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arXiv:1003

.5744v3[math.MG]6Apr2011

FIXED POINTS AND LINES IN 2-METRIC SPACES

ABDELKRIM ALIOUCHE AND CARLOS SIMPSON

Abstract. We consider bounded 2-metric spaces satisfying anadditional axiom, and show that a contractive mapping has eithera fixed point or a fixed line.

1. Introduction

Gahler introduced in the 1960s the notion of 2-metric space [10] [11][12], and several authors have studied the question of fixed point the-orems for mappings on such spaces. A 2-metric is a function d(x, y, z )symmetric under permutations, satisfying the tetrahedral inequality

d(x,y,z) d(x,y,a)+ d(x,a,z)+ d(a,y,z) for all x, y, z, aX.

as well as conditions (Z) and (N) which will be recalled below. In theprototypical example, d(x,y,z) is the area of the triangle spanned byx, y, z .

This notion has been considered by several authors (see [9]), whohave notably generalized Banachs principle to obtain fixed point theo-

rems, for example White [31], Iseki[14], Rhoades [28], Khan [16], Singh,Tiwari and Gupta [30], Naidu and Prasad[25], Naidu [26] and Zhang[18], Abd El-Monsef, Abu-Donia, Abd-Rabou[2], Ahmed [3]and oth-ers. The contractivity conditions used in these works are usually of theform

d(F(x), F(y), a) . . .

for any a X. We may think of this as meaning that d(x,y,a) is afamily of distance-like functions ofx and y, indexed by a X. Thisinterpretation intervenes in our transitivity condition (Trans) below.However, Hsiao has shown that these kinds of contractivity conditions

dont have a wide range of applications, since they imply colinearityof the sequence of iterates starting with any point [13]. We thank B.Rhoades for pointing this out to us.

There have also been several different notions of a space togetherwith a function of 3-variables. For example, Dhage [8] introduced the

2000 Mathematics Subject Classification. Primary 54H25; Secondary 47H10.Key words and phrases. 2-metric space, D-metric, Fixed point.

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2 A. ALIOUCHE AND C. SIMPSON

concept ofD-metric space and proved the existence of a unique fixedpoint of a self-mapping satisfying a contractive condition. Dhages def-

inition uses the symmetry and tetrahedral axioms present in Gahlersdefinition, but includes the coincidenceaxiom thatd(x, y, z ) = 0 if andonly ifx = y = z.

A sequence {xn} in a D-metric space (X, d) is said by Dhage to beconvergent to an element x X (or d-convergent)[8] if given >0,there exists an N Nsuch that d(xm, xn, x)< for all m, n N. Hecalls a sequence {xn}in aD-metric space (X, d) Cauchy (ord-Cauchy)[8] if given >0, there exists anN Nsuch thatd(xn, xm, xp)< forall n, m, p N.

These definitions, distinct from those used by Gahler et al, motivatethe definition of the propertyLIM(y, (xi)) in Definition4.4and studied

in Theorem4.8below.The question of fixed-point theorems on such spaces has proven to

be somewhat delicate [22]. Mustafa and Sims introduced a notion ofG-metric space [23][24], in which the tetrahedral inequality is replacedby an inequality involving repetition of indices. In their point of viewthe function d(x, y, z ) is thought of as representing the perimeter of atriangle.

The question of fixed points for mappings on G-metric spaces hasbeen considered by Abbas-Rhoades [1], Mustafa and co-authors [20],[21]. This is not an exhaustive description of the large literature onthis subject.

In the present paper, we return to the notion of 2-metric space. Thebasic philosophy is that since a 2-metric measures area, a contraction,that is a map F such that

(1.1) d(F(x), F(y), F(z)) kd(x, y, z )

for some k

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FIXED POINTS AND LINES 3

their results concern the euclidean case or the case of a 2-normed linearspace.

In order to obtain a treatment which applies to more general 2-metric spaces, yet always assuming that d is globally bounded (B), weadd an additional quadratic axiom (Trans) to the original definition of2-metric. Roughly speaking this axiom says that ifx, y,zare approxi-mately colinear, ify , z , ware approximately colinear, and ify and zarefar enough apart, then x,z,w and x, y, w are approximately colinear.The axiom (Trans) will be shown to hold in the example X=S2 whered(x,y,z) is given by a determinant (Section 5), which has appeared in[19], as well as for the standard area 2-metric on Rn. The abbreviationcomes from the fact that (Trans) implies transitivity of the relation ofcolinearity, see Lemma4.2. This axiom allows us to consider a notion

offixed lineof a mapping Fwhich is contractive in the sense of (1.1).With these hypotheses on d and under appropriate compactness as-sumptions we prove that such a mapping has either a fixed point or afixed line.

