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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 $p, q$ .

1.1

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

A4algebra

1.2

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

A1A2

2.1

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

step 1.1step 1.2A3A5A6

3.1

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 $p (T) = q (T)$ .

step 2.1

4.1

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

step 2.1step 3.1∎

Depends on

Used by

Continuous functional calculus for bounded self adjoint operators Theorem

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, Theorem 5.54, printed pp.250–262 (standard reference, not scraped)
Dana P. Williams, Lecture Notes on the Spectral Theorem, §4, pp.10–13 (standard reference, not scraped)