On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area
The author expresses his sincere gratitude to V.I. Diskant for his useful discussions of the considered problem.
Збережено в:
| Дата: | 2011 |
|---|---|
| Автор: | |
| Формат: | Стаття |
| Мова: | English |
| Опубліковано: |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2011
|
| Назва видання: | Журнал математической физики, анализа, геометрии |
| Онлайн доступ: | http://dspace.nbuv.gov.ua/handle/123456789/106670 |
| Теги: |
Додати тег
Немає тегів, Будьте першим, хто поставить тег для цього запису!
|
| Назва журналу: | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| Цитувати: | On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area / A.I. Shcherba // Журнал математической физики, анализа, геометрии. — 2011. — Т. 7, № 2. — С. 158-175. — Бібліогр.: 11 назв. — англ. |
Репозитарії
Digital Library of Periodicals of National Academy of Sciences of Ukraine| id |
irk-123456789-106670 |
|---|---|
| record_format |
dspace |
| spelling |
irk-123456789-1066702016-10-02T03:02:33Z On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area Shcherba, A.I. The author expresses his sincere gratitude to V.I. Diskant for his useful discussions of the considered problem. 2011 Article On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area / A.I. Shcherba // Журнал математической физики, анализа, геометрии. — 2011. — Т. 7, № 2. — С. 158-175. — Бібліогр.: 11 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106670 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| institution |
Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| collection |
DSpace DC |
| language |
English |
| description |
The author expresses his sincere gratitude to V.I. Diskant for his useful discussions of the considered problem. |
| format |
Article |
| author |
Shcherba, A.I. |
| spellingShingle |
Shcherba, A.I. On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area Журнал математической физики, анализа, геометрии |
| author_facet |
Shcherba, A.I. |
| author_sort |
Shcherba, A.I. |
| title |
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area |
| title_short |
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area |
| title_full |
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area |
| title_fullStr |
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area |
| title_full_unstemmed |
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area |
| title_sort |
on stability of a unit ball in minkowski space with respect to self-area |
| publisher |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| publishDate |
2011 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/106670 |
| citation_txt |
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area / A.I. Shcherba // Журнал математической физики, анализа, геометрии. — 2011. — Т. 7, № 2. — С. 158-175. — Бібліогр.: 11 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT shcherbaai onstabilityofaunitballinminkowskispacewithrespecttoselfarea |
| first_indexed |
2025-07-07T18:50:33Z |
| last_indexed |
2025-07-07T18:50:33Z |
| _version_ |
1837015220306837504 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2011, vol. 7, No. 2, pp. 158–175
On Stability of a Unit Ball in Minkowski Space
with Respect to Self-Area
A.I. Shcherba
Cherkassy State Technological University
460 Shevchenko Blvd., Cherkassy, 18006, Ukraine
E-mail: shcherba anatoly@mail.ru
Received February 23, 2010
The main results of the paper are the following two statements. If the
length of the unit circle ∂B = {||x|| = 1} on Minkowski plane M2 is equal
to O(B) = 8(1− ε), 0≤ ε ≤ 0.04, then there exists a parallelogram which is
centrally symmetric with respect to the origin o and the sides of which lie
inside an annulus (1+18ε)−1 ≤ ||x|| ≤ 1. If the area of the unit sphere ∂B in
the Minkowski space Mn, n ≥ 3, is equal to O(B) = 2n ·ωn−1 ·(1−ε), where
ε is a sufficiently small nonnegative constant and ωn is a volume of the unit
ball in Rn, then in the globular layer (1 + εδ)−1 ≤ ||x|| ≤ 1, δ = 2−n · (n!)−2
it is possible to place a parallelepiped symmetric with respect the origin o.
Key words: Minkowski space, self-perimeter, self-area, stability.
Mathematics Subject Classification 2000: 52A38, 52A40.
Let B be a normalizing body of the n-dimensional Minkowski space Mn,
n ≥ 2. This body is usually called a unit ball, and its boundary ∂B is called a
unit sphere in Mn. Denote by Rn a Euclidean space adjoined to Mn the distance
function of which is used as an auxiliary metric [1, 2]. In its turn, the auxiliary
metric is chosen in such a way that the Euclidean n-dimensional volume Vn(B)
of B equals the volume of the n-dimensional unit ball in Rn,
Vn(B) = ωn :=
π
n
2
Γ
(
n
2 + 1
) .
We identify the points in Mn with their position vectors from the origin o.
Following Busemann [3], we define an (n − 1)-dimensional area of the surface
of nonempty compact convex body K. Let Mm be an m-dimensional plane
in Mn. Then the m-dimensional Minkowski volume in Mm (1 ≤ m ≤ n) is an
m-dimensional Lebesgue measure of V B
m in Mm normalized such that
V B
m (B ∩Mm
o ) = ωm,
c© A.I. Shcherba, 2011
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area
where Mm
o is a translant (i.e., a result of some translation) of Mm which passes
through the origin o. For any compact convex set K in Mm,
V B
m (K) = ωm · Vm(K)/Vm(B ∩Mm
o ), 1 ≤ m ≤ n,
where Vm is an arbitrary taken (affine) m-dimensional Lebesgue measure.
Isoperimetrix I in Mn is an o centrally symmetric compact convex body with
the support function hI given on the unit sphere Ω = {< u, u >= 1} ⊂ Rn by
hI(u) = ωn−1 · V −1
n−1(B ∩Ao(u)), (1)
where Vn−1 is a Euclidean (n− 1)-dimensional volume and Ao(u) is a hyperplane
having the normal u and passing through the origin o .
Notice that the isoperimtrix I in Mn depends only on the normalizing body
B and does not depend on the choice of the auxiliary metric [1, p. 279].
Let K0 and K1 be convex bodies in Rn. Consider a segment Kθ = (1 − θ) ·
K0 + θ · K1 (0 ≤ θ ≤ 1) connecting the bodies K0 and K1. In [4], Minkowski,
introducing the notion of the mixed volumes, expressed the volume V (Kθ) as
V (Kθ) =
n∑
υ=0
Cυ
n · (1− θ)n−υ · θυ · Vυ(K0,K1), (2)
where Vυ(K0,K1) is a mixed volume of the bodies K0 and K1 which corresponds
to the parameter υ. Here we use the standard notations [5, p. 113]. By Minkowski,
the value
OB(K) = n · V1(K, I)
is called a surface area of the body K.
By a self-area of the surface of the unit ball B we understand the value
O(B) = OB(B) = n · V1(B, I). (3)
In the case of n = 2, the value O(B) is called a self-perimeter of the unit circle.
In 1932, Golab S. [6] found optimal estimations for the perimeter: 6 ≤ O(B) ≤ 8.
In 1956, Busemann H. and Petti K. [7] obtained the following result.
Theorem A. If B is a unit ball in the n-dimensional Minkowski space Mn,
then O(B) ≤ 2n · ωn−1, and the equality holds only when B is a parallelepiped.
In this paper we study a stability of the unit ball B in the case when the
self-area O(B) is close to the greatest possible value 2n · ωn−1. There are proved
the following theorems.
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2 159
A.I. Shcherba
Theorem 1. Let the self-perimeter of a unit ball B on Minkowski plane M2 be
equal to O(B) = 8 · (1− ε), where 0 ≤ ε ≤ 1
25 . Then there exists a parallelogram
P which is centrally symmetric with respect to the origin o and for which the
inclusions
P ⊂ B ⊂ (1 + 18 · ε) · P (4)
hold.
Theorem 2. Let the self-area O(B) of a unit sphere ∂B in Minkowski space
Mn, n ≥ 3, be equal to O(B) = 2n · ωn−1 · (1 − ε). Then there exists a positive
constant ε0 depending only on the dimension n and the centrally symmetric w.r.
to the origin o parallelepiped P for which the inclusions
P ⊂ B ⊂ (1 + εδ) · P, (5)
hold, where 0 ≤ ε ≤ ε0 and δ = 2−n · (n!)−2.
The main results of the paper can be formulated in terms of the metric ‖x‖ of
Minkowski space Mn. For example, Theorem 1 can be reformulated as follows: if
the self-area of a unit sphere is equal to 2nωn−1 ·(1−ε), where ε is a small enough
nonnegative constant, then in the globular layer (1+εδ)−1 ≤ ‖x‖ ≤ 1 of the space
Mn (n ≥ 3) it is possible to place some parallelepiped P symmetric w.r. to the
origin o. And also the area of P satisfies (1 + εδ)1−n · O(B) ≤ OB(P ) ≤ O(B)
that follows at once from definition (3) and monotonicity of the mixed volume.
Studying the possibility of the equality O(B) = 2n · ωn−1, Busemann H. and
Petti K. used the fact that the body B, being a cylindrical one, possesses n linearly
independent one-dimensional generators. Discussing the results obtained in this
paper, Diskant V.I. drew my attention that I used only one such a generator in
the proof of Theorem 2. In fact, it is proved by induction over the dimension
m of Mm (n ≥ m ≥ 2) by constructing a cylinder in Minkowski space, which
approximates a unit ball with a given accuracy. In our opinion, this construction
is of independent interest.
If K is a convex body in Mn, then there are two supporting hyperplanes H+
K
and H−
K parallel to any given (n− 1)-dimensional hyperplane H. By Minkowski,
the value
∆B(K, H) = min
{‖x1 − x2‖ : x1 ∈ H+
K , x2 ∈ H−
K
}
is called the width of the convex body K in Mn w.r. to H [2, p. 106], [8]. Since
the isoperimetrix I is symmetric w.r. to the origin o, its width satisfies the
equality ∆B(I, H) = 2 · min {‖x‖ : x ∈ HI}, where HI is one of two supporting
hyperplanes. Consider the body B as the one located in some adjoint space Rn
and specify a unit vector u normal to HI = HI(u). Let hI(u) and hB(u) be the
supporting numbers of I and B. Then ∆B(I, H) = 2·hI(u)·h−1
B (u). There follows
the theorem on the stability of the unit ball B w.r. to the width of isoperimetrix.
160 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area
Theorem 3. If ∆B(I, H) = 4(1 − ε) · ωn−1/ωn, 0 ≤ ε ≤ 10−4n3
, then there
exists a cylinder Cn(D) with one-dimensional generators such that:
1. Cn(D) is centrally symmetric w.r. to the origin o;
2. Cn(D) cross-section D is parallel to H;
3. Cn(D) ⊂ B ⊂ Cn(D) ·
(
1 + ε
1
2n2
)
. (6)
This result is close to that obtained by Diskant V.I. on the estimation from above
for the width of the isoperimetrix ∆B(I, H) ≤ 4ωn−1 · ω−1
n , where the equality
holds only when B is a cylinder [8].
