TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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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 $\int z d E^{'} (z) = T$ .

1.1A1A2

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 $q (λ) = r > 0$ .

1.2A1A5

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 $f$ .

2.1step 1.2A2A3A5

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 $E^{'} (Λ ∖ σ (T)) = 0$ .

3.1step 2.1A2A4

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 $B \subseteq σ (T)$ .

4.1step 1.1step 3.1A6∎

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 $T$ .

Support and uniqueness of the spectral measure

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources