Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 4.
WEIGHTED SHIFTS
then c(n, i) satisfy a double-indexed recursive formula, i.e.,
⎧
c(1, 1) = u 1 v 0 + v 1 u 0 − w 0
⎪⎨
c(n, 0) = u 0 · · · u n
c(n, n + 1) = v 0 · · · v n
⎪⎩
c(n + 2, i) = u n+2 c(n + 1, i) + v n+2 c(n + 1, i − 1) − w n+1 c(n, i − 1).
Theorem 4.3.6. (Outer propagation) Let T be a weighted shift with weight sequence
{α n } ∞ n=0. If T is quadratically hyponormal then
α n = α n+1 = α n+2 for some n =⇒ α n = α n+1 = α n+2 = α n+3 = · · · .
Proof. We may assume that n = 0 and α 0 = α 1 = α 2 = 1. We want to show that
α 3 = 1. A straightforward calculation shows that
d 0 = 1 + t
d 1 = t 2
d 2 = ( α 2 3 − 1 ) t 3
d 3 = ( α 2 3 − 1 ) ( α 2 3α 2 4 − 1 ) t 4
d 4 = q 4 d 3 − γ 2 3d 2
So
= [( α 2 4 − α 2 3)
+ t
(
α
2
4 α 2 5 − α 2 3)] (
α
2
3 − 1 ) ( α 2 3α 2 4 − 1 ) t 4 − tα 2 3
(
α
2
4 − 1 ) 2 (
α
2
3 − 1 ) t 3 .
which implies that α 3 = 1.
d 4
lim
t→0 + t 4 = ( −α2 4 α
2
3 − 1 ) 3
≥ 0,
Theorem 4.3.7. (Inner Propagation) Let T be a weighted shift with weight sequence
{α n } ∞ n=0. If T is quadratically hyponormal then
α n = α n+1 = α n+2 for some n =⇒ α 1 = · · · = α n .
Proof. Withou loss of generality we may assume n = 2, i.e., α 2 = α 3 = α 4 = 1. We
want to show that α 1 = 1. We consider d 3 . Now,
Therefore
d 3 (0) = q 3 (0)d 2 (0) = 0 since q 3 (0) = α 2 3 − α 2 2 = 0
d ′ 3(0) = q ′ 3(0)d 2 (0) − α 2 2(α 2 3 − α 2 1) 2 α 1 (0) = · · · = 0
d ′′
3(0) = 2q ′ 3(0)d ′ 2(0) − 2(1 − α 2 1)α ′ 1(0) = · · · = −2α 4 0(1 − α 2 1) 3 .
Since d 3 ≥ 0 (all t ≥ 0), it follows α 1 = 1.
d 3 (t) = −α 4 0(1 − α 2 1) 3 t 2 + · · · .
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