diploma_thesis_mhenze the mahler conjecture

Otto-von-Guericke-Universitat Magdeburg

Fakultat fur Mathematik

Institut fur Algebra und Geometrie

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Diplomarbeit

The Mahler Conjecture

Matthias Henze

March 26, 2008

Supervisor: Prof. Dr. Martin Henk

3

Contents

Preface 4

1 Introduction 6

2 An overview of the current state 82.1 The Blaschke-Santalo inequality . . . . . . . . . . . . . . . . . 92.2 Mahler’s proof in dimension 2 . . . . . . . . . . . . . . . . . . 14

3 Kuperberg’s “low-technology” approach 193.1 p-sums, p-intersections and p-products . . . . . . . . . . . . . 193.2 Kuperberg’s arguments . . . . . . . . . . . . . . . . . . . . . . 23

4 Hanner polytopes and beyond 284.1 Hanner polytopes . . . . . . . . . . . . . . . . . . . . . . . . . 284.2 The transference property of ×p . . . . . . . . . . . . . . . . . 29

5 Mahler’s Conjecture and zonoids 335.1 Zonotopes and zonoids . . . . . . . . . . . . . . . . . . . . . . 335.2 The Mahler volume of zonoids . . . . . . . . . . . . . . . . . . 38

6 Polytopes with at most 2n+ 2 facets 446.1 Equivalent description . . . . . . . . . . . . . . . . . . . . . . 446.2 Transformation to a finite search . . . . . . . . . . . . . . . . 45

7 Mahler volume of 1-unconditional bodies 497.1 Definition and preparation . . . . . . . . . . . . . . . . . . . . 497.2 Details of the proof . . . . . . . . . . . . . . . . . . . . . . . . 53

Index 58

References 59

4

Preface

Kurt Mahler was born in 1903 into a Jewish middle-class family in Krefeld,Germany. In his youth, he early began to read mathematical textbooks andsoon wrote small articles about the content and formulated implications ofthe material. Without knowledge of his son, Kurt Mahler’s father sent thesenotes to a mathematician at the local school. After a short while, the ma-terial got into the office of Carl Siegel, who later invited him to study inFrankfurt am Main. Mahler attended his first lectures there at the age of 20and his mathematical career started with his doctoral thesis, that was ap-proved in 1927. Due to the political situation in Germany, he had to changehis working place several times, spending years in the United Kingdom, theNetherlands and in Canberra, Australia, where he died in 1988 at the ageof 85. Mahler mainly worked in transcendental number theory, Diophantineapproximation, p-adic analysis and geometry of numbers. A detailed descrip-tion of his contributions to mathematics and more biographical backgroundis presented in an obituary by Cassels [5].

During his time in Manchester, Mahler submitted his famous work “EinUbertragunsprinzip fur konvexe Korper” [15]. In this article, Mahler raisedthe problem, that the present work is devoted to. On pages 96 and 97 hewrote

“Wir werden jetzt zeigen, daß ∆ fur jedes konvexe Korperpaar zwischenzwei positiven, allein von der Dimension n abhangigen Schranken bleibt.Wahrscheinlich lauten die exacten Ungleichungen

4n

n!5 JJ ′ 5

πn

Γ(n2

+ 1)2 ,

wobei die linke Seite zutrifft fur das polare Korperpaar von Parallelepipedund Oktaeder [..] und die rechte Seite fur die polaren Ellipsoide.”

The quantity ∆ denotes the product of the volume J of an arbitrary centrallysymmetric convex body and the volume J ′ of its polar. Mahler’s Conjectureturned out to be a persistent problem, yet no general proof is known fordimensions greater than two. Mahler [14] himself showed the validity in theplane and applied his arguments also to the non-symmetric case, where theconjectured lower bound is given by simplices. But also in this case, theproblem remains open for bodies of more than two dimensions.

However, the present work only covers the centrally symmetric case ofMahler’s Conjecture and is organised as follows. In Chapter 1 we introducethe reader to the notations and conventions from Convex Geometry that

5

we use throughout. It is followed by a survey of the current state of theknowledge. Mahler’s original proof in the plane and an investigation onthe upper bound are also discussed in Chapter 2. The third chapter dealswith an estimation to the lower bound that is the best known result, whichcan be obtained without use of deep theory. Since Hanner polytopes areconjectured to characterise the lower bound, Chapter 4 takes a detailed lookat these bodies. Additionally, we give an approach to combine the knownresults by proving a transference property of the p-product relation that isintroduced in Chapter 3. The second part of this work, that consists of theChapters 5, 6 and 7, discusses the validity of the Mahler Conjecture for theclasses of zonoids, polytopes with at most 2n+ 2 facets and 1-unconditionalbodies.

We are grateful to Prof. Dr. Martin Henk for proposing that interestingproblem as the topic of this work, for valuable suggestions and indispensablediscussions. We also thank Dr. Gennadiy Averkov for useful hints concern-ing the POV-Ray programme and his remarks on the style of writing, andfurthermore Prof. Dr. Shlomo Reisner, who provided the main literature forChapter 7.

6 1 Introduction

1 Introduction

The purpose of this introductory chapter is to provide definitions from Con-vex Geometry that will be used throughout the present work without furtherreference. Additionally, basic properties of some of these concepts are stated.

Convex bodies

A convex body in n-dimensional Euclidean space Rn is a compact convex setwith nonempty interior. We denote with Kn the set of all convex bodies inRn and with Kn0 all those being additionally centrally symmetric, i.e., forx ∈ K it also holds −x ∈ K. A centrally symmetric convex body K is incorrespondence to a norm on Rn which is given by

||x||K = min{t ≥ 0 | x ∈ tK}, x ∈ Rn.

It holds K = {x ∈ Rn | ||x||K ≤ 1}. We omit the subscript, when we consider

the Euclidean unit ball Bn, whose norm is given by ||x|| = (∑n

i=1 x2i )

12 , x ∈

Rn. Other frequently used examples of centrally symmetric convex bodies arethe unit cube Cn = [−1, 1]n and the crosspolytope C?

n = conv{±e1, . . . ,±en},where ei, 1 ≤ i ≤ n, denote the i-th unit normal vectors in Rn. The boundaryof some convex body K will be denoted with ∂K and the special case Sn−1 =∂Bn = {x ∈ Rn | ||x|| = 1} is called the unit n-sphere.

Polarity of convex bodies is of major interest in this work. Thus, wedefine the polar body of K ∈ Kn as

K? = {y ∈ Rn | xᵀy ≤ 1 for all x ∈ K}.

The polar operation is inclusion inversive, since A ⊂ B implies B? ⊂ A?.If K is centrally symmetric, we have (K?)? = K and furthermore, for anyinvertible linear transformation T on Rn it holds (TK)? = (T ᵀ)−1K?.

The family Kn is equipped with a metric. The Hausdorff distance of twoconvex bodies K1 and K2 is defined by

δ(K1, K2) = min{ε ≥ 0 | K1 ⊂ K2 + εBn and K2 ⊂ K1 + εBn}.

Two common geometric quantities of a convex body K are circumradiusand inradius, that are given by R(K) = inf{R > 0 | K ⊂ RBn} andr(K) = sup{r > 0 | rBn ⊂ K}, respectively. Additionally, the eccentricity

of K is said to be the quotient e(K) = R(K)r(K)

of these numbers.

7

Let χA denote the characteristic function of a subset A of Rn. If A is mea-surable, its volume vol(A) =

∫Rn χA(x) dx =

∫Adx is declared by the usual

Lebesgue measure on Rn. The volume of the n-dimensional Euclidean unitball is abbreviated by κn = vol(Bn).

The orthogonal complement of a linear subspace S is defined by S⊥ ={x ∈ Rn | xᵀy = 0 for all y ∈ S}. For fixed x ∈ Rn there exist unique y ∈ Sand z ∈ S⊥ such that x = y + z. The operator that maps x to y ∈ S iscalled the orthogonal projection of x onto S. The image of A ⊂ Rn underthis mapping is called the projection of A onto S and is denoted with A|S.For a centrally symmetric convex body K, we have (K|S)? = K? ∩ S.

Support and radial function

A useful concept in Convex Geometry is the support function hK : Rn → Rof a convex body K which is given by

hK(x) = maxy∈K

xᵀy, x ∈ Rn.

It is continuous and positive homogeneous of degree one, i.e., for c > 0 wehave hK(cx) = c · hK(x). The radial function ρK : Rn \ {0} → R of K is thedual concept and defined by

ρK(x) = max{c ∈ R | cx ∈ K}, x ∈ Rn \ {0}.

Support and radial function determine the convex body K and moreover, Kis centrally symmetric if and only if hK(−u) = hK(u) [ρK(−u) = ρK(u)], forall unit vectors u ∈ Sn−1. Let the origin be an interior point of K. Then,for all u ∈ Sn−1 we have ρK?(u) = 1

hK(u), which affirms the duality of both

concepts. Furthermore, if K is centrally symmetric, the following relationshold

||x||K =1

ρK(x), x ∈ Rn \ {0}, and ||y||K? = hK(y), y ∈ Rn.

These two functionals enable us to construct new bodies from a given cen-trally symmetric K ∈ Kn0 . Let n ≥ 2 and define the projection body of K asthe convex body ΠK such that

hΠK(u) = voln−1(K|u⊥), for all u ∈ Sn−1,

and the intersection body of K as the convex body IK such that

ρIK(u) = voln−1(K ∩ u⊥), for all u ∈ Sn−1.

Both constructions ΠK and IK yield centrally symmetric convex bodies,and for a detailed investigation we refer to Gardner [6].

8 2 An overview of the current state

2 An overview of the current state

Let K be a convex body in Rn that contains the origin in its interior. Thefunctional M, that maps K to its volume product M(K) = vol(K)vol(K?),is called the Mahler volume of K. This mapping M is affine invariant, i.e., forany invertible linear transformation T : Rn → Rn, we have M(TK) = M(K).This can be seen by (TK)? = (T ᵀ)−1K? and vol(TK) = | det(T )|vol(K).Moreover, for centrally symmetric K, it holds M(K) = M(K?).

In 1939, Mahler [15] proved, that if we restrict the functional M to cen-trally symmetric convex bodies K, it is bounded from below as well as fromabove by constants which only depend on the dimension n. Precisely, heshowed

4n

(n!)2≤M(K) ≤ 4n, K ∈ Kn0 . (2.1)

Furthermore, he conjectured that the sharp inequalities are given by

4n

n!≤M(K) ≤ πn

Γ(n2

+1)2, K ∈ Kn0 , (2.2)

where the lower bound is attained for the cube Cn and its polar, the crosspoly-tope C?

n, and the upper estimate holds for the Euclidean unit ball Bn. Onthe right hand side, it appears the well-known Gamma function Γ(x) =∫∞

0sx−1e−s ds, x > 0.The upper bound in (2.2) was shown by Blaschke [3] and Santalo [24], and

equality holds only for ellipsoids. Details are discussed in Section 2.1 below.However, the lower bound remains to be an open problem and is called theMahler Conjecture. There is an analogue conjecture for the general case ofconvex bodies, which are not necessary centrally symmetric. It states

M(K) ≥ (n+ 1)n+1

(n!)2, K ∈ Kn. (2.3)

It is also conjectured that the case of equality is characterised by simplicesin Rn. The current state of the art is given in [17], where (2.3) is shown tobe true for polytopes with at most n + 3 vertices. Mahler [14] verified thisinequality in dimension n = 2.

This work exclusively deals with the centrally symmetric case (2.2), andbefore we work through any details, we survey the current knowledge on

2.1 The Blaschke-Santalo inequality 9

Mahler’s Conjecture. Mahler himself gave a proof in dimension two, thatis discussed in Section 2.2, but for dimensions n ≥ 3 no general argumentwas found yet. In the literature there are two different approaches to dealwith the problem. On the one hand, special classes of bodies in Kn0 areinvestigated. Contributions in this direction were made by Saint-Raymond[23] for 1-unconditional bodies, by Reisner [21] for zonoids and by Lopezand Reisner [13] for polytopes with at most 2n + 2 facets. On the otherhand, one is interested in a general estimation of M(K) from below up to acertain factor. Those estimates are either given in terms of M(Cn) or M(Bn).Mahler’s Conjecture says

M(K) ≥M(Cn) and M(K) ≥ mnM(Bn), K ∈ Kn0 , (2.4)

where m = 4π

(Γ( n

2+1)2

n!

) 1n

tends to 2π

from above, when n goes to infinity.

Mahler initiated this approach, since (2.1) yields M(K) ≥ 1n!

M(Cn), K ∈ Kn0 .In 1987, Bourgain and Milman [4] obtained

2.1 Theorem (Bourgain, Milman [4]).There is a universally constant c > 0 such that for every body K ∈ Kn0

M(K) ≥ cnM(Bn).

The original arguments heavily rely on the local theory of Banach spaces. Asimplified proof can be found in Chapter 7 of Pisier’s book [18]. Kuperberg[12] used more elementary geometric reasoning to get a weaker result thanthe Bourgain-Milman Theorem, which states that the constant m in (2.4)can be chosen as 1

log2 n, for dimensions greater than or equal to four. The

best known lower bound on M(K) is also due to Kuperberg [11]. In 2006,he published a paper with the following outcome

2.2 Theorem (Kuperberg [11]).For any centrally symmetric convex body K it holds

M(K) ≥(π

4

)nγnM(Cn),

where γn ≥ 4π

only depends on the dimension and tends to√

2, when n goesto infinity.

2.1 The Blaschke-Santalo inequality

This section deals with the upper bound in (2.2). In 1923, it was shown byBlaschke [3], that it is true in dimensions n = 2, 3, and 26 years later Santalo


[24] found a proof for arbitrary dimensions. The literature contains severaldifferent approaches on the inequality (see [16], [23]). These two papers alsotreat the case of equality, which is characterised by ellipsoids.

What follows, is a sketch of the proof by Meyer and Pajor [16], who utiliseSteiner symmetrisations.

2.3 Definition (Steiner symmetrisation).Let K ∈ Kn and let H ⊂ Rn be a hyperplane. The set

stH(K) =

{x+

1

2(v − w)

∣∣∣∣ x ∈ K|H, x+ v, x+ w ∈ K}

is called the Steiner symmetrisation of K with respect to H.

Figure 1 illustrates the Steiner symmetrisation of a polygon with respect toH = {(x, y) ∈ R2 | y = 0}.

K

stH(K)x

y

Figure 1: Steiner symmetrisation of K

Two essential properties of stH(K) are given in the following proposition.

2.4 Proposition. Let K ∈ Kn and let H ⊂ Rn be a hyperplane. Steinersymmetrisation does not change the volume, i.e., it holds

vol(stH(K)) = vol(K).

Furthermore, stH(·) is a continuous function on {K ∈ Kn | dimK = n}.

2.5 Lemma. Let K ∈ Kn with dimK = n. Furthermore, write SK for thefamily of all convex bodies, that can be derived of K by finitely many Steinersymmetrisations with respect to hyperplanes, that contain the origin. Then


there is a sequence (Ki)i∈N ⊂ SK that converges to

(vol(K)

vol(Bn)

) 1n

Bn, with

respect to the Hausdorff metric.

