# The General Dual Orlicz Geominimal Surface Area

Authors: Ni Li; Shuang Mou
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we study the general dual Orlicz geominimal surface area by the general dual Orlicz mixed volume which was introduced by Gardner et al. (2019). We find the conditions to the existence of the general dual Orlicz-Petty body and hence prove the continuity of the general geominimal surface area in the Orlicz setting (2010 Mathematics Subject Classification: 52A20, 53A15).


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## Body

## 1. Introduction

In the Euclidean spaceℝn, we call a compact and convex subset K⊂ℝn a convex body if K has nonempty interior. Let Kn be the family of all convex bodies in ℝn and K0n be the family of the convex bodies containing the origin in their interiors. We use the standard notations ⋅,⋅ and ⋅ to denote the inner product and the Euclidean norm in ℝn. Let K denote the volume of a set K and B2n denote the unit ball in ℝn, i.e., B2n=x∈ℝn:x≤1; the volume of B2n is denoted by ωn. Let Sn−1=x∈ℝn:x=1 be the unit sphere and σ denote the spheral measure.The geominimal surface area and its extension play important roles in connecting the relative differential geometry with Minkowski geometry. The geominimal surface area was firstly introduced by Petty in 1974 (see [1]). The Lp geominimal surface area was investigated by Lutwak (see [2]) and others (see, e.g., [3–5]). Recently, Lutwak et al. (see [6, 7]) developed the Orlicz–Brunn–Minkowski theory, a new extension of Brunn–Minkowski theory. Simultaneously, the geominimal surface area in Orlicz setting was studied in [8, 9]. The dual Lp Brunn–Minkowski theory was introduced in [10, 11]. In 2011, Wang and Qi [12] introduced the Lp geominimal surface area in dual theory which is dual to the Lp geominimal surface area. Moreover, they established some affine isoperimetric inequalities, for example, the Blaschke Santaló-type inequality for the Lp dual geominimal surface area. The dual Brunn–Minkowski theory in Orlicz setting has already been introduced by Gardner et al. [13, 14], Ye [15], and Zhu et al. [16]. For more results on the geominimal surface area, one can refer to [3, 4, 12, 17–20] and so on.The Orlicz–Brunn–Minkowski theory for the general volume was established by Gardner et al. [21], and the general dual mixed volume V˜ϕ,φK,L in Orlicz setting was also introduced (see [21], p13), i.e., for two star bodies K and L:(1)V˜ϕ,φK,L=1n∫Sn−1ϕρKuuφρLuρKuρKundσu,where ϕ:ℝn\0⟶0,∞ is a continuous function, φ:0,∞⟶0,∞ is also a continuous function, and ρK denotes the radial function of K (see Section 2 for undefined and unexplained notations). There is a natural problem: whether there is a convex body T∈K0n with T∘=ωn, such that it is a solution to the following problem:(2)infnV˜ϕ,φK,M:M∈K0nandM∘=ωn.Here,M∘ is the polar of M (see Section 2).Ifϕx=xq−n for q∈ℝ and x∈ℝn, we show this problem is solvable. For instance, we obtain one of the main results.Theorem 1.
LetK∈L0n be a star body. If the function φ:0,∞⟶0,∞ is increasing and q∈ℝ, there is a convex body T∈K0n with T∘=ωn, such that(3)nV˜q,φK,T=infnV˜q,φK,M:M∈K0nandM∘=ωn.In addition, ifφ is convex, the solution T is unique.Here,V˜q,φK,L can be formulated by (1) and ϕx=xq−n for q∈ℝ, i.e.,(4)V˜q,φK,L=1n∫Sn−1φρLuρKuρKuqdσu.


