Nakajima's problem: convex bodies of constant width and constant brightness

0.11 MB·English

by  Ralph Howard

NAKAJIMA’S PROBLEM: CONVEX BODIES OF CONSTANT WIDTH AND
6 CONSTANT BRIGHTNESS
0
0
2 RALPHHOWARDANDDANIELHUG

n
DedicatedtoRolfSchneiderontheoccasionofhis65thbirthday
a
J

0 ABSTRACT. ForaconvexbodyK⊂Rn,thekthprojectionfunctionofKassignstoany
2 k-dimensionallinearsubspaceofRnthek-volumeoftheorthogonalprojectionofK to
thatsubspace.LetKandK0beconvexbodiesinRn,andletK0becentrallysymmetric
G] andsatisfyaweakregularityandcurvaturecondition(whichincludesallK0with∂K0of
classC2withpositiveradiiofcurvature). AssumethatKandK0haveproportional1st
M projectionfunctions(i.e.,widthfunctions)andproportionalkthprojectionfunctions. For
2≤k<(n+1)/2andfork=3,n=5weshowthatKandK0arehomothetic.Inthe
.
h specialcasewhereK0isaEuclideanball,wethusobtaincharacterizations ofEuclidean
t ballsasconvexbodiesofconstantwidthandconstantk-brightness.
a
m
[

1 1. INTRODUCTION AND STATEMENTOF RESULTS
v
9 LetK beaconvexbody(acompact,convexsetwithnonemptyinterior)inRn,n 3.
≥
9 Assumethat, foranyline, the lengthofthe projectionofK to the lineis independentof
4 thatline and, foranyhyperplane,the volumeof theprojectionofK to the hyperplaneis
1 independentofthathyperplane.MustK thenbeaEuclideanball?
0
In dimensionthree, thisproblemhasbecomeknownasNakajima’sproblem[11]; see
6
0 [1],[2],[3],[4],[5],[6].ItiseasytocheckthattheanswertoitisintheaffirmativeifKisa
/ convexbodyinR3ofclassC2.ForgeneralconvexbodiesinR3,theproblemismuchmore
h
difficultandasolutionhasonlybeenfoundrecently.LetG(n,k)denotetheGrassmannian
t
a ofk-dimensionallinearsubspacesofRn. AconvexbodyK inRnissaidtohaveconstant
m
k-brightness,k 1,...,n 1 ,ifthek-volumeV (K U)oftheorthogonalprojection
k
: ofK tothelinea∈rs{ubspaceU− }G(n,k)isindependento|fthatsubspace.Themap
v ∈
Xi πk: G(n,k) R, U Vk(K U),
→ 7→ |
r is referredto as the kth projection functionof K. Hence a convexbodyK has constant
a
width(i.e.constant1-brightness)ifithasconstant1stprojectionfunction(widthfunction).
1.1.Theorem ([7]). Let K be a convexbody in Rn having constantwidth and constant
2-brightness.ThenK isaEuclideanball.
ThistheoremprovidesacompletesolutionoftheNakajimaprobleminR3 forgeneral
convex bodies. In the present paper, we continue this line of research. Our main result
complementsTheorem1.1bycoveringthecasesofconvexbodiesofconstantwidthand
constantk-brightnesswith2 k <(n+1)/2ork =3,n=5.
≤
Date:December11,2005.
2000MathematicsSubjectClassification. 52A20.
Key wordsand phrases. Constant width, constant brightness, projection function, characterization ofEu-
clideanballs,umbilics.
1
2 RALPHHOWARDANDDANIELHUG
1.2. Theorem. Let K be a convex body in Rn having constant width and constant k-
brightnesswith2 k <(n+1)/2,ork =3,n=5. ThenK isaEuclideanball.
≤
Theprecedingtwotheoremscanbegeneralizedto pairsofconvexbodiesK,K0 hav-
ing proportionalprojectionfunctions, providedthat K0 is centrally symmetric and has a
minimalamountofregularity.
