Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 4.
WEIGHTED SHIFTS
4.5 The Extensions
In [Sta3], J. Stampfli showed that given α : √ a, √ b, √ c with 0 < a < b < c, there
always exists a subnormal completion of α, but that for α : √ a, √ b, √ c, √ d (a < b <
c < d) such a subnormal completion may not exist.
There are instances where k-hyponormality implies subnormality for weighted
shifts. For example, in [CuF3], it was shown that if α(x) : √ x, ( √ a, √ b, √ c) ∧ (a <
b < c) then W α(x) is 2-hyponormal if and only if it is subnormal: more concretely,
W α(x) is 2-hyponormal if and only if
√ x ≤ H2 ( √ a, √ √
b, √ ab(c − b)
c) :=
(b − a) 2 + b(c − b) ,
in which case W α(x) is subnormal. In this section we extend the above result to weight
sequences of the form α : x n , · · · , x 1 , (α 0 , · · · , α k ) ∧ with 0 < α 0 < · · · < α k . We here
show:
Extensions of Recursively Generated Weighted Shifts.
If α : x n , · · · , x 1 , (α 0 , · · · , α k ) ∧ then
{
W α is ([ k+1
W α is subnormal ⇐⇒
2
] + 1)-hyponormal (n = 1)
W α is ([ k+1
2
] + 2)-hyponormal (n > 1).
In particular, the above theorem shows that the subnormality of an extension of
the recursive shift is independent of its length if the length is bigger than 1.
Given an initial segment of weights
α : α 0 , · · · , α 2k (k ≥ 0),
suppose ˆα ≡ (α 0 , · · · , α 2k ) ∧ , i.e., ˆα is recursively generated by α. Write
⎡ ⎤
γ ṇ
⎢
v n := ⎣ .
⎥
⎦ (0 ≤ n ≤ k + 1).
γ n+k
Then {v 0 , · · · , v k+1 } is linearly dependent in R k+1 . Now the rank of α is defined
by the smallest integer i (1 ≤ i ≤ k + 1) such that v i is a linear combination of
v 0 , · · · , v i−1 . Since {v 0 , · · · , v i−1 } is linearly independent, there exists a unique i-
tuple φ ≡ (φ 0 , · · · , φ i−1 ) ∈ R i such that v i = φ 0 v 0 + · · · + φ i−1 v i−1 , or equivalently,
γ j = φ i−1 γ j−1 + · · · + φ 0 γ j−i
(i ≤ j ≤ k + i),
which says that (α 0 , · · · , α k+i ) is recursively generated by (α 0 , · · · , α i ). In this case,
W α is said to be i-recursive (cf. [CuF3, Definition 5.14]).
We begin with:
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