P r o o f of the Theorem 1. Let Q2 be a parallelogram of the smallest
area and let it be centered at o and circumscribed around B. The midpoints
of the Q2 sides necessarily lie on ∂B [1, p. 121]. On M2, chose an auxiliary
Euclidean metric such that on the adjoint plane R2 with the Cartesian system
xoy the parallelogram Q2 becomes a square abcd with the vertices a(−1; 1), b(1; 1),
c(1;−1), d(−1;−1). The points e(0; 1), f(1; 0), g(0;−1), p(−1; 0) lie on ∂Q2 and
efgp ⊂ B. Denote by n and m the points of intersection of straight lines y = x
and y = −x with ∂B in a half-plane y > 0. Let 0 < ξ < 1
2 and 0 < η < 1
2 be the
parameters that determine n and m by n(1−ξ, 1−ξ) and m(−1+η; 1−η). From
the symmetry B = −B, the points −n(−1 + ξ;−1 + ξ) and −m(1 − η;−1 + η)
lie on ∂B. Draw the straight lines (pm), (ab) and denote their intersection by
a2 = (pm)∩ (ab); draw the straight lines (em), (da) and denote their intersection
by a1 = (em) ∩ (da). Set b2 = (en) ∩ (bc), b1 = (fn) ∩ (ab), c1,2 = −a1,2,
d1,2 = −b1,2. Since B is convex, its line of support at m crosses the segments
[a2e] and [pa1], and hence the segment [a1a2] does not have common points with
the interior
◦
B. Therefore, B ⊂ a1a2b1b2c1c2d1d2, and it follows then that
8 · (1− ε) ≤ O(B) ≤ OB(a1a2b1b2c1c2d1d2) ≤ OB(Q2) = 8. (7)
Denote by ‖x‖ the length of a vector x on M2 with a normalizing body B
and by |x| , its Euclidean length on R2. Taking into account (7), we have
‖pa1‖+ ‖a1a2‖+ ‖a2e‖ ≤ ‖ap‖+ ‖ae‖ = 2,
‖eb1‖+ ‖b1b2‖+ ‖b2f‖ ≤ 2,
4− 4ε ≤ (‖pa1‖+ ‖a1a2‖+ ‖a2e‖) + (‖eb1‖+ ‖b1b2‖+ ‖b2f‖) ≤ 4.
Hence, {
0 ≤ 2− (‖pa1‖+ ‖a1a2‖+ ‖a2e‖) ≤ 4ε,
0 ≤ 2− (‖eb1‖+ ‖b1b2‖+ ‖b2f‖) ≤ 4ε.
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2 161
A.I. Shcherba
By calculating
|aa2| = |a1a| = η
1− η
,
we can see that
‖a2e‖ = ‖pa1‖ = 1− η
1− η
and
‖a1a2‖ =
|a1a2|
|on| =
|aa2|
nx
=
η
(1− η)(1− ξ)
.
Consequently,
2− 4ε ≤ ‖pa1‖+ ‖a1a2‖+ ‖a2e‖ = 2− η
1− η
(
2− 1
1− ξ
)
.
After the similar calculations for n, compose the system
{
η(1− 2ξ) ≤ 4ε(1− η)(1− ξ),
ξ(1− 2η) ≤ 4ε(1− η)(1− ξ),
where 0 ≤ ξ, η ≤ 1
2 .
Combining the inequalities, we get
(1 + 8ε)(ξ + η) ≤ (1 + 2ε)4ξη + 8ε.
Since 4ξη ≤ (ξ + η)2, the value z = ξ + η satisfies the square inequality
(1 + 2ε)z2 − (1 + 8ε)z + 8ε ≥ 0.
It is obvious that either
0 ≤ ξ + η ≤ 1 + 8ε−√1− 16ε
2(1 + 2ε)
or
1 + 8ε +
√
1− 16ε
2(1 + 2ε)
≤ ξ + η ≤ 1.
As a consequence, either
max {ξ; η} ≤ 1 + 8ε−√1− 16ε
2(1 + 2ε)
or max
{
1
2
− ξ;
1
2
− η
}
≤ 1− 4ε−√1− 16ε
2(1 + 2ε)
.
If 0 ≤ ε ≤ 1
25 , then
√
1− 16ε ≥ 1− 10ε. There are two cases:
1) max {ξ; η} ≤ 9ε
1 + 2ε
≤ 9ε; 2) max
{
1
2
− ξ;
1
2
− η
}
≤ 3ε
1 + 2ε
≤ 3ε.
Consider each case separately. Suppose (1) holds. Chose a square r1r2r3r4
with the vertices at points r1(−1+9ε; 1−9ε), r2(1−9ε; 1−9ε), r3(1−9ε;−1+9ε),
162 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area
r4(−1 + 9ε;−1 + 9ε) to be a parallelogram P in (4). By the construction, P ⊂
B ⊂ Q2. Since Q2 = 1
1−9εP , we have Q2 ⊂ (1 + 18ε)P .
Suppose (2) holds. Chose a square efgp to be P in (4). As noticed above,
[a1a2] ∩
o
B = ∅. The points a1(−1; 1− η
1−η ) and a2(−1 + η
1−η ; 1) lie on a straight
line y = x + 2− η
1−η . For 1
2 − η ≤ 3ε we have
2− η
1− η
≤ 1 +
12ε
1 + 6ε
≤ 1 + 12ε,
and hence the figure B is under a straight line y = x+ 1+ 12ε. For the segments
[b1b2], [c1c2], [d1d2] we draw the straight lines y = −x+1+12ε, y = x−1−12ε, y =
−x−1−12ε. Denote by S2 a square with vertices at e1(0; 1+12ε), f1(1+12ε; 0),
g1(0;−1 − 12ε, p1(−1 − 12ε; 0). Then B ⊂ S2 = (1 + 12ε) · P . The proof is
complete.
To prove Theorem 3 we need some auxiliary statements. Without loss of
generality, further we will consider a proper convex compact body B symmetric
w.r. to the origin o and located in the corresponding adjoint Euclidean space Rn
(n ≥ 2).
Proposition 1. Let K0 and K1 be convex compact bodies in Rm, m ≥ 2, with
the m-dimensional Euclidean volumes satisfying V (K0) ≤ V (K1). Let V0 be a
constant such that V (Kθ) ≤ V0, 0 ≤ θ ≤ 1. Then
V1(K0,K1)− V (K0) ≤ e(V0 − V (K0)). (8)
P r o o f. The Brunn inequality implies
V
1
m (Kθ) ≥ (1− θ)V
1
m (K0) + θV
1
m (K1) ≥ V
1
m (K0),
and hence V (Kθ) ≥ V (K0).
