Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2014
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| Цитувати: | Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies / V.I. Diskant // Журнал математической физики, анализа, геометрии. — 2014. — Т. 10, № 3. — С. 309-319. — Бібліогр.: 8 назв. — англ. |
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irk-123456789-1068002016-10-06T03:02:25Z Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies Diskant, V.I. 2014 Article Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies / V.I. Diskant // Журнал математической физики, анализа, геометрии. — 2014. — Т. 10, № 3. — С. 309-319. — Бібліогр.: 8 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106800 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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English |
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Article |
| author |
Diskant, V.I. |
| spellingShingle |
Diskant, V.I. Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies Журнал математической физики, анализа, геометрии |
| author_facet |
Diskant, V.I. |
| author_sort |
Diskant, V.I. |
| title |
Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies |
| title_short |
Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies |
| title_full |
Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies |
| title_fullStr |
Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies |
| title_full_unstemmed |
Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies |
| title_sort |
refinement of isoperimetric inequality of minkowski with the account of singularities in boundaries of intrinsic parallel bodies |
| publisher |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| publishDate |
2014 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/106800 |
| citation_txt |
Refinement of Isoperimetric Inequality of Minkowski with the Account of Singularities in Boundaries of Intrinsic Parallel Bodies / V.I. Diskant // Журнал математической физики, анализа, геометрии. — 2014. — Т. 10, № 3. — С. 309-319. — Бібліогр.: 8 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT diskantvi refinementofisoperimetricinequalityofminkowskiwiththeaccountofsingularitiesinboundariesofintrinsicparallelbodies |
| first_indexed |
2025-07-07T19:02:12Z |
| last_indexed |
2025-07-07T19:02:12Z |
| _version_ |
1837016054606331904 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2014, vol. 10, No. 3, pp. 309–319
Refinement of Isoperimetric Inequality of Minkowski
with the Account of Singularities in Boundaries of
Intrinsic Parallel Bodies
V.I. Diskant
Cherkasy State Technologic University
460 Shevchenko Blvd., Cherkasy 18006, Ukraine
E-mail: diskant@chiti.uch.net
Received May 14, 2013, revised December 23, 2013
The following inequalities are proved:
Sn(A,B) ≥ nn
k−1∑
i=0
V (BAi)
(
V n−1(Ai)− V n−1(Ai+1)
)
+ Sn(A−T (B), B),
Sn(A,B) ≥ nn
T∫
0
g(t)df(t) + Sn(A−T (B), B),
Sn(A,B) ≥ nn
q∫
0
g(t)df(t) + Sn(A−q(B), B),
where V (A), V (B) stand for the volumes of convex bodies A and B in Rn
(n ≥ 2), S(A,B) denotes the area of the surface of the body A relative to
the body B, q is the capacity factor of the body B with respect to the body
A, Ai = A−ti(B) = A/(tiB) is the inner body parallel to the body A with
respect to the body B at a distance ti, 0 = t0 < t1 < . . . < ti < . . . < tk−1 <
tk = T < q, BAi is a shape body of Ai relative to B, g(t) = V (BA−t(B)),
f(t) = −V n−1(A−t(B)),
T∫
0
g(t)df(t) is the Riemann–Stieltjes integral of the
function g(t) by the function f(t), and
q∫
0
g(t)df(t) = lim
T→q
T∫
0
g(t)df(t).
Key words: convex body, isoperimetric inequality, Minkowski inequality.
Mathematics Subject Classification 2010: 53B50.
c© V.I. Diskant, 2014
V.I. Diskant
By a convex body in the n-dimensional Euclidean space Rn (n ≥ 2) we mean
a convex compact set with non-empty interior.
Let A and B be convex bodies in Rn, A+λB with λ ≥ 0 stand for their linear
combination in Minkowski’s sense, E be a unit ball whose center coincides with
the origin ō of the coordinate system in Rn, Ω denote the boundary sphere of the
ball E.