1.3.Theorem. LetK,K0beconvexbodiesinRn,andletK0becentrallysymmetricwith
positive principal radii of curvature on some Borel subset of the unit sphere of positive
measure. Let2 k < (n+1)/2,orletk = 3,n = 5inwhichcaseassumethesurface
≤
areameasureS4(K0, )ofK0 isabsolutelycontinuouswithpositivedensity. Assumethat
·
thereareconstantsα,β >0suchthat
π1(K)=απ1(K0) and πk(K)=βπk(K0).
ThenK andK0arehomothetic.
As the naturalmeasure on the unit sphere, Sn−1, we use the invariantHaar probabil-
ity measure (i.e. spherical Lebesgue measure), or what is the same thing the (n 1)-
dimensionalHausdorffmeasure, n−1,normalizedsothatthetotalmassisone. We−view
H
the principal radii of curvature as functions of the unit normal, despite the fact that the
unit normal map is in general a set valued function (cf. the beginning of Section 2 be-
low). Theassumptionthattheprincipalradiiofcurvaturearepositiveonasetofpositive
measuremeansthatthereisaBorelsubsetofSn−1 ofpositivemeasuresuchthatonthis
setthereverseGaussmapissinglevalued,differentiable(inageneralizedsense)andthe
eigenvaluesofthedifferentialarepositive.Explicitly,thisconditioncanbestatedinterms
of second orderdifferentiabilitypropertiesof the supportfunction(againsee Section 2).
In particular, it is certainly satisfied if K0 is of class C+2, and therefore letting K0 be a
EuclideanballrecoversTheorem1.2. TherequiredconditionallowsforpartsofK0 tobe
quiteirregular.Forexampleif∂K0hasapointthathasasmallneighborhoodwhere∂K0is
C2 withpositiveGauss-Kroneckercurvature,thentheassumptionwillhold,regardlessof
howroughtherestoftheboundaryis. Forexamplea“sphericalpolyhedron”constructed
byintersectinga finite numberofEuclideanballsin Rn willsatisfy the condition. More
generallyiftheconvexbodyK0 isanintersectionofafinitecollectionofbodiesofclass
C2,itwillsatisfythecondition.
+
Theorem1.3extendsthe mainresultsin[8]forthe rangeofdimensionsk,nwhereit
appliesbyreducingtheregularityassumptiononK0 anddoingawaywithanyregularity
assumptions on K. However, the classical Nakajima problem, which concerns the case
n=3andk =2,isnotcoveredbythepresentapproach.
Despite recentprogressontheNakajimaproblemvariousquestionsremainopen. For
instance, can Euclidean balls be characterized as convex bodies having constant width
and constant (n 1)-brightnessif n 4? This question is apparently unresolvedeven
− ≥
for smooth convex bodies. A positive answer is available for smooth convex bodies of
revolution(cf.[8]). Fromtheargumentsofthepresentpaperthefollowingpropositionis
easytocheck.
1.4. Proposition. Let K,K0 Rn be convex bodies that have a common axis of revo-
⊂
lution. Let K0 be centrally symmetric with positive principal radii of curvature almost
everywhere. AssumethatK andK0 haveproportionalwidth functionsandproportional
kthprojectionfunctionsforsomek 2,...,n 2 . ThenK andK0arehomothetic.
∈{ − }
ItisapleasurefortheauthorstodedicatethispapertoRolfSchneider. ProfessorRolf
Schneiderhasbeenalargesourceofinspirationforcountlessstudentsandcolleaguesall
NAKAJIMA’SPROBLEM:CONVEXBODIESOFCONSTANTWIDTHANDCONSTANTBRIGHTNESS 3
over the world. His willingness to communicate and share his knowledge make contact
withhimapleasurableandmathematicallyrewardingexperience. Thesecondnamedau-
thorhasparticularlybeenenjoyingmanyyearsof support, personalinteractionandjoint
research.
