Convexity of asymptotic geodesics in Hilbert Geometry
Charalampos Charitos, Ioannis Papadoperakis, Georgios Tsapogas
aa r X i v : . [ m a t h . M G ] M a r Convexity of asymptotic geodesics in HilbertGeometry
Charalampos Charitos, Ioannis Papadoperakisand Georgios TsapogasAgricultural University of AthensMarch 24, 2020
Abstract
If Ω is the interior of a convex polygon in R and f, g two asymptoticgeodesics, we show that the distance function d ( f ( t ) , g ( t )) is convex for t sufficiently large. The same result is obtained in the case ∂ Ω is of class C and the curvature of ∂ Ω at the point f ( ∞ ) = g ( ∞ ) does not vanish.An example is provided for the necessity of the curvature assumption. Let Ω be a bounded convex (open) domain in R n and h the Hilbert metric on Ωdefined as follows: for any distinct points p, q in Ω let p ′ and q ′ be the intersectionsof the line through p and q with ∂ Ω closest to p and q respectively. Then h ( p, q ) = log | p ′ − q | · | q ′ − p || p ′ − p | · | q ′ − q | where | z − w | denotes the usual Euclidean distance. The quantity | p ′ − q |·| q ′ − p || p ′ − p |·| q ′ − q | is the cross ratio of the colinear points p, q, q ′ , p ′ denoted by [ p, q, q ′ , p ′ ] and isinvariant under projective transformations of R n . We refer to [2], [7] and [6] forthe basic properties of the distance h as well as a presentation of classic andcontemporary aspects of Hilbert Geometry.1t is well known that, contrary to non-positively curved Riemannian geometry,the distance between two points moving at unit speed along two geodesics is notnecessarily convex. The behavior near infinity of the distance function t → h ( f ( t ) , g ( t ))when f, g are two intersecting geodesics is studied in detail in [8]. In this notewe are concerned with the case of asymptotic geodesics.Each geodesic line f determines two points at infinity denoted by f ( −∞ )and f (+ ∞ ) which are distinct points in ∂ Ω . Two geodesics f, g are said to be asymptotic if f (+ ∞ ) = g (+ ∞ ) . We first show that, up to re-parametrization,the distance function h ( f ( t ) , g ( t )) tends to 0 when t → ∞ , provided that ∂ Ω is C at the point f (+ ∞ ) = g (+ ∞ ) . Then we show that, near infinity, the distancefunction is convex when Ω is the interior of a convex polytope in R n as well aswhen Ω is a convex domain in R n with C boundary such that the curvature of ∂ Ω at f (+ ∞ ) = g (+ ∞ ) along the plane determined by f and g is not zero. Theprecise statement is the following Theorem 1
Suppose f, g are two asymptotic geodesic lines in a convex boundeddomain Ω with common boundary point ξ = f (+ ∞ ) = g (+ ∞ ) ∈ ∂ Ω and P theplane determined by f and g. If Ω is either(a) the interior of a convex polytope in R n , or(b) a convex domain in R n with C boundary and the curvature of ∂ Ω at ξ along P is not zero,then, there exists T > such that the function t → h ( f ( t ) , g ( t )) is convex for t > T. In the last Section and for the case of non-asymptotic geodesics, two sim-ple examples are provided demonstrating non-convexity for either intersectingor, disjoint geodesics. Finally, an example demonstrating the necessity of thecurvature condition in Theorem 1b above is provided (see Example 3).Dynamical properties of the geodesic flow on Ω / Γ equipped with the Hilbertmetric, where Γ is a torsion free discrete group which divides the strictly convexdomain Ω, have been studied by Y. Benoist (see [1]) using the Anosov propertiesof the flow. In view of Eberlein’s approach in the study of the geodesic flow (see[4], [5] and [3]) which is based on the convexity of the distance function as well as2 ( y n ) θ ( x n ) ℓ ξ ξ Ω g ( −∞ ) f ( −∞ ) LL L t n x ′ n f (0) g (0) x n y n y x y ′ n g ( t n ) f ( t n )Figure 1: Overview of notation for the proof of Proposition 2.the zero distance of asymptotic geodesics, mixing of geodesic flow in the Hilbertgeometry setting can be established using Theorem 1b and Proposition 2 below. We will always work with a pair of geodesic lines which determine a plane P in R n . As the distance function only depends on the affine section P ∩ Ω we willassume for the rest of this paper that Ω is a bounded convex domain in R . Recall that the Euclidean line ℓ ξ is called a support line for Ω at the point ξ ∈ ∂ Ω if ∂ Ω ∩ ℓ ξ ∋ ξ and Ω ∩ ℓ ξ = ∅ . Note that if ∂ Ω is smooth at ξ, then thesupport line is the unique tangent line at ξ. Proposition 2
Let f, g be two asymptotic geodesic lines with common boundarypoint ξ = f (+ ∞ ) = g (+ ∞ ) ∈ ∂ Ω . Assume that ∂ Ω is C at ξ. Then there existsa (geodesic) re-parametrization of f such that lim t →∞ h ( f ( t ) , g ( t )) = 0 . Proof.
