arXiv:0812.4550v1 [math.MG] 24 Dec 2008
Inequalities for mixed paffine surface area ∗
Elisabeth Werner† Deping Ye
Abstract
We prove new AlexandrovFenchel type inequalities and new affine isoperimetric
inequalities for mixed paffine surface areas. We introduce a new class
of bodies, the illumination surface bodies, and establish some of their properties.
We show, for instance, that they are not necessarily convex. We give
geometric interpretations of Lp affine surface areas, mixed paffine surface areas
and other functionals via these bodies. The surprising new element is that
not necessarily convex bodies provide the tool for these interpretations.
1 Introduction
This article deals with affine isoperimetric inequalities and AlexandrovFenchel
type inequalities for mixed paffine surface area. Mixed paffine surface area was
introduced by Lutwak for p ≥ 1 in [28]. It has the dual mixed volume [25] and
the Lp affine surface area [28] as special cases. Lp affine surface area is at the core
of the rapidly developing Lp BrunnMinkowski theory. Contributions here include
the study of solutions of nontrivial ordinary and, respectively, partial differential
equations (see e.g. Chen [10], Chou and Wang [11], Stancu [39, 40]), the study
of the Lp ChristoffelMinkowski problem by Hu, Ma and Shen [18], extensions of
Lp affine surface area to all p (see e.g., [34, 37, 38, 45]), a new proof by Fleury,
Gu´edon and Paouris [12] of a result by Klartag [19] on concentration of volume,
results on approximation of convex bodies by polytopes (e.g., [16, 24, 38]), results
on valuations (e.g., Alesker [2, 3], and Ludwig and Reitzner [22, 23]) and the affine
Plateau problem solved in R3 by Trudinger and Wang [41], and Wang [43].
∗Keywords: mixed paffine surface area, affine isoperimetric inequality, Lp affine surface area,
Lp Brunn Minkowski theory. 2000 Mathematics Subject Classification: 52A20, 53A15
†Partially supported by an NSF grant, a FRGNSF grant and a BSF grant
1
The classical affine isoperimetric inequality, which gives an upper bound for
the affine surface area in terms of volume, is fundamental in many problems (e.g.
[14, 15, 29, 36]). In particular, it was used to show the uniqueness of selfsimilar
solutions of the affine curvature flow and to study its asymptotic behavior by
Andrews [4, 5], Sapiro and Tannenbaum [35]. More general Lp affine isoperimetric
inequalities were proved in [28] for p > 1 and in [45] for all p. These Lp affine
isoperimetric inequalities generalize the celebrated BlaschkeSantal´o inequality and
inverse Santal´o inequality due to Bourgain and Milman [6] (see also Kuperberg
[20]). We also refer to related works by Lutwak, Yang and Zhang [30] and Campi
and Gronchi [9].
For mixed paffine surface area, AlexandrovFenchel type inequalities (for p =
1,±∞) and affine isoperimetric inequalities (for 1 ≤ p ≤ n) were first established by
Lutwak in [25, 26, 28]. Here we derive new AlexandrovFenchel type inequalities for
mixed paffine surface area for all p ∈ [−∞,∞] and new mixed paffine isoperimetric
inequalities for all p ∈ [0,∞]. Classification of the equality cases for all p in the
AlexandrovFenchel type inequalities for mixed paffine surface area is related to the
uniqueness of solutions of the Lp Minkowski problem (e.g., [10, 11, 27, 29, 31, 32, 39,
40]), which is unsolved for many cases. This is similar to the classical Alexandrov
Fenchel inequalities for mixed volume, where the complete classification of the
equality cases is also an unsolved problem.
We also give new geometric interpretations for functionals on convex bodies. In
particular, for Lp affine surface area, mixed paffine surface area, and ith mixed
paffine surface area (see below for the definitions). To do so, we construct a new
class of bodies, the illumination surface bodies, and study the asymptotic behavior
of their volumes. We show that the illumination surface bodies are not necessarily
convex, thus introducing a novel idea in the theory of geometric characterizations
of functionals on convex bodies, where to date only convex bodies where used (e.g.
[34, 37, 38, 45]).
From now on, we will always assume that the centroid of a convex body K in
Rn is at the origin. We write K ∈ C2+
, if K has C2 boundary with everywhere
strictly positive Gaussian curvature. For real p ≥ 1, the mixed paffine surface
area, asp(K1, · · · ,Kn), of n convex bodies Ki ∈ C2+
was introduced in [28] by
asp(K1, · · · ,Kn) =
Z
Sn−1
hK1(u)1−pfK1(u) · · · h1−p
Kn
fKn(u)
1
n+p
dσ(u). (1.1)
2
Here Sn−1 is the boundary of the Euclidean unit ball Bn
2 in Rn, σ is the usual
surface area measure on Sn−1, hK(u) is the support function of the convex body
K at u ∈ Sn−1, and fK(u) is the curvature function of K at u, i.e., the reciprocal
of the Gauss curvature κK(x) at this point x ∈ ∂K, the boundary of K, that has
u as its outer normal.
We propose here to extend the definition (1.1) for mixed paffine surface area
to all p 6= −n. We also propose a definition for the (−n) mixed affine surface area
(see Section 2).
We show that mixed paffine surface areas are affine invariants for all p. Note
that for p = ±∞,
as±∞(K1, · · · ,Kn) =
Z
Sn−1
1
hK1(u) · · ·
1
hKn(u)
dσ(u)
= n ˜ V (K◦1 , · · · ,K◦n ) (1.2)
where K◦ = {y ∈ Rn, hx, yi ≤ 1, ∀x ∈ K} is the polar body of K, and ˜ V (K◦1 , · · · ,K◦n)
is the dual mixed volume of K◦1 , · · · ,K◦n , introduced by Lutwak in [25].
When all Ki coincide with K, then for all p 6= −n
asp(K, · · · ,K) =
Z
Sn−1
fK(u)
n
n+p
hK(u)
n(p−1)
n+p
dσ(u) = asp(K). (1.3)
asp(K) is the Lp affine surface area of K, which is defined for a general convex
body K as in [28] (p > 1) and in [38] (p < 1) by
asp(K) =
Z
@K
κK(x)
p
n+p
hx,NK(x)i
n(p−1)
n+p
dμK(x). (1.4)
NK(x) is the outer unit normal vector at x to ∂K, μK denotes the usual surface
area measure on ∂K, and h·, ·i is the standard inner product on Rn which induces
the Euclidian norm k · k. If K ∈ C2+
, (1.4) can be rewritten as (1.3). We show
in Section 2 that the corresponding formula (1.3) for p = −n also holds, where
as−n(K) is the L−n affine surface area of K introduced in [34].
Note further that the surface area of K can be written as (−1)th mixed 1affine
surface area of K and the Euclidean ball Bn
2 (see Section 2).
Thus, mixed paffine surface area is an extension of dual mixed volume, surface
area, and Lp affine surface area.
3
Further notations. For sets A and B, [A,B] = conv(A,B) := {λx+(1−λ)y :
λ ∈ [0, 1], x, y ∈ A∪B} is the convex hull of A∪B. A subset K of Rn is star convex
if there exists x0 ∈ K such that the line segment [x0, x], from x0 to any point x in
K, is contained in K. A convex body K is said to be strictly convex if ∂K does
not contain any line segment.
For a convex body K in Rn, K stands for the ndimensional volume of K.
More generally, for a set M, M denotes the Hausdorff content of its appropriate
dimension.
For u ∈ Sn−1, H(x, u) is the hyperplane through x with outer normal vector u,
H(x, u) = {y ∈ Rn, hy, ui = hx, ui}. The two halfspaces generated by H(x, u) are
H−(x, u) = {y ∈ Rn, hy, ui ≥ hx, ui} and H+(x, u) = {y ∈ Rn, hy, ui ≤ hx, ui}. For
f : ∂K → R+ ∪ {0}, μf is the measure on ∂K defined by μf (A) =
R
A fdμK.
The paper is organized as follows. In Section 2, we prove new Alexandrov
Fenchel type inequalities and new isoperimetric inequalities for mixed paffine surface
areas. We show monotonicity behaviour of the quotients
asp(K1, · · · ,Kn)
as∞(K1, · · · ,Kn)
n+p
and
asp(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+p
p
.
We prove BlaschkeSantal´o type inequalities for mixed paffine surface areas. Similar
results for the ith mixed paffine surface areas are also proved in Section 2.
In Section 3, we introduce the illumination surface body and describe some of its
properties. In Section 4, we derive the asymptotic behavior of the volume of the
illumination surface body, and geometric interpretations of Lp affine surface areas,
mixed paffine surface areas, and other functionals on convex bodies.
2 Mixed paffine surface area and related inequalities
2.1 Inequalities for mixed paffine surface area
We begin by proving that mixed paffine surface area is affine invariant for all p.
For p ≥ 1, this was proved by Lutwak [28]. We will first treat the case p 6= −n.
All the results concerning the case p = −n are at the end of this subsection.
It will be convenient to use the notation
fp(K, u) = h1−p
K (u)fK(u) (2.5)
4
for a convex body K in Rn and u ∈ Sn−1. We will also write asmp
(K1, · · · ,Kn)
for
asp(K1, · · · ,Kn)
m, and det(T) for the absolute value of the determinant of
linear transform T.
Lemma 2.1 Let T : Rn → Rn be an invertible linear transform. Then for all
p 6= −n,
asp(TK1, · · · , TKn) = det(T)
n−p
n+p asp(K1, · · · ,Kn).
In particular, if det(T) = 1, then asp(K1, · · · ,Kn) is affine invariant:
asp(TK1, · · · , TKn) = asp(K1, · · · ,Kn).
Proof.
Since K ∈ C2+
, for any u ∈ Sn−1, there exists a unique x ∈ ∂K such that u = NK(x)
and fK(u) = 1
K(x) . By Lemma 12 of [38]
fK(u) =
1
κK(x)
=
fTK (v)
det2(T) kT−1t(u)kn+1 , (2.6)
where v = T−1t(u)
kT−1t(u)k ∈ Sn−1 and where for an operator A, At denotes its usual
adjoint. On the other hand,
hK(u) = hx, ui = hTx, T−1t(u)i = kT−1t(u)k hTx, vi = kT−1t(u)k hTK(v).
