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Diagonal spectrum

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space (Hilbert space), let $(e_{n})_{n \in N}$ be a complete orthonormal family of $H$ (Orthonormal families, complete orthonormal systems and Hilbert bases), let $(λ_{n})_{n \in N} \subseteq R$ , and let $A$ be the self-adjoint diagonal operator of Diagonal adjoint and graph-core lemma:

$D (A) = {x \in H : \sum_{n} λ_{n}^{2} ∣ ⟨ x, e_{n} ⟩ ∣^{2} < \infty}, A x = \sum_{n} λ_{n} ⟨ x, e_{n} ⟩ e_{n} .$

Then, with $ρ (A)$ , $σ (A)$ and $R (λ, A)$ as in Resolvent and spectrum of a closed operator on a Banach space:

(a) $A e_{n} = λ_{n} e_{n}$ for every $n$ ; every $λ_{n}$ is an eigenvalue of $A$ , hence belongs to $σ (A)$ ;

(b) $σ (A) = \overline{{λ_{n} : n \in N}}$ , the closure in $C$ ; equivalently $λ \in ρ (A)$ exactly when $d : = \inf_{n} ∣ λ - λ_{n} ∣ > 0$ , and then

$R (λ, A) x = \sum_{n} (λ - λ_{n})^{- 1} ⟨ x, e_{n} ⟩ e_{n}, ∥ R (λ, A) ∥ = d^{- 1} = (\inf_{n} ∣ λ - λ_{n} ∣)^{- 1};$

(c) if $λ \in σ (A) ∖ {λ_{n} : n \in N}$ , then $λ$ is an approximate eigenvalue: for some subsequence $(n_{k})$ , $∥ (λ I - A) e_{n_{k}} ∥ \to 0$ with $∥ e_{n_{k}} ∥ = 1$ . Moreover, $\ker (λ I - A) = {0}$ , so $λ$ is not an eigenvalue of $A$ ; no further identification of spectral types is made.

Facts & Assumptions

Given: the Hilbert space $H$ , the complete orthonormal family $(e_{n})$ , real numbers $λ_{n}$ , and the self-adjoint diagonal operator $A$ of Diagonal adjoint and graph-core lemma.

[F1]

$D (A) = {x : \sum_{n} λ_{n}^{2} ∣ ⟨ x, e_{n} ⟩ ∣^{2} < \infty}$ , $A x = \sum_{n} λ_{n} ⟨ x, e_{n} ⟩ e_{n}$ with $∥ A x ∥^{2} = \sum_{n} λ_{n}^{2} ∣ ⟨ x, e_{n} ⟩ ∣^{2}$ , $A e_{n} = λ_{n} e_{n}$ , and $A$ is self-adjoint (Diagonal adjoint and graph-core lemma(a),(b),(e)); for every $x \in H$ the expansion $x = \sum_{n} ⟨ x, e_{n} ⟩ e_{n}$ converges and $\sum_{n} ∣ ⟨ x, e_{n} ⟩ ∣^{2} = ∥ x ∥^{2}$ (Parseval equivalences for an orthonormal family).

[F2]

$λ \in ρ (A)$ exactly when $λ I - A : D (A) \to H$ is bijective with bounded inverse, and then that inverse is $R (λ, A)$ ; $σ (A) = C ∖ ρ (A)$ (Resolvent and spectrum of a closed operator on a Banach space).

[F3]

$∣ \sum_{n} a_{n} \overline{b_{n}} ∣ \leq (\sum_{n} ∣ a_{n} ∣^{2})^{1 / 2} (\sum_{n} ∣ b_{n} ∣^{2})^{1 / 2}$ and $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors).

Proof

1.1givenF1F2

[Eigenvalues are spectral] By [F1], $A e_{n} = λ_{n} e_{n}$ with $e_{n} \neq 0$ , so $λ_{n}$ is an eigenvalue of $A$ . If $λ_{n} \in ρ (A)$ , then $λ_{n} I - A$ is injective by [F2], contradicting $(λ_{n} I - A) e_{n} = 0$ . Hence ${λ_{n} : n \in N} \subseteq σ (A)$ .

