Question on charaterization of the parts of the spectrum of an operator.

Let $\mathcal H$ be a separable Hilbert space and $\{e_n\}_{n \geq 1}$ be an orthonormal basis for $\mathcal H.$ Let $D = \sum\limits_{n \geq 1} d_n \left |e_n \rangle \langle e_n \right |,$ where $\left |e_n \rangle \langle e_n \right |$ is a bounded linear operator on $\mathcal H$ defined by $x \mapsto \left \langle e_n,x \right \rangle e_n$ of norm $1.$ Then show that

$(1)$ $\text {ran} (D - \lambda)$ is closed iff $\lambda$ is not a limit point of $\sigma_p (D) = \left \{\lambda_n\ |\ n \in \mathbb N \right \},$ the point spectrum of $D.$

$(2)$ $\sigma_{\text {ess}} (D)$ (the essential spectrum of $D$) is the set of all limit points of $D.$

$(3)$ $\lambda \in \sigma_c (D)$ (the compression spectrum of $D$) iff $\lambda$ is a limit point of $\sigma_p (D)$ and $\lambda \notin \sigma_p (D)$ i.e. $\sigma_{c} (D) = \sigma_{\text {ess}} (D) \setminus \sigma_p (D).$

$\textbf {My Attempt} :$ If we can show $(1)$ we are done with $(2).$ Because if $\lambda \in \sigma_{\text {ess}} (D)$ then either $\dim \text {ker} (D - \lambda) = \infty$ or $\text {ran} (D - \lambda)$ is not closed. But $\dim \text {ker} (D - \lambda) \neq \infty$ since each eigenvalue of $D$ will have finite multiplicity (in fact $\text {ker} (D - d_n) = \text {span} \{e_n \}$). So $\text {ran} (D - \lambda)$ can't be closed and hence by $(1)$ it follows that $\lambda$ is a limit point of $\{d_n\ |\ n \in \mathbb N \}.$ Conversely, if $\lambda$ is a limit point of $\{d_n\ |\ n \in \mathbb N \}$ then again by $(1)$ we can conclude that $\text {ran} (D - \lambda)$ is not closed and hence $\lambda \in \sigma_{\text {ess}} (D).$

For $(3)$ I first noticed that $D$ is normal. So if $\lambda \in \sigma_c (D)$ i.e. if $\text {ran} (D - \lambda)$ is not dense then $\text {ker} (D - \lambda) \neq 0.$ So $\lambda = d_n$ for some $n \in \mathbb N$ i.e. $\lambda \in \sigma_p (D).$ But that contradicts $(3).$

Also I have some issue in proving $(1).$ If $\text {ran} (D - \lambda)$ is closed then two cases may arise (normality of $D$ allows only two cases). Either $D - \lambda$ is invertible in which case we are through. Otherwise $\text {ker} (D - \lambda) \neq 0$ and $\text {ran} (D - \lambda) \subsetneq \mathcal H.$ In this case $\lambda \in \sigma_p (D)$ but I can't conclude that $\lambda$ is a limit point of $\sigma_p (D).$ Also I have no idea how to prove the other way round.

Any help would be a boon for me at this stage.

Thanks a lot.

Solution 1:

I'm not too familiar with the essential spectrum, but it looks like you already handled $(2)$ and $(3)$ so here goes $(1)$.

Lemma 1.

We have $\sigma_p(D) = \{d_n : n \in \Bbb{N}\}.$

Proof.

Clearly $(D-d_n I) e_n = 0$ so $d_n \in \sigma_p(D)$ for all $n \in \Bbb{N}$. Conversely, let $\lambda \notin \{d_n : n \in \Bbb{N}\}$ and assume $(D-\lambda I) x = 0$. It follows $$Dx = \lambda x \implies d_n\langle x, e_n\rangle = \lambda\langle x, e_n\rangle, \forall n \in \Bbb{N} \implies \langle x, e_n\rangle = 0, \forall n \in \Bbb{N} \implies x= 0$$ so $D-\lambda I$ is injective and hence $\lambda \notin \sigma_p(D).$

Lemma 2.

We have $\sigma(D) = \overline{\{d_n : n \in \Bbb{N}\}}.$

Proof.

