A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
V. I. O LIKER
The boundary value Minkowski problem. The parametric case
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 9, n
o3 (1982), p. 463-490
<http://www.numdam.org/item?id=ASNSP_1982_4_9_3_463_0>
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The Parametric Case (*).
V. I. OLIKER
0. - Introduction.
0.1. The celebrated Minkowski
problem
consists offinding
a closedconvex
hypersurface
in whose Gaussian curvature at thepoint
with aunit
normal ~
isprescribed
in advance as apositive
continuous functionK($).
The latter isgiven
on a unithypersphere E
and shouldsatisfy
thenecessary condition
where du
is the n-volume element of 27.Minkowski himself solved first the
problem
for convexpolyhedral
sur-faces ;
in this case instead of the Gaussian curvature oneprescribes
theareas of the faces and the normals to them of the
polyhedron
which is tobe found.
Passing
to a limit frompolyhedrons
togeneral
convex surfacesone finds the
generalized (weak)
solution of theproblem
in terms of certain associated measures. The very difficultproblem
of existence of aregular
solution under various
assumptions
on .g has been solved for n = 2by
H.
Lewy [7]
in theanalytic
case,Pogorelov [12]
andNirenberg [9]
in theclass of smooth
functions,
and forarbitrary n by Pogorelov [14] (see
alsoCheng
and Yau[4]).
(*) Partially supported by
NSF Grant MCS80-02779. Part of research for this paper was done while the author was aguest
of theSonderforschungsbereich
Theoretische Mathematik of the Universitat Bonn in the summer of 1980. The final version was written when the author was
visiting
theHeidelberg University
in
July
of 1981. The author would likeespecially
to thank Prof. W.Jager
ofHeidelberg University
for hishospitality.
Pervenuto alla Redazione l’l l Novembre 1981 ed in forma definitiva il 5 Marzo 1982.
0.2. The
original problem
admits variousgeneralizations.
Gluck[6]
studied the situation when the Gaussian curvature is
prescribed
on Z butthe
hypersurface
inquestion
does not have to be animbedding
via theinverse of the Gauss map, as is the case in the classical formulation. A
generalization
moredirectly
related to theoriginal problem
can be statedas follows.
Let cv be a domain on the unit
hypersphere E
and .K apositive
con- tinuous function in roo Find a convexhypersurface
withboundary
whosespherical image
is m and at thepoint
with the unitnormal $
its Gaussian curvature isK(~).
It is natural toimpose
someboundary
conditions. It turns out thatprescribing
theboundary
of thehypersurface
leads to anover-determined
problem;
this follows from our results in[11].
The
analytic
formulation of theproblem suggests naturally
toprescribe only
the normalcomponent
of theposition
vectoralong
theboundary
of c~.In such a form
(actually, slightly
morerestricted)
theproblem
wasproposed by
Aleksandrov[1 ],
p.319,
and alsoby Pogorelov [13],
p. 657. When cc~is convex and lies
strictly
inside ahemisphere,
theproblem
has been in-vestigated extensively [1, 2, 5, 12, 15]. However,
very little is known in the case when co is not contained in ahemisphere. Geometrically
the lattercase means that we are
dealing
withhypersurfaces
inparametric form, and,
since the closedness is notassumed,
serious difficulties arise.Already
thequestion
ofuniqueness
isquite
nontrivial. Forexample,
the maximumprinciple,
which isapplicable
in the case when co is inside ahemisphere
does not hold anymore for the
corresponding
linearizedequation, and,
infact,
the linearizedproblem
may havegeometrically
nontrivial solutions[10, 11]. Apparently,
forarbitrary given
data theproblem
may not be solvable at all.0.3. The purpose of this paper is an
investigation
of thesolvability
ofthe
boundary
value Minkowskiproblem
in thefollowing setting.
Fix a domain co on the unit
hypersphere 27
such that itscomplement
is a convex set
lying strictly
inside ahemisphere,
and letq($)
be a
positive
continuous function in c5. We want to find a convexhyper-
surface S which is arelatively compact
subset of some closed convexhyper-
surfaceS,
itsspherical image
isw,
the Gaussian curvature at thepoint
withthe unit
normal ~
is1 jg~(~), ~
E cv, and the normalcomponent h
of theposi-
tion vector of S on the
boundary
as coincides with a linear function in Rn+z. The last condition has asimple geometric meaning. Namely,
if weconsider the
envelope
of the set ofhyperplanes tangent
to Salong
its bound-ary, then the
boundary
condition on h means that thisenvelope
has avertex. In other
words,
thehypersurface
« sits » on a cone.Since no
assumptions
on smoothness or 8m are made sofar,
all theterms and conditions above should be understood in a
generalized
senseusually given
to thoseconcepts
whendealing
withgeneral
convexhyper-
surfaces.