## 2. Background and Notation

For any setK∈Kn, it can be determined by its support function hKu=maxx,u:x∈K for u∈Sn−1. Obviously, hK≥0 if K∈Kn and hK>0 if K∈K0n. A set K will be called a star set about the origin 0 if every line passes through 0 and meets K with only a line segment. Moreover, if K is a star set about 0, its radial function ρK:Sn−1⟶0,∞ can be defined by the formula ρK=maxt>0:tu∈K for all u∈Sn−1. If λ>0, one obtains(5)ρλKu=λρKuandρKλu=λ−1ρKu,for allu∈Sn−1.If the radial functionρK of a star set K about the origin 0 is continuous and positive, then K is called a star body about the origin 0. Denote by L0n the family of star bodies in ℝn endowed with radial metric; that is, the distance between K,L∈L0n is defined as ρK−ρL∞, where ⋅∞ is max-norm in Hausdorff metric space. Suppose there exist a sequence Kii≥1⊂L0n and a set K∈L0n such that Ki⟶K, then the radial function ρKiu is uniformly convergent to ρKu.For a convex bodyK∈Kn, the polar K∘ of K is also a convex body in ℝn, that is,(6)K∘=x∈ℝn:x,y≤1for eachy∈K.It has been proved thatK∘∘=K if K∈K0n. Moreover, if K∈K0n, the following formulas hold:(7)hK∘u=1ρKu,ρK∘u=1hKu,for allu∈Sn−1.We use the notationGLn to denote the set of all invertible and linear transforms on ℝn and detA to denote the determinant of A∈GLn. Especially, SLn=A:A∈GLnanddetA=±1. By At and A−t, we mean the transpose of A and the inverse of the transpose of A, respectively. From definitions of the radial function and the support function, we observe immediately that for A∈SLn, L∈L0n, and K∈K0n, one has(8)ρALu=ρLA−1u,hAKu=hKAtu,for allu∈Sn−1.By formulas (7) and (8), if K∈K0n, we have(9)AK∘=A−tK∘.For anyK,L∈L0n and λ,μ>0, the harmonic radial combination λ∘K+˜−1μ∘L is defined by (see [22])(10)ρλ∘K+˜−1μ∘L−1u=λρK−1u+μρL−1ufor allu∈Sn−1,where λ∘K defines the radial multiplication, i.e., λ∘K=λ−1K. Obviously, the harmonic radial combination of two convex bodies K,L∈K0n is still a convex body according to formulas (7) and (10). For q∈ℝ, the qth dual mixed volume of two star bodies K,L∈L0n is given by (see [2])(11)V˜qK,L=1n∫Sn−1ρKn−quρLqudσu.LettingL=B2n, then V˜qK,B2n=1/n∫Sn−1ρKn−qudσu. By the Hölder inequality, one has for q>0 (see [10])(12)V˜qK,B2n≤VKq/nωnn−q/n,with equality iff K is a ball. If q<0, the reverse inequality of (12) holds. Obviously,(13)V˜qK,K=K,where K=1/n∫Sn−1ρKnudσu is the volume of star body K∈L0n.In convex geometry, the Blaschke selection theorem is very useful (see, e.g., [23, 24]), which means that if a sequence Kn of convex bodies is bounded, there is a subsequence Knm of Kn and a convex set K such that Knm converges to K.Lemma 1 (see [2]).
SupposeKii≥1⊂K0n is a convergent sequence such that Ki⟶K0 in the Hausdorff distance. If Ki∘i≥1 is bounded, then K0∈K0n.Lemma 2 (see [25]).
IfKii≥1⊂K0n is a bounded sequence and the sequence Ki∘i≥1 is also bounded, there is a subsequence Kijj≥1 and a body K∈K0n such that Kij⟶K. In addition, if Ki∘=ωn, then K∘=ωn.


