Notes on isotropic convex bodies
Warsaw, October 2003
Contents
Notation
v
1 Isotropic position and the slicing problem
1
1.1 Isotropic position: existence and uniqueness 1
1.2 Isotropic constant 3
1.3 Connection with Sylvester’s problem 6
1.4 Binet ellipsoid of inertia 10
1.5 Sections of an isotropic convex body 10
1.6 Connection with the M-position 15
1.7 Connection with the Busemann-Petty problem 19
2 An
2.1
2.2
2.3
2.4
2.5
upper bound for the isotropic constant
Khintchine type inequalities for polynomials
Ψα -estimates 26
Reduction to small diameter 28
Upper bound for the isotropic constant 30
Alternative proof 31
23
23
3 Isotropic convex bodies with small diameter
37
3.1 Lq -centroid bodies 37
3.2 Isotropic convex bodies with small diameter 40
3.3 An example 44
4 Reduction to bounded volume ratio
51
4.1 Symmetrization of isotropic convex bodies 51
4.2 Monotonicity in the dimension 56
4.3 Generalizations of Busemann’s inequality 58
4.4 Reduction to bounded volume ratio 61
5 The
5.1
5.2
5.3
5.4
5.5
1-unconditional case
67
Bound for the isotropic constant 67
Tail estimates for the Euclidean norm
Linear functionals on the `n1 -ball 73
The comparison theorem 77
Ψ2 -behavior of linear functionals 78
71
iv · Contents
6 Concentration of volume on isotropic convex bodies
6.1 Formulation of the problem 81
6.2 A first reduction 82
6.3 Average decay of hyperplane sections 83
6.4 Reduction to the average sectional decay 85
7 Random points and random polytopes
91
7.1 Random points in isotropic convex bodies 91
7.2 Random polytopes in isotropic convex bodies 96
8 The
8.1
8.2
8.3
8.4
central limit problem 107
Concentration property for p-balls 107
The ε-concentration hypothesis 110
The variance hypothesis 118
Consequences of the variance hypothesis
120
A Entropy estimates 123
A.1 Subgaussian processes 123
A.2 Metric entropy - the case of Gaussian processes 124
A.3 Dudley’s bound for subgaussian processes 125
A.4 Majorizing measures 126
A.5 Translation to the language of convex bodies 127
References
129
81
Notation
We work on Rn , which is equipped with a Euclidean structure h·, ·i. We denote
by | · | the corresponding Euclidean norm, and write B2n for the Euclidean unit
ball and S n−1 for the unit sphere. We fix a coordinate system defined by an
orthonormal basis {e1 , . . . , en }. Volume (n-dimensional Lebesgue measure) and
the cardinality of a finite set are also denoted by |·|. We write ωn for the volume
of B2n .
We write L(Rn ) for the family of all linear transformations T : Rn → Rn . The
class of invertible T ∈ L(Rn ) is denoted by GL(n), and SL(n) denotes the
subclass of volume preserving transformations.
In these notes, convex body is a compact convex subset K of Rn with 0 ∈ int(K).
A convex body K is called symmetric if x ∈ K ⇒ −x ∈ K. We say that K has
center of mass at the origin if
Z
(1)
hx, θidx = 0
K
n−1
for every θ ∈ S
.
The radial function ρK : Rn \{0} → R+ of K is defined by
(2)
ρK (x) = max{λ > 0 : λx ∈ K}.
The support function hK : Rn → R of K is defined by
(3)
hK (x) = max{hx, yi : y ∈ K}.
The width of K in the direction of θ ∈ S n−1 is the quantity w(K, θ) = hK (θ) +
hK (−θ), and the mean width of K is defined by
Z
Z
1
w(K) =
w(K, θ)σ(dθ) =
hK (θ)σ(dθ),
(4)
2 S n−1
S n−1
where σ is the rotationally invariant probability measure on S n−1 . We write µ
for the Haar probability measure on O(n). With Gn,k we denote the Grassmann
manifold of k-dimensional subspaces of Rn . Then, O(n) equips Gn,k with a Haar
probability measure νn,k .
The circumradius of K is the quantity
(5)
R(K) = max{|x| : x ∈ K}.
The polar body K ◦ of K is
(6)
K ◦ := {y ∈ Rn : hx, yi ≤ 1 for all x ∈ K}.
vi · Contents
The Brunn-Minkowski inequality describes the effect of Minkowski addition to
volume: If A and B are two non empty compact subsets of Rn , then
(7)
|A + B|1/n ≥ |A|1/n + |B|1/n ,
where A + B := {a + b | a ∈ A, b ∈ B}. It follows that, for every λ ∈ (0, 1)
(8)
|λA + (1 − λ)B|1/n ≥ λ|A|1/n + (1 − λ)|B|1/n ,
and, by the arithmetic-geometric means inequality,
(9)
|λA + (1 − λ)B| ≥ |A|λ |B|1−λ .
Let K be a symmetric convex body in Rn . The function
(10)
kxkK = min{λ ≥ 0 : x ∈ λK}
is a norm on Rn . The normed space (Rn , k · kK ) will be denoted by XK .
Conversely, if X = (Rn , k · k) is a normed space, then the unit ball KX = {x ∈
Rn : kxk ≤ 1} of X is a symmetric convex body in Rn .
The dual norm k · k∗ of k · k is defined by kyk∗ = max{|hx, yi| : kxk ≤ 1}.
From the definition it is clear that |hx, yi| ≤ kyk∗ kxk for all x, y ∈ Rn . If
◦
X ∗ = (Rn , k · k∗ ) is the dual space of X, then KX ∗ = KX
. Note that if
n−1
θ∈S
then ρK (θ) = 1/kθkK and hK (θ) = kθkK ◦ .
Whenever we write a ' b, we mean that there exist absolute constants c1 , c2 > 0
such that c1 a ≤ b ≤ c2 a (also, b ¿ a means that a exceeds Cb for some (large)
absolute constant C > 1). The letters c, c0 , C, c1 , c2 etc. denote absolute positive
constants which may change from line to line.
Basic references on classical and asymptotic convex geometry are the books [17],
[45], [51] and [53].
Chapter 1
Isotropic position and the
slicing problem
1.1
Isotropic position: existence and uniqueness
A convex body K in Rn is called isotropic if it has volume |K| = 1, center of
mass at the origin, and there is a constant α > 0 such that
Z
(1.1.1)
hx, yi2 dx = α2 |y|2
K
for all y ∈ Rn .
Remark 1: (i) It is not hard to check that the isotropic condition (1.1.1) is
equivalent to each one of the following statements:
1. For every i, j = 1, . . . , n,
Z
(1.1.2)
xi xj dx = α2 δij ,
K
where x1 , . . . , xn are the coordinates of x with respect to some orthonormal
basis.
2. For every T ∈ L(Rn ),
Z
(1.1.3)
hx, T xidx = α2 (trT ).
K
(ii) Note that if K satisfies the isotropic condition (1.1.1) then
Z
(1.1.4)
|x|2 dx = nα2 .
K
(ii) If K is an isotropic convex body in Rn and U ∈ O(n), then U (K) is also
isotropic.
Our first result proves the existence of an isotropic body in every linear class.
Proposition 1.1.1. Let K be a convex body in Rn with center of mass at the
origin. There exists T ∈ GL(n) such that T (K) is isotropic.
2 · Isotropic position and the slicing problem
R
Proof. The operator M ∈ L(Rn ) defined by M (y) = K hx, yixdx has a symmetric and positive square root S. Consider the linear image K̃ = S −1 (K) of
K. Then, for every y ∈ Rn we have
Z
Z
hx, yi2 dx = |detS|−1
hS −1 x, yi2 dx
K̃
K
Z
−1
hx, S −1 yi2 dx
= |detS|
K
Z
­
®
= |detS|−1
hx, S −1 yixdx, S −1 y
K
= |detS|−1 hM S −1 y, S −1 yi = |detS|−1 |y|2 .
Normalizing volume, we get the result.
Let K be a convex body in Rn with volume 1 and center of mass at the
origin. We say that K̃ is a position of K if K̃ = T K for some T ∈ SL(n). The
next Theorem shows that the isotropic position of a convex body is uniquely
determined (if we ignore orthogonal transformations) and arises as a solution of
a minimization problem.
Theorem 1.1.2. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. Define
½Z
¾
2
(1.1.5)
B(K) = inf
|x| dx : T ∈ SL(n) .
TK
Then, a position K1 of K is isotropic if and only if
Z
(1.1.6)
|x|2 dx = B(K).
K1
If K1 and K2 are isotropic positions of K, there exists U ∈ O(n) such that
K2 = U (K1 ).
Proof. (1) Fix an isotropic position K1 of K. Remark 1 shows that there exists
α > 0 such that
Z
(1.1.7)
hx, T xidx = α2 (trT )
K1
for every T ∈ L(Rn ). Then, for every T ∈ SL(n) we have
Z
Z
Z
2
2
|x| dx =
|T x| dx =
hx, T ∗ T xidx
T K1
K1
K1
Z
2
∗
2
= α tr(T T ) ≥ nα =
|x|2 dx,
K1
where we have used the arithmetic-geometric means inequality in the form
(1.1.8)
tr(T ∗ T ) ≥ n[det(T ∗ T )]1/n .
This shows that K1 satisfies (1.1.6). In particular, the infimum in (1.1.5) is a
minimum.
1.2 Isotropic constant · 3
(2) Note also that if we have equality in (1.1.8) then T ∗ T = I, that is T ∈
O(n). This shows that any other position K̃ of K which satisfies (1.1.6) is an
orthogonal image of K1 , therefore it is isotropic.
(3) Finally, if K2 is some other isotropic position of K then the first part of
the proof shows that K2 satisfies (1.1.6). By (2) we must have K2 = U (K1 ) for
some U ∈ O(n).
Remark 2: An alternative way to see that if K is a solution of the minimization
problem above then K is isotropic, is using the following simple variational
argument. Let T ∈ L(Rn ). For small ε > 0, I + εT is invertible, and hence
(I + εT )/[det(I + εT )]1/n preserves volume. Consequently,
Z
Z
|x + εT x|2
(1.1.9)
|x|2 dx ≤
dx.
2/n
K
K [det(I + εT )]
Noting that |x + εT x|2 = |x|2 + 2εhx, T xi + O(ε2 ) and [det(I + εT )]2/n =
2
+
1 + 2ε trT
n + O(ε ) and letting ε → 0 , we see that
Z
Z
trT
2
(1.1.10)
|x| dx ≤
hx, T xidx.
n K
K
Since T was arbitrary, a similar inequality holds for −T , therefore
Z
Z
trT
2
(1.1.11)
|x| dx =
hx, T xidx
n K
K
for all T ∈ L(Rn ). By Remark 1(i), this condition implies that K is isotropic.
1.2
Isotropic constant
The preceding discussion shows that, for every convex body K in Rn with center
of mass at the origin, the constant
½
¾
Z
¯
1
1
2
¯
(1.2.1)
L2K = min
|x|
dx
T
∈
GL(n)
2
n
|T K|1+ n T K
is well-defined and depends only on the linear class of K. Also, if K is isotropic
then for all θ ∈ S n−1 we have
Z
(1.2.2)
hx, θi2 dx = L2K .
K
The constant LK is called isotropic constant of K.
Conjecture 1 (isotropic constant): There exists an absolute constant C > 0
such that
(1.2.3)
LK ≤ C
for every convex body K with center of mass at the origin. Equivalently, if K
is an isotropic convex body in Rn , then
Z
(1.2.4)
hx, θi2 dx ≤ C 2
K
4 · Isotropic position and the slicing problem
for every θ ∈ S n−1 .
A reverse estimate is available and quite simple. Therefore, what the conjecture
states is that the isotropic constants of all convex bodies are (uniformly with
respect to n) of the order of 1.
Proposition 1.2.1. For every isotropic convex body K in Rn ,
(1.2.5)
LK ≥ LB2n ≥ c,
where c > 0 is an absolute constant.
−1/n
Proof. If rn = ωn
, then |rn B2n | = 1 and rn B2n is isotropic. Let K be an
isotropic convex body. Observe that |x| > rn on K \ rn B2n and |x| ≤ rn on
rn B2n \ K. It follows that
Z
|x|2 dx
nL2K =
ZK
Z
=
|x|2 dx +
|x|2 dx
Z
K∩rn B2n
K\rn B2n
Z
|x|2 dx +
≥
Z
K∩rn B2n
=
rn B2n
|x|2 dx
rn B2n \K
|x|2 dx = nL2B2n .
A simple computation shows that
(1.2.6)
L2B2n
1
=
n
Z
−2/n
|x|2 dx =
rn B2n
1 nωn n+2
ωn
r
=
≥ c2 ,
nn+2 n
n+2
where c > 0 is an absolute constant, therefore LK ≥ LB2n ≥ c.
In the symmetric case, a first upper bound for the isotropic constant follows
from John’s theorem.
Proposition
1.2.2. Let K be a symmetric convex body in Rn . Then, LK ≤
√
c n.
Proof. There exist r > 0 and T ∈ GL(n) such that rB2n is the ellipsoid of
n
maximal
√ volume inscribed in T K and |T K| = 1. Since rB2 ⊆ T K we have
r ≤ c n, and Theorem 1.1.2 shows that
Z
(1.2.7)
nL2K ≤
|x|2 dx ≤ R2 ,
TK
where R = R(K) is the circumradius of T K. Therefore,
(1.2.8)
LK ≤
R r
R
·√ ≤c .
r
r
n
Finally, from John’s theorem (see [29]) we have R ≤
√
nr.
Let us also note the following consequence of the reverse Santaló inequality [15].
1.2 Isotropic constant · 5
Proposition 1.2.3. Let K be a symmetric convex body in Rn . Then,
L2K L2K ◦ ≤ cnφ(K),
(1.2.9)
where
(1.2.10)
φ(K) =
1
|K| · |K ◦ |
√
In particular, LK LK ◦ ≤ c1 n.
Z Z
hx, yi2 dydx.
K◦
K
Proof. Observe that φ(K) = φ(T K) for every T ∈ GL(n). So, we may assume
that K ◦ is isotropic. Then, Theorem 1.1.2 shows that
(1.2.11)
1
|K|1+2/n · |K ◦ |1+2/n
Z Z
K
K◦
hx, yi2 dydx = L2K ◦
1
Z
|K|1+2/n
|x|2 dx ≥ nL2K L2K ◦ .
K
Therefore,
¡
¢−2/n
nL2K L2K ◦ ≤ |K| · |K ◦ |
φ(K),
(1.2.12)
and (1.2.9) follows from the reverse Santaló inequality:
¡
¢1/n
(1.2.13)
|K| · |K ◦ |
≥ c2 ωn2/n ≥ c3 /n,
where c3 > 0 is an absolute constant. Since φ(K) is trivially bounded by 1, the
proof is complete.
Finally, we give an estimate for the diameter of an isotropic convex body K
in terms of the isotropic constant LK (this proof is from [33]).
Theorem 1.2.4. Let K be an isotropic convex body in Rn . Then, R(K) ≤
(n + 1)LK .
Proof. Let x ∈ K. We define h : S n−1 → R by
(1.2.14)
h(u) = max{t > 0 : x + tu ∈ K}.
We can express the volume of K as
Z
1 = |K| =
=
nωn
Z
ωn
Z
S n−1
h(u)
tn−1 dtσ(du)
0
hn (u)σ(du).
S n−1
Since K is isotropic, for every θ ∈ S n−1 we may write
Z
L2K
hy, θi2 dy
=
K
=
Z
=
Z
Z
S n−1
≥
=
Z
tn−1 hx + tu, θi2 dtσ(du)
0
¡
¢
tn−1 hx, θi2 + 2tn hx, θihu, θi + tn+1 hu, θi2 dtσ(du)
0
µ n
¶
Z
h (u)
hn+1 (u)
hn+2 (u)
nωn
hx, θi2 + 2
hx, θihu, θi +
hu, θi2 σ(du)
n
n+1
n+2
S n−1
Z
hn (u)
2
hx, θi σ(du)
nωn
2
S n−1 n(n + 1)
Z
hx, θi2
hx, θi2
n
h
(u)σ(du)
=
ω
.
n
(n + 1)2
(n + 1)2
S n−1
nωn
S n−1
=
h(u)
nωn
h(u)
6 · Isotropic position and the slicing problem
This means that for every x ∈ K and θ ∈ S n−1 ,
(1.2.15)
|hx, θi| ≤ (n + 1)LK .
Therefore,
(1.2.16)
|x| = max
|hx, θi| ≤ (n + 1)LK .
n−1
θ∈S
Since x ∈ K was arbitrary, the proof is complete.
1.3
Connection with Sylvester’s problem
Let K be a convex body of volume 1 in Rn . We choose (n+1) points x1 , . . . , xn+1
independently and uniformly from K. Their convex hull co(x1 , . . . , xn+1 ) is a
random simplex in K. For every p > 0 we define
µZ
(1.3.1)
mp (K) =
¶1/p
Z
p
...
|co(x1 , . . . , xn+1 )| dxn+1 . . . dx1
K
.
K
If we do not want to assume that |K| = 1, the correct normalization is
µ
(1.3.2) mp (K) =
1
Z
p
...
|K|n+p+1
¶1/p
Z
|co(x1 , . . . , xn+1 )| dxn+1 . . . dx1
K
.
K
Then, mp (K) is invariant under non-degenerate affine transformations: If T ∈
GL(n) and u ∈ Rn , then mp (K) = mp (T K + u) for all p > 0. The quantity
m1 (K) is the expectation of the normalized volume of a random simplex inside
K.
Sylvester’s problem is the following question: describe the affine classes of
convex bodies for which mp (K) is minimized or maximized. It is known that,
for every p > 0,
mp (K) ≥ mp (B2n )
(1.3.3)
with equality if and only if K is an ellipsoid. In the other direction, the problem
is open if n ≥ 3.
Conjecture 2 (simplex conjecture): For every convex body K in Rn ,
(1.3.4)
m1 (K) ≤ m1 (Sn ),
where Sn is a simplex in Rn .
The conjecture is correct when n = 2. In this Section we will see that
Sylvester’s problem is connected with the isotropic constant problem: If the
simplex conjecture is correct, then LK ≤ C for every convex body K.
To see this, we define the variant of mp (K)
µ
(1.3.5)
Sp (K) =
1
|K|n+p
Z
K
¶1/p
Z
|co(0, x1 , . . . , xn )|p dxn . . . dx1
...
K
.
1.3 Connection with Sylvester’s problem · 7
Proposition 1.3.1. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. Then, for every p ≥ 1 we have
(1.3.6)
Sp (K) ≤ mp (K) ≤ (n + 1)Sp (K).
Proof. For every x ∈ K we define
µZ
¶1/p
Z
p
(1.3.7)
Sp (K; x) =
...
|co(x, x1 , . . . , xn )| dxn . . . dx1
.
K
K
We know that |co(x, x1 , . . . , xn )| = |det(x̃, x˜1 , . . . , x˜n )|/n! where z̃ = (z, 1) ∈
Rn+1 if z ∈ Rn , and this determinant is an affine function of x. It follows
that |co(x, x1 , . . . , xn )|p is a convex function on K. Integrating with respect to
x1 , . . . , xn we see that Spp (K; x) is also convex. Since 0 is the center of mass of
K and |K| = 1, we conclude that
Z
(1.3.8)
Spp (K; 0) ≤
Spp (K; x)dx,
K
which gives
Spp (K) ≤ mpp (K).
(1.3.9)
This proves the left hand side inequality. For the right hand side inequality we
observe that if x1 , . . . , xn+1 ∈ K, then
(1.3.10)
|co(x1 , . . . , xn+1 )| =
1
|det(x̃1 . . . , x̃n+1 )|
n!
where x̃j = (xj , 1) ∈ Rn+1 , and if we develop the determinant in the column
(1, . . . , 1) and apply the triangle inequality we get
(1.3.11)
|co(x1 , . . . , xn+1 )| ≤
n+1
X
|co(0, xi : i 6= j)|.
j=1
It follows that
µZ
|co(x1 , . . . , xn+1 )|p dxn+1 . . . dx1
...

K
K
Z
≤ 
Z
K
n+1
X µZ
j=1


...
K
≤
¶1/p
Z
mp (K) =
K
n+1
X
p
1/p
|co(0, xi : i 6= j)| dxn+1 . . . dx1 
j=1
¶1/p
Z
p
...
|co(0, xi : i 6= j)| dxn+1 . . . dx1
K
= (n + 1)Sp (K).
2
Remark 1: The function fi : K → R defined by xi 7→ det(x1 , . . . , xn ) for fixed
xj , j 6= i in K, is a linear functional. This leads to the following Proposition
(we postpone its proof, which is a simple consequence of the reverse Hölder
inequalities of §2.1).
8 · Isotropic position and the slicing problem
Proposition 1.3.2. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. Then,
S2 (K) ≤ cn S1 (K),
(1.3.12)
where c > 0 is an absolute constant.
2
Fix an orthonormal basis in Rn and write M (K) for the matrix with entries
Z
xi xj dx.
(1.3.13)
[M (K)]ij =
K
This is the matrix of inertia of K. The connection of mp (K) and Sp (K) with
the isotropic constant of K becomes clear by the next Proposition.
Proposition 1.3.3. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. Then,
det(M (K))
.
n!
S22 (K) =
(1.3.14)
Proof. By definition,
Z
S22 (K) =
(1.3.15)
Z
|co(0, x1 , . . . , xn )|2 dxn . . . dx1 .
...
K
K
We write xi = (xij ), j = 1, . . . , n. Then,
Z
Z
2 2
(1.3.16)
(n!) S2 (K) =
...
|det(x1 , . . . , xn )|2 dxn . . . dx1 ,
K
K
and expanding the determinant we get
Z
(n!)2 S22 (K)
=
...
K
K
Z
=
...
K
...
K
=
X
σ,ϕ
=
²ϕ
i=1
²σ ²τ
xi,σ(i)
¶µ X
²τ
τ
n
Y
!
xi,σ(i) xi,τ (i)
i=1
²ϕ
σ,ϕ
n µZ
Y
n
Y
i=1
σ,τ
Z ÃX
K
²σ
σ
Z ÃX
K
Z
=
Z µX
n
Y
¶
xi,τ (i) dxn . . . dx1
i=1
dxn . . . dx1
!
xi,σ(i) xi,ϕ(σ(i))
i=1
n
Y
dxn . . . dx1
¶
xi xϕ(i) dx
K
n! det(M (K)).
2
n
Remark 2: Let K be a convex body in R with volume 1 and center of mass at
the origin. Since S2 (K) is invariant under invertible linear transformations, the
identity (1.3.14) shows that if T ∈ SL(n) then
(1.3.17)
det(M (T K)) = det(M (K)).
If we choose T so that T K will be isotropic, then M (T K) = L2K I. So,
(1.3.18)
det(M (K)) = L2n
K.
Thus, we have proved the following.
1.3 Connection with Sylvester’s problem · 9
Theorem 1.3.4. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. Then,
2
L2n
K = n! S2 (K).
(1.3.19)
√
Corollary 1.3.5. LK ≤ c n for every convex body K in Rn with center of
mass at the origin.
Proof. We may assume that K is isotropic. Since S2 (K) is obviously bounded
by 1, we get
(1.3.20)
and
√
2n
LK ≤
√
2n
n!,
√
n! ≤ c n for some absolute constant c > 0.
Corollary 1.3.6. If the simplex conjecture is correct, then
(1.3.21)
LK ≤ C
for every convex body K in Rn with center of mass at the origin.
Proof. Consider the simplex
½
(1.3.22)
Sn =
x ∈ Rn : −
¾
n
X
n
1
1
≤ xi ≤
,
xi ≤
.
n+1
n + 1 i=1
n+1
Then, Sn0 = (n!)1/n Sn has volume 1 and center of mass at the origin. A simple
computation shows that
Z
(1.3.23)
0
Sn
2
x2i dx <
(n!)1+ n
,
(n + 2)!
and since M (K) is symmetric and positive definite, Hadamard’s inequality gives
det(M (Sn0 ))
1
S22 (Sn0 ) =
≤
n!
n!
Ã
2
(n!)1+ n
(n + 2)!
!n
≤
1
.
n!
Now, if K is an isotropic convex body in Rn we have m1 (K) ≤ m1 (Sn0 ), and
combining Propositions 1.3.1, 1.3.2 and Theorem 1.3.4 we obtain
LnK
=
≤
≤
≤
√
√
n! cn S1 (K)
√
n! cn m1 (K) ≤ n! cn m1 (Sn0 )
√
√
n! cn (n + 1)S1 (Sn0 ) ≤ n! cn (n + 1)S2 (Sn0 )
√
n! S2 (K) ≤
(n + 1)cn .
It follows that LK ≤ 2c.
10 · Isotropic position and the slicing problem
1.4
Binet ellipsoid of inertia
Let K be a convex body in Rn with volume 1 and center of mass at the origin.
The Binet ellipsoid EB (K) of K is defined by
Z
2
(1.4.1)
kykEB (K) =
hx, yi2 dx = hM y, yi.
K
n
Therefore, K is isotropic if and only if EB (K) = L−1
K B2 . In particular, if K is
n
an isotropic convex body in R , we have
|EB (K)| = ωn L−n
K .
(1.4.2)
Lemma 1.4.1. Let K be a convex body in Rn with volume 1 and center of mass
at the origin. Then,
|EB (K)| = ωn L−n
K .
(1.4.3)
Proof. We have already seen that |det(M (K)| = | det M (T K)| for every T ∈
SL(n). This shows that
(1.4.4)
|EB (T K)| = ωn | det M (T K)|−1/2 = ωn | det M (K)|−1/2 = |EB (K)|
for every T ∈ SL(n). We now choose T ∈ SL(n) so that T (K) is isotropic and
recall (1.4.2).
Corollary 1.4.2. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. There exists θ ∈ S n−1 such that
Z
(1.4.5)
hx, θi2 dx ≤ L2K .
K
Proof. Note that
Z
(1.4.6)
L−n
K = |EB (K)|/ωn =
S n−1
kθk−n
EB (K) σ(dθ).
It follows that minθ∈S n−1 kθkEB (K) ≤ LK .
1.5
Sections of an isotropic convex body
Let K be a convex body in Rn and fix a direction θ ∈ S n−1 . We define a
function f = fK,θ : R → R+ by
(1.5.1)
f (t) = |K ∩ (θ⊥ + tθ)|.
Here, | · | denotes (n − 1)-dimensional volume. So, f (t) gives the area of the
section of K with the hyperplane which is perpendicular to θ, at a distance t
from θ⊥ .
Theorem 1.5.1. Let K be a convex body in Rn , θ ∈ S n−1 , and f (t) = |K ∩
1
(θ⊥ + tθ)|. Then, f n−1 is concave on its support.
1.5 Sections of an isotropic convex body · 11
Proof. We may assume that θ = en , and identify θ⊥ with Rn−1 . For every t ∈ R
we set
K(t) = {x ∈ Rn−1 : (x, t) ∈ K}.
(1.5.2)
Let I = {t : K(t) 6= ∅}. Then, K(t) is convex for every t ∈ I, and if t, s ∈ I,
λ ∈ (0, 1), then
(1.5.3)
λK(t) + (1 − λ)K(s) ⊆ K(λt + (1 − λ)s).
Applying the Brunn-Minkowski inequality on Rn−1 , we get
(1.5.4)
1
1
1
|K(λt + (1 − λ)s)| n−1 ≥ λ|K(t)| n−1 + (1 − λ)|K(s)| n−1 .
Since f (t) = |K ∩ (θ⊥ + tθ)| = |K(t)|, the proof is complete.
Corollary 1.5.2. In the notation of Theorem 1.5.1, f is a log-concave function.
2
Corollary 1.5.3. If K is a symmetric convex body, then kf k∞ = f (0).
2
If we do not assume the symmetry of K, we still have that the hyperplane
section passing through the center of mass of K is comparable to the maximal
section in this direction.
Proposition 1.5.4. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. If θ ∈ S n−1 and f (t) = |K ∩ (θ⊥ + tθ)|, then
µ
(1.5.5)
kf k∞ ≤
n+1
n
¶n−1
f (0) ≤ e|K ∩ θ⊥ |.
Proof. Let [−a, b] be the support of f , where a, b > 0, and assume that kf k∞ =
f (t0 ). We will show that
µ
(1.5.6)
f (0) ≥
n
n+1
¶n−1
f (t0 ) ≥
1
f (t0 ).
e
Replacing θ by −θ if needed, we may assume that 0 < t0 ≤ b (we can also
assume that f (0) < f (t0 ); otherwise, there is nothing to prove). Since K has
its center of mass at the origin, we have
Z
Z
b
(1.5.7)
tf (t)dt =
−a
hx, θidx = 0.
K
Therefore,
Z
Z
0
(1.5.8)
(−t)f (t)dt =
−a
1
Z
b
t0
tf (t)dt ≥
0
tf (t)dt.
0
We set h = f n−1 and consider the linear function g defined by the equations
g(0) = h(0) and g(t0 ) = h(t0 ). By the Brunn-Minkowski inequality, h is concave,
which implies that g ≥ h on [−a, 0] and g ≤ h on [0, t0 ]. Since g(0) < g(t0 ),
12 · Isotropic position and the slicing problem
we see that g is strictly increasing and g(−γ) = 0 for some −γ ≤ −a. In other
words, g(t) = c(t + γ) for some c > 0 and γ ≥ α. Then,
Z 0
Z 0
Z 0
(−t)[g(t)]n−1 dt ≥
(−t)[g(t)]n−1 dt ≥
(−t)[h(t)]n−1 dt
−γ
−a
t0
Z
≥
Z
t0
t(t + γ)n−1 dt
−γ
γ+t0
= c
t
n−1
µ
n−1
(t − γ)dt = c
0
which implies t0 ≤
(1.5.9)
γ
n.
t[g(t)]n−1 dt.
0
This means that
Z t0
Z
0 ≥
t[g(t)]n−1 dt = cn−1
n−1
t0
t[h(t)]n−1 dt ≥
0
−γ
−a
Z
(γ + t0 )n+1
(γ + t0 )n
−γ
n+1
n
¶
,
Therefore,
f (0)
=
f (t0 )
µ
g(0)
g(t0 )
¶n−1
µ
=
γ
γ + t0
¶n−1
µ
≥
n
n+1
¶n−1
.
2
Our next observation is that the volume of the (n − 1)-dimensional section
|K ∩ θ⊥ | is closely related to integrals of the form
µZ
(1.5.10)
¶1/2
hx, θi dx
.
2
I2 (K, θ) :=
K
This connection becomes clear if we write I2 (K, θ) in the form
Z
Z
(1.5.11)
hx, θi2 dx =
t2 fK,θ (t)dt,
K
R
and will provide a link between the volume of sections and the isotropic position.
Proposition 1.5.5. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. For every θ ∈ S n−1 ,
µZ
¶1/2
1
1
(1.5.12)
hx, θi2 dx
≥ √
.
2 3e |K ∩ θ⊥ |
K
R +∞
Proof. Let f := fK,θ . We set B = 0 f (t)dt and define
(1.5.13)
g(t) = kf k∞ χ[0,B/kf k∞ ] (t).
Since g ≥ f on the support of g, we have
Z s
Z
f (t)dt ≤
0
s
g(t)dt
0
for every 0 ≤ s ≤ B/kf k∞ . The integrals of f and g on [0, +∞) are both equal
to B. So,
Z ∞
Z ∞
(1.5.14)
g(t)dt ≤
f (t)dt
s
s
1.5 Sections of an isotropic convex body · 13
for every s ≥ 0. It follows that
Z
∞
Z
t2 f (t)dt =
0
∞
Z
2sf (t)dsdt
µZ ∞
¶
2s
f (t)dt ds
0
s
¶
Z ∞ µZ ∞
2s
g(t)dt ds
s
Z0 ∞
t2 g(t)dt
Z0 ∞
=
≥
=
t
0
0
Z
B/kf k∞
=
t2 kf k∞ dt =
0
In the same way, if A =
R0
B3
.
3kf k2∞
f (t)dt, we see that
−∞
Z
0
t2 f (t)dt ≥
(1.5.15)
−∞
A3
.
3kf k2∞
Summing the estimates we conclude that
Z
B 3 + A3
,
3kf k2∞
hx, θi2 dx ≥
(1.5.16)
K
and since A + B = |K| = 1, it follows that
µZ
¶1/2
1
1
≥ √
.
2 3 kf k∞
hx, θi2 dx
(1.5.17)
K
Taking into account Proposition 1.5.4, we get the result.
Proposition 1.5.6. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. For every θ ∈ S n−1 we have
µZ
¶1/2
hx, θi2 dx
(1.5.18)
≤c
K
1
,
|K ∩ θ⊥ |
where c > 0 is an absolute constant.
Proof. We define f, A and B as in Proposition 1.5.5 and distinguish two cases.
Assume first that there exists s > 0 such that f (s) = f (0)/2. Since f is logconcave on [0, s] and f (0) ≥ f (s), we see that f (t) ≥ f (s) for all t ∈ [0, s].
Therefore,
Z
(1.5.19)
Z
∞
1≥B=
s
f (t)dt ≥
0
f (t)dt ≥ sf (s) = sf (0)/2.
0
s
s
If t > s, using the fact that f is log-concave, we have f (s) ≥ [f (0)]1− t [f (t)] t ,
14 · Isotropic position and the slicing problem
which implies f (t) ≤ f (0)2−t/s . We write
Z s
Z ∞
Z ∞
2
2
t f (t)dt =
t f (t)dt +
t2 f (t)dt
0
0
s
Z s
Z ∞
2
≤ kf k∞
t dt +
t2 f (0)2−t/s dt
0
s
µ 3
¶
Z ∞
s
3
2 −u
≤ f (0) e + s
u 2 du
3
1
≤
(c/f (0))2 ,
where we have used Proposition 1.5.4 and the estimate s ≤ 2/f (0).
On the other hand, if for all s > 0 on the support of f we have f (s) > f (0)/2,
then the role of s is played by s0 = max (suppf ∩ R+ ). We have 1 ≥ B ≥
f (0)s0 /2 and
Z ∞
Z s0
2
(1.5.20)
t f (t)dt =
t2 f (t)dt ≤ ef (0)s30 /3,
0
0
which leads to the same upper bound (c/f (0))2 .
The same reasoning applies to (−∞, 0]. It follows that
Z
Z ∞
Z 0
hx, θi2 dx =
t2 f (t)dt +
t2 f (t)dt
K
0
≤
−∞
(c1 /f (0))2 ,
where c1 > 0 is an absolute constant.
Assume that K is isotropic. Then, Propositions 1.5.5 and 1.5.6 show that
all (n − 1)-dimensional sections K ∩ θ⊥ of K have “the same volume”.
Theorem 1.5.7. Let K be an isotropic convex body in Rn . For every θ ∈ S n−1
we have
c1
c2
(1.5.21)
≤ |K ∩ θ⊥ | ≤
,
LK
LK
where c1 , c2 > 0 are absolute constants.
2
This establishes the connection to the slicing problem.
Conjecture 3 (slicing problem): There exists an absolute constant c > 0
with the following property: if K is a convex body in Rn with volume 1 and
center of mass at the origin, there exists θ ∈ S n−1 such that
(1.5.22)
|K ∩ θ⊥ | ≥ c.
