ESI
The Erwin Schrödinger International
Institute for Mathematical Physics
Boltzmanngasse 9
A-1090 Wien, Austria
Typical Faces of Best Approximating Three–Polytopes
Károly Böröczky, Jr.
Péter Tick
Gergely Wintsche
Vienna, Preprint ESI 1692 (2005)
Supported by the Austrian Federal Ministry of Education, Science and Culture
Available via http://www.esi.ac.at
September 8, 2005
Typical faces of best approximating
three-polytopes
Károly Böröczky, Jr., Péter Tick, Gergely Wintsche
October 7, 2005
Dedicated to Rolf Schneider on his 65th birthday
AMS classification: 52A27, 52A40
Key words: polytopal approximation, extremal problems
Abstract
For a given convex body K in R3 with C 2 boundary, let Pni be the
inscribed polytope of maximal volume with at most n vertices, and let
c
P(n)
be the circumscribed polytope of minimal volume with at most n
faces. We prove that typical faces of Pni are close to regular triangles in
c
are close to regular hexagons
a suitable sense, and typical faces of P(n)
in a suitable sense. We prove the analogous statements if not the
volume difference from K is to be minimized but the mean width
difference or the Hausdorff metric. In addition we discuss the cases
when there is no restriction like being inscribed or circumscribed on
the position of the polytopes.
We note that if the Gauß–Kronecker curvature is positive on ∂K
then many of these statements were verified earlier by P.M. Gruber;
namely, [10] considers the case of Hausdorff metric, and [11] discusses
the cases of circumscribed polytopes and volume difference, and of
inscribed polytopes and mean width difference.
Acknowledgement: The first named author is supported by OTKA grants
T 042769, 043520 and 049301, and by the Marie Curie Research Training
Network PhD. We are grateful to Endre Makai who pointed out various
mistakes in an earlier version, and whose remarks considerably improved the
paper.
1
1
Introduction
First we introduce some notions. For functions f and g of positive integers,
we write f(n) = O(g(n)) if there exists an absolute constant c such that
(n)
= o. In
|f(n)| ≤ c · g(n) for all n ≥ 1, and f(n) = o(g(n)) if limn→∞ fg(n)
(n)
addition we write f(n) ∼ g(n) if limn→∞ fg(n)
= 1.
For a compact convex set C, we write aff C to denote its affine hull, and
if it is of dimension two then we call C a convex disc. In addition if C is
circular disc of radius one then we simply call C a unit disc. If C is any
convex disc then we write relint C to denote its relative interior. We write
o to denote the origin in R2, B 2 to denote the unit disc in R2 centred at o,
and | · | to denote area. Similarly o also denotes the origin in R3 , and B 3
denotes the unit ball in R3 centred at the origin. For any two points p and
q, we write [p, q] to denote the segment connecting p and q.
Concerning additional notions for convex bodies and polytopes in this
paper, consult the beautiful monograph R. Schneider [18]. For any convex
bodies K and M in R3 , we write δS (K, M) to denote the volume of the
symmetric difference of K and M. We observe that δS is a metric on the
space of convex bodies.
Now let K be a convex body in R3 with C 2 boundary. We always integrate
on ∂K with respect to the two–dimensional Hausdorff–measure. For any
x ∈ ∂K, we write Qx to denote the second fundamental form at x, hence Qx
is positive semi definite. Its two eigenvalues are the principal curvatures at
x, and its determinant κ(x) is the Gauß–Kronecker curvature at x. Readily
κ(x) ≥ 0 for any x ∈ ∂K.
We define Pn to be the polytope with at most n vertices such that
δS (K, Pn ) is minimal, and P(n) to be the polytope with at most n faces such
that δS (K, P(n) ) is minimal. In addition let Pni be the polytope inscribed
c
be the
into K with at most n vertices and of maximal volume, and let P(n)
polytope circumscribed around K with at most n faces and of minimal volume. Completing a task initiated by L. Fejes Tóth [5], and developed by
P.M. Gruber and M. Ludwig, K. Böröczky, Jr. [2] proved
2
Z
1
1
1/4
∼ √
κ(x) dx · ;
n
4 3
∂K
Z
2
5
1
c
√
δS (K, P(n)
) ∼
κ(x)1/4dx · ;
n
36 3
∂K
δS (K, Pni )
2
(1)
(2)
Z
2
1
1
1
1/4
√ −
κ(x) dx · ;
δS (K, Pn ) ∼
n
12 3 16π
∂K
2
Z
1
1
5
√ −
κ(x)1/4dx ·
δS (K, P(n) ) ∼
n
36 3 8π
∂K
(3)
(4)
as n tends to infinity. Under the assumption that κ(x) > 0 for all x ∈ ∂K,
(1) is due to P.M. Gruber [7], (2) is due to P.M. Gruber [8], moreover (3)
and (4) follow from combining M. Ludwig [15] and K. Böröczky, Jr. and M.
Ludwig [3].
The goal of this paper is to continue a task initiated by P.M. Gruber
[10] and [11]; namely, to describe the typical faces of the extremal polytopes
above. For any x ∈ ∂K, we write u(x) to denote the exterior unit normal to
∂K at x. Let F be a face of one of the extremal polytopes above such that
there is a unique point xF ∈ ∂K such that u(x) is an exterior normal also to
F . In this case, let QF = QxF . If QF is positive definite and ε > 0 then we
say that F is ε–close to some polygon Π if there exists y ∈ F such that
(1 + ε)−1 · Π ⊂ F − y ⊂ (1 + ε) · Π.
We note that under the assumption that κ(x) > 0 for all x ∈ ∂K, the
c
following Theorem 1.1 for circumscribed polytopes (namely, for P(n)
) is due
to P.M. Gruber [11]. We also note that according to Euler’s theorem, a
polytope with at most n vertices has at most 2n − 4 faces.
THEOREM 1.1 Given a convex body K in R3 with C 2 boundary, there
exist ν(n) > 0 with limn→∞ ν(n) = 0 and the following properties where we
use the notation as above:
(i) If f(n) is number of faces F of Pn such that QF is positive definite, F
is a triangle, and F is ν(n)–close
to some triangle that is regular with
R
respect to QF and of area
1/4 dx
∂K κ(x)
,
2n·κ(xF )1/4
f (n)
n→∞ 2n
then lim
= 1.
(ii) If g(n) is number of faces F of P(n) such that QF is positive definite, F
is a hexagon, and F is ν(n)–close
to some hexagon that is regular with
R
1/4 dx
∂K κ(x)
respect to QF and of area n·κ(x )1/4 , then lim g(n)
= 1.
n
F
n→∞
c
In addition the analogous statements hold for Pni in place of Pn , and P(n)
in
place of P(n) .
3
c
Remark: In the case of Pni , all vertices lie on ∂K, in the case of P(n)
, the
faces touch K in their centroids, and in the case of P(n) , |F ∩ K| = 21 |F |
holds for each face F . Finally in the case of Pn , if F is a typical face then
1 + ν(n)
1 − ν(n)
· |F | ≤ |F ∩ K| ≤
· |F |.
(5)
2
2
We write h·, ·i to denote the scalar product in R3, and S 2 to denote the
surface of the unit ball centred at the origin. Given a convex body C in R3,
its support function hC is defined by
hC (u) = maxhx, ui.
x∈C
If M is another convex body then the L1 metric of C and M is
Z
δ1(C, M) =
|hC (u) − hM (u)| du.
S2
In particular if M ⊂ C then δ1(C, M) is proportional to the difference of
the mean widths of C and M, hence to the difference of the first intrinsic
volumes of C and M.
We note that under the assumption that κ(x) > 0 for all x ∈ ∂K, the
following Theorem 1.2 for inscribed polytopes (namely, for Mni ) is due to
P.M. Gruber [11].
THEOREM 1.2 Given a convex body K in R3 with C 2 boundary, there
exist ν(n) > 0 with limn→∞ ν(n) = 0 and the following properties:
(i) If Mn is a polytope with at most n vertices such that δ1(K, Mn ) is minimal, and f(n) is number of faces F of Mn such that QF is positive
definite, F is a triangle, and F is ν(n)–close
to some triangle that is
R
regular with respect to QF and of area
3/4 dx
∂K κ(x)
2n·κ(xF )3/4
f (n)
n→∞ 2n
then lim
= 1.
(ii) If M(n) is a polytope with at most n faces such that δ1 (K, M(n) ) is minimal, and g(n) is number of faces F of P(n) such that QF is positive
definite, F is a hexagon, and F is ν(n)–close
to some hexagon that is
R
regular with respect to QF and of area
κ(x)3/4 dx
n·κ(xF )3/4
∂K
then lim
n→∞
g(n)
n
= 1.
Finally let Mni be an inscribed polytope with at most n vertices such that
c
δ1(K, Mni ) is minimal, and let M(n)
be a circumscribed polytope with at most
c
n faces such that δ1 (K, M(n) ) is minimal. Then the analogous statements
c
hold for Mni in place of Mn , and M(n)
in place of M(n) .
