Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 4.
WEIGHTED SHIFTS
Write f m+1 := u m v m+1 −w m . If we choose α m+2 such that f m+1 ≥ 0, then d m+1 (t) ≥
0 for all t > 0. In particular we can choose α m+2 so that f m+1 = 0. i.e., u m v m+1 =
w m , or
α 2 m+2 := α2 m(α 2 m+1 − α 2 m−1) 2 + α 2 m−1α 2 m(α 2 m − α 2 m−1)
α 2 m+1 (α2 m − α 2 m−1 ) ,
or equivalently,
α 2 m+2 := α 2 m+1 + α2 m−1(α 2 m+1 − α 2 m) 2
α 2 m+1 (α2 m − α 2 m−1 ) .
In this case, d m+1 (t) ≥ u m+1 d m (t) ≥ 0. Repeating the argument (with α m+3 to be
chosen later), we obtain
d m+2 (t) = q m+2 (t)d m+1 (t) − |r m+1 (t)| 2 d m (t)
≥ 1 [
u m+1 q m+2 (t) − |r m+1 (t)|
]d 2 m+1 (t)
u m+1
= 1 [
]
u m+1 u m+2 + (u m+1 v m+2 − w m+1 ) t d m+1 (t)
u m+1
= u m+2 d m+1 (t) + t
u m+1
(u m+1 v m+2 − w m+1 ) d m+1 (t).
Write f m+2 := u m+1 v m+2 − w m+1 . If we choose α m+3 such that f m+2 = 0, i.e.,
α 2 m+3 := α 2 m+2 + α2 m(α 2 m+2 − α 2 m+1) 2
α 2 m+2 (α2 m+1 − α2 m) ,
then d m+2 (t) ≥ u m+2 d m+1 (t) ≥ 0. Continuing this process with the sequence {α n } ∞ n=m+2
defined recursively by
and
φ 1 := α2 m(αm+1 2 − αm−1)
2
αm 2 − αm−1
2 , φ 0 := − α2 m−1αm(α 2 m+1 2 − αm)
2
αm 2 − αm−1
2
α 2 n+1 := φ 1 + φ 0
α 2 n
(n ≥ m + 1),
we obtain that d n (t) ≥ 0 for all t > 0 and all n ≥ m + 2. On the other hand, by
an argument of [Sta3, Theorem 5], the sequence {α n } ∞ n=m+2 is bounded. Therefore,
a quadratically hyponormal completion {α n } ∞ n=0 is obtained. The above recursive
relation shows that the sequence {α n } ∞ n=m+2 is obtained recursively from α m−1 , α m
and α m+1 , that is, {α n } ∞ n=m−1 = (α m−1 , α m , α m+1 ) ∧ . This completes the proof.
Given four weights α : α 0 < α 1 < α 2 < α 3 , it may not be possible to find
a 2-hyponormal completion. In fact, by the preceding criterion for subnormal and
k-hyponormal completions, the following statements are equivalent:
(i) α has a subnormal completion;
(ii) α has a 2-hyponormal completion;
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