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Borel functional calculus for bounded normal operators

Statement

Assume AC. Let $T$ be a bounded normal operator on a nonzero complex Hilbert space $H$ , let $E$ be its spectral projection valued measure on $σ (T)$ , and let $f \mapsto f (T)$ be the bounded Borel functional calculus of Borel functional calculus for a bounded normal operator. Then:

$f (T)$ extends the continuous calculus: for continuous $f$ it agrees with the operator denoted $f (T)$ by Continuous functional calculus for bounded normal operators, and $1_{B} (T) = E (B)$ for every Borel $B \subseteq σ (T)$ ;
the calculus is linear, unital, multiplicative and star-preserving: $(a f + b g) (T) = a f (T) + b g (T)$ , $1 (T) = I$ , $(f g) (T) = f (T) g (T)$ and $\overline{f} (T) = f (T)^{*}$ ;
$∥ f (T) x ∥^{2} = \int ∣ f ∣^{2} d E_{x}$ for every $x \in H$ and $∥ f (T) ∥ = ∥ f ∥_{E, \infty} \leq ∥ f ∥_{\infty}$ , the exact norm being the $E$ -essential supremum of $f$ ; in particular $f (T)$ is normal with $f (T)^{*} = \overline{f} (T)$ ;
if $(f_{n})$ are uniformly bounded Borel functions, $f$ is bounded Borel, and $f_{n} \to f$ pointwise $E$ -almost everywhere, then $f_{n} (T) \to f (T)$ in the strong operator topology;
every $S \in B (H)$ commuting with $T$ and $T^{*}$ commutes with every $f (T)$ .

Facts & Assumptions

[A1]

The spectral PVM $E$ of $T$ is the unique regular PVM on the Borel $σ$ -algebra of $σ (T)$ with $\int z d E = T$ ; it is obtained from the continuous calculus $π (f) = f (T)$ by the construction that represents each continuous functional $f \mapsto ⟨ π (f) x, y ⟩$ by a unique finite regular complex measure $μ_{x, y}$ with $\int f d μ_{x, y} = ⟨ f (T) x, y ⟩$ , and then $⟨ E (B) x, y ⟩ = μ_{x, y} (B)$ and $⟨ Φ_{E} (h) x, y ⟩ = \int h d μ_{x, y}$ for every bounded Borel $h$ (Continuous functional calculus produces a regular PVM, Spectral theorem for bounded normal operators pvm form).

[A2]

For continuous $f$ the calculus satisfies $Φ_{E} (f) = f (T)$ , and the map $f \mapsto f (T)$ on $C (σ (T))$ is a unital star-homomorphism that commutes with every $S$ satisfying $S T = T S$ and $S T^{*} = T^{*} S$ (Continuous functional calculus for bounded normal operators, Continuous functional calculus properties, Borel functional calculus for a bounded normal operator).

[A3]

$Φ_{E}$ is linear, unital, multiplicative and star-preserving on bounded Borel functions, $∥ Φ_{E} (f) ∥ = ∥ f ∥_{E, \infty}$ , $∥ Φ_{E} (f) x ∥^{2} = \int ∣ f ∣^{2} d E_{x}$ , and uniformly bounded pointwise $E$ -almost everywhere convergence implies strong convergence; by definition $f (T) = Φ_{E} (f)$ (Pvm integral is a star homomorphism, Bounded borel pvm integral, Borel functional calculus for a bounded normal operator).

[A4]

An operator $S$ commutes with $T$ and $T^{*}$ ; the adjoint satisfies $⟨ S x, y ⟩ = ⟨ x, S^{*} y ⟩$ and normal means $N N^{*} = N^{*} N$ (Hilbert-adjoint identities, Self-adjoint, positive, unitary and normal operators, Hilbert space).

[A5]

If two finite regular complex measures on a compact metric space have equal integrals against every continuous function then they are equal, because bounded complex functionals on $C (K; C)$ have a unique representing regular complex measure (The bounded complex dual of C_0(X) is regular complex measures).

[A6]

AC is the declared choice hypothesis of this page from the construction item onward (The Axiom of Choice).

Proof

technique · direct

Given: A nonzero complex Hilbert space $H$ , a bounded normal operator $T$ with spectral PVM $E$ and Borel calculus $f \mapsto f (T) = Φ_{E} (f)$ , and a bounded operator $S$ commuting with $T$ and $T^{*}$ .

1.1A2A3

Clauses 1 to 4: the agreement with the continuous calculus and $1_{B} (T) = E (B)$ are the definition of the Borel calculus and the identification $Φ_{E} (1_{B}) = E (B)$ ; linearity, unitality, multiplicativity and conjugation preservation, the quadratic identity, the norm formula and the strong-convergence property are the corresponding properties of $Φ_{E}$ ; normality follows because $f (T) f (T)^{*} = Φ_{E} (f \overline{f}) = Φ_{E} (\overline{f} f) = f (T)^{*} f (T)$ .

1.2A1A2A4

The measures $μ_{S x, y}$ and $μ_{x, S^{*} y}$ coincide: for every continuous $f$ one has $\int f d μ_{S x, y} = ⟨ f (T) S x, y ⟩ = ⟨ S f (T) x, y ⟩ = ⟨ f (T) x, S^{*} y ⟩ = \int f d μ_{x, S^{*} y}$ , using the commutant clause of the continuous calculus and the adjoint identity; both are finite regular complex measures, so equality of their integrals against all continuous functions gives $μ_{S x, y} = μ_{x, S^{*} y}$ .

2.1A1step 1.2A3A4A5

The commutant clause: for every bounded Borel $h$ and all $x, y \in H$ one has $⟨ h (T) S x, y ⟩ = \int h d μ_{S x, y} = \int h d μ_{x, S^{*} y} = ⟨ h (T) x, S^{*} y ⟩ = ⟨ S h (T) x, y ⟩$ , so $S h (T) = h (T) S$ ; in particular $S$ commutes with every spectral projection $E (B) = 1_{B} (T)$ and with every Borel calculus operator.

3.1step 1.1step 2.1A6∎

All five clauses hold: the Borel calculus extends the continuous calculus, is a unital star-homomorphism with $E$ -essential-supremum norm, is strongly continuous under uniformly bounded pointwise $E$ -almost everywhere convergence, and every operator commuting with $T$ and $T^{*}$ commutes with all of it.

Borel functional calculus for bounded normal operators

Statement

Facts & Assumptions

Proof

Depends on

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Sources