Using the identity
1 =
m∑
υ=0
Cυ
m(1− θ)m−υθυ,
rewrite (2) in the form of
V (Kθ)− V (K0) =
m∑
υ=0
Cυ
m(1− θ)m−υθυ [Vυ(K0,K1)− V (K0)] . (9)
Write down the inequality for the mixed volumes
V m
υ (K0,K1) ≥ V m−υ(K0)V υ(K1),
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2 163
A.I. Shcherba
which is a consequence of a more general A.D. Aleksandrov’s inequality [9, p. 78].
Then V m
υ (K0,K1) ≥ V m(K0) and Vυ(K0,K1) − V (K0) ≥ 0. Since all terms in
the right-hand side of (9) are nonnegative, then
m(1− θ)m−1θ [V1(K0,K1)− V (K0)] ≤ V (Kθ)− V (K0) ≤ V0 − V (K0).
The inequality holds for all 0≤ θ ≤ 1. For θ = 1
m we get
(
1− 1
m
)m−1
[V1(K0,K1)− V (K0)] ≤ V0 − V (K0).
Since the Euler sequence an =
(
1 + 1
n
)n
< e is monotonously increasing, then
(
1− 1
m
)m−1
=
(
1 +
1
m− 1
)1−m
>
1
e
.
Therefore,
1
e
[V1(K0, K1)− V (K0)] ≤ V0 − V (K0),
which completes the proof of Proposition 1.
Further we will use a method suggested by V.I. Diskant [10, 11] for studying
a stability in the theory of convex bodies. Denote by q = q(K0,K1) a capacity
coefficient of K1 w.r. K0, i.e., the greatest of γ’s for which the body γ · K1 is
embedded into K0 by a translation. Recall one of Diskant’s inequalities for the
mixed volumes [10, p. 101]:
V
m
m−1
1 (K0,K1)− V (K0)V
1
m−1 (K1) ≥
[
V
1
m−1
1 (K0,K1)− qV
1
m−1 (K1)
]m
. (10)
Proposition 2. Let the bodies K0 and K1 meet the requirements of Proposi-
tion 1. Set α = 3(V0/V (K0)− 1) ≤ 1
4 . Then the capacity coefficient q satisfies
q(K0,K1) ≥ 1− 2α
1
m . (11)
P r o o f. To estimate q(K0,K1) from below, we use inequality (10) (see
formula (2.1) in [10, p. 110])
q ≥
[
V1(K0,K1)
V (K1)
] 1
m−1
−
[
V
m
m−1
1 (K0,K1)− V (K0)V
1
m−1 (K1)
] 1
m · V −1
m−1 (K1).
164 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area
Transform this inequality
q ≥
[
V1(K0,K1)
V (K1)
] 1
m−1
−
[
V1(K0,K1)
V (K1)
] 1
m
{(
V1(K0,K1)
V (K1)
) 1
m−1
− V (K0)
V1(K0,K1)
} 1
m
. (12)
The inequality V1(K0,K1) ≥ V (K0) implies
V1(K0, K1)
V (K1)
≥ V (K0)
V (K1)
≥ V (K0)
V0
=
1
1 + α
3
≥ 1− α
3
. (13)
By (8), we have V1(K0,K1)− V (K0) ≤ 3 · (V0 − V (K0)), and hence
V1(K0,K1)
V (K1)
≤ V1(K0,K1)
V (K0)
≤ 1 + 3(
V0
V (K0)
− 1) = 1 + α. (14)
Besides,
V (K0)
V1(K0,K1)
≥ 1
1 + α
≥ 1− α. (15)
Substituting (13), (14), (15) into (12), we obtain
q ≥
(
1− α
3
) 1
m−1 − (1 + α)
1
m
{
(1 + α)
1
m−1 − (1− α)
} 1
m
.
For p ≥ 1 we have
(1) (1 + x)
1
p ≤ 1 + x
p , 0 ≤ x ≤ 1;
(2) (1− x)
1
p ≥ 1− 12
11x, 0 ≤ x ≤ 1
12 .
Therefore,
q ≥ 1− 4
11
α−
(
1 +
α
m
){
m
m− 1
α
} 1
m
≥ 1− 4
11
α− 9
8
(
m
m− 1
) 1
m
α
1
m .
The conditions m ≥ 2 and 0≤ α ≤ 1
4 provide
α ≤ 1
2
α
1
m and
(
m
m− 1
) 1
m
≤
√
2.
Finally,
q ≥ 1− 2
11
α
1
m − 9
8
√
2α
1
m ≥ 1− 2α
1
m .
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2 165
A.I. Shcherba
Denote by At(u) a hyperplane in Rn which is parallel to Ao(u) and is at the
distance t in the direction of the vector u. If t < 0, then At(u) is at the same
distance from Ao(u) in the direction of the vector −u.
We denote by hB = h(u) (u ∈ Ω) a supporting function of the normalizing
body B. Denote by H(u) the hyperplanes of support that correspond to h(u).
Let Bt(u) = B ∩ At(u). If −h(u) ≤ t ≤ h(u), then Bt(u) 6= ∅. The central
symmetry of the unit ball B = −B provides the equalities B−t(u) = −Bt(u).
Consider the function
φu(t) = V
1
n−1
n−1 (Bt(u)), t ∈ [−h(u);h(u)] .
The function is even, φu(−t) = φu(t), and by the Brunn inequality it is convex
upwards. Then max
t
φu(t) = φu(0), and this provides the estimation
Vn(B) ≤ 2h(u) · Vn−1(B0(u)).
Denote by ∆V (u) the difference
∆V (u) = 2h(u)Vn−1(B0(u))− Vn(B).
