5
$\begingroup$

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:

  1. Is $\sigma(\Delta)=\{\mu+\frac{1}{\mu}:|\mu|=1\}-2$ true?

  2. Does the fact that $\sigma(S)$ is purely continuous imply that $\sigma(\Delta)$ is also continuous?

$\endgroup$
8
$\begingroup$

Yes. Let $f(z) = z + z^{-1} - 2$. Since $\Delta = f(S)$, $\sigma(\Delta) = f(\sigma(S))$. And yes, $\sigma(S) = \{z: |z| = 1\}$. Now note that if $z = e^{i\theta}$, $f(z) = 2 \cos(\theta) - 2$, so $\sigma(\Delta) = [-4, 0]$.

The spectrum of $\Delta$ is all continuous: it is easy to see that $\Delta$ has no eigenvalues in $\ell^2({\mathbb Z})$. In fact, it is absolutely continuous, and this follows from the fact that the inverse image under $f$ of any set of measure 0 in $\mathbb R$ has measure 0 in the unit circle.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.