ANALYTIC METHODS IN CONVEX GEOMETRY

Transcription

1 ANALYTIC METHODS IN CONVEX GEOMETRY September 21, 2013 DMITRY RYABOGIN AND ARTEM ZVAVITCH Dedicated to the memory of Joram Lindenstrauss, and Aleksander Pelczynski Contents 1. The General Plan 2 2. Short Introduction Main definitions and facts Brunn-Minkowski inequality 5 3. Duality and Volume, the first look R 2 -case Unconditional case Local Minimum Zonoids and Zonotopes Mahler conjecture: case of zonoids Radon and Cosine Transforms Local and Equatorial characterizations of Zonoids A motivating example Local equatorial characterization Local characterization What information can uniquely define a convex body? Questions of Bonnensen and Klee Isomorphic version of the Mahler conjecture Introduction Part I. Main Idea: the application of the Paley-Wiener Theorem Part II. The adjustment of the main idea to the context of the Bergman space A 2 (T K ) The Rothaus-Korányi-Hsin formula for the reproducing kernel in a tube domain T K Estimates related to the Bergman Kernel in the tube domain Part III. The auxiliary construction The construction of the the pluri-sub-harmonic φ on the tube domain T K The construction of the Mexican-hat function g supported on K C Part IV. The construction of the analytic function F that helps to obtain the sharp lower bound on v 2n (K C )K(0, 0) 58 Appendix A. The Paley-Wiener Theorem 59 Appendix B. The Bergman Spaces 61 The authors were supported in part by the Institute of Mathematics, Polish Academy of Sciences and U.S. National Science Foundation grant DMS

2 B.1. Reproducing kernel 61 Appendix C. The Hörmander Theorem and some of its applications 61 C.1. Convexity and Pseudo-Convexity 62 C.2. Formulation of the Hörmander theorem 65 C.3. From sub-harmonic to analytic. Examples in the case n = 1 66 C.4. The example of a non-trivial analytic function F = F (φ, g) obtained after the application of the Hörmander Theorem in the strip Ω = T K 68 Appendix D. Proof of Hörmander s Theorem 70 References The General Plan These lecture notes are devoted to duality phenomena in Convex Geometry and different analytic methods for its study. The notion of duality is one of the most important ones in mathematics. It plays a central role in Functional Analysis and Convex Geometry. It is hard to believe, but there are still many classical open problems tightly connected to this notion. Our main goal here is to discuss some of these problems. In these lectures we will frequently use theorems, facts and general techniques from [Ba4, Ga1, Ga2, K5, KY, MS, Pi, Sc2, Web]. At first we will review some classical background and refresh some facts from Convex Geometry. In particular, we will remind the reader the F. John Theorem and the Brunn-Minkowski Inequality. Next, we will discuss the notion of duality, the volume product and present a proof of the Santalo inequality. We continue our discussion with the Mahler conjecture, asking about the minimum of the volume product of a convex body and its polar. The conjecture is open (even in the three-dimensional case) since 1939 and has been confirmed only for some particular classes of bodies. We will present the solution in the two-dimensional case and in the case of unconditional bodies. We will also discuss some new results and show that the unit cube is a local minimizer for the volume product. After that we will define Zonotopes and Zonoids. This class of bodies will be our main playground to test different conjectures for correctness, for example, we will verify the Mahler s conjecture for Zonoids. Harmonic Analysis is an extremely helpful tool to understand the behavior of volume under duality: we will discuss Koldobsky s method of representing the Spherical Radon and the Cosine transforms using the language of the Fourier transform of distributions. As an application we will present results on the local characterization of Zonoids. We will also talk about questions of the unique determination of convex bodies by the volumes of their projections and sections. In particular, we will discuss recent results related to the problems of Bonnensen and the solutions of the problems of Klee. We will finish the lectures by presenting Nazarov s proof of the Bourgain-Milman inequality, which is an isomorphic version of the Mahler conjecture. In many cases, writing down all details would take us away from the main point. Therefore, when the details become too tedious, we prefer to provide the reader with the 2

3 exact reference to the book or the paper, where the rigorous proof of the corresponding fact can be found. To help the reader with the understanding of the material, we include in the text many exercises of various difficulty. Acknowledgment. We would like to thank Radoslaw Adamczak, Rafal Latala, Piotr Mankiewicz, Krzysztof Oleszkiewicz, Pawel Wolff and other colleagues in Warsaw for organization of these lectures and for the wonderful time we had in Warsaw. We are very grateful to Fedor Nazarov for a number of suggestions and for letting us use his published (and unpublished!) results. We would also like to thank Jaegil Kim for helpful advice during the preparation of these lecture notes. 2. Short Introduction 2.1. Main definitions and facts. As usual, S n 1 denotes the unit sphere, B n 2 the unit Euclidean ball, 0 is the origin, and the norm in Euclidean n-space R n. If x, y R n, then x y is the inner product of x and y and [x, y] denotes the line segment with endpoints x and y. If X is a set, dim(x) is its dimension, that is, the dimension of its affine hull, and X is its boundary. A set is o-symmetric if it is centrally symmetric, with center at the origin. If X and Y are sets in R n, then X + Y = {x + y : x X, y Y } is the Minkowski sum of X and Y. A body is a compact set equal to the closure of its interior. A body K R n is called star-shaped if every straight line through the origin intersects the boundary of K at exactly two points different from the origin. The Minkowski functional of K is defined as x K = min{a > 0 : x ak}, which is a continuous function on R n. A radial function ρ K (x) : R n \ {0} R + is defined by ρ K (x) = max{a > 0 : ax K}. Notice that ρ K (x) = x 1 K. For a k-dimensional set A R n we write A for the k-dimensional Lebesgue measure in R n, k = 1,..., n. If K is a k-dimensional body in R n, then we refer to K as its volume. A set K R n is called convex, if [x, y] K, for all x, y K. A set in R n is called a convex body if it is convex and compact with nonempty interior. We denote by conv(a) the closed convex hull of a set A R n, and conv(a, B, C,... ) the closed convex hull of A B C,.... Exercise 2.1. Show that K is a convex symmetric body iff x K is a norm on R n. Show that x K iff x K 1. Exercise 2.2. Show that if K and L are star-shaped bodies, then K L iff x K x L, or, equivalently ρ K (x) ρ L (x), for every x S n 1. 3

