Summary: Spectrum vs. Numerical Range This thread is only Q&A!
Given a Hilbert space $\mathcal{H}$.
Consider operators:
$$T:\mathcal{D}(T)\to\mathcal{H}$$
Denote for shorthand:
$$\Omega\subseteq\mathbb{C}:\quad\langle\Omega\rangle:=\operatorname{conv}(\Omega)$$
Then one has:
$$\langle\sigma(T)\rangle=\overline{\mathcal{W}(T)}$$
Does this hold true?
 A: Bounded Operators
For bounded operators:
$$A\in\mathcal{B}(\mathcal{H}):\quad\langle\sigma(A)\rangle=\overline{\mathcal{W}(A)}\subseteq\mathbb{C}$$
For nilpotent operators:
$$N^{K+1}=0:\quad\langle\sigma(N)\rangle=(0)\subseteq\overline{\mathcal{W}(N)}$$
As standard example:
$$\left\langle\sigma\begin{pmatrix}0&1\\0&0\end{pmatrix}\right\rangle=(0)\subsetneq\frac12\mathbb{D}=\overline{\mathcal{W}\begin{pmatrix}0&1\\0&0\end{pmatrix}}$$
But inclusion always holds!
Unbounded Operators
For normal operators:
$$N^*N=NN^*:\quad\langle\sigma(N)\rangle=\overline{\mathcal{W}(N)}\subseteq\mathbb{C}$$
For symmetric operators:
$$S\subseteq\overline{S}\subsetneq S^*:\quad\langle\sigma(S)\rangle\nsubseteq\mathbb{R}\quad\overline{\mathcal{W}(S)}\subseteq\mathbb{R}$$
For unclosed operators:
$$T\subsetneq\overline{T}=T^{**}:\quad\langle\sigma(T)\rangle=\mathbb{C}\supseteq\overline{\mathcal{W}(T)}$$
It may happen even:*
$$S=\overline{S}\subsetneq S^*:\quad\langle\sigma(S)\rangle=\mathbb{C}\supsetneq\mathbb{R}\supseteq\overline{\mathcal{W}(S)}$$
And even worse also:*
$$T=\overline{T}=T^{**}:\quad\langle\sigma(T)\rangle=\varnothing\subsetneq\overline{\mathcal{W}(T)}\subseteq\mathbb{C}$$
So inclusion mostly fails!
References
For examples see: Closed Operators
For the proof see: Normal Operators, Spectral Measures
