Geom Dedicata https://doi.org/10.1007/s10711-018-0352-3 ORIGINAL PAPER

L ∞ Loomis–Whitney inequalities Songjun Lv1

Received: 4 August 2017 / Accepted: 12 April 2018 © Springer Science+Business Media B.V., part of Springer Nature 2018

Abstract By proving a “weighted” reverse affine isoperimetric inequality and its dual, we establish a sharp L ∞ Loomis–Whitney inequality and its dual both associated with even isotropic measures. This complements the recently-discovered family of L p Loomis– Whitney inequalities and their dual inequalities for p ≥ 1. Keywords Reverse affine isoperimetric inequality · Loomis–Whitney inequality · Isotropic measure · L ∞ zonoid · Weighted np -balls Mathematics Subject Classification (2000) 52A40

1 Introduction The classical Loomis–Whitney inequality [18] states that if Pe⊥ K is the orthogonal projection i

of any compact set K onto the hyperplane ei⊥ through the origin perpendicular to the unit vector ei , then n |K |n−1 ≤ i=1 |Pe⊥ K |,

(1.1)

i

n forms the standard basis of Rn , and | · | denotes the Lebesgue measure of where {ei }i=1 measurable sets of the appropriate dimension. The Loomis–Whitney inequality is sharp when the set K is a cube. It is usually viewed as an n-parameter isoperimetric inequality, and in fact the classical isoperimetric inequality (without the sharp constant) in Rn can be easily derived from it. Over decades, the Loomis–Whitney inequality has been generalized and found extensive applications in other mathematical areas, see, e.g., [4,8,10–13,30].

Research supported partly by NSFC under Grant 10801140, CSTC under Grant 2013-JCYJ-A00005, CQNU Foundation under Grant 13XLZ05.

B 1

Songjun Lv [email protected] School of Mathematical Sciences, Chongqing Normal University, Chongqing 401331, China

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One of the most important extensions of the Loomis–Whitney inequality is the following n Ball–Loomis–Whitney inequality. Suppose positive real numbers {ci }i=1 and unit vectors n {u i }i=1 satisfy the John condition (see e.g., [3,14]) m ci u i ⊗ u i = In , i=1

(1.2)

m |Pu ⊥ K |ci , |K |n−1 ≤ i=1

(1.3)

then i

where u ⊗ u is the rank 1 linear operator on Rn that takes x to (x · u)u, and In is the identity map on Rn . The case, where m = n and ci = 1 in (1.2), reduces (1.3) to the classical Loomis–Whitney inequality (1.1). The John condition (1.2) naturally leads to the notion of isotropic measure. A Borel measure μ on the unit sphere, S n−1 , of Rn is said to be isotropic provided  u ⊗ udμ(u) = In . (1.4) S n−1

m c (δ +δ m m n−1 , Evidently, setting μ = 21 i=1 i ui −u i ) in (1.4) with positive {ci }i=1 and {u i }i=1 ⊂ S n−1 one recovers the John condition (1.2). Here δu denotes the delta measure defined on S . A Borel measure on S n−1 is said to be even if it assumes the same mass on antipodal Borel subsets of S n−1 . Two important examples of even isotropic measures on S n−1 are the spherical Lebesgue measure and the cross measures. The basic cross measure can be n (δ + δ expressed by μn = 21 i=1 ei −ei ). A cross measure is just a rotation of the basic cross measure, and it is equally concentrated on O{± e1 , ± e2 . . . , ± en } for some O ∈ O(n). If the Lebesgue measurable set K is a convex body (compact, convex subset of Rn with non-empty interior), then for u ∈ S n−1 , the (n − 1)-dimensional volume, |Pu ⊥ K |, of Pu ⊥ K actually defines the projection body K of K , which is an origin-symmetric convex body determined by

h K (u) = |Pu ⊥ K |, u ∈ S n−1 ,

(1.5)

where h M represents the support function of convex body M defined as h M (x) = max{x · y : y ∈ M} for x ∈ Rn \{0}. By combining the notions presented above, one may anticipate the following “continuous” Ball–Loomis–Whitney inequality. Suppose that μ is an even isotropic measure on S n−1 and K is a convex body in Rn , then  |K |

n−1

≤ exp

S n−1

 log h K (u)dμ(u) .

m c (δ + δ m m n−1 Obviously, if we set μ = 21 i=1 i ui −u i ), with positive {ci }i=1 and {u i }i=1 ⊂ S satisfying the John condition (1.2), in the above inequality, then in light of (1.5) we obtain the (discrete) Ball–Loomis–Whitney inequality (1.3). Based on this fundamental observation, recently Li and Huang [15] investigated “continuous” Ball–Loomis–Whitney inequalities within the framework of the L p Brunn–Minkowski theory. See, for example, [19–23,25,29] for the recent enormous progress in the L p Brunn– Minkowski theory.