4 We denote B n = {x R n : max 1 i n x i 1} and B n p = {x R n : Exercise 2.3. Show that also show that S n 1 = n B n 2. B n p = Hint: Use two ways to compute R n e ( ( )) n 2Γ p Γ(1 + n), p n i=1 n x i p 1} for 1 p <. i=1 x i p dx, see ([Pi], page 11). The set of convex bodies is pretty large. For many problems we would like to make it smaller, so sometimes it makes sense to study the convex bodies up to linear transformations. Definition 2.1. Let K, L R n be convex symmetric bodies. Then the Banach-Mazur distance between these bodies is defined by d BM (K, L) = min{d > 0 : K T L dk; T GL(n)}. Exercise 2.4. Show that d BM (T K, L) = d BM (K, L) for all T GL(n). d BM (K, L) d BM (K, M)d BM (M, L). We will frequently use the following theorem, which will help us to approximate a convex symmetric body by an ellipsoid. One of the consequences of this theorem is that the set of convex symmetric bodies in R n is compact with respect to the Banach-Mazur distance. Theorem 2.1. (F. John) For any convex symmetric body K R n, we get d BM (K, B n 2 ) n. Proof. Let E be an ellipsoid of maximal volume inscribed in K, i.e. E K and E = max{ D : D is an ellipsoid and D K}. Our goal is to show that K ne. The Banach-Mazur distance is invariant under linear transformations, so we may assume that E = B n 2. If K nb n 2, then there exists p K so that p > n. Applying convexity and symmetry of K we get that D = conv (B n 2, p, p) K. We would like to show that D (and thus K) contains an ellipsoid of the volume which is larger than the volume of B2 n, this would contradict an assumption that B2 n is the ellipsoid of the maximal volume in K. Without loss of generality we may assume that p = (d, 0,..., 0) and d > n. Consider an ellipsoid E = {x R n : x2 1 a 2 + n i=2 x 2 i b 2 1}. 4

5 Then E = ab n 1 B2 n and if a2 + (1 1 )b 2 1, then E D. Indeed, E and D are d 2 d 2 rotation invariant around x 1, thus to check E D we need to make two dimensional calculation in (x 1, x 2 ). It is easy to check that a = d/ n, b = 1 1/n/ 1 1/d 2 satisfy this last condition and ab n 1 > 1. Exercise 2.5. Prove that d BM (B, n B2 n ) = n. Hint: It is easy to see that d BM (B, n B2 n ) n. To prove the reverse inequality consider T GL(n) such that T B2 n B n dt B2 n, check the conditions you will get on the diagonal elements of T, from the fact that e i T B2 n. Also check the conditions on the elements of dt, it will help to notice that the vertices of B n must be outside of T B2 n, and to study them as a random vector with ±1 coordinates. Definition 2.2. The support function h K : R n R + of a convex body K R n is defined by h K (x) = max y K x y. Lemma 2.1. If K and L are convex bodies in R n then Exercise 2.6. Prove Lemma 2.1. h K+L = h K + h L. Exercise 2.7. Show that for a convex body K, x K iff for all y R n we have x y h K (y). Also prove that if K, L are convex bodies then K L iff h K (x) h L (x), for all x R n. Exercise 2.8. Consider a convex body K R n. Show that h K (ξ) exists for almost all ξ S n 1 and, moreover, if it exists, then h K (ξ) K and ξ is a normal vector to K at h K (ξ) Brunn-Minkowski inequality. Theorem 2.2. (Brunn-Minkowski inequality) Let, A, B be two measurable, non-empty sets in R n. Then, (1) A + B 1/n A 1/n + B 1/n. Equivalently, for all λ [0, 1] (2) λa + (1 λ)b A λ B 1 λ, for all λ [0, 1]. Proof. We will sketch the proof of (1) due to Lusternik [Lus]. We refer to [Ga2] for other proofs. We first consider the case when A, B are n-dimensional boxes. The volume is invariant under the translation, so we may assume that one of the vertices of A and B is the origin: n n A = [0, a i ] and B = [0, b i ] i=1 i=1 5

6 Thus A + B = n [0, a i + b i ]. But then i=1 ( ) 1 ( A n + A + B B A + B ) 1 n ( n a i = a i=1 i + b i ( n 1 n i=1 a i a i + b i ) 1 ( n n + ) + 1 n i=1 b i a i + b i ( n i=1 ) 1 n b i a i + b i Now suppose that A is the union of m 1 disjoint rectangular boxes and B is the union of m 2 disjoint rectangular boxes. We will prove (1) by the induction on the number m 1 + m 2. The case m 1 + m 2 = 2 is checked, so we may assume that m 1 + m 2 > 2 and that A contains at least 2 disjoint boxes. Again, (1) is invariant under the translation. Hence, we may assume that at least two boxes of A are separated by the coordinate plane x n = 0. Let A + = {x A : x n 0} and A = {x A : x 0}. By the construction, the number of boxes in each A + and A is smaller than m 1. Next, translate B so that A + A = B+ B and, thus, A A = B B, where B + = {x B : x n 0} and B = {x B : x 0}. Notice that the sets A + + B + and A + B are disjoint (up to a set of volume zero) and ( A + + B +) ( A + B ) A + B. Moreover, we can apply the induction hypothesis to A + + B + and A + B to get A + B A + + B + + A + B ( A + 1/n + B + 1/n) n ( + A 1/n + B 1/n) n ) = A + (1 + B+ 1/n n ) + A (1 + B 1/n n A + 1/n A 1/n ) n ) n = A + (1 + B 1/n + A (1 + B 1/n = ( A 1/n + B 1/n) n. A 1/n A 1/n Exercise 2.9. Finish the proof of the Brunn-Minkowski inequality, i.e. show that it is enough to prove (1) for A and B that are disjoint union of boxes. Exercise Note that (2) easily follows from (1). Prove that the reverse direction is, unexpectedly, also true. Hint: use (2) with A = K/ K 1/n, B = L/ L 1/n and λ = K 1/n /( K 1/n + L 1/n ). Using the Brunn-Minkowski inequality one can obtain a number of properties of convex bodies. Below we consider several applications of this inequality that will be very useful for us. For direction ξ S n 1, we denote by ξ the hyperplane perpendicular to ξ, i.e. ξ = {x R n : x ξ = 0}. ) = 1. 6

