1. PSEUDODIFFERENTIAL CALCULUS AND SUMMABILITY 9
Notation. Given a semifinite von Neumann algebra N with faithful normal
semifinite trace τ, we let
L , τ), 1 ≤ p ∞, denote the set of τ-measurable
operators T aﬃliated to N with τ(|T
∞. We do not often use this no-
tion of p-integrable elements, preferring to use the bounded analogue,
, τ) :=
L , τ) ∩ N , normed with T → τ(|T |p)1/p + T .
Remarks. (1) If
, τ) for all (s) p ≥ 1, then B2(D,p) =
N , since then the weights ϕs, s p, are bounded and the norms Qn are all
equivalent to the operator norm.
(2) The triangle inequality for Qn follows from the Cauchy-Schwarz inequal-
ity applied to the inner product T, S
along with the equality
Qn(T )2 = T 2 + T, T
+ T ∗,T ∗
. In concrete terms, an element T ∈ N belongs
to B2(D,p) if and only if for all s p, both T (1 +
, τ), the ideal of τ-Hilbert-Schmidt operators.
(3) The norms Qn are increasing, in the sense that for n ≤ m we have Qn ≤ Qm.
We leave this as an exercise, but observe that this requires the cyclicity of the trace.
The following result of Brown and Kosaki gives the strongest statement on this
cyclicity. By the preceding Remark (2), we do not need the full power of this result
here, but record it for future use.
Proposition 1.4. [8, Theorem 17] Let τ be a faithful normal semifinite trace
on a von Neumann algebra N , and let A, B be τ-measurable operators aﬃliated to
N . If AB, BA ∈
L , τ) then τ(AB) = τ(BA).
Another important result that we will frequently use comes from Bikchentaev’s
Proposition 1.5. [6, Theorem 3] Let N be a semifinite von Neumann algebra
with faithful normal semfinite trace τ. If A, B ∈ N satisfy A ≥ 0, B ≥ 0, and are
such that AB is trace class, then
are also trace class,
with τ(AB) =
Next we show that the topological algebra B2(D,p) is complete and thus is a
Fr´ echet algebra. The completeness argument relies on the Fatou property for the
trace τ, .
Proposition 1.6. The ∗-algebra B2(D,p) ⊂ N is a Fr´ echet algebra.
Proof. Showing that B2(D,p) is a ∗-algebra is routine with the aid of the
following argument. For T, S ∈ B2(D,p), the operator inequality
and, therefore, Qn(TS) ≤ Qn(T ) Qn(S).
For the completeness, let (Tk)k≥1 be a Cauchy sequence in B2(D,p). Then
(Tk)k≥1 converges in norm, and so there exists T ∈ N such that Tk → T in N .
For each norm Qn we have | Qn(Tk) − Qn(Tl) | ≤ Qn(Tk − Tl), so we see that the
numerical sequence (Qn(Tk))k≥1 possesses a limit. Now since
(1+D2)−p/4−1/4nTk ∗ Tk(1+D2)−p/4−1/4n