≤ kGk
p
Z
n+1 n
kJ
r,n+1
k
2
d r
p 2
Z
n+1 n
|h
r
|
2
d r
p 2
≤ kGk
p
Z
n+1 n
kJ
r,n+1
k
p
d r |||h|||
p n
, and then applying the first bound.
In addition to these first Malliavin derivatives, we will need the control of the derivative of various objects involving the Malliavin derivative. The following lemma gives control over two objects related
to the second Malliavin derivative:
Lemma 5.13 For all p ∈ [0, ¯p2], one has the bounds
sup
s,r ∈[n,n+1]
E kD
i s
J
r,n+1
k
p
≤ exp2pηη
′ n
V u + 2pκC
L
+ pC
J
+ pC
2 J
, sup
s ∈[n,n+1]
E kD
i s
A
n
k
p
≤ |||G|||
p
exp2p ηη
′ n
V u + 2pκC
L
+ pC
J
+ pC
2 J
. Proof. For this, we note that by 35 one has the identities
D
i s
J
r,n+1
ξ =
J
2 s,n+1
J
r,s
ξ, g
i
for r ≤ s,
J
2 r,n+1
J
s,r
g
i
, ξ for s ≤ r,
D
i s
A
n
v = Z
n+1 n
D
i s
J
r,n+1
G v
r
d r . Hence if p
∈ [0, ¯p2] which by the way also ensures that 2pκ 1 it follows from Proposition 5.12 that
E kD
i s
J
r,n+1
k
p
≤ EkJ
2 s,n+1
k
2p
E kJ
r,s
k
2p
1 2
≤ E exp2pηV u
n
+ pC
J
+ pC
2 J
≤ exp2pηη
′ n
V u + 2pκC
L
+ pC
J
+ pC
2 J
for r ≤ s and similarly for s ≤ r. Since, for p ≥ 1, we can write
E kD
i s
A
n
k
p
≤ kGk
p
Z
n+1 n
E kD
i s
J
r,n+1
k
p
d r , the second estimate then follows from the first one.
5.5 Controlling the error term
ρ
t
ρ
t
ρ
t
The purpose of this section is to show that the “error term” ρ
t
= D
ξ
u
t
− D
h
ξ
u
t
goes the zero as t
→ ∞, provided that the “control” h
ξ
is chosen as explained in Section 5.3. We begin by observing that for even integer times,
ρ
n
is given recursively by ρ
2n+2
= J
2n+1
ρ
2n+1
= J
2n+1
R
β
2n
2n
J
2n
ρ
2n
, 54
692
where R
β k
is the operator R
β k
def
= 1 − M
k
β + M
k −1
= ββ + M
k −1
. Observe that
R
β k
measures the error between M
k
β + M
k −1
and the identity, which we will see is small for
β very small. This recursion is of the form ρ
2n+2
= Ξ
2n+2
ρ
2n
, with the random operator Ξ
2n+2
: H → H defined by Ξ
2n+2
= J
2n+1
R
β
2n
2n
J
2n
. Notice that Ξ
2n
is F
2n
-measurable and that Ξ
k
is defined only for even integers k. Define the n-fold product of the Ξ
2k
by Ξ
2n
=
n
Y
k=1
Ξ
2k
, so that
ρ
2n
= Ξ
2n
ξ. It is our aim to show that it is possible under the assumptions of Section 5 to choose the sequence
β
n
in an adapted way such that for a sufficiently small constant ¯ η and p ∈ [0, ¯p2] one has
E kρ
2n
k
p
≤ E kΞ
2n
k
p
kρ k
p
≤ exp p ¯ ηV u
− pn˜κ kρ
k
p
. 55
for some ˜ κ 0. This will give the needed control over the last term in 48.
By Assumption B.1, we have a bound on the Malliavin covariance matrix of the form
P inf
kΠϕk≥αkϕk
〈ϕ, M
k
ϕ〉 ≤ ǫkϕk
2
F
k
≤ Cα, p U
p
u
k
ǫ
p
. 56
Here, by the Markov property, the quantities ǫ and α do not necessarily need to be constant, but are
allowed to be F
k
-measurable random variables. In order to obtain 55, the idea is to decompose Ξ
2n+2
as Ξ
2n+2
= J
2n+1
R
β
2n
2n
J
2n
= J
2n+1
Π
⊥
R
β
2n
2n
J
2n
+ J
2n+1
ΠR
β
2n
2n
J
2n
def
= I
2n+2,1
+ I
2n+2,2
. 57
The crux of the matter is controlling the term Π R
β
2n
2n
since J
2n+1
Π
⊥
is controlled by Assumption B.4 and we know that
kR
β
2n
2n
k ≤ 1. To understand and control the I
2n+2,2
term, we explore the properties of a general operator of the form of
R
β 2n
.
Lemma 5.14 Let Π be an orthogonal projection on H and M be a self-adjoint, positive linear operator
on H satisfying for some γ 0 and δ ∈ 0, 1]
inf
ξ∈Λ
δ
〈Mξ, ξ〉 kξk
2
≥ γ , 58
where Λ
δ
= {ξ : kΠξk ≥ δkξk}. Then, defining R = 1 − Mβ + M
−1
= ββ + M
−1
for some β 0,
one has kΠRk ≤ δ ∨
p βγ.
