Birkhoff billiards are insecure

We prove that every compact plane billiard, bounded by a smooth curve, is insecure: there exist pairs of points $A,B$ such that no finite set of points can block all billiard trajectories from $A$ to $B$.

Two points A and B of a Riemannian manifold M are called secure if there exists a finite set of points S ⊂ M − {A, B} such that every geodesic connecting A and B passes through a point of S. One says that the set S blocks A from B. A manifold is called secure (or has the finite blocking property) if any pair of its points is secure. For example, every pair of nonantipodal points of the Euclidean sphere is secure, but a pair of antipodal points is not secure, so the sphere is insecure. A flat torus of any dimension is secure.
In the recent years, the notion of security has attracted a considerable attention, see [1,2,3,4,9,10,11,12]. This notion extends naturally to Riemannian manifolds with boundary, in which case one considers billiard trajectories from A to B with the billiard reflection off the boundary.
In this note we consider a compact plane billiard domain M bounded by a smooth curve and prove that M is insecure. More specifically, one has the following local insecurity result. Consider a sufficiently short outward convex arc γ ⊂ ∂M with end-points A and B (such an arc always exists). Proof. Denote by T n the polygonal line A = P 0 , P 1 , . . . P n−1 , P n = B, P i ∈ γ, of minimal length; this is a billiard trajectory from A to B. If n is large then T n lies in a small neighborhood of γ.
Working toward contradiction, assume that a finite set of points S ⊂ M − {A, B} blocks every billiard trajectory from A to B. Decompose S as S ′ ∪ S ′′ where the points of S ′ lie on the boundary and the points of S ′′ lie inside the billiard table. For n large enough, the trajectory T n is disjoint from S ′′ . We want to show that there is a sufficiently large n such that the set P n = {P 1 , . . . P n−1 } is disjoint from S ′ .
Let s be the arc-length parameter and k(s) the curvature of γ. Let σ be a new parameter on the arc γ such that dσ = (1/2)k 2/3 ds. By rescaling the arc γ, we may assume that the range of σ is [0, 1] with σ(0) = A and σ(1) = B. Let Q 0 = A, Q 1 , . . . , Q n−1 , Q n = B be the points that divide the σ-measure of γ into n equal parts, that is, Q m = σ(m/n).
Remark 3 This Claim is consistent with Theorem 6 (iii) of [6] which describes the limit distribution of the vertices of the inscribed polygons that best approximate a convex curve relative the deviation of the perimeter length.
To prove Proposition 2, we use the theory of interpolating Hamiltonians, see [7,8] and especially [5]. Recall the relevant facts from this theory.
First, some generalities about plane billiards (see, e. g., [13,14]). The phase space X of the billiard ball map consists of inward unit tangent vectors (x, v) to M with the foot point x on the boundary ∂M; x is the position of the billiard ball and v is its velocity. The billiard ball map F takes (x, v) to the vector obtained by moving x along v until it hits ∂M and then elastically reflecting v according to the law "angle of incidence equals angle of reflection". Let φ be the angle made by v with the positive direction of ∂M.
In a nutshell, the theory of interpolating Hamiltonians asserts that the billiard ball map equals an integrable symplectic map, modulo smooth symplectic maps that fix the boundary of the phase space X to all orders. More specifically, one can choose new symplectic coordinates H and Z near the boundary φ = 0 such that ω = dH ∧ dZ, H is an integral of the map F , up to all orders in φ, and also up to all orders in φ. The function H is given by a series in even powers of φ, namely, and this series is uniquely determined by the above conditions on H and Z.

Lemma 4 One may choose the coordinate Z in such a way that
Proof. Let Z = f (s) + g(s)φ + O(φ 2 ). We have: ω = dH ∧ dZ. Equating the coefficients of φ dφ ∧ ds and of φ 2 dφ ∧ ds and using (2) we obtain the equations: The first equation implies that df = dσ and the second that g = Ck −1/3 where C is a constant. We can choose f (0) = 0. Since Z is defined up to summation with functions of H, it follows from (2) that the term g(s)φ can be eliminated by subtracting CH 1/2 . 2 Now we can prove Proposition 2. The billiard trajectory T n corresponds to a phase orbit x 0 , . . . , x n , F (x i ) = x i+1 . Since H is an integral of the map F , the orbit x 0 , . . . , x n lies on a level curve H = c n . Due to (1), we have: n √ c n = O(1), and hence c n = O(1/n 2 ) which, in view of (2), implies that Consider σ and Z as functions on the phase space X. Since σ(x m ) = P m and the σ-coordinate of Q m is m/n, we need to show that Since F is a shift in Z-coordinate, see (1), one has: the second equality due to Lemma 4 and (3). This proves Proposition 2. 2 From now on, we identify the arc γ with the segment [0, 1] using the parameter σ; the points P 1 , . . . , P n−1 are considered as reals between 0 and 1. Assume that a finite set S ′ = {t 1 , . . . , t k } ⊂ (0, 1) is blocking, that is, for all sufficiently large n, one has P n ∩ S ′ = ∅.
Some of the numbers t i ∈ S ′ may be rational; denote them by p i /q i , i = 1, . . . , l (fractions in lowest terms), and let Q = q 1 · · · q l . Set n i = 1 + (N + i)Q, i = 0, . . . , k.
Proposition 5 For N sufficiently large, at least one of the sets P n i is disjoint from S ′ .
Proof. Assume not. Then, by the Pigeonhole Principle, there exist l, i, j such that t l ∈ P n i ∩ P n j . According to Proposition 2, there is a constant C (independent of n) such that, for P m ∈ P n , one has: Therefore for some m 1 , m 2 .

Lemma 6
If N sufficiently large then t i / ∈ Q.
Proof. First, we claim that, given a fraction p/q and a constant C, if for all sufficiently large n then m/n = p/q. Indeed, if m/n = p/q then 1 ≤ |pn − qm|, hence which cannot hold for n > Cq.
Next, we claim that, for all M, N ∈ Z and each i = 1, . . . , l, Indeed, if the equality holds then Mq i = p i (1 + NQ). The right hand side is divisible by q i but 1 + NQ is coprime with q i ; this contradicts the assumption that q i and p i are coprime. The two claims combined imply the lemma. 2 Next, (5) and the triangle inequality imply that for some m 1 , m 2 . It follows that The expression in the parentheses on the right hand side has limit 2, as N → ∞, hence one has, for sufficiently great N, Denote by M the (finite) set of fractions with the denominators jQ, j ∈ {1, 2, . . . , k}, and let δ > 0 be the distance between the sets S ′ − Q and M.
Lemma 7 For sufficiently large N, one has: Proof. For N large enough, it follows from (5) that t l − m 1 n i < δ 2 .
Since t l / ∈ Q, it follows that the distance from m 1 /n i to M is greater than δ/2. One has: as claimed. 2 Finally, for large N, Lemma 7 contradicts inequality (6), and Proposition 5 follows. 2 This Proposition implies Theorem 1, and we are done. 2 Remark 8 Theorem 1, along with its proof, can be extended to billiards in higher dimensional Euclidean spaces: the role of the curve γ is played by the shortest geodesic on the boundary of the billiard table connecting A and B.