The Dual Orlicz–Aleksandrov–Fenchel Inequality

mathematics

Article The Dual Orlicz–Aleksandrov–Fenchel Inequality

Chang-Jian Zhao

Department of Mathematics, China Jiliang University, Hangzhou 310018, China; [email protected] or [email protected]

Received: 20 October 2020; Accepted: 4 November 2020; Published: 10 November 2020

Abstract: In this paper, the classical dual mixed volume of star bodies Ve(K1, ··· , Kn) and dual Aleksandrov–Fenchel inequality are extended to the Orlicz space. Under the framework of dual Orlicz-Brunn-Minkowski theory, we put forward a new affine geometric quantity by calculating first order Orlicz variation of the dual mixed volume, and call it Orlicz multiple dual mixed volume. We generalize the fundamental notions and conclusions of the dual mixed volume and dual Aleksandrov-Fenchel inequality to an Orlicz setting. The classical dual Aleksandrov-Fenchel inequality and dual Orlicz-Minkowski inequality are all special cases of the new dual Orlicz-Aleksandrov-Fenchel inequality. The related concepts of Lp-dual multiple mixed volumes and Lp-dual Aleksandrov-Fenchel inequality are first derived here. As an application, the dual Orlicz–Brunn–Minkowski inequality for the Orlicz harmonic addition is also established.

Keywords: dual mixed volume; dual Aleksandrov–Fenchel inequality; Orlicz harmonic radial addition; Orlicz dual mixed volume; Orlicz dual Minkowski inequality; dual Orlicz–Brunn–Minkowski theory

1. Introduction It is well known that vector addition is one of the important operators in convex geometry. As an operation between sets K and L, deﬁned by

K + L = {x + y : x ∈ K, y ∈ L}, it is called Minkowski addition and plays an important role in the convex geometry. During the last few decades, the theory has been extended to Lp–Brunn–Minkowski theory. Lp addition of K and L was introduced by Firey in [1,2], denoted by +p, and deﬁned by

p p p h(K +p L, x) = h(K, x) + h(L, x) , for p ≥ 1, x ∈ Rn and compact convex sets K and L in Rn containing the origin. Here, function h(K, ·) denotes the support function of K. If K is a nonempty closed (not necessarily bounded) convex set in Rn, then h(K, x) = max{x · y : y ∈ K},

n for x ∈ R . A nonempty closed convex set is uniquely determined by its support function. Lp-addition is the fundamental and core content in the Lp–Brunn–Minkowski theory. For recent important results and more information from this theory, we refer to [3–23] and the references therein. In recent years, the study turned to an Orlicz–Brunn–Minkowski theory, initiated by Lutwak, Yang, and Zhang [24,25]. Gardner, Hug, and Weil [26] introduced a corresponding Orlicz addition and established ﬁrst the Orlicz–Minkowski, and Orlicz–Brunn–Minkowski inequalities. The same concepts and inequalities are derived by Xi, Jin and Leng [27] using a new geometric symmetry technique. Other articles on this theory can be found in the literature [28–35].

Mathematics 2020, 8, 2005; doi:10.3390/math8112005 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 2005 2 of 23

The radial addition K+e L of star sets (compact sets that are star-shaped at o and contain o) K and L can be deﬁned by ρ(K+e L, ·) = ρ(K, ·) + ρ(L, ·), where ρ(K, ·) denotes the radial function of star set K. The radial function of star set K is deﬁned by

ρ(K, u) = max{c ≥ 0 : cu ∈ K}, for u ∈ Sn−1. The origin and history of the radial addition can be referred to [36], p. 235. When ρ(K, ·) is positive and continuous, K will be called a star body. Let Sn denote the set of star bodies about the origin in Rn. The radial addition and volume are the core and essence of the classical dual Brunn–Minkowski theory and played an important role in the theory (see, e.g., [20,37–42] for recent important contributions). Lutwak [43] introduced the concept of dual mixed volumes that laid the foundation of the dual Brunn–Minkowski theory. What is particularly important is that this theory plays a very important and key role in solving the Busemann–Petty problem in [38,44–46]. For any p 6= 0, the Lp-radial addition K+e p L deﬁned by (see [47] and [48])

p p p ρ(K+e p L, x) = ρ(K, x) + ρ(L, x) ,

n n for x ∈ R and K, L ∈ S . Obviously, when p = 1, the Lp-radial addition +e p becomes the radial addition +e . The Lp-harmonic radial addition was deﬁned by Lutwak [9]: If K, L are star bodies, the Lp-harmonic radial addition, deﬁned by

−p −p −p ρ(K+b p L, x) = ρ(K, x) + ρ(L, x) , (1)

n for p ≥ 1 and x ∈ R . The Lp-harmonic radial addition of convex bodies was first studied by Firey [1]. The operation of the Lp-harmonic radial addition and Lp-dual Minkowski, Brunn–Minkwski inequalities are the basic concept and inequalities in the Lp-dual Brunn–Minkowski theory. The latest information and important results of this theory can be referred to [32,37,39,40,47–51] and the references therein. For a systematic investigation on the concepts of the addition for convex body and star body, we refer the reader to [26,48,50]. Lp-dual Brunn–Minkowski theory has been extended to dual Orlicz–Brunn–Minkowski theory. The dual Orlicz–Brunn–Minkowski theory has also attracted attention, see [52–57]. The Orlicz harmonic radial addition K+b φ L of two star bodies K and L, defined by (see [57]) ρ(K, u) ρ(L, u) ρ(K+b φ L, u)) = sup λ > 0 : φ + φ ≤ φ(1) , (2) λ λ where u ∈ Sn−1, and φ : (0, ∞) → (0, ∞) is a convex and decreasing function such that φ(0) = ∞, limt→∞ φ(t) = 0 and limt→0 φ(t) = ∞. Let C denote the class of the convex and decreasing functions φ −p with φ(0) = ∞, limt→∞ φ(t) = 0, and limt→0 φ(t) = ∞. Obviously, if φ(t) = t and p ≥ 1, then the Orlicz harmonic radial addition becomes the Lp-harmonic radial addition. The dual Orlicz mixed volume, denoted by Veφ(K, L), defined by

0 φr(1) V(K+b φε · L) − V(K) Veφ(K, L) := lim n ε→0+ ε

1 Z ρ(L, u) = φ ρ(K, u)ndS(u), (3) n Sn−1 ρ(K, u) Mathematics 2020, 8, 2005 3 of 23

where K+b φε · L is the Orlicz harmonic linear combination of K and L (see Section3), and the right 0 −p derivative of a real-valued function φ is denoted by φr. When φ(t) = t and p ≥ 1, the dual Orlicz mixed volume Veφ(K, L) becomes the Lp-dual mixed volume Ve−p(K, L), deﬁned by (see [9])

Z 1 n+p −p Ve−p(K, L) = ρ(K, u) ρ(L, u) dS(u). (4) n Sn−1

n If K1, ... , Kn ∈ S , the dual mixed volume of star bodies K1, ... , Kn, denoted by Ve(K1, ... , Kn), deﬁned by Lutwak (see [43])

