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volume 5, issue 3, article 54, 2004.

Received 25 September, 2003;

accepted 02 April, 2004.

Communicated by:J.M. Borwein

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Journal of Inequalities in Pure and Applied Mathematics

A METRIC INEQUALITY FOR THE THOMPSON AND HILBERT GEOMETRIES

ROGER D. NUSSBAUM AND CORMAC WALSH

Mathematics Department Rutgers University New Brunswick NJ 08903.

EMail:nussbaum@math.rutgers.edu INRIA Rocquencourt

B.P. 105,

78153 Le Chesnay Cedex France.

EMail:cormac.walsh@inria.fr

c

2000Victoria University ISSN (electronic): 1443-5756 131-03

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A Metric Inequality for the Thompson and Hilbert

Geometries

Roger D. Nussbaum and Cormac Walsh

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J. Ineq. Pure and Appl. Math. 5(3) Art. 54, 2004

http://jipam.vu.edu.au

Abstract

There are two natural metrics defined on an arbitrary convex cone: Thompson’s part metric and Hilbert’s projective metric. For both, we establish an inequality giving information about how far the metric is from being non-positively curved.

2000 Mathematics Subject Classification:53C60

Key words: Hilbert geometry, Thompson’s part metric, Cone metric, Non-positive curvature, Finsler space.

1. Introduction

Let C be a cone in a vector spaceV. Then C induces a partial ordering onV given byx ≤ yif and only ify−x ∈ C. For eachx ∈ C\{0},y ∈ V, define M(y/x) := inf{λ ∈ R :y ≤ λx}. Thompson’s part metric onC is defined to be

dT(x, y) := log max (M(x/y), M(y/x)) and Hilbert’s projective metric onC is defined to be

d_H(x, y) := log (M(x/y)M(y/x)).

Two points inC are said to be in the same part if the distance between them is finite in the Thompson metric. IfC is almost Archimedean, then, with respect to this metric, each part of C is a complete metric space. Hilbert’s projective metric, however, is only a pseudo-metric: it is possible to find two distinct points which are zero distance apart. Indeed it is not difficult to see thatd_H(x, y) = 0 if and only ifx=λyfor someλ >0. Thusd_H is a metric on the space of rays of the cone. For further details, see Chapter 1 of the monograph [23].

Suppose C is finite dimensional and let S be a cross section of C, that is S :={x ∈ C : l(x) = 1}, wherel : V → Ris some positive linear functional with respect to the ordering on V. Suppose x, y ∈ S are distinct. Let a and b be the points in the boundary of S such that a, x, y, and b are collinear and are arranged in this order along the line in which they lie. It can be shown that the Hilbert distance betweenxandyis then given by the logarithm of the cross ratio of these four points:

d_H(x, y) = log|bx| |ay|

|by| |ax|.

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Indeed, this was the original definition of Hilbert. IfS is the open unit disk, the Hilbert metric is exactly the Klein model of the hyperbolic plane.

An interesting feature of the two metrics above is that they show many signs of being non-positively curved. For example, when endowed with the Hilbert metric, the Lorentz cone {(t, x₁, . . . , x_n) ∈ Rⁿ⁺¹ : t² > x²₁ +· · ·+ x²_n} is isometric ton-dimensional hyperbolic space. At the other extreme, the positive cone Rⁿ+ := {(x₁, . . . , x_n) : x_i ≥0for1≤i≤n} with either the Thompson or the Hilbert metric is isometric to a normed space [11], which one may think of as being flat. In between, for Hilbert geometries having a strictly-convexC² boundary with non-vanishing Hessian, the methods of Finsler geometry [28]

apply. It is known that such geometries have constant flag curvature−1. More general Hilbert geometries were investigated in [17] where a definition was given of a point of positive curvature. It was shown that no Hilbert geometries have such points.

However, there are some notions of non-positive curvature which do not apply. For example, a Hilbert geometry will only be a CAT(0) space (see [6]) if the cone is Lorentzian. Another notion related to negative curvature is that of Gromov hyperbolicity [15]. In [2], a condition is given characterising those Hilbert geometries that are Gromov hyperbolic. This notion has also been investigated in the wider context of uniform Finsler Hadamard manifolds, which includes certain Hilbert geometries [12].

Busemann has defined non-positive curvature for chord spaces [7]. These are metric spaces in which there is a distinguished set of geodesics, satisfying certain axioms. In such a space, denote by m_xy the midpoint along the distinguished geodesic connecting the pair of pointsxandy. Then the chord space is

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non-positively curved if, for all pointsu,x, andyin the space,

(1.1) d(m_ux, m_uy)≤ 1

2d(x, y), wheredis the metric.

