10. Commutative Banach algebras - Aarhus Universitet

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The proof of Theorem 11.3 Proof. (a) : Let J be a proper ideal in A. The proper ideals in A which contain J constitute a partially ordered set when ordered by inclusion. By Hausdorff’s maximality theorem there is a maximal totally ordered set Ω in this partially ordered set. Let I be the union of the ideals in Ω. It follows I is an ideal since Ω is totally ordered. If 1 ∈ J there must be an ideal from Ω which contains 1. By (a) of Lemma 2 this is not possible since the ideals in Ω are proper. This shows that J is a proper ideal. Note that J ⊆ I. By maximality of Ω there can not be a proper ideal J ′ in A such that J ⊆ J ′ and J = J ′ , i.e. J is a maximal ideal. Klaus Thomsen 10. Commutative Banach algebras
The proof of Theorem 11.3 Proof. (b) Let J be a maximal ideal in A. Then J is an ideal in A by (b) of Lemma 2. If 1 ∈ J there is an element x ∈ J such that 1 − x < 1. But then x = 1 − (1 − x) is invertible, which is not possible since J is a proper ideal, cf. (a) of Lemma 2. Thus 1 /∈ J, proving that J is a proper ideal. Since J ⊆ J and J is maximal, we conclude that J = J, i.e. J is closed. 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: Commutative Banach algebras - ideal
Page 9 and 10: Quotients Lemma Let X be a Banach s
Page 11 and 12: Quotients Proof. First we observe t
Page 13 and 14: Theorem 11.5 Let A be a Banach alge
Page 15 and 16: Theorem 11.5 Proof. It follows that
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
Page 23: The Gelfand transform - the proof o

10. Commutative Banach algebras - Aarhus Universitet

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