Let ℓ ξ be the tangent line at ξ and L the line containing f ( −∞ ) and g ( −∞ ) . We first treat the case where ℓ ξ ∩ ∂ Ω = { ξ } , the other case being when ℓ ξ ∩ ∂ Ω is an Euclidean segment containing ξ. If ℓ ξ ∩ L is a (finite) point A, we may compose with a projective transformationwhich sends A to ∞ and, thus, we may assume that ℓ ξ , L are parallel. For each3 ∈ R , using the line L t containing g ( t ) and parallel to L we obtain a newparametrization for f by setting f ( t ) := L t ∩ Im f. For all t ∈ R , by similarity ofthe (Euclidean) triangles (cid:0) ξ, f ( t ) , g ( t ) (cid:1) and (cid:0) ξ, f (0) , g (0) (cid:1) , we have h ( f (0) , f ( t )) = h ( g (0) , g ( t )) = t hence f is re-parametrized by arc length.Pick an arbitrary sequence { t n } of positive reals converging to infinity. Foreach n ∈ N , the geodesic segment [ f ( t n ) , g ( t n )] determines two points in ∂ Ωdenoted by x n and y n so that [ x n , g ( t n )] contains f ( t n ) and does not contain y n . For each n > ξ, x n ] and denote by x ′ n itsintersection with the line L . All the above notation is displayed in Figure 1.Denote by T n (resp. T ′ n ) the triangle with vertices ξ, x n and y n (resp. x ′ n and y ′ n ). Denote by h T n and h T ′ n the corresponding Hilbert metrics. We have h ( f ( t n ) , g ( t n )) = h T n ( f ( t n ) ,g ( t n )) = log[ f ( t n ) , g ( t n ) , y n , x n ]= log[ f (0) , g (0) , y ′ n , x ′ n ] = d T ′ n ( f (0) , g (0)) . Hence, it suffices to show that h T ′ n ( f (0) , g (0)) → n → ∞ . As ∂ Ω is smooth at ξ and ℓ ξ ∩ ∂ Ω = { ξ } the angle θ ( x n ) (resp. θ ( y n )) formedby ℓ ξ and the segment [ x n , ξ ] (resp. [ y n , ξ ]) is well defined. Clearly, θ ( x n ) → θ ( y n ) → n → ∞ which implies that | x ′ n − f (0) | → ∞ and | y ′ n − g (0) | → ∞ . Therefor, both fractions | x ′ n − g (0) || x ′ n − f (0) | = | x ′ n − f (0) | + | f (0) − g (0) || x ′ n − f (0) | , | y ′ n − f (0) || y ′ n − g (0) | = | y ′ n − g (0) | + | f (0) − g (0) || y ′ n − g (0) | converge to 1 and hence h T ′ n ( f (0) , g (0)) = log (cid:18) | x ′ n − g (0) | | y ′ n − f (0) | x ′ n − f (0) | | y ′ n − g (0) | (cid:19) → . We now treat the case where ξ is contained in a segment σ ⊂ ∂ Ω . We may assumethat σ and the line L containing f ( −∞ ) and g ( −∞ ) are parallel, otherwise, we4ay compose by a projective transformation sending the intersection point atinfinity. As in the previous case, define a new geodesic parametrization of f bysetting f ( t ) := L t ∩ Im f where L t is the line containing g ( t ) and parallel to L. Pick a simple close C curve τ with the following properties:(1) τ bounds a convex domain Ω ′ ( Ω(2) τ contains ξ, f ( −∞ ) , g ( −∞ )(3) the tangent line to τ at ξ contains σ. Denote by h ′ the Hilbert distance in Ω ′ . By the previous case, h ′ ( f ( t ) , g ( t )) → t → ∞ . Since Ω ′ ⊂ Ω , we have h ( f ( t ) , g ( t )) ≤ h ′ ( f ( t ) , g ( t ))for all t, which completes the proof of the proposition. Remark 3 If ∂ Ω is not smooth at ξ then the distance function is bounded awayfrom . To see this, pick two distinct support lines ℓ , ℓ at ξ. One of them, say ℓ , intersects L at a point, say A. Then, a fraction involving the sines of the anglesformed by the segments [ A, ξ ] , [ f ( −∞ ) , ξ ] and [ g ( −∞ ) , ξ ] at ξ is a lower boundfor the distance function. Corollary 4
Let f, g be two asymptotic geodesic lines with common boundarypoint ξ = f (+ ∞ ) = g (+ ∞ ) ∈ ∂ Ω . Assume that ∂ Ω is C at ξ. Then lim t →∞ h ( f ( t ) , g ( t )) = C for some non-negative real C. Proof.
In the proof of Proposition 2, a new parametrization for f was defined.Denote by f the same geodesic line with the new parametrization and let C bethe unique real number so that f ( C ) = f (0) . Clearly, f ( t + C ) = f ( t ) , hence, h ( f ( t ) , g ( t )) ≤ h ( f ( t ) , f ( t + C )) + h ( f ( t + C ) , g ( t )) ≤ | C | + h (cid:0) f ( t ) , g ( t ) (cid:1) and h ( f ( t ) , g ( t )) ≥ h ( f ( t ) , f ( t + C )) − h ( f ( t + C ) , g ( t ))= | C | − h (cid:0) f ( t ) , g ( t ) (cid:1) where h (cid:0) f ( t ) , g ( t ) (cid:1) → → ∞ . Convexity of the distance function
We start with an elementary lemma concerning the Hilbert distance.
