Table Of ContentA COMPLETE COMPLEX HYPERSURFACE IN THE BALL OF CN
4 Josip Globevnik
1
0
2 Abstract In 1977 P. Yang asked whether there exist complete immersed complex
n submanifolds ϕ: Mk CN with bounded image. A positive answer is known for holomor-
→
a phic curves (k = 1) and partial answers are known for the case when k > 1. The principal
J
result of the present paper is a construction of a holomorphic function on the open unit
4
1 ball B of CN whose real part is unbounded on every path in B of finite length that
N N
ends on bB . A consequence is the existence of a complete, closed complex hypersurface
] N
V in B . This gives a positive answer to Yang’s question in all dimensions k, N, 1 k < N,
N
≤
C by providing properly embedded complete complex manifolds.
.
h
t 1. Introduction and the main result
a
m
Denote by ∆ the open unit disc in C and by B the open unit ball in CN, N 2.
[ N
≥
In 1977 P. Yang asked whether there exist complete immersed complex submanifolds
1
v ϕ:Mk CN with bounded image [Y1, Y2]. The first answer was obtained by P. Jones
→
5 [J] who constructed a bounded complete immersion ϕ:∆ C2 and a complete proper
3 →
1 holomorphic embedding ϕ:∆ B4. Since then there has been a series of results on
→
3 bounded complete holomorphic curves (k = 1) immersed in C2 [MUY, AL1, AF] the most
.
1 recent being that every bordered Riemann surface admits a complete proper holomorphic
0
immersion to B and a complete proper holomorphic embedding to B [AF]. The more
4 2 3
1 difficult complete embedding problem for k = 1 and N = 2 has been solved only very
v: recently by A. Alarc´on and F. J. L´opez [AL2] who proved that every convex domain in C2
Xi contains a complete, properly embedded complex curve.
r In the present paper we are interested primarilyin the higher dimensional case (k > 1)
a
where there are partial answers which are easy consequences of the results for complete
curves. For instance, it is known that for any k IN there are complete bounded embedded
∈
complex k-dimensional submanifolds of C2k and it is an open question whether, in this
case, N = 2k is the minimal possible dimension [AL2]. In the present paper we consider
the case where ϕ is a proper holomorphic embedding. In this case ϕ(Mk) is a closed
submanifold. We restate the definition of completeness for this case:
DEFINITION 1.1 A closed complex submanifold M of B is complete if every path
N
p:[0,1) M such that p(t) 1 as t 1 has infinite length.
→ | | → →
Note that this coincides with the standard definition of completeness since the paths
p:[0,1) M such that p(t) 1 as t 1 are precisely the paths that leaveevery compact
→ | | → →
subset of M as t 1 .
→
1
Here is our main result:
THEOREM 1.1 Let N 2. There is a holomorphic function f on B such that f is
N
≥ ℜ
unbounded on every path of finite length that ends on bB .
N
So our function f has the property that if p:[0,1] B is a path of finite length such
N
→
that p(t) < 1 (0 t < 1) and p(1) = 1 then t f(p(t) is unbounded on [0,1).
| | ≤ | | → ℜ
The following corollary answers the question of Yang in all dimensions k and N by
(cid:0) (cid:1)
providing properly embedded complete complex manifolds.
COROLLARY 1.2 For each k,N, 1 k < N, there is a complete, closed, k-dimensional
≤
complex submanifold of B .
N
Proof. We first prove the corollary for k = N 1 (that is, we first prove the existence
−
of the hypersurface, mentioned in the title). Let f be the function given by Theorem 1.1.
By Sard’s theorem one can choose c C such that the level set M = z B :f(z) = c
N
∈ { ∈ }
is a closed submanifold of B . Let p:[0,1) M be a path such that p(t) bB as
N N
→ →
t 1. Assume that p has finite length. Then there is a point w on bB such that
N
→
lim p(t) = w. By the properties of f, f is unbounded on p([0,1)). On the other hand,
t→1
ℜ
f((p(t)) = c (0 t < 1), a contradiction. So p must have infinite length. This proves that
≤
M is complete and so completes the proof of the corollary for k = N 1. Assume now that
−
1 k N 2. By the first part of the proof there is a complete, closed, k dimensional
≤ ≤ − −
complex submanifold M of B B . Clearly M is a complete, closed k dimensional
k+1 N
⊂ −
manifold of B . This completes the proof.
