Topics in Convex Geometry and Phenomena in High Dimension

TOPICS IN CONVEX GEOMETRY AND PHENOMENA IN HIGH DIMENSION

DEPING YE

Submitted in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy

Dissertation Advisors: Dr. Stanislaw J. Szarek and Dr. Elisabeth M. Werner

Department of Mathematics

CASE WESTERN RESERVE UNIVERSITY

August, 2009 CASE WESTERN RESERVE UNIVERSITY

SCHOOL OF GRADUATE STUDIES

We hereby approve the thesis/dissertation of

______

candidate for the ______degree *.

(signed)______(chair of the committee)

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(date) ______

*We also certify that written approval has been obtained for any proprietary material contained therein. Table of Contents

Table of Contents ...... iii Dedication ...... v Acknowledgement ...... vi Abstract ...... vii

1 Introduction and Background 1 1.1 Introduction and Overview of Results ...... 1 1.2 Background on convex geometry and asymptotic geometric analysis . 7 1.2.1 Preliminaries on Convex Bodies ...... 7 1.2.2 Lp affine surface area and mixed p-affine surface area . . . . . 12 1.2.3 Inequalities related to the volume radius ...... 14 1.2.4 Symmetrization and Rogers-Shephard inequality ...... 19 1.2.5 John ellipsoid and Löwnerellipsoid ...... 20 1.2.6 Tensor products of convex bodies ...... 21 1.3 Background on Geometry of Quantum States ...... 23 1.3.1 Mathematical framework ...... 23 1.3.2 Peres-Horodecki’ positive partial transpose criterion ...... 24 1.3.3 Bures metric on D ...... 26 1.3.4 Hilbert-Schmidt and Bures measures on D ...... 32

2 New Lp Affine Isoperimetric Inequalities 40 2.1 L n affine surface area of the polar body ...... 40 − n+2 2.2 Lp affine surface areas ...... 47 2.3 Lp affine isoperimetric inequalities ...... 53

3 Inequalities for mixed p-affine surface area 61 3.1 Mixed p-affine surface area and related inequalities ...... 62 3.1.1 Inequalities for mixed p-affine surface area ...... 62 3.1.2 i-th mixed p-affine surface area and related inequalities . . . . 73 3.2 Illumination surface bodies ...... 80 3.3 Geometric interpretation of functionals on convex bodies ...... 86

iii 4 On the Bures Volume of Separable Quantum States 99 4.1 Hilbert-Schmidt volume of separable quantum states ...... 100 4.2 Bures volume of separable quantum states ...... 106 4.3 Optimality of the bounds ...... 116 4.3.1 Optimality of the lower bound ...... 116 4.3.2 Optimality of the upper bound ...... 117 4.4 Conclusion and Comments ...... 119 Bibliography ...... 121

iv Dedication

To my new born baby boy: Joseph Hengrui Ye. He is the best gift for my 30-years birthday and the completion of my Ph. D. degree in mathematics. I hope that he will have a bright future, a happy and healthy life.

To my lovely wife: Minglu Qian. My dissertation was started right after our marriage. Her strong support and encouragement were essential for the completion of my Ph.D. dissertation.

v ACKNOWLEDGEMENTS

I am greatly indebted to my advisors Dr. Stanislaw J. Szarek and Dr. Elisa- beth M. Werner, who are not only great advisors but also true mentors, for their patience, kindness and many beneﬁcial discussions. Their guidance made me become more and more mature, both academically and as a person. Their achievements and professionalism as mathematician were and will always be my models.

My thanks also go to the Mathematics Department at Case Western Reserve University. I would like to thank the Institut Henri Poincar´eand the Mathematics Department at Texas A&M University for providing excellent research environments during my visits. Special thanks go to my defense committee members Dr. Harsh Mathur and Dr. Mark Meckes.

My research has been partially supported by grants from the National Science Foundation (U.S.A.).

Finally, I would like to give my deepest gratitude to my wife Minglu for her encouragement and understanding during the past few years. My appreciation also goes to my supportive family: my parents Jinhe and Baihua, my sister Zhongying, my grandmother-in-law Yu’e, and my parents-in-law Guangzhen and Jihong for their dedication and ongoing support.

vi Topics in Convex Geometry and Phenomena in High Dimension

Abstract

DEPING YE

This dissertation deals with topics in convex geometry and phenomena in high dimension. Convex geometry studies geometry of convex bodies of ﬁxed dimension, while phenomena in high dimension aims to understand the structure of large dimensional objects.

Our objects of study in convex geometry are related to Lp aﬃne surface areas

and mixed p-affine surface areas. In particular, we prove new Lp affine isoperimetric inequalities for all p ∈ [−∞, 1). For p ≥ 1, such inequalities had been established earlier. We generalize these inequalities to the analogous inequalities for mixed p-affine surface area. We also prove new Alexandrov-Fenchel type inequalities for mixed p- affine surface area. To generalize the Lp affine surface and mixed p-affine surface area to all p, we study the asymptotic behavior of the volume of the illumination surface bodies and the polar bodies of the surface bodies. Properties of the illumination surface bodies are established, for instance, we show that the illumination surface bodies are star-convex, but not necessarily convex. As applications of our asymptotic results, we give geometric interpretations of functionals associated with convex bodies, such as Lp affine surface area, surface area, and mixed p-affine surface area. Moreover, we establish, for all p 6= −n, a duality formula which shows that Lp affine surface area

vii ◦ of a convex body K equals L n2 aﬃne surface area of the polar body K . p In phenomena in high dimension, we will study the relative sizes of various sets of quantum states. We obtain two sided estimates for the Bures volume of an arbitrary subset of the set of N × N density matrices in terms of the Hilbert-Schmidt volume of that subset. For general subsets, our results are essentially optimal (for large N). As applications, we derive, in particular, nontrivial lower and upper bounds for the Bures volume of sets of separable states and for sets of states with positive partial transpose.

Key words: affine isoperimetric inequality, Lp affine surface area, Lp Brunn- Minkowski theory, mixed p-affine surface area, Bures metric, Bures volume, separable states, positive partial transpose.

viii Chapter 1

Introduction and Background

1.1 Introduction and Overview of Results

Convex geometry studies the geometry of convex bodies in ﬁxed dimension, while asymptotic geometric analysis strives to understand structures in high dimensions by using tools of convex geometry, functional analysis and probability. Asymptotic geometric analysis provides “isomorphic” rather than “isometric” solutions of many open problems in geometry and such a relaxation often makes these problems solvable.

One part of the dissertation will focus on Lp aﬃne surface area and mixed p- aﬃne surface area. In particular, we prove new isoperimetric inequalities and new Alexandrov-Fenchel type inequalities. Another part of the dissertation is about ob- taining nontrivial bounds for the Bures volume of sets of quantum separable states for large dimension.

Isoperimetric inequalities play an important role in many areas of mathematics. For instance, the classical isoperimetric inequality, which compares the area of a convex body K with its volume, is an extremely powerful tool in geometry and related areas. Many isoperimetric inequalities have an affine invariant flavor. As an example we mention the celebrated Blaschke-Santalóinequality which can be obtained as a consequence of the classical affine isoperimetric inequality [78]. This classical affine

1 2 isoperimetric inequality provides an upper bound of as(K), the affine surface area of a convex body K, in terms of its volume. More generally, affine isoperimetric inequalities compare two functionals associated with convex bodies (or more general sets) where the ratio of the functionals is invariant under non-degenerate linear or affine transformations.

The affine surface area, as(K), of a convex body K was introduced by Blaschke in 1923 for dimension 2 and 3 in [13]. Originally a basic concept from the field of affine differential geometry, it has recently attracted increasing attention (e.g. [4, 5, 10, 48, 53, 55, 56, 60, 69, 84]). Affine surface area has many nice and useful properties. Aside from the fact that there is an affine isoperimetric inequality associate with it, it is invariant under affine maps and it is a valuation. It-in a sense-“measures” the boundary behavior of a convex body and therefore appears naturally and plays crucial roles in the theory of approximation of convex bodies by polytopes. We refer the readers to [33, 86, 57] for the details and more references. Also it is the subject of the affine Plateau problem solved in R3 by Trudinger and Wang [102], and Wang [108].

In the ground breaking paper [62], Lutwak introduced a generalization of the

classical affine surface area, Lp affine surface area for all p ≥ 1. When p = 1, L1 affine

surface area is just the classical affine surface area. The notion of Lp affine surface area was further extended to all p and to general convex bodies in, e.g., [70, 85, 86]. In fact, extensions were obtained by investigating the asymptotic behavior of the volume of certain families of (convex) bodies. These ideas were already successfully used in the study of the geometric interpretation of the classical affine surface area via the convex floating body by Schüttand Werner [84] and the convex illumination body by Werner [109], (see also [51, 69].)

Lp affine surface area is now at the core of the rapidly developing Lp Brunn- Minkowski theory. Contributions here include the discovery of new ellipsoids [54, 64], the study of solutions of nontrivial ordinary and, respectively, partial differential 3 equations (see e.g. Chen [18], Chou and Wang [20], Stancu [96, 97]), and the study of the Lp Christoffel-Minkowski problem by Hu, Ma and Shen [44].

A generalization of Lp affine surface area is mixed p-affine surface area which, for p ≥ 1, was also introduced by Lutwak in [62]. Aside from Lp affine surface area, other special cases of mixed p-affine surface area are the dual mixed volume and the surface area. In this dissertation, we will extend the definition of the mixed p-affine surface area to all p ∈ [−∞, ∞]. To do so, we will provide new geometric interpretations for functionals on convex bodies. In particular, for Lp affine surface area, mixed p- affine surface area, and i-th mixed p-affine surface area (see below for the definitions). In Chapter 2, we investigate the asymptotic behavior of the volume of the polar of surface bodies, and give a new geometric interpretation of Lp affine surface area. As ◦ ◦ an application, we show the duality formula as n2 (K ) = asp(K) for p 6= −n where K p is the polar body of K. This duality formula was proved by Hug [47] for p > 0 using a different method. In Chapter 3, we construct a new class of bodies, the illumination surface bodies, and study the asymptotic behavior of their volumes. We show that the illumination surface bodies are not necessarily convex, thus introducing a novel idea in the theory of geometric characterizations of functionals on convex bodies, where to date only convex bodies were used (e.g. [70, 85, 86]).

The classical affine isoperimetric inequality is fundamental in many problems. Besides the ones already mentioned earlier, see also, e.g., [30, 31, 63, 82]. More recently, it was used, e.g., by Andrews [6, 7], Sapiro and Tannenbaum [79] to show the uniqueness of self-similar solutions of the affine curvature flow and to study its

asymptotic behavior. Lp aﬃne isoperimetric inequalities for all p ≥ 1 were ﬁrst proved

by Lutwak in [62]. He gave upper bounds of the Lp aﬃne surface area of K for all p ≥ 1

in terms of the volume of K. We derive new Lp aﬃne isoperimetric inequalities for all

p ∈ [−∞, 1) in Chapter 2. These Lp affine isoperimetric inequalities can be viewed as extensions of the Blaschke-Santalóinequality and inverse Santalóinequality (due to Bourgain and Milman [14] (see also Kuperberg [49])). We also provide examples 4 to show that these inequalities cannot be improved.

In Chapter 3, we prove isoperimetric inequalities for mixed p-affine surface area . For mixed p-affine surface area, Alexandrov-Fenchel type inequalities (for p = 1, ±∞) and affine isoperimetric inequalities (for 1 ≤ p ≤ n) were first established by Lutwak in [58, 59, 62]. We derive new Alexandrov-Fenchel type inequalities for mixed p-affine surface area for all p ∈ [−∞, ∞] and new mixed p-affine isoperimetric inequalities for all p ∈ [0, ∞]. Classification of the equality cases for all p in the Alexandrov-Fenchel type inequalities for mixed p-affine surface area is related to the uniqueness of solutions

of the Lp Minkowski problem (e.g., [18, 20, 61, 63, 65, 66, 96, 97]), which is unsolved for many cases. Thus the situation is similar to the classical Alexandrov-Fenchel inequalities for mixed volume, where the complete classiﬁcation of the equality cases is also an unsolved problem.

Phenomena in high dimension studies geometric objects in dimension n, as n becomes large. The ultimate goal is to obtain uniform estimations independent of dimension n, or to obtain estimations with best dependence on dimension n. A typical result is the Bourgain-Milman’s version of inverse Santal´oinequality, which says that the volume radius product of K and K◦ is essentially a constant. This inverse Santal´o inequality provides an “isomorphic” solution of the Mahler conjecture (see Section 1.2.3 for details).

The techniques from phenomena in high dimension have been used to study the relative size of the set of separable quantum states within the set of all quantum states. In [8, 99], the authors obtained rather striking results: in large dimension, all but extremely few quantum states are entangled (not separable), in the sense of Hilbert-Schmidt measure. Moreover, they proved that the powerful positive partial transpose criterion is not precise in large dimension as a tool to detect the separability, in the sense of Hilbert-Schmidt measure. Their results rely on the Euclidean structure of the Hilbert-Schmidt metric. An arguably more important measure is the Bures measure, which is induced by the non-Euclidean Bures metric. In Chapter 4, we 5 deal with the relative size of the set of separable quantum states (and other sets of interest) within the set of all quantum states in terms of the Bures measure.

Quantum entanglement was discovered in 1930’s [25, 83] and is now at the heart of quantum computation and quantum information. The key ingredients in quantum algorithms such as Shor’s algorithm for integer factorization [87] or Deutsch-Jozsa algorithm (see e.g. [72]), are entangled quantum states, i.e., those states which can not be represented as a mixture of tensor products of states on subsystems. Fol- lowing [112], states that can be so represented are called separable states. Since determining whether a state is entangled or separable is in general a difficult problem [34], sufficient and/or necessary conditions for separability are very important in quantum computation and quantum information theory, and have been studied extensively in the literature (see e.g. [38, 39, 40, 41, 42, 43, 73]). One well-known tool is the Peres-Horodecki’s positive partial transpose (PPT) criterion [38, 73], that is, if a state on H = CD1 ⊗ CD2 · · · ⊗ CDn is separable then its partial transpose must be positive. Equivalently, if a state on H does not have positive partial transpose, it must be entangled. This criterion works perfectly, namely, the set of separable states S = S(H) equals to the set of states with positive partial transpose PPT = PPT (H) for H = C2 ⊗ C2 (two-qubits), H = C2 ⊗ C3 (qubit-qutrit), and H = C3 ⊗ C2 (qutrit- qubit) [38, 79, 113]. However, entangled states with positive partial transpose appear in the composite Hilbert space H = C2 ⊗ C4 and H = C3 ⊗ C3 [39] (and of course in all “larger” composite spaces). The results in [8, 99] show that, in some measures, Qn the positive partial transpose criterion becomes less and less precise as N = i=1 Di grows to infinity. This is inferred by comparing the Hilbert-Schmidt volumes of S and PPT . The methods in [8, 99], which rely on the special geometric properties of the Hilbert-Schmidt metric and tools of phenomena in high dimension, can also be employed to derive tight estimates for the Hilbert-Schmidt volume of D = D(H) (the set of all states on H). However, a closed expression for the exact value of this volume is known; it was found in [115] via random matrix theory and calculating some nontrivial multivariate integrals. 6

Compared with the Hilbert-Schmidt metric, the Bures metric on D [16, 103] is, in some sense, more natural. It has attracted considerable attention (see e.g. [24, 23, 22, 46, 45, 106, 105, 104]). The Bures metric is Riemannian [74] but not flat. It is monotone, i.e., it does not increase under the action of any completely positive, trace preserving map. It induces the Bures measure [12, 35, 95], which has singularities on the boundary of D. The Bures volume of D has been calculated exactly in [95] and 2 1 happens to be equal to the volume of an (N − 1)-dimensional hemisphere of radius 2 [12, 95]. (This mysterious fact does not seem to have a satisfactory explanation.) On the other hand, the precise Bures (or Hilbert-Schmidt) volumes of S and PPT are rather difficult to calculate since the geometry of these sets is not very well understood and the relevant integrals seem quite intractable. These quantities can be used to measure the priori Bures probabilities of separability and of positive partial transpose within the set of all quantum states. (Here priori means that the state is selected randomly according to the Bures measure and no further information about it is available.) For small N, e.g., N = 2 × 2 and N = 2 × 3, the Bures volume of S (hence of PPT ) has been extensively studied by numerical methods in [93, 92, 91, 90, 89, 88]. For large N, the asymptotic behavior of the Hilbert-Schmidt volume of S and PPT was successfully studied in [8, 99]. Based on that work, we derive qualitatively similar “large N” results for the Bures volume. In summary, our results state that the relative size of S within D is extremely small for large N (see Corollaries 7 and 8 for detail). On the other hand, the corresponding relative size for PPT within D is, in the Bures volume radius sense (see Section 1.3.4 for a precise definition), bounded from below by a universal (independent of N) positive constant (see Corollary 9). In conclusion, the priori Bures probability of finding a separable state within PPT is exceedingly small is when N is large. In other words, we have shown that, as a tool to detect separability, the positive partial transpose criterion for large N is not precise in the priori Bures probability sense. Its effectiveness to detect entanglement is less clear (see comments following Corollary 9). 7 1.2 Background on convex geometry and asymptotic geometric analysis

1.2.1 Preliminaries on Convex Bodies

A convex body K in Rn is a convex, compact subset of Rn with nonempty interior. Unless stated otherwise, here we will always assume that the centroid of a convex n n n body K in R is at the origin. We use B2 to denote the Euclidean ball in R and n−1 n n−1 S to denote the unit sphere, the boundary of B2 . For u ∈ S , the support function of a convex body K is deﬁned as

hK (u) = maxhx, ui x∈K where h·, ·i is the standard inner product on Rn which induces the Euclidian norm k · k. The radial function of K at u ∈ Sn−1 is

ρK (u) = max{λ : λu ∈ K}.

n−1 + Both hK (·) and ρK (·) are functions from S to R and both determine a convex body uniquely. The volume of K can be calculated as Z 1 n |K| = ρK (u) dσ(u) n Sn−1 where σ is the usual surface area measure on Sn−1. More generally, for a set M, we use |M| to denote the Hausdorﬀ content of its appropriate dimension.

For any convex body K with the origin in its interior, K◦ = {y ∈ Rn, hx, yi ≤ 1, ∀x ∈ K} is the polar body of K. K◦ is a convex body with the origin in its interior. The bipolar theorem (see [82]) states that (K◦)◦ = K. It is easy to check that for u ∈ Sn−1, 1 1 ρK (u) = , and ρK◦ (u) = . hK◦ (u) hK (u) Hence the volume of K◦ is Z Z ◦ 1 n 1 1 |K | = ρK◦ (u) dσ(u) = n dσ(u). n Sn−1 n Sn−1 hK (u) 8

n Let K be a convex body in R with the origin in its interior. The gauge k · kK of K is deﬁned as

kxkK = inf{λ > 0 : x ∈ λK}.

Then one has for all u ∈ Sn−1,

1 kukK = = hK◦ (u). ρK (u)

If K is 0-symmetric, i.e., K = −K, k · kK is the norm which has K as its unit ball.

If K and L are two convex bodies in Rn with the origin in their interiors, we define the geometric distance between K and L as the product αβ where α and β are the smallest numbers such that β−1L ⊂ K ⊂ αL. This quantifies how the “shapes” of K and L differ.

2 We use ∂K to denote the boundary of convex body K. We write K ∈ C+, 2 if ∂K is C with everywhere strictly positive Gaussian curvature κK (x). We use

NK (x) to denote the unit outer normal vector of x ∈ ∂K. Thus, the hyperplane n H(x, NK (x)) = {y ∈ R : hNK (x), yi = hNK (x), xi} passing through x with normal 2 vector NK (x) is the tangent hyperplane of ∂K at x. In the case K ∈ C+, the Gauss n−1 1 map NK (·): ∂K → S is C . The hyperplane H(x, NK (x)) can be identified with n−1 R with (any) orthonormal basis, say {e˜1, ··· , eñ−1}. Then {e˜1, ··· , eñ−1,NK (x)} forms an orthonormal basis of Rn with the origin at x. In a neighborhood of x ∈ ∂K, ∂K can be represented by a convex function f : Rn−1 → R with f(0) = 0 as following: for any point z ∈ ∂K in a neighborhood of x,

z = x + (t1, ··· , tn−1, −f(t1, ··· , tn−1)).

Then κK (x), the Gaussian curvature of K at x ∈ ∂K, is deﬁned as the determinant of 2 the Hessian matrix of f at 0. K ∈ C+ implies that κK (·) is a continuous function on ∂K. Moreover, the spherical measure σ is absolutely continuous with respect to the

pushforward of the surface area measure µK (or the (n − 1)-dimensional Hausdorﬀ 9 measure) on ∂K along NK , and vice versa. Hence, by the Radon-Nikodym theorem, n−1 there is a function fK (·): S → R, such that

dµ (x) K = f (u) dσ(u) K and also dσ(u) = κK (x) dµK (x)

1 where u = NK (x). Consequently, κK (x) = . The function fK (u) is the curvature fK (u) 2 function of the convex body K ∈ C+ at the direction u and fK (·) is a continuous n−1 function on S . In fact, the curvature function fK (·) is closely related to the support function and can be determined as follows (see Corollary 2.5.3 in [82]).

Corollary 1 fK (u) equals to the product of all nonzero eigenvalues of the Hessian n n−1 matrix (with respect to any orthonormal basis of R ) of hK at u ∈ S .

For any convex body K, its volume can be calculated as follows:

1 Z |K| = hx, NK (x)i dµK (x). n ∂K

2 If K ∈ C+, the above formula is equivalent to

1 Z |K| = fK (u)hK (u) dσ(u). n Sn−1

For two convex bodies K and L with the origin in their interiors, and λ,η ≥ 0 (not both zero), the Minkowski linear combination λK + ηL is the convex body with n−1 support function hλK+ηL where for u ∈ S

hλK+ηL(u) = λhK (u) + ηhL(u).

A fundamental inequality in convex geometry is the Brunn-Minkowski inequality. This inequality also holds for more general compact sets instead of convex bodies. 10

Theorem 1 (Brunn-Minkowski inequality) Let K and L be two convex bodies in Rn. Then the volume measure is log-concave. That is, for all t ∈ [0, 1],

|tK + (1 − t)L| ≥ |K|t |L|1−t,

with equality for all t ∈ [0, 1] if and only if K = L + a for some a ∈ Rn. Equivalently,

1 1 1 |K + L| n ≥ |K| n + |L| n ,

with equality if and only if K = cL + a for some a ∈ Rn and c ∈ R.

The mixed volume of K and L, denoted by V1(K,L), is deﬁned by

|K + L| − |K| V1(K,L) = lim . →0 n

n It is easy to verify that nV1(K,B2 ) coincides with the usual surface area of K, i.e., n nV1(K,B2 ) = µK (K). Moreover, V1(K,K) = |K|. Alexandrov [2] and Fenchel and Jessen [26] have shown that for each convex body K there is a positive Borel measure S(K, ·) on Sn−1, such that, for all convex bodies L,

1 Z V1(K,L) = hL(u) dS(K, u). n Sn−1

As noted in [60], the measure S(K, ·) has the following simple geometric description: n−1 for any Borel subset A of S , S(K,A) is just the µK measure of the set of points of ∂K which have an unit outer normal vector in A, namely

S(K,A) = µK ({x ∈ ∂K : ∃u ∈ A, s.t., H(x, u) is a support hyperplane of ∂K at x}).

2 Note that S(K, ·) is in general diﬀerent from µK . For K ∈ C+, the measure pushfor-

ward of µK by NK coincides with S(K, ·), and hence S(K, ·) is absolutely continuous with respect to the spherical measure σ on Sn−1. The Radon-Nikodym theorem then implies that dS(K, u) = f (u). dσ(u) K 11

Good general references for this material are [17, 50].

The Minkowski linear combination and the mixed volume V1(·, ·) can be generalized to all p ≥ 1. For convex bodies K and L in Rn with the origin in their interiors and λ, η ≥ 0 (not both zeros), the Firey p-sum λK +p ηL for p ≥ 1 [28] is the convex body with support function

p p p hλK+pηL(u) = λ hK (u) + η hL(u) .

If p = 1, this is the Minkowski sum of two convex bodies λK and ηL.

The p-mixed volume of K and L, denoted by Vp(K,L), was deﬁned by Lutwak [61]

1 |K +p L| − |K| Vp(K,L) = lim . (1.1) p →0 n

It has been proved by Lutwak [61] that analogous to above, for each convex body K n in R there is a measure Sp(K, u) such that

Z 1 p Vp(K,L) = hL(u) dSp(K, u). n Sn−1

The measure Sp(K, ·) is absolutely continuous with respect to the measure S(K, ·) and the Radon-Nikodym theorem implies that for all p ≥ 1,

dS (K, u) p = h1−p(u). dS(K, u) K

2 If K ∈ C+, dS (K, u) p = f (u) h1−p(u). dσ(u) K K

The Minkowski mixed volume inequality says that for any two convex bodies K and L, n n−1 V1(K,L) ≥ |K| |L| 12

with equality if and only if K and L are dilates. The analogous inequality for p ≥ 1 is proved in [61]: for any two convex bodies K and L, and any p ≥ 1,

n n−p p Vp(K,L) ≥ |K| |L|

with equality if and only if K and L are dilates. We refer readers to, e.g., [61, 62] for

more properties of the p-mixed volume Vp(·, ·).

1.2.2 Lp aﬃne surface area and mixed p-aﬃne surface area

n A subset L of R is star-shaped (about x0 ∈ L) if there exists x0 ∈ L such that the

line segment [x0, x], from x0 to any point x in L, is contained in L.

Let K be a convex body with its center at the origin and let L be a star-convex body (about the origin), i.e., L is a compact star-shaped subset (about the origin) ◦ with continuous radial function ρL(·). For all p ≥ 1, deﬁne Vp(K,L ) by

Z ◦ 1 −p Vp(K,L ) = ρL(u) dSp(K, u). n Sn−1

−1 For a convex body L, ρL(u) = hL◦ (u) , and then this deﬁnition agrees with the previous one.

For p ≥ 1, Lutwak deﬁned the Lp aﬃne surface area of a convex body K [62], by

n ◦ n p o asp(K) = inf n Vp(K,L ) n+p |L| n+p (1.2) where the infimum runs over all star-convex bodies (about the origin). This definition generalizes the definition of affine surface area (for p = 1) of [52].

If the boundary of K is sufficiently smooth, then the Lp affine surface area defined in (1.2) can be written as an integral over Sn−1 (see Theorem 4.4 in [62]),

n Z n+p fK (u) asp(K) = n(p−1) dσ(u). n−1 S hK (u) n+p 13

The deﬁnition of Lp aﬃne surface area was further generalized to all convex body and all p 6= −n in [86] by

p Z n+p κK (x) asp(K) = n(p−1) dµK (x) (1.3) ∂K hx, NK (x)i n+p and Z κK (x) as±∞(K) = n dµK (x) (1.4) ∂K hx, NK (x)i provided the above integrals exist. In particular, the volume of K corresponds to p = 0, Z as0(K) = hx, NK (x)i dµK (x) = n|K|, ∂K

◦ 2 and the volume of K corresponds to p = ±∞ if K ∈ C+,

Z 1 ◦ as±∞(K) = n dσ(u) = n|K |. (1.5) Sn−1 hK (u)

Moreover the classical aﬃne surface area of K, as(K), is the case p = 1, namely

Z 1 as(K) = κK (x) n+1 dµK (x). ∂K as(K) is independent of the position of K in space. This is not the case when p 6= 1.

One important property of the Lp aﬃne surface area is that it is aﬃne invariant for all p 6= −n. This follows from the fact that for any convex body K in Rn and any liner transformation T , we have [62, 86]

For all real p ≥ 1, the mixed p-aﬃne surface area, asp(K1, ··· ,Kn), of n convex 2 bodies Ki ∈ C+ was introduced in [62]

1 Z n+p as (K , ··· ,K ) = h (u)1−pf (u) ··· h1−pf (u) dσ(u). (1.6) p 1 n K1 K1 Kn Kn Sn−1

We propose here to extend the definition (1.6) for mixed p-affine surface area to all p 6= −n. In Section 3.2, we also give a definition for (−n)- mixed affine surface area.

