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Spherical thin-shell concentration for convex
measures
Matthieu Fradelizi, Olivier Gu´edon and Alain Pajor
Abstract
We prove that for s < 0, s-concave measures on Rn satisfy a spherical thin shell concentration similar to the log-concave one. It leads to a
Berry-Esseen type estimate for most of their one dimensional marginal
distributions. We also establish sharp reverse H¨older inequalities for
s-concave measures.
1
Introduction
Let s ∈ [−∞, 1]. A measure µ is called s-concave on Rn whenever
µ ((1 − λ)A + λB) ≥ ((1 − λ)µ(A)s + λµ(B)s )1/s
for every compact sets A, B ⊂ Rn such that µ(A)µ(B) > 0. When s = 0,
this inequality should be read as
µ ((1 − λ)A + λB) ≥ µ(A)1−λ µ(B)λ
and it defines µ as a log-concave measure. When s = −∞, the measure is
said to be convex and the inequality is replaced by
µ ((1 − λ)A + λB) ≥ min (µ(A), µ(B)) .
The purpose of this paper is to prove a spherical thin shell concentration
for s-concave measures in the case s ≤ 0, which we consider from now on.
By measure, we always mean probability measure. Notice that the class of
s-concave measures on Rn is decreasing in s. In particular a log-concave
1
measure is s-concave for any s < 0 and any s-concave measure with s ≤ 0 is
(−∞)-concave, thus a convex measure.
The class of s-concave measures was introduced and studied in [9, 10],
where the following complete characterization was established. An s-concave
measure is supported on some convex subset of an affine subspace where it
has a density. If the support of s-concave measure µ generates the whole
space, then µ is absolutely continuous with respect to the Lebesgue measure
and has a density h which is log-concave when s = 0 and when s < 0, is of
the form
1
h = f −α with α = n − ≥ n,
s
n
where f : R → (0, ∞] is a convex function.
A random vector with an s-concave distribution is called s-concave. The
linear image of a s-concave measure is also s-concave. It is known that
any semi-norm of an n-dimensional vector with a s-concave distribution has
moments up to the order p < −1/s (see [9] and [1]). Since we are interested
in comparison of moments with the moment of order 2, we will always assume
that −1/2 < s ≤ 0, which corresponds to α > n + 2.
We say that a random vector X is isotropic if EX = 0 and for every
θ ∈ S n−1 , EhX, θi2 = 1. For every −1/2 < s ≤ 0, we can always find an
affine transformation A such that AX is isotropic.
Our main result is the following
Theorem 1. Let r > 2. Let X be a (−1/r)-concave isotropic random vector
on Rn . Then for any p ∈ R such that 0 < |p| ≤ c min(r, n1/3 ), we have
3/5
C |p − 2|
(E|X|p2 )1/p
C |p − 2|
− 1 ≤
+
αp (X) := (E|X|22 )1/2
r
n1/3
where C and c are universal constants.
In the general case, let A be the affine transformation such that AX is
1/3
isotropic. Then for any p ∈ R such that 0 < |p| ≤ c min r, kAk2/3nkA−1 k2/3 ,
we have
3/5
(E|X|p2 )1/p
C |p − 2|
C |p − 2|(kAkkA−1 k)2/3
αp (X) := − 1 ≤
+
(E|X|22 )1/2
r
n1/3
where C and c are universal constants.
2
√
Remark 2. For r > n + n, we have a better inequality which recovers the
log-concave estimate from [16].
To present the connections between moment inequalities, concentration
in a spherical thin shell property and the Berry-Esseen theorem for one dimensional marginals, let us introduce some notations.
Let X ∈ Rn be an isotropic random vector. Thus E|X|2 = n. Define
ε(X) to be the smallest number ε > 0 such that
|X|
(1)
P √ − 1 ≥ ε ≤ ε.
n
If ε(X) = o(1) with respect to the dimension n, we say that X is concentrated
in a spherical thin shell. It was shown in [2] (see also [12, 13]) that if an
isotropic random vector X uniformly distributed on a convex body in Rn is
such that ε(X) = o(1), then almost all marginal distributions of X satisfy a
Berry-Esseen theorem. More generally, let X ∈ Rn be an isotropic random
vector, it was proved in [6] that
4
n−1
σn−1 θ ∈ S
: sup |P(hX, θi ≤ t) − Φ(t)| ≥ 4ε(X) + δ ≤ 4n3/8 e−cnδ
t∈R
where σn−1 denotes the rotation invariant probability measure on the unit
sphere S n−1 , Φ is the standard normal distribution function and c > 0 is
a universal constant. Therefore if ε(X) is small then almost all the one
dimension marginal distributions of X are approximately Gaussian. The
fact that indeed for all log-concave random vector ε(X) = o(1) was proved
later in [18, 15] and the best estimate at this date [16] is that
ε(X) = O(n−1/6 log n).
