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Advances in Mathematics
www.elsevier.com/locate/aim
Existence of solutions to the Orlicz–Minkowski problem
✩Huaiyu Jiana,Jian Lub,∗
a DepartmentofMathematics,TsinghuaUniversity,Beijing100084,China
b DepartmentofAppliedMathematics,ZhejiangUniversityof Technology, Hangzhou310023,China
a r t i c l e i n f o a bs t r a c t
Article history:
Received18January2018 Receivedinrevisedform14 December2018
Accepted18December2018 Availableonline10January2019 CommunicatedbyErwinLutwak
MSC:
35J96 35J75 34C40
Keywords:
Orlicz–Minkowskiproblem Monge–Ampèreequation Alexandrovbody Variationalmethod
InthispapertheOrlicz–Minkowskiproblem,ageneralization of the classical Minkowski problem, is studied. Using the variational method, we obtain a new existence result of solutionstothisproblemforgeneralmeasures.
©2019ElsevierInc.Allrightsreserved.
1. Introduction
In recent years, the Orlicz–Brunn–Minkowski theory in convex geometry has been built up gradually and is developing rapidly. It can be viewed as the recent develop-
✩ TheauthorsweresupportedbyNaturalScienceFoundationofChina(11771237,11871432,11401527).
* Correspondingauthor.
E-mailaddresses:[email protected](H. Jian),[email protected](J. Lu).
https://doi.org/10.1016/j.aim.2019.01.004 0001-8708/©2019ElsevierInc. Allrightsreserved.
mentof the classicalBrunn–Minkowskitheory, and hasattracted great attention from manyscholars,seeforexample[4,7–9,11,13,14,16,17,21,22,27,29,30,32–35,37–40,44] and references therein. In the Brunn–Minkowskitheory, it is well known thatthe classical Minkowski problem is of central importance, and has many applications. In the new Orlicz–Brunn–Minkowski theory, the corresponding Minkowski problem is called the Orlicz–Minkowskiproblem.
Letϕ: (0,+∞)→(0,+∞) beagivencontinuousfunction.ForaconvexbodyK⊂Rn withtheorigin0∈K,theOrliczsurfaceareameasureisdefinedasϕ(hK)dSK.HerehK
isthesupportfunctionofK,andSK isthesurfaceareameasure.TheOrlicz–Minkowski problem,firstproposedin[10],askswhatarethenecessaryandsufficientconditionsfor aBorelmeasure μ onthe unitsphere Sn−1 to be amultipleof theOrlicz surfacearea measure of aconvex body K. Namely,this problem isto finda convex bodyK ⊂Rn suchthat
c ϕ(hK)dSK=dμonSn−1 (1.1) forsomepositiveconstant c. Sincethe Orlicz surfaceareameasure dependson ϕ,it is alsocalled Lϕ-surfaceareameasure.Correspondingly,theOrlicz–Minkowskiproblem is sometimescalled theLϕ-Minkowskiproblem.
When ϕ is a constant function, Eq. (1.1) is just the classical Minkowski problem.
When ϕ(s) = s1−p, Eq. (1.1) reduces to the Lp-Minkowski problem, which has been extensively studied, see e.g. [1–3,12,15,17–20,23–26,28,31,41,43] and Schneider’s book [36],andcorresponding referencestherein.
When the Radon–Nikodym derivative of μ with respect to the spherical measure on Sn−1 exists,namely dμ =f dx for a non-negative integrable function f, the equa- tion (1.1) canbewrittenas
c ϕ(hK) det(∇2hK+hKI) =f onSn−1, (1.2) where∇2hK = (∇ijhK) istheHessianmatrixofcovariantderivativesofhKwithrespect to an orthonormal frame on Sn−1, and I is the unit matrix of order n−1. This is a Monge–Ampèretypeequation.
Eq. (1.1) has a variational structure, which can be used to prove the existence of solutions[10,14,37]. Haberl, Lutwak,Yang and Zhang [10] considered theeven Orlicz–
Minkowskiproblem undertheassumption
(A)ϕ: (0,+∞)→(0,+∞)isacontinuousfunctionsuchthat φ(t)=t
01/ϕ(s)dsexists forevery t>0andisunbounded ast→+∞.
Theyprovedthefollowing
Theorem 1.1 ([10, Theorem 2]). Suppose (A) is satisfied. If μ is an even finite Borel measureonSn−1 whichisnotconcentratedon anygreat sub-sphere ofSn−1,then there
exists an originsymmetric convex body K in Rn anda number c >0satisfying (1.1).
Moreover, onecan requirethat theOrlicz-norm ofhK isequalto1.
