Verification of the values in the continuous spectrum of an operator

Question

I am needing help with the continuous spectrum.

Let $(\lambda_j)$ be a sequence of real numbers with $\lambda_j \neq 1$ for all $j$ and $\lambda_j \rightarrow 1$. Consider $T: \ell^2 \rightarrow \ell^2$ defined for $\xi_j \in \ell^2$ by $$T(\xi_j)=(\lambda_j \xi_j)$$

Questions: Find the point spectrum, residual spectrum, and continuous spectrum.

My thoughts...

I believe that the point spectrum is the set of the $\lambda_j$. Since $$Te_j=\lambda_j e_j$$ for $e_j$ having a 1 in the $jth$ position and zeroes elsewhere.

The operator is bounded and self adjoint since the $\lambda_j$ are real (I will omit the details), so we know the residual spectrum is empty.

This is the part I am not sure about.

EDIT: I now think that the continuous spectrum only contains 1. I think this because that is the only value we can get arbitrarily close to for $\lambda \neq \lambda_j)$. If someone could verify this that would be great. Thanks!

Since the spectrum is closed, we know that the closure of the set of $\lambda_j \subset $ the spectrum.

We can show that the set that is the closure of the set of $\lambda_j$ without the $\lambda_j$ is in the continuous spectrum since the resolvent would be unbounded. I think it is similar to this question here, (see the last paragraph). But I am not sure because we don't have that the $\lambda_j$ are dense.

All other values are in the resolvent set.

Answer 1

In general, for any complex sequence $(\lambda_n)_n$ the spectra for the diagonal operator $T$ are:

$$\sigma(T) = \overline{\{\lambda_n : n \in \mathbb{N}\}}$$ $$\sigma_p(T) = \{\lambda_n : n \in \mathbb{N}\}$$ $$\sigma_r(T) = \emptyset$$ $$\sigma_c(T) = \overline{\{\lambda_n : n \in \mathbb{N}\}} \setminus \{\lambda_n : n \in \mathbb{N}\}$$

So in your case we have $$\sigma(T)= \{\lambda_n : n \in \mathbb{N}\} \cup \{1\}$$ $$\sigma_p(T) = \{\lambda_n : n \in \mathbb{N}\}$$ $$\sigma_r(T) = \emptyset$$ $$\sigma_c(T) = \{1\}$$

Answer 2

If $\lambda \notin \{1,\lambda_1,\lambda_2,...\}$ ther exists $a>0$ such that $|\lambda -c|>a$ for all $c \in \{1,\lambda_1,\lambda_2,...\}$ (This is because the set here is compact). Now solve the equarion $Tx-\lambda x=y$ for any $y \in l^{2}$. You will get a unique solution. This implies that $T-\lambda I$ is invertible. Hence $\sigma (T) \subset \{1,\lambda_1,\lambda_2,...\}$. Since $\{\lambda_1,\lambda_2,...\}$ is contained in the spectrum and the spectrum is closed we get $\sigma (T) = \{1,\lambda_1,\lambda_2,...\}$.