Now let p > 2 and assume that |X|2 has finite p moment. For an isotropic
random vector X,
αp (X) =
(E|X|p2 )1/p
(E|X|p2 )1/p
√
−
1
=
− 1.
(E|X|22 )1/2
n
In the particular case p = 4, it is easy to deduce from Markov inequality
that ε(X) ≤ (24 α4 (X))1/3 . Thus if α4 (X) is small, then so is ε(X) and
from this discussion, it follows that almost all one dimensional marginal
3
distributions of X are approximately Gaussian. Hence Theorem 1 ensures
that if r → +∞ with the dimension n then any isotropic (−1/r)-concave
vector satisfies a spherical thin shell concentration and therefore almost all its
marginals verify a Berry-Esseen theorem. As a matter of fact, this condition
on r is necessary. If r is fixed and does not depend on the dimension n,
we give below an example of an isotropic (−1/r)-concave random vector
X ∈ Rn which does not satisfy a spherical thin shell property. This shows
the sharpness of Theorem 1 for small values of r.
Fact: Let r > 2. For every n ≥ 1, there exists an isotropic (−1/r)-concave
random vector Xn ∈ Rn such that
ε(Xn ) ≥ c(r) > 0
where c(r) > 0 depends only on r. Moreover, when r tends to ∞, one has
lim αp (Xn ) ∼
n→∞
p−2
.
r−p
(2)
Proof. Let r > 2 and 2 < p < r and let Xn ∈ Rn be an isotropic random
vector with density
c1
fn,r (x) =
(1 + c2 |x|2 )r+n
where c1 and c2 are normalization factors. Such a random vector is (−1/r)concave. An immediate computation gives that
(E|Xn |p2 )1/p
=
(E|Xn |22 )1/2
B(n + p, r − p)
B(n, r)
1/p B(n + 2, r − 2)
B(n, r)
−1/2
.
For fixed r and 2 < p < r, we have
(E|Xn |p2 )1/p
lim
=
1/2
n→+∞ (E|Xn |2
2)
Γ(r − p)
Γ(r)
1/p Γ(r − 2)
Γ(r)
Therefore, when r → +∞
lim αp (Xn ) ∼
n→+∞
4
p−2
.
r−p
−1/2
6= 1.
(3)
From Paley-Zygmund inequality we get that for any 2 < p < s < r
P
|Xn |
√ ≥1+ε
n
(αp + 1)p − (1 + ε)p
≥
(αs + 1)sp
1/(s−p)
.
For fixed r, 2 < p < r and n large enough, choosing for instance p = (2+r)/2
and s = (p + r)/2, we get from (3) that
|Xn |
P √ ≥ 1 + ε ≥ c(r) > 0
n
for every 0 < ε ≤ ε(r), where ε(r) and c(r) > 0 depend only on r. Therefore
ε(Xn ) ≥ min(ε(r), c(r)) > 0.
To build the proof of Theorem 1, we need to extend to the case of sconcave measures several tools coming from the study of log-concave measures. This is the purpose of Section 2. Some of them were already achieved
by Bobkov [7], like the analogous of the Ball’s bodies [5] in the s-concave
setting. Some others were also noticed previously (see e.g. [7], [1]) but not
with the most accurate point of view. These new ingredients are analogous
to the results of [11] in the log-concave setting and are at the heart of our
proof. As in the approach of [14] or [16], an important ingredient is the LogSobolev inequality on SO(n). It allows to get reverse H¨older inequalities, see
[21, 14]: for every f : SO(n) → R, if L is the Log-Lipschitz constant of f
then for every q > r > 0,
2
cL
q 1/q
(q − r) (EU |f |r )1/r .
(4)
(EU |f | ) ≤ exp
n
According to the characterization of −1/r-concave measures in [10], let X
be a (−1/r)-concave random vector in Rn with full dimensional support and
distributed according to a measure with a (−1/α)-concave density function
w : Rn → R+ with α = r + n. For any linear subspace E, denote by PE the
orthogonal projection onto E and for any x ∈ E denote by
Z
πE w(x) =
w(y)dy
x+E ⊥
5
the marginal of w on E. Given an integer k between 1 and n, a real number
p ∈ (−k, r), a linear subspace E0 ∈ Gn,k and θ0 ∈ S(E0 ), define the function
hk,p : SO(n) → R+ :
Z ∞
k−1
tp+k−1 πU (E0 ) w(tU (θ0 ))dt.
(5)
hk,p (U ) := Vol(S )
0
Following the approach of [19, 14], we observe that for any p ∈ (−k, r)
E|X|p =
Γ((p + n)/2)Γ(k/2)
EU hk,p (U )
Γ(n/2)Γ((p + k)/2)
(6)
where U is uniformly distributed on SO(n). In view of (6) and the definition
of hk,p , we notice that it is of importance to work with family of measures
which are stable after taking the marginals. For any subspace E of dimension
k, since X is (−1/r)-concave with density w, we know that PE X is (−1/r)concave and πE w is (−1/αk )-concave with αk = r + k.
