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A convex body whose centroid and Santalo´ point are far apart ∗ 0 1 0 Mathieu Meyer, Carsten Schu¨tt and Elisabeth M. Werner † 2 n a J Abstract 5 We give an example of a convex body whose centroid and Santaló point ] A are “far apart”. F . h t a m [ 1 v 4 1 7 0 . 1 0 0 1 : v i X r a ∗Keywords: Santaló point, . 2000 Mathematics Subject Classification: 52A20, 53A15 †Partially supported by an NSF grant, a FRG-NSF grant and a BSF grant 1 1 Introduction. The main theorem. Inhissurveypaper[7],Gru¨nbaumintroducedmeasuresofasymmetryforconvexbodies in Rn, i.e. compact, convex sets in Rn with nonempty interior. Measures of asymmetry determine the degree of non-symmetricity of a non symmetric convex body. Recall that a convex set K in Rn is centrally symmetric with respect to the origin if x K implies ∈ x K. More generally, K is centrally symmetric with respect to the point z if x K − ∈ ∈ implies 2z x K. − ∈ Gru¨nbaum argues that a measure of asymmetry should be a function that is 0 for centrallysymmetricconvexbodiesandmaximalforthesimplexandonlyforthesimplex. However, the simplex has enough symmetries to ensure that all affine invariant points are identical. This leads to the question: How far apart can specific affine invariant points be? This distance can be used as a measure for the asymmetry of the convex body. Of particular interest are the centroid g(K) and the Santaló point s(K) of a convex bodyK inRn. Thesepointsareofmajorimportanceandhavebeenwidelystudied. For the (less well known) Santaló point see e.g. [4, 12, 14, 13, 20]. Recent developments in this direction include e.g. the works [1, 2, 3, 5, 6, 10, 15]. We show here that the points can be very far apart, namely Theorem 1 There is an absolute constant c > 0 and n N such that for all n n 0 0 there is a convex body C =C in Rn with ∈ ≥ n g(C) s(C) g(C) s(C) c k − k2 = k − k2 . (1) ≤ vol (ℓ C) w (u) 1 C ∩ is the Euclidean norm, l is the line through g =g(C) and s=s(C) and w (u)= 2 C hkC·k(u)+hC(−u) is the width of C in directon of the unit vector u= kgg−−ssk2. The proof actually shows that we can asymptotically determine the constant c of the theorem. g(C) s(C) 2 k − k vol (ℓ C) 1 ∩ is asymptotically with respect to the dimension, greater than or equal to 1 √eπ 2 1 − =0.142673... − e √eπ+ 2 (cid:18) (cid:19) e 1 − and is asymptotically with respect to the dimension, less than or equal to 1 √e 1 1 − =0.18383... − e √e+ 1 (cid:18) (cid:19) e 1 − Throughout the paper we use the following notations. Bn(a,r) is the Euclidean ball in 2 Rn centered at a with radius r. We denote Bn = Bn(0,1). Sn 1 is the boundary ∂Bn 2 2 − 2 2 of the Euclidean unit ball. For x Rn, let x = ( n x p)p1, if 1 p < and x = max x , if p = an∈d let Bn =k kxp Rn :i=1x| i| 1 be th≤e unit∞ball of tkhek∞space ln.1≤i≤n| i| ∞ p { ∈ P k kp ≤ } p For a convex body K in Rn we denote the volume by vol (K) (if