In the case of the Hilbert and Thompson geometries on a part of a closed cone C, there will not necessarily be a unique minimal geodesic connecting each pair of points. However, it is known that, setting β := M(y/x;C) and α := 1/M(x/y;C), the curveφ : [0,1]→C :

(1.2) φ(s;x, y) :=







β^s−α^s β−α

y+

βα^s−αβ^s β−α

x, ifβ 6=α,

α^sx, ifβ =α

is always a minimal geodesic from x to y with respect to both the Thompson and Hilbert metrics. We view these as distinguished geodesics. If the cone C is finite dimensional, then each part ofC will be a chord space under both the Thompson and Hilbert metrics. Notice that the geodesics above are projective straight lines. If the cone is strictly convex, these are the only geodesics that are minimal with respect to the Hilbert metric. For Thompson’s metric, if two points are in the same part of C and are linearly independent, then there are infinitely many minimal geodesics between them.

In this paper we investigate whether inequalities similar to (1.1) hold for the Hilbert and Thompson geometries with the geodesics given in (1.2). We prove the following two theorems.

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Theorem 1.1. Let C be an almost Archimedean cone. Suppose u, x, y ∈ C are in the same part. Also suppose that 0 < s < 1 and R > 0, and that d_H(u, x) ≤ R and d_H(u, y) ≤ R. If the linear span of {u, x, y} is 1- or 2- dimensional, thend_T(φ(s;u, x), φ(s;u, y))≤sd_T(x, y). In general

(1.3) dT φ(s;u, x), φ(s;u, y)

≤

2(1−e^−Rs) 1−e^−R −s

dT(x, y).

Note that the bracketed value on the right hand side of this inequality is strictly increasing in R. As R → 0, this value goes to s, which reflects the fact that in small neighborhoods the Thompson metric looks like a norm. As R → ∞, the bracketed value goes to2−s.

Theorem 1.2. Let C be an almost Archimedean cone. Suppose u, x, y ∈ C are in the same part. Also suppose that 0 < s < 1 and R > 0 and that d_H(u, x) ≤ R and d_H(u, y) ≤ R. If the linear span of {u, x, y} is 1- or 2- dimensional, thend_H(φ(s;u, x), φ(s;u, y))≤sd_H(x, y). In general

(1.4) d_H φ(s;u, x), φ(s;u, y)

≤

1−e^−Rs 1−e^−R

d_H(x, y).

Again, the bracketed value on the right hand side increases strictly with in- creasingR. This time, it ranges betweensasR→0and1asR → ∞.

Our method of proof will be to first establish the results whenC is the positive coneR^N+, with N ≥ 3. It will be obvious from the proofs that the bounds given are the best possible in this case. A crucial lemma will state that any finite set of n elements of a Thomson or Hilbert geometry can be isometrically

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embedded in Rⁿ⁽ⁿ⁻¹⁾+ with, respectively, its Thompson or Hilbert metric. This lemma will allow us to extend the same bounds to more general cones, although in the general case the bounds may no longer be tight.

A special case of Theorem1.2was proved in [29] using a simple geometrical argument. It was shown that if two particles start at the same point and travel along distinct straight-line geodesics at unit speed in the Hilbert metric, then the Hilbert distance between them is strictly increasing. This is equivalent to the special case of Theorem 1.2when d_H(u, x) = d_H(u, y)andR approaches infinity.

A consequence of Theorems 1.1 and 1.2 is that both the Thompson and Hilbert geometries are semihyperbolic in the sense of Alonso and Bridson [1].

Recall that a metric space is semihyperbolic if it admits a bounded quasi-geodesic bicombing. A bicombing is a choice of path between each pair of points. We may use the one given by

ζ_(x,y)(t) :=





 φ

t

d(x, y);x, y

, ift ∈[0, d(x, y)]

y, otherwise

for each pair of pointsxandyin the same part ofC. Heredis either the Thomp- son or Hilbert metric. This bicombing is geodesic and hence quasi-geodesic. To say it is bounded means that there exist constantsM andsuch that

d(ζ_(x,y)(t), ζ_(w,z)(t))≤Mmax(d(x, w), d(y, z)) + for eachx, y, w, z ∈Candt∈[0,∞).

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Corollary 1.3. Each part of C is semihyperbolic when endowed with either Thompson’s part metric or Hilbert’s projective metric.

It should be pointed out that for some cones there are other good choices of distinguished geodesics. For example, for the cone of positive definite symmet- ric matrices Sym(n), a natural choice would be

φ(s;X, Y) :=X^1/2(X^−1/2Y X^−1/2)^sX^1/2

for X, Y ∈ Sym(n) and s ∈ [0,1]. It can be shown that, with this choice, Sym(n) is non-positively curved in the sense of Busemann under both the Thompson and Hilbert metrics. This result has been generalized to both sym- metric cones [16] and to the cone of positive elements of aC^∗-algebra [10].

