Hilbert geometry for convex polygonal domains

HILBER T GEOMETR Y F OR CONV E X POL YGONAL DOMAINS BRU NO COLBOIS, CONST ANTIN VERNICOS, AND P A TRICK VEROVIC Abstract. W e prov e in this paper that the Hilb ert geometry ass o c ia ted with a n o pen conv ex po lygonal set is Lipschitz equiv alent to Euclidea n plane. 1. Intr oduction A H ilb ert domain in R m is a metric space ( C , d C ), where C is an op en b ounde d c on v ex set in R m and d C is the distance function on C — called t he Hilb ert metric — defined as fo llo ws. Giv en t wo distinct p oints p and q in C , let a and b be the in tersection p oin ts of the straig ht line defined by p and q w ith ∂ C so that p = (1 − s ) a + sb a nd q = (1 − t ) a + tb with 0 < s < t < 1. Then d C ( p, q ) : = 1 2 ln[ a, p, q , b ] , where [ a, p, q , b ] : = 1 − s s × t 1 − t > 1 is the cross ratio of the 4-t uple of ordered collinear p oin ts ( a, p, q , b ) ( see Figure 1). W e complete the definition b y setting d C ( p, p ) : = 0. a b p q ∂ C Figure 1. The Hilb ert metric d C The metric spac e ( C , d C ) th us obtained is a comple te non-compact g eo desic me tric space whose top ology is the one induced b y the canonical top ology of R m and in which the affine op en segmen ts joining tw o p oin ts of the b oundary ∂ C are geo desics that are isometric to ( R , | · | ). It is to b e men tioned here that in general the a ffine segmen t b etw een tw o p oin ts in C may no t b e the unique g eo desic joining these p oints (for example, if C is a square). Nev ertheless, this uniqueness holds whenev er C is s trictly conv ex. Date : O ctobe r 31 , 201 8 . 2000 Mathema tics Subje ct Classific ation. Primar y: global Finsler geometry , Secondary: conv exity . 1 2 BRUNO COLBOIS, CON ST ANTIN VERNICOS, A ND P A TRICK VEROVIC F or further information ab out Hilb ert geometry , w e refer to [1, 2 , 3, 6, 8, 12] and the excellen t in tro duction [11] b y So ci ´ e-M ´ ethou. The t w o fundamen tal examples of Hilb ert doma ins ( C , d C ) in R m corresp ond to the case when C is an ellipsoid, whic h giv es the Klein mo del of m -dimensional h yp erb o lic geometry (see for example [11, first c hapter]), and the case when C is a m -simplex, for whic h there exists a norm k·k C on R m suc h that ( C , d C ) is isometric to the normed v ector space ( R m , k·k C ) (see [5, pag es 110–113] or [10, pages 2 2–23]). Therefore, it is natural to study the Hilb ert domains ( C , d C ) in R m for whic h C is close to either an ellipsoid or a m -simplex. The first a nd last autho rs th us prov ed in [4] that an y Hilb ert do main ( C , d C ) in R m suc h t ha t the b oundary ∂ C is a C 2 h yp ersurface with non-v a nishing Gaussian curv a ture is Lipsc hitz equiv alen t to m -dimensional h yp erb o lic space H m . On the o ther hand, F¨ ortsc h and Karlsson sho w ed in [7] that a Hilb ert domain in R m is isometric to a normed v ector space if and only if it is giv en b y a m -simplex. In addition, Lins establishe d in his PhD thesis [9, Lemma 2.2.5] that the Hilb ert geometry asso ciated with a n op en con ve x p olygonal set in R 2 can b e isometrically em b edded in the normed v ector space ( R N 2 , k·k ∞ ), where N is the num ber of v ertices of the p olygon. The aim of t his pap er is to prov e that the Hilb ert g eometry asso ciated with an op en con v ex p olygonal set P in R 2 is Lipsch itz equiv alent to Euclidean plane (Theorem 3 .1 in the last section). A straighforward consequence o f this result is that al l the Hilb ert p olygonal domains in R 2 are Lipsc hitz equiv alen t to eac h other, whic h is a fa ct t hat is far f rom b eing ob vious at a first glance. The idea of the pro of is to decompo se a give n op en conv ex n -sided p olygo n P into n tria ng les ha ving o ne common v ertex in P and whose o pp osite edges to that v ertex are the sides of P , and then to sho w that eac h of these triangles is Lipsc hitz equiv alent to the cone it defines with that v ertex. This second p oint is the most tec hnical part of the pap er and is based on Prop osition 2.2. Remark. It seems tha t our result migh t b e extended to higher dimensions to prov e more generally that an y Hilb ert domain in R m giv en b y a p olytop e is Lipsc hitz equiv alen t to m - dimensional Euclidean s pace. Nev ertheless , computations in that case app ear to b e m uc h more difficult since they in volv e not only the edges of the p olytop e but also it s f aces. 2. Preliminaries This section is dev oted to some tec hnical prop erties w e will need for the pro of o f Theorem 3.1 in Section 3. The k ey results are con tained in Prop o sition 2.1 and Prop osition 2.2 . Let us first recall that the distance function d C is asso ciated with the Finsler metric F C on C giv en, fo r a n y p ∈ C a nd any v ∈ T p C = R m (tangen t vec to r space to C at p ), b y F C ( p, v ) : = 1 2  1 t − + 1 t +  if v 6 = 0 , where t − = t − C ( p, v ) and t + = t + C ( p, v ) are the unique p ositive n umbers suc h that p − t − v ∈ ∂ C and p + t + v ∈ ∂ C , and F C ( p, 0) : = 0 (see Figure 2). 