Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 2.
WEYL THEORY
Moreover, by (2.17), every Toeplitz operator with quasicontinuous symbol is essentially
normal. The (normal) Laurent operator M f◦φ on L 2 (T) has its spectrum contained
within the spectrum of the (essentially normal) Toeplitz operator T f◦φ . Thus
there is the following relationship involving the essentially normal operators f(T φ )
and M f◦φ ⊕ f(T φ ):
σ e
(
f(Tφ ) ⊕ M f◦φ
)
= σe (f(T φ )) and SP(f(T φ )) = SP ( f(T φ ) ⊕ M f◦φ
)
,
where SP(T ) denotes the spectral picture of an operator T . (The spectral picture
SP(T ) is the structure consisting of the set σ e (T ), the collection of holes and pseudoholes
in σ e (T ), and the Fredholm indices associated with these holes and pseudoholes.)
Thus it follows from the Brown-Douglas-Fillmore theorem [Pe] that f(T φ ) is compalent
to f(T φ ) ⊕ M f◦φ , in the sense that there exists a unitary operator W and a
compact operator K such that W ( f(T φ ) ⊕ M f◦φ
)
W ∗ + K = f(T φ ).
Corollary 2.2.9 (1) can be viewed as saying that σ(f(T φ )) \ σ(T f◦φ ) consists of
holes with winding number zero.
We consider the following question ([Ob2]):
if T φ is a Toeplitz operator, then does Weyl’s theorem hold for T 2 φ (2.19)
To answer the above question, we need a spectral property of Toeplitz operators with
continuous symbols.
Lemma 2.2.10. Suppose that φ is continuous and that f ∈ H(σ(T φ )). Then
σ(T f◦φ ) ⊆ f(σ(T φ )) , (2.20)
and equality occurs if and only if Weyl’s theorem holds for f(T φ ).
Proof. By Corollary 2.2.9, σ(T f◦φ ) = ω(f(T φ )) ⊆ σ(f(T φ )) = f(σ(T φ )). Because
σ(T φ ) is connected, so is f(σ(T φ )) = σ(f(T φ )); therefore the set π 00 (f(T φ )) is
empty. Again by Corollary 2.2.9, ω(f(T φ )) = σ(T f◦φ ) and so ω(f(T φ )) = σ(f(T φ )) \
π 00 (f(T φ )) if and only if σ(T f◦φ ) = f(σ(T φ )).
If φ is not continuous, it is possible for Weyl’s theorem to hold for some f(T φ )
without σ(T f◦φ ) being equal to f(σ(T φ )). One example is as follows. Let φ(e i θ ) =
e i θ
3 (0 ≤ θ < 2π), a piecewise continuous function. The operator T φ is invertible but
T φ 2 is not; hence 0 ∈ σ(T φ 2) \ {σ(T φ )} 2 . However ω(Tφ) 2 = σ(Tφ), 2 and π 00 (Tφ) 2 is
empty (see Figure 2); therefore Weyl’s theorem holds for Tφ.
2
We can now answer the question (2.19): the answer is no.
Example 2.2.11. There exists a continuous function φ ∈ C(T) such that σ(T φ 2) ≠
{σ(T φ )} 2 .
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