It is now not hard to see that this conjecture is equivalent to the conjecture
about the isotropic constant. Assume that the slicing problem has an affirmative
answer. If K is isotropic, Theorem 1.5.7 shows that all sections K ∩ θ⊥ have
volume bounded by c2 /LK . Since (1.5.22) must be true for at least one θ ∈ S n−1 ,
we get LK ≤ c2 /c.
Conversely, if there exists an absolute bound C 2 for the isotropic constant,
then the slicing conjecture follows. An easy way to see this is through the Binet
1.6 Connection with the M-position · 15
ellipsoid of inertia. Assume that K is a convex body in Rn with volume 1 and
center of mass at the origin. According to Corollary 1.4.2, there exists θ ∈ S n−1
such that
Z
(1.5.23)
hx, θi2 dx ≤ L2K ≤ C 2 .
K
Now, Proposition 1.5.5 shows that
(1.5.24)
1.6
1
|K ∩ θ⊥ | ≥ c := √
.
2 3eC
Connection with the M-position
The following theorem of Milman ([40], see also [41]) establishes the existence
of “M -ellipsoids” associated to any convex body.
Theorem 1.6.1. There exists an absolute constant c > 0 with the following
property: For every convex body K in Rn with center of mass at the origin,
there exists an origin symmetric ellipsoid MK such that |K| = |MK | and for
every convex body T in Rn
(1.6.1)
¯1/n ¯
¯1/n
¯
¯1/n
1 ¯¯
MK + T ¯
≤ ¯K + T ¯
≤ c¯MK + T ¯
c
and
(1.6.2)
1 ◦
◦
|M + T |1/n ≤ |K ◦ + T |1/n ≤ c|MK
+ T |1/n .
c K
A consequence of Theorem 1.6.1 is that for every convex body K in Rn there
exists a position K̃ = uK (K) of volume |K̃| = |K| such that for every pair of
convex bodies K1 and K2 in Rn and for all t1 , t2 > 0,
³
´
(1.6.3)
|t1 K̃1 + t2 K̃2 |1/n ≤ c t1 |K̃1 |1/n + t2 |K̃2 |1/n ,
where c > 0 is an absolute constant. This statement is the reverse BrunnMinkowski inequality.
Recall the definition of the covering number N (A, B) of two convex bodies
A and B: this is the least integer N for which there exist N translates of B
whose union covers A. It is quite easy to check that |A + B| ≤ N (A, B)|2B|
and if B is symmetric, |A + B/2| ≥ N (A, B)|B/2|.
Interchanging the roles of K and MK , let us assume that the assertion of
Theorem 1.6.1 is satisfied by MK = B2n . This is always possible if we apply a
linear transformation to K. Then,
(1.6.4)
N (K, B2n ) ≤
2n |K + B2n |
≤ exp(c1 n),
|B2n |
where c1 > 0 is a constant depending only on c. This condition on N (K, B2n )
implies (see e.g. [44]) that
(1.6.5)
max{N (B2n , K), N (K ◦ , B2n ), N (B2n , K ◦ )} ≤ exp(c2 n)
for some constant c2 which again depends only on c1 . In other words, we have
the following.
16 · Isotropic position and the slicing problem
Proposition 1.6.2. There exists an absolute constant β > 0 such that every
convex body K in Rn with center of mass at the origin has a linear image K̃
with |K̃| = |B2n | which satisfies
(1.6.6)
max{N (K̃, B2n ), N (B2n , K̃), N (K̃ ◦ , B2n ), N (B2n , K̃ ◦ )} ≤ exp(βn).
We say that a convex body K in Rn which has volume |K| = |B2n |, center of
mass at the origin and satisfies (1.6.6) is in M -position with constant β. If K1
and K2 are two such convex bodies, then it is easily checked that
³
´
(1.6.7)
|K1 + K2 |1/n ≤ C |K1 |1/n + |K2 |1/n
and
(1.6.8)
³
´
|K1◦ + K2◦ |1/n ≤ C |K1◦ |1/n + |K2◦ |1/n
where C is a constant depending only on β. If K is in M -position with constant
β, setting K1 = K, K2 = B2n and using the reverse Santaló inequality, we get
(1.6.9)
cn |K| · |K ◦ | ≤ |K ∩ B2n | · |co(K ◦ ∪ B2n )| ≤ |K ∩ B2n | · |K ◦ + B2n |,
which, combined with (1.6.8), gives |K ∩ B2n | ≥ cn |K|.
Pisier (see [51], Chapter 7) has given a different approach to these results, which
provides a construction of special M -ellipsoids with regularity estimates on the
covering numbers. The precise statement is as follows.
Theorem 1.6.3. For every α > 1/2 and every convex body K there exists an
affine image K̃ of K which satisfies |K̃| = |B2n | and
(1.6.10)
µ
¶
c(α)n
max{N (K̃, tB2n ), N (B2n , tK̃), N (K̃ ◦ , tB2n ), N (B2n , tK̃ ◦ )} ≤ exp
t1/α
for¡ every t ≥ ¢1, where c(α) is a constant depending only on α, with c(α) =
O (α − 21 )−1/2 as α → 12 .
2
We then say that K̃ is in M -position of order α (or α-regular M -position).
The next Theorem shows that if the isotropic constant is uniformly bounded,
then isotropic convex bodies are in M -position of order 1. This would give a
description of the M -position in classical terms. Our method is based on an
estimate of the covering numbers N (K, tB2n ) in terms of the quantity
Z
n
(1.6.11)
M (K, B2 ) =
|x|dx.
K
Note that
µZ
(1.6.12)
M (K, B2n )
¶1/2
2
≤
|x| dx
=
√
nLK .
K
Theorem 1.6.4. Let K be an isotropic convex body in Rn . For every t > 0 we
have
¶
µ 3/2
6n LK
.
(1.6.13)
N (K, tB2n ) ≤ 2 exp
t
1.6 Connection with the M-position · 17
Proof. Consider the Minkowski functional pK (x) = inf{λ > 0 : x ∈ λK}. It
is clear that pK is subadditive and positively homogeneous. We define a Borel
probability measure on Rn by
Z
1
e−pK (x) dx,
(1.6.14)
µ(A) =
cK A
R
where cK = Rn exp(−pK (x))dx. Let {x1 , . . . , xN } be a subset of K which is
maximal with respect to the condition
(1.6.15)
i 6= j =⇒ |xi − xj | ≥ t.
S
Then, the balls xi + (t/2)B2n have disjoint interiors, and K ⊆ i≤N (xi + tB2n ).
Consequently, N (K, tB2n ) ≤ N .
We choose b > 0 so that µ(bB2n ) ≥ 1/2. If we set yi = (2b/t)xi , then
Z
Z
1
1
µ(yi + bB2n ) =
e−pK (x+yi ) dx ≥
e−pK (x) e−pK (yi ) dx
cK bB2n
cK bB2n
Z
2b
1
e−pK (x) dx = e− t pK (xi ) µ(bB2n )
= e−pK (yi )
cK bB2n
e−2b/t µ(bB2n ),
≥
since pK (xi ) ≤ 1, i = 1, . . . , N . The balls yi + bB2n have disjoint interiors,
therefore
ÃN
!
N
[
X
−2b/t
n
n
n
(1.6.16)
Ne
µ(bB2 ) ≤
µ(yi + bB2 ) = µ
(yi + bB2 ) ≤ 1.
i=1
i=1
It follows that
N (K, tB2n ) ≤ e2b/t (µ(bB2n ))−1 ≤ 2e2b/t .
(1.6.17)
What remains is to estimate b. We first compute the constant
Z
Z Z ∞
−pK (x)
cK =
e
dx =
e−s dsdx
Rn
∞
Rn
Z
=
pK (x)
Z
0
It follows that
J
Z
|x|µ(dx) =
Rn
1
n!
sn e−s ds = n!.
0
:=
=
∞
e−s |{x : pK (x) ≤ s}|ds =
Z
∞
1
cK
Z
Z
Rn
K
e−s dsdx
pK (x)
Z
sn+1 e−s ds ·
0
∞
|x|
|x|dx = (n + 1)M (K, B2n ).
n
From Markov’s inequality, µ(x ∈ R : |x| > 2J) ≤ 1/2, which shows thar
µ(2JB2n ) ≥ 1/2. If we choose b = 2J, we get
N (K, tB2n ) ≤ 2 exp(4J/t) ≤ 2 exp(4(n + 1)M (K, B2n )/t)
≤ 2 exp(6n3/2 LK /t),
which is the assertion of the Theorem.
18 · Isotropic position and the slicing problem
Note that |K| = 1, and hence |K| = |rn B2n | for some rn > 0 with rn '
Therefore,
µ
¶
cLK n
(1.6.18)
N (K, t(rn B2n )) ≤ 2 exp
t
√
n.
for all t > 0, where c > 0 is an absolute constant. This proves that if LK is
uniformly bounded, then isotropic convex bodies are in M -position of order 1.
In particular, every pair of isotropic convex bodies K and T we would satisfy
the reverse Brunn-Minkowski inequality. We will show that the converse is also
true.
Theorem 1.6.5. Assume that there exists a constant A > 0 such that for every
n the following holds true: if K and T are isotropic convex bodies in Rn , then
³
´
(1.6.19)
|K + T |1/n ≤ 2A = A |K|1/n + |T |1/n .
Then, for every convex body in Rn we have
LK ≤ cA4 ,
(1.6.20)
where c > 0 is an absolute constant.
We first prove a simple Lemma.
Lemma 1.6.6. Let K and T be two
isotropic convexn bodies in Rn and Rm rem
n+m
K × (LK /LT ) n+m T is an isotropic convex
spectively. Then, W := (LT /LK )
n+m
body in R
, and
n
m
n+m
LK×T = LK
LTn+m .
(1.6.21)
Proof. Let E be the subspace spanned by the first n standard unit vectors in
Rn+m . We define W = aK × bT where an bm = 1 so that |W | = 1, and write M
for the operator MW ∈ L(Rn+m ) defined by
Z
(1.6.22)
M (z) =
hw, zizdw.
W
It is clear that if z ∈ E then M (z) ∈ E. Also,
(1.6.23)
Z
Z
2
m
hM (z), zi =
hw, zi dw = b
hx, zi2 dx = bm an+2 L2K |z|2 = a2 L2K |z|2 .
W
aK
The same argument shows that if z ∈ E ⊥ then M (z) ∈ E ⊥ and
(1.6.24)
hM z, zi = b2 L2T |z|2 .
Since M acts as a multiple of the identity on both E and E ⊥ , we see that W
will be isotropic provided that
(1.6.25)
n m
Since a b
(1.6.26)
aLK = bLT .
= 1 this condition gives
µ
¶ m
¶ n
µ
LT n+m
LK n+m
a=
and b =
.
LK
LT
Also, LK×T = LW = aLK and inserting the value of a we complete the proof.
1.7 Connection with the Busemann-Petty problem · 19
Proof of Theorem 1.6.5: Let Ds = rs B2s be the Euclidean ball of volume 1 in
Rs . Let K be an isotropic convex body in Rn . According to Lemma 1.6.6, the
body W = (LDn /LK )1/2 K × (LK /LDn )Dn is an isotropic convex body in R2n
(and LW = (LK LDn )1/2 ). Applying (1.6.19) to W and D2n we get
|W + D2n |1/n ≤ 2A.
(1.6.27)
On the other hand,
(1.6.28) W + D2n ⊇ (LK /LDn )1/2 Dn + D2n ⊇ (LK /LDn )1/2 Dn × (r2n /rn )Dn .
Since r2n /rn ≥ c1 for some absolute constant c1 > 0, we get
µ
(1.6.29)
LK
LDn
¶1/4
1/2
c1
≤ 2A.
Since LDn ≤ c2 for some absolute constant c2 > 0, we have LK ≤ cA4 (where
2
c = 16c2 /c21 ).
1.7
Connection with the Busemann-Petty problem
Let K be a convex body in Rn . Assume that 0 ∈ int(K). For every θ ∈ S n−1
consider the section K ∩ θ⊥ of K, and the normalization of S1 (·)
(1.7.1)
Z
Z
1
⊥
S1 (K ∩ θ ) =
...
|co(0, x1 , . . . , xn−1 )|dxn−1 . . . dx1 .
|K ∩ θ⊥ |n K∩θ⊥
K∩θ ⊥
Busemann proved a formula which connects the volume of K with the areas of
the (n − 1)-dimensional sections K ∩ θ⊥ , θ ∈ S n−1 :
Theorem 1.7.1. If K is a convex body in Rn with 0 ∈ int(K), then
Z
n! ωn
|K ∩ θ⊥ |n S1 (K ∩ θ⊥ )σ(dθ).
(1.7.2)
|K|n−1 =
2
n−1
S
Proof. Consider Rn(n−1) = Rn1 ×· · ·×Rnn−1 , and set K̃ = K ×· · ·×K ⊂ Rn(n−1) .
We will write the coordinates of any xi ∈ Rni in the form xi = (x1i , . . . , xni ).
Then,
Z
Z
Z
n−1
dx1n−1 . . . dxnn−1 . . . dx11 . . . dxn1 .
(1.7.3) |K|
=
...
dxn−1 . . . dx1 =
K
K̃
K
Given x1 , . . . , xn−1 , let a1 , . . . , an−1 be the solution of the linear system
(1.7.4)
xni =
n−1
X
aj xji ,
i = 1, . . . , n − 1,
j=1
and consider the new coordinates x11 , . . . , x1n−1 , a1 , . . . , x1n−1 , . . . , xn−1
n−1 , an−1 on
Rn(n−1) . The Jacobian of this change of variables is
(1.7.5)
J = det[(xji )i,j=1,...,n−1 ].
20 · Isotropic position and the slicing problem
The set where J = 0 has measure 0 and outside this set the transformation is
well-defined.
P
If b = (1 + aj2 )−1/2 , then the unit vector θ normal to the hyperplane xn =
a1 x1 + · · · + an−1 xn−1 is described either by
θj = baj (j = 1, . . . , n − 1) and θn = −b,
(1.7.6)
or by
θj = −baj (j = 1, . . . , n − 1) and θn = b.
(1.7.7)
If φ is the angle formed by θ and the xn -axis, then b = | cos φ|, and hence,
dθ = 1b dθ1 . . . dθn−1 . Again, we may ignore those hyperplanes that are parallel
to the xn -axis. We can then check that
det[(∂θj /∂ai )i,j=1,...,n−1 ] = bn+1 ,
(1.7.8)
which gives
(1.7.9)
dθ =
1 1
dθ . . . dθn−1 = bn da1 . . . dan−1 = | cosn φ|da1 . . . dan−1 .
b
Note also that for the hyperplane xnj = a1 x1j + · · · + an−1 xn−1
we have
j
dxj = | sec φ|dx1j . . . dxn−1
.
j
(1.7.10)
Taking into account all the above, we write
Z
n−1
. . . dx1n−1 , . . . , dxn−1
|K|
=
|J|dx11 . . . dxn−1
n−1 da1 . . . dan−1
1
J −1 K̃
Z
Z
Z
nωn
=
...
|J sec φ|dx1 . . . dxn−1 σ(dθ),
2 S n−1 K∩θ⊥
K∩θ ⊥
where x1 , . . . , xn−1 are considered as points of θ⊥ . Observe that the projection
of the simplex co(0, x1 , . . . , xn−1 ) onto xn = 0 has volume |J|/(n − 1)!. Since
all these points belong to θ⊥ , we get
(1.7.11)
|co(0, x1 , . . . , xn−1 )|| cos φ| =
|J|
.
(n − 1)!
It follows that
|K|n−1
=
=
Z
Z
Z
n! ωn
|co(0, x1 , . . . , xn−1 )|dxn−1 . . . dx1 σ(dθ)
...
2
K∩θ ⊥
S n−1 K∩θ ⊥
Z
n! ωn
|K ∩ θ⊥ |n S1 (K ∩ θ⊥ )σ(dθ).
2
S n−1
2
If we assume that K is symmetric, then the results of §1.3 show that
p
(1.7.12)
S1 (K ∩ θ⊥ ) ' S2 (K ∩ θ⊥ ) = Ln−1
/ (n − 1)!.
K∩θ ⊥
Therefore, Theorem 1.7.1 has the following immediate consequences.
1.7 Connection with the Busemann-Petty problem · 21
Corollary 1.7.2. Let K be a symmetric convex body in Rn . Then,
Z
n−1
n−1
n
⊥ n
(1.7.13)
|K|
≤c
LK∩θ
⊥ |K ∩ θ | σ(dθ),
S n−1
where c > 0 is an absolute constant.
Corollary 1.7.3. Let K be a symmetric convex body in Rn . Then,
¶1/n
|K ∩ θ | σ(dθ)
,
µZ
(1.7.14)
|K|
n−1
n
≥ c1
⊥ n
S n−1
where c1 > 0 is an absolute constant.
These last results bring us to the Busemann-Petty problem which was originally
formulated as follows:
Assume that K1 and K2 are symmetric convex bodies in Rn and satisfy
(1.7.15)
|K1 ∩ θ⊥ | ≤ |K2 ∩ θ⊥ |
for all θ ∈ S n−1 . Does it follow that |K1 | ≤ |K2 |?
The answer is positive if n ≤ 4 and negative for all higher dimensions. What
remains open is the asymptotic version of the problem.
Conjecture 4 (Busemann-Petty problem): There exists an absolute constant c > 0 such that if K1 and K2 are symmetric convex bodies in Rn and
|K1 ∩ θ⊥ | ≤ |K2 ∩ θ⊥ | for all θ ∈ S n−1 , then |K1 | ≤ c|K2 |.
The results of this Section show that this conjecture is equivalent to the boundedness of the isotropic constant, at least if we restrict ourselves to the class of
symmetric convex bodies. Let us first assume that there is a constant C > 0
such that LW ≤ C for every symmetric convex body W . If K1 and K2 satisfy
(1.7.16)
|K1 ∩ θ⊥ | ≤ |K2 ∩ θ⊥ |
for all θ ∈ S n−1 , then Corollaries 1.7.1 and 1.7.2 show that
Z
n−1
n n−1
|K1 |
≤ c C
|K1 ∩ θ⊥ |n σ(dθ)
S n−1
Z
n n−1
≤ c C
|K2 ∩ θ⊥ |n σ(dθ)
n
≤ (c/c1 ) C
S n−1
n−1
|K2 |n−1 ,
which gives
(1.7.17)
|K1 | ≤ c3 |K2 |
for an absolute constant c3 > 0. Conversely, let us assume that Conjecture 4 is
correct and let K be an isotropic symmetric convex body in Rn . Let θ0 ∈ S n−1
be such that
(1.7.18)
|K ∩ θ0⊥ | = max
|K ∩ θ⊥ |.
n−1
θ∈S
22 · Isotropic position and the slicing problem
We choose r > 0 so that ωn−1 rn−1 = |K ∩ θ0⊥ |. Then,
(1.7.19)
|K ∩ θ⊥ | ≤ rn−1 ωn−1 = |(rB2n ) ∩ θ⊥ |
for all θ ∈ S n−1 , therefore
(1.7.20)
|K|n−1 ≤ cn−1 |rB2n |n−1 =
cn−1 ωnn−1
|K ∩ θ0⊥ |n ≤ cn1 |K ∩ θ0⊥ |n ,
n
ωn−1
for some absolute constant c1 > 0. Since |K| = 1, we see that
(1.7.21)
|K ∩ θ0⊥ | ≥ 1/c1 .
Since K is isotropic, we have |K ∩ θ⊥ | ' 1/LK for every θ ∈ S n−1 . It follows
that LK ≤ C for some absolute constant C > 0.
Notes and References
A basic reference on the isotropic position is the paper of Milman and Pajor [43].
It contains more information on the history of this topic (see also [3], [27] and
[9]), on basic properties of isotropic convex bodies and on the connection of the
isotropic constant conjecture with other questions (slicing problem, Sylvester’s
problem, Busemann-Petty problem) The fact that if LK is bounded then K is in
M -position was observed in [42], [43]. The argument in these notes is from [28]
(see also [36]) and works for not necessarily symmetric convex bodies. Theorem
1.6.5 is recent (see [14]).
Chapter 2
An upper bound for the
isotropic constant
2.1
Khintchine type inequalities for polynomials
Let K be a convex body in Rn with volume |K| = 1. For every p > 0 and
θ ∈ S n−1 we define
µZ
(2.1.1)
¶1/p
p
Ip (K, θ) =
|hx, θi| dx
.
K
Proposition 2.1.1. There exists an absolute constant c > 0 with the following
property: If K is a convex body in Rn with volume 1, then for all θ ∈ Rn and
all p > 1 we have
(2.1.2)
Ip (K, θ) ≤ cpI1 (K, θ).
The proof is based on Borell’s lemma.
Lemma 2.1.2. Let K be a convex body of volume 1 in Rn . Assume that A is
a symmetric convex subset of Rn such that |K ∩ A| = θ > 1/2. Then, for every
t > 1 we have
µ
n
(2.1.3)
|(R \ tA) ∩ K| ≤ θ
1−θ
θ
¶ t+1
2
.
Proof. Observe that
Rn \ A ⊇
2
t−1
(Rn \ tA) +
A,
t+1
t+1
therefore
(Rn \ A) ∩ K ⊇
¢ t − 1¡
¢
2 ¡ n
(R \ tA) ∩ K +
A∩K .
t+1
t+1
Then, apply the Brunn-Minkowski inequality.
24 · An upper bound for the isotropic constant
Proof of Proposition 2.1.1: We set I := I1 (K, θ) and
A = {x ∈ Rn : |hx, θi| ≤ 3I}.
(2.1.4)
Then, A is a symmetric convex subset of Rn , and Markov’s inequality shows
that
(2.1.5)
|A ∩ K| ≥ 2/3.
¡
¢
We observe that {x ∈ K : |hx, θi| ≥ t} = K ∩ Rn \ (t/3I)A , and write
Z
Z
3I
|hx, θi|p dx =
K
0
Z
ptp−1 |K ∩ (Rn \ (t/3I)A)|dt
∞
+
ptp−1 |K ∩ (Rn \ (t/3I)A)|dt.
3I
The first integral is bounded by
Z
(2.1.6)
3I
ptp−1 dt = (3I)p ,
0
while for the second one, making the change of variables t = 3Is and using
Lemma 2.1.2 we get
Z ∞
Z ∞
ptp−1 |K ∩ (Rn \ (t/3I)A)|dt = (3I)p
psp−1 |K ∩ (Rn \ sA)|ds
3I
Z1 ∞
p
≤ (3I)
psp−1 2−s/2 ds.
1
Combining the above and estimating the last integral, we see that
Z
£
¤
(2.1.7)
|hx, θi|p dx ≤ (3I)p 1 + (c1 p)p
K
for some absolute constant c1 > 0.
2
Remark 1: The same argument works if we replace the function x 7→ |hx, θi| by
any seminorm.
Proposition 2.1.1 can be equivalently stated as follows.
Proposition 2.1.3. Let K be a convex body of volume 1 in Rn . For every
θ ∈ S n−1 we have
Z
(2.1.8)
exp(|hx, θi|/CI1 (K, θ))dx ≤ 2,
K
where C > 0 is an absolute constant.
2
Corollary 2.1.4. Let K be a convex body of volume 1 in Rn . For every θ ∈
S n−1 and every s > 0,
(2.1.9)
Prob (x ∈ K : |hx, θi| ≥ CI1 (K, θ)s) ≤ 2e−s ,
where C > 0 is the constant in Proposition 2.1.3.
2
2.1 Khintchine type inequalities for polynomials · 25
More generally, write Pd,n for the space of polynomials f : Rn → R of degree
less than or equal to d. Bourgain [10] (see also Bobkov [5]) proved that for every
1 ≤ q, r ≤ ∞ there exists a constant cq,r,d > 0 depending only on q, r and d,
such that kf kLq (K) ≤ cq,r,d kf kLr (K) for every f ∈ Pd,n and every convex body
K of volume 1 in Rn . Carbery and Wright (see [19]) have recently established
the best possible dependence of the constant cq,r,d on q, r and d. In order to
formulate their result, for every f ∈ Pd,n we define f # (x) = |f (x)|1/d .
Theorem 2.1.5. There exists an absolute constant C > 0 such that for every
convex body K of volume 1 in Rn and for every f ∈ Pd,n the following hold true:
(i) If n ≤ r ≤ q ≤ ∞, then
kf # kq ≤ Ckf # kr .
(2.1.10)
(ii) If 1 ≤ r ≤ n ≤ q ≤ ∞, then
n
kf # kq ≤ C kf # kr .
r
(2.1.11)
(iii) If 1 ≤ r ≤ q ≤ n, then
q
kf # kq ≤ C kf # kr .
r
(2.1.12)
In particular,
q
kf # kq ≤ C kf # kr
r
(2.1.13)
whenever 1 ≤ r ≤ q < ∞.
2
These inequalities allow us to obtain tail estimates for polynomials f ∈ Pd,n .
Lemma 2.1.6. Let K be a convex body of volume 1 in Rn and let f ∈ Pd,n .
Then,
¡
¢
(2.1.14)
Prob x ∈ K : f # (x) ≥ 3Ckf # kq · s ≤ e−qs
for all q ≥ 1 and s ≥ 1, where C is the constant in Theorem 2.1.5.
Proof. From Theorem 2.1.5 we have
Z
(2.1.15)
f # (x)qp dx ≤ (Cp)qp kf # kqp
q .
K
for every p ≥ 1. Markov’s inequality gives
¡
¢
(2.1.16) (3Ckf # kq s)qp Prob x ∈ K : f # (x) ≥ 3Ckf # kq · s ≤ (Cp)qp kf # kqp
q ,
which shows that
(2.1.17)
¡
¢ ³ p ´qp
.
Prob x ∈ K : f # (x) ≥ 3Ckf # kq · s ≤
3s
If we choose p = 3s/e ≥ 1, we get
¡
¢
(2.1.18)
Prob x ∈ K : f # (x) ≥ 3Ckf # kq · s ≤ e−3qs/e ≤ e−qs
which is the assertion of the Lemma.
26 · An upper bound for the isotropic constant
For every p > 0 we define
µZ
(2.1.19)
¶1/p
p
Ip (K) =
|x| dx
.
K
Applying Lemma 2.1.6 to linear functionals f (x) = hx, θi and to the polynomial
f (x) = |x|2 , we have the following immediate consequence.
Proposition 2.1.7. Let K be a convex body of volume 1 in Rn . If q ≥ 1, then
(2.1.20)
Prob (x ∈ K : |hx, θi| ≥ 3CIq (K, θ)s) ≤ e−qs
for all θ ∈ S n−1 and s ≥ 1, and
(2.1.21)
Prob (x ∈ K : |x| ≥ 3CIq (K)s) ≤ e−qs
for all s ≥ 1, where C is the constant in Theorem 2.1.5.
2.2
2
Ψα -estimates
Let K be a convex body of volume 1 in Rn . If f : K → R is a bounded
measurable function and if α ≥ 1, the Orlicz norm kf kψα of f is defined by
½
¾
Z
α
(2.2.1)
kf kψα = inf t > 0 :
exp ((|f (x)|/t) ) dx ≤ 2 .
K
An equivalent description is the following.
Lemma 2.2.1. Let K be a convex body of volume 1 in Rn , f : K → R be a
bounded measurable function and α ≥ 1. Then,
½
¾
kf kp
(2.2.2)
kf kψα ' sup
:p≥α .
p1/α
Let α ≥ 1 and y 6= 0 in Rn . We say that K satisfies a ψα -estimate with constant
bα in the direction of y if
(2.2.3)
kh·, yikψα ≤ bα kh·, yik1 .
Proposition 2.1.3 is then taking the following form.
Proposition 2.2.2. There exists an absolute constant C > 0 such that for every
convex body K of volume 1 in Rn and every y 6= 0,
(2.2.4)
kh·, yikψ1 ≤ Ckh·, yik1 .
Next, we concentrate on isotropic convex bodies and give some first results on
the ψα -behavior of linear functionals and of the Euclidean norm. First, note
that in the case of an isotropic convex body Proposition 2.2.2 states that
(2.2.5)
kh·, θikψ1 ≤ CLK
for all θ ∈ S n−1 . We say that “every isotropic convex body is a ψ1 -body with
constant C”.
2.2 Ψα -estimates · 27
Proposition 2.2.3. Let α ≥ 1. Every isotropic convex body K is a ψα -body
with constant
¡
¢1− α1
.
(2.2.6)
bα ≤ c R(K)/LK
Proof. For every θ ∈ S n−1 and every p ≥ α we have
µZ
¡
¢ ¶1/p
p
1
|hx, θi| α |hx, θi|p 1− α dx
Ip (K, θ) =
K
1
1
R(K)1− α I αp (K, θ) α
´ α1
³ p
1
R(K)1− α c I1 (K, θ)
α
³ p
´ α1
1
1− α
R(K)
c LK
α
≤
≤
≤
1
1
1
α
c1 R(K)1− α LK
pα .
≤
The result now follows from Lemma 2.2.1.
The next Theorem (see [1]) asserts that the Euclidean norm satisfies a ψ2 estimate on isotropic convex bodies.
Theorem 2.2.4. Let K be an isotropic convex body in Rn . If f (x) = |x|, then
√
(2.2.7)
kf kψ2 ≤ c nLK ,
where c > 0 is an absolute constant.
We will use the following Lemma.
Lemma 2.2.5. Let K be a convex body in Rn with volume 1 and center of mass
at the origin. For every q ≥ 1,
µZ
¶1/q
Z
q
(2.2.8)
|hx, θi| dxσ(dθ)
S n−1
r
'
K
q
Iq (K).
q+n
Proof. For every q ≥ 1 and x ∈ Rn , we check that
µZ
(2.2.9)
¶1/q
√
q
|hx, θi| σ(dθ)
'√
|x|.
q+n
S n−1
q
A simple application of Fubini’s theorem gives the result.
Proof of Theorem 2.2.4: It suffices to prove that for every q > 1
µZ
¶1/q
|x|q dx
(2.2.10)
√ √
≤ c1 q nLK
K
for some absolute constant c1 > 0. From Proposition 2.1.1 we know that for
every θ ∈ S n−1
Z
(2.2.11)
|hx, θi|q dx ≤ cq2 q q LqK .
K
28 · An upper bound for the isotropic constant
Integrating on the sphere we get
Z
Z
(2.2.12)
|hx, θi|q dxσ(dθ) ≤ cq2 q q LqK .
S n−1
K
Taking into account Lemma 2.2.5, we see that
µZ
r
¶1/q
|x|q dx
(2.2.13)
≤ c3 q
K
n+q
√ √
LK ≤ c4 q nLK ,
q
provided that q ≤ n. On the other hand, if q > n, using the fact that R(K) ≤
(n + 1)LK , we get
µZ
¶1/q
√ √
|x| dx
≤ cnLK ≤ c q nLK .
q
(2.2.14)
K
Combining the above we see that there is an absolute constant c1 > 0 such that
µZ
¶1/q
√ √
|x| dx
≤ c1 q nLK
q
(2.2.15)
K
for all q > 1.
2
Corollary 2.2.6. There exists an absolute constant c > 0 such that: if K is an
isotropic convex body in Rn then
(2.2.16)
√
Prob(x ∈ K : |x| ≥ c nLK s) ≤ 2 exp(−s2 )
for every s > 0.
2.3
2
Reduction to small diameter
Proposition 2.3.1. Let K be an isotropic convex body in Rn . Then,√ there
exists an isotropic convex body Q in Rn with LQ ' LK and R(Q) ≤ c nLQ ,
where c > 0 is an absolute constant.
For the proof we need a simple observation about the stability of the isotropic
position
Lemma 2.3.2. Let K be a convex body of volume 1 in Rn . Assume that for
some constant L > 0 and some a, b > 0 we have
Z
−2 2
(2.3.1)
a L ≤
hx, θi2 dx ≤ b2 L2
K
for all θ ∈ S n−1 . Then,
(2.3.2)
a−1 L ≤ LK ≤ bL.
2.3 Reduction to small diameter · 29
Proof. Consider the Binet ellipsoid EB (K) which is defined by
Z
(2.3.3)
kθk2EB (K) =
hx, θi2 dx.
K
Our assumption implies that
a
1 1/n
ω
≤ |EB (K)|1/n ≤ ωn1/n .
bL n
L
(2.3.4)
On the other hand,
1/n
|EB (K)|1/n = L−1
K ωn
(2.3.5)
from Lemma 1.4.1. Combining the above we conclude the proof.
Proof of Proposition 2.3.1: Since K is isotropic, we have
Z
(2.3.6)
|x|2 dx = nL2K .
K
For every r > 0 we define
√
Kr = {x ∈ K : |x| ≤ r nLK }.
(2.3.7)
Markov’s inequality shows that |Kr | ≥ 1 − r−2 . Then, for every θ ∈ S n−1 ,
Z
Z
Z
hx, θi2 dx =
hx, θi2 dx −
hx, θi2 dx
Kr
K
K\Kr
µZ
≥
L2K − |K \ Kr |1/2
≥
L2K − r−1 (4c)2 L2K ,
¶1/2
hx, θi4 dx
K
where c > 0 is the constant in Proposition 2.1.1. If we choose r = 32c2 , we have
Z
L2K
(2.3.8)
≤
hx, θi2 dx ≤ L2K
2
Kr
for all θ ∈ S n−1 . We find a ≥ 1 with an ≤ c1 := 1/(1−r−2 ), such that W = aKr
has volume |W | = 1. Then,
Z
L2K 2
(2.3.9)
|y| ≤
hx, yi2 dx ≤ c21 L2K |y|2
2
W
for all y ∈ Rn , and Lemma 2.3.2 shows that LW ' LK .
Let T ∈ SL(n) be such that Q = T (W ) is isotropic. From (2.3.9), for every
θ ∈ S n−1 we have
Z
Z
(2.3.10)
L2W =
hx, θi2 dx =
hx, T ∗ θi2 dx ' |T ∗ θ|2 L2K .
Q
W
Since LQ = LW ' LK , we obtain
c2 ≤ |T ∗ θ| ≤ c3
(2.3.11)
for every θ ∈ S n−1 . Therefore,
(2.3.12)
√
√
√
R(Q) ≤ kT : `n2 → `n2 kR(W ) ≤ c3 ar nLK ≤ c5 nLK ≤ c nLQ
where c > 0 is an absolute constant.
2
30 · An upper bound for the isotropic constant
2.4
Upper bound for the isotropic constant
Bourgain’s argument (see [10]) may now be described for isotropic convex bodies
with small diameter.
Theorem 2.4.1. Let K be an isotropic convex body in Rn . Then,
√
(2.4.1)
LK ≤ c 4 n log n,
where c > 0 is an absolute constant.
n
Proof. From Proposition 2.3.1
√ we can find an isotropic convex body Q in R
with LQ ' LK and R(Q) ≤ c nLQ . Recall that
Z
(2.4.2)
hx, T xidx = (trT ) · L2Q
Q
for every T ∈ L(Rn ). The extremal property (Theorem 1.1.2) of the isotropic
position, combined with the arithmetic-geometric means inequality, shows that
for every symmetric positive definite T ∈ SL(n) we have
Z
Z
2
2
(2.4.3)
nLQ ≤ (trT ) · LQ =
hx, T xidx ≤
max hy, xidx.
