= Solution
For $x\in X$,
$$
\|Tx\|^2=\sum_{n\in\mathbb Z}|x_{n+1}|^2
=\sum_{n\in\mathbb Z}|x_n|^2=\|x\|^2.
$$
The map $S$ defined by $(Sx)_n=x_{n-1}$ satisfies $ST=TS=I$, so $T$ is a surjective isometry and hence a <unitary operator>.
If $Tx=\lambda x$, then
$$
x_{n+1}=\lambda x_n.
$$
For $\lambda\ne0$, this gives $x_n=\lambda^nx_0$ for every integer $n$. When $|\lambda|>1$ the positive tail is not square summable, when $|\lambda|<1$ the negative tail is not square summable, and when $|\lambda|=1$ neither tail is square summable unless $x_0=0$. The case $\lambda=0$ also forces every coordinate to vanish. Thus
$$
\boxed{\sigma_p(T)=\varnothing.}
$$
The reverse triangle inequality gives, for every $x$,
$$
\|(T-\lambda I)x\|
\geq\bigl|\|Tx\|-|\lambda|\|x\|\bigr|
=|1-|\lambda||\,\|x\|.
$$
Hence $T-\lambda I$ is bounded below whenever $|\lambda|\ne1$, so such $\lambda$ do not belong to the <approximate point spectrum>.
Now let $|\lambda|=1$ and define the unit vector
$$
x^{(N)}_n=
\begin{cases}
\lambda^n/\sqrt{N+1},&0\leq n\leq N,\\
0,&\text{otherwise}.
\end{cases}
$$
The equation $x^{(N)}_{n+1}=\lambda x^{(N)}_n$ fails only at the two endpoints, so
$$
\|(T-\lambda I)x^{(N)}\|^2=\frac2{N+1}\longrightarrow0.
$$
Therefore the <bilateral shift operator> has
$$
\boxed{\sigma_{\mathrm{ap}}(T)=\{\lambda\in\mathbb C:|\lambda|=1\}.}
$$
Back to article page