In the contractivity condition (1.1), the function F is applied toall three variables. Consequently, it turns out that there exist manymappings satisfying our contractivity condition, but not the trivialityobserved by Hsiao [13]. Some examples will be discussed in Section7. In the example ofS2 with the norm of determinant 2-metric, onecan take a neighborhood of the equator which contracts towards theequator, composed with a rotation. This will have the equator as fixed

line, but no fixed point, and the successive iterates of a given pointwill not generally be colinear. Interesting examples of 2-metrics onmanifolds are obtained from embedding in Rn and pulling back thestandard area 2-metric. The properties in a local coordinate chartdepend in some way on the curvature of the embedded submanifold.We consider a first case of patches on S2 in the euclidean R3. Thesesatisfy an estimate (Lemma 5.5) which allows to exhibit a flabbyfamily of contractible mappings depending on functional parameters(Proposition7.2). This shows that in a strong sense the objection of[13] doesnt apply.

The first section of the paper considers usual metric-like functions

of two variables, pointing out that the classical triangle inequality maybe weakened in various ways. A bounded 2-metric leads naturally tosuch a distance-like function (x, y) but we also take the opportunityto sketch some directions for fixed point results in this general context,undoubtedly in the same direction as [27] but more elementary.

As a small motivation to readers more oriented towards abstractcategory theory, we would like to point out that a metric space (in

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the classical sense) may be considered as an enriched category: theordered set (R0, ) considered as a category has a monoidal struc-

ture +, and a metric space is just an (R0, , +)-enriched category.We have learned this observation from Leinster and Willerton [17] al-though it was certainly known before. An interesting question is, whatcategorical structure corresponds to the notion of 2-metric?

2. Asymmetric triangle inequality

SupposeXis a set together with a function (x, y) defined forx, yXsuch that:(R)(x, x) = 0;(S)(x, y) = (y, x);(AT)for a constant C1 saying

(x, y) (x, z) + C(z, y).

In this case we say that (X, ) satisfies the asymmetric triangle in-equality.

It follows that (x, y) 0 for all x, y X. Furthermore, if weintroduce a relation x y when (x, y) = 0, then the three axiomsimply that this is an equivalence relation, and furthermore when x x

and y y then (x, y) = (x, y). Thus, descends to a functionon the quotient X/ and on the quotient it has the property that(x, y) = 0 x = y. In view of this discussion it is sometimesreasonable to add the strict reflexivityaxiom

(SR)if(x, y) = 0 then x= y.B. Rhoades pointed out to us that the asymmetric triangle inequality

implies the property (x, y) ((x, z) + (z, y)) of a quasidistanceused by Peppo [27] and it seems likely that the following discussioncould be a consequence of her fixed point result for (,i,j,k)-mappings,although that deduction doesnt seem immediate.

It is easy to see for (X, ) satisfying the asymmetric triangle inequal-ity, that the notion of limit for the distance function makes sense,similarly the notion of Cauchy sequence for makes sense, and we cansay that (X, ) is complete if every Cauchy sequence has a limit. If asequence has a limit then it is Cauchy. The function is continuous,

i.e., it transforms limits into limits. If furthermore the strictness axiom(SR) satisfied, then limit is unique.

A point of accumulationof a sequence (xi)iN is a limit of a subse-quence, that is to say a point y such that there exists a subsequence(xi(j))jN with i(j) increasing, such that y = limj xi(j). A set Xprovided with a distance function satisfying the asymmetric triangleinequality (i.e. (R), (S) and (AT)), is compactif every sequence has a

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point of accumulation. In other words, every sequence admits a conver-gent subsequence. This notion should perhaps be called sequentially

compact but it is the only compactness notion which will be used inwhat follows.

Lemma 2.1. Suppose(X, )satisfies the asymmetric triangle inequal-ity with the constant C. Suppose F : X X is a map such that(F x , F y) k(x, y) with k < (1/C). Then for any x X the se-quence{Fi(x)} is Cauchy. If (X, ) is complete and strictly reflexivethen its limit is the unique fixed point ofF.

Proof. Letx0be an arbitrary point inXand {xn} the sequence definedbyxn+1= F(xn) =F

n(x0) for all positive integer n. We have

(xn+1, xn) = (F xn, F xn1) k(xn, xn1).By induction, we obtain

(xn+1, xn)kn(x0, x1)

Using the asymmetric triangle inequality several times we get for allpositive integers n, msuch that m > n

(xn, xm) (xn, xn+1)+C(xn+1, xn+2)+C2(xn+2, xn+3)+...

...+ Cmn1(xm1, xm).

Then

(xn, xm) kn(x0, x1) + Ck

n+1(x0, x1) + C2kn+2(x0, x1)+

...+ Cmn1km1(x0, x1).

Therefore

(xn, xm) (1 + Ck + C2k2 + .... + Cmn1kmn1)kn(x0, x1)

and so

(xn, xm)< kn

1 Ck(x0, x1)

Hence, the sequence {xn} is Cauchy. Since (X, ) is complete, it con-verges to some x X. Now, we show that z is a fixed point of F.Suppose not. Then

(F z , F xn) k(z, xn1)

As ntends to infinity we get z=F zusing (SR). The uniqueness ofzfollows easily.

