The minimal Orlicz surface area Advances in Applied Mathematics

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Advances in Applied Mathematics

www.elsevier.com/locate/yaama

The minimal Orlicz surface area

Du Zou,Ge Xiong^∗

DepartmentofMathematics,ShanghaiUniversity,Shanghai,200444,PRChina

a r t i c l e i n f o a b s t r a c t

Articlehistory:

Received27October2013 Receivedinrevisedform30July 2014

Accepted7August2014

Availableonline17September2014

MSC:

52A40

Keywords:

OrliczBrunn–Minkowskitheory Minimalsurfacearea

Reverseisoperimetricinequality

Petty proved that a convex body in Rⁿ has the minimal surface area amongst its SL(n) images, if, and only if, its surfaceareameasureisisotropic.Byintroducinganewnotion of minimal Orlicz surface area, wegeneralize thisresult to theOrliczsetting.TheanalogofBall’sreverse isoperimetric inequalityisestablished.

© 2014ElsevierInc.All rights reserved.

1. Introduction

AclassicalandusefulresultprovedbyPetty[41]istheminimalsurfaceareatheorem, which states: A convex body (i.e., a compact convex set with non-empty interior) in Euclidean n-space Rⁿ has the minimal surface areaamongst its SL(n) images, if, and only if, its surface area measure is isotropic on the unit sphere Sⁿ⁻¹. Its importance was rediscovered in the 1990s. In [9], Clack generalized it to Minkowski space. Later,

✩ Research of the authors was supported by Innovation Program of Shanghai Municipal Education CommissionNo.11YZ11andNSFCNo.11471206.

* Correspondingauthor.

E-mailaddress:[email protected](G. Xiong).

http://dx.doi.org/10.1016/j.aam.2014.08.006 0196-8858/© 2014ElsevierInc.All rights reserved.

GiannopoulosandPapadimitrakis[14]usedisotropicsurfaceareameasuretostudy the hyperplane projectionsof convexbodies.

As theBrunn–Minkowskitheory[44] wasextendedto theLp Brunn–Minkowskithe- ory (see,e.g.,[7,8,10,20,25–34,40,43,45–47,50,49,51,53]),thenotionsofsurfaceareaand surface area measure were extended to those of Lp surface area and Lp surface area measure,respectively.SeetheinitialworksofLutwak[28,29].In[34],Lutwak,Yangand ZhangshowedthatPetty’stheoremhasanaturalLpgeneralization:TheLpsurfacearea ofaconvexbodyisminimalamongstitsSL(n) images,if,andonlyif,itsLpsurfacearea measure isisotropiconSⁿ⁻¹.

Beginningwith aseriesof ground-breaking articles[26,35,36,19]andthevery recent work[12],amorewideextension oftheL_p Brunn–Minkowskitheoryemerged,whichis now called theOrlicz Brunn–Minkowski theory. In this context, the main goal of this paper istoseekanOrliczextension oftheminimalL_p surfacearea.

Throughout thispaper,letϕ: [0,∞)→[0,∞) beconvexandstrictlyincreasingwith ϕ(0)= 0,anddenote byΦtheclassof thoseϕ.

Suppose K isa convexbody inRⁿ with theorigin inits interior.Its Orlicz surface area Sϕ(K) withrespecttoϕisdefinedby

Sϕ(K) =

∂K

ϕ 1

x·ν(x)

x·ν(x)dHⁿ⁻¹(x).

Here, ν(x), x·ν(x) and Hⁿ⁻¹ denote the outer unit normal of ∂K at x ∈ ∂K, the standardinner product ofxandν(x), andthe(n−1)-dimensionalHausdorffmeasure, respectively.Notethatν(x) existsforHⁿ⁻¹-almost allx∈∂K.

Ifϕ(t)=t^p,1≤p<∞,thenSϕ(K) ispreciselytheLp surfacearea[28,29]ofK.

In Section3, we demonstratethat moduloorthogonal transformations, the body K has a unique SL(n) image with minimal Orlicz surface area. In view of this fact, we define theminimalOrliczsurface areaofK withrespect toϕby

Aϕ(K) = min

Sϕ(T K) :T ∈SL(n) .

For ϕ ∈Φ∩C¹(0,∞),i.e., forsmooth functions ϕin Φ, we introducethe transfor- mation, ϕ → ϕ, defined by ϕ(t) = tϕ(t). Then, for each Borel set ω ⊆ Sⁿ⁻¹, we write

S_ϕ(K, ω) =

ν⁻¹(ω)

ϕ 1

x·ν(x)

dHⁿ⁻¹(x).

Ifϕ(t)=t^p,1≤p<∞,thenSϕ(K,·)/pisjusttheLp surfaceareameasureofK.

InSection4,weextendPetty’sresulttotheOrliczsetting.

