DUALITY PHENOMENA AND VOLUME INEQUALITIES IN CONVEX GEOMETRY

A dissertation submitted to Kent State University in partial fulfillment of the requirements for the degree of Doctor of Philosophy

by

Jaegil Kim

August, 2013 Dissertation written by

Jaegil Kim

B.S., Pohang University of Science and Technology, 2005

M.S., Pohang University of Science and Technology, 2008

Ph.D., Kent State University, 2013

Approved by

Chair, Doctoral Dissertation Committee Dr. Artem Zvavitch Members, Doctoral Dissertation Committee Dr. Richard M. Aron Members, Doctoral Dissertation Committee Dr. Fedor Nazarov Members, Doctoral Dissertation Committee Dr. Dmitry Ryabogin Members, Outside Discipline Dr. Fedor F. Dragan Members, Graduate Faculty Representative Dr. David W. Allender

Accepted by

Chair, Department of Mathematical Sciences Dr. Andrew Tonge

Associate Dean Raymond Craig Dean, College of Arts and Sciences

ii TABLE OF CONTENTS

ACKNOWLEDGEMENTS ...... v

Introduction and Preliminaries 1

0.1 Basic concepts in Convex Geometry ...... 2

0.2 Summary and Notation ...... 5

0.3 Bibliographic Note ...... 7

I Mahler conjecture 8

1 Introduction to the volume product ...... 9

1.1 The volume product and its invariant properties ...... 9

1.2 Mahler conjecture and related results ...... 13

2 More on Hanner polytopes ...... 16

2.1 Connection with perfect graphs ...... 16

2.2 Direct sums of polytopes ...... 20

3 Local minimality in the symmetric case ...... 31

3.1 Construction of Polytopes ...... 31

3.2 Computation of volumes of polytopes ...... 41

3.3 A gap between K and a polytope ...... 60

4 Local minimality in the nonsymmetric case ...... 79

4.1 Continuity of the Santal´omap ...... 80

iii 4.2 Construction of polytopes for the simplex case ...... 86

5 Stability results for unconditional bodies ...... 94

5.1 Stability for unconditional bodies ...... 95

5.2 Minimality near unconditional convex bodies ...... 103

6 Proposed further research ...... 106

II ‘Convexity’ of intersection bodies 108

7 Introduction to intersection bodies ...... 109

7.1 Classical theorems and questions ...... 110

7.2 Several ways to measure ‘convexity’ ...... 111

8 Quasi-convexity for intersection bodies ...... 114

8.1 Quasi-convexity and related results ...... 114

8.2 Generalization to log-concave measures ...... 122

8.3 Non-symmetric cases and s-concave measures ...... 127

9 Local convexity for bodies of revolution ...... 132

9.1 Equatorial power type for bodies of revolution ...... 133

9.2 Equatorial power type 2 for intersection bodies ...... 136

9.3 Double intersection bodies of revolution in high dimension ...... 149

10 Uniform convexity for intersection bodies ...... 155

10.1 The Proof of Theorem ...... 157

11 Proposed further research ...... 166

BIBLIOGRAPHY ...... 167

iv ACKNOWLEDGEMENTS

First of all I would like to thank my advisor Dr. Artem Zvavitch for his constant support and guidance throughout my graduate study. He has been a great mentor not only in my academic pursuits but also in my life at Kent State. Next I would like to thank Dr. Richard

Aron, Dr. Fedor Nazarov, and Dr. Dmitry Ryabogin for their help in class or seminar and also serving as my dissertation committee members. Lastly I would like to thank my family, my wife Sieun and my son Sonu for their support and love.

v Introduction and Preliminaries

1 2

We start with the review of some classical background and general techniques, containing

John’s Theorem and the Brunn-Minkowski Inequality, in Convex Geometry. We refer the reader to the books [22], [47], [48], [68], [74] for more information on classical results and their application in Convex Geometry and Geometric Tomography.

0.1 Basic concepts in Convex Geometry

n We denote by R the Euclidean space of dimension n with the inner product , and h· ·i n the standard basis e1, . . . , en . A nonempty subset S of R is said to be { }

convex if the midpoint of any two points in S belongs to S. •

symmetric if it is symmetric at the origin, i.e. x S for all x S. • − ∈ ∈

unconditional (with respect to a basis e1, . . . , en ) if it is symmetric with respect to • { } n each coordinate hyperplane, i.e., tj x, ej S for any x S and any choice of 1 h i ∈ ∈

signs t1, . . . , tn 1, 1 . P ∈ {− }

star-shaped at a point p S if the intersection with any straight line containing p is • ∈ always a line segment.

n For a star-shaped set K at the origin, the Minkowski functional of K is defined in R by

n x K = min λ > 0 : x λK , x R k k { ∈ } ∈ and the radial function of K is defined on the sphere Sn−1 by

ρ (u) = max λ > 0 : λu K = u −1, u Sn−1. K { ∈ } k kK ∈

n n A body in R is a compact set in R which is the closure of its interior. In particular, a convex body is a body which is convex, or, equivalently, a compact convex set with nonempty 3

interior. A star body is a star-shaped body at the origin whose radial function is positive and continuous. For example, the n-dimensional `p ball, which is defined by

n n n p Bp = x R : x, ej 1 when 0 < p < ∈ j=1 | h i | ≤ ∞ n o n n X B∞ = x R : x, ej 1 for 1 j n when p = , ∈ |h i| ≤ ≤ ≤ ∞ n o is a star body if p > 0, and is a convex body if p 1. ≥ n Let K be a convex body in R containing the origin in its interior. The polar body of K is defined by

◦ n K = y R x, y 1 for all x K . ∈ h i ≤ ∈ 

This concept of duality (polarity) in convex bodies can be generalized by taking another

n interior point instead of the origin. If K R is a convex body and z is an interior point ⊂ of K, then the polar body of K with respect to z, denoted by Kz, is given as

z n K = y R : y z, x z 1 for all x K . { ∈ h − − i ≤ ∈ }

It can be expressed, in terms of the usual polarity, as Kz = (K z)◦ + z. Moreover we have − the following basic properties of polarity.

n n n Proposition 0.1. Let K, L be convex bodies in R . Let T : R R be an invertible → linear operator, and fix u Sn−1. ∈ 1. (K◦)◦ = K (Kz)z = K

2. K◦ L◦ whenever K LKz Lz whenever K L ⊃ ⊂ ⊃ ⊂ 3. (K L)◦ = conv(K◦,L◦)(K L)z = conv(Kz,Lz) ∩ ∩ 4. (K u⊥)◦ = K◦ u⊥ where K u⊥ denotes the projection of K to u⊥. ∩ | | 5. (TK)◦ = (T ∗)−1K◦ where T ∗ denotes the adjoint operator of T .

See [22], [47], [48], and [79] for the proof of the above properties. 4

One of fundamental results in Convex Geometry is the Brunn-Minkowski inequality. It will be useful for us to prove some of results related to the volume of a convex body.

Theorem 0.1 (Brunn-Minkowski inequality). Let, A, B be non-empty compact subsets of

n R . Then,

A + B 1/n A 1/n + B 1/n, | | ≥ | | | | or, equivalently, for all λ [0, 1], ∈

λA + (1 λ)B A λ B 1−λ, | − | ≥ | | | |

n where A + B = a + b R : a A, b B is the Minkowski sum of A and B, and λA = { ∈ ∈ ∈ } n λa R : a A . { ∈ ∈ }

In order to study local properties of convex bodies, we may consider two different metrics among convex bodies: the Banach-Mazur distance and the Hausdorff distance. The classical

Banach-Mazur distance dBM between two symmetric bodies is defined by

dBM (K,L) = min r 1 : L TK rL for T GL(n) , ≥ ⊂ ⊂ ∈ n o

n where GL(n) is the set of all invertible linear transformations on R . More generally, in case that we do not have symmetry condition on bodies K and L, it can be defined as

dBM (K,L) = min r 1 : AL BK rAL for A, B AL(n) , ≥ ⊂ ⊂ ∈ n o

n where AL(n) is the set of all invertible affine transformations on R . In addition, the

Hausdorff distance dH between two bodies K and L is defined by

dH(K,L) = max max min x y , max min x y , x∈K y∈L | − | y∈L x∈K | − |   5

or, equivalently,

n n dH(K,L) = min ε 0 : K L + εB ,L K + εB . ≥ ⊂ 2 ⊂ 2 n o

n We note that the volume function of bodies in R is continuous with respect to both the

Banach-Mazur distance and the Hausdorff distance. Moreover the Banach-Mazur distance satisfies the following classical theorem by F. John.

n Theorem 0.2 (John). For any convex symmetric body K R , we get ⊂

n dBM (K,B ) √n. 2 ≤

Here the equality holds if and only if K is an ellipsoid.

One of the consequences of the theorem is that the set of symmetric convex bodies in

n R is compact with respect to the Banach-Mazur distance.

0.2 Summary and Notation

In this dissertation we investigate two topics related to the volume in Convex Geometry; one is about duality phenomena, and the other is about intersection bodies.

First, the volume product of a symmetric convex body is defined as the product of volumes of the body and its polar body. The Blaschke-Santalo inequality says that the maximum of the volume product is attained at the Euclidean balls. On the other hand, the exact minimum of the volume product is not known in general. The Mahler’s conjecture, posted in 1939, is asking whether the cube/simplex is a minimizer for the volume product in the class of symmetric/general convex bodies in a fixed dimension. As a partial answer of the conjecture, we study the local minimality of the volume product at the cube, the simplex, and the Hanner polytopes.

Second, the intersection body of a star body is the star body whose radial function in 6

each direction is equal to the volume of the central section of the body perpendicular to the direction. A celebrated theorem of Busemann states that the intersection body of a symmetric convex body is convex. In this sense it is natural to ask how much of convexity is preserved under the intersection body operation. We provide quantitative versions of

Busemann’s theorem for a class of quasi-convex bodies, a class of uniformly convex bodies, and a class of bodies of revolution. More precisely we prove that the intersection body of a quasi-convex body is again quasi-convex for a proper choice of quasi-convex type, and also shown that every uniformly convex body with modulus of convexity of power type p has the intersection body of the same power type. Finally we apply our results to present partial answers to Lutwaks question on the uniqueness of the fixed point of the intersection operator.

List of Notation

n Let A be a non-empty subset of R . Denote by , the usual inner product. h· ·i conv(A) : the minimal convex set containing A (convex hull) • span(A) : the minimal linear subspace containing A (linear span) • aff(A) : the minimal affine subspace containing A (affine hull) • n n int(A) = a R :(a + εB2 ) aff(A) A for some ε > 0 (relative interior) • { ∈ ∩ ⊂ } ◦ n A = b R : a, b 1, a A (polar) • { ∈ h i ≤ ∀ ∈ } ⊥ n A = b R : a, b = 0, a A (orthogonal complement) • { ∈ h i ∀ ∈ } n A + B = a + b R : a A, b B (Minkowski sum) • { ∈ n ∈ ∈ } A B = a b R : a A, b B − { − ∈ ∈ ∈ } dim(A) : the dimension of span(A A)(dimension) • − diam(A) = sup a : a A A (diameter) • {| | ∈ − }

In addition, by c we denote a constant depending on dimension only, which may change

f(ε) from line to line, and write f(ε) = O(ε) if sup ε < for some ε0 > 0, and f(ε) = o(ε) 0<ε<ε0 ∞

7

f(ε) if lim ε = 0. Denote by ∂K the boundary of a convex body K. The origin in the ε→0+

Euclidean space n of any dimension is always denoted by 0. If A is a measurable set of R n dimension k in R , we denote by A the k-dimensional volume (Lebesgue measure) of A. | | n There should be no confusion with the notation for the Euclidean norm of a vector x R , ∈ n which is x = x, x . We work in the Euclidean space R of a fixed dimension n with the | | h i p standard basis e1, , en . { ··· }

0.3 Bibliographic Note

Many of the results presented in this dissertation have been published or submitted for publication. The results in Chapter 2 and 3 are has been submitted for publication

[40]. Chapter 4 is joint work with Shlomo Reisner appeared in [41], Mathematika 57, 2011.

Chapter 5 is joint work with Artem Zvavitch submitted for publication [43]. The results of Chapter 7 are joint work with Vladyslav Yaskin and Artem Zvavitch appeared in [42],

Advances in Mathematics 226, 2011. Chapter 8 is joint work with M.Angeles Alfonseca submitted for publication. Part I

Mahler conjecture

8 Chapter 1. Introduction to the volume product

The notion of duality is one of the most important notions in mathematics, especially, in

Functional Analysis and Convex Geometry. In this chapter we introduce the concept of the volume product of a convex body, which is related to the duality (polarity) among convex bodies. The maximum of the volume product in the class of (symmetric) convex bodies is well-known from the Blaschke-Santalo inequality. On the other hand, the minimum of the volume product is not known in general. We will study basic properties of the volume product and discuss some results and open problems related to the volume product.

1.1 The volume product and its invariant properties

n z Let K be a convex body in R . We denote by K the polar body of K with respect to a point z in the interior of K as defined in Section 0.1.

n Definition 1.1. Let K be a convex body in R . The volume product of K is defined by

(K) = min K Kz : z int(K) . P | || | ∈ n o

It turns out that the function z Kz on the interior of K is strictly convex and tends 7→ | | to infinity as z approaches to the boundary of K (see [81] or [65, 82]). Thus the minimum

of the function z Kz is attained at a single point, called the Santal´opoint of K. 7→ | |

Definition 1.2. The Santal´opoint of a convex body K is the unique point z = s(K) at

which the volume of the polar of K with respect to z attains its minimum, i.e.,

Ks(K) = min Kz : z int(K) . | | | | ∈ n o 9 10

If K is a symmetric convex body, i.e., K = K, then the Santal´opoint of K should be − the origin because the function z Kz is even. So the volume product of K is equal to 7→ | |

K K◦ , if K is symmetric, (K) = | | | | P   K Ks(K) , in general. | | | |   Directly from the formula (TK)◦ = (T ∗)−1K◦ in Proposition 0.1, we have the following invariant property of the volume product.

n Proposition 1.1. Let K be a convex body in R . Then the volume product of K is invariant

n under any invertible affine transformation on R , that is,

n n (TK) = (K) for all invertible affine T : R R . (1.1) P P →

Remark 1.1. For the local behavior of the volume product, it is natural to consider the

Banach-Mazur distance between symmetric convex bodies because the volume product is invariant under linear transformations and the polarity.

To see another invariant property of the volume product, we start with the definition for the `p-sums of two convex bodies.

n Definition 1.3. For 1 p < , the `p-sum A +p B of two subsets A, B of R is defined as ≤ ∞ the set of all points of the form x, y, or λ1/px + (1 λ)1/py for x A, y B and λ (0, 1). − ∈ ∈ ∈

In particular, if A, B are convex, then their `1-sum and `∞-sum are given by

A +1 B = conv(A B) and A +∞ B = A + B. ∪

We write A p B instead of A +p B if span(A) span(B) = 0 . ⊕ ∩ { }

In case that both A and B are star-shaped at the origin, the `p-sum of A and B can be 11

given in terms of their Minkowski functionals as follows.

1/p a p + b p , if 1 p <  k kA k kB ≤ ∞ (a, b) A⊕ B = k k p   max a , b , if p = . {k kA k kB} ∞   The classical results on the `p-sums of two Banach spaces in Functional Analysis say that

n n the polar body of the `p-sum of two symmetric convex bodies K R 1 and L R 2 is ⊂ ⊂ ◦ ◦ equal to the `q-sum of K and L for q with 1/p + 1/q = 1, i.e.,

◦ ◦ ◦ (K p L) = K q L . (1.2) ⊕ ⊕

The next lemma is a standard computation.

n n Lemma 1.1. Let K R 1 , L R 2 be symmetric convex bodies. Then the volume of the ⊂ ⊂ n n n `p-sum of K and L in R = R 1 R 2 is given by ⊕

n1 n2 Γ( p + 1)Γ( p + 1) K L = K L , p n1+n2 | ⊕ | Γ( p + 1) | || | where Γ( ) is the gamma function (see e.g. [47]). ·

Proof. Note that

∞ ∞ −kzkp −t −t e A dz = e dtdz = e dzdt n n kzkp 0 z∈t1/pA ZR ZR Z A Z Z ∞ ∞ = t1/pA e−tdt = A tn/pe−tdt = A Γ(n/p + 1). | | | | | | Z0 Z0

Since

p −k(x,y)k −kxkp −kykp e K⊕pL dxdy = e K dx e L dy, n n n n ZR 1 ×R 2 ZR 1 ZR 2 12

we have

K p L Γ(n/p + 1) = K Γ(n1/p + 1) L Γ(n2/p + 1), | ⊕ | | | · | | which completes the proof.

The above lemma gives another invariant property for the volume product as follows.

Proposition 1.2 ([80, 49]). The volume product of the `p-sum of two symmetric convex

bodies is the same as that of the `q-sum for p, q [1, ] with 1/p + 1/q = 1, that is, ∈ ∞

(K p L) = (K q L) (K 1 L). P ⊕ P ⊕ ≥ P ⊕

Moreover the lemma makes it possible to calculate the exact volume of of the `p-balls

and its volume product .

Corollary 1.1. The volume of the n-dimensional `p ball is equal to

[2Γ(1 + 1/p)]n Bn = , p (0, ]. (1.3) p Γ(1 + n/p) ∈ ∞

It gives that for p, q [1, ] with 1/p + 1/q = 1 ∈ ∞

[2Γ(1 + 1/p)]n [2Γ(1 + 1/q)]n (Bn) = (Bn) = . P p P q Γ(1 + n/p) · Γ(1 + n/q)

Especially, for p = 1 or , ∞ 4n (Bn ) = (Bn) = . (1.4) P ∞ P 1 n!

Remark 1.2. We recall from the Mahler conjecture that the simplex is a candidate for the minimizers of the volume product in the non-symmetric setting. The volume product of the 13

n-dimensional simplex can be calculated directly from the regular simplex ∆n as follows.

(n + 1)n+1 (∆n) = . (1.5) P (n!)2

1.2 Mahler conjecture and related results

It follows from John’s theorem [36] and the continuity of the volume function in the

Banach-Mazur compactum that the volume product attains its maximum and minimum.

n It turns out that the maximum of the volume product is attained at the Euclidean ball B2 .

The corresponding inequality is known as the Blaschke-Santal´oinequality [81, 73, 66].

n Theorem 1.1 (Blaschke-Santal´oinequality). For any convex body K R , ⊂

(K) (Bn). P ≤ P 2

Here the equality holds if and only if K is an ellipsoid.

For the minimum of the volume product, it was conjectured by Mahler in [60, 61] that

n n (K) is minimized at the cube B∞ in the class of symmetric convex bodies in R , and P n n minimized at the simplex ∆ in the class of general convex bodies in R .

Conjecture 1.1 (Mahler conjecture). It is asked whether the followings are true:

n 1. For any symmetric convex body K in R ,

4n (K) (Bn ) = (1.6) P ≥ P ∞ n!

n 2. For any convex body K in R ,

(n + 1)n+1 (K) (∆n) = . (1.7) P ≥ P (n!)2 14

The case of n = 2 was proved by Mahler [60]. It was also proved in several special cases,

like, e.g., unconditional bodies [80, 64, 77, 5], zonoids [76, 26, 10], bodies of revolution [63]

and bodies with some positive curvature assumption [84, 78, 27]. An isomorphic version of

the conjectures was proved by Bourgain and Milman [12]: there is a universal constant c > 0 such that (K) cn (Bn); see also different proofs in [50, 69, 23]. Functional versions of P ≥ P 2 the Blaschke-Santal´oinequality and the Mahler conjecture in terms of log-concave functions were investigated by Ball [4], Artstein, Klartag, Milman [2], and Fradelizi, Meyer [17, 19, 18].

For more information on Mahler’s conjecture, see expository articles [85, 62, 79].

From Proposition 1.2 we recall that the volume product is invariant between the `1-sum and the `∞. It suggests other candidates, different from the cube or the cross-polytope, for the minimizers of the volume product in the symmetric case.

Definition 1.4. A symmetric convex body H is called a Hanner polytope if H is one-

dimensional, or it is the `1 or `∞ sum of two (lower dimensional) Hanner polytopes.

n In other words every Hanner polytope in R can be obtained from n symmetric intervals by taking the `1 or `∞ sums. For example, H = ((I1 1 I2) ∞ I3) 1 (I4 ∞ I5) is ⊕ ⊕ ⊕ ⊕ 5 5 a Hanner polytope in R obtained from symmetric intervals I1,...,I5 R . Note that ⊂ every summand in the representation of H in terms of symmetric intervals is also a (lower

n dimensional) Hanner polytope. Note also that every Hanner polytope in R has the same

n volume product as the cube B∞ by Proposition 1.2.

For the study of a local property of the volume product, it is natural to consider the

Banach-Mazur distance among convex bodies, as see in Proposition 1.1 and the remark after

it. The local minimality of the volume product was first investigated in [70] by Nazarov,

Petrov, Ryabogin, and Zvavitch. Namely, they proved that the cube is a strict local min-

imizer of the volume product in the class of symmetric convex bodies endowed with the

Banach-Mazur distance. It turns out that the basic procedure of the proof used in [70] can 15

be applied for other polytopes such as the simplex and the Hanner polytopes. In case of the simplex, the technique of [70] can be adapted to the non-symmetric setting by showing the stability of the Santal´opoint, which leads to the local minimality of the simplex in non-symmetric setting given in Chapter 4 (see also [41]).

In case of Hanner polytopes, however, it is not so simple to get the same conclusion as the cube because the structure of a Hanner polytope may be much more complicated than that of the cube. To illustrate the structure of Hanner polytopes, we notice that

Hanner polytopes are in one-to-one correspondence with the (perfect) graphs which do not contain any induced path of edge length 3 (see Chapter 2). In this correspondence, the n-dimensional cube is associated with the graph of n vertices without any edges. Thus, a

Hanner polytope may have more delicate combinatorial structure that the cube, especially in large dimension. To overcome such difficulties, we employ and analyze the combinatorial representation of Hanner polytopes, which helps us prove the local minimality for Hanner polytopes given in Chapter 3 or [40].

In the class of unconditional convex bodies, Saint Raymond confirmed the Mahler con- jecture in [80], and Meyer and Reisner, independently, characterized the equality case. In

Chapter 5 (see also [43]) we present a stability version of these results and also show that any symmetric convex body, which is sufficiently close to an unconditional body, satisfies the the reverse Blaschke-Santal´oinequality (1.6). Chapter 2. More on Hanner polytopes

In this chapter we first investigate the combinatorial structure of Hanner polytopes associated with perfect graphs, and then we describe the facial structure of Hanner polytopes as a part of a general theory of the facial structure of the `1 or `∞ sum of two polytopes.

2.1 Connection with perfect graphs

n We recall that every Hanner polytope in R can be obtained from n symmetric intervals

n n in R by taking the `1 or `∞ sums. In particular, a Hanner polytope in R is called standard if it is obtained from the intervals [ e1, e1],..., [ en, en] by taking the `1 or `∞ sums. It − − is easy to see that every Hanner polytope is a linear image of a standard Hanner polytope.

Moreover, each coordinate of any vertex of a standard Hanner polytope is 0 or 1. The ± polytopes with this property as in standard Hanner polytopes are defined as follows.

n Definition 2.1. An unconditional polytope P in R is called a dual 0-1 polytope if each

coordinate of any vertex of P and P ◦ is 0 or 1. The support of a vertex v of a dual 0-1 ± polytope is defined by

supp(v) = j 1, . . . , n : v, ej = 0 . ∈ { } h i 6 n o

The definition of a dual 0-1 polytope was given in [75]; the term ‘a dual 0-1 space’ is used there, as a normed space with the unit ball being a dual 0-1 polytope.

16 17

Next we need the following basic notions from Graph Theory to give some connection between polytopes and graphs.

The complement G of a graph G is a graph with the same vertex set but whose edge • set consists of the edges not present in G.

A subgraph S of a graph G is said to be induced if two vertices of S are connected by • an edge of H if and only if they are connected by an edge of G.

A subset I of the vertex set of a graph G is called an independent set of G if no two • points in I are connected by an edge.

A subset J of the vertex set of a graph G is called a clique of G if any two points in • J are connected by an edge of G.

The chromatic number of a graph G is the minimum number of colors needed to color • the vertices of G in such a way that no two adjacent vertices have the same color.

A perfect graph is a graph in which the chromatic number of every induced subgraph • equals the size of the largest clique of that subgraph.

In [75], Reisner gave a connection between dual 0-1 polytopes and perfect graphs, which

n states that every dual 0-1 polytope P in R can be associated with the graph G = G(P )

with the vertex set 1, , n and the edge set defined as follows: { ··· }

i, j 1, . . . , n are connected by an edge of G, (denoted by i j), if ei + ej P . ∈ { } ∼ 6∈

3 Some examples for standard Hanner polytopes in R and their perfect graphs are provided

in the end of this section.

We note that for each i, j 1, . . . , n both the section and the projection of a dual 0-1 ∈ { }

polytope P by span ei, ej are dual 0-1 polytopes, that is, 2-dimensional `1- or `∞-balls. It { } ◦ ◦ gives that ei + ej P if and only if ei + ej P , and thus G(P ) = G(P ) where G denotes 6∈ ∈ the compliment of G. 18

n Theorem 2.1 ([75]). The dual 0-1 polytopes in R are in one-to-one correspondence with

perfect graphs with vertex set 1, . . . , n . { }

In particular, the standard Hanner polytopes, as a special family of dual 0-1 polytopes,

have one-to-one correspondence in the following way. See [83] or Lemma 3.5 in [75].

n Theorem 2.2. The standard Hanner polytopes in R are in one-to-one correspondence

with the graphs with vertex set 1, . . . , n not containing any induced path of edge length 3. { }

The following result gives a way to recover the dual 0-1 polytope P from a given perfect

graph G such that G = G(P ).

Proposition 2.1 (Lemma 2.5, [75]). Let G be the graph associated with a dual 0-1 polytope

n P . Then a point v R is a vertex of P if and only if each coordinate of v is 1, 0, or 1, ∈ − and the support of v is a maximal independent set of G.

Here, the maximality of cliques and independent sets comes from the partial order of inclu-

? n ◦ ? sion. Similarly, a point v R is a vertex of H if and only if each coordinate of v is 1, ∈ − 0, or 1, and the support of v? is a maximal clique of G.

In addition, it is known [32, 34, 75] that every Hanner polytope H satisfies CL-property:

v, v? = 1 for every vertex v of H and every vertex v? of H◦. |h i|

Equivalently it can be stated in the following way.

Proposition 2.2. Let G be the perfect graph associated with a standard Hanner polytope.

Then there exists a (unique) common element between any maximal independent set and

any maximal clique in G. 19

Hanner polytopes Perfect graph Max. indep. Vertices

3 H R G = G(H) sets of G of H ⊂

e3

1 e2

e1

H = (I I) I 2 3 1, 2, 3 ( 1, 1, 1) ⊕∞ ⊕∞ { } ± ± ±

e3

e2 1 e1 1 ( 1, 0, 0) { } ± 2 (0, 1, 0) { } ± H = (I 1 I) 1 I 2 3 3 (0, 0, 1) ⊕ ⊕ { } ±

e3

e 2 1

e1

1, 3 ( 1, 0, 1) { } ± ± H = (I 1 I) I 2 3 2, 3 (0, 1, 1) ⊕ ⊕∞ { } ± ±

e3

1

e2

e1 3 (0, 0, 1) { } ± H = (I I) 1 I 2 3 1, 2 ( 1, 1, 0) ⊕∞ ⊕ { } ± ± 20

2.2 Direct sums of polytopes

In this section we investigate basic properties of the direct sum (mostly the `1 or `∞ sum) of two polytopes to get the general form of a face and of its centroid. In the end of this section we provide the facial structure of the `1 or `∞ sum of two polytopes in terms of the summands.

A subset F of P is called a face of P if there exists a supporting hyperplane A of P such that F = P A. In particular, a face P is called a vertex if its dimension is 0, and a facet if ∩ its dimension is n 1. The empty set ∅ and the entire polytope P are always considered − as improper faces of P of dimensions 1 and n, respectively. −

Definition 2.2. The dual face F ∗ of a face F is defined by

F ∗ = y P ◦ : x, y = 1, x F . ∈ h i ∀ ∈ n o

Note that F ∗ is a face of P ◦ with dim F ∗ = n 1 dim F (see [29, Ch. 3]). In case of − − ∗ ∗ ◦ ∗ improper faces, we get P = ∅ and ∅ = P ; hence the formula dim F = n 1 dim F − − works even for improper faces F .

0 k n−1 Definition 2.3. An ordered set F = F ,...,F , ,F consisting of n faces of P is ··· called a flag over P if each F k is of dimension k and F 0 F 1 F n−1. ⊂ ⊂ · · · ⊂

n n We recall that the `1/`∞-sum of convex sets A R 1 and B R 2 is given by ⊂ ⊂

A +1 B = conv(A B) and A +∞ B = A + B. ∪

We write A p B instead of A +p B if span(A) span(B) = 0 . In addition, in case of ⊕ ∩ { } p = , we mostly use the notation A + B, A B without subscript. ∞ ⊕

n n Lemma 2.1. Let A, B be non-empty convex subsets of R 1 , R 2 respectively, and let 21

n n n R = R 1 R 2 . Then ⊕ dim(A B) = dim A + dim B. ⊕

If aff(A) aff(B) = ∅ (i.e., at least one of aff(A) and aff(B) does not contain the origin), ∩ then

dim(A 1 B) = dim A + dim B + 1. ⊕

In addition, the second formula holds even if one of the summands is the empty set under the convention dim(∅) = 1. −

Proof. If one of the summands is the empty set, say B = ∅, then dim(A 1 ∅) = dim A = ⊕ dim A + dim ∅ + 1. If the dimensions of non-empty sets A, B are k1, k2 respectively, then there exist affinely independent sets a0, a1, . . . , ak A and b0, b1, . . . , bk B. { 1 } ⊂ { 2 } ⊂ Consider the sets

S1 = a0, a1, . . . , ak , b0, b1, . . . , bk A 1 B 1 2 ⊂ ⊕ n o and

S∞ = a0 + b0, a0 + b1, . . . , a0 + bk , a1 + b0, . . . , ak + b0 A ∞ B. 2 1 ⊂ ⊕ n o

Then, aff(S1) contains A, B, and also their convex hull, so aff(S1) A 1 B. In addition, ⊃ ⊕

aff(S∞) contains a0 + aff(B), aff(A) + b0 and their convex hull, so aff(S∞) A ∞ B. So, ⊃ ⊕

it remains to prove that each of the sets S1 and S∞ is affinely independent. To verify the

independence of S1, assume that 0 aff(A) and 6∈

k1 k2 λjaj µjbj = 0 − j=0 j=0 X X

k1 k2 k1 k2 where λj’s and µj’s satisfy j=0 λj = j=0 µj. It implies j=0 λjaj = j=0 µjbj = 0

because the origin is the onlyP common pointP of span(A) and span(P B). Since aff(P A) does not

Pk1 k1 j=0 λj aj contain the origin, then j=0 λj = 0; otherwise, 0 = Pk1 aff(A). Thus j λj = 0 j=0 λj ∈ P P 22

and µj = 0. The affine independence of each of a0, . . . , ak and b0, . . . , bk gives j { 1 } { 2 }

λ0 =P = λk = 0 and µ0 = = µk = 0. We now prove the affine independence of S∞. ··· 1 ··· 2 Assume that k1 k2 α(a0 + b0) + λj(aj + b0) + µj(a0 + bj) = 0 (2.1) j=1 j=1 X X

k1 k2 where α, λj’s and µ’s satisfy α + j=1 λj + j=1 µj = 0. Projecting the equality (2.1) onto

n n R 1 and R 2 respectively, we getP P

k2 k1 k1 k2 α + µj a0 + λjaj = 0 and α + λj b0 + µjbj = 0.     j=1 j=1 j=1 j=1 X X X X    

Finally, the affine independence of each of a0, . . . , ak and b0, . . . , bk gives λ1 = = { 1 } { 2 } ···

λk = 0, µ1 = = µk = 0, and α = 0. 1 ··· 2

n n Lemma 2.2. Let A, B be non-empty convex subsets in R 1 , R 2 respectively, and let

n n n R = R 1 R 2 . Then the centroid of the `∞-sum of A, B is given by ⊕

c(A B) = c(A) + c(B), ⊕

where c(A) denotes the centroid of A. If aff(A) aff(B) = ∅ (i.e., at least one of aff(A) ∩

and aff(B) does not contain the origin), then the centroid of the `1-sum of A, B is

dim A + 1 dim B + 1 c(A 1 B) = c(A) + c(B). ⊕ dim A + dim B + 2 dim A + dim B + 2

In addition, the second formula holds even if one of the summands is the empty set ∅ under the conventions dim(∅) = 1 and c(∅) = 0. − n Proof. Since c(TA) = T c(A) for any invertible linear transformation T on R , we may

n n n n n assume that R = R 1 R 2 is the orthogonal sum of R 1 and R 2 . Let k1 = dim A, ⊕ k2 = dim B. First, consider the `∞-sum case. If, say, k1 = 0, then c(A B) = c(c(A)+B) = ⊕ 23

c(A) + c(B). So, we may assume k1, k2 1. The convex sets A, B can be viewed as subsets ≥ k k of R 1 , R 2 respectively. Then

1 1 c(A B) = (x + y) dxdy = (x + y) dxdy ⊕ A B A B | ⊕ | ZA ZB | || | ZA ZB 1 1 = x dx + y dy = c(A) + c(B), A B | | ZA | | ZB

k k where dx, dy are the Lebesgue measures on R 1 , R 2 .