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Corollary 2.2. Suppose (X, ) satisfies the asymmetric triangle in-equality, is strictly reflexive and complete. If F : X X is a map

such that(F x , F y) k(x, y) withk

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Proof. Letx0be an arbitrary point inXand {xn} the sequence definedbyxn+1= F(xn) =F

n(x0) for all positive integer n. We have

(xn, xm) ((xn, xn+1) + (xn+1, xm))e(xn,xm,xn+1)

kn(x0, x1)e(xn,xm,xn+1) + (xn+1, xm)e

(xn,xm,xn+1)

kn(x0, x1)e(xn,xm,xn+1)+

((xn+1, xn+2) + (xn+2, xm))e(xn+1,xm,xn+2)+(xn,xm,xn+1)


kn+1(x0, x1)e(xn+1,xm,xn+2)+(xn,xm,xn+1)+

(xn+2, xm)e

(xn+1,xm,xn+2)+(xn,xm,xn+1)



kn+2(x0, x1)e(xn+2,xm,xn+3)+(xn+1,xm,xn+2)+(xn,xm,xn+1)+

(xn+3, xm)e(xn+2,xm,xn+3)+(xn+1,xm,xn+2)+(xn,xm,xn+1)



kn+2(x0, x1)e(xn+2,xm,xn+3)+(xn+1,xm,xn+2)+(xn,xm,xn+1) + ...+

km1(x0, x1)e(xn,xm,xn+1)+(xn+1,xm,xn+2)+...+(xm2,xm,xm1)

kn(x0, x1)(ekn(x0,xp,x1) + kek

n(x0,xp,x1)+kn+1(x0,xp,x1)+

k2ekn(x0,xp,x1)+kn+1(x0,xp,x1)+kn+2(x0,xp,x1)+

. . . +

kmn1ekn(x0,xp,x1)+kn+1(x0,xp,x1)+...+km1(x0,xp,x1))

kn(x0, x1)(eMkn + keM(k

n+kn+1) + k2eM(kn+kn+1+kn+2)+

. . . +

kmn1eM(kn+kn+1+kn+2+...+km1))

since is bounded. Hence, the sequence {xn} is Cauchy. Since (X, )is complete, it converges to some x X. The rest of the proof followsas in Lemma2.1.

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3. Bounded 2-metric spaces

Gahler defined the notion of 2-metric space to be a set Xwith func-tion d: X3 R denoted (x, y, z )d(x, y, z ) satisfying the followingaxioms [10][11][12]:(Sym)thatd(x, y, z ) is invariant under permutations of the variablesx, y, z .(Tetr)for alla, b, c, xwe have

d(a,b,c) d(a,b,x) + d(b,c,x) + d(a,c,x).

(Z)for all a, bwe have d(a,b,b) = 0.(N)for all a, bthere exists c such that d(a,b,c)= 0.

One can think ofd(x, y, z ) as measuring how far are x, y,zfrom beingaligned or colinear.

The 2-metric spaces (X, d) have been the subject of much study, see[1] and [2]for example. The prototypical example of a 2-metric space isobtained by setting d(x, y, z ) equal to the area of the triangle spannedbyx, y,z.

Assume that the 2-metric isbounded, and by rescaling the bound canbe supposed equal to 1:(B)the function is bounded byd(x, y, z )1 for all x, y, z X.

Define the associated distanceby

(x, y) := supzX

d(x,y,z).

Lemma 3.1. We have d(x,y,z) 0 and hence (x, y) 0. Also(x, x) = 0 and(x, y) =(y, x).

Proof. Applying the axiom (Tetr) with b= c, we get

d(a,b,b) d(a,b,x) + d(b,a,x) + d(a,b,x).

By the axiom (Z) and the symmetry ofd we obtain d(a,b,x) 0 andso d(x,y,z) 0. Then, (x, y) 0. Symmetry of follows frominvariance ofd under permutations (Sym).

Lemma 3.2. We have the triangle inequality with cost (2.1)

(x, y) (x, z) + (z, y) + d(x,y,z).Therefore

(x, y) (x, z) + (z, y) + min((x, z), (z, y))

and hence the asymmetric triangle inequality (AT)

(x, y) (x, z) + 2(z, y).

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Proof. We have

d(x,y,z0) d(x, y, z ) + d(y, z0, z) + d(x, z0, z)

(x, z) + (z, y) + d(x, y, z ).

For the next statement, note that by definition

d(x,y,z) min((x, z), (z, y)),

and for the last statement, min((x, z), (z, y)) (z, y).

In particular the distance satisfies the axioms (R), (S) and (AT) ofSection2. This allows us to speak of limits, Cauchy sequences, points ofaccumulation, completeness and compactness, see also [27]. For clarityit will usually be specified that these notions concern the function .Axiom (N) for d is equivalent to strict reflexivity (SR) for ; if this is

not assumed from the start, it can be fixed as follows.3.1. Nondegeneracy. It is possible to start without supposing thenondegeneracy axiom (N), define an equivalence relation, and obtaina 2-metric on the quotient satisfying (N). For the next lemma and itscorollary, we assume that d satisfies all of (Sym), (Tetr), (Z), (B), butnot necessarily (N).

Lemma 3.3. Ifa, b, x, y are any points then

|d(a,b,x) d(a,b,y)| 2(x, y).

