Show spectrum of a bilateral shift is contained in a donut

Question

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!

Answer 1

Assuming that $We_n:=w_ne_{n+1}$ and the space is $\ell^\infty(\mathbb{Z})$, then $W=LD$ where $L$ is the right-shift operator and $D$ is the multiplier operator with sequence $(w_n)$. Note that both $L$ and $D$ are invertible.

Then $\|W\|\le\|L\|\|D\|=\sup_n|w_n|\le1$; hence $\sigma(W)$ is bounded by $1$.

Now repeat for $W^{-1}$: $\|W^{-1}\|\le\|D^{-1}\|\|L^{-1}\|=\sup_n|w_n^{-1}|\le\frac{1}{\delta}$; hence $\sigma(W^{-1})$ is bounded by $1/\delta$, and thus $\sigma(W)$ is bounded below by $\delta$.

Answer 2

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.