Proof. For L ∈ Kn, define ρ(L) = min{γ ≥ 0 | L ⊂ γBn} and let ρ bethe infimum of ρ(S) taken over all S ∈ SK . Thus, there exists a sequence(Si)i∈N ⊂ SK , such that ρ(Si) tends to ρ, when i tends to infinity. Due tothe fact that we only consider Steiner symmetrisations of K with respect tohyperplanes that contain the origin, and since, taking the Steiner symmetraldoes not increase the diameter of K, we have ρ(S) ≤ ρ(K) for all S ∈ SK .Therefore, all the bodies S ∈ SK are contained in ρ(K)Bn. Under theseconditions, Blaschke’s selection theorem (see [25], Theorem 1.8.6) yields aconvergent subsequence of (Si)i∈N. Without loss of generality, let Si tend toM , for i→∞. Note, that ρ(·) is continuous on SK and so we have ρ(M) = ρ.This means, that M ⊂ ρBn.

Suppose, that we have M 6= ρBn. Then there is an x0 ∈ ρSn−1\M . Since,M is compact, we can find by the separation theorem (see [10], Chapter 3)a strict separating hyperplane H(a, α) = {x ∈ Rn | aᵀx = α} of M andx0 with aᵀx0 > α and aᵀx ≤ α, for all x ∈ M . Next, we consider forv ∈ ρSn−1 the spherical cap around v with “radius” α, which is given byC(v, α) = {x ∈ ρSn−1 | vᵀx > α}. Since, Sn−1 is compact, we find pointsx1, . . . , xm ∈ ρSn−1 such that

ρSn−1 =m⋃i=0

C(xi, α).

Now, write Hi = H(xi − x0, 0), for 1 ≤ i ≤ m, and Cj = C(xj, α), for0 ≤ j ≤ m. An important observation is, that the cap Ci is the reflectionof C0 at Hi. To see this, it is sufficient to show, that xi is mapped to x0 bythat reflection Ri. We have

Ri(xi) = xi −2(xi − x0)ᵀxi

(xi − x0)ᵀ(xi − x0)(xi − x0) = xi − (xi − x0) = x0,

because 2(xi−x0)ᵀxi

(xi−x0)ᵀ(xi−x0)=

2xᵀi xi−2xᵀ

0xi

xᵀi xi−2xᵀ

0xi+xᵀ0x0

= 1, since x0 and xi are of the same

length. By definition of C0, we get C0 ∩M = ∅ and therefore, stH1(M) ∩(C0 ∪C1) = ∅. Similarly, we obtain stH2(stH1(M)) ∩ (C0 ∪C1 ∪C2) = ∅ andinductively

M = stHm(stHm−1(. . . (stH1(M)) . . .)) ∩ ρSn−1 = ∅.

This means that M ⊂ int(ρBn). Let us now denote

Si = stHm(stHm−1(. . . (stH1(Si)) . . .)).


Then, clearly Si ∈ SK and hence ρ(Si) ≥ ρ. The continuity of the Steinersymmetrisation (see Proposition 2.4) yields, that Si converges to M , since Siconverges to M , for i tending to infinity. But this creates a contradiction,because we have ρ(Si) → ρ(M) < ρ. Thus, we must drop the assumptionand get M = ρBn.

All bodies in SK have the same volume, as Steiner symmetrisation doesnot change it. So, we are done, because vol(K) = vol(M) = ρnvol(Bn), orequivalently,

ρ =

(vol(K)

vol(Bn)

) 1n

.

�

2.6 Lemma. Let K ∈ Kn0 with dimK = n and let H ⊂ Rn be a hyperplane,that contains the origin. Then

vol((stH(K))?) ≥ vol(K?).

Proof. First of all, for all affine transformations A : Rn → Rn we have

vol((AK)?) = vol(A−ᵀK?) = | detA|−1vol(K?).

Since, H contains the origin, we can therefore find a suitable rotation A(| detA| = 1) of K, to assume that H = {x ∈ Rn | xn = 0}. For this specialhyperplane H we get

stH(K) =

{(x, s)ᵀ

∣∣∣∣ x ∈ K|H, s =1

2(a− b), (x, a)ᵀ, (x, b)ᵀ ∈ K

}and

K? = {(y, t)ᵀ | xᵀy + st ≤ 1, x ∈ K|H, (x, s)ᵀ ∈ K}.Combining these two identities yields

(stH(K))? =

{(y, t)ᵀ

∣∣∣∣ xᵀy +1

2(a− b)t ≤ 1, x ∈ K|H, (x, a)ᵀ, (x, b)ᵀ ∈ K

}.

Next, for A ⊂ Rn and t ∈ R, let A(t) = {x ∈ Rn−1 | (x, t)ᵀ ∈ A}.Claim: For t ∈ R it holds 1

2(K?(t) +K?(−t)) ⊂ (stH(K))?(t).

To see this, pick some fixed y1 ∈ K?(t) and y2 ∈ K?(−t). Then, for all(x, a)ᵀ, (x, b)ᵀ ∈ K we have(

x,1

2(a− b)

)ᵀ(1

2y1 +

1

2y2, t

)=

1

2xᵀy1 +

1

2xᵀy2 +

1

2at− 1

2bt

=1

2(xᵀy1 + at︸︷︷︸

≤1

) +1

2(xᵀy2 + b(−t)︸︷︷︸

≤1

)

≤ 1.


Therefore, by the above identity, this implies 12(y1 + y2) ∈ (stH(K))?(t).

Since, K ∈ Kn0 , the polar K? is also centrally symmetric, which leads toK?(t) = −K?(−t). Thus, we obtain voln−1(K?(t)) = voln−1(K?(−t)) andan application of the famous Brunn-Minkowski inequality (see [25], Theorem6.1.1), yields, for all real t,

voln−1((stH(K))?(t)) ≥ voln−1

(1

2K?(t) +

1

2K?(−t)

)≥

(1

2voln−1(K?(t))

1n−1 +

1

2voln−1(K?(−t))

1n−1

)n−1

= voln−1(K?(t)).

Eventually, we use Cavalieri’s principle and derive

vol((stH(K))?) =

∫ ∞−∞

voln−1((stH(K))?(t)) dt

≥∫ ∞−∞

voln−1(K?(t)) dt = vol(K?),

and the lemma is shown. �

We are now well-prepared to prove the main theorem.

2.7 Theorem (Blaschke [3], Santalo [24]).Let K be a centrally symmetric convex body in Rn. Then M(K) ≤M(Bn).

Proof. As a consequence of Lemma 2.5, we can derive, that there is a sequenceof hyperplanes Hi, i ∈ N, such that

Ki := stHi. . . stH1(K)→

(vol(K)

vol(Bn)

) 1n

Bn.

This yields that the “polar sequence” (K?i )i∈N converges to

(vol(K)

vol(Bn)

)− 1n

Bn.

By Proposition 2.4 and Lemma 2.6, we have

M(K) ≤M(K1) ≤ . . . ≤M(Ki),

for all i ∈ N. The continuity of the Mahler volume, together with the abovestatements, imply that this sequence of inequalities converges to M(Bn). �


2.2 Mahler’s proof in dimension 2

In a previous work to his famous paper “Ein Ubertragungsprinzip fur kon-vexe Korper” [15], Kurt Mahler showed in [14], that in dimension two thefunctional M(·) on K2

0 is bounded from below by the constant 8 = 42

2!. Prob-

ably, this result was the main aspect that lead him to state his conjectureafterwards.

Preparatory, we have to introduce a definition for polarity, that Mahlerused in his proof and which differs from the one we denote with ? in thiswork. A vector p ∈ R2 \ {0} is said to be polar to the line lp = {x ∈R2 | p1x1 + p2x2 = 1} and vice versa. Consider a convex polygon P =conv{x1, . . . , xm} and collect for all edges [xi, xj] of P the polar point vij tothe line through xi and xj. The polar set P ′ of P is given by the convex hullof {vij | [xi, xj] is an edge of P}.

1

1

T

x1

x2

x3

T ′

x′1

x′2

x′3

x

y

Figure 2: Triangle and its polar

If P contains the origin in its interior, it holds P ′ = P ?. But if 0 /∈P , then P ? is unbounded, while P ′ is surely not, being the convex hullof finitely many points. Figure 2 stresses that difference by illustrating atriangle T = conv{x1, x2, x3}, which does not contain the origin, and itspolar T ′ = conv{x′1, x′2, x′3}. The dashed lines enlighten the fact that say x′1lies on the line through the origin perpendicular to the edge [x1, x2] and the

2.2 Mahler’s proof in dimension 2 15

section of the unit circle shows that the length of x′1 is given by the reciprocalof the distance from [x1, x2] to 0.

Next, we write for x, y ∈ R2

dxy = det

(x1 y1

x2 y2

)= x1y2 − x2y1.

The following proposition relates the area of a triangle with that of its polar.

2.8 Proposition. Let T = conv{x, y, z} be a triangle in R2 and let T ′ =conv{x′, y′, z′} be its polar. Furthermore, assume that no two of x, y and zare linearly dependent. Then, it holds

vol(T ′) =2vol(T )2

dxydyzdzx. (2.5)

Proof. First of all, note that the assumption on say x and y not to be linearlydependent prevents dxy from being zero. Now, x′ is given as the intersectionpoint of the lines lx and ly polar to x and y. Therefore, by solving a system

of two linear equations we obtain x′ = 1dxy

(y2 − x2

x1 − y1

). Similarly, we have

y′ = 1dyz

(z2 − y2

y1 − z1

)and z′ = 1

dzx

(x2 − z2

z1 − x1

). Elementary calculations

yield, that the area of T ′ is given by

vol(T ′) =1

2det(y′ − x′, z′ − x′) =

1

2(dx′y′ + dy′z′ + dz′x′)

=1

2

(1

dxydyz

∣∣∣∣ y2 − x2 z2 − y2

x1 − y1 y1 − z1

∣∣∣∣+1

dyzdzx

∣∣∣∣ z2 − y2 x2 − z2

y1 − z1 z1 − x1

∣∣∣∣+

1

dzxdxy

∣∣∣∣ x2 − z2 y2 − x2

z1 − x1 x1 − y1

∣∣∣∣ )=

1

2dxydyzdzx

(d2xy + d2

yz + d2zx + 2dxydyz + 2dyzdzx + 2dzxdxy

)=

(dxy + dyz + dzx)2

2dxydyzdzx=

2vol(T )2

dxydyzdzx.

�

Let no two of x, y and z be linearly dependent. Suppose, that 0 /∈ T =conv{x, y, z}. Then, dxy, dyz, dzx are not all of the same sign, so we canassume without loss of generality, that dxy, dyz > 0 and dzx < 0. Next, fixx, z and vol(T ). The equation

dxy + dyz − |dzx| = 2vol(T )


defines a line g parallel to [x, z] from which we pick y arbitrarily. Due tothe positivity of dxy, we get all points on g, when 0 < dxy < 2vol(T ) + |dzx|.Bounding dxy by 0 < α ≤ dxy ≤ β < 2vol(T ) + |dzx| yields a stretch S of g.By equation (2.5) we get

vol(T ′) =2vol(T )2

dxy(2vol(T ) + |dzx| − dxy)|dzx|.

Thus, vol(T ′) depends continuously on y ∈ S and attains a maximum onthat stretch. Furthermore, we have(

vol(T ) +|dzx|

2

)2

−(

vol(T ) +|dzx|

2− dxy

)2

=

(vol(T ) +

|dzx|2

)2

−

((vol(T ) +

|dzx|2

)2

− 2dxy

(vol(T ) +

|dzx|2

)+ d2

xy

)

= dxy(2vol(T ) + |dzx| − dxy) =2vol(T )2

|dzx|vol(T ′).

The left hand side attains a maximum at dxy = vol(T ) + |dzx|2

and decreasesthereof strongly monotonically in both directions. So, vol(T ′) is maximal,when y is a boundary point of S. Precisely,

◦ if vol(T ) + |dzx|2

< α, the maximum is attained at dxy = β,

◦ if vol(T ) + |dzx|2

> β, the maximum is attained at dxy = α,

◦ if α ≤ vol(T ) + |dzx|2≤ β, the maximum is attained at dxy ∈ {α, β},

depending on which point the area of T ′ is bigger.

The subsequent lemma contains the crucial construction, from that the proofwill be obtained.

2.9 Lemma. Let P ∈ K20 be a polygon with 2m ≥ 6 vertices. Then, there

exists a polygon Q ∈ K20 with 2(m− 1) vertices, such that

vol(Q)vol(Q′) < vol(P )vol(P ′).

Proof. Let v1, . . . , vm,−v1, . . . ,−vm be the clockwise ordered vertices ofP and let P ′ = conv{±v′1, . . . ,±v′m}, where ±v′i corresponds to the edge[±vi,±vi+1], for 1 ≤ i ≤ m− 1, and ±v′m corresponds to [±vm,∓v1].

Due to the assumption, that 2m ≥ 6, we can find a denotation thatthe triangle T = conv{v1, v2, v3} does not contain the origin. Addition-ally, let R = conv{±v1,±v3, . . . ,±vm}, so we have P = R ∪ T ∪ −T .

2.2 Mahler’s proof in dimension 2 17

−v6

v1

v2v3 v4

v5

v6

−v1

−v2−v3−v4−v5

R

R′

T

−T

T ′

−T ′

−v′6

v′1

w′

v′2

v′3v′4

v′5

v′6

−v′1

−w′

−v′2−v′3

−v′4

−v′5

x

y

Figure 3: Cutting the vertices ±v2

The polar triangles T ′ = conv{v′1, w′, v′2} and −T ′ are contained in R′ =conv{±w′,±v′3, . . . ,±v′m} and it holds P ′ = R′ \ (T ′ ∪ −T ′). The just de-scribed situation is depicted in Figure 3.

Consider the line g through v2 parallel to the edge [v1, v3]. By extendingthe edges [−vm, v1] and [v3, v4] towards g we resect a stretch S of that line. Asshown in Figure 4, we can move v2 arbitrarily on S and will always maintainthe convexity of P and its volume, since the volume of T would not change.Let us call T (v) = conv{v1, v, v3}, for v ∈ S. By the considerations afterProposition 2.8, we choose v as the boundary point of S such that the areaof T (v)′ becomes maximal over all v ∈ S.

Now, write Q = conv{±v1,±v,±v3, . . . ,±vm} and we get vol(Q) =vol(P ) and

vol(Q′) = vol(P ′) + vol(T ′)− vol(T (v)′) < vol(P ′).

Notice, that Q indeed has 2(m−1) vertices, since the choice of v ensures thatit lays on the extension of either [−vm, v1] or [v3, v4]. In conclusion, we thushave found the desired polygon fulfilling vol(Q)vol(Q′) < vol(P )vol(P ′). �


−vm

v1

v2Sv

v3

v4

P

T

g

Figure 4: Varying v2 on S

2.10 Theorem (Mahler [14]).Let P ∈ K2

0 be a polygon. Then we have M(P ) ≥ 8 and equality is attainedif and only if P is a parallelogram.