Thus, with notation (2.5), for all p,
fp(K, u) =
fTK (v) h1−p
TK (v) kT−1t(u)k1−p
det2(T) kT−1t(u)kn+1 =
fp(TK, v)
det2(T) kT−1t(u)kn+p . (2.7)
Lemma 10 and its proof in [38] show that up to a small errorfTK(
v) dσ(v) = det(T) kT−1t(u)kfK(u) dσ(u).
Together with (2.6), one gets that (again up to a small error) kT−1t(u)k−n dσ(u) =
det(T) dσ(v). Therefore, up to a small error,
[fp(K1, u) · · · fp(Kn, u)]
1
n+p dσ(u) = det(T)
p−n
n+p [fp(TK1, v) · · · fp(TKn, v)]
1
n+p dσ(v).
The lemma then follows by integrating over Sn−1.
5
A general version of the classical AlexandrovFenchel inequalities for mixed
volumes (see [1, 8, 36]) can be written as
mY−1
i=0
V (K1, · · · ,Kn−m,K n−i, ·{·z· ,Kn−}i
m
) ≤ V m(K1, · · · ,Kn).
Here we prove the analogous inequalities for mixed paffine surface area. For p =
±∞ and p = 1, the inequalities were proved by Lutwak [25, 26]. For p ≥ 1,
inequality (2.8) was proved by Lutwak in [28], with equality if and only if the
associated Ki are dilates of each other.
Proposition 2.1 Let all Ki be convex bodies in C2+
with centroid at the origin. If
p 6= −n, then for 1 ≤ m ≤ n
asmp
(K1, · · · ,Kn) ≤
mY−1
i=0
asp(K1, · · · ,Kn−m,K n−i, ·{·z· ,Kn−}i
m
).
Equality holds if the Kk, for k = n − m + 1, · · · , n are dilates of each other. If
m = 1, equality holds trivially.
In particular, if m = n,
asnp
(K1, · · · ,Kn) ≤ asp(K1) · · · asp(Kn). (2.8)
Proof. Put g0(u) = [fp(K1, u) · · · fp(Kn−m, u)]
1
n+p and for i = 0, · · · ,m − 1, put
gi+1(u) = [fp(Kn−i, u)]
1
n+p . By H¨older’s inequality (see [17])
asp(K1, · · · ,Kn) =
Z
Sn−1
g0(u)g1(u) · · · gm(u) dσ(u)
≤
mY−1
i=0
Z
Sn−1
g0(u)gm
i+1(u) dσ(u)
1
m
=
mY−1
i=0
as
1
m
p (K1, · · · ,Kn−m, Kn−i ·{·z·Kn−}i
m
).
As Ki ∈ C2+
, fp(Ki, u) > 0 for all i and all u ∈ Sn−1. Therefore, equality in
H¨older’s inequality holds if and only if g0(u)gm
i+1(u) = λmg0(u)gm
j+1(u) for some
λ > 0 and all 0 ≤ i 6= j ≤ m − 1. This is equivalent to hKn−i(u)1−pfKn−i(u) =
λhKn−j (u)1−pfKn−j (u) for all 0 ≤ i 6= j ≤ m − 1. This condition holds true if the
Kk, for k = n −m + 1, · · · , n are dilates of each other.
6
Remark. It is an unsolved problem for many p whether fp(K, u) = λfp(L, u) guarantees
that K and L are dilates of each other. This is equivalent to the uniqueness
of the solution of the Lp Minkowski problem: for fixed α ∈ R, under which conditions
on a continuous function γ : Sn−1 → (0,∞), there exists a (unique) convex
body K such that hK(u)fK(u) = γ(u) for all u ∈ Sn−1. In many cases, the uniqueness
of the solution is an open problem. We refer to e.g., [11, 27, 29, 32, 39, 40]
for detailed information and more references on the subject. For p ≥ 1, p 6= n, the
solution to the Lp Minkowski problem is known to be unique and for p = n, the
solution is unique modulo dilates [27]. Therefore, we have the characterization of
equality in Proposition 2.1 for p ≥ 1.
Next, we prove affine isoperimetric inequalities for mixed paffine surface areas.
Proposition 2.2 Let all Ki be convex bodies in C2+
with centroid at the origin.
(i) For p ≥ 0,
asnp
(K1, · · · ,Kn)
asnp
(Bn
2 , · · · ,Bn
2 ) ≤
K1
Bn
2  · · · Kn
Bn
2 
n−p
n+p
,
with equality if the Ki are ellipsoids that are dilates of one another.
(ii) For 0 ≤ p ≤ n,
asp(K1, · · · ,Kn)
asp(Bn
2 , · · · ,Bn
2 ) ≤
V (K1, · · · ,Kn)
V (Bn
2 , · · · ,Bn
2 )
n−p
n+p
with equality if the Ki are ellipsoids that are dilates of one another.
In particular, for p = n
asn(K1, · · · ,Kn) ≤ asn(Bn
2 , · · · ,Bn
2 ),
with equality if and only if the Ki are ellipsoids that are dilates of one another.
(iii) For p ≥ n,
asp(K1, · · · ,Kn)
asp(Bn
2 , · · · ,Bn
2 ) ≤
˜V (K1, · · · ,Kn)
˜ V (Bn
2 , · · · ,Bn
2 )
!n−p
n+p
,
with equality if and only if the Ki are ellipsoids that are dilates of one another.
7
In particular, for p = ±∞
˜ V (K1, · · · ,Kn) ˜ V (K◦1 , · · · ,K◦n) ≤ Bn
2 2,
with equality if and only if Ki are ellipsoids that are dilates of one another.
Remark. For 1 ≤ p ≤ n, inequality (ii) (with equality if and only if the Ki are
ellipsoids that are dilates of one another) was proved by Lutwak in [28]. If Ki = K
for all i, one recovers the Lp affine isoperimetric inequality proved in [45].
Remark. We cannot expect to get strictly positive lower bounds in Proposition
2.2. As in [45], we consider the convex body K(R, ε) ⊂ R2, obtained as the intersection
of four Euclidean balls with radius R centered at (±(R−1), 0), (0,±(R−1)), R
arbitrarily large. We then “round” the corners by putting there arcs of Euclidean
balls of radius ε, ε arbitrarily small. To obtain a body in C2+
, we “bridge” between
the Rarcs and εarcs by C2+
arcs on a set of arbitrarily small measure. Then
asp(K(R, ε)) ≤ 16
R
p
2+p
+ 4π ε
2
2+p , which goes to 0 as R → ∞ and ε → 0. Choose
now Ri and εi, 1 ≤ i ≤ n, such that Ri → ∞ and εi → 0, and let Ki = K(Ri, εi)
for i = 1, 2 · · · , n. By inequality (2.8), asnp
(K1, · · · ,Kn) ≤
Qn
i=1 asp(K(Ri, εi)) and
thus asp(K1, · · · ,Kn) → 0 for p > 0. A similar construction can be done in higher
dimensions.
Proof of Proposition 2.2.
(i) Clearly asp(Bn
2 , · · · ,Bn
2 ) = asp(Bn
2 ) = nBn
2  for all p 6= −n. By inequality
(2.8), one gets for all p ≥ 0
asnp
(K1, · · · ,Kn)
asnp
(Bn
2 , · · · ,Bn
2 ) ≤
asp(K1)
asp(Bn
2 ) · · ·
asp(Kn)
asp(Bn
2 ) ≤
K1
Bn
2  · · · Kn
Bn
2 
n−p
n+p
. (2.9)
The second inequality follows, for p ≥ 0, from the Lp affine isoperimetric inequality
in [45]. Equality holds true in the Lp isoperimetric inequality [45] if and only if
the Ki are all ellipsoids, and equality holds true in inequality (2.8) if the Ki are
dilates of one another. Thus, equality holds in (2.9) if the Ki are ellipsoids that
are dilates of one another.
(ii) A direct consequence of the classical AlexandrovFenchel inequality for mixed
volume (see e.g. [7, 21]) is that
K1 · · · Kn ≤ V n(K1, · · · ,Kn).
8
If 0 ≤ p ≤ n, then n−p
n+p ≥ 0. Thus
K1 · · · Kn
n−p
n+p ≤
V n(K1, · · · ,Kn)
n−p
n+p .
As V (Bn
2 , · · · ,Bn
2 ) = Bn
2 , one gets together with (2.9)
asp(K1, · · · ,Kn)
asp(Bn
2 , · · · ,Bn
2 ) ≤
V (K1, · · · ,Kn)
V (Bn
2 , · · · ,Bn
2 )
n−p
n+p
,
with equality if the Ki are ellipsoids that are dilates of one another.
(iii) The analogous inequality for dual mixed volume [25] is
K1 · · · Kn ≥ ˜ V n(K1, · · · ,Kn),
with equality if and only the Ki are dilates of one another. p > n implies n−p
n+p < 0.
Thus
K1 · · · Kn
n−p
n+p ≤
˜ V n(K1, · · · ,Kn)
n−p
n+p .
Together with (2.9) and ˜ V (Bn
2 , · · · ,Bn
2 ) = Bn
2 , one gets
asp(K1, · · · ,Kn)
asp(Bn
2 , · · · ,Bn
2 ) ≤
˜V (K1, · · · ,Kn)
˜ V (Bn
2 , · · · ,Bn
2 )
!n−p
n+p
.
As for p ≥ 1 equality in (2.8) holds if and only if the Ki are dilates of one another,
equality holds true here if and only if the Ki are ellipsoids that are dilates of one
another.
Proposition 2.3 Let E be a centered ellipsoid. If either all Ki ∈ C2+
are subsets
of E, for 0 ≤ p < n, or E is subset of all Ki for p > n, then
asp(K1, · · · ,Kn) ≤ asp(E).
For p = n, the inequality holds for all Ki in C2+
by Proposition 2.2 (ii).
Remark. This proposition was proved by Lutwak [28] if Ki = K for all i.
Proof. It is enough to prove the proposition for E = Bn
2 . For 0 ≤ p < n, one has
n−p
n+p > 0 and hence
Ki
Bn
2 
n−p
n+p ≤ 1 as Ki ⊂ Bn
2 . Similarly, p > n implies n−p
n+p < 0
and therefore
Ki
Bn
2 
n−p
n+p ≤ 1 as Bn
2 ⊂ Ki for all i. In both cases, the proposition
follows by inequality (2.9).
