[proofplan]
We represent the normal operator $N$ by its spectral measure and compare its norm with the size of the coordinate function on the spectrum. The spectral integral immediately gives the upper bound $\|N\|_{\mathcal L(H)} \le \sup_{\lambda \in \sigma(N)} |\lambda|$. For the reverse inequality, the fact that the spectrum is the support of the spectral measure ensures that every neighbourhood of a spectral point has a nonzero spectral projection, and testing $N$ on a vector in that spectral subspace forces the norm to be at least the modulus of that point. Compactness of the spectrum turns the supremum into a maximum.
[/proofplan]
[step:Represent $N$ as the spectral integral of the coordinate function]
By the [spectral theorem for bounded normal operators](/theorems/8412), [citetheorem:8412], there exists a projection-valued measure $E$ on $\sigma(N)$ such that
\begin{align*}
N = \int_{\sigma(N)} z\,dE(z).
\end{align*}
Define the continuous coordinate function by
\begin{align*}
\iota:\sigma(N)\to\mathbb C,\quad z\mapsto z.
\end{align*}
Then the displayed identity is precisely
\begin{align*}
N=\int_{\sigma(N)}\iota\,dE.
\end{align*}
[/step]
[step:Use the spectral integral norm bound to obtain the upper estimate]
Since $\iota$ is a bounded Borel function on the compact set $\sigma(N)$, the norm estimate for spectral integrals from the spectral theorem, [citetheorem:8412], applies to $\iota$ and gives
\begin{align*}
\|N\|_{\mathcal L(H)} = \left\|\int_{\sigma(N)} \iota\,dE\right\|_{\mathcal L(H)} \le \|\iota\|_\infty.
\end{align*}
By the definition of the supremum norm on $C(\sigma(N))$,
\begin{align*}\|\iota\|_\infty = \sup_{\lambda \in \sigma(N)} |\iota(\lambda)| = \sup_{\lambda \in \sigma(N)} |\lambda|.\end{align*}
Thus
\begin{align*}
\|N\|_{\mathcal L(H)}
\le
\sup_{\lambda \in \sigma(N)} |\lambda|.
\end{align*}
[guided]
The spectral theorem has converted the operator $N$ into an operator-valued integral of the scalar function $\iota(z)=z$. This is useful because the norm of a spectral integral is controlled by the supremum norm of the function being integrated.
More precisely, the function
\begin{align*}
\iota: \sigma(N) \to \mathbb C
\end{align*}
is defined by $\iota(z)=z$. Since $\sigma(N)$ is compact for a bounded operator on a complex [Banach space](/page/Banach%20Space), $\iota$ is bounded; since $\iota$ is continuous, it is Borel measurable. The spectral integral norm estimate from the spectral theorem, [citetheorem:8412], therefore applies to $\iota$ and gives
\begin{align*}
\left\|\int_{\sigma(N)} \iota\,dE\right\|_{\mathcal L(H)} \le \|\iota\|_\infty.
\end{align*}
Using the representation of $N$ from the spectral theorem, the left-hand side is $\|N\|_{\mathcal L(H)}$. On the right-hand side, the supremum norm is computed directly:
\begin{align*}\|\iota\|_\infty = \sup_{\lambda \in \sigma(N)} |\iota(\lambda)| = \sup_{\lambda \in \sigma(N)} |\lambda|.\end{align*}
Therefore
\begin{align*}
\|N\|_{\mathcal L(H)}
\le
\sup_{\lambda \in \sigma(N)} |\lambda|.
\end{align*}
This proves that the operator norm cannot exceed the largest spectral modulus.
[/guided]
[/step]
[step:Use spectral support to force the reverse estimate]
Let $\lambda_0 \in \sigma(N)$ and let $\varepsilon>0$. Define the open neighbourhood
\begin{align*}
U_\varepsilon := \{z \in \sigma(N) : |z-\lambda_0|<\varepsilon\}.
\end{align*}
By the spectral theorem and the support characterisation for normal operators, [citetheorem:8414], we have
\begin{align*}
\sigma(N)=\operatorname{supp}E.
\end{align*}
where $\operatorname{supp}E$ denotes the support of the projection-valued measure $E$. Hence $E(U_\varepsilon)\ne 0$, because $U_\varepsilon$ is a relatively open neighbourhood of the support point $\lambda_0$.
Choose $x \in \operatorname{Range}(E(U_\varepsilon))$ with $\|x\|_H=1$, where $\operatorname{Range}(E(U_\varepsilon))$ denotes the range of the projection $E(U_\varepsilon)$. Since $E(U_\varepsilon)x=x$, the spectral integral representation gives
\begin{align*}Nx = \int_{\sigma(N)} z\,dE(z)x = \int_{U_\varepsilon} z\,dE(z)x.\end{align*}
The same projection reduction gives
\begin{align*}\lambda_0 x = \int_{U_\varepsilon} \lambda_0\,dE(z)x.\end{align*}
Subtracting these two spectral integrals and applying the spectral integral norm estimate, [citetheorem:8412], to the bounded Borel function $z\mapsto z-\lambda_0$ restricted to $U_\varepsilon$, we obtain, with $I_H:H\to H$ denoting the identity operator,
\begin{align*}
\|(N-\lambda_0 I_H)x\|_H \le \sup_{z\in U_\varepsilon}|z-\lambda_0|\,\|x\|_H \le \varepsilon.
\end{align*}
Therefore
\begin{align*}
|\lambda_0| = \|\lambda_0 x\|_H \le \|Nx\|_H+\|(N-\lambda_0 I_H)x\|_H \le \|N\|_{\mathcal L(H)}+\varepsilon.
\end{align*}
Letting $\varepsilon \downarrow 0$ gives
\begin{align*}
|\lambda_0| \le \|N\|_{\mathcal L(H)}.
\end{align*}
Since $\lambda_0\in\sigma(N)$ was arbitrary,
\begin{align*}
\sup_{\lambda\in\sigma(N)}|\lambda| \le \|N\|_{\mathcal L(H)}.
\end{align*}
[/step]
[step:Convert the supremum over the spectrum into a maximum]
Combining the two estimates gives
\begin{align*}\|N\|_{\mathcal L(H)} = \sup_{\lambda\in\sigma(N)}|\lambda|.\end{align*}
The spectrum $\sigma(N)$ is a nonempty compact subset of $\mathbb C$ by the standard compactness and nonemptiness theorem for spectra in a unital complex Banach algebra, applied to the element $N\in\mathcal L(H)$. The function
\begin{align*}
\sigma(N) \to \mathbb R, \qquad \lambda \mapsto |\lambda|
\end{align*}
is continuous, so the extreme value theorem applies on the nonempty compact set $\sigma(N)$ and gives that it attains its supremum. Hence
\begin{align*}\sup_{\lambda\in\sigma(N)}|\lambda| = \max_{\lambda\in\sigma(N)}|\lambda|.\end{align*}
Therefore
\begin{align*}
\|N\|_{\mathcal L(H)} = \max_{\lambda\in\sigma(N)}|\lambda|.
\end{align*}
This is the desired norm equality.
[/step]
