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

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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 $x_{j} \in H_{j}$ .

1.1A1A3

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 $H_{x_{j}} = H_{j}$ .

2.1step 1.1A2

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 $U_{j} M_{z} = T U_{j}$ .

3.1step 2.1A1A4

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 $H$ .

3.2step 2.1A2A4

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 $U (⨁_{j} M_{z}) U^{- 1} = T$ .

3.3step 2.1A1

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.

4.1step 3.1step 3.2step 3.3A5∎

$T$ 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.

Depends on

Used by

Dependency tree · two levels

54 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, §5.7 and Theorem 5.84, printed pp.293–296 (standard reference, not scraped)
Andreas Kriegl, Funktionalanalysis, §8.61, printed pp.196–198 (standard reference, not scraped)