(A
brief review of the necessary information isgiven
in section 1of this
paper.)
Because of the fact that S must be a
portion
of a closed convexhyper- surface,
y the functionq($)
shouldsatisfy
some necessary condition similar to(0.1)
butexpressed
in terms of co.Analytically,
y such a condition canbe
given
in the formwhere
A > 0
andgeometrically
it means that the n-volume of theprojection
of S(taken
with thesign)
on anyhyperplane
with the normalcorresponding
to the «missing » part
of S isnonnegative.
It turns out that the
requirement A
> 0 in(0.2) along
with the otherconditions stated before are sufficient to
produce
ageneralized
solution ofthe
boundary
valueproblem posed
above. This solution isunique.
0.4. The
proof
consistsessentially
of twoparts.
Atfirst,
in section 2.the
problem
is formulated and solved in the class of convexpolyhedrons,
and then a
generalized
solution is obtainedby
a passage to a limit. Thisstep
is described in section 3. The reader familiar with the existenceproof
in the case of closed
hypersurfaces
willrecognize
here the method which goes back to Minkowski and which has been usedsuccessively by
A. D.Aleksandrov, Pogorelov,
and other mathematicians.However,
its concreterealization for the
boundary
valueproblem
is far frombeing
a standardrepetition
of thearguments
used in the case of closedhypersurfaces.
Inparticular,
the information that theproblem
in the latter case is solvable does not seem to behelpful,
and the condition(0.2)
is used in a much moresubtle way than
(0.1)
in the classical situation.0.5. In the last section of the paper we prove interior
regularity
ofthe
generalized
solution. The results here are basedessentially
on the apriori
estimates forMonge-Ampére equations
obtainedby Pogorelov [14]
and Calabi
[3].
1. - Generalized solutions.
1.1. In the
(n + 1)-dimensional
Euclidean space we fix a unithyper-
sphere E
with the center 0 which coincides with theorigin
of asystem
ofCartesian coordinates. On 27 we introduce smooth local coordinates
u1,
...,The standard metric on 27 will be denoted
by (gii), (g’5)
standsfor the inverse matrix. Let 8 be a
piece
of a C2hypersurface
inwhose
Gauss map y is adiffeomorphism
on a domain m c E.Obviously, S
can beconsidered as an immersion r : m -
Rn+’;
moreover, we canidentify
theposition
vector r withy-1.
Under such circumstances thecorrespondence
between S and m is established via
correspondence
betweenpoints
withparallel
unit normals. Theexpression
for r,given
asy-i,
can be found interms of the
support
function h ofS,
which is defined aswhere ( , )
is the innerproduct
inRn+,, and $
is apoint
in co.Here,
andin the rest of the paper we make the obvious identification of the vector with its
endpoint
on ~.Since
r(~)
=y-~(~),
we findby differentiating h(~)
The vectors ...,
8$/8un, $
form a basis in Rn+l and we can expressr($)
in terms of h andgrad
h. The result iswhere
(The
summation convention is in effect here and in thesequel.)
The second fundamental form
~dr, d~ ~
==bij
can also be expres- sed in terms of h.Namely,
if we differentiate(1.2)
weget
By
the Gauss formulas for 27 we havewhere
7~
are the Christoffelsymbols
of the metric g;; .Substituting
it intothe last
expression
andtaking
into account(1.1)
and(1.2)
we obtainwhere
The
principal
radii of curvature of 8 are theeigenvalues
of the differen- tialof y (by
theRodrigues formula),
and therefore are the rootsRiy
...,R.
of the
equation
detBgij)
=== 0.Since y
is adiffeomorphisni, Ri:71-- 0, i
=1,
...1 n, and we can considern B
the Gaussian curvature
/
view of(1.3)
wehave
Let dv denote the n-dimensional volume element of
S and du
the n-vo-lume element of ~. Then the
following relationship
holds:For an
arbitrary
Borel set G on E such that G c m the sety-1(O)
is also aBorel set and its n-volume is
given by
1.2. Assume now that
/§
is a closed convexhypersurface
of class C2with
positive
Gaussian curvature. Let co be asimply
connected domainon
E,
different fromE,
and S aportion of S corresponding
to co = (J) aco.The
previous
discussion shows that in order to solve theproblem posed
in section 0.3 of the introduction we have to consider the differential
equation
subject
to theboundary
conditionwhere is the
position-vector
of the vertex of theenveloping
conealong
as.