3 This means that for ev ery p, q ∈ C and v ∈ T p C = R m , we hav e F C ( p, v ) = d d t   t =0 d C ( p, p + tv ) and d C ( p, q ) is the infim um of the length Z 1 0 F C ( σ ( t ) , σ ′ ( t ))d t with resp ect to F C when σ : [0 , 1 ] − → C ranges o ve r all the C 1 curv es joining p to q . Remark. F or p ∈ C and v ∈ T p C = R m with v 6 = 0 , w e will define p − = p − C ( p, v ) : = p − t − C ( p, v ) v and p + = p + C ( p, v ) : = p + t + C ( p, v ) v . Then, given any a rbitrary norm k·k on R m , w e can write F C ( p, v ) = 1 2 k v k  1 k p − p − k + 1 k p − p + k  . p p + p − v v ∂ C Figure 2. The Finsler metric F C Notations. Let S : = ] − 1 , 1[ × ] − 1 , 1[ ⊆ R 2 b e the standard op en square, ∆ : = { ( x, y ) ∈ R 2 | | y | < x < 1 } ⊆ S the op en triangle whose v ertices are 0 = (0 , 0), (1 , − 1) and (1 , 1), and Z : = { ( X , Y ) ∈ R 2 | | Y | < X } ⊆ R 2 the op en cone asso ciated with ∆ (see Figure 3). The canonical basis of R 2 will b e denoted b y ( e 1 , e 2 ). The usual ℓ 1 -norm on R 2 and its asso ciated distance will b e denoted respectiv ely b y k·k and d . Definition 2.1. F or an y pair ( V 1 , V 2 ) of vectors in R 2 r { 0 } , the set S ( V 1 , V 2 ) : = { sV 1 + tV 2 | s > 0 and t > 0 } will b e called the se ctor asso ciated with this pair. Remark. The sector S ( V 1 , V 2 ) is the con ve x h ull of the set ( R + V 1 ) ∪ ( R + V 2 ). Let us b egin with the following useful lemma: Lemma 2.1. Given a b asis ( V 1 , V 2 ) of R 2 and a ve ctor V ∈ R 2 , we have V ∈ S ( V 1 , V 2 ) ⇐ ⇒  det ( V 1 ,V 2 ) ( V 1 , V ) > 0 and det ( V 1 ,V 2 ) ( V , V 2 ) > 0  . Pr o of. The lemma is a mere consequence of the fact that the co ordinate system ( s, t ) of an y v ector V in R 2 with resp ect to a basis ( V 1 , V 2 ) of R 2 is equal to  det ( V 1 ,V 2 ) ( V , V 2 ) , det ( V 1 ,V 2 ) ( V 1 , V )  .  4 BRUNO COLBOIS, CON ST ANTIN VERNICOS, A ND P A TRICK VEROVIC No w, w e ha ve Prop osition 2.1. The ma p Φ : S − → R 2 define d by Φ( x, y ) = ( X , Y ) : = (atanh( x ) , atanh( y )) is a s mo oth diff e omorph i s m s uch that (1) Φ(∆) = Z , and (2) for al l m ∈ ∆ and V ∈ T m S = R 2 , F S ( m, V ) 6 k T m Φ · V k 6 2 F S ( m, V ) . Before pro ving this result, w e will need t he f ollo wing (see F igure 3) : Lemma 2.2. L et m = ( x, y ) ∈ ∆ ⊆ S , and define in T m S = R 2 the ve ctors V 1 : = (1 , 1 ) − m = (1 − x , 1 − y ) , V 2 : = m − (1 , − 1) = ( − 1 + x , 1 + y ) , V 3 : = ( − 1 , 1) − m = ( − 1 − x , 1 − y ) and V 4 : = ( − 1 , − 1) − m = ( − 1 − x , − 1 − y ) . Then we have the inclusions (1) S ( V 1 , V 2 ) ⊆  V = ( λ, µ ) ∈ R 2     µ > 0 and | λ | 1 − x 2 6 µ 1 − y 2  , (2) S ( V 2 , V 3 ) ⊆ { V = ( λ, µ ) ∈ R 2 | λ < 0 < µ } , (3) S ( V 3 , V 4 ) ⊆  V = ( λ, µ ) ∈ R 2     λ < 0 and | µ | 1 − y 2 6 − λ 1 − x 2  , and (4) S ( V 4 , − V 1 ) ⊆ { V = ( λ, µ ) ∈ R 2 | λ < 0 and µ < 0 } . 0 − 1 1 − 1 1 S x y V 1 − V 2 V 4 V 3 − V 1 V 2 ∆ V m Figure 3. The six zones for the v ector V 5 Pr o of. First of all, w e ha ve det ( e 1 ,e 2 ) ( V 1 , V 2 ) = 2(1 − x ) > 0, det ( e 1 ,e 2 ) ( V 2 , V 3 ) = 2( x + y ) > 0, det ( e 1 ,e 2 ) ( V 3 , V 4 ) = 2(1 − x ) > 0 a nd det ( e 1 ,e 2 ) ( V 4 , − V 1 ) = 2( x − y ) > 0 (since x > y ). This sho ws that ( V 1 , V 2 ), ( V 2 , V 3 ), ( V 3 , V 4 ) and ( V 4 , − V 1 ) are all bases of R 2 ha ving the same orien tation as ( e 1 , e 2 ). Then, let V = ( λ, µ ) b e an a r bitrary v ector in T m S = R 2 . • P oint (1): If V ∈ S ( V 1 , V 2 ), then, according to Lemma 2.1, w e ha ve 0 6 det ( e 1 ,e 2 ) ( V 1 , V ) = (1 − x ) µ − (1 − y ) λ and 0 6 det ( e 1 ,e 2 ) ( V , V 2 ) = (1 + y ) λ + (1 − x ) µ since ( V 1 , V 2 ) is a basis of R 2 ha ving the same orien tation as ( e 1 , e 2 ). This writes λ 1 − x 6 µ 1 − y and − µ 1 + y 6 λ 1 − x , and hence, m ultiplying b oth inequalities by 1 1 + x > 0, w e g et (2.1) λ 1 − x 2 6 µ (1 + x )(1 − y ) and − µ (1 + x )(1 + y ) 6 λ 1 − x 2 . On the other hand, writing V = sV 1 + tV 2 with s : = det ( V 1 ,V 2 ) ( V , V 2 ) > 0 and t : = det ( V 1 ,V 2 ) ( V 1 , V ) > 0 , the second co ordinate µ of V with resp ect to the canonical basis ( e 1 , e 2 ) of R 2 equals µ = det ( e 1 ,e 2 ) ( e 1 , V ) = s det ( e 1 ,e 2 ) ( e 1 , V 1 ) + t det ( e 1 ,e 2 ) ( e 1 , V 2 ) = s (1 − y ) + t (1 + y ) > 0 . This yields µ (1 + x )(1 − y ) 6 µ 1 − y 2 since 0 < 1 + y 6 1 + x , and hence (2.2) λ 1 − x 2 6 µ 1 − y 2 from the first part of Equation 2.1. Moreo v er, w e a lso hav e − µ 1 − y 2 6 − µ (1 + x )(1 + y ) since 0 < 1 − y 6 1 + x . Th us, (2.3) − µ 1 − y 2 6 λ 1 − x 2 from the second part of Equation 2.1 . Finally , summarizing Equations 2.2 and 2 .3 , we obtain | λ | 1 − x 2 6 µ 1 − y 2 . • P oint (2): If V ∈ S ( V 2 , V 3 ), let us write V = sV 2 + tV 3 with s : = det ( V 2 ,V 3 ) ( V , V 3 ) > 0 and t : = det ( V 2 ,V 3 ) ( V 2 , V ) > 0 . Then the first co ordinate λ of V with resp ect to the canonical basis ( e 1 , e 2 ) of R 2 equals λ = det ( e 1 ,e 2 ) ( V , e 2 ) = s det ( e 1 ,e 2 ) ( V 2 , e 2 ) + t det ( e 1 ,e 2 ) ( V 3 , e 2 ) = − s (1 − x ) − t (1 + x ) < 0 . On the other hand, the second co or dinate µ of V with resp ect to ( e 1 , e 2 ) is equal to µ = det ( e 1 ,e 2 ) ( e 1 , V ) = s det ( e 1 ,e 2 ) ( e 1 , V 2 ) + t det ( e 1 ,e 2 ) ( e 1 , V 3 ) = s (1 + y ) + t (1 − y ) > 0 . • P oint (3): If V ∈ S ( V 3 , V 4 ), then, according to Lemma 2.1, w e ha ve 0 6 det ( e 1 ,e 2 ) ( V 3 , V ) = − (1 + x ) µ − (1 − y ) λ and 0 6 det ( e 1 ,e 2 ) ( V , V 4 ) = − (1 + y ) λ + (1 + x ) µ 6 BRUNO COLBOIS, CON ST ANTIN VERNICOS, A ND P A TRICK VEROVIC since ( V 3 , V 4 ) is a basis of R 2 ha ving the same orien tation as ( e 1 , e 2 ). This writes µ 1 − y 6 − λ 1 + x and λ 1 + x 6 µ 1 + y , and hence, m ultiplying the first inequalit y by 1 1 + y > 0 and the second one by 1 1 − y > 0, w e get (2.4) µ 1 − y 2 6 − λ (1 + x )(1 + y ) and λ (1 + x )(1 − y ) 6 µ 1 − y 2 . On the other hand, writing V = sV 3 + tV 4 with s : = det ( V 3 ,V 4 ) ( V , V 4 ) > 0 and t : = det ( V 3 ,V 4 ) ( V 3 , V ) > 0 , the first co ordinate µ of V with resp ect to the canonical basis ( e 1 , e 2 ) of R 2 equals λ = det ( e 1 ,e 2 ) ( V , e 2 ) = s det ( e 1 ,e 2 ) ( V 3 , e 2 ) + t det ( e 1 ,e 2 ) ( V 4 , e 2 ) = − s (1 + x ) − t (1 + x ) < 0 . This yields − λ (1 + x )(1 + y ) 6 − λ 1 − x 2 since 0 < 1 − x 6 1 + y , a nd hence (2.5) µ 1 − y 2 6 − λ 1 − x 2 from the first part of Equation 2.4. Moreo v er, w e a lso hav e λ 1 − x 2 6 λ (1 + x )(1 − y ) since 0 < 1 − x 6 1 − y . Th us, (2.6) λ 1 − x 2 6 µ 1 − y 2 from the second part of Equation 2.4 . Finally , summarizing Equations 2.5 and 2 .6 , we obtain | µ | 1 − y 2 6 − λ 1 − x 2 . • P oint (4): If V ∈ S ( V 4 , − V 1 ), let us write V = sV 4 − tV 1 with s : = det ( V 4 , − V 1 ) ( V , V 4 ) > 0 a nd t : = det ( V 4 , − V 1 ) ( − V 1 , V ) > 0 . Then the first co ordinate λ of V with resp ect to the canonical basis ( e 1 , e 2 ) of R 2 equals λ = det ( e 1 ,e 2 ) ( V , e 2 ) = s det ( e 1 ,e 2 ) ( V 4 , e 2 ) − t det ( e 1 ,e 2 ) ( V 1 , e 2 ) = − s (1 + x ) − t (1 − x ) < 0 . On the other hand, the second co or dinate µ of V with resp ect to ( e 1 , e 2 ) is equal to µ = det ( e 1 ,e 2 ) ( e 1 , V ) = s det ( e 1 ,e 2 ) ( e 1 , V 4 ) − t det ( e 1 ,e 2 ) ( e 1 , V 1 ) = − s (1 + y ) − t (1 − y ) < 0 .  Pr o of of Pr op osition 2.1. Only the second p oint has to b e pr ov ed since the first one is obvious. So, fix m = ( x, y ) ∈ ∆ ⊆ S and V = ( λ, µ ) ∈ T m S = R 2 suc h that V 6 = 0. A straigh tfo rw ard computation shows that k T m Φ · V k =  λ 1 − x 2 , µ 1 − y 2  , and th us k T m Φ · V k = | λ | 1 − x 2 + | µ | 1 − y 2 . 7 No w, let us define the vec to rs V 1 , V 2 , V 3 and V 4 in T m S = R 2 as in Lemma 2.2. Since R 2 is equal t o the union of the sectors S ( V 1 , V 2 ), S ( V 2 , V 3 ), S ( V 3 , V 4 ), S ( V 4 , − V 1 ) and their images b y the symmetry ab out the origin 0, a nd since the Finsler metric F S on S is rev ersible, there are four cases to b e considered. • Case 1: V ∈ S ( V 1 , V 2 ). The unique p ositiv e n um b ers τ − and τ + suc h that m − τ − V ∈ ∂ S and m + τ + V ∈ ∂ S satisfy y − τ − µ = − 1 a nd y + τ + µ = 1. So, τ − = (1 + y ) /µ a nd τ + = (1 − y ) /µ , and hence F S ( m, V ) = 1 2  1 τ − + 1 τ +  = µ 1 − y 2 . But k T m Φ · V k > | µ | 1 − y 2 = µ 1 − y 2 since µ > 0 by p oin t (1) in Lemma 2.2. Therefore, w e hav e F S ( m, V ) 6 k T m Φ · V k . On the other hand, p oint (1) in Lemma 2 .2 yields k T m Φ · V k = | λ | 1 − x 2 + µ 1 − y 2 6 2 µ 1 − y 2 , whic h show s that k T m Φ · V k 6 2 F S ( m, V ) . • Case 2: V ∈ S ( V 2 , V 3 ). The unique p ositiv e n um b ers τ − and τ + suc h that m − τ − V ∈ ∂ S and m + τ + V ∈ ∂ S satisfy x − τ − λ = 1 and y + τ + µ = 1. So, τ − = − (1 − x ) /λ and τ + = (1 − y ) /µ , and hence F S ( m, V ) = 1 2  − λ 1 − x + µ 1 − y  . But p oint (2) in Lemma 2.2 implies k T m Φ · V k = − λ 1 − x 2 + µ 1 − y 2 =  1 1 + x   − λ 1 − x  +  1 1 + y   µ 1 − y  with − λ 1 − x > 0 a nd µ 1 − y > 0. Therefore, since 1 2 6 1 1 + x 6 1 and 1 2 6 1 1 + y 6 1, w e ha ve F S ( m, V ) 6 k T m Φ · V k 6 2 F S ( m, V ) . • Case 3: V ∈ S ( V 3 , V 4 ). The unique p ositiv e n um b ers τ − and τ + suc h that m − τ − V ∈ ∂ S and m + τ + V ∈ ∂ S satisfy x − τ − λ = 1 and x + τ + λ = − 1. So, τ − = − (1 − x ) /λ and τ + = − (1 + x ) /λ , and hence F S ( m, V ) = − λ 1 − x 2 . But k T m Φ · V k > | λ | 1 − x 2 = − λ 1 − x 2 since λ < 0 by p oin t (3) in Lemma 2.2 . 8 BRUNO COLBOIS, CON ST ANTIN VERNICOS, A ND P A TRICK VEROVIC Therefore, w e hav e F S ( m, V ) 6 k T m Φ · V k . On the other hand, p oint (3) in Lemma 2 .2 yields k T m Φ · V k = − λ 1 − x 2 + | µ | 1 − y 2 6 − 2 λ 1 − x 2 , whic h show s that k T m Φ · V k 6 2 F S ( m, V ) . • Case 4: V ∈ S ( V 4 , − V 1 ). The unique p ositiv e n um b ers τ − and τ + suc h that m − τ − V ∈ ∂ S and m + τ + V ∈ ∂ S satisfy x − τ − λ = 1 and y + τ + µ = − 1. So, τ − = − (1 − x ) /λ and τ + = − (1 + y ) /µ , and hence F S ( m, V ) = 1 2  − λ 1 − x + − µ 1 + y  . But p oint (2) in Lemma 2.2 implies k T m Φ · V k = − λ 1 − x 2 + − µ 1 − y 2 =  1 1 + x   − λ 1 − x  +  1 1 + y   − µ 1 − y  with − λ 1 − x > 0 a nd − µ 1 − y > 0. Therefore, since 1 2 6 1 1 + x 6 1 and 1 2 6 1 1 + y 6 1, w e ha ve F S ( m, V ) 6 k T m Φ · V k 6 2 F S ( m, V ) .  Remark. It is to b e p oin ted out that the Lipsch itz constants 1 and 2 obtained in Prop osition 2.1 are o ptimal. Indeed, taking m : = (1 / 2 , 0) ∈ ∆ and V : = (0 , 1) ∈ S ( V 1 , V 2 ) ⊆ T m S = R 2 , w e get k T m Φ · V k = F S ( m, V ). On the other hand, w e hav e k T m Φ · V k F S ( m, V ) − → 2 when m − → (0 , 0) and V − → (1 , 1) with m ∈ ∆ and V ∈ S ( V 1 , V 2 ) ⊆ T m S = R 2 . Giv en real n umbers a ∈ (0 , 1) and c > b > 1 , let T ⊆ R 2 b e the triangle defined as the op en con v ex h ull of the p o in ts (1 , − 1), (1 , 1 ) and ( − a, 0), and let Q ⊆ R 2 b e the quadrilatera l defined as the op en conv ex h ull of the p oin ts (1 , − 1), (1 , 1), ( − b, c ) and ( − b, − c ) (see Figure 4). Then w e hav e ∆ ⊆ T ⊆ Q and Prop osition 2.2. Ther e exists a c onstant A = A ( a, b, c ) ∈ (0 , 1] such that AF T ( m, V ) 6 F Q ( m, V ) 6 F T ( m, V ) for al l m ∈ ∆ and V ∈ T m T = T m Q = R 2 . Remark. This is the k ey result of this section, but also the most tec hnical one of t he pap er. So, the reader ma y skip it in a first reading without an y loss of k eeping trac k of the ideas that lead to the final theorem in Section 3. Before provin g Prop osition 2 .2 , we will need the fo llowing simple but ve ry useful fact (see