# A New Result on Tricyclic Mappings

Authors: Taoufik Sabar; Abdelhafid Bassou; Mohamed Aamri
Journal: Journal of Mathematics
 (2020)

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## 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.


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## Body

## 1. Introduction and Preliminaries

Back in 2017, the current authors introduced the class of tricyclic mappings and best proximity points thereof. A mappingT:A∪B∪C⟶A∪B∪C is said to be tricyclic provided that TA⊆B, TB⊆C, and TC⊆A, where A,B,andC 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 Dx,Tx,T2x=DA,B,C where the mapping D:X×X×X⟶0,+∞ is defined by Dx,y,z:=dx,y+dy,z+dz,x and DA,B,C:=infDx,y,z:x∈A,y∈Bandz∈C. Moreover, the mapping T is said to be a tricyclic contraction if it is tricyclic and verifies DTx,Ty,Tz≤kDx,y,z+1−kδA,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]).
LetA,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. Then, T has a best proximity point.Given three subsetsA,B, and C of a metric space X,d, we set(1)E≕x,y∈A×B:Δx,yC≤λ,where Δx,yC:=supDx,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. 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 subclassesCAT0spaces.A geodesic spaceX,d is a metric space in which every two points xandy are joined by a geodesic, that is, a map c:0,l⟶X such that c0=x and cl=y and dct,ct′=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 dx,z=λdx,y and dy,z=1−λdx,y and it is written as z=1−tx⊕ty.The geodesic spaceX is said to be reflexive if for every decreasing chain Ci⊂X, where i∈I, Ci is nonempty, bounded, closed, and convex for all i∈I such that we have ∩i∈ICi≠∅.A geodesic triangleΔp,q,r in X,d consists of three points p,q, and rvertices ofΔ connected by three geodesic segments sides ofΔ. A comparison triangle for Δp,q,r is a triangle Δ¯p,q,r:=Δp¯,q¯,r¯ 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 c1:0,a1⟶X and c2:0,a2⟶X; the following inequality is satisfied:(2)dc1ta1,c2ta2≤1−tdc10,c20+tdc1a1,c2a2for allt∈0,1.Equivalently,(3)d1−tx1⊕tx2,1−ty1⊕ty2≤1−tdx1,y1+tdx2,y2,for all x1,x2,y1,y2∈X and t∈0,1.Besides Busemann convex spaces, CAT0 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, CAT0 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 CAT0 spaces. Complete CAT0 spaces are called Hadamard spaces and are reflexive [4].


