5.1 Proof of Theorem 2.1
We prove Theorem 2.1 in the case σ = 1. The general case follows by dividing the random variables
by σ. Since ζ
r
P
aX
, P
aY
= |a|
r
ζ
r
P
X
, P
Y
, it is enough to bound up ζ
r
P
S
n
, G
n
. We first give an upper bound for
ζ
p,N
:= ζ
p
P
S
2N
, G
2
N
.
Proposition 5.1. Let X
i i
∈Z
be a stationary martingale differences sequence in L
p
for p in ]2, 3]. Let M
p
= E|X |
p
. Then for any natural integer N , 2
−2Np
ζ
2 p
p,N
≤ M
p
+ 1
2 p
2
N
X
K=0
2
Kp 2−2
kZ
K
k
1,Φ,p 2
p
+ 2
p ∆
N
, 5.1
where Z
K
= ES
2 2
K
|F − ES
2 2
K
and ∆
N
= P
N −1
K=0
2
−2Kp
kZ
K
k
p 2
.
Proof of Proposition 5.1. The proof is done by induction on N . Let Y
i i
∈N
be a sequence of N 0, 1-distributed independent random variables, independent of the sequence X
i i
∈Z
. For m 0,
let T
m
= Y
1
+ Y
2
+ · · · + Y
m
. Set S = T
= 0. For any numerical function f and m ≤ n, set f
n −m
x = E f x + T
n
− T
m
. Then, from the independence of the above sequences,
E f S
n
− f T
n
=
n
X
m=1
D
m
with D
m
= E f
n −m
S
m −1
+ X
m
− f
n −m
S
m −1
+ Y
m
. 5.2
For any two-times differentiable function g, the Taylor integral formula at order two writes gx + h
− gx = g
′
xh + 1
2 h
2
g
′′
x + h
2
Z
1
1 − tg
′′
x + th − g
′′
xd t. 5.3
Hence, for any q in ]2, 3], |gx + h − gx − g
′
xh − 1
2 h
2
g
′′
x| ≤ h
2
Z
1
1 − t|th|
q −2
|g|
Λ
q
d t ≤
1 qq
− 1 |h|
q
|g|
Λ
q
. 5.4
Let D
′ m
= E f
′′ n
−m
S
m −1
X
2 m
− 1 = E f
′′ n
−m
S
m −1
X
2 m
− Y
2 m
From 5.4 applied twice with g = f
n −m
, x = S
m −1
and h = X
m
or h = Y
m
together with the martingale property,
D
m
− 1
2 D
′ m
≤ 1
pp − 1
| f
n −m
|
Λ
p
E |X
m
|
p
+ |Y
m
|
p
. Now E
|Y
m
|
p
≤ p − 1 ≤ p − 1M
p
. Hence |D
m
− D
′ m
2| ≤ M
p
| f
n −m
|
Λ
p
5.5 Assume now that f belongs to Λ
p
. Then the smoothed function f
n −m
belongs to Λ
p
also, so that | f
n −m
|
Λ
p
≤ 1. Hence, summing on m, we get that E
f S
n
− f T
n
≤ nM
p
+ D
′
2 where D
′
= D
′ 1
+ D
′ 2
+ · · · + D
′ n
. 5.6
Suppose now that n = 2
N
. To bound up D
′
, we introduce a dyadic scheme. 994
Notation 5.2. Set m = m − 1 and write m
in basis 2: m =
P
N i=0
b
i
2
i
with b
i
= 0 or b
i
= 1 note that b
N
= 0. Set m
L
= P
N i=L
b
i
2
i
, so that m
N
= 0. Let I
L,k
=]k2
L
, k + 12
L
] ∩ N note that I
N ,1
=]2
N
, 2
N +1
], U
k L
= P
i ∈I
L,k
X
i
and ˜ U
k L
= P
i ∈I
L,k
Y
i
. For the sake of brevity, let U
L
= U
L
and ˜
U
L
= ˜ U
L
. Since m
N
= 0, the following elementary identity is valid D
′ m
=
N −1
X
L=0
E f
′′ n
−1−m
L
S
m
L
− f
′′ n
−1−m
L+1
S
m
L+1
X
2 m
− 1 .
