Problem
Given a Hilbert space $\mathcal{H}$.
Denote for readability: $$\Omega\subseteq\mathbb{C}:\quad\|\Omega\|:=\|\omega\|_{\omega\in\Omega}:=\sup_{\omega\in\Omega}|\omega|$$
Denote for shorthand: $$\Omega\subseteq\mathbb{R}:\quad\Omega_+:=\sup_{\omega\in\Omega}\omega\quad \Omega_-:=\inf_{\omega\in\Omega}\omega$$
For bounded operators: $$A\in\mathcal{B}(\mathcal{H}):\quad\sigma(A)\subseteq\overline{\mathcal{W}(A)}$$
For normal operators: $$N^*N=NN^*:\quad\|\sigma(N)\|=\|\mathcal{W}(N)\|$$
But one has even: $$H=H^*:\quad\sigma(H)_\pm=\mathcal{W}(H)_\pm$$
How can I prove this?
Attempt
The argument goes as: $$|\langle N\hat{\varphi},\hat{\varphi}\rangle|\leq\|N\|\cdot\|\hat{\varphi}\|^2=\|\sigma(N)\|$$ (It exploits normality!)
Example
As standard example: $$\left\langle\sigma\begin{pmatrix}0&1\\0&0\end{pmatrix}\right\rangle=(0)\quad\overline{\mathcal{W}\begin{pmatrix}0&1\\0&0\end{pmatrix}}=\tfrac12\mathbb{D}$$
(It exploits nilpotence!)