Proof. By condition (Tetr),

d(a,b,y) d(a,b,x)+d(b,y,x)+d(a,y,x) d(a,b,x)+2(x, y).The same in the other direction gives the required estimate.

Corollary 3.4. If x, y are two points with (x, y) = 0 then for anya, b we have d(a,b,x) = d(a,b,y). Therefore, if is the equivalencerelation considered in the second paragraph of Section2, the functiond descends to a function (X/ )3 R satisfying the same propertiesbut in addition its associated distance function is strictly reflexive andd satisfies (N).

Proof. For the first statement, apply the previous lemma. This invari-ance applies in each of the three arguments since d is invariant underpermutations, which in turn yields the descent of d to a function on(X/)3. The associated distance function is the descent of which isstrictly reflexive.

In view of this lemma, we shall henceforth assume that satisfies(SR) or equivalently d satisfies (N) too. In particular the limit of asequence is unique if it exists.

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3.2. Surjective mappings. If F is surjective, then a boundednesscondition for d implies the same for . Since we are assuming that d

is globally bounded (condition (B)), a surjective mapping cannot bestrictly contractive:

Lemma 3.5. SupposeF :XXis a map such that

d(F(x), F(y), F(z)) kd(x, y, z )

for some constant k > 0. If F is surjective then (F(x), F(y)) k(x, y). The global boundedness condition implies thatk 1 in thiscase.

Proof. Suppose x, y X. For any z X, choose a preimage w Xsuch thatF(w) = zby surjectivity ofF. Then

d(F(x), F(y), z) = d(F(x), F(y), F(w)) kd(x, y, w) k(x, y).

It follows that

(F(x), F(y)) = supzX

d(F(x), F(y), z) k(x, y).

Suppose now that k

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Definition 4.1. Say that(x, y, z ) arecolinear ifd(x,y,z) = 0. A lineis a maximal subsetY Xconsisting of colinear points, that is to say

satisfying(4.1) x,y,z Y, d(x, y, z ) = 0.

The colinearity condition is symmetric under permutations by (Sym).

Lemma 4.2. Using all of the above axioms including (N) and assump-tion (Trans), colinearity satisfies the following transitivity property: ifx, y, z are colinear, y , z , w are colinear, and y = z, thenx, y, w andx,z,w are colinear.

Proof. By (N), (y, z) = 0, then use the above version of (Trans)rewritten after some permutations as

d(x,y,w) (d(x, y, z ) + d(y, z, w))(y, z)1.

This shows that x,y,w are colinear. Symmetrically, the same forx,z,w.

A line is nonempty, by maximality since d(y , y , y) = 0 by (Z).

Lemma 4.3. If x = y are two points then there is a unique line Ycontainingxandy, andY is the set of pointsa colinear withx andy ,i.e. such thatd(a,x,y) = 0.

Proof. The set {x, y} satisfies Condition (4.1), so there is at least onemaximal such set Y containing xand y. Choose one such Y. Ifa Ythen automaticallyd(a,x,y) = 0.

Suppose d(a,x,y) = 0. By Lemma4.2,d(a,x,u) = 0 for anyu Y.Now suppose u, v Y. If u = x then the preceding shows that

d(a,u,v) = 0. If u = x then, since d(x,u,v) = 0 and d(a,x,u) = 0,Lemma4.2 implies that d(a,u,v) = 0. This shows thata is colinearwith any two points ofY. In particular,Y{a} also satisfies Condition(4.1) so by maximality, a Y. This shows that Y is the set of pointsasuch that d(a,x,y) = 0, which characterizes it uniquely.

The notions of colinearity and lines come from the geometric exam-ples of 2-metrics which will be discussed in Section5 below. It should

be pointed out that there can be interesting examples of 2-metric spaceswhich dont satisfy the transitivity condition of Lemma4.2 and whichtherefore dont satisfy Axiom (Trans). The remainder of our discussiondoesnt apply to such examples.

We assume Axiom (Trans) from now on. It allows us to look at thequestion of fixed subsets of a contractive mapping F when F is notsurjective. In addition to the possibility of having a fixed point, there

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will also be the possibility of having a fixed line. We see in examplesbelow that this can happen.

Definition 4.4. Consider a sequence of pointsxi X. The propertyLIM(y, (xi)) is defined to mean:

>0 a, i, j a, d(y, xi, xj)< .

Suppose LIM(y, (xi)) and LIM(y, (xi)). We would like to show

that d(y, y, xi) 0. However, this is not necessarily true: if (xi) isCauchy then the properties LIM are automatic (see Proposition 4.9below). So, we need to include the hypothesis that our sequence is notCauchy, in the following statements.

Lemma 4.5. Suppose (xi) is not Cauchy. If both LIM(y, (xi)) and

LIM(y

, (xi)) hold, thend(y, y

, xi) 0 asi .Proof. The sequence (xi) is supposed not to be Cauchy for , so thereexists0 > 0 such that for anym 0 there arei, j mwith(xi, xj)0. Therefore, in view of the definition of , for any m there existi(m), j(m) mand a pointz(m) Xsuch thatd(xi(m), xj(m), z(m))0/2.