Proof. We proceed by induction on the number of vertices of P . Let’s makea start with P = conv{v, w,−v,−w}, i.e., P is a parallelogram. SubdividingP into four triangles ∆(0, v, w),∆(0, w,−v),∆(0,−v,−w) and ∆(0,−w, v),which are all of the same area, yields

vol(P ) =1

2dvw +

1

2dw,−v +

1

2d−v,−w +

1

2d−w,v = 2dvw.

Note, that we do not have to distinguish P ′ from P ? here, since 0 ∈ int(P ).Similarly to the reasoning in the proof of Proposition 2.8, we obtain P ? =

conv{±v′,±w′}, where v′ = 1dvw

(w2 − v2

v1 − w1

)and w′ = 1

dvw

(−v2 − w2

v1 + w1

),

which implies vol(P ?) = 4dvw

and therefore M(P ) = 8.For the induction step let P = conv{±v1, . . . ,±vm} with m ≥ 3. Lemma

2.9 states the existence of a polygon Q with 2(m− 1) vertices and

M(P ) > M(Q) ≥ 8.

The latter comes from the induction hypothesis. And the strict inequalityalso yields the characterisation of the case of equality. �

Mahler also remarked, that for an arbitrary centrally symmetric convex bodyK in R2 we can find a sequence of polygons in K2

0 that converges to K. Since,the Mahler volume is a continuous functional on K2

0, we also have M(K) ≥ 8,but the characterisation of the equality case is lost.

19

3 Kuperberg’s “low-technology” approach

In this chapter we deal with a work by Kuperberg [12] from 1992. He provesa weaker result than Theorem 2.1, but the arguments are more elementary.

3.1 Theorem (Kuperberg [12]).Let n ≥ 4 and K ∈ Kn0 . Then

M(K) ≥ (log2 n)−nM(Bn).

In order to get this result, Kuperberg considers special constructions of con-vex bodies that turn out to be very useful. The definitions and importantproperties of those convex sets are given in the subsequent part. Afterwards,the details of the arguments are discussed.

3.1 p-sums, p-intersections and p-products

In the following, we declare that 1∞ = 0.

3.2 Definition (p-sum, p-intersection, p-product).Let K,L ∈ Kn0 , K ′ ∈ Km0 and 1 ≤ p ≤ ∞. We define the p-sum set of K andL as

K +p L :=⋃

0≤t≤1

((1− t)

1pK + t

1pL),

the p-intersection set of K and L as

K ∩p L :=⋃

0≤t≤1

((1− t)

1pK ∩ t

1pL),

and the p-product set of K and K ′ as

K ×p K ′ :=⋃

0≤t≤1

((1− t)

1pK × t

1pK ′

).

The above constructions are generalisations of very common operations, itholds K +∞ L = K + L, K ∩∞ L = K ∩ L, K ×∞ K ′ = K × K ′ andK +1 L = conv{K ∪ L}.

3.3 Proposition. The operations +p, ∩p and ×p are associative and theresult is always centrally symmetric.

20 3 Kuperberg’s “low-technology” approach

Proof. The argument to prove the associativity is independent on the op-eration +p, ∩p and ×p and it therefore suffices to show for p ≥ 1 andK,L,M ∈ Kn0 that (K +p L) +pM = K +p (L+pM). We have

(K +p L) +pM =⋃

0≤t≤1

((1− t)

1p

⋃0≤s≤1

((1− s)

1pK + s

1pL)

+ t1pM

)=

⋃0≤t≤1

⋃0≤s≤1

((1− t)

1p (1− s)

1pK + (1− t)

1p s

1pL+ t

1pM).

Now, we consider the following substitutions:

t = s+ t− st and s =t

s+ t− stfor s+ t 6= 0 and s = 1 if s = t = 0.

The maps (t, s) 7→ t and (t, s) 7→ s are both surjective on [0, 1]× [0, 1] and itholds 1− t = (1− t)(1− s), t(1− s) = (1− t)s and ts = t. This leads to

(K +p L) +pM =⋃

0≤t≤1

⋃0≤s≤1

((1− t)

1pK + t

1p (1− s)

1pL+ t

1p s

1pM)

=⋃

0≤t≤1

((1− t)

1pK + t

1p

⋃0≤s≤1

((1− s)

1pL+ s

1pM))

= K +p (L+pM).

The property that the three described constructions always lead to centrallysymmetric sets simply inherits from the assumption on the original bodiesto be centrally symmetric. �

3.4 Proposition. Let K,L ∈ Kn0 , K ′ ∈ Km0 and 1 ≤ p <∞. Then

i) K +p L = {x+ y | x, y ∈ Rn and ||x||pK + ||y||pL ≤ 1}.

ii) For x ∈ Rn we have ||x||pK∩pL= ||x||pK + ||x||pL.

iii) For (x, y) ∈ Rn × Rm we have ||(x, y)||pK×pK′= ||x||pK + ||y||pK′ .

iv) By identifying K with K × {0}m ∈ Kn+m0 and K ′ with {0}n × K ′ ∈

Kn+m0 , we have K ×p K ′ = K +p K

′ ⊂ Rn × Rm.

Proof. i) To see the inclusion from left to right, let w = (1 − t)1px + t

1py ∈

K +p L, where t ∈ [0, 1], x ∈ K and y ∈ L. We get

||(1− t)1px||pK + ||t

1py||pL = (1− t)||x||pK + t||y||pL ≤ 1.

3.1 p-sums, p-intersections and p-products 21

Conversely, let x, y ∈ Rn, s = ||x||K and t = ||y||L such that sp + tp ≤ 1.Therefore, since s, t ≤ 1, there are x ∈ K and y ∈ L with x = sx and y = ty.Without loss of generality we assume, that s is not equal to 1, since otherwiset would be zero and x+ y = sx ∈ K ⊂ K +p L. We obtain

x+ y = sx+ ty = sx+ (1− sp)1p (t(1− sp)−

1p y︸︷︷︸

=:y∈L

) = r1p x+ (1− r)

1p y,

where r = sp ∈ [0, 1]. This means, that x+ y ∈ K +p L.ii) We have to show that K ∩p L = {x ∈ Rn | ||x||pK + ||x||pL ≤ 1} =: D.

On the one hand, for x = (1− t)1py = t

1p z ∈ K ∩pL, it holds ||x||K ≤ (1− t)

1p

and ||x||L ≤ t1p . Thus, ||x||pK + ||x||pL ≤ 1− t+ t = 1 and x ∈ D. On the other

hand, let x ∈ D and s = ||x||K , t = ||x||L, so sp + tp ≤ 1. Therefore, thereexist y ∈ K and z ∈ L such that x = sy = tz. Analoguesly to i) we define

r = sp and z = t(1− sp)1p z ∈ L and get x = r

1py = (1− r)

1p z ∈ K ∩p L.

The statement iii) can be proven similarly to the second one and iv)directly follows from the definitions. �

Figure 5: C?3 ∩3 C3 and C1 ×1.2 C1 ×4 C1

Since, we wish to investigate the Mahler volume of sets coming from p-constructions, it is inevitable to study the relation to the polar operation,which is stated without proof.

3.5 Lemma. Let K,L ∈ Kn0 , K ′ ∈ Km0 and 1 ≤ p, q ≤ ∞ with 1p

+ 1q

= 1.Then the following relations hold.

i) (K +p L)? = K? ∩q L?, ii) (K ∩p L)? = K? +q L?,

iii) (K ×p K ′)? = K? ×q (K ′)?.


3.6 Remark. Proposition 3.4 and the preceding lemma show that p-sums,p-intersections and p-products of centrally symmetric convex bodies alwayscorrespond to a norm, which means that the outcome of such constructionsis always in Kn0 .

We utilise the properties above to compute some examples for bodies thatcome from p-constructions. The pictures in Figure 5 were rendered withPOV-Ray for Windows v3.5 [19].

The following formula enables us to derive the volume of a p-product setfrom that of the involved bodies.

3.7 Lemma. For K ∈ Kn0 and 1 ≤ p <∞ we have∫Rn

e−||x||pKdx = Γ(n

p+1)vol(K).

Proof. The statement is shown by the following computation.∫Rn

e−||x||pKdx =

∫Rn

(−e−s

∣∣∣∞||x|pK

)dx =

∫Rn

∫p√s>0:x∈ p√sK

e−sds dx

=

∫Rn

∫t>0:x∈tK

ptp−1e−tp

dt dx

(Fubini) =

∫ ∞0

∫tK

ptp−1e−tp

dx dt

=

∫ ∞0

ptp−1e−tp

vol(tK) dt = vol(K)

∫ ∞0

ptn+p−1e−tp

dt

= vol(K)

∫ ∞0

snp e−sds = Γ(n

p+1)vol(K).

�

3.8 Theorem (Product formula).Let K ∈ Kn0 , K ′ ∈ Km0 and 1 ≤ p ≤ ∞. It holds

vol(K ×p K ′) =Γ(n

p+1)vol(K)Γ(m

p+1)vol(K ′)

Γ(n+mp

+1).

Proof. For the case p =∞, we use that K ×∞ K ′ = K ×K ′. So, we get

vol(K ×∞ K ′) = vol(K ×K ′) = vol(K)vol(K ′).

Assume that 1 ≤ p <∞. By combining Lemma 3.7 with the representationof the norm of K ×p K ′ in Proposition 3.4 iii) we obtain

Γ(n+mp

+1)vol(K ×p K ′) =

∫Rn×Rm

e−||(x,y)||p

K×pK′d(x, y)

3.2 Kuperberg’s arguments 23

=

∫Rn×Rm

e−||x||pK−||y||

p

K′d(x, y)

=

(∫Rn

e−||x||pKdx

)(∫Rm

e−||y||p

K′dy

)= Γ(n

p+1)vol(K)Γ(m

p+1)vol(K ′),

and the product formula follows. �

3.2 Kuperberg’s arguments

We start by stating formulas for the Mahler volume of p-products. They arededuced with the help of Lemma 3.5, Theorem 3.8 and the associativity ofthe ×p relation.

3.9 Proposition. Let Ki ∈ Kni0 , 1 ≤ i ≤ s, n = n1 + . . . + ns and further-

more 1 ≤ p, q ≤ ∞ such that 1p

+ 1q

= 1. Then, K1×pK2 and K1×qK2 havethe same Mahler volume and it holds

M(K1 ×p K2) =Γ(n1

p+1)Γ(n1

q+1)Γ(n2

p+1)Γ(n2

q+1)

Γ(n1+n2p

+1)Γ(n1+n2q

+1)M(K1)M(K2)

and

M(K1 ×p . . .×p Ks) =

∏si=1 Γ(ni

p+1)Γ(ni

q+1)M(Ki)

Γ(np

+1)Γ(nq

+1).

Examples: Consider the interval I = [−1, 1]. We have I? = I and therefore

M(I) = 4. For 1 ≤ p < ∞ the p-norm is defined by ||x||p =

(n∑i=1

|xi|p) 1

p

and for p = ∞ by ||x||∞ = max1≤i≤n

|xi|, x ∈ Rn. We denote the unit ball of

the p-norm with Bnp = {x ∈ Rn | ||x||p ≤ 1}. In Proposition 3.4 iii) we have

already seen that for K ∈ Kn0 and K ′ ∈ Km0 the norm of K ×p K ′ is relatedto that of K and K ′ by

||(x, y)||pK×pK′= ||x||pK + ||y||pK′ , (x, y) ∈ Rn × Rm for 1 ≤ p <∞.

This can be extended to the p-product of more than two convex bodies. ForKi ∈ Kni

0 , 1 ≤ i ≤ s, it holds

||(x1, . . . , xs)||pK1×p...×pKs=

s∑i=1

||xi||pKi, (x1, . . . , xs) ∈ Rn1 × . . .× Rns .


Now, we have ||x||I = |x|, x ∈ R, which leads to

||(x1, . . . , xn)||pI×p...×pI=

n∑i=1

|xi|p = ||x||pp, x ∈ Rn.

This ends in a very useful description of the unit balls of the p-norm:

Bnp = I ×p . . .×p I, for 1 ≤ p <∞.

Additionally, we also have Bn∞ = (Bn

1 )? = (I ×1 . . .×1 I)? = I ×∞ . . .×∞ I.With help of Proposition 3.9 we now compute the Mahler volume of the

Euclidean unit ball Bn = Bn2 , the cube Cn = Bn

∞ and the crosspolytopeC?n = Bn

1 . We have

M(Bn) = M(I ×2 . . .×2 I) =πn

Γ(n2

+1)2

and

M(Cn) = M(C?n) = M(I ×1 . . .×1 I) =

4n

Γ(n+ 1)=

4n

n!.

Note, that we used some properties of the Gamma function, i.e., Γ(12) =√

π, Γ(n+ 1) = n!, for n ∈ N0, and Γ(x+ 1) = xΓ(x), for x > 0.

3.10 Proposition. For every centrally symmetric convex body K ∈ Kn0 wehave

i) e(K)−nM(Bn) ≤M(K) ≤ e(K)nM(Bn) and

ii) n−n2 M(Bn) ≤M(K) ≤ n

n2 M(Bn).

Proof. i) The definitions of in- and circumradius of K yield

r(K)Bn ⊂ K ⊂ R(K)Bn

and by taking polars we get

1

R(K)Bn ⊂ K? ⊂ 1

r(K)Bn.

This leads to

M(K) = vol(K)vol(K?) ≥ vol(r(K)Bn)vol(1

R(K)Bn) = e(K)−nM(Bn)

and analoguesly, we obtain M(K) ≤ e(K)nM(Bn).ii) By John’s Theorem (see [6], Theorem 4.2.12) there is an invertible lineartransformation T : Rn → Rn such that Bn ⊂ TK ⊂

√nBn and therefore, it

holds e(TK) ≤√n. Due to the affine invariance of the Mahler volume this

and i) yield the desired inequalities. �


3.11 Lemma. For any body K ∈ Kn0 we have M(K) ≥ 2−nM(K ∩2 K?).

Proof. The key for the proof will be to investigate the body C := T (K×2K?),

where T : R2n → R2n is the linear map defined by T (x, y) := (x, x+y), x, y ∈Rn, which has determinant 1. By using the product formula for the volumeof p-products (see Theorem 3.8), we compute the volume of C as

vol(C) = vol(K ×2 K?) =

Γ(n2

+1)vol(K)Γ(n2

+1)vol(K?)

Γ(n+ 1)=

M(K)(nn2

) . (3.1)

Note, that we adopt the notation of the binomial coefficients for arbitraryreals by writing

(rs

)= Γ(r+1)

Γ(s+1)Γ(r−s+1), for 0 ≤ s ≤ r.

Next, we consider the projection map π : Rn × Rn → Rn, π(x, y) = y.Claim 1: i) C ∩ π−1(0) = (K ∩2 K

?)× {0} and ii) π(C) = K +2 K?.