4
Remark: In the case of Mni , all vertices lie on ∂K. In addition if F is a
c
typical face of M(n)
then the point where F touches K is of distance at most
ν(n) · n
−1
2
from the centroid of F .
We do not discuss the proof Theorem 1.2 because it consists of combining
the proof of Theorem 1.1 with a well–known duality argument for δ1 (see S.
Glasauer and P.M. Gruber [6] and M. Ludwig [15] for the case when the curvature is positive, and K. Böröczky, Jr. [2] for the case when the curvature
is allowed to be zero).
We note that the Hausdorff metric δH (C, M) of two convex bodies C and
M in R3 is
δH (C, M) = max2 |hC (u) − hM (u)|;
u∈S
namely, the minimal d such that any point of C is of distance at most d from
M, and any point of M is of distance at most d from C. For a convex body
K in R3 with C 2 boundary, let Wn be the polytope with at most n vertices
such that δH (K, Wn ) is minimal, and let W(n) be the polytope with at most
n faces such that δH (K, W(n) ) is minimal. In addition we write Wni to denote
an extremal polytope with at most n vertices if the polytopes are assumed to
c
be inscribed into K, and W(n)
to denote an extremal polytope with at most n
faces if the polytopes are assumed to be circumscribed around K. Following
the work of R. Schneider [16], [17] and P.M. Gruber [9], who considered the
case when κ(x) > 0 for all x ∈ ∂K, K. Böröczky, Jr. [2] proved
Z
1
1
1
δH (K, Wn ), δH (K, W(n) ) ∼ √
κ(x) 2 dx · .
(6)
n
6 3 ∂K
Under the additional assumption that the polytope is inscribed or circumscribed, the optimal approximation is twice the value in (6) (see the references
above). We note that under the assumption that κ(x) > 0 for all x ∈ ∂K,
the following Theorem 1.3 is due to P.M. Gruber [10].
THEOREM 1.3 Given a convex body K in R3 with C 2 boundary, there
exist ν(n) > 0 with limn→∞ ν(n) = 0 and the following properties where we
use the notation as above:
(i) If f(n) is number of faces F of Wn such that QF is positive definite, F
is a triangle, and F is ν(n)–close
to some triangle that is regular with
R
respect to QF and of area
κ(x)1/2 dx
2n·κ(xF )1/2
∂K
5
f (n)
n→∞ 2n
then lim
= 1.
(ii) If g(n) is number of faces F of W(n) such that QF is positive definite,
F is a hexagon, and F is ν(n)–close
to some hexagon that is regular
R
with respect to QF and is of area
κ(x)1/2 dx
n·κ(xF )1/2
∂K
then lim
n→∞
g(n)
n
= 1.
c
In addition, the analogous statements hold for Wni in place of Wn , and W(n)
in place of W(n) .
Remark: In the case of Wni , all vertices lie on ∂K. Let F be a typical face of
c
either of the other three extremal polytopes. In the case of W(n)
, the point
−1
where F touches K is of distance at most ν(n) · n 2 from the centroid of F .
In the cases of W(n) and Wn , we have the following: If v is a vertex of F ,
√
√
·(v−p) ∈ K and p+ 1+ν(n)
·(v−p) 6∈ K.
and p is the centroid of F then p+ 1−ν(n)
2
2
We do not discuss the proof Theorem 1.3 because it runs as the proof of
Theorem 1.1, only one uses ideas in P.M. Gruber [10] to handle the ”round
part” of ∂K.
Naturally if K = B 3 then the metric induced by a second fundamental
form is always the Euclidean metric in R3 . Therefore in all the three cases
(approximation with respect to volume, L1 –metric and Hausdorff–metric),
the typical faces of extremal polytopes are close to be regular triangles if
the number of vertices is restricted, and to regular hexagons if the number
of faces is restricted. P.M. Gruber [11] proved that if K is a ball then in
some cases, extremal polytopes with respect to the Hausdorff metric are
asymptotically optimal with respect to the symmetric difference metric and
the L1 metric; namely, using the notations of Theorems 1.1, 1.2 and 1.3, if n
tends to infinity then
c
c
);
) ∼ δS (K, W(n)
δS (K, P(n)
δ1(K, Mni ) ∼ δ1(K, Wni ).
We note without any further argument that combining Theorems 1.1, 1.2
and 1.3 also yield
COROLLARY 1.4 Using the notations of Theorems 1.1, 1.2 and 1.3, if
K is a ball and n tends to infinity then
δS (K, Pni ) ∼ δS (K, Wni );
c
c
δ1(K, M(n)
) ∼ δ1(K, W(n)
).
6
We note that W(n) and Wn are asymptotically optimal neither with respect to δS nor with respect to δ1 .
2
Some notation and basic observations
For t ∈ (0, π2 ), let R(t) be the triangle with a right angle such that o is a
vertex, the angle at o is t, and the longest side is of length one.
We will need that certain type of functions are concave or monotonic:
PROPOSITION 2.1 Let ω ∈ [ 31 , 3], and let f(t) = tan t +
(0, π2 ).
ω
tan t
for t ∈
);
(i) f(t)−1 is concave on (0, π2 ), and (f(t)−1 )” < −0.0001 if t ∈ ( π7 , 5π
12
(ii) t · f(t) is increasing on (0, π2 ), and (t · f(t))′ ≥ 0.1 if t ∈ ( π6 , π2 ).
Proof: (i) follows by
(f(t)−1 )” = −(tan2 t · (3ω − 1) + 3ω − ω 2 ) ·
For (ii), we have by tan t > t + 31 t3 that
(t · f(t))
′
2 tan t(1 + tan2 t)
.
(ω + tan2 t)3
1
t
= t + tan t + t · tan t + ω
−
−t
tan t tan2 t
t3
2
−t .
≥ t + tan t + t · tan t + ω
3 tan 2 t
2
Therefore we may assume that ω = 3, hence x + x12 − 2 > 1 − x for x > 1
yields
t2
tan t
′
+
− 2 + t · tan2 t
(t · f(t)) ≥ t ·
t
tan2 t
≥ t − tan t + t · tan2 t.
Since t − tan t + t · tan2 t is increasing, we obtain (ii). 2
If f is a C 2 function on (a, b) and t, t0 ∈ (a, b) then the Taylor formula
says that
f(t) = f(t0) + f ′ (t0 ) · (t − t0) + 21 f”(t0 + s(t − t0)) · (t − t0)2
7
where s ∈ (0, 1). We deduce the following properties of convex and concave
functions:
PROPOSITION 2.2 Let f be a positive convex function on [a, b]. If ω > 0,
and f”(t) ≥ ω f(t0 ) for all t ∈ [a, b] with |t − t0 | < ε0, and t1, . . . , tn ∈ [a, b]
n
= t0 and the number of ti with |ti − t0 | ≥ ε is m for
satisfy that t1 +...+t
n
ε ∈ (0, ε0) then
f(t1 ) + . . . + f(tn )
m
≥ 1 + 2n
· ωε2 · f(t0).
n
In addition the analogous result holds if f is positive concave and f”(t) ≤
−ω f(t0 ) for all t ∈ [a, b] with |t − t0| ≤ ε0 ; namely, in this case we conclude
f(t1) + . . . + f(tn )
m
≤ 1 − 2n
· ωε2 · f(t0 ).
n
We will also use the following consequence of Cauchy–Schwarz inequality:
If γi , Ai > 0 for i = 1, . . . , m then
!−1 m
!2
m
m
X
X
X
1
γi A2i ≥
Ai .
(7)
γi
i=1
i=1
i=1
Finally we prove a consequence of the Euler theorem:
PROPOSITION 2.3 For any convex polygon Π there exists δ > 0 with the
following property: If Π is tiled by the convex polygons Π1, . . . , Πn such that
each Πi is of diameter at most δ then writing ki to denote the number of sides
of Πi , we have k1 + . . . + kn < 6n.
Proof: We write m to denote the number of sides of Π, and p to denote the
p
.
perimeter of Π, and define δ = 2m
Let Π1 , . . . , Πn be the tiling of Π as in Proposition 2.3. If e is the number
of edges in the tiling, and v is the number of vertices in the tiling then the
Euler theorem says
v − e + n = 1 > 0.
(8)
Since at least two edges of the tiling meet at any vertex of Π, and at least
three edges of the tiling meet at any other vertices of the tiling, summing up
the degrees of the vertices of the tiling leads to 3v ≤ 2e + m. It follows by (8)
that e < 3n + m. In addition let b be the number of segments that are sides
of some Πi and are contained in some side of Π, hence b ≥ 2m. Therefore
k1 + . . . + kn = 2e − b < 6n + 2m − b ≤ 6n.
