Large Poisson-Voronoi Cells and Crofton Cells

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1 Large Poisson-Voronoi Cells an Crofton Cells Daniel Hug Matthias Reitzner Rolf Schneier Abstract. It is prove that the shape of the typical cell of a stationary Poisson-Voronoi tessellation in Eucliean space, uner the conition that the volume of the typical cell is large, must be close to spherical shape, with high probability. The same hols if the volume is replace by the surface area or other suitable functionals. Similar results are establishe for the zero cell of a stationary an isotropic Poisson hyperplane tessellation. Keywors: Poisson-Voronoi tessellation; typical cell; Poisson hyperplane tessellation; Crofton cell; asymptotic shape; D.G. Kenall s conjecture; intrinsic volume AMS 2000 Subject Classification: Primary 60D05 Seconary 52A22, 52A40 1 Introuction In this paper, we continue a line of research that began with a problem of D.G. Kenall (see the forewor to the first eition of [17]). He conjecture that the shape of the zero cell (or Crofton cell) of a stationary an isotropic Poisson line process in the plane, given that the area of the cell tens to infinity, must become circular. Contributions to Kenall s question are ue to Miles [11] an Golman [3], an the conjecture was prove by Kovalenko [8], [10]. In [7], Kovalenko s result was extene to higher imensions an to stationary, but not necessarily isotropic Poisson hyperplane processes. It was also strengthene, by estimating the probability of large eviations from spherical shape, given that the volume of the zero cell lies in a prescribe interval. In the present paper, we prove an analogous result for the typical cell of a stationary Poisson-Voronoi tessellation (mosaic) of -imensional space. Thus we exten an strengthen a result of Kovalenko [9] in the planar case. We further exten this result by consiering, in aition to the volume functional, also the kth intrinsic volume, k = 1,..., 1. This inclues cells of large surface area or of large mean with. The result from [7] on Crofton cells of stationary Poisson hyperplane processes with large volume is also extene to the kth intrinsic volume, but only for k 2 an uner the aitional assumption of isotropy. For both types of ranom polytopes, Poisson-Voronoi cells an isotropic Crofton cells, we can also replace (somewhat easier) the conition of large volume by the conition of Postal aress: Mathematisches Institut, Albert-Luwigs-Universität, Eckerstr. 1, D Freiburg, Germany. aress: aniel.hug@math.uni-freiburg.e Postal aress: Institut für Analysis un Technische Mathematik, Technische Universität Wien, Wiener Hauptstrasse 8-10, A-1040 Wien, Austria. aress: mreitzne@mail.zserv.tuwien.ac.at Postal aress: Mathematisches Institut, Albert-Luwigs-Universität, Eckerstr. 1, D Freiburg, Germany. aress: rolf.schneier@math.uni-freiburg.e 1

2 large inraius. This is suggeste by consierations of Miles [11] on Crofton cells in the plane an by the work of Calka [1] on planar Poisson-Voronoi tessellations. Finally, we mention here that cells of large volume in Poisson-Delaunay tessellations were treate in [6]; such cells ten to be regular simplices. Let A be a locally finite point set in Eucliean space R (with scalar prouct, an norm ), where 2. For x A, the Voronoi cell of x with respect to A is efine by C(x, A) := y R : y x y a for all a A. Let X be a stationary Poisson point process of intensity λ > 0 in R. (In treating simple point processes, we conveniently ientify a simple counting measure with its support.) Then X := C(x, X) : x X is the Poisson-Voronoi tessellation erive from X. Let Z enote the typical cell of X (we recall its efinition in Section 2). For a convex boy (non-empty, compact, convex set) K R, we enote the volume by v (K). The (conveniently renormalize) intrinsic volumes v 0 (K),..., v 1 (K) can be efine by means of the Steiner formula v (K + ɛb ) = ( ) ɛ k v k (K), ɛ 0. k k=0 Here B := x R : x 1 is the unit ball. Equivalently, v i (K) is the mixe volume V (K,..., K, B,..., B ), where K appears i times an B appears i times. The functional W j = v j is known as the jth quermassintegral. In particular, v 1 is the surface area, an (2/κ )v 1 is the mean with; here κ := v (B ). More information is foun in [15]. Let K R be a compact set with o K an containing more than one point. In orer to measure the eviation of K from a ball with centre o, we efine ϑ(k) := R o ρ o R o + ρ o, where R o is the raius of the smallest ball with centre o containing K an ρ o is the raius (possibly zero) of the largest ball with centre o containe in K. By P we enote the unerlying probability, an P( ) is a conitional probability. Theorem 1. Let X enote the Poisson-Voronoi tessellation erive from a stationary Poisson point process with intensity λ > 0 in R ; let Z be its typical cell. Let k 1,...,. There is a positive constant c 0 epening only on the imension such that the following is true. If ɛ (0, 1) an I = [a, b) is any interval (possibly b = ) with a /k λ σ 0, for some constant σ 0 > 0, then P(ϑ(Z) ɛ v k (Z) I) c exp c 0 ɛ (+3)/2 a /k λ, where c is a constant epening only on, ɛ an σ 0. In particular, lim P(ϑ(Z) ɛ v k(z) a) = 0 for every ɛ > 0, (1) a 2

3 but Theorem 1 provies much stronger information. The relation (1) can equivalently be formulate as follows (see Section 2 for further explanation). Corollary. The conitional law for the shape of Z, given a lower boun for v k (Z), converges weakly, as that lower boun tens to infinity, to the law concentrate at the shape of a ball. Theorem 1 will be prove in Section 6, after preliminary explanations in Section 2 an preparations in Sections 3 to 5. In [7], a similar result was obtaine for the volume of the zero cell (also calle Crofton cell) of the tessellation generate by a stationary Poisson hyperplane process. We will inicate in Section 7 how, uner the aitional assumption of isotropy, this result can be extene to the kth intrinsic volume, k = 2,...,. As in [7], we measure the eviation of the shape of a convex boy K R with interior points from spherical shape by r B, which we abbreviate by r, thus r (K) := mins/r 1 : rb + z K sb + z, z R, r, s > 0. Theorem 2. Let Z o be the zero cell of the tessellation inuce by a stationary isotropic Poisson hyperplane process in R with intensity λ > 0. Let k 2,...,. There is a positive constant c 0 epening only on the imension such that the following is true. If ɛ (0, 1) an I = [a, b) is any interval (possibly b = ) with a 1/k λ σ 0, for some constant σ 0 > 0, then P(r (Z o ) ɛ v k (Z o ) I) c exp c 0 ɛ (+3)/2 a 1/k λ, where c is a constant epening only on, ɛ an σ 0. The case of the volume is inclue here for k =. We remark that in this case the inequality of Theorem 2 is sharper (in its epenence on ɛ) than Theorem 1 of [7], specialize to the isotropic case. The reason for this improvement lies in the fact that in the isotropic case sharper stability estimates from convex geometry are available. Unfortunately, our metho of proof oes not permit us to treat the case k = 1, which in the plane is the case of the perimeter, alreay stuie by Miles [11] in his heuristic approach. In aition to ρ o (K) efine above, we enote by ρ(k) the raius of the largest ball containe in the convex boy K. Theorem 3. Let Z an Z o be efine as in Theorems 1 an 2, respectively. There is a positive constant c 0 epening only on the imension such that the following is true. If ɛ (0, 1) an I = [a, b) is any interval (possibly b = ) with a λ σ 0 in the case of Z, respectively aλ σ 0 in the case of Z o, with some constant σ 0 > 0, then P(ϑ(Z) ɛ ρ o (Z) I) c exp c 0 ɛ (+1)/2 a λ an P(r (Z o ) ɛ ρ(z o ) I) c exp c 0 ɛ (+1)/2 aλ, where c is a constant epening only on, ɛ an σ 0. 3

