Asymptotic shape of a random polytope in a convex body

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1 Asymptotic shape of a random polytope in a convex body N. Dafnis, A. Giannopoulos, and A. Tsolomitis Abstract Let K be an isotropic convex body in R n and let Z qk be the L q centroid body of K. For every N > n consider the random polytope K N := conv{x 1,..., x N } where x 1,..., x N are independent random points, uniformly distributed in K. We prove that a random K N is asymptotically equivalent to Z [lnn/n] K in the following sense: there exist absolute constants ρ 1, ρ 2 > 0 such that, for all β 0, 1 ] and all N Nn, β, one 2 has: i K N cβz qk for every q ρ 1 lnn/n, with probability greater than 1 c 1 exp c 2N 1 β n β. ii For every q ρ 2 lnn/n, the expected mean width E [wk N ] of K N is bounded by c 3 wz qk. As an application we show that the volume radius K N 1/n of a random ln2n/n K N satisfies the bounds c 4 K N 1/n ln2n/n c 5L K N expn. for all Keywords: convex body; isotropic body; isotropic constant; random polytope; centroid bodies; mean width; volume radius MSC: Primary 52A21; Secondary 46B07, 52A40, 60D05. 1 Introduction Let K be a convex body of volume 1 in R n. For every q 1 we define the L q centroid body Z q K of K by its support function: 1.1 h ZqKx =, x q := K y, x q dy 1/q. The aim of this article is to provide some precise quantitative information on the asymptotic shape of a random polytope K N = conv{x 1,..., x N } spanned by N independent random points x 1,..., x N uniformly distributed 1

2 in K. Our approach is to compare K N with the L q -centroid body Z q K of K for q lnn/n. The origin of our work is in the study of the behavior of symmetric random ±1 polytopes, the absolute convex hulls of random subsets of the discrete cube E2 n = { 1, 1} n. The natural way to produce these random polytopes is to fix N > n and to consider the convex hull K n,n = conv { ± X 1,..., ± X } N of N independent random points X1,..., X N, uniformly distributed over E2 n. It turns out see [10] that a random polytope K n,n has the largest possible volume among all ±1 polytopes with N vertices, at every scale of n and N. This is a consequence of the following fact: If n n 0 and if N nln n 2, then 1.2 K n,n c lnn/nb n 2 B n with probability greater than 1 e n, where c > 0 is an absolute constant, B2 n is the Euclidean unit ball in Rn and B n = [ 1, 1] n. In [17], Litvak, Pajor, Rudelson, and Tomczak-Jaegermann worked in a more general setting which contains the previous Bernoulli model and the Gaussian model; let K n,n be the absolute convex hull of the rows of the random matrix Γ n,n = ξ ij 1 i N, 1 j n, where ξ ij are independent symmetric random variables satisfying certain conditions ξ ij L 2 1 and ξ ij L ψ 2 ρ for some ρ 1, where L ψ 2 is the Orlicz norm corresponding to the function ψ 2 t = e t2 1. For this larger class of random polytopes, the estimates from [10] were generalized and improved in two ways: the paper [17] provides estimates for all N 1 + δn, where δ > 0 can be as small as 1/ ln n, and establishes the following inclusion: for every 0 < β < 1, 1.3 K n,n cρ β lnn/nb n 2 B n with probability greater than 1 exp c 1 n β N 1 β exp c 2 N. The proof in [17] is based on a lower bound of the order of N for the smallest singular value of the random matrix Γ n,n with probability greater than 1 exp cn. In a sense, both works correspond to the study of the size of a random polytope K N = conv{x 1,..., x N } spanned by N independent random points x 1,..., x N uniformly distributed in the unit cube Q n := [ 1/2, 1/2] n. The connection of the estimates 1.2 and 1.3 with L q -centroid bodies comes from the following observation. Remark. For x R n and t > 0, define 1.4 K 1,2 x, t := inf { u 1 + t x u 2 : u R n }. 2