N
Remark If we want to have a connected, complete closed complex submanifold of B
N
then we simply take a connected component of M as above. Note also that the same
function f gives many complete closed complex manifolds of B since, by Sard’s theorem,
N
one can use the same reasoning for almost every c in the range of f.
2. Outline of the proof of Theorem 1.1
Let M IN. For x IRM 0 and α IR, write
∈ ∈ \{ } ∈
H(x,α) = y IRM:< y x >= α , K(x,α) = y IRM:< y x > α .
{ ∈ | } { ∈ | ≤ }
Assume that x IRM 0 (1 i n) and that
i
∈ \{ } ≤ ≤
n
P = K(x ,1) (2.1)
i
i=1
\
is a bounded set. Then P is a convex polytope, that is, the convex hull of a finite set. So
P is a compact convex set that contains the origin in its interior. A convex subset F of
P is called a face of P if any closed segment with endpoints in P whose relative interior
meets F is contained in F. A k-face is a face F with dimF = k, that is, the affine hull of
F is k-dimensional. A face of dimension M 1 is called a facet of P. Let P be a convex
−
polytope such that the representation (2.1) is irreducible, that is,
n
P = K(x ,1) for each k, 1 k n.
i
6 ≤ ≤
i=1,i6=k
\
2
Then
n
bP = H(x ,1) P
i
∩
i=1
[
and the sets F = H(x ,1) P, 1 i n, are precisely the facets of P. See [B] for the
i i
∩ ≤ ≤
details.
Given a convex set G, denote by ri(G) the relative interior of G in the affine hull of
G. What remains of the boundary of a convex polytope P after we have removed relative
interiors of all facets F , 1 i n, we call the skeleton of P (or more precisely, the
i
≤ ≤
(M 2)-skeleton of P, the union of all (M 2)-dimensional faces of P) and denote by
− −
skel(P). Thus
n
skel(P) = F ri(F ) .
i i
\
i=1
[(cid:2) (cid:3)
To prove Theorem 1.1 we first prove
THEOREM 2.1 Let B be the open unit ball of IRM, M 3. There is a sequence of
≥
convex polytopes P , n IN, such that
n
∈
∞
P IntP P IntP B, P = B,
1 2 2 3 j
⊂ ⊂ ⊂ ⊂ ··· ⊂
j=1
[
such that if w skel(P ) (j IN) then
j j
∈ ∈
∞
w w = (2.2)
j+1 j
| − | ∞
j=1
X
that is, the series in (2.2) diverges.
In the proof of Theorem 1.1 we shall use the following
COROLLARY 2.2 Let P , n IN be the sequence of convex polytopes from Theorem
n
∈ ∞
2.1. Let θ be a decreasing sequence of positive numbers such that θ < . For
n n=1 n ∞
each n IN, let bP be the θ -neighbourhood of skel(P ) in bP , that is, = w
n n n n n n
∈ U ⊂ P U { ∈
bP : dist(w,skel(P )) < θ . Let p: [0,1) B be a path such that p(t) 1 as t 1,
n n n
} → | | → →
and such that for all sufficiently large n IN, p([0,1)) meets bP only at . Then p has
n n
∈ U
infinite length.
Once we have proved Corollary 2.2 we prove Theorem 1.1 as follows. Let B be the
N
open unit ball of CN, N 2. Let P , n IN, be a sequence of convex polytopes as in
n
≥ ∈
Theorem 2.1 with M = 2N, and let , n IN, be as in Corollary 2.2. Given ε > 0 and
n n
U ∈
L < we use an idea from [GS] to construct a function f , holomorphic on B , such
n n N
∞
that f < ε on P and such that f > L on bP . By choosing L and ε
n n n−1 n n n n n n
| | ℜ ∞ \ U
inductively in the right way we then see that f = f has all the required properties.
n=1 n
P
3. Beginning of the proof of Theorem 2.1
3
Let w be a sequence in Bsuch that w 1 as n . If w does not converge then
n n n
| | → → ∞
(2.2) holds so to prove Theorem 2.1 it is enough to consider only the convergent sequences
w .
n
First, we try to explain the idea of the most important part of the proof. Suppose for
a moment that we have a sequence P of convex polytopes with the desired properties and
n
that there is an increasing sequence R of positive numbers converging to 1 such that
n
bP R B R B (n IN).
n n n−1
⊂ \ ∈
Let W = U (1 ν,1 + ν) be a small open neighbourhood of z = (0,0, 0,1) in IRM
× − ···
where U is a small open ball in IRM−1 centered at the origin and ν > 0 is small. Assume
that U 1 ν R B.