Lemma 5
Let Ω be a convex domain and f a geodesic line in Ω such that | f ( −∞ ) − f (0) | = | f (+ ∞ ) − f (0) | = 1 / . Then for any t > the Euclideandistance E ( t ) = | f ( t ) − f (+ ∞ ) | is given by the formula E ( t ) = 1 e t + 1 . The proof is straightforward by the definition of the Hilbert metric: h ( f (0) , f ( t )) = log | f ( −∞ ) − f ( t ) | · | f (+ ∞ ) − f (0) || f ( −∞ ) − f (0) | · | f (+ ∞ ) − f ( t ) | = log (1 − E ( t )) · · E ( t )hence e t = e h ( f (0) ,f ( t )) = 1 − E ( t ) E ( t ) and E ( t ) = 1 e t + 1 . We also need an elementary calculus lemma.
Lemma 6
There exists
T > such that the following function is convex φ : [ T, + ∞ ) → R : φ ( t ) = log β + (cid:0) α + (cid:1) E ( t ) β + (cid:0) α − (cid:1) E ( t ) where α, β are real numbers with β > and E ( t ) = e t +1 . Proof.
An elementary calculation shows that φ ′ ( t ) = βE ′ ( t ) (cid:2) β + (cid:0) α + (cid:1) E ( t ) (cid:3) (cid:2) β + (cid:0) α − (cid:1) E ( t ) (cid:3) and the second derivative of φ is φ ′′ ( t ) = Φ (cid:2) β + (cid:0) α + (cid:1) E ( t ) (cid:3) (cid:2) β + (cid:0) α − (cid:1) E ( t ) (cid:3) where the numerator Φ isΦ = β (cid:2) β + (cid:0) α + (cid:1) E ( t ) (cid:3) (cid:2) β + (cid:0) α − (cid:1) E ( t ) (cid:3) E ′′ ( t )+ βE ′ ( t ) (cid:0) α + (cid:1) E ′ ( t ) (cid:2) β + (cid:0) α − (cid:1) E ( t ) (cid:3) − βE ′ ( t ) (cid:0) α − (cid:1) E ′ ( t ) (cid:2) β + (cid:0) α + (cid:1) E ( t ) (cid:3) . ξξ ξξg ( −∞ ) g ( −∞ ) g ( −∞ ) g ( −∞ ) f ( −∞ ) f ( −∞ ) f ( −∞ ) f ( −∞ )Figure 2: The four generic positions of the sides of ∂ Ω with respect to ξ and thecoordinate axes (Theorem 1a). 7s the asymptotic behavior of E ( t ) , E ′ ( t ) and E ′′ ( t ) is e t , − e t and e t respectively,the dominant summand of Φ is β E ′′ ( t ) = β e t − e t ( e t + 1) . Since β > , this completes the proof. Proof of Theorem 1(a).
Let Ω be the interior of a convex polygon in R and f, g two asymptoticgeodesic lines with common boundary point ξ = f (+ ∞ ) = g (+ ∞ ) ∈ ∂ Ω . Recallthat a projective transformation preserves straight lines, convexity and cross ratioof four colinear points. Moreover, a projective transformation is uniquely deter-mined by its image on four points provided that no three of them are colinear.As the latter property is satisfied by the points f ( −∞ ) , g ( −∞ ) , f (0) and g (0) , we may assume, after composing by the appropriate projective transformationthat the coordinates of the four points mentioned above are: f ( −∞ ) ≡ (cid:16) , √ (cid:17) , g ( −∞ ) ≡ (cid:16) − , √ (cid:17) , f (0) ≡ (cid:16) , √ (cid:17) and g (0) ≡ (cid:16) − , √ (cid:17) . In particular, the point ξ = f (+ ∞ ) = g (+ ∞ ) is the point (0 ,
0) and the points ξ, f ( −∞ ) , g ( −∞ ) form an equilateral triangle with side length 1 . There are fourgeneric cases to examine depending on whether the point ξ ≡ (0 ,
0) is a ver-tex of the polygon ∂ Ω or not and whether the side containing ξ intersects the x − axis only at ξ or not. In Figure 2 the thick segments represent sides of ∂ Ωdemonstrating the four cases.Clearly, in Case IV the distance h ( f ( t ) , g ( t )) is constant because for all suffi-ciently large t = t ′ the lines containing f ( t ) , g ( t ) and f ( t ′ ) , g ( t ′ ) are parallel. Wewill deal in detail with Case I and the arguments will suffice for the remainingcases II and III.By Lemma 5 we have | f ( t ) − ξ | = | g ( t ) − ξ | = 1 e t + 1 ≡ E ( t ) . Let (0 , y ( t )) be the intersection point of the line containing f ( t ) , g ( t ) with the y − axis. As the triangle formed by ξ, g ( t ) , f ( t ) is equilateral we have y ( t ) = E ( t ) √