Making
aparallel
translation of we can assume that this vertex is located at theorigin
0.Note that is
simply
a restriction to E of a linear function inAlso,
a directcomputation
shows that-f-
= 0. Thisactually
be-comes obvious if we notice that
a, ~>
is thesupport
function of the constant vector a. That also means that theequation (1.7)
remains invariant underparallel
translations of theorigin.
In order not to introduce newsymbols,
we leave the notation h for the
support
function of~S’
withrespect
to thenew
position
of theorigin
in relative to which theboundary
condi-tion assumes the form
In view of
(1.6)
and(1.7)
we havefor any Borel subset of 1: contained in co. The last
formula, however,
makes sense if we assume
only
that S is aportion
of a closed convexhyper-
surface not
necessarily
of class C2 and p is apositive
continuous function in w(and, obviously,
y even under moregeneral assumptions).
The
appropriate generalization
is made in thefollowing
way. Let~S
be a closed convex
hypersurface
which is understood to be theboundary
of an
arbitrary
finite convexbody
in Rn+l. Then the convexhypersurface S
is a
simply
connectedrelatively compact
subset of~.
If x is an interiorpoint
of S then for any
supporting hyperplane
at x with the exterior unitnormal ~
we
put
incorrespondence
apoint
on 27 which is theendpoint
of a unit vec-tor
parallel to ~
andstarting
at the center 0 of 27. Themapping
definedin such a way is
obviously
ageneralization
of the standard Gauss map,and,
ingeneral,
it is a set valued function. We map all interiorpoints
of ~S’into
~,
and then take the relative closure of theimage;
the latter is called thespherical image
of S.Let G be a subset of 27. Define
on 9
a set G’ as the set ofpoints
whosespherical images
are in G. If G is a Borel subset of 27 then G’ is also aBorel set on
~S’;
we denoteby v(G)
the n-volume of the set G’. The func- tion v is called the areafunction.
It is anonnegative completely
additivemeasure on the
a-algebra
of Borel subsets of E. Since/§
isfinite, v
is alsofinite. If the
Radon-Nikodym
derivative of v withrespect
to the measure it on 27 exists at apoint ~
and it is different from zero then itsreciprocal
iscalled the Gaussian curvature. When
~S’
is smooth then in view of(1.5)
this derivative is
equal
to~ 1(~) (see [2],
section8).
For a convex
hypersurface c S, were 8
is not assumed to besmooth,
by
its area function we mean the area function of the closedhypersurface ~S‘.
A
generalized
solution of(1.7)
in c~ c 1; is now defined as a convexhyper-
surface contained in some closed convex
hypersurface 9
and such thatits
spherical image
isc5y
and for any Borel subset contained in cv,The
support
functionh(~), ~
E~,
of S is a function of a unitvector ~ giving
the oriented distance from theorigin
0 to asupporting hyperplane tao 8
with the exterior unitnormal ~.
It is known(see [2],
section6)
thath(~)
is a continuous function on the entireS, and,
inparticular, h(~)
isdefined on 8m.
Thus, (1.8)
is also defined.2. - Solution of the
problem
in the class ofpolyhedrons.
2.1. Let .~ be a closed convex
polyhedron
in Take anarbitrary
4 true» vertex
(i.e.,
different from an interiorpoint
of any ’fl,- k dimen- sionalface, 0 k n - 1)
of .~f and suppose aparallel
translation is madeso that this vertex coincides with
origin
0 of a Cartesian coordinatesystem
in Rn+l. We
distinguish
twotypes
of faces and exterior unit normals of M.Namely,
denoteby
gi , ..., 7 g. the faces for which 0 n g, =0, i
=1,...,
m.Their
corresponding
exterior unit normals we denoteby ~l’
...,7 ~..
