By Thill M.

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1), we have: (ab)(k) e ikt = τ (ab) = k∈Z = l∈Z a(l) b(m) e ikt k∈Z l+m=k a(l) e ilt · b(m) e imt m∈Z = τ (a) · τ (b). As a multiplicative linear functional is determined by its value at δ1 , it follows that ∆(ℓ 1 Z) can be identified with R/2πZ. Also, a function on R/2πZ is the Gelfand transform of an element of ℓ 1 (Z) if and only if it has an absolutely convergent Fourier expansion. 24). Ì ÓÖ Ñ (Wiener)º Let f be a function on R/2πZ which has an absolutely convergent Fourier expansion. If f (t) = 0 for all t ∈ R/2πZ, then also 1/f has an absolutely convergent Fourier expansion.

2). Furthermore, the map π factors to an injective ∗-algebra homomorphism π1 from C to B. The mapping π1 is contractive in the auxiliary ∗-norm on C, cf. 2). This implies that the involution in C has closed graph, cf. 7). But then the involution is continuous, as C is complete. Hence the auxiliary ∗-norm on C and the quotient norm on C are equivalent, cf. 6). It follows that π1 is continuous in the quotient norm on C, which is enough to prove the theorem. 28 1. 9). ÈÖÓÔÓ× Ø ÓÒº Let A, B be normed algebras and let π : A → B be a continuous algebra homomorphism.

It shall next be shown that (K, φ) is equivalent to the compactification associated with A := AK as in (i). For x ∈ K, consider θx : f → g(x), (f ∈ AK ), where g is the unique function in C(K) such that f = g ◦ φ, see above. Then θx is a multiplicative linear functional on AK because it is non-zero as AK is unital in Cb (Ω). Now the map θ : K → ∆(AK ), x → θx is the desired homeomorphism K → ∆(AK ). 11) because AK is essentially the same C*-algebra as C(K), see above. It is also clear that θ ◦ φ = ε.