Abstract
The main purpose of this paper is first to introduce the concept of total asymptotically nonexpansive mappings and to prove a -convergence theorem for finding a common fixed point of the total asymptotically nonexpansive mappings and the asymptotically nonexpansive mappings. The demiclosed principle for this kind of mappings in CAT(0) space is also proved in the paper. Our results extend and improve many results in the literature.
1. Introduction
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. Fixed point theory in a CAT(0) space was first studied by Kirk [1, 2]. He showed that every nonexpansive 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 various mappings in CAT(0) space has been developed rapidly and many papers have appeared [3–10]. On the other hand, Browder [11] introduced the demiclosed principle which states that if is a uniformly convex Banach space, is a nonempty closed convex subset of , and if is nonexpansive mapping, then is demiclosed at each , that is, for any sequence in conditions weakly and strongly imply that (where is the identity mapping of ). Xu [12] proved the demiclosed principle for asymptotically nonexpansive mappings in the setting of a uniformly convex Banach space. Nanjaras and Panyanak [13] proved the demiclosed principle for asymptotically nonexpansive mappings in CAT(0) space and obtained a -convergence theorem for the Krasnosel’skii-Mann iteration.
Motivated and inspired by the researches going on in this direction, especially inspired by Nanjaras and Panyanak, and so forth [13], the purpose of this paper is to introduce a general mapping, namely, total asymptotically nonexpansive mapping and to prove its demiclosed principle in CAT(0) space. As a consequence, we construct a hierarchical iterative algorithm to study the fixed point of the total asymptotically nonexpansive mappings and obtain a -convergence theorem.
2. Preliminaries and Lemmas
Let be a metric space and with . A geodesic path from to is a isometry such that , . The image of a geodesic path is called geodesic segment. A space is a (uniquely) geodesic space if every two points of are joined by only one geodesic segment. A geodesic triangle in a geodesic metric space consists of three points 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 space such that for .
A geodesic space is said to be a CAT(0) space if for each geodesic triangle in and its comparison triangle in , the CAT(0) inequality is satisfied for all and .
In this paper, we write for 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 said to be convex if for all .
Lemma 2.1 (see [14]). A geodesic space is a CAT(0) space, if and only if the following inequality is satisfied for all and . In particular, if are points in a CAT(0) space and , then
Let be a bounded sequence in a CAT(0) space . For , one sets The asymptotic radius of is given by the asymptotic radius of with respect to is given by the asymptotic center of is the set the asymptotic center of with respect to is the set
Recall that a bounded sequence in is said to be regular if for every subsequence of .
Proposition 2.2 (see [15]). If is a bounded sequence in a complete CAT(0) space and is a closed convex subset of , then(1)there exists a unique point such that (2) and are both singleton.
Lemma 2.3 2.3 (see [16]). If is a closed convex subset of a complete CAT(0) space and if is a bounded sequence in , then the asymptotic center of is in .
Definition 2.4 (see [17]). A sequence in a CAT(0) space is said to -converge to if is the unique asymptotic center of for every subsequence of . In this case one writes and call the -limit of .
Lemma 2.5 (see [17]). Every bounded sequence in a complete CAT(0) space always has a -convergent subsequence.
Let be a bounded sequence in a CAT(0) space and let be a closed convex subset of which contains . We denote the notation where .
Now one gives a connection between the “” convergence and -convergence.
Proposition 2.6 2.6 (see [13]). Let be a bounded sequence in a CAT(0) space and let be a closed convex subset of which contains . Then(1) implies that ;(2) and is regular imply that ;
Let be a closed subset of a metric space . Recall that a mapping is said to be nonexpansive if is said to be asymptotically nonexpansive if there is a sequence with such that
is said to be closed if, for any sequence with and , then .
is called uniformly Lipschitzian, if there exists a constant such that
Definition 2.7. Let be a metric space and let be a closed subset of . A mapping is said to be total asymptotically nonexpansive if there exist nonnegative real sequences with , and a strictly increasing continuous function with such that
Remark 2.8. (1) It is obvious that If is uniformly Lipschitzian, then is closed.