8
2
3
An extremal property of regular polygons
related to general polytopes
This section builds on K. Böröczky, Jr. and M. Ludwig [3]. We only present
what should be modified in the corresponding argument in K. Böröczky, Jr.
and M. Ludwig [3]. For t ∈ (0, π2 ), we define, letting x2 = hx, xi,
R
minα∈R R(t) |x2 − α| dx
.
γ(t) =
|R(t)|2
According to K. Böröczky, Jr. and M. Ludwig [3],
γ(t) =
tan t
1
1
+
−
if t ∈ (0, 1.05] where π/3 < 1.05.
tan t
3
2t
(9)
LEMMA 3.1 If q is a positive definite quadratic form, α ∈ R and Π is a
polygon of at most k sides then
Z
p
γ( π )
|q(x) − α| dx ≥ k · |Π|2 det q.
2k
Π
In addition there exist positive absolute constants ε0 and ϑ with the following
R
√
(1+ε)γ( π
)
k
properties: Let us assume that k ≤ 6, and Π |q(x)−α| dx ≤
|Π|2 det q
2k
for ε ∈ (0, ε0). In this case Π is a k–gon, and there exists some k–gon Π0,
which is regular with respect to q, whose centroid is o, and satisfies
√
Π0 ⊂ Π ⊂ (1 + ϑ 4 ε) · Π0.
Moreover for E = {z : q(z) ≤ α}, we have
√
1−ϑ ε
2
· |Π| ≤ |Π ∩ E| ≤
√
1+ϑ ε
2
· |Π|
Remark: If k = 3, 6 then it follows by (9) that
γ( π3 )
1
1
= √ −
;
6
3 3 4π
γ( π6 )
1
5
√ − .
=
12
18 3 4π
To prove Lemma 3.1, we need several auxiliary statements.
9
(10)
(11)
PROPOSITION 3.2 t γ(t) is increasing on (0, 1.05], and if π/7 ≤ t < 1.05
then (t γ(t))′ > 0.002.
Proof: For E(t) = 3 tan t + tan3 t − 2t tan2 t + t tan4 t − 3t, we have (t γ(t))′ =
E(t)
. Readily
3 tan2 t
E ′ (t) = 4 tan2 t − 4t tan t + 4 tan4 t + 4t tan5 t > 0,
hence
E(t)
3 tan2 t
≥
E(π/7)
3 tan2 1.05
> 0.002 if π/7 ≤ t ≤ 1.05. 2
We define θ̃(t) = γ(t)−1 for t ∈ (0, π/2), and
γ(t)−1
for t ∈ (0, 1.05];
θ(t) =
′
θ̃(1.05) + θ̃ (1.05)(t − 1.05) for t ∈ [1.05, π/2).
PROPOSITION 3.3 Let t ∈ (0, π/2). Then θ(t) ≥ γ(t)−1 , θ(t) is concave,
and actually θ”(t) < −0.1 for t ∈ (π/7, 1.05).
Proof: There exists a unique t1 ∈ (0, π2 ) such that tan t1 = 2t1 . In particular
t1 = 1.1655 . . . > 1.05 >
p
3/5 = 0.7745 . . . > π/7.
For t ∈ (0, t1], K. Böröczky, Jr. and M. Ludwig [3] calculates that (9) still
holds, and the proof of Lemma 4 in [3] and tan t ≤ 2t yield that
16
tan t
(1 + tan2 t) −
(t + 2 tan t + t tan2 t)
2
9
3t
10t2 + 10t2 tan2 t − 6 tan2 t
.
≥
9t2
p
Since tant t is increasing with t and tan 3/5 < 1, we deduce that 10t2 −
p
6 tan2 t > 0 for t ∈ (0, 3/5), which in turn yields
p
10 tan2 t
if t ∈ (0,
′
2
9
p 3/5)
γ(t)γ”(t) − 2γ (t) ≥
10
if t ∈ [ 3/5, t1).
9
γ(t)γ”(t) − 2γ ′ (t)2 ≥
Therefore θ̃”(t) =
2γ ′ (t)2 −γ(t)γ”(t)
3
γ(t)
p
on (0, π/2). In addition
< 0 for t ∈ (0, t1), hence θ(t) is concave
3/5 > π/7, γ(π/7) > γ(1.05), and differentiation
10
shows that γ(t) itself is convex on (0, t1). We conclude that if t ∈ ( π7 , 1.05)
then
10 tan2 π7
θ”(t) < −
< −0.1.
9γ( π7 )3
Finally the concavity of θ̃ on (0, t1] yields that θ(t) ≥ γ(t)−1 for t ∈
[1.05, t1], hence all we need to prove is that θ(t) ≥ γ(t)−1 for t > t1. The
proof of Lemma 5 in [3] shows that
γ(t) ≥ γ(t1 ) + t − t1
for t ∈ (t1 , π/2).
Since θ̃′ (1.05) < −1 by numerical calculations, it is sufficient to prove
1
1
− (t − 1.05)
≤
γ(t1 ) + t − t1
γ(1.05)
for t ∈ (t1, π/2), which in turn is equivalent with
1
ψ(t) = (γ(t1) + t − t1)
− t + 1.05 ≥ 1.
γ(1.05)
(12)
(13)
Now ψ(t) is quadratic with negative main coefficient, and ψ(1.05), ψ(π/2) > 1
by numerical evaluation, completing the proof of Lemma 3.3. 2
PROPOSITION 3.4 If α ∈ R and T is a triangle with an obtuse angle
such that o is a vertex and the angle at o is t ∈ (0, π/2) then
Z
|x2 − α| dx > γ(t) · |T |2.
T
Proof: Let T be the family of triangles such that the area is |T |, o is a vertex,
there exists an angle that is at least π/2, and the angle at o is t. According
to the proof of Lemma 3 in K. Böröczky, Jr. and M. Ludwig [3], there exists
S ∈ T and r > 0 such that if M ∈ T and β ∈ R then
Z
Z
2
|x − β| dx ≥
|x2 − r2 | dx.
M
S
Suppose by contradiction that S has an obtuse angle. Again the proof of
Lemma 3 in K. Böröczky, Jr. and M. Ludwig [3] shows that after fixing r,
11
R
|x2 −r 2 | dt
|S|2
can be decreased by rotating the side of S opposite to o around
the midpoint of that side. After rescaling the resulting triangle, we deduce
that S is not an extremal triangle in T , which contradiction completes the
proof Lemma 3.4. 2
S
Combining Lemma 2 in K. Böröczky, Jr. and M. Ludwig [3] and Proposition 3.2 leads to
PROPOSITION 3.5 If α ∈ R, Π is a polygon with at most k sides, and
o 6∈ relint Π then
Z
γ( π )
|x2 − α| dx ≥ 1.1 · k · |Π|2.
2k
Π
Proof of Lemma 3.1: We may assume that q(z) = z 2 , and according to
Proposition 3.5, we may also assume that o ∈ relint Π. Let T1, . . . , Tm be all
the (non degenerate) triangles obtained in the following way: One vertex is
o, the second vertex is the closest point of some side s of Π to o, and the third
vertex is an endpoint of s. Then T1 , . . . , Tm tiles Π, and m ≤ 2k. For each
Ti , we write ti to denote the (acute) angle of Ti at o, and observe that Ti has
an angle that is at least π/2. We deduce by Proposition 3.3, Proposition 3.4
and (7) that
Z
Π
|x2 − α| dx ≥
≥
m
X
i=1
γ(ti)|Ti |2 ≥
m
X
i=1
θ(ti )
!−1
m
X
i=1
1
γ(ti )
!−1
|Π|2
|Π|2 ≥ m · θ( 2π
)
m
−1
|Π|2.
Since all ti are acute, we deduce that m ≥ 5. If m = 5 then direct calculation
and Proposition 3.2 yield
Z
−1
γ( πk )
γ( π3 )
2
2
·
|Π|
≥
1.1
·
· |Π|2. (14)
)
|Π|
>
1.1
·
|x2 − α| dx ≥ 5 · θ( 2π
5
6
2k
Π
If m ≥ 6 then
2π
m
< 1.05, hence m ≤ 2k and Proposition 3.2 imply
Z
Π
|x2 − α| dx ≥
γ( 2π
)
γ( π )
m
· |Π|2 ≥ k · |Π|2.
m
2k
12
(15)
In turn we conclude the first half of Lemma 3.1.
π
Pm
(1+ε)γ( )
k · |Π|2 . If
Therefore let k ≤ 6, and assume that i=1 γ(ti )|Ti|2 ≤
2k
ε is small enough then we deduce that m = 2k by (14), (15) and Proposition 3.2. In particular all triangles Ti have a right angle.
During the rest of the argument, we write ϑ1 , ϑ2, . . . to denote positive
absolute constants. It follows by Proposition 2.2 and Proposition 3.3 that
√
√
for any i = 1, . . . , 2k, we have |ti − πk | < ϑ1 ε, hence γ(ti) ≥ (1 − ϑ2 ε)γ( πk ).