1 Z Ve(K1,..., Kn) = ρ(K1, u) ··· ρ(Kn, u)dS(u). (5) n Sn−1

n Lutwak’s dual Aleksandrov–Fenchel inequality is the following: If K1, ... , Kn ∈ S and 1 ≤ r ≤ n, then

r 1/r Ve(K1, ··· , Kn) ≤ ∏ Ve(Ki ..., Ki, Kr+1,..., Kn) , i=1 with equality if and only if K1,..., Kr are all dilations of each other. As we all know, the dual mixed volume Ve−1(K, L) of star bodies K and L has been extended to the Lp-space. Following this, the Lp-dual mixed volume Ve−p(K, L) has been extended to the Orlicz space and becomes dual Orlicz mixed volume Veφ(K, L). However, the classical dual mixed volume Ve(K1, ... , Kn) has not been extended to the Orlicz space, and this question becomes a difficult research in convex geometry. Why? We all know that the history of geometric research has always followed the order from general convex geometric space to Lp-space, and then from Lp-space to Orlicz space. The dual mixed volume Ve(K1, ... , Kn) has not been extended to Lp-space. In other words, there is nothing in the Lp-space about the dual mixed volume Ve(K1, ... , Kn), which can be used as the basis for our further study. As a result, directly extend it to the Orlicz space. Its difficulty can be imagined. In this paper, our main aim is to generalize direct the classical dual mixed volumes Ve(K1, ... , Kn) and dual Aleksandrov–Fenchel inequality to the Orlicz space without passing through Lp-space. Amazingly, all the corresponding concepts and inequalities of the Lp-space of the dual mixed volume Ve(K1, ... , Kn) are all derived, which subverts the order of historical research on the issue, directly deriving the results of Orlicz space, saving a lot of time and resources. This is also unimaginable. Under the framework of dual Orlicz–Brunn–Minkowski theory, we introduce the affine geometric quantity by calculating the first order Orlicz variation of the dual mixed volumes, and call it Orlicz multiple dual mixed volumes, denoted by Veφ(K1, ... , Kn, Ln), which involves (n + 1) star bodies in n R . The fundamental notions and conclusions of the dual mixed volume Ve(K1, ... , Kn) and the dual Minkowski, and Aleksandrov–Fenchel inequalities are extended to an Orlicz setting. The related concepts and conclusions of Lp-multiple dual mixed volume Ve−p(K1, ... , Kn, Ln) and Lp-dual Aleksandrov–Fenchel inequality are first derived here. The new dual Orlicz–Aleksandrov–Fenchel inequality in special cases yields the dual Aleksandrov–Fenchel inequality and the Orlicz dual Minkowski inequality for the dual quermassintegrals, respectively. As an application, a new dual Orlicz–Brunn–Minkowski inequality for the Orlicz harmonic radial addition is established, which implies the dual Orlicz–Brunn–Minkowski inequality for the dual quermassintegrals. Complying with the spirit of introduction of Aleksandrov, Fenchel and Jessen’s mixed quermassintegrals, and introduction of Lutwak’s Lp-mixed quermassintegrals, we calculate the first order Orlicz variational of dual mixed volumes. If convex bodies K2, ... , Kn are given, we often use the n n abbreviations = := K2, ... , Kn; = ∈ S := K2, ... , Kn ∈ S and ρ= := ρ(K2, u) ··· ρ(Kn, u). In Section4, we prove that the first order Orlicz variation of the dual mixed volumes can be expressed as:

d 1 Ve(L1+b φε · K1, =) = 0 · Veφ(L1, K1, =), dε ε=0+ φr(1) Mathematics 2020, 8, 2005 4 of 23

n where L1, K1, = ∈ S , φ ∈ C and ε > 0. In the above first order variational equation, we find a new geometric quantity. Based on this, we extract the required geometric quantity, denoted by Veφ(L1, K1, =) and call it Orlicz multiple dual mixed volume of (n + 1) star bodies L1, K1, =, defined by

0 d Veφ(L1, K1, =) := φr(1) · Ve(L1+b φε · K1, =). dε ε=0+

We also prove the new afﬁne geometric quantity Veφ(L1, K1, =) has an integral representation.

Z 1 ρ(K1, u) Veφ(L1, K1, =) = φ ρ(L1, u)ρ=dS(u), (6) n Sn−1 ρ(L1, u)

Obviously, the dual mixed volume Ve(K1, ... , Kn) and dual Orlicz mixed volume Veφ(K, L) are −p all special cases of Veφ(L1, K1, =). When φ(t) = t , p ≥ 1, Orlicz multiple dual mixed volume Veφ(L1, K1, =) becomes a new dual mixed volume in Lp-place, denoted by Ve−p(K1, ... , Kn, Ln), call it Lp multiple dual mixed volume. From (6), we have

Z p 1 ρ(L1, u) Ve−p(L1, K1, =) = ρ(L1, u)ρ=dS(u). (7) n Sn−1 ρ(K1, u)

The following harmonic mixed p-quermassintegral We −p,i(K, L) is a special case of Ve−p(L1, K1, =), deﬁned by (see Section2)

Z 1 n−i+p −p We −p,i(K, L) = ρ(K, u) ρ(L, u) dS(u). (8) n Sn−1

In Section5, we establish the following dual Orlicz–Aleksandrov–Fenchel inequality for the Orlicz multiple dual mixed volumes. n The dual Orlicz–Aleksandrov–Fenchel inequality If L1, K1, = ∈ S , φ ∈ C and 1 ≤ r ≤ n, then

r 1/r ! ∏i=1 Ve(Ki ..., Ki, Kr+1,..., Kn) Veφ(L1, K1, =) ≥ Ve(L1, =) · φ . (9) Ve(L1, =)

If φ is strictly convex, equality holds if and only if L1, K1,..., Kr are all dilations of each other. n Obviously, Lutwak’s dual Aleksandrov–Fenchel inequality is a special case of (9). If K1, = ∈ S and 1 ≤ r ≤ n, then r 1/r Ve(K1, ··· , Kn) ≤ ∏ Ve(Ki ..., Ki, Kr+1,..., Kn) , (10) i=1 −p with equality if and only if K1, ... , Kr are all dilations of each other. When φ(t) = t , p ≥ 1, the dual Orlicz–Aleksandrov–Fenchel inequality (9) becomes the following Lp-dual Aleksandrov–Fenchel inequality. n The Lp-dual Aleksandrov–Fenchel inequality If L1, K1, = ∈ S , p ≥ 1 and 1 ≤ r ≤ n, then

r p+1 −p/r Ve−p(L1, K1, =) ≥ Ve(L1, =) · ∏ Ve(Ki ..., Ki, Kr+1,..., Kn) . (11) i=1

If φ is strictly convex, equality holds if and only if L1, K1,..., Kr are all dilations of each other. The following dual Orlicz–Minkowski inequality (see [57]) is a special case of the dual Orlicz–Aleksandrov–Fenchel inequality (9). If K, L ∈ Sn and φ ∈ C, then ! V(L) 1/n Veφ(K, L) ≥ V(K) · φ . (12) V(K) Mathematics 2020, 8, 2005 5 of 23

If φ is strictly convex, equality holds if and only if K and L are dilates. In Section5, we show also the Orlicz–Aleksandrov–Fenchel inequality (9) in special case yields also the following result. If K, L ∈ Sn, 0 ≤ i < n and φ ∈ C, then

 !1/(n−i) We i(L) We φ,i(K, L) ≥ We i(K)φ   . (13) We i(K)

If φ is strictly convex, equality holds if and only if K and L are dilates. Here, We i(K) is the usually dual quermassintegral of K, and We φ,i(K, L) is the Orlicz dual mixed quermassintegral of K and L, deﬁned by (see Section4) Z 1 ρ(L, u) n−i We φ,i(K, L) = φ ρ(K, u) dS(u). (14) n Sn−1 ρ(K, u) In Section6, we establish the following dual Orlicz Brunn–Minkowski type inequality. n If L1, K1, = ∈ S and φ ∈ C, then ! ! Ve(K , =) Ve(L , =) φ(1) ≥ φ 1 + φ 1 . (15) Ve(K1+b φ L1, =) Ve(K1+b φ L1, =)

If φ is strictly convex, equality holds if and only if L1, K1, = are all dilations of each other. A special case of (15) is the following inequality.

