A stability result for mean width of -centroid bodies

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Advances in Mathematics 214 (2007) 865877www.elsevier.com/locate/aimA stability result for mean width of Lp-centroid bodiesB. Fleury a, O. Gudon a,, G. Paouris b,1a Universit Pierre et Marie Curie, Institut de Mathmatiques de Jussieu, bote 186, 4 Place Jussieu,75252 Paris Cedex 05, Franceb Universit de Marne-la-Valle, Laboratoire dAnalyse et de Mathmatiques Appliques, 5 Bd Descartes,Champs-sur-Marne, 77454 Marne-la-Valle Cedex 2, FranceReceived 16 October 2006; accepted 22 March 2007Available online 30 March 2007Communicated by Michael J. HopkinsAbstractWe give a different proof of a recent result of Klartag [B. Klartag, A central limit theorem for convexsets, Invent. Math. 168 (1) (2007) 91131] concerning the concentration of the volume of a convex bodywithin a thin Euclidean shell and proving a conjecture of Anttila, Ball and Perissinaki [M. Anttila, K. Ball,I. Perissinaki, The central limit problem for convex bodies, Trans. Amer. Math. Soc. 355 (12) (2003) 47234735]. It is based on the study of the Lp-centroid bodies. We prove an almost isometric reverse Hlderinequality for their mean width and a refined form of a stability result. 2007 Elsevier Inc. All rights reserved.Keywords: Lp-centroid bodies; Reverse Hlder inequality; Concentration hypothesis; Variance hypothesis1. IntroductionIn this paper we study how the volume of a symmetric convex body concentrates within a verythin Euclidean shell. Let K be an isotropic convex body in Rn, i.e. a symmetric convex body ofvolume 1 such that for some fixed LK > 0,* Corresponding author.E-mail addresses: fleury_bruno@yahoo.fr (B. Fleury), guedon@math.jussieu.fr (O. Gudon),grigoris_paouris@yahoo.co.uk (G. Paouris).1 Research supported by a Marie Curie Intra-European Fellowship (EIF), Contract MEIF-CT-2005-025017.0001-8708/$ see front matter 2007 Elsevier Inc. All rights reserved.doi:10.1016/j.aim.2007.03.008866 B. Fleury et al. / Advances in Mathematics 214 (2007) 865877 Sn1,Kx, 2 dx = L2K.It is known that every symmetric convex body has an affine image which is isotropic. We denoteby |x|2 the Euclidean norm of x Rn. In the paper [1], Anttila, Ball and Perissinaki asked ifevery isotropic convex body in Rn satisfy an -concentration hypothesis namely:Concentration hypothesis. Does there exist n such that limn n = 0 and{x K, |x|2LKn 1 n} n?We will prove the following.Theorem 1. There exists c and c such that for every isotropic convex body K in Rn, and everyp (logn)1/3,1(K|x|p2 dx)1/p/(K|x|2 dx) 1 + cp/(logn)1/3.In particular, for every (0,1),{x K, |x|2nLK 1 } 2ec(logn)1/12 . (1)This implies that the concentration hypothesis holds with n = c(log logn)2/(logn)1/6. Thisresult has been very recently obtained in full generality by Klartag [12], where he proved that (1)holds true with 2e2 logn for every isotropic convex body with center of mass at the origin. Ourgoal is to present a different approach via the notion of Lp-centroid bodies. To any star-shapebody with respect to the origin, L Rn, we associate its Lp-centroid body Zp(L) which is asymmetric convex body defined by its support function:y Rn, hZp(L)(y) =(Lx, yp dx)1/p.This body is homothetic to the Lp-centroid body defined by Lutwak and Zhang in [15] (seealso [16]). For any symmetric convex body C, we define the pth mean