3 If we write x j j n for the decreasing rearrangement of x j j n we have Holmstedt s approximation formula [t 1 2 1/2 ] 1.5 c K n 1,2x, t x j + t x j 2 K 1,2 x, t j=1 j=[t 2 ]+1 where c > 0 is an absolute constant see [15]. Now, for any α 1 define Cα = αb n 2 Bn. Then, 1.6 h Cα θ = K 1,2 θ, α for every θ S n 1. On the other hand, 1.7, θ L q Q n θj + q θj 2 j q q<j n for every q 1 see, for example, [7]. In other words, 1.8 C q Z q Q n where Z q K is the L q -centroid body of K. This shows that 1.3 or 1.2 can be written in the form 1.9 K n,n cρz β lnn/n Q n. This observation leads us to consider a random polytope K N = conv{x 1,..., x N } spanned by N independent random points x 1,..., x N uniformly distributed in an isotropic convex body K and try to compare K N with Z q K for a suitable value q = qn, n lnn/n. Our first main result states that an analogue of 1.9 holds true in full generality. Theorem 1.1 Let β 0, 1/2] and γ > 1. If 1.10 N Nγ, n = cγn, where c > 0 is an absolute constant, for every isotropic convex body K in R n we have 1.11 K N c 1 Z q K for all q c 2 β lnn/n, with probability greater than exp c 3 N 1 β n β P Γ : l n 2 l N 2 γl K N, where Γ : l n 2 ln 2 is the random operator Γy = x 1, y,... x N, y defined by the vertices x 1,..., x N of K N. 3 1/2

4 The proof of Theorem 1.1 is given in Section 2, where we also collect what is known about the probability P Γ : l n 2 ln 2 γl K N which appears in It should be emphasized that a reverse inclusion of the form K N c 4 Z q K cannot be expected with probability close to 1, unless q is of the order of n. This follows by a simple volume argument which makes use of the upper estimate of Paouris see [21] for the volume of Z q K and is presented in Section 3. However, one can easily see that K N is weakly sandwiched between Z qi K i = 1, 2, where q i lnn/n, in the following sense: Proposition 1.2 For every α > 1 one has 1.13 E [ σθ : h KN θ αh ZqKθ ] Nα q. This shows that if q c 5 ln N then for most K N, for most θ S n 1 we have h KN θ c 6 h ZqKθ. It follows that several geometric parameters of K N, e.g. the mean width, are controlled by the corresponding parameter of Z [lnn/n] K. As an application, we discuss the volume radius of K N : Let K be a convex body of volume 1 in R n. The question to estimate the expected volume radius 1.14 EK, N = K convx 1,..., x N 1/n dx N dx 1 K of K N was studied in [13] where it was proved that for every isotropic convex body K in R n and every N n + 1, 1.15 EBn, N EK, N cl K ln2n/n, where Bn is a ball of volume 1. This estimate is rather weak for large values of N: a strong conjecture is that { } ln2n/n 1.16 EK, N min, 1 n L K for every N n + 1. This was verified in [11] in the unconditional case, where it was also shown that the general problem is related to the ψ 2 - behavior of linear functionals on isotropic convex bodies. Using a recent result of G. Paouris [22] on the negative moments of the support function of h ZqK we can settle the question for the full range of values of N. 4

5 Theorem 1.3 For every N expn, one has 1.17 c 4 ln2n/n K N 1/n c 5 L K ln2n/n with probability greater than 1 1 N, where c 4, c 5 > 0 are absolute constants. Notation and terminology. We work in R n, which is equipped with a Euclidean structure,. We denote by 2 the corresponding Euclidean norm, and write B2 n for the Euclidean unit ball, and Sn 1 for the unit sphere. Volume is denoted by. We write ω n for the volume of B2 n and σ for the rotationally invariant probability measure on S n 1. We also write A for the homothetic image of volume 1 of a convex body A R n, i.e. A := A/ A 1/n. A convex body is a compact convex subset C of R n with non-empty interior. We say that C is symmetric if x C whenever x C. We say that C has center of mass at the origin if C x, θ dx = 0 for every θ Sn 1. The support function h C : R n R of C is defined by h C x = max{ x, y : y C}. The mean width of C is defined by 1.18 wc = h C θσdθ. S n 1 The radius of C is the quantity RC = max{ x 2 : x C}, and the polar body C of C is 1.19 C := {y R n : x, y 1 for all x C}. Whenever we write a b, we mean that there exist universal constants c 1, c 2 > 0 such that c 1 a b c 2 a. The letters c, c, c 1, c 2 > 0 etc., denote universal positive constants which may change from line to line. A convex body K in R n is called isotropic if it has volume K = 1, center of mass at the origin, and there is a constant L K > 0 such that 1.20 x, θ 2 dx = L 2 K K for every θ in the Euclidean unit sphere S n 1 2. For every convex body K in R n there exists an affine transformation T of R n such that T K is isotropic. Moreover, if we ignore orthogonal transformations, this isotropic image is unique, and hence, the isotropic constant L K is an invariant of the affine class of K. We refer to [19] and [9] for more information on isotropic convex bodies. 5