 !1/n  !1/n V(K ) ··· V(K ) V(L )V(K ) ··· V(K ) φ(1) ≥ φ 1 n + φ 1 2 n . (16)  n   n  Ve(K1+b φ L1, =) Ve(K1+b φ L1, =)

If φ is strictly convex, equality holds if and only if L1, K1, ... , Kn are all dilations of each other. Putting K1 = K, L1 = L and K2 = ··· = Kn = K1+b φ L1 in (16), it follows the Orlicz dual Brunn–Minkowski inequality established in [57]. In Section6, we show also the dual Orlicz–Brunn–Minkowski inequality (16) in a special case yields the following result. If K, L ∈ Sn, φ ∈ C and 0 ≤ i < n − 1, then

 !1/(n−i)  !1/(n−i) We i(K) We i(L) φ(1) ≥ φ   + φ   . (17) We i(K+b φ L) We i(K+b φ L)

If φ is strictly convex, equality holds if and only if K and L are dilates.

2. Preliminaries

The setting for this paper is n-dimensional Euclidean space Rn. A body in Rn is a compact set equal to the closure of its interior. For a compact set K ⊂ Rn, we write V(K) for the (n-dimensional) Lebesgue measure of K and call this the volume of K. The unit ball in Rn and its surface are denoted by B and Sn−1, respectively. Let Kn denote the class of nonempty compact convex subsets containing the origin in their interiors in Rn. Associated with a compact subset K of Rn, which is star-shaped with respect to the origin and contains the origin, its radial function is ρ(K, ·) : Sn−1 → [0, ∞), deﬁned by

ρ(K, u) = max{λ ≥ 0 : λu ∈ K}.

Two star bodies K and L are dilates if ρ(K, u)/ρ(L, u) is independent of u ∈ Sn−1. If λ > 0, then

ρ(λK, u) = λρ(K, u). Mathematics 2020, 8, 2005 6 of 23

From the deﬁnition of the radial function, it follows immediately that for A ∈ GL(n) the radial function of the image AK = {Ay : y ∈ K} of K is given by (see e.g., [36])

ρ(AK, u) = ρ(K, A−1u), for all u ∈ Sn−1. Namely, the radial function is homogeneous of degree −1. Let δ˜ denote the radial Hausdorff metric, as follows, if K, L ∈ Sn, then (see e.g., [58])

δ˜(K, L) = |ρ(K, u) − ρ(L, u)|∞.

2.1. Dual Mixed Volumes The polar coordinate formula for volume of a compact set K is

1 Z V(K) = ρ(K, u)ndS(u). (18) n Sn−1

The ﬁrst dual mixed volume, Ve1(K, L), deﬁned by

1 V(K+e ε · L) − V(K) Ve1(K, L) = lim , n ε→0+ ε where K, L ∈ Sn. The integral representation for ﬁrst dual mixed volume is proved: For K, L ∈ Sn,

Z 1 n−1 Ve1(K, L) = ρ(K, u) ρ(L, u)dS(u). (19) n Sn−1

The Minkowski inequality for ﬁrst dual mixed volume is the following: If K, L ∈ Sn, then

n n−1 Ve1(K, L) ≤ V(K) V(L),

n with equality if and only if K and L are dilates. (see [45]) If K1, ... , Kn ∈ S , the dual mixed volume Ve(K1,..., Kn) is deﬁned by (see [43])

1 Z Ve(K1,..., Kn) = ρ(K1, u) ··· ρ(Kn, u)dS(u). (20) n Sn−1

If K1 = ··· = Kn−i = K, Kn−i+1 = ··· = Kn = L, the dual mixed volume Ve(K1, ... , Kn) is written as Vei(K, L). If L = B, the dual mixed volume Vei(K, L) = Vei(K, B) is written as We i(K) and called dual quermassintegral of K. For K ∈ Sn and 0 ≤ i < n,

Z 1 n−i We i(K) = ρ(K, u) dS(u). (21) n Sn−1

If K1 = ··· = Kn−i−1 = K, Kn−i = ··· = Kn−1 = B and Kn = L, the dual mixed volume Ve(K,..., K, B,..., B, L) is written as We i(K, L) and called dual mixed quermassintegral of K and L. | {z } | {z } n−i−1 i For K, L ∈ Sn and 0 ≤ i < n, it is easy that ([33])

Z We i(K+e ε · L) − We i(K) 1 n−i−1 We i(K, L) = lim = ρ(K, u) ρ(L, u)dS(u). (22) ε→0+ ε n Sn−1

The fundamental inequality for dual mixed quermassintegral stated that: If K, L ∈ Sn and 0 ≤ i < n, then n−i n−1−i We i(K, L) ≤ We i(K) We i(L), (23) Mathematics 2020, 8, 2005 7 of 23 with equality if and only if K and L are dilates. The Brunn–Minkowski inequality for dual quermassintegral is the following: If K, L ∈ Sn and 0 ≤ i < n, then

1/(n−i) 1/(n−i) 1/(n−i) We i(K+e L) ≤ We i(K) + We i(L) , (24) with equality if and only if K and L are dilates.

2.2. Lp-dual Mixed Volume

The dual mixed volume Ve−1(K, L) of star bodies K and L is deﬁned by ([9])

V(K) − V(K+b ε · L) Ve−1(K, L) = lim , (25) ε→0+ ε where +b is the harmonic addition. The following is a integral representation for the dual mixed volume Ve−1(K, L): Z 1 n+1 −1 Ve−1(K, L) = ρ(K, u) ρ(L, u) dS(u). (26) n Sn−1 The dual Minkowski inequality for the dual mixed volume states that

n n+1 −1 Ve−1(K, L) ≥ V(K) V(L) , (27) with equality if and only if K and L are dilates. (see ([42])) The dual Brunn–Minkowski inequality for the harmonic addition states that

−1/n −1/n −1/n V(K+b L) ≥ V(K) + V(L) , (28) with equality if and only if K and L are dilates (This inequality is due to Firey [1]). The Lp-dual mixed volume Ve−p(K, L) of K and L is deﬁned by ([9])

p V(K+b pε · L) − V(K) Ve−p(K, L) = − lim , (29) n ε→0+ ε where K, L ∈ Sn and p ≥ 1. n The following is an integral representation for the Lp-dual mixed volume: For K, L ∈ S and p ≥ 1, Z 1 n+p −p Ve−p(K, L) = ρ(K, u) ρ(L, u) dS(u). (30) n Sn−1 n Lp-dual Minkowski and Brunn-Minkowski inequalities were established by Lutwak [9]: If K, L ∈ S and p ≥ 1, then n n+p −p Ve−p(K, L) ≥ V(K) V(L) , (31) with equality if and only if K and L are dilates, and

−p/n −p/n −p/n V(K+b p L) ≥ V(K) + V(L) , (32) with equality if and only if K and L are dilates.

