A New Result on Tricyclic Mappings

Back in 2017, the current authors introduced the class of tricyclic mappings and best proximity points thereof. A mapping T: A∪B∪C⟶ A∪B∪C is said to be tricyclic provided that T(A)⊆B, T(B)⊆C, and T(C)⊆A, where A, B, andC are nonempty subsets of ametric space (X, d), and a best proximity point of T is a point x ∈ A∪B∪C such that D(x, Tx, T2x) � D(A, B, C) where the mapping D: X × X × X⟶ [0, +∞) is defined by D(x, y, z): � d(x, y) + d(y, z) + d(z, x) and D (A, B, C): � inf D(x, y, z): x ∈ A, y ∈ B and z ∈ C 􏼈 􏼉. Moreover, the mapping T is said to be a tricyclic contraction if it is tricyclic and verifies D(Tx, Ty, Tz)≤ kD(x, y, z) + (1 − k)δ (A, B, C), for some k ∈ (0, 1) and for all (x, y, z) ∈ A × B × C. 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.


Introduction and Preliminaries
Back in 2017, the current authors introduced the class of tricyclic mappings and best proximity points thereof. A mapping T: A ∪ B ∪ C ⟶ A ∪ B ∪ C is said to be tricyclic provided that T(A) ⊆ B, T(B) ⊆ C, and T(C) ⊆ A, where A, B, and C are nonempty subsets of a metric space (X, d), and a best proximity point of T is a point x ∈ A ∪ B ∪ C such that D(x, Tx, T 2 x) � D(A, B, C) where the mapping D: X × X × X ⟶ [0, +∞) is defined by D(x, y, z): � d(x, y) + d(y, z) + d(z, x) and D (A, B, C): � inf D(x, y, z): x ∈ A, y ∈ B and z ∈ C . Moreover, the mapping T is said to be a tricyclic contraction if it is tricyclic and verifies D(Tx, Ty, Tz) ≤ kD(x, y, z) + (1 − k)δ (A, B, C), for some k ∈ (0, 1) and for all (x, y, z) ∈ A × B × C. 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 A, B, and C be nonempty, closed, bounded, and convex subsets of a reflexive Banach space X, and let T: A ∪ B ∪ C ⟶ A ∪ B ∪ C be a tricyclic contraction map. en, T has a best proximity point.
Given three subsets A, B, and C of a metric space (X, d), we set where Δ (x,y) (C): � sup D(x, y, z): z ∈ C and λ 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 E is convex. It is easy to see that E is convex in linear normed spaces. But how far from linear metric spaces can we go? Later on, we show that the set E 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. en, 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 CAT(0) spaces.
A geodesic space (X, d) is a metric space in which every two points x and y are joined by a geodesic, that is, a map c: [0, l] ⟶ X such that c(0) � x and c(l) � y and d(c(t), c(t′)) � |t − t ′ | for all t, t ′ ∈ [0, l]. A subset C of a geodesic space X is said to be convex if every pair of points x, y ∈ C are joined by a geodesic whose image is contained in C. A point z ∈ X belongs to a geodesic segment [x, y] if and only if there exists λ ∈ [0, 1] such that d(x, z) � λd(x, y) and d(y, z) � (1 − λ)d(x, y) and it is written as e geodesic space X is said to be reflexive if for every decreasing chain (C i ) ⊂ X, where i ∈ I, C i is nonempty, bounded, closed, and convex for all i ∈ I such that we have ∩ i∈I C i ≠ ∅.
A geodesic triangle Δ(p, q, r) in (X, d) consists of three points p, q, and r (vertices of Δ) connected by three geodesic segments (sides of Δ). A comparison triangle for Δ(p, q, r) is a triangle Δ(p, q, r): � Δ(p, tqn, qr) 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 c 1 : [0, a 1 ] ⟶ X and c 2 : [0, a 2 ] ⟶ X; the following inequality is satisfied: Equivalently, for all x 1 , x 2 , y 1 , y 2 ∈ X and t ∈ [0, 1]. Besides Busemann convex spaces, CAT(0) spaces is a very important subclass of geodesic spaces. ose 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(0) 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(0) spaces. Complete CAT(0) spaces are called Hadamard spaces and are reflexive [4].

Main Results
Direct usage of the norm properties shows that E is convex if X is a normed linear space and the subsets A and B are convex. e same goes for CAT(0) spaces.

Lemma 1. If A and B are nonempty convex subsets of a CAT(0) space X, then so is E.
Proof. Let (x 1 , y 1 ), (x 2 , y 2 ) ∈ E, and the map given by is a geodesic joining (x 1 , y 1 ) and (x 2 , y 2 ) in the set X × X endowed with the metric d((x, x ′ ), (y, y ′ )) � d(x, y) + d(x ′ , y ′ ), so we must verify that Given z ∈ C, Hence, E is convex. Now, we can assert our best proximity point existence result. en, T 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 E is also convex in the setting of hyperbolic spaces in the sense of Kohlenbach [5]. And, in [2], it is proved that E is convex in the setting of (S) − convex metric spaces, if A, B, and C verify supplementary conditions. Now, besides the metric spaces mentioned previously in this article, does E 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.
e geodesic space (X, d) is of nonnegative (resp., nonpositive) curvature if the following property is true: let Δ(p, q, r) be a triangle in X and Δ(p, tqn, qr) be its comparison triangle. en, for points x and y in the sides [ d(x, y) ≤ d(x, ty)) holds. e following observation is natural, yet it has been very useful to our work. And, by the nonnegativity of the curvature e inequality from the previous proposition implies that and, inspired by the last inequality, we may claim that there are cases where the inequality in the last proposition becomes for some t ∈ (0, 1), just like illustrated in the following ( Figure 1). Now, let (X, d) be a space with nonnegative curvature, and assume there exist three geodesic segments of X, And, from the previous remark, we assume that there exist x ∈ (p, q) and y ∈ (p, r) such that d(x, y) > d(q, r) and d(x, s) � d(y, s).
Since (p, p), (q, r) ∈ E, if E is convex, then it must contain (x, y). We now show the opposite, in this order, and we suppose that x and y verify the extra condition that, if en, which means that E is not convex. Now, we support our claim with a concrete example where all the previous suppositions hold true. Example 1. Let X: � S 2 be the unit sphere in R 3 endowed with the function d: X × X ⟶ R that assigns to each pair (x, y) ∈ X × X the unique real number d(x, y) � where <x, y > denotes the Euclidean scalar product. It is known that d is a metric on a X. Moreover, d(x, y) is the length of the shortest curve on S 2 connecting x and y. Such a curve is a segment on a great circle through x and y ( at is the intersection of the sphere with the plane through x and y and the centre of the sphere.). A natural way to parameterize arcs of great circles is as follows: given a point x ∈ S 2 , a unit vector u in the Euclidean space R 3 such that 〈x, u〉 � 0, and a number a ∈ [0, π], consider the path c: [0, a] ⟶ S 2 given by c(t) � (cos t)A + (sin t)u. We obtain and the image of c is contained in the great circle where the vector subspace spanned by x and u meets S 2 . e image of c is said to be a minimal great arc, more precisely, the great arc with initial vector u and joining x to c(a). Let p � (0, 0, 1), q � (− ( We obtain d(p, q) � d(p, r) � d(s, q) � d(s, r) and d(p, s) � d(q, r).

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And, if we consider C to be the major arc joining p and s, we plainly see that Δ (q,r) (C) � D(p, q, r) � D(s, q, r).
Consider the vectors , 0 and y ≕ � 3 5 , and the following paths   Journal of Mathematics