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Advances
in
Applied
Mathematics
www.elsevier.com/locate/yaama
The
Orlicz
version
of
the
L
pMinkowski
problem
for
−n
< p
< 0
✩Gabriele Bianchia,∗, KárolyJ. Böröczkyb,c, Andrea Colesantia aDipartimentodiMatematicaeInformatica“U.Dini”,UniversitàdiFirenze,
VialeMorgagni67/A,Firenze,I-50134,Italy
b
AlfrédRényiInstituteofMathematics,HungarianAcademyofSciences, Reltanodau.13-15,H-1053Budapest,Hungary
cDepartmentofMathematics,CentralEuropeanUniversity,Nadoru9,H-1051,
Budapest,Hungary
a r t i c l e i n f o a b s t r a c t
Articlehistory:
Received28June2019 Accepted9August2019 Availableonline30August2019
MSC:
primary52A38,35J96
Keywords:
LpMinkowskiproblem
OrliczMinkowskiproblem Monge-Ampèreequation
Given a function f on the unit sphere Sn−1, the Lp
Minkowski problem asks for a convex body K whose Lp
surface area measure has density f with respect to the standard(n−1)-HausdorffmeasureonSn−1.Inthispaperwe dealwiththegeneralizationof thisproblemwhich arisesin theOrlicz-Brunn-MinkowskitheorywhenanOrliczfunction ϕ substitutestheLpnormandp isintherange(−n,0).This
problemisequivalenttosolvetheMonge-Ampereequation ϕ(h) det(∇2h + hI) = f on Sn−1,
whereh isthesupportfunctionoftheconvexbodyK. ©2019ElsevierInc.Allrightsreserved.
✩ Firstandthirdauthorsare supportedinpartbythe GruppoNazionaleper l’Analisi Matematica,la
Probabilità e le loro Applicazioni(GNAMPA) of the Istituto Nazionale di AltaMatematica (INdAM). SecondauthorissupportedinpartbyNKFIHgrants116451,121649and129630.
* Correspondingauthor.
E-mailaddresses:gabriele.bianchi@unifi.it(G. Bianchi),boroczky.karoly.j@renyi.mta.hu
(K.J. Böröczky),andrea.colesanti@unifi.it(A. Colesanti).
https://doi.org/10.1016/j.aam.2019.101937
1. Introduction
Weworkinthen-dimensionalEuclideanspaceRn,n≥ 2.Aconvex body K inRn is
acompactconvexsetthathasnon-emptyinterior.Given aconvexbodyK,forx∈ ∂K we denote by νK(x) ⊂ Sn−1 the family of all unit exteriornormal vectors to K at x
(theGaußmap).Wecanthendefinethesurfaceareameasure SK ofK,whichisaBorel
measure ontheunitsphereSn−1 ofRn, asfollows:foraBorelsetω⊂ Sn−1 weset SK(ω) =Hn−1
νK−1(ω)=Hn−1({x ∈ ∂K : νK(x)∩ ω = ∅}) (see, e.g.,Schneider[38]).
TheclassicalMinkowskiproblemcanbeformulatedasfollows:givenaBorelmeasure
μ onSn−1,findaconvexbodyK suchthatμ= SK.Thereaderisreferredto[38,Chapter
8] foranexhaustivepresentationof thisproblemanditssolution.
Throughoutthispaperwewillconsider(eitherfortheclassicalMinkowskiproblemor foritsvariants)thecaseinwhichμ hasadensityf withrespecttothe(n−1)-dimensional HausdorffmeasureonSn−1.UnderthisassumptiontheMinkowskiproblemisequivalent tosolve(intheclassicorintheweaksense)adifferentialequationonthesphere.Namely:
det(∇2h + hI) = f, (1)
where:h isthesupportfunctionofK,∇2h isthematrixformedbythesecondcovariant
derivativesofh withrespecttoalocalorthonormalframeonSn−1 andI istheidentity matrixoforder (n− 1).
Many different types of variations of the Minkowski problem have been considered (wereferforinstanceto [38,Chapters8and9]).Ofparticularinterestforourpurposes is the so called Lp version of theproblem (see [1–11,14–34,36,37,39–50]).At theorigin ofthis newproblem thereisthereplacementoftheusualMinkowskiadditionofconvex bodiesbythep-addition.Asaneffect,thecorresponding differentialequationtakes the form
h1−pdet(∇2h + hI) = f, (2)
(see[38,Section9.2]).ThestudyoftheLpMinkowskiproblemsdevelopedinasignificant wayinthelastdecades,asapartofthesocalledLpBrunn-Minkowskitheory,which rep-resentsnowasubstantialareaofConvexGeometry.Oneofthemostinterestingaspects ofthisproblemisthatseveralthresholdvaluesoftheparameterp canbeidentified,e.g. p= 1, p= 0, p=−n, across whichthe natureof theproblem changesdrastically. For an account on the literatureand on the stateof the art ofthe Lp Minkowskiproblem (especially forthevaluesp< 1)wereferthereaderto[1] and[2].
Of particular interest here is the range −n < p < 0. In this case Chou and Wang (see [10]) solved thecorresponding problem when the measure μ hasadensity f , and
f isbounded and bounded awayfrom zero.This result wasslightlygeneralised bythe authorsincollaborationwithYang in[1],where f isallowed tobe inL n
n+p.
Theorem1.1(ChouandWang; Bianchi,Böröczky,Colesanti andYang). Forn≥ 1 and
−n< p< 0, if thenon-negative and non-trivialfunction f is in L n
n+p(S
n−1) then (2)
has a solution in the Alexandrov sense; namely, f dHn−1 = dSK,p for a convex body
K∈ Kn
0.In addition, iff is invariantunder aclosed subgroup G ofO(n), then K can
bechosentobe invariantunder G.
AsafurtherextensionoftheLpMinkowskiproblem,onemayconsider itsOrlicz ver-sion.Formally,thisproblem arisesinthecontextoftheOrlicz-Brunn-Minkowskitheory ofconvexbodies(see[38,Chapter 9]).Inpractice,therelevantdifferentialequationis
ϕ(h) det(∇2h + hI) = f,
where ϕ is a suitable Orlicz function. The Lp Minkowski problem is obtained when ϕ(t)= t1−p, fort≥ 0.
Whenϕ : (0,∞)→ (0,∞) iscontinuousandmonotonedecreasing,thisproblem(under
asymmetryassumption) hasbeen consideredbyHaberl, Lutwak,Yang,Zhang in[17]. Comparingthepreviousassumptionsonϕ withtheLpcase,weseethatthiscorresponds tothevaluesp≥ 1.
Weareinterestedinthecaseinwhichthemonotonicityassumption isreversed, cor-respondingto thevaluesp< 1.Henceweassumethatϕ : [0,∞)→ R iscontinuousand monotoneincreasing,havingtheexampleϕ(t)= t1−p,p< 1,asaprototype.Tocontrol inamorepreciseformthebehaviourofϕ withthatofapowerfunction,weassumethat thereexists p< 1 such that
lim inf t→0+
ϕ(t)
t1−p > 0. (3)
Concerningthebehaviourofϕ at∞ weimposethecondition: ∞
1
1
ϕ(t)dt <∞. (4)
The corresponding Minkowski problem in this setting can be called the Orlicz Lp Minkowskiproblem.Thesolutionofthis problemintherangep∈ (0,1) isdueto Jian, Lu[25]. We alsonote thatOrlicz versionsof theso called Lp dual Minkowskiproblem havebeenconsideredrecentlybyGardner,Hug,Weil,Xing,Ye[13],Gardner,Hug,Xing, Ye[14],Xing,Ye,Zhu [43] andXing, Ye[44].
In this paper we focus on the range of values p ∈ (−n,0). As an extension of the resultscontainedin[1],weestablishthefollowingexistencetheorem(notethat,asusual
inthecaseofOrliczversionsofMinkowskitypeproblems,wecanonlyprovideasolution upto aconstantfactor).