7 Definition 2.3. Consider a body K R n, and a direction ξ S n 1 then we define a function A K,ξ (t) : R R + such that A K,ξ (t) = K {ξ + tξ}. Corollary 2.1. (Brunn s Theorem) Consider a convex body K R n, then A 1 n 1 K,ξ (t) is concave on its support. Moreover, if K is symmetric then A K,ξ (0) A K,ξ (t), for all t R. Proof. It is enough to note that from convexity of the body K we get λ ( K {ξ + t 1 ξ} ) + (1 λ) ( K {ξ + t 2 ξ} ) ( K {ξ + [λt 1 + (1 λ)t 2 ]ξ} ). Exercise Is it possible to guarantee the convexity of K from the concavity of A 1 n 1 K,ξ (t)? More precisely: Show that if K R 2 is a star shaped body and A K,ξ (t) is concave on its support, then K is convex. Show that the above statement is not true in R 3, i.e. construct an example of a non-convex star-shaped origin-symmetric body K R 3 such that its section function A 1 2 K,ξ (t) is concave on it s support for every fixed direction ξ S 2. Hint: Use rotation invariant bodies, see more formulas in [GR]. Definition 2.4. Assume that K R n and that the boundary K of K is smooth. Then the Minkowski volume of K is defined by K + tb2 n K K = lim, t 0 t provided that the limit exists. Exercise Consider a convex polytope P R n with (n 1)-dimensional faces {F i } m i=1. Show that the Minkowski volume of K is F i. Corollary 2.2. Assume K = B n 2, then K S n 1. Proof. Corollary follows from the definition of K and the Brunn-Minkowski inequality: ( K + tb2 n K ( K 1/n + tb2 n 1/n) n K (3) K = lim lim t 0 t t 0 t = n K n 1 n B n (4) 2 1/n = S n 1. Here we used that S n 1 = n B n 2. A consequence of the Minkowski theorem, (Theorem in [Sc2]), tells us that if K, L R n are convex bodies, then K + tl is a polynomial in t 0, with positive coefficients: n ( ) n K + tl = V k (K, L)t k. k k=0 7

8 Here V i (K, L) is called the i-th mixed volume of K and L. We will not need the above identity in these lecture notes, but we will use analytic properties of V 1 (K, L), for which we may give an alternative definition: Definition 2.5. If K and L are convex bodies in R n, the mixed volume V 1 (K, L) is equal to V 1 (K, L) = 1 n lim K + tl K. t 0 t Applying the Brunn-Minkowski inequality, as in the proof of Corollary 2.2, we get: Corollary 2.3. (First Minkowski Inequality) For any convex bodies K, L R n, V 1 (K, L) K n 1 n L 1 n. Consider a convex polytope P R n with (n 1)-dimensional faces {F i } m i=1 and the corresponding normal vectors u i S n 1 to F i. Using the formula for the volume of the pyramid and the definition of the support function h P, we get (5) P = 1 m h P (u i ) F i. n i=1 Let K θ be the orthogonal projection of K onto hyperplane perpendicular to the unit vector θ. It is easy to see that then F i θ = u i θ F i (indeed, u i θ is just a cosine of the angle between the normal vector u i and the direction θ). This gives the well-known Cauchy formula: (6) P θ = 1 m F i θ = 1 m u i θ F i. 2 2 i=1 Exercise Find among B n ξ, ξ S n 1, the one with maximal volume; minimal volume. Do the same for the projections of B n 1. Finally, we can also create a formula for V 1 (P, L). Indeed, it is enough to understand how the volume of P is changing under the addition of tl, for very small t. Moreover, we only need to consider the rate of change of the volume of the order t (the rate t 2 will be canceled after taking the limit). When we add tl to P, each face F i moves outward by th L (u i ) in the direction of u i. Adding up these changes we get (7) V 1 (P, L) = 1 m h L (u i ) F i. n Formulas (5), (6) and (7) are extremely useful for our lecture notes. We will also need their generalizations from polytopes to the case of general convex bodies. In the general case the formulas can be obtained by the approximation argument; but at first we need to generalize the notion of the volume measure of the face. Definition 2.6. The surface area measure S(K, ) of a convex body K in R n is a finite Borel measure on S n 1, such that for every Borel set E S n 1, S(K, E) is the volume of the part of K where normal vector belongs to E. If S(K, ) is absolutely continuous with respect to the Lebesque measure on S n 1, then the density f K of S(K, ) is called the curvature function. i=1 i=1 8

9 Exercise Consider a polytope P, compute S(P, ). Exercise Let K be a convex body with C 2 -smooth boundary. Assume also that the gaussian curvature at each point of K is positive. Prove that S(K, ) is absolutely continuous. Hint: in this case the surface area measure is the reciprocal of the Gaussian curvature, viewed as a function of the unit normal vector. Exercise Compute f B n p for p (1, ). Hint: see [KRZ]. Using the approximation argument and the above definition we get the following theorem: Theorem 2.3. Consider a convex body in K R n, then (1) K = 1 h n K (u)ds(k, u). S n 1 (2) K θ = 1 u θ ds(k, u). 2 S n 1 (3) V 1 (K, L) = 1 h n L (u)ds(k, u). S n 1 3. Duality and Volume, the first look Consider a convex symmetric body K R n. We define a polar body K of K as K = {y R n : y x 1 for all x K}. Clearly h K = x K = ρ 1 K (x). One can also give the definition of the duality for a non-symmetric case: polar body K z of a convex body K with center of polarity z is defined by K z = {y R n : (y z) (x z) 1 for all x K}. If the center of polarity is taken to be the origin, we denote by K the polar body of K, thus K z = (K z) + z. A well known result of Santaló [Sa] states that in every convex body K in R n, there exists a unique point s(k), called the Santaló point of K, such that K s(k) = min z int(k) Kz. Exercise 3.1. Prove the existence and uniqueness of s(k) (Hint: see [Sc2], p. 419). Also prove that s(k) = 0, provided K is symmetric. Definition 3.1. The volume product of K is defined by P(K) = inf{ K K z : z int(k)} = K K s(k), thus if K is a symmetric convex body then P(K) = K K. Lemma 3.1. Here we list the main properties of duality and volume product for convex bodies containing the origin in their interior. (1) We have (K ) = K. (2) If K L, then L K. 9

10 (3) The volume product is invariant under the non-degenerate linear transformations, that is, (T K) = (T ) 1 K and thus P(T (K)) = P(K), for all T GL(n). (4) We have ( K ξ ) = K ξ. Remark 3.1. We note that the property ( K ξ ) = K ξ shows that the operation of duality transforms sections of the given body into projections of the dual one. It is the origin for understanding of statements of duality between subspaces and quotient spaces in Functional Analysis. Exercise 3.2. Let K = [ 5, 4], find (K ). Exercise 3.3. Prove Lemma 3.1. Note that (3) follows from the fact that T x (T ) 1 y = x y, for T GL(n). Exercise 3.4. Show that for convex, symmetric bodies K, L R n (K L) = conv (K, L ). Exercise 3.5. Let K, L R n be a convex symmetric bodies. Show that d BM (K, L) = d BM (K, L ). Exercise 3.6. Show that the map K P(K) is a continuous on the set of convex symmetric bodies equipped with the Banach-Mazur distance (Hint: see [FMZ]). Since the set of all convex bodies in R n is compact with the respect to the Banach- Mazur distance and the map K P(K) is continuous (this follows from the F. John theorem and the above exercise), it is very natural to ask for a maximal and minimal values of P(K). We start with the maximum, which is given by the following Blaschke- Santaló inequality Theorem 3.1. For any convex symmetric body K R n, P(K) P(B n 2 ). Proof. We will present the proof from [MePa] and [Ba]. The main tool is the Stiener Symmetrization which (in the one co-dimensional case) can be described as follows. Consider a convex body K R n and a hyperplane u, where u S n 1. To get S u (K) perform the following algorithm: for each point p u let l p be a line through p perpendicular to u (i.e. l p = p + tu, t R). If K l p = - do nothing, otherwise translate the segment K l p along l p until its midpoint belongs to u (see the Figure 1). 10