For the `1-sum case, let E = aff(A)+aff(B) c(B), and θ = c(B) c(A). Then, θ E E − − 6∈ − because E E θ = aff(A) + aff(B) c(A) c(B) (c(B) c(A)) = aff(A) aff(B) does − − − − − − − not contain the origin, and

tθ + E = t(c(B) c(A)) + aff(A) + aff(B) c(B) − − = (1 t) aff(A) + t aff(B). −

Choose ϕ Sn−1 span(E θ ) which is orthogonal to E. Thus, from ∈ ∩ ∪ { }

A 1 B (tθ + E), ⊕ ⊂ 0≤t≤1 [

we get 1 A 1 B = (A 1 B) (tθ + E) θ, ϕ dt. | ⊕ | ⊕ ∩ |h i| Z0

Since

(A 1 B) (tθ + E) = ((1 s)A + sB) ((1 t)aff(A) + taff(B)) ⊕ ∩ − ∩ − 0≤s≤1 [ = (1 t)A + tB, − 24

we get

1 1 A 1 B = (1 t)A + tB θ, ϕ dt = θ, ϕ (1 t)A tB dt | ⊕ | 0 − |h i| |h i| 0 − Z 1 Z k1!k2! θ, ϕ A B = θ, ϕ A B (1 t)k1 tk2 dt = |h i| | || | . |h i| | || | − (k + k + 1)! Z0 1 2

Similarly, 1 z dz = z dz θ, ϕ dt. |h i| ZA⊕1B Z0 Z(1−t)A+tB ! Note that (1 t)A + tB −1 x dx is the centroid of (1 t)A + tB, which is equal | − | (1−t)A+tB − to (1 t)c(A) + tc(B) by theR `∞-case. Thus −

1 x dx = θ, ϕ (1 t)c(A) + tc(B) (1 t)A + tB dt |h i| − − ZA⊕1B Z0  1  1 = θ, ϕ A B c(A) (1 t)k1+1 tk2 dt + c(B) (1 t)k1 tk2+1 dt |h i| | || | − −  Z0 Z0  k1!k2! θ, ϕ A B k1 + 1 k2 + 1 = |h i| | || | c(A) + c(B) (k + k + 1)! k + k + 2 k + k + 2 1 2  1 2 1 2 

Dividing the above by A 1 B , we get the formula for c(A 1 B). The case that a summand | ⊕ | ⊕ contains the empty set is clear.

The description of extreme points of the `1 or `∞ sum of two convex sets is well known

(e.g., [86]). By ext(A) we denote the set of extreme points of a convex set A. Then the

extreme points of the `1 or `∞ sum of two convex sets are described in the following way:

n n n n n if A R 1 , B R 2 are convex sets and R = R 1 R 2 , then ⊂ ⊂ ⊕

ext(A 1 B) = ext(A) ext(B) and ext(A B) = ext(A) + ext(B). (2.2) ⊕ ∪ ⊕

In particular, if A, B are polytopes, then it gives a characterization of the vertices (zero- dimensional faces) of the sum A 1 B or A B. Moreover, it can be generalized to the faces ⊕ ⊕ 25

of any dimension as follows.

n n Lemma 2.3. Let P1 R 1 , P2 R 2 be convex polytopes containing the origin, and let ⊂ ⊂ n n n R = R 1 R 2 . Then, F is a face of P1 1 P2 if and only if F is of the form ⊕ ⊕

F1 1 ∅, ∅ 1 F2, or F1 1 F2, ⊕ ⊕ ⊕

where F1, F2 are faces of P1, P2. In case of the `∞-sum, F is a face of P1 ∞ P2 if and ⊕ only if F is of the form

F1 P2,P1 F2, or F1 F2, ⊕ ⊕ ⊕

where F1, F2 are faces of P1, P2.

Proof. First, consider the `1-sum case. Let F be a face of P1 1 P2. Then there exists a ⊕

supporting hyperplane A of P1 1 P2 such that F = A (P1 1 P2). By (2.2), every face ⊕ ∩ ⊕

F of P1 1 P2 can be expressed as F = F1 1 F2 where Fj is the convex hull of some of ⊕ ⊕

extreme points of Pj for j = 1, 2. In addition, for each j = 1, 2,

nj nj nj nj (A R ) Pj = A (R (P1 1 P2)) = R (A (P1 1 P2)) = R F ∩ ∩ ∩ ∩ ⊕ ∩ ∩ ⊕ ∩

nj = R (F1 1 F2) = Fj. ∩ ⊕

n n n If Fj = ∅, then A R j is a hyperplane in R 1 with (A R j ) Pj = Fj, which means that 6 ∩ ∩ ∩

Fj is a face of Pj.

For the `∞-sum case, let F be a face of P1 ∞ P2. Then there exists a hyperplane A of ⊕ ⊥ P1 ∞ P2 with F = A (P1 ∞ P2). Write A = x1 + x2 + θ for x1 P1, x2 P2 with ⊕ ∩ ⊕ ∈ ∈

x1 + x2 F . Note that y1 + x2, θ x1 + x2, θ and x1 + y2, θ x1 + x2, θ for each ∈ h i ≤ h i h i ≤ h i y1 P1, y2 P2. It implies y1, θ x1, θ and y2, θ x2, θ for each y1 P1, y2 P2. ∈ ∈ h i ≤ h i h i ≤ h i ∈ ∈ 26

Thus, y1 + y2, θ = x1 + x2, θ if and only if y1, θ = x1, θ and y1, θ = x1, θ . In other h i h i h i h i h i h i words,

⊥ F = (P1 ∞ P2) A = (P1 + P2) (x1 + x2 + θ ) ⊕ ∩ ∩ ⊥ ⊥ = P1 (x1 + θ ) + P2 (x2 + θ ). ∩ ∩

⊥ n Let Fj = Pj (xj + θ ) for each j = 1, 2. If θ R j for j = 1 or 2, then Fj is a face ∩ ∈

of Pj and the other summand is an improper face. Otherwise, each Fj is a face of Pj for

j = 1, 2.

n n Lemma 2.4. Let P1 R 1 , P2 R 2 be convex polytopes containing the origin in their ⊂ ⊂ n n n interiors, and let R = R 1 R 2 . Then, the dual face of a face of the `1 or `∞ sum of P1 ⊕ and P2 is given by

∗ ∗ ◦ ∗ ◦ ∗ ∗ ∗ ∗ (F1 1 ∅) = F1 P2 , (∅ 1 F2) = P1 F2 , (F1 1 F2) = F1 F2 , ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ∗ ∗ ∗ ∗ ∗ ∗ ∗ (F1 P2) = F1 1 ∅, (P1 F2) = ∅ 1 F2 , (F1 F2) = F1 1 F2 , ⊕ ⊕ ⊕ ⊕ ⊕ ⊕

∗ where F1, F2 are faces of polytopes P1, P2 respectively, and F denotes the dual face of a face F .

Proof. Let F = F1 1 F2 where each Fj is ∅ or a face of Pj for j = 1, 2. Then, by Lemma ⊕ 2.3, F ∗ F ∗ is a face of P ◦ P ◦, and clearly F ∗ F ∗ F ∗. Moreover, it follows from 1 ⊕ 2 1 ⊕ 2 1 ⊕ 2 ⊂ Lemma 2.1 and the formula dim F ∗ = n 1 dim F that the dimension of F ∗ F ∗ is the − − 1 ⊕ 2 same as that of F ∗. So, aff(F ∗ F ∗) = aff(F ∗). It implies 1 ⊕ 2

∗ ∗ ∗ ◦ ∗ ∗ ◦ ◦ (F1 1 F2) = F = aff(F ) (P1 1 P2) = aff(F F ) (P P ) ⊕ ∩ ⊕ 1 ⊕ 2 ∩ 1 ⊕ 2 = (aff(F ∗) P ◦) (aff(F ∗) P ◦) = F ∗ F ∗. 1 ∩ 1 ⊕ 2 ∩ 2 1 ⊕ 2 27

∗ ∗ whenever each Fj is ∅ or a face Pj for j = 1, 2. In addition, replacing F1 , F2 with G1, G2,

∗ ∗ ∗ ∗ ∗ ∗ we get (G G ) = G1 1 G2. It implies (G1 1 G2) = G G whenever each Gj is Pj 1 ⊕ 2 ⊕ ⊕ 1 ⊕ 2 or a face Pj.

Remark 2.1 (More on the faces). We describe the faces of the sum of two polytopes.

Lemma 2.1 and Lemma 2.3 make it possible to describe the faces over the sum with the

n n n dimensions of summands. Let P1 R 1 , P2 R 2 be convex polytopes, and let R = ⊂ ⊂ n n k−1 R 1 R 2 . For each k = 1, . . . , n, every face F of dimension k 1 of P1 1 P2 is of the ⊕ − ⊕ form

k−1 k1−1 k2−1 F = F 1 F , (2.3) 1 ⊕ 2

n−k and every face F of dimension n k of P1 ∞ P2 is of the form − ⊕

F n−k = F n1−k1 F n2−k2 , (2.4) 1 ⊕ 2

d where 0 k1 n1, 0 k2 n2, k1 + k2 = k, and each F denotes a face or an improper ≤ ≤ ≤ ≤ j face of Pj of dimension d for j = 1, 2.

If a face F = F1 1 F2 (respectively F1 F2) is contained in another face G = G1 1 G2 ⊕ ⊕ ⊕

(respectively G1 G2), then it follows from (2.2) that F1 G1 and F2 G2. Furthermore, ⊕ ⊂ ⊂ if F G with dim G = dim F + 1, then we have, by Lemma 2.1, either ⊂

F1 = G1,F2 ( G2, dim G2 = dim F2 + 1, or F2 = G2,F1 ( G1, dim G1 = dim F1 + 1.

We introduce the following notion to describe a flag over the sum of two polytopes.

n n n n Definition 2.4. Let P1, P2 be convex polytopes in R 1 , R 2 respectively, and R = R 1 ⊕ n 0 n−1 n R 2 . A flag F = F ,...,F over P1 1 P2 is of type (j1, . . . , jn) 1, 2 if each ⊕ ∈ { } element of F is obtained in the following way: 28

0 1. The face F of dimension 0 is the `1-sum of a zero-dimensional face of Pj1 and ∅.

k−1 2. For 2 k n, the face F of dimension k 1 is obtained from the `1-decomposition ≤ ≤ − k−2 of F by increasing the dimension of the jk-th summand (a face of Pjk ), i.e., if

k−2 F = F1 1 F2, then ⊕

F1 1 F2, if jk = 1 F k−1 =  ⊕   Fe1 1 F2, if jk = 2 ⊕   e where each Fj is a face of Pj of dimension 1 + dim Fj containing Fj for j = 1, 2.

e 0 n−1 In case of the `∞ sum, we say that a flag F = F ,...,F over P1 ∞ P2 is of type ⊕ ∗ n−1 ∗ 0 ∗ (j1, . . . , jn) if its dual flag F = (F ) ,..., (F ) is of type (j1, . . . , jn), as a flag over

◦ ◦ P 1 P .  1 ⊕ 2

n Remark 2.2 (Description of flags). Note that every flag over the `1 or `∞ sum in R =

n n n n R 1 R 2 of P1 R 1 and P2 R 2 has a type as an n-tuple consisting of n1 copies of 1 ⊕ ⊂ ⊂ and n2 copies of 2, that is, an element of

n #{k:jk=1}=n1 J = (j1, . . . , jn) 1, 2 . (2.5) ∈ { } #{k:jk=2}=n2 n o

In addition, for p = 1 or , every flag F over the `p-sum of P1 and P2 gives two flags F1, ∞ F2 in lower dimensions, defined by

F1 = F1 F1 p F2 F for some F2, 0 dim F1 < n1 ⊕ ∈ ≤ (2.6) n o

F2 = F2 F1 p F2 F for some F1, 0 dim F2 < n2 . ⊕ ∈ ≤ n o

Every flag F is uniquely determined by lower dimensional flags F1, F2 and a type σ J. ∈ 0 n1−1 0 n2−1 Moreover, two flags F1 = F1 ,...,F , F2 = F2 ,...,F in lower dimensions and { 1 } { 2 } 29

0 n−1 a type σ = (j1, . . . , jn) J can be associated with the flag F = F ,...,F defined by, ∈ { } for 1 k n, ≤ ≤

k−1 σ1(k)−1 σ2(k)−1 F = F 1 F (`1 -sum case) 1 ⊕ 2

n−k n1−σ1(k) n2−σ2(k) F = F F (`∞-sum case) 1 ⊕ 2

where σj(k) denotes the number of j’s among the first k entries j1, . . . , jk of σ = (j1, . . . , jn) for each j = 1, 2. We can also see that this flag constructed from F1, F2 and σ is of type

σ and has the lower dimensional flags F1, F2. Consequently, every flag F over the `1 or `∞ sum of P1 and P2 can be considered as an triple of a type σ J and two lower dimensional ∈ flags F1, F2 over P1, P2.

0 1 2 3 0 1 2 Example 2.1. Let F1 = F1 ,F1 ,F1 ,F1 and F2 = F2 ,F2 ,F2 be flags over a polytope 4  3  P1 R and a polytope P2 R respectively. Then ⊂ ⊂

0 1 1 0 2 0 2 1 2 2 3 2 F = F1 1 ∅,F1 1 ∅,F1 1 F2 ,F1 1 F2 ,F1 1 F2 ,F1 1 F2 ,F1 1 F2 ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ n o is a flag over the `1-sum of type (1, 1, 2, 1, 2, 2, 1).

Example 2.2. Consider the following flags over Hanner polytopes in dimension 2 or 3. In

0 1 2 the first figure the flag F ,F ,F over (I 1 I) ∞ I is given as ⊕ ⊕ 

2 F = [ e2, e1] ∞ [ e3, e3] − ⊕ − 1 F = e1 ∞ [ e3, e3] { } ⊕ − 0 F = e1 ∞ e3 . { } ⊕ { }

Thus the first flag is of type (1, 1, 2). 30

e3

1 0 F cF 1 F F 0 F 2

0 e2 e1 0 cF 2 cF 1

F 1 e1 e − 2

H = (I 1 I) I H = I I ⊕ ⊕∞ ⊕∞

0 1 In the second figure the flag F ,F over I ∞ I is given as ⊕ 

1 F = [ e1, e1] ∞ e2 − ⊕ { } 0 F = e1 ∞ e2 , { } ⊕ { }

which is of type (2, 1). Chapter 3. Local minimality in the symmetric case

n In this chapter we prove that every Hanner polytope in R is a strict local minimizer of the volume product in the class of symmetric convex bodies endowed with the Banach-

Mazur distance. More precisely, we will prove the following result.

n Theorem 3.1. Let K R be a symmetric convex body close enough to one of Hanner ⊂ polytopes in the sense that

n δ = min dBM (K,H) 1 : H is a Hanner polytope in R − n o is small enough. Then

(K) (Bn ) + c(n)δ P ≥ P ∞

where c(n) > 0 is a constant depending on the dimension n only.

3.1 Construction of Polytopes

n We fix a convex polytope P in R containing the origin in its interior, and we restate

the key steps from [70] for the polytope P . For each face F of P (resp. P ◦), choose an

affine subspace AF satisfying

◦ (a) AF intersects P (resp. P ) at a single point, denoted by cF , in the interior of F .

(b) dim A + dim F = n 1. F −

In addition we assume that every pair of a face F and its dual face F ∗ satisfies

(c) x, y = 1 for every x A , y A ∗ . h i ∈ F ∈ F

31 32

F F cF yF AF xF

AF

K span(A ) ∩ F span(AF ) H = (I 1 I) I ⊕ ⊕∞

n Let K be a convex body in R containing the origin in its interior. In order to construct

the points xF , yF for each face F of P , we move the affine subspace AF in the direction

of cF (or equivalently, dilate AF from the origin) until it is tangent to K. On the affine

subspace tangent to K obtained by moving AF , take the point xF on the boundary of K

that is nearest to cF , and the point yF on the line containing the origin and cF . In other

words, the points xF , yF are taken so that

x (t A ) ∂K and y = t c , (3.1) F ∈ F F ∩ F F F

where t = tF > 0 is chosen so that the affine subspace tAF is tangent to K. For the dual

◦ ∗ ◦ case, we get the points xF ∗ , yF ∗ by replacing P , F , AF , K with P , F , AF ∗ K , respectively.

0 k n−1 We recall that a flag over a polytope P is an ordered set F = F ,...,F , ,F ··· consisting of n faces of P such that each F k is of dimension k andF 0 F 1 F n−1 . ⊂ ⊂ · · · ⊂

n Definition 3.1. Let P be a convex polytope in R containing the origin in its interior, F the set of all faces of P , and N the number of all faces of P .

n N 1. The volume function V :(R ) R+ is defined by →

n N V (Z) = Z for Z =(zF )F ∈F (R ) , | F| ∈ : flag over P F X 33

where ZF denotes the simplex defined by

0 n−1 Z = conv 0, z 0 , . . . , z n−1 for Z = (zF )F ∈F , F = F , ,F . (3.2) F F F ···  

n 2. Let K be a convex body in R containing the origin. Suppose that the affine subspaces

AF , AF ∗ for each face F satisfy the conditions (a), (b), (c). Then we consider the points

nN in R defined by

X =(xF )F ∈F ,Y =(yF )F ∈F ,C =(cF )F ∈F ,

∗ ∗ ∗ X =(xF ∗ )F ∈F ,Y =(yF ∗ )F ∈F ,C =(cF ∗ )F ∈F .

where cF denotes the unique common point of F and AF , and xF , yF , xF ∗ , yF ∗ are the

points constructed in (3.1) from K, K◦.

The point C = (cF ) depends on P and AF ’s only, but X = (xF ), Y = (yF ) also depend on K. With the notation of (3.2), each flag F gives the simplices XF, YF, CF. Then, the polytopes P , P ◦ can be written as

◦ ∗ P = CF and P = CF∗ . (3.3) [F [F

Indeed, we can use the induction on n to prove that every polytope P of dimension n containing the origin satisfies P = C whenever c int(F ) for each face F of P . Assume F F F ∈ that it is true for any (n 1)-dimensionalS polytope containing the origin. Let P be a polytope − of dimension n containing the origin. Then, every facet G of P is an (n 1)-dimensional − polytope, and each flag over G (as a polytope of dimension n 1) can be viewed as a flag − over P containing G. Thus, the inductive assumption gives G = conv c : F F F3G { F ∈ } by viewing each facet G as an (n 1)-dimensional polytope and theS point c as the origin. − G 34

The convex hull of the origin and G is the same as that of the origin and cF ’s for any face

F F. Thus, conv( 0 G) = conv( 0 c : F F ) because the right hand ∈ F3G { } ∪ F3G { } ∪ { F ∈ } sideS is already convex. Finally, S

P = conv( 0 cF : F F ) = C = C . { } ∪ { ∈ } F F G 3G G 3G [ F[ [ F[ [F

Moreover, the interiors of any two simplices CF, CF0 in the above union are disjoint. In-

0 deed, if the facet in a flag F is different from that of another flag F , then the intersection of

0 conv c : F F and conv c : F F is at most (n 2)-dimensional. Thus, the intersec- { F ∈ } { F ∈ } −

tion of C and C 0 is at most (n 1)-dimensional, which implies that the interiors of C and F F − F 0 CF0 are disjoint. On the other hand, if F and F contain a common facet, then the facet can be viewed as an (n 1)-dimensional polytope and the c -point as the origin. Using the same − F

(inductive) argument as above, we conclude that the interiors of any different simplices CF,

CF0 are disjoint. Furthermore, if every xF -point is close enough to the cF -point, then the

above argument can be applied to prove that the interiors of any two simplices XF, XF0 are also disjoint. In that case, V (X) can be viewed as the volume of the (star-shaped, not

necessary convex) polytope, defined by the union of simplices XF’s over all flags.

n Proposition 3.1. Let P be a convex polytope, and K a convex body in R such that 0 ∈

int(P K). Let A , A ∗ be affine subspaces satisfying the conditions (a), (b), (c), and let ∩ F F

xF , yF , cF , xF ∗ , yF ∗ , cF ∗ be the points obtained from K and AF , AF ∗ , as in (3.1). Then

x , x ∗ = y , y ∗ = 1. h F F i h F F i

Moreover, if δ = dH(K,P ) is small enough, then x c cδ and y c cδ where | F − F | ≤ | F − F | ≤

c > 0 does not depend on K, but may depend on P , F and AF . 35

Proof. Note first that

n AF ∗ = y R : x, y = 1 x AF . (3.4) ∈ h i ∀ ∈ n o Indeed, if we denote the set on the right hand side by B, then the condition (c) implies

A ∗ B. Moreover, since c , y = 1 and x c , y = 0 for each x A , y B, we have F ⊂ h F i h − F i ∈ F ∈ ⊥ ⊥ B (c ∗ + c ) (A c ) . So, the dimension of B is at most n 1 dim A which is ⊂ F F ∩ F − F − − F ∗ equal to n 1 (n 1 dim F ) = dim F = n 1 dim F = dim A ∗ by the condition (b) − − − − − − F ∗ n and the formula dim F = n 1 dim F . Thus A ∗ = B = y R : x, y = 1 x A . − − F { ∈ h i ∀ ∈ F }

Let t = tF be such that tAF is tangent to K at xF . Consider a hyperplane A containing tA that is tangent to K at x . Let z be the dual point of A, i.e., y, z = 1 for every y A. F F h i ∈ ◦ Then z K , and ty, z = 1 for all y A , which implies tz A ∗ by (3.4). Moreover, ∈ h i ∈ F ∈ F 1 ◦ 1 A ∗ is tangent to K at z. Indeed, every p A ∗ satisfies tp, q = 1 for each q A and t F ∈ t F h i ∈ F 1 ◦ 1 hence p, x = 1, so every p A ∗ is not in the interior of K . It implies that A ∗ is h F i ∈ t F t F ◦ 1 tangent to K at z. Therefore, we get t ∗ = , which gives 1 = t t ∗ = y , y ∗ = x , x ∗ . F t F F h F F i h F F i For the second part, replacing δ by cδ if necessary, we may assume that (1 δ)P − ⊂ K (1 + δ)P . Then x tA ∂K and y = tc for some t with 1 δ t 1 + δ. So, ⊂ F ∈ F ∩ F F − ≤ ≤

y c = t 1 c c1δ. Consider P1 = P span(A ). Then c is a vertex of P1 | F − F | | − | · | F | ≤ ∩ F F

because cF is the unique common point of P1 and AF . It implies that there exists a constant

c2, depending on P , F , AF , such that for every δ > 0,

diam(P (1 δ)A ) = diam(P1 (1 δ)A ) c2δ. ∩ − F ∩ − F ≤

Finally, since xF , yF are contained in (1 + δ)P and also lie on tAF ,

t x y diam (1 + δ)P tA (1 + δ) diam P A | F − F | ≤ ∩ F ≤ · ∩ 1 + δ F     t c2(1 + δ) 1 2c2δ, ≤ − 1 + δ ≤   36

which implies x c (c1 + 2c2)δ. | F − F | ≤

Proposition 3.2. Suppose that , ◦ are divided into simplices , ∗ of equal volume, P P CF CF i.e.,

∗ ∗ 0 C = C 0 and C = C 0 for any two flags F, F . (3.5) | F| | F | | F| | F |

Then V (Y ) V (Y ∗) V (C) V (C∗) = P P ◦ . · ≥ · | | · | |

0 n−1 Proof. For each flag F = F ,...,F over P , 

∗ 1 Y Y = det y 0 , . . . , y n−1 det y 0 ∗ , . . . , y n−1 ∗ | F| | F | n!2 F F · (F ) (F ) 1

= det t 0 c 0 , . . . , t n− 1 c n−1 det t 0 ∗ c 0 ∗ , . . . , t n−1 ∗ c n−1 ∗ n!2 F F F F · (F ) (F ) (F ) (F ) n−1

1 = t k t k ∗ det c 0 , . . . , c n−1 det c 0 ∗ , . . . , c n−1 ∗ , n!2 F (F ) F F · (F ) (F ) k=0  Y 

∗ which is equal to C C because t t ∗ = y , y ∗ = 1 for any F by Proposition 3.1. From | F| | F| F F h F F i Cauchy-Schwarz inequality and the assumption (3.5),

2 2 1 1 1 1 ∗ ∗ 2 ∗ 2 2 ∗ 2 V (Y )V (Y ) = YF YF YF YF = CF CF . | |! | |! ≥ " | | | | # " | | | | # XF XF XF XF

∗ Since C C is constant for each flag F, we have | F| | F|

V (Y )V (Y ∗) C C∗ = P P ◦ . ≥ | F| | F| | | | | XF XF

The directional derivative of the function V along a vector Z = (zF ) at a point A = (aF ), which we denote by V 0(A),Z , is defined as h i

V (A + tZ) V (A) V 0(A),Z = lim − . t→0 t

37

n Proposition 3.3. Let P be a convex polytope in R containing the origin in its interior.

For each face F of P , choose an affine subspace AF satisfying the conditions (a), (b), (c), and consider the point C = (cF ) given in Definition 3.1. Suppose that

V 0(C),Z = 0 for every Z = z with all z A A . (3.6) F F ∈ F − F

2 Then V (X) V (Y ) cδ , provided that δ = dH(K,P ) is small enough, where c > 0 is a | − | ≤ constant depending on P and AF ’s.

Proof. The Talyor series expansion of V around the point C = (cF ) gives

V (X) = V (C) + V 0(C),X C + O( X C 2) − | − |

and

V (Y ) = V (C) + V 0(C),Y C + O( Y C 2) − | − |

By subtracting the above two equations, the assumption (3.6) gives

V (X) V (Y ) = V 0(C),X Y + O( X C 2) + O( Y C 2) − − | − | | − |

= O( X C 2) + O( Y C 2), | − | | − |

which is O(δ2) by Proposition 3.1.

Remark 3.1 (Plan for the proof). Note that the conditions (3.5), (3.6) in Propositions 3.2 and 3.3 depend on the polytope P only. If a polytope P satisfies these two conditions, then

Propositions 3.1, 3.2, and 3.3 give

V (X)V (X∗) P P ◦ cδ2, ≥ | || | − 38

where δ = dH(K,P ) is small enough and c > 0 is a constant depending on P and AF ’s only.

In Section 3.2, we fix a Hanner polytope H, and define affine subspaces AF ’s associated

with H that satisfy the conditions (a), (b), (c). The special choice of AF is not essential

to prove the condition (3.5), but is necessary for the proof of the condition (3.6). In fact,

the condition (3.5) is always true whenever the centroid of each face F of H is chosen as

the unique common point cF of AF and H. To complete the proof, we prove in Section ??

that there exists a symmetric convex body K˜ with dBM (K,K˜ ) = 1 + o(δ) such that

K˜ K˜ ◦ V (X˜)V (X˜ ∗) + c0δ, | || | ≥

˜ ˜ ∗ ˜ ˜ ◦ where X, X are the sets of the xF -points constructed from K, K , as in Definition 3.1.

For each Hanner polytope, we now define very particular affine subspaces AF satisfying

(a), (b), (c) in Section 3.1 to guarantee the conditions (3.5), (3.6) in Propositions 3.2, 3.3.

When a representation of a Hanner polytope H in terms of symmetric intervals is given, we

define the affine subspaces AF ’s inductively for all lower dimensional Hanner polytopes of

H which appear as a summand in the representation of H.

Definition 3.2. Define the affine subspaces AF ’s inductively for any Hanner polytope H

in the following way:

1. If H is one-dimensional, H has the only two faces (of dimension zero). For each face

F of a one-dimensional Hanner polytope H, define the affine subspace AF by AF = F.

2. Let H be a Hanner polytope obtained from two lower dimensional Hanner polytopes

H1 and H2 by taking the `1 or `∞-sum. Then, for the `1 sum case, the affine subspace 39

A for each face of the form F1 1 ∅, ∅ 1 F2, or F1 1 F2 is defined by F ⊕ ⊕ ⊕

AF1⊕1 ∅ = AF1 + span(H2),A∅ ⊕1F2 = span(H1) + AF2 ,

dim F1 + 1 dim F2 + 1 AF1⊕1F2 = AF1 + AF2 . (3.7) dim F1 + dim F2 + 2 dim F1 + dim F2 + 2

For the `∞ sum case, it is defined by

AF1⊕H2 = AF1 ,AH1⊕F2 = AF2

cF1 cF2 AF1⊕F2 = AF1 + AF2 + span , (3.8) dim H1 dim F1 − dim H2 dim F2 n − − o

where each cFj is the centroid of a face Fj of Hj for j = 1, 2.

n Lemma 3.1. Let H be any Hanner polytope in R , and F a face of H. Consider the affine

subspace AF defined as above. Then AF satisfies the conditions (a), (b), (c) from Section

3.1. Moreover, the unique common point cF of H and AF is the centroid of F .

Proof. We use the induction on the dimension n of H. The statement is trivial when

n dim H = 1. Let H R be the `1 or `∞ sum of lower dimensional Hanner polytopes ⊂ n n H1 R 1 , H2 R 2 . Assume that the affine subspaces A for any face Fj of Hj satisfy ⊂ ⊂ Fj the conditions (a), (b), (c) in Section 3.1, and H A = c for j = 1, 2. First consider ∩ Fj { Fj } the case that a face F of H contains an improper face as a summand. If, say, F = F1 1 ∅ ⊕ n or F1 H2, then the affine subspace A is given by A = A + R 2 or A = A . ⊕ F F1⊕1∅ F1 F1⊕H2 F1 We can see that all conditions (a), (b), (c) hold under the induction hypothesis. For

n (a), note that H A = (H1 1 H2) (A + R 2 ) = H1 A and H A = ∩ F1⊕1∅ ⊕ ∩ F1 ∩ F1 ∩ F1⊕H2

(H1 ∞ H2) A = H1 A , both of which have the unique element c . For (b), note ⊕ ∩ F1 ∩ F1 F1 that dim A = dim A + n2 = (n1 1 dim F1) + n2 = n 1 dim F1 1 ∅ and F1⊕1∅ F1 − − − − ⊕

dim A = dim A = n1 1 dim F1 = n 1 dim(F1 H2). For (c), we have F1⊕H2 F1 − − − − ⊕ n2 x1 + z2, y1 = x1, y1 = 1 for every x1 +z2 AF ⊕ = AF +R and y1 AF ∗⊕H◦ = AF ∗ . h i h i ∈ 1 1∅ 1 ∈ 1 2 1 40

Now, it remains to consider the cases: F = F1 1 F2 or F1 F2 where each Fj is a face of ⊕ ⊕ dimension kj of Hj.