We show that mixed p-aﬃne surface areas are aﬃne invariants for all p. Note that for p = ±∞,

Z 1 1 as±∞(K1, ··· ,Kn) = ··· dσ(u) Sn−1 hK1 (u) hKn (u) ˜ ◦ ◦ = nV (K1 , ··· ,Kn) (1.7)

˜ ◦ ◦ ◦ ◦ where V (K1 , ··· ,Kn) is the dual mixed volume of K1 , ··· ,Kn, introduced by Lutwak in [58]. If all Ki coincide with K, then for all p 6= −n

n Z n+p fK (u) asp(K, ··· ,K) = n(p−1) dσ(u) = asp(K). (1.8) n−1 S hK (u) n+p

Note further that the surface area of K can be written as (−1)-th mixed 1-affine n surface area of K and the Euclidean ball B2 (see Section 3.2). Thus, mixed p-affine surface area is an extension of dual mixed volume, surface area, and Lp affine surface area. A formula corresponding to (1.8) also holds for p = −n.

1.2.3 Inequalities related to the volume radius

For any convex body K, the volume ratio of K, vrad(K), is deﬁned as

1 |K| n vrad(K) = n . |B2 |

This quantity equals to the radius of the ball whose volume is the same as |K|, and hence it is a natural measure of the size of K in terms of volume. 15

As an immediate consequence of the Brunn-Minkowski inequality (see Theorem 1), one has the following isoperimetric inequality.

Theorem 2 (Isoperimetric Inequality) Let K be a convex body in Rn. Then

1 1 n n−1 |K| µK (K) n ≤ n |B2 | n|B2 | with equality if and only if K is a ball.

This classical isoperimetric inequality is one example of isoperimetric inequalities. It compares the surface area of a convex body K to its volume. In Chapter 2 and

Chapter 3, we will study related isoperimetric inequalities for Lp affine surface area and mixed p-affine surface area. Note that µK (K) is a special case of mixed p-affine surface area (see Chapter 3 for details).

Theorem 3 (Blaschke-Santal´oinequality) Let K be a convex body with centroid at the origin. Then vrad(K) · vrad(K◦) ≤ 1 with equality if and only if K is a 0- ◦ n 2 centered ellipsoid. Equivalently, one has |K| |K | ≤ |B2 | with equality if and only K is a 0-centered ellipsoid.

Remark. The Blaschke-Santal´oinequality says that the volume product |K| |K◦| attains its maximum at 0-centered ellipsoids. We refer the reader, for instance, to [68] for a proof based on Steiner symmetrization and the Brunn-Minkowski inequality. On the other hand, the precise lower bound (or minimum) of the volume product |K| |K◦| is still open, and this problem is known as Mahler Conjecture.

Mahler Conjecture (symmetric case) Let K be a 0-symmetric convex body in n ◦ 4n R , i.e. K = −K. Is the volume product of |K| |K | at least n! (with equality if, for n instance, K = B∞)?

Remark. It is known that if the Mahler conjecture is true, there are many minimiz- ers, at least for n ≥ 4. 16

Mahler Conjecture (nonsymmetric case) Let K be any convex body in Rn. Is ◦ (n+1)n+1 the volume product of |K| |K | at least (n!)2 with equality if and only if K is a simplex?

Bourgain and Milman in [14] proved an asymptotic version of these conjectures, which solved the Mahler conjecture up to a constant.

Theorem 4 (Inverse Santal´oinequality) There exists a universal constant c > 0, such that, for every n ≥ 1, and for every convex body K in Rn, one has

vrad(K) vrad(K◦) ≥ c.

Equivalently, ◦ n n 2 |K| |K | ≥ c |B2 | .

Together with the Blaschke-Santal´oinequality, one gets that the volume radius product of K and K◦ is essentially a constant (independent of dimension n). See also Kuperberg [49] for an improvement of constant c in the inverse Santal´oinequality.

Lemma 1 Let K be any convex body in Rn with the origin in its interior. Then

1 1 n Z n |K| −n n = kykK dy |B2 | Sn−1

n−1 dσ with dy the normalized standard measure on the sphere S , i.e., dy = n . n|B2 |

The proof of this lemma is simple. By the polar coordinates, we have

1 Z Z kyk Z n K n−1 n −n |K| = |B2 | nr dr dy = |B2 | kykK dy Sn−1 0 Sn−1

where dy is the normalized measure of the sphere Sn−1. 17

Theorem 5 (Urysohn Inequality) Let K be any convex body in Rn with the origin in its interior. Then

1 |K| n Z vrad(K) = n ≤ kykK◦ dy |B2 | Sn−1

where dy is the normalized standard measure on the sphere Sn−1.

Remark. By Lemma 1, one has

1 Z n Z −n kykK dy ≤ kykK◦ dy. Sn−1 Sn−1

Lemma 2 Suppose f(x): → satisﬁes R |f(x)| dx < ∞ and R |f(x)|−n dx < ∞, R R R R then n 1 Z 1+n Z n+1 1 ≤ |f(x)| dx |f(x)|−n dx . R R Equivalently, one has

− 1 Z n Z |f(x)|−n dx ≤ |f(x)| dx. R R

Proof. This lemma follows H¨older inequality:

Z n 1 = |f(x)| · |f(x)|−1 n+1 dx R 1 n Z n+1 n+1 Z n+1 n+1 − n n n ≤ |f(x)| n+1 |f(x)| n+1 R R 1 n Z n+1 Z n+1 ≤ |f(x)|−n dx |f(x)| dx . R R Proof of Theorem 5. If K = −K is a 0-symmetric convex body, then the Blaschke- Santal´oinequality shows that

1 − 1 |K| n |K◦| n n ≤ n . |B2 | |B2 | 18

Lemma 1 implies that

1 − 1 − 1 Z n ◦ n Z n −n |K | −n vrad(K) = kykK dy ≤ n = kykK◦ dy . Sn−1 |B2 | Sn−1

Employing Lemma 2 with f(y) = kykK◦ , we get the Urysohn inequality for symmetric convex body K. The Blaschke-Santal´oinequality holds also for not-necessarily- symmetric convex body after an appropriate translation, and hence the above inequality also holds for any bounded convex body.

The mean width of K is the mean of the support functions in all directions, that is, Z Z ω(K) := hK (y) dy = maxhx, yi dy Sn−1 Sn−1 x∈K where dy is the normalized Lebesgue measure on the sphere Sn−1. The mean width can be rewritten as Z ω(K) = kykK◦ dy. (1.9) Sn−1 So the Urysohn inequality is equivalent to vrad(K) ≤ ω(K) for any convex body K with the origin in its interior.

A more convenient quantity related to the mean width ω(K) is the so-called

Gaussian mean width, denoted by ωG(K) and deﬁned as

ωG(K) = maxhx, Gi E x∈K

with G ∼ N (0, Idn) Gaussian vector, and E is the expectation with respect to G.A

simple and routine calculation by passing to polar coordinates shows that ωG(K) = √ 1+n √ Γ( 2 ) τn ω(K) with τn = 2 n . This constant is close to n. More precisely, τn ∈ Γ( 2 ) √ √ ( n − 1, n).

2 Theorem 6 Let X1, ··· ,Xv be Gaussian random variables with EXi ≤ 1, for any i = 1, ··· , v, then p max Xi ≤ 2 log n. E 1≤i≤n 19

We refer the reader to [101] for a nice proof. From this theorem, one immedi- √ n ately has that ωG(K) ≤ 2 ln v, where K = conv({x1, ··· , xv}) ⊂ B2 is a poly- v n v n tope with v-vertices {xi}i=1 ⊂ B2 . Moreover, if K = conv({±xi}i=1) ⊂ B2 is a 0-symmetric polytope with 2v > 2 vertices, this inequality also holds, that is, √ v ωG(conv({±xi}i=1)) ≤ 2 ln v (see e.g. [8, 27]). Hence, combining with Urysohn’s lemma, one gets √ v v 1 v 1 vrad(conv({±xi}i=1)) ≤ ω(conv({±xi}i=1)) = ωG(conv({±xi}i=1)) < 2 ln v. τn τn (1.10)

1.2.4 Symmetrization and Rogers-Shephard inequality

Let H be an aﬃne hyperplane in Rn+1, not containing the origin, and K be any convex body in H. We deﬁne the symmetric convex body Ω as: Ω := conv(−K ∪ K). The following theorem gives an estimation of the volume of Ω based on the volume of K. One reason to consider this symmetrization is that much more information is available about centrally symmetric convex bodies as opposed to nonsymmetric convex bodies.

Theorem 7 Let H be a hyperplane in Rn+1, not containing the origin and with distance h from H to the origin. Suppose K be a convex body in H and Ω = conv(−K ∪ K). Then 2n 2h |K| ≤ |Ω| ≤ 2h |K|. n + 1 Remark. The left hand inequality is an immediate consequence of the Brunn- Minkowski inequality. The equality holds for the left hand inequality if K is centrally symmetric (with respect to some point x0 ∈ K). The right hand inequality is the so-called Rogers-Shephard inequality [77]. The equality holds for the right hand 2n inequality if K is a simplex. Although n+1 looks large, its n-th root is smaller than 2. This n-th root is more important here because we are ultimately interested in comparing the volume radii. 20

1.2.5 John ellipsoid and L¨owner ellipsoid

T (x) = |vihu|xi = hu, xi|vi.

Theorem 8 (John’s Theorem) There is a unique maximal ellipsoid contained in the convex body K ⊂ Rn, called the John ellipsoid, which we denote J (K). The John n ellipsoid coincides with B2 if and only if

n 1. B2 ⊂ K,

n−1 2. There exists a ﬁnite family (ui, ci), where ui ∈ S ∩ ∂K and ci are positive P numbers, such that, i ciui = 0, and the identity Idn can be decomposed as P n P 2 2 Idn = i ci|uiihui|, or ∀x ∈ R , ci|hx, uii| = kxk .

Similarly, there exists a unique ellipsoid of minimal volume containing K, called Löwner ellipsoid, which we denote L(K). Moreover, the Löwnerellipsoid of K equals n n to B2 if and only if K ⊂ B2 , and there exists a finite family (ui, ci) satisfy the conditions in the John’s theorem.

P For a proof of this theorem, see [9]. The sequence ci also satisfies that i ci = n P by taking trace from both sides of Idn = i ci|uiihui|. The John ellipsoid and the Löwner ellipsoid are affine invariant, namely, for any affine map T , J (TK) = T J (K) and L(TK) = T L(K). Moreover, for any symmetric convex body K, J (K◦) = L(K)◦ ◦ ◦ ◦ n 2 and L(K ) = J (K) which is a consequence of the formula |E| |E | = |B2 | for any 0− symmetric ellipsoid E. 21

1.2.6 Tensor products of convex bodies

0 If K and K0 are two convex bodies, respectively, in Rn and Rn , their projective tensor product (e.g., [8]) is deﬁned as

K⊗ˆ K0 = conv{x ⊗ x0, x ∈ K, x0 ∈ K0}.

0 Clearly K⊗ˆ K0 is a convex body in Rn ⊗ Rn .

n 0 n0 For E1 = SB2 and E2 = S B2 , their Hilbertian tensor product is the ellipsoid 0 nn0 E1 ⊗2 E2 := (S ⊗ S )B2 . This deﬁnition does not depend on the choice of operators S and S0. The following lemma shows that the L¨ownerellipsoid behaves well with respect to the projective tensor product. Its proof can be found in [8] and we omit the proof here.

0 Lemma 3 Let K ⊂ Rn and K0 ⊂ Rn be two convex bodies. Then the L¨ownerellipsoid of their projective tensor product is the Hilbertian tensor product of the respective L¨ownerellipsoids.

0 In terms of scalar product, for every x, y ∈ Rn, and x0, y0 ∈ Rn ,

0 0 0 0 hx ⊗ x , y ⊗ y iL(K⊗ˆ K0) = hx, yiL(K)hx , y iL(K0).

The following Gordon-Chevet type lemma [19, 32] is related to the mean widths of convex bodies to that of their projective tensor product.

n 0 0 n0 Lemma 4 For any two convex bodies K ⊂ rB2 and K ⊂ r B2 , one has

0 0 0 ωG(K⊗ˆ K ) ≤ r ωG(K) + r ωG(K ).

In terms of the mean width, one has

0 0 r τ rτ 0 r r ω(K⊗ˆ K0) ≤ n ω(K) + n ω(K0) ≤ √ ω(K) + √ ω(K0). τnn0 τnn0 n0 n 22

When r = r0 = 1, this is Lemma 2 in [8] (here we can remove the condition: at least one of the two convex bodies is the convex hull of a subset of the corresponding unit √ √τn sphere). The second inequality follows from n increasing and τn ∼ n.

To prove the above lemma, one needs the Slepian’s lemma [94].

Lemma 5 (Slepian’s Lemma) Let {Xj}j∈T and {Yj}j∈T be two families of joint Gaussian random variables with ﬁnite index set T , such that, ∀j, k ∈ T ,

kXj − Xkk ≤ kYj − Ykk.

Then

max Xj ≤ max Yj. E j∈T E j∈T

Here if X is a (Gaussian) random vector, kXk = EhX,Xi.

Proof of Lemma 4. Deﬁne φ : Rn × Rm → Rn ⊗ Rm as

φ(x, y) = x ⊗ y and ϕ : Rn × Rm → Rn ⊗ Rm as

ϕ(x, y) = (kykx, kxky).

Then for any x, x0 ∈ Rn and y, y0 ∈ Rm, one has

kφ(x, y) − φ(x0, y0)k ≤ kϕ(x, y) − ϕ(x0, y0)k.

Let ext(K) be the extreme point set of K. Deﬁne the index set T to be T = ext(K)× ext(K0), and the Gaussian process families as

0 X 0 = hG n n0 , t ⊗ t i, (t,t ) R ⊗R 0 0 Y 0 = hG n , kt kti + hG n0 , ktkt i, (t,t ) R R

l where G l refers to the standard Gaussian random vector in . Hence, by the R R Slepian’s lemma, one has

max Xa ≤ max Ya. E a∈T E a∈T 23

0 0 Considering ktk2 ≤ r, and kt k2 ≤ r , one has

0 0 max Ya ≤ r ωG(K) + r ωG(K ). E a∈T

Hence, we have proved that

0 0 0 ωG(K⊗ˆ K ) ≤ r ωG(K) + r ωG(K ).

1.3 Background on Geometry of Quantum States

1.3.1 Mathematical framework

We now introduce the mathematical framework and some notation. Let H be the

D1 D2 Dn (complex) Hilbert space C ⊗C · · ·⊗C with (complex) dimension N = D1D2 ··· Dn.

Here we always assume n ≥ 2 and Di ≥ 2 for all i = 1, 2, ··· , n. Recall that Di = 2

for all i corresponds to n-qubits, and Di = 3 for all i corresponds to n-qutrits. n = 2 corresponds to bipartite quantum systems and n > 2 corresponds to multipartite quantum systems. Denote by B(H) the space of linear maps on H. Deﬁne the † Hilbert-Schmidt inner product on space B(H) as hA, BiHS = tr(A B). The subspace of B(H) consisting of all self-adjoint operators is Bsa(H). It inherits a (real) Euclidean structure from the scalar product h·, ·iHS on B(H). (This is because if A, B ∈ Bsa(H), then hA, BiHS must be a real number.) D denotes the set of all states on H (more precisely, states on B(H)), i.e., positive (semi) deﬁnite trace one operator in Bsa(H):

D = D(H) := {ρ ∈ Bsa(H), ρ ≥ 0, tr ρ = 1}.

A state in D is said to be separable if it is a convex combination of tensor products of n states (otherwise, it is called entangled). Denote the set of separable states by S, then

Di S = S(H) := conv{ρ1 ⊗ · · · ⊗ ρn, ρi ∈ D(C )}.

2 Both D and S are convex subsets of Bsa(H) of (real) dimension d = N − 1. We also denote the set of entangled states on H by E(H). 24

Indent: The notation S(H) is in principle ambiguous: separability of a state on B(H) is not an intrinsic property of the Hilbert space H nor of the algebra B(H); it depends on the particular decomposition of H as a tensor product of (smaller) Hilbert spaces. However, this will not be an issue here since our study focuses on ﬁxed decompositions.

It is well-known that any quantum state on H with (complex) dimension N can be represented as a density matrix with size N × N, i.e., an N × N positive (semi) deﬁnite matrix with trace 1. Here and after, we will work on D, the set of all N × N density matrices. The separable quantum states can be represented as the convex hull of the tensor product of density matrices on subsystem CDi .

1.3.2 Peres-Horodecki’ positive partial transpose criterion

For a bipartite system H = CD1 ⊗ CD2 , any state ρ on H can be expressed uniquely as D D X1 X2 ρ = ρiα,jβ|ei ⊗ fαihej ⊗ fβ| i,j=1 α,β

D1 D2 D1 D2 where {ei}i=1 and {fα}α=1 are the canonical bases of C and C respectively. Deﬁne the partial transpose T (ρ) with respect to the ﬁrst subsystem as

D D X1 X2 T (ρ) = ρjα,iβ|ei ⊗ fαihej ⊗ fβ|. i,j=1 α,β

t It is clear that T (·) is linear, tr(T (ρ)) = 1 for any state ρ on H, and T (ρ1⊗ρ2) = ρ1⊗ρ2

Dj t for all ρj ∈ D(C ), j = 1, 2 where ρ1 is the transpose of ρ1. The transpose operator t preserves the positivity of matrices, i.e, if ρ1 ≥ 0, then ρ1 ≥ 0. Hence, T (ρ1 ⊗ ρ2) ≥ 0

for any ρj ≥ 0, j = 1, 2. We write PPT for the set of states ρ such that T (ρ) is also positive, i.e., PPT = {ρ ∈ D : T (ρ) is positive} ⊂ D. 25

We can also view that PPT is the intersection of D and T (D) = T −1(D), i.e., PPT = D ∩ T (D). It is clear that PPT is a convex body. Note PPT is basis- independent because eigenvalues of T (ρ) do not depend on basis. The following theorem by Peres [73] gives a way to detect the separability (see also [38]).

Theorem 9 Let H = CD1 ⊗ CD2 be any bipartite system. For any ρ ∈ S, one has T (ρ) is positive. Equivalently S ⊂ PPT ⊂ D.

Pm 1 2 Pm Proof. Any ρ ∈ S takes the form ρ = i=1 ζiρi ⊗ ρi where ζi ≥ 0, i=1 ζi = 1, and j ρi are positive (semi) deﬁnite matrix with size Dj × Dj, j = 1, 2. By the linearity of T (·), one immediately has

m m X 1 2 X 1 t 2 T (ρ) = ζiT (ρi ⊗ ρi ) = ζi(ρi ) ⊗ ρi , i=1 i=1 and hence T (ρ) ≥ 0.

This theorem only gives the necessary condition for separability. If a quantum state ρ ∈ D does not have positive partial transpose, i.e., one or more eigenvalues of T (ρ) are strictly negative, then ρ is an entangled quantum state. We refer the reader to reference [39] for the examples where a quantum state ρ on H with dimension N ≥ 8 has positive T (ρ) but ρ∈ / S. In Chapter 4, we will discuss how precise is the positive partial transpose criterion for “large N” in the Hilbert-Schmidt and the Bures volume sense. It is well-known that, for qubit-qubit system C2 ⊗ C2 with N = 4 and qubit-qutrit (or qutrit-qubit) system C2 ⊗C3 (or C3 ⊗C2) with N = 6, the positive partial transpose criterion provides a suﬃcient condition as well. This can be stated as the following theorem. We refer the reader to the references [38, 98, 113] for its proof.

Theorem 10 If H is a bipartite system with complex dimension N = 4 or N = 6, then any quantum state on H with positive partial transpose is a separable quantum state on H, i.e., S = PPT . 26

An interesting feature of the geometric structure of PPT in the Euclidean metric is revealed in Szarek, Bengtsson, and Zyczkowski in [100]. In fact, the authors show that PPT is a body of constant height, that is, every boundary point of PPT is contained in a face tangent to the inscribed ball of PPT .

1.3.3 Bures metric on D

For any two density matrices ρ and σ, the Hilbert-Schmidt distance between ρ and σ is defined as q † kρ − σkHS = tr (ρ − σ) (ρ − σ) where A† is the conjugate transpose of the matrix A. This metric induces the Eu- clidean structure of D. A nontrivial, but arguably more important, metric in the present context is the so-called Bures metric [16], dB(·, ·), whose definition may be formulated as r q√ √ dB(ρ, σ) = 2 − 2tr ρσ ρ (1.11) √ for any two states ρ and σ. The matrix ρ is the usual square root of ρ, defined as √ U ΛU † when ρ has the eigenvalue decomposition ρ = UΛU † for some unitary matrix

U and diagonal matrix Λ. Hereafter, we denote IdN the N × N identity matrix, and † † U(N) the N × N unitary group, i.e., ∀U ∈ U(N), U U = UU = IdN . √ √ The quantity F (ρ, σ) = (trp ρσ ρ)2 is called the Fidelity. It is not obvious from formula (1.11) to see that dB(·, ·) is a metric. The following theorem gives equivalent deﬁnitions for Bures metric.

Theorem 11 For any two quantum states ρ and σ on H, one has

r q √ √ † † dB(ρ, σ) = 2 − 2tr ρσ ρ = inf{kA − BkHS : A A = ρ, and B B = σ}.

One can further get √ † √ † dB(ρ, σ) = inf{kA − σkHS : A A = ρ} = inf{kB − ρkHS : B B = σ}. 27

Remark. Clearly, by Theorem 11, dB(ρ, ρ) = 0 for all state ρ and dB(ρ, σ) = dB(σ, ρ) p√ √ for all ρ, σ. If dB(ρ, σ) = 0, then tr ρσ ρ = 1. On the other hand,

q√ √ q√ √ √ √ √ √ tr ρσ ρ = tr ρ σ σ ρ = tr(| ρ σ|).

By Cauchy-Schwarz inequality, √ √ pF (ρ, σ) = tr(| ρ σ|) ≤ ptr(ρ) tr(σ) = 1

with equality if and only if ρ = σ. Therefore, dB(ρ, σ) = 1 if and only if ρ = σ.

To verify the triangle inequality, for any states ρ, σ, δ, one has √ √ † † dB(ρ, δ) + dB(σ, δ) = inf{kA − δkHS : A A = ρ} + inf{kB − δkHS : B B = σ} √ √ † † = inf{kA − δkHS + kB − δkHS : A A = ρ, and B B = σ}

† † ≥ inf{kA − BkHS : A A = ρ, and B B = σ}

= dB(ρ, σ) where the inequality follows the triangular inequality for k · kHS.

To prove Theorem 11, we need the following theorem (see [12]).

Theorem 12 (Uhlmann’s Fidelity) The Fidelity of any two states ρ and σ on H takes the following form:

tr(A†B) + tr(B†A) pF (ρ, σ) = sup : A†A = ρ and B†B = σ . 2

Remark. An immediate consequence of this theorem is the symmetry of F (·, ·), i.e., for all two states ρ, σ ∈ D, one has F (ρ, σ) = F (σ, ρ). It is also easy to see that 0 ≤ pF (ρ, σ) ≤ 1. pF (ρ, σ) = 0 if and only if ρ and σ have orthogonal supports, namely, the eigenvectors of ρ with respect to nonzero eigenvalues sits in the eigenspace spanned by the eigenvectors of σ with respect to eigenvalue 0, and vice versa. The ﬁdelity (and hence the Bures metric) is unitarily invariant, that 28

is, pF (UρU †, UσU †) = pF (ρ, σ) for any unitary matrix U ∈ U(N). It can be proved that the Fidelity F (ρ, σ) ≤ F Φ(ρ), Φ(σ) for any completely positive, trace preserving map Φ (see [12] for details). This is equivalent to dB Φ(ρ), Φ(σ) ≤

dB(ρ, σ) for any completely positive, trace preserving map Φ.

Proof. Any positive (semi)-deﬁnite matrix ρ can be decomposed as the product of A with its Hermitean conjugate, that is, ρ = A†A for some matrix A. Moreover, any A √ is of the form A = V ρ for some unitary matrix V ∈ U(N). Similarly, for σ = B†B, √ we can take B = W σ for some unitary matrix W ∈ U(N). Therefore, one has

tr(A†B) + tr(B†A) F˜(A, B) : = sup : A†A = ρ and B†B = σ 2 √ √ √ √ tr( ρV †W σ) + tr( σW †V ρ) = sup : V,W ∈ U(N) 2 √ √ √ √ tr( σ ρV †W ) + tr( ρ σW †V ) = sup : V,W ∈ U(N) . 2

Notice V˜ = W †V is still a unitary matrix, then the above formula can be rewritten as ( √ √ √ √ ) tr( σ ρV˜ † + tr( ρ σV˜ ) F˜(A, B) = sup : V˜ ∈ U(N) . (1.12) 2 √ √ √ √ √ √ √ √ A direct computation shows that ρ σ ρ = ρ σ σ ρ. This implies that ρ σ = √ √ p ρσ ρW˜ for some unitary matrix W˜ . Substitute this into formula (1.12), and redeﬁne V˜ as W˜ V˜ , one gets ( √ √ √ √ ) tr(p ρσ ρV˜ †) + tr(p ρσ ρV˜ ) F˜(A, B) = sup : V˜ ∈ U(N) . (1.13) 2 √ Recall that sup{|tr(AU)| : U ∈ U(N)} = tr( A†A) for any matrix A [80], with √ equality if and only if AU = eiθ A†A for some θ ∈ R. Therefore, formula (1.13) implies that q√ √ F˜(A, B) = tr ρσ ρ = pF (ρ, σ) 29

˜ with equality if and only if V = IdN , the identity matrix with size N × N.

Proof of Theorem 11. By Theorem 12, one has

r q√ √ dB(ρ, σ) = 2 − 2tr ρσ ρ r = 2 − sup {tr(A†B) + tr(B†A)} A†A=ρ,B†B=σ

= inf{ptr(A†A) + tr(B†B) − tr(A†B) − tr(B†A): A†A = ρ and B†B = σ}

† † = inf{kA − BkHS : A A = ρ and B B = σ}. √ If B†B = σ, then B = U σ for some unitary matrix U ∈ U(N). Therefore, √ dB(ρ, σ) = inf kA − U σkHS A†A=ρ,U∈U(N)

† √ = inf kU A − σkHS A†A=ρ,U∈U(N) √ ˜ = inf kA − σkHS A˜†A˜=ρ

where we replace U †A by A˜, and A˜ satisﬁes A˜†A˜ = ρ. Similarly, one can prove

√ dB(ρ, σ) = inf k ρ − BkHS. B†B=σ

ρσ = σρ means that, up to unitary equivalence, ρ = diag(a1, ··· , aN ) and σ =

diag(b1, ··· , bN ) are diagonal matrices with ρii = ai and σii = bi for all i = 1, ··· ,N. PN PN Here ai, bi ≥ 0, and i=1 ai = i=1 bi = 1. Then

v v u N u N √ u X p uX p 2 dB(ρ, σ) = t2 − 2 aibi = t ( ai − bi) . i=1 i=1

This coincides with the so-called Hellinger distance between the discrete probability

distribution P (Xi) = ai and the discrete probability distribution Q(Xi) = bi. In this 30

case, the root of the Fidelity coincides with the Bhattacharyya distance between the distribution P and Q, i.e.,

N p X p F (ρ, σ) = ai bi = B(P,Q). i=1

In fact, the Hellinger distance or the Bhattacharyya distance measures the statistical distinguishability between two probability distribution by statistical sampling (see e.g. [12]).

In the rest of this subsection, we will link the Bures distance between any two states ρ and σ with the positive operator valued measurement and the projection measurement. The positive operator valued measure (POVM) is deﬁned by any par-

tition of the identity operator into a set of k positive operator Ei (not necessary rank

1 projection) acting on the N-dimensional Hilbert space H (see e.g., [12]). These Ei satisfy k X † Ei = IdN , and Ei = Ei , Ei ≥ 0, i = 1, ··· , k. i=1

k Employing the POVM measurement {Ei}i=1 to a state ρ results in the i-th outcome

with probability p(Ei, ρ) = tr(Eiρ), i = 1, ··· , k. The projective measurement (PM) N {Pi}i=1 is a special POVM measurement with all Pi, i = 1, ··· ,N being rank 1 orthogonal projection operators. The following theorem was proved in [11]. Because of its importance in explaining the signiﬁcance of ﬁdelity and the Bures distance, we include a proof for completeness.