11 p + t 1u p + t 2u K l p u S u (K) p + t1 t2 2 u u p p t1 t2 2 u Figure 1. Steiner Symmetrization in R 2. More precisely, let K R n be a convex body, U be a subspace of R n and U its orthogonal complement. We define the Stiener symmetrization S U (K) of K with respect to U to be the set of points x R n for which there are p U and v, w U such that x = p + v w 2 p + v, p + w K. We will need the following useful facts: Assume that K, L are convex symmetric bodies in R n, then (a) S U (K) is also a symmetric, convex body. (b) S U ( ) is a monotone operation, i.e. S U (L) S U (K), for all L K. (c) S U (λk) = λs U (K), for all λ > 0. (d) S U (B2 n ) = B2 n. (e) S U (K) K. (f) S U (K ) (S U (K)). (g) P(K) P(S U (K)). It is easy to check that S U (K) is convex and moreover (8) [S U (K) p] U = 1 [(K p) U (K p) U]. 2 We see that (e) follows from Fubini Theorem, equation (8) and the Brunn-Minkowski inequality (Theorem 2.2). Now consider x S U (K) and y S U (K ). Then x = p (v w), y = z + 1 (s t), 2 11

12 where p, s, t U ; z, v, w U; p + v, p + w K and z + s, z + t K. Moreover, p z = 0, (v w) (s t) = 0 and x y = 1 2 p (s t) (v w) z = 1 2 (p + v) (z + s) 1 (p + w) (z + t). 2 This gives x y 1, which implies (f). To prove (g) we apply (e) and consider Steiner symmetrization of K and K with respect to U and U. We also use (f): P(K) = K K S U (K) S U (K ) S U (K) (S U (K)) = P(S U (K)). Another well known fact (see Theorem [Web]) is that for any convex body K R n there is a sequence of unit vectors ξ 1, ξ 2,... such that S ξ i S ξ 1 (K) converges to the Euclidian ball of volume K as i. Thus, Theorem 3.1 follows immediately from (f). Exercise 3.7. Show that for two convex bodies K, L R n we have S U (K) + S U (L) S U (K + L). Use the above inclusion to prove Brunn-Minkowski inequality for convex bodies. The minimality of P(K) turns out to the much harder question which is still open! It is often called the Mahler s conjecture [Ma1, Ma2], and it states that for every convex body K in R n, (9) P(K) P( n ) = (n + 1)n+1 (n!) 2, where n is an n-dimensional simplex. It is also conjectured that equality in (9) is attained only if K is a simplex. Exercise 3.8. Show that P( n ) = (n + 1)n+1 (n!) 2. In our lectures we will concentrate on the symmetric case of the Mahler conjecture, which states that for every convex symmetric body K R n : (10) P(K) P(B n ) = P(B n 1 ) = 4n n!. Exercise 3.9. Try to prove directly, using Exercise 2.3, that P(B n p ) P(B n ), for p 1. Note, that this fact follows immediately from Theorem 3.3 (see below). The conjecture is confirmed in R 2. We will sketch this fact in the next section as well as give a complete proof as a part of the local minima result (see Theorem 3.4 below). 12

13 3.1. R 2 -case. We would like to show that Theorem 3.2. (Mahler) P(P ) P(B 2 ) for all symmetric convex polygons P R 2. Proof. (Sketch) The main idea is to remove/glue vertices. Let P k be any given symmetric polygon with k vertices. Following Mahler, we will construct a sequence of transformations of symmetric polygon P k : P k P k 2 P k 4... P 4, such that P(P k ) P(P k 2 ) P(P k 4 )... P(P 4 ) = P(B 2 ). Here the last equality follows from invariance of P under linear transformations (Lemma 3.1). On each step of the construction, we will have P k = P k 2 and P k P k 2. In fact, the procedure is just to reduce the question to a comparison of the area of a triangle and the dual triangle. The duality operation K K does not work well on polygons (bodies) that do not contain the origin (see Exercise 3.2). To overcome this difficulty, Mahler introduced another duality like operation: Definition 3.2. Let K be a convex polytope with faces f 1,..., f m. Then K is a polytope that has vertices ṽ 1,..., ṽ m such that ṽ m f m and dist(ṽ m, 0) dist(aff(f m ), 0) = 1, where aff(f m ) denotes the minimal affine subspace (line in this case) containing f m. Exercise Show that if a polytope K contains the origin, then K = K. Exercise Let v 1, v 2, v 3 be the vertices of a triangle conv(v 1, v 2, v 3 ) such that 0 conv(v 1, v 2, v 3 ). Find ṽ 1, ṽ 2 and ṽ 3, the vertices of triangle conv(v1, v 2, v 3 ). Consider a polygon P k with vertices v 1, v 2, v 3,..., v k, written in the clockwise order. Let l be a line passing through v 2 parallel to the segment v 1, v 3, and let v be any point on l. Then Exercise As long as v stays on l between lines l v3,v 4, l vk,v 1 (passing through v 3, v 4 and v k, v 1 respectively), we have P k = P k (v), where P k (v) := conv{v 1, v, v 3,..., v k }. See Figure 2. The main observation in this construction is Exercise Prove that (11) P k (v) = P k 1 (v), Pk (v) = P k 1 \ (v), where (12) (v) = conv{v 1, v, v 3 }, and P k 1 = conv{v 1, v 3,..., v k }. Also prove that (11), (12) imply (13) P k (v) = P k 1 + (v), (14) P k (v) = P k 1 (v). and thus (v) (v 2 ) gives P k (v) P k (v 2 ) = P k. 13

14 l l vk,v1 v k v 1 v l v 2 v l v3,v4 v r 0 v 3 v 4 Figure 2. Moving vertex v 2 to vertex v along the line l, keeping P k (v) constant. Let v l = l l vk,v 1 and v r = l l v3,v 4 (See Figure 2). To finish the proof it is enough to show that (15) max{ (v l ), (v r ) } (v 2 ) Indeed P k (v l ) and P k (v r ) have k 1 vertices, and it follows from Exercise 3.13 and (15) that P(P k ) min{p(p k (v l )), P(P k (v r ))}. To prove (15) we need to show that the function f(v) = (v) attains its maximum at the end points of the interval [v l, v r ]. This can be done by using Exercise 3.11 and computing the exact formula for f(v) Unconditional case. Let us also note that if the conjecture is true in the symmetric case, then for n 4, B n and B1 n is not the only minimal pair. For sets K R n 1 and L R n 2 we denote by K L their direct sum, i.e. K L = {(x, y) R n 1+n 2 : x K, y L}. Lemma 3.2. Consider two convex symmetric bodies K R n 1 In particular, P(K L) = n 1!n 2! (n 1 + n 2 )! P(K)P(L). P(B n 1 1 B n 2 1 ) = P(B n 1+n 2 1 ). and L R n 2. Then 14