## 2. Main Results

Direct usage of the norm properties shows thatE is convex if X is a normed linear space and the subsets A and B are convex. The same goes for CAT0 spaces.Lemma 1.
IfA and B are nonempty convex subsets of a CAT0 space X, then so is E.Proof.
Letx1,y1,x2,y2∈E, and the map given by(4)t↦1−tx1⊕tx2,1−ty1⊕ty2is a geodesic joining x1,y1 and x2,y2 in the set X×X endowed with the metric dx,x′,y,y′=dx,y+dx′,y′, so we must verify that(5)1−tx1⊕tx2,1−ty1⊕ty2∈E.
Givenz∈C,(6)D1−tx1⊕tx2,1−ty1⊕ty2,z=d1−tx1⊕tx2,1−ty1⊕ty2+d1−tx1⊕tx2,z+d1−ty1⊕ty2,z,≤1−tdx1,y1+tdx2,y2+1−tdx1,z+tdx2,z+1−tdy1,z+tdy2,z,=1−tDx1,y1,z+tDx2,y2,z≤maxDx1,y1,z,Dx2,y2,z≤maxΔx1,y1C,Δx2,y2C≤λ.
Hence,E is convex.
Now, we can assert our best proximity point existence result.Theorem 2.
LetA,B, and C be nonempty, closed, bounded, and convex subsets of Hadamard space X, and let T:A∪B∪C⟶A∪B∪C be a tricyclic contraction map. Then, 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 thatE 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.Proposition 1 (see [6], Exercise 4.3.8).
The geodesic spaceX,d is of nonnegative (resp., nonpositive) curvature if the following property is true: let Δp,q,r be a triangle in X and Δp¯,q¯,r¯ be its comparison triangle. Then, for points xandy in the sides p,q and q,r and points x¯andy¯ in the sides p¯,q¯ and q¯,r¯ such that dx,q=dx¯,q¯ and dy,q=dy¯,q¯, the inequality dx,y≥dx¯,y¯ (resp., dx,y≤dx¯,y¯) holds.The following observation is natural, yet it has been very useful to our work.Proposition 2.
LetX be a space with nonnegative curvature, given any pair of geodesics c:0,1⟶X and c′:0,1⟶X emanating from the same point p=c0=c′0; the following inequality holds for all t∈0,1:(7)dct,c′t≥tdc1,c′1.Proof.
Consider a comparison triangleΔ¯⊂E2 for the triangle Δp,q,r:=Δc0,c1,c′1. Given t∈0,1, elementary geometry tells us that(8)dct¯,c′t¯=tdc1¯,c′1¯=tdc1,c′1.And, by the nonnegativity of the curvature dct,c′t≥dct¯,c′t¯; thus, dct,c′t≥tdc1,c′1.Remark 1.
The inequality from the previous proposition implies that(9)sup0<t<1dct,c′t≥dc1,c′1.and, inspired by the last inequality, we may claim that there are cases where the inequality in the last proposition becomes(10)dct,c′t>dc1,c′1,for some t∈0,1, just like illustrated in the following (Figure 1).
Now, letX,d be a space with nonnegative curvature, and assume there exist three geodesic segments of X, A=p,q, B=p,r, and C=p,s such that dp,q=dp,r=ds,q=ds,r and dp,s=dq,r. Suppose(11)Δq,rC=Dp,q,r=Ds,q,r.And, from the previous remark, we assume that there exist x∈p,q and y∈p,r such that(12)dx,y>dq,randdx,s=dy,s.
Put(13)E=x,y∈A×B:Δx,yC≤Dp,q,r.
Sincep,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 ds,q>ds,x, we have(14)dx,y−dq,r>2ds,q−ds,x.
Then,(15)Dx,y,s=dx,y+2ds,x,>dq,r+2ds,q,=Dq,r,s=Dq,r,p.which means that E is not convex.
Now, we support our claim with a concrete example where all the previous suppositions hold true.Figure 1
An illustrative figure.Example 1.
LetX:=S2 be the unit sphere in ℝ3 endowed with the function d:X×X⟶ℝ that assigns to each pair x,y∈X×X the unique real number dx,y=cos−1<x,y>∈0,π where <x,y> denotes the Euclidean scalar product. It is known that d is a metric on a X. Moreover, dx,y is the length of the shortest curve on S2 connecting x and y. Such a curve is a segment on a great circle through x and y (That is the intersection of the sphere with the plane through xandy and the centre of the sphere.). A natural way to parameterize arcs of great circles is as follows: given a point x∈S2, a unit vector u in the Euclidean space ℝ3 such that x,u=0, and a number a∈0,π, consider the path c:0,a⟶S2 given by ct=costA+sintu. We obtain(16)dct,ct′≤t−t′for allt,t′∈0,a,and the image of c is contained in the great circle where the vector subspace spanned by x and u meets S2. The image of c is said to be a minimal great arc, more precisely, the great arc with initial vector u and joining x to ca. Let p=0,0,1, q=−3/3,2/3,−2/3, r=3/3,2/3,−2/3, and s=0,−22/3,1/3. Since (Figure 2),(17)p,q=p,r=s,q=s,r=−23andp,s=q,r=13,
We obtain(18)dp,q=dp,r=ds,q=ds,randdp,s=dq,r.
And, if we considerC to be the major arc joining p and s, we plainly see that(19)Δq,rC=Dp,q,r=Ds,q,r.
Consider the vectors(20)x≕−35,25,0andy≕35,25,0,and the following paths(21)c:0,cos−1−23⟶S2,t↦−35sint,25sint,cost,c′:0,cos−1−23⟶S2,t↦35sint,25sint,cost.
Clearly, the image ofcc′,resp. is the great arc with initial vector x and joining p to qyand joiningptor,resp.. And,(22)x,y=−15andx,s=y,s=−4515,implying that(23)dx,y>dq,randdx,y−dq,r>2ds,q−ds,x.
Now, all of the conditions previously mentioned are furnished.Figure 2
A counter example in three dimension.Remark 2.
Retain the same notations as in the previous example. We emphasize that the major great arcC is not a geodesic segment in the sense of geodesic metric spaces. Indeed, an arc of a great circle on S2 fails to be a geodesic segment as long as it is longer than half of the equator. However, if S2 is seen as a two-dimensional manifold in the sense of differential geometry, C is a geodesic and is geodesically convex.



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*Source: 9525890-2020-10-05.xml*