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Spectral Measures and Borel Functional Calculus

1 · Prerequisites

2 · Summary

This page refines the continuous functional calculus of the preceding pair into the measurable calculus of a bounded normal operator. The route is the projection valued measure: a PVM assigns an orthogonal projection to every measurable set, multiplicatively on intersections and strongly countably additively, and its scalar pairings $E_{x, y} (B) = ⟨ E (B) x, y ⟩$ are finite complex measures of total variation at most $∥ x ∥ ∥ y ∥$ . The page first fixes those conventions, proves that weak and strong countable additivity coincide for projection values, and constructs the integral of a simple function as a finite sum of projection values, which is shown to be independent of the disjoint presentation. Uniform approximation by simple functions then extends the integral to every bounded measurable function, with $∥ Φ_{E} (f) ∥ \leq ∥ f ∥_{\infty}$ , the scalar pairing identity, the quadratic identity $∥ Φ_{E} (f) x ∥^{2} = \int ∣ f ∣^{2} d E_{x}$ , and exact norm the $E$ -essential supremum, the supremum over unit vectors of the $E_{x}$ -essential supremum of $f$ . The measurable integral is a unital $*$ -homomorphism, and uniformly bounded pointwise $E$ -almost everywhere convergence of functions implies strong convergence of operators.

The spectral theorem is proved rather than assumed. For a compact space $K$ and a unital star-homomorphism $π : C (K) \to B (H)$ , the scalar functionals $f \mapsto ⟨ π (f) x, y ⟩$ are represented by unique finite regular complex measures $μ_{x, y}$ with $∣ μ_{x, y} ∣ (K) \leq ∥ x ∥ ∥ y ∥$ , the polarised family is sesquilinear, and the Hilbert space Riesz representation theorem builds bounded operators $E (h)$ with $⟨ E (h) x, y ⟩ = \int h d μ_{x, y}$ . Multiplicativity is obtained by testing the densities $f μ_{x, y}$ against continuous functions and invoking uniqueness of the representing measure; consequently $B \mapsto E (1_{B})$ is a regular PVM with $π (f) = \int f d E$ , and it is unique because two regular PVMs with the same continuous integrals have the same scalar measures. Applied to the continuous calculus of $T$ on $σ (T)$ , this yields the unique regular spectral PVM with $\int z d E (z) = T$ ; conversely the coordinate integral of any regular PVM on a compact set is a bounded normal operator whose spectrum lies in that set. The Borel functional calculus $f (T) = Φ_{E} (f)$ therefore extends the continuous calculus, is a unital $*$ -homomorphism with $E$ -essential-supremum norm and strong limits, and every operator commuting with $T$ and $T^{*}$ commutes with all of it. The spectral projections reduce $T$ , the eigenspace at $λ$ is exactly $E ({λ}) H$ , and for self-adjoint $T$ the half-line projections form an increasing strongly right continuous family with limits $0$ and $I$ at the two infinities. The support of the spectral measure is all of $σ (T)$ , and the PVM is determined by $T$ among regular PVMs on compact sets.

The second half develops the cyclic and multiplicity theory that makes the multiplication model canonical. A vector is cyclic when the closed span of ${f (T) x}$ is everything; the map $f \mapsto f (T) x$ extends from continuous functions to a unitary $L^{2} (σ (T), E_{x}) \to H_{x}$ intertwining multiplication by $z$ with $T$ and every bounded Borel multiplier with the Borel calculus. Zorn's lemma produces a maximal orthogonal family of cyclic reducing subspaces, and in the separable case a dense sequence produces a finite or countable one, so every bounded normal operator is unitarily equivalent to multiplication by the coordinate on an orthogonal sum $⨁_{j} L^{2} (σ (T), μ_{j})$ of cyclic summands. Choosing a common dominating measure $μ$ and writing $μ_{j} = h_{j} μ$ gives the multiplicity function $m (z) = # {j : h_{j} (z) > 0}$ , the fibre dimension of the standard measurable-field model with fibres $span {e_{1}, \dots, e_{m (z)}}$ ; the model is unitarily equivalent to the sum of the cyclic $L^{2}$ -spaces through the rank enumeration of the active coordinates. Unitary intertwiners are shown to preserve both the class of $μ$ and the fibre dimension almost everywhere: they commute with every bounded Borel multiplier, so the two scalar measures have the same null sets, and after localising to a set on which both multiplicities are constant the constant-fibre commutant argument forces the two constants to be equal. Hence two bounded normal operators on nonzero separable spaces are unitarily equivalent exactly when their spectra agree, their scalar spectral measure classes agree on that common set, and their multiplicity functions agree almost everywhere; there is no change of spectral coordinate, and the zero space is the separate trivial class. The page closes with Stone's resolvent formula: the strong limit of $(2 π i)^{- 1} \int_{a}^{b} [(T - (t + i ε))^{- 1} - (T - (t - i ε))^{- 1}] d t$ as $ε ↓ 0$ is $E ((a, b)) + \frac{1}{2} (E ({a}) + E ({b}))$ , with the half-masses at the endpoints stated explicitly.

The declared choice strength is uniform and explicit: the early PVM infrastructure inherits Countable Choice through the Hilbert projection and adjoint suppliers, while the construction of the spectral PVM, the Borel calculus and the whole multiplicity classification assume AC, the general cyclic decomposition using it through Zorn's lemma and the regular-measure representation theorem. The separable alternative to the Zorn decomposition is stated separately, and the nonseparable multiplicity theory is orientation only and is not used as a supplier.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-09-22Open item page →

Projection valued measure

Definition

Assume Countable Choice. Let $(X, Σ)$ be a measurable space (Measurable spaces and measurable sets, Sigma-algebras) and let $H$ be a complex Hilbert space (Hilbert space). A projection valued measure (PVM) on $(X, Σ)$ is a map

$E : Σ ⟶ B (H), B ⟼ E (B),$

such that:

$E (B)$ is an orthogonal projection for every $B \in Σ$ , that is, a bounded operator with $E (B)^{2} = E (B) = E (B)^{*}$ ; there is no finite-dimensional restriction;
$E (\emptyset) = 0$ and $E (X) = I$ ;
$E (B \cap C) = E (B) E (C)$ for all $B, C \in Σ$ ;
for every pairwise disjoint sequence $(B_{n})_{n \in N}$ in $Σ$ with union $B$ and every $x \in H$ , the series $\sum_{n} E (B_{n}) x$ converges in norm to $E (B) x$ , that is $E (B) x = \sum_{n = 0}^{\infty} E (B_{n}) x .$

Clause 4 is strong countable additivity. A PVM is called regular when $X$ is a locally compact Hausdorff space (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space), the $σ$ -algebra is the Borel $σ$ -algebra, and every one of the finite positive measures

$E_{x} (B) : = ⟨ E (B) x, x ⟩, x \in H, B \in Σ,$

is a regular Borel measure (Regular Borel measure on an LCH space). Inner products are linear in the first variable and conjugate-linear in the second (Real and complex inner-product spaces and their induced length); this fixed convention is the one used for every pairing $E_{x, y} (B) = ⟨ E (B) x, y ⟩$ on this page.

Well-definedness: the projection clause. The conditions $P^{2} = P = P^{*}$ are simultaneously meaningful and describe exactly the Hilbert orthogonal projections, with no hidden finite-dimensional hypothesis. Indeed, for a bounded $P$ with $P^{2} = P = P^{*}$ the range $ran P$ is a linear subspace, and for $y = P w$ one has $⟨ x - P x, y ⟩ = ⟨ P (x - P x), w ⟩ = ⟨ P x - P^{2} x, w ⟩ = 0$ , so $x - P x ⊥ ran P$ ; conversely $⟨ P z, P z ⟩ = ⟨ z, P^{*} P z ⟩ = ⟨ z, P z ⟩ = 0$ for $z ⊥ ran P$ , so $z \in \ker P$ , while $⟨ z, P x ⟩ = ⟨ P z, x ⟩ = 0$ for $z \in \ker P$ gives $\ker P = (ran P)^{⊥}$ (Hilbert-adjoint identities, Real and complex inner-product spaces and their induced length). Hence $ran P = (\ker P)^{⊥}$ is closed, and the two defining properties $P x \in ran P$ , $x - P x \in (ran P)^{⊥}$ of The Hilbert orthogonal projection onto a closed subspace show that $P = P_{ran P}$ is the Hilbert orthogonal projection onto a closed subspace, and conversely every such $P_{M}$ is idempotent and self-adjoint by Hilbert projections are linear, self-adjoint and contractive and Orthogonal decomposition by a closed subspace. Finally $P$ is contractive: from $⟨ P x, x ⟩ = ⟨ P x, P x ⟩ = ∥ P x ∥^{2}$ and Cauchy–Schwarz, $∥ P x ∥^{2} \leq ∥ P x ∥ ∥ x ∥$ , so $∥ P x ∥ \leq ∥ x ∥$ for all $x$ , and $⟨ P x, x ⟩ = ∥ P x ∥^{2} \geq 0$ is a nonnegative real number.

Well-definedness: the regularity clause. For an orthogonal projection value the pairing $E_{x} (B) = ⟨ E (B) x, x ⟩$ is a nonnegative real number and $E_{x} (X) = ⟨ x, x ⟩ = ∥ x ∥^{2} < + \infty$ , so every $E_{x}$ is a finite nonnegative set function and the regularity requirement is a meaningful condition on it; that each $E_{x}$ is genuinely a countably additive measure of total mass $∥ x ∥^{2}$ , and that the polarized pairings $E_{x, y}$ are finite complex measures, is proved as Scalar and complex measures from a pvm before either is used.

LemmaStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Weak and strong additivity of orthogonal projections

Statement

Assume Countable Choice. Let $(X, Σ)$ be a measurable space, let $H$ be a complex Hilbert space, and let $E : Σ \to B (H)$ take values in orthogonal projections and satisfy $E (\emptyset) = 0$ and $E (B \cap C) = E (B) E (C)$ for all $B, C \in Σ$ . Then the following two properties are equivalent:

(weak countable additivity) for every pairwise disjoint sequence $(B_{n})_{n \in N}$ in $Σ$ with union $B$ and all $x, y \in H$ , $⟨ E (B) x, y ⟩ = \sum_{n = 0}^{\infty} ⟨ E (B_{n}) x, y ⟩;$
(strong countable additivity) for every such sequence and every $x \in H$ , $E (B) x = \sum_{n = 0}^{\infty} E (B_{n}) x$ with the series converging in norm.

Facts & Assumptions

[A1]

An orthogonal projection value $P = E (B)$ satisfies $P^{2} = P = P^{*}$ and $⟨ P x, x ⟩ = ∥ P x ∥^{2} \geq 0$ , and it is contractive, $∥ P u ∥ \leq ∥ u ∥$ for all $u$ (Projection valued measure, Hilbert projections are linear, self-adjoint and contractive).

[A2]

The Hilbert adjoint satisfies $⟨ S u, v ⟩ = ⟨ u, S^{*} v ⟩$ for all $u, v$ , and conjugate symmetry gives $⟨ S^{*} u, v ⟩ = \overline{⟨ v, S^{*} u ⟩} = \overline{⟨ S v, u ⟩} = ⟨ u, S v ⟩$ ; for a self-adjoint $S$ the two pairings with $S$ coincide (The Hilbert-space adjoint of a bounded operator, Real and complex inner-product spaces and their induced length).

[A3]

Multiplicativity on intersections and the empty-set value hold: $E (B \cap C) = E (B) E (C)$ and $E (\emptyset) = 0$ ; if $C \subseteq B$ then $E (C) = E (B \cap C) = E (B) E (C) = E (C) E (B)$ (Projection valued measure).

[A4]

Cauchy–Schwarz gives $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[A5]

The pairing is linear in the first argument and conjugate-linear in the second, and $∥ u ∥^{2} = ⟨ u, u ⟩$ (Real and complex inner-product spaces and their induced length).

[A6]

Countable Choice is the declared standing hypothesis of this block of the page (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: A measurable space $(X, Σ)$ , a complex Hilbert space $H$ , a map $E$ with orthogonal projection values, $E (\emptyset) = 0$ , $E (B \cap C) = E (B) E (C)$ , a pairwise disjoint sequence $(B_{n})$ in $Σ$ with union $B$ , vectors $x, y \in H$ , and partial sums $Q_{N} : = \sum_{n \leq N} E (B_{n})$ .

1.1

Strong implies weak: for every $N$ the difference of the two sides of the weak identity is $⟨ E (B) x - Q_{N} x, y ⟩$ , so by Cauchy–Schwarz $∣ ⟨ E (B) x, y ⟩ - \sum_{n \leq N} ⟨ E (B_{n}) x, y ⟩ ∣ \leq ∥ E (B) x - Q_{N} x ∥ ∥ y ∥$ , which tends to $0$ because $E (B) x - Q_{N} x \to 0$ in norm by hypothesis.

A1A4A5

1.2

Weak implies strong: expanding $∥ E (B) x - Q_{N} x ∥^{2} = ⟨ E (B) x - Q_{N} x, E (B) x - Q_{N} x ⟩$ and using $P^{2} = P = P^{*}$ for each projection value gives $∥ E (B) x - Q_{N} x ∥^{2} = ⟨ E (B) x, x ⟩ - 2 \sum_{n \leq N} ⟨ E (B_{n}) x, x ⟩ + \sum_{n, m \leq N} ⟨ E (B_{m} \cap B_{n}) x, x ⟩$ , because $⟨ E (B) x, E (B) x ⟩ = ⟨ E (B)^{2} x, x ⟩$ , because $⟨ E (B_{n}) x, E (B) x ⟩ = ⟨ E (B) E (B_{n}) x, x ⟩$ with $B_{n} \subseteq B$ , and because $⟨ E (B_{n}) x, E (B_{m}) x ⟩ = ⟨ E (B_{m}) E (B_{n}) x, x ⟩ = ⟨ E (B_{m} \cap B_{n}) x, x ⟩$ .

A1A2A3A5

2.1

In the last sum the off-diagonal terms are $⟨ E (\emptyset) x, x ⟩ = 0$ and the diagonal terms are $⟨ E (B_{n}) x, x ⟩$ , so $∥ E (B) x - Q_{N} x ∥^{2} = ⟨ E (B) x, x ⟩ - \sum_{n \leq N} ⟨ E (B_{n}) x, x ⟩$ , which tends to $0$ by weak countable additivity applied with $y = x$ ; hence $Q_{N} x \to E (B) x$ in norm.

step 1.2A3A5

3.1

Both implications hold for an arbitrary pairwise disjoint sequence, so weak and strong countable additivity of $E$ are equivalent.

step 1.1step 2.1A6∎

LemmaStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Scalar and complex measures from a pvm

Statement

Assume Countable Choice. Let $(X, Σ)$ be a measurable space, let $H$ be a complex Hilbert space, let $E$ be a projection valued measure on $(X, Σ)$ , and for $x, y \in H$ define

$E_{x} (B) : = ⟨ E (B) x, x ⟩, E_{x, y} (B) : = ⟨ E (B) x, y ⟩ (B \in Σ) .$

Then:

$E_{x}$ is a positive measure on $(X, Σ)$ with $E_{x} (X) = ∥ x ∥^{2}$ and $0 \leq E_{x} (B) \leq ∥ x ∥^{2}$ for every $B$ ;
$E_{x, y}$ is a finite complex measure on $(X, Σ)$ , the map $(x, y) \mapsto E_{x, y}$ is linear in $x$ and conjugate-linear in $y$ , and $E_{y, x} = \overline{E_{x, y}}$ , meaning $E_{y, x} (B) = \overline{E_{x, y} (B)}$ for every $B$ ;
$∣ E_{x, y} ∣ (X) \leq ∥ x ∥ ∥ y ∥$ , so $∣ E_{x, y} (B) ∣ \leq ∥ x ∥ ∥ y ∥$ for every $B$ , and the polarization identity $E_{x, y} = \frac{1}{4} \sum_{k = 0}^{3} i^{k} E_{x + i^{k} y}$ holds as an identity of complex measures, where $i^{k}$ are the fourth roots of unity $1, i, - 1, - i$ .

Facts & Assumptions

[A1]

Projection values satisfy $P^{2} = P = P^{*}$ , $⟨ P x, x ⟩ = ∥ P x ∥^{2} \geq 0$ and $∥ P x ∥ \leq ∥ x ∥$ (Projection valued measure, Hilbert projections are linear, self-adjoint and contractive).

[A2]

$E (\emptyset) = 0$ , $E (X) = I$ , $E (B \cap C) = E (B) E (C)$ , and for every pairwise disjoint sequence $(B_{n})$ with union $B$ one has $E (B) x = \lim_{N} \sum_{n \leq N} E (B_{n}) x$ in norm (Projection valued measure).

[A3]

$⟨ S u, v ⟩ = ⟨ u, S^{*} v ⟩$ for all $u, v$ , and for self-adjoint $S$ one has $⟨ S u, v ⟩ = ⟨ u, S v ⟩$ and $⟨ S u, v ⟩ = \overline{⟨ S v, u ⟩}$ (The Hilbert-space adjoint of a bounded operator, Real and complex inner-product spaces and their induced length).