These two statements are consequences of Proposition 3.4. It holds

T (K ×2 K?) ∩ π−1(0) = {(x, x+ y) | ||x||2K + ||y||2K? ≤ 1, x+ y = 0}

= {(x, 0) | ||x||2K + ||x||2K? ≤ 1}= (K ∩2 K

?)× {0}

and

π(T (K ×2 K?)) = π

({(x, x+ y) | ||x||2K + ||y||2K? ≤ 1}

)= {x+ y | ||x||2K + ||y||2K? ≤ 1}= K +2 K

?.

Now, we identify a dilated copy of C ∩ π−1(0) in every affine sectionC ∩ π−1(y), y ∈ π(C), of C.Claim 2: For all t ∈ [0, 1] and y ∈ tπ(C) there exists a z ∈ Rn×Rn such thatz + (1− t)(C ∩ π−1(0)) ⊂ C ∩ π−1(y).

For fixed t ∈ [0, 1] and (x, y) ∈ C, let y = tπ(x, y) = ty. Now, for anarbitrary (x, 0) ∈ C ∩ π−1(0) we have t(x, y) + (1 − t)(x, 0) ∈ C, since C isconvex. Moreover, t(x, y) + (1− t)(x, 0) = (tx+ (1− t)x, y) ∈ π−1(y), whichverifies the claim with z = t(x, y).

Having these relations, we can estimate the volume of C in another way.The second equality comes from Cavalieri’s principle.

vol(C) =

∫C

d(x, y) =

∫π(C)

(∫C∩π−1(y)

dx

)dy

=

∫ 1

0

∫π(C)

∫C∩π−1(y)

dx dy dt


≥∫ 1

0

∫tπ(C)

∫C∩π−1(y)

dx dy dt

(Claim 2) ≥∫ 1

0

∫tπ(C)

∫(1−t)(C∩π−1(0))

dx dy dt

= vol(π(C)×1 (C ∩ π−1(0)))

(Claim 1) = vol((K +2 K?)×1 (K ∩2 K

? × {0}))= vol((K +2 K

?)×1 (K ∩2 K?))

(Theorem 3.8) =Γ(n+ 1)2

Γ(2n+ 1)vol(K +2 K

?)vol(K ∩2 K?)

(Lemma 3.5) =vol(K ∩2 K

?)vol((K ∩2 K?)?)(

2nn

) =M(K ∩2 K

?)(2nn

) .

When we look back to our first computation (3.1) of vol(C), the lemma willbe shown as long as we have seen that(

2n

n

)≤ 2n

(nn2

), n ∈ N.

It is well known that(

2nn

)=∑n

k=0

(nk

)2,∑n

k=0

(nk

)= 2n and

(nbn

2c

)≥(nk

), for

all 0 ≤ k ≤ n. Thus, we estimate(2n

n

)=

n∑k=0

(n

k

)2

≤(n

bn2c

) n∑k=0

(n

k

)= 2n

(n

bn2c

).

Note, that

(n

bn2c

)≤(nn2

)is equivalent to Γ(n

2+1)2 ≤ Γ(bn

2c+1)Γ(n−bn

2c+1).

The Gamma function is known to be log-convex, which means that for allt ∈ [0, 1] and x, y > 0 we have

ln Γ((1− t)x+ ty) ≤ (1− t) ln Γ(x) + t ln Γ(y)

and soΓ((1− t)x+ ty) ≤ Γ(x)1−tΓ(y)t.

If we pick t = 12, x = bn

2c+1 and y = n−bn

2c+1, we have (1−t)x+ty = n

2+1

and therefore Γ(n2

+1) ≤ Γ(bn2c+1)

12 Γ(n−bn

2c+1)

12 . Eventually, we proved the

desired statement, since

M(K) ≥(nn2

)M(K ∩2 K

?)(2nn

) ≥ 2−nM(K ∩2 K?).

�

We are now able to show the main theorem of this section.


3.12 Theorem. Let n ≥ 4 and K ∈ Kn0 . Then

M(K) ≥ (2 log2 e(K))−nM(Bn).

In particular, from John’s Theorem and the affine invariance of the Mahlervolume, we have

M(K) ≥ (log2 n)−nM(Bn).

Proof. We will proceed by induction onm ∈ N such that 22m−1< e(K) ≤ 22m

.At first, we may consider K ∈ Kn0 with e(K) ≤ 2. In this case, the inequalityin Proposition 3.10 i), i.e., e(K)−nM(Bn) ≤M(K), is sharper than the claimabove, since n ≥ 4 and therefore e(K)−n ≥ 2−n ≥ (log2 n)−n.

Now, we assume that for all bodies K ∈ Kn0 with e(K) ≤ 22m−1the

theorem is proven and consider K ∈ Kn0 with 22m−1< e(K) ≤ 22m

. Bydefinition we have r(K)Bn ⊂ K ⊂ R(K)Bn, and so

r(K)

r(K)12R(K)

12︸︷︷︸

=e(K)−12

Bn ⊂ 1

r(K)12R(K)

12

K ⊂ R(K)

r(K)12R(K)

12︸︷︷︸

=e(K)12

Bn

Since e(tK) = e(K), for positive t, we can assume that e(K)−nBn ⊂ K ⊂e(K)nBn. Taking polars, yields e(K)−nBn ⊂ K? ⊂ e(K)nBn. Define K :=K ∩2 K

?. We see that

K ⊃⋃t∈[0,1]

((1− t)12 e(K)−

12Bn ∩ t

12 e(K)−

12Bn) ⊃ 1√

2e(K)−

12Bn.

This means, that r(K) ≥ 1√2e(K)−

12 . For x ∈ ((1− t) 1

2K ∩ t 12K?) it holds

||x||2 = xᵀx ≤ (1− t)12 t

12 ≤ 1

2,

which leads to R(K) ≤ 1√2. Combining these two inequalities yields

e(K) =R(K)

r(K)≤ e(K)

12 ≤ (22m

)12 = 22m−1

,

and by induction hypothesis we get

M(K) ≥ (2 log2 e(K))−nM(Bn) ≥ (log2 e(K))−nM(Bn).

From Lemma 3.11 we know M(K) ≥ 2−nM(K) and the claim follows. �

28 4 Hanner polytopes and beyond

4 Hanner polytopes and beyond

This chapter primary deals with Hanner polytopes which first occurred ina work by Olof Hanner [8]. These polytopes are related to the problem ofMahler, because they satisfy the conjectured lower bound in (2.2) with equal-ity, which is shown below. We will also see, that the p-product constructionof Kuperberg (see Definition 3.2) admits another description of a Hannerpolytope. Basing ourselves on that, we investigate in the second section howthe Mahler volume changes under taking p-products. The result is, thatK ×p L satisfies the Mahler Conjecture for all choices of p ∈ [1,∞], if bothK and L do so, in their respective dimensions.

4.1 Hanner polytopes

Considering a given convex body K ∈ Kn, Hanner [8] was interested in theminimal number I(K) of vectors ui ∈ Rn, such that the bodiesK+ui pairwisemeet, but the section of them all is empty. He proved that I(K) is either3, 4 or nonexistent - we let I(K) be infinity in this case. The polytopes ofinterest occurred in his work as examples of convex bodies with I(K) > 3. In1981, Hansen and Lima [9] showed that Hanner polytopes are characterisedby that property.

The free sum of two convex sets K ∈ Kn and L ∈ Km is defined as theconvex hull of K × {0} and {0} × L.

4.1 Definition (Hanner polytope).A Hanner polytope is obtained by successively applying Cartesian productsand free sums to centered line segments in arbitrary order.

In the sequel, we will restrict to the interval I = [−1, 1] as the involved linesegment, since the Mahler volume is preserved under affine transformations.The p-product notation permits us to rewrite the Cartesian product of K ∈Kn and L ∈ Km as K ×∞ L = K × L and the free sum as K ×1 L ∼=(K × {0}) +1 ({0} × L) = conv{(K × {0}) ∪ ({0} × L)}. Therefore, everyHanner polytope in Rn is an affine image of I ×p1 . . . ×pn−1 I, where pi iseither 1 or infinity, for 1 ≤ i ≤ n− 1.

The following proposition is a direct consequence of I? = I, the associa-tivity of the p-product relation and Lemma 3.5.

4.2 The transference property of ×p 29

4.2 Proposition. Let H = I×p1 . . .×pn−1 I ∈ Kn0 be a Hanner polytope anddefine pi = 1, if pi =∞, and pi =∞, if pi = 1, for 1 ≤ i ≤ n− 1. Then thepolar body H? is also a Hanner polytope and it holds H? = I×p1 . . .×pn−1 I.

Next, we show that Hanner polytopes attain the conjectured minimal Mahlervolume.

4.3 Lemma. Let H ∈ Kn0 be a Hanner polytope. Then, M(H) = 4n

n!.

Proof. Let H = I×p1 . . .×pn−1 I with pi ∈ {1,∞}, 1 ≤ i ≤ n−1. Proposition3.9 states that M(K×1L) = M(K×∞L) and together with the associativityof the p-product relation we obtain

M(H) = M(I ×p1 (I ×p2 . . .×pn−1 I)) = M(I ×1 (I ×p2 . . .×pn−1 I)).

Inductively, this leads to M(H) = M(I ×1 . . .×1 I) = M(C?n) = 4n

n!. �

To conclude our consideration, we inspect Hanner polytopes in small dimen-sions up to affine transformations. In dimension one, there is only the linesegment I = [−1, 1].

For n = 2, we also have only one Hanner polytope, since the linear

transformation given by the matrix

(1 1−1 1

)maps the diamond C?

2 =

I ×1 I to the square C2 = I ×∞ I.This affine transformation of square and diamond implies, that in dimen-

sion n = 3 the polytope I ×∞ I ×1 I is affinely isomorphic to the octahedronC?

3 . Thus, there are two 3-dimensional Hanner polytopes, namely C3 andC?

3 .Similarly, one gets four 4-dimensional Hanner polytopes - the cube C4,

the crosspolytope C?4 , I ×∞ I ×∞ I ×1 I and I ×1 I ×1 I ×∞ I.

Observe, that when we add another ×1 relation to H = I ×p1 . . .×pn−1 I,where pi ∈ {1,∞}, we end up with an (n+ 1)-dimensional Hanner polytopewith two additional vertices. If we use another ×∞ relation, the number ofvertices is doubled. For example, there is no 4-dimensional Hanner polytopewith 14 vertices.

4.2 The transference property of ×p

Suppose, we have two convex bodies K ∈ Kn0 and L ∈ Km0 , that fulfill theMahler Conjecture. Is this property transferee to the p-product of K and L?A proof for the positive answer to this question is given in the sequel of thissection.


In upcoming lemma, we write collections of elements, where the sameelement can occur various times as [x1, . . . , xm]. Moreover, we adopt theusual notation for the union of sets and write [x1, . . . , xm] ∪ [y1, . . . , yk] =[x1, . . . , xm, y1, . . . , yk].

4.4 Lemma. Let k, l ∈ N and Mk =[jk| j = 1, . . . , k

]. There exists a

bijection f : Mk ∪Ml → Mk+l, such that m ≥ f(m), for all m ∈ Mk ∪Ml.Furthermore, there has to be an element m′ ∈Mk ∪Ml with m′ > f(m′).

Proof. Let M = Mk∪Ml = [m1, . . . ,mk+l] and N = Mk+l = [n1, . . . , nk+l] begiven in ascending order. Suppose mi ∈ Mk, i < k + l, so there is an integerz ∈ {1, . . . , k} with mi = z

k. Recall the ceiling function dne of an integer

n as the smallest integer greater than or equal to n. By definition we haved lkze − 1 < l

kz ≤ d l

kze and therefore,

d lkze − 1

l<z

k≤d lkzel.

Apparently, there are z − 1 occurrences of elements from Mk in M whichprecede z

k. Thus, we identify the index i as the number d l

kze+z−1. It holds

lkz ≤ d l

kze < l

kz + 1, and equivalent transformations lead to

k

⌈l

kz

⌉+ zk − k < zk + zl ≤ k

⌈l

kz

⌉+ zk.

This means, that

ni =i

k + l=d lkze+ z − 1

k + l<z

k= mi ≤

d lkze+ z

k + l=i+ 1

k + l= ni+1.

Similarly, we derive for mj ∈ Ml, j < k + l, that nj < mj ≤ nj+1. Togetherthis yields

n1 < m1 ≤ n2 < m2 ≤ . . . ≤ nk+l−2 < mk+l−2 ≤ nk+l−1 < mk+l−1 ≤ nk+l,

and furthermore we have nk+l = mk+l = 1. Thus, we can define a bijectionf : M → N with the stipulated property, by mapping mi to f(mi) = ni, for1 ≤ i ≤ k+l. The additional statement follows immediately by the definitionof f , since k + l ≥ 2. �

This lemma enables us to extend an argument by Saint-Raymond [23], whichhe used to verify the Mahler Conjecture for the class of unit balls of p-norms.We get the following statement, that turns out to be the main ingredient forthe proof of the transference property of ×p.

4.2 The transference property of ×p 31

4.5 Lemma. Let k, l ∈ N, and 1 ≤ p, q ≤ ∞, such that 1p

+ 1q

= 1. Thenthe following inequalities hold

k! · l!(k + l)!

≤Γ( k

p+1)Γ( l

p+1)Γ( k

q+1)Γ( l

q+1)

Γ( k+lp

+1)Γ( k+lq

+1)≤ Γ( k

2+1)2Γ( l

2+1)2

Γ( k+l2

+1)2.

Equality is attained on the left hand side, if and only if p = 1 or q = 1, andon the right hand side, if and only if p = q = 2.

Proof. Consider the well-known digamma function ψ(x) = ddx

ln Γ(x) forx > 0. For the first derivative of ψ there is the following series representation(see [2])

ψ′(x) =d2

dx2ln Γ(x) =

∞∑k=0

1

(x+ k)2.

For x ≥ 0 define the function g(x) := Γ(1+kx)Γ(1+lx)Γ(1+(k+l)x)

. We obtain

d2

dx2ln g(x) = k2ψ′(1 + kx) + l2ψ′(1 + lx)− (k + l)2ψ′(1 + (k + l)x)

=∞∑i=1

k2

(kx+ i)2+∞∑i=1

l2

(lx+ i)2−∞∑i=1

(k + l)2

((k + l)x+ i)2

=∞∑i=1

1

(x+ ik)2

+∞∑i=1

1

(x+ il)2−∞∑i=1

1

(x+ ik+l

)2

=∞∑i=0

(k∑j=1

1

(x+ i+ jk)2

+l∑

j=1

1

(x+ i+ jl)2−

k+l∑j=1

1

(x+ i+ jk+l

)2

)︸︷︷︸

=:sk,l(i)

.

The sign of sk,l(i) is independent of x and i as long as x+i ≥ 0 which is alwaysfulfilled. By Lemma 4.4, we get sk,l(i) < 0, for all i ∈ N0. Thus, x 7→ ln g(x)is concave in [0, 1], and so is ϕ(x) = ln(g(x)g(1 − x)). Additionally, ϕ issymmetric around x = 1

2, it attains its maximum at x = 1

2and its minima

at x = 0 and x = 1, which gives us

ϕ(0) = ϕ(1) ≤ ϕ(x) ≤ ϕ(1

2).