Theorem 13 Let ρ and σ be any two states on the Hilbert space H. Then

p X p p F (ρ, σ) = inf tr(ρEi) tr(σEi) { i} E i where the inﬁmum runs over all the POVM measurements {Ei}. 31 p p Proof. For convenience, let pi = tr(ρEi) and qi = tr(σEi). Then

X X p p piqi = tr(ρEi) tr(σEi) i i q q √ √ X √ p p √ † p p = tr(U ρ Ei Ei ρU ) tr( σ Ei Ei σ) i where U ∈ U(N) is any unitary matrix. The noncommutative Cauchy-Schwarz inequality implies that |tr(AB†)| ≤ ptr(AA†)tr(BB†) with equality if and only A = ϑB or B = ϑA for some constant ϑ. Then for any unitary matrix U ∈ U(N)

and any POVM {Ei}, one has

X X √ √ piqi ≥ |tr(U ρEi σ)| i i X √ √ ≥ | tr( σU ρEi)| i √ √ = |tr(U ρ σ)|.

The equality holds if and only if

√ p √ p †√ p √ p U ρ Ei = ϑi σ Ei, or, U σ Ei = ϑi ρ Ei (1.14) for some constant ϑi ≥ 0. Hence

X √ √ p min piqi ≥ max |tr(U ρ σ)| = F (ρ, σ). (1.15) { i} U E i

√ √ p√ √ The equality holds if and only if U0 ρ σ = ρσ ρ. Combining with formula (1.14), the equality is attained at

p p M Ei = ϑi Ei √ √ √ √ with M = σ−1p σρ σ σ−1 if σ is invertible. One way to satisfy this condition is to take Ei = |iihi| (projection operators) where the vector |ii are an orthonormal eigenbasis of M with ϑi the eigenvalue of |ii. Similarly, if ρ is invertible, one can 32 √ √ replace M by M = pρ−1p ρσ ρpρ−1 and repeat the above process to find (projection operators) Ei. When both ρ and σ are not invertible, one can first choose, for instance, Ei, i = 1, ··· ,N − rank(σ) to be an orthonormal basis of the null space of σ. In the support of σ (the orthocomplement of its null space), we can construct the analogous M as above and find other rank(σ) many (projection operators) Ei. Here rank(σ) means the rank of σ. Or similarly, one chooses Ei, i = 1, ··· ,N − rank(ρ) to be an orthonormal basis of the null space of ρ and add other rank(ρ) many Ei as above in the support of ρ.

Remark. From this theorem, one can deduce that the Bures distance between ρ and σ is equal to

1 2! 2 X p p dB(ρ, σ) = sup tr(Eiρ) − tr(Eiσ) {Ei} i where the supremum runs over all the POVM measurements {Ei}. The proof of this theorem also shows that

N p X p p F (ρ, σ) = inf tr(ρPi) tr(σPi) { i} P i=1

N where the inﬁmum runs over all the projective measurements {Pi}i=1. Thus, the Bures distance between ρ and σ is also equal to

1 N 2! 2 X p p dB(ρ, σ) = sup tr(Piρ) − tr(Piσ) {Pi} i=1

N where the supremum runs over all the projection measurements {Pi}i=1.

1.3.4 Hilbert-Schmidt and Bures measures on D

Any quantum state on H can be represented as a density matrix, i.e., the N ×N positive (semi) deﬁnite matrix whose diagonal elements sum up to 1. Therefore, any quantum state ρ ∈ D has eigenvalue decomposition ρ = UΛU † for some unitary matrix 33

U ∈ U(N) and some diagonal matrix Λ = diag(λ1, ··· , λN ) with (λ1, ··· , λN ) ∈ ∆. We denote by ∆ the regular simplex in RN , i.e.,

( N ) N X ∆ = (λ1, ··· , λN ) ∈ R : λi ≥ 0, λi = 1 . i=1

The Weyl chamber of ∆, deﬁned by the constraint λ1 ≥ · · · ≥ λN , is denoted by ∆1. Clearly, for any ρ = UΛU † as above and for any diagonal matrix B ∈ U(N), i.e., B =

iθ1 iθN † † † diag(e , ··· , e ) with θj ∈ R for all j = 1, ··· ,N, we have UΛU = UBΛB U . Thus, to have unique parameterization of generic states ρ = UΛU † ∈ D, we have to restrict (λ1, ··· , λN ), for instance, to ∆1 and select one speciﬁc point in the coset space F N = U(N)/[U(1)]N (the ﬂag manifold).

We will be interested in various measures on D. A natural restriction is to require invariance with respect to unitary rotations. For most problems, the interesting class of measures are those that are invariant under conjugation by a unitary matrix. Such measures can normally be represented as the product of some measure on ∆1 and the invariant measure on F N (see [12, 37] for more on this and for the background on the discussion that follows). The unique (up to a multiplicative constant) invariant measure γ on F N is induced by the Haar measure on the unitary group U(N) and has the form Y −1 −1 dγ = 2Re(U dU)ijIm(U dU)ij, 1≤i

(2π)N(N−1)/2 Z = , where E(N) = QN Γ(j). (1.16) N E(N) j=1

R ∞ x−1 −t Here Γ(x) = 0 t e dt is the Gamma function.

The Hilbert-Schmidt measure VHS(·) on D, induced by the Hilbert-Schmidt met- 34

ric, may be expressed as

√ N−1 Y 2 Y dVHS = N (λi − λj) dλi dγ, (1.17) 1≤i

dρ = dU Λ U † + U dΛ U † + U Λ dU †.

† † † U U = IdN implies that dU U +U dU = 0. Then the inﬁnitesimal distance, deﬁned 2 as dHS(ρ, ρ + dρ), equals to

N 2 X 2 X 2 † 2 dHS(ρ, ρ + dρ) = (dλi) + 2 (λi − λj) |(U dU)ij| i=1 1≤i

here | · | is the length of complex number. Since the density matrices satisfy with the PN PN PN−1 condition i=1 λi = 1, one has i=1 dλi = 0. Therefore, dλN = − i=1 dλi which implies that

N N−1 N−1 !2 N−1 X 2 X 2 X X (dλi) = (dλi) + dλi = dλigijdλj. i=1 i=1 i=1 i,j=1

Here gii = 2, i = 1, ··· ,N − 1 and gij = 1 for i 6= j. It is easy to check that the determinant of the matrix g equals to N, and hence

√ N−1 Y 2 Y dVHS = N (λi − λj) dλi dγ. 1≤i

That is, the Hilbert-Schmidt measure takes the product form

dVHS = dνHS(λ1, ··· , λN ) × dγ 35 √ Q 2 QN−1 with the measure dνHS(λ1, ··· , λN ) = N 1≤i

For any α, β > 0, one deﬁnes the following integral:

Z ∞ N ! N N 1 X Y α−1 Y β Y α,β := δ λi − 1 λi |λi − λj| dλi. (1.18) CN 0 i=1 i=1 1≤i

Here δ denotes the dirac measure. By using the Laguerre ensemble formula (see [12]), one obtains that

N 1 Y Γ[1 + jβ/2]Γ[α + (j − 1)β/2] Cα,β = N Γ[αN + βN(N − 1)/2] Γ[1 + β/2] j=1

Therefore, to obtain the Hilbert-Schmidt volume of D, one has to calculate the following integral [115]:

Z √ N−1 Y 2 Y VHS(D) = N (λi − λj) dλi dγ N ∆1 × F 1≤i

1 where N! appears because ∆1 is only one chamber of N! similar chambers of ∆.

We deﬁne vradHS(K), the Hilbert-Schmidt volume radius of K ⊂ D, to be the radius of d-dimensional Euclidean ball which has the same volume as the Hilbert- Schmidt volume of K. In other words,

1 d VHS(K) vradHS(K) = , σd

πd/2 where σd = Γ(1+d/2) is the volume of d-dimensional Euclidean ball. For later con- 1/d VHS (K) vradHS (K) venience, we also denote VRHS(K, L) as VRHS(K, L) = = . It VHS (L) vradHS (L) amounts to comparing the Hilbert-Schmidt volume radii of K and L. 36

− 1 − 1 It is known that vradHS(D) ∼ e 4 d 4 [99] by Stirling approximation r 2π z z 1 Γ(z) = 1 + O . (1.20) z e z √ 1 − 1 Stirling approximation (1.20) also implies that (σd) d ∼ 2eπ d 2 and therefore (after lengthy but routine calculation)

1 2 1 − 3 (VHS(D)) d ∼ (4π e) 4 d 4 . (1.21)

Here a(n) ∼ b(n) means limn→∞ a(n)/b(n) = 1.

The Bures measure, induced by the Bures distance dB(·, ·), can be written as

2 2−N−N 2 N−1 2 2 Y (λi − λj) Y dVB = √ dλi dγ, (1.22) λ ··· λ λi + λj 1 N 1≤i

where (λ1, ··· , λN ) ∈ ∆1. The Bures measure has singularities (with respect to the Hilbert-Schmidt measure) on the boundary of D. (The boundary corresponds to at least one of the λi’s being 0, and if two or more of them are 0, then some denominators in (1.22) vanish.)

To obtain formula (1.22), one can compute the Bures inﬁnitesimal distance between ρ+dρ and ρ, i.e., dB(ρ, ρ+dρ), with the valuation dρ at ρ such that ρ+dρ ∈ D.

First, we compute the case when ρ is a diagonal matrix, say ρ = diag(λ1, ··· , λN ). Let q√ √ ρ(ρ + dρ) ρ = ρ + X + Y

where the matrix X is of order 1 and Y is of order 2. Squaring the equation, we obtained the ﬁrst and second order √ √ ρ dρ ρ = Xρ + ρX, −X2 = Y ρ + ρY. (1.23)

Considering ρ = diag(λ1, ··· , λN ) with λi > 0 for all i = 1, ··· ,N, one can easily get that p λiλj 2 1 Xij = dρij,Yij = −(X )ij . (1.24) λi + λj λi + λj 37

Equation (1.23) implies that

p √ √ p dρ = ρ−1 X ρ + ρX ρ−1

trdρ and hence trX = 2 = 0 as trdρ = 0. Similarly,

−1 2 2 −tr(ρ X ) X 1 X 1 |dρij| trY = = − |X |2 = − . 2 2λ ij 4 λ + λ i,j i i,j i j

Hence the inﬁnitesimal of Bures metric at ρ = diag(λ1, ··· , λN ) is

2 2 X 1 |dρij| d (ρ, ρ + dρ) = . (1.25) B 2 λ + λ i,j i j

Now let us compute the inﬁnitesimal of the Bures metric at general state ρ with eigenvalue decomposition ρ = UΛU †. The valuation of dρ at ρ takes the form

dρ = dU Λ U † + U dΛ U † + U Λ dU †.

Hence U †(ρ + dρ)U = Λ + dΛ + U † dU Λ − Λ U † dU.

Letρ ˜ = Λ and dρ˜ = dΛ + U † dU Λ − Λ U † dU. Employing formula (1.25) toρ ˜ and dρ˜, and by the unitary invariance of the Bures distance, one gets

2 2 2 X 1 |dρ˜ij| d (ρ, ρ + dρ) = d (˜ρ, ρ˜ + dρ˜) = B B 2 λ + λ i,j i j

N 2 2 X (dλi) X (λi − λj) † 2 = + |(U dU)ij| . 4λi λi + λj i=1 1≤i

PN Note terms where the denominator vanishes have to be excluded. i=1 λi = 1 implies PN that i=1 dλi = 0. Eliminating dλN , one obtains that

2 N 2 N−1 2 N−1 ! 1 X (dλi) 1 X (dλi) 1 X 1 = + dλ . 4 λ 4 λ 4 i λ i=1 i i=1 i i=1 N 38

This metric in the (N − 1)-dimensional simplex is of the form

1 1 1 gii = + , and gij = , ∀i 6= j. 4λi 4λN 4λN

1−N QN −1 The determinant of the matrix g is equal to 4 i=1 λi . Thus, the Bures volume element is 2 2−N−N 2 N−1 2 2 Y (λi − λj) Y dVB = √ dλi dγ. λ ··· λ λi + λj 1 N 1≤i

It also takes the product form

dVB = dνB(λ1, ··· , λN ) × dγ

2 2−N−N 2 2 2 Q (λi−λj ) QN−1 with the measure dνB(λ1, ··· , λN ) = √ dλi on ∆1. λ1···λN 1≤i

2 Z 2−N−N 2 N−1 2 2 Y (λi − λj) Y VB(D) = √ dλi dγ N λ ··· λ λi + λj ∆1 × F 1 N 1≤i

N 2/2 2 π = 21−N . (1.26) Γ(N 2/2)

As mentioned earlier, this value happens to be the d-dimensional volume of the d- 1 dimensional hemisphere with radius 2 . We deﬁne vradB(K), the Bures volume radius of K ⊂ D, to be

1 d VB(K) vradB(K) = . σd While comparing the Bures volume of K with the Hilbert-Schmidt volume of the Euclidean ball does not have immediate geometric meaning, we ﬁnd this way of describing the size of K in the Bures volume sense convenient in our calculations. 39

1 By formula (1.20) and (1.26), one has vradB(D) ∼ 2 and hence

r 1 eπ 1 d − V (D) ∼ d 2 . (1.27) B 2

For later convenience, we also deﬁne the (relative) Bures volume radii ratio of K to L 1/d VB (K) vradB (K) as VRB(K, L) = = . This can be used as a measure of the relative VB (L) vradB (L) size of K to L in the Bures volume sense, and clearly does have geometric meaning.

We refer the reader to the references [12, 35, 37, 95, 115] for more detailed background and for motivation with regards to the issues addressed in this section. In

Chapter 4, we will study the (asymptotical) behavior of VRB(K, D) in terms of its relative VRHS(K, D). Chapter 2

New Lp Aﬃne Isoperimetric Inequalities

This chapter, based on the work [111], deals with isoperimetric inequalities for Lp affine surface area, and geometric interpretations of Lp affine surface area via the convex floating body and the surface body. This chapter is organized as follows. In ◦ Section 1, we investigate the asymptotic behavior of (Kδ) , the polar body of the convex floating body, and provide a new geometric interpretation of the L n affine − n+2 ◦ ◦ surface area of K . In Section 2, we investigate the asymptotic behavior of (Kf,s) , the polar body of the surface body, and provide new geometric interpretations of the

Lp aﬃne surface area of K for all p 6= −n. As an application, the duality formula ◦ asp(K) = as n2 (K ) for p 6= −n is proved. Section 3 will be dedicated to Lp aﬃne p isoperimetric inequalities for p ∈ [−∞, ∞]. We also show a monotonicity behavior of

n+r asr(K) r the quotient n|K| and obtain Santal´otype inequalities.

2.1 L n aﬃne surface area of the polar body −n+2

2 It was proved in [70] that for a convex body K ∈ C+

40 41

◦ ◦ Z Z n+2 |(Kδ) | − |K | dσ(u) κK (x) n+1 lim cn 2 = 1 = dµK (x) δ→0 n+1 n+1 δ n+1 Sn−1 fK (u) n+1 hK (u) ∂K hx, NK (x)i

= as−n(n+2)(K), (2.1)

2 n−1 n+1 |B2 | where cn = 2 n+1 and Kδ is the convex floating body [84]: the intersection of all halfspaces H+ whose defining hyperplanes H cut off a set of volume δ from K, that is,

\ + Kδ = H . H:|H−∩K|≤δ

Hereafter, we use H(x, u) to denote the hyperplane through x with outer normal vector u ∈ Sn−1, i.e., H(x, u) = {y ∈ Rn, hy, ui = hx, ui}. The two half-spaces generated by H(x, u) are H−(x, u) = {y ∈ Rn, hy, ui ≥ hx, ui} and H+(x, u) = {y ∈ Rn, hy, ui ≤ hx, ui}.

Assumptions on the boundary of K are needed in order that (2.1) holds. n n n To see that, consider B∞ = {x ∈ R : max1≤i≤n|xi| ≤ 1}. As κB∞ (x) = 0 a.e. on n ∂B∞, n+2 Z n n+1 κB∞ (x) dµ n (x) = 0. n+1 B∞ ∂Bn n ∞ x, NB∞ (x) However n ◦ n ◦ |((B∞)δ) | − |(B∞) | lim cn 2 = ∞. (2.2) δ→0 δ n+1

n Indeed, writing K for B∞, we will construct a 0-symmetric convex body K1 such ◦ ◦ ◦ that Kδ ⊆ K1 ⊆ K. Then K ⊆ K1 ⊆ Kδ . Therefore, to show (2.2), it is enough to show that ◦ ◦ |K1 | − |K | lim cn 2 = ∞. δ→0 δ n+1 42

+ n + + Let R = {(xj)j=1 : xj ≥ 0, 1 ≤ j ≤ n}. It is enough to consider K = R ∩ K and + + to construct (K1) = K1 ∩ R .

+ + + We deﬁne (K1) to be the intersection of R with the half-spaces Hi , 1 ≤ i ≤ n n + 1, where Hi = {(xj)j=1 : xi = 1}, 1 ≤ i ≤ n, and

( n ) 1 n X n Hn+1 = (xj)j=1 : xj = n − (n!δ) , j=1

δ > 0 suﬃciently small. Notice that the hyperplane Hn+1 (orthogonal to the vector

(1,..., 1)) cuts oﬀ a set of volume exactly δ from K and therefore Kδ ⊂ K1. ◦ Moreover, K1 can be written as a convex hull:

1 K◦ = conv {±e , 1 ≤ i ≤ n} ∪ (ε , . . . , ε ), ε = ±1, 1 ≤ j ≤ n , 1 i s 1 n j

1 where s = n − (n!δ) n . Hence

n ◦ 2 n |K1 | = · 1 n! n − (n!δ) n

and therefore

◦ ◦ n 1 |K | − |K | 2 −2 (n!δ) n 1 n+1 lim 2 = lim δ 1 = ∞. δ→0 δ n+1 n! δ→0 (n − (n!δ) n )

Now we show

2 Theorem 14 Let K be a convex body in C+ such that 0 ∈ int(K). Then

◦ ◦ |(Kδ) | − |K | ◦ lim c = as n (K ). n 2 − n+2 δ→0 δ n+1

2 As a corollary of (2.1) and Theorem 14 we get that for a convex body K ∈ C+

◦ as (K) = as n (K ). (2.3) −n(n+2) − n+2 43

◦ This is a special case for p = −n(n + 2) of the formula asp(K) = as n2 (K ) proved p in [47] for p > 0. We will show in the next section that this formula holds for all p < 0, p 6= −n for convex bodies with suﬃciently smooth boundary. For p = 0 (and 2 ◦ K ∈ C+) the formula holds trivially as as0(K) = n|K| and as∞(K ) = n|K| (see [86]).

For the proof of Theorem 14 we need the following lemmas.

2 ◦ Lemma 6 Let K ∈ C+. Then for any x ∈ ∂K , we have

n 2 hx, NK◦ (x)i kxδk hx, NK◦ (x)i lim 2 − 1 = 1 δ→0 n δ n+1 kxk cn (κK◦ (x)) n+1

◦ where xδ ∈ ∂(Kδ) is in the ray passing through 0 and x.

Proof 2 ◦ ◦ 2 Since K, and hence also Kδ, are in C+ one has that K and (Kδ) are in C+. Therefore, for x ∈ ∂K◦ there exists a unique y ∈ ∂K, such that, hx, yi = 1, namely y =

NKo (x) x 1 . y has outer normal vector NK (y) = and kxk = hy, NK (y)i. hNKo (x), xi kxk

◦ Similarly, for xδ ∈ ∂(Kδ) there exists a unique yδ in ∂Kδ such that hxδ, yδi = 1,

o N(Kδ) (xδ) xδ x namely yδ = , yδ has outer normal vector NKδ (yδ) = = and o hN(Kδ) (xδ), xδi kxδk kxk 1 = hyδ,NK (yδ)i. kxδk δ

0 Let y = [0, y] ∩ ∂Kδ ([z1, z2] denotes the line segment from z1 to z2) and let 0 0 yδ ∈ ∂K be such that yδ = [0, yδ] ∩ Kδ.

We have

1 x = hy, N (y)i ≥ hy0 ,N (y)i = hy0 , i, kxk K δ K δ kxk

1 0 0 x = hyδ,NKδ (yδ)i ≥ hy ,NKδ (yδ)i = hy , i. kxδk kxk 44

Hence

kx kn hy, N (y)i n δ − 1 = K − 1 kxk hyδ,NKδ (yδ)i n " hy0 , x i! # 0 n δ kxk kyδk ≥ x − 1 = − 1 , hyδ, kxk i kyδk

n n " x !n # kxδk hy, NK (y)i hy, kxk i − 1 = − 1 ≤ 0 x − 1 kxk hyδ,NKδ (yδ)i hy , kxk i

kyk n = − 1 (2.4) ky0k

and therefore

0 n n hx, N ◦ (x)i ky k hx, N ◦ (x)i kx k K δ − 1 ≤ K δ − 1 n kyδk n kxk n hx, N ◦ (x)i kyk ≤ K − 1 . n ky0k

We ﬁrst consider the lower bound.

n 0 0 0 n ◦ ◦ hx, NK (x)i kxδk hx, NK (x)i hyδ,NKδ (yδ)i kyδk lim 2 − 1 ≥ lim 2 − 1 . δ→0 δ→0 0 0 n δ n+1 kxk hyδ,NKδ (yδ)i n δ n+1 kyδk

0 2 0 As δ → 0, yδ → y. As K is in C+, NKδ (yδ) → NK (y) as δ → 0. 0 0 Therefore limδ→0hyδ,NKδ (yδ)i = hy, NK (y)i. By Lemma 7 and Lemma 10 of [84],

n 1 0 0 0 n+1 hyδ,NKδ (yδ)i kyδk (κK (y)) lim 2 − 1 = . δ→0 n δ n+1 kyδk cn

Hence

n 1 2 hx, NK◦ (x)i kxδk hx, NK◦ (x)i (κK (y)) n+1 hx, NK◦ (x)i lim 2 − 1 ≥ = 1 . δ→0 n δ n+1 kxk hy, NK (y)i cn cn (κK◦ (x)) n+1 45

2 The last equation follows from the fact that if K ∈ C+, then, for any y ∈ ∂K, there is a unique point x ∈ ∂K◦ such that hx, yi = 1 and [47]

1 hy, NK (y)ihx, NK◦ (x)i = (κK (y)κK◦ (x)) n+1 . (2.5)

Similarly, one gets for the upper bound

n 2 hx, NK◦ (x)i kxδk hx, NK◦ (x)i lim 2 − 1 ≤ 1 , δ→0 n δ n+1 kxk cn (κK◦ (x)) n+1

hence altogether

n 2 hx, NK◦ (x)i kxδk hx, NK◦ (x)i lim 2 − 1 = 1 . δ→0 n δ n+1 kxk cn (κK◦ (x)) n+1

2 Lemma 7 Let K ∈ C+. Then we have

n hx, NK◦ (x)i kxδk 2 − 1 ≤ c(K, n), n δ n+1 kxk

where c(K, n) is a constant (depending on K and n only) and x and xδ are as in Lemma 6.

Proof By (2.4)

n n hx, NK◦ (x)i kxδk hx, NK◦ (x)i hy, NK (y)i kyk 2 − 1 ≤ 2 0 − 1 n δ n+1 kxk hy, NK (y)i n δ n+1 ky k n 0 n hx, NK◦ (x)i kyk hy, NK (y)i ky k ≤ 0 2 1 − . hy, NK (y)i ky k n δ n+1 kyk

Since Kδ is increasing to K as δ → 0, there exists δ0 > 0 such that for all δ < δ0, n n 1 0 ∈ int(Kδ). Therefore there exits α > 0 such that B2 (0, α) ⊂ Kδ ⊂ K ⊂ B2 (0, α ) n for all δ < δ0. B2 (0, r) is the n-dimensional Euclidean ball centered at 0 with radius r. 46

Hence for δ < δ0

n 0 n hx, NK◦ (x)i kxδk hy, NK (y)i ky k n−1 −2(n+1) 0 − n+1 2 − 1 ≤ α 2 1 − ≤ C r(y) n δ n+1 kxk n δ n+1 kyk due to Lemma 6 in [84]. Here r(y) is the radius of the biggest Euclidean ball contained in K and touching ∂K at y. 2 Since K is C+, by the Blaschke rolling theorem (see [82]) there is r0 > 0 such that − n−1 0 n+1 r0 ≤ min r(y). We put c(K, n) = C r0 . y∈∂K

Proof of Theorem 14.

◦ ◦ Z n |(Kδ) | − |K | 1 kxδk ◦ ◦ 2 = 2 hx, NK (x)i − 1 dµK (x). δ n+1 n δ n+1 ∂Ko kxk Combining Lemma 6, Lemma 7 and Lebesgue’s dominated convergence theorem, gives Theorem 14:

◦ ◦ Z n |(Kδ) | − |K | 1 kxδk ◦ ◦ lim 2 = lim 2 hx, NK (x)i − 1 dµK (x) δ→0 δ n+1 δ→0 n δ n+1 ∂Ko kxk Z n 1 kxδk ◦ ◦ = lim 2 hx, NK (x)i − 1 dµK (x) ∂Ko δ→0 n δ n+1 kxk Z 2 hx, NK◦ (x)i ◦ = 1 dµK (x) ∂Ko cn (κK◦ (x)) n+1

1 ◦ = as −n (K ). cn n+2

Remark The proof of Theorem 14 provides a uniform method to evaluate

◦ ◦ |(Kt) | − |K | lim 2 t→0 t n+1 where Kt is a family convex bodies constructed from the convex body K such that

Kt ⊂ K or- similarly- such that K ⊂ Kt. In particular, we can apply this method 47

to prove the analog statements as in (2.1) and Theorem 14 if we take as Kt the illumination body of K [109], or the Santal´obody of K [69], or the convolution body of K [81] - and there are many more.

2.2 Lp aﬃne surface areas

2 ◦ We now prove that for all p 6= −n and all K ∈ C+, asp(K) = as n2 (K ). To do so, p we use the surface body of a convex body which was introduced in [85, 86]. We also

give a new geometric interpretation of Lp aﬃne surface area for all p 6= −n.

Deﬁnition 1 Let s ≥ 0 and f : ∂K → R be a nonnegative, integrable function. The + surface body Kf,s is the intersection of all the closed half-spaces H whose deﬁning

hyperplanes H cut oﬀ a set of fµK -measure less than or equal to s from ∂K. More precisely,

\ + Kf,s = H . R ∂K∩H− fdµK ≤s

2 Theorem 15 Let K be a convex body in C+ and such that 0 is the center of gravity of K. Let f : ∂K → R be an integrable function such that f(x) > c for all x ∈ ∂K 2 n−1 n−1 and some constant c > 0. Let βn = 2 |B2 | . Then

◦ ◦ Z |(Kf,s) | − |K | dσ(u) lim βn 2 = 2 s→0 1 −1 s n−1 Sn−1 n+1 n−1 n−1 hK (u) fK (u) f(NK (u))

n−1 where NK : ∂K → S , x → NK (x) = u is the Gauss map.

Proof n−1 Let u ∈ S . Let x ∈ ∂K be such that NK (x) = u and let xs ∈ ∂Kf,s be such that NKf,s (xs) = u. Let H∆ = H(x − ∆u, u) be the hyperplane through x − ∆u with 48

outer normal vector u. Since K has everywhere strictly positive Gaussian curvature, by Lemma 21 in [86] almost everywhere on ∂K,

1 Z lim − |f(x) − f(y)| dµK (y) = 0. ∆→0 |∂K ∩ H | − ∆ ∂K∩H∆

This implies that

1 Z lim − f(y) dµ∂K (y) = f(x). (2.6) ∆→0 |∂K ∩ H | − ∆ ∂K∩H∆

Let bs = hK (u) − hKf,s (u). As H(x − bsu, u) = H(xs, u) (the hyperplane through xs

with outer normal u) and as bs → 0 as s → 0, (2.6) implies

1 Z lim − f(y) dµK (y) = f(x). (2.7) s→0 − |∂K ∩ H (xs, u)| ∂K∩H (xs,u)

Hence there exists s1 small enough, such that for all s < s1,

Z − s ≤ f(y) dµK (y) ≤ (1 + ε)f(x)|∂K ∩ H (xs, u)|. (2.8) − ∂K∩H (xs,u)

As ∂K has everywhere strictly positive Gaussian curvature, the indicatrix of Dupin exists everywhere on ∂K and is an ellipsoid. It then follows from (2.8) with Lemmas

1.2, 1.3 and 1.4 in [85] that there exists 0 < s2 < s1 such that for all 0 < s < s2

n−1 n−1 p 2 s ≤ (1 + ε) f(x) |B2 | fK (u) (2bs) , or, equivalently

bs 1 − c1 ε 2 ≥ 2 1 , (2.9) n−1 −1 n−1 n−1 s βn f(NK (u)) fK (u) where c1 is an absolute constant.

0 0 Let now xs ∈ [0, x] ∩ ∂Kf,s. Then hxs, ui ≤ hKf,s (u). Therefore bs = hK (u) − 0 hKf,s (u) ≤ hx − xs, ui. 49

Hence for s suﬃciently small

0 0 0 bs hx − xs, ui hx, ui kxsk hx, ui kxs − xk 2 ≤ 2 ≤ 2 1 − ≤ 2 s n−1 s n−1 s n−1 kxk s n−1 kxk

0 n hx, NK (x)i kxsk ≤ (1 + ε) 2 1 − . (2.10) n s n−1 kxk

0 n 0 kxsk n kxs−xk The last inequality follows as 1 − kxk ≥ (1 − ε) kxk for suﬃciently small s. By Lemma 23 in [86]

0 n 1 kxsk 1 lim 2 hx, NK (x)i 1 − = 2 1 . (2.11) s→0 n−1 kxk −1 n−1 n−1 ns βnf(NK (u)) fK (u)

Thus we get from (2.9), (2.10) and (2.11) that

bs 1 lim 2 = 2 1 . (2.12) s→0 n−1 −1 n−1 n−1 s βnf(NK (u)) fK (u)

As (1 − t)−n ≥ 1 + nt for 0 ≤ t < 1 and by (2.12),

" −n # βn −n −n βn −n bs lim 2 [hKf,s (u)] − [hK (u)] = lim 2 [hK (u)] 1 + − 1 s→0 ns n−1 s→0 ns n−1 hK (u)

βn bs ≥ lim n+1 2 s→0 [hK (u)] s n−1 1 = 2 1 . (2.13) n+1 −1 n−1 n−1 [hK (u)] f(NK (u)) fK (u)

0 As hKf,s (u) ≥ hxs, ui,

0 0 hK (u) hx , ui kx k f,s ≥ s = s . (2.14) hK (u) hx, ui kxk

2 Since K ∈ C+, hKf,s (u) → hK (u) as s → 0. Therefore,

h (u)n βn −n −n βn −n Kf,s lim 2 [hKf,s (u)] − [hK (u)] = lim 2 [hKf,s (u)] 1 − s→0 ns n−1 s→0 ns n−1 hK (u) 50

0 n 0 n βn −n kxsk βn −n hx, ui kxsk ≤ lim 2 [hKf,s (u)] 1 − = lim 2 [hKf,s (u)] 1 − s→0 ns n−1 kxk s→0 ns n−1 hx, ui kxk

0 n 1 βn kxsk = lim lim 2 hx, NK (x)i 1 − s→0 n s→0 hK (u)[hKf,s (u)] ns n−1 kxk 1 = 2 1 (2.15) n+1 −1 n−1 n−1 [hK (u)] f(NK (u)) fK (u) where the last equality follows from (2.11).

Altogether, (2.13) and (2.15) give

βn −n −n 1 lim 2 [hKf,s (u)] − [hK (u)] = 2 1 . s→0 n−1 n+1 −1 n−1 n−1 ns [hK (u)] f(NK (u)) fK (u)

Therefore

◦ ◦ Z n n |(Kf,s) | − |K | βn 1 1 lim βn 2 = lim 2 − dσ(u) s→0 s→0 s n−1 n s n−1 Sn−1 hKf,s (u) hK (u) Z n n βn 1 1 = lim 2 − dσ(u) s→0 Sn−1 n s n−1 hKf,s (u) hK (u) Z dσ(u) = , 1 2 Sn−1 n+1 n−1 −1 n−1 hK (u) fK (u) f(NK (u)) provided we can interchange integration and limit.

We show this next. To do so, we show that for all u ∈ Sn−1 and all suﬃciently small s > 0, 1 1 n 1 n 2 − ≤ g(u) n s n−1 hKf,s (u) hK (u) R with Sn−1 g(u) dσ(u) < ∞. As 0 ∈ int(K), the interior of K, there exists α > 0 n n 1 such that for all s suﬃciently small B2 (0, α) ⊂ Kf,s ⊂ K ⊂ B2 (0, α ). Therefore, 1 1 1 1 α ≤ hKf,s (u) ≤ hK (u) ≤ α and α ≤ h (u) ≤ h (u) ≤ α . K Kf,s 51

With (2.14), we thus get for all s > 0,

1 −n −n 2 hKf,s (u) − hK (u) n s n−1 n 1 hK (u) = h (u)−n 1 − f,s 2 Kf,s n n s n−1 hK (u) −n 0 n α kxsk ≤ 2 1 − n s n−1 kxk 0 n −(n+1) hx, ui kxsk ≤ α 2 1 − . n s n−1 kxk

By Lemma 17 in [86] there exists s3 such that for all s ≤ s3

0 n hx, ui kxsk C 2 1 − ≤ 2 , s n−1 kxk (Mf (x)) n−1 r(x)

where C is an absolute constant and, as in the proof of Lemma 7, r(x) is the biggest 2 Euclidean ball contained in K that touches ∂K at x. Thus, as ∂K is C+, by Blaschke’s

rolling theorem (see [82]) there is r0 such that r(x) ≥ r0.

R ∂K∩H−(x ,N (x )) fdµK s Kf,s s Mf (x) = inf 0

~~is the minimal function. It was introduced in [86]. By the assumption on f, Mf (x) ≥ c. Thus altogether~~

~~ −(n+1) 1 −n −n α C 2 hKf,s (u) − hK (u) ≤ 2 = g(u), n s n−1 n c n−1 r0~~

~~which, as a constant, is integrable.~~

~~2 Theorem 16 Let K be a convex body in C+ and such that 0 is the center of gravity of K. Let f : ∂K → R be an integrable function such that f(y) > c for all y ∈ ∂K 52~~

~~2 n−1 n−1 and some constant c > 0. Let βn = 2 |B2 | . Then~~

~~◦ ◦ Z 1 ! |(Kf,s) | − |K | hx, NK◦ (x)i κK (y(x)) n−1 ◦ lim βn 2 = 2 dµK (x) s→0 s n−1 ∂K◦ hy(x),NK (y(x))i f(y(x)) n−1~~

~~Here y(x) ∈ ∂K is such that hy(x), xi = 1.~~

~~Proof We follow the pattern of the proof of Theorem 15 integrating now over ∂K◦ instead of Sn−1.~~

~~As a corollary we get the following geometric interpretation of Lp aﬃne surface area.~~

~~2 Corollary 2 Let K ∈ C+ be a convex body. For p ∈ R, p 6= −n, let fp : ∂K → R be~~

n2+p −(n−1)(n2+2n+p) deﬁned by fp(y) = κK (y) 2(n+p) hy, NK (y)i 2(n+p) . Then (i) ◦ ◦ |(Kfp,s) | − |K | ◦ lim βn 2 = as n2 (K ). s→0 s n−1 p (ii) ◦ ◦ |(Kfp,s) | − |K | lim βn 2 = asp(K). s→0 s n−1 (iii) ◦ asp(K) = as n2 (K ). p

~~Proof~~

~~Notice ﬁrst that fp(y) veriﬁes the conditions of Theorems 15 and 16. (i) For x ∈ ∂K◦, let now y(x) be the (unique) element in ∂K such that hx, y(x)i =~~

~~1. Then, by Theorem 16, with f(y(x)) = fp(y(x)), and with (2.5) 53~~

~~◦ n(n+1) ◦ Z n+p | Kfp,s | − |K | hy(x),NK (y(x))i ◦ ◦ lim βn 2 = hx, NK (x)i n dµK (x) s→0 s n−1 ∂K◦ κK (y(x)) n+p~~

~~n Z n+p κK◦ (x) ◦ = dµ ◦ (x) = as 2 (K ). n2−p K n p ∂K◦ n+p hx, NK◦ (x)i~~

~~n−1 −1 (ii) For u ∈ S , let now y ∈ ∂K be such that NK (y) = u. Then fp(NK (u)) =~~

~~n2+p −(n−1)(n2+2n+p) − 2(n+p) 2(n+p) −1 −1 fK (u) hK (u) . By Theorem 15 with f(NK (u)) = fp(NK (u))~~

~~◦ n ◦ Z n+p | Kfp,s | − |K | fK (u) lim βn 2 = n(p−1) dσ(u) = asp(K). s→0 n−1 n−1 s S hK (u) n+p~~

~~(iii) follows from (i) and (ii).~~

~~2.3 Lp aﬃne isoperimetric inequalities~~

Theorem 17 Let s 6= −n, r 6= −n, t 6= −n be real numbers. Let K be a convex body n in R with centroid at the origin and such that µK {x ∈ ∂K : κK (x) = 0} = 0. (n+r)(t−s) (i) If (n+t)(r−s) > 1, then

~~(r−s)(n+t) (t−r)(n+s) (t−s)(n+r) (t−s)(n+r) asr(K) ≤ ast(K) ass(K) .~~

~~(n+r)t (ii) If (n+t)r > 1, then~~

~~r(n+t) as (K) as (K) t(n+r) r ≤ t . n|K| n|K|~~

~~Proof 54~~

~~(n+r)(s−t) (i) By H¨older’s inequality -which enforces the condition (n+t)(s−r) > 1~~

~~Z r κK (x) n+r asr(K) = n(r−1) dµK (x) ∂K hx, NK (x)i n+r~~

~~(r−s)(n+t) (t−r)(n+s) Z t ! (t−s)(n+r) s ! (t−s)(n+r) κK (x) n+t κK (x) n+s = n(t−1) n(s−1) dµK (x) ∂K hx, NK (x)i n+t hx, NK (x)i n+s~~

~~(r−s)(n+t) (t−r)(n+s) (t−s)(n+r) (t−s)(n+r) ≤ ast(K) ass(K) .~~

~~(ii) Similarly, again using H¨older’s inequality -which now enforces the condition (n+r)t (n+t)r > 1,~~

~~Z r κK (x) n+r asr(K) = n(r−1) dµK (x) ∂K hx, NK (x)i n+r~~

~~r(n+t) Z t ! t(n+r) κK (x) n+t dµK (x) = n(t−1) (r−t)n ∂K n+t (n+r)t hx, NK (x)i hx, NK (x)i~~

~~r(n+t) (t−r)n t(n+r) (n+r)t ≤ ast(K) n |K| .~~

(n+r)(t−s) Condition (n+t)(r−s) > 1 implies 8 cases: −n < s < r < t, s < −n < t < r, r < t < −n < s, t < r < s < −n, s < r < t < −n, r < s < −n < t, t < −n < s < r and −n < t < r < s.

~~n+r asr(K) r Note also that (ii) describes a monotonicity condition for n|K| : if 0 < r < t, or r < t < −n, or −n < r < t < 0 then~~

~~n+r n+t as (K) r as (K) t r ≤ t . n|K| n|K|~~

~~n(s−t) We now analyze various subcases of Theorem 17 (i) and (ii). For r = 0, if s(n+t) > 1 55~~

~~s(n+t) t(n+s) n(s−t) n(t−s) n|K| ≤ ast(K) ass(K) .~~

~~t(n+r) For s = 0, if r(n+t) > 1,~~

~~n(t−r) r(n+t) t(n+r) asr(K) ≤ (n|K|) t(n+r) ast(K) . (2.18)~~

~~n+r For s → ∞, if n+t > 1,~~

~~r−t n+t n+r n+r asr(K) ≤ as∞(K) ast(K) . (2.19)~~

~~t−s 2 For r → ∞, if n+t > 1 and if K is in C+,~~

~~n+t n+s ◦ t−s s−t as∞(K) = n|K | ≤ ast(K) ass(K) . (2.20)~~

~~◦ As for all convex bodies K, as∞(K) ≤ n|K | (see [86]), it follows from (2.18) that, for all convex body K with centroid at origin,~~

~~n ◦ r asr(K) ≤ (n|K|) n+r (n|K |) n+r , r > 0 (2.21) and from (2.19),~~

~~t n+t ◦ n n n|K| n|K | ≤ ast(K) , −n < t < 0. (2.22)~~

~~2 Similarly, (2.20) implies that, if in addition K is in C+,~~

~~n n+t ◦ t t n|K | n|K| ≤ ast(K) , t < −n (2.23)~~

~~(2.21) can also be obtained from Proposition 4.6 of [62] and Theorem 3.2 of [47].~~

~~Inequalities (2.21), (2.22) and (2.23) yield the following Corollary which was proved by Lutwak [62] in the case p ≥ 1.~~

~~Corollary 3 Let K be convex body in Rn with centroid at the origin. 56~~

~~(i) For all p ≥ 0 ◦ 2 ◦ asp(K) asp(K ) ≤ n |K| |K |.~~

~~(ii) For −n < p < 0,~~

~~◦ 2 ◦ asp(K)asp(K ) ≥ n |K| |K |.~~

~~2 If K is in addition in C+, inequality (ii) holds for all p < −n.~~

◦ n 2 Thus, using Santalóinequality in (i), for p ≥ 0, asp(K)asp(K ) ≤ asp(B2 ) , and ◦ n n 2 inverse Santalóinequality in (ii), for −n < p < 0, asp(K)asp(K ) ≥ c asp(B2 ) . c is the constant in the inverse Santalóinequality [14, 49].

~~Proof (i) follows immediately from (2.21). (ii) follows from (2.22) if −n < p < 0 and from (2.23) if p < −n.~~

~~Lutwak [62] proved for p ≥ 1~~

~~n−p n+p asp(K) |K| n ≤ n asp(B2 ) |B2 | with equality if and only if K is an ellipsoid. We now generalize these Lp-aﬃne isoperimetric inequalities to p < 1.~~

~~Theorem 18 Let K be a convex body with centroid at the origin. (i) If p ≥ 0, then~~

~~n−p n+p asp(K) |K| n ≤ n , asp(B2 ) |B2 | with equality if and only if K is an ellipsoid. For p = 0, equality holds trivially for all K. 57~~

~~(ii) If −n < p < 0, then~~

~~n−p n+p asp(K) |K| n ≥ n , asp(B2 ) |B2 | with equality if and only if K is an ellipsoid. 2 (iii) If K is in addition in C+ and if p < −n, then~~

~~n−p n+p asp(K) np |K| n+p n ≥ c n . asp(B2 ) |B2 | where c is the constant in the inverse Santal´oinequality [14, 49].~~

We cannot expect to get a strictly positive lower bound in Theorem 18 (i), even 2 2 if K is in C+: Consider, in R , the convex body K(R, ε) obtained as the intersection of four Euclidean balls with radius R centered at (±(R − 1), 0), (0, ±(R − 1)), R 2 arbitrarily large. To obtain a body in C+, we “round” the corners by putting there 16 Euclidean balls with radius ε, ε arbitrarily small. Then asp(K(R, ε)) ≤ p + R 2+p 2 4π ε 2+p . A similar construction can be done in higher dimensions. This example also shows that, likewise, we cannot expect ﬁnite upper bounds in

~~3(p+1) −p Theorem 18 (ii) and (iii). If −2 < p < 0, then asp(K(R, ε)) ≥ 2 2+p R 2+p . If p < −2, 4 then −2 < p < 0 and thus~~

~~◦ −2 12+3p asp(K(R, ε) ) = as 4 (K(R, ε)) ≥ R p+2 2 4+2p . p~~

~~np Note also that in part (iii) we cannot remove the constant c n+p . Indeed, if p → n n 2 ◦ −∞, the inequality becomes c |B2 | ≤ |K||K |.~~

~~Proof of Theorem 18 (i) The case p = 0 is trivial. We prove the case p > 0. Combining inequality n q n n n+q n n+q (2.21), the Blaschke Santal´oinequality, and asq(B2 ) = n|B2 | |B2 | , one obtains~~

~~p n n−p ◦ n+p n+p n+p asp(K) |K | |K| |K| n ≤ n n ≤ n . asp(B2 ) |B2 | |B2 | |B2 | 58~~

~~This proves the inequality. The equality case follows from the equality case for the Blaschke Santal´oinequality.~~

~~n+p p n n n n n (ii) Combining inequality (2.22) and asp(B2 ) = n|B2 | n|B2 | , one gets, for −n < p < 0,~~

~~n+p p n−p n ◦ n n asp(K) |K| |K | |K| n ≥ n n ≥ n asp(B2 ) |B2 | |B2 | |B2 |~~

~~p where the last inequality follows from the Blaschke Santal´oinequality. As n < 0,~~

~~p p ◦ n n n n (|K| |K |) ≥ (|B2 | |B2 |) .~~

~~As n + p > 0,~~

~~n−p n+p asp(K) |K| n ≥ n . asp(B2 ) |B2 | The equality case follows from the equality case for the Blaschke Santal´oinequality.~~

~~n+p n+1 n n p−1 n 1−p (iii) Similarly, combining (2.23), n|B2 | = (asp(B2 )) (as(B2 )) , and the In- verse Santal´oinequality, we get, for p < −n,~~

~~n+p n n−p p ◦ p p asp(K) |K | |K| n |K| n ≥ n n ≥ c n . asp(B2 ) |B2 | |B2 | |B2 |~~

~~n+p As p > 0,~~

~~n−p n+p asp(K) np |K| n+p n ≥ c n . asp(B2 ) |B2 |~~

~~2 The L−n aﬃne surface area was deﬁned in [70] for convex bodies K in C+ and with centroid at the origin by~~

~~1 n+1 as−n(K) = max fK (u) 2 hK (u) 2 . u∈Sn−1 59~~

More generally, one could define the L−n affine surface area for any convex body n+1 hx,NK (x)i 2 K with centroid at the origin by as−n(K) = supx∈∂K 1 . But as in most κK (x) 2 2 cases then as−n(K) = ∞, it suffices to consider K in C+.

~~A statement similar to Theorem 17 holds.~~

~~2 Proposition 1 Let K be a convex body in C+ with centroid at the origin. Let p 6= −n and s 6= −n be real numbers. n(s−p) (i) If (n+p)(n+s) ≥ 0, then~~

~~2n(s−p) (n+p)(n+s) asp(K) ≤ as−n(K) ass(K).~~

~~n(s−p) (ii) If (n+p)(n+s) ≤ 0, then~~

~~2n(s−p) (n+p)(n+s) asp(K) ≥ as−n(K) ass(K).~~

~~(iii) The L−n aﬃne isoperimetric inequality holds~~

~~as−n(K) |K| n ≥ n . as−n(B2 ) |B2 |~~

~~Proof (i) and (ii)~~

~~p Z n+p κK (x) asp(K) = n(p−1) dµK (x) ∂K hx, NK (x)i n+p~~

2n(s−p) Z s ! n+1 ! (n+p)(n+s) κK (x) n+s hx, NK (x)i 2 = n(s−1) 1 dµK (x) 2 ∂K hx, NK (x)i n+s κK (x) which is 2n(s−p) (n+p)(n+s) n(s−p) ≤ as−n(K) ass(K), if (n+p)(n+s) ≥ 0, and 2n(s−p) (n+p)(n+s) n(s−p) ≥ as−n(K) ass(K), if (n+p)(n+s) ≤ 0. 60

~~n(s−p) (iii) Note that (n+p)(n+s) > 0 implies that s > p > −n or p < s < −n or s < −n < p. If p = 0 and s → ∞, then~~

~~s |K| as (K) ≥ . (2.24) −n |K◦|~~

~~This gives the L−n aﬃne isoperimetric inequality~~

~~s s s 2 2 as−n(K) |K| |K| |K| |K| n = as−n(K) ≥ ◦ ≥ ◦ ≥ n 2 = n . as−n(B2 ) |K | |K| |K | |B2 | |B2 |~~

~~Analogous to corollary 3, an immediate consequence of (2.24) is the following corollary. It can also be proved directly using (2.5).~~

~~2 Corollary 4 Let K be a convex body in C+ with centroid at the origin. Then~~

~~◦ n 2 as−n(K )as−n(K) ≥ as−n(B2 ) . Chapter 3~~

~~Inequalities for mixed p-aﬃne surface area~~

In this chapter, we study mixed p-aﬃne surface areas, and inequalities related to them, such as, related isoperimetric inequalities and Alexandrov-Fenchel type inequalities. The chapter is based on the work [110] and is organized as follows. In Section 2, we prove new Alexandrov-Fenchel type inequalities and new isoperimetric inequalities for mixed p-aﬃne surface areas. We show a monotonicity behaviour of the quotients

~~n+p n+p as (K , ··· ,K ) as (K , ··· ,K ) p p 1 n and p 1 n . as∞(K1, ··· ,Kn) as0(K1, ··· ,Kn)~~

We prove Blaschke-Santalótype inequalities for mixed p-affine surface areas. Similar results for the i-th mixed p-affine surface areas are also proved in Section 2. In Section 3, we introduce the illumination surface body and describe some of its properties. In Section 4, we derive the asymptotic behavior of the volume of the illumination surface body, and geometric interpretations of Lp affine surface areas, mixed p-affine surface areas, and other functionals on convex bodies.

~~61 62 3.1 Mixed p-aﬃne surface area and related inequalities~~

~~3.1.1 Inequalities for mixed p-aﬃne surface area~~

We begin by proving that mixed p-affine surface area is affine invariant for all p. For p ≥ 1, this was proved by Lutwak [62]. We will first treat the case p 6= −n. All the results concerning the case p = −n are at the end of this subsection. It will be convenient to use the notation

~~1−p fp(K, u) = hK (u)fK (u) (3.1)~~

~~n n−1 m for a convex body K in R and u ∈ S . We will also write asp (K1, ··· ,Kn) for m asp(K1, ··· ,Kn) , and |det(T )| for the absolute value of the determinant of a linear transform T .~~

~~Lemma 8 Let T : Rn → Rn be an invertible linear transform. Then for all p 6= −n,~~

~~n−p asp(TK1, ··· ,TKn) = |det(T )| n+p asp(K1, ··· ,Kn).~~

~~In particular, if |det(T )| = 1, then asp(K1, ··· ,Kn) is aﬃne invariant:~~

~~asp(TK1, ··· ,TKn) = asp(K1, ··· ,Kn).~~

~~Proof. 2 n−1 Since K ∈ C+, for any u ∈ S , there exists a unique x ∈ ∂K such that u = NK (x) 1 and fK (u) = . By Lemma 12 of [86] κK (x)~~

~~1 fTK (v) fK (u) = = 2 −1t n+1 , (3.2) κK (x) det (T ) kT (u)k~~

~~T −1t(u) n−1 t where v = kT −1t(u)k ∈ S and where for an operator A, A denotes its usual adjoint. On the other hand,~~

~~−1t −1t −1t hK (u) = hx, ui = hT x, T (u)i = kT (u)k hT x, vi = kT (u)k hTK (v). 63~~

~~Thus, with notation (3.1), for all p,~~

~~f (v) h1−p(v) kT −1t(u)k1−p f (T K, v) f (K, u) = TK TK = p . (3.3) p det2(T ) kT −1t(u)kn+1 det2(T ) kT −1t(u)kn+p~~

~~Lemma 10 and its proof in [86] show that -up to a small error-~~

~~−1t fTK (v) dσ(v) = |det(T )| kT (u)kfK (u) dσ(u).~~

~~Together with (3.2), one gets that (again up to a small error) kT −1t(u)k−n dσ(u) = |det(T )| dσ(v). Therefore, up to a small error,~~

~~1 p−n 1 [fp(K1, u) ··· fp(Kn, u)] n+p dσ(u) = |det(T )| n+p [fp(TK1, v) ··· fp(TKn, v)] n+p dσ(v).~~

~~The lemma then follows by integrating over Sn−1.~~

~~A general version of the classical Alexandrov-Fenchel inequalities for mixed volumes (see [3, 17, 82]) can be written as~~

~~m−1 Y m V (K1, ··· ,Kn−m,Kn−i, ··· ,Kn−i) ≤ V (K1, ··· ,Kn). | {z } i=0 m~~

~~Here we prove the analogous inequalities for mixed p-aﬃne surface area. For p = ±∞ and p = 1, the inequalities were proved by Lutwak [58, 59]. For p ≥ 1, inequality~~

~~(3.4) was proved by Lutwak in [62], with equality if and only if the associated Ki are dilates of each other.~~

~~2 Proposition 2 Let all Ki be convex bodies in C+ with centroid at the origin. If p 6= −n, then for 1 ≤ m ≤ n~~

~~m−1 m Y asp (K1, ··· ,Kn) ≤ asp(K1, ··· ,Kn−m,Kn−i, ··· ,Kn−i). | {z } i=0 m~~

~~Equality holds if the Kk, for k = n − m + 1, ··· , n are dilates of each other. If m = 1, equality holds trivially. 64~~

~~In particular, if m = n,~~

~~n asp (K1, ··· ,Kn) ≤ asp(K1) ··· asp(Kn). (3.4)~~

~~1 Proof. Put g0(u) = [fp(K1, u) ··· fp(Kn−m, u)] n+p and for i = 0, ··· , m − 1, put~~

~~1 gi+1(u) = [fp(Kn−i, u)] n+p . By H¨older’s inequality (see [36]) Z asp(K1, ··· ,Kn) = g0(u)g1(u) ··· gm(u) dσ(u) Sn−1~~

~~m−1 1 Z m Y m ≤ g0(u)gi+1(u) dσ(u) n−1 i=0 S~~

~~m−1 1 Y m = asp (K1, ··· ,Kn−m,Kn−i ··· Kn−i). | {z } i=0 m~~

2 n−1 As Ki ∈ C+, fp(Ki, u) > 0 for all i and all u ∈ S . Therefore, equality in H¨older’s m m m inequality holds if and only if g0(u)gi+1(u) = λ g0(u)gj+1(u) for some λ > 0 and all 1−p 1−p 0 ≤ i 6= j ≤ m−1. This is equivalent to hKn−i (u) fKn−i (u) = λhKn−j (u) fKn−j (u)

~~for all 0 ≤ i 6= j ≤ m−1. This condition holds true if the Kk, for k = n−m+1, ··· , n are dilates of each other.~~

~~Remark. It is an unsolved problem for many p whether fp(K, u) = λfp(L, u) guarantees that K and L are dilates of each other. This is equivalent to the uniqueness of~~

the solution of the Lp Minkowski problem: for ﬁxed α ∈ R, under which conditions on a continuous function γ : Sn−1 → (0, ∞), there exists a (unique) convex body K α n−1 such that hK (u) fK (u) = γ(u) for all u ∈ S . In many cases, the uniqueness of the solution is an open problem. We refer to e.g., [20, 61, 63, 66, 96, 97] for detailed information and more references on the subject. For p ≥ 1, p 6= n, the solution to

~~the Lp Minkowski problem is known to be unique and for p = n, the solution is unique modulo dilates [61]. Therefore, we have the characterization of equality in Proposition 2 for p ≥ 1.~~

~~Next, we prove aﬃne isoperimetric inequalities for mixed p-aﬃne surface areas. 65~~

~~2 Proposition 3 Let all Ki be convex bodies in C+ with centroid at the origin.~~

~~(i) For p ≥ 0,~~

~~n−p n n+p asp (K1, ··· ,Kn) |K1| |Kn| n n n ≤ n ··· n , asp (B2 , ··· ,B2 ) |B2 | |B2 |~~

~~with equality if the Ki are ellipsoids that are dilates of one another.~~

~~(ii) For 0 ≤ p ≤ n,~~

~~n−p n+p asp(K1, ··· ,Kn) V (K1, ··· ,Kn) n n ≤ n n asp(B2 , ··· ,B2 ) V (B2 , ··· ,B2 )~~

~~with equality if the Ki are ellipsoids that are dilates of one another.~~

~~In particular, for p = n~~

~~n n asn(K1, ··· ,Kn) ≤ asn(B2 , ··· ,B2 ),~~

~~with equality if and only if the Ki are ellipsoids that are dilates of one another.~~

~~(iii) For p ≥ n,~~

n−p ! n+p as (K , ··· ,K ) V˜ (K , ··· ,K ) p 1 n ≤ 1 n , n n ˜ n n asp(B2 , ··· ,B2 ) V (B2 , ··· ,B2 ) with equality if and only if the Ki are ellipsoids that are dilates of one another.

