I'm reading this script and I got stuck on the last Let $X$ be the set of all non-zero multiplicative linear functionals on the unital commutative Banach algebra $\mathcal{A}$ then $\mathrm{sp}(\hat{A})=\hat{A}(X)$.
In the script (Theorem page 7) the author says that it follows from the elementary fact that $\mathrm{sp}(f)=f(X)$ for any $f\in C(X)$.
I agree that it would follow but to me it's not that elementary.
Any kind of help is appreciated.
$X$ is a compact Hausdorff space. If $f\in C(X), \lambda\in\Bbb{C},\lambda\not\in f(X)$ then $\operatorname{dist}(\lambda,f(X)) = d>0$ because $f$ is continuous, so $f(X)$ is a compact subset of $\Bbb{C}$ and its complement $\Bbb{C}\setminus f(X)$ is open. Therefore $(f-\lambda)^{-1}$ is a continuous function bounded in absolute value by $1/d$ and this function is the inverse of $f-\lambda$ in $C(X)$. Therefore $\lambda\not\in\operatorname{sp}(f)$.
If $\lambda\in f(X)$, for a similar reason, $(f-\lambda)^{-1}$ is unbounded near a point $z$ for which $f(z)=\lambda$, and since $(f-\lambda)^{-1}$ can be the only inverse for $f-\lambda$ where the operation is pointwise multiplication of functions, it follows that $f-\lambda$ is not invertible, so $\lambda\in\operatorname{sp}(f)$.
Note: $(f-\lambda)^{-1}$ denotes $\frac{1}{f-\lambda}$, not an inverse function in the sense of composition of functions.