TITLE: Spectrum of the right shift operator on $\ell^2({\bf Z})$
QUESTION [15 upvotes]: Here is the question:

Considering the right shift operator $S$ on $\ell^2({\bf Z})$, what can one know about ran$(S-\lambda)$? 

Here is what I thought:

If one wants to prove that the operator $S-\lambda$ is onto when $\lambda$ satisfies some conditions, does one have to construct the solution?
One needs to find the solution $(S-\lambda)x=y$ where $x,y\in \ell^2({\bf Z})$. Using the standard basis $e_n=(\delta_{nk})_{k=-\infty}^{\infty}$, one has to solve
$\lambda x_{k}-x_{k-1}=y_{k}, (k\in {\bf Z})$.
Intuitively, when $|\lambda|=1$, there may be no way for $S-\lambda:\ell^2({\bf Z})\to \ell^2({\bf Z})$ to be onto. However, when $|\lambda|\neq 1$, can one explicitly find $x=(x_{k})_{k=-\infty}^{\infty}$? 


[ADDED] Thanks to a recent editing of my question, I have learned that the right and left shift operators acting on two-sided infinite sequences are also called bilateral shifts.

REPLY [3 votes]: To address the final part of your question: you can write down the inverse of $S-\lambda$ for $|\lambda|\ne1$, so
in this sense you can "explicitly find $x$". (Although the spectral argument is a more efficient way to answer the range question).
If $\|T\|<1$ then $1-T$ is invertible, and
$$ (1-T)^{-1}=\sum_{k\ge0}T^k.$$
Wikipedia calls this the Neumann series of $T$.
Since $S^{-1}$ is the backward shift, we have $\|S^{-1}\|=1$. So if $0<|\lambda|<1$ then $\|\lambda S^{-1}\|<1$ and
$$(S-\lambda)^{-1}=-S^{-1}(1-\lambda S^{-1})^{-1}=S^{-1}\sum_{k\ge0}(\lambda S^{-1})^k.$$
Since $\|S\|=1$, if $|\lambda|>1$ then $\|\lambda^{-1}S\|<1$ so
$$ (S-\lambda)^{-1}=-\lambda^{-1}(1-\lambda^{-1} S)^{-1}=-\lambda^{-1}\sum_{k\geq 0}(\lambda^{-1}S)^k.$$