Q y∈T Q
Q
√
Since R(Q) ≤ c nLQ , Proposition 2.2.3 shows that
°
°
°
°
h·, yi
°
° ≤ 1,
(2.4.4)
° c1 √
4
nLQ |y| °ψ2
for every y 6= 0, where c1 > 0 is an absolute constant. It follows that
√
(2.4.5)
Prob({x ∈ Q : hx, yi ≥ c1 4 nLQ t}) ≤ 2 exp(−t2 /|y|2 )
for every y 6= 0 and every t > 0.
We define a process X = (Xy )y∈T Q , where Xy : Q → R is given by
(2.4.6)
Xy (x) =
hx, yi
√
.
c1 4 nLQ
Then, for every y 6= z ∈ T Q and every t > 0, we have
Prob(|Xy − Xz | ≥ t)
√
= Prob({x ∈ Q : |hy − z, xi| ≥ c1 4 nLQ t})
¡ 2
¢
≤ 2 exp −t /|y − z|2 ,
that is, X is subgaussian with respect to the Euclidean metric on T Q.
Let g1 , . . . , gn be independent standard Gaussian random variables on some
probability space Ω, and consider the Gaussian process Z = (Zy )y∈T Q with
(2.4.7)
Zy (ω) = hG(ω), yi,
where G = (g1 , . . . , gn ). Then, X and Z satisfy the assumptions of Talagrand’s
comparison theorem (see Appendix A) . Therefore,
(2.4.8)
E sup Xy ≤ C · E sup Zy ,
y∈T Q
y∈T Q
2.5 Alternative proof · 31
where C > 0 is an absolute constant. A simple computation (see Appendix A)
shows that
√
(2.4.9)
E sup Zy ' nw(T Q).
y∈T Q
Therefore,
√
√
√
nL2Q ≤ c1 4 nLQ · E sup Xy ≤ c1 C 4 nLQ · nw(T Q).
(2.4.10)
y∈T Q
In other words,
√
LQ ≤ c2 w(T Q)/ 4 n,
(2.4.11)
where c2 > 0 is an absolute constant. Well-known results of Lewis [34] , Figiel
and Tomczak-Jaegermann [23], combined with Pisier’s inequality [50] on the
norm of the Rademacher projection, show that
√ there exists a symmetric positive
definite T ∈ SL(n) for which w(T Q) = O( n log n). This completes the proof.
The argument above makes use of the simple ψ2 -estimate from Proposition
2.2.3. It is natural to ask if linear functionals on an isotropic convex body with
small diameter exhibit better ψ2 -behavior. One may ask for the worst direction
θ ∈ S n−1 or, at least, for probability estimates on kh·, θikψ2 with respect to σ.
We study these two questions in the next Chapter.
2.5
Alternative proof
√
We sketch one more proof of the bound LK = O( 4 n log n), which is based on
the ψ1 -behavior of linear functionals on convex bodies. This argument is more
elementary, since it involves simpler entropy considerations.
1. Expectation of the maximum of linear functionals on convex bodies
Let K be an isotropic convex body in Rn , which satisfies the ψα -estimate
(2.5.1)
kh·, θikψα ≤ Bkh·, θik1 ≤ BLK
for all θ ∈ S n−1 , where α ∈ [1, 2] and B > 0.
Proposition 2.5.1. If θ1 , . . . , θN ∈ S n−1 , then
Z
(2.5.2)
max |hx, θi i| dx ≤ CBLK (log N )1/α ,
K 1≤i≤N
where C > 0 is an absolute constant.
Proof. For every t > 0,
µ
¶
Prob x ∈ K : max |hx, θi i| ≥ t
≤
1≤i≤N
N
X
Prob (x ∈ K : |hx, θi i| ≥ t)
i=1
≤
2N exp (−(t/BLK )α ) .
32 · An upper bound for the isotropic constant
Then, given A > 0 we may write
µ
¶
Z
Z ∞
max |hx, θi i|dx =
Prob x ∈ K : max |hx, θi i| ≥ t dt
1≤i≤N
K 1≤i≤N
0
µ
¶
Z ∞
≤ A+
Prob x ∈ K : max |hx, θi i| ≥ t dt
1≤i≤N
A
Z ∞
exp (−(t/BLK )α ) dt.
≤ A + 2N
A
Choosing A = 4BLK (log N )1/α we get
Z ∞
Z
exp (−(t/BLK )α ) dt = 4BLK (log N )1/α
∞
exp(−4α s log N )ds
1
Z ∞
1/α
4BLK (log N )
exp(−2 log N )
e−s ds
A
≤
1
1/α
≤
4BLK (log N )
N
−2
,
where we have used the fact that
(2.5.3)
exp(−4α s log N ) ≤ exp(−2 log N ) · e−s
is valid for all s ≥ 1. It follows that
Z
(2.5.4)
max |hx, θi i|dx ≤ CBLK (log N )1/α ,
K 1≤i≤N
with C = 12.
2. Dudley-Fernique decomposition
Let K be a convex body in Rn . We assume that 0 ∈ K and write R for the
circumradius of K. For every j ∈ N we may find a subset Nj of K such that
|Nj | = N (K, (R/2j )B2n ),
(2.5.5)
and
(2.5.6)
K⊆
[
(y + (R/2j )B2n ).
y∈Nj
Sudakov’s inequality (see Appendix A) shows that
µ
(2.5.7)
log |Nj | ≤ cn
2j w(K)
R
¶2
.
We define N0 = {0} and
(2.5.8)
for every j ≥ 1.
¯
Wj = Nj − Nj−1 = {y − y 0 ¯ y ∈ Nj , y 0 ∈ Nj−1 }
2.5 Alternative proof · 33
Lemma 2.5.2. For every x ∈ K and any m ∈ N we can find zj ∈ Wj ∩
(3R/2j )B2n , j = 1, . . . , m and wm ∈ (R/2m )B2n such that
(2.5.9)
x = z1 + · · · + zm + wm .
Proof. Let x ∈ K. By the definition of Nj , we can find yj ∈ Nj , j = 1, . . . , m,
such that
(2.5.10)
|x − yj | ≤
R
.
2j
We write
(2.5.11)
x = 0 + (y1 − 0) + (y2 − y1 ) + · · · + (ym − ym−1 ) + (x − ym ).
We set y0 = 0 and wm = x − ym , zj = yj − yj−1 for j = 1, . . . , m. Then,
|wm | = |x − ym | ≤ R/2m , and zj ∈ Nj − Nj−1 = Wj . Also,
(2.5.12)
|zj | ≤ |x − yj | + |x − yj−1 | ≤
R
R
3R
+ j−1 = j .
j
2
2
2
Finally, x = z1 + · · · + zm + wm .
We set Zj = Wj ∩ (3R/2j )B2n . Then, taking into account (2.5.7), we may state
the following.
Theorem 2.5.3. Let K be a convex body in Rn , with 0 ∈ K and circumradius
equal to R. There exist Zj ⊆ (3R/2j )B2n , j ∈ N, such that
µ
(2.5.13)
log |Zj | ≤ cn
2j w(K)
R
¶2
,
with the following property: for every x ∈ K and any m ∈ N we can find zj ∈ Zj ,
2
j = 1, . . . , m and wm ∈ (R/2m )B2n such that x = z1 + · · · + zm + wm .
Theorem 2.5.4. Let K be an isotropic convex body in Rn and let B > 0.
Assume that, for some 1 ≤ α < 2, K satisfies the ψα -estimate
(2.5.14)
kh·, θikψα ≤ Bkh·, θik1 ≤ BLK
for every θ ∈ S n−1 . Then,
α
α
1
α
LK ≤ CB 2 (2 − α)− 2 n 2 − 4 log n
(2.5.15)
where C > 0 is an absolute constant.
Proof. There exists a symmetric and positive T ∈ SL(n) such that
√
(2.5.16)
w(T K) ≤ c n log n.
We write
Z
(2.5.17)
nL2K
trT
=
|x| dx ≤
n
K
Z
2
Z
2
|x| =
K
hx, T xidx.
K
34 · An upper bound for the isotropic constant
Therefore,
Z
nL2K ≤
(2.5.18)
max |hy, xi|dx.
K y∈T K
We now use Theorem 2.5.3 for T K. If R is the circumradius of T K, for every
j = 1, . . . , m, m ∈ N we can find Zj ⊂ (3R/2j )B2n such that
µ
(2.5.19)
log |Zj | ≤ cn
w(T K)2j
R
¶2
,
and every y ∈ T K is written in the form y = z1 + · · · + zm + wm , where zj ∈ Zj
and wm ∈ (R/2m )B2n . This implies that
m
X
max |hy, xi| ≤
y∈T K
j=1
m
X
≤
j=1
max |hz, xi| +
z∈Zj
max
w∈(R/2m )B2n
|hw, xi|
3R
R
max |hz, xi| + m |x|,
j
2 z∈Zj
2
where z denotes the unit vector in the direction of z. Noting that
√
nLK and using the above, we see that
Z
Z
m
X
3R
R
2
nLK ≤
max |hz, xi|dx + m
|x|dx
2j K z∈Zj
2
K
j=1
Z
m
X
3R
R√
≤
max |hz, xi|dx + m nLK .
j
z∈Z
2
2
j
K
j=1
R
K
From Proposition 2.5.1 we get
(2.5.20)
nL2K ≤
m
X
3R
j=1
2j
µ
1
c00 n α LK B
w(T K)2j
R
¶ α2
+
The sum on the right is bounded by
(2.5.21)
2
2
1
2
c1
LK Bn α w(T K) α 2m( α −1) R1− α .
(2 − α)
Solving the equation
(2.5.22)
B
¢
¡
2
1
2
n α w(T K) α 2m α −1
2
R α −1 (2 − α)
√
R n
= m
2
we see that the optimal value of m satisfies the equation
1
(2.5.23)
α
1
n 2 − 4 w(T K) α
=
B2.
α
2m
(2 − α) 2 R
Going back to (2.5.20), we obtain
α
α
α
nL2K ≤ c2 (2 − α)− 2 n1− 4 w(T K)LK B 2 .
√
Since w(T K) ≤ c3 n log n, we get the result.
(2.5.24)
R√
nLK .
2m
|x|dx ≤
2.5 Alternative proof · 35
Notes and References
Bourgain’s theorem 2.4.1 (see [10]) is the best known general upper bound on the
isotropic constant. Reduction to small diameter is implicit in his argument (and
explicitly described in [48]). Originally, the Theorem was proved for symmetric
convex bodies. An alternative proof, based on the same ideas, was given by Dar
(see√[20]). A modification of Dar’s argument (see [46]) leads to the upper bound
n
O( 4 n log
√ n) for every convex body in R (the previously known general estimate
was O( n); see e.g. [21]). Bourgain [12] has also proved that LK ≤ cb log b for
ψ2 -bodies with constant b.
Chapter 3
Isotropic convex bodies
with small diameter
3.1
Lq -centroid bodies
Let K be a convex body of volume 1 in Rn . For every q ≥ 1 we define the
q-centroid body Zq (K) of K by
µZ
(3.1.1)
hZq (K) (y) = Iq (K, y) :=
¶1/q
|hx, yi|q dx
.
K
For simplicity we will sometimes write Hq (y) := hZq (K) (y). Since |K| = 1, it is
easy to check that
(3.1.2)
Z1 (K) ⊆ Zp (K) ⊆ Zq (K) ⊆ Z∞ (K)
for every 1 ≤ p ≤ q ≤ ∞, where Z∞ (K) = K̂ := co{K, −K}. Another simple
observation is that
(3.1.3)
Zq (T K) = T (Zq (K))
for every T ∈ SL(n) and q ∈ [1, ∞].
If K has its center of mass at the origin, then Zq (K) ' K̂ for all q ≥ n.
This follows from the next Lemma.
Lemma 3.1.1. Let K be a convex body in Rn with volume 1 and center of mass
at the origin. Then,
Z
©
ª
Γ(q + 1)Γ(n)
(3.1.4)
|hx, θi|q dx ≥
max hqK (θ), hqK (−θ)
2eΓ(q
+
n
+
1)
K
for every θ ∈ S n−1 . In particular,
(3.1.5)
Zq (K) ⊇ cK̂
for all q ≥ n, where c > 0 is an absolute constant.
38 · Isotropic convex bodies with small diameter
1
Proof. Consider the function f (t) := fθ (t) = |K ∩ (θ⊥ + tθ)|. Since f n−1 is
³
´n−1
concave, we have f (t) ≥ 1 − hKt(θ)
f (0) for all t ∈ [0, hK (θ)]. It follows
that
Z
Z hK (θ)
Z hK (−θ)
|hx, θi|q dx =
tq f (t)dt +
tq f−θ (t)dt
K
0
Z
hK (θ)
≥
tq 1 −
0
t
hK (θ)
¶n−1
f (0)dt
¶n−1
t
+
f (0)dt
hK (−θ)
0
Z
¡
¢ 1 q
q+1
f (0) hq+1
(θ)
+
h
(−θ)
s (1 − s)n−1 ds
K
K
Z
=
0
µ
hK (−θ)
µ
t 1−
q
0
=
≥
³
´
Γ(q + 1)Γ(n)
q+1
f (0) hq+1
(θ)
+
h
(−θ)
K
K
Γ(q + n + 1)
Γ(q + 1)Γ(n)
f (0)(hK (θ) + hK (−θ)) max{hqK (θ), hqK (−θ)}.
2Γ(q + n + 1)
Since K has its center of mass at the origin, we have kf k∞ ≤ ef (0), therefore
Z
hK (θ)
1 = |K| =
f (t)dt ≤ e (hK (θ) + hK (−θ)) f (0).
−hK (−θ)
This completes the proof of (3.1.4). Now, if q ≥ n we have
hZq (K) (θ)
≥ hZn (K) (θ)
·
¸1/n
©
ª
Γ(n + 1)Γ(n)
≥
max hK (θ), hK (−θ)
2eΓ(2n + 1)
©
ª
≥ c max hK (θ), hK (−θ) ,
©
ª
where c > 0 is an absolute constant. Since hK̂ (θ) = max hK (θ), hK (−θ) , this
proves that hZq (K) ≥ chK̂ for all q ≥ n.
Let V be a symmetric convex body in Rn , and let k · k be the norm induced by
V to Rn . For every p ≥ 1 we define
µZ
(3.1.6)
Mp := Mp (V ) =
¶1/p
kθk σ(dθ)
.
p
S n−1
The parameters Mp were studied by Litvak, Milman and Schechtman in [37].
Theorem 3.1.2. Let V be a symmetric convex body in Rn and let k · k be the
corresponding norm on Rn . We denote by b the smallest constant for which
kxk ≤ b|x| holds true for every x ∈ Rn . Then,
½
½
√ ¾
√ ¾
b p
b p
(3.1.7)
max M1 , c1 √
≤ Mp ≤ max 2M1 , c2 √
n
n
for all p ∈ [1, n], where c1 , c2 > 0 are absolute constants.
3.1 Lq -centroid bodies · 39
Proof. The function kxk : S n−1 → R is Lipschitz continuous with constant b.
By the spherical isoperimetric inequality (see [22] and [45]) it follows that
¯
¯
¡
¢
(3.1.8)
σ x ∈ S n−1 : ¯ kxk − M1 ¯ > t ≤ 2 exp(−ct2 n/b2 )
for all t > 0. Then,
Z
¯
¯
¯ kxk − M1 ¯p σ(dx) ≤
S n−1
=
≤
Z ∞
2p
tp−1 exp(−ct2 n/b2 )dt
µ 0 ¶p Z ∞
b
√
2p
sp−1 exp(−s2 )ds
cn
0
µ √ ¶p
b p
,
C√
n
for some absolute constant C > 0. The triangle inequality on Lp (S n−1 ) implies
that
√
b p
(3.1.9)
Mp − M1 ≤ k kxk − M1 kp ≤ C √ .
n
In other words,
(3.1.10)
½
√ ¾
b p
Mp ≤ 2 max M1 , C √
.
n
For the left hand side inequality we observe that 1/b is the minimal width of A,
so there exists z ∈ S n−1 such that A ⊂ {y : |hy, zi| ≤ 1/b}. It follows that
(3.1.11)
{x ∈ S n−1 : kxk ≥ t} ⊃ Ct := {x ∈ S n−1 : |hx, zi| ≥ t/b}
for every t > 0. A simple computation shows that
√ t
(3.1.12)
σ(Ct ) ≥ c n exp(−cnt2 /b2 )
b
if t ≤ b/3, where c > 0 is an absolute constant. Then,
µ Z ∞
¶1/p
p−1
Mp =
p
t σ(Ct )dt
≥ s[σ(Cs )]1/p
0
√
n
≥ cs
exp(−cns2 /pb2 )
b
√
√
for all s ≤ b/3, and the choice s = b p/3 n completes the proof.
The change of behavior of Mp happens when p ' n(M1 /b)2 . This value is equivalent to the largest integer k = k(V ) for which the majority of k-dimensional
sections of V are 4-Euclidean (the Dvoretzky number of V ; see [22]). It is clear
that Mp ≤ Mr if p ≤ r. Therefore, Mr ' b if r ≥ n. In other words, we have a
second change of behavior of Mp at the point p = n.
We shall apply Theorem 3.1.2 to the symmetric convex body V = [Zq (K)]◦ .
Then, b = R(Zq (K)) and M1 = w(Zq (K)). For every p, q ≥ 1 we define
µZ
¶1/p
p
(3.1.13)
wp (Zq (K)) =
hZq (K) (θ)σ(dθ)
.
S n−1
With this notation, we may rephrase Theorem 3.1.2 as follows.
40 · Isotropic convex bodies with small diameter
Theorem 3.1.3. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. For every p ∈ [1, n] and q ≥ 1 we have
½
√ ¾
R(Zq (K)) p
√
max w(Zq (K)), c1
≤ wp (Zq (K))
n
½
√ ¾
R(Zq (K)) p
√
≤ max 2w(Zq (K)), c2
.
n
If p ≥ n, then wp (Zq (K)) ' R(Zq (K)).
2
The quantities Iq (K) are related to the q-centroid bodies of K through the
following Proposition.
Proposition 3.1.4. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. For every q ≥ 1,
r
q
(3.1.14)
wq (Zq (K)) '
Iq (K).
q+n
Proof. Since
µZ
(3.1.15)
wq (Zq (K)) =
Z
¶1/q
|hx, θi| dxσ(dθ)
,
q
S n−1
K
this follows by Lemma 2.2.5.
3.2
Isotropic convex bodies with small diameter
n
In this Section
√ we assume that K is an isotropic convex body in R with radius
R(K) = α nLK , where α ≥ 1 is a positive constant. Our main result will
be that most directions θ ∈ S n−1 are “good ψ2 -directions” if α is uniformly
bounded. The precise statement is as follows.
√
Theorem 3.2.1. Let K be an isotropic convex body in Rn with R(K) = α nLK
for some α ≥ 1. Then,
¡
¢
¡
¢
√
(3.2.1)
σ θ ∈ S n−1 : kh·, θikψ2 ≥ c1 αLK t ≤ exp −c2 nt2 /α2
for all t ≥ 1, where c1 , c2 > 0 are absolute constants.
The proof starts with the following observation. By the definition of Zq (K), for
every θ ∈ S n−1 we have
(3.2.2)
Hq (θ)
kh·, θikψ2 ' sup √ .
q
q≥1
So, our first task will be to fix q ≥ 1 and give an upper bound for probabilities
of the form
¢
¡
√
σ θ ∈ S n−1 : Hq (θ) ≥ c1 α qLK t ,
where the constant c1 > 0 will be specified. Here, we are using for the first time
the fact that K has small diameter.
3.2 Isotropic convex bodies with small diameter · 41
Lemma 3.2.2. For every q ≥ 1 we have
(3.2.3)
√
w(Zq (K)) ≤ wq (Zq (K)) ≤ c1 α qLK ,
where c1 > 0 is an absolute constant.
Proof. The first inequality is a simple consequence of Hölder’s inequality, while
for the second one we may assume that q ≤ n. From Proposition 3.1.4 we have
r
r
√
q
q
√
Iq (K) ≤ c1
· α nLK = c1 α qLK ,
(3.2.4)
wq (Zq (K)) ≤ c1
n
n
√
since Iq (K) ≤ R(K) = α nLK for every q ≥ 1.
Lemma 3.2.2 shows that
(3.2.5)
¡
¢
¡
¢
√
σ θ ∈ S n−1 : Hq (θ) ≥ c1 α qLK t ≤ σ θ ∈ S n−1 : Hq (θ) ≥ w(Zq (K))t ,
for every q ≥ 1 and t ≥ 1. At this point, the spherical isoperimetric inequality
may be used through the following Lemma.
Lemma 3.2.3. Let M be a convex body in Rn and assume that for some R > 0
we have hM (θ) ≤ R for all θ ∈ S n−1 . Then,
µ
¶
¡
¢
w2 (M )t2 n
n−1
(3.2.6)
σ θ∈S
: hM (θ) ≥ 3tw(M ) ≤ exp −c
R2
for all t ≥ 1, where c > 0 is an absolute constant.
Proof. Let m be the Lévy mean of hM on S n−1 . Since hM is Lipschitz with
constant R, the spherical isoperimetric inequality gives
µ
¶
¡
¢
w2 (M )t2 n
(3.2.7)
σ θ ∈ S n−1 : hM (θ) ≥ m + tw(M ) ≤ exp −c
R2
for all t > 0. On observing that m ≤ 2w(M ), we conclude the proof.
Applying this for M = Zq (K) and taking into account (3.2.5), we have proved
the following fact.
Proposition
3.2.4. Let K be an isotropic convex body in Rn with R(K) =
√
α nLK for some α ≥ 1. Then,
µ
¶
¢
¡
w(Zq (K))2
√
(3.2.8)
σ θ ∈ S n−1 : Hq (θ) ≥ c1 α qLK t ≤ exp −ct2 n
,
R(Zq (K))2
for every q ≥ 1 and t ≥ 1, where c, c1 > 0 are absolute constants.
2
It is now natural to study the quantity nw(Zq (K))2 /R(Zq (K))2 as a function of q. To this end, we specify an absolute constant β > 0 such that
R√2 (V ) < βnw2 (V ) for every symmetric convex body V in Rn (note that R(M ) ≤
c nw(M ) in general), and for every convex body K in Rn with volume 1 and
center of mass at the origin, we define a function mK : [1, ∞) → (1, ∞) by
(3.2.9)
mK (q) = βn
w(Zq (K))2
.
R(Zq (K))2
42 · Isotropic convex bodies with small diameter
It is clear that mK takes values in the interval (1, βn]. We consider the sets
PK
NK
=
=
{q ∈ [1, 2βn] : q ≥ mK (q)}
{q ∈ [1, 2βn] : q ≤ mK (q)}
Observe that [βn, 2βn] ⊆ PK and 1 ∈ NK . Since mK is continuous and the sets
PK , NK are non-empty, q0 := min PK is well-defined, and mK (q0 ) = q0 . We
shall repeatedly use the following observation.
Lemma 3.2.5. If p, q ∈ [1, 2βn] and p ≤ mK (q), then
(3.2.10)
w(Zq (K)) ≥ cwp (Zq (K)).
Proof. Since p ≤ mK (q), we have
√ p
w(Zq (K)) ≥ R(Zq (K)) p/ βn.
(3.2.11)
The result follows from Theorem 3.1.3.
Some first observations on the function mK follow.
Proposition 3.2.6. There exist absolute constants c1 , c2 , c3 > 0 such that: for
every convex body K in Rn with volume 1 and center of mass at the origin,
(i) If q ∈ NK , then
(3.2.12)
mK (q) ≥ c1
mK (2)
.
q
(ii) If mK (2) > 1/c3 , then
(3.2.13)
[1, c3
p
mK (2)] ⊆ NK .
(iii) If q ∈ PK , then
√
R(Zq (K)) ≥ c2 nw(Z2 (K)).
(3.2.14)
Proof. (i) Let q ∈ NK . Since q ≤ mK (q), Lemma 3.2.5 and Proposition 3.1.4
give
r
r
q
q
0
0
w(Zq (K)) ≥ cwq (Zq (K)) ≥ c
Iq (K) ≥ c
I2 (K)
n
n
√
√
≥ c00 qw2 (Z2 (K)) ≥ c00 qw(Z2 (K)).
By Proposition 2.1.1 we have Hq (θ) ≤ cqH2 (θ) for all θ ∈ S n−1 , therefore
(3.2.15)
R(Zq (K)) ≤ cqR(Z2 (K)).
It follows that
(3.2.16)
mK (q) = βn
w(Zq (K))2
qw(Z2 (K))2
mK (2)
≥
c
βn
= c1
.
1
R(Zq (K))2
q 2 R(Z2 (K))2
q
(ii) If q0 = min PK we have mK (q0 ) = q0 and the previous claim shows that
(3.2.17)
q02 ≥ c1 mK (2).
3.2 Isotropic convex bodies with small diameter · 43
p
√
Therefore, [1, c3 mK (2)] ⊆ NK , where c3 = c1 .
(iii) For the value q0 , using Lemma 3.2.5 and Proposition 3.1.4 we have
(3.2.18)
p w(Zq0 (K))
√ wq (Zq (K))
R(Zq0 (K)) = βn
≥c n 0 √0
' Iq0 (K) ≥ cI2 (K).
√
q0
q0
√
By Proposition 3.1.4 we have I2 (K) ' nw2 (Z2 (K)), and hence,
√
(3.2.19)
R(Zq0 (K)) ≥ c nw(Z2 (K)).
If q ∈ PK then R(Zq (K)) ≥ R(Zq0 (K)), which completes the proof.
If we assume that K is isotropic, we have R(Z2 (K)) = w(Z2 (K)) = LK .
Then, Proposition 3.2.6 takes the following form.
Proposition 3.2.7. Let K be an isotropic convex body in Rn . Then,
(i) For every q ∈ NK we have mK (q) ≥ c1 n/q, where c1 > 0 is an absolute
constant.
√
(ii) [1, c2 n] ⊂ NK , where c2 > 0 is an absolute constant.
√
(iii) For every q ∈ PK we have R(Zq (K)) ≥ c3 nLK , where c3 > 0 is an
absolute constant.
2
We now use again the additional assumption that K has small diameter.
Proposition
3.2.8. Let K be an isotropic convex body in Rn with R(K) =
√
α nLK for some α ≥ 1. For every q ≥ 1 we have
√
(3.2.20)
mK (q) ≥ c3 n/α2 ,
where c3 > 0 is an absolute constant.
√
Proof. From Proposition 3.2.7 there exists q1 ' n with the property [1, q1 ] ⊂
NK . From Lemma 3.2.5 we have
(3.2.21)
√
√
√
√
√
w(Zq1 (K)) ≥ cwq1 (Zq1 (K)) ' q1 Iq1 (K)/ n ≥ q1 I2 (K)/ n = q1 LK .
Let q ≥ q1 . Then,
(3.2.22)
√
√
w(Zq (K)) ≥ w(Zq1 (K)) ≥ c q1 LK ≥ c0 4 nLK .
On the other hand,
(3.2.23)
√
R(Zq (K)) ≤ R(K) ≤ α nLK ,
therefore
(3.2.24)
mK (q) = βn
√
w(Zq (K))2
≥ c n/α2 .
2
R(Zq (K))
If q ≤ q1 then q ∈ NK , and Proposition 3.2.7 shows that
(3.2.25)
mK (q) ≥ c1
√
n
n
≥ c1 ≥ c01 n.
q
q1
The result follows from (3.2.24) and (3.2.25).
44 · Isotropic convex bodies with small diameter
Proof of Theorem 3.2.1: Proposition 3.2.4 shows that for every q, t ≥ 1 we have
¡
¢
√
(3.2.26)
σ θ ∈ S n−1 : Hq (θ) ≥ c1 α qLK t ≤ exp(−c2 mK (q)t2 ).
√
From Proposition 3.2.8, mK (q) ≥ c3 n/α2 for all q ≥ 1. Therefore,
¢
¡
√
√
(3.2.27)
σ θ ∈ S n−1 : Hq (θ) ≥ c1 α qLK t ≤ exp(−c4 nt2 /α2 ).
Lemma 3.1.1 and (3.2.2) show that
Hq (θ)
√ .
q
1≤q≤n
(3.2.28)
kh·, θikψ2 ' sup
It easily follows that there exists an absolute constant c5 > 0 such that
(3.2.29)
n
[
{θ ∈ S n−1 : kh·, θikψ2 ≥ c5 αtLK ) ⊆
Jq ,
q=2
where, for the integer q ∈ {2, . . . , n} we define
(3.2.30)
√
Jq := {θ ∈ S n−1 : Hq (θ) ≥ 2c1 αt qLK }.
Taking into account (3.2.27) we see that
(3.2.31)
√
σ(θ ∈ S n−1 : kh·, θikψ2 ≥ c5 αtLK ) ≤ n exp(−c6 nt2 /α2 )
for some absolute constant c6 > 0. This completes the proof for n ≥ n0 (α). 2
3.3
An example
We consider symmetric convex bodies in Rn of the form
(3.3.1)
K = {(x, t) ∈ Rn−1 × R : |t| ≤ R, |x| ≤ f (|t|)},
where f : [0, R] → R+ is a decreasing linear function. We may write f (t) = a−bt,
where a > 0 and 0 ≤ b ≤ a/R. We assume that |K| = 1, which gives the
condition
Z R
(3.3.2)
2ωn−1
(a − bt)n−1 dt = 1.
0
We also write I for the quantity
Z
Z
(3.3.3)
I := I1 (K, en ) =
|t|d(x, t) = 2ωn−1
K
R
t(a − bt)n−1 dt.
0
We shall show that the right choice of the parameters R, a and b leads to
very simple examples of bad ψ2 -behavior. It is convenient to assume from the
beginning that
(3.3.4)
I¿R
and
a
a
≤ b ≤ c3 .
nI
nI
where ci > 0 are absolute constants. One can actually show that (3.3.4) implies
(3.3.5). In the sequel, K denotes the convex body defined by (3.3.1), where
f (t) = a − bt.
(3.3.5)
c2
3.3 An example · 45
Lemma 3.3.1. Assume that K satisfies (3.3.2), (3.3.4) and (3.3.5). If s ≤
c0 min{R, a/b}, then
(3.3.6)
Prob(|t| ≥ s) ≥ c4 exp(−c5 s/I),
where c0 , c4 , c5 > 0 are absolute constants.
Proof. The probability is equal to
Z
Prob(|t| ≥ s)
R
(a − bt)n−1 dt
µµ
¶n µ
¶n ¶
bR
a
bs
= 2ωn−1
1−
− 1−
.
nb
a
a
=
2ωn−1
s
n
Since |K| = 1, Propositions 1.5.5 and 2.1.1 show that
ωn−1 an−1 = |K ∩ e⊥
n | ≥ c/I.
(3.3.7)
Taking into account the assumption b ≤ c3 a/(nI), we see that
(3.3.8)
2ωn−1
an
≥c
nb
for some absolute constant c > 0. If s ≤ a/2b, using the numerical inequality
1 − x ≥ e−2x for x ∈ [0, 1/2], we obtain
µ
¶n
bs
1−
≥ exp(−2bns/a).
a
(3.3.9)
On the other hand,
µ
¶n
bR
1−
≤ exp(−bRn/a).
a
(3.3.10)
If s ¿ R, then
(3.3.11)
exp(−bRn/a) ≤
1
exp(−2bns/a).
2
Here, we also use the fact that I ¿ R, which gives exp(bRn/a) ≥ exp(cR/I) À
1. It follows that if s ≤ c0 min{R, a/b}, then
(3.3.12)
Prob(|t| ≥ s) ≥ (c/2) exp(−2bns/a) ≥ (c/2) exp(−2c2 s/I).
This gives the result, with c4 = c/2 and c5 = 2c2 .
Lemma 3.3.2. For every j = 1, . . . , n − 1 we have
√
√
n
n−1
(3.3.13)
c6
R,
≤ |K ∩ e⊥
j | ≤ c7
a
where c6 , c7 > 0 are absolute constants.
46 · Isotropic convex bodies with small diameter
Proof. For the upper bound we use Hölder’s inequality and (3.3.2):
Z
|K ∩ e⊥
j |
= 2ωn−2
R
ÃZ
R
(a − bt)n−2 dt ≤ 2ωn−2
0
√
n−2 n−1
= 2ωn−2 (2ωn−1 )− n−1
R ≤ c7
√
n−1
! n−2
n−1
(a − bt)n−1 dt
√
n−1
R
0
R,
where c7 > 0 is an absolute constant. For the lower bound we observe that
Z R
Z R
ωn−2
n−2
|K ∩ e⊥
|
=
2ω
(a
−
bt)
dt
≥
2ω
(a − bt)n−1 dt
n−2
n−1
j
aω
n−1
0
0
√
n
ωn−2
≥ c6
,
=
aωn−1
a
where c6 > 0 is an absolute constant.
√
√
Lemma 3.3.3. There exist a ' n and b ' 1/ n such that the symmetric
convex body
(3.3.14)
W = {y = (x, t) : |t| ≤ a, |x| ≤ a − b|t|}
has volume 1 and satisfies
Z
(3.3.15)
hy, θi2 dy ≤ c9
c8 ≤
W
for every θ ∈ S n−1 , where c8 , c9 > 0 are absolute constants.
Proof. Let r be the solution of the equation ωn−1 rn−1 = 1, and consider the
body
√
(3.3.16)
K = {(x, t) : |t| ≤ r, |x| ≤ r − |t|/ n}.
Then,
(3.3.17)
|K| = 2ωn−1 rn−1 ·
√
nr ·
³
1− 1−
n
√1
n
´n
' 1,
√
since r ' n. Let s > 0 be such that the body W√:= sK has volume
1. Then,
√
sn ' 1, and W is of the form (3.3.14), where a ' n and b = 1/ n. Note that
(3.3.18)
n−1 n−1
I −1 ' |W ∩ e⊥
s
'1
n | = ωn−1 r
where I = I1 (W, en ), and hence
Z
(3.3.19)
hy, en i2 dy ' I 2 ' 1.
W
Since W is symmetric with respect to the coordinate subspaces, (3.3.15) will
hold for every θ ∈ S n−1 provided that
Z
(3.3.20)
c8 ≤
hy, ej i2 dy ≤ c9
W
3.3 An example · 47
for every j = 1, . . . , n. Because of (3.3.19), we only need to check the case
j ≤ n − 1. From Lemma 3.3.2, for every j = 1, . . . , n − 1 we have
√
√
n
n−1
(3.3.21)
c06 ≤ c6
a ≤ c07 ,
≤ |W ∩ e⊥
j | ≤ c7
a
which implies
Z
(3.3.22)
W
−2
hy, ej i2 dy ' |W ∩ e⊥
' 1.
j |
This proves the Lemma.
Starting with W , we may easily pass to a “similar” isotropic body.
Theorem 3.3.4. There exist a1 , R1 '
metric convex body
(3.3.23)
√
√
n and b1 ' 1/ n such that the sym-
Q = {y = (x, t) : |t| ≤ R1 , |x| ≤ a1 − b1 |t|}
is isotropic.
Proof. Consider the body W of the previous Lemma. There exists a diagonal
operator T = diag(u, . . . , u, v) such that Q = T (W ) is isotropic. Lemma 2.3.2
and the proof of Proposition 2.3.1 show that u, v ' 1. Then, Q can be written
in the form (3.3.23) with R1 = av, a1 = au and b1 = bu/v.
The next two Lemmas describe two “contradictory” properties of Q.
Lemma 3.3.5. There exist absolute constants c, C > 0 such that
(3.3.24)
√
√
c nB2n ⊆ Q ⊆ C nB2n .
Proof. The problem is two-dimensional. For every y = (x, t) ∈ Q we have
(3.3.25)
kyk22 = |x|2 + t2 ≤ a21 + R12 ≤ C 2 n,
√
where C√ > 0 is an absolute constant, because a1 , R1 ' n. This shows that
n
Q ⊆ C nB2 . For the other inclusion, we observe that the inradius of Q is
equal to min{R1 , d}, where d is the distance from (0, 0) to the line y = a1 − b1 t
in R2 . We have
(3.3.26)
a1
d= p
1+
b21
'
√
n,
√
and hence Q ⊇ c nB2n for some absolute constant c > 0.
Lemma 3.3.6. There exists an absolute constant c > 0 such that
(3.3.27)
√
kh·, en ikψ2 ≥ c 4 n.
48 · Isotropic convex bodies with small diameter
Proof. For every q ≥ 1 we have
(3.3.28)
Iq := Iq (Q, en ) ≤ c1 qI1 (Q, en ) ≤ c2 q,
where c2 > 0 is an absolute constant. Proposition 2.1.7 shows that
(3.3.29)
Prob(y ∈ Q : |hy, en i| ≥ 3CIq ) ≤ e−q
for every q ≥ 1. If 3C · c2 q ≤ c0 min{R1 , a1 /b1 } where c0 is the constant in
Lemma 3.3.1, we have
(3.3.30)
Prob(y ∈ Q : |hy, en i| ≥ 3CIq ) ≥ exp(−3c5 CIq /I).
For these values of q it follows that qI ≤ 3c5 CIq , and since I ' 1, we conclude
that
Iq
√
√ ≥ c00 q,
q
(3.3.31)
√
where c00 > 0 is an absolute constant. Since min{R1 , a1 /b1 }√= R1 ' n, the
largest value of q for which (3.3.31) holds
√ is of the order of n (note that this
is the largest possible, since R(Q) = O( n)). It follows that
½
¾
√
kh·, en ikq
(3.3.32)
kh·, en ikψ2 = sup
:q≥1 ≥c4n
√
q
for some absolute constant c > 0.
We summarize in the following Theorem.
Theorem 3.3.7. There exists an isotropic convex body of revolution Q in Rn
with the following properties:
√
√
(3.3.33)
c1 nB2n ⊆ Q ⊆ c2 nB2n
and
(3.3.34)
√
kh·, en ikψ2 ≥ c3 4 n
where c1 , c2 , c3 > 0 are absolute constants.
2
Remark 1: Theorem 3.3.7 shows that the simple ψ2 -estimate of Proposition 2.2.3
cannot be improved, even for bodies which have uniformly bounded geometric
distance to a Euclidean ball. We can also check that Proposition 3.2.8 is sharp:
if we apply it to Q we have
√
(3.3.35)
mQ (q) ≥ c4 n
for every q ≥ 1. If q0 is the largest q ≥ 1 for which (3.3.31) holds, then
(3.3.36)
R(Zq0 (Q)) ≥ cq0
and Lemma 3.2.2 implies
(3.3.37)
√
w(Zq0 (Q)) ≤ C q0 .
3.3 An example · 49
Therefore,
(3.3.38)
Since q0 '
mQ (q0 ) = βn
√
w(Zq0 (Q))2
C 2 q0
≤
βn
≤ C1 n/q0 .
R(Zq0 (Q))2
c2 q02
n, we conclude that
(3.3.39)
inf mQ (q) '
q≥1
√
n.
In other words, all the basic estimates of §3.2 are exact.
We can actually give a complete description of the Lq -centroid bodies of Q,
using the following consequence of the “Lq -affine isoperimetric inequality” of
Lutwak, Yang and Zhang (see [39], also [18]).
Proposition 3.3.8. Let K be a convex body of volume 1 in Rn . Then,
p
(3.3.40)
|Zq (K)|1/n ≥ c q/n
for every 1 ≤ q ≤ n, where c > 0 is an absolute constant.
2
Theorem 3.3.9.
Let Q be the isotropic convex body in Theorem 3.3.7. There
√
exists q0 ' n such that:
(i) For every 1 ≤ q ≤ n,
w(Zq (Q)) '
√
q.
(ii) If 1 ≤ q ≤ q0 then R(Zq (Q)) ' q, and if q ≥ q0 then R(Zq (Q)) '
√
n.
(iii) If 1 ≤ q ≤ q0 then mQ (q) ' n/q, and if q0 ≤ q ≤ n then mQ (q) ' q.
Proof. Let q0 be the largest q ≥ 1 for which (3.3.31) holds.
1. From Proposition 3.3.8 and Urysohn’s inequality, for every 1 ≤ q ≤ n we
have
√
√
(3.3.41)
w(Zq (Q)) ≥ c n|Zq (Q)|1/n ≥ c0 q.
√
On the other hand, since R(Q) = O( n), we see that
p
√
(3.3.42)
w(Zq (Q)) ≤ wq (Zq (Q)) ' q/nIq (Q) ≤ c00 q.
2. In the case q ≤ q0 , (3.3.31) shows that
(3.3.43)
R(Zq (Q)) ' q.
In particular, R(Zq0 (Q)) ' q0 . It follows that if q ≥ q0 , then
√
√
(3.3.44)
n ' q0 ' R(Zq0 (Q)) ≤ R(Zq (Q)) ≤ R(Q) ' n.
3. Since we have determined R(Zq (Q)) and w(Zq (Q)) for every value of q ∈
[1, n], we can compute the parameter mQ (q).
Notes and References
Lq -centroid bodies were introduced by Lutwak and G. Zhang in [38]. The idea
to use q-centroid bodies (and the result of [37]) for ψ2 -estimates comes from the
thesis of Paouris. All the results of §3.2 and §3.3 can be found in [48].
Chapter 4
Reduction to bounded
volume ratio
4.1
Symmetrization of isotropic convex bodies
Let K be a convex body in Rn . Suppose that E is a k-dimensional subspace of
Rn and T is a convex body in E with volume 1 and center of mass at the origin.
The (T, E)-symmetrization of K is the unique convex body K(T, E) which has
the following two properties:
1. If y ∈ E ⊥ then |K ∩ (y + E)| = |K(T, E) ∩ (y + E)|.
2. If y ∈ PE ⊥ (K), then (K(T, E) − y) ∩ E is homothetic to T and has its center
of mass at 0.
Lemma 4.1.1. K(T, E) is a convex body.
Proof. We set K1 := K(T, E). It suffices to prove that for every y1 , y2 ∈
PE ⊥ (K1 ) = PE ⊥ (K) and every λ ∈ (0, 1),
(4.1.1) λ[K1 ∩ (y1 + E)] + (1 − λ)[K1 ∩ (y2 + E)] ⊆ K1 ∩ (λy1 + (1 − λ)y2 + E).
Since (K1 − y) ∩ E is homothetic to T for every y ∈ PE ⊥ (K), it suffices to prove
that
1
1
1
(4.1.2) λ|K1 ∩(y1 +E)| k +(1−λ)|K1 ∩(y2 +E)| k ≤ |K1 ∩(λy1 +(1−λ)y2 +E)| k .
By the definition of K1 ,
(4.1.3)
1
1
|K1 ∩ (y + E)| k = |K ∩ (y + E)| k
for every y ∈ PE ⊥ (K), so the Lemma follows from the Brunn-Minkowski inequality applied to K.
Definition: Let A be a convex body in Rn with volume 1 and center of mass
at the origin. We say that e ∈ S n−1
R is an axis of inertia of A if it is an
eigenvector of the operator MA (y) = A hx, yixdx. Since MA is symmetric and
52 · Reduction to bounded volume ratio
positive, there exists an orthonormal basis consisting of eigenvectors of MA .
Since det MA = L2n
A , it is clear that if {e1 , . . . , en } is such a basis, then
L2n
K
(4.1.4)
n Z
Y
=
i=1
hx, ei i2 dx.
A
We will say that such a basis is a basis of inertia for A.
Remark 1: Fubini’s theorem shows that if θ1 ∈ Rn then
Z
Z
Z
2
(4.1.5)
hx, θi =
hy + z, θi2 dzdy.
K1
PE ⊥ (K)
(K1 −y)∩E
If θ ∈ E ⊥ , then the inner integral is independent of z, and since |(K1 − y) ∩ E| =
|(K − y) ∩ E| for every y ∈ PE ⊥ (K), we get
Z
Z
(4.1.6)
hx, θi2 =
hx, θi2 dx.
K1
K
If θ ∈ E, then (4.1.5) gives
Z
Z
2
hx, θi =
K1
P
Z
hz, θi2 dzdy
ÃZE
⊥ (K)
(K1 −y)∩E
! µZ
2
1+ k
|(K − y) ∩ E|
=
¶
2
dy
hz, θi dz .
PE ⊥ (K)
T
It follows that: if θ1 , θ2 ∈ E ⊥ then
(4.1.7)
hMK1 (θ1 ), θ2 i = hMK (θ1 ), θ2 i,
and if θ1 , θ2 ∈ E then
(4.1.8)
hMK1 (θ1 ), θ2 i = c(K, E)hMT (θ1 ), θ2 i,
where
(4.1.9)
Z
2
|K ∩ (y + E)|1+ k dy.
c(K, E) =
PE ⊥ (K)
Lemma 4.1.2. Let K be an isotropic convex body in Rn . Let E be a kdimensional subspace of Rn and let T be a convex body in E with volume 1
and center of mass at the origin. If e1 , . . . , ek are axes of inertia of T and
{ek+1 , . . . , en } is any orthonormal basis of E ⊥ , then {e1 , . . . , en } is a basis of
inertia for K(T, E).
Proof. Set K1 := K(T, E). We first observe that the restriction of MK(T,E)
onto E ⊥ is a multiple of the identity: from (4.1.7) we have
(4.1.10)
hMK1 (θ1 ), θ2 i = L2K hθ1 , θ2 i
for all θ1 , θ2 ∈ E ⊥ , which shows that MK1 |E ⊥ = L2K IE ⊥ . Then, any orthonormal basis {ek+1 , . . . , en } of E ⊥ consists of axes of inertia of K1 . Also, since
MK1 is symmetric, we have that E is invariant under MK1 .
4.1 Symmetrization of isotropic convex bodies · 53
Now, from (4.1.8) we have
hMK1 (ei ), θi = c(K, E)hMT (ei ), θi
for every θ ∈ E and every i = 1, . . . , k. Since e1 , . . . , ek are eigenvectors of MT ,
this shows that e1 , . . . , ek are axes of inertia of K1 .
Proposition 4.1.3. Let f : Rn → R+ be a functionRwith compact support, such
that f 1/k is concave on supp(f ). Assume also that f (x)dx = 1. Then,
Z
(k + 1)(k + 2)
(4.1.11)
kf k2/k
≤
f (x)1+2/k dx ≤ kf k2/k
∞
∞ .
(n + k + 1)(n + k + 2)
Remark 2: The right hand side inequality is trivial:
Z
Z
1+2/k
2/k
(4.1.12)
f (x)
dx ≤ kf k∞
f (x)dx = kf k2/k
∞ .
For the left hand side inequality, we will use the following Lemma.
Lemma 4.1.4. Let f : R+ → R+ be a function with compact support, such that
f 1/k is concave on supp(f ) and a = f (0) > 0. Let n ∈ N and choose b > 0 such
that
Z ∞
Z ∞
(4.1.13)
f (x)xn dx =
(a1/k − bx)k+ xn dx,
0
0
where x+ = max{x, 0}. Then,
Z ∞
Z
(4.1.14)
f (x)p xn dx ≥
0
0
∞
n
(a1/k − bx)pk
+ x dx
for every p > 1.
R∞
Proof. The existence of b is clear because 0 f (x)xn dx < +∞ (f has compact
support). Define h(x) = a1/k − f (x)1/k . Then, h is a convex function with
h(0) = 0. Therefore, h1 (x) = h(x)/x is increasing. Writing (4.1.13) in the form
Z ∞
Z ∞
(4.1.15)
(a1/k − xh1 (x))k+ xn dx =
(a1/k − bx)k+ xn dx,
0
0
we see that h1 (x) cannot be everywhere less or everywhere greater than b. Since
h1 is increasing, there exists x0 ∈ [0, ∞) such that h1 ≤ b on [0, x0 ] and h1 ≥ b
on [x0 , ∞). It follows that
(4.1.16)
(g(x) − f (x))(x − x0 ) ≥ 0
for all x ≥ 0, where g(x) = (a1/k − bx)k+ . Since g p−1 is a decreasing function,
we have
Z f (x)
Z f (x)
Z x0
Z x0
g(x0 )p−1 dydx
y p−1 dydx ≥
xn
(4.1.17)
xn
g(x)
0
g(x)
0
and
Z
∞
(4.1.18)
x0
Z
xn
g(x)
f (x)
Z
y p−1 dydx ≤
∞
x0
Z
xn
g(x)
f (x)
g(x0 )p−1 dydx.
54 · Reduction to bounded volume ratio
Subtracting these two inequalities, we get
Z
∞
(4.1.19)
Z
f (x)
xn
0
Z
∞
y p−1 dydx ≥ g(x0 )p−1
g(x)
(f (x) − g(x))xn dx = 0.
0
This shows that
Z
∞
(4.1.20)
0
Z
xn
f (x)
Z
∞
py p−1 dydx ≥
0
Z
g(x)
xn
0
py p−1 dydx,
0
which is equivalent to (4.1.14).
Proof of Proposition 4.1.3: Since the inequality is translation invariant, we may
assume that kf k∞ = f (0). We write
Z
Z
Z ∞
1+2/k
(4.1.21)
f (x)
dx = nωn
f (rθ)1+2/k rn−1 drσ(dθ),
S n−1
0
we fix θ ∈ S n−1 and define g(r) = f (rθ) on [0, ∞) (note that g 1/k is concave).
Lemma 4.1.4 shows that
Z ∞
Z ∞
n−1
(4.1.22)
g(x)1+2/k xn−1 dx ≥
(a1/k − bx)k+2
dx
+ x
0
0
where b > 0 is chosen so that
Z ∞
Z
n−1
(4.1.23)
g(x)x
dx =
0
0
∞
(a1/k − bx)k+ xn−1 dx
and a = g(0). Direct computation shows that
Z ∞
Z
n−1
2/k
(4.1.24)
(a1/k − bx)k+2
x
dx
=
c
a
n,k
+
0
∞
0
(a1/k − bx)k+ xn−1 dx,
where
(4.1.25)
cn,k =
(k + 1)(k + 2)
.
(n + k + 1)(n + k + 2)
Combining with (4.1.22) and (4.1.23) we get
Z
Z ∞
(4.1.26)
g(x)1+2/k xn−1 dx ≥ cn,k g(0)2/k
0
∞
g(x)xn−1 dx.
0
In other words,
Z
∞
(4.1.27)
0
for every θ ∈ S
n−1
Z
(4.1.28)
Z
f (rθ)1+2/k rn−1 dr ≥ cn,k kf k2/k
∞
∞
f (rθ)rn−1 dr
0
. Integrating this last inequality with respect to θ we get
Z
1+2/k
2/k
f (x)
dx ≥ cn,k kf k∞
f (x)dx = cn,k kf k2/k
∞ .
We will also need the following generalization of Proposition 1.5.4 (see [24]).
4.1 Symmetrization of isotropic convex bodies · 55
Proposition 4.1.5. Let K be a convex body in Rn with center of mass at the
origin. Let k < n and let E be a k-dimensional subspace of Rn . Then,
µ
(4.1.29)
max |K ∩ (E + y)| ≤
y∈E ⊥
n+1
k+1
¶k
|K ∩ E|.
2
Proposition 4.1.6. Let K be a convex body in Rn with volume 1 and center
of mass at the origin. Let E be a k-dimensional subspace of Rn and let T be
a convex body in E with volume 1 and center of mass at the origin. Then, for
every θ ∈ E,
µ
(4.1.30)
k+1
n+1
¶2
Z
|K ∩ E|
Z
2
k
2
hx, θi2 dx
hz, θi dz ≤
T
K(T,E)
and
µ
Z
hx, θi2 dx ≤
(4.1.31)
K(T,E)
n+1
k+1
Proof. From Remark 1 we have
Z
Z
2
(4.1.32)
hx, θi dx =
K(T,E)
¶2
Z
2
hz, θi2 dz.
|K ∩ E| k
T
Z
|K ∩ (E + w)|
2
1+ k
hz, θi2 dz.
dw ·
PE ⊥ (K)
T
Let g(y) = |K ∩R (E + y)| on PE ⊥ (K) . By the Brunn-Minkowski inequality, g 1/k
is concave and g(y)dy = |K| = 1. Proposition 4.1.3 shows that
(4.1.33)
Z
Z
Z
2
2
(k + 1)(k + 2)
k
k
kgk∞
hz, θi2 dy ≤
hx, θi2 dx ≤ kgk∞
hz, θi2 dy.
(n + 1)(n + 2)
T
K(T,E)
T
Since K has its center of mass at the origin, Proposition 4.1.5 shows that
µ
(4.1.34)
g(0) ≤ kgk∞ ≤
n+1
k+1
¶k
g(0).
On observing that g(0) = |K ∩ E| and
(k + 1)(k + 2)
≥
(n + 1)(n + 2)
µ
k+1
n+1
¶2
,
we conclude the proof.
Theorem 4.1.7. Let K be an isotropic convex body in Rn . Let E be a kdimensional subspace of Rn and let T be a convex body in E with volume 1 and
center of mass at the origin. Then,
µ
(4.1.35)
k+1
n+1
¶ nk
k
1− n
LK(T,E) ≤ LK
k
µ
1
LTn |K ∩ E| n ≤
n+1
k+1
¶ nk
LK(T,E) .
56 · Reduction to bounded volume ratio
Proof. We choose an orthonormal basis {e1 , . . . , en } as in Lemma 4.1.2. Then,
(4.1.36)
L2n
K(T,E) =
n Z
Y
i=1
2(n−k)
K(T,E)
hx, ei i2 dx = LK
k Z
Y
i=1
hx, ei i2 dx.
K(T,E)
By the left hand side inequality of Proposition 4.1.6, this gives
µ
(4.1.37)
L2n
K(T,E)
≥
k+1
n+1
¶2k
2(n−k)
LK
|K
∩ E|
2
k Z
Y
i=1
hx, ei i2 dx,
T
and since {e1 , . . . , ek } is a basis of axes of inertia of T ,
µ
(4.1.38)
L2n
K(T,E) ≥
k+1
n+1
¶2k
2(n−k)
LK
2
L2k
T |K ∩ E| .
It follows that
µ
(4.1.39)
LK(T,E) ≥
k+1
n+1
¶k/n
k
1− n
LK
k
1
LTn |K ∩ E| n .
Using the right hand side inequality of Proposition 4.1.6 and following the same
argument, we get
µ
(4.1.40)
LK(T,E) ≤
n+1
k+1
¶k/n
k
1− n
LK
k
1
LTn |K ∩ E| n .
Note that [(k + 1)/(n + 1)]k/n ≥ c for all k < n, where c > 0 is an absolute
constant.
4.2
Monotonicity in the dimension
For every n ∈ N we define
(4.2.1)
Ln = sup{LK : K is an isotropic convex body in Rn }.
In §1.6 we saw that if K and T are isotropic convex bodies in Rn and Rm
respectively, then
(4.2.2)
n
m
n+m
LK×T = LK
LTn+m .
Choosing K and T so that LK = Ln and LT = Lm , we readily see that
(4.2.3)
n+m
Ln+m
≥ Lnn Lm
m
for all n, m ∈ N. In particular, we have the following.
Lemma 4.2.1. Let k, n ∈ N. If k divides n, then Lk ≤ Ln .
2
In this Section we will prove that the sequence Ln is “monotone” in the following
precise sense.
4.2 Monotonicity in the dimension · 57
Theorem 4.2.2. There exists an absolute constant C > 0 such that: if m, n ∈ N
and m < n then Lm ≤ CLn .
The proof will be based on Theorem 4.1.7 and on the following fact.
Proposition 4.2.3. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. For every k < n there exists a k-dimensional subspace E of
Rn such that
|K ∩ E|1/n ≥ c,
(4.2.4)
where c > 0 is an absolute constant.
Proof. There exists an ellipsoid E such that |K| = |E| and
(4.2.5)
N (K, E) ≤ exp(βn),
where β > 0 is an absolute constant (in the terminology of §1.6 this is an
M -ellipsoid with constant β).
Averaging over all (n − k)-dimensional “coordinate subspaces” with respect
to the axes of E we see that there exists a (n − k)-dimensional subspace F such
that
(4.2.6)
|PF (E)| ≤
µ
ωn−k
(n−k)/n
ωn
≤
n
n−k
¶ n−k
2
.
From (4.2.5) we see that
µ
(4.2.7)
|PF (K)|1/n ≤ eβ |PF (E)|1/n ≤ eβ
n
n−k
¶ n−k
2n
≤ C = C(β).
Define E = F ⊥ . Using Proposition 4.1.5 and Fubini’s theorem we get
µ
(4.2.8)
1 = |K| ≤ max |K ∩ (y + E)||PF (K)| ≤
y∈E ⊥
n+1
k+1
¶k
|K ∩ E|.
Then, (4.2.7) shows that
(4.2.8)
1/n
|K ∩ E|
1
≥
C
µ
k+1
n+1
¶k/n
≥ c,
where c > 0 is an absolute constant.
Proof of Theorem 4.2.2: Let m, n ∈ N with m < n. Consider the largest integer
s for which 2s m ≤ n. Lemma 4.2.1 shows that
(4.2.9)
L2s m ≥ Lm .
Set k = 2s m. Let K be an isotropic convex body in Rn such that LK = Ln .
From Proposition 4.2.3 there exists a k-dimensional subspace E satisfying
(4.2.10)
|K ∩ E|1/n ≥ c.
58 · Reduction to bounded volume ratio
Let T be an isotropic convex body in E such that LT = Lk . Since K has
extremal isotropic constant, using Theorem 4.1.7 and (4.2.10) we get
(4.2.11)
LK ≥ LK(T,E) ≥
1− k k
LK n LTn |K
µ
∩ E|
1
n
n+1
k+1
¶− nk
,
which shows that
(4.2.12)
LT ≤ LK ·
n+1
k+1
µ ¶ nk
1
≤ CLK
c
because n < 2k. By the definition of K and T we have
(4.2.13)
Lm ≤ Lk = LT ≤ CLK = CLn .
2
4.3
Generalizations of Busemann’s inequality
Let K be a symmetric convex body in Rn . Busemann’s inequality states that
the function
(4.3.1)
x 7→
|x|
|K ∩ x⊥ |
is a norm. This fact is a special case (take k = 2) of the following Theorem.
Theorem 4.3.1. Let K be a symmetric convex body in Rn . Let 2 ≤ k ≤ n − 1
and let E be a k-codimensional subspace of Rn . Let F = E ⊥ and, for every
z ∈ F define E(z) = {x + tz : x ∈ E, t > 0}. Then, the function
(4.3.2)
z 7→
|z|
|K ∩ E(z)|
is a norm on F .
Proof. Let z1 and z2 be linearly independent vectors in F . Set z3 = z1 + z2 and
define
¯
¡
¢¯
(4.3.3)
fi (t) = ¯K ∩ t(zi /|zi |) + E ¯
for all t > 0, i = 1, 2, 3. Then,
Z
(4.3.4)
∞
fi (t)dt.
Fi = |K ∩ E(zi )| =
0
We will prove that
(4.3.5)
|z1 | |z2 |
|z3 |
≤
+
.
F3
F1
F2
Let t1 , t2 > 0 and let yi = ti zi /|zi |, i = 1, 2. The segment [y1 , y2 ] intersects the
ray in the direction of z3 at the point y3 = t3 z3 /|z3 |. Writing y3 = αy1 +(1−α)y2 ,
we see that
(4.3.6)
α=
t2 /|z2 |
.
t1 /|z1 | + t2 /|z2 |
4.3 Generalizations of Busemann’s inequality · 59
Then, from the equation
αt1
(1 − α)t2
t3
z1 +
z2 =
z3
|z1 |
|z2 |
|z3 |
(4.3.7)
we get
|z3 |
|z1 | |z2 |
=
+
.
t3
t1
t2
(4.3.8)
For every s ∈ [0, 1] we define t1 (s) and t2 (s) by the equations
(4.3.9)
s=
1
F1
Z
t1 (s)
f1 (u)du =
0
1
F2
Z
t2 (s)
f2 (u)du.
0
We have
dti
Fi
=
,
ds
fi (ti (s))
(4.3.10)
and differentiating (4.3.8) we see that
|z3 | dt3
|z1 |
F1
|z2 |
F2
= 2
+ 2
.
2
t3 (s) ds
t1 (s) f1 (t1 (s)) t2 (s) f2 (t2 (s))
(4.3.11)
Applying the Brunn-Minkowski inequality (in log-concave form) we see that
f3 (t3 (s)) ≥ f1 (t1 (s))α f2 (t2 (s))1−α .
(4.3.12)
We write
F3
≥
|z3 |
(4.3.13)
Z
1
0
f3 (t3 (s)) dt3
ds.
|z3 |
ds
Now, the integrand is greater than or equal to
µ
¶
t23 (s) |z1 |
F1
|z2 |
F2
(4.3.14)
+
f1 (t1 (s))α f2 (t2 (s))1−α .
|z3 |2 t21 (s) f1 (t1 (s)) t22 (s) f2 (t2 (s))
If we set a = |z1 |/t1 and b = |z2 |/t2 , from (4.3.6) and (4.3.8) we may write the
last expression in the form
(4.3.15)
µ
¶
a
b
1
F1
F2
2
2
a
+
b
f1 (t1 (s)) a+b f2 (t2 (s)) a+b .
(a + b)2
|z1 |f1 (t1 (s))
|z2 |f2 (t2 (s))
By the arithmetic-geometric means inequality,
µ
a
aF1
|z1 |f1 (t1 )
¶
µ
+b
bF2
|z2 |f2 (t2 )
¶
µ
≥ (a + b)
aF1
|z1 |f1 (t1 )
a
µ
¶ a+b
so the integrand in (4.3.13) is greater than
(4.3.16)
1
(a + b)
µ
aF1
|z1 |
a
¶ a+b
µ
bF2
|z2 |
b
¶ a+b
.
bF2
|z2 |f2 (t2 )
b
¶ a+b
,
60 · Reduction to bounded volume ratio
Applying once again the arithmetic-geometric means inequality, we see that
µ
¶−1
a |z1 |
b |z2 |
+
≥
a + b aF1
a + b bF2
µ
¶−1
|z1 | |z2 |
= (a + b)
+
.
F1
F2
³
´−1
This shows that the integrand in (4.3.13) is greater |zF11| + |zF22|
, which proves
(4.3.5).
aF1
|z1 |
a
¶ a+b
µ
bF2
|z2 |
b
¶ a+b
µ
A generalization of Theorem 4.3.1 (whose proof follows the same lines) is given
in the next Theorem.
Theorem 4.3.2. Let K be a symmetric convex body in Rn . Let 2 ≤ k ≤ n − 1
and let E be a k-codimensional subspace of Rn . Let F = E ⊥ and, for every
z ∈ F define E(z) = {x + tz : x ∈ E, t > 0}. Then, for every p ≥ 0 the function
p
(4.3.17)
|z|1+ p+1
kzkp = ³
R
|hx, zi|p dx
K∩E(z)
1
´ p+1
is a norm on F .
2
Theorem 4.3.3. Let K be an isotropic symmetric convex body in Rn . Let
2 ≤ k ≤ n − 1 and let E be a k-codimensional subspace of Rn . Let F = E ⊥
and write W for the unit ball of (F, k · kk+1 ), where k · kp is the norm defined in
Theorem 4.3.2. Then,
(4.3.18)
c1
LW
LW
≤ |K ∩ E|1/k ≤ c2
LK
LK
where c1 , c2 > 0 are absolute constants.
Proof. Let y ∈ F . For every x ∈ Rn we write x = x1 + ρθ where x1 ∈ E, ρ ≥ 0
and θ ∈ S(F ). Then,
ÃZ
!
Z
Z
hx, yi2 dx = kωk
(4.3.19)
K
hθ, yi2
|hw, θi|k+1 dw σF (dθ).
S(F )
K∩E(θ)
In other words,
Z
L2K
(4.3.20)
= kωk
1
hθ, yi
σF (dθ) = (k + 2)
kθkk+2
S(F )
W
Z
2
hx, yi2 dx.
W
This means that W/|W |1/k is isotropic, and
2
L2K = (k + 2)|W |1+ k L2W .
(4.3.21)
Observe that

(4.3.22)
2
1+ k
|W |
Z
ÃZ
k
! k+2
= ωk
|hx, θi|k+1 dx
S(F )
K∩E(θ)
 k+2
k
σF (dθ)
,
4.4 Reduction to bounded volume ratio · 61
while
Z
(4.3.23)
Z
|hx, θi|k−1 dxσF (dθ).
1 = |K| = ωk
S(F )
K∩E(θ)
Using the results of §2.1 we check that
2
2
(k + 2)|W |1+ k ' |K ∩ E|− k .
(4.3.24)
The Theorem is a consequence of (4.3.21) and (4.3.24).
Remark 1: All the results of this Section can be stated for not necessarily
symmetric convex bodies. For example, in Theorem 4.3.2 the conclusion would
be that
1
ÃZ
! p+1
(4.3.25)
hx, θip dx
ρp (θ) =
K∩E(θ)
defined on F \ {0} is the radial function of a convex body. In fact, the only
assumption needed is that 0 ∈ int(K). Then, Theorem 4.3.3 is also true (with
the obvious modification).
4.4
Reduction to bounded volume ratio
Recall the definition of volume ratio: If K is a convex body in Rn with center
of mass at the origin, then
½µ
¶1/n
¾
|K|
(4.4.1)
vr(K) = inf
: E is an 0-symmetric ellipsoid in K .
|E|
For every α > 1 define
(4.4.2)
Ln (α) = sup{LK : K is isotropic in Rn and vr(K) ≤ α}.
In this Section we will prove the following reduction of the slicing problem.
Theorem 4.4.1. There exist two constants c > 0 and α > 1 such that
(4.4.3)
Ln ≤ c[Ln (α)]4
for all n.
In the rest of this Section we assume that K is an isotropic convex body in Rn
such that
(4.4.4)
LK = Ln = sup{LK : K is an isotropic convex body in Rn }.
As a consequence of Theorem 4.3.3 and of the monotonicity of the sequence
{Ls }, we have the following property of K.
Proposition 4.4.2. Let K be an isotropic convex body in Rn with LK = Ln .
For every k-codimensional subspace E of Rn ,
(4.4.5)
|K ∩ E|1/k ≤ C1 ,
where C1 > 0 is an absolute constant.
62 · Reduction to bounded volume ratio
Proof. Theorem 4.3.3 shows that there exists a k-dimensional symmetric convex
body W such that
|K ∩ E|1/k ≤ c2
(4.4.6)
LW
LK
where c2 > 0 is an absolute constant. On the other hand, by Theorem 4.2.2 we
have
(4.4.7)
LW ≤ Lk ≤ CLn = CLK
for some absolute constant C > 0, and the result follows.
Proposition 4.4.2 imposes strong conditions on the axes of any M -ellipsoid of
K.
Proposition 4.4.3. Let K be an isotropic convex body in Rn with LK = Ln .
Let E be an ellipsoid with |K| = |E| and N (K, E) ≤ exp(βn). If
½
(4.4.8)
E=
x ∈ Rn :
n
X
hx, ei i2
i=1
nλ2i
¾
≤1 ,
where {e1 , . . . , en } is an orthonormal basis of Rn and λ1 ≤ . . . ≤ λn , then
(4.4.9)
λ[n/2]+1 ≤ C2 (β)
for some constant C2 (β) ≤ c exp(2β).
Proof. Let k = [n/2] and let E = span{e1 , . . . , ek }. In the proof of Proposition
4.2.3 we saw that
µ
¶k
n+1
(4.4.10)
1 = |K| ≤
|K ∩ E ⊥ | · |PE (K)|.
k+1
Proposition 4.4.2 shows that
|K ∩ E ⊥ | ≤ C1k .
(4.4.11)
Also, PE (K) is covered by at most exp(βn) translates of PE (E), so
(4.4.12)
|PE (K)| ≤ exp(βn)ωk nk/2
k
Y
λi .
i=1
It follows that
µ
(4.4.13)
1≤
n+1
k+1
¶k
exp(βn)ωk nk/2 C1k
i=1
On the other hand, since |E| = 1, we have
(4.4.14)
n
Y
i=1
k
Y
λi =
1
nn/2 ωn
.
λi .
4.4 Reduction to bounded volume ratio · 63
Therefore,
λn−k
k+1
(4.4.15)
≤
n
Y
µ
λi ≤
i=k+1
n+1
k+1
¶k
C1k exp(βn)
ωk nk/2
,
ωn nn/2
which implies
(4.4.16)
λk+1 ≤ C(β)
for some constant depending only on β.
The last ingredient of the proof is the fact that every convex body which is in
M -position has orthogonal projections of proportional dimension with bounded
volume ratio. For the proof we need the “volume ratio theorem” (see [30], [55]
and [56]).
Theorem 4.4.4. Let K be a convex body in Rn such that B2n ⊆ K and |K| =
αn |B2n | for some α > 1. For every 1 ≤ k ≤ n, a random subspace E ∈ Gn,k
satisfies with probability greater than 1 − e−n :
n
(4.4.17)
BE ⊆ K ∩ E ⊆ (cα) n−k BE ,
where c > 0 is an absolute constant.
For completeness we give a proof of the volume ratio theorem which was recently
given by Klartag (see [31]).
Lemma 4.4.5. Let K be a convex body in Rn . For every 1 ≤ k ≤ n,
Z
Z
kωk
(4.4.18)
|x|n−k dxνn,k (dE) =
|K|.
nω
n
Gn,k K∩E
Proof. Let χK be the indicator function of K. Integrating in polar coordinates
we get
Z
Z
Z
|x|n−k dxdνn,k
Gn,k
Z
=
K∩E
Gn,k
S(E)
Z ∞
Z
=
Z
∞
kωk
χK (rθ)rn−1 drdσE dνn,k
0
χK (rθ)rn−1 σ(dθ)
kωk
S n−1
=
=
0
Z
kωk
χK (x)dx
nωn Rn
kωk
|K|.
nωn
2
Lemma 4.4.6. Let K be a convex body in Rn with 0 ∈ int(K) and let R(K) =
max{|x| : x ∈ K}. Then,
(4.4.19)
|{x ∈ K : |x| > R(K)/2}| ≥ cn |K|,
where c > 0 is an absolute constant.
64 · Reduction to bounded volume ratio
Proof. We may assume that |K| = 1. There exists θ ∈ S n−1 such that hK (θ) =
r = R(K). Since
(4.4.20)
|{x ∈ K : |x| > R(K)/2}| ≥ |{x ∈ K : hx, θi > r/2}|,
it is enough to prove that
Z
3r/4
(4.4.21)
f (t)dt ≥ cn ,
r/2
where f = fK,θ .
Let y0 ∈ [−r, r] with f (y0 ) = kf k∞ . Suppose that r/2 ≤ x ≤ 3r/4.
