Function Spaces as Dirichlet Spaces (About a Paper by Maz'ya and ...

16 N. Jacob and R. L. Schilling
Again we use that s ↦→ g(s)/s is decreasing to find
∫ ∞
1
x −n/2−1 h(x) dx
c ′ n
{ ∫ ∞
1
1
[ ∫ 1
2
0
g( x 2 )
x
2
] ∫ dx ∞
[ ∫ 1
t n/2−1 dt
x +
{( ∫ ∞
g( x
˜c ) )( ∫ 1 )
2
2
n ( x 2
dx t
2) n/2−1 dt +
0
1
1
2
∫ ∞
1
g( x 2 )
x
2
( ∫ 1
2
0
] } dx
dt
x
g(xs)
xs
ds ) dx
x
}
.
By assumption, the first term is finite, and for the second term we find
∫ ∞
( ∫ 1
( ∫
2
∞
1
as µ( 1 ) µ(1) < ∞.
2
0
) ∫ 1
g(xs) dx
xs ds x = 2
0
s
g(y)
y
dy
y
)
ds = µ ( 1
2)
< ∞
Corollary 4.2. The Bernstein function f associated with s ↦→ g(s −1 )s −n/2 is
given by
f(r) = 1 π n/2 ∫ ∞
(
4 Γ ( 1 − e
)
−rλ ∫ ∞
[ ∫ 1 (
exp − 1 − t ) ] ) ρ(ds)
t n/2−1 dt dλ .
n
2 0 λ 2 0+ 0 4sλ
s
Proof. Applying Lemma 2.1 to the measure η(dx) = h(x) dx, x > 0, we find
f(r) =
∫ ∞
0+
(1 − e −rx )(4πx) n/2 Φ(η)(dx)
where Φ(x) = 1/4x. Hence
∫ ∞
( ∫ ∞
[ ∫ 1
] )
f(r) = πn/2
ρ(ds)
Γ ( )
n
(1 − e −r/4x ) x −n/2 x n/2 e −x/s t n/2−1 e −tx/s dt dx
2 0
0+
0
s
∫ ∞
( ∫ ∞
[ ∫ 1
] )
= πn/2
ρ(ds)
Γ ( )
n
(1 − e −r/4x )e −x(1−t)/s t n/2−1 dt dx
2 0 0+ 0
s
∫ ∞
( ∫ ∞
[ ∫ 1
= πn/2
(
4 Γ ( )
n
(1 − e −rλ ) exp − 1 − t ) ] ) ρ(ds) dλ
t n/2−1 dt
2 0 0+ 0
4sλ
s λ 2 ,
where we used the substitution λ = 1/4x.
It is by no means obvious to find out the cases where f and µ are equal.
However, we can always prove that f ∼ µ holds.
Proposition 4.3. Assume that f is as in the above corollary, µ as in Lemma 3.3
and that the assumptions of Lemma 4.1 are satisfied. Then there are two constants
κ 0 , κ 1 > 0 such that
κ 0 f(r) µ(r) κ 1 f(r), r > 0.