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Woo Young Lee Lecture Notes on Operator Theory

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CHAPTER 4.<br />

WEIGHTED SHIFTS<br />

Therefore if W α is 2-hyp<strong>on</strong>ormal then by Lemma 4.4.7, the sequences<br />

{ α<br />

2<br />

n+1 − α 2 n<br />

α 2 n − α 2 n−1<br />

} ∞<br />

n=2<br />

and<br />

{ α<br />

2<br />

n−1 − α 2 n−2<br />

α 2 n − α 2 n−1<br />

} ∞<br />

n=2<br />

are both bounded, so that {k n } ∞ n=2 is bounded. This proves Claim 2.<br />

Write k := sup n k n . Without loss of generality we assume k < 1 (this is possible<br />

from the observati<strong>on</strong> that cα induces {c 2 k n }). Choose a sufficiently small perturbati<strong>on</strong><br />

α ′ of α such that if we let<br />

h :=<br />

sup<br />

0≤l≤N+2;0≤m≤1<br />

N+4<br />

∑<br />

∣<br />

k=4<br />

p ′ k−2<br />

⎛<br />

∏<br />

⎝<br />

N+3<br />

v j<br />

′<br />

j=k<br />

⎞<br />

⎠ c ′ (k − 3, l) + v N+3 ′ · · · v 3 ′ ρ ′ m<br />

∣<br />

(4.21)<br />

then<br />

c ′ (N + 3, i) − 1 h > 0 (0 ≤ i ≤ N + 3) (4.22)<br />

1 − k<br />

(this is always possible because by Theorem 4.4.2, we can choose a sufficiently small<br />

|p ′ i | such that<br />

c ′ (N + 3, i) > v 0 · · · v i−1 u i · · · u N+3 − ϵ and |h| < (1 − k) ( v 0 · · · v i−1 u i · · · u N+3 − ϵ )<br />

for any small ϵ > 0).<br />

Claim 3. For j ≥ 4 and 0 ≤ i ≤ N + j,<br />

)<br />

∑j−3<br />

c ′ (N + j, i) ≥ u N+j · · · u N+4<br />

(c ′ (N + 3, i) − k n h . (4.23)<br />

Proof of Claim 3. We use inducti<strong>on</strong>. If j = 4 then by Claim 1 and (4.21),<br />

c ′ (N + 4, i) = u ′ N+4c ′ (N + 3, i) + p ′ N+3c ′ (N + 2, i − 1)<br />

⎛ ⎞<br />

N+4<br />

∑<br />

N+3<br />

∏<br />

+ v N+4<br />

′ p ′ ⎝<br />

k−2 v j<br />

′ ⎠ c ′ (k − 3, i − N + k − 6) + v N+4 ′ · · · v 3ρ ′ ′ i−(N+3)<br />

k=4<br />

j=k<br />

n=1<br />

≥ u ′ N+4c ′ (N + 3, i) + p ′ N+3c ′ (N + 2, i − 1) − v ′ N+4h<br />

≥ u N+4<br />

(<br />

c ′ (N + 3, i) − k N+4 h )<br />

≥ u N+4<br />

(<br />

c ′ (N + 3, i) − k h )<br />

because u ′ N+4 = u N+4, v ′ N+4 = v N+4 and p ′ N+3 = p N+3 ≥ 0. Now suppose (4.23)<br />

131

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