9
The next proposition gives a BlaschkeSantal´o type inequality for pmixed affine
surface area. When Ki = K for all i, the proposition was proved in [45].
Proposition 2.4 Let all Ki be convex bodies in C2+
with centroid at the origin.
For all p ≥ 0,
asnp
(K1, · · · ,Kn)asnp
(K◦1 , · · · ,K◦n ) ≤ n2nK1K◦1  · · · KnK◦n . (2.10)
Furthermore, asp(K1, · · · ,Kn)asp(K◦1 , · · · ,K◦n) ≤ as2
p(Bn
2 ) with equality if the Ki
are ellipsoids that are dilates of one another.
Proof. It follows from (2.8) that for all p 6= −n,
asnp
(K1, · · · ,Kn)asnp
(K◦1 , · · · ,K◦n ) ≤ asp(K1)asp(K◦1 ) · · · asp(Kn)asp(K◦n ),
with equality if the Ki are dilates of one another. By Corollary 4.1 in [45], for
p ≥ 0,
asnp
(K1, · · · ,Kn)asnp
(K◦1 , · · · ,K◦n ) ≤ n2nK1K◦1  · · · KnK◦n .
BlaschkeSantal´o inequality states that KK◦ ≤ Bn
2 2 with equality if and only
if K is a 0centered ellipsoid. We apply it to inequality (2.10), and obtain that for
p ≥ 0,
asp(K1, · · · ,Kn)asp(K◦1 , · · · ,K◦n ) ≤ as2
p(Bn
2 , · · · ,Bn
2 ).
Equality holds if the Ki are ellipsoids that are dilates of one another.
Theorem 2.1 Let s 6= −n, r 6= −n, p 6= −n be real numbers. Let all Ki be convex
bodies in C2+
with centroid at the origin.
(i) If (n+p)(r−s)
(n+r)(p−s) > 1, then
asp(K1, · · · ,Kn) ≤
asr(K1, · · · ,Kn)
(p−s)(n+r)
(r−s)(n+p)
ass(K1, · · · ,Kn)
(r−p)(n+s)
(r−s)(n+p) .
(ii) If n+p
n+r > 1, then
asp(K1, · · · ,Kn) ≤ (asr(K1, · · · ,Kn))
n+r
n+p
n ˜ V (K◦1 , · · · ,K◦n )
p−r
n+p . (2.11)
10
Remark. When all Ki coincide with K, (i) of Theorem 2.1 was proved in [45].
Proof.
(i) By H¨older’s inequality which enforces the condition (n+p)(r−s)
(n+r)(p−s) > 1,
asp(K1, · · · ,Kn) =
Z
Sn−1
fp(K1, u) · · · fp(Kn, u)
1
n+p dσ(u)
=
Z
Sn−1
fr(K1, u) · · · fr(Kn, u)
1
n+r
(n+r)(p−s)
(n+p)(r−s)
fs(K1, u) · · · fs(Kn, u)
1
n+s
(n+s)(r−p)
(n+p)(r−s)
dσ(u)
≤
asr(K1, · · · ,Kn)
(p−s)(n+r)
(r−s)(n+p)
ass(K1, · · · ,Kn)
(r−p)(n+s)
(r−s)(n+p) .
(ii) Similarly, again using H¨older’s inequality which now enforces the condition
n+p
n+r > 1
asp(K1, · · · ,Kn) =
Z
Sn−1
fp(K1, u) · · · fp(Kn, u)
1
n+p dσ(u)
=
Z
Sn−1
fr(K1, u) · · · fr(Kn, u)
1
n+r
n+r
n+p
1
hK1(u) · · · hKn(u)
p−r
n+p
dσ(u)
≤ (asr(K1, · · · ,Kn))
n+r
n+p (as∞(K1, · · · ,Kn))
p−r
n+p .
Together with (1.2), this completes the proof.
Remark. The condition (n+p)(r−s)
(n+r)(p−s) > 1 implies 8 cases: −n < s < p < r, s <
−n < r < p, p < r < −n < s, r < p < s < −n, s < p < r < −n, p < s < −n < r,
r < −n < s < p and −n < r < p < s.
In [45], we proved monotonicity properties of
asr(K)
nK
n+r
r . Here we prove
similar results for mixed paffine surface area.
Proposition 2.5 Let all Ki ∈ C2+
be convex bodies with centroid at the origin.
(i) If −n < r < p or r < p < −n, one has
asp(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n )
!n+p
≤
asr(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n)
!n+r
.
11
(ii) If 0 < p < r, or p < r < −n, or r < −n < 0 < p, or −n < p < r < 0, one has
asp(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+p
p
≤
asr(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+r
r
.
Proof.
(i) We divide both sides of inequality (2.11) by n ˜ V (K◦1 , · · · ,K◦n), and get for n+p
n+r >
1,
asp(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n ) ≤
asr(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n)
!n+r
n+p
. (2.12)
Condition n+p
n+r > 1 implies that −n < r < p or p < r < −n. If −n < r < p,
n + p > 0 and therefore, inequality (2.12) implies inequality (i). If p < r < −n,
n + p < 0 and therefore,
asp(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n )
!n+p
≥
asr(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n)
!n+r
.
Switching r and p, one obtains the inequality in (i): for r < p < −n,
asp(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n )
!n+p
≤
asr(K1, · · · ,Kn)
n ˜ V (K◦1 , · · · ,K◦n)
!n+r
.
(ii) Let s = 0 in inequality (i) of Theorem 2.1. Then for r(n+p)
p(n+r) > 1,
asp(K1, · · · ,Kn) ≤
asr(K1, · · · ,Kn)
p(n+r)
r(n+p)
as0(K1, · · · ,Kn)
(r−p)n
r(n+p) .
We divide both sides of the inequality by as0(K1, · · · ,Kn) and get
asp(K1, · · · ,Kn)
as0(K1, · · · ,Kn) ≤
asr(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
p(n+r)
r(n+p)
.
The condition r(n+p)
p(n+r) > 1 implies that 0 < p < r, or p < r < −n, or −n < r < p <
0, or r < −n < 0 < p. In the cases 0 < p < r, or p < r < −n, or r < −n < 0 < p,
one has n+p
p > 0 and therefore inequality (ii) holds true. On the other hand, if
−n < r < p < 0, then n+p
p < 0 and hence,
asr(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+r
r
≤
asp(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+p
p
.
12
Switching r and p, one gets inequality (ii): if −n < p < r < 0, then
asp(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+p
p
≤
asr(K1, · · · ,Kn)
as0(K1, · · · ,Kn)
n+r
r
.
Now we treat the case p = −n. The mixed (−n)affine surface area of K1, · · · ,Kn
is defined as
as−n(K1, · · · ,Kn) = max
u∈Sn−1
fK1(u)
1
2n hK1(u)
n+1
2n · · · fKn(u)
1
2n hKn(u)
n+1
2n
. (2.13)
It is easy to verify that as−n(K, · · · ,K) equals to as−n(K), the L−n affine surface
area of K [34]. We have the following proposition.
Proposition 2.6 Let all Ki be convex bodies in C2+
with centroid at the origin.
Let p 6= −n and s 6= −n be real numbers.
(i) Let T : Rn → Rn be an invertible linear transform. Then
as−n(TK1, · · · , TKn) = det(T) as−n(K1, · · · ,Kn).
In particular, if det(T) = 1, then as−n(K1, · · · ,Kn) is affine invariant:
as−n(TK1, · · · , TKn) = as−n(K1, · · · ,Kn).
(ii) AlexandrovFenchel type inequalities
asm−
n(K1, · · · ,Kn) ≤
mY−1
i=0
as−n(K1, · · · ,Kn−m,K n−i, ·{·z· ,Kn−}i
m
),
with equality if the Kj , for j = n − m + 1, · · · , n are dilates.
In particular, if m = n,
asn
−n(K1, · · · ,Kn) ≤ as−n(K1) · · · as−n(Kn).
(iii) If n(s−p)
(n+p)(n+s) ≥ 0, then
asp(K1, · · · ,Kn) ≤
as−n(K1, · · · ,Kn)
2n(s−p)
(n+p)(n+s) ass(K1, · · · ,Kn).
(iv) If n(s−p)
(n+p)(n+s) ≤ 0, then
asp(K1, · · · ,Kn) ≥
as−n(K1, · · · ,Kn)
2n(s−p)
(n+p)(n+s) ass(K1, · · · ,Kn).
13
Proof.
(i) By formula (2.7), f−n(TK, v) = det(T)2 f−n(K, u). Therefore
as−n(TK1, · · · , TKn) = maxv∈Sn−1 [f−n(TK1, v) · · · f−n(TKn, v)]
1
2n
= det(T) maxu∈Sn−1 [f−n(K1, u) · · · f−n(Kn, u)]
1
2n
= det(T) as−n(K1, · · · ,Kn).
(ii) Let ˜g0(u) = fK1(u)
1
2n hK1(u)
n+1
2n · · · fKn−m(u)
1
2n hKn−m(u)
n+1
2n and ˜gi+1(u) =
fKn−i(u)
1
2n hKn−i(u)
n+1
2n , for i = 0, · · · ,m − 1. Then
as−n(K1, · · · ,Kn) = max
u∈Sn−1
˜g0(u)˜g1(u) · · · ˜gm(u)
≤
mY−1
i=0
max
u∈Sn−1
˜g0(u)˜gm
i+1(u)
1
m
=
mY−1
i=0
as
1
m
−n(K1, · · · ,Kn−m, Kn−i · ·{·z ,Kn−}i
m
).
Equality holds if and only if for all i, 0 ≤ i ≤ m − 1, ˜g0(u)˜gm
i+1(u) attain their
maximum at the same direction u0. This condition holds true if the Kj , for j =
n − m + 1, · · · , n, are dilates.
(iii) and (iv)
asp(K1, · · · ,Kn) =
Z
Sn−1
[fp(K1, u) · · · fp(Kn, u)]
1
n+p dσ(u)
=
Z
Sn−1
[fs(K1, u) · · · fs(Kn, u)]
1
n+s
h
n+1
2n
K1
(u)f
1
2n
K1
(u) · · · h
n+1
2n
Kn
(u)f
1
2n
Kn
(u)
2n(s−p)
(n+p)(n+s)
dσ(u)
which is
≤
as−n(K1, · · · ,Kn)
2n(s−p)
(n+p)(n+s) ass(K1, · · · ,Kn), if
n(s − p)
(n + p)(n + s) ≥ 0,
and
≥
as−n(K1, · · · ,Kn)
2n(s−p)
(n+p)(n+s) ass(K1, · · · ,Kn), if
n(s − p)
(n + p)(n + s) ≤ 0.
