Funk geometry

Athanase Papadopoulosand Marc Troyanov

Institut de Recherche Mathematique Avancee,Universite de Strasbourg and CNRS,

7 rue Rene Descartes,67084 Strasbourg Cedex, France.

email: athanase.papadopoulos@math.unistra.fr

Ecole Polytechnique Federale de Lausanne,SB MATHGEOM GR-TR MA B3 495 (Batiment MA) Station 8,

CH-1015 Lausanneemail: marc.troyanov@epfl.ch

Abstract. We survey some basic geometric properties of the Funk metric of aconvex set in Rn. In particular, we study its geodesics, its topology, its metric balls,its convexity properties and its perpendicularity theory. Most of the results wereknown to Busemann.The Hilbert metric is a symmetrization of the Funk metric in the sense that if H(x, y)and F (x, y) denote the Hilbert and the Funk distance functions associated to a convexset then, we have H(x, y) = 1

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

2000 Mathematics Subject Classification:

Keywords: Funk metric, convexity, Hilbert metric, Menelaus Theorem.

Contents

E En cours de revision, ne pas toucher !

The first author was partially supported by the French ANR project FINSLER

2 Athanase Papadopoulos and and Marc Troyanov

1 Introduction

The Funk metric is a weak metric in the sense that it does not satisfy all theaxioms of a metric (it is not symmetric, and we shall also allow the distancebetween two points to be zero); see [?] in this volume. Weak metrics oftenoccur in the calculus of variations and in Finsler geometry and the studyof such metrics has been revived recently in low-dimensional topology andgeometry by Thurston who introduced an asymmetric metric on Teichmullerspace, which became the subject of intense research. In this chapter, we shalluse the expression metric instead of weak metric in order to simplify.

The Funk metric is associated to an open convex subset of a Euclideanspace Rn. It is important in the setting of his Handbook because of its relationto the Hilbert metric. Indeed, the Hilbert metric H of a convex set is thearithmetic symmetrization of its Funk metric F. More precisely, for any xand y in , we have

H(x, y) =1

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

The Funk metric is a nice example of a weak metric, because there aremany natural questions for a given convex subset that one can ask and solvefor such a metric, regarding its geodesics, its balls, its isometries, its boundarystructure, its parallelism theory, and so on. Of course, the answer depends onthe shape of the convex set , and it is an interesting aspect of the theory tosee the influence of the properties of the boundary (its degree of smoothness,the fact that it is a polyhedron, a stricly convex hypersurface, etc.) on theFunk geometry. There is a section on the Funk metric in Busemanns book[?],

E which section ?

althought the name Funk metric is not used there, but it is used byBusemann in his later papers and books, see e.g. [?], and in the memoir byZaustinsky [?].

We studied some aspects of this metric in [?], following Busemanns ideas.But a systematic study of this metric is something which seems to be missingin the literature.

Ecette phrase est un peu en contradiction avec lidee que Busemann

connaissait tout ca on va ameliorer...

In this chapter, we shall study several of the topological and geometricproperties of this metric. Euclidean segments in are geodesics for the Funk

Funk geometry 3

metrics, and in the case where the domain is strictly convex, the Euclideansegments are the unique geodesic segments. The property for a metric of hav-ing the Euclidean segments geodesics is the subject of Hilberts Problem IV(see [?] in this Handbook). One version of this problem asks for a character-ization of non-symmetric metrics on subsets of Rn such that the Euclideansegments are geodesics.

In the paper [?], we introduced the notions of tautological and of reversibletautological Finsler structure of the domain . The Funk metric F is thedistance function induced by the tautological weak Finsler structure, and theHilbert metricH is the distance function induced by the reversible tautologicalFinsler structure of the domain . This is in fact how Funk introduced hismetric in 1929, see [?, ?]. The reversible Finsler structure is obtained bythe process of harmonic symmetrization at the level of the convex sets inthe tangent spaces that define these structures. This gives another relationbetween the Hilbert and the Funk metrics. The Finsler geometry of the Funkmetric is studied in some detail in the survey [?] in this Handbook.