693
Proof. Since kRk ≤ 1, for Rξ ∈ Λ
c δ
one has kΠRξk
kξk ≤
kΠRξk kRξk
≤ δ . Now for R
ξ ∈ Λ
δ
, we have by assumption 58 γ
kΠRξk
2
kξk
2
≤ γ kRξk
2
kξk
2
≤ 〈MRξ, Rξ〉
kξk
2
≤ 〈M + βRξ, Rξ〉
kξk
2
= β 〈ξ, Rξ〉
kξk
2
≤ β . Combining both estimates gives the required bound.
This result can be applied almost directly to our setting in the following way:
Corollary 5.15 Let M ω be a random operator satisfying the conditions of Lemma 5.14 almost surely
for some random variable γ. If we choose β such that, for some deterministic δ ∈ 0, 1 , p ≥ 1 and
C
0, one has the bound Pβ ≥ δ
2
γ ≤ Cδ
p
, then
E kΠRk
p
≤ 1 + Cδ
p
. In particular, for any
δ ∈ 0, 1, setting β
k
= δ
3
Uu
k
Cδ, ¯p
1 ¯
p
, 59
where C is the constant from 56, produces the bound E
kΠR
β
2n
2n
k
p
|F
2n
≤ 2δ
p
, valid for every p ≤ ¯p.
Proof. To see the first part define Ω = {ω : βω ≤ δ
2
γω}. It the follows from Lemma 5.14, the fact that
kΠRk ≤ 1 and the assumption PΩ
c
≤ Cδ
p
, that
E kΠRk
p
≤ E δ ∨
r β
γ
p
1
Ω
+ 1
Ω
c
≤ δ
p
+ PΩ
c
≤ 1 + Cδ
p
, 60
as required. To obtain the second statement, it is sufficient to consider 56 with
ǫ = β
2n
δ
2
, so that one can take for
γ the random variable equal to ǫ on the set for which the bound 56 holds and 0 on its complement. It then follows from the choice 59 for
β
2n
that the assumption for the first part are satisfied with C = 1 and p = ¯
p, so that the statement follows. We now introduce a “compensator”
χ
2n+2
= exp ηV u
2n+1
+ ηV u
2n
, and, in analogy to before, we set
χ
2n
= Q
n k=1
χ
2k
. Proposition 5.11 implies that for any p ∈ [0, ¯p]
E χ
2n p
≤ exppκV u + pκC
L
2n , 61
where κ = η1 − η
′
. Note that Assumption B.3 made sure that η is sufficiently small so that κ¯ p
≤ 1. With these preliminaries complete, we now return to the analysis of 57.
694
Lemma 5.16 For any ǫ 0 and p ∈ [0, ¯p2], there exists a δ 0 sufficiently small so that if one
chooses β
n
as in Corollary 5.15 and η such that κ¯
p ≤ 1, one has
E kΞ
2n+2
k
p
χ
−p 2n+2
|F
2n
≤ exppC
J
− pC
Π
+ ǫp . Proof.
Since for every ǫ 0 there exists a constant C
ǫ
such that |x + y|
p
≤ e
p ǫ2
|x|
p
+ C
p ǫ
| y|
p
, recalling the definition of I
2n+2,1
and I
2n+2,2
from 57 we have that
E
kΞ
2n+2
k
p
χ
−p 2n+2
|F
2n
≤ e
ǫp2
E
kI
2n+2,1
k
p
χ
−p 2n+2
|F
2n
+ C
p ǫ
E kI
2n+2,2
k
p
χ
−p 2n+2
|F
2n
. We begin with the first term since it is the most straightforward one. Using the fact that
kR
β
2n
2n
k ≤ 1 and that ¯
p η 1 − η
′
by the assumption on η, we obtain from Assumptions B.2 and B.3 that
E kI
2n+2,1
k
p
χ
−p 2n+2
|F
2n
≤ exp−pηV u
2n
E E
kJ
2n+1
Π
⊥
k
p
|F
2n+1
× exp−pηV u
2n+1
kJ
2n
k
p
F
2n
≤ exppC
J
− pC
Π
. Turning to the second term, we obtain for any
δ ∈ 0, 1 the bound
E kI
2n+2,2
k
p
χ
−p 2n+2
|F
2n
≤ exp−pηV u
2n
q
E kΠR
β
2n
2n
k
2p
|F
2n
× Ç
E E
kJ
2n+1
k
2p
|F
2n+1
exp −2pηV u
2n+1
kJ
2n
k
2p
F
2n
≤ expp2C
J
δ
p
p 2 ,
provided that we choose β
n
as in Corollary 5.15. Choosing now δ sufficiently small it suffices to
choose it such that δ
p
≤
ǫp 2
p 2
C
−p ǫ
e
−pC
J
−pC
Π
for every p ≤ ¯p2 we obtain the desired bound.
Combining Lemma 5.16 with 61, we obtain the needed result which ensures that the “error term” ρ
t
from 48 goes to zero.
Lemma 5.17 For any p ∈ [0, ¯p4] and ˜κ ∈ [0, C
Π
− C
J
− 2κC
L
there exists a choice of the β
n
of the form 59 so that
E kΞ
2n+2
k
p
≤ exp pκV u − p˜κn
, for all u
∈ H . Proof. Since
E kΞ
2n+2
k
p
≤ E
kΞ
2n+2
k
2p
χ
2n+2 −2p
1 2
E χ
2n+2 2p
1 2
, the result follows by combining Lemma 5.16 with 61.
695
5.6 Controlling the size of the variation h