The volume V (A + λE), λ ≥ 0, is expressed by the Steiner formula
V (A + λE) =
n∑
k=0
Ck
nVn−k(A)λk,
where Vn−k(A) is the (n − k)-th basic measure of the body A. In particular,
Vn(A) = V (A), whereas nVn−1(A) = S(A) is the area of the boundary surface of
the body A [1, p. 176]. It follows from the Steiner formula that
S(A) = lim
λ→+0
V (A + λE)− V (A)
λ
.
Minkowski [2, p. 57] obtained a generalization of the Steiner formula, which
reads as follows:
V (A + λB) =
n∑
k=0
Ck
nVk(A,B)λk,
where Vk(A,B) is the k-th mixed volume of the bodies A and B, V0(A, B) =
V (A). It follows from the Minkowski formula that
nV1(A,B) = lim
λ→+0
V (A + λB)− V (A)
λ
.
Since nV1(A,E) = S(A), it is natural to call nV1(A,B) the area of the surface
of the body A relative to the body B and denote it by S(A,B).
If A is a polyhedron with k facets, then by Minkowski (see [3, p. 61]), its first
mixed volume is expressed by the formula
V1(A,B) =
1
n
k∑
i=1
S(Ai)hB(ûi), (1)
where ūi (i = 1, 2, . . . , k) are outside unit vectors normal to the facets of A, S(Ai)
is the area of the i-th facet of A, hB(ûi) is the support value of the body B with
respect to the vector ūi.
310 Journal of Mathematical Physics, Analysis, Geometry, 2014, vol. 10, No. 3
Refinement of Isoperimetric Inequality of Minkowski...
In the general case, an expression for the first mixed volume, obtained by
A.D. Aleksandrov [4, p. 39], has the form
V1(A,B) =
1
n
∫
Ω
hB(ū)F (A, dω),
where F (A,ω) is the surface function of the body A, ω denotes a domain in Ω.
Recall the first inequality of Minkowski for mixed volumes (see [3, p. 65]),
V n
1 (A,B) ≥ V (B)V n−1(A). (2)
The equality holds if and only if A is positively homothetic to the body B, i.e.,
A = kB, k > 0.
The isoperimetric inequality of Minkowski,
Sn(A,B) = (nV1(A,B))n ≥ nnV (B)V n−1(A), (3)
is virtually the same as (2). The equality holds in (3) if and only if it holds in
(2). Besides, if B = E, then (3) leads to the following classical isoperimetric
inequality:
Sn(A) ≥ nnV (E)V n−1(A). (4)
Let us recall some notions. The Minkowski difference D = A/B of the convex
bodies A and B is defined as the set of all points d̄ ∈ Rn such that d̄ + B ⊂ A
(see [1, p. 83]). It is known that the body D is convex. Moreover, if the origin
in Rn is moved, then D is subject to a parallel translation. The capacity factor
q = q(A,B) of the body B with respect to the body A is defined as the greatest
number α such that the body αB can be placed inside A by a parallel translation
[5, p. 100]. Given 0 ≤ σ ≤ q, the body A−σ(B) = A/(σB) is called the intrinsic
body parallel to the body A relative to the body B at the distance σ.
In [4, p. 97], A.D. Alexandrov introduced the notion of a convex body with
a given domain of definition of the support function. Namely, let Ω′ be a closed
subset of the unit sphere Ω which does not belong to any closed hemisphere of Ω.
Let H∗(ū) be a continuous positive function defined in Ω′. Given a vector ū ∈ Ω′,
consider a hyperplane T (ū) in Rn orthogonal to ū and lying at the distance H∗(ū)
from the origin ō in the direction of ū. Denote by T (ū) the closed half-space in
Rn bounded by T (ū) and containing the origin. By A.D. Alexandrov, the convex
body N = ∩ū∈Ω′T (ū) is called a convex body with the definition domain Ω′ of
the support function. In what follows, we will use the notation N = (Ω′,H∗(ū)).
When dealing with functionals invariant under parallel translations, the choice
of the origin in Rn does not matter. Hence we will assume that qB ⊂ A and that
the origin ō belongs to the interior of B. Let HA(ū) denote the support function
of the body A. Then A = (Ω,HA(ū)) since HA(ū) is continuous and positive in
Journal of Mathematical Physics, Analysis, Geometry, 2014, vol. 10, No. 3 311
V.I. Diskant
Ω. It is shown in [5, p. 107] that for A there exists a minimal domain Ω′ = ΩA
such that A = (ΩA, H∗
A(ū)), where H∗
A(ū) is the restriction to ΩA of the support
function HA(ū) of the body A. For example, if A is a polyhedron in Rn, then
ΩA is just a set of outward unit vectors normal to the facets of A. The body
BA = (ΩA,H∗
B(ū)) is called the shape body of the body A relative to the body
B (see [5, p. 108]). Note that B ⊂ BA. Remarkably, the body BA takes into
account the singularities in the boundary of the body A relative to the body B.
G. Hadwiger [6, p. 368] obtained the following refinement of the classical
isoperimetric inequality (4):
Sn(A) ≥ nnV (EA)V n−1(A), (5)
where EA = (ΩA,H∗
E(ū)). In [5, 108], inequality (5) was generalized by taking
into account singularities in the boundary of the body A relative to the body B,
Sn(A,B) ≥ nnV (BA)V n−1(A). (6)
In [7, p. 43], inequality (6) was refined by taking into account the non-
degeneracy of A−q(B),
Sn(A,B) ≥ nnV (BA)V n−1(A) + Sn(A−q(B), B). (7)
Now let T be an arbitrary number satisfying 0 < T < q. Divide the segment
[0, T ] into k parts by points 0 = t0 < t1 < . . . < ti < ...tk = T . Denote
Ai = A−ti(B), 1 ≤ i ≤ k − 1, so A0 = A, Ak = A−T (B).
The aim of the paper is to prove the following.
Theorem 1. The following inequality holds true, which refines inequality
(6) by taking into account singularities in the boundaries of the bodies A1, A2,
. . . , Ak−1:
Sn(A,B) ≥ nn
k−1∑
i=0
V (BAi)
(
V n−1(Ai)− V n−1(Ai+1)
)
+ Sn(A−T (B), B).
Theorem 2. The following inequality holds true:
Sn(A,B) ≥ nn
T∫
0
g(t)df(t) + Sn(A−T (B), B),
where g(t) = V (BA−t(B)), f(t) = −V n−1(A−t(B)),
T∫
0
g(t)df(t) is the Riemann–
Stieltjes integral of g(t) by f(t).
312 Journal of Mathematical Physics, Analysis, Geometry, 2014, vol. 10, No. 3
Refinement of Isoperimetric Inequality of Minkowski...
Theorem 3. The following inequality holds true:
Sn(A,B) ≥ nn
q∫
0
g(t)df(t) + Sn(A−q(B), B), (8)
where
q∫
0
g(t)df(t) = lim
T→q
g(t)df(t).
P r o o f of Theorem 1. To prove Theorem 1, we will apply Lemma 1 obtained
in [7, p. 43].
Lemma 1. The following inequality holds true for any 0 < ρ < q:
V n
1 (A, B)− V (BA)V n−1(A) ≥ V n
1 (A−ρ(B), B)− V (BA)V n−1(A−ρ(B)).
Putting ρ = t1 in this inequality, we rewrite it as follows:
V n
1 (A,B) ≥ V (BA)
(
V n−1(A)− V n−1(A1(B))
)
+ V n
1 (A1(B), B). (9)
Next, replacing A by A1 and A1 by A2, we have
V n
1 (A1, B) ≥ V (BA1)
(
V n−1(A1)− V n−1(A2(B))
)
+ V n
1 (A2(B), B).
Substituting this lower bound for V n
1 (A1, B) in (9), we obtain the inequality
V n
1 (A,B) ≥ V (BA)
(
V n−1(A)− V n−1(A1(B))
)
+
+V (BA1)
(
V n−1(A1)− V n−1(A2(B))
)
+ V n
1 (A2(B), B). (10)
Similarly, using (9), one can obtain lower bounds for V n
1 (A2, B), V n
1 (A3, B), . . . ,
and substitute these bounds step by step into (10). After k steps, this iterative
procedure results into the desired inequality given in Theorem 1, q.e.d.