That implies for 1p

+ 1q

= 1,

k! · l!(k + l)!

≤Γ( k

p+1)Γ( l

p+1)Γ( k

q+1)Γ( l

q+1)

Γ( k+lp

+1)Γ( k+lq

+1)≤ Γ( k

2+1)2Γ( l

2+1)2

Γ( k+l2

+1)2,

where equality is attained if and only if p = 1 or q = 1 on the left hand sideand if and only if p = q = 2 on the right hand side. �


4.6 Theorem. Let Ki ∈ Kni0 , 1 ≤ i ≤ k, and n = n1 + . . . + nk. If all Ki

satisfy Mahler’s Conjecture in their respective dimension, i.e., M(Ki) ≥ 4ni

ni!,

then for all p1, . . . , pk−1 ∈ [1,∞], we have

M(K1 ×p1 . . .×pk−1Kk) ≥

4n

n!. (4.1)

Proof. We proceed by induction on the number k of involved bodies. There isnothing to show for k = 1, since we have n = n1 and therefore, M(K1) ≥ 4n

n!.

Let k ≥ 2 and write K = K1 ×p1 . . .×pk−1Kk. Proposition 3.9 gives us

M(K) =Γ(n1

p1+1)Γ(n−n1

p1+1)Γ(n1

q1+1)Γ(n−n1

q1+1)

Γ( np1

+1)Γ( nq1

+1)M(K1)M(K2 ×p2 . . .×pk−1

Kk),

where q1 satisfies 1p1

+ 1q1

= 1. Lemma 4.5 and the induction hypothesis imply

M(K) ≥ n1!(n− n1)!

n!· 4n1

n1!· 4n−n1

(n− n1)!=

4n

n!,

and we are done. �

4.7 Remark. The equality characterisation in Lemma 4.5 yields the fol-lowing addition to the above statement. Assume, that all the bodies fulfillM(Ki) = 4ni

ni!, then equality holds in (4.1), if and only, if pj is either 1 or

infinity, for 1 ≤ j ≤ k − 1.

Theorem 4.6 arises the question, whether we get new classes of centrally sym-metric convex bodies, when applying p-products to bodies, that are knownto satisfy the Mahler Conjecture. The following chapters deal with familiesof centrally symmetric convex bodies, for which the validity of Mahler’s Con-jecture is known. An interesting task for further work, would be to studythe bodies that are given by p-products of members of such families.

33

5 Mahler’s Conjecture and zonoids

In this chapter we investigate an approach on the Mahler Conjecture for thespecial class of zonoids. In [20] Reisner used probabilistic arguments involv-ing random hyperplanes and a result on the number of vertices of certainpolyhedra, to show that Mahler’s Conjecture is valid for zonoids. Addition-ally, he characterised the equality case in [21]. In joint work with Gordonand Meyer [7] from 1987, he found a much more self-contained proof, thatmakes no use of probability and which is presented in the sequel.

The upcoming section will introduce the concept of zonoids. Severalequivalent descriptions are given and a formula for the computation of thevolume of such a convex body will be deduced. Subsequently, we will followthe details of the proof by Gordon, Meyer and Reisner.

5.1 Zonotopes and zonoids

5.1 Definition (Zonotope).A zonotope Z ∈ Kn is a finite Minkowski sum of line segments, i.e., there areai, bi ∈ Rn, 1 ≤ i ≤ m, such that Z = [a1, b1] + . . .+ [am, bm].

By the definition of the support function we derive, that for K,L ∈ Kn andt, s > 0 it holds

htK+sL(x) = thK(x) + shL(x), for all x ∈ Rn. (5.1)

This property implies the following lemma.

5.2 Lemma. Let Z ∈ Kn0 be a centrally symmetric zonotope. The supportfunction of Z is given by

hZ(u) =m∑i=1

ai|uᵀvi|, for u ∈ Sn−1,

where ai > 0, vi ∈ Sn−1, 1 ≤ i ≤ m.

Proof. Pick ai > 0, vi ∈ Sn−1 and wi ∈ Rn, 1 ≤ i ≤ m, such that

Z =m∑i=1

(ai[−vi, vi] + wi).

34 5 Mahler’s Conjecture and zonoids

The equation (5.1) yields

hZ(u) =m∑i=1

(aih[−vi,vi](u) + hwi(u)) =

m∑i=1

ai|uᵀvi|+m∑i=1

uᵀwi.

Since, Z is centrally symmetric, we have hZ(u) = hZ(−u), for all u ∈ Sn−1.This leads to

∑mi=1 u

ᵀwi = 0 and therefore to the desired form of hZ . �

5.3 Lemma. The support function gives an alternative way to compute theHausdorff distance of K1, K2 ∈ Kn. We have

δ(K1, K2) = supu∈Sn−1

|hK1(u)− hK2(u)| = ||hK1 − hK2||∞.

Proof. Let u ∈ Sn−1. Then

hK2+||hK1−hK2

||∞Bn(u) = hK2(u) + ||hK1 − hK2||∞hBn(u)

= hK2(u) + ||hK1 − hK2||∞≥ hK2(u) + |hK1(u)− hK2(u)| ≥ hK1(u).

This means K1 ⊂ K2 + ||hK1 − hK2||∞Bn and by changing the roles of K1

and K2, we also get K2 ⊂ K1 + ||hK1 − hK2||∞Bn. In other words, it holdsδ(K1, K2) ≤ ||hK1 − hK2||∞.

To see the converse inequality, let ε > 0 such that δ(K1, K2) = ε. Bydefinition of the Hausdorff distance we have K1 ⊂ K2 + εBn and K2 ⊂K1 + εBn, or equivalently, hK1(u) ≤ hK2(u) + ε and hK2(u) ≤ hK1(u) + ε, forall u ∈ Sn−1. Thus, |hK1(u) − hK2(u)| ≤ ε = δ(K1, K2), for u ∈ Sn−1, andthe lemma is proven. �

5.4 Definition (Zonoid).A compact convex set Z ⊂ Rn is called a zonoid, if it is the Hausdorff limitof a sequence of zonotopes, that is, there are zonotopes Zj ∈ Kn, j ∈ N, suchthat lim

j→∞δ(Zj, Z) = 0.

5.5 Remark. The class of zonotopes is invariant under Minkowski linearcombinations and affine transformations. In particular, projections of zono-topes (zonoids) are again zonotopes (zonoids).Moreover, zonotopes are symmetric to a certain point in Rn and since thisproperty is preserved under taking Hausdorff limits, zonoids are as well.

In the following, we state a characterisation of zonoids in terms of Borelmeasures in Sn−1.

5.1 Zonotopes and zonoids 35

5.6 Definition (Borel measure).The smallest σ-algebra containing all open sets in Rn is said to be the familyof Borel sets. A signed Borel measure on S ⊂ Rn is a real-valued function µdefined on all Borel sets in S which is countably additive, i.e., for all pairwise

disjoint Borel sets Aj ⊂ S, j ∈ N, we have µ

(⋃j∈N

Aj

)=∑j∈N

µ(Aj).

If the function µ is furthermore nonnegative, it is called a Borel measure onS. A signed Borel measure µ on S ⊂ Rn is said to be even [odd], if for allBorel sets A in S we have µ(−A) = µ(A) [= −µ(A)].

In the subsequent theorem we use a special case of Cauchy’s projection for-mula (see [6], Appendix A.5), which states that for all v ∈ Sn−1 and K ∈ Kn0it holds

hΠK(v) = voln−1(K|v⊥) =

∫Sn−1

|uᵀv| dSn−1(K, u), (5.2)

where Sn−1(K, ·) is a finite Borel measure on the unit n-sphere Sn−1, thesurface area measure of K. The surface area measure of K = Bn coincideswith the usual Lebesgue measure on Sn−1. Thus, we have

hΠBn(v) = voln−1(Bn|v⊥) =

∫Sn−1

|uᵀv| du. (5.3)

5.7 Theorem. A compact K ∈ Kn is a centrally symmetric zonoid if andonly if

hK(u) =

∫Sn−1

|uᵀv| dµ(v), for u ∈ Sn−1, (and so for u ∈ Rn), (5.4)

where µ is an even Borel measure on Sn−1.

Proof. As a start, let the support function of K be of the form (5.4). Then,K is centrally symmetric, since hK(u) = hK(−u), for all u ∈ Sn−1. Fix ε > 0and let δ = ε

µ(Sn−1)> 0. Next, we pick a partition of the unit sphere, that

is, a family Ei ⊂ Sn−1, 1 ≤ i ≤ m, such that Ei ∩ Ej = ∅, whenever i 6= j,and Sn−1 =

⋃mi=1Ei. Furthermore, we assume that for 1 ≤ i ≤ m we have

maxx,y∈Ei

||x− y|| < δ. Fix some vi ∈ Ei, 1 ≤ i ≤ m. Then, for all u ∈ Sn−1,

∣∣∣∣∣∫Sn−1

|uᵀv| dµ(v)−m∑i=1

|uᵀvi|µ(Ei)

∣∣∣∣∣ =

∣∣∣∣∣m∑i=1

∫Ei

(|uᵀv| − |uᵀvi|) dµ(v)

∣∣∣∣∣


≤m∑i=1

∫Ei

∣∣∣|uᵀv| − |uᵀvi|∣∣∣ dµ(v) ≤

m∑i=1

∫Ei

|uᵀ(v − vi)| dµ(v)

≤m∑i=1

∫Ei

||v − vi|| dµ(v) < δ

m∑i=1

∫Ei

dµ(v)

= δ

m∑i=1

µ(Ei) = δµ(Sn−1) = ε.

Note, that we used the famous Cauchy-Schwarz inequality here. This com-putation shows that the zonotope Z =

∑mi=1 µ(Ei)[−vi, vi], with support

function hZ(u) =∑m

i=1 µ(Ei)|uᵀvi|, and K have Hausdorff distance

δ(K,Z) = ||hK − hZ ||∞ = supu∈Sn−1

|hK(u)− hZ(u)| < ε.

Since ε > 0 was chosen arbitrarily, K is the Hausdorff limit of zonotopes andtherefore a zonoid.

In order to show, that the condition (5.4) on hK is necessary for K tobe a centrally symmetric zonoid, consider a sequence (Zj)j∈N of zonotopes inRn converging to K. Without loss of generality we can assume that all Zjare centrally symmetric, since K is. By Lemma 5.2 the support function ofa Zj, j ∈ N, has the form

hZj(u) =

m∑i=1

ai|uᵀvi| =∫Sn−1

|uᵀv| dµ(v), u ∈ Sn−1,

where µ is the sum of point masses of weight ai

2at ±vi, 1 ≤ i ≤ m. Therefore,

it suffices to show, that the set of compact Z ∈ Kn0 whose support functionshave the desired form (5.4), is closed in Kn.

To this end, let Kj ∈ Kn0 , j ∈ N, be compact with support function

hKj(u) =

∫Sn−1

|uᵀv| dµj(v), ∀u ∈ Sn−1,

and µj even Borel measures on Sn−1. Suppose that Kj → K, j → ∞, inthe Hausdorff metric. By the boundedness of K, there is an r > 0 anda j0 ∈ N such that Kj ⊂ rBn, for j ≥ j0. Consider the projection bodyΠBn of the Euclidean unit ball. Since, any projection of Bn on an (n − 1)-dimensional subspace of Rn is an (n − 1)-dimensional Euclidean unit ball,we have hΠBn(u) = voln−1(Bn|u⊥) = κn−1, for all u ∈ Sn−1. Combiningequation (5.3) with Fubini’s theorem leads to

2κn−1µj(Sn−1) = 2

∫Sn−1

hΠBn(v) dµj(v) =

∫Sn−1

∫Sn−1

|uᵀv| du dµj(v)

5.1 Zonotopes and zonoids 37

=

∫Sn−1

∫Sn−1

|uᵀv| dµj(v) du =

∫Sn−1

hKj(u) du

≤∫Sn−1

hrBn(u) du = r

∫Sn−1

du = rnκn, for j ≥ j0.

This means, that the sequence (µj(Sn−1))j∈N is bounded, which implies the

existence of a subsequence (µjk)k∈N ⊂ (µj)j∈N that converges weakly to aneven Borel measure µ on Sn−1 (see [25], the proof of Theorem 5.5.2). Thatis, for all bounded, continuous, real-valued functions on Sn−1∫

Sn−1

f(u) dµjk(u)→∫Sn−1

f(u) dµ(u), k →∞.

Since the scalar product is continuous, this leads to

hKjk(u) =

∫Sn−1

|uᵀv| dµjk(v)→∫Sn−1

|uᵀv| dµ(v), k →∞.

We also have Kj → K, so by Lemma 5.3, the sequence (hKj)j∈N converges

uniformly to hK . Finally, by the uniqueness of the limit, we get hK(u) =∫Sn−1 |uᵀv| dµ(v) for all u ∈ Sn−1. �

5.8 Remark. The even Borel measure µ in the previous theorem is calledthe generating measure of the zonoid K. It is essentially unique, since frommeasure theory µ = 0, if for all u ∈ Sn−1 it holds

∫Sn−1 |uᵀv| dµ(v) = 0. A

proof can be found in [6], Appendix C.

5.9 Proposition. Let K ∈ Kn0 be a zonoid with generating measure µ.Then the volume of K is given by

vol(K) =1

n

∫Sn−1

voln−1(K|u⊥) dµ(u).

Proof. The surface area measure admits a formula (see [6], Theorem A.3.1)to compute the volume of K with the help of its support function hK . Itstates that

vol(K) =1

n

∫Sn−1

hK(u) dSn−1(K, u).

With Theorem 5.7 and equation (5.2) we derive

vol(K) =1

n

∫Sn−1

hK(v) dSn−1(K, v)

=1

n

∫Sn−1

(1

2

∫Sn−1

|vᵀu| dµ(u)

)dSn−1(K, v)


=1

n

∫Sn−1

(1

2

∫Sn−1

|vᵀu| dSn−1(K, v)

)dµ(u)

=1

n

∫Sn−1


�

As the following statement shows, projection bodies and zonoids describeexactly the same class of convex bodies, when we restrict to the centrallysymmetric case.

5.10 Theorem. Any projection body is a centrally symmetric zonoid. Con-versely, every n-dimensional centrally symmetric zonoid in Rn is the projec-tion body of a unique centrally symmetric convex body.

Proof. Let K ∈ Kn0 and let A be a subset of Sn−1. The surface area mea-sure fulfills Sn−1(K,−A) = Sn−1(−K,A). Hence, from Cauchy’s projectionformula (5.2), we obtain

hΠK(u) =1

2

∫Sn−1

|uᵀv| dSn−1(K, v)

=1

4

(∫Sn−1

|uᵀv| dSn−1(K, v) +

∫Sn−1

|uᵀv| dSn−1(−K, v)

)=

∫Sn−1

|uᵀv| dµ(v),

where µ = 14(Sn−1(K, ·) + Sn−1(−K, ·)) is a Borel measure on Sn−1. We are

up to show, that µ is even. Again, by the above property of Sn−1(K, ·), weget

µ(−A) =1

4(Sn−1(K,−A) + Sn−1(−K,−A))

=1

4(Sn−1(−K,A) + Sn−1(K,A)) = µ(A).

