Functional Analysis – Spectrum of the Discrete Laplacian Operator

functional-analysisnumerical methodsspectral-theory

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?

Best Answer

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.

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