10. Commutative Banach algebras - Aarhus Universitet

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Theorem 11.5 Proof. a) Let J be a maximal ideal in A. Then J is closed by b) of Theorem 11.3. Hence J is a closed two-sided ideal in A, and we conclude that A/J is a Banach algebra by Lemma 5. We assert that every non-zero element of A/J is invertible. To see this, let z ∈ A/J be a non-zero element. Then z (A/J) = {zξ : ξ ∈ A/J} is an ideal in A/J, and it follows that q −1 (z (A/J)) is an ideal in A. Note that J ⊆ q −1 (z (A/J)) and that J = q −1 (z (A/J)). Since J is maximal ideal, q −1 (z (A/J)) can then not be a proper ideal in A, i.e. A = q −1 (z (A/J)). In particular, 1 = q(1) ∈ z (A/J), proving that z is invertible. Klaus Thomsen 10. Commutative Banach algebras
Theorem 11.5 Proof. It follows that A/J = C1 by the Gelfand-Mazur theorem. Define ω : A → C such that ω(a)1 = q(a), where q : A → A/J is the quotient map. Then ω ∈ ∆ and J = ker ω. b): Let µ ∈ ∆. Then ker µ is clearly (!) an ideal in A. Let I ⊆ A be a proper ideal such that ker µ ⊆ I and ker µ = I. Let x ∈ I be an element such that µ(x) = 0. Then λµ(x) = 1 for some λ ∈ C, and it follows that µ (x − µ(x)1) = 0. Hence x − µ(x)1 ∈ ker µ ⊆ I, and we deduce that µ(x)1 ∈ I. This contradicts the properness of I, and we conclude that ker µ is a maximal ideal. Klaus Thomsen 10. Commutative Banach algebras
Page 1 and 2: 10. Commutative Banach algebras Kla
Page 3 and 4: The Gelfand-Mazur theorem Let A and
Page 5 and 6: Commutative Banach algebras - ideal
Page 7 and 8: The proof of Theorem 11.3 Proof. (b
Page 9 and 10: Quotients Lemma Let X be a Banach s
Page 11 and 12: Quotients Proof. First we observe t
Page 13: Theorem 11.5 Let A be a Banach alge
Page 17 and 18: Theorem 11.5 Proof. (e) : If λ ∈
Page 19 and 20: The Gelfand transform - Theorem 11.
Page 21 and 22: The Gelfand transform - the proof o
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10. Commutative Banach algebras - Aarhus Universitet

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