WenotethattheOrlicz-normofhK inthistheorem isdefinedwithrespect toφ(t)= t
01/ϕ(s)dsandμ.DenotingitbyhKφ,μ,wehavethat hKφ,μ= inf
⎧⎨
⎩λ >0 : 1 μ(Sn−1)
Sn−1
φ hK
λ
dμ≤φ(1)
⎫⎬
⎭.
One can consult[10, Section4] for moreproperties about this norm. We will use this notation throughoutthispaper.
HuangandHe[14] studiedthegeneral(notnecessarilyeven)Orlicz–Minkowskiprob- lem andobtainedthefollowingresult.
Theorem 1.2 ([14, Theorem 1.2]).In addition to (A), further suppose that ϕ(s) tends to +∞as s→0+.If μ isa finite Borelmeasureon Sn−1 which isnot concentratedin any closed hemisphere of Sn−1,then thereexists aconvex bodyK in Rn andanumber c>0satisfying (1.1). Moreover,one can requirethat theOrlicz-norm hKφ,μ is equal to 1.
We note that Theorem 1.1 includes the even Lp-Minkowski problem for p > 0, and Theorem 1.2 includes the general Lp-Minkowski problem for p > 1. There is no result about the general Orlicz–Minkowski problem which can include the general Lp-Minkowskiproblem for0< p<1.Inthispaper,we willfillthisgap.Weobtainthe following
Theorem1.3.Inadditionto(A),furthersupposethatϕisnon-decreasingandϕ(s)tends to0ass→0+.Ifμ isafinite BorelmeasureonSn−1 whichisnotconcentratedinany closed hemisphereofSn−1,thenthereexistsaconvexbodyK inRn andanumberc>0 satisfying (1.1).Moreover, onecan requirethatthevolumeof hK isequalto1.
Onecanseethatϕ(s)=s1−pwith0< p<1 satisfiestheassumptionsofTheorem1.3.
Therefore this theorem includes the general Lp-Minkowski problem for 0 < p < 1.
ActuallytherewasevennoexistenceresultaboutthegeneralLp-Minkowskiproblemfor 0< p<1 whenourpaperwascompleted,whilenow thereisoneexistenceresultabout this Lp-Minkowskiproblem[5].
The method of proving Theorem 1.3 is the variational method, which was used to studytheOrlicz–Minkowskiproblemin[10,14] andtheLp-Minkowskiproblemin[6,42].
Howeverourmethodisnotadirectgeneralizationofthesepreviousmethods,since[10]
isfortheoriginsymmetriccase,whilein[6,14,42] extremumproblemswereconsideredin aclassofsupportfunctionsofconvexbodies,whichadditionallyneedtoanalyzerelated propertiesof extremumconvex bodieswhencomputing variations,see Lemmas 5.5–5.6
in[6] orLemmas3.5–3.6in[42].Toovercomethesedifficulties,inthispaperweuseanew techniquecombiningthefunctionalsgivenin[6] and[10],andmakinguseofAlexandrov bodies to compute an extremum problem inthe class of positive continuous functions on Sn−1. No additional properties of extremum convex bodies will be needed in our method.
Thepaperisorganizedas follows. Insection 2, weprovidesomepreliminaries about convexbodies.Insection3,westateTheorem3.1whichismoregeneralthanTheorem1.3 and prove a special discrete case of the theorem. Then we complete the proof of the theoreminsection4.
2. Preliminaries
In this section we state some notations and basicfacts about convex bodies which willbeusedthroughoutthispaper.Forgeneralreferencesaboutconvexbodies,onecan consult[36].
A convex body is a compact convex subset of Rn with non-empty interior. For a convex body K, we use intK to denote the interior of K. The support function of a convexbody,denotedbyhK,isgivenby
hK(x) := max
ξ∈Kξ·x, x∈Sn−1,
where“·”denotestheinner productintheEuclideanspace Rn. Itiswell knownthata convex bodyis uniquely determined by itssupport function,and the convergence ofa sequenceofconvexbodiesisequivalenttotheuniformconvergenceofthecorresponding support functions on Sn−1. The Blaschke selection theorem says that every bounded sequenceofconvexbodieshasasubsequencethatconvergestoaconvexbody.
Denote the set of positive continuous functions on Sn−1 by C+(Sn−1). For g ∈ C+(Sn−1) and a closed subset ω ⊂ Sn−1 not lying in any closed hemisphere, define theAlexandrovbody associatedwith(g,ω) as
K:=
x∈ω
{ξ∈Rn:ξ·x≤g(x)}.
OnecanseethatK isaboundedconvex bodyand0∈K.Note that hK(x)≤g(x), x∈ω.
Wewritevol(g,ω) forthevolumeoftheAlexandrovbodyassociatedwith(g,ω).Forthe conceptofAlexandrovbody,thereisausefulvariationalformuladuetoAlexandrov,see e.g.[36, Lemma7.5.3].