## 3. Properties for the General Dual Mixed VolumeV˜q,φK,L

In this section, we will discuss some properties of the general dual Orlicz mixed volume, which is defined as follows [21]: Let K,L∈L0n; if ϕ:ℝn\0⟶0,∞ is continuous and φ∈C+0,∞, the general dual Orlicz mixed volume is(14)V˜ϕ,φK,L=1n∫Sn−1ϕρKuuφρLuρKuρKnudσu.Here,CI denotes the family of all the continuous functions on interval I⊂ℝ, and C+I is the set of f∈CI such that f>0. Let(15)ℐ=φ∈C+0,∞:φtis increasing function withlimt⟶∞φt=∞andlimt⟶0φt=0,D=φ∈C+0,∞:φtis decreasing function withlimt⟶∞φt=0andlimt⟶0φt=∞.In this paper, we will consider the functionφ∈ℐ∪D and ϕx=xq−n for q∈ℝ, that is(16)V˜q,φK,L=1n∫Sn−1φρLuρKuρKqudσu.We will now prove the continuity of the general dual Orlicz mixed volume as follows.Proposition 1.
LetKii=1∞⊂L0n and Lii=1∞⊂L0n be two sequences of star bodies such that Ki⟶K∈L0n and Li⟶L∈L0n as i⟶∞. If φ∈ℐ∪D and q∈ℝ, then(17)V˜q,φKi,Li⟶V˜q,φK,Lasi⟶∞.Proof.
SinceKi⟶K∈L0n and Li⟶L∈L0n as i⟶∞, then ρKi⟶ρK and ρLi⟶ρL uniformly on Sn−1, and thus, there are two constants m,M>0 such that(18)mB2n⊂Ki,K,Li,L⊂MB2n.
Then, for anyi≥1 and u∈Sn−1, we have(19)mM≤ρLiuρKiu,ρLuρKu≤Mm.
From the continuity ofφ on interval m/M,M/m, we have(20)φρLiuρKiuρKiuq⟶φρLuρKuρKuquniformly onSn−1,
Hence, one has(21)limi⟶∞V˜q,φKi,Li=limi⟶∞1n∫Sn−1φρLiuρKiuρKiqudσu=1n∫Sn−1φρLuρKuρKqudσu=V˜q,φK,L.
The next proposition is necessary to prove our main results.Proposition 2.
LetKii≥1⊂L0n be a sequence and K∈L0n such that Ki⟶K as i⟶∞. If Lii≥1⊂L0n is a sequence such that V˜q,φKi,Lii≥1 is bounded for φ∈ℐ and q∈ℝ, then Lii≥1 is bounded.Proof.
The boundedness ofV˜q,φKi,Lii≥1 shows that there is a M>0 such that V˜q,φKi,Li<M for all i≥1. We now assume Lii≥1 is unbounded, that is, supi≥1Ri=∞, where Ri=maxρLiu:u∈Sn−1. Hence, there is a constant D>0 such that Ri>D. Let R=maxρKu:u∈Sn−1 and r=minρKu:u∈Sn−1. Combining with continuity of the radial function, φ∈ℐ, and Fatou’s lemma, one has(22)M>limi⟶∞V˜q,φKi,Li=limi⟶∞1n∫Sn−1φρLiuρKiuρKiqudσu≥limi⟶∞1n∫Sn−1φDρKiuρKiqudσu≥1n∫Sn−1liminfi⟶∞φDρKiuρKiqudσu=φDRrqωn.
LetD⟶∞; then, one has a contradiction M≥∞. Thus, Lii≥1 is bounded.