2. PRELIMINARIES
Let K be a convexbody in Rn, and let h : Rn R be the supportfunctionof K,
K
whichisaconvexfunction.Forx Rnlet∂h (x)be→thesubdifferentialofh atx. This
K K
isthesetofvectorsv Rn sucht∈hatthefunctionh v, achievesitsminimumatx.
K
Itiswellknownthat,f∈orallx Rn,∂h (x)isanon−emhpty·icompactconvexsetandisa
K
∈
singletonpreciselyatthosepointswhereh isdifferentiableintheclassicalsense(cf.[13,
K
pp. 30–31]). For u Sn−1 the set ∂h (u) is exactly the set of x ∂K such that u is
K
∈ ∈
anoutwardpointingnormaltoK atx(cf.[13,Thm1.7.4]). Butthisisjustthedefinition
ofthereverseGaussmap(whichingeneralisnotsinglevalued,butasetvaluedfunction)
andso the functionu ∂h (u) givesa formulaforthe reverseGaussmapin termsof
K
7→
thesupportfunction.
Inthefollowing,by“almosteverywhere”ontheunitsphereorby“foralmostallunit
vectors” we mean for all unit vectors with the possible exclusion of a set of spherical
Lebesgue measure zero. A theorem of Aleksandrov states that a convex function has a
generalizedsecondderivativealmosteverywhere,whichwewillviewasapositivesemi-
definite symmetric linear map rather than a symmetric bilinear form. This generalized
derivativecaneitherbedefinedintermsofasecondorderapproximatingTaylorpolyno-
mialatthepoint,orintermsofthesetvaluedfunctionx ∂h (x)beingdifferentiable
K
7→
inthesenseofsetvaluedfunctions(boththesedefinitionsarediscussedin[13,p.32]).At
pointswheretheAleksandrovsecondderivativeexists∂h issinglevalued. Becauseh
K K
is positivelyhomogeneousof degreeone, if it is Aleksandrovdifferentiableat a pointx,
thenitisAleksandrovdifferentiableatallpointsλxwith λ > 0. ThenFubini’stheorem
impliesthatnotonlyish Aleksandrovdifferentiableat nalmostallpointsofRn,butit
K
isalsoAleksandrovdifferentiableat n−1almostallpoinHtsofSn−1. Forpointsu Sn−1
whereitexists,letd2h (u)denotetHheAleksandrovsecondderivativeofh . Let∈u⊥ de-
K K
notetheorthogonalcomplementofu. Thenthe restrictiond2h (u)u⊥ isthe derivative
K
ofthereverseGaussmapatu. Theeigenvaluesofd2h (u)u⊥ aret|heprincipalradiiof
K
curvatureatu. Asthediscussionaboveshowstheseexistata|lmostallpointsofSn−1.
Ausefultoolforthestudyofprojectionfunctionsofconvexbodiesarethesurfacearea
measures. An introductionto these Borelmeasuresonthe unitsphereisgivenin [13], a
morespecializedreference(forthepresentpurpose)iscontainedintheprecedingwork[8].
The topordersurface area measure Sn−1(K, ) of the convexbodyK Rn can be ob-
tainedasthe(n 1)-dimensionalHausdorffm·easure n−1oftherevers⊂esphericalimage
ofBorelsetsof−theunitsphereSn−1. TheRadon-NikHodymderivativeofSn−1(K, )with
·
respecttothesphericalLebesguemeasureistheproductoftheprincipalradiiofcurvature
ofK. Sinceforalmosteveryu Sn−1,theradiiofcurvatureofK atu Sn−1 arethe
eigenvalues of d2hK(u)u⊥, the∈Radon-Nikodym derivative of Sn−1(K,∈) with respect
tosphericalLebesguem|easureisthefunctionu det d2h (u)u⊥ , w·hichisdefined
K
7→ |
almost everywhere on Sn−1. In particular, if Sn−1(K,(cid:0)) is absolutel(cid:1)y continuous with
·
respect to spherical Lebesgue measure, the density function is just the Radon-Nikodym
derivative. For explicit definitions of these and other basic notions of convex geometry
neededhere,wereferto[13]and[8].