Proposition 3. Let u0 be a unit normal vector of some hyperplane of support
H0 = HI(u0) for the isoperimetrix I. If a Minkowski width of I in the direction
u0 is equal to ∆B(I,H0) = 4(1− ε)ωn−1ω
−1
n , 0≤ ε < 1, then
∆V (u0) = ε2h(u0)Vn−1(B0(u0)). (16)
P r o o f. Indeed, from the expression in the terms of supporting numbers
for the Minkowski width of the body I in the adjoint space Rn and the explicit
expression for the isoperimetrix I supporting function hI given by (1), we get
∆B(I, H0) = 2
hI(u0)
hB(u0)
= 2
ωn−1
h(u0)Vn−1(B0(u0))
.
Taking into account the normalization Vn(B) = ωn, we have
∆B(I, H0) = 4
ωn−1
ωn
Vn(B)
2h(u0)Vn−1(B0(u0))
.
Together with the condition imposed on ∆B by the hypothesis, the latter equality
provides (16).
Set V0 = Vn−1(B0(u0)), h0 = hB(u0), φ0(t) = φu0(t) and ∆V (u0) = 2h0V0ε.
Denote by B∗ a Schwartz-symmetrized body B w.r. to a straight line L(u0)
which is parallel to u0 and passes through the origin o. By the construction,
166 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area
Vn(B∗) = Vn(B). By the Brunn theorem, the body of rotation B∗ is convex [5,
p. 89]. On R2 with the Cartesian coordinates xoy, define the function
x(y) = φ0(y)ω
− 1
n−1
n−1 , −h0 ≤ y ≤ h0.
Set for brevity x(0) = r. The function x = x(y) defines the radii of the (n− 1)-
dimensional balls that generate B∗. On the graph of this function, mark the
point M0(x0; y0) which is an intersection point of the graph and a straight line
y = h0
r x. We have 0 < x0 ≤ r, 0 < y0 ≤ h0. It is convenient to use a parameter
τ = r − x0. Then M0(r − τ, h0 − h0
r τ).
Proposition 4. If the conditions of Proposition 3 hold, then
τ ≤ r
√
ε
2
. (17)
P r o o f. If τ = 0, then inequality (17) is trivial. Notice that by the
Minkowski–Brunn theorem, the equality τ = 0 holds only when the body B is a
cylinder with the generators parallel to u0.
Suppose τ > 0. Draw a supporting straight line to x = x(y) at M0. The
intersection points of this line and straight lines y = h0 and x = r denote by P
and Q, respectively. The points P1 = (0; h0), Q1 = (r; 0) and the whole segment
[P1Q1] are to the left of the convex curve x = x(y), 0 ≤ y ≤ h0. Therefore,
0 < τ ≤ r
2 . Rewrite the coordinates of P and Q in the terms of a and b, namely,
P = (r − a;h0) and Q = (r; h0 − b).
Define the function r1 = r1(y), y ∈ [−h0; h0] by
r1(y) =
r, if b− h0 ≤ y ≤ h0 − b;
r − a− a
b (y − h0), if h0 − b ≤ y ≤ h0;
r − a + a
b (y + h0), if − h0 ≤ y ≤ b− h0.
In Rn, construct a rotation body B̂ with the axis L(u0) and the radii of the
(n−1)-dimensional spheres given by the function r1 = r1(y). By the construction,
x(y) ≤ r1(y), which provides B∗ ⊂ B̂. Estimate from below a difference ∆V (u0)
in the terms of Vn(B̂)
∆V (u0) = 2h0V0 − Vn(B∗) ≥ 2h0V0 − Vn(B̂)
= 2ωn−1r
n−1b− 2ωn−1
b∫
0
(r − a
b z)n−1dz
= 2ωn−1
b
na
[
(r − a)n − rn + narn−1
]
.
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2 167
A.I. Shcherba
It is easy to verify that the function
φ(s) = (1− s)n − 1 + ns− n
2
s2, 0 ≤ s ≤ 1, n ≥ 2,
is monotonously increasing. Multiplying the inequality
(1− s)n − 1 + ns ≥ n
2
s2
by rn and denoting rs = a, we obtain
(r − a)n − rn + nrn−1a ≥ n
2
rn−2a2.
Thus,
∆V (u0) ≥ ωn−1r
n−2ab. (18)
The chosen point M0(r−τ ;h0− h0
r τ) lies on the supporting straight line, therefore
a and b are connected by the equation
τ
a
+
h0
r
τ
b
= 1.
The product ab in the right-hand side of (18) can be expressed in the terms of b
ab = τ(b +
h0
r
a) =
rb2τ
rb− h0τ
= f(b).
Estimate ab from below by min f(b) = f(b0), where b0 = 2h0
r τ, a0 = 2τ . Then
ab ≥ a0b0 = 4
h0
r
τ2.
The hypotheses of Proposition 3, (16) and (18) imply
ε2h0V0 = ∆V (u0) ≥ 4ωn−1r
n−3h0τ
2 = 4
V0
r2
h0τ
2, or ε ≥ 2
(τ
r
)2
.
Corollary 1. A cross-section Bt = B ∩ At(u0), which corresponds to M0, is
defined by T = h0 − h0
r · τ . Besides, the distance between AT (u0) and A0(u0) is
T ≥ t0 = h0
(
1−
√
ε
2
)
. (19)
The Euclidean (n− 1)-dimensional volume of the section BT satisfies
Vn−1(BT ) = ωn−1(r − τ)n−1
≥ ωn−1r
n−1
(
1−
√
ε
2
)n−1
= V0
(
1−
√
ε
2
)n−1
. (20)
168 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area
The following theorem (the analog of Theorem 3 for the plane) illustrates the
importance of inequalities (19) and (20).
Theorem 4. If the width of the isoperimetrix I on M2, which corresponds to
a straight line H0, satisfies ∆B(I,H0) = 8
π (1− ε), 0≤ ε ≤ 1
6 , then there exists a
symmetric w.r. to the origin o parallelogram P with a side parallel to H0 such
that P ⊂ B ⊂ (1 + 2
√
ε) · P .