[A4]

Cauchy–Schwarz gives $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[A5]

A measure is a $[0, + \infty]$ -valued countably additive set function vanishing at $\emptyset$ (Measures on sigma-algebras); a complex measure is a $C$ -valued countably additive set function vanishing at $\emptyset$ (A complex measure is a finite-valued countably additive set function); its total variation is the supremum of $\sum_{n} ∣ ν (E_{n}) ∣$ over countable measurable partitions (The total variation |nu|(E) from countable measurable partitions).

[A6]

The pairing is linear in the first argument, conjugate-linear in the second, and $∥ u ∥^{2} = ⟨ u, u ⟩$ (Real and complex inner-product spaces and their induced length).

[A7]

Countable Choice is the declared standing hypothesis of this block of the page (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: A measurable space $(X, Σ)$ , a complex Hilbert space $H$ , a projection valued measure $E$ on it, vectors $x, y \in H$ , and a pairwise disjoint sequence $(B_{n}) \subseteq Σ$ with union $B$ .

1.1

$E_{x}$ is countably additive and vanishes at $\emptyset$ : using strong additivity, $E_{x} (B) = ⟨ \lim_{N} \sum_{n \leq N} E (B_{n}) x, x ⟩ = \lim_{N} \sum_{n \leq N} ⟨ E (B_{n}) x, x ⟩ = \sum_{n} E_{x} (B_{n})$ , with the limit pulled through the linear functional $⟨ \cdot, x ⟩$ , continuous since $∣ ⟨ v, x ⟩ - ⟨ w, x ⟩ ∣ \leq ∥ v - w ∥ ∥ x ∥$ , while $E_{x} (\emptyset) = ⟨ 0, x ⟩ = 0$ and $E_{x} (X) = ⟨ x, x ⟩ = ∥ x ∥^{2}$ .

A2A4A6

1.2

$E_{x, y}$ is a finite complex measure and the map is linear in $x$ and conjugate-linear in $y$ : countable additivity holds by the same limit argument with the continuous functional $⟨ \cdot, y ⟩$ , $E_{x, y} (\emptyset) = 0$ , and $⟨ E (B) (x + u), y ⟩ = ⟨ E (B) x, y ⟩ + ⟨ E (B) u, y ⟩$ , $⟨ E (B) (λ x), y ⟩ = λ ⟨ E (B) x, y ⟩$ , $⟨ E (B) x, λ y ⟩ = \overline{λ} ⟨ E (B) x, y ⟩$ ; finiteness follows from $∣ E_{x, y} (B) ∣ = ∣ ⟨ E (B) x, y ⟩ ∣ \leq ∥ E (B) x ∥ ∥ y ∥ \leq ∥ x ∥ ∥ y ∥$ .

A1A2A3A4A5A6

1.3

Conjugate symmetry: $\overline{E_{x, y} (B)} = \overline{⟨ E (B) x, y ⟩} = ⟨ y, E (B) x ⟩ = ⟨ E (B) y, x ⟩ = E_{y, x} (B)$ , using that $E (B)$ is self-adjoint.

A1A3A6

1.4

Polarization: for fixed $B$ the form $Λ (u, v) : = ⟨ E (B) u, v ⟩$ is sesquilinear, so expanding the four terms gives $\frac{1}{4} \sum_{k = 0}^{3} i^{k} Λ (u + i^{k} v, u + i^{k} v) = Λ (u, v)$ for all $u, v$ (the $∣ u ∣^{2}$ and $∣ v ∣^{2}$ coefficients cancel and the mixed terms add to $4 Λ (u, v)$ ); reading the identity at $B$ and letting $B$ vary gives $E_{x, y} = \frac{1}{4} \sum_{k = 0}^{3} i^{k} E_{x + i^{k} y}$ .

A6algebra

2.1

$E_{x}$ is nonnegative and bounded by its total mass: $E_{x} (B) = ⟨ E (B) x, E (B) x ⟩ = ∥ E (B) x ∥^{2} \geq 0$ and $∥ E (B) x ∥ \leq ∥ x ∥$ , so $0 \leq E_{x} (B) \leq ∥ x ∥^{2}$ ; hence $E_{x}$ is a positive measure with $E_{x} (X) = ∥ x ∥^{2}$ .

step 1.1A1A5A6

3.1

Variation bound: let $(B_{j})_{j \geq 0}$ be a countable measurable partition of $X$ . For every $N$ , Cauchy--Schwarz gives $\sum_{j = 0}^{N} ∣ E_{x, y} (B_{j}) ∣ = \sum_{j = 0}^{N} ∣ ⟨ E (B_{j}) x, E (B_{j}) y ⟩ ∣ \leq (\sum_{j = 0}^{N} ∥ E (B_{j}) x ∥^{2})^{1 / 2} (\sum_{j = 0}^{N} ∥ E (B_{j}) y ∥^{2})^{1 / 2} \leq ∥ x ∥ ∥ y ∥$ . Indeed, orthogonality and strong additivity give $\sum_{j \geq 0} ∥ E (B_{j}) x ∥^{2} = \sum_{j \geq 0} ⟨ E (B_{j}) x, x ⟩ = E_{x} (X) = ∥ x ∥^{2}$ , because $E (B_{j})^{2} = E (B_{j}) = E (B_{j})^{*}$ and the $B_{j}$ partition $X$ ; the same holds with $y$ . Taking $N \to \infty$ yields $\sum_{j \geq 0} ∣ E_{x, y} (B_{j}) ∣ \leq ∥ x ∥ ∥ y ∥$ .

step 1.1step 2.1A1A2A3A4A5algebra

4.1

Hence every countable partition contributes at most $∥ x ∥ ∥ y ∥$ to the defining supremum of $∣ E_{x, y} ∣ (X)$ , so $∣ E_{x, y} ∣ (X) \leq ∥ x ∥ ∥ y ∥$ and $∣ E_{x, y} (B) ∣ \leq ∣ E_{x, y} ∣ (X) \leq ∥ x ∥ ∥ y ∥$ for every $B$ , since $(B, X ∖ B, \emptyset, \dots)$ is one of those countable partitions.

step 3.1A5

5.1

All asserted properties of $E_{x}$ , $E_{x, y}$ hold for arbitrary $x, y$ , so the scalar pairings of a projection valued measure are a positive measure of mass $∥ x ∥^{2}$ and a family of finite complex measures of variation at most $∥ x ∥ ∥ y ∥$ , conjugate symmetric and recovered by polarization.

step 1.1step 2.1step 1.2step 1.3step 4.1step 1.4A7∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-09-22Open item page →

Integral of a simple function against a pvm

Definition

Assume Countable Choice. Let $(X, Σ)$ be a measurable space, let $H$ be a complex Hilbert space, let $E$ be a projection valued measure on $(X, Σ)$ , and let $s : X \to C$ be a complex simple function (Complex simple functions as finite sums of measurable indicators, Measurable spaces and measurable sets). Present $s$ in disjoint normal form with a zero-coefficient complement, that is

$s = \sum_{j = 1}^{m} a_{j} 1_{B_{j}},$

where $m \geq 1$ , and $B_{1}, \dots, B_{m} \in Σ$ are pairwise disjoint with $B_{1} \cup \dots \cup B_{m} = X$ and $a_{1}, \dots, a_{m} \in C$ ; a representation over a disjoint family whose union misses some measurable set is completed by adding that set with the coefficient $0$ , and any coefficient is allowed to vanish. In particular the zero function on the empty space uses $m = 1$ , $B_{1} = \emptyset$ and $a_{1} = 0$ ; an empty presentation is not used. Then define

$\int s d E : = \sum_{j = 1}^{m} a_{j} E (B_{j}) \in B (H) .$

Here $1_{B}$ is the indicator of $B$ and $E (B_{j})$ is the value of the projection valued measure on $B_{j}$ (Projection valued measure).

Well-definedness. Because the $B_{j}$ are pairwise disjoint and cover $X$ , the operator $\sum_{j} a_{j} E (B_{j})$ is a finite sum of bounded operators and hence a bounded operator; the sum is meaningful in $B (H)$ with the operator norm, and $∥ \sum_{j} a_{j} E (B_{j}) ∥ \leq \max_{j} ∣ a_{j} ∣$ , since the projection values are contractive and pairwise orthogonal. The value displayed a priori depends on the chosen disjoint presentation of $s$ ; that it does not, $\int s d E$ is independent of the disjoint presentation of $s$ , is proved as Simple pvm integral is representation independent immediately below, before the symbol is used. The normal-form convention $s = \sum_{j} a_{j} 1_{B_{j}}$ with $B_{1} \cup \dots \cup B_{m} = X$ is the one used throughout this page, and the identity $\int 1_{B} d E = E (B)$ holds for every $B \in Σ$ .

LemmaStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Simple pvm integral is representation independent

Statement

the operator $\int s d E$ of Integral of a simple function against a pvm is independent of the disjoint normal form of $s$ ;
for all $x, y \in H$ , $⟨ (\int s d E) x, y ⟩ = \int s d E_{x, y}$ , the scalar integral against the complex measure $E_{x, y}$ ;
for every $x \in H$ , $∥ (\int s d E) x ∥^{2} = \int ∣ s ∣^{2} d E_{x}$ , and $∥ \int s d E ∥ \leq M_{s}$ .

Here $M_{s} : = \max ({0} \cup {∣ s (t) ∣ : t \in X})$ ; thus $M_{s} = \max_{t \in X} ∣ s (t) ∣$ when $X \neq \emptyset$ , and $M_{s} = 0$ when $X = \emptyset$ .

Facts & Assumptions

[A1]

For a disjoint normal form $s = \sum_{j = 1}^{m} a_{j} 1_{B_{j}}$ with $B_{1}, \dots, B_{m}$ pairwise disjoint and covering $X$ , the integral is $\int s d E = \sum_{j} a_{j} E (B_{j})$ (Integral of a simple function against a pvm).

[A2]

$E (\emptyset) = 0$ , $E (X) = I$ , $E (B \cap C) = E (B) E (C)$ , each $E (B)$ satisfies $E (B)^{2} = E (B) = E (B)^{*}$ and is contractive, and for pairwise disjoint $(D_{n})$ with union $D$ one has $E (D) x = \sum_{n} E (D_{n}) x$ in norm (Projection valued measure).

[A3]

$E_{x, y} (B) = ⟨ E (B) x, y ⟩$ is a finite complex measure, $E_{x} (B) = ⟨ E (B) x, x ⟩ = ∥ E (B) x ∥^{2}$ is a positive measure of mass $∥ x ∥^{2}$ , $∣ E_{x, y} ∣ (X) \leq ∥ x ∥ ∥ y ∥$ , and $E_{y, x} = \overline{E_{x, y}}$ (Scalar and complex measures from a pvm).

[A4]

For a complex measure $ν$ and a complex simple function whose nonzero level sets have finite total variation, presented over the nonzero level sets, the scalar simple integral is $\int s d ν = \sum_{j} c_{j} ν (E_{j})$ , and the value is unchanged by deleting empty level sets (The simple integral against a signed or complex measure, Complex simple functions as finite sums of measurable indicators).

[A5]

The pairing is linear in the first argument and conjugate-linear in the second, so $⟨ \sum_{j} z_{j}, y ⟩ = \sum_{j} ⟨ z_{j}, y ⟩$ and $⟨ z, w ⟩ = \overline{⟨ w, z ⟩}$ (Real and complex inner-product spaces and their induced length). The adjoint identity is $⟨ P u, v ⟩ = ⟨ u, P^{*} v ⟩$ (The Hilbert-space adjoint of a bounded operator).

[A6]

Countable Choice is the declared standing hypothesis of this block of the page (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[A7]

The operator norm is the unit-ball supremum (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A8]

Total variation is the supremum of the nonnegative sums over countable measurable partitions (The total variation |nu|(E) from countable measurable partitions).

Proof

technique · direct

Given: A measurable space $(X, Σ)$ , a complex Hilbert space $H$ , a projection valued measure $E$ , a complex simple function $s$ with two disjoint normal forms $s = \sum_{j = 1}^{m} a_{j} 1_{B_{j}} = \sum_{k = 1}^{n} b_{k} 1_{C_{k}}$ covering $X$ , and vectors $x, y \in H$ .

1.1

Finite additivity follows by padding a finite disjoint family with empty sets in strong countable additivity. The scalar measures also have finite additivity. Every measurable subset has finite $∣ E_{x, y} ∣$ -variation: a countable partition of that subset extends to one of $X$ by adding its complement, so its sum is at most $∣ E_{x, y} ∣ (X) \leq ∥ x ∥ ∥ y ∥$ . In particular the scalar simple integrals below are defined; for $E_{x} = E_{x, x}$ the same argument applies.

A2A3A4A8

2.1

The intersections $B_{j} \cap C_{k}$ form a disjoint cover of $X$ . If an intersection is nonempty then $a_{j} = b_{k}$ , and if empty its projection value is zero. Finite additivity therefore gives $\sum_{j} a_{j} E (B_{j}) = \sum_{j, k} a_{j} E (B_{j} \cap C_{k}) = \sum_{j, k} b_{k} E (B_{j} \cap C_{k}) = \sum_{k} b_{k} E (C_{k})$ . This proves representation independence.

step 1.1A1A2algebra

3.1

Expanding the pairing gives $⟨ (\int s d E) x, y ⟩ = \sum_{j} a_{j} E_{x, y} (B_{j})$ . Discard empty cells and regroup the remaining indices by $c \in s (X)$ . Finite additivity gives $\sum_{j} a_{j} E_{x, y} (B_{j}) = \sum_{c \in s (X)} c E_{x, y} (s^{- 1} ({c})) = \int s d E_{x, y}$ , where the zero-value term is zero. If $X$ is empty all cells and sums contribute zero.

step 1.1step 2.1A1A3A4A5algebra

3.2

The adjoint identity and the projection rules give $⟨ E (B_{j}) x, E (B_{k}) x ⟩ = ⟨ E (B_{k})^{*} E (B_{j}) x, x ⟩ = ⟨ E (B_{k} \cap B_{j}) x, x ⟩$ . Thus expansion of the squared norm leaves only diagonal terms: $∥ (\int s d E) x ∥^{2} = \sum_{j} ∣ a_{j} ∣^{2} E_{x} (B_{j})$ . Regrouping the nonempty cells by the value $d = ∣ a_{j} ∣^{2}$ , finite additivity identifies this sum with $\sum_{d \in ∣ s ∣^{2} (X)} d E_{x} ((∣ s ∣^{2})^{- 1} ({d})) = \int ∣ s ∣^{2} d E_{x}$ ; the zero term vanishes.

step 1.1step 2.1A1A2A3A4A5algebra

4.1

Empty cells contribute zero to the sum in step 3.2. On every nonempty $B_{j}$ , $∣ a_{j} ∣ \leq M_{s}$ because $a_{j}$ is a value of $s$ . Positivity and finite additivity give $∥ (\int s d E) x ∥^{2} \leq M_{s}^{2} \sum_{j} E_{x} (B_{j}) = M_{s}^{2} ∥ x ∥^{2}$ . Taking nonnegative square roots and then the unit-ball supremum gives $∥ \int s d E ∥ \leq M_{s}$ . When $X = \emptyset$ , all projection values are zero and the integral is zero, so the same bound with $M_{s} = 0$ holds.

step 1.1step 3.2A2A3A7algebra

5.1

The integral is independent of the presentation, has the asserted scalar pairings and squared-norm identity, and satisfies the stated bound, including the empty-space case.

step 2.1step 3.1step 3.2step 4.1A6∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Bounded borel pvm integral

Statement

Assume Countable Choice. Let $(X, Σ)$ be a measurable space, let $H$ be a nonzero complex Hilbert space, let $E$ be a projection valued measure on $(X, Σ)$ , and let $f : X \to C$ be bounded and $Σ$ -measurable (A measurable function between measurable spaces). Then:

there is a unique operator $Φ_{E} (f) \in B (H)$ with $⟨ Φ_{E} (f) x, y ⟩ = \int f d E_{x, y} (x, y \in H),$ and for every sequence $(s_{n})$ of complex simple functions with $∥ f - s_{n} ∥_{\infty} \to 0$ one has $∥ Φ_{E} (f) - \int s_{n} d E ∥ \to 0$ : the integral is obtained from uniform simple approximations and is independent of the approximating sequence;
$∥ Φ_{E} (f) ∥ \leq ∥ f ∥_{\infty}$ , and for every $x \in H$ $⟨ Φ_{E} (f) x, x ⟩ = \int f d E_{x}, ∥ Φ_{E} (f) x ∥^{2} = \int ∣ f ∣^{2} d E_{x};$
the exact norm is the $E$ -essential supremum $∥ f ∥_{E, \infty} : = \sup {∥ f ∥_{\infty, E_{x}} : x \in H, ∥ x ∥ = 1}, ∥ Φ_{E} (f) ∥ = ∥ f ∥_{E, \infty},$ where $∥ f ∥_{\infty, E_{x}}$ denotes the essential supremum of the real measurable function $∣ f ∣$ with respect to the finite measure $E_{x}$ (The essential supremum of a measurable function with respect to a measure).

Facts & Assumptions

[A1]

For a complex simple function $s$ the operator $\int s d E$ satisfies $⟨ (\int s d E) x, y ⟩ = \int s d E_{x, y}$ , $∥ (\int s d E) x ∥^{2} = \int ∣ s ∣^{2} d E_{x}$ and $∥ \int s d E ∥ \leq \max ∣ s ∣ \leq ∥ s ∥_{\infty}$ ; the construction is linear on simple functions presented over a common refinement (Simple pvm integral is representation independent).