4 The proof will be sketche in Section 8. A few wors about the choices of the shape parameters ϑ(k) an r (K) seem in orer. For the formulation of our estimates, we want a simple, similarity invariant measure for the eviation of the shape of a convex boy from spherical shape. Such a measure appears implicitly in relation (2) of Miles [11], an explicitly in the paper of Kovalenko [10]. The number r (K) use above is the extension of the latter to higher imensions. For an interior point z of the compact convex set K, let Rz be the raius of the smallest ball with centre z containing K, an ρz the raius of the largest ball with centre z containe in K. Then r (K) is the minimum, over all z in the interior of K, of the quotient (Rz ρz)/ρz. In the case of the typical cell of a Poisson-Voronoi tessellation, for which the nucleus o is a istinguishe point, we get a sharper result if we use instea the eviation measure (Ro ρo)/ρo, or, which for boies close to a ball with centre o is essentially equivalent, (Ro ρo)/(ro +ρo) = ϑ(k). Here we have chosen the latter, since for this the crucial stability estimate (13) below takes a simple form. Finally, we mention that Golman [3], in the planar case, consiers the first eigenvalue of the Laplacian for the Dirichlet problem an fins the same asymptotic behaviour for large Crofton cells an their incircles, but this oes apparently not lea to explicit geometric estimates for the eviation from circular shape. 2 The typical cell of a Poisson-Voronoi tessellation In general, the notion of the typical cell of a stationary ranom tessellation requires the choice of a centroi function (e.g., see Møller [12, Section 3.2]), but for Voronoi cells there is a canonical choice, the nucleus. Let X be the Poisson-Voronoi tessellation generate by a stationary Poisson point process X. Let K o enote the space of convex boies K in R with o K, equippe with the Hausorff metric an corresponing Borel structure. The istribution Q of the typical cell of X can be efine by Q(A) = 1 λ E x X 1 A (C(x, X) x)1 B (x) (E enotes mathematical expectation) for Borel sets A Ko, where B R is an arbitrary Borel set with λ (B) = 1; here λ enotes Lebesgue measure in R. The istribution Q is commonly interprete as the conitional istribution of C(o, X) given that o X. An intuitive interpretation follows from the fact that stationary Poisson-Voronoi tessellations are mixing ([16, Satz 6.4.1]) an hence ergoic. This implies that hols with probability one. Q(A) = lim r car x X rb : C(x, X) x A car ( X rb ) Since X is a stationary Poisson process, it follows from Slivnyak s theorem that the typical cell of the Poisson-Voronoi tessellation X is equal in istribution to the ranom polytope Z = C(o, X o) (see [12, Remark 4.1.1]; see also [13]). Hence, we can consier Z as the typical cell of X, an for this we obtain a convenient representation. For x R, we efine H(x) := y R : y, x = x 2 /2, H (x) := y R : y, x x 2 /2, 4

5 so that H(x) is the mi hyperplane of o an x. Then Z = H (x), thus Z is the zero cell of the tessellation inuce by the Poisson hyperplane process x X Y := H(x) : x X. The intensity measure E Y ( ) of this process can be represente as follows (recall that Y enotes a ranom simple counting measure on the space of hyperplanes as well as its support). For a Borel set A in the space of hyperplanes, we have Writing E Y (A) = E car (A Y ) = E car x X : H(x) A = λ λ (x R : H(x) A). H(u, t) := y R : y, u = t, H (u, t) := y R : y, u t for u S 1 := x R : x = 1 an t R, an introucing polar coorinates, we get E Y ( ) = 2 λ 1H(u, t) t 1 t σ(u), (2) S 1 where σ enotes spherical Lebesgue measure on the unit sphere S 1. 0 In particular, for K K o let H K be the set of all hyperplanes H R with H K. Then (2) gives E Y (H K ) = 2 λu(k) (3) with U(K) := 1 where h(k, ) is the support function of K. Writing S 1 h(k, u) σ(u), (4) Φ(K) := y R : H(2y) K, we have U(K) = λ (Φ(K)). The star boy Φ(K) is the union of all close balls having a iameter segment [o, x] with x K. The relation (3), together with the Poisson property of the hyperplane process Y, implies that P(Y (H K ) = n) = [2 U(K)λ] n exp 2 U(K)λ (5) n! for K K o an n N 0. It is now clear that we are in a similar situation as in [7]. There, the zero cell of a stationary (not necessarily isotropic) Poisson hyperplane process, with intensity measure given by [7, (2)], was stuie. This process is now replace by the isotropic, non-stationary Poisson hyperplane process Y, with intensity measure given by (2). The functional U(K), efine by (4), will play the role of the mixe volume V 1 (B, K) in [7] (up to imensional factors). In aition to the volume functional consiere in [7], we now treat general intrinsic volumes. All these ifferences require a number of changes an new arguments, but also some parallel 5