For the condition (a), we can see from Lemma 2.2 that the centroid of any face F belongs to AF . So, we need to show the uniqueness of the common point of AF and H. For the

k1+1 k2+1 `1-sum case, let p + q for p A , q A be a common point of H and k1+k2+2 k1+k2+2 ∈ F1 ∈ F2

AF . Then, 1 k1 + 1 1 k2 + 1 p H1 and q H2 λ · k1 + k2 + 2 ∈ 1 λ · k1 + k2 + 2 ∈ −

for some λ (0, 1) because every point in H is a convex combination of two points in H1, ∈ H . Then 1 k1+1 = 1 = 1 k2+1 . Otherwise, either 1 k1+1 > 1 or 1 k2+1 > 1. 2 λ k1+k2+2 1−λ k1+k2+2 λ k1+k2+2 1−λ k1+k2+2

1 k1+1 If r := > 1, then rA H1 must be empty because A intersects H at a single λ k1+k2+2 F1 ∩ F1

point by the induction hypothesis. However, it is impossible because rp rA H1. Thus, ∈ F1 ∩

r = 1 gives p H1 A and q H1 A . By the induction hypothesis, we get p = c ∈ ∩ F1 ∈ ∩ F2 F1 and q = cF2 , which give the uniqueness of the common point of AF and H. Now, for the 1 1 `∞-sum case, suppose that p + q + t c c for p A , q A is a common n1−k1 F1 − n2−k2 F2 ∈ F1 ∈ F2 t t point of H A . Then p + c H1 and q
c H2 because H = H1 ∞ H2. ∩ F n1−k1 F1 ∈ − n2−k2 F2 ∈ ⊕ t t It implies that 1 + A intersects H1, and 1 A intersects H2. But one of n1−k1 F1 − n2−k2 F2 them is impossible unless
t = 0 by the same reason as in the
`1-sum case. So, t = 0 gives p H1 A , q H2 A , i.e., p = c , q = c . ∈ ∩ F1 ∈ ∩ F2 F1 F2 To verify the condition (b), we use (3.7), (3.8), and Lemma 2.1. Then

dim AF1 + dim AF2 = (n1 1 k1) + (n2 1 k2), if F = F1 1 F2 dim A =  − − − − ⊕ F   dim A + dim A + 1 = (n1 1 k1) + (n2 1 k2) + 1, if F = F1 F2 F1 F2 − − − − ⊕  = n 1 dim F − − 41

To verify the condition (c), it suffices to show x, y = 1 for x AF ⊕ F , y AF ∗⊕F ∗ . Write h i ∈ 1 1 2 ∈ 1 2

k1 + 1 k2 + 1 x = p + q AF1⊕1F2 , k1 + k2 + 2 k1 + k2 + 2 ∈

∗ ∗ 1 1 y = p + q + cF ∗ cF ∗ AF ∗⊕F ∗ , k1 + 1 1 − k2 + 1 2 ∈ 1 2

∗ ∗ where p AF , q AF and p AF ∗ , q AF ∗ . Thus, ∈ 1 ∈ 2 ∈ 1 ∈ 2

k1 + 1 k2 + 1 ∗ ∗ 1 1 x, y = p + q, p + q + cF ∗ cF ∗ h i k + k + 2 k + k + 2 k + 1 1 − k + 1 2  1 2 1 2 1 2  k1 + 1 ∗ k2 + 1 ∗ 1 = p, p + q, q + p, cF ∗ q, cF ∗ k1 + k2 + 2 h i k1 + k2 + 2 h i k1 + k2 + 2 h 1 i − h 2 i   k1 + 1 k2 + 1 1 = + + (1 1) = 1 k1 + k2 + 2 k1 + k2 + 2 k1 + k2 + 2 − because p, p∗ = q, q∗ = 1 by the induction hypothesis. h i h i

3.2 Computation of volumes of polytopes

To compute the volume of the simplex CF or XF defined in (3.2) from C = (cF ), X =

(xF ), and a flag F, we need the following lemma.

Lemma 3.2. Let σ = (j1, . . . , jn) be an n-tuple consisting of n1 copies of 1 and n2 copies

of 2 for n = n1 + n2. For j = 1, 2, denote by σj(k) the number of j’s among the first k

entries j1, . . . , jk of σ. Consider the n n matrix M whose rows consist of ×

pσ1(k) + qσ2(k) + ξkz for k = 1, . . . , n

n where ξ1, . . . , ξn R, p0 = q0 = 0, p1, . . . , pn1 , q1, . . . , qn2 , z R . Then the absolute value ∈ ∈ 42

of the determinant of M is the same as that of the matrix M 0 whose rows consist of

σ pk1 + φ1 (k1)z for k1 = 1, . . . , n1,

σ qk2 + φ2 (k2)z for k2 = 1, . . . , n2,

σ σ where φ1 , φ2 are functions determined by ξ1, . . . , ξn R and σ. More precisely, for j = 1, 2, ∈ the function φj is defined by φj(0) = 0 and

σ −1 φ (`) = (Φj σ )(`), for ` = 1, . . . , nj, (3.9) j ◦ j where

−1 σj (`) = min k : σj(k) = ` for ` = 1, . . . , nj, n o j+j` Φj(k) = ξk + ( 1) ξ` for k = 1, . . . , n. − `

Proof. It is enough to show that the matrix M can be obtained from M 0 through the Gauss elimination. For each k = 1, . . . , n, replace one of the rows

σ pσ1(k) + φ1 (σ1(k))z, (if jk = 1)

σ qσ2(k) + φ2 (σ2(k))z, (if jk = 2)

by the row

σ σ pσ1(k) + qσ2(k) + [φ1 (σ1(k)) + φ2 (σ2(k))] z.

σ σ We claim that φ1 (σ1(k)) + φ2 (σ2(k)) = ξk for each k = 1, . . . , n; if it is true, then every

row of M can be obtained by the above replacement of rows in M 0. First, consider the case

that either σ1(k) or σ2(k) is zero; assume σ2(k) = 0. This case implies j1 = = jk = 1 ··· 43

−1 and σ1 (k) = k. So,

σ σ σ σ φ1 (σ1(k)) + φ2 (σ2(k)) = φ1 (k) + φ2 (0) = Φ1(k) = ξk.

Assume that both σ1(k) and σ2(k) are positive integers. Then, for j = 1, 2,

σ j+j` −1 φ (σj(k)) = Φj(κj) = ξκ + ( 1) ξ`, for κj = σ (k) j j − j `<κ : j 6=j jX` `+1 j = ξκ + ( 1) ξ` ξ` , j − − Iσ ∩[1,κ ) Iσ ∩[1,κ )  12X j 21X j 

σ σ where I = ` : j` = 1, j`+1 = 2 and I = ` : j` = 2, j`+1 = 1 . If jk = 1, then κ1 = 12 { } 21 { } −1 −1 σ σ σ (k) = k and jκ = 2, jκ +1 = = jk = 1 for κ2 = σ (k). Thus, I [1, κ2) = I [1, k) 1 2 2 ··· 2 12 ∩ 12 ∩ σ σ and κ2 I [1, κ2) = I [1, k). In case of jk = 1, we have { } ∪ 21 ∩ 21 ∩

σ φ (σ1(k)) = ξk + ξ` ξ`, (3.10) 1 − Iσ ∩[1,k) Iσ ∩[1,k) 12X 21X σ φ (σ2(k)) = ξ` ξ`. (3.11) 2 − Iσ ∩[1,k) Iσ ∩[1,k) 21X 12X

−1 −1 In case of jk = 2, we get κ2 = σ (k) = k and jκ = 1, jκ +1 = = jk = 2 for κ1 = σ (k). 2 1 1 ··· 1 Thus,

σ φ (σ1(k)) = ξ` ξ`, (3.12) 1 − Iσ ∩[1,k) Iσ ∩[1,k) 12X 21X σ φ (σ2(k)) = ξk + ξ` ξ`. (3.13) 2 − Iσ ∩[1,k) Iσ ∩[1,k) 21X 12X

σ σ In both cases, we have φ1 (σ1(k)) + φ2 (σ2(k)) = ξk, which completes the proof.

n Proposition 3.4. Let H be a Hanner polytope in R . Consider the set C = (cF ) of the

centroids of faces of H. Then, the simplex CF, defined in (3.2), has the same volume for 44

every flag F of H.

n n Proof. Let H be the `1 or `∞-sum of two Hanner polytopes H1 R 1 and H2 R 2 . Since ⊂ ⊂ the volume of the linear image of C by T GL(n) is always equal to det T C for each F ∈ | | | F| n n n n n flag F, we may assume that R = R 1 R 2 is the orthogonal sum of R 1 and R 2 . ⊕ 0 n−1 Let F = F ,...,F be a flag over H. Consider the lower dimensional flags, defined in (2.6), 

0 n1−1 0 n2−1 F1 = F1 ,...,F1 and F2 = F2 ,...,F2 . n o n o

Let σ = (j1, . . . , jn) J be the type of F (Definition 2.4). For j = 1, 2, denote by σj(k) the ∈ −1 nj number of j’s among the first k entries j1, . . . , jk of σ, and let Fj = ∅, Fj = Hj be the improper faces of Hj.

CASE (H = H1 1 H2). It follows from Remark 2.2 that each element of a flag F = ⊕ F 0,...,F n−1 of type σ is expressed by 

k−1 σ1(k)−1 σ2(k)−1 F = F 1 F , for 1 k n. 1 ⊕ 2 ≤ ≤

By Lemma 2.2, the centroid of F k−1 is given by

σ1(k) σ2(k) c k−1 = c σ (k)−1 + c σ (k)−1 F F 1 F 2 σ1(k) + σ2(k) 1 σ1(k) + σ2(k) 2 1 = σ1(k) c σ1(k)−1 + σ2(k) c σ2(k)−1 . k F1 F2   45

Thus, the volume of the simplex CF is equal to the absolute value of

cF 0  .  . . .   1   1   det c  = det σ (k) c σ (k)−1 + σ (k) c σ (k)−1 . n!  F k−1  n!2 1 F 1 2 F 2    1 2   .   .   .   .               cF n−1     

Applying Lemma 3.2 with ξ1 = = ξn = 0, the absolute value of the determinant of the ··· matrix whose rows are

σ1(k) c σ1(k)−1 + σ2(k) c σ2(k)−1 , k = 1, . . . , n F1 F2

is equal to that of the matrix whose rows are

k1 c k1−1 , k1 = 1, . . . , n1 F1

k2 c k2−1 , k2 = 1, . . . , n2. F2

The determinant of the above matrix is, up to a multiple of 1, equal to ±

1 cF 0 0 · 1  . .  . .     cF 0 cF 0   1 2  n1 cF n1−1 0   .   .  det · 1  = (n1! n2!) det . det . ,        0 1 cF 0       · 2        cF n1−1 cF n2−1  . .   1   2   . .                 0 n2 cF n2−1   · 2    46

Thus, for the `1-sum case, we have

2 n1!n2! C = (C1) (C2) (3.14) | F| n! F1 · F2  

where each Cj = (cFj ) is the set of centroids of faces of Hj for j = 1, 2.

0 n−1 CASE (H = H1 ∞ H2). From Remark 2.2, each element of the flag F = F ,...,F ⊕ of type σ can be expressed by 

n −σ (k) n −σ (k) F n−k = F 1 1 F 2 2 . 1 ⊕ 2

By Lemma 2.2, the volume of the simplex CF is equal to the absolute value of

cF 0  .  . . .   1   1   det c  = det c n −σ (k) + c n −σ (k) . n!  F n−k  n! F 1 1 F 2 2    1 2   .   .   .   .               cF n−1     

Applying Lemma 3.2 with ξ1 = = ξn = 0, the absolute value of the determinant of the ··· matrix whose rows are

c n1−σ1(k) + c n2−σ2(k) , k = 1, . . . , n F1 F2 is equal to that of the matrix whose rows are

c n1−k1 , k1 = 1, . . . , n1 F1

c n2−k2 , k2 = 1, . . . , n2. F2 47

Then, by a similar argument to the `1-sum case, the volume of CF for the `∞-sum case is given by

n1!n2! C = (C1) (C2) , (3.15) | F| n! F1 · F2

where each Cj = (cFj ) is the set of centroids of faces of Hj for j = 1, 2. Finally, we use the induction on the dimension of H to conclude from (3.14), (3.15) that

the simplices CF have equal volumes for each flag F over H.

n Corollary 3.1. Let H be a Hanner polytope in R and C = (cF ) the set of centroids of

faces of H. Then, the volume function V ( ) for H given in Definition 3.1 is infinitely ·

differentiable around the point C = (cF ).

Note that the function V ( ) is a sum of the volumes of simplices, and the volume of · each simplex can be expressed by the absolute value of a determinant function. Since every

simplex C has equal volume which is non-zero, the volume function V ( ) around the point F · C = (c ) is just a sum of the determinant functions. Thus V ( ) is infinitely differentiable F ·

in a neighborhood of the point C = (cF ).

In this section we prove that the condition (3.6) in Proposition 3.3,

0 V (C),Z = 0 for any Z = zF with all z A A F ∈ F − F
holds for Hanner polytopes. By linearity it is enough to take Z = zF with

z, if F = G  zF =  0, otherwise

  48

for a fixed face G and z AG AG. Since ∈ −

V (C + tZ) V (C) V 0(C),Z = lim − t→0 t 1 = lim (C + tZ) C , t→0 t | F| − | F| 3G 3G ! FX FX

0 we need to show (C + tZ)F = CF for small t > 0 to get V (C),Z = 0. F3G | | F3G | | h i P P n Lemma 3.3. Let H be a Hanner polytope in R and C = (cF ) the set of centroids of faces of

n H. For z, ξ = (ξ1, . . . , ξn) R , consider the point Wξ,z = (w ) defined by w = ξdim F +1 z ∈ F F for each face F of H. Then, for any ξ close to the origin, V (C + Wξ,z) = V (C), i.e.,

(C + Wξ,z) = C . | F| | F| XF XF

n n Proof. Let z = (z1, . . . , zn) R . Consider the function from R to R+, defined by ∈

T ξ = (ξ1, . . . , ξn) det(M + ξ z), (3.16) 7−→

T where M is a n n matrix and ξ z is the n n matrix with ξizj in the (i, j)-entry. Note × × T that ∂(det(M+ξ z)) is equal to the determinant of the matrix obtained from M by replacing ∂ξi1 ···∂ξik the i1, . . . , ik-th rows of M with the same row z. It follows that all second-order derivatives are zero, so the function (3.16) is affine. Moreover, since the volume function V (C + Wξ,z)

can be expressed by a sum of determinant functions similar to (3.16), the function

ξ V (C + Wξ,z) (3.17) 7−→ 49

is also affine. Also, (3.17) is an even function. Indeed,

V (C + W−ξ,z) = (C Wξ,z) = (C Wξ,z)− | − F| | − F| XF XF

= ( C Wξ,z) = V (C + Wξ,z). | − − F| XF

Thus, it should be a constant. It implies that V (C + Wξ,z) = V (C) when ξ is close to the origin.

n Theorem 3.2. Let H be a Hanner polytope in R and C = (cF ) the set of centroids of

faces of H. Fix a face G of H. Take any ξ1, . . . , ξn R with small absolute values, and ∈

z AG AG where AG is the affine subspace defined in Section 3.2. Consider W = (w ) ∈ − F with

ξdim F +1 z, if F G or F G  ⊃ ⊂ wF =  0, otherwise.

 Then 

(C + W ) = C . | F| | F| 3G 3G FX FX n n Proof. Let H be the `1 or `∞-sum of two Hanner polytopes H1 R 1 and H2 R 2 . ⊂ ⊂ n n n n n Assume that R = R 1 R 2 is the orthogonal sum of R 1 and R 2 Fix a face G of H, ⊕ 0 n−1 and consider a flag F = F ,...,F over H containing G. Then F induces two lower

0 n1−1 0 n2−1 dimensional flags F1 = F1 ,...,F , F2 = F2 ,...,F defined in (2.6). Denote by { 1 } { 2 } −1 nj −1 Fj = ∅ and Fj = Hj the improper faces of Hj for each j = 1, 2. Let

1 + dim G, for the `1-sum case m =    n dim G, for the `∞-sum case. −   50

From (2.3), (2.4), the face G can be written by

m−1 m1−1 m2−1 F = F1 1 F2 , for the `1-sum case G =  ⊕ (3.18)   n−m n1−m1 n2−m2 F = F F , for the `∞-sum case 1 ⊕ 2   where m1, m2 are integers satisfying 0 m1 n1 0 m2 n2 and m1 + m2 = m. ≤ ≤ ≤ ≤ Suppose that F is of type σ = (j1, . . . , jn). Since F contains G, it follows from the above

representation of G and Remark 2.2 that the number of j’s among the first m entries of σ

is mj for each j = 1, 2. Thus, the type σ of a flag F containing the face G can be viewed as

an element of

n σ1(n)=n1, σ1(m)=m1 Jm = σ = (j1, . . . , jn) 1, 2 (3.19) ∈ { } σ2(n)=n2, σ2(m)=m2 n o

where σj(k) denotes the number of j’s among the first k entries j1, . . . , jk of σ = (j1, . . . , jn)

for j = 1, 2. As in Remark 2.2, every flag over the `1 or `∞ sum of H1, H2 containing the

face G is associated with an triple consisting of an element of Jm and two lower dimensional

flags over H1, H2 containing the corresponding summands in the decomposition (3.18) of G.

CASE (H = H1 1 H2). It follows from Remark 2.2 that each element of a flag F = ⊕ F 0,...,F n−1 of type σ is expressed by 

k−1 σ1(k)−1 σ2(k)−1 F = F 1 F , (1 k n). 1 ⊕ 2 ≤ ≤

By Lemma 2.2, the centroid of F k−1 is given by

1 cF k−1 = σ1(k) c σ1(k)−1 + σ2(k) c σ2(k)−1 . k F1 F2   51

So, the volume of the simplex (C + W )F is equal to the absolute value of

cF 0 + ξ1z  .  . . .   1   1   det c + ξ z  = det σ (k) c σ (k)−1 + σ (k) c σ (k)−1 + (kξ )z . n!  F k−1 k  n!2 1 F 1 2 F 2 k    1 2   .   .   .   .               cF n−1 + ξnz      By Lemma 3.2, the absolute value of the determinant of the matrix whose rows are

σ1(k) c σ1(k)−1 + σ2(k) c σ2(k)−1 + (kξk)z, k = 1, . . . , n F1 F2 is equal to that of the matrix whose rows are

¯ k1 c k1−1 + k1φ1(k1)z, k1 = 1, . . . , n1 F1 ¯ k2 c k2−1 + k2φ2(k2)z, k2 = 1, . . . , n2 F2

where φ¯1, φ¯2 are some functions depending on σ Jm and ξ1, . . . , ξn. The exact formulas ∈ for φ¯1, φ¯2 are not needed in this proof of the `1-sum case. Thus, the volume of the simplex

(C + W )F is equal to the absolute value of

. . . .     ¯ ¯ 1 k1 c k1−1 + k1φ1(k1)z n ! n ! c k1−1 + φ1(k1)z F1 1 2 F1 2 det  = 2 det  . (3.20) n!  ¯  n!  ¯   k2 c k2−1 + k2φ2(k2)z   c k2−1 + φ2(k2)z   F2   F2       .   .   .   .          52

Note that z AG AG, and G is written from (3.18) by ∈ −

m1−1 m2−1 G = G1 1 G2 for G1 = F and G2 = F . ⊕ 1 2

Also, from (3.7) we have

(AG1 AG1 ) + (AG2 AG2 ), if G1 = ∅,G2 = ∅  − − 6 6  AG AG =  n2 (AG1 AG1 ) + R , if G1 = ∅,G2 = ∅ −  − 6  n1 R + (AG2 AG2 ), if G1 = ∅,G2 = ∅.  − 6    Without loss of generality, we may assume one of the following three cases.

1. z AG1 AG1 when G1 = ∅, G2 = ∅. ∈ − 6 6

2. z AG1 AG1 when G1 = ∅, G2 = ∅. ∈ − 6

n 3. z R 1 when G1 = ∅, G2 = ∅. ∈ 6

n Since z R 1 in any cases, from (3.20) we have ∈

¯ cF 0 + φ1(1)z 0  1  . . . .     ¯  c n1−1 + φ1(n1)z 0  n1!n2!  F1  (C + W )F = 2 det  | | n!  ¯   φ2(1)z c 0   F2     . .   . .       ¯   φ2(n2)z cF n2−1   2  2   n1!n2 ! = (C1 + Wσ) (C2) (3.21) n! F1 · F2  

53

where Wσ = (wF ) is defined by

φ¯1(dim F + 1) z, if F G1 or F G1  ⊃ ⊂ wF =  0, otherwise.

  Consider the first case that z AG1 AG1 when G1 = ∅ and G2 = ∅. Since G = G1 1 G2, ∈ − 6 6 ⊕

(C + W ) = (C + W ) , | F| | F| 3G σ∈J 3G 3G FX Xm FX1 1 FX2 2 where the flag F on the right hand side means the flag induced by lower dimensional flags

F1, F2 and a type σ as in Remark 2.2. It follows from (3.21) that

(C + W ) = (Const) (C1 + Wσ) , (3.22) | F| | F1 | 3G σ∈J 3G FX Xm FX1 1

2 n1!n2! where (Const) = (C2) . Similarly, letting ξ1 = = ξn = 0 we get n! F23G2 | F2 | ···   P

C = (Const) (C1) . (3.23) | F| | F1 | 3G σ∈J 3G FX Xm FX1 1

Since

(C1 + Wσ) = (C1) | F1 | | F1 | 3G 3G FX1 1 FX1 1 by the induction hypothesis, the right hand side of (3.22) is the same as that of (3.23),

which implies (C + W ) = C . Similarly, in the second case, we have the F3G | F| F3G | F| 2 n1!n2! same conclusionP with a different constantP (Const) = (C2) for (3.22). n! F2 | F2 | n   Consider the third case that z R 1 when G1 = ∅ and G2P= ∅. In this case, note that ∈ 6 ¯ Wσ = (wF ) is the point with wF = φ1(dim F +1) z for each face F of H1 (without any other 54

restrictions on F ). So, Lemma 3.3 gives

(C + Wσ) = C . | F| | F| XF1 XF1

Therefore,

(C + W ) = (Const) (C + Wσ) = (Const) C = C , | F| | F| | F| | F| 3G σ∈J σ∈J 3G FX Xm XF1 Xm XF1 FX which completes the proof for the `1-sum case.

0 n−1 CASE (H = H1 ∞ H2). From Remark 2.2, each element of the flag F = F ,...,F ⊕ of type σ can be expressed by 

n −σ (k) n −σ (k) F n−k = F 1 1 F 2 2 . 1 ⊕ 2

Note that z AG AG, and by (3.18) ∈ −

n1−m1 n2−m2 G = G1 G2 for G1 = F and G2 = F . ⊕ 1 2

In addition, AG AG can be written from (3.8) by −

AG AG , if G1 = H1,G2 = H2 1 − 1 6

AG AG , if G1 = H1,G2 = H2 2 − 2 6 and

1 1 (AG AG ) + (AG AG ) + span c c if G1 = H1,G2 = H2. 1 − 1 2 − 2 m G1 − m G2 6 6  1 2  55

Without loss of generality, we may consider the following three cases:

1. z AG AG when G1 = H1, G2 = H2. ∈ 1 − 1 6

2. z AG AG when G1 = H1, G2 = H2. ∈ 1 − 1 6 6

1 1 3. z span c c when G1 = H1, G2 = H2. ∈ m1 G1 − m2 G2 6 6 n o The volume of the simplex (C + W ) is, up to a multiple of 1, F ±

cF 0 + ξ1z  .  . . .   1   1   det c + ξ z  = det c n −σ (k) + c n −σ (k) + ξ z . n!  F n−k k  n! F 1 1 F 2 2 k    1 2   .   .   .   .               cF n−1 + ξnz     

Here, for simplicity, ξ1, . . . , ξn are rearranged by ξn−dim F instead of ξdim F +1. By Lemma

3.2, the determinant of the matrix whose rows are

c n1−σ1(k) + c n2−σ2(k) + ξkz, k = 1, . . . , n F1 F2

is equal to that of the matrix whose rows are

σ c k1−1 + φ1 (k1)z, k1 = 1, . . . , n1 F1

σ c k2−1 + φ2 (k2)z, k2 = 1, . . . , n2. F2 56

σ σ where φ1 , φ2 are the functions defined in (3.9). Thus

. .   σ 1 c n1−k1 + φ1 (k1)z F1 (C + W )F = det  . (3.24) | | n!  σ   c n2−k2 + φ2 (k2)z   F2     .   .     

For the first two cases z AG AG when G1 = H1, we can conclude by the same ∈ 1 − 1 6 argument as in the `1-sum case to get (C + W ) = C . For the third one, F3G | F| F3G | F| let z = 1 c 1 c . In this case, theP 2 n sub-matrix fromP (3.24) m1 G1 − m2 G2 ×

σ σ σ φ1 (m1) φ1 (m1) cG + φ1 (m1)z 1 + m cG m cG 1 = 1 1 − 2 2    σ σ  σ h φ2 (m2i) φ2 (m2) cG + φ2 (m2)z m cG + 1 m cG  2   1 1 − 2 2     h i  is, under the Gauss elimination, equivalent to

λ1 cG1   λ2 cG  2   

σ σ σ σ σ φ1 (m1) φ2 (m2) φ1 (m1) φ2 (m2) φ1 (m1) −1 where λ1 = 1 + and λ2 = 1 + 1 + . So, m1 − m2 m2 m1 m1   σ σ φ1 (m1) φ2 (m2) λ1λ2 = 1 + . m1 − m2

Since z is a linear combination of two rows cG1 and cG2 , the z-term in each row of the 57

matrix in (3.24) disappears through the Gauss elimination, that is, the absolute value of

. .

 σ  c n −k + φ (k1)z F 1 1 1  1   σ   c n −k + φ (k )z   F 2 2 2 2   2  .  .  .  .  .       σ σ σ  c  cF n1−k1 + φ1 (k1)z φ (m1) φ (m2)  G1  1 1 2   det  = 1 + det   σ  m1 − m2 ·  n −k   c  cF 2 2 + φ2 (k2)z    G2   2     .   .   .   .   .          σ 0  c 0 + φ (k )z   n1−k1 1 1   F1     σ 0  c 0 + φ (k )z  n2−k2 2 2   F2     .   .      is equal to

c 0 c 0 F1 F2 σ σ     φ1 (m1) φ2 (m2) 1 . . (C + W )F = 1 + det . det . | | m1 − m2 · n!   ·          c   c   F n1−1   F n2−1   1   2  σ σ     φ1 (m1) φ2 (m2) n1! n2! = 1 + (C1) 1 (C2) 2 . (3.25) m − m · n! | F | | F |  1 2 

From G = G1 G2, we get ⊕

(C + W ) = (C + W ) , | F| | F| 3G σ∈J 3G 3G FX Xm FX1 1 FX2 2

where the flag F on the right hand side is the flag induced by lower dimensional flags F1, 58

F2 and a type σ as in Remark 2.2. The formula (3.25) implies

σ σ φ1 (m1) φ2 (m2) (C + W )F = (Const) 1 + (3.26) | | m1 − m2 3G σ∈J FX Xm 

n1!n2! where (Const) = (C1) (C2) . Similarly, letting ξ1 = = ξn = n! F13G1 F23G2 | F1 | | F2 | ··· σ Pσ P 0, we get φ1 (k) = 0 = φ2 (k) for each k and hence

C = (Const) 1. (3.27) | F| 3G σ∈J FX Xm

To complete the proof from (3.26) and (3.27), we claim that

σ σ φ (m1) φ (m2) 1 2 = 0. (3.28) m1 − m2 σ∈J Xm 

It follows from (3.10), (3.11), (3.12), (3.13) that

σ σ 1+jm φ1 (m1) φ2 (m2) ( 1) 1 1 = − ξm + + ξ` ξ` , m1 − m2 mj m1 m2 − m Iσ ∩[1,m) Iσ ∩[1,k)    12X 21X 

σ σ where I = ` : j` = 1, j`+1 = 2 and I = ` : j` = 2, j`+1 = 1 . Therefore, 12 { } 21 { }

σ σ 1 2 m−1 φ1 (m1) φ2 (m2) µm µm 1 1 12 21 = ξm + + (µ` µ` )ξ`, m1 − m2 m1 − m2 m1 m2 − σ∈J `=1 Xm      X 59

where µ1 , µ2 and µ12, µ21 for ` = 1, . . . , m 1 are the numbers given by m m ` ` −

1 m 1 n m µm = # σ = (j1, . . . , jn) Jm : jm = 1 = − − ∈ m1 1 · n1 m1 n o  −   −  2 m 1 n m µm = # σ = (j1, . . . , jn) Jm : jm = 2 = − − ∈ m2 1 · n1 m1 n o  −   −  12 m 2 n m µ` = # σ = (j1, . . . , jn) Jm : j` = 1, j`+1 = 2 = − − ∈ m1 1 · n1 m1 n o  −   −  21 m 2 n m µ` = # σ = (j1, . . . , jn) Jm : j` = 2, j`+1 = 1 = − − . ∈ m2 1 · n1 m1 n o  −   − 

1 2 We can easily see that µm = µm and µ12 = µ21 from n + n = n and m + m = m, which m1 m2 ` ` 1 2 1 2 implies (3.28) and completes the proof for the `∞-sum case.

n Corollary 3.2. Let H be a Hanner polytope in R , C = (cF ) the set of centroids of faces

of H, and AF ’s the affine subspaces defined from H as in Section 3.2. Then

V 0(C),Z = 0 for any Z = z with all z A A . F F ∈ F − F

Proof. By linearity it is enough to take Z = zF with

z, if F = G  zF =  0, otherwise

  for a face G and z AG AG. Letting all ξ1, . . . , ξn be zero except ξ1+dim G in Theorem ∈ − 3.2, we have

V (C + tZ) V (C) 1 V 0(C),Z = lim − = lim (C + tZ) C = 0, t→0 t t→0 t | F| − | F| 3G 3G ! FX FX which completes the proof. 60

3.3 A gap between K and a polytope

n Let K be a symmetric convex body in R with dBM (K,H) = 1+δ. Taking an appropriate linear transformation on K, we may assume that H is a standard Hanner polytope and dH(K,H) = δ. From the xF -points constructed in Section 3.1 we obtained the star-shaped, not necessary convex, polytope PK = F XF. Note that the volume of PK is equal to V (X) by the definition of the function V ( ).S · Propositions 3.3 and Corollary 3.2 give • V (X) V (Y ) = O(δ2) = V (X∗) V (Y ∗) . | − | | − | Propositions 3.2 and 3.4 give • V (Y )V (Y ∗) H H◦ . ≥ | || |

Therefore we now have V (X) (H) O(δ2). ≥ P − In this section, after taking an appropriate linear transformation on K, we will show

that (K) V (X)V (X∗) + c(n)δ, which is given in Theorem 3.3. We start with some P ≥ preparatory lemmas and propositions.

