[proofplan]
We use the continuous functional calculus for the normal element $a$: it identifies $C(\sigma_A(a))$ with the unital commutative $C^*$-subalgebra $C^*(a,1_A)\subset A$ by a unital isometric $*$-isomorphism. The multiplicative, adjoint, and norm identities are exactly the multiplicativity, $*$-preservation, and isometry of this isomorphism. For the spectral mapping formula, we compute the spectrum of the function $f$ inside $C(\sigma_A(a))$ directly as its range, transfer that spectrum through the isomorphism, and then use spectral permanence to regard the result as the spectrum in the ambient algebra $A$.
[/proofplan]
[step:Introduce the functional calculus isomorphism]
Let $B:=C^*(a,1_A)$ denote the unital $C^*$-subalgebra of $A$ generated by $a$ and $1_A$. Since $A$ is a unital complex Banach algebra, $\sigma_A(a)$ is a nonempty compact subset of $\mathbb C$, so $C(\sigma_A(a))$ is a unital commutative complex $C^*$-algebra under pointwise operations, pointwise conjugation, and the supremum norm. Since $a$ is normal, the continuous functional calculus theorem for normal elements applies. Thus there is a unital isometric $*$-isomorphism
\begin{align*}
\Phi_a:C(\sigma_A(a))\to B
\end{align*}
such that, for every polynomial $p$ in one complex variable and its conjugate variable, $\Phi_a(p)=p(a,a^*)$. By definition of the notation used in the theorem, for every $h\in C(\sigma_A(a))$ we write
\begin{align*}
h(a):=\Phi_a(h).
\end{align*}
[/step]
[step:Read multiplicativity and adjoints from the $*$-isomorphism]
The product $fg\in C(\sigma_A(a))$ is the pointwise product defined by $(fg)(\lambda)=f(\lambda)g(\lambda)$ for $\lambda\in\sigma_A(a)$. Since $\Phi_a$ is multiplicative,
\begin{align*}
(fg)(a)=\Phi_a(fg)=\Phi_a(f)\Phi_a(g)=f(a)g(a).
\end{align*}
The function $\bar f\in C(\sigma_A(a))$ is the pointwise complex conjugate of $f$. The involution in $C(\sigma_A(a))$ is pointwise conjugation, so $f^*=\bar f$ in the $C^*$-algebra $C(\sigma_A(a))$. Since $\Phi_a$ is $*$-preserving,
\begin{align*}
(\bar f)(a)=\Phi_a(\bar f)=\Phi_a(f^*)=\Phi_a(f)^*=f(a)^*.
\end{align*}
[guided]
The continuous functional calculus is designed so that algebraic operations on continuous functions become the corresponding algebraic operations on the element $a$. We make this precise through the map
\begin{align*}
\Phi_a:C(\sigma_A(a))\to C^*(a,1_A).
\end{align*}
This map is a unital $*$-isomorphism, so it preserves multiplication and the involution.
First consider multiplication. The function $fg:\sigma_A(a)\to\mathbb C$ is defined pointwise by
\begin{align*}
(fg)(\lambda)=f(\lambda)g(\lambda)
\end{align*}
for every $\lambda\in\sigma_A(a)$. Multiplicativity of $\Phi_a$ gives
\begin{align*}
(fg)(a)=\Phi_a(fg)=\Phi_a(f)\Phi_a(g)=f(a)g(a).
\end{align*}
The last equality is only notation: by definition, $\Phi_a(f)=f(a)$ and $\Phi_a(g)=g(a)$.
Now consider the adjoint. In the commutative $C^*$-algebra $C(\sigma_A(a))$, the involution is pointwise complex conjugation. Therefore the adjoint of $f$ as an element of $C(\sigma_A(a))$ is the function $\bar f:\sigma_A(a)\to\mathbb C$ defined by
\begin{align*}
\bar f(\lambda)=\overline{f(\lambda)}.
\end{align*}
Because $\Phi_a$ is $*$-preserving, applying $\Phi_a$ after taking the adjoint is the same as taking the adjoint after applying $\Phi_a$. Hence
\begin{align*}
(\bar f)(a)=\Phi_a(\bar f)=\Phi_a(f^*)=\Phi_a(f)^*=f(a)^*.
\end{align*}
[/guided]
[/step]
[step:Read the norm identity from isometry]
The norm on $C(\sigma_A(a))$ is the supremum norm
\begin{align*}
\|f\|_\infty=\sup_{\lambda\in\sigma_A(a)}|f(\lambda)|.
\end{align*}
Since $\Phi_a$ is isometric and $f(a)=\Phi_a(f)$,
\begin{align*}
\|f(a)\|_A=\|\Phi_a(f)\|_A=\|f\|_\infty.
\end{align*}
[/step]
[step:Compute the spectrum of a continuous function as its range]
Let $X:=\sigma_A(a)$, viewed as a compact subset of $\mathbb C$. Let $1_{C(X)}:X\to\mathbb C$ denote the constant function with value $1$. We claim that the spectrum of $f$ as an element of the unital $C^*$-algebra $C(X)$ is
\begin{align*}
\sigma_{C(X)}(f)=f(X).
\end{align*}
Indeed, fix $\lambda\in\mathbb C$. The element $f-\lambda 1_{C(X)}$ is invertible in $C(X)$ if and only if the [continuous function](/page/Continuous%20Function)
\begin{align*}
x\mapsto f(x)-\lambda
\end{align*}
has no zero on $X$. If it has no zero, then the reciprocal function
\begin{align*}
h:X\to\mathbb C,\qquad x\mapsto \frac{1}{f(x)-\lambda}
\end{align*}
is continuous and satisfies $h(f-\lambda 1_{C(X)})=1_{C(X)}$. Conversely, if $f(x_0)=\lambda$ for some $x_0\in X$, then every product $h(f-\lambda 1_{C(X)})$ vanishes at $x_0$, so $f-\lambda 1_{C(X)}$ cannot be invertible. Therefore $\lambda\notin\sigma_{C(X)}(f)$ exactly when $\lambda\notin f(X)$, which proves the claim.
[/step]
[step:Transfer the spectrum back to the ambient algebra]
Since $\Phi_a:C(\sigma_A(a))\to B$ is a unital algebra isomorphism, it preserves invertibility and therefore preserves spectra:
\begin{align*}
\sigma_B(\Phi_a(f))=\sigma_{C(\sigma_A(a))}(f).
\end{align*}
Using $\Phi_a(f)=f(a)$ and the preceding step,
\begin{align*}
\sigma_B(f(a))=f(\sigma_A(a)).
\end{align*}
Finally, $B=C^*(a,1_A)$ is a unital $C^*$-subalgebra of the unital $C^*$-algebra $A$ with the same unit, and $f(a)\in B$. By spectral permanence for unital $C^*$-subalgebras,
\begin{align*}
\sigma_A(f(a))=\sigma_B(f(a)).
\end{align*}
Combining the last two displayed equalities gives
\begin{align*}
\sigma_A(f(a))=f(\sigma_A(a)).
\end{align*}
This is the desired spectral mapping formula, and the proof is complete.
[/step]