The proofs of Theorems 2 and 3 will be preceded by five lemmas.
Lemma 2 [7]. If the inequality
HL(ū) ≤ H∗(ū)
holds true for any ū ∈ Ω′, then L ⊂ L̄ ⊂ N .
Next, let C = A−σ(B) = (Ω′,H∗
C(ū)), where σ ∈ [0, q] is a fixed number.
Consider the body (Ω′,H∗
B(ū)) and the function H∗
t (ū) = H∗
C(ū)−tH∗
B(ū), ū ∈ Ω′,
t ∈ (0, q − σ). Since the support function of a convex body is continuous in Ω,
the function H∗
t (ū) is continuous in ω̄ ∈ Ω′. Moreover, the inclusion qB ⊂ A,
the choice of the origin o and the inequality σ + t < q together imply that
Journal of Mathematical Physics, Analysis, Geometry, 2014, vol. 10, No. 3 313
V.I. Diskant
H∗
t (ū) = H∗
C(ū)− th∗B(ū) > 0 holds for any ū ∈ Ω′. Therefore the function H∗
t (ū)
generates a convex body Nt = (Ω′,H∗
t (ū)).
Lemma 3. Given the convex bodies C and B, the following equality holds
true for any 0 < t < q − σ:
Nt = C−t(B̄) = C−t(B).
P r o o f. Let us show that Nt ⊂ C−t(B̄) = C/(tB̄). Let ā be a point of Nt.
Then Hā(ū) ≤ H∗
t (ū) holds for any ū ∈ Ω′. The support function Hā+tB̄(ū) of
the body ā + tB̄ satisfies the inequality
Hā+tB̄(ū) = Hā(ū) + tHB̄(ū) ≤ H∗t(ū) + tHB̄(ū) = H∗
C(ū), ū ∈ Ω′.
Then it follows from Lemma 2 that ā + tB̄ ⊂ C. Therefore, ā ∈ C/(tB̄). Thus,
Nt ⊂ C−t(B̄) = C/(tB̄).
Now let us show that C−t(B) ⊂ Nt. Let b̄ be a point not belonging to
Nt. Then there exists ū0 ∈ Ω′ such that Hb̄(ū0) > H∗
t (ū0). Consequently,
Hb̄(ū0) + tHB(ū0) = Hb̄(ū0) + tH∗
B(ū0) > H∗
C(ū0). This means that the body
b̄ + tB does not belong to C. Hence b̄ does not belong to C−t(B) = C/(tB).
Thus, C−t(B) ⊂ Nt.
The inclusions Nt ⊂ C−t(B̄) ⊂ C−t(B) ⊂ Nt result in the desired equalities
Nt = C−t(B̄) = C−t(B), q.e.d.
Lemma 4. The value of V
(
BA−σ(B)
)
is finite positive for any 0 ≤ σ < q.
P r o o f. Consider the shape body BA−σ(B) of the body A−σ(B) relative to
the body B. The support function of this shape body is defined in the domain
ΩA−σ(B), which is the minimal domain of definition of the support function of the
body A−σ(B). The minimal domain of definition ΩA for the support function
of the convex body A contains the set of all outward unit vectors normal to the
support planes of the body A at regular points of the surface of A. Recall that
a point in the boundary of A is called regular if there exists a unique support
plane of A passing through this point. In [6, p. 368], it is shown that ΩA can not
belong to any closed hemisphere of the unit sphere Ω. Hence the shape body of
the convex body A relative to the convex body B is a convex body and it has a
finite positive volume, q.e.d.
Lemma 5. Let 0 ≤ σ1 < σ2 < q. Then the following inclusions hold true:
ΩA−σ1 (B) ⊃ ΩA−σ2 (B), BA−σ1(B) ⊂ BA−σ2(B). Moreover, the function g(σ) =
V (BA−σ(B)) is increasing for any σ ∈ [0, q).