Case 1: y0 ≤ x ≤ r. Then, we write
(4.4.22)
x=
r−x
x − y0
y0 +
r,
r − y0
r − y0
and applying the Brunn-Minkowski inequality we get
(4.4.23)
1
f (x) n−1 ≥
1
1
r−x
1
n−1
f (y0 ) n−1 ≥ kf k∞
,
r − y0
8
because r − x ≥ r/4 and r − y0 ≤ 2r.
Case 2: 0 ≤ x ≤ y0 . We write
(4.4.24)
1
f (x) n−1 ≥
1
1
1
x
y0 − x
1
n−1
f (y0 ) n−1 +
f (0) n−1 ≥ kf k∞
y0
y0
2
because x ≥ y0 /2.
Rr
Now, observe that 1 = −r f (t)dt ≤ 2rkf k∞ , therefore
Z
3r/4
(4.4.25)
f (t)dt ≥
r/2
1
8n−1
r
1
kf k∞ ≥ n .
4
8
This proves the Lemma, with c = 1/8.
Proposition 4.4.7. Let K be a convex body in Rn with 0 ∈ int(K). For every
1 ≤ k ≤ n, a random subspace E ∈ Gn,k satisfies with probability greater than
1 − e−n :
(4.4.26)
|K ∩ E|
|K|
(R(K ∩ E))n−k ≤ cn
,
ωk
ωn
where c > 0 is an absolute constant.
Proof. By Lemma 4.4.6, for every E ∈ Gn,k we have
Z
1 R(K ∩ E)n−k
(4.4.27) k
|K
∩
E|
≤
|x|n−k dx ≤ R(K ∩ E)n−k |K ∩ E|.
8
2n−k
K∩E
Integrating over Gn,k and using Lemma 4.4.5, we get
Z
|K|
|K ∩ E|
R(K ∩ E)n−k νn,k (dE) ≤ 16n
.
(4.4.28)
ωk
ωn
Gn,k
The result now follows from Markov’s inequality.
4.4 Reduction to bounded volume ratio · 65
Proof of Theorem 4.4.4: Let E ∈ Gn,k be a subspace satisfying (4.4.26). Since
B2n ⊆ K, we have |K ∩ E| ≥ ωk . Therefore,
(R(K ∩ E))n−k ≤ (cα)n ,
(4.4.29)
n
which means: K ∩ E ⊆ (cα) n−k BE .
2
Proposition 4.4.8. Let A be a convex body in Rk . Assume that |A| = |B2k |
and N (A, B2k ) ≤ exp(βk) for some constant β > 0. Then, for any 1 ≤ s ≤ k, a
random orthogonal projection PF (A) of A onto an s-dimensional subspace F of
Rk has volume ratio bounded by a constant C(β, s/k).
Proof. Recall that |co(A◦ ∪B2k )|1/k ≤ C|B2k |1/k where C depends on β. In other
words, W = co(A◦ ∪ B2k ) has bounded volume ratio, and Theorem 4.4.4 shows
that for a random E ∈ Gk,s ,
k
A◦ ∩ F ⊆ W ∩ F ⊆ C(β) k−s BF .
(4.4.30)
By duality, this means that
(4.4.31)
PF (A) ⊇ rBF ,
k
where r = C(β)− k−s . Since
(4.4.32)
|PF (A)| ≤ N (PF (A), BF )|BF | ≤ N (A, B2k )|BF | ≤ exp(βs)|BF |,
this shows that
(4.4.33)
¡
¢1/s
k
|PF (A)|/|rBF |
≤ C(β, s/k) = exp(βk/s)C(β) k−s .
2
Proof of Theorem 4.4.1: There exists an absolute constant β > 0 and an ellipsoid
E described as in Proposition 4.4.3, such that |K| = |E| = 1 and N (K, E) ≤
exp(βn). Let k = [n/2] + 1. If E = span{e1 , . . . , ek }, from Proposition 4.4.2
and (4.4.10) we have
(4.4.34)
|PE (K)| ≥ ck1 .
Now,
(4.4.35)
N (PE (K), PE (E)) ≤ N (K, E) ≤ exp(2βk)
and Proposition 4.4.3 shows that
(4.4.36)
√
PE (E) ⊆ c kBE .
Let √
ρ > 0 be defined so that |PE (K)| = |ρBE |. From (4.4.34) we see that
ρ ' k. This means that
(4.4.37)
ρBE ⊇ c1 PE (E),
and (4.4.35) gives
(4.4.38)
N (PE (K), ρBE ) ≤ exp(c2 βk).
66 · Reduction to bounded volume ratio
So, we may apply Proposition 4.4.8 to find a subspace F of E with dimension
dimF = s = [k/2] + 1, such that
(4.4.39)
vr(PF (K)) = vr(PF (PE (K))) ≤ C = C(β).
Actually, Proposition 4.4.8 shows that PF (K) contains a ball rBF such that
|PF (K)| ≤ C(β)s |rBF |.
(4.4.40)
Consider the (DF ⊥ , F ⊥ )-symmetrization of K, where DF ⊥ is the Euclidean ball
of volume 1 in F ⊥ . We denote this body by K1 .
Claim: vr(K1 ) ≤ α, where α > 1 is an absolute constant.
Proof of the claim: We define the ellipsoid
(4.4.41)
F = {ax + by : a2 + b2 ≤ 1, x ∈ rBF , y ∈ DF ⊥ }.
Note that DF ⊥ = K1 ∩ F ⊥ . Then, it is not hard to check that
1
√ F ⊆ co{rBF , K1 ∩ F ⊥ } ⊆ K1 .
2
(4.4.42)
√
On the other hand, | 2F| ≥ |rBF | · |K1 ∩ F ⊥ | and (4.4.40) shows that
(4.4.43)
|F|1/n ≥ √
1
1
|PF (K1 )|1/n |K1 ∩ F ⊥ |1/n ≥ √
.
2C(β)s/n
2C 0 (β)
This shows that vr(K1 ) ≤ α, where α = 2C 0 (β).
2
We now use Theorem 4.1.7: Since s ≥ n/4, we have
(4.4.43)
1
3
1
4
4
LK
LD
F⊥
|K ∩ F ⊥ | n ≤ c1 LK1 .
Also, from (4.4.38) we have |PF (K)| ≤ exp(cβn)|ρBF |, which gives
|PF (K)|1/n ≤ C 00 (β).
(4.4.44)
This means that
(4.4.45)
1
1
|K ∩ F ⊥ | n ≥ c|PF (K)|− n ≥ c(β).
Going back to (4.4.43) and taking into account the fact that LDF ⊥ ≥ c, we get
(4.4.46)
LK ≤ c1 (β)L4K1 .
Since LK = Ln and LK1 ≤ Ln (α), the proof is complete.
2
Notes and References
All the results of this Chapter come from [14] (see also [13]). See [3] and [43]
for complete proofs of the generalizations of Busemann’s inequality which are
outlined in §4.3.
Chapter 5
The 1-unconditional case
5.1
Bound for the isotropic constant
In this Chapter we study the case of symmetric convex bodies which generate a
norm with 1-unconditional basis. After a linear transformation, we may assume
that the canonical orthonormal basis {e1 , . . . , en } is 1-unconditional basis for
k · kK . That is, for every choice of real numbers t1 , . . . , tn and every choice of
signs εi = ±1,
(5.1.1)
°
°
°ε1 t1 e1 + · · · + εn tn en °
K
°
°
= °t1 e1 + · · · + tn en °K .
Geometrically, this means that if x = (x1 , . . . , xn ) ∈ K then the whole rectangle
[−|x1 |, |x1 |] × · · · × [−|xn |, |xn |] is contained in K.
Note that the matrix of inertia of such a body is diagonal, therefore one can
bring it to the isotropic position by a diagonal operator. This explains that for
every 1-unconditional convex body K in Rn there exists a linear image K̃ of K
which has the following properties:
1. The volume of K̃ is equal to 1.
2. If x = (x1 , . . . , xn ) ∈ K̃ then [−|x1 |, |x1 |] × · · · × [−|xn |, |xn |] ⊆ K̃.
3. For every j = 1, . . . , n,
Z
(5.1.2)
K̃
x2j dx = L2K .
This last condition implies that K̃ is in isotropic position, because
Z
(5.1.3)
xi xj dx = 0 for all i 6= j
K̃
by Property 2.
In the rest of this Chapter we assume that K has these three properties. It
is not hard to prove that LK ≤ C for some absolute constant C > 0. One way
to see this is using the Loomis-Whitney inequality (which even holds without
the convexity assumption) .
68 · The 1-unconditional case
Lemma 5.1.1. For every convex body K in Rn ,
(5.1.4)
|K|n−1 ≤
n
Y
|Pi (K)|,
i=1
where Pi (K) is the orthogonal projection of K onto e⊥
i .
2
Now, since K has Property 2, it is clear that
Pi (K) = K ∩ e⊥
i
(5.1.5)
for all i = 1, . . . , n. The Loomis-Whitney inequality implies that
|K ∩ e⊥
i |≥1
(5.1.6)
for some i ≤ n. From Proposition 1.5.6,
Z
(5.1.7)
L2K =
x2i dx ≤
K
c
2
|K ∩ e⊥
i |
where c > 0 is an absolute constant. This proves that LK ≤ C, with C =
√
c.
Theorem 5.1.2. There exists an absolute constant C > 0 such that LK ≤ C
for every n and every 1-unconditional convex body K in Rn .
2
We shall give an independent proof of Theorem 5.1.2. It will be convenient
to consider the normalized part
K + = 2K ∩ Rn+
(5.1.8)
of K in Rn+ = [0, +∞)n . In other words, if x = (x1 , . . . , xn ) is uniformly
distributed in K, then (2|x1 |, . . . , 2|xn |) is uniformly distributed in K + . It is
easy to check that K + has the following three properties:
4. The volume of K + is equal to 1.
5. If x = (x1 , . . . , xn ) ∈ K + and 0 ≤ yj ≤ xj for all 1 ≤ j ≤ n, then
y = (y1 , . . . , yn ) ∈ K + .
6. For every j = 1, . . . , n,
Z
(5.1.9)
x2j dx = 4L2K .
K+
Theorem 5.1.3. Let K be an isotropic 1-unconditional convex body in Rn .
Then, L2K ≤ 1/2.
Proof. Property 5 shows that if x = (x1 , . . . , xn ) ∈ K + , then [0, x1 ] × · · · ×
[0, xn ] ⊆ K + . It follows that
n
Y
(5.1.10)
xj ≤ 1
j=1
for every x ∈ K + . Define
½
(5.1.11)
V =
x ∈ Rn+ :
n
Y
j=1
¾
xj ≥ 1 .
5.1 Bound for the isotropic constant · 69
Then, V is convex and the sets K + and V have disjoint interiors. So, there
exists a hyperplane H of the form
(5.1.12)
λ1 x 1 + · · · + λn x n = α
+
with λj > 0, which touches V and separates
Qn it from K . Since H touches V
we can choose the λi ’s so that α = n and j=1 λj = 1.
Since H separates V from K + , we have
©
ª
(5.1.13)
K + ⊆ x ∈ RN
+ : λ1 x1 + · · · + λn xn ≤ n .
The arithmetic-geometric means inequality gives
(5.1.14)
n µZ
Y
j=1
K+
¶λj /n Z
xj dx
≤
K+
λ1 x 1 + · · · + λn x n
dx ≤ 1.
n
Now, Khintchine type inequalities for linear functionals (see §2.1) show that
µZ
Z
4L2K =
(5.1.15)
K+
x2j dx ≤ 2
K+
¶2
xj dx
for all j = 1, . . . , n (for the exact constant 2 one needs to use a different version
of the proof of Proposition 2.1.1, in the spirit of Lemma 4.1.4). If we set d =
λ1 +···+λn
, then
n
(5.1.16)
(2L2K )d =
n
Y
(2L2K )λj /n ≤
n µZ
Y
j=1
j=1
¶λj /n
K+
xj dx
≤ 1,
which shows that L2K ≤ 1/2.
Combining with Proposition 1.5.5 we have:
Corollary 5.1.4. Let K be an isotropic 1-unconditional convex body in Rn .
Then,
1
|K ∩ θ⊥ | ≥ √
6
(5.1.17)
for every θ ∈ S n−1 .
Proof. Let θ ∈ S n−1 . Proposition 1.5.5 shows that
µZ
(5.1.18)
¶1/2
2
LK =
hx, θi dx
K
1
1
.
≥ √
2 3 |K ∩ θ⊥ |
Note that if K is symmetric, then the factor e is not needed in the statement
of Proposition 1.5.5.
The basic source of all the results in this Chapter is the distributional inequality
in the next Theorem.
70 · The 1-unconditional case
Theorem 5.1.5. Let K be an isotropic 1-unconditional convex body in Rn .
Then,
µ
¶n
α1 + · · · + αn
+
√
(5.1.19)
|{x ∈ K : x1 ≥ α1 , . . . , xn ≥ αn }| ≤ 1 −
,
6n
for all (α1 , . . . , αn ) ∈ K + .
Proof. We define a function u : K + → R+ by
(5.1.20)
u(α1 , . . . , αn ) = |{x ∈ K + : x1 ≥ α1 , . . . , xn ≥ αn }|.
1
The Brunn-Minkowski inequality shows that the function h = u n is concave on
K + . Observe that u(0) = 1 and
∂u
1
(0) = −|K ∩ e⊥
j | ≤ −√ ,
∂αj
6
(5.1.21)
where the last inequality comes from Corollary 5.1.4. Let α ∈ K + and consider
the function hα : [0, 1] → R defined by hα (t) = h(αt). Note that
(5.1.22)
h0α (0) =
n
X
j=1
n
αj
X
∂h
1 ∂u
α1 + · · · + αn
√
(0) =
αi ·
(0) ≤ −
∂αj
n
∂α
6n
j
i=1
by (5.1.21). Since h is concave, hα is concave on [0, 1]. This implies that h0α is
decreasing on [0, 1], and hence,
(5.1.23)
h(α) − 1 = hα (1) − hα (0) ≤ h0α (0) ≤ −
α1 + · · · + αn
√
6n
for all α ∈ K + . This proves the Theorem.
As a direct consequence we get the following statement, which is valid for all
αj ≥ 0.
Corollary 5.1.6. Let K be an isotropic 1-unconditional convex body in Rn .
Then,
|{x ∈ K + : x1 ≥ α1 , . . . , xn ≥ αn }| ≤ exp(−c(α1 + · · · + αn )),
√
for all α1 , . . . , αn ≥ 0, where c = 1/ 6.
(5.1.24)
Proof. If (α1 , . . . , αn ) ∈ K + we just note that 1 − x ≤ e−x for all x ≥ 0. If not,
then the quantity on the left is equal to zero.
Another consequence of Theorem 5.1.5 is that an interior point (α1 , . . . , αn ) of
K + necessarily satisfies
√
(5.1.25)
α1 + · · · + αn < 6n.
This observation can be equivalently stated as follows.
Proposition
isotropic 1-unconditional convex body in Rn .
p 5.1.7.n Let K be an
n
Then K ⊆ 3/2nB1 , where B1 = {x ∈ Rn : |x1 | + · · · + |xn | ≤ 1} is the unit
ball of `n1 .
2
5.2 Tail estimates for the Euclidean norm · 71
One last observation is that, on the other hand, K contains a large cube:
Proposition 5.1.8. Let K be an isotropic 1-unconditional convex body in Rn .
Then,
√
√
(5.1.26)
K ⊇ [−LK / 2, LK / 2]n .
Proof. The center of mass v = (v1 , . . . , vn ) of K + is in K + , therefore the rectangle [0, v1 ] × · · · × [0, vn ] is contained in K + , where
Z
√
xj dx ≥ 2LK .
(5.1.26)
vj =
K+
Then,
(5.1.27)
[−v1 /2, v1 /2] × · · · × [−vn /2, vn /2] ⊆ K,
√
√
√
and this proves the Proposition. Note that LK / 2 ≥ LB2n / 2 ≥ 1/(2 πe).
5.2
Tail estimates for the Euclidean norm
Let x = (x1 , . . . , xn ) ∈ Rn+ . We write X1 , . . . , Xn for its coordinates in decreasing order. That is,
(5.2.1)
max xj = X1 ≥ X2 ≥ · · · ≥ Xn = min xj .
+
Let µ+
K denote the uniform distribution on K . Corollary 5.1.6 has the following
consequence.
Proposition 5.2.1. Let K be an isotropic 1-unconditional convex body in Rn .
Then,
µ ¶
n −ckα
n
(5.2.2)
µ+
({x
∈
R
:
X
≥
α})
≤
e
k
+
K
k
√
for all α ≥ 0 and 1 ≤ k ≤ n, where c = 1/ 6.
Proof. Let 1 ≤ j1 < · · · < jk ≤ n. From Corollary 5.1.6 we have
(5.2.3)
n
µ+
K ({x ∈ R+ : xj1 ≥ α, . . . , xjk ≥ α}) ≤ exp(−ckα).
Since
(5.2.4)
{x ∈ Rn+ : Xk ≥ α} =
[
{x ∈ Rn+ : xj1 ≥ α, . . . , xjk ≥ α},
1≤j1 <···<jk ≤n
we get
µ+
K ({x : Xk ≥ α}) ≤
X
1≤j1 <···<jk ≤n
≤
µ ¶
n −ckα
e
.
k
µ+
K ({x : xj1 ≥ α, . . . , xjk ≥ α})
72 · The 1-unconditional case
Theorem 5.2.2. Let K be an isotropic 1-unconditional convex body in Rn .
Then, for every t > 0
µ
¶
n
(5.2.5)
|{x ∈ K : kxk1 ≥ c1 (1 + t)n} ≤ n exp −2t
,
log n + 1
√
where c1 = 6.
Proof. Let α1 , . . . , αn ≥ 0. From Proposition 5.2.1 we have
ý
¯½
¾¯
¾!
n
n
n
X
X
X
¯
¯
+
n
¯ x ∈ K : kxk1 ≥
¯
αk ¯ = µK
x ∈ R+ :
xk ≥ 2
αk
¯
k=1
k=1
k=1
ý
¾!
n
n
X
X
= µ+
x ∈ Rn+ :
Xk ≥ 2
αk
K
k=1
≤
≤
n
X
k=1
n
µ+
K ({x ∈ R+ : Xk ≥ 2αk })
k=1
n µ
X
k=1
¶
n
exp(−2ckαk ),
k
√
where c = 1/ 6. Since
µ ¶ ³ ´
n
en k
≤
,
k
k
(5.2.6)
we get
¯½
¾¯
n
n
³ ³
X
¯
en ´´
αk ¯¯ X
¯ x ∈ K : kxk1 ≥
.
≤
exp
−k
2α
−
log
k
¯
c ¯
k
(5.2.7)
k=1
k=1
We choose
(5.2.8)
Since
αk =
Pn
(5.2.9)
1
en
t
n
log
+
.
2
k
k log n + 1
1
k=1 k
≤ log n + 1 and n! ≥ (n/e)n , we get
µ n n¶
n
n
X
X
1
n e
n
1
αk = log
+t
≤ n + nt.
2
n!
log n + 1
k
k=1
k=1
Going back to (5.2.7) we get
(5.2.10)
µ
√
¯©
ª¯
¯ x ∈ K : kxk1 ≥ 6(1 + t)n ¯ ≤ n exp −2t
n
log n + 1
¶
.
Theorem 5.2.3. Let K be an isotropic 1-unconditional convex body in Rn .
Then, for every t ≥ 4
µ √ ¶
√
t n
,
(5.2.11)
|{x ∈ K : |x| ≥ c2 t n}| ≤ exp −
2
√
where c2 = 6.
5.3 Linear functionals on the `n1 -ball · 73
Proof. Let α1 , . . . , αn ≥ 0. From Proposition 5.2.1 we have
ý
¯½
¾¯
¾!
n
n
n
X
X
X
¯
¯
+
2 ¯
n
2
2
¯ x ∈ K : |x|2 ≥
x ∈ R+ :
xk ≥ 4
αk
αk ¯ = µK
¯
k=1
k=1
k=1
ý
¾!
n
n
X
X
+
= µK
x ∈ Rn+ :
Xk2 ≥ 4
αk2
k=1
≤
≤
n
X
k=1
n
µ+
K ({x ∈ R+ : Xk ≥ 2αk })
k=1
n µ
X
k=1
¶
n
exp(−2cαk ).
k
This shows that
¯½
¾¯
n
n
³ ³
X
¯
en ´´
αk2 ¯¯ X
¯ x ∈ K : |x|2 ≥
(5.2.12)
≤
exp
−k
2α
−
log
,
k
¯
c2 ¯
k
k=1
k=1
√
where c = 1/ 6. We choose
√
en
n
1
+t
.
(5.2.13)
αk = log
2
k
k
Pn
We check that if t ≥ 2 then k=1 αk2 ≤ 4nt2 , and going back to (5.2.13) we
have
√ √ ª¯
¯©
√
√
¯ x ∈ K : |x| ≥ 2 6t n ¯ ≤ n exp(−2t n) ≤ exp(−t n)
(5.2.14)
for every t ≥ 2. This proves the Theorem.
5.3
Linear functionals on the `n1 -ball
Let B1n , the unit ball of `n1 , be equipped with the uniform distribution µn . The
density of µn is given by
(5.3.1)
dµn (x)
n!
= n χB1n (x).
dx
2
We also define ∆n = {x ∈ Rn+ : x1 + · · · + xn ≤ 1}.
Let θ = (θ1 , . . . , θn ) ∈ Rn and consider the linear functional fθ (x) = θ1 x1 +
· · · + θn xn .
Lemma 5.3.1. For every q ∈ N,
Z
n!(2q)! X 2q1
(5.3.2)
|fθ (x)|2q dµn (x) =
θ1 . . . θn2qn ,
(n + 2q)!
where the summation is over all non-negative integers q1 , . . . , qn with q1 + · · · +
qn = q.
74 · The 1-unconditional case
Proof. Using the fact that k · k1 is 1-unconditional, we write
Z
Z
X
(2q)!
2qn
1
(5.3.3)
|fθ |2q dµn =
θ12q1 . . . θn2qn x2q
1 . . . xn dµn (x),
(2q1 )! . . . (2qn )!
where the summation is over all non-negative integers q1 , . . . , qn with q1 + · · · +
qn = q. Now, a simple computation shows that
Z
Z
n!(2q1 )! . . . (2qn )!
2qn
2qn
1
1
.
(5.3.4)
x2q
.
.
.
x
dµ
(x)
=
n!
x2q
n
n
1
1 . . . xn dx =
(n + 2q)!
∆n
This proves the Lemma.
Theorem 5.3.2. For every θ ∈ Rn ,
√
(5.3.5)
c1 kθk∞ ≤ nkfθ kLψ2 (µn ) ≤ c2 kθk∞ ,
√
√
where c1 = 1/ 6 and c2 = 2 2.
√
Proof. If α = nkθk∞ , we have θ12q1 . . . θn2qn ≤ α2q /nq for ¡all q1 ,¢. . . , qn ≥ 1
with q1 + · · · + qn = q. Since the sum in (5.3.2) consists of n+q−1
terms, we
n−1
have
Z
n!(2q)! (n + q − 1)! α2q
|fθ (x)|2q dµn (x) ≤
(n + 2q)! (n − 1)!q! nq
µ ¶
n
2q q!α2q
=
(n + q) · · · (n + 2q) q
nq
4q q!α2q
≤
.
n2q
Therefore, if |t| < 1/(2α), we have
Z
¡
¢
exp (tnfθ (x))2 dµn (x)
=
≤
1+
1+
∞ 2q 2q Z
X
t n
q=1
∞
X
q!
|fθ (x)|2q dµn (x)
(2tα)2q
q=1
=
1
.
1 − 4t2 α2
√
If we choose t = 1/(2 2α), the last quantity becomes equal to 2. This shows
that
√ √
(5.3.6)
nkfθ kLψ2 (µn ) ≤ 2 2 nkθk∞ .
We now turn to the lower bound. We may clearly assume that θj ≥ 0 for
all 1 ≤ j ≤ n, and Lemma 5.3.1 shows that for every q ≥ 1 the function
F2q (θ) := kfθ k2q is increasing with respect to each coordinate on Rn+ . It follows
that Fψ2 (θ) := kfθ kLψ2 (µn ) has the same property. So,
(5.3.7)
kfθ kLψ2 (µn ) ≥ kθk∞ kfe1 kLψ2 (µn ) .
5.3 Linear functionals on the `n1 -ball · 75
Since g1 := f1 /kf1 kLψ2 (µn ) has ψ2 -norm equal to 1, we have
Z
(5.3.8)
2≥
¡
¢
1
1
exp [g1 (x)]2 dµn (x) ≥
n! kfe1 k2n
Lψ2 (µn )
Z
|fe1 (x)|2n dµn (x).
Taking into account (5.3.4), we get
kfe1 k2n
Lψ2 (µn ) ≥
(5.3.9)
1
1 n!(2n)!
≥
.
2n! (3n)!
2(3n)n
This shows that
√
(5.3.10)
√
n
√ √ kθk∞ ,
nkfθ kLψ2 (µn ) ≥ 2n
2 3n
which is the left hand side inequality of the Theorem.
Note that the proof of Theorem 5.3.2 gives the following estimate for kfθ kLp (µn ) ,
p ≥ 1.
Proposition 5.3.3. For every θ ∈ Rn and every p ≥ 1,
√
√
nkfθ kp ≤ c pkθk∞ ,
(5.3.11)
√
where c = 2 2.
Proof. For p = 2q this follows immediately (with c = 2) from the estimate
Z
|fθ (x)|2q dµn (x) ≤
(5.3.12)
where α =
√
4q q!α2q
n2q
nkθk∞ . It is then easily extended to all values of p ≥ 1.
We will prove a stronger statement. Define
p
(5.3.13)
Cn (θ) = kθk∞ n/ log n
for θ ∈ Rn and n ≥ 2. Since the expectation of kθk∞ on S n−1 is of the order of
p
log n/n, for a random θ ∈ S n−1 we have Cn (θ) ' 1.
Theorem 5.3.4. For every θ ∈ S n−1 and every p ≥ 2,
p
©√
ª
(5.3.14)
nkfθ kLp (µn ) ≤ C max
p, Cn (θ) p log p ,
where C > 0 is an absolute constant.
Remark 1: Observe that if θ ∈ S n−1 then
p
©√
ª √
(5.3.15)
max
p, Cn (θ) p log p ≤ npkθk∞
for all 1 ≤ p ≤ n. This shows that Theorem 5.3.4 gives stronger information
than Proposition 5.3.3 in the range p ∈ [1, n].
76 · The 1-unconditional case
Lemma 5.3.5. Let q ≥ 1 be an integer and set
X
(5.3.16)
Pq (y) =
y1q1 · · · ynqn
q1 +···+qn =q
Rn+
on
(as always in this Section, the summation is over all non-negative integers
q1 , . . . , qn with q1 + · · · + qn = q). If y ∈ Rn+ and y1 + · · · + yn = 1, then
q
(5.3.17)
Pq (y) ≤ (2e max{1/q, kyk∞ }) .
Proof. Let y ∈ Rn+ with y1 + · · · + yn = 1 and let 0 ≤ t ≤ 1/(2kyk∞ ). If
a = ty, then 0 ≤ ai ≤ 1/2 and hence, 1/(1 − ai ) ≤ e2ai for every i = 1, . . . , n.
Consequently,
(5.3.18)
n
X
X q
Y
1
q1
qn
qn
1
Pq (a) =
a1 · · · an ≤
a1 · · · an =
≤ e2(a1 +···+an ) .
(1
−
a
)
i
q +···+q =q
i=1
1
qi ≥0
n
In other words, for every t ∈ [0, 1/(2kyk∞ )] we have
(5.3.19)
Since t−q e2t
(5.3.20)
e2t
.
tq
has a minimum at q/2, we get
µ ¶q
2e
Pq (y) ≤
q
Pq (y) ≤
if kyk∞ ≤ 1/q, and
(5.3.21)
Pq (y) ≤ (2kyk∞ )q exp(1/kyk∞ )
if kyk∞ ≥ 1/q.
Proof of Theorem 5.3.4: Define y = (θ12 , . . . , θn2 ). From Lemma 5.3.1 and Lemma
5.3.5, for every q ≥ 1 we have
¢q
n!(2q)! ¡
2
(5.3.22)
kfθ k2q
.
L2q (µn ) ≤ (n + 2q)! 2e max{1/q, kθk∞ }
Since (n + 2q)! ≥ n!n2q and (2q)! ≤ (2q)2q , this gives
√
√
© 1
ª
√
(5.3.23) nkfθ kL2q (µn ) ≤ 2 2e max √ , kθk∞ = 2 2e max{ q, qkθk∞ }.
q
Therefore, if p ≥ 2 we find an integer q ≥ 1 such that q ≤ p ≤ 2q and write
√
√
nkfθ kLp (µn ) ≤ nkfθ kL2q (µn ) ≤ 2 2e max{ q, qkθk∞ }
√
√
≤ 2 2e max{ p, pkθk∞ }
p
√
√
= 2 2e max{ p, Cn (θ)p log n/n}.
Now, it is easy to check that if 2 ≤ p ≤ n we have logn n ≤ 2 logp p , and hence,
p
√
√
(5.3.24)
nkfθ kLp (µn ) ≤ 4 2e max{ p, Cn (θ) p log p},
while, if p > n then
(5.3.25) nkfθ kLp (µn ) ≤ nkfθ k∞ = nkθk∞ ≤ Cn (θ)
This proves the Theorem.
p
p
n log n ≤ Cn (θ) p log p.
2
5.4 The comparison theorem · 77
5.4
The comparison theorem
In Section 5.1 we saw that there exists an absolute constant C > 0 with the
following property: if K is an isotropic 1-unconditional convex body in Rn , then
µ
¶n
α1 + · · · + αn
+
(5.4.1)
|{x ∈ K : x1 ≥ α1 , . . . , xn ≥ αn }| ≤ 1 −
,
Cn
√
for all (α1 , . . . , αn ) ∈ K + . In particular, K ⊆ (C/2)nB1n (the constant C = 6
works). We fix such a constant C and define V = (C/2)nB1n . We also denote
by µT the uniform distribution on a convex body T .
We say that a function F : Rn → R+ belongs to the class Fn if it satisfies
the following:
1. F is symmetric with respect to each coordinate: we have
F (x1 , . . . , xn ) = F (ε1 x1 , . . . , εn xn )
for every x ∈ Rn and all choices of signs εi .
2. There exists a positive Borel measure ν on Rn+ which is finite on all compact
sets, such that
(5.4.2)
F (x) = ν([0, x1 ] × · · · × [0, xn ])
for every x = (x1 , . . . , xn ) ∈ Rn+ . If F is absolutely continuous with respect to
Lebesgue measure, this means that there exists a measurable function q : Rn+ →
R+ such that
Z x1
Z xn
(5.4.3)
F (x) =
···
q(z)dz
0
0
Rn+ .
for every x = (x1 , . . . , xn ) ∈
In this Section we prove the following comparison theorem.
Theorem 5.4.1. Let F ∈ Fn . For every isotropic 1-unconditional convex body
K in Rn and every t ≥ 0,
µK ({x ∈ Rn : F (x) ≥ t}) ≤ µV ({x ∈ Rn : F (x) ≥ t}).
(5.4.4)
In particular,
Z
(5.4.5)
Z
F (x)dµK (x) ≤
F (x)dµV (x).
Proof. We will prove that for every α1 , . . . , αn ≥ 0,
(5.4.6)
µK (|x1 | ≥ α1 , . . . , |xn | ≥ αn ) ≤ µV (|x1 | ≥ α1 , . . . , |xn | ≥ αn ).
We may assume that α = (α1 , . . . , αn ) is in the interior of K (otherwise, the
quantity on the left is equal to zero). Then, we also have α ∈ V .
Observe that (5.4.1) is equivalent to
¶n
µ
2(α1 + · · · + αn )
.
(5.4.7)
µK (|x1 | ≥ α1 , . . . , |xn | ≥ αn ) ≤ 1 −
Cn
78 · The 1-unconditional case
On the other hand,
µV (|x1 | ≥ α1 , . . . , |xn | ≥ αn )
=
=
=
|{x ∈ V : |x1 | ≥ α1 , . . . , |xn | ≥ αn }|
|V |
|[(C/2)n − (α1 + · · · + αn )]B1n |
|(C/2)nB1n |
µ
¶n
2(α1 + . . . + αn )
1−
.
Cn
This proves (5.4.6). Observe that this is equivalent to (5.4.4) for the function
Fα = χ{|x1 |≥α1 ,...,|xn |≥αn } . Note that Fα corresponds to the Dirac measure να
which gives a unit mass to the point α. Since the class {Fα : α ∈ Rn+ } coincides
with the extreme points of the cone Fn , the result follows.
5.5
Ψ2 -behavior of linear functionals
Let K be an isotropic 1-unconditional convex body in Rn . For every θ =
(θ1 , . . . , θn ) ∈ Rn , consider the linear functional fθ (x) = θ1 x1 + · · · + θn xn .
Note that for every q ∈ N,
Z
(5.5.1)
2q
|fθ (x)| dµK (x) = (2q)!
X
θ12q1 . . . θn2qn
(2q1 )! · · · (2qn )!
Z
2qn
1
x2q
1 · · · xn dµK (x),
where the summation is over all non-negative integers q1 , . . . , qn with q1 + · · · +
qn = q. From (5.4.6) it is clear that
Z
Z
(5.5.2)
|x1 |p1 · · · |xn |pn dµK (x) ≤ |x1 |p1 · · · |xn |pn dµV (x)
for all p1 , . . . , pn ≥ 0. This gives immediately the following.
Proposition 5.5.1. Let K be an isotropic 1-unconditional convex body in Rn .
For every θ ∈ Rn and every integer q ≥ 1,
Z
Z
(5.5.3)
|fθ (x)|2q dµK (x) ≤ |fθ (x)|2q dµV (x)
and
(5.5.4)
kfθ kLψ2 (µK ) ≤ kfθ kLψ2 (µV ) .
Proof. The second
follows from the first if we take the Taylor ex¡ inequality
¢
pansion of exp (fθ /r)2 with r = kfθ kLψ2 (µV ) and integrate with respect to
µK .
Observe that
(5.5.6)
kfθ kLψ2 (µV ) =
Cn
kfθ kLψ2 (µn ) .
2
Then, the next Theorem follows from Theorem 5.3.2.
5.5 Ψ2 -behavior of linear functionals · 79
Theorem 5.5.2. Let K be an isotropic 1-unconditional convex body in Rn . For
every θ ∈ Rn ,
√
(5.5.7)
kfθ kLψ2 (µK ) ≤ c nkθk∞ ,
where c > 0 is an absolute constant.
2
We also have a generalization of Theorem 5.3.4.
Theorem 5.5.3. Let K be an isotropic 1-unconditional convex body in Rn . For
every θ ∈ Rn and every p ≥ 2,
p
(5.5.8)
kfθ kLp (µK ) ≤ c max{1, Cn (θ)} p log p,
where c > 0 is an absolute constant and Cn (θ) = kθk∞
p
n/ log n.
2
Notes and References
The results of this Chapter were proved by Bobkov and Nazarov (see [7] and
[8]).
Chapter 6
Concentration of volume on
isotropic convex bodies
6.1
Formulation of the problem
In this Chapter we discuss the following question about the concentration of
volume on isotropic convex bodies.
