Successive radii and Orlicz Minkowski addition

0.19 MB·English

by  Fangwei Chen

SUCCESSIVE RADII AND ORLICZ MINKOWSKI ADDITION
1 2 2
FANGWEI CHEN , CONGLI YANG , MIAO LUO ,
Abstract. In this paper, we deal with the successive inner and outer radii
5
with respect to Orlicz Minkowski sum. The upper and lower bounds for the
1
radii of the Orlicz Minkowski sum of two convex bodies are established.
0
2

n
Let n denotethesetofconvexbodies(compact, convexsubsetswithnonempty
a
J interioKrs) in Euclidean n-space, Rn. Let , and denote the standard inner
2
7 product and the Euclidean norm in Rn, rhe·sp·iectivekly·.kDenote by e (i = 1, ,n),
i
the orthogonal unit vectors in Rn. The n-dimensional unit ball and its bou··n·dary,
]
G i.e., the (n 1)-dimensional unit sphere is denoted by B and Sn 1, respectively.
n −
−
M The volume of a set K n, i.e., its n-dimensional Lebesgue measure is
. denoted by K . The set of a∈ll iK-dimensional linear subspaces of Rn is denoted by
h | |
n. For L n, L denotes its orthogonal complement. Let K n, L n,
at Li ∈ Li ⊥ ∈ K ∈ Li
the projection of K onto L is denoted by K L.
m
|
Following the traditional notations, we use D(K), ω(K), R(K) and r(K) to
[
denote the diameter, minimal width, circumradius and inradius of a convex body

1
K, respectively. The behavior of the diameter, minimal width, circumradius and
v
0 inradius with respect to the Minkowski sum is well known(see [25]), namely
9
4 D(K +K ) D(K)+D(K ), ω(K +K ) ω(K)+ω(K ),
′ ′ ′ ′
6 ≤ ≤
0 R(K +K ) R(K)+R(K ), r(K +K ) r(K)+r(K ).
′ ′ ′ ′
. ≤ ≤
2
0 Let K n, and i = 1,2, ,n, the successive outer and inner radii of K are
5 ∈ K ···
defined as
1
:
v R (K) = min R(K L), r (K) = maxmaxr(K (x+L) : x+L).
i i
Xi L∈Lni | L∈Lni x∈L⊥ ∩
r Notice that R (K) is the smallest radius of a solid cylinder with i-dimensional
a i
spherical cross section containing K, and r (K) is the radius of the greatest i-
i
dimensional ball contained in K. It is clear that the outer radii are increasing in
i, whereas the inner radii are decreasing in i, and the following hold (see [3]).
ω(K) D(K)
R (K) = R(K), R (K) = , r (K) = r(K), r (K) = .
n 1 n 1
2 2
2010 Mathematics Subject Classification. 52A20, 52A40, 52A38.
Key words and phrases. Orlicz Minkowski sum, asymmetric Orlicz zonotopes, Shadow sys-
tem, volume product, volume ratio.
The work is supported in part by CNSF (Grant No. 11161007, Grant No. 11101099),
Guizhou (Unite) Foundation for Science and Technology (Grant No. [2014] 2044, No. [2012]
2273, No. [2011] 16), Guizhou Technology Foundation for Selected Overseas Chinese Scholar
and Doctor foundation of Guizhou Normal University.
1
2 F. CHEN, C. YANG,M. LUO
The first systematic study of the successive radii was developed in [1], and one
can refer [2,7,13,14,16,19–21] and references within for more details.
TheradiiofconvexbodieswhichconnectedtheMinkowski sum(orL -Minkowski
p
sum) are studied by Gonz´alez and Hern´andez Cifre [13,14].
Beginning with the articles [18,23,24] of Haberl, Lutwak, Yang ang Zhang, a
more wide extension of the L Brunn-Minkowski theory emerged. Recently, in
p
a paper of Gardner, Hug and Weil [12], a systematic studies are made on the
Orlicz Minkowski addition, the Orlicz Brunn-Minkowski inequality and Orlicz
Minkowski inequality are obtained. The Orlicz Brunn-Minkowski theory are es-
tablished. See, e.g., [4–6,9,10,12,18,22,26,27] about the Orlicz Brunn-Minkowski
theory. In this context, the main goal of this paper is to seek the relations of the
radii for Orlicz Minkowski sum.