2
Preliminary results for s-concave measures
Theorem 3. Let α > 0 and f : [0, ∞) → [0, ∞) be (−1/α)-concave and
integrable. Define Hf : [0, α) → R+ by

Z +∞
1

tp−1 f (t)dt, 0 < p < α
Hf (p) =
B(p, α − p) 0

f (0)
p=0
R1
R +∞
where B(p, α − p) = 0 up−1 (1 − u)α−p−1 du = 0 tp−1 (t + 1)−α dt. Then Hf
is log-concave on [0, α).
Since the proof is identical to the well known 1/n-concave case [11], we
postpone it to the Appendix. We present several consequences of this result
like some reverse H¨older inequalities with sharp constants in the spirit of
Borell’s [11] and Berwald’s [3] inequalities.
Corollary 4. Let r > 0 and µ be a (−1/r)-concave measure on Rn . Let
φ : Rn → R+ be concave on its support. Then
Z
1
p 7→
φ(x)p dµ(x)
pB(p, r − p)
6
is log-concave on [0, r). Moreover, for any 0 < p ≤ q < r,
1/q
1/p
Z
(qB(q, r − q))1/q
p
φ(x) dµ(x)
≤
φ(x) dµ(x)
(pB(p, r − p))1/p
Rn
Rn
Z
q
and for every θ ∈ S n−1 ,
Z
1/q
Z
1/p
(qB(q, r − q))1/q
q
p
hx, θi+ dµ(x)
≤
hx, θi+ dµ(x)
(pB(p, r − p))1/p
Rn
Rn
where
hx, θi+ =
hx, θi
0
(7)
if hx, θi ≥ 0
otherwise.
Proof. By concavity of φ, for every u, v ≥ 0, and any λ ∈ [0, 1]
(1 − λ){φ ≥ u} + λ{φ ≥ v} ⊂ {φ ≥ (1 − λ)u + λv}.
By −1/r-concavity of µ, the function f (t) = µ({φ > t}) is −1/r-concave.
Observe by Fubini that for any p > 0,
Z
Z +∞
p
φ(x) dµ(x) =
ptp−1 f (t)dt.
Rn
0
The result follows from Theorem 3. Since µ is a probability we have f (0) =
1 = Hf (0) and the moreover part follows from the fact that p 7→ Hf (p)1/p is
a non increasing function. For the last inequality, we notice that for every
θ ∈ S n−1 , the function φ(x) = hx, θi+ is affine on its support.
The second corollary concerns the function hk,p .
Corollary 5. Let r > 0. For any −1/(r +n)-concave function w : Rn → R+ ,
and any subspace F of dimension k ≤ n,

hk,p (U )

, p > −k + 1,
p 7→
B(p + k, r − p)

Vol(S k−1 )πU (E0 ) w(0), p = −k + 1
is log-concave on [−k + 1, r).
Proof. Since w is −1/(r + n)-concave, we note that t 7→ πU (E0 ) w(tU (θ0 )) is
−1/(r + k)-concave. Theorem 3 proves the result.
7
We finish with some geometric properties of a family of bodies introduced
by K. Ball in [5] in the log-concave case.
Corollary 6. Let α > 0. Let w : Rn → R+ be a −1/α-concave function such
that w(0) > 0. For 0 < a < α let
Z +∞
a−1
n
t w(tx)dt ≥ w(0) .
Ka (w) = x ∈ R ; a
0
Then for any 0 < a ≤ b < α
1−1
w(0) a b
(bB(b, α − b))1/b
Ka (w) ⊂ Kb (w) ⊂
Ka (w).
kwk∞
(aB(a, α − a))1/a
Proof. Notice that the sets Ka are star-shaped with respect to the origin.
For any x ∈ Rn , let f be the −1/α-concave function defined on R+ by
f (t) = w(tx)/w(0). By Theorem 3, the function Hf : [0, α) → R+ is logconcave on [0, α) with Hf (0) = 1. For 0 < a ≤ b < α, we have thus
Hf (b)1/b ≤ Hf (a)1/a . The right hand side inclusion follows. The left hand
side is a simple consequence of H¨older’s inequality in the even case. The
non-even case may be easily treated as Lemma 2.1 in [22].
3
Thin shell for convex measures
The purpose of this section is to prove Theorem 1. We follow the strategy
from the log-concave case initiated in [18, 15, 19] and further developed in
[14, 16].
To any random vector X in Rn and any p ≥ 1, we associate its Zp+ -body
defined by its support function
1/p
hZp+ (X) (θ) = E hX, θip+
.