We deduce that
!2
P2k
P2k
2
√
|T
|
|T
|
i
i
i=1
i=1
≤ (1 + ϑ3 ε)
,
2k
2k
therefore Proposition 2.2 applied to the function f(s) = s2 implies that the
√
1
|Π| is at most ϑ4 4 ε · |Π| for i = 1, . . . , 2k. In turn
absolute value of |Ti | − 2k
we conclude the existence of Π0 .
For any β > 0, let Eβ = {z : z 2 ≤ β}, hence
Z
d
|x2 − β| dx = |Π ∩ Eβ | − |Π\Eβ |.
(16)
dβ Π
R
Therefore Π |x2 − β| dx attains its minimum for β = α0 where |Π ∩ Eα0 | =
1
|Π|. Since ̺0 B 2 ⊂ Π ⊂ ̺1 B 2 where ̺̺01 < 1 + ϑ5 , and the first half of
2
Lemma 3.1 yields
Z
γ( πk )
2
|x − α0 | dx ≥
· |Π|2,
2k
Π
√
α
we deduce that | α0 − 1| < ϑ6 ε, completing the proof of Lemma 3.1. 2
4
An extremal property of regular polygons
related to circumscribed polytopes
Most of the results of this section are contained in P.M. Gruber [11] or in G.
Fejes Tóth [4]. Still we provide proof because some of the statements are not
stated exactly as we need. For t ∈ (0, π2 ), we define
R
x2 dx
R(t)
c
γ (t) =
,
|R(t)|2
13
which readily satisfies
γ c (t) =
tan t
1
+
.
tan t
3
(17)
LEMMA 4.1 If q is a positive definite quadratic form, α ≤ 0 and Π is a
polygon of at most k sides then
Z
p
γ c ( πk )
{q(x) − α} dx ≥
· |Π|2 det q.
2k
Π
R
√
(1+ε)γ c ( π
)
2
k
In addition if k ≤ 6 and Π {q(x)−α}
dx
≤
|Π|
det q for ε ∈ (0, ε0 )
2k
√
4
then Π is a k–gon, and is ϑ ε–close to some k–gon that is regular with respect
to q where ε0 and ϑ are positive absolute constants.
Remark: If k = 6 then it follows by (17) that
γ c ( π6 )
5
= √ .
12
18 3
(18)
To prove Lemma 4.1, we need several auxiliary statements. Propositions 4.2 and 4.3 are consequences of Proposition 2.1.
PROPOSITION 4.2 t γ c (t) is increasing on (0, π2 ), and (t γ c (t))′ > 0.1 for
t ∈ ( π6 , π2 ).
PROPOSITION 4.3 γ c (t)−1 is concave on (0, π2 ), and actually (γ c (t)−1 )” <
).
−0.0001 for t ∈ ( π7 , 5π
12
Next we consider triangles with obtuse angles:
PROPOSITION 4.4 If T is a triangle with an obtuse angle such that o is
a vertex and the angle at o is t ∈ (0, π/2) then
Z
x2 dx > γ c (t) · |T |2.
T
Proof: We may assume that |T | = |R(t)|, and T is positioned in a way that
T and R(t) share their angle t at o, and their longest sides are collinear. In
this
of R(t)\T are closer to o than any point of T \R(t), hence
R 2case allR points
2
x
dx.
2
x
dx
>
R(t)
T
Finally we show that the ”centre of the moment” should be inside to get
a good estimate.
14
PROPOSITION 4.5 If Π is a convex disc with o 6∈ relint Π, and k ≥ 3
then
Z
γ c ( πk )
2
· |Π|2.
x dx ≥ 1.1 ·
2k
Π
Proof: Since there exists a half plane containing Π such that o lies on the
boundary of the half plane, we may assume that Π is a semi circular disc
centred at o. In this case direct calculations and Proposition 4.2 yield
Z
γ c ( π3 )
γ c ( πk )
2
2
x dx ≥ 1.1 ·
· |Π| ≥ 1.1 ·
· |Π|2. 2
6
2k
Π
Finally Lemma 4.1 can be proved analogously to Lemma 3.1. 2
5
An extremal property of regular polygons
related to inscribed polytopes
For t ∈ (0, π2 ), we define
i
γ (t) =
R
R(t)
{1 − x2} dx
|R(t)|2
,
which by (17) satisfies
γ i (t) =
1
5 tan t
+
.
tan t
3
(19)
LEMMA 5.1 If q is a positive definite quadratic form, α > 0 and Π is a
triangle such that q(x) ≤ α for x ∈ Π then
Z
p
γ i ( π3 )
{α − q(x)} dx ≥
· |Π|2 det q.
6
Π
R
√
(1+ε)γ i ( π3 )
In addition if Π {α−q(x)} dx ≤
|Π|2 det q for ε ∈ (0, ε0 ) then there
6
exists some triangle Π0, which is regular with respect to q, whose centroid is
o, and satisfies
√
Π0 ⊂ Π ⊂ (1 + ϑ 4 ε) · Π0
where ε0 and ϑ are positive absolute constants.
15
Remark: If follows by (19) that
γ i ( π3 )
1
=√ .
6
3
(20)
To prove Lemma 5.1, we need several auxiliary statements. Propositions 5.2 and 5.3 are consequences of Proposition 2.1.
PROPOSITION 5.2 t γ i (t) is increasing on (0, π2 ), and (t γ i (t))′ > 0.1 for
t ∈ ( π6 , π2 ).
PROPOSITION 5.3 γ i (t)−1 is concave on (0, π2 ), and actually (γ i (t)−1 )” <
−0.0001 for t ∈ ( π7 , 5π
).
12
Next we consider triangles with obtuse angles:
PROPOSITION 5.4 If T ⊂ rB 2 is a triangle with an obtuse angle such
that o is a vertex and the angle at o is t ∈ (0, π/2) then
Z
{r2 − x2} dx > γ i (t) · |T |2.
T
Proof: Let b the vertex of T with the obtuse angle, and let c be the third
vertex, hence kxk ≤ kck for x ∈ T . It follows that
Z
Z
2
2
{r − x } dx ≥ {c2 − x2} dx.
T
T
Next we define the points a ∈ [o, c] and b′ with the following properties:
∠ab′c = π2 , [a, b′] is parallel to [o, b] and intersects [b, c] in some d, and the
triangle T ′ = ab′c satisfies |T ′| = |T |. Since kxk < kdk < kyk if x ∈ T \T ′
and y ∈ T ′\T , we deduce
Z
Z
2
2
{c − x } dx >
{c2 − x2} dx.
T′
T
Now for any point x ∈ T ′, we have c2 − x2 > (c − a)2 − (x − a)2, therefore
Z
Z
2
2
{c − x } dx >
{(c − a)2 − (x − a)2} dx = γ i (t) · |T ′|2. 2
T′
T′
16
Finally we show that the ”centre of the moment” should be inside to get
a good estimate (see Proposition 5.6). Unfortunately this fact is harder to
prove than in the previous cases. The reason is that Proposition 5.6 does not
hold for any convex disc as Π; for example, if Π is a semi circular disc with
R
γi ( π )
centre o and radius one then Π {1 − x2 } dx < 63 · |Π|2. Therefore first we
need an auxiliary statement:
PROPOSITION 5.5 If T ⊂ rB 2 is a triangle such that o is a vertex, and
the angle at o is 2t for t ∈ (0, π2 ), and Π ⊂ T is a triangle, then
Z
{r2 − x2 } dx ≥ 21 γ i (t) · |Π|2.
Π
Proof: We may assume that r = 1, and T is the isosceles triangle whose two
other vertices are a = (sin t, cos t) and b = (− sin t, cos t). In particular, the
“horizontal direction” is parallel to the side ab, and the “vertical direction”
is orthogonal to it. We may assume that Π 6= T since
Z
{1 − x2 } dx = 21 γ i (t) · |T |2.
(21)
T
In the proof we will use that if C1 , C2 ⊂ T are two convex discs with
|C1| = |C2| then
Z
Z
Z
Z
2
2
2
{1− x } dx ≤
{1− x } dx if and only if
x dx ≥
x2 dx. (22)
C1
C2
C1
C2
We apply the so-called Blaschke–Schüttelung (see T. Bonnesen, W. Fenchel
[1]) twice to Π: If l is any vertical line intersecting Π then let us translate
l ∩Π upwards along l until it hits the side ab but is still inside T . This way we
obtain a triangle Π1, which is the union of the new positions of the vertical
segments (it is a well–known fact that Blaschke–Schüttelung does not lead
out of the class of triangles). Next if l is any horizontal line intersecting Π1
then we translate l ∩ Π1 to the right along l until it hits the side ao but is
still inside T , and write Π2 to denote the resulting triangle. Therefore only
one side of Π2 intersects relint T . We observe that |Π2| = |Π| and
Z
Z
2
x dx ≥
x2 dx.