An interesting variational descriptions of the Funk metric is studied in [?].There are non-Euclidean versions, see [?] in this volume.

2 The Funk metric

In this section, is a nonempty open convex subset of Rn. We denote by the closure of and by the topological boundary of , that is, the set \ . For any two points x 6= y in Rn, we denote by [x, y] (resp. [x, y), etc.)the closed (respectively closed at x and open at y, etc.) segment joining thetwo points. We also denote by R(x, y) the Euclidean ray starting at x andpassing through y.

In the next definition, if x 6= y are in and if R(x, y) 6 , then we let a+

denote the intersection point R(x, y) .

Definition 2.1 (The Funk metric). The Funk metric on , denoted by F,is defined, for x and y in , by the formula

F(x, y) =

log

|x a+|

|y a+|if x 6= y and R(x, y) 6=

0 ifR(x, y) 6 .

(In particular, if the point a+ does not exist, then F(x, y) = 0.)

From the definition, it follows that if = Rn, then F 0. We shallhenceforth assume that 6= Rn whenever we shall deal with the Funk metricof a nonempty open convex subset of Rn.

4 Athanase Papadopoulos and and Marc Troyanov

The function F satisfies the triangle inequality. A proof is given in Zaustin-sky [?] p. 85 using arguments that are identical to those of the classical proofof the triangle inequality for the Hilbert metric, as given by D. Hilbert in[?]. This proof is based on the cross ratio invariance and on the theorem ofMenelaus that gives a necessary and sufficient condition on the alignment ofthree points situated on three given lines in the Euclidean plane, each pointon its corresponding line. Yamada gave in [?] a proof of the triangle inequalityfor the Funk metric using a variational formula for this metric. The trian-gle inequality also follows from the fact that this metric is associated to thetautological Finsler structure on , see [?].

For the convenience of the reader, we reproduce the classical proof. It usesthe theorem of Menelaus, which we recall in the appendix.

Let x, y, z be three points in . We may assume that they are not collinear,otherwise they satisfy a degenerate triangle inequality (see Proposition ??below). Let a, b, c, d, e, f be the intersections with of the lines xz, yx andzy, using the notation of Figure ?? concerning the order of intersections.

ba

a

b

c

d e

f

p

gx

y

z

Figure 1.

From the invariance of the cross ratio, we have

|x c|

|y c||d y|

|d x|=|x b|

|g b||a g|

|a x|

and

|y e|

|z e||f z|

|f y|=|g b|

|z b||a z|

|a g|.

Multiplying both sides of these two equations, we get

|x c|

|y c||y e|

|z e|=|x b|

|z b||a z|

|a x||d x|

|d y||f y|

|f z|.

Funk geometry 5

By the Theorem of Menelaus (Theorem ??) applied to the triangle faz, wehave

|d x|

|d y||f y|

|f z|=|a x|

|a z|.

This gives

|x c|

|y c||y e|

|z e|=|x b|

|z b||x b|

|z b|,

and the inequality is strict unless b = b. From this the triangle inequalityfollows, and the inequality is strict for all x, y, z unless contains a Euclideansegment. This gives the desired result.

Proposition 2.2. The Funk metric is unbounded.

Proof. Recall that we always assume 6= Rn, and therefore 6= . Let xbe a point in and b a point in and consider the open Euclidean segment(x, b), contained in . For any sequence xn in this segment converging to b

(with respect to the Euclidean metric), we have F(x, xn) =|x b|

|xn b| as

n.

Notice that, in the preceding proof, the sequence F(xn, x) is bounded.This is one indication of the fact that the metric F defined by F

(x, y) =F(y, x) is very different from the Funk metric F. We call F

the reverseFunk metric of .

The following proposition is also easy to prove from the definitions.

