Let $$H=\ell^2=\{(x_n)_{n=1}^\infty:\sum\limits_{n=1}^\infty |x|^2<\infty\},$$ so that $S:\ell^2\rightarrow \ell^2,$ $$ S(x_1, x_2, \cdots) = (0, x_1, x_2,\cdots) . $$
Can you help me how to prove:
If $S \in B(H)=\{ \text{bounded linear operators on } H\}$, and $ \| S\|= 1 $, then $$\sigma(S) = \{ \lambda \in \mathbb{C} : \lambda\leqslant 1\}$$
That $||S||=1$ is easy to see (show that $||Sx||=||x||$ for all $x \in H$.
We get: $\sigma(S) \subseteq\{ \lambda \in \mathbb{C} : \mid\lambda\mid\leqslant 1\}$.
Next consider $T:=S^{*}$. We have $T(x_{1}, x_{2}, \ldots) = (x_{2} , x_{3},\ldots) $.
Observe that $ \sigma (T)= \sigma (S)$
It should be clear, that $0$ is an eigenvalue of $T$.
Let $ 0<|\mu|<1$ and $x=(1, \mu, \mu^2,...)$
Show that $x \in H$ and that$Tx=\mu x$
We get: $ \{\mu: |\mu|<1\} \subseteq \sigma (T) \subseteq \{\mu: |\mu| \le1\}$
Since $ \sigma (T)$ is closed: $\sigma (T) =\{\mu: |\mu| \le1\}$