular,
(2.14)

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
(2.15)

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
(2.16)
By induction, we have
(2.17)

for all . This implies that is bounded and so is the sequence .

Next, we show that . To see this, we consider the following:

(2.18)
By the conditions (C2) and (C3), we have
(2.19)
Consequently, by the condition (C1),
(2.20)
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:
(2.21)
By Lemma 2.2, we have for all Banach limits . Moreover, since ,
(2.22)
It follows from and Lemma 1.2 that
(2.23)

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
(3.1)

where is a sequence in satisfying

(C1) ;

(C2) ;

(C3) or .

Suppose, in addition, that
(3.2)

Then converges to which is nearest .

Here the function takes values in .

Proof.

By [16, Theorem ], we have
(3.3)

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

(3.4)

for each bounded subset of of .

If is a closed subset and satisfies AKTT-condition, then we can define such that

(3.5)

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
(3.6)

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
(3.7)
By conditions (C2), (C3), AKTT-condition, and Lemma 1.3, we have
(3.8)
Consequently, and hence
(3.9)
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
(3.10)
and this implies that
(3.11)

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,
(3.12)

This implies . As , we have , as desired.

Proof of Theorem 3.3.

We first define a family of mappings by
(3.13)
By Lemma 3.4, each is a nonexpansive mapping satisfying . Notice that, for fixed ,
(3.14)
From the estimation above, we have
(3.15)
for each bounded subset of . In particular, is a Cauchy sequence for each . We now define the nonexpansive mapping by
(3.16)
Finally, we prove that
(3.17)
The latter equality is clearly verified and holds. On the other hand, let and . We consider the following:
(3.18)
Then
(3.19)
Letting yields
(3.20)

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
(4.1)

(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
(4.2)

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
(4.3)

where is a sequence in satisfying

(C1) ;

(C2) ;

(C3) or .

Then converges to which is nearest .

Declarations

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.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Khon Kaen University
(2)
Centre of Excellence in Mathematics, CHE

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© Satit Saejung. 2010

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