In $\ell^2(\mathbb{Z})$, what is the spectrum of the right-shift operator?
I have looked at relevant posts on the website and noticed that someone had mentioned in hints that $\ell^2(\mathbb{Z})$ is $L^2(\mathbb{T})$ and hence the right-shift operator can be transformed into a multiplication operator. I am interested in this approach and could anyone show how this is achieved?
Indeed, $\ell^2(\mathbb{Z})$ is isometric to $L^2(\mathbb{T})$ through the Fourier's transform, namely the following map is an isometry: $$f\mapsto\left(\frac{1}{2\pi}\int_0^{2\pi}f(t)e^{-int}\mathrm{d}t\right)_{n\in\mathbb{Z}}.$$ This is partially a consequence of Parseval's equality.
Now, notice that the right-shift operator in $\ell^2(\mathbb{Z})$ becomes the multiplication by $x\mapsto e^{-ix}$ in $L^2(\mathbb{T})$ through the above isometry. Therefore, since $\mathbb{S}^1=\{e^{it},t\in\mathbb{R}\}$, the spectrum of the multiplication by $x\mapsto e^{-ix}$ is $\mathbb{S}^1$ and so is the spectrum of the right-shift operator.
I let you work out why the spectrum is unchanged through the isometry.