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

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24 N. Jacob and R. L. Schilling = 2 · ∫ ∞ ( ∫ ∞ 4−n/2 Γ ( ) n 2 0+ 0+ 2 · ( ∫ ∞ 4−n/2 Γ ( ) n 2 0+ < ∞ . Therefore, the function m(|y| 2 ) := ∫ ∞ 0+ exp ) st 2 ( ) 1 + st τ(ds) exp − t2 t n−1 dt 2 4 )( ∫ s 1 1 + s τ(ds) 0+ t n−1 e −t2 /4 dt + ( ) − |y|2 τ(ds) 4s (4πs) = 1 ∫ ∞ n/2 π n/2 0+ ∫ ∞ 1 ) t n+1 e −t2 /4 dt e −s|y|2 s n/2 Φ(τ)(ds) is the Laplace transform of the measure s n/2 Φ(τ)(ds) on (0, ∞) where Φ : s ↦→ (4s) −1 . Moreover, the above calculation shows that m(|y| 2 ) dy has the integrability properties of a Lévy measure. Since τ is the representing measure of the Bernstein function g, we find ∫ 1 0+ s −n/2 s n/2 Φ(τ)(ds) + = ∫ 1/4 0+ 4s τ(ds) + ∫ ∞ 1 ∫ ∞ 1/4 s −n/2−1 s n/2 Φ(τ)(ds) τ(ds) < ∞ . Conversely, assume that ψ is of the form (14) with Lévy measure m(|y| 2 ) dy where ∫ m(r) is the Laplace transform of a measure ν on (0, ∞) such that 1 0+ s−n/2 ν(ds) + ∫ ∞ s −n/2−1 ν(ds) < ∞. Using calculations similar to the ones 1 used above it is enough to check that the function given by (15) is indeed a Bernstein function. This, however, follows from ∫ ∞ 0+ ∫ ∞ s 1 1 + s sn/2 Φ(ν)(ds) = 0+ 1 + 4s (4s)−n/2 ν(ds) ( ∫ 1 ∫ ∞ ) (4s) −n/2 ν(ds) + (4s) −n/2−1 ν(ds) < ∞. 0+ 1 The proof of Theorem 2.2 uses similar techniques to the ones used by Harzallah [8] to prove that the Bernstein functions are the only class of functions with the property that f ◦ ψ is a c.n.d.f. for all continuous negative definite functions ψ. Lemma. Denote by O the family of functions f : [0, ∞) → R such that f(|ξ| 2 ) is a c.n.d.f. on every R n , n ∈ N. Then (a) O is closed under locally uniform convergence, and for all f, g ∈ O (b) f 0
(c) λf + (1 − λ)g ∈ O for all 0 λ 1 (d) f| (0,∞) is continuous (e) τ c f := f(• + c 2 ) − f(c 2 ) ∈ O for all c ∈ R (f) f − τ c f ∈ O for all c ∈ R (g) f is increasing, subadditive and concave. Function Spaces as Dirichlet Spaces 25 Proof. The properties (a) – (d) follow immediately from the corresponding properties for the family of c.n.d.fs. To see (e), we use criterion (7). Set ψ(ξ) := τ c f(|ξ| 2 ). Obviously, ψ is real valued and ψ(0) 0. Now fix vectors ξ 1 , . . . , ξ N ∈ R n , numbers λ 1 , . . . , λ N ∈ C such that ∑ j λ j = 0 and assume that ∑ j,k f(|ξ j − ξ k | 2 + c 2 )λ j¯λk > 0. We define M vectors η l ∈ R M by η l := √ c 2 e l (e l is the lth unit vector in R M ). Then |η l − η m | 2 = c 2 if l ≠ m and = 0 if l = m. By our assumption we can choose M so large that for µ l = µ m := 1 , 1 l, m M, we find M N∑ ∑ N∑ f(|ξ j − ξ k | 2 + c 2 )λ j¯λk µ l¯µ m > − f(|ξ j − ξ k | 2 )λ j¯λk j,k=1 l≠m j,k=1 M ∑ l=1 |µ l | 2 . Thus, 0 < ∑ N∑ M∑ N∑ f(|ξ j − ξ k | 2 + c 2 )λ j¯λk µ l¯µ m + f(|ξ j − ξ k | 2 )λ j¯λk |µ l | 2 l≠m j,k=1 l=1 j,k=1 = ∑ ∑ f(|(ξ j , η l ) − (ξ k , η m )| 2 )(λ j µ l )(λ k µ m ). (l,j) (m,k) Since R N+M ∋ (ξ, η) ↦→ f(|(ξ, η)| 2 ) is negative definite and since ∑ (l,j) λ jµ l = 0, this contradicts (7). For (f) we use again the criterion (7). Choose ξ 1 , . . . , ξ N ∈ R n and λ 1 , . . . , λ N ∈ C with ∑ j λ j = 0. For c ∈ R we introduce the auxiliary function φ(c) := N∑ [ f(|ξj − ξ k | 2 + c 2 ) − f(|ξ j − ξ k | 2 ) ] λ j¯λk j,k=1 and show that c ↦→ φ(c) is negative definite. Obviously, φ is real-valued, φ(0) = 0 and for all c 1 , . . . , c M ∈ R and µ 1 , . . . , µ M ∈ C with ∑ l µ l = 0 we see M∑ l,m=1 φ((c l − c m ) 2 )µ l¯µ m = M∑ l,m=1 j,k=1 = ∑ (j,l) N∑ f(|ξ j − ξ k | 2 + (c l − c m ) 2 )λ j¯λk µ l¯µ m ∑ f(|(ξ j , c l ) − (ξ k , c m )| 2 )(λ j µ l )(λ k µ m ) 0. (k,m)
Page 1 and 2: Zeitschrift für Analysis und ihre
Page 3 and 4: Function Spaces as Dirichlet Spaces
Page 9 and 10: Hence, µ(1) < ∞ implies G < ∞.
Page 13 and 14: Since 1 − e −λ λ ∧ 1 2λ/
Page 17 and 18: and so J ′ (λ) = a λ 2 ∫ λ 0
Page 21: Function Spaces as Dirichlet Spaces
Page 25 and 26: and for all R > 1 we have by the co

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