I am reading this expository work Numerical Range of Operators by Paul Skoufranis (see this). Here is the theorem and proof I'm working through:
Theorem 2.11 Let $N \in B(\mathcal{H})$ be a normal operator. Then $\overline{W(N)} = conv(\sigma(N))$
Proof: By the Weyl-von Neumann Berg-Theorem, there exists a diagonal normal operator $D$ such that $N$ and $D$ are approximately unitarily equivalent and $\sigma (N) = \sigma (D)$. Therefore, by theorem 2.10, $\overline{W(N)} = \overline{W(D)}$. However, by example 2.7, $W(D)$ is the convex hull of the eigenvalues $\{a_n\}_{n \ge 1}$ of $D$. Since the eigenvalues of $D$ are dense in the spectrum of $D$, it is clear that $$\overline{W(D)} = \overline{conv (\{a_n\}_{n \ge 1})} = \overline{conv(\sigma (N))} = conv(\sigma (N))$$ as $\sigma(N)$ is compact subset of $\Bbb{C}$ so its convex hull is closed. Hence the result is complete.
First, does anyone know of a nice statement of the Weyl-von Neumann Berg-Theorem and proof of it (references would be appreciated). Second, why are the eigenvalues of $D$ dense in $\sigma (D)$?
For the first question, see Weyl–von Neumann theorem and references therein. For the second, if $D$ is a diagonal operator on a Hilbert space with eigenvalues $(\lambda_i)_i\in I$, then its spectrum $\sigma(D)$ is exactly the closure of the set $\Lambda\subset \mathbb C$ of its eigenvalues. Indeed, as the spectrum of a bounded operator is closed, $\sigma(D)\supseteq \overline\Lambda$. For the other inclusion, let $(e_i)_{i\in I}$ be its eigenbasis. If $\lambda\not\in \overline{\Lambda}$, then the operator defined by $$ R_\lambda e_i:= \frac{1}{\lambda_i-\lambda}\cdot e_i $$ is clearly the inverse of $D-\lambda\cdot\mathrm{id}$, so $\lambda\not\in\sigma(D)$.