Spectrum: Boundary 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!)

Define the value:
$$\omega:=\omega_-:=\mathcal{W}(H)_-$$
By linearity of range:
$$\mathcal{W}(H-\omega)=\mathcal{W}(H)-\omega\geq0$$
For bounded operators:
$$\mathcal{W}(H-\omega)\geq0\implies\sigma(H-\omega)\geq0$$
For normal operators:
$$\sigma(H-\omega)_+=\|\sigma(H-\omega)\|=\|\mathcal{W}(H-\omega)\|=\mathcal{W}(H-\omega)_+$$
By linearity of both:
$$\sigma(H)_+=\sigma(H-\omega)_++\omega=\mathcal{W}(H-\omega)_++\omega=\mathcal{W}(H)_+$$
$$\sigma(H)_-=-\sigma(-H)_+=-\mathcal{W}(-H)_+=\mathcal{W}(H)_-$$
Concluding the assertion.
