Consider the family of operators $\mathcal F$ on $\mathcal H = \ell^2(\Bbb Z)$ given by $\mathcal F = \{A_k\}_{k=1}^\infty \cup \{A_\infty\}$ where $$A_k e_n = \begin{cases}e_{n+1} & n \ne 0\\ \frac{e_{n+1}}{k} & n = 0 \end{cases}$$ and $1/\infty := 0$. What are the spectra of operators in $\mathcal F$? Is the spectrum "continuous"?
I presume this amounts to finding $\sigma(A_k)$ for $k\in \Bbb N$, and $\sigma(A_\infty)$ separately, and taking the limit (in some sense) as $k\to\infty$ to see that the spectrum (as a map from $\mathcal B(\mathcal H)$ to compact subsets of $\Bbb C$ is not continuous.) For $x = \sum_n x_n e_n \in \mathcal H$ and $\lambda \in \Bbb C$, we have $$(A_k - \lambda I) x = \left(\frac{x_0}{k} - x_1\lambda\right)e_1 + \sum_{n\ne 0} (x_n - \lambda x_{n+1})e_{n+1}$$ and $$(A_\infty - \lambda I) x = -x_1\lambda e_1 + \sum_{n\ne 0} (x_n - \lambda x_{n+1})e_{n+1}$$ How should I proceed? I do not have a guess for the spectra of operators in $\mathcal F$. Thank you!
Let $S$ denote the shift operator defined as $Se_n=e_{n+1}.$ Let $M_k$ be the operator defined by $$M_ke_n=\begin{cases} e_n & n\ge 1\\ ke_n & n\le 0 \end{cases} $$ Then $A_k=M_kSM_k^{-1}.$ Therefore $\sigma(A_k)=\sigma(S).$ It is well known that $\sigma(S)=\{z\,:\,|z|=1\}.$ Hence $\sigma(A_k)=\{z\,:\,|z|=1\}$ for $k\in \mathbb{N}.$
For the operator $A_\infty$ we have the orthogonal decomposition $\ell^2(\mathbb{Z})=\ell^2(\mathbb{N})\oplus\ell^2(-\mathbb{N}\cup\{0\}),$ where both subspaces are invariant for the operator $A_\infty.$ Therefore $\sigma(A_\infty)$ is equal to the union of the spectra of $A_\infty$ restricted to each of the subspaces. The operator $A_\infty$ restricted to the first subspace is the unilateral shift whose spectrum coincides with $\{z\,:\,|z|\le 1\}.$ Similarly the operator $A_\infty$ restricted to the second subspace is the adjoint of the unilateral shift, hence its spectrum also coincides with $\{z\,:\,|z|\le 1\}.$ Summarizing $\sigma(A_\infty)=\{z\,:\,|z|\le 1\}.$