Information Technology Reference
In-Depth Information
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
Search WWH ::




Custom Search