Let $T(x_n)_{n \in \mathbb Z} = (x_{n+1})_{n \in \mathbb Z}$, proof that $\sigma(T) = S^1$.

Question

Consider the unit operator

$T: \ell^2(\mathbb Z) \to \ell^2(\mathbb Z), \, T(x_n)_{n \in \mathbb Z} = (x_{n+1})_{n \in \mathbb Z}$. I'm trying to understand the proof that $\sigma(T) = S^1$ given here.

First we have the following results:

$\forall \lambda \in \mathbb C, T - \lambda I$ is injective.

If $E$ is a Banach Space, $S \in \mathcal B(E), z \in \mathbb C, z \neq 0, \, \sigma(zS) = z\sigma(S)$.

If $S,U \in \mathcal B(E), \, \sigma(U^{-1}SU) = \sigma(S)$.

Given $z \in S^1$, we define the bounded operator in $\ell^2(\mathbb Z)$, $$M_z(x_n)_{n \in \mathbb Z} = (...,\overline z ^{2}x_{-2}, \overline z ^{-1}x_{-1},x_0, zx_1, z^2 x_2,...).$$

Then, we have that $M_z^{-1}T M_z= zT$. So, for each $z \in S^1$,

$$ \sigma(T) = \sigma(M_z ^{-1}T M_z) = \sigma(zT) = z\sigma(T). $$

In the proof given in the above link, Jonas Meyer used that exists $z_0 \in \sigma(T)$ with $|z_0| = 1.$ However, I couldn't see why this happens.

QUESTION: If $|\lambda| < 1$, how one can proof that $T - \lambda I$ is invertible?

In particular, the above question implies in $\sigma(T) S^1$, since $\sigma(T) \subset B[0, 1]$. With this, I can conclude the argument, as follows:

Since $\sigma(T) \neq \emptyset$, then $\exists z_0 \in \sigma(T)$. Given $z \in S^1$, we have that $\sigma(T) = z\overline{z_0}\sigma(T)$. Hence,

$$ z_0 \in \sigma(T) \Rightarrow z = z(\overline{z_0}z_0) (z\overline{z_0})z_0 \in z\overline{z_0}\sigma(T) = \sigma(T) \Rightarrow S^1\subset \sigma(T).$$

Help?

Answer 1

You just need to note that, if $0<|\lambda|<1$, then $T-\lambda = T(Id - \lambda T^{⁻1})$ and $\|\lambda T^{-1}\|\leq |\lambda| <1$. Now, observe that $(Id - \lambda T^{⁻1})^{-1} = \sum_{n=0}^\infty (\lambda T^{-1})^n$ and: $$(T-\lambda)^{-1}= (Id-\lambda T^{-1})^{-1}T^{-1}.$$

It proves that $\sigma(T) \subset S^1$.

Answer 2

Since $T$ is a unitary ($T^*T=TT^*=I$), it follows that $\sigma(T)\subset S^1$: indeed, since $T^*=T^{-1}$, $$ \overline{\sigma(T)}=\sigma(T^*)=\sigma(T^{-1})=\{\lambda^{-1}:\ \lambda\in\sigma(T)\}. $$ So, if $\lambda\in\sigma(T)$, then $\overline{\lambda^{-1}}\in\sigma(T)$. As $\|T\|=1$, we have $|\lambda|\leq1$ for all $\lambda\in\sigma(T)$, so it follows that $|\lambda|=1$ for all $\lambda\in\sigma(T)$.

In particular, if $|\lambda|\ne1$, then $T-\lambda I$ is invertible.

As for the existence of $z_0$, it is just a consequence that the spectrum is never empty.