## 4. Continuity of the General Dual Orlicz-Petty Body

In this section, we firstly define the general dual Orlicz geominimal surface area, denoted byG˜q,φK, and prove the existence and uniqueness of the general dual Orlicz-Petty body. Moreover, we will also provide some properties of the general dual Orlicz-Petty body. Finally, the continuity of G˜q,φK will be established.Definition 1.
SupposeK∈L0n is a star body. For φ∈ℐ and q∈ℝ, define the general dual Orlicz geominimal surface area of K by(23)G˜q,φK=infnV˜q,φK,L:L∈K0nwithL∘=ωn.
The affine invariance of the general dual Orlicz geominimal surface area can be shown in the following consequence.Proposition 3.
LetK∈L0n and A∈SLn. If q∈ℝ and φ∈ℐ, then(24)G˜q,φAK=G˜q,φK.Proof.
For anyx∈ℝn\0, let x¯=x/x. From Definition 1, together with formulas (8), (9), and (16), we have(25)G˜q,φAK=infnV˜q,φAK,L:L∈K0nwithL∘=ωn=inf∫Sn−1φρLuρAKuρAKqudσu:L∈K0nwithL∘=ωn=inf∫Sn−1φρA−1LA−1uρKA−1uρKqA−1udσu:L∈K0nwithL∘=ωn=inf∫Sn−1φρA−1LA−1uρKA−1uρKqA−1udσAA−1u:A−1L∈K0nwithA−1L∘=ωn=inf∫Sn−1φρA−1LA−1u¯ρKA−1u¯ρKqA−1u¯dσA−1u¯:A−1L∈K0nwithA−1L∘=ωn=G˜q,φK.
We now prove the existence of the general dual Orlicz-Petty body as follows.Theorem 2.
SupposeK∈L0n is a star body. If φ∈ℐ and q∈ℝ, then there is a convex body T∈K0n such that(26)G˜q,φK=nV˜q,φK,TandT∘=ωn.
In addition, ifφ is convex, the body T is unique.Proof.
From Definition 1, there is a sequenceTii≥1⊂Kon with Ti∘=ωn, such that V˜q,φK,B2n≥V˜q,φK,Ti for each i≥1 and nV˜q,φK,Ti⟶G˜q,φK. We now prove that the sequence Tii≥1 is bounded. Similar to Proposition 2, assume that Tii≥1 is unbounded, let Ri=maxρTiu:u∈Sn−1, and thus supi≥1Ri=∞; there exists a constant sequence Dii≥1 and a constant D such that Ri>Di and Di⟶D as i⟶∞. Let R=maxρKu:u∈Sn−1 and r=minρKu:u∈Sn−1. By formula (16) and Fatou’s lemma, we have(27)V˜q,φK,B2n=limi⟶∞V˜q,φK,Ti=limi⟶∞1n∫Sn−1φρTiuρKuρKqudσu≥limi⟶∞1n∫Sn−1φDiρKuρKqudσu≥1n∫Sn−1liminfi⟶∞φDiρKuρKqudσu=φDRrqωn.
This will not hold ifD⟶∞ as i⟶∞. Hence, Tii≥1⊂K0n is bounded; then by the Blaschke selection theorem, there is one subsequence Tijj≥1⊂Tii≥1 and one compact convex set T so that Tij⟶T as j⟶∞; it also implies that Tij∘⟶T∘. Since Tij∘=ωn, by Lemma 2, we have T∘=ωn and by Lemma 1 we have T∈K0n. This together with Proposition 1 shows that(28)G˜q,φK=limi⟶∞nV˜q,φK,Ti=nV˜q,φK,TwithT∘=ωn.
Ifφ is convex, assume that there are two convex bodies T1,T2∈K0n such that T1∘=T2∘=ωn and G˜q,φK=nV˜q,φK,T1=nV˜q,φK,T2. Define a body T by(29)T=12∘T1+˜−112∘T2.
Obviously(30)T∘=12T1∘+12T2∘.
Thus,T is a convex body and T∈K0n by formulas (7) and (10). By the Brunn–Minkowski inequality, we have T∘≥ωn with equality iff T1=T2. Since both ft=t−1 and φ∈ℐ are convex functions, we have(31)nV˜q,φK,T=∫Sn−1φρTuρKuρKqudσu=∫Sn−1φρ1/2∘T1+˜−11/2∘T2uρKuρKqudσu=∫Sn−1φ1/2ρT1u−1+1/2ρT2u−1−1ρKuρKqudσu≤∫Sn−1φ1/2ρT1u+1/2ρT2uρKuρKqudσu≤∫Sn−112φρT1uρKu+12φρT2uρKuρKqudσu=12nV˜q,φK,T1+12nV˜q,φK,T2=nV˜q,φK,T1=nV˜q,φK,T2.