Although Hilbert’s projective metric arose in geometry, it has also been of great interest to analysts. This is because many naturally occurring maps in analysis, both linear and non-linear, are either non-expansive or contractive with respect to it. Perhaps the first example of this is due to G. Birkhoff [3,4], who noted that matrices with strictly positive entries (or indeed integral operators with strictly positive kernels) are strict contractions with respect to Hilbert’s metric. References to the literature connecting this metric to positive linear operators can be found in [14, 13]. It has also been used to study the spec- tral radii of elements of Coxeter groups [20]. Both metrics have been applied to questions concerning the convergence of iterates of non-linear operators [8,16,23,24,25]. The two metrics have been used to solve problems in- volving non-linear integral equations [27,30], linear operator equations [8, 9], and ordinary differential equations [5,25,31,32]. Thompson’s metric has also been usefully applied in [24, 26] to obtain “DAD theorems”, which are scal- ing results concerning kernels of integral operators. Another application of this

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metric is in Optimal Filtering [19], while Hilbert’s metric has been used in Er- godic Theory [18] and Fractal Diffusions [21].

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2. Proofs

A cone is a subset of a (real) vector space that is convex, closed under multiplication by positive scalars, and does not contain any vector subspaces of dimension one. We say that a cone is almost Archimedean if the closure of its restriction to any two-dimensional subspace is also a cone.

The proofs of Theorems1.1 and 1.2will involve the use of some infinitesimal arguments. We recall that both the Thompson and Hilbert geometries are Finsler spaces [22]. IfC is a closed cone inR^N with non-empty interior, then intC can be considered to be anN-dimensional manifold and its tangent space at each point can be identified withR^N. If a norm

|v|^T_x := inf{α >0 :−αx≤v ≤αx}

is defined on the tangent space at each pointx ∈ intC, then the length of any piecewiseC¹ curveα: [a, b]→ intCcan be defined to be

L^T(α) :=

Z b a

|α⁰(t)|^T_α(t)dt.

The Thompson distance between any two points is recovered by minimizing over all paths connecting the points:

d_T(x, y) = inf{L^T(α) :α∈P C¹[x, y]},

where P C¹[x, y] denotes the set of all piecewise C¹ paths α : [0,1] → intC with α(0) = x and α(1) = y. A similar procedure yields the Hilbert metric when the norm above is replaced by the semi-norm

|v|^H_x :=M(v/x)−m(v/x).

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HereM(v/x)is as before and m(v/x) := sup{λ ∈ R: v ≥λx}. The Hilbert geometry will be Riemannian only in the case of the Lorentz cone. The Thomp- son geometry will be Riemannian only in the trivial case of the one-dimensional coneR+.

Our strategy will be to first prove the theorems for the case of the positive coneR^N+, and then extend them to the general case. The proof in the case ofR^N+

will involve investigation of the mapg : intR^N+ → intR^N+: g(x) := φ(s;1, x)

(2.1)

=







b^s−a^s b−a

x+

ba^s−ab^s b−a

1, ifb 6=a,

a^s1, ifb =a,

whereb :=b(x) := max_ix_i anda :=a(x) := min_ix_i. Heres ∈(0,1)is fixed and we are using the notation1:= (1, . . . ,1). The derivative ofgatx∈ intR^N+

is a linear map fromR^N → R^N. Taking| · |^T_x as the norm on the domain and

| · |^T_g(x)as the norm on the range, the norm ofg⁰(x)is

||g⁰(x)||_T := sup{|g⁰(x)(v)|^T_g(x) :|v|^T_x ≤1}.

If, instead, we take the appropriate infinitesimal Hilbert semi-norms on the domain and range, then the norm ofg⁰(x)is given by

||g⁰(x)||_H := sup{|g⁰(x)(v)|^H_g(x) :|v|^H_x ≤1}.

For each pair of distinct integersI andJ contained in{1, . . . , N}, let U_I,J :=n

x∈ intR^N+ : 0< x_I < x_i < x_J for alli∈ {1, . . . , N}\{I, J}o .

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On each setU_I,J, the mapg isC¹and is given by the formula g(x) =

x^s_J−x^s_I xJ−xI

x+

x_Jx^s_I−x_Ix^s_J xJ −xI

1.

Let U denote the union of the sets U_I,J; I, J ∈ {1, . . . , N}, I 6= J. If x ∈ R^N+\U, then there must exist distinct integers m, n ∈ {1, . . . , N} with either xn =xm = maxixi orxn =xm = minixi. The setx∈ R^N+ withxn=xmhas (N-dimensional) Lebesgue measure zero, so the complement of U in R^N+ has Lebesgue measure zero.

We recall the following results from [22]. The first is a combination of Corol- laries 1.3 and 1.5 from that paper.

Proposition 2.1. Let C be a closed cone with non-empty interior in a finite dimensional normed space V. SupposeGis an open subset of intC such that φ(s;x, y)∈Gfor allx, y ∈Gands∈[0,1]. Suppose also thatf :G→ intC is a locally Lipschitzian map with respect to the norm onV. Then

inf{k≥0 :d_T(f(x), f(y))≤kd_T(x, y)for allx, y ∈G}

= ess sup_x∈G||f⁰(x)||T. It is useful in this context to recall that every locally Lipschitzian map is Fréchet differentiable Lebesgue almost everywhere. The next proposition is a special case of Theorem 2.5 in [22].