Theorem 1.2. For n ≥ 2, −n < p < 0 and monotone increasing continuous function ϕ: [0,∞)→ [0,∞) satisfyingϕ(0)= 0, andconditions (3) and(4), if thenon-negative
non-trivial function f is in L n
n+p(S
n−1), then there exists λ > 0 and a convex body
K∈ Kn
0 withV (K)= 1 such that
λϕ(h) det(∇2h + hI) = f
holdsforh= hK intheAlexandrovsense;namely,λϕ(hK)dSK= f dHn−1.Inaddition,
iff isinvariantunderaclosedsubgroupG ofO(n),thenK canbechosentobeinvariant
under G.
Wenote thattheoriginmaylieon∂K forthesolutionK inTheorem 1.2.
We observe that Theorem 1.2 readily yields Theorem 1.1. Indeed if −n < p < 0,
f ∈ L n
n+p(S
n−1), f ≥ 0,f ≡ 0 andλh1−p
K dSK = f dHn−1 for K∈ Kn0 and λ> 0,then
h1−p
K dSK= f dHn−1 forK = λ
1
n−pK.
In Section3 we sketchthe proof of Theorem 1.2 and describe the structure of the paper.
2. Notation
ThescalarproductonRn isdenotedby ·,· ,andthecorrespondingEuclideannorm isdenotedby· .Thek-dimensionalHausdorffmeasurenormalizedinsuchawaythat it coincides with theLebesgue measure on Rk is denotedby Hk. The angle (spherical distance) ofu,v∈ Sn−1 isdenotedby∠(u,v).
We write Kn
0 (K(0)n ) to denote thefamily of convex bodies with o ∈ K (o∈ int K).
Given aconvex bodyK, fora Borelset ω ⊂ Sn−1,νK−1(ω) is theBorel set of x∈ ∂K withνK(x)∩ω = ∅.Apointx∈ ∂K iscalledsmoothifνK(x) consistsofauniquevector, and in this case, we use νK(x) to denote this unique vector, as well. It is well-known thatHn−1-almosteveryx∈ ∂K issmooth(see,e.g.,Schneider[38]);let∂K denotethe familyofsmoothpointsof ∂K.
Foraconvexcompactset K inRn, lethK be itssupportfunction: hK(u) = max{ x, u : x ∈ K} for u ∈ Rn.
Note that if K ∈ Kn0, then hK ≥ 0. If p ∈ R and K ∈ Kn0, then the Lp-surface area
measure isdefinedby
where forp> 1 the right-handside isassumed to be afinite measure.Inparticular, if p= 1,thenSK,p= SK,andifp< 1 andω⊂ Sn−1 Borel,then
SK,p(ω) = ν−1K(ω)
x, νK(x) 1−pdHn−1(x).
3. Sketchofthe proofofTheorem 1.2
To sketch the argument leading to Theorem 1.2, first we consider the case when
−n< p≤ −(n− 1) and ϕ(t)= t1−p, andτ
1≤ f ≤ τ2 for someconstants τ2 > τ1> 0.
Wesetψ(t)= 1/ϕ(t)= tp−1 fort> 0,anddefine Ψ: (0,∞)→ (0,∞) by
Ψ(t) = ∞ t ψ(s) ds =−1 pt p,
whichisastrictly convexfunction. GivenaconvexbodyK inRn,weset
Φ(K, ξ) = Sn−1
Ψ(hK−ξ)f dHn−1;
this is astrictly convex functionof ξ∈ int K. As f > τ1 and p≤ −(n− 1),there is a
(unique)ξ(K)∈ int K suchthat
Φ(K, ξ(K)) = min
ξ∈int KΦ(K, ξ).
ThisstatementisprovedinProposition5.2,buttheconditionsf > τ1andp≤ −(n− 1)
areactuallyusedinthepreparatorystatementLemma5.1.
Usingp>−n andtheBlaschke-Santalóinequality(seeLemma5.4and the prepara-torystatementLemma4.3),oneverifiesthatthereexists aconvexbodyK0 inRn with
V (K0)= 1 maximizingΦ(K,ξ(K)) overallconvexbodiesK inRn withV (K)= 1.
Finallyavariationalargument proves thatthere exists λ0> 0 such thatf dHn−1 =
λ0ϕ(hK0)dSK0. A crucial ingredient (see Lemma6.2)is that,as ψ is C
1 and ψ < 0, Φ(Kt,ξ(Kt)) isadifferentiablefunctionofKtforasuitablevariationKtofK0.
Inthegeneralcase,whenstill keepingtheconditionτ1≤ f ≤ τ2 butallowingany ϕ
whichsatisfiestheassumptionsofTheorem1.2,wemeettwomainobstacles.Ontheone hand,evenifϕ(t)= t1−p but0< t<−(n− 1),it mayhappen thatfor aconvex body
K inRn,theinfimumofΦ(K,ξ) forξ∈ int K isattainedwhenξ tendstotheboundary ofK.Ontheotherhand,thepossiblelackofdifferentiabilityofϕ (orequivalentlyofψ) destroysthevariationalargument.
Therefore, we approximateψ bysmooth functions, andalso make surethatthe ap-proximating functions are large enough near zero to ensure that the minimum of the analogues ofΦ(K,ξ) as afunctionofξ∈ int K existsforany convexbodyK.
Section4provessomepreparatorystatements,Section5introducesthesuitable ana-logueoftheenergyfunctionΦ(K,ξ(K)),andSection6providesthevariationalformula foranextremalbodyfortheenergyfunction.WeproveTheorem1.2iff isboundedand bounded awayfrom zeroinSection7,andfinally infullstrength inSection8.
4. Somepreliminaryestimates
Inthissection,weprovethesimplebuttechnicalestimatesLemmas4.1and4.3that will beusedinvarioussettings duringthemainargument.
Lemma 4.1. For δ ∈ (0,1), A,ℵ > 0 and˜ q ∈ (−n,0), let ψ : (0, ∞) → (0,∞) satisfy that ψ(t) ≤ ˜ℵtq−1 fort∈ (0,δ] and∞
δ ψ≤ A.If t∈ (0,δ) and ˜ℵ0= max{ ˜ ℵ |q|,δAq},then Ψ(t)=t∞ψ satisfies Ψ(t)≤ ˜ℵ0tq.
Proof. We observethatift∈ (0,δ),then
Ψ(t)≤ δ t ˜ ψ(s) ds + A≤ ˜ℵ δ t sq−1ds + A = ℵ˜ |q|(tq− δq) + A≤ tqmax ˜ ℵ |q|, A δq . 2
We write Bn to denote theEuclidean unit ballin Rn, and set κ
n =Hn(Bn).For a convex bodyK inRn,letσ(K) denoteitscentroid,whichsatisfies (seeSchneider[38])
−(K − σ(K)) ⊂ n(K − σ(K)). (5)
Next,ifo∈ int K thenthepolarofK is
K ={x ∈ Rn : x, y ≤ 1 ∀y ∈ K} = {tu : u ∈ Sn−1 and 0≤ t ≤ hK(u)−1}.
In particular, the Blaschke-Santaló inequality V (K)V ((K − σ(K))∗) ≤ V (Bn)2 (see Schneider [38])yieldsthat
Sn−1
h−nK−σ(K)dHn−1≤ nV (B n)2
V (K) . (6)
As apreparationfortheproofofLemma4.3,weneedthefollowingstatementabout absolutely continuous measures.Fort ∈ (0,1) andv ∈ Sn−1, we considerthe spherical strip
Ξ(v, t) ={u ∈ Sn−1 : | u, v | ≤ t}. Lemma4.2. If f ∈ L1(Sn−1) and f(t) = sup v∈Sn−1 Ξ(v,t) |f| dHn−1 fort∈ (0,1),thenwehave limt→0+ f(t)= 0.
Proof. We observe that f(t) is decreasing, therefore the limit limt→0+ f(t) = δ ≥ 0
exists.Wesuppose thatδ > 0,andseekacontradiction.