Theorem 1.1. Suppose K is a convex body in Rⁿ with the origin in its interior, and ϕ∈Φ∩C¹(0,∞).ThenAϕ(K)=Sϕ(K)ifandonly if Sϕ(K,·) isisotropiconSⁿ⁻¹.

In the last section, we provide bounds for the minimal Orlicz surfacearea Aϕ(K).

When the volume of K is fixed, origin-symmetric ellipsoids attain the minimum; the volumeof L_∞ Johnellipsoid introduced in[34] dominates it from above.Especially, if theJohn point (i.e., center ofthe John ellipsoid) ofbody K isat the origin,we prove thefollowing

Theorem1.2. AmongstallconvexbodiesinRⁿ with thesamevolumeandJohnpointsat theorigin,moduloSL(n)transformations,then-dimensionalsimplexuniquelymaximizes the minimal Orlicz surface area. If the involved bodies are origin-symmetric, then the n-dimensionalparallelotopeistheuniquemaximizer.

2. Preliminaries

Inordertokeepthepaperself-contained,wecollectheresomebasicfactsfromConvex Geometry.GoodreferencesonthetheoryofconvexbodiesarethebooksbyGardner[11], Gruber[16], Pisier[42],Schneider[44], andThompson[48],etc.

Asusual, x·y denotes thestandardinner productof xandy inRⁿ; B ={x∈Rⁿ : x·x≤1}andSⁿ⁻¹=∂B denotetheunitball andunitsphereinRⁿ,respectively.The volumeofB isπ^n/2/Γ(1 +n/2).

Accordingtothecontext,onecancatchclearlythatthenotation|·|hasseveraldiffer- entmeanings:theabsolutevalue,thestandardEuclideannormonRⁿ,then-dimensional volume,theabsolute valueof determinantof ann×n matrix,and thetotalmass ofa finitemeasure.

Forbrevity,wewrite x=|x|⁻¹x,forx∈Rⁿ\ {0}.

GivenaconvexbodyK inRⁿ,itssupportfunction hK :Rⁿ →Risdefinedby h_K(x) = max{x·y:y∈K}.

Thedefinitionimmediately givesthatforT ∈GL(n) andx∈Rⁿ, hT K(x) =hK

T^tx

. (2.1)

Throughoutthispaper,KoⁿdenotestheclassofconvexbodiesinRⁿ thatcontainthe origin intheir interiors. Koⁿ is often equipped with the Hausdorff metric δ_H, which is definedforK1,K2∈ Kⁿo,byδH(K1,K2)= max_Sn−1|hK1−hK2|.

The classical surface area measure SK, of a convex body K, is the unique Borel measureonSⁿ⁻¹ suchthat

Sⁿ⁻¹

f(u)dS_K(u) =

∂K

f ν_K(y)

dHⁿ⁻¹(y),

foreachcontinuousf :Sⁿ⁻¹→R.

Thecone-volumemeasureVK,ofaconvexbodyK,isaBorelmeasureonSⁿ⁻¹defined foraBorelsetω⊆Sⁿ⁻¹ by

VK(ω) = 1 n

ω

hKdSK.

It is convenient to use thenormalized cone-volumemeasure V_K = ^V_|K|^K,of K. Observe that VK is a probability measure on Sⁿ⁻¹. Also, VK is GL(n)-invariant, i.e., for T ∈ GL(n) andaBorelsubsetω⊆Sⁿ⁻¹,ityields

VT^tK(ω) =VK

T ω

, (2.2)

where T ω={ T u:u∈ω}.

Thecone-volumemeasurehasbeenappearedandinvestigatedwidelyinvariouscon- texts recently,see e.g.,[5,7,8,15,21,20,26,27,38–40,46,47,51].

What followsrecallssomefundamentalfactsestablishedin[12].

ForK∈ Kⁿo,ϕ∈Φ,andε≥0,definethefunctionhϕ,ε:Rⁿ →[0,∞) by

hϕ,ε(x) = inf λ >0 :ϕ

hK(x) λ

+εϕ

|x| λ

≤ϕ(1)

.

Observe that hϕ,ε is both sublinear and positive definite. Thus, there exists a unique convex bodyK_ϕ,ε∈ Kⁿo suchthatitssupportfunctionispreciselyh_ϕ,ε.

Accordingto Lemmas8.2and 8.4in[12],wehaveKϕ,ε→K,as ε→0⁺, and

∂

∂ε

ε=0⁺

h_ϕ,ε= h_K(u) ϕ₋(1)ϕ

1 hK(u)

, uniformly foru∈Sⁿ⁻¹. Here,ϕ₋(1) denotestheleft derivativeofϕat 1.

Note thath(ε,u)=hϕ,ε(u): (0,∞)×Sⁿ⁻¹ →(0,∞) is continuous ([54]). Thus, by Aleksandrov’svariationalprinciple(seeLemma3.1in[8],Lemma8.3in[12]orLemma 1 in[19]),ityieldsthat

d dε

ε=0⁺

|Kϕ,ε|= n ϕ₋(1)

Sⁿ⁻¹

ϕ 1

hK

dVK.