[A2]

$E_{x, y}$ is a finite complex measure on $(X, Σ)$ with $∣ E_{x, y} ∣ (X) \leq ∥ x ∥ ∥ y ∥$ , the integral $\int g d E_{x, y}$ of a bounded measurable $g$ satisfies $∣ \int g d E_{x, y} ∣ \leq ∥ g ∥_{\infty} ∣ E_{x, y} ∣ (X)$ (Scalar and complex measures from a pvm, Integrals against signed or complex measures are bounded by total variation, Integration against a signed or complex measure, and the class L^1(nu) = L^1(|nu|)).

[A3]

$E_{x} (B) = ⟨ E (B) x, x ⟩ = ∥ E (B) x ∥^{2}$ is a positive measure with $E_{x} (X) = ∥ x ∥^{2}$ . Moreover, if $E (A) y = y$ , then $E (X ∖ A) y = E (X ∖ A) E (A) y = E (\emptyset) y = 0$ , so $E_{y} (X ∖ A) = 0$ (Scalar and complex measures from a pvm, Projection valued measure).

[A4]

$B (H)$ is complete for the operator norm (If (Y) is Banach then (\mathcal B(X,Y)) is Banach, Hilbert space).

[A5]

A bounded measurable complex function is integrable against every finite measure and dominated convergence holds: if $g_{n} \to g$ pointwise and $∣ g_{n} ∣ \leq C$ with $C$ integrable, then $\int g_{n} d μ \to \int g d μ$ (Dominated convergence).

[A6]

For a finite measure $μ$ , $∥ g ∥_{\infty, μ}$ is the least essential bound of $∣ g ∣$ : $∣ g ∣ \leq ∥ g ∥_{\infty, μ}$ $μ$ -almost everywhere, and if $∣ g ∣ \leq M$ almost everywhere then $∥ g ∥_{\infty, μ} \leq M$ ; if $∥ g ∥_{\infty, μ} > c$ then $μ ({∣ g ∣ > c}) > 0$ (The essential supremum is attained as the least essential bound, The essential supremum of a measurable function with respect to a measure).

[A7]

A complex simple function is a finite linear combination of indicators of pairwise disjoint measurable sets; for a measurable $f$ and $ε > 0$ the square $[- M, M]^{2}$ containing the range of a bounded $f$ can be cut into finitely many Borel squares of diameter $< ε$ whose inverse images refine to a disjoint measurable cover of $X$ (Complex simple functions as finite sums of measurable indicators, A measurable function between measurable spaces).

[A8]

Countable Choice is the declared standing hypothesis of this block of the page (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: A measurable space $(X, Σ)$ , a nonzero complex Hilbert space $H$ , a projection valued measure $E$ on $(X, Σ)$ , and a bounded measurable $f : X \to C$ with $M : = ∥ f ∥_{\infty} < + \infty$ .

1.1

Uniform simple approximation: for each $n \geq 0$ choose finitely many pairwise disjoint measurable sets $D_{1}, \dots, D_{r}$ covering $X$ and complex numbers $c_{i}$ with $∣ f - c_{i} ∣ \leq 1 / (n + 1)$ on $D_{i}$ (cut a square containing the range of $f$ into finitely many squares of diameter $< 1 / (n + 1)$ and take inverse images), so $s_{n} : = \sum_{i} c_{i} 1_{D_{i}}$ is a complex simple function with $∥ f - s_{n} ∥_{\infty} \leq 1 / (n + 1)$ .

1.2

Difference of simple integrals: if $s, t$ are complex simple functions, presenting both over the common refinement of their disjoint normal forms gives $\int (s - t) d E = \int s d E - \int t d E$ , hence $∥ \int s d E - \int t d E ∥ \leq ∥ s - t ∥_{\infty}$ ; in particular the sequence $\int s_{n} d E$ is Cauchy, since $∥ \int s_{n} d E - \int s_{m} d E ∥ \leq ∥ s_{n} - s_{m} ∥_{\infty} \leq \frac{1}{n + 1} + \frac{1}{m + 1}$ .

2.1

By completeness of $B (H)$ the sequence $\int s_{n} d E$ has a norm limit $Φ_{E} (f)$ , and for any other uniformly approximating sequence $(t_{n})$ one has $∥ \int s_{n} d E - \int t_{n} d E ∥ \leq ∥ s_{n} - t_{n} ∥_{\infty} \to 0$ , so the limit does not depend on the sequence; the same argument applies to the difference of two candidate limits.

step 1.2A4

3.1

Pairing identity: for all $x, y \in H$ , $⟨ Φ_{E} (f) x, y ⟩ = \lim_{n} ⟨ (\int s_{n} d E) x, y ⟩ = \lim_{n} \int s_{n} d E_{x, y} = \int f d E_{x, y}$ , because $∣ \int (s_{n} - f) d E_{x, y} ∣ \leq ∥ s_{n} - f ∥_{\infty} ∣ E_{x, y} ∣ (X) \leq \frac{1}{n + 1} ∥ x ∥ ∥ y ∥ \to 0$ .

step 2.1A1A2

4.1

Norm identities: $∥ Φ_{E} (f) x ∥^{2} = \lim_{n} ∥ (\int s_{n} d E) x ∥^{2} = \lim_{n} \int ∣ s_{n} ∣^{2} d E_{x} = \int ∣ f ∣^{2} d E_{x}$ by dominated convergence applied to the finite measure $E_{x}$ with the constant dominating function $(M + 1)^{2}$ , since $s_{n} \to f$ and $∣ s_{n} ∣ \leq M + 1$ ; taking $y = x$ in the pairing identity gives $⟨ Φ_{E} (f) x, x ⟩ = \int f d E_{x}$ .

step 1.1step 2.1step 3.1A1A3A5

4.2

Norm bound and uniqueness: $∣ ⟨ Φ_{E} (f) x, y ⟩ ∣ \leq ∥ f ∥_{\infty} ∥ x ∥ ∥ y ∥$ by the pairing identity just proved and the variation bound, so $∥ Φ_{E} (f) ∥ \leq ∥ f ∥_{\infty}$ ; and any operator $T$ with $⟨ T x, y ⟩ = \int f d E_{x, y}$ for all $x, y$ equals $Φ_{E} (f)$ , since $T - Φ_{E} (f)$ has all pairings zero, whence $(T - Φ_{E} (f)) x = 0$ for every $x$ by positive definiteness of the pairing.

step 3.1A2

5.1

Upper bound for the norm: for $x \neq 0$ one has $E_{x} = ∥ x ∥^{2} E_{x / ∥ x ∥}$ and hence $∥ f ∥_{\infty, E_{x}} = ∥ f ∥_{\infty, E_{x / ∥ x ∥}}$ , so $∥ Φ_{E} (f) x ∥^{2} = \int ∣ f ∣^{2} d E_{x} \leq ∥ f ∥_{\infty, E_{x}}^{2} ∥ x ∥^{2} \leq ∥ f ∥_{E, \infty}^{2} ∥ x ∥^{2}$ by the least-essential-bound property; therefore $∥ Φ_{E} (f) ∥ \leq ∥ f ∥_{E, \infty}$ .

step 4.1A3A6

5.2

Lower bound for the norm: write $a = ∥ f ∥_{E, \infty}$ . If $a = 0$ , the lower bound follows from nonnegativity of the operator norm. If $a > 0$ , given $0 \leq c < a$ choose a unit vector $x$ with $∥ f ∥_{\infty, E_{x}} > c$ , so $A : = {∣ f ∣ > c}$ satisfies $E_{x} (A) > 0$ by the least-essential-bound property; put $y : = E (A) x \neq 0$ , so $E_{y} (X ∖ A) = 0$ and hence $∥ Φ_{E} (f) y ∥^{2} = \int ∣ f ∣^{2} d E_{y} \geq c^{2} E_{y} (X) = c^{2} ∥ y ∥^{2}$ , giving $∥ Φ_{E} (f) ∥ \geq c$ ; since $0 \leq c < a$ was arbitrary, $∥ Φ_{E} (f) ∥ \geq ∥ f ∥_{E, \infty}$ .

step 4.1A3A6

6.1

The integral $Φ_{E} (f)$ is well defined, obtained from uniform simple approximations, satisfies the pairing and quadratic identities and the bound $∥ Φ_{E} (f) ∥ \leq ∥ f ∥_{\infty}$ , and its exact norm is the $E$ -essential supremum $∥ f ∥_{E, \infty}$ .

step 2.1step 3.1step 4.1step 4.2step 5.1step 5.2A8∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Pvm integral is a star homomorphism

Statement

Assume Countable Choice. Let $(X, Σ)$ be a measurable space, let $H$ be a nonzero complex Hilbert space, let $E$ be a projection valued measure on $(X, Σ)$ , and for a bounded measurable $f$ write $Φ_{E} (f) = \int f d E$ for the operator of Bounded borel pvm integral. Then:

$Φ_{E}$ is linear and unital: $Φ_{E} (1_{X}) = I$ and $Φ_{E} (a f + b g) = a Φ_{E} (f) + b Φ_{E} (g)$ for bounded measurable $f, g$ and $a, b \in C$ ;
$Φ_{E}$ is multiplicative: $Φ_{E} (f g) = Φ_{E} (f) Φ_{E} (g)$ ;
$Φ_{E}$ preserves conjugation: $Φ_{E} (\overline{f}) = Φ_{E} (f)^{*}$ ;
if $(f_{n})$ are bounded measurable with $\sup_{n} ∥ f_{n} ∥_{\infty} < \infty$ , $f$ is bounded measurable, and $f_{n} \to f$ pointwise $E$ -almost everywhere, meaning $E_{x}$ -almost everywhere for every $x \in H$ , then $Φ_{E} (f_{n}) \to Φ_{E} (f)$ in the strong operator topology.

Facts & Assumptions

[A1]

$Φ_{E} (g)$ is the unique operator with $⟨ Φ_{E} (g) x, y ⟩ = \int g d E_{x, y}$ for all $x, y$ , it satisfies $∥ Φ_{E} (g) ∥ \leq ∥ g ∥_{\infty}$ and $∥ Φ_{E} (g) x ∥^{2} = \int ∣ g ∣^{2} d E_{x}$ , and it is the norm limit of $\int s_{n} d E$ for any complex simple $s_{n} \to g$ uniformly (Bounded borel pvm integral).

[A2]

The simple integral of $s = \sum_{j} c_{j} 1_{D_{j}}$ over a disjoint measurable cover is $\sum_{j} c_{j} E (D_{j})$ (Integral of a simple function against a pvm), independently of the presentation; it has the scalar pairing and quadratic identities (Simple pvm integral is representation independent). Complex simple functions have finite measurable range (Complex simple functions as finite sums of measurable indicators).

[A3]

$E_{y, x} = \overline{E_{x, y}}$ , $E_{x}$ is a positive measure of mass $∥ x ∥^{2}$ , and $E_{x} (B) = ⟨ E (B) x, x ⟩$ (Scalar and complex measures from a pvm).

[A4]

Each $E (B)$ is self-adjoint and idempotent, $E (\emptyset) = 0$ , $E (X) = I$ and $E (B \cap C) = E (B) E (C)$ (Projection valued measure).

[A5]

Dominated convergence: if $g_{n} \to g$ pointwise almost everywhere and $∣ g_{n} ∣ \leq C$ for an integrable constant $C$ , then $\int g_{n} d μ \to \int g d μ$ (Dominated convergence).

[A6]

The adjoint is conjugate-linear on operator sums and norm-preserving, $∥ S^{*} ∥ = ∥ S ∥$ , and operator multiplication is norm-continuous, $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ (Hilbert-adjoint identities, Composition satisfies |ST|\le|S|,|T|).

[A7]

Countable Choice is the declared standing hypothesis of this block of the page (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: A measurable space $(X, Σ)$ , a nonzero complex Hilbert space $H$ , a projection valued measure $E$ , bounded measurable functions $f, g$ with uniformly approximating complex simple functions $s_{n} \to f$ , $t_{n} \to g$ , and scalars $a, b$ .

1.1

Unitality: $1_{X}$ is simple, and $Φ_{E} (1_{X}) = E (X) = I$ by the definition of the simple integral and $E (X) = I$ .

A1A2A4

1.2

Linearity on simple functions: presenting $s$ and $t$ over a common refinement of their disjoint normal forms, $\int (a s + b t) d E = a \int s d E + b \int t d E$ because both sides are the corresponding coefficient-weighted sum of the same projection values.

1.3

Multiplicativity on simple functions: over a common disjoint normal form $s = \sum_{j} c_{j} 1_{D_{j}}$ , $t = \sum_{j} d_{j} 1_{D_{j}}$ one has $s t = \sum_{j} c_{j} d_{j} 1_{D_{j}}$ and $(\int s d E) (\int t d E) = \sum_{j, k} c_{j} d_{k} E (D_{j}) E (D_{k}) = \sum_{j} c_{j} d_{j} E (D_{j}) = \int s t d E$ , because $E (D_{j}) E (D_{k}) = E (D_{j} \cap D_{k})$ vanishes for $j \neq k$ and equals $E (D_{j})$ for $j = k$ .

A2A4

1.4

Conjugation on simple functions: self-adjointness of the projection values and conjugate-linearity of the adjoint give $(\int s d E)^{*} = (\sum_{j} c_{j} E (D_{j}))^{*} = \sum_{j} \overline{c_{j}} E (D_{j}) = \int \overline{s} d E$ . The bounded integral agrees with the simple integral by taking a constant approximating sequence.

A1A2A4A6

2.1

Linearity, multiplicativity and conjugation pass to uniform limits: if $s_{n} \to f$ and $t_{n} \to g$ uniformly then $a s_{n} + b t_{n} \to a f + b g$ , $s_{n} t_{n} \to f g$ and $\overline{s_{n}} \to \overline{f}$ uniformly, and A1 gives convergence of the simple integrals to the integrals of each of these limits, so $Φ_{E} (a f + b g) = a Φ_{E} (f) + b Φ_{E} (g)$ , $Φ_{E} (f g) = Φ_{E} (f) Φ_{E} (g)$ and $Φ_{E} (\overline{f}) = Φ_{E} (f)^{*}$ by taking norm limits and using norm continuity of the adjoint.

A1A6step 1.2step 1.3step 1.4

3.1

Strong convergence: if $\sup_{n} ∥ f_{n} ∥_{\infty} \leq C < \infty$ and $f_{n} \to f$ $E$ -almost everywhere, then for each $x$ the functions $∣ f_{n} - f ∣^{2}$ converge to $0$ $E_{x}$ -almost everywhere and are dominated by the constant $(C + ∥ f ∥_{\infty})^{2}$ , which is integrable for the finite measure $E_{x}$ ; hence $∥ Φ_{E} (f_{n}) x - Φ_{E} (f) x ∥^{2} = \int ∣ f_{n} - f ∣^{2} d E_{x} \to 0$ .

step 2.1A1A3A5

4.1

$Φ_{E}$ is a unital star homomorphism on the bounded measurable functions, and bounded pointwise $E$ -almost everywhere convergence with a uniform bound implies strong convergence of the operators.

step 1.1step 2.1step 3.1A7∎

LemmaStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Continuous functional calculus produces a regular PVM

Statement

Assume AC. Let $K$ be a nonempty compact Hausdorff space, let $H$ be a nonzero complex Hilbert space, and let $π : C (K; C) \to B (H)$ be a unital star-homomorphism: $π$ is complex-linear, $π (1) = I$ , $π (f g) = π (f) π (g)$ and $π (\overline{f}) = π (f)^{*}$ for all continuous $f, g$ . Then there is a unique regular projection valued measure $E$ on the Borel $σ$ -algebra of $K$ such that

$π (f) = \int f d E for every f \in C (K; C),$

where $\int f d E = Φ_{E} (f)$ is the bounded Borel integral of Bounded borel pvm integral and regularity is the requirement that each finite measure $E_{x} (B) = ⟨ E (B) x, x ⟩$ is a regular Borel measure (Regular Borel measure on an LCH space). The PVM constructed satisfies $⟨ E (B) x, y ⟩ = μ_{x, y} (B)$ for the scalar measures $μ_{x, y}$ built from $π$ below.

Facts & Assumptions

[A1]

Hypothesis on $π$ : $π$ is complex-linear and unital with $π (f g) = π (f) π (g)$ and $π (\overline{f}) = π (f)^{*}$ for continuous $f, g$ ; in particular $π (1) = I$ , where $1$ is the constant function.

[A2]

A unital star-homomorphism between complex C*-algebras maps positive elements to positive elements: if $g = \overline{h} h$ in $C (K; C)$ then $π (g) = π (h)^{*} π (h)$ ; for $g \geq 0$ pointwise there is $h = \sqrt{g}$ continuous with $g = \overline{h} h$ (C star algebra, Self-adjoint positive unitary and normal elements).

[A3]

The pairing is linear in the first argument and conjugate-linear in the second, $∥ u ∥^{2} = ⟨ u, u ⟩$ , and for a bounded operator $S$ one has $⟨ S u, u ⟩ = ⟨ u, S^{*} u ⟩$ , so $⟨ S^{*} S u, u ⟩ = ∥ S u ∥^{2} \geq 0$ (Hilbert-adjoint identities, Real and complex inner-product spaces and their induced length, Hilbert space).

[A4]

Every bounded complex linear functional $L$ on $C_{0} (X; C)$ for LCH $X$ has a unique representation $L (f) = \int f d μ$ by a finite regular complex Borel measure $μ$ , with $∥ L ∥ = ∣ μ ∣ (X)$ ; a positive functional's representing measure is a positive measure, and two Radon measures with equal integrals of all continuous functions coincide (The bounded complex dual of C_0(X) is regular complex measures, Positive C_0(X) functionals have finite regular representing measures, Uniqueness of the RMK representing measure among Radon measures).

[A5]

A finite complex measure $ν$ satisfies $∣ \int g d ν ∣ \leq ∥ g ∥_{\infty} ∣ ν ∣ (X)$ for bounded measurable $g$ , and for a measurable set $B$ and countable measurable partition $(B_{j})$ of $B$ one has $∣ \int_{B_{j}} g d ν ∣ \leq \int_{B_{j}} ∣ g ∣ d ∣ ν ∣$ ; a complex measure $σ$ with $∣ σ ∣ \leq C ρ$ for a finite regular Borel measure $ρ$ is regular, because inner and outer approximation transfer from $ρ$ to $∣ σ ∣$ with the factor $C$ (Integrals against signed or complex measures are bounded by total variation, The total variation |nu|(E) from countable measurable partitions, Regular Borel measure on an LCH space, Regular complex Borel measures).

[A6]

For $x \in H$ and a bounded conjugate-linear functional $φ$ on $H$ there is a unique $z \in H$ with $φ (y) = ⟨ z, y ⟩$ and $∥ z ∥ = ∥ φ ∥$ (Hilbert Riesz representation for the first-variable-linear convention; Riesz representation for Hilbert spaces).

[A7]

Polarization for a sesquilinear form $Λ$ : $Λ (x, y) = \frac{1}{4} \sum_{k = 0}^{3} i^{k} Λ (x + i^{k} y, x + i^{k} y)$ , and if $P$ is an orthogonal projection then $⟨ P x, x ⟩ = ∥ P x ∥^{2}$ (Jordan–von Neumann: a norm is induced by an inner product exactly when it satisfies the parallelogram law, Projection valued measure).

[A8]

For a complex measure $ν$ , $μ \mapsto \int h d μ$ is linear in $μ$ and $\int 1_{B} d μ = μ (B)$ ; and $E_{x} = μ_{x, x}$ below is a regular measure because it is the representing measure of a bounded functional on $C (K; C)$ (Integration against a signed or complex measure, and the class L^1(nu) = L^1(|nu|), A complex measure is a finite-valued countably additive set function).

[A9]

AC is the declared choice hypothesis of this page from this item onward, and it entails the Countable Choice hypothesis of the Hilbert Riesz supplier (The Axiom of Choice).

Proof

technique · direct

Given: A nonempty compact Hausdorff space $K$ , a nonzero complex Hilbert space $H$ and a unital star-homomorphism $π : C (K; C) \to B (H)$ .

1.1

Contractivity of $π$ : if $∥ f ∥_{\infty} \leq 1$ then $1 - \overline{f} f \geq 0$ pointwise, so $1 - \overline{f} f = \overline{g} g$ with $g = \sqrt{1 - \overline{f} f}$ continuous and hence $I - π (f)^{*} π (f) = π (g)^{*} π (g)$ ; for every $x$ this gives $∥ x ∥^{2} - ∥ π (f) x ∥^{2} = ∥ π (g) x ∥^{2} \geq 0$ , so $∥ π (f) x ∥ \leq ∥ x ∥$ and $∥ π (f) ∥ \leq 1$ ; for general $f \neq 0$ apply this to $f / ∥ f ∥_{\infty}$ .

A1A2A3

2.1

For all $x, y \in H$ the map $L_{x, y} (f) : = ⟨ π (f) x, y ⟩$ is a bounded complex linear functional on $C (K; C)$ with $∥ L_{x, y} ∥ \leq ∥ x ∥ ∥ y ∥$ , because $π$ is linear and $∣ ⟨ π (f) x, y ⟩ ∣ \leq ∥ π (f) x ∥ ∥ y ∥ \leq ∥ f ∥_{\infty} ∥ x ∥ ∥ y ∥$ .

step 1.1A3

3.1

Since $K$ is compact, $C (K; C) = C_{0} (K; C)$ , so the representation theorem applies: for each pair $x, y$ there is a unique finite regular complex Borel measure $μ_{x, y}$ with $⟨ π (f) x, y ⟩ = \int f d μ_{x, y}$ for all continuous $f$ , and $∣ μ_{x, y} ∣ (K) = ∥ L_{x, y} ∥ \leq ∥ x ∥ ∥ y ∥$ .

step 2.1A4

4.1

Sesquilinearity: for all $x, z, y \in H$ , $λ \in C$ and continuous $f$ one has $\int f d μ_{x + z, y} = ⟨ π (f) (x + z), y ⟩ = \int f d μ_{x, y} + \int f d μ_{z, y}$ and $\int f d μ_{λ x, y} = λ \int f d μ_{x, y}$ , while $\int f d μ_{x, y + z} = \int f d μ_{x, y} + \int f d μ_{x, z}$ and $\int f d μ_{x, λ y} = \overline{λ} \int f d μ_{x, y}$ ; both sides in each identity are finite regular complex measures with equal integrals against every continuous $f$ , so they coincide by uniqueness in the representation theorem.

step 3.1A4

5.1

For a bounded Borel $h$ the form $Λ_{h} (x, y) : = \int h d μ_{x, y}$ is sesquilinear by the previous step, and $∣ Λ_{h} (x, y) ∣ \leq ∥ h ∥_{\infty} ∣ μ_{x, y} ∣ (K) \leq ∥ h ∥_{\infty} ∥ x ∥ ∥ y ∥$ ; hence for each $x$ the map $y \mapsto Λ_{h} (x, y)$ is a bounded conjugate-linear functional and Hilbert Riesz representation gives a unique vector $E (h) x$ with $⟨ E (h) x, y ⟩ = \int h d μ_{x, y}$ for all $y$ , where $∥ E (h) x ∥ \leq ∥ h ∥_{\infty} ∥ x ∥$ ; the map $x \mapsto E (h) x$ is linear by sesquilinearity, so $E (h) \in B (H)$ and $∥ E (h) ∥ \leq ∥ h ∥_{\infty}$ .

A5A6step 4.1

6.1

For bounded Borel $h_{1}, h_{2}$ and scalars $a, b$ one has $E (a h_{1} + b h_{2}) = a E (h_{1}) + b E (h_{2})$ because the defining pairings agree, and $E (1_{K}) = I$ because $⟨ E (1_{K}) x, y ⟩ = \int 1_{K} d μ_{x, y} = μ_{x, y} (K) = ⟨ π (1) x, y ⟩ = ⟨ x, y ⟩$ ; moreover $E (f) = π (f)$ for every continuous $f$ , since their pairings are equal by the defining property of $μ_{x, y}$ .

step 3.1step 5.1A1A8

6.2

Conjugate symmetry of the scalar measures: for continuous $f$ one has $\int f d μ_{y, x} = ⟨ π (f) y, x ⟩ = \overline{⟨ π (\overline{f}) x, y ⟩} = \overline{\int \overline{f} d μ_{x, y}} = \int f d \overline{μ_{x, y}}$ , so $μ_{y, x} = \overline{μ_{x, y}}$ by uniqueness; consequently for bounded Borel $h$ and all $x, y$ , $⟨ E (h)^{*} x, y ⟩ = \overline{⟨ E (h) y, x ⟩} = \overline{\int h d μ_{y, x}} = \overline{\int h d \overline{μ_{x, y}}} = \int \overline{h} d μ_{x, y} = ⟨ E (\overline{h}) x, y ⟩$ , where the third expression inserts $μ_{y, x} = \overline{μ_{x, y}}$ and the fourth uses $\int h d \overline{μ_{x, y}} = \overline{\int \overline{h} d μ_{x, y}}$ ; hence $E (h)^{*} = E (\overline{h})$ .

step 3.1step 5.1A1A4

7.1

Multiplicativity with a continuous factor: fix continuous $f$ ; for the measures $ν : = \int_{\cdot} f d μ_{x, y}$ and $ρ : = μ_{x, E (\overline{f}) y}$ and every continuous $g$ one computes $\int g d ν = \int f g d μ_{x, y} = ⟨ E (f g) x, y ⟩ = ⟨ E (f) E (g) x, y ⟩ = ⟨ E (g) x, E (\overline{f}) y ⟩ = \int g d ρ$ , where the middle identity uses $E (f) = π (f)$ , the multiplicativity of $π$ and $E (g) = π (g)$ for continuous $g$ ; here $ν$ is regular because $∣ ν ∣ \leq ∥ f ∥_{\infty} ∣ μ_{x, y} ∣$ and $ρ$ is regular by construction, so uniqueness in the representation theorem gives $ν = ρ$ and hence $\int h d (f μ_{x, y}) = \int h d μ_{x, E (\overline{f}) y}$ for every bounded Borel $h$ ; therefore $E (f h) = E (f) E (h)$ for every bounded Borel $h$ .

step 3.1step 5.1step 6.1A4A5

8.1

Measure identity for a Borel density: for every bounded Borel $f$ and all $x, y$ the finite complex measures $f μ_{x, y} : B \mapsto \int_{B} f d μ_{x, y}$ and $μ_{x, E (\overline{f}) y}$ are equal, because for every continuous $g$ one has $\int g d (f μ_{x, y}) = \int f g d μ_{x, y} = ⟨ E (f g) x, y ⟩ = ⟨ E (f) E (g) x, y ⟩ = ⟨ E (g) x, E (\overline{f}) y ⟩ = \int g d μ_{x, E (\overline{f}) y}$ , using multiplicativity with a continuous second factor, which follows from the continuous-factor case together with $E (h)^{*} = E (\overline{h})$ .

step 7.1step 6.2A5

9.1

Full multiplicativity: for bounded Borel $f, h$ and all $x, y$ , $⟨ E (f h) x, y ⟩ = \int f h d μ_{x, y} = \int h d (f μ_{x, y}) = \int h d μ_{x, E (\overline{f}) y} = ⟨ E (h) x, E (\overline{f}) y ⟩ = ⟨ E (h) x, E (f)^{*} y ⟩ = ⟨ E (f) E (h) x, y ⟩$ , so $E (f h) = E (f) E (h)$ .

step 5.1step 6.2step 8.1A3

10.1

The set function $B \mapsto E (B) : = E (1_{B})$ takes values in orthogonal projections: $E (B)^{2} = E (1_{B}^{2}) = E (1_{B}) = E (B)$ by multiplicativity and $E (B)^{*} = E (1_{B}) = E (B)$ by conjugation symmetry; moreover $E (\emptyset) = 0$ , $E (K) = I$ and $E (B \cap C) = E (B) E (C)$ for all Borel $B, C$ .

step 6.1step 6.2step 9.1A7

11.1

Strong countable additivity and regularity: for pairwise disjoint Borel sets $B_{n}$ with union $B$ and $C_{N} : = ⋃_{n \leq N} B_{n}$ one has $E (B) - E (C_{N}) = E (1_{B ∖ C_{N}})$ by linearity, so $∥ E (B) x - E (C_{N}) x ∥^{2} = ⟨ E (1_{B ∖ C_{N}}) x, x ⟩ = μ_{x, x} (B ∖ C_{N})$ ; the sets $B ∖ C_{N}$ decrease to $\emptyset$ and $μ_{x, x}$ is a finite complex measure, so its values on them tend to $0$ , giving $E (B) x = \lim_{N} E (C_{N}) x = \sum_{n} E (B_{n}) x$ in norm; moreover $E_{x} (B) = ⟨ E (B) x, x ⟩ = μ_{x, x} (B)$ is a regular Borel measure, since $μ_{x, x}$ is regular by construction.

step 3.1step 6.1step 9.1step 10.1A5A7A8

12.1

Uniqueness: if $E^{'}$ is a regular PVM on the Borel $σ$ -algebra of $K$ with $π (f) = \int f d E^{'}$ for every continuous $f$ , then for each $x$ the finite regular positive measures $E_{x}$ and $E_{x}^{'}$ have $\int f d E_{x} = ⟨ π (f) x, x ⟩ = \int f d E_{x}^{'}$ for every continuous $f$ , so $E_{x} = E_{x}^{'}$ by the uniqueness theorem for Radon measures; the polarization formula $⟨ E (B) x, y ⟩ = \frac{1}{4} \sum_{k} i^{k} E_{x + i^{k} y} (B)$ for the sesquilinear form $(x, y) \mapsto ⟨ E (B) x, y ⟩$ then gives $⟨ E (B) x, y ⟩ = ⟨ E^{'} (B) x, y ⟩$ for all $B, x, y$ , hence $E (B) = E^{'} (B)$ .

step 10.1step 11.1A4A7

13.1

Consequently $E (B) = E (1_{B})$ defines a regular PVM on the Borel $σ$ -algebra of $K$ with $\int f d E = π (f)$ for every continuous $f$ , and it is the unique such regular PVM; for bounded Borel $h$ the operator $\int h d E$ of Bounded borel pvm integral coincides with $E (h)$ , since both have the pairings $\int h d μ_{x, y}$ .

step 5.1step 6.1step 9.1step 10.1step 11.1step 12.1A9∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Spectral theorem for bounded normal operators pvm form

Statement

Assume AC. Let $T$ be a bounded normal operator on a nonzero complex Hilbert space $H$ , with operator spectrum $σ (T)$ (Spectrum and resolvent of a bounded operator). Then:

there is a unique regular projection valued measure $E$ on the Borel $σ$ -algebra of the nonempty compact set $σ (T) \subseteq C$ such that $\int z d E (z) = T$ and, equivalently, $Φ_{E} (f) = f (T) for every continuous f : σ (T) \to C,$ where $f (T)$ is the continuous functional calculus of Continuous functional calculus for bounded normal operators and $Φ_{E}$ is the bounded Borel integral of the projection valued measure $E$ ; this $E$ is the spectral projection valued measure of $T$ ;
conversely, if $Λ \subseteq C$ is nonempty and compact and $E$ is a regular projection valued measure on the Borel $σ$ -algebra of $Λ$ , then $T_{E} : = \int z d E (z) = Φ_{E} (z)$ is a bounded normal operator with $σ (T_{E}) \subseteq Λ$ and $∥ T_{E} ∥ \leq \max_{z \in Λ} ∣ z ∣$ .

Facts & Assumptions

[A1]

For normal $T$ the map $π (f) : = f (T)$ is the unique isometric unital star-isomorphism $C (σ (T)) \to C^{*} (I, T)$ with $π (z) = T$ ; it is complex-linear, multiplicative, unital and star-preserving, and its range consists of the continuous-calculus operators (Continuous functional calculus for bounded normal operators, Continuous functional calculus properties, C star algebra generated by a normal operator).

[A2]

$σ (T)$ is a nonempty compact subset of $C$ : the operator spectrum coincides with the spectrum of $T$ in the unital C*-algebra $C^{*} (I, T)$ by spectral permanence and the bounded inverse theorem, and the spectrum of an element of a nonzero unital complex Banach algebra is nonempty and compact (Spectral permanence for unital c star subalgebras, Bounded inverse theorem, Spectrum is nonempty compact and norm bounded, Spectrum and resolvent of a bounded operator).

[A3]

For a nonempty compact Hausdorff $K$ and a unital star-homomorphism $π : C (K; C) \to B (H)$ there is a unique regular PVM $E$ on the Borel $σ$ -algebra of $K$ with $Φ_{E} (f) = π (f)$ for every continuous $f$ , and for bounded Borel $h$ the operator $Φ_{E} (h)$ satisfies $∥ Φ_{E} (h) ∥ \leq ∥ h ∥_{\infty}$ (Continuous functional calculus produces a regular PVM, Bounded borel pvm integral).

[A4]

For every PVM the map $Φ_{E}$ is linear, unital, multiplicative and star-preserving on bounded Borel functions; consequently $T_{E} : = Φ_{E} (z)$ is normal because $T_{E} T_{E}^{*} = Φ_{E} (∣ z ∣^{2}) = T_{E}^{*} T_{E}$ (Pvm integral is a star homomorphism).

[A5]

For an orthogonal projection value $E (Λ) = I$ and a continuous bounded $g$ on $Λ$ the operators $Φ_{E} (g)$ are bounded by $∥ g ∥_{\infty}$ , and for $∣ λ - z ∣ > 0$ on $Λ$ the function $z \mapsto (λ - z)^{- 1}$ is continuous (Bounded borel pvm integral, Projection valued measure).

[A6]

The $*$ -polynomials in $z, \overline{z}$ are uniformly dense in $C (Λ; C)$ for compact $Λ \subseteq C$ , and the image of a compact set under a continuous map is compact (Complex Stone–Weierstrass dichotomy for separating self-adjoint algebras; the unital case is dense, The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

[A7]

A bounded operator with a two-sided bounded inverse at $λ$ has $λ \notin σ (T)$ ; normality is $T^{*} T = T T^{*}$ (Spectrum and resolvent of a bounded operator, Self-adjoint, positive, unitary and normal operators, Hilbert space).

[A8]

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 $H$ and a bounded normal operator $T \in B (H)$ ; for the converse a nonempty compact $Λ \subseteq C$ and a regular PVM $E$ on its Borel $σ$ -algebra.

