Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 4.
WEIGHTED SHIFTS
In fact,
⎧
x 1 = H i (α 0 , · · · , α 2i−2 )
⎪⎨
· · · · · · · · ·
x n−1 = H i (x n−2 , · · · , · · · , α 2i−n )
⎪⎩
x n ≤ H i (x n−1 , · · · , α 2i−n−1 ),
where H i is the modulus of i-hyponormality (cf. [CuF3, Proposition 3.4 and (3.4)]),
i.e.,
H i (α) := sup{x > 0 : W xα is i-hyponormal}.
Therefore, W α = W xn (x n−1 ,··· ,α 2i−n−1 ) ∧.
Proof. Consider the (k + 1) × (l + 1) “Hankel” matrix A(n; k, l) by
⎡
⎤
γ n γ n+1 . . . γ n+l
γ n+1 γ n+2 . . . γ n+1+l
A(n; k, l) := ⎢
⎥
⎣ . .
. ⎦
γ n+k γ n+k+1 . . . γ n+k+l
(n ≥ 0).
Case 1 (α : x 1 , (α 0 , · · · , α k ) ∧ ): Let Â(n; k, l) and A(n; k, l) denote the Hankel
matrices corresponding to the weight sequences (α 0 , · · · , α k ) ∧ and α, respectively.
Suppose W α is ([ k+1
2 ] + 1)-hyponormal. Then by Lemma 4.5.2, W (α 0,··· ,α k ) ∧ is subnormal.
Observe that
A(n + 1; m, m) = x 2 1 Â(n; m, m) for all n ≥ 0 and all m ≥ 0.
Thus it suffices to show that A(0; m, m) ≥ 0 for all m ≥ [ k+1
2
] + 2. Also note that
if ˜B denotes the (k − 1) × k matrix obtained by eliminating the first row of a k × k
matrix B then
Ã(0; m, m) = x 2 1
+ 1
Â(0; m − 1, m) for all m ≥ [k ] + 2.
2
Therefore for every m ≥ [ k+1
k+1
2
] + 2, A(0; m, m) is a flat extension of A(0; [
2 ] +
1, [ k+1
k+1
2
] + 1). This implies A(0; m, m) ≥ 0 for all m ≥ [
2 ] + 2 and therefore W α is
subnormal.
Case 2 (α : x n , · · · , x 1 , (α 0 , · · · , α k ) ∧ )): As in Case 1, let Â(n; k, l) and A(n; k, l)
denote the Hankel matrices corresponding to the weight sequences (α 0 , · · · , α k ) ∧ and
k+1 k+1
α, respectively. Observe that det Â(n; [
2
] + 1, [
2
] + 1) = 0 for all n ≥ 0. Suppose
W α is ([ k+1
2
] + 2)-hyponormal. Observe that
so that
A(n + 1; [ k + 1
2
] + 1, [ k + 1 ] + 1) = x 2 1 · · · x
2
det A(n + 1; [ k + 1
2
2 + 1
nÂ(1; [k
2
] + 1, [ k + 1 ] + 1),
2
] + 1, [ k + 1 ] + 1) = 0. (4.29)
2
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