Proposition 2.2. Let C be a closed cone with non-empty interior in a normed space V of finite dimension N. Let l be a linear functional on V such that

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l(x) > 0for all x ∈ intC, and defineS := {x ∈ C : l(x) = 1}. Let Gbe a relatively-open convex subset of S. Suppose thatf : G → intC is a locally Lipschitzian map with respect to the norm onV. Then

inf{k≥0 :dH(f(x), f(y))≤kdH(x, y)for allx, y ∈G}

= ess sup_x∈G||f⁰(x)||H˜, where ||f⁰(x)||H˜ := sup{|f⁰(x)(v)|^H_f_(x) : |v|^H_x ≤1,l(v) = 0}. Here the essen- tial supremum is taken with respect to theN−1-dimensional Lebesgue measure onS.

Since we wish to apply Propositions2.1and2.2to the mapg, we must prove that it is locally Lipschitzian.

Lemma 2.3. The map g : int(R^N+) → int(R^N+) defined by (2.1) is locally Lipschitzian.

Proof. We use the supremum norm||x||∞:= max_i|x_i|onR^N. Clearly,|b(x)− b(y)| ≤ ||x−y||_∞ and |a(x)−a(y)| ≤ ||x−y||_∞ for all x, y ∈ int(R^N+).

Therefore bothaandbare Lipschitzian with Lipschitz constant 1.

Letγ : [0,∞)→[0,∞)be defined by γ(t) :=





 t^s−1

t−1, fort 6= 1, s, fort = 1.

Thengmay be expressed as

g(x) =a^s−1γ(b/a)x+a^s

1−γ(b/a) 1.

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The Binomial Theorem gives that γ(t) =

∞

X

k=1

s k

(t−1)^k for|t−1|<1

and so γ isC^∞ on a neighborhood of 1. Hence it isC^∞ on [0,∞), and thus locally Lipschitzian. It follows thatg is also locally Lipschitzian.

2.1. Thompson’s Metric

We have the following bound on the norm ofg⁰(x)with respect to the Thompson metric.

Lemma 2.4. Consider the Thompson metric on intR^N+. Letx∈U1,N. IfN = 1 orN = 2then the norm ofg⁰ atxis given by||g⁰(x)||_T =s. IfN ≥3, then (2.2) ||g⁰(x)||_T = x_N −xN−1

x_N −x₁ θ x_N

x₁

x^s+1₁

E_N₋₁ +(x^s_N −x^s₁)xN−1

E_N−1 + xN−1−x₁ xN −x1

θ x₁

xN

x^s+1_N EN−1

whereθ(t) := (1−s)−t^s+standE_i(x) :=E_i :=x_i(x^s_N−x^s₁) +x_Nx^s₁−x₁x^s_N. Proof. IfN = 1andx > 0, then g(x) = x^s. We leave the proof in this case to the reader and assume thatN ≥2.

Forx∈U_1,N, g(x) =

x^s_N −x^s₁ x_N −x₁

x+

x_Nx^s₁−x₁x^s_N x_N −x₁

1.

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Let

h_ij(x) := x_j g_i(x)

∂g_i

∂x_j(x).

Straightforward calculation gives, for eachj ∈ {1, . . . , N}, h_1j(x) =sδ_1j

and h_{N j}(x) =sδ_{N j}.

Hereδ_ij is the Kronecker delta function which takes the value1ifi=jand the value 0 ifi 6= j. Clearly,h_ij(x) = 0for 1 < i < N andj 6∈ {1, i, N}. For 1< i < N,

h_i1(x) = _x^x^N^−xⁱ

N−x₁ θ

xN

x1

_xs+1 1

Ei ≥0, (2.3)

h_ii(x) = ^x^s^N_E^−x^s¹

i x_i ≥0,

(2.4)

hiN(x) =−_x^xⁱ^−x¹

N−x₁ θ x1

xN

_xs+1 N

Ei ≤0.

(2.5)

Inequalities (2.3) – (2.5) rely on the fact thatθ(t) ≥ 0fort ≥ 0. This may be established by observing thatθ(1) =θ⁰(1) = 0andθ⁰⁰(t)>0fort ≥0.

Let

B˜^T :=

v ∈R^N : max_j|v_j| ≤1 . We wish to calculate

(2.6) ||g⁰(x)||_T = sup (

X

j

h_ijv_j

: 1≤i≤N,v ∈B˜^T )

.

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Fori = 1ori = N, we have|P

jh_ijv_j| ≤ sfor any choice ofv ∈ B˜^T. If N = 2, then it follows that||g⁰(x)||_T =sfor allx∈U_1,N.