Π2
Π
17
Now we enlarge the class of triangles satisfying the above property of Π2
to include certain quadrilaterals. Let Q be the family of intersections H ∩ T ,
where H is a closed half plane such that
o 6∈ relint H and |H∩T | ≥ |Π|. There
R
{1−x2 } dx
is minimal. In particular, even
exists some Q0 ∈ Q satisfying that Q0 |Q0|2
if possibly Π 6∈ Q, we have
R
R
2
{1 − x2 } dx
{1
−
x
}
dx
Q0
Π
≥
.
|Π|2
|Q0|2
If Q0 = T then we are done by (21), hence we assume that Q0 6= T . In
addition we may assume by symmetry that a is a vertex of Q0 .
First we suppose that neither b nor o are vertices of Q0, hence Q0 is a
triangle. Let l be the supporting line of Q0 parallel to the line ob, and let
c = l ∩ [o, a]. We observe that if x ∈ Q0 \{a} then hx − a, ci < 0, thus
(a − c)2 − (x − c)2 < a2 − x2. It follows that
Z
Z
2
2
{(a − c) − (x − c) } dx <
{1 − x2} dx.
Q0
Q0
Since Q′ ∈ Q for Q′ = a +
R
Q′
{1 − x2} dx
|Q′|2
=
R
1
ka−ck
(Q0 − a), and
{(a − c)2 − (x − c)2 } dx
Q0
|Q0|2
<
R
Q0
{1 − x2 } dx
|Q0|2
,
we obtain a contradiction with the extremality of Q0 . In particular Q0 and
T share two vertices.
The condition Q0 6= T yields that Q0 has exactly one side s that intersects
relint T . We write p and q to denote the endpoints of s, and prove
kpk = kqk.
(23)
We suppose that the claim (23) does not hold, and seek a contradiction.
We write m to denote the midpoint of s, and f to denote the foot of the
perpendicular from o to aff s, hence m 6= f.
We write H0 to denote the half plane such that H0 contains Q0 and s lies
on the boundary of H0 . For small ϕ > 0, we rotate H0 around m by ϕ in a
way that the resulting half plane Hϕ intersects the segment [m, f] only in m.
We define Qϕ = Hϕ ∩ T , Sϕ+ to be the triangle that is the closure of Qϕ \Q0,
and Sϕ− to be the closure of Q0\Qϕ . In the following we write O(f(ϕ)) to
18
denote any real function g(ϕ) such that |g(ϕ)| ≤ c · f(ϕ) where c depends on
Q0. We observe that
−(1 − O(ϕ)) · (Sϕ+ − m) ⊂ Sϕ− − m ⊂ −(1 + O(ϕ)) · (Sϕ+ − m),
and
|Sϕ+ | = |Sϕ− | =
1
8
kp − qk2 · ϕ + O(ϕ2 ), hence |Qϕ | = |Q0| + O(ϕ2 ).
(24)
In addition if x ∈ Sϕ+ and y ∈ Sϕ− are symmetric with respect to m (ie.,
m = x+y
) then simple calculations yield
2
x2 − y 2 = 4kf − mk · kx − mk + O(ϕ).
We deduce that there exists η > 0 depending only on Q0 such that if ϕ > 0
is small enough then
Z
Z
Z
Z
2
2
2
x dx −
x dx =
x dx −
x2 dx ≥ η · ϕ.
(25)
−
Sϕ
+
Sϕ
Q0
Qϕ
eϕ be the translate of Hϕ such that |Q
eϕ | = |Q0 | holds for Q
eϕ =
Next let H
eϕ ∩ T , hence Q
e ϕ ∈ Q. It follows by (24) and (25) that if ϕ > 0 is small
H
enough then
Z
Z
eϕ
Q
x2 dx >
x2 dx,
Q0
which contradicts the extremality of Q0 according to (22). In turn we conclude the claim (23).
Since T and Q0 shares two vertices, it follows by (23) that Q0 is a trapezoid whose two parallel sides are ab and a0b0 where a0 ∈ [o, a] and b0 ∈ [o, b].
We write T0 to denote the triangle oa0b0 , and define (compare (17))
sin t
1 cos t
1 c
,
+
γ = 2 γ (t) =
2 sin t 3 cos t
which satisfies
Z
Q0
2
x dx =
Z
2
T
x dx −
Z
T0
x2 dx = γ · (|T |2 − |T0|2 ).
19
We deduce by |T | = sin t cos t,
R
R
2
{1 − x2} dx
{1
−
x
}
dx
|T | − |T0| − γ(|T |2 − |T0|2)
Q0
Π
≥
=
|Π|2
|Q0 |2
(|T | − |T0|)2
2
sin2 t
1 − 2γ |T |
+γ = 3
+γ
=
|T | − |T0|
|T | − |T0|
R
2
{1 − x2 } dx
sin2 t
3
>
+γ = T
.
|T |
|T |2
Therefore we conclude Proposition 5.5 by (21). 2
PROPOSITION 5.6 If Π ⊂ rB 2 is a triangle with o 6∈ relint Π then
Z
γ i ( π3 )
{r2 − x2 } dx ≥ 1.1 ·
· |Π|2.
6
Π
Proof: It follows by (19) and (20) that if t ∈ (0, π2 ) then
q
γ i ( π3 )
i
5
.
(26)
γ (t) ≥ 2 3 > 4 · 1.1 ·
6
We write a, b, c to denote the vertices of Π, and distinguish two cases: If
the triangle say obc contains a, and ∠boc = 2t then Proposition 5.5 yields
directly that
Z
γ i ( π3 )
2
2
2
1 i
{r − x } dx ≥ 2 γ (t) · |Π| > 1.1 ·
· |Π|2.
6
Π
Otherwise we may assume that [o, c] intersects [a, b]. We write Ta to denote
the triangle oca, and Tb to denote the triangle ocb. Moreover let 2ta = ∠coa
and 2tb = ∠cob, and let Πa = Π ∩ Ta and Πb = Π ∩ Tb . We deduce by
Proposition 5.5, the Cauchy–Schwarz inequality (7), and the concavity of
γ i (t)−1 (see Proposition 5.3) that
Z
{r2 − x2} dx ≥ 21 γ i (ta) · |Πa|2 + 12 γ i (tb) · |Πb |2
Π
−1
1 i
γ (ta)−1 + γ i (tb )−1
|Π|2
2
γ i ( π3 )
2
1 i ta +tb
· |Π|2. 2
· |Π| > 1.1 ·
≥ 4γ
2
6
≥
20
Finally Lemma 5.1 can be proved analogously to Lemma 3.1. 2
6
A transfer lemma
As usual in polytopal approximation, we plan to transfer the original problem
in R3 into a planar problem where certain integral expressions based on the
second moment are investigated.
LEMMA 6.1 Let ε ∈ (0, 2−21 ), let q be a positive definite quadratic form
in two variables, and let C be a convex polygon tiled by the convex polygons
Π1, . . . , Πk . We assume that for i = 1, . . . , k, there exist ai ∈ R2, αi , α′i ∈ R
and a convex function gi on C − ai such that for any i, j = 1, . . . , k and
y ∈ Πi , we have
gi (y − ai ) − αi ≤ gj (y − aj ) − αj .
For some convex polygon C ′ ⊂ C, we number the tiles in a way that Π1, . . . , Πk′
are the ones intersecting relint C ′, and we assume the following for any y ∈ C
and i = 1, . . . , k ′: We have
(1 + ε)−1 q(y − ai ) ≤ gi (y − ai ) ≤ (1 + ε) q(y − ai),
and if gi (y − ai ) ≤ αi then y ∈ relint C. Moreover α′i = αi if αi ≤ 0, and
αi ≤ α′i ≤ (1 + ε)αi if αi > 0. Then
k Z
X
i=1
k Z
X
′
21
Πi
|gi (y − ai) − αi | dy ≥ (1 − 2 ε) ·
i=1
Πi
|q(y − ai ) − α′i | dy.
Proof: We may assume that q(z) = hz, zi. For i = 1, . . . , k ′ , if αi ≤ 0 then
√
we define Di = ∅, and if αi > 0 then we define ri = αi and
Di = y ∈ R2 : gi (y − ai ) ≤ α2i ,
which by the conditions in the Lemma satisfy
ai +
In addition let
ri
2
B 2 ⊂ Di ⊂ C.
′
C ∗ = ∪ki=1 (Πi ∪ Di ) ,
21
(27)
and for any i = 1, . . . , k ′ , let
Ωi = {y ∈ C ∗ : ∀j = 1, . . . , k ′ , gi (y − ai ) − αi ≤ gj (y − aj ) − αj }.
First we prove the estimates
X Z
i=1,...,k′
k Z
X
′
20
Ωi
gi (y − ai ) dy ≤ 2
X Z
1≤i≤k′
αi >0
i=1
k Z
X
|gi (y − ai ) − αi | dy;
(28)
Ωi
|gi (y − ai ) − αi | dy.