When the distribution of X has a density g, we write Zp+ (g) = Zp+ (X).
Extending a theorem of Ball [5] on log-concave functions, Bobkov proved in
[7] that if w is a −1/(r + n)-concave on Rn such that w(0) > 0, then
Ka (w) is convex for any 0 < a ≤ r + n − 1.
In the case of log-concave measures [23, 24, 16, 17], several relations between
the Zp+ bodies and the convex sets Ka are known. We need their analogue in
the setting of s-concave measures for negative s. We start with two technical
lemmas. We postpone their proofs to the Appendix.
8
Lemma 7. Let x, y ≥ 1, then
x
x
c
≤ (xB(x, y))1/x ≤ C
,
x+y
x+y
(8)
where c, C are positive universal constants. Moreover for k, r > 1, we have
B(k + p, r − p)
1
1
d 1
log
≤
+
(9)
0≤
dp p
B(k, r)
r−1 k−1
for |p| ≤ (min(r, k) − 1)/2.
Lemma 8. Let r, m and p be such that m is an integer, r ≥ m + 1 and
− m2 ≤ p ≤ r − 1, Let F be a subspace of dimension m and let g be a
−1/(r + m)-concave density of a probability measure on F with barycenter at
the origin. Then we have
+
d(Km+p (g), B2 (F )) ≤ c d(Zmax(m,p)
(g), B2 (F ))
where c is a universal constant.
An important ingredient in the proof of a thin-shell concentration inequality is an estimate from above of the log-Lipschitz constant of U 7→ hk,p (U ).
Proposition 9. Let n ≥ 1, r > 10 and w be a −1/(r + n)-concave density
of a probability measure on Rn . Let k be an integer such that 2k + 1 ≤ n and
2k + 2 ≤ r. Let p such that − k2 ≤ p ≤ r − 1. Denote by Lk,p the log-Lipschitz
constant of U 7→ hk,p (U ). Then
+
Lk,p ≤ C max(k, p)d(Zmax(k,p)
(w), B2n )
where C is a universal constant.
Proof. Since Ka (w) are appropriate convex bodies, the proof of Theorem 2.1
in [16] works identically and lead to the upper bound :
max{(m + p) d(Km+p (πF (w)), B2 (F ))}
F
over all subspaces F of dimension m = k, k + 1, 2k + 1. From Lemma 8, we
have
+
d(Km+p (πF (w)), B2 (F )) ≤ c d(Zmax(p,m)
(πF (w)), B2 (F )).
By integration in polar coordinates, Zp+ (πF (w)) = PF (Zp+ (w)) and the distance to the Euclidean ball cannot increase after projections. Hence
+
d(Km+p (πF (w)), B2 (F )) ≤ cd(Zmax(p,m)
(w), B2n ).
9
We define the q-condition number of a random vector X to be
1/q
sup|θ|2 =1 E hX, θiq+
ρq (X) =
1/q .
inf |θ|2 =1 E hX, θiq+
Obviously, if w is the density of X, ρq (X) = d(Zq+ (w), B2n ).
Proposition 10. With the same assumptions as in Proposition 9, if the
random vector X with density w is isotropic then
Lk,p ≤ C max(k, p)2 .
More generally if A is such that AX is isotropic then
Lk,p ≤ C max(k, p)2 kAkkA−1 k.
(10)
Proof. Let q = max(k, p) then one has 1 ≤ q ≤ r − 1. If X is isotropic then
from Corollary 4 and equation (8) we get that for q ≥ 2
ρq (X) ≤
(qB(q, r − q))1/q
1/2
(2B(2, r − 2))
≤
Cq/r
≤ c0 q,
c/r
for some universal constant c0 > 0. The same inequality holds for q ∈ [1, 2].
The conclusion follows from Proposition 9. In the general case, notice that
Zq+ (AX) = AZq+ (X) and d(AB2n , B2n ) = kAkkA−1 k thus
ρq (X) ≤ ρq (AX)kAkkA−1 k.
Proof of Theorem 1. We start by presenting a complete argument following
[14]. This will give a complete proof of Theorem 1 with a slightly weaker
result. In the second part, we just indicate the needed modifications of the
argument of [16] to get the complete conclusion.
From now on, we assume without loss of generality that r > 10 (otherwise
the inequality of Theorem 1 is obvious, see [1]) and |p| ≤ (r − 2)/4. Let k
be an integer such that −p < k ≤ n. We will optimize the choice of k at the
end of the proof. Recall that by (6),
E|X|p =
Γ((p + n)/2)Γ(k/2)
EU hk,p (U )
Γ(n/2)Γ((p + k)/2)
10
where U is uniformly distributed on SO(n). Then for all 0 < p < r and
n ≥ k > p we have
B((n + p)/2, (n − p)/2)
EU hk,p (U )EU hk,−p (U )
B((k + p)/2, (k − p)/2)
≤ EU hk,p (U )EU hk,−p (U )
(11)
E|X|p E|X|−p =
since B((n+p)/2, (n−p)/2) ≤ B((k+p)/2, (k−p)/2). Applying Log-Sobolev
inequality (4) to hk,p and hk,−p we get
EU h2k,p ≤ e
c L2
k,p
n
(EU hk,p )2 and EU h2k,−p ≤ e
c L2
k,−p
n
(EU hk,−p )2 .
