ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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Functional calculus for a diagonal operator

Example

Assume AC. Let $(λ_{n})_{n \in N}$ be a bounded complex sequence and let $T$ be the diagonal operator $T e_{n} = λ_{n} e_{n}$ on $ℓ^{2} (N; C)$ , where $(e_{n})$ is the standard orthonormal basis. Then $σ (T) = \overline{{λ_{n} : n \in N}}$ and $f (T) e_{n} = f (λ_{n}) e_{n}$ for every continuous $f$ on $σ (T)$ .

Facts & Assumptions

[A1]

$ℓ^{2} (N; C)$ is the space of square-summable families with $⟨ x, y ⟩ = \sum_{n} x_{n} \overline{y_{n}}$ ; the vectors $e_{n}$ form an orthonormal family with $⟨ x, e_{n} ⟩ = x_{n}$ , and the space is complete, hence a Hilbert space (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ , A Hilbert space with a given orthonormal basis is $ℓ^{2}$ of the index set, Orthonormal families, complete orthonormal systems and Hilbert bases, Hilbert space).

[A2]

$z \in ρ (T)$ exactly when $z I - T$ is bijective with bounded inverse; a bounded operator that is not bounded below has no bounded inverse (Spectrum and resolvent of a bounded operator, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A3]

For normal $T$ with a nonzero eigenvector $x$ satisfying $T x = λ x$ , one has $λ \in σ (T)$ and, for every $f \in C (σ (T))$ , $f (T) x = f (λ) x$ (Continuous functional calculus properties, Continuous functional calculus for bounded normal operators).

[A4]

AC is the hypothesis of the calculus supplier (The Axiom of Choice).

Verification

technique · direct

Given: A bounded complex sequence $(λ_{n})$ with $M : = \sup_{n} ∣ λ_{n} ∣ < \infty$ and the diagonal operator $T x = (λ_{n} x_{n})$ on $ℓ^{2} (N; C)$ .

1.1A1A2

The formula $T x = (λ_{n} x_{n})$ defines a linear bounded operator with $∥ T x ∥_{2}^{2} = \sum_{n} ∣ λ_{n} x_{n} ∣^{2} \leq M^{2} ∥ x ∥_{2}^{2}$ , so $∥ T ∥ \leq M$ and $T e_{n} = λ_{n} e_{n}$ ; moreover $∥ T ∥ \geq \sup_{n} ∣ λ_{n} ∣ = M$ because $∥ T e_{n} ∥ = ∣ λ_{n} ∣$ , so $∥ T ∥ = M$ .

1.2A1algebra

$T$ is normal: $T^{*} y = (\overline{λ_{n}} y_{n})$ , since $⟨ T x, y ⟩ = \sum_{n} λ_{n} x_{n} \overline{y_{n}} = \sum_{n} x_{n} \overline{λ_{n} y_{n}}$ , and therefore $T^{*} T = T T^{*}$ is the diagonal operator with entries $∣ λ_{n} ∣^{2}$ .

2.1step 1.1step 1.2A2

$σ (T) = \overline{{λ_{n}}}$ : if $z \notin \overline{{λ_{n}}}$ then $δ : = \inf_{n} ∣ z - λ_{n} ∣ > 0$ , the diagonal operator $S$ with entries $1 / (z - λ_{n})$ is bounded with $∥ S ∥ \leq 1 / δ$ , and $S (z I - T) = (z I - T) S = I$ , so $z \in ρ (T)$ ; if $z \in \overline{{λ_{n}}}$ choose $n_{k}$ with $λ_{n_{k}} \to z$ , so $∥ (T - z I) e_{n_{k}} ∥ = ∣ λ_{n_{k}} - z ∣ \to 0$ , whence $T - z I$ is not bounded below and lies in $σ (T)$ .

3.1step 2.1A3

For $f \in C (σ (T))$ and each $n$ , the basis vector $e_{n}$ is an eigenvector of the normal operator $T$ at $λ_{n} \in σ (T)$ , so $f (T) e_{n} = f (λ_{n}) e_{n}$ .

4.1step 2.1step 3.1A4∎

Hence $σ (T) = \overline{{λ_{n}}}$ and the calculus acts diagonally on the standard basis, as asserted.

Depends on

Used by

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Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §5.3, printed pp.235–245 (standard reference, not scraped)
Dana P. Williams, Lecture Notes on the Spectral Theorem, §4, pp.10–15 (standard reference, not scraped)