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Let K be a convex body in Rn with the origin in its interior, for p ≥ 1 the L p projection body  p K of K can be defined by  1/ p  1 1− p |u · v|h K (v)d S K (v) , u ∈ S n−1 , h  p K (u) = |B np∗ | p/n S n−1 where B np∗ is the unit ball of np∗ , p ∗ is the Hölder conjugate of p, and S K (·) is the surface area of K (see Sect. 2 for definitions). Note that if p = 1, then from Cauchy’s projection formula (see e.g. [29]), one has  1 h K (u) = |u · v|d S K (v) = |Pu ⊥ K |, u ∈ S n−1 . 2 S n−1 In [15], Li and Huang actually obtained the following L p Loomis–Whitney inequality associated with even isotropic measures on S n−1 for 1 ≤ p < ∞. Suppose that K is a convex body in Rn with the origin in its interior and μ is an even isotropic measure on S n−1 . Then   n− p p |K | ≤ exp log h  p K (u)dμ(u) , (1.6) S n−1

with equality if and only if μ is a cross measure with support {± u 1 , . . . , ± u n and for n 2 = p > 1 there exist positive {αi }i=1 such that   1/ p∗  n n p∗ ≤1 ; K = x ∈ R : i=1 αi |x · u i | n such that and for p = 1 there exists a vector x0 ∈ Rn and positive {αi }i=1 n K = x0 + i=1 αi [− u i , u i ].

Another important isoperimetric type inequality due to Meyer [27] is the so-called dual Loomis–Whitney inequality, which gives a lower bound of |K |n−1 in terms of the product of (n − 1)-dimensional volumes of sections of K with hyperplanes through the origin perpendicular to ei , i = 1, . . . , n. This was also extended to the L p setting very recently in [16], where Li and Huang established a dual L p Loomis–Whitney inequality associated with even isotropic measures on S n−1 . The goal of this paper is to investigate the limiting case p = ∞ of the L p Loomis– Whitney inequality (1.6) and its dual inequality associated with even isotropic measures on S n−1 . The main tool is a “weighted” reverse affine isoperimetric inequality, as well as its dual inequality. Our “weighted” reverse affine isoperimetric inequalities are substantive extensions of volume inequalities for subspaces of L p (see [2,22]) when p → ∞. Just as having been done in [17,22–24,28], we shall use a direct optimal transportation approach, which does not employ the Brascamp–Lieb inequality [1,2,4–7,9,26], to prove the “weighted” reverse affine isoperimetric inequalities. The L ∞ Ball–Loomis–Whitney inequality associated with even isotropic measures on S n−1 can be stated as follows. Theorem 1.1 Let K be an origin-symmetric convex body in Rn and μ an even isotropic measure on S n−1 , then   2n exp log ρ K (u)dμ(u) , (1.7) |K | ≥ n! S n−1

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with equality if and only if μ is a cross measure with support {± u 1 , . . . , ± u n } and there n exist positive numbers {wi }i=1 such that n K = {x ∈ Rn : i=1 |x · u i |wi ≤ 1}. m c (δ + δ m m n−1 When taking μ = 21 i=1 i ui −u i ) in (1.7) with positive {ci }i=1 and {u i }i=1 ⊂ S satisfying the John condition (1.2), we get a discrete version of the L ∞ Loomis–Whitney inequality

2n m ci  ρ (u i ). n! i=1 K This shows that the inequality (1.7) is actually of Loomis–Whitney (alternatively, isoperimetric) type. The dual L ∞ Ball–Loomis–Whitney inequality associated with even isotropic measures on S n−1 , which is the complement to the limiting case p = ∞ of the dual L p Ball–Loomis– Whitney inequality formulated in [16], states the following |K | ≥

Theorem 1.2 Let K be an origin-symmetric convex body in Rn and μ an even isotropic measure on S n−1 , then   log h K (u)dμ(u) , (1.8) |K | ≤ 2n exp S n−1

with equality if and only if μ is a cross measure with support {± u 1 , . . . , ± u n } and there n exist positive numbers {wi }i=1 such that K = {x ∈ Rn : |x · u i |wi ≤ 1, for all i = 1, . . . , n}. m c (δ + δ m m Similarly, if we set μ = 21 i=1 i ui −u i ) in (1.8) with positive {ci }i=1 and {u i }i=1 ⊂ S n−1 satisfying the John condition (1.2), then m |K | ≤ 2n i=1 h cKi (u i ),

which asserts that the dual L ∞ Loomis–Whitney inequality is also of isoperimetric type.

2 Preliminaries The setting for this paper is the n-dimensional Euclidean space, Rn . We shall write x · y for the standard inner product of x, y ∈ Rn . Let B2n and S n−1 denote the standard unit ball and the unit sphere in Rn , respectively. The most fundamental functional on convex body in Rn is volume (Lebesgue measure), denoted by | · |. The notation | · | will also be used to denote the standard Euclidean norm · 2 on Rn , the absolute value of a real number and of the determinant of an n × n matrix, as well as the total mass of a finite Borel measure on the unit sphere S n−1 . Denote by GL(n) the group of invertible linear transformations in Rn and by SL(n) the subgroup of GL(n) whose elements have determinant ± 1. For T ∈ GL(n), let trT and det T denote the trace and determinant of T . Also write T t for the transpose of T ∈ GL(n), and T −t for the transpose of the inverse of T , respectively. The identity in GL(n) will be denoted by In . A convex body in Rn is understood to be a compact, convex subset of Rn with nonempty interior. The support function of a convex body K in Rn is defined by h K (x) = max{x · y : y ∈ K },

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for x ∈ Rn \{0}. Obviously, for A ∈ GL(n), the support function of the image AK = {Ax : x ∈ K } is given by h AK (x) = h K (At x),

(2.1)

for x ∈ Rn . The set of convex bodies in Rn that contain the origin in their interiors will be denoted by n Ko . For K ∈ Kon , the polar body K ∗ of K is defined by K ∗ = {x ∈ Rn : x · y ≤ 1 for all y ∈ K }. From the definition, we see that K ∗∗ = K and that for Q ∈ GL(n), (Q K )∗ = Q −t K ∗ .