If ϕ(t)=t,then K_ϕ,ε=K_ε:={x+y:x∈K,y∈εB}isprecisely theouterparallel body[44] ofK.Correspondingly,theaboveformulareducesto theclassicalversion

d dε

ε=0⁺

|Kε|=|SK|=Hⁿ⁻¹(∂K).

Inthissense,we introducethefollowing

Definition 2.1. TheOrlicz surface area Sϕ(K),of aconvex bodyK ∈ Kⁿo withrespect toafunctionϕ∈Φ,isdefinedby

S_ϕ(K) =n

Sⁿ⁻¹

ϕ 1

h_K

dV_K.

Ifϕ(t)=t^p,1≤p<∞,then Sϕ(K) turnsto Sp(K),theLp surfaceareaofK.

RecallthattheOrlicz mixedvolumeVϕ(K,L),ofK,L∈ Kⁿo,isdefinedby

V_ϕ(K, L) = 1 n

Sⁿ⁻¹

ϕ h_L

hK

dV_K.

(See,e.g.,[12,52,54]).Thus,Sϕ(K)=nVϕ(K,B).

3. TheminimalOrliczsurfacearea

InordertodemonstratetheexistenceanduniquenessofminimalOrliczsurfacearea, webeginbyprovingtwo lemmas.

Lemma3.1. SupposeK∈ Kⁿo,ϕ∈Φand T ∈GL(n).Then

S_ϕ(T K) =n|T|

Sⁿ⁻¹

ϕ

h_T−1B

hK

dV_K.

Proof. From Definition 2.1,(2.1)and(2.2),wehave

Sϕ(T K) =n|T K|

Sⁿ⁻¹

ϕ 1

h_{T K}(u)

dVT K(u)

=n|T||K|

Sⁿ⁻¹

ϕ

|T^−t T^tu| hK( T^tu)

dV_K

T^tu

=n|T||K|

Sⁿ⁻¹

ϕ

h_T−1B( T^tu) h_K(T^tu)

dVK

T^tu

=n|T|

Sⁿ⁻¹

ϕ

h_T−1B

h_K

dVK,

asdesired. 2

Given anorigin-symmetricellipsoidE inRⁿ,letdE= max{hE(u):u∈Sⁿ⁻¹}.Then there existsavE∈Sⁿ⁻¹ suchthatforallu∈Sⁿ⁻¹,

dE|u·vE| ≤hE(u). (3.1)

Lemma 3.2.Suppose K∈ Kⁿo,ϕ∈ΦandT ∈SL(n).Then

d_T−1B ≤ n|K|ϕ⁻¹(^Sϕ_n|K|^{(T K)}) min_v∈Sn−1

Sⁿ⁻¹|u·v|dS_K(u), where ϕ⁻¹ istheinverse functionofϕ.

Proof. FromLemma 3.1,thestrictmonotonicityofϕtogetherwith(3.1),theconvexity of ϕtogether withJensen’sinequality,wehave

Sϕ(T K) n|K| =

Sⁿ⁻¹

ϕ

h_T−1B

h_K

dVK

≥

Sⁿ⁻¹

ϕ

d_T−1B|u·v_T−1B| hK(u)

dVK(u)

≥ϕ

Sⁿ⁻¹

d_T−1B|u·v_T−1B|

hK(u) dV_K(u)

≥ϕ

d_T−1B

n|K| min

v∈Sⁿ⁻¹

Sⁿ⁻¹

|u·v|dSK(u)

.

Since ϕ⁻¹ is strictly increasing in [0,∞), and min_v∈Sn−1

Sⁿ⁻¹|u·v|dSK(u)>0 by the fact thatSK is notconcentrated on any great subsphere,the desired inequalityis derived. 2

Withthepreviouslemmasinhand,wecanshowthattheminimalOrliczsurfacearea is well-defined.

Theorem 3.3. Suppose K ∈ Kⁿo and ϕ ∈ Φ. Then modulo orthogonal transformations, there existsauniquesolution totheminimization problem

min

T∈SL(n)Sϕ(T K).

Alternatively, by using Orlicz mixed volume, we can reformulate this theorem as follows: Theunitball B hasauniqueSL(n) imageE0 suchthat

Vϕ(K, E0) = min

Vϕ(K, T B) :T ∈SL(n) .

Proof. Let{Tk}k ⊂SL(n) beaminimizingsequence fortheproblem,thatis,

klim→∞S_ϕ(T_kK) = inf

S_ϕ(T K) :T ∈SL(n) .