0
×{ − } ⊂
Let π be the orthogonal projection onto IRM−1, so π(x , ,x ) = (x , ,x ).
1 M 1 M−1
··· ···
For each n, consider C , the part of bP W consisting of the facets of P contained in
n n n
∩
W. The projection π is one to one on C and for each of these facets its image under π is
n
a convex polytope in U that is a cell of a partition of π(C ) into convex polytopes. Call
n
this partition and notice that as n , π(C ) tends to U. If we remove from each
n n
L → ∞
cell of its relative interior then we get what we call the skeleton of and denote by
n n
L L
skel( ). Clearly π(skel(P ) C ) = skel( ). Since, by our assumption at the moment,
n n n n
L ∩ L
every sequence w contained in W which meets skel(P ) for all sufficiently large n must
n n
satisfy (2.2), looking at z = π(w ) we conclude that every sequence z U such that
n n n
∞ ∈
z skel( ) for all sufficiently large n must satisfy z z = . The idea now
n ∈ Ln n=1| n+1− n| ∞
is to reverse the direction of reasoning. Let R be so close to 1 that U 1 ν R B.
0 0
P ×{ − } ⊂
In a typical induction step of constructing our polytopes the data will be a partition of
L
IRM−1 into convex polytopes and ρ and r, R < ρ < r < 1. Denote by the union of those
0
C
cells of the partition that are contained in U and let be the set of their vertices. We
L V
will ”lift” to b(rB) by putting V = (π W b(rB))−1( ). We want V to be the set of
V | ∩ V
vertices of a convex polyhedral surface C such that π(C) = and such that π maps the
C
facets of C precisely onto the cells of . We will do this in such a way that C stays out
C
of ρB - for this, the cells of C, and consequently the cells of will have to be sufficiently
C
small, of size proportional to √r ρ. Then we will construct a convex polytope P such
−
that C will be a part of its boundary bP and such that ρB IntP P rB.
⊂ ⊂ ⊂
There is a potential problem already at the first step. Namely, the points of V need
not be the vertices of a convex surface C. For this to happen we will need two things:
will have to be a true Delaunay partition of IRM−1 and the ball U in the definition
L
of W will have to be sufficiently small so that the part of b(rB) contained in W will be
sufficiently flat.
4. A Delaunay tessellation of IRM−1
Perturb the canonical orthonormal basis in IRM−1 a little to get an (M 1)-tuple of
−
vectors e ,e , ,e in general position so that the lattice
1 2 M−1
···
M−1
Λ = n e : n Z, 1 i M 1 (4.1)
i i i
∈ ≤ ≤ −
i=1
(cid:8)X (cid:9)
4
will be generic, and, in particular, no more than M points of Λ will lie on the same sphere.
For each point x Λ there is the Voronei cell V(x) consisting of those points of
∈
IRM−1 that are at least as close to x as to any other y Λ, so
∈
V(x)) = y IRM−1: dist(y,x) dist(y,z) for all z Λ .
{ ∈ ≤ ∈ }
M−1
In our case it is easy to see how to get V(0). Consider the finite set E = n e :
{ j=1 i i
1 n 1,1 i M 1 and for each x E 0 , look at K(x, x 2/2), that is, at
− ≤ i ≤ ≤ ≤ − } ∈ \{ } | | P
the halfspace which contains the origin and is bounded by the hyperplane passing through
x/2 which is perpendicular to x. Then
V(0) = K(x, x 2/2).
| |
x∈E\{0}
\
This is a convex polytope. It is known that the Voronei cells form a tessellation of RM−1
and in our case they are all congruent, of the form V(0)+x, x Λ [CS].
∈
There is a Delaunay cell for each point that is a vertex of a Voronei cell. It is the
convex polytope that is the convex hull of the points in Λ closest to that point - these
points are all on a sphere centered at this point. In our case, when there are no more than
M points of Λ on a sphere, Delaunay cells are (M 1)-simplices. Delaunay cells form a
−
tessellation of RM−1 [CS]. It is a true Delaunay tessellation , that is, for each cell, the
circumsphere of each cell S contains no other points of Λ than the vertices of S. We shall
denote by (Λ) the family of all simplices - cells of the Delaunay tessellation for the lattice
D
Λ.