For the rest of the proof we shall therefore assume thatN ≥3. For1< i <

N, it is clear from inequalities (2.3) – (2.5) that|P

jh_ijv_j|is maximized when v1 =vi = 1andvN =−1. In this case

X

j

h_ijv_j

= 1 E_i

x_N −x_i x_N −x₁θ

x_N x₁

x^s+1₁ (2.7)

+(x^s_N −x^s₁)x_i+ x_i−x₁ x_N −x₁θ

x₁ x_N

x^s+1_N

= c₁x_i+c₂ c₃x_i+c₄, (2.8)

wherec₁,c₂,c₃, andc₄depend onx₁ andx_N but not onx_i. Observe thatc₃x_i+ c4 6= 0forx1 ≤ xi ≤ xN. Given this fact, the general form of expression (2.8) leads us to conclude that it is either non-increasing or non-decreasing when regarded as a function ofx_i. When we substitutex_i =x₁, we get|P

jh_ijv_j|= s. When we substitutexi =xN, we get

(2.9)

X

j

h_ijv_j

= 2

1−(x1/xN)^s

1−(x1/xN) −s.

Now, writingΓ(t) := 2(1−t^s)/(1−t)−s, we haveΓ⁰(t) = −2t^sθ(t⁻¹)/(1− t)² < 0, in other wordsΓ is decreasing on (0,1). In particular, Γ(x₁/x_N) ≥ limt→1Γ(t) = s. Therefore expression (2.7) is non-decreasing in x_i. So, the

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supremum in (2.6) is attained whenv is as above and i = N −1. Recall that xN−1is the second largest component ofx. The conclusion follows.

Corollary 2.5. Let R > 0. IfN = 1orN = 2, then ess sup{||g⁰(x)||_T : x ∈ intR^N+}=s. IfN ≥3, then

ess sup{||g⁰(x)||_T :d_H(x,1)≤R}= 2(1−e^−Rs) 1−e^−R −s.

Proof. Note that ifσ :R^N+ →R^N+ is some permutation of the components, then g◦σ(x) =σ◦g(x)for allx∈R^N+. Furthermore,σwill be an isometry of both the Thompson and Hilbert metrics. It follows that, given any x ∈ U_I,J with I, J ∈ {1, . . . , N}, I 6=J, we may reorder the components ofxto find a point y in U_1,N such that ||g⁰(y)||_T = ||g⁰(x)||_T. Recall, also, that the complement of U in intR^N+ has N-dimensional Lebesgue measure zero. From these two facts, it follows that the essential supremum of||g⁰(x)||_T overB_R(1) := {x∈ intR^N+ :d_H(x,1)≤R}is the same as its supremum overB_R(1)∩U_1,N.

In the case whenN = 1orN = 2, the conclusion follows immediately.

ForN = 3, we must maximize expression (2.2) under the constraintsx₁ <

x_N₋₁ < x_N andx₁/x_N ≥ exp(−R). First, we maximize overx_N₋₁, keeping x₁ andx_N fixed. In the proof of the previous lemma, we showed that expression (2.2) is non-decreasing inx_N−1, and so it will be maximized whenxN−1

approachesx_N. Here it will attain the value

(2.10)

2

1−(x₁/x_N)^s

1−(x₁/x_N) −s= Γ(x₁/x_N).

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We also showed that Γis decreasing on (0,1). Therefore (2.10) will be maximized whenx1/xN = exp(−R), where it takes the value

2(1−e^−Rs) 1−e^−R −s.

Lemma 2.6. Let C be an almost Archimedean cone and let {x_i : i ∈ I} be a finite collection of elements of C of cardinalityn, all lying in the same part.

Denote by W the linear span of {xi : i ∈ I} and write CW := C ∩ W. Denote by intC_W the interior ofC_W as a subset ofW, using onW the unique Hausdorff linear topology. Then each of the points x_i;i ∈ I is contained in intCW. Furthermore, there exists a linear map F : W → Rⁿ⁽ⁿ⁻¹⁾ such that F( intC_W)⊂ intRⁿ⁽ⁿ⁻¹⁾+ and

(2.11) M(xi/xj;C) = M(F(xi)/F(xj);Rⁿ⁽ⁿ⁻¹⁾+ ) for eachi, j ∈I.

Proof. Since the points{x_i : i ∈ I}all lie in the same part of C, they also all lie in the same part ofC_W. Therefore there exist positive constantsa_ij such that xj−aijxi ∈CW for alli, j ∈I. If we definea:= min{aij :i, j ∈I}it follows thatx_j +δx_i ∈ C_W whenever|δ| ≤ aandi, j ∈ I. Now selecti₁, . . . , i_m ∈ I such that{x_i_k : 1 ≤ k ≤ m}form a linear basis forW. For eachy ∈ W, we define ||y|| := max{|bk| : 1 ≤ k ≤ m}, wherey = Pm

k=1bkxi_k is the unique representation ofyin terms of this basis. The topology onW generated by this

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norm is the same as the one we have been using. If||y|| ≤a/mandj ∈I, then xj +mbkxi_k ∈CW for1≤k≤m. It follows that

xj+y= 1 m

m

X

k=1

(xj+mbkxi_k)∈CW

whenever||y|| ≤a/m. This proves thatx_j ∈ intC_W for allj ∈I.