Throughoutthispaper, let betheclassofconvex, strictlyincreasing functions
C
ϕ : [0, ) [0, ) satisfying ϕ(0) = 0 and ϕ(1) = 1. Here the normalization is
∞ → ∞
a matter of convenience and other choices are possible.
Let n denote the class of convex bodies containing the origin. K,L n,
Ko ∈ Ko
ϕ , the Orlicz Minkowski sum of K and L is the convex body K + L with
ϕ
∈ C
support function
h (x) h (x)
K L
h (x) = inf λ > 0 : ϕ +ϕ 1 .
K+ϕL λ λ ≤
(cid:26) (cid:18) (cid:19) (cid:18) (cid:19) (cid:27)
If ϕ(t) = tp, p 1, then K + L is precisely the L Minkowski sum K + L.
ϕ p p
≥
For the successive outer radii of Orlicz Minkowski sum, we establish the fol-
lowing theorem.
Theorem 0.1. Let ϕ and K,K n. Then
∈ C ′ ∈ Ko
2ϕ 1(1/2)R (K + K ) R (K)+R (K ), (0.1)
− 1 ϕ ′ 1 1 ′
≥
2√2ϕ 1(1/2)R (K + K ) R (K)+R (K ), i = 2, ,n. (0.2)
− i ϕ ′ i i ′
≥ ···
All inequalities are best possible.
We also prove that there is non-existence the reverse inequalities for the suc-
cessive outer radius excepted R . That is
n
Proposition 0.2. Let ϕ and K,K n, for i = 1, n 1, there exists no
∈ C ′ ∈ Ko ··· −
constant c > 0 such that
cR (K + K ) R (K)+R (K ).
i ϕ ′ i i ′
≤
Similarly, for the successive inner radii of Orlicz Minkowski sum, we obtain
Theorem 0.3. Let ϕ and K,K n. Then
∈ C ′ ∈ Ko
2ϕ 1(1/2)r (K + K ) r (K)+r (K ), (0.3)
− 1 ϕ ′ 1 1 ′
≥
2√2ϕ 1(1/2)r (K + K ) r (K)+r (K ), i = 1, ,n 1. (0.4)
− i ϕ ′ i i ′
≥ ··· −
All inequalities are best possible.
The analogous Proposition 0.2 for the successive inner radii are
SUCCESSIVE RADII AND ORLICZ MINKOWSKI ADDITION 3
Proposition 0.4. Let ϕ and K,K n, for i = 2, ,n, there exists no
∈ C ′ ∈ Ko ···
constant c > 0 such that
cr (K + K ) r (K)+r (K ).
i ϕ ′ i i ′
≤
Ifwe takeϕ(t) = tp, p 1,these results areproved by Gonz´alezandHern´andez
≥
Cifre (see [14]). Specially, if p = 1, it was shown in [13].
The second part of our result is regard the Orlicz difference body. We obtain
the following:
Theorem 0.5. Let ϕ , and K n, for all i = 1,2 ,n, then
∈ C ∈ Ko ···
√2 i+1
R (K) R (K + ( K)) 2R (K), (0.5)
2ϕ 1(1/2) i i ≤ i ϕ − ≤ i
− r
1
r (K) r (K + ( K)) < 2(i+1)r (K). (0.6)
ϕ 1(1/2) i ≤ i ϕ − i
−
Ifwe takeϕ(t) = tp, p 1, these results areproved byGonz´alezandHern´andez
≥
Cifre (see [13,14]).
The paper is organized as follows. In section 1, we introduce the Orlicz
Minkowski sum and show some of their properties. The proof of the results
of successive outer and inner radii for Orlicz Minkowski sum are given in Section
2. Section 3 deals with the successive radii for Orlicz difference body.
1. Orlicz Minkowski addition
In this section, some basic definitions and notations about Orlicz Minkowski
sum and some of their properties are introduced.
Let ϕ , x = (x ,x , ,x ) Rn, the Orlicz norm x of a point x Rn
1 2 n ϕ
∈ C ··· ∈ k k ∈
is defined as
n
x
x = inf λ > 0 : (x ,x , ,x ) Rn, ϕ | i| 1 . (1.1)
ϕ 1 2 n
k k ··· ∈ λ ≤
( )
i=1 (cid:18) (cid:19)
X
Note that, if take ϕ(t) = tp, then is precisely the L norm , if p = 2, it
ϕ p p
k·k k·k
is the Euclidean norm .