We now use condition (Trans) with x = xi(m) and y = xj(m) andc= z(m), for a = y and b = y . This says

d(y, y, xi(m))0/2 d(y, xi(m), xj(m)) + d(y, xi(m), xj(m)).

IfLI M(y, (xi)),LIM(y, (xi)), then for any we can assume m is big

enough so that

d(y, xi(m), xj(m)) 0/4

and

d(y, xi(m), xj(m)) 0/4.

Putting these together gives d(y, y, xi(m)) .Choose m so that for all j, k m we have d(y, xj, xk) and the

same for y . Then we have by (Tetr), for any j m

d(y, y , xj) d(xi(m), y , xj)+d(y, xi(m), xj)+d(y

, y , xi(m)) 3.

Changingby a factor of three, we obtain the following statment: forany >0 there existsm such that for alli mwe haved(y, y , xi) .This is the required convergence.

Corollary 4.6. If the sequence(xi) is not Cauchy for the distance,then the following property holds:ifLIM(y, (xi)),LI M(y

, (xi)), andLIM(y, (xi))then(y, y

, y)arecolinear.

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Proof. We use the fact that

d(y, y, y) d(y, y, xi) + d(y, y, xi) + d(y, y

, xi).

By Lemma4.5, all three terms on the right approach 0 as i . Thisproves that d(y, y, y) = 0.

Lemma 4.7. Suppose that (xi) is not Cauchy for the distance . IfLIM(y, (xi)), LIM(y

, (xi)), (y, y) > 0 and y is a point such that

(y, y, y) are colinear, then also LIM(y, (xi)).

Proof. By Lemma 4.5, for any > 0 there exists m such that for alli m we have d(y, y , xi) .

Letu, v denote some xi or xj . By hypothesis (y, y)>0 so there is

a pointzsuch thatd(y, y, z) =1 > 0. Then condition (Trans) applied

witha= y , b= u, x= y , c= z, y = y givesd(y, u , y)d(z, y, y) d(y, y, y) + d(u, y, y).

Hence

d(y, u , y) d(u, y, y)/1.

We can do the same for v , and also interchanging y and y , to get

d(y, v , y) d(v, y, y)/1,

d(y, u , y) d(u , y, y)/1,

d(y, v , y) d(v , y, y)/1.

We have

d(y, u , v) d(y, u , v) + d(y, y , v) + d(y, u , y)

d(y,u,v) + (d(u , y, y) + d(v , y, y))/1.

For u = xi and v = xj with i, jm as previously we get

d(y, xi, xj) d(y, xi, xj) + (d(xi, y , y) + d(xj , y , y

))/1.

By choosing m big enough this can be made arbitrarily small, thusgiving the condition LI M(y, (xi)).

Theorem 4.8. Suppose that(X, d) is a bounded2-metric space satis-fying axiom (Trans) as above. Suppose(xi) is a sequence. Then thereare the following possibilities (not necessarily exclusive):there is no pointy withLIM(y, (xi));there is exactly one pointy withLIM(y, (xi));the sequence(xi) is Cauchy for the distance; orthe subsetY Xof pointsy such thatLIM(y, (xi)), is a line.

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Proof. Consider the subset Y Xof points y such that LIM(y, (xi))holds. We may assume that there are two distinct points y1=y2 inY,

for otherwise one of the first two possibilities would hold. Suppose that(xi) is not Cauchy for ; in particular, Lemmas4.5,4.7 and Corollary4.6apply.

Ify , y, y are any three points in Y, then by Corollary4.6, they arecolinear. ThusY is a subset satisfying Condition (4.1) in the definitionof a line; to show that it is a line, we have to show that it is a maximalsuch subset.

SupposeY Y1and Y1also satisfies (4.1). Sincey1=y2, and we areassuming that satisfies strict reflexivity (SR), we have (y1, y2)= 0.By Lemma3.1, (y1, y2)> 0. Ify Y1 then by (4.1), d(y, y1, y2) = 0.By Lemma 4.7, y must also satisfy LIM(y, (xi)), thus y Y. This

shows that Y1 Y, giving maximality ofY. Thus, Y is a line. The following proposition shows that the case when (xi) Cauchy has

to be included in the statement of the theorem.

Proposition 4.9. If(xi),(yj)and(zk)are Cauchy sequences, then thesequenced(xi, yj, zk)is Cauchy in the sense that for any >0 there ex-istsMsuch that fori, j, k,p, q,r Mthen|d(xi, yj, zk)d(xp, yq, zr)|0, m, i, j , k m d(xi, xj, xk)< .

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Lemma 4.10. Suppose (xi) is a tri-Cauchy sequence, and y X isan accumulation point of the sequence with respect to the distance.

ThenLIM(y, (xi)).