4 RALPHHOWARDANDDANIELHUG
The following lemma contains more precise information about the Radon-Nikodym
derivative of the top order surface area measure. We denote the support function of a
convexbodyK byh,ifK isclearfromthecontext. Forafixedunitvectoru Sn−1 and
i N, we also putω := v Sn−1 : v,u 1 (2i2)−1 , wheneveru is∈clear from
i
∈ ∈ h i≥ −
thecontext. Henceωi u(cid:8),asi ,inthesenseofHaus(cid:9)dorffconvergenceofclosed
↓ { } → ∞
sets.
2.1. Lemma. Let K Rn be a convex body. If u Sn−1 is a point of second order
⊂ ∈
differentiabilityofthesupportfunctionhofK,then
lim Sn−1(K,ωi) =det d2h(u)u⊥ .
i→∞ Hn−1(ωi) (cid:0) | (cid:1)
Proof. ThisisimplicitlycontainedintheproofofHilfssatz2in[10]. Asimilarargument,
inaslightlymoreinvolvedsituation,canbefoundin[9]. (cid:3)
AnanalogueofLemma2.1forcurvaturemeasuresisprovidedin[12,(3.6)Hilfssatz].
As another ingredient in our approach to Nakajima’s problem, we need two simple
algebraic lemmas. Here we write M for the cardinality of a set M. If x1,...,xn are
| |
realnumbersand I = {i1,...,ik} ⊆ {1,...,n} we set xI := xi1...xik. We also put
x∅ :=1.
2.2. Lemma. Let b > 0 be fixed. Let x1,...,xn−1,y1,...,yn−1 be nonnegative real
numberssatisfying
x +y =2 and x +y =2b
i i I I
foralli=1,...,n 1andallI 1,...,n 1 with I =k,wherek 2,...,n 2 .
− ⊂{ − } | | ∈{ − }
Then x1,...,xn−1 2and y1,...,yn−1 2.
|{ }|≤ |{ }|≤
Proof. Wecanassumethatx1 xn−1. Thenwehavey1 yn−1.
≤···≤ ≥···≥
Ifx1 = 0,theny1 = 2. Further,forI′ 2,...,n 1 with I′ = k 1,wehave
⊂ { − } | | −
y1yI′ = 2b, hence yI′ = b. Since k 2, we get y2,...,yn−1 > 0. Moreover, since
≥
k 1 n 3,weconcludethaty2 = =yn−1. Thisshowsthatalsox2 = =xn−1,
− ≤ − ··· ···
andthus x1,...,xn−1 2and y1,...,yn−1 2.
|{ }|≤ |{ }|≤
Ifyn−1 =0,thesameconclusionisobtainedbysymmetry.
If x1 > 0 and yn−1 > 0, then x1,...,xn−1,y1,...,yn−1 > 0. Now we fix any set
J 1,...,n 1 with J = k +1. The argumentat the beginning of the proof of
⊆ { − } | |
Lemma4.2in [8] showsthat x : i J 2. Since k+1 3, we first obtainthat
i
x1,...,xn−1 2,andthe|n{also ∈y1,.}.|.,≤yn−1 2. ≥ (cid:3)
|{ }|≤ |{ }|≤
2.3. Lemma. Let n 4, and let b > 0 be fixed. Let x1,...,xn−1,y1,...,yn−1 be
≥
nonnegativerealnumberssatisfying
x +y =2 and x +y =2b
i i I I
foralli=1,...,n 1andallI 1,...,n 1 with I =n 2. Then
− ⊂{ − } | | −
(2.1) x = y =b
l l
lY6=i,j lY6=i,j
wheneveri,j 1,...,n 1 aresuchthatx =x .
i j
∈{ − } 6
NAKAJIMA’SPROBLEM:CONVEXBODIESOFCONSTANTWIDTHANDCONSTANTBRIGHTNESS 5
Proof. Fortheproof,wemayassumethati = 1andj = n 1,tosimplifythenotation.