6 reasoning is possible. In the latter cases, we will be brief an just list how the arguments of [7] have to be moifie. In analogy to Kenall s original conjecture, of which a more general version was treate in [7], we have formulate the Corollary after Theorem 1. We make this more precise. As a space of shapes, suitable for our purpose, we may take the space S (with the Hausorff metric) of convex boies containing o an with circumraius one. The ilate version of Z containe in S is enote by Z. The conitional law for the shape of Z, given the lower boun a for v k (Z), can be efine as the probability measure on S given by µ a ( ) := P(Z v k (Z) a). We are asserting that lim a µ a = δ B weakly, where δ B is the Dirac measure on S concentrate at the ball B (as we are prescribing the centre of the ball, this is a slightly stronger assertion than formulate in the Corollary). Thus, we have to prove that lim sup µ a (C) δ B (C) (6) a for every close set C S. This is trivial if B C. Let B / C. Since the eviation measure ϑ is continuous an positive on C, there exists ɛ > 0 such that C K S : ϑ(k) ɛ. This implies µ a (C) µ a (K S : ϑ(k) ɛ) = P(ϑ(Z) ɛ v k (Z) a) 0 for a, by (1). This proves the Corollary. Conversely, the assertion of the Corollary implies (1): given ɛ > 0, let C := K S : ϑ(k) ɛ, then C is close an B / C, hence (6) yiels (1). As the etails of the proofs for Theorems 1 3 are a bit technical, we want to sketch here the main lines of the reasoning, taking Theorem 1 as an example. Let k 1,...,. In the first instance, we are intereste in bouning the probability P(ϑ(Z) ɛ v k (Z) a) from above, for given ɛ > 0 an large a. This conitional probability is a quotient. A lower boun for the enominator P(v k (Z) a) follows immeiately from (5): the ball B a := (a/κ ) 1/k B satisfies v k (B a ) = a, hence P(v k (Z) a) P(Y (H Ba ) = 0) = exp 2 U(B a )λ = exp 2 κ 1 /k a /k λ. (7) To obtain an upper boun for the numerator P(ϑ(Z) ɛ, v k (Z) a), we use the geometric inequality U(K) κ 1 /k v k (K) /k, (8) in a strengthene form. Equality in (8) hols if an only if K is a ball with centre o. Let K K o be a convex boy satisfying ϑ(k) ɛ > 0. It can be prove (Lemma 1) that with f(ɛ) > 0. If now K satisfies this gives U(K) (1 + f(ɛ))κ 1 /k v k (K) /k ϑ(k) ɛ an v k (K) a, (9) P(Y (H K ) = 0) = exp 2 U(K)λ exp (1 + f(ɛ))2 κ 1 /k a /k λ. Since a convex boy containe in the interior of the cell Z oes not meet any hyperplane of Y, one might now hope that this estimate remains essentially true if the fixe convex boy K 6

7 is replace by the cell Z, in the cases where the latter satisfies the inequalities (9). Although this replacement is not legitimate, a similar an slightly weaker inequality of the form P(ϑ(Z) ɛ, v k (Z) a) c exp (1 + g(ɛ))2 κ 1 /k a /k λ, (10) with a constant c > 0 an g(ɛ) > 0, might be true. If this hols, then iviing (10) by (7) immeiately gives P(ϑ(Z) ɛ v k (Z) a) c exp g(ɛ)2 κ 1 /k a /k λ, which is of the require type. An estimate of type (10) can inee be prove, but at first not for v k (Z) in intervals [a, ), but in intervals a(1, 2). This is one in Lemmas 4 an 6. The istinction of two cases is necessary since it turns out that cells which are in a sense too elongate nee an extra treatment. The convex boies are classifie by an integer parameter m such that increasing m means increasing elongation. The auxiliary geometric Lemma 3 provies inner an outer inclusion estimates for boies of given elongation. For large m, the estimate of Lemma 4 can be use; here the conition ϑ(z) ɛ oes not play a role. Its proof uses only elementary geometric arguments. For small m, the conition ϑ(z) ɛ is essential. Lemma 6 contains the relevant estimate. It is here that the geometric stability result of Lemma 1 is neee. Moreover, the approximation result for polytopes expresse in Lemma 5 is require for the reuction to a situation involving only a fixe number of hyperplanes. The further Lemmas 7, 8, 9 are neee to pass from intervals a(1, 2) to intervals a(1, 1 + h), with fixe h; the extension is achieve by a transformation. The obtaine estimates are then combine into Lemma 10, which gives an upper estimate for the probability P(ϑ(Z) ɛ, v k (Z) a(1, 1+h)). Since this upper boun contains h as a factor, it is necessary to estimate the enominator P(v k (Z) a(1, 1 + h)) from below by a suitable boun which is also linear in h. This is achieve in Lemma 2. Its proof is essentially constructive, exhibiting sufficiently many realizations of Y for which the event v k (Z) a(1, 1 + h) occurs. In both, Lemmas 2 an 10, the number h must be sufficiently small. The final proof of Theorem 1 extens the estimates from the special intervals a(1, 1 + h), with small h, to general intervals [a, b), by a covering argument, as in the proof of Theorem 1 in [7]. 3 A stability estimate For K K o, we trivially have K Φ(K), hence U(K) v (K). Here equality hols if an only if K is a ball with centre o (this follows from the consierations below, but can also be shown irectly). A similar inequality can be obtaine for the other intrinsic volumes. In the following, we write h(k, ) =: h K for the support function of K. Integrations with respect to σ exten over S 1. By Höler s inequality, U(K) 1 (κ ) 1 ( h K σ), an since h K σ = v 1 (K), we get U(K) 1/ κ (1 )/ v 1 (K). 7

8 A well-known inequality (e.g., [15, p. 334]) says that for k = 1,...,. Hence, v 1 (K) k κ k 1 v k (K) (11) U(K) κ 1 /k v k (K) /k. (12) Equality for a number k 1,..., hols if an only if K is a ball with centre o. We will nee an improve version of (12), in the form of a stability estimate. The following proof combines techniques evelope for obtaining stability results relate to Höler s inequality an to isoperimetric inequalities such as (11). Lemma 1. For K K o an k 1,...,, U(K) (1 + γϑ(k) (+3)/2 )κ 1 /k v k (K) /k, (13) where γ is a positive constant epening only on the imension. Proof. We assume that K contains more than one point; otherwise the assertion is trivial. In orer to improve Höler s inequality, we use Lemma 4.2 of Garner an Vassallo [2]. There we put m = 2, f 0 = f 2 = 1, f 1 = h K, w 1 = 1/, w 2 = ( 1)/ (hence w = 1/), an obtain hk σ 1 ( h K σ ) 1/ ( ) ( 1)/ 1 σ 1 [ h /2 K ( h K σ ) 1/2 1 ( 1 σ ) 1/2 ] 2 σ =: β(k). Using (1 β)(1 + β) 1, we euce that U(K) 1/ (1 + β(k))κ (1 )/ v 1 (K). (14) Next, we establish an estimate of the form β(k) cϑ(k) α with α > 0. For this, we argue similarly as in the proof of Lemma 1 in [5] (which is reprouce in [4], see inequality (2.3.3)). From now on in this paper, c 1, c 2,... enote constants epening only on the imension, except in those cases where other epenences are explicitly inicate. Let K Ko be given. Since β an ϑ are invariant uner ilatations, we can normalize K an assume that h K σ = 1 σ = κ, (15) which permits us to obtain the following relations. From (12) we get κ = U(K) κ 1 v 1 (K) c 1 D(K), where D enotes the iameter, hence D(K) c 2 an therefore h K c 2. (16) Moreover, β(k) = c 3 ( h /2 K 1 ) 2 σ. 8

