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Proof. 30. For the second statement, let a ∈ A. We need to prove that a(1 + D2 )−s/2 is trace p +1 class for a = bc with b ∈ B2 (D, p) and c ∈ B2 (D, p). Thus, for all k ≤ p + 1 44 A. L. CAREY, V. GAYRAL, A. RENNIE, and F. A. SUKOCHEV and all s > p we have b(1 + D2 )−s/4 , (1 + D 2 )−s/4 δ k (c) ∈ L2 (N , τ ). We start from the identity (−1)k Γ(s + k) 1 (1 + |D|)−s−k = Γ(s)Γ(k + 1) 2πi λ−s (λ − 1 − |D|)−k−1 dλ, (λ)=1/2 and then by induction we have p [(λ − 1 − |D|)−1 , c] = (−1)k+1 (λ − 1 − |D|)−k−1 δ k (c) k=1 + (−1) p (λ − 1 − |D|)− p −1 δ p +1 (c)(λ − 1 − |D|)−1 .
A. SUKOCHEV for all k ∈ N0 and s > p. 10) (1 + D 2 )−s/4 |δ k (a)|(1 + D2 )−s/4 ∈ L1 (N , τ ), for all k ∈ N0 and s > p. 10) is satisﬁed if |δ k (a)|(1 + D2 )−s/2 ∈ L1 (N , τ ), for all k ∈ N0 and s > p, which in turn is equivalent to δ k (a)(1 + D2 )−s/2 ∈ L1 (N , τ ), for all k ∈ N0 and s > p. 12) (1 + D2 )−s/4 δ k (a)(1 + D2 )−s/4 ∈ L1 (N , τ ), for all k ∈ N0 and s > p. 12) is equivalent to (1 + D 2 )−s/4 Lk (a)(1 + D2 )−s/4 ∈ L1 (N , τ ), for all k ∈ N0 and s > p. In an entirely similar way, we see that δ k ([D, a]) ∈ B1 (D, p) if (1 + D2 )−s/4 Lk ([D, a])(1 + D2 )−s/4 ∈ L1 (N , τ ), for all k ∈ N0 and s > p.
17. Let (A, H, D) be a ﬁnitely summable semiﬁnite spectral triple of spectral dimension p. Then A is a subalgebra of B1 (D, p). Proof. Since A is a ∗-algebra, it suﬃces to consider self-adjoint elements. For a = a∗ ∈ A, we have by assumption that a(1 + D2 )−s/2 ∈ L1 (N , τ ), for all s > p. Now let a = v|a| = |a|v ∗ be the polar decomposition. Observe that neither v nor |a| need be in A. However, |a|(1 + D2 )−s/2 = v ∗ a(1 + D2 )−s/2 ∈ L1 (N , τ ) for all s > p. 5, from [6, Theorem 3], implies that |a|1/2 (1 + D2 )−s/4 ∈ L2 (N , τ ), for all s > p, and so |a|1/2 ∈ B2 (D, p).