Santalo's formula and stability of trapping sets of positive measure

0.18 MB·

by  Luchezar Stoyanov

Santalo’s formula and stability of trapping sets
of positive measure
Luchezar Stoyanov1
School of Mathematics and Statistics, University of Western Australia, Crawley 6009 WA, Australia
6 Abstract
1
Billiard trajectories (broken generalised geodesics) are considered in the exterior of an obstacle K
0
2 with smooth boundary on an arbitrary Riemannian manifold. We prove a generalisation of the well-
knownSantalo’sformula. Asaconsequence,itisestablishedthatifthe setoftrappedpointshaspositive
t
c measure, then for all sufficiently small smooth perturbations of the boundary of K the set of trapped
O ′
pointsforthenewobstacleK obtainedinthiswayalsohaspositivemeasure. Moregenerallythemeasure
3 ofthesetoftrappedpointsdependscontinuouslyonperturbationsoftheobstacleK. Someconsequences
1 are derived in the case of scattering by an obstacle K in IRn. For example, it is shown that, for a large
classofobstaclesK,thevolumeofK isuniquelydeterminedbytheaveragetravellingtimesofscattering

] rays in the exterior of K.
S
D MSC: 37D20, 37D40, 53D25, 58J50
.
h Keywords: geodesic, billiard trajectory, Santalo’s formula, trapped point, Liouville measure, travelling
t
a time, scattering by obstacles
m
[
1 Introduction

2
v
8 Let g be the geodesic flow on the unit sphere bundle SM of a Ck (k ≥ 2) Riemannian
t
2
8 manifold M without boundary, and let n = dim(M) = n ≥ 2. Consider an arbitrary
3 compact subset M of M with non-empty smooth (Ck)fboundary ∂M and non-empty
0
. interior Mf\∂M. f
1
0 Next, suppose that Kfis a compact subset of the interior M \∂M of M with smooth
6 boundary ∂K and non-empty interior K \∂K. Consider the compact subset
1
:
v Ω = M \K
i
X
r of M. Then the boundary of Ω in M is ∂Ω = ∂M ∪ ∂K. Define the billiard flow φ in
a t
S(Ω) in the usual way: it coincides with the geodesic flow g in the interior of S(Ω), and
t
when a geodesic hits the boundary f∂Ω it reflect following the law of geometrical optics.
Let dq be the measure on M determined by the Riemannian metric and let dv be the
Lebesgue measure on the unit sphere S M. It is well-known (see Sect. 2.4 in [CFS]) that
q
the geodesic flow g preservefs the Liouville measure dqdv on SM, and so the billiard flow
t
φ preserves the restriction λ of dqdv to Sf(Ω). Finally, denote by µ the Liouville measure
t
on S+(∂Ω) defined by f
dµ = dρ(q)dω |hν(q),vi|,
q
where ρ is the measure on ∂Ω determined by the Riemannian metric on ∂Ω and ω is the
q
Lebesgue measure on the (n − 1)-dimensional unit sphere S (M) (see e.g. Sect. 6.1 in
q
[CFS]).
1E-mail: [email protected]
1
For any q ∈ ∂Ω, denote by ν(q) ∈ SM the unit normal vector to ∂Ω pointing into the
interior of Ω. Set
f
S+(∂Ω) = {x = (q,v) : q ∈ ∂Ω,hv,ν(q)i ≥ 0},
and in a similar way define S+(∂M) ⊂ S+(∂Ω). For any x = S+(∂Ω) denote by τ(x) ≥ 0
the maximal number (or ∞) such that φ (x) = g (x) is in the interior of Ω for all 0 < t <
t t
τ(x). Set τ(x) = 0 if hν(q),vi = 0. The function τ thus defined is called the first return
time function for the flow φ on Ω.
t
In the present paper we will assume that the geodesic flow g has no trapped trajec-
t
tories in M, i.e.
(A): For every x = (q,v) ∈ S+(∂M) with hv,ν(q)i > 0 ∃t > 0 with pr (g (x)) ∈ ∂M,
1 t
where we set pr (q,v) = q. In particular, it follows from this assumption that τ(x) < ∞
1
for every x ∈ S+(∂Ω).
Then Santalo’s formula (see e.g. pp. 336-338 in [Sa] or [Cha]) gives
τ(x)
f(x) dλ(x) = f(g (x)) dt dµ(x) (1.1)
t
ZS(Ω) ZS+(∂Ω) Z0 !
for every λ-integrable function f on S(Ω).
Santalo’s formula has been widely used in geometry and dynamical systems (including
billiards) – see e.g. [CULV], [Ch], [CaG], [C] and the references there.
By assumption (A), the geodesic flow g has no trapped trajectories in M. However,
t
the billiard flow φ may have trapped trajectories in M with respect to the obstacle K.
t
More precisely, we will now consider the billiard trajectories in Ω as scattering trajectories
in M reflecting at the boundary ∂K when they hit K. Given x = S(Ω), let t(x) ≥ 0 be
the maximal number (or ∞) such that φ (x) is in the interior of M for all 0 < t < t(x).
t
For x = (q,v) ∈ S+(∂Ω) with hν(q),vi = 0 set t(x) = 0. Then t(x) is called the travelling
time function on Ω. Given x ∈ S+(∂M), consider the billiard trajectory
γ+(x) = {φ (x) : 0 ≤ t ≤ t(x)}
t
which starts at a point x ∈ S+(∂M) and, if t(x) < ∞, it ends at another point y ∈
S+(∂M). Between x and y the trajectory may hit ∂K several times.
Problems related to recovery of information about the manifold M from travelling
times τ(x) of geodesics in the interior of M have been considered in Riemannian geometry
for a rather long time – see [SUV], [SU], [CULV] and the references there for some general
information. Similar problems have been studied when an obstacle K is present; then
one deals with travelling times t(x) of billiard trajectories (generalised geodesics) in the
exterior of K – see [NS1], [NS2], [St3] and the references there (see also [LP] and [M] for
some general information on inverse scattering problems). For some classes of obstacles
K the travelling time function t(x) = t (x) completely determines K ([NS1], [NS2]). For
K
example, it was recently established ([NS1]) that this is so in the class of obstacles K in
Euclidean spaces that are finite disjoint unions of strictly convex domains with smooth
boundaries.
2
Let Trap+(∂M) be the set of the trapped points x = (q,v) ∈ S+(∂M) of the billiard
Ω
flow φ in Ω, i.e. the points for which t(x) = ∞. In general, it may happen that
t
Trap+(∂M) 6= ∅, however it follows from Theorem 1.6.2 in [LP] (see also Proposition 2.4
Ω
in [St1]) that
µ(Trap+(∂M)) = 0. (1.2)
Ω
Next, let Trap(Ω) be the set of all trapped points x = (q,v) ∈ S(Ω) of the billiard flow
φ in Ω, i.e. the points for which t(q,v) = ∞ or t(q,−v) = ∞. As Livschits’ example
t
shows (see Figure 1), in general Trap(Ω) may have positive Liouville measure and
even a non-empty interior.
Generalising Santalo’s formula (1.1), we prove the following.
Theorem 1.1. Assume that M and the geodesic flow g in M satisfy the condition (A).
t
Let f : S(Ω)\Trap(Ω) −→ C be a λ-measurable function. Assume that either:
(i) Trap(Ω) = ∅ and f is integrable on S(Ω), or
(ii) |f| is integrable.
Then we have
t(x)
f(x) dλ(x) = f(φ (x)) dt dµ(x). (1.3)
t
ZS(Ω)\Trap(Ω) ZS+(∂M)\Trap+Ω(∂M) Z0 !
Clearly, in the case (ii) above Trap(Ω) is allowed to have positive measure.
We will now describe the main consequence of Theorem 1.1 derived in this paper.
Let k ≥ 3 and let Ck(∂K,M) be the space of all smooth embedding F : ∂K −→ M
endowed with the Whitney Ck topology (see [Hi]). Given F ∈ Ck(∂K,M), let K be the
F
obstacle inM withboundary ∂K = F(∂K) so that K ∩∂M = ∅, andlet Ω = M \K .
F F F F
Our second main result in this paper is the following.
