Information Technology Reference
In-Depth Information
C
0
∈
R
Remark 10.4.1
Let
ϕ,φ
(
)
. Then, by the definition of the bilinear form
a
J
(
·
·
,
)
and (
10.17
), we find
a
J
(ϕ, φ)
J
ϕ)(x)φ(x)
d
x
ψ(ξ)e
ixξ
=
(
−
A
=
ϕ(ξ)
d
ξφ(x)
d
x
R
R
R
ϕ(ξ)
ϕ(ξ)φ(ξ)
d
ξ.
e
ixξ
φ(x)
d
x
d
ξ
=
=
ψ(ξ)
ψ(ξ)
R
R
R
We want to show that
a
J
(
·
,
·
)
is continuous (3.8) and satisfies a Gårding inequality
H
ρ
(
V
=
R
(3.9)on
)
. By Remark
10.4.1
, it is sufficient to study the characteristic
exponent
ψ
. We do this in the following lemma.
Lemma 10.4.2
Let X be a Lévy process with characteristic triplet (σ
2
,ν,
0
) where
the Lévy measure satisfies
(
10.12
), (
10.13
).
Then
,
there exist constants C
i
>
0,
i
=
1
,
2
,
3,
for all ξ
∈ R
holds
2
ρ
,
2
ρ
ψ(ξ)
≥
C
1
|
ξ
|
|
ψ(ξ)
|≤
C
2
|
ξ
|
+
C
3
.
1
0 and denote
k
sym
Proof
Consider first
σ
=
=
2
(k(z)
+
k(
−
z))
.Using(
10.13
) and
2sin
(z/
2
)
2
C
1
z
2
1
−
cos
(z)
=
≥
for
|
z
|≤
1, we obtain
cos
(ξ z))k
sym
(z)
d
z
≥
C
1
1
/
|
ξ
|
ξ
2
z
2
k
sym
(z)
d
z
ψ(ξ)
=
(
1
−
R
0
1
/
|
ξ
|
C
1
C
−
2
α
.
For an upper bound, we first consider
α<
1. Then, with
ξ
2
z
1
−
α
d
z
=
≥
C
1
C
−
α
|
ξ
|
−
0
|
z
|≤
1
|
z
|
ν(
d
z) <
∞
and
(
10.12
),
−
e
iξz
ν(
d
z)
≤
1
1
/
|
ξ
|
|
e
iξz
|
ψ(ξ)
|=
−
1
|
ν(
d
z)
+
C
2
R
−
1
/
|
ξ
|
2
1
/
|
ξ
|
≤
|
ξz
|
ν(
d
z)
+
C
2
−
1
/
|
ξ
|
1
/
|
ξ
|
4
C
+
1
|
−
α
d
z
α
≤
|
||
+
C
2
≤
α
|
|
+
4
C
+
ξ
z
ξ
C
2
.
−
0
Similarly, we obtain for 1
≤
α<
2 that
iξz
ν(
d
z)
≤
1
1
/
|
ξ
|
e
iξz
e
iξz
|
|=
−
+
|
−
−
|
+
C
3
+
C
4
|
|
ψ(ξ)
1
iξz
ν(
d
z)
ξ
R
−
1
/
|
ξ
|
1
/
|
ξ
|
1
/
|
ξ
|
2
ν(
d
z)
2
1
−
α
d
z
≤
|
ξz
|
+
C
3
+
C
4
|
ξ
|≤
2
C
|
ξ
|
|
z
|
+
C
3
+
C
4
|
ξ
|
+
−
1
/
|
ξ
|
0
α
C
6
.
For
σ>
0, we immediately have
≤
C
5
|
ξ
|
+
Search WWH ::
Custom Search