[proofplan]
We view spectral integration against $E$ as the bounded [Borel functional calculus](/theorems/2696) associated to the projection-valued measure. The coordinate function $g(z)=z$ is bounded and Borel on the compact set $K$, so the operator $T$ is $\int_K g\,dE$. The $*$-homomorphism rules for spectral integration identify adjoints and products with conjugates and products of scalar functions, which gives normality, commutation with every $\int_K f\,dE$, and self-adjointness in the real case. The norm estimate follows from the general operator-norm bound for spectral integrals.
[/proofplan]
[step:Declare the coordinate function and invoke the spectral integral calculus]
Define the coordinate function $g:K\to\mathbb C$ by $g(z)=z$ for every $z\in K$. Since $K$ is compact, $g$ is continuous and bounded; in particular, $g$ is a bounded Borel function on $K$. With this notation,
\begin{align*}
T=\int_K g\,dE.
\end{align*}
Let $B_b(K)$ denote the complex algebra of bounded Borel functions $u:K\to\mathbb C$, equipped with pointwise addition, pointwise multiplication, pointwise complex conjugation, and the supremum norm $\|u\|_\infty=\sup_{z\in K}|u(z)|$. We use the bounded Borel spectral integral map $\Phi_E:B_b(K)\to\mathcal L(H)$ defined by
\begin{align*}
\Phi_E(u)=\int_K u\,dE
\end{align*}
for every $u\in B_b(K)$. Since $K\subset\mathbb C$ is compact, the metric inherited from $\mathbb C$ makes $K$ a compact [metric space](/page/Metric%20Space), and the stated map $E:\mathcal B(K)\to\mathcal L(H)$ is a projection-valued measure on its Borel $\sigma$-algebra. By [citetheorem:8407], $\Phi_E$ is a $*$-homomorphism and satisfies the norm estimate
\begin{align*}
\|\Phi_E(u)\|_{\mathcal L(H)}\le \|u\|_\infty
\end{align*}
for every bounded Borel function $u:K\to\mathbb C$.
[guided]
The first task is to put the notation into the language where the algebraic rules for spectral integrals apply. Define $g:K\to\mathbb C$ by $g(z)=z$ for every $z\in K$. This is the coordinate function restricted to $K$. Because $K$ is compact and $g$ is continuous, the image $g(K)=K$ is compact in $\mathbb C$, hence bounded. Therefore $g$ is a bounded Borel function on $K$.
The operator in the theorem is precisely the spectral integral of this function:
\begin{align*}
T=\int_K z\,dE(z)=\int_K g\,dE.
\end{align*}
Let $B_b(K)$ denote the complex algebra of bounded Borel functions $u:K\to\mathbb C$, equipped with pointwise addition, pointwise multiplication, pointwise complex conjugation, and the supremum norm $\|u\|_\infty=\sup_{z\in K}|u(z)|$. We now invoke the spectral integration construction for projection-valued measures, [citetheorem:8407]. It applies because $K\subset\mathbb C$ is compact, hence is a compact metric space with the inherited metric, because $E:\mathcal B(K)\to\mathcal L(H)$ is a projection-valued measure on the Borel $\sigma$-algebra of that compact metric space, and because the functions considered below lie in $B_b(K)$. The theorem gives a map $\Phi_E:B_b(K)\to\mathcal L(H)$ defined by
\begin{align*}
\Phi_E(u)=\int_K u\,dE
\end{align*}
for each $u\in B_b(K)$. The two properties we will use are multiplicativity and compatibility with adjoints:
\begin{align*}
\Phi_E(uv)=\Phi_E(u)\Phi_E(v)
\end{align*}
and
\begin{align*}
\Phi_E(\overline{u})=\Phi_E(u)^*
\end{align*}
for bounded Borel functions $u,v:K\to\mathbb C$. We also use the corresponding norm estimate
\begin{align*}
\|\Phi_E(u)\|_{\mathcal L(H)}\le \|u\|_\infty.
\end{align*}
These are exactly the properties that turn scalar identities between functions on $K$ into operator identities on $H$.
[/guided]
[/step]
[step:Compute the adjoint and prove normality]
The complex conjugate function $\overline g:K\to\mathbb C$ is defined by $\overline g(z)=\overline z$. It is bounded and Borel. By the $*$-property of $\Phi_E$,
\begin{align*}
T^*=\Phi_E(g)^*=\Phi_E(\overline g)=\int_K \overline z\,dE(z).
\end{align*}
Using multiplicativity of $\Phi_E$,
\begin{align*}
T^*T=\Phi_E(\overline g)\Phi_E(g)=\Phi_E(\overline g\,g)=\int_K |z|^2\,dE(z).
\end{align*}
Similarly,
\begin{align*}
TT^*=\Phi_E(g)\Phi_E(\overline g)=\Phi_E(g\,\overline g)=\int_K |z|^2\,dE(z).
\end{align*}
Since $\overline g\,g=g\,\overline g$, the two products are equal:
\begin{align*}
T^*T=TT^*.
\end{align*}
Thus $T$ is normal.
[/step]
[step:Apply the norm estimate to the coordinate function]
Applying the norm estimate from [citetheorem:8407] to the bounded Borel function $g:K\to\mathbb C$ gives
\begin{align*}
\|T\|_{\mathcal L(H)}=\|\Phi_E(g)\|_{\mathcal L(H)}\le \|g\|_\infty.
\end{align*}
By the definition of the supremum norm on $K$,
\begin{align*}
\|g\|_\infty=\sup_{z\in K}|g(z)|=\sup_{z\in K}|z|.
\end{align*}
Therefore
\begin{align*}
\|T\|_{\mathcal L(H)}\le \sup_{z\in K}|z|.
\end{align*}
[/step]
[step:Show every bounded Borel spectral integral commutes with $T$]
Let $f:K\to\mathbb C$ be a bounded Borel function. Since $f$ and $g$ are bounded Borel functions, the pointwise products $fg$ and $gf$ are bounded Borel functions on $K$. By multiplicativity of $\Phi_E$,
\begin{align*}
f(T;E)T=\Phi_E(f)\Phi_E(g)=\Phi_E(fg).
\end{align*}
Also,
\begin{align*}
Tf(T;E)=\Phi_E(g)\Phi_E(f)=\Phi_E(gf).
\end{align*}
For every $z\in K$, complex multiplication gives $f(z)g(z)=g(z)f(z)$, so $fg=gf$. Hence
\begin{align*}
f(T;E)T=Tf(T;E).
\end{align*}
Thus $f(T;E)$ commutes with $T$.
[/step]
[step:Use reality of the coordinate function to obtain self-adjointness]
Assume now that $K\subset\mathbb R$. Then for every $z\in K$,
\begin{align*}
g(z)=z=\overline z=\overline g(z).
\end{align*}
Thus $g=\overline g$ as bounded Borel functions on $K$. Using the $*$-property of $\Phi_E$,
\begin{align*}
T^*=\Phi_E(g)^*=\Phi_E(\overline g)=\Phi_E(g)=T.
\end{align*}
Therefore $T$ is self-adjoint. This completes the proof.
[/step]