6 2 The main inclusion In this Section we prove Theorem 1.1. Let K be an isotropic convex body in R n. For every q 1 consider the L q -centroid body Z q K of K; recall that 1/q 2.1 h ZqKx =, x q := y, x dy q. Since K = 1, we readily see that Z 1 K Z p K Z q K Z K for every 1 p q, where Z K = conv{k, K}. On the other hand, one has the reverse inclusions 2.2 Z q K cq p Z pk for every 1 p < q <, as a consequence of the ψ 1 -behavior of y y, x. Observe that Z q K is always symmetric, and Z q T K = T Z q K for every T SLn and q [1, ]. Also, if K has its center of mass at the origin, then Z q K cz K for all q n, where c > 0 is an absolute constant. We refer to [9] for proofs of these assertions and further information. Lemma 2.1 Let 0 < t < 1. For every θ S n 1 one has 2.3 P {x K : x, θ t, θ q } 1 tq 2 C q. Proof. We apply the Paley-Zygmund inequality 2.4 P gx t q E g 1 t q 2 [E g]2 E g 2 for the function gx = x, θ q. Since, by 2.2, 2.5 E g 2 = E x, θ 2q C q E x, θ q 2 = C q [E g] 2 K for some absolute constant C > 0, the lemma is proved. Lemma 2.2 For every σ {1,..., N} and any θ S n 1 one has 2.6 P { X = x 1,..., x N K N : max x j, θ 1 j σ 2, θ q} exp σ /4C q, where C > 0 is an absolute constant. 6

7 Proof. Applying Lemma 2.1 with t = 1/2 we see that P max x j, θ 1 j σ 2, θ q = P x j, θ 1 2, θ q j σ 1 1 σ 4C q exp σ /4C q, since 1 v < e v for every v > 0. Proof of Theorem 1.1. Let Γ : l n 2 ln 2 by be the random operator defined 2.7 Γy = x 1, y,..., x N, y. We modify an idea from [17]. Define m = [5N/n β ] + 1 and k = [N/m]. Fix a partition σ 1,..., σ k of {1,..., N} with m σ i for all i = 1,..., k and define the norm 2.8 u 0 = 1 k k P σi u. Since 2.9 h KN z = max 1 j N x j, z P σi Γz for all z R n and i = 1,..., k, we observe that 2.10 h KN z Γz 0. If z R n and Γz 0 < 1 4, z q, then, Markov s inequality implies that there exists I {1,..., k} with I > k/2 such that P σi Γz < 1 2, z q, for all i I. It follows that, for fixed z S n 1 and α 1, 7

8 P Γz 0 < 1 4, z q Choosing I >k/2 I >k/2 i I P P σi Γz < 1 2, z q, for all i I I >k/2 i I k j>k/2 P P σi Γz < 1 2, z q exp σ i /4C q k j exp km/8c q exp k ln 2 km/8c q q β lnn/n we see that 2.12 P Γz 0 < 1 4, z q exp c 2 N 1 β n β. Let S = {z :, z q /2 = 1} and consider a δ-net U of S with cardinality U 3/δ n. For every u U we have 2.13 P Γu 0 < 1 exp c 2 N 1 β n β, 2 and hence, 2.14 P u U Fix γ > 1 and set { Γu 0 < 2} 1 exp n ln3/δ c 2 N 1 β n β Ω γ = {Γ : Γ : l n 2 l N 2 γl K N}. Since Z q K cl K B2 n, we have 2.16 Γz 0 1 k Γz 2 cγl K N/k z 2 cγ N/k, z q 8