2.3. Mixed p-harmonic Quermassintegral From (1), it is easy to see that if K, L ∈ Sn, 0 ≤ i < n and p ≥ 1, then

Z p We i(K+b pε · L) − We i(L) 1 − lim = ρ(K.u)n−i+pρ(L.u)−pdS(u). (33) n − i ε→0+ ε n Sn−1 Mathematics 2020, 8, 2005 8 of 23

Let K, L ∈ Sn, 0 ≤ i < n and p ≥ 1, the mixed p-harmonic quermassintegral of star K and L, denoted by We −p,i(K, L), deﬁned by (see [59])

Z 1 n−i+p −p We −p,i(K, L) = ρ(K, u) ρ(L, u) dS(u). (34) n Sn−1

Obviously, when K = L, the p-harmonic quermassintegral We −p,i(K, L) becomes the dual quermassintegral We i(K). The Minkowski and Brunn–Minkowski inequalities for the mixed p-harmonic quermassintegral are following (see [59]): If K, L ∈ Sn, 0 ≤ i < n, and p ≥ 1, then

n−i n−i+p −p We −p,i(K, L) ≥ We i(K) We i(L) , (35) with equality if and only if K and L are dilates. If K, L ∈ Sn, 0 ≤ i < n, and p ≥ 1, then

−p/(n−i) −p/(n−i) −p/(n−i) We i(K+b p L) ≥ We i(K) + We i(L) , (36) with equality if and only if K and L are dilates. Inequality (36) is a Brunn–Minkowski type inequality for the p-harmonic addition. For different variants of dual Brunn–Minkowski inequalities, we refer to [16,46,60–65] and the references therein.

3. Orlicz Harmonic Linear Combination

n Throughout the paper, the standard orthonormal basis for R will be {e1, ... , en}. Let Cm, m ∈ N, denote the set of convex function φ : [0, ∞)m → (0, ∞) that are strictly decreasing in each variable and satisfy φ(0) = ∞. When m = 1, we shall write C instead of C1. Orlicz harmonic radial addition is deﬁned below.

n Definition 1. Let m ≥ 2, φ ∈ Cm, Kj ∈ S , and j = 1, ... , m, define the Orlicz harmonic addition of K1,..., Km, denoted by +b φ(K1,..., Km), defined by

ρ(K , x) ρ(K , x) ρ(+ (K ,..., K ), x) = sup λ > 0 : φ 1 ,..., m ≤ φ(1) , (37) b φ 1 m λ λ for x ∈ Rn.

Equivalently, the Orlicz multiple harmonic addition +b φ(K1, ... , Km) can be deﬁned implicitly by ! ρ(K , x) ρ(K , x) φ 1 ,..., m = φ(1), (38) ρ(+b φ(K1,..., Km), x) ρ(+b φ(K1,..., Km), x) for all x ∈ Rn. An important special case is obtained when

m φ(x1,..., xm) = ∑ φ(xj), j=1 for φ(t) ∈ Cm. We then write +b φ(K1, ... , Km) = K1+b φ ··· +b φKm. This means that K1+b φ ··· +b φKm is deﬁned either by

( m ) ρ(Kj, x) ρ(K1+b φ ··· +b φKm, x) = sup λ > 0 : ∑ φ ≤ φ(1) , (39) j=1 λ Mathematics 2020, 8, 2005 9 of 23 for all x ∈ Rn, or by the corresponding special case of (37). From (39), it follows easy that

m ρ(Kj, x) ∑ φ = φ(1), j=1 λ if and only if λ = ρ(K1+b φ ··· +b φKm, x). (40) Next, deﬁne a new Orlicz dual harmonic linear combination on the case m = 2.

Deﬁnition 2. The Orlicz dual harmonic linear combination is denoted +b φ(K, L, α, β), deﬁned by ! ! ρ(K, x) ρ(L, x) αφ + βφ = φ(1), (41) ρ(+b φ(K, L, α, β), x) ρ(+b φ(K, L, α, β), x) for K, L ∈ Sn, x ∈ Rn, and α, β ≥ 0 (not both zero).

−p When φ(t) = t and p ≥ 1, then Orlicz harmonic linear combination +b φ(K, L, α, β) changes to the Lp-harmonic linear combination α · K+b pβ · L. We shall write K+b φε · L instead of +b φ(K, L, 1, ε), for ε ≥ 0 and assume throughout that this is deﬁned by (41), where α = 1, β = ε, and φ ∈ C. It is easy that +b φ(K, L, 1, 1) = K+b φ L.

4. Orlicz Multiple Dual Mixed Volumes Let us introduce the Orlicz multiple dual mixed volumes.

n Deﬁnition 3. For L1, K1, = ∈ S and φ ∈ C, the Orlicz multiple dual mixed volume of L1, K1, =, denoted by Veφ(L1, K1, =), deﬁned by

Z 1 ρ(K1, u) Veφ(L1, K1, =) =: φ ρ(L1, u) · ρ=dS(u). n Sn−1 ρ(L1, u)

n Lemma 1 ([57]). If K1, L1 ∈ S and φ ∈ C, then

L1+b φε · K1 → L1 (42) as ε → 0+.

n Lemma 2. If L1, K1, = ∈ S and φ ∈ C, then

Z d 1 ρ(K1, u) Ve(L1+φε · K1, =) = φ ρ(L1, u)ρ=dS(u). (43) b 0 n−1 dε ε=0+ nφr(1) S ρ(L1, u)

n n−1 Proof. Suppose ε > 0, K1, L1 ∈ S , and u ∈ S , let

ρε = ρ(L1+b φε · K1, u). Mathematics 2020, 8, 2005 10 of 23

From Lemma 1, and noting that φ is a continuous function, we obtain

(ρε − ρ(L1, u))ρ= ρε ρε − ρ(L1, u) lim = ρ= lim · ε→0+ ε ε→0+ ε ρε ρ(K1, u) = ρ= lim ρε · φ ε→0+ ρε ρ(K , u) 1 − φ−1 φ(1) − εφ 1 ρε × . ρ(K , u) φ(1) − φ(1) − εφ 1 ρε

Noting that y → 1+ as ε → 0+, we have (ρε − ρ(L1, u))ρ= ρ(K1, u) 1 − y lim = ρ(L1, u)ρ=φ lim ε→0+ ε ρ(L1, u) y→1+ φ(1) − φ(y) 1 ρ(K1, u) = 0 φ ρ(L1, u)ρ=, (44) φr(1) ρ(L1, u) where ρ(K , u) y = φ−1 φ(1) − εφ 1 . ρε The Equation (43) follows immediately from (20) with (44).

Second proof Since

− dφ 1(y) ρ(K , u) ρ(L , u) φ 1 dρ 1 dy ρε ε = , (45) dε ρ(K , u) 2 ρ(K , u)ρ(L , u) dφ−1(y) dφ(z) −1 ( ) − 1 + · 1 1 φ φ 1 εφ ε 2 ρε ρε dy dz where ρ(K , u) y = φ(1) − εφ 1 , ρε and ρ(K , u) z = 1 . ρε Hence, (ρε − ρ(L1, u))ρ= ρε − ρ(L1, u) lim = ρ= lim ε→0+ ε ε→0+ ε

dρε = ρ= lim . (46) ε→0+ dε On the other hand,

dφ−1(y) φ−1(φ(1) + 4y) − 1 lim = lim ε→0+ dy 4y→0+ 4y

ω − 1 = lim ω→1+ φ(ω) − φ(1) 1 = 0 , (47) φr(1) where ω = φ−1(φ(1) + 4y). Mathematics 2020, 8, 2005 11 of 23

From (45), (46), (47), and Lemma 1, we obtain

(ρ − ρ(L , u))ρ 1 ρ(K , u) lim ε 1 = = · φ 1 ρ(L , u)ρ . (48) + 0 1 = ε→0 ε φr(1) ρ(L1, u)

From (20) and (48), the Equation (43) follows easy. n For any L1, K1, = ∈ S , and φ ∈ C, the integral on the right-hand side of (43) denoting by Veφ(L1, K1, =), and hence this new Orlicz multiple dual mixed volume Veφ(L1, K1, =) has been born.

n Lemma 3. If L1, K1, = ∈ S and φ ∈ C, then

0 d Veφ(L1, K1, =) = φr(1) · Ve(L1+b φε · K1, =). (49) dε ε=0+

Proof. This yields immediately from the Deﬁnition 3 and the variational formula of volume (43).

Lemma 4. Let K, L ∈ Sn and φ ∈ C, then

Ve1(K, K+b φε · L) − V(K) 1 V(K+b φε · L) − V(K) lim = lim . (50) ε→0+ ε n ε→0+ ε

Proof. Suppose ε > 0, K, L ∈ Sn, and u ∈ Sn−1, let

ρ¯ε = ρ(K+b φε · L, u).

From (3), (18), (19), and (45), we obtain

Z n−1 n Ve1(K, K+b φε · L) − V(K) 1 ρ(K+b φε · L), u)ρ(K, u) − ρ(L, u) lim = lim dS(u) ε→0+ ε n Sn−1 ε→0+ ε 1 Z dρ¯ = ρ(K, u)n−1 lim ε dS(u) n Sn−1 ε→0+ dε Z 1 ρ(L, u) n = 0 φ ρ(K, u) dS(u) nφr(1) Sn−1 ρ(K, u) 1 V(K+b φε · L) − V(K) = lim . n ε→0+ ε

n Lemma 5. Let L1, K1, = ∈ S , and φ ∈ C, then

Veφ(L1, K1, =) = Veφ(K, L), (51) if K2 = ··· = Kn = K,L1 = K and K1 = L.

Proof. On the one hand, putting K2 = ··· = Kn = K, L1 = K and K1 = L in (49), and noting Lemma 4 and (3), it follows that

0 d Veφ(L1, K1, =) = φr(1) Ve(L1+b φε · K1, =) dε ε=0+

0 Ve1(K, K+b φε · L) − V(K) = φr(1) lim ε→0+ ε 0 φ (1) V(K+b φε · L) − V(K) = r lim n ε→0+ ε Mathematics 2020, 8, 2005 12 of 23

= Veφ(K, L). (52)

On the other hand, let K2 = ··· = Kn = K, L1 = K, and K1 = L, from Deﬁnition 3 and (3), then Z 1 ρ(K1, u) Veφ(L1, K1, λ=) = φ ρ(L1, u)ρ=dS(u) n Sn−1 ρ(L1, u) 1 Z ρ(L, u) = φ ρ(K, u)ndS(u) n Sn−1 ρ(K, u)

= Veφ(K, L). (53) Combining (52) and (53), this shows that

Veφ(L1, K1, =) = Veφ(K, L), if K2 = ··· = Kn = K, L1 = K and K1 = L.

n Lemma 6 ([57]). If Ki, Li ∈ S , and Ki → K,Li → L as i → ∞, then

a · Ki+b φb · Li → a · K+b φb · L, as i → ∞, (54) for all a and b.

n Lemma 7. If L1, K1, =, K, L ∈ S , λ1, ··· , λn ≥ 0, and φ ∈ C, then (1) Veφ(L1, K1, =) ≥ 0. (2) Veφ(K1, K1, =) = φ(1)Ve(K1, =). (3) Veφ(K, K, ··· , K) = φ(1)V(K). (4) Veφ(L1, K1, λi=) = λ2 ··· λnVeφ(L1, K1, =), where λi= denotes λ2K2, ··· , λnKn. (5) Veφ(L1, K1, λ1K+b λ2L, K3 ··· , Kn)

= λ1Vφ(L1, K1, K, K3, ··· , Kn) + λ2Vφ(L1, K1, L, K3, ··· , Kn).

This shows the Orlicz multiple mixed volume Vφ(L1, K1, =) is linear in its back (n − 1) variables. (6) Veφ(L1, K1, =) is continuous.

Proof. From Deﬁnition 3, it immediately gives (1), (2), (3), and (4). From Deﬁnition 3, combining the following fact

ρ(λ1K+e λ2L, ·) = λ1ρ(K, ·) + λ2ρ(L, ·), it yields (5) directly. Suppose Li1 → L1, Kij → Kj as i → ∞ where j = 1, ... , n, combining Deﬁnition 3 and Lemma 6 with the following facts Ve(Li1+b φε · Ki1, =i) → Ve(L1+b φε · K1, =) and Ve(Li1, =i) → Ve(L1, =) as i → ∞, where =i denotes Ki2, ··· , Kin. It yields (6) directly. Mathematics 2020, 8, 2005 13 of 23

Lemma 8. ([57]) Suppose K, L ∈ Sn and ε > 0. If φ ∈ C, then for A ∈ GL(n)

A(K+b φε · L) = AK+b φε · AL. (55)

We easily ﬁnd that Orlicz multiple dual mixed volume Veφ(L1, K1, =) is invariant under simultaneous unimodular centro-afﬁne transformation.

n Lemma 9. If L1, K1, = ∈ S and φ ∈ C, then for A ∈ SL(n),

Veφ(AL1, AK1, A=) = Veφ(L1, K1, =), (56) where A= denotes AK2,..., AKn.

Proof. From (49) and Lemma 8, we have, for A ∈ SL(n),

0 Ve(AL1+b φε · AK1, A=) − Ve(AL1, A=) Veφ(AL1, AK1, A=) = φr(1) lim ε→0+ ε

0 Ve(A(L1+b φε · K1), A=) − Ve(L1, =) = φr(1) lim ε→0+ ε

0 Ve(L1+b φε · K1, =) − Ve(L1, =) = φr(1) lim ε→0+ ε

= Veφ(L1, K1, =).

This completes the proof.