−
Thenwehave
x1 xn−2+y1 yn−2 =2b,
··· ···
x2 xn−1+y2 yn−1 =2b,
··· ···
whichimpliesthat
x2 xn−2(xn−1 x1)+y2 yn−2(yn−1 y1)=0.
··· − ··· −
Moreover,x1+y1 =2=xn−1+yn−1yields
xn−1 x1 =y1 yn−1 =0,
− − 6
andthus
x2 xn−2 =y2 yn−2.
··· ···
Hence
2x2 xn−2 =(x1+y1)x2 xn−2 =x1x2 xn−2+y1x2 xn−2
··· ··· ··· ···
=x1x2 xn−2+y1y2 yn−2 =2b,
··· ···
andthus
b=x2 xn−2 =y2 yn−2.
··· ···
(cid:3)
3. PROOFS
First,bypossiblydilatingK,wecanassumethatα=1. Hencetheassumptioncanbe
statedas
(3.1) π1(K)=π1(K0) and πk(K)=βπk(K0)
forsomek 2,...,n 2 . LetK∗ denotethereflectionofK intheorigin. Then(3.1)
∈{ − }
yieldsthat
K+K∗ =2K0 and Vk(K U)=βVk(K0 U)
| |
forallU G(n,k). Minkowski’sinequality(cf.[13])thenimpliesthat
∈
Vk(2K0 U)=Vk(K U +K∗ U)
| | |
Vk(K U)k1 +Vk(K∗ U)k1 k
≥ (cid:16) | | (cid:17)
1 k
= 2Vk(K U)k
(cid:16) | (cid:17)
=βVk(2K0 U).
|
EqualityinMinkowski’sinequalitywillholdifandonlyifK∗ U andK U arehomothetic.
| |
Astheyhavethesamevolumethisisequivalenttotheirbeingtranslatesofeachother,in
whichcaseK U iscentrallysymmetric. Henceβ 1withequalityifandonlyifK U is
centrallysymm| etricforalllinearsubspacesU G≤(n,k). Sincek 2,thisistheca|seif
∈ ≥
andonlyif K is centrallysymmetric(cf. [4, Thm.3.1.3]). So if β = 1, then K and K0
mustbehomothetic.
Inthefollowing,weassumethatβ (0,1). Thiswillleadtoacontradictionandthus
∈
provethetheorem.
Wewriteh,h0 forthesupportfunctionsofK,K0. Hereandinthefollowing,“almost
all”or“almostevery”referstothenaturalHaarprobabilitymeasureonSn−1. Moreover
a linear subspace “E” as an upper index indicates that the corresponding functional or
6 RALPHHOWARDANDDANIELHUG
measure is considered with respect to E as the surroundingspace. By assumption there
is a Borel subset P Sn−1 with positive measure such that for all u P all the radii
⊆ ∈
ofcurvatureof K0 in the directionu existandare positive. As K0 is symmetricwe can
assume thatu P if andonlyif u P. Let N be the set of pointsu Sn−1 where
∈ − ∈ ∈
theprincipalradiiofcurvatureofK donotexist. SinceN isthe setofpointswherethe
Alexandrovsecondderivativeofhdoesnotexist,itisasetofmeasurezero. Byreplacing
P byP r(N ( N))wecanassumethattheradiiofcurvatureofbothK0 andK exist
∪ −
atallpointsofP. AsbothN and N havemeasurezerothissetwillstillhavepositive
−
measure.