Letμ betheabsolutelycontinuousmeasuredμ= |f|dHn−1onSn−1.Accordingtothe definitionof f,foranyk≥ 2,thereexistssomevk ∈ Sn−1suchthatμ(Ξ(vk,k1))≥ δ/2. Let v ∈ Sn−1 be an accumulation point of the sequence {vk}. For any m ≥ 2, there exists αm > 0 such thatΞ(u,2m1 ) ⊂ Ξ(v,m1) if u∈ Sn−1 and ∠(u,v) ≤ αm. Since for any m≥ 2, there exists somek≥ 2m such that∠(vk,v)≤ αm, we haveμ(Ξ(v,m1))≥
μ(Ξ(vk,1k)) ≥ δ/2. We deduce that μ(v⊥ ∩ Sn−1) = μ ∩m≥2Ξ(v,m1) ≥ δ/2, which contradictsμ(v⊥∩ Sn−1)= 0. 2
Lemma4.3. Forδ∈ (0,1),ℵ > 0 and˜ q∈ (−n,0),letΨ : (0, ∞)→ (0,∞) beamonotone decreasingcontinuous functionsuchthat Ψ(t) ≤ ˜ℵtq fort∈ (0,δ] andlim
t→∞Ψ(t) = 0,
and let f be˜ a non-negative function in L n
n+p(S
n−1). Then for any ζ > 0, there exists
a Dζ depending on ζ, Ψ, δ, ℵ,˜ q and f such˜ that if K is a convex body in Rn with
V (K)= 1 anddiam K≥ Dζ then
Sn−1
( Ψ◦ hK−σ(K)) ˜f dHn−1 ≤ ζ.
Proof. We mayassume that σ(K)= o. Let R = maxx∈Kx, and let v ∈ Sn−1 such thatRv∈ K.Itfollowsfrom (5) that −R
nv∈ K.
Sincelimt→∞Ψ(t) = 0 andf is˜ inL1(Sn−1) bytheHölderinequality,wecanchoose
r≥ 1 suchthat Ψ(r) Sn−1 ˜ f dHn−1< ζ 2. (7)
WepartitionSn−1 intothetwomeasurable parts
Ξ0={u ∈ Sn−1: hK(u)≥ r}
Letus estimatetheintegralsoverΞ0 andΞ1.Wededucefrom(7) that Ξ0 ( Ψ◦ hK) ˜f dHn−1≤ ζ 2. (8)
Nextweclaimthat
Ξ1⊂ Ξ v,nr R . (9)
Forany u∈ Ξ1,wechooseη∈ {−1,1} suchthat u,ηv ≥ 0,thus ηRn v ∈ K yieldsthat
r > hK(u)≥ u,ηRn v .Inturn,weconclude(9).Itfollowsfrom(9) andLemma4.2that fortheL1functionf = ˜f
n n+q,wehave Ξ1 ˜ fn+qn ≤ f nr R . (10)
To estimatethedecreasingfunctionΨ on (0,r), weclaimthatift∈ (0,r) then
Ψ(t)≤ ℵδ˜ q
rq t
q. (11)
Werecallthatr≥ 1> δ.Inparticular,ift≤ δ,thenΨ(t) ≤ ˜ℵtqyields(11).Ift∈ (δ,r), then usingthatΨ is decreasing,(11) followsfrom
Ψ(t)≤ Ψ(δ)≤ ˜ℵδ q tq t q≤ ℵδ˜ q rq t q.
Applying first (11), then the Hölder inequality, after that the Blaschke-Santaló in-equality (6) withV (K)= 1 andfinally (10),wededuce that
Ξ1 ( Ψ◦ hK) ˜f dHn−1 ≤ ˜ ℵδq rq Ξ1 h−|q|K f d˜ Hn−1 ≤ℵδ˜ q rq ⎛ ⎝ Ξ1 h−nK dHn−1 ⎞ ⎠ |q| n ⎛ ⎝ Ξ1 ˜ fn−|q|n dHn−1 ⎞ ⎠ n−|q| n ≤ℵδ˜rqq nV (Bn)2 |q| n f nr R n+q n .
Therefore afterfixingr≥ 1 satisfying(7),wemaychooseR0> r suchthat
˜ ℵδq rq n |q| nV (Bn) 2|q| n f nr R0 n+q n <ζ 2
byLemma4.3. Inparticular, ifR≥ R0,then Ξ1 ( Ψ◦ hK) ˜f dHn−1 ≤ ζ 2.
Combining this estimatewith (8) shows that settingDζ = 2R0, ifdiam K ≥ Dζ, then R≥ R0, andhence
Sn−1( Ψ◦ hK)f d˜ Hn−1 ≤ ζ. 2
5. Theextremalproblemrelatedto Theorem1.2whenf is boundedandbounded awayfromzero
For0< τ1< τ2,letthereal functionf onSn−1 satisfy
τ1< f (u) < τ2 for u∈ Sn−1. (12)
Inaddition,letϕ: [0,∞)→ [0,∞) be acontinuousmonotone increasingfunction satis-fyingϕ(0)= 0, lim inf t→0+ ϕ(t) t1−p > 0 and ∞ 1 1 ϕ(t)dt <∞.
It will be more convenient to work with the decreasingfunction ψ = 1/ϕ : (0,∞) → (0,∞),whichhastheproperties
lim sup t→0+ ψ(t) tp−1, <∞ (13) ∞ 1 ψ(t) dt <∞. (14)
WeconsiderthefunctionΨ: (0,∞)→ (0,∞) definedby
Ψ(t) = ∞
t
ψ(s) ds,
whichreadilysatisfies
Ψ=−ψ, and hence Ψ is convex and strictly monotone decreasing, (15)
lim
t→∞Ψ(t) = 0. (16)
ψ(t) <ℵtp−1 for t∈ (0, δ). (17) AswepointedoutinSection3,wesmoothenψ usingconvolution.Letη : R→ [0,∞) be anon-negative C∞ “approximationof identity”withsupp η⊂ [−1,0] andRη = 1. Foranyε∈ (0,1),weconsiderthenon-negativeηε(t)=1εη(tε) satisfyingthat
Rηε= 1, supp ηε⊂ [−ε,0],anddefine θε: (0,∞)→ (0,∞) by
θε(t) = R ψ(t− τ)ηε(τ ) dτ = 0 −ε ψ(t− τ)ηε(τ ) dτ.
As ψ ismonotonedecreasingandcontinuouson(0,∞),thepropertiesofηεyield
θε(t) ≤ ψ(t) for t > 0 and ε ∈ (0, 1)
θε(t1) ≥ θε(t2) for t2> t1> 0 and ε∈ (0, 1)
θεtends uniformly to ψ on any interval with positive endpoints as ε tends to zero. Next,forany t0> 0,thefunctionlt0 onR defined by
lt0(t) =
ψ(t) if t≥ t0
0 if t < t0
is bounded, and hence locally integrable. For the convolution lt0 ∗ ηε, we have that
(lt0∗ ηε)(t)= θε(t) fort> t0 andε∈ (0,1),thus
θε is C1 for each ε∈ (0, 1).
As it is explained in Section 3, we need to modify ψ in a way such that the new functionisoforder atleastt−(n−1) ift> 0 is small.Weset
q = min{p, −(n − 1)},
and hence(17) andδ∈ (0,1) yieldsthat
θε(t)≤ ψ(t) < ℵtq−1 for t∈ (0, δ) and ε ∈ (0, δ). (18)
Nextweconstructθ˜ε: (0,∞)→ (0,∞) satisfying ˜ θε(t) = θε(t)≤ ψ(t) for t ≥ ε and ε ∈ (0, δ) ˜ θε(t) ≤ ℵtq−1 for t∈ (0, δ) and ε ∈ (0, δ) ˜ θε(t) = ℵtq−1 for t∈ (0, ε 2] and ε∈ (0, δ) ˜
Itfollowsthat ˜
θεtends uniformly to ψ on any interval with positive endpoints as ε tends to zero. Toconstructsuitableθ˜ε,firstweobservethattheconditionsabovedetermineθ˜εoutside theinterval (ε
2,ε), and θ˜ε(ε)<ℵεq−1. WritingΔ to denote the degreeonepolynomial
whosegraphisthetangentto thegraphoft→ ℵtq−1 att= ε/2,wehaveΔ(t)<ℵtq−1 fort> ε/2 andΔ(ε)< 0.Thereforewecanchooset0∈ (ε2,ε) suchthatθ˜ε(ε)< Δ(t0)<
ℵεq−1. Wedefine θ˜ε(t) = Δ(t) fort ∈ (ε
2,t0),and construct θ˜ε on(t0,ε) inaway that
˜
θεstaysC1 on(0,∞).Itfollows fromtheway θ˜εisconstructedthatθ˜ε(t)≤ ℵtq−1 also fort∈ [ε
2,ε].
