Let $T : l^2 (\mathbb{C}) \longrightarrow l^2(\mathbb{C})$ be the bounded linear operator defined by $T(x_1,x_2,x_3,x_4, \ldots)=(0,4x_1,x_2,4x_3,x_4,\ldots)$. I want to prove that the spectrum $\sigma(T)=\left\{{ \lambda \in \mathbb{C}: T-\lambda I \textrm{ is not invertible} }\right\}$ is equal to $\overline{B_{2}(0)}=\left\{{ \mu: |\mu| \leq 2 }\right\}$ . The first step proposed by the textbook (linear functional analysis by Bryan Rynne) is to show that if $ |\mu|<4 $ then $\mu$ is an eigenvalue of $(T^*)^2$. I agree with this procedure since this prove that $B_{4}(0) \subset \sigma((T^*)^2)$ and therefore, since $ \sigma((T^*)^2)$ is a closed set contained in the closed ball $\overline{B_{4}(0)}$ (since $(T^*)^2$ has norm equal to $4$ ) it follows that $\overline{B_{4}(0)} = \sigma((T^*)^2)$. Hence, if I prove that $\overline{B_{2}(0)} = \sigma(T^*)$ it follows that $\overline{B_{2}(0)} = \sigma(T)$.
Now, since also $\sigma((T^*)^2)=\left\{{ \mu^2: \mu \in \sigma(T^*) }\right\}=p(\sigma(T^*))$ where $p(z)=z^2$, that is, $\sigma((T^*)^2)$ agrees with the image of $\sigma(T^*)$under $p$, we have that $$ \begin{equation} \label{eq1} \begin{split} \overline{B_{2}(0)} & = p^{-1}(\overline{B_{4}(0)}) \\ & =p^{-1}(p(\sigma(T^*)) \\ & \supset\sigma(T^*) \end{split} \end{equation} $$ Therefore, it remains to prove that $\overline{B_{2}(0)} \subset \sigma(T^*)$. Now, at the outset I thought in the following argument: let $\mu \in \overline{B_{2}(0)}$, if $\mu \notin \sigma(T^*)$ then $\mu \notin \sigma((T^*)^2)$ and hence $| \mu |^2 >4$ which contradicts the fact that $|\mu| \leq 2$. However, there is something wrong with this argument, in general, if $f$ is a function, the fact that $x \notin B$ doesn't mean that $f(x) \notin f(B)$. For instance, if $B=[0,1]$ then $-1 \notin B$ but $p(-1)=(-1)^2=1=p(1) \in p(B)$.
Any help in order to correct this proof?
In advance thank you very much.
For $|u|=1$ let $U(x_1,x_2,\ldots )=(ux_1,u^2x_2,\ldots ).$ Then $U$ is a unitary operator and $UTU^{-1}=u T.$ We thus have $$\sigma(T)=\sigma(UTU^{-1})=\sigma(uT)=u\sigma(T)$$ Therefore the set $\sigma(T)$ is invariant under rotations centered at $0.$ By OP we have $\sigma(T)^2=\sigma(T^2)=\{|z|\le 4\}.$ Thus for $0\le r\le 4$ at least one of the numbers $\pm r^{1/2}$ belongs to $\sigma(T).$ By the invarance under rotations we obtain $\sigma(T)=\{|z|\le 2\}.$
I prefer another solution, not following the hint from the textbook. Namely let $$A(x_1,x_2,x_3,x_4,\ldots )=(2x_1,x_2,2x_3,x_4,\ldots )$$ and $S(x_1,x_2,\ldots )=(0,x_1,x_2,\ldots ).$ It is well known that $\sigma(S)=\{|z|\le 1\}.$ Observe that $T=2A^{-1}SA.$ Hence $$\sigma(T)=2\sigma(A^{-1}SA)=2\sigma(S)=\{|z|\le 2\}$$