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

26 N. Jacob and R. L. Schilling
The latter follows from (7) and the fact that R n+1 ∋ (ξ, c) ↦→ f(|(ξ, c)| 2 ) is
negative definite. Thus, φ(c) φ(0) 0 and we conclude that
N∑
(f − τ c f)(|ξ j − ξ k | 2 )λ j¯λk = −φ(c) 0
j,k=1
which, in view of (7), means that f(|ξ| 2 ) − τ c f(|ξ| 2 ) is negative definite.
To see (g) we note that because of (e) τ c f ∈ O for all c ∈ R, thus f(|ξ| 2 +
c 2 ) f(c 2 ) which proves that f is increasing. By (f), f − τ c f ∈ O, hence
0 f(|ξ| 2 ) + f(c 2 ) − f(|ξ| 2 + c 2 ), proving subadditivity. Finally using that
τ c (f − τ d f) ∈ O we get
0 τ c (f − τ d f)(d 2 ) = 2f(c 2 + d 2 ) − f(c 2 + 2d 2 ) − f(c 2 )
which can be rearranged to give 1 2 (f(c2 + 2d 2 ) + f(c 2 )) f(c 2 + d 2 ). Since f is
continuous, this mid-point property implies concavity.
We are now ready for the proof of Theorem 2.2.
Proof of Theorem 2.2. The above Lemma shows that O is a closed convex
cone contained in the space L = L 1 ((0, ∞), e −r dr). The set
{ ∫ ∞
}
B = f ∈ O : f(r)e −r dr = 1 ⊂ L
is a convex base of O. For f ∈ B we have
f(r)e −r = f(r)
∫ ∞
r
0
e −s ds
∫ ∞
r
f(s)e −s ds 1. (25)
Let us show that B is a relative compact subset of L. We check the conditions
of Kolmogorov’s compactness criterion for L p . First,
sup ‖f‖ L = 1. (26)
f∈B
Moreover, for all x, y 0 contained in a bounded set we have
‖f(• + x) − f(• + y)‖ L =
∫ ∞
0
|f(r + x) − f(r + y)|e −r dr
∫ ∞
= e x∧y |f(r + x ∨ y − x ∧ y) − f(r)|e −r dr
x∧y
∫ ∞
e x∧y |f(r + |x − y|) − f(r)|e −r dr
0
= e x∧y (
e |x−y| ∫ ∞
|x−y|
)
f(r)e −r dr − 1
e x∧y( e |x−y| − 1 ) |x−y|→0
−−−−−→ 0,
(27)