10 A. L. CAREY, V. GAYRAL, A. RENNIE, and F. A. SUKOCHEV
in norm, it also converges in measure, and so we may apply the Fatou Lemma,
[26, Theorem 3.5 (i)], to deduce that
τ
(
(1 +
D2)−p/4−1/4nT ∗T
(1 +
D2)−p/4−1/4n
)
lim inf
k→∞
τ
(
(1 +
D2)−p/4−1/4nTk ∗Tk(1
+
D2)−p/4−1/4n
)
.
Since the same conclusion holds for TT

in place of T
∗T
, we see that
Qn(T ) lim inf
k→∞
Qn(Tk) = lim
k→∞
Qn(Tk) ∞,
and so T B2(D,p). Finally, fix ε 0 and n 1. Now choose N large enough so
that Qn(Tk Tl) ε for all k, l N. Applying the Fatou Lemma to the sequence
(Tk)k≥1, we have Qn(T Tl) lim infk→∞ Qn(Tk Tl) ε. Hence Tk T in the
topology of B2(D,p).
We now give some easy but useful stability properties of the algebras B2(D,p).
Lemma 1.7. Let T B2(D,p), S N and let f L∞(R).
(1) The operators Tf(D), f(D)T are in B2(D,p). If moreover T = T , then
Tf(T ) B2(D,p). In all these cases,
Qn(Tf(D)), Qn(f(D)T ), Qn(Tf(T )) f

Qn(T ).
(2) If
S∗S
T
∗T
and
SS∗
TT
∗,
then S B2(D,p) with Qn(S) Qn(T ).
(3) We have S B2(D,p) if and only if |S|,
|S∗|
B2(D,p).
(4) The real and imaginary parts (T ), (T ) belong to B2(D,p).
(5) If T = T
∗,
let T = T+ T− be the Jordan decomposition of T into pos-
itive and negative parts. Then T+, T− B2(D,p). Consequently B2(D,p) =
span{B2(D,p)+}.
Proof. (1) Since T (1 +
D2)−s/4,
T
∗(1
+
D2)−s/4

L2(N
, τ), we immediately
see that
Tf(D)(1 +
D2)−s/4
= T (1 +
D2)−s/4f(D),
¯(D)T
f
∗(1
+
D2)−s/4

L2(N
, τ),
and when T is self-adjoint, we also have
Tf(T )(1 +
D2)−s/4
= f(T )T (1 +
D2)−s/4,
¯(T
f )T (1 +
D2)−s/4

L2(N
, τ).
To prove the inequality we use the trace property to see that
τ((1 +
D2)−s/4
¯(D)T
f
∗Tf(D)(1
+
D2)−s/4)
= τ(T (1 +
D2)−s/4|f|2(D)(1
+
D2)−s/4T ∗)
f
2

τ((1 +
D2)−s/4T ∗T
(1 +
D2)−s/4),
and similarly for Tf(D) and Tf(T ) when T = T .
(2) Clearly, ϕs(S∗S) ϕs(T ∗T ) and ϕs(SS∗) ϕs(TT ∗). The assertion follows
immediately.
(3) This follows from Qn(T ) = (Qn(|T |) + Qn(|T
∗|))/2.
Item (4) follows since
B2(D,p) is a ∗-algebra, and then item (5) follows from (2), since for a self-adjoint
element T B2(D,p):
T
∗T
=
|T|2
= (T+ +
T−)2
= T+
2
+ T−
2
T+,
2 T−.2
This completes the proof.
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