Lemma 2.1.Let >0. AssumeGt(x)=G(t,x): (−,)×ω →(0,+∞)is continuous.
If thereisacontinuous functiong onω such that
t→lim0+
Gt−G0
t =g uniformly onω, then
tlim→0+
vol(Gt, ω)−vol(G0, ω)
t =
ω
g(x)dSK(x),
where KistheAlexandrovbodyassociatedwith(G0,ω).The sameassertionholdsifthe one-sidedlimit limt→0+ isreplacedby limt→0− orby limt→0.
ForafiniteBorelmeasureμonSn−1,denoteits supportsetbysupp(μ),anditstotal massμ(Sn−1) by|μ|.
3. Aspecialdiscretecase
Inthispaper, insteadofprovingTheorem1.3directly,wewillprovethefollowing Theorem 3.1.Suppose φ : (0,+∞) → (0,+∞) is an increasing concave C1 function satisfying that limt→+∞φ(t)= +∞,φ(t)>0and limt→0+φ(t)= +∞.If μis afinite BorelmeasureonSn−1whichisnotconcentratedinanyclosedhemisphereofSn−1,then there existsaconvexbody K inRn andanumberc>0satisfying
c
φ(hK)dSK=dμ onSn−1. (3.1) Moreover, onecan requirethevolumeof hK isequaltoany givennumber v >0.
OnecaneasilyproveTheorem 1.3byvirtueofTheorem3.1.
Proof of Theorem1.3. Given ϕas inTheorem 1.3,wedefine φas
φ(t) = t
0
1/ϕ(s)ds, ∀t >0.
Byassumption(A), φisanincreasing C1 functionin(0,+∞) satisfying
t→+∞lim φ(t) = +∞.
Note φ= 1/ϕ>0 isnon-increasing,thenφisaconcave function.Also
tlim→0+φ(t) = lim
t→0+
1
ϕ(t) = +∞.
So φ satisfies all assumptions of Theorem 3.1. Bythis theorem, there exists a convex bodyK inRn andanumberc>0 satisfying
c
φ(hK)dSK=dμonSn−1, namely
c ϕ(hK)dSK=dμonSn−1,
whichisjustequation(1.1).Theorem3.1 alsosaysthatthevolumeofhK canbeequal to any given positivenumber v. We choose v = 1. Now Theorem 1.3 is proved to be true. 2
Fromnowon,we onlyfocus onTheorem3.1.
Inthis section,wemainlyprovethefollowing lemma,whichisaspecialdiscretecase ofTheorem3.1.
Lemma 3.2.Suppose φ: (0,+∞)→(0,+∞) isan increasing concave C2 functionsat- isfyingthat limt→+∞φ(t)= +∞, φ(t)>0, limt→0+φ(t)= +∞, and φ(t)<0. If μ isafinitediscretemeasureon Sn−1which isnotconcentratedinany closedhemisphere of Sn−1,then there existsa convex body K in Rn containing theorigin in itsinterior, andanumberc>0satisfyingEq.(3.1).Moreover,one canrequirethevolumeof hK is equaltoany givennumberv >0.
WeuseavariationalmethodtoprovethisLemma.Foranyfixedpositiveconstantv, weconsiderthefollowing minimizingproblem:
inf
ξ∈Ksupg
J[g(x)−ξ·x] :g∈C+(Sn−1), vol(Kg) =v
, (3.2)
whereKg istheAlexandrovbodyassociatedwith(g,supp(μ)),and J[g] =
Sn−1
φ(g(x))dμ(x) =
supp(μ)
φ(g(x))dμ(x). (3.3)
Notewhenξ∈Kg andx∈supp(μ),there is
g(x)−ξ·x≥hKg(x)−ξ·x≥0.
Bytheassumptions of φ,we candefine φ(0) aslimt→0+φ(t) which exists and isfinite.
Noteφ(0)≥0.ThereforeJ[g(x)−ξ·x] in (3.2) iswell-defined.TheproofofLemma3.2 willbe carriedoutinthefollowingLemmas3.3–3.6andfinished afterLemma3.6.
Lemma 3.3. Assume that φ: (0,+∞)→(0,+∞)is an increasing concave C1 function satisfying limt→0+φ(t)= +∞,andthatμisafinitediscretemeasureonSn−1 whichis not concentratedinany closedhemisphereof Sn−1.Thenforeverynon-negative contin- uous function g on Sn−1 with Kg havingnonempty interior, there isat leastone point of Kg,denoted byξg,suchthat
J[g(x)−ξg·x] = sup
ξ∈Kg
J[g(x)−ξ·x]. (3.4)
And forany such point, wehave ξg ∈intKg.If φ is additionallystrictly concave, then ξg is unique,and dependscontinuouslyon g wheng∈C+(Sn−1).