Theorem 1.2. Assume that M and the geodesic flow g in M satisfy the condition (A).
t
(a) If λ(Trap(Ω)) > 0, then there exists an open neighbourhood U of id in Ck(∂K,M)
such that for every F ∈ U we have λ(Trap(Ω )) > 0.
F
(b) More generally, for every ǫ > 0 there exists an open neighbourhood U of id in
Ck(∂K,M) such that for every F ∈ U we have |λ(Trap(Ω ))−λ(Trap(Ω))| < ǫ.
F
We prove Theorem 1.1 in Sect. 2 below, and then use it in Sect.3 to derive Theorem
1.2. Let us note that in the case Trap(Ω) = ∅ the formula (1.3) with f = 1 was mentioned
without proof and used by Plakhov and Roshchina in [PlaR] (see also [Pla]).
Formula (1.3) with f = 1 implies the following.
Corollary 1.3. Under the assumptions of Theorem 1.1 we have
λ(Trap(Ω)) = λ(S(Ω))− t(x)dµ(x).
ZS+(∂M)
So, if we know the travelling time function t(x) and have enough information about Ω
to determine its volume, then we can determine the measure of the set of trapped points
in Ω, as well.
3
Some other consequences of Theorem 1.1 concerning scattering by obstacles are dis-
cussed in Sects. 4 and 5 below.
E
P Q
F F
1 2
U V
Figure 1: Livshits’ Example (Ch. 5 in [M])
E is a half-ellipse with end points P and Q and foci F1 and F2. Any trajectory entering the interior of
the ellipse between the foci must exit between the foci after reflection. So, no trajectory ‘coming from
infinity’ has a common point with the parts U and V of the boundary and the nearby regions.
Similar examples in higher dimensions are described in [NS3].
2 A generalised Santalo’s formula
Given x ∈ S(Ω), we will say that γ+(x) contains a gliding segment on the boundary ∂Ω
if there exist 0 ≤ t < t ≤ t(x) such that φ (x) = g (x) ∈ S(∂Ω) for all t ∈ [t ,t ] (i.e.
1 2 t t 1 2
γ+(x) contains a non-trivial geodesic segment lying entirely in ∂Ω).
It follows from results2 in [MS2] that for λ-almost all x ∈ S(Ω), the billiard trajectory
γ+(x) does not contain any gliding segments on the boundary ∂Ω, and γ+(x) has only
finitely many reflections.
First, we prove a special case of Theorem 1.1.
Lemma 2.1. Assume that M and the geodesic flow g in M satisfy the condition (A).
t
Let V be an open subset of S(Ω) containing Trap(Ω) and such that φ (x) ∈ V for any
t
x ∈ V and any t ∈ [0,t(x)]. Assume that f : S(Ω) −→ C is an integrable function such
that f = 0 on V. Then (1.3) holds.
2See Sect. 3 in [MS2]; see also [MS1], [H] or [PS] for general information about generalisedgeodesics.
4
Proof of Lemma 2.1. Let V and f satisfy the assumptions of the lemma. Then we
have
f(x) dλ(x) = f(x) dλ(x) (2.1)
ZS(Ω) ZS(Ω)\V
and
t(x) t(x)
f(φ (x)) dt dµ(x) = f(φ (x)) dt dµ(x). (2.2)
t t
ZS+(∂M) Z0 ! ZS+(∂M)\V Z0 !
We will prove that the right-hand-sides of (2.1) and (2.2) are equal.
Given an integer k ≥ 0 denote by Γ the set of those x ∈ S+(∂M)\V such that γ+(x)
k
contains no gliding segments on the boundary ∂Ω, and γ+(x) has exactly k reflections at
∂K. Clearly Γ are disjoint, measurable subsets of S+(∂M) and, as remarked earlier, it
k
follows from [MS2] that
µ S+(∂M)\(V ∪∪∞ Γ ) = 0. (2.3)
k=0 k
The billiard ball map B is(cid:0)defined in the usual way(cid:1): given x = (q,v) ∈ S+(∂Ω) with
g (x) ∈ S(∂Ω) for some t ∈ (0,∞), take the smallest t > 0 with this property, and let
t
g (x) = (p,w) for some p ∈ ∂Ω, w ∈ Sn−1. Set B(x) = (p,σ (w)) ∈ S+(∂Ω), where
t p
σ : T (M) −→ T (M)
p p p
is the symmetry through the tangent plane T (∂Ω), i.e.
f p f
σ (ξ) = ξ −2hν(p),ξi ν(p).
p
It follows from the condition (A) that B is well-defined on a set Λ of full µ-measure in
S+(∂Ω) and the Liouville measure µ is invariant with respect to B (see e.g. Lemma 6.6.1
in [CFS]).