Thus, ΠK is a centrally symmetric zonoid by Theorem 5.7.The remaining implication will not be shown here, since it uses some more

concepts that are out of the focus of this work. The arguments can be foundin Chapter 4 of the book by Gardner [6]. �

5.2 The Mahler volume of zonoids

In this section we illustrate that Mahler’s Conjecture is true for the class ofzonoids. As mentioned in the introduction to this chapter, we are followinga work from 1988 by Gordon, Meyer and Reisner [7].

5.2 The Mahler volume of zonoids 39

5.11 Lemma. Let K ∈ Kn0 be a zonoid with generating measure µ. Then

(n+1)vol(K)

∫Sn−1

∫K?

|uᵀx| dx dµ(u) = 2vol(K?)

∫Sn−1


Furthermore, for some u0 ∈ Sn−1 we have

(n+ 1)vol(K)

∫K?

|uᵀ0x| dx ≥ 2vol(K?)voln−1(K|u⊥0 ).

Proof. Using Fubini’s theorem and hK(·) = || · ||K? , we get∫Sn−1

∫K?

|uᵀx| dx dµ(u) =

∫K?

∫Sn−1

|uᵀx| dµ(u) dx

(Theorem 5.7) = 2

∫K?

hK(x) dx = 2

∫K?

||x||K?dx

= 2

∫K?

∫ ||x||K?

0

dt dx.

For the integration domain we have {(x, t) ∈ Rn+1 | x ∈ K?, 0 ≤ t ≤||x||K?} = {(x, t) ∈ Rn+1 | 0 ≤ t ≤ ||x||K? ≤ 1}, and we continue thesequence of equalities by

= 2

∫ 1

0

vol({x ∈ Rn | t ≤ ||x||K? ≤ 1}) dt

= 2

∫ 1

0

(vol(K?)− vol(tK?)) dt

= 2vol(K?)

∫ 1

0

(1− tn) dt =2n

n+ 1vol(K?)

(Proposition 5.9) =2vol(K?)

(n+ 1)vol(K)

∫Sn−1

voln−1(K|u⊥) dµ(u),

and the claim follows. The existence of a desired u0 ∈ Sn−1, such that

(n+ 1)vol(K)

∫K?

|uᵀ0x| dx ≥ 2vol(K?)voln−1(K|u⊥0 ),

follows from the non-negativity of the generating measure µ of K. If therewould be no such u0, equality in the above formula could not hold. �

The next step is to prove a technical lemma about estimating a certain inte-gral.


5.12 Lemma. Let f : R≥0 → R≥0 be such that 0 <∫∞

0f(t) dt < ∞ and

f(0) = 1. Moreover, for some p > 0 let f1p be concave. Then∫ ∞

0

tf(t) dt ≤ p+ 1

p+ 2

(∫ ∞0

f(t) dt

)2

with equality if and only if there is an a > 0 such that for all real t ≥ 0 wehave f(t) = (1− at)p+ = max{(1− at)p, 0}.

Proof. Consider h : R>0 → R>0 given by h(a) = 1a(p+1)

. Clearly, h is

surjective. By positivity of∫∞

0f(t) dt there exists an a > 0 such that 1

a(p+1)=∫∞

0f(t) dt. Hence∫ ∞

0

(1− at)p+ dt =

∫ 1a

0

(1− at)p dt = − 1

a(p+ 1)(1− at)p+1

∣∣∣10

=1

a(p+ 1)=

∫ ∞0

f(t) dt.

Now, define g(t) := f(t) − (1 − at)p+, where t ≥ 0. Thus, g(0) = 0 and∫∞0g(t) dt = 0. Let us show that there exists an x0 ≥ 0 such that g(t) ≥ 0

for all t ∈ [0, x0] and g(t) ≤ 0, for t ≥ x0.In order to see this, we assume that there is an ε > 0 with g(t) < 0,

whenever t ∈ (0, ε]. This implies for such t ∈ (0, ε] that f(t) < (1 − at)p+,

or equivalently, f(t)1p < (1 − at)+. The concavity of f

1p and the fact, that

(1− at)+ is tangent on f1p in t = 0, also lead to f(t)

1p ≤ (1− at)+ for t ≥ ε.

We arrive at a contradiction, since, then∫∞

0f(t) dt <

∫∞0

(1 − at)p+ dt, butthe definition of g yields

∫∞0g(t) dt = 0. We come up with an x ≥ 0 fulfilling

g(t) ≥ 0, whenever t ∈ [0, x]. The same argument shows, that g(t) ≥ 0cannot hold everywhere, thus, there has to be a desired x0 ≥ 0 with thestated properties.

That statement implies∫∞xg(t) dt ≤ 0, for all x ≥ 0. Therefore,∫ ∞

0

tf(t) dt =

∫ ∞0

∫ t

0

f(t) ds dt =

∫ ∞0

∫ ∞s

f(t) dt ds

≤∫ ∞

0

∫ ∞s

(1− at)p+ dt ds =

∫ 1a

0

∫ 1a

s

(1− at)p dt ds

=

∫ 1a

0

(1− as)p+1

a(p+ 1)ds

=1

a2(p+ 1)(p+ 2)=

p+ 1

p+ 2

(∫ ∞0

f(t)dt

)2

.


Equality is attained if and only if for any nonnegative s it holds∫∞sf(t) dt =∫∞

s(1− at)p+ dt, that is, if and only if f(t) = (1− at)p+, for t ≥ 0. �

5.13 Lemma. Let K ∈ Kn0 and u ∈ Sn−1. Then∫K

|uᵀx| dx ≤ nvol(K)2

2(n+ 1)voln−1(K ∩ u⊥)(5.5)

with equality if and only if K = conv{y,−y,K ∩ u⊥}, for some y ∈ K.

Proof. At first, we evaluate∫K

|uᵀx| dx = 2

∫ ∞0

∫K∩{uᵀx=t}

|uᵀx| dx dt (5.6)

= 2

∫ ∞0

tvoln−1(K ∩ {uᵀx = t}) dt = 2

∫ ∞0

tg(t) dt,

where g(t) = voln−1(K ∩ {uᵀx = t}), for t ∈ R. Since K is centrally sym-metric, g is even and we have g(t) = 0, for |t| > hK(u) = ||u||K? . Moreover,it holds g(0) = voln−1(K ∩ u⊥). The Brunn-Minkowski inequality (see [25],Theorem 6.1.1) yields

g(λt+ (1− λ)s)1

n−1 = voln−1(K ∩ {uᵀx = λt+ (1− λ)s})1

n−1

≥ voln−1(λ(K ∩ {uᵀx = t}) + (1− λ)(K ∩ {uᵀx = s}))1

n−1

≥ λvoln−1(K ∩ {uᵀx = t})1

n−1 + (1− λ)voln−1(K ∩ {uᵀx = s})1

n−1

= λg(t)1

n−1 + (1− λ)g(s)1

n−1 ,

for arbitrary λ ∈ [0, 1] and s, t ∈ (−||u||K? , ||u||K?). This means, that g1

n−1

is concave on (−||u||K? , ||u||K?). Next, Fubini’s theorem and the evenness ofg give

vol(K) =

∫ ∞−∞

voln−1(K ∩ {uᵀx = t}) dt = 2

∫ ∞0

g(t) dt. (5.7)

Applying Lemma 5.12 to f(t) = g(t)g(0)

, t ≥ 0, and p = n− 1, we derive∫ ∞0

tg(t)

g(0)dt ≤ n

n+ 1

(∫ ∞0

g(t)

g(0)

)2

.

By definition of g this combines with (5.6) and (5.7) to∫K

|uᵀx| dx ≤ nvol(K)2

2(n+ 1)g(0)=

nvol(K)2

2(n+ 1)voln−1(K ∩ u⊥).


Let us now consider the case of equality. Lemma 5.12 states, that equality

holds, if and only if, for all real t ≥ 0, we have g(t)g(0)

=(

1− t||u||K?

)n−1

+. We

can pick a y ∈ K, such that

||u||K? = hK(u) = maxx∈K|uᵀx| = |uᵀy|.

Clearly, it holds, that conv{y,−y,K ∩ u⊥} ⊂ K. Assume, that we haveequality in (5.5). Then

1

g(0)

∫ ∞0

g(t) dt =

∫ ||u||K?

0

(1− t

||u||K?

)n−1

dt =||u||K?

n,

and thus

vol(K) ≥ vol(conv{y,−y,K ∩ u⊥}) =2

n||u||K?voln−1(K ∩ u⊥)

= 2

∫ ∞0

g(t) dt = vol(K).

Therefore, equality must hold throughout and K = conv{y,−y,K ∩ u⊥}, asdesired.

Conversely, let K = conv{y,−y,K ∩ u⊥} and write Kt = K ∩ {uᵀx = t}.For t ∈ [0, 1], the bodies K0 = K ∩ u⊥ and Kt are homothetic, i.e., there isa v ∈ Rn and a nonnegative constant s, such that Kt = sK0 + v. Since, Kis a double pyramid, the dilatation factor s is |u

ᵀy|−t|uᵀy| = (1 − t

||u||K?) and we

obtain

g(t)

g(0)=

voln−1(Kt)

voln−1(K0)=

voln−1(sK0 + v)

voln−1(K0)= sn−1 =

(1− t

||u||K?

)n−1

.

Because y is an apex of the double pyramid K, we have g(t) = 0, whenever

t > |uᵀy| and finally it holds g(t)g(0)

=(

1− t||u||K?

)n−1

+, for all t greater than or

equal to zero, which yields equality in (5.5). �

We are now ready to prove the main theorem of this chapter.

5.14 Theorem. Let K ∈ Kn0 be a zonoid. Then M(K) ≥ 4n

n!, with equality

if and only if K is an n-cube.

Proof. We proceed by induction on n. For n = 1 the result is clear, sinceevery K ∈ K1

0 is just a line segment.


Assume that n > 1. From Lemma 5.11 and 5.13 we derive the existenceof u0 ∈ Sn−1 such that

2vol(K?)voln−1(K|u⊥0 ) ≤ (n+ 1)vol(K)

∫K?

|uᵀ0x| dx ≤

nvol(K)vol(K?)2

2voln−1(K? ∩ u⊥0 ).

For the polar body of the projection it holds (K|u⊥0 )? = K? ∩ u⊥0 , whichimplies

M(K) = vol(K)vol(K?) ≥ 4

nvol(K|u⊥0 )vol((K|u⊥0 )?) =

4

nM(K|u⊥0 ).

As mentioned in Remark 5.5, projections of zonoids are again zonoids. There-fore, by induction hypothesis we have

M(K) ≥ 4

nM(K|u⊥0 ) ≥ 4

n

4n−1

(n− 1)!=

4n

n!.

In the case of equality, M(K) = 4n

n!, it holds M(K|u⊥0 ) = 4n−1

(n−1)!and the

induction hypothesis states that K|u⊥0 is an (n − 1)-cube. The equalitycharacterisation in Lemma 5.13 yields, that there is a y0 ∈ Rn such thatK? = conv{y0,−y0, K

? ∩ u⊥0 }, an n-dimensional crosspolytope. Thus, K isan n-cube. �

44 6 Polytopes with at most 2n+ 2 facets

6 Polytopes with at most 2n + 2 facets

In 1998, Lopez and Reisner [13] published a work on centrally symmetricpolytopes in Rn with at most 2n + 2 facets. They relied on a note by Ball[1], who stated that asking for the minimal Mahler volume among thesepolytopes is equivalent to find the minimal Mahler volume in all sections ofthe (n+ 1)-dimensional cube Cn+1 by n-dimensional hyperplanes. By basingon that, Lopez and Reisner transformed the problem to a finite search infixed dimension n. They used the computer to verify Mahler’s Conjecture indimensions n ≤ 8 for these special polytopes.

Throughout this chapter we denote by P a centrally symmetric polytopein Rn, having at most 2n + 2 facets. Note, that P ? is a polytope with atmost 2n+ 2 vertices.

The illustration of the proof of the following result is given in the twosubsequent sections. We start by enlightening the mentioned equivalence forthe polytopes of interest, and afterwards working through the arguments in[13] to transform the problem to a finite search.

6.1 Theorem. Let 2 ≤ n ≤ 8 and let P ∈ Kn0 be a polytope with at most2n+ 2 facets. Then, we have M(P ) ≥ 4n

n!.

Note, that Lopez and Reisner also characterised the equality cases in theconcerning dimensions. Again, the Hanner polytopes show up here. Theyfound out that the minimum is attained if and only if up to affine transfor-mations P is either the n-cube or P = C?

3 × Cn−3 for 3 ≤ n ≤ 8. Since, C?3

has 8 facets and Cn−3 has 2(n−3) = 2n−6, the latter body has 2n+2 facetsfor all n ≥ 3.

6.1 Equivalent description

We introduce the n-dimensional hyperplane Hu = {x ∈ Rn+1 | uᵀx = 0}perpendicular to u ∈ Sn.

6.2 Lemma. Let P ∈ Kn0 be a polytope with at most 2m facets. Then thereexists a direction vector u ∈ Sn and a linear map A : Rm → Rn such thatP = A(Cm ∩Hu).

Proof. There are vectors vi ∈ Rn \ {0}, 1 ≤ i ≤ m, such that

P = {x ∈ Rn | −1 ≤ vᵀi x ≤ 1, for 1 ≤ i ≤ m}.

6.2 Transformation to a finite search 45

Consider the linear mapping L : Rn → Rm with L(x) =

vᵀ1...vᵀm

x. We have

Cm ∩ L(Rn) =

vᵀ

1...vᵀm

x∣∣∣ x ∈ Rn, |vᵀ

i x| ≤ 1, 1 ≤ i ≤ m

= L(P ).

Thus, the inverse map L−1 of L : Rn → L(Rn) enables us to write thepolytope as P = L−1(Cm ∩ L(Rn)) and since L(Rn) is an n-dimensionalsubspace of Rm, there is a vector u ∈ Sn with L(Rn) = Hu. �

Putting m = n+1 in Lemma 6.2 and noticing, that (Cn+1∩Hu)? = C?

n+1|Hu,implies

6.3 Proposition. For the class of polytopes P ∈ Kn0 with at most 2n + 2facets, Mahler’s Conjecture is equivalent to

minu∈Sn

vol(Cn+1 ∩Hu) · vol(C?n+1|Hu) ≥

4n

n!. (6.1)

Note, that the minimum is independent on the length of the direction vectoru ∈ Sn.

6.2 Transformation to a finite search

At first, we further investigate the minimisation problem (6.1). Consider thebodies E = (ΠC?

n+1)? and F = ICn+1 ∈ Kn+10 . By definition of projection

and intersection body, we have for v ∈ Rn+1 \ {0}

vol(C?n+1|Hv) = hΠC?

n+1

(v

||v||2

)=∣∣∣∣∣∣ v

||v||2

∣∣∣∣∣∣(ΠC?

n+1)?=||v||E||v||2

,

and similarly

vol(Cn+1 ∩Hv) = ρICn+1

(v

||v||2

)=||v||2||v||F

.

