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2
σ
2
s
2
ρ
+
2
μ
∂
s
ϕ∂
s
φ
d
s
+
σ
2
(ρ
+
μ)
R
R
1
a
CEV
s
2
ρ
+
2
μ
−
1
∂
s
ϕφ
d
s
ρ,μ
(ϕ, φ)
:=
0
0
r
r
R
R
s
1
+
2
μ
∂
s
ϕφ
d
s
s
2
μ
ϕφ
d
s.
−
+
(4.28)
0
0
Note that
a
CEV
ρ,
0
a
CEV
ρ
and
a
CEV
1
a
BS
. We introduce the spaces
W
ρ,μ
as closures
=
=
of
C
0
(G)
with respect to the norm
R
s
2
ρ
+
2
μ
2
d
s,
2
2
s
2
μ
ϕ
ρ,μ
:=
|
∂
s
ϕ
|
+
|
ϕ
|
(4.29)
0
compare with (
4.20
). Note that
ϕ
ρ
=
ϕ
ρ,
0
. We now show the analog to Propo-
sition
4.5.1
,for
ρ
∈[
1
/
2
,
1
]
.
Proposition 4.5.3
Assume
0
≤
ρ
≤
1
and select
μ
=
≤
ρ<
2
,ρ
=
0
if
0
1
,
(4.30)
1
2
<μ<
2
−
1
−
ρ
if
2
≤
ρ<
1
.
Assume also r>
0.
Then there exist C
1
,C
2
>
0
such that
∀
ϕ
,
φ
∈
W
ρ,μ
the follow-
ing holds
:
a
CEV
|
|≤
C
1
ρ,μ
ρ,μ
,
ρ,μ
(ϕ, φ)
ϕ
φ
(4.31)
a
CEV
2
ρ,μ
(ϕ, ϕ)
≥
C
2
ϕ
ρ,μ
.
(4.32)
Proof
The continuity (
4.31
)of
a
CEV
W
ρ,μ
(
0
,R)
follows from the
Cauchy-Schwarz inequality and by Hardy's inequality (
4.26
) with
ε
in
W
ρ,μ
(
0
,R)
×
ρ,μ
=
2
(ρ
+
μ)
=
1
ρ,μ
s
2
ρ
+
2
μ
−
2
φ
2
d
s
1
/
2
R
1
2
σ
2
a
CEV
σ
2
(ρ
|
ρ,μ
(ϕ, φ)
|≤
ϕ
ρ,μ
φ
ρ,μ
+
+
μ)
ϕ
0
ρ,μ
s
2
+
2
μ
−
2
ρ
φ
2
d
s
1
/
2
R
+
r
ϕ
0
σ
2
2
+
rR
1
−
ρ
2
σ
2
(ρ
+
μ)
≤
|
+
ϕ
ρ,μ
φ
ρ,μ
.
|
2
ρ
+
2
μ
−
1
Let
ϕ
∈
C
0
(G)
. We calculate
2
σ
2
(ρ
+
μ)
R
1
2
σ
2
1
a
CEV
s
ρ
+
μ
∂
s
ϕ
2
s
2
ρ
+
2
μ
−
1
∂
s
(ϕ
2
)
d
s
ρ,μ
(ϕ, ϕ)
=
L
2
(G)
+
0
2
r
R
1
s
1
+
2
μ
∂
s
(ϕ
2
)
d
s
s
μ
ϕ
2
L
2
(G)
−
+
r
0
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