Proof. DefineG:Kg→Ras
G(ξ) :=J[g(x)−ξ·x] =
Sn−1
φ(g(x)−ξ·x)dμ(x).
WeclaimthatGisconcavewithrespecttoξ.Infact,forλ1,λ2∈(0,1) withλ1+λ2= 1, and ξ1,ξ2∈Kg, wehave
G(λ1ξ1+λ2ξ2) =
Sn−1
φ[g(x)−(λ1ξ1+λ2ξ2)·x]dμ(x)
=
Sn−1
φ[λ1(g(x)−ξ1·x) +λ2(g(x)−ξ2·x)]dμ(x)
≥
Sn−1
[λ1φ(g(x)−ξ1·x) +λ2φ(g(x)−ξ2·x)]dμ(x)
=λ1G(ξ1) +λ2G(ξ2).
Here we have used the concavity of φ. If φ is additionally strictly concave, when the aboveequalityholds, theremustbe
g(x)−ξ1·x=g(x)−ξ2·x, ∀x∈supp(μ), namely
(ξ1−ξ2)·x= 0, ∀x∈supp(μ).
Recall μ is notconcentrated on any closed hemisphere,then supp(μ) spans the whole space Rn.Thusξ1=ξ2,whichimpliesGisstrictly concaveonKg.
Note GiscontinuousontheconvexbodyKg,there existsatleastonepointξg∈Kg suchthat
G(ξg) = sup
ξ∈Kg
G(ξ),
whichisjust(3.4).Weneedtoproveξg∈intKg.Otherwise,supposeξg∈∂Kg.Wewill provethatforsomee∈Sn−1andsmallλ>0,ξg+λe∈intKgandG(ξg+λe)> G(ξg), whichleadsto acontradiction.
Recallthedefinitionof Kg:
Kg=
x∈supp(μ)
{ξ∈Rn :ξ·x≤g(x)},
theremustexist onex∈supp(μ) suchthat ξg·x=g(x),
sinceotherwise ξg·x+δ < g(x) for someδ > 0 andevery x∈supp(μ), which would implyξg ∈intKg.
Nowwewritesupp(μ) astheunionoftwodisjoint nonemptysets:
supp(μ) =A∪B, (3.5)
where
A:={x∈supp(μ) :ξg·x=g(x)}, B:={x∈supp(μ) :ξg·x < g(x)}.
Byvirtueoftheassumption thatKg hasnonempty interior,onecanfindaunitvector e∈Sn−1 suchthat
e·x <0 for everyx∈A. (3.6) Since supp(μ) isadiscrete set bythe assumption,B is aclosed subset of Sn−1. Then thereexists aλ0>0 satisfying
ξg·x+ 2λ0< g(x), ∀x∈B.
Thusforany0< λ<2λ0, wehave
(ξg+λe)·x < g(x), ∀x∈supp(μ).
BythedefinitionofKg,
ξ(λ) :=ξg+λe∈intKg.
Wewanttoprove
G(ξ(λ))> G(ξg) forsufficiently smallλ,whichisacontradiction.
Recalling (3.5) andφ(0) iswelldefined,wehave G(ξ(λ))−G(ξg) =
A∪B
φ[g(x)−ξ(λ)·x]dμ(x)−
A∪B
φ[g(x)−ξg·x]dμ(x)
=
A
φ[g(x)−ξ(λ)·x]−φ(0)
dμ(x) (3.7)
+
B
φ[g(x)−ξ(λ)·x]−φ[g(x)−ξg·x]
dμ(x).
Note that
g(x)−ξ(λ)·x=−λe·x, ∀x∈A.
Andwe canstrengthen(3.6) as
e·x <−δ0<0 ∀x∈A
forsomeconstantδ0>0.Thenthefirstintegralintheendof(3.7)
A
φ[g(x)−ξ(λ)·x]−φ(0)
dμ(x) =
A
φ(−λe·x)−φ(0) dμ(x)
≥
A
φ(δ0λ)−φ(0)
dμ(x) (3.8)
= [φ(δ0λ)−φ(0)]μ(A).
Here we haveused thatφ is increasing. To estimate thelast integralin(3.7), we note thatwhen0< λ< λ0 andx∈B,there is
g(x)−ξ(λ)·x=g(x)−ξg·x−λe·x
>2λ0−λ
> λ0.
Recalling φisconcave,wehave
|φ[g(x)−ξ(λ)·x]−φ[g(x)−ξg·x]| ≤φ(λ0)| −λe·x| ≤λφ(λ0),
whichimpliesthat
B
|φ[g(x)−ξ(λ)·x]−φ[g(x)−ξg·x]|dμ(x)≤λφ(λ0)μ(B). (3.9)
By(3.8) and(3.9),when 0< λ< λ0wesimplify(3.7) as
G(ξ(λ))−G(ξg)≥[φ(δ0λ)−φ(0)]μ(A)−λφ(λ0)μ(B).
