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
By 3.1, we get that see Volný 1993 X
= D + Z
− Z ◦ T,
5.34 where
Z =
∞
X
k=0
E X
k
|F
−1
−
∞
X
k=1
X
−k
− EX
−k
|F
−1
and D
= X
k ∈Z
E X
k
|F − EX
k
|F
−1
. Note that Z
∈ L
p
, D ∈ L
p
, D is
F -measurable, and ED
|F
−1
= 0. Let D
i
= D ◦ T
i
, and Z
i
= Z ◦ T
i
. We obtain that S
n
= M
n
+ Z
1
− Z
n+1
, 5.35
where M
n
= P
n j=1
D
j
. We first bound up E f S
n
− f M
n
by using the following lemma
Lemma 5.2. Let p ∈]2, 3] and r ∈ [p − 2, p]. Let X
i i
∈Z
be a stationary sequence of centered random variables in L
2 ∨r
. Assume that S
n
= M
n
+ R
n
where M
n
− M
n −1
n 1
is a strictly stationary sequence of martingale differences in L
2 ∨r
, and R
n
is such that ER
n
= 0. Let nσ
2
= EM
2 n
, nσ
2 n
= ES
2 n
and α
n
= σ
n
σ. 1. If r
∈ [p − 2, 1] and E|R
n
|
r
= On
r+2−p2
, then ζ
r
P
S
n
, P
M
n
= On
r+2−p2
. 2. If r
∈]1, 2] and kR
n
k
r
= On
3−p2
, then ζ
r
P
S
n
, P
M
n
= On
r+2−p2
. 3. If r
∈]2, p], σ
2
0 and kR
n
k
r
= On
3−p2
, then ζ
r
P
S
n
, P
α
n
M
n
= On
r+2−p2
. 4. If r
∈]2, p], σ
2
= 0 and kR
n
k
r
= On
r+2−p2r
, then ζ
r
P
S
n
, G
n σ
2 n
= On
r+2−p2
.
Remark 5.1. All the assumptions on R
n
in items 1, 2, 3 and 4 of Lemma 5.2 are satisfied as soon as sup
n
kR
n
k
p
∞. 1001
Proof of Lemma 5.2. For r ∈]0, 1], ζ
r
P
S
n
, P
M
n
≤ E|R
n
|
r
, which implies item 1. If f
∈ Λ
r
with r ∈]1, 2], from the Taylor integral formula and since ER
n
= 0, we get E
f S
n
− f M
n
= E R
n
f
′
M
n
− f
′
0 + Z
1
f
′
M
n
+ tR
n
− f
′
M
n
d t .
Using that | f
′
x − f
′
y| ≤ |x − y|
r −1
and applying Hölder’s inequality, it follows that E
f S
n
− f M
n
≤ kR
n
k
r
k f
′
M
n
− f
′
0k
r r−1
+ kR
n
k
r r
≤ kR
n
k
r
kM
n
k
r −1
r
+ kR
n
k
r r
. Since
kM
n
k
r
≤ kM
n
k
2
= p
n σ, we infer that ζ
r
P
S
n
, P
M
n
= On
r+2−p2
. Now if f
∈ Λ
r
with r ∈]2, p] and if σ 0, we define g by
gt = f t − t f
′
0 − t
2
f
′′
02 . The function g is then also in Λ
r
and is such that g
′
0 = g
′′
0 = 0. Since α
2 n
E M
2 n
= ES
2 n
, we have
E f S
n
− f α
n
M
n
= EgS
n
− gα
n
M
n
. 5.36
Now from the Taylor integral formula at order two, setting ˜ R
n
= R
n
+ 1 − α
n
M
n
, E
gS
n
− gα
n
M
n
= E˜ R
n
g
′
α
n
M
n
+ 1
2 E
˜ R
n 2
g
′′
α
n
M
n
+ E ˜ R
n 2
Z
1
1 − tg
′′
α
n
M
n
+ t ˜ R
n
− g
′′
α
n
M
n
d t . 5.37
Note that, since g
′
0 = g
′′
0 = 0, one has E
˜ R
n
g
′
α
n
M
n
= E ˜ R
n
α
n
M
n
Z
1
g
′′
tα
n
M
n
− g
′′
0d t Using that
|g
′′
x − g
′′
y| ≤ |x − y|
r −2
