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Cyclic spectral representation

Statement

Assume AC. Let $T$ be a bounded normal operator on a nonzero complex Hilbert space $H$ with spectral projection valued measure $E$ , let $x \in H$ , and let $H_{x}$ be the cyclic subspace of Cyclic vector and cyclic normal operator. Then:

the formula $U [f] : = f (T) x$ , initially defined for continuous $f$ , extends uniquely to a unitary operator $U : L^{2} (σ (T), E_{x}) \to H_{x}$ satisfying $∥ U [f] ∥^{2} = \int ∣ f ∣^{2} d E_{x}$ , and the class $[f]$ of a continuous function is mapped to $f (T) x$ ;
$U$ intertwines multiplication by the coordinate function with the restriction of $T$ : $U M_{z} = T U$ on $L^{2} (σ (T), E_{x})$ , where $(M_{z} f) (z) = z f (z)$ ;
more generally $U M_{h} = h (T) U$ for every bounded Borel function $h$ on $σ (T)$ , where $M_{h}$ is multiplication by $h$ ;
if $x$ is cyclic, that is $H_{x} = H$ , then $T$ is unitarily equivalent to multiplication by the coordinate on $L^{2} (σ (T), E_{x})$ .

Facts & Assumptions

[A1]

$E_{x} (B) = ⟨ E (B) x, x ⟩$ is a positive measure on the Borel $σ$ -algebra of $σ (T)$ with $E_{x} (σ (T)) = ∥ x ∥^{2} < + \infty$ , and for every bounded Borel $f$ one has $∥ f (T) x ∥^{2} = \int ∣ f ∣^{2} d E_{x}$ (Scalar and complex measures from a pvm, Bounded borel pvm integral).

[A2]

$E$ is regular, $σ (T) \subseteq C$ is compact and locally compact Hausdorff, $C (σ (T))$ is dense in $L^{p} (E_{x})$ for $1 \leq p < \infty$ , and $L^{2} (E_{x})$ is complete (Regular Borel measure on an LCH space, C_c(X) is dense in L^p(mu) for a Radon measure, Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ ).

[A3]

Elements of $L^{2} (σ (T), E_{x})$ are almost-everywhere classes of measurable functions, and a measurable function is unchanged as a class by modification on an $E_{x}$ -null set (The space $L^{p} (μ)$ as the quotient by null functions).

[A4]

The Borel calculus is multiplicative, $Φ_{E} (f g) = f (T) g (T)$ , and $T = Φ_{E} (z)$ ; for continuous $f$ one has $T f (T) x = (z f) (T) x$ and $T^{*} f (T) x = (\overline{z} f) (T) x$ (Borel functional calculus for bounded normal operators, Borel functional calculus for a bounded normal operator, Continuous functional calculus for bounded normal operators).

[A5]

$H_{x}$ is the closed span of ${f (T) x : f \in C (σ (T))}$ ; an isometry from a complete space has closed image, and a linear isometry with dense image into a Hilbert space is unitary onto its codomain (Cyclic vector and cyclic normal operator).

[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 $E$ , a vector $x \in H$ with cyclic subspace $H_{x}$ , and the map $U_{0} [f] : = f (T) x$ on $C (σ (T))$ .

1.1

Isometry of $U_{0}$ : for $f \in C (σ (T))$ one has $∥ U_{0} [f] ∥^{2} = ∥ f (T) x ∥^{2} = \int ∣ f ∣^{2} d E_{x}$ , so $U_{0}$ preserves norms and, by linearity of the calculus, is complex-linear; if $f$ and $g$ agree $E_{x}$ -almost everywhere then $∥ U_{0} [f] - U_{0} [g] ∥^{2} = \int ∣ f - g ∣^{2} d E_{x} = 0$ , so $U_{0}$ is well defined on classes.

A1A3A4

2.1

$U_{0}$ extends uniquely to an isometry $U$ on the closure of the continuous classes, which is $L^{2} (σ (T), E_{x})$ : for any Cauchy sequence of continuous functions the images are Cauchy, and the limit is independent of the approximating sequence because two uniformly- $L^{2}$ equivalent choices differ by a sequence with vanishing norm.

step 1.1A2A3

3.1

The image of $U$ is $H_{x}$ : the continuous classes map onto the set ${f (T) x : f \in C (σ (T))}$ , whose closed span is $H_{x}$ by definition, and an isometry with complete domain has closed image, so the image of the extension is exactly $H_{x}$ ; hence $U : L^{2} (σ (T), E_{x}) \to H_{x}$ is a unitary.

step 2.1A2A5

3.2

Borel multipliers: first let $q$ be bounded Borel and choose continuous $q_{n} \to q$ in $L^{2} (E_{x})$ . Then $U [q] = \lim_{n} q_{n} (T) x = q (T) x$ , because $∥ (q_{n} (T) - q (T)) x ∥^{2} = \int ∣ q_{n} - q ∣^{2} d E_{x} \to 0$ . Now, for bounded Borel $h$ and continuous $f$ , the product $h f$ is bounded Borel, so $U (M_{h} f) = (h f) (T) x = h (T) f (T) x = h (T) U (f)$ by multiplicativity. Since $M_{h}$ and $h (T)$ are bounded, density extends this equality to all of $L^{2} (σ (T), E_{x})$ .

step 2.1A1A2A4

4.1

Intertwining with the coordinate: for continuous $f$ one has $U (M_{z} f) = (z f) (T) x = T f (T) x = T U (f)$ , because $T = Φ_{E} (z)$ and the calculus is multiplicative; both $U M_{z}$ and $T U$ are bounded linear maps agreeing on the dense set of continuous classes, so $U M_{z} = T U$ on $L^{2} (σ (T), E_{x})$ .

step 2.1step 3.1A2A4

5.1

If $x$ is cyclic then $H_{x} = H$ and the unitary $U : L^{2} (σ (T), E_{x}) \to H$ satisfies $U M_{z} U^{- 1} = T$ , so $T$ is unitarily equivalent to multiplication by the coordinate.

step 4.1A5

6.1

The cyclic representation $U$ is a unitary onto the cyclic subspace, intertwines the coordinate multiplication with $T$ , intertwines every bounded Borel multiplier with the Borel calculus, and is a unitary equivalence between $T$ and multiplication by the coordinate when $x$ is cyclic.

step 3.1step 4.1step 3.2step 5.1A6∎

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Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, Theorem 5.84, printed pp.294–296 (standard reference, not scraped)
John B. Conway, A Course in Functional Analysis, 2nd ed., Chapter IX §10, printed pp.293–299 (standard reference, not scraped)