(12)
Since Varf = Ef 2 − (Ef )2 we deduce that
Varhk,p ≤ (e
c L2
k,p
n
− 1) (EU hk,p )2 , Varhk,−p ≤ (e
c L2
k,−p
n
− 1) (EU hk,−p )2 . (13)
By Corollary 5, we know that p 7→ hk,p (U )/B(k + p, r − p) is log-concave on
[−k + 1, r) hence
B(k + p, r − p) B(k − p, r − p)
hk,p (U ) hk,−p (U ) ≤
h2k,0 (U ).
B(k, r)
B(k, r)
Taking the expectation with respect to SO(n), we get that
B(k + p, r − p) B(k − p, r − p)
EU h2k,0 (U ).
EU hk,p (U )hk,−p (U ) ≤
B(k, r)
B(k, r
Since EU hk,0 (U ) = 1 we deduce from (12) that
EU h2k,0 (U ) ≤ e
c L2
k,0
n
.
By (9), we know that for p ≤ (k − 1)/2,
B(k + p, r − p) B(k − p, r − p)
B(k, r)
B(k, r)
hence
EU hk,p (U )hk,−p (U ) ≤ e
11
1/p
c L2
k,0
n
≤ e4p(1/k+1/r)
+4p2 ( k1 + r1 )
.
(14)
Moreover
EU hk,p (U ) hk,−p (U ) = EU hk,p (U ) EU hk,−p (U ) + Cov(hk,p , hk,−p )
p
≥ EU hk,p (U ) EU hk,−p (U ) − Varhk,p Varhk,−p
s
c L2
!
c L2
k,p
k,−p
e n −1
e n −1
≥ EU hk,p (U ) EU hk,−p (U ) 1 −
where the last inequality follows from (13). For p ≤ (k−1)/2, we can evaluate
Lk,p , Lk,−p and Lk,0 from Proposition 10 since the assumptions are full filled.
We get that if X is isotropic then max(Lk,p , Lk,−p , Lk,0 ) ≤ Ck 2 . Combining
the previous estimate with (14) we get that if k ≤ cn1/4 and p ≤ (k − 1)/2
then
2
4
e4p (1/k+1/r)+ck /n
r
c k4
EU hk,p (U )hk,−p (U )
c k4
1−
e n −1 e n −1
k4
1 1
2
+
≤
1 + Cp
+C
.
k r
n
EU hk,p (U ) EU hk,−p (U ) ≤
We conclude from (11) that
p
−p 1/p
(E|X| E|X| )
≤
1 + Cp
1 1
+
k r
k4
+C
n
.
It remains to optimize the choice of k. If 0 < p < n−1/2 then we choose k = 5
and get
C
C
p
−p 1/p
.
(15)
(E|X| E|X| ) ≤ 1 + + √
r
n
If p ≥ n−1/2 we can choose k such that k 5 = p2 n with the restriction 2p + 1 ≤
k ≤ cn1/4 . This proves that
3/5 !
Cp
Cp
.
∀n−1/2 ≤ p ≤ c n1/8 , (E|X|p E|X|−p )1/p ≤ 1 +
+
r
n1/3
Therefore for any p such that |p| ≤ min(c n1/8 , (r − 2)/4),
3/5 !
(E|X|p )1/p
C(1
+
|p|)
C(1
+
|p|)
,
+
(E|X|2 )1/2 − 1 ≤
r
n1/3
12
which is already enough to get a spherical thin shell concentration.
In the second part, we follow the argument developed in [16] to get a
better conclusion. We continue to assume that |p| ≤ (r − 2)/4. Most of the
computation of section 3.2 in [16] is identical and it required the use of the
reverse H¨older inequalities (12). We estimate
1
Γ((p + n)/2)Γ(k/2)
d
d
d 1
p p1
p
log((E|X| ) ) =
log((EU hk,p (U )) )+
log
.
dp
dp
dp p
Γ(n/2)Γ((p + k)/2)
By Corollary 5
p 7→
hk,p (U )
,
B(p + k, r − p)
is log-concave on [−k + 1, r). This can be written as
hk,0 (U )
d 1
hk,p (U )
− log
log
≤ 0.
dp p
B(p + k, r − p)
B(k, r)
We know from Lemma 7 that for all p ∈ [− k−1
, r−1
]
2
2
B(k + p, r − p)
d 1
1 1
log
+
≤C
.