(2.2)

The Minkowski functional · K of K ∈ Kon is defined by x K = min{λ > 0 : x ∈ λK }, for x ∈ Rn . It is easily seen that · K = h K ∗ .

(2.3)

Note that for K , L ∈ Kon , K ⊆L

⇔

· L ≤ · K .

(2.4)

Further, by applying the polar coordinate formula, one can obtain the following volume formula for all p > 0:  p 1 e− x K d x. (2.5) |K | = (1 + n/ p) Rn A compact set K in Rn is star-shaped about the origin if the intersection of K with each straight line through the origin is a line segment. Associated with a star-shaped K in Rn is the radial function ρ K : Rn \{0} → R which is defined for x = 0 by ρ K (x) = max{λ ≥ 0 : λx ∈ K }. Obviously, for x = 0 and A ∈ SL(n), ρ AK (x) = ρ K (A−1 x).

(2.6)

If ρ K is positive and continuous, K is called a star body. Throughout, the set of star bodies in Rn will be denoted by Son . It is easily seen that if K ∈ Kon , then ρ K ∗ = 1/ h K , and h K ∗ = 1/ρ K .

(2.7)

If, in addition, K is origin-symmetric then ρ K (u) = min{h K (u)/|u · v| : v ∈ S n−1 }, for all u ∈ S n−1 .

(2.8)

For a convex body K in Rn , the classical surface area measure, S K = S(K , ·), of K can be defined as the unique Borel measure on S n−1 such that for each continuous f : S n−1 → R,   f (u)d S(K , u) = f (ν K (x))d Hn−1 (x), S n−1

∂K

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where ν K : ∂  K → S n−1 is the Gauss map of K , defined on ∂  K , the set of points of ∂ K that have a unique outer normal, and Hn−1 is the (n − 1)-dimensional Hausdorff measure. Note that ν K exists almost everywhere for x ∈ ∂ K with respect to Hn−1 on ∂ K . In the remarkable papers [19,20], Lutwak introduced the notion of L p -surface area measure of a convex body. If K ∈ Kon , then for each real p, the L p -surface area measure S p (K , ·) of K is a Borel measure on the unit sphere S n−1 defined on a Borel set ω ⊂ S n−1 by  (x · ν K (x))1− p d Hn−1 (x). S p (K , ω) = −1 x∈ν K (ω)

Note that when p = 1, the measure S1 (K , ·) is the classical surface area measure of K ∈ Kon . If K , L ∈ Kon , then for real p > 0 define the L p -mixed volume of K and L by (see, e.g., [19,21])  1 p h (u)d S p (K , u). (2.9) V p (K , L) = n S n−1 L For 1 < p < ∞, the L p Minkowski mixed volume inequality states that V p (K , L) ≥ |K |(n− p)/n |L| p/n ,

(2.10)

with equality if and only if K and L are dilates. To facilitate the formulation of our problem for the case p → ∞, we also require the volume-normalized L p mixed volumes introduced by Lutwak et al. [21]. For K , L ∈ Kon , and for each real p > 0, define V p (K , L) =

V p (K , L) |K |

1/ p

=

1 n|K |



 S n−1

h L (u) h K (u)

p

1/ p h K (u)d S(K , u)

,

(2.11)

and for p = ∞ define V ∞ (K , L) = lim V p (K , L) = p→∞

max

u∈suppS(K ,·)

h L (u)/ h K (u),

(2.12)

where suppS(K , ·) is the support of the measure S(K , ·). From Jensen’s inequality we see that, for 0 < p < q ≤ ∞, V p (K , L) ≤ V q (K , L),

(2.13)

with equality if and only if h L / h K is constant on suppS(K , ·). Combining (2.10) with (2.13), we obtain V ∞ (K , L) ≥ |K |−1/n |L|1/n ,

(2.14)

with equality if and only if h L / h K is constant on suppS(K , ·). For K ∈ Kon and p > 1, the volume-normalized L p projection body (with an abuse of notation, we still denote it by  p K ) of K is a convex body defined by (see, e.g., [21])  1 p |u · v| p d S p (K , v), h  p K (u) = |K | S n−1 for p > 1, and h ∞ K (u) = for p = ∞.

123

max

v∈suppS(K ,·)

{|u · v|/ h K (v)},

(2.15)

Geom Dedicata

It was shown in [21] that, for A ∈ GL(n) and for all p ∈ [1, ∞],  p (AK ) = A−t  p K .

(2.16)

We shall denote by ∗p K the polar body of  p K for 1 < p ≤ ∞. If, in addition, K is origin-symmetric, then (cf. (2.8)) ∗∞ K = K and ∞ K = K ∗ . If K , L ∈ e.g., [20])

Son ,

(2.17)

then for real p > 0 define the dual L p mixed volume of K and L by (see, 1 Vˇ− p (K , L) = n



n+ p

S n−1

ρK

−p

(u)ρ L (u)d S(u).