P r o o f. Let Ω′ be a domain of definition for the support function of the
body A−σ1(B). It follows from Lemma 3 that Ω′ is a domain of definition for the
support function of the body A−σ2(B) since σ2 = σ1 + t, where t = σ2 − σ1 > 0.
314 Journal of Mathematical Physics, Analysis, Geometry, 2014, vol. 10, No. 3
Refinement of Isoperimetric Inequality of Minkowski...
Replace Ω′ by the minimal domain of definition ΩA−σ1 (B) of the support function
of the body A−σ1(B). It follows from Lemma 3 that ΩA−σ1 (B) is a domain of
definition for the support function of the body A−σ2(B). Therefore, ΩA−σ1 (B) ⊃
ΩA−σ2 (B).
Let us prove the inclusion BA−σ1(B) ⊂ BA−σ2(B). Actually, BA−σ1 (B) (respec-
tively, BA−σ2 (B)) is the intersection of the closed half-spaces supporting the body
B whose outward unite normal vectors, translated to ō, belong to ΩA−σ1 (B) (re-
spectively, to ΩA−σ2(B)). Since ΩA−σ2 (B) ⊂ ΩA−σ1(B), then BA−σ1 (B) ⊂ BA−σ2 (B).
Moreover, V (BA−σ1 (B)) ≤ V (BA−σ2 (B)). Therefore, the function g(σ) =
V (BA−σ(B)), defined for σ ∈ [0, q), is increasing, q.e.d.
Set f1(σ) = V (A−σ(B)).
Lemma 6. The function f1(σ), defined for σ ∈ [0, q), is continuous and
decreasing everywhere in [0, q).
P r o o f. Along with the body N = (Ω′,H∗(ū)), let us consider a family
of bodies Nt = (Ω′,H∗(ū) + tδH∗(ū)), where δH∗(ū) is a continuous function
defined for ū ∈ Ω′. A.D. Alexandrov proved that the first variation of the volume
V (N), i.e.,
δV (N) = lim
t→0
V (Nt)− V (N)
t
,
is equal to
δV (N) =
∫
Ω′
δH∗(ū)F (N, dω),
where F (N, dω) is a surface measure of the body N which satisfies
F (N,Ω − Ω′) = 0 (see [4, p. 100–101]). It is shown in [7, p. 44] that Nσ =
(Ω′,H∗
σ(ū)) = A−σ(B). Applying the above to the body Nσ and to the family of
bodies (Nσ)t =
(
Ω′, H∗
σ+t(ū)
)
= (Ω′,H∗
σ(ū)− tH∗
B(ū)), we get
dV (A−σ(B))
dσ
= −
∫
Ω′
H∗
B(ū)F (A−σ(B), dω) = −nV1(A−σ(B), B), σ ∈ [0, q).
Therefore, the function f1(σ) has a finite derivative at the interval (0, q) and
a finite right-hand side derivative at σ = 0, which is actually equal to nV1(A, B).
This leads to the continuity of f1(σ) in [0, q).
It follows from (2) that the first mixed volume V1(A−σ(B), B) of A−σ(B) and
B is positive. Hence, dV (A−σ(B))
dσ < 0 for σ ∈ [0, q), q.e.d.
Journal of Mathematical Physics, Analysis, Geometry, 2014, vol. 10, No. 3 315
V.I. Diskant
P r o o f of Theorem 2. Let us show that the integral I =
T∫
0
g(t)df(t),
where 0 < T < q, g(t) = V (BA−t(B)), f(t) = −V n−1(A−t(B)), i.e., the Riemann–
Stieltjes integral of g(t) relative to f(t) (see [8, p. 201]), exists.
By Lemma 5, the function g(t) = V (BA−t(B)) is increasing in the interval
t ∈ [0, T ]. Therefore, g(t) is a function of finite variation in [0, T ]. Besides, in
view of Lemma 6, the function f1(t) = V (A−t(B)) is continuous in the interval
t ∈ [0, q). Thus the function −V n−1(A−t(B)) is continuous in [0, T ].