Letu Sn−1 besuchthathandh0 aresecondorderdifferentiableatuandat uand
∈ −
thattheradiiofcurvatureofK0atuarepositive.Thisistrueofallpointsu P,whichis
notemptyasithaspositivemeasure. LetE G(n,k+1)besuchthatu ∈E. Thenthe
∈ ∈
assumptionimpliesthatalso
πkE(K|E)=βπkE(K0|E).
Henceweconcludeasin[8]that
SkE(K|E,·)+SkE(K∗|E,·)=2βSkE(K0|E,·).
Since h(K|E,·) = hK|E andh(K0|E,·) = hK0|E are secondorderdifferentiableat u
andat uwithrespecttoE, Lemma2.1appliedwith respectto thesubspaceE implies
−
that
det d2hK|E(u)E u⊥ +det d2hK∗|E(u)E u⊥
| ∩ | ∩
(cid:0) (cid:1) (cid:0) =2β(cid:1)det d2h (u)E u⊥ .
K0|E | ∩
(cid:0) (cid:1)
Sincehandh0aresecondorderdifferentiableatuandat u,thelinearmaps
−
L(h)(u): T Sn−1 T Sn−1, v d2h(u)(v),
u u
→ 7→
L(h0)(u): TuSn−1 TuSn−1, v d2h0(u)(v),
→ 7→
are well defined and positive semidefinite. Since the radii of curvature of K0 at u are
positive,wecandefine
Lh0(h)(u):=L(h0)(u)−1/2◦L(h)(u)◦L(h0)(u)−1/2
asin[8]inthesmoothcase.
Inthissituation,theargumentsin[8]canberepeatedtoyieldthat
L (h)(u)+L (h)( u)=2id
(3.2) h0 h0 −
kL (h)(u)+ kL (h)( u)=2β kid,
∧ h0 ∧ h0 − ∧
where id is the identity map on T Sn−1. Lemma 3.4 in [8] shows that L (h)(u) and
u h0
Lh0(h)(−u) have a common orthonormalbasis of eigenvectors e1,...,en−1, with cor-
respondingeigenvalues(relativeprincipalradiiof curvature)x1,...,xn−1 at u and with
eigenvaluesy1,...,yn−1at u. Afterachangeofnotation(ifnecessary),wecanassume
−
that0 x1 x2 xn−1. By(3.2)wethusobtain
≤ ≤ ≤···≤
(3.3) x +y =2 and x +y =2β
i i I I
fori=1,...,n 1andI 1,...,n 1 with I =k.
− ⊂{ − } | |
Proof of Theorem 1.3 when 2 k < (n + 1)/2. From (3.3) and Lemma 2.2 we
≤
concludethatthereissomeℓ 0,...,n 1 suchthat
∈{ − }
x1 = =xℓ <xℓ+1 = =xn−1 and y1 = =yℓ >yℓ+1 = =yn−1.
··· ··· ··· ···
NAKAJIMA’SPROBLEM:CONVEXBODIESOFCONSTANTWIDTHANDCONSTANTBRIGHTNESS 7
(a)Ifk ℓ,then
≤
x1+y1 =2 and xk1 +y1k =2β.
Hence
1= x1+y1 k xk1 +y1k =β,
(cid:18) 2 (cid:19) ≤ 2
contradictingtheassumptionthatβ <1.
(b)Letk > ℓ. Sincek < (n+1)/2wehave2k < n+1ork < n+1 k. Hence
−
k n k <n ℓ,andthusk n 1 ℓ. Butthen
≤ − − ≤ − −
xℓ+1+yℓ+1 =2 and xkℓ+1+yℓk+1 =2β,
and we arrive at a contradiction as before. This proves Theorem 1.3 when 2 k <
(n+1)/2 ≤ (cid:3)
Proof of Theorem 1.3 when k = 3,n = 5. In this case we are assuming that K0
haspositiveradiiofcurvatureatalmostallpointsofSn−1. Ash hasAlexandrovsecond
derivativesatalmostallpoints,foralmostallu Sn−1 theradiiofcurvatureofK exist
∈
atbothuand uandattheseunitvectorsK0 haspositiveradiiofcurvature. Recallthat
−
x1 ≤···≤x4aretheeigenvaluesofLh0(h)(u). Wedistinguishthreecaseseachofwhich
willleadtoacontradiction.
