Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 5.
TOEPLITZ THEORY
From (5.18) we have
e n = φ
(a n e n+1 + √ )
√
1 − a 2 n η n = a 2 ne n + a n 1 − a
2 n ξ n + √ 1 − a 2 n φ η n . (5.20)
Then (5.20) is equivalent to
φη n = −a n ξ n + √ 1 − a 2 n e n . (5.21)
Set d n := η n
t
and ρ n := ξ n
t
(|t| = 1). Then () is equivalent to
φd n = −a n ρ n + √ 1 − a 2 n
e n
t . (5.22)
Since en t
∈ (H 2 ) ⊥ and {d n } ∞ n=0 is an orthonormal basis for H 2 , we can see that
{
||T φ d 0 || = a 0 = inf ||x||=1 ||T φ x|| = ||T φ e 0 ||
(5.23)
||T φ d l || = a l = ||T φ e l || .
Then (5.17) and (5.23) imply
d n = r n e n (|r n | = 1). (5.24)
Substituting (5.24) into (5.23) and comparing it with (5.18) gives
a n e n+1 + √ 1 − a 2 n η n = φe n = − a n
r n
ρ n +
√
1 − a
2 n
r n
e n
t ,
which implies {
−r n ρ n = e n+1
r n
e nt
= η n .
Therefore (5.18) is reduced to:
{
φe n = a n e n+1 + √ 1 − a 2 n r n
e nt
φe n+1 = a n e n − √ 1 − a 2 n r n
e n+1
t
(5.25)
(5.26)
Put e −(n+1) := e n
t
∈ (H 2 ) ⊥ (n ≥ 0). We now claim that
φe 0 = re −1 (|r| = 1) : (5.27)
( )
indeed, T φe0
φ t
= P ( e0 t ) = 0, so e 0 = r φe0
t
for |r| = 1, and hence φe 0 = re −1 . From
(5.26) we have
√
φe 0 = a 0 e 1 + r 0
√1 − a 2 0 e −1 = a 0 e 1 + r 0 r 1 − a 2 0 φe 0, (5.28)
or, equivalently, (
)
φ − r 0 r
√1 − a 2 0 φ e 0 = a 0 e 1 . (5.29)
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