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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.

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    related: http://math.stackexchange.com/q/617601/1731472015-06-08

2 Answers 2

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The right shift is unitary, so its spectrum is contained in the unit circle. It has no eigenvalues, so $S-\lambda$ is always injective. The spectrum is nonempty, so there exists $\lambda_0$ with $|\lambda_0|=1$ such that $S-\lambda_0$ is not invertible. For each $\lambda$ in the unit circle, the operator $\lambda S$ is unitarily equivalent to $S$, via the diagonal unitary operator $(\ldots,x_{-2},x_{-1},x_0,x_1,x_2,\ldots)\mapsto(\ldots,\overline{\lambda}^2x_{-2},\overline{\lambda}x_{-1},x_0,\lambda x_1,\lambda^2x_2,\ldots).$ Thus for each $\lambda$ in the unit circle, $\sigma(S)=\sigma((\lambda\cdot\overline{\lambda_0})S)=(\lambda\cdot\overline{\lambda_0})\sigma(S)$, and therefore $\sigma(S)$ contains $\lambda$. This shows that the spectrum of $S$ is the unit circle.

For each $\lambda\in\sigma(S)$, $S-\lambda$ is not surjective, because $S-\lambda$ is injective but not invertible. However, $S-\lambda$ has dense range, which follows from the fact that the left shift $S^*$ also has no eigenvalues.

Feel free to ask for elaboration on any of the claims I've made.

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    @Jack: If $T$ is an operator on$a$Hilbert space $H$ and $T^*$ is its adjoint, then $(TH)^\perp=\mathrm{ker}T^*$. So for all $\lambda\in\mathbb{C}$, $((S-\lambda)\ell^2(\mathbb{Z}))^\perp=\mathrm{ker}(S^*-\overline{\lambda})=\{0\}$, which implies that $S-\lambda$ has dense range.2011-04-13
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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.$