Spectrum of a bilateral shift 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 $\sigma(u)=\Bbb S$? 

Hint: Let $e_n$ $(n\in \mathbb Z)$ be the standard basis in $\ell^2(\mathbb Z)$. For $\lambda \in {\mathbb T}$, let
$$ \xi_n=\frac{1}{2n+1}\bigl( \lambda^{-n} e_{-n}+\lambda^{-n+1} e_{-n+1}+\cdots+e_0+\cdots+\lambda^{n} e_n\bigr)\qquad (n\in \mathbb Z). $$
Note that $(\xi_n)$ is a sequnce of vectors with norm $1$. Consider
$$ u\xi_n-\lambda \xi_n\quad \text{when}\quad n\to \infty. $$
