Suppose $A$ is a commutative Banach algebra without a unit, let $\Delta$ be the set of all complex homomorphisms of $A$ which are not identically $0$. Each $x\in A$ defines a function $\hat x$ on $\Delta$, given by $$\hat x(h)=h(x)\quad (h\in\Delta).$$ If we give $\Delta$ the Gelfand topology, then $\Delta$ is a locally compact Hausdorff space, i.e., the maximal ideal space of $A$. Then how to show that each $\hat x$ is a member of $C_0(\Delta)$?
We know that if $A$ is unital, then $\Delta$ is actually compact and it follows that $C_0(\Delta)=C(\Delta)$ which implies each $\hat x$ is in $C_0(\Delta)$. But I cannot find a proof when $\Delta$ is not unital. Also, I do not have a solid background in Banach algebra. I just want to make sense of this statement. Thank you!
Or why can we conclude that $\hat x\in C_0(\Delta(\Lambda))$ in the first definition between Proposition 23 and Theorem 24 in this document: http://www.karlin.mff.cuni.cz/~kalenda/data/fa116e-14.pdf
Fix an $x\in A$. For every $\varepsilon \gt o$, the set $\{h\in\Delta\mid |h(x)|\geq\varepsilon\}$ is weak-star closed(by definition of the weak-star topology) and bounded($\|h\|\leq1$ for all $h\in\Delta$). Thus by Banach-Alaoglu, $\{h\in\Delta\mid |h(x)|\geq\varepsilon\}$ is weak-star compact. That's exactly what you needed to prove.