Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 5.
TOEPLITZ THEORY
Proof. Suppose A 0 is k-hyponormal. We now use induction on k. If k = 2 then A 0 is
2-hyponormal, and so D 0 := [A ∗ 0, A 0 ] ≥ 0. By Theorem 5.3.26 (i), D 1 2
0 A 0 D − 1 2
0 | ran (D0)
is bounded. Let A 1 be the bounded extension of D 1 2
0 A 0 D − 1 2
0[
from Ran
]
(D 0 ) to H 1 :=
Ran (D 0 ) and D 1 := D 0 | H1 + [A ∗ 1, A 1 ]. Writing Â0 :=
A 0 D 1 2
0 , we have
0 A Â0 =
1
m.p.n.e. (A 0 ), which is hyponormal by Theorem 5.3.26(ii). Thus
[Â0∗
, Â 0 ] =
[ ]
0 0
0 D 0 | H1 + [A ∗ ≥ 0.
1, A 1 ]
and hence D 1 ≥ 0. Also by [CuL2, Lemma 2.2], A 0 (ker D 0 ) ⊆ ker D 0 whenever A 0 is
2-hyponormal. Thus (I n ), (II n ), and (III n ) hold for n = 0, 1. Assume now that if A 0
is k-hyponormal then (I n ),(II n ) and (III n ) hold for all 0 ≤ n ≤ k − 1. Suppose A 0 is
(k + 1)-hyponormal. We must show that (I n ),(II n ) and (III n ) hold for n = k. Define
⎡
⎤
A 0 D 1 2 0 0
A 1 D 1 2
1
S :=
. .. . ..
⊕k−1
⊕k−1
: H i −→ H i .
⎢
⎣
. ..
1 ⎥ i=0
i=0
D
2 ⎦
k−2
0 A k−1
By our inductive assumption, D k−1 ≥ 0. Writing ̂T (n) := m.p.n.e.( ̂T (n−1) ) when it
exists, we can see by our assumption that S = Â0(k−1)
: indeed, if
⎡
⎤
A 0 D 1 2 0 0
A 1 D 1 2
1
S l :=
. .. . ..
⎢
⎣
. ⎥ ..
D 1 2 ⎦
l−2
0 A l−1
then since by assumption [Sl ∗, S l] = 0 ⊕ D l and A l = D 1 2
l−1
A l−1 D − 1 2
l−1 | Ran (D l−1 ), it
follows that S l is the minimal partially normal extension of S l−1 (1 ≤ l ≤ k − 1). But
since by our assumption A 0 is (k + 1)-hyponormal, it follows from Lemma 5.3.26(ii)
that S is 2-hyponormal. Thus by Theorem 5.3.26(i), [S ∗ , S] 1 2 S[S ∗ , S] − 1 2 | Ran ([S∗ ,S]) is
bounded, which says that D 1 2
k−1
A k−1 D − 1 2
k−1 | Ran (D k−1 ) is bounded, proving (III n ) for
]
n = k. Observe that A k , H k and D k are well-defined. Writing
[S Ŝ := D 1 2
k−1 ,
0 A k
we can see that Ŝ = m.p.n.e.(S), [ ] which is hyponormal, again by Theorem 5.3.26(ii).
0 0
Thus, since [Ŝ∗ , Ŝ] = ≥ 0, we have D
0 D k ≥ 0, proving (I n ) for n = k. On the
k
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