width asWp(C) =( Sn1hC()p d())1/p.The main result of this paper compares the mean width of the Lp-centroid bodies of an isotropicconvex body to the mean width of the Lp-centroid bodies of the Euclidean unit ball of volume 1.B. Fleury et al. / Advances in Mathematics 214 (2007) 865877 867Theorem 2. There exists a constant c such that for any n, for every isotropic convex body Kin Rn, if D denotes the Euclidean unit ball in Rn of volume 1, for every p (logn)1/3W1(Zp(K))W1(Z1(K))W1(Z1(D))W1(Zp(D)) 1 + cp/(logn)1/3.Regarding K as a probability space, these techniques were used by the third named author[20] to prove that the Lq -norms of the Euclidean norm are almost constant for any q n, i.e.(see Theorem 1.2 in [20])C 1, q cn,(K|x|q2 dx)1/q C(K|x|22 dx)1/2= CnLK. (2)Theorem 1 is in fact an almost isometric version of this result (although it does not recover thefull isomorphic one). It is also related to a weak form of Kannan, Lovsz and Simonovits [11]conjecture about the Cheeger-type isoperimetric constant for convex bodies: does there existc > 0 such that for any isotropic convex body K , 2K :=Var(|X|22)nL4K cwhere X is a random vector uniformly distributed on K? We refer to the paper of Bobkov [4]for more details between the full KLS-conjecture and this weaker form. Theorem 1 implies thatlimn 2K/n = 0. Up to now, the only known upper bound was the trivial one, K cn.On the way, we will need a new type of stability result for the Lp-centroid bodies. Let K andL be symmetric convex bodies of volume 1 in Rd , if Zp(L) is close to Zp(K) for the geometricdistance, what can we say about the geometric distance between K and L? This type of questionhas been studied by Bourgain and Lindenstrauss [6] in the case of projection bodies, i.e. p = 1.We will prove a more precise result when one of the bodies is the Euclidean unit ball D. Thegeometric distance between two symmetric convex bodies K and L is defined byd(K,L) = inf{ab | a, b > 0 and 1/aK L bK}.Theorem 3. There exists c > 0 such that for every integer d greater than 3 and any odd integerp d , we have the following property: if K is a symmetric convex body in Rd such that for some > 1 and (0, (c)2d3)d(K,D) and d(Zp(D),Zp(K)) 1 + ,where K = |K|1/dK and D = |D|1/dD thend(K,D) 1 + h() and (1 h())Zp(D) Zp(K) (1 + h())Zp(D)where h() = (c)d+p+11/d2 .868 B. Fleury et al. / Advances in Mathematics 214 (2007) 865877It was proved in [1] that the concentration hypothesis implies some type of central limittheorem. The conjecture about a central limit theorem for convex sets stated by Anttila, Ball,Perissinaki [1] and Brehm, Voigt [7] has been recently proved by Klartag [12] and we refer tothat paper for more precise references on this subject.The paper is organized as follows. In Section 2, we shall explain how we reduce the study ofconcentration of the volume of an isotropic convex body to the study of its Lp-centroid bodies.We shall prove the main Theorem 2 in Section 3. The proof of Theorem 3 is given in Section 4and uses standard tools coming from the theory of spherical harmonics.Notations. Throughout this paper, D will be the Euclidean unit ball in Rn and Sn1 the unitsphere. The volume is denoted by | |. We write n for the volume of D and for the rotationallyinvariant probability measure on Sn1. By L we denote the convex body that is homothetic toL Rn and has volume 1, that is L = |L|1/nL