we want to emphasize n the dimension) or by K . co[A,B] = λa+(1 λ)b : a A,b B,0 λ 1 is the | | { − ∈ ∈ ≤ ≤ } convex hull of A and B. h (u)=max u,x is the supportfunction of K in direction u. K x K For x K, Kx∈=h(Ki x) = y Rn : y,z x 1 for all z K is the polar ◦ ∈ − { ∈ h − i ≤ ∈ } body of K with respect to x. Let ξ Rn with ξ = 1 and s R. Then the (n 1)-dimensional section of K 2 ∈ k k ∈ − orthogonal to ξ through sξ is K(s,ξ)= x K ξ,x =s . { ∈ |h i } We just write K(s) if it is clear which direction ξ is meant. The centroid g(K) of a convex body K in Rn is the point 1 g(K)= xdx. vol (K) n ZK The Santaló point s(K) of a convex body K is the unique point s(K) for which vol (K)vol (Kx) n n attains its minimum. It is more difficult to compute the Santaló point as it is defined implicitly. Before we go into the construction of the convex bodies that fulfill the properties of Theorem 1, we discuss an example that shows that a more elaborate construction is necessary to get the statement of Theorem 1. For instance, it is not enough to take one half of a Euclidean ball B = x Rn x 1,x 0 . 2 1 { ∈ |k k ≤ ≥ } We compute the centroid of B. The coordinates of the centroid are g(B)(2) = = ··· g(B)(n)=0 and g(B)(1) = |B22n|Z01tvoln−1(cid:16)p1−t2B2n−1(cid:17)dt= 2v(onln+−11(cid:0))|BB2n2n−|1(cid:1) . NowwecomputetheSantalópointofB. ThepolarbodyoftheEuclideanballBn(λe ,1) 2 1 with center λe , 0 λ<1, and radius 1 is 1 ≤ λ 2 x Rn (1 λ2)2 x + +(1 λ2)x2+ +(1 λ2)x2 1 . (2) ( ∈ (cid:12)(cid:12) − (cid:18) 1 1−λ2(cid:19) − 2 ··· − n ≤ ) (cid:12) Therefore, the(cid:12) polar body of the half ball B with respect to λe is the convex hull of (cid:12) 1 the vector −eλ1 and the ellipsoid (2). We estimate the volume of (B −λe1)◦. Since (B λe ) contains the ellipsoid (2) 1 ◦ − (1−λ2)−n+21 voln(B2n)≤voln((B−λe1)◦). 3 On the other hand, as (B λe ) is contained in the union of the ellipsoid (2) and a 1 ◦ − cone with height 1 and with a base that is a Euclidean ball with radius (1+λ2) 1, λ − voln((B−λe1)◦)≤ (1volnλ(2B)2nn+2)1 + nvλol(n1−+1(λB22n)−n−1)1 − For λ= 1 we get √n voln((B−λe1)◦)≤ 1− n1 −n+21 voln(B2n)+ √vonln(1−1+(B12n)−n1)1 . (cid:18) (cid:19) n − Since voln 1(B2n−1) πn − vol (Bn) ∼ 2 n 2 r we get 1 π vol ((B λe ) ). √e+ vol (Bn) . n − 1 ◦ e 2 n 2 (cid:18) r (cid:19) On the other hand, for λ= γ √n n+1 γ2 − 2 vol ((B λe ) ) 1 vol (Bn). n − 1 ◦ ≥ − n n 2 (cid:18) (cid:19) With 1 t e t − − ≤ voln((B−λe1)◦)≥eγ2n2+n1 voln(B2n) . Therefore,thereisaγ suchthatforalln Nwehaves(B)(1) γ . Thus kg(B)−s(B)k2 is of order 1 . ∈ ≤ √n vol1(ℓ∩B) √n The next lemma is well known (see [20]). Lemma 2 For any convex body K, an interior point x of K is the Santalo´ point if and only if 0 is the centroid of (K x) . ◦ − This lemma can be rephrased as follows: Let K be a convex body. Then 0 is the Santalo´ point of Kg(K). Indeed, Kg(K) =(K g(K))◦ and(K g(K))◦◦ =K g(K). Since0isthecentroidof − − − (K