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

functional-analysishilbert-spacesoperator-theoryspectral-theory

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)?

Best Answer

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.

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