#### Abstract

In this note, we present a best proximity point theorem for tricyclic contractions in the setting of CAT(0) spaces. In the same context, we give an elaborate counterexample.

#### 1. Introduction and Preliminaries

Back in 2017, the current authors introduced the class of tricyclic mappings and best proximity points thereof. A mapping is said to be tricyclic provided that , , and , where are nonempty subsets of a metric space , and a best proximity point of is a point such that where the mapping is defined by and . Moreover, the mapping is said to be a tricyclic contraction if it is tricyclic and verifies , for some and for all . For detailed information, we refer to [1, 2].

In the same research paper where they defined tricylic mappings, the following best proximity point result was obtained.

Theorem 1 (see [1]). *Let , and be nonempty, closed, bounded, and convex subsets of a reflexive Banach space , and let be a tricyclic contraction map. Then, has a best proximity point.*

Given three subsets , and of a metric space , we setwhere and is a nonnegative real number.

Taking heed in the proof of the main result of [1], one becomes aware that the proof of such result in a metric space, where some sort of convexity and reflexivity is defined of course, comes down to proving that the setting is convex. It is easy to see that is convex in linear normed spaces. But how far from linear metric spaces can we go? Later on, we show that the set loses its convexity in an arbitrary geodesic metric space.

In this paper, we extend the previous theorem to a wider class of spaces, and it is that of geodesic metric spaces. Then, we give an answer to the tricyclic mapping-related question that remained open for a couple of years in the form of a counterexample.

Preparatory to giving our main result, we hark back to geodesic spaces and one of their most important subclasses .

A geodesic space is a metric space in which every two points are joined by a geodesic, that is, a map such that and and for all . A subset of a geodesic space is said to be if every pair of points are joined by a geodesic whose image is contained in . A point belongs to a geodesic segment if and only if there exists such that and and it is written as .

The geodesic space is said to be if for every decreasing chain , where , is nonempty, bounded, closed, and convex for all such that we have .

A geodesic triangle in consists of three points , and connected by three geodesic segments . A comparison triangle for is a triangle in the Euclidean plane with the same lengths of sides.

We also recall that a geodesic space is called convex in the sense of Buesmann [3] if given any pair of geodesics and ; the following inequality is satisfied:

Equivalently,for all and .

Besides Busemann convex spaces, CAT spaces is a very important subclass of geodesic spaces. Those are metric spaces that are, of course, geodesically connected, and their geodesic triangles are at least as “thin” as their comparison triangles in the Euclidean plane.

Needless to say, CAT spaces are convex in the sense of Busemann. As a matter of fact, that is the most fundamental of the properties defining the nature of CAT spaces. Complete CAT spaces are called Hadamard spaces and are reflexive [4].

#### 2. Main Results

Direct usage of the norm properties shows that is convex if is a normed linear space and the subsets and are convex. The same goes for CAT spaces.

Lemma 1. *If and are nonempty convex subsets of a CAT space , then so is .*

*Proof. *Let , and the map given byis a geodesic joining and in the set endowed with the metric , so we must verify thatGiven ,Hence, is convex.

Now, we can assert our best proximity point existence result.

Theorem 2. *Let , and be nonempty, closed, bounded, and convex subsets of Hadamard space , and let be a tricyclic contraction map. Then, has a best proximity point.**Evidently, the previous theorem generalizes the one of [1], since the latter was proven in the setting of reflexive Banach spaces.**Furthermore, it is easy to see that is also convex in the setting of hyperbolic spaces in the sense of Kohlenbach [5]. And, in [2], it is proved that is convex in the setting of metric spaces, if , and verify supplementary conditions. Now, besides the metric spaces mentioned previously in this article, does stay convex in a more general framework, say, geodesic spaces or any metric spaces, with some sense of convexity?**Next, we claim that it is not the case and we give a subtle counterexample. First, we mention a couple of propositions.*

Proposition 1 (see [6], Exercise 4.3.8). *The geodesic space is of nonnegative (resp., nonpositive) curvature if the following property is true: let be a triangle in and be its comparison triangle. Then, for points in the sides and and points in the sides and such that and , the inequality (resp., ) holds.*

The following observation is natural, yet it has been very useful to our work.

Proposition 2. *Let be a space with nonnegative curvature, given any pair of geodesics and emanating from the same point ; the following inequality holds for all :*

*Proof. *Consider a comparison triangle for the triangle . Given , elementary geometry tells us thatAnd, by the nonnegativity of the curvature ; thus, .

*Remark 1. *The inequality from the previous proposition implies thatand, inspired by the last inequality, we may claim that there are cases where the inequality in the last proposition becomesfor some , just like illustrated in the following (Figure 1).

Now, let be a space with nonnegative curvature, and assume there exist three geodesic segments of , , , and such that and . SupposeAnd, from the previous remark, we assume that there exist and such thatPutSince , if is convex, then it must contain . We now show the opposite, in this order, and we suppose that and verify the extra condition that, if , we haveThen,which means that is not convex.

Now, we support our claim with a concrete example where all the previous suppositions hold true.

*Example 1. *Let be the unit sphere in endowed with the function that assigns to each pair the unique real number where denotes the Euclidean scalar product. It is known that is a metric on a . Moreover, is the length of the shortest curve on connecting and . Such a curve is a segment on a great circle through and (That is the intersection of the sphere with the plane through and the centre of the sphere.). A natural way to parameterize arcs of great circles is as follows: given a point , a unit vector in the Euclidean space such that , and a number , consider the path given by . We obtainand the image of is contained in the great circle where the vector subspace spanned by and meets . The image of is said to be a minimal great arc, more precisely, the great arc with initial vector and joining to . Let , , , and . Since (Figure 2),We obtainAnd, if we consider to be the major arc joining and , we plainly see thatConsider the vectorsand the following pathsClearly, the image of is the great arc with initial vector and joining to . And,implying thatNow, all of the conditions previously mentioned are furnished.

*Remark 2. *Retain the same notations as in the previous example. We emphasize that the major great arc is not a geodesic segment in the sense of geodesic metric spaces. Indeed, an arc of a great circle on fails to be a geodesic segment as long as it is longer than half of the equator. However, if is seen as a two-dimensional manifold in the sense of differential geometry, is a geodesic and is geodesically convex.

#### Data Availability

No data were used to support this study.

#### Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

#### Acknowledgments

This research was supported by the National Centre of Scientific and Technological Research grant.