For the convenience of writing, when K1 = ··· = Ki = K, Ki+1 = ··· = Kn = L, Ln = M, the Orlicz multiple dual mixed volume Veφ(K, ··· , K, L, ··· , L, M), with i copies of K, n − i copies of L, and 1 copy of M, will be denoted by Veφ(K [i], L [n − i], M).

Lemma 10. If K, L ∈ Sn and φ ∈ C, and 0 ≤ i < n then

Z 1 ρ(L, u) n−i Veφ(K, L, K [n − i − 1], B [i]) = φ ρ(K, u) dS(u). (57) n Sn−1 ρ(K, u)

Proof. On the one hand, putting L1 = K, K1 = L,K2 = ··· = Kn−i = K, and Kn−i+1 = ··· = Kn = B in (49), from (21), (22), (45), and (47), we obtain for ϕ1, ϕ2 ∈ Φ

0 We i(K, K+b φε · L) − We i(K) Veφ(K, L, K [n − i − 1], B [i]) = φr(1) lim ε→0+ ε n−i−1 n−i 1 0 Z ρ(K+b φε · L)ρ(K, u) − ρ(K, u) = φr(1) lim dS(u) n Sn−1 ε→0+ ε Z 1 0 n−i−1 dρε = φl (1) ρ(K, u) lim dS(u) n Sn−1 ε→0+ ε

1 Z ρ(L, u) = φ ρ(K, u)n−idS(u). (58) n Sn−1 ρ(K, u)

On the other hand, putting L1 = K, K1 = L,K2 = ··· = Kn−i = K, and Kn−i+1 = ··· = Kn = B in Deﬁnition 3, we have

Z 1 ρ(L, u) n−i Veφ(K, L, K [n − i − 1], B [i]) = φ ρ(K, u) dS(u). (59) n Sn−1 ρ(K, u)

Combining (58) and (59), (57) yields easy. Mathematics 2020, 8, 2005 14 of 23

Here, we denote the Orlicz multiple dual mixed volume Veφ(K, L, K [n − i − 1], B [i]) by We φ,i(K, L), and call We φ,i(K, L) as Orlicz dual quermassintegral of star bodies K and L. When i = 0, Orlicz dual quermassintegral We φ,i(K, L) becomes Orlicz dual mixed volume Veφ(K, L).

−p 0 Remark 1. When φ(t) = t , p = 1, and L1 = K1, from (49) and noting that φr(1) = −1, hence

Ve(K1, =) − Ve(K1+b ε · K1, =) Ve(K1, =) = lim . (60) ε→0+ ε

This is very interesting for the usually dual mixed volume of this form.

−p Remark 2. When φ(t) = t , p ≥ 1, write the Orlicz multiple dual mixed volume Veφ(L1, K1, =) as Ve−p(L1, K1, =) and call it the Lp-multiple dual mixed volume, from Deﬁnition 3, it easily yields

Z 1 −p 1+p Ve−p(L1, K1, =) = ρ(K1, u) ρ(L1, u) ρ=dS(u). (61) n Sn−1

−p When ϕ(t) = t and p ≥ 1, from (49), we get the following expression of Lp-multiple dual mixed volume. 1 Ve(L1+b pε · K1, =) − Ve(L1, =) Ve−p(L1, K1, =) = lim . −p ε→0+ ε

When K1 = L and L1 = K2 = ··· = Kn = K, the Orlicz multiple dual mixed volume Veφ(L1, K1, =) becomes the usual dual Orlicz mixed volume Veφ(K, L). Putting L1 = K1 in (61), the Lp multiple dual mixed volume Ve−p(L1, K1, =) becomes the usual dual mixed volume Ve(K1, =). Putting K1 = L and L1 = K2 = ··· = Kn = K in (61), Ve−p(L1, K1, =) becomes the Lp dual mixed volume Ve−p(K, L). Putting K1 = L, L1 = K2 = ··· = Kn−i = K, and Kn−i+1 = ··· = Kn = B in (61), Ve−p(L1, K1, =) becomes the harmonic mixed p-quermassintegral We −p,i(K, L),

Lemma 11. (Jensen’s inequality) Let µ be a probability measure on a space X and g : X → I ⊂ R is a µ-integrable function, where I is a possibly inﬁnite interval. If ψ : I → R is a convex function, then

Z Z ψ(g(x))dµ(x) ≥ ψ g(x)dµ(x) . (62) X X

If ψ is strictly convex, equality holds if and only if g(x) is constant for µ-almost all x ∈ X (see [63]).

5. The Dual Orlicz–Aleksandrov–Fenchel Inequality

n Theorem 1. If L1, K1,..., Kn ∈ S and φ ∈ C, then ! Ve(K1,..., Kn) Veφ(L1, K1,..., Kn) ≥ Ve(L1, K2 ..., Kn)φ . (63) Ve(L1, K2 ..., Kn)

If φ is strictly convex, equality holds if and only if K1 and L1 are dilates.

n n−1 ρ(K1, u)ρ= Proof. For K1, = ∈ S and any u ∈ S , it is not difﬁcult to see that S(u) is a probability nVe(K1, =) measure on Sn−1. Mathematics 2020, 8, 2005 15 of 23

From Deﬁnition 3 and Jensen’s inequality (43) and (20), it follows that

Z Veφ(L1, K1, =) 1 ρ(K1, u) = φ ρ(L1, u)ρ=dS(u) n−1 Ve(L1, =) nVe(L1, =) S ρ(L1, u) ! 1 Z ≥ φ ρ(K1, u)ρ=dS(u) n−1 nVe(L1, =) S ! Ve(K , =) = φ 1 . (64) Ve(L1, =) If φ is strictly convex, from the equality condition of Jensen’s inequality, it follows that the equality in (64) holds if and only if K1 and L1 are dilates.

n Theorem 2. (The dual Orlicz–Aleksandrov–Fenchel inequality) If L1, K1, ... , Kn ∈ S , 1 ≤ r ≤ n, and φ ∈ C, then

r 1/r ! ∏i=1 Ve(Ki ..., Ki, Kr+1,..., Kn) Veφ(L1, K1,..., Kn) ≥ Ve(L1, K2 ..., Kn) · φ . (65) Ve(L1, K2 ..., Kn)

If φ is strictly convex, equality holds if and only if L1, K1,..., Kr are all dilations of each other.

Proof. This follows immediately from Theorem 1 with the dual Aleksandrov–Fenchel inequality.

Obviously, putting L1 = K1 in (65), (65) becomes the Lutwak’s dual Aleksandrov–Fenchel inequality (11) stated in the introduction.

n Corollary 1. If L1, K1, = ∈ S and φ ∈ C, then

 !1/n V(K1) ··· V(Kn) Vφ(L , K , =) ≥ V(L , =)φ . (66) e 1 1 e 1  n  Ve(L1, =)

If φ is strictly convex, equality holds if and only if L1, K1,..., Kn are all dilations of each other.

Proof. This follows immediately from Theorem 2 with r = n.

Corollary 2. If K, L ∈ Sn, 0 ≤ i < n and φ ∈ C, then

 !1/(n−i) We i(L) We φ,i(K, L) ≥ We i(K)φ   . (67) We i(K)

If φ is strictly convex, equality holds if and only if K and L are dilates.

