Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 4.
WEIGHTED SHIFTS
(iii)
α 2 n
α 2 n+2
u n+2
u n+3
≤ un+1
u n+2
(n ≥ 0);
(iv)
p n ≥ 0 (n ≥ 0).
Proof. This follows from a straightforward calculation.
We are ready for:
Proof of Theorem 4.4.2. If α is not strictly increasing then α is flat, by the argument
of [Cu2, Corollary 6], i.e., α 0 = α 1 = α 2 = · · · . Then
[ ]
α
2
D n (s) = 0 + |s| 2 α0 4 ¯sα0
3
sα0 3 |s| 2 α0
4 ⊕ 0 ∞
(cf. (4.7)), so that (4.10) is evident. Thus we may assume that α is strictly increasing,
so that u n > 0, v n > 0 and w n > 0 for all n ≥ 0. Recall that if we write d n (t) :=
∑ n+1
i=0 c(n, i)ti then the c(n, i)’s satisfy the following recursive formulas (cf. (4.9)):
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) (n ≥ 0, 1 ≤ i ≤ n).
(4.14)
Also, c(n, n + 1) = v 0 · · · v n (again by (4.9) and p n := u n v n+1 − w n ≥ 0 (n ≥ 0), by
Lemma 4.4.6. A straightforward calculation shows that
d 0 (t) = u 0 + v 0 t;
d 1 (t) = u 0 u 1 + (v 0 u 1 + p 0 ) t + v 0 v 1 t 2 ;
d 2 (t) = u 0 u 1 u 2 + (v 0 u 1 u 2 + u 0 p 1 + u 2 p 0 ) t + (v 0 v 1 u 2 + v 0 p 1 + v 2 p 0 ) t 2 + v 0 v 1 v 2 t 3 .
(4.15)
Evidently,
c(n, i) ≥ 0 (0 ≤ n ≤ 2, 0 ≤ i ≤ n + 1). (4.16)
Define
β(n, i) := c(n, i) − v 0 · · · v i−1 u i · · · u n
(n ≥ 1, 1 ≤ i ≤ n).
For every n ≥ 1, we now have
⎧
⎪⎨ u 0 · · · u n ≥ 0 (i = 0)
c(n, i) = v 0 · · · v i−1 u i · · · u n + β(n, i) (1 ≤ i ≤ n)
⎪⎩
v 0 · · · v n ≥ 0 (i = n + 1).
For notational convenience we let β(n, 0) := 0 for every n ≥ 0.
Claim 1. For n ≥ 1,
c(n, n) ≥ u n c(n − 1, n) ≥ 0.
(4.17)
Proof of Claim 1. We use mathematical induction. For n = 1,
c(1, 1) = v 0 u 1 + p 0 ≥ u 1 c(0, 1) ≥ 0,
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