dp p
B(k, r)
k r
Following Section 3.2 in [16], we arrive at
1
c
d
C C
log((E|X|p ) p ) ≤ 2 (2L2k,p + 3L2k,0 ) + + .
dp
pn
k
r
Assume that X is isotropic. By Proposition 10, we know that for any k ≥ 2|p|
and k ≤ (r − 2)/2, Lk,p , Lk,−p , Lk,0 are smaller than Ck 2 . We get that
4
d
k
1 1
p p1
log((E|X| ) ) ≤ C
+ +
.
dp
p2 n k r
We have to minimize this expression for k ∈ [2|p|, (r − 2)/2]. For |p| ≤ cn1/3 ,
we set k = min((r − 2)/2, c(p2 n)1/5 ) so that k ∈ [2|p|, (r − 2)/2] and we get
d
c
1
p p1
log((E|X| ) ) ≤ C
+
.
dp
(p2 n)1/5 r
Let p0 = n−1/2 , we get after integration that ∀p ∈ [p0 , c min(r, n1/3 )],
3/5
(E|X|p )1/p
C |p − 2|
C |p − 2|
+
(E|X|2 )1/2 − 1 ≤
r
n1/3
13
and ∀p ∈ [−c min(r, n1/3 ), −p0 ],
3/5
C |p − p0 |
(E|X|p )1/p
C |p − p0 |
+
.
(E|X|−p0 )−1/p0 − 1 ≤
r
n1/3
For p ∈ [−p0 , p0 ], we use (15) and this concludes the proof of Theorem 1.
If X is such that AX is isotropic, we know from Proposition 10 that for
any k ≥ 2|p|,
max(Lk,p , Lk,−p , Lk,0 ) ≤ Ck 2 kAkkA−1 k.
The proof is identical to the previous one replacing n by
n
.
kAk2 kA−1 k2
Remark 11. In [16], a preprocessing step was to add a Gaussian isotropic
vector to get from the very beginning a better information on the Zp+ bodies
associated to the measure. In [19], [14], this convolution argument played a
role of regularization. It is natural to ask if such a process could be done in
the situation of s-concave measure. Nothing is doable by adding a Gaussian
vector because for s < 0, the new vector does not belong to any class of
s-concave vectors. However, we can build a similar argument for r > n,
taking the addition of X with a random vector Z uniformly distributed on
the isotropic Euclidean ball, see also [8]. Since Z is 1/n-concave and X
1
√
will be a − r−n
-concave random
is −1/r-concave, the new vector Y = X+Z
2
isotropic vector. For any p ≥ 1, we have (see inequality (4.7) in [16])
αp (X) ≤ α2p (Y ) (2 + α2p (Y ))
so that it remains to bound α2p (Y ). It is easy to see that Y is such that for
1/q
√
every q ≥ 2 and every θ ∈ S n−1 , E hY, θiq+
≥ c q. Adapting the proof of
Proposition 10, we get Lk,p ≤ C max(k, p)3/2 . As in [16], this improvement
leads to the √
estimate : if r − n > 2, then for any p such that 1 ≤ p ≤
c min(r − n, n)
1/2
C(2p − 2)
C(2p − 2)
√
+
.
α2p (Y ) ≤
r−n
n
√
For r > n + n, we recover the same spherical thin shell concentration
as in the log-concave case. It would be interesting to understand in which
precise sense the s-concave measures are close to the log-concave measures
for s ∈ (−1/n, 1/n). An other question is to know what kind of preprocessing
argument like in [20] would enable to recover the small ball estimates from
[1].
14
4
Appendix
Proof of Theorem 3. Since f is −1/α-concave, there exists a convex function
ϕ : [0, ∞) → (0, ∞) such that f = ϕ−α . Since f is integrable it follows that
ϕ tends to +∞ at +∞. From the convexity of ϕ, one deduces that for some
constant c > 0, ϕ(t) ≥ (1 + t)/c. Thus f (t) ≤ c(1 + t)−α , for every t ≥ 0.
Therefore, tp−1 f is integrable for every p < α, which means that Hf (p) < +∞
for every 0 ≤ p < α. For any p ∈ (0, α) and any m, M > 0, we have
−α
Z 1
Z +∞
t
p
p−1
v p−1 (1 − v)α−p−1 dv (16)
1+
dt = mM
mt
M
0
0
= mM p B(p, α − p).
Take 0 < a < b < c < α. Choose m and M such that mM a = Hf (a)
and mM b = Hf (b) so that Hg (a) = Hf (a) and Hg (b) = Hf (b), where g :
−α
R+ → R+ is the function defined by g(t) = m 1 + Mt
. Then we observe
from (16) that Hg ((1 − λ)a + λc) = Hg (a)1−λ Hg (c)λ where λ is such that
b = (1 − λ)a + λc. Our goal is to prove that
Z +∞
tc−1 (g − f )(t)dt ≥ 0.