Proposition 2.3. The metric F is separating (that is, we have, for every xand y in , x 6= y F(x, y) > 0) if and only if is bounded.

Il is useful to notice that computing the Funk distance between two pointsin is a one-dimensional operation. More precisely, for x and y in , withx 6= y, we consider the line in Rn joining x and y. It intersects in someinterval I (bounded or not) which has its proper Funk metric, and we have,for all x and y in I, F(x, y) = FI(x, y). This remark immediately implies thefollowing more general statement:

Eexpliquer pourquoi (il faut utiliser quun convexe non borne contient un

rayon)

Proposition 2.4. Let be a nonempty open convex subset of Rn, let be the intersection of with an affine subspace of Rn, and suppose that 6= .Then, F is the metric induced by F on

as a subspace of (, F).

6 Athanase Papadopoulos and and Marc Troyanov

Now we prove the following:

Proposition 2.5. Let be an open convex bounded subset of Rn. Then, wehave, for x in and for a sequence xn in ,

F(x, xn) 0 |x xn| 0 F(xn, x) 0.

Note that if is unbounded, the conclusion of Proposition ?? may fail,since we may have points x and y satisfying F(x, y) = 0 while F(y, x) 6= 0.

E verifier cette preuve

Proof. We first prove that F(x, xn) 0 |xxn| 0. We take a sequencexn in and a point x in such that |xxn| 6 0. Up to taking a subsequence,we assume that there exists a real number > 0 such that F(x, xn) 0 and|x xn| > . Consider the sphere S(x, ) of center x and radius . For anyx S(x, ), let ax be the intersection of the ray R(x, a) with . The function

a 7 log|x ax|

|a ax|defined on the compact set S(x, ) is positive and therefore

bounded from below by a positive number . Now we take any point xn inour sequence and we let a S(x, ) be the intersection of the ray R(x, xn)

with the sphere S(x, ). We have log|x ax|

|xn ax|> log

|x ax|

|a ax|> . Therefore

F(x, xn) 6 0.The implications |x xn| 0 F(xn, x) 0 and |xn x| 0

F(xn, x) 0 follow from the next proposition.

Proposition 2.6. Let 1 and 2 be two open convex subsets of Rn. Then,

for every x and y in 1 2, we have

F12(x, y) = max (F1(x, y), F1 (x, y)) .

Proof. The set 1 2 is convex. Let x and y be two distinct points in1 2, and assume first that the ray R(x, y) has nonempty intersection with(1 2), say, at the point a0. We assume without loss of generality thata0 2. In this case, if a1 denotes the intersection point of R(x, y) with2, then, the four points x, y, a0, a1 are aligned in that order (we may havea0 = a1, or a1 at infinity), and therefore we have

|x a0|

|y a0||x a1|

|y a1|.

This implies F12(x, y) = F2 (x, y) F1(x, y).

Funk geometry 7

The remaining case is when R(x, y) is contained in 1 2. In this case,we have F12(x, y) = F1 (x, y) = F1(x, y) = 0.

3 Balls and the topology of the Funk metric

Since we are dealing with non-symmetric distances, we need to distinguishbetween right and left open balls. For x in and > 0, we set

B+(x, ) = {y B,F(x, y) < } (3.1)

and we call it the open right open ball centered at x and of radius . Likewise,we set

B(x, ) = {y B,F(y, x) < } (3.2)

and we call it the open left open ball centered at x and of radius .We define closed right and left open balls by replacing the inequalities in

(??) and (??) by large inequalities, and in the same way we defineright andleft spheres by replacing the inequalities in (??) and (??) by qualities.

In Funk geometry, the left balls and the left right balls have generallydifferent shapes and different properties, see for instance in Example ?? below.

If is a bounded convex open set of Rn equipped with its Funk metric,then it follows from Proposition ?? that the collections of left and of right openballs are sub-bases of the same topology on , and this topology coincides withthe topology induced from the inclusion of in Rn.