By utilising these two identities we rewrite (6.1) as

1

max||v||E=1

||v||F=

1

maxv∈Rn+1\{0}

||v||F||v||E

= minv∈Rn+1\{0}

||v||E||v||F

≥ 4n

n!.

Therefore, the minimum is attained at an extreme point of E, that is, thepolar of the projection body of the (n+ 1)-dimensional crosspolytope.


6.4 Proposition. Let v ∈ Rn+1 \ {0}. Then we have

vol(C?n+1|Hv) =

1

2n!

∑ε∈{−1,1}n+1

|vᵀε|.

Proof. The projection of the surface of C?n+1 onto Hv covers almost all of

C?n+1|Hv exactly twice. Furthermore, the boundary ∂C?

n+1 is composed of2n+1 facets Fε, indexed by their outer normal vectors ε ∈ {−1, 1}n+1. Thesefacets are all regular simplices and hence of the same volume.

Denote with 1 = (1, . . . , 1) and let S = conv{0, e1, . . . , en+1} be thestandard (n+ 1)-simplex. We compute

1

(n+ 1)!= vol(S) =

vol(F1)

n+ 1

1√n+ 1

.

Thus, vol(Fε) =√n+1n!

, for all ε ∈ {−1, 1}n+1. Rescaling identifies 1√n+1

ε asthe outer unit normal vector of the facet Fε and hence, the volume of theprojection of Fε onto Hv is given by

vol(Fε|Hv) = vol(Fε)1√n+ 1

|vᵀε| = 1

n!|vᵀε|.

In conclusion, this yields vol(C?n+1|Hv) =

1

2n!

∑ε∈{−1,1}n+1

|vᵀε|, as desired. �

The next statement ensures, that the search for extreme points of E is indeedfinite. Write E = {ε ∈ {−1, 1}n+1 | ε1 = 1} and consider E ′ = n!E which isno restriction since we are only interested in direction vectors of extremepoints of E.

6.5 Proposition. A vector v ∈ Rn+1 is an extreme point of E ′ if and onlyif∑

ε∈E |vᵀε| = 1 and there are linearly independent vectors ε1, . . . , εn ∈ E ,such that vᵀεi = 0, for 1 ≤ i ≤ n.

Proof. The definition of E ′ and Proposition 6.4 yield

||x||E′ =∑ε∈E

|xᵀε|, x ∈ Rn+1. (6.2)

Next, consider the linear map ι : Rn+1 → R2nthat takes a vector x to

ι(x) = (xᵀε)ε∈E . Since, E surely contains n linear independent vectors, ιis injective, and we identify Rn+1 with its image M = ι(Rn+1) under thisembedding. From (6.2) we obtain ||x||E′ = ||ι(x)||C?

2n, and therefore v is

6.2 Transformation to a finite search 47

an extreme point of E ′ if and only if ||v||E′ =∑

ε∈E |vᵀε| = 1 and v isan intersection point of the (n + 1)-dimensional subspace M of R2n

and a(2n − (n+ 1))-dimensional face of C?

2n .An m-dimensional crosspolytope is given by C?

m = conv{±e1, . . . ,±em},so we can identify each (m − i)-dimensional face of C?

m as the set of allvectors x ∈ Rm with norm ||x||C?

m= 1 and equal sign pattern. Furthermore,

exactly i − 1 fixed coordinates of all such x always vanish. Therefore, anydirection of an extreme point of E ′ is contained in the intersection of Mwith an (2n − n)-dimensional subspace N = {x ∈ R2n | xεi

= 0, 1 ≤ i ≤ n}of R2n

, where ε1, . . . , εn are distinct elements in E . This property remainstrue, if we restrict to linearly independent subsets {ε1, . . . , εn} ⊂ E . In thiscase, the intersection M ∩N consists of all solutions v ∈ Rn+1 of the systemvᵀεi = 0, 1 ≤ i ≤ n, and hence is one-dimensional.

Summarising, we state that v is an extreme point of E ′ if and onlyif∑

ε∈E |xᵀε| = 1 and v fulfills vᵀεi = 0, for some linearly independent{ε1, . . . , εn}, which we wished to show. �

Utilising this proposition we are now able to formulate an algorithm to findthe minimum in (6.1). Note, that we neglect the norm condition, since weare only interested in the direction vectors.

(1) for all subsets S ⊂ E of size n do

if S = {ε1, . . . , εn} is linearly independent then collect S

(2) for all sets S collected in (1) do

compute a nontrivial solution v ∈ Rn+1 of the linear system

vᵀεi = 0, εi ∈ S

(3) for all v computed in (2) do

calculate vol(Cn+1 ∩Hv) · vol(C?n+1|Hv)

(4) return the minimum of the values calculated in (3)

Step (1) of this algorithm is very costly. Therefore, the search should bereduced by proving additional sufficient conditions for v to be a directionvector of an extreme point of E ′.

At first, the symmetry of cube and crosspolytope to all coordinate hy-perplanes implies that it is sufficient to consider only nonnegative vectorsv ∈ Rn+1, whose entries form a non-increasing sequence. A further reduc-tion step assumes, that the inequality is valid for dimensions smaller than


n. Lopez and Reisner show, that a vector v which satisfies the conditionsin Proposition 6.5 is parallel to an integer vector w, such that for everycoordinate 1 ≤ i ≤ n+ 1, it holds

1 ≤ wi ≤n

n2

2n−1. (6.3)

Now, one has to find all non-increasing integer vectors w that satisfy (6.3),and check whether there are n linearly independent elements in E , that areperpendicular to w. For the latter define a matrix U , whose rows are givenby all vectors ε ∈ E , such that wᵀε = 0 and check if the rank of U is n. Note,that the rank cannot exceed n, since all elements in E have the same firstcoordinate.

The number of candidates still grows exponentially with the dimension,but up to n = 8 Lopez and Reisner could do the computations to obtainTheorem 6.1.

49

7 Mahler volume of 1-unconditional bodies

In this chapter, we prove Mahler’s Conjecture for a third class of centrallysymmetric convex bodies, namely those which admit a 1-unconditional basis.This result was obtained by Saint-Raymond [23] in 1980.

7.1 Theorem (Saint-Raymond [23]).Let K ∈ Kn0 be 1-unconditional, then M(K) ≥ 4n

n!and equality holds if K is

a Hanner polytope.

Reisner [22] showed in a work from 1987, that the equality case is indeedcharacterised by Hanner polytopes.

The upcoming section deals with the definition and some examples of1-unconditional bodies. Furthermore, we introduce a construction which theproof of Theorem 7.1 relies on and discuss its properties that will be used toderive the result in the second part.

7.1 Definition and preparation

7.2 Definition (1-unconditional body).A centrally symmetric convex body K ∈ Kn0 is called 1-unconditional, if thereexists a basis {x1, . . . , xn} of Rn, such that for all scalars ai ∈ R and signs

εi ∈ {−1, 1}, 1 ≤ i ≤ n, we have∣∣∣∣∣∣ n∑

i=1

aixi

∣∣∣∣∣∣K

=∣∣∣∣∣∣ n∑

i=1

εiaixi

∣∣∣∣∣∣K

.

Let {x1, . . . , xn} be a basis of the 1-unconditional convex body K ∈ Kn0 .Such a basis is called 1-unconditional basis of K. Consider the unit vectorsei ∈ Rn, 1 ≤ i ≤ n, and the linear transformation A that maps xi to ei, for1 ≤ i ≤ n. Then, {e1, . . . , en} is a 1-unconditional basis of AK. To see this,we note that ||x||AK = ||A−1x||K , for x ∈ Rn, and get

∣∣∣∣∣∣ n∑i=1

aiei

∣∣∣∣∣∣AK

=∣∣∣∣∣∣A−1

(n∑i=1

aiei

)∣∣∣∣∣∣K

=∣∣∣∣∣∣ n∑

i=1

aiA−1ei

∣∣∣∣∣∣K

=∣∣∣∣∣∣ n∑

i=1

aixi

∣∣∣∣∣∣K

=∣∣∣∣∣∣ n∑

i=1

εiaixi

∣∣∣∣∣∣K

=∣∣∣∣∣∣ n∑

i=1

εiaiei

∣∣∣∣∣∣AK.

Due to the affine invariance of the Mahler volume, we therefore can restrict tosuch 1-unconditional bodies with basis {e1, . . . , en}. In this case the definition

50 7 Mahler volume of 1-unconditional bodies

yields

||(y1, . . . , yn)||K = ||(|y1|, . . . , |yn|)||K , y ∈ Rn,

which means that the body K is symmetric with respect to all coordinatehyperplanes.

Examples:

◦ By definition, the p-norm ||x||p only depends on the absolute value ofthe coordinates of x ∈ Rn. Thus, every p-ball Bn

p is a 1-unconditionalbody with basis {e1, . . . , en}.

◦ Let K ∈ Kn and L ∈ Km be 1-unconditional bodies with basis{x1, . . . , xn} and {y1, . . . , ym}, respectively. The relation

||(x, y)||pK×pL= ||x||pK + ||y||pL, for (x, y) ∈ Rn × Rm

implies, that K ×p L is also 1-unconditional with basis

{(x1, 0), . . . , (xn, 0), (0, y1), . . . , (0, ym)}.

In particular, any Hanner polytope is a 1-unconditional body.

Next, we consider the mapping

Tn : Rn>0 → Rn, Tn(t1, . . . , tn) := (− ln t1, . . . ,− ln tn)

and for K ∈ Kn0 denote T (K) := Tn(K ∩ Rn>0). In order to get an idea how

the resulting set T (K) looks like, we consider the unit ball Bnp of the p-norm,

for 1 ≤ p <∞. We have

T (Bnp ) = {y ∈ Rn | ∃(t1, . . . , tn) ∈ Bn

p with y = (− ln t1, . . . ,− ln tn)}= {y ∈ Rn | (e−y1 , . . . , e−yn) ∈ Bn

p )}

=

{y ∈ Rn

∣∣∣ n∑i=1

e−p·yi ≤ 1

}=

{1

py ∈ Rn

∣∣∣ n∑i=1

e−yi ≤ 1

}=

1

pT (C?

n).

This means, that all the sets T (Bnp ) are dilatations of T (C?

n). Figure 6illustrates the octahedron C?

3 and its image under T .The following three statements, are essential properties of T (K) and im-

portant ingredients for the proof of the main result.

7.1 Definition and preparation 51

Figure 6: Octahedron C?3 and T (C?

3)

7.3 Proposition. Let K ∈ Kn0 be 1-unconditional with basis {e1, . . . , en}.Then T (K) is convex.

Proof. Pick some s, t ∈ K ∩Rn>0 with s ≤ t and let x = (− ln s1, . . . ,− ln sn)

and y = (− ln t1, . . . ,− ln tn) ∈ T (K). For all λ ∈ [0, 1] and 1 ≤ i ≤ n wehave on the one hand

si ≤ sλi t1−λi ≤ ti (7.1)

and one the other hand

λxi + (1− λ)yi = −λ ln si − (1− λ) ln ti = − ln(sλi t1−λi ). (7.2)

Since, K is symmetric with respect to all coordinate hyperplanes, the cube

Qt = conv{(ε1t1, . . . , εntn) | ε ∈ {−1, 1}n}

is contained in K. By (7.1) this yields

(sλ1 t1−λ1 , . . . , sλn t

1−λn ) ∈ [s1, t1]× . . .× [sn, tn] ⊂ Qt ⊂ K.

Thus, with (7.2) we obtain λx+(1−λ)y ∈ T (K), which shows the convexityof T (K). �

7.4 Lemma. For every K ∈ Kn0 it holds T (K) + T (K?) = T (C?n).

Proof. First, observe that T (C?n) = {u ∈ Rn |

∑ni=1 e

−ui ≤ 1} and T (K) ={u ∈ Rn | e−u = (e−u1 , . . . , e−un) ∈ K}.


In order to see that T (K) + T (K?) ⊂ T (C?n), let u ∈ T (K), v ∈ T (K?)

and w = u + v. Thus, we have e−u ∈ K and e−v ∈ K?. This gives us∑ni=1 e

−wi =∑n

i=1 e−uie−vi = (e−u)ᵀ(e−v) ≤ 1 and so w ∈ T (C?

n).For the converse, pick a w ∈ T (C?

n) and write αi = e−wi , for 1 ≤ i ≤ n.We clearly have

∑ni=1 αi ≤ 1. Next, define a mapping L : K ∩ Rn

>0 → Rby L(x) =

∑ni=1 αi lnxi. For all 1 ≤ j ≤ n we have ∂2

∂x2iL(x) = −αj

x2j< 0,

which means that L is strictly concave. Observe, that the unique supremumof L on K ∩ Rn

>0 is attained and does not lie “near” some of the coordinatehyperplanes, because ln xi → −∞, if xi → 0, and L is strictly increasing.Therefore, it exists an x ∈ K ∩Rn

>0 such that λ := L(x) = supx∈K∩Rn>0L(x).

This implies, that we can find a u ∈ T (K) with x = e−u.Now, define an open set U = {x ∈ Rn

>0 | L(x) > λ}. It is convex due tothe concavity of L. By the definition of U , it has no point in common withK ∩ Rn

>0, and the well-known separation theorem for distinct convex bodies(see [10], Chapter 3) yields a supporting hyperplane H = {x ∈ Rn | aᵀx = 1}at x, such that U ⊂ {x ∈ Rn | aᵀx > 1} and K ∩Rn

>0 ⊂ {x ∈ Rn | aᵀx ≤ 1}.Note, that a is forced to be strictly positive.

Write f(x) = aᵀx. Due to the maximality of x there exists a γ > 0such that gradL(x) = γgradf(x). This is equivalent to (α1

x1, . . . , αn

xn) =

γ(a1, . . . , an), or αi = γaie−ui , for 1 ≤ i ≤ n. From here, we obtain

1 ≥n∑i=1

αi = γn∑i=1

aie−ui = γaᵀx = γ

and, since a > 0, we have

||γa||K? = γ||a||K? = γhK(a) = γ supx∈K

aᵀx = γ supx∈K∩Rn

>0

aᵀx = γ ≤ 1.

In conclusion, we can find an element v of T (K?) with γai = e−vi , 1 ≤ i ≤ n,and we get e−wi = αi = γaie

−ui = e−(ui+vi). Thus, we end up with w =u+ v ∈ T (K) + T (K?) and the lemma is proven. �

7.5 Proposition. For all K ∈ Kn0 there are a, b ∈ Rn such that

a+ Rn≥0 ⊂ T (K) ⊂ b+ Rn

≥0.

Proof. Recall that T (K) = {u ∈ Rn | e−u ∈ K} = {u ∈ Rn | ||e−u||K ≤ 1}.The equivalence of all norms on Rn gives us constants α, β ∈ R such that

e−β||x||∞ ≤ ||x||K ≤ e−α||x||∞, ∀x ∈ Rn.

7.2 Details of the proof 53

Putting a = −(α, . . . , α) and b = −(β, . . . , β) ∈ Rn, yields

a+ Rn≥0 = a+ T (Cn) = {a+ u ∈ Rn | ||e−u||∞ ≤ 1}

= {u ∈ Rn | ||e−(u−a)||∞ ≤ 1} = {u ∈ Rn | e−α||e−u||∞ ≤ 1}⊂ {u ∈ Rn | ||e−u||K ≤ 1} = T (K),

and similarly T (K) ⊂ b+ Rn≥0. �

7.2 Details of the proof

Let f : Rn → R be an integrable function. Then the Laplace transformL(f) : Rn → R of f is defined by

L(f)(t) =

∫Rn

e−uᵀtf(u) du, t ∈ Rn.

7.6 Proposition. Let f, g : Rn → R be integrable functions and f ∗ g be itsconvolution which is given by

(f ∗ g)(t) =

∫Rn

f(x)g(t− x) dx, t ∈ Rn.

Then, we have L(f)L(g) = L(f ∗ g).

Proof. Pick a t ∈ Rn. Then

L(f ∗ g)(t) =

∫Rn

e−uᵀt(f ∗ g)(u) du =

∫Rn

e−uᵀt

∫Rn

f(x)g(u− x) dx du

(Fubini) =

∫Rn

∫Rn

e−xᵀtf(x)e−(u−x)ᵀtg(u− x) du dx

=

∫Rn

e−xᵀtf(x)

∫Rn

e−(u−x)ᵀtg(u− x) du︸︷︷︸=L(g)(t)

dx

= L(f)(t) · L(g)(t).

�

The subsequent lemma provides a connection between the Mahler volume ofK and the Laplace transforms of the characteristic functions of T (K) andT (K?).

7.7 Lemma. Let K ∈ Kn0 be 1-unconditional with basis {e1, . . . , en}. Then

M(K) = 4nL(χT (K) ∗ χT (K?))(1, . . . , 1).


Proof. The symmetry of K with respect to all coordinate hyperplanes givesus

vol(K) = 2nvol(K ∩ Rn>0).

Let k ∈ Nn. By substituting xi = e−ui , 1 ≤ i ≤ n, we compute∫K∩Rn

>0

xk1−11 · . . . · xkn−1

n dx =

∫Tn(K∩Rn

>0)

e−u1k1 · . . . · e−unkn du1 . . . dun

=

∫T (K)

e−uᵀk du =

∫Rn

e−uᵀkχT (K)(u) du = L(χT (K))(k).

Choosing k = (1, ..., 1) yields

M(K) = 4nvol(K ∩ Rn>0)vol(K? ∩ Rn

>0)

= 4n

(∫K∩Rn

>0

dx

)(∫K?∩Rn

>0

dx

)= 4nL(χT (K))(1, . . . , 1) · L(χT (K?))(1, . . . , 1)

(Proposition 7.6) = 4nL(χT (K) ∗ χT (K?))(1, . . . , 1),

as desired. �

The next step, is to relate the characteristic functions of two convex sets C1

and C2 to that of their Minkowski sum C1 + C2.

7.8 Lemma. Let C1, C2 ⊂ Rn be two closed convex sets, such that thereare a, b ∈ Rn fulfilling a+ Rn

≥0 ⊂ Ci ⊂ b+ Rn≥0 for i = 1, 2. Furthermore, we

define the cross-section of Ci, i = 1, 2, at altitude t ∈ R as the set Ci(t) ={y ∈ Rn−1 | (y, t) ∈ Ci}. Then, for all y ∈ Rn−1 and r ∈ R it holds

δ(y, r) :=

∫RχC1(t)+C2(r−t)(y) dt ≥

∫ r

−∞χ(C1+C2)(t)(y) dt =: h(y, r).

Proof. First of all, we illustrate that we can restrict to C1 and C2 such thatthere is a b ∈ Rn with Rn

≥0 ⊂ Ci ⊂ b + Rn≥0. From the definition of the

cross-section we get

(C1 − a)(t) + (C2 − a)(r− t) = C1(t+ an) +C2(r + 2an − (t+ an))− 2a(n−1)

and(C1 − a+ C2 − a)(t) = (C1 + C2)(t+ 2an)− 2a(n−1),


where a(n−1) = (a1, . . . , an−1) ∈ Rn−1. This leads to∫Rχ(C1−a)(t)+(C2−a)(r−t)(y) dt =

∫RχC1(t+an)+C2(r+2an−(t+an))(y + 2a(n−1)) dt

=

∫RχC1(t)+C2(r+2an−t))(y + 2a(n−1)) dt

= δ(y + 2a(n−1), r + 2an)

and ∫ r

−∞χ(C1−a+C2−a)(t)(y) dt =

∫ r

−∞χ(C1+C2)(t+2an)(y + 2a(n−1)) dt

=

∫ r+2an

−∞χ(C1+C2)(t)(y + 2a(n−1)) dt

= h(y + 2a(n−1), r + 2an),

for all y ∈ Rn−1 and r ∈ R.Next, we fix some (y, r) ∈ Rn−1 × R and abbreviate δ = δ(y, r) and

h = h(y, r). Observe that, for t ∈ R, we have y ∈ C1(t) + C2(r − t) if andonly if there exists a y1 ∈ Rn−1 such that (y1, t) ∈ C1 and (y−y1, r− t) ∈ C2.This is equivalent to (C1 ∩ [(y, r)− C2])(t) 6= ∅ and therefore

δ = vol({t ∈ R | y ∈ C1(t) + C2(r − t)})= vol({t ∈ R | (C1 ∩ [(y, r)− C2])(t) 6= ∅})= vol(C1 ∩ [(y, r)− C2] | Hn),

where Hn = {x ∈ Rn | x1 = . . . = xn−1 = 0}. So, δ is the length of aninterval in R, since Ci is convex for i = 1, 2.

In order to get δ ≥ h, it suffices to show y /∈ (C1 + C2)(t), whenevert < r − δ, or equivalently (y, t) /∈ C1 + C2. Suppose it holds the contrary(y, t) ∈ C1 + C2. Then there exists (u, s) ∈ C1 such that (y − u, t − s) =(y − u, r + t − s − r) ∈ C2. From the assumption Rn

≥0 ⊂ Ci, we get Ci +Rn≥0 ⊂ Ci, i = 1, 2. And since r − t > 0, we obtain (u, s + r − t) ∈ C1 and

(y − u, t− s+ r − t) = (y − u, r − s) ∈ C2, which implies

(u, s) ∈ C1 ∩ [(y, r)− C2] and (u, s+ r − t) ∈ C1 ∩ [(y, r)− C2].

Thus, s and s + r − t belong to the projection (C1 ∩ [(y, r)− C2]) | Hn andwe have s+ r− t− s = r− t ≤ δ, which contradicts our choice of t < r− δ.�

7.9 Lemma. Let C1, C2 ⊂ Rn be two closed convex sets, such that thereare a, b ∈ Rn fulfilling a+ Rn

≥0 ⊂ Ci ⊂ b+ Rn≥0, for i = 1, 2. Then we have

χC1 ∗ χC2 ≥ χC1+C2 ∗ χRn≥0.


Proof. We proceed by induction on the dimension n. In the case n = 1, thereare αi ∈ R such that Ci = [αi,∞), i = 1, 2. This yields C1+C2 = [α1+α2,∞)and for all x ∈ R

χC1 ∗ χC2(x) =

∫RχC1(t)χC2(x− t) dt =

∫RχC1(t)χx−C2(t) dt

=

∫RχC1∩(x−C2)(t) dt =

∫Rχ[x−α2,α1](t) dt

= max{α1 + α2 − x, 0}.

Similarly, we get

χC1+C2 ∗ χR≥0(x) =

∫Rχ[x,α1+α2](t) dt = max{α1 + α2 − x, 0}.

Thus, in dimension one it even holds equality.Next, we assume n > 1. Recall from Lemma 7.8, that for t ∈ R and

i = 1, 2 we write Ci(t) = {y ∈ Rn−1 | (y, t) ∈ Ci}. Pick (x, r) ∈ Rn−1 × Rand denote y(n−1) = (y1, . . . , yn−1). It holds

χC1 ∗ χC2(x, r) =

∫Rn

χC1(y)χC2((x, r)− y) dy

=

∫Rn

χC1(y(n−1), yn)χC2(x− y(n−1), r − yn) dy

=

∫R

∫Rn−1

χC1(yn)(y(n−1))χC2(r−yn)(x− y(n−1)) dy(n−1) dy

=

∫RχC1(yn) ∗ χC2(r−yn)(x) dyn

≥∫

RχC1(yn)+C2(r−yn) ∗ χRn−1

≥0(x) dyn

=

∫R

∫Rn−1

χC1(yn)+C2(r−yn)(y(n−1))χRn−1≥0

(x− y(n−1)) dy(n−1) dyn

=

∫Rn−1

(∫RχC1(yn)+C2(r−yn)(y(n−1)) dyn

)χRn−1≥0

(x− y(n−1)) dy(n−1)

≥∫

Rn−1

(∫ r

−∞χ(C1+C2)(yn)(y(n−1)) dyn

)χRn−1≥0

(x− y(n−1)) dy(n−1)

=

∫Rn−1

∫Rχ(C1+C2)(yn)(y(n−1))χRn

≥0(r−yn)(x− y(n−1)) dyn dy(n−1)

=

∫Rχ(C1+C2)(yn) ∗ χRn

≥0(r−yn)(x) dyn


=

∫Rn

χC1+C2(y)χRn≥0

((x, r)− y) dy

= χC1+C2 ∗ χRn≥0

(x, r).

The first inequality above comes from the induction hypothesis and the sec-ond one from Lemma 7.8. �

Eventually, we are well prepared to show the main theorem.

Proof of Theorem 7.1. Since, M(C?n) = 4n

n!and because of Lemma 7.7 it

suffices to show that

L(χT (K) ∗ χT (K?))(1, . . . , 1) ≥ L(χT (C?n) ∗ χT (Cn))(1, . . . , 1). (7.3)

We have

T (Cn) = {y ∈ Rn | ∃(t1, . . . , tn) ∈ Cn with y = (− ln t1, . . . ,− ln tn)}= {y ∈ Rn | (e−y1 , . . . , e−yn) ∈ Cn)}= {y ∈ Rn | e−yi ≤ 1, 1 ≤ i ≤ n} = Rn

≥0,

and by Lemma 7.4, it holds T (K) + T (K?) = T (C?n). In Proposition 7.3

we have seen that the set T (K) is always convex for 1-unconditional K. Bydefinition it is also closed and with Lemma 7.9 we obtain

χT (K) ∗ χT (K?) ≥ χT (K)+T (K?) ∗ χRn≥0

= χT (C?n) ∗ χT (Cn).

This verifies (7.3) and therefore, we have indeed M(K) ≥ 4n

n!. �

Index

Bn, 6Bnp , 23

Cn, 6C?n, 6

Sn−1, 6M(·), 8Kn, 6Kn0 , 6χA, 7κn, 7d·e, 30vol(·), 7p-intersection, 19p-product, 19p-sum, 19|| · ||p, 231-unconditional

basis, 49body, 49

Blaschke, W., 9Borel measure, 35

even, 35odd, 35signed, 35

Borel set, 35Bourgain, J., 9

Cauchy projection formula, 35circumradius, 6convex body, 6cross-section, 54

digamma function, 31

eccentricity, 6

gamma function, 8generating measure, 37

Gordon, Y., 33

Hanner polytope, 28Hanner, O., 28Hausdorff

distance, 6limit, 34

homothetic bodies, 42

inradius, 6intersection body, 7

Kuperberg, G., 19

Laplace transform, 53Lopez, M. A., 44

Mahler Conjecture, 8Mahler volume, 8Mahler, K., 14Meyer, M., 10, 33Milman, V. D., 9

orthogonal projection, 7

Pajor, A., 10polar body, 6projection body, 7

radial function, 7Reisner, S., 33, 44, 49

Saint-Raymond, J., 30, 49Santalo, L. A., 9Steiner symmetrisation, 10support function, 7surface area measure, 35

zonoid, 34zonotope, 33

58

59

References

[1] K. Ball, Mahler’s Conjecture and Wavelets, Discrete Comput. Geom. 13(1995), 271-277.

[2] N. Batir, On some properties of digamma and polygamma functions, J.Math. Anal. Appl. 328 (2007), 452-465.

[3] W. Blaschke, Vorlesungen uber Differentialgeometrie II, Berlin - Heidel-berg - New York, 1923.

[4] J. Bourgain, V. D. Milman, New volume ratio properties for convexsymmetric bodies in Rn, Invent. Math. 88 (1987), 319-340.

[5] J. W. S. Cassels, Obituary of Kurt Mahler, Acta Arith. 58 (1991), 215-228.

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[8] O. Hanner, Intersections of translates of convex bodies, Math. Scand. 4(1956), 65-87.

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magdeburg.de/∼henk/lecture notes/notes convex discrete geo-

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[11] G. Kuperberg, From the Mahler conjecture to Gauss linking integrals,(2006), arXiv:math.MG/0610904v2.

[12] G. Kuperberg, A low-technology estimate in convex geometry, (1992),arXiv:math.MG/9211216v1.

[13] M. A. Lopez, S. Reisner, A Special Case of Mahler’s Conjecture, DiscreteComput. Geom. 20 (1998), 163-177.

[14] K. Mahler, Ein Minimalproblem fur konvexe Polygone, Mathematica(Zutphen) B7 (1938), 118-127.

60

[15] K. Mahler, Ein Ubertragungsprinzip fur konvexe Korper, Casopis Pest.Mat. Fys. 68 (1939), 93-102.

[16] M. Meyer, A. Pajor, On the Blaschke-Santalo inequality, Arch. Math.55 (1990), 82-93.

[17] M. Meyer, S. Reisner, Inequalities involving integrals of polar-conjugateconcave functions, Monatsh. Math. 125 (1998), 219-227.

[18] G. Pisier, The volume of convex bodies and Banach space geometry,Cambridge University Press, 1989.

[19] POV-Ray, http://www.povray.org/.

[20] S. Reisner, Random polytopes and the volume product of symmetric con-vex bodies, Math. Scand. 57 (1985), 386-392.

[21] S. Reisner, Zonoids with minimal volume product, Math. Z. 192 (1986),339-346.

[22] S. Reisner, Minimal volume-product in Banach spaces with a 1-uncondi-tional basis, J. London Math. Soc. 2 (1987), no. 36, 126-136.

[23] J. Saint-Raymond, Sur le volume de corps convexes symetriques, Semi-naire d’initation a l’analyse, 20e Annee, (1980/81), Exp. 11.

[24] L. A. Santalo, Un invariante afin para los cuerpos convexos del espaciode n dimensiones, Portugal. Math. 8 (1949), 155-161.

[25] R. Schneider, Convex bodies: The Brunn-Minkowski theory, CambridgeUniversity Press, 1993.

[26] T. Tao, Santalo’s inequality, http://www.math.ucla.edu/∼tao/pre-prints/Expository/santalo.dvi.

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