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1
2
σ
2
ξ
2
1
2
σ
2
ξ
2
,
ψ(ξ)
=
+
(
1
−
cos
(ξ z))ν(
d
z)
≥
R
1
2
σ
2
ξ
2
e
iξz
C
7
ξ
2
|
ψ(ξ)
|≤
+
|
−
1
−
iξz
|
ν(
d
z)
≤
+
C
8
.
R
Lemma
10.4.2
shows that
ψ
satisfies the so-called sector condition
|
ψ(ξ)
|≤
C
ψ(ξ)
, for all
ξ
∈ R
. Using this, we can show
Theorem 10.4.3
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,
such that for all ϕ,φ
∈
H
ρ
(
R
) one has
|
a
J
(ϕ, φ)
|≤
C
1
ϕ
H
ρ
(
R
)
φ
H
ρ
(
R
)
,
J
(ϕ, ϕ)
≥
C
2
ϕ
2
2
H
ρ
(
R
)
−
C
3
ϕ
L
2
(
R
)
.
Proof
Using Lemma
10.4.2
, there exist positive constants
C
1
,C
2
>
0 such that
|
ψ(ξ)
|≤
C
2
|
ξ
|
1
.
2
ρ
,
2
ρ
ψ(ξ)
≥
C
1
|
ξ
|
+
Then, we have, using Hölder's inequality and Remark
10.4.1
,
ϕ(ξ)φ(ξ)
d
ξ
a
J
(ϕ, φ)
|
|=
ψ(ξ)
R
≤
C
2
2
ρ
)
|
ϕ(ξ)φ(ξ)
|
(
1
+|
ξ
|
d
ξ
R
≤
C
2
)
2
ρ
ϕ(ξ)φ(ξ)
+|
|
|
|
(
1
ξ
d
ξ
R
≤
C
2
2
d
ξ
1
/
2
2
d
ξ
1
/
2
)
2
ρ
)
2
ρ
|
φ(ξ)
(
1
+|
ξ
|
|
ϕ(ξ)
|
(
1
+|
ξ
|
|
R
R
=
C
2
ϕ
H
ρ
(
R
)
φ
H
ρ
(
R
)
,
where we used the fact that there exists a
c>
0 such that
)
2
ρ
(
1
+|
ξ
|
1
c
<
0
<c
≤
≤
∞
,
∀
ξ
∈ R
.
1
+|
ξ
|
2
ρ
Furthermore, to prove the Gårding inequality, one finds
C
1
a
J
(ϕ, ϕ)
2
d
ξ
2
d
ξ
2
d
ξ,
=
ψ(ξ)
|
ϕ(ξ)
|
=
(C
1
+
ψ(ξ))
|
ϕ(ξ)
|
−
|
ϕ(ξ)
|
R
R
R
and
C
1
2
d
ξ
2
ρ
)
2
d
ξ
(C
1
+
|
|
≥
+|
|
|
|
ψ(ξ))
ϕ(ξ)
(
1
ξ
ϕ(ξ)
R
R
≥
C
1
)
2
ρ
2
d
ξ.
(
1
+|
ξ
|
|
ϕ(ξ)
|
R
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