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Let $\mathcal{A}$ be a non-unital commutative Banach algebra. Consider the Gelfand transform \begin{align*}\Gamma_{\mathcal{A}}:\mathcal{A}&\to C(\sigma(\mathcal{A}))\\ x&\mapsto \hat{x} \end{align*} where \begin{align*}\hat{x}:\sigma(\mathcal{A})&\to \mathbb{C}\\ h&\mapsto \hat{x}(h)=h(x) \end{align*} In Folland's A course in Abstract Harmonic Analysis it is stated that in fact $\Gamma_{\mathcal{A}}(\mathcal{A})\subset C_0(\sigma(\mathcal{A}))$, i.e. if $x\in \mathcal{A}$ then for all $\varepsilon>0$ there is a compact $K\subset \sigma(\mathcal{A})$ such that $|\hat{x}(h)|\leq \varepsilon$ for all $h\in \sigma(\mathcal{A})\setminus K$. It is not well explained why this is true.

The basic idea would be to try to get back to the unital case. We can consider the unital extension $\tilde{A}=\mathcal{A}\times \mathbb{C}$ of $\mathcal{A}$, where $\mathcal{A}\cong \mathcal{A}\times \left\{0\right\}\subset \tilde{A}$. We then have a homeomorphism (I think) \begin{align*}\sigma(\tilde{\mathcal{A}})&\cong \sigma(\mathcal{A})\cup \left\{0\right\}\\ 0&\mapsto \tilde{0}:(x,\lambda)\mapsto 0\\ \sigma(\mathcal{A})\ni h&\mapsto \tilde{h}:(x,\lambda)\mapsto h(x)+\lambda \end{align*} both are compact Hausdorff spaces in the weak*-topology induced by $\mathcal{A}^*$ and $\mathcal{A}$ respectively, while $\sigma(\mathcal{A})$ is only locally compact in general.

The above homeomorphism in turn induces a Banach algebra isomorphism $C(\sigma(\tilde{\mathcal{A}}))\cong C(\sigma(\mathcal{A})\cup\left\{0\right\})$. With this identification we have $\Gamma_{\mathcal{A}}=r\circ \left.\Gamma_{\tilde{\mathcal{A}}}\right|_{\mathcal{A}}$ where $r:C(\sigma(\mathcal{A})\cup\left\{0\right\})\to C(\sigma(\mathcal{A}))$ is the restriction map. Now I would like to use the topological/algebraic properties I know about $r$ and $\Gamma_{\tilde{\mathcal{A}}}$ to show that $\Gamma_{\mathcal{A}}(\mathcal{A})\subset C_0(\sigma(\mathcal{A}))$.

I am aware that $\mathcal{A}$ is a closed maximal ideal of $\tilde{A}$ so since $\tilde{\mathcal{A}}$ is unital, $\Gamma_{\tilde{\mathcal{A}}}(\mathcal{A})$ is itself a closed non-trivial subalgebra of $C(\sigma(\mathcal{A})\cup\left\{0\right\})$.

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  • $\begingroup$ In regards to your second paragraph, $C_0(\sigma(\mathcal{A}))$ isn't the algebra of compactly supported functions, it's the algebra of functions vanishing at infinity. $\endgroup$ – Aweygan May 1 '18 at 15:50
  • $\begingroup$ True, thanks for pointing it out. I actually had the correct space in mind, just wrote down the wrong definition, so my issues remain. $\endgroup$ – Lorenzo Quarisa May 1 '18 at 16:00
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The following may or may not answer the specific questions you have in mind, but it does I think explain what's really going on.

Say $A$ is a Banach algebra. Usually in this context "complex homomorphism" means "non-zero complex homomorphism", but here I'm going to include the map $h(x)=0$ as a "complex homomorphism".

Let $K$ be the set of all complex homomorphisms of $A$, and let $\sigma(A)=K\setminus\{0\}$ be the set of non-zero complex homomorphisms.

Let $A^*$ be the Banach-space dual of $A$. Then $K$ is a subset of the unit ball of $A^*$, and it's easy to see that $K$ is weak* closed; hence $K$ is weak* compact. The Gelfand toplogy is just the weak* topology, so $K$ is compact.

Hence $\sigma(A)=K\setminus\{0\}$ is locally compact. (Of course $K$ is the one-point compactification of $\sigma(A)$, which means that the "point at infinity" for $\sigma(A)$ is given by the amusing formula $$\infty=0.)$$

Now if $A$ has an identity $e$ then $\sigma(A)=\{h\in K:h(e)=1\}$; hence $\sigma(A)$ is a closed subset of $K$, hence $\sigma(A)$ is compact. But in general, whether $A$ has an identity or not, we have, writing $\hat x$ for the Gelfand transform,

If $x\in A$ then $\hat x\in C_0(\sigma(A))$.

Proof: Let $\epsilon>0$. We need to show that $\{h\in\sigma(A):|\hat x(h)|\ge\epsilon\}$ is compact. But this set is equal to$\{h\in K:|h(x)|\ge\epsilon\}$, which is a closed subset of $K$.

Note: I'm pretty sure that if $\sigma(A)$ is compact then $A$ has an identity. That's less trivial than the rest of this - right now I don't recall the proof.

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  • $\begingroup$ Thanks for your answer. Just for reference on your note, Folland notes that if $0$ is an isolated point of $\sigma(\mathcal{A})\cup\left\{0\right\}$ then $\sigma(\mathcal{A})$ is still compact (when talking of non-unital algebras), although he does claim that in 'the most common case' $\sigma(\mathcal{A})$ is non-compact. $\endgroup$ – Lorenzo Quarisa May 1 '18 at 17:13
  • $\begingroup$ @LorenzoQ. He doesn't actually give an example of an algebra with no identity where $\sigma(A)$ is compact, does he? $\endgroup$ – David C. Ullrich May 1 '18 at 17:25
  • $\begingroup$ No, he does not! $\endgroup$ – Lorenzo Quarisa May 1 '18 at 17:34
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Ok, I solved it.

Let $x\in \mathcal{A}$. With the usual identification $\mathcal{A}\cong \mathcal{A}\times \left\{0\right\}\subset \tilde{\mathcal{A}}$, we have that $\hat{x}=\left.(x,0)^{\hat{}}\right|_{\sigma(\mathcal{A})}$ and $\sigma(\tilde{\mathcal{A}})\cong \sigma(\mathcal{A})\cup\left\{0\right\}$. For any $\varepsilon>0$, $$\left\{h\in \sigma(\mathcal{A})\cup\left\{0\right\}:|(x,0)^{\hat{}}(\tilde{h})|\geq \varepsilon\right\}=\left((x,0)^{\hat{}}\right)^{-1}(\mathbb{C}\setminus B_0(\varepsilon))=:S_{\varepsilon} $$ Now, the key insight is to notice that $0\notin S_{\varepsilon}$ for all $\varepsilon >0$, because $(x,0)^{\hat{}}(\tilde{0})=0(x,0)=0$. Since moreover $\hat{x}$ is continuous, $S_{\varepsilon}$ is a closed subset of $\sigma(\mathcal{A})$, and hence compact because it is also contained in $\sigma(\mathcal{A})\cup\left\{0\right\}$ which is compact.

Thus $S_{\varepsilon}\subset \sigma(\mathcal{A})$ is compact, and $|\hat{x}(h)|<\varepsilon$ outside of $S_{\varepsilon}$. This proves that $\hat{x}=\left.(x,0)^{\hat{}}\right|_{\sigma(\mathcal{A})}\in C_0(\sigma(\mathcal{A}))$.

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