Figure 5): 9 0 − 1 − b ( − b, c ) ( − b, − c ) 1 x y − a T ∆ Q (1 , 1) (1 , − 1) Figure 4. The triangle T and t he quadrila t era l Q Lemma 2.3. L et ω , q 1 and q 2 b e non-c ol line ar p oints in R 2 , and c onsider p 1 ∈ ] ω , q 1 [ and p 2 ∈ ] ω , q 2 [ such that the lines ( p 1 p 2 ) and ( q 1 q 2 ) interse ct in a p oint ω 0 . If q 1 ∈ ] ω 0 , q 2 [ , then we have ω q 2 ω p 2 > ω q 1 ω p 1 . p 1 p 2 q 1 q 2 ω ω 0 Figure 5. In tersecting pairs of half-lines Pr o of. Let π : R 2 − → R 2 b e t he pro jection of R 2 on to t he line ( ω q 2 ) along the direction of ( ω 0 p 2 ). Since π is affine, it is barycen tre-preserving, and hence q 1 ∈ ] ω 0 , q 2 [ necessarily implies π ( q 1 ) ∈ ] π ( ω 0 ) , π ( q 2 )[ = ] p 2 , q 2 [. So , ω q 2 > ω π ( q 1 ). But, π b eing affine with π ( ω ) = ω and π ( p 1 ) = p 2 , w e also hav e ω π ( q 1 ) ω p 2 = ω q 1 ω p 1 , whic h prov es Lemma 2.3 thanks to the previous inequality .  10 BRUNO COLBOIS, CON ST ANTIN VERNICOS, A ND P A TRICK VEROVIC Pr o of of Pr op osition 2.2. Since T ⊆ Q , we already hav e the second inequalit y . So, the v ery thing to b e pro ved here is the first inequalit y . Recall tha t k·k and d denote resp ectiv ely the usual ℓ 1 -norm on R 2 and it s asso ciated distance. Define κ 0 : = diam d ( Q ) > 0, the diameter of Q with resp ect to d , and let θ 0 b e t he in tersection p oin t of the line R (1 , 1) with the line passing through the p oin ts (1 , − 1) and ( − b, c ); in other w ords, θ 0 = ( α 0 , α 0 ) with α 0 : = c − b c + b + 2 ∈ (0 , 1). Next consider ∆ + : = { ( x, y ) ∈ R 2 | 0 6 y < x < 1 } ⊆ ∆ and Σ : = { ( x, y ) ∈ R 2 | α 0 6 y < x < 1 } ⊆ ∆ + , fix m = ( x, y ) ∈ Σ, and define in T m T = T m Q = R 2 the v ectors V 1 : = (1 , 1) − m = (1 − x , 1 − y ), V 2 : = m − (1 , − 1) = ( − 1 + x , 1 + y ), V 3 : = ( − b, c ) − m = ( − b − x , c − y ) and V 4 : = ( − a, 0) − m = ( − a − x , − y ). Then w e hav e det ( e 1 ,e 2 ) ( V 1 , V 2 ) = det ( e 1 ,e 2 ) ( − V 2 , V 1 ) = det ( e 1 ,e 2 ) ( − V 1 , − V 2 ) = 2(1 − x ) > 0 , det ( e 1 ,e 2 ) ( V 2 , V 3 ) = det ( e 1 ,e 2 ) ( − V 2 , − V 3 ) = (1 + c ) x + (1 + b ) y + b − c > (2 + b + c ) y + b − c > 0 (since x > y > α 0 ) , det ( e 1 ,e 2 ) ( V 1 , V 3 ) = det ( e 1 ,e 2 ) ( − V 3 , V 1 ) = (1 − c ) x − (1 + b ) y + b + c > ( b + c )(1 − x ) > 0 (since x > y ) , det ( e 1 ,e 2 ) ( V 3 , V 4 ) = ( b − a ) y + c ( x + a ) > 0 (since b > 1 > a ) , det ( e 1 ,e 2 ) ( V 2 , V 4 ) = det ( e 1 ,e 2 ) ( V 4 , − V 2 ) = det ( e 1 ,e 2 ) ( − V 2 , − V 4 ) = (1 + a ) y + x + a > 0 , and det ( e 1 ,e 2 ) ( V 1 , V 4 ) = det ( e 1 ,e 2 ) ( V 4 , − V 1 ) = det ( e 1 ,e 2 ) ( − V 4 , V 1 ) = a (1 − y ) + x − y > a (1 − y ) > 0 (since x > y ) . This sho ws that ( V 1 , V 3 ), ( V 1 , V 4 ), ( V 3 , V 4 ), ( V 4 , − V 2 ) and ( − V 2 , V 1 ) are bases of R 2 ha ving the same orientation as ( e 1 , e 2 ) with V 2 ∈ S ( V 1 , V 3 ) ∩ S ( V 1 , V 4 ), V 3 ∈ S ( V 1 , V 4 ), − V 1 ∈ S ( V 4 , − V 2 ) and − V 3 , − V 4 ∈ S ( − V 2 , V 1 ). Giv en an arbitra r y v ector V = ( λ, µ ) ∈ T m T = T m Q = R 2 suc h that V 6 = 0, there are now four cases to b e dealt with. • Case 1: V ∈ S ( V 1 , V 2 ) (see Figure 6). F rom V 2 ∈ S ( V 1 , V 3 ), the half-line m + R + V 2 in tersects with the segmen t [(1 , 1) , ( − b, c )] ⊆ ∂ Q , and hence the same holds for the half-line m + R + V since V ∈ S ( V 1 , V 2 ) ⊆ S ( V 1 , V 3 ). Moreo v er, since V 2 ∈ S ( V 1 , V 4 ), w e ha v e V ∈ S ( V 1 , V 2 ) ⊆ S ( V 1 , V 4 ), and this implies tha t the half-lines m + R + V 2 and m + R + V also intersec t with the segmen t [(1 , 1) , ( − a, 0)] ⊆ ∂ T . Therefore, if V 6∈ { V 1 , V 2 } , Lemma 2.3 with ω : = m , p 1 : = p + T ( m, V ), q 1 : = p + Q ( m, V ), p 2 : = p + T ( m, V 2 ), q 2 : = p + Q ( m, V 2 ) and ω 0 : = (1 , 1) giv es (2.7) t + Q ( m, V 2 ) t + T ( m, V 2 ) > t + Q ( m, V ) t + T ( m, V ) , and this still holds when V : = V 1 or V : = V 2 . 11 0 − 1 − b ( − b, c ) = q 0 ( − b, − c ) 1 x − a T Q Σ (1 , 1) = ω 0 (1 , − 1) = m 0 m θ 0 p 0 P Q ( m, V 2 ) P T ( m, V 2 ) V P + Q ( m, V ) P + T ( m, V ) P − T ( m, V ) P − Q ( m, V ) V 2 − V 2 V 1 V 3 V 4 Figure 6. The case when V ∈ S ( V 1 , V 2 ) On the other hand, if m 0 : = (1 , − 1), q 0 : = ( − b, c ) and p 0 is the inters ection p o in t o f the line pass- ing through (1 , 1) and ( − a, 0) with the line ( m 0 q 0 ), Lemma 2.3 with ω : = m 0 , p 1 : = p + T ( m, V 2 ), q 1 : = p + Q ( m, V 2 ), p 2 : = p 0 , q 2 : = q 0 and ω 0 : = (1 , 1) yields (2.8) m 0 q 0 m 0 p 0 > m 0 p + Q ( m, V 2 ) m 0 p + T ( m, V 2 ) . But t + Q ( m, V 2 ) = mp + Q ( m, V 2 ) > mp + T ( m, V 2 ) = t + T ( m, V 2 ) > 0 (since T ⊆ Q ) and m 0 m > 0, whic h implies (2.9) m 0 p + Q ( m, V 2 ) m 0 p + T ( m, V 2 ) > 1 > m 0 p + T ( m, V 2 ) m 0 p + Q ( m, V 2 ) = m 0 m + t + T ( m, V 2 ) m 0 m + t + Q ( m, V 2 ) > t + T ( m, V 2 ) t + Q ( m, V 2 ) . Then, com bining Equations 2.7, 2.8 and 2.9 , we get (2.10) m 0 q 0 m 0 p 0 > t + Q ( m, V ) t + T ( m, V ) . F urthermore, since m ∈ Σ and p − T ( m, V ) ∈ [( − a, 0) , (1 , − 1)] ⊆ R × ( −∞ , 0] (indeed, − V 1 ∈ S ( V 4 , − V 2 ) implies − V ∈ S ( − V 1 , − V 2 ) ⊆ S ( V 4 , − V 2 )), w e ha v e   m − p − T ( m, V )   > d ( m, R × { 0 } ) > α 0 , a nd th us t − T ( m, V ) =   m − p − T ( m, V )   k V k > α 0 k V k . 