15 Proof. We first notice that K L = K L. For convex bodies K R n 1 and L R n 2 define a new convex body K 1 L R n 1+n 2 using its Minkowsky functional: Exercise Prove that: K 1 L = conv(k, L). (K L) = K 1 L. (x 1, x 2 ) K 1 L = x 1 K + x 2 L, for x 1 R n 1, x 2 R n 2. Finally, Lemma 3.2 follows from (16) K = 1 e x K dx, for K R n, K is convex, symmetric body. n! R n Indeed, e x K dx = R n R n x K e t dtdx = 0 x K t e t dxdt = 0 tk e t dt = n! K. Definition 3.3. A symmetric convex body is called a Hanner polytope if it is onedimensional, or if it is the or the 1 sum of two (lower dimensional) Hanner polytopes. If we apply Lemma 3.2 to Hanner polytopes we immediately get that P(H) = P(B ) n for every Hanner polytope H in R n. Thus, starting from dimension 4 there are a lot of possible minimizers of the volume product. Exercise Please, check the above statement, more precisely: Prove that P(H) = P(B n ) for every Hanner polytope H R n. Show that every Hanner polytope H R 3 is, up to a linear transformation, is B 3 or B 3 1. Show that there exists a Hanner polytope H R 4 such that it is not a linear transformation of B 4 or B 4 1. Hint: you may use (I 1 I 1 I) I, where I = [ 1, 1]. Open Problem 3.1. Does there exist a convex symmetric body K R 3, such that but K is not a linear image of B 3 1 or B 3? P(K) = P(B 3 ), Open Problem 3.2. Does there exist a convex body K R n, n 3 such that but K is not an affine image of n? P(K) = P( n ), Open Problem 3.3. Assume that convex symmetric body K is a minimizer for a volume product over the set of convex symmetric bodies in R n. Is it true that K is necessary a polytope? (see [St], [RSW] and [GM] for a related discussion) 15

16 Definition 3.4. K R n is called unconditional if for any x = (x 1, x 2,..., x n ) K we have (±x 1, ±x 2,..., ±x n ) K. Exercise Show that there is a convex symmetric body K R 2 (and thus in R n ) such that T K is not unconditional for all T GL(n). The following theorem is due to Saint-Raymond [SR]. We present here a proof due to Meyer [Me]: Theorem 3.3. Let K be an unconditional convex body in R n, then P(K) P(B n ). Proof. It is clear that the theorem (as well as the Mahler conjecture) is true in R. Indeed, then K = [ a, a] and K = [ 1/a, 1/a] for a > 0, from this we get P(K) = 4 for all origin-symmetric K R. We will prove the theorem by induction. Define K + = K R n + = {x K : x i 0, for i = 1,..., n}. Consider x K + and consider n pyramids created as the convex hull of x and intersection of K + with coordinate planes. More precisely, let K + i = conv{x, K + e i }, where e 1,... e n is the standard basis of R n. Note that the intersection of K + i has a zero volume for i j (see Figure 3). and K + j e2 K + x1 x K + 1 x2 K + 2 e1 K Figure 3. Construction of K + i. Then, using the unconditionality and convexity of K: n n K = 2 n K + 2 n K + i = 1 2n n x K e i i = 2 n 1 i=1 i=1 n i=1 ( ) 2 x i n K e i. 16

17 The above inequality can be rewritten as n ( ) 2 K e (17) x i i 1, for all x K +, and thus for all x K. n K i=1 Now (17) immediately yields the fact that the vector { } 2 K e n (18) i K. n K Applying the same argument to K we get { 2 K e i (19) n K i=1 i=1 } n i=1 K. From the definition of polarity, we know, that x y 1, for all x K and y K, thus from (18) and (19) we get: n 2 K e i (20) 2 K e i 1. n K n K Next, we notice that property (4) of Lemma 3.1 gives (21) K e i = (K e i ) = (K e i ), where the second equality follows from the fact that K is unconditional. Finally, we use (20), (21) and the inductive hypothesis to obtain K K 4 n K e n 2 i (K e i ) 4 n n 4n 1 2 (n 1)! = 4n n!. i= Local Minimum. In his blog on the Mahler conjecture T. Tao [T] asked whether the cube B n is a local minimizer. More precisely, is it possible to select δ > 0 such that if U δ (B n ) := {K convex, symmetric body : d BM (B n, K) 1 + δ}, then the Mahler conjecture is true for all K U δ (B n ). The following theorem from [NPRZ] answers his question: Theorem 3.4. Let K R n be an origin-symmetric convex body. Then P(K) P(B n ), provided that d BM (K, B n ) 1 + δ, and δ = δ(n) > 0 is small enough. Moreover, the equality holds only if d BM (K, B n ) = 1, i.e., if K is a parallelepiped. The proof of the above theorem is quite involved. We will split it into several lemmata. The first difficulty in proving local minimality of the unit cube is that there are plenty of small (linear) perturbations with the same volume product, namely all close parallelepipeds. We will overcome this difficulty by choosing a canonical representative in each class of affinely equivalent convex bodies. More precisely, we consider only the bodies K for which the unit cube is a parallelepiped of the least volume containing K. 17

18 In addition to taking care of all close parallelepipeds, it allows us to fix 2n points on the boundary of K and K (the centers of the (n 1)-dimensional faces of B n ). Lemma 3.3. Let P be a parallelepiped of minimal volume containing a convex originsymmetric body K. Let T : R n R n be a linear transformation such that P = T B n. Then T 1 K B n and ±e j T 1 K, j = 1,..., n. Proof. Note that B n is a parallelepiped of minimal volume containing T 1 K. Assume that e j T 1 K. Then it is quite easy to see that we can construct a parallelepiped of smaller volume than B, n still containing T 1 K. For this. just rotate and push towards the origin a facet of B, n containing e j. More precisely, if e j T 1 K then there exists an affine hyperplane H e j such that H T 1 K =. Note that the volume of the parallelepiped bounded by H, H, and the affine hyperplanes {x : x e i = ±1}, i j, equals B, n and that this parallelepiped still contains K. But then we can shift H and H towards K a little bit and a get a new parallelepiped of smaller volume containing K. Exercise Let S be a simplex of minimal volume containing convex (not necessary symmetric) body K. Show that all the centroids of facets of S belong to K (see [Kl] for a solution). From the statement of the theorem we see that we need to use the parallelepiped closest to K in Banach-Mazur distance, which is not necessary a parallelepiped of minimal volume. Lemma 3.5 below solves this problem. But first we will prove a technical lemma, which is quite useful for our computations: Lemma 3.4. Let Π R n be a star-shaped (with respect to the origin) polytope such that every (n 1)-dimensional face F of Π has area at least A and satisfies dist(aff(f ), 0) r, where aff(f ) denotes the minimal affine subspace containing F. Let x (1 + µ)π for some µ > 0. Then conv(π, x) Π + µra n. Proof. Let y = Π [0, x]. Let F be a face of Π containing y. Then conv(π, x) \ Π contains the pyramid with base F and apex x. The assumptions of the lemma imply that the height of this pyramid is at least µ dist(aff(f ), 0) µr, so its volume is at least µra n. Now we will show that if K is sufficiently close to B n, then K is also close to any parallelepiped of minimal volume containing K. Lemma 3.5. Let K be a convex body satisfying (22) (1 δ)b n K B n. Then there exist a constant C and a linear operator T such that and ±e i T 1 K. (1 Cδ)B n T 1 K B n, 18

