10. Commutative Banach algebras - Aarhus Universitet

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Theorem 11.5 Proof. c) If x ∈ A is invertible, xx −1 = 1 and it follows from Proposition 10.6 that 1 = µ(1) = µ(x)µ x −1 , and we conclude that µ(x) = 0. Conversely, if x is not invertible, the set xA = {xa : a ∈ A} is a proper ideal in A. By (a) of Theorem 11.3 it is then contained in a maximal ideal in A, and it follows from (a), proved above, that it is contained in the kernel of an element ω ∈ ∆. d) : If x ∈ A is invertible, it is not an element of any proper ideal by (a) of Proposition 11.2 in Rudin. Conversely, if x lies in no proper ideal, it does, in particular, not lie in any maximal ideal. Hence ω(x) = 0 for all ω ∈ ∆ by (a) and thus x is invertible by c). Klaus Thomsen 10. Commutative Banach algebras
Theorem 11.5 Proof. (e) : If λ ∈ σ(x), λ1 − x is not invertible, and hence µ (λ1 − x) = 0 for some µ ∈ ∆. It follows that λ ∈ {ω(x) : ω ∈ ∆} since µ(1) = 1 by Proposition 10.6. If λ = ω(x) for some ω ∈ ∆, we conclude from (c) that λ1 − x is not invertible (since ω(λ1 − x) = 0), and it follows that λ ∈ σ(x). 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 and 14: Theorem 11.5 Let A be a Banach alge
Page 15: Theorem 11.5 Proof. It follows that
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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