Bytheassumptionsof φ,
t→lim0+
φ(t)−φ(0)
t = lim
t→0+φ(t) = +∞. Hence,wecanchoosepositivenumbersδand
M > φ(λ0)μ(B) δ0μ(A) ,
suchthat
φ(t)−φ(0)> M t, ∀0< t < δ.
Nowfor0< λ<min{λ0, δ/δ0},wehave
G(ξ(λ))−G(ξg)≥[M δ0μ(A)−φ(λ0)μ(B)]λ >0, whichisimpossible.Soξg cannotbeon∂Kg, namelyξg∈intKg.
Ifφisadditionallystrictlyconcave,thenGisalsostrictlyconcaveonKg.Soξg must be unique. Let g ∈ C+(Sn−1), and {gk} ⊂ C+(Sn−1) be any sequence of functions uniformly converging to g on Sn−1. We wantto provethatξgk converges to ξg in Rn. Note that Kgk → Kg and ξgk ∈ Kgk, therefore {ξgk}is bounded. For any convergent subsequence
ξgki
⊂ {ξgk},weneedtoproveitslimit,sayξ0,equalsξg.
Observethatfor any ξ∈Kg, there exists asequenceof ξki ∈Kgki whichconverges toξ. Then
G(ξ) =J[g(x)−ξ·x]
= lim
ki
J[gki(x)−ξki·x]
≤lim
ki
J[gki(x)−ξgki·x]
=J[g(x)−ξ0·x]
=G(ξ0),
whichimpliesthat
G(ξ0) = sup
ξ∈Kg
G(ξ).
Bytheuniquenessof ξg,wehave
ξ0=ξg. Theproof ofthislemmaiscompleted. 2
Lemma 3.4. Under the assumptionsof Lemma 3.2,theminimizing problem (3.2)has a solution h.
Proof. Letmbetheinfimumof(3.2),namely m= inf
ξ∈Ksupg
J[g(x)−ξ·x] :g∈C+(Sn−1), vol(Kg) =v
.
Byφ≥0,weseem≥0.
Let {gk} ⊂C+(Sn−1),vol(Kgk)=v be aminimizingsequence. Denote the support functionofKgk byhk.Then
hk(x)≤gk(x), ∀x∈supp(μ),
and Khk =Kgk.Since 0∈intKgk, hk ispositiveonSn−1.For anyξ∈Khk =Kgk,by themonotonicityofφ,
J[hk(x)−ξ·x] =
supp(μ)
φ(hk(x)−ξ·x)dμ(x)
≤
supp(μ)
φ(gk(x)−ξ·x)dμ(x)
=J[gk(x)−ξ·x], whichimplies
sup
ξ∈Khk
J[hk(x)−ξ·x]≤ sup
ξ∈Kgk
J[gk(x)−ξ·x].
Therefore
k→lim+∞ sup
ξ∈Khk
J[hk(x)−ξ·x] =m.
Namely {hk} is also a minimizing sequence of (3.2). Recalling Lemma 3.3, we have ξhk ∈intKhk suchthat
J[hk(x)−ξhk·x] = sup
ξ∈Khk
J[hk(x)−ξ·x].
Note hk is alsothe supportfunctionofKhk, byatranslation transformwe canalways assumeξhk = 0.Thisfactwillbe usedafewtimes.
Weclaimthat{hk}isuniformly boundedonSn−1.Ifnot,wecanassume
k→lim+∞ max
x∈Sn−1hk(x) = +∞.
WriteRk= maxx∈Sn−1hk(x).Foreachk,thereexistsxk∈Sn−1suchthathk(xk)=Rk. Since {xk} ⊂ Sn−1, there exists aconvergent subsequence. Without loss of generality, weassume
xk→x0∈Sn−1 whenk→+∞.
Recallsupp(μ) isnotconcentratedonanyclosedhemisphere,thereissomex¯∈supp(μ) suchthatx¯·x0>0.Writeδ=12x¯·x0,thenδ >0 andforsufficientlylargek,e.g.k≥k0, wehave
¯
x·xk> δ.
Bythedefinitionofsupportfunction,there is
hk(¯x)≥Rk(¯x·xk)> Rkδ, k≥k0.
Noteφisincreasing,limt→+∞φ(t)= +∞andμisafinitediscretemeasure,wehave m= lim
k→+∞J[hk]
= lim
k→+∞
Sn−1
φ(hk(x))dμ(x)
≥ lim
k→+∞φ(hk(¯x))μ(¯x)
≥ lim
k→+∞φ(Rkδ)μ(¯x)→+∞.
However,byLemma3.3,mmustbefinite.Thisisacontradiction.Thus{hk}isuniformly bounded.