P r o o f. As above, denote by u0 a normal vector of the isoperimetrix
I supporting the straight line HI = HI(u0), HI ||H0 in the adjoint plane R2.
If n = 2, then the section BT is a segment [ab]. Set c = −a, d = −b. Then
B−T = −BT = [cd]. Denote B0 = [ef ]. Then |oe| = |of | = r. We assume that
the points a, b, f, c, d, e are on ∂B in the cyclic order and clockwise.
Show that P = abcd can be taken as a required parallelogram. The inclusion
P ⊂ B is obvious. Prove now the inclusion B ⊂ (1 + 2
√
ε)P . Denote a1 =
(de) ∩ (ab); d1 = (ae) ∩ (cd); b1 = (cf) ∩ (ab); c1 = (bf) ∩ (dc). The segments
[ea1], [ed1], [fb1], [fc1] do not have any common points with
◦
B. Therefore,
the figure B is in a strip bounded by the two parallel straight lines (d1a1) and
(c1b1). Denote a2 = (d1a1) ∩H(u0), b2 = (c1b1) ∩H(u0), c2 = (c1b1) ∩H(−u0),
d2 = (d1a1)∩H(−u0). Mark also the points f1 = (cb)∩(ef) and e1 = (da)∩(ef).
Due to (20), we have
|ab| = 2 |of1| = 2(r − τ) ≥ 2r
(
1−
√
ε
2
)
and |f1f | = τ ≤ r
√
ε
2
.
A similarity of the triangles ∆cf1f and ∆cbb1 implies |bb1| = 2|f1f | ≤ r
√
2ε.
Besides, |cc1| = |dd1| = |aa1| = |bb1| ≤ r
√
2ε. It is easy to see that
|a2b2|
|ab| ≤ |ab|+ 2r
√
2ε
|ab| ≤ 1 +
√
2ε
1−√
ε
2
≤ 1 + 2
√
ε.
The estimation (19) provides
|c2b2|
|cb| ≤ h0
h0
(
1−√
ε
2
) ≤ 1 +
√
2ε.
By the construction, the parallelogram a2b2c2d2 ⊃ B, hence B ⊂ (1+2
√
ε)P .
The theorem is proved.
To prove Theorem 3 for the case of n ≥ 3 we need an estimation from below
for the capacity coefficient of BT w.r. to B−T .
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2 169
A.I. Shcherba
Corollary 2. If 0 ≤ ε ≤ 10−2, then the capacity coefficient satisfies
q(−BT ; BT ) ≥ 1− 5ε
1
2(n−1) , n ≥ 3. (21)
P r o o f. By means of translation, place the convex bodies BT and B−T
in the (n − 1)-dimensional hyperplane Ao(u0). Denote by B′
T = BT − Tu0,
B′
−T = B−T + Tu0 the corresponding traslants. The equalities Vn−1(B′
T ) =
Vn−1(B′
−T ) and q(−BT , BT ) = q(B′
−T , B′
T ) are obvious. The line segment Kθ =
(1− θ)BT + θB−T , 0≤ θ ≤ 1 is in the section B(1−2θ)T = B∩A(1−2θ)T (u0). Thus,
for K ′
θ = (1− θ)B′
T + θB′
−T , we have
Vn−1(K ′
θ) = Vn−1(Kθ) ≤ V0 = Vn−1(B0(u0)).
Take K0 = B′
−T and K1 = B′
T from the hypothesis of Proposition 2. Then
from (20) we get
α = 3
(
V0
V (K0)
− 1
)
≤ 3
((
1−
√
ε
2
)1−n
− 1
)
,
and for the capacity coefficient
q(−BT ; BT ) ≥ 1− 2× 3
1
n−1
(
1−
(
1−
√
ε
2
)n−1
) 1
n−1 (
1−
√
ε
2
)−1
.
For m ≥ 2, 0 ≤ x ≤ 1
2 , 0 ≤ ε ≤ 10−2 the inequalities
(
3m√
2
) 1
m
≤ 9
4
, (1− (1− x)m)
1
m ≤ m
1
m x
1
m , 1−
√
ε
2
≥ 10
11
hold. The estimation (21) follows at once from the above.
P r o o f of the Theorem 3. Let BT be defined as in Corollary 1 and
the relations (19)–(21) be fulfilled. Let B′
T and B′
−T respectively denote the
tranlants of BT and B−T after a translation on Ao(u). Notice that B′
−T = −B′
T .
Let γ = q(B′
T ,−B′
T ). Then there is a vector a in the hyperplane Ao(u0) such that
a + γ(−B′
T ) ⊂ B′
T . On Ao(u) consider a mapping ϕ(x) = a − γx, x ∈ Ao(u).
Evidently, φ(B′
T ) = a−γB′
T = a+γ(−B′
T ) ⊂ B′
T . Since Vn−1(B′
T ) = Vn−1(B′
−T ),
then γ ≤ 1. If γ = 1, then the bodies B′
T and B′
−T coincide after a translation
on the vector a. In general, γ < 1. Denote by x0 a solution of the equation
x0 = a− γx0, i.e., x0 = (1 + γ)−1a.
By the choice, ϕ(x0) = x0. Let B̃T = B′
T − x0, B̃−T = B′
−T + x0. Then
B̃−T = −B̃T . It is easy to check that after this replacement we have γB̃−T ⊂ B̃T .
170 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area
Hence, the inclusions γ(−B̃T ) ⊂ B̃T ⊂ 1
γ
(
−B̃T
)
hold. Moreover, q(B̃T ,−B̃T ) =
q(−B̃T , B̃T ) = γ. On the hyperplane Ao(u0), construct a body D = B̃T ∩
(
−B̃T
)
which is centrally symmetric w.r. to the origin o. Set ν = x0 + Tu0. By the
construction, DT ≡ D + v ⊂ BT and D−T ≡ D − v ⊂ B−T . Notice also that
D ⊃ γ(B̃T ∪ B̃−T ). (22)
Denote by Cn(D) ⊂ Rn a cylinder whose cross-sections coincide with D and
whose 1-dimensional generators are parallel to v and bounded by the hyperplanes
AT (u0) and A−T (u0). This cylinder is symmetric w.r. to the origin o: Cn(D) =
−Cn(D). Since the symmetric body B is convex, the inclusion Cn(D) ⊂ B holds.