12 BRUNO COLBOIS, CON ST ANTIN VERNICOS, A ND P A TRICK VEROVIC In addition, since m, p − Q ( m, V ) ∈ Q and κ 0 = diam d  Q  , we also ha ve t − Q ( m, V ) =   m − p − Q ( m, V )   k V k 6 κ 0 k V k . Hence, (2.11) t − Q ( m, V ) t − T ( m, V ) 6 κ 0 α 0 . Finally , if K 1 = K 1 ( a, b, c ) : = min { α 0 /κ 0 , m 0 p 0 /m 0 q 0 } ∈ (0 , 1], Equations 2.10 and 2.11 lead to 1 t − Q ( m, V ) + 1 t + Q ( m, V ) > K 1  1 t − T ( m, V ) + 1 t + T ( m, V )  , or equiv alen tly F Q ( m, V ) > K 1 F T ( m, V ) . • Case 2: V ∈ S ( V 2 , V 3 ) (see Figure 7). 0 − 1 − b ( − b, c ) ( − b, − c ) 1 x − a T Q Σ (1 , 1) (1 , − 1) m θ 0 P + Q ( m, V ) P + T ( m, V ) P − T ( m, V ) = P − Q ( m, V ) V V 2 − V 2 V 1 V 3 V 4 Figure 7. The case when V ∈ S ( V 2 , V 3 ) Since V 2 ∈ S ( V 1 , V 3 ), w e hav e V ∈ S ( V 2 , V 3 ) ⊆ S ( V 1 , V 3 ), and hence the half- line m + R + V in tersects with the segmen t [(1 , 1) , ( − b, c )] ⊆ ∂ Q . On the other hand, this half-line also intersec ts with the segmen t [(1 , 1) , ( − a, 0)] ⊆ ∂ T since V ∈ S ( V 2 , V 3 ) ⊆ S ( V 1 , V 4 ) (indeed, w e hav e V 2 , V 3 ∈ S ( V 1 , V 4 )). 13 Then, writing that the p oin t p + T ( m, V ) = ( x + t + T ( m, V ) λ , y + t + T ( m, V ) µ ) (resp. p + Q ( m, V ) = ( x + t + Q ( m, V ) λ , y + t + Q ( m, V ) µ )) b elongs t o the line pa ssing thro ugh (1 , 1) and ( − a, 0) (resp. ( − b, c )) whose equation is X − (1 + a ) Y + a = 0 (resp. (1 − c ) X − (1 + b ) Y + ( b + c ) = 0), we get (1 + a ) µ − λ > 0 and ( c − 1) λ + (1 + b ) µ > 0 together with (2.12) t + T ( m, V ) = x − (1 + a ) y + a (1 + a ) µ − λ and t + Q ( m, V ) = (1 − c ) x − (1 + b ) y + ( b + c ) ( c − 1) λ + (1 + b ) µ . Next, since − V 3 ∈ S ( − V 2 , V 1 ), w e hav e − V ∈ S ( − V 2 , − V 3 ) ⊆ S ( − V 2 , V 1 ), and hence the half-line m − R + V inters ects with t he segmen t [(1 , − 1) , (1 , 1)] ⊆ ∂ T ∩ ∂ Q . The po in ts p − T ( m, V ) = ( x − t − T ( m, V ) λ , y − t − T ( m, V ) µ ) and p − Q ( m, V ) = ( x − t − Q ( m, V ) λ , y − t − Q ( m, V ) µ )) t hus lie on the line passing thro ugh (1 , − 1) and (1 , 1) whose equation is X − 1 = 0, whic h gives λ < 0 and (2.13) t − T ( m, V ) = t − Q ( m, V ) = x − 1 λ . No w, from Equations 2.1 2 and 2.13, one obtains 2 F Q ( m, V ) = 1 t − Q ( m, V ) + 1 t + Q ( m, V ) = 1 + b x − 1 × λ (1 − y ) − µ (1 − x ) (1 − c ) x − ( 1 + b ) y + ( b + c ) together with 2 F T ( m, V ) = 1 t − T ( m, V ) + 1 t + T ( m, V ) = 1 + a x − 1 × λ (1 − y ) − µ (1 − x ) x − (1 + a ) y + a , and hence (2.14) F Q ( m, V ) F T ( m, V ) = 1 + b 1 + a × x − (1 + a ) y + a (1 − c ) x − (1 + b ) y + ( b + c ) . As y 6 x < 1, w e ha v e both x − (1 + a ) y + a > a (1 − y ) > 0 a nd 0 < c (1 − x ) + b (1 − y ) + ( x − y ) = (1 − c ) x − (1 + b ) y + ( b + c ) 6 ( b + c )(1 − y ) ( indeed, (1 − c ) x 6 (1 − c ) y since c > 1) , from whic h Equation 2.1 4 finally yields F Q ( m, V ) > K 2 F T ( m, V ) , where K 2 = K 2 ( a, b, c ) : = a (1 + b ) (1 + a )( b + c ) ∈ (0 , 1 ]. • Case 3: V ∈ S ( V 3 , V 4 ) (see Figure 8). Since − V 3 , − V 4 ∈ S ( − V 2 , V 1 ), we ha v e − V ∈ S ( − V 3 , − V 4 ) ⊆ S ( − V 2 , V 1 ), a nd hence the half-line m − R + V inters ects with t he segmen t [(1 , − 1) , (1 , 1)] ⊆ ∂ T ∩ ∂ Q . So, w e get ag ain λ < 0 and (2.15) t − T ( m, V ) = t − Q ( m, V ) = x − 1 λ . On the other hand, let V 5 : = ( − b, − c ) − m = ( − b − x , − c − y ). W e hav e det ( e 1 ,e 2 ) ( V 3 , V 4 ) > 0, det ( e 1 ,e 2 ) ( V 4 , V 5 ) = cx + ( a − b ) y + ac > ( a + c − b ) y + ac > 0 (since x > y and c > b ) and det ( e 1 ,e 2 ) ( V 3 , V 5 ) = 2 c ( b + x ) > 0, whic h sho ws that ( V 3 , V 5 ) is a basis of R 2 ha ving the same orientation as ( e 1 , e 2 ) with V 4 ∈ S ( V 3 , V 5 ). Therefore, V ∈ S ( V 3 , V 4 ) ⊆ S ( V 3 , V 5 ), and hence the half-line m + R + V interse cts with the segmen t [( − b, c ) , ( − b, − c )] ⊆ ∂ Q . F urthermore, this half-line also in tersects with the segmen t [(1 , 1) , ( − a, 0)] ⊆ ∂ T s ince V ∈ S ( V 3 , V 4 ) ⊆ S ( V 1 , V 4 ) (indeed, w e hav e V 3 ∈ S ( V 1 , V 4 )). 14 BRUNO COLBOIS, CON ST ANTIN VERNICOS, A ND P A TRICK VEROVIC 0 − 1 − b ( − b, c ) ( − b, − c ) 1 x − a T Q Σ (1 , 1) (1 , − 1) m θ 0 P + Q ( m, V ) P + T ( m, V ) P − T ( m, V ) = P − Q ( m, V ) V − V 2 V 1 V 3 V 4 V 5 Figure 8. The case when V ∈ S ( V 3 , V 4 ) Then, writing that the p oin t p + T ( m, V ) (resp. p + Q ( m, V )) b elongs to the line passing through (1 , 1 ) and ( − a, 0) (resp. ( − b, c ) and ( − b, − c )) whose equation is X − (1 + a ) Y + a = 0 (resp. X + b = 0), w e compute (1 + a ) µ − λ > 0 together with (2.16) t + T ( m, V ) = x − (1 + a ) y + a (1 + a ) µ − λ and t + Q ( m, V ) = x + b − λ . Equations 2.15 and 2.16 then yield λ (1 − y ) − µ (1 − x ) < 0 and (2.17) F Q ( m, V ) F T ( m, V ) = 1 + b 1 + a × x − (1 + a ) y + a x + b × λ λ (1 − y ) − µ (1 − x ) . Since V ∈ S ( V 3 , V 4 ) and ( V 3 , V 4 ) is a basis of R 2 with the same orien tation as ( e 1 , e 2 ), w e ha ve det ( e 1 ,e 2 ) ( V 3 , V ) = ( λy − µx ) − ( cλ + bµ ) > 0 and det ( e 1 ,e 2 ) ( V , V 4 ) = ( λy − µx ) − aµ 6 0, and th us − ( b − a ) µ > cλ . But y 6 x < 1, b > a and λ < 0 then imply 0 > ( b − a )( λ ( 1 − y ) − µ (1 − x )) = ( b − a ) λ (1 − y ) − ( b − a ) µ (1 − x ) > ( b − a ) λ (1 − y ) + cλ (1 − x ) > ( b − a ) λ (1 − y ) + cλ (1 − y ) = λ ( b − a + c )(1 − y ) , and hence λ λ (1 − y ) − µ (1 − x ) = ( b − a ) λ ( b − a )( λ (1 − y ) − µ (1 − x )) > b − a ( b − a + c )(1 − y ) > 0 . 15 Finally , using x − (1 + a ) y + a > a (1 − y ) > 0 tog ether with 0 < x + b 6 1 + b , Equation 2.17 giv es F Q ( m, V ) > K 3 F T ( m, V ) , where K 3 = K 3 ( a, b, c ) : = a ( b − a ) (1 + a )( b − a + c ) ∈ (0 , 1 ]. • Case 4: V ∈ S ( V 4 , − V 1 ) (see Figure 9). 