Bythe Blaschke selection theorem,there is a subsequence of {hk}which uniformly converges to somesupportfunction hon Sn−1. CorrespondinglyKhk converges to Kh
whichistheconvexbodydeterminedbyh.Obviouslyh≥0 onSn−1,vol(Kh)=v,and
J[h] =m.
Forany ξ∈Kh,thereexists ξk ∈Khk suchthatξk →ξask→+∞.Then J[h(x)−ξ·x] = lim
k→+∞J[hk(x)−ξk·x]
≤ lim
k→+∞J[hk(x)−ξhk·x]
= lim
k→+∞J[hk(x)]
=J[h(x)], whichimpliesthat
J[h] = sup
ξ∈Kh
J[h(x)−ξ·x].
ByLemma3.3,0∈intKh.Thereforeh>0 onSn−1.Hence,weseethathisasolution to theminimizingproblem (3.2),andhisthesupportfunctionofKh. 2
InthefollowingweprovethatthesolutionhobtainedinLemma3.4isalsoasolution to (3.1) forsomec>0.
Forany givenη∈C(Sn−1),let
qt=h+tη fort≥0.
Byh∈C+(Sn−1),qt∈C+(Sn−1) forsufficiently smallt.ByLemma2.1,wehave Lemma 3.5.
tlim→0+
vol(Kqt)−vol(Kh)
t =
supp(μ)
ηdSKh(x).
Letgt(x)=β(t)qt(x) where
β(t) = vol(Kqt)−1/nv1/n.
Then gt∈C+(Sn−1),andvol(Kgt)=v.Noteg0(x)=h(x),and
t→0lim+
gt(x)−g0(x)
t =η(x) +β(0)h(x) uniformly onSn−1. (3.10) AlsobyLemma3.5,
β(0) =−v1/n
n vol(Kqt)−1/n−1dvol(Kqt) dt
t=0
=− 1 nv
Sn−1
ηdSKh(x). (3.11)
Foreachgt,ξ(t):=ξgt ∈Rn iswell definedbyLemma3.3.
Lemma3.6. ξ(t) isLipschitzcontinuous withrespecttot.
Proof. Sincesupξ∈K
gtJ[gt(x)−ξ·x] isattainedat ξ=ξ(t),wehave
Sn−1
φ(gt(x)−ξ(t)·x)xdμ(x) = 0. (3.12)
Recallingξ(0)=ξh= 0,and takingt= 0 intheaboveequality,wehave
Sn−1
φ(h(x))xdμ(x) = 0. (3.13)
Recallingthatφ<0 in(0,+∞),andsubtracting(3.13) from(3.12),weget
Sn−1
φ(τ)[gt(x)−ξ(t)·x−h(x)]xdμ(x) = 0,
whereτ :Sn−1×(0,+∞)→Randτ(x,t) isbetweengt(x)−ξ(t)·xandh(x).Then
Sn−1
φ(τ)[gt(x)−h(x)]xdμ(x) =
Sn−1
φ(τ)(ξ(t)·x)xdμ(x).
Takingtheinnerproductwith ξ(t),wehave
Sn−1
φ(τ)[gt(x)−h(x)](ξ(t)·x)dμ(x) =
Sn−1
φ(τ)(ξ(t)·x)2dμ(x). (3.14)
Notethatwhentissmall, sup
x∈Sn−1|gt(x)−h(x)|= sup
x∈Sn−1|(β(t)−1)h(x) +tβ(t)η(x)|
≤C[|β(t)−1|+tβ(t)]
≤Ct
forsomepositiveconstantC whichisindependentofxandt. Sinceh>0,andgt(x)− ξ(t)·xconvergestoh(x) uniformlyonSn−1whent→0+,wecanassume 12minxh(x)<
τ <2maxxh(x).ThereforethereexisttwopositiveconstantsC1andC2dependingonly onhandφ suchthat
−C1≤φ(τ)≤ −C2. Thus wecanestimate(3.14) as
C2
Sn−1
(ξ(t)·x)2dμ(x)≤
Sn−1
(−φ(τ))[gt(x)−h(x)](ξ(t)·x)dμ(x)
≤C1Ct
Sn−1
|ξ(t)·x|dμ(x) (3.15)
≤C1C|μ| · |ξ(t)|t.
Recall μ is notconcentrated on any closed hemisphere,there exists C3 >0 depending onlyonμ,suchthat
Sn−1
(y·x)2dμ(x)≥C3, ∀y∈Sn−1.
Then
Sn−1
(ξ(t)·x)2dμ(x)≥C3|ξ(t)|2.
Therefore, itfollowsfrom(3.15) that
|ξ(t)| ≤ C1C|μ| C2C3
t,
whichisthedesiredresult. 2
Now, wearegoingtofinishtheproofof Lemma3.2.Let J(t) :=J[gt(x)−ξ(t)·x].