Now m
L
6= m
L+1
only if b
L
= 1, then in this case m
L
= k2
L
with k odd. It follows that D
′
=
N −1
X
L=0
X
k ∈IN−L,0
k odd
E f
′′ n
−1−k2
L
S
k2
L
− f
′′ n
−1−k−12
L
S
k−12
L
X
{m:m
L
=k2
L
}
X
2 m
− σ
2
. 5.7
Note that {m : m
L
= k2
L
} = I
L,k
. Now by the martingale property, E
k2
L
X
i ∈I
L,k
X
2 i
− σ
2
= E
k2
L
U
k L
2
− EU
k L
2
:= Z
k L
. Consequently
D
′
=
N −1
X
L=0
X
k ∈IN−L,0
k odd
E f
′′ n
−1−k2
L
S
k2
L
− f
′′ n
−1−k−12
L
S
k−12
L
Z
k L
=
N −1
X
L=0
X
k ∈IN−L,0
k odd
E f
′′ n
−1−k2
L
S
k2
L
− f
′′ n
−1−k2
L
S
k−12
L
+ T
k2
L
− T
k−12
L
Z
k L
, 5.8
since X
i i
∈N
and Y
i i
∈N
are independent. By using 1.2, we get that D
′
≤
N −1
X
L=0
X
k ∈IN−L,0
k odd
E |U
k−1 L
− ˜ U
k−1 L
|
p −2
|Z
k L
| . From the stationarity of X
i i
∈N
and the above inequality, D
′
≤ 1
2
N −1
X
K=0
2
N −K
E |U
K
− ˜ U
K
|
p −2
|Z
1 K
|. 5.9
Now let V
K
be the N 0, 2
K
-distributed random variable defined from U
K
via the quantile transfor- mation, that is
V
K
= 2
K 2
Φ
−1
F
K
U
K
− 0 + δ
K
F
K
U
K
− F
K
U
K
− 0 where F
K
denotes the d.f. of U
K
, and δ
K
is a sequence of independent r.v.’s uniformly distributed on [0, 1], independent of the underlying random variables. Now, from the subadditivity of x → x
p −2
, |U
K
− ˜ U
K
|
p −2
≤ |U
K
− V
K
|
p −2
+ |V
K
− ˜ U
K
|
p −2
. Hence E
|U
K
− ˜ U
K
|
p −2
|Z
1 K
| ≤ kU
K
− V
K
k
p −2
p
kZ
1 K
k
p 2
+ E|V
K
− ˜ U
K
|
p −2
|Z
1 K
| . 5.10
995
By definition of V
K
, the real number kU
K
− V
K
k
p
is the so-called Wasserstein distance of order p between the law of U
K
and the N 0, 2
K
normal law. Therefrom, by Theorem 3.1 of Rio 2007 which improves the constants given in Theorem 1 of Rio 1998, we get that, for p
∈]2, 3], kU
K
− V
K
k
p
≤ 22p − 1ζ
p,K 1
p
≤ 24ζ
p,K 1
p
. 5.11
Now, since V
K
and ˜ U
K
are independent, their difference has the N 0, 2
K+1
distribution. Note that if Y is a N 0, 1-distributed random variable, Q
|Y |
p −2
u = Φ
−1
1 − u2
p −2
. Hence, by Fréchet’s inequality 1957 see also Inequality 1.11b page 9 in Rio 2000, and by definition of the norm
k . k
1,Φ,p
, E
|V
K
− ˜ U
K
|
p −2
|Z
1 K
| ≤ 2
K+1p2−1
kZ
K
k
1,Φ,p
. 5.12
From 5.10, 5.11 and 5.12, we get that E
|U
K
− ˜ U
K
|
p −2
|Z
1 K
| ≤ 2
p −4p
ζ
p−2p p,K
kZ
K
k
p 2
+ 2
K+1p2−1
kZ
K
k
1,Φ,p
. 5.13
Then, from 5.6, 5.9 and 5.13, we get 2
−N
ζ
p,N
≤ M
p
+ 2
p 2−3
∆
′ N
+ 2
p −2−4p
N −1
X
K=0
2
−K
ζ