0 − 1 − b ( − b, c ) ( − b, − c ) 1 x − a T Q Σ (1 , 1) (1 , − 1) m θ 0 P + Q ( m, V ) P + T ( m, V ) P − T ( m, V ) = P − Q ( m, V ) V − V 1 − V 2 V 1 V 4 Figure 9. The case when V ∈ S ( V 4 , − V 1 ) Since − V 4 ∈ S ( − V 2 , V 1 ), w e hav e − V ∈ S ( − V 4 , V 1 ) ⊆ S ( − V 2 , V 1 ), and hence the half-line m − R + V inters ects with the segmen t [(1 , − 1) , (1 , 1)] ⊆ ∂ T ∩ ∂ Q . So, t − T ( m, V ) = t − Q ( m, V ), whic h means that (2.18) t − Q ( m, V ) t − T ( m, V ) = 1 . On the other hand, since m ∈ Σ and p + T ( m, V ) ∈ [( − a, 0) , (1 , − 1)] ⊆ R × ( −∞ , 0] (indeed, − V 1 ∈ S ( V 4 , − V 2 ) implies V ∈ S ( V 4 , − V 1 ) ⊆ S ( V 4 , − V 2 )), w e ha ve   m − p + T ( m, V )   > d ( m, R × { 0 } ) > α 0 , a nd th us t + T ( m, V ) =   m − p + T ( m, V )   k V k > α 0 k V k . 16 BRUNO COLBOIS, CON ST ANTIN VERNICOS, A ND P A TRICK VEROVIC In addition, since m, p + Q ( m, V ) ∈ Q and κ 0 = diam d  Q  , we also ha ve t + Q ( m, V ) =   m − p + Q ( m, V )   k V k 6 κ 0 k V k . Hence, (2.19) t + Q ( m, V ) t + T ( m, V ) 6 κ 0 α 0 . Finally , if K 4 = K 4 ( a, b, c ) : = min { 1 , α 0 /κ 0 } ∈ (0 , 1], Equations 2.18 and 2.19 lead to F Q ( m, V ) > K 4 F T ( m, V ) . A t this stage of the pro of , defining K = K ( a, b, c ) : = min { K 1 , K 2 , K 3 , K 4 } ∈ (0 , 1] and summing up the results obtained in the fo ur cases discussed ab o ve, w e can write (2.20) F Q ( m, V ) > K F T ( m, V ) for all m ∈ Σ and V ∈ T m T = T m Q = R 2 . No w, the only thing to b e done is to establish a similar inequalit y as in Equation 2.2 0 for m ∈ ∆ + r Σ, fr o m which w e will get Prop o sition 2.2 since b o th T and Q ar e preserv ed b y the reflection ab out the x -axis. So, let δ 0 : = d (∆ + r Σ , [(1 , 1) , ( − a, 0)]) > 0 and consider a p oin t m ∈ ∆ + r Σ together with a v ector V = ( λ, µ ) ∈ R 2 suc h that V 6 = 0 . First of all, since m, p + Q ( m, V ) ∈ Q and κ 0 = diam d  Q  , w e hav e t + Q ( m, V ) =   m − p + Q ( m, V )   k V k 6 κ 0 k V k . Next, the F insler metrics F T and F Q b eing rev ersible, w e can a ssume that λ 6 0, and hence p + T ( m, V ) ∈ ∂ T r ( { 1 } × [ − 1 , 1]). This implies that   m − p + T ( m, V )   > δ 0 and g iv es t + T ( m, V ) =   m − p + T ( m, V )   k V k > δ 0 k V k . Therefore, (2.21) t + Q ( m, V ) t + T ( m, V ) 6 κ 0 δ 0 . On the other hand, as regards t − T ( m, V ) and t − Q ( m, V ), w e ha v e t wo cases to lo ok at. • First case: p − T ( m, V ) ∈ { 1 } × [ − 1 , 1] ⊆ ∂ T . In that case, w e also hav e p − Q ( m, V ) ∈ { 1 } × [ − 1 , 1] ⊆ ∂ Q , and hence t − T ( m, V ) = t − Q ( m, V ), or equiv alen tly (2.22) t − Q ( m, V ) t − T ( m, V ) = 1 . So, if K 5 = K 5 ( a, b, c ) : = min { 1 , δ 0 /κ 0 } ∈ (0 , 1], Equations 2.21 and 2.22 yield (2.23) F Q ( m, V ) > K 5 F T ( m, V ) . • Second case: p − T ( m, V ) ∈ ∂ T r ( { 1 } × [ − 1 , 1]). 17 Then w e hav e t − T ( m, V ) =   m − p − T ( m, V )   k V k > δ 0 k V k . But t − Q ( m, V ) =   m − p − Q ( m, V )   k V k 6 κ 0 k V k since m, p + Q ( m, V ) ∈ Q and κ 0 = diam d  Q  . Therefore, (2.24) t − Q ( m, V ) t − T ( m, V ) 6 κ 0 δ 0 . If K 6 = K 6 ( a, b, c ) : = δ 0 /κ 0 ∈ (0 , 1 ], Equations 2.21 and 2.24 th us lead to (2.25) F Q ( m, V ) > K 6 F T ( m, V ) . Conclusion: combining Equations 2.20, 2.23 and 2.25, and defining A = A ( a, b, c ) : = min { K, K 5 , K 6 } ∈ (0 , 1] , w e hav e finally obtained that F Q ( m, V ) > AF T ( m, V ) for all m ∈ ∆ + and V ∈ T m T = T m Q = R 2 , whic h ends the pro of of Prop osition 2.2.  F rom Prop osition 2.2, w e can then deduce Prop osition 2.3. L et C 1 and C 2 b e op en b ounde d c onvex sets in R 2 such that (1) the se gment { 1 } × [ − 1 , 1] is include d in b oth b oundaries ∂ C 1 and ∂ C 2 , (2) (1 , − 1) and (1 , 1) ar e c orner p oints of C 1 and C 2 , (3) the origin 0 lies in C 1 ∩ C 2 , and (4) ∆ ⊆ C 1 ∩ C 2 . Then ther e exis ts a c onstant B = B ( C 1 , C 2 ) > 1 that satisfies 1 B F C 1 ( m, V ) 6 F C 2 ( m, V ) 6 B F C 1 ( m, V ) for al l m ∈ ∆ and V ∈ T m C 1 = T m C 2 = R 2 . Pr o of. Since C 1 ∩ C 2 is an op en set in R 2 that con tains the origin 0 by p oint (3 ), its in tersection I with R × { 0 } is an o p en set in R × { 0 } whic h also contains 0, a nd hence there exists a n umber a ∈ (0 , 1) suc h that [ − a, a ] × { 0 } ⊆ I . This implies that ( − a, 0) ∈ I ⊆ C 1 ∩ C 2 , and therefore T ⊆ C 1 ∩ C 2 , where T ⊆ R 2 is the triangle define d as the o p en conv ex h ull of the p oints (1 , − 1), (1 , 1) a nd ( − a, 0) (indeed, C 1 and C 2 are con v ex sets in R 2 whose b oundaries con ta in (1 , − 1) and (1 , 1) b y p oin t (1)). On the o ther hand, since the sets C 1 and C 2 are b o unded, there exists a num b er b > 1 suc h that they are b oth inside [ − b, b ] × [ − b, b ]. So, C 1 and C 2 are included in the op en half-plane of R 2 whose b oundary is the line {− b } × R and which contains the or igin 0. Next, the conv exity of C 1 (resp. C 2 ) together with p oints (1) and (3) show that C 1 (resp. C 2 ) lies inside the op en half-plane of R 2 whose b o undar y is the line R (( 1 , − 1) − (1 , 1)) = { 1 } × R and whic h contains the o r ig in 0. Moreo v er, p o in t (2 ) implies that C 1 (resp. C 2 ) has supp or t lines L − 1 (resp. L − 2 ) and L + 1 (resp. L + 2 ) at resp ectiv ely (1 , − 1) a nd (1 , 1) whic h are not equal to the line { 1 } × R . Therefore, C 1 (resp. C 2 ) 18 BRUNO COLBOIS, CON ST ANTIN VERNICOS, A ND P A TRICK VEROVIC lies inside the op en half-planes of R 2 whic h con tain the origin 0 and whose b o undar ies are the lines L − 1 (resp. L − 2 ) and L + 1 (resp. L + 2 ). So, if w