In the case where the convex open set is unbounded, the left and theright open balls of the Funk metric are always noncompact.

We shall return to the study of balls in ??.

4 Examples of Funk metrics

Example 4.1 (The interval). Let a be a positive real number and considerthe open interval Ia = (0, a) R. Then, for x and y in Ia, we have

FIa(x, y) =

log(x/y) if x > y

loga x

a yif x < y

0 if x = y.

Note that for fixed x < y, the derivative of the function a 7 FIa(x, y), whichis equal to 1

ax 1

ay, is negative, and therefore this function is decreasing.

8 Athanase Papadopoulos and and Marc Troyanov

This remark was already used in proposition ?? and it will be useful in theproof of the next corollary.

Corollary 4.2. Let and be two open convex subsets of Rn satisfying . For any x and y in , we have F(x, y) F (x, y).

Proof. By restricting the result to the Euclidean line containing x and y, theresult to prove becomes one-dimensional, and this one-dimensional case followsfrom the remark at the end of Example ??.

Note that Corollary ?? also follows from Proposition ??.In the next two examples, we consider the Funk metrics of the open half-

plane and of the open strip. We note that these sets are the only open convexsubsets of the Euclidean plane that contain complete Euclidean straight lines.

Example 4.3 (The open strip). Let = B R2 be the open strip definedby

B = {(x1, x2) R2 |0 < x2 < 1}.

Then, for x = (x1, x2) and y = (y1, y2) in B, we have

FB(x, y) =

log(x2/y2) if x2 > y2

log1 x21 y2

if x2 < y2

0 if x2 = y2.

(Compare this formula with the formula for the Funk metric of the interval,Example ??).

The following is easy to prove:

(1) Let x be an arbitrary point in B and let be a positive real num-ber. Then, the right open ball B+(x, ) = {y B,FB(x, y) < } is anopen strip contained by a union of two Euclidean lines contained in Band which are parallel to the sides of B. Likewise, the left open ballB(x, ) = {y B,FB(y, x) < } is an open strip contained by a unionof two Euclidean lines parallel to the sides of B. For small , the sidesof B(x, ) are both contained in B, and otherwise, one side of B(x, )is contained in B and the other side coincides with a side of B.

(2) Given any three points pi = (xi, yi) i = 1, 2, 3 in B, satisfying eithery1 < y2 < y3 or y3 < y2 < y1, then we have FB(p1, p3) = FB(p1, p2) +FB(p2, p3). We deduce that the three points are aligned. We deducethat any curve in B joining the two boundaries and with monotonicy-coordinate is the image of a geodesic.

Funk geometry 9

Example 4.4 (The Euclidean half-plane). Let = H R2 be the upperhalf-plane, that is, the set

H = {(x1, x2) R2 |x2 > 0}.

Consider two points x = (x1, x2) and y = (y1, y2) in H . The ray R(x, y)intersects H if and only if x2 > y2. In Figure ??, we have represented thetwo possibilities:(1) R(x, y) intersects H ;

(2) R(x, y) does not intersect H .By using the intercept theorem, we have

FH(x, y) =

{log(x2/y2) if x2 > y2

0 otherwise.

In all cases, we have

FH(x, y) = max(log(x2/y2), 0).

x

xy

y

R(x, y)

R(x, y)

Figure 2. In the case where is the upper half-plane (Example ??) the distancefrom x to y is zero if the altitude of the point y is at least equal to that of x(the case represented on the left hand side).

x

x

Figure 3. In the case where is the upper half-plane (Example ??), the twoballs B+(x, ) (the case represented on the right hand side of the figure) andB(x, ) (the case represented on the left hand side of the figure).

Let be a positive number and let x be a point in H . The following is easyto prove:(1) The right closed ball B+(x, ) = {y H,FH(x, y) < } is a Euclidean

half-plane bounded by a line parallel to boundary of H (represented onthe left hand side of Figure ??).