Proof. The condition thaty is an accumulation point means that thereexists a subsequence (xu(k)) such that (xu(k))y with respect to thedistance. We have by (Tetr)

d(y, xi, xj) d(xu(k), xi, xj) + d(y, xu(k), xj) + d(y, xi, xu(k))

d(xu(k), xi, xj) + 2(y, xu(k)),

and both terms on the right become small, for i, j big in the originalsequence and k big in the subsequence. Hence d(y, xi, xj) 0 as

i, j 0, which is exactly the condition LI M(y, (xi)).

We say that (X, d) is tri-complete if, for any tri-Cauchy sequence,the set Yof points satisfying LIM(y, (xi)) is nonempty. By Theorem4.8, Yis either a single point, a line, or (in case (xi) is Cauchy) all ofX.

Lemma 4.11. Suppose (X, ) is compact. Then it is tri-complete,and for any tri-Cauchy sequence(xi) we have one of the following twopossibilities:(xi) has a limit; or

the subsetY of pointsy withLIM(y, (xi)) is a line.

Proof. Suppose (xi) is a tri-Cauchy sequence. By compactness thereis at least one point of accumulation, so the set Y of points y withLIM(y, (xi)) is nonempty by Lemma 4.10. This rules out the firstpossibility of Theorem4.8.

Suppose Yconsists of a single point y. We claim then that xi y .Suppose not: then there is a subsequence which doesnt contain y in itsclosure, but since X is compact after going to a further subsequencewe may assume that the subsequence has a limit point y = y. Butagain by Lemma 4.10, we would have LIM(y, (xi)), a contradiction.

So in this case, the sequence (xi) is Cauchy for and hasy as its limit;thus we are also in the situation of the third possibility. Note howeverthat, since (xi) is Cauchy, the set of points Y consists of all ofX byProposition4.9, so the second possibility doesnt occur unless X is asingleton.

From Theorem 4.8 the remaining cases are that (xi) is Cauchy, inwhich case it has a limit by compactness; or that Y is a line.

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5. Some examples

The classic example of a 2-metric is the function on R2 defined bysettingd(x, y, z ) to be the area of the triangle spanned by x, y, z . Beforediscussing this example we look at a related example on RP2 which isalso very canonical.

5.1. A projective area function. Let X := S2 := {(x1, x2, x3) R3, x21+ x

22+ x

23 = 1}. Define the function d(x, y, z ) by taking the

absolute value of the determinant of the matrix containing x,y,z ascolumn vectors:

d

x1x2x3

,

y1y2y3

,

z1z2z3

:=

det

x1 y1 z1x2 y2 z2x3

y3

z3

.

This has appeared in Example 2.2 of [19].

Proposition 5.1. This function satisfies axioms (Sym), (Tetr), (Z),(B), (Trans).

Proof. Invariance under permutations (Sym) comes from the corre-sponding fact for determinants. Condition (Z) comes from vanishingof a determinant with two columns which are the same. Condition (B)comes from the fact that the determinant of a matrix whose columnshave norm 1, is in [1, 1]. We have to verify (Tetr) and (Trans).

For condition (Tetr), suppose given vectorsx, y, z S2 as above, and

suppose thatd(x,y,z)> 0 i.e. they are linearly independent. Suppose

given another vectora =

a1a2a3

too. Then the determinantsd(a,y,z),

d(x,a,y) and d(x, y, a) are the absolute values of the numerators ap-pearing in Cramers rule. This means that if we write

a= x + y + z

then

d(a,y,z)

d(x,y,z)

=||,

d(x,a,z)

d(x,y,z)=||,

and

d(x,y,a)

d(x, y, z )=||.

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Now by the triangle inequality in R3 we have

1 =a || + || + ||

which gives exactly the relation (Tetr).To prove (Trans), notice that it is invariant under orthogonal trans-

formations ofR3 so we may assume that

x=

100

, y =

uv0

, u2 + v2 = 1.

In this case, supcS2d(c,x,y) = |v|, so we are reduced to considering

a=

a1

a2a3

, b=

b1

b2b3

.

Nowd(a,x,y) = |a3v| and d(b,x,y) = |b3v|, so we have to show that

d(a,b,x)|v| |a3v| + |b3v|.

But d(a,b,x) = |a2b3 a3b2| so this inequality is true (since |a2| 1and|b2| 1). This completes the proof of (Trans).

Corollary 5.2. If X S2 then with the same function d(x,y,z), itstill satisfies the axioms.

The lines for the function d defined above, are the great circles orgeodesics onS2.

Remark 5.3. The distance function is not strictly reflexive in thisexample, indeed the associated equivalence relation identifies antipodalpoints. The quotient S2/ is the real projective plane. The corre-sponding function onRP2 is a bounded2-metric satisfying (Trans).

5.2. The Euclidean area function. We next go back and considerthe standard area function on Euclidean space, which we denote by.

Proposition 5.4. LetU Rn be a ball of diameter 1. Let(x, y, z )be the area of the triangle spanned by x,y,z U. Then satisfiesaxioms (Sym), (Tetr), (Z), (N), (B), (Trans).

Proof. Conditions (Sym), (Tetr), (Z), (N) are classical. Condition (B)comes from the bound on the size of the ball. It remains to showCondition (Trans). By invariance under orthogonal transformations,we may assume that x= (0, . . . , 0) and y = (y1, 0, . . . , 0) with y1 >0.