9 Suppose that ρ o B K R o B, where ρ o is maximal an R o is minimal. It follows from (15) that ρ o 1 R o. Case 1: R o 1 1 ρ o. We put R o = 1 + h, then ϑ(k) = R o ρ o R o + ρ o 2h. (17) There exists a vector u 0 S 1 such that h K (u 0 ) = R o = 1 + h, an the point p = (1 + h)u 0 belongs to K. For u S 1, let E(u) be the hyperplane through p an orthogonal to u. Let ω enote the angle between u an u 0, an let ω 0 be the angle between u 0 an any v such that E(v) is tangent to B. Then cos ω 0 = 1/(1 + h) an (with σ 1 := ( 1)κ 1 ) ( h /2 K 1 ) 2 σ σ 1 ω0 0 [(1 + h) cos ω] /2 1 2 (sin ω) 2 ω. We set (1 + h) /2 1 =: α an substitute [(1 + h) cos ω] /2 1 = αx, to obtain ( ) h /2 2 K 1 σ 2σ 1α 3 (1 + h) 2 By (16), we can estimate an, for 0 x 1, This gives The function 1 0 x 2 [(1 + h) 2 (αx + 1) 4/ ] ( 3)/2 (αx + 1) (2 )/ x. (1 + h) 2 c 4 (αx + 1) (2 )/ (α + 1) (2 )/ = (1 + h) (2 )/2 c 1/ β(k) c 5 α 3 x 2 [(1 + h) 2 (αx + 1) 4/ ] ( 3)/2 x. 0 f(x) := (1 + h) 2 (αx + 1) 4/, 0 x 1, satisfies f(0) = h(h + 2), f(1) = 0, an f (1) = (4/)α(1 + h) (4 )/2. It is convex for 4 an concave for = 2, 3. For 4 we euce that f(x) 4 α(1 + h)(4 )/2 (1 x) c 6 α(1 x), an for = 2, 3 we get Together with this yiels From (17) we now get f(x) h(h + 2)(1 x) h(1 x). α = (1 + h) /2 1 2 h β(k) c 7 h (+3)/2. β(k) c 8 ϑ(k) (+3)/2. 9

10 Case 2: R o 1 < 1 ρ o. We put ρ o = 1 h, then R o < 1 + h, hence an K (1 + h)b (18) ϑ(k) 2h. (19) There is a vector u 0 S 1 such that h K (u 0 ) = ρ o = 1 h, an the hyperplane G through ρ o u 0 orthogonal to u 0 supports K. Let p ( (1 + h)b ) G. Let u 1 S 1 be the vector, positively spanne by u 0 an p, that is orthogonal to a support plane of B through p. Let ω 0 be the angle between u 0 an u 1. For u S 1, let ω u be the angle between u 0 an u. If ω u ω 0, then h K (u) 1, hence 1 h K (u) /2 1 h K (u) 0 an thus β(k) c 3 (1 h K (u)) 2 1ω u ω 0 σ(u). Now exactly the argument of Case 2 in [5, pp ] leas to which together with (19) gives β(k) c 9 h (+3)/2, β(k) c 10 ϑ(k) (+3)/2. In each case, we conclue from (14) an (11) the inequality (13). 4 Probabilities involving small intervals In this section we prove an inequality which replaces the easily obtaine estimate (7) in the case where v k (Z) is containe in a boune interval. For A R we efine Z(A) := x A H (x) an set Z(x 1,..., x n ) := Z(x 1,..., x n ). For the ranom polytope Z( X), the typical cell of X, we retain the notation Z. Let k 1,..., be fixe. A main feature of the following estimate is the linear epenence of the lower boun on the length h of the interval (1, 1 + h). A similar result, for the zero cell of a stationary Poisson hyperplane process an the function v, is given in [7] as Lemma 3.2. It is an important avantage of the present argument that it merely uses the monotoneity, continuity an homogeneity properties of the function v k. Lemma 2. For each β > 0, there are constants h 0 > 0, N N an c 11 > 0, epening only on β an, such that for a > 0 an 0 < h < h 0, ( ) N P(v k (Z) a(1, 1 + h)) c 11 h a /k λ exp (1 + β) 2 κ 1 /k a /k λ. 10

11 Proof. Let β > 0 an a > 0 be given. For n 2 we efine ( Q n := (x 1,..., x n 1, u) (1 + β/2)b ) n 1 S 1 : Z(x 1,..., x n 1, u) 2 1 (1 + β/2)b, v k (Z(x 1,..., x n 1, u)) 2 k κ. A continuity argument shows that we can choose N = N(β, ) N (sufficiently large) so that... 1(x 1,..., x N 1, u) Q N x 1... x N 1 σ(u) =: c 12 (, β) > 0. S 1 R R If (x 1,..., x N 1, u) Q N, then We efine h 0 = h 0 (β) > 0 by v k (Z(x 1,..., x N 1, u)) (1 + β/2) k 2 k κ. (20) (1 + h 0 ) 1/k (1 + β/2) = 1 + β (21) an suppose that 0 < h < h 0. As before, we put B a := (a/κ ) 1/k B. We start with the trivial estimate (again we use that a realization of X enotes a simple counting measure an also its support) P(v k (Z) a(1, 1 + h)) P( X(2(1 + β)ba ) = N, Z( X 2(1 + β)b a ) (1 + β)b a, v k (Z( X ) 2(1 + β)b a )) a(1, 1 + h) ) ( = P( X(2(1 + β)ba ) = N P Z( X 2(1 + β)b a ) (1 + β)b a, v k (Z( X 2(1 + β)b a )) a(1, 1 + h) ) X(2(1 + β)ba ) = N. Since X is a stationary Poisson process with intensity λ, we obtain (using Satz 3.2.3(b) of [16]) P(v k (Z) a(1, 1 + h)) λn N! exp v (2(1 + β)b a )λ... R 1 i : x i 2(1 + β)b a 1 Z(x 1,..., x N ) (1 + β)b a R 1 v k (Z(x 1,..., x N )) a(1, 1 + h) x 1... x N. Assume that x 1,..., x N R satisfy the conitions (i) (x 1,..., x N ) x N Q N ; (ii) v k (Z(x 1,..., x N )) a(1, 1 + h). Then, using (ii), (i) an the efinition of Q N, we get a(1 + h) x N k v k(z(x 1,..., x N )) x N k 2 k κ, 11

