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

Hence, µ(1) < ∞ implies G < ∞.
Now suppose that G < ∞. Integration by parts yields
µ(t) =
=
= t
∫ t
( ∫ ∞
0 r
∫ ∞
[
r
r
∫ ∞
t
g(s) ds
s 2 )
dr
g(s) ds
s 2 ] r=t
g(s)
s
r=0
∫
ds t
s +
0
+
Function Spaces as Dirichlet Spaces 11
∫ t
0
sg(s) ds
s 2
g(s) ds
s − lim
r→0
( ∫ ∞
g(s)
r
r s
)
ds
.
s
Since g(rt)/t g(t)/t if 0 r 1, and since by assumption t ↦→ g(t)/t is
integrable at +∞, the dominated convergence theorem applies and
( ∫ ∞
) ∫
g(s) ds
∞
lim r
= lim g(rt) dt
r→0
r s s r→0
1 t = 0,
where we used that g(0) = 0.
Lemma 3.6. Let g(r) = ∫ ∞ r
ρ(ds) be a complete Bernstein function and
0+ s+r
let
∫
µ(t) be as in Lemma 3.3. Then µ(1) < ∞ – and, by Lemma 3.5, G =
1
g(s)/s ds + ∫ ∞
g(s)/s 2 ds < ∞ – if, and only if, the measure ρ representing
0 1
g satisfies
∫ 1
0+
log(1 + s −1 ) ρ(ds) +
∫ ∞
1
log(1 + s) ρ(ds)
s
< ∞.
Proof. Because of Tonelli’s Theorem we find
∫ 1
g(s) ds ∫ 1
( ∫ ∞
)
0 s = ρ(dr)
ds
0 0+ s + r
∫ ∞
( ∫ 1
)
ds
=
ρ(dr)
0+ 0 s + r
∫ ∞
s=1
= log(s + r)
∣ ρ(dr)
=
0+
∫ 1
0+
s=0
log(1 + r −1 )ρ(dr) +
∫ ∞
1
log(1 + r −1 ) ρ(dr).
Since log(1 + r −1 ) r −1 and since, by assumption, ∫ ∞
1
r −1 ρ(dr) < ∞ we find
that
∫ 1
0
g(s) ds
∫ 1
s < ∞ if, and only if, log(1 + r −1 ) ρ(dr) < ∞.
0+