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Again by the bound on the size of the ball, for any cwe have(x, y, c) y1. Thus, we need to show

(x,a,b)y1 (x,y,a) + (x, y, b).

Write a = (a1, . . . , an) andb = (b1, . . . , bn). Then

(x, y, a) = y1a

where a= (0, a2, . . . , an) is the projection ofa on the plane orthogonalto (xy). Similarlyd(x,y,b) =y1b. To complete the proof it sufficesto show that (x,a,b) a + b.

Writea = a1e1 + aand b = b1e1+ bwheree1 is the first basis vector.Then

(x,a,b) =1

2a b

=1

2a1e1 b b1e1 a + a b

a + b

sincea1 1, b11, a 1, and b 1.

5.3. Euclidean submanifolds. IfU Rn is any subset contained ina ball of diameter 1 but not in a single line, then the function d(x, y, z )onU3 induced by the standard Euclidean area function on Rn is againa bounded isometric 2-metric. This is interesting in caseU is a patch

inside a submanifold. The induced 2-metric will have different prop-erties depending on the curvature of the submanifold. We considerUS2 R3 as an example.

Choose a patchUonS2, contained in the neighborhood of the south

pole of radius r < 1/4. Let p : U = V R2 be the vertical projec-

tion. Leth(x, y, z ) := (p1(x), p1(y), p1(z)) be the 2-metric on Vinduced from the standard Euclidean area function ofS2 R3 viathe projection p. This satisfies upper and lower bounds which mirrorthe convexity of the piece of surface U:

Lemma 5.5. Keep the above notations for h(x, y, z ). Let 2 denotethe standard area function onV R2. Define a function

(x,y,z) :=x y x z y z.

Then there is a constantC >1 such that

(5.1)1C

(2(x,y,z) + (x, y, z )) h(x, y, z )h(x, y, z ) C(2(x,y,z) + (x, y, z ))

forx, y, z V.

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Proof. If any two points coincide the estimate holds, so we may assumethe three points x,y,z are distinct, lying on a unique plane P. In

particular, they lie on a circleS2

Pof radius 1. A simple calculationon the circle gives a bound of the form h(x,y,z) (1/C)(x, y, z ).The vertical projection from P to R2 has Jacobian determinant 1 soh(x, y, z ) 2(x, y, z ). Together these give the lower bound.

For the upper bound, consider the unit normal vector to P and letu3be the vertical component. If|u3| 1/4 then the plane is somewhathorizontal and h(x, y, z ) C2(x, y, z ). If |u3| 1/4 then the planeis almost vertical, and in view of the fact that the patch is near thesouth pole, the intersection P S2 is a circle of radius 1/4. In thiscaseh(x,y,z) C(x,y,z). From these two cases we get the requiredupper bound.

6. Fixed points of a map

Suppose that (X, ) is compactmeaning that every sequence has aconvergent subsequence. Suppose F : X X is a d-decreasing mapi.e. one with d(F(x), F(y), F(z)) kd(x,y,z) for 0< k

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Proof. Pick x0 Xand define the sequence of iterates (xi) as above.By the previous corollary, either xi y or else the set Y of points y

withLIM(y, (xi)) is a line.Suppose we are in the second possibility but not the first; thus (xi)has more than one accumulation point. Suppose y is a point withpropertyLI M(y, (xi)), which means that d(y, xi, xj)< for i, jm.Hence ifi, jm+ 1 then

d(F(y), xi, xj) =d(F(y), F(xi1), F(xj1))< k.

This shows the property LIM(F(y), (xi)). Thus, F(Y) Y.Suppose Y1 is a line with F(Y) Y1. If F(Y) contains at least

two distinct points then there is at most one line containing F(Y)by Lemma 4.3 and we obtain the second conclusion of the theorem.

Suppose F(Y) = {y0} consists of a single point. Then, y0 Y soF(y0) F(Y), which shows that F(y0) = y0; in this case F admits afixed point.

We may now assume that we are in the first case of the first para-graph: xi y . IfF(y) =y we have a fixed point, so assume

1 F(y)=y.Let Ybe the unique line containing y and F(y) by Lemma4.3whichalso says Y is the set of points colinear with y and F(y).

We claim that F(X) Y. Ifz X then LIM(z, (xi)) by the lastpart of Proposition4.9. For a given there ismsuch thati, jm d(xi, xj, z) < . However, for i fixed we have xj y by hypothesis,and the continuity ofd (Proposition 4.9) implies that d(xi, y , z ) < .

ApplyF, givingi m, d(xi+1, F(y), F(z))< k.

Again using continuity ofd, we have

d(xi+1, F(y), F(z)) d(y, F(y), F(z))

and the above then implies that d(y, F(y), F(z)) < k for any > 0.Henced(y, F(y), F(z)) = 0 which means F(z) Y. This proves thatF(X) Y a fortioriF(Y) Y. IfF(F(y)) is distinct from F(y) thenY is the unique line containing F(Y), otherwise F(y) is a fixed pointofF. This completes the proof of the theorem.