2
Let x Rn, then k·k
∈
x 0,
ϕ
k k ≥
and for c > 0, we have
cx = c x .
ϕ ϕ
k k k k
The Orlicz ball is defined as
Bϕ = x = (x , ,x ) Rn : x 1 .
n { 1 ··· n ∈ k kϕ ≤ }
We have the following Lemma.
Lemma 1.1. Let ϕ ,ϕ , x Rn. If ϕ ϕ , then
1 2 1 2
∈ C ∈ ≤
x x , and Bϕ2 Bϕ1.
k kϕ1 ≤ k kϕ2 n ⊆ n
4 F. CHEN, C. YANG,M. LUO
Proof. Let x Rn, set x = λ (i = 1, 2), by the definition (1.1), we have
∈ k kϕi i
ni=1ϕi |λxii| ≤ 1. Since ϕ is strictly increasing, then λ → ni=1ϕ |xλi| is
strictly d(cid:16)ecre(cid:17)asing, so x = λ if and only if (cid:16) (cid:17)
P k kϕ i P
n
x
i
ϕ | | = 1.
i
λ
i=1 (cid:18) i (cid:19)
X
Note that ϕ ϕ , then
1 2
≤
n
x
i
ϕ | | 1.
2
λ ≥
i=1 (cid:18) 1 (cid:19)
X
By the definition of x we have x x .
k kϕ k kϕ1 ≤ k kϕ2
Let x ∂Bϕ2, then x = 1, by (1.1), we have
∈ n k kϕ2
x x = 1.
k kϕ1 ≤ k kϕ2
Then we have x Bϕ1, so we complete the proof. (cid:3)
∈ n
Let K,L n, ϕ , the Orlicz Minkowski sum of K and L is the convex
∈ Ko ∈ C
body K + L, with support function
ϕ
h (x) h (x)
K L
h (x) = inf λ > 0 : ϕ +ϕ 1 .
K+ϕL λ λ ≤
(cid:26) (cid:18) (cid:19) (cid:18) (cid:19) (cid:27)
Since ϕ is strictly increasing, then
h (x) h (x)
K L
λ ϕ +ϕ ,
→ λ λ
(cid:18) (cid:19) (cid:18) (cid:19)
is strictly decreasing. So, equivalently, h (u ) = λ if and only if
K+ϕL 0 0
h (u ) h (u )
K 0 L 0
ϕ +ϕ = 1. (1.2)
λ λ
(cid:18) 0 (cid:19) (cid:18) 0 (cid:19)
For the body K + L, we have the following results.
ϕ
Theorem 1.2. Let ϕ and ϕ ,ϕ in , for K,L n, then
1 2 C ∈ Ko
(i). If ϕ ϕ for all x [0,1], then K + L K + L,
1 ≤ 2 ∈ ϕ1 ⊆ ϕ2
(ii). For ϕ , then 1 K +L K + L K +L,
∈ C 2ϕ−1(1/2) ⊆ ϕ ⊆
(iii). conv(K L) K + L,
ϕ
∪ ⊆
(iv). K + L 1 conv(K L).
ϕ ⊆ ϕ−1(1/2) ∪
Proof. In order to prove (i), we only need to show
h (u) h (u), for u Sn 1.
K+ϕ1L ≤ K+ϕ2L ∈ −
By formula (1.2), we have
h (u) h (u)
K L
ϕ +ϕ = 1,
1 1
h (u) h (u)
(cid:18) K+ϕ1L (cid:19) (cid:18) K+ϕ1L (cid:19)
h (u) h (u)
K L
ϕ +ϕ = 1.
2 2
h (u) h (u)
(cid:18) K+ϕ2L (cid:19) (cid:18) K+ϕ2L (cid:19)
SUCCESSIVE RADII AND ORLICZ MINKOWSKI ADDITION 5
Since ϕ ϕ , so
1 2
≤
h (u) h (u)
K L
ϕ +ϕ 1,
2 2
h (u) h (u) ≥
(cid:18) K+ϕ1L (cid:19) (cid:18) K+ϕ1L (cid:19)
which means
h (u) h (u).
K+ϕ1L ≤ K+ϕ2L
So K + L K + L .