e denote b y c 1 (resp. c 2 ) the maxim um of the absolute v alues of the second co ordinates of the in tersection p oints of the lines L − 1 (resp. L − 2 ) and L + 1 (resp. L + 2 ) with the line { − b } × R , then C 1 and C 2 are included in the op en half-planes of R 2 whic h con ta in the origin 0 and whose b oundaries a re the line s R ((1 , − 1) − ( − b, − c )) and R (( − b, c ) − (1 , 1 )), where c : = max { c 1 , c 2 } + b + 1 > b . Conclusion: w e hav e C 1 ⊆ Q and C 2 ⊆ Q , where Q ⊆ R 2 is the quadrilateral defined as the op en con ve x h ull of the p oints (1 , − 1), (1 , 1), ( − b, c ) and ( − b, − c ). No w, for all m ∈ ∆ and V ∈ T m T = T m C 1 = T m C 2 = T m Q = R 2 , w e can write AF C 1 ( m, V ) 6 AF T ( m, V ) (since T ⊆ C 1 ) 6 F Q ( m, V ) (by the first inequalit y in Prop osition 2.2) 6 F C 2 ( m, V ) (since C 2 ⊆ Q ) 6 F T ( m, V ) (since T ⊆ C 2 ) 6 1 A F Q ( m, V ) (by the first inequalit y in Prop osition 2.2) 6 1 A F C 1 ( m, V ) (since C 1 ⊆ Q ) , whic h prov es Prop osition 2.3 with B : = 1 / A > 1.  3. Lipschitz equiv alence to Euclidean plane In this section, we build a homeomorphism fro m an op en con v ex p olygonal set to Euclidean plane, and pro ve t ha t it is bi-Lipsc hitz with resp ect to the Hilb ert metric of the p olygonal set and the Euclidean distance of the plane. This is the statemen t of Theorem 3.1. So, let P b e an o p en con v ex p olygonal set in R 2 that con ta ins the origin 0. Let v 1 , . . . , v n b e t he v ertices of P ( i.e. , the corner p oints of the conv ex set P ) that w e assume to b e cyclically ordered in ∂ P (notice w e hav e n > 3 ). Define v 0 : = v n and v n +1 : = v 1 . Let f : P − → R 2 b e the map defined as follows . F or eac h k ∈ { 1 , . . . , n } , let ∆ k : = { sv k + tv k +1 | s > 0 , t > 0 and s + t < 1 } ⊆ P , and consider the unique linear transformatio n L k of R 2 suc h that L k ( v k ) : = (1 , − 1) and L k ( v k +1 ) : = (1 , 1). Then, giv en an y p ∈ ∆ k , w e define f ( p ) : = L − 1 k (Φ( L k ( p ))) , where Φ : S − → R 2 is the map considered in Prop osition 2.1. In other w ords, f mak es t he f ollo wing diagr a m commute (see Figure 10): ∆ k f − − − → S ( v k , v k +1 ) L k   y   y L k ∆ r ( { 1 } × [ − 1 , 1]) − − − → Φ Z , 19 where we recall that S : = ] − 1 , 1[ × ] − 1 , 1[ ⊆ R 2 , ∆ : = { ( x, y ) ∈ R 2 | | y | < x < 1 } ⊆ S and Z : = { ( X , Y ) ∈ R 2 | | Y | < X } ⊆ R 2 . This mak es sense since n [ i =1 ∆ i = P , L k (∆ k ) = ∆ r ( { 1 } × [ − 1 , 1]) ⊆ S and fo r all p ∈ [0 , 1) v k = ∆ k − 1 ∩ ∆ k , w e hav e L − 1 k − 1 (Φ( L k − 1 ( p ))) = L − 1 k (Φ( L k ( p ))) ∈ R v k . 0 ∆ k v n = v 0 P v 1 = v n +1 v 2 v k v k +1 p v 0 S ( v k , v k +1 ) v n = v 0 v 1 = v n +1 v 2 v k v k +1 f ( p ) T p f · v 0 − 1 − 1 1 1 S ∆ r  { 1 } × [ − 1 , 1]  m V x y 0 X Y Z Φ( m ) T m Φ · V f Φ L k L − 1 k Figure 10. The bi-Lipsc hitz homeomorphism f With this definition, k eeping in mind that k·k and d denote resp ectiv ely the usual ℓ 1 -norm on R 2 and it s asso ciated distance, we get Theorem 3.1. The map f satisfies the fol lowin g pr op e rties: (1) f is a hom e omorphism. (2) I f U and V ar e the op en sets in R 2 define d by U : = P r n [ k =1 [0 , 1) v k ! and V : = R 2 r n [ k =1 R v k ! , then f ( U ) = V a nd f induc es a sm o oth diffe omorphism f r om U onto V . 20 BRUNO COLBOIS, CON ST ANTIN VERNICOS, A ND P A TRICK VEROVIC (3) The r e exists a c onstant C > 1 such that 1 C d P ( p, q ) 6 d ( f ( p ) , f ( q )) 6 C d P ( p, q ) for al l p, q ∈ P . Before pro ving this theorem, let us establish the following: Lemma 3.1. Given any r e al numb er α > 0 , ther e is a c onstant M = M ( α ) > 1 such that 1 M ln  1 − s 1 − t × 1 + α t 1 + αs  6 ln  1 − s 1 − t × 1 + t 1 + s  6 M ln  1 − s 1 − t × 1 + α t 1 + αs  for al l 0 6 s < t < 1 . Pr o of. Consider D : = { ( s, t ) ∈ R 2 | 0 6 s < t 6 1 } and let ϕ : D − → R b e the function defined by ϕ ( s, t ) : = ln  1 + t 1 + s   ln  1 + αt 1 + αs  . Giv en λ ∈ [0 , 1], w e ha v e for all ( s, t ) ∈ D , ϕ ( s, t ) ∼  1 + t 1 + s − 1    1 + αt 1 + αs − 1  = 1 + αs α (1 + s ) − → 1 + αλ α (1 + λ ) as ( s, t ) − → ( λ, λ ). Hence, b y contin uit y o f ϕ , the function ˆ ϕ : D − → R defined b y ˆ ϕ ( s, t ) : = ϕ ( s, t ) if ( s, t ) ∈ D and ˆ ϕ ( s, t ) : = 1 + αs α (1 + s ) if s = t is con tin uous. Then, compactness of D implies that ˆ ϕ has a minim um and a maxim um. But these latters a re p ositiv e since one can easily chec k that ˆ ϕ ( s, t ) > 0 for all ( s, t ) ∈ D , and this imp lies t ha t there is a constan t M > 1 such t hat (3.1) 1 M ln  1 + αt 1 + αs  6 ln  1 + t 1 + s  6 M ln  1 + αt 1 + αs  for all ( s, t ) ∈ D . 21 Finally , for all 0 6 s < t < 1, w e ha v e 1 M ln  1 − s 1 − t × 1 + αt 1 + αs  = 1 M ln  1 − s 1 − t  + 1 M ln  1 + αt 1 + αs  6 ln  1 − s 1 − t  + 1 M ln  1 + αt 1 + αs  (since 1 / M 6 1) 6 ln  1 − s 1 − t  + ln  1 + t 1 + s  = ln  1 − s 1 − t × 1 + t 1 + s  (b y t he first inequalit y in Equation 3 .1) 6 ln  1 − s 1 − t  + M ln  1 + αt 1 + αs  (b y t he second inequalit y in Equation 3.1) 6 M ln  1 − s 1 − t  + M ln  1 + αt 1 + αs  (since M > 1) = M ln  1 − s 1 − t × 1 + α t 1 + αs  This pro ves Lemma 3.1 .  