Since the spectrum is closed and contains $\sigma_p(D)$, clearly we have $$\overline{\{d_n : n \in \Bbb{N}\}} = \overline{\sigma_p(D)} \subseteq \sigma(D).$$ Conversely, assume that $\lambda \notin \overline{\{d_n : n \in \Bbb{N}\}}$. Then there exists $\varepsilon>0$ such that $|\lambda - d_n| \ge \varepsilon$ for all $n \in \Bbb{N}$. For any $x \in \mathcal{H}$ we then have $$\|(D-\lambda I)x\|^2 = \left\|\sum_{n=1}^\infty (d_n-\lambda)\langle x,e_n\rangle e_n\right\|^2 = \sum_{n=1}^\infty |d_n-\lambda|^2|\langle x,e_n\rangle|^2 \ge \varepsilon^2 \sum_{n=1}^\infty|\langle x,e_n\rangle|^2 = \varepsilon^2\|x\|^2$$ so $D-\lambda I$ is bounded from below. Also, for all $n \in \Bbb{N}$ we have $$(D-\lambda I)e_n = \underbrace{(d_n-\lambda)}_{\ne 0}e_n$$ so $R(D-\lambda I)$ contains all $(e_n)_{n=1}^\infty$ and is therefore dense in $H$. Operator which is bounded from below and has dense range is invertible so it follows that $D-\lambda I$ is invertible and therefore $\lambda \notin \sigma(D)$.

Now we can prove $(1)$.

Assume that $\lambda$ is a limit point of $\sigma_p(D)$.

First assume that $\lambda \notin \sigma_p(D)$. Since $\lambda$ is a limit point of $\sigma_p(D)$ it follows $\lambda \in \overline{\sigma_p(D)} = \sigma(D)$ so $D-\lambda I$ is injective but not invertible. Therefore, $D - \lambda I$ is not surjective. However, for all $n \in \Bbb{N}$ we have $$(D-\lambda I)e_n = \underbrace{(d_n-\lambda)}_{\ne 0}e_n$$ so $R(D-\lambda I)$ contains all $(e_n)_{n=1}^\infty$ and is therefore dense in $H$, but not equal to $\mathcal{H}$. It follows that $R(D-\lambda I)$ is not closed.

Now assume that $\lambda \in \sigma_p(D)$ (and is a limit point of $\sigma_p(D)$). Let $\lambda = d_m$ for some $m \in \Bbb{N}$. Denote the Hilbert subspace $$\mathcal{K} := \overline{\operatorname{span}\{e_n : n \in \Bbb{N}, d_n \ne d_m\}}.$$ We can look at $D|_\mathcal{K} : \mathcal{K} \to \mathcal{K}$ which is again a diagonal operator induced by $\{d_n : n \in \Bbb{N}, d_n \ne d_m\}$ and we now have that $\lambda = d_m$ is a limit point of this set but not contained in it. By the first case it follows that the range $R(D|_\mathcal{K} - \lambda I_{\mathcal{K}})$ is not closed in $\mathcal{K}$, so $$R(D-\lambda I) = R(D|_\mathcal{K} - \lambda I_{\mathcal{K}})$$ cannot be closed in $\mathcal{H}$.

Conversely, assume that $\lambda$ is not a limit point of $\sigma_p(D)$.

First assume $\lambda \notin \sigma_p(D)$. Then $\lambda \notin \overline{\sigma_p(D)} = \sigma(D)$ so $D-\lambda I$ is invertible. Therefore $R(D-\lambda I) = \mathcal{H}$ and hence closed.

Now assume that $\lambda \in \sigma_p(D)$ (and is not a limit point of $\sigma_p(D)$). Let $\lambda = d_m$ for some $m \in \Bbb{N}$. Again, defining $\mathcal{K}$ as above, we have that $D|_\mathcal{K} : \mathcal{K} \to \mathcal{K}$ is a diagonal operator induced by $\{d_n : n \in \Bbb{N}, d_n \ne d_m\}$ and we now have that $\lambda = d_m$ is not a limit point of this set nor is it contained in it. Hence $$\lambda \notin \overline{\{d_n : n \in \Bbb{N}, d_n \ne d_m\}} = \sigma(D|_\mathcal{K})$$ so $D|_\mathcal{K} - \lambda I_{\mathcal{K}}$ is invertible and hence surjective so we have $$R(D-\lambda I) = R(D|_\mathcal{K} - \lambda I_{\mathcal{K}}) = \mathcal{K}$$ which is closed in $\mathcal{H}$.