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 $\sigma(u)=\Bbb S$?
2
$\begingroup$
$\endgroup$
3
$\begingroup$
$\endgroup$
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 $$ u\xi_n-\lambda \xi_n\quad \text{when}\quad n\to \infty. $$
2
$\begingroup$
$\endgroup$
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 $\frac{1}{(z-\lambda)}\in L^2(\mathbb{T})$.