Notethat

inf

S_ϕ(T K) :T ∈SL(n)

≤S_ϕ(K)<∞,

this implies {Sϕ(TkK)}k, therefore {ϕ⁻¹(^Sϕ_n|K|^(TkK))}k, is bounded from above. So, by Lemma 3.2, {T_k⁻¹B}k is bounded with respect to the Hausdorff metric. From the Blaschkeselectiontheorem,{T_k⁻¹B}k hasaconvergentsubsequence{T_k⁻¹

j B}j thatcon- vergestoabodyE.SincevolumefunctionaliscontinuouswithrespecttotheHausdorff metric, and |T_k⁻_j¹B| = ωn for each j, it yields that |E| = ωn; Since the conver- genceof {T_k⁻_j¹B}j is equivalent to theuniform convergence of{h_T−1

kj B}j onSⁿ⁻¹,and h_T−1

kjB(u)=h_T−1

kj B(−u) forallu∈Sⁿ⁻¹,ityieldsthath_E(u)=h_E(−u) forallu∈Sⁿ⁻¹. Thus,E isanon-degeneratedorigin-symmetricellipsoid.

Consequently, there exists atransformation T0 ∈ SL(n) such thatE =T₀⁻¹B. This demonstratestheexistenceof solutionsto theconsideredproblem.

Now, we prove the uniqueness by contradiction. Assume there are two solutions T₁,T₂ ∈ SL(n) to the considered problem, and they don’t differ only by an orthogo- naltransformation. Itis knownthatT_i⁻¹ canbe representedin theform T_i⁻¹ =PiQi, i= 1,2,wherePiisasymmetricpositivedefinitematrixandQiisanorthogonalmatrix.

Theabove assumption implies thatP1 = λP2, for allλ >0. Hence,the Minkowski inequalityforsymmetricpositivedefinitematricesshowsthat

det

P1+P2

2 _n¹

> 1

2det (P1)ⁿ¹ +1

2det (P2)ⁿ¹ = 1.

Let

T3=

det

P₁+P₂ 2

−n¹(P₁+P₂) 2

−1

. Then, T₃ ∈SL(n) andh_T−1

3 B(u)< h¹

2(P1+P2)B(u), forall u∈Sⁿ⁻¹. Sinceϕ isstrictly increasingandconvexin[0,∞),wehave

ϕ h_T−1

3 B

h_K

< ϕ h¹

2(P1+P2)B

h_K

≤ 1 2ϕ

hP1B

h_K

+1 2ϕ

hP2B

h_K

.

Thus,byLemma 3.1, wehave Sϕ(T3K) =n

Sⁿ⁻¹

ϕ h_T−1

3 B

h_K

dVK

<n 2

Sⁿ⁻¹

ϕ hP1B

h_K

dVK+n 2

Sⁿ⁻¹

ϕ hP2B

h_K

dVK

=1

2Sϕ(T1K) +1

2Sϕ(T2K)

=S_ϕ(T₁K)

=S_ϕ(T₂K).

Thatis,

Sϕ(T3K)< Sϕ(T1K) =Sϕ(T2K).

However,bythepreviousassumptiononT1 andT2,wehave Sϕ(T3K)≥Sϕ(T1K) =Sϕ(T2K), whichisacontradiction.Theproof iscomplete. 2

InviewofTheorem 3.3, naturally,weintroducethefollowing Definition 3.4.SupposeK∈ Koⁿ andϕ∈Φ.Thequantity

Aϕ(K) = min

Sϕ(T K) :T ∈SL(n)

is calledtheminimalOrlicz surfaceareaof theconvex bodyK withrespect toϕ.

Obviously, A_ϕ(K) isSL(n) invariantand ageneralizationof Petty’sminimal surface area. If ϕ(t)=t^p,1≤p< ∞,then the notionof minimal Orlicz surface areareduces to thatofminimal Lp surfaceareaandparticularlythatofminimaltotalLp curvature, whenevertheboundary∂Kisof classC₊². See[34].

4. AcharacterizationoftheminimalOrliczsurfacearea

Throughoutthissection,weimposeaconditiononϕ∈Φ,thatϕissmoothin(0,∞).

SupposeK∈ Kⁿo andϕ∈Φ∩C¹(0,∞).Recallthat Sϕ(K, ω) =

ω

ϕ 1

hK

dSK,

foreachBorelsubsetω⊆Sⁿ⁻¹.

For further discussion, we introduce the important notion of isotropy of measures.

A nonnegativeBorelmeasureμonSⁿ⁻¹ issaidtobe isotropicif

Sⁿ⁻¹

(u·v)²dμ(u) = |μ|

n , for allv∈Sⁿ⁻¹. Thedefinitionimmediately yields

Sⁿ⁻¹

u²_idμ(u) = |μ| n ,

whereuidenotestheithcomponentofthecoordinateofu.Formoreinformationabout theisotropy,wereferto[1,2,6,13,14,37].