Hereand in the
sequel,
unless otherwisestated,
we understand n-dimensional faces.By
q1, ..., q, we denote the faces for which 0 is aboundary point;
..., are their
corresponding
exterior unit normals. Let 7 ..., ag be thehyperplanes containing
faces q~, ... , qs . The intersection of the half- spaces determinedby
andcontaining M,
is an infinite convexbody
whoseboundary
is an infinite convexpolyhedron.
Lethi , ... , h~
bethe
support
numbers of the faces gl, ..., 7 g., thatis,
the oriented distances from 0 to thehyperplanes containing
thecorresponding
faces.Obviously,
the
support
numbers of ql, ... , qs areequal
to zero.Now consider a convex
polyhedron
.~.~’ which also has a true vertex at 0.Suppose
it has the same number of faces as M and(1) gi and g’
areparallel
for all i =1, ... ,
m ; the latter means thatthey
have the same ex-terior unit
normals; (2)
thehyperplanes
’containing
the facescoincide with (Xi,...,o~, and for every
q~ , ~ = 1, ... , s,
0 is aboundary point. Geometrically, (2)
means that the infinitepolyhedral
conesformed
by
thehyperplanes
of the faces of M and .M’adjacent
to 0 are thesame.
2.2. PROPOSITION. Let M and M’ be as above.
Assume,
inaddition,
thatthe n-volum8S
Fi
and~’$ of
gi andg’,
i =1, ... ,
m, arecorrespondingly equal.
Then M coincides ’With M’.
PROOF. We will use a method of Minkowski based on his
theory
ofmixed volumes. Details
concerning
mixed volumestheory
can be foundin
[2,
section6].
The mixed volume of the
polyhedrons
and M’ is defined aswhere hj
are thesupport
numbers of the facesqi, j
=1,
..., 8, and.~’~
arethe n-volumes of
q~ .
Since q, passthrough 0, h¡= 0, j
=1,
..., s, and the second sum inV(M, M’ )
vanishes.Replacing
inV(M, .M’ )
thepolyhedron
M’
by
M weget
the usual(n + I)-volume
of the convexbody
boundedby
M.Similarly,
yV(M’, M~)
=(n + 1 ) -volume
of the convexbody
boundedby
M’.According
to the Minkowskiinequality (see [2],
section7),
and the
equality
holds if andonly
if and M’ are homothetic.Thus,
Since
.I’~
_ =1, ... ,
m, we haveChanging
the roles of M andM’,
weget
Thus,
M and .~C’ must be homothetic. Since the n-volumes of Ui and g,are
equal,
.M’ must coincide with M’. Theproposition
isproved.
2.3. LEMMA. Lot
tJ
be a closed convex set on the unithypersphere
Elying
strictly
inside ahemisphere,
and P a convex cone with vertex at 0 at the centerof
E and withspherical intage D. (By de f inition
thespherical image o f
P oon-sists
of
all exterior unit normals tohyperplanes supporting
toP.)
Also let R be the
infinite
convexbody
whoseboundary
isP,
and -r = R r1 1:.Denote
by D’c
1: the setsymmetric
to,~
withrespect
to 0 and -, -r ~’1Q’.
Then int
Je =1= for
anypoint
q E int Je there exists apositive
number c
(depending
on such that2.3.1. PROOF. Denote
by fo*
thehypersphere containing strictly
inside and
by
a, theequatorial hyperplane determining 1:;. Let ~o
be thepole
of2:’0.
We consider twopossibilities: a) b)
We start with the case
a).
Consider the infinite convex cone I~formed by
the raysoriginating
at 0 and of direction~, where ~
E,~.
Since~Qf/:
intQ,
there exists a
hyperplane
0"1containing $0
andsupporting
toK.
The lattermeans
that K
lies in one of the dihedralangles
formedby
0"0 and Note also that(ao
r) r1D
=0,
since 0"0n D
=0.