and R(L) will be the circumradius of L, i.e. thesmallest real number such that L R(L)D. The letter c will always be used as being a universalconstant and it can change from line to line.2. Reduction to Lp centroid bodiesFor any isotropic convex body K , we define Ip(K) = (K|x|p2 dx)1/p. It is easy to check thatthere exists a constant cn,p such that for every Sn1cpn,pSn1, xp d() = |x|p2 , i.e. cn,p =( (p+n2 )(p+12 )(n2 ))1/p.Note that cn,p is of the same magnitude than(n+ p)/p. By the Fubini theorem and the def-inition of Wp(Zp(K)), Ip(K) = cn,p Wp(Zp(K)). We first need some precise computations inthe case of the Euclidean ball of volume 1.Lemma 1. Let D be the Euclidean unit ball in Rn, then for any p n,Ip(D)/I1(D) 1 + cp/n2. (3)Let k be an integer and p k n and denote by DF the Euclidean unit ball of volume 1 in anyk-dimensional subspace F of Rn then(W1(Z1(D))/W1(Zp(D)))(W1(Zp(DF ))/W1(Z1(DF ))) 1 + c p/k.Proof. For any 1 p n, we havecn,pcn,1Wp(Zp(D))W1(Z1(D))=(D|x|p2 dx)1/p/D|x|2 dx = (1 + 1/n)(1 + p/n)1/p 1 + cp/n2.Since for any p 1, W1(Zp(D)) = Wp(Zp(D)) and x(x) = (x + 1), we getW1(Z1(D))W1(Zp(DF ))W1(Zp(D))W1(Z1(DF ))=((1 + n+p2 )(1 + k2 )(1 + n2 )(1 + k+p2 ))1/p (1 + n2 )(1 + k+12 )(1 + n+12 )(1 + k2 ).Easy computations involving the function give the stated estimate when p k. B. Fleury et al. / Advances in Mathematics 214 (2007) 865877 869For any fixed symmetric convex body L, Litvak, Milman and Schechtman [14] studied thebehavior of Wp(L) as a function of p.Lemma. (See [14].) Let L be a symmetric convex body of Rn. ThenWp(L)W1(L) hL(u)W1(L)p c2pnR(L), (4)for any p c1n(W1(L)/R(L))2, where c1 and c2 are universal constants.The next lemma was essentially proved in [20].Lemma 2. There exists c > 0 such that for every isotropic convex body K Rn, for every 1p cn,R(Zp(K)) cpW1(Zp(K)). (5)Proof. We briefly indicate a proof. In isotropic position, R(Zp(K)) cpR(Z2(K)) = cpLK .Corollary 3.11 in [20] means that if p cn, Wp(Zp(K)) is similar up to universal constantsto W1(Zp(K)). Observe that Wp(Zp(K)) cp/nIp(K) cpLK andpW1(Zp(K)) cpLK cR(Zp(K)). Proof of Theorem 1. We writeIp(K)I1(D)I1(K)Ip(D)= Wp(Zp(K))W1(Zp(K))(W1(Zp(K))W1(Z1(K))W1(Z1(D))W1(Zp(D))). (6)From (4) and (5), we get 1Wp(Zp(K))/W1(Zp(K)) 1+c pn if p cn. Hence Theorem 1is proved using (3), (6) and Theorem 2. In particular,K( |x|22nL2K 1)2dx = I44 (K)I 42 (K) 1 c/(logn)1/3. (7)The function f (x) = ( |x|22nL2K 1) is a polynomial of degree 2 and we can use the results ofBobkov [3] about Lr -norms of polynomials. Indeed, Theorem 1 of [3] states that there existsa universal constant c > 0 such thatKef (x)/cK f (x) dx dx 2 where f = |f |1/2. For every (0,1), since Kf (x) dx (Kf 2(x) dx)1/4, we get by (7) and by the Chebychev inequality,{ |x|2 } { |x|22 } c(logn)1/12 x K, nLK 1 x K, nL2K 1 2e . 870 B. Fleury et al. / Advances in Mathematics 214 (2007) 8658773. Proof of Theorem 2We now introduce some notations and recall some well-known facts from local theory ofBanach spaces. For a given subspace F Rn, denote by E the orthogonal subspace to F and forevery SF , the Euclidean sphere in F , we define E() to be {x span{E,}, x, 0}. Forany q 0, define the star body Bq by its radial function SF , rBq () =( KE()x,q dx)1/(q+1).A theorem of Ball [2] asserts that when K is a symmetric convex body in Rn, this radial functiondefines a symmetric convex body in F . These balls are related