12 hence By (21) an (22), x N 2(a/κ ) 1/k (1 + h) 1/k. (22) x N 2(1 + β)(a/κ ) 1/k. Further, using (i), the efinition of Q N, (22) an (21), we fin that, for i = 1,..., N 1, x i x N (1 + β/2) 2(a/κ ) 1/k (1 + h) 1/k (1 + β/2) 2(1 + β)(a/κ ) 1/k, thus x i 2(1 + β)b a. Finally, (i), the efinition of Q N, (22) an (21) imply that Z(x 1,..., x N ) 2 1 x N (1 + β/2)b (a/κ ) 1/k (1 + h) 1/k (1 + β/2)b (1 + β)b a. Hence, introucing polar coorinates, we obtain P(v k (Z) a(1, 1 + h)) λn N! exp v (2(1 + β)b a )λ... 1 (x 1,..., x N ) x N Q N R R 1v k (Z(x 1,..., x N )) a(1, 1 + h)x 1... x N = λn N! exp v (2(1 + β)b a )λ... 1 (x 1,..., x N 1, ru) rq N S 1 0 R R 1v k (Z(x 1,..., x N 1, ru)) a(1, 1 + h)r 1 x 1... x N 1 r σ(u). Substituting x i = ry i for i = 1,..., N 1, we get P(v k (Z) a(1, 1 + h)) λn N! exp v (2(1 + β)b a )λ... 1(y 1,..., y N 1, u) Q N S 1 0 R R 1r k v k (Z(y 1,..., y N 1, u)) a(1, 1 + h)r N 1 y 1... y N 1 r σ(u) = λn N! exp v (2(1 + β)b a )λ S 1... R 1(y 1,..., y N 1, u) Q N R an/k ) N v k(z(y 1,..., y N 1, u)) N/k (1 + h) N/k 1 y 1... y N 1 σ(u) λn N! exp v (2(1 + β)b a )λ an/k ( ) N/k N 2 k (1 + β/2) k κ 1 N k h c 12(, β) ( N h a λ) /k exp v (2(1 + β)b a )λ c 13 (, β), where also (20) was use. assertion. Since v (2(1 + β)b a ) = (1 + β) 2 κ 1 /k a /k, this proves the 12

13 5 Probabilities involving elongate cells We will later nee estimates showing that typical cells which, compare to their value of v k, are too long, occur only with small probability. This requires the following preparations, in the course of which convex boies are classifie accoring to their egree of elongation. We enote by P o K o the subset of convex polytopes an by G(, k) the Grassmannian of k-imensional linear subspaces of R. For K K o an L G(, k), the set K L is the image of K uner orthogonal projection to L. For k 1,...,, we efine η k (K) := mind(k)/ (K, L) : L G(, k), where D(K) enotes the iameter of K an (K, L) is the with of K L evaluate in L. The simplest cases are k =, where η (K) is the ratio of the iameter an the with of K, an k = 1, where η 1 (K) = 1. Let a > 0 be given. For m N, we set K,k a (m) := K K o : v k (K) a(1, 2), η k (K) [m k, (m + 1) k ) The information that K Ka,k (m) can be use to erive estimates for the size of K from above an below.. Lemma 3. Let m N an k 1,...,. Then (a) K K,k a (m) implies that K c 14 m k a 1/k B =: C; (b) there exists a measurable map Ka,k (m) Po P v(p ) such that v(p ) is a vertex of P with v(p ) c 15 ma 1/k. Proof. We use repeately that v k (K L) is a constant multiple of the k-imensional volume of K L. (a) A special case of equation (5.3.23) in [15] an the monotoneity of mixe volumes imply that v k (K L) c 16 v k (K) hols for all K Ko an L G(, k). Let K Ka,k (m), an choose L 0 G(, k) such that η k (K) = D(K)/ (K, L 0 ). Then 2a v k (K) c 1 16 v k(k L 0 ) c 17 (K, L 0 ) k 1 D(K L 0 ), where the estimate (16) from [7] was use. Thus an therefore Since o K, this implies (a). 2a c 17 (K, L 0 ) k c 17 (m + 1) k2 D(K) k D(K) c 18 m k a 1/k. (b) For any L G(, k), we enclose K L in a rectangular parallelepipe in L with one ege length equal to (K, L) an the other ege lengths at most D(K L). Then v k (K L) c 19 (K, L)D(K L) k 1 c 19 m k D(K)D(K) k 1 13

14 an hence, by an integral-geometric projection formula ([15], (5.3.27)), a v k (K) c 20 m k D(K) k. Therefore, K has a point at istance at least 2 1 c 1/k 20 ma 1/k from the origin. If K is a polytope, such a point can be chosen as a vertex. That a measurable selection is possible, follows as in [7, Lemma 4.3(c)]. Remark. We have η 1 (K) = 1; moreover, Ka,1 (m) only for m = 1. Therefore, some of the subsequent arguments simplify consierably, or can be omitte, in the case k = 1. Let a > 0, ɛ > 0 be given. For m N, we efine an K,k a,ɛ (m) := K K,k a (m) : ϑ(k) ɛ (23) q k a,ɛ(m) := P(Z K,k a,ɛ (m)). (24) Similarly as in [7], we prove two estimates concerning the ecay of q k a,ɛ(m) as a /k λ. The epenence on ɛ will not play a role until Lemma 6. For u 1,..., u n S 1 an t 1,..., t n (0, ) we introuce the abbreviation n H (u i, t i ) =: P (u (n), t (n) ). i=1 Our next lemma correspons to Lemma 5.1 in [7]; its proof is a moifie an slightly simplifie version of the proof of the latter. Lemma 4. For m N an a /k λ σ 0, where σ 0 > 0 is a constant, q k a,ɛ(m) c 21 (, σ 0 ) exp c 22 m a /k λ. (25) Proof. Let C be the ball efine in Lemma 3(a). Then Here, q k a,ɛ(m) = N=+1 P(Y (H C ) = N)P(Z K,k a,ɛ (m) Y (H C ) = N). (26) p N := P(Z Ka,ɛ,k (m) Y (H C ) = N) = 1 U(C) N... S 1 0 S P (u (N), t (N) ) Ka,ɛ,k (m) (27) 1 i : H(u i, t i ) C (t 1 t N ) 1 t 1 σ(u 1 )... t N σ(u N ). Suppose that u 1,..., u N, t 1,..., t N are such that the inicator functions occurring in the multiple integral are all equal to one; then P := P (u (N), t (N) ) Ka,ɛ,k (m) has a vertex v(p ) 14