It is easy to see thatβ_ij := M(x_i/x_j;C) = M(x_i/x_j;C_W)for alli, j ∈ I, i 6= j. Observe thatβ_ijx_j −x_i ∈ ∂C_W. Since intC_W is a non-empty open convex set which does not containβ_ijx_j−x_i, the geometric version of the Hahn- Banach Theorem implies that there exists a linear functionalf_ij : W → Rand a real numberr_ij such thatf_ij(β_ijx_j −x_i) ≤ r_ij < f_ij(z)for all z ∈ intC_W. Because0is in the closure of intC_W andf_ij(0) = 0, we haver_ij ≤0. On the other hand, iff_ij(z)<0for somez ∈ intC_W, then consideringf_ij(tz)we see thatf_ij would not be bounded below on intC_W. It follows thatr_ij = 0. Since β_ijx_j −x_iis in the closure of intC_W, we must havef_ij(β_ijx_j−x_i) = 0.

Now, define

F :W →Rⁿ⁽ⁿ⁻¹⁾ :z 7→(f_ij(z))i,j∈I, i6=j,

so that f_ij(z);i, j ∈ I, i 6= j are the components ofF(z). Clearly, F is linear and maps intCW into intRⁿ⁽ⁿ⁻¹⁾+ . Also, for alli, j ∈I,i6=j,

M(F(x_i)/F(x_j);Rⁿ⁽ⁿ⁻¹⁾+ ) = inf{λ >0 :f_kl(λx_j −x_i)≥0for allk, l ∈I,k6=l}.

For λ ≥ β_ij, we have λx_j − x_i ∈ clC_W and so f_kl(λx_j −x_i) ≥ 0 for all k, l ∈I,k 6=l. On the other hand, forλ < β_ij, we havef_ij(λx_j−x_i)<0since fij(xj)>0. We conclude thatM(F(xi)/F(xj);Rⁿ⁽ⁿ⁻¹⁾+ ) = βij.

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Lemma 2.7. Theorem1.1holds in the special case whenC =R^N+ withN ≥3.

Proof. Each part of R^N+ consists of elements ofR^N+ all having the same components equal to zero. Thus each part can be naturally identified with intRⁿ+, wherenis the number of strictly positive components of its elements. We may therefore assume initially that{x, y, u} ⊂ intR^N+.

DefineL :R^N →R^N byL(z) := (u₁z₁, . . . , u_Nz_N). Its inverse is given by L⁻¹(z) := (u⁻¹₁ z₁, . . . , u⁻¹_N z_N). BothL andL⁻¹ are linear maps which leave R^N+ invariant. It follows that L and L⁻¹ are isometries of R^N+ with respect to both the Thompson and Hilbert metrics. Therefore, foru, z ∈ intR^N+,

L⁻¹(φ(s;u, z)) =φ(s;L⁻¹(u), L⁻¹(z)).

Thus, we may as well assume thatu=1.

We now wish to apply Proposition2.1 with f := g and G := BR+(1) = {z ∈ R^N+ : d_H(z,1) < R + }. It was shown in [23] that G is a convex cone, in other words that it is closed under multiplication by positive scalars and under addition of its elements. Since φ(s;w, z) is a positive combination of w andz, it follows that φ(s;w, z) is inGif w andz are. If we now apply Lemma 2.3, Proposition 2.1, and Corollary 2.5, and let approach zero, we obtain the desired result.

Lemma 2.8. Theorem 1.1 holds in the special case when the linear span of {x, y, u}is one- or two-dimensional.

Proof. Let W denote the linear span of {x, y, u}, in other words the smallest linear subspace containing these points. By Lemma 2.6, x, y, anduare in the

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interior ofC∩W inW. It is easy to see thatM(z/w;C) = M(z/w;C∩W) for allw, z ∈ int(C∩W). Therefore, we can work in the coneC∩W.

It is not difficult to show [14] that ifm := dimW is either one or two, then there is a linear isomorphismF fromW toR^mtaking int(C∩W)to intR^m+. It follows thatF is an isometry of both the Thompson and Hilbert metrics and F(φ(s;z, w)) = φ(s;F(z), F(w))for all z, w ∈ int(C∩W). We may thus assume thatC =R^m+ andu, x, y ∈ intC.

As in the proof of Lemma2.7, we may assume thatu=1.