9 for all z R n and all Γ in Ω γ. Let z S. There exists u U such that 1 2, z u q < δ, which implies that 2.17 Γu 0 Γz 0 + cγδ N/k on Ω γ. Now, choose δ = k/n/4cγ. Then, P {Γ Ω γ : z R n : Γz 0, z q /8} = P {Γ Ω γ : z S : Γz 0 1/4} P {Γ Ω γ : u U : Γu 0 1/2} exp n ln12cγ N/k c 2 N 1 β n β exp c 3 N 1 β n β provided that N is large enough. Since h KN z Γz 0 for every z R n, we get that K N cz q K with probability greater than 1 exp c 4 N 1 β n β P Γ : l n 2 ln 2 γl K N. We now analyze the restriction for N; we need n ln12c 4 γ N/k CN 1 β n β for some suitable constant C > 0. Assuming 2.18 N 12cγn, and since β 0, 1 2 ], using the definitions of k and m we see that it is enough to guarantee lnn/n C N/n, which is valid if N/n c 5 for a suitable absolute constant c 5 > 0. We get the result taking 2.18 into account. Remark 2.3 The statement of Theorem 1.1 raises the question to estimate the probability 2.19 PΩ γ = P Γ : l n 2 l N 2 γl K N. In [17] it was proved that if Γ n,n = ξ ij 1 i N, 1 j n is a random matrix, where ξ ij are independent symmetric random variables satisfying ξ ij L 2 1 and ξ ij L ψ 2 ρ for some ρ 1, then PΩ γ exp cρ, γn. In our case, Γ is a random N n matrix whose rows are N uniform random points from 9

10 an isotropic convex body K in R n. Then, the question is to estimate the probability that, N random points x 1,..., x N from K satisfy N N x j, θ 2 γ 2 L 2 K j=1 for all θ S n 1. This is related to the following well-studied question of Kannan, Lovász and Simonovits [16] which has its origin in the problem of finding a fast algorithm for the computation of the volume of a given convex body: given δ, ε 0, 1, find the smallest positive integer N 0 n, δ, ε so that if N N 0 then with probability greater than 1 δ one has εl 2 K 1 N N x j, θ εl 2 K j=1 for all θ S n 1. In [16] it was proved that one can have N 0 cδ, εn 2, which was later improved to N 0 cδ, εnln n 3 by Bourgain [3] and to N 0 cδ, εnln n 2 by Rudelson [25]. One can actually check see [12] that this last estimate can be obtained by Bourgain s argument if we also use Alesker s concentration inequality. For subsequent developments, see see, for instance, [21], [14], [18] and [2]. Here, we are only interested in the upper bound of 2.21; actually, we need an isomorphic version of this upper estimate, since we are allowed to choose an absolute constant γ 1 in An application of the main result of [18] to the isotropic case gives such an estimate: If N c 1 n ln 2 n, then 1/ P Γ : l n 2 l N 2 γl K N exp c 2 γ N ln Nn ln n A slightly better estimate can be extracted from the work of Guédon and Rudelson in [14]. It should be emphasized that this term does not allow us to fully exploit the second term exp c 3 N 1 β n β in the probability estimate of Theorem 1.1. However, it is not clear if it is optimal. Recently see [1], Theorem 3.5 it was proved that for every n N e c n we have p e c n. Remark 2.4 G. Paouris and E. Werner [23] have recently studied the relation between the family of L q -centroid bodies and the family of floating bodies of a convex body K. Given δ 0, 1 2 ], the floating body K δ of K. 10