Pr o of of The or em 3.1 . • Point (1): L et g : R 2 − → P b e the map giv en b y g ( P ) : = L − 1 k (Φ − 1 ( L k ( P ))) for all k ∈ { 1 , . . . , n } and P ∈ S ( v k , v k +1 ), this definition making sense since n [ i =1 S ( v i , v i +1 ) = R 2 , Φ − 1 ( L k ( S ( v k , v k +1 ))) = ∆ r ( { 1 } × [ − 1 , 1]) (using L k ( S ( v k , v k +1 )) = Z and p oint (1) in Prop o- sition 2.1 ) a nd L − 1 k − 1 (Φ − 1 ( L k − 1 ( P ))) = L − 1 k (Φ − 1 ( L k ( P ))) ∈ [0 , 1) v k whenev er P ∈ R v k = S ( v k − 1 , v k ) ∩ S ( v k , v k +1 ). Then it is easy to c hec k that f ◦ g = I R 2 and g ◦ f = I P (iden tit y maps), whic h sho ws that f is bijectiv e with f − 1 = g . In addition, f a nd g are contin uous since L 1 , . . . , L n and Φ are homeomorphisms. • P oint (2): F or each k ∈ { 1 , . . . , n } , w e ha v e f ([0 , 1) v k ) = R v k , and therefore f ( U ) = f ( P ) r n [ k =1 f ([0 , 1) v k ) ! = R 2 r n [ k =1 R v k ! = V . Moreo v er, since U = n [ k =1 ◦ ∆ k and V = n [ k =1 ◦ z }| { S ( v k , v k +1 ) together with the fact that L 1 , . . . , L n are smoot h b y linearity and Φ is a smo oth diffeomorphism b y Prop osition 2 .1, w e get that f |U and g |V are smo oth. • P oint (3): F ixing k ∈ { 1 , . . . , n } a nd applying Prop osition 2.3 with C 1 : = S and C 2 : = L k ( P ), w e get a constan t B k > 1 suc h that for all m ∈ ∆ and V ∈ T m S = T m ( L k ( P )) = R 2 , 1 B k F S ( m, V ) 6 F L k ( P ) ( m, V ) 6 B k F S ( m, V ) , and hence 1 B k F L k ( P ) ( m, V ) 6 k T m Φ · V k 6 2 B k F L k ( P ) ( m, V ) 22 BRUNO COLBOIS, CON ST ANTIN VERNICOS, A ND P A TRICK VEROVIC b y Prop osition 2 .1. But, since L k induces an isometry from ( P , d P ) on to ( R 2 , d ) (b eing affine, L k preserv es the cross ratio), this is equiv alent to saying that for all p ∈ ◦ ∆ k and v ∈ T p P = R 2 (writing m = L k ( p ) and V = L k ( v )), w e ha ve 1 B k F P ( p, v ) 6   T L k ( p ) Φ · L k ( v )   6 2 B k F P ( p, v ) , whic h yields 1 B k | | | L k | | | F P ( p, v ) 6   L − 1 k ( T L k ( p ) Φ · L k ( v ))   = k T p f · v k 6 2 B k       L − 1 k       F P ( p, v ) , where | | | ·| | | denotes the op erator nor m on End( R 2 ) asso ciated with k·k . No w, if K : = max  B k | | | L k | | | + 2 B k       L − 1 k       + 1   1 6 k 6 n  > 1, then (3.2) 1 K F P ( p, v ) 6 k T p f · v k 6 K F P ( p, v ) for all p ∈ n [ k =1 ◦ ∆ k = U and v ∈ T p P = R 2 . W e will then prov e Theorem 3.1 using the fact that ( P , d P ) and ( R 2 , d ) are geo desic metric spaces in whic h a ffine segmen ts are geo desics (see In tro duction). Let p, q ∈ P and γ : [0 , 1] − → P defined b y γ ( t ) : = (1 − t ) p + tq . Assume tha t γ | [0 , 1) ⊆ U and q = γ (1 ) ∈ n [ k =1 [0 , 1) v k . The second inequalit y in Equation 3.2 then implies that for all t ∈ [0 , 1), w e hav e d ( f ( p ) , f ( γ ( t ))) 6 Z t 0   T γ ( s ) f · γ ′ ( s )   d s 6 K Z t 0 F P ( γ ( s ) , γ ′ ( s ))d s = K d P ( p, γ ( t )) , and th us d ( f ( p ) , f ( γ ( t ))) 6 K d P ( p, γ ( t )), whic h gives (3.3) d ( f ( p ) , f ( q )) 6 K d P ( p, q ) as t − → 1. If no w p : = sv k and q : = tv k for some 0 6 s < t < 1 and k ∈ { 1 , . . . , n } , a stra ig h tforward calculation giv es f ( p ) = L − 1 k (Φ( sL k ( v k ))) = L − 1 k (Φ( s (1 , − 1))) = L − 1 k (Φ( s, − s )) = L − 1 k (atanh( s ) , − atanh( s )) = atanh( s ) L − 1 k (1 , − 1) = atanh( s ) v k , and hence (3.4) d ( f ( p ) , f ( q )) = k f ( q ) − f ( p ) k = (atanh( t ) − a t a nh( s )) k v k k = k v k k 2 ln  1 − s 1 − t × 1 + t 1 + s  , together with e 2 d P ( p,q ) =  v k , q , p , − t − P (0 , v k ) v k  = 1 − s 1 − t × t + t − P (0 , v k ) s + t − P (0 , v k ) , or equiv alen tly (3.5) d P ( p, q ) = 1 2 ln  1 − s 1 − t × 1 + tα k 1 + sα k  , where α k : = 1 /t − P (0 , v k ) > 0. 23 Then, using Lemma 3.1 with α : = α k and denoting Λ k : = M ( α k ) × max {k v k k , 1 / k v k k} > 1, Equations 3.4 and 3.5 yield (3.6) 1 Λ k d P ( p, q ) 6 d ( f ( p ) , f ( q )) 6 Λ k d P ( p, q ) . If no w p and q are arbitrary c hosen in P , the closed affine segmen t joining p and q either meets n [ k =1 [0 , 1) v k in at most n p oints, o r ha s at least t w o distinct points in common with some [0 , 1) v k for k ∈ { 1 , . . . , n } . Therefore, it follows from Equation 3 .3 and t he second inequalit y in Equation 3.6 that d ( f ( p ) , f ( q )) 6 C d P ( p, q ) holds with C : = max { Λ k | 1 6 k 6 n } + K > 1. Finally , using the first inequalities in Equations 3.2 and 3.6, the same argumen ts for f − 1 as those for f lead to d P ( p, q ) 6 C d ( f ( p ) , f ( q )) for an y p, q ∈ P . This ends the pro of of Theorem 3.1.  Reference s [1] Benoist, Y. A survey on con vex divisible sets. Lecture notes, International Conference and Instructional W orkshop o n Discrete Groups in Beijing, 200 6 . [2] Busemann, H. The ge ometry of ge o desics . Academic Pre s s, 1955. [3] Busemann, H., and Kell y, P. Pr oje ctive ge ometry and pr oje ctive metrics . Academic Press, 1 9 53. 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PhD thesis, Uni- versit y of Strasbour g, 2000 . [12] Vernicos, C. In tro duction aux g´ eom ´ etr ies de Hilbert. S ´ emin. Th ´ eor. Sp e ctr. G ´ eom. 23 (2005 ), 145 –168. Bruno Colbois, Universit ´ e de N euch ˆ atel, Institut de ma th ´ ema tique, Rue ´ Emile Argand 11, Case post ale 15 8, CH–2009 Neuch ˆ atel, Switzerla n d E-mail addr ess : br uno.co lbois@ unine.ch Const antin Vernicos, UMR 5149 du CNRS & Universit ´ e Mo n tpellier I I, I nstitut de m a th ´ e- ma tique et de mod ´ elisa tion de Montpellier, Pl ace Eug ` ene Ba t aillon, Case cou rrier 5 1 , F–34095 Mo n tpellier Cedex, France E-mail addr ess : co nstant in.ver nicos@math.univ-montp2.fr P a trick Vero vic, UMR 5127 du CNRS & U niversit ´ e de Sa voie, Labora toire de ma th ´ ema tique, Campus scientifique, F–73376 Le Bourget-du-La c Cedex, France E-mail addr ess : ve rovic@ univ-s avoie.fr