Let $W$ be a bilateral shift with weight sequence $(w_n) \in \ell^{\infty}(\mathbb{Z})$, for $w_n \geq 0$ for each $n \in \mathbb{Z}$.
i.e. we have that $We_n = we_{n+1}$ for $n\in \mathbb{Z}$, where $(e_n)$ is an orthonormal basis.
Suppose that there exists $\delta >0$ such that $\delta \leq w_n \leq 1$.
Show that $\sigma(W) \subset \{ z\in \mathbb{C}: \delta \leq |z| \leq 1\}$.
Things I've done:
I showed that if $\alpha \in \sigma(W)$, then we have $\{z\in \mathbb{C}: |z|=\alpha \} \subset \sigma(W)$.
But I want to show that $\alpha \in \{ z\in \mathbb{C}: \delta \leq |z| \leq 1\}$.
Any help is appreciated!
Best Answer
I guess you mean $We_n = w_ne_{n+1}$ and that your space is $\ell^2(\mathbb Z)$. Then $W = SD_w$, where $S$ is the usual bilateral shift and $D_w$ is the diagonal matrix with the $w_n$ on the diagonal. So, $\|Wx\| = \|SD_wx\| = \|D_wx\|$, which implies $$ \delta\|x\|\le\|Wx\|\le\|x\|. $$ Similarly, $\|W^*x\|\le\|x\|$ and $\|W^*x\| = \|D_wS^*x\|\ge\delta\|x\|$.
From here you should be able to see that the spectrum of $W$ is in that annulus.