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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

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