Function Spaces as Dirichlet Spaces (About a Paper by Maz'ya and ...
Function Spaces as Dirichlet Spaces (About a Paper by Maz'ya and ...
Function Spaces as Dirichlet Spaces (About a Paper by Maz'ya and ...
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Since 1 − e −λ λ ∧ 1 2λ/(1 + λ) for all λ > 0, we find<br />
∫ 1<br />
0<br />
x −n/2 h(x) dx 2 ∫ ∞<br />
( ∫ 1<br />
Γ ( 1<br />
)<br />
n<br />
2 0+ 0 1 − t<br />
= 2 ∫ 1<br />
( ∫ ∞<br />
Γ ( )<br />
n<br />
2 0 0+<br />
= 2 ∫ 1<br />
Γ ( g(1 − t)<br />
)<br />
t n/2−1 dt<br />
n<br />
2 0 1 − t<br />
= 2 ( ∫ 1/2<br />
Γ ( g(1 − t)<br />
)<br />
t n/2−1 dt +<br />
n<br />
2 0 1 − t<br />
( ∫ 1/2<br />
c ′ g( 1)<br />
∫ 1<br />
2<br />
n<br />
t n/2−1 dt +<br />
0<br />
1<br />
2<br />
<strong>Function</strong> <strong>Spaces</strong> <strong>as</strong> <strong>Dirichlet</strong> <strong>Spaces</strong> 15<br />
1−t<br />
)<br />
s<br />
1 + 1−t t n/2−1 dt ρ(ds)<br />
s<br />
)<br />
1<br />
s + 1 − t ρ(ds) t n/2−1 dt<br />
1/2<br />
∫ 1<br />
1/2<br />
g(1 − t)<br />
1 − t<br />
g(1 − t)<br />
1 − t<br />
)<br />
t n/2−1 dt ,<br />
)<br />
t n/2−1 dt<br />
where we use that s ↦→ g(s)/s is decre<strong>as</strong>ing. Thus,<br />
∫ 1<br />
0<br />
( ∫ 1/2<br />
x −n/2 h(x) dx ˜c n t n/2−1 dt +<br />
0<br />
( ∫ 1/2<br />
c n t n/2−1 dt +<br />
0<br />
∫ 1<br />
1/2<br />
∫ 1<br />
0<br />
g(1 − t)<br />
1 − t<br />
)<br />
t n/2−1 dt<br />
g(s)<br />
s ds )<br />
< ∞.<br />
To show the finiteness of the second integral in the statement, we use the elementary<br />
estimate e −λ 1/(1 + λ), λ 0, to get<br />
∫ ∞<br />
1<br />
x −n/2−1 h(x) dx<br />
=<br />
∫ ∞<br />
x −n/2−1 xn/2<br />
1 Γ ( n<br />
2<br />
∫ ∞<br />
( ∫ 1<br />
= 1<br />
Γ ( )<br />
n<br />
2<br />
1 )<br />
Γ ( n<br />
2<br />
= 1<br />
Γ ( )<br />
n<br />
2<br />
= 1<br />
Γ ( )<br />
n<br />
2<br />
1<br />
∫ ∞<br />
1<br />
∫ ∞<br />
1<br />
∫ ∞<br />
1<br />
0<br />
( ∫ ∞<br />
[ ∫ 1<br />
) e −x/s t n/2 e<br />
0+<br />
[ ∫ ∞<br />
e −x(1−t)/s t<br />
0+<br />
( ∫ 1<br />
[ ∫ ∞<br />
0<br />
0+<br />
( ∫ 1<br />
[ ∫ ∞<br />
0<br />
( ∫ 1<br />
0<br />
0+<br />
1<br />
1 + x(1−t)<br />
s<br />
g(x(1 − t))<br />
x(1 − t)<br />
0<br />
tx/s dt<br />
n/2−1 ρ(ds)<br />
s<br />
1<br />
s tn/2−1 ρ(ds)<br />
] ρ(ds)<br />
t s<br />
] ) dx<br />
dt<br />
x<br />
]<br />
]<br />
1<br />
s + x(1 − t) ρ(ds) t n/2−1 dt<br />
t n/2−1 dt) dx<br />
x .<br />
) dx<br />
dt<br />
x<br />
) dx<br />
x<br />
)<br />
dx