⊥ Lemma 3.4. For j = 1, . . . , n, consider Ej = H (ej + e ), which is a face of H with ∩ j

centroid ej. Then,

aff(Ej) = ej + span ei H : i  j (3.29) { ∈ }

A = ej + span ei H : i j ,. (3.30) Ej { ∈ ∼ }

Proof. The equality (3.29) can be obtained from

conv ei + ej H : i  j Ej aff ei + ej H : i  j . {± ∈ } ⊂ ⊂ {± ∈ }

Here, the first inclusion follows directly from definitions of Ej and i  j. We can get 61

⊥ the second inclusion by showing that Ej e whenever i j. Indeed, if x Ej, then ⊂ i ∼ ∈ the orthogonal projection of x to span ei, ej is x, ei ei + x, ej ej H. If i j, then { } h i h i ∈ ∼

x, ei + x, ej 1 because the section of H by span ei, ej is the 2-dimensional `1-ball. |h i| |h i| ≤ { }

Since x, ej = 1, we get x, ei = 0. h i h i To prove (3.30), we use the induction on the dimension n = dim H. If n = 1, then (3.30) is trivial. Now assume that (3.30) is true for all standard Hanner polytopes of dimension

n less than n. Let H R be the `1 or `∞ sum of two standard Hanner polytopes H1 and ∈

H2. Without loss of generality, we fix j = 1 and assume e1 span(H1). Consider F = ∈ ⊥ ⊥ H (e1+e1 ) and F1 = H1 (e1+e1 ). From (3.29), we get aff(F ) = e1+span ei H : i  1 ∩ ∩ { ∈ } and aff(F1) = e1+span ei H1 : i  1 . So, if H = H1 1H2, then we get aff(F ) = aff(F1), { ∈ } ⊕ which implies F = F1 1 ∅ by Lemma 2.3. If H = H1 ∞ H2, then we get aff(F ) = ⊕ ⊕ aff(F1) + span(H2), which implies F = F1 H2 by Lemma 2.3. It follows from Definition ⊕ 3.2 that

AF1 + span(H2), if H = H1 1 H2, A =  ⊕ F   A , if H = H1 ∞ H2. F1 ⊕   Since A = e1 + span ei H1 : i 1 by the induction hypothesis, in any case we have F1 { ∈ ∼ }

A = e1 + span ei H : i 1 . F { ∈ ∼ }

n Proposition 3.5. Let H be a standard Hanner polytope in R , and K0 be a symmetric

n convex body in R with dH(K0,H) = δ for small δ > 0. Consider the polytope P0 defined by the convex hull of all the xF -points of K0. Then, there exists a polytope K with dBM (K,P0) =

1 + O(δ2) such that

n n dH(K,H) = O(δ) and B K B . 1 ⊂ ⊂ ∞

⊥ Proof. Let j 1, . . . , n . Consider Ej = H (ej + e ), which is a face of H with centroid ∈ { } ∩ j ej, and let xj = x be the x -point for F = Ej. Then, xj ej = O(δ) by Proposition 3.1. Ej F | − | 62

⊥ Choose a hyperplane ϕj + ϕj tangent to K0 at xj and parallel to AEj . First, we will prove

ϕj ej = O(δ). (3.31) | − |

Indeed, if i j, then ei is parallel to A by (3.30), which implies ∼ Ej

ei, ϕj = 0 whenever i j. (3.32) h i ∼

± ± If i  j, then ei + ej H gives t (ej ei) K0 for some t = 1 + O(δ). So, the ∈ ± ∈ ⊥ points ej ei are bounded by the hyperplanes (1 + O(δ))(ϕj + ϕ ). In addition, note ± ± j

that xj, ϕj = ϕj, ϕj , xj ej = O(δ), and c1 ϕj c2 for constants c1, c2 (by h i h i | − | ≤ | | ≤ 2 ◦ ϕj/ ϕj ∂K0 ). Thus, if i  j, then | | ∈

ei, ϕj = ej ei, ϕj xj, ϕj + xj ej, ϕj h± i h ± i − h i h − i

= ej ei, ϕj ϕj, ϕj + O(δ) O(δ). (3.33) h ± i − h i ≤

To estimate ej, ϕj , note that h i

n 2 2 2 2 2 ϕj, ϕj = ei, ϕj = ej, ϕj + ei, ϕj = ej, ϕj + O(δ ) h i h i h i h i h i i=1 i j X X

⊥ is equal to xj, ϕj = ej, ϕj + O(δ) because xj ϕj + ϕ . It implies ej, ϕj = 1 + O(δ), h i h i ∈ j h i which, together with (3.32) and (3.33), completes the proof of (3.31).

n−1 Choose θj S which is orthogonal to ∈

⊥ span ej ei : i  j ϕj + span xi : i j , (3.34) { } ∪ { } ∩ ∼   n o

and satisfies ej, θj 0. Here, the dimension of the right hand side of (3.34) is at most h i ≥ 63

n 1 because ϕj span( ej ei : i  j ) by (3.32), so such θj exists. Moreover, − ∈ { } ∪ { }

θj ej = O(δ). (3.35) | − |

Indeed, if i j, then xi, θj = 0, which implies ei, θj = O(δ). Assume i  j. Let ∼ h i h i ⊥ Pϕ⊥ (ei) be the orthogonal projection of ei onto ϕj . Then, Pϕ⊥ (ei) ei is parallel to ϕj, j j − 2 i.e., Pϕ⊥ (ei) ei = αϕj where α = ei, ϕj / ϕj is obtained from Pϕ⊥ (ei), ϕj = 0. j − − h i | | h j i

So, Pϕ⊥ (ei) ei = αϕj = ei, ϕj / ϕj = O(δ). Moreover, (3.32) implies Pϕ⊥ (ei) = | j − | | | |h i| | | j ⊥ ei + αϕj ϕj span( ej ek : k  j ). Thus, we have Pϕ⊥ (ei), θj = 0, which implies ∈ ∩ { } ∪ { } j 2 D 2 E 2 2 ei, θj = O(δ). To estimate ej, θj , note that θj = ei, θj = ej, θj + O(δ ). From h i h i | | h i h i 2 θj = 1 and ej, θj > 0, we get ej, θj = 1 + O(δ ), whichP completes the proof of (3.35). | | h i h i ⊥ Consider the hyperplane xj + θ . We will prove that P0 = conv x is bounded by j { F } 2 ⊥ hyperplanes (1 + O(δ ))(xj + θ ). Note that a face F is contained in Ej if and only if ± j ⊥ c ej + e . Indeed, since c int(F ), it can be written as c = λvv where v runs over F ∈ j F ∈ F v P all vertices of H contained in F , and all λv’s are positive numbers satisfying v λv = 1.

It implies that c , ej = 1 if and only if v, ej = 1 for each vertex v of H in PF , which is h F i h i ⊥ equivalent to F Ej. Thus, the point x is O(δ)-close to xj + θ if and only if F Ej. It ⊂ F j ⊂ 2 suffices to prove x , θj xj, θj + O(δ ) whenever F Ej. For any face F Ej, h F i ≤ h i ⊂ ⊂

x , θj xj, θj = x xj, ϕj + x xj, θj ϕj h F i − h i h F − i h F − − i

= x xj, ϕj + sc ϕj, θj ϕj + x sc , θj ϕj (3.36) h F − i h F − − i h F − F − i

2 where s = ϕj / c , ϕj = 1 + O(δ). The first term in (3.36) satisfies x xj, ϕj | | h F i h F − i ≤ ⊥ 0 because ϕj + ϕj is tangent to K0 at xj. For the second term in (3.36), note that

⊥ sc ϕj ϕ by choice of s. Moreover, since ei, ϕj = 0 = ei, c for each i j F − ∈ j h i h F i ∼ ⊥ by (3.32) and (3.29), the point sc ϕj belongs to ϕ span( ej ei : i  j ), which F − j ∩ { } ∪ { } 64

⊥ gives sc ϕj θ . Thus, sc ϕj, θj ϕj = 0. The last term in (3.36) satisfies F − ∈ j h F − − i 2 2 x sc , θj ϕj x sc θj ϕj = O(δ ). Finally, x , θj xj, θj + O(δ ). h F − F − i ≤ | F − F | · | − | h F i ≤ h i ⊥ ⊥ Take a linear transformation T1 which maps xj + θj to ej + ej for each j = 1, . . . , n.

Indeed, to see the existence of such a linear transformation, consider the parallelepiped Q

⊥ bounded by the hyperplanes xj + θ for j = 1, . . . , n. If T1 is the linear transformation ± j ⊥ which maps the centroid of the facet Q (xj + θ ) of Q to ej for j = 1, . . . , n, then it ∩ j

is a desired linear transformation. Here, both xj and θj are O(δ)-close to ej, so is the

⊥ centroid of Q (xj + θ ). Thus, T1 Id = O(δ) where Id is the identity operator ∩ j k − k n 0 on R . Let P1 = T1P0, and xj = T1xj for each j. Then, dH(P1,P0) = O(δ). Also,

n 2 2 ⊥ dH(P1,P1 B ) = O(δ ) because P0 is bounded by hyperplanes (1 + O(δ ))(xj + θ ) for ∩ ∞ ± j ⊥ 0 each j. Moreover, if i j, then xi is parallel to xj + θ by (3.34). So, x is parallel to ∼ j i ⊥ ej + e whenever i j, that is, j ∼

0 x , ej = 0 whenever i j. (3.37) i ∼

Let (1 cδ)H P1 (1 + cδ)H for some constant c > 0. By (3.37) we can write − ⊂ ⊂

0 x1 = e1 + λjej j∈J X

where J = j : j  1 and λj = O(δ). Consider the point { }

1 cδ z = − λj e1 sign(λj) ej , λ | | − j∈J X

1 where λ = λj = O(δ). Since e1 ej H for each j J, we get λj (e1 | | ± ∈ ∈ λ j∈J | | − P P 65

sign(λj)ej) H. So, z P1. Moreover, consider the point ∈ ∈

0 λ x1 + 1−cδ z 1 + λ λ = λ e1 =: r1e1, 1 + 1−cδ 1 + 1−cδ

2 2 where r1 = 1 cδλ+O(δ ) = 1+O(δ ). Then, r1e1 P1 because it is a convex combination of − ∈ 0 x P1 and z P1. Repeat the above procedure to get r2e2, . . . , rnen P1 for j = 2, . . . , n 1 ∈ ∈ ∈ 2 where rj = 1 O(δ ) for each j. −

Take a linear transformation T2 which maps rjej to ej for each j. Then T2 Id = k − k 2 2 n O(δ ) because rj = 1 O(δ ). Let T = T2T1 and K = TP0 B . Then, dBM (K,P0) = − ∩ ∞ 2 n n 1 + O(δ ), dH(K,H) = O(δ) and B K B . 1 ⊂ ⊂ ∞

n (v) For each vertex v of a standard Hanner polytope in R , consider the subspace R , the (v) (v) lower dimensional cube B∞ , and the lower dimensional cross-polytope B1 , defined by

(v) (v) n (v) (v) n (v) R = span ej : j supp(v) ,B∞ = B∞ R ,B = B1 R . { ∈ } ∩ 1 ∩

Then, we can see that

(v) (v) (v) B∞ = H R = H R , for every vertex v of H, ∩ | (v?) (v?) (v?) ? ◦ B = H R = H R , for every vertex v of H , 1 ∩ |

(v) (v) where H R denotes the orthogonal projection of H to R . |

n Proposition 3.6. Let H be a standard Hanner polytope in R and K a symmetric convex

n body in R with dH(K,H) = δ for small δ > 0. Then, there exists either a vertex v of H ? (v) (v) ? ◦ ◦ (v?) (v ) with dH K R ,B∞ cδ, or a vertex v of H with dH K R ,B∞ cδ, where | ≥ | ≥     c = c(n) is a positive constant. 66

For the proof by contradiction, we assume that

(v) (v) dH K R ,B∞ = o(δ) for every vertex v of H, (3.38) |  (v?) (v?) ? ◦ dH K R ,B = o(δ) for every vertex v of H . (3.39) ∩ 1  

? ◦ (v?) (v ) Here, the equality in (3.39) is equivalent to dH K R ,B∞ = o(δ) because the polar | (v?) (v?) ◦ (v?)   of K R in R is the same as K R . We need the following two lemmas to prove ∩ | Proposition 3.6.

n Lemma 3.5. Let v be a vertex of a standard Hanner polytope H in R , and Av the affine subspace for the (zero-dimensional) face v defined in Definition 3.2. Then,

(v) (v) Av R = Av R . ∩ |

Proof. We use the induction on n = dim H. The statement is trivial if n = 1. Assume that Lemma 3.5 is true for all standard Hanner polytopes of dimension less than n. Let

n n n n H R = R 1 R 2 be the `1 or `∞ sum of standard Hanner polytopes H1 R 1 , ⊂ ⊕ ⊂ n H2 R 2 . Let v be a vertex of H. First, consider the case H = H1 1 H2. Then, it follows ⊂ ⊕ from Remark 2.1 that v = v1 1 ∅ or ∅ 1 v2 where v1, v2 are vertices of H1, H2 respectively. ⊕ ⊕ (v) (v1) Say, if v = v1 1 ∅, then R = R , and Definition 3.2 gives Av = Av1 + span(H2). So, ⊕ (v) (v1) (v1) (v1) (v) Av R = (Av1 + span(H2)) R = Av1 R is equal to Av1 R = Av R by the ∩ ∩ ∩ | | induction hypothesis. Now, consider the case H = H1 ∞ H2. In this case, v can be written ⊕ (v) (v ) (v ) as v = v1 v2 where v1, v2 are vertices of H1, H2 respectively. So, R = R 1 + R 2 and ⊕ (v) Av = Av1 + Av2 + span v1/n1 v2/n2 . Since v1/n1 v2/n2 R , we get { − } − ∈

(v) (v) (v) Av R Av1 R + Av2 R + span v1/n1 v2/n2 ∩ ⊃ ∩ ∩ { − }

(v1) (v2) = Av1 R + Av2 R + span v1/n1 v2/n2 , ∩ ∩ { − } 67

and

(v) (v) (v) Av R Av1 R + Av2 R + span v1/n1 v2/n2 | ⊂ | | { − }

(v1) (v2) = Av1 R + Av2 R + span v1/n1 v2/n2 | | { − }

(v1) (v1) (v2) (v2) Since Av1 R = Av1 R and Av2 R = Av2 R by the induction hypothesis, we ∩ | ∩ | (v) (v) have Av R Av R . The opposite inclusion is trivial. ∩ ⊃ |

Under the assumption (3.38), Lemma 3.5 implies that the orthogonal projection of the

(v) point xv (the xF -point for F = v) to R is o(δ)-close to v. In other words, for any vertex

v, the assumption (3.38) gives

xv, ej = v, ej + o(δ) for every j supp(v). (3.40) h i h i ∈

(v) To prove (3.40), letx ˜v, A˜v, K˜ be the orthogonal projections to R of xv, Av, K respectively.

(v) Then A˜v = Av R by Lemma 3.5. It gives that tA˜v is tangent to K˜ atx ˜v whenever tAv ∩ (v) is tangent to K at xv. Moreover, A˜v satisfies the conditions (a), (b) for P = B∞ given

in Section 3.1. By Proposition 3.1, the assumption (3.38) implies x˜v v = o(δ), which | − | complete the proof of (3.40).

The next lemma gives an estimate for the j-th coordinates of the xv-points when j 6∈ supp(v).

n Lemma 3.6. Let H be a standard Hanner polytope in R and K a symmetric convex body

n in R with (1 δ)H K (1 + δ)H. Let v, w be vertices of H with supp(v) = supp(w). − ⊂ ⊂ Suppose that 1 / supp(v) and ∈

v, ei = w, ei i  1. (3.41) h i h i ∀ 68

Then, under the assumptions (3.38) and (3.39), we have xv, e1 = xw, e1 + o(δ). h i h i

Proof. It suffices to consider the case that two vertices v, w have different values in only

one coordinate and the same values in the other coordinates; for the general case, consider

a sequence of vertices v = v1, v2, . . . , vk = w such that any two consecutive vertices vj, vj+1

have different values in only one coordinate. Thus, we may assume that v, ej = w, ej for h i h i

any j = 1, . . . , n 1, and v, en = w, en . Let I = supp(v). Then v and w can be written − h i 6 h i as

v = en + ej and w = en + ej. − j∈XI\{n} j∈XI\{n}

Under the assumption (3.38), as in (3.40), the points xv, xw K can be expressed by ∈

xv = en + ej + ajej + o(δ) j∈XI\{n} Xj∈ /I

xw = en + ej + bjej + o(δ) − j∈XI\{n} Xj∈ /I

where aj = O(δ) and bj = O(δ) follow from xv v = O(δ). So, | | | | | − |

xv xw aj bj − = en + ξjej + o(δ), ξj = − = O(δ). (3.42) 2 2 Xj∈ /I

Here, we may assume that each ξj is nonnegative; if ξj < 0, then replace ej with ej. Then − we claim that

ξ1 = o(δ). (3.43)

To prove (3.43), letx ¯ = en + ξjej. Then, (1 ε)¯x K for some ε = o(δ) by (3.42). j∈ /I − ∈ Take a maximal clique J containingP 1, n in the graph G associated with H, and let { } ? ◦ v = ej, which is a vertex of H with support J. First, if I J = 1, . . . , n , then j∈J ∪ { } (v?) x¯ = enP+ ξjej R . By (3.39), the `1-norm of (1 ε)¯x satisfies (1 ε)¯x 1+o(δ), j∈ /I ∈ − k − k1 ≤ P 69

which gives ξ1 = o(δ). Thus, it remains to consider the case that M := 1, . . . , n (I J) { }\ ∪

is not empty. It follows from maximality of J that for every k M, there exists jk J ∈ ∈ with jk  k. Consider the point

ejk + en y = (1 δ) λ + λk(ej ek) , − 2 k − k∈M X   where λ and all λk’s are non-negative and obtained from k∈M (λ + λk) = 1 and λk =

ejk +en ξk/(√δ + δ). Then, y K because it is a convex combinationP of points (1 δ) K ∈ − 2 ∈ and (1 δ)(ej ek) K for k M. Let z = √δy + (1 √δ)¯x. That is, − k − ∈ ∈ −

ejk + en z = √δ(1 δ) λ + λk(ej ek) + (1 √δ) en + ξjej − 2 k − − k∈M X    Xj∈ /I 

= √δ(1 δ) M λ/2 + (1 √δ) en + √δ(1 δ) (λ/2 + λk)ej + (?) − | | − − k k∈M   X where

(?) = √δ(1 δ) λkek + (1 √δ) ξjej − − − k∈M X Xj∈ /I 1 √δ = √δ(1 δ) λk − ξk ek + (1 √δ) ξjej = (1 √δ) ξjej. − − − √ − − k∈M δ(1 δ) ! X − j∈XJ\{n} j∈XJ\{n}

Thus,

z = √δ(1 δ) M λ/2 + 1 √δ en+√δ(1 δ) (λ/2+λk)ej +(1 √δ) ξjej. (3.44) − | | − − k − k∈M   X j∈XJ\{n}

(v?) (v?) Note that (1 ε)z K R because (1 ε)y, (1 ε)¯x K and z R . So, (3.39) gives − ∈ ∩ − − ∈ ∈ (1 ε)z 1 + o(δ), that is, z 1 + o(δ). Since all coefficients of the terms in (3.44) k − k1 ≤ k k1 ≤ 70

are non-negative by the assumption ξj 0, the `1-norm of z is ≥

z = √δ(1 δ) M λ/2 + 1 √δ + √δ(1 δ) (λ/2 + λk) + (1 √δ) ξj k k1 − | | − − − k∈M X j∈XJ\{n}

= 1 √δ + √δ(1 δ) (λ + λk) + (1 √δ) ξj = 1 δ√δ + (1 √δ) ξj − − − − − k∈M X j∈XJ\{n} j∈XJ\{n}

Therefore, z 1 + o(δ) implies k k1 ≤

δ√δ ξj + o(δ) = O(δ√δ) + o(δ) = o(δ), ≤ 1 √δ j∈XJ\{n} −

which completes the proof.

Proof of Proposition 3.6. Suppose that K and H satisfy (3.38), (3.39). It follows from

◦ dH(K,H) = δ that there exist a constant c = c(n) and a vertex v of H or H with xv v | − | ≥ ◦ cδ. Otherwise, xv v = o(δ) for every vertex v of H or H , which gives dH(K,H) = o(δ); | − | contradiction. In addition, since xv, ej = v, ej + o(δ) for every j supp(v) by (3.40), h i h i ∈ there exists j supp(v) such that xv, ej cδ. Without loss of generality, we may assume 6∈ |h i| ≥ that v is a vertex of H,

1 / supp(v), and xv, e1 cδ. (3.45) ∈ h i ≥

Let I = supp(v). Take a maximal clique J containing 1 in the graph G associated with H,

and let m be the unique common point between I and J. Without loss of generality, we

write J = 1, 2, , m , I J = m , and v = ei. For k = 1, , m, define the subset { ··· } ∩ { } i∈I ···

Ik of I by P

I1 = i I : i  1 ∈ n o Ik = i I (I1 Ik−1): i  k , if k = 2, . . . , m. ∈ \ ∪ · · · ∪ n o 71

The maximality of J gives that for each i I m there exists j J that is not connected ∈ \{ } ∈ to i, so

I = (I1 Im) m . (3.46) ∪ · · · ∪ ∪ { }

Im I1 I Ik 2 I = supp(v) m

1 I1 e I ? 2 J = supp(v ) e

k 2 Ik = supp(vk±) = supp(wk±) e

Since each Ik k is an independent set, we can choose a maximal independent set Ik ∪ { } + − containing Ik k for k = 1, , m 1. Consider the vertices v , v with support Ik for ∪ { } ··· − k k e k = 1 . . . , m 1 defined by − e

± v = ek ei ei, k − ± I Xk Iek\(XIk∪{k})

+ − and the vertices vm, vm with support I defined by

± v = em + ei ei. m ± I ∪···∪I I 1 Xm−1 Xm

For j = 1, . . . , m, let

tj = sign xv+ + xv− , ej . (3.47)  k k  k∈XJ\{j} D E   Let v? be the vertex of H◦ with support J defined as

m ? v = tjej. j=1 X 72

+ − + − Consider the vertices w1 , w1 , . . . , wm, wm defined by

± w = tkek ei ei for k = 1, . . . , m 1 k − ± − I Xk Iek\(XIk∪{k}) ± w = tmem + ei ei. m ± I ∪···∪I I 1 Xm−1 Xm

Since j supp(w±) for each j, (3.40) implies ∈ j

± x ± , ej = w , ej + o(δ) = tj + o(δ) for j = 1, . . . , m. (3.48) wj j D E D E

± ± Note that wk , vk have the same support and may have different values only in the k-the coordinate. If k J is different from j, then j k and by Lemma 3.6 we get ∈ ∼

xw± , ej = xv± , ej + o(δ) for each k J j . k k ∈ \{ } D E D E

Together with (3.47), it implies

xw+ + xw− , tjej = tj xv+ + xv− , ej + o(δ) o(δ). (3.49) k k k k ≥ k∈XJ\{j} D E k∈XJ\{j} D E

± ? From the construction of wk ’s and v , we can see that

m w+ + w− k k = v?. (3.50) 2 k=1 X

Let F be the dual face of v?, i.e., F = (v?)∗ = x H : x, v? = 1 . Consider the point { ∈ h i }

m 1 xw+ + xw− y = k k , m 2 k=1 X which belongs to K. To get a lower bound of y, v? , let us split it into three parts as h i 73

follows:

m m m m x + + x − x + + x − ? 1 wk wk 1 wk wk y, v = , tjej = , tjej h i m 2 m 2 * k=1 j=1 + j=1 k=1 X X X X   = (I) + (II) + (III)

where

m x + + x − 1 wj wj (I) = , tjej , m 2 j=1 * + X x + + x − t1 wk wk (II) = , e1 , m 2 k∈XJ\{1}   m x + + x − 1 wk wk (III) = tj , ej . m 2 j=2 X k∈XJ\{j}  

1 m 2 The first term satisfies (I) = m j=1 tj + o(δ) = 1 + o(δ) by (3.48), and the third term (III) is non-negative or o(δ)-small byP (3.49). For (II), note that 1 supp(w±) for k = 2, . . . , m 6∈ k

because k 1, and every element of I I1 is connected to 1 due to the choice of I1. The ∼ \ ± vertices wm and v have the same support I and may have different values only in the

coordinates i Im m I I1. Thus, Lemma 3.6 gives ∈ ∪ { } ⊂ \

x ± , e1 = xv, e1 + o(δ) (3.51) wm h i D E

+ − For k = 2, . . . , m 1, the vertices w and w have the same support Ik and different − k − k

values only in the coordinates i Ik k I I1. By Lemma 3.6, we get ∈ ∪ { } ⊂ \ e

xw+ , e1 = x−w− , e1 + o(δ) for k = 2, . . . , m 1, (3.52) k k − D E D E 74

which gives x + + x − , e1 = o(δ) by symmetry of K. Thus, by (3.51) and (3.52), wk wk D E m x + + x − wk wk , e1 = xv, e1 + o(δ) cδ + o(δ). 2 h i ≥ k=2 X  

c It gives t1 = 1 and hence the second term (II) is at least m δ + o(δ). Finally we have

c y, v? 1 + δ + o(δ). (3.53) h i ≥ m

1 ? On the other hand, y is a point of K which is O(δ)-close to the point m v on the facet F

1 ? by (3.50). In fact, m v is an interior point of F . Indeed,

1 ? 1 1 1 v +v ¯ 1 v = tjej = = v, m J J Vj 2 J Vj j∈J j∈J v∈V j∈J v∈V | | X | | X | | Xj  X Xj | | | |

1 ? where Vj = v ext(F ): j supp(v) for j J andv ¯ = 2tjej v Vj for v Vj. So, v { ∈ ∈ } ∈ − ∈ ∈ m 1 ? can be written as m v = v∈ext(F ) λvv where all λv’s are positive and satisfy λv = 1. It 1 ? P P implies m v is an interior point of F . In addition, under the assumption (3.39), the point (v?) 1 v∗ is o(δ)-almost on the boundary of K because 1 v? ∂B . This contradicts (3.53). m m ∈ 1 t ? Indeed, write y = y0 + v for y0 F and t > 0, and take a point z on the boundary m ∈ 1 ? 0 0 of F meeting with the ray from y0 to v . Then, (1 c δ)z K for some c > 0, and m − ∈ 1 ? 1 ? v = (1 µ)y0 + µz for some µ = O(δ) because v is an interior point of the facet F . m − m Then, the point

µ 0 (1 µ)y + 1−c0δ (1 c δ)z 1 + (1 µ)t 1 ? 2 1 ? − µ − = − µ v = 1 + (1 µ)t + O(δ ) v (3.54) 1 µ + 0 1 µ + 0 · m − · m − 1−c δ − 1−c δ

(v?) belongs to K R because it is a convex combination of two points y and (1 δ)z in K. ∩ −

So, the `1-norm of (3.54) is at most 1 + o(δ) by the assumption (3.39), which gives t = o(δ).

It implies y, v? = 1 + t = 1 + o(δ) which contradicts (3.53) and completes the proof. h i 75

n Proposition 3.7. Let H be a standard Hanner polytope in R and K a symmetric convex

n n (v) (v) body satisfying B1 K B∞. If dH(K R ,B∞ ) = δ for small δ > 0 and some vertex v ⊂ ⊂ | of H, then

K K◦ V (X)V (X∗) + c(n)δ, | || | ≥

∗ ◦ where X = (xF ), X = (xF ∗ ) are the xF -points obtained from K, K as in Definition 3.2.

(v) (v) (v) Proof. Without loss of generality, we may assume that (1 δ)B∞ K R B∞ and − ⊂ | ⊂ (v) (1 δ)v is on the boundary of K R . In addition, taking an appropriate coordinate system, − | (v) m we may assume that R = span e1, . . . , em = R and xv, e1 = min1≤j≤m xv, ej . { } h i | h i | Then,

xv, e1 1 δ. h i ≤ −

m Indeed, if xv, ej > 1 δ for all j = 1, . . . , m, then the orthogonal projection of xv to R is h i − m m m in the interior of v + δB∞ by Lemma 3.5, but it is impossible because K R (1 δ)B∞ | ⊃ − m and (1 δ)v is on the boundary of K R . − |

Let ε = 1 max c , c ∗ : F G, faces of H . Then ε > 0. Indeed, since c int(F ), − h F G i 6⊂ F ∈ n o the centroid can be written as cF = v λvv where v runs over all vertices of H contained

in F , and all λv’s are positive numbersP satisfying λv = 1. So, if c , c ∗ = 1 then we v h F G i have v, w = 1 for any vertex v in F and any vertexP w in G∗. It implies G∗ F ∗; hence h i ⊂

F G. Thus, max c , c ∗ : F G < 1. ⊂ h F G i 6⊂ n o Let E1 = y H : y, e1 = 1 be the face of H with centroid e1. Then, v E1, and { ∈ h i } ⊂ n xE∗ = e1 because K B∞ and e1 K. Consider the point 1 ⊂ ∈

? x = (1 δ)xE∗ + (1 + ε)δxv∗ − 1

= (1 δ)e1 + (1 + ε)δx ∗ , − v 76

where v∗ is the dual face of (a zero-dimensional face) v. Then

? xv, x = (1 δ) xv, e1 + (1 + ε)δ h i − h i (1 δ)2 + (1 + ε)δ = 1 (1 ε)δ + δ2 ≤ − − − and, for any face F of H different from v,

? x , x = (1 δ) x , e1 + (1 + ε)δ x , x ∗ h F i − h F i h F v i

(1 δ) + (1 + ε)δ [ c , c ∗ + O(δ)] ≤ − h F v i (1 δ) + (1 + ε)δ(1 ε) + O(δ2) = 1 ε2δ + O(δ2). ≤ − − −

Therefore, for small δ > 0, ◦ ? 2 x (1 + ε δ) XF ∈ F h [ i

where XF is the simplex defined from X = (xF ) as in (3.2). First, consider the case that K (1 + ε2δ)( X ). Then x? K◦. In addition, ⊂ F F ∈ ? ∗ S 0 n−1 the point x is outside of the polytope F XF∗ . If G = F ,...,F is a flag over H S  ? ◦ containing v and E1, then the simplex by x(F 0)∗ , . . . , x(F n−1)∗ and x is contained in K , not

∗ in the polytope F XF. So S

◦ ∗ ? K V (X ) + conv x , x 0 ∗ , . . . , x n−1 ∗ | | ≥ (F ) (F ) n o = V (X∗) + εδ X∗ , | G|

which implies

K K◦ K V (X∗) + ε K X∗ δ | | | | ≥ | | | | | G| 1 Bn V (X)V (X∗) + ε Bn | 1 | δ. ≥ 2 · · | 1 | · n!2n   77

2 On the other hand, if K (1 + ε δ)( X ), then there exist a flag F and a point x 6⊂ F F ∈ 2 2 conv x : F F such that (1 + ε δ)xS K. Since it gives K V (X) + ε δ X , { F ∈ } ∈ | | ≥ | F|

K K◦ V (X) K◦ + ε2 X K◦ δ | | | | ≥ | | | F| | | 1 Bn V (X)V (X∗) + ε2 | 1 | Bn δ. ≥ 2 · · n!2n · | 1 |  

2 n 2 ◦ ∗ ε B1 2n−1ε2 Consequently, we have K K V (X)V (X ) + c(n)δ where c(n) = | n+1| = 3 . | | | | ≥ n!2 n!

n Theorem 3.3. Let H be a standard Hanner polytope in R and K0 a symmetric convex

n body in R with dBM (K0,H) = 1+δ for small δ > 0. Then, there exists a symmetric convex

body K of with dH(K,K0) = 1 + o(δ) such that

K K◦ V (X)V (X∗) + c(n)δ (3.55) | || | ≥

∗ ◦ where X = (xF ), X = (xF ∗ ) are the sets of the xF -points obtained from K, K .

n Proof. Let H be a standard Hanner polytope in R , and K a symmetric convex body in

n R with dH(K,H) = δ for small δ > 0. As mentioned in Section 3.1, note that V (X) can

be viewed as the volume of a star-shaped, not necessary convex, polytope P = F XF.. First, consider the case that the (Hausdorff) distance between K and P is comparableS to

δ. Then, there exist a flag F and a point x conv x : F F with (1 + cδ)x K. The ∈ { F ∈ } ∈ convex hull of (1 + cδ)x and all x for F F is contained in K int(P ), and its volume is F ∈ \

cδ times the volume of XF because the distance between x and (1 + cδ)x is cδ times the distance between x and the origin. Thus, in this case

K K◦ (V (X) + cδ X ) V (X∗) | | | | ≥ | F| c Bn 2 V (X)V (X∗) + | 1 | δ. ≥ n!2n 78

To prove the other case dH(K,P ) = o(δ), we apply Propositions 3.5, 3.6, and 3.7. So, there

2 exists a polytope K˜ with dBM (K,˜ conv(P )) = 1 + O(δ ) such that

K˜ K˜ ◦ V (X˜)V (X˜ ∗) + c(n)δ, | || | ≥

˜ ˜ ∗ ˜ ˜ ◦ where X = (xF ), X = (xF ∗ ) are the sets of the xF -points obtained from K, K as in

Definition 3.2. Moreover, we get dBM (K,K˜ ) dBM (K,P˜ ) dBM (K,P ) = 1 + o(δ) in this ≤ · case, which complete the proof.

Finally we are ready to prove the local minimality of the volume product at Hanner polytopes.