32 = √ / e t + 1 . − , √ (cid:17) = g ( −∞ ) f ( −∞ ) = (cid:16) , √ (cid:17)(cid:16) − , √ (cid:17) = g (0) f (0) = (cid:16) , √ (cid:17) w ( t ) z ( t ) y ( t ) f ( t ) g ( t )(0 , ≡ ξ Figure 3: The calculation of the Hilbert distance h ( f ( t ) , g ( t )) (Theorem 1a)9hen the unique side of ∂ Ω not containing ξ and intersecting the x − axis has theform z ( t ) = ( αy ( t ) + β, y ( t ))and the side containing ξ has the form w ( t ) = ( − α ′ y ( t ) , y ( t ))for α ∈ R , β > α ′ > √ . The latter holds because the angle formed by thesegments [ ξ, w ( t )] and [ ξ, (0 , y ( t ))] at ξ belongs to (cid:0) π , π (cid:1) . All the above notationis visualized in Figure 3.We next compute the Euclidean distances involved in the definition of thedistance h ( f ( t ) , g ( t )) : | z ( t ) − f ( t ) | = αy ( t ) + β − | g ( t ) − f ( t ) | == α √ / e t +1 + β −
12 1 e t +1 = (cid:16) α √ − (cid:17) E ( t ) + β | z ( t ) − g ( t ) | = (cid:16) α √ + (cid:17) E ( t ) + β | w ( t ) − g ( t ) | = α ′ y ( t ) − | g ( t ) − f ( t ) | == α ′ √ / e t +1 −
12 1 e t +1 = (cid:16) α ′ √ − (cid:17) E ( t ) | w ( t ) − f ( t ) | = (cid:16) α ′ √ + (cid:17) E ( t )It follows that h ( f ( t ) , g ( t )) = log (cid:16) α √ + (cid:17) E ( t ) + β (cid:16) α √ − (cid:17) E ( t ) + β + log (cid:18) A ′ E ( t ) (cid:19) where A ′ = α ′ √ + α ′ √ − . The second summand is convex for all t because α ′ > √ , hence, A ′ > . Thereexists, by Lemma 6,
T > t > T asrequired. This completes the proof of part 1a.
Proof of Theorem 1(b).
Let Ω be a convex domain in R with C boundary and f, g two asymptotic geodesic lines with common boundary point ξ = f (+ ∞ ) = g (+ ∞ ) ∈ ∂ Ω . We will need the following well known Lemma whose proof isincluded for the reader’s convenience.
Lemma 7
Let p be a projective transformation of R . sending ∂ Ω to a boundedcurve. If the curvature of ∂ Ω at the point ξ ∈ ∂ Ω is not zero then the same holdsfor the point p ( ξ ) ∈ p ( ∂ (Ω)) . roof of Lemma. Identify R with the plane Π = { ( x, y, z ) | z = 1 } in the realprojective space R P = R − { (0 , , } / ∼ whose points are rays emanating fromthe origin. The image of ∂ Ω under a projective transformation can be taken asthe composition of • an invertible linear transformation of R , and • the projection of A ( ∂ Ω) ⊂ A (Π) onto Π along the rays through the origin.Since an invertible linear transformation of R sends C curves to C curves andpreserves the non-vanishing curvature property, it suffices to check the desiredproperty for the projection A (Π) −→ Π . To see this, let E , E be two hyperplanes intersecting a C cone K throughthe origin and denote by σ i the simple closed convex C curve determined by theintersection E i ∩ K, i = 1 , . Moreover, as K is convex, the curvature of σ i at anypoint is ≥ . Let ℓ be a line through the origin contained in K intersecting σ i atthe point ξ i , i = 1 , E ′ the hyperplane containg ξ and parallel to E . Denoteby κ ( ξ i ) the curvature of σ i at ξ i and by κ ′ ( ξ ) the curvature of the curve E ′ ∩ K at ξ . Assume κ ( ξ ) = 0 and, clearly, κ ′ ( ξ ) = 0 . For, if κ ( ξ ) = 0 then, since theline ℓ contains ξ , both principal curvature at ξ would be 0 contradicting thefact that κ ′ ( ξ ) = 0 . Returning to the proof of Theorem 1(b), the points f ( −∞ ) , g ( −∞ ) , f (0) and g (0) form a non trivial quadrilateral. After composing by a projective transfor-mation we may assume that the coordinates of the four points mentioned aboveare: f ( −∞ ) ≡ √ , √ ! , g ( −∞ ) ≡ − √ , √ ! ,f (0) ≡ √ , √ ! and g (0) ≡ − √ , √ ! . In particular, the point ξ = f (+ ∞ ) = g (+ ∞ ) is the point (0 ,
0) and the points ξ, f ( −∞ ) , g ( −∞ ) form a right equilateral triangle. There are three cases toexamine depending on the intersection of ∂ Ω with the x − axis:Case 1: ∂ Ω ∩ { y = 0 } = { ξ } . Case 2: ∂ Ω ∩ { y = 0 } = (cid:8) ξ, ( λ, (cid:12)(cid:12) λ > (cid:9) . Case 3: ∂ Ω ∩ { y = 0 } = (cid:8) ξ, ( λ, (cid:12)(cid:12) λ < (cid:9) . In Figure 4, these cases are demonstrated with the additional consideration, in11ubcase 2bCase 1 Case 3Subcase 2a ξ ξ ξξg ( −∞ ) g ( −∞ ) g ( −∞ ) − λλ λg ( −∞ ) f ( −∞ ) f ( −∞ ) f ( −∞ ) f ( −∞ )Figure 4: Cases of intersection of ∂ Ω with the x − axis (Theorem 1b).12ase 2, of two subcases (2a and 2b defined below) depending on the tangent lineat