g(K)) ,itfollowsbyLemma2that0istheSantalópointof(K g(K)) =Kg(K). ◦◦ ◦ − − For convex bodies K and L in Rn and natural numbers k, 0 k n, the coefficients ≤ ≤ V (K,L)=V(K,...,K,L,...,L) n k,k − n k k − | {z } | {z } 4 in the expansion n n voln(λ1K+λ2L)= k λ1n−kλk2Vn−k,k(K,L) (3) k=0(cid:18) (cid:19) X are the mixed volumes of K and L (see [20]). Now we introduce the convex bodies which will serve as candidates for Theorem 1. For convex bodies K and L in Rn and real numbers a > 0 and b > 0, we construct a convex body M in Rn+1 n M =co[(K, a),(L,b)]= t(x, a)+(1 t)(y,b)x K,y L,0 t 1 . (4) n − { − − | ∈ ∈ ≤ ≤ } The bodies we are using in Theorem 1 will be the polar bodies to M . The polar of M n n can be described as follows: For 1 s 1, the sections of M orthogonal to e −a ≤ ≤ b n◦ n+1 and containing se are n+1 M (s)=M (s,e )=(1+s a)K (1 s b)L . (5) n◦ n◦ n+1 ◦∩ − ◦ We show this. M = (z,s) Rn R x K :<z,x> sa 1and y L:<z,y >+sb 1 n◦ { ∈ × |∀ ∈ − ≤ ∀ ∈ ≤ } = (z,s) Rn Rz (1+sa)K (1 bs)L ◦ ◦ { ∈ × | ∈ ∩ − } ThebodyM istheconvexhulloftwon-dimensionalfacesK andL. Inthefollowing n proposition we choose specific bodies for those faces. We choose them in such a way that their volume product differs greatly, which has as effect that the centroid and the Santaló point of Mg(Mn) are “far apart”. One face is the Euclidean ball and the other n the unit ball of ℓn . We normalize both balls so that their volume is 1. ∞ Proposition 3 Let Bn 1 K = 2 L= Bn . |B2n|n1 2 ∞ and a=1 and b= 1 . Let M be the convex body in Rn+1 defined in (4). Then e 1 n − (i) lim g(M )=0. n n →∞ (ii) Let √e 1 2 √eπ s = − = 0.290815... s = − = 0.225705... 0 −√e+ 1 − 1 √eπ+ 2 − e 1 e 1 − − Then for every ε>0 there is n such that for all n n the n+1-st coordinate of the 0 0 ≥ centroid g(M ) of M satisfies n◦ n◦ s ε g(M )(n+1) s +ε. 0− ≤ n◦ ≤ 1 5 The Santalo´ point of Mg(Mn), s = s(Mg(Mn)) = 0 and therefore the centroid g and n ∗n n n∗ the Santalo´ point s of Mg(Mn) satisfy ∗n n √e 1 2 √eπ √e+−e−11 ≥linm→s∞up|gn∗ −s∗n|≥linm→i∞nf|gn∗ −s∗n|=linm→i∞nf|gn∗|≥ √e−π+ e−21 . As for the proof of Proposition 3, it follows from Lemma 2 that s(Mg(Mn))=0. For the n centroid g we actually estimate the coordinates of g(M ) and not g(Mg(Mn)). Since n∗ n◦ n limg(M ) = 0 it suffices to use a perturbation argument. The remaining part of the n proof of the proposition is in the next section. Proof of Theorem 1. The proof follows immediately from Proposition 3: For every n, let C =M . n n◦ 2 Proof of Proposition 3 This section is devoted to the proof of Proposition 3. We will need several lemmas and then give the proof of the proposition at the end of the section. Lemma 4 Let K and L be convex bodies in Rn such that their centroid are at 0. Let c>0 and let M be the convex body in Rn+1 n M =co[(K,0),(L,c)]. n Then the (n+1)-coordinate of the centroid g(M )(n+1) of M satisfies n n c n (k+1)V (K,L) g(Mn)(n+1)=hg(Mn),en+1i= n+2 k=0n V n−(kK,k,L) . P k=0 n−k,k P Proof. By