Let $A:\ell^2(\mathbb Z) \longmapsto \ell^2(\mathbb Z)$ be defined as following $A(x)_{n} = x_{n−1} − x_{n+1}$
Find A's spectrum. $\sigma(A)=\sigma_c(A)=\sigma_{ess}(A)=[-2i,2i]$
Operator is not self-adjoined, $A=S_r-S_l$ and $||A||=2$.
The space $\ell^2(\mathbb Z)$ is isomorphic with $L^2(\mathbb T)$, via the identification $\delta_n\longmapsto (t\mapsto e^{in\pi\,t})$.
$S_r$ mapped to $M_z$, $S_l$ mapped to $M_z^{-1}$ So the spectrum of $S_r-S_l$ is the same as that of $M_{z^{-1} - z}$.
I don't understand why $$\sigma_c(M_f)=\sigma(M_f)=\overline{f(\mathbb R)}.$$ Why for a multiplication operator, its spectrum is the closure of the range? And spectrum continuous?
$$\sigma(S_r-S_l)=\overline{\{z-z^{-1}:\ z\in\mathbb T\}} =\overline{\{z -\bar z:\ |z|=1\}}=\overline{\{2 \text{Im}\,z:\ |z|=1\}}$$ How to prove from that that $\sigma_c=[-2i;2i]$?