(2) From the definitions, it is to know that, each nonexpansive mapping is a asymptotically nonexpansive mapping with sequence , and each asymptotically nonexpansive mapping is a total asymptotically nonexpansive mapping with , , for all , and , .
Lemma 2.9 (demiclosed principle for total asymptotically nonexpansive mappings). Let be a closed and convex subset of a complete CAT(0) space and let be a uniformly Lipschitzian and total asymptotically nonexpansive mapping. Let be a bounded sequence in such that and . Then .
Proof. By the definition, if and only if . By Lemma 2.3, we have .
Since , by induction we can prove that
In fact, it is obvious that, the conclusion is true for . Suppose the conclusion holds for , now we prove that the conclusion is also true for . In fact, since is a uniformly Lipschitzian mapping, we have
Equation (2.16) is proved. Hence for each and from (2.16) we have
In (2.18) taking , we have
Let and taking superior limit on the both sides, it gets that
Furthermore, for any it follows from inequality (2.3) with that
Let and taking superior limit on the both sides of the above inequality, for any we get
Since , we have
which implies that
By (2.20) and (2.24), we have . This implies that . Since is uniformly Lipschitzian, is uniformly continuous. Hence we have . This completes the proof of Lemma 2.9.
The following proposition can be obtained from Lemma 2.9 immediately which is a generalization of Kirk and Panyanak [17] and Nanjaras and Panyanak [13].
Proposition 2.10. Let be a closed and convex subset of a complete CAT(0) space and let be an asymptotically nonexpansive mapping. Let be a bounded sequence in such that and . Then .
Definition 2.11 (see [18]). Let be a CAT(0) space then is uniformly convex, that is, for any given and , there exists a such that, for all , where the function is called the modulus of uniform convexity of CAT(0).
Lemma 2.12 (see [14]). If is a bounded sequence in a complete CAT(0) space with , is a subsequence of with , and the sequence converges, then .
Lemma 2.13. Let , and be sequences of nonnegative real numbers satisfying the inequality If and , then is bounded and exists.
Lemma 2.14 (see [13]). Let be a CAT(0) space, be a given point and be a sequence in with and . Let and be any sequences in such that for some . Then
3. Main Results
In this section, we will prove our main theorem.
Theorem 3.1. Let be a nonempty bounded closed and convex subset of a complete CAT(0) space . Let be a asymptotically nonexpansive mapping with sequence , and be a uniformly L-Lipschitzian and total asymptotically nonexpansive mapping such that . From arbitrary , defined the sequence as follows: for all , where is a sequence in (0, 1). If the following conditions are satisfied:(i);(ii)there exists a constant such that , ;(iii)there exist constants with such that ;(iv) for each bounded subset of .Then the sequence -converges to a fixed point of .
Proof. We divide the proof of Theorem 3.1 into four steps. (I) First we prove that for each the following limit exists
In fact, for each , we have
It follows from Lemma 2.13 that is bounded and exists. Without loss of generality, we can assume . (II) Next we prove that
In fact, since
for all and , we have
On the other hand, since
by Lemma 2.14, we have
From condition (iv), we have
Hence from (3.8) and (3.9) we have that
By (3.9) and (3.10) it gets that
Hence from (3.10) and (3.11) we have that
Again since is uniformly -Lipschitzian, from (3.10) and (3.12) we have that
Equation (3.4) is proved. (III) Now we prove that
and consists exactly of one point.
In fact, let , then there exists a subsequence of such that . By Lemmas 2.5 and 2.3, there exists a subsequence of such that . By Lemma 2.9, we have . By Lemma 2.12, . This shows that .
Let be a subsequence of with and let . Since and converges, by Lemma 2.12, we have . This shows that consists of exactly one point. (IV) Finally we prove -converges to a point of .
In fact, it follows from (3.2) that is convergent for each . By (3.4) . By (3.14) and consists of exactly one point. This shows that -converges to a point of .
This completes the proof of Theorem 3.1.
Conflict of Interests
The authors declare that they have no competing interests.
Authors’ Contribution
All the authors contributed equally to the writing of the present article. And they also read and approved the final paper.
Acknowledgments
The authors would like to express their thanks to the referees for their helpful suggestions and comments. This study was supported by the Scientific Research Fund of Sichuan Provincial Education Department (12ZB346).