14
2.2 ith mixed paffine surface area and related inequalities
For all p ≥ 1 and all real i, the ith mixed paffine surface area of K,L ∈ C2+
is
defined as [26, 42]
asp,i(K,L) =
Z
Sn−1
fp(K, u)
n−i
n+p fp(L, u)
i
n+p dσ(u).
Recall that fp(K, u) = fK(u)h1−p
K (u). Here we further generalize this definition to
all p 6= −n and all i. An analogous definition for the ith mixed (−n)affine surface
area of K and L is
as−n,i(K,L) = max
u∈Sn−1
fK(u)
n−i
2n hK(u)
(n+1)(n−i)
2n fL(u)
i
2n hL(u)
(n+1)i
2n
.
When i ∈ N, 0 ≤ i ≤ n, then, for all p, the ith mixed paffine surface area of K
and L is
asp,i(K,L) = asp(K , ·{·z· ,K}
n−i
, L, ·{·z· ,L}
i
).
Clearly, for all p, asp,0(K,L) = asp(K), and asp,n(K,L) = asp(L). When L = Bn
2 ,
we write asp,i(K) for asp,i(K,Bn
2 ). Thus
asp,i(K) =
Z
Sn−1
fp(K, u)
n−i
n+p dσ(u), for p 6= −n,
asp,i(K) = max
u∈Sn−1
fK(u)
n−i
2n hK(u)
(n+1)(n−i)
2n
, for p = −n.
In particular, as1,−1(K) =
R
Sn−1 fK(u) dσ(u) is the surface area of K.
The next proposition and its proof is similar to Proposition 2.1 and its proof.
Therefore we omit it.
Proposition 2.7 Let K and L be convex bodies in C2+
with centroid at the origin.
Let i ∈ R and s 6= −n, r 6= −n, and p 6= −n be real numbers.
(i) If (n+p)(r−s)
(n+r)(p−s) > 1, then
asp,i(K,L) ≤
asr,i(K,L)
(p−s)(n+r)
(r−s)(n+p)
ass,i(K,L)
(r−p)(n+s)
(r−s)(n+p) .
(ii) If n+p
n+r > 1, then
asp,i(K,L) ≤ (asr,i(K,L))
n+r
n+p
n ˜ Vi(K◦,L◦)
p−r
n+p
15
where ˜Vi(K◦,L◦) = 1
n
R
Sn−1
1
hK(u)n−i hL(u)i dσ(u) for all i.
(iii) If n(s−p)
(n+p)(n+s) ≥ 0, then
asp,i(K,L) ≤
as−n,i(K,L)
2n(s−p)
(n+p)(n+s) ass,i(K,L).
(iv) If n(s−p)
(n+p)(n+s) ≤ 0, then
asp,i(K,L) ≥
as−n,i(K,L)
2n(s−p)
(n+p)(n+s) ass,i(K,L).
The following proposition was proved in [26, 42] for p ≥ 1.
Proposition 2.8 Let K and L be convex bodies in C2+
with centroid at the origin.
If j < i < k or k < i < j (equivalently, k−j
k−i > 1 ), then for all p,
asp,i(K,L) ≤ asp,j(K,L)
k−i
k−j asp,k(K,L)
i−j
k−j ,
with equality if K and L are dilates of each other.
In particular,
asp,i(K) ≤ asp,j(K)
k−i
k−j asp,k(K)
i−j
k−j ,
with equality if K is a ball.
For p 6= −n, the proof is the same as the proof in [26, 42]. For p = −n, it is similar
to the proof of Proposition 2.1. Note that for i ∈ N, 0 < i < m, m = j and k = 0,
the proposition is a direct consequence of Proposition 2.1.
In Proposition 2.8, if j = 0 and k = n, then for all p and 0 ≤ i ≤ n
asnp,i(K,L) ≤ asn−i
p (K)asi
p(L). (2.14)
If we let i = 0 and j = n, then for all k ≤ 0 and for all p
asnp
,k(K,L) ≥ asn−k
p (K)askp
(L). (2.15)
Let i = n, j = 0 and k > n. Then inequality (2.15) also holds true for k ≥ n and
all p. In both (2.14) and (2.15), equality holds for all p if K and L are dilates.
16
From inequality (2.14) and Corollary 4.1 in [45], one gets that
asnp
,i(K,L)asnp
,i(K◦,L◦) ≤
asp(K)asp(K◦)
n−i
asp(L)asp(L◦)
i
≤ n2n(KK◦)n−i(LL◦)i (2.16)
holds true for all p ≥ 0 and 0 < i < n. The inequality also holds if i = 0 and i = n
[45]. We apply BlaschkeSantal´o inequality to inequality (2.16) and get
asp,i(K,L)asp,i(K◦,L◦) ≤ as2
p(Bn
2 )
for all p ≥ 0 and 0 ≤ i ≤ n. Equality holds true if K and L are ellipsoids that are
dilates of each other. Hence we have proved the following proposition, which, for
p ≥ 1, was proved in [42].
Proposition 2.9 Let K and L be convex bodies in C2+
with centroid at the origin.
If p ≥ 0 and 0 ≤ i ≤ n, then
asp,i(K,L)asp,i(K◦,L◦) ≤ as2
p(Bn
2 ),
with equality if K and L are ellipsoids that are dilates of each other.
We now establish isoperimetric inequalities for asp,i(K).
Proposition 2.10 Let K ∈ C2+
be a convex body with centroid at the origin.
(i) If p ≥ 0 and 0 ≤ i ≤ n, then
asp,i(K)
asp,i(Bn
2 ) ≤
K
Bn
2 
(n−p)(n−i)
(n+p)n
with equality if K is a ball. Moreover, asp,i(K)asp,i(K◦) ≤ as2
p(Bn
2 ) with equality
if K is a ball.
(ii) If p ≥ 0 and i ≥ n, then
asp,i(K)
asp,i(Bn
2 ) ≥
K
Bn
2 
(n−p)(n−i)
(n+p)n
with equality if K is a ball. Moreover, asp,i(K)asp,i(K◦) ≥ as2
p(Bn
2 ) with equality
if K is a ball.
17
(iii) If −n < p < 0 and i ≤ 0, then
asp,i(K)
asp,i(Bn
2 ) ≥
K
Bn
2 
(n−p)(n−i)
(n+p)n
with equality if K is a ball. Moreover, asp,i(K)asp,i(K◦) ≥ cn−ias2
p(Bn
2 ) where c is
the universal constant in the inverse Santal´o inequality [6, 20].
(iv) If p < −n and i ≤ 0, then
asp,i(K)
asp,i(Bn
2 ) ≥ c
(n−i)p
n+p
K
Bn
2 
(n−p)(n−i)
(n+p)n
.
Moreover, asp,i(K)asp,i(K◦) ≥ cn−ias2
p(Bn
2 ) where c is the same constant as in
(iii).
(v) If i ≤ 0, then
as−n,i(K)
as−n,i(Bn
2 ) ≥
K
Bn
2 
n−i
n
.
Moreover, as−n,i(K)as−n,i(K◦) ≥ as2
−n,i(Bn
2 ).
Proof.
(i) For i = n, the equality holds trivially. For i = 0, the inequality was proved in
[45]. We now prove the case 0 < i < n. L = Bn
2 in inequality (2.14) gives
asp,i(K)
asp,i(Bn
2 )
n
≤
asp(K)
asp(Bn
2 )
n−i
(2.17)
for all p 6= −n and 0 ≤ i ≤ n. We also use that asp,i(Bn
2 ) = asp(Bn
2 ). Then, as
asp(Bn
2 ) = nBn
2 , we get for all p ≥ 0 and 0 ≤ i ≤ n, the following isoperimetric
inequality as a consequence of the Lp affine isoperimetric inequality in [45]
asp,i(K)
asp,i(Bn
2 ) ≤
asp(K)
asp(Bn
2 )
n−i
n
≤
K
Bn
2 
(n−p)(n−i)
(n+p)n
,
with equality if K is a ball. The inequality asp,i(K)asp,i(K◦) ≤ as2
p,i(Bn
2 ) follows
from Proposition 2.9 with L = Bn
2 .
(ii) For i = n, the equality holds trivially. Similarly, let L = Bn
2 in inequality
(2.15), then for all p 6= −n, and i ≥ n or i ≤ 0,
asp,i(K)
asp,i(Bn
2 )
n
≥
asp(K)
asp(Bn
2 )
n−i
. (2.18)
18
Hence for i ≥ n and p ≥ 0, the Lp affine isoperimetric inequality in [45] implies
that
asp,i(K)
asp,i(Bn
2 ) ≥
asp(K)
asp(Bn
2 )
n−i
n
≥
K
Bn
2 
(n−p)(n−i)
(n+p)n
with equality if K is a ball. Moreover, by Corollary 4.1 (i) in [45] and the remark
after it, one has for all i ≥ n
asp,i(K)asp,i(K◦)
as2
p,i(Bn
2 )
!n
≥
asp(K)asp(K◦)
as2
p(Bn
2 )
n−i
≥ 1,
or equivalently, asp,i(K)asp,i(K◦) ≥ as2
p,i(Bn
2 ), with equality if K is a ball.
(iii) If i ≤ 0 and −n < p < 0, inequality (2.18) and Theorem 4.2 (ii) of [45] imply
that
asp,i(K)
asp,i(Bn
2 ) ≥
asp(K)
asp(Bn
2 )
n−i
n
≥
K
Bn
2 
(n−p)(n−i)
(n+p)n
with equality if K is a ball. By Corollary 4.1 (ii) of [45] and the remark after it,
asp,i(K)asp,i(K◦)
as2
p,i(Bn
2 )
!n
≥
asp(K)asp(K◦)
as2
p(Bn
2 )
n−i
≥ cn(n−i),
or equivalently, asp,i(K)asp,i(K◦) ≥ cn−ias2
p,i(Bn
2 ) where c is the constant in the
inverse Santal´o inequality [6, 20].
(iv) If i ≤ 0 and p < −n inequality (2.18) and Theorem 4.2 (iii) of [45] imply that
asp,i(K)
asp,i(Bn
2 ) ≥
asp(K)
asp(Bn
2 )
n−i
n
≥ c
(n−i)p
n+p
K
Bn
2 
(n−p)(n−i)
(n+p)n
.
The proof of asp,i(K)asp,i(K◦) ≥ cn−ias2
p,i(Bn
2 ) is same as in (iii).
(v) Inequality (2.15) implies that as−n,i(K)n ≥ as−n(K)n−i for i ≤ 0. As as−n,i(Bn
2 ) =
1 for all i,
as−n,i(K)
as−n,i(Bn
2 )
n
≥
as−n(K)
as−n(Bn
2 )
n−i
≥
K
Bn
2 
n−i
.
The second inequality follows from the L−n affine isoperimetric inequality in [45].
Moreover, by Corollary 4.2 in [45], for i ≤ 0,
as−n,i(K)as−n,i(K◦)
as2
−n,i(Bn
2 )
!n
≥
as−n(K)as−n(K◦)
as2
−n(Bn
2 )
n−i
≥ 1,
19
or equivalently, as−n,i(K)as−n,i(K◦) ≥ as2
−n,i(Bn
2 ).
Remark. The example K(R, ε) mentioned in the remarks after Proposition 2.2
shows that we cannot expect to get strictly positive lower bounds in (i) of Proposition
2.10 for p > 0 and 0 ≤ i < n. In fact, by inequality (2.17), one has
asp,i(K(R, ε))
asp,i(Bn
2 )
n
≤
asp(K(R, ε))
asp(Bn
2 )
n−i
.
As in [45], asp(K(R, ε)) → 0 for p > 0 as R → ∞ and ε → 0. 0 ≤ i < n implies
that n − i > 0, and therefore asp,i(K(R, ε)) → 0.
This example also shows that, likewise, we cannot expect finite upper bounds
in (ii) (for p > 0 and i > n), (iii) (for −n < p < 0 and i ≤ 0), and (iv) (for p < −n
and i ≤ 0), of Proposition 2.10. For instance, if i ≤ 0, by inequality (2.18), one has
asp,i(K(R, ε))
asp,i(Bn
2 )
n
≥
asp(K(R, ε))
asp(Bn
2 )
n−i
.
For −2 < p < 0, one has asp(K(R, ε)) → ∞ as R → ∞ and ε → 0. Therefore, if
i ≤ 0, we obtain that asp,i(K(R, ε)) → ∞ as R → ∞ and ε → 0, i.e., there are no
finite upper bounds in (iii).
Remark. In (iv), if p = −∞, then for all i ≤ 0,
Z Z
Sn−1×Sn−1
hK(u)hK◦ (v)
i−n dσ(u) dσ(v) ≥ cn−ias2
p,i(Bn
2 )
or equivalently, for all i ≤ 0,
Z Z
Sn−1×Sn−1
ρK(u)ρK◦(v)
n−i dσ(u) dσ(v) ≥ cn−ias2
p,i(Bn
2 ).
In particular, if i = 0, this is equivalent to the inverse Santal´o inequality [6].
3 Illumination surface bodies
We now define a new family of bodies associated with a given convex body K.
These new bodies are a variant of the illumination bodies [44] (compare also [38]).
Definition 3.1 (Illumination surface body) Let s ≥ 0 and f : ∂K → R be a
nonnegative, integrable function. The illumination surface body Kf,s is defined as
Kf,s =
n
x : μf (∂K ∩ [x,K]\K) ≤ s
o
.
20
Obviously, K ⊆ Kf,s for any s ≥ 0 and any nonnegative, integrable function f.
Moreover, Kf,s ⊆ Kf,t for any 0 ≤ s ≤ t.
Notice also that Kf,s needs to be neither bounded nor convex:
Example 3.1 Let K = B2∞
= {x ∈ R2 : max1≤i≤2xi ≤ 1} and
f(x) =
(
1
12 , x ∈ [(−1, 1), (1, 1)] ∪ [(1, 1), (1,−1)]
1
6 , otherwise
Kf,s is calculated as follows. If s < 1
6 , then Kf,s = K.
If s ∈ [ 1
6 , 1
3 ), then
Kf,s = {(x1, x2) : x1 ≥ −1, x2 ∈ [−1, 1]; or x2 ≥ −1, x1 ∈ [−1, 1]}.
If s ∈ [ 1
3 , 1
2 ), then
Kf,s = {(x1, x2) : x1, x2 ≥ −1; or x1 ≤ −1, x2 ∈ [−1, 1]; or x2 ≤ −1, x1 ∈ [−1, 1]}.
If s ∈ [ 1
2 , 2
3 ), then Kf,s = {(x1, x2) : x1 ≥ −1 or x2 ≥ −1}.
Except for s < 1/6, all of them are neither bounded nor convex.
If s ≥ 2
3 , then Kf,s = R2
The following lemmas describe some of the properties of the bodies Kf,s.
Lemma 3.1 Let s ≥ 0 and f : ∂K → R be a nonnegative, integrable function.
Then
(i) Kf,s =
T
>0 Kf,s+.
(ii) Kf,s is star convex, i.e., for all x ∈ Kf,s: [0, x] ⊂ Kf,s.
Proof.
(i) We only need to show that Kf,s ⊇
T
>0 Kf,s+. Let x ∈
T
>0 Kf,s+. Then for
all δ > 0, μf (∂K ∩ [x,K]\K) ≤ s+δ. Thus, letting δ → 0, μf (∂K ∩ [x,K]\K) ≤ s.
(ii) Let x ∈ Kf,s. We claim that [0, x] ⊆ Kf,s. Let y ∈ [0, x]. Since y ∈ [0, x] ⊂
[x,K], we have [y,K] \K ⊆ [x,K] \K and thus ∂K ∩[y,K] \ K ⊆ ∂K ∩ [x,K]\K.
This implies that
μf (∂K ∩ [y,K] \ K) ≤ μf (∂K ∩ [x,K] \ K) ≤ s
21
and hence y ∈ Kf,s.
Remark. We can not expect Kf,s to be convex, even for K = Bn
2 and f smooth.
Indeed, let K = B2
2 and s = 1
64 . Define
f(x) =
1
4 , x is in the first and third quadrant
1
16 , x is in the fourth quadrant
23
16 , x is in the second quadrant
Then Kf,s is not convex. In fact, Kf,s contains the arc from the point (tan(
32 ), 1)
to the point (1, tan(
32 )) of the Euclidean ball centered at 0 with radius r = sec(
32 ).
Moreover, the point (sec(
20 ), 0) is on the boundary of Kf,s. The tangent line at
(1, tan(
32 )) of B2
2(0, r) is y = −x+r2
√r2−1
. This tangent line intersects the xaxis at
(r2, 0) = (sec2(
32 ), 0). Since sec2(
32 ) ∼ 1.009701 < sec(
20 ) ∼ 1.01264, Kf,s is not
convex.
We can modify f so that it becomes smooth also at the points (±1, 0) and
(0,±1) and ∂Kf,s still intersects the positive xaxis at the point (sec(
20 ), 0). Therefore,
Kf,s is not convex, even if f is smooth.
Lemma 3.2 Let s ≥ 0 and f : ∂K → R be an integrable, μKalmost everywhere
strictly positive function. Then
(i) K = Kf,0 .
(ii) There exists s0 > 0, such that for all 0 ≤ s ≤ s0, Kf,s is bounded.
Proof. (i) It is enough to prove that Kf,0 ⊆ K. Suppose this is not the case.
Then there is x ∈ Kf,0 but x /∈ K. Since 0 ∈ int(K), there is α > 0 such that
Bn
2 (0, α) ⊆ K ⊆ Bn
2 (0, 1/α). (3.19)
Let y ∈ [0, x] ∩ ∂K and Con(x, α) = [x,Bn
2 (0, α)] be the convex hull of x and
Bn
2 (0, α). H(y,NK(y)) ∩ Con(x, α) contains a (n − 1)dimensional Euclidean ball
with radius (at least) r1 = αkx−yk
kxk
> 0.
Hence μK(∂K ∩ [x,K] \ K) ≥ H(y,NK(y)) ∩ Con(x, α) ≥ rn−1
1 Bn−1
2  > 0. Let
Ej =
z ∈ ∂K ∩ [x,K]\K : f(z) ≥
1
j
, j = 1, 2, · · · .
As μK ({z ∈ ∂K : f(z) = 0}) = 0 and Ej ⊆ Ej+1 for all j,
μK
∂K ∩ [x,K] \ K
= μK
∞[
j=1
Ej
= lim
j→∞
μK(Ej).
22
Therefore there exists j1 such that μK(Ej1) > 0. Thus
μf (∂K ∩ [x,K]\K) ≥ μf (Ej1) ≥
μK(Ej1)
j1
> 0
which contradicts that x ∈ Kf,0.
(ii) is an immediate consequence of Lemma 3.1 (i) and Lemma (3.2) (i). Indeed,
these lemmas imply that K = Kf,0 =
T
s>0Kf,s. So, also using (3.19), there
exists s0 > 0 such that for all 0 ≤ s ≤ s0, Kf,s ⊂ 2K ⊂ Bn
2 (0, 2
). In particular,
Kf,s0 ⊂ Bn
2 (0, 2
).
Remark. The assumption that f is μKalmost everywhere strictly positive is
necessary in order that Kf,0 = K. To see that, let K = B2
2 and
f(x, y) =
(
0 x =
p
1 − y2, y ∈ [−1, 1],
1
otherwise.
Then Kf,0 = K ∪ {(x, y) : x ≥ 0, y ≤ 1}.
This example also shows that there is no s0 such that Kf,s is bounded for all
0 ≤ s ≤ s0 unless f is μKalmost everywhere strictly positive.
Let K be a convex body with 0 ∈ int(K). Let f : ∂K → R be an integrable,
μKalmost everywhere strictly positive function. For x /∈ K, let t0 = t0(x) be the
strictly positive real number such that t0x = ∂K ∩ [0, x]. Define hx(t) to be
hx(t) = μf (∂K ∩ [tx,K] \ K), t ≥ t0.
Clearly hx(t0) = 0. Moreover hx(t) ≤ s if tx ∈ Kf,s, and hx(t) > s if tx /∈ Kf,s.
