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Find the Spectrum of the following operator in $l^{2}$ : $$ (Ax)_{n}= x_{n-2}+x_{n+2}-2x_{n}$$

I found this problem in a list of many exercises for my course, and I am quite lost. I have solved many excersices with operators with functions, and a few with sequences, so Im not that strong with the later. This time I don't know how to work with the $(Ax)_{n}$, should I proceed with something like $((Ax)_{n}-\lambda)=y$ or how should I proceed?. I feel this problem it is not that difficult, but I am unsure of what to do. Thanks in advance for the help.

AdrinMI49
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  • How is the operator defined for $n=0,1$? Also, if ($\lambda$, $x$) is an eigenpair, then $x_{n-2} + x_{n+2} - 2x_n = \lambda x_n$ and you can analyze the solutions of this difference equation for various $\lambda$. Here is a related problem https://math.stackexchange.com/questions/139735/eigenstructure-of-discrete-laplacian-on-uniform-grid – whpowell96 Nov 27 '23 at 01:28
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    Is the space $l^2(\mathbb{N})$ or $l^2(\mathbb{Z})$? – user8469759 Nov 27 '23 at 01:30
  • I think the space is $l^2(\mathbb{Z})$ – AdrinMI49 Nov 27 '23 at 05:54
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    The space can be decomposed into two orthogonal invariant subspaces $\ell^(2\mathbb{F})$ and $\ell^2(2\mathbb{F}+1),$ where $\mathbb{F}=\mathbb{Z}$ or $\mathbb{F}=\mathbb{N}.$ The operator $A$ restricted to either space leads to two operators unitarily equiavlent to $Bx_n=x_{n-1}+x_{n+1}-2x_n.$ The spectrum of the latter can be determined by applying the shift operator $(Sx)n=x{n+1},$ as $B=S+S^*-2I.$ – Ryszard Szwarc Nov 27 '23 at 11:18
  • Hi, @RyszardSzwarc I have been trying this problem for the past week, with what you told me, but Im not getting anywhere, could say some extra things of how to solve it? So maybe that way I can understand better the idea of the solution? Thank you very much – AdrinMI49 Dec 03 '23 at 21:05

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For $\ell^2(\mathbb{Z})$ consider the operator $(Sx)_n=x_{n+2}.$ It is a unitary operator, hence $\sigma(S)\subset \{z\,:\,|z|=1\}.$ Actually $\sigma(S)= \{z\,:\,|z|=1\}.$ Indeed, for $|u|=1$ let $$v_n=\begin{cases}u^n & 1\le n\le N\\ 0 & {\rm otherwise}\end{cases}$$ Then $$(Sv)_n=\begin{cases}u^n & -1\le n\le N-2\\ 0 & {\rm otherwise}\end{cases}$$ Thus $$\|(u^2I-S)v\|_2=2,\quad \|v\|_2=\sqrt{N}$$ Hence $${\|(u^2I-S)v\|_2\over \|v\|_2}={2\over \sqrt{N}}\underset{N}{\longrightarrow} 0 $$ which shows that $u^2\in\sigma(S).$ Any number $z$ in the unit circle can be represented as $z=u^2,$ $|u|=1,$ thus $ \{z\,:\,|z|=1\}\subset \sigma(S).$ Next observe that $A=S^{-1}+S-2I,$ i.e. $A=f(S),$ where $f(z)=z+z^{-1}-2.$ The function $f$ is holomorphic on the unit circle therefore $$\sigma(A)=\sigma(f(S))=f(\sigma(S))=f\{z\,:|z|=1\}=[-4,0]$$ Since the function $f(z)$ is pretty simple, the second (crucial) equality can be proved straightforward.

Remark The spectrum is the same for $\ell^2(\mathbb{N}_0)$ but this time we use the fact $A=S^*+S-2I$ and $\sigma(S)=\{z\,:\,|z|\le 1\}.$ The latter follows from the fact that any sequence $\{u^n\}_{n=0}^\infty$, $|u|<1,$ is an eigenvector of $S$ corresponding to the eigenvalue $u^2.$ The operator $A$ is selfadjoint.

Ryszard Szwarc
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