A note on constructive methods in Banach algebras

Abstract

proof. We let I denote the ideal generated by fi, i.e. the set of all functions of the form Efigi. We wish to show that 1 belongs to I. We assume that ISA. First, we extend I to a maximal ideal J, invoking the axiom of choice. Second, we observe that J is closed. Third, we form the quotient Banach algebra A/J. Fourth, we use the Gelfand-Mazur theorem to deduce that there is a homomorphism of A onto the complexes sending J, and hence fi, into zero. Finally, we show that any such homomorphism is given by evaluation at some point of the disc, which is a contradiction. We note that step two above is accomplished by the following simple device, which we can expect to occur in any classical proof. LEMMA. If IIf 1 1, then 1/(z-X) belongs to A and so we may take h equal to it. If lXi _1, then we have say f1(X) O0. Now the polynomials in z are dense in A, a fact which is used in the last step of the abstract proof, and may be proved by taking Fejer means. Thus there is a polynomial f(z), such that If1-fill < f(X) 1 /4. For some k in A we clearly have (1) (z -) k (z) = f(z) -f(X),