Proof of Theorem 3.1. After taking a linear transformation and a small o(δ)-perturbation on K, by Theorem 3.3 we have

K K◦ V (X)V (X∗) + c(n)δ. | | | | ≥

Propositions 3.3 and Corollary 3.2 imply

V (X) V (Y ) = O(δ2) = V (X∗) V (Y ∗) . | − | | − |

Thus

K K◦ V (X)V (X∗) + c(n)δ | | | | ≥ V (Y )V (Y ∗) + c0(n)δ. ≥

Finally, since V (Y )V (Y ∗) V (C)V (C∗) = H H◦ by Propositions 3.2 and 3.4, we have ≥ | || | (K) (H) + c(n)δ. P ≥ P Chapter 4. Local minimality in the nonsymmetric case

We now consider the general case of the Mahler conjecture. In this case the simplex is the only known candidate for a minimizer of the volume product in the class of convex

n bodies in R . In this chapter we prove that the simplex is a strict local minimizer of the volume product. More precisely we will prove the following result.

n n Theorem 4.1. Let K R be a convex body close enough to the simplex ∆ in the sense ⊂ n of dBM (K, ∆ ) = 1 + ε for small ε > 0. Then

(K) (∆n) + c(n)ε, P ≥ P

where c(n) > 0 is a constant depending on the dimension n only.

n n Throughout this chapter we fix a regular simplex ∆ with n + 1 vertices v0, . . . , vn R ∈ satisfying n, if i = j, vi, vj = (4.1) h i   1, if i = j.  − 6 n  n ◦ n Under this assumption ∆ = conv v0, . . . , vn , we have (∆ ) = ∆ . { } −

n Lemma 4.1. Let ∆ be the regular simplex defined from vertices v0, . . . , vn as above. Let

n ∗ F be a face of ∆ and F its dual face; denote their centroids by cF , cF ∗ . Then

∗ 1. vj F if and only if vj F . ∈ − 6∈

⊥ 2. F = span vi vj : vi, vj F . { − 6∈ }

1+dim F 3. cF ∗ = n−dim F cF .

79 80

Proof. For simplicity, assume that F = conv v0, . . . , vk for k = dim F < n. By (4.1), { }

∗ n ◦ F = y (∆ ) : x, y = 1 x F = conv vk+1,..., vn , ∈ h i ∀ ∈ − − n o n o

which implies the first and the third properties. For the second one, we see that vi vj − ⊥ ⊥ for vi, vj F belongs to F , and both F and span vi vj : vi, vj F have the same 6∈ { − 6∈ } dimension n 1 k. − −

4.1 Continuity of the Santal´omap

The following volume formula is known (using polar coordinates). For every interior point z of K,

z 1 −n K = (hK (u) z, u ) du (4.2) | | n n−1 − h i ZS where hK (x) = max x, y : y K is the support function of K. By the minimum property {h i ∈ } of Kz at the Santal´opoint s(K), it turns out that z = s(K) is a unique point satisfying | | the condition (see [82], p. 419)

−n−1 (hK (u) z, u ) u du = 0. (4.3) n−1 − h i ZS

This is equivalent to the fact that the centroid of (K z)◦ is the origin. − n n Denote by the space of convex bodies in R endowed with the Hausdorff metric dH. K n n−1 The space ( , dH) is isometrically embedded in the space C(S ) of continuous functions K n−1 on the sphere S by the isometry K hK , that is, dH(K,L) = hK hL ∞ for every 7→ k − k n K,L in where hK is the support function (see [82], p. 37). K

n n Proposition 4.1. The Santal´omap s :( , dH) R mapping a convex body K to K → its Santal´opoint s(K) is continuous. Furthermore, for every convex body K, there exist 81

constants cK , δK > 0 such that for each convex body L with dH(K,L) δK , ≤

s(K) s(L) cK dH(K,L). | − | ≤

Proof. The continuity of the Santal´omap s( ) is proved using a standard argument (the · dominated convergence theorem and the uniqueness of s(L) in (4.3)).

For the second part, fix a convex body K and let L be any convex body which is close to K in the Hausdorff metric. Since s(K) is in the interior of K, there is a rK > 0 such that the ball B(s(K), rK ) with center s(K) and radius rK is contained in K. Then, since

◦ ◦ (K s(K)) B(0, rK ) = B(0, 1/rK ), − ⊂ we have, for every u Sn−1, ∈

hK (u) s(K), u = u ◦ rK . − h i k k(K−s(K)) ≥

Define three functions a, x, y on Sn−1 by

a(u) = hK (u) s(K), u , − h i x(u) = s(L), u s(K), u , h i − h i

y(u) = hL(u) hK (u). −

n−1 Note that hL(u) s(L), u = a(u) x(u) + y(u) for every u S and − h i − ∈

a(u) rK , x(u) s(L) s(K) , y(u) dH(K,L). (4.4) | | ≥ | | ≤ | − | | | ≤ 82

Note also that the Taylor theorem gives

x y −n−1 (a x + y)−n−1 = a−n−1 1 − − − a h i x y x y = a−n−1 1 + (n + 1) − + f − a a h   i where the function f(t) := (1 t)−n−1 1 (n + 1)t is O(t2) for small t > 0. Then (4.3) − − − implies

−n−1 −n−1 0 = (hL(u) s(L), u ) u du = (a x + y) u du Sn−1 − h i Sn−1 − Z Zx−y −n−1 (n + 1)(x y) f a = a + n+2 − + n+1 u du Sn−1 a a Z 
 = (a) + (n + 1) (b) (c) + (d). (4.5) − h i

Here (a), (b), (c) and (d) are given as:

−n−1 −n−1 (a) = a(u) u du = (hK (u) s(K), u ) u du n−1 n−1 − h i ZS ZS is equal to 0, (i.e., the origin), by (4.3). The Euclidean norm of

x(u) s(L) s(K), u (b) = n+2 u du = h − in+2 u du Sn−1 a(u) Sn−1 [hK (u) s(K), u ] Z Z − h i is greater than or equal to

s(L) s(K), u s(K) s(L) u, v 2 du v, h − in+2 u du = | − || h i |n+2 , Sn−1 [hK (u) s(K), u ] Sn−1 [hK (u) s(K), u ]  Z − h i  Z − h i where v = s(L)−s(K) Sn−1. Thus |s(K)−s(L)| ∈

n−1 s(K) s(L) 2 S s(K) s(L) (b) | − n+2| u, v du = | −n+2 |. | | ≥ diam(K) Sn−1 |h i| n diam( K) | | Z | | 83

The Euclidean norm of y(u) (c) = n+2 u du n−1 a(u) ZS is less than or equal to, by (4.4),

y(u) dH(K,L) n−1 | n+2| du n+2 S . n−1 a(u) ≤ r ZS K | |

That is, Sn−1 (c) n+2 dH(K,L). | | ≤ rK !

If c > 0 is an absolute constant such that f(t) c t 2 for small t > 0, then the Euclidean | | ≤ | | norm of x−y f a (d) = n+1 u du n−1 a ZS
is less than or equal to

x(u) y(u) 2 1 x(u) 2 + y(u) 2 c − n+1 du 2c | | n|+3 | du. n−1 a(u) a(u) ≤ n−1 a(u) ZS ZS | | | |

By (4.4), n−1 2c S 2 2 (d) n+3 s(K) s(L) + dH(K,L) . | | ≤ r K | − |   In summary,

(a) = 0, (b) c1 s(K) s(L) , (c) c2dH(K,L), | | ≥ | − | | | ≤ and

2 2 (d) c3 s(K) s(L) + dH(K,L) , | | ≤ | − |
where Sn−1 Sn−1 2c Sn−1 c1 = , c2 = , and c3 = . n diam(K) n+2 n+2 n+3 rK r K | | 84

Since (b) (c) 1 ((a) + (d)) by (4.5), we have ≤ − n+1

c2 c3 2 2 s(K) s(L) dH(K,L) + s(K) s(L) + dH(K,L) . | − | ≤ c1 (n + 1)c1 | − |

By continuity of the Santal´omap s( ), the Santal´opoint of L converges to that of K as a · body L approaches to K in the sense of dH(K,L) 0. Thus the two quadratic terms in → above inequality can be ignored if dH(K,L) is small enough. Therefore, there exists δK > 0 such that for every L with dH(K,L) δK ≤

s(L) s(K) cK dH(K,L) | − | ≤

n+2 where cK is a constant greater than c2/c1 =(diam(K)/rK ) .

n Proposition 4.2. Let K be a convex body in R , and let rK be the distance from the

Santal´opoint to the boundary of K, i.e., rK = dist(s(K), ∂K). Then, for each z with

rK z s(K) < 2n , | − | 2 z s(K) 2n 2 K K 1 + 2 z s(K) . | | ≤ rK | − |  

Proof. First assume that s(K) = 0. Then, by (4.3),

y dy = y dy = 0. ◦ ◦ ZK Z(K−s(K))

The formula (4.2) can be written as

∞ z 1 −n −n K = (hK (u) z, u ) du = (t z, u ) dt du | | n n−1 − h i n−1 − h i ZS ZS ZhK (u) ρ (u) ρ (u) K 1/r2 dr du K rn−1dr du = n+1 = n+1 , n−1 (1/r z, u ) n−1 (1 z, ru ) ZS Z0 − h i ZS Z0 − h i 85

which gives

z dy K = n+1 . | | ◦ (1 z, y ) ZK − h i Since ∞ (n + 1) (n + j) (1 t)−n−1 = 1 + (n + 1)t + ··· tj, − j! j=1 X we have

∞ (n + 1) (n + j) Kz = 1 + (n + 1) z, y + ··· z, y j dy | | h i j! h i j=2 KZ◦  X  ∞ (n + 1) (n + j) j = dy + ··· n z, y dy. j!nj h i Z◦ Z◦ j=2 K K X

◦ Note that K is contained in the ball of radius 1/rK and center 0 because K contains the

◦ ◦ ball of radius rK and center s(K) = 0. So, y 1/rK for y K . Thus, for every y K | | ≤ ∈ ∈ and every z with z s(K) < rK , | − | 2n

r 1 1 n z, y n K = . | h i| ≤ · 2n · rK 2

We get

∞ 1 j−2 Kz K◦ + n z, y 2 dy | | ≤ | | ◦ | h i | 2 K j=2 Z X  2n2 K◦ 1 + z 2 . ≤ | | r2 | |  K 

Substitute K and z by K s(K) and z s(K) in above inequality to get the general case. − − Then 2n2 Kz = (K s(K))z−s(K) Ks(K) 1 + z s(K) 2 . | | | − | ≤ | | r2 | − |  K  86

4.2 Construction of polytopes for the simplex case

For the proof of Theorem 4.1, we follow the construction given in Chapter 3; Replace

n the polytope P in Section 3.1 by the regular simplex ∆ = conv v0, . . . , vn , and for each { }

face F define the affine subspace AF by

⊥ AF = cF + F ,

where cF denotes the centroid of a face F . Then the affine subspaces satisfy three conditions in Section 3.1, that is, for each face F and its dual face F ∗,

n n ◦ (a) A ∆ = c and A ∗ (∆ ) = c ∗ . F ∩ { F } F ∩ { F }

(b) dim A = dim F ⊥ = n 1 dim F . F − −

(c) x, y = 1 for each x A and y A ∗ . h i ∈ F ∈ F

Proof. For simplicity, assume F = conv v0, . . . , vm for m = dim F . By Lemma 4.1, we { } ∗ have F = conv vm+1, . . . , vn and { }

v0 + + vm A = ··· + span vi vj : i, j > m , F m + 1 − n o vm+1 + + vn A ∗ = ··· + span vi vj : i, j m , F n m − ≤ − n o which gives the first and third properties. The second one is trivial.

Now, as in Definition 3.1, from the body K and the polytope ∆n we construct the volume

∗ ∗ function V ( ) and the points X = (x ), Y = (y ), C = (c ), X = (x ∗ ), Y = (y ∗ ), · F F F F F ∗ n N n C = (cF ∗ ), as a point in (R ) for the number N of faces of ∆ .

Lemma 4.2. The simplex ∆n satisfies the condition (3.5) in Proposition 3.2, i.e., the

simplex CF for each flag F has the equal volume. 87

0 n−1 n Proof. Let F = F ,...,F be a flag over ∆ = conv v0, . . . , vn . Without loss of { } generality, we may assume that

k F = conv v0, . . . , vk , for k = 0, 1, . . . , n 1. { } −

For each k, the centroid of F k can be computed as

1 c k = (v0 + + vk). F k + 1 ···

Thus, the volume of the simplex CF is equal to the absolute value of

cF 0 v0 v0       1 cF 1 1 v0 + v1 1 v1 det  = det  = det  , n!  .  (n!)2  .  (n!)2  .   .   .   .                     c n−1   v0 + v1 + + vn−1   vn−1   F   ···          which is a constant.

Lemma 4.3. The condition (3.6) from Proposition 3.3 holds, i.e.,

V 0(C),Z = 0 for every Z = z with all z A A . F F ∈ F − F

n Proof. For z R and a face F , we denote by zF the point zF = (z ) with z = z and ∈ G F z = 0 for G = F . G 6 b b n Fix a face F of ∆ = conv v0, . . . , vn ; write F = conv v0, . . . , vm for m = dim F . By { } { } Lemma 4.1,

⊥ A A = F = span vi vj : i, j > m . F − F { − }

0 n By linearity, it suffices to show V (C), (vi vj)F = 0 for i, j > m. By ‘symmetry’ of ∆ , − D E b 88

0 we have V (C + tviF ) = V (C + tvjF ) for any t > 0, which implies V (C), (vi vj)F = 0. − D E b b b

n Lemma 4.4. Let K be a convex body in R with dBM (K,L) = 1 + ε for small ε > 0. Then

n n there are a constant c = c(n) and an affine transformation A : R R such that every → centroid of facets of ∆n belongs to AK and (1 cε)∆n AK ∆n. − ⊂ ⊂

n n n Proof. By definition, there exist an affine isomorphism A : R R and a ∆ such that → ∈ (1 ε)(∆n a) AK ∆n a. Adding the point a in both inequalities, we have − − ⊂ ⊂ −

(1 ε)∆n + εa AK + a ∆n. − ⊂ ⊂

Furthermore, from ∆n n∆n we have that the simplex (1 ε)∆n + εa contains a smaller − ⊂ − simplex (1 (n + 1)ε)∆n. Indeed, −

(1 ε)∆n + εa = (1 ε nε)∆n + nε∆n + εa − − −   (1 ε nε)∆n + ε( a) + εa = (1 (1 + n)ε)∆n. ⊂ − − − −

Thus

(1 cε)∆n AK + a ∆n, for c = n + 1. − ⊂ ⊂

Let S be a simplex of minimal volume containing AK + a. In [46], Klee prove that all the

centroids of facets of S should belong to AK+a. On the other hand, if x S (1+κ)(1 ε)∆n ∈ \ − for some κ > 0, then, (as in Lemma 2 of [70]),

1 1 n2 S (1 cε)∆n + κ(1 cε)n ∆n | | ≥ | − | n · − · n · n + 1 | | κ = ∆n (1 cε)n 1 + . | | − n + 1   89

−n n κ If κ > κ0 = (n + 1) [(1 cε) 1], then (1 cε) 1 + > 1 and hence the existence − − − n+1 n   of such x implies S > ∆ which is a contradiction. Thus, for κ > κ0, it should be | | | | S (1 + κ)(1 ε)∆n, which implies ⊂ −

(1 κ)S (1 κ) (1 + κ) 1 cε ∆n (1 cε)∆n. − ⊂ − − ⊂ − 


Therefore,

(1 κ)S AK + a S. − ⊂ ⊂

−n 2 Taking κ = 2κ0 = 2(n + 1)(1 cε) 1) 2n(n + 1) ε completes the proof. − − ≈

Lemma 4.5. Let K be a convex body contained in ∆n and containing all the centroids of facets of ∆n. Suppose that δ = min d > 0 : (1 d)∆n K ∆n is small enough. Then, { − ⊂ ⊂ }

K K◦ V (X)V (X∗) + c(n)δ, | || | ≥

∗ n where X = (x ), X = (x ∗ ), and V ( ) come from Definition 3.1 for K and ∆ . F F ·

Proof. By minimality of δ, there is a vertex v of ∆n such that (1 δ)v is on the boundary − of K. Then,

xv, v n δ. (4.6) h i ≤ − 90

n Indeed, if xv, v > n δ, the point w = 2(1 δ)v xv is an interior point of ∆ because h i − − −

w, v = 2(1 δ) v, v xv, v h − i − h − i − h − i 2(1 δ) v, v v, v = n + 2nδ < 1, ≤ − h − i − h − i −

w, vj = 2(1 δ) v, vj xv, vj for vj = v h − i − h − i − h − i 6

= 2(1 δ) xv, v + vi = 2(1 δ) xv, v + xv, vi − − * + − − h i h − i vi6∈{Xv,vj } vi6∈{Xv,vj } 2(1 δ) (n δ) + (n 1) = 1 δ < 1. ≤ − − − − −

1 n So, xv, v > n δ implies that (1 δ)v = (w+xv) is an interior point of ∆ ; contradiction! h i − − 2

We may assume that max xv, vj : vj = v = xv, v0 . Consider the point {h i 6 } h i

? x = (1 δ)xV ∗ + (1 + δ)δxv∗ = (1 δ)v0 + (1 + δ)δxv∗ , − 0 − −

where V0 = conv v1, . . . , vn and 0 < δ < 1/n. Note that − { }

0 = xv, v0 + + vn xv, v + n xv, v0 , h ··· i ≤ h i h i

1 n−δ which implies xv, v0 xv, v . By (4.6), we have xv, v0 , which implies h − i ≤ n h i h − i ≤ n

? xv, x = (1 δ) xv, v0 + (1 + δ)δ h i − h − i (1 δ)(1 δ/n) + (1 + δ)δ = 1 (1/n δ)δ + δ2. ≤ − − − − 91

For any face F different from v,

? x , x = (1 δ) x , v0 + (1 + δ)δ x , x ∗ h F i − h F − i h F v i

(1 δ) + (1 + δ)δ [ c , c ∗ + O(δ)] ≤ − h F v i 1 1 δ (1 δ) + (1 + δ)δ + O(δ2) = 1 − δ + O(δ2). ≤ − 2 − 2

Therefore, for small δ > 0,

◦ ? 1 1 δ x (1 + cδ) XF , c = max δ, − ∈ F n − 2 h [ i   where XF is the simplex defined from X = (xF ) as in (3.2). First, consider the case that K (1+cδ)( X ). Then x? K◦. In addition, the point ⊂ F F ∈ ? ∗ S 0 n−1 x is outside of the polytope F XF∗ . If G = F ,...,F is a flag over H containing S  ? ◦ v and E1, then the simplex by x(F 0)∗ , . . . , x(F n−1)∗ and x is contained in K , not in the

∗ polytope F XF. So S

◦ ∗ ? K V (X ) + conv x , x 0 ∗ , . . . , x n−1 ∗ | | ≥ (F ) (F ) n o = V (X∗) + δδ X∗ , | G|

which implies

K K◦ K V (X∗) + δ K X∗ δ | | | | ≥ | | | | | G| 1 Bn V (X)V (X∗) + δ Bn | 1 | δ. ≥ 2 · · | 1 | · n!2n  

On the other hand, if K (1 + cδ)( X ), then there exist a flag F and a point x 6⊂ F F ∈ S 92

conv x : F F such that (1 + cδ)x K. Since it gives K V (X) + cδ X , { F ∈ } ∈ | | ≥ | F|

K K◦ V (X) K◦ + c X K◦ δ | | | | ≥ | | | F| | | c Bn V (X)V (X∗) + | 1 | Bn δ. ≥ 2 · n!2n · | 1 |  

n 2 ◦ ∗ 0 0 c B1 2n−1c Consequently, we have K K V (X)V (X ) + c (n)δ where c (n) = | n+1| = 3 . | | | | ≥ n!2 n!

n Proof of Theorem 4.1. Let K be a convex body with dBM (K, ∆ ) = 1 + ε for small

ε > 0. By Lemma 4.4, after taking an affine transformation we may assume that all the

n n n centroids of facets of ∆ belong to K and (1 c1ε)∆ K ∆ for some c1 = c1(n) > 0. − ⊂ ⊂ n n n Let δ = min d > 0 : (1 d)∆ K ∆ . Then dH(K, ∆ ) = O(δ) and δ c1ε. By { − ⊂ ⊂ } ≥

Lemma 4.5, there exists c2 = c2(n) > 0 such that

◦ ∗ K K V (X)V (X ) + c2δ. | | | | ≥

Proposition 3.3 and Lemma 4.3 imply

V (X) V (Y ) = O(δ2) = V (X∗) V (Y ∗) . | − | | − |

Proposition 3.2 and Lemma 4.2,

V (Y )V (Y ∗) V (C)V (C∗) = ∆n (∆n)◦ . ≥ | || |

Thus there exists c3 = c3(n) > 0 such that

◦ ∗ ∗ K K V (X)V (X ) + c2δ V (Y )V (Y ) + c3δ | || | ≥ ≥ n n ◦ n ∆ (∆ ) + c3δ = (∆ ) + c3δ. ≥ | || | P 93

n By Proposition 4.1, dH(K, ∆ ) = O(δ) implies s(K) c4δ for some c4 = c4(n) > 0. Let | | ≤

rK be the distance from s(K) to the boundary of K. Then

n rK = dist(s(K), ∂K) dist(0, (1 δ)∆ ) s(K) ≥ − − | | 1 n dist 0, ∆ =: c5 ≥ 2  

By Proposition 4.2, we get

2 2 2 ◦ s(K) 2n 2 s(K) 2 2n c4 K K 1 + s(K) K (1 + c6δ ), for c6 = . | | ≤ | | r2 | | ≤ | | c2  K  5

The volume product of K satisfies

s(K) ◦ 2 (K) = K K K K (1 c6δ ) P | || | ≥ | | | | − n 2 n ( (∆ ) + c3δ) (1 c6δ ) (∆ ) + (c3/2)δ ≥ P − ≥ P (∆n) + cε, ≥ P

for a constant c = c(n) = c1c3/2 > 0. Chapter 5. Stability results for unconditional bodies

In the class of unconditional convex bodies, Saint Raymond confirmed the conjecture, and Meyer and Reisner, independently, characterized the equality cases. In this section we present a stability version of these results and also show that any symmetric convex body, which is sufficiently close to an unconditional body, satisfies the the reverse Blaschke-Santal´o inequality.

We will prove that if the volume product of an unconditional convex body is sufficiently close to that of the cube, then the body must be close to a Hanner polytope:

n Theorem 5.1. Let K be an unconditional convex body in R . If

(K) (Bn ) ε |P − P ∞ | ≤

n for small ε > 0, then there exists a Hanner polytope H R such that ⊂

dBM (K,H) 1 + c(n)ε, ≤

where c(n) > 0 is a constant depending on n only.

2 We would like to notice that the R -case of Theorems 5.1 was proved as a part of more

2 general stability result in R by B¨or¨oczky, Makai, Meyer and Reisner in [11], so our goal is

to concentrate on the case n 3. ≥ It is shown in Chapter 3 that a Hanner polytope is a local minimizer of the volume

product in the symmetric setting (see Theorem 3.1). We will use local minimality for

Hanner polytopes and Theorem 5.1 to prove that any convex symmetric convex body,

94 95

which is sufficiently close to an unconditional body, satisfies the reverse Blaschke-Santal´o inequality.

n Theorem 5.2. Let K be a symmetric convex body in R with

n min dBM (K,L): L R unconditional convex body = 1 + ε, ⊂ n o for small ε > 0. Then,

(K) (1 + c(n)ε) (Bn ), P ≥ ·P ∞ where c(n) > 0 is a constant depending on dimension n only.

5.1 Stability for unconditional bodies

n n + n Let R+ = x R : xi 0, i = 1, . . . , n and K = K R+. To prove Theorem 5.1 { ∈ ≥ ∀ } ∩ we first prove the following lemma which is based on the inductive argument of Meyer (see

[64] or [79]):

n Lemma 5.1. Consider ε 0 and unconditional convex body K R , n 2 such that ≥ ⊂ ≥

(K) (1 + ε) (Bn ). P ≤ P ∞

Then

(K e⊥) (1 + nε) (Bn−1), j = 1, . . . , n. (5.1) P ∩ j ≤ P ∞

Proof. Consider x K+ and n pyramids created by taking the convex hull of x and the ∈ intersection of K+ with each coordinate hyperplane. More precisely, let K+ = conv x, K+ i { ∩ e⊥ for i = 1, . . . , n. Then, using symmetry of K with respect to coordinate hyperplanes, i }

n n ⊥ n ⊥ n + n + n 1 K ei 2 K ei K = 2 K 2 K = 2 x, ei | ∩ | = x, ei | ∩ |, | | | | ≥ | i | n h i 2n−1 h i · n i=1 i=1 i=1 [ X X 96

or equivalently, n ⊥ 2 K ei + x, | ∩ | ei 1, for all x K . n K ≤ ∈ * i=1 + X | | Note that, by the unconditionality of K, the above inequality holds for all x K, so ∈

n ⊥ 2 K ei ◦ | ∩ | ei K . n K ∈ i=1 X | |

Applying the same argument to K◦ we get

n ◦ ⊥ 2 K ei | ∩ | ei K. n K◦ ∈ i=1 X | |

Thus, using the definition of polarity we get:

n 2 K e⊥ 2 K◦ e⊥ | ∩ i | | ∩ i | 1. (5.2) n K × n K◦ ≤ i=1 X | | | |

Next notice that

K◦ e⊥ = (K e⊥)◦ = (K e⊥)◦, (5.3) ∩ i | i ∩ i where the last equality follows from the unconditionality of K. Finally from (5.2), (5.3) we get n n 4 4 (K) K e⊥ (K e⊥)◦ = (K e⊥). (5.4) P ≥ n2 | ∩ i | × | ∩ i | n2 P ∩ j i=1 j=1 X X Now, we will use (5.4) to prove our lemma. Since (K) (1 + ε) (Bn ) and (Bn ) = P ≤ P ∞ P ∞ 4 n (Bn−1), we get n2 j=1 P ∞ P n n 4 4 (1 + ε) (Bn−1) = (1 + ε) (Bn ) (K) (K e⊥). n2 P ∞ P ∞ ≥ P ≥ n2 P ∩ j j=1 j=1 X X

It implies that

(K e⊥) (1 + nε) (Bn−1), j = 1, . . . , n. P ∩ j ≤ P ∞ 97

Indeed, if (K e⊥) > (1 + nε) (Bn−1), then P ∩ 1 P ∞

n n2 (K e⊥) > (1 + nε) (Bn−1) + (n 1) (Bn−1) = (1 + ε) (Bn ), (5.5) P ∩ j P ∞ − P ∞ 4 P ∞ j=1 X

⊥ where in the first inequality we used our assumption for the e1 -section and the reverse

⊥ Blaschke-Santal´oinequality for unconditional convex bodies for the sections by ei , i =

2, . . . , n. Finally note that (5.5) together with (5.4) gives (K) > (1 + ε) (Bn ); contra- P P ∞ diction!

n Next lemma will help us to treat the case of a convex body K R whose sections by ⊂ coordinate hyperplanes are close to the (n 1)-dimensional cube. −

n n Lemma 5.2. Let K B∞ be a convex body in R , for n 3, satisfying ⊂ ≥

K e⊥ = Bn e⊥, j = 1, . . . , n. (5.6) ∩ j ∞ ∩ j ∀

n Let p = (t, . . . , t) ∂K R+. Then ∈ ∩

K K◦ (1 + c(1 t)) Bn Bn , | || | ≥ − | 1 || ∞| where c = c(n) > 0 is a constant depending on n only.

Remark: The proof of Lemma 5.2 does not require the assumption of unconditionality of

K. Such assumption would make the proof a bit shorter and would improve constant c(n).

Proof. Using (5.6) we claim that K contains n vectors of the form (1,..., 1, 0, 1,..., 1) (i.e. n 1 coordinates are equal to 1 and one coordinate is equal to 0). Moreover, K must contain − n the convex hull of those points. Thus x : x, e1 + + en = n 1 B K, which { h ··· i − } ∩ ∞ ⊂ gives t (n 1)/n. Next we choose q ∂K◦ such that p, q = 1. Here, (5.6) guarantees ≥ − ∈ h i 98

n n that the normal vector to ∂K at p belongs to R+, and thus we may assume q R+. Let ∈

n n P = conv p x B∞ R+ : x, e1 + + en n 1 , { } ∪ ∈ ∩ h ··· i ≤ −  n n
Q = conv q B1 R+ . { } ∪ ∩

n n ◦ n ◦ n Then K R+ P . Also from K B∞ we get K B1 and thus K R+ Q. Next ∩ ⊃ ⊂ ⊃ ∩ ⊃ we notice that q belongs to the hyperplane with normal vector (1/√n, . . . , 1/√n) whose distance from the origin is 1/(t√n). Thus

1 1 t√n n−1 √n 1 t P = 1 + − n = 1 − , | | − n! n! · 1/√n − (n 1)! − 1 1/(√nt) 1 Q = = . | | n! · 1/√n n!t

It gives

1 1 1 t −n n n P Q = 1 + 1 − 4 B B (1 + cn(1 t)), (5.7) | || | n! − (n 1)! t ≥ | ∞|| 1 | −   −  

1 where cn = 1 is a positive constant in case of n 3. Note also that we have P Q − (n−1)! ≥ | || | ≥ 4−n Bn Bn , independently of the position of p (i.e. independently of the lower/upper | ∞|| 1 | bounds on t).

If we would assume that K is unconditional then we would be able to finish the proof

by simply multiplying (5.7) by 4n. In all other cases we must split K and K◦ into 4n parts

each depending on the choice of signs of coordinates. Construct P and Q corresponding to

each part, compute the corresponding volumes, and take the sum. We may apply (5.7) to

n the part corresponding to R+. In all other parts we are not guaranteed the position of the

point on the boundary, so we can estimate the volume using the remark after (5.7). More 99

precisely,

2 ◦ ◦ K K = Kδ K Pδ Qδ Pδ Qδ , | || | | | | δ | ≥ | | | | ≥ | || | δ ! δ ! δ ! δ ! δ ! X X X X X p where the sum is taken over all possible choices of n signs δ and Kδ is a subset of K corresponding to δ. Next

2 ◦ −n n n n −n n n K K 4 B B (1 + cn(1 t)) + (2 1) 4 B B | || | ≥ | ∞|| 1 | − − | ∞|| 1 |   q 2q n n −n = B B 1 + 2 1 + cn(1 t) 1 | ∞|| 1 | − − n n h −n−1p i B B 1 + 2 cn(1 t) . ≥ | ∞|| 1 | −

Note that if K is unconditional then we may apply a diagonal transformation to K and

assume that Bn K Bn . Indeed, this follows immediately from the fact that a tangent 1 ⊂ ⊂ ∞

hyperplane with a normal vector ei touches K at point yi which is a dilate of ei. Thus to

prove Theorem 5.1 it is enough to prove the following statement:

n Equivalent form of Theorem 5.1: Let K be an unconditional convex body in R

satisfying Bn K Bn . If (K) (1 + ε) (Bn ), then there exists a standard Hanner 1 ⊂ ⊂ ∞ P ≤ P ∞ n polytope H in R such that dH(K,H) = O(ε).

Proof of Theorem 5.1. We use induction on the dimension n to prove the above state-

ment. It is trivial if n = 1, and the case n = 2 was proved in [11]. Assume that the

statement is true for all unconditional convex bodies of dimension (n 1), n 3. For the − ≥ n inductive step, consider an unconditional convex body K R satisfying ⊂

Bn K Bn and (K) (1 + ε) (Bn ). 1 ⊂ ⊂ ∞ P ≤ P ∞ 100

We apply Lemma 5.1 and the inductive hypothesis to get standard Hanner polytopes H1 ⊂ n ⊥ n ⊥ R e1 , ... , Hn R en such that ∩ ⊂ ∩

⊥ dH(Hj,K e ) = O(ε), for j = 1, . . . , n. (5.8) ∩ j

Now we would like to show that we can ‘glue’ these (n 1)-dimensional Hanner polytopes − to create an n-dimensional Hanner polytope which is O(ε)-close to K. For each j, let

Gj = G(Hj) be the graph associated with Hj. Note that 1, . . . , n j is the vertex set { }\{ } of Gj and that Gj does not contain an induced path of edge-length 3.

Consider the graph G with vertex set 1, . . . , n and containing all edges of G1,...,Gn. { } Our goal is to show that the polytope P corresponding to G is a Hanner polytope which is

O(ε)- close to K.

We claim that the graph G satisfies the following three properties.

1. Each Gj is an induced subgraph of G (i.e. for any k, l 1, . . . , n , k and l are ∈ { }

connected by an edge of Gj if and only if they are connected by an edge of G).