the point ( λ, . The Euclidean line containing f ( t ) and g ( t ) intersects the y − axis at the point(0 , y ( t )) where, by Lemma 5, | f ( t ) − ξ | = e t +1 , hence, y ( t ) = √ / e t + 1 . (1)We also have that when t → + ∞ ye t = √ / e t + 1 e t −→ √ / , y ′ e t = − ( √ / e t ( e t + 1) e t −→ −√ / y ′′ e t = √ e t − e t ( e t + 1) e t −→ √ / f ( t ) and g ( t ) also intersects ∂ Ω at two points withcoordinates ( x ( t ) , y ( t )) , x ( t ) > x ( t ) , y ( t )) , x ( t ) < . For t large enough, y is a function of x, say, y ( t ) = K ( x ( t )) for some 1-1 and C function K and,hence, x is a function of y, namely, x ( t ) = K − ( y ( t )) . Similarly, for t largeenough, y ( t ) = K ( x ( t )) for some 1-1 and C function K and x is a function of y, x ( t ) = K − ( y ( t )) . The Hilbert distance of f ( t ) , g ( t ) is given by h ( f ( t ) , g ( t )) = log x ( t ) + y ( t ) x ( t ) − y ( t ) + log | x ( t ) | + y ( t ) | x ( t ) | − y ( t ) (3)and we denote by φ ( t ) and φ ( t ) the first and second summand respectively. Were-write the above mentioned Cases using the notation just introduced:Case 1: Both x ( t ) , x ( t ) −→ t → + ∞ . Case 2: x ( t ) → λ > x ( t ) −→ t → + ∞ . Case 3: x ( t ) → x ( t ) −→ λ ′ < t → + ∞ . It suffices to deal only with the convexity of φ ( t ) in all three cases: in Case 1 theproof for the convexity of φ is identical with that of φ and convexity of φ in Case2 (resp. Case 3) follows from convexity of φ in Case 3 (resp. Case 2).We will suppress the parameter t and we will be writing dydx instead of dKdx and dxdy instead of d ( K − ) dy . By the following calculation d xdy = ddy (cid:18) dxdy (cid:19) = ddy (cid:0) dydx (cid:1) ! = − (cid:0) dydx (cid:1) d ydx dxdy = − d ydx (cid:18) dydx (cid:19)
13e have the formula d xdy = − d ydx (cid:18) dxdy (cid:19) (4)First and second derivatives of φ ( t ) are as follows: φ ′ ( t ) = 2 (cid:16) yx (cid:17) ′ − (cid:0) yx (cid:1) and φ ′′ ( t ) = 2 (cid:16) yx (cid:17) ′′ − (cid:0) yx (cid:1) + 2 (cid:16) yx (cid:17) ′ yx (cid:0) yx (cid:1) ′ (cid:16) − (cid:0) yx (cid:1) (cid:17) . As the slope of the geodesic line f is 1 , it suffices to show that for t large enough (cid:0) yx (cid:1) ′′ > . We have the following calculations (cid:16) yx (cid:17) ′ = y ′ x − yx ′ x = y ′ x − dxdy y ′ yx = x − dxdy yx y ′ (cid:16) yx (cid:17) ′′ x dydx = (cid:20) − dydx d xdy y − yx dxdy (cid:21) ( y ′ ) + (cid:18) x dydx − y (cid:19) y ′′ (5a) (cid:16) yx (cid:17) ′′ x dydx by ( ) = "(cid:18) dydx (cid:19) − d ydx y − yx dxdy ( y ′ ) + (cid:18) x dydx − y (cid:19) y ′′ (5b)Case 1: In this case the x − axis is the tangent line to ∂ Ω at ξ = (0 ,
0) and usingLemma 7 and our curvature hypothesis we have that d ydx (cid:12)(cid:12)(cid:12)(cid:12) x =0 = 0 and dydx (cid:12)(cid:12)(cid:12)(cid:12) x =0 = 0 . Multiplying both sides of equation (5b) by e t we have (cid:16) yx (cid:17) ′′ x dydx e t = " (cid:18) dydx (cid:19) − d ydx y | {z } term1 − yx dxdy | {z } term2 ( y ′ ) e t + (cid:18) x dydx e t | {z } term3 − ye t (cid:19) y ′′ e t . (6)We will show that term1 and term2 both converge to 1 / t → + ∞ andterm3 converges to 2 . Then using (2) the right hand side of (6) converges to (cid:0) − (cid:1) (cid:0) −√ / (cid:1) + (cid:0) √ − √ / (cid:1) (cid:0) √ / (cid:1) = > . This shows that (cid:0) yx (cid:1) ′′ > φ ( t ) is convex for large enough t. In the followingcalculations limits are always taken as t → + ∞ or, equivalently, x, y → ∼ between two functions indicates that the their limits as t → + ∞ are equal. term1 = d ydx y (cid:0) dydx (cid:1) ∼ d ydx y ′ dydx (cid:0) dydx (cid:1) ′ = d ydx y ′ dydx d ydx dxdy y ′ −→ . yx dxdy ∼ (cid:0) yx (cid:1) ′ (cid:0) dydx (cid:1) ′ = (cid:16) x − y dxdy (cid:17) y ′ x − d ydx dxdy y ′ = (cid:18) d ydx (cid:19) − x dydx − yx ∼ (cid:18) d ydx (cid:19) − (cid:0) x dydx − y (cid:1) ′ ( x ) ′ = (cid:18) d ydx (cid:19) − dxdy y ′ dydx + x d ydx dxdy y ′ − y ′ x dxdy y ′ = (cid:18) d ydx (cid:19) − d