10 Athanase Papadopoulos and and Marc Troyanov

(2) The left closed ball B(x, ) = {y H,FH(y, x) < } is a Euclideanstrip (represented to the right hand side of Figure ??).

(3) Either ball is non-compact.

xx

xyy

y

x1 x1x1

x2

x2

x2

y1 y1y1

y2

y2

y2

Figure 4.

Example 4.5 (The quarter plane). Let = Q R2 be the quarter planedefined by

Q = {(x1, x2) R2 |x1 > 0, x2 > 0}

and consider two points x = (x1, x2) and y = (y1, y2) in Q. In Figure ??, thethree cases represented are the following:

(1) R(x, y) cuts the horizontal axis. In this case we have F(x, y) = log(x2/y2).

(2) R(x, y) cuts the vertical axis. In this case we have F(x, y) = log(x1/y1)

(3) R(x, y) does not intersect the boundary of the quarter plane. In thiscase we have F(x, y) = 0.

The following formula is valid in all cases:

FQ(x, y) = max(log(x2/y2), log(x1/y1), 0).

Now we study balls in Q.Using the intercept theorem, one can easily see that the right closed ball

centered at any point P in Q is a quarter plane bounded by two rays parallelto boundary rays of Q. This quarter plane is represented on the left hand sideof Figure ??.

In the same way one can see that the left sphere centered at a point P in Qis a Euclidean rectangle; this is the rectangle OBCD represented on the righthand side of Figure ??.

Example 4.6 (Polygon). Let R2 be the interior of a polygon. Thefollowing is a ruler and compass construction of a right sphere centered ata point, illustrated in Figure ??. Let O and A be two distinct points in .

Funk geometry 11

PP

Figure 5. A right and left ball (shaded regions) in the case where is aquarter-plane (Example ??)

O

AA1 A2

Figure 6. In the case where is a polygon, a right sphere centered at a pointO.

12 Athanase Papadopoulos and and Marc Troyanov

Assume first that the ray OA does not contain a vertex of the polygon .To construct the sphere of center O passing through A, we take the Euclideanparallel through A to the edge a that intersects the ray OA. The segment onthis parallel with endpoints on the two rays that join O to the vertices of theedge a is contained in the boundary of the required sphere. Now we continuethe construction of this boundary by starting at the endpoints A1 or A2 of thissegment (see Figure ??), taking the parallel to the next edge of , etc. Thisgives the required construction of the sphere. The construction in the casewhere the ray OA contains a vertex of the polygon has also been handledin this construction.

The construction of left spheres is different, and left spheres may be non-compact. An example is given in Figure ??. In this example, the space is the triangle. For some value of r, a left sphere of radius r is the union ofthe three segments in dashed lines. This sphere is not connected and non-compact. The ball of radius r (the shaded region in Figure ??) is bounded bythe hexagon having these three dashed lines as three non-consecutive edges.It is non compact.

.P

Figure 7. A left sphere centered at a point x in the case where is a triangle.

It should also be clear from the construction in theis example that when is the interior of a convex polygon, then both left and right spheres arepolygons, and that the left spheres are not always compact.

Example 4.7 (The Euclidean unit ball). The following is the formula for theFunk metric in the case where is the unit ball in Rn (see [?]):

F(x, y) =

(1 |x|2)|y|2 + x, y2

1 |x|2

x, y

1 |x|2.

E je pense que cest faux (il y a une confusion)

Funk geometry 13

In the last two examples, the right closed balls around a point in arealways homothetic to the boundary of . This is a general fact for the Funkmetric and we shall prove it below (Proposition ??).

Example 4.8 (Parabola). Let R2 be the interior of a parabola. Here, is unbounded and for any x , there is a unique ray starting at x andcontained in . The direction of this ray is the same for all points x (it isthe axis of the parabola). In this sense, has a unique point at infinity.

A convex subset of the plane bounded by one branch of a hyperbola issuch that for each point there is an open set in the set of directions where theopen ray starting at this point is contained in . This convex set has a lineat infinity.