By periodicity there are finitely many simplices S , ,S such that every other sim-
1 ℓ
···
plex of (Λ) is of the form S +w where w Λ and 1 i ℓ. It isthen clear by periodicity
i
D ∈ ≤ ≤
that there is an η > 0 such that for every simplex S (Λ) in η-neighbourhood of the
∈ D
closed ball bounded by the circumsphere of S there are no other points of Λ than the
vertices of S.
We shall typically replace the lattice Λ by the lattice Λ+q = x+q: x Λ where
{ ∈ }
q IRM−1, or, more generally, by the lattice σ(Λ+ q) where σ > 0 is small. Again, we
∈
shall denote by (σ(Λ+q)) the family of all simplices - cells of the Delaunay tessellation
D
for σ(Λ+q). These are the simplices of the form σ(S +q) where S (Λ). Passing from
∈ D
Λ to σ(Λ + q) everything in the reasoning will change proportionally. In particular, for
every simplex S (σ(Λ+q)) in (ση)-neighbourhood of the closed ball bounded by the
∈ D
circumsphere of S there will be no other points of σ(Λ+ q) than the vertices of S. We
shall also need the notion of the skeleton of the Delaunay tessellation for σ(Λ+q). This is
what remains after we remove the interiors of all S (σ(Λ+q)), hence
∈ D
skel (σ(Λ+q)) = S IntS = IRM−1 IntS .
D \ \
S∈D(σ(Λ+q)) S∈D(σ(Λ+q))
(cid:0) (cid:1) [ (cid:2) (cid:3) (cid:2) [ (cid:3)
The author is gratefull to John M.Sullivan who suggested the use of a generic lattice
for our purpose here.
5
5. Lifting the lattice from IRM−1 to the sphere
Let z, W = U (1 ν,1+ν) and π be as in Section 3. Let Λ IRM−1 be as in (4.1).
× − ⊂
Fix R , 0 < R < 1, so largethat U 1 ν R Band assume that R < ρ < r < 1.
0 0 0 0
×{ − } ⊂
Thepart ofthe sphere b(rB)inW cannow be writtenasa graphofa real analyticfunction,
call it ψ , so
r
b(rB) W = (x,ψ (x)): x U
r
∩ { ∈ }
where
M−1 1/2
ψ (x) = ψ (x , ,x ) = r2 x2 . (5.1)
r r 1 ··· M−1 − j
(cid:18) j=1 (cid:19)
X
Note that (grad ψ )(0) = 0, R < r < 1.
r 0
The map π maps W b(rB) in a one to one way onto U. We shall ”lift” (σΛ) from U
∩
to b(rB) W by the inverse of this map, that is, by the map x (x,ψ (x)). We want to
r
∩ 7→
get a convex polyhedral surface C with vertices w = (v,ψ (v)) where v are the vertices of
r
those cells of the Delaunay tessellation for σΛ which are contained in U and we want that
π maps the facets of the surface C precisely onto the Delaunay cells of σΛ contained in
U. Let us describe the conditions for this to happen. Let S be a simplex of the Delaunay
tessellation for σΛ. Let v , ,v be the vertices of S. We want that the simplex with
1 M
···
vertices w = (v ,ψ(v )), 1 j M, is a facet of a convex poyhedral surface. For this
j j j
≤ ≤
to happen, all other points w = (v,ψ (v)), v σΛ U, v = v , ,v , must lie in the
r 1 M
∈ ∩ 6 ···
open halfspace bounded by the hyperplane Π through w , 1 j M, which contains the
j
≤ ≤
origin, that is, they must lie on b(rB) outside the ”small” sphere Γ = Π b(rB). Since
∩
π W b(rB) is one to one, this happens if and only if the points v σΛ which are the
| ∩ ∈
vertices of the Delaunay cells of σΛ contained in U and are different from v , ,v , are
1 M
···
outside the projection π(Γ), an ellipsoid in IRM−1.
Aswe shallsee, thiswillhappen for allsuch simplicesS iftheball U IRM−1 centered
⊂
at the origin will be small enough so that the the gradient of ψ and thus the Lipschitz
r
constant of ψ will be small enough on U. The choice of U will depend only on η from
r
Section 4 and the same reasoning will work for any σ > 0.