For p > 0, the dual L p Minkowski mixed volume inequality is Vˇ− p (K , L) ≥ |K |(n+ p)/n |L|− p/n ,

(2.18)

with equality if and only if K and L are dilates. It will be helpful to introduce the volume-normalized dual L p mixed volume of star bodies. For p > 0 and K , L ∈ Son , define  1/ p

1/ p   Vˇ− p (K , L) 1 ρ K (u) p

− p (K , L) = = ρ K (u)n d S(u) , V |K | n|K | S n−1 ρ L (u) and for p = ∞ define

−∞ (K , L) = lim V

− p (K , L) = max ρ K (u)/ρ L (u). V p→∞

u∈S n−1

(2.19)

From (2.18) and Jensen’s inequality, it follows that, for 0 < p < ∞,

− p (K , L) ≤ V

−∞ (K , L), |K |1/n |L|−1/n ≤ V

(2.20)

with equality, in either inequality, if and only if K and L are dilates. For p ≥ 1 and K ∈ Son , the L p centroid body, p K , of K is the convex body defined by   n+p 1 p n+ p p |x · y| p d x = |x · v| p ρ K (v)d S(v) = x ∗ K , h p K (x) = p |K | K |K | S n−1 for all x ∈ Rn , where ∗p K is the polar of p K . Define ∞ K , as well as its polar body, ∗ K , by ∞ ∗ K, h ∞ K (x) = lim h p K (x) = max |x · v|ρ K (v) = x ∞

p→∞

v∈S n−1

(2.21)

for all x ∈ Rn . It is easily seen that when K is origin-symmetric, ∞ K = K .

(2.22)

It was shown in [25] that for A ∈ GL(n) and for all p ∈ [1, ∞], p (AK ) = A p K . Recall that a Borel measure μ on S n−1 is isotropic if  u ⊗ udμ(u) = In .

(2.23)

(2.24)

S n−1

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Taking the trace in (2.24) leads to μ(S n−1 ) = n.

(2.25)

It is easily seen that (2.24) is equivalent to  |x · u|2 dμ(u) = |x|2 ,

(2.26)

S n−1

for all x ∈ Rn . Associated with isotropic measures is the following continuous version of the Ball–Barthe inequality proved by Lutwak et al. [22] as a consequence of the Hölder inequality. Lemma 2.1 If μ is an isotropic measure on S n−1 and the function t : S n−1 → (0, ∞) is continuous, then    det t (u)u ⊗ udμ(u) ≥ exp log t (u)dμ(u) , S n−1

S n−1

with equality if and only if t (u 1 ) · · · t (u n ) is constant for linearly independent u 1 , . . . , u n ∈ suppμ. Let ν be a cross measure with support suppν = {± u 1 , ± u 2 , . . . , ± u n } = O{± e1 , ± n be n positive numbers. Then the weighted e2 , . . . , ± en } for some O ∈ O(n), and let {wi }i=1 np -ball formed by ν, B np,w = B np,w (ν), is defined by   1/ p  n |x · wi u i | p ≤1 , (2.27) B np,w = x ∈ Rn : i=1 for 1 ≤ p < ∞ and n B∞,w = {x ∈ Rn : |x · u i |wi ≤ 1, for all i = 1, . . . , n},

(2.28)

for p = ∞. Obviously, if w ≡ 1 on S n−1 , then B np,w reduces to the standard np -ball B np in Rn for all p ∈ [1, ∞]. Moreover, for all p ∈ [1, ∞], direct calculations (cf. [15,16]) lead to n |B np,w | = |B np |i=1 wi−1 ;

(B np,w )∗ |(B np,w )∗ |

= =

(2.29)

B np∗ ,1/w ; n |B np∗ |i=1 wi .

(2.30) (2.31)

Suppose that μ is an even isotropic measure on S n−1 and w : S n−1 → (0, ∞) is an even positive continuous function. For x ∈ Rn , define the weighted L p zonoid Z p,w = Z p,w (μ) by  p p |w(u)u · x| p dμ(x) = x Z ∗ , (2.32) h Z p,w (x) = p,w

S n−1

for p ∈ [1, ∞), and ∗ h Z ∞,w (x) = lim h Z p,w (x) = max |x · u|w(u) = x Z ∞,w ,

p→∞

u∈suppμ

(2.33)

for p = ∞. If we set p = 2 and w ≡ 1 in (2.32), then in view of (2.26) we have μ is isotropic if and only if Z 2,1 (μ) = B2n .

123

(2.34)

Geom Dedicata

It follows from (2.3) that if μ is a cross measure (denote it by ν) with suppν = {± u 1 , . . . , ± u n }, then for all p ∈ [1, ∞], the weighted L p zonoids Z p,w (ν) reduce to the weighted np∗ -balls formed by ν; i.e., Z p,w (ν) = B np∗ ,1/w . Consequently, if we set wi = w(u i ), then from (2.30) and (2.31) we have n wi . |Z ∞,w (ν)| = |B1n |i=1

(2.35)

∗ n = B∞,w and then At the same time, we have Z ∞,w ∗ n n |Z ∞,w (ν)| = |B∞ |i=1 wi−1 .

(2.36)

3 An L ∞ Loomis–Whitney inequality associated with even isotropic measures To establish the L ∞ Loomis–Whitney inequality for even isotropic measures on S n−1 , the following lemma will be needed. The proof is basically similar to that of [22, Lemma 3.1]. For completeness, we present the proof below. Lemma 3.1 Let w : S n−1 → (0, ∞) be an even continuous positive function and let μ be an even Borel measure on S n−1 . If f ∈ L 1 (S n−1 , μ), then        u f (u)w(u)dμ(u) ≤ | f (u)|dμ(u). (3.1)  S n−1

Z ∞,w

Proof For f ∈ L 1 (S n−1 , μ), define f =

S n−1



o

S n−1

and denote

u f (u)w(u)dμ(u),

(3.2)

 | f : μ|1 =

S n−1

| f (u)|dμ(u).