(a) x1 = x2. Then Lemma 2.2 yields that x1 < x2 = x3 = x4 and therefore also
6
y2 =y3 =y4. Hence
x32+y23 =2β and x2+y2 =2,
andthus
1= x2+y2 3 x32+y23 =β,
(cid:18) 2 (cid:19) ≤ 2
contradictingthatβ <1. Sothiscasecannotarise.
(b)x1 =x2 andx1 =x3,i.e.x1 =x2 =x3. Thenalsoy1 =y2 =y3,andweget
x31+y13 =2β and x1+y1 =2,
which,asbefore,leadstoacontradictionandthusthiscasecannotarise.
(c) x1 = x2 and x1 = x3, i.e. x1 = x2 < x3 = x4 by Lemma2.2. Since x1 = x3,
6 6
Lemma2.3impliesthat
(3.4) x2x4 =β =y2y4.
Inaddition,wehave
(3.5) x2+y2 =2=x4+y4.
Weshowthattheseequationsdeterminex2,x4,y2,y4asfunctionsofβ. Substituting(3.4)
into(3.5),weget
β β
+y2 =2, x4+ =2.
x4 y2
Combiningthesetwoequations,wearriveat
β
y2+ =2,
2 β
− y2
8 RALPHHOWARDANDDANIELHUG
whereweusedthatx4 =2− yβ2 6=0. Thisequationfory2canberewrittenas
y22 2y2+β =0.
−
Hence,wefindthat(recallthat0<β <1)
y2 =1 1 β.
± −
p
Consequently,
x2 =2 y2 =1 1 β.
− ∓ −
From(3.4),wealsoget p
β β
x4 = = =1 1 β,
x2 1∓√1−β ±p −
andfinallyagainby(3.4)
β β
y4 = = =1 1 β.
y2 1±√1−β ∓p −
Sincex1 =x2 <x3 =x4,thisshowsthat
(3.6) x1 =x2 =1 1 β, x3 =x4 =1+ 1 β.
− − −
p p
By assumption the surface area measure S4(K0, ) of K0 is absolutely continuous with
density function u det(d2h0(u)u⊥). Since K· +K∗ = 2K0, the non-negativityof
7→ |
themixedsurfaceareameasuresS(K[i],K∗[4 i], )andthemultilinearityofthesurface
− ·
areameasuresyieldsthat
4
4
S4(K, ) S(K[i],K∗[4 i], )
· ≤ (cid:18)i(cid:19) − ·
Xi=0
=S4(K+K∗, )=24S4(K0, ).
· ·
This implies that S4(K, ) is absolutely continuous as well, with density function u
det(d2h(u)u⊥). Nowob·servethatthecases(a)and(b)havealreadybeenexcludeda7→nd
|
therefore the present case (c) is the only remaining one. Hence, using the definition of
L (h)(u),
h0
det(d2h(u)u⊥)
det(d2h0(u)|u⊥) =det(Lh0(h)(u))=x1x2x3x4 =β2,
|
foralmostallu S4. Thuswededucethat
∈
S4(K, )=β2S4(K0, ).
· ·
Minkowski’suniquenesstheoremnowimpliesthatK andK0arehomothetic,henceK is
centrallysymmetric.Symmetricconvexbodieswiththesamewidthfunctionaretranslates
ofeachother. Butthenagainβ =1,acontradiction.
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129(1967),26–37.
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DEPARTMENTOFMATHEMATICS,UNIVERSITYOFSOUTHCAROLINA,COLUMBIA,S.C.29208,USA
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