~~In particular, for p = ±∞~~

~~˜ ˜ ◦ ◦ n 2 V (K1, ··· ,Kn)V (K1 , ··· ,Kn) ≤ |B2 | ,~~

~~with equality if and only if Ki are ellipsoids that are dilates of one another.~~

Remark. For 1 ≤ p ≤ n, inequality (ii) (with equality if and only if the Ki are ellipsoids that are dilates of one another) was proved by Lutwak in [62]. If Ki = K for all i, one recovers the Lp aﬃne isoperimetric inequality proved in Chapter 2. 66

Remark. We cannot expect to get strictly positive lower bounds in Proposition 3. As in Chapter 2, we consider the convex body K(R, ε) ⊂ R2, obtained as the intersection of four Euclidean balls with radius R centered at (±(R − 1), 0), (0, ±(R − 1)), R arbitrarily large. We then “round” the corners by putting there arcs of Euclidean balls 2 of radius ε, ε arbitrarily small. To obtain a body in C+, we “bridge” between the R- 2 arcs and ε-arcs by C+-arcs on a set of arbitrarily small measure. Then asp(K(R, ε)) ≤ 16 2 p + 4π ε 2+p , which goes to 0 as R → ∞ and ε → 0. Choose now Ri and R 2+p

εi, 1 ≤ i ≤ n, such that Ri → ∞ and εi → 0, and let Ki = K(Ri, εi) for i = n Qn 1, 2 ··· , n. By inequality (3.4), asp (K1, ··· ,Kn) ≤ i=1 asp(K(Ri, εi)) and thus asp(K1, ··· ,Kn) → 0 for p > 0. A similar construction can be done in higher dimensions.

~~Proof of Proposition 3. n n n n (i) Clearly asp(B2 , ··· ,B2 ) = asp(B2 ) = n|B2 | for all p 6= −n. By inequality (3.4), one gets for all p ≥ 0~~

~~n−p n n+p asp (K1, ··· ,Kn) asp(K1) asp(Kn) |K1| |Kn| n n n ≤ n ··· n ≤ n ··· n . (3.5) asp (B2 , ··· ,B2 ) asp(B2 ) asp(B2 ) |B2 | |B2 |~~

~~The second inequality follows, for p ≥ 0, from the Lp aﬃne isoperimetric inequality~~

~~in Chapter 2. Equality holds true in the Lp isoperimetric inequality if and only if the~~

~~Ki are all ellipsoids, and equality holds true in inequality (3.4) if the Ki are dilates~~

~~of one another. Thus, equality holds in (3.5) if the Ki are ellipsoids that are dilates of one another.~~

~~(ii) A direct consequence of the classical Alexandrov-Fenchel inequality for mixed volume (see e.g. [15, 50]) is that~~

~~n |K1| · · · |Kn| ≤ V (K1, ··· ,Kn).~~

~~n−p If 0 ≤ p ≤ n, then n+p ≥ 0. Thus~~

~~n−p n−p n+p n n+p |K1| · · · |Kn| ≤ V (K1, ··· ,Kn) . 67~~

~~n n n As V (B2 , ··· ,B2 ) = |B2 |, one gets together with (3.5)~~

~~n−p n+p asp(K1, ··· ,Kn) V (K1, ··· ,Kn) n n ≤ n n , asp(B2 , ··· ,B2 ) V (B2 , ··· ,B2 ) with equality if the Ki are ellipsoids that are dilates of one another.~~

~~(iii) The analogous inequality for dual mixed volume [58] is~~

~~˜ n |K1| · · · |Kn| ≥ V (K1, ··· ,Kn),~~

~~n−p with equality if and only the Ki are dilates of one another. p > n implies n+p < 0. Thus n−p n−p n+p ˜ n n+p |K1| · · · |Kn| ≤ V (K1, ··· ,Kn) .~~

~~˜ n n n Together with (3.5) and V (B2 , ··· ,B2 ) = |B2 |, one gets~~

~~n−p ! n+p as (K , ··· ,K ) V˜ (K , ··· ,K ) p 1 n ≤ 1 n . n n ˜ n n asp(B2 , ··· ,B2 ) V (B2 , ··· ,B2 )~~

~~As for p ≥ 1 equality in (3.4) holds if and only if the Ki are dilates of one another, equality holds true here if and only if the Ki are ellipsoids that are dilates of one another.~~

~~2 Proposition 4 Let E be a centered ellipsoid. If either all Ki ∈ C+ are subsets of E, for 0 ≤ p < n, or E is subset of all Ki for p > n, then~~

~~asp(K1, ··· ,Kn) ≤ asp(E).~~

~~2 For p = n, the inequality holds for all Ki in C+ by Proposition 3 (ii).~~

~~Remark. This proposition was proved by Lutwak [62] if Ki = K for all i.~~

n Proof. It is enough to prove the proposition for E = B2 . For 0 ≤ p < n, one has n−p n+p n−p |Ki| n n−p > 0 and hence n ≤ 1 as Ki ⊂ B2 . Similarly, p > n implies < 0 and n+p |B2 | n+p 68

~~n−p n+p |Ki| n therefore n ≤ 1 as B2 ⊂ Ki for all i. In both cases, the proposition follows |B2 | by inequality (3.5).~~

~~The next proposition gives a Blaschke-Santal´otype inequality for p-mixed aﬃne sur-~~

~~face area. When Ki = K for all i, the proposition was proved in Chapter 2.~~

~~2 Proposition 5 Let all Ki be convex bodies in C+ with centroid at the origin. For all p ≥ 0,~~

~~n n ◦ ◦ 2n ◦ ◦ asp (K1, ··· ,Kn)asp (K1 , ··· ,Kn) ≤ n |K1||K1 | · · · |Kn||Kn|. (3.6)~~

~~◦ ◦ 2 n Furthermore, asp(K1, ··· ,Kn)asp(K1 , ··· ,Kn) ≤ asp(B2 ) with equality if the Ki are ellipsoids that are dilates of one another.~~

~~Proof. It follows from (3.4) that for all p 6= −n,~~

~~n n ◦ ◦ ◦ ◦ asp (K1, ··· ,Kn)asp (K1 , ··· ,Kn) ≤ asp(K1)asp(K1 ) ··· asp(Kn)asp(Kn),~~

~~with equality if the Ki are dilates of one another. By Corollary 3, for p ≥ 0,~~

~~n n ◦ ◦ 2n ◦ ◦ asp (K1, ··· ,Kn)asp (K1 , ··· ,Kn) ≤ n |K1||K1 | · · · |Kn||Kn|.~~

~~◦ n 2 Blaschke-Santal´oinequality states that |K||K | ≤ |B2 | with equality if and only if K is a 0-centered ellipsoid. We apply it to inequality (3.6), and obtain that for p ≥ 0,~~

~~◦ ◦ 2 n n asp(K1, ··· ,Kn)asp(K1 , ··· ,Kn) ≤ asp(B2 , ··· ,B2 ).~~

~~Equality holds if the Ki are ellipsoids that are dilates of one another.~~

~~Theorem 19 Let s 6= −n, r 6= −n, p 6= −n be real numbers. Let all Ki be convex 2 bodies in C+ with centroid at the origin. (n+p)(r−s) (i) If (n+r)(p−s) > 1, then~~

~~(p−s)(n+r) (r−p)(n+s) (r−s)(n+p) (r−s)(n+p) asp(K1, ··· ,Kn) ≤ asr(K1, ··· ,Kn) ass(K1, ··· ,Kn) . 69~~

~~n+p (ii) If n+r > 1, then~~

~~p−r n+r n+p n+p ˜ ◦ ◦ asp(K1, ··· ,Kn) ≤ (asr(K1, ··· ,Kn)) nV (K1 , ··· ,Kn) . (3.7)~~

~~Remark. When all Ki coincide with K, (i) of Theorem 19 was proved in Chapter 2.~~

~~Proof.~~

~~(n+p)(r−s) (i) By H¨older’s inequality -which enforces the condition (n+r)(p−s) > 1,~~

~~Z 1 n+p asp(K1, ··· ,Kn) = fp(K1, u) ··· fp(Kn, u) dσ(u) Sn−1~~

~~(n+r)(p−s) Z 1 (n+p)(r−s) n+r = fr(K1, u) ··· fr(Kn, u) Sn−1~~

~~(n+s)(r−p) 1 (n+p)(r−s) n+s fs(K1, u) ··· fs(Kn, u) dσ(u)~~

~~(p−s)(n+r) (r−p)(n+s) (r−s)(n+p) (r−s)(n+p) ≤ asr(K1, ··· ,Kn) ass(K1, ··· ,Kn) .~~

~~n+p (ii) Similarly, again using H¨older’sinequality -which now enforces the condition n+r > 1~~

~~Z 1 n+p asp(K1, ··· ,Kn) = fp(K1, u) ··· fp(Kn, u) dσ(u) Sn−1~~

~~n+r Z 1 n+p n+r = fr(K1, u) ··· fr(Kn, u) Sn−1~~

~~p−r 1 n+p dσ(u) hK1 (u) ··· hKn (u)~~

~~n+r p−r ≤ (asr(K1, ··· ,Kn)) n+p (as∞(K1, ··· ,Kn)) n+p .~~

~~Together with (1.7), this completes the proof.~~

(n+p)(r−s) Remark. The condition (n+r)(p−s) > 1 implies 8 cases: −n < s < p < r, s < −n < r < p, p < r < −n < s, r < p < s < −n, s < p < r < −n, p < s < −n < r, r < −n < s < p and −n < r < p < s. 70

~~n+r asr(K) r In Chapter 2, we proved monotonicity properties of n|K| . Here we prove similar results for mixed p-aﬃne surface area.~~

~~2 Proposition 6 Let all Ki ∈ C+ be convex bodies with centroid at the origin.~~

~~(i) If −n < r < p or r < p < −n, one has~~

~~ as (K , ··· ,K ) n+p as (K , ··· ,K ) n+r p 1 n ≤ r 1 n . ˜ ◦ ◦ ˜ ◦ ◦ nV (K1 , ··· ,Kn) nV (K1 , ··· ,Kn)~~

~~(ii) If 0 < p < r, or p < r < −n, or r < −n < 0 < p, or −n < p < r < 0, one has~~

~~n+p n+r as (K , ··· ,K ) p as (K , ··· ,K ) r p 1 n ≤ r 1 n . as0(K1, ··· ,Kn) as0(K1, ··· ,Kn)~~

~~Proof.~~

~~˜ ◦ ◦ n+p (i) We divide both sides of inequality (3.7) by nV (K1 , ··· ,Kn), and get for n+r > 1,~~

~~n+r as (K , ··· ,K ) as (K , ··· ,K ) n+p p 1 n ≤ r 1 n . (3.8) ˜ ◦ ◦ ˜ ◦ ◦ nV (K1 , ··· ,Kn) nV (K1 , ··· ,Kn)~~

~~n+p Condition n+r > 1 implies that −n < r < p or p < r < −n. If −n < r < p, n + p > 0 and therefore, inequality (3.8) implies inequality (i). If p < r < −n, n + p < 0 and therefore,~~

~~ as (K , ··· ,K ) n+p as (K , ··· ,K ) n+r p 1 n ≥ r 1 n . ˜ ◦ ◦ ˜ ◦ ◦ nV (K1 , ··· ,Kn) nV (K1 , ··· ,Kn)~~

~~Switching r and p, one obtains the inequality in (i): for r < p < −n,~~

~~ as (K , ··· ,K ) n+p as (K , ··· ,K ) n+r p 1 n ≤ r 1 n . ˜ ◦ ◦ ˜ ◦ ◦ nV (K1 , ··· ,Kn) nV (K1 , ··· ,Kn)~~

~~r(n+p) (ii) Let s = 0 in inequality (i) of Theorem 19. Then for p(n+r) > 1,~~

~~p(n+r) (r−p)n r(n+p) r(n+p) asp(K1, ··· ,Kn) ≤ asr(K1, ··· ,Kn) as0(K1, ··· ,Kn) . 71~~

~~We divide both sides of the inequality by as0(K1, ··· ,Kn) and get~~

~~p(n+r) as (K , ··· ,K ) as (K , ··· ,K ) r(n+p) p 1 n ≤ r 1 n . as0(K1, ··· ,Kn) as0(K1, ··· ,Kn)~~

r(n+p) The condition p(n+r) > 1 implies that 0 < p < r, or p < r < −n, or −n < r < p < 0, or r < −n < 0 < p. In the cases 0 < p < r, or p < r < −n, or r < −n < 0 < p, n+p one has p > 0 and therefore inequality (ii) holds true. On the other hand, if n+p −n < r < p < 0, then p < 0 and hence,

~~n+r n+p as (K , ··· ,K ) r as (K , ··· ,K ) p r 1 n ≤ p 1 n . as0(K1, ··· ,Kn) as0(K1, ··· ,Kn)~~

~~Switching r and p, one gets inequality (ii): if −n < p < r < 0, then~~

~~n+p n+r as (K , ··· ,K ) p as (K , ··· ,K ) r p 1 n ≤ r 1 n . as0(K1, ··· ,Kn) as0(K1, ··· ,Kn)~~

~~Now we treat the case p = −n. The mixed (−n)-aﬃne surface area of K1, ··· ,Kn is deﬁned as~~

~~1 n+1 1 n+1 2n 2n 2n 2n as−n(K1, ··· ,Kn) = max fK1 (u) hK1 (u) ··· fKn (u) hKn (u) . (3.9) u∈Sn−1~~

~~It is easy to verify that as−n(K, ··· ,K) equals to as−n(K), the L−n aﬃne surface area of K [70]. We have the following proposition.~~

~~2 Proposition 7 Let all Ki be convex bodies in C+ with centroid at the origin. Let p 6= −n and s 6= −n be real numbers. (i) Let T : Rn → Rn be an invertible linear transform. Then~~

~~as−n(TK1, ··· ,TKn) = |det(T )| as−n(K1, ··· ,Kn).~~

~~In particular, if |det(T )| = 1, then as−n(K1, ··· ,Kn) is aﬃne invariant:~~

~~as−n(TK1, ··· ,TKn) = as−n(K1, ··· ,Kn). 72~~

~~(ii) Alexandrov-Fenchel type inequalities~~

~~m−1 m Y as−n(K1, ··· ,Kn) ≤ as−n(K1, ··· ,Kn−m,Kn−i, ··· ,Kn−i), | {z } i=0 m with equality if the Kj, for j = n − m + 1, ··· , n are dilates. In particular, if m = n,~~

~~n as−n(K1, ··· ,Kn) ≤ as−n(K1) ··· as−n(Kn).~~

~~n(s−p) (iii) If (n+p)(n+s) ≥ 0, then~~

~~2n(s−p) (n+p)(n+s) asp(K1, ··· ,Kn) ≤ as−n(K1, ··· ,Kn) ass(K1, ··· ,Kn).~~

~~n(s−p) (iv) If (n+p)(n+s) ≤ 0, then~~

~~2n(s−p) (n+p)(n+s) asp(K1, ··· ,Kn) ≥ as−n(K1, ··· ,Kn) ass(K1, ··· ,Kn).~~

~~Proof. 2 (i) By formula (3.3), f−n(T K, v) = det(T ) f−n(K, u). Therefore~~

~~1 as−n(TK1, ··· ,TKn) = maxv∈Sn−1 [f−n(TK1, v) ··· f−n(TKn, v)] 2n~~

~~1 = |det(T )| maxu∈Sn−1 [f−n(K1, u) ··· f−n(Kn, u)] 2n~~

~~= |det(T )| as−n(K1, ··· ,Kn).~~

~~1 n+1 1 n+1 2n 2n 2n 2n (ii) Letg ˜0(u) = fK1 (u) hK1 (u) ··· fKn−m (u) hKn−m (u) and 1 n+1 2n 2n g˜i+1(u) = fKn−i (u) hKn−i (u) , for i = 0, ··· , m − 1. Then~~

~~as−n(K1, ··· ,Kn) = max g˜0(u)˜g1(u) ··· g˜m(u) u∈Sn−1~~

~~m−1 1 m Y m ≤ max g˜0(u)˜gi+1(u) u∈Sn−1 i=0~~

~~m−1 1 Y m = as−n(K1, ··· ,Kn−m,Kn−i ··· ,Kn−i). | {z } i=0 m 73~~

~~m Equality holds if and only if for all i, 0 ≤ i ≤ m−1,g ˜0(u)˜gi+1(u) attain their maximum~~

~~at the same direction u0. This condition holds true if the Kj, for j = n−m+1, ··· , n, are dilates.~~

~~(iii) and (iv)~~

~~Z 1 asp(K1, ··· ,Kn) = [fp(K1, u) ··· fp(Kn, u)] n+p dσ(u) Sn−1~~

~~Z 1 = [fs(K1, u) ··· fs(Kn, u)] n+s Sn−1~~

~~2n(s−p) n+1 1 n+1 1 (n+p)(n+s) h 2n (u)f 2n (u) ··· h 2n (u)f 2n (u) dσ(u) K1 K1 Kn Kn which is~~

~~2n(s−p) n(s − p) ≤ as (K , ··· ,K ) (n+p)(n+s) as (K , ··· ,K ), if ≥ 0, −n 1 n s 1 n (n + p)(n + s) and~~

~~2n(s−p) n(s − p) ≥ as (K , ··· ,K ) (n+p)(n+s) as (K , ··· ,K ), if ≤ 0. −n 1 n s 1 n (n + p)(n + s)~~

~~3.1.2 i-th mixed p-aﬃne surface area and related inequalities~~

~~2 For all p ≥ 1 and all real i, the i-th mixed p-aﬃne surface area of K,L ∈ C+ is deﬁned as [59, 107]~~

~~Z n−i i asp,i(K,L) = fp(K, u) n+p fp(L, u) n+p dσ(u). Sn−1~~

~~1−p Recall that fp(K, u) = fK (u)hK (u). Here we further generalize this definition to all p 6= −n and all i. An analogous definition for the i-th mixed (−n)-affine surface area of K and L is~~

~~n−i (n+1)(n−i) i (n+1)i as−n,i(K,L) = max fK (u) 2n hK (u) 2n fL(u) 2n hL(u) 2n . u∈Sn−1 74~~

~~When i ∈ N, 0 ≤ i ≤ n, then, for all p, the i-th mixed p-aﬃne surface area of K and L is~~

~~asp,i(K,L) = asp(K, ··· ,K, L, ··· ,L). | {z } | {z } n−i i~~

~~n Clearly, for all p, asp,0(K,L) = asp(K), and asp,n(K,L) = asp(L). When L = B2 , we n write asp,i(K) for asp,i(K,B2 ). Thus~~

Z n−i asp,i(K) = fp(K, u) n+p dσ(u), for p 6= −n, Sn−1 n−i (n+1)(n−i) asp,i(K) = max fK (u) 2n hK (u) 2n , for p = −n. u∈Sn−1 R In particular, as1,−1(K) = Sn−1 fK (u) dσ(u) is the surface area of K.

~~The next proposition and its proof is similar to Theorem 19 and its proof. There- fore we omit it.~~

~~2 Proposition 8 Let K and L be convex bodies in C+ with centroid at the origin. Let i ∈ R and s 6= −n, r 6= −n, and p 6= −n be real numbers.~~

~~(n+p)(r−s) (i) If (n+r)(p−s) > 1, then~~

~~(p−s)(n+r) (r−p)(n+s) (r−s)(n+p) (r−s)(n+p) asp,i(K,L) ≤ asr,i(K,L) ass,i(K,L) .~~

~~n+p (ii) If n+r > 1, then~~

~~p−r n+r n+p n+p ˜ ◦ ◦ asp,i(K,L) ≤ (asr,i(K,L)) nVi(K ,L )~~

~~˜ ◦ ◦ 1 R 1 where Vi(K ,L ) = n−1 n−i i dσ(u) for all i. n S hK (u) hL(u) n(s−p) (iii) If (n+p)(n+s) ≥ 0, then~~

~~2n(s−p) (n+p)(n+s) asp,i(K,L) ≤ as−n,i(K,L) ass,i(K,L).~~

~~n(s−p) (iv) If (n+p)(n+s) ≤ 0, then~~

~~2n(s−p) (n+p)(n+s) asp,i(K,L) ≥ as−n,i(K,L) ass,i(K,L). 75~~

~~The following proposition was proved in [59, 107] for p ≥ 1.~~

~~2 Proposition 9 Let K and L be convex bodies in C+ with centroid at the origin. If k−j j < i < k or k < i < j (equivalently, k−i > 1 ), then for all p,~~

~~k−i i−j asp,i(K,L) ≤ asp,j(K,L) k−j asp,k(K,L) k−j , with equality if K and L are dilates of each other. In particular,~~

~~k−i i−j asp,i(K) ≤ asp,j(K) k−j asp,k(K) k−j ,~~

~~with equality if K is a ball.~~

For p 6= −n, the proof is the same as the proof in [59, 107]. For p = −n, it is similar to the proof of Proposition 2. Note that for i ∈ N, 0 < i < m, m = j and k = 0, the proposition is a direct consequence of Proposition 2.

~~In Proposition 9, if j = 0 and k = n, then for all p and 0 ≤ i ≤ n~~

~~n n−i i asp,i(K,L) ≤ asp (K)asp(L). (3.10)~~

~~If we let i = 0 and j = n, then for all k ≤ 0 and for all p~~

~~n n−k k asp,k(K,L) ≥ asp (K)asp(L). (3.11)~~

~~Let i = n, j = 0 and k > n. Then inequality (3.11) also holds true for k ≥ n and all p. In both (3.10) and (3.11), equality holds for all p if K and L are dilates.~~

~~From inequality (3.10) and Corollary 3, one gets that~~

~~n n ◦ ◦ ◦ n−i ◦ i asp,i(K,L)asp,i(K ,L ) ≤ asp(K)asp(K ) asp(L)asp(L ) ≤ n2n(|K||K◦|)n−i(|L||L◦|)i (3.12)~~

~~holds true for all p ≥ 0 and 0 < i < n. The inequality also holds if i = 0 and i = n (see Chapter 2). We apply Blaschke-Santal´oinequality to inequality (3.12) and get~~

~~◦ ◦ 2 n asp,i(K,L)asp,i(K ,L ) ≤ asp(B2 ) 76~~

for all p ≥ 0 and 0 ≤ i ≤ n. Equality holds true if K and L are ellipsoids that are dilates of each other. Hence we have proved the following proposition, which, for p ≥ 1, was proved in [107].

~~2 Proposition 10 Let K and L be convex bodies in C+ with centroid at the origin. If p ≥ 0 and 0 ≤ i ≤ n, then~~

~~◦ ◦ 2 n asp,i(K,L)asp,i(K ,L ) ≤ asp(B2 ),~~

~~with equality if K and L are ellipsoids that are dilates of each other.~~

~~We now establish isoperimetric inequalities for asp,i(K).~~

~~2 Proposition 11 Let K ∈ C+ be a convex body with centroid at the origin. (i) If p ≥ 0 and 0 ≤ i ≤ n, then~~

~~(n−p)(n−i) (n+p)n asp,i(K) |K| n ≤ n asp,i(B2 ) |B2 |~~

~~◦ 2 n with equality if K is a ball. Moreover, asp,i(K)asp,i(K ) ≤ asp(B2 ) with equality if K is a ball. (ii) If p ≥ 0 and i ≥ n, then~~

~~(n−p)(n−i) (n+p)n asp,i(K) |K| n ≥ n asp,i(B2 ) |B2 |~~

~~◦ 2 n with equality if K is a ball. Moreover, asp,i(K)asp,i(K ) ≥ asp(B2 ) with equality if K is a ball. (iii) If −n < p < 0 and i ≤ 0, then~~

~~(n−p)(n−i) (n+p)n asp,i(K) |K| n ≥ n asp,i(B2 ) |B2 |~~

~~◦ n−i 2 n with equality if K is a ball. Moreover, asp,i(K)asp,i(K ) ≥ c asp(B2 ) where c is the universal constant in the inverse Santal´oinequality [14, 49]. 77~~

~~(iv) If p < −n and i ≤ 0, then~~

~~(n−p)(n−i) (n+p)n asp,i(K) (n−i)p |K| n+p n ≥ c n . asp,i(B2 ) |B2 |~~

~~◦ n−i 2 n Moreover, asp,i(K)asp,i(K ) ≥ c asp(B2 ) where c is the same constant as in (iii). (v) If i ≤ 0, then~~

~~n−i n as−n,i(K) |K| n ≥ n . as−n,i(B2 ) |B2 |~~

~~◦ 2 n Moreover, as−n,i(K)as−n,i(K ) ≥ as−n,i(B2 ).~~

~~Proof.~~

~~(i) For i = n, the equality holds trivially. For i = 0, the inequality was proved in n Chapter 2. We now prove the case 0 < i < n. L = B2 in inequality (3.10) gives~~

~~ n n−i asp,i(K) asp(K) n ≤ n (3.13) asp,i(B2 ) asp(B2 )~~

n n for all p 6= −n and 0 ≤ i ≤ n. We also use that asp,i(B2 ) = asp(B2 ). Then, as n n asp(B2 ) = n|B2 |, we get for all p ≥ 0 and 0 ≤ i ≤ n, the following isoperimetric inequality as a consequence of the Lp aﬃne isoperimetric inequality in Chapter 2

~~n−i (n−p)(n−i) n (n+p)n asp,i(K) asp(K) |K| n ≤ n ≤ n , asp,i(B2 ) asp(B2 ) |B2 |~~

~~◦ 2 n with equality if K is a ball. The inequality asp,i(K)asp,i(K ) ≤ asp,i(B2 ) follows from n Proposition 10 with L = B2 .~~

~~n (ii) For i = n, the equality holds trivially. Similarly, let L = B2 in inequality (3.11), then for all p 6= −n, and i ≥ n or i ≤ 0,~~

~~ n n−i asp,i(K) asp(K) n ≥ n . (3.14) asp,i(B2 ) asp(B2 ) 78~~

Hence for i ≥ n and p ≥ 0, the Lp aﬃne isoperimetric inequality in Chapter 2 implies that n−i (n−p)(n−i) n (n+p)n asp,i(K) asp(K) |K| n ≥ n ≥ n asp,i(B2 ) asp(B2 ) |B2 | with equality if K is a ball. Moreover, by Corollary 3 and the remark after it, one has for all i ≥ n

~~ ◦ n ◦ n−i asp,i(K)asp,i(K ) asp(K)asp(K ) 2 n ≥ 2 n ≥ 1, asp,i(B2 ) asp(B2 )~~

~~◦ 2 n or equivalently, asp,i(K)asp,i(K ) ≥ asp,i(B2 ), with equality if K is a ball.~~

~~(iii) If i ≤ 0 and −n < p < 0, inequality (3.14) and Theorem 18 imply that~~

~~n−i (n−p)(n−i) n (n+p)n asp,i(K) asp(K) |K| n ≥ n ≥ n asp,i(B2 ) asp(B2 ) |B2 | with equality if K is a ball. By Corollary 3 and the remark after it,~~

~~ ◦ n ◦ n−i asp,i(K)asp,i(K ) asp(K)asp(K ) n(n−i) 2 n ≥ 2 n ≥ c , asp,i(B2 ) asp(B2 )~~

~~◦ n−i 2 n or equivalently, asp,i(K)asp,i(K ) ≥ c asp,i(B2 ) where c is the constant in the inverse Santal´oinequality [14, 49].~~

~~(iv) If i ≤ 0 and p < −n inequality (3.14) and Theorem 18 imply that~~

~~n−i (n−p)(n−i) n (n+p)n asp,i(K) asp(K) (n−i)p |K| n+p n ≥ n ≥ c n . asp,i(B2 ) asp(B2 ) |B2 |~~

~~◦ n−i 2 n The proof of asp,i(K)asp,i(K ) ≥ c asp,i(B2 ) is same as in (iii).~~

n n−i n (v) Inequality (3.11) implies that as−n,i(K) ≥ as−n(K) for i ≤ 0. As as−n,i(B2 ) = 1 for all i, n n−i n−i as−n,i(K) as−n(K) |K| n ≥ n ≥ n . as−n,i(B2 ) as−n(B2 ) |B2 | 79

~~The second inequality follows from the L−n aﬃne isoperimetric inequality in Chapter 2.~~

~~Moreover, by Corollary 4, for i ≤ 0,~~

~~ ◦ n ◦ n−i as−n,i(K)as−n,i(K ) as−n(K)as−n(K ) 2 n ≥ 2 n ≥ 1, as−n,i(B2 ) as−n(B2 )~~

~~◦ 2 n or equivalently, as−n,i(K)as−n,i(K ) ≥ as−n,i(B2 ).~~

Remark. The example K(R, ε) mentioned in the remarks after Proposition 3 shows that we cannot expect to get strictly positive lower bounds in (i) of Proposition 11 for p > 0 and 0 ≤ i < n. In fact, by inequality (3.13), one has

~~ n n−i asp,i(K(R, ε)) asp(K(R, ε)) n ≤ n . asp,i(B2 ) asp(B2 )~~

~~As in Chapter 2, asp(K(R, ε)) → 0 for p > 0 as R → ∞ and ε → 0. 0 ≤ i < n~~

~~implies that n − i > 0, and therefore asp,i(K(R, ε)) → 0.~~

This example also shows that, likewise, we cannot expect ﬁnite upper bounds in (ii) (for p > 0 and i > n), (iii) (for −n < p < 0 and i ≤ 0), and (iv) (for p < −n and i ≤ 0), of Proposition 11. For instance, if i ≤ 0, by inequality (3.14), one has

~~ n n−i asp,i(K(R, ε)) asp(K(R, ε)) n ≥ n . asp,i(B2 ) asp(B2 )~~

~~For −2 < p < 0, one has asp(K(R, ε)) → ∞ as R → ∞ and ε → 0. Therefore, if~~

~~i ≤ 0, we obtain that asp,i(K(R, ε)) → ∞ as R → ∞ and ε → 0, i.e., there are no ﬁnite upper bounds in (iii).~~

Remark. In (iv), if p = −∞, then for all i ≤ 0, ZZ i−n n−i 2 n hK (u)hK◦ (v) dσ(u) dσ(v) ≥ c asp,i(B2 ) Sn−1×Sn−1 or equivalently, for all i ≤ 0, ZZ n−i n−i 2 n ρK (u)ρK◦ (v) dσ(u) dσ(v) ≥ c asp,i(B2 ). Sn−1×Sn−1 80

~~In particular, if i = 0, this is equivalent to the inverse Santal´oinequality [14].~~

~~3.2 Illumination surface bodies~~

We now deﬁne a new family of bodies associated with a given convex body K. These new bodies are a variant of the illumination bodies [109] (compare also [86]). For sets A and B,[A, B] = conv(A, B) := {λx + (1 − λ)y : λ ∈ [0, 1], x, y ∈ A ∪ B} is the convex hull of A ∪ B. For f : ∂K → R+ ∪ {0}, µf is the measure on ∂K deﬁned by R µf (A) = A fdµK .