Estimate from below the capacity coefficient q(Cn(D), B). By formula (22),
Vn(B) ≥ Vn(Cn(D)) = 2TVn−1(D) ≥ 2TVn−1(γBT ) = 2Tγn−1Vn−1(BT ).
Using estimations (19)–(21), we conclude
Vn(B) ≥ 2h0V0
(
1− 5ε
1
2(n−1)
)n−1
(
1−
√
ε
2
)n
.
For further calculations to be substantial, we assume 0 ≤ ε ≤ (10(n− 1))−2(n−1).
Initially, 2h0V0 ≥ Vn(B) (see, for example, equality (16)). Hence,
Vn(B) ≥ Vn(Cn(D)) ≥ Vn(B)
(
1− 5(n− 1)ε
1
2(n−1)
)(
1− n
√
ε
2
)
,
or
1 ≤ Vn(B)
Vn(Cn(D))
≤
(
1− 7(n− 1)ε
1
2(n−1)
)−1
≤ 1 + 14(n− 1)ε
1
2(n−1) .
Consider a segment Kθ = (1−θ)Cn(D)+θ·B, 0≤ θ ≤ 1 inside B. In Proposition 2
assume that K0 = Cn(D), K1 = B, V0 = Vn(B), where Vn(Kθ) ≤ Vn(B). Then
α ≤ 50(n− 1)ε
1
2(n−1) , and the capacity coefficient q(Cn(D), B) can be estimated
by (11),
q1 = q(Cn(D), B) ≥ 1− 10ε
1
2n(n−1) , n ≥ 3.
The bodies Cn(D) and B being centrally symmetric w.r. to the origin o, we
will get q1B ⊂ Cn(D) and B ⊂ 1
q1
Cn(D). Since 1
1−x ≤ 1+2x, 0≤ x ≤ 1
2 , we have
Cn(D) ⊂ B ⊂
(
1 + 20ε
1
2n(n−1)
)
Cn(D). (23)
Finally, if 0≤ ε ≤ 10−4n3
, then the inclusions (6) hold. The theorem is proved.
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2 171
A.I. Shcherba
P r o o f of the Theorem 2. The proof is based on the idea of Busemann–
Petti (see Theorem 7.4.1 [2]) and on the properties of a superficial function of
convex body introduced by Aleksandrov A.D. [9, p. 39].
For a convex body B, the superficial function F (B, ω) on a unit sphere Ω is
defined by the following construction. Let a Lebesgue measurable set ω be given
on Ω. Denote by σ(ω) a set of all points on the surface of the convex body B
having a normal u directed to ω. The superficial function F (σ(ω)) is the area of
σ(ω).
Write down the first mixed volume from definition (3) in the terms of the
Stieltjes–Radon integral for the continuous isoperimetrix I supporting function
hI(u) over a unit sphere,
O(B) =
∫
Ω
hI(u)F (B, dω).
Since the origin o is inside of B, then hB(u) > 0, and hence the ratio
hI(u)/hB(u) is a continuous function on Ω. By the integral mean value theo-
rem, there is a vector u0 on Ω such that
O(B) =
∫
Ω
hI(u)
hB(u)
hB(u)F (B, dω)
=
hI(u0)
hB(u0)
∫
Ω
hB(u)F (B, dω) =
hI(u0)
hB(u0)
nVn(B).
The plane of support H0 = HI(u0) for I is given by the supporting number
hI(u0). By Theorem 2, the area O(B) = 2nωn−1(1− ε), and hence the width
∆B(I,H0) = 2hI(u0)/hB(u0) = 4(1− ε)ωn−1/ωn.
By Theorem 3, in Mn there is a cylinder with the cross-section perpendicular to
u0 for which (6) holds.
Now we study the cross-section D of the cylinder Cn = Cn(D). Show that
the body D, by analogy with (6), can be approximated by some ”(n − 1) -
dimensional” cylinder Cn−1 = Cn−1(Dn−2) with the cross-section Dn−2. Denote
by Q = 1 + 20 · ε 1
2n(n−1) the factor from (23). Without loss of generality, assume
that the generators of the cylinder Cn(D) are perpendicular to the cross-section
D, i.e., v ‖ u0. The latter is based on the affine invariancy of the definition of
self-area of the surface O(B) and on the free choosing of the auxiliary metric
172 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area
in Mn. Notice that the inclusions (23) provide
O(B) ≤ OB(QCn) =
∫
Ω
ωn−1
Vn−1(B ∩A0(u))
F (QCn, dω)
≤ Qn−1
∫
Ω
ωn−1
Vn−1(Cn ∩A0(u))
F (Cn, dω) = Qn−1OCn(Cn(D)).
From the conditions imposed on O(B) in Theorem 2, we have
OCn(Cn) ≥ Q1−nO(B) = Q1−n(1− ε)2nωn−1 ≥ Q−n2nωn−1.
