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Show that $I-\mu S_r$ and $I-\mu S_\ell$ are not invertible when $\left | \mu\right | = 1$, where $S_\ell, S_r$ are the left and right shift operators in $\ell_2$ respectively.

Me and a friend of mine are stuck on this question for a long time. Their kernels are both $\{0\}$, and since $(I-\mu S_\ell)^* = I-\overline{\mu} S_r$, this means that we need to show that their images are not $\ell_2$, which is quite difficult to do as they are dense in $\ell_2$.

Any suggestions?

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We have $S_l(x_1,x_2,x_3,...)=(x_2,x_3,..)$ for $x=(x_n) \in \ell_2$.

Now let $ \lambda $ such that $| \lambda|<1$ und put $x=(1, \lambda, \lambda^2, ...)$. Then we have

$$S_lx= \lambda x,$$

hence $ \lambda \in \sigma(S_l)$. Since $\sigma(S_l)$ is closed, we have

$$\{ \mu: | \mu| \le 1\} \subseteq \sigma(S_l).$$

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  • $\begingroup$ I'm afraid I can't use the spectrum because it's not part of the material for the assignment. $\endgroup$ – Bary12 May 8 '18 at 12:54

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