Forx∈Rⁿ\ {0},themapx⊗x:Rⁿ →Rⁿ istherank1 linearoperatory→(x·y)x.

So,itfollowsthattr(x⊗x)=|x|².

Thenexttheorem characterizestheconvexbodywithminimalOrliczsurfacearea.

Theorem4.1. SupposeK ∈ Kⁿo, ϕ∈Φ∩C¹(0,∞) andT₀ ∈SL(n).Then thefollowing assertionsare equivalent:

(1) Aϕ(K)=Sϕ(T0K).

(2) The measureSϕ(T0K,·) isisotropiconSⁿ⁻¹. (3) Forallx∈Rⁿ,thetransformationT0 satisfies

|x|²

ω

T₀^−tuϕ

|T₀^−tu| hK(u)

dSK(u) =n

Sⁿ⁻¹

|x·T₀^−tu|²

|T₀^−tu| ϕ

|T₀^−tu| hK(u)

dSK(u).

Proof. First,weprovetheequivalenceof (1)and(2).

Suppose that(1) holds. Since Aϕ(K) isSL(n) invariant, we mayassume thatT0 is then×nidentitymatrixIn.

Let T be alinear transformation. Then there exists ε0 > 0 such thatfor each ε ∈ (0,ε0),thematricesIn+εT andIn−εT arestillpositivedefinite.Forε∈(0,ε0),define

T_ε= In+εT

|In+εT|ⁿ¹.

Ifεissufficientlysmall,fromLemma 3.1,thesmoothnessofϕ,togetherwiththetwo equalities

|In+εT|¹ⁿ = 1 + ε

ntrT+o ε²

, (I_n+εT)^−tu= 1−εu·T^tu+o

ε² ,

andthedefinitionofS_ϕ,wehave Sϕ(TεK) =

Sⁿ⁻¹

ϕ

|T_ε^−tu| h_K(u)

hK(u)dSK(u)

=

Sⁿ⁻¹

ϕ

|In+εT|¹ⁿ|(In+εT)^−tu| hK(u)

hK(u)dSK(u)

=S_ϕ(K) +ε

Sⁿ⁻¹

trT

n −u·T^tu

ϕ 1

hK(u)

dS_K(u) +o ε²

=Sϕ(K) +ε

Sⁿ⁻¹

trT

n −u·T^tu

dSϕ(u) +o ε²

.

BytheassumptionthatS_ϕ(T_εK)≥S_ϕ(K),itimmediatelyyields

Sⁿ⁻¹

trT

n −u·T^tu

dS_ϕ(u)≥0.

Replacing T by−T intheaboveinequality,weget

Sⁿ⁻¹

trT

n −u·T^tu

dS_ϕ(u)≤0.

Hence,

Sⁿ⁻¹

u·T^tudSϕ(K, u) =trT

n Sϕ(K,·).

Taking T =x⊗x,with x∈Rⁿ\ {0},andusing thefactsu·(x⊗x)^tu=|x·u|² and tr(x⊗x)=|x|²,wehave

Sⁿ⁻¹

(x·u)²dSϕ(K, u) = |S_ϕ(K,·)| n |x|², whichshows theisotropyofSϕ(K,·).

Next, we show the implication “(2) =⇒ (1)”. The proof will be completed by two steps.

Firstly,forapoint a= (a₁,· · ·,a_n)∈[0,∞)ⁿ,let F(a) =

Sⁿ⁻¹

ϕ

|diag(a₁,· · ·, a_n)u| hK(u)

dV_K(u),

where diag(a1,· · ·,an) denotes thediagonalmatrixwithdiagonalelements a1,· · ·,an. Weaimto showthat

F(a)≥F(e), whenevera1a2· · ·an= 1. (4.1) Here,edenotesthepoint(1,· · ·,1).

Itcanbe checked thatF : [0,∞)ⁿ →[0,∞) is continuousand convex, andF(λa) is strictly increasing inλ∈[0,∞), forany a∈(0,∞)ⁿ. Thus, F⁻¹([0,F(e)]) is compact, convexandofnon-emptyinterior. Precisely,itisaconvexbody.

By the smoothness of ϕ and that |diag(a1,· · ·,an)u| is smooth in (a1,· · ·,an) uni- formlyforu∈Sⁿ⁻¹,wehave

∂

∂a_j

a=e

F(a) =

Sⁿ⁻¹

∂

∂a_j

a=e

ϕ

|diag(a1,· · ·, an)u| h_K(u)

dV_K(u)

= 1 n

Sⁿ⁻¹

u²_jdS_ϕ(K, u).