In [8, p. 204], it is proved that any function of finite variation is integrable
with respect to any continuous function. Therefore the integral I =
T∫
0
g(t)df(t)
exists.
Consider a Riemann sum of the integral in question which corresponds to a
partition of the segment [0, T ] by points 0 = t0 < t1 < . . . < Tk−1 < tk = T with
ξi = ti. This Riemann sum has the form
σ =
k−1∑
i=0
V (BAi)
(−V n−1(Ai+1)− (−V n−1(Ai))
)
.
Hence the statement of Theorem 1 can be rewritten as follows:
Sn(A,B) ≥ nnσ + Sn(A−T (B), B).
Applying an appropriate choice both of a sufficiently fine partition of the
segment [0, T ] into segments [ti, ti+1] and of points ξi ∈ [ti, ti+1], one can get
|I − σ| < ε for any arbitrary ε > 0. Thus Theorem 2 is proved.
P r o o f of Theorem 3. Let ϕ(T ) =
T∫
0
g(t)df(t), ψ(T ) = Sn(A−T (B), B).
Then the inequality in Theorem 2 can be rewritten as follows:
Sn(A,B) ≥ nnϕ(T ) + ψ(T ), 0 ≤ T < q. (11)
Given 0 ≤ t1 < t2 ≤ q, we have A−t2(B) ⊂ A−t1(B). Indeed, if a point
a2 belongs to A−t2(B), then a2 + t2B ⊂ A. Moreover, since t1B ⊂ t2B, then
a2 + t1B ⊂ A. Therefore, a2 belongs to A−t1(B), so A−t2(B) ⊂ A−t1(B).
Because the mixed volume is monotone with respect to any of its arguments
and non-negative [2, p. 49], the function ψ(T ) is decreasing for 0 ≤ T < q, and
its minimal value is equal to S(A−q(B), B).
From Lemma 4, it follows that g(t) = V (BA−t(B)) is a finite positive number
for any t ∈ [0, T ]. Moreover, it follows from Lemma 6 that f(t) = −V n−1(A−t(B)
is increasing for any t ∈ [0, T ]. Therefore, the integral sum ϕ(T ) is non-negative
and it increases if T increases. Because the summands in the right-hand side of
316 Journal of Mathematical Physics, Analysis, Geometry, 2014, vol. 10, No. 3
Refinement of Isoperimetric Inequality of Minkowski...
(11) are non-negative, any of them is less than or equal to the left-hand side of
(11). Therefore, Sn(A,B) ≥ nnϕ(T ), and hence ϕ(T ) has a limit value at t → q.
Denote this limit value by lim
T→q
ϕ(T ) =
q∫
0
g(t)df(t). Then, passing to the limit in
the right-hand side of (11), we get (8), q.e.d.
Let us give an example when
q∫
0
g(t)df(t) exists and (8) refines (7).
E x a m p l e. We will consider convex polygons in the plane (see Fig. 1)
where n = 2. Let B be an isosceles right triangle, k = 3, assuming that the origin
ō belongs to B. Let A be a hexagon bounded by the broken line abcdef . Clearly,
A = A0(B) = A/(0B) = A/ō. Moreover, A and B are chosen in such a way that
q = 4.
Fig. 1. The isosceles right triangle B, hexagon A,
and the sequence of shape bodies.
To provide a partition of the segment [0, q], we chose t1 = 2, t2 = 3, t3 = 4.
Then A1 = A−2(B) = A/(2B) = A/(2BA) is a pentagon bounded by the broken
line ab1c1d1e1. Next, A2 = A−3(B) = A/(3B) = A1/B = A1/BA1 is a parallel-
ogram bounded by a broken line a2b2c2d2. Finally, A3 = A−4(B) = A/(4B) =
A2/B = A2/BA2 is a segment o1o2.
Journal of Mathematical Physics, Analysis, Geometry, 2014, vol. 10, No. 3 317
V.I. Diskant
Let us describe the minimal domains of definition for support functions of the
planar convex polygons in question. All these domains belong to the unite circle
Ω centered at the origin o. We have the following.
