Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 5.
TOEPLITZ THEORY
Proof. If kerH φ ≠ {0} then since H φ f = 0 ⇒ (1 − P )φf = 0 ⇒ φf = P φf := g, we
have
∃ f, g ∈ H 2 s.t. φf = g.
Hence φ = g f . Remembering that if 1 φ ∈ L∞ then φ is outer if and only if 1 φ ∈ H∞
and dividing the outer part of f into g gives
φ = ψ θ
(ψ ∈ H ∞ , θ inner).
Conversely, if φ = ψ θ
(ψ ∈ H∞ , θ inner), then θ ∈ kerH φ because φθ = ψ ∈ H ∞ ⇒
(1 − P )φθ = 0 ⇒ H φ θ = 0.
From Theorem 5.3.1 we can see that
φ = ψ θ (θ, ψ inner), T φ subnormal ⇒ T φ normal or analytic (5.15)
The following proposition strengthen the conclusion of (5.15), whereas weakens
the hypothesis of (5.15).
Proposition 5.3.4. If φ = ψ θ (θ, ψ inner) and if T φ is hyponormal, then T φ is
analytic.
Proof. Observe that
1 = ||θ|| = ||P (θ)|| = ||P (φθφ)|| = ||P (φψ)||
= ||T φ (ψ)|| ≤ ||T φ (ψ)|| = ||P ( ψ2
θ
)|| ≤ ||ψ2 || = 1,
θ
which implies that ψ2
θ
∈ H 2 , so θ divides ψ 2 . Thus if one choose ψ and θ to be
relatively prime (i.e., if φ = ψ θ
is in lowest terms), then θ is constant. Therefore T φ
is analytic.
Proposition 5.3.5. If A is a weighted shift with weights a 0 , a 1 , a 2 , · · · such that
0 ≤ a 0 ≤ a 1 ≤ · · · < a N = a N+1 = · · · = 1,
then A is not unitarily equivalent to any Toeplitz operator.
Proof. Note that A is hyponormal, ||A|| = 1 and A attains its norm. If A is unitarily
equivalent to T φ then by a result of Brown and Douglas [BD], T φ is hyponormal and
φ = ψ θ (θ, ψ inner). By Proposition 5.3.4, T φ ≡ T ψ is an isometry, so a 0 = 1, a
contradiction.
√
Recall that the Bergman shift (whose weights are given by
The following question arises naturally:
n+1
n+2
) is subnormal.
Is the Bergman shift unitarily equivalent to a Toeplitz operator (5.16)
180