RATE OF CONVERGENCE IN CLT ON STRATIFIED GROUPS
GYULA PAP
∗
Lajos Kossuth University, Debrecen, Hungary
It is given a uniform estimate of the rate of convergence in CLT
on stratified groups for variational distance when some pseudomoment
is sufficiently small. Moreover it is given a nonuniform estimate on the
system of homogeneous balls when some pseudomoment is finite.
1. Introduction. Let G be a stratified group of step s, that is a simply connected
nilpotent Lie group whose Lie algebra G has a vector space decomposition G = ⊕ sj=1 Vj
such that [Vi , Vj ] ⊂ Vi+j when i + j ≤ s and [Vi , Vj ] = 0 when i + j > s, and V1
generates G as an algebra. Let exp : G 7→ G be the exponential mapping (which is now
a diffeomorphism). We equip G as well as G with the natural dilations by extending
◦
◦
δ t (X) = tj X , t > 0, X ∈ Vj by linearity to G and putting δt (exp X) = exp(δ t X). The
family (δt )t>0 is a continuous one-parameter semigroup of automorphisms of G.
Let {X1 , . . . , Xm } be a basis of G adapted to the vector space decomposition G =
that is {Xk : dk = j} is a basis for Vj where dk = j when Xk ∈ Vj . We denote
by {ξ1 , . . . , ξm } the basis for the linear forms on G dual to the basis {X1 , . . . , Xm }, and
we set ηj = ξj ◦ exp−1 . The functions {η1 , . . . , ηm } form a global coordinate system on G.
⊕sj=1 Vj ,
An element X ∈ G can be regarded as a (left-invariant) differential operator on G:
for f ∈ C 1 we put
f (x exp tX) − f (x)
Xf (x) = lim
.
t→0
t
The the following central limit theorem on G (cf. [4], [14]) is known . Let µ be
a centered probability measure on G with finite second homogeneous moment. (That is,
R
R
ηi (x)µ(dx) = 0 when di = 1 and
|ηi (x)|2/di µ(dx) < ∞ for i = 1, . . . , m.) Then
δ1/√n (µn ) → ν1
(1)
where (νt )t>0 denotes the continuous convolution semigroup of Gaussian measures on G
AMS 1980 subject classifications: Primary 60B15; secondary 60F05
Key words and phrases: Stratified groups, Gauss semigroups of measures on Lie groups,
pseudomoments
∗
Research work supported by the Alexander von Humboldt Foundation and completed
at the University of Tübingen, Federal Republic of Germany
1
whose infinitesimal generator is
X
1X
aij Xi Xj
di =dj =1
di =2
2
R
R
where ai = ηi (x)µ(dx) for di = 2 and aij = ηi (x)ηj (x)µ(dx) for di = dj = 1. (For
information on Gauss semigroups cf. [9], [16]).
a i Xi +
The aim of this paper is to examine the rate of convergence in (1). For the sake of
simplicity it will be considered only the special case when ai = 0 for di = 2 and aij = δij
for di = dj = 1. We obtain the optimal speed of convergence n−1/2 . We use the
assumption of the finiteness of some pseudomoment and give estimates in terms of different
kinds of moments and pseudomoments. We apply the method of composition. This has been
used in this context for Euclidean spaces or Banach spaces; see for example [1], [2], [3], [11],
[13] and [15]. For Heisenberg group see [12]; for that special group the results are a little bit
better then the ones of the present paper.
2. Preliminaries. The letter c (respectively c(.) ) with or without indices will
denote positive constants (respectively positive constants depending only on the quantities
in parentheses); the same symbol may stand for different constants.
◦
◦
It is known that if λ denotes the Lebesgue measure on G then λ = λ ◦ exp−1 is an
invariant Haar measure on G.
Let (νt )t>0 be the continuous convolution semigroup of Gaussian measures on G whose
P
infinitesimal generator is the sub-Laplacian (1/2) dk =1 Xk2 . Let R∗+ = {r ∈ R : r > 0}.
The differential operator
1X
∂
Xk2
−
dk =1
∂t 2
is hypoelliptic on R∗+ × G by the Hörmander’s hypoellipticity criterion since {Xk : dk = 1}
generates the whole Lie algebra G (cf. [10]; see also 6.3.7 in [9]). Thus the Gauss semigroup
(νt )t>0 on G is absolutely continuous with respect to λ and there exists an infinitely
differentiable function p > 0 on R∗+ × G such that pt (·) = p(t, ·) is a λ-density of the
measure νt (cf. [16]). We shall write simply ν and p instead of ν1 and p1 , respectively.
It is known that
p(r 2 t, δr x) = r −Q p(t, x)
(2)
Pm
for all x ∈ G, t > 0, r > 0, where Q = i=1 di is the homogeneous dimension of G.
Taking into account that
λ(δr (B)) = r Q λ(B)
for all B ∈ B(G) we obtain νt = δ√t (ν1 ) for all t > 0. In particular, ν1n = νn = δ√n (ν1 ).
Hence the measure ν is stable in the sense of Baldi and also stable with respect to the
one-parameter semigroup (δ√t )t>0 of automorphisms of G in the sense of Hazod (see [7]).
2
A homogeneous norm on G is a function x → |x| from G to [0, ∞) satisfying
(i) x → |x| is continuous on G and C ∞ on G \ {e};
(ii) |x| = 0 if and only if x = e;
(iii) |δt x| = t|x| for t > 0, x ∈ G.
Homogeneous norms always exist (cf. [5]). Moreover it is known that for a homogeneous
norm | · | on G there is a constant c > 0 such that |xy| ≤ c(|x| + |y|) for all x, y ∈ G.
A function f : G \ {0} 7→ R will be called homogeneous of degree α (α ∈ R)
if f ◦ δr = rα f for r > 0. Clearly a homogeneous norm is a homogeneous function of
degree 1. A linear differential operator D on G will be called homogeneous of degree
α if D(f ◦ δr ) = r α (Df ) ◦ δr for any f ∈ C1 and r > 0. It can be observed that if D is
homogeneous of degree α1 and f is homogeneous of degree α2 then Df is homogeneous of
degree α2 − α1 . For example the differential operator Xk ∈ G is homogeneous of degree dk .
LEMMA 1. Let | · | be a homogeneous norm on G and let f : G \ {0} 7→ R be a
continuous function which is homogeneous of degree d ∈ R. Then there exists a constant
c1 > 0 such that
|f (x)| ≤ c1 |x|d
for all x ∈ G \ {0}. If moreover f (x) 6= 0 for x ∈ G \ {0} then there exists a constant
c2 > 0 such that
|f (x)| ≥ c2 |x|d
for all x ∈ G \ {0}.
PROOF. Both sides of the desired inequalities are homogeneous of degree d, so it is
suffices to assume that |x| = 1. The set {x ∈ G : |x| = 1} is compact (see 1.4 Lemma in
[5]), and does not contain e ∈ G. Hence the assertion. #
Let us define
%(x) =
Xm
i=1
|ηi (x)|1/di
for x ∈ G. The function % satisfies the assumptions of Lemma 1, thus for an arbitrary
homogeneous norm | · | on G there exist c1 , c2 > 0 such that
c1 %(x) ≤ |x| ≤ c2 %(x).
(3)
We adopt the following multiindex notation. Let Z+ = N ∪ {0}. If I = (i1 , . . . , im ) ∈
i1
I
im
im
Zm
and η I (x) = η1i1 (x) . . . ηm
(x), x ∈ G.
+ is a multiindex, we set X = X1 . . . Xm
Further, we set |I| = i1 + . . . + im and d(I) = i1 d1 + . . . + im dm . Thus |I| is the order of
the differential operator X I , while d(I) is its degree of homogeneity (see [5]).
A function P : G 7→ R will be called a polynomial if P ◦ exp is a polynomial on G.
Every polynomial on G can be written uniquely as
X
P =
aI η I
I
3
where all but finitely many of the coefficients aI vanish. The homogeneous degree of a
P
polynomial P = I aI η I is max{d(I) : aI 6= 0}. Clearly η I is a homogeneous function
of degree d(I), thus by Lemma 1
|η I (x)| ≤ cI |x|d(I)
(4)
for all x ∈ G with a suitable constant cI > 0. A polynomial P will be called homogeneous
if it is homogeneous as a function. If P is a homogeneous polynomial which is homogeneous
of degree d then it can be written uniquely as
X
P =
aI η I
d(I)=d
and Lemma 1 implies that there is a constant cP > 0 such that
|P (x)| ≤ cP |x|d
for all x ∈ G.
