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Polynomial calculus is isometric for self adjoint operators

Statement

Assume AC. If $T = T^{*}$ and $p$ is a complex polynomial, then $∥ p (T) ∥ = \max_{λ \in σ (T)} ∣ p (λ) ∣$ ; hence polynomial restriction classes on $σ (T)$ give a well-defined isometric calculus.

Facts & Assumptions

[A1]

For $T = T^{*}$ the spectrum satisfies $σ (T) \subseteq R$ ; a self-adjoint operator is normal (Spectrum of a self adjoint operator is real, Self-adjoint, positive, unitary and normal operators).

[A2]

$σ (p (a)) = p (σ (a))$ for every polynomial $p$ and every element $a$ of a unital complex Banach algebra (Polynomial spectral mapping).

[A3]

For a normal element $a$ of a unital complex C*-algebra one has $r (a) = ∥ a ∥$ , and $B (H)$ is such a C*-algebra (C star spectral radius equals norm for normal elements, Bounded Hilbert operators form a C star algebra).

[A4]

$(a T + b S)^{*} = \overline{a} T^{*} + \overline{b} S^{*}$ and $(S T)^{*} = T^{*} S^{*}$ , so for a polynomial $p (z) = \sum_{k} c_{k} z^{k}$ and $T = T^{*}$ one has $p (T)^{*} = \overline{p} (T)$ with $\overline{p} (z) = \sum_{k} \overline{c_{k}} z^{k}$ ; the maps $p \mapsto p (T)$ and $p \mapsto \overline{p} (T)$ are ring homomorphisms (Hilbert-adjoint identities).

[A5]

For a normal operator the norm equals the spectral radius and the maximum is attained: $∥ T ∥ = r (T) = \max {∣ λ ∣ : λ \in σ (T)}$ (Normal operator norm equals spectral radius).

[A6]

AC is the hypothesis of the spectral-radius and Gelfand-theoretic suppliers (The Axiom of Choice).

Proof

technique · direct

Given: A nonzero complex Hilbert space $H$ , a bounded self-adjoint $T \in B (H)$ and complex polynomials $p, q$ .

1.1A4algebra

$p (T)^{*} = \overline{p} (T)$ , and $p (T)$ and $\overline{p} (T)$ are polynomials in $T$ , hence commute; therefore $p (T)$ is normal.

1.2A1A2

The spectrum of $T$ is contained in $R$ , and polynomial spectral mapping gives $σ (p (T)) = p (σ (T))$ .

2.1step 1.1step 1.2A3A5A6

$∥ p (T) ∥ = \max_{λ \in σ (T)} ∣ p (λ) ∣$ : by normality of $p (T)$ its norm is the spectral radius, the spectral radius is the maximum of $∣ z ∣$ over $σ (p (T))$ , and $σ (p (T)) = p (σ (T))$ .

3.1step 2.1

If $p$ and $q$ agree on $σ (T)$ , then $p - q$ vanishes there, so $∥ p (T) - q (T) ∥ = \max_{λ \in σ (T)} ∣ p (λ) - q (λ) ∣ = 0$ and $p (T) = q (T)$ .

4.1step 2.1step 3.1∎

The assignment $[p] \mapsto p (T)$ is therefore well defined on restriction classes, and it preserves the supremum norm because $∥ p (T) ∥ = \max_{σ (T)} ∣ p ∣ = ∥ p ∣_{σ (T)} ∥_{\infty}$ .

Polynomial calculus is isometric for self adjoint operators

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources