[Math] Spectrum of a bilateral shift

Question

Let $u$ be a bilateral shift on Hilbert space $\ell^2(\Bbb Z)$. As for unilateral shifts, the spectrum of $u$ does not contain any eigenvalue. Also $u$ is unitary, so $\sigma(u) \subset \Bbb S$ ($\Bbb S$ means unit circle). How can I show that $\sigma(u)=\Bbb S$?

Answer 1

We can solve it in an isomorphic Hilbert space. Consider $L^2(\mathbb{T})$, with the length measure on $\mathbb{T}$. The functions $z^n$, $n\in\mathbb{Z}$ form a basis and we see that $Wf=zf$ is the bilateral shift.

We see now that $(W-\lambda I)f=(z-\lambda)f$. Therefore $\lambda\notin Spec(W)$ iff $\frac{1}{(z-\lambda)}\in L^2(\mathbb{T})$.