Question: Do there exist an absolute constant c > 0 and a function φ : N → R+
with φ(n) → ∞ as n → ∞, such that for every isotropic convex body K in Rn
the inequality
¡©
ª¢
¡
¢
√
(6.1.1)
Prob x ∈ K : |x| ≥ c nLK t ≤ exp − φ(n)t
holds true for every t ≥ 1?
If we restrict ourselves to the class of isotropic 1-unconditional convex bodies,
then Theorem
5.2.3 shows that the answer is positive, and one can even take
√
φ(n) = n:
Fact 1: (Theorem 5.2.3) There exists an absolute constant c > 0 such that if
K is an isotropic 1-unconditional convex body in Rn , then
¡©
¡ √ ¢
√ ª¢
(6.1.2)
Prob x ∈ K : |x| ≥ c nt ≤ exp − nt
for every t ≥ 1.
2
Let us recall some general results of this kind. If K is an isotropic convex body
in Rn , then
Z
(6.1.3)
|x|2 dx = nL2K .
K
√
Applying Markov’s inequality we see that |K ∩(3 nLK )B2n | ≥ 8/9, and Borell’s
lemma proves the following:
Fact 2: If K is an isotropic convex body in Rn , then
¡©
ª¢
√
(6.1.4)
Prob x ∈ K : |x| ≥ 3 nLK t ≤ exp(−t)
for every t ≥ 1.
82 · Concentration of volume on isotropic convex bodies
Alesker (see Theorem 2.2.4) showed that if K is isotropic, then the Euclidean
norm f (x) = |x| satisfies the ψ2 -estimate
√
kf kψ2 ≤ c nLK ,
(6.1.5)
where c > 0 is an absolute constant. In particular, we have the following
improvement of the estimate in Fact 2:
Fact 3: There exists an absolute constant c > 0 such that if K is an isotropic
convex body in Rn , then
(6.1.6)
Prob
¡©
ª¢
√
x ∈ K : |x| ≥ c nLK t ≤ 2 exp(−t2 )
for every t ≥ 1.
Note that LK ' 1 in the case of 1-unconditional convex bodies. Since the
circumradius R(K) of an isotropic convex body K in Rn is always bounded by
cnLK , the estimate of Bobkov and Nazarov is stronger than the previous ones
for all t ≥ 1.
6.2
A first reduction
Proposition 6.2.1. Let K be an isotropic convex body in Rn which satisfies
(6.2.1)
Prob
¡©
ª¢
¡
¢
√
x ∈ K : |x| ≥ γ nLK t ≤ exp − φ(n)t
for every t ≥ 1, where γ ≥ 1 and φ(n) À 1 are two constants. Then,
√
Iq (K) ≤ 2γ nLK
(6.2.2)
for all 2 ≤ q ≤ c1 φ(n), where c1 > 0 is an absolute constant.
Proof. We write
Z
Iqq (K)
Z
K
Z γ √nLK
≤
≤
qsq−1 Prob (x ∈ K : |x| ≥ s) ds
0
qs
0
=
∞
|x|q dx =
=
q−1
Z
ds +
γ
∞
√
nLK
qs
q−1
µ
¶
φ(n)s
√
exp −
ds
γ nLK
µ √
¶q Z ∞
¡ √
¢q
γ nLK
γ nLK +
qtq−1 e−t dt
φ(n)
φ(n)
µ
µ
¶q ¶
cq
γ q nq/2 LqK 1 +
,
φ(n)
where c > 0 is an absolute constant. If cq ≤ φ(n), then
(6.2.3)
√
Iq (K) ≤ 21/q γ nLK ,
so we get the result for all 2 ≤ q ≤ c1 φ(n), where c1 := 1/c.
6.3 Average decay of hyperplane sections · 83
Proposition 6.2.2. Let K be an isotropic convex body in Rn and let γ ≥ 1,
ψ(n) > 2 be two constants. If
√
(6.2.4)
Iq (K) ≤ γ nLK
for all 2 ≤ q ≤ ψ(n), then
¡
¢
¡
¢
√
(6.2.5)
Prob x ∈ K : |x| ≥ cγ nLK t ≤ exp − ψ(n)t
for every t ≥ 1, where c > 0 is an absolute constant.
Proof. From Proposition 2.1.7 we have
Prob (x ∈ K : |x| ≥ 3CIq (K)t) ≤ e−qt
(6.2.6)
for every t ≥ 1, where C > 0 is an absolute constant. Setting q = ψ(n) and
using (6.2.4) we get
¡
¢
√
(6.2.7)
Prob x ∈ K : |x| ≥ 3Cγ nLK t ≤ exp(−ψ(n)t)
for every t ≥ 1, and the result follows with c := 3C.
p
From Proposition 3.1.4 we have wq (Zq (K)) ' q/nIq (K) for every isotropic
convex body K in Rn and every q ≤ n. Taking into account the previous two
Propositions we may state the following.
Theorem 6.2.3. Let φ(n) À 1 and let K be an isotropic convex body in Rn .
For every γ ≥ 1 the following are equivalent:
(i) For every t ≥ 1,
¡
¢
¡
¢
√
(6.2.8)
Prob x ∈ K : |x| ≥ γ nLK t ≤ exp − φ(n)t .
(ii) For every 2 ≤ q ≤ c1 φ(n),
√
Iq (K) ≤ c2 (γ) nLK ,
(6.2.9)
where c2 (γ) ' γ.
(iii) For every 2 ≤ q ≤ c3 φ(n),
(6.2.10)
√
wq (Zq (K)) ≤ c4 (γ) qLK ,
where c4 (γ) ' γ.
6.3
2
Average decay of hyperplane sections
Let K be a convex body in Rn with volume 1 and center of mass at the origin.
Recall that
(6.3.1)
fK,θ (t) = |K ∩ (θ⊥ + tθ)|
for every θ ∈ S n−1 and t ≥ 0. We define a function fK , which measures the
average decay of the volume of hyperplane sections of K, by
Z
(6.3.2)
fK (t) =
fK,θ (t)σ(dθ).
S n−1
The next Proposition gives an integral formula for fK (this proof is from [6];
see also [16]).
84 · Concentration of volume on isotropic convex bodies
Proposition 6.3.1. Let K be a convex body in Rn with volume 1 and center of
mass at the origin. For every t ≥ 0,
Z
(6.3.3)
fK (t) = cn
UK (t)
1
|x|
µ
¶ n−3
2
t2
1− 2
dx,
|x|
where UK (t) = {x ∈ K : |x| ≥ t} and
¡ ¢
Γ n2
cn = √ ¡ n−1 ¢ .
πΓ 2
(6.3.4)
Proof. We denote by λθ,t the Lebesgue measure on the hyperplane Nθ (t) = {x :
hx, θi = t}. Consider the measure
Z
(6.3.5)
λt =
λθ,t σ(dθ).
S n−1
Then, λt is a positive measure on Rn and
Z
Z
(6.3.6)
fK (t) =
χK (x)dλθ,t (x)σ(dθ) = λt (K).
S n−1
Nθ (t)
The density of λt is invariant under orthogonal transformations, therefore
dλt
= pt (|x|)
dx
(6.3.7)
where pt : [t, +∞) → [0, +∞). In order to find pt , for every r > t we compute
λt (B(0, r)) in two different ways. First, we have
Z
Z r
(6.3.8)
λt (B(0, r)) =
pt (|x|)dx = nωn
pt (s)sn−1 ds.
B(0,r)
t
On the other hand, since the intersection
√ of B(0, r) with the hyperplane Nθ (t)
is an (n − 1)-dimensional ball of radius r2 − t2 for every θ ∈ S n−1 , we get
Z
n−1
(6.3.9)
λt (B(0, r)) =
λθ,t (B(0, r))σ(dθ) = ωn−1 (r2 − t2 ) 2 .
S n−1
Differentiating with respect to r ≥ t, we see that
(6.3.10)
ωn−1
n−3
n−1 2
(r − t2 ) 2 2r = nωn pt (r)rn−1 .
2
This shows that
(6.3.11)
pt (r) =
(n − 1)ωn−1 (r2 − t2 )
nωn
rn−2
n−3
2
.
Therefore,
Z
(6.3.12)
fK (t) =
pt (|x|)dx =
UK (t)
(n − 1)ωn−1
nωn
Z
UK (t)
(|x|2 − t2 )
|x|n−2
n−3
2
dx,
and the result follows.
Corollary 6.3.2. fK is a decreasing function.
2
6.4 Reduction to the average sectional decay · 85
6.4
Reduction to the average sectional decay
In this Section we present one more reduction of the question, this time to the
behavior of the function fK . Our main tool will be some inequalities relating fK
to the generalized mean widths of the Lq -centroid bodies. It will be convenient
to simplify the notation of Chapter 3.
Definition. Let K be an isotropic convex body in Rn . For every q > 0 and
t > 0 we define
¶1/q
µZ
Z
|hx, θi|q dx σ(dθ)
(6.4.1)
Z(q) = wq (Zq (K)) =
S n−1
K
and
ÃZ
(6.4.2)
!1/q
Z
q
|hx, θi| dx σ(dθ)
Z(q, t) =
S n−1
BK,θ (t)
where
(6.4.3)
BK,θ (t) = {x ∈ K : |hx, θi| ≤ t}.
Note that Z(q, t) ≤ Z(q) for every t > 0. Also, Z(q) is a continuous and
increasing function of q.
Lemma 6.4.1. Let K be a convex body in Rn with volume 1 and center of mass
at the origin. For every t ≥ 0 we have the identity
Z t
(6.4.4)
Z q (q, t) = 2
rq fK (r)dr.
0
Proof. It is an immediate consequence of Fubini’s theorem:
Z
Z t
Z t Z
Z q (q, t) = 2
rq fK,θ (r)drσ(dθ) = 2
rq
fK,θ (r)σ(dθ)dr
S n−1 0
0
S n−1
Z t
= 2
rq fK (r)dr,
0
by the definition of fK .
Since Z(q) = wq (Zq (K)), Proposition 3.1.4 takes the following form.
Lemma 6.4.2. Let K be a convex body in Rn with volume 1 and center of mass
at the origin. Then,
r
q
(6.4.5)
Z(q) '
Iq (K)
q+n
for every q ≥ 1.
2
For every θ ∈ S n−1 and q ≥ 1 we write Hq (θ) = kh·, θikq . Let C ≥ 1 be the
absolute constant from Theorem 2.1.5. The next Lemma shows that integration of the function |h·, θi|q on the strip BK,θ (3αHq (θ)s) - for suitable s ' 1 essentially captures the value of Hqq (θ).
86 · Concentration of volume on isotropic convex bodies
Lemma 6.4.3. Let K be a convex body in Rn with volume 1 and center of mass
at the origin. Then, for every θ ∈ S n−1 and every q, s ≥ 1,
Z
³
´
−qs/2
q
q
(6.4.6)
1−e
(2C) Hq (θ) ≤
|hx, θi|q dx,
BK,θ (3CHq (θ)s)
where C > 0 is the constant in Theorem 2.1.5.
Proof. Proposition 2.1.7 shows that
(6.4.7)
|K \ BK,θ (3CHq (θ)s)| ≤ exp(−qs)
for all q, s ≥ 1. We write
Z
q
Hq (θ) =
Z
q
|hx, θi|q dx
|hx, θi| dx +
BK,θ (3CHq (θ)s)
K\BK,θ (3CHq (θ)s)
µZ
Z
q
≤
|hx, θi| dx + exp(−qs/2)
BK,θ (3CHq (θ)s)
BK,θ (3CHq (θ)s)
|hx, θi| dx
K
Z
≤
¶1/2
2q
|hx, θi|q dx + exp(−qs/2)(2C)q Hqq (θ),
where we have used (6.4.7), Cauchy-Schwarz inequality and Theorem 2.1.5 (for
the pair q, 2q).
Our main Lemma is the next one: it shows that Z(q) ' Z(q, t) when t becomes
of the order of Z(q).
Proposition 6.4.4. There exists an absolute constant β > 0 with the following
property: if K is a convex body K in Rn with volume 1 and center of mass at
the origin, then for every q ≥ 1,
(6.4.8)
Z q (q) ≤ 2Z q (q, βZ(q)).
Proof. For every t > 0 we set Ut = {θ ∈ S n−1 : Hq (θ) ≥ tZ(q)}. Markov’s
inequality shows that σ(Ut ) ≤ t−q . Using Lemma 6.4.3, for every s ≥ 1 we
write
Z
Z
−qs/2
q
q
(1 − e
(2C) )Z (q) ≤
|hx, θi|q dxσ(dθ)
S n−1 \Ut
Z
Z
BK,θ (3CHq (θ)s)
|hx, θi|q dxσ(dθ)
+
Z
Ut
BK,θ (3CHq (θ)s)
Z
|hx, θi|q dxσ(dθ)
≤
S n−1
+σ(Ut )
≤
≤
BK,θ (3CtsZ(q))
µZ
1/2
¶1/2
Z
2q
|hx, θi| dxσ(dθ)
S n−1
K
−q/2
Z q (q, 3CtsZ(q)) + t
Z q (2q)
q
q −q/2 q
Z (q, 3CtsZ(q)) + (2C) t
Z (q),
because Z(2q) ≤ 2CZ(q) (this follows from the fact that I2q (K, θ) ≤ 2CIq (K, θ)
for every θ ∈ S n−1 , by Theorem 2.1.5 applied to the functional f (x) = hx, θi).
6.4 Reduction to the average sectional decay · 87
We now choose s, t so that
√
t = 8C and es/2 = 8C. Then,
(1 − 4−q )Z q (q) ≤ Z q (q, 3CtsZ(q)) + 4−q Z q (q).
(6.4.9)
Inserting the values of t, s in (6.4.9) we compute the value of β.
Remark 1: The argument gives
¡
¢
(6.4.10)
1 − (2C)q (e−qs/2 + t−q/2 ) Z q (q) ≤ Z q (q, 3CtsZ(q))
for all t, s ≥ 1.
Proposition 6.4.5. There exists an absolute constant c > 0 with the following
property: if K is a convex body in Rn with volume 1 and center of mass at the
origin, then for every q ≥ 1,
(6.4.11)
fK (cZ(q)) ≤
1
exp(−q).
Z(q)
Proof. We set
G(t) = Z q (q, t).
(6.4.12)
From (6.4.10) and the observation that Z(q, t) ≤ Z(q) we have
¡
¢
(6.4.13)
1 − (2C)q (e−qs/2 + t−q/2 ) Z q (q) ≤ G(3CtsZ(q)) ≤ Z q (q)
for all t, s ≥ 1. Lemma 6.4.1 and the fact that fK is decreasing show that, if
u < v then
Z v
v q+1 − uq+1
(6.4.14)
G(v) − G(u) = 2
rq fK (r)dr ≥ 2fK (v)
.
q+1
u
It follows that for every t, s ≥ 1 we have
(6.4.15)
fK (6CtsZ(q)) ≤
q+1
(2C)q (e−qs/2 + t−q/2 )
Z q (q).
2 (6CtsZ(q))q+1 − (3CtsZ(q))q+1
Choosing s = 1 and t = e we have
(6.4.16)
fK (6CeZ(q)) ≤
q+1
1
1 −q
(3e/2)−q e−q/2 ≤
e ,
2q+1 − 1 Z(q)
Z(q)
which proves the Proposition with c := 6Ce.
We can now see the exact relation of our question to the behavior of fK .
Theorem 6.4.6. Let γ ≥ 1 and let K be an isotropic convex body in Rn . If
1 ¿ φ(n) ¿ n and
ª¢
¡
¢
¡©
√
(6.4.17)
Prob x ∈ K : |x| ≥ γ nLK t ≤ exp − φ(n)t
for every t ≥ 1, then
(6.4.18)
for all 0 < t ≤ c3 γ
fK (t) ≤
p
φ(n)LK .
¡
¢
c1
exp −c2 t2 /(γ 2 L2K )
LK
88 · Concentration of volume on isotropic convex bodies
Proof. We will use the integral formula for fK . We assume that n > 3 and write
Z
(6.4.19)
fK (t) = cn
gt (|x|)dx
UK (t)
for all t > 0, where gt is defined by
(6.4.20)
gt (s) =
1
s
µ
¶ n−3
2
t2
1− 2
s
√
on [t, ∞), and cn ' n. Differentiating gt wepsee that it is increasing on
√
[t, t n − 2] and then decreasing. Let 0 < t ≤ c3 γ φ(n)LK , where
√ the absolute
√
constant c3 > 0√is to be chosen. Assume first that γ nLK ≤ t n − 2 (this is
satisfied if t ≥ 2γLK ). Then, we write
Z
fK (t)
=
Z
cn
K∩{t≤|x|≤γ
≤
=
≤
≤
√
nLK }
gt (|x|)dx + cn
UK (γ
√
gt (|x|)dx
nLK )
√
√
cn gt (γ nLK ) + exp(−φ(n))cn gt (t n − 2)
µ
µ
¶ n−3
¶ n−3
2
2
cn
t2
1
cn
√
1− 2 2
1−
+ exp(−φ(n)) √
γ nLK
n−2
γ nLK
t n−2
c01
c00
exp(−c2 t2 /γ 2 L2K ) + 1 exp(−φ(n))
LK
LK
c1
2
2 2
exp(−c2 t /γ LK ),
LK
√
because φ(n) ≥ c2 t2 /γ 2 L2K if we choose c3 = 1/ c2 .
p
√
If 0 < t ≤ min{ 2γLK , c3 γ φ(n)LK }, then
(6.4.21)
fK (t) ≤
c5
2c5
≤
exp(−c7 t2 /γ 2 L2K ),
LK
LK
because fK,θ (t) ≤ c5 /LK for all θ ∈ S n−1 and exp(−c7 t2 /γ 2 L2K ) ≥ exp(−2c7 ) ≥
1/2 if c7 >p0 is suitably chosen. It follows that (6.4.18) holds true for all
0 < t ≤ c3 γ φ(n)LK .
Second proof: From Theorem 6.2.3 we have
(6.4.22)
√
Z(q) = wq (Zq (K)) ≤ c2 γ qLK
for all 2 ≤ q ≤ ψ(n) := c1 φ(n).
Let c > 0 be the constant in Proposition 6.4.5. The function D : [2, ψ(n)] → R
defined
by D(q) = cZ(q) is continuous and increasing. Since D(q) = cZ(q) '
p
√
q/n Iq (K) ≥ c3 qLK for every 2 ≤ q ≤ n, we have
p
(6.4.23)
D([2, ψ(n)]) ⊇ [cLK , c3 ψ(n)LK ].
p
Then, for every t ∈ [cLK , c3 ψ(n)LK ] there exists q(t)
p ∈ [2, ψ(n)] such that
cZ(q(t)) = D(q(t)) = t. Moreover, Z(q(t)) ≤ c2 γ q(t)LK from (6.4.22).
Therefore,
p
(6.4.24)
t ≤ c4 γ q(t)LK .
6.4 Reduction to the average sectional decay · 89
Now, Proposition 6.4.5 shows that
(6.4.25)
fK (t) ≤
1
c
exp(−q(t)) ≤ exp(−t2 /c24 γ 2 L2K ).
Z(q(t))
t
Since c/t ≤ 1/LK , we obtain
(6.4.26)
fK (t) ≤
c5
exp(−c6 t2 /γ 2 L2K ).
LK
The same inequality is clearly correct for 0 < t < cLK : it suffices to change the
value of the (absolute) constants c5 , c6 if needed.
2
The other direction is a consequence of Proposition 6.4.1.
Theorem 6.4.7. Let γ ' 1 and let K be an isotropic convex body in Rn .
Assume that
¡
¢
c1
(6.4.27)
fK (t) ≤
exp −t2 /γ 2 L2K
LK
for all 0 < t ≤ γψ(n)LK . Then, for every 2 ≤ q ≤ c2 ψ 2 (n) we have
√
(6.4.28)
Iq (K) ≤ c3 γ nLK .
Proof. Note that Z(2) = LK and Z(n) ≥ c0 R(K) for some absolute constant
c0 > 0.
We first assume that β ≤ γψ(n) < βc0 R(K)/LK . Then, there exists 2 ≤ s ≤ n
such that βZ(s) = γψ(n)LK . From Lemma 6.4.1 and Proposition 6.4.4 we see
that
Z βZ(s)
s
s
Z (s) ≤ 2Z (s, βZ(s)) = 4
rs fK (r)dr
0
Z
=
γψ(n)LK
4
rs fK (r)dr
0
Z
4c1 γψ(n)LK s
r exp(−r2 /γ 2 L2K )dr
LK 0
Z
4c1 ∞ s
r exp(−r2 /γ 2 L2K )dr
LK 0
√
(c4 γ sLK )s .
≤
≤
≤
In other words,
√
Z(s) ≤ c4 γ sLK .
(6.4.29)
Lemma 6.4.2 implies that
r
(6.4.30)
Is (K) ≤ c5
√
n
Z(s) ≤ c3 γ nLK ,
s
and Hölder’s inequality gives
(6.4.31)
√
Iq (K) ≤ Is (K) ≤ c3 γ nLK
90 · Concentration of volume on isotropic convex bodies
for all q ≤ s. On the other hand, by the definition of s and (6.4.29),
(6.4.32)
s≥
Z 2 (s)
c24 γ 2 L2K
=
ψ 2 (n)
=: c2 ψ 2 (n),
c24 β 2
which gives the result in this case.
To conclude the proof, observe that the range β ≤ γψ(n) ≤ βc0 R(K)/LK
is the interesting one for the parameter ψ(n). If 0 < γψ(n) < β, then the
conclusion of the Theorem is trivially true. If γψ(n) ≥ βc0 R(K)/LK , then we
have (6.4.27) for every
√ t > 0. Following the previous argument, we check that
In (K) ' Z(n) ≤ cγ nLK . But In (K) ' R(K), and this implies (6.4.28) for
every q ≥ 2.
Theorems 6.4.6 and 6.4.7, combined with the reduction in §6.2 prove the following.
Theorem 6.4.8. Let 1 ¿ φ(n) ¿ n and let K be an isotropic convex body in
Rn . For every γ ' 1 the following are equivalent:
(i) For every t ≥ 1,
√
(6.4.33)
Prob(x ∈ K : |x| ≥ γ nLK t) ≤ exp(−φ(n)t).
p
(ii) For every 0 < t ≤ c1 (γ) φ(n)LK ,
(6.4.34)
fK (t) ≤
¡
¢
c2
exp − t2 /(c3 (γ)LK )2 ,
LK
where ci (γ) ' γ.
Notes and References
The results of this Chapter come from the thesis of Paouris: see also [49].
2
Chapter 7
Random points and random
polytopes
7.1
Random points in isotropic convex bodies
The purpose of this Section is to prove the following Theorem about independent
random points which are uniformly distributed in an isotropic convex body.
Theorem 7.1.1. Let K be an isotropic convex body in Rn and let δ, ε ∈ (0, 1).
If m ≥ c(δ, ε)ε−2 n(log n)2 , then m random points x1 , . . . , xm which are chosen independently and uniformly from K satisfy with probability > 1 − δ the
following: For every θ ∈ S n−1 ,
m
(1 − ε)L2K ≤
(7.1.1)
1 X
hxj , θi2 ≤ (1 + ε)L2K .
m j=1
Remark
¡ 21:¢ The constant c(δ, ε) may be assumed to be bounded by a fixed power
.
of log εδ
We will use the following facts from Chapter 2: First, K satisfies a ψ1 -estimate
with constant C for some absolute constant C > 0. That is,
Z
¡
¢
(7.1.2)
exp |hx, θi|/CLK dx ≤ 2
K
for every θ ∈ S
that
n−1
. Equivalently, there exists an absolute constant c1 > 0 such
µZ
(7.1.3)
LK ≤
¶1/p
|hx, θi|p dx
≤ c1 pLK ,
K
n−1
for every p ≥ 2 and θ ∈ S
. Second, we have a ψ2 -estimate for the Euclidean
norm: there exists an absolute constant c2 > 0 such that
µ
¶
Z
|x|2
(7.1.4)
exp 2 2 dx ≤ 2.
c2 nLK
K
The important step for the proof is done in the next Theorem.
92 · Random points and random polytopes
Theorem 7.1.2. Let δ ∈ (0, 1) and let x1 , . . . , xm be random points independently and uniformly distributed in K. With probability greater than 1 − δ we
have
p
√
(7.1.5)
|xj | ≤ c1 (δ) nLK log m
for all j ∈ {1, . . . , m}, and
p
p √
¯X ¯
¯
(7.1.6)
xi ¯ ≤ c2 (δ)LK log m |E| n + c3 (δ)LK (log m)|E|
i∈E
q
log 2δ , i = 1, 2, 3.
for all E ⊆ {1, . . . , m}, where ci (δ) '
Proof. From (7.1.4) we have
µ
¶
√
δ
(7.1.7)
Prob max |xj | ≥ c2 t nLK ≤ m exp(−t2 ) <
1≤j≤m
2
p
provided that t ≥ log(2m/δ). So, with probability greater than 1 − (δ/2) we
have
p
√
(7.1.8)
|xj | ≤ c1 (δ) nLK log m
for every j = 1, . . . , m.
Let E ⊆ {1, . . . , m}. We write
(7.1.9)
X
X
¯ X ¯2 X
¯
hxi , xj i.
xi ¯ =
hxi , xj i ≤ c21 (δ)L2K n(log m)|E| +
|xi |2 +
i∈E
i∈E
i6=j∈E
i6=j∈E
If δj takes the values 0 or 1 with probability 1/2, then
(7.1.10)
E~δ
m
­X
δi xi ,
i=1
m
X
® 1 X
(1 − δj )xj =
hxi , xj i.
4
j=1
i6=j∈E
Therefore, we can find E1 , E2 ⊂ E with |E1 | ≥ |E2 |, E1 ∩ E2 = ∅, E1 ∪ E2 = E,
such that
X
X ®
­X
hxi , xj i ≤ 4
xi ,
xj
i∈E1
i6=j∈E
≤
j∈E2
X ¯­
X ®¯
¯ xi ,
4
x j ¯.
i∈E1
j∈E2
Rewrite this last sum in the form
X ¯­
X ®¯ ¯ X ¯ X
¯ xi ,
(7.1.11)
xj ¯ = ¯
xj ¯ ·
|hxi , yE2 i|,
i∈E1
j∈E2
j∈E2
i∈E1
where
P
(7.1.12)
j∈E xj
yE2 = ¯¯ P 2 ¯¯ ,
j∈E2 xj
7.1 Random points in isotropic convex bodies · 93
and |yE2 | = 1. Observe that the set {xi }i∈E1 is independent from yE2 , since
E1 ∩ E2 = ∅. If we fix |E1 | = k, applying (7.1.2) and Fubini’s theorem we see
that
!
Ã
Z
1 X
|hxi , yE2 i| dx ≤ 2k .
(7.1.13)
exp
CLK
Kk
i∈E1
Therefore,
Ã
(7.1.14)
Prob ~x ∈ K
m
:
X
!
< 2k e−kt ≤ e−kt/2
|hxi , yE2 i| > CtkLK
i∈E1
if t ≥ 2. The number of possible E1 ’s is bounded by mk , and hence, for every
k≤m
(7.1.15)Ã
!
X
m
Prob ~x ∈ K : ∃E1 ⊂ E, |E1 | = k,
|hxi , yE2 i| > CtkLK < mk e−kt/2 .
i∈E1
This
£ m ¤ probability will be smaller than δ/m ifm t 'δ log m. Doing this for k =
satisfies with probability greater
2 + 1, . . . , m, we see that (x1 , . . . , xm ) ∈ K
than 1 − 2δ the following: For every E ⊆ {1, . . . , m},
(7.1.16)
X
¯
¯¾
¯ X
¯
|E1 | ¯¯
xj ¯¯ .
½
hxi , xj i ≤ c2 (δ)LK (log m) max
E1 ⊂E
i6=j∈E
j∈E\E1
To finish the proof, consider any m-tuple which satisfies (7.1.9) and (7.1.16),
and for every s ≤ m write
¯X ¯
(7.1.17)
As = max ¯
x j ¯,
j∈F
where the maximum is over all F ⊆ {1, . . . , m} with |F | ≤ s. For every E ⊆
{1, . . . , m} we have
(7.1.18)
¯ X ¯2
¯
xi ¯ ≤ c21 (δ)L2K n(log m)|E| + c2 (δ)LK (log m)|E|A|E| ,
i∈E
therefore
(7.1.19)
A2|E| ≤ c21 (δ)L2K n(log m)|E| + c2 (δ)LK (log m)|E|A|E| ,
which implies
(7.1.20)
p
√ p
A|E| ≤ c2 (δ)LK n log m |E| + c3 (δ)LK (log m)|E|.
q
From the argument we easily check that ci (δ) '
log 2δ , i = 1, 2, 3.
We will also use a version of Bernstein’s inequality:
94 · Random points and random polytopes
Lemma 7.1.3. Let {fj }j≤m be independent random variables with mean 0 on
some probability space (Ω, µ). If kfj k1 ≤ 2 and kfj k∞ ≤ M , then, for every
ε ∈ (0, 1),


m
¯X
¯
(7.1.21)
Prob ¯
fj ¯ > εm ≤ 2 exp(−ε2 m/8M ).
j=1
Lemma 7.1.4. Let K be an isotropic convex body in Rn . Fix δ, ζ, ε ∈ (0, 1). If
µ ¶
2
LK
(7.1.22)
B ≥ c4 log
ε
and
µ
(7.1.23)
m ≥ c5 ε−2 n
B
LK
¶2
µ
log
2
δζ
¶
,
then m random points x1 , . . . , xm which are chosen independently and uniformly
from K satisfy with probability greater than 1 − δ the following: for all θ in a
ζ-net for S n−1 ,
³
³
X
ε´ 2
ε´ 2
1
hxj , θi2 ≤ 1 +
(7.1.24)
1−
LK ≤
LK
2
m
2
{j:|hxj ,θi|≤B}
Proof. There exists a ζ-net N for S n−1 with cardinality |N | ≤ (3/ζ)n .
We fix θ ∈ N and define fθ : K → R by
(7.1.23)
fθ (x) =
1
hx, θi2 χ{z∈K:|hz,θi|≤B} (x).
L2K
From (7.1.2) and (7.1.3) we see that
1
L2K
Z
hx, θi2 dx
≤
|{z ∈ K : |hz, θi| ≥ B}|1/2 ·
≤
c3 exp(−B/CLK ),
{|h·,θi|≥B}
1
L2K
µZ
hx, θi4 dx
¶1/2
K
where c3 > 0 is an absolute constant. This will be less than ε/4 if
µ ¶
2
(7.1.25)
B ≥ c4 log
LK
ε
where c4 > 0 is an absolute constant. If B satisfies (7.1.25), then
Z
1
ε
(7.1.26)
1 − Efθ = 2
hx, θi2 dx ≤ ,
LK {z∈K:|hz,θi|>B}
4
for every θ ∈ S n−1 . We define fθ,j (x1 , . . . , xm ) = fθ (xj ) − Efθ on K m . We
easily check that
¶2
µ
B
.
(7.1.27)
kfθ,j k1 ≤ 2 , Efθ,j = 0 and kfθ,j k∞ ≤
LK
7.1 Random points in isotropic convex bodies · 95
Then, Lemma 7.1.3 gives


¶
µ
m
¯ ε
¯1 X
ε2 m
δ
¯
¯


fθ (xj ) − Efθ >
(7.1.28) Prob
≤ 2 exp − 0
<
2
m j=1
4
c3 (B/LK )
|N |
if we choose
µ
−2
(7.1.29)
m ≥ c5 ε
n
B
LK
¶2
µ
log
2
δζ
¶
.
Therefore, with probability greater than 1 − δ the points x1 , . . . , xm satisfy
¯
¯
X
¯1
¯ εL2K
2
2
¯
¯≤
(7.1.30)
hx
,
θi
−
L
Ef
,
j
θ
K
¯m
¯
4
{j:|hxj ,θi|≤B}
for every θ ∈ N . Taking into account (7.1.26) we get the assertion of the
Lemma.
Proof of Theorem 7.1.1: We may assume that x1 , . . . , xm satisfy the conclusion
of Theorem 7.1.2 with probability > 1 − 2δ . We fix ζ ∈ (0, 1) (which will be later
chosen of the order of ε). Let
(7.1.31)
B = 4c3 (δ)LK log m,
where c3 (δ) is the constant in Theorem 7.1.2. If log m > c6 log
and
µ ¶
µ ¶
2
2
(7.1.32)
m ≥ c7 ε−2 n(log m)2 log
log
,
δ
δζ
¡2¢
ε
/ log1/2
¡2¢
δ
then, we know that with probability greater than 1 − 2δ , the random points
x1 , . . . , xm ∈ K satisfy
³
³
X
ε´ 2
ε´ 2
1
hxj , θi2 ≤ 1 +
(7.1.33)
1−
LK ≤
LK
2
m
2
{j:|hxj ,θi|≤B}
for all θ in a ζ-net N of S n−1 .
For every β ≥ B and every θ ∈ S n−1 we define
(7.1.35)
Eβ (θ) = {j ≤ m : |hxj , θi| > β}.
Using Theorem 7.1.2 we can estimate the cardinality of Eβ (θ) as follows:
X
|hxj , θi|
β|Eβ | ≤
j∈Eβ
≤
≤
≤
≤
max |
εj =±1
X
2 max |
F ⊆Eβ
εj xj |
j∈Eβ
X
j∈F
xj |
√ q
log m n |Eβ | + 2c3 (δ)LK (log m)|Eβ |
p
√ q
β
2c2 (δ)LK log m n |Eβ | + |Eβ |.
2
2c2 (δ)LK
p
96 · Random points and random polytopes
This gives: for every θ ∈ S n−1 ,
β 2 |Eβ (θ)| ≤ 16c22 (δ)L2K n log m.
(7.1.36)
It follows that
X
hxj , θi2
kX
0 −1
=
X
hxj , θi2
k=0 {j:2k B<|hxj ,θi|≤2k+1 B}
{j:|hxj ,θi|>B}
kX
0 −1
≤
k=0
≤
c log
|E2k B |(2k+1 B)2
µ ¶
2
L2K n(log m)k0 ,
δ
where the summation is over all non-empty E, and k0 is the least integer for
which R(K) ≤ 2k0 B. Since K is isotropic, we have R(K) ≤ (n + 1)LK . So,
recalling the definition of B, we get
µ
¶
nLK
(7.1.37)
k0 ≤ c log
≤ c0 log n.
B
Now, if m satisfies (7.1.32), we have
(7.1.38)
1
m
X
2
hxj , θi ≤
{j:|hxj ,θi|>B}
c log
¡2¢
δ
L2K n(log n)(log m)
εL2K
<
.
m
20
Combining with (7.1.33) we conclude the proof for every θ ∈ N . Finally, choosing ζ = ε/10 and using a standard successive approximation argument we get a
similar estimate for every θ ∈ S n−1 .
2
7.2
Random polytopes in isotropic convex bodies
Let K be a convex body in Rn with volume 1. We fix N ≥ n + 1 and consider
random points x1 , . . . , xN independently and uniformly distributed in K. Let
C(x1 , . . . , xN ) be their convex hull. For every p > 0 we consider the quantity
µZ
(7.2.1)
¶1/pn
Z
Ep (K, N ) =
|C(x1 , . . . , xN )|p dxN . . . dx1
...