Meanwhile, since the boundary of F⁻¹([0,F(e)]) is given by the equation F(a) = F(e) with a ∈ Rⁿ+, so the vector

Sⁿ⁻¹(u²₁,· · ·,u²_n)dSϕ(K,u) is an outer normal of F⁻¹([0,F(e)]) at theboundarypointe.Note thatS_ϕ(K,·) isisotropic,ityields

Sⁿ⁻¹

u²₁,· · ·, u²_n

dS_ϕ(K, u) = |S_ϕ(K,·)|

n (1,· · ·,1) = |S_ϕ(K,·)| n e.

Thus,eisanouternormalofF⁻¹([0,F(e)]) attheboundarypointe. Consequently, F⁻¹

0, F(e)

⊂

a∈Rⁿ:a·e≤n .

Thatis to say, for alla ∈ [0,∞)ⁿ, if F(a)≤ F(e), then a·e≤ n. In contrast, for all b= (b1,· · ·,bn)∈(0,∞)ⁿ withb1· · ·bn = 1,theAM-GMinequalityyieldsthatb·e≥n, withequalityifandonlyifb=e.Thus,(4.1)isderived.

Secondly,with(4.1)inhand,weaimtoshowthatforallT ∈SL(n),S_ϕ(T K)≥S_ϕ(K), withequalityifandonlyifT isorthogonal.

ItisknownthatT^−tcanberepresentedintheformT^−t=P AQ,whereP andQare n×n orthogonal matrices, and A = diag(a₁,· · ·,a_n) is diagonal and positivedefinite.

So,byLemma 3.1, (2.2), (4.1), andLemma 3.1again, wehave

Sϕ(T K) =n

Sⁿ⁻¹

ϕ

|Au| h_QK(u)

dVQK(u)

=n

Sⁿ⁻¹

ϕ

|diag(a₁,· · ·, a_n)u| hQK(u)

dV_QK(u)

≥n

Sⁿ⁻¹

ϕ

|diag(1,· · ·,1)u| hQK(u)

dVQK(u)

=n

Sⁿ⁻¹

ϕ 1

h_QK(u)

dVQK(u)

=S_ϕ(QK)

=S_ϕ(K),

equality holds if and only if (a1,· · ·,an) = (1,· · ·,1), equivalently, if and only if T is orthogonal. Thus,theimplication“(2) =⇒(1)” isshown.

Intherest, weprovetheequivalenceof(2)and (3).

From the definitions of Sϕ(T0K,·) and cone-volume measure, (2.1), and (2.2), we have

dSϕ(T0K, u) =nϕ 1

h_T0K(u) 1

h_T0K(u)dVT0K(u)

=nϕ

|T₀^−t T₀^tu| h_K( T₀^tu)

|T₀^−t T₀^tu| h_K( T₀^tu)dV_K

T₀^tu

=ϕ

|T₀^−t T₀^tu| hK( T₀^tu)

T₀^−t

T₀^tudSK

T₀^tu ,

whichimmediately yieldsthat

Sⁿ⁻¹

T₀^−tvϕ

|T₀^−tv| hK(v)

dSK(v) =Sϕ(T0K,·).

Meanwhile, forx∈Rⁿ we have

Sⁿ⁻¹

|x·u|²dSϕ(T0K, u)

=

Sⁿ⁻¹

x·T₀^−t

T₀^tu²T₀^tu²ϕ

|T₀^−t T₀^tu| hK( T₀^tu)

T₀^−t

T₀^tudSK

T₀^tu

=

Sⁿ⁻¹

x·T₀^−tv²T₀^−tv⁻²ϕ

|T₀^−tv| hK(v)

T₀^−tvdSK(v)

=

Sⁿ⁻¹

x·T₀^−tv²T₀^−tv⁻¹ϕ

|T₀^−tv| hK(v)

dSK(v).

Withthese,theequivalenceof(2)and(3)isshown.

Theproof iscomplete. 2

5. BoundsfortheminimalOrliczsurfacearea

Inthissection, weestimate theminimalOrlicz surfaceareaAϕ(K).Lemma 5.1 and Theorem 5.2givelowerbounds. Theorem 5.3andTheorem 5.7giveupperbounds.

WriteA(K)= min{Hⁿ⁻¹(∂(T K)):T ∈SL(n)} for theminimal surface areaof K.

Note thatforour purpose,this quantityis alittledifferent from thatof Giannopoulos andPapadimitrakis[14].

Thenextlemmashows therelationshipbetweenA_ϕ(K) andA(K),and isneededin theproofsofTheorem 5.2, Lemma 5.4andLemma 5.5.

Lemma5.1. SupposeK∈ Kⁿo andϕ∈Φ.Then

Aϕ(K)≥n|K|ϕ A(K)

n|K|

. (5.1)

If K has an SL(n) image K such that: (1) SK is isotropic; (2) hK|suppSK, that is, the restriction of h_K to the support set of S_K, is constant, then equality holds in(5.1).

Conversely,ifϕisstrictlyconvex,thenequalityholdsin(5.1)onlyifKhasanSL(n) image K whichsatisfies(1) and(2).

