Gelfand transform of $a \in A$ vanishes at infinity?

Question

Let $A$ be a non-unital commutative Banach algebra defined over the field of complex numbers.

Given $a \in A$, one defines the function $\widehat{a}:\Phi_A\to{\mathbb C}$ by $\widehat{a}(\varphi)=\varphi(a)$, where $\Phi_A$ is endowed with the Gelfand topology (i.e. the restriction of the $w^*$-topology on $A^*$ to $\Phi_A$).

The definition of $\Phi_A$ and of its topology ensure that $\widehat{a}$ is continuous. According to wiki (https://en.wikipedia.org/wiki/Gelfand_representation) it seems that $\widehat{a} \in \mathcal{C}_0(\Phi_A)$, that is to say that $\widehat{a}$ vanishes at infinity. I struggle to see why this latter fact is true.

Answer 1

Since nobody posted an answer, I will write down my attempt. Could anyone check is this is correct please ?

Note that $\big\{ \varphi \in \Phi_A :~~ \big| \varphi(a)\big| \geq \varepsilon \big\}$ is $w^*$-closed in the closed unit ball of $A^*$ for all $\varepsilon > 0$ because

$$\big\{ \varphi \in \Phi_A :~~ \big| \varphi(a)\big| \geq \varepsilon \big\} = \widehat{a}^{-1}\Big( \{z : |z| \geq \varepsilon\}\Big)$$ and $\{z : |z| \geq \varepsilon\}$ is closed and $\widehat{a}$ is continuous when $\Phi_A$ has the Gelfand topology.

Hence Banach-Alaoglu implies that $\big\{ \varphi \in \Phi_A :~~ \big| \varphi(a)\big| \geq \varepsilon \big\}$ is $w^*$-compact.