Let $\ell^2 =\ell^2(\mathbb{Z})$. Choose $\theta \in ]0,1[$ and set:
$Tx=(\theta x_{n-1} +(1-\theta)x_{n+1})_{n\in \mathbb{Z}}$
for each $x=(x_n)_{n\in \mathbb{Z}}\in \ell^2$ (thus $T$ is a convex combination of the right and left shift operators).
It is easy to prove that, for every $\theta$, $T$ is a bounded linear operator of $\ell^2$ into itself, that $\lVert T\rVert =1$ and that $T$ is selfadjoint iff $\theta =\ frac{1}{2}$. Moreover $T$ is not compact: in fact, if $e^m:=(\delta_n^m)$ (so $e^m$ is a vector of the canonical base of $\ell^2$), one has:
$|Te^m -Te^p|^2=\begin{cases} 0 &\text{, if } p=m \\ \theta^2 +(1-\theta)^2+1 &\text{, if } m=p+2 \text{ or } p=m+2 \\ 2\theta^2+2(1-\theta)^2 &\text{, otherwise} \end{cases} \; ,$
thus $|Te^m-Te^p|^2> \theta^2+(1-\theta)^2>0$ for $m\neq p$; therefore the sequence $\{ Te^m\}_{m\in \mathbb{N}}$ does not contain any Cauchy's subsequence.
The problem is:
I am not able to find the spectrum of $T$.
About the eigenvalues, the only thing I know for sure is that $1$ is not in the point spectrum of $T$ for any value of $\theta$: in fact if $1$ were in the point spectrum $\sigma_P(T)$, then the eigenvectors would satisfy the linear recurrence:
$x_n=\theta x_{n-1}+(1-\theta) x_{n+1} \; ,$
hence they have to be sequences of the type:
$x_n=A \left( \frac{\theta}{1-\theta}\right)^n +B$
($A,B$ suitable constants); but a sequence like this doesn't belong to $\ell^2$ except in the trivial case $A=B=0$, which however doesn't give a valid eigenvector. Therefore $1\notin \sigma_P(T)$.
But now, what about other eigenvalues? And what about the residue and continuous spectra of $T$?
Any hint is welcome.