Proof. ForT ∈SL(n),recallthat Sϕ(T K)

n|K| =

Sⁿ⁻¹

ϕ 1

h_{T K}

dV_{T K},

and

Hⁿ⁻¹(∂(T K)) n|K| =

Sⁿ⁻¹dS_{T K} n|K| =

Sⁿ⁻¹

1 hT K

dVT K.

SinceϕisconvexandV_{T K} isaprobabilitymeasure,byJensen’sinequality,wehave Sϕ(T K)

n|K| ≥ϕ

Sⁿ⁻¹

1

h_{T K}dVT K

=ϕ

Hⁿ⁻¹(∂(T K)) n|K|

,

whichyields(5.1)bytheexistenceofAϕ(K) andA(K).

Weproceedtoprovetheequalitycondition.

Ononehand,byPetty’sminimalsurfaceareatheoremand (1),wehave A(K) =Hⁿ⁻¹

∂K .

ByTheorem 4.1,(1)and(2),wehave Aϕ(K) =Sϕ

K

=n|K|ϕ

Hⁿ⁻¹(∂K) n|K|

.

Thus, Aϕ(K)=n|K|ϕ(^A(K)_n|K|).

Conversely,theequalityA_ϕ(K)=n|K|ϕ(^A(K)_n|K|),aswellastheexistenceofA(K) and Aϕ(K),impliesthatK hastwoSL(n) imagesK1and K2 whichsatisfythefollowing:

(3) S_ϕ(K₁)=A_ϕ(K).

(4) Sϕ(K1)=n|K|ϕ(^Hn−1_n|K|^(∂K2)).

(5) For allT ∈ SL(n), Hⁿ⁻¹(∂(T K2))≥ Hⁿ⁻¹(∂K2), with equality ifand onlyif T is orthogonal.

TheprovedinequalitySϕ(K1)≥n|K|ϕ(^Hn−1_n|K|^(∂K1)) togetherwith(4), yields

ϕ

Hⁿ⁻¹(∂K2) n|K|

≥ϕ

Hⁿ⁻¹(∂K1) n|K|

.

Since ϕis strictlyincreasing,wehave

Hⁿ⁻¹(∂K2)≥ Hⁿ⁻¹(∂K1).

Withthisand(5),weconcludethatK1differsfromK2onlybyanorthogonaltransfor- mation. Thus, byPetty’sminimal surfaceareatheorem,weknow thatS_K1 isisotropic onSⁿ⁻¹.Moreover,bytheorthogonalinvarianceofHⁿ⁻¹and(4), wehave

Sϕ(K1) =n|K|ϕ

Hⁿ⁻¹(∂K1) n|K|

.

Thatis,

Sⁿ⁻¹

ϕ 1

h_K1

dV_K1=ϕ

Sⁿ⁻¹

1 h_K1dV_K1

.

SinceVK1 isaprobabilitymeasureandϕisstrictlyconvex,bytheequalitycondition of Jensen’sinequality,itfollowsthath_K1|suppVK1,i.e.,h_K1|suppSK1,isconstant. 2 Theorem 5.2.SupposeK∈ Kⁿo andϕ∈Φ.Then

Aϕ(K)≥n|K|ϕ |B|

|K| ¹_n

. (5.2)

If ϕ is strictly convex, then equality holds in (5.2) if and only if K is an origin- symmetricellipsoid.

Proof. ApplyingtheclassicalisoperimetricinequalitytoA(K),wehave A(K)

n|K| ≥ |B|

|K| _n¹

,

withequalityifandonlyifKis anellipsoid.Withthisand(5.1),(5.2)isestablished.

Supposeϕisstrictlyconvex.SinceK∈ Kⁿo,byLemma 5.1andtheequalitycondition in the above inequality, it follows that equality holds in (5.2) if and only if K is an origin-symmetricellipsoid. 2

In what follows, we use the L_∞ John ellipsoid discovered in [34] to estimate the minimalOrliczsurfaceareaAϕ(K).

Suppose K ∈ Kⁿo. Recall that the L_∞ John ellipsoid, E_∞K, is the unique origin- symmetricellipsoid contained inK with maximal volume.Thatis, amongst allorigin- symmetricellipsoidsE,E_∞Kistheuniqueonethatsolvestheconstrainedmaximization problem

maxE |E| subject toV_∞(K, E)≤1,

whereV_∞(K,E)= max{^h_h^E_K^(u)_(u) :u∈suppSK}.Indeed,E_∞K necessarilysatisfies V_∞(K,E_∞K) = 1.

WriteE_∞K for(_|E^|B|

∞K|)^1/nE_∞K. Asitwasshownin[34],E_∞K istheuniqueSL(n) imageofB whichsatisfies

V_∞(K,E_∞K) = min

V_∞(K, T B) :T ∈SL(n) .