ydx = 12 . (7)For the calculation of term3 first observe that d ydx x dxdy ∼ d ydx x ′ (cid:0) dydx (cid:1) ′ = d ydx dxdy y ′ d ydx dxdy y ′ = 1 . (8)Then we haveterm3 = x dxdy e t ∼ (cid:16) x dxdy (cid:17) ′ ( e − t ) ′ = dxdy y ′ dydx + x d ydx dxdy y ′ − e − t = − y ′ e t (cid:18) x d ydx dxdy (cid:19) by ( ) , ( ) −→ − (cid:16) −√ / (cid:17) (1 + 1) = √ . This completes the proof for the convexity of φ in Case 1.Case 2: In this case the x ( t ) → λ > t → + ∞ and we have two sub-casesdepending on whether dxdy (cid:12)(cid:12)(cid:12)(cid:12) y =0 = 0 equivalently dydx (cid:12)(cid:12)(cid:12)(cid:12) x = λ = ∞ (Subcase 2a)or dxdy (cid:12)(cid:12)(cid:12)(cid:12) y =0 = 0 = dydx (cid:12)(cid:12)(cid:12)(cid:12) x = λ (Subcase 2b)For Subcase 2a, multiply both sides of equation (5b) by e t dxdy to get (cid:16) yx (cid:17) ′′ x e t = "(cid:18) dydx (cid:19) − d ydx y − dxdy + 2 yx (cid:18) dxdy (cid:19) ( y ′ ) e t + (cid:18) x − y dxdy (cid:19) y ′′ e t . (9)By (2), y ′′ e t → √ / (cid:16) x − y dxdy (cid:17) → λ. Moreover, the quantity inside the square bracket is easily seen to be boundedand, since by (2) ( y ′ ) e t → , it follows that (cid:16) yx (cid:17) ′′ x e t −→ λ √ ( −∞ ) g ( −∞ ) f ( t ) g ( t ) − − ≡ g ( ∞ ) 1 ≡ f ( ∞ ) 2 3Figure 5: The thick segments are equal of Euclidean length E ( t ) = e t +1 . hence, (cid:0) yx (cid:1) ′′ is positive for large enough t. For Subcase 2b, observe that dydx (cid:12)(cid:12) x = λ may be negative. For this reason, we multi-ply both sides of equation (5b) by e t to get (cid:16) yx (cid:17) ′′ x dydx e t = "(cid:18) dydx (cid:19) − d ydx y − yx dxdy ( y ′ ) e t + (cid:18) x dydx − y (cid:19) y ′′ e t . (10)In a similar manner as in the previous subcase we obtain (cid:16) yx (cid:17) ′′ x dydx e t −→ λ dydx √ . This completes the proof of the convexity of φ in Case 2.Case 3: In this case x ( t ) → t → + ∞ and dydx (cid:12)(cid:12) x =0 ∈ (0 ,
1) because the slopeof the geodesic line f is 1 . We have the following preliminary calculations x dydx − ye − t ∼ x d ydx dxdy y ′ − e − t = − d ydx dxdy (cid:0) e t x (cid:1) (cid:0) e t y ′ (cid:1) ∼ − d ydx dxdy dxdy √ ! − √ ! = 14 d ydx (cid:18) dxdy (cid:19) > e t x ∼ dxdy √ and e t y ′ → − √ . In a similar manner we obtain x − dxdy yx ∼ − (cid:18) dydx (cid:19) d xdy (12)16e now multiply both sides of equation (5a) by e t to get (cid:16) yx (cid:17) ′′ x dydx e t = " − dydx d xdy ye t + 2 − yx dxdy e − t ( y ′ ) e t + x dydx − ye − t y ′′ e t . (13)By (11) and the fact ( y ′ ) e t → (cid:0) √ / (cid:1) , it suffices to show that the term in thesquare bracket converges to 0 as t → ∞ . For the first summand inside the squarebracket we have − dydx d xdy ye t → − dydx (cid:12)(cid:12)(cid:12)(cid:12) d xdy √
22 (14)For the second summand inside the square bracket we have2 − yx dxdy e − t ∼ (cid:0) yx (cid:1) ′ dxdy + yx d xdy y ′ − e − t = − y ′ e t " yx d xdy + x − dxdy yx dxdy by ( ) ∼ − y ′ e t " yx d xdy − (cid:18) dydx (cid:19) d xdy dxdy → √ (cid:20) dydx (cid:12)(cid:12)(cid:12)(cid:12) d xdy (cid:21) (15)By (14) and (15) the term in the square bracket on the right hand side of (13)converges to 0 as required. This completes the proof in Case 3 and the proof ofTheorem 1(b). Remark 8
The convexity result posited in Theorem 1 also holds for boundedconvex domains Ω with piece wise C boundary which consists of either segmentsor, C curves with nno-vanishing curvature. This follows by combining parts (a)and (b) of Theorem 1 and the fact that the distance funtion studied in the aboveproof has two summands each of which was treated separetely and shown to beconvex. We will construct an example which demonstrates the necessity of the curvaturecondition in Theorem 1b. Apart from asymptotic geodesics, there are two morecases: intersecting and disjoint geodesics. In the case Ω is a convex polytope,we provide below examples showing that convexity does not hold neither forintersecting nor for disjoint geodesics.