15 accoring to Lemma 3(b). This vertex is the intersection of facets of P. Hence, there exists an inex set J 1,..., N with elements such that v(p ) = i J H(u i, t i ). The segment S := [o, v(p )] satisfies relint S H(u j, t j ) = for j 1,..., N \ J, where relint enotes the relative interior. Since S C, we have 1H(u, t) C, H(u, t) S = t 1 t σ(u) S 1 0 = U(C) U(S) = U(C) 2 κ S, (28) where S enotes the length of S. Similarly as in the proof of [7, Lemma 5.1] we obtain ( ) N 1 p N U(C) N... 1H(u i, t i ) C, i = 1,..., S 1 0 S 1 [U(C) c 23 m a /k ] N (t 1 t ) 1 t 1 σ(u 1 )... t σ(u ) 0 = ( N ) ( ) N 1 c 23m a /k. (29) U(C) This leas to the estimate qa,ɛ(m) k [2 U(C)λ] N N! N=+1 ( ) ( N exp 2 U(C)λ 1 c 23m a /k U(C) ) N = 1! [2 U(C)λ] exp 2 U(C)λ N=+1 1 [ ] N 2 U(C)λ c 24 m a /k λ (N )! 1! [2 U(C)λ] exp c 24 m a /k λ ( ) c 25 m k2 a /k λ exp c 24 m a /k λ c 27 (, σ 0 ) exp c 26 m a /k λ, which completes the proof. The following result allows us to approximate a given convex polytope P by a polytope L P with a restricte number of vertices such that U(L) is not much smaller than U(P ). For a polytope P, let extp be the set of vertices an f 0 (P ) the number of vertices of P. Lemma 5. Let α > 0 be given. There is a number ν N epening only on an α such that the following is true. For P P o there exists a polytope L = L(P ) P o satisfying 15

16 extl extp, f 0 (L) ν, an U(L) (1 α)u(p ). Moreover, there exists a measurable selection P L(P ). Proof. The following can be extracte from the proof of Lemma 4.2 in [7]. There exist numbers k 0 = k 0 () an b 0 = b 0 () such that the following is true. Let P P o be a polytope an let P RB, where R is minimal. Let k k 0. There is a measurable map P L(P ) such that L = L(P ) is the convex hull of at most (k + 1) vertices of P, o L, an P L + κrb with κ = b 0 k 2/( 1). There is a unit vector u such that R[o, u] P an hence U(P ) U(R[o, u]) = 2 κ R. By a suitable choice of k, epening only on an α, we can achieve that κ 1 an 4 κ α. Since h L R, we get U(P ) 1 [h L + κr] σ U(L) + κ 2 κ R U(L) + 4 κu(p ), S 1 thus U(L) (1 α)u(p ), an ν = (k + 1) is the require number. For the probabilities q k a,ɛ, we now state another upper boun, which is base on the stability estimate in Lemma 1 an on the preceing approximation result. Lemma 6. For m N, ɛ (0, 1) an a, λ > 0, ( qa,ɛ(m) k c 28 (, ɛ)m k2ν exp where ν epens only on an ɛ. 1 + c 29 ɛ (+3)/2) 2 κ 1 /k a /k λ, Proof. We efine C as in Lemma 3(a) an use (26) an (27) again. Assume that u 1,..., u N, t 1,..., t N are such that the inicator functions in (27) are all equal to one. Then, by Lemma 1, U ( P (u (N), t (N) ) ) (1 + γɛ (+3)/2 )κ 1 /k a /k. (30) Let α := γɛ (+3)/2 /(2 + γɛ (+3)/2 ); then (1 α)(1 + γɛ (+3)/2 ) = 1 + α. Put c 30 := γ/(2 + γ); then α > c 30 ɛ (+3)/2. By Lemma 5, there are ν = ν(, ɛ) vertices of P (u (N), t (N) ) such that the convex hull L = L ( P (u (N), t (N) ) of these vertices satisfies The inequalities (30) an (31) imply that U(L) (1 α)u ( P (u (N), t (N) ) ). (31) U(L) (1 + α) κ 1 /k a /k. (32) Now the same argument as in the proof of Lemma 5.2 in [7], with the obvious moifications, yiels P(Z K,k a,ɛ (m) Y (H C ) = N)[U(C)] N ν j=+1 ( N j )( ) j ν [ U(C) (1 + α)κ 1 /k a /k] N j [U(C)] j. 16

17 Here j enotes the number of hyperplanes generating the vertices of L, an ( j ) bouns the number of points of intersection of these hyperplanes; thus ( j ν ) estimates the possibilities to choose the vertices of L. The probability that the other N j hyperplanes intersecting C o not meet L is given by [U(C) U(L)] N j U(C) N+j, which is estimate using (32). Inserting the inequality in (26), we can continue as in [7], finally using U(C) = (c 14 m k a 1/k ) κ an α > c 30 ɛ (+3)/2. This completes the proof. 6 Proof of Theorem 1 From now on, the proofs follow essentially the lines of those given in [7]. We will, therefore, state only the necessary lemmas in their moifie forms an refer to the corresponing proofs in [7]. thus with Let a > 0 an ɛ (0, 1) be given. For h (0, 1] an m N we efine K,k a,ɛ,h K (m) := Ko : v k (K) a(1, 1 + h), η k (K) [m k, (m + 1) k ), ϑ(k) ɛ ; (33) Moreover, we put P(v k (Z) a(1, 1 + h), ϑ(z) ɛ) = m=1 q k,h a,ɛ (m) ( ) qa,ɛ k,h (m) := P Z K,k a,ɛ,h (m). (34) ( ) qa,ɛ k,h (m, n) := P Z K,k a,ɛ,h (m), f 1(Z) = n for n N; here f 1 (P ) enotes the number of facets of a polytope P. Then we have P(v k (Z) a(1, 1 + h), ϑ(z) ɛ) = m=1 n=+1 q k,h a,ɛ (m, n). (35) Finally, we efine R k,h a,ɛ (m, n) := (u 1,..., u n, t 1,..., t n ) (S 1 ) n (0, ) n : P (u (n), t (n) ) K,k a,ɛ,h (m), f ( 1 P (u(n), t (n) ) ) = n, H(u i, t i ) C for i = 1,..., n, where the ball C is again efine as in Lemma 3(a), for the given a, ɛ, m. Lemma 7. For m, n N, n + 1 an h (0, 1], qa,ɛ k,h (m, n) = (2 λ) n n! R k,h a,ɛ (m,n) exp 2 U ( P (u (n), t (n) ) ) λ (t 1 t n ) 1 t 1... t n σ(u 1 )... σ(u n ). 17