to the Lp-centroid bodies by thefollowing proposition (see Proposition 4.3 in [20]).Proposition. (See [20].) Let K be a symmetric convex body in Rn and let 1 k n 1. Forevery subspace F of Rn of dimension k and every q 1, we havePF(Zq(K))= (k + q)1/qZq(Bk+q1) = (k + q)1/q |Bk+q1|1/k+1/qZq(Bk+q1). (8)Moreover, an application of a result of Borell [5] gives comparison between these norms.Lemma. (See [5].) For f being a log-concave non-increasing function on [0,+), defineF : t 1(t)+0xt1f (x)dx, G : t t+0xt1f (x)dxthen F is log-concave and G is log-convex on (0,+).Proposition 3. Let K be a symmetric convex body in Rn, let F be a k-dimensional subspaceof Rn, and for any t 1, define the symmetric convex body Bt1 in F as before. For every SF and every 1 s t u, we havetBt1 (s)(1)(u)(t)(1)sBs1 uBu1 and tBt1 ts(1)u(1)sBs1 uBu1where t = (1 )s + u.Proof. Let f(y) = |{K(E+y)}| for y R+ then by the BrunnMinkowski inequality, f isa log-concave function and non-increasing. By Fubini, for every SF ,tBt1 =+0yt1f(y) dy = t1G(t) = (t)F (t)and the conclusion follows easily by the above lemma. B. Fleury et al. / Advances in Mathematics 214 (2007) 865877 871We will also use a refinement of Dvoretzkys theorem proved by Milman [17] (see also [18]).Theorem. (See [17].) There exist constants c1, c2 such that for any n, any > 0 and any symmet-ric convex body L Rn, if k c1(2/ log(1/))n(W1(L)/R(L))2, the set of subspaces F Gn,ksuch that(1 )W1(L)DF PFL (1 + )W1(L)DF(where DF is the Euclidean unit ball of F ) has Haar measure greater than 1 ec2k .It was proved by Gordon [9] that we may take 2 instead of 2/ log(1/).Proof of Theorem 2. Let K be an isotropic convex body in Rn. Hence from (5), for every1 q cn, R(Zq(K)) cq W1(Zq(K)). Without loss of generality, we can assume that pis an odd integer. Let k and (0,1/3) (to be chosen later) be such that k2 c2n and k p.Since Dvoretzkys theorem holds with high probability, we can choose a subspace F of Rn ofdimension k such that five conditions hold simultaneously: for every q {1,p, k,2k p,2k},(1 ) W1(Zq(K))W1(Zq(DF ))Zq(DF ) PFZq(K) (1 + ) W1(Zq(K))W1(Zq(DF ))Zq(DF ).Indeed, observe that q {1,p, k,2k p,2k}, k c2n/q c12n(W1(Zq(K))/R(Zq(K)))2.From (8), these inclusions mean that for every q {1,p, k,2k p,2k},(1 )qZq(DF ) Zq(Bk+q1) (1 + )qZq(DF ) (9)whereq = W1(Zq(K))(k + q)1/q |Bk+q1|1/k+1/qW1(Zq(DF )). (10)The first step is to prove the following.Claim. There is a universal constant c such that, for q {1,p}, d(Bk+q1,DF ) c.Indeed, since Bk+q1 is a symmetric convex body in a k-dimensional space, it is well knownthat there exists a universal constant c such that cBk+q1 Zq(Bk+q1) Bk+q1 for q k(see for example Lemma 4.1 in [19] or Lemma 3.1.1 in [8]). For q {k,2k p,2k}, we de-duce from (9) that d(Bk+q1,DF ) c where c is a universal constant. Now, for q {1,p},Proposition 3 with s = k + q, t = 2k,u = 3k q (i.e. t = (1 )s + u with = 1/2) gives2kB2k1 (k + q)1/2(3k q)1/2(2k)(k+q)/2Bk+q1 (3kq)/2B3kq1 ,2kB2k1 2k(k + q)1/2 (3k q)1/2 (k+q)/2Bk+q1 (3kq)/2B3kq1for every SF . Since q p k, it is easy to conclude the proof of the claim.872 B. Fleury et al. / Advances in Mathematics 214 (2007) 865877In the second step, we apply Theorem 3. Indeed, for q {1,p}, we get from (9) thatd(Zq(Bk+q1),Zq(DF )) (1 + )/(1 ) 1 + 3 and we have seen that d(Bk+q1,DF ) ctherefore, Theorem 3 (since q is a non-even number) states that there exists a universal