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Continuous functional calculus properties

Statement

Assume AC. The continuous functional calculus on a nonzero complex Hilbert space preserves sums, products, conjugation and positivity, is norm-continuous, obeys continuous composition, sends eigenvectors at $λ$ to scalar evaluation at $λ$ , and commutes with every $S$ satisfying $S T = T S$ and $S T^{*} = T^{*} S$ .

Facts & Assumptions

[A1]

The normal calculus $f \mapsto f (T)$ is an isometric unital star-isomorphism $C (σ (T)) \to C^{*} (I, T)$ , hence complex-linear, multiplicative, unital, star-preserving and isometric; when $T$ is self-adjoint, this normal calculus agrees function-by-function with the self-adjoint calculus, which has the same properties and range (Continuous functional calculus for bounded normal operators, Continuous functional calculus for bounded self adjoint operators).

[A2]

$σ (f (T)) = f (σ (T))$ and $f (T)$ is normal (Spectral mapping for continuous normal functional calculus).

[A3]

$*$ -polynomials are uniformly dense in the continuous functions on a compact subset of $C$ (Complex Stone–Weierstrass dichotomy for separating self-adjoint algebras; the unital case is dense).

[A4]

$C^{*} (I, T)$ is the norm closure of the unital $*$ -algebra of $*$ -polynomials in $T$ , and multiplication in $B (H)$ is continuous (C star algebra generated by a normal operator, Hilbert-adjoint identities).

[A5]

For normal $T$ and $T x = λ x$ one has $T^{*} x = \overline{λ} x$ : $∥ (T^{*} - \overline{λ}) x ∥^{2} = ⟨ (T^{*} - \overline{λ}) (T - λ) x, x ⟩ = ⟨ (T - λ) x, (T - λ) x ⟩ = 0$ by normality (Self-adjoint, positive, unitary and normal operators, Hilbert-adjoint identities).

[A6]

If $X \geq 0$ is a bounded positive operator then $σ (X) \subseteq [0, + \infty)$ (Spectrum of a positive operator is nonnegative).

[A7]

AC is the hypothesis of the calculus and spectral suppliers (The Axiom of Choice).

Proof

technique · direct

Given: A nonzero complex Hilbert space $H$ , a bounded normal $T \in B (H)$ , functions $f, k \in C (σ (T))$ , a function $h \in C (σ (f (T)))$ , and an operator $S$ commuting with $T$ and $T^{*}$ .

1.1A1

Linearity, multiplicativity, conjugation and norm-continuity hold: $f (T)$ is the image of $f$ under an isometric unital $*$ -isomorphism, so $(f + k) (T) = f (T) + k (T)$ , $(f k) (T) = f (T) k (T)$ , $(λ f) (T) = λ f (T)$ , $\overline{f} (T) = f (T)^{*}$ and $∥ f (T) ∥ = ∥ f ∥_{\infty}$ .

1.2A1A2A6algebra

If $f \geq 0$ is real-valued, write $f = ∣ r ∣^{2}$ with $r = \sqrt{f}$ continuous; then $f (T) = r (T)^{*} r (T)$ and $⟨ f (T) x, x ⟩ = ∥ r (T) x ∥^{2} \geq 0$ , so $f (T)$ is a positive operator; conversely, if $f (T) \geq 0$ as a quadratic form, then $σ (f (T)) \subseteq [0, + \infty)$ and [A2] gives $f (σ (T)) \subseteq [0, + \infty)$ , so $f \geq 0$ .

1.3A4A5algebra

For every $*$ -polynomial $q$ in $z, \overline{z}$ and every eigenvector $x$ of $T$ at $λ$ , one has $T^{*} x = \overline{λ} x$ and hence $q (T, T^{*}) x = q (λ, \overline{λ}) x$ .

1.4A4algebra

If $S$ commutes with $T$ and $T^{*}$ , then $S$ commutes with every $*$ -polynomial in $T$ , and hence with every element of $C^{*} (I, T)$ by continuity of multiplication.

2.1step 1.3A1A3

Eigenvector identity: for eigenvectors $x$ of $T$ at $λ$ and $f$ continuous, approximate $f$ uniformly on $σ (T)$ by $*$ -polynomials $q_{n}$ ; then $f (T) x = \lim q_{n} (T) x = \lim q_{n} (λ, \overline{λ}) x = f (λ) x$ .

2.2step 1.1A1A2A3algebra

Composition: since $f (T)$ is normal and $σ (f (T)) = f (σ (T))$ , choose $*$ -polynomials $q_{n}$ converging uniformly to $h$ on $σ (f (T))$ . Multiplicativity and conjugation give $q_{n} (f (T), f (T)^{*}) = (q_{n} (f, \overline{f})) (T)$ . The left side converges to $h (f (T))$ by the isometry of the calculus for $f (T)$ , while the right side converges to $(h \circ f) (T)$ because $q_{n} \circ f \to h \circ f$ uniformly on $σ (T)$ . Thus $h (f (T)) = (h \circ f) (T)$ .

2.3step 1.4A4

Commutant: since $f (T) \in C^{*} (I, T)$ and $S$ commutes with all of $C^{*} (I, T)$ , $S f (T) = f (T) S$ .

3.1step 1.1step 1.2step 2.1step 2.2step 2.3A7∎

The calculus therefore preserves sums, products, conjugation and positivity, is norm-continuous, obeys continuous composition $h (f (T)) = (h \circ f) (T)$ , sends eigenvectors at $λ$ to the scalar $f (λ)$ , and commutes with every $S$ commuting with $T$ and $T^{*}$ .

Continuous functional calculus properties

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources