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!)
 A: Let $H=H^{\star}$ be a bounded selfadjoint operator and let $\rho=\inf_{\|x\|=1}(Hx,x)$. Then $\rho_{\epsilon}(x,y)=((H-\rho I+\epsilon I)x,y)$ is an inner product on $\mathcal{H}$ for any $\epsilon > 0$. So,
$$
       |\rho_{\epsilon}(x,y)| \le \rho_{\epsilon}(x,x)^{1/2}\rho_{\epsilon}(y,y)^{1/2}.
$$
Letting $\epsilon \downarrow 0$ gives
$$
       |((H-\rho I)x,y)| \le ((H-\rho I)x,x)^{1/2}((H-\rho I)y,y)^{1/2}.
$$
Let $y = (H-\rho I)x$ in order to obtain
\begin{align}
     \|(H-\rho I)x\|^{2} & \le ((H-\rho I)x,x)^{1/2}((H-\rho I)(H-\rho I)x,(H-\rho I)x)^{1/2} \\
       & \le ((H-\rho I)x,x)^{1/2}\|H-\rho I\|^{1/2}\|(H-\rho I)x\|
\end{align}
Hence,
$$
    \|(H-\rho I)x\| \le \|H-\rho I\|^{1/2}((H-\rho I)x,x)^{1/2}.
$$
By the definition of $\rho$, there is a sequence of unit vectors $\{ x_{n} \}$ such that $((H-\rho I)x_{n},x_{n})\rightarrow 0$. Therefore,
$$
                \lim_{n}\|(H-\rho I)x_{n} \|=0,
$$
which means $\rho \in \sigma(H)$ because it is in the point spectrum or approximate point spectrum of $H$. By the way, you also see that a unit vector $x$ minimizes $(Hx,x)$ over all unit vectors iff $Hx = \rho x$.
The opposite direction is not so difficult, which is to say that every $\rho'$ such that $\rho ' < \rho=\inf_{\|x\|=1}(Hx,x)$ is in the resolvent of $H$. I've shown that to you before.
A: 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.
