The Stieltjes convolution and a functional calculus for non-negative ...

( ⃗ λ ∈ R n +), then we have for ∀x ∈ X and ∀x ∗ ∈ X ∗
S(ϕ x,x ∗) = 〈R R (S)x,x ∗ 〉.
Proof: The function ϕ x,x ∗ is in H(R n +) because
∣ ⃗ λ ⃗k+1 D ⃗ 〈( ∏ n k
R i (λ i ) ) x,x ∗〉∣ ∣〈( ∣ = ⃗ n k! ∣∣ ∏
λ k i+1
i R i (λ i ) ) k i+1
x,x ∗〉∣ ∣ i=1
for ∀ ⃗ λ ∈ R n +, and we have
✷
S(ϕ x,x ∗) = ∑
| ⃗ k|≤K
= ∑
| ⃗ k|≤K
∫
(−1) |⃗ k|
⃗ k!
∫
R n +
= 〈R R (S)x,x ∗ 〉.
R n +
Lemma 7 R R is well-defined.
i=1
≤ ⃗ k!M |⃗ k|+n ||⃗x|| ||x ∗ ||
D ⃗ 〈( ∏ n k
R i (λ i ) ) x,x ∗〉 dµ ⃗k ( ⃗ λ)
i=1
〈( ∏ n R i (λ i ) ) k i+1
x,x ∗〉 dµ ⃗k ( ⃗ λ)
i=1
Proof: Assume there are two representations of the same distribution S ∈
D S ∗(R n +) ′ so that the operators assigned to them are not the same. Then
the difference of them is a non-trivial representation of the zero-distribution
O whose corresponding operator R R (O) is not the zero-operator. Therefore
there are x ∈ X and x ∗ ∈ X ∗ with 〈R R (O)x,x ∗ 〉 ≠ 0. But according to Lemma
6 we have 〈R R (O)x,x ∗ 〉 = O(ϕ x,x ∗) = 0. Contradiction. ✷
Note that R R is well-defined even if the resolvent families R i do not commute.
Fixing an order for the operators in the expression ∏ n
i=1 R i (λ i ), this can be
useful to prove some non-trivial operator equations of the form R R (S) = 0 for
which the upcoming theorem is not needed.
Theorem 4 If S 1 ,S 2 ∈ D S ∗(R n +) ′ and the resolvent families R i , i = 1,...,n,
commute, then R R (S 1 ∗ S 2 ) = R R (S 1 )R R (S 2 ).
Proof: Let x ∈ X and x ∗ ∈ X ∗ , and define ϕ x,x ∗ as in Lemma 7. Based on
the resolvent equation we can generalize Lemma 3 to
(∆ n ϕ x,x ∗)( ⃗ λ 1 , ⃗ λ 2 ) = 〈 2∏ n∏
R i (λ j,i )x,x ∗〉 ( λ ⃗ 1 , λ ⃗ 2 ∈ R n +).
j=1 i=1
Using two representations (8) for S 1 and S 2 , equation (20) and Lemma 7, we
find that
20