Strong Convergence of Modified Halpern Iterations in CAT(0) Spaces

Abstract

Strong convergence theorems are established for the modified Halpern iterations of nonexpansive mappings in CAT(0) spaces. Our results extend and improve the recent ones announced by Kim and Xu (2005), Hu (2008), Song and Chen (2008), Saejung (2010), and many others.

1. Introduction

Let be a nonempty subset of a metric space . A mapping is said to be nonexpansive if

(11)

A point is called a fixed point of if . We will denote by the set of fixed points of . In 1967, Halpern [1] introduced an explicit iterative scheme for a nonexpansive mapping on a subset of a Hilbert space by taking any points and defined the iterative sequence by

(12)

where . He pointed out that the control conditions: (C1) and (C2) are necessary for the convergence of to a fixed point of . Subsequently, many mathematicians worked on the Halpern iterations both in Hilbert and Banach spaces (see, e.g., [2–11] and the references therein). Among other things, Wittmann [7] proved strong convergence of the Halpern iteration under the control conditions (C1), (C2), and (C4) in a Hilbert space. In 2005, Kim and Xu [12] generalized Wittmann's result by introducing a modified Halpern iteration in a Banach space as follows. Let be a closed convex subset of a uniformly smooth Banach space , and let be a nonexpansive mapping. For any points , the sequence is defined by

(13)

where and are sequences in . They proved under the following control conditions:

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that the sequence converges strongly to a fixed point of .

The purpose of this paper is to extend Kim-Xu's result to a special kind of metric spaces, namely, CAT(0) spaces. We also prove a strong convergence theorem for another kind of modified Halpern iteration defined by Hu [13] in this setting.

2. CAT(0) Spaces

A metric space is a CAT(0) space if it is geodesically connected and if every geodesic triangle in is 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 CAT(0) space. Other examples include Pre-Hilbert spaces (see [14]), -trees (see [15]), Euclidean buildings (see [16]), the complex Hilbert ball with a hyperbolic metric (see [17]), 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 Haefliger [14].

Fixed point theory in CAT(0) 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 CAT(0) space always has a fixed point. Since then, the fixed point theory for single-valued and multivalued mappings in CAT(0) spaces has been rapidly developed, and many papers have appeared (see, e.g., [20–31] and the references therein). It is worth mentioning that fixed point theorems in CAT(0) spaces (specially in -trees) can be applied to graph theory, biology, and computer science (see, e.g., [15, 32–35]).

Let be a metric space. A geodesic path joining to (or, more briefly, a geodesic from to ) is a map from a closed interval to such that and for all . In particular, is an isometry and . The image of is called a geodesic (or metric) segment joining and . When it is unique, this geodesic segment is denoted by . The space is said to be a geodesic space if every two points of are joined by a geodesic, and is said to be uniquely geodesic if there is exactly one geodesic joining and for each . A subset is said to be convex if includes every geodesic segment joining any two of its points.

A geodesic triangle in a geodesic metric space consists of three points in (thevertices of ) and a geodesic segment between each pair of vertices (the edges of ). A comparison triangle for the geodesic triangle in is a triangle in the Euclidean plane such that for .

A geodesic space is said to be a CAT(0) space if all geodesic triangles satisfy the following comparison axiom.

CAT(0): let be a geodesic triangle in , and let be a comparison triangle for . Then, is said to satisfy the CAT(0) inequality if for all and all comparison points ,

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Let , and by Lemma  2.1 (iv) of [23] for each , there exists a unique point such that

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From now on, we will use the notation for the unique point satisfying (2.2). We now collect some elementary facts about CAT(0) spaces which will be used in the proofs of our main results.

Lemma 2.1.

Let be a space. Then,

• (i) (see [23, Lemma  2.4]) for each and , one has

  

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• (ii) (see [21]) for each and , one has

  

  (24)

 (iii) (see [19, Lemma  3]) for each and , one has

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 (iv) (see [23, Lemma  2.5]) for each and , one has

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Recall that a continuous linear functional on , the Banach space of bounded real sequences, is called a Banach limit if and for all .

Lemma 2.2 (see [8, Proposition  2]).

Let be such that for all Banach limits and . Then, .

Lemma 2.3 (see [28, Lemma  2.1]).

Let be a closed convex subset of a complete space , and let be a nonexpansive mapping. Let be fixed. For each , the mapping defined by

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has a unique fixed point , that is,

(28)

Lemma 2.4 (see [28, Lemma  2.2]).

Let and be as the preceding lemma. Then, if and only if given by (2.8) 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 .

Lemma 2.5 (see [10, Lemma  2.1]).

Let be a sequence of nonnegative real numbers satisfying the condition

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where and are sequences of real numbers such that

(1) and ,

• (2) either or .

Then, .

Lemma 2.6 (see [27, 36]).

Let and be bounded sequences in a space , and let be a sequence in with . Suppose that for all and

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Then, .

3. Main Results

The following result is an analog of Theorem  1 of Kim and Xu [12]. They prove the theorem by using the concept of duality mapping, while we use the concept of Banach limit. We also observe that the condition in [12, Theorem  1] is superfluous.

Theorem 3.1.

Let be a nonempty closed convex subset of a complete space , and let be a nonexpansive mapping such that . Given a point and sequences and in , the following conditions are satisfied:

(A1) and ,

(A2) , and .

Define a sequence in by arbitrarily, and

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Then, converges to a fixed point which is nearest .

Proof.

For each , we let . We divide the proof into 3 steps. (i) We will show that , , and are bounded sequences. (ii) We show that . Finally, we show that (iii) converges to a fixed point which is nearest .

• (i) As in the first part of the proof of [12, Theorem  1], we can show that is bounded and so is and . Notice also that

  

  (32)

 (ii) It suffices to show that

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Indeed, if (3.3) holds, we obtain

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By using Lemma 2.1, we get

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Hence,

(36)

where is a constant such that for all . By assumptions, we have

(37)

Hence, Lemma 2.5 is applicable to (3.6), and we obtain .

• (iii) From Lemma 2.3, let , where is given by (2.8). Then, is the point of which is nearest . We observe that

  

  (38)

By Lemma 2.4, we have for all Banach limit . Moreover, since ,

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It follows from and Lemma 2.2 that

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Hence, the conclusion follows from Lemma 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 also [37, 38] for subsequence comments).

Theorem 3.2.

Let C be a nonempty closed and convex subset of a complete space , and let be a nonexpansive mapping such that . Given a point and an initial value . The sequence is defined iteratively by

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Suppose that both and are sequences in satisfying

(B1) ,

(B2) ,

(B3) .

Then, converges to a fixed point which is nearest .

Proof.

Let . We divide the proof into 3 steps.

Step 1.

We show that , , and are bounded sequences. Let , then we have

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Now, an induction yields

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Hence, is bounded and so are and .

Step 2.

We show that . By using Lemma 2.1, we get

(314)

This implies that

(315)

Since and are bounded and , it follows that

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Hence, by Lemma 2.6, we get

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On the other hand,

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Using (3.17) and (3.18), we get

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Step 3.

We show that converges to a fixed point of . Let , where is given by (2.8), then . Finally, we show that

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By Lemma 2.4, we have for all Banach limit . Moreover, since

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it follows from condition ,  and Lemma 2.2 that

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Hence, the conclusion follows by Lemma 2.5.