Example 1 (disjoint geodesics): let Ω be the interior of the trapezoid withvertices (2 , , (3 , , ( − ,
1) and ( − , . Let f be the geodesic whose image is17 ( −∞ ) g ( −∞ ) g ( t ) f ( t ) − − ≡ f ( ∞ ) 1 ≡ g ( ∞ ) 2 3Figure 6: The thick segments are equal of Euclidean length √ | f ( t ) − f (+ ∞ ) | = √ √ e t +1 = E ( t ) . the intersection of Ω with the line x = 1 and f (0) = (cid:0) , (cid:1) , having arc lengthparametrization. Similarly, let g be the geodesic with image the intersection of Ωwith the line x = 0 and g (0) = (cid:0) , (cid:1) , see Figure 5. By Lemma 5, the Euclideandistance | g ( t ) − g (+ ∞ ) | is e t +1 ≡ E ( t ) , in other words, the horizontal lines y = e t +1 , t ∈ (0 , ∞ ) intersect the images of the geodesics f and g at the points f ( t ) and g ( t ) respectively. The distance function is given by h ( f ( t ) , g ( t )) = log 2 + E ( t )1 + E ( t ) + log 2 + E ( t )1 + E ( t ) . An elementary calculation shows that h ′′ = 2 Φ[2 + E ( t )] [1 + E ( t )] where the dominant summand of Φ is − E ′′ ( t ) = − e t − e t ( e t + 1) . Thus, for large enough t, the function h ( f ( t ) , g ( t )) is not convex. Example 2 (intersecting geodesics): let Ω be as above. Let f, g be thegeodesics with Im f = Ω ∩ { y = x } and Im g = Ω ∩ { y = − x + 1 } respectivelyand f (0) = g (0) = (cid:0) , (cid:1) , see Figure 6. By the same procedure as in the previous18xample we obtain h ( f ( t ) , g ( t )) = log 21 + 2 E ( t ) + log 21 + 2 E ( t ) . An analogous elementary calculations shows that h ′′ = 2 Φ[1 + 2 E ( t )] where the dominant summand of Φ is − E ′′ ( t ) e t − e t ( e t +1) . Example 3:
We will construct a (convex) C curve whose curvature is zeroat exactly one point, positive at every other point and two geodesics asymptoticat the point of zero curvature such that their distance function is not convex.Consider the convex domain Ω bounded below by the function y = (cid:12)(cid:12) x (cid:12)(cid:12) , x ∈ [ − , . Let f (resp. g ) be the geodesic line whose image is the intersection of Ω with theline y = x (resp. y = − x ). We may assume that Ω is a bounded convex domaincontaining the above mentioned geodesics. Although Theorem 1b does not applybecause ∂ Ω has curvature 0 at the point (0 , , it can be shown that the distancefunction D ( t ) := d ( f ( t ) , g ( t )) = 2 log x ( t ) + y ( t ) x ( t ) − y ( t )is in fact convex for sufficiently large t. We will alter Ω by replacing a subarc of itsboundary by a segment so that the distance function will no longer be convex atthe corresponding time interval. Then we will repeat the same process infinitelymany times to ensure that convexity does not hold for large t and we will takeappropriate care for the C property.By symmetry, we will restrict our attention to x ( t ) > . Lemma 9
Let y ( t ) = ce t + 1 , c > and y ( t ) = Ax ( t ) + B with A ∈ (0 , and B < . Then the first and second derivative of D ( t ) = 2 log x ( t ) + y ( t ) x ( t ) − y ( t ) are asfollows: D ′ ( t ) = 4 Ax − yx − y A (cid:18) y c − y (cid:19) D ′′ ( t ) C = 2 ( Ay − x ) (cid:18) − y c + y (cid:19) + (cid:0) x − y (cid:1) (cid:18) − yAc + A (cid:19) ( t ) y ( t ) x ( t ) x ax x b x y = x y = (2 x ) x − x y = h (cid:0) x (cid:1) i x − (cid:0) x (cid:1) I I Figure 7: The segments I n for n = 0 , . here C is the following positive real number A ( x − y ) − B (cid:16) − y c + y (cid:17) . For x ∈ (0 ,
