I am looking for an example of a bilateral weighted shift $T\in B(\ell^2(\mathbb{Z})) $ with $\sigma(T) = \mathbb{T}$, which is the unit circle.
NOTE: what I'm looking for a bilateral weighted shift such that $Te_n=t_ne_{n+1}$ with $t_n$ a non-constant.
I know that a bilateral shift $S$ with $Se_n=e_{n+1}$, $n\in \mathbb{Z}$, has spectrum $\sigma(S)=\mathbb{T}$.
So I started trying out things.
Here are some ideas:
Let $T$ be defined by $Te_n =t_ne_{n+1}$, where
$ t_n= \begin{cases} \frac{1}{1+n},& \text{if } n\geq 0\\ \frac{1}{1-n}, & \text{if } n <0 \end{cases} $
First we know that the spectrum is always closed.
But I'm not sure how to find out if the spectrum of this is $\mathbb{T}$.
To show $\sigma(T) \subset \mathbb{T}$, some facts that might be needed are:
The spectrum is always closed.
Could someone please let me know if my example works?
If it does, why $\sigma(T) = \mathbb{T}$?
If not, could I get some suggestions on the other possible weighted shifts? Thank you!
Any bilateral shift with weights in $\mathbb T$ will have spectrum $\mathbb T$, so you can choose $t_n$ to be whatever you want, as long as $|t_n|=1$.
This is because if you define $Ve_n=\lambda^ne_n$ for some $\lambda\in\mathbb T$, then $$ VTe_n=t_nVe_{n+1}=t_n\lambda^{n+1}e_{n+1},\qquad\qquad TVe_n=\lambda^nTe_n=t_n\lambda^ne_n. $$ So $TV=\lambda VT$. As $V$ is a unitary, $V^*TV=\lambda T$. Then $$ \sigma(T)=\sigma(V^*TV)=\lambda\sigma(T). $$ That is, $\sigma(T)$ is invariant for rotation, as we can do this for any $\lambda\in\mathbb T$. So $\sigma(T)=\mathbb T$.