11 is the intersection of all halfspaces whose defining hyperplanes cut off a set of volume δ from K. It was observed in [19] that K δ is isomorphic to an ellipsoid as long as δ stays away from 0. In [23] it is proved that 2.23 c 1 Z ln1/δ K K δ c 2 Z ln1/δ K where c 1, c 2 > 0 are absolute constants. From Theorem 1.1 it follows that if K is isotropic and if, for example, N n 2 then 2.24 K N c 3 K 1/N with probability greater than 1 o n 1, where c 3 > 0 is an absolute constant. This fact should be compared with the following well-known result from [4]: for any convex body K in R n one has c K 1/N E K N c n K 1/N where the constant on the left is absolute and the right hand side inequality holds true with a constant c n depending on the dimension, for N large enough; the critical value of N is exponential in n. 2.1 Unconditional case In this subsection we consider separately the case of unconditional convex bodies: we assume that K is centrally symmetric and that, after a linear transformation, the standard orthonormal basis {e 1,..., e n } of R n is a 1- unconditional basis for K, i.e. for every choice of real numbers t 1,..., t n and every choice of signs ε j = ±1, 2.25 ε1 t 1 e ε n t n e K n = t1 e t n e K n. Then, a diagonal operator brings K to the isotropic position. It is also known that the isotropic constant of an unconditional convex body K satisfies L K 1. Bobkov and Nazarov have proved that K c 2 Q n, where Q n = [ 1 2, 1 2 ]n see [5]. The following argument of R. Latala private communication shows that the family of L q -centroid bodies of the cube Q n is extremal in the sense that Z q K cz q Q n for all q 1, where c > 0 is an absolute constant: Let ε 1, ε 2,..., ε n be independent and identically distributed ±1 random variables, defined on some probability space Ω, F, P, with distribution Pε i = 1 = Pε i = 1 = 1 2. For every θ Sn 1, by the unconditionality of K, Jensen s inequality and the contraction principle, one 11

12 has, θ L q K = K Ω Q n n q 1/q n q 1/q θ i x i dx = θ i ε i x i dx dpε Ω K n q 1/q n q θ i ε i x i dx dpε = t i θ i ε i dpε K Ω n q 1/q t i θ i y i dy =, tθ L q Q n, where t i = K x i dx and tθ = t 1 θ 1,..., t n θ n. Since t i 1 for all i = 1,..., n, from 1.7 we readily see that, θ L q K, tθ L q Q n c, θ L q Q n. In view of 1.8, this observation and Theorem 1.1 show that, if K is unconditional, then a random K N contains Z lnn/n Q n : Theorem 2.5 Let β 0, 1/2] and γ > 1. There exists an absolute constant c > 0 so that if 2.26 N Nγ, n = cγn, and if K N = conv{x 1,..., x N } is a random polytope spanned by N independent random points x 1,..., x N uniformly distributed in an unconditional isotropic convex body K in R n, then we have 2.27 K N c 1 Cα = c 1 αb n 2 B n for all α c 2 β lnn/n, with probability greater than exp c 3 N 1 β n β P Γ : l n 2 l N 2 γ N, where Γ : l n 2 ln 2 is the random operator Γy = x 1, y,... x N, y defined by the vertices x 1,..., x N of K N. Next, we outline a direct proof of Theorem 2.5 in which L q -centroid bodies are not involved: For k N and y R n, define 2.29 y P k := sup k j Bi y 2 j 1/2 : k B i = [n], B i B j = i j, 1/q 12

13 where we write [n] for the set {1, 2,..., n}. Montgomery Smith has shown see [20] that: For any y R n and k N, one has n 1 k 2.30 P ε i y i λ y P k 1 2λ 2 2k 0 λ 1/ 2. 3 Also, for y R n, one has 2.31 y P t 2 K 1,2 y, t 2 y P t 2 when t 2 N, from where one concludes the following: Lemma 2.6 There exists a constant c > 0 such that, for all y R n and any t > 0, n 2.32 P ε i y i λk 1,2 y, t e φλt2, where φλ = 4 ln 31 2λ 2 2 for 0 < λ < 1/ 2. P. Pivovarov [24] has recently obtained the following result: There exists an absolute constant C 1 such that for any unconditional isotropic convex body K in R n, the spherical measure of the set of θ S n 1 such that P x, θ t exp Ct 2 whenever C t n C ln n, is at least 1 2 n. The proof of the next Lemma follows more or less the same lines. Lemma 2.7 Let K be an isotropic unconditional convex body in R n. For every θ S n 1 and any α 1 we have 2.33 P x x, θ c0 h Cα θ c 1 e c 2α 2, where c 0 and c 1 are absolute positive constants. Proof. For θ = θ i n Sn 1 and 0 < s < 1/ 2 define the set 2.34 K s θ = {x K : K 1,2 θ, α sk 1,2 xθ, α}, 13

