I've come across the fact that the Fourier transform (or, more general, the Gelfand transform) vanishes at $\infty$. See for example "Principles of Harmonic Analysis" by Deitmar and Echterhoff, page 47 or Wikipedia. I also found the following two related posts: (1) Gelfand transform of $a \in A$ vanishes at infinity?, (2) Range of the Gelfand transform on a non-unital Banach algebra. But I am not satisfied with the arguments and I am looking for a convincing proof. It would be great if someone could give me some hints!
I'm going to restrict myself to the case where $A = L^1(G)$, i.e., I will just look at the Fourier transform, but the general case is similar. Let $f \in L^1(G)$ and $G$ be a locally compact abelian group.
First, about (1) and (2): Both use that $B = \widehat{f}^{-1}(\mathbb{C} \setminus B_\varepsilon(0))$ is a weak* closed subset of $\sigma(L^1)$. Since $\sigma(L^1) \cup \{0\}$ is weak* compact, they claim that $B$ must be weak* compact in $\sigma(L^1)$. But since the set is not necessarily closed in the superset, I don't see why that should hold. Is there a way to fix this argument?
About the approach in the book, which I also found in some other books: There, they use that $\sigma(L^1) \cup \{0\}$ is the one-point-compactification of $\sigma(L^1)$ and that vanishing at infinity is the same as the continuous extension of $\widehat{f}$ satisfying $\widehat{f}(0) =0$. All of them just state that $\widehat{f}(0) =0$ is obvious. My question here is: How can I prove that this is the actual continuous extension? I tried to show that it is continuous in $0$ but I somehow got stuck. I'm working with the formula $\widehat{f}(\xi) = h_\xi(f)= \int_G \overline{\langle x, \xi \rangle} f(x) \ dx$, where $\xi \in \widehat{G}$. I know that we can use $\widehat{G}$ with the topology of compact convergence and $\sigma(L^1)$ with the weak* topology interchangeably. For $\varepsilon > 0$, we have
$$\widehat{f}^{-1}(B_\varepsilon(\widehat{f}(0))) = \widehat{f}^{-1}(B_\varepsilon(0)) = \{\eta \in \widehat{G} \cup \{0\} \mid |\widehat{f}(\eta)| < \varepsilon\}\\ \cong \{h_\eta \in \sigma(L^1) \cup \{0\} \mid |h_\eta(f)| < \varepsilon\} = \{h_\eta \in \sigma(L^1) \mid |h_\eta(f)| < \varepsilon\} \cup \{0\}. $$ But why is the latter open in the one-point-compactification $\sigma(L^1) \cup \{0\}$?
I am thankful for any hints!
Here is my attempt on fixing the first approach:
Let $f \in L^1(G)$. We extend $\hat{f}: \widehat{G} \rightarrow \mathbb{C}$ to a function $\hat{f}_{\text{ext}}: \widehat{G}\cup \{0\} \rightarrow \mathbb{C}$ by defining $\hat{f}_{\text{ext}}(0) = h_0(f) = 0$. We claim that this extension is continuous.
Indeed, let $U \subset \mathbb{C}$ be open and choose some $\xi \in \hat{f}_{\text{ext}}^{-1}(U) \subset \widehat{G}\cup \{0\} $. Since $U$ is open, there exists some $\varepsilon > 0$ such that the ball $B_\varepsilon(\hat{f}_{\text{ext}}(\xi))$ lies in $U$. Now, it suffices to prove that $\hat{f}_{\text{ext}}^{-1}(B_\varepsilon(\hat{f}_{\text{ext}}(\xi))) \subset \widehat{G} \cup \{0\}$ is open. We have $$\hat{f}_{\text{ext}}^{-1}(B_\varepsilon(\widehat{f}_{\text{ext}}(\xi))) = \big\{\eta \in \widehat{G}\cup\{0\} \mid \big\vert \hat{f}_{\text{ext}}(\xi)- \hat{f}_{\text{ext}}(\eta)\big\vert < \varepsilon\big\} \cong \big\{h_\eta \in \sigma(L^1(G))\cup \{0\} \mid \left\vert h_\xi(f)- h_\eta(f)\right\vert < \varepsilon\big\}. $$ The latter is weak*-open in $\sigma(L^1(G))\cup \{0\}$, so the left-hand side is open in $\widehat{G}\cup \{0\}$. Thus, $\hat{f}_{\text{ext}}$ is continuous.
Now, let $\varepsilon >0$. Since $\hat{f}_{\text{ext}}$ is continuous, the set $\hat{f}_{\text{ext}}^{-1}(\mathbb{C} \setminus B_\varepsilon(0))$ is closed. As a subset of the compact set $\widehat{G}\cup \{0\}$ it is compact. Since $\hat{f}_{\text{ext}}(0) = 0$, we have $\hat{f}^{-1}(\mathbb{C} \setminus B_\varepsilon(0)) = \hat{f}_{\text{ext}}^{-1}(\mathbb{C} \setminus B_\varepsilon(0))$. Since these sets are compact in $\widehat{G} \cup \{0\}$, they are also compact in $\widehat{G}$. Therefore, $\hat{f}$ vanishes at infinity.
Any concerns?