Proof. This follows immediately from Theorem 2 with r = n − i, L1 = K, K1 = L, K2 = ··· = Kn−i = K, and Kn−i+1 = ··· = Kn = B.

The following inequality follows immediately from (67) with φ(t) = t−p and p ≥ 1. If K, L ∈ Sn, 0 ≤ i < n, and p ≥ 1, then n−i n−i+p −p We −p,i(K, L) ≥ We i(K) We i(L) , (68) Mathematics 2020, 8, 2005 16 of 23

with equality if and only if K and L are dilates. Taking i = 0 in (68), this yields Lutwak’s Lp-dual Minkowski inequality: If K, L ∈ Sn and p ≥ 1, then

n n+p −p Ve−p(K, L) ≥ V(K) V(L) , (69) with equality if and only if K and L are dilates.

Theorem 3. (Orlicz dual isoperimetric inequality) If K ∈ Sn and φ ∈ C, and 0 ≤ i < n then

 !1/(n−i) Veφ(K, B, K [n − i − 1], B [i]) V(B) ≥ φ   . (70) We i(K) We i(K)

If φ is strictly convex, equality holds if and only if K is a ball.

Proof. This follows immediately from (65) with r = n − i, L1 = K, K1 = B, K2 ··· = Kn−i = K, and Kn−i+1 = ··· = Kn = B.

−p When φ(t) = t , p ≥ 1, the Orlicz isoperimetric inequality (70) becomes the following Lp-dual isoperimetric inequality. If K is a star body, p ≥ 1 and 0 ≤ i < n, then

!n−i !n−i+p nVe−p(K, B) We (K) ≥ i , (71) ωn κn with equality if and only if K is a ball, and where κn denotes volume of the unit ball B, and its surface area by ωn. Putting p = 1 and i = 0 in (71), (71) becomes the following dual isoperimetric inequality. If K is a star body, then !n n+1 nVe− (K, B) V(K) 1 ≥ , ωn κn with equality if and only if K is a ball.

n Theorem 4. If L1, K1, = ∈ M ⊂ S , and φ ∈ C be strictly convex, and if either

Veφ(Q, K1, =) = Veφ(Q, L1, =), f or all Q ∈ M, (72) or Veφ(K1, Q, =) Veφ(L1, Q, =) = , f or all Q ∈ M, (73) Ve(K1, =) Ve(L1, =) then K1 = L1.

Proof. Suppose (72) holds. Taking K1 for Q, then from Deﬁnition 3 and Theorem 1, we obtain ! Ve(L1, =) φ(1)Ve(K1, =) = Veφ(K1, L1, =) ≥ Ve(K1, =)φ , Ve(K1, =) with equality if and only if K1 and L1 are dilates. Hence, ! Ve(L , =) φ(1) ≥ φ 1 , Ve(K1, =) Mathematics 2020, 8, 2005 17 of 23

with equality if and only if K1 and L1 are dilates. Since ϕ is a decreasing function on (0, ∞), it follows that Ve(K1, =) ≤ Ve(L1, =), with equality if and only if K1 and L1 are dilates. On the other hand, if taking L1 for Q, we similarly get Ve(K1, =) ≥ Ve(L1, =), with equality if and only if K1 and L1 are dilates. Hence, Ve(K1, =) = Ve(L1, =), and K1 and L1 are dilates, it follows that K1 and L1 must be equal. Suppose (73) holds. Taking K1 for Q, then from Deﬁnition 3 and Theorem 1, we obtain ! Veφ(L1, K1, =) Ve(K , =) φ(1) = ≥ φ 1 , Ve(L1, =) Ve(L1, =) with equality if and only if K1 and L1 are dilates. Since ϕ is an increasing function on (0, ∞), this follows that Ve(L1, =) ≤ Ve(K1, =), with equality if and only if K1 and L1 are dilates. On the other hand, if taking L1 for Q, we similar get Ve(L1, =) ≥ Ve(K1, =), with equality if and only if K1 and L1 are dilates. Hence, Ve(L1, =) = Ve(K1, =), and K1 and L1 are dilates, it follows that K1 and L1 must be equal.

Corollary 3. Let K, L ∈ M ⊂ Sn, 0 ≤ i < n, and φ ∈ C be strictly convex, and if either

We φ,i(Q, K) = We φ,i(Q, L), for all Q ∈ M, or We φ,i(K, Q) We φ,i(L, Q) = , for all Q ∈ M, We i(K) We i(L) then K = L.

Proof. This yields immediately from Theorem 4 and Lemma 10.

Remark 3. When φ(t) = t−p and p = 1, the dual Orlicz Aleksandrov–Fenchel inequality (65) becomes the n following inequality. If L1, K1, = ∈ S and 1 ≤ r ≤ n, then

2 Ve(L1, =) Ve−1(L1, K1, =) ≥ r , (74) 1/r ∏ Ve(Ki ..., Ki, Kr+1,..., Kn) i=1 with equality if and only if L1, K1,..., Kr are all dilations of each other.

Putting L1 = K1 in (74) and noting that Ve−1(K1, K1, =) = Ve(K1, =), (74) becomes the dual Aleksandrov–Fenchel inequality (11). Putting r = n in (74), (74) becomes the following inequality.

n 2n −1 Ve−1(L1, K1, =) ≥ Ve(L1, =) (V(K1) ··· V(Kn)) , (75) with equality if and only if L1, K1, = are all dilations of each other. Putting L1 = K, K1 = L and n K2 = ··· = Kn = K in (75), (75) becomes the well-known Minkowski inequality. If K, L ∈ S , then

n n+1 −1 Ve−1(K, L) ≥ V(K) V(L) , (76) Mathematics 2020, 8, 2005 18 of 23 with equality if and only if K and L are dilates. Obviously, inequality (74) in a special case yields also n the following result. If K1, = ∈ S and 0 ≤ i < n, then

n−i n−i+1 −1 We −1,i(K, L) ≥ We i(K) Wi(L) , (77) with equality if and only if K and L are dilates. When i = 0, (77) becomes (76). On the other hand, n putting L1 = K1 in (75), (75) becomes the well-known inequality. If K1, = ∈ S , then

n Ve(K1,..., Kn) ≤ V(K1) ··· V(Kn), with equality if and only if K1, = are all dilations of each other.

6. The Dual Orlicz–Brunn–Minkowski Inequality

n Lemma 12. If L1, K1, = ∈ S and φ ∈ C, then

φ(1)Ve(K1+b φ L1, =) = Veφ(K1+b φ L1, K1, =) + Veφ(K1+b φ L1, L1, =). (78)

n Proof. Suppose ε > 0, K1, L1 ∈ S , let

Q = K1+b φε · L1.

From Deﬁnition 3, (20), and (40), we have

Z 1 ρ(K1, u) ρ(L1, u) φ(1)Ve(Q, =) = φ + φ ρ(Q, u)ρ=dS(u) n Sn−1 ρ(Q, u) ρ(Q, u)

= Veφ(Q, K1, =) + Veφ(Q, L1, =). (79) This completes the proof.