Inordertoensureanegative derivative,weconsiderψε: (0,∞)→ (0,∞) definedby ψε(t) = ˜θε(t) +
ε
1 + t2 (19)
forε∈ (0,δ) andt> 0.This C1 functionψ
εhasthefollowingproperties:
ψε(t) ≤ ψ(t) +1+t12 for t≥ ε and ε ∈ (0, δ)
ψε(t) < 0 for t > 0 and ε∈ (0, δ) ψε(t) < 2ℵtq−1 for t∈ (0, δ) and ε ∈ (0, δ) ψε(t) > ℵtq−1 for t∈ (0,ε2) and ε∈ (0, δ)
ψεtends uniformly to ψ on any interval with positive endpoints as ε tends to zero. (20)
Forε∈ (0,δ),wealsoconsider theC2functionΨε: (0,∞)→ (0,∞) definedby
Ψε(t) = ∞
t
ψε(s) ds,
andhence(20) yields lim
t→∞Ψε(t) = 0 (21)
Ψε=−ψε, thus Ψεis strictly decreasing and strictly convex. (22)
Forε∈ (0,δ),Lemma4.1and(20) implythatsetting
A = ∞ δ ψ(t) + 1 1 + t2dt, wehave
Ψε(t)≤ ℵ0tq forℵ0= max{
2ℵ |q|,
A
δq} and t ∈ (0, δ). (23)
Ontheother hand,ifε∈ (0,δ) and t∈ (0,ε
4),then Ψε(t)≥ ε/2 t ℵsq−1ds = ℵ |q|(tq− (ε/2)q)≥|q|ℵ (tq− (2t)q) =ℵ1tq forℵ1= (1− 2q)ℵ |q| > 0. (24)
According to (20), we have limε→0+ψε(t) = ψ(t) and ψε(t) ≤ ψ(t)+ 1
1+t2 for any
t> 0,thereforeLebesgue’sDominatedConvergenceTheoremimplies
lim
ε→0+Ψε(t) = Ψ(t) for any t > 0. (25)
It alsofollowsfrom (20) that ift≥ ε,then
Ψε(t) = ∞ t ψε≤ ∞ t ψ(s) + 1 1 + s2ds≤ Ψ(t) + π 2. (26)
Forany convexbodyK andξ∈ int K,weconsider Φε(K, ξ) = Sn−1 (Ψε◦ hK−ξ)f dHn−1= Sn−1
Ψε(hK(u)− ξ, u )f(u) dHn−1(u).
Naturally,Φε(K) dependsonψ andf ,aswell,butwedonotsignalthesedependences. Weequip Kn
0 withtheHausdorffmetric,whichistheL∞ metriconthespaceof the
restrictions ofsupportfunctionstoSn−1.Forv ∈ Sn−1 and α∈ [0,π2],we consider the sphericalcap
Ω(v, α) ={u ∈ Sn−1 u, v ≥ cos α}. Wewrite π : Rn\{o}→ Sn−1 theradialprojection:
π(x) = x
x.
Inparticular,ifπ isrestrictedtotheboundaryofaK∈ Kn
(0),thenthismapisLipschitz.
Another typical application of the radial projection is to consider, for v ∈ Sn−1, the compositionx→ π(x+ v) asamapv⊥→ Sn−1 where
ThefollowingLemma5.1isthestatementwhereweapplydirectlythatψ ismodified tobe essentiallytqift is verysmall.
Lemma 5.1. Let ε ∈ (0,δ), and let {Ki} be a sequence of convex bodies tending to a
convexbody K inRn,andletξi∈ int Ki suchthat limi→∞ξi= x0∈ ∂K. Then
lim
i→∞Φε(Ki, ξi) =∞.
Proof. Wemayassumethatlimi→∞ξi= x0= o.Letv∈ Sn−1 beanexteriornormalto
∂K at o, andchoosesomeR > 0 such thatK ⊂ RBn. Thereforewe mayassumethat
Ki− ξi ⊂ (R + 1)Bn,hKi(v)< ε/8 andξi< ε/8 forall ξi, thushKi−ξi(v)< ε/4 for
alli.
Foranyζ∈ (0,ε8),thereexistsIζ suchthatifi≥ Iζ,thenξi≤ ζ/2 and y,v ≤ ζ/2 for all y ∈ Ki, and hence y,v ≤ ζ for all y ∈ Ki− ξi. For i ≥ Iζ, any y ∈ Ki− ξi canbe written inthe form y = sv + z where s ≤ ζ andz ∈ v⊥∩ (R + 1)Bn, thus if ∠(v,u)= α∈ [ζ,π2) foru∈ Sn−1,then wehave
hKi−ξi(u)≤ (R + 1) sin α + ζ cos α ≤ (R + 2)α. (28)
Wesetβ = ε
4(R+2),andforζ∈ (0,β),wedefine
Ωζ = Ω(v, β)\Ω(v, ζ).
Inparticular,asΨε(t)≥ ℵ1tqfort∈ (0,ε4) accordingto(24),ifu∈ Ωζ,then(28) implies Ψε(hKi−ξi(u))≥ γ(∠(v, u))
q
forγ =ℵ1(R + 2)q.
Thefunctionx→ π(x+ v) maps Bζ = v⊥∩
(tan β)Bn\(tan ζ)Bnbijectivelyonto Ωζ,whileβ < 18 and(27) yieldthattheJacobianofthismap isatleast2−nonBζ.
Sincef > τ1 and∠(v,π(x+ v))≤ 2x forx∈ Bζ,ifi≥ Iζ,then
Φε(Ki, ξi) = Sn−1 Ψε(hKi−ξi(u))f (u) dH n−1(u)≥ Ωζ τ1γ(∠(v, u))qdHn−1(u) ≥ τ1γ 2n+|q| Bζ xqdHn−1(x) = (n− 1)κn−1τ1γ 2n+|q| tan β tan ζ tq+n−2dt.
Asζ > 0 isarbitrarilysmallandq≤ 1−n,weconcludethatlimi→∞Φε(Ki,ξi)=∞. 2 Nowwesingleouttheoptimalξ∈ int K.
Proposition5.2. Forε∈ (0,δ) andaconvexbodyK inRn,thereexistsauniqueξ(K)∈ int K such that
Φε(K, ξ(K)) = min
ξ∈int KΦε(K, ξ).
In addition, ξ(K) and Φε(K,ξ(K)) are continuous functionsof K, and Φε(K,ξ(K)) is
translation invariant.
Proof. The firstpartofthis proof,theoneregardingtheexistenceofξ(K)∈ int K and its uniqueness, isvery similar to the proofof [1, Proposition3.2] given by theauthors and Yang for the Lp Minkowski problem. It is very short and we rewrite it here for completeness.
Let ξ1,ξ2 ∈ int K,ξ1= ξ2,and letλ∈ (0,1).Ifu∈ Sn−1\(ξ1− ξ2)⊥,then u,ξ1 =
u,ξ2 ,andhencethestrictconvexityofΨε (see(22))yieldsthat
Ψε(hK(u)− u, λξ1+ (1− λ)ξ2 ) > λΨε(hK(u)− u, ξ1 ) + (1 − λ)Ψε(hK(u)− u, ξ2 ),
thus Φε(K,ξ) isastrictlyconvexfunctionofξ∈ int K byf > τ1.
Letξi∈ int K suchthat
lim
i→∞Φε(K, ξi) =ξ∈int Kinf Φε(K, ξ).