1.1

Existence: $σ (T)$ is a nonempty compact subset of $C$ and $π (f) : = f (T)$ is a unital star-homomorphism $C (σ (T); C) \to B (H)$ ; applying the construction lemma to $K = σ (T)$ gives a regular PVM $E$ on the Borel $σ$ -algebra of $σ (T)$ with $Φ_{E} (f) = π (f) = f (T)$ for every continuous $f$ ; taking $f$ to be the coordinate function $z$ gives $\int z d E = π (z) = T$ .

A1A2A3

1.2

Converse, boundedness and normality: for a regular PVM $E$ on a nonempty compact $Λ$ the coordinate function is bounded by $\max_{z \in Λ} ∣ z ∣$ and measurable, so $T_{E} = Φ_{E} (z)$ is a bounded operator with $∥ T_{E} ∥ \leq \max_{z \in Λ} ∣ z ∣$ , its adjoint is $T_{E}^{*} = Φ_{E} (\overline{z})$ by conjugation preservation, and $T_{E} T_{E}^{*} = Φ_{E} (z \overline{z}) = Φ_{E} (\overline{z} z) = T_{E}^{*} T_{E}$ by multiplicativity, so $T_{E}$ is normal.

A4A5

2.1

Converse, spectrum: if $λ \in C ∖ Λ$ then $Λ$ is closed and $g (z) : = (λ - z)^{- 1}$ is continuous on $Λ$ with $(λ - z) g (z) = 1$ ; hence $(λ I - T_{E}) Φ_{E} (g) = Φ_{E} (λ 1 - z) Φ_{E} (g) = Φ_{E} ((λ - z) g) = Φ_{E} (1) = I$ and likewise $Φ_{E} (g) (λ I - T_{E}) = I$ , so $λ I - T_{E}$ has a two-sided bounded inverse and $λ \notin σ (T_{E})$ ; therefore $σ (T_{E}) \subseteq Λ$ .

step 1.2A4A5A7

2.2

Uniqueness for the direct statement: if $E^{'}$ is a regular PVM on the Borel $σ$ -algebra of $σ (T)$ with $\int z d E^{'} = T$ , then for every $*$ -polynomial $p (z, \overline{z})$ one has $Φ_{E^{'}} (p) = p (T_{E}^{'}, T_{E}^{' *}) = p (T, T^{*}) = p (T) = Φ_{E} (p)$ , using multiplicativity and conjugation preservation of $Φ_{E^{'}}$ together with $T_{E}^{'} = T$ and the identification of $T^{*}$ with the calculus value of $\overline{z}$ ; since $*$ -polynomials are uniformly dense in $C (σ (T))$ and both $Φ_{E^{'}}$ and $Φ_{E}$ are bounded linear maps agreeing there, they agree on every continuous function, so $E^{'} = E$ by the uniqueness clause of the construction lemma.

step 1.1A1A3A6

3.1

The spectral PVM $E$ of $T$ therefore exists, is unique among regular PVMs on $σ (T)$ whose coordinate integral is $T$ , and satisfies $Φ_{E} (f) = f (T)$ for all continuous $f$ ; conversely every coordinate integral of a regular PVM on a compact set is a bounded normal operator with spectrum inside that set.

step 1.1step 1.2step 2.1step 2.2A8∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-09-22Open item page →

Borel functional calculus for a bounded normal operator

Definition

Assume AC. Let $T$ be a bounded normal operator on a nonzero complex Hilbert space $H$ and let $E$ be its spectral projection valued measure on the Borel $σ$ -algebra of $σ (T)$ , the unique regular projection valued measure with $\int z d E (z) = T$ (Spectral theorem for bounded normal operators pvm form). For a bounded Borel function $f : σ (T) \to C$ , that is a bounded $Σ$ -measurable function on the Borel $σ$ -algebra of $σ (T)$ (A measurable function between measurable spaces), define the Borel functional calculus of $T$ at $f$ by

$f (T) : = Φ_{E} (f) = \int_{σ (T)} f d E,$

the bounded Borel integral of $f$ against $E$ from Bounded borel pvm integral. The map $f \mapsto f (T)$ from bounded Borel functions on $σ (T)$ to $B (H)$ is the bounded Borel functional calculus of $T$ .

Well-definedness and consistency. The operator $f (T)$ is well defined because the spectral projection valued measure $E$ of $T$ is unique: if $E^{'}$ were another regular projection valued measure with $\int z d E^{'} = T$ , then $E^{'} = E$ . The construction agrees with the continuous calculus on continuous functions, $Φ_{E} (f) = f (T)$ for $f \in C (σ (T))$ , by the displayed clause of the spectral theorem. It inherits the algebraic behaviour of the projection valued measure integral: $f (T)$ is complex-linear in $f$ , unital with $1 (T) = I$ , multiplicative, star-preserving with $\overline{f} (T) = f (T)^{*}$ , norm bounded by $∥ f (T) ∥ \leq ∥ f ∥_{\infty}$ , strongly continuous for bounded Borel $f_{n}, f$ when $f_{n} \to f$ pointwise $E$ -almost everywhere and $\sup_{n} ∥ f_{n} ∥_{\infty} < \infty$ , as supplied by Pvm integral is a star homomorphism. In particular $1_{B} (T) = E (B)$ for every Borel set $B \subseteq σ (T)$ , and the norm of $f (T)$ is the $E$ -essential supremum of $f$ (Bounded borel pvm integral).

Commutation. Every bounded $S$ commuting with $T$ and $T^{*}$ commutes with every $f (T)$ . Write $π (g) = g (T)$ for the continuous calculus. The construction in Continuous functional calculus produces a regular PVM provides finite regular complex measures $μ_{x, y}$ with $μ_{x, y} (B) = ⟨ E (B) x, y ⟩$ and $\int g d μ_{x, y} = ⟨ π (g) x, y ⟩$ for continuous $g$ ; its PVM is the present $E$ by uniqueness. The continuous commutant property (Continuous functional calculus properties) and the defining adjoint identity (The Hilbert-space adjoint of a bounded operator) give $\int g d μ_{S x, y} = ⟨ π (g) S x, y ⟩ = ⟨ S π (g) x, y ⟩ = ⟨ π (g) x, S^{*} y ⟩ = \int g d μ_{x, S^{*} y} .$ Uniqueness of the finite regular complex representing measure on the compact spectrum, where $C_{0} (σ (T)) = C (σ (T))$ , gives $μ_{S x, y} = μ_{x, S^{*} y}$ (The bounded complex dual of C_0(X) is regular complex measures). For bounded Borel $f$ , the bounded-integral pairing identity therefore yields $⟨ f (T) S x, y ⟩ = \int f d μ_{S x, y} = \int f d μ_{x, S^{*} y} = ⟨ S f (T) x, y ⟩ .$ Testing the difference against itself proves $f (T) S = S f (T)$ . These properties are collected in Borel functional calculus for bounded normal operators.

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

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 $T^{*}$ .

1.1

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 $f (T) f (T)^{*} = Φ_{E} (f \overline{f}) = Φ_{E} (\overline{f} f) = f (T)^{*} f (T)$ .

A2A3

1.2

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 $μ_{S x, y} = μ_{x, S^{*} y}$ .

A1A2A4

2.1

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.

A1step 1.2A3A4A5

3.1

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.

step 1.1step 2.1A6∎

CorollaryStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Spectral projections and resolution of the identity

Statement

Assume AC. Let $T$ be a bounded normal operator on a nonzero complex Hilbert space $H$ , with spectral projection valued measure $E$ on $σ (T)$ and Borel calculus $f \mapsto f (T)$ . For every Borel set $A \subseteq C$ , use the zero-extension convention

$E (A) : = E (A \cap σ (T)) .$

Then:

every spectral projection $E (B) = 1_{B} (T)$ reduces $T$ : it commutes with $T$ and $T^{*}$ , so its range and its kernel are invariant under $T$ and $T^{*}$ ;
$E ({λ}) H = \ker (T - λ I)$ for every $λ \in C$ : the eigenspace of $T$ at $λ$ is the range of the spectral projection of the singleton ${λ}$ , and it is nonzero precisely when $E ({λ}) \neq 0$ ;
if $T$ is self-adjoint, then $F (t) : = E (σ (T) \cap (- \infty, t])$ , $t \in R$ , is an increasing family of orthogonal projections which is strongly right continuous, $\lim_{s ↓ t} F (s) = F (t)$ in the strong operator topology, and $\lim_{t \to - \infty} F (t) = 0$ , $\lim_{t \to + \infty} F (t) = I$ strongly.

Facts & Assumptions

[A1]

The Borel calculus satisfies $(f g) (T) = f (T) g (T) = g (T) f (T)$ , $\overline{f} (T) = f (T)^{*}$ , $f (T)$ is bounded by $∥ f ∥_{\infty}$ , and $1_{B} (T) = E (B)$ ; in particular $B \mapsto E (B)$ satisfies $E (B \cap C) = E (B) E (C)$ , $E (\emptyset) = 0$ , $E (σ (T)) = I$ and finite additivity on disjoint measurable sets. Uniformly bounded Borel functions converging pointwise have strongly convergent calculus images (Borel functional calculus for bounded normal operators, Borel functional calculus for a bounded normal operator, Projection valued measure).

[A2]

$T = Φ_{E} (z) = \int z d E (z)$ and $⟨ f (T) x, y ⟩ = \int f d E_{x, y}$ for bounded Borel $f$ , with $∥ f (T) x ∥^{2} = \int ∣ f ∣^{2} d E_{x}$ and $E_{x, y} = ⟨ E (\cdot) x, y ⟩$ a finite regular complex measure and $E_{x} = ⟨ E (\cdot) x, x ⟩$ a positive measure of mass $∥ x ∥^{2}$ ; two finite regular complex measures with equal integrals against all continuous functions are equal (Borel functional calculus for a bounded normal operator, The bounded complex dual of C_0(X) is regular complex measures, Scalar and complex measures from a pvm, Borel functional calculus for bounded normal operators).

[A3]

For normal $T$ and $f$ continuous one has $f (T) x = f (λ) x$ whenever $T x = λ x$ (Continuous functional calculus properties).

[A4]

For self-adjoint $T$ one has $σ (T) \subseteq R$ , $\min σ (T) I \leq T \leq \max σ (T) I$ and $∥ T ∥ = \max {∣ \min σ (T) ∣, ∣ \max σ (T) ∣}$ , so $σ (T) \subseteq [- ∥ T ∥, ∥ T ∥]$ (Spectrum of a self adjoint operator is real, Self adjoint norm and spectrum extrema).

[A5]

The adjoint product rule $(R S)^{*} = S^{*} R^{*}$ and involution $S^{* *} = S$ and the definition of the spectrum via $λ I - T$ (Hilbert-adjoint identities, Spectrum and resolvent of a bounded operator, Self-adjoint, positive, unitary and normal operators).

[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 $f \mapsto f (T)$ , a Borel set $B \subseteq σ (T)$ and a scalar $λ \in C$ .

1.1

$E (B)$ commutes with $T$ and $T^{*}$ : $E (B) T = Φ_{E} (1_{B}) Φ_{E} (z) = Φ_{E} (1_{B} z) = T E (B)$ and $E (B) = Φ_{E} (1_{B}) = Φ_{E} (\overline{1_{B}}) = Φ_{E} (1_{B})^{*}$ is self-adjoint, so taking adjoints gives $E (B) T^{*} = T^{*} E (B)$ ; hence $ran E (B)$ and $\ker E (B)$ are invariant under $T$ and $T^{*}$ .

A1A2A5

1.2

Eigenvectors of spectral projections: if $x = E ({λ}) x$ then $T x = Φ_{E} (z) x = Φ_{E} (z 1_{{λ}}) x = Φ_{E} (λ 1_{{λ}}) x = λ E ({λ}) x = λ x$ , so $ran E ({λ}) \subseteq \ker (T - λ I)$ .

A1A2

1.3

For self-adjoint $T$ define $F (t) : = E (σ (T) \cap (- \infty, t])$ for real $t$ ; for $s \leq t$ the set $A_{s} : = σ (T) \cap (- \infty, s]$ is contained in $A_{t} : = σ (T) \cap (- \infty, t]$ , so $F (s) = E (A_{s}) = E (A_{s} \cap A_{t}) = F (s) F (t) = F (t) F (s)$ , and $F (t) - F (s) = E (A_{t} ∖ A_{s})$ is the image of an indicator and therefore an orthogonal projection: $F$ is increasing in the projection order.

A1A4

2.1

Conversely, suppose $T x = λ x$ . For $x = 0$ the identity $x = E ({λ}) x$ holds by linearity, with no point evaluation. For $x \neq 0$ , the operator $λ I - T$ has nonzero kernel and is not invertible, so $λ \in σ (T)$ . Thus evaluation at $λ$ and the Dirac measure $δ_{λ}$ on $σ (T)$ are defined. This Dirac measure is regular: a set containing $λ$ contains the compact singleton, and a set omitting it has the open superset $σ (T) ∖ {λ}$ of zero mass. Now for every continuous $f$ one has $\int f d E_{x} = ⟨ f (T) x, x ⟩ = f (λ) ∥ x ∥^{2}$ , so the positive measure $E_{x}$ and the mass $∥ x ∥^{2} δ_{λ}$ are finite regular measures with equal integrals against all continuous functions and hence are equal; therefore $E_{x} (σ (T) ∖ {λ}) = 0$ , which gives $∥ E (σ (T) ∖ {λ}) x ∥^{2} = 0$ and, since $E ({λ}) + E (σ (T) ∖ {λ}) = E (σ (T)) = I$ , the identity $x = E ({λ}) x$ ; with the preceding step this proves $E ({λ}) H = \ker (T - λ I)$ .

step 1.2A1A2A3A5

2.2

Strong right continuity: fix $t$ and put $t_{n} = t + 1 / (n + 1)$ for $n \geq 0$ . The Borel indicators of $σ (T) \cap (t, t_{n}]$ are bounded by one and converge pointwise to zero. Hence $F (t_{n}) - F (t)$ converges strongly to zero. For $t < s \leq t_{n}$ , the norm-square formula for indicators gives $∥ (F (s) - F (t)) x ∥^{2} = E_{x} (σ (T) \cap (t, s]) \leq E_{x} (σ (T) \cap (t, t_{n}]) = ∥ (F (t_{n}) - F (t)) x ∥^{2}$ . This proves the full right-hand strong limit as $s ↓ t$ , for every $x$ .

step 1.3A1A2

2.3

Limits at infinity: since $σ (T) \subseteq [- ∥ T ∥, ∥ T ∥]$ , for $t < - ∥ T ∥$ the set $σ (T) \cap (- \infty, t]$ is empty and $F (t) = 0$ , while for $t \geq ∥ T ∥$ it is all of $σ (T)$ and $F (t) = E (σ (T)) = I$ ; hence the strong limits at the two infinities are $0$ and $I$ .

step 1.3A1A4

3.1

The spectral projections reduce $T$ , the eigenspace at $λ$ is exactly $E ({λ}) H$ , and for self-adjoint $T$ the family $F (t) = E (σ (T) \cap (- \infty, t])$ is increasing, strongly right continuous, with strong limits $0$ and $I$ at the two infinities.

step 1.1step 2.1step 2.2step 2.3A5A6∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Support and uniqueness of the spectral measure

Statement

Assume AC. Let $T$ be a bounded normal operator on a nonzero complex Hilbert space $H$ with spectral projection valued measure $E$ on the Borel $σ$ -algebra of $σ (T)$ . Then:

support: $E (U) \neq 0$ for every nonempty relatively open subset $U \subseteq σ (T)$ ; equivalently the support of $E$ is $σ (T)$ ;
uniqueness: if $Λ \subseteq C$ is nonempty and compact and $E^{'}$ is a regular projection valued measure on the Borel $σ$ -algebra of $Λ$ with $\int z d E^{'} (z) = T$ , then $E^{'} (Λ ∖ σ (T)) = 0$ and $E^{'} (B) = E (B)$ for every Borel set $B \subseteq σ (T)$ ; in particular every scalar pairing $⟨ E (\cdot) x, y ⟩$ is determined by $T$ .

Facts & Assumptions

[A1]

For every continuous $f$ on $σ (T)$ one has $∥ f (T) ∥ = ∥ f ∥_{\infty}$ and $f (T) = Φ_{E} (f) = \int f d E$ ; for every bounded Borel $h$ one has $⟨ h (T) x, y ⟩ = \int h d E_{x, y}$ with $E_{x, y} (B) = ⟨ E (B) x, y ⟩$ . For every PVM $F$ , its bounded integral $Φ_{F}$ is a unital star-homomorphism (Continuous functional calculus for bounded normal operators, Spectral theorem for bounded normal operators pvm form, Borel functional calculus for bounded normal operators, Pvm integral is a star homomorphism).

[A2]

The Borel calculus is multiplicative: $(f g) (T) = f (T) g (T)$ for bounded Borel $f, g$ , and $E (B) = 1_{B} (T)$ ; in particular $Φ_{E} (g) = 0$ whenever $g$ vanishes on a Borel set carrying the full projection (Borel functional calculus for bounded normal operators, Projection valued measure).