~~Deﬁnition 2 (Illumination surface body) Let s ≥ 0 and f : ∂K → R be a nonnegative, integrable function. The illumination surface body Kf,s is deﬁned as~~

~~f,s n o K = x : µf (∂K ∩ [x, K]\K) ≤ s .~~

~~Obviously, K ⊆ Kf,s for any s ≥ 0 and any nonnegative, integrable function f. Moreover, Kf,s ⊆ Kf,t for any 0 ≤ s ≤ t.~~

~~Notice also that Kf,s needs to be neither bounded nor convex:~~

~~2 2 Example 1 Let K = B∞ = {x ∈ R : max1≤i≤2|xi| ≤ 1} and~~

~~ 1 12 , x ∈ [(−1, 1), (1, 1)] ∪ [(1, 1), (1, −1)] f(x) = 1 6 , otherwise~~

~~f,s 1 f,s K is calculated as follows. If s < 6 , then K = K. 1 1 If s ∈ [ 6 , 3 ), then~~

~~f,s K = {(x1, x2): x1 ≥ −1, x2 ∈ [−1, 1]; or x2 ≥ −1, x1 ∈ [−1, 1]}.~~

~~1 1 If s ∈ [ 3 , 2 ), then~~

~~f,s K = {(x1, x2): x1, x2 ≥ −1; or x1 ≤ −1, x2 ∈ [−1, 1]; or x2 ≤ −1, x1 ∈ [−1, 1]}.~~

~~1 2 f,s If s ∈ [ 2 , 3 ), then K = {(x1, x2): x1 ≥ −1 or x2 ≥ −1}. Except for s < 1/6, all of them are neither bounded nor convex. 2 f,s 2 If s ≥ 3 , then K = R 81~~

~~The following lemmas describe some of the properties of the bodies Kf,s.~~

~~Lemma 9 Let s ≥ 0 and f : ∂K → R be a nonnegative, integrable function. Then f,s T f,s+δ (i) K = δ>0 K . (ii) Kf,s is star convex, i.e., for all x ∈ Kf,s: [0, x] ⊂ Kf,s.~~

~~Proof. f,s T f,s+δ T f,s+δ (i) We only need to show that K ⊇ δ>0 K . Let x ∈ δ>0 K . Then for~~

all δ > 0, µf (∂K ∩ [x, K]\K) ≤ s + δ. Thus, letting δ → 0, µf (∂K ∩ [x, K]\K) ≤ s. (ii) Let x ∈ Kf,s. We claim that [0, x] ⊆ Kf,s. Let y ∈ [0, x]. Since y ∈ [0, x] ⊂ [x, K], we have [y, K] \ K ⊆ [x, K] \ K and thus ∂K ∩ [y, K] \ K ⊆ ∂K ∩ [x, K]\K. This implies that

~~µf (∂K ∩ [y, K] \ K) ≤ µf (∂K ∩ [x, K] \ K) ≤ s~~

~~and hence y ∈ Kf,s.~~

~~f,s n Remark. We can not expect K to be convex, even for K = B2 and f smooth. 2 1 Indeed, let K = B2 and s = 64 . Deﬁne~~

~~ 1 4π , x is in the ﬁrst and third quadrant  1 f(x) = 16π , x is in the fourth quadrant  23 16π , x is in the second quadrant~~

f,s f,s π Then K is not convex. In fact, K contains the arc from the point (tan( 32 ), 1) π π to the point (1, tan( 32 )) of the Euclidean ball centered at 0 with radius r = sec( 32 ). π f,s Moreover, the point (sec( 20 ), 0) is on the boundary of K . The tangent line at 2 (1, tan( π )) of B2(0, r) is y = −√x+r . This tangent line intersects the x-axis at (r2, 0) = 32 2 r2−1 2 π 2 π π f,s (sec ( 32 ), 0). Since sec ( 32 ) ∼ 1.009701 < sec( 20 ) ∼ 1.01264, K is not convex. We can modify f so that it becomes smooth also at the points (±1, 0) and (0, ±1) f,s π f,s and ∂K still intersects the positive x-axis at the point (sec( 20 ), 0). Therefore, K is not convex, even if f is smooth. 82

Lemma 10 Let s ≥ 0 and f : ∂K → R be an integrable, µK -almost everywhere strictly positive function. Then (i) K = Kf,0 . f,s (ii) There exists s0 > 0, such that for all 0 ≤ s ≤ s0, K is bounded.

~~Proof. (i) It is enough to prove that Kf,0 ⊆ K. Suppose this is not the case. Then there is x ∈ Kf,0 but x∈ / K. Since 0 ∈ int(K), there is α > 0 such that~~

~~n n B2 (0, α) ⊆ K ⊆ B2 (0, 1/α). (3.15)~~

~~n n Let y ∈ [0, x]∩∂K and Con(x, α) = [x, B2 (0, α)] be the convex hull of x and B2 (0, α).~~

~~H(y, NK (y)) ∩ Con(x, α) contains a (n − 1)-dimensional Euclidean ball with radius kx−yk (at least) r1 = α kxk > 0.~~

~~n−1 n−1 Hence µK (∂K ∩ [x, K] \ K) ≥ |H(y, NK (y)) ∩ Con(x, α)| ≥ r1 |B2 | > 0. Let~~

~~ 1 E = z ∈ ∂K ∩ [x, K]\K : f(z) ≥ , j = 1, 2, ··· . j j~~

~~As µK ({z ∈ ∂K : f(z) = 0}) = 0 and Ej ⊆ Ej+1 for all j,~~

~~∞ [ µK ∂K ∩ [x, K] \ K = µK Ej = lim µK (Ej). j→∞ j=1~~

~~Therefore there exists j1 such that µK (Ej1 ) > 0. Thus~~

~~µK (Ej1 ) µf (∂K ∩ [x, K]\K) ≥ µf (Ej1 ) ≥ > 0 j1~~

which contradicts that x ∈ Kf,0. (ii) is an immediate consequence of Lemma 9 (i) and Lemma (10) (i). Indeed, these f,0 T f,s lemmas imply that K = K = s>0 K . So, also using (3.15), there exists s0 > 0

~~f,s n 2 f,s0 n 2 such that for all 0 ≤ s ≤ s0, K ⊂ 2K ⊂ B2 (0, α ). In particular, K ⊂ B2 (0, α ). 83~~

~~Remark. The assumption that f is µK -almost everywhere strictly positive is neces- f,0 2 sary in order that K = K. To see that, let K = B2 and~~

~~ 0 x = p1 − y2, y ∈ [−1, 1], f(x, y) = 1 π otherwise.~~

~~Then Kf,0 = K ∪ {(x, y): x ≥ 0, |y| ≤ 1}.~~

~~f,s This example also shows that there is no s0 such that K is bounded for all 0 ≤ s ≤ s0~~

~~unless f is µK -almost everywhere strictly positive.~~

~~Let K be a convex body with 0 ∈ int(K). Let f : ∂K → R be an integrable, µK -~~

~~almost everywhere strictly positive function. For x∈ / K, let t0 = t0(x) be the strictly~~

~~positive real number such that t0x = ∂K ∩ [0, x]. Deﬁne hx(t) to be~~

~~hx(t) = µf (∂K ∩ [tx, K] \ K), t ≥ t0.~~

~~f,s f,s Clearly hx(t0) = 0. Moreover hx(t) ≤ s if tx ∈ K , and hx(t) > s if tx∈ / K .~~

~~n Lemma 11 Let K be a convex body in R and f : ∂K → R be an integrable, µK - almost everywhere strictly positive function.~~

~~(i) hx(t) is increasing and left continuous on [t0, ∞).~~

~~f,s (ii) K is closed for all s ≥ 0. In particular, it is compact for all 0 ≤ s ≤ s0. If K is in addition strictly convex, i.e., ∂K does not contain any line segment. then~~

~~(iii) hx(t) is continuous on [t0, ∞).~~

~~f,s (iv) For any 0 ≤ s ≤ s0 and x ∈ ∂K , one has µf (∂K ∩ [x, K] \ K) = s.~~

~~Proof.~~

~~(i) If t1 ≤ t2, then ∂K ∩ [t1x, K] \ K ⊆ ∂K ∩ [t2x, K] \ K. Thus hx(t1) ≤ hx(t2).~~

~~Let now t > t0 and (tm)m∈N be a sequence, increasing to t. Then, by monotonicity~~

~~of hx, hx(tm) ≤ hx(t) for all m and thus limmhx(tm) ≤ hx(t). We have to show~~

~~that limmhx(tm) ≥ hx(t). Let y ∈ relint∂K (∂K ∩ [tx, K] \ K), where relintB(A) is 84~~

the relative (with respect to B) interior of a set A ⊆ B, i.e., relintB(A) = {x ∈ A : there is a δ > 0, such that B(x, δ) ∩ B ⊂ A}. Then y ∈ int ([tx, K]), and therefore there exists m0(y) ∈ N, such that y ∈ int([tm0(y)x, K]). This implies that y ∈ [tm0(y)x, K] \ K ∩ ∂K and thus

~~[ relint∂K ([tx, K] \ K ∩ ∂K) ⊂ [tmx, K] \ K ∩ ∂K. m≥1~~

~~By continuity of the measure µf from below, one has~~

~~ hx(t) = µf [tx, K] \ K ∩ ∂K = µf relint∂K [tx, K] \ K ∩ ∂K [ ≤ µf [tmx, K] \ K ∩ ∂K m≥1 = lim µf ∂K ∩ [tmx, K] \ K = limmhx(tm) m~~

f,s (ii) It will follow from Lemma 10 (ii) that K is compact for 0 ≤ s ≤ s0, once we have proved that Kf,s is closed. To that end, we show that (Kf,s)c, the complement of Kf,s in Rn, is open for all s ≥ 0. Suppose this is not the case. Then there exists x ∈ (Kf,s)c and a sequence f,s (xm)m∈N, such that xm → x as m → ∞ but xm ∈ K for all m. Without loss of

~~generality, we can assume that xm are not in the ray of {tx : t ≥ 0}. Otherwise, if~~

~~f,s kxmk kxmk xm ∈ K are in the ray, then hx kxk ≤ s and by (i), limm hx kxk = hx(1) ≤ s.~~

~~This contradicts with hx(1) > s.~~

~~Now we let Km = [xm,K]. For suﬃciently big m, ∂Km ∩ [0, x] 6= ∅. Suppose not, then x ∈ Km implies that [x, K] ⊂ Km, and hence [x, K] \ K ∩ ∂K ⊂ Km \ K ∩ ∂K.~~

Since µf ([x, K] \ K ∩ ∂K) > s, one gets µf (Km \ K ∩ ∂K) > s, a contradiction with f,s xm ∈ K . Let ym = ∂Km ∩ [0, x]. Thus µf ([ym,K] \ K ∩ ∂K) ≤ s. Let α be as in n (3.15). Similarly, ∂ [xm,B2 (0, α)] ∩ [0, x] 6= ∅ for suﬃciently big m and we denote n zm = ∂ [xm,B2 (0, α)] ∩ [0, x]. 85

It is easy to check that 0 ≤ kxk − kymk ≤ kxk − kzmk for any m. As α ≤ kzmk ≤ kxk and α ≤ kx−xmk , one has kxk − kz k ≤ kxk kx−xmk . Thus z → x, and hence kzmk kxk−kzmk m α m

~~also ym → x, as m → ∞. Therefore we can choose a subsequence (ymk )k∈N that is~~

~~kymk k monotone increasing to x. By (i) with tmk = kxk , hx(tmk ) % hx(1) as k → ∞.~~

~~Since for all k, hx(tmk ) ≤ s, one has hx(1) ≤ s, a contradiction.~~

~~(iii) It is enough to prove that hx(t) is right continuous on [t0, ∞). To do so, let t ≥ t0~~

and let (tm)m∈N be a sequence decreasing to t. By (i), hx(tm) ≥ hx(t) for all m, thus limmhx(tm) ≥ hx(t) and we have to show that limmhx(tm) ≤ hx(t). We claim that if K is strictly convex, then

~~∞ \ ∂K ∩ [tx, K] \ K = ∂K ∩ [tmx, K] \ K . (3.16) m=1~~

~~T∞ We only need to prove that m=1 ∂K ∩ [tmx, K] \ K ⊆ ∂K ∩ [tx, K] \ K. Let T∞ z0 ∈ m=1 ∂K ∩ [tmx, K] \ K . Thus z0 ∈ ∂K. Let l(z0, tx) be the line passing~~

~~through tx and z0. We have two cases.~~

~~Case 1: l(z0, tx) is in a tangent hyperplane of K. Then l(z0, tx)∩∂K = {z0} by strict~~

convexity of K. Therefore, {z0} = [z0, tx] \ K ∩ ∂K ⊆ [tx, K] \ K ∩ ∂K. T∞ Case 2: l(z0, tx) ∩ ∂K consists of two points, z0 and z1. As z0 ∈ m=1 ∂K ∩ [tmx, K] \ K , we must have ktx − z0k < ktx − z1k. Therefore, {z0} = [z0, tx] \ K ∩ ∂K ⊆ [tx, K] \ K ∩ ∂K.

~~Hence by (3.16) and continuity of the measure µf from above,~~

~~∞ \ h i hx(t) = µf ∂K ∩ [tmx, K] \ K m=1 = lim µf ∂K ∩ [tmx, K] \ K = lim hx(tm). m m~~

f,s (iv) Let 0 ≤ s ≤ s0, and x ∈ ∂K which implies that hx(1) ≤ s. Deﬁne Φx(s) = n 3 {t : hx(t) = s}. Then Φx(s) 6= ∅. Indeed, let tax = ∂B2 (0, α ) ∩ Tx, where α is as in (3.15) and Tx = {tx : t ≥ t0(x) > 0}. The proof of Lemma 10 (ii) shows that 86

~~f,s0 n 2 f,s0 K ⊂ B2 (0, α ). It is clear that tαx∈ / K , and hence hx(tα) > s0. In fact, if~~

~~f,s0 n 2 n 3 tαx ∈ K , then tαx ∈ B2 (0, α ), but by deﬁnition of tα, tαx ∈ ∂B2 (0, α ). This is a contradiction.~~

By continuity of hx(·), there must exist t ∈ [t0, tα], such that hx(t) = s. This also f,s shows that t¯ = sup Φx(s) ≤ tα. Clearly hx(t¯) = s and thus tx¯ ∈ K . This implies f,s that t¯ ≤ 1 because x ∈ ∂K . Suppose t¯ < 1. Then s = hx(t¯) ≤ hx(1) ≤ s by monotonicity of hx(·), a contradiction with t¯= sup Φ(s). Thus t¯= 1 and hx(1) = s.

~~Remark. Strict convexity is needed in (iii) and (iv). Indeed, let x = (0, 2) and~~

~~K = conv{(1, 1), (−1, 1), (−2, 0), (2, 0)}.~~

~~Then ∂K ∩ [x, K]\K = [(−1, 1), (1, 1)]. However for any point tx with t > 1,~~

~~[tx, K] \ K ∩ ∂K = ∂K \ [(−2, 0), (2, 0)] ) ∂K ∩ [x, K]\K.~~

~~Thus, for any function f with f > 0 on [(−2, 0), (−1, 1)] and/or [(1, 1), (2, 0)], hx(·) is not right continuous on [1, ∞).~~

~~To see that strict convexity is needed also in (iv), observe that Kf,1/12 = K in Example 1. Thus, for x ∈ ∂Kf,1/12 = ∂K, we have 1 µ ([x, K] \ K ∩ ∂K) = 0 6= . f 12~~

~~3.3 Geometric interpretation of functionals on convex bodies~~

We now give geometric interpretations of functionals on convex bodies, such as Lp affine surface area and mixed p-affine surface area for all p 6= −n using the non convex illumination surface bodies. While there are no geometric interpretations for mixed p-affine surface area, many geometric interpretations of Lp affine surface area have been discovered in the last years, all based on using convex bodies (e.g., [69, 85, 86, 111]). The remarkable new fact here is that now the bodies involved in the geometric interpretation are not necessarily convex. 87

~~2 Theorem 20 Let K be a convex body in C+. Let c > 0 be a constant, and f : ∂K →~~

~~R be an integrable function such that f ≥ c µK -almost everywhere. Then~~

~~f,s Z 1 |K | − |K| κK (x) n−1 lim cn 2 = 2 dµK (x), (3.17) s→0 s n−1 ∂K f(x) n−1~~

~~2 n−1 n−1 where cn = 2|B2 | .~~

~~Remark. As dµK = fK dσ, we also have~~

~~f,s Z n−2 |K | − |K| fK (u) n−1 lim cn 2 = 2 dσ(u), (3.18) s→0 n−1 n−1 −1 n−1 s S f(NK (u))~~

~~−1 where NK is the inverse of the Gauss map NK (·).~~

The geometric interpretation of Lp aﬃne surface area is then a corollary to Theorem 20. The theorem also gives geometric interpretations of other known functionals on convex bodies, e.g. the surface area and the mixed p-aﬃne surface area. Notice that these geometric interpretations can also be obtained using e.g. the (convex) surface body [86, 111].

~~Deﬁne n−2 1−n ˜ −1 2 2(n+p) f(NK (u)) = fK (u) [fp(K1, u) ··· fp(Kn, u)] ,~~

~~1−p where fp(K, u) = hK (u) fK (u).~~

~~2 Corollary 5 Let K and Ki, i = 1, ··· , n, be convex bodies in C+. Then~~

~~|Kf,s˜ | − |K| lim cn 2 = asp(K1, ··· ,Kn). s→0 s n−1~~

~~In particular, if all Ki coincide with K, then asp(K1, ··· ,Kn) = asp(K) and we get~~

~~a geometric interpretation of asp(K)~~

~~|Kgp,s| − |K| cn lim 2 = asp(K), s→0 s n−1~~

~~n+2p−np n(n−1)(p−1) where gp : ∂K → R is deﬁned by gp(x) = κK (x) 2(n+p) hx, NK (x)i 2(n+p) . 88~~

~~2 p Corollary 6 Let K be a convex body in C+ and g(x) = κK (x). Then~~

~~|Kg,s| − |K| lim cn 2 = µK (∂K). s→0 s n−1~~

~~The proof of the corollaries follows immediately from Theorem 20. To prove Theorem 20, we need several other concepts and lemmas.~~

2 As K is in C+, for any x ∈ ∂K, the indicatrix of Dupin is an ellipsoid. As in [86], we apply an aﬃne transform T : Rn → Rn to K so that the indicatrix of Dupin is transformed into an (n−1)-dimensional Euclidean ball. T has the following properties:

~~T (x) = x T (NK (x)) = NK (x) det(T ) = 1 (3.19)~~

and T maps a measurable subset of a hyperplane orthogonal to NK (x) onto a subset of the same (n − 1)-dimensional measure. It was also shown in [86] that for any > 0 − there is ∆1 = ∆1(ε) > 0 such that for all measurable subsets A of ∂K ∩ H (x −

~~∆1NK (x),NK (x))~~

~~(1 − ) µK (A) ≤ |T (A)| ≤ (1 + ) µK (A). (3.20)~~

~~T (K) can be approximated at x = T (x) by a n-dimensional Euclidean ball: For any > 0 there is ∆2 = ∆2(ε) such that~~

~~n − B2 (x − rNK (x), r) ∩ H (x − ∆2NK (x),NK (x)) − ⊆ T (K) ∩ H (x − ∆2NK (x),NK (x)) (3.21)~~

~~n − ⊆ B2 (x − RNK (x),R) ∩ H (x − ∆2NK (x),NK (x)) ,~~

~~− 1 where r = r(x) = κK (x) n−1 and R = R(x) with r ≤ R ≤ (1 + )r . We put~~

~~∆ = ∆(ε) = min{∆1, ∆2}. (3.22)~~

~~Moreover, for x ∈ ∂K, let~~

~~f,s xs ∈ ∂K be such that x ∈ [0, xs] ∩ ∂K (3.23) 89~~

~~and deﬁnex ˜s to be the orthogonal projection of xs onto the ray {y : y = x + tNK (x), t ≥ 0}. Clearly T (˜xs) =x ˜s, and the distance from T (xs) to the hyperplane~~

~~H(x, NK (x)) is the same as the distance from xs to this hyperplane.~~

~~We say that a family of sets Es ⊆ ∂K, 0 < s ≤ s0 shrinks nicely to a point x ∈ ∂K (see [29]) if~~

~~(SN1) diamEs → 0, as s → 0.~~

~~(SN2) There is a constant β > 0 such that for all s ≤ s0 there exists ts with~~

~~µK (∂K ∩ B(x, ts)) ≥ µK (Es) ≥ β µK (∂K ∩ B(x, ts)) .~~

~~2 Lemma 12 Let K be a convex body in C+ and f : ∂K → R an integrable, µK -almost everywhere strictly positive function. Let x ∈ ∂K and let xs and x˜s be as above (3.23). Then~~

~~(i) The family ∂K ∩ [˜xs,K] \ K, 0 < s ≤ s0 shrinks nicely to x.~~

~~(ii) The family ∂K ∩ [xs,K] \ K, 0 < s ≤ s0 shrinks nicely to x. (iii)~~

~~µf (∂K ∩ [˜xs,K] \ K) lim = f(x) µK -a.e. (3.24) s→0 µK (∂K ∩ [˜xs,K] \ K) (iv)~~

~~µf (∂K ∩ [xs,K] \ K) lim = f(x) µK - a.e. (3.25) s→0 µK (∂K ∩ [xs,K] \ K)~~

~~Proof. Formulas (3.24) and (3.25) in (iii) and (iv) follow from the Lebesgue diﬀer-~~

~~entiation theorem (see [29]) once we have proved that ∂K ∩ [x ˜s,K] \ K and ∂K ∩~~

~~[xs,K] \ K shrink nicely to x. Therefore it is enough to prove (i) and (ii).~~

~~(i) For x ∈ ∂K, let r = r(x) and R = R(x) be as in (3.21). We abbreviate B(r) = n n B2 (x − rNK (x), r) and B(R) = B2 (x − RNK (x),R). Let~~

~~ x ∆(x, s) = ,N (x) kx − xk = hx − x, N (x)i kxk K s s K 90~~

be the distance from xs to H x, NK (x) . This is the same as the distance fromx ˜s (deﬁned after formula (3.23)) to H x, NK (x) . R ∆(x,s) Let hR = hR(s) = R+∆(x,s) be the height of the cap of B(R) that is “illuminated” by x˜s. Then

~~− H x − hRNK (x),NK (x) ∩ ∂B(R) = [T (˜xs),B(R)] \ B(R) ∩ ∂B(R). (3.26)~~

~~Let ∆ be as in (3.22). Since ∆(x, s) → 0 as s → 0, one can choose s1 ≤ s0, such that~~

~~for all 0 < s ≤ s1, h = 2hR < ∆. Therefore (3.21) holds:~~

~~− − H (x − hNK (x),NK (x)) ∩ B(r) ⊂ H x − hNK (x),NK (x) ∩ T (K) − ⊂ H x − hNK (x),NK (x) ∩ B(R). (3.27)~~

~~(3.27) and (3.26) imply that for all small enough s ≤ s2 ≤ s1 T [˜xs,K] \ K ∩ ∂K = [T (˜xs),T (K)] \ T (K) ∩ ∂T (K)~~

~~− − ⊆ H x − hNK (x),NK (x) ∩ ∂T (K) ⊆ B(R) ∩ H x − hNK (x),NK (x) .~~

~~q 4R2 ∆(x,s) Let ts = kx−zk = R+∆(x,s) where z is any point in H x−hNK (x),NK (x) ∩∂B(R). − n As B(R) ∩ H x − hNK (x),NK (x) ⊆ B2 (x, ts),~~

~~n T [˜xs,K] \ K ∩ ∂T (K) = T [˜xs,K] \ K ∩ ∂K ⊆ B2 (x, ts) ∩ ∂T (K)~~

~~and ts → 0 as s → 0. This shows that condition (SN1) is satisﬁed for T [˜xs,K] \ K ∩ ∂T (K) to shrink nicely to T (x) = x.~~

~~We now show that condition (SN2) also holds true.~~

~~First, [T (˜xs),B(r)] \ B(r)∩H(x, NK (x)) is a (n −1)-dimensional Euclidean ball with r∆(x,s) radius √ . Then for any 0 < s < s2, 2r∆(x,s)+∆2(x,s)~~

~~|T ([˜xs,K] \ K ∩ ∂K)| ≥ |[T (˜xs),B(r)] \ B(r) ∩ H(x, NK (x))| !n−1 r∆(x, s) ≥ |Bn−1| . 2 p2r∆(x, s) + ∆2(x, s) 91~~

~~∆(x,s) We can choose ( a new, smaller) s2 such that r ≤ 2. Then~~

~~n−1 1−n n−1 2 |T ([˜xs,K] \ K ∩ ∂K)| ≥ 2 |B2 | (r∆(x, s)) . (3.28)~~

~~On the other hand, for ε small enough, there exists s3 < s2, such that, for all 0 < s ≤ − s3 and for any subset A of H (x − 2hNK (x),NK (x)) ∩ ∂T (K) [86]~~