We may assume that limi→∞ξi = x0 ∈ K, and Lemma 5.1 yields x0 ∈ int K. Since
Φε(K,ξ) is a strictly convex and continuous function of ξ ∈ int K, x0 is the unique
minimumpointofξ→ Φε(K,ξ),whichwedenotebyξ(K) (notsignallingthedependence
onε,ψ andf ).
Readilyξ(K) istranslationequivariant,andΦε(K,ξ(K)) istranslationinvariant. Forthecontinuityofξ(K) andΦε(K,ξ(K)),letusconsiderasequence{Ki} ofconvex bodiestendingtoaconvexbodyK inRn.Wemayassumethatξ(K
i) tendstoax0∈ K.
For any y ∈ int K,there exists anI such thaty ∈ int Ki fori ≥ I.SincehKi tends
uniformly tohK onSn−1,wehavethat lim sup
i→∞ Φε(Ki, ξ(Ki))≤ limi→∞i≥I Φε(Ki, y) = Φε(K, y).
AgainLemma5.1impliesthatx0∈ int K.ItfollowsthathKi−ξi(Ki) tendsuniformlyto
hK−x0,thus
Φε(K, x0) = lim
i→∞Φε(Ki, ξ(Ki))≤ limi→∞ i≥I
Φε(Ki, y) = Φε(K, y).
In particular, Φε(K,x0) ≤ Φε(K,y) for any y ∈ int K, thus x0 = ξ(K). In turn, we
Sinceξ→ Φε(K,ξ)=
Sn−1Ψε(hK(u)− u,ξ )f(u)dH
n−1(u) ismaximalat ξ(K)∈ int K andΨε=−ψε,wededuce
Corollary5.3. Forε∈ (0,δ) and aconvexbodyK in Rn,wehave Sn−1 u ψε hK(u)− u, ξ(K) f (u) dHn−1(u) = o.
For aclosed subgroup G ofO(n), we write Kn,G(0) to denote the family of K ∈ Kn
(0)
invariantunderG.
Lemma5.4. Forε∈ (0,δ),there existsaKε∈ Kn(0) with V (Kε)= 1 such that Φε(Kε, ξ(Kε))≥ Φε(K, ξ(K)) for any K∈ Kn(0) with V (K) = 1.
Inaddition,iff isinvariantunderaclosedsubgroupG ofO(n),thenKε canbechosen
tobeinvariantunder G.
Proof. WechooseasequenceKi∈ K(0)n withV (Ki)= 1 fori≥ 1 suchthat lim
i→∞Φ(Ki, ξ(Ki)) = sup{Φ(K, ξ(K)) : K ∈ K
n
(0) with V (K) = 1}.
WritingB1= κ−1/nn Bn todenotetheunitballcentredattheoriginandhavingvolume 1,we mayassumethateachKi satisfies
Φε(Ki, σ(Ki))≥ Φε(Ki, ξ(Ki))≥ Φε(B1, ξ(B1)). (29)
AccordingtoProposition5.2,wemayalsoassumethatσKi= o for eachKi.
Wededuce from Lemma4.3, (21), (23) and(29) that there exists someR > 0 such that Ki ⊂ RBn for any i ≥ 1. According to the Blaschke selection theorem, we may assume that Ki tends to a compact convex set Kε with o ∈ Kε. It follows from the continuity of the volume that V (Kε) = 1, and hence int Kε = ∅. We conclude from Lemma5.2 thatΦε(Kε,ξ(Kε))= limi→∞Φε(Ki,ξ(Ki)).
Iff isinvariantunderaclosedsubgroupG ofO(n),thenweapplythesameargument toconvexbodiesinKn,G(0) insteadofKn(0). 2
SinceΦ(5)< Φ(4),(25) yields some˜δ∈ (0,δ) such thatΨε(4)≥ Φ(5) forε∈ (0,δ).˜ For futurereference, the monotonicityof Ψε, diamκ−1/nn Bn ≤ 4 and(29) yield thatif ε∈ (0,δ),˜ then Φε(Kε, σ(Kε))≥ Φε κ−1/nn Bn, ξ(κ−1/nn Bn) ≥ Sn−1 Ψε(4)f dHn−1≥ Ψ(5) Sn−1 f dHn−1. (30)
6. Variationalformulae andsmoothnessoftheextremalbodywhenf isboundedand boundedawayfromzero
In this section, again let 0 < τ1 < τ2 and let the real function f on Sn−1 satisfy
τ1 < f < τ2. Inaddition, letϕ bethecontinuous functionof Theorem1.2,and weuse
thenotationdevelopedinSection5, sayψ : (0,∞)→ (0,∞) isdefinedbyψ = 1/ϕ. Now thatwe have constructed an extremal body Kε, we want to show thatit sat-isfies therequired differential equation inthe Alexandrov sense by using a variational argument.Thissectionprovidestheformulaethatwewillneed,andensuretherequired smoothnessofKε.
Concerningthevariationofvolume,akeytoolisAlexandrov’sLemma6.1(seeLemma 7.5.3in[38]).Tostatethis,foranycontinuoush: Sn−1→ (0,∞),wedefinethe Alexan-drov body
[h] ={x ∈ Rn: x, u ≤ h(u) for u ∈ Sn−1}
whichisaconvexbodycontainingtheorigininitsinterior.Obviously,ifK∈ Kn
(0) then
K = [hK].
Lemma 6.1 (Alexandrov). For K ∈ Kn
(0) and a continuous function g : Sn−1 → R,
K(t)= [hK+ tg] satisfies lim t→0 V (K(t))− V (K) t = Sn−1 g(u) dSK(u).
TohandlethevariationofΦε(K(t),ξ(K(t))) forafamilyK(t) isamoresubtle prob-lem. The next lemma shows essentially that ifwe perturba convex body K in a way such that the support function is differentiable as a function of the parameter t for Hn−1-almostallu∈ Sn−1,thenξ(K) changesalsoinadifferentiableway.Lemma6.2is thepoint oftheproof whereweusethatψε isC1andψε < 0.
Lemma6.2. Forε∈ (0,δ),letc> 0 andt0> 0,andletK(t) beafamilyofconvexbodies
with supportfunctionhtfort∈ [0,t0). Assumethat
(i) |ht(u)− h0(u)|≤ ct foreach u∈ Sn−1 andt∈ [0,t0),
(ii) limt→0+ht(u)−h0(u)
t existsforHn−1-almostallu∈ Sn−1.
Then limt→0+ξ(K(t))−ξ(K(0))
t exists.
Proof. We set K = K(0). We may assume thatξ(K) = o, and hence Proposition5.2 yields that
lim
ThereexistssomeR > r > 0 suchthatr≤ ht(u)− u,ξ(K(t)) = hK(t)−ξ(K(t))(u)≤ R foru∈ Sn−1 andt∈ [0,t0).SinceψεisC1 on[r,R],wecanwrite
ψε(t)− ψε(s) = ψε(s)(t− s) + η0(s, t)(t− s)
for t,s ∈ [r,R] where limt→sη0(s,t) = 0. Let g(t,u) = ht(u)− hK(u) for u ∈ Sn−1
and t ∈ [0,t0). Since hK(t)−ξ(K(t)) tends uniformly to hK on Sn−1, we deduce that if
t∈ [0,t0),then ψε ht(u)− u, ξ(K(t)) − ψε(hK(u)) = ψε(hK(u)) g(t, u)− u, ξ(K(t)) + e(t, u) (31) where
|e(t, u)| ≤ η(t)|g(t, u) − u, ξ(Kt) | and η(t) = η0(hK(u), ht(u)− u, ξ(K(t)) ). Notethatlimt→0+η(t)= 0 uniformlyinu∈ Sn−1.
Inparticular,(i)yieldsthate(t,u)= e1(t,u)+ e2(t,u) where
|e1(t, u)| ≤ cη(t)t and |e2(t, u)| ≤ η(t)ξ(K(t)). (32)
Itfollowsfrom (31) andfromapplyingCorollary 5.3toK(t) andK that
Sn−1
uψε(hK(u))
g(t, u)− u, ξ(K(t)) + e(t, u) f (u)dHn−1(u) = o,
whichcanbe writtenas
Sn−1
u ψε(hK(u)) g(t, u) f (u)dHn−1(u)
+ Sn−1 u e1(t, u) f (u)dHn−1(u) = Sn−1
u u, ξ(Kt) ψε(hK(u)) f (u)dHn−1(u)
− Sn−1
u e2(t, u) f (u)dHn−1(u).
