Spectrum of a operator in $l^2$

Question

Given a sequence of number $\{ a_{n}\} $, define an operator on $l^2$ as follows: $$A\left( x_{1},x_{2},...,x_{n},...\right) =\left( a_{1}x_{1},a_{2}x_{2},...,a_{n}x_{n},...\right) $$ If $A$ is a bounded operator, determine the spectral point type of $A$.

My attempt:

It's easy to see that $\{ a_{n}\} $ is bounded if $A$ is a bounded operator, but I think this condition is not enough to determine the whole spectrum, it depends on the property of $\{ a_{n}\} $, for example, $a_{n}=\frac{1}{n} $, we can determine the whole spectrum.

Answer 1

I claim first that $\{a_j:j\in\mathbb{N}\}$ is the set of eigenvalues of $A$. Indeed it is clear that each $a_j$ is an eigenvalue, as if $e_j\in\ell^2$ is $1$ in the $j$'th position, and $0$ everywhere else, then

$$Ae_j=a_je_j.$$

To see that any eigenvalue is in the set, suppose $\lambda$ is an eigenvalue of $A$ with eigenvector $x=\{x_j\}_{j\in\mathbb{N}}$, then

$$Ax=\{a_jx_j\}_{j\in\mathbb{N}}=\{\lambda x_j\}_{j\in\mathbb{N}},$$

so $\lambda x_j=a_j x_j$ for all $j\in\mathbb{N}$, and so in particular, as $x\neq0$, $\lambda=a_j$ for some $j\in\mathbb{N}$. This proves the claim.

We show now that $\sigma(A)=\overline{\{a_j:j\in\mathbb{N}\}}$. It is clear that

$$\overline{\{a_j:j\in\mathbb{N}\}}\subseteq\sigma(A)$$

as $\sigma(A)$ is a closed set containing all eigenvalues of $A$. We show now that

$$\mathbb{C}\setminus\overline{\{a_j:j\in\mathbb{N}\}}\subseteq\rho(A).$$

Let $\lambda\in\mathbb{C}\setminus\overline{\{a_j:j\in\mathbb{N}\}}$, and find $\varepsilon>0$ such that $\lvert\lambda-a_j\rvert\geq\varepsilon$ for all $j\in\mathbb{N}$. Then

$$(\lambda I-A)\{x_j\}_{j\in\mathbb{N}}=\{(\lambda-a_j)x_j\}_{j\in\mathbb{N}},$$

so we have the inverse

$$(\lambda I-A)^{-1}\{x_j\}_{j\in\mathbb{N}}=\left\{\frac{1}{\lambda-a_j}x_j\right\}_{j\in\mathbb{N}},$$

which is a well defined operator $\ell^2\to\ell^2$ as $\frac{1}{\lambda-a_j}\leq\frac{1}{\varepsilon}$ for all $j\in\mathbb{N}$, and so $\lambda\in\rho(A)$. It follows that

$$\sigma(A)=\overline{\{a_j:j\in\mathbb{N}\}}.$$

Answer 2

$A$ is bounded iff $(a_n)$ is in $\ell_\infty$, in which case he spectrum of the (multiplication) operator $A$ is $\overline { \{a_n \colon n \ge 1 \}}$. Clearly each number $a_n$ is an eigenvalue of $A$ and hence $\{a_n \colon n \ge 1 \} \subset \sigma (A)$, whence $$\overline { \{a_n \colon n \ge 1 \}} \subset \sigma (A). $$ On the other hand if $z \notin \overline { \{a_n \colon n \ge 1 \}}$ then there exists $δ>0$ such that $|z-a_n| \ge δ$ for all $n$. Then the bounded operator $(x_n)_{n \ge 1} \mapsto (\frac{1}{z-a_n} x_n)_{n \ge 1}$ is the invese of $z-A$ so that $z \notin \sigma (A)$.