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.
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.