1. Introduction
Hadamard metric spaces, also known as complete
metric spaces, play an important role when dealing with the geometry of Bruhat–Tits building, metric trees, Hadamard manifolds, or simply connected nonpositively curved symmetric spaces. In fact, the power of Hadamard spaces goes beyond geometry. For example,
geometry was used to solve an interesting problem in Dynamical billiards [
1]. In a very simplistic way, Hadamard metric spaces are the nonlinear version of Hilbert vector spaces.
A metric Hadamard space
is characterized by the inequality (
1) [
2,
3], known as the inequality of Bruhat and Tits, i.e., for any
, there exists
such that
It is easy to check that
c is a metric midpoint of
a and
b, i.e.,
. Note that in the linear Hilbert spaces, the inequality (
1) becomes an equality. A Hadamard space for which the inequality (
1) is an equality are known as flat Hadamard spaces. They are isomorphic to closed convex subsets of Hilbert spaces.
To understand this inequality, one has to look at the formal definition of a
space with the comparison triangles taken in the Euclidean plane
. In this case, the triangles in
M are kind of slimmer than the comparison triangles in
. This fundamental property motivated the authors of [
4] to consider metric spaces for which the corresponding triangles are taken in general Banach spaces. The most natural example is to take these comparison triangles in
, for
. The authors of [
4] called these more general metric spaces “generalized
”. In this work, we continue investigating the properties of these metric spaces and establish some existence fixed point results and their approximations.
For readers interested in metric fixed point theory, we recommend the book [
5]. For more on geodesic metric spaces, we recommend the excellent book [
2].
2. Basic Definitions and Preliminaries
Most of the terminology of geodesic metric spaces is taken from the work in [
2]. Consider a metric space
. A geodesic function
is any function that satisfies
, for every
.
is said to be a geodesic space if, for every two points
, there exists a function
such that
and
and is geodesic. Throughout, we will use the notation
, for
.
is said to be uniquely geodesic if any two points in
M are connected by a unique geodesic. In this case, the range of the unique geodesic function connecting
a and
b will be denoted by
, i.e.,
.
Normed vector spaces are natural examples of geodesic metric spaces. Complete Riemannian manifolds, and polyhedral complexes of piecewise constant curvature are examples of nonlinear geodesic metric spaces. In these two examples, it is not obvious to show the existence of geodesics and show that they are unique. To determine when such spaces are uniquely geodesic is also a very hard task.
Geodesic triangles are naturally introduced in geodesic metric spaces. Indeed, let
be a geodesic metric space. Any three points—
—will define a geodesic triangle
, which consists of the three given points called its vertices and the geodesic segments between each pair of vertices also known as the edges of
. Comparison triangles are crucial to the definition of
spaces [
6]. Given a geodesic triangle
in a geodesic metric space
, a triangle
in the Euclidean plane
is said to be a comparison triangle to
whenever
holds for any
. The point
is called a comparison point for
, for any
and
.
Throughout stands for a uniquely geodesic metric space.
Definition 1. [2] Let be a geodesic triangle in M and be a comparison triangle for Δ
in . We say that Δ
satisfies the inequality if for any and their comparison points , the following holds, is said to be a space if all geodesic triangles satisfy the inequality. A Hadamard metric space is any complete space [7]. Let
be a
space. Let
be in
M. If
is the midpoint of
and
, then the
inequality implies:
Strictly convex Banach spaces are obviously uniquely geodesic. It is well known that a normed vector space is a
space if and only if it is a pre-Hilbert space [
2].
A recent extension to
spaces was initiated in [
4]. It is based on the idea that comparison triangles belong to a general Banach spaces instead of the Euclidean plane.
Definition 2. [4] be a Banach space. The geodesic metric space is said to be a space if for any geodesic triangle Δ
in M, there exists a comparison triangle in such thatfor any and their comparison points . If , for , we say M is a space. It is obvious that is a space. If is not a pre-Hilbert space, then is not a space. In other words, Definition 2 gives a larger class of hyperbolic metric spaces.
Throughout our work, we mainly focus on metric spaces for . It is obvious that space is exactly the classical space, which has been extensively studied.
The classical inequality (
1) gives information about the middle point of two points. Except that many successive iterations, which present some interesting behavior, do not involve the middle point, but a convex combination of the given two points. Therefore, it is of utmost importance to prove or discover some metric properties of this kind of convex combinations.