K
.
K
Observe that Ep (K, N ) is an affinely invariant quantity, so we may also assume
that K has its center of mass at the origin. When N = n + 1, these quantities
are exact functions of the isotropic constant of K. To see this, recall that
Z
Z
(7.2.2)
S22 (K) :=
...
|co(0, x1 , . . . , xn )|2 dxn . . . dx1
K
K
satisfies the identity
(7.2.3)
2
L2n
K = n!S2 (K)
7.2 Random polytopes in isotropic convex bodies · 97
and
2 2
S22 (K) ≤ E2n
2 (K, n + 1) ≤ (n + 1) S2 (K).
(7.2.4)
It follows that
LK
LK
c1 √ ≤ E2 (K, n + 1) ≤ c2 √
n
n
(7.2.5)
where c1 , c2 > 0 are absolute constants. Moreover, using Khintchine type inequalities for linear functionals
on convex bodies (see §2.1) one can show that
√
Ep (K, n + 1) ≥ cLK / n for every p > 0, where c > 0 is an absolute constant.
In this Section, we give estimates for the volume radius E1/n (K, N ) of a
random N -tope C(x1 , . . . , xN ) in K.
1. Upper bounds: As it turns out, a generalization of the upper bound in
(7.2.5) is possible.
Theorem 7.2.1. Let K be a convex body in Rn with volume 1. For every
N ≥ n + 1, we have
(7.2.6)
E1/n (K, N ) ≤ cLK
log(2N/n)
√
n
where c > 0 is an absolute constant.
Remark 1: Let α ∈ [1, 2]. We will say that a convex body K in Rn is a ψα -body
with constant bα if
µZ
¶1/p
Z
(7.2.7)
|hx, yi|p dx
≤ bα p1/α
|hx, yi|dx
K
K
n
for every y ∈ R and p ≥ 1. Our method shows
p that if K is√a ψ2 -body then one
has the stronger estimate E1/n (K, N ) ≤ cLK log(2N/n)/ n. This is optimal
and might be the right dependence for every convex body K in Rn .
We shall prove the following.
Theorem 7.2.2. Let K be an isotropic convex body in Rn . Assume that for
some α ∈ [1, 2], K is a ψα -body with constant bα . Then, for every N ≥ n + 1,
¡
¢1/α
log(2N/n)
√
(7.2.8)
E1/n (K, N ) ≤ cbα LK
.
n
This implies Theorem 7.2.1 because every convex body of volume 1 has an
affine image which is isotropic, and every isotropic convex body is a ψ1 -body
with constant b1 ≤ C. For the proof, we will use a result of Ball and Pajor on
the volume of symmetric convex bodies which are intersections of symmetric
strips in Rn .
Lemma 7.2.3. Let x1 , . . . , xN ∈ Rn \ {0} and let 1 ≤ q < ∞. If
(7.2.9)
W = {z ∈ Rn : |hz, xj i| ≤ 1, j = 1, . . . , N },
then

(7.2.10)
|W |1/n
−1/q
Z
N
X
n
+
q
1
≥ 2
|hz, xj i|q dz 
.
n j=1 |Bqn | Bqn
98 · Random points and random polytopes
2
Proof of Theorem 7.2.2: In the sequel we will write KN for the absolute convex
hull co{±x1 , . . . , ±xN } of N random points from K. By the Blaschke-Santaló
inequality,
◦ −1/n
E1/n (K, N ) ≤ E|KN |1/n ≤ ωn2/n · E|KN
|
(7.2.11)
◦
where KN
is the polar body of KN . Lemma 7.2.3 shows that
◦ −1/n
|KN
|
(7.2.12)
1/q

Z
N
1 n + q X 1
≤
|hz, xj i|q dz 
2
n j=1 |Bqn | Bqn
for every q ≥ 1. Consider the convex body W = K × · · · × K (N times) in
RN n . We apply Hölder’s inequality, change the order of integration and use the
ψα -property of K:

1/q
Z
N
X
1 n + q
1
|hz, xj i|q dz  dxN . . . dx1
2
n j=1 |Bqn | Bqn
Z
◦ −1/n
E|KN
|
2/n
Since ωn
(7.2.13)
≤
W
≤

1/q
Z Z
N
1 n + q X 1
|hz, xj i|q dxN . . . dx1 dz 
2
n j=1 |Bqn | Bqn W
≤
1
2
Ã
Z
´q
n + q ³ 1/α
1
q bα LK N n
|z|q dz
n
|Bq | Bqn
!1/q
.
≤ c1 /n and |z| ≤ n1/2−1/q for all z ∈ Bqn when q ≥ 2, we get
c
E1/n (K, N ) ≤ √ bα LK q 1/α
n
µ
N
n
¶1/q µ
n+q
n
¶1/q
for every q ≥ 2. Choosing q = log(e2 N/n) we complete the proof.
2
2. The 1-unconditional case: Let K be an isotropic 1-unconditional convex
body in Rn . We will prove an optimal upper bound.
Theorem 7.2.4. Let K be an isotropic 1-unconditional convex body in Rn .
Then, for every N ≥ n + 1,
p
log(2N/n)
√
(7.2.14)
E1/n (K, N ) ≤ C
,
n
where C > 0 is an absolute constant.
We will use the following properties of K (which were proved in Chapter 5):
(i) For every y ∈ Rn and every p ≥ 1,
µZ
¶1/p
√ √
|hx, yi| dx
≤ c1 p nkyk∞ .
p
(7.2.15)
K
7.2 Random polytopes in isotropic convex bodies · 99
In other words,
√
kh·, yikψ2 ≤ c nkyk∞ .
(7.2.16)
(ii) There exists an absolute constant c2 > 0 such that for every t ≥ 1,
¯©
¡ √ ¢
√ ª¯
¯ x ∈ K : |x| ≥ c2 t n ¯ ≤ exp −t n .
(7.2.17)
We will also use a result of Bárány and Füredi which is in the spirit of Lemma
7.2.3.
Lemma 7.2.5. There exists an absolute constant c3 > 0 such that: if N ≥ n+1
and x1 , . . . , xN ∈ Rn , then
p
log(2N/n)
1/n
(7.2.18)
|C(x1 , . . . , xN )|
.
≤ c3 · max |xi | ·
i≤N
n
Proof of Theorem 7.2.4: We distinguish two cases (small and large N ):
Case 1: N ≤ n2 : Fix t ≥ 1, which will be suitably chosen. We know that
¡ √ ¢
¡
√ ¢
(7.2.19) Prob (x1 , . . . , xN ) : ∃i ≤ N s.t |xi | ≥ c2 t n ≤ N · exp −t n .
If A is the event in (7.2.19), using Lemma 7.2.5 we write
Z
Z
1/n
E1/n (K, N ) =
|C(x1 , . . . , xN )|
+
|C(x1 , . . . , xN )|1/n
A
Ac
p
√
log(2N/n)
c
≤ Prob(A) + Prob(A ) · c3 c2 t n ·
n
p
¡
√ ¢
log(2N/n)
√
.
≤ N · exp −c1 t n + c4 t ·
n
We have assumed that N ≤ n2 , which implies
(7.2.20)
¡ √ ¢ c6 t6 n3
n·N
√
exp c1 t n ≥ 1
≥
6!
c4 t log 2
if t ≥ 1 is chosen large enough (independently from n and N ). Then,
p
log(2N/n)
√
(7.2.21)
E1/n (K, N ) ≤ (2c4 t) ·
.
n
Case 2: N ≥ n2 : Repeating the argument of the proof of Theorem 7.2.2, we
arrive at
Ã
!1/q
Z Z
2/n
N (n + q) 1
ωn
q
|hz, xi| dx dz
.
(7.2.22)
E1/n (K, N ) ≤
2
n
|Bqn | Bqn K
Now,
Z
¡ √ √
¢q
|hz, xi|q dx ≤ C q nkzk∞
(7.2.23)
K
100 · Random points and random polytopes
for every z ∈ Bqn . Therefore,
2/n
(7.2.24)
ωn
√ √
E1/n (K, N ) ≤
C q n
2
Ã
N (n + q) 1
n
|Bqn |
!1/q
Z
Bqn
kzkq∞ dz
.
Observe that kzk∞ ≤ kzkq and
Z
Z
Bqn
kzkqq dz
1
=
Z
0
1
=
0
qtq−1 |{z ∈ Bqn : kzkq ≥ t}|dt
qtq−1 |Bqn \tBqn |dt
= |Bqn |
=
Z
1
qtq−1 (1 − tn )dt
0
n
|B n |.
n+q q
It follows that
(7.2.25)
1
|Bqn |
Z
Bqn
kzkq∞ dz ≤
1
|Bqn |
Z
Bqn
kzkqq dz =
n
.
n+q
Combining the above we get
(7.2.26)
√
q
E1/n (K, N ) ≤ C2 √ · N 1/q
n
for every q ≥ 1. Choose q = log(2N/n). Since N ≥ n2 , we have
µ
¶
µ
¶
log N
log N
1/q
√
(7.2.27)
N
= exp
≤ exp
≤ e2 .
log(2N/n)
log(2 N )
Therefore,
p
(7.2.28)
E1/n (K, N ) ≤ C3
log(2N/n)
√
.
n
2
3. Lower bounds: Our lower bound is based on an extension of a result of
Groemer.
Theorem 7.2.6. Let K be a convex body in Rn with volume 1. Write B for a
ball in Rn with volume 1. Then,
(7.2.29)
Ep (K, N ) ≥ Ep (B, N )
for every p > 0. In particular, the expected volume radius E1/n (K, N ) of a
random N -tope in K is minimal when K = B.
The method is to show that Steiner symmetrization decreases Ep (K, N ): Let H
be an (n − 1)-dimensional subspace of Rn . We identify H with Rn−1 and write
7.2 Random polytopes in isotropic convex bodies · 101
x = (y, t), y ∈ H, t ∈ R for a point x ∈ Rn . If K is a convex body in Rn with
|K| = 1 and P (K) is the orthogonal projection of K onto H, then
Z
Z
(7.2.30)
Epn
(K,
N
)
=
.
.
.
Mp,K (y1 , . . . , yN )dyN . . . dy1
p
P (K)
where
(7.2.31)
Z
P (K)
Z
Mp,K (y1 , . . . , yN ) =
|C((y1 , t1 ), . . . , (yN , tN ))|p dtN . . . dt1
...
`(K,y1 )
`(K,yN )
and `(K, y) = {t ∈ R : (y, t) ∈ K}.
We fix y1 , . . . , yN ∈ H and consider the function FY : RN → R defined by
(7.2.32)
FY (t1 , . . . , tN ) = |C((y1 , t1 ), . . . , (yN , tN ))|,
where Y = (y1 , . . . , yN ). The key observation is the following:
Lemma 7.2.7. For any y1 , . . . , yN ∈ P (K), the function FY is convex.
2
We now also fix r1 , . . . , rN > 0 and define Q = {U = (u1 , . . . , uN ) : |ui | ≤
ri , i = 1, . . . , N }. For every N -tuple W = (w1 , . . . , wN ) ∈ RN we set
(7.2.33)
GW (u1 , . . . , uN ) = FY (w1 + u1 , . . . , wN + uN ),
and write
GW (U ) = FY (W + U ).
This is the volume of the polytope which is generated by the points (yi , wi + ui ).
Finally, for every W ∈ RN and α > 0, we define
(7.2.34)
A(W, α) = {U ∈ Q : GW (U ) ≤ α}.
With this notation, we have
Lemma 7.2.8. Let α > 0 and λ ∈ (0, 1). If W, W 0 ∈ RN , then
(7.2.35)
|A(λW + (1 − λ)W 0 , α)| ≥ |A(W, α)|λ |A(W 0 , α)|1−λ .
Proof. Let U ∈ A(W, α) and U 0 ∈ A(W 0 , α). Then, using the convexity of FY
we see that
GλW +(1−λ)W 0 (λU + (1 − λ)U 0 )
= FY (λ(W + U ) + (1 − λ)(W 0 + U 0 ))
≤
=
λFY (W + U ) + (1 − λ)FY (W 0 + U 0 )
λGW (U ) + (1 − λ)GW 0 (U 0 )
≤
α.
Therefore,
(7.2.36)
A(λW + (1 − λ)W 0 ) ⊇ λA(W, α) + (1 − λ)A(W 0 , α)
and the result follows from the Brunn-Minkowski inequality.
102 · Random points and random polytopes
Observe that the polytopes C((yi , wi + ui )i≤N ) and C((yi , −wi − ui )i≤N )
have the same volume since they are reflections of each other with respect to
H. It follows that
(7.2.37)
A(−W, α) = −A(W, α)
for every α > 0. Taking W 0 = −W and λ = 1/2 in Lemma 7.2.8, we obtain the
following:
Lemma 7.2.9. Let y1 , . . . , yN ∈ H. For every W ∈ RN and every α > 0,
(7.2.38)
|A(O, α)| ≥ |A(W, α)|,
where O is the origin in RN .
2
For every y ∈ P (K), we denote by w(y) the midpoint and by 2r(y) the length
of `(K, y). Let S(K) be the Steiner symmetral of K. By definition, P (S(K)) =
P (K) = P and for every y ∈ P the midpoint and length of `(S(K), y) are
w0 (y) = 0 and 2r0 (y) = 2r(y) respectively.
Lemma 7.2.10. Let y1 , . . . , yN ∈ P (K) = P (S(K)). Then,
(7.2.39)
Mp,K (y1 , . . . , yN ) ≥ Mp,S(K) (y1 , . . . , yN )
for every p > 0.
Proof. In the notation of the previous Lemmas, we have
Z
Mp,K (y1 , . . . , yN ) =
[GW (U )]p dU
Z
Q
∞
Z
0
=
=
∞
|{U ∈ Q : GW (U ) ≥ t1/p }|dt
³
´
|Q| − |A(W, t1/p )| dt.
0
By the definition of S(K),
Z
(7.2.40)
Z
∞
p
Mp,K (y1 , . . . , yN ) =
[GO (U )] dU =
Q
³
´
|Q| − |A(O, t1/p )| dt,
0
and the result follows from Lemma 7.2.9.
It is now clear that Ep (K, N ) decreases under Steiner symmetrization.
Theorem 7.2.11. Let K be a convex body with volume |K| = 1 and let H be
an (n − 1)-dimensional subspace of Rn . If SH (K) is the Steiner symmetral of
K with respect to H, then
(7.2.41)
for every p > 0.
Ep (S(K), N ) ≤ Ep (K, N )
7.2 Random polytopes in isotropic convex bodies · 103
Proof. We may assume that H = Rn−1 . Since P (S(K)) = P (K), Lemma 7.2.10
and (7.2.30) show that
Z
Z
Epn
(K,
N
)
=
.
.
.
Mp,K (y1 , . . . , yN )dyN . . . dy1
p
PK
ZP K
Z
≥
...
Mp,SK (y1 , . . . , yN )dyN . . . dy1
P (SK)
=
P (SK)
Epn
p (S(K), N ).
Proof of Theorem 7.2.6: Since the ball B of volume 1 is the Hausdorff limit of
a sequence of successive Steiner symmetrizations of K, Theorem 7.2.11 shows
that the expected volume radius is minimal in the case of B.
2
Remark 2: The argument shows that a more general fact holds true:
Theorem 7.2.12. Let K be a convex body in Rn with volume 1. Write B for
a ball in Rn with volume 1. Then,
(7.2.42)
Z
Z
Z
Z
...
f (|C(x1 , . . . , xN )|)dxN . . . dx1 ≥
...
f (|C(x1 , . . . , xN )|)dxN . . . dx1
K
K
B
+
B
+
for every increasing function f : R → R .
2
Next, we give a lower bound for E1/n (B, N ). We will actually prove that
the
p convex hull√ of N random points from K = B contains a ball of radius
c log(2N/n)/ n.
Lemma 7.2.13. Let B = rB2n be the centered ball of volume 1 in Rn . If
θ ∈ S n−1 , then
Prob (x ∈ B : hx, θi ≥ εr) ≥ exp(−4ε2 n)
√
for every ε ∈ (c1 / n, 1/4), where c1 > 0 is an absolute constant.
(7.2.43)
Proof. A simple calculation shows that
Z
Prob (x ∈ B : hx, θi ≥ εr)
= ωn−1 rn
1
(1 − t2 )(n−1)/2 dt
ε
ωn−1
ε(1 − 4ε2 )(n−1)/2
ωn
≥ exp(−4(n − 1)ε2 ) ≥ exp(−4ε2 n)
≥
since
√
nωn ≤ c1 ωn−1 for some absolute constant c1 > 0.
Lemma 7.2.14. There exist c > 0 and n0 ∈ N such that: if n ≥ n0 and
n(log n)2 ≤ N ≤ exp(cn), then
p
log(N/n)
√
(7.2.44)
C(x1 , . . . , xN ) ⊇
B
6 n
with probability greater than 1 − exp(−n).
104 · Random points and random polytopes
Proof. By Lemma 7.2.13, for every θ ∈ S n−1 we have
µ
¶
≤ (1 − exp(−4ε2 n))N
Prob (x1 , . . . , xN ) : maxhxj , θi ≤ εr
j≤N
≤
exp(−N exp(−4ε2 n))
√
for every ε ∈ (c1 / n, 1/4). Let N be a ρ-net for S n−1 with cardinality |N | ≤
exp(log(1 + 2/ρ)n). If
¡
¢
(7.2.45)
exp n log(1 + 2/ρ) − N exp(−4ε2 n) ≤ exp(−n),
then with probability greater than 1 − exp(−n) we have maxj≤N hxj , θi > εr for
all θ ∈ N . For every u ∈ S n−1 we find θ ∈ N with |θ − u| < ρ. Then,
(7.2.46)
maxhxj , ui ≥ maxhxj , θi − maxhxj , θ − ui ≥ (ε − ρ)r.
j≤N
j≤N
We choose ε = 2a ((log(N/n)/n)
mined) and ρ = ε/2. Then,
1/2
j≤N
(a > 0 is an absolute constant to be deterµ
(7.2.47)
n log(1 + 2/ρ) + n ≤ 2n log(3/ρ) ≤ n log
9n
a2 log(N/n)
¶
≤ n log n,
if a2 ≥ 9/ log(N/n). Therefore, (7.2.45) will be a consequence of
(7.2.48)
exp(16a2 log(N/n)) ≤
N
,
n log n
which can be written equivalently in the form
µ ¶1−16a2
N
(7.2.49)
≥ log n.
n
If N ≥ n(log n)2 and a = 1/6, then (7.2.49) is clearly satisfied. Finally, our
choice of a should be such that a2 ≥ 9/(2 log log n), which is also satisfied when
n ≥ n0 , for a suitable (absolute) n0 ∈ N.
Theorem 7.2.15. Let B be a ball of volume 1 in Rn . If n(log n)2 ≤ N ≤
exp(cn), then
p
log(N/n)
√
(7.2.50)
E1/n (B, N ) ≥ c
n
where c > 0 is an absolute constant.
p
√
Proof. Let f (N, n) = log(N/n)/(6 n) and
A = {(x1 , . . . , xN ) : C(x1 , . . . , xn ) ⊇ f (N, n)B}.
By Lemma 7.2.14 we have Prob(A) ≥ 1 − exp(−n), and hence
Z
E1/n (B, N ) ≥
|C(x1 , . . . , xN )|1/n dx1 . . . dxN
A
≥
(1 − exp(−n))f (N, n)|B|
≥
f (N, n)/2
if n exceeds some (absolute) n0 ∈ N. This proves the Theorem.
7.2 Random polytopes in isotropic convex bodies · 105
Combining Theorem 7.2.15 with Theorem 7.2.6 we have:
Theorem 7.2.16. Let K be a convex body of volume 1 in Rn . If n(log n)2 ≤
N ≤ exp(cn), then
p
log(N/n)
√
(7.2.51)
E1/n (K, N ) ≥ c
n
where c > 0 is an absolute constant.
2
Notes and References
Theorem 7.1.1 was proved by Bourgain with n(log n)3 dependence of m on n
(see [11]). Rudelson [52] gave a different proof with n(log n)2 dependence. The
proof we present follows Bourgain’s ideas and comes from [25] (where several
extensions of this result are obtained). The material of §7.2 comes from [26]
(Theorem 7.2.7 is an observation which appears in [28]).
Chapter 8
The central limit problem
8.1
Concentration property for p-balls
Let 1 ≤ p ≤ ∞ and let rp,n > 0 be a constant such that |rp,n Bpn | = 1. We
write Lp,n for the isotropic constant of Bpn and µp,n for the Lebesgue measure
on rp,n Bpn .
n
As the next Theorem shows, most of the volume
√ of the normalized `p -ball lies
in a very thin spherical shell around the radius nLp,n :
Theorem 8.1.1. For every t > 0,
¯
µ¯ 2
¶
¯
¯ |x|
CL4p,n
2 ¯
¯
(8.1.1)
µp,n ¯
− Lp,n ¯ ≥ t ≤
,
n
nt2
where C > 0 is an absolute constant.
The proof is based on the fact that normalized `np -balls have the following
subindependence property.
Theorem 8.1.2 (Subindependence). Let K := rp,n Bpn and P := µp,n . If
t1 ,. . . ,tn are non-negative numbers, then
à n
!
n
\©
Y
ª
¡
¢
(8.1.2)
P
|xi | ≥ ti
≤
P {|xi | ≥ ti } .
i=1
i=1
Proof. The Theorem will follow by induction if we show that
à n
!
à n
!
\©
\©
ª
ª
(8.1.3)
P
≤ P (|x1 | ≥ t1 ) P
.
|xi | ≥ ti
|xi | ≥ ti
i=1
i=2
Set
(8.1.4)
S=
n
\
©
ª
|xi | ≥ ti .
i=2
Then, we need to prove that
(8.1.5)
|K ∩ {|x1 | ≥ t1 }| |K ∩ S|
|K ∩ S ∩ {|x1 | ≥ t1 }|
≤
·
.
|K|
|K|
|K|
108 · The central limit problem
We will apply the following simple fact: if µ is a positive measure on [0, 1] and
f : [0, 1] → R is increasing, then
Z s
Z 1
(8.1.6)
µ([0, 1])
f dµ ≤ µ([0, s])
f dµ
0
0
for all s ∈ [0, 1]. If
(8.1.7)
|K ∩ S ∩ {|x1 | = 1 − u}|
,
|K ∩ {|x1 | = 1 − u}|
f (u) =
it is not hard to check that f is increasing. Let µ be the probability measure
with density
(8.1.8)
|K ∩ {|x1 | = 1 − u}|
.
|K|
g(u) =
Then,
Z
(8.1.9)
Z
1
1
0
|K ∩ S|
|K ∩ S ∩ {|x1 | = 1 − u}| |K ∩ {|x1 | = 1 − u}|
du =
|K ∩ {|x1 | = 1 − u}|
|K|
|K|
Z
1−t
f (u)dµ =
0
and
(8.1.10)
f (u)dµ =
0
|K ∩ S ∩ {|x1 | ≥ t}|
.
|K|
Applying (8.1.6) for s = 1 − t1 we get the result.
Theorem 8.1.2 immediately implies an anti-correlation inequality for the coordinate functions.
Corollary 8.1.3. Let K := rp,n Bpn . Then,
Z
Z
Z
(8.1.11)
x2i x2j dx ≤
x2i dx ·
x2j dx
K
K
K
for all i 6= j in {1, . . . , n}.
Proof. We write
Z
x2i x2j dx
Z
= 4
K∩{xi ≥0,xj ≥0}
Z ∞Z ∞
K
= 4
x2i x2j dx
4ti tj P (xi ≥ ti , xj ≥ tj )dti dtj
Z
0
∞
0
Z
∞
≤ 4
4ti tj P (xi ≥ ti )P (xj ≥ tj )dti dtj
¶ µZ ∞
¶
∞
≤ 4
2ti P (xi ≥ ti )dti
2tj P (xj ≥ tj )dtj
0
Z 0
Z
= 4
x2i dx
x2j dx
0
µZ
Z
0
K∩{xi ≥0}
K∩{xj ≥0}
Z
=
K
x2i dx
K
x2j dx.
8.1 Concentration property for p-balls · 109
Proof of Theorem 8.1.1: From the Cauchy-Schwarz inequality we have
µZ
2
(8.1.12)
n
L4p,n
¶2
Z
2
=
|x| dx
|x|4 dx.
≤
K
K
On the other hand, using Corollary 8.1.3 we have
Z
4
|x| dx =
Z ÃX
n
K
K
Z
≤
n
K
!2
x2i
i=1
x41 dx +
n
K
K
i=1
XZ
i6=j
Z
=
n Z
X
dx =
K
x4i dx +
i6=j
Z
x2i dx
K
XZ
K
x2i x2j dx
x2j dx
x41 dx + n(n − 1)L4p,n .
Since
µZ
Z
(8.1.13)
K
x41 dx
≤C
K
¶2
x21 dx
= CL4p,n
for some absolute constant C > 0, we get
µ
¶
Z
1
C
4
4
(8.1.14)
Lp,n ≤ 2
|x| dx ≤ 1 +
L4p,n .
n K
n
This implies that
Z µ
(8.1.15)
K
|x|2
− L2p,n
n
¶2
1
dx = 2
n
Z
K
|x|4 dx − L4p,n ≤
C 4
L .
n p,n
Then, Chebyshev’s inequality gives
(8.1.16)
¯
µ¯ 2
¶2
¶ Z µ 2
¯ |x|
¯
C
|x|
t2 µp,n ¯¯
− L2p,n ¯¯ ≥ t ≤
− L2p,n dx ≤ L4p,n
n
n
n
K
for every t > 0, which is exactly the assertion of the Theorem.
Corollary 8.1.4. For every t > 0,
(8.1.17)
¯
µ¯
¶
¯ |x|
¯
CL2p,n
¯
¯
µp,n ¯ √ − Lp,n ¯ ≥ t ≤
.
nt2
n
Proof. Let t > 0. We have
¯
¡¯
√ ¢
√
µp,n ¯ |x| − nLp,n ¯ ≥ t n
¯
¡¯
¢
≤ µp,n ¯ |x|2 − nL2p,n ¯ ≥ tnLp,n
≤
by Theorem 8.1.1.
CL2p,n
CL4p,n
=
t2 nL2p,n
t2 n
2
110 · The central limit problem
8.2
The ε-concentration hypothesis
Let K be an isotropic convex body in Rn . We view K as a probability space
(with the Lebesgue measure µK on K) and for every θ ∈ S n−1 we consider the
random variable Xθ (x) = hx, θi. Since K is isotropic, we have
EXθ = 0 and Var(Xθ ) = L2K
(8.2.1)
for every θ ∈ S n−1 .
It is conjectured that most of these random variables have to be very close to
a Gaussian random variable γ with mean 0 and variance L2K . In this Section
we will see that this is true, at least for isotropic symmetric convex bodies,
under the following general √
hypothesis which states that the Euclidean norm
concentrates near the value nLK as a function on K.
Concentration hypothesis: Let 0 < ε < 21 . We say that K satisfies the
ε-concentration hypothesis if
¯
µ¯
¶
¯ |x|
¯
¯
¯
(8.2.2)
µK ¯ √ − LK ¯ ≥ εLK ≤ ε.
n
Remark 1: The results of §8.1 (see Corollary 8.1.4) show that the class of lpn
1
balls satisfies the ε-concentration hypothesis with ε ' n1/3
.
Before stating the Theorem, we need to introduce some notation. We denote
by g(s) the density of the Gaussian random variable γ with variance L2K and
for simplicity we write gθ (s) for the density of Xθ . Note that
gθ (s) = fK,θ (s) = |K ∩ (θ⊥ + sθ)|
(8.2.3)
and
µ
¶
1
s2
g(s) = √
exp − 2 .
2LK
2πLK
(8.2.4)
Theorem 8.2.1. Let K be an isotropic symmetric convex body in Rn which
satisfies the ε-concentration hypothesis for some 0 < ε < 12 . Then, for every
δ>0
µ½
(8.2.5)
σ
¯Z
¯
θ : ¯¯
Z
t
gθ (s) ds −
−t
¯
¾¶
¯
c1
g(s) ds¯¯ ≤ δ + 4ε + √ for every t ∈ R
n
−t
t
≥ 1 − n e−c2 δ
2
n
,
where c1 , c2 > 0 are absolute constants.
The proof is divided three steps. We first consider the average function
Z
Z t
(8.2.6)
A(t) =
gθ (s) ds σ(dθ)
S n−1
−t
and show that, under the ε-concentration hypothesis,
¯
¯
Z t
¯
¯
c1
¯A(t) −
¯ ≤ 4ε + √
(8.2.7)
g(s)
ds
¯
¯
n
−t
for every t > 0.
8.2 The ε-concentration hypothesis · 111
Lemma 8.2.2. Let K be an isotropic convex body in Rn . For every t > 0,
√
¯
¯
n
Z Z t|x|
¯
¯
2
2
c1
¯
¯
− u2
(8.2.8)
e
du dx¯ ≤ √ ,
¯A(t) − √
¯
¯
n
2π K 0
where c1 > 0 is an absolute constant.
Proof. From Proposition 6.3.1 we have
Z
Z
(8.2.9)
fK (s) =
gθ (s)σ(dθ) = cn
S n−1
{x∈K:|x|≥s}
1
|x|
µ
¶ n−3
2
s2
dx,
1− 2
|x|
¡ ¢ √ ¡
¢
where cn = Γ n2 / πΓ n−1
. Note that
2
Z
1
=
2cn
(8.2.10)
1
(1 − u2 )
n−3
2
du.
0
Now, Fubini’s theorem gives
Z
A(t) =
Z tZ
t
2
fK (s)ds = 2cn
0
0
Z Z
t
min{1, |x|
}
= 2cn
K
{x∈K:|x|≥t}
(1 − u2 )
n−3
2
1
|x|
µ
¶ n−3
2
s2
1− 2
dx ds
|x|
du dx.
0
We will prove that
√
¯
¯
n
t
Z t|x|
Z min{1, |x|
¯
¯
}
u2
2
c1
¯
¯
2 n−3
−
(8.2.11)
(1 − u ) 2 du − √
e 2 du¯ ≤ √
¯2cn
¯
¯
n
2π 0
0
for every t > 0 and x ∈ K, where c1 > 0 is an absolute constant. This will
prove the Lemma since the volume of K is equal to 1.
Case 1: If |x| ≤ t, then by (8.2.10) we have to show that
√
¯
¯
n
Z t|x|
¯
¯
2
2
c1
¯
¯
− u2
(8.2.12)
du¯ ≤ √ .
e
¯1 − √
¯
¯
n
2π 0
But the left hand side is equal to
Z ∞
Z ∞
2
u2
n
2
2
2
2
− u2
√
e
e− 2 du ≤ √
e− 2 ≤ √
.
(8.2.13)
du ≤ √
√
√
t
n
2π |x|
2π
2πn
2πn
n
Case 2: If |x| ≥ t, we write
Z
(8.2.14)
2cn
0
t
|x|
2
(1 − u )
n−3
2
2cn
du = √
n
Z
√
t n
|x|
0
µ
¶ n−3
2
u2
1−
du.
n
Then, the left hand side of (8.2.11) is bounded by
√
n
¯
¯ Z t√n
Z t|x|
¯ 2cn
¯ |x| − u2
2
2c
n
¯
¯
2
e
(8.2.15) ¯ √ − √ ¯
du + √
n
n 0
2π 0
¯µ
¯
¶ n−3
¯
¯
2
u2 ¯
u2
¯
− e− 2 ¯ du.
¯ 1−
¯
¯
n
112 · The central limit problem
√
¡
Using the asymptotic formula Γ(x) = xx−1 e−x 2πx 1 +
+∞, we see that
(8.2.16)
2
2cn
√ =√
n
2π
µ
n
n−1
¶ n2 −1
1
√
e
1
12x
¢
+ O(x−2 ) as x →
µ
µ ¶¶
1
1+O
.
n
It follows that the first term in (8.2.15) is bounded by c/n as n → ∞.
³
´ n−3
2
2
u2
For the second term, consider the function h(u) = e− 2 − 1 − un
√
√
on [0, n]. Note that h(0) = 0 and h( n) = exp(−n/2). If there is a point
³
´ n−5
√
2
2
v2
n
v ∈ [0, n] such that h0 (v) = 0, then 1 − vn
= n−3
e− 2 , and hence,
2
√
2
−3 − v2
h(v) = vn−3
e
Since this last expression is O(1/n) on [0, n], we have
Z
(8.2.17)
√
t n
|x|
0
¯µ
¯
¶ n−3
¯
¯
2
2
u2
c0
¯
− u2 ¯
−e
¯ 1−
¯ du ≤ √ .
¯
¯
n
n
√
On the other hand we have
√ cn ' n, which shows that the second term in
(8.2.15) is bounded by c/ n as n → ∞.
Theorem 8.2.3. Let K be an isotropic convex body in Rn . If K satisfies the
ε-concentration hypothesis, then
¯
¯
Z t
¯
¯
c1
¯A(t) −
¯ ≤ 4ε + √
(8.2.18)
g(s)ds
¯
¯
n
−t
for every t > 0.
Proof. Let t > 0 and set
(8.2.19)
2
Ft (s) = √
2π
Z
t
s
e−
u2
2
du.
0
In this notation, Lemma 8.2.2 states that
¯
µ
¶ ¯
Z
¯
¯
|x|
c1
¯
(8.2.20)
Ft √
dx¯¯ ≤ √ .
¯A(t) −
n
n
K
Note that
2
(8.2.21) Ft (LK ) = √
2π
Z
t
LK
0
e
2
− u2
1
du = √
2πLK
Z
t
−
e
s2
2L2
K
Z
−t
So, the Theorem states that
(8.2.22)
t
ds =
c1
|A(t) − Ft (LK )| ≤ 4ε + √ .
n
Because of (8.2.20), the Theorem will be proved if we check that
¯
¯Z
¶
µ
¯
¯
|x|
¯
¯
(8.2.23)
¯ Ft √n dx − Ft (LK )¯ ≤ 4ε.
K
g(s)ds.
−t
8.2 The ε-concentration hypothesis · 113
We divide K into two subsets:
¯
½¯
¾
¯ |x|
¯
¯
¯
K1 = K ∩ ¯ √ − LK ¯ ≤ εLK
n
(8.2.24)
and
¯
½¯
¾
¯ |x|
¯
K2 = K ∩ ¯¯ √ − LK ¯¯ ≥ εLK .
n
(8.2.25)
Then,
¯ X
¯Z
¶
µ
2 Z
¯
¯
|x|
¯ Ft √
¯
dx
−
F
(L
)
≤
t
K ¯
¯
n
(8.2.26)
K
i=1
Ki
¯ µ
¯
¶
¯
¯
|x|
¯F t √
¯
−
F
(L
)
t
K ¯ dx.
¯
n
³ ´
|x|
To estimate the integral on K2 , we just use the fact that Ft √
and Ft (LK )
n
are bounded by 1. Since K satisfies the ε-concentration hypothesis,
¯
¶
Z ¯ µ
¯
¯
|x|
¯
¯
(8.2.27)
¯Ft √n − Ft (LK )¯ dx ≤ 2|K2 | ≤ 2ε.