19 Proof. Let as before P = T B n be a parallelepiped of minimal volume containing K. Then K P and our goal is to show that there is C > 0 such that (1 Cδ)P K. Applying (22) we see that it is enough to show that (1 Cδ)P (1 δ)b n or that there is a small κ 0 such that P (1 + κ 0 )(1 δ)b. n Note that P 2 n (indeed, K B, n so, by minimality, P B ). n On the other hand, if there exists a κ > 0 such that x P \ (1 + κ)(1 δ)b, n then, by Lemma 3.4, with Π = (1 δ)b, n r = 1 δ, A = (1 δ)b n 1 and µ = κ, P (1 δ)b n κ(1 δ) + (1 δ)b n 1 = 2 n (1 δ) n + κ 2n 1 n n (1 δ)n. The right hand side is greater than 2 n if κ > κ 0 = 2n((1 δ) n 1). Thus, P (1 + κ 0 )(1 δ)b, n and thereby (1 κ 0 )P (1 δ)b n K. It remains to note that κ 0 4n 2 δ for sufficiently small δ > 0. Thus, replacing K by its suitable linear image we may assume everywhere below that K B n, ±e j K, j = 1,..., n. Let δ > 0 be the minimal number such that (1 δ)b n K. Exercise Study the geometric and combinatorial properties of a polytope and its polar: Let F i be a face of B, n dim(f i ) = i, i = 0,..., n 1. Show that there is a face Fi of B1 n such that x y = 1 for all x F i, y Fi and dim(fi ) = n 1 i. Generalize this statement to the case of general convex polytopes P and P in R n, such that 0 is interior point of P. More precisely show that if F i is i-dimensional face of P, then F i = {y P : x y = 1, for all x F i } is n 1 i dimensional face of P. Moreover, the mapping Ψ(F i ) = Fi is one-to-one mapping between faces of P and P and Ψ(Ψ(F i )) = F i (see Chapter 3.4 in [Gru]). The next construction and the lemma will give the first idea of how we will approximate K and K by polytopes. Consider flag F = {F 0, F 1,..., F n 1 } such that F 0 F 1 F n 1 are faces of B n, dimf i = i, i.e. F 0 is a vertex (0-dimensional face) and F n 1 is a facet (n 1-dimensional face). We will denote by c(f i ) a center of face F i and by c (F i ) a center of the dual face to F i. Exercise Show that c (F i ) = (n dimf i ) 1 c(f i ). For each i 0,..., n 1, fix two vectors y i, y i having the same direction as c(f i ) such that y i y i = 1 (i.e. y i = α i c(f i ) and y i = α i c (F i ), where α i, α i > 0 and α i α i = 1) and consider two simplexes: S y (F) = conv{0, y 0,..., y n 1 } and S y (F) = conv{0, y 0,..., y n 1}. Now consider Q = S y (F) and Q = S y (F), F F where the union is taken over all 2 n n! flags of B. n 19

20 Lemma 3.6. Q Q P(B n ). Proof. We first notice that S y (F) S y (F) = 1/(n!) 3. Indeed, consider a simplex = conv{0, c(f 0 ),..., c(f n 1 )}, then = 1/n!. To get the volume of S y (F) or S y (F) we apply to a linear transformation (see the definition of those simplexes). We finish the proof of the lemma applying: l i=1 d i l i=1 Indeed, then ( ) ( ) Q Q = S y (F) S y (F) F F d 1 i l 2, for any d 1,..., d l > 0. (2n n!) 2 (n!) 3 ( ) ( ) = 1 S (n!) 3 y (F) S y (F) 1 F F = P(B n ). We are ready to start the construction of polytopes approximating K and K and having properties similar to the polytopes used in Lemma 3.6. Let us start with a simple two-dimensional case (which will not require K to be close to B 2 and actually gives yet another solution to the Mahler conjecture in R 2 ): Two dimensional case. Assume that B 2 is a parallelogram of maximal area containing K. Then from Lemma 3.3 we get that ±e 1, ±e 2 K B 2. x 1 = c(f 1 ) = e 2 x 0 S F l 0 e 1 Figure 4. Construction of polytope P. We will now construct polytopes P and P approximating K and K respectively. Consider a flag of faces F = {F 0 F 1 }, where F 1 is just a side of B 2 and F 0 is one of the vertices belonging to F 1. Define x 1 = c(f 1 ) {±e 1, ±e 2 }. At first we construct a point x 0 corresponding to F 0. Consider a tangent line l to K perpendicular to c(f 0 ). Let x 0 be a tangent point of l and K (see Figure 4). Notice that x 0 and c(f 0 ) need 20

21 not to be parallel! This is why we make things look harder and do not immediately use notation of Lemma 3.6. Let S F = conv{0, x 0, x 1 } and P = F S F, where P is the union of all (eight) simplexes defined by flags of B. 2 Notice that P K and that in the two dimensional case P is convex (this will not be the case in higher dimension!). Now we will create P K. Again, P will be the union of simplexes: as before, consider a flag of faces F of B, 2 also consider points x 0, x 1. Let x i be a dual point to x i. More precisely, x i is parallel to the normal vector to K at point x i and x i x i = 1. Thus, x i K. We also note that x 1 = x 1 {±e 1, ±e 2 } (see Figure 5). x 1 = x1 = c(f1) = e2 c(f0) x0 x 0 l 0 e1 Figure 5. Construction of polytopes P and P. Let S F = conv{0, x 0, x 1} and P = F S F. By construction, we have that P K and P K and thus P(K) P P. We could finish the proof if we would be able to apply Lemma 3.6 to P and P as to Q and Q. The only obstacle is that for each flag F points x 0 from above may not be parallel to c(f 0 )! (notice that x 0 is in the right place). The solution is to move them! This reminds the original Mahler s proof of the two-dimensional case, see Theorem 3.2. Indeed, moving those points along tangent line l will not change the volume of P. This follows from the simple observation described in Exercise 3.12: consider triangle ABC; moving vertex C along the line parallel to AB would not change the area of the triangle. In our case we take C = x 0 and A, B {±e 1, ±e 2 }, so that A, B are vertices of P adjacent to x 0 and notice that the line l is parallel to AB (See Figure 6). 21

