$U$ is defined to be a bilateral shift operator such that $Ue_n= u_ne_{n+1}$ for $n\in \mathbb{Z}$, where $u_n \ne 1$.
I am looking for an example of $U$ and a rank-one operator $T \in \mathcal{B}(\ell^2(\mathbb{Z}))$ so that $\sigma (U) = \{ z \in \mathbb{C} : |z| = 1 \}$ and $\sigma(U+T)= \{ z \in \mathbb{C} : |z| \le 1 \}$.
Some thoughts:
I showed that for any bilateral shift $U$ with $Ue_n=u_ne_{n+1}$ for $n \in \mathbb{Z}$ such that $|u_n|=1$, we have $ \sigma (U) = \{ z \in \mathbb{C} : |z| = 1 \}$.
Also, I saw it from here The spectrum of the operators $\sigma(U+T) = \sigma(U) \cup \sigma(T) $.
So I wonder if I could find some rank one operator to satisfy the condition based on that.
But other suggestions will also be great.
Thank you.
Best Answer
Let $V$ be the standard bilateral shift on $\ell ^2(\mathbb Z)$, namely the unitary operator $V$ such that $Ve_n=e_{n+1}$, and let $T$ be the rank-one operator given by $$ T(x)= -\langle x,e_0\rangle e_1, \quad\forall x\in \ell ^2(\mathbb Z). $$ Then $R:= V+T$ becomes the operator given by $$ R(e_n) = \left\{\matrix{ e_{n+1}, & \text { if } n\neq 1, \cr 0 , & \text { if } n=0. }\right. $$ One then sees that $R$ leaves invariant the decomposition $$ \ell ^2(\mathbb Z)=\ell ^2(\mathbb Z_-)\oplus \ell ^2(\mathbb Z_+^*), $$ where $\mathbb Z_-$ stands for the non-positive integers and $\mathbb Z_+^*$, for the strictly positive ones.
The restriction of $R$ to $\ell ^2(\mathbb Z_+^*)$ is then the so called unilateral shift $S$, while the restriction of $R$ to $\ell ^2(\mathbb Z_-)$ is seen to be unitarily equivalent to $S^*$. We then conclude that $R$ is unitarily equivalent to $S\oplus S^*$, whose spectrum is given by $$ \sigma (R) = \sigma (S) \cup \sigma (S^*) = \sigma (S) \cup \overline{ \sigma (S)}. $$ Since $\sigma (S)$ is the unit disk, we see that $\sigma (R)$ is the unit disk as well.
Now, the OP has asked for a version of the bilateral shift where $Ue_n=u_ne_{n+1}$, with $u_n\neq 1$, so we may take $U=-V$, i.e. take all of the $u_n=-1$, the requested rank-one operator being $-T$.