Notice that each of the sets Γ is contained in Λ and moreover
k
Γ = Bj(Γ ) ⊂ Λ
k,j k
for all j = 0,1,...,k. Also, by the definition of the sets Γ we have
k
Bj(Γ ) ⊂ S+(∂K) , 1 ≤ j ≤ k. (2.4)
k
The sets Γ are clearly measurable, and moreover
k,j
Γ ∩Γ = ∅ whenever k 6= m or j 6= i. (2.5)
k,j m,i
Indeed, assumethaty ∈ Γ ∩Γ forsomenon-negativenumbersk,j,m,iwith0 ≤ j ≤ k
k,j m,i
and 0 ≤ i ≤ m. Then y = Bj(x) for some x ∈ Γ and y = Bi(z) for some z ∈ Γ . Assume
k m
e.g. j > i. Then Bi(z) = y = Bj(x) = Bi(Bj−i(x)) implies z = Bj−i(x). Now (2.4) gives
z ∈ S+(∂K) which is a contradiction with z ∈ Γ ⊂ S+(∂M). Thus, we must have j = i
m
and therefore Bi(z) = y = Bi(x), so z = x ∈ Γ ∩Γ . The latter is non-empty only when
k m
k = m. This proves (2.5).
5
Finally, notice that
µ S+(∂Ω)\(V ∪∪∞ ∪k Γ ) = 0. (2.6)
k=0 j=0 k,j
(cid:0) (cid:1)
Indeed, as mentioned earlier, it follows from results in [MS2] that for λ-almost all x ∈
S(Ω), the billiardtrajectory γ+(x) does not containany gliding segments ontheboundary
∂Ω, and γ+(x) has only finitely many reflections. Since Trap(Ω) ⊂ V, for almost all
y ∈ S+(∂K) \V there exist t ≥ 0 and x ∈ S+(∂M) such that y = φ (x), t(x) < ∞, the
t
billiardtrajectoryγ+(x)doesnotcontainanyglidingsegments on∂Ω andhasonlyfinitely
many reflections. Let k be the number of those reflections; then for some j = 0,1,...,k
we have y = Bj(x). Thus, x ∈ Γ and y ∈ Γ . This proves (2.6).
k k,j
Next, given k > 0 and x ∈ Γ , clearly we have
k
t(x) = τ(x)+τ(B(x))+τ(B2(x))+...+τ(Bk(x)).
Set
T (x) = τ(x)+τ(B(x))+...+τ(Bj(x))
j
for j = 0,1,...,k. Clearly φ (x) = Bj+1(x) for 0 ≤ j ≤ k − 1. Hence for any
Tj(x)
j = 1,...,k we have
φ (x) = φ (φ (x)) = g (Bj(x)),
s+Tj−1(x) s Tj−1(x) s
and therefore, using the substitution t = s+T (x) below we get
j−1
Tj(x) τ(Bj(x))
f(φ (x)) dt = f(φ (Bj(x))) ds.
t s
ZTj−1(x) Z0
Thus,
t(x) τ(x) k Tj(x)
f(φ (x)) dt = f(φ (x)) dt+ f(φ (x)) dt
t t t
Z0 Z0 j=1 ZTj−1(x)
X
τ(x) k τ(Bj(x))
= f(φ (x)) dt+ f(φ (Bj(x))) ds
t s
Z0 j=1 Z0
X
k τ(Bj(x))
= f(φ (Bj(x))) dt.
t
j=0 Z0
X
Now, using the above and the fact that Bj : Γ −→ Bj(Γ ) is a measure preserving
k,j k,j
bijection, we get
τ(Bj(x)) τ(y)
f(φ (Bj(x))) dt dµ(x) = f(φ (y)) dt dµ(y),
t t
ZΓk Z0 ! ZΓk,j Z0 !
6
which implies
t(x) k τ(Bj(x))
f(φ (x)) dt dµ(x) = f(φ (Bj(x))) dt dµ(x)
t t
ZΓk Z0 ! ZΓk j=0 Z0 !
X
k τ(Bj(x))
= f(φ (Bj(x))) dt dµ(x)
t
j=0 ZΓk Z0 !
X
k τ(y)
= f(φ (y)) dt dµ(y).
t
j=0 ZΓk,j Z0 !
X
Using (2.2), (2.3), (2.4) and (2.6), we derive
t(x) t(x)
f(φ (x)) dt dµ(x) = f(φ (x)) dt dµ(x)
t t
ZS+(∂M) Z0 ! ZS+(∂M)\V Z0 !
∞ t(x)
= f(φ (x)) dt dµ(x)
t
k=0ZΓk Z0 !
X
∞ k τ(y)
= f(φ (y)) dt dµ(y)
t
k=0 j=0 ZΓk,j Z0 !
XX
τ(y)
= f(φ (y)) dt dµ(y)
t
ZS+(∂Ω)\V Z0 !
τ(y)
= f(g (y)) dt dµ(y).
t
ZS+(∂Ω) Z0 !
By Santalo’s formula (1.1), the latter is equal to f(x) dλ(x).
ZS(Ω)
Proof of Theorem 1.1. If Trap(Ω) = ∅, taking V = ∅ in Lemma 2.1, proves the case
(i).
To prove the case (ii), assume that Trap(Ω) 6= ∅. Let f be a measurable function of
S(Ω)\Trap(Ω) such that |f| is integrable. Extend3 f as 0 on Trap(Ω). Since Trap+(∂M)
Ω
is closed in S+(∂M), there exists a sequence of compact sets
F ⊂ F ⊂ ... ⊂ F ⊂ ... ⊂ S+(∂M)\Trap+(∂M)
1 2 m Ω
such that
∪∞ F = S+(∂M)\Trap+(∂M). (2.7)
m=1 m Ω
Choose such a sequence, and for every m set
G = {φ (x) : x ∈ F , 0 ≤ t ≤ t(x)}.
m t m
3This is already assumed in the right-hand-side of (1.3).
7
Then we have
∪∞ G = S(Ω)\Trap(Ω). (2.8)
m=1 m
Indeed, if x ∈ G for some m, then x = φ (y) for some y ∈ F ⊂ S+(∂M)\Trap+(∂M)
m t m Ω
and some t ∈ [0,t(x)]. Then y ∈/ Trap+(∂M) gives t(y) < ∞ and so t(x) < ∞, too.
Ω
Also, if x = (q,v), then t(q,−v) = t < ∞. Thus, x ∈/ Trap(Ω). This proves that
G ⊂ S(Ω) \ Trap(Ω) for all m. Next, let x = (q,v) ∈ S(Ω) \ Trap(Ω). Then t =
m
t(q,−v) < ∞, so x = φ (y) for some y ∈ S+(∂M). Moreover, t(y) = t(x) + t < ∞, so
t
y ∈ S+(∂M) \ Trap+(∂M), and by (2.7) we have y ∈ F for some m ≥ 1. Moreover
Ω m
0 ≤ t ≤ t(y). Thus, x ∈ G . This proves (2.8).
m
Notice that all sets G are closed in S(Ω). Indeed, given m, a standard argument (see
m
e.g. the proof of Lemma 3.1 below) shows that the function t(x) is uniformly bounded
on the compact set F . Let {x }∞ be a sequence in G with x → x as p → ∞. Then
m p p=1 m p
x = φ (y ) for some t ∈ [0,T] and some y ∈ F , where T = sup t(y). Taking an
p tp p p p m y∈Fm
appropriate sub-sequence, we may assume y → y and t → t as p → ∞. Since F is
p p m
compact, y ∈ F . Now a simple continuity argument gives x = φ (y), so x ∈ G .
m t m
Thus, G is closed in S(Ω) and so V = S(Ω)\G is open in S(Ω). Moreover, it is
m m m
clear that φ (x) ∈ V for all x ∈ V and all t ∈ [0,t(x)]. Let χ be the characteristic
t m m Gm
function of G in S(Ω). Applying Lemma 2.1 to V = V and f replaced by f = f·χ ,
m m m Gm
we get
t(x)
f (x) dλ(x) = f (φ (x)) dt dµ(x).
m m t
ZS(Ω)\Trap(Ω) ZS+(∂M) Z0 !