Sinceψε(s)< 0 foralls> 0,thesymmetricmatrix A =
Sn−1
u⊗ u ψε(hK(u)) f (u)dHn−1(u)
vTAv = Sn−1
u, v 2ψ
ε(hK(u)) f (u) dHn−1(u) < 0.
Inaddition, A satisfies Sn−1
u u, ξ(Kt) ψε(hK(u)) f (u)dHn−1(u) = A ξ(Kt).
It followsfrom(32) that ift≥ 0 issmall, then
A−1 Sn−1
u ψε(hK(u)) g(t, u) f (u)dHn−1(u) + ˜e1(t) = ξ(Kt)− ˜e2(t), (33)
where ˜e1(t) ≤ α1η(t)t and ˜e2(t) ≤ α2η(t)ξ(Kt) for constants α1,α2 > 0. Since
η(t) tends to 0 with t, if t ≥ 0 is small, then ξ(K(t))− ˜e2(t)≥ 12ξ(Kt). Adding the estimate g(t,u)≤ ct, we deduce thatξ(K(t))≤ β t for aconstantβ > 0, which in turn yields that limt→0+ ˜ei(t)
t = 0 and ˜ei(0) = 0 for i = 1,2. Since there exists limt→0+g(t,u)−g(0,u)
t = ∂1g(0,u) forH
n−1 almostallu∈ Sn−1, and g(t,u)−g(0,u)
t < c for
allu∈ Sn−1 andt> 0,weconcludethat d dtξ(K(t)) t=0+ = A−1 Sn−1
u ψε(hK(u)) ∂1g(0, u) f (u) dHn−1(u). 2
Corollary 6.3.Under theconditions of Lemma6.2,and settingK = K(0), wehave d dtΦε(K(t), ξ(K(t))) t=0+ =− Sn−1 ∂ ∂thK(t)(u) t=0+ ψε hK(u)− u, ξ(K) f (u) dHn−1(u).
We omit theproof of thisresult sinceit isvery similar to thatof [1, Corollary 3.6], givenbytheauthorsandYangfortheLp Minkowskiproblem,withf (u)dHn−1(u),Ψε, −ψε,Lemma6.2andCorollary5.3replacingrespectivelydμ(u),ϕε,ϕε,Lemma3.5and Corollary 3.3.
Given afamilyK(t) of convex bodiesfor t∈ [0,t0), t0 > 0, to handle thevariation
ofΦε(K(t),ξ(K(t))) atK(0)= K viaapplyingCorollary6.3,weneedtheproperty(see Lemma6.2)thatthereexistsc> 0 suchthat
|hK(t)(u)− hK(u)| ≤ c |t| for any u ∈ Sn−1 and t∈ [0, t0) (34)
lim t→0+
hK(t)(u)− hK(u)
t exists forH
However,evenifK(t)= [hK+ thC] forK,C∈ Kn(0)andfort∈ (−t0,t0),K mustsatisfy
somesmoothness assumption inorder to ensurethat (35) holds also for thetwo sided limits(problemsoccursayifK isapolytopeandC issmooth).
Werecallthat∂K denotes thesetofsmoothpointsof ∂K.WesaythatK is
quasi-smooth if Hn−1(Sn−1\νK(∂K)) = 0; namely, the set of u ∈ Sn−1 that are exterior
normalsonlyatsingularpointshasHn−1-measurezero.ThefollowingLemma6.4,taken fromBianchi,Böröczky,Colesanti,Yang[1],showsthat(34) and(35) aresatisfiedevenif t∈ (−t0,t0) atleastforK(t)= [hK+ thC] witharbitraryC∈ Kn(0)ifK isquasi-smooth. Lemma 6.4.Let K,C ∈ Kn
(0) be suchthat rC⊂ K for some r > 0.Fort ∈ (−r,r) and
K(t)= [hK+ thC],
(i) if K ⊂ R C for R > 0, then |hK(t)(u)− hK(u)|≤ Rr|t| for any u∈ Sn−1 and
t∈ (−r,r);
(ii) ifu∈ Sn−1 istheexteriornormal atsomesmoothpointz∈ ∂K, then
lim t→0
hK(t)(u)− hK(u)
t = hC(u).
We will need the condition (35) in the following rather special setting taken from Bianchi,Böröczky,Colesanti,Yang[1].
Lemma6.5.LetK beaconvexbodywithrBn⊂ int K forr > 0, letω⊂ Sn−1 beclosed, andif t∈ [0,r),thenlet
K(t) = [hK− 1ω] ={x ∈ K : x, u ≤ hK(u)− t for every u ∈ ω}.
(i) Wehave limt→0+hK(t)(u)−hK(u)
t existsandisnon-positiveforallu∈ S
n−1,and ifu∈ ω,theneven limt→0+
hK(t)(u)−hK(u)
t ≤ −1.
(ii) IfSK(ω)= 0,then limt→0+V (K(t))−V (K)
t = 0.
Proposition6.6. Forε∈ (0,δ),Kε isquasi-smooth.
Proof. We suppose that Kε is not quasi-smooth, and seek a contradiction. It follows thatHn−1(X)> 0 forX = Sn−1\νKε(∂Kε),thereforethereexistsaclosedω⊂ X such
thatHn−1(ω)> 0.SinceνK−1ε(ω)⊂ ∂Kε\∂Kε,wededucethatSKε(ω)= 0.
We may assume that ξ(Kε) = o and rBn ⊂ Kε ⊂ RBn for R > r > 0. As in Lemma6.5,ift∈ [0,r),thenwedefine
K(t) = [hKε− 1ω] ={x ∈ Kε: x, u ≤ hK(u)− t for every u ∈ ω}.
Clearly, K(0) equals Kε. We define α(t) = V (K(t))−1/n, and hence α(0) = 1, and Lemma6.5 (ii)yieldsthatα(0)= 0.
Weset K(t) = α(t)K(t),andhenceK(0) = Kεand V ( K(t))= 1 for allt∈ [0,r). In addition,weconsiderh(t,u)= hK(t)(u) andh(t,˜ u)= hK(t)(u)= α(t)h(t,u) foru∈ S
n−1 andt∈ [0,r).Since[hKε−thBn]⊂ K(t),Lemma6.4(i)yieldsthat|h(t,u)−h(0,u)|≤ R
r t foru∈ Sn−1 andt∈ [0,r).Henceα(0)= 0 impliesthatthereexistc> 0 andt
0∈ (0,r)
such that |˜h(t,u)− ˜h(0,u)| ≤ ct for u ∈ Sn−1 and t ∈ [0,t0). Applying α(0) = 1,
α(0)= 0 andLemma6.5(i),wededucethat
∂1˜h(0, u) = lim t→0+ ˜ h(t, u)− ˜h(0, u) t = limt→0+ h(t, u)− h(0, u)
t ≤ 0 exists for all u ∈ S
n−1, ∂1˜h(0, u)≤ −1 for all u ∈ ω.
As ψε is positive and monotone decreasing, f > τ1 and Hn−1(ω) > 0, Corollary 6.3
impliesthat d dtΦε( K(t), ξ( K(t))) t=0+ =− Sn−1
∂1h(0, u)˜ · ψε(hK(u)) f (u) dHn−1(u)
≥ −
ω
(−1)ψε(R)τ1dHn−1(u) > 0.
Therefore Φε( K(t),ξ( K(t))) > Φε(Kε,ξ(Kε)) for small t > 0. This contradicts the definitionofKεand concludestheproof. 2
Forε∈ (0,δ),wedefine λε= 1 n Sn−1 hKε−ξ(Kε)· ψε(hKε−ξ(Kε))· f dHn−1. (36)
Proposition6.7. Forε∈ (0,δ),ψε(hKε−ξ(Kε))·f dHn−1= λεdSKε asmeasuresonSn−1.