1) consider the line determined by the points ( x , x ) and (cid:0) x , (cid:1) whose equation is y = (2 x ) x − x . This line determines a segment I withendpoints ( x , x ) and ( b, b ) for some b < x which can be computed explicitly(see Figure 7). Using the previous Lemma it is easy to see that for t such that x ( t ) = x we have A = 2 x , y ( t ) = x and, thus, D ′′ ( t ) C = − c x + 2 x which is negative for sufficiently small x . Clearly, if the subarc of ∂ Ω determinedby the points ( b, b ) and ( x , x ) is replaced by the segment I then D ( t ) will notbe convex near t . The same non-convexity property can be obtained by replacingthe above mentioned subarc of ∂ Ω by a C arc σ : [ b, x ] −→ (cid:2) b , x (cid:3) of constant and sufficiently small curvature.Using x n = x n , n ∈ N as starting point we obtain the corresponding intervals I n with endpoints on ∂ Ω and we perform the same replacement for all n ∈ N using C arcs σ n of constant and sufficiently small curvature. Moreover, we mayarrange so that the curvature of each σ n → n → ∞ . This guarantees that the distance function D ( t ) with respect to the new (altered)convex domain, denoted again by Ω , cannot be convex for t large enough.The endpoint ( b, b ) of the interval I can be computed explicitly but we will onlyneed the fact that b ∈ (cid:18) x , x (cid:19) . (16)Denote by a the point x = x . Our Final Step is to replace the subarc of ∂ Ω with endpoints ( a, a ) and ( b, b )by a C curve σ , : [ a, b ] −→ (cid:2) a , b (cid:3) so that the first and second derivatives of σ , matches those of σ and σ at theappropriate points. 21 emma 10 Let [ α, β ] be an interval. Let α (0) , α (1) , α (2) and β (0) , β (1) , β (2) bepositive real numbers satisfying < α (0) < β (0) , < α (1) < β (1) and α (1) ( β − α ) < β (0) − α (0) < β (1) ( β − α ) (17) Then there exists a C function σ : [ α, β ] −→ (cid:2) α (0) , β (0) (cid:3) satisfying • σ ( α ) = α (0) , σ ′ ( α ) = α (1) , σ ′′ ( α ) = α (2) • σ ( β ) = β (0) , σ ′ ( β ) = β (1) , σ ′′ ( β ) = β (2) • σ ′′ ( x ) > for all x ∈ ( α, β ) . Proof.
First, we may find a strictly increasing differentiable function σ (1) : [ α, β ] −→ (cid:2) α (1) , β (1) (cid:3) satisfying lim t → α ddt σ (1) ( t ) = α (2) and lim t → β ddt σ (1) ( t ) = β (2) Set σ ( t ) = α (0) + R tα σ (1) ( s ) ds and we need the following equality to hold Z βα σ (1) ( t ) dt = σ ( β ) − σ ( α ) = β (0) − α (0) . (18)As α (1) ( β − α ) < Z βα σ (1) ( t ) dt < β (1) ( β − α )equation (18) can be achieved provided that (17) holds.In order to use the above Lemma to find σ , we need to check that theboundary values of σ and σ which correspond to the segments I and I satisfythe assumptions of the above Lemma.The slope of I (resp. I ) is 2 x (resp. 2( x / ) so the first two inequalitiesrequired by Lemma 10 clearly hold:0 < (cid:16) x (cid:17) < b and 0 < (cid:16) x (cid:17) < x . For condition (17) of Lemma 10 we need to check that2 (cid:16) x (cid:17) (cid:16) b − x (cid:17) < b − (cid:16) x (cid:17) < x (cid:16) b − x (cid:17) . (cid:18) x (cid:19) − (cid:16) x (cid:17) < x (cid:16) x − x (cid:17) which is equivalent to x < x which holds. For the left hand side inequalityand using again (16) it suffices to check that2 (cid:16) x (cid:17) (cid:18) x − x (cid:19) < (cid:16) x (cid:17) − (cid:16) x (cid:17) which is equivalent to x < x which holds.As we have the liberty to choose the arcs σ n arbitrarily close to the segments I n , it follows that the inequalities required in Lemma 10 hold for any pair ofsegments I n , I n +1 . Therefor, the above procedure can be followed in an identicalway to construct the curves σ n,n +1 joining the curves σ n σ n +1 for all n. The finalcurve obtained in this way is clearly a C curve with zero curvature only at thepoint (0 ,
0) and, by construction, the distance function between the geodesics f, g which are asymptotic at (0 ,
0) is not convex.
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