Convex Bodies with Minimal Mean Width

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1 Covex Boies with Miimal Mea With A.A. Giaopoulos 1, V.D. Milma 2, a M. Ruelso 3 1 Departmet of Mathematics, Uiversity of Crete, Iraklio, Greece 2 School of Mathematical Scieces, Tel Aviv Uiversity, Tel Aviv 69978, Israel 3 Departmet of Mathematics, Uiversity of Missouri, Columbia, MO 65211, USA 1 Itrouctio Let K be a covex boy i R, a {T K T SL(} be the family of its positios. I [GM] it was show that for may atural fuctioals of the form T f(t K, T SL(, the solutio T 0 of the problem mi{f(t K T SL(} is isotropic with respect to a appropriate measure epeig o f. The purpose of this ote is to provie applicatios of this poit of view i the case of the mea with fuctioal T w(t K uer various costraits. Recall that the with of K i the irectio of u S 1 is efie by w(k, u = h K (u + h K ( u, where h K (y = max x K x, y is the support fuctio of K. The with fuctio w(k, is traslatio ivariat, therefore we may assume that o it(k. The mea with of K is give by w(k = w(k, uσ(u = 2 h K (uσ(u, S 1 S 1 where σ is the rotatioally ivariat probability measure o the uit sphere S 1. We say that K has miimal mea with if w(t K w(k for every T SL(. The followig isotropic characterizatio of the miimal mea with positio was prove i [GM]: Fact. A covex boy K i R has miimal mea with if a oly if h K (u u, θ 2 σ(u = w(k S 2 1 for every θ S 1. Moreover, if U SL( a UK has miimal mea with, we must have U O(. Research of the seco ame author partially supporte by the Israel Sciece Fouatio foue by the Acaemy of Scieces a Humaities. Research of the thir ame author was supporte i part by NSF Grat DMS

2 2 A.A. Giaopoulos et al. Our first result is a applicatio of this fact to a reverse Urysoh iequality problem: The classical Urysoh iequality states that w(k ( K /ω 1/ where ω is the volume of the Eucliea uit ball D, with equality if a oly if K is a ball. A atural questio is to ask for which boies K a := max mi w(t K K =1 T SL( is attaie, a what is the precise orer of growth of a as. Examples such as the regular simplex or the cross-polytope show that a c log( + 1. O the other ha, it is kow that every symmetric covex boy K i R has a image T K with T K = 1 for which w(t K c 1 log[(xk, l 2 + 1], where X K = (R, K a eotes the Baach-Mazur istace. This statemet follows from a iequality of Pisier [Pi], combie with work of Lewis [L], Figiel a Tomczak-Jaegerma [FT]. Joh s theorem [J] implies that mi w(t K c 2 log( + 1, T SL( for every symmetric covex boy K with K = 1, a a simple argumet base o the ifferece boy a the Rogers-Shephar iequality [RS] shows that the same hols true without the symmetry assumptio. Therefore, c log( + 1 a c 3 log( + 1. Here, we shall give a precise estimate for the miimal mea with of zoois (this is the class of symmetric covex boies which ca be approximate by Mikowski sums of lie segmets i the Hausorff sese: Theorem A. Let Z be a zooi i R with volume Z = 1. The, where Q = [ 1/2, 1/2]. mi w(t Z w(q = 2ω 1, T SL( ω For our seco applicatio, we cosier the class of origi symmetric covex boies i R. Every symmetric boy K iuces a orm x K = mi{λ 0 : x λk} o R, a we write X K for the orme space (R, K. The polar boy of K is efie by x K = max y K x, y = h K (x, a will be eote by K. Wheever we write (1/a x x K b x, we assume that a, b are the smallest positive umbers for which this iequality hols true for every x R. We cosier the average M(K = x K σ(x S 1