ϕ1 ⊆ ϕ2
(ii). Let Id denote the identity function on [0,1], by the convexity of ϕ on
[0,1], for x [0,1], we have
∈
ϕ(x) = ϕ(x 1+(1 x)0) xϕ(1)+(1 x)ϕ(0),
· − ≤ −
sowe haveϕ(x) x, which meansthatϕ Id. Ontheother hand, whenϕ = Id,
≤ ≤
the Orlicz Minkowski sum K + L is precisely the Minkowski sum K +L. Now
ϕ
by (i), we have K + L K +L.
ϕ
⊆
For the left hand inclusion, let u Sn 1, by the definition of Orlicz Minkowski
−
∈
sum we have
h (u) h (u)
K L
ϕ +ϕ = 1.
h (u) h (u)
(cid:18) K+ϕL (cid:19) (cid:18) K+ϕL (cid:19)
The convexity of ϕ implies
h (u)+h (u)
K L
2ϕ 1.
2h (u) ≤
(cid:18) K+ϕL (cid:19)
Then, we have
1
(h (u)+h (u)) h (u).
2ϕ 1(1/2) K L ≤ K+ϕL
−
By the definition of Minkowski sum we have h (u) + h (u) = h (u), so we
K L K+L
obtain
1
h (u) h (u).
2ϕ 1(1/2) K+L ≤ K+ϕL
−
Which means 1 K +L K + L.
2ϕ−1(1/2) ⊆ ϕ
(iii). Note that h (u) = max h (u),h (u) . Then
conv(K L) K L
∪ { }
a), If h (u) = h (u), for some u Ω, then
conv(K L) K
∪ ∈
h (u) h (u) h (u)
K L L
ϕ +ϕ = ϕ(1)+ϕ 1.
h (u) h (u) h (u) ≥
(cid:18) conv(K L) (cid:19) (cid:18) conv(K L) (cid:19) (cid:18) K (cid:19)
∪ ∪
b), If h (u) = h (u), for some u Sn 1 Ω, then
conv(K L) L −
∪ ∈ \
h (u) h (u) h (u)
K L K
ϕ +ϕ = ϕ(1)+ϕ 1.
h (u) h (u) h (u) ≥
(cid:18) conv(K L) (cid:19) (cid:18) conv(K L) (cid:19) (cid:18) L (cid:19)
∪ ∪
So we obtain
h (u) h (u)
K L
ϕ +ϕ 1,
h (u) h (u) ≥
(cid:18) conv(K L) (cid:19) (cid:18) conv(K L) (cid:19)
∪ ∪
for u Sn 1. By the definition of Orlicz Minkowski sum we obtain
−
∈
h (u) h (u).
conv(K∪L) ≤ K+ϕL
6 F. CHEN, C. YANG,M. LUO
So we have conv(K L) K + L.
ϕ
∪ ⊆
(iv). Since
h (u) h (u)
K L
ϕ +ϕ = 1.
h (u) h (u)
(cid:18) K+ϕL (cid:19) (cid:18) K+ϕL (cid:19)
The increasing of ϕ implies
max h (u),h (u) max h (u),h (u)
K L K L
1 ϕ { } +ϕ { } .
≤ h (u) h (u)
(cid:18) K+ϕL (cid:19) (cid:18) K+ϕL (cid:19)
Then,
1
h (u) max h (u),h (u) .
K+ϕL ≤ ϕ 1(1/2) { K L }
−
So we have K + L 1 conv(K L). We complete the proof. (cid:3)
ϕ ⊆ ϕ−1(1/2) ∪
Let L n, and K n, the orthogonal projection of K onto L is denoted by
∈ Li ∈ K
K L.
|
For the Orlicz Minkowski sum we have the following lemma.
Lemma 1.3. Let ϕ , K,K n, and L n, then
∈ C ′ ∈ Ko ∈ Li
(K + K ) L = K L+ K L.
ϕ ′ ϕ ′
| | |
Proof. Let x Rn, the following fact is obvious,
∈
h (x L) = h (x). (1.3)
K+ϕK′ | (K+ϕK′)|L
By the definition of Orlicz Minkowski sum we have
h (x L) h (x L)
K K′
ϕ | +ϕ | = 1.
h (x L) h (x L)
(cid:18) K+ϕK′ | (cid:19) (cid:18) K+ϕK′ | (cid:19)
By (1.3) we have,
h (x) h (x)
K L K′L
ϕ | +ϕ | = 1.
h (x L) h (x L)
(cid:18) K+ϕK′ | (cid:19) (cid:18) K+ϕK′ | (cid:19)
On the other hand,
h (x) h (x)
K L K′L
ϕ | +ϕ | = 1.
h (x) h (x)
(cid:18) (K L+ϕK′L) (cid:19) (cid:18) (K L+ϕK′L) (cid:19)
| | | |
Comparing with the above two formulas shows
h (x) = h (x L) = h (x).