(17)
0
We will then deduce that
Hf (b) = Hg (b) = Hg (a)1−λ Hg (c)λ ≥ Hf (a)1−λ Hf (c)λ
and this will prove the log-concavity of H on (0, α).
Let h = (g − f ). Let
Z +∞
Z +∞
a−1
sb−a−1 H1 (s)ds.
s h(s)ds, and H2 (t) =
H1 (t) =
t
t
α
Using that h(t) ≤ c/(1 + t) , on R+ , one can see that for every t ≥ 0
c
c
H1 (t) ≤
and
H
(t)
≤
.
2
(1 + t)α−a
(1 + t)α−b
R +∞
We have 0 ta−1 h(t)dt = 0 thus H1 (∞) = H1 (0) = 0. Obviously H2 (∞) =
0. We also observe
Z +∞
Z +∞
Z +∞
b−1
b−a a−1
0=
t h(t)dt =
t t h(t)dt = −
tb−a H10 (t)dt
0
0
0
Z +∞
= [tb−a H1 (t)]+∞
+ (b − a)
tb−a−1 H1 (t)dt = (b − a)H2 (0),
0
0
15
whence H2 (∞) = H2 (0) = 0. Since
H20 (t) = −tb−a−1 H1 (t),
the function H1 changes sign at least once (if not, then H20 ≥ 0 or H20 ≤ 0, but
since H2 (0) = H2 (∞) = 0, we have H2 ≡ 0, but then H1 ≡ 0, h ≡ 0 and there
is nothing to do). Since H1 changes sign at least once and H1 (0) = H1 (∞) =
0, therefore H10 changes of sign at least twice. Since H10 (t) = −ta−1 h(t), we
have that h changes sign at least twice, but since g −α is affine and f −α is
convex, h changes sign exactly twice and we have g −α (0) < f −α (0) so that
h(0) > 0. We deduce that H2 ≥ 0. Therefore,
Z +∞
Z +∞
Z +∞
c−1
c−a a−1
t h(t)dt =
t t h(t)dt = −
tc−a H10 (t)dt
0
0
Z0 +∞
= [−tc−a H1 (t)]+∞
+ (c − a)
tc−a−1 H1 (t)dt
0
0
Z +∞
= (c − a)
tc−b tb−a−1 H1 (t)dt
0
Z +∞
c−b
+∞
= [−t H2 (t)]0 + (c − a)(c − b)
tc−b−1 H2 (t)dt ≥ 0.
0
This proves (17). Thus the log-concavity of H on (0, α). To get it on [0, α),
it is enough to prove that H is continuous at 0. This is classical, but let
us recall the argument for completeness. Since f is −1/α-concave, for all
0 ≤ s ≤ kf k∞ , its level sets {f ≥ s} are convex in [0, +∞) thus there exists
0 ≤ a(s) ≤ b(s) ≤ kf k∞ such that
Z +∞
Z kf k∞ Z b(s)
p−1
pt f (t)dt =
ptp−1 dtds.
0
0
a(s)
Since a(s) = inf{t ≥ 0; f (t) ≥ s} is non increasing on [0, kf k∞ ], one has
a(s) = 0 for 0 ≤ s ≤ f (0), thus
Z +∞
Z f (0) Z b(s)
Z kf k∞ Z b(s)
p−1
p−1
pt f (t)dt =
pt dtds +
ptp−1 dtds
0
0
Z
=
0
f (0)
f (0)
b(s)p ds +
0
Z
a(s)
kf k∞
(b(s)p − a(s)p )ds.
f (0)
R +∞
We conclude that 0 ptp−1 f (t)dt tends to f (0) when p tends to 0. Applied
to both f and (1 + t)−α this proves that Hf is continuous at 0.
16
Proof of Lemma 7. Equation (8) follows easily from the classical bounds for
the Gamma function (see [4]), valid for x ≥ 1:
√
√
1
1
1
2πxx− 2 e−x ≤ Γ(x) ≤ 2πxx− 2 e−x+ 12 .
For equation (9), we write that
B(k + p, r − p)
Γ(k + p)Γ(r − p)
=
.
B(k, r)
Γ(k)Γ(r)
P
Denote G(p) = log Γ(p), for p > 0. We know that G00 (p) = i≥0 1/(p + i)2
hence G00 is non increasing and 0 ≤ G00 (p) ≤ 1/(p − 1), for p > 1. Denote
R1
Fk (p) = G(k+p)−G(k)
, for k > 0 and p > −k. We have Fk (p) = 0 G0 (k+up)du.
p
Using that G00 is non increasing, we get that for k > 1 and p ≥ −(k − 1)/2,
Z 1
Z 1
k+1
1 00 k + 1
1
00
00
0
G (k+up)udu ≤ G
udu = G
Fk (p) =
≤
.
