In numerical analysis, the discrete Laplacian operator $\Delta$ on $\ell^2({\bf Z})$ can be written in terms of the shift operator
$\Delta=S+S^*-2I$
where $S$ is the right shift operator. Since it is self-adjoint, the spectrum should be in the real line. On the other hand, simple calculation show that one can write the operator $\Delta-\lambda$ as the following
$\Delta-\lambda=-\frac{1}{\mu}(S-\mu)(S^*-\mu)$ (*)
where $\mu$ is such that $\mu+\frac{1}{\mu}=2+\lambda$.
(*) can give the intuition that the spectrum
$\sigma(\Delta)=\{\mu+\frac{1}{\mu}:|\mu|=1\}-2$
However, for proving it, (*) seems not work.
Here are my questions:
Is $\sigma(\Delta)=\{\mu+\frac{1}{\mu}:|\mu|=1\}-2$ true?
Does the fact that $\sigma(S)$ is purely continuous imply that $\sigma(\Delta)$ is also continuous?