14 where by xθ we mean the vector with coordinates x i θ i and s is to be chosen. We have: P x n x i θ i c 0 h Cα θ n = P x ε i x i θ i c 0 h Cα θ n = P ε ε i x i θ i c 0 h Cα θ dx K n = P ε ε i x i θ i c 0 K 1,2 θ, α dx K n ε i x i θ i sk 1,2 xθ, α dx K c 1 0 sθ P ε e φsα2 K c 1 0 sθ, by Lemma 2.6. Assume first that m := α 2 is an integer and let B 1, B 2,..., B m be a partition of the set {1, 2,..., n} so that 2.35 K 1,2 θ, α = m θ j 2 j Bi 1/2 =: A. Consider the seminorm 2.36 fx = m x j θ j 2 j Bi 1/2, and observe that fx K 1,2 xθ, α. Then, using the reverse Hölder inequality c 1 f L 2 K f L 1 K and the fact that L K 1, we get K m fx dx = 1/2 x j θ j 2 K j Bi m c 1 θ j 2 x j 2 K j Bi c 2 A. 1/2 14

15 We now apply the Paley-Zygmund inequality to get 2.37 K c 1 0 sθ P x f > c0 s 1 A E f 2 c 0 s 1 A 2 2 E [f 4. ] Choosing s = and c 0 > 0 so that c 0 < c 2 s we get ca4 K c 1 0 sθ E [f 4 ], for a suitable new absolute constant c > 0. On the other hand, we can estimate E [f 4 ] from above, by the reverse Hölder inequality: As a result, K c 1 0 E [f 4 ] 1/4 4cE f = 4c m E x j θ j 2 j Bi m 4cL K θ j 2 j Bi 1/2 sθ c. Returning to the estimate 2.38 P x n x i θ i c 0 h Cα θ we get: 2.39 P x n x i θ i c 0 h Cα θ 4cA. 1/2 e φsα2 K c 1 0 sθ, ce cα2. This proves the Lemma for α 2 N and the result follows easily for every α. Proof of Theorem 2.5. Now, using the procedure of the proof of Theorem 1.1 we complete the proof of Theorem 2.5. Remark 2.8 Regarding the probability P Γ : l n 2 ln 2 γ N, in the unconditional case Aubrun has proved in [2] that for every ρ > 1 and N ρn, one has 2.40 P Γ : l n 2 l N 2 c 1 ρ N exp c 2 ρn 1/5. 15

16 In particular, one can find c, C > 0 so that, if N Cn, then 2.41 P Γ : l n 2 l N 2 C N exp cn 1/5. This allows us to use Theorem 2.5 with a probability estimate 1 exp cn c for values of N which are proportional to n. 3 Weakly sandwiching K N We proceed to the question whether the inclusion given by Theorem 1.1 is sharp. It was already mentioned in the Introduction that we cannot expect a reverse inclusion of the form K N c 4 Z q K with probability close to 1, unless if q is of the order of n. To see this, observe that, for any α > 0, P K N αz q K = P x 1, x 2,..., x N αz q K = P x αz q K N αz q K N. It was proved in [21] that, for every q n, the volume of Z q K is bounded by c q/nl K n. This leads immediately to the estimate 3.1 P K N αz q K cα q/nl K nn, where c > 0 is an absolute constant. Assume that K has bounded isotropic constant and we want to keep α 1. Then, 3.1 shows that, independently from the value of N, we have to choose q of the order of n so that it might be possible to show that P K N αz q K is really close to 1. Actually, if q n then this is always the case, because Z n K ck. Lemma 3.1 Let K be a convex body of volume 1 in R n and let N > n. Fix α > 1. Then, for every θ S n 1 one has 3.2 P h KN θ αh ZqKθ Nα q. Proof. Markov s inequality shows that 3.3 Pα, θ := P x K : x, θ α, θ q α q. Then, P h KN θ αh ZqKθ = P max j N x j, θ α, θ q N Pα, θ and the result follows. 16