K|L+ϕK′|L (K+ϕK′) | (K+ϕK′)|L
Which means
(K + K ) L = K L+ K L.
ϕ ′ ϕ ′
| | |
(cid:3)
So we complete the proof.
If ϕ(t) = tp (p 1), these results is obtained by Firey [11].
≥
Let L n, write B = B L the unit ball contained in L. The following
∈ Li i,L n ∩
Lemma will be useful in the proof of the main results.
SUCCESSIVE RADII AND ORLICZ MINKOWSKI ADDITION 7
Lemma 1.4. Letϕ , e ,e be theorthogonalunit vectorsin Rn, L˜ = span e ,e ,
i j i j
∈ C { }
then
√2
[ e ,e ]+ [ e ,e ] B . (1.4)
− i i ϕ − j j ⊆ 2ϕ 1(1/2) 2,L˜
−
Proof. To prove (1.4), we only need to show
√2
h (u) , for u Sn 1.
[−ei,ei]+ϕ[−ej,ej] ≤ 2ϕ−1(1/2) ∈ −
For write simply, let h (u) = λ . By the Orlicz Minkowski sum we
[ ei,ei]+ϕ[ ej,ej] u
− −
have
h (u) h (u)
ϕ [−ei,ei] +ϕ [−ej,ej] = 1.
λ λ
(cid:18) u (cid:19) (cid:18) u (cid:19)
The symmetry of [ e ,e ]+ [ e ,e ] ensure us it is enough to discuss it’s support
i i ϕ j j
− −
function with the parameter θ on interval (0, π]. Let ω(θ) = eiθ, then
4
cosθ sinθ
ϕ +ϕ = 1,
λ λ
(cid:18) ω(θ)(cid:19) (cid:18) ω(θ)(cid:19)
where 0 < θ π. The increasing of ϕ and sinθ cosθ on interval (0, π] implies
≤ 4 ≤ 4
cosθ sinθ cosθ
1 = ϕ +ϕ 2ϕ .
λ λ ≤ λ
(cid:18) ω(θ)(cid:19) (cid:18) ω(θ)(cid:19) (cid:18) ω(θ)(cid:19)
Then we have
cosθ √2
λ .
ω(θ) ≤ ϕ 1(1/2) ≤ 2ϕ 1(1/2)
− −
For other intervals, by the symmetry of [ e ,e ]+ [ e ,e ], we have
i i ϕ j j
− −
ii:i:IIffπ4π<<θθ≤ π2π,,tthheennλλω(θ) ==λλω(π2−θ)..
2 ≤ ω(θ) ω(π−θ)
iii: If π < θ 2π, then λ = λ .
ω(θ) ω(2π θ)
≤ −
So we have λ √2 for 0 < θ 2π. We complete the proof. (cid:3)
ωθ ≤ 2ϕ−1(1/2) ≤
If ϕ(t) = tp, this result is obtained by Gordon and Junge [15].
2. Proof of main results
In this section, we give the proofs of the main results.
Proof of Theorem 0.1: By Theorem 1.2 we have 1 K +K K + K ,
2ϕ−1(1/2) ′ ⊆ ϕ ′
then
K +K 2ϕ 1(1/2)(K + K ).
′ − ϕ ′
⊆
So we have
2ϕ 1(1/2)R (K + K ) R (K +K ) R (K)+R (K ),
− 1 ϕ ′ 1 ′ 1 1 ′
≥ ≥
1
2ϕ 1(1/2)R (K + K ) R (K +K ) (R (K)+R (K )),
− i ϕ ′ i ′ i i ′
≥ ≥ √2
8 F. CHEN, C. YANG,M. LUO
i = 2, ,n, which shows the inequalities (0.1) and (0.2). Where we use the
···
fact R (K + K ) R (K) + R (K ) and R (K + K ) 1 (R (K) + R (K )),
1 ′ ≥ 1 1 ′ i ′ ≥ √2 i i ′
i = 2, ,n (see [13]).
···
To show theinequalities (0.1) and(0.2) are best possible, we find convex bodies
satisfy the equality conditions.