LEMMA 5.1 Let π:IRM IRM−1 be the standard projection, π(x , ,x ) = (x ,
1 M 1
→ ··· ···
,x ). Let Λ be the lattice in IRM−1 as in (4.1) and let η > 0. There is a constant
M−1
ω > 0 such that for every σ > 0 the following holds. Let S IRM−1 be a simplex belonging
⊂
to (σΛ). Suppose that ψ is a Lipschitz function in a neighbourhood of S with Lipschitz
D
constant ω. Let v , ,v be the vertices of S and let w , ,w be the points in IRM
1 M 1 M
≤ ··· ···
given by w = (v ,ψ(v )), 1 j M. Let Π be the hyperplane in IRM containing the
j j j
≤ ≤
points w , ,w and let Γ be the sphere in Π, containing these points (that is, let Γ be
1 M
···
the circumsphere of the (M 1)-simplex in Π with vertices w , ,w ). Then π(Γ) is
1 M
− ···
contained in the (ση)-neighbourhood of the circumsphere of the simplex S.
6. Proof of Lemma 5.1
Let S (Λ) and let η > 0. If we replace ψ with ψ +c where c is a constant, Π will
∈ D
change to Π+(0,c), Γ to Γ+(0,c) and consequently π(Γ) will not change. Thus, π(Γ)
remains unchanged if we subtract ψ(v ) from each ψ(v ), 1 j M. Thus, π(Γ) will be
M j
≤ ≤
6
determined precisely once we know β = ψ(v ) ψ(v ), ,β = ψ(v ) ψ(v ).
1 1 M M−1 M−1 M
− ··· −
We shall show that π(Γ) changes continuously with (β , ,β ) near (0,0, ,0) if
1 M−1
··· ···
w = (v ,β ), ,w = (v ,β ) and w = (v ,0). Note that when β = =
1 1 1 M−1 M−1 M−1 M M 1
··· ···
β = 0, then Γ = π(Γ) is the circumsphere of S in IRM−1. Let w = (w , w ,1)
M−1 0 01 0,M−1
···
be the vector in IRM perpendicular to Π whose last component equals 1. So w must be
0
perpendicular to w w , 1 j M 1, so < w w w >= 0 (1 j M 1)
j M j M 0
− ≤ ≤ − − | ≤ ≤ −
which, if v = (v , ,v ), 1 j M 1, is the system of linear equations
j j1 j,M−1
··· ≤ ≤ −
(v v )w + +(v v )w = β (1 j M 1).
j1 M1 01 j,M−1 M,M−1 0,M−1 j
− ··· − − ≤ ≤ −
This is a system of M 1 linear equations for M 1 unknowns w , ,w whose
01 0,M−1
− − ···
matrix is nonsingular, since, S being a (M 1)-simplex, the vectors v v , 1 j
j M
− − ≤ ≤
M 1, are linearly independent. Its solution depends linearly on (β , ,β ). When
1 M−1
− ···
β = = β = 0 the solution is the zero vector. In this case w = (0, ,0,1). Let
1 M−1 0
··· ···
z = (z , ,z ) be the center of the sphere in Π that contains w , ,w . Then z is in
1 M 1 M
··· ···
Π so
< z w w >= 0. (6.1)
M 0
− |
Further, for each i, 1 i M 1, z isat equal distance from w and w , so z iscontained
i M
≤ ≤ −
in the hyperplane in IRM that passes through the midpoint of the segment joining w and
i
w , and is perpendicular to this segment, so z must satisfy
M
< [z (w +w )/2] [w w ] >= 0.
i M i M
− | −
Thus,
< z [w w ] >= (1/2) < [w +w ] [w w ] > (1 i M 1).
i M i M i m
| − | − ≤ ≤ −
Together with (6.1) this becomes the following system of linear equations for z , ,z :
1 M
···
z (v v )+ +z (v v )+z β = ( w 2 w 2)/2 (1 i M 1)
1 i1 M1 M−1 i,M−1 M,M−1 M i i M
− ··· − | | −| | ≤ ≤ −
z w + +z w +z = v w + +v w .
