Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 3.
HYPONORMAL AND SUBNORMAL THEORY
Proposition 3.3.11. If S is a subnormal operator then the following hold:
(a) (Halmos, 1952) σ n (S) ⊆ σ(S).
(b) σ ap (S) ⊆ σ n (S) and ∂σ(S) ⊆ ∂σ n (S).
(c) (Bram, 1955) If U is a bounded component of C\σ n (S), then either U ∩σ(S) = ∅
or U ⊆ σ(S).
Proof. (a) We want to show that S is invertible ⇒ N is invertible.
If N = ∫ zdE(z) is the spectral decomposition of N, ε > 0, and M = E (B(0; ε)) K
then we claim that
||N k f|| ≤ ε k ∥f∥ for k = 1, 2, 3, · · · and f ∈ M.
To see this let △ := B(0; ε). Then
∫
NE(△) = zχ △ (z)dE(z) = ϕ(N), where ϕ = zχ △ .
We thus have
∥NE(△)∥ = ∥ϕ(N)∥ ≤ ∥ϕ∥ = sup|ϕ(z)| = sup{|z| : z ∈ △} ≤ ε.
So if f ∈ M then E(△)f = f. Therefore
∥Nf∥ = ∥NE(△)f∥ ≤ ∥NE(△)∥∥f∥ ≤ ε∥f∥.
So if f ∈ M and h ∈ H,
|⟨f, h⟩| = ∣ ∣⟨f, S k S −k h⟩ ∣ = ∣ ∣⟨f, N k S −k h⟩ ∣ ∣
= ∣⟨N ∗k f, S −k h⟩ ∣
Letting k → ∞ shows that
≤ ∥ ∥N ∗k f ∥ ∥ · ∥∥S −k h ∥ ∥ ≤ ε k ∥f∥ ∥ ∥S −k∥ ∥ ≤ ε k ∥ ∥S −1∥ ∥ k ∥f∥ ∥h∥.
ε < 1
∥S −1 ∥
=⇒ ⟨f, h⟩ = 0,
so that H ⊆ M ⊥ . Since M is a reducing subspace for N, N| M ⊥ is a normal extension
of S. By the minimality of N, M = {0} and so N is invertible because N = N φ and
|φ(x) ≥ ε a.e.
(b) Observe that
λ ∈ σ ap (S) =⇒ ∃ unit vectors h n ∈ H such that ∥(λ − S)h n ∥ −→ 0.
But (λ − S)h n = (λ − N)h n .
=⇒ σ ap (S) ⊆ σ ap (N) = σ(N) = σ n (S).
λ ∈ ∂σ(S) =⇒ λ ∈ σ ap (S) =⇒ λ ∈ σ n (S) =⇒ λ /∈ int σ n (S) =⇒ λ ∈ ∂σ n (S).
(c) (Due to S. Parrot) Let U be a bounded component of σ n (S) c and put
U + = U \ σ(S) and U − = U ∩ σ(S).
So U = U − ∪ U + , U + ∩ U − = ∅ and U + is open. By (b), U − = U ∩ int σ(S), so that
U − is open. By the connectedness of U, either U + = ∅ or U − = ∅.
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