Abstract
Viscosity approximation methods for nonexpansive mappings in CAT(0) spaces are studied. Consider a nonexpansive self-mapping of a closed convex subset of a CAT(0) space . Suppose that the set Fix of fixed points of is nonempty. For a contraction on and , let be the unique fixed point of the contraction . We will show that if is a CAT(0) space satisfying some property, then converge strongly to a fixed point of which solves some variational inequality. Consider also the iteration process , where is arbitrary and for , where . It is shown that under certain appropriate conditions on converge strongly to a fixed point of which solves some variational inequality.
1. 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 [1], R-trees [2], Euclidean buildings [3], the complex Hilbert ball with a hyperbolic metric [4], 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 [1].
Fixed-point theory in CAT(0) spaces was first studied by Kirk (see [5, 6]). 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 [2, 7–17].
The purpose of this paper is to study the iterative scheme defined as follows. Consider a nonexpansive self-mapping of a closed convex subset of a CAT(0) space . Suppose that the set Fix of fixed points of is nonempty. For a contraction on and , let be the unique fixed point of the contraction . Consider the iteration process , where is arbitrary and for , where . We show that converge strongly to a fixed point of under certain appropriate conditions on , and the fixed point of solves some variational inequality.
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 geodesicmetric space consists of three points , and in (the vertices 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 ,
Let , and by Lemma (iv) of [18] for each , there exists a unique point such that From now on, we will use the notation for the unique point satisfying the above equation.
We now collect some elementary facts about CAT(0) spaces which will be used in the proofs of our main results.
Lemma 1.1. Let be a CAT(0) space. Then,(i)(see [18, Lemma 2.4]) for each and , one has(ii) (see [7]) for each and , one has(iii) (see [6, Lemma 3]) for each and , one has(iv) (see [18, Lemma 2.5]) for each and , one has
Let be a complete CAT(0) space, let be a bounded sequence in a complete , and for , set
The asymptotic radius of is given by
and the asymptotic center of is the set
It is known (see, e.g., [11, Proposition 7]) that in a CAT(0) space, consists of exactly one point.
A sequence in is said to -converge to if is the unique asymptotic center of for every subsequence of . In this case, we write - and call the -limit of .
Lemma 1.2. Assume that is a CAT(0) space. Then,(i) (see [14]) every bounded sequence in has a -convergent subsequence,(ii) (see [14, Proposition 3.7]) if is a closed convex subset of , and is a nonexpansive mapping, then the conditions -converge to and and imply and .
Lemma 1.3 (see [19, Proposition 3.5]). Assume that is a CAT(0) space, is a closed convex subset of . Then the metric (nearest point) projection is a nonexpansive mapping. one calls a CAT(0) space satisfying property if for ,
Remark 1.4. The property in Hilbert space corresponds to the inequality
Recall that a continuous linear functional on , the Banach space of bounded real sequences, is called a Banach limit if and for all .
Lemma 1.5 (see [20, Proposition 2]). Let be such that for all Banach limits and . Then .
Lemma 1.6 (see [21, Lemma 2.3]). 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. Viscosity Iteration for a Single Mapping
In this section, we prove the main results of this paper.
Lemma 2.1. Let be a closed convex subset of a complete CAT(0) space , and let be a nonexpansive mapping. Let be a contraction on with coefficient . For each , the mapping defined by has a unique fixed point , that is,
Proof. For , according to Lemma 1.1, we have the following: This implies that is a contraction mapping, and hence, the conclusion follows.
The following result is to prove that the net converge strongly to a fixed point of .
Theorem 2.2. Let be a closed convex subset of a complete CAT(0) space satisfying the property , and let be a nonexpansive mapping. Let be a contraction on with coefficient . For each , let be given by Then one has and .
Proof. We first show that is bounded. Indeed choose a , and using Lemma 1.1 and the nonexpansive of , we derive that
It follows that
Hence,
and is bounded, so are and . As a result, we can get that
Assume that is such that as . Put . We will show that contains a subsequence converging strongly to , where .
Since is bounded, by Lemma 1.2(i),(ii), we may assume that -converges to a point , and .
Next we will prove that converge strongly to .
Indeed, according to Lemma 1.1 and the property of and , we can get that
It follows that
Since , we can get that
Let be a comparison triangle for in . Then,
Hence,
Let be a comparison triangle for in . For each , let be the point of the segment which is nearest to , and let be the point of the segment for which .
By passing to subsequences again, we may suppose that converges to , converges to .
Since -converges to a point , we have
Thus, . This implies that by uniqueness of the asymptotic center. Hence, . That is to say, converges to , and converges to .
Moreover, since satisfies the property , we can get that
It follows that
Since converges to , we obtain that , that is, converge strongly to . Since is such that as is arbitrarily selected, we can get that .
Finally, we will prove that satisfy the equation .
Indeed, for any ,
It follows that
Since , we can get that
Hence,
That is to say, .
Consider now the iteration process where satisfies(H1),(H2),(H3)either or .
Theorem 2.3. Let be a CAT(0) space satisfying the property , a closed convex subset of , a nonexpansive mapping with , and a contraction with coefficient . Let , be generated by . Then under the hypotheses (H1 )–(H3 ), , where .
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 and .
We claim that . Indeed, we have
By the conditions H2 and H3, we have
Consequently, by the condition H1,
Since is bounded, we may assume that -converges to a point . By Lemma 1.2, we have .
Next we will prove that converge strongly to and . Indeed, according to Lemma 1.1 and the property of and , we can get that
With a minor modification of the proof of the analogous statement in Theorem 2.2, we can get that
and .
Thus,
where . Since and , we obtain that
According to Lemma 1.6, we can get .
Finally, we prove that .
Indeed, for any ,
Let be a Banach limit. Then,
Since , we obtain that
It follows that
that is to say, . Since is a contraction and , we know that .
Acknowledgment
This paper was supported by NSFC Grant no. 11071279.