09.01.2014 Views

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 ...

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

<strong>Function</strong> <strong>Spaces</strong> <strong>as</strong> <strong>Dirichlet</strong> <strong>Spaces</strong> 9<br />

Nagel discussed in [19, 20] the rotationally invariant analogue of Lemma 3.2.<br />

Lemma 3.3. Let g : [0, ∞] → [0, ∞] be a monotone incre<strong>as</strong>ing concave function<br />

<strong>and</strong> define µ(t) := ∫ t ∫ ∞<br />

g(s)/s 2 ds dr ∈ [0, ∞]. Then<br />

0 r<br />

( ∫∫<br />

( 1<br />

) ) 1<br />

dx dy<br />

|u(x + y) − u(x)| 2 2<br />

g<br />

R n ×R |y| n 2 |y| n<br />

( ∫<br />

) (19)<br />

1<br />

∼ |û(ξ)| 2 µ(|ξ| 2 2<br />

) dξ<br />

R n<br />

are equivalent (semi-)norms on S(R n ).<br />

Let us now consider the rotationally invariant c<strong>as</strong>e. Comparing (16) with<br />

(19) we would like to relate<br />

( 1<br />

) 1<br />

m(t) with g<br />

<strong>and</strong> f(|ξ| 2 ) with µ(|ξ| 2 ).<br />

t t n/2<br />

Lemma 3.4. Let g(t) = ∫ ∞ t<br />

ρ(ds) be a complete Bernstein function. Then<br />

0+ s+t<br />

the functions r ↦→ g(r −1 ) <strong>and</strong> r ↦→ r −n/2 g(r −1 ) are completely monotone. In<br />

particular, we have the representations<br />

( 1<br />

)<br />

g =<br />

r<br />

( 1<br />

)<br />

r −n/2 g =<br />

r<br />

∫ ∞<br />

0<br />

∫ ∞<br />

0<br />

( ∫ ∞<br />

e −xr −x/s ρ(ds)<br />

e<br />

s<br />

0+<br />

( 1<br />

e −xr Γ ( )<br />

n<br />

2<br />

∫ ∞<br />

0+<br />

)<br />

dx (20)<br />

s n/2−1 e −x/s ∫ x/s<br />

0<br />

)<br />

y n/2−1 e y dy ρ(ds) dx. (21)<br />

Proof. Since the formulae (20) <strong>and</strong> (21) prove that r ↦→ r −n/2 g(r −1 ), n 0, is<br />

completely monotone, it is enough to establish these two representations.<br />

From the definition of g we have<br />

Since we can write<br />

0+<br />

( 1<br />

)<br />

g =<br />

r<br />

∫ ∞<br />

0+<br />

1<br />

r<br />

s + 1 r<br />

ρ(ds) =<br />

∫<br />

1<br />

∞<br />

rs + 1 = e −t(rs+1) dt =<br />

0<br />

0<br />

∫ ∞<br />

0+<br />

∫ ∞<br />

0<br />

0<br />

1<br />

rs + 1 ρ(ds).<br />

e −x/s e −xr dx s<br />

we can apply Tonelli’s Theorem to find<br />

( 1<br />

) ∫ ∞ ∫ ∞<br />

g = e −x/s e −xr dx ∫ ∞<br />

( ∫ ∞<br />

)<br />

r s ρ(ds) = e −xr −x/s ρ(ds)<br />

e dx<br />

s<br />

which proves (20). To prove (21), we use the elementary identity<br />

r −n/2 = 1<br />

Γ ( )<br />

n<br />

2<br />

∫ ∞<br />

0<br />

0+<br />

x n/2−1 e −rx dx, n > 0,

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!