(3.3)

Evidently, for real λ > 0 (λ f )o = λ f o . Define M∞ as the closure of the set { f o ∈ Rn : | f : μ|1 ≤ 1}, that is,   M∞ = cl f o ∈ Rn : | f : μ|1 ≤ 1 .

(3.4)

(3.5)

Then M∞ is a convex body in Rn . By (3.4) and the fact that the Minkowski functional is homogeneous of degree 1, we see that once the desired inequality is confirmed for f = f 0 , then it must hold for f = λ f 0 for all λ > 0. Hence, it suffices to show (3.1) for the special case where | f : μ|1 = 1. From (3.5), (3.2), and (2.33), it follows that h M∞ (u) = ≤

sup |u · f o |

| f :μ|1 ≤1

  sup 

| f :μ|1 ≤1

S n−1

  (u · v) f (v)w(v)dμ(v)

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≤

sup |u · v|w(v)

v∈suppμ

= h Z ∞,w (u). Thus, M∞ ⊆ Z ∞,w . This, together with (2.4), yields the desired inequality (3.1).

 

The following “weighted” reverse isoperimetric inequality for even isotropic measures, which is an extension of a result due to Lutwak et al. [22] (see [2,5] for discrete measures), plays an important role in the proof of the L ∞ Loomis–Whitney inequality. Theorem 3.2 Let w : S n−1 → (0, ∞) be an even continuous positive function and let μ be an even isotropic measure on S n−1 . Then   2n |Z ∞,w | ≥ log w(u)dμ(u) , (3.6) exp n! S n−1 with equality if and only if μ is a cross measure. Proof Define the strictly increasing function φ : R → R by  φ(t)  t 3 −s 2 e ds = e−|s| ds. 2 −∞ −∞ Differentiating both sides with respect to t gives

3 −|φ(t)|  −t 2 = φ (t). e e 2 Thus, φ  > 0 and for all t ∈ R, − t 2 = log Define T : Rn → Rn by



3 − |φ(t)| + log φ  (t). 2

 Tx =

S n−1

uφ(x · u)w(u)dμ(u),

for each x ∈ Rn . Now, (3.8) and Lemma 3.1 show that for each x ∈ Rn ,  |φ(x · u)|dμ(u). T x Z ∞,w ≤ S n−1

Note that the differential of T is given by  u ⊗ uφ  (x · u)w(u)dμ(u), dT (x) = S n−1

for each x ∈

Rn .

Thus v · dT (x)v =

123

(3.7)

 S n−1

|u · v|2 φ  (x · u)w(u)dμ(u).

(3.8)

(3.9)

Geom Dedicata

Since φ  > 0, w > 0, and μ is not concentrated on a great sub-sphere of S n−1 , we see that dT (x) is positive definite for each x ∈ Rn . Hence, by the mean value theorem, we conclude that T : Rn → Rn is injective. For the sake of brevity, below we denote  −1 c = exp log w(u)dμ(u) . S n−1

From (2.26), (3.7) with t = x · u, Lemma 2.1, (3.9), (2.5) with p = 1, we have n  1 2 = e−|x| d x 2 Rn     = exp − |x · u|2 dμ(u) d x Rn S n−1 

  3  exp log = − |φ(x · u)| + log φ (x · u)dμ(u) d x 2 Rn S n−1    n    3 exp − |φ(x · u)|dμ(u) exp log φ  (x · u)w(u)dμ(u) = c 2 Rn S n−1    n  3 exp − |φ(x · u)|dμ(u) |dT (x)|d x ≤ c 2 Rn S n−1 n    3 ≤ c exp − T x Z ∞,w |dT (x)|d x 2 Rn n    3 ≤ c exp − y Z ∞,w dy n 2 R n 3 = c n!|Z ∞,w |. 2 This proves the desired inequality (3.6). If μ is a cross measure with support {± u 1 , . . . , ± u n }, then from (2.35) we see that the inequality (3.6) becomes an equality. Conversely, assume that equality holds in (3.6). Since there exist linearly independent {u 1 , . . . , u n } ⊂ suppμ and μ is even, we have {± u 1 , . . . , ± u n } ⊂ suppμ. We argue by contradiction and assume that a unit vector v ∈ suppμ exists such that v ∈ / {± u 1 , . . . , ± u n }. Write v = λ1 u 1 + · · · + λn u n . At least one coefficient, say λ1 , is not zero. Observing that {u 1 , u 2 , . . . , u n } ⊂ suppμ and {v, u 2 , . . . , u n } ⊂ suppμ are both linearly independent sets, the equality condition of the Ball–Barthe inequality implies that φ  (x · u 1 )w(u 1 )φ  (x · u 2 )w(u 2 ) · · · φ  (x · u n )w(u n ) = φ  (x · v)w(v)φ  (x · u 2 )w(u 2 ) · · · φ  (x · u n )w(u n ), for all x ∈ Rn . This together with φ  > 0, w > 0 gives that φ  (x · u 1 )w(u 1 ) = φ  (x · v)w(v), for all x ∈ Rn . Differentiating this on both sides with respect to x yields that φ  (x · u 1 )w(u 1 )u 1 = φ  (x · v)w(v)v,

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Geom Dedicata

for all x ∈ Rn . Since φ  is not constant, there exists an x ∈ Rn such that φ  (x · u 1 ) = 0. The desired contradiction now follows from the fact that w > 0 on S n−1 . Therefore, we must have v = ±u 1 and suppμ = {± u 1 , . . . , ± u n }. Hence, for all x ∈ Rn , we must have n μ({±u i })|x · u i |2 = |x|2 . i=1

Substituting x = u j , we get n μ({±u i })|u j · u i |2 = 1. i=1

This shows that μ({±u j }) ≤ 1. j = 1, 2, . . . , n; and from (3.10) that μ is a cross measure.

