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Maximal ideal space is compact Hausdorff

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $A$ be a nonzero commutative unital complex Banach algebra (Unital Banach algebra) and let $Δ (A)$ be its character space (Character and maximal ideal space). Then:

each character of $A$ is a bounded linear functional of norm one, so that $Δ (A)$ is a subset of the closed dual unit ball $B_{A^{*}} = {f \in A^{*} : ∥ f ∥ \leq 1}$ ;
the pointwise-evaluation topology of $Δ (A)$ is the subspace topology induced by the weak-star topology $σ (A^{*}, A)$ ;
$Δ (A)$ is a weak-star closed subset of $B_{A^{*}}$ ;
$Δ (A)$ is nonempty, compact and Hausdorff, hence a compact Hausdorff space in the weak-star topology, and it is a nonempty compact Hausdorff subset of the dual unit ball in that topology.

The Axiom of Choice enters exactly through the ultrafilter lemma (for Banach–Alaoglu) and through the existence of maximal ideals (for nonemptiness); no further selection is made.

Facts & Assumptions

Given: An assumed Axiom of Choice, a nonzero commutative unital complex Banach algebra $A$ , its dual $A^{*}$ with weak-star topology $σ (A^{*}, A)$ , and $Δ (A)$ with the pointwise-evaluation topology.

[L1]

Characters of $A$ are unital, bounded and satisfy $∣ χ (a) ∣ \leq ∥ a ∥$ for all $a \in A$ ; in particular $χ \in A^{*}$ with $∥ χ ∥ = 1$ and the evaluation maps $χ \mapsto χ (a)$ are the linear functionals of $A^{*}$ evaluated at $a$ (Characters on a unital Banach algebra are continuous).

[L2]

$χ \in Δ (A)$ means that $χ : A \to C$ is nonzero, complex-linear and multiplicative; the pointwise-evaluation topology is the coarsest topology making all evaluations $χ \mapsto χ (a)$ continuous, and every $χ$ satisfies $χ (1) = 1$ by [L1] (Character and maximal ideal space).

[L3]

Under the Axiom of Choice the ultrafilter lemma holds, and under the ultrafilter lemma the closed dual unit ball $B_{X^{*}}$ of a real or complex normed space $X$ is weak-star compact (The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter, Banach–Alaoglu, The Axiom of Choice).

[L4]

Since $A$ is nonzero, the zero ideal is proper, so by the maximal ideal theorem applied under the Axiom of Choice there is a maximal ideal of $A$ , and every maximal ideal is the kernel of some character (Maximal ideals and characters of a commutative Banach algebra).

Proof

technique · direct

1.1

By [L1] every character has $∥ χ ∥ = 1$ , so $Δ (A) \subseteq B_{A^{*}}$ , and the weak-star topology on $A^{*}$ is by definition the topology of pointwise convergence on $A$ , so on $Δ (A)$ it induces exactly the pointwise-evaluation topology of [L2].

L1L2

1.2

Write $C : = B_{A^{*}} \cap {f : f (1) = 1} \cap ⋂_{a, b \in A} {f : f (a b) = f (a) f (b)}$ , a subset of $B_{A^{*}}$ . Each set displayed is weak-star closed: ${f : f (1) = 1}$ and the sets ${f : f (a b) = f (a) f (b)}$ are preimages of the closed sets ${1}$ and ${0}$ under the continuous functions $f \mapsto f (1)$ and $f \mapsto f (a b) - f (a) f (b)$ . Hence $C$ is weak-star closed.

L1L2algebra

1.3

The ultrafilter lemma follows from the Axiom of Choice by [L3], so $B_{A^{*}}$ is weak-star compact by Banach–Alaoglu.

1.4

$Δ (A) \neq \emptyset$ : by [L4] there is a maximal ideal, and it is the kernel of a character.

1.5

$Δ (A)$ is Hausdorff in the pointwise-evaluation topology: if $χ \neq ψ$ then there is $a \in A$ with $χ (a) \neq ψ (a)$ , and with $ε : = ∣ χ (a) - ψ (a) ∣ / 2 > 0$ the basic evaluation-open sets ${φ : ∣ φ (a) - χ (a) ∣ < ε}$ and ${φ : ∣ φ (a) - ψ (a) ∣ < ε}$ are disjoint.

L2algebra

2.1

$C = Δ (A)$ : a bounded linear functional $f$ with $f (1) = 1$ and $f (a b) = f (a) f (b)$ is a nonzero linear multiplicative map, that is, a character, and conversely every character lies in $B_{A^{*}}$ and satisfies these two equations, by [L1] and [L2].

step 1.1step 1.2L1L2

3.1

By [step 1.2] and [step 2.1] the set $Δ (A)$ is weak-star closed, and by [step 1.3] $B_{A^{*}}$ is weak-star compact; a closed subset of a compact space is compact, so $Δ (A)$ is compact in the weak-star topology, and by [step 1.1] the same topology on $Δ (A)$ is the pointwise-evaluation topology.

step 1.1step 1.2step 1.3step 2.1

4.1

By [step 3.1] $Δ (A)$ is compact in the pointwise-evaluation topology, by [step 1.5] it is Hausdorff, and by [step 1.4] it is nonempty; together with [step 1.1] and [step 1.2] this proves all four claims.

step 1.1step 1.4step 1.5step 3.1∎

Remarks

Compactness is a weak-star statement. Banach–Alaoglu is applied to $B_{A^{*}}$ with no completeness hypothesis on $A$ ; the only use of completeness is through the continuity and norm bound of characters, and the only use of commutativity is through the maximal ideal theorem.
Nonemptiness is not automatic. For a commutative unital Banach algebra over $C$ it is a consequence of Zorn; the proof does not construct a character explicitly, and the case of the zero algebra is excluded by the hypothesis that $A$ is nonzero.
The two topologies agree on $Δ (A)$ only because characters are bounded. Before the automatic-continuity theorem the pointwise-evaluation topology of $Δ (A)$ is not a weak-star subspace topology, since $Δ (A)$ is not a subset of the dual.

Depends on

Used by

Gelfand transform Definition
Character space of generated normal algebra is operator spectrum Lemma
Spectral permanence for unital c star subalgebras Lemma
Character space of the unitization is one-point compactification Theorem
Commutative Gelfand duality Theorem
Commutative Gelfand Naimark Theorem
Continuous functional calculus for bounded normal operators Theorem

Dependency tree · two levels

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Sources

Vahid Shirbisheh, Lectures on C-star Algebras, v2 — Proposition 3.1.12 and Remark 3.1.13, printed pp. 60–61 (standard reference, not scraped)
Theo Bühler and Dietmar A. Salamon, Functional Analysis — Lemmas 5.61–5.62, printed pp. 262–265 (standard reference, not scraped)
Dana P. Williams, Lecture Notes on the Spectral Theorem — Theorem 3.6, printed pp. 8–9 (standard reference, not scraped)