5 Geodesics and convexity in Funk geometry

Given a (weak) metric space (X, d) and an interval I of the real line, a mapf : I X is said to be geodesic if for any three points x1, x2, x3 in X satisfyingx1 < x2 < x3 we have d(x1, x3) = d(x1, x2) + d(x2, x3). Note that in the casewhere the metric d is non-symmetric, the order of the arguments in d(., .) isimportant. The (weak) metric space is said to be geodesic if every two pointscan be joinded by a geodesic map. The length of a curve is defined as usualas the supremum over all subdivisions of the image set of a segment of the setof lengths of polygonal curves joining the vertices of the subdivision. (Again,we note that in computing lengths, the order of the arguments is important.)

We also recall that a subset K of a (weak) geodesic metric space is saidto be geodesically convex if given any two points in K, any geodesic joiningthem if contained in K. This notion of convexity was thoroughly studied byBusemann (and it was probably introduced by him). We prove that for theFunk metric associated to a strictly convex subset of Rn, right open ballsare geodesically convex. In fact, for the Funk metric on a such a domain,the geodesic lines are the affine lines (see Proposition ?? and Corollary ??).Therefore, for such a metric, geodesic convexity of balls is equivalent to affineconvexity. Geodesic convexity of open balls is an important property in ametric space. For instance, Minkowski spaces (see [?] in this volume) andsimply connected spaces of nonpositive curvature have this property. Spheresequipped with their canonical metrics are examples of metric spaces where thisproperty is not satisfied.

In this section, is again a nonempty open subset of Rn. We study thegeodesics and the geometric balls of the Funk metric of .

Proposition 5.1. Let x, y and z be three points in lying in that order ona Euclidean line. Then, we have F(x, y) + F(y, z) = F(x, z).

14 Athanase Papadopoulos and and Marc Troyanov

Proof. We may assume that the three points are distinct, otherwise the proofis trivial. As before, we denote by R(x, y) the Euclidean ray starting at xand passing through y. We have the equivalence R(x, y) R(x, z) R(y, z) , and this holds if and only if the three quantities F(x, y),F(y, z) and F(x, z) are equal to 0. Thus, the conclusion also holds triviallyin this case. Therefore, we can assume that R(x, y) 6 . In this case, we havea+(x, y) = a+(x, z) = a+(y, z). Denoting this common point by a+, we have

|x a+|

|y a+|

|y a+|

|z a+|=|x a+|

|z a+|,

which implies

log|x a+|

|y a+|+ log

|y a+|

|z a+|= log

|x a+|

|z a+|,

which completes the proof.

Corollary 5.2. The Euclidean segments in are geodesic segments for theFunk metric on .

Corollary ?? implies that (, F) is a geodesic weak metric space. Metricson subsets of Euclidean space for which the Euclidean segments are geodesicsegments are important, in particular because of Hilberts Problem IV whichprecisely asks for a characterization of such metrics.

The Funk metric is not always a (weak version of a) Desarguesian space inthe sense of Busemann (see [?] in this volume for the definition), because ingeneral geodesics between two distinct points are not unique. The followingproposition implies that there exist other types of geodesic segments in ,provided there exists a Euclidean segment of nonempty interior contained inthe boundary of .

Proposition 5.3 (Non-unique geodesics). Let be an open convex subsetof Rn such that contains some Euclidean segment [p, q] and let x and zbe two points in such that R(x, z) [p, q] 6= . Let be the intersectionof with the affine subspace of Rn spanned by {x} [p, q]. Then, for anypoint y in satisfying R(x, y) [p, q] 6= and R(y, z) [p, q] 6= , we haveF(x, y) + F(y, z) = F(x, z).

Proof. It suffices to work in the space . Let x, y and z denote the feet ofthe perpendiculars from x, y and z respectively on the Euclidean line joiningthe points p and q (see Figure ??). Let b = R(x, z) [p, q]. Since the trianglesbxx and bzz are similar, we have

F(x, z) = log|x b|

|z b|= log

|x x|

|z z|.