To obtain the required result, we apply Lemma2.3, Corollary2.5, and Propo- sition2.1withf :=g andG:= intR^m+.

of Theorem1.1. Let W denote the linear span of {x, y, u}. Lemma 2.8 han- dles the case when these three points are not linearly independent; we will therefore assume that they are. Thus the five points x, y, u, φ(s;u, x), and φ(s;u, y)are distinct. We apply Lemma2.6and obtain a linear mapF :W → R²⁰+ with the specified properties. From (2.11), it is clear that d_T(z, w) = d_T⁰(F(z), F(w)) for eachz, w ∈ {x, y, u, φ(s;u, x), φ(s;u, y)}. Here we are using dT0

to denote the Thompson metric on R²⁰+. Note that φ(s;u, x) is a positive combination of u and x and that the coefficients of u and x depend only on s, M(u/x;C), and M(x/u;C). The latter two quantities are equal to M(F(u)/F(x);R²⁰+)and M(F(x)/F(u);R²⁰+) respectively. We conclude that F(φ(s;u, x)) = φ(s;F(u), F(x)). A similar argument gives F(φ(s;u, y)) = φ(s;F(u), F(y)). Inequality (1.3) follows by applying Lemma2.7to the points F(x),F(y), andF(u)in the coneR²⁰+.

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2.2. Hilbert’s Metric

We shall continue to use the same notation. Thus, for a given N ∈ N and s ∈ (0,1), we useg to denote the function in (2.1) and U to denote the union of setsU_I,J withI, J ∈ {1, . . . , N}, I 6=J. We also use the functionsθ(t) :=

(1−s)−t^s+standE_i(x) :=E_i :=x_i(x^s_N −x^s₁) +x_Nx^s₁ −x₁x^s_N, and write h_ij(x) := (x_j/g_i(x))∂g_i/∂x_j(x). As was noted earlier, θ(t) > 0 ift > 0 and t 6= 1. Also, γ(t) := (1 − t^s)/(1− t), γ(1) := s is strictly decreasing on [0,∞). We shall also use the simple but useful observation that if c₁, c₂, c₃, and c₄ are constants such that c₃t+c₄ 6= 0 for a ≤ t ≤ b, then the function t 7→ (c₁t +c₂)/(c₃t+c₄) is either increasing on [a, b] (if c₁c₄ −c₂c₃ ≥ 0) or decreasing on [a, b](if c₁c₄ −c₂c₃ ≤ 0). Either way, the function attains is maximum over[a, b]ataorb.

Recall that ifgis Fréchet differentiable atx∈ intR^N+ then||g⁰(x)||_H denotes the norm ofg⁰(x)as a linear map from(R^N,|| · ||^H_x)to(R^N,|| · ||^H_g(x)), although, of course,|| · ||^H_x and|| · ||^H_g(x)are semi-norms rather than norms.

Lemma 2.9. Consider the Hilbert metric on intR^N+ withN ≥2. Letx∈U1,N. IfN = 2then the norm ofg⁰ atxis given by||g⁰(x)||_H =s. IfN ≥3, then

(2.12) ||g⁰(x)||_H = x_N −x_N−1

x_N −x₁ θ x_N

x₁

x^s+1₁ EN−1

+(x^s_N −x^s₁)x_N−1

E_N−1

.

Proof. The norm ofg⁰(x)as a map from(R^N,|| · ||^H_x)to(R^N,|| · ||^H_g(x))is given by

||g⁰(x)||_H = sup

v∈B˜^H

max

i,k

X

j

(h_ij −h_kj)v_j,

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where

B˜^H :=

v ∈R^N : maxjvj−minjvj ≤1 .

To calculate||g⁰(x)||_H we will need to determine the sign ofh_ij −h_kj for each i, j, k ∈ {1, . . . , N}. We introduce the notation

(2.13) L_ik := sup

v∈B˜^H

X

j

(h_ij −h_kj)v_j.

Note thatgis homogeneous of degrees, in other wordsg(λx) =λ^sg(x)for allx∈R^N+ andλ >0. Therefore,

X

j

x_j∂g_i

∂x_j(x) = sg_i(x) for each i∈ {1, . . . , N}. ThusP

jh_ij =sfor eachi ∈ {1, . . . , N}, a fact that could also have been obtained by straightforward calculation. It follows that

(2.14) X

j

(h_ij −h_kj)v_j =X

j

(h_ij −h_kj)(v_j+c)

for any constantc∈R.

It is clear that an optimal choice ofv in (2.13) would be to takev_j := 1for each componentj such thathij−hkj >0andvj := 0for each component such thath_ij −h_kj < 0. Alternatively, we may choosev_j := 0whenh_ij −h_kj > 0 andv_j := −1whenh_ij −h_kj < 0. That the optimal value is the same in both cases follows from (2.14). Also, it is easy to see thatLik =Lki.

Fixi, k ∈ {1, . . . , N}so thati < k. There are four cases to consider.

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• Case 1.1< i < k < N. Recall thath_1j(x) =sδ_1j andh_{N j}(x) =sδ_{N j}. A calculation using equations (2.3) – (2.5) gives

E_i(x)E_k(x)(h_i1(x)−h_k1(x)) =x^s_Nx^s+1₁ (x_k−x_i)θx_N x₁

≥0

and

(2.15) E_i(x)E_k(x)(h_iN(x)−h_kN(x)) = x^s₁x^s+1_N (x_k−x_i)θx₁ x_N

≥0.