3 Covex Boies with Miimal Mea With 3 of the orm K o S 1, a efie M (K = M(K. Thus, M (K is half the mea with of K. We will say that K has miimal M if M(K M(T K for every T SL(. Equivaletly, if K has miimal mea with. Our purpose is to show that if K has miimal M, the the volume raius of K is boue by a fuctio of b a M. Actually, it is of the orer of b/m. The precise formulatio is as follows: Theorem B. Let K be a symmetric covex boy i R with miimal M, such that (1/a x x K b x, x R. The, ( 1/ b K M c b ( 2b (1/bD M log, M where c > 0 is a absolute costat. Our last result cocers optimizatio of the with fuctioal uer a ifferet coitio. We say that a -imesioal symmetric covex boy K is i the Gauss-Joh positio if the miimum of the fuctioal E g T K uer the costrait T K D is attaie for T = I. That is, K has miimal mea with uer the coitio T K D (it miimizes M uer the coitio a(t K 1. We ca cosier this optimizatio problem oly for positive self-ajoit operators T. Sice the orm of T shoul be boue to guaratee that T K D a the orm of T 1 shoul be boue as well, there exists T for which the miimum is attaie. Deote by γ the staar Gaussia measure i R. The, we have the followig ecompositio. Theorem C. Let K be i the Gauss-Joh positio. The there exist: m (+1/2, cotact poits x 1,..., x m K S 1 a umbers c 1,..., c m > 0 such that m c i = 1 a ( m (x x I x K γ(x = x K γ(x c i x i x i. R R The Gauss-Joh positio is ot equivalet to the classical Joh positio. Examples show that, whe K is i the Gauss-Joh positio, the istace betwee D a the Joh ellipsoi may be of orer / log. 2 Reverse Urysoh Iequality for Zoois The proof of Theorem A will make use of a characterizatio of the miimal surface positio, which was give by Petty [Pe] (see also [GP]: Recall that the area measure σ K of a covex boy K is efie o S 1 a correspos

4 4 A.A. Giaopoulos et al. to the usual surface measure o K via the Gauss map: For every Borel V S 1, we have σ K (V = ν ({x b(k : the outer ormal to K at x is i V }, where ν is the ( 1-imesioal surface measure o K. If A(K is the surface area of K, we obviously have A(K = σ K (S 1. We say that K has miimal surface area if A(K A(T K for every T SL(. With these efiitios, we have: 2.1 Theorem. A covex boy K i R has miimal surface area if a oly if u, θ 2 σ K (u = A(K S 1 for every θ S 1. Moreover, if U SL( a UK has miimal surface area, we must have U O(. Recall also the efiitio of the projectio boy ΠK of K: it is the symmetric covex boy whose support fuctio is efie by h ΠK (θ = P θ (K where P θ (K is the orthogoal projectio of K oto θ, θ S 1. It is kow that Z is a zooi i R if a oly if there exists a covex boy K i R such that Z = ΠK. By the formula for the area of projectios, this ca be writte i the form h Z (x = 1 x, u σ K (u. 2 S 1 The, the characterizatio of the miimal mea with positio a Theorem 2.1 imply the followig: 2.2 Lemma. Let Z = ΠK be a zooi. The, Z has miimal mea with if a oly if K has miimal surface area. Proof. The proof (moulo the characterizatio of the miimal mea with positio may be fou i [Pe]: By Cauchy s surface area formula, A(K = ω h Z (θσ(θ. ω 1 S 1 If f 2 is a spherical harmoic of egree 2, the Fuk-Hecke formula shows that S 1 f 2 (u u, τ σ(u = c f 2 (τ for all u, τ S 1, where c is a costat epeig oly o the imesio. Therefore, f 2 (uh Z (uσ(u = 1 f 2 (u u, τ σ(uσ K (τ S 2 1 S 1 S 1 = c f 2 (τσ K (τ. 2 S 1

5 Covex Boies with Miimal Mea With 5 Sice u u, θ 2 is homogeeous of egree 2, this implies h Z (u u, θ 2 σ(u = c u, θ 2 σ K (u S 2 1 S 1 for every θ S 1. The characterizatios of the miimal mea with a the miimal surface area positios make it clear that Z has miimal mea with if a oly if K has miimal surface area. Our ext lemma is a well-kow fact, prove by K. Ball [B]: 2.3 Lemma Let {u j } j m be uit vectors i R a {c j } j m be positive umbers satisfyig m I = c j u j u j. j=1 If Z = m j=1 α j[ u j, u j ] for some α j > 0, the Z 2 m ( αj j=1 c j cj. We apply this result to the projectio boy of a covex boy with miimal surface area. 2.4 Lemma If K has miimal surface area, the A(K ΠK 1/. Proof. We may assume that K is a polytope with facets F j a ormals u j, j = 1,..., m. The, Theorem 2.1 is equivalet to the statemet I = m c j u j u j j=1 where c j = F j /A(K (see [GP]. O the other ha, ΠK = A(K 2 m c j [ u j, u j ]. j=1 We ow apply Lemma 2.3 for Z = ΠK, with α j = A(K 2 c j: m ( A(K ΠK 2. 2 j=1 Remark. I the previous argumet, equality ca hol oly if (u j j m is a orthoormal basis of R (see [Ba]. This meas that if K is a polytope the equality i Lemma 2.3 ca hol oly if K is a cube.