[A3]

$E_{x} (B) = ⟨ E (B) x, x ⟩ = ∥ E (B) x ∥^{2}$ is a positive measure; for a nonnegative measurable $g$ one has $\int g d E_{x} = 0$ if and only if $g = 0$ $E_{x}$ -almost everywhere (Scalar and complex measures from a pvm, A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[A4]

$E$ is the unique regular PVM on the Borel $σ$ -algebra of $σ (T)$ whose coordinate integral is $T$ , and $Λ ∖ σ (T)$ is a countable union of compact subsets of $Λ$ (Spectral theorem for bounded normal operators pvm form, Continuous functional calculus produces a regular PVM, Regular Borel measure on an LCH space).

[A5]

The $*$ -polynomials are uniformly dense in $C (Λ; C)$ for compact $Λ \subseteq C$ , and for compact $Λ \subseteq C$ the distance function $z \mapsto d (z, σ (T))$ is continuous, nonnegative, and vanishes exactly on $σ (T)$ (Complex Stone–Weierstrass dichotomy for separating self-adjoint algebras; the unital case is dense).

[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 $E$ on $σ (T)$ , a nonempty relatively open $U \subseteq σ (T)$ , and a regular PVM $E^{'}$ on a nonempty compact $Λ \subseteq C$ with $\int z d E^{'} (z) = T$ .

1.1

Suppose $U \subseteq σ (T)$ is nonempty and relatively open with $E (U) = 0$ ; choose $λ \in U$ and $r > 0$ with $σ (T) \cap {∣ z - λ ∣ < r} \subseteq U$ and put $q (z) : = \max {0, r - ∣ z - λ ∣}$ on $σ (T)$ . Then $q$ is continuous, $q \geq 0$ , $q (λ) = r = ∥ q ∥_{\infty}$ because $λ \in σ (T)$ , and $q$ vanishes outside $U$ , so $q 1_{U} = q$ and $Φ_{E} (q) = Φ_{E} (q 1_{U}) = Φ_{E} (q) Φ_{E} (1_{U}) = q (T) E (U) = 0$ , whence $∥ q (T) ∥ = 0$ ; by the isometry of the continuous calculus $∥ q ∥_{\infty} = ∥ q (T) ∥ = 0$ , contradicting $q (λ) = r > 0$ .

A1A2

1.2

For a *-polynomial $p$ in $z, \overline{z}$ on $Λ$ one has $Φ_{E^{'}} (p) = p (T, T^{*})$ , while the continuous calculus on $σ (T)$ gives $p (T, T^{*}) = (p ∣_{σ (T)}) (T) = Φ_{E} (p ∣_{σ (T)})$ ; hence the bounded linear maps $f \mapsto Φ_{E^{'}} (f)$ and $f \mapsto Φ_{E} (f ∣_{σ (T)})$ on $C (Λ; C)$ agree on the uniformly dense family of *-polynomials and therefore on all continuous $f$ .

A1A5

2.1

Consequently $Φ_{E^{'}} (q) = 0$ for $q (z) : = d (z, σ (T))$ , since $q$ is continuous on $Λ$ with $q ∣_{σ (T)} = 0$ ; then $\int q d E_{x}^{'} = ⟨ Φ_{E^{'}} (q) x, x ⟩ = 0$ with $q \geq 0$ , so $q = 0$ $E_{x}^{'}$ -almost everywhere and $E_{x}^{'} (Λ ∖ σ (T)) = 0$ ; as $∥ E^{'} (Λ ∖ σ (T)) x ∥^{2} = E_{x}^{'} (Λ ∖ σ (T)) = 0$ for every $x$ , one gets $E^{'} (Λ ∖ σ (T)) = 0$ .

step 1.2A2A3A5

3.1

The restriction $E^{''} (B) : = E^{'} (B)$ for Borel $B \subseteq σ (T)$ is a regular PVM on $σ (T)$ with $E^{''} (σ (T)) = E^{'} (Λ) - E^{'} (Λ ∖ σ (T)) = I$ and $Φ_{E^{''}} (z) = Φ_{E^{'}} (z) - Φ_{E^{'}} (z 1_{Λ ∖ σ (T)}) = T - 0 = T$ , because functions supported in the $E^{'}$ -null set $Λ ∖ σ (T)$ integrate to $0$ ; by the uniqueness clause of the spectral theorem $E^{''} = E$ , so $E^{'} (B) = E (B)$ for every Borel $B \subseteq σ (T)$ .

step 2.1A2A4

4.1

The support of $E$ is all of $σ (T)$ , and any regular PVM on a compact set whose coordinate integral is $T$ agrees with $E$ on $σ (T)$ and vanishes off it; in particular all pairings of $E$ are determined by $T$ .

step 1.1step 3.1A6∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-09-22Open item page →

Cyclic vector and cyclic normal operator

Definition

Assume AC. Let $T$ be a bounded normal operator on a nonzero complex Hilbert space $H$ , with continuous functional calculus $f \mapsto f (T)$ , a unital star-isomorphism $C (σ (T)) \to C^{*} (I, T)$ (Continuous functional calculus for bounded normal operators).

A vector $x \in H$ is cyclic for $T$ when the closed linear span of the set ${f (T) x : f \in C (σ (T))}$ is all of $H$ ; equivalently, when $\overset{}{span} {f (T) x : f \in C (σ (T))} = H .$ For a normal operator this is the same as cyclicity for the unital $*$ -algebra generated by $T$ : because the calculus is a unital $*$ -isomorphism onto $C^{*} (I, T)$ and $C^{*} (I, T)$ is the norm closure of the unital $*$ -algebra generated by $T$ and $T^{*}$ (C star algebra generated by a normal operator), the set ${f (T) x : f \in C (σ (T))}$ is precisely the set ${a x : a \in C^{*} (I, T)}$ , and cyclicity uses the normal operator $T$ together with its adjoint, never only the nonnegative powers of $T$ . A normal operator is called cyclic when it has a cyclic vector.

For an arbitrary vector $x \in H$ define its cyclic subspace $H_{x} : = \overset{}{span} {f (T) x : f \in C (σ (T))} .$

Well-definedness and elementary properties. $H_{x}$ is a closed linear subspace by definition, it contains $x = 1 (T) x$ , and it is invariant under $T$ and $T^{*}$ : for $f$ continuous one has $T f (T) x = (z f) (T) x$ and $T^{*} f (T) x = (\overline{z} f) (T) x$ , both again of the form $g (T) x$ with $g$ continuous (Continuous functional calculus properties). Hence $H_{x}$ is a reducing subspace for $T$ . If $x \neq 0$ , then $H_{x} \neq {0}$ and the restriction $R : = T ∣_{H_{x}}$ is a bounded normal operator. It is cyclic with cyclic vector $x$ : every $f (T) \in C^{*} (I, T)$ is a norm limit of star-polynomials $q_{n} (T, T^{*})$ , and restriction to $H_{x}$ gives $q_{n} (R, R^{*}) \to f (T) ∣_{H_{x}}$ , so $f (T) ∣_{H_{x}} \in C^{*} (I, R)$ . The continuous calculus for $R$ is onto $C^{*} (I, R)$ , so its orbit of $x$ contains the original orbit ${f (T) x : f \in C (σ (T))}$ , whose closed span is $H_{x}$ by definition; therefore $x$ is cyclic for $R$ . If $x = 0$ , then $H_{x} = {0}$ ; the restriction to this zero space is not fed to the library's nonzero-space functional calculus, and $0$ is not cyclic for the original nonzero $H$ .

Finally, $x$ is cyclic for $T$ if and only if no nonzero $y \in H$ is orthogonal to every $f (T) x$ ; this is the standard description of a closed span as the orthogonal complement of its annihilator (Orthogonality and the orthogonal complement, Hilbert space, Self-adjoint, positive, unitary and normal operators).

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

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∎

LemmaStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Maximal orthogonal family of cyclic reducing subspaces

Statement

Assume AC. Let $T$ be a bounded normal operator on a nonzero complex Hilbert space $H$ . Then:

there is a family $(H_{j})_{j \in J}$ of pairwise orthogonal nonzero closed subspaces of $H$ , each reducing $T$ and cyclic for $T ∣_{H_{j}}$ , whose Hilbert sum is all of $H$ : the closed span of $⋃_{j} H_{j}$ equals $H$ ;
if in addition $H$ is separable and $(v_{n})_{n \geq 0}$ is a dense sequence in $H$ , put $S_{n} = H_{0} \oplus \dots \oplus H_{n - 1}$ with $S_{0} = {0}$ . Project $v_{n}$ onto $S_{n}^{⊥}$ to obtain $x_{n}$ , and take $H_{n}$ to be its cyclic subspace for the original operator $T$ on $H$ . The $H_{n}$ are closed and reducing, zero summands are allowed, and the nonzero summands are cyclic for their restrictions. Their closed span is $H$ , denoted $H = ⨁_{n \geq 0} H_{n}$ . After discarding zero summands this is a finite or countable family with the properties in claim 1.

Facts & Assumptions

[A1]

A closed subspace reduces $T$ if it is invariant under $T$ and $T^{*}$ . For every $y \in H$ the ambient cyclic subspace $H_{y}$ is closed, reducing and contains $y$ ; it is the closed span of the unital polynomial orbit in $T, T^{*}$ (Cyclic vector and cyclic normal operator). If $y = 0$ it is ${0}$ . On a nonzero reducing subspace $M$ , the restriction of $T^{*}$ is the adjoint of $T ∣_{M}$ by the defining pairing, so $T ∣_{M}$ is normal. A closed subspace is complete because Cauchy sequences converge in $H$ and their limits stay in the subspace. For nonzero $y$ , restricting the same polynomial orbit to $H_{y}$ shows $y$ is cyclic for $T ∣_{H_{y}}$ . No spectrum or calculus on a zero space is used. The adjoint pairing and involution are supplied by The Hilbert-space adjoint of a bounded operator and Hilbert-adjoint identities, and bounded operators are continuous by For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent.

[A2]

If $M$ reduces $T$ , then $M^{⊥}$ reduces $T$ : for $y \in M^{⊥}$ and $m \in M$ one has $⟨ T y, m ⟩ = ⟨ y, T^{*} m ⟩ = 0$ and $⟨ T^{*} y, m ⟩ = ⟨ y, T m ⟩ = 0$ , since $T^{*} M \subseteq M$ and $T M \subseteq M$ (Orthogonality and the orthogonal complement, The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[A3]

Orthogonal decompositions: for a closed subspace $M$ one has $H = M \oplus M^{⊥}$ , and the orthogonal complement of a closed subspace is closed; a vector orthogonal to a closed subspace $N$ lies in $N^{⊥}$ (Orthogonal decomposition by a closed subspace, Orthogonal complements are closed, Orthogonality and the orthogonal complement).

[A4]

Zorn's lemma: a nonempty poset in which every chain has an upper bound has a maximal element (Zorn's lemma, The Axiom of Choice).

[A5]

If a closed linear subspace $L$ has $L^{⊥} = {0}$ , then $H = L \oplus L^{⊥} = L$ . Equivalently, double orthogonal complementation of any linear subspace gives its closure (Orthogonal decomposition by a closed subspace, The double orthogonal complement of a subspace is its closure). A closed set containing a dense subset is the whole space (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets).

[A6]

Separability means existence of an at most countable dense subset (Separability: the existence of an at most countable dense subset). Finite sums of pairwise orthogonal closed subspaces are closed: their orthogonal projections $P_{j}$ are bounded, are the identity on their own subspace and vanish on the others (The Hilbert orthogonal projection onto a closed subspace, Hilbert projections are linear, self-adjoint and contractive). Thus $P = \sum_{j < n} P_{j}$ satisfies $P^{2} = P$ and has range $S = \sum_{j < n} H_{j}$ , so $S = \ker (I - P)$ . This kernel is closed: for $x \notin S$ , a ball of radius $∥ x - P x ∥ / (2 (1 + ∥ P ∥))$ misses it, by $∥ (I - P) (y - x) ∥ \leq (1 + ∥ P ∥) ∥ y - x ∥$ . For $n = 0$ , $P = 0$ and $S = {0}$ . If each summand reduces $T$ , their finite sum does too by linearity. No closedness of an infinite algebraic sum is asserted.

[A7]

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 $H$ and a bounded normal operator $T \in B (H)$ ; in the separable case also a dense sequence $(v_{n})$ .

1.1

Every chain in the poset $P \subseteq P (P (H))$ of sets ${M_{j}}_{j \in J}$ of pairwise orthogonal nonzero closed $T$ -reducing subspaces, cyclic for $T ∣_{M_{j}}$ , ordered by inclusion, has an upper bound (the empty family belongs to this poset): the union of the chain is again such a family, because any two of its members lie in a common family of the chain and are therefore orthogonal, and each is reducing and cyclic by membership.

A1A4

1.2

Separable construction: start with $S_{0} = {0}$ , $K_{0} = H$ , $x_{0} = v_{0}$ and the ambient cyclic subspace $H_{0} = H_{x_{0}}$ . Recursively, once $H_{j}$ for $j < n$ are defined, their finite orthogonal sum $S_{n}$ is closed and reducing by [A6]. Its orthogonal complement $K_{n} = S_{n}^{⊥}$ is closed and reducing by [A2, A3]. Let $x_{n} = P_{K_{n}} v_{n}$ and define $H_{n} = H_{x_{n}}$ using the original $T$ on $H$ . Since $K_{n}$ is closed and invariant under $T, T^{*}$ , the whole polynomial orbit of $x_{n}$ and its closure lie in $K_{n}$ . Thus $H_{n}$ is closed, reducing and orthogonal to all earlier summands. If $x_{n} = 0$ , set $H_{n} = {0}$ with no restriction calculus.

A1A2A3A6

2.1

By Zorn's lemma $P$ has a maximal element $(H_{j})_{j \in J}$ ; its members are pairwise orthogonal, nonzero, $T$ -reducing and cyclic for the restrictions.

step 1.1A4

2.2

The family $(H_{n})$ is pairwise orthogonal and every $H_{n}$ reduces $T$ and each nonzero $H_{n}$ is cyclic for the restriction, by construction and by the preceding paragraph; the nonzero members form a pairwise orthogonal family of cyclic reducing subspaces.

step 1.2A1

2.3

Spanning in the separable case: each $v_{n}$ splits as $v_{n} = p_{n} + x_{n}$ with $p_{n} \in H_{0} \oplus \dots \oplus H_{n - 1}$ and $x_{n} \in K_{n}$ , and $x_{n} \in H_{n}$ by the ambient cyclic-subspace construction, including $x_{n} = 0$ ; hence $v_{n} \in H_{0} \oplus \dots \oplus H_{n}$ for every $n$ , so the closed sum $⨁_{n} H_{n}$ contains the dense sequence $(v_{n})$ and therefore equals $H$ .

step 1.2A5A6

3.1

Maximality forces $K : = (span ⋃_{j} H_{j})^{⊥} = {0}$ : $K$ is the orthogonal complement of the closed span of a family of cyclic reducing subspaces, hence closed and $T$ -reducing: the algebraic span is invariant under $T, T^{*}$ , its closure stays invariant by their continuity, and [A2] applies; if $y \in K$ is nonzero, its ambient cyclic subspace $M : = H_{y}$ is nonzero, closed and $T$ -reducing, and lies in $K$ by invariance of $K$ under the polynomial orbit and orthogonal to every $H_{j}$ , and it is cyclic for $T ∣_{M}$ , so $(H_{j})_{j \in J} \cup {M}$ is a strictly larger element of $P$ , contradicting maximality.

step 2.1A1A2A3

4.1

Hence no nonzero vector is orthogonal to the closed span of $⋃_{j} H_{j}$ , so [A5] gives that this closed span equals $H$ , which is the first assertion.

step 3.1A3A5

5.1

The separable construction therefore yields a finite or countable family of nonzero members of $(H_{n})_{n \geq 0}$ , consisting of pairwise orthogonal cyclic reducing subspaces with $H = ⨁_{n} H_{n}$ , as claimed.

step 2.2step 2.3A7∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Multiplication operator form of the bounded normal spectral theorem

Statement

Assume AC. Let $T$ be a bounded normal operator on a nonzero complex Hilbert space $H$ , with spectral projection valued measure $E$ on $σ (T)$ . Then there is a family of nonzero finite positive regular Borel measures $(μ_{j})_{j \in J}$ on $σ (T)$ and a unitary operator

$U : ⨁_{j \in J} L^{2} (σ (T), μ_{j}) ⟶ H$

such that $U (⨁_{j} M_{z}) U^{- 1} = T$ , where $M_{z}$ is multiplication by the coordinate function on each summand. Concretely one may take a family $(x_{j})_{j \in J}$ of vectors with $H = ⨁_{j} H_{x_{j}}$ and $μ_{j} = E_{x_{j}} = ⟨ E (\cdot) x_{j}, x_{j} ⟩$ , and $U = ⨁_{j} U_{j}$ with $U_{j} [f] = f (T) x_{j}$ the cyclic representation of Cyclic spectral representation. If in addition $H$ is separable, the index set may be taken finite or countable.

Facts & Assumptions

[A1]

There is a family $(H_{j})_{j \in J}$ of pairwise orthogonal nonzero closed $T$ -reducing subspaces, each cyclic for $T ∣_{H_{j}}$ , with the closed span of the union equal to $H$ ; in the separable case the family may be taken finite or countable (Maximal orthogonal family of cyclic reducing subspaces, Separability: the existence of an at most countable dense subset).