1 01 M−1 0,M−1 M M1 01 M,M−1 0,M−1
··· ···
Its matrix
v v , ,v v ,β
11 M1 1,M−1 M,M−1 1
− ··· −
....
v v , ,v ,β
M−1,1 M1 M−1,M−1 M−1
− ···
w , w , 1
01 ··· 0,M−1
is nonsingular for β = = β = 0 when w = = w = 0. The matrix de-
1 M−1 01 0,M−1
··· ···
pends continuously on(β , β ) and so do the right sides (1/2)( v 2 v 2+β2), 1
1 ··· M−1 | i| −| M| i ≤
i, M 1, and, since w depends continuously on (β , ,β ), also v w + +
0 1 M−1 M1 01
≤ − ··· ···
v w depends continuously on (β , ,β ). So the solution z = (z , ,z ),
M,M−1 0,M−1 1 M−1 1 M
··· ···
the center of the sphere Γ, depends continuously on (β , ,β ) near (0,0, ,0) and
1 M−1
··· ···
so does its radius z w = (z v )2+ +(z v )2+z2 )1/2. Recall that Π
| − M| 1− 1 ··· M−1 − M−1 M
(cid:0)
7
passes through w = (v ,0)anditsperpendicular directionw changes continuously with
M M 0
(β , ,β ) so Π changes continuously with(β , ,β ). We have seen that the cen-
1 M−1 1 M−1
··· ···
ter z of the sphere Γ in Π and its radius also change continuously with (β , ,β ) near
1 M−1
···
(0,0, 0). Thus, π(Γ) changes continuously with (β , ,β ) near the origin where
1 M−1
··· ···
π(Γ) = Γ is the circumsphere of S when β = β = β = 0. Thus, π(Γ) is contained
1 2 M−1
···
in the η-neighbourhood of the circumsphere of of the simplex S in IRM−1 provided that
ψ(v ) ψ(v ), ,ψ(v ) ψ(v ) are small enough. If ψ is a Lipschitz function with
1 M M−1 M
− ··· −
the Lipschitz constant ω then ψ(v ) ψ(v ) ω v v , 1 i M 1, so there
i M i M
| − | ≤ | − | ≤ ≤ −
is an ω such that if ψ is a Lipschitz function with the Lipschitz constant not exceeding
ω then π(Γ) is contained in the η-neighbourhood of the circumsphere of the simplex S.
Recall that every simplex in (Λ) is of the form S +x, 1 i ℓ, x Λ. Repeating the
i
D ≤ ≤ ∈
reasoning above for each S , 1 i ℓ, we get the Lipschitz constant that works for every
i
≤ ≤
simplex S in (Λ). This completes the proof for σ = 1.
D
Now, let σ > 0 be arbitrary and let S IRM−1 be a simplex in (σΛ). Let ψ be a
⊂ D
Lipschitz function with Lipschitz constant not exceeding ω in a neighbourhood of S, so its
graph is given by x = ψ(x , ,x ). Introduce new coordinates X , ,X in IRM
M 1 M−1 1 M
··· ···
by x = σX , 1 j M. In new coordinates we have σX = ψ(σX , ,σX ) so
j j M 1 M−1
≤ ≤ ···
X = Ψ(X , X ) = (1/σ)ψ(σX , σX ). Bothψ andΨareLipschitzfunctions
M 1 M−1 1 M−1
··· ···
with the same Lipschitz constants, so in new coordinates Ψ is a Lipschitz function in a
neighbourhood of S, which, in new coordinates, belongs to (Λ). Thus, applying the first
D
part of the proof we see that in new coordinates π(Γ) is contained in the η-neighbourhood
of the circumsphere of S. In follows that in old coordinates π(Γ) is contained in the
(ση)-neighbourhood of the circumsphere of S. This completes the proof.
7. Polyhedral convex surface contained in a spherical shell
Let η > 0 be as in Section 4 and let ω be the one given by Lemma 5.1. Let again
W = U (1 ν,1 + ν) where ν > 0 is small, U is a small open ball centered at the
× −
origin in IRM−1 and let R < 1 be so large that U 1 ν R B. For every r, R <
0 0 0
×{ − } ⊂
r < 1, W b(rB) = (x,ψ (x): x U where the function ψ is as in (5.1). We have
r r
∩ { ∈ }
(grad(ψ ))(x) = (r2 x 2)−1/2x (x U) so we may, passing to a smaller U if necessary,
r
− −| | ∈
assume that (gradψ )(x) ω (x U, R < r < 1) so that for each r, R < r < 1, ψ is
r 0 0 r
| | ≤ ∈
a Lipschitz function on U with Lipschitz constant not exceeding ω.
Let Λ be as in (4.1), let σ > 0 be small and let R < r < 1. Let ψ be as in (5.1).