17 Lemma 3.2 Let K be a convex body of volume 1 in R n and let N > n. For every α > 1 one has 3.4 E [ σθ : h KN θ αh ZqKθ ] Nα q. Proof. Immediate: observe that E [ σθ : h KN θ αh ] ZqKθ = P h KN θ αh ZqKθ dσθ S n 1 by Fubini s theorem. The estimate of Lemma 3.2 is already enough to show that if q c ln N then, on the average, h KN θ ch ZqKθ with probability greater than 1 N c. In particular, the mean width of a random K N is bounded by the mean width of Z lnn/n K: Proposition 3.3 Let K be an isotropic convex body in R n. If q 2 ln N then 3.5 E [ wk N ] c wz q K, where c > 0 is an absolute constant. Proof. Using the fact that K N K n + 1L N B2 n for the latter inclusion see [9], we write 3.6 wk N h KN θ dσθ + cσa c NnL K, A N where A N = {θ : h KN θ αh ZqKθ}. Then, 3.7 wk N α h ZqKθ dσθ + cσa c NnL K, A N and hence, by Lemma 3.2, 3.8 E wk N αwz q K + cnnα q L K. Since wz q K wz 2 K = L K, we get 3.9 E wk N α + cnnα q wz q K. The result follows if we choose α = e. 17

18 3.1 Volume radius of K N Next, we discuss the volume radius of K N. A lower bound follows by comparison with the Euclidean ball. It was proved in [13, Lemma 3.3] that if K is a convex body in R n with volume 1, then 3.10 P K N t P [B n 2 ] N t for every t > 0. Therefore, it is enough to consider the case of B2 n. In [11] it is shown that there exist c 1 > 1 and c 2 > 0 such that if N c 1 n and x 1,..., x N are independent random points uniformly distributed in B n 2, then { } 3.11 [B n ln2n/n 2 ] N c 2 min, 1 B n n 2 with probability greater than 1 exp n. It follows that if N c 1 n then, with probability greater than 1 exp n we have { } ln2n/n 3.12 K N 1/n c 2 min, 1, n where c 1 > 1 and c 2 > 0 are absolute constants. The case n < N < c 1 n was studied in [8] where it was proved that 3.11 continues to hold true with probability greater than 1 exp cn/ ln n, where c > 0 is an absolute constant. Combining this fact with 3.10, we see that 3.12 is valid for all N > n. We now pass to the upper bound; Proposition 3.3, combined with Urysohn s inequality, yields the following: Proposition 3.4 Let K be an isotropic convex body in R n. If N > n and q 2 ln N, then 3.13 EK, N c 1E [wk N ] where c 1, c 2 > 0 are absolute constants. c 2 wz q K, Proposition 3.4 reduces, in a sense, the question to that of giving upper bounds for wz q K. It is proved in [21] that, if q = ln N then wz q K c ql K. It follows that for N > n 1+δ 3.14 EK, N c lnn/nlk, 18

19 which is the conjectured estimate for N e n. Actually the restriction N > n 1+δ is not needed for 3.14 as shown in [8], and N needs only to satisfy n + 1 N ne n. For q = ln N > we know that wz q K ql K / 4 since Z q K q/ Z K. This is most probably a nonoptimal bound. However, we can further exploit the simple estimate of Lemma 3.1 to obtain a sharp estimate for larger values of N. We will make use of the following facts: Fact 1. Let A be a symmetric convex body in R n. For any 1 q < n, set 1 1/q 3.15 w q A = S n 1 h q dσθ. A θ An application of Hölder s inequality shows that 3.16 A 1/n 1 1/n 1 1/q B2 n = S n 1 h n dσθ A θ S n 1 h q dσθ = A θ From the Blaschke Santaló inequality, it follows that 3.17 A 1/n B n 2 1/n w q A c 1w q A. 1 w q A. Fact 2. A recent result of G. Paouris see [22, Proposition 5.4] shows that if A is an isotropic convex body in R n then, for any 1 q < n/2, q 3.18 w q Z q A I q A n where 1/p 3.19 I p A = x p 2 dx, A p > n. Fact 3. Let K be an isotropic convex body in R n, let N > n 2 and q = 2 ln2n. We write 2 [w q/2 Z q K] q = 1 S n 1 h q/2 dσθ Z qk θ S n 1 1 h q K N θ dσθ 19 q h S n 1 h q Z qk K N θ dσθ. θ