Let a be the dihedral
angle containing
1~’ andEO’
theportion
of1:0
whichis contained in a. Consider a
point ;lE ,~~
such that the vector isorthogonal
to the
subspace
ui .Again
it mayhappen
that int S~. In this casewe
repeat
the above constructionby taking
ahyperplane
o~2containing ~l
andidentifying
the trihedralangle containing
,~2. This process is continued until either after ksteps, k n,
we construct or after nsteps
weconstruct
a
hyperoctant containing
the cone K. In the first case thehyperplane
or, withthe normal-
~k
will beclearly strictly supporting
toK,
thatis,
thehemisphere I:k
determinedby
O"k will containQ strictly
inside.Since $w
is an interiorpoint
of.~,
the normals tosupporting hyperplanes
of K form with8~ angles greater
than~/2 ;
thatis,
thehyperplane a,
is alsostrictly supporting
tothe cone R.
(Note
that is thespherical image
of~. )
On the other hand the int Je consists ofpoints corresponding
tohyperplanes strictly supporting
to both R and K.
Thus,
int 3C is notempty,
and theinequality (*)
followsimmediately.
Now consider the situation when after n
steps
we obtained ahyperoc-
tant T
containing
K.Let ~
E int Q. Since T containsK,
thehyperplane ~’
with the unit
normal ~
isstrictly supporting
toboth K
and T.Obviously,
the same
hyperplane
will bestrictly supporting
to.~,
since is thespherical image
of K.Thus, - ~
E int í, and therefore int3C 0 0.
Asbefore,
the
inequality ( ~ )
follows from the openness of int K. The lemma isproved.
2.4.
LEMMA. Let D be a
closed convex set on thehypersphere
f withpolyhedral boundary lying strictly
inside a and P a convex conewith vertex 0 and
spherical image D.
Denoteby
0153l’..., as thehyperplanes containing
the n-dimensionalfaces of P. Finally,
let co = and~1, ..., ~~ be
unit vectors uxathorigins
at 0 andendpoints
in co.Then there exists a convex
polyhedron
M with a true vertex at 0 and such that it has sfaces
q,, ..., qsadjacent
to0,
eachof
which lies on thecorresponding hyperplane (Xj, j = 1,
... , s. M also has m otherfaces
gl , ..., gm with exterior normals81,
... ,~m
there exists aneighborhood of
0 on Pfree of points of
thef aees
g1, ... , gm . Inaddition, if
the... , ~~,
are alldirected in the
hemisphere containing ,~
then M is anin f inite polyhedron;
otherwise,
it is a closed convexpolyhedron.
2.5. PROOF. As in Lemma 2.3 we denote
by R
the infinite convexbody
whose
boundary
isP,
and 7: = .l~ r1 .~.Correspondingly, D’
denotes the setsymmetric to D
withrespect
to 0 and H = -r nQ’. Also,
we denoteby K
the infinite convex cone formedby
the raysoriginating
at 0 andpassing through ~, where $
E,~. Finally, put
C = 8K.By
Lemma2.3,
intJC 0,
and wechoose (
E int K. Let l be astraight
line
passing through
0 andcontaining ~,
and H ahyperplane passing through
0 andperpendicular
to l. Denoteby
0’ anarbitrary point
on Idifferent from 0 and taken in the direction
~.
Let C’ be the cone obtainedfrom C
by parallel
translation so that in the newposition
its vertex coin- cides with 0’. The intersection of the infinite convexbody
boundedby
C’with the infinite convex
body
.Rgives
a finite convexbody
whoseboundary
is a closed convex
polyhedron.
Let T denote thispolyhedron.
We note that every
point
of the intersection 0’ f1 P is apositive
dis-tance away from 0. In
fact,
let ~ E 0’ f1 P and be such that so _--__ distx)
== min
dist(0,
~x).
Also let arc cos~).
Since H isstrictly supporting, 7&/2
y n, and a calculation shows that dist(0,
---_= dist
(0, 0’) sin y
> 0 for any x E 0’ f1 P.Next we consider the
hyperplanes
ai, ..., amsupporting
to T whoseexterior unit normals are
~1, ... , ~m .
None of thosehyperplanes
passesthrough 0,
since$i c co, i
=1,
..., m.Among
al, ..., am we take an arbi-.trary hyperplane,
say ai, and move itparallel
to itselfby
a small distancein the direction
- ~i .
If this distance issufficiently
small thenclearly
noneof the faces of T will
disappear.
