Looking for spectrum of an operator Let's define the following Hilbert space on the field $\mathbb{C}$:
$$ \ell^2(\mathbb{Z}) = \{(...,x_{-1},x_0,x_1,...): \sum_{n \in \mathbb{Z}} |x_n|^2 < \infty   \}  $$
with the inner product $\langle x,y \rangle = \sum_{n \in \mathbb{Z}} x_n \overline{y_n}$ and define operator $T \in L(\ell^2(\mathbb{Z}))$ by
$$T(...,x_{-1},x_0,x_1,...) = (...,\frac{x_{-2} + x_0}{2},\frac{x_{-1} + x_1}{2},\frac{x_{0} + x_2}{2},...). $$
I would like to classify the spectrum of this operator.
First of all, this operator is self-adjoint, $\lVert T \rVert = 1$ so from few properties of spectra of such operators we know that:

*

*$\sigma(T)$ is real,

*$\sigma(T) \subset [-1,1]$

*$r_{\sigma}(T) = \lVert T \rVert$ hence at least one of $-1,1$ belong to spectrum,

*residual spectrum of this operator is empty.

I wanted now to check what is the continuous spectrum and the point spectrum of this operator, since then $\sigma(T) = \sigma_P(T) \cup \sigma_C(T)$.
Let's fix $\lambda \in \sigma_P(T)$. We look for such $x \neq 0$ that
$$Tx - \lambda x = 0.$$
I noticed the following: suppose that there exists (a finite) $N \in \mathbb{Z}$ such that $N$ is the smallest integer satisfying $x_n \neq 0$. Then, from the definition of $T$ we get
$$\frac{x_{N-2} + x_{N}}{2} = \lambda x_{N-1}$$
and since by assumption $x_{N-2} = 0 = x_{N-1}$ hence
$$\frac{0 + x_N}{2} = \lambda 0$$
which means that $x_N = 0$; contradiction.
Similarly, if there exists (a finite) $N \in \mathbb{Z}$ being the largest integer satisfying $x_N \neq 0$ then
$$\frac{x_N + x_{N+2}}{2} = \lambda x_{N+1}$$
and since $x_{N+1} = 0 = x_{N+2}$ hence $x_{N} = 0$; contradiction.
This means that we cannot have a situation, where $x$ is equal $0$ starting from some point.
Unfortunately here I'm stuck. I would appreciate if anyone could share some thoughts here.
Thanks!
 A: The solution is straightforward but the details tedious.
Suppose such a (real) eigenvalue $\lambda$ exists. Then $x$ satisfies the difference equation
$x_{k+2}=2\lambda x_{k+1}-x_k$ in both 'directions' (positive & negative $k$).
The roots of the difference equation are ${ 1\over 2}(2\lambda \pm \sqrt{4\lambda^2-4})$, or $\lambda \pm \sqrt{\lambda^2-1}$.
Since $|\lambda| \le 1$ we see that the real part of the roots is $\lambda$.
If $(x_0,x_1) \ne0$, then if $\lambda \ge 0$ then $x_k \not\to 0$ and if $\lambda <0$ then $x_{-k} \not\to 0$, and so we must have
$(x_0,x_1) =0$ and so $x=0$.
In particular, no such $\lambda$ can exist. Hence $T$ has no point spectrum.
I claim that the continuous spectrum is $[-1,1]$. Since we know the spectrum is closed, it is sufficient to show that a dense subset of $[-1,1]$ is in the continuous spectrum.
It is straightforward to show that if $\lambda$ is not in the point or residual spectrum then it is in the continuous spectrum iff for any $\epsilon>0$ there is a unit vector $x$ such that $\|(T-\lambda I)v\| < \epsilon$.
Let $T_n$ be the $n\times n$ dimensional tridiagonal Toepliz matrix that 'corresponds' to $T$.
From https://mathoverflow.net/a/188489/31729 we see that the eigenvalues are $\lambda_{n,k} = \cos ({k \pi \over n+1})$, with $k=1,...,n$. Let $v_{n,k}$ be a corresponding unit eigenvector.
It is straightforward to see that $\lambda_{n,k}$ are dense in $[-1,1]$.
Very roughly speaking, the idea is to 'stack' a finite number of the $v_{n,k}$ eigenvectors together so that we can find a unit $v$ with $\|(T-\lambda I)v\| < \epsilon$. Then we see that the $\lambda_{n,k}$ are in the continuous spectrum which finishes the proof.
Define the permutation $P$ on $\mathbb{R}^n$ by $P e_k = e_{n-k+1}$, note that $P^2=I$ and it is straightforward to check that if $T_n v_{n,k} = \lambda_{n,k} v_{n,k}$ then $T_n Pv_{n,k} = \lambda_{n,k} Pv_{n,k}$. In particular, 'reversing' the eigenvector $v_{n,k}$ gives another eigenvector.
Define the linear injection $\sigma_k:\mathbb{R}^n \to l_2$ by $\sigma_k(e_j) = e_{k+j-1}$ where the $e_k$ are the usual basis vectors in their respective spaces.
Note that $T \sigma_k(x) = \sigma_k(T_n x) + {1 \over 2} x_1 e_{k-1}+ {1 \over 2} x_n e_{k+n}$ and in particular,
$(T -\lambda I)\sigma_k(v_{n,k}) = {1 \over 2} [v_{n,k}]_1 e_{k-1}+ {1 \over 2} [v_{n,v}]_n e_{k+n}$ and
\begin{eqnarray}
(T -\lambda I) \sigma_{k+n+1}(-Pv_{n,k}) &=&  {1 \over 2} [-Pv_{n,k}]_1 e_{k+n}+ {1 \over 2} [-Pv_{n,k}]_n e_{k+2n+1} \\
&=& - {1 \over 2} [v_{n,k}]_n e_{k+n}- {1 \over 2} [v_{n,k}]_1 e_{k+2n+1}
\end{eqnarray}
Adding together gives
$(T -\lambda I) (\sigma_k(v_{n,k})+ \sigma_{k+n+1}(-Pv_{n,k})) = {1 \over 2} [v_{n,k}]_1 e_{k-1} - {1 \over 2} [v_{n,k}]_1 e_{k+2n+1}$.
Let $v_m = \sum_{i=0}^m \sigma_{k+i(n+1)} ( (-P)^{i} v_{n,k})$ (that is, $m+1$ copies of $v_{n,k}$ or $-P v_{n,k}$ stacked together but separated by a $0$). It is not too hard to see (assuming I have not made an error) that
$(T -\lambda I) v_m = {1 \over 2} [v_{n,k}]_1 e_{k-1} + {1 \over 2} (-1)^m [v_{n,k}]_{l_m} e_{k+n+ m(n+1)}$,
where $l_m$ is $1$ if $m$ is odd and $n$ is $m$ is even.
Note that $\|v_m\| = \sqrt{m+1}$, and we see that
$\lim_{m \to \infty} {1 \over \sqrt{m+1}} \|(T -\lambda I) v_m\| = 0$.
