Spectrum of operator $T: \ell^{2}\supset\text{dom}(T) \rightarrow \ell^{2}$ defined by $(x_{n})_{n\in \mathbb{N}} \mapsto (nx_{n})_{n\in \mathbb {N}}$ What is the point, continuous and residual spectrum of the following operator?
$T: \ell^{2}\supset\text{dom}(T) \rightarrow \ell^{2}$, $(x_{n})_{n\in \mathbb{N}} \mapsto (nx_{n})_{n\in \mathbb {N}}$, where $\text{dom}(T) = \left\{(x_{n})_{n\in \mathbb{N}} \in \ell^{2} | 
 (nx_{n})_{n\in \mathbb {N}} \in \ell^{2} \right\}$
My thoughts so far: Since $T$ is self-adjoint it must be that $\sigma_{r}(T) = \emptyset$. Now considering the point spectrum, I tried to find the eigenvalues of $T$: $Tx = \lambda x \iff (n-\lambda) x_{n} = 0$. But then I would have $x_{n}=1 $ for $n = \lambda$ and $x_{n}=0$ otherwise, so the corresponding eigenvectors are not in $\text{dom}(T)$, right? So $\sigma_{p}(T) = \emptyset$? Can anyone provide hints for $\sigma_r(T)$ or $\sigma_c(T)$, please?
 A: Let $e_1,e_2,e_3,....$ be the usual orthonormal basis of $\ell^2.$ Then $e_1,e_2,e_3,....\in dom(T)$ and $Te_k=ke_k$ for all $k \in \mathbb N.$ This gives
$$\mathbb N \subseteq \sigma_p(T).$$
If $ \lambda \in \sigma_p(T)$, then there is $x=(x_1,x_2,...) \in dom(T)$ such that $Tx= \lambda x$ and $x \ne 0.$ It follows that
$$kx_k = \lambda x_k$$
for all $k \in \mathbb N.$ This shows that $\lambda= m$ for some $m \in \mathbb N $, $x_m \ne 0$ and $x_k=0$ for $k \ne m$, hence $x=x_me_m.$
Conclusion:
$$\mathbb N =\sigma_p(T).$$
A: You are correct that $T$ is self-adjoint and thus $\sigma_r(T) = \emptyset$ (and also $\sigma(T) \subset \mathbb R$).
Your reasoning for $\sigma_p(T) = \emptyset$ is only partially correct (as the other answer indicates) because it only works if $\lambda \in \mathbb N$ (otherwise $n = \lambda$ doesn't work!).
But the sequence
$$
x_n = \begin{cases}
1, & \text{for } n = \lambda, \\
0, & \text{else.}
\end{cases}
$$
is in $\text{dom}(T)$, as
$$
n x_n = \begin{cases}
\lambda, & \text{for } n = \lambda, \\
0, & \text{else.}
\end{cases}
$$
and thus
$$
\| (n x_n)_{n \in \mathbb N} \|_{\ell^2}
= \sqrt{\sum_{n = 0}^{\infty} (n x_n)^2}
= \lambda
< \infty.
$$
This shows that $\mathbb N \subset \sigma_p(T)$ and equality if you argue that for $\lambda \not\in \mathbb N$ there is no solution to $(n - \lambda) x_n = 0$ for all $n \in \mathbb N$ and the solution above is the only one for $\lambda \in \mathbb N$.
We have
$$
\sigma_c(T)
:= \{ \lambda \in \mathbb R: T - \lambda \text{ injective}, \text{ ran}(T - \lambda) \subsetneq \ell^2 \text{ dense} \}.
$$
Inorder for $T - \lambda$ to be injective, $(T - \lambda) x = ((n - \lambda) x_n)_{n \in \mathbb N} = 0$ has to imply $x = 0$.
We have seen above that this is true for all $\lambda \in \mathbb R \setminus \mathbb N$.