2. If G is not the path of edge-length 3, then G does not contain any path of edge-length

3 as an induced subgraph.

If G is the path of edge-length 3, then it is a perfect graph. Otherwise, the second property implies that G does not contain any path of edge-length 3, that is, G is the graph associated with a Hanner polytope. Thus, in any cases G must be a perfect graph.

Let P be the 0-1 polytope associated with G, i.e., G = G(P ).

3. If G is neither the complete graph nor its complement, then dH(K,P ) = O(ε).

Indeed, to show the first property consider different i, j, k 1, . . . , n . Note that the ∈ { } existence of ij edge depends only on the properties of K span ei, ej , so i, j are connected ∩ { } 101

by an edge of Gk if and only if they are connected by an edge of G. It implies that each Gk is an induced subgraph of G.

To prove the second property, assume that G is not the path of edge-length 3. If G contains a path of edge-length 3, then there is j 1, . . . , n which does not belong to this ∈ { } path. Thus, Gj, as an induced subgraph of G, must also contain the path of edge-length 3, which contradict with the definition of Gj.

To show the third property, we first note that if G is not the complete graph then every maximal independent set of G is a proper subset of 1, . . . , n . The vertices of P have { } coordinates 0, 1 or 1 and the support of a vertex of P is a maximal independent set of G. − This implies that every vertex of P has at least one zero coordinate, i.e. every vertex of P is

contained in one of coordinate hyperplanes. This, together with the inductive hypothesis,

gives that each vertex of P is O(ε)-close to the boundary of K, which means that P is

contained in O(ε)-neighborhood of K. It gives K (1 + O(ε))P . Repeating the same ⊃ ◦ ◦ argument for G we get K (1 + O(ε))P and thus dH(K,P ) = O(ε). ⊃ From the above three properties, we finish the proof of Theorem 5.1, modulo two patho-

logical cases:

I. G or G is the complete graph.

II. G is the path of edge-length 3.

Therefore, it remains to consider those cases:

CASE I: G (respectively G) is the complete graph, so all H1,...,Hn are the (n 1)- − dimensional cubes (respectively (n 1)-dimensional cross-polytopes). Taking the polar if −

necessary, we may assume, without loss of generality, that all H1,...,Hn are the (n 1)- − dimensional cubes. Let K˜ = 1 K Bn , where 1−δ ∩ ∞  

⊥ δ = min d : (1 d)Hj K e Hj, j = 1, . . . , n . − ⊂ ∩ j ⊂ ∀ n o 102

Notice that δ = O(ε) and hence dH(K, K˜ ) = O(ε). By (linear order) continuity of the volume product in the Hausdorff metric, we get (K˜ ) (1 + O(ε)) (K). Thus, the P ≤ P assumption (K) (1 + ε) (Bn ) gives P ≤ P ∞

(K˜ ) (1 + O(ε)) (Bn ). (5.9) P ≤ P ∞

n ⊥ Also, using Hj = B e and the definition of δ we get ∞ ∩ j

K˜ e⊥ = Bn e⊥, j = 1, . . . , n. (5.10) ∩ j ∞ ∩ j ∀

Consider p = (t, . . . , t) ∂K˜ . Then Lemma 5.2 gives ∈

(K˜ ) (1 + c(1 t)) (Bn ). P ≥ − P ∞

Together with (5.9) the above implies that 1 t = O(ε). Thus p is in O(ε)-neighborhood of − n ˜ ˜ n the vertex of B∞. the unconditionality of K and (5.10) give that dH(K,B∞) = O(ε) and

n hence dH(K,B∞) = O(ε).

CASE II: G is the path of edge-length 3. The direct computation below will show that this

4 case contradicts with assumption on the volume product of K. Indeed in this case K R ⊂ is O(ε)-close to the polytope P which is associated with the path of edge length 3. The

4 volume product of P is strictly greater than that of the cube B∞. Thus selecting ε4 small enough we will be able to contradict the assumption of the Theorem 5.1. More precisely, if

G is represented by the vertex set 1, 2, 3, 4 and the edge set 12, 23, 34 , then G has the { } { } same vertex set and the edge set 24, 41, 13 . Applying the characterization of vertices of { } a polytope by maximal independent sets, we can get

P = conv ( 1, 0, 1, 0), (0, 1, 0, 1) and P ◦ = conv ( 1, 1, 0, 0), (0, 0, 1, 1) . ± ± ± ± ± ± ± ± n o n o 103

Then we have P = P ◦ and | | | |

P = conv ( 1, 0, 1, 0), (0, 1, 0, 1) | | ± ± ± ± n o 4 = x R : x1 + x2 1, x2 + x3 1, x3 + x4 1 ∈ | | | | ≤ | | | | ≤ | | | | ≤ n 1 o 3 = 2 x R : x1 + x2 1, x2 + x3 1, x3 1 s ds 0 ∈ | | | | ≤ | | | | ≤ | | ≤ − Z 1 n1−s o 3 = 4 x R : x1 + x2 1, x2 1 t dtds 0 0 ∈ | | | | ≤ | | ≤ − Z Z n o 1 1−s 10 = 8 (1 t2)dtds = . − 3 Z0 Z0

2 4 Thus (P ) = 10 > 4 = (B4 ). On the other hand, note that in this case G and G P 3 4! P ∞ are not complete
graphs, so we may apply the Property (3) from above to claim that K is O(ε)-close to P . Therefore, (K) > (B4 ), which contradicts the assumption (K) P P ∞ P ≤ (1 + ε) (B4 ) for ε > 0 small enough. P ∞

5.2 Minimality near unconditional convex bodies

The local minimality result, proved in Chapter 3, is the main tool of this section.

Let γn > 0 be a common threshold of ε to satisfy both Theorem 5.1 and Theorem 3.1.

More precisely, the constant γn satisfies the following:

n If K is a symmetric convex body in R with dBM (K,H) = 1 + ε, ε γn, for some • ≤ Hanner polytope H, then

n (K) (1 + αnε) (B ). (5.11) P ≥ P ∞

n If K is an unconditional convex body in R with dBM (K,H) 1 + ε, ε γn, for • ≥ ≤ every Hanner polytope H, then

n (K) (1 + βnε) (B ). (5.12) P ≥ P ∞ 104

where αn, βn > 0 are constants depending on n only.

Proof of Theorem 5.2. We will use γn to select the constant εn, a threshold of ε in the

n n statement of the Theorem 5.2. Consider a convex symmetric body K R . Let L R be ⊂ ⊂ an unconditional convex body with smallest, among unconditional bodies, Banach-Mazur

distance to K and let ε 0 be such that dBM (K,L) = 1 + ε. Also consider a Hanner ≥

polytope H0 with smallest, among Hanner polytopes, Banach-Mazur distance to L and let

δ 0 be such that dBM (L, H0) = 1 + δ. ≥

We will consider two cases: δ γn/3 and δ > γn/3. ≤

γn γn 1. Assume δ and thus dBM (L, H0) . Then ≤ 3 ≤ 3

dBM (K,H0) dBM (K,L)dBM (L, H0) = (1 + ε)(1 + δ) ≤

1 + ε + (1 + ε)γn/3 1 + γn/3 + 2γn/3 = 1 + γn, ≤ ≤

γn where, to guarantee the above inequalities, we select εn min , 1 . By (5.11), ≤ { 3 }

n (K) (1 + αnε) (B ). P ≥ P ∞

γn γn 2. Now assume that δ > 3 and thus dBM (L, H) > 3 for every Hanner polytope H. We

βnγn 1 will require εn min , . Since L TK (1 + ε)L for some T GL(n), ≤ 12n 2n ⊂ ⊂ ∈ n o

(K) = (TK) L (1 + ε)−1L◦ = (1 + ε)−n (L) P P ≥ | || | P (1 ε)n (L) (1 nε) (L). ≥ − P ≥ − P 105

n Moreover, since (L) (1 + βnγn/3) (B ) by (5.12), P ≥ P ∞

n (K) (1 nε) (L) (1 nε)(1 + βnγn/3) (B ) P ≥ − P ≥ − P ∞ (1 nε)(1 + 4nε) (Bn ) = (1 + 3nε 4n2ε2) (Bn ) ≥ − P ∞ − P ∞ (1 + nε) (Bn ). ≥ P ∞

Finally we combine the above two cases by taking

γn βnγn 1 εn = min , , and τn = min αn, n . 3 12n 2n { }  

n Then, (K) (1 + τnε) (B ) whenever ε εn. P ≥ P ∞ ≤ Chapter 6. Proposed further research

Functional versions of the Blaschke-Santal´oinequality and Mahler’s conjecture in terms of log-concave functions were investigated by Ball [3], Artstein-Avidan, Klartag, Milman

[2], and Fradelizi, Meyer [16, 17, 18].

To define the duality in the class of log-concave functions, we recall that the Legendre

z n n transform ϕ of a function ϕ : R R with respect to z R is given by L → ∪ {∞} ∈

z n ϕ(y) = sup x z, y z ϕ(x), for y R . L x∈Rn h − − i − ∈

If ϕ is convex, lower semi-continuous and ϕ(z) , then z zϕ = ϕ. ≤ ∞ L L n Let f : R [0, ) be a log-concave function. Then φ = log f is a convex and lower → ∞ − n z semi-continuous function from R to R . The polar f of for a log-concave function ∪ {∞} n n f : R R+ with respect to z R is defined by → ∈

−hx−z,y−zi z e f z(y) = e−L ϕ(y) = inf . z∈Rn f(x)

As an analogue of the volume product of a convex body, the functional (volume) product

n of a log-concave function f : R [0, ) is defined by → ∞

(f) = inf f(x)dx f z(y)dy. P z∈ n n n R ZR ZR

Then the functional Blaschke-Santal´oinequality [2, 16, 51] states that

2 (f) (e−| · | /2). P ≤ P 106 107

As in the case of convex bodies, the minimum of the functional volume product is not known in general. It was conjectured (see [16]) that

n −k·k (f) 4 = (e 1 ) P ≥ P

n n for any even log-concave function f : R [0, ), and (f) e for any log-concave → ∞ P ≥ n function f : R [0, ). → ∞ As a further study of the local minimality of the volume product, it would be good to try for the local minimality of the functional volume product of log-concave functions. Part II

‘Convexity’ of intersection bodies

108 Chapter 7. Introduction to intersection bodies

In this chapter we introduce basic notions and classical results on intersection bodies.

First let us start with the definition of the intersection body of a star body. In [58], Lutwak introduced the notion of the intersection body.

n Definition 7.1. Let K be a star body in R . The intersection body IK of K is defined by

its radial function

ρ (u) = K u⊥ , for u Sn−1, IK | ∩ | ∈

where u⊥ denotes the central hyperplane perpendicular to u.

K IK n 1 u u S − K u⊥ ∈ | ∩ |

K u⊥ ∩

From the volume formula in polar coordinates for the section K u⊥, the following ∩ analytic definition of the intersection body of a star body can be derived: the radial function of the intersection body IK of a star body K is given by

1 n−1 1 n−1 n−1 ρIK (u) = ρK (v) dv = ρK (u), u S . n 1 n−1 ⊥ n 1 R ∈ − ZS ∩u −

Here stands for the spherical Radon transform. The more general class of intersection R bodies is defined in the following way (see [22], [47]). A star body K is an intersection body

109 110

if its radial function ρK is the spherical Radon transform of an even non-negative measure

µ. We refer the reader to the books [22], [47], [48] for more information on the definition and properties of intersection bodies, and their applications in convex geometry and geometric tomography.

7.1 Classical theorems and questions

One of the fundamental results regarding intersection bodies is Busemann’s theorem [13].

n Theorem 7.1 (Busemann). Let K be an symmetric convex body in R . Then its intersec- tion body IK is convex.

The symmetric condition on the body K in the theorem is essential. It may not be true without symmetry (see the book [22] for a counterexample). Recently, a new proof of Busemann’s theorem was established by Berck [8], who also generalized the theorem to the case of Lp intersection bodies (see [21], [31], [87] and [47] for more information on the theory of Lp intersection bodies).

In view of Busemann’s theorem it is natural to ask how much of convexity is preserved or improved under the intersection body operation. As a way to measure ‘convexity’ of a body, we may consider the Banach-Mazur distance from the Euclidean ball. In [35], Hensley provided an interesting result regarding the Banach-Mazur distance between intersection bodies of symmetric convex bodies and the Euclidean ball.

Theorem 7.2 (Hensley). The Banach-Mazur distance between the intersection body of any symmetric convex body and the ball is bounded by an absolute constant, that is, for every

n symmetric convex body K R , ⊂

n dBM (IK, B ) C, 2 ≤ where C > 1 is an absolute constant. 111

n Compared with John’s classical result (Theorem 0.2), dBM (K,B ) √n for any sym- 2 ≤ metric convex body K, we see that in most cases the intersection body operation improves convexity in the sense of the Banach-Mazur distance from the ball.

Given that the intersection body of a Euclidean ball is again a Euclidean ball, another question about the intersection body operator I comes from works of Lutwak [59] and

Gardner [22, Prob. 8.6-7] (see also [28]):

Open Question Are there other fixed points of the intersection body operator?

As a partial answer to this question, it is shown in [15] that in a sufficiently small neigh-

borhood of the ball in the Banach-Mazur distance there are no other fixed points of the

intersection body operator. However, in general this question is still open.

We will provide several quantitative versions of Busemann’s theorem to show the (strict) improvement of convexity under the intersection body operation.

7.2 Several ways to measure ‘convexity’

In this section we study several ways to measure convexity of a body to provide quan- titative information on convexity of intersection bodies.

Quasi-convexity. First, the development of the theory of intersection bodies shows that

it is not natural to restrict ourselves to the class of convex bodies, and in fact, in many

situations one has to deal with bodies which are not necessarily convex. The following

notion helps us to describe ‘convexity’ for quasi-convex bodies.

Definition 7.2. Let q (0, 1]. A star body K is quasi-convex if there exists a q (0, 1] ∈ ∈ n such that for all x, y R , ∈ x + y q x q + y q , k kK ≤ k kK k kK or, equivalently t1/qx+(1 t)1/qy K whenever x and y are in K and t (0, 1). Especially, − ∈ ∈ a quasi-convex body with above constant q (0, 1] is said to be q-convex. ∈ 112

One can see that q-convex bodies with q = 1 are just convex. Note also that a q1-convex body is q2-convex for all 0 < q2 q1. ≤ Example 7.1. Consider the 2-dimensional star body

2 q q K = (x, y) R : x + y 1 , 0 < q 1. ∈ | | | | ≤ ≤ 

q = 1/8 q = 1/4 q = 1/2 q = 3/4 q = 1 (convex)

Note that the q-convex body K becomes to be convex as q tends to 1.

There is an extensive literature on p-convex bodies as well as the closely related concept of quasi-convex bodies, see for example [1], [7], [6], [14], [24], [25], [30], [37], [39], [38], [57],

[56], [53], [54], [55], [67] and others.

Modulus of convexity. The next notion is the modulus of convexity for symmetric convex bodies.

n Definition 7.3. Let K be a symmetric convex body in R . The modulus of convexity of

K is defined by

x + y δK (ε) = inf 1 : x K, y K, x y ε , ε (0, 2]. − 2 ∈ ∈ k − kK ≥ ∈  K 

It is non-decreasing on (0, 2] and also invariant for any invertible linear transformation.

x

K x+y 2 K δK

O 1 y 113

n For example, the modulus of convexity of the Euclidean ball B2 can be given by the paral- lelogram identity as follows:

2 ε 1 2 2 n δB2 (ε) = 1 1 = ε + o(ε ). − r − 4 8

Nordlander [71] proved that the ellipsoids have the largest modulus of convexity in the class of symmetric convex bodies, i.e., for any symmetric K,

δK (ε) δBn (ε). ≤ 2

In view of this result, we may say that the convexity of a body increases when the modulus

of convexity becomes bigger.

Banach-Mazur distance from the ball. As another notion to measure convexity, the

result of Hensley (Theorem 7.2) and the above Nordlander’s result suggest to consider the

Banach-Mazur distance from the Euclidean ball. Moreover, we note that the intersection

bodies of linearly equivalent star bodies are linearly equivalent (see Theorem 8.1.6 in [22]),

in the sense that for any T GL(n) ∈

I(TK) = det T (T ∗)−1IK. | |

This gives that, for any T GL(n), ∈

dBM (I(TK),I(TL)) = dBM (IK, IL).

Thus the Banach-Mazur distance is fitted very well to measure convexity of intersection

bodies. Chapter 8. Quasi-convexity for intersection bodies

In this chapter we provide a quantitative version of Busemann’s Theorem 7.1 for quasi- convex bodies, and we discuss the Banach-Mazur distance from the Euclidean ball in the class of quasi-convex bodies. Moreover we extend these results to some general measure spaces with log-concave and s-concave measures.

In Section 8.1 we answer to the following question: Is the intersection body IK of a

symmetric p-convex body necessarily q-convex for some q? We prove that such a constant

q satisfies 1 −1 q 1 (n 1) + 1 , ≤ p − −  
and moreover this upper bound of q is asymptotically sharp. In addition, it is shown that there are quasi-convex bodies for which the intersection body is farther from the Euclidean ball. It means that Hensley’s Theorem 7.2 does not hold in the class of quasi-convex bodies.

In recent times a lot of attention has been attracted to the study of log-concave measures.

These are measures whose densities are log-concave functions. The interest to such measures stems from the Brunn-Minkowski inequality, and many results known for convex bodies are now generalized to log-concave measures, see for example [3], [4], [2], [45], [17], [72] and the references cited therein.

In Section 8.2 we study intersection bodies in spaces with log-concave measures. We define the intersection body IµK of a star body K with respect to µ, and extend the theorems of Section to a log-concave measure space.

8.1 Quasi-convexity and related results

Here we prove a version of Busemann’s theorem for quasi-convex bodies.

114 115

n Theorem 8.1. Let K be a symmetric p-convex body in R , p (0, 1], and E a (k 1)- ∈ − n dimensional subspace of R for 1 k n. Then the map ≤ ≤

u u | | , u E⊥ 7−→ K span(u, E) ∈ ∩ k

defines the Minkowski functional of a q-convex body in E⊥ with q = [(1/p 1) k + 1]−1. −

Proof. We follow the general idea of the proof from [66] (see also [22, p.311]). Let u1, u2 ∈ ⊥ E 0 be nonparallel vectors. Denote u = u1 + u2, and \{ }

∞ K span u1,E ρ(u1) = | ∩ { }| = K (ru1 + E) dr, u1 −∞ | ∩ | | | Z ∞ K span u2,E ρ(u2) = | ∩ { }| = K (ru2 + E) dr. u2 | ∩ | | | Z−∞

Define the functions r1 = r1(t) and r2 = r2(t) by

1 r1 t = K (ru1 + E) dr ρ(u ) | ∩ | 1 Z0 1 r2 = K (ru2 + E) dr, t [0, 1/2]. ρ(u ) | ∩ | ∈ 2 Z0

− 1 −p −p −p −p p r1 r2 Let r = r1 + r2 , λ1 = −p −p and λ2 = −p −p . Then r1 +r2 r1 +r2  

− 1 −1 dr 1 p dr1 dr2 = r−p + r−p p r−p−1 p r−p−1 dt −p 1 2 − 1 dt − 2 dt     1 dr1 1 dr2 = r λ1 + λ2 r dt r dt   1   2  λ1 λ2 λ1 λ2 1 dr1 1 dr2 1/p dr1 1/p dr2 r = λ λ ≥ r dt r dt 1 dt 2 dt  1   2      1 λ1 λ2 λ1 λ2 p ρ(u1) ρ(u2) = λ1 λ2 . K (r1u1 + E) K (r2u2 + E)   | ∩ | | ∩ | 116

1 1 1 −p −p − p p p On the other hand, since r u = (r1 + r2 ) (u1 + u2) = λ1 r1u1 + λ2 r2u2, we have

1 1 p p K (r u + E) λ (K (r1 u1 + E)) + λ (K (r2 u2 + E)) ∩ ⊃ 1 ∩ 2 ∩ 1 1 p −1 p −1 = λ1 λ K (r1 u1 + E) + λ2 λ K (r2 u2 + E) . 1 ∩ 2 ∩    

Thus, by the Brunn-Minkowski inequality (see, for example, [20]), we get

1 λ1 1 λ2 p −1 p −1 K (ru + E) λ K (r1u1 + E) λ K (r2u2 + E) | ∩ | ≥ 1 ∩ 2 ∩

( 1 −1)(k−1) λ1 λ2 p λ1 λ2 = λ λ K (r1u1 + E) K (r2u2 + E) . 1 2 | ∩ | | ∩ |  

Finally, we have the following:

∞ 1/2 dr ρ(u1 + u2) = K (ru + E) dr = 2 K (ru + E) dt | ∩ | | ∩ | dt Z−∞ Z0 1/2 ( 1 −1)(k−1)+ 1 λ1 λ2 p p λ1 λ2 2 λ1 λ2 ρ(u1) ρ(u2) dt ≥ 0 Z   1/2 λ λ 1/q q 1 q 2 2 λ1 [ρ(u1)] λ2 [ρ(u2)] dt ≥ Z0   1/2    −1/q λ1 λ2 2 + dt ≥ λ [ρ(u )]q λ [ρ(u )]q Z0  1 1 2 2  −q −q −1/q = [ρ(u1)] + [ρ(u2)] .  

Therefore ρ(u) defines a q-convex body.

As an immediate corollary of the previous theorem we obtain the following.

n Theorem 8.2. Let K be a symmetric p-convex body in R for p (0, 1]. Then the inter- ∈ section body IK of K is q-convex for q = [(1/p 1) (n 1) + 1]−1. − −

Proof. Let L = IK be the intersection body of K. Let v1, v2 span u1, u2 be orthogonal ∈ { } 117

⊥ to u1 and u2 correspondingly. Denote E = span u1, u2 . Then { }

ρL(v1) = K span u1,E = ρ(u1), | ∩ { }|

ρL(v2) = K span u2,E = ρ(u2). | ∩ { }|

Theorem 8.1 implies, for k = n 1, that L is q-convex for q = [(1/p 1) (n 1) + 1]−1. − − −

Remark 8.1. Note that the previous theorem does not hold without the symmetry as- sumption. To see this, use the idea from [22, Thm 8.1.8], where it is shown that IK is not necessarily convex if K is not symmetric.

A natural question is to see whether the value of q in Theorem 8.2 is optimal. Unfortu- nately, we were unable to construct a body that gives exactly this value of q, but our next result shows that the bound is asymptotically correct.

n Theorem 8.3. There exists a p-convex body K R such that IK is q-convex with ⊂

−1 q [(1/p 1)(n 1) + 1 + gn(p)] , ≤ − −

where gn(p) is a function that satisfies

1. gn(p) log (n 1), ≥ − 2 −

2. lim gn(p) = 0. p→1−

n Proof. Consider the following two (n 1)-dimensional cubes in R : −

C1 = x1 1, ..., xn−1 1, xn = 1 {| | ≤ | | ≤ }

C−1 = x1 1, ..., xn−1 1, xn = 1 . {| | ≤ | | ≤ − } 118

n For a fixed 0 < p < 1, let us define a set K R as follows: ⊂

n 1/p 1/p K = z R : z = t x + (1 t) y, x C1, y C−1, 0 t 1 . ∈ − ∈ ∈ ≤ ≤ n o

e1

e1+e2 √2

We claim that K is p-convex. To show this let us consider two arbitrary points z1, z2 K, ∈

1/p 1/p 1/p 1/p z1 = t x1 + (1 t1) y1, z2 = t x2 + (1 t2) y2, 1 − 2 −

where x1, x2 C1, y1, y2 C−1, and t1, t2 [0, 1]. We need to show that for all s (0, 1) ∈ ∈ ∈ ∈ 1/p 1/p the point w = s z1 + (1 s) z2 belongs to K. Assume first that t1 and t2 are neither −

both equal to zero nor both equal to one. Since C1 and C−1 are convex sets, it follows that

the points

1/p 1/p 1/p 1/p 1/p 1/p 1/p 1/p s t1 x1 + (1 s) t2 x2 s (1 t1) y1 + (1 s) (1 t2) y2 x¯ = 1/p − 1/p andy ¯ = 1/p − 1/p − 1/p − 1/p s1/pt + (1 s)1/pt s (1 t1) + (1 s) (1 t2) 1 − 2 − − − belong to C1 and C−1 correspondingly. Then w = αx¯ + βy¯, where

1/p 1/p 1/p 1/p 1/p 1/p 1/p 1/p α = s t + (1 s) t and β = s (1 t1) + (1 s) (1 t2) . 1 − 2 − − −

p p Note that α + β st1 + (1 s)t2 + s(1 t1) + (1 s)(1 t2) = 1. Therefore, there exists ≤ − − − − 119

µ 0 such that (α + µ)p + (β + µ)p = 1 and ≥

βy¯ + µ( x¯) w = (α + µ)¯x + (βy¯ + µ( x¯)) = (α + µ)¯x + (β + µ) − . − β + µ

Sincey ¯ C−1 and x¯ C−1, it follows that ∈ − ∈

βy¯ + µ( x¯) y = − C−1. β + µ ∈ e Therefore w is a p-convex combination ofx ¯ C1 and y C−1. ∈ ∈ If t and t are either both zero or one, then either α = 0 or β = 0. Without loss of 1 2 e

generality let us say α = 0, then w = βy.¯ Now choosex ¯ C1 arbitrarily, and apply the ∈ considerations above to the point w = 0¯x + βy.¯ The claim follows.

Note that K can be written as

1 1 K = r p x + (1 r) p y : x C1, y C−1, 0 r 1 − ∈ ∈ ≤ ≤ n 1 1 1 1 o n−1 = r p v + (1 r) p w + r p (1 r) p en : v, w B , 0 r 1 − − − ∈ ∞ ≤ ≤ n 1 1 h 1 1i n−1 o = r p + (1 r) p z + r p (1 r) p en : z B , 0 r 1 − − − ∈ ∞ ≤ ≤ nh i n−h1 i o = f(t)z + ten : z B , 1 t 1 , (8.1) ∈ ∞ − ≤ ≤ n o

n−1 n−1 n−1 where B∞ = [ 1, 1] R and f is a function on [ 1, 1] defined as the solution − ⊂ − s = f(t) of s + t p s t p + − = 1, s t , 1 t 1. 2 2 ≥ | | − ≤ ≤     Let L = IK be the intersection body of K. Then

n−1 ⊥ n−1 4 ρL(en) = K e = (2f(0)) = . | ∩ n | 21/p   120

In order to compute the volume of the central section of K orthogonal to (e1 + en)/√2,

⊥ use (8.1) to notice that its projection onto x1 = 0 coincides with K e . Therefore ∩ 1

1 n−2 ρL((e1 + en)/√2) = √2ρL(e1) = 2√2 [2f(t)] dt. Z0

Let L = IK be q-convex. In order to estimate q, we will use the inequality

q q q q en + e1 en e1 en + e1 √2en + − = 2 , L ≤ √ √ √ 2 L 2 L 2 L

that is √2 21/q . ρL(en) ≤ ρL((e1 + en)/√2)

Thus, we have n−1 21/p 1 21/q 2 f(t)n−2dt. ≥ 2 ! Z0 We now estimate the latter integral. ¿From the definition of f it follows that f(t) 2 ≥ 21/p and f(t) t for t [0, 1]. Taking the maximum of these two functions, we get ≥ ∈

2 2 1/p , for 0 t 1/p , f(t) 2 ≤ ≤ 2 ≥   t, for 2 t 1.  21/p ≤ ≤  Therefore,

1 2 n−2 1 n−2 21/p 2 n−2 f(t) dt 1/p dt + t dt 0 ≥ 0 2 2 Z Z   Z 21/p 2 n−1 1 2 n−1 = + 1 . (8.2) 21/p n 1 − 21/p   − "   #

Hence, 1 n−1 21/q 2 1 + 21/p−1 1 , ≥ n 1 −  −    121

which implies

n−1 −1 1 (n 2) 21−1/p + 1 q 1 (n 1) + 1 + log − . ≤ p − − 2 n 1 "  −
#

Denoting n−1 (n 2) 21−1/p + 1 gn(p) = log − , 2 n 1 −
we get the statement of the theorem.

We will use the above example to show that in general the intersection body operator

n does not improve the Banach-Mazur distance to the Euclidean ball B2 .

Theorem 8.4. Let p (0, 1) and let c be any constant satisfying 1 < c < 21/p−1. Then for ∈ n all large enough n, there exists a p-convex body K R such that ⊂

n n n c dBM (K,B2 ) < dBM (IK, B2 ).

Proof. We will consider K from the previous theorem. One can see that K Bn √nBn. ⊂ ∞ ⊂ 2 n Also note that for any a B , there exist x C1, y C−1 and λ [0, 1] such that ∈ ∞ ∈ ∈ ∈ p p−1 a = λx + (1 λ)y. Then we have a λp + (1 λ)p 21−p. Therefore, K 2 p Bn − k kK ≤ − ≤ ⊃ ∞ ⊃ p−1 p n 2 B2 , and thus 1−p n dBM (K,B ) 2 p √n. (8.3) 2 ≤

n Next we would like to provide a lower bound for dBM (IK, B2 ). Let E be an ellipsoid such

that E IK dE, for some d. Then ⊂ ⊂

1 IK conv(IK) dE and conv(IK) E IK. ⊂ ⊂ d ⊂ ⊂ 122

Therefore, 1/d 1/r, where r = min t : conv(IK) tIK . Thus, ≤ { ⊂ }

n ρconv(IK)(θ) n−1 ρconv(IK)(en) dBM (IK, B ) r = max , θ S . 2 ≥ ρ (θ) ∈ ≥ ρ (e )  IK  IK n

The convexity of conv(IK) gives

1 en + e1 en + e1 en e1 en e1 ρconv(IK)(en) ρIK + ρIK − − ≥ 2 √2 √2 √2 √2       2

1 en + e1 = ρIK . √2 √2   Combining the above inequalities with inequality (8.2) from the previous theorem, we get

en+e1 n−1 ρIK ( √ ) 1/p n 2 2 1 dBM (IK, B2 ) . ≥ √2ρ (e ) ≥ 2 n 1 IK n ! −

Comparing this with (8.3) we get the statement of the theorem.

8.2 Generalization to log-concave measures

n n A measure µ on R is called log-concave if for any measurable A, B R and 0 < λ < 1, ⊂

µ(λA + (1 λ)B) µ(A)λµ(B)(1−λ) − ≥ whenever λA + (1 λ)B is measurable. − n Borell [9] has shown that a measure µ on R whose support is not contained in any affine hyperplane is a log-concave measure if and only if it is absolutely continuous with respect to the Lebesgue measure, and its density is a log-concave function.

n To extend Busemann’s theorem to log-concave measures on R , we need the following

theorem of Ball [3], [4].

n Theorem 8.5. Let f : R [0, ) be an even log-concave function satisfying 0 < n f < → ∞ R R 123

and let k 1. Then the map ∞ ≥

1 ∞ − k x f(rx)rk−1dr 7−→ Z0 

n defines a norm on R .

An immediate consequence of Ball’s theorem is a generalization of the classical Buse-

n mann theorem to log-concave measures on R .

n Let µ be a measure on R , absolutely continuous with respect to the Lebesgue measure m, and f its density function. If f is locally integrable on k-dimensional affine subspaces

n of R , then we denote by µk = fmk the restriction of µ to k-dimensional subspaces, where mk is the k-dimensional Lebesgue measure.

Definition 8.1. Define the intersection body IµK of a star body K with respect to µ by

⊥ n−1 ρI K (u) = µn−1(K u ), u S . µ ∩ ∈

n Let µ be a symmetric log-concave measure on R and K a symmetric convex body in

n R . Let f be the density of the measure µ. If we apply Theorem 8.5 to the log-concave

function 1K f, we get a symmetric convex body L whose Minkowski functional is given by

1 ∞ − n−1 n−2 x L = (n 1) (1K f)(rx)r dr . k k −  Z0 

Then for every u Sn−1, ∈

∞ ⊥ n−2 µn−1(K u ) = (1K f)(rθ)r drdθ ∩ n−1 ⊥ ZS ∩u Z0 1 −n+1 ⊥ = θ L dθ = L u . n 1 n−1 ⊥ k k | ∩ | − ZS ∩u

Using Theorem 7.1 for the convex body L, one immediately obtains the following version 124

of Busemann’s theorem for log-concave measures.

n Theorem 8.6. Let µ be a symmetric log-concave measure on R and K a symmetric convex

n body in R . Then the intersection body IµK is convex.