constantc such that1 hk() q 1 + hk() (11)and for every , 0 SF ,(1 + hk())10Bk+q1 Bk+q1 (1 + hk())0Bk+q1 , (12)where hk() = c2k(3)1/k2 . We want that this last quantity goes to 0 when k goes to infinity hencewe choose = (2c)2k3 in such a way that hk() ek . In order to use Dvoretzkys theorem,k has been chosen such that k2 = c2n which means that k c(logn)1/3. By (10) and (11),W1(Zp(K))W1(Z1(K))W1(Z1(DF ))W1(Zp(DF )) (1 + ek)(k + p)1/p|Bk+p1|1/k+1/p(1 ek)(k + 1)|Bk|1/k . (13)To conclude, it is left to observe that |K| = 1 can be written as1 = |K| = kkSFKE()x, k1 dx dF () = kk SFkBk1 dF ()so that there exists a 0 SF such that 1 = kk0kBk1 . Using relation (12),(k + p)1/p|Bk+p1|1/k+1/p(k + 1)|Bk|1+1/k =(k + p)1/p(kSFkBk+p1 dF ())1/k+1/p(k + 1)(kSFkBk dF ())1+1/k(1 + ek)k+2+k/p(k + p)1/p0k+1Bk(k + 1)11/pk 01+k/pBk+p1.Proposition 3 with s = k, t = k + 1, u = k + p (i.e. t = (1 )s + u with = 1/p) gives0k+1Bk (k)11/p(k + p)1/p(k + 1) 0k(11/p)Bk1 01+k/pBk+p1 .Since 0kBk1 = kk and p k, easy computations involving the function gives(k + p)1/p|Bk+p1|1/k+1/p(k + 1)|Bk|1+1/k (1 + ek)2k (1 + p/k)1/p(1 + 1/k)(k)11/p(k + p)1/p(k + 1)( )1/p= (1 + ek)2k 1k + 1(k + p + 1)(k + 1) 1 + cp/k.B. Fleury et al. / Advances in Mathematics 214 (2007) 865877 873Combining this last inequality with (13) and with Lemma 1, we conclude that if p kW1(Zp(K))W1(Z1(K))W1(Z1(D))W1(Zp(D)) 1 + cp/(logn)1/3for a universal constant c. 4. Stability result for Lp-centroid bodiesIn Theorem 3, the equality case (i.e. = 0) may be treated via the use of the FunckHecketheorem. This is why we will follow an approach using the decomposition in spherical harmonicsand we refer to Chapter 3 of the book of Groemer [10] for more detailed explanation. Thistechnique was also used by Bourgain and Lindenstrauss [6].Let p be an odd integer with p d , we consider the function :R R defined by (t) = |t |pand we define the operator J on L2(Sd1) byJ(F )(u) =Sd1(u,v)F(v)d(v)for any u Sd1. By the FunckHecke theorem, for every harmonic polynomial H homoge-neous of degree l on the sphere Sd1 we have J(F ),H = d,l()F,H , where , denotesthe usual scalar product in L2(Sd1) andd,l() = (1)l(d1)/22l1(l + d12 )11(t)dldt l(1 t2)(l+ d32 ) dt.These coefficients are known, see [21] or [13, Lemma 1]. Hence, for any odd values of l,d,l() = 0 and for any even values of l,d,l() = d/21(p + 1) sin((l p)/2)((l p)/2)2p1((l + d + p)/2) .Standard computations involving the function give a universal constant c such that for anyeven integer l,1d,l()1/(p+d/2) cmax(d, l). (14)For a continuous function F :Sd1 R such that F :Rd R defined by F(x) = F(x/|x|2)is differentiable on Rd \ {0}, we set for any u Sd1, 0F(u) = F(u). The next propositionis a standard trick using spherical harmonics [10].Proposition 4. There exists a universal constant c such that for any continuous even functionF :Sd1 R such that 0F exists,F2 cJ(F )2/(d+2p+2)2 (0F22 + d2F22) 12 (12/(d+2p+2)).874 B. Fleury et al. / Advances in Mathematics 214 (2007) 865877Proof. Let F Ql(F ) be the decomposition in spherical harmonics of F (with Ql(F ) spher-ical harmonics of degree l) then by Corollary 3.2.12 in [10]0F22 =l0l(l + d 2)Ql(F )22.For any odd l, d,l() = 0 and since F is even, Ql(F ) = 0. Hence from the Parseval equalityF22 =l evenQl(F )22 = l even(d,l()Ql(F )2)Ql(F )22 d,l(),where (0,2) is chosen such that 2/(2 ) = 2/(p + d/2). By the Hlder inequality,F22 ( l evend,l()2Ql(F )22)/2( l evenQl(F )22d,l()2/(2))1/2.By the FunckHecke theorem, J(F )22 =l even d,l()2Ql(F )22 and by the inequal-ity (14),l evenQl(F )22d,l()2/(p+d/2) c2 l evenmax(d2, l2)Ql(F )22 c2(d2l even, ldQl(F )22 + l even, ldl2Ql(F )22) c2(d2F22 +0(F )22).This proves that F2 cJ(F )2/(d+2p+2)2 (0F22 + d2F22)12 (12/(d+2p+2)). We will also need the following simple lemma.Lemma 5. Let F :Sd1 R be a Lipschitz function and let M = max(F2,FLip) thenF 5M(d1)/(d+1)F2/(d+1)2 .Proof. Let u Sd1 such that |F(u)| = F and let C(u,R) be the spherical cap of radius Rcentered at u. For any 1, define A = {v Sd1, |F(v)| F2} then by the Chebychevinequality, (A) 11/2. For any R (0,2), it is well known that (C(u,R)) 12 (R2 )d1. IfR is chosen such that 12 (R2 )d1 = 12then A C(u,R) = . In that case, take v A C(u,R)thenF(u) F(u) F(v)+ F(v) FLip|u v|2 + F2 RM + F2.Since R = 2(2/2)1/(d1), we get the estimate taking = (M/F2)(d1)/(d+1) 1. B. Fleury et al. / Advances in Mathematics 214 (2007) 865877 875Proof of Theorem 3. Using the support functions, d(Zp(K),Zp(D)) 1 + implies that thereexists > 0 such that hZp(D) hZp(K) (1 + ) hZp(D). (15)For any symmetric convex body L Rd , by integration in polar coordinates,hZp(L)(u)p =Lx,up dx = ddd + pSd1v,up 1vd+pLd(v)hence applying it for L = K and L = D, we get for any u Sd1,Sd1v,up( 1vd+p p1+p/dd)d(v) ((1 + )p 1) p1+p/ddSd1v,up d(v),where is the norm with unit ball K. For every u Sd1, let F(u) = 1+p/dd pup+d 1. Sinceu Sd1, Sd1 |v,u|p d(v) 1, we getJ(F )2 J(F ) ((1 + )p 1). (16)Since d(K,D) , there exists a, b > 1 such that 1/aD K bD and ab = . For ally Sd1, p(1 + )phpZp(D)(y) hpZp(K)(y)D/ax, yp dx = hpZp(D)(y)/ad+p,therefore 1/ p ad+p(1 + )p . For any x Rd , 1/dd b1|x|2 x a1/dd |x|2 and foru Sd1,F(u) = 1+p/dd p(p + d)up+d(u (u)u),therefore0F2 0F (p + d)bp+d p(1 + ab) 4dd+p+1(1 + )p. (17)We also have F2 F 1 + bp+d/ p 2p+d(1 + )p . Using Proposition 4 with (16)and (17), we getF2 c((1 + )p 1)2/(d+2p+2)(6dd+p+1(1 + )p)12/(d+2p+2) c2/(d+2p+2)(4)d+p+1.876 B. Fleury et al. / Advances in Mathematics 214 (2007) 865877Moreover, for any u,v Sd1, F(u) F(v) = 1+p/dd / p(1/up+d 1/vp+d) andF(u) F(v) 1+p/dd pu vd+p1i=0u(d+pi)v(i+1) 2dd+p+1(1 + )p|u v|2.Therefore max(F2,FLip) (4)d+p+1 and by Lemma 5,F c(4)d+p+14/(d+1)(d+2p+2) c(4)d+p+11/d2 := f ().Recalling the definition of F , F(u) = 1 +1+p/dd / pup+d , u Sd1, we have proved(1 f ())1/d+p p/d+pD K (1 + f ())1/d+p p/d+pD. (18)Since |K| = |D| = 1, (1 + f ())1 p (1 f ())1 and choosing (c)2d3 , (15) and(18) prove the assertions of Theorem 3. AcknowledgmentThe authors would like to thank K. Ball for several useful discussions.References[1] M. Anttila, K. Ball, I. Perissinaki, The central limit problem for convex bodies, Trans. Amer. Math. Soc. 355 (12)(2003) 47234735.[2] K. Ball, Logarithmically concave functions and sections of convex sets in Rn , Studia Math. 88 (1) (1988) 6984.[3] S.G. Bobkov, Remarks on the growth of Lp-norms of polynomials, in: Geometric Aspects of Functional Analysis,in: Lecture Notes in Math., vol. 1745, Springer, Berlin, 2000, pp. 2735.[4] S.G. Bobkov, On isoperimetric constants for log-concave probability distributions, Geometric Aspects of FunctionalAnalysis, Springer, Berlin, 2006, in press.[5] C. Borell, Complements of Lyapunovs inequality, Math. Ann. 205 (1973) 323331.[6] J. Bourgain, J. 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