Lemma 3.3 Let K be a convex body in Rn and f : ∂K → R be an integrable,
μKalmost everywhere strictly positive function.
(i) hx(t) is increasing and left continuous on [t0,∞).
(ii) Kf,s is closed for all s ≥ 0. In particular, it is compact for all 0 ≤ s ≤ s0.
If K is in addition strictly convex, then
(iii) hx(t) is continuous on [t0,∞).
(iv) For any 0 ≤ s ≤ s0 and x ∈ ∂Kf,s, one has μf (∂K ∩ [x,K] \ K) = s.
23
Proof.
(i) If t1 ≤ t2, then ∂K ∩ [t1x,K] \ K ⊆ ∂K ∩ [t2x,K] \ K. Thus hx(t1) ≤ hx(t2).
Let now t > t0 and (tm)m∈N be a sequence, increasing to t. Then, by monotonicity
of hx, hx(tm) ≤ hx(t) for all m and thus limmhx(tm) ≤ hx(t). We have to show
that limmhx(tm) ≥ hx(t). Let y ∈ relint@K(∂K ∩ [tx,K] \ K), where relintB(A)
is the relative (with respect to B) interior of a set A ⊆ B, i.e., relintB(A) =
{x ∈ A : there is a δ > 0, such that B(x, δ) ∩ B ⊂ A}. Then y ∈ int ([tx,K]), and
therefore there exists m0(y) ∈ N, such that y ∈ int([tm0(y)x,K]). This implies that
y ∈ [tm0(y)x,K] \ K ∩ ∂K and thus
relint@K([tx,K] \ K ∩ ∂K) ⊂
[
m≥1
[tmx,K] \ K ∩ ∂K.
By continuity of the measure μf from below, one has
hx(t) = μf
[tx,K] \ K ∩ ∂K
= μf
relint@K
[tx,K] \ K ∩ ∂K
≤ μf
[
m≥1
[tmx,K] \ K ∩ ∂K
= lim
m
μf
∂K ∩ [tmx,K] \ K
= limmhx(tm)
(ii) It will follow from Lemma 3.2 (ii) that Kf,s is compact for 0 ≤ s ≤ s0, once we
have proved that Kf,s is closed.
To that end, we show that (Kf,s)c, the complement of Kf,s in Rn, is open for all
s ≥ 0. Suppose this is not the case. Then there exists x ∈ (Kf,s)c and a sequence
(xm)m∈N, such that xm → x as m → ∞ but xm ∈ Kf,s for all m. Without loss of
generality, we can assume that xm are not in the ray of {tx : t ≥ 0}. Otherwise,
if xm ∈ Kf,s are in the ray, then hx
kxmk
kxk
≤ s and by (i), limm hx
kxmk
kxk
=
hx(1) ≤ s. This contradicts with hx(1) > s.
Now we let Km = [xm,K]. For sufficiently big m, ∂Km∩[0, x] 6= ∅. Suppose not,
then x ∈ Km implies that [x,K] ⊂ Km, and hence [x,K] \ K∩∂K ⊂ Km \ K∩∂K.
Since μf ([x,K] \ K ∩ ∂K) > s, one gets μf (Km \ K ∩ ∂K) > s, a contradiction
with xm ∈ Kf,s. Let ym = ∂Km ∩ [0, x]. Thus μf ([ym,K] \ K ∩ ∂K) ≤ s. Let α
be as in (3.19). Similarly, ∂
[xm,Bn
2 (0, α)]
∩ [0, x] 6= ∅ for sufficiently big m and
we denote zm = ∂
[xm,Bn
2 (0, α)]
∩ [0, x].
24
It is easy to check that 0 ≤ kxk−kymk ≤ kxk−kzmk for any m. As α ≤ kzmk ≤
kxk and
kzmk ≤ kx−xmk
kxk−kzmk
, one has kxk − kzmk ≤ kxk kx−xmk . Thus zm → x, and
hence also ym → x, as m → ∞. Therefore we can choose a subsequence (ymk )k∈N
that is monotone increasing to x. By (i) with tmk = kymk k
kxk
, hx(tmk ) ր hx(1) as
k → ∞. Since for all k, hx(tmk ) ≤ s, one has hx(1) ≤ s, a contradiction.
(iii) It is enough to prove that hx(t) is right continuous on [t0,∞). To do so, let
t ≥ t0 and let (tm)m∈N be a sequence decreasing to t. By (i), hx(tm) ≥ hx(t) for all
m, thus limmhx(tm) ≥ hx(t) and we have to show that limmhx(tm) ≤ hx(t). We
claim that if K is strictly convex, then
∂K ∩ [tx,K] \ K =
∞\
m=1
∂K ∩ [tmx,K] \ K
. (3.20)
We only need to prove that
T
∞m=1
∂K ∩ [tmx,K] \ K
⊆ ∂K ∩ [tx,K] \ K. Let
z0 ∈
T
∞m=1
∂K ∩ [tmx,K] \ K
. Thus z0 ∈ ∂K. Let l(z0, tx) be the line passing
through tx and z0. We have two cases.
Case 1: l(z0, tx) is in a tangent hyperplane of K. Then l(z0, tx) ∩ ∂K = {z0} by
strict convexity of K. Therefore, {z0} = [z0, tx] \ K ∩ ∂K ⊆ [tx,K] \ K ∩ ∂K.
Case 2: l(z0, tx) ∩ ∂K consists of two points, z0 and z1. As z0 ∈
T
∞m=1
∂K ∩
[tmx,K] \ K
, we must have ktx−z0k < ktx−z1k. Therefore, {z0} = [z0, tx] \ K∩
∂K ⊆ [tx,K] \ K ∩ ∂K.
Hence by (3.20) and continuity of the measure μf from above,
hx(t) = μf
∞\
m=1
h
∂K ∩ [tmx,K] \ K
i
= lim
m
μf
∂K ∩ [tmx,K] \ K
= lim
m
hx(tm).
(iv) Let 0 ≤ s ≤ s0, and x ∈ ∂Kf,s which implies that hx(1) ≤ s. Define x(s) =
{t : hx(t) = s}. Then x(s) 6= ∅. Indeed, let tax = ∂Bn
2 (0, 3
) ∩ Tx, where α is as
in (3.19) and Tx = {tx : t ≥ t0(x) > 0}. The proof of Lemma 3.2 (ii) shows that
Kf,s0 ⊂ Bn
2 (0, 2
). It is clear that tx /∈ Kf,s0, and hence hx(t) > s0. In fact, if
tx ∈ Kf,s0, then tx ∈ Bn
2 (0, 2
), but by definition of t, tx ∈ ∂Bn
2 (0, 3
). This is
a contradiction.
By continuity of hx(·), there must exist t ∈ [t0, t], such that hx(t) = s. This
also shows that ¯t = supx(s) ≤ t. Clearly hx(¯t) = s and thus ¯tx ∈ Kf,s. This
implies that ¯t ≤ 1 because x ∈ ∂Kf,s. Suppose ¯t < 1. Then s = hx(¯t) ≤ hx(1) ≤ s
25
by monotonicity of hx(·), a contradiction with ¯t = sup(s). Thus ¯t = 1 and
hx(1) = s.
Remark. Strict convexity is needed in (iii) and (iv). Indeed, let x = (0, 2) and
K = conv
{(1, 1), (−1, 1), (−2, 0), (2, 0)}
.
Then ∂K ∩ [x,K]\K = [(−1, 1), (1, 1)]. However for any point tx with t > 1,
[tx,K] \ K ∩ ∂K = ∂K \ [(−2, 0), (2, 0)] ) ∂K ∩ [x,K]\K.
Thus, for any function f with f > 0 on [(−2, 0), (−1, 1)] and/or [(1, 1), (2, 0)], hx(·)
is not right continuous on [1,∞).
To see that strict convexity is needed also in (iv), observe that Kf,1/12 = K in
Example 3.1. Thus, for x ∈ ∂Kf,1/12 = ∂K, we have
μf ([x,K] \ K ∩ ∂K) = 0 6=
1
12
.
4 Geometric interpretation of functionals on convex
bodies
We now give geometric interpretations of functionals on convex bodies, such as Lp
affine surface area and mixed paffine surface area for all p 6= −n using the non
convex illumination surface bodies. While there are no geometric interpretations
for mixed paffine surface area, many geometric interpretations of Lp affine surface
area have been discovered in the last years, all based on using convex bodies (e.g.,
[33, 37, 38, 45]). The remarkable new fact here is that now the bodies involved in
the geometric interpretation are not necessarily convex.
Theorem 4.1 Let K be a convex body in C2+
. Let c > 0 be a constant, and
f : ∂K → R be an integrable function such that f ≥ c μKalmost everywhere.
Then
lim
s→0
cn Kf,s − K
s
2
n−1
=
Z
@K
κK(x)
1
n−1
f(x)
2
n−1
dμK(x), (4.21)
where cn = 2Bn−1
2 
2
n−1 .
26
Remark. As dμK = fKdσ, we also have
lim
s→0
cn Kf,s − K
s
2
n−1
=
Z
Sn−1
fK(u)
n−2
n−1
f(N−1
K (u))
2
n−1
dσ(u), (4.22)
where N−1
K is the inverse of the Gauss map NK(·).
The geometric interpretation of Lp affine surface area is then a corollary to Theorem
4.1. The theorem also gives geometric interpretations of other known functionals
on convex bodies, e.g. the surface area and the mixed paffine surface area. Notice
that these geometric interpretations can also be obtained using e.g. the (convex)
surface body [38, 45].
Define
˜ f(N−1
K (u)) = fK(u)
n−2
2 [fp(K1, u) · · · fp(Kn, u)]
1−n
2(n+p) ,
where fp(K, u) = hK(u)1−pfK(u).
Corollary 4.1 Let K and Ki, i = 1, · · · , n, be convex bodies in C2+
. Then
lim
s→0
cn K ˜ f,s − K
s
2
n−1
= asp(K1, · · · ,Kn).
In particular, if all Ki coincide with K, then asp(K1, · · · ,Kn) = asp(K) and we
get a geometric interpretation of asp(K)
cn lim
s→0
Kgp,s − K
s
2
n−1
= asp(K),
where gp : ∂K → R is defined by gp(x) = κK(x)
n+2p−np
2(n+p) hx,NK(x)i
n(n−1)(p−1)
2(n+p) .