K2
For the integral on K1 we shall use a Lipschitz estimate for Ft . Observe that
2 t − t22
1
|F 0 (s)| = √
e 2s ≤ ,
2
s
2π s
√
since x exp(−x2 /2) ≤ 1/ e on [0, +∞). Also, since ε < 1/2, for every x ∈ K1
we have
(8.2.28)
|x|
L
√ > K.
2
n
(8.2.29)
It follows that
Z
(8.2.30)
K1
¯ µ
¯
¶
Z
¯
¯
|x|
¯F t √
− Ft (LK )¯¯ dx ≤
¯
n
K1
¯
¯
Z
¯
2 ¯¯ |x|
√ − LK ¯¯ dx ≤
2ε dx ≤ 2ε.
¯
LK
n
K1
Combining the above we get (8.2.23), and taking into account (8.2.20) we conclude the proof.
In the second step we use the estimate of Theorem 8.2.3 for the average A(t) to
obtain
θ ∈ S n−1 . The idea is to show
R t a similar estimate for “most directions”
n−1
that −t gθ (s)ds is the (restriction on S
of the) radial function of a symmetric
convex body in Rn and then use the spherical isoperimetric inequality in the
context of Lipschitz continuous functions on the sphere.
Here, we make use of Busemann’s inequality (see §4.3) which states the following:
Lemma 8.2.4. Let W be a symmetric convex body in Rm and define r(u) =
|W ∩ u⊥ | for all u ∈ S m−1 . Then, r is the (the restriction of the) radial function
of a symmetric convex body in Rm .
2
114 · The central limit problem
Proposition 8.2.5. Let K be a symmetric convex body in Rn . Fix t > 0 and
define
(8.2.31)
|x|
kxkt = R t
x (s)ds
g |x|
−t
.
Then, k · kt is a norm on Rn .
¯
¯
Proof. Recall that gθ (s) = ¯K ∩ (θ⊥ + sθ)¯ for every θ ∈ S n−1 . We define
Z
(8.2.32)
t
v(x, t) =
−t
x (s)ds
g |x|
and prove that for all x, y ∈ Rn ,
1
2
(8.2.33)
µ
|x|
|y|
+
v(x, t) v(y, t)
¶
¯ x+y ¯
¯
¯
2
¢.
≥ ¡ x+y
v 2 ,t
We may clearly assume that x and y are linearly independent. Consider the
convex body K 0 = K × [−1, 1] in Rn+1 . Lemma 8.2.4 shows that |K 0|θ|
∩θ ⊥ |
defines a norm on Rn+1 /{0}. That is,
¯
¯
¯ θ+φ ¯
µ
¶
¯ 2 ¯
1
|θ|
|φ|
(8.2.34)
+
≥¯
¡
¢
2 |K 0 ∩ θ⊥ | |K 0 ∩ φ⊥ |
¯K 0 ∩ θ+φ ⊥ |
2
for all linearly independent θ, φ ∈ Rn+1 .
√
Let r ∈ (0, 1) be defined by the equation tr = 1 − r2 . We observe that if
z ∈ Rn \ {0} and
µ
¶
z p
(8.2.35)
u(z) = r , 1 − r2 ,
|z|
then the projection of K 0 ∩ u(z)⊥ onto the first n coordinates is {w ∈ K :
|hw, zi| ≤ t|z|}. It follows that
p
(8.2.36)
v(z, t) = 1 − r2 |K 0 ∩ u(z)⊥ |.
We define η(z) = |z| u(z). Then, |K 0 ∩ u(z)⊥ | = |K 0 ∩ η(z)⊥ | and |η(z)| = |z|.
If we set θ = η(x) and φ = η(y), Lemma 8.2.4 shows that
¯ θ+φ ¯
µ
¶
¯
¯
|x|
|y|
1
1
2
+
≥√
(8.2.37)
¯
¡ θ+φ ¢⊥ ¯ .
2
2 v(x, t) v(y, t)
¯
1 − r ¯K 0 ∩
2
Observe that
θ+φ
2
=
=
µ
¶
x+y p
(|x| + |y|)
2
r
, 1−r
2
2
µ
¶
p
|x + y|
x+y
|x| + |y|
2
r
, 1−r
.
2
|x + y|
|x + y|
8.2 The ε-concentration hypothesis · 115
Then, the projection of K 0 ∩
perpendicular to
x+y
2 ,
³
θ+φ
2
´⊥
onto the first n coordinates is a strip
with width
(8.2.38)
s=
|x| + |y|
t,
|x + y|
and this gives
(8.2.39)
v
¡ x+y
¯
¯K 0 ∩
¢
2 ,s
¡ θ+φ ¢⊥ ¢¯
¯
2
√
=
1 − r2 |x| + |y|
¯ θ+φ ¯ .
¯
¯
2
2
Then, (8.2.37) takes the form
µ
¶
|x|
|y|
1 |x| + |y|
1
¡
¢.
+
≥
(8.2.40)
2 v(x, t) v(y, t)
2 v x+y
2 ,s
Observe that if a ≥ 1, then for every z ∈ Rn we have
(8.2.41)
v(z, at) ≤ a v(z, t)
(this follows from the fact that gθ (as) ≤ gθ (s) for all θ ∈ S n−1 and s > 0). So,
¯ x+y ¯
¯
¯
|x| + |y|
|x| + |y|
|x| + |y|
2
¡
¢
¡
¢.
(8.2.42)
= ¡ x+y |x|+|y| ¢ ≥ |x|+|y| ¡ x+y ¢ = x+y
x+y
2v 2 , s
v 2 ,t
2v 2 , |x+y| t
2 |x+y| v 2 , t
Going back to (8.2.40) we conclude the proof.
Lemma 8.2.6. Let K be a symmetric convex body in Rn with volume 1 . For
every θ ∈ S n−1 and every t > 0,
Z ∞
1
(8.2.43)
gθ (s)ds ≤ e−2gθ (0)t .
2
t
Proof. Consider the function
Z ∞
Z
(8.2.44)
H(t) =
gθ (s)ds =
t
∞
0
χ[t,∞) (s)gθ (s)ds.
Using the fact that gθ is log-concave and applying the Prékopa-Leindler inequality, we may easily check that H is log-concave. It follows that
(8.2.45)
(log H)(t) − (log H)(0) ≤ (log H)0 (0)t
for every t > 0. Observe that H(0) = 1/2 by the symmetry of K, and
(8.2.46)
(log H)0 (0) = −
gθ (0)
= −2gθ (0).
H(0)
It follows that
(8.2.47)
¡
¢ 1
H(t) ≤ H(0) exp (log H)0 (0)t = exp(−2gθ (0)t),
2
as stated in the Lemma.
116 · The central limit problem
Lemma 8.2.7. Let K be an isotropic symmetric convex body in Rn . For every
t > 0, the norm
(8.2.48)
kxkt = R t
−t
|x|
x (s) ds
g |x|
satisfies
(8.2.49)
a|x| ≤ kxkt ≤ b|x|
for every x ∈ Rn , where a, b are two positive constants such that a ≥ 1 and
b/a ≤ c for some absolute constant c > 0.
n−1
Proof. Since K is isotropic, we know that gθ (0) ' L−1
.
K for every θ ∈ S
Then, by the symmetry of K we have
½
¾
Z t
c1 t
(8.2.50)
gθ (s) ds ≤ min {2t gθ (0), 1} ≤ min
,1 .
LK
−t
Also, Lemma 8.2.6 shows that
Z t
Z
(8.2.51)
gθ (s) ds = 1 − 2
−t
∞
c t
− L2
gθ (s) ds ≥ 1 − e−2gθ (0)t ≥ 1 − e
K
.
t
We easily check that
(8.2.52)
1−e
c t
− L5
K
≥
c2 t
2LK
if c2 t ≤ LK . In any case,
½
¾
Z t
c3 t
(8.2.53)
gθ (s) ds ≥ min
, 1 − e−1 .
LK
−t
In other words
(8.2.54)
½
a := max
¾
LK
,1
c4 t
½
≤ kθkt ≤ b := max
LK
e
,
c3 t e − 1
¾
for every θ ∈ S n−1 . Note that a ≥ 1 and b/a is bounded independently of t and
LK .
Theorem 8.2.8. Let K be an isotropic symmetric convex body in Rn . If K
satisfies the ε-concentration hypothesis, then for every t > 0 and δ > 0,
¯
¾¶
µ½ ¯Z t
Z t
¯
¯
2
c3
¯
¯
≤ 2e−c4 δ n ,
(8.2.55) σ
θ:¯
gθ (s)ds −
g(s)ds¯ ≥ δ + 4ε + √
n
−t
−t
where c3 , c4 > 0 are absolute constants.
Proof. Let t > 0 and δ > 0 be fixed. We will use the spherical isoperimetric
inequality through the following fact: If f : S n−1 → R is d-Lipschitz and M(f)
is its mean, then
¾¶
µ½
δ2 n
c2
≤ 2e− 2d2 .
(8.2.56)
σ
θ : |f − M (f )| ≥ δ + √
n
8.2 The ε-concentration hypothesis · 117
Rt
We shall apply this to the function f (θ) = −t gθ(s) ds, Observe that
¯Z t
¯
¯
¯
Z t
¯
¯
¯ 1
1 ¯¯
¯
¯
¯
g
(s)
ds
−
g
(s)
ds
−
=
θ
φ
¯
¯
¯ ||θ|| ||φ|| ¯
−t
−t
||θ − φ||
≤
||θ|| ||φ||
b
≤
|θ − φ|
a2
≤ c|θ − φ|,
where c is the absolute constant in Lemma 8.2.7. Also, note that M (f ) = A(t).
It follows that
¯
µ½ ¯Z t
¾¶
µ 2 ¶
¯
¯
c2
δ n
¯
¯
(8.2.57)
σ
θ:¯
gθ (s) ds − A(t)¯ ≥ δ + √
≤ 2 exp − 2 .
2c
n
−t
Combining this with Theorem 8.2.3, we get
µ½
(8.2.58)
σ
¯Z
¯
θ : ¯¯
Z
t
gθ (s)ds −
−t
¯
¾¶
¯
¡
¢
c3
g(s)ds¯¯ ≥ δ + 4ε + √
≤ 2 exp −c4 δ 2 n ,
n
−t
t
where c3 = c1 + c2 (c1 is the constant in Theorem 8.2.3) and c4 = 1/(2c2 ).
Proof of Theorem 8.2.1: First, fix some θ ∈ S n−1 . Since
(8.2.59)
c1
LK
gθ (s) ≤ gθ (0) ≤
and
(8.2.60)
g(s) = √
1
1
exp(−s2 /(2L2K )) ≤ √
2πLK
2πLK
for every s > 0, the function
(8.2.61)
¯Z
¯
H(t) = ¯¯
Z
t
t
gθ (s)ds −
−t
is Lipschitz continuous with constant d ≤
−t
¯
¯
g(s)ds¯¯
c2
LK ,
where c2 is an absolute constant.
√
Also, there is an absolute constant c3 > 0 such that H(t) ≤ 1/ n for every
t ≥ c3 LK log n. This is a consequence of the equality
¯Z ∞
¯
Z ∞
¯
¯
(8.2.62)
H(t) = 2 ¯¯
gθ (s)ds −
g(s)ds¯¯
t
t
and of Lemma 8.2.6: if c3 > 0 is chosen large enough, when t ≥ c3 LK log n we
have
½Z ∞
¾
Z ∞
1
(8.2.63)
max
gθ (s)ds,
g(s)ds < √ .
2
n
t
t
√
√
Define tk = kα, where α = LK / n and k = 1, . . . , k0 = [c3 n log n] + 1. From
Theorem 8.2.8, for every δ > 0 we have
√
2
(8.2.64)
σ (A) ≤ 2c3 n(log n)e−c6 δ n ,
118 · The central limit problem
where
¯Z
¯
θ : ∃k ≤ k0 s.t. ¯¯
¯
¾
¯
c5
¯
gθ (s)ds −
g(s)ds¯ ≥ δ + 4ε + √
n
−tk
−tk
½
(8.2.65) A =
Z
tk
tk
and c5 , c6 > 0 are absolute constants. If θ is not in A, then
(8.2.66)
c5
H(tk ) ≤ δ + 4ε + √
n
2
for all k = 1, . . . , k0 . Since H is LcK
-Lipschitz, we get a similar estimate for
√
H(t), t ∈ [0, c3 LK log n]. Finally, if t > c3 LK log n, we know that H(t) < 1/ n.
2
8.3
The variance hypothesis
Let K be an isotropic convex body in Rn . In this Section we study the parameter
σK of K which is defined by
2
σK
=
(8.3.1)
Var(|x|2 )
nL4K
and its connections with the problems we discussed in this Chapter as well as
in Chapter 6.
It is useful to write σK in the form
nVar(|x|2 )
2
σK
= ¡
¢2 .
E|x|2
(8.3.2)
In this way the quantity becomes invariant under homotheties, and hence, easier
to compute.
A simple computation shows that if K = B2n then
(8.3.3)
E|x|4 =
n
n
and E|x|2 =
.
n+4
n+2
Therefore,
Ã
(8.3.4)
2
σB
n = n
2
!
E|x|4
¡
¢2 − 1
E|x|2
=
4
.
n+4
Actually, Euclidean balls have minimal σK as the next Theorem shows.
Theorem 8.3.1. Let K be an isotropic convex body in Rn . Then,
(8.3.5)
σK ≥ σB2n .
Proof. Let x be uniformly distributed in K. The distribution function F (r) =
|{x ∈ K : |x| ≤ r}| has density
(8.3.6)
F 0 (r) = nωn rn−1 σ
¡1 ¢
K
r
8.3 The variance hypothesis · 119
¡
¢
for r > 0. We define q(r) = nωn σ 1r K . Observe that q is increasing and can
be assumed absolutely continuous. Therefore, we can write q in the form
Z ∞
p(s)
(8.3.7)
q(r) = n
ds,
sn
r
where p : (0, +∞) → R is a non-negative measurable function. Then, Fubini’s
theorem shows that
µZ s
¶
Z ∞
Z ∞
Z ∞
p(s)
n−1
p(s)ds = n
r
rn−1 q(r)dr = 1,
(8.3.8)
dr ds =
sn
0
0
0
0
which means that p is the density of some positive random variable ξ.
Also, for every α > −n,
Z ∞
Z ∞
n
n
(8.3.9)
E|x|α =
sα p(s)ds =
Eξ α .
rα+n−1 q(r)dr =
n+α 0
n+α
0
We can now compute
Var(|x|2 ) =
=
≥
µ
¶2
n
n
Eξ 4 −
Eξ 2
n+4
n+2
¡ 2 ¢2
4n
n
Eξ
+
Var(ξ 2 )
(n + 4)(n + 2)2
n+4
¡ 2 ¢2
4n
Eξ .
2
(n + 4)(n + 2)
It follows that
(8.3.10)
2
σK
¡ 2 ¢2
4n
Var(|x|2 )
4
(n+4)(n+2)2 Eξ
=n¡
=
,
¢2 ≥ n
³
´2
2
n
+
4
n
E|x|
Eξ 2
n+2
and the Theorem follows from (8.3.4).
Remark 1: Simple computations show that
(8.3.11)
2
σB
n = 1 −
1
2(n + 1)
→ 1 as n → ∞
(n + 3)(n + 4)
and
(8.3.12)
n =
σB∞
4
for every n.
5
In the next Section we briefly discuss various consequences of the following
hypothesis.
Variance hypothesis: There exists an absolute constant C > 0 such that
2
σK
≤ C for every isotropic convex body.
2
Let us first note that σK
is uniformly bounded for all `np . This follows by the
subindependence theorem of §8.1. Actually, the argument is inside the proof of
Theorem 8.1.1.
120 · The central limit problem
2
Proposition 8.3.2. There exists an absolute constant C > 0 such that σB
n ≤ C
p
for every n and every p ∈ [1, ∞].
Proof. In the proof of Theorem 8.1.1 we saw that
Z
2 4
(8.3.13)
n Lp,n ≤
|x|4 dx ≤ (n2 + Cn)L4p,n
K
for some absolute constant C > 0. Then,
µ
¶
E|x|4
2
(8.3.14)
σB
=
n
−
1
≤C
n
p
n2 L4p,n
for all p and n.
8.4
Consequences of the variance hypothesis
8.4.1 The ε-concentration hypothesis
2
Theorem 8.4.1. Let K be an isotropic convex body in Rn with 8σK
< n. Then,
³ 2 ´1/3
σK
K satisfies the ε-concentration hypothesis with ε = n
.
Proof. For every ε > 0 we have
¯
¯
¢
¡
√
√
µK x ∈ K : ¯ |x| − nLK ¯ ≥ ε nLK
≤
¯
¯
¡
¢
µK x ∈ K : ¯ |x|2 − nL2K ¯ ≥ εnL2K
≤
Var(|x|2 )
.
ε2 n2 L4K
2
By the definition of σK
,
2
2
Var(|x|2 )
σK
nL4K
σK
=
≤
.
ε2 n2 L4K
ε2 n2 L4K
ε2 n
(8.4.1)
³
If we define ε =
2
σK
n
´1/3
, then
2
σK
ε2 n
= ε and the result follows.
Corollary 8.4.2. Assume that there is an absolute constant C > 0 such that
2
σK
≤ C for every isotropic convex body K. Then, every isotropic convex body
¡ ¢1/3
satisfies the ε-concentration hypothesis with ε = C
.
2
n
Then, Theorem 8.1.1 has the following consequence.
Corollary 8.4.3. Assume that there is an absolute constant C > 0 such that
2
σK
≤ C for every isotropic convex body K. Then, for every isotropic symmetric
convex body K in Rn we have
µ½
(8.4.2) σ
¯Z
¯
θ : ¯¯
Z
t
gθ (s) ds −
−t
√
¯
¾¶
√
¯
3
c1 3 C
for
all
t
∈
R
≥ 1 − e−c2 n ,
g(s) ds¯¯ ≤ √
3
n
−t
t
where c1 , c2 > 0 are absolute constants.
2
8.4 Consequences of the variance hypothesis · 121
8.4.2 Tail estimates for the Euclidean norm
Let K be an isotropic convex body in Rn . The proof of Theorem 8.4.1 shows
that
(8.4.3)
√
σ2
µK (x ∈ K : |x| ≥ (1 + s) nLK ) ≤ 2K .
s n
Therefore,
(8.4.4)
√
1
µK (x ∈ K : |x| ≥ (1 + 2σK ) nLK ) ≤
.
4n
Applying Borell’s lemma we get:
Theorem 8.4.4. Let K be an isotropic convex body in Rn . Then,
√
t
(8.4.5)
|{x ∈ K : |x| ≥ (1 + 2σK ) nLK t}| ≤ n− 2
for every t ≥ 1.
2
Corollary 8.4.5. Assume that there is an absolute constant C > 0 such that
2
σK
≤ C for every isotropic convex body K. Then, the Question of Chapter 6
has an affirmative answer with φ(n) ' log n.
2
8.4.3 Average sectional decay
Bobkov and Koldobsky (see [6]) proved the following.
Theorem 8.4.6. Let K be an isotropic convex body in Rn . Then,
¯
µ
¶
2 ¯
¯
− t2 ¯
1
σK LK
1
2L
¯
¯
K
(8.4.6)
¯ ≤ c1 t2 √n + n
¯fK (t) − √2πL e
K
√
for all 0 < t ≤ c2 n, where c1 , c2 > 0 are absolute constants.
2
2
If we assume that there is an absolute constant C > 0 such that σK
≤ C for
every isotropic convex body K, this result shows that: for every isotropic convex
body, fK is very close to the Gaussian density with mean zero and variance L2K .
8.4.4 Further reductions
Let K be an isotropic convex body in Rn . For every x ∈ K we denote by χ(x)
the length of the longest line segment which is contained in K and has x as its
midpoint. Let
Z
(8.4.7)
χ(K) =
χ(x)dx.
K
Using the localization lemma, Kannan, Lovasz and Simonovits (see [33]) proved
the following.
Theorem 8.4.7. If K is an isotropic convex body, then σK LK ≤ 8χ(K).
One should also mention the following result of Voigt [58]:
2
122 · The central limit problem
Theorem 8.4.8. Let ε > 0. Consider the quantity
©
ª
(8.4.8)
A(ε, n) := sup µK (x ∈ K : | |x|2 − nL2K | ≥ εnL2K ) ,
where the supremum is over all isotropic convex bodies in Rn . If
(8.4.9)
lim A(ε, n) = 0
n→∞
for every ε > 0, then
(Z
(8.4.10)
lim sup
n→∞
µZ
2
|x| dx
K
K
1
dx
|x|
¶2 )
= 1,
where the supremum is again over all isotropic convex bodies in Rn .
2
Proposition 6.3.1 shows that
Z
(8.4.11)
fK (0) = cn
K
1
dx
|x|
and, in the proof of Lemma 8.2.2, we saw that
√
cn 2π
(8.4.12)
lim √
= 1.
n→∞
n
Therefore, (8.4.10) is equivalent to
½
¾
fK (0)
n
√
(8.4.13)
lim sup
: K isotropic in R
= 1.
n→∞
2πLK
This has been verified only in special cases by Koldobsky and Lifshits (see [32]).
Notes and References
The results of Sections §8.1 and §8.2 are due to Anttila, Ball and Perissinaki (see
[2] and [4]). The variance hypothesis is discussed in [6]. It appears implicitly in
various articles (see the discussion in §8.4).
Appendix A
Entropy estimates
A.1
Subgaussian processes
Let (T, d) be a metric space, let (Ω, A, P ) be a probability space, and let Y = (Yt )t∈T
be a family of real valued random variables with indices from T . We say that the
process Y = (Yt )t∈T is subgaussian with respect to d if
(A.1)
EYt = 0
for all t ∈ T and, for all t, s ∈ T and every u > 0,
µ
Prob(|Yt − Ys | ≥ u) ≤ 2 exp −
(A.2)
u2
2
d (t, s)
¶
.
Typical example: Let E2n = {−1, 1}n be equipped with the uniform probability
measure. Write ε = (ε1 , . . . , εn ) for the points of E2n . The Rademacher functions are
the random variables ri : E2n → {−1, 1} (1 ≤ i ≤ n) defined by ri (ε) = εi .
For every t = (t1 , . . . , tn ) ∈ T ⊆ Rn we define
(A.3)
Yt (ε) = ht, εi = t1 r1 (ε) + · · · + tn rn (ε).
The inequality
(A.4)
¡
¢
Prob(ε ∈ E2n : |t1 r1 (ε) + · · · + tn rn (ε)| ≥ u) ≤ 2 exp − u2 /2(t21 + · · · + t2n )
which is - for example - a consequence of the classical Khintchine inequality, shows
that Y = (Yt )t∈Rn is subgaussian with respect to the Euclidean metric.
We write g for a standard Gaussian random variable and G = (g1 , . . . , gn ) for the
standard Gaussian random vector in Rn . The distribution of G is the Gaussian measure
γn , with density (2π)−n/2 exp(−|x|2 /2).
Let T be a non empty set. A family Z = (Zt )t∈T of real valued random variables on
(Ω, A, P ) is called a Gaussian process if every linear combination a1 Zt1 + · · · + am Ztm
of Zt ’s is a Gaussian random variable with mean 0. We may view Z as a subset of
L2 (Ω), and then it induces on T the metric
(A.5)
d(t, s) = kZt − Zs kL2 (Ω) .
By the definition of a Gaussian process, for every t, s ∈ T , Zt − Zs is a Gaussian
random variable with mean 0 and variance E(Zt − Zs )2 = d2 (t, s). Consequently, for
every u > 0 we have
(A.6)
¶
µ
¶
µ
Z ∞
u2
r2
2
√
dr ≤ 2 exp − 2
,
exp − 2
Prob(|Zt − Zs | ≥ u) =
2d (t, s)
d (t, s)
d(t, s) 2π u
124 · Entropy estimates
which implies that Z is subgaussian with respect to the metric d it induces to T .
Typical examples: (i) If g1 , . . . , gN are independent standard Gaussian random
variables on (Ω, A, P ), then Z = {g1 , . . . , gN } is a Gaussian process.
(ii) Consider n independent standard Gaussian random variables g1 , . . . , gn . For every
non empty T ⊆ Rn we define a process Z = (Zt )t∈T , by
(A.7)
n
n
­ X
® X
Zt (ω) = ht, G(ω)i = t,
gi ei =
ht, ei igi ,
i=1
i=1
n
where {e1 , . . . , en } is an orthonormal basis of R and G = (g1 , . . . , gn ). Then, Z is
a Gaussian process and the induced metric is the Euclidean metric on Rn : for all
t, s ∈ T ,
(A.8)
d(t, s) = kZt − Zs kL2 (Ω) = |t − s|.
If Y = (Yt )t∈T is a subgaussian process, we define
©
ª
(A.9)
E sup Yt = sup E max Yt : F ⊆ T, 0 < |F | < ∞ .
t∈T
t∈F
The results of the next three Sections concern the relation of E sup Yt with the geometry
of (T, d).
A.2
Metric entropy - the case of Gaussian processes
Let (T, d) be a metric space. For every ε > 0 we define
(A.10)
½
¾
N
[
N (T, d, ε) = min N : t1 , . . . , tN ∈ T : T ⊆
B(ti , ε) ,
i=1
where B(t, ε) = {s ∈ T : d(t, s) < ε}. The function ε 7→ H(T, d, ε) = log N (T, d, ε) is
the metric entropy function of T .
Consider as an
, gN }. We easily check that
√ example the Gaussian process Z = {g1 , . . .√
√
kgi − gj k2 = 2 if i 6= j, and hence, N (ε) = N if 0 < ε ≤ 2 and N (ε) = 1 if ε > 2.
Also, using the fact that gi are independent we may check that
p
(A.11)
E max gi ' log N .
1≤i≤N
Let Z = (Zt )t∈T be a Gaussian process. We view T as a metric space with the induced
metric d. The next Theorem gives upper and lower bounds for E sup Zt in terms of
the metric entropy function of (T, d).
Theorem A.1 There exist constants c1 , c2 > 0 with the following property: if Z =
(Zt )t∈T is a Gaussian process and d is the induced metric, then
Z ∞p
p
(A.12)
c1 sup ε log N (T, d, ε) ≤ E sup Zt ≤ c2
log N (T, d, ε)dε.
ε>0
t∈T
0
The left hand side inequality is Sudakov’s inequality [54], while the right hand side
inequality is Dudley’s inequality. In the example of Z = {g1 , . . . , gN }, both bounds
give the right order of E sup gi . The proof of Sudakov’s inequality is based on a classical
comparison lemma of Slepian.
Theorem A.2 If (X1 , . . . , XN ) and (Y1 , . . . , YN ) are two N -tuples of Gaussian random
variables with mean 0 which satisfy the condition
(A.13)
kXi − Xj k2 ≤ kYi − Yj k2
A.3 Dudley’s bound for subgaussian processes · 125
for all i 6= j, then
(A.14)
E max Xi ≤ E max Yi .
i≤N
i≤N
We now use Slepian’s lemma as follows: Let Z = (Zt )t∈T be a Gaussian process and
let d be the induced metric. Given ε > 0 we consider a subset {t1 , . . . , tNS
} of T which
is maximal with respect to the condition “d(t, s) ≥ ε if t 6= s”. Then T ⊆ N
i=1 B(ti , ε),
which implies N (T, d, ε) ≤ N .
¡ 1
¢
δg
√ , . . . , √N , where gi are indeIf δ = min kZti − Ztj k2 , we consider the N -tuple δg
2
2
pendent standard Gaussian random variables. If i 6= j then
°
°
° δgi
j°
° √ − δg
° = δ ≤ kZt − Zt k2 ,
√
(A.15)
i
j
° 2
2 °2
so we can apply Slepian’s lemma. It follows that
(A.16)
Thus, E supt∈T
A.3
p
δ
E sup Zt ≥ E max Zti ≥ √ E max gi ≥ c1 ε log N .
i≤N
i≤N
2
t∈T
p
Zt ≥ c1 supε>0 ε log N (T, d, ε).
Dudley’s bound for subgaussian processes
Dudley’s inequality is more generally valid for subgaussian processes Y = (Yt )t∈T .
The proof uses a successive approximation argument which we briefly describe:
We consider a non empty finite subset F of T and fix t0 ∈ F . We set R = max{d(t, t0 ) :
t ∈ F } and rk = R/2k for all k ≥ 0.
We define A0 = {t0 } andSfor every k ≥ 1 we find Ak ⊆ F with cardinality |Ak | =
N (F, d, rk ) such that F ⊆ t∈Ak B(t, rk ). Finally, for every t ∈ F and k ≥ 0 we choose
πk (t) ∈ Ak with the property d(t, πk (t)) ≤ rk . Since F is finite, for every t ∈ F we
eventually have πk (t) = t. Note also that
(A.17)
d(πk (t), πk−1 (t)) ≤ rk + rk−1 = 3rk .
For every t ∈ F we write
(A.18)
Yt − Yt0 =
∞
X
¡
¢
Yπk (t) − Yπk−1 (t)
k=1
and, using EYt0 = 0,
Z
(A.19)
E max Yt = E max(Yt − Yt0 ) =
t∈F
t∈F
∞
µ
¶
Prob max(Yt − Yt0 ) ≥ u du.
t∈F
0
P
We fix αk > 0 (which will be suitably chosen) with S :=
αk < ∞ and set Bk =
{(w, z) ∈ Ak × Ak−1 : d(w, z) ≤ 3rk }. Using the subgaussian assumption we write
̰
!
µ
¶
∞
X
X
Prob max(Yt − Yt0 ) ≥ uS
≤ Prob
max(Yπk (t) − Yπk−1 (t) ) ≥
uαk
t∈F
k=1
≤
≤
∞
X
k=1
∞
X
k=1
t∈F
µ
Prob
k=1
¶
max (Yw − Yz ) ≥ uαk
(w,z)∈Bk
exp(−u2 αk2 /9rk2 )|Ak | · |Ak−1 |.
126 · Entropy estimates
We now choose αk = 3rk
p
log(2k |Ak |2 ). For every u ≥ 1 and every k we have
³
´−u2
2
exp(−u2 αk2 /9rk2 )|Ak | · |Ak−1 | ≤ |Ak |2 2k |Ak |2
≤ 2−u k ,
(A.20)
and hence.
µ
¶ X
∞
2
2
Prob max(Yt − Yt0 ) ≥ uS ≤
2−u k ≤ c2−u .
(A.21)
t∈F
This shows that
(A.22)
Z
E max Yt = S
t∈F
∞
0
k=1
µ
¶
Z
Prob max(Yt − Yt0 ) ≥ uS du ≤ S + S
t∈F
∞
2
c2−u du ≤ c0 S.
1
Going back to the definition of S we see that
S
=
'
∞
∞
´ X
X
p
3R ³ p
Rp
k) '
k
log
2
+
log
N
(F,
d,
R/2
log N (F, d, R/2k )
2k
2k
k=1
k=1
Z ∞p
log N (F, d, ε)dε.
0
Since N (F, d, ε) ≤ N (T, d, ε), the proof is complete.
A.4
Majorizing measures
Dudley’s bound is not always sharp, as one can see by the following example: Consider
an infinite sequence {gn } of independent standard Gaussian random variables, fix
a = (an ) ∈ `2 and define the ellipsoid
½
¾
∞
X
(A.23)
E = t = (tn ) ∈ `2 :
t2n /a2n ≤ 1 .
n=1
If we set Zt =
P
n tn gn ,
then Z = (Zt )t∈E is a Gaussian process and
Ã
(A.24)
E sup Zt '
t∈E
∞
X
!1/2
a2n
< ∞.
n=1
On the other hand, one can choose a ∈ `2 so that “Dudley’s integral” will diverge.
A second approach to the problem is through majorizing measures: If (T, d) is a
metric space, for every Borel probability measure µ on T we consider the quantity
¶
µ
Z ∞s
1
dt.
(A.25)
γ(T, µ) = sup
log
µ(B(t, ε))
t∈T 0
In a sense, 1/µ(B(t, ε)) replaces the entropy number N (T, d, ε) and the integral (which
depends on t ∈ T ) is the analogue of Dudley’s integral.
Theorem A.3 There exists a constant C > 0 with the following property. If (T, d) is
a metric space and Y = (Yt )t∈T is a subgaussian (with respect to d) process, then
(A.26)
E sup Yt ≤ C · γ(T ),
t∈T
where
(A.27)
γ(T ) = inf γ(T, µ).
µ
A.5 Translation to the language of convex bodies · 127
In the case of Gaussian processes, Talagrand [57] proved that the bound of Theorem
A.3 gives always the right order for E supt∈T Zt .
Theorem A.4 There exists a constant C > 0 with the following property. If Z =
(Zt )t∈T is a Gaussian process and d is the induced metric, then
1
· γ(T ) ≤ E sup Zt .
C
t∈T
(A.28)
A direct consequence of the above is the following comparison theorem.
Theorem A.5 Let Z = (Zt )t∈T be a Gaussian process and let d be the induced metric.
If the process Y = (Yt )t∈T is subgaussian with respect to d, then
(A.29)
E sup Yt ≤ C · E sup Zt ,
t∈T
t∈T
where C > 0 is an absolute constant.
A.5
Translation to the language of convex bodies
Let K be a convex body in Rn . We assume that 0 ∈ int(K) and denote by hK the
support function of K which is defined by
(A.30)
hK (y) = maxht, yi.
t∈K
The width of K in the direction of θ ∈ S n−1 is the quantity w(K, θ) = hK (θ)+hK (−θ),
and the mean width of K is defined by
Z
Z
1
(A.31)
w(K) =
w(K, θ)σ(dθ) =
hK (θ)σ(dθ).
2 S n−1
S n−1
We can associate to K the Gaussian process Z = (Zt )t∈K , where Zt (ω) = ht, G(ω)i
(G = (g1 , . . . , gn ) is the standard Gaussian random vector). We have already seen
that the induced metric is the Euclidean metric on K: kZt − Zs k2 = |t − s| for all
t, s ∈ K.
A basic observation, which connects the preceding theory with convex geometric analysis, is that
E sup Zt
=
t∈K
=
'
E supht, Gi = E hK (G)
t∈K
Z
2
1
hK (x)e−|x| /2 dx
(2π)n/2 Rn
Z
√
√
hK (θ)σ(dθ) ' nw(K).
n
S n−1
We just used the fact that the distribution of G is the standard Gaussian measure on
Rn and polar integration.
In this context, Sudakov’s inequality takes the following form: if K is a convex body
in Rn , then for every ε > 0 we have
µ
¶
w2 (K)
(A.32)
N (K, εB2n ) ≤ exp cn
,
ε2
where N (K, εB2n ) is the least number of balls of radius ε whose union covers K.
Dudley’s inequality takes an analogous form.
128 · Entropy estimates
Mean width plays an important role in convex geometric analysis. From Urysohn’s
inequality, for every convex body K in Rn we have the inequality
µ
(A.33)
w(K) ≥
|K|
|B2n |
¶1/n
.
(among all bodies of the same volume, Euclidean ball has minimal mean width).
Fundamental results of Lewis [34], Figiel and Tomczak-Jaegermann [23], Pisier [50]
establish the following “reverse Urysohn inequality”.
Theorem A.6 There exists an absolute constant c > 0 with the following property.
For every convex body K in Rn we can find an affine image T (K) such that |T (K)| = 1
and
√
(A.34)
w(T (K)) ≤ c n log n.
Notes and References
Complete proofs of the above can be found in the book of Ledoux and Talagrand [35].
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