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My argument below is hacked together from pages 5-6 of Davidson's "$C^*$ algebras by example".

Theorem: The multiplicative linear functionals on a unital abelian Banach algebra are continuous of norm 1.

Proof: Let $\varphi$ be a multiplicative linear functional on $\mathfrak A$, and $A \in \mathfrak A$. Suppose towards a contradiction that $\|A\| < \varphi(A)$; by considering instead $A'=A/\varphi(A)$ if necessary, we can assume $\varphi(A)=1$. Let $B = \sum_{n \geq 1} A^n$; then $A+AB=B$, and as a result, $\varphi(B)=\varphi(A)+\varphi(A)\varphi(B)=1+\varphi(B)$, which is nonsense. So $\|\varphi\| \leq 1$. Since $\varphi(I)=1$, that's an equality.

If $\mathfrak A$ is non-unital, it seems to me the early part of that argument still applies, so that $\|\varphi\| \leq 1$. In either case $\mathfrak{A}$ is a subspace of the unit ball in $\mathfrak{A}^*$. We topologize the set of multiplicative linear functionals $\mathcal{M_{\mathfrak A}}$ as a subspace of the dual space $\mathfrak A^*$.

Now weak-* convergence of nets is just pointwise convergence. As a result it's clear that the weak-* limit of multiplicative linear functionals is a multiplicative linear functional. (As is standard in mathematics, the part where I say 'it's clear' is where I think there's likeliest to be a problem.) So $\mathcal{M_{\mathfrak A}}$ is a closed subset of the unit ball of $\mathfrak A^*$; Banach-Alaoglu says that it's compact.

On the other hand, I know that $\mathcal{M_{\mathfrak A}}$ is not compact for $\mathfrak A$ non-unital, but rather only locally compact. So something in the above argument is wrong. What is it?

As mentioned, I think it's the "clear" bit. Because we can extend the 0 functional in a unique way, and every other functional extends uniquely, it seems likely to me that there's a net of multiplicative functionals that weak-* converge to the 0 functional. I'd like to see such a net. (It actually is clear in the unital case: since the functionals have norm 1, they're bounded away from the 0 functional, so this can't happen. They're not bounded away from the 0 functional in the non-unital case, so this is almost certainly the failure of this argument.

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    Careful: the norm is not weak-star continuous. It's possible for a sequence of functionals to have norm bounded away from 0 yet still converge weak star to 0. – Nate Eldredge Apr 04 '14 at 07:09
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    @NateEldredge Damn, thanks for correcting me. I must admit I'm very weak on my weak-* everything. I can modify, however, by noting that $\phi(I)=1$ for all $\phi \in \mathcal{M}_{\tilde{\mathfrak A}}$, so that no net can converge to the 0 functional. –  Apr 04 '14 at 07:13

3 Answers3

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Here is a simple example. Let $X$ be a non-compact LCH space. Every multiplicative linear functional has the form $\phi_x(f) := f(x)$ for $f \in C_0(X)$, but if you choose a sequence (or net) $x_n$ which goes off to $\infty$ then $\phi_{x_n}(f) = f(x_n) \to 0$ for every $f$ which means $\phi_{x_n} \to 0$ in the weak*-topology.

Paul Siegel
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Gelfand spectrum is a set of non-zero characters. For non-unital algebra you can not prove that the limit of the net of characters is non-zero character, that is the place where you argument fails. In fact, Alexandrov's compactification of the spectrum of non-ubital Banach algebra is just an addition of zero character.

Norbert
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  • I believe that one can prove that the limit of this net is the non-zero characters (though I'm not sure how to direct this set), but I don't know how. Can you point me in the right direction? –  Apr 04 '14 at 07:15
  • See theorem 2.8 in my answer here – Norbert Apr 04 '14 at 08:16
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Consider the Banach algebra $C_0(Z)$ and the evaluation functionals $l_n(f)=f(n)$. They all have norm 1 but converge to 0 in weak-star.

The fact that this can't happen in the unital case is because all the functionals send the identity to 1, hence so must their limit, so the limit can't be 0.

Nate Eldredge
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