0 r
Then x Ψ = (x,ψ (x)) is a one to one map from U onto W b(rB). We now look at
r r
7→ ∩
the points Ψ (x),x (σΛ) U and want to see them as vertices of a convex polyhedral
r
∈ ∩
hypersurface in IRM.
Consider a simplex S (σΛ) which is contained in U. Let v , ,v be its vertices.
1 M
∈ D ···
We can extend the restriction of the function ψ to this set of vertices to a function ϕ on
r r
all S by putting
M M M
ϕ α v = α ψ (v ) (0 α 1, 1 j M, α = 1)
r j j j r j j j
≤ ≤ ≤ ≤
j=1 j=1 j=1
(cid:0)X (cid:1) X X
to get an affine function ϕ on S so that x Φ (x) = (x,ϕ (x)) is an affine map mapping
r r r
7→
S to Φ (S), the simplex with vertices Ψ (v ), ,Ψ (v ). We do this for every simplex
r r 1 r M
···
8
S (σΛ) that is contained in U. Thus, we get a piecewise linear function ϕ on the
r
∈ D
union of the simplices S (σΛ)contained in U and so the union C (σ) of all these Φ (S),
r r
∈ D
the graph of the function ϕ , is then a polyhedral surface in IRM. We shall show that the
r
function ϕ is convex so that C (σ) is a convex polyhedral surface. Later we shall show
r r
that the part of C (σ) contained in W = U (1 ν,1+ν) with U being a ball in RM−1
r 0 0 0
∩ −
centered at the origin, strictly smaller than U, is a part of the boundary bP of a suitable
convex polytope P.
Given S (σΛ), S U, let Π be the hyperplane in IRM that contains Φ (S).
r
∈ D ⊂
Then Π b(rB) is the sphere in Π and which is the circumsphere of Φ (S) and which was
r
∩
denoted by Γ in Section 5. By Lemma 5.1, π(Γ) is contained in the (ση)-neighbourhood
of the circumsphere of S in IRM−1. We know that the ση-neighbourhood of the closed
ball in IRM−1 bounded by the circumsphere of S contains no other points of σΛ than the
vertices of S which implies that all points of Ψ (U (σΛ)) other than the vertices of Φ (S)
r r
∩
lie outside of the small ”spherical cap” that Π cuts out of b(rB), that is, outside of the
”small” part of b(rB) bounded by Γ. This shows that all other vertices of the simplices
in C (σ) that are not the vertices of Φ (S) are contained in the open halfspace of IRM
r r
bounded by Π that contains the origin. Thus, Φ (S) is a facet of C (σ). Since this holds
r r
for every S (σΛ),S U, it follows that the surface C (σ) is convex.
r
∈ D ⊂
The simplices Φ (S) where S (σΛ), S U, have all their vertices on b(rB). We
r
∈ D ⊂
want to estimate how far into rB they reach. To do this, we need the following
PROPOSITION 7.1 Let 0 < r < 1, let a b(rB) and let A b(rB) be a set such
∈ ⊂
that x a γ for all x A where γ < r. Then the convex hull of A misses ρB where
| − | ≤ ∈
ρ = r γ2/r.
−
Proof. A is contained in x b(rB): x a γ . With no loss of generality assume that
{ ∈ | − | ≤ }
a = (r,0, ,0). Then A x b(rB): (x r)2+x2+ +x2 γ2 x b(rB): r2
··· ⊂ { ∈ 1− 2 ··· M ≤ } ⊂ { ∈ −
2x r + r2 γ2 = x b(rB): 2r2 2x r < 2γ2 = x b(rB): x > (r2 γ2)/r
1 1 1
≤ } { ∈ − } { ∈ − }
x rB: x > (r2 γ2)/r . The last set is a convex set that contains A and misses ρB
1
⊂ { ∈ − }
which completes the proof.
Denote by d the length of the longest edge of simplices in (Λ) so that σd is the length
D
of the longest edge of the simplices in (σΛ). Since ψ is a Lipschitz function with the
r
D
Lipschitz constant not exceeding ω the length of the longest edge of the simplices Φ (S)
r
where S (σΛ), S U, does not exceed √1+ω2σd. Now, we use Proposition 7.1. If
∈ D ⊂
R < r < 1 then r γ2/r > r γ2/R . Thus, putting
0 0
− −
(1+ω2)d2
λ =
R
0
we get the following
PROPOSITION 7.2 If R < r < 1 then the simplices Φ (S) where S (σΛ), S U,
0 r
⊂ D ⊂
miss ρB where ρ = r σ2λ.
−
8. A convex polytope with a prescribed part of the boundary
We keep the meaning of R ,U,d and λ. Recall that U is an open ball in IRM−1
0
9
centered at the origin. Let µ be its radius. Let 0 < µ < µ < µ < µ < µ and let
0 1 2 3
U = x IRM−1: x < µ , W = U (1 ν,1+ν), 0 i 3.
i i i i
{ ∈ | | } × − ≤ ≤
Choose σ > 0 so small that
0
σ d < min µ µ ,µ µ ,µ µ ,µ µ . (8.1)
0 3 3 2 2 1 1 0
{ − − − − }
Then, since the maximal edge length of simplices in (σΛ) equals σd it follows that if
D
0 < σ < σ then
0
- the simplices S (σΛ) that meet U are contained in U
0 1
∈ D
- the simplices S (σΛ) that are contained in U cover U .
2
∈ D
PROPOSITION 8.1 There is a κ > 0 such that whenever R R 1 and R < R′ <
0
≤ ≤
R + κ then each hyperplane in IRM which meets W R′B RB and misses W RB
2 3
∩ \ ∩
misses RB.
(cid:0) (cid:1)
Proof Suppose that there is no such κ > 0. Then there are a sequence R , R
n 0
≤
R 1 (n IN), and a sequence x W , such that x > R (n IN) and such that
n n 2 n n
≤ ∈ ∈ | | ∈
x R 0 as n , and for each n a hyperplane H through x which misses
n n n n
| | − → → ∞
W R B and meets R B W . Since x IR 0 as n we may, passing to
3 n n 3 n n
∩ \ | | − → → ∞
subsequences if necessary, with no loss of generality assume that R converges to an R
n
and x converges to x b(RB) W . Since for each n, H misses W R B it follows
n 2 n 3 n
∈ ∩ ∩
that H converges to H, the hyperplane through x tangent to b(RB) at x. In particular,
n
H (RB W ) is empty, so for sufficiently large n, H (R B W ) must be empty, a
3 n n 3
∩ \ ∩ \
contradiction. This completes the proof.
With no loss of generality, passing to a smaller σ if necessary, we may assume that
0
σ2λ < κ. Suppose now that 0 < σ < σ and let R ρ < r < 1 where ρ = r σ2λ.
0 0 0 ≤ −
We know that the union C (σ) of the simplices Φ (S) where S (σΛ),S U, is a
r r
∈ D ⊂
convex polyhedral surface which, by Proposition 7.2, is contained in rB ρB. Each of these
\
simplices Φ (S) is contained in a hyperplane H. We want that these hyperplanes miss
r
ρB. Note that by (8.1) the simplices in (σΛ), contained in U cover U . So the function
3
D
ϕ is well defined on U and its graph C (σ) W is contained in W (rB ρB). The
r 3 r 3 3
∩ ∩ \
function ϕ is piecewise linear and convex. Thus, if S (σΛ) meets U then, by (8.1),
r 2
∈ D
S U and by the convexity of ϕ , the graph of ϕ U lies on one side of the hyperplane
3 r r 3
⊂ |
H that contains Φ (S) which, in particular, implies that H misses W ρB and thus, by
r 3
∩
Proposition 8.1, H misses ρB. This shows that the part of C (σ) contained in W can
r 2
be described in terms of the hyperplanes that miss ρB. So we find x , ,x bB and
1 n
··· ∈
α , ,α , ρ < α r (1 i n), such that
1 n i
··· ≤ ≤ ≤
G = x B: < x x > α ,1 i n .
1 i i
{ ∈ | ≤ ≤ ≤ }
is a convex set containing ρB in its interior, and is such that W bG = W C (σ).
2 1 2 r
∩ ∩
PROPOSITION 8.2 Let R < r < 1, let 0 < σ < σ , and let ρ = r σ2λ > R , There
0 0 0
−
is a convex polytope P which contains ρB in its interior, such that bP rB ρB, and such
⊂ \
that every Φ (S) where S (σΛ), S U , is a facet of P.
r 1
∈ D ⊂
Proposition 8.2 implies in particular, that
W skel(P) = Φ U skel( (σΛ))
0 r 0
∩ ∩ D
(cid:0) (cid:1)
10