Let k ∈ N, f ∈ C k and x ∈ G. The left Taylor polynomial of f at x of
homogeneous degree k is the unique polynomial P of homogeneous degree k such
that X I P (e) = X I f (x) for I ∈ Zm
+ with d(I) ≤ k (see [5]). We shall use the following
stratified Taylor inequality.
THEOREM A. Let | · | be a homogeneous norm on G. For each k ∈ N there is a
constant ck such that for all f ∈ C k and all x, y ∈ G,
|f (xy) − Px(k) (y)| ≤ ck |y|k+1 supd(I)=k+1,|z|≤bk+1 |y| |X I f (xz)|
where Pxk is the left Taylor polynomial of f at x of homogeneous degree k and b ≥ 1
is a constant (depending only on G and | · |) (cf. 1.44 Corollary in [5]).
We shall apply also the following result:
THEOREM B. Let | · | be a homogeneous norm on G.
constants {CI , I ∈ Zm
+ } and C such that
Then there exist positive
|X I pt (x)| ≤ CI t−(d(I)+Q)/2 exp{−C|x|2 /t}
for every t > 0, x ∈ G and I ∈ Zm
+ (cf. [8]).
COROLLARY 1. For arbitrary k ∈ Z+ and I ∈ Zm
+ we have
Z
(1)
|x|k |X I pt (x)|λ(dx) = ckI t(k−d(I))/2
Z
∂ I
(2)
|x|k pt (x) λ(dx) = ckI t(k−d(I))/2
∂η
Z
(3)
|x|k |X I pt (x)|λ(dx) ≤ ckI
|x|>1
Z
∂ I
(4)
k
pt (x) λ(dx) ≤ ckI .
|x| ∂η
|x|>1
4
(i)
with some positive constants ckI , i = 1, 2, 3, 4.
PROOF. From the homogeneity (2) we obtain X I pt (x) = t−(Q+d(I))/2 X I p1 (δ1/√t (x)).
The integrability of the function x → |x|k |X I pt (x)| with respect to the measure λ follows
from Theorem B and Corollary 1.17 in [5]. Substituting x = δ√t (y) and using λ(δ√t (dy)) =
tQ/2 λ(dy) we obtain the first equation.
It is known that
∂
∂η
I
=
X
|J|≤|I|,d(J)≥d(I)
PIJ X J
(5)
where PIJ is a homogeneous polynomial of degree d(J) − d(I) (cf. p.25 in [5]). Thus the
inequality |PIJ (x)| ≤ cIJ |x|d(J)−d(I) , Theorem B and the homogeneity (2) imply the second
equation.
Applying again Theorem B and (2) we have
Z
Z
k
I
(k−d(I))/2
k
I
|x| |X pt (x)|λ(dx) = t
√ |y| |X p1 (y)|λ(dy) ≤
|y|>1/ t
|x|>1
Z
(3)
(d(I)+k)/2
|X I p1 (y)|λ(dy) ≤ ckI .
≤
√ |y|
|y|>1/ t
One can prove the last inequality in the same way.
#
For the sake of simplicity we shall write ν and p instead of ν1 and p1 respectively,
if this does not cause misunderstanding.
3. Taylor polynomial of homogeneous degree 2.
LEMMA 2. Let f ∈ C 2 and x ∈ G. Then the left Taylor polynomial of f at x of
homogeneous degree 2 is
X
1X
ηi (y)ηj (y)Xi Xj f (x).
Px(2) (y) = f (x) +
ηi (y)Xi f (x) +
di =dj =1
di =1,2
2
(2)
PROOF. The Taylor polynomial Px can be written uniquely as
X
1X
aij ηi (y)ηj (y).
Px(2) (y) = a0 +
ai ηi (y) +
di =dj =1
di =1,2
2
(2)
(2)
We have X I Px (e) = X I f (x) for all I ∈ Zm
Clearly Px (e) = a0 ,
+ with d(I) ≤ 2.
I
thus a0 = f (x). If d(I) = 1, 2 then X = Xk with dk = 1, 2 or X I = Xk X` with
1 ≤ k ≤ ` ≤ m, dk = d` = 1.
The Campbell-Hausdorff formula implies that
X
ηk (xy) = ηk (x) + ηk (y) +
d(I)+d(J)=dk ,I6=0,J6=0
5
I
J
cIJ
k η (x)η (y)
(6)
for all x, y ∈ G, k = 1, . . . , m (see p. 23 in [5]). Therefore
Xk =
X
∂
∂
I[k]
+
c` η I
d` ≥dk +1,d(I)=d` −dk
∂ηk
∂η`
(7)
where [k] is the multiindex with 1 in the k th place and zeros elsewhere (cf. 1.26 Proposition
in [5]). Thus
δki
if di ≤ dk
I[k] I
(8)
Xk ηi (y) = P
c
η
(y)
if
d i ≥ dk + 1
d(I)=di −dk i
Consequently Xk ηi (e) = δki and (Xk ηi ηj )(e) = 0.
k ∈ {1, . . . , m} with dk = 1, 2, thus ak = Xk f (x).
(2)
We conclude Xk Px (e) = ak for
Let now k, ` ∈ {1, . . . , m} with dk = d` = 1. If i ∈ {1, . . . , m} with di = 1 then
we have X` ηi (y) = δ`i , thus Xk X` ηi (e) = 0. If di = 2 then (8) implies X` ηi (y) =
P
[j][`]
ηj (y), thus
dj =1 ci
Xk X` ηi (y) =
X
[j][`]
c
Xk ηj (y)
dj =1 i
[k][`]
= ci
.
Obviously ηk (x2 ) = 2ηk (x) since exp X exp X = exp(X +X) for all X ∈ G. Thus applying
(6) with x = y we obtain
X
[k][`]
dk =d` =1
ci
ηk (x)η` (x) = 0
for i ∈ {1, . . . , m} with di = 2. Since x ∈ G can be arbitrary we conclude
[k][`]
ci
[`][k]
+ ci
= 0.
(9)
Applying (7) it can be shown that if di = dj = dk = d` = 1 then
(Xk X` ηi ηj )(e) =
(
2 if i = j = k = `
1 if i 6= j and i = k, j = ` or i = `, j = k
0 else.
Let 1 ≤ k ≤ ` ≤ m with dk = d` = 1. We have
Xk X` Px(2) (e)
(2)
=
2akk P
[k][`]
ak` + di =2 ai ci
if k = `
if k =
6 `
thus Xk X` Px (e) = Xk X` f (x) implies akk = (1/2)Xk2 f (x) and
ak` = (Xk X` −
X
[k][`]
c
Xi )f (x).
di =2 i
6
(10)
Now X` Xk is a right-invariant differential operator of order 2 and homogeneous degree 2, so
it is a linear combination of the monomials X I such that d(I) ≤ 2 (see [5]). Consequently
(2)
we have again X` Xk Px (e) = X` Xk f (x). On the other hand
2akk
if k = `
(2)
[`][k]
X` Xk Px (e) = a + P
ac
if k 6= `
k`
thus
ak` = (X` Xk −
di =2
X
i i
[`][k]
c
Xi )f (x).
di =2 i
(11)
Adding (10) and (11) and applying (9) we obtain ak` = (1/2)(Xk X` + X` Xk )f (x). Thus
the assertion. #
4. Homogeneous moments and pseudomoments. Let µ be a probability
measure on G. For k ∈ N consider the k th homogeneous moment of µ:
Mk (µ) =
m Z
X
i=1
|ηi (x)|k/di µ(dx).
It should be noted that Mk (µ) < ∞ if and only if the measure µ has finite k th moment
on G in the sense of [6], [14]. We use also the k th homogeneous pseudomoment of µ
and ν defined by
m Z
X
βk (µ, ν) =
|ηi (x)|k/di |µ − ν|(dx)
i=1
where |µ − ν| denotes the total variation of the signed measure µ − ν. The inequality
(3) implies that for k ∈ N and for a homogeneous norm | · | on G there are constants
(1) (2)
ck , ck > 0 such that
Z
Z
(2)
(1)
k
|x|k µ(dx)
|x| µ(dx) ≤ Mk (µ) ≤ ck
ck
(1)
ck
Z
k
|x| |µ − ν|(dx) ≤ βk (µ, ν) ≤
(2)
ck
Z
|x|k |µ − ν|(dx).