Using the inequalities 1
1+x ≥ 1− x and (1− x)n ≥ 1− nx for
0 ≤ x = 20ε
1
2n(n−1) ≤ 1
2n
,
we obtain
OCn(Cn) ≥ 2nωn−1
(
1− 20nε
1
2n(n−1)
)
. (24)
The surface of the cylinder Cn(D) consists of two bases DT , D−T that are equal
to D and of a lateral surface C ′
n. Hence,
OCn(Cn) = 2OCn(D) + OCn(C ′
n) = 2ωn−1 + OCn(C ′
n). (25)
Denote by Ω′ an intersection of the unit sphere Ω and the (n − 1) dimensional
hyperplane Rn−1 which corresponds to A0(u0). Recall the equalities
{
Vn−1(Cn ∩A0(w)) = 2h0Vn−2(D ∩A0(w)), w ∈ Ω′;
Fn−1(C ′
n, dω) = 2h0Fn−2(D, d′ω),
,
where d′ω is a restriction of dω on Ω′. Thus,
OCn(C ′
n) =
∫
Ω
ωn−1
Vn−1(Cn ∩A0(w))
Fn−1(C ′
n, dω)
=
∫
Ω′
ωn−1
Vn−2(D ∩A0(w))
Fn−2(D, d′ω) =
ωn−1
ωn−2
OD(D).
Imposing the condition ε ≤ (20n)−2n3
and taking into account (24) and (25), we
obtain
OD(D) ≥ 2(n− 1)ωn−2
(
1− 20n2
n− 1
ε
1
2n(n−1)
)
≥ 2(n− 1)ωn−2
(
1− ε
1
2n2
)
.
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2 173
A.I. Shcherba
Set ε1 = ε
1
2n2 . Then for the compact proper central symmetric (n − 1)-
dimensional body D the estimate
O(D) ≥ 2(n− 1)ωn−2(1− ε1)
holds.
Remark. Calculations in the proof of Theorem 3 up to formula (23) remain
valid for the dimension (n− 1) ≥ 3.
Taking initially a body D instead of B, which is in the space Rn−1 adjoint to
A0(u0), we can construct a centrally symmetric cylinder Cn−1 = Cn−1(Dn−2)
with the cross-section Dn−2 ⊂ Rn−2 satisfying the inclusions similar to (6).
Namely,
Cn−1(Dn−2) ⊂ D ⊂
(
1 + ε
1
2(n−1)2
1
)
Cn−1(Dn−2).
In Rn consider a cylinder Cn(Cn−1(Dn−2)) whose cross-sections coincide with
the ”(n− 1)-dimensional” cylinder Cn−1(Dn−2); the one-dimensional generators
are parallel to u0 and bounded by the hyperplanes AT (u0) and A−T (u0). The
cylinder possesses a specific property
Cn(Cn−1(Dn−2)) ⊂ B ⊂
(
1 + ε
1
2n2
)(
1 + ε
1
22n2(n−1)2
)
Cn(Cn−1(Dn−2)). (26)
Using recurrently (n − 2) times the specified above constructions that cor-
respond to the pass from formula (23) to formula (26), we get a cylinder C̃ =
Cn(Cn−1(. . . (C3(D2)) . . .)) which approximates the initial normalizing body B
as follows:
C̃ ⊂ B ⊂
(
1 + ε
1
2n2
)(
1 + ε
1
22n2(n−1)2
)
. . .
(
1 + ε
1
2n−2n2(n−1)2 ... 32
)
C̃.
The cylinder C2 on the plane M2 is a parallelogram. Approximate a figure
D2 by the parallelogram; the approximation order is defined on the (n− 1)-step
by εn−1 = ε24−n(n!)−2
. Recall that on the plane M2 there is formula (4) from
Theorem 1, where εn−1 appears to be in the first degree. Thus, it is possible to
approximate the body B by the parallelepiped P for which the inclusions
P ⊂ B ⊂
(
1 + ε
1
2n2
)(
1 + ε
1
22n2(n−1)2
)
× . . .
×
(
1 + ε
1
2n−4(n!)2
)(
1 + 18ε
1
2n−4t(n!)2
)
P (27)
174 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2
On Stability of a Unit Ball in Minkowski Space with Respect to Self-Area
hold. There is such a sufficiently small positive ε0(n) depending only on the
dimension n that the inequalities
(
1 + 18ε24−n(n!)−2
)n−1
≤ 1 + 18nε24−n(n!)−2 ≤ 1 + ε2−n(n!)−2
(28)
hold for 0≤ ε ≤ ε0. Put δ = 2−n(n!)−2. Then from (27) and (28) we derive
formula (5). The theorem is proved.
The author expresses his sincere gratitude to V.I. Diskant for his useful dis-
cussions of the considered problem.
References
[1] K. Leichtweiß, Konvexe Mengen. VEB Deutscher Verlag, Berlin (1980).
[2] A.C. Thompson, Minkowski Geometry. Cambridge University Press, Cambridge
(1996).
[3] H. Busemann, Intrinsic Area. — Ann. Math. 48 (1947), 234–267.
[4] H. Minkowski, Volume und Oberfläche. — Ann. Math. 75 (1903), 447–495.
[5] T. Bonnesen and W. Fenchel, Teorie der Konvexen Körper. Springer (1934); New
York (1948).
[6] S. Golab, Quelques Problèmes Métriques de la Géometrie de Minkowski. Trav.
l’Acad. Mines Cracovie 6 (1932), 1–79.
[7] H. Busemann and C.M. Petty, Problems on Convex Bodies. — Math. Scand. 4
(1956), 88–94.
[8] V.I. Diskant, Estimates for Diameter and Width for the Isoperimetrix in Minkowski
Geometry. — J. Math. Phys., Anal., Geom. 2 (2006), No. 4, 388–395.
[9] A.D. Aleksandrov, Geometry and Applications. Selected Papers. V. 1. Nauka,
Novosibirsk (2006). (Russian)
[10] V.I. Diskant, Refinement of Isoperimetric Inequality and Stability Theorems in the
Theory of Convex Bodies. — Tr. Inst. Mat., Siberian Branch USSR Acad. Sci. 14
(1989), 98–132. (Russian)
[11] V.I. Diskant, Stability of Solutions of Minkowski and Brunn Equations. — Mat.
Fiz., Anal., Geom. 6 (1999), No. 3/4, 245–252. (Russian)
Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 2 175
|