Numerical range and essential spectrum The following question is inspired by the proof of Proposition $\S 40.4$ (the Glimm's Lemma) in A Course in Operator Theory written by John Conway.
Let $H$ be a separable Hilbert space and $\mathcal{A}\subseteq B(H)$ be a unital separable $C^*$-Algebra that contains $\mathcal{K}(H)$. Fix a self-adjoint $A\in\mathcal{A}$, and we use $\pi$ to denote the canonical quotient mapping from $B(H)$ to $B(H)/\mathcal{K}(H)$, and $\sigma_e(A)$ (the essential spectrum) to denote the spectrum of $\pi(A)$. If we suppose $\alpha = \min\sigma_e(A)$ and $\beta = \max\sigma_e(A)$. My questions are below:

*

*Is it true that there exists an orthonormal sequence $\{e_n\}\subseteq H$ such that a subsequence of $\{\langle A e_n, e_n\rangle\}$ converges to $\alpha$ or $\beta$?


*Given a sequence of unit vectors $\{h_n\}\subseteq H$ which weakly converges to zero, there is always a subsequence of $\{\langle A h_n, h_n\rangle\}$ that is converging and we use $\{h_n\}-\operatorname{LIM}(A)$ to denote the limit of that subsequence. Once $\{h_n\}$ is given, $\{h_n\}-\operatorname{LIM}$ is a well defined states on $\mathcal{A}$ and we use $\Sigma_1$ to denote the set of states having this form. If we define $\Sigma_1(A) = \{\psi(A): \psi\in\Sigma_1\}$, is $\Sigma_1(A) = [\alpha, \beta]$?
It is shown in this post that whenever $A$ is normal, the numerical range of $A$, $W(A)$ defined as the set $\{\langle Av, v\rangle: \|v\|=1\}$, satisfies that $\overline{W(A)} = \overline{\operatorname{conv}(\sigma(A))}$. For the second questions, if we define:
$$
W_e(A) = \Big\{\{h_n\}-\operatorname{LIM}(A): h_n\overset{w}{\rightarrow}0\text{  and  }\|h_n\|=1 \Big\}
$$
then the ultimate question is that if $\overline{W_e(A)} = \overline{\operatorname{conv}(\sigma_e(A))}$. I am not sure where I need the separability of $\mathcal{A}$ either, and also wonder if $\overline{W_e(A)} = \overline{\operatorname{conv}(\sigma_e(A))}$ is true when $H$ is not separable.
A result that may help with the second question is $\S 7.8$ in the same textbook:

If $\mathfrak{A}$ is a $C^*$-Algebra and $a\in\mathfrak{A}$ is self-adjoint with $\alpha = \min\sigma(a), \beta = \max\sigma(a)$, then $[\alpha, \beta] = \{\phi(a): \phi\text{  is a state on  }\mathfrak{A}\}$


 A: $\def\e{\varepsilon}$

*

*Yes. One characterization of the essential spectrum is that $E_A(\alpha-\e,\alpha+\e)$ is infinite for any $\e>0$. Then we can choose an orthonormal sequence $\{e_n\}$ with $e_n\in E_A(\alpha-\frac1n,\alpha+\frac1n)$. Since
$$
\big(\alpha-\frac1n\big)E_A(\alpha-\frac1n,\alpha+\frac1n)\leq A\,E_A(\alpha-\frac1n,\alpha+\frac1n)\leq (\alpha+\frac1n)\,E_A(\alpha-\frac1n,\alpha+\frac1n),
$$
we have
\begin{align}
\alpha-\frac1n
&=\langle (\alpha-\frac1n)E_A(\alpha-\frac1n,\alpha+\frac1n)e_n,e_n\rangle\\[0.3cm]
&\leq \langle Ae_n,e_n\rangle\\[0.3cm]
&\leq \langle (\alpha+\frac1n)E_A(\alpha-\frac1n,\alpha+\frac1n)e_n,e_n\rangle\\[0.3cm]
&=\alpha+\frac1n.
\end{align}
Thus $\langle Ae_n,e_n\rangle\to\alpha$. The same can be done for $\beta$ and for any other $\lambda\in\sigma_e(A)$.


*You say

Once $\{h_n\}$ is given, $\{h_n\}-\operatorname{LIM}$ is a well defined states on $\mathcal{A}$

As stated, that's not true because the sequence $\{\langle Ah_n,h_n\rangle\}$ can have many accumulation points. For instance, fix an orthonormal basis $\{h_n\}$ and let $A$ be the selfadjoint operator that does $Ae_{2n}=e_{2n}$, $Ae_{2n-1}=2e_{n-1}$. Then both $1$ and $2$ are accumulation points for the sequence $\{\langle Ah_n,h_n\rangle\}$. Tweaking this idea one can get even an interval as the set of accumulation points.
In any case, the answer is yes. By 1) we have that both $\alpha$ and $\beta$ can be obtained in this fashion. Because the numerical range is convex, for each $t\in[\alpha,\beta]$ there exists a state $\phi_t$ on the Calkin algebra such that $\phi(\pi(A))=t$. The state $\phi\circ\pi$ on $B(H)$ is zero on $K(H)$ and hence Glimm's Lemma applies to give us an orthonormal sequence $\{h_n\}$ with $t=\phi_t(\pi(A))=\lim\langle Ah_n,h_n\rangle$.