(29)
Ωi
′
20
Ωi
αi dy ≤ 2
i=1
If αi ≤ 0 for all i = 1, . . . , k ′ then (28) and (29) readily follow, therefore we
assume that α1 ≥ . . . ≥ αk′ , and αm > 0 for some 1 ≤ m ≤ k ′ , and αi ≤ 0 if
e i = ai + 8ri B 2 .
i > m. For i ≤ m, we define Di′ = ai + 2ri B 2 and D
Now we define a subsequence l1 < . . . < lm′ of 1, . . . , m. Let l1 = 1. If lj
is known and all Di′ , i ≤ m, intersect at least one of Dl′1 , . . . , Dl′j then j = m′.
Otherwise let lj+1 be the smallest index such that Dl′j+1 does not intersect
Dl′1 , . . . , Dl′j . It follows that
′
′
m e
∪m
i=1 Di ⊂ ∪j=1 Dlj .
If i = 1, . . . , m and y ∈ Ωi \Di′ then readily αi ≤
1
2
gi (y − ai), hence
αi ≤ gi (y − ai) − αi ;
gi (y − ai ) ≤ 2 · [gi (y − ai) − αi ].
(30)
el ,
However if y ∈ Ωi ∩ Di′ then let j be the smallest index such that y ∈ D
j
hence lj ≤ i. We deduce
αi ≤ αlj ,
(31)
which fact combining with gi (y − ai ) − αi ≤ glj (y − alj ) − αlj leads to
gi (y − ai ) ≤ gi (y − ai ) − αi + αlj ≤ glj (y − alj ) ≤ 2 · (y − alj )2 .
It follows by using (27), as well, that
m Z
X
i=1
m Z
X
′
Ωi ∩Di′
gi (y − ai ) dy ≤
j=1
22
el
D
j
2 · (y − alj )2 dy
m Z
X
′
17
≤ 2
j=1
XZ
(y − alj )2 dy
Dlj
m′
18
≤ 2
j=1
m Z
X
glj (y − alj ) dy
Dlj
′
18
≤ 2
j=1
|glj (y − alj ) − αlj | dy.
Dlj
In addition if y ∈ Dlj ∩ Ωt for some j = 1, . . . , m′ and t = 1, . . . , k ′ then
gt (y − at ) − αt ≤ gt (y − alj ) − αlj < 0, hence
m Z
X
′
Ωi ∩Di′
i=1
k Z
X
18
gi (y − ai ) dy ≤ 2
Ωt
t=1
|gt (y − at ) − αt | dy.
Therefore we deduce by (30) that
m Z
X
i=1
Ωi
k Z
X
′
gi (y − ai ) dy ≤ 219
Ωt
t=1
|gt (y − at) − αt | dy,
which in turn yields (28). Turning to (29), we use the notation as above. It
follows by (31) that
m Z
X
i=1
Ωi ∩Di′
m Z
X
αi dy ≤
j=1
m Z
X
′
′
el
D
j
αlj dy ≤
j=1
el
D
j
2 · (y − alj )2 dy,
hence the rest of the argument for (29) is similar to the proof (28).
For i = 1, . . . , k ′, assume that y ∈ Ωi \Πi , hence αi > 0 and y ∈ Di .
Naturally y ∈ Πj for some j = 1, . . . , k, and gj (y−aj )−αj ≤ gi (y−ai )−αi < 0
yields |gj (y − aj ) − αj | ≥ |gi (y − ai ) − αi |. Defining α∗i = max{αi , 0}, we
deduce by (28) and (29) that
k Z
X
′
i=1
k Z
X
′
2
Ωi
|(y − ai ) −
α′i | dy
≤
i=1
Ωi
|gi (y − ai ) − αi | dy +
k Z
X
′
ε·
i=1
23
Ωi
{gi (y − ai ) + α∗i } dy
k Z
X
′
21
≤ (1 + 2 ε) ·
≤ (1 + 221 ε) ·
i=1
Ωi
|gi (y − ai ) − αi | dy
i=1
Πi
|gi (y − ai ) − αi | dy.
k Z
X
In turn we conclude Lemma 6.1 by Πi ⊂ Ωi for i = 1, . . . , k ′ . 2
7
The idea of the proof of Theorem 1.1
Let us present the main idea how to prove Theorem 1.1 where we start
with the case when the number of faces is restricted, say with P(n) . The
rough idea is that if F is a face of P(n) and F is a k–gon then we compare the
contribution of F towards the symmetric difference metric to the contribution
of some regular k–gon of the same area. If k 6= 6 then we may actually
think F regular by the results of Section 3. On the one hand, the average
number of sides of the faces of P(n) is at most six according to the Euler
theorem. On the other hand, the estimate of the asymptotic formula for
the symmetric difference metric is attained by some polytopes whose typical
faces are approximately regular hexagons in the suitable sense. Therefore
convexity of various functions involved yields that the typical faces also of
P(n) are approximately regular hexagons (in the suitable sense).
Since the Gauß–Kronecker curvature of ∂K is allowed to be zero, we meet
some additional technical difficulties. First we fix a small positive ν, and
then consider some arbitrary small positive µ < ν 12 . We separate the ”round
part” X of ∂K (X depends on ν and µ), and will actually consider only
faces of P(n) “near X”. We construct an auxiliary circumscribed polytope
M depending also on ν and µ, and write M to denote the faces of M “near
X”. For any C ∈ M, we project the faces of of P(n) near C into aff C. With
the help of Lemma 6.1, we transfer our problem of polytopal approximation
in R3 into estimating the second moment corresponding to some tiling of a
slightly dilated copy of C. Since the average number of sides of the tiles in
C is at most six, we can apply the estimates of Section 3. In particular we
may assume that nearly all faces of P(n) are hexagons. Now if QF exists for
a hexagonal face F of P(n) , and F is not ν–close to a regular hexagon with
respect QF then F generates an error of order ν 4 |F |, and using this fact we
24
conclude Theorem 1.1.
In the case of circumscribed polytopes the proof is the same. When the
number of vertices is restricted then the proof is simpler because we may
essentially assume that all faces are triangles.
We only prove Theorem 1.1 for P(n) in detail, and sketch the necessary
changes for the other cases afterwards.
8
The proof of Theorem 1.1 for P(n)
It is sufficient to prove the following statement:
For a given convex body K in R3 with C 2 boundary, let P(n) be a polytope
with at most n faces such that δS (K, P(n) ) is minimal. For ν ∈ (0, ν0 ) and
µ ∈ (0, ν 12), if g(n) is number of faces F of P(n) such that QF is positive
definite, F is a hexagon, and F Ris ϑν–close to some hexagon that is regular
κ(x)1/4 dx
then
with respect to QF and of area ∂K
n·κ(x )1/4
F
√
g(n) > (1 − ϑ̃ 6 µ)n for n > n0
(32)
where ϑ and ϑ̃ are positive absolute constants, and ν0 > 0 and n0 depend on
K.
During the proof of (32), ϑ1, ϑ2 , . . . denote positive absolute constants,
and ω1 , ω2 , . . . positive constants depending on K.
For any x ∈ ∂K, let u(x) be the exterior unit normal to ∂K at x. It is
well–known (see say K. Leichtweiß [14]) that there exists η > 0 such that
balls of radius η roll from inside on ∂K; namely, for any x ∈ ∂K, the three–
ball of radius η and of centre x − η u(x) is contained in K. Let K−η be the
family of points z such that z + η B 3 ⊂ K. Now if y ∈ R3 \K−η then there
exists a unique closest point of ∂K to y, and we write π(y) to denote this
point.
We write cl Y to denote the closure of any Y ⊂ R3 , and consider ∂K
with the subspace topology as a subset of R3 . We say that Y ⊂ ∂K is Jordan measurable if the relative boundary of Y on ∂K is of two–dimensional
Hausdorff–measure zero. Let X0 , X ′ and X be relatively open Jordan measurable subsets of ∂K such that cl X0 ⊂ X, cl X ⊂ X ′ , κ(x) > 0 for x ∈ cl X ′ ,
and
Z
Z
1/4
4
κ(x) dx ≥ (1 − µν )
κ(x)1/4dx.
X0
∂K
25
Now we construct a circumscribed polytope M whose typical faces are
hexagonal, and the faces near X are small enough. A useful tool is the Taylor
formula that we use in the following form: Let f be a convex C 2 function on a
convex disc C. For y ∈ C, we write ly to denote the linear form representing
the derivative of f at y, and qy to denote the quadratic form representing
the second derivative of f at y. Now if a, z ∈ C then there exists t ∈ (0, 1)
satisfying
f(y) = f(a) + la(y)(y − a) + 21 qa+t (y−a) (y − a).