We omit the proof of this result since it is very similar to that of [1, Proposition 6.1], given by the authors and Yang for the Lp Minkowski problem, with −λε, −ψε, Lemma6.1,Lemma6.4,Corollary6.3,and[38] replacingrespectivelyλε,ϕε,Lemma 5.2, Lemma2.3,Corollary 3.6and[35].
7. TheproofofTheorem1.2 whenf isboundedandboundedawayfromzero
Inthissection,againlet0< τ1< τ2,lettherealfunctionf onSn−1 satisfyτ1< f <
τ2,andletϕ bethecontinuousfunctionon[0,∞) ofTheorem 1.2.Weusethenotation
developedinSection5,andhenceψ : (0,∞)→ (0,∞) andψ = 1/ϕ.
To ensurethataconvex bodyis“fat” enoughinLemma7.2and later,thefollowing observationisuseful:
Lemma 7.1.If K is a convex body in Rn with V (K) = 1 and K ⊂ σ(K)+ RBn for R > 0,then
σ(K) + rBn⊂ K for r = 1
cκn−1n
−3/2R−(n−1).
Proof. Letz0+ r0Bn bealargest ballinK.Accordingto theSteinhagentheorem [12,
Theorem50], thereexistsv∈ Sn−1 suchthat
| x − z0, v | ≤ c√nr0 for x∈ K,
where c is apositiveuniversal constant.It follows that1= V (K) ≤ c√nr0κn−1Rn−1,
thusr0≥ cκn−11 n−1/2R−(n−1).Sinceσ(K)+rn0 Bn⊂ K by−(K−σ(K))⊂ n(K−σ(K)),
wemaychooser = 1
cκn−1n
−3/2R−(n−1). 2
Werecall(compare(36))thatifε∈ (0,δ) andξ(Kε)= o,then λ
ε isdefinedby λε= 1 n Sn−1 hKεψε(hKε)f dHn−1. (37)
Lemma 7.2. There existR0 > 1, r0 > 0 and λ˜2 > ˜λ1 > 0 depending on f,q,ψ,ℵ such
that ifε∈ (0,δ0) forδ0= min{˜δ,r20} whereδ comes˜ from (30),then ˜λ1≤ λε≤ ˜λ2 and
σ(Kε) + r0Bn⊂ Kε⊂ σ(Kε) + R0Bn.
Proof. Accordingto(23),thereexistsℵ0> 0 dependingonq,ψ,ℵ suchthatifε∈ (0,δ)
and t∈ (0,δ),then Ψε(t) ≤ ℵ0tq.In addition, limt→∞Ψε(t)= 0 by(21), therefore we mayapply Lemma4.3.Since (30) provides thecondition
Sn−1 Ψε(hKε−σ(Kε))f dHn−1≥ Ψ(5) Sn−1 f dHn−1
forany ε∈ (0,δ),˜ wededuce fromLemma4.3 theexistence ofR0> 0 such thatKε⊂
σ(Kε)+ R
0Bn foranyε∈ (0,δ).˜ Inaddition, theexistenceofr0independent ofε such
thatσ(Kε)+ r
0Bn ⊂ Kε followsfromLemma7.1.
Toestimateλε,weassumeξ(Kε)= o.Letwε∈ Sn−1and ε≥ 0 besuchthatσ(Kε)= εwε,andhencer0wε∈ Kε.ItfollowsthathKε(u)≥ r0/2 holdsforu∈ Ω(wε,π
3),while
Kε⊂ 2R0Bn,R0> 1 and themonotonicityofψε implythatψε(hKε(u))≥ ψε(2R0)=
ψ(2R0) forallu∈ Sn−1.
Wededucefrom(37) that λε= 1 n Sn−1 hKεψε(hKε)f dHn−1≥ 1 n· r0 2 · ψ(2R0)· τ1· H n−1Ωw ε, π 3 = ˜λ1.
To have asuitable upper bound on λε, we define α∈ (0,π2) with cos α = 2Rr00, and hence Ω(−wε, α) = u∈ Sn−1: u, wε ≤ −r0 2R0 . A key observation is that if u ∈ Sn−1\Ω(−w
ε,α), then u,wε >−2Rr00 and ε ≤ R0
imply hKε(u)≥ u, wε+ r0u ≥ r0−r0 ε 2R0 ≥ r 0/2, therefore ε<r0 2 yields ψε(hKε(u))≤ ψε(r0/2) = ψ(r0/2). (38)
AnotherobservationisthatKε⊂ 2R
0Bn implies
hKε(u) < 2R0 for any u∈ Sn−1. (39)
It followsdirectlyfrom(38) and(39) that
Sn−1\Ω(−wε,α)
hKεψε(hKε)f dHn−1≤ (2R0)ψ(r0/2)τ2nκn. (40)
However, if u ∈ Ω(−wε,α), then ψε(hKε(u)) can be arbitrary large as ξ(Kε) can
be arbitrary closeto ∂Kε ifε > 0 is small, and hence we transferthe problem to the previous caseu∈ Sn−1\Ω(−wε,α) using Corollary 5.3. Firstapplying u,−wε ≥ 2Rr00 foru∈ Ω(−wε,α),thenCorollary5.3,andafterthat u,wε ≤ 1,f ≤ τ2and(38) implies
Ω(−wε,α)
ψε(hKε(u))f (u) dHn−1(u)≤ 2R0
r0
Ω(−wε,α)
u, −wε ψε(hKε(u))f (u) dHn−1(u)
= 2R0 r0
Sn−1\Ω(−wε,α)
u, wε ψε(hKε(u))f (u) dHn−1(u)
≤ 2R0 r0 · ψ r 0 2 τ2nκn. Now (39) yields Ω(−wε,α) hKψε(hK)f dHn−1≤ (2R0)2 r0 · ψ r0 2 τ2nκn,
which estimate combinedwith (40) leadsto λε< (2R0)2 r0 + 2R0 ψ(r0 2)τ2nκn. Inturn, we concludeLemma7.2. 2
NowweproveTheorem1.2iff isboundedandbounded awayfrom zero.
Theorem7.3.For0< τ1< τ2,lettherealfunctionf onSn−1satisfyτ1< f < τ2,andlet
ϕ: [0,∞)→ [0,∞) beincreasing andcontinuoussatisfyingϕ(0)= 0,lim inft→0+ ϕ(t)
t1−p >
0, and 1∞ϕ1 < ∞. Let Ψ(t) = t∞ϕ1. Then there exist λ > 0 and a K ∈ Kn
0 with V (K)= 1 suchthat f dHn−1= λϕ(hK) dSK, asmeasureson Sn−1,and Sn−1 Ψ(hK−σ(K))f dHn−1≥ Ψ(5) Sn−1 f dHn−1. (41)
Inaddition, iff is invariantunder aclosedsubgroup G of O(n),then K can bechosen
tobeinvariantunder G.
Proof. We assume that ξ(Kε) = o for all ε ∈ (0,δ
0) where δ0 ∈ (0,δ] comes from
Lemma7.2.UsingtheconstantR0ofLemma7.2,wehavethat
Kε⊂ 2R0Bn and hKε(u) < 2R0 for any u∈ Sn−1 and ε∈ (0, δ0). (42)
We consider the continuous increasing function ϕε : [0,∞)] → [0,∞) defined by ϕε(0)= 0 andϕε(t)= 1/ψε(t) forε∈ (0,δ).Weclaimthat
ϕεtends uniformly to ϕ on [0, 2R0] as ε > 0 tends to zero. (43)
Forany smallζ > 0,there existsδζ > 0 suchthatϕ(t)≤ ζ/2 fort∈ [0,δζ]. Wededuce from (20) thatifε> 0 is small,then |ϕε(t)− ϕ(t)|≤ ζ/2 fort∈ [δζ,2R0]. Howeverϕε ismonotone increasing,thereforeϕε(t),ϕ(t)∈ [0,ζ] for t∈ [0,δζ], completingthe proof of(43).