20 Observe that K N K n + 1L K and Z q K Z 2 K L K B n 2, and hence, h KN θ n + 1h ZqKθ for all θ S n 1. Therefore, 3.20 q h S n 1 h q Z qk K N θ dσθ = θ n+1 0 qt q 1 [ σ θ : h KN θ th ZqKθ ] dt. Fact 4. Taking expectations in 3.20 and using Lemma 3.2, we see that, for every a > 1, [ ] q hk E N θ n+1 S n 1 h q dσθ a q + qt q 1 Nt q dt Z qk θ a n + 1 = a q + qn ln. a Choosing a = 2e and using the fact that e q = 2N 2 by the choice of q, we see that [ ] q hk 3.21 E N θ dσθ c q 2 θ S n 1 h q Z qk where c 2 > 0 is an absolute constant. Then, Markov s inequality implies that q hk 3.22 N θ h q Z θ dσθ c 2e q qk S n 1 with probability greater than 1 e q. Going back to Fact 3, we conclude that [w q/2 Z q K] q c q 3 [w qk N ] q, i.e w q K N c 4 w q/2 Z q K with probability greater than 1 e q. Proof of Theorem 1.3. Assume that K N satisfies 3.23 and set S N = K N K N. From Fact 1 we have 3.24 K N 1/n S N 1/n c 1 w q S N = 2c 1 w q K N. Now, Fact 4 shows that 3.25 K N 1/n c 5 w q/2 Z q K 20

21 with probability greater than 1 e q. Since Z q K cz q/2 K, using Fact 2 we write 3.26 w q/2 Z q K c 6 w q/2 Z q/2 K c 7 q I n q/2 K. Since K is isotropic, we have I q/2 K I 2 K = L K, which implies 3.27 w q/2 Z q K c 7 q LK. Putting everything together, we have 3.28 K N 1/n c q L K lnn/nlk, with probability greater than 1 e q 1 1 N. This completes the proof. Acknowledgement. We would like to thank Grigoris Paouris for bringing to our attention his formula on negative moments of the support function of centroid bodies. References [1] R. Adamczak, A.E. Litvak, A. Pajor, and N. Tomczak-Jaegermann, Quantitative estimates of the convergence of the empirical covariance matrix in log-concave ensembles preprint, arxiv: v1. [2] G. Aubrun, Sampling convex bodies: a random matrix approach, Proc. Amer. Math. Soc , [3] J. Bourgain, Random points in isotropic convex bodies, in Convex Geometric Analysis Berkeley, CA, 1996 Math. Sci. Res. Inst. Publ , [4] I. Bárány and D. G. Larman, Convex bodies, economic cap coverings, random polytopes, Mathematika , [5] S. G. Bobkov and F. L. Nazarov, On convex bodies and log-concave probability measures with unconditional basis, Geom. Aspects of Funct. Analysis Milman- Schechtman eds., Lecture Notes in Math , [6] S. G. Bobkov and F. L. Nazarov, Large deviations of typical linear functionals on a convex body with unconditional basis, Stochastic Inequalities and Applications, Progr. Probab. 56, Birkhäuser, Basel, 2003,

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23 [24] P. Pivovarov, Volume of caps and mean width of isotropic convex bodies, Preprint. [25] M. Rudelson, Random vectors in the isotropic position, J. Funct. Anal , N. Dafnis: Department of Mathematics, University of Athens, Panepistimiopolis , Athens, Greece. A. Giannopoulos: Department of Mathematics, University of Athens, Panepistimiopolis , Athens, Greece. A. Tsolomitis: Department of Mathematics, University of the Aegean, Karlovassi , Samos, Greece. 23