Extending f as 0 on V , we get a sequence of measurable functions on S(Ω) with
m m
|f | ≤ |f| for all m and f (x) → f(x) for all x ∈ S(Ω). Since |f| is integrable, Lebesgue’s
m m
Theorem now implies
t(x)
f(x) dλ(x) = f(φ (x)) dt dµ(x),
t
ZS(Ω)\Trap(Ω) ZS+(∂M) Z0 !
which proves the theorem.
3 Proof of Theorem 1.2
It is enough to prove the more general part (b).
Let M and Ω satisfy the assumptions of Theorem 1.2, so (A) holds. Let ǫ > 0. Assume
that there is no open neighbourhood U of id in Ck(∂K,M) such that for every F ∈ U
we have |λ(Trap(Ω )) − λ(Trap(Ω)| < 2ǫ. Then there exists a sequence {F }∞ ⊂
m m m=1
Ck(∂K,M) with F → id as m → ∞ in the Ck Whitney topology such that
m
|λ(Trap(Ω ))−λ(Trap(Ω)| ≥ 2ǫ (3.1)
m
for all m, where we set Ω = Ω for brevity. Set K = M \Ω ; then ∂K = F (∂K).
m Fm m m m m
Since all obstacles K arein the interior of M, we have Ω ⊂ M, so the Liouville measure
m m
8
λ = dqdv is well-defined on S(Ω ). The function t(x) and the billiard trajectory γ+(x)
m
for Ω will be denoted by t (x) and γ+(x), respectively. Set
m m m
T ′ = Trap+(∂M)∪∪∞ Trap+ (∂M).
Ω m=1 Ωm
Then µ(T′) = 0 by (1.2). Theorem 1.1 with f = 1 implies
λ(S(Ω)\Trap(Ω)) = t(x)dµ(x) = t(x)dµ(x),
ZS+(∂M) ZS+(∂M)\T′
and
λ(S(Ω )\Trap(Ω )) = t (x)dµ(x) = t (x)dµ(x).
m m m m
ZS+(∂M) ZS+(∂M)\T′
Since λ(S(Ω )) → λ(S(Ω)) as m → ∞, we may assume that |λ(S(Ω ))−λ(S(Ω))| < ǫ
m m
for all m ≥ 1. Combining this with (3.1) and the above equalities, we get
|I −I| ≥ ǫ (3.2)
m
for all m ≥ 1, where we set for brevity
I = t (x)dµ(x) , I = t(x)dµ(x).
m m
ZS+(∂M)\T′ ZS+(∂M)\T′
For every m ≥ 1 denote by T′′ the set of those x ∈ S+(∂M) such that γ+(x) has a
m m
tangent point to ∂Ω , and let T′′ be the set of those x ∈ S+(∂M) such that γ+(x) has
m 0
a tangent point to ∂Ω. As remarked earlier, we have µ(T′′) = 0 for all m ≥ 0. Thus, for
m
the set
T′′ = ∪∞ T ′′
m=0 m
we have µ(T ′′) = 0, and therefore µ(T ′ ∪T′′) = 0. Let
D ⊂ D ⊂ ... ⊂ D ⊂ ... ⊂ S+(∂M)\(T′ ∪T ′′)
1 2 r
be compact sets so that µ(D ) ր µ(S+(∂M)) as r → ∞. Set
r
I(r) = t (x)dµ(x) , I(r) = t(x)dµ(x)
m m
ZDr ZDr
for brevity. Then
I(r) ր I , I(r) ր I (3.3)
m m
as r → ∞, for any fixed m in the first limit. It follows from this that there exists an
integer r so that
0
ǫ
I − < I(r) ≤ I (3.4)
3
for all r ≥ r .