~~|PH(x−2hNK (x),NK (x))(A)| ≤ |A| ≤ (1 + ε)|PH(x−2hNK (x),NK (x))(A)| (3.29)~~

~~where PH (A) is the orthogonal projection of A onto the hyperplane H. We apply this n to A = B2 (x, ts) ∩ ∂T (K):~~

~~n n |B2 (x, ts) ∩ ∂T (K)| ≤ (1 + ε)|PH(x−2hNK (x),NK (x))(B2 (x, ts) ∩ ∂T (K))|~~

~~≤ (1 + ε) |B(R) ∩ H(x − 2hNK (x),N)| √ !n−1 2 2RpR∆(x, s) ≤ (1 + ε) |Bn−1| 2 R + ∆(x, s)~~

~~n−1 n n−1 2 ≤ 4 |B2 | (r∆(x, s)) (3.30)~~

The last inequality follows as r ≤ R < (1 + ε)r. It now follows from (3.28) and (3.30) that also condition (SN2) holds true for e.g. β = 8−n. Hence the family T [˜xs,K] \ K ∩ ∂T (K) shrinks nicely to T (x) = x and therefore,

~~−1 −1 as T exists, the family [˜xs,K] \ K ∩ ∂K = T T [˜xs,K] \ K ∩ ∂T (K) shrinks nicely to x.~~

(ii) Let v1 = xs − (x − rNK (x)) and v2 = xs − (x − RNK (x)). θ denotes the angle between NK (x) and x and φi = φi(x, s), i = 1, 2 is the angle between NK (x) and vi, i = 1, 2. These angles can be computed as follows

~~∆(x, s) tan(θ) tan(φ ) = 1 r + ∆(x, s) ∆(x, s) tan(θ) tan(φ ) = . 2 R + ∆(x, s) 92~~

~~2 Then, for i = 1, 2, φi → 0 as s → 0. Since K is in C+, this means that for any ε > 0 there iss ¯ε ≤ s0 such that for all s ≤ s¯ε~~

~~µ ([x ,K] \ K ∩ ∂K) 1 − ε ≤ K s ≤ 1 + ε. (3.31) µK ([˜xs,K] \ K ∩ ∂K)~~

~~˜ ˜ 3R ∆(x,s) By (3.21) and as φi → 0, i = 1, 2 as s → 0, one can choose hR = hR(s) = R+∆(x,s) so small that T [xs,K] \ K ∩ ∂K = [T (xs),T (K)] \ T (K) ∩ ∂T (K)~~

~~− ˜ ⊂ H x − hRNK (x),NK (x) ∩ B(R).~~

~~q 2 ˜ 6R ∆(x,s) ˜ Let ts = R+∆(x,s) be the distance from x to any point in H x − hRNK (x),NK (x) ∩ ∂B(R). Then n ˜ T [xs,K] \ K ∩ ∂K ⊆ B2 (x, ts) ∩ ∂T (K).~~

~~(3.20), (3.31) and Lemma 12 (i) then give~~

~~ 3 |T [xs,K] \ K ∩ ∂K | ≥ (1 − ε) |T [˜xs,K] \ K ∩ ∂K |~~

~~3 n ≥ (1 − ε) β|B2 (x, ts) ∩ ∂T (K)|. (3.32)~~

~~Furthermore, by (3.29), one has~~

~~n |B2 (x, ts) ∩ ∂T (K)| ≥ |H(x − hNK (x),NK (x)) ∩ T (K)|~~

~~≥ |H(x − hNK (x),NK (x)) ∩ B(r)|~~

~~n−1 4R2r∆(x, s) + 4Rr∆(x, s)2 − 4R2∆(x, s)2 2 = |Bn−1|. (R + ∆(x, s))2 2~~

~~Since ∆(x, s) → 0 as s → 0, fors ¯ε small enough, and any 0 < s < s¯ε, one get~~

~~ 2R n−1 |Bn(x, t ) ∩ ∂T (K)| ≥ pr∆(x, s) − ∆(x, s)2 |Bn−1| 2 s R + ∆(x, s) 2~~

~~n−1 −n 2 n−1 ≥ 2 (r∆(x, s)) |B2 |. (3.33) 93~~

~~A computation similar to (3.30) shows that for all 0 < s ≤ s¯ε with (a possibly new)~~

~~s¯ε small enough~~

~~n ˜ ˜ |B2 (x, ts) ∩ ∂T (K)| ≤ (1 + ε) |B(R) ∩ H(x − hNK (x),N)| √ !n−1 6RpR∆(x, s) = (1 + ε) |Bn−1| 2 R + ∆(x, s)~~

~~n−1 n n−1 2 ≤ 3 |B2 | (r∆(x, s)) (3.34)~~

~~(3.32), (3.33) and (3.34) imply that~~

~~ −n−1 n ˜ |T [xs,K] \ K ∩ ∂K | ≥ (48) β|B2 (x, ts) ∩ ∂T (K)|.~~

~~ This shows that T [xs,K] \ K ∩ ∂K shrinks nicely to x. Therefore also [xs,K] \ K∩ ∂K shrinks nicely to x.~~

~~2 Lemma 13 Let K be a convex body in C+ and f : ∂K → R an integrable, µK -almost~~

~~everywhere strictly positive function. Then for µK -almost all x ∈ ∂K one has~~

~~h n i kxsk 1 hx, NK (x)i − 1 kxk κK (x) n−1 lim cn 2 = 2 , (3.35) s→0 s n−1 f(x) n−1~~

~~f,s where xs ∈ ∂K is such that x ∈ [0, xs].~~

~~Proof. It is enough to consider x ∈ ∂K such that f(x) > 0. As x and xs are collinear and as (1 + t)n ≥ 1 + tn for t ∈ [0, 1), one has for small enough s,~~

~~hx, N (x)i kx kn hx, N (x)i kx − xkn K s − 1 = K 1 + s − 1 ≥ ∆(x, s). n kxk n kxk~~

~~D x E Recall that ∆(x, s) = kxk ,NK (x) kxs − xk = hxs − x, NK (x)i is the distance from xs to H (x, NK (x)). 94~~

~~Similarly, as (1 + t)n ≤ 1 + nt + 2nt2 for t ∈ [0, 1), one has for s small enough,~~

~~hx, N (x)i kx kn 2n kx − xk K s − 1 ≤ ∆(x, s) 1 + s . (3.36) n kxk n kxk~~

~~Hence for ε > 0 there exists sε ≤ s0 such that for all 0 < s ≤ sε~~

~~n h kxsk i hx, NK (x)i kxk − 1 1 ≤ ≤ 1 + ε. n∆(x, s)~~

~~2 K is strictly convex as K ∈ C+. Thus, µf (∂K ∩ [xs,K] \ K) = s by Lemma 11 (iv). Therefore~~

~~n 2 h kxsk i n−1 hx, NK (x)i kxk − 1 µf (∂K ∩ [xs,K] \ K) 1 ≤ 2 ≤ 1 + ε. n s n−1 ∆(x, s)~~

~~By Lemma 12 (iv) and (3.31), it then follows that we can choose (a new) sε so small that we have for all s ≤ sε~~

~~n 2 h kxsk i n−1 hx, NK (x)i kxk − 1 f(x) µK (∂K ∩ [x ˜s,K] \ K) 1 − c1ε ≤ 2 ≤ 1 + c2ε. n s n−1 ∆(x, s) (3.37)~~

with absolute constants c1 and c2. Let T be as in (3.19) and let r = r(x) and R = R(x) be as in (3.21). We abbreviate n n again B(r) = B2 (x − rNK (x), r) and B(R) = B2 (x − RNK (x),R). Let hr = hr(s) = r∆(x,s) r+∆(x,s) . As hr → 0 as s → 0, we have for all s suﬃciently small that hr < ∆ where ∆ is as in (3.22). Hence by (3.21)

~~− − H (x − hrNK (x),NK (x)) ∩ B(r) ⊂ H (x − hrNK (x),NK (x)) ∩ T (K).~~

~~If we denote by PH the orthogonal projection onto the hyperplane~~

~~− H (x − hrNK (x),NK (x)) , 95~~

~~this then implies that~~

~~−1 −1 −1 PH ∂K ∩ [˜xs,K] \ K ⊃ PH ∂T (B(r)) ∩ [˜xs,T (B(r))] \ T (B(r))~~

~~−1 = H (x − hrNK (x),NK (x)) ∩ T (B(r)).~~

~~Hence for s suﬃciently small~~

~~ µK (∂K ∩ [˜xs,K] \ K) ≥ PH ∂K ∩ [˜xs,K] \ K~~

~~−1 ≥ H (x − hrNK (x),NK (x)) ∩ T (B(r)) −1 = T H (x − hrNK (x),NK (x)) ∩ T (B(r))~~

~~= H (x − hrNK (x),NK (x)) ∩ B(r)~~

~~n−1 ∆(x,s) 2 1 − n−1 = 2r (2r∆(x, s)) 2 |Bn−1| ∆(x,s) n−1 2 1 + r~~

~~n−1 2 (2∆(x, s)) n−1 ≥ (1 − c3ε) p |B2 | (3.38) κK (x) where c3 > 0 is an absolute constant.~~

~~By (3.21), one has~~

~~−1 −1 −1 PH ∂K ∩ [˜xs,K] \ K ⊆ PH ∂ T (B(R)) ∩ [˜xs,T (B(R))] \ T (B(R))~~

~~−1 = H (x − hRNK (x),NK (x)) ∩ T (B(R))~~

~~R∆(x,s) where H = H (x − hRNK (x),NK (x)). The equality follows as hR = R+∆(x,s) .~~

~~Together with (3.29), for sε small enough, whenever 0 < s < sε, one has~~

~~µK (∂K ∩ [˜xs,K] \ K) ≤ (1 + ε)|PH (∂K ∩ [˜xs,K] \ K)|~~

~~−1 ≤ (1 + ε) H (x − hRNK (x),NK (x)) ∩ T (B(R)) .~~

~~A calculation similar to (3.38) then shows that with an absolute constant c4~~

~~n−1 (2∆(x, s)) 2 n−1 µK (∂K ∩ [˜xs,K] \ K) ≤ (1 + c4ε) p |B2 |. (3.39) κK (x) 96~~

~~Combining (3.37), (3.38) and (3.39), we prove the formula (3.35), i.e., h n i kxsk 1 hx, NK (x)i − 1 kxk κK (x) n−1 lim cn 2 = 2 . s→0 n s n−1 f(x) n−1~~

~~2 Lemma 14 Let K be a convex body in C+. Let c > 0 be a constant and f : ∂K → R~~

~~an integrable function with f ≥ c µK -almost everywhere. Then there exists s ≤ s0, such that for all s ≤ s,~~

~~n h kxsk i hx, NK (x)i kxk − 1 2 ≤ c(K, n), s n−1~~

~~where c(K, n) is a constant (depending on K and n only), and x and xs are as in Lemma 13.~~

Proof. 2 As K ∈ C+, by the Blaschke rolling theorem [82], there exists r0 > 0 such that n for all x ∈ ∂K, B2 (x − r0NK (x), r0) ⊆ K. Let γ be such that 0 < γ ≤ min{1, r0}. By f,0 T f,s Lemmas 9 (i) and 10 (ii), K = K = s>0 K . Therefore there exists s = sγ ≤ s0, f,s f,s such that for all s ≤ s, K ⊆ (1 + γ)K. Hence for xs ∈ ∂K and x = [0, xs] ∩ ∂K,

~~kxsk kxk ≤ 1 + γ, or equivalently -as x and xs are collinear-~~

~~kx k kx − xk s − 1 = s ≤ γ ≤ 1. (3.40) kxk kxk~~

~~Together with (3.36), one has for all s ≤ s (with a possibly smaller s)~~

~~kx kn hx, N (x)i s − 1 ≤ ∆(x, s)[n + 2n] . (3.41) K kxk~~

~~2 As K ∈ C+, K is strictly convex. Hence, by Lemma 11 (iv), (3.21) and as f ≥ c on ∂K Z s = µf (∂K ∩ [xs,K] \ K) = fdµK ≥ c µK ∂K ∩ [xs,K] \ K ∂K[xs,K]\K)~~

~~−1 ≥ c H (x, NK (x)) ∩ [xs,T (B(r))] = c H (x, NK (x)) ∩ [T (xs),B(r)] 97~~

~~where T is as in (3.19) and r = r(x) is as in (3.21).~~

~~As in the proof of Lemma 10 (i), H (x, NK (x)) ∩ [T (xs),B(r)] contains a (n − 1)- dimensional Euclidean ball of radius at least~~

~~rp2r∆(x, s) + ∆2(x, s) α ≥ p2r ∆(x, s). 2r + ∆(x, s) 1 + 2α 0~~

~~kxs−xk γ 1 The inequality follows as ∆(x, s) = kxk hx, NK (x)i ≤ α ≤ α , which is a direct consequence of (3.15) and (3.40).~~

~~n−1 α p n−1 Hence s ≥ c 1+2α 2r0∆(x, s) |B2 | and with (3.41) we get that~~

~~ n 2 hx, NK (x)i kxsk 1 + 2α 2 2 −1 n n−1 n−1 n−1 2 − 1 ≤ (n + 2 ) 2r0 c |B2 | . s n−1 kxk α~~

~~Finally, we also need the following lemma. It is well known and we omit the proof.~~

~~Lemma 15 Let K be a convex body and L be a star-convex body in Rn. (i) If 0 ∈ int(L) and L ⊂ K, then~~

~~1 Z kx0kn |K| − |L| = hx, NK (x)i 1 − dµK (x) n ∂K kxk~~

~~where x ∈ ∂K and x0 ∈ ∂L ∩ [0, x]. (ii) If 0 ∈ int(K) and K ⊂ L, then~~

~~1 Z kx0kn |L| − |K| = hx, NK (x)i − 1 dµK (x) n ∂K kxk~~

~~where x ∈ ∂K, x0 ∈ ∂L and x = ∂K ∩ [0, x0].~~

~~Proof of Theorem 20. 98~~

~~2 As K ∈ C+, K is strictly convex. By Lemmas 15, 13, 14 and Lebegue’s Dominated Convergence theorem~~

n h kxsk i f,s |K | − |K| Z hx, NK (x)i kxk − 1 cn lim 2 = cn lim 2 dµK (x) s→0 s n−1 s→0 ∂K n s n−1 n h kxsk i Z hx, NK (x)i kxk − 1 = cn lim 2 dµK (x) ∂K s→0 n s n−1 Z 1 κK (x) n−1 = 2 dµK (x). ∂K f(x) n−1 Chapter 4

~~On the Bures Volume of Separable Quantum States~~

Geometry of quantum states is one of the fundamental objects of quantum information theory, in particular, geometry of the set of entangled quantum states. In this chapter, we will investigate two very important questions (1): what is the relative size of S (and hence E(H)) in D? (2): how precise is the positive partial transpose criterion as a tool to detect the separability? We show that the relative size of S in D for large N is extremely small in the sense of Bures volume. Moreover, our results show that the powerful positive partial transpose criterion is imprecise as a tool to detect separability for large N. It is still unclear whether the positive partial transpose criterion is precise as a tool to detect the entanglement.

This chapter is based on the work [114], and is organized as follows. In section 1, we review the results by Szarek [99] and by Aubrun and Szarek [8]. In section 2, we will compare the Bures volume of any subset K of D with its Hilbert-Schmidt volume. As an application, we employ this result to obtain estimates of the Bures volume of S and PPT . Section 3 explains why our estimates are essentially optimal for general subsets of quantum states. Section 4 contains conclusions, comments and ﬁnal remarks.

~~99 100 4.1 Hilbert-Schmidt volume of separable quantum states~~

D D In this section, we will focus on the Hilbert Space H = C ⊗ · · · ⊗ C with n ≥ 2 | {z } n and D ≥ 2. Hence, N = dim(H) = Dn ≥ 4. Theorem 21 and 22 are dealing with the Hilbert-Schmidt volume of S, and Theorem 23 provides rough estimate of the Hilbert-Schmidt volume of PPT .

~~Theorem 21 (Large number of small subsystems): For system H = (CD)⊗n, ˜ there exist universal constants c˜2, C2 > 0, such that for all D, n ≥ 2,~~

~~1/2 c˜2 ˜ (Dn ln n) ≤ VRHS(S, D) ≤ C2 N 1/2+αD N 1/2+αD~~

~~1 1 1 where αD = 2 logD(1 + D ) − 2D2 logD(D + 1).~~

~~1 27 In particular, if D = 2, this recovers the results in [99], with α2 = 8 log2( 16 ).~~

~~Theorem 22 (Small number of large subsystems): For system H = (CD)⊗n, ˜ there exist universal constants c˜3, C3 > 0, such that for all D, n ≥ 2,~~

~~c˜n (n ln n)1/2 3 ≤ VR (S, D) ≤ C˜ . N 1/2−1/(2n) HS 3 N 1/2−1/(2n)~~

~~Theorem 23 (Asymptotic roughness of PPT criterion): There exists an ab- D D solute constant c˜0 > 0 such that for any bipartite system H = C ⊗ C ,~~

~~c˜0 ≤ VRHS(PPT , D) ≤ 1.~~

~~Here, we only provide the proof of Theorem 21 for D = 2 based on the approach~~

in [8]. This proof presents how to estimate VHS(S) by techniques from asymptotic geometric analysis. Interested readers should read [8, 99] for more general results of D > 2 and similar calculations for other sets of states. 101

~~Recall that the precise value of the Hilbert-Schmidt volume of D is (see formula (1.19))~~

N(N−1) √ E(N) V (D) = (2π) 2 N HS Γ(N 2) where E(N) is deﬁned in formula (1.16). Moreover, as we already pointed out in formula (1.21) the Hilbert-Schmidt volume radius of D is (asymptotically) vradHS(D) ∼ − 1 − 1 e 4 d 4 . Hence, to prove Theorem 21, one only needs appropriate upper and lower estimates for VHS(S) (or vradHS(S)).

~~Recall that the set of separable quantum state is~~

~~2 2 ⊗n S = conv{ρ1 ⊗ · · · ⊗ ρn, ρi ∈ D(C )} ⊂ Bsa((C ) ).~~

It is easy to verify that D(C2) coincides with the Bloch ball, which geometrically is a √ Id2 3-dimensional Euclidean ball with center at 2 and radius r = 1/ 2. Its boundary is the so-called Bloch sphere which consists of pure states on B(C2). Accordingly, Ξ(C2) := conv(D(C2) ∪ −D(C2)) is a 4-dimensional cylinder whose base is the Bloch ball and whose axis is the segment [−Id2/2, Id2/2]. Deﬁne the Pauli matrices

~~0 1 0 −i 1 0 σ = , σ = , σ = . x 1 0 y i 0 z 0 −1~~

~~√ √ √ √ 2 Via the orthonormal basis {Id2/ 2, σx/ 2, σy/ 2, σz/ 2}, one can identify Bsa(C ) with R4.~~

For any δ > 0, a δ-net of a set K is a subset N ⊂ K such that for each x ∈ K, there exists y ∈ N whose distance to x is smaller than δ. If N ⊂ C2, we write P (N ) 2 for the polytope conv({±|xihx|}x∈N ) ⊂ Bsa(C ).

~~p √ Lemma 16 Let δ < 2 − 2 ≈ 0.765, and let N be a δ-net of the unit sphere of C2. Then 2 4 2 2 (1 − 2δ + δ /2)Ξ(C ) ⊂ P (N ) ⊂ Ξ(C ). 102~~

~~Proof. The second inclusion holds trivially, so we will only show the ﬁrst one. For 2 2 ◦ any A ∈ Bsa(C ), its norm measured by the gauge of Ξ(C ) is~~

~~kAkΞ( 2)◦ = max hA, ±|xihx|iHS = max |hx|A|xi| = kAkop. C 2 2 x∈C ,kxk=1 x∈C ,kxk=1~~

Here kAkop is the operator norm of A: the maximum of the singular values of A. Its ◦ norm measured by the gauge of P (N ) is kAkP (N )◦ = maxy∈N |hy|A|yi|. We need to show that 2 4 kAkP (N )◦ ≥ (1 − 2δ + δ /2)kAkop.

~~2 To this end, by homogeneity, it is enough to consider A ∈ Bsa(C ) such that kAkop and the largest eigenvalues of A are both equal to 1. Hence, there exists a vector x ∈ C2~~

~~such that Ax = x. Choose y0 ∈ N verifying kx − y0k ≤ δ. By writing y0 = y1 + y2 2 2 δ2 with y1 = hy0|xix, one can easily check that hy1|A|y1i ≥ |hy0|xi| ≥ 1 − 2 ,~~

~~2 2 δ2 |hy2|A|y2i| ≤ ky2k ≤ δ 1 − 4 and hy1|A|y2i = 0. It then easily follows that 4 hy0|A|y0i ≥ 1 − 2δ2 + δ /2 which implies the assertion.~~

~~2 2 ˆ Now, let Υ(H) = conv(S ∪ −S) with H = C ⊗ · · · ⊗ C . Notice that Ξ(C2)⊗n = | {z } n Υ(H). Tensoring the conclusion of the preceding lemma yields an inclusion~~

~~(1 − 2δ2 + δ4/2)nΥ(H) ⊂ P (N )⊗ˆ n.~~

~~p √ Therefore, for any 0 < δ < 2 − 2, one has~~

~~1 |Υ(H)| d+1 1 − 2δ2 + δ4/2n ≤ vrad(P (N )⊗ˆ n). |BHS|~~

~~It is well known [75] that for δ ≤ 1, we can ﬁnd δ-nets in the sphere S3 of~~

2 4 2 4 ⊗ˆ n cardinality not exceeding 1 + δ , that is #N ≤ 1 + δ . Observe that P (N ) n 2 4n is a symmetric polytope with at most 2(#N ) ≤ 2 1 + δ vertices. Combin- ing this estimate with the Urysohn inequality (Theorem 5), one obtains that for all 103 p √ 0 < δ < 2 − 2,

1 d+1 2 4 n |Υ(H)| ⊗ˆ n 1 ⊗ˆ n 1 − 2δ + δ /2 ≤ ω(P (N ) ) = ωG P (N ) |BHS| τd+1 s 1 24n ≤ 2 ln 1 + τd+1 δ where the second inequality follows from Theorem 6 and the comments following it. Hence,

~~q 1 2 4n |Υ(H)| d+1 2 ln 1 + ≤ inf δ . √ √ 2 4 n |BHS| 0<δ< 2− 2 τd+1 (1 − 2δ + δ /2)~~

~~This approach can be easily generalized to other Euclidean structures; the only constraint is that the vertices of the polytope are of norm less than or equal to 1. That is,~~

~~q 1 2 4n |Υ(H)| d+1 2 ln 1 + ≤ inf δ (4.1) √ √ 2 4 n |E| 0<δ< 2− 2 τd+1 (1 − 2δ + δ /2)~~

where E is any ellipsoid containing Υ(H). When E is taken as the L¨ownerellipsoid of Υ(H), i.e., the ellipsoid containing Υ(H) with minimal volume, then inequality (4.1) gives a tight estimate of Υ(H) (and hence of S(H)). To determine the L¨owner ellipsoid of Υ(H), one needs the following lemma.

m−1 ⊥ Lemma 17 Let u ∈ S , and H0 = {u} be the orthogonal hyperplane. Let h > 0 be a real number and H = H0 + hu be a translate of H0. W is a convex body in H, and its symmetrization is Ω = conv{W ∪ −W }. Then the following assertions are equivalent. m (i) Ω is in Löwner position (i.e., the unit ball B2 is the Löwnerellipsoid of Ω). √ m (ii) h = 1/ m and B2 ∩ H is the Löwnerellipsoid of W . 104

~~m Proof of Lemma 17. Assume that Ω is in L¨owner position, so Ω ⊂ B2 . By~~

~~Theorem 8 (John’s theorem), this means that there exists a sequence (xi, ci) with~~

m2−1 P ci > 0 and xi ∈ Ω ∩ S such that i ci|xiihxi| = Idm. Since xi are extreme points of Ω, they must be either in W or in −W ; and since Ω is symmetric, we can assume that all xi are in W . Let P be the orthogonal projection on H0. It follows √ P 2 0 2 that i ci|P xiihP xi| = P ; note kP xik = 1 − h . With ci = ci(1 − h ), one can 0 verify that (P xi, ci) satisfies the conditions in Theorem 8 (John’s theorem). That is 2 −1/2 m (1 − h ) PW is in Löwnerposition; equivalently, B2 ∩ H is the Löwner ellipsoid √ P P 0 of W . Since i ci = 1 and i ci = n − 1, one gets h = 1/ m.