With equality iffT1=T2. Thus, if T1≠T2, we have(32)nV˜q,φK,T<nV˜q,φK,T1=nV˜q,φK,T2,which is a contradiction with the definition of G˜q,φK. This finishes the proof.Remark 1.
The bodyT in Theorem 2 is usually called as the general dual Orlicz-Petty body and is denoted by(33)T˜q,φK=T:G˜q,φK=nV˜q,φK,TwithT∘=ωn.
SinceT is unique if the function φ is convex by Theorem 2, the mapping T˜q,φK contains only one element.
In the next theorem, we prove the continuity of the general geominimal surface areaG˜q,φK.Theorem 3.
LetKii=1∞⊂L0n and K∈L0n such that Ki⟶K as i⟶∞. If φ∈ℐ and q∈ℝ, then G˜q,φKi⟶G˜q,φK.Proof.
LetKii=1∞⊂L0n and K∈L0n such that Ki⟶K as i⟶∞. For any real number ϵ>0, from Definition 1 and Proposition 1, there is a convex body Tε∈K0n so that Tε∘=ωn and(34)G˜q,φK+ε>nV˜q,φK,Tε=limi⟶∞nV˜q,φKi,Tε=limsupi⟶∞nV˜q,φKi,Tε≥limsupi⟶∞G˜q,φKi.
Sinceε>0 is arbitrary,(35)G˜q,φK≥limsupi⟶∞G˜q,φKi.
LetTi∈T˜q,φKi for each Ki∈L0n, that is, G˜q,φKi=nV˜q,φKi,Ti. This, together with formula (35), implies that the sequence V˜q,φKi,Ti is bounded and thus the sequence Tii≥1 is bounded by Proposition 2. Suppose Kijkk≥1⊂Kii≥1 is a subsequence so that(36)limk⟶∞G˜q,φKijk=liminfi⟶∞G˜q,φKi.
SinceTijj≥1 is uniformly bounded, then by Lemma 1, there is a subsequence Tijkk≥1⊂Tijj≥1 and a convex body T∈K0n such that limk⟶∞Tijk=T. According to Definition 1, Proposition 1, and Theorem 2, one has(37)G˜q,φK≤nV˜q,φK,T=nlimk⟶∞V˜q,φKijk,Tijk=limk⟶∞G˜q,φKijk=liminfi⟶∞G˜q,φKi.
Combining (35) and (37), one obtains(38)limi⟶∞G˜q,φKi=G˜q,φK.
Similar to the statement of Remark1, the mapping T˜q,φK contains only one element if the function φ is convex, so it can define an operator. Moreover, we can prove this operator is continuous as follows.Theorem 4.
LetKii=1∞⊂L0n and K∈L0n such that Ki⟶K as i⟶∞. If q∈ℝ and φ∈ℐ are convex, then T˜q,φKi⟶T˜q,φK.Proof.
LetT=T˜q,φK and Ti=T˜q,φKi by the uniqueness of T˜q,φK and T˜q,φKi when φ∈ℐ is convex. By the subsequence Kijj≥1 of Kii≥1 such that Kij⟶K as j⟶∞ and Theorem 3, then there is a subsequence Tijj≥1⊂Tii≥1 such that(39)G˜q,φK=limj⟶∞G˜q,φKij=limj⟶∞nV˜q,φKij,Tij.
This implies thatV˜q,φKij,Tij is bounded, and hence, Tijj≥1 is bounded by Proposition 2. Thus, according to Lemma 2, there are a subsequence Tijkk≥1⊂Tijj≥1 and a body T0∈K0n so that limk⟶∞Tijk=T0 and T0∘=ωn. By Proposition 1 and formula (18), we have(40)G˜q,φK=limk⟶∞G˜q,φKijk=limk⟶∞nV˜q,φKijk,Tijk=nV˜q,φK,T0.
On the other hand, sinceT∈T˜q,φK, T satisfies T∘=ωn and G˜q,φK=nV˜q,φK,T. The convexity of φ∈ℐ, by Theorem 2, implies T=T0; that is, limi⟶∞Ti=T. Hence, T˜q,φKi⟶T˜q,φK.



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