But  nj=1 μ({±u j }) we see that |u j · u i |

(3.10)

= n, hence μ({±u j }) = 1 for all = 0 for j = i. Hence, we conclude  

We are now in a position to establish the L ∞ Loomis–Whitney inequality associated with even isotropic measures. Theorem 3.3 If K is a convex body in Rn with the origin in its interior, and μ is an even isotropic measure on S n−1 , then  −1 2n exp log h ∞ K (u)dμ(u) , (3.11) |K | ≥ n! S n−1 with equality if and only if μ is a cross measure and K is a weighted n1 -ball formed by μ. Proof From (2.14), (2.12), (2.33), and (2.15), we have |K | ≥ |Z ∞,w |V ∞ (K , Z ∞,w )−n  −n = |Z ∞,w | max h Z ∞,w (u)/h K (u) u∈suppS(K ,·)

 = |Z ∞,w |

max

−n max |u · v|w(v)/h K (u)

u∈suppS(K ,·) v∈suppμ

 = |Z ∞,w |

max

v∈suppμ u∈suppS(K ,·)

 = |Z ∞,w |

max

−n |u · v|/h K (u)w(v)

−n max h ∞ K (v)w(v) .

v∈suppμ

For each v ∈ suppμ, letting w(v) = h ∞ K (v)−1 ,

(3.12)

|K | ≥ |Z ∞,w |.

(3.13)

we conclude that

Combining (3.6) with (3.13), we arrive at    2n log h ∞ K (u)dμ(u) . |K | ≥ exp − n! S n−1 That proves the desired inequality. If equality in (3.11) holds, then the equality conditions of (3.6) imply that μ is a cross n measure, and then Z ∞,w is a weighted n1 -ball formed by μ, which is exactly B1,1/w . Meanwhile, the equality conditions of (2.14) yield that K and Z ∞,w are dilates. Therefore, K is a n dilation of B1,1/w , which is still a weighted n1 -ball formed by μ.

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Geom Dedicata

Conversely, we shall show that if μ is a cross measure with suppμ = {± u 1 , . . . , ± u n }, n and K is a weighted n1 -ball formed by μ, that is, there are positive numbers {wi }i=1 such that n K = {x ∈ Rn : i=1 |x · u i |wi ≤ 1},

(3.14)

then equality holds in (3.11). Under these assumptions, it suffices to show that K and Z ∞,w are dilates. Observe that K = O A−1 B1n ,

(3.15)

where O ∈ O(n) is an orthogonal transform such that u i = Oei for all i = 1, . . . , n; and A = diag{w1 , . . . , wn } is a diagonal matrix. From (3.12), (3.15), (2.16), and (2.1), we see that −1 w(u i ) = h −1 ∞ K (u i ) = h 

∞ (O A

−1 B n ) 1

(u i ) = h −1 O At ∞ B n (u i ) 1

−1 −1 −1 = h −1 ∞ B n (AO u i ) = h ∞ B n (Aei ) = h ∞ B n (wi ei ) 1

1

1

−1 = h −1 ∞ B n (ei )wi , 1

for each i = 1, . . . , n. This, together with the fact that h −1 ∞ B n (ei ) is a positive constant 1

(denoted by c0 ), shows that w(u i ) = c0 wi−1 , for each i = 1, . . . , n. Recalling that μ is a cross measure, we obtain from (3.14) that n n Z ∞,w = B1,1/w = {x ∈ Rn : i=1 |x · u i |w(u i )−1 ≤ 1} n |x · u i |wi ≤ c0 } = {x ∈ Rn : i=1

= c0 K . This proves that K and Z ∞,w are dilates and, thus, shows that there is equality in (3.11).   If K is in addition origin-symmetric in Theorem 3.3, then from (2.17) and (2.7), we conclude the following even version of the L ∞ Loomis–Whitney inequality formulated as Theorem 1.1. Corollary 3.4 Let K be an origin-symmetric convex body in Rn and μ an even isotropic measure on S n−1 , then   2n log ρ K (u)dμ(u) , |K | ≥ exp n! S n−1 with equality if and only if μ is a cross measure with support {± u 1 , . . . , ± u n } and K is a weighted n1 -ball formed by μ. If we replace (3.12) by w(v) ≡ 1, then the above argument yields the following isoperimetric type inequality (cf. [2,22]). Theorem 3.5 If K is a convex body in Rn with the origin in its interior, and μ is an even isotropic measure on S n−1 , then

−n 2n , (3.16) max h  K (v) |K | ≥ n! v∈suppμ ∞ with equality if and only if μ is a cross measure and K is a dilation of O B1n for some O ∈ O(n).

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Geom Dedicata

If K is in addition origin-symmetric in Theorem 3.5, then we immediately get Corollary 3.6 If K is an origin-symmetric convex body in Rn , and μ is an even isotropic measure on S n−1 , then

n 2n min ρ K (v) , |K | ≥ n! v∈suppμ with equality if and only if μ is a cross measure and K is a dilation of O B1n for some O ∈ O(n).