Corollary 6.3. SupposeXS2

is a closed subset. Define the functiond(x,y,z) by a determinant as in Proposition5.1. IfF : X X is afunction such thatd(F(x), F(y), F(z)) kd(x,y,z)for0 < k

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Proof. Recall that we should really be working with the image ofX inthe real projective space RP2 =S2/(Remark5.3). On this quotient,

the previous theorem applies. Note that by a fixed great circle wemean a subsetY Xwhich is the intersection ofXwith a great circle,and such that Y is the intersection ofXwith the unique great circlecontainingF(Y).

7. Examples of contractive mappings

Here are some basic examples which show that the phenomenonpointed out by Hsiao [13] doesnt affect our notion of contractive map-ping. He showed that for a number of different contraction condi-tions for a mapping Fon a 2-metric space, the iterates xi defined byxi+1:= F(xi) starting with anyx0, are all colinear i.e. d(xi, xj , xk) = 0for all i, j, k.

7.1. Consider first the standard 2-metric given by the area of thespanned triangle in Rn, see Proposition 5.4. Suppose U is a convexregion containing 0, contained in a ball of diameter 1. Choose 0 1. Such d exists by Lemma 5.5. Given CA > 0,there is a constantC >0 such that the following holds.

SupposeA is a fixed2 2 matrix with det(A) = 0 andA CA.Then there is a constantc >0 depending onA, such that ifF :V Vis a C2 mapping withJ(F, x) A c anddJ(F) c overV,then

d(F(x), F(y), F(z)) C|det(A)|d(x,y,z)

for anyx, y,z V.

Proof. In what follows the constants Ci and C will depend only onCand the bound CA for A, assuming c

to be chosen small enoughdepending on A. By choosing c small enough we may assume thatF is locally a diffeomorphism, as J(F, x) is invertible, being close tothe invertible matrixA. By Rolles theorem there is a point y on thesegment joiningx to y, and a positive real number u, such that

F(y) F(x) = uJ(F, y)(y x).

Furthermore u < C1, ifc is small enough. We thank N. Mestrano for

this argument.Similarly there is a point z on the segment joining x to z, and a

positive real number v < C1 such that

F(z) F(x) =vJ(F, z)(z x).

PutS:= J(F, y)J(F, x) andT :=J(F, z)J(F, x). Now the bounddJ(F) c implies that

S 4cy x, T 4cz x.

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We have

2(F(x), F(y), F(z)) = |(F(y) F(x)) (F(z) F(x))|

=|(uJ(F, y)(y x)) (vJ(F, z)(z x))|

=uv |(J(F, x) + S)(y x) (J(F, x) + T)(z x)|

uv|detJ(F, x)|2(x, y, z )

+ uv[(S + T)J(F, x) + ST] y x z x.

Note that

yx2zx+yxzx2+yx2zx2 C2(x, y, z ),

and J(F, x) C3. By chosing c small enough depending on A,

we may assume that |detJ(F, x)| C4|detA|. So again possibly byreducingc, we get

2(F(x), F(y), F(z)) C4|det(A)|2(x,y,z)+C1C2C3(x, y, z ).

It is also clear that (F(x), F(y), F(z)) C5(x,y,z). Using the con-vexity estimate in the hypothesis of the proposition, we get

d(F(x), F(y), F(z)) C|det(A)|d(x,y,z)

for a constant C which depends onCandCA but not onA itself.

Suppose V is a disc centered at the origin in R2 with a 2-metricd satisfying the convexity estimate of the form (5.1). For example Vcould be the projection of a patch in the Euclidean S2 as in Lemma5.5. Fix a bound CA = 2 for example. We get a constant C

from the

previous proposition. Choose then a matrixAsuch thatC

|det(A)|< 1and A V V. Then there is c given by the previous proposition.There exist plenty ofC2 mappingsF :V V withJ(F, x) A c

and dJ(F) c. By the conclusion of the proposition, these arecontractive.

This example shows in a strong sense that there are no colinearityconstraints such as [13] for mappings which are contractive in our sense.

As a 2-metric moves from a convex one towards the flat Euclideanone (for example for patches of the same size but on spheres with biggerand bigger radii) we have some kind of a deformation from a nonrigidsituation to a rigid one. This provides a model for investigating this

general phenomenon in geometry.The question of understanding the geometry of contractive map-

pings, or even mappings F with d(F(x), F(y), F(z)) K d(x,y,z) forany constantK, seems to be an interesting geometric problem. Follow-ing the extensive literature in this subject, more complicated situationsinvolving several compatible mappings and additional terms in the con-tractivity condition may also be envisioned.

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A. Aliouche, Department of Mathematics, University of Larbi Ben

MHidi, Oum-El Bouaghi, 04000, Algeria

E-mail address: [email protected]

C. Simpson, CNRS, Laboratoire J. A. Dieudonne, UMR 6621, Univer-

site de Nice-Sophia Antipolis, 06108 Nice, Cedex 2, France

E-mail address: [email protected]

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