ItisknownthattheclassicalJohnellipsoidJKofK,istheuniqueellipsoidcontained inK withmaximal volume.NotethatiftheJohn pointofK, i.e.,thecenterof JK, is attheorigin,thenE_∞KispreciselyJK.FormoreinformationabouttheJohnellipsoid, wereferto[3,13,18,17,22–24].

Theorem5.3. SupposeK∈ Koⁿ andϕ∈Φ.Then

Aϕ(K)≤n|K|ϕ

|B|

|E_∞K| _n¹

.

Proof. Suppose T ∈SL(n) and1≤p<∞. From Lemma 3.1,Jensen’s inequality,and thedefinitionofV_∞,wehave

Sϕ(T K) n|K| =

Sⁿ⁻¹

ϕ

h_T−1B

h_K

dVK

≤

Sⁿ⁻¹

ϕ

h_T−1B

hK

p

dVK

1/p

≤ lim

p→∞

Sⁿ⁻¹

ϕ

h_T−1B

hK

p

dV_K 1/p

= max ϕ

h_T−1B(u) h_K(u)

:u∈suppSK

=ϕ

max h_T−1B(u)

hK(u) :u∈suppSK

=ϕ V_∞

K, T⁻¹B .

Thatis,

Sϕ(T K) n|K| ≤ϕ

V_∞

K, T⁻¹B

. (5.3)

Now, from(5.3)andthedefinitionsofA_ϕ(K),E_∞K,V_∞ andE_∞K,itfollowsthat A_ϕ(T K)

n|K| ≤min ϕ

V_∞

K, T⁻¹B

:T ∈SL(n)

=ϕ min

V_∞

K, T⁻¹B

:T ∈SL(n)

=ϕ

V_∞(K,E_∞K)

=ϕ

|B|

|E_∞K| 1/n

V_∞(K,E_∞K)

=ϕ

|B|

|E_∞K| 1/n

,

as desired. 2

If theJohn pointof K is atthe origin,two precise upperbounds forAϕ(K) canbe obtained.

Lemma 5.4.Suppose Tn isann-dimensionalregularsimplex inRⁿ.Then (1) The surfaceareameasure S_Tn isisotropicon Sⁿ⁻¹.

(2) A(Tn)=Hⁿ⁻¹(∂Tn).

(3) If theJohnpointof Tn isattheorigin, thenforallϕ∈Φ, Aϕ(Tn) =Sϕ(Tn) =n|Tn|ϕ

A(Tn) n|Tn|

.

Proof. Lete1,· · ·,en bethestandardorthonormalbasisofRⁿandtakeRⁿ⁺¹asRⁿ×R. Letu⁽ⁿ⁾₁ ,· · ·,u⁽ⁿ⁾_n+1betheouterunitnormalvectorsofTn.Withoutlossofgenerality,we mayassumeu⁽ⁿ⁾_n+1=−e_n.ThenS_Tn canbe representedby

S_Tn= |S_Tn| n+ 1

n+1

j=1

δ_u(n) j

,

whereδ_u(n)

j denotestheDiracmeasureat u⁽ⁿ⁾_j .

Weprove(1)byinductiononthedimensionn. Ifn= 1,itisobvious.

Assume(1)holdsforn.Then,forallx₁∈Rⁿ,fromthedefinitionofisotropy,wehave

n+1

j=1

x1·u⁽ⁿ⁾_j ²

= n+ 1

|S_Tn| · |STn| n+ 1

n+1

j=1

x1·u⁽ⁿ⁾_j ²

= n+ 1

|S_Tn| ·|STn|

n |x1|²= n+ 1 n |x1|².

ThefactthatthecentroidofSTn is attheorigin,i.e.,n+1 j=1 |S_Tn|

n+1u⁽ⁿ⁾_j = 0,impliesthat forallx1∈Rⁿ,

n+1

j=1

x₁·u⁽ⁿ⁾_j = 0.

Notethatforeachj= 1,· · ·,n+ 1, u⁽ⁿ⁺¹⁾_j = 1

n+ 1

n²+ 2nu⁽ⁿ⁾_j ,1 .

Hence,forallx= (x1,x2)∈Rⁿ×R,

|S_Tn+1| n+ 2

n+2

j=1

x·u⁽ⁿ⁺¹⁾_j ²=|S_Tn+1| n+ 2

n²+ 2n (n+ 1)²

n+1

j=1

x₁·u⁽ⁿ⁾_j 2

+n+ 2 n+ 1x₂²

=|S_Tn+1| n+ 2

n²+ 2n (n+ 1)² ·n+ 1

n |x1|²+n+ 2 n+ 1x²₂

=|STn+1| n+ 1 |x|²,

whichshowstheisotropyofS_Tn+1,andthereforeconcludes(1).

Petty’sminimal surfaceareatheoremimpliesthat(1)and(2)areequivalent.