Note ξ(0)= 0 andJ(0)=J[h]. Sincehisaminimizerof(3.2),wehave J(t)≥J(0)
forallsmall t≥0.Thus
lim
tk→0+
J(tk)−J(0)
tk ≥0 (3.16)
for any convergent subsequence {tk}. By Lemma 3.6, we can assume without loss of generalitythat
lim
tk→0+
ξ(tk)−ξ(0)
t =γ.
Recalling(3.10),weseethat(3.16) issimplifiedas
Sn−1
φ(h)[η(x) +β(0)h(x)−γ·x]dμ(x)≥0, (3.17)
which,togetherwith(3.13),implies
Sn−1
φ(h)[η(x) +β(0)h(x)]dμ(x)≥0.
By(3.11),weobtain
Sn−1
φ(h)ηdμ−c
Sn−1
ηdSKh ≥0,
where
c= 1 nv
Sn−1
φ(h)hdμ.
Replacingη by−η,weseethat
Sn−1
φ(h)ηdμ−c
Sn−1
ηdSKh= 0
forallη∈C(Sn−1).Thus
φ(h)dμ−c dSKh = 0, namely
c
φ(h)dSKh =dμ,
whichmeansthathsolvesequation (3.1).Obviouslyc>0.Theproof ofLemma3.2 is completed.
4. Thegeneralcase
In the previous section, we have proved Lemma 3.2. This lemma says that Theo- rem3.1 holdsundertheadditionalassumptions:φ∈C2(0,+∞) withφ<0 andμisa discrete measure onSn−1. Inthissection, wewill removethese additionalassumptions to complete theproof ofTheorem 3.1.Wewill useapproximations toachieve thisaim.
First we remove the assumptions: φ ∈ C2(0,+∞) with φ < 0. Namely we prove the following:
Lemma 4.1. Suppose φ: (0,+∞)→(0,+∞)is an increasing concave C1 functionsat- isfying that limt→+∞φ(t) = +∞, φ(t) > 0 and limt→0+φ(t) = +∞. If μ is a finite discrete measureon Sn−1 whichis not concentratedin any closed hemisphere of Sn−1, then thereexistsaconvex bodyK in Rn andac>0satisfying
c
φ(hK)dSK=dμ onSn−1.
Moreover, one can require the volume of hK is equal to any given number v > 0. And hK ∈C+(Sn−1)isa minimizerof
inf
sup
ξ∈Kg
J[g(x)−ξ·x] :g∈C+(Sn−1), vol(Kg) =v
. (4.1)
HereKg is theAlexandrov bodyassociated with(g,supp(μ)),andJ isgivenby (3.3).
Proof. Assumeρ∈ C∞(R) isanon-negative smoothfunction compactly supportedin [−1,0],and
R
ρ(t)dt= 1.
Let ρ(t)=−1ρ(t/) for >0.Then ρ isan approximation to theidentity. Letφ˜be theextension ofφfrom(0,+∞) toR,givenby
φ(t) :=˜
⎧⎪
⎪⎨
⎪⎪
⎩
φ(t), if t >0,
t→0lim+φ(t), if t= 0, 0, if t <0.
Thenφ˜isnon-decreasingandnon-negativeinR.Letφ˜ betheconvolutionproductof ˜φ andρ,namelyforany t∈R,
φ˜(t) := ( ˜φ∗ρ)(t)
=
R
φ(t˜ −τ)ρ(τ)dτ
= 0
−
φ(t˜ −τ)ρ(τ)dτ.
Thenφ˜ isanon-negative C∞ functioninR.Forany t2> t1,wehave φ˜(t2)−φ˜(t1) =
R
[ ˜φ(t2−τ)−φ(t˜ 1−τ)]ρ(τ)dτ ≥0,
wheretheinequalityisduetothemonotonicity ofφ˜inR.Thereforeφ˜isnon-decreasing inR,andthen φ˜≥0.Foranyt>0,wealso have
φ˜(t) = 0
−
φ(t˜ −τ)ρ(τ)dτ
≥ 0
−
φ(t)ρ˜ (τ)dτ
= ˜φ(t) =φ(t), whichimpliesthat
t→lim+∞φ˜(t)≥ lim
t→+∞φ(t) = +∞.
Nextwe showφ˜is alsoconcavein(0,+∞).Infact,foranyt2> t1>0,thereis
φ˜t1+t2 2
= 0
−
φ˜t1+t2 2 −τ
ρ(τ)dτ
= 0
−
φ˜
t1−τ+t2−τ 2
ρ(τ)dτ
≥ 0
−
1
2[ ˜φ(t1−τ) + ˜φ(t2−τ)]ρ(τ)dτ
=1
2[ ˜φ(t1) + ˜φ(t2)],
where theinequalityistruesinceφ˜=φ isconcave in(0,+∞).Soφ˜ isanon-negative, non-decreasing concave C∞ functionin (0,+∞) satisfying thatlimt→+∞φ˜(t)= +∞, φ˜(t)≥0,andφ˜(t)≤0.