Corollary 4.2 Let K be a convex body in C2+
and g(x) =
p
κK(x). Then
lim
s→0
cn Kg,s − K
s
2
n−1
= μK(∂K).
The proof of the corollaries follows immediately from Theorem 4.1. To prove
Theorem 4.1, we need several other concepts and lemmas.
As K is in C2+
, for any x ∈ ∂K, the indicatrix of Dupin is an ellipsoid. As in [38],
we apply an affine transform T : Rn → Rn to K so that the indicatrix of Dupin
27
is transformed into an (n − 1)dimensional Euclidean ball. T has the following
properties:
T(x) = x T(NK(x)) = NK(x) det(T) = 1 (4.23)
and T maps a measurable subset of a hyperplane orthogonal to NK(x) onto a
subset of the same (n − 1)dimensional measure. It was also shown in [38] that
for any ǫ > 0 there is 1 = 1(ε) > 0 such that for all measurable subsets A of
∂K ∩ H−(x − 1NK(x),NK(x))
(1 − ǫ) μK(A) ≤ T(A) ≤ (1 + ǫ) μK(A). (4.24)
T(K) can be approximated at x = T(x) by a ndimensional Euclidean ball: For
any ǫ > 0 there is 2 = 2(ε) such that
Bn
2 (x − rNK(x), r) ∩ H− (x − 2NK(x),NK(x))
⊆ T(K) ∩ H− (x − 2NK(x),NK(x)) (4.25)
⊆ Bn
2 (x − RNK(x),R) ∩ H− (x − 2NK(x),NK(x)) ,
where r = r(x) = κK(x)− 1
n−1 and R = R(x) with r ≤ R ≤ (1 + ǫ)r . We put
= (ε) = min{1,2}. (4.26)
Moreover, for x ∈ ∂K, let
xs ∈ ∂Kf,s be such that x ∈ [0, xs] ∩ ∂K (4.27)
and define ˜xs to be the orthogonal projection of xs onto the ray {y : y = x +
tNK(x), t ≥ 0}. Clearly T(˜xs) = ˜xs, and the distance from T(xs) to the hyperplane
H(x,NK(x)) is the same as the distance from xs to this hyperplane.
We say that a family of sets Es ⊆ ∂K, 0 < s ≤ s0 shrinks nicely to a point
x ∈ ∂K (see [13]) if
(SN1) diamEs → 0, as s → 0.
(SN2) There is a constant β > 0 such that for all s ≤ s0 there exists ts with
μK (∂K ∩ B(x, ts)) ≥ μK(Es) ≥ β μK (∂K ∩ B(x, ts)) .
28
Lemma 4.1 Let K be a convex body in C2+
and f : ∂K → R an integrable, μKalmost
everywhere strictly positive function. Let x ∈ ∂K and let xs and ˜xs be as
above (4.27). Then
(i) The family ∂K ∩ [˜xs,K] \ K, 0 < s ≤ s0 shrinks nicely to x.
(ii) The family ∂K ∩ [xs,K] \ K, 0 < s ≤ s0 shrinks nicely to x.
(iii)
lim
s→0
μf (∂K ∩ [˜xs,K] \ K)
μK(∂K ∩ [˜xs,K] \ K)
= f(x) μKa.e. (4.28)
(iv)
lim
s→0
μf (∂K ∩ [xs,K] \ K)
μK(∂K ∩ [xs,K] \ K)
= f(x) μK a.e. (4.29)
Proof. Formulas (4.28) and (4.29) in (iii) and (iv) follow from the Lebesgue
differentiation theorem (see [13]) once we have proved that ∂K ∩ [ ˜ xs,K] \ K and
∂K ∩ [xs,K] \ K shrink nicely to x. Therefore it is enough to prove (i) and (ii).
(i) For x ∈ ∂K, let r = r(x) and R = R(x) be as in (4.25). We abbreviate
B(r) = Bn
2 (x − rNK(x), r) and B(R) = Bn
2 (x − RNK(x),R). Let
(x, s) =
x
kxk
,NK(x)
kxs − xk = hxs − x,NK(x)i
be the distance from xs to H
x,NK(x)
. This is the same as the distance from ˜xs
(defined after formula (4.27)) to H
x,NK(x)
.
Let hR = hR(s) = R (x,s)
R+(x,s) be the height of the cap of B(R) that is “illuminated”
by ˜xs. Then
H−
x − hRNK(x),NK(x)
∩ ∂B(R) = [T(˜xs),B(R)] \ B(R) ∩ ∂B(R). (4.30)
Let be as in (4.26). Since (x, s) → 0 as s → 0, one can choose s1 ≤ s0, such
that for all 0 < s ≤ s1, h = 2hR < . Therefore (4.25) holds:
H−(x − hNK(x),NK(x)) ∩ B(r) ⊂ H−
x − hNK(x),NK(x)
∩ T(K)
⊂ H−
x − hNK(x),NK(x)
∩ B(R). (4.31)
(4.31) and (4.30) imply that for all small enough s ≤ s2 ≤ s1
T
[˜xs,K] \ K ∩ ∂K
= [T(˜xs), T(K)] \ T(K) ∩ ∂T(K)
29
⊆ H−
x − hNK(x),NK(x)
∩ ∂T(K) ⊆ B(R) ∩ H−
x − hNK(x),NK(x)
.
Let ts = kx − zk =
q
4R2 (x,s)
R+(x,s) where z is any point in H
x − hNK(x),NK(x)
∩
∂B(R). As B(R) ∩ H−
x − hNK(x),NK(x)
⊆ Bn
2 (x, ts),
T
[˜xs,K] \ K
∩ ∂T(K) = T
[˜xs,K] \ K ∩ ∂K
⊆ Bn
2 (x, ts) ∩ ∂T(K)
and ts → 0 as s → 0.
This shows that condition (SN1) is satisfied for T
[˜xs,K] \ K
∩ ∂T(K) to shrink
nicely to T(x) = x.
We now show that condition (SN2) also holds true.
First, [T(˜xs),B(r)] \ B(r) ∩ H(x,NK(x)) is a (n − 1)dimensional Euclidean ball
with radius r(x,s) √2r(x,s)+2(x,s)
. Then for any 0 < s < s2,
T([˜xs,K] \ K ∩ ∂K) ≥ [T(˜xs),B(r)] \ B(r) ∩ H(x,NK(x))
≥ Bn−1
2 
r(x, s) p
2r(x, s) + 2(x, s)
!n−1
.
We can choose ( a new, smaller) s2 such that (x,s)
r ≤ 2. Then
T([˜xs,K] \ K ∩ ∂K) ≥ 21−n Bn−1
2  (r(x, s))
n−1
2 . (4.32)
On the other hand, for ε small enough, there exists s3 < s2, such that, for all
0 < s ≤ s3 and for any subset A of H−(x − 2hNK(x),NK(x)) ∩ ∂T(K) [38]
PH(x−2hNK(x),NK(x))(A) ≤ A ≤ (1 + ε)PH(x−2hNK(x),NK(x))(A) (4.33)
where PH(A) is the orthogonal projection of A onto the hyperplane H. We apply
this to A = Bn
2 (x, ts) ∩ ∂T(K):
Bn
2 (x, ts) ∩ ∂T(K) ≤ (1 + ε)PH(x−2hNK(x),NK(x))(Bn
2 (x, ts) ∩ ∂T(K))
≤ (1 + ε) B(R) ∩ H(x − 2hNK(x),N)
≤ (1 + ε) Bn−1
2 
2√2R
p
R(x, s)
R + (x, s)
!n−1
≤ 4n Bn−1
2  (r(x, s))
n−1
2 (4.34)
The last inequality follows as r ≤ R < (1 + ε)r. It now follows from (4.32) and
(4.34) that also condition (SN2) holds true for e.g. β = 8−n.
30
Hence the family T
[˜xs,K] \ K
∩∂T(K) shrinks nicely to T(x) = x and therefore,
as T−1 exists, the family [˜xs,K] \ K∩∂K = T−1
T
[˜xs,K] \ K
∩∂T(K)
shrinks
nicely to x.
(ii) Let v1 = xs − (x − rNK(x)) and v2 = xs − (x − RNK(x)). θ denotes the angle
between NK(x) and x and φi = φi(x, s), i = 1, 2 is the angle between NK(x) and
vi, i = 1, 2. These angles can be computed as follows
tan(φ1) =
(x, s) tan(θ)
r + (x, s)
tan(φ2) =
(x, s) tan(θ)
R + (x, s)
.
Then, for i = 1, 2, φi → 0 as s → 0. Since K is in C2+
, this means that for any
ε > 0 there is ¯s" ≤ s0 such that for all s ≤ ¯s"
1 − ε ≤
μK([xs,K] \ K ∩ ∂K)
μK([˜xs,K] \ K ∩ ∂K) ≤ 1 + ε. (4.35)
By (4.25) and as φi → 0, i = 1, 2 as s → 0, one can choose ˜hR = ˜hR(s) = 3R (x,s)
R+(x,s)
so small that
T
[xs,K] \ K ∩ ∂K
= [T(xs), T(K)] \ T(K) ∩ ∂T(K)
⊂ H−
x −˜hRNK(x),NK(x)
∩ B(R).
Let ˜ts =
q
6R2 (x,s)
R+(x,s) be the distance from x to any point in H
x −˜hRNK(x),NK(x)
∩
∂B(R). Then
T
[xs,K] \ K ∩ ∂K
⊆ Bn
2 (x, ˜ts) ∩ ∂T(K).
(4.24), (4.35) and Lemma 4.1 (i) then give
T
[xs,K] \ K ∩ ∂K
 ≥ (1 − ε)3T
[˜xs,K] \ K ∩ ∂K

≥ (1 − ε)3βBn
2 (x, ts) ∩ ∂T(K). (4.36)
Furthermore, by (4.33), one has
Bn
2 (x, ts) ∩ ∂T(K) ≥ H(x − hNK(x),NK(x)) ∩ T(K)
≥ H(x − hNK(x),NK(x)) ∩ B(r)
=
4R2r(x, s) + 4Rr(x, s)2 − 4R2(x, s)2
(R + (x, s))2
n−1
2
Bn−1
2 .
