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

18 N. Jacob and R. L. Schilling
Setting
∫ 1
J(λ) := λe −(1−t)λ t n/2−1 dt and λ := r
0
4y
we see
∫ r
I 1 = 4J(λ) g(y) dy
0 y ,
and it remains to show that 0 < c J(λ) C < ∞ for all λ 1 . This will be
4
done in Lemma 4.4 below. From this we conclude that
From Lemma 3.5 we know that
µ(r) =
∫ r
0
I 1 ∼
∫ r
0
g(y) dy
y .
g(y) dy ∫ ∞
y + r
and collecting all of the above calculations gives
Lemma 4.4. The function
J(λ) := λ
µ(r) ∼ I 1 + I 2 ∼ f(r).
∫ 1
0
r
g(y) dy
y 2 ,
e −(1−t)λ t n/2−1 dt, λ > 0,
satisfies 0 < min{1 − e 1/4 , J( 1 4 )} J(λ) 1 for all λ 1 4 .
Proof. For n 2 we have
J(λ) λ
∫ 1
and for n = 1 we set s = t/λ to find
0
e −(1−t)λ dt = 1 − e −λ 1,
J(λ) = √ ∫ λ
λ e −λ e −s s 1/2−1 ds √ ∫ ∞
λ e −λ
0
0
e −s s 1/2−1 ds = √ λ e −λ Γ ( 1
2)
.
Clearly, √ λ e −λ Γ ( √
1
2)
= πλ e −λ < 1.
To get the lower bounds for n = 1, 2 observe that 0 t 1 implies that
t n/2−1 1, and so
J(λ) λ
∫ 1
For n 3 we have with a := n 2 − 1 > 0
0
J(λ) =
e −(1−t)λ dt = 1 − e −λ 1 − e −1/4 .
∫ λ
0
e −t (
1 − t λ) a
dt,