Let (νt )t>0 be the Gauss semigroup on G whose infinitesimal generator is the subP
Laplacian (1/2) dk =1 Xk2 . We write simply ν instead of ν1 .
R
R
LEMMA 3. We have
ηi (x)ν(dx) = 0 for i = 1, ..., m and
ηi (x)ηj (x)ν(dx) = δij
for di = dj = 1.
PROOF. Theorem B implies that all moments of ν exist. The measure ν is symmetric,
R
hence
ηi (x)ν(dx) = 0 for all i = 1, ..., m. The central limit theorem (1) holds for the
R
measure µ = ν, thus
ηi (x)ηj (x)ν(dx) = δij for di = dj = 1. #
7
The variational distance between ν and a probability measure µ on G can be
estimated by the help of their pseudomoments in the following way.
LEMMA 4. For all k ∈ N there exists a constant ck > 0 such that for every probability
measure µ on G one has
Q/(Q+k)
|µ − ν|(G) ≤ ck βk
(µ, ν).
PROOF. We use the ideas of Lemma 1, p.12 in [15]. Suppose first that
0 < v = |µ − ν|(G) < 2.
Let G = D + ∪ D− be a Hahn decomposition of G with respect to the signed measure
µ − ν, i.e. D + and D − are disjoint Borel sets such that (µ − ν)(B) ≥ 0 if B ⊂ D + and
(µ − ν)(B) ≤ 0 if B ⊂ D − . Define a measure ζ on B(G) by
ζ(B) = ν(B ∩ D + ) + µ(B ∩ D − ).
Then
βk (µ, ν) ≥ ck
Z
k
% (x)|µ − ν|(dx) ≥ ck
Z
%k (x)(ν − ζ)(dx).
(12)
Note that ζ(G) = 1 − v/2. Put Aa,r = {x ∈ G : %(a−1 x) < r} for a ∈ G, r > 0.
Let r > 0 such that ν(Ae,r ) = v/2. Then ν(Ace,r ) = 1 − v/2 = ζ(Ae,r ) + ζ(Ace,r ) thus
ζ(Ae,r ) = (ν − ζ)(Ace,r ). Consequently we have
Z
k
Ae,r
k
k
% (x)ζ(dx) ≤ r ζ(Ae,r ) = r (ν −
and it follows that
Z
Z
k
% (x)(ν − ζ)(dx) ≥
ζ)(Ace,r )
k
Ae,r
% (x)(ν − ζ)(dx) +
Now (12) and (13) imply
βk (µ, ν) ≥ ck
Z
Z
≤
Z
Ace,r
k
%k (x)(ν − ζ)(dx)
% (x)ζ(dx) =
Ae,r
Z
%k (x)ν(dx).
(13)
Ae,r
%k (x)ν(dx).
Ae,r
Let h ∈ (0, r) such that ν(Ae,h ) = ν(Ae,r \ Ae,h ) = v/4. Using that the density function
of ν is bounded we have
Z
Z
Q
λ(dy) ≤ c, hQ .
ν(Ae,h ) ≤ cλ(Ae,h ) = c
λ(dx) = ch
%(y)<1
%(x)<h
8
Consequently h ≥ c,, v 1/Q and
Z
βk (µ, ν) ≥ ck
%k (x)ν(dx) ≥ ck hk ν(Ae,r \ Ae,h ) ≥ ck hk v/4 ≥ c,k v (Q+k)/Q .
Ae,r \Ae,h
Q/(Q+k)
From this it follows v = |µ − ν|(G) ≤ c,,k βk
(µ, ν).
If |µ − ν|(G) = 0 then the statement is obvious. If |µ − ν|(G) = 2 then µ and ν
are orthogonal, and we have
Z
k
% (x)|µ − ν|(dx)
Hence the assertion.
Q/(Q+k)
≥
Z
k
% (x)ν(dx)
Q/(Q+k)
= ck = ck |µ − ν|(G)/2.
#
The variational distance between ν and the normalized convolution power δ 1/√n (µn )
can be estimated as follows.
LEMMA 5. For all n ∈ N there exists a constant cn > 0 such that if µ is a probability
R
R
measure on G with
ηi (x)µ(dx) = 0 for di = 1, 2 and
ηi (x)ηj (x)µ(dx) = δij for
di = dj = 1 then for all n ∈ N
|δ1/√n (µn ) − ν|(G) ≤ (|µ − ν|(G))n + cn
X3s
j=3
βj (µ, ν).
(14)
PROOF. We use again the ideas of Lemma 1, p.12 in [15]. Let us first observe that
|(µ − ν)n (B)| ≤
1
(|µ − ν|(G))n
2
(15)
for all n ∈ N and B ∈ B(G). Indeed this is true for n = 1 and for n ≥ 2 using induction
on n we have
Z
n
n−1 −1
|(µ − ν) (B)| = (µ − ν)
(x B)(µ − ν)(dx) ≤
Z
1
1
n−1
|µ − ν|(dx) = (|µ − ν|(G))n .
≤ (|µ − ν|(G))
2
2
Let us now estimate |ν ∗ (µ − ν)(B)| for B ∈ B(G). We have
ZZ
ZZ
ν ∗ (µ − ν)(B) =
χB (yx)p(y)λ(dy)(µ − ν)(dx) =
χB (u)p(ux−1 )λ(du)(µ − ν)(dx).
Now we shall use Taylor inequality for the function
Z
f (x) = χB (u)p(ux−1 )λ(du),
9
x∈G
at e ∈ G. Put gu (x) = p(ux−1 ) for x, u ∈ G. First we show that f ∈ C ∞ and
Z
I
X f (x) = χB (u)X I gu (x)λ(du)
for all x ∈ G and I ∈ Zm
+ . We have
I
X gu (x) =
X
|J|≤|I|,d(J)≥d(I)
PIJ (x)
∂
∂η
J
(16)
gu (x)
where PIJ is a homogeneous polynomial of degree d(J) − d(I) (cf. [5, p.25]). Moreover,
from (6)
J
K
X
∂
∂
gu (x) =
p(ux−1 )
QJK (x, u)
|K|≤|J|,d(K)≥d(J)
∂η
∂η
where QJK is a polynomial on G × G which is jointly homogeneous of degree d(K) −
d(J) (that is, QJK (δr (x), δr (u)) = r d(K)−d(J) QJK (x, u)). By (4), Corollary 1 and by the
inequality |vx| ≤ c(|v| + |x|)
J
Z ∂
gu (x) λ(du)
χB (u)
∂η
K
Z X
∂
−1 ≤
p(ux ) λ(du)
QJK (x, u)
|K|≤|J|,d(K)≥d(J)
∂η
K
Z X
(17)
∂
≤
p(v) λ(dv)
QJK (x, vx)
|K|≤|J|,d(K)≥d(J)
∂η
Z
∂ K
X
≤ cJ
(|v| + |x|)d(K)−d(J) p(v) λ(dv)
|K|≤|J|,d(K)≥d(J)
∂η
≤ c(J, R)
if |x| ≤ R. Thus
∂
∂η
J
f (x) =
Z
χB (u)
∂
∂η
J
gu (x)λ(du)
which implies (16).
Denote P (2) the left Taylor polynomial of f at e ∈ G of homogeneous degree 2.
The moment conditions, Lemma 2 and Lemma 3 imply
Z
P (2) (x)(µ − ν)(dx) = 0.
By Theorem A we have
|f (x) − P (2) (x)| ≤ |x|3 supd(I)=3,|z|≤b3 |x| |X I f (z)|
10
for all x ∈ G. For |z| ≤ b3 |x| by (16) and (17)
K
Z ∂
I
p(v) λ(dv)
|X f (z)| ≤
|PIJ (z)| QJK (z, vz)
|K|≤|J|≤|I|
∂η
d(K)≥d(J)≥d(I)
X
≤ c(I)
|x|d(J)−d(I) 1 + |x|d(K)−d(J) .
|K|≤|J|≤|I|
X
d(K)≥d(J)≥d(I)
Consequently
Z
Z
(2)
|ν ∗ (µ − ν)(B)| = f (x)(µ − ν)(dx) = (f (x) − P (x))(µ − ν)(dx)
Z
Z
3
≤ c |x| supd(I)=3,|z|≤b3 |x| |X I f (z)||µ − ν|(dx)
Z
X
X
|x|d(J)−d(I) + |x|d(K)−d(I) |µ − ν|(dx)
≤ c |x|3
|K|≤|J|≤|I|
d(I)=3
≤c
X
3≤j≤3s
Z
d(K)≥d(J)≥d(I)
|x|j |µ − ν|(dx)
since d(I) = 3 and |K| ≤ |I| imply d(K) ≤ s|K| ≤ s|I| ≤ sd(I) = 3s. Thus we conclude
|ν ∗ (µ − ν)(B)| ≤ c
X3s
βj (µ, ν).
(18)
|(µ − ν) ∗ ν(B)| ≤ c
X3s
βj (µ, ν).
(19)
Similarly
j=3
j=3
Now we prove (14). It is true for n = 1. Suppose that it is true for 1, 2, . . . , n − 1. Then
for all Borel sets B ∈ B(G) we have
|(δ1/√n (µn ) − ν)(B)| = |(µn − ν n )(δ√n (B))| ≤ |(µ − ν)n (δ√n (B))| + S1 + S2
where
n−1
X
(µ − ν)k ∗ ν ∗ µn−k−1 (δ√n (B))
S1 = k=1
n−1
X
S2 = ν k ∗ (µ − ν) ∗ µn−k−1 (δ√n (B)) .
k=1
From (15), (18), (19) and (20) we obtain (14).
#
5. Rate of convergence in CLT for variational distance.
11
(20)
TEOREM 1. There exist two constants C1 , C2 > 0 with the following property: if µ is
R
R
a probability measure on G such that ηi (x)µ(dx) = 0 when di = 1, 2, ηi (x)ηj (x)µ(dx) =
δij when di = dj = 1 and β3s (µ, ν) ≤ C1 then for all n ≥ 4 one has
∆n (µ, ν) =
X3s−3
sup δ1/√n (µn )(B) − ν(B) ≤ C2
β3+j (µ, ν)n−(1+j)/2 .
j=0
B∈B(G)
(21)
REMARK. Lemma 4 and Lemma 5 give an estimate for ∆n (µ, ν) when n = 1, 2, 3.
PROOF. We use induction on n and the method of composition, and follow the ideas
of the proof of Lemma 2 in [2]. By the assumption β3s ≤ C1 , Lemma 4 and Lemma 5
X3s
1
1 √ 4
4
(|µ − ν|(G)) + c
βj ≤
∆4 (µ, ν) ≤ δ1/ 4 (µ ) − ν (G) ≤
j=3
2
2
X3s
X3s
3(Q−s)/(Q+3s)
4Q/(Q+3s)
≤ c β3s
+
βj ≤ c 1 + C 1
βj
j=3
j=3
since Q ≥ m ≥ s. We conclude that (21) is true for n = 4 if we choose C2 ≥ 4(1+3s)/2 c(1 +
3(Q−s)/(Q+3s)
C1
).
Now we assume that if β3s (µ, ν) ≤ C1 then
X3s−3
√ k
β3+j (µ, ν)k −(1+j)/2 = Sk
sup δ1/ k (µ )(B) − ν(B) ≤ C2
j=0
B∈B(G)
(22)
for k = 2, 3, . . . , n − 1 and B ∈ B(G).
Consider the decomposition
δ1/√n (µn )
where
−ν =
n
X
(γk + γ̃k )
(23)
k=1
γk = (δ1/√n ν)k−1 ∗ (δ1/√n µ − δ1/√n ν) ∗ ((δ1/√n µ)n−k − (δ1/√n ν)n−k )
γ̃k = (δ1/√n ν)k−1 ∗ (δ1/√n µ − δ1/√n ν) ∗ (δ1/√n ν)n−k .
First we give an estimate for γ1 (B), B ∈ B(G) :
γ1 (B) =
Z
(δ1/√n µn−1 − δ1/√n ν n−1 )(x−1 B)(δ1/√n µ − δ1/√n ν)(dx).
Using the inductive hypothesis (22) for n − 1 we have
|(δ1/√n µn−1 − δ1/√n ν n−1 )(x−1 B)| = |(δ1/√n−1 µn−1 − ν)(δ√n/(n−1) (x−1 B))| ≤ Sn−1
12
thus using Sn−1 ≤ 2(3s−2)/2 Sn we obtain by Lemma 4
Q/(Q+3s)
|γ1 (B)| ≤ Sn−1 |δ1/√n µ − δ1/√n ν|(G) ≤ cSn |µ − ν|(G) ≤ c, β3s
Sn .
(24)
Now we give an estimate of γk (B) for 2 ≤ k ≤ [n/2] :
γk (B) =
ZZ
(δ1/√n µn−k − δ1/√n ν n−k )(u−1 B)pt (ux−1 )λ(du)(δ1/√n µ − δ1/√n ν)(dx)
where t = (k − 1)/n.
We use now the Taylor inequality for the function f : G 7→ R,
f (x) =
Z
(δ1/√n µn−k − δ1/√n ν n−k )(u−1 B)pt (ux−1 )λ(du)
as in the proof of Lemma 5. Put gut (x) = pt (ux−1 ) for x, u ∈ G and t > 0, and
l(u) = δ1/√n (µn−k − ν n−k )(u−1 B) for u ∈ G. From the inductive hypothesis (22) for n − k
we have
|l(u)| ≤ Sn−k ≤ cSn .
(25)
As in the proof of Lemma 5 one can show that f ∈ C ∞ and
I
X f (x) =
Z
l(u)X I gut (x)λ(du).
Denote P (2) the left Taylor polynomial of f at e ∈ G of homogeneous degree 2. By the
moment conditions we have again
Z
P (2) (x)δ1/√n (µ − ν)(dx) = 0.
By Theorem A we have
|f (x) − P (2) (x)| ≤ |x|3 supd(I)=3,|z|≤b3 |x| |X I f (z)|
(26)
for all x ∈ G. By (25) we have as in the proof of Lemma 5 for |z| ≤ b3 |x|
|X I f (z)| ≤
≤ c I Sn
≤ c I Sn
X
X
|K|≤|J|≤|I|
d(K)≥d(J)≥d(I)
|K|≤|J|≤|I|
d(K)≥d(J)≥d(I)
∂ K
d(J)−d(I)
d(K)−d(J)
d(K)−d(J) pt (v) λ(dv)
|x|
|x|
+ |v|
∂η
|x|d(K)−d(I) t−d(K)/2 + |x|d(J)−d(I) t−d(J)/2
Z
13
taking into account Corollary 1. Thus by (26)
Z
Z
|γk (B)| = f (x)δ1/√n (µ − ν)(dx) = (f (x) − P (2) )δ1/√n (µ − ν)(dx) ≤
Z
X3s
−j/2
≤ cSn
|x|j |δ1/√n (µ − ν)|(dx) ≤
t
j=3
≤ cSn
Consequently
X3s
j=3
((k − 1)/n)−j/2 βj n−j/2 = cSn
X[n/2]
k=2
|γk (B)| ≤ cSn
X3(s−1)
X3s−3
j=0
j=0
β3+j (k − 1)−(3+j)/2 .
β3+j .
(27)
If [n/2] < k ≤ n then instead of (25) we use the simple estimate
|(δ1/√n µn−k − δ1/√n ν n−k )(u−1 B)| ≤ 1
(since both δ1/√n µn−k (u−1 B) and δ1/√n ν n−k )(u−1 B) are in [0, 1]), and t = (k − 1)/n ≥
1/2 implies
X3(s−1)
|γk (B)| ≤ c
β3+j n−(3+j)/2 .
j=0
The same estimate is valid for γ̃k . Thus
Xn
k=[n/2]+1
Xn
k=1
|γk (B)| ≤ c
|γ̃k (B)| ≤ c
j/k
X3(s−1)
j=0
X3(s−1)
j=0
β3+j n−(1+j)/2 ≤ cC2−1 Sn
(28)
β3+j n−(1+j)/2 ≤ cC2−1 Sn .
(29)
Hölder inequality implies βj ≤ βk
for 1 ≤ j ≤ k. Collecting the estimates (24) and
(27)—(29) we obtain from (23) that
X3s−3
Q/(Q+3s)
−1
n
|(δ1/√n µ − ν)(B)| ≤ c β3s
+
β3+j + C2
Sn
j=0
X3s−3 (3+j)/3s
Q/(Q+3s)
−1
≤ c C1
+
C1
+ C2
Sn
j=0
≤ Sn
taking C1 sufficiently small and C2 sufficiently large. The proof is complete.