~~The reverse implication follows by retracing the above argument in the opposite direction.~~

~~2 Proof of Theorem 21 Recall that the following is an orthonomal basis of Bsa(C ) √ √ √ √ {Id2/ 2, σx/ 2, σy/ 2, σz/ 2}.~~

2 Define the linear map A on Bsa(C ) which is diagonal in that basis and whose action is √ p defined by AId2 = Id2/ 2, and Aσi = 3/2σi for all i = x, y, z. Hence, A maps the √ 2 2 hyperplane T1 containing D(C ) = S(C ) into the hyperplane T with Id2/ 8 ∈ T 1 and with distance to the origin 2 . One verifies conditions in Lemma 17 (with m = 4) for AD(C2), and hence one concludes that AΞ(C2) is in Löwner position. That is, 2 2 BHS(C ), the unit ball of Hilbert-Schmidt metric of Bsa(C ), is the Löwnerellipsoid of AΞ(C2). Lemma 3 implies that the Löwnerellipsoid of

~~⊗ˆ n 2 2 Υ(˜ H) = A Υ(H) = (AΞ(C ))⊗ˆ · · · ⊗ˆ (AΞ(C )) | {z } n~~

~~2 is the Hilbertian tensor product of BHS(C ), which equals to BHS(H), the unit ball of~~

the Hilbert-Schmidt metric of Bsa(H). Recall that for any affine map T , the Löwner ellipsoid of TK, L(TK), equals to the image of the Löwnerellipsoid of K under T , i.e., L(TK) = T L(K). Hence, the Löwnerellipsoid of Υ(H) is

~~−1 ⊗ˆ n ˜ −1 ⊗ˆ n L(Υ(H)) = (A ) L(Υ(H)) = (A ) BHS(H). 105~~

~~The volume radius of L(Υ(H)) may be calculated as~~

~~1 2 N 1 n4n−1 |L(Υ(H))| −1 ⊗ˆ n 2 −1 vrad(L(Υ(H))) = = det((A ) ) N = (det(A )) N2 . |BHS(H)|~~

~~q 27 n Clearly det(A) = 16 and since N = 2 , it follows that~~

~~n 16 8 vrad(L(Υ(H))) = = 2−nα2 = N −α2 (4.2) 27~~

~~1 27 by the deﬁnition of α2, i.e., α2 = 8 log2( 16 ).~~

~~Put the L¨owner ellipsoid of Υ(H) into the inequality (4.1), one gets~~

~~q 2 4n vrad(Υ(H)) 2 ln 1 + ≤ inf δ . √ √ 2 4 n vrad(L(Υ(H))) 0<δ< 2− 2 τd+1 (1 − 2δ + δ /2)~~

~~p Let δ = 1/ n log2(2n), we obtain~~

~~q 2 ln 1 + 2 4n vrad(Υ(H)) ≤ vrad(L(Υ(H))) inf δ √ √ 2 4 n 0<δ< 2− 2 τd+1 (1 − 2δ + δ /2)~~

~~p4n log (4n) ≤ 2 . N 1+α2~~

~~The Rogers-Shephard inequality (Theorem 7) implies that the volume radius of~~

S(H) equals to the volume radius of Υ(H) (up to a factor 2). As vradHS(D) ∼ − 1 − 1 − 1 − 1 ˜ e 4 d 4 ∼ e 4 N 2 , there exists a universal (independent of N) constant C2 > 0, such that √ ˜ 2n ln n VRHS(S, D) ≤ C2 . (4.3) N 1/2+α2

~~The lower bound of Theorem 21 directly follows from the classical fact [9] that for √ a centrally symmetric convex body K in Rn, we have the inclusion L(K) ⊂ nK. 106~~

~~Hence, vrad(Υ(H)) 1 ≥ . vrad(L(Υ(H))) N~~

~~1 1 1 − 4 − 2 Equivalently vrad(Υ(H)) ≥ N 1+α2 . Combining with vradHS(D) ∼ e N and~~

~~Rogers-Shephard inequality (Theorem 7), there exists a universal constantc ˜2, such that c˜2 VRHS(S, D) ≥ N 1/2+α2 which completes the proof of Theorem 21.~~

~~4.2 Bures volume of separable quantum states~~

In this section, K will be an arbitrary (Borel) subset of D. We will estimate the Bures volume of K, in terms of the Hilbert-Schmidt volume of K, both from below and from above. The following lemma is our main tool to study the asymptotical behavior of

~~VRB(K, D). We point out that these estimates are independent of the possible tensor product structure of H.~~

~~Lemma 18 For any subset K in D and any p > 1, one has~~

~~1 2p N−N2 N2−1 N2+N−2 VHS(K) 2p−1 2 2 N 2 VHS(K) ≤ 2 2 VB(K) ≤ √ I(p) 2p N where I(p) is deﬁned as~~

~~ N Γ 1 + j(p−1) Γ j(p−1) N(N−1) 1 Y 2p−1 2p−1 (2π) 2 I(p) := . (4.4) (p−1)N 2 3p−2 E(N) N!Γ 2p−1 j=1 Γ 2p−1~~

1 Remark. I(p) can be deﬁned for all p∈ / [ 2 , 1] (irrespective of N; since the Gamma 1 function has poles at nonpositive integers, there are singularities in [ 2 , 1] whose exact V (D) locations depend on N.) In particular, I(0) = HS√ . The quantity E(N) was deﬁned N in (1.16). 107

~~Proof. First of all, we estimate VB(K) from below. To that end, deﬁne h : ∆ → R as N Y Y 2 h(λ1, ··· , λN ) = λi (λi + λj) . i=1 1≤i~~

Lagrange multiplier method implies that (1/N, ··· , 1/N) is the only critical point of h(λ1, ··· , λN ) in the interior of simplex ∆. Clearly h(λ1, ··· , λN ) is always 0 on the boundary of the simplex ∆, which consists of sequences for which one or more of the

λi’s equal 0, and strictly positive in the interior of the simplex ∆. By compactness, h(λ1, ··· , λN ) must have a maximum inside, and the critical point (1/N, ··· , 1/N) must be the (only) maximizer of h(λ1, ··· , λN ) on ∆. Therefore,

~~1 1 Y 1 N−N2 N2 p = √ ≥ 2 2 N 2 . (4.5) λ ··· λ λi + λj h(λ1, ··· , λN ) 1 N 1≤i~~

~~R By formula (1.22), the Bures volume of K equals to K dVB, i.e.,~~

~~N−1 Z 2 2 2−N−N 1 Y (λi − λj) Y VB(K) = 2 2 √ dλi dγ. λ ··· λ λi + λj K 1 N 1≤i~~

~~Considering inequality (4.5) and formula (1.17), one gets~~

~~N−1 2 Z 1−N 2 N Y 2 Y VB(K) ≥ 2 N 2 (λi − λj) dλi dγ K 1≤i~~

~~2 1−N 2 N −1 = 2 N 2 VHS(K).~~

Next, we will derive the upper bound, which is more involved (and more important for our results). The subset ∂∆, the boundary of ∆, consists of sequences for which some λi = 0 and has zero N −1 dimensional measure. Thus, without loss of generality,

~~λi−λj we can assume λi > 0 for all i = 1, ··· ,N and, in particular, < 1 for all i 6= j. λi+λj 108~~

~~This implies~~

~~N−1 2 Z 2 N +N−2 1 Y (λi − λj) Y 2 2 VB(K) = √ dλi dγ λ ··· λ λi + λj K 1 N 1≤i~~

~~N−1 Z 1 Y Y < √ |λ − λ | dλ dγ λ ··· λ i j i K 1 N 1≤i~~

~~N−1 Z Y = f g dλi dγ (4.6) K i=1 where, to reduce the clutter, we denoted~~

~~1 Y 1− 1 g(λ , ··· , λ ) = √ |λ − λ | p , 1 N λ ··· λ i j 1 N 1≤i~~

~~Y 1 f(λ1, ··· , λN ) = |λi − λj| p . 1≤i~~

~~1 For any p > 2 (so that 2p > 1), we employ the H¨olderinequality to (4.6) and get~~

~~N−1 1 N−1 2p−1 2 Z 2p Z 2p N +N−2 2p Y 2p Y 2 2 VB(K) ≤ f dλi dγ g 2p−1 dλi dγ . (4.7) K i=1 K i=1~~

~~Substituting f into the ﬁrst integral of (4.7) and by (1.17), one has~~

~~N−1 1 N−1 1 1 Z 2p Z 2p 2p 2p Y Y 2 Y VHS(K) f dλi dγ = (λi − λj) dλi dγ = √ (4.8). K i=1 K 1≤i~~

~~Substituting g into the second integral of (4.7) leads to~~

~~N−1 N N−1 Z 2p Z 2p−2 ( p−1 −1) 2p−1 Y Y 2p−1 Y 2p−1 Y g dλi dγ = |λi − λj| λi dλi dγ K i=1 K 1≤i~~

~~N N−1 Z 2p−2 ( p−1 −1) Y 2p−1 Y 2p−1 Y ≤ |λi − λj| λi dλi dγ, (4.9) D 1≤i~~

~~the inequality following just from K ⊂ D. By (1.16) and the Fubini’s theorem, the last integral in (4.9) equals to~~

~~N(N−1)/2 N p−1 N−1 (2π) Z Y 2p−2 Y ( −1) Y |λ − λ | 2p−1 λ 2p−1 dλ . (4.10) E(N) i j i i ∆1 1≤i~~

~~p−1 Under the condition 2p−1 > 0 (i.e., p > 1 or p < 1/2), one has (see e.g. [67, 115])~~

~~N N−1 Z 2p−2 p−1 −1 Y 2p−1 Y 2p−1 Y |λi − λj| λi dλi ∆ 1≤i~~

~~ N Γ 1 + j(p−1) Γ j(p−1) 1 Y 2p−1 2p−1 = . (p−1)N 2 3p−2 Γ 2p−1 j=1 Γ 2p−1~~

~~Taking into account that ∆ consists of N! Weyl chambers, we conclude that the expression in (4.10) is then equal to I(p). In other words, we have shown that~~

~~N N−1 Z 2p−2 ( p−1 −1) Y 2p−1 Y 2p−1 Y I(p) = |λi − λj| λi dλi dγ. D 1≤i~~

~~Combining this with (4.7), (4.8), and (4.9), we conclude that if p > 1, then~~

~~1 2p N2+N−2 VHS(K) 2p−1 2 2 VB(K) ≤ √ I(p) 2p , N which is the upper estimate from Lemma 18.~~

~~Theorem 24 There is a universal computable constant c1 > 0, such that for any Hilbert space H and any subset K ⊂ D,~~

~~c1 VRHS(K, D) ≤ VRB(K, D). 110~~

~~Proof. Recall d = N 2 − 1. From the lower bound of Lemma 18, one has~~

~~1 1 1 1 1 1 1 V (K) d ≥ (d + 1) 4 V (K) d > d 4 V (K) d . B 2 HS 2 HS~~

~~1 Dividing both sides by VB(D) d , one obtains~~

~~1 1 1 d d d VB(K) 1 1 VHS(K) VHS(D) > d 4 VB(D) 2 VHS(D) VB(D)~~

~~1 d 1 1 vradHS(D) VHS(K) = d 4 . (4.11) 2 vradB(D) VHS(D)~~

~~1 − 1 Formulas (1.21) and (1.27) imply that d 4 vradHS(D) ∼ 2e 4 vradB(D), i.e., 1 1 vradHS(D) − 1 lim d 4 = e 4 . d→∞ 2 vradB(D)~~

~~Therefore, there is a (computable) universal constant c1 > 0, such that, for any N,~~

~~1 1 vradHS(D) d 4 ≥ c1. 2 vradB(D)~~

~~Together with (4.11), this shows that for any K ⊂ D and for any N,~~

~~VRB(K, D) ≥ c1VRHS(K, D).~~

−1/4 Remark. If N is relatively large, the optimal constant c1 = c1(N) is close to e 1 d 4 vradHS (D) −1/4 because c1(N) = → e as N → ∞. For speciﬁc values of N, one can 2 vradB (D) compute c1 precisely. For instance, c1(4) ≈ 0.7572. Actually, c1(6) ≈ 0.7686, c1(8) ≈ −1/4 0.7728 and c1(10) ≈ 0.7748 which are very close to the e = 0.7788. It appears that the sequence c1(N) is increasing and so c1 = c1(4) should work for all N, however, we do not have a rigorous proof.

~~Theorem 25 There is a universal computable constant C1 > 0 such that, for any Hilbert space H and any K ⊂ D,~~

~~√ ln ln(e/α) VR (K, D) ≤ C α exp B 1 2N 111~~

~~where α = VRHS(K, D).~~

~~2 Proof. Recall N = D1D2 ··· Dn and d = N − 1. For any p > 1, Lemma 18 implies that~~

~~d N2+N−2 d 1 2p−1 2 2 VB(K) ≤ 2 2 VB(K) ≤ α 2p (VHS(D)) 2p I(p) 2p .~~

~~p−1 Let β = 2p−1 . Replacing VHS(D) and I(p) by formula (1.19) and formula (4.4), one has~~

~~1 1 1 −1 QN 1− 2p d d N2−N [E(N)] p N 4p [ j=1 Γ(βj)Γ(1 + βj) ] 2 2 VB(K) ≤ α 2p (2π) 2 . 1− 1 1 1− 1 [N!] 2p [Γ(N 2)] 2p [Γ(βN 2)Γ(1 + β)N ] 2p~~

~~1 −1 1 1 −1 Clearly [E(N)] p ≤ 1 and N 4p (N!) 2p ≤ 1 if p > 1. Hence,~~

~~1 QN 1− 2p d d [ j=1 Γ(βj)Γ(1 + βj) ] VB(K) ≤ α 2p π 2 . (4.12) 1 1− 1 [Γ(N 2)] 2p [Γ(βN 2)Γ(1 + β)N ] 2p~~

Since xΓ(x) = Γ(x + 1), it is easy to see that for all x ∈ (0, 1) the upper estimate 1 1 of Γ(x) is x and (somewhat less easy that) the lower estimate of Γ(x) is ϑx , where ϑ ≈ 1.12917 [21]. That is 1 1 1 ≤ Γ(x) ≤ , or ≤ Γ(1 + x) ≤ 1, for all x ∈ (0, 1). (4.13) ϑx x ϑ

N 2 ln(e/α)−1 p−1 1 Pick p = p(N, α) := N 2 ln(e/α)−2 as a function of N and α, so that β = 2p−1 = N 2 ln(e/α) . 2 1 2 Equivalently, βN = ln(e/α) . Since α ≤ 1 for all K ⊂ D, then βN < 1 and hence βj ≤ 1 for all j = 1, 2, ··· N 2. Taking inequality (4.13) into account, one has

~~1− 1 Γ(1 + βj) 2p ≤ ϑ, for all j = 1, 2, ··· N. (4.14) Γ(1 + β)~~

~~Consequently, again by inequality (4.13), and N ≥ 4,~~

~~1 " N #1− 2p 1 Q 1− 2p j=1 Γ(βj) ϑ N (1−N)(1− 1 ) (1−N)(1− 1 ) ≤ β 2p ≤ β 2p . (4.15) Γ(βN 2) (N − 1)! 112~~

~~Combining inequality (4.12) with inequalities (4.14) and (4.15), one has~~

~~d d N (1−N)(1− 1 ) 2 −1 VB(K) ≤ α 2p π 2 ϑ β 2p [Γ(N )] 2p .~~

~~Equivalently, taking d-th root from both sides,~~

~~1 1 1 1 −1 (1− 1 ) 2 −1 (VB(K)) d ≤ α 2p π 2 ϑ N−1 β N+1 2p [Γ(N )] 2pd .~~

~~1 1 1 Note N ≥ 4, and hence ϑ N−1 ≤ ϑ 3 = 1.0413. Now dividing VB(D) d from both sides of the above inequality, one gets~~

~~1 1 2 1 (1− 1 ) 2 1 2 −1 VRB(K, D) ≤ 2ϑ 3 α 2p [N ln(e/α)] N+1 2p [Γ N /2 ] d [Γ(N )] 2dp . (4.16)~~

~~It is then easy to verify that~~

~~1 ln N ln ln(e/α) √ ln ln(e/α) [N 2 ln(e/α)] 2(N+1) ≤ exp exp ≤ 2 exp , N 2N 2N~~

~~ 2 2 1 (1− 1 ) ln(N ln(e/α)) 1 1 [N ln(e/α)] 2(N+1) p = exp ≤ exp ≤ e 4 . 2(N + 1)(N 2 ln(e/α) − 1) N~~

~~Therefore,~~

~~√ 2 1 (1− 1 ) 1 ln ln(e/α) [N ln(e/α)] N+1 2p ≤ 2 e 4 exp (4.17) 2N~~

~~Also, we can verify that~~

~~ 1 1 ln(1/α) 1 1 1 √ α 2p = α 2 exp ≤ α 2 exp ≤ e 32 α.(4.18) 2[N 2(1 + ln(1/α)) − 1] 2N 2~~

~~Since Γ(N 2) ≤ (N 2)! ≤ exp(N 2 ln N 2), one has~~

~~ 2 2 1 − 1 ln(Γ(N )) Γ(N ) 2d 2pd = exp 2(N 2 − 1) (N 2 ln(e/α) − 1) 2 ln N 2 1 ≤ exp ≤ exp ≤ e 2 . (4.19) N 2 ln(e/α) N 113~~

~~Stirling approximation formula (1.20) implies that~~

~~2 1 Γ(N /2) d 1 lim 1 = √ . (4.20) N→∞ [Γ(N 2)] 2d 2~~

~~Together with inequalities (4.16), (4.17),(4.18), and (4.19), this shows that there exists a universal (independent of N, α) constant C1 > 0, such that, VRB(K, D) ≤~~

~~√ ln ln(e/α) C1 α exp 2N as asserted.~~

~~Remark. A slightly more precise calculation shows that~~

~~√ ln ln(e/α) ln N VR (K, D) ≤ 2α exp 1 + O . B 2N N~~

~~The calculation yields explicit (not necessarily optimal) values of C1 in the theorem.~~

~~For small dimensions, one can take C1(4) ≈ 2.5164 if N = 4, C1(6) ≈ 2.2137, and~~

C1(8) ≈ 2.0478. As the dimension N becomes large, the value of C1 given by the √ argument tends to 2 ≈ 1.4142. The Legendre duplication formula (see [1]) says that √ Γ(z)Γ(z + 1/2) = 21−2z π Γ(2z).

~~By taking z = N 2/2, one has~~

~~1 √ 1 Γ(N 2/2) Γ(N 2/2) 2(N2−1) 1 π Γ(N 2/2) 2(N2−1) = √ . (4.21) Γ(N 2) 2 Γ(N 2/2 + 1/2)~~

~~Gamma function [71] is log-convex, and hence, one has~~

~~Γ(N 2/2 + 1/2)2 Γ(N 2/2)2 ≤ Γ(N 2/2 − 1/2)Γ(N 2/2 + 1/2) = . N 2/2 − 1/2~~

Equivalently r Γ(N 2/2 + 1/2) N 2 − 1 ≥ , Γ(N 2/2) 2 √ √ which is greater than π iﬀ N > 2π + 1 ≈ 2.7. Together with formula (4.21), one shows that the asymptotic relative (4.20) is in fact an upper bound for all N ≥ 3, it

~~follows that C1 ≈ 2.5164 works for all N ≥ 4, C1 ≈ 2.2137 works for all N ≥ 6, etc. 114~~

~~ ln ln(e/α) Remark. In most cases of interest α is such that the factor exp 2N is bounded~~

~~a2N by a universal numerical constant. For instance, if ln(1/α) ≤ a1 e for some constants a1 > 0, a2 > 0, then~~

~~ √ p ln N VR (K, D) ≤ 2ea2 VR (K, D) 1 + O . B HS N~~

~~While our argument doesn’t give similar estimates for general α, other ways of writing the estimates in more transparent ways are possible. For example, for any ﬁxed p > 1~~

~~there is a constant Cp > 0 depending on p (but independent of N and α), such that~~

~~1 2p VRB(K, D) ≤ Cp VRHS(K, D) .~~

1 k In the cases of S and PPT , α is bounded from above by N for some (ﬁxed) integer ˜ p ˜ k [8, 99]. Therefore, VRB(S, D) ≤ C1 VRHS(S, D) where C1 > 0 is a universal ¯ p constant independent of N. Similarly, VRB(PPT , D) ≤ C1 VRHS(PPT , D) where ¯ C1 > 0 is a universal constant independent of N.

1 Remark. We point out that there is a lot of ﬂexibility in the choice of β = N 2 ln(e/α) 1 (hence the choice of p(N, α)). For example, one can choose β = eN ln(e/α) , and proves Theorem 25 with diﬀerent (larger) constants. However, formula (4.18) does suggest that the factor ln(e/α) in β is essentially optimal in general.

~~As applications of Theorems 24 and 25, and the estimates for VRHS(S, D) implicit in [8], one immediately has the following corollaries.~~

~~Corollary 7 (Large number of small subsystems) For system H = (CD)⊗n,~~

~~there exist universal computable constants c2,C2 > 0, such that for all D, n ≥ 2,~~

~~r 1/2 c2 (Dn ln n) ≤ VRB(S, D) ≤ C2 . N 1/2+αD N 1/2+αD~~

~~1 1 1 where αD = 2 logD(1 + D ) − 2D2 logD(D + 1). 115~~

~~Corollary 8 (Small number of large subsystems) For system H = (CD)⊗n,~~

~~there exist universal computable constants c3,C3 > 0, such that for all D, n ≥ 2,~~

~~r cn (n ln n)1/2 3 ≤ VR (S, D) ≤ C . N 1/2−1/(2n) B 3 N 1/2−1/(2n)~~

Remark. Recall that if H = (CD)⊗n, then the dimension of H is N = Dn and so the expressions in the the numerators of the estimates in the Corollaries above are of smaller order than the denominators. Hence, for any ﬁxed small D, Corollary 7 shows that VRB(S, D) goes to 0 exponentially as n → ∞. On the other hand, for H = (CD)⊗n and for ﬁxed small n, Corollary 8 shows that “the order of decay” of 1 − n 1 − n VRB(S, D) is between D 2 2 and D 4 4 as D → ∞. In both cases, the priori Bures probability of separability is extremely small for large (and even for moderate) N. It is possible to provide (not necessarily optimal) estimates on the constants appearing in both corollaries. In the case of the Hilbert-Schmidt volume, estimates on the relevant constants were supplied in [8]. Feeding those estimates into our arguments shows that, √ for instance, in Corollary 7, one can take c2(4) = 0.2272,C2(4) = 4.4 C1(4) = 5.2785 √ if N = 4, c2(6) = 0.2306,C2(6) = 4.4 C1(6) = 4.6436, and c2(8) = 0.2318,C2(8) = √ 4.4 C1(8) = 4.2955. Similarly, if the dimension N is large (particularly for large n),

the relevant asymptotic behaviors of c2 and C2 given by the proofs are: c2 tends to p e 3/4 1/8 2π ≈ 0.6577, and C2 tends to 2 e ≈ 1.9057. In Corollary 8, c3 can be taken as √ −1/4 c3 = e / 6 ≈ 0.3179 and C3 = C2.

~~The following corollary, which is a direct consequence of Theorems 24, 25, and of Theorem 4 of [8], gives the estimation of the Bures volume of PPT .~~

~~Corollary 9 (Bures Volume of PPT ) There exists an absolute computable con- D D stant c0 > 0, such that, for any bipartite system H = C ⊗C , c0 ≤ VRB(PPT , D) ≤ 1.~~

~~Indent: An unsolved question is: does there exist a universal constant 0 < C0 < 1 such~~

that VRB(PPT , D) ≤ C0 < 1? Answering this question would help us understand 116 the eﬀectiveness of positive partial transpose criterion as a tool to detect quantum entanglement for all D ≥ 3. (The answer to the analogous question about VRHS is not known, either.)

An immediate consequence of Corollaries 8 and 9 is that, for H = CD ⊗ CD and large D, there exist universal constants c4,C4 > 0 (independent of D), such that, − 1 − 1 c4D 2 ≤ VRB(S, PPT ) ≤ C4D 4 . The upper bound decreases to 0 as D → ∞. In other word, the conditional priori Bures probability of separability given positive partial transpose condition is exceedingly small. Hence, for large N, the PPT criterion is not precise as a tool to detect separability.

~~4.3 Optimality of the bounds~~

In this section, we will prove that, in general, the bounds in Theorems 24 and 25 are essentially optimal. The Bures volume has singularities close to the boundary of D, so the optimal upper bound is intuitively attained by the subsets close to the boundary of D. On the other hand, for the subsets located near the maximal state

~~IdN ρmax = N , we can achieve the lower bound.~~

~~4.3.1 Optimality of the lower bound~~

~~For 0 < t < 1, let Kt = tD + (1 − t)ρmax, i.e.,~~

~~ 1 − t 1 − t K = UXU † : X = diag + tλ , ··· , + tλ t N 1 N N~~

~~ with (λ1, ··· λN ) ∈ ∆ and U ∈ U(N)~~

~~Let ZN be as in (1.16). 117~~

~~We now estimate VB(Kt) from above. By formula (1.22),~~

~~2 N−1 2 Z N −1 2 N +N−2 ZN t Y (λi − λj) Y 2 2 VB(Kt) = q 2−2t dλi. N 1−t ( + tλ + tλ ) ∆1 Q 1≤i~~

~~1−t 1−t As N + tλi ≥ N for all i, one obtains~~

~~2 2 N−1 2 N −1 N /2 Z N +N−2 t N Y 2 Y 2 2 V (K ) ≤ Z (λ − λ ) dλ B t N 2(N 2−N)/2 (1 − t)N 2/2 i j i ∆1 1≤i~~

~~tN 2−1 N (N 2−1)/2 = V (D) 2(N 2−N)/2 (1 − t)N 2/2 HS where the equality follows the formula (1.17). By Lemma 18, one gets~~

~~tN 2−1 tN 2−1 V (K ) ≤ V (D) ≤ V (D). B t N2 B N 2−1 B (1 − t) 2 (1 − t)~~

t 3 Hence, VRB(Kt, D) ≤ 1−t ≤ 4t for all t ≤ 4 . On the other hand, VRHS(Kt, D) = t holds trivially because of the homogeneity of the Hilbert-Schmidt measure. We have 3 proved that VRB(Kt, D) ≤ 4VRHS(Kt, D) for all Kt such that VRHS(Kt, D) ≤ 4 .

Theorem 24 guarantees that the lower bound of VRB(Kt, D) is at least (up to a multiplicative constant) VRHS(Kt, D). So the lower bound of VRB(Kt, D) in Theorem 24 can be obtained, and hence is optimal in general.

~~4.3.2 Optimality of the upper bound~~

~~For 0 < t < 1, we consider Kt as~~

~~t † K = {UXU : X = diag(1 − t + tλ1, tλ2, ··· , tλN )~~

~~with (λ1, ··· λN ) ∈ ∆1 and U ∈ U(N)}.~~

~~Recall ∆1 is the chamber of ∆ with order λ1 ≥ · · · ≥ λN . 118~~

~~The Hilbert-Schmidt volume of Kt can be calculated by the following integral~~

~~√ Z N N−1 t (N−1)2 Y 2 Y 2 Y VHS(K ) = ZN t N (λi − λj) (tλ1 − tλk + 1 − t) dλi. ∆1 2≤i~~

~~As 0 ≤ tλ1 − tλk + 1 − t ≤ 1, one has~~

~~√ Z N−1 t (N−1)2 Y 2 Y VHS(K ) ≤ ZN t N (λi − λj) dλi ∆1 2≤i~~

~~r Z 1 ZN (N−1)2 N N 2−2N = t VHS(DN−1) (1 − λ1) dλ1 ZN−1 N − 1 0 √ ZN (N−1)2 ≤ 2 t VHS(DN−1). (4.22) ZN−1~~

~~t −2 By Stirling approximation (1.20), VRHS(K , D) ≤ C4 t t N+1 holds for some universal −2 c4(−1−N) constant C4 > 0. If t N+1 is bounded from above, e.g., t > e for some constant~~

~~c4 > 0, then t VRHS(K , D) ≤ C5 t (4.23) holds for a new universal constant C5 > 0.~~

~~Next, we estimate the Bures volume of Kt from below. By formula (1.22),~~

~~2 N +N−2 t 2 2 VB(K )~~

~~2 Z (N−1) 2 N 2 N−1 ZN t 2 Y (λi − λj) Y (tλ1 − tλk + 1 − t) Y = p dλi λi + λj tλ1 + tλk + 1 − t ∆1 (tλ1 + 1 − t)λ2 ··· λN 2≤i~~

~~2 Z 2N−2 (N−1) 2 N−1 (1 − t) t 2 Y (λi − λj) Y ≥ ZN √ dλi λ ··· λ λi + λj ∆1 2 N 2≤i~~

where the inequality is because of 0 ≤ (tλ1 + 1 − t) ≤ 1, 0 ≤ (tλ1 + tλk + 1 − t) ≤ 1 and (tλ1 − tλk + 1 − t) ≥ 1 − t. The last integral can be computed as in (4.22) and 119 leads to

~~2 N +N−2 t 2 2 VB(K )~~

~~2 2 Z 1 2 ZN 2N−2 (N−1) N −N−2 N −2N−1 ≥ (1 − t) t 2 2 2 VB(DN−1) (1 − λ1) 2 dλ1 ZN−1 0~~

2 2 N −N 2N−2 (N−1) ZN 2 2 (1 − t) t 2 = 2 VB(DN−1). ZN−1 (N − 1) √ t Employing the Stirling approximation (1.20) one gets VRB(K , D) ≥ c5 t for some 4 universal constant c5 > 0 if, say, t < 5 . Together with (4.23), we have thus proved

~~t p t VRB(K , D) ≥ c¯5 VRHS(K , D)~~

4 c4(−1−N) for some universal constantc ¯5 > 0, if t < 5 and t > e . Theorem 25 guarantees t p t that VRB(K , D) ≤ C1 VRHS(K , D). Therefore, the upper bound in Theorem 25 can also be achieved, and is optimal in general.

~~4.4 Conclusion and Comments~~

In summary, we proved that if K is a Borel subset of D, then the priori Bures probability of K can be estimated from above and from below in terms of the priori Hilbert-Schmidt probability of K. Speciﬁcally, under some mild conditions on K the relative Bures volume radius VRB(K, D) can be (approximately) bounded from below by the relative Hilbert-Schmidt volume radius VRHS(K, D), and from above p by VRHS(K, D). We employ these results to estimate the Bures volume of S and PPT and the relevant priori Bures probabilities. We deduce that positive partial transpose criterion becomes less and less precise as the dimension of H becomes larger and larger, at least if the goal is to detect separability. We also give examples showing that, for general subsets, our bounds are essentially optimal.

When N is small, for instance when N = 4 or 6, our estimates for VB(S) are less precise than Slater’s numerical results. However, our methods overcome the big 120 disadvantage of the numerical approach, which works only for small N. Moreover, our results are independent of the structure of H. (In applications, of course, the information on the structure of H will be hidden in the calculation of VRHS(K, D).) Proceeding along similar lines one can obtain similar results for real Hilbert spaces, and then estimate the Bures volume of sets analogous to S or PPT on a real Hilbert space. As is well known, for sets in a Euclidean space (in particular, for sets of matrices endowed with the Hilbert-Schmidt metric) the volumetric information is roughly equivalent to the metric entropy information such as covering and packing numbers (see, e.g., [76]). However, for the Bures geometry the parallels are not so immediate. Consequently, further work is required to answer (even approximately) questions of the type: Given ε > 0, what is the maximal cardinality of a subset of D (or S), every two elements of which are at least ε apart in the Bures metric? 121 Bibliography

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