Spectrum of double infinite shift using isometry to Fourier series I'm trying to find the spectrum of the operator $T: l^2(\mathbb{Z}) \to l^2(\mathbb{Z})$ given by right shift but I am having some difficulties.
I can show that $l^2$ is isomorphic to $L^2(\mathbb{T})$ via $\phi: (\alpha_n) \to \sum \alpha_n z^n$ where $z = e^{it}$ and then under this map $T$ acts on $L^2(\mathbb{T})$ by $f \to zf$.
I'm now trying to find the spectrum of this multiplication operator. It's easy to show that the point spectrum is empty. The operator is unitary and so the spectrum must lie in the unit circle and it's normal so the spectrum must be entirely approximate point.
I'm struggling to find the approximate point spectrum. In particular I'm not quite sure how we should write the norm on $L^2(\mathbb{T})$. Usually we should look for functions $f_n$ with $||(\lambda I - T)f_n|| = \int_{0} ^{2 \pi} |\lambda - e^{it}||f(t)| \to 0$ but this seems no easier than finding the approximate point spectrum of $T$ as an operator on $l^2(\mathbb{Z})$ so I feel like I'm missing something - perhaps there is a better way to write the norm in terms of $z = e^it$ to take advantage of being able to write the operator as multiplication?
Thanks
 A: Whether you use $L^{2}(\mathbb{T})$ or $L^{2}[0,2\pi]$ is a matter of preference. I prefer the classical Fourier setting
$$
                 U : L^{2}[0,2\pi]\rightarrow \ell^{2}(\mathbb{Z})
$$
defined by
$$
                 Uf = \left\{ \int_{0}^{2\pi}f(t)\frac{e^{-int}}{\sqrt{2\pi}}dt\right\}_{n=--\infty}^{\infty}
$$
The inverse of $U$ is
$$
                   U \{ c_{n} \}_{n=-\infty}^{\infty} = \sum_{n=-\infty}^{\infty}c_{n} \frac{e^{int}}{\sqrt{2\pi}}
$$
$U$ is unitary because $\{ \frac{e^{int}}{\sqrt{2\pi}}\}_{n=-\infty}^{\infty}$ is a complete orthonormal basis of $L^{2}[0,2\pi]$ with inner product $(f,g)=\int_{0}^{2\pi}f(t)\overline{g(t)}dt$. If $S$ is the right shift on $\ell^{2}(\mathbb{Z})$, then
$$
                      U^{-1}SUf = e^{it}f(t).
$$
Every $|\lambda|$ is an approximate eigenvalue of $Mf = e^{it}f$. For example, if $0 < \theta < 2\pi$, consider $\chi_{[\theta-\delta,\theta+\delta]}$. Then
\begin{align}
       \|(M-e^{i\theta}I)\chi_{[\theta-\delta,\theta+\delta]}\|^{2} & =       \int_{0}^{2\pi}|(M-e^{i\theta}I)\chi_{[\theta-\delta,\delta+\delta]}|^{2}dt \\
   & = \int_{\theta-\delta}^{\theta+\delta}|e^{it}-e^{i\theta}|^{2}dt \\
   & \le |e^{i\delta}-1|^{2}\|\chi_{[\theta-\delta,\theta+\delta]}\|^{2} \\
   & = [(1-\cos\delta)^{2}+\sin^{2}\delta]\|\chi_{[\theta-\delta,\theta+\delta]}\|^{2} \\
   & = 2(1-\cos\delta)\|\chi_{[\theta-\delta,\theta+\delta]}\|^{2}
\end{align}
The conclusion is that every $\lambda$ for which $|\lambda|=1$ is an approximate eigenvalue of $M$ and, hence, also of the shift $S$.
A: You are sort of on the right track. Let's write things out a little bit more explicitly. Consider
$$(\lambda I-T)f = \lambda f - Tf = \lambda f - zf = \sum_{n=-\infty}^{\infty} \lambda \alpha_n z^n - \sum_{n=-\infty}^{\infty} \alpha_n z^{n+1}.$$
Reindexing, the second sum becomes $\sum\limits_{n=-\infty}^{\infty} \alpha_{n-1} z^n$ so the above becomes
$$(\lambda I - T)f = \sum_{n=-\infty}^{\infty} (\lambda\alpha_n-\alpha_{n-1})z^n.$$
As you note, we want to find $f_i$ such that $\|(\lambda I - T)f_i\|\to 0$. To do this, it's best to sort of reduce yourself to the case of $\ell^2(\Bbb N_0)$ since there the right shift operator is easily understood. This is accomplished by considering those $f_i$ such that $f_i = (\ldots, 0, a_{i0},a_{i1},\ldots)$, where $a_{i0}$ occurs at the zeroth spot and eventually the $a_{ij}$ are zero.
To make things as simple on ourselves as possible, we'd like to set up a recursion of sorts with the $\alpha_i$ (so that there is cancellation to make the computation easy) and so we'd naively pick $\alpha_n = \frac{1}{\lambda}\alpha_{n-1}$. This gives the following candidate elements:
$$f_i = (\ldots, 0, 1, \lambda^{-1},\lambda^{-2},\ldots,\lambda^{1-n},0,\ldots).$$
Can you take it from there?