2
2
2
k−1
0
0
Therefore, if p ≤ (min(r, k) − 1)/2 then
B(k + p, r − p)
d 1
d
log
(Fk (p) − Fr (−p))
=
dp p
B(k, r)
dp
= Fk0 (p) + Fr0 (−p) ≤
1
1
+
.
k−1 r−1
Proof of Lemma 8. We present here a similar but simpler proof than in the
Appendix of [16]. By integration in polar coordinates, it is well known [23]
(see also [17]) that we have the following relation between the Zq+ -bodies
associated with g and the Zq+ -bodies associated with one of the convex bodies
Ka (g): for any 0 < q < r
Zq+ (g) = g(0)1/q Zq+ (Km+q (g)).
(18)
Let us prove that for any convex body K ⊂ Rm containing 0 in its interior,
for any q ∈ (0, +∞), and for every θ ∈ Rn
hK (θ) ≥
hZq+ (K) (θ)
Vol(K ∩ {hx, θi ≥
0})1/q
17
≥ (qB(q, m + 1))1/q hK (θ).
(19)
Indeed, observe that
hqZ + (K) (θ)
q
Z
=
hx, θiq+ dx
Z
+∞
=q
tq−1 f (t)dt,
0
K
where f (t) = Vol({x ∈ K : hx, θi ≥ t}) is a 1/m-concave function on
(0, +∞). The analogous of Theorem 3 in the 1/m-concave case [11] (see
also [17] for a modern presentation) tells that the function Gf : [0, ∞) → R+
defined by
 R +∞ q−1
 0 t f (t)dt
q>0
Gf (q) =
 B(q, m + 1)
f (0)
q=0
1/q
Gf (q)
is log-concave on [0, +∞). Hence Gf (0)
is nonincreasing. Since
hZq+ (K) (θ) = (qB(q, m + 1))1/q Gf (q)1/q
and Gf (0) = f (0) = Vol(K ∩ {hx, θi ≥ 0}), we get for any q ∈ (0, +∞),
hZq+ (K) (θ)
(qB(q, m + 1)Vol(K ∩ {hx, θi ≥ 0}))1/q
is greater than the limit of the same quantity when q goes to +∞. Since 0
is in the interior of K, we have
lim
q→+∞
hZq+ (K) (θ)
Vol(K ∩ {hx, θi ≥ 0})1/q
= max hx, θi = hK (θ).
x∈K
A computation gives that (qB(q, m + 1))1/q ∼q→+∞ 1 and this proves (19).
First assume that p ≥ m. Take q = max(p, m) = p and K = Km+p . By
Lemma 7, for any p ≥ m ≥ 1, (pB(p, m + 1))1/p ≥ cp/(m + p + 1) ≥ c/3 and
we deduce from (19) that
c Km+p ⊂
1
Zp+ (Km+p ) ⊂ Km+p
1/p
Vol(Km+p ∩ {hx, θi ≥ 0})
where c is a universal constant. We conclude from (18) that
d(Km+p , Zp+ (g)) ≤ c
18
for a universal constant c. This gives the conclusion.
In the case when − m2 < p < m, we get from Corollary 6
g(0)
kgk∞
1
1
m+p
− 2m
Km+p ⊂ K2m ⊂
((2m)B(2m, r − m))1/2m
Km+p .
((m + p)B(m + p, r − p))1/m+p
From Lemma 7, we have
((2m)B(2m, r − m))1/2m
2m
≤ 4c
≤c
1/m+p
((m + p)B(m + p, r − p))
m+p
since p ≥ − m2 .
For the other inclusion, we use the fact that g has barycentre at the origin
and the sharp lower bound of the ratio g(0)/kgk∞ , proved in Lemma 20 of
[1],
r+m
r−1
g(0)
≥
≥ e−2m
kgk∞
r+m−1
for m ≤ r − 1. Since p ≥ − m2 , we have (m − p)/(m + p) ≤ 3 hence
g(0)
kgk∞
1
1
m+p
− 2m
≥ e−3 .
Therefore d(Km+p , K2m ) ≤ c where c is a universal constant. We conclude
+
(g), B2 ) and this finthat d(Km+p (g), B2 (F )) ≤ cd(K2m (g), B2 (F )) ≤ c0 d(Zm
ishes the proof.
Acknowledgement. The research of the three authors is partially supported by the ANR GeMeCoD, ANR 2011 BS01 007 01.
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Matthieu Fradelizi, Olivier Gu´edon, Alain Pajor
Universit´e Paris-Est
Laboratoire d’Analyse et Math´ematiques Appliqu´ees (UMR 8050).
UPEMLV, F-77454, Marne-la-Vall´ee, France
e-mail: [email protected], [email protected],
[email protected]
21