Next, we will discuss a property of the convex combinations that holds in spaces for .
Lemma 1. Let be a metric space, with . Then, for any in M and , we have Proof. Note that this inequality is valid in
, for
. Indeed, Lim [
8] proved the following inequality,
for any
and
, where
and
, for
is the unique solution to
In particular, we have
, for any
. Therefore,
for any
and
. Next, we turn our attention to the proof of Lemma 1. Consider the geodesic triangle
, where
, and
. Ase
M is a
space, there exists a comparison geodesic triangle
in
. The comparison axiom implies that
The inequality (
2) implies that
which implies the conclusion of Lemma 1. □
Throughout, we will use the notation , for .
In the next section, we extend some known fixed point results in Banach spaces and spaces to the case of , for .
3. Fixed Point Results in
Next, we investigate the fixed point problem for the class of asymptotically nonexpansive mappings. Note that this family of mappings was introduced by Goebel and Kirk [
9] as a family of mappings that sits between the family of nonexpansive mappings [
10] and the family of uniformly Lipschitzian mappings [
11].
As we said before, throughout we consider to be a geodesic metric space.
Definition 3. [9,10,11] Let be a map. - (1)
J is asymptotically nonexpansive if there exists such that andfor any and . We can always assume that , for any . - (2)
J is uniformly Lipschitzian if there exists such thatfor any and . - (3)
A point is a fixed point of J if holds. will denote the set of fixed points of J.
The fixed point problem for this class of mappings was extensively investigated [
12,
13,
14,
15]. It followed two directions: The first deals with the existence of a fixed point. The second deals with the approximation of the fixed points based on algorithms initiated by Schu [
16]. In this work, we will follow the same directions as well.
A powerful tool used in investigating the existence of fixed points is the concept of type functions, which plays a major role in the study of metric fixed point theory in Banach spaces. Historically, it is also known as the asymptotic center.
Definition 4. Let be a metric space. A function is a type function if there exists a bounded sequence in M such thatfor any . A sequence in M is said to be a minimizing sequence of θ whenever The following technical lemma shows why type functions are a powerful tool.
Lemma 2. Let be a complete metric space, with . Let C be a nonempty closed convex subset of M and be a type function generated by a bounded sequence . Then, the following hold.
- (1)
Any minimizing sequence of θ is convergent.
- (2)
All minimizing sequences of θ converge to the same limit .
- (3)
z is a minimum point of θ, i.e., .
Proof. Set
. Without loss of generality, we assume
. Let
be a minimizing sequence of
. Assume that
is not Cauchy. As any subsequence of
is also a minimizing sequence of
, we may assume there exists
such that
, for any
. As
C is convex, then
, for any
. Lemma 1 implies
for any
. If we let
, we get
which implies
for any
. If we let
, we get
This contradiction shows that is Cauchy, which shows that holds. To prove , let and be two minimizing sequences of . Consider the sequence defined by and , for any . Then, is also a minimizing sequence of . From we conclude that is convergent. As both and are subsequence of we conclude that and have the same limit. The conclusion of follows from the simple fact that type functions are continuous. □
In the first result, we discuss the existence of a fixed point for asymptotically nonexpansive mappings in spaces.
Theorem 1. Let be a complete metric space, with . Let C be a nonempty closed bounded convex subset of M. Let be an asymptotically nonexpansive mapping. Then, J has a fixed point. Moreover, is closed and convex.
Proof. Let
be the Lipschitz sequence associated to
J. Fix
. Consider the type function
generated by
. Let
z be the minimum point of
which exists by using Lemma 2. Therefore,
for any
. If we let
, we will get
, for any
. If we let
, and using
, we get
, i.e.,
is a minimizing sequence of
. Using Lemma 2, we conclude that
converges to
z. As
J is continuous, we conclude that
. The fact that
is closed is obvious from the continuity of
J. Let us prove that
is convex. Let
be different. As
is closed, we only need to prove that
. Note that
for
and
. Therefore,
for any
. As
, we conclude that
Similarly, we can show that
Using Lemma 1, we get
for any
. If we let
, we get
. As
J is continuous, then
as claimed, which completes the proof of Theorem 1. □
Note that we may refine the boundedness assumption of C by assuming that an orbit of J is bounded. In this case, the above proof still holds. The convexity of the set of fixed points is a useful information, because it will allow us to prove the existence of a common fixed point for this class of mappings for example.