0
Next, consider an arbitrary point x = (q,v) ∈ S+(∂M) \ (T′ ∪ T ′′). Then x ∈/
Trap+(∂M) gives t(x) < ∞, while x ∈/ T ′′ shows that γ+(x) is a simply (transversally)
Ω 0
9
reflecting trajectory in Ω with no tangencies to ∂Ω. Thus, the number j (x) of reflection
0
points of γ+(x) is finite, and γ+(x) has no other common points with ∂Ω.
Lemma 3.1. Let r ≥ 1. There exists an integer m = m (r) ≥ 1 such that for every
0 0
x ∈ D and every integer m ≥ m the trajectory γ+(x) has at most j (x) common points4
r 0 m 0
with ∂Ω .
m
Proof of Lemma 3.1. Fix r ≥ 1 and assume there exist arbitrarily large m such that
γ+(x )hasatleastj (x )+1reflectionpointsforsomex ∈ D . Choosingasubsequence,
m m 0 m m r
we may assume that the latter is true for all m ≥ 1 and that x → x ∈ D as m → ∞.
m r
Since both γ+(x ) and γ+(x) are simply reflecting, it follows that j (x ) = j (x) for all
m 0 m 0
sufficiently large m; we will assume this is true for all m. Set j = j (x). Thus, for every
0 0
m the trajectory γ+(x ) has at least j +1 reflection points. Let q ,...,q ∈ ∂K be the
m m 0 1 j0
successive reflection points of γ+(x). Set q = g for convenience. Notice that some
j0+1 1
of the points q may coincide, however q 6= q for all j. For each j = 1,...,j choose
j j+1 j 0
a small open neighbourhood V of q on ∂K so that V ∩ V = ∅ for all j = 1,...,j .
j j j j+1 0
Since F → id in the Ck Whitney topology, a simple continuity argument shows that for
m
sufficiently large m we have F (V )∩F (V ) = ∅ for all j = 1,...,j . Assuming that
m j m j+1 0
the neighbourhoods V are chosen sufficiently small and m is sufficiently large, it follows
j
that for each j = 1,...,j , γ+(x ) has an unique reflection point q(m) in F (V ). Let
0 m m j m j
q(m) be a reflection point of γ+(x ) which is not in ∪j0 F (V ); such a point exists, since
j0+1 m m j=1 m j
by assumption the trajectory γ+(x ) has at least j + 1 reflection points. Choosing an
m m 0
(m)
appropriate sub-sequence again, we may assume that q → q ∈ ∂K as m → ∞.
j0+1 j0+1
It is now clear that q is a common point of γ+(x) and ∂K which does not belong to
j0+1
∪j0 V , so q 6= q for all j = 1,...,j . This is a contradiction which proves the lemma.
j=1 j j0+1 j 0
We now continues with the proof of Theorem 1.2.
According to (3.2), we either have I ≤ I −ǫ for infinitely many m, or I ≥ I +ǫ for
m m
infinitely many m.
Case 1. I ≥ I +ǫ for infinitely many m. Considering an appropriate subsequence, we
m
may assume I ≥ I +ǫ for all m ≥ 1.
m
Let r ≥ r so that (3.4) holds for r, and let m = m (r) be as in Lemma 3.1. Then
0 0 0
the compactness of D and a simple continuity argument show that
r
i = sup j (x) < ∞.
0 0
x∈Dr
This and Lemma 3.1 now imply
t (x) ≤ Di , m ≥ m , x ∈ D .
m 0 0 r
Another simple continuity argument shows that for x ∈ D the only possible limit point
r
of the sequence {t (x)}∞ is t(x), so we must have lim t (x) = t(x) for all x ∈ D .
m m=1 m→∞ m r
This is true for all r, so lim t (x) = t(x) for all x ∈ S+(∂M)\(T′ ∪T ′′). Setting
m→∞ m
t˜m(x) = sup tm′(x) , x ∈ S+(∂M)\(T ′ ∪T′′),
m′≥m
4Clearly all of them must be simple reflection points, since x∈/ T′ ∪T′′.
m m
10