6 6 A.A. Giaopoulos et al. Proof of Theorem A. Let Z be a zooi with miimal mea with a volume Z = 1. By Lemma 2.2, Z is the projectio boy ΠK of some covex boy K with miimal surface area. We have w(z = 2 h Z (uσ(u = 2 S 1 P u (K σ(u = 2ω 1 S ω 1 A(K. By Lemma 2.4, the area of K is boue by Z 1/ =. We have equality whe K is a cube, a this correspos to the case Z = Q. Therefore, w(z w(q = 2ω 1 ω. Remark. Urysoh s iequality a Theorem A show that if Z is a zooi with Z = 1, the where α, β 1 as. 2 2 α mi πe w(t Z β, T SL( π 3 Volume Ratio of Symmetric Covex Boies with Miimal M For the proof of Theorem B we will ee the followig fact which was prove i [GM]: 3.1 Theorem. Let K be a symmetric covex boy i R with miimal M. The, for every λ (0, 1 there exists a [(1 λ]-imesioal subspace E of R such that b r(λ x x K b x, x E, (3.1 b where r(λ c log( 2b Mλ 1/2 Mλ, a c > 0 is a absolute costat. Actually, the proof of Theorem 3.1 shows that the statemet hols true for a raom [(1 λ]-imesioal subspace E of R. Oe ca assume that for every k we have the result with probability greater tha c log (this formulatio is correct whe 0, where 0 N is absolute. This assumptio o the measure of subspaces satisfyig (3.1 implies that there is a icreasig sequece of subspaces E 1 E 2... E k0, where k 0 = [ c log 2 ] a ime k = k, so that (3.1 hols for each E k with r = r(k/. We will also ee the followig

7 Covex Boies with Miimal Mea With Lemma. Let K be a symmetric covex boy i R, such that (1/a x x K b x. If E is a k-imesioal subspace of R, the ( K D Ca( where C > 0 is a absolute costat. k 1/2 k K E, D k Proof. Let E be a k-imesioal subspace of R. Replacig K by (1/aK, we may assume that a = 1, so K D. Usig Bru s theorem we see that K = K (E + y y P E (K K E This shows that P E (K K E D k. K D D k D k D K E D k Proof of Theorem B. We first observe that ( C k ab C log. ( k/2 K E. D k Iee, let e S 1 be such that e K = b a let γ be a staar ormal variable. The cb = E γe E g K. Similarly, ca E g K. Multiplyig these iequalities, we obtai ab CE g K E g K C log. The last iequality follows from the fact that M is miimal for K, a Pisier s iequality [Pi]. Assume ow that b = 1. Let t = [log ]. For s = 1, 2,..., t put k s = [(1 1/s] a let E s = E ks be a subspace from our flag. The by Theorem 3.1 we have (1/a s x x K x o E s, where a s r(( k s / c M s1/2 log ( 2s M s c M log ( 2 =: s c(m. M Now, Lemma 3.2 shows that K ( D Ca kt K Et t (C log 2 / log K E t D kt D kt C K E t D kt

8 8 A.A. Giaopoulos et al. a K E s+1 D ks+1 (Ca s s ks+1 ks K E s D ks ( Cc(Ms 2 k s+1 k s K E s, D ks for all s = 1, 2,..., t. Sice a 1 c(m, we have K E 1 / D k1 c(m k1. Hece, multiplyig the iequalities above, we get K D C (Cc(M kt k1 c(m k1 By the efiitio of k s, ( t s 2(ks+1 ks exp c therefore s=2 t s=2 t s 2(ks+1 ks. s=2 ( 1/ K C 1 c(m. (1/bD log s s 2 e c, The left ha sie iequality is a immeiate cosequece of Höler s iequality: ( 1/ ( K = b (1/bD ( b S 1 x K σ(x 1/ S 1 x K 1 = b M. 4 Gauss-Joh Positio We prove Theorem C. Cosier the followig optimizatio problem: F (T = T 1 x K γ(x mi (4.1 R uer the costrait H x (T = T x for x K. Assume that the boy K is i the Gauss-Joh positio, amely the miimum i (4.1 is attaie for T = I. Let W be the set of the cotact poits of K: W = K S 1. First we apply a argumet of Joh [J] to show that we

9 Covex Boies with Miimal Mea With 9 ca cosier oly fiitely may costraits. Sice the paper [J] is ot easily available, we shall sketch the argumet. Let T be a self-ajoit operator a let T s = I + st. We shall prove that if for every x W, the Iee, assume that s H x(t s s=0 < 0 s F (T s s=0 0. a = sup x W s H x(t s s=0 < 0. Let W ε be a ε-eighborhoo of W : W ε = {x K ist(x, W < ε}. There exists a ε > 0 such that s H x(t s s=0 < a 2 for every x W ε. So, there exists s 0 > 0 such that for ay 0 < s < s 0 a ay x W ε H x (T s < H x (I 0. O the other ha, if x K \ W ε the Here, a H x (T s H x (T s H x (I + H x (I. H x (T s H x (I = T s x 2 x 2 T (1 + T s sup H x (I = sup x 2 1 < 0 x K\W ε x K\W ε sice K \ W ε is compact. Thus for a sufficietly small s we have H x (T s < 0 for all x K. So, sice I is the solutio of the miimizatio problem (4.1, s F (T s s=0 0. Sice s H x(t s s=0 = H x (I, T a s F (T s s=0 = F (I, T, this meas that the vector F (I caot be separate from the set { H x (I x W } by a hyperplae. By Carathéoory s theorem, there exist M (+1/2 cotact poits x 1... x M W a umbers λ 1... λ M > 0 such that M M F (I = λ i H xi (I = λ i x i x i. (4.2