p−2p p,K
kZ
K
k
p 2
, where ∆
′ N
= P
N −1
K=0
2
Kp 2−2
kZ
K
k
1,Φ,p
. Consequently we get the induction inequality 2
−N
ζ
p,N
≤ M
p
+ 1
2 p
2 ∆
′ N
+
N −1
X
K=0
2
−K
ζ
p−2p p,K
kZ
K
k
p 2
. 5.14
We now prove 5.1 by induction on N . First by 5.6 applied with n = 1, one has ζ
p,0
≤ M
p
, since D
′ 1
= f
′′
0EX
2 1
− 1 = 0. Assume now that ζ
p,L
satisfies 5.1 for any L in [0, N − 1]. Starting
from 5.14, using the induction hypothesis and the fact that ∆
′ K
≤ ∆
′ N
, we get that 2
−N
ζ
p,N
≤ M
p
+ 1
2 p
2 ∆
′ N
+
N −1
X
K=0
2
−2Kp
kZ
K
k
p 2
M
p
+ 1
2 p
2 ∆
′ N
2 p
+ 2
p ∆
K p
2−1
. Now 2
−2Kp
kZ
K
k
p 2
= ∆
K+1
− ∆
K
. Consequently 2
−N
ζ
p,N
≤ M
p
+ 1
2 p
2 ∆
′ N
+ Z
∆
N
M
p
+ 1
2 p
2 ∆
′ N
2 p
+ 2
p x
p 2−1
d x , which implies 5.1 for
ζ
p,N
. In order to prove Theorem 2.1, we will also need a smoothing argument. This is the purpose of the
lemma below.
Lemma 5.1. Let S and T be two centered and square integrable random variables with the same variance. For any r in ]0, p],
ζ
r
P
S
, P
T
≤ 2ζ
r
P
S
∗ G
1
, P
T
∗ G
1
+ 4 p
2.
996
Proof of Lemma 5.1. Throughout the sequel, let Y be a N 0, 1-distributed random variable, inde- pendent of the
σ-field generated by S, T . For r
≤ 2, since ζ
r
is an ideal metric with respect to the convolution, ζ
r
P
S
, P
T
≤ ζ
r
P
S
∗ G
1
, P
T
∗ G
1
+ 2ζ
r
δ , G
1
≤ ζ
r
P
S
∗ G
1
, P
T
∗ G
1
+ 2E|Y |
r
which implies Lemma 5.1 for r ≤ 2. For r 2, from 5.4, for any f in Λ
r
, f S
− f S + Y + f
′
SY − 1
2 f
′′
SY
2
≤ 1
rr − 1
|Y |
r
. Taking the expectation and noting that E
|Y |
r
≤ r − 1 for r in ]2, 3], we infer that E
f S − f S + Y −
1 2
f
′′
S ≤
1 r
. Obviously this inequality still holds for T instead of S and
− f instead of f , so that adding the so obtained inequality,
E f S − f T ≤ E f S + Y − f T + Y +
1 2
E f
′′
S − f
′′
T + 1. Since f
′′
belongs to Λ
r −2
, it follows that ζ
r
P
S
, P
T
≤ ζ
r
P
S
∗ G
1
, P
T
∗ G
1
+ 1
2 ζ
r −2
P
S
, P
T
+ 1. Now r
− 2 ≤ 1. Hence ζ
r −2
P
S
, P
T
= W
r −2
P
S
, P
T
≤ W
r
P
S
, P
T r
−2
. Next, by Theorem 3.1 in Rio 2007, W
r
P
S
, P
T
≤ 32ζ
r
P
S
, P
T 1
r
. Furthermore 32ζ
r
P
S
, P
T 1
−2r
≤ ζ
r
P
S
, P
T
as soon as ζ
r
P
S
, P
T
≥ 2
5r2−5
. This condition holds for any r in ]2, 3] if ζ
r
P
S
, P
T
≥ 4 p
2. Then, from the above inequalities
ζ
r
P
S
, P
T
≤ ζ
r
P
S
∗ G
1
, P
T
∗ G
1
+ 1
2 ζ
r
P
S
, P
T
+ 1, which implies Lemma 5.1.