18 The proof is the same as that for Lemma 6.1 in [7], with the obvious necessary changes. Lemma 8. Let w > 0, h (0, 1/2) an r 1. Then k 1+h 1 x r exp wx x c 20 hw[1 + (expw/2 1) 1 ] k 2 1 x r exp wx x. After the substitution x = y, one can imitate the proof of Lemma 6.2 in [7] to obtain the result. The next (technical) lemma states that each boun for qa,ɛ k,1 yiels a boun for qa,ɛ k,h which is linear in h. This shoul be compare to the boun in Lemma 2 which is also linear in h. Lemma 9. For m N, h (0, 1/2) an a /k λ σ 0 > 0, q k,h a,ɛ (m) c 31 (, σ 0 )h a /k λ m k q k,1 a,ɛ (m). Again, the proof is obtaine by aapting the corresponing one from [7], namely that of Lemma 6.3. After applying Lemmas 7 an 8, we arrive at the inequality qa,ɛ k,h (m, n) (2 λ) n t(ζ) n c 32 hu(k(ζ, t(ζ)))λ n! U(m,n) ( ( ) ) exp 2 1 U(K(ζ, t(ζ)))λ 1 k 2 1 s n 1 exp 2 U(K(ζ, t(ζ)))λs s (t 1 t n 1 ) 1 t 1... t n 1 σ(u 1 )... σ(u n ), where U(m, n), t(ζ) an K(, ) are efine as in [7], with the obvious changes. Now we have to observe that v k (K(ζ, t(ζ))) = a, hence K(ζ, t(ζ)) K,k a,ɛ (m), which implies that c 33 m a /k U(K(ζ, t(ζ))) c 34 m k a /k, by Lemma 3. The estimation can now be complete as in [7]. The following lemma establishes an upper estimate for an unconitional probability. Lemma 10. Let ɛ (0, 1), h (0, 1/2) an a /k λ σ 0 > 0. Then P(v k (Z) a(1, 1 + h), ϑ(z) ɛ) ( c 35 (, ɛ, σ 0 )h exp 1 + (c 29 /2)ɛ (+3)/2) 2 κ 1 /k a /k λ. Here c 29 is the constant appearing in Lemma 6. The proof of Lemma 10 follows the one of Proposition 7.1 in [7] an uses Lemmas 9, 4 an 6, in this orer. 18

19 The choice (1 + β) = 1 + (c 29 /4)ɛ (+3)/2 in Lemma 2 immeiately proves Theorem 1 with b = a(1 + h) in the case h min(h 0, 1/2). As to arbitrary b a, we observe that Lemmas 2 an 10 have the same structure as Lemma 3.2 an Proposition 7.1, respectively, in [7]; they iffer only by the values of some parameters. It is, therefore, clear that Theorem 1 now follows precisely in the same way as Theorem 1 of [7] was prove. 7 Proof of Theorem 2 In this section, Y enotes a stationary isotropic Poisson hyperplane process in R with intensity λ > 0. For a convex boy K R an for n N 0, we have P(Y (H K ) = n) = [2κ 1 v 1(K)λ] n n! exp 2κ 1 v 1(K)λ, (36) by [7, (4)], where B is now the ball with surface area 1, thus B = (κ ) 1/( 1) B. Let Z o be the zero cell of the tessellation inuce by Y. Lemma 11. Let k 1,...,. For each β > 0, there are constants h 0 > 0, N N an c 36 > 0, epening only on β an, such that for a > 0 an 0 < h < h 0, ( ) N P(v k (Z o ) a(1, 1 + h)) c 36 h a 1/k λ exp (1 + β)2κ 1/k a 1/k λ. Proof. In orer to be able to essentially copy the proof of Lemma 2, we efine a measure ψ on R by ψ(a) := 1 1 A (tu) t σ(u) κ S 1 0 for Borel sets A R. Let X be the Poisson process in R with intensity measure λψ, an let Y be the hyperplane process efine by Y := H(x) : x X. Then Y is a Poisson process in the space H of hyperplanes, an for a Borel set A H we have E Y (A) = λψ(x R : H(x) A) = 2λ 1 A (H(u, t)) t σ(u). κ S 1 This shows that Y has the same intensity measure as Y. Since Y an Y are Poisson processes, they are stochastically equivalent. We can now repeat the proof of Lemma 2, where we replace Z by Z o an x by ψ(x). Further, we observe that ψ((ru)) = (κ ) 1 r σ(u) for u = 1 an ψ((ry)) = rψ(y) for r > 0. In the exponential, v (2(1 + β)b a ) (where B a = (a/κ ) 1/k B ) has to be replace by ψ(2(1 + β)b a ) = 2(1 + β)(a/κ ) 1/k. With these changes, the proof of Lemma 2 yiels the assertion of Lemma We will nee a stability version of the inequality (11). Lemma 12. There is a positive constant γ, epening only on the imension, such that for ɛ (0, 1) an every convex boy K R with r (K) ɛ, the inequality ( v 1 (K) 1 + γɛ (+3)/2) κ 1 1/k v k (K) 1/k (37) 19

20 hols for k = 2,...,. Proof. Without loss of generality, we assume that K has interior points, mean with 2 (the same as the unit ball), an Steiner point o. We put v i := v i (K) for i = 0,..., (v 0 = κ ) an use the Aleksanrov-Fenchel inequalities v 2 i v i 1v i+1 for i = 1,..., 1 (see [15]). Theorem an Lemma of [15] provie an improvement of the first of these inequalities, namely v 2 1 v 0 v 2 c 37 δ(k, B ) (+3)/2, where δ is the Hausorff istance. This can be rewritten as v 1 v ( c ) 37 δ(k, B ) (+3)/2 v 2 (1 + α) v 0 v 1 v 0 v 2 v 1 with α := c 38 δ(k, B ) (+3)/2, since v 0 v 2 v1 2 = c 39 (v 1 being a constant multiple of the mean with). From v 1 (1 + α) v 2 (1 + α) v k v 0 v 1 v k 1 we get an thus ( v1 If δ := δ(k, B ) 1/2, then (since ɛ < 1) v 0 ) k 1 (1 + α) k 1 v k v 1 v k 1 v k 1 0 v k 1 + c 38 δ(k, B ) (+3)/2. v k 1 v k 1 0 v k 1 + c 40 ɛ (+3)/2. If δ 1/2, let rb K sb where r is maximal an s is minimal. Then (by the efinition of the Hausorff metric) s 1 + δ an r 1 δ 1/2, hence ɛ r (K) (s/r) 1 4δ an thus v1 k v0 k c 41 ɛ (+3)/2. v k Both cases together give ( v c 42 ɛ (+3)/2) 1/k ( 1 1/k v 0 v 1/k k 1 + c 43 ɛ (+3)/2) κ 1 1/k 0 v 1/k k. Theorem 2 can now be prove in essentially the same way as Theorem 1, an we list only the necessary changes in Sections 5 an 6, in aition to those alreay mentione in the proof of Lemma 11. Definitions (23) an (24) are replace by K,k a,ɛ (m) := K K,k a (m) : r (K) ɛ, q k a,ɛ(m) := P(Z o K,k a,ɛ (m)). Lemma 13. Let k 1,...,. For m N an a 1/k λ σ 0, where σ 0 > 0 is a given constant, qa,ɛ(m) k c 44 (, σ 0 ) exp c 45 ma 1/k λ. (38) 20