For the equality of (0.1), let K = K , we have
′
1
K + K = K. (2.1)
ϕ ϕ 1(1/2)
−
In fact, since
h (u) h (u)
K K
ϕ +ϕ = 1,
h (u) h (u)
(cid:18) K+ϕK (cid:19) (cid:18) K+ϕK (cid:19)
then,
h (u) = ϕ 1(1/2)h (u),
K − K+ϕK
so (2.1) holds. Then we obtain,
1
2ϕ 1(1/2)R (K + K) = 2ϕ 1(1/2)R K = R (K)+R (K).
− 1 ϕ − 1 ϕ 1(1/2) 1 1
(cid:18) − (cid:19)
Which mean the equality in (0.1) holds.
Next, for i = 2, ,n 1, we consider the convex bodies,
··· −
n n
K = [ e ,e ]+ [ e ,e ], K = [ e ,e ]+ [ e ,e ],
1 1 k k ′ 2 2 k k
− − − −
k=i+1 k=i+1
X X
i.e. the 0-symmetric (n i+1)-cubes with edges parallel to the coordinate axes
−
and with length 2 in the subspaces L n (j=1, 2). For i = n, we take
K = [ e ,e ], K = [ e ,e ]. j ⊆ Ln−i+1
1 1 ′ 2 2
− −
Clearly, for L n, R(K L) 1, R(K L) 1. Specially, if L n is
∈ Li | ≥ ′| ≥ 0 ∈ Li
generated by e , ,e , then
1 i
{ ··· }
K L = [ e ,e ], K L = [ e ,e ].
0 1 1 ′ 0 2 2
| − | −
Then we have
R (K) = min R(K L) = R([ e ,e ]) = 1,
i 1 1
L n | −
∈Li
R (K ) = min R(K L) = R([ e ,e ]) = 1.
i ′ ′ 2 2
L n | −
∈Li
Since dim(L L) 1, then there exist x K L and x K L with
j ′ ′
∩ ≥ ∈ ∩ ∈ ∩
x , x 1. By the symmetry of K and K , we may assume that x x > 0,
2 ′ 2 ′ ′
k k k k ≥ h · i
then
x+x ( x 2 + x 2)1/2 21/2 √2
′ k k2 k ′k2 = .
2ϕ 1(1/2) ≥ 2ϕ 1(1/2) ≥ 2ϕ 1(1/2) 2ϕ 1(1/2)
(cid:13) − (cid:13)2 − − −
(cid:13) (cid:13)
Since x(cid:13)+x′ (K +(cid:13) K ) L. Using the fact that for K n and L n,
2ϕ−(cid:13)1(1/2) ∈ (cid:13)ϕ ′ ∩ ∈ Ko ∈ Li
K L K L,
∩ ⊆ |
SUCCESSIVE RADII AND ORLICZ MINKOWSKI ADDITION 9
we obtain,
√2
R((K + K ) L) R((K + K ) L) . (2.2)
ϕ ′ | ≥ ϕ ′ ∩ ≥ 2ϕ 1(1/2)
−
On the other hand, by Lemma 1.3 we have
(K + K ) L = K L + K L = [ e ,e ]+ [ e ,e ].
ϕ ′ 0 0 ϕ ′ 0 1 1 ϕ 2 2
| | | − −
Lemma 1.4 shows that
√2
(K + K ) L = [ e ,e ]+ [ e ,e ] B .
ϕ ′ | 0 − 1 1 ϕ − 2 2 ⊆ 2ϕ 1(1/2) 2,L˜
−
Where L˜ = span e ,e . So we have
1 2
{ }
√2
R((K + K ) L) . (2.3)
ϕ ′
| ≤ 2ϕ 1(1/2)
−
Together with (2.2) and (2.3) we have
√2
R((K + K ) L) = .
ϕ ′ | 2ϕ 1(1/2)
−
Moreover,
√2
R (K + K ) = min R((K + K ) L) = (R (K)+R (K )).
i ϕ ′ L∈Lni ϕ ′ | 4ϕ−1(1/2) i i ′
(cid:3)
Which gives the equality of (0.2). So we complete the proof.
Notice that, when i = n, namely, the circumradius R(K + K ). By (ii) of
ϕ ′
Theorem 1.2 and the fact R(K+K ) R(K)+R(K ), the reverse inequality for
′ ′
≤
the circumradius holds
R(K + K ) R(K)+R(K ).