Funk geometry 15

p

q

x

y

z

x

y

z

b

Figure 8.

Similar formulas hold for F(x, y) and F(y, z). Therefore,

F(x, z) = log|x x|

|z z|

= log

(|x x|

|y y|

|y y|

|z z|

)

= log

(|x x|

|y y|

)+ log

(|y y|

|z z|

)= F(x, y) + F(y, z).

Corollary 5.4. For any triple of points x, y, z as in Proposition ??, the unionof the two segments [x, y] [y, z] is a geodesic segment.

Proof. Let x, y, z be three points on the topological segment [x, y] [y, z]such that the points x, x, y, z, z are in that order. In the case where the threepoints x, y, z lie in that order on a Euclidean segment, then, by Proposition??, they satisfy F(x

, y)+F(y, z) = F(x

, z). Otherwise, it is easy to seeby elementary Euclidean geometry that the triple of points x, y, z satisfiesthe properties of the triple x, y, z of Proposition ??, and in that case, we alsohave F(x

, y) + F(y, z) = F(x

, z).

16 Athanase Papadopoulos and and Marc Troyanov

Remark 5.5. Given any open convex set Rn containing a nonempty opensegment in this boundary, Corollary ?? allows us to construct polygonal pathsthat are not Euclidean segments but that are geodesics for the Funk metric of. By taking limits of polygonal paths, we can construct, from Proposition??, geodesics which are smooth ands which are not Euclidean segments.

We state the following:

Corollary 5.6. For any open convex subset Rn such that con-tains some Euclidean segment, there are Funk geodesic segments in that arenot Euclidean segments. In particular, the Funk metric of is not uniquelygeodesic.

Now that we know that there is a class of distinguished geodesics in a Funkgeometry, namely, the Euclidean segments, a natural question is whether thisclass is preserved by an isometry of the metric. The answer will turn out to beyes, but by an indirect method; it follows from the fact which we prove below,that the isometries of the Funk metric are the affine maps.

In the proof of the next proposition, we use the notion of a support hy-perplane for , and we recall it here for the convenience of the reader. Asupport hyperplane for an open convex set Rn at a point a is anaffine hyperplane in Rn that intersects the boundary of and that does notintersect . If the boundary of is smooth, then a tangent plane to ata is a support hyperplane. Support hyperplanes exist at every point of theboundary of a convex set, but they may be not unique. If the convex set isstrictly convex, then the support hyperplane at any boundary point is unique.For these classical result, the reader can consult to the books by Fenchel [?],Eggleston [?] or Valentine [?].

Proposition 5.7. Let be an open convex subset of Rn. Let x and z be twodistinct points in such that R(x, z) 6= and such that at the pointb = R(x, z) , there is a support hyperplane whose intersection with isreduced to b. Let y be a point in such that the three points x, y, z in donot lie on the same affine line. Then, F(x, z) < F(x, y) + F(y, z).

Proof. To prove the proposition, it suffices to work in the affine plane spannedby x, y and z. In this affine plane, we can compute the three Funk distancesthat are involved in the proposition. In other words, we can assume withoutloss of generality that n = 2.

We assume that the intersection points of R(x, y) and R(y, z) with are not empty, and we let a and c be respectively these points. From the

Funk geometry 17

hypothesis, there is a support line of (which we callD) at b whose intersectionwith is reduced to the point b.

For the proof, we distinguish three cases.

D

a

a

x

y

z

c

b

c

Figure 9.

Case 1. The two rays R(x, y) and R(y, z) intersect the line D (see Figure??).

Let a and c be respectively these intersection points. Note that the threepoints a, b and c are in that order on D. By reasoning with the projectionson the line D and arguing as we did in the proof of Proposition ??, we have

|x b|

|z b|=|x a|

|y a|

|y c|

|z c|.

Since we have

|x a|

|y a|