Next, we discuss the behavior of the successive iterations introduced by Schu [
16] for asymptotically nonexpansive. In this case, Lemma 1 will prove to be crucial.
Recall that for a bounded nonempty subset
C of a metric space
,
denotes the diameter of
C and is defined by
Lemma 3. Let be a complete metric space, with . Let C be a nonempty closed bounded convex subset of M. Let be an asymptotically nonexpansive mapping with as its associated Lipschitz constants. Let , such that . The modified Mann iteration process [16] is defined byfor any , where is a fixed arbitrary point. If , then exists. Proof. First, note that
which implies
, for any
. In particular, we have
, for any
. Therefore,
for any
. If we let
, we get
for any
. Using the assumption
, we obtain
Therefore, , i.e., is convergent as claimed. □
Remark 1. In the original work of Schu [16], the conclusion of Lemma 3 is obtained under the stronger assumption . One may be confused to how we can assume a weaker assumption that involves the sequence . In fact, the construction of the sequence is done during the computation of the sequence at the same time making sure we have the convergence of the series . Moreover, if we assume that , for some , then is convergent if and only if is convergent. In the next result, we show that the sequence generated by (
3) almost gives a fixed point.
Theorem 2. Let be a complete metric space, with . Let C be a nonempty closed bounded convex subset of M. Let be an asymptotically nonexpansive mapping with as its associated Lipschitz constants. Assume . Let , where . Fix and consider the sequence generated by the iteration (3). Then,for any , i.e., the sequence is said to be an approximate fixed point sequence of T. Proof. First, let us prove that
. Using Theorem 1,
J has a fixed point
. Lemma 3 implies that
exists. Using Lemma 1, we get
since
, which implies
for any
. As
we conclude that
. Next, we prove that for any
, we have
. As
is convergent, it is bounded. Set
. Therefore,
for any
. As
, we conclude that
Finally, fix
. As we have
for any
. Therefore, we have
, for any
. □
From a computational point of view, the algorithm (
3) almost generated a fixed point of
J. However, from a mathematical point of view, we still need to look at the convergence of
. First, we have a strong convergence as did Schu [
16]. To obtain this, we need some kind of compactness assumption. Recall that
is said to be compact if the closure of
is compact.
Theorem 3. Let be a complete metric space, with . Let C be a nonempty closed bounded convex subset of M. Let be an asymptotically nonexpansive mapping such that is compact for some . Assume , where is the Lipschitz sequence associated to J. Let , where . Fix and consider the sequence generated by the iteration (3). Then, converges strongly to a fixed point z of J, i.e., . Proof. Let such that is compact. Then, there exists a subsequence such that converges to some point . Using Theorem 2, we know that , which implies also converges to z. Again, using Theorem 2, we know that , which implies that also converges to z. As J is continuous, we conclude that , i.e, . Moreover, Lemma 3 implies exists. As , we conclude that , i.e., . In other words, the sequence converges to z as claimed. □
Therefore, we wonder whether a weaker convergence is happening if we relax the compactness assumption in Theorem 3. In the original work of Schu [
16], the setting is a Banach space. Therefore, we may consider naturally the weak topology. In the nonlinear setting, it is still unknown what the weak topology looks like. Lim [
17] introduced a convergence concept he called
-convergence based on the asymptotic center of a sequence. Except that this convergence does not capture the weak topology once we restrict ourselves to Banach spaces. It only happens if the Banach space enjoys the Opial property [
18]. In the next result, we discard the compactness assumption.
Theorem 4. Let be a complete metric space, with . Let C be a nonempty closed bounded convex subset of M. Let be an asymptotically nonexpansive mapping. Assume , where is the Lipschitz sequence associated to J. Let , where . Fix and consider the sequence generated by the iteration (3). For any subsequence , consider the type on C. Then, the minimum point z of is independent of the subsequence and is a fixed point of J. We say that Δ
-converges to z. Proof. Consider the type function
defined by
, i.e.,
, for any
. According to Lemma 2, the type function
has a unique minimum point
z, which is a fixed point of
J. Let us prove that
z is also the minimum point of any type function
generated by a subsequence
of
. Again according to Lemma 2, there exists a unique minimum point
of
, which is also a fixed point of
J. Lemma 3 implies that
and
exist. As
z is the minimum point of
, we get
, which implies
or
i.e.,
. The uniqueness of the minimum point of
implies that
, which completes the proof of Theorem 4. □