(33)
We have δ > 0 with the following properties: (X0 + 2δ B 3 ) ∩ ∂K ⊂ X
and (X + 2δ B 3 ) ∩ ∂K ⊂ X ′ . Moreover if C is a convex disc that touches K
in x ∈ X and C is of diameter at most δ then
(i) writing C ′ to denote the orthogonal projection of π(C) into aff C, we
have
x + (1 − µν 4 )(C − x) ⊂ C ′ ⊂ x + (1 − µν 4 )−1 (C − x);
(ii) if w ∈ π(C) then hu(w), u(x)i ≥ 1 − µν 4 ;
(iii) if f is the convex function on C such that its graph is the part of ∂K,
and qy is the quadratic form representing the second derivative of f at
y ∈ C (hence Qx = qx) then
(1 + µν 4 )−1 Qx ≤ qy ≤ (1 + µν 4 )Qx .
There exists a convex polytope M circumscribed around K such that each
face that touch ∂K in a point of X is of diameter less than δ. We write M
to denote the family of faces C of M that touch K in a point x(C) of X. If
ν0 , δ and ̺ are small enough then
Z
X
1/4
4
κ(x(C)) |C| ≥ (1 − ϑ1µν )
κ(x)1/4dx.
(34)
∂K
C∈M
We start to investigate P(n) . We define
1 γ( π6 )
5
1
= ·
.
γ̃ = √ −
2 12
36 3 8π
26
According to (4), if n is large then
4
δS (K, P(n) ) < (1 + µν ) · γ̃ ·
Z
κ(x)
1/4
dx
∂K
2
·
1
.
n
(35)
It follows by (35) and the existence of the rolling ball of radius η that
δH (K, P(n) ) ≤ ω1 n−1/2.
(36)
Therefore if n0 is large enough then K−η ⊂ intP(n) . Since the infimum of the
principal curvatures at the points of X ′ is positive, we deduce that if F is a
face of P(n) such that π(F ) ⊂ X ′ then
diam F ≤ ω2 n−1/4 .
(37)
f to denote the family of convex discs of the form
Next we write M
(1 − 2µν 4 )(H − x(H)) + x(H)
f we write x(C) to denote
as H runs through the elements of M. For C ∈ M,
the point where C touches K, and define
C ′ = (1 − µν 4 )(C − x(C)) + x(C).
In addition let FC to denote the family of faces of P(n) near C whose orthogonal projection to affC intersects relint C. We deduce by (i) and (37) that if
f are pairwise disjoint, and
n0 is large enough then the families FC for C ∈ M
by (34) that
Z
X
1/4
′
4
κ(x(C)) |C | ≥ (1 − ϑ2µν )
κ(x)1/4dx.
(38)
∂K
f
C∈M
For any plane L in R3 , we write pL to denote the orthogonal projection
f We write F ′ to denote the family of all F ∈ FC such
into L. Let C ∈ M.
C
that paff C (F ) intersects relint C ′. Again if n0 is large enough then (36) yields
for any F ∈ FC′ that
paff C (K ∩ aff F ) ⊂ relint C.
(39)
We write fC to denote the convex function on the face of P(n) containing
C whose graph is part of ∂K. We recall that for any F ∈ FC , xF denotes
27
the point of ∂K such that u(xF ) is an exterior unit normal to F , and write
aF = paff C (xF ). Then writing lF to denote the linear function representing
the derivative of fC at aF , the part aff F above C is the graph of
fC (aF ) + lF (y − aF ) + αF
for y ∈ C where αF ∈ R. In addition there exists a convex function gF on
C − aF such that if y ∈ C then
fC (y) = fC (aF ) + lF (y − aF ) + gF (y − aF ),
and the Taylor formula (33) and (iii) yield that
(1 + µν 4 )−1 21 Qx(C)(y − aF ) ≤ gF (y − aF ) ≤ (1 + µν 4 ) 21 Qx(C) (y − aF ).
We also observe that according to (39), if F ∈ FC′ , and y ∈ aff C satisfies
gF (y − aF ) ≤ αF then y ∈ relint C. For any F ∈ FC′ , we define
ΠF = C ′ ∩ paff C (F ).
Therefore we may apply Lemma 6.1, and obtain using the Fubini theorem
that
X X Z
4
δS (K, P(n) ) ≥ (1 − ϑ3 µν )
| 21 Qx(C) (y − aF ) − αF | dy. (40)
′
f F ∈FC
C∈M
ΠF
For any F ∈ FC′ , we define k(F ) to be the number of sides of ΠF , and
I(F ) = κ(x(C))1/4|ΠF |.
′
Next we decompose ∪C∈M
fFC into the families F1 , F2 , F3 and F4 . Let
F ∈ FC′ . We put F into F4 if k(F ) 6= 6, and into F3 if ΠF is a hexagon that
is not ν–close to any hexagon that is regular with respect Qx(C). Therefore
F ∈ F1 ∪ F2 if ΠF is a hexagon, which is ν–close to some hexagon that is
regular with respect Qx(C) . Assuming this, we have
R
1/4
∂K κ(x) dx
F ∈ F1 if − 1 ≤ ν;
n · I(F )
R
1/4
∂K κ(x) dx
− 1 > ν.
F ∈ F2 if n · I(F )
28
We write nj to denote the cardinality of Fj . It follows by Lemma 3.1, (35)
and (38) that

(1 + ϑ4µν 4 ) · γ̃ · 
X
F ∈∪4j=1 Fj
2
I(F )

1
≥ γ̃ · 
n

X
I(F )2
F ∈∪3j=1 Fj
X
+ϑ5ν 4
I(F )2
F ∈F3
π
)
γ( k(F
)
!
1 X
· I(F )2.
2 F ∈F 2k(F )
+
4
We claim that last term above satisfies
π
1 X γ( k(F ) )
· I(F )2 ≥ (1 + ϑ6 ) · γ̃ ·
2 F ∈F 2k(F )
4
X
I(F )
F ∈F4
!2
·
1
.
n4
(41)
It follows by the Cauchy–Schwarz inequality (7) that
π
X γ( k(F
)
)
F ∈F4
2k(F )
· I(F )
2
=
X
F ∈F4
≥
2k(F ) ·
X
F ∈F4
π
γ( k(F
)
)
·
I(F )
2k(F )
!−1
π
2k(F ) · γ( k(F
)−1
)
·
2
X
F ∈F4
I(F )
!2
.
As C ′ is tiled by ΠF as F runs through FC′ , combining Proposition 2.3 and
(37) yields that the average number of sides of all ΠF , F ∈ FC′ , is at most
six if n0 is large enough. In particular the average of all k(F ), F ∈ F4, is at
most six. If the average is at least 5.5 then we use Proposition 2.2 to γ(t)−1
(compare Proposition 3.3) with error term, and use the monotonicity of tγ(t)
(compare Proposition 3.2) to obtain
X
F ∈F4
π
)−1
2k(F ) · γ( k(F
)
!−1
≥
≥
1 + ϑ7
n4
· γ P n4 πk(F )
·P
F ∈F4
n4
F ∈F4 2k(F )
1 + ϑ7 γ( π6 )
1 + ϑ7
·
· 2γ̃.
=
n4
12
n4
29
If the average all k(F ), F ∈ F4 , is less than 5.5 then we use the error term
for the monotonicity of tγ(t) (compare Proposition 3.2, and observe that the
average is at least three), and obtain
X
F ∈F4
π
2k(F ) · γ( k(F
)−1
)
!−1
≥
n4
1
· γ P n4 πk(F )
·P
F ∈F4
n4
F ∈F4 2k(F )
1 + ϑ8
1 + ϑ8 γ( π6 )
=
·
· 2γ̃.
≥
n4
12
n4
In turn we deduce the claim (41).
P
Now by applying the inequality for quadratic mean to F ∈F3 I(F )2, we
deduce

2


X
X
1
(1 + ϑ4 µν 4 ) · γ̃ 
I(F ) ·
≥ γ̃ · 
I(F )2
(42)
n
4
3
F ∈∪j=1 Fj
F ∈∪j=1 Fj
X
+ϑ5ν 4
I(F )
F ∈F3
+(1 + ϑ6) · γ̃ ·
!2
X
·
1
n3
I(F )
F ∈F4
!2
·
1
.
n4
First we show that the contribution coming from faces in F3 and F4 is negligible. Applying the inequality for quadratic mean and the Cauchy–Schwarz
inequality (7) in (42) leads to

(1 + ϑ4 µν 4 ) · γ̃ 
X
F ∈∪4j=1 Fj
2
I(F ) ·

1
≥ γ̃ · 
n
X
F ∈∪4j=1 Fj
+ϑ9ν 4
I(F ) · P4
X
F ∈F3 ∪F4
Since
P4
j=1
nj ≤ n, it follows that
X
F ∈F3 ∪F4
√
I(F ) ≤ ϑ10 µ ·
30
X
F ∈∪4j=1 Fj
2
I(F ).
I(F )
1
j=1
!2
nj
1
.
n3 + n4
(43)
Thus (38) and (42) yield
!2
!
X
X
1
√
I(F ) ·
≥ (1 − O( µ))
I(F )2 ;
n
F ∈F1 ∪F2
F ∈F ∪F
Z 1 2
X
√
I(F ) = (1 + O( µ)) ·
κ(x)1/4dx.