#
COROLLARY 2. There exist two constants C1 , C2 > 0 such that for every probability
measure µ on G satisfying the assumptions of Theorem 1 and for all n ≥ 4
∆n (µ, ν) ≤ C2 (β3 (µ, ν)n−1/2 + β3s (µ, ν)n−(3s−2)/2 )
∆n (µ, ν) ≤ C2 (m3 (µ)n−1/2 + m3s (µ)n−(3s−2)/2 ).
14
6. Homogeneous balls. Let | · | be a homogeneous norm on G. For a ∈ G and
r > 0 we define
B(a, r) = {x ∈ G : |a−1 x| < r}
and we call B(a, r) the homogeneous ball of radius r about a. We observe that
δt (bB(a, r)) = B(δt (ba), rt)
for all a, b ∈ G and r, t > 0. In other words, the system {B(a, r) : a ∈ G, r > 0} of
homogeneous balls is closed with respect to the dilations (δt )t>0 and translation from the
left.
Let us consider now the homogeneous norm on G defined by
|x| =
m
X
i=1
|ηi (x)|
d/di
!1/d
(30)
for x ∈ G, where d ∈ N and d/di are even numbers for i = 1, . . . , m. We shall show
that the system of homogeneous balls with respect to the above homogeneous norm has two
important properties.
LEMMA 6. Let | · | be the homogeneous norm on G defined by (30). Then there exists
a constant c > 0 such that
ν(B(a, r + ε) \ B(a, r)) ≤ cε(1 + |a|)s−1
for all a ∈ G, r > 0 and ε > 0.
PROOF. To avoid complicated notation we write xi = ηi (x) for x ∈ G, i = 1, . . . , m
in the proof.
For r > 0, i = 1, . . . , m consider the sets
n
X
Ai = x ∈ G :
1≤j≤m,j6=i
|xj |d/dj ≤ c0 rd
o
where (m − 1)/m < c0 < 1 is a fixed constant. It is easy to notice that for
r > ((c0 m/(m − 1))1/d − 1)−1 ε = c1 ε
we have
m
\
i=1
thus
Aci ⊂ (B(e, r + ε))c
B(e, r + ε) \ B(e, r) ⊂
m
[
i=1
(Ai ∩ (B(e, r + ε) \ B(e, r))).
15
Applying the translation invariance of the homogeneous balls we can estimate in the following
way:
Z
ν(B(a, r + ε) \ B(a, r)) =
=
where
Ii =
Z
Z
p(x)λ(dx) =
Z
B(a,r+ε)\B(a,r)
B(e,r+ε)\B(e,r)
p(ay)λ(dy) ≤
m
X
Ii
i=1
p(ay)λ(dy)
Ai ∩(B(e,r+ε)\B(e,r))
= P
j6=i
|yj |d/dj ≤c0 r d
Z
p(ay)λ(dy).
{yi :r≤|y|<r+ε}
Let us fix y1 , . . . , yi−1 , yi+1 , . . . , ym ∈ R. We remark that
{yi : r ≤ |y| < r + ε} = (−bi , −ai ] ∪ [ai , bi )
where
ai (y) = (r d −
X
|yj |d/dj )di /d
X
bi (y) = ((r + ε)d −
|yj |d/dj )di /d .
j6=i
j6=i
For y ∈ Ai ∩ (B(e, r + ε) \ B(e, r)) we have (taking also into account r ≥ c1 ε):
bi (y) − ai (y) ≤ ε supr≤z≤r+ε di z d−1 (z d −
X
j6=i
|yj |d/dj )di /d−1
≤ εdi (r + ε)d−1 (1 − c0 )di /d−1 rdi −d
≤ cεr di −1 .
Applying Theorem B we obtain
Z
bi
ai
p(ay)dyi ≤ cεr di −1 supai ≤yi ≤bi exp{−C|ay|2 }.
The supremum can be estimated in the following way:
Z
2
supai ≤yi ≤bi exp{−C|ay| } ≤
|∂i exp{−C|ay|2 }|dyi .
yi ≥ai
By formula (6) we obtain
∂i |ay| = |ay|1−d
We have
X
dj ≥di
d/dj −1
d−1
j (ηj (ay))
∂i =
X
dk ≥di
16
X
d(I)+d(J)=dj
Pik Xk
I
J
cIJ
j η (a)∂i η (y).
where Pik is a homogeneous polynomial of degree dk − di (cf. p.25 in [5]). Since Xk η J is
homogeneous of degree d(J) − dk Lemma 1 implies
X
X
|∂i η J (y)| ≤
|Pik (y)||Xk η J (y)| ≤ c
|y|dk −di |y|d(J)−dk ≤ c, |y|d(J)−di .
dk ≥di
dk ≥di
If d(J) < di then clearly ∂i η J (y) = 0 for all y ∈ G. Using also (4) we obtain
X
X
∂i |ay| ≤ c
|ay|1−dj
|a|d(I) |y|d(J)−di .
dj ≥di
d(I)+d(J)=dj
Collecting the above estimates we have
Z
Z
di −1
Ii ≤ cεr
P
≤ cε
= cε
Z
Z
j6=i
|yj |
d/dj
≤c0
rd
|y|di −1 exp{−C|ay|2 }
exp{−C|x|2 }
≤ c, ε(1 + |a|s−1 )
X
|yi |≥ai
X
dj ≥di
|∂i exp{−C|ay|2 }|λ(dy)
dj ≥di
|x|2−dj
|ay|2−dj
X
X
d(I)+d(J)=dj
d(I)+d(J)=dj
|a|d(I) |y|d(J)−di λ(dy)
|a|d(I) |a−1 x|d(J)−1 λ(dy)
where the following integrability condition was used: if f is a measurable function on G
such that |f (x)| = O(|x|α−Q ) where α > 0 then f is integrable near 0 (cf. p.15 in [5]).
In case r < c1 ε we can simply use the boundedness of the density function p:
ν(B(a, r + ε) \ B(a, r)) ≤ cλ(B(a, r + ε) \ B(a, r)) = c, ((r + ε)Q − rQ ) ≤ c, εQ ≤ c, ε
if ε ≤ 1. In case ε > 1 the proposition is trivial, since we can choose c = 1.
#
Let us fix an infinitely differentiable function ψ : R 7→ [0, 1] with ψ(z) = 1 for z ≤ 0
and ψ(z) = 0 for z ≥ 1. Let r, ε > 0. Put
( d
|x| −(r−ε)d
if r > ε
(31)
Ψ(x) = ψ rd −(r−ε)d
0
if r ≤ ε
for x ∈ G. It is easy to show that we have
χB(e, r − ε) ≤ Ψ ≤ χB(e, r)
where χA denotes the indicator function of the set A.
LEMMA 7. Let | · | be the homogeneous norm on G defined by (30). Let r, ε > 0.
Let Ψ be defined by (31). Then for all I ∈ Zm
+ there exists a constant cI > 0 such that
∂ I
Ψ(x) ≤ cI ε−|I| r|I|−d(I)
∂η
17
for all x ∈ G.
PROOF. Let I = (i1 , . . . , im ) ∈ Zm
+ . We can assume r > ε. Since Ψ(x) = 0 for
x ∈ G with |x| ≤ r − ε or |x| ≥ r it can be assumed r − ε ≤ |x| ≤ r. We have
∂ I
X
Ym |ηk (x)|jk d/dk −ik
Ψ(x) ≤ c(I)
.
k=1 (r d − (r − ε)d )jk
il dl /d≤jl ≤il , 1≤l≤m
∂η
For jk ≤ ik we have 1 − (1 − ε/r)d ≥ ε/r ≤ (ε/r)ik /jk , thus (r d − (r − ε)d )jk ≥ εik rjk d−ik .
Applying also the inequality |ηk (x)| ≤ |x|dk ≤ rdk we obtain the assertion. #
7. Rate of convergence in CLT on homogeneous balls.
following ‘smoothing inequality’ (cf. [11], [13, Chapter 5, Lemma 1.1]).
We shall use the
LEMMA 8. Let µ1 and µ2 be probability measures on a measurable space (Ω, F).