4 Dual L ∞ Loomis–Whitney inequality associated with even isotropic measures We now consider the dual L ∞ Loomis–Whitney inequality associated with even isotropic measures on S n−1 , for the proof of which a dual “weighted” reverse isoperimetric inequality is of prime importance. To establish such a dual “weighted” inequality, the next lemma which is a special case of Lemma 4.1 from [22] will be required. Lemma 4.1 Suppose that μ is an even Borel measure on S n−1 . If f ∈ L 2 (S n−1 , μ), then 2       u f (u)dμ(u) ≤ f (u)2 dμ(u).   S n−1

Z 2,1

S n−1

Theorem 4.2 Let w : S n−1 → (0, ∞) be an even continuous positive function and let μ be an even isotropic measure on S n−1 . Then  −1 ∗ | ≤ 2n exp log w(u)dμ(u) , (4.1) |Z ∞,w S n−1

with equality if and only if μ is a cross measure. Proof For every u ∈ suppμ, define the strictly increasing function φu : (−w(u)−1 , w(u)−1 ) → R by  w(u)t  φu (t) 3 2 1[−1,1] (τ )dτ = e−τ dτ. 2 −1 −∞ Then φu > 0, and



3 2 w(u)1[−1,1] (w(u)t) = e−φu (t) φu (t), 2

(4.2)

for all t ∈ (−w(u)−1 , w(u)−1 ). Observe that φu is an even function that is strictly decreasing on the interval (−w(u)−1 , 0) and strictly increasing on the interval (0, w(u)−1 ). From (2.33) we have   ∗ int Z ∞,w = x ∈ Rn : max |x · u|w(u) < 1 . (4.3) u∈suppμ

Therefore, for each x ∈

∗ , int Z ∞,w



exp suppμ

123

 log 1[−1,1] (x · w(u)u)dμ(u) = 1.

(4.4)

Geom Dedicata ∗ Define T : int Z ∞,w → Rn by



Tx =

S n−1

uφu (x · u)dμ(u).

(4.5)

It now follows from (4.5), (2.34), and Lemma 4.1 with f (u) = φu (x · u) that  φu (x · u)2 dμ(u). |T x|2 ≤

(4.6)

S n−1

Meanwhile, (4.5) gives  dT (x) =

S n−1

u ⊗ uφu (x · u)dμ(u).

(4.7)

Since φu > 0, w > 0, μ is isotropic and thus not concentrated on a great sub-sphere of S n−1 , ∗ the matrix dT (x) is positive definite for each x ∈ int Z ∞,w . Therefore, the transformation ∗ T : int Z ∞,w → Rn is injective. For the sake of brevity, we denote    d = exp − log w(u)dμ(u) . (4.8) S n−1

From (4.4), (4.2), (2.25), (4.8), Lemma 2.1, (4.6), and (2.34), we have    ∗ exp log 1[−1,1] (x · w(u)u)dμ(u) d x |Z ∞,w | = ∗ int Z ∞,w

suppμ



 = =

∗ int Z ∞,w



1  3 n 2

exp

log suppμ



∗ int Z ∞,w



exp

≤ ≤



2

d  3 n 2

d  3 n 2

2

 log w(u)−1 dμ(u)

suppμ

suppμ

≤





  1 −1 −φu (x·u)2  φu (x · u) dμ(u) d x  3  w(u) e

  −φu (x · u)2 dμ(u) exp

× exp d  3 n



∗ int Z ∞,w



Rn

  = 2n exp −

e

suppμ

e

− T x 2Z

− y 2Z

S n−1

 −φu (x · u) dμ(u) |dT (x)|d x 2

exp



∗ int Z ∞,w

suppμ





 log φu (x · u)dμ(u) d x

2,1

2,1

|dT (x)|d x

dy

 log w(u)dμ(u) .

This concludes the proof of the inequality (4.1). If μ is a cross measure with support {± u 1 , . . . , ± u n }, then from (2.36) we see that equality holds in (4.1). Conversely, assume that equality holds in (4.1). Since μ is even and there exist linearly independent {u 1 , . . . , u n } ⊂ suppμ, we have {± u 1 , . . . , ± u n } ⊂ suppμ.

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Geom Dedicata

We argue by contradiction and assume that a vector v ∈ suppμ exists such that v ∈ / {± u 1 , . . . , ± u n }. Write v = λ1 u 1 + · · · + λn u n . At least one coefficient, say λ1 , is not zero. Observing that {u 1 , u 2 , . . . , u n } ⊂ suppμ and {v, u 2 , . . . , u n } ⊂ suppμ are both linearly independent sets, the equality condition of the Ball–Barthe inequality implies that there exists a δ > 0 such that for all fixed x satisfying |x| < δ, it follows that φu (x · u 1 )φu (x · u 2 ) · · · φu (x · u n ) = φu (x · v)φu (x · u 2 ) · · · φu (x · u n ). This together with φu > 0 gives that φu (x · u 1 ) = φu (x · v), whenever |x| < δ. Differentiating this on both sides with respect to x yields that φu (x · u 1 )u 1 = φu (x · v)v, whenever |x| < δ. Since φu is not constant, there exists an x satisfying |x| < δ such that φu (x · u 1 ) = 0. That yields the desired contradiction. Thus, it follows that v = ±u 1 and suppμ = {± u 1 , . . . , ± u n }. That is to say, whenever |x| < δ, we must have n i=1 μ({±u i })|x · u i |2 = |x|2 .

Substituting x = δ1 u j with 0 < δ1 < δ, then we obtain n i=1 μ({±u i })|u j · u i |2 = 1.