21 Proof. In the proof of Lemma 4, the number U(C) is replace by v 1 (C). In the integrations, t 1 t is replace by t. Equation (28) now reas 1H(u, t) C, H(u, t) S = t σ(u) S 1 0 = v 1 (C) v 1 (S) = v 1 (C) κ 1 S, hence (29) reas p N ( N ) ( ) N 1 c 46ma 1/k. v 1 (C) Continuing as in the proof of Lemma 4, we arrive at (38). Lemma 14. Let k 2,...,. For m N, ɛ (0, 1) an a, λ > 0, ( qa,ɛ(m) k c 47 (, ɛ)m kν exp 1 + c 48 ɛ (+3)/2) 2κ 1/k a 1/k λ, where ν epens only on an ɛ. Proof. We use (26) an (27), with the changes alreay mae for the proof of Lemma 13. Assume that u 1,..., u N, t 1,..., t N are such that the inicator functions in (27) are all equal to one. By Lemma 12, ( v 1 (P (u (N), t (N) )) 1 + γɛ (+3)/2) κ 1 1/k a 1/k. We efine α as in the proof of Lemma 6 an put c 49 := γ/(2 + γ), so that α > c 49 ɛ (+3)/2. By Lemma 4.2 of [7] (with B = (κ ) 1/( 1) B ), there are ν = ν(, ɛ) vertices of P (u (N), t (N) ) such that their convex hull L = L(P (u (N), t (N) )) satisfies This gives v 1 (L) (1 α)v 1 (P (u (N), t (N) )). v 1 (L) (1 + α)κ 1 1/k a 1/k. As in the proof of Lemma 6 (an of Lemma 5.2 in [7]) we euce that By Lemma 3, P(Z o K,k a,ɛ (m) Y (H C ) = N)[v 1 (C)] N ν j=+1 ( N j )( ) j ν [ v 1 (C) (1 + α)κ 1 1/k a 1/k] N j [v1 (C)] j. v 1 (C) = c 50 m k a 1/k. The assertion is now obtaine as in the proofs quote above. The further proof again follows the lines of [7] an of the proof of Theorem 1. Definitions (33), (34) an (35) are replace by K,k a,ɛ,h K (m) := Ko : v k (K) a(1, 1 + h), η k (K) [m k, (m + 1) k ), r (K) ɛ, 21

22 Lemma 7 is replace by q k,h a,ɛ (m, n) = q k,h a,ɛ (m) := P q k,h a,ɛ (m, n) := P (2λ) n (κ ) n n! ( Z o K,k a,ɛ,h (m) ), ( Z o K,k a,ɛ,h (m), f 1(Z o ) = n R k,h a,ɛ (m,n) t 1... t n σ(u 1 )... σ(u n ). ). exp 2κ 1 v ( 1 P (u(n), t (n) ) ) λ This is a special case of Lemma 6.1 in [7]. Instea of Lemma 8, we use Lemma 6.2 from [7], with replace by k 2. Lemma 9 is replace by the following assertion. For m N, h (0, 1/2) an a 1/k λ σ 0 > 0, q k,h a,ɛ (m) c 51 (, σ 0 )h a 1/k λ m k q k,1 a,ɛ (m). The proof is obtaine by aapting the proof of Lemma 6.3 in [7]. In the course of the proof, one has to use that Lemma 3 implies c 52 ma 1/k v 1 (K) c 53 m k a 1/k for K K,k a,ɛ (m). The counterpart of Lemma 10 now reas as follows. Let ɛ (0, 1), h (0, 1/2) an a 1/k λ σ 0 > 0. Then P(v k (Z o ) a(1, 1 + h), r (Z o ) ɛ) ( c 54 (, ɛ, σ 0 )h exp 1 + (c 48 /2)ɛ (+3)/2) 2κ 1/k a 1/k λ. With these preliminaries, the proof of Theorem 2 can now be complete in the same way as that of Theorem 1. 8 Proof of Theorem 3 In large parts of the proofs of Theorems 1 an 2, only the following properties of the functional v k are use: it is monotone uner set inclusion, i.e., v k (K 1 ) v k (K 2 ) if K 1 K 2, positively homogeneous of egree k, i.e., v k (rk) = r k v k (K) for r 0, an continuous with respect to the Hausorff metric. These properties are share, with k = 1, by the inraius functionals ρ o an ρ. Hence, the corresponing parts of the proofs apply also to Theorem 3. Aitional properties of the function v k were only neee for the stability estimates of Lemmas 1 an 12 an for Lemma 3. We replace these lemmas by the following ones. Lemma 15. There is a positive constant γ, epening only on the imension, such that for ɛ (0, 1) an every convex boy K K o with ϑ(k) ɛ the inequality hols. U(K) (1 + γɛ (+1)/2 )κ ρ o (K) 22

23 Proof. Without loss of generality, we assume that ρ o (K) = 1. Let K R o B, where R o is minimal. First we assume that R o 3. Put R o = 1 + h, then ɛ ϑ(k) h 2. Proceeing similarly as in Case 1 of the proof of Lemma 1, we fin that U(K) U(B ) c 55 h (+1)/2 c 55 ɛ (+1)/2. Now suppose that R o > 3. Then there is a spherical cap A S 1 with σ(a) = c 56 > 0 on which h K 2. It follows that U(K) U(B ) 1 ( ) h K 1 σ c 57 ɛ (+1)/2, since ɛ < 1. Hence, both cases yiel U(K) U(B ) + c 58 ɛ (+1)/2 = A ( 1 + c 59 ɛ (+1)/2) κ ρ o (K). Lemma 16. There is a positive constant γ, epening only on the imension, such that for ɛ (0, 1) an every convex boy K K o with r (K) ɛ the inequality hols. v 1 (K) (1 + γɛ (+1)/2 )κ ρ(k) The proof is similar to that of Lemma 15. For a replacement of Lemma 3, let a > 0 be given. For m N we set K a(m) := K K o : ρ(k) a(1, 2), D(K)/ρ(K) [m, m + 1). Lemma 17. Let m N. Then (a) K K a(m) implies that K c 60 mab ; (b) there exists a measurable map K a(m) P o P v(p ) such that v(p ) is a vertex of P with v(p ) c 61 ma. A similar result hols with ρ o instea of ρ(k). The proof is immeiate. With these changes, the proofs of Theorems 1 an 2 yiel the proof of Theorem 3. References [1] Calka, P. (2002) The istributions of the smallest isks containing the Poisson-Voronoi typical cell an the Crofton cell in the plane. Av. Appl. Prob. 34, [2] Garner, R.J. an Vassallo, S. (1999) Stability of inequalities in the ual Brunn- Minkowski theory. J. Math. Anal. Appl. 231, [3] Golman, A. (1998) Sur une conjecture e D.G. Kenall concernant la cellule e Crofton u plan et sur sa contrepartie brownienne. Ann. Probab. 26,

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