and applying Hölder’s inequality in 5.37, it follows that E
gS
n
− gα
n
M
n
≤ 1
r − 1
E |˜R
n
||α
n
M
n
|
r −1
+ 1
2 k˜R
n
k
2 r
kg
′′
α
n
M
n
k
r r−2
+ 1
2 k˜R
n
k
r r
≤ 1
r − 1
α
r −1
n
k˜R
n
k
r
kM
n
k
r −1
r
+ 1
2 α
r −2
n
k˜R
n
k
2 r
kM
n
k
r −2
r
+ 1
2 k˜R
n
k
r r
. Now
α
n
= O1 and k˜R
n
k
r
≤ kR
n
k
r
+ |1 − α
n
|kM
n
k
r
. Since |kS
n
k
2
− kM
n
k
2
| ≤ kR
n
k
2
, we infer that
|1 − α
n
| = On
2−p2
. Hence, applying Burkhölder’s inequality for martingales, we infer that k˜R
n
k
r
= On
3−p2
, and consequently ζ
r
P
S
n
, P
α
n
M
n
= On
r+2−p2
. If
σ
2
= 0, then S
n
= R
n
. Let Y be a N 0, 1 random variable. Using that E
f S
n
− f p
n σ
n
Y = EgR
n
− g p
n σ
n
Y and applying again Taylor’s formula, we obtain that
sup
f ∈Λ
r
|E f S
n
− f p
n σ
n
Y | ≤
1 r
− 1 k¯R
n
k
r
k p
n σ
n
Y k
r −1
r
+ 1
2 k¯R
n
k
2 r
k p
n σ
n
Y k
r −2
r
+ 1
2 k¯R
n
k
r r
, 1002
where ¯ R
n
= R
n
− p
n σ
n
Y . Since p
n σ
n
= kR
n
k
2
≤ kR
n
k
r
and since kR
n
k
r
= On
r+2−p2r
, we infer that
p n
σ
n
= On
r+2−p2r
and that k¯R
n
k
r
= On
r+2−p2r
. The result follows. By 5.35, we can apply Lemma 5.2 with R
n
:= Z
1
− Z
n+1
. Then for p − 2 ≤ r ≤ 2, the result follows
if we prove that under 3.1 and 3.2, M
n
satisfies the conclusion of Theorem 2.1. Now if 2 r ≤ p
and σ
2
0, we first notice that ζ
r
P
α
n
M
n
, G
n σ
2 n
= α
r n
ζ
r
P
M
n
, G
n σ
2
. Since
α
n
= O1, the result will follow by Item 3 of Lemma 5.2, if we prove that under 3.1 and 3.2, M
n
satisfies the conclusion of Theorem 2.1. We shall prove that X
n ≥1
1 n
3 −p2
kEM
2 n
|F − EM
2 n
k
p 2
∞ . 5.38
In this way, according to Remark 2.1, both 2.1 and 2.2 will be satisfied. Suppose that we can show that
X
n ≥1
1 n
3 −p2
kEM
2 n
|F − ES
2 n
|F k
p 2
∞ , 5.39
then by taking into account the condition 3.2, 5.38 will follow. Indeed, it suffices to notice that 5.39 also entails that
X
n ≥1
1 n
3 −p2
|ES
2 n
− EM
2 n
| ∞ , 5.40
and to write that kEM
2 n
|F − EM
2 n
k
p 2
≤ kEM
2 n
|F − ES
2 n
|F k
p 2
+kES
2 n
|F − ES
2 n
k
p 2
+ |ES
2 n
− EM
2 n
| . Hence, it remains to prove 5.39. Since S
n
= M
n
+ Z
1
− Z
n+1
, and since Z
i
= Z ◦ T
i
is in L
p
, 5.39 will be satisfied provided that
X
n ≥1
1 n
3 −p2
kS
n
Z
1
− Z
n+1
k
p 2
∞ . 5.41
Notice that kS
n
Z
1
− Z
n+1
k
p 2
≤ kM
n
k
p
kZ
1
− Z
n+1
k
p
+ kZ
1
− Z
n+1
k
2 p
. From Burkholder’s inequality,
kM
n
k
p
= O p
n and from 3.1, sup
n
kZ
1
− Z
n+1
k
p
∞. Conse- quently 5.41 is satisfied for any p in ]2, 3[.
5.3 Proof of Theorem 3.2