10 10 A.A. Giaopoulos et al. Now we have to calculate F (I. We have F (T = (2π /2 so, 2 /2 x T 1 x K e x R = (2π /2 T x et T x K e 2 /2 x, R F (I = ((2π /2 x K e x 2 /2 x I R (2π /2 x K e x 2 /2 x xx R = (I x x x K γ(x. R Combiig it with (4.2 we obtai R (I x x x K γ(x + Takig the trace, we get ( Tr (I x x x K γ(x R = x K γ(x x 2 x K γ(x R R = r e 2 /2 r ω K m(ω 0 S 1 0 = x K γ(x. R M λ i x i x i = 0. r +2 e 2 /2 r ω K m(ω S 1 Fially, puttig λ i = c i R x K γ(x, we obtai the ecompositio ( M (I x x x K γ(x = x K γ(x c i x i x i, R R where M c i = 1. This completes the proof of Theorem C. We procee to compare D with the Joh ellipsoi i the Gauss-Joh positio: Propositio. Let K be a symmetric covex boy i R which is i the Gauss- Joh positio. The, (i (2/π 1/2 1 D K D ;

11 Covex Boies with Miimal Mea With 11 (ii It may happe that c log D is ot cotaie i K. Proof. (i Let T 0 be a operator which puts K ito the maximal volume positio. The mi F (T F (T 0. From the other sie, if there exists y S 1 such that y K < (2/π 1/2 1 the x K γ(x x, y/ y K γ(x (2/π 1/2 1/ y K >. R R (ii Let K = B1 1 + [ e, e ]. Let T be a positive self-ajoit operator such that T K is i the Gauss-Joh positio. We first prove that T is a iagoal operator. Let G O( be the group geerate by the operators U i = I 2e i e i, i = 1,..., a let m be the uiform measure o G. Notice that U i K = K for every i. The, x T K γ(x = Ux T U(K γ(xm(u R G R ( U 1 T 1 U m(u x γ(x. R G Put We claim that W = Iee, sice for ay i a G U 1 T 1 Um(U = iag(t 1. W ( iag(t 1. e i, iag(t 1 e i = e i, (T 1 e i e i, ( iag(t 1 ei = e i, T e i 1, the claim follows from the fact that for ay θ S 1 θ, T 1 θ θ, T θ 1. Let S = iag(t. Sice W S 1, we have F (T = x T K γ(x R W x K γ(x R S 1 x K γ(x = F (S. R [Notice that sice T K D, ( SK = U 1 T U m(u (K D, G so the restrictios of the optimizatio problem (4.1 are satisfie.]

12 12 A.A. Giaopoulos et al. Let ow G O( be the group geerate by the operators U ij = I e i e i e j e j + e i e j + e j e i for i, j = 1,..., 1, i j. Arguig the same way we ca show that there exist a, b > 0 such that F (S F (T 0, where ( 1 T 0 = a e i e i + be e. Sice the vertices of T 0 K are cotact poits, We have x T0K = max a 2 + b 2 = 1. ( x i, b 1 x. a 1 1 Deote x 1 = 1 x i a let t = t(x = (b/a x 1. The, ψ(b = x T0Kγ(x R ( 1 t = a 1 x 1 e x2 /2 x + 2 b 1 x e x2 /2 x γ(x 2π 2π = R 1 R 1 t ( 2 a x 1Φ(t + 2 b 1 e t2 /2 γ(x, 2π where Φ(t = (1/ 2π t /2 u. We have to show that b c log 0 e u2. We may assume that b c/. Puttig a = (1 b 2 1/2 a ifferetiatig, we get after some calculatios ( b ψ(b = 2a 3 b x 1 Φ(t 2 b 2 e t2 /2 γ(x. 2π R 1 Sice b c/ a x 1 C with probability at least 1/2, we have Φ(t > c with probability 1/2, for some absolute costat c > 0. So, which is positive whe b c log /. Remark. The ual problem f(t = uer the costrait b ψ(b c Cb 2 exp( c 2 b 2, R sup x, y γ(x max y T K h x (T = T x for x K is very ifferet. The examples suggest that the matrix T for which the maximum is attaie may be sigular. t

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