In order to generalize Theorem 8.2 to log-concave measures, we will first prove a version of Ball’s theorem (Thm 8.5) for p-convex bodies.

n Theorem 8.7. Let f : R [0, ) be an even log-concave function, k 1, and K a → ∞ ≥ n p-convex body in R for 0 < p 1. Then the body L defined by the Minkowski functional ≤

− 1 kxk−1 k K k−1 n x L = f(rx)r dr , x R , k k ∈ "Z0 #

is p-convex.

n Proof. Fix two non-parallel vectors x1, x2 R and denote x3 = x1 + x2. We claim ∈ p p p that x3 x1 + x2 . Consider the following 2-dimensional bodies in the plane k kL ≤ k kL k kL

E = span x1, x2 , { }

¯ t1x1 t2x2 p p K = + : t1, t2 0, t1 + t2 1 x1 K x2 K ≥ ≤ k k k k 

and 1 −1 − k kxk ¯ ¯ n K k−1 L = x R : x L¯ = f(rx)r dr 1 .  ∈ k k " 0 # ≤   Z  One can see that the boundary of K¯ consists of a p-arc connecting the points x1 kx1kK and x2 , and two straight line segments connecting the origin with these two points. kx2kK

¯ ¯ x1 Clearly K is p-convex and K K. Also note that xi K¯ = xi K for i = 1, 2, since ⊂ k k k k kx1kK x2 ¯ ¯ and are on the boundary of K, and x3 K¯ x3 K since K K. It follows that kx2kK k k ≥ k k ⊂

xi ¯ = xi L (i = 1, 2), and x3 ¯ x3 L. k kL k k k kL ≥ k k 125

kx1kL¯ kx2kL¯ y Consider the point y = x1 + x2 in the plane E. The point lies on the kx1kK¯ kx2kK¯ kykK¯ p-arc connecting x1 and x2 . Consider the tangent line to this arc at the point y . kx1kK¯ kx2kK¯ kykK¯

tixi This line intersects the segments [0, xi/ xi K¯ ], i = 1, 2, at some points with ti (0, 1). k k kxikK¯ ∈ Since t1x1 , t2x2 and y are on the same line, it follows that the coefficients of t1x1 kx1kK¯ kx2kK¯ kykK¯ kx1kK¯ and t2x2 in the equality kx2kK¯

y 1 x1 ¯ t1x1 x2 ¯ t2x2 = k kL + k kL y ¯ y ¯ t1 · x1 ¯ t2 · x2 ¯ k kK k kK  k kK k kK 

have to add up to 1. Therefore,

x1 L¯ x2 L¯ y K¯ = k k + k k . k k t1 t2

Note also that the line between t1x1 and t2x2 separates x3 from the origin, which kx1kK¯ kx2kK¯ kx3kK¯ means that the three points t1x1 , t2x2 and x3 are in the “convex position”. Applying kx1kK¯ kx2kK¯ kx3kK¯ Ball’s theorem on log-concave functions (Thm 8.5) to these three points, we have

1 1 1 t − t − 1 − k 1 k 2 k kx3kK¯ k−1 kx1kK¯ k−1 kx2kK¯ k−1 f(rx3)r dr f(rx1)r dr + f(rx2)r dr . ≤ "Z0 # "Z0 # "Z0 #

1 ti k kxikK¯ k−1 If we let si = xi L¯ 0 f(rxi)r dr for each i = 1, 2, the above inequality becomes k k " # R

x1 L¯ x2 L¯ x3 L¯ k k + k k . k k ≤ s1 s2

By a change of variables, we get

1 1 1 k 1 k kxikK¯ k−1 kxikK¯ k−1 si = ti xi ¯ f(tirxi)r dr ti xi ¯ f(rxi)r dr = ti k kL ≥ k kL "Z0 # "Z0 # 126

for each i = 1, 2. The above inequality comes from the fact that an even log-concave function has to be non-increasing on [0, ). Indeed, ∞

1 + ti 1 ti 1+ti 1−ti f(tirxi) = f rxi − rxi f(rxi) 2 f( rxi) 2 = f(rxi). 2 · − 2 · ≥ −  

Putting all together, we have

x1 L¯ x2 L¯ x1 L¯ x2 L¯ x3 L x3 L¯ k k + k k k k + k k = y K¯ . k k ≤ k k ≤ s1 s2 ≤ t1 t2 k k

Using the p-convexity of K¯ , we have

p p p x1 L¯ x2 L¯ p p p p y K¯ k k x1 + k k x2 = x1 L¯ + x2 L¯ = x1 L + x2 L, k k ≤ x ¯ ¯ x ¯ ¯ k k k k k k k k 1 K K 2 K K k k k k

p p p and therefore x3 x1 + x2 . k kL ≤ k kL k kL

Corollary 8.1. Let µ be a symmetric log-concave measure and K a symmetric p-convex

n body in R for p (0, 1]. Then the intersection body IµK of K is q-convex with q = ∈ [(1/p 1)(n 1) + 1]−1. − −

Proof. Let f be the density function of µ. By Theorem 8.7, the body L with the Minkowski

functional −1 kxk−1 n−1 K n−2 n x L = (n 1) f(rx)r dr , x R , k k − ∈ " Z0 # is p-convex.

On the other hand, the intersection body IµK of K is given by the radial function

kuk−1 ⊥ K n−2 ρIµK (u) = µn−1(K u ) = f(rv)r drdv ∩ n−1 ⊥ ZS ∩u Z0 1 −n+1 ⊥ = v L dv = L u n−1 = ρIL(u), n 1 n−1 ⊥ k k | ∩ | − ZS ∩u 127

−1 which means IµK = IL. By Theorem 8.1, IL is q-convex with q = [(1/p 1)(n 1) + 1] , − − and therefore so is IµK.

We conclude this section with an example that shows that the condition on f to be even in Theorem 8.7 cannot be dropped.

n Example 1. Let µ be a log-concave measure on R with density

1−1/p 1, if x1 + x2 2 , ≥ f(x1, . . . , xn) =   0, otherwise.

 Consider the p-convex body K = Bn for p (0, 1). If L is the body defined in Theorem 8.7, p ∈

then e1 + e2 L = 0 and e1 L = e2 L > 0, which means L is not q-convex for any q > 0. k k k k k k

8.3 Non-symmetric cases and s-concave measures

Note that Ball’s theorem (Thm 8.5) remains valid even if f is not even, as was shown by

Klartag [44]. On the other hand, as we explained above, Theorem 8.7 does not hold for non-

symmetric log-concave measures. However, if we restrict ourselves to the class of s-concave

measures, s > 0, then it is possible to give a version of Theorem 8.7 for non-symmetric

measures.

n Borell [9] introduced the classes Ms(Ω), ( s ,Ω R open convex) of s- −∞ ≤ ≤ ∞ ⊂ concave measures, which are Radon measures µ on Ω satisfying the following condition: the

inequality

s s 1 µ(λA + (1 λ)B) [λµ(A) + (1 λ)µ(B) ] s − ≥ − holds for all nonempty compact A, B Ω and all λ (0, 1). In particular, s = 0 gives the ⊂ ∈ class of log-concave measures.

Let us consider the case 0 < s < 1/n. According to Borell, µ is s-concave if and only if

s the support of µ is n-dimensional and dµ = fdm for some f L1 (Ω) such that f 1−ns is a ∈ loc 128

concave function on Ω.

n Theorem 8.8. Let µ be an s-concave measure on Ω R with density f, for 0 < s < 1/n, ⊂ and K a p-convex body in Ω, for p (0, 1]. If k 1, then the body L whose Minkowski ∈ ≥ functional is given by

1 ∞ − k k−1 n x L = 1K (rx)f(rx)r dr , x R k k ∈ Z0 

−1 is q-convex with q = 1 1 1 n 1 + 1 . p − s − k p h

i n Proof. Let x1, x2 R and x3 = x1 + x2. Then, for i = 1, 2, ∈

∞ ∞ −k k−1 1 − 1 − 1 − k −1 xi = 1K (rxi)f(rxi)r dr = 1K (s p xi)f(s p xi)s p ds k kL p Z0 Z0 1 ∞ 1 ∞ = Fi(s)ds = s (0, ): Fi(s) > t dt, p 0 p 0 { ∈ ∞ } Z Z

− 1 − 1 − k −1 where Fi(s) = 1K (s p xi)f(s p xi)s p for each i = 1, 2, 3. We claim that

k +1 λ1 λ2 2 q F3(s3) F1(s1) F2(s2) ≥

whenever s = s + s and λ = si for i = 1, 2. Indeed, since 3 1 2 i s1+s2

1 1 1 p 1 p 1 − p s1 − p s2 − p s x3 = s x1 + s x2 3 s + s 1 s + s 2  1 2   1 2  1 1 1 1 p −1 − p p −1 − p = λ1(λ1 s1 x1) + λ2(λ2 s2 x2), 129

γ s the concavity of f , γ = 1−ns , gives

− 1 1 −1 − 1 1 −1 − 1 γ p γ p p γ p p f (s x3) λ1f (λ s x1) + λ2f (λ s x2) 3 ≥ 1 1 2 2 λ λ 1 −1 − 1 1 1 −1 − 1 2 γ p p γ p p f (λ s x1) f (λ s x2) ≥ 1 1 2 2     λ λ 1 −1 − 1 1 1 −1 − 1 2 p γ p p γ p λ f (s x1) λ f (s x2) ≥ 1 1 2 2     γ 1 1 λ1 1 1 λ2 γ ( p −1) 1 γ ( p −1) 1 s1 − p s2 − p = f(s x1) f(s x2) ,  s 1 s 2  " 3  # " 3  #   that is, 1 1 1 2 1 1 1 λi γ ( p −1) − p γ ( p −1) − p s f(s x3) s f(s xi) . 3 3 ≥ i i i=1 Y   On the other hand, note that

λ1 λ2 2 λ1 λ2 1 1 = + s s s ≥ s s 3 1 2  1   2  and 1 1 1 − p − p − p 1K (s x3) 1K (s x1)1K (s x2), 3 ≥ 1 2

1 1 1 1 1 − p p − p p − p since s3 x3 = λ1 (s1 x1) + λ2 (s2 x2). Thus

2 λ − 1 − 1 − k −1 i λ1 λ2 p p p F1(s1) F2(s2) = 1K (si xi)f(si xi)si i=1 Y   2 1 1 k λi 1 1 1 1 γ ( p −1)+ p +1 − p γ ( p −1) − p 1 1K (s x3) s f(s xi) ≤ 3 i i · s i=1 " i # Y   1 1 1 1 k k − p − p ( −1)+ +1 − p −1 1K (s x3)f(s x3)2 γ p p s ≤ 3 3 3 k +1 2 q F3(s3). ≤ 130

It follows that for every t > 0

k +1 s3 : 2 q F3(s3) > t s1 : F1(s1) > t + s2 : F2(s2) > t . { } ⊃ { } { }

Applying the Brunn-Minkowski inequality, we have

∞ −k −k 1 x1 + x2 L = x3 L = F3(s)ds k k k k p 0 ∞Z 1 1 k +1 = s3 (0, ) : 2 q F3(s3) > t dt k +1 q · p 0 { ∈ ∞ } 2 Z 1 1 ∞ s1 : F1(s1) > t + s2 : F2(s 2) > t dt k +1 ≥ 2 q · p 0 { } { } Z   1 −k −k = ( x1 + x2 ). k +1 L L 2 q k k k k

Thus,

q 1 1 − q − k k k −k −k k −q q −q q 1 x1 L + x2 L 1 ( x1 L ) + ( x1 L ) x1 + x2 L 2 q k k k k = k k k k k k ≤ 2 ! 2 2 ! 

1 1 −q −q − q  q q −1 − q  1 x1 + x2 1 x1 + x2 k kL k kL k kL k kL ≤ 2 2 ≤ 2 2 " !# "   # 1 q q q = x1 + x2 , k kL k kL
which means that L is q-convex.

The following example shows that the value of q in the above theorem is sharp.

n Example 2. Let µ be an s-concave measure on Ω = (x1, . . . , xn) R : x1 0 for s > 0 { ∈ ≥ } with density

1/s−n f(x1, . . . , xn) = x1 | | 131

and let

p p x1 + x2 x1 x2 K = (x1, . . . , xn): x1 0, + − 1, xi 1 i = 3, . . . , n . ≥ 2 2 ≤ | | ≤ ∀  

1−1/p Note that e1 K = 2 and e1 + e2 K = e1 e2 K = 1. If L is the body defined by K k k k k k − k in the above theorem, then

1 1   − 1−1/p − 1 1 k 2 k 1− p ( s −n+k) 1/s−n k−1 2 e1 L = r r dr = 1 k k 0  n + k  "Z # s −   and − 1 1 1 k k 1/s−n k−1 1 e1 + e2 L = r r dr = n + k . k k s − Z0    q q 1/q If L is q-convex for some q, then the inequality 2e1 L e1 + e2 + e1 e2 k k ≤ k kL k − kL implies
  − 1 1− 1 ( 1 −n+k) k 1 2 p s 1 1 k 2 2 q n + k  1 n + k  ≤ s − s −     that is, 1 1 1 1 −1 q 1 n + . ≤ p − s − k p     Note that in our construction Ω is not open, as opposed to what we said in the beginning of Section 4. This is done for the sake of simplicity of the presentation. To be more

n precise one would need to define Ω = (x1, . . . , xn) R : x1 >  and f(x1, . . . , xn) = { ∈ − } 1/s−n + x1 +  , for  > 0, and then send  0 . | | → Chapter 9. Local convexity for bodies of revolution

In this chapter we concentrate on the symmetric bodies of revolution to provide several results on the (strict) improvement of convexity under the intersection body operation.

It is shown that the intersection body of a symmetric convex body of revolution has the same asymptotic behavior near the equator as the ball. We apply this result to show that the double intersection body of a symmetric convex body of revolution in sufficiently high dimension is very close to an ellipsoid in the Banach-Mazur distance. We also prove results on the local convexity at the equator of intersection bodies in the class of star bodies of revolution.

Following [22, Theorem C.2.9], the radial function ρIK (θ) of IK at the angle θ from the axis of revolution is given by, for θ (0, π/2], ∈

n−4 π/2 2 cn cos ϕ 2 ρ (θ) = ρ (ϕ)n−1 1 sin ϕ dϕ (9.1) IK sin θ K − sin2 θ Zπ/2−θ   and, if θ = 0,

n−1 n−1 ρIK (0) = ρK (π/2) B2 = cndn π/(2n) ρK (π/2) , (9.2) p n−2 2πn/2−1 n−1 √ n/2 Γ(n/2) where cn = and dn = 1 for large n. Since a dilation of the n−1 · Γ(n/2) n−2 · Γ((n+1)/2) ≈ body does not change its regularity or convexity, in this chapter we replace cn by 1.

In Section 9.1 we introduce several concepts containing the equatorial power type to describe quantitative information about convexity of bodies of revolution.

In Section 9.2 we investigate the equatorial behavior of symmetric intersection bodies of revolution under the convexity assumption. We prove that if K is a symmetric convex body of revolution, then the intersection body of K has uniform equatorial power type 2, which

132 133

means that its boundary near the equator is asymptotically the same as the ball. Using this result, we prove in Section 9.3 that if K is a symmetric convex body of revolution in sufficiently high dimension, then its double intersection body is close, in the Banach-Mazur distance, to the Euclidean ball.

9.1 Equatorial power type for bodies of revolution

We assume that the axis of revolution for any body of revolution is the e1-axis. The equator of a body K of revolution is the boundary of the section of K by the central hyperplane perpendicular to the axis of revolution, i.e., ∂K e⊥. For a star body K of ∩ 1 revolution, we denote by ρK (θ) the radial function of K at a direction whose angle from the e1-axis is θ. The goal of this section is to introduce parameters that could measure convexity at the equator for bodies of revolution.

n Definition 9.1. Let K be a body of revolution in R about the e1-axis, and let 1 p < . ≤ ∞ + Then the function ψK : R R is defined by →

−1 ψK (x) = ρ (θ) sin θ for θ = tan (1/x). (9.3) K | |

In particular, if K is a symmetric convex body, then ψK is a continuous even function which is non-increasing in [0, + ) and ψK (x) = O(1/x) as x tends to infinity. ∞ A body K of revolution is said to have equatorial power type p if there exist constants

p c1, c2 > 0, depending on K, such that c1 < ψK (x) ψK (0) /x < c2 for every x R. | − | ∈

The classical modulus of convexity of a symmetric convex body K is given in Defi- nition 7.3. However, since we focus on the convexity around the equator for bodies of revolution, it would be better to consider the following notion.

n Definition 9.2. Let K be a convex body of revolution in R about the e1-axis. The 134

1

ψK(x)

ψK K φ x

x = cot φ

Figure 9.1: The function ψK modulus of convexity of K at the equator is defined by

e x + y δ (ε) = inf 1 : x, y K, x y 2ε, x y span e1 , K − 2 ∈ k − kK ≥ − ∈ { }  K 

where denotes the Minkowski functional of K. In particular, if K is symmetric, then k·kK e δK can be expressed by

e ρK (π/2) ρK (θ) sin θ ψK (0) ψK (cot θ) δK (ε) = − = − , (9.4) ρK (π/2) ψK (0)

ρK (θ) where the angle θ is obtained from ε = ρ (0) cos θ. K ·

It follows from (9.4) that a convex body K of revolution has equatorial power type p if

e p and only if there exist constants c1, c2 > 0 such that c1 < δ (ε)/ε < c2 for all ε (0, 1]. K ∈

Moreover, differently from the function ψK for a (star) body K of revolution, we notice

that the modulus of convexity for a convex body of revolution is invariant for any dilations

on the axis of revolution or its orthogonal complement.

n For example, if K is the body of revolution in R obtained by rotating a 2-dimensional

`p-ball with respect to the axis e1, then it has equatorial power type p; more precisely

e p p δK (ε) = ε /p + o(ε ). 135

Definition 9.3. For 1 p < , a collection of convex bodies of revolution is said to ≤ ∞ C have uniform equatorial power type p if every convex body in has equatorial power type C

p, and moreover there exist uniform constants c1, c2 > 0 such that

e δK (ε) c1 < < c2 for every ε (0, 1] and K . εp ∈ ∈ C

e First, let us show some relation between δK and ψK when K is a symmetric convex body. Fix ε (0, 1] and choose the angle θ (0, π/2) so that ε = (ρ (θ)/ρ (0)) cos θ. ∈ ∈ K K · Then,

(1 ε)ρ (π/2) ρ (θ) sin θ ρ (π/2), (9.5) − K ≤ K ≤ K which can be obtained by applying convexity property of K to three points on the boundary of K with angles 0, θ, and π/2. Notice that (9.5) gives

ρ (0) ε ρ (0) cot θ = K · = K ε + O(ε2). ρ (θ) sin θ ρ (π/2) K  K 

Next it follows from (9.4) that

e ψK (0) ψK (δ) ρK (0) 2 δK (ε) = − , for δ = ε + O(ε ). (9.6) ψK (0) ρK (π/2) ·

e In particular, we have δ (ε) 1 ψK (ε) under the assumption ρ (0) = ρ (π/2) = 1. K ≈ − K K

Secondly, the formulas (9.1) and (9.2) provide a very nice relation between ψK and ψIK . 136

Indeed, (9.1) implies that

n−4 n−4 θ 2 2 n−1 1 2 sin φ dφ ρIK (θ) sin θ = [ρK (π/2 φ) cos φ] 2 1 2 2 0 − cos φ − sin θ cos φ Z   n−4  θ tan2 φ 2 dφ = [ρ (π/2 φ) cos φ]n−1 1 K − − tan2 θ cos2 φ Z0   tan θ n−4 n−1 2 2 2 = ψK (t) 1 t cot θ dt. − Z0  

Thus we have

∞ n−1 ψIK (0) = ρIK (π/2) = ψK (t) dt, (9.7) Z0 1/x n−4 n−1 2 2 2 ψIK (x) = ψK (t) 1 x t dt, x (0, ). (9.8) − ∈ ∞ Z0  

We are now ready to investigate the equatorial behavior of intersection bodies of revolution.

9.2 Equatorial power type 2 for intersection bodies

In this section we prove that the class of all intersection bodies of symmetric convex bodies of revolution has uniform equatorial power type 2. Namely, if K is a symmetric con- vex body of revolution, then IK has equatorial power type 2 and, moreover, the coefficient

e of the quadratic term in the expansion of δIK (ε) is bounded (above and below) by absolute constants.

First, we need a specific formula for the function ψK in the case that K is a symmetric body of revolution obtained by rotating line segments.

n Lemma 9.1. Let La,b R be the symmetric body of revolution whose boundary is deter- ⊂ mined by a line segment (x, y): ax + by = 1, 0 x 1/a for a, b 0. Namely, the body { ≤ ≤ } ≥

La,b can be given by

n n−1 La,b = (x, y) R = R R : a x + b y 1 . ∈ × | | | | ≤  137

Then the function ψLa,b , defined in (9.3), is given by

1 ψL (x) = . (9.9) a,b a x + b | |

n Moreover, if K R is a symmetric convex body of revolution with ρ (0) = ρ (π/2) = 1, ⊂ K K then 1 1 ψK (x) min 1, . (9.10) x + 1 ≤ ≤ x | |  | |

Proof. Let x = cot θ > 0, and write L = La,b to shorten the notation. Then the point

2 (ρL(θ) cos θ, ρL(θ) sin θ) R ∈

2 lies on the straight line (x, y) R : ax + by = 1 . Thus we have ∈  ρ (θ) cos θ (1 bρ (θ) sin θ)/a x = cot θ = L = − L ρL(θ) sin θ ρL(θ) sin θ 1 b ψL(x) = − , a ψL(x)

which gives ψL(x) = 1/(ax + b).

n Now, if K R is a symmetric convex body of revolution with ρ (0) = ρ (π/2) = 1, ⊂ K K then B1 K B∞ where ⊂ ⊂

n n−1 B1 = (x, y) R = R R : x + y 1 (double cone) ∈ × | | | | ≤  n n−1 B∞ = (x, y) R = R R : x 1, y 1 (cylinder) ∈ × | | ≤ | | ≤ 

We also see that ψB ψK ψB . Here ψB and ψB can be obtained from (9.9): 1 ≤ ≤ ∞ 1 ∞

1 ψB (x) = ψL (x) = 1 1,1 x + 1 138

and

ψL0,1 (x) = 1, if 0 x 1 ψ (x) =  ≤ ≤ B∞   ψL (x) = 1/x, if x 1, 1,0 ≥  which imply (9.10). 

The inequalities in (9.10) of Lemma 9.1 give easy upper/lower bounds for the function

ψK . However, we will need better bounds in high dimensions. These are given by the following Lemma.

n Lemma 9.2. Let K R be a convex body of revolution with ρ (π/2) = 1. For every ⊂ K σ > 0 and t > 1, 1 −1 ψK (σt) 1 + t 1 . (9.11) ≤ ψ (σ) −   K 

−1 −1 Proof. Let φ1 = tan (1/σ) and φ2 = tan (1/σt). Choose three points P0,P1,P2 ∈

∂K span e1, e2 whose angles from the e1-axis are π/2, φ1 and φ2, respectively. That is, ∩ { }

P0 = (0, 1)

P1 = (ρK (φ1) cos φ1, ρK (φ1) sin φ1) =: (x1, y1)

P2 = (ρK (φ2) cos φ2, ρK (φ2) sin φ2) =: (x2, y2)

Since 1−y2 x2 by convexity of K, 1−y1 ≥ x1

1 ρ (φ2) sin φ2 ρ (φ2) cos φ2 ρ (φ2) sin φ2 cot φ2 − K K = K · , 1 ρ (φ1) sin φ1 ≥ ρ (φ1) cos φ1 ρ (φ1) sin φ1 cot φ1 − K K K · which implies

1 ψK (σt) ψK (σt)σt − . 1 ψK (σ) ≥ ψK (σ)σ − 139

Simplifying the above inequality, we have the inequality (9.11).

The next Lemma will be helpful to bound the integral in (9.7), and to control its tail.

n Lemma 9.3. Let K R , for n 4, be a convex body of revolution with ρ (π/2) = 1, and ⊂ ≥ K −1 let σK = ψ (1 1/n). Then K −

∞ 1 n−1 c1 ψK (t) dt c2, (9.12) ≤ σ ≤ K Z0 where c1, c2 > 0 are absolute constants. In addition, for every R > 1,

∞ n−1 2−n ψK (σK t) dt = O [1 + R/n] . R Z  

Proof. For any t R, Lemma 9.2 gives ≥

t −1 t −1 ψK (σK t) 1 + t = 1 + . ≤ ψK (σK ) − n 1    − 

Thus

∞ ∞ 1−n 2−n n−1 t n 1 R ψK (σK t) dt 1 + dt = − 1 + . ≤ n 1 n 2 n 1 ZR ZR  −  −  − 

Next we will show an upper bound in (9.12),

∞ σK ∞ n−1 n−1 n−1 ψK (t) dt = ψK (t) dt + ψK (t) dt Z0 Z0 ZσK ∞ n−1 σK + σK ψK (σK t) dt ≤ Z1 n 1 1 2−n = σK + σK − 1 + (1 + 1/e)σK . n 2 n 1 . −  −  140

For a lower bound,

σK σK n−1 n−1 1 σK ψK (t) dt 1 dt . ≥ − n ≈ e Z0 Z0  

Thus, ∞ σK n−1 ψK (t) dt (1 + 1/e)σK . e . . Z0

Next, Lemma 9.4 will allow us to estimate the integral in (9.8).

n Lemma 9.4. Let K R , for n 4, be a symmetric convex body of revolution with ⊂ ≥ −1 1 ρ (π/2) = 1. Fix R > 1 and let σK = ψ (1 1/n). Then, for each x , K K − ≤ RσK

R n−4 ψIK (x) n−1 2 2 2 2 2−n = ψK (σK t) 1 σ x t dt + O [1 + R/n] . σ − K K Z0    

Proof. Apply Lemma 9.3. If x = 0, then 6

1/x n−4 n−1 2 2 2 ψIK (x) = ψK (t) 1 x t dt − Z0 R   n−4 ∞ n−1 2 2 2 2 n−1 σK ψK (σK t) 1 σ x t dt + σK ψK (σK t) dt ≤ − K Z0 ZR R   n−4 n−1 2 2 2 2 2−n = σK ψK (σK t) 1 σ x t dt + O [1 + R/n] . − K Z0     

If x = 0, then

∞ R n−1 n−1 2−n ψIK (0) = ψK (t) dt = σK ψK (σK t) dt + O [1 + R/n] . 0 0 Z Z  

Now we are ready to prove the main result of this section. 141

Theorem 9.1. The class of intersection bodies of symmetric convex bodies of revolution in dimension n 4 has uniform equatorial power type 2. Namely, if K is a symmetric convex ≥ n body of revolution in R for n 4, then its intersection body IK has modulus of convexity ≥ of the form

e 2 3 δIK (ε) = cK ε + O(ε )

where cK is a constant bounded by absolute constants.

Proof. The modulus of convexity at the equator is invariant for any dilations on the axis of revolution or its orthogonal complement, so we may start with ρK (π/2) = ρK (0) = 1. Fix a small number ε > 0 and choose the angle θ such that

ρ (θ) IK cos θ = ε. ρIK (0)

−1 Let δ = cot θ and σK = ψ (1 1/n). By Lemma 9.3, we have K −

∞ n−1 ψIK (0) := ψK (t) dt = dK σK , (9.13) Z0 where c1 dK c2 for absolute constants c1, c2 > 0. By convexity of IK, as in (9.5), ≤ ≤

(1 ε)ρ (π/2) ρ (θ) sin θ ρ (π/2). − IK ≤ IK ≤ IK

Moreover, it follows from (9.2) that ρ (0) π , so IK ≈ 2n p π/2 ε π/2 ε . σK δ . . (9.14) dK · √n dK (1 ε) · √n p p −

First consider the case n 14. The formula (9.14) and Lemma 9.4 give that, for any R ≥ 142

with σK R 1/δ, ≤

R ψIK (0) n−1 2−n = ψK (σK t) dt + O [1 + R/n] σK 0 Z   and

R n−4 ψIK (δ) n−1 2 2 2−n = ψK (σK t) 1 (σK δt) dt + O [1 + R/n] σ − K Z0 R     n−1 n 4 2 4 4 2−n = ψK (σK t) 1 − (σK δt) + O(ε R ) dt + O [1 + R/n] . (9.15) 0 − 2 Z    

−1/4 Since σK δ is comparable to ε/√n by absolute constants (see (9.14)), we can take R = ε

to control error terms of above equation. Then we have ε4R4 = ε3, and for n 14, ≥

n 12 (1 + R/n)2−n (1 + R/n)−12 = ε3 ≤ 1 + nε1/4   −1/4 (1 + R/n)2−n e−R = e−ε as n , → → ∞ so the remainder part of (9.15) is O(ε3) for n 14. Thus, ≥

R R ψIK (δ) n−1 1 2 n−1 2 3 = ψK (σK t) dt (n 4)(σK δ) ψK (σK t) t dt + O(ε ) σK 0 − 2 − 0 Z R Z ψIK (0) π 2 n−1 2 3 = ε ψK (σK t) t dt + O(ε ). (9.16) σ − 4d2 K K Z0

The formula (9.6), together with (9.13) and (9.16), gives the modulus of convexity at the equator as follows.

R e ψIK (0) ψIK (δ) π 2 n−1 2 3 δ (ε) = − = ε ψK (σK t) t dt + O(ε ). IK ψ (0) 4d3 IK K Z0

R n−1 2 Now it is enough to compute 0 ψK (σK t) t dt. For a upper bound, apply Lemma 9.2. R 143

Then, for any t 1, ≥

t −1 t −1 ψK (σK t) 1 + t = 1 + . ≤ ψK (σK ) − n 1    − 

Thus

R 1 ∞ 1−n ∞ n−1 2 2 t 3 2 1−n ψK (σK t) t dx dt + t 1 + dt = 1 + (n 1) (s 1) s ds 0 ≤ 0 1 n 1 − n − Z Z Z  −  Z n−1 1 n−1 5n3 15n2 + 12n 5 = 1 + 1 − 1 + . − n (n 2)(n 3)(n 4) ≈ e   − − −

For a lower bound,

R 1 1 n−1 n−1 2 n−1 2 1 2 1 ψK (σK t) t dt ψK (σK t) t dt 1 t dt , 0 ≥ 0 ≥ 0 − n ≈ 3e Z Z Z h i which completes the proof for n 14. ≥

Now consider the case of 4 n < 14. It follows from (9.7) and (9.8) that ≤

∞ ∞ n−1 n−1 2 2 n−4 ψIK (0) ψIK (δ) = ψK (t) dt ψK (t) (1 δ t ) 2 dt − 0 − 0 − Z ∞ Z 1/δ n−1 n−1 2 2 n−4 = ψK (t) dt + ψK (t) 1 (1 δ t ) 2 dt 1/δ 0 − − Z Z h i = (I) + (II).

For (I), use the inequalities (9.10) from Lemma 9.1. Then

∞ 1 δn−2 (I) dt = = O(δn−2) ≤ tn−1 n 2 Z1/δ − 144

and ∞ 1 (1 + 1/δ)2−n (I) dt = = O(δn−2). ≥ (t + 1)n−1 n 2 Z1/δ − Since δ is comparable to ε by (9.14), we have that (I) is O(εn−2), which is at most O(ε3) if

n 5. ≥ To get an upper bound of (II), use (9.10) again.

1 1/δ 2 2 n−4 n−1 2 2 n−4 (II) 1 (1 δ t ) 2 dt + (1/t) 1 (1 δ t ) 2 dt ≤ 0 − − 1 − − Z h i Z h i = (II-1) + (II-2), where 1 n 4 n 4 (II-1) = 1 (1 − δ2t2) dt + O(δ4) = − δ2 + O(δ4) − − 2 6 Z0   and

1/δ 1/δ 2 2 n−4 (1 δ t ) 2 dt (II-2) = (1/t)n−1dt − − tn−4 t3 Z1 Z1 n−2 1 n−2 2 n−2 1 δ 1 2 n−4 1 δ (1 δ ) 2 = − (s δ ) 2 ds = − − − n 2 − 2 2 − n 2 − Zδ − 1 = δ2 + O(δn−2). 2

A lower bound of (II) is given by

1/δ 2 2 n−4 1 (1 δ t ) 2 (II) − − dt ≥ (t + 1)n−1 Z0 1 2 2 n−4 1/δ n−1 2 2 n−4 1 (1 δ t ) 2 t 1 (1 δ t ) 2 = − − dt + − − dt (t + 1)n−1 t + 1 tn−1 Z0 Z0   1 1/δ 2 2 n−4 1 n−4 1 1 (1 δ t ) 2 2 2 2 n−1 1 (1 δ t ) dt + n−1 − −n−1 dt ≥ 2 0 − − 2 0 t Z h i Z n 4 1 n 4 + 3 2n−1 = − δ2 + (II-2) = − · δ2 + O(δn−2). 3 2n 2n−1 3 2n ·  ·  145

In summary, if n 5, then (I) is at most O(ε3) and (II) is asymptotically cδ2 + O(δ3). ≥ 2 3 e In addition, if n = 4, then (II) disappears and (I) is cδ + O(δ ). Note that δIK (ε) =

ψIK (0)−ψIK (δ) and c1 < ψIK (0) < c2 by Lemma 9.3. Finally, we get ψIK (0)

0 e 2 0 c1 < δIK (ε)/ε < c2,

0 0 where c1, c2 are positive absolute constants.

Remark 9.1. In general, Theorem 9.1 is not true in dimension 3. For example, the inter-

3 section body of the double cone B1 R , ⊂

3 2 2 3 K = B1 = (x, y1, y2) R : x + y + y 1 , (double cone in R ) ∈ | | 1 2 ≤  q  does not have equatorial power type 2.

Proof. It follows from Lemma 9.1 that the function ψK for the double cone K is given by

1 ψK (x) = x+1 . Let ε =(ρIK (θ)/ρIK (0)) cos θ for some angle θ and let δ = cot θ. Then

∞ ∞ 2 −2 ψIK (0) = ψK (t) dt = (t + 1) dt = 1 0 0 Z 1/δ Z 1/δ 2 2 2 −1/2 dt ψIK (δ) = ψK (t) (1 δ t ) dt = . 2 2 2 0 − 0 (t + 1) √1 δ t Z Z −

So,

∞ 1/δ e 1 dt δ (ε) = 1 ψIK (δ) = dt IK 2 2 2 2 − 0 (t + 1) − 0 (t + 1) √1 δ t Z Z − ∞ 1 1/δ 1 1 = dt + 1 dt 2 2 2 2 1/δ (t + 1) 0 (t + 1) − √1 δ t Z Z  −  δ2 1 t2dt = δ δ 2 2 2 − 1 + δ − 0 (t + δ) √1 t (1 + √1 t )   Z − − = δ δf(δ) + O(δ2), − 146

where 1 t2dt f(δ) = . 2 2 2 0 (t + δ) √1 t (1 + √1 t ) Z − − Note that 1 dt f(0) = = 1 2 2 0 √1 t (1 + √1 t ) Z − − and f(0) f(δ) 1 dt lim − = = , 2 2 δ→0 δ 0 t√1 t (1 + √1 t ) ∞ Z − − Since δ is comparable to ε, we conclude that δe (ε) = o(ε), but δe (ε) = O(ε2). IK IK 6

Remark 9.2. The convexity condition of K in Theorem 9.1 is crucial to get the uniform boundedness of the constant cK . For t > 0, consider the star body of revolution Kt, defined as the union Kt = Lt B∞ of two cylinders ∪

n n−1 −1/t Lt = (x, y) R = R R : x e , y 1/t ∈ × | | ≤ | | ≤ n o and

n n−1 B∞ = (x, y) R = R R : x 1, y 1 . ∈ × | | ≤ | | ≤  If t > 0 is small enough, then the intersection body of Kt is almost the same as that of B∞

−1/t around the equator. In other words, ψIKt (0) = ψIB∞ (0)+O(e ) and ψIKt (ε) = ψIB∞ (ε)+

O(e−1/t) for small ε > 0. Nevertheless, note that ρ (0) = 1, but ρ (0) = 1/tn−1, i.e., IB∞ IKδ they have quite different radial functions on the axis as t approaches to zero. So,

n−1 e ψIKt (0) ψIKt (δ) ψIB∞ (0) ψIB∞ (δ/t ) −1/t δIKt (ε) = − = − + O(e ) ψIKt (0) ψIB∞ (0)

e n−1 −1/t = δIB∞ (ε/t ) + O(e ),

ρ (0) ε ρ (0) ε IKt IB∞ 1 −1/t e e 2n−2 where δ = ρ (π/2) = ρ (π/2) tn−1 + O(e ). Thus, δIKt (ε)/δIB∞ (ε) (1/t) , which IKt IB∞ · ≈ 147

tends to infinity as t tends to zero. Therefore, this example shows that the constant cK in

Theorem 9.1 may be unbounded in case of star bodies.

n Theorem 9.2. Let K be a symmetric star body of revolution in R , for n 4, such that ≥ the origin is an interior point of K. Then IK has equatorial power type 2.

Proof. Since the origin is an interior point, we may assume that ρ (0) = 1 and rB∞ K ⊂

K RB∞ for some constants r, R > 0 depending on K where B∞ is the cylinder defined in ⊂ the proof of Lemma 9.1. For θ [0, π/2], consider the symmetric convex body Kθ defined ∈ by

ρ (ϕ), 0 ϕ θ  K ≤ ≤ ρKθ (ϕ) =  ρ (ϕ), θ ϕ π/2 Lθ ≤ ≤   where Lθ is a body of revolution obtained by rotating the line containing two points of angles 0, θ on the boundary of K, i.e.,

n n−1 1 ρK (θ) cos θ Lθ = (x, y) R = R R : x + b y = 1 for b = b(θ) = − . ∈ × | | | | ρK (θ) sin θ 

Then, by (9.9) in Lemma 9.1, we have

1 ψK (x) = ψL (x) = , for every x cot θ, (9.17) θ θ x + b ≥

and moreover, from rB∞ K RB∞, ⊂ ⊂

r R ψK (x) , for every x 1. (9.18) x ≤ θ ≤ x ≥ 148

We need to compute ψIK (0) ψIK (δ) for small δ: −

∞ 1/δ n−4 n−1 n−1 2 2 2 ψIKθ (0) ψIKθ (δ) = ψKθ (t) dt + ψKθ (t) 1 (1 δ t ) dt − 1/δ 0 − − Z Z h i = (I) + (II).

If θ π/2 and δ < tan θ, then (9.18) gives upper/lower bounds of the firs term: ≤

∞ ∞ (r/t)n−1dt (I) (R/t)n−1dt. ≤ ≤ Z1/δ Z1/δ

So, the first term is bounded by (rn−1/(n 2))δn−2 and (Rn−1/(n 2))δn−2, which are − − independent of θ. If n = 4, then (I) is asymptotically equivalent to δ2 and the second term

(II) is equal to zero. Assume n 5. Then the second term (II) is divided into two parts as ≥ follows.

cot θ 1/δ n−4 n−1 2 2 2 (II) = + ψKθ (t) 1 (1 δ t ) dt 0 cot θ − − Z Z h i = (II-1) + (II-2), where

cot θ n−1 n 4 2 2 4 (II-1) = ψK (t) 1 (1 − δ t + O(δ )) dt θ − − 2 Z0   cot θ n 4 n−1 2 2 = − ψK (t) t dt δ , 2 θ  Z0  149

and

1/δ 2 2 n−4 1/δ 1/δ 2 2 n−4 1 (1 δ t ) 2 1 (1 δ t ) 2 dt (II-2) − − dt = dt −  tn−1 tn−1 − tn−4 t3 Zcot θ Zcot θ Zcot θ n−2 n−2 tan2 θ (tan θ) δ 1 2 n−4 = − (s δ ) 2 ds n 2 − 2 2 − − Zδ n−2 n−2 2 2 n−2 n−4 (tan θ) δ (tan θ δ ) 2 (tan θ) = − − − = δ2 + O(δn−2). n 2 2 −

Furthermore, when n 5, the integral of (II-1) is bounded above and below by positive ≥ constants independent of θ:

cot θ 1 ∞ n−1 2 n−1 2 n−1 2 n 1 n−1 ψK (t) t dt R t dt + (R/t) t dt = − R θ ≤ 3(n 4) Z0 Z0 Z1 − and cot θ 1 n−1 2 n−1 2 1 n−1 ψK (t) t dt r t dt = r . θ ≥ 3 Z0 Z0 Finally, letting θ go to zero, we have equatorial power type 2 for the body K.

9.3 Double intersection bodies of revolution in high dimension

Recently, Fish, Nazarov, Ryabogin, and Zvavitch [15] proved that the iterations of the intersection body operator, applied to any symmetric star body sufficiently close to

n n a Euclidean ball B2 in the Banach-Mazur distance, converge to B2 in the Banach-Mazur

n n distance. Namely, if K is a star body in R with dBM (K,B2 ) = 1 + ε for small ε > 0, then

m n lim dBM (I K,B ) = 1. m→∞ 2

In case of bodies of revolution in high dimension, it turns out that it is enough to apply the intersection body operator twice to get close to the Euclidean ball in the Banach-Mazur distance, which will be shown in this section. The uniform boundedness of the constant cK 150

by absolute constants in Theorem 9.1 plays an important role in the following result.

n Theorem 9.3. Let K be a symmetric convex body of revolution in R . Then the double intersection body I2K is close to an ellipsoid if the dimension n is large enough. More precisely, for every ε > 0 there exists an integer N > 0 such that for every n N and any ≥ n body K R of revolution, ⊂ 2 n dBM (I K,B ) 1 + ε, 2 ≤

n n n where B2 = x R : x 1 is the unit Euclidean ball in R . { ∈ | | ≤ }

n n Proof. Denote the unit ball in R by B instead of B2 . It follows from Theorem 9.1 that

e 2 3 δIK (ε) = cK ε + O(ε ) where c1 < cK < c2 for absolute constants c1, c2 > 0. Also note

e 2 3 δB(ε) = ε /2 + O(ε ) for the unit ball B. Consider a linear transformation T (dilation) which gives ρT (IK)(π/2) = 1 and ρT (IK)(0) = 1/√2cK . Denote L := T (IK). Then,

e 3 2 3 ψL(t) = 1 δ t/√2cK + O(t ) = 1 t /2 + O(t ). (9.19) − L −

Also, it is not hard to compute the function ψB for the ball B,

1 2 3 ψB(t) = = 1 t /2 + O(t ). (9.20) √1 + t2 −

Let

−1 −1/2 σL = ψ (1 1/n) = 2/n + o(n ) L − −1 p −1/2 σB = ψ (1 1/n) = 2/n + o(n ). B − p

Fix ε > 0, and let R = 4 log ε, N = R2/ε4. Then, we claim that for every n N, − ≥

ρ (θ) IL 1 ε θ [0, π/2]. (9.21) ρ (θ) − ≤ ∀ ∈ IB

151

Note that ρ (0) = ρ (0) π by (9.2). In case that θ is close to 0, the claim (9.21) can IL IB ≈ 2n be obtained from the convexityp of IL. Thus, we may assume that

tan θ ε. ≥

2−n 2−n R 1−R/2 2 For n N, since (1 + R/n) e 2n e = eε , we have ≥ ≤ ≤   R 2−n 1 + = O(ε2). (9.22) n  

2 Note also that for n N, since σL, σB are bounded by 2/N = √2ε /R, ≥ p

2 2 σLR = O(ε ) and σBR = O(ε ). (9.23)

Lemma 9.4 and (9.22) give

n−4 R 2 2 2 ρIL(θ) sin θ n−1 σLt 2 = ψL(σLt) 1 dt + O(ε ). σ − tan2 θ L Z0  

π Note that ρ (π/2) is comparable to σL by Lemma 9.3, and ρ (0) is also comparable IL IL ≈ 2n to σL. So, by convexity of IL, the radial function for IK at any anglep is comparable to σL.

Moreove, since

2 2 σL ε σL σLε = ρIL(θ) sin θ 2ε ρIL(θ) sin θ ρIL(θ) · sin θ · ≤ ρIL(θ) · ·

= O(ε) ρ (θ) sin θ, · IL 152

we have

σLR n−4 n−1 2 2 2 2 ρ (θ) sin θ = ψL(t) 1 t / tan θ dt + O(σLε ) IL − Z0 σLR
n−4 n−1 2 2 2 = (1 + O(ε)) ψL(t) 1 t / tan θ dt. − Z0

Similarly, we have the same equality for IB. Without loss of generality, we may assume

σL σB. Then, ≥

σLR n−4 n−1 2 2 2 ρ (θ) sin θ = (1 + O(ε)) ψL(t) 1 t / tan θ dt (9.24) IL − Z0 σLR
n−4 n−1 2 2 2 ρ (θ) sin θ = (1 + O(ε)) ψB(t) 1 t / tan θ dt (9.25) IB − Z0

Moreover, (9.19) and (9.20) give

n−1 ψ (t) n−1 L = 1 + O(σ3 R3) = 1 + O(nσ3 R3). ψ (t) L L  B   

2 2 2 2 Here, since nσ 3, εR = 16ε(log ε) 1, and σLR = O(ε ), we get L ≤ ≤

3 3 2 2 nσLR = (nσL)(εR )(σLR/ε) = O(ε).

Finally, we have ρ (θ) IL = 1 + O(ε) θ tan−1 ε, ρIB(θ) ∀ ≥ which completes the proof.

Remark 9.3. Theorem 9.3 says that the double intersection body of any body of revolution becomes close to an ellipsoid as the dimension increases to the infinity. However, it is not 153

true for the single intersection body, in general. For example, consider the cylinder

n n−1 B∞ = (x, y) R = R R : x 1, y 1 . (cylinder) ∈ × | | ≤ | | ≤ 

n Then the Banach-Mazur distance between IB∞ and B2 does not converge to 1 as n tends to the infinity.

Proof. The function ψB∞ for the cylinder B∞ is given by ψB∞ (t) = min(1, 1/t), as in the

proof of Lemma 9.1. Note that ρIB∞ (0) = π/(2n) by (9.2), and p ∞ 1 ∞ n−1 1−n n 1 ρ (π/2) = ψB (t) = 1dt + t dt = − . (9.26) IB∞ ∞ n 2 Z0 Z0 Z1 −

ρ (π/2) IB∞ Choose the angle θ with tan θ = ρ (0) , and let x = cot θ. Then IB∞

ρ (0) π n 2 x = IB∞ = − = O(1/√n). (9.27) ρ (π/2) 2n · n 1 IB∞ r −

In addition,

1/x n−1 2 2 n−4 ρ (θ) sin θ = ψIB (x) = ψB (t) (1 x t ) 2 dt = (I) + (II), IB∞ ∞ ∞ − Z0

where 1 1/x 2 2 n−4 1−n 2 2 n−4 (I) = (1 x t ) 2 dt and (II) = t (1 x t ) 2 dt. − − Z0 Z1 For the first term, note that

n−4 1 π (n 2)2 t2 2 lim (I) = lim 1 − dt n→∞ n→∞ − 2 (n 1)2 · n 0   1 Z 1 − − π t2 2 = e 4 dt (1 πt /4)dt = 1 π/12. ≥ − − Z0 Z0 154

The second term

n−4 1/x 1/x 2 1−n 2 2 n−4 1 1 2 (II) = t (1 x t ) 2 dt = x dt − t3 t2 − Z1 Z1   2 n−4 1−x 2 2 1 n−4 1 2 n−4 1 π (n 2) 1 = s 2 ds = (1 x ) 2 = 1 − 2 n 2 − n 2 − 2 (n 1)2 · n Z0 − −  −  converges to zero as n tends to infinity. Let L be the body of revolution obtained by

⊥ shrinking IB∞ by ρ (0) on span e1 and by ρ (π/2) on e . That is, ρ (0) = ρ (π/2) = IB∞ { } IB∞ 1 L L 1. Then

ρIB∞ (θ) sin θ ρL(π/4) sin(π/4) = , ρIB∞ (π/2) and it limit as n is given by → ∞

(I) + (II) lim 1 π/12. n→∞ (n 1)/(n 2) ≥ − − −

Thus, we get ρ (π/4) √2(1 π/12) > 1 for large n. Note that L is a symmetric body L ≥ − of revolution about the axis e1 satisfying ρL(0) = ρL(π/2) = 1 and ρL(π/4) = c > 1, which

n implies that dBM (L, B ) c > 1 for large n. Therefore 2 ≥

n n lim dBM (IB∞,B ) = lim dBM (L, B ) c > 1. n→∞ 2 n→∞ 2 ≥ Chapter 10. Uniform convexity for intersection bodies

In this chapter we provide a quantitative version of the Busemanns theorem for uniformly convex bodies. More precisely, we prove that if a symmetric convex body K has modulus

p of convexity of power type p, i.e., δK (ε) cε , then its intersection body has modulus of ≥ convexity of power type p or smaller.

A convex body K is uniformly convex if its modulus of convexity is positive, i.e., δK (ε) >

0 for each ε (0, 2]. Furthermore we define the power type for uniformly convex bodies in ∈ the following way.

n Definition 10.1. Let K be a uniformly convex body in R . We say that K has modulus of convexity of power type p if

p δK (ε) cε ≥ for some c > 0 (see, e.g. [52, 1.e]).

Nordlander [71] proved that δK (ε) δBn (ε) for any symmetric K. It implies that δK ≤ 2 cannot be of power type p < 2. For example, the unit balls of Lp or `p are of power type 2 if 1 < p 2 and power type p if p 2 (see [33]). ≤ ≥ In this chapter we prove the following:

n Theorem 10.1. Let K be a symmetric convex body in R and E a k-dimensional subspace

n ⊥ of R for 0 k n 1. Suppose that a body L in E is defined by ≤ ≤ −

x x = | | , x E⊥. (10.1) k kL K span(x, E) ∈ | ∩ |

Then, if K has convexity of power type p for p 2, the body L is also of power type p. ≥ 155 156

As an immediate corollary of the theorem we obtain the following.

n Corollary 10.1. Let K be a symmetric convex body in R and p 2. Then the intersection ≥ body IK of K has convexity of power type p whenever K does.

Proof. Let IK be the intersection body of K and fix u1, u2. Let v1, v2 span u1, u2 be ∈ { } ⊥ orthogonal to u1 and u2 correspondingly. Denote E = span v1, v2 . Apply the above { }

theorem to K, E, v1, and v2.

The following example shows that the statement of Theorem is optimal: the uniform convexity in the theorem cannot be improved.

Remark 10.1. Consider K = Bn for p 2. Then K is of power type p. Moreover the p ≥ intersection body IK of K contains a 2-dimensional `p-section which is exactly of power type p, not power type q for any q < p. Thus IK is also of power type p, not power type q for any q < p.

n n−1 ⊥ Proof. Let K = B and E = span(e3, e4, , en). Fix u S E . Note that K p ··· ∈ ∩ ∩

span(u, E) is the `p-sum of K E and [ ρ (u)u, ρ (u)u]. So we have ∩ − K K

n−2 1 Γ p + 1 Γ p + 1 K span(u, E) = K E ρK (u). | ∩ |  n−1   | ∩ | · Γ p + 1   (It can be induced by Fubini’s theorem). It means ρ (u) = c ρ (u) for each u E⊥ = IK · K ∈ ⊥ span(e1, e2) where c > 0 is a constant. Namely, IK E is a 2-dimensional `p-ball. ∩

n Remark 10.2. There exists a uniformly convex body in R which does not have any power type.

Proof. It is enough to consider the 2-dimensional case. Let

2 2 2 − 1 K = (x, y) R : x + cy 1, y 1 e |x| , ∈ ≤ | | ≤ − n o 157

where c = 0.999. We can see that K is convex. Indeed, the boundary of K around the point

− 1 (0, 1) is determined by the graph of y = 1 e |x| which is ‘almost flat’ and concave near − 0. Note also that K does not contain a line segment on the boundary, so K is uniformly

convex. Moreover, taking two points (ε, 1 e−1/ε) and ( ε, 1 e−1/ε) on the boundary of − − − K, we have, for 0 < ε √1 c, ≤ − −1/ε δK (ε) e ≤ which cannot be of power type p for any p 2. ≥

10.1 The Proof of Theorem

n ⊥ Let E be a k-dimensional subspace of R and u1, u2 E be nonparallel vectors. Denote ∈ u1+u2 u = 2 . For each i = 1, 2, define the functions ri = ri(t) by

ri 0 K (ξui + E) dξ t = ∞ | ∩ | , t [0, 1]. (10.2) K (ξui + E) dξ ∈ R0 | ∩ | R Take r = r(t) such that ru is on the line segment between r1u1 and r2u2, that is,

ru = λ1r1u1 + λ2r2u2

where

r2 r1 2r1r2 λ1 = , λ2 = and r = . (10.3) r1 + r2 r1 + r2 r1 + r2

n Lemma 10.1. Let K be a symmetric convex body in R and E be a k-dimensional subspace

n n−k ⊥ n−k ⊥ of R . Suppose that c1B K E c2B . Then, for each x E , 2 ⊂ | ⊂ 2 ∈

2c1 K E K span(x, E) 2c2 K E . k + 1 | ∩ | ≤ | ∩ | ≤ | ∩ |

Proof. Note that K span(x, E) contains the convex hull of K E and a point in K whose ∩ ∩ 158

maximal distance from E is greater than c1. It means

1 K span(x, E) K E (2c1). | ∩ | ≥ k + 1 | ∩ | ·

Since the maximal distance of a point in K from E is c2,

K span(x, E) K E (2c2). | ∩ | ≤ | ∩ | ·

(By Brunn’s theorem, the central section gives a maximal area!).

Lemma 10.2. There exists T GL(n) such that ∈

1 TK E = | ∩ | 2 and 1 Bn−k TK E⊥ Bn−k. √n k 2 ⊂ | ⊂ 2 − Proof. Consider a dilation on E to get K E = 1/2 and a transformation on E⊥ which | ∩ | gives the ellipsoid of minimal volume containing K E⊥. |

Lemma 10.3. The function ri defined in (10.2) for each i = 1, 2 satisfies

1 ri(t) 1 (1 t) k+1 t. − − ≤ ri(1) ≤

Proof. Put a = ri(1). By concavity, for any s [0, ri], ∈

1 a ri 1 ri s 1 K (riui + E) k − K (sui + E) k + − K (aui + E) k , | ∩ | ≥ a s | ∩ | a s | ∩ | − − 159

which gives

K (riui + E) k K (sui + E) | ∩ k |(a s) . | ∩ | ≤ (a ri) − −

So, by integrating on s [0, ri], ∈

ri k+1 k+1 K (riui + E) a (a ri) K (sui + E) ds | ∩ k | − − . (10.4) | ∩ | ≤ (a ri) · k + 1 Z0 −

On the other hand, since K contains the convex hull of K (riui + E) and a point in ∩

K (aui + E), ∩

a 1 K (sui + E) ds K (riui + E) (a ri). (10.5) | ∩ | ≥ k + 1 | ∩ | − Zri

Dividing (10.5) by (10.4), we get

ri k+1 k+1 t 0 K (sui + E) ds a (a ri) = a | ∩ | − −k+1 . 1 t K (sui + E) ds ≤ (a ri) − R ri | ∩ | − R By simplifying the above, we have a lower bound,

ri 1 1 (1 t) k+1 . a ≥ − −

For a upper bound, note that the map s K (sui + E) is decreasing. So 7→ | ∩ |

ri K (sui + E) ds ri K (riui + E) | ∩ | ≥ | ∩ | Z0

and a K (sui + E) ds (a ri) K (riui + E) . | ∩ | ≤ − | ∩ | Zri Dividing these two inequalities, we get t ri , which gives the upper bound ri t. 1−t ≥ a−ri a ≤ 160

Lemma 10.4. Suppose that K, L, E, r1, r2 and r are defined as (10.1), (10.2) and (10.3).

dr −1 λ1 K (r1u1 + E) + λ2 K (r2u2 + E) u1 + u2 . ∩ ∩ dt ≥ k kL k kL      

Proof. For the proof of this lemma, we follow the general idea from the proof of Busemann’s

theorem [13]. Note that (10.2) implies that, for any t [0, 1] and i = 1, 2, ∈

dri 1 K (riui + E) = . (10.6) | ∩ | dt 2 ui k kL

The Brunn-Minkowski inequality gives

λ1 λ2 λ1 K (r1u1 + E) + λ2 K (r2u2 + E) K (r1u1 + E) K (r2u2 + E) . ∩ ∩ ≥ | ∩ | | ∩ |     (10.7)

On the other hand, it follows from (10.3) that

λ1 λ2 dr 2 dr1 2 dr2 dr1 dr2 = 2 λ + λ 2λ1 2λ2 . (10.8) dt 1 dt 2 dt ≥ dt · dt      

From (10.6), (10.7) and (10.8), we get

dr λ1 K (r1u1 + E) + λ2 K (r2u2 + E) ∩ ∩ dt    λ1  λ2 dr1 dr2 2λ1 K (r1u1 + E) 2λ 2 K (r2u2 + E) ≥ | ∩ | dt · | ∩ | dt     λ1 λ2 λ1 λ2 −1 = u1 L + u2 L . u1 L · u2 L ≥ k k k k k k  k k    161

Proof of the theorem. Since the modulus of convexity is invariant for any linear trans- formation, by Lemma 10.2 we may assume that

1 1 K E = and Bn−k K E⊥ Bn−k. | ∩ | 2 √n k 2 ⊂ | ⊂ 2 − √ 2/ n−k Then, by Lemma10.1, K span(x, E) K E = 1√ for each x E⊥. In | ∩ | ≥ k+1 | ∩ | (k+1) n−k ∈ other words, for each x E⊥, ∈

1. x ⊥ x , k kK|E ≥ | |

2. x (k + 1)√n k x . k kL ≤ − | |

⊥ Now fix u1, u2 E such that u1 = 1 = u2 and u1 u2 ε. We need to ∈ k kL k kL k − kL ≥ u1+u2 u1+u2 compute 1 to estimate the modulus of convexity δL of L. Let u = and − 2 L 2 take any

z := λ1(r1u1 + y1) + λ2(r2u2 + y2) λ1 K (r1u1 + E) + λ2 K (r2u2 + E) . ∈ ∩ ∩    

Note that z can be written as the midpoint of two elements in K: if λ := min(λ1, λ2) = λ1,

2λ(r1u1 + y1) + (1 2λ)(r2u2 + y2) + r2u2 + y2 z = − .  2   

Thus,

z K 1 δK 2λ(r1u1 + y1) + (1 2λ)(r2u2 + y2) r2u2 + y2 k k ≤ − − − K      

= 1 δK 2λ r1u1 r2u2 + y1 y2 − k − − kK   1 δK 2λ r1u1 r2u2 ⊥ . ≤ − k − kK|E   162

Estimation for λ = min(λ1, λ2):

By Lemma 10.1, we have

2 ri(1) · K E ui 2 ri(1) K E , k + 1 | ∩ | ≤ k kL ≤ · | ∩ | which gives, under the above assumptions,

1 ri(1) k + 1. ≤ ≤

By Lemma 10.3,

1 1 (1 t) k+1 ri(t) (k + 1)t. (10.9) − − ≤ ≤

Thus

r1 r2 λ = min , r1 + r2 r1 + r2  1  1 (1 t) k+1 1 min − − = . ≥ t∈[0,1] 2(k + 1)t 2(k + 1)2

Estimation for r1u1 r2u2 ⊥ : k − kK|E By Lemma 10.2(or the above property 1),

r1u1 r2u2 ⊥ r1u1 r2u2 . k − kK|E ≥ | − |

Moreover, the triangle inequality and the arithmetic/harmonic mean inequality imply

r1u1 r2u2 + r2u1 r1u2 r1 + r2 r1u1 r2u2 = | − | | − | u1 u2 r u1 u2 . | − | 2 ≥ 2 | − | ≥ | − | 163

ku1−u2kL By Lemma 10.2 (or the above property 2) again, u1 u2 √ . Thus | − | ≥ (k+1) n−k

εr r1u1 r2u2 ⊥ . k − kK|E ≥ (k + 1)√n k −

From above two lower bounds, it follows

z 1 δK 2λ r1u1 r2u2 ⊥ k kK ≤ − k − kK|E  1 εr 1 δK 2 ≤ − · 2(k + 1)2 · (k + 1)√n k  − 

= 1 δK (Cεr) , −

1√ where C = (k+1)3 n−k . Thus, we have

[1 + δK (Cεr)] z K. ∈

In other words,

K [1 + δK (Cεr)] λ1 K (r1u1 + E) + λ2 K (r2u2 + E) . (10.10) ⊃ ∩ ∩ h    i

Let s(t) = [1 + δK (Cεr(t))] r(t). Then

−1 1 u1 + u2 = u −1 = 2 K (su + E) s0(t)dt, 2 k kL | ∩ | L Z0

0 s(t)−s(t−h) where s (t) = limh→0+ (δK ( ) is nondecreasing and r( ) is differentiable). Using h · · 164

the fact t δK (t)/t is non-decreasing [52, 1.e.8], we can get 7→

s(t) s(t h) s0(t) = lim − − h→0+ h r(t) r(t h) δK (Cεr(t))r(t) δK (Cεr(t h))r(t h) = lim − − + − − − h→0+ h h r(t−h) dr δK (Cεr(t))r(t) δK (Cεr(t)) r(t) r(t h) + lim − − ≥ dt h→0+ h dr = [1 + 2δK (Cεr(t))] . dt

By (10.10),

k K (su + E) 1 + δK (Cεr) λ1 K (r1u1 + E) + λ2 K (r2u2 + E) . | ∩ | ≥ | | ∩ ∩    

It follows from Lemma 10.4

k 0 dr 2 K (su + E) s 2 1 + δK (Cεr) K (ru + E) 1 + 2δK (Cεr) | ∩ | ≥ | ∩ | · dt    −1  2 1 + (k + 2)δK (Cεr) u1 + u2 ≥ · k kL k kL     1 + (k + 2)δK (Cεr). ≥

By integrating on t, we get

−1 1 u1 + u2 = 2 K (su + E) s0dt 2 | ∩ | L Z0 1

1 + (k + 2) δK (Cεr)dt, ≥ Z0 that is, 1 u1 + u2 1 (k + 2) δK (Cεr)dt. − 2 ≥ L Z0

165

Finally 1 δL(ε) (k + 2) δK (Cεr)dt > 0 ≥ Z0 1 p where C = √ . Moreover, if K is of power type p, δK (ε) cε for p 2, then (k+1)3 n−k ≥ ≥

1 1 p p p δL(ε) (k + 2) δK (Cεr)dt (k + 2)cC ε r dt. ≥ ≥ Z0 Z0

2r (t)r (t) 1 Since r(t) = 1 2 1 (1 t) k+1 by (10.9), r1(t)+r2(t) ≥ − −

p 1 (k + 2)cε 1 p δ (ε) 1 (1 t) k+1 dt L 3 p ≥ (k + 1) √n k 0 − − − Z   p 1 p/2  (k + 2)cε  1 2 1 (1 t) k+1 dt 3 p ≥ (k + 1) √n k 0 − − − Z    cn −4pεp  ≥ which means that L is also power type p. Chapter 11. Proposed further research

1. As mentioned in Section 7.1 we are interested in proving or disproving that there exist

other fixed points except Euclidean balls of the intersection body operator. More

precisely, is it true that IK = K implies that K is a Euclidean ball? As the first

step to this question we may try to see whether it is true in the class of bodies of

revolution, which turns out to be true in dimension 4 or 6, but not known in general.

2. It is shown in Chapter 10 that the intersection body IK is of power type p whenever K

is. In other words we may say that ‘convexity’ in the sense of power type is preserved

under the (single) intersection body operation. Moreover this result should be optimal

as seen in Remark 10.1, so we cannot say that the ‘convexity’ increases strictly under

the (single) intersection body operation. However, if we take the double intersection

body operator I2 instead of the single intersection body operator I, then we may have

a more strong result. As my further research on intersection bodies, I would like to

study the ‘convexity’ of the double intersection body I2K in the sense of power type

or the modulus of convexity δK , in order to get a strict improvement on its ‘convexity’.

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