Lemma 13. ([53]) Let K, L ∈ Sn, ε > 0 and φ ∈ C. (1) If K and L are dilates, then K and K+b φε · L are dilates. (2) If K and K+b φε · L are dilates, then K and L are dilates.

n Theorem 5. (The dual Orlicz–Brunn–Minkowski inequality) If L1, K1, ... , Kn ∈ S and φ ∈ C, then for ε > 0 ! ! Ve(K ,..., Kn) Ve(L , K ..., Kn) φ(1) ≥ φ 1 + ε · φ 1 2 , (80) Ve(K1+b φε · L1, K2 ..., Kn) Ve(K1+b φε · L1, K2 ..., Kn)

If φ is strictly convex, equality holds if and only if K1 and L1 are dilates.

Proof. From Theorem 1 and Lemma 12, we have

φ(1)Ve(K1+b φε · L1, =) = Veφ(K1+b φε · L1, K1, =)

+ ε · Veφ(K1+b φε · L1, L1, =) ( ! Ve(K1, =) ≥ Ve(K1+b φε · L1, =) φ Ve(K1+b φε · L1, =) !) Ve(L , =) + φ 1 . Ve(K1+b φε · L1, =) Mathematics 2020, 8, 2005 19 of 23

If φ is strictly convex, from the equality condition of Theorem 1, the equality in (80) holds if and only if K1 and K1+b φ L1, and L1 and K1+b φ L1 are dilates. Further, from Lemma 13, it follows that if φ is strictly convex, the equality in (80) holds if and only if K1 and L1 are dilates.

n Theorem 6. (The dual Orlicz–Brunn–Minkowski type inequality) If L1, K1, = ∈ S , 0 ≤ i, j < n, 1 < r ≤ n, and φ ∈ C, then

r 1/r ! r 1/r ! ∏ Ve(K ,..., K , K + ,..., Kn) M(r) ∏j=2 ·Ve(Kj,..., Kj, Kr+1,..., Kn) φ(1) ≥ φ i=1 i i r 1 + φ , (81) Ve(K1+b φL1, =) Ve(K1+b φL1, =)

1/r where M(r) = Ve(L1, ... , L1, Kr+1, ... , Kn) . If φ is strictly convex, equality holds if and only if L1, K1,..., Kr are all dilations of each other.

Proof. This follows immediately from Theorem 5 and the dual Aleksandrov–Fenchel inequality.

n Corollary 4. (Lp-dual Brunn-Minkowski inequality) If L1, K1, = ∈ S , 0 ≤ i, j < n, 1 < r ≤ n, and p ≥ 1, then

r − −p p Ve(K1+b p L1, =) ≥ ∏ Ve(Ki,..., Ki, Kr+1,..., Kn) r i=1

r − −p p + ∏ M(r) · Ve(Kj,..., Kj, Kr+1,..., Kn) r , (82) j=1 with equality if and only if L1, K1,..., Kr are all dilations of each other, and M(r) is as in Theorem 6.

Proof. This follows immediately from (81) with φ(t) = t−p and p ≥ 1.

Corollary 5. If K, L ∈ Sn, φ ∈ C and 0 ≤ i < n − 1, then

 !1/(n−i)  !1/(n−i) We i(K) We i(L) φ(1) ≥ φ   + φ   . (83) We i(K+b φ L) We i(K+b φ L)

If φ is strictly convex, equality holds if and only if K and L are dilates.

Proof. This follows immediately from Theorem 6 with r = n − i, K2 = ··· = Kn−i = K+b φ L, Kn−i+1 = ··· = Kn = B.

The following inequality follows immediately from (83) with φ(t) = t−p and p ≥ 1. If K, L ∈ Sn, 0 ≤ i < n, and p ≥ 1, then

−p/(n−i) −p/(n−i) −p/(n−i) We i(K+b p L) ≥ We i(K) + We i(L) , with equality if and only if K and L are dilates.

n Corollary 6. If L1, K1, = ∈ S and φ ∈ C, then

 !1/n  !1/n V(K ) ··· V(K ) V(L )V(K ) ··· V(K ) φ(1) ≥ φ 1 n + φ 1 2 n . (84)  n   n  Ve(K1+b φ L1, =) Ve(K1+b φ L1, =)

If φ is strictly convex, equality holds if and only if L1, K1, = are all dilations of each other. Mathematics 2020, 8, 2005 20 of 23

Proof. This follows immediately from Theorem 6 with r = n.

n Corollary 7. If L1, K1, = ∈ S and p ≥ 1, then

−p −p/n −p/n Ve(K1+b p L1, =) ≥ (V(K1) ··· V(Kn)) + (V(L1)V(K2) ··· V(Kn)) , (85) with equality if and only if L1, K1, = are all dilations of each other.

Proof. This follows immediately from (84) with φ(t) = t−p and p ≥ 1.

Putting K2 = ··· = Kn = K1+b p L1 in (85), (85) becomes Lutwak’s Lp-dual Brunn–Minkowski inequality −p/n −p/n −p/n V(K+b p L) ≥ V(K) + V(L) , with equality if and only if K and L are dilates.

n Corollary 8. If L1, K1, = ∈ S , 1 ≤ r ≤ n, and φ ∈ C, then

r 1/r ! ∏i=1 Ve(Ki ..., Ki, Kr+1,..., Kn) Veφ(L1, K1, =) ≥ Ve(L1, =)φ . (86) Ve(L1, =)

If φ is strictly convex, equality holds if and only if L1, K1,..., Kr are all dilations of each other.

Proof. Let Kε = L1+b φε · K1. From (49), dual Orlicz–Brunn–Minkowski inequality (80), and dual Aleksandrov–Fenchel inequality, we obtain

1 d 0 · Veφ(L1, K1, =) = Ve(Kε, =) φ+(1) dε ε=0+ Ve(Kε, =) − Ve(L , =) = lim 1 ε→0+ ε ! Ve(L , =) Ve(L1, =) 1 − 1 φ(1) − φ Ve(Kε, =) Ve(Kε, =) = lim ! · · Ve(Kε, =) ε→0+ Ve(L , =) ε φ(1) − φ 1 Ve(Kε, =) ! Ve(L , =) φ(1) − φ 1 1 − t Ve(Kε, =) = lim · lim · lim Ve(Kε, =) t→0+ φ(1) − (t) ε→0+ ε ε→0+ ! 1 Ve(K , =) ≥ · lim φ 1 · V(L , =) 0 + e 1 φ+(1) ε→0 Ve(Kε, =) ! 1 Ve(K1, =) = 0 · φ · Ve(L1, =) φ+(1) Ve(L1, =)

r 1/r ! 1 ∏i=1 Ve(Ki ..., Ki, Kr+1,..., Kn) ≥ 0 · φ · Ve(L1, =). (87) φ+(1) Ve(L1, =) From (87), inequality (86) easily follows. From the equality conditions of the dual Orlicz–Brunn–Minkowski inequality (80) and dual Aleksandrov–Fenchel inequality, it follows that if φ is strictly convex, the equality in (87) holds if and only if L1, K1, = are all dilations of each other. Mathematics 2020, 8, 2005 21 of 23

This proof is complete.

From the proof of (87) and (80), it is not difﬁcult to see that inequalities (63) and (80) are equivalent.

Funding: This research is supported by National Natural Sciences Foundation of China (10971205, 11371334). Conﬂicts of Interest: The author declares that he has no competing interests.

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