1.2givenF1F2F3algebra

[The resolvent outside the closure] Let $λ \in C$ with $d = \inf_{n} ∣ λ - λ_{n} ∣ > 0$ , and define $R x : = \sum_{n} (λ - λ_{n})^{- 1} ⟨ x, e_{n} ⟩ e_{n}$ for $x \in H$ . The series converges and $∥ R x ∥^{2} = \sum_{n} ∣ λ - λ_{n} ∣^{- 2} ∣ ⟨ x, e_{n} ⟩ ∣^{2} \leq d^{- 2} ∥ x ∥^{2}$ by [F1], so $R \in B (H)$ with $∥ R ∥ \leq d^{- 1}$ ; testing on $e_{n}$ gives $∥ R ∥ \geq \sup_{n} ∣ λ - λ_{n} ∣^{- 1} = d^{- 1}$ , so $∥ R ∥ = d^{- 1}$ . For $x \in H$ , the coefficients of $R x$ satisfy $\sum_{n} λ_{n}^{2} ∣ (λ - λ_{n})^{- 1} ⟨ x, e_{n} ⟩ ∣^{2} \leq (1 + ∣ λ ∣ d^{- 1})^{2} \sum_{n} ∣ ⟨ x, e_{n} ⟩ ∣^{2} < \infty$ because $∣ λ_{n} ∣ \leq ∣ λ - λ_{n} ∣ + ∣ λ ∣ \leq ∣ λ - λ_{n} ∣ (1 + ∣ λ ∣ d^{- 1})$ for every $n$ ; hence $R x \in D (A)$ by [F1]. Computing coefficients with [F1] gives $(λ I - A) R x = \sum_{n} ⟨ x, e_{n} ⟩ e_{n} = x$ , and symmetrically $R (λ I - A) x = \sum_{n} (λ - λ_{n})^{- 1} (λ - λ_{n}) ⟨ x, e_{n} ⟩ e_{n} = x$ for $x \in D (A)$ . So $λ I - A$ is bijective with bounded inverse $R$ , that is, $λ \in ρ (A)$ and $R (λ, A) = R$ by [F2].

2.1step 1.1F1F2algebra

[Closure points are spectral] Let $λ \in \overline{{λ_{n}}}$ . If $λ = λ_{m}$ for some $m$ , then $λ \in σ (A)$ by step 1.1. Otherwise pick a subsequence with $λ_{n_{k}} \to λ$ and $λ_{n_{k}} \neq λ$ for all $k$ , and suppose for contradiction that $λ \in ρ (A)$ . Then $1 = ∥ e_{n_{k}} ∥ = ∥ R (λ, A) (λ I - A) e_{n_{k}} ∥ = ∥ R (λ, A) (λ - λ_{n_{k}}) e_{n_{k}} ∥ \leq ∥ R (λ, A) ∥ ∣ λ - λ_{n_{k}} ∣ \to 0$ , which is impossible. Hence $λ \in σ (A)$ .

3.1step 1.2step 2.1step 1.1F1∎

[Conclusion] Steps 1.1 and 2.1 give $\overline{{λ_{n}}} \subseteq σ (A)$ , and step 1.2 gives $C ∖ \overline{{λ_{n}}} \subseteq ρ (A)$ ; taking complements yields $σ (A) \subseteq \overline{{λ_{n}}}$ . Hence $σ (A) = \overline{{λ_{n}}}$ , and the equivalence " $λ \in ρ (A)$ exactly when $\inf_{n} ∣ λ - λ_{n} ∣ > 0$ " together with the resolvent formula and its norm are exactly the contents of step 1.2. For (c), if $λ \in σ (A) ∖ {λ_{n}}$ , the spectrum equality gives $λ \in \overline{{λ_{n}}}$ , so a subsequence with $λ_{n_{k}} \to λ$ satisfies $∥ (λ I - A) e_{n_{k}} ∥ = ∣ λ - λ_{n_{k}} ∣ \to 0$ and $∥ e_{n_{k}} ∥ = 1$ . If $x \in D (A)$ and $A x = λ x$ , then comparing coefficients in the complete orthonormal family gives $(λ_{n} - λ) ⟨ x, e_{n} ⟩ = 0$ for every $n$ ; because $λ \neq λ_{n}$ , every coefficient vanishes, and Parseval gives $x = 0$ . Thus $\ker (λ I - A) = {0}$ and $λ$ is not an eigenvalue. No further spectral-type classification is asserted.

Diagonal spectrum

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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