22 y 1 = y 1 = x 1 = x 1 = c(f 1 ) = e 2 x 0 y 0 c(f 0 ) l 0 e 1 Figure 6. Moving x 0 to y 0 (triangle e 1, x 0, e 2 to triangle e 1, y 0, e 2 ). Thus, we may move x 0 to y 0 which is the point of intersection of l and [0, c(f 0 )]: y 1 = y1 = x 1 = x1 = c(f1) = e2 c(f0) x0 y0 x 0 = y 0 l 0 e1 Figure 7. Now we have Q and Q as in Lemma 3.6. Note that this move does not change the position of x 0. Finally, we apply Lemma 3.6 with P = Q and Q is P with vertices x 0 moved to y 0 (see Figure 7). This concludes the proof of two-dimensional case. The main difficulty of the case n > 2 is that S F will have more than one vertices that needs to be moved and some of them will have to be moved not exactly parallel to the base. Now, to make a move and not to change the volume of P (and P ) by much we are required to assume that K is close to B n. 22

23 General case. We first outline the plan. We use Lemma 3.5 to get that ±e 1, ±e n K B n. As before we need to select other points on the boundary of K and K and to construct two (not necessarily convex!) polytopes P K and P K such that P P P(B n ) Cδ 2, where δ is the least positive number for which (1 δ)b n K. We conclude that B n is a lower semi-stationary point for the volume product functional P. This means that the perturbation of B n by δ in the Banach-Mazur distance may result in decreasing the product volume only by δ 2, i.e., in the second order rather than in the first. Our last step is to show that either K contains a point outside (1 + cδ)p or K contains a point outside (1 + cδ)p for some small positive c. This allows us to apply Lemma 3.4 to conclude that P(K) exceeds P P by at least cδ and get the final estimate P(K) P(B n ) + cδ Cδ 2 from which the strict local minimality follows immediately. We first prove two lemmata which would allow us to move vertices of a simplex similarly as we did in the dimension 2. As before, given a set F R n, we define aff(f ) to be an affine subspace of minimal dimension containing F. Lemma 3.7. Let D R n be a compact convex set with a non-empty interior, and let dim(aff(d)) = n 1. If f(x) = conv(x, D), then f(x) is parallel to the unit normal vector n D of aff(d). Proof. We can assume that D {y R n : y n = 0} and x n > 0. Then f(x) = x n D n, D f(x) = (0,..., 0, n ). It is clear that f(x) = f(y), provided (x y) n D = 0. The next lemma will allow us to move the vertices of the simplex along an affine hyperplane almost parallel to the base opposite to the moved vertex, so that the change in the volume is of the order δ 2. Lemma 3.8. Let f be as above, and let δ > 0 be small enough. If y x = O(δ), and n = 1 is such that n n D = O(δ), (x y) n = 0, then f(x) f(y) = O(δ 2 ). Proof. By the previous lemma, f(y)/ f(y) = n D. Hence, f(x) f(y) = f(y) (n D n) (x y) + f(y) n (x y) + O( x y 2 ) = f(y) (n D n) (x y) + O( x y 2 ). Applying the Cauchy-Schwartz inequality, we get the result. Next we will define the vertices of the polytopes P K and P K and show that they can be moved to the right place. Let F be a face of B n and F B1 n be a corresponding dual face (see Exercise 3.18). As before let c F be a center of F and c F = 1 c n dimf F be a center of F. Let F = {y R n : x y = 0, for all x F }, where dim(f ) = n 1 dim(f ). Consider α F > 0 such that α F c F + F is tangent to K. We define x F to be a corresponding 23

24 tangent point. We also set y F = α F c F. Notice that from (1 δ)b n K B n we get 1 δ α F 1. Now we switch to the dual face F. The same way as above, we define points x F and yf = α F c F by replacing F, K and Bn by F, K and B1 n, respectively. In Lemma 3.9 below we will present properties of x F, y F, x F and y F which are essential for our construction, but before let us suggest the following continuation of Exercise 3.18: Exercise Let P R n be a convex polytope such that 0 is an interior point of P. Consider a face F of P and the dual face F of P. Let l(f ) be the linear subspace parallel to a face F of P, such that dim(f ) = dim(l(f )). Prove the following relations: x l(f ), for all x F and thus l(f ) F. x l(f ), for all x F and thus l(f ) (F ). F = l(f ). (F ) = l(f ). l(f ) l(f ). Lemma 3.9. Let F be a face of B n. Suppose that (1 δ)b n K B n. Then (1) (x F y F ) c F and (x F y F ) c F, (2) x F x F = 1 and y F y F = 1, (i.e. α F α F = 1), (3) x F y F = O(δ) and x F y F = O(δ). Proof. We note that x F = y F + h F and x F = y F + h F, where h F F and h F (F ). Hence, (1) follows immediately from c F F and c F F. Next we will prove (2). Consider a hyperplane H tangent to K at point x F, note that α F c F + F H. Consider the point w dual to H, i.e. such that w z = 1 for every z H. Thus, w K. Also note that α F w u = 1 for all u c F + F. Using Exercise 3.9 and the fact that c F is parallel to c F we notice that c F + (F ) = {q R n : q u = 1, u c F + F }. This implies α F w c F + (F ). Thus, w ( 1 α F c F + (F ) ) K, which gives αf = 1/α F. To prove (3) we denote by U a minimal subspace containing α F c F + F. Then dimu = dimf + 1 and B n U is a dimu-dimensional unit cube with a vertex c F. Thus, using the fact that the distance from α F c F + F to c F is O(δ), we get x F y F diam ( (α F c F + F ) B n ) = diam ( (α F c F + F ) U B n ) = O(δ). Similarly, using B n 1 K 1 1 δ Bn 1 and α F = 1/α F, we get x F y F = O(δ). Remark 3.2. Notice that the estimate in (3) from Lemma 3.9 is exactly the reason why we used F and not the tangent hyperplane in order to define x F. Consider a flag F = {F 0, F 1,..., F n 1 } of faces of B. n Since ±e j K and ±e j K (see Lemma 3.5), we will choose x Fn 1 = y Fn 1 = x F n 1 = yf n 1 = c Fn 1 = c F n 1. For each other face F F, we consider four points x F, x F, y F and yf defined in Lemma 3.9. These points induce the following four polytopes (in general, not convex!): P = F conv(0, x F0,..., x Fn 1 ) and P = F conv(0, x F 0,..., x F n 1 ), 24

25 Q = F conv(0, y F0,..., y Fn 1 ) and Q = F conv(0, y F 0,..., y F n 1 ). Thus, by Lemmata 3.9, 3.8 (applying them to each simplex in construction of P and P ) we get P Q Cδ 2 and P Q Cδ 2, whence, by Lemma 3.6 (23) P P Q Q Cδ 2 P(B n ) Cδ 2. To prove the theorem we need to show that there is a reasonable gap in volumes of K and P or in volumes of K and P, more precisely (24) K P + c δ or K P + c δ, for some c > 0. Indeed, (24) together with (23) yields P(K) P(B n ) + c δ Cδ 2 > P(B n ), provided that δ > 0 is small enough. Since K P and K P, to prove (24) we need to show that for some c > 0, either K (1 + cδ)p, or K (1 + cδ)p. Then, by Lemma 3.4, we will get (24). The conclusion of the proof: Note that at least one of the coordinates of one of the x F, where F is a vertex of B, n is at most 1 δ. Indeed, assume that all coordinates are greater then (1 δ ) in absolute value with some 0 < δ < δ. Define D = conv{x F : F is a vertex of B } n K. Consider z D, then 1 x F z for all vertices F. Choose the vertex F so that (x F ) j z j 0 for all j = 1,..., n. Then, 1 x F z (1 δ ) j z j. Thus, D (1 δ ) 1 B n 1 and D (1 δ )B n, contradicting the minimality of δ. Due to symmetry, we may assume without loss of generality that F = (1,..., 1) and that (x F ) 1 1 δ. Let us now explain the idea of the next construction. Note that, as we mentioned above, if K (1 + cδ)p, then there is a gap of volume cδ between K and P (see Figure 8). 25

26 x F n 1 = x Fn 1 = c(f n 1 ) = e n K δ x F F =(1,1,..., 1) Pyramid of volume cδ P Figure 8. K (1 + cδ)p. Thus we need to consider the case K (1 + cδ)p and to show that there is a cδ-gap between the volumes of K and P. In this case we notice that K is close to P and thus has a part of the boundary which looks like a face of P. Such a part, containing x F, will produce a dual point which belongs to K but far from P (see Figure 9). x F n 1 = x Fn 1 = c(f n 1 ) = e n K δ x F F =(1,1,..., 1) A part of K is close to the face of P This gives x K, which is far from P P Figure 9. K (1 + cδ)p. 26

27 Now we are ready to present a detailed proof. Assume that 1 K (1 + cδ)p, and thus 1 + cδ P K. We will now find a point x such that x K but x (1 + cδ)p. We will consider x to be a point close to vertex (1, 0,..., 0) of B1 n. More precisely, let x = (1 δ, c δ,..., c δ), where c = 1/(n 5). Then x (1 4 c δ)p, where c = 1/(4n 5). Indeed, it is enough to check that x x F 1 c δ for all vertices x F of P. If F (1,..., 1), then all coordinates of x F do not exceed 1 and at least one does not exceed 1/2 (otherwise, x F will be to far from F ). Thus, if δ is small enough, we get x x F (1 δ) + (n 2)c δ + c δ 2 = 1 δ + (n 3 2 )c δ = 1 c δ. If F has the vertex (1,..., 1), then x x F (1 δ) 2 + (n 1)c δ = 1 2δ + n 1 n 5 δ + δ 2 1 2δ δ + δ2 1 c δ, 4 provided that δ > 0 is small enough. Therefore if c < c, we get x (1 c δ)p cδ P K. Now we will show that we may select c small enough such that x (1 + cδ)p. Note that for every x P, we have x 1 + (1 C δ) j 2 x j 1, provided C is chosen large enough. Indeed, again it is enough to check this for the vertices x F of P. From P K 1 1 δ Bn 1 we get that (x F ) j 1 + Cδ. j 1 If c F (±1, 0,..., 0) we have (x F ) j 1/3, so j 2 (x F ) 1 + (1 C δ) j 2 (x F ) j j 1 (x F ) j C δ j 2 (x F ) j 1 + Cδ C δ 3 1, provided that C 3C. If c F = (±1, 0..., 0), then x F = ±e 1 and the inequality is trivial. Now it remains to note that x 1 + (1 C δ) j 2 x j = 1 δ + (1 C δ)(n 1)c δ = 1 + c δ C (n 1)c δ 2 > 1 + cδ, provided that c < c /2 and δ is small enough, whence x (1 + cδ)p. Remark 3.3. It was shown in Lemma 3.2 and Exercise 3.15 that Hanner polytopes have the same volume product as the cube. Recently, Jaegil Kim [K] proved that any Hanner polytope is a local minimizer for P(K). 27

28 4. Zonoids and Zonotopes Definition 4.1. A Minkowski sum of line segments in R n is called a zonotope, i.e. a symmetric convex body Z R n is a zonotope if there exists a set of segments {[ v i, v i ]} m i=1 R n, such that m Z = [ v i, v i ]. i=1 Zonoid is a limit of zonotopes in Hausdorff metric. Exercise 4.1. Show that any face of a zonotope is again a zonotope and thus must be symmetric. (Hint: See [Bl], [Sc2].) Exercise 4.2. Show that a projection of a zonotope (zonoid) is again a zonotope (zonoid). Exercise 4.3. Show that any symmetric polygon in R 2 is a zonotope (and thus any symmetric convex body in R 2 is a zonoid). We will give a Harmonic Analysis proof of this fact later, (see Corollary 5.1). Assume that α i = v i > 0 and define u i = v i / v i S n 1. Then, from the definition and the properties of the support function we get m m h Z (x) = h [ vi,v i ](x) = α i x u i. Taking the limit we obtain i=1 Lemma 4.1. Z is a zonoid iff there exists an even measure µ (called the generating measure of Z) on S n 1 such that h Z (x) = x u dµ(u). S n 1 Exercise 4.4. Check which of Bp n, p 1, are zonoids: Show that B n is a zonoid. Show that B2 n is a zonoid. Show that B1 n is not a zonotope for n 3. Show that a zonotope is a linear image of B, N for N big enough. Show that a polytope P is the polar of a zonoid if and only if it is a central section of B1 N, for N big enough. It is also true that Bp n is a zonoid for any p 2 and not a zonoid for any p [1, 2) and n 3, but this fact requires a non-trivial use of Harmonic Analysis, see [K5]. Lemma 4.1 allows us to give a quite useful formula for the volume of zonoids. Indeed, from Theorem 2.3 we get Z = 1 h Z (θ)ds(z, θ) = 1 ξ θ dµ(ξ)ds(z, θ) n S n n 1 S n 1 S n 1 = 1 ξ θ ds(z, θ)dµ(ξ) = 2 Z ξ dµ(ξ). n S n 1 S n n 1 S n 1 i=1 28