Let A, A1 , A2 ∈ F such that A1 ⊂ A ⊂ A2 and let ϕ1 and ϕ2 be measurable functions on
(Ω, F) with χA ≤ ϕ1 ≤ χA ≤ ϕ2 ≤ χA . Then
1
2
Z
|µ1 (A) − µ2 (A)| ≤ max ϕi (x)(µ1 − µ2 )(dx) + min µi (A2 \ A1 ).
i=1,2
i=1,2
Ω
TEOREM 2. Let | · | be a homogeneous norm on G with the following two properties:
(i) There exists a constant c > 0 such that
ν(B(a, r + ε) \ B(a, r)) ≤ cε(1 + |a|)s−1
(32)
for all a ∈ G, r > 0 and ε > 0;
(ii) for r > ε > 0 there exists a function Ψr,ε : G 7→ [0, 1] such that
χB(e, r − ε) (x) ≤ Ψr,ε (x) ≤ χB(e, r) (x)
for all x ∈ G, and
∂ I
Ψr,ε (x) ≤ cI ε−|I| r|I|−d(I)
∂η
for all x ∈ G and for all I ∈ Zm
+ whith d(I) ≤ 3s.
(33)
(34)
Then there exist a constant C > 0 with the following property: if µ is a probability
R
R
measure on G such that
ηi (x)µ(dx) = 0 when di = 1, 2,
ηi (x)ηj (x)µ(dx) = δij when
R
3s/di
di = dj = 1,
|ηi (x)|
µ(dx) < ∞ when i = 1, . . . , m, then for all n ≥ 4, a ∈ G and
r > 0 one has
δ1/√n (µn )(B(a, r)) − ν(B(a, r)) ≤ Cκ(a, r)(β3 (µ, ν) + β3s (µ, ν))n−1/2
(35)
18
where
κ(a, r) = (1 + |a|)s−1 (1 + (1 + |a|)/r})3(s−1) .
REMARK. Lemma 6 and Lemma 7 show that there exist always homogeneous norms
with properties (i) and (ii) on arbitrary stratified groups.
PROOF. We use again induction on n and the method of composition and follow the
ideas of the proof of Theorem 1 in [2]. If β3s > 1 then using Lemma 4 we have
|(µ − ν)(B)| ≤
1
Q/(Q+3s)
|µ − ν|(G) ≤ cβ3s
≤ cβ3s
2
for all B ∈ B(G), thus (35) is true for n = 1 (choosing C ≥ c), and we begin the induction
with n = 1.
If β3s ≤ 1 then as at the beginning of the proof of Theorem 1
X3s
4Q/(Q+3s)
|(δ1/√4 (µ4 ) − ν)(B)| ≤ c(β3s
+
βj )
j=3
X3s
βj ≤ c,, (β3 + β3s )
≤ c,
j=3
for all B ∈ B(G), since βj ≤ cj (β3 + β3s ) for 3 ≤ j ≤ 3s. Thus (35) is true for n = 4
(choosing C ≥ 2c), and we begin the induction with n = 4.
Now we assume that
√ k
δ1/ k (µ )(B(a, r)) − ν(B(a, r)) ≤ Cκ(a, r)(β3 (µ, ν) + β3s (µ, ν))k −1/2 = Sk
(36)
for a ∈ G, r > 0 and k ≤ n − 1.
Let now a ∈ G and r, ε > 0. Denote by Ψr,ε the function on G with properties
(33) and (34) when r > ε and put Ψr,ε (x) ≡ 0 for x ∈ G when r ≤ ε. Put Ψ1 = Ψr,ε
and Ψ2 = Ψr+ε,ε . Put ϕi (x) = Ψi (a−1 x) for x ∈ G, i = 1, 2. Then
χB(a, r − ε) ≤ ϕ1 ≤ χB(a, r) ≤ ϕ2 ≤ χB(a, r + ε) .
(37)
Applying Lemma 8 and the assumption (32) we obtain
δ1/√n (µn )(B(a, r)) − ν(B(a, r)) ≤
Z
n
√
≤ max ϕi (x)(δ1/ n µ − ν)(dx) + cε(1 + |a|)s−1 .
i=1,2
We shall write from now on ϕ and Ψ instead of ϕi and Ψi .
19
(38)
We use again the decomposition (22). First we suppose that ε < min{1, r/2} and we
R
shall give an estimate for
ϕ(x)(γk + γ̃k )(dx) if 1 ≤ k ≤ [n/2]. We have
Z
ZZZ
ϕ(x)(γk + γ̃k )(dx) =
ϕ(yuz)δ1/√n ν k−1 (dy)δ1/√n (µ − ν)(du)δ1/√n µn−k (dz).
Now we use the Taylor inequality for the function
ZZ
f (u) =
ϕ(yuz)δ1/√n ν k−1 (dy)δ1/√n µn−k (dz),
u∈G
at e ∈ G. Put gyz (u) = ϕ(yuz) = Ψ(a−1 yuz) for y, u, z ∈ G. First we show that f ∈ C 3
and
ZZ
I
X f (u) =
X I gyz (u)δ1/√n ν k−1 (dy)δ1/√n µn−k (dz)
(39)
for all u ∈ G and I ∈ Zm
with d(I) ≤ 3. Since the map (y, u, z) → ηk (yuz)
+
is a polynomial on G × G × G which is jointly homogeneous of degree dk (that is,
ηk (δr (y)δr (u)δr (z)) = r dk ηk (yuz)) we have
X
ηk (a−1 yuz) =
η I (a−1 y)η J (u)η K (z)
cIJK
k
d(I)+d(J)+d(K)=dk
for some constants cIJK
. Therefore
k
∂
∂η
J
gyz (u) =
X
|K|≤|J|,d(K)≥d(J)
PJK (a
−1
y, u, z)
∂
∂η
K
Ψ(a−1 yuz)
(40)
where PJK is a polynomial on G×G×G which is jointly homogeneous of degree d(K)−d(J).
Moreover,
J
X
∂
I
QIJ (u)
X gyz (u) =
gyz (u)
(41)
|J|≤|I|,d(J)≥d(I)
∂η
where QIJ is a homogeneous polynomial on G of degree d(J) − d(I).
Let us consider the case gyz (u) = ϕ2 (yuz) = Ψ2 (a−1 yuz). The case gyz (u) =
ϕ1 (yuz) = Ψ1 (a−1 yuz) can be handled similarly. The assumption (33) implies that
Ψ2 (a−1 yuz) = 0
if |a−1 yuz| > r + ε. If |a−1 yuz| ≤ r + ε we have
|z| ≤ c(|u−1 y −1 a| + |a−1 yuz|) ≤ c(|u| + |a−1 y| + r + ε).
(42)
For K ∈ Zm
+ with |K| ≤ |I| ≤ d(I) ≤ 3 we have d(K) ≤ 3s, and assumption (34) implies
∂ K
−1
Ψ2 (a yuz) ≤ cK ε−|K| (r + ε)|K|−d(K) .
(43)
∂η
20
By (40)—(43) and applying Theorem B we obtain
ZZ
|X I gyz (u)|δ1/√n ν k−1 (dy)δ1/√n µn−k (dz) ≤ c(I, ε, r, R)
for u ∈ G with |u| ≤ R, which implies (39).
Denote P (2) the left Taylor polynomial of f at e ∈ G of homogeneous degree 2. By
the moment conditions we have again
Z
P (2) (x)δ1/√n (µ − ν)(dx) = 0.
Thus by Lemma 2 we obtain
Z
where
ϕ(x)(γk + γ̃k )(dx) = I0 − I1 − I2 + I3
Z
(f (u) − f (e))δ1/√n (µ − ν)(du)
|u|>1
Z
X
I1 =
ηi (u)Xi f (e)δ1/√n (µ − ν)(du)
di =1,2 |u|>1
Z
X
I2 =
ηi (u)ηj (u)Xi Xj f (e)δ1/√n (µ − ν)(du)
di =dj =1 |u|>1
Z
I3 =
(f (u) − P (2) (u))δ1/√n (µ − ν)(du).
I0 =
|u|≤1
Using |ϕ(yuz) − ϕ(yz)| ≤ 1 for y, u, z ∈ G we obtain |f (u) − f (e)| ≤ 1 for u ∈ G thus
|I0 | ≤ |δ1/√n (µ − ν)|({u ∈ G : |u| > 1}) ≤ cβ3 n−3/2 .
Now we shall estimate
I
X gyz (e) =
∂
∂η
I
gyz (e).
