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Spectral mapping for continuous normal functional calculus

Statement

Assume AC. For a bounded normal operator $T$ on a nonzero complex Hilbert space and $f \in C (σ (T))$ , the operator $f (T)$ is normal and $σ_{B (H)} (f (T)) = f (σ (T))$ . This includes constant functions and disconnected spectra.

Facts & Assumptions

[A1]

The normal calculus $f \mapsto f (T)$ is an isometric unital star-isomorphism $C (σ (T)) \to C^{*} (I, T)$ with $z \mapsto T$ , so $f (T)^{*} = \overline{f} (T)$ and $f (T) g (T) = (f g) (T)$ (Continuous functional calculus for bounded normal operators).

[A2]

The algebra $C^{*} (I, T)$ is commutative when $T$ is normal, and it is a unital C*-subalgebra of $B (H)$ with the same identity (C star algebra generated by a normal operator as used by the calculus, C star algebra).

[A3]

For a unital C*-subalgebra $B \subseteq A$ with the same identity one has $σ_{B} (b) = σ_{A} (b)$ for every $b \in B$ (Spectral permanence for unital c star subalgebras).

[A4]

For a nonzero commutative unital complex Banach algebra $C$ and $c \in C$ one has $σ_{C} (c) = {χ (c) : χ \in Δ (C)}$ (Spectrum as character values).

[A5]

For normal $T$ the map $Φ (χ) = χ (T)$ is a homeomorphism $Δ (C^{*} (I, T)) \to σ (T)$ . More precisely, $χ \circ Ψ$ is the evaluation character at the unique point $Φ (χ)$ , so $χ (f (T)) = f (Φ (χ))$ (Character space of generated normal algebra is operator spectrum, Characters of continuous functions are evaluations).

[A6]

The operator spectrum of $f (T) \in C^{*} (I, T) \subseteq B (H)$ is the spectrum in $B (H)$ (Spectrum and resolvent of a bounded operator for the spectrum convention).

[A7]

For a constant function $f \equiv c$ , unitality and linearity give $f (T) = c I$ ; moreover $λ I - c I = (λ - c) I$ is boundedly invertible exactly when $λ \neq c$ , so $σ (c I) = {c}$ (Continuous functional calculus for bounded normal operators, Spectrum and resolvent of a bounded operator).

[A8]

AC is the hypothesis of the permanence and character-space suppliers (The Axiom of Choice).

Proof

technique · direct

Given: A nonzero complex Hilbert space $H$ , a bounded normal $T \in B (H)$ and $f \in C (σ (T))$ , with $f (T)$ the image of $f$ under the normal calculus.

1.1A1A2

$f (T)$ lies in the commutative C*-algebra $C^{*} (I, T)$ and $f (T)^{*} = \overline{f} (T)$ also lies there, so the two commute and $f (T)$ is normal.

1.2A4A5

The spectrum of $f (T)$ computed in $C^{*} (I, T)$ is the set of character values. For $χ \in Δ (C^{*} (I, T))$ , A5 gives $χ (f (T)) = f (Φ (χ))$ ; hence $σ_{C^{*} (I, T)} (f (T)) = f (Φ (Δ (C^{*} (I, T)))) = f (σ (T))$ .

2.1step 1.1step 1.2A3A6A8

Spectral permanence for the unital C*-subalgebra $C^{*} (I, T) \subseteq B (H)$ gives $σ_{B (H)} (f (T)) = σ_{C^{*} (I, T)} (f (T)) = f (σ (T))$ .

3.1step 2.1A7∎

Hence $f (T)$ is normal and its operator spectrum is the image of the spectrum of $T$ under $f$ ; constant functions give $f (T) = c I$ and $f (σ (T)) = {c}$ , and no connectedness of $σ (T)$ is used, so disconnected spectra are covered by the same pointwise argument.

Spectral mapping for continuous normal functional calculus

Statement

Facts & Assumptions

Proof

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