[A2]

If $x \in H$ has cyclic subspace $H_{x}$ , then $U_{x} [f] : = f (T) x$ extends to a unitary $U_{x} : L^{2} (σ (T), E_{x}) \to H_{x}$ with $U_{x} M_{z} = T U_{x}$ , where $E_{x}$ is the finite positive regular measure $⟨ E (\cdot) x, x ⟩$ (Cyclic spectral representation, Scalar and complex measures from a pvm).

[A3]

If $H_{j}$ reduces $T$ and $R_{j} : = T ∣_{H_{j}}$ , then $T^{*} ∣_{H_{j}} = R_{j}^{*}$ and every star-polynomial restricts by $q (T, T^{*}) ∣_{H_{j}} = q (R_{j}, R_{j}^{*})$ . The normal calculus maps onto $C^{*} (I, R_{j})$ , the norm closure of those restricted star-polynomials (Cyclic vector and cyclic normal operator, Continuous functional calculus for bounded normal operators, C star algebra generated by a normal operator).

[A4]

Orthogonal direct sums of Hilbert spaces: vectors with pairwise orthogonal component subspaces have squares of norms summing, and a direct sum of unitaries between corresponding summands is a unitary between the Hilbert sums; the direct sum of multiplication operators acts componentwise (Hilbert space, The space $L^{p} (μ)$ as the quotient by null functions, Self-adjoint, positive, unitary and normal operators).

[A5]

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 $H$ and a bounded normal operator $T$ with spectral PVM $E$ ; a family $(H_{j})$ as in the maximal-orthogonal-family lemma, with cyclic vectors $x_{j} \in H_{j}$ .

1.1

For each $j$ put $R_{j} : = T ∣_{H_{j}}$ . Since $H_{j}$ reduces $T$ , every star-polynomial in $T, T^{*}$ preserves $H_{j}$ , and norm approximation in $C^{*} (I, T)$ shows $f (T) x_{j} \in H_{j}$ for every $f \in C (σ (T))$ ; hence $H_{x_{j}} \subseteq H_{j}$ . Conversely, if $g \in C (σ (R_{j}))$ , choose star-polynomials $q_{n} (R_{j}, R_{j}^{*}) \to g (R_{j})$ using the range description of the calculus. Then $q_{n} (R_{j}, R_{j}^{*}) x_{j} = q_{n} (T, T^{*}) x_{j} \in H_{x_{j}}$ , so closedness gives $g (R_{j}) x_{j} \in H_{x_{j}}$ . Since $x_{j}$ is cyclic for $R_{j}$ , these vectors have dense span in $H_{j}$ , whence $H_{j} \subseteq H_{x_{j}}$ and therefore $H_{x_{j}} = H_{j}$ .

A1A3

2.1

For each $j$ the cyclic representation gives a unitary $U_{j} : L^{2} (σ (T), E_{x_{j}}) \to H_{j}$ with $U_{j} [f] = f (T) x_{j}$ on continuous $f$ and $U_{j} M_{z} = T U_{j}$ .

step 1.1A2

3.1

The direct sum $U : = ⨁_{j} U_{j}$ maps $⨁_{j} L^{2} (σ (T), E_{x_{j}})$ onto the closed span $⨁_{j} H_{j} = H$ : it is isometric because $∥ (f_{j})_{j} ∥^{2} = \sum_{j} ∥ f_{j} ∥^{2} = \sum_{j} ∥ U_{j} f_{j} ∥^{2}$ , and it is surjective because each $U_{j}$ is onto $H_{j}$ and the Hilbert sum of the $H_{j}$ is $H$ .

step 2.1A1A4

3.2

Intertwining: $U (⨁_{j} M_{z}) (f_{j})_{j} = U (M_{z} f_{j})_{j} = (U_{j} M_{z} f_{j})_{j} = (T U_{j} f_{j})_{j} = T U (f_{j})_{j}$ , so $U (⨁_{j} M_{z}) U^{- 1} = T$ .

step 2.1A2A4

3.3

In the separable case the family from the maximal-orthogonal-family lemma may be chosen finite or countable, and the construction above then exhibits $H$ as a finite or countable orthogonal sum of cyclic $L^{2}$ summands.

step 2.1A1

4.1

$T$ is therefore unitarily equivalent to the coordinate multiplication on an orthogonal sum of $L^{2}$ -spaces over the scalar spectral measures of cyclic vectors, with a finite or countable index set in the separable case.

step 3.1step 3.2step 3.3A5∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-09-22Open item page →

Spectral multiplicity function in the separable case

Definition

Assume AC. Let $T$ be a bounded normal operator on a nonzero separable complex Hilbert space $H$ (Separability: the existence of an at most countable dense subset) with spectral projection valued measure $E$ on $σ (T)$ and spectral measure class to be described below.

Step 1: a countable cyclic decomposition. By the multiplication-operator form of the spectral theorem applied to a dense sequence, fix once and for all a finite or countable family $(x_{j})_{j \in J}$ of nonzero vectors with $H = ⨁_{j \in J} H_{x_{j}}, μ_{j} : = E_{x_{j}} = ⟨ E (\cdot) x_{j}, x_{j} ⟩,$ where $H_{x_{j}}$ is the cyclic subspace of $x_{j}$ and each $μ_{j}$ is a nonzero finite positive regular Borel measure on $σ (T)$ (Multiplication operator form of the bounded normal spectral theorem, Cyclic vector and cyclic normal operator, Scalar and complex measures from a pvm). The index set $J$ is a finite or countable subset of $N$ , listed in increasing order, and summands with $x_{j} = 0$ are omitted.

Step 2: a common dominating measure. Put $μ : = \sum_{j \in J} \frac{2^{- j}}{1 + μ_{j} (σ (T))} μ_{j} .$ Then $μ$ is a finite positive measure with $μ_{j} ≪ μ$ for every $j$ ; let $h_{j} : = \frac{d μ_{j}}{d μ}, F_{j} : = {z \in σ (T) : h_{j} (z) > 0}$ be the Radon–Nikodym derivative and its positivity set (A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density).

Step 3: the multiplicity function. Define the multiplicity function of the decomposition by $m (z) : = \sum_{j \in J} 1_{F_{j}} (z) \in {0, 1, 2, \dots} \cup {\infty} (z \in σ (T)) .$ It is a Borel function, and $z \mapsto m (z)$ is the fiber dimension function of the model below. The set $S (z) : = {j \in J : z \in F_{j}}$ of active coordinates has cardinality $m (z)$ , and for $r \geq 1$ the $r$ -th active coordinate is $k_{r} (z) : = \min {k \in J : # (S (z) \cap {j \in J : j \leq k}) = r}, z \in A_{r} : = {z : m (z) \geq r},$ a Borel function on the Borel set $A_{r}$ ; the sets $A_{1} \supseteq A_{2} \supseteq \dots$ are decreasing and $⋃_{r} A_{r}$ is $μ$ -conull.

Step 4: the standard measurable-field model. The standard model of the decomposition is the space $L^{2} (σ (T), μ, m) : = {f = (f_{r})_{r \geq 1} : f_{r} Borel, f_{r} = 0 μ -a.e. on {m < r}, \sum_{r \geq 1} \int ∣ f_{r} ∣^{2} d μ < \infty}$ of measurable fields with componentwise operations and inner product $⟨ f, g ⟩ = \sum_{r} \int f_{r} \overline{g_{r}} d μ$ (A measurable function between measurable spaces, The space $L^{p} (μ)$ as the quotient by null functions), where fields are identified when they agree $μ$ -almost everywhere in every component. Its fiber at $z$ is $span {e_{1}, \dots, e_{m (z)}} \subseteq ℓ^{2} (N)$ , so the fiber dimension is exactly $m (z)$ , and the model is the direct sum $⨁_{r \geq 1} L^{2} (μ ∣_{A_{r}})$ of the ordinary $L^{2}$ -spaces of the restrictions of $μ$ to the decreasing sets $A_{r}$ (Hilbert space).

Claim of the definition (identification with the operator model). The rank enumeration of active coordinates together with the Radon–Nikodym derivatives identifies the standard model unitarily with the orthogonal sum of the cyclic $L^{2}$ -summands: the map $W : L^{2} (σ (T), μ, m) ⟶ ⨁_{j \in J} L^{2} (σ (T), μ_{j}), (W f)_{j} (z) : = \frac{f_{r (z, j)} (z)}{\sqrt{h_{j} (z)}} for z \in F_{j}, (W f)_{j} : = 0 off F_{j},$ where $r (z, j)$ is the rank of $j$ in $S (z)$ , is a well-defined unitary operator intertwining the multiplications by every bounded Borel function and, in particular, intertwining $M_{z}$ with $M_{z}$ .

Well-definedness. (1) $μ$ is finite and nonzero and $μ_{j} ≪ μ$ for each $j$ , since the $j$ -th summand of the sum dominates $\frac{2^{- j}}{1 + μ_{j} (σ (T))} μ_{j}$ . (2) $h_{j}$ exists, is nonnegative $μ$ -almost everywhere and is unique up to $μ$ -null sets (Radon–Nikodym); $F_{j} = {h_{j} > 0}$ is Borel and $μ_{j} (σ (T) ∖ F_{j}) = \int_{{h_{j} = 0}} h_{j} d μ = 0$ . (3) $⋃_{j} F_{j}$ is $μ$ -conull: since $h_{j} = 0$ off $F_{j}$ one has $μ_{j} (σ (T) ∖ ⋃_{j} F_{j}) = \int_{σ (T) ∖ ⋃_{j} F_{j}} h_{j} d μ = 0$ for every $j$ , so the dominating sum vanishes there, and therefore $m \geq 1$ $μ$ -almost everywhere. (4) $m$ is Borel as a countable sum of indicators, and $k_{r}$ is Borel on $A_{r}$ because ${k_{r} = k} = F_{k} \cap {\sum_{ℓ < k, ℓ \in J} 1_{F_{ℓ}} = r - 1}$ is a Borel set. (5) $L^{2} (σ (T), μ, m)$ is a Hilbert space: it is the orthogonal direct sum $⨁_{r} L^{2} (μ ∣_{A_{r}})$ , whose components are complete by Riesz– Fischer and whose direct sum is complete because a Cauchy sequence in the sum has summable component norms, its components converge in the complete spaces $L^{2} (μ ∣_{A_{r}})$ , and the componentwise limit has finite norm and is the norm limit (dominated convergence for the counting measure) (Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ , Dominated convergence). (6) $W$ is well defined: $(W f)_{j}$ is Borel because on each Borel set ${r (z, j) = r}$ it equals $f_{r} / \sqrt{h_{j}}$ with $h_{j} > 0$ there, it vanishes off $F_{j}$ , and it is $μ_{j}$ -square-integrable because $\sum_{j \in J} \int ∣ (W f)_{j} ∣^{2} d μ_{j} = \sum_{j \in J} \int_{F_{j}} ∣ f_{r (z, j)} ∣^{2} d μ = \sum_{r \geq 1} \int_{A_{r}} ∣ f_{r} ∣^{2} d μ = ∥ f ∥^{2} < + \infty,$ using $d μ_{j} = h_{j} d μ$ on $F_{j}$ (Integrating against a Radon-Nikodym derivative recovers integration against the measure, The total variation of an absolutely continuous signed or complex measure has density the absolute value of the Radon-Nikodym derivative); the same display shows $W$ is isometric, and $W$ is onto with inverse $(g_{j})_{j} \mapsto f$ , $f_{r} (z) : = g_{k_{r} (z)} (z) \sqrt{h_{k_{r} (z)} (z)}$ on $A_{r}$ and $f_{r} : = 0$ otherwise, which is Borel by the same rank-measurability and is square-integrable by the same computation; multiplicativity is componentwise and gives $W M_{φ} = M_{φ} W$ for every bounded Borel $φ$ , hence $W M_{z} = M_{z} W$ . (7) The data $(μ, h_{j}, F_{j}, m)$ depend on the chosen decomposition; the measure class of $μ$ and the almost-everywhere class of $m$ are independent of the choice by the intertwiner lemma and the classification theorem proved immediately below on this page. The multiplicity function is the fiber dimension $z \mapsto \dim H_{z}$ of the model, equal to $m (z)$ ; at a non-atomic point of $σ (T)$ with respect to $E$ it is not the dimension of the eigenspace $\ker (T - z I)$ , which is the fiber of the atoms carried by $E ({z})$ .

LemmaStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Unitary intertwiners preserve direct-integral fiber dimension

Statement

Assume AC. Let $Λ \subseteq C$ be nonempty and compact, let $μ$ and $ν$ be nonzero finite positive regular Borel measures on $Λ$ , let $m, m^{'} : Λ \to {1, 2, \dots} \cup {\infty}$ be Borel functions, and consider the standard measurable-field models $L^{2} (μ, m)$ , $L^{2} (ν, m^{'})$ of Spectral multiplicity function in the separable case, recalled below. If $U : L^{2} (μ, m) ⟶ L^{2} (ν, m^{'})$ is a unitary operator with $U M_{z} = M_{z} U$ , where $M_{z}$ is multiplication by the coordinate function on $Λ$ , then:

$μ$ and $ν$ are mutually absolutely continuous;
$m = m^{'}$ $ν$ -almost everywhere, equivalently $μ$ -almost everywhere, where the two functions are compared after their common domain $Λ$ and the equivalence classes are pushed forward along the class equality of clause 1.

Facts & Assumptions

[A1]

The standard model is the orthogonal sum of the ordinary $L^{2}$ -spaces of the level sets: $L^{2} (μ, m) = ⨁_{r \geq 1} L^{2} (μ ∣_{A_{r}})$ with $A_{r} = {m \geq r}$ , and multiplication by a bounded Borel $h$ acts componentwise; likewise $L^{2} (ν, m^{'}) = ⨁_{s \geq 1} L^{2} (ν ∣_{B_{s}})$ , $B_{s} = {m^{'} \geq s}$ ; and $m \geq 1$ $μ$ -a.e., $m^{'} \geq 1$ $ν$ -a.e. (Spectral multiplicity function in the separable case).

[A2]

If a bounded linear functional on $C (Λ; C)$ is represented by two finite regular complex measures, then the two measures coincide, and $∣ \int g d σ ∣ \leq ∥ g ∥_{\infty} ∣ σ ∣ (Λ)$ ; the total variation of a measure with density $σ \in L^{1} (μ)$ is $∣ σ ∣ d μ$ . If finite positive measures $ν ≪ μ$ , the Radon--Nikodym theorem supplies an integrable density $h = d ν / d μ$ ; positivity forces $h \geq 0$ almost everywhere, and if also $μ ≪ ν$ then $h > 0$ almost everywhere (The bounded complex dual of C_0(X) is regular complex measures, Integrals against signed or complex measures are bounded by total variation, The total variation of an absolutely continuous signed or complex measure has density the absolute value of the Radon-Nikodym derivative, A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density).

[A3]

For a finite regular Borel measure $μ$ on the compact metric space $Λ$ and $σ \in L^{1} (μ)$ , the measure $E \mapsto \int_{E} σ d μ$ is regular: continuous densities are reducible to the regular case by domination, $C (Λ)$ is dense in $L^{1} (μ)$ , and regularity is preserved by total-variation limits; the identification $∣ σ d μ ∣ = ∣ σ ∣ d μ$ holds (C_c(X) is dense in L^p(mu) for a Radon measure, Regular complex Borel measures, The total variation |nu|(E) from countable measurable partitions).

[A4]

$L^{2}$ of a finite measure is complete and $L^{\infty}$ is dense in it; a bounded operator on it commuting with every multiplication $M_{h}$ is itself a multiplication: it is $M_{g}$ with $g : = T 1_{Λ}$ , because $T (h) = h T (1)$ for every $h \in L^{\infty}$ , and $∥ M_{g} ∥ = ∥ g ∥_{\infty}$ (Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ , C_c(X) is dense in L^p(mu) for a Radon measure, The space $L^{p} (μ)$ as the quotient by null functions).

[A5]

Adjoints: $(U M_{z})^{*} = M_{z}^{*} U^{*}$ and $M_{z}^{*} = M_{\overline{z}}$ ; a unitary satisfies $U^{*} U = I$ and $U U^{*} = I$ (Hilbert-adjoint identities, Hilbert space).

[A6]

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

Proof

technique · direct

Given: Nonempty compact $Λ \subseteq C$ , nonzero finite regular positive measures $μ, ν$ , Borel multiplicities $m, m^{'}$ , and a unitary $U : L^{2} (μ, m) \to L^{2} (ν, m^{'})$ with $U M_{z} = M_{z} U$ ; write $X : = ⨁_{r} L^{2} (μ ∣_{A_{r}})$ , $Y : = ⨁_{s} L^{2} (ν ∣_{B_{s}})$ .

1.1

$U$ preserves all multiplications: $U M_{z} = M_{z} U$ gives $U M_{z}^{*} = M_{z}^{*} U$ by taking adjoints, so $U$ commutes with every $*$ -polynomial in $z, \overline{z}$ ; $*$ -polynomials are uniformly dense in the continuous functions and both sides are bounded, so $U M_{h} = M_{h} U$ for continuous $h$ ; for fixed vectors $p, q \in X$ the measure $ρ_{p, q} (E) : = ⟨ M_{1_{E}} p, q ⟩ = \int_{E} σ_{p, q} d μ$ with $σ_{p, q} = \sum_{r} p_{r} \overline{q_{r}} \in L^{1} (μ)$ is a finite regular complex measure, and for continuous $h$ one has $\int h d ρ_{p, q} = ⟨ M_{h} p, q ⟩ = ⟨ U M_{h} p, U q ⟩ = ⟨ M_{h} U p, U q ⟩ = \int h d ρ_{U p, U q}$ , so the two regular measures coincide and hence $⟨ M_{h} p, q ⟩ = ⟨ M_{h} U p, U q ⟩$ for every bounded Borel $h$ and all $p, q$ ; therefore $U M_{h} = M_{h} U$ for every bounded Borel $h$ .