(44)
(45)
∂K
F ∈F1 ∪F2
Applying the inequality for quadratic mean in (44) leads to n1 + n2 = (1 +
√
O( µ)) · n, hence (45) shows that
R
P
κ(x)1/4dx
√
F ∈F1 ∪F2 I(F )
∂K
I0 =
.
= (1 + O( µ)) ·
n1 + n2
n
Therefore if ν0 after (32) is small enough then Proposition 2.2 implies
!2
4
X
X
ϑ
ν
n
1
11
2
I(F )2 ≥ 1 +
I(F ) ·
.
n1 + n2
n1 + n2
F ∈F1 ∪F2
F ∈F1 ∪F2
√
√
µ
O( ν 4 ) = O( 6 µ),
2
and hence n1 ≥
Comparing to (44) leads to n1n+n
=
2
√
6
(1 − ϑ12 µ) · n.
We are not ready because some ΠF is not the projection of F . We call
f ΠF meets the relaF ∈ F1 a border face if assuming F ∈ FC , C ∈ M,
′
tive boundary of C . Otherwise we call F ∈ F1 an inner face; namely, if
ΠF ⊂ relint C ′. We observe that if F is an inner face then ΠF is the projection of F , hence F is ϑ13ν–close to some hexagon
that is regular with
R
1/4 dx
∂K κ(x)
respect to the positive definite QF , and is of area n·κ(x )1/4 . However if
F
F is a border face and F ∈ FC then ΠF lies in a ω3 n−1/2 neighbourhood of
the relative boundary of C ′ in aff C. Since any border face F is √
in F1 , we
ω4
have |ΠF | > n , therefore the number of border faces is at most ω5 n. After
√
choosing n0 large enough, the number of border faces is less than 6 µ · n,
√
hence g(n) ≥ (1 − ϑ14 6 µ) · n. In turn, we conclude (32), and hence Theorem 1.1 for P(n) . 2
9
The proof of Theorem 1.1 for Pn
The proof is essentially the same as for P(n) , therefore we only sketch the
changes. We recall that according to Euler’s theorem, Pn has at most 2n
31
faces. The main changes in (32) is that now g(n) counts the number of
√
regular triangles, not of regular hexagons, and we prove g(n) > (1− ϑ̃ 6 µ)2n.
We define X, M and M̃ as in Section 8. Instead of γ̃, we use
γ∗ =
1
1 γ( π )
1
√ −
= · 3 .
4
6
12 3 16π
Here we have factor 14 unlike the factor 12 in the definition of γ̃ because the
asymptotically optimal upper bound on the number of faces of Pn is 2n, not
n as in the case of P(n) .
An essential change in the argument that first we triangulate ∂Pn by
triangulating any non–triangular face by diagonals from a fixed vertex of the
face. We write Σ to denote the resulting triangular complex, that has the
same family of vertices as Pn . For C ∈ M̃, we write FC to denote the family
of all faces of Σ that lies near C and paff C (F ) intersects relint C, and FC′ to
denote the family of all F ∈ FC such that paff C (F ) intersects relint C ′. For
any F ∈ FC , we define aF , αF and I(F ) analogously as in Section 8.
Other changes compared to the argument in Section 8 are concerned with
the definitions of ΠF and Fj . If F ∈ FC′ then ΠF = paff C (F ) (no need to
′
intersect with C ′ ). We decompose ∪C∈M
fFC into only three families F1 , F2
R
κ(x)1/4 dx
and F3. Let F ∈ FC′ . We put F into F1 if ∂Kn·I(F )
− 1 ≤ ν, and there
exists a triangle T whose centroid is aF , which is regular with respect Qx(C),
and
(1 + ν)−1 (T − aF ) ⊂ ΠF − aF ⊂ (1 + ν)(T − aF ).
R
κ(x)1/4 dx
We put F into F2 if such a T exists but ∂Kn·I(F )
− 1 > ν. Finally F ∈ F3
if no such T exists.
The part of the argument in Section 8 that is concerned with F1 , F2 and
√
F3 proves that n1 ≥ (1 − ϑ∗ 6 µ) · 2n where ϑ∗ is a positive absolute constant.
We observe that if a face G of Pn contains an element F ∈ F1 then G is near
f and paff C (x(G)) = aF ∈ relintΠF . Therefore any face of Pn
some C ∈ M,
contains only at most one element of F1 . Since Pn has at most 2n faces, we
√
deduce that g(n) ≥ (1 − 2ϑ∗ 6 µ) · 2n. 2
Remark: To prove the Remark about Pn after Theorem 1.1, we put the
following extra condition on a face F ∈ FC′ that belongs to F1 ∪ F2: Let LC
be the tangent plane at x(C), and let EF = {z ∈ LC : Qx(C) (z − aF ) ≤ αF }.
Then F should satisfy
1−ν
2
· |F | ≤ |F ∩ EF | ≤
32
1+ν
2
· |F |.
10
c
The proof of Theorem 1.1 for P(n)
and Pni
c
The proof in the case of P(n)
runs closely as for P(n) , only one uses the results
of Section 4 instead of Section 3.
The proof in the case of Pni runs closely as for Pn , the main difference
is that one uses the results of Section 5 instead of Section 3. There is one
additional change in the argument. For each F ∈ FC′ , we define
α′F = (1 + µν 4 ) · αF .
1
2
Qx(C)(y − aF ) ≤ α′F for any y ∈ ΠF , and (40) is replaced by
X X Z i
4
α′F − 21 Qx(C)(y − aF ) dy. 2
δS (K, Pn ) ≥ (1 − ϑ3µν )
Therefore
′
f F ∈FC
C∈M
ΠF
References
[1] T. Bonnesen, W. Fenchel: Theory of convex bodies. BCS. Assoc., Moscow
(Idaho), 1987. Translated from German: Theorie der konvexen Körper.
Springer-Verlag, 1934.
[2] K. Böröczky Jr.: Approximation of smooth convex bodies. Adv. Math.,
153 (2000), 325–341.
[3] K. Böröczky, Jr., M. Ludwig: Approximation of convex bodies and a
momentum lemma for power diagrams. Monats. Math., 127 (1999), 101–
110.
[4] G. Fejes Tóth: A stability criterion to the moment theorem. Studia Sci.
Math. Hungar., 38 (2001), 209–224.
[5] L. Fejes Tóth: Lagerungen in der Ebene, auf der Kugel und im Raum.
Springer-Verlag, Berlin, 2nd edition, 1972.
[6] S. Glasauer, P.M. Gruber: Asymptotic estimates for best and stepwise
approximation of convex bodies III. Forum Math., 9 (1997), 383–404.
[7] P.M. Gruber: Volume approximation of convex bodies by inscribed polytopes. Math. Ann., 281 (1988), 229–245.
33
[8] P. M. Gruber: Volume approximation of convex bodies by circumscribed
polytopes. In: Applied geometry and discrete mathematics, DIMACS
Ser. Discrete Math. Theoret. Comput. Sci. 4, Amer. Math. Soc., 1991,
309–317.
[9] P.M. Gruber: Asymptotic estimates for best and stepwise approximation
of convex bodies I. Forum Math., 5 (1993), 281–297.
[10] P.M. Gruber: Asymptotic estimates for best and stepwise approximation
of convex bodies IV. Forum Math., 10 (1998), 665-686.
[11] P.M. Gruber: Optimal configurations of finite sets in Riemannian 2manifolds. Geom. Dedicata, 84 (2001), 271–320.
[12] P.M. Gruber: Optimale Quantisierung. Math. Semesterber., 49 (2002),
227–251.
[13] P.M. Gruber: Optimum quantization and its applications. Adv. Math.,
186 (2004), 456–497.
[14] K. Leichtweiß: Affine Geometry of Convex Bodies. Johann Ambrosius
Barth Verlag, Heidelberg, Leipzig, 1998.
[15] M. Ludwig: Asymptotic approximation of smooth convex bodies by
general polytopes. Mathematika, 46 (1999), 103–125.
[16] R. Schneider: Zur optimalen Approximation konvexer Hyperflächen
durch Polyeder. Math. Ann., 256 (1981), 289–301.
[17] R. Schneider: Polyhedral approximation of smooth convex bodies. J.
Math. Anal. Appl. 128 (1987), 470–474.
[18] R. Schneider: Convex Bodies – the Brunn–Minkowski theory. Cambridge
Univ. Press, 1993.
Károly Böröczky, Jr.
Alfréd Rényi Institute of Mathematics, Budapest,
PO Box 127, H–1364, Hungary, [email protected]
and
Department of Geometry, Roland Eötvös University, Budapest,
34
Pázmány Péter sétány 1/C, H–1117, Hungary
Gergely Wintsche
Teacher Training Department, Roland Eötvös University, Budapest,
Pázmány Péter sétány 1/C, H–1117, Hungary, [email protected]
Péter Tick
Budapest, Gyűrű utca 24., H–1039, Hungary, [email protected]
35