ϕ ′ ′
≤
However, for R (K + K ) (i = 1,2, ,n 1) there is no chance to get reverse
i ϕ ′
··· −
inequalities.
Proposition 2.1. Let ϕ and K,K n, for i = 1, n 1, there exists no
∈ C ′ ∈ Ko ··· −
constant c > 0 such that
cR (K + K ) R (K)+R (K ).
i ϕ ′ i i ′
≤
Proof. To show the non-existence of a reverse inequality, for i = 1, ,n 1.
··· −
Take the convex bodies
n i
−
K = [ e ,e ] and K = [ e ,e ].
n i+1 n i+1 ′ k k
− − − −
k=1
X
LetL¯ andL¯ in n spannedby e ,e , ,e and e ,e , ,e ,
0 ′0 Li { n−i n−i+2 ··· n} { n−i+1 n−i+2 ··· n}
respectively. Then
K L¯ = K L¯ = 0 .
| 0 ′| ′0 { }
10 F. CHEN, C. YANG,M. LUO
So, R (K) = R (K ) = 0, i.e., R (K)+R (K ) = 0. On the other hand,
i i ′ i i ′
n i+1
1 1 −
K + K (K +K ) = [ e ,e ],
ϕ ′ ⊇ 2ϕ 1(1/2) ′ 2ϕ 1(1/2) − k k
− −
k=1
X
that is K + K contains an (n i + 1) dimensional cube, which implies that
ϕ ′
− −
dim((K + K ) L) 1 for L n. Then, R (K + K ) > 0, so there exists no
ϕ ′ | ≥ ∈ Li i ϕ ′
constant c > 0 such that
cR (K + K ) R (K)+R (K ).
i ϕ ′ i i ′
≤
(cid:3)
We complete the proof.
If take ϕ(t) = tp, we obtain the following corollaries, which is obtained by
Gonz´alez and Hern´andez Cifre [14].
Corollary 2.2. Let K,K n and p 1. Then
′ ∈ Ko ≥
p−1
2 p R1(K +p K′) R1(K)+R1(K′), for p 1,
≥ ≥
3p−2
2 2p Ri(K +p K′) Ri(K)+Ri(K′), for 1 p 2, i = 2, ,n,
≥ ≤ ≤ ···
R (K + K ) max R (K),R (K ) , for p 2, i = 2, ,n.
i p ′ i i ′
≥ { } ≥ ···
All inequalities are best possible.
Corollary 2.3. Let K,K n and p 1. Then
′ ∈ Ko ≥
R (K + K ) R (K)+R (K )
n p ′ n n ′
≤
which is tight, and for any i = 1, ,n 1, there exists no constant c > 0 such
··· −
that cR (K + K ) R (K)+R (K ).
i p ′ i i ′
≤
More specially, if p = 1, it was shown in [13].
Now we deal with the inner radii r . The proof of Theorem 0.3 is similar with
i
Theorem 0.1.
Proof of Theorem 0.3: By Theorem 1.2, we have 1 K +K K + K ,
2ϕ−1(1/2) ′ ⊆ ϕ ′
we obtain
2ϕ 1(1/2)r (K + K ) r (K +K ) r (K)+r (K ),
− n ϕ ′ n ′ n n ′
≥ ≥
1
2ϕ 1(1/2)r (K + K ) r (K +K ) (r (K)+r (K )),
− i ϕ ′ i ′ i i ′
≥ ≥ √2
i = 1,2, ,n 1, which shows the inequalities (0.3) and (0.4).
··· −
Similarly, we will show inequalities (0.3) and (0.4) are the best possible. Let
K = K , by (2.1), we have
′
2ϕ 1(1/2)r (K + K) = r (K)+r (K).
− n ϕ n n
Fori = 1,2, ,n 1,letj = 2i n,if2i n, andj = 0otherwise. Weconsider
the i dimensio·n·a·l li−near subspace−s L˜ = sp≥an e , ,e ,e , ,e and L˜ =
− 0 { 1 ··· j j+1 ··· i} ′0
span e , ,e ,e , ,e in n. Let B and B be i-dimensional unit
{ 1 ··· j i+1 ··· 2i−j} Li i,L˜0 i,L˜′0
balls. It is clear that, r (B ) = r (B ) = 1.
i i,L˜0 i i,L˜′0