definition cw vol (M (w))dw g(M ),e = 0 n n . h n n+1i cvol (M (w))dw R 0 n n Note that R w w co[(K,0),(L,c)]= (1 )K+ L,w 0 w c . − c c ≤ ≤ n(cid:16) (cid:17)(cid:12) o Therefore (cid:12) (cid:12) cw vol ((1 w)K+ wL)dw g(M ),e = 0 n − c c h n n+1i cvol ((1 w)K+ wL)dw R 0 n − c c = cR01t voln((1−t)K+tL)dt. R 01voln((1−t)K+tL)dt R 6 Now we use the mixed volume formula (3). Thus 1 n n tk+1 (1 t)n k V (K,L) dt 0 k=0 k − − n−k,k g(M ),e = c (cid:18) (cid:19) h n n+1i R 1P n (cid:0) (cid:1)n tk (1 t)n k V (K,L) dt 0 k=0 k − − n−k,k (cid:18) (cid:19) cR Pn (k(cid:0)+(cid:1)1)V (K,L) = n+2 k=0n V n−(kK,k,L) , P k=0 n−k,k where we have also used the BetafunctPion 1 Γ(k)Γ(l) B(k,l)= tk 1(1 t)l 1dt= , k,l>0. − − − Γ(k+l) Z0 (cid:3) The next lemma is well known ([8], p. 216, formula 54). Lemma 5 For all n N and t 0 ∈ ≥ n n voln(B2n+tB∞n )= k 2kvoln−k B2n−k tk, k=0(cid:18) (cid:19) X (cid:0) (cid:1) with the convention that vol (B0)=1. Therefore, for 0 k n, 0 2 ≤ ≤ Vn k,k(B2n,Bn )=2k voln k B2n−k . − ∞ − (cid:0) (cid:1) Lemma 6 The following formula holds n k+1 1 k=0 Γ(1+n)nkΓ(1+n−k) 1 lim 2 2 =1 . n n+2 Pn 1 − e →∞ k=0 Γ(1+n)nkΓ(1+n−k) 2 2 P Proof. One has n k+1 n n k+1 1 k=0 Γ(1+n)nkΓ(1+n−k) 1 k=0 Γ(1+n)−n−nkΓ(1+k) 2 2 = 2 2 n+2 Pn 1 n+2 Pn 1 k=0 Γ(1+n2)nkΓ(1+n−2k) k=0 Γ(1+n2)n−nkΓ(1+k2) P P 1 nk=0 (n−k+Γ(11)+Γ(k1+) n2)nk n+1 1 nk=0kΓΓ(1(1++n2k))nk = 2 = 2 . n+2P n Γ(1+n)nk n+2 − n+2Pn Γ(1+n)nk 2 2 k=0 Γ(1+k) k=0 Γ(1+k) 2 2 It is thus needed to proPve that P n kΓ(1+n2)nk 1 k=0 Γ(1+k) 1 lim 2 = . (6) n→∞nPn Γ(1+n2)nk e k=0 Γ(1+k) 2 P 7 For every x 0, ≥ n kxk n kxk n xk 1 = =2x − Γ(1+ k) kΓ(k) Γ(k) k=0 2 k=1 2 2 k=1 2 X X X = 2x 1 + n xk−1 =2x 1 +xn−2 xk . Γ(1) Γ(k) Γ(1) Γ(1+ k) (cid:16) 2 kX=2 2 (cid:17) (cid:16) 2 kX=0 2 (cid:17) Thus for x=Γ(1+ n)n1 2 n kΓ(1+ n2)nk =2Γ 1+ n n1 1 +Γ 1+ n n1 n−2Γ(1+ n2)nk . (7) Γ(1+ k) 2 Γ(1) 2 Γ(1+ k) ! kX=0 2 (cid:16) (cid:17) 2 (cid:16) (cid:17) kX=0 2 Let n Γ(1+ n)nk n Γ(1+ n)nk A = k 2 and B = 2 . n Γ(1+ k) n Γ(1+ k) k=0 2 k=0 2 X X (6) is equivalent to A 1 lim n = . n→∞nBn e By Stirling’s formula n n n n n 2 e−n2√πn Γ 1+ 2 e−n2√πne6n−112 2 ≤ 2 ≤ 2 (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) and thus n n 1 n (πn)21n Γ 1+ n (πn)21nen(6n1−12) 2e ≤ 2 ≤ 2e r r (cid:16) (cid:17) Moreover, Γ(1+ n)n2 n B 2 . n ≥ Γ(2) ≥ 2e (In fact, it can be proved that Bn 2ec2n 2e2ne, where we write a(n) b(n), to mean ∼ ∼ ∼ that there are absolute constants c ,c >0 such that c a(n) b(n) c a(n). But for 1 2 1 2 ≤ ≤ our purposes the above estimate is enough.) Also, for n 3 ≥ Γ(1+ n)nn−1 Γ(1+ n) 2 = 2 Γ(1+ n−21) Γ(1+ n2)n1Γ(1+ n−21) (n2)n2e−n2√πn e6n−112 ≤ (πn)21n 2ne(n−21)n−21e−n−21 π(n−1) p(n)n2√n ep 21n ≤ √n(n 1)n−21 (n 1) πn − − (cid:16) (cid:17) n n = e 21n np 2 e21n 1+ 1 2 exp n+1 e. πn n 1 ≤ n 1 ≤ 2(n 1) ≤ (cid:16) (cid:17) (cid:18) − (cid:19) (cid:18) − (cid:19) (cid:18) − (cid:19) 8 1 Therefore, by (7) and with c = Γ(1+ n) n, n 2 (cid:0) (cid:1) Γ(1+n2)n−n1 + Γ(1+n2)nn lim An = lim 2cn 1 +c 1 Γ(1+n−21) Γ(1+n2) n→∞nBn n→∞ n Γ(12)Bn n − Bn     2c2  2 n 2 1  = lim n = lim = . n n n n 2e e →∞ →∞ (cid:18)r (cid:19) (cid:3) Lemma 7 Let a,b R, a>0, b>0 and ∈ Bn Bn K = voln(B22n)n1 L= 2∞. Let M = co[(K, a),(L,b)] be defined as in (4). Then the center of gravity g(M ) n n − satisfies 1 1 lim g(M )= ( a)+ 1 b. n n e − − e →∞ (cid:18) (cid:19) Proof. By symmetry, the centroid of M is an element of the (n+1)-axis. Therefore n we only need to compute its (n+1)-th coordinate. Instead of M = co[(K, a),(L,b)], we consider M˜ = co[(K,0),(L,a+b)] with n n − centroid g˜=g˜ . The centroid g of M is then given by g =g˜ a. n n n n n − We use Lemma 4 with c=a+b to get a+b n (k+1)V (K,L) hg˜(Mñ),en+1i= n+2 k=0n V n−(kK,k,L) . P k=0 n−k,k By Lemma 5 and the linearity of the mixedPvolumes in each component, we get hg˜(Mñ),en+1i = na++b2 Pnk=0(nkk=0+v1o)lnv(oBln2n()Bnk2nv)onklnvokln−Bk2n(cid:0)−Bk2n−k(cid:1) − n k+1 a+b k=P0 k (cid:0) (cid:1) = (Γ(1+n/2)n Γ(1+(n−k)/2). (8) n+2 Pn 1 k=0 k (Γ(1+n/2)n Γ(1+(n k)/2) − P Now we apply Lemma 6. (cid:3) Eventually we will have to investigate expressions of the form Bn Bn vol 2 t 1 , t 0. n voln(B2n)n1 ∩ voln(B1n)n1 ! ≥ 9 Schechtman, Schmuckenschläger and Zinn established asymptotic formulas for large n forthevolumesofBn tBn [17,18,19]. Todoso, theyconsideredindependentrandom variables hp,...,hp pwi∩th dqensity 1 n p e tp (9) 2Γ(1) −| | p Here, we need uniform estimates instead of asymptotic ones. Lemma 8 Let 0<p,q < and let hp be a random variable with density (9). Then ∞ Ehp q = p ∞ tqe tpdt= p ∞tqe tpdt= 1 Γ q+1 (10) | | 2Γ(1) | | −| | Γ(1) − Γ(1) p p Z−∞ p Z0 p (cid:18) (cid:19) In particular, for p=1 and q =1 respectively q =2 we get Eh1 =1 respectively Eh1 2 =2. (11) | | | | For p=2 and q =2 we get 1 Eh2 2 = . | | 2 The next lemma treats the Gaußian case of a more general statement which we will prove below. Lemma 9 Let (Ω,P) be a probability space. Let g :Ω R, 1 i n, be independent i → ≤ ≤ N(0,1)-random variables. Then, for all γ >0 there is n such that for all n n 0 0 ≥ 1 n g (ω) 2 1 P ω n i=1| i | γ (12)  (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n1 Pni=1|gi(ω)|2 21 −rπ(cid:12)(cid:12)(cid:12)≤ ≥ 2 (cid:12)(cid:12) (cid:12)  (cid:12)(cid:12)(cid:12)(cid:12)(cid:0) P (cid:1) (cid:12)(cid:12)  Proof. By Chebyshev’s inequality. (cid:3) Lemma 10 Let (Ω,P) be a probability space. Let h1 : Ω Rn, 1 i n, be indepen- i → ≤ ≤ dent random variables with density e t. Then for all γ >0 there is n such that for all −| | 0 n n 0 ≥ P ω √2 γ (n1 ni=1|h1i(ω)|2)12 √2+γ 1 (13) ( (cid:12)(cid:12) − ≤ n1P ni=1|h1i(ω)| ≤ )≥ 2 (cid:12) (cid:12) P (cid:12) Proof. Again, by Chebyshev’s inequality. (cid:3) Lemma 11 Let (Ω,P) be a probability space. Let g :Ω R, 1 i n, be independent i → ≤ ≤ N(0,1)-random variables. Then 1 n g t voln(B2n∩t B1n)=n voln(B2n)Z0 rn−1P ( Pni=i=11|g|i|i2|)21 ≤ r! dr (14) P 10