31
Since (x, s) → 0 as s → 0, for ¯s" small enough, and any 0 < s < ¯s", one get
Bn
2 (x, ts) ∩ ∂T(K) ≥
2R
R + (x, s)
p
r(x, s) − (x, s)2
n−1
Bn−1
2 
≥ 2−n (r(x, s))
n−1
2 Bn−1
2 . (4.37)
A computation similar to (4.34) shows that for all 0 < s ≤ ¯s" with (a possibly new)
¯s" small enough
Bn
2 (x, ˜ts) ∩ ∂T(K) ≤ (1 + ε) B(R) ∩ H(x −˜ hNK(x),N)
= (1 + ε) Bn−1
2 
√6R
p
R(x, s)
R + (x, s)
!n−1
≤ 3n Bn−1
2  (r(x, s))
n−1
2 (4.38)
(4.36), (4.37) and (4.38) imply that
T
[xs,K] \ K ∩ ∂K
 ≥ (48)−n−1βBn
2 (x, ˜ts) ∩ ∂T(K).
This shows that T
[xs,K] \ K ∩ ∂K
shrinks nicely to x. Therefore also [xs,K] \ K∩
∂K shrinks nicely to x.
Lemma 4.2 Let K be a convex body in C2+
and f : ∂K → R an integrable, μKalmost
everywhere strictly positive function. Then for μKalmost all x ∈ ∂K one
has
lim
s→0
cn
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
s
2
n−1
=
κK(x)
1
n−1
f(x)
2
n−1
, (4.39)
where xs ∈ ∂Kf,s is such that x ∈ [0, xs].
Proof. It is enough to consider x ∈ ∂K such that f(x) > 0. As x and xs are
collinear and as (1 + t)n ≥ 1 + tn for t ∈ [0, 1), one has for small enough s,
hx,NK(x)i
n
kxsk
kxk
n
− 1
= hx,NK(x)i
n
1 + kxs − xk
kxk
n
− 1
≥ (x, s).
Recall that (x, s) =
D
x
kxk
,NK(x)
E
kxs−xk = hxs−x,NK(x)i is the distance from
xs to H (x,NK(x)).
32
Similarly, as (1 + t)n ≤ 1 + nt + 2nt2 for t ∈ [0, 1), one has for s small enough,
hx,NK(x)i
n
kxsk
kxk
n
− 1
≤ (x, s)
1 +
2n
n
kxs − xk
kxk
. (4.40)
Hence for ε > 0 there exists s" ≤ s0 such that for all 0 < s ≤ s"
1 ≤
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
n(x, s) ≤ 1 + ε.
K is strictly convex as K ∈ C2+
. Thus, μf (∂K ∩ [xs,K] \ K) = s by Lemma 3.3
(iv). Therefore
1 ≤
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
μf (∂K ∩ [xs,K] \ K)
2
n−1
n s
2
n−1 (x, s) ≤ 1 + ε.
By Lemma 4.1 (iv) and (4.35), it then follows that we can choose (a new) s" so
small that we have for all s ≤ s"
1 − c1ε ≤
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
f(x) μK(∂K ∩ [ ˜ xs,K] \ K)
2
n−1
n s
2
n−1 (x, s) ≤ 1 + c2ε.
(4.41)
with absolute constants c1 and c2.
Let T be as in (4.23) and let r = r(x) and R = R(x) be as in (4.25). We abbreviate
again B(r) = Bn
2 (x−rNK(x), r) and B(R) = Bn
2 (x−RNK(x),R). Let hr = hr(s) =
r(x,s)
r+(x,s) . As hr → 0 as s → 0, we have for all s sufficiently small that hr <
where is as in (4.26). Hence by (4.25)
H− (x − hrNK(x),NK(x)) ∩ B(r) ⊂ H− (x − hrNK(x),NK(x)) ∩ T(K).
If we denote by PH the orthogonal projection onto the hyperplane H− (x − hrNK(x),NK(x)),
this then implies that
PH
∂K ∩ [˜xs,K] \ K
⊃ PH
∂T−1(B(r)) ∩ [˜xs, T−1(B(r))] \ T−1(B(r))
= H (x − hrNK(x),NK(x)) ∩ T−1(B(r)).
33
Hence for s sufficiently small
μK(∂K ∩ [˜xs,K] \ K) ≥
PH
∂K ∩ [˜xs,K] \ K
≥
H (x − hrNK(x),NK(x)) ∩ T−1(B(r))
=
T
H (x − hrNK(x),NK(x)) ∩ T−1(B(r))
=
H (x − hrNK(x),NK(x)) ∩ B(r)
=
1 − (x,s)
2r
n−1
2
1 + (x,s)
r
n−1 (2r(x, s))
n−1
2 Bn−1
2 
≥ (1 − c3ε)
(2(x, s))
n−1
2 p
κK(x) Bn−1
2  (4.42)
where c3 > 0 is an absolute constant.
By (4.25), one has
PH
∂K ∩ [˜xs,K] \ K
⊆ PH
∂
T−1(B(R))
∩ [˜xs, T−1(B(R))] \ T−1(B(R))
= H (x − hRNK(x),NK(x)) ∩ T−1(B(R))
where H = H (x − hRNK(x),NK(x)). The equality follows as hR = R(x,s)
R+(x,s) .
Together with (4.33), for s" small enough, whenever 0 < s < s", one has
μK(∂K ∩ [˜xs,K] \ K) ≤ (1 + ε)PH(∂K ∩ [˜xs,K] \ K)
≤ (1 + ε)
H (x − hRNK(x),NK(x)) ∩ T−1(B(R))
.
A calculation similar to (4.42) then shows that with an absolute constant c4
μK(∂K ∩ [˜xs,K] \ K) ≤ (1 + c4ε)
(2(x, s))
n−1
2 p
κK(x) Bn−1
2 . (4.43)
Combining (4.41), (4.42) and (4.43), we prove the formula (4.39), i.e.,
lim
s→0
cn
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
n s
2
n−1
=
κK(x)
1
n−1
f(x)
2
n−1
.
Lemma 4.3 Let K be a convex body in C2+
. Let c > 0 be a constant and f :
∂K → R an integrable function with f ≥ c μKalmost everywhere. Then there
34
exists s ≤ s0, such that for all s ≤ s,
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
s
2
n−1 ≤ c(K, n),
where c(K, n) is a constant (depending on K and n only), and x and xs are as in
Lemma 4.2.
Proof.
As K ∈ C2+
, by the Blaschke rolling theorem [36], there exists r0 > 0 such that
for all x ∈ ∂K, Bn
2 (x − r0NK(x), r0) ⊆ K. Let γ be such that 0 < γ ≤ min{1, r0}.
By Lemmas 3.1 (i) and 3.2 (ii), K = Kf,0 =
T
s>0Kf,s. Therefore there exists
s = s
≤ s0, such that for all s ≤ s, Kf,s ⊆ (1 + γ)K. Hence for xs ∈ ∂Kf,s and
x = [0, xs] ∩ ∂K, kxsk
kxk ≤ 1 + γ, or equivalently as x and xs are collinear
kxsk
kxk − 1 = kxs − xk
kxk ≤ γ ≤ 1. (4.44)
Together with (4.40), one has for all s ≤ s (with a possibly smaller s)
hx,NK(x)i
kxsk
kxk
n
− 1
≤ (x, s) [n + 2n] . (4.45)
As K ∈ C2+
, K is strictly convex. Hence, by Lemma 3.3 (iv), (4.25) and as
f ≥ c on ∂K
s = μf (∂K ∩ [xs,K] \ K) =
Z
@K[xs,K]\K)
fdμK ≥ c μK
∂K ∩ [xs,K] \ K
≥ c
H (x,NK(x)) ∩ [xs, T−1(B(r))]
= c
H (x,NK(x)) ∩ [T(xs), B(r)]
where T is as in (4.23) and r = r(x) is as in (4.25).
As in the proof of Lemma 3.2 (i), H (x,NK(x)) ∩ [T(xs), B(r)] contains a (n−1)
dimensional Euclidean ball of radius at least
r
p
2r(x, s) + 2(x, s)
2r + (x, s) ≥
α
1 + 2α
p
2r0(x, s).
The inequality follows as (x, s) = kxs−xk
kxk hx,NK(x)i ≤
≤ 1
, which is a direct
consequence of (3.19) and (4.44).
35
Hence s ≥ c
1+2
p
2r0(x, s)
n−1
Bn−1
2  and with (4.45) we get that
hx,NK(x)i
s
2
n−1
kxsk
kxk
n
− 1
≤ (n + 2n)
1 + 2α
α
2
2r0 c
2
n−1 Bn−1
2 
2
n−1
−1.
Finally, we also need the following lemma. It is well known and we omit the proof.
Lemma 4.4 Let K be a convex body and L be a starconvex body in Rn.
(i) If 0 ∈ int(L) and L ⊂ K, then
K − L =
1
n
Z
@Khx,NK(x)i
1 −
kx′k
kxk
n
dμK(x)
where x ∈ ∂K and x′ ∈ ∂L ∩ [0, x].
(ii) If 0 ∈ int(K) and K ⊂ L, then
L − K =
1
n
Z
@Khx,NK(x)i
kx′k
kxk
n
− 1
dμK(x)
where x ∈ ∂K, x′ ∈ ∂L and x = ∂K ∩ [0, x′].
Proof of Theorem 4.1.
As K ∈ C2+
, K is strictly convex. By Lemmas 4.4, 4.2, 4.3 and Lebegue’s Dominated
Convergence theorem
cn lim
s→0
Kf,s − K
s
2
n−1
= cn lim
s→0
Z
@K
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
n s
2
n−1
dμK(x)
= cn
Z
@K
lim
s→0
hx,NK(x)i
h
kxsk
kxk
n
− 1
i
n s
2
n−1
dμK(x)
=
Z
@K
κK(x)
1
n−1
f(x)
2
n−1
dμK(x).
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Elisabeth Werner
Department of Mathematics Universit´e de Lille 1
Case Western Reserve University UFR de Math´ematique
Cleveland, Ohio 44106, U. S. A. 59655 Villeneuve d’Ascq, France
elisabeth.werner@case.edu
Deping Ye
Department of Mathematics
Case Western Reserve University
Cleveland, Ohio 44106, U. S. A.
dxy23@case.edu
39
Friday, December 26, 2008
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