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

1.1

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∎

Depends on

Used by

Spectral theorem for unbounded self-adjoint operators (PVM form) Theorem

Dependency tree · two levels

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

Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, Theorem 5.74 and §5.7, printed pp.288–296 (standard reference, not scraped)
Andreas Kriegl, Funktionalanalysis, §8.61, printed pp.196–198 (standard reference, not scraped)