The spectrum of the right shift operator on $\ell^2(\mathbb{Z}$)

Question

The right shift operator $S$ on $\ell^2(\mathbb{Z})$ is defined such that for $x=\left(x_n\right)_{n \in \mathbb{Z}} \in \ell^2(\mathbb{Z})$, we have $S(x)_k=x_{k-1}$ for all $k \in \mathbb{Z}$.

1. Prove that $\lambda \in \rho(S)$, the resolvent set of $S$, for all $\lambda \in \mathbb{C}$ such that $|\lambda| \neq 1$. Start with the case $|\lambda|>1$ and next treat the case $|\lambda|<1$. In both cases use a contraction principle (e.g. contraction mapping theorem).

2. Show that the spectrum of $S$ is purely continuous and given by the unit circle $|\lambda|=1$. [Hint: consider vectors of the form $x_k=\varphi_k e^{i \theta k}$ for some $0 \leq \theta<2 \pi$ and for instance $\varphi_k=1$ for $|k| \leq K$ and $\varphi_k=0$ otherwise. Also, we recall that a space is not dense in $\mathcal{H}$ when we can find a non-vanishing element in $\mathcal{H}$ that is orthogonal to its closure.]

I've managed to do (1) without using the contraction principle. I did this by considering the adjoint left-shift operator, then using that the resolvent set should contain the complement of the unit ball of radius $||S||$. For (2), I've done this by showing that $(\dots,0,1,0,\dots)$ is not in the range, but again I wasn't able to use the hint. Can anyone help me understand how to use the hints to solve the problem the way it was intended (with the hints)?

Answer 1

We will base on the fact that if $T\in B(X),$ where $X$ is a Banach space, and $\|T\|<1$ then $I-T$ is invertible. The simplest way is to show that the inverse operator is given by the absolutely convergent power series $\sum_{n=0}^\infty T^n.$

The operator $S$ maps $\ell^2$ isometrically onto $\ell^2.$ Therefore $\|S\|=\|S^{-1}\|=1.$ For $|\lambda|>1$ the operator $\lambda I-S$ is invertible, as $\lambda I-S=\lambda (I-\lambda^{-1}S)$ and $\|\lambda^{-1}S\|=|\lambda|^{-1}<1.$ Similarly for $0<|\lambda|<1$ we have $\lambda I-S=-S(I-\lambda S^{-1}).$ Hence it is invertible since $\|\lambda S^{-1}\|=|\lambda|<1.$

Assume $|\lambda|=1.$ Let the vector $v^{(K)}$ be defined by $$v^{(K)}_k=\begin{cases} \lambda^{-k} & |k|\le K\\ 0 & {\rm otherwise} \end{cases} $$ Then $$[(\lambda I-S)v^{(K)}]_k=\begin{cases} 0 & |k|\le K-1,\ |k|\ge K+1\\ \lambda^{k+1} & k=-K\\ -\lambda^{-k} & k=K+1 \end{cases} $$ Then $${\|(\lambda I-S)v^{(K)}\|\over \|v^{(K)}\|} ={2\over 2K+1}\underset{K\to \infty}{\longrightarrow}0$$ Therefore $\lambda\in \sigma(S).$ Assume a vector $u$ is orthogonal to ${\rm Im}(\lambda I-S).$ Then $u\perp (\lambda I-S)\delta_k$ for any $k,$ where $\delta_k$ denote the elements of the standard basis. Hence $u\perp \lambda \delta_k-\lambda \delta_{k+1}$ and consequently $u_k=\lambda u_{k+1}.$ We get $|u_k|=|u_{k+1}|$ for any $k,$ therefore the sequence $|u_k|$ is constant. It is possible only if $u=0.$ Summarizing the orthogonal complement of ${\rm Im}\,(\lambda I-S)$ is trivial, i.e. the space ${\rm Im}\,(\lambda I-S)$ is dense.