 Solution

For $x\in X$,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 $Tx=\lambda x$, thenFor $\lambda \ne 0$, this gives $x_n={\lambda }^n{x}_0$ 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

The reverse triangle inequality gives, for every $x$,Hence $T-\lambda I$ is bounded below whenever $|\lambda |\ne 1$, so such $\lambda$ do not belong to the approximate point spectrum.

Now let $|\lambda |=1$ and define the unit vectorThe equation $x_{n+1}^{(N)}=\lambda x_n^{(N)}$ fails only at the two endpoints, soTherefore the bilateral shift operator has