(4.9)

Now the same argument as in the proof of Theorem 3.2 shows that μ is a cross measure.   When w ≡ 1, the above inequality was established by Lutwak et al. [22]. For discrete measures, it was proved by Ball [2]. The equality conditions for discrete measures are due to Barthe [5]. We are now in a position to establish the dual L ∞ Loomis–Whitney inequality associated with even isotropic measures. Theorem 4.3 If K is a convex body in Rn with the origin in its interior, and μ is an even isotropic measure on S n−1 , then   log h ∞ K (u)dμ(u) , (4.10) |K | ≤ 2n exp S n−1

with equality if and only if μ is a cross measure and K is a weighted n∞ -ball formed by μ. Proof From (2.20), (2.19), (2.7), (2.33), and (2.21), it follows that ∗ ∗

−∞ (K , Z ∞,w |K | ≤ |Z ∞,w |V )n  n ∗ ∗ = |Z ∞,w | max ρ K (u)/ρ Z ∞,w (u) u∈S n−1  n ∗ = |Z ∞,w | max ρ K (u)h Z ∞,w (u) u∈S n−1  n ∗ = |Z ∞,w | max max |u · v|w(v)ρ K (u) u∈S n−1 v∈suppμ

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Geom Dedicata

n max |u · v|ρ K (u)w(v) v∈suppμ u∈S n−1  n ∗ = |Z ∞,w | max h ∞ K (v)w(v) . ∗ = |Z ∞,w |



max

v∈suppμ

Setting w(v) = h −1 ∞ K (v),

(4.11)

and combining the above estimate with (4.1), we obtain   ∗ | ≤ 2n exp log h ∞ K (u)dμ(u) . |K | ≤ |Z ∞,w S n−1

This proves the desired inequality. If equality holds in (4.10), then the equality conditions of (4.1) imply that μ is a cross ∗ n measure, and then Z ∞,w is a weighted n∞ -ball formed by μ, B∞,w . Meanwhile, the equality ∗ n conditions of (2.20) yield that K and Z ∞,w are dilates. Therefore, K is a dilation of B∞,w , n which is still a weighted ∞ -ball formed by μ. Conversely, we shall show that if μ is a cross measure with suppμ = {± u 1 , . . . , ± u n }, n and K is a weighted n∞ -ball formed by μ, that is, there exist positive numbers {wi }i=1 such that K = {x ∈ Rn : |x · u i |wi ≤ 1 for all i = 1, . . . , n},

(4.12)

then equality holds in (4.10). ∗ It suffices to prove that K and Z ∞,w are dilates. Note that under our assumptions n , K = O A−1 B∞

(4.13)

where O ∈ O(n) is an orthogonal transform such that u i = Oei for all i = 1, . . . , n; and A = diag{w1 , . . . , wn } is a diagonal matrix. From (4.11), (4.13), (2.23) with p = ∞, and (2.1), we see that −1 −1 w(u i ) = h −1 ∞ K (u i ) = h (O A−1 B n ) (u i ) = h O A−1 ∞

n ∞ B∞

∞

(u i )

−1 −1 −1 −t −1 −t = h −1 ∞ B n (A O u i ) = h ∞ B n (A ei ) = h ∞ B n (wi ei ) ∞

= wi h −1 ∞ B n (ei ),

∞

∞

∞

for each i = 1, . . . , n. This, together with the fact that h −1 n (ei ) is a positive constant ∞ B∞ (denoted as c1 ), shows that w(u i ) = c1 wi , for each i = 1, . . . , n. Recalling that μ is a cross measure, we obtain from (4.12) that ∗ n Z ∞,w = B∞,w = {x ∈ Rn : |x · u i |w(u i ) ≤ 1 for all i = 1, . . . , n}

= {x ∈ Rn : |x · u i |wi ≤ c1−1 for all i = 1, . . . , n} = c1−1 K . ∗ are dilates and, thus, yields equality in (4.10). This proves that K and Z ∞,w

 

If K in Theorem 4.3 is origin-symmetric, then from (2.22) we obtain the following corollary, which is an even version of the so-called dual L ∞ Loomis–Whitney inequality (Theorem 1.2).

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Geom Dedicata

Corollary 4.4 Let K be an origin-symmetric convex body in Rn . If μ is an even isotropic measure on S n−1 , then   log h K (u)dμ(u) , |K | ≤ 2n exp S n−1

with equality if and only if μ is a cross measure with support {± u 1 , . . . , ± u n } and K is a weighted n∞ -ball formed by μ. If we replace (4.11) by w(v) ≡ 1, then the above argument leads to the following isoperimetric type inequality (cf. [5,22]). Theorem 4.5 If K is a convex body in Rn with the origin in its interior, and μ is an even isotropic measure on S n−1 , then

n |K | ≤ 2 max h ∞ K (v) , (4.14) v∈suppμ

n for some with equality if and only if μ is a cross measure and K is a dilation of O B∞ O ∈ O(n).

If K is in addition origin-symmetric in Theorem 4.5, then we obtain Corollary 4.6 If K is an origin-symmetric convex body in Rn , and μ is an even isotropic measure on S n−1 , then

n |K | ≤ 2 max h K (v) , v∈suppμ

n for some with equality if and only if μ is a cross measure and K is a dilation of O B∞ O ∈ O(n).

Acknowledgements This work was carried out at the Chern Institute of Mathematics (CIM) of Nankai University and the author is grateful for the hospitality and the support provided by the CIM during his stay. He also would like to thank the anonymous referee for his/her very careful reading and helpful comments on improving the presentation of this paper.

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