For any ε ∈ (0,δ0), it follows from Lemma 6.7 thatψε(hKε)f dHn−1 = λεdSKε as
measuresonSn−1.Integratinggϕε(hKε) foranycontinuousrealfunctiong onSn−1,we
deducethat
f dHn−1 = λεϕε(hKε) dSKε (44)
asmeasures onSn−1.
Sinceλ˜1≤ λε≤ ˜λ2forsomeλ˜2> ˜λ1 independentofε accordingtoLemma7.2,(42)
yieldsthe existenceofλ> 0,K ∈ Kn
0 withV (K)= 1 and sequence {ε(m)} tendingto
0 suchthatlimm→∞λε(m)= λ andlimm→∞Kε(m)= K.As hKε(m) tends uniformlyto
hK onSn−1, wededuce thatλε(m)ϕε(m)(hKε(m)) tendsuniformly to λϕ(hK) on Sn−1.
f dHn−1 = λϕ(hK) dSK.
Wenote thatiff isinvariantunderaclosed subgroupG ofO(n),theneachKεcanbe chosento beinvariantunderG accordingtoLemma5.4,thereforeK isinvariantunder G in thiscase.
To prove(41),ifε∈ (0,δ0),then(30) providesthecondition
Sn−1 Ψε(hKε−σ(Kε))f dHn−1≥ Ψ(5) Sn−1 f dHn−1. (45)
Now Lemma7.2 yields thatthere exists r0 > 0 such that ifε ∈ (0,δ0), then σ(Kε)+
r0Bn ⊂ Kε where0< δ0≤ r20.Inparticular, ifu∈ Sn−1,thenhKε−σ(Kε)(u)≥ r0,and
hencewededucefrom (26) that
Ψε(hKε−σ(Kε)(u))≤ Ψ(hKε−σ(Kε)(u)) +
π
2. (46)
Since Kε(m)− σ(Kε(m)) tendsto K− σ(K),(25) impliesthatifu∈ Sn−1,then lim
ε→0+Ψε(hK
ε−σ(Kε)(u)) = Ψ(hK−σ(K)(u)). (47)
Combining (45), (46) and (47) with Lebesgue’s Dominated Convergence Theorem,we conclude(41),andinturnTheorem7.3. 2
8. TheproofofTheorem1.2
Let−n< p< 0,letf beanon-negative non-trivialfunctioninL n n+p(S
n−1),and let ϕ: [0,∞)→ [0,∞) beamonotoneincreasingcontinuousfunctionsatisfyingϕ(0)= 0,
lim inf t→0+ ϕ(t) t1−p > 0 (48) ∞ 1 1 ϕ(t)dt <∞. (49)
Weassociatecertainfunctionstof andϕ.Foranyintegerm≥ 2,wedefinefmonSn−1 as follows: fm(u) = ⎧ ⎪ ⎨ ⎪ ⎩ m if f (u)≥ m, f (u) if m1 < f (u) < m, 1 m if f (u)≤ 1 m.
In particular,fm≤ ˜f wherethefunctionf : S˜ n−1 → [0,∞) inL n n+p(S
n−1),and hence inL1(Sn−1),isdefinedby
˜ f (u) =
f (u) if f (u) > 1, 1 if f (u)≤ 1. AsinSection5, using(49),wedefinethefunction
Ψ(t) = ∞ t 1 ϕ for t > 0.
Accordingto(48),thereexist someδ∈ (0,1) andℵ> 1 suchthat 1
ϕ(t) <ℵt
p−1 for t∈ (0, δ). (50)
WededucefromLemma4.1thatthere existsℵ0> 1 depending onϕ suchthat
Ψ(t) <ℵ0tp for t∈ (0, δ). (51)
Form≥ 2,Theorem7.3yieldsaλm> 0 andaconvexbodyKm∈ Kn0 withξ(Km)= o∈ int Km,V (Km)= 1 suchthat
λmϕ(hKm) dSKm = fmdH n−1 (52) Sn−1 Ψ(hKm−σ(Km))fmdH n−1 ≥ Ψ(5) Sn−1 fmdHn−1. (53)
Inaddition,iff isinvariantunderaclosedsubgroupG ofO(n),thenfmisalsoinvariant underG,and henceKmcanbechosento beinvariantunderG.
Sincefm≤ ˜f ,andfmconverges pointwiseto f ,Lebesgue’sDominated Convergence theoremyieldstheexistenceofm0> 2 suchthatifm> m0,then
1 2 Sn−1 f < Sn−1 fm< 2 Sn−1 f. (54) Inparticular,(53) implies Sn−1 Ψ(hKm−σ(Km)) ˜f dH n−1≥Ψ(5) 2 Sn−1 f dHn−1. (55)
WededucefromV (Km)= 1,limt→∞Ψ(t)= 0,(51),(55) andLemma4.3 thatthere existsR0> 0 independentofm suchthat
Km⊂ σ(Km) + R0Bn⊂ 2R0Bn for all m > m0. (56)
σ(Km) + r0Bn ⊂ Km for all m > m0. (57)
To estimateλm frombelow,(56) impliesthat
Sn−1
ϕ(hKm) dSKm ≤ ϕ(2R0)H
n−1(∂K
m)≤ ϕ(2R0)(2R0)n−1nκn,
and henceitfollows from (52) and (54) the existenceofλ˜1 > 0 independent ofm such
that λm= Sn−1fmdHn−1 Sn−1ϕ(hKm) dSKm ≥ ˜λ1 for all m > m0. (58)
To haveasuitableupper bound onλmfor any m> m0, wechoosewm∈ Sn−1 and m≥ 0 such thatσ(Km)= mwm.WesetB#m= wm⊥∩ int Bn and considertherelative open set Ξm= (∂Km)∩ mwm+ r0B#m+ (0,∞)wm .
Ifu is anexteriorunitnormalatanx∈ Ξmform> m0,thenx= ( m+ s)wm+ rv for s> 0,r∈ [0,r0) andv∈ wm⊥∩ Sn−1,andhence mwm+ rv∈ Kmyields
u, ( m+ s)wm+ rv = hKm(u)≥ u, mwm+ rv ,
implying that u,wm ≥ 0; or in other words, u ∈ Ω(wm,π2). Since the orthogonal projectionofΞm ontowm⊥ isBm# form> m0,wededucethat
SKm Ωwm, π 2 ≥ Hn−1(Ξ m)≥ Hn−1(Bm#) = r0n−1κn−1. (59) Ontheother hand,ifu∈ Ω(wm,π2) form> m0,then mwm+ r0u∈ Km yields
hKm(u)≥ u, mwm+ r0u ≥ r0. (60)
Combining (54),(59) and(60) implies
λm= Ω(wm,π2) fmdHn−1 Ω(wm,π2)ϕ(hKm) dSKm ≤2 Sn−1f dHn−1 ϕ(r0)r0n−1κn−1 = ˜λ2 for all m > m0. (61)
Since Km ⊂ 2R0Bn and λ˜1 ≤ λm ≤ ˜λ2 for m > m0 by (56), (58) and (61), there
exists subsequence{Km}⊂ {Km} suchthatKm tendstosomeconvexcompactset K and λm tendsto someλ> 0. Aso∈ Km and V (Km)= 1 forallm, wehaveo ∈ K
Weclaimthatforanycontinuous functiong : Sn−1 → R,we have Sn−1 gλϕ(hK) dSK = Sn−1 gf dHn−1. (62)
Asϕ is uniformlycontinuous on[0,2R0] andhKm tends uniformlyto hK on S
n−1, we deducethatλmϕ(hKm) tendsuniformlyto λϕ(hK) onS
n−1.SinceS Km tendsweakly toSK,wehave lim m→∞ Sn−1 gλmϕ(hKm) dSKm = Sn−1 gλϕ(hK) dSK.
Ontheotherhand,|gfm|≤ ˜f· maxSn−1|g| forallm≥ 2,andgfmtendspointwisetogf
as m tends to infinity.Therefore Lebesgue’s Dominated Convergence Theorem implies
that lim m→∞ Sn−1 gfmdHn−1= Sn−1 gf dHn−1,
whichinturn yields(62) by(52).Inturn,we concludeTheorem 1.2by(62). 2 References
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