TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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Spectrum as character values

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $A$ be a nonzero commutative unital complex Banach algebra (Unital Banach algebra), let $a \in A$ , and let $σ_{A} (a)$ be the spectrum of $a$ in $A$ (Spectrum and resolvent set in a Banach algebra), with spectral radius $r (a)$ (Spectral radius). Then

$σ_{A} (a) = {χ (a) : χ \in Δ (A)},$

and consequently every character $χ \in Δ (A)$ satisfies $∣ χ (a) ∣ \leq r (a) \leq ∥ a ∥$ for every $a \in A$ .

The spectrum is taken in the ambient algebra $A$ ; the statement is not a claim about the spectrum computed in a subalgebra, and no injectivity or surjectivity of the Gelfand transform is asserted.

Facts & Assumptions

Given: An assumed Axiom of Choice, a nonzero commutative unital complex Banach algebra $A$ , and an element $a \in A$ .

[L1]

Every proper ideal of $A$ is contained in a maximal ideal, and the maximal ideals of $A$ are exactly the kernels of the characters of $A$ (Maximal ideals and characters of a commutative Banach algebra, The Axiom of Choice).

[L2]

$z \in σ_{A} (a)$ exactly when $z 1 - a$ is not invertible in $A$ ; an element of a proper ideal is never invertible (Spectrum and resolvent set in a Banach algebra).

[L3]

Every character of $A$ is unital, so $χ (1) = 1$ , and satisfies $∣ χ (x) ∣ \leq ∥ x ∥$ for all $x$ (Characters on a unital Banach algebra are continuous).

[L4]

The spectral radius is $r (a) = \max {∣ z ∣ : z \in σ_{A} (a)}$ and satisfies $0 \leq r (a) \leq ∥ a ∥$ (Spectral radius, The Axiom of Choice).

Proof

technique · direct

1.1

For $χ \in Δ (A)$ the element $a - χ (a) 1$ lies in $\ker χ$ , because $χ (a - χ (a) 1) = χ (a) - χ (a) χ (1) = 0$ by linearity and [L3]; as $\ker χ$ is a proper ideal, $a - χ (a) 1$ is not invertible, so $χ (a) \in σ_{A} (a)$ by [L2].

L1L2L3

1.2

If $λ \in σ_{A} (a)$ then the ideal $I : = (a - λ 1) A$ generated by $a - λ 1$ is proper: were $I = A$ , there would be $b \in A$ with $b (a - λ 1) = 1$ , making $a - λ 1$ invertible, contrary to [L2].

2.1

By [L1] the proper ideal $I$ of [step 1.2] is contained in a maximal ideal $M$ , and $M = \ker χ$ for some character $χ$ ; since $a - λ 1 \in I \subseteq M$ we get $χ (a - λ 1) = 0$ , that is, $χ (a) = λ$ by linearity and [L3].

step 1.2L1L3

3.1

Steps [step 1.1] and [step 2.1] give $σ_{A} (a) = {χ (a) : χ \in Δ (A)}$ . For each character, $χ (a) \in σ_{A} (a)$ , so $∣ χ (a) ∣ \leq \max {∣ z ∣ : z \in σ_{A} (a)} = r (a)$ by [L4], and $r (a) \leq ∥ a ∥$ by [L4]; the displayed consequence follows.

step 1.1step 2.1L4∎

Remarks

AC is spent once, in the maximal-ideal extension. Both inclusions are otherwise algebraic: the forward inclusion only tests the character on a coset representative, and the reverse inclusion only extends an ideal.
Consequences for the Gelfand transform. Since $\hat{a} (χ) = χ (a)$ , the equality of the statement says that the range of $\hat{a}$ is exactly $σ_{A} (a)$ , which is how the norm formula $∥ \hat{a} ∥_{\infty} = r (a)$ is proved in Gelfand transform is a contractive unital homomorphism.
No isometry claim. The inequality chain $∣ χ (a) ∣ \leq r (a) \leq ∥ a ∥$ is all that the spectrum identity yields; for a general commutative Banach algebra the first inequality can be strict, as cex-gelfand-transform-of-a-banach-algebra-need-not-be-isometric records.
Reading order. The example items named by ID above are homed on later pages of the plan, so they are named rather than hyperlinked: a body link to later material must be declared as a forward reference, and Step-5b closure removes every such declaration. Rehoming those items to an earlier page (an owner-only reading-order change) would make the citations backward and restore the links.

Depends on

Used by

Gelfand transform Definition
Character space of generated normal algebra is operator spectrum Lemma
Gelfand transform is a contractive unital homomorphism Theorem
Spectral mapping for continuous normal functional calculus Theorem

Dependency tree · two levels

17 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

Vahid Shirbisheh, Lectures on C-star Algebras, v2 — Corollary 3.1.8 and §3.1, printed pp. 54–61 (standard reference, not scraped)
Theo Bühler and Dietmar A. Salamon, Functional Analysis — Theorem 5.63, printed pp. 262–266 (standard reference, not scraped)