The intersection of T with ahalfspace
determined
by
ai in the direction- ~i gives
a new convexpolyhedron
withadditional face whose normal is
~i . However,
it couldhappen
that oci al-ready
contained an n-dimensional face of T before we moved it. For con-venience,
we assume in this case that ai was « moved » butby
a distanceequal
to zero. Now we move each ai in the fashion describedabove,
andthen consider the intersection of the
halfspaces
determinedby
0153l, ..., am inthe directions -
~1’
... , -$m
and the infinitebody
1~. Thisintersection
is a convexbody,
whoseboundary
is a closed convexpolyhedron
with allproperties
stated in theProposition 2.4, except
for the last one which is îyet
to be verified.Let .~ be the convex
polyhedron
constructed above.Suppose ~1, ..., ~m
are all directed in the
hemisphere Z’ containing 17. Let ~
EE+
bethe
unitvector
perpendicular
to theequatorial hyperplane bounding ~+.
Thenarc cos
«, ~ i) yr/2y ~ = I , ... ,
m, and arc cos~, ~> ~/2 for ,E li.
Ob-viously, the
ray with theorigin
at 0 and of direction- I
intersects neithernor .P which means that .~ is infinite. The
remaining
case istreated
similarly.
The lemma isproved.
2.6. REMARK. In the
following the
cone P in Zmw 2.4 will bereferred
to as cone
of
M.2.7. We will use the
mapping
lemma of Aleksandrov to prove the ex-istence of a
polyhedron
withprescribed boundary
cone,normals,
and n-vol-umes of the
corresponding
faces. For convenience of the reader wequote
it here. More details can be found in
[1 ],
p. 90.Let A and B be two m-dimensional manifolds without boundaries
(not necessarily connected)
and 7: amapping
of .A. into Bsatisfying
the conditions:1)
each connectedcomponent
of B containsimages
ofpoints
inA ; 2) -r
is a one-to-onecorrespondence;
3)
z iscontinuous;
4)
if a sequencebk E B, 1~ = 1, 2, ... ,
is such that there exists a se- quenceakE A.
for which-r(ak)= = 1, ~, ...,
andbk
convergesthen there exists a
E A,
for which7:(a) = b,
and one can select asubsequence converging
to a.Under the conditions
1)-4), r(A)
= B.2.8.
THEOREM. Let lil be
as in Lemma2.4, i. e. , a
closed convex set onthe unit
hypersphere E
withpolyhedral boundary lying strictly
inside a hemi-sphere,
P a convex cone with vertex 0 at the centero f
E andspherical image iil,
and
~1, ..., ~m points in w. Further,
let.Fl,
...,Pm
bepositive
-numbers such that
E Q and A > 0.
Then exists a closed convex
polyhedron
M with a true vertex at 0for
P is a
boundary
cone, andwhich,
in addition tofaces generated by P,
has m
other faces
UI, ..., Um with exterior unit normals~1, ... , ~~
and n-volume$F, 7 7 Fn
-,2.9. In order to prove the theorem we need some
preparatory
statements.Consider
the
set 77 of closed convexpolyhedrons
which have 0 as their truevertex
and P as theboundary
cone. Assume alsothat besides
the faceslying
on Z’they
haveonly m
other faces with exterior unit normals$1, ... , $~. By
«defi-7aition)>
of theboundary
cone the faces withthe
nor-mals
~1,
...,~n
cannot beadjacent
to the vertex 0. It follows from Lem-ma 2.4 that
0.
Each polyhedron
is definedby
itssupport
numbershl, ... , hm corresponding
to the normals~1, ... , ~m . The support
numbers of the faceslying
on P areequal
to zero.Clearly, 07 i
=1,
..., m, andtherefore,
with
each
we can associate apoint
..., inthe positive
coor-dinate
angle
ofthe m-dimensional Euclidean
space Em. Denoteby A
theset of
points
in Emcorresponding
topolyhedrons
fromII.
The set A is open.
Indeed, for
any .~ EH
if we make aparallel
transla-tion
by
asufficiently
smalldistance
of ahyperplane
aicontaining
a facewith
the normal~a,
none ofthe
other faces willdisappear
and P willstill
remain a
boundary
cone.Therefore,
the newpolyhedron
obtained from .1~will also lie in II. Of course, we can move all ai
simultaneously
and stillremain in ff.
Consider now an
arbitrary polyhedron
Since P is aboundary
cone for
M,
there exists aneighborhood
TT of the vertex 0 on P free ofm
points
of the faces gi with normals =1,
..., m. LetM = U
Thet-i
orthogonal projection
of2ff
on thehyperplane H,
constructed in theproof
of Lemma
2.4, obviously
covers theprojection
of V on H.Therefore,
thereexists e > 0 such that
where f is
the n-volume of the face ga . The left hand side in(2.2)
is then-volume of the
projection
ofM
on H.In
general,
the number e in(2.2) depends
onthe particular polyhedron.
We define the set as the set of
polyhedrons
in II whichsatisfy (2.2).
It is clear that for any 8 > 0 the set77~
is notempty.
This followsfrom Lemma 2.4 if we take
the polyhedron
constructed there and make ahomothetic transformation with the center at 0 and
appropriate
coefficient.Correspondingly
to the set-U".
we have the open setAs.
2.1U. Next we consider the set
$ð
ofpositive
numbersf 1, ... , 1m
whichsatisfy
theinequality
for some fixed
positive ð.
In the m-dimensional Euclidean space with coor- dinates(2.3)
defines an open convex subset of thepositive
coor-m
dinate
angle.
This subset lies above the; = i
We choose 6 so that contains the
point (1fl, ..., 1f m) given
in Theo-rem 2.8. Let us show that it can be done. It follows from the construction of
vector n
that « « max0,
since thehyperplane H
isstrictly supporting
to the cone C made up of rays whichoriginate
at 0 and go intopoints
of 8Q. On the otherhand,
because of andsince E Q,
we haveThus,
if wetake ð
= - then will contain thepoint
..., Wefix this
67
and also in(2.2)
take E = 6.2.11. The
proof
of Theorem 2.8 consists of verification of the condi- tions1)-4)
of theMapping
Lemma with .A =A,5 B
= As for the map-ping
T one takes it asSince is an open convex
set,
it hasonly
onecomponent.
It containsimages
ofpoints
in~
becauseby
Lemma 2.4 and the discussion in2.9, A6
is notempty. Thus,
condition1)
is verified.Condition
2)
holds because ofProposition
2.2.To
verify 3)
one shouldjust
note that under a smallchange
ofsupport
numbers
hi ,
...,hm
thecorresponding
n-volumesf 1,
... ,1m change continuously.
2.12.
Finally,
we establish4).
Letil =-- f’), k == 1, 2, ...,
be asequence of
points
from~3~ converging
tof
=(11’ ... , f m),
and is thecorresponding
sequence ofpolyhedrons
from77~
withsupport
numbersof the faces
= 1,
..., m. We will prove now that thesequence is bounded. ,
Fix an
arbitrary polyhedron
Mk from the sequence~~k~.
LetQ
be apoint
on Mk such that the dist(0, Q)
is maximal. In order to prove thatis bounded it suffices to show that dist
(0, Q)
is boundedby
a constantindependent
on .~k.Let H be
the hyperplane
with the unitnormal n
constructed in theproof
of Lemma 2.4. In view of the condition(2.2)
and because of theway
IIa
was constructed there exists aneighborhood U
of 0 on P free ofpoints
of the faces with normals~1, ... , ~m
of anypolyhedron
in There-fore,
we can move thehyperplane
.H~parallel
to itself in thedirection iji
sothat in the new
position
H will still have commonpoints only
with U.Denote
by
.H’ the translatedhyperplane
H. The intersection of .b~’ with the convexbody
boundedby
P is a closed convexbody 0
whose n-volumeis bounded away from zero, say
by
aconstant y
> 0. Consider apyramid
with the base 0 and vertex
Q.
It liesentirely
within the convexbody Mk
bounded
by
and therefore its(n + 1)-volume
TT does not exceed the(n + 1)-VOIUMe
offlk. Thus,
we havewhere s is the distance from
Q
to H’.By
theisoperimetric inequality
theright
hand side of(2.4)
is boundedprovided
the n-volume of jM~ is bounded.m m
The total n-volume of is
equal to z ff where ejk
denote the4=1 1=1
n-volumes of
the
faces ofadjacent
to 0. Let us show that the n-volu nese§’
are bounded. Since thepolyhedron
Mk isclosed,
where q; are the unit normals to the faces of the cone P. Then
On the other
hand - ~
is an interiorpoint
of9,
and(in
>0,
since none of the faces of ~kadjacent
to 0degenerate).
