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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<strong>Function</strong> <strong>Spaces</strong> <strong>as</strong> <strong>Dirichlet</strong> <strong>Spaces</strong> 13<br />
<strong>and</strong> we get (23) if we define ν <strong>by</strong> (24). Since<br />
∫<br />
t ∞<br />
s + t = (1 − e −λt ) se −λs dλ,<br />
0<br />
we find<br />
µ(t) =<br />
=<br />
∫ ∞<br />
0<br />
∫ ∞<br />
( ∫ ∞<br />
)<br />
(1 − e −λt ) se −λs ν(ds) dλ<br />
0+<br />
( ∫ ∞<br />
)<br />
(1 − e −λt ) se −λs ν(ds) dλ .<br />
0<br />
0+<br />
Finally,<br />
∫ ∞<br />
0<br />
se −λs ν(ds)<br />
∫ 1<br />
( ∫ ∞<br />
[ ∫ ∞<br />
] ) dy dx<br />
=<br />
se −λs ρ(xy · ds)<br />
0 1 0+<br />
y 2 x<br />
∫ 1<br />
( ∫ ∞<br />
[ ∫ ∞<br />
z<br />
(<br />
=<br />
0 1 0+ xy exp − λz ) ] ) dy dx<br />
ρ(dz)<br />
xy y 2 x<br />
∫ 1<br />
( ∫ ∞<br />
[ ∫ ∞<br />
] )<br />
=<br />
zr e −λzξr ρ(dz) dξ dr<br />
0 1 0+<br />
∫ ∞<br />
( ∫ 1<br />
[ ∫ ∞<br />
] )<br />
=<br />
e −λzξr dξ zr dr ρ(dz)<br />
0+ 0 1<br />
∫ ∞<br />
( ∫ 1<br />
)<br />
e −λzr<br />
=<br />
λ dr ρ(dz)<br />
0+<br />
0<br />
= 1 ∫ ∞<br />
1 − e −λz<br />
ρ(dz),<br />
λ 2 z<br />
0+<br />
(z = xys)<br />
(<br />
ξ =<br />
1<br />
, r = )<br />
1<br />
x y<br />
<strong>and</strong> the proof is finished.<br />
4. Equivalent Seminorms<br />
Let ψ : R n → R be a c.n.d.f. of the form<br />
∫<br />
ψ(ξ) = (1 − cos yξ) m(|y| 2 ) dy<br />
y≠0<br />
where<br />
m(r) =<br />
∫ ∞<br />
0+<br />
e −rs ν(ds)