We also have that h_ii(x)−h_ki(x) = h_ii(x) > 0 andh_ik(x)−h_kk(x) =

−h_kk(x)<0. So an optimal choice ofv ∈B˜^H in equation (2.13) is given byv_j :=−δ_jk. We conclude thatL_ik =h_kkin this case.

• Case 2. 1 = i < k < N. We will show that h_k1(x) ≤ h₁₁(x) = s.

Considerx₁ and x_N as fixed and x_k as varying in the range x₁ ≤ x_k ≤ x_N. From equation (2.3), h_k1(x) = (c₁x_k +c₂)/(c₃x_k +c₄), wherec₁, c₂, c₃, and c₄ depend on x₁ and x_N, and both c₃ and c₄ are positive. A simple calculation shows that c₁c₄ −c₂c₃ = −θ(x_N/x₁)x^s+1₁ x^s_N, which is negative. Hence h_k1 is decreasing in x_k and takes its maximum value whenx_k =x₁. Here it achieves the value

x₁ xN −x1

θx_N x1

=s− x^1−s₁ (x^s_N −x^s₁) xN −x1

< s.

Thus we conclude thath11(x)−hk1(x)> 0. We also have thath1k(x)− h_kk(x) = −h_kk(x)≤ 0andh_1N(x)−h_kN(x) = −h_kN(x)≥0. Thus the optimal choice ofv ∈ B˜^H is given by v_j := −δ_jk. We conclude that in this caseL1k(x) = hkk(x).

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• Case 3. 1 < i < k = N. Here h_i1 ≥ h_N₁ = 0, h_ii ≥ h_{N i} = 0, and hiN ≤ hN N =s. So the optimalv ∈ B˜^H is given byvj :=δj1+δji. We conclude thatL_iN =h_i1+h_ii.

• Case 4. i = 1and k = N. Here s = h₁₁ ≥ h_N1 = 0and 0 = h_1N ≤ hN N =s. Thus the optimalv ∈ B˜^H is given byvj :=δ1j. We conclude thatL_1N =s.

IfN = 2then Case 4 is the only one possible, and so||g⁰(x)||_H =s. So, for the rest of the proof, we will assume thatN ≥3.

We know thath_i1(x) +h_ii(x) = s−h_iN(x)≥sso Case 3 dominates Case 4, that is to sayL_iN(x)≥L_1N(x)fori >1. Sinceh_i1(x)≥0fori∈ {1, . . . , N}, Case 3 also dominates Cases 1 and 2, meaning thatL_iN(x)≥L_ik(x)fork < N, i < k.

The final step is to maximizeL_iN(x) = h_i1(x) +h_ii(x) = s−h_iN(x)over i ∈ {2, . . . , N −1}. From (2.15), h_mN(x) ≥ h_nN(x) form < n. Thus the maximum occurs wheni=N−1. Recall that we have ordered the components ofxin such a way thatxN−1is the second largest component ofx. We conclude that

||g⁰(x)||H = max

i,k:i<kLik =hN−1,1+hN−1,N−1

By substituting the expressions in (2.3) and (2.4), we obtain the required formula.

Corollary 2.10. LetR >0andN ≥2. Letlbe a linear functional onR^N such that l(x) > 0 for allx ∈ intR^N+ and define S := {x ∈ R^N+ : l(x) = 1}. If

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N = 2, then ess sup{||g⁰(x)||_H :x∈S}=s. IfN ≥3, then ess sup{||g⁰(x)||_H :d_H(x,1)≤R,x∈S}= 1−e^−Rs

1−e^−R.

In both cases, the essential supremum is taken with respect to the N − 1- dimensional Lebesgue measure onS.

Proof. Note that the complement ofU∩SinShasN−1-dimensional Lebesgue measure zero. Using the reordering argument in the proof of Corollary2.5, we deduce the result in the case whenN = 2.

The case whenN ≥ 3reduces to maximizing the right hand side of (2.12) subject to the constraints x1 < xN−1 < xN and x1/xN ≥ exp(−R). We can write the expression in (2.12) in the form s+ (c₁x_N−1 +c₂)/(c₃x_N−1 +c₄), wherec₁,c₂,c₃, andc₄ depend only onx₁ andx_N andc₁ ≥ 0,c₂ ≤ 0,c₃ ≥0, c4 ≥0. It follows that, if we viewx1andxN as fixed andxN−1 as variable, the expression is maximized whenx_N−1 =x_N. The value obtained there will be

1−(x₁/x_N)^s

1−(x₁/x_N) =γ(x₁/x_N).

If we recall thatγis decreasing on[0,1)andx₁/x_N ≥exp(−R), we see that

||g⁰(x)||_H ≤ 1−e^−Rs 1−e^−R.

Ifx₁/x_N = exp(−R), then, by choosingx ∈ U_1,N withx_N−1 close tox_N, we can arrange that||g⁰(x)||_H is as close as desired to this value.