We go back to the proof of Theorem 2.1. Let n ∈]2
N
, 2
N +1
] and ℓ = n − 2
N
. The main step is then to prove the inequalities below: for r
≥ p − 2 and r, p 6= 1, 3, for some εN tending to zero as N tends to infinity,
ζ
r
P
S
n
, G
n
≤ c
r,p
2
N r −p2
ζ
p
P
S
ℓ
, G
ℓ
+ C2
N r+2 −p2
+ 2
N r −p2+2p
εN ζ
p
P
S
ℓ
, G
ℓ p−2p
5.15 and for r = 1 and p = 3,
ζ
1
P
S
n
, G
n
≤ CN + 2
−N
ζ
3
P
S
ℓ
, G
ℓ
+ 2
−N3
ζ
3
P
S
ℓ
, G
ℓ 1
3
. 5.16
997
Assuming that 5.15 and 5.16 hold, we now complete the proof of Theorem 2.1. Let ζ
∗ p,N
= sup
n ≤2
N
ζ
p
P
S
n
, G
n
, we infer from 5.15 applied to r = p that ζ
∗ p,N +1
≤ ζ
∗ p,N
+ C2
N
+ 2
2N p
εN ζ
∗ p,N
p−2p
. Let N
be such that C εN ≤ 12 for N ≤ N
, and let K ≥ 1 be such that ζ
∗ p,N
≤ K2
N
. Choosing K large enough such that K
≥ 2C, we can easily prove by induction that ζ
∗ p,N
≤ K2
N
for any N ≥ N
. Hence Theorem 2.1 is proved in the case r = p. For r in [p
− 2, p[, Theorem 2.1 follows by taking into account the bound
ζ
∗ p,N
≤ K2
N
, valid for any N ≥ N
, in the inequalities 5.15 and 5.16. We now prove 5.15 and 5.16. We will bound up
ζ
∗ p,N
by induction on N . For n ∈]2
N
, 2
N +1
] and ℓ = n − 2
N
, we notice that ζ
r
P
S
n
, G
n
≤ ζ
r
P
S
n
, P
S
ℓ
∗ G
2
N
+ ζ
r
P
S
ℓ
∗ G
2
N
, G
ℓ
∗ G
2
N
. Let
φ
t
be the density of the law N 0, t
2
. With the same notation as in the proof of Proposition 5.1, we have
ζ
r
P
S
ℓ
∗ G
2
N
, G
ℓ
∗ G
2
N
= sup
f ∈Λ
r
E f
2
N
S
ℓ
− f
2
N
T
ℓ
≤ | f ∗ φ
2
N 2
|
Λ
p
ζ
p
P
S
ℓ
, G
ℓ
. Applying Lemma 6.1, we infer that
ζ
r
P
S
n
, G
n
≤ ζ
r
P
S
n
, P
S
ℓ
∗ G
2
N
+ c
r,p
2
N r −p2
ζ
p
P
S
ℓ
, G
ℓ
. 5.17
On the other hand, setting ˜ S
ℓ
= X
1 −ℓ
+ · · · + X , we have that S
n
is distributed as ˜ S
ℓ
+ S
2
N
and, S
ℓ
+ T
2
N
as ˜ S
ℓ
+ T
2
N
. Let Y be a N 0, 1-distributed random variable independent of X
i i
∈Z
and Y
i i
∈Z
. Using Lemma 5.1, we then derive that ζ
r
P
S
n
, P
S
ℓ
∗ G
2
N
≤ 4 p
2 + 2 sup
f ∈Λ
r
E f ˜
S
ℓ
+ S
2
N
+ Y − f ˜S
ℓ
+ T
2
N
+ Y . 5.18
Let D
′ m
= E f
′′ 2
N
−m+1
˜ S
ℓ
+ S
m −1
X
2 m
− 1. We follow the proof of Proposition 5.1. From the Taylor expansion 5.3 applied twice with g = f
2
N
−m+1
, x = ˜ S
ℓ
+ S
m −1
and h = X
m
or h = Y
m
together with the martingale property, we get that
E f ˜
S
ℓ
+ S
2
N
+ Y − f ˜S
ℓ
+ T
2
N
+ Y =
2
N
X
m=1
E f
2
N
−m+1
˜ S
ℓ
+ S
m −1
+ X
m
− f
2
N
−m+1
˜ S
ℓ
+ S
m −1
+ Y
m
= D
′ 1
+ · · · + D
′ 2
N
2 + R
1
+ · · · + R
2
N
, 5.19
where, as in 5.5, R
m
≤ M
p
| f
2
N
−m+1
|
Λ
p
. 5.20
In the case r = p − 2, we will need the more precise upper bound
R
m
≤ E X
2 m
k f
′′ 2
N
−m+1
k
∞
∧ 1
6 k f
3 2
N
−m+1
k
∞
|X
m
| +
1 6
k f
3 2
N
−m+1
k
∞
E |Y
m
|
3
, 5.21
998
which is derived from the Taylor formula at orders two and three. From 5.20 and Lemma 6.1, we have that
R := R
1
+ · · · + R
2
N
= O2
N r −p+22
if r p − 2, and R = ON if r, p = 1, 3 .
5.22 It remains to consider the case r = p
− 2 and r 1. Applying Lemma 6.1, we get that for i ≥ 2, k f
i 2
N
−m+1
k
∞
≤ c
r,i
2
N
− m + 1
r−i2
. 5.23
It follows that
2
N
X
m=1
E X
2 m
k f
′′ 2
N
−m+1
k
∞
∧ k f
3 2
N
−m+1
k
∞
|X
m
| ≤ C
∞
X
m=1
1 m
1 −r2
E X
2
1 ∧
|X |
p m
≤ CE
[X
2
]
X
m=1
X
2
m
1 −r2
+
∞
X
m=[X
2
]+1
|X |
3
m
3−r2
. Consequently for r = p
− 2 and r 1, R
1
+ · · · + R
2
N
≤ CM
p
+ E|Y |
3
. 5.24
We now bound up D
′ 1
+ · · · + D
′ 2
N
. Using the dyadic scheme as in the proof of Proposition 5.1, we get that
D
′ m
=
N −1
X
L=0
E f
′′ 2
N
−m
L
˜ S
ℓ
+ S
m
L
− f
′′ 2
N
−m
L+1
˜ S
ℓ
+ S
m
L+1
X
2 m
− 1 + E f
′′ 2
N
˜ S
ℓ
X
2 m
− 1 :=
D
′′ m
+ E f
′′ 2
N
˜ S
ℓ
X
2 m
− 1 . Notice first that
2
N
X
m=1
E f
′′ 2
N
˜ S
ℓ
X
2 m
− 1 = E f
′′ 2
N
˜ S
ℓ
− f
′′ 2
N
T
ℓ
Z
N
. 5.25
Since f belongs to Λ
r
i.e. | f |
Λ
r
≤ 1, we infer from Lemma 6.1 that | f
i
|
Λ
p
≤ C i
r−p2
which means exactly that
| f
′′ i
x − f
′′ i
y| ≤ C i
r−p2
|x − y|
p −2
. 5.26
Starting from 5.25 and using 5.26 with i = 2
N
, it follows that
2
N
X
m=1
E f
′′ 2
N
˜ S
ℓ
X
2 m
− 1 ≤ C2
N r −p2
E |˜S
ℓ
− T
ℓ
|
p −2
|Z
N
| . Proceeding as to get 5.13 that is, using similar upper bounds as in 5.10, 5.11 and 5.12, we
obtain that E
|˜S
ℓ
− T
ℓ
|
p −2
|Z
N
| ≤ 2
p −4p
ζ
p
P
S
ℓ
, G
ℓ p−2p
kZ
N
k
p 2
+ 2ℓ
p 2−1
kZ
N
k
1,Φ,p
.