ΩA consists of six points in Ω, which are the end-points of outward normal
unit vectors to the sides of the hexagon abcdef .
ΩA1 consists of five points in Ω, which are the end-points of outward normal
unit vectors to the sides of the pentagon ab1c1d1e1. ΩA1 is the same as ΩA
excluding the end-point of the outward normal unit vector to the side ab.
ΩA2 differs from ΩA1 by one point, the end-point of the outward normal unit
vector to the side c1d1.
Ωo1o2 consists of two points in Ω, which are the end-points of outward normal
unit vectors to the segment o1o2.
ΩA−t(B) = ΩA for any t ∈ [0, 2). Hence the shape body for any ΩA−t(B),
t ∈ [0, 2), is BA.
ΩA−t(B) = ΩA1 for any t ∈ [2, 3). Hence the shape body for any ΩA−t(B),
t ∈ [2, 3), is BA1 .
ΩA−t(B) = ΩA2 for any t ∈ [3, 4). Hence the shape body for any ΩA−t(B),
t ∈ [3, 4), is BA2 .
Since n = 2, replace S by l and V by S in (8) to get
S(A) = 38, S(A1) = 15, S(A2) = 6,
S(BA) = 3
4 , S(BA1) = 1, S(BA2) = 2, |o1o2| = 4.
Applying (1) and the equality S(A,B) = nV1(A,B), we have
S(A,B) = 2 V1(A,B) = 2
(
1
2
6∑
I=1
aihB(ūi)
)
= |ab| · 0 + |bc| ·
√
2
2
+ |cd| · 1 + |de| ·
√
2
2
+ |ef | ·
√
2
2
+ |fa| · 0 = 13,
S(A−q(B), B) = |o1o2| · 1 + |o2o1| · 0 = 4.
Therefore, the left-hand side of (8) is equal to l2(A, B) = 132 = 169.
On the other hand, g(t) is a step-like function,
g(t) =
S(BA) = 3
4 , t ∈ [0, 2);
S(BA1) = 1, t ∈ [2, 3);
S(BA2) = 2, t ∈ [3, 4).
318 Journal of Mathematical Physics, Analysis, Geometry, 2014, vol. 10, No. 3
Refinement of Isoperimetric Inequality of Minkowski...
Hence the right-hand side of (8) is equal to
4
4∫
0
g(t)df(t) + S2(A−q(B), B)
= 4
2∫
0
g(t)df(t) +
3∫
2
g(t)df(t) +
4∫
4
g(t)df(t)
+ S2(A−q(B), B)
= 4
(
3
4
(38− 15) + 1 · (15− 6) + 2 · 6
)
+ 42 = 169.
The left-hand side of (7) also equals 169, whereas the right-hand side of (7)
is equal to 130. Hence (8) refines (7).
References
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[2] T. Bonnesen and W. Fenchel, Theory of Convex Bodies. BSC Associates, Moscow,
Idaho, 1987.
[3] H. Busemann, Convex Surfaces. Interscience Tracts in Pure and Applied Mathe-
matics. 6. New York–London: Interscience Publishers, 1958.
[4] A.D. Alexandrov, Selected Works, Volume 1. Geometry and Applications Nauka,
Novosibirsk, 2006. (Russian)
[5] V.I. Diskant, Refinements of Isoperimetric Inequality and Stability Theorems in
the Theory of Convex Bodies. — Modern Problems of Geometry and Analysis 14
(1989), 98–132. (Russian)
[6] H. Hadwiger, Vorlesungen uber Inhalt, Oberflache und Isoperimetrie.
Die Grundlehren der Mathematischen Wissenschaften. 93. Berlin–Gottingen–
Heidelberg, Springer–Verlag, 1957. (German).
[7] V.I. Diskant, The Behavior of Izoperemetric Difference at the Transition to a Parallel
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Anal., Geom. 10 (2003), No. 1, 40–48.
[8] I.P. Natanson, Theory of Functions of a Real Variable. F. Unger Publishing Co,
New York, 1964.
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