A1A2A3A5

2.1

Absolute continuity: the field $e : = (1_{A_{1}}, 0, 0, \dots) \in X$ has scalar measure $⟨ M_{h} e, e ⟩ = \int h 1_{A_{1}} d μ = \int h d μ$ because $A_{1}$ is $μ$ -conull, while the scalar measure of $U e$ in $Y$ is $\int h w d ν$ with $w : = \sum_{s} ∣ (U e)_{s} ∣^{2} \in L^{1} (ν)$ ; the identity $⟨ M_{h} U e, U e ⟩ = ⟨ U M_{h} e, U e ⟩ = ⟨ M_{h} e, e ⟩$ for all bounded Borel $h$ gives $\int_{E} w d ν = μ (E)$ for every Borel $E$ , hence $μ ≪ ν$ .

step 1.1A1

2.2

Localisation to constant multiplicity: let $B \subseteq Λ$ be Borel with $m \equiv k$ and $m^{'} \equiv k^{'}$ on $B$ for constants $k, k^{'} \in {1, 2, \dots} \cup {\infty}$ and $μ (B) > 0$ ; since $U M_{1_{B}} = M_{1_{B}} U$ , the unitary restricts to a unitary $U_{B}$ between the localised spaces $X_{B} : = {p \in X : p_{r} = 0 μ -a.e. off B} = ⨁_{r \leq k} L^{2} (μ ∣_{B})$ and $Y_{B} = ⨁_{s \leq k^{'}} L^{2} (ν ∣_{B})$ .

step 1.1A1

3.1

The same argument applied to the unitary $U^{*}$ , which also intertwines the multiplications, shows $ν ≪ μ$ ; hence $μ$ and $ν$ are mutually absolutely continuous.

step 1.1step 2.1A5

3.2

Transferring $Y_{B}$ to the measure $μ ∣_{B}$ by the Radon–Nikodym factor: because $ν ∣_{B} \sim μ ∣_{B}$ , A2 supplies a positive almost-everywhere density $h = d ν / d μ$ . The map $(g_{s})_{s} \mapsto (g_{s} \sqrt{h})_{s}$ is an isometry $⨁_{s \leq k^{'}} L^{2} (ν ∣_{B}) \to ⨁_{s \leq k^{'}} L^{2} (μ ∣_{B})$ by the defining integral identity, and it is onto because $h > 0$ almost everywhere and its inverse is multiplication by $h^{- 1 / 2}$ . It commutes with all multiplications, so composing it with $U_{B}$ gives a unitary $V : ⨁_{r \leq k} L^{2} (μ ∣_{B}) \to ⨁_{s \leq k^{'}} L^{2} (μ ∣_{B})$ commuting with all multiplications.

step 2.2A2A5

4.1

Constant-fibre rigidity: writing $P_{r}, Q_{s}$ for the coordinate projections and $V_{s r} : = Q_{s} V P_{r}$ , each $V_{s r}$ commutes with all multiplications, so $V_{s r} = M_{g_{s r}}$ for a bounded Borel function $g_{s r}$ ; hence $V$ is given fibrewise by the measurable matrix field $φ (z) : = (g_{s r} (z))$ , and $V^{*} V = I$ , $V V^{*} = I$ force $φ (z)^{*} φ (z) = I_{k}$ and $φ (z) φ (z)^{*} = I_{k^{'}}$ for $μ$ -almost every $z$ .

step 3.2A4

5.1

A measurable field of linear maps satisfying both identities exists only if $k = k^{'}$ : if $k < \infty$ then $φ (z)^{*} φ (z) = I_{k}$ shows the range of $φ (z)^{*}$ spans a $k$ -dimensional space, so $φ (z) φ (z)^{*}$ , whose rank is at most $k$ , cannot equal $I_{k^{'}}$ when $k^{'} = \infty$ ; symmetrically $k^{'} < \infty < k$ is impossible; and $k = k^{'} \in {1, 2, \dots} \cup {\infty}$ is consistent. Hence $k = k^{'}$ .

step 4.1

6.1

The Borel sets ${m = k} \cap {m^{'} = k^{'}}$ over $k, k^{'}$ cover a conull set, and by the rigidity steps above every one of them with positive measure satisfies $k = k^{'}$ ; therefore $m = m^{'}$ $μ$ -almost everywhere, and by the class equality also $ν$ -almost everywhere.

step 3.1step 5.1A6∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Unitary equivalence classified by measure class and multiplicity

Statement

Assume AC. Let $T$ be a bounded normal operator on a nonzero separable complex Hilbert space $H$ and let $T^{'}$ be a bounded normal operator on a nonzero separable complex Hilbert space $H^{'}$ , with multiplicity data $(μ, m)$ and $(ν, m^{'})$ obtained from countable cyclic decompositions as in Spectral multiplicity function in the separable case. Then $T$ and $T^{'}$ are unitarily equivalent — there is a unitary $V : H \to H^{'}$ with $V T V^{*} = T^{'}$ — if and only if

$σ (T) = σ (T^{'})$ as subsets of $C$ , and on this common compact set the scalar measures are in the same class, $[μ] = [ν]$ (mutual absolute continuity);
the multiplicity functions agree almost everywhere for that class, $m = m^{'}$ $μ$ -almost everywhere, equivalently $ν$ -almost everywhere.

No change of spectral coordinate is allowed: the identification of the two scalar measure classes is an equality of measures on the common set $σ (T) = σ (T^{'})$ , not an identification after a homeomorphism of spectra. The zero Hilbert space is a separate trivial class: it carries the zero operator, whose spectrum is empty, and no regular projection valued measure on the empty set is used anywhere above.

Facts & Assumptions

[A1]

The multiplicity data $(μ, m)$ live on $σ (T)$ , with $μ$ a nonzero finite positive regular Borel measure and $m \geq 1$ $μ$ -almost everywhere; the standard model $L^{2} (μ, m)$ is unitarily equivalent to $⨁_{j} L^{2} (σ (T), μ_{j})$ for the cyclic decomposition with scalar measures $μ_{j} = E_{x_{j}}$ and the identification intertwines the multiplications on both sides (Spectral multiplicity function in the separable case).

[A2]

$T$ is unitarily equivalent to multiplication by the coordinate on the orthogonal sum of the cyclic summands, hence, via $[A 1]$ , to $M_{z}$ on $L^{2} (μ, m)$ (Multiplication operator form of the bounded normal spectral theorem, Cyclic spectral representation, Spectral multiplicity function in the separable case).

[A3]

If $λ$ is any finite positive measure equivalent to $μ$ that dominates all $μ_{j}$ , the same construction applies with $λ$ in place of $μ$ and produces a model canonically unitarily equivalent to $L^{2} (μ, m)$ : the Radon–Nikodym ratio $d μ / d λ$ is positive $λ$ -almost everywhere and multiplication by its square root is a unitary intertwining all multiplications (A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density, Integrating against a Radon-Nikodym derivative recovers integration against the measure, Spectral multiplicity function in the separable case).

[A4]

A unitary intertwiner between two standard models preserves the class of the dominating measure and the multiplicity function almost everywhere (Unitary intertwiners preserve direct-integral fiber dimension).

[A5]

Unitary equivalence preserves the spectrum: $λ \notin σ (T)$ exactly when $λ I - T$ is bijective with bounded inverse, and conjugating by a unitary carries this property to $T^{'}$ ; more generally conjugating by a unitary is an isometric isomorphism of $B (H)$ onto $B (H^{'})$ (Spectrum and resolvent of a bounded operator, Self-adjoint, positive, unitary and normal operators, Hilbert space).

[A6]

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

Proof

technique · direct

Given: Bounded normal operators $T, T^{'}$ on nonzero separable complex Hilbert spaces with multiplicity data $(μ, m)$ on $σ (T)$ and $(ν, m^{'})$ on $σ (T^{'})$ .

1.1

Unitary invariance of the spectrum: if $V : H \to H^{'}$ is unitary with $V T V^{*} = T^{'}$ , then for every $λ$ the operator $λ I - T$ is bijective with bounded inverse exactly when $λ I - T^{'} = V (λ I - T) V^{*}$ is, so $ρ (T) = ρ (T^{'})$ and $σ (T) = σ (T^{'})$ ; in particular the two spectra are compared on a common compact subset of $C$ .

1.2

Converse: suppose $σ (T) = σ (T^{'}) = : Λ$ , $[μ] = [ν]$ and $m = m^{'}$ $μ$ -almost everywhere, and put $λ : = μ$ . Then $ν \sim λ$ and, by clause (2), $m = m^{'}$ $λ$ -almost everywhere, so the standard models $L^{2} (λ, m)$ and $L^{2} (λ, m^{'})$ are the same space with the same multiplication operators: the level sets $A_{r} = {m \geq r}$ and $A_{r}^{'} = {m^{'} \geq r}$ differ by $λ$ -null sets, so each $L^{2} (λ ∣_{A_{r}})$ equals $L^{2} (λ ∣_{A_{r}^{'}})$ as a subspace of $L^{2} (λ)$ .

A1A3

2.1

Direct implication: suppose $V T V^{*} = T^{'}$ . Since $m \geq 1$ $μ$ -almost everywhere and $m^{'} \geq 1$ $ν$ -almost everywhere, replace their values by $1$ on the respective null sets where they are $0$ , obtaining Borel representatives $\tilde{m}, {\tilde{m}}^{'} : Λ \to {1, 2, \dots} \cup {\infty}$ . Their level sets differ from those of $m, m^{'}$ only by null sets, so the resulting $L^{2}$ models and coordinate multiplications are unchanged. By the definitional identification, $T$ is unitarily equivalent to $M_{z}$ on $L^{2} (μ, \tilde{m})$ and $T^{'}$ to $M_{z}$ on $L^{2} (ν, {\tilde{m}}^{'})$ ; composing these equivalences with $V$ gives a unitary $U : L^{2} (μ, \tilde{m}) \to L^{2} (ν, {\tilde{m}}^{'})$ with $U M_{z} = M_{z} U$ .

step 1.1A1A2

2.2

The model of $T^{'}$ with dominating measure $ν$ is canonically unitarily equivalent to the model with dominating measure $λ$ ; hence $T$ and $T^{'}$ are both unitarily equivalent to $M_{z}$ on $L^{2} (λ, m)$ , and composing one equivalence with the inverse of the other gives a unitary $H \to H^{'}$ conjugating $T$ to $T^{'}$ .

step 1.2A1A2A3

3.1

Applying the intertwiner lemma to the everywhere-positive representatives in step 2.1 gives $[μ] = [ν]$ and $\tilde{m} = {\tilde{m}}^{'}$ almost everywhere for that class. Since each representative differs from the original multiplicity only on a null set, $m = m^{'}$ almost everywhere as well; together with step 1.1 this proves the "only if" implication.

step 1.1step 2.1A4

4.1

Therefore $T$ and $T^{'}$ are unitarily equivalent exactly when the spectra agree and, on the common spectrum, the scalar measure classes agree and the multiplicity functions agree almost everywhere; the zero space is the excluded trivial case carrying the zero operator and no PVM.

step 3.1step 2.2A6∎

TheoremStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-22Open item page →

Stone resolvent formula for spectral projections

Statement

Assume AC. Let $T$ be a bounded self-adjoint operator on a nonzero complex Hilbert space $H$ with spectral projection valued measure $E$ on the compact set $σ (T) \subseteq R$ , and let $a < b$ be real. For every Borel set $A \subseteq R$ , write $E (A) : = E (A \cap σ (T))$ . Then, with the resolvents $(T - (t \pm i ε))^{- 1}$ defined for $ε > 0$ by Spectrum and resolvent of a bounded operator and the integral of a continuous $B (H)$ -valued function understood in the Bochner sense (Bochner-integrable function),

$\frac{1}{2 π i} \int_{a}^{b} [(T - (t + i ε))^{- 1} - (T - (t - i ε))^{- 1}] d t ⟶ E ((a, b)) + \frac{E ({a}) + E ({b})}{2}$

in the strong operator topology as $ε ↓ 0$ . In particular, if $a, b \notin σ (T)$ then the limit is the spectral projection $E ((a, b))$ , and the half-masses at $a$ and $b$ appear exactly when these points are atoms of the spectrum.

Facts & Assumptions

[A1]

For $z \notin R$ the function $g_{z} (λ) : = (λ - z)^{- 1}$ is bounded and Borel on $σ (T)$ , with $∥ g_{z} ∥_{\infty} \leq ∣ Im z ∣^{- 1}$ , and its Borel calculus value satisfies $(T - z I)^{- 1} = Φ_{E} (g_{z})$ : indeed $(T - z I) Φ_{E} (g_{z}) = Φ_{E} ((λ - z) g_{z}) = Φ_{E} (1) = I$ and $Φ_{E} (g_{z}) (T - z I) = I$ by linearity and multiplicativity of the Borel calculus (Borel functional calculus for bounded normal operators, Borel functional calculus for a bounded normal operator).

[A2]

$σ (T) \subseteq R$ for self-adjoint $T$ , so the functions $g_{z}$ are defined on the spectrum (Spectrum of a self adjoint operator is real).

[A3]

A continuous function on the compact interval $[a, b]$ with values in the Banach space $B (H)$ is Bochner integrable, and a bounded linear map $Φ$ satisfies $Φ (\int_{E} f d μ) = \int_{E} Φ f d μ$ (Bochner integrability criterion, Bounded linear maps commute with Bochner integration).

[A4]

The Borel calculus is a unital star-homomorphism: $Φ_{E} (1_{B}) = E (B)$ , and uniformly bounded pointwise $E$ -almost everywhere convergence implies strong convergence (Pvm integral is a star homomorphism, Projection valued measure).

[A5]

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

Proof

technique · direct

Given: A bounded self-adjoint operator $T$ with spectral PVM $E$ on $σ (T) \subseteq R$ , real numbers $a < b$ , and $ε > 0$ .

1.1

Resolvent identity: since $σ (T) \subseteq R$ , the function $g_{z}$ is bounded Borel for $z = t \pm i ε$ and (A1) identifies $(T - (t \pm i ε))^{- 1} = Φ_{E} (g_{t \pm i ε})$ , so the integrand $t \mapsto Φ_{E} (g_{t + i ε}) - Φ_{E} (g_{t - i ε})$ is a continuous $B (H)$ -valued function on the compact interval and the Bochner integral converges.

A1A2A3

1.2

Scalar kernel: for real $λ$ and $t$ one computes $(λ - (t + i ε))^{- 1} - (λ - (t - i ε))^{- 1} = \frac{2 i ε}{(λ - t)^{2} + ε^{2}}$ , hence the bounded Borel function $K_{ε} (λ) : = \frac{1}{2 π i} \int_{a}^{b} [(λ - (t + i ε))^{- 1} - (λ - (t - i ε))^{- 1}] d t$ equals $\frac{1}{π} \int_{a}^{b} \frac{ε d t}{(λ - t)^{2} + ε^{2}} = \frac{1}{π} [\arctan \frac{b - λ}{ε} - \arctan \frac{a - λ}{ε}]$ , with $∣ K_{ε} ∣ \leq 1$ and, for every real $λ$ , $K_{ε} (λ) \to 1_{(a, b)} (λ) + \frac{1}{2} 1_{{a, b}} (λ)$ as $ε ↓ 0$ .

A2algebra

2.1

The operator integral is the calculus value of the kernel: by linearity of $Φ_{E}$ and commutation of the bounded linear map $Φ_{E}$ with Bochner integration, $\frac{1}{2 π i} \int_{a}^{b} [(T - (t + i ε))^{- 1} - (T - (t - i ε))^{- 1}] d t = \frac{1}{2 π i} \int_{a}^{b} [Φ_{E} (g_{t + i ε}) - Φ_{E} (g_{t - i ε})] d t = Φ_{E} (\frac{1}{2 π i} \int_{a}^{b} [g_{t + i ε} - g_{t - i ε}] d t) = Φ_{E} (K_{ε})$ .

step 1.1step 1.2A3

3.1

Strong limit: $∣ K_{ε} ∣ \leq 1$ and $K_{ε} \to 1_{(a, b)} + \frac{1}{2} 1_{{a, b}}$ pointwise, so the strong-convergence clause of the calculus gives $Φ_{E} (K_{ε}) \to Φ_{E} (1_{(a, b)} + \frac{1}{2} 1_{{a, b}}) = E ((a, b)) + \frac{1}{2} (E ({a}) + E ({b}))$ in the strong operator topology.

step 2.1A4

4.1

Therefore the resolvent expression converges strongly to $E ((a, b)) + \frac{1}{2} (E ({a}) + E ({b}))$ as $ε ↓ 0$ ; if $a, b \notin σ (T)$ the endpoint atoms vanish and the limit is the open-interval spectral projection $E ((a, b))$ .

step 3.1A5∎

5 · Examples, counterexamples and false statements

None yet.