999
Using Remark 2.6, 2.1 and 2.2 entail that kZ
N
k
p 2
= o2
2N p
and kZ
N
k
1,Φ,p
= o2
N 2 −p2
. Hence, for some
εN tending to 0 as N tends to infinity, one has
2
N
X
m=1
D
′ m
≤
2
N
X
m=1
D
′′ m
+ CεN 2
N r −p2+2p
ζ
p
P
S
ℓ
, G
ℓ p−2p
+ 2
N r+2 −p2
. 5.27
Next, proceeding as in the proof of 5.8, we get that
2
N
X
m=1
D
′′ m
≤
N −1
X
L=0
X
k ∈IN−L,0
k odd
E f
′′ 2
N
−k2
L
˜ S
ℓ
+ S
k2
L
− f
′′ 2
N
−k2
L
˜ S
ℓ
+ S
k−12
L
+ T
k2
L
− T
k−12
L
Z
k L
. 5.28
Let r p − 2 or r, p = 1, 3. Using 5.26 with i = 2
N
− k2
L
, 5.28, and the stationarity of X
i i
∈N
, we infer that
2
N
X
m=1
D
′′ m
≤ C
N −1
X
L=0
X
k ∈IN−L,0
k odd
2
N
− k2
L r−p2
E |U
L
− ˜ U
L
|
p −2
Z
1 L
.
It follows that
2
N
X
m=1
D
′′ m
≤ C2
N r+2−p2 N
X
L=0
2
−L
E U
L
− ˜ U
L p
−2
Z
1 L
if r p − 2,
5.29
2
N
X
m=1
D
′′ m
≤ C N
N
X
L=0
2
−L
E U
L
− ˜ U
L
Z
1 L
if r = 1 and p = 3. 5.30
In the case r = p − 2 and r 1, we have
2
N
X
m=1
D
′′ m
≤ C
N −1
X
L=0
X
k ∈IN−L,0
k odd
E k f
′′ 2
N
−k2
L
k
∞
∧ k f
′′′ 2
N
−k2
L
k
∞
U
L
− ˜ U
L
Z
1 L
. Applying 5.23 to i = 2 and i = 3, we obtain
2
N
X
m=1
D
′′ m
≤ C
N
X
L=0
2
r−2L2
E Z
1 L
2
N −L
X
k=1
k
r−22
1 ∧
1 2
L 2
p k
U
L
− ˜ U
L
, Proceeding as to get 5.24, we have that
2
N −L
X
k=1
k
r−22
1 ∧
1 2
L 2
p k
U
L
− ˜ U
L
≤
∞
X
k=1
k
r−22
1 ∧
1 2
L 2
p k
U
L
− ˜ U
L
≤ C2
−Lr2
|U
L
− ˜ U
L r
. It follows that
2
N
X
m=1
D
′′ m
≤ C
N
X
L=0
2
−L
E U
L
− ˜ U
L r
Z
1 L
if r = p − 2 and r 1.
5.31 1000
Now by Remark 2.6, 2.1 and 2.2 are respectively equivalent to X
K ≥0
2
Kp 2−2
kZ
K
k
1,Φ,p
∞ , and X
K ≥0
2
−2Kp
kZ
K
k
p 2
∞ . Next, by Proposition 5.1,
ζ
p,K
= O2
K
under 2.1 and 2.2. Therefrom, taking into account the inequality 5.13, we derive that under 2.1 and 2.2,
2
−L
E U
L
− ˜ U
L p
−2
Z
1 L
≤ C2
−2Lp
kZ
L
k
p 2
+ C2
Lp 2−2
kZ
L
k
1,Φ,p
. 5.32
Consequently, combining 5.32 with the upper bounds 5.29, 5.30 and 5.31, we obtain that
2
N
X
m=1
D
′′ m
= ¨
O2
N r+2 −p2
if r ≥ p − 2 and r, p 6= 1, 3
ON if r = 1 and p = 3.
5.33 From 5.17, 5.18, 5.19, 5.22, 5.24, 5.27 and 5.33, we obtain 5.15 and 5.16.
5.2 Proof of Theorem 3.1
Kategori