We consider again only the case gyz (u) = ϕ2 (yuz) = Ψ2 (a−1 yuz). We have
∂
∂η
K
Ψ2 (a−1 yz) = 0
if |a−1 yz| ≤ r or |a−1 yz| ≥ r + ε, that is, if z ∈
/ B(y −1 a, r + ε) \ B(y −1 a, r). Thus
δ1/√n µ
n−k
(
z∈G:
∂
∂η
I
gyz (e) 6= 0
)!
≤ δ1/√n µn−k (B(y −1 a, r + ε) \ B(y −1 a, r)).
21
Using ε < r/2, k ≤ [n/2] and the inductive hypothesis (36) for n − k we have
|δ1/√n (µn−k − ν n−k )(B(y −1 a, r + ε))|
= |δ1/√n−k (µn−k − ν n−k )(δ√n/(n−k) (B(y −1 a, r + ε)))|
p
≤ Cκ δ√n/(n−k) (y −1 a), (r + ε) n/(n − k) (β3 + β3s )(n − k)−1/2
≤ cC(1 + |a| + |y|)s−1 (1 + (1 + |a| + |y|)/r)3(s−1) (β3 + β3s )n−1/2 .
In the same way we obtain the same estimate for |δ1/√n (µn−k − ν n−k )(B(y −1 a, r))|. From
the assumption (32) follows
ν δ√n/(n−k) B(y −1 a, r + ε) \ B(y −1 a, r) ≤ cε(1 + |a| + |y|)s−1 .
Consequently for fixed a, y ∈ G we have
δ1/√n µn−k z ∈ G : X I gyz (e) 6= 0
≤ c(1 + |a| + |y|)s−1 ε + C(1 + (1 + |a| + |y|)/r)3(s−1) (β3 + β3s )n−1/2 .
For z ∈ B(y −1 a, r + ε) \ B(y −1 a, r) we have |z| ≤ c(|a−1 y| + r + ε) consequently by (40)
and (43)
∂ I
|X I gyz (e)| = gyz (e) ≤
∂η
X
≤ c(I)
|a−1 y|i |a−1 y|j + (r + ε)j ε−|J| (r + ε)|J|−d(J) .
|J|≤|I|,i+j=d(J)−d(I)≥0
We have ε−|J| (r + ε)|J|−d(J) ≤ ε−d(I) (r + ε)d(I)−d(J) , since |J| − d(I) ≤ |J| − |I| ≤ 0.
Theorem B and the above estimates imply
ZZ
I
|X f (e)| ≤
|X I gyz (e)|δ1/√n ν k−1 (dy)δ1/√n µn−k (dz) ≤
≤ c(I)(1 + |a|)s−1 ε + C(1 + (1 + |a|)/r)3(s−1) (β3 + β3s )n−1/2 ε−d(I) ×
X
×
(1 + |a|i ) 1 + |a|j + (r + ε)j (r + ε)d(I)−d(J)
|J|≤|I|,i+j=d(J)−d(I)≥0
≤ c(I)ε−d(I) (1 + (1 + |a|)/r)(s−1)d(I) (Sn + ε(1 + |a|)s−1 )
since |J| ≤ |I| implies d(J) ≤ s|J| ≤ s|I| ≤ sd(I). We have for I ∈ Zm
+ with d(I) ≤ 3
Z
I
|u|>1
≤
Z
|η (u)| |δ1/√n (µ − ν)|(du) ≤
3
|u| |δ1/
√
n (µ
Z
− ν)|(du) ≤ cβ3 n
22
|u|>1
−3/2
.
|u|d(I) |δ1/√n (µ − ν)|(du) ≤
(44)
Collecting the above estimates we obtain for I ∈ Zm
+ with d(I) ≤ 3
Z
η I (u)X I f (e)δ1/√n (µ − ν)(du) ≤
|u|>1
≤ c(I)ε−d(I) (1 + (1 + |a|)/r)(s−1)d(I) (Sn + ε(1 + |a|)s−1 )β3 n−3/2 .
Thus using ε < 1 we can conclude
|Ii | ≤ cε−3 (1 + (1 + |a|)/r)3(s−1) (Sn + ε(1 + min{|a|, r})s−1 )β3 n−3/2
for i = 1, 2. By Theorem A we have
|f (u) − P (2) (u)| ≤ |u|3 supd(I)=3,|x|≤b3 |u| |X I f (x)|
for all u ∈ G. Applying ε < min{1, r/2} it can be proved as above that for fixed a, y, u ∈ G
with |u| ≤ 1 we have
δ1/√n µn−k z ∈ G : X I gyz (u) 6= 0
≤ δ1/√n µn−k (B(u−1 y −1 a, r + ε) \ B(u−1 y −1 a, r))
3(s−1)
≤ c(1 + |a| + |y| + |u|)s−1 ε + C (1 + (1 + |a| + |y| + |u|)/r)
(β3 + β3s )n−1/2
≤ c(1 + |a| + |y|)s−1 ε + C(1 + (1 + |a| + |y|)/r)3(s−1) (β3 + β3s )n−1/2 .
Thus using (40)—(43) one can show
|I3 | ≤ cε−3 (1 + (1 + |a|)/r)3(s−1) (Sn + ε(1 + |a|)s−1 )β3 n−3/2
Consequently
Z
ϕ(x)(γk + γ̃k )(dx) ≤ |I0 | + |I1 | + |I2 | + |I3 | ≤
≤ cε
−3
(1 + (1 + |a|)/r)
3(s−1)
(Sn + ε(1 + |a|)
s−1
)β3 n
(45)
−3/2
+ cβ3 n
−3/2
if ε < min{1, r/2} and 1 ≤ k ≤ [n/2].
R
Now we give another estimate for
ϕ(x)(γk + γ̃k )(dx) if ε < min{1, r/2} and
1 ≤ k ≤ [n/2]. First we observe
Z
ZZ
ϕ(x)γk (dx) =
l(v)pt (vu−1 )λ(dv)δ1/√n (µ − ν)(du)
R
where t = (k − 1)/n and l(v) = ϕ(vz)δ1/√n (µn−k − ν n−k )(dz), v ∈ G. We use the
R
Taylor inequality for the function f (u) = l(v)pt (vu−1 )λ(dv), u ∈ G at e ∈ G. Denote
P (2) (u) the left Taylor polynomial of f at e ∈ G of homogeneous degree 2. We have again
Z
ϕ(x)γk (dx) = I0 − I1 − I2 + I3
23
where
Z
(f (u) − f (e))δ1/√n (µ − ν)(du)
Z
X
I1 =
ηi (u)Xi f (e)δ1/√n (µ − ν)(du)
di =1,2 |u|>1
Z
X
I2 =
ηi (u)ηj (u)Xi Xj f (e)δ1/√n (µ − ν)(du)
di =dj =1 |u|>1
Z
I3 =
(f (u) − P (2) (u))δ1/√n (µ − ν)(du).
I0 =
|u|>1
|u|≤1
From |ϕ| ≤ 1 it follows |l| ≤ 1 and |f | ≤ 1, thus
Z
|I0 | ≤ 2
|δ1/√n (µ − ν)|(du) ≤ cβ3 n−3/2 .
|u|>1
Put gvt (u) = pt (vu−1 ) for u, v ∈ G and t > 0. We have
Z
I
X f (u) = l(v)X I gvt (u)λ(dv)
for all x ∈ G, I ∈ Zm
+.
First we shall estimate
I
X f (e) =
We have
∂
∂η
I
gvt (e) =
Z
X
l(v)
∂
∂η
I
gvt (e)λ(dv).
|J|≤|I|,d(J)≥d(I)
PIJ (v)
∂
∂η
J
pt (v)
(46)
where PIJ is a homogeneous polynomial of degree d(J)−d(I). Using |l| ≤ 1 and Corollary
1 we obtain
Z
I
∂
l(v)
gvt (e)λ(dv) ≤
|v|>1
∂η
Z
∂ J
X
d(J)−d(I) |v|
≤ cI
pt (v) λ(dv)
|J|≤|I|,d(J)≥d(I) |v|>1
∂η
≤ cI .
For |v| ≤ 1 we have
|l(v)| ≤ c(Sn + ε(1 + |a|)s−1 )
applying the inductive hypothesis (36) for n − k and using ε < min{1, r/2}.
Corollary 1 and (46)
Z
I
∂
gvt (e)λ(dv) ≤ cI (Sn + ε(1 + |a|)s−1 )t−d(I)/2 .
l(v)
|v|≤1
∂η
24
(47)
Thus by
Consequently by (44) we have for I ∈ Zm
+ with d(I) ≤ 3
Z
η I (u)X I f (e)δ1/√n (µ − ν)(du) ≤
|u|>1
≤ cβ3 n−3/2 + cβ3 (k − 1)−3/2 (Sn + ε(1 + |a|)s−1 ).
Thus we can conclude
|Ii | ≤ cβ3 (k − 1)−3/2 (Sn + ε(1 + |a|)s−1 ) + cβ3 n−3/2
for i = 1, 2. By Theorem A we have
|f (u) − P (2) (u)| ≤ |u|3 supd(I)=3,|x|≤b3 |u| |X I f (x)|
(48)
for all u ∈ G. Using |l| ≤ 1 and Corollary 1 we have for u ∈ G with |u| ≤ 1, x ∈ G
with |x| ≤ b3 |u| and for a sufficiently large constant c > 0
Z
l(v)X I gvt (x)λ(dv) ≤
|v|>c
Z
∂ K
X
d(K)−d(J) −1 ≤ cI
1
+
|v|
p
(vx
)
λ(dv)
t
|K|≤|J|≤|I|
∂η
d(K)≥d(J)≥d(I) |v|>c
Z
∂ K
X
d(K)−d(J) ≤ cI
1
+
|w|
p
(w)
λ(dw)
t
|K|≤|J|≤|I|
∂η
d(K)≥d(J)≥d(I) |wx|>c
≤ cI
since |x| ≤ b3 and |wx| > c imply |wx| ≤ c0 (|w| + |x|) ≤ c0 (|w| + b3 ) thus |w| ≥
−1
3
3
c−1
0 |wx| − b > c0 c − b ≥ 1 if c is sufficiently large. Moreover, for |v| ≤ c we have again
(47), thus by Corollary 1 we have for u ∈ G with |u| ≤ 1 and x ∈ G with |x| ≤ b3 |u| as
in the proof of Theorem 1
Z
I
l(v)X gvt (x)λ(dv) ≤
|v|≤c
X
d(K)−d(I) −d(K)/2
d(J)−d(I) −d(J)/2
.
|x|
t
+
|x|
t
≤ cI (Sn + ε(1 + |a|)s−1 )
|K|≤|J|≤|I|
d(K)≥d(J)≥d(I)
Consequently by (44) and (48)
Z
|I3 | ≤
|f (u) − P (2) (u)| |δ1/√n (µ − ν)|(du)
|u|≤1
≤ cβ3 n−3/2 + c(Sn + ε(1 + |a|)s−1 )
25
X3(s−1)
j=0
β3+j (k − 1)−(3+j)/2 .
Collecting the estimates we conclude
Z
ϕ(x)γk (dx) ≤ c(Sn + ε(1 + |a|)s−1 )(β3 + β3s )(k − 1)−3/2 + cβ3 n−3/2 .
(49)
We obtain as in the proof of Theorem 1 (using |l| ≤ 1)
Z
X3s−3
ϕ(x)γ̃k (dx) ≤ c
β3+j n−(3+j)/2 ≤ c(β3 + β3s )n−3/2
j=0
for 1 ≤ k ≤ n. The same estimate is valid for
R
(50)
ϕ(x)(γk + γ̃k )(dx) in case [n/2] ≤ k ≤ n.
Let m ∈ {1, 2, . . . , [n/2]}. For k = 1, 2, . . . , m we use the estimate (45) :
Xm Z
ϕ(x)(γk + γ̃k )(dx) ≤ c(β3 + β3s )n−1/2 +
k=1 + cε−3 (1 + (1 + |a|)/r)
3(s−1)
(Sn + ε(1 + |a|)s−1 )(β3 + β3s )mn−3/2 .
For k = m + 1, . . . , [n/2] we use (49) :
Z
ϕ(x)(γk + γ̃k )(dx) ≤
k=m+1
X[n/2]
≤ c(Sn + ε(1 + |a|)s−1 )(β3 + β3s )m−1/2 + c(β3 + β3s )n−1/2 .
For k = [n/2] + 1, . . . , n we use (50) :
Z
ϕ(x)(γk + γ̃k )(dx) ≤ c(β3 + β3s )n−1/2 .
k=[n/2]+1
Xn
Let K > 0 be a large enough constant (we shall specify it later) and
ε = K(1 + (1 + |a|)/r)s−1 (β3 + β3s )n−1/2 .
Using (38) and collecting the above estimates we obtain
δ1/√n (µn )(B(a, r)) − ν(B(a, r)) ≤
c (K −1 + C −1 )(K −2 (β3 + β3s )−2 m + K(β3 + β3s )m−1/2 ) + C −1 (K + 1) Sn .
From Theorem 1 it follows that if β3 + β3s ≤ C1 then
δ1/√n (µn )(B(a, r)) − ν(B(a, r)) ≤ C2 (β3 + β3s )n−1/2 ≤ C2 C −1 Sn
for all n ≥ 4.
26
(51)
If C1 ≤ (β3 + β3s ) < K −1 then we choose m = 1 and we have
δ1/√n (µn )(B(a, r)) − ν(B(a, r)) ≤ c (K −1 + C −1 )(K −2 C −2 + 1) + C −1 (K + 1) Sn .
1
(52)
If β3 + β3s > K −1 [n/2]1/2 then we choose m = n and we have
δ1/√n (µn )(B(a, r)) − ν(B(a, r)) ≤ c K −1 + C −1 (K + 2) Sn .
(53)
If K −1 < β3 + β3s ≤ K −1 [n/2]1/2 then there exists m ∈ {1, 2, . . . , [n/2]}. such that
m − 1 < K 2 (β3 + β3s )2 ≤ m
and then
δ1/√n (µn )(B(a, r)) − ν(B(a, r)) ≤ c 3K −1 + C −1 (K + 4) Sn .
(54)
If ε = K(1 + (1 + |a|)/r)s−1 (β3 + β3s )n−1/2 ≥ r/2 then we need a new estimate of the
R
integral
ϕ(x)(γk + γ̃k )(dx). We use the inequality (38) with ε = 1 :
Z
δ1/√n (µn )(B(a, r)) − ν(B(a, r)) ≤ max ϕi (x)(δ1/√n µn − ν)(dx) + c(1 + |a|)s−1 ,
i=1,2 noting that now in the definition of Ψi one should replace ε with 1. Using again the Taylor
inequality for the function
ZZ
f (u) =
ϕ(yuz)δ1/√n ν k−1 (dy)δ1/√n µn−k (dz),
u∈G
at e ∈ G we obtain
Z
Z
ϕ(x)(γk + γ̃k )(dx) ≤ |f (u) − P (2) (u)| |δ1/√n (µ − ν)|(du)
where P (2) is the left Taylor polynomial of f at e ∈ G of homogeneous degree 2. We
obtain
Z
ϕ(x)(γk + γ̃k )(dx) ≤ c(1 + (1 + |a|)/r)3(s−1) (β3 + β3s )n−3/2 .
Consequently from 1 < 2r −1 ε = 2r −1 K(1 + (1 + |a|)/r)s−1 (β3 + β3s )n−1/2 we conclude
δ1/√n (µn )(B(a, r)) − ν(B(a, r)) ≤ cC −1 (K + 1)Sn .
(55)
If ε = K(1 + (1 + |a|)/r)s−1 (β3 + β3s )n−1/2 ≥ 1 then we have
δ1/√n (µn )(B) − ν(B) ≤ 1 ≤ K(1 + (1 + |a|)/r)s−1 (β3 + β3s )n−1/2 ≤ C −1 KSn
27
(56)
From (51)—(56) it follows that δ1/√n (µn )(B(a, r)) − ν(B(a, r)) ≤ Sn if we first fix a
sufficiently large K > 0, then choose a sufficiently large C > 0. The induction is complete.
#
COROLLARY 3. The statement of Theorem 2 remains true if we replace β 3 (µ, ν) +
β3s (µ, ν) in (35) by M3 (µ, ν) + M3s (µ, ν).
ACKNOWLEDGEMENT. The author is grateful to Professor Herbert Heyer and to Professor Eberhard Siebert for supporting his work.
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MATHEMATICAL INSTITUTE
LAJOS KOSSUTH UNIVERSITY
PF. 12
H-4010 DEBRECEN, HUNGARY
29