In the second
paragraph
of 2.10 it was shown thatmin
> 0.Therefore,
Thus,
the n-volume of Mk is boundedby
aquantity depending only
m
_
on Z ff
and the domainD. Finally,
y(2.4)
and theisoperimetric inequality
show that the distance s is bounded. Since the distance between H and H’
is taken to be bounded and does not
depend
on theparticular polyhedron
from the distance
from Q
to .TC is bounded. On the other hand H isstrictly supporting
to the cone .R.Therefore,
any rayoriginating
at 0 andgoing
in R formswith
anangle
less thana/2. Then, obviously
the bound-edness of the distance from
Q
to .b~implies
boundedness of the distancefrom
Q
to 0. mconverges to
f,
thesums fk
can beuniformly
bounded inm
terms of
f;, I
=1,
..., m, for allsufficiently large
k.Therefore,
the sequencefhkl
isbounded,
and we can select from it aconverging subsequence. Correspondingly,
y we will have asubsequence
ofconvex
polyhedrons converging
to some convexpolyhedron
M. The vol-umes of the faces of
polyhedrons
of thatsubsequence
converge tothe
num-bers
f 1, ... , f ~, . Thus,
the condition4)
of theMapping
Lemma also holds.From this we conclude that to
any f
therecorresponds
apolyhedron
from
~d .
On the otherhand,
it has been shown in 2.10 that thepoint
.F’ ==
(F1, ... , .F’,~) given
in Theorem 2.8belongs
to%,,.
Thiscompletes
theproof
of Theorem 2.8.3. - Generalized solution of the
boundary
valueMinkowski problem.
3.1. THEOREM. let co be a on a unit
hypersphere
I such that4li = eonvex domains
lying strictly
inside ahemisphere. Let
be a continuous
function defined
in Co and such thatwhere  > 0
and
E Q.Then there exists a convex
hypersurface S,
contained in a closed convexhypersurface 9,
such that thespherical image of
S isCo,
the Gaussian curvature(that is,
theRadon-Nikodym
derivativeof
the areaf unction)
at an interiorpoint ~ of
w is and thesupport function
vanishes at theboundary
aco.the
hypersurface S
can be constructed in such a way that thesupport f unetion o f ~S’
vanishes on the entireD.
In order to prove the theorem we construct a
converging
sequence ofconvex
polyhedrons
withprescribed
set of unitnormals,
n-volumes of thecorresponding faces,
and certainboundary
cones. Existence of suchpoly- hedrons
is assuredby Theorem
2.8. The sequence is constructed in such a.way that the area functions of the
polyhedrons
convergeweakly
to thearea function
generated by
the function99(~).
Also theboundary
conesof
these polyhedrons
converge to a cone which will be theboundary
coneof the
hypersurface
inquestion.
3.1.1.
Consider
a convexpolyhedral
domain whoseboundary
consists of k
(n -1 )-dimensional faces,
and is circumscribed about 8Q.The
(n-1)-dimensional
faces arepieces
of(%- 1)-dimensionai spheres
formed
by intersections
of Z with n-dimensionalhyperplanes passing through
the center 0 of Z.
Suppose k
islarge
and the diameters of the faces ofas2k
are small
enough
so that liestrictly
inside thehemisphere containing
Q.If k ~oo and diameters of the faces of converge
uniformly
to zero, then 8Q.Convergence
is understood in the standard sensegiven
toconvergence of sets in Euclidean space, which is
applicable here,
since weare
dealing
with convex setslying
inside ahemisphere.
To each
Qk
therecorresponds
a closed domain which is contained in the domain co. Wepartition C-0 k
into smalldomains @)
anddefine
positive
numbersPi
and unitvectors $) by
the conditionA certain care is needed in order to assure that
vectors $§
lie inside This condition will be satisfied if onetakes,
forexample,
atriangulation
of (Ok into convex sets. The latter can be done in several
quite
obviousways,
though describing
the actual constructions is rather tedious. For that reason we omit it here.It follows from conditions
1), 2)
in the theorem and(3.1)
that for sufh-ciently large k
there exist > 0 and unitvectors 8;
E Q such thatMoreover,
since the diameters off3k
tend to zero when k and c~/, -~ o),,we have -
A$-
Denote
by Pk
the convex cone with vertex 0 andspherical image
Then in view of
(3.2)
it follows from Theorem 2.8 that there exists a convexpolyhedron Mk
with a true vertex at 0 for whichPk
is aboundary
coneand