Now defineφ as
φ(t) := ˜φ(t) +α(t), ∀t >0, where
α(t) =
√t 1 +√
t.
Directcomputations showthat
α(t) = 1 2(1 +√
t)2√ t >0, α(t) =− 1 + 3√
t 4(1 +√
t)3t3/2 <0.
Recalling the above properties about φ˜, we see that φ is a positive, increasing and concave C∞ functionin(0,+∞) withlimt→+∞φ(t)= +∞, φ(t)>0 and φ(t)<0.
Observing φ˜ issmoothinR,andlimt→0+α(t)= +∞,weobtain
t→lim0+φ(t) = lim
t→0+
φ˜(t) + lim
t→0+α(t)
= ˜φ(0) + lim
t→0+α(t)
= +∞.
Henceφ satisfiesalltheassumptionsonφinLemma3.2.
Now applyingLemma3.2 onφ,there existsh∈C+(Sn−1) whichisaminimizerof
inf
ξ∈Ksupg
J[g(x)−ξ·x] :g∈C+(Sn−1), vol(Kg) =v
,
where
J[g] =
Sn−1
φ(g(x))dμ(x).
Moreoverh satisfiesthefollowing c
φ(h)dSKh =dμ, (4.2)
where
c= 1 nv
Sn−1
φ(h)hdμ. (4.3)
Andh isthesupportfunctionofKh. For0< <1,let
m=J[h].
Weclaimmisuniformlyboundedfromabove.Infact,denoteKμtheAlexandrovbody associatedwith(1,supp(μ)).Here1 means theconstantfunctiononSn−1. Let
¯ g≡
v vol(Kμ)
1/n
,
wehaveKg¯= ¯gKμ,andthenvol(K¯g)=gnvol(Kμ)=v.Bydefinition,there is m≤ sup
ξ∈K¯g
J[¯g−ξ·x]
= sup
ξ∈K¯g
supp(μ)
φ(¯g−ξ·x)dμ(x)
≤
supp(μ)
φ(diam(Kg¯))dμ(x)
=φ(diam(K¯g))|μ|. Notethatwhen0< <1
φ(t)< φ(t+) + < φ(t+ 1) + 1, wehave
m<[φ(diam(K¯g) + 1) + 1]· |μ|. (4.4) Next, we prove{h} is uniformly bounded on Sn−1. If not, there exists a sequence {k}suchthat
k→lim+∞ max
x∈Sn−1hk(x) = +∞.
Write Rk = maxx∈Sn−1hk(x). For each k, there exists xk ∈ Sn−1 such that hk(xk) =Rk. Since {xk} ⊂ Sn−1, there exists a convergent subsequence. Without lossofgenerality, weassume
xk→x0∈Sn−1 whenk→+∞.
Recallsupp(μ) isnotconcentratedonanyclosedhemisphere,thereissomex¯∈supp(μ) suchthatx¯·x0>0.Writeδ=12x¯·x0>0,then forsufficientlylargek,e.g.k≥k0,we have
¯
x·xk> δ. (4.5)
Bythedefinitionofsupportfunction,there is
hk(¯x)≥Rk(¯x·xk)> Rkδ, k≥k0. (4.6) Note φisincreasing, φ(t)> φ(t) fort>0,andlimt→+∞φ(t)= +∞, wehave
mk=
Sn−1
φk(hk(x))dμ(x)
≥φk(hk(¯x))μ(¯x)
≥φk(Rkδ)μ(¯x)
≥φ(Rkδ)μ(¯x)→+∞.
(4.7)
However, by (4.4), mk is uniformly bounded. This is a contradiction. Thus {h} is uniformly bounded,namelythere existssomepositiveconstantC1suchthat
max
x∈Sn−1h(x)≤C1, ∀0< <1. (4.8) By the Blaschke selection theorem, we can assume h converges to some support function h uniformly on Sn−1 when → 0+. Correspondingly Kh converges to Kh. Note thatvol(h)=v andh≥0 on Sn−1.Weclaimthatifnon-negativeg convergesto someg uniformlyonSn−1,then
→0lim+ sup
ξ∈Kg
J[g(x)−ξ·x] = sup
ξ∈Kg
J[g(x)−ξ·x]. (4.9)
Infact,letξˆ beapointinKg suchthat J[g(x)−ξˆ·x] = sup
ξ∈Kg
J[g(x)−ξ·x].
Since Kg convergesto Kg, onecanassume ξˆ converges to someξˆ∈Kg.By ourcon- struction,φ convergestoφuniformly onanyclosedintervalof[0,+∞),then
