Continuous and residual spectrum

Note: Please do not give a solution; I would prefer guidance to help me complete the question myself. Thank you.

I am having trouble understanding and finding the continuous and residual spectrum. I am working through the following problem:

Let $$\alpha = (\alpha_{i})\in\ell^{\infty}$$ and let $$T_{\alpha}:\ell^{2}\rightarrow\ell^{2}$$ with $$T_{\alpha}x=(\alpha_{1}x_{1},\alpha_{2}x_{2},\ldots)$$.

(i) Compute the spectrum, $$\sigma(T_{\alpha})$$, of $$T_{\alpha}$$.

(ii) Identify the point spectrum, $$\sigma_{p}(T_{\alpha})$$, the continuous spectrum, $$\sigma_{c}(T_{\alpha})$$, and the residual spectrum, $$\sigma_{r}(T_{\alpha})$$, of $$T_{\alpha}$$.

My Solution

(i) Suppose $$\lambda\in\rho(T_{\alpha})$$ (where $$\rho(T_{\alpha})$$ is the resolvent set of $$T_{\alpha}$$) then $$\lambda I-T_{\alpha}$$ is bijective. Hence, for every $$y\in\ell^{2}$$ $$\exists ! x\in\ell^{2}$$ such that, \begin{align} (\lambda I-T_{\alpha})x = y\implies x = (\lambda I-T_{\alpha})^{-1}y, \end{align} and \begin{align} x_{n} = \frac{y_{n}}{\lambda - \alpha_{n}}, \end{align} for each element, $$n\in\mathbb{N}$$. Since $$x\in\ell^{2}$$, \begin{align} |x|_{2}^{2} = \sum_{n=1}^{\infty}|x_{n}|^{2}=\sum_{n=1}^{\infty}\frac{|y_{n}|^{2}}{|\lambda-\alpha_{n}|^{2}}<\infty. \end{align} Therefore, if $$\lambda\in\overline{\{\alpha_{n}\}}$$ then $$|x|_{2} = \infty$$. Hence $$\sigma(T_{\alpha}) = \{\lambda\in\mathbb{C}|\lambda\in\overline{\{\alpha_{n}\}},n\in\mathbb{N}\}$$.

(ii) Using the definition of point spectrum, \begin{align} \sigma_{p}(T_{\alpha}):=\{\lambda\in\mathbb{C}|\lambda I-T_{\alpha} \text{is not injective}\}, \end{align} then there exists $$x_{1},x_{2}\in\ell^{2}$$, with $$x_{1}\neq x_{2}$$, such that $$(\lambda I-T_{\alpha})x_{1} = (\lambda I-T_{\alpha})x_{2}$$. This occurs when $$|x|_{2}=\infty$$, that is, $$\lambda = \alpha_{n}$$ for some $$n\in\mathbb{N}$$. Hence, $$\sigma_{p}(T_{\alpha})=\{\lambda\in\mathbb{C}|\lambda = \alpha_{n}\text{ for some }n\in\mathbb{N}\}$$.

At this stage I do not know how to continue. Particularly, I don't understand how to use the definitions of continuous and residual spectrum: \begin{align} \sigma_{c}(T_{\alpha})&:=\{\lambda\in\mathbb{C}|\lambda I-T_{\alpha}\text{ is injective, }\overline{\text{im}(\lambda I-T_{\alpha})}=\ell^{2},(\lambda I-T_{\alpha})^{-1}\text{ is unbounded}\},\\ \sigma_{r}(T_{\alpha})&:=\{\lambda\in\mathbb{C}|\lambda I-T_{\alpha}\text{ is injective, }\overline{\text{im}(\lambda I-T_{\alpha})}\neq\ell^{2}\} \end{align}

Finding continuous and residual spectrum

Assume $$\overline{\text{im}(\lambda I-T_{\alpha})}\neq\ell^{2}$$ then $$\text{im}(\lambda I-T_{\alpha})^{\perp}\neq\{0\}$$. Hence there is $$y\in\text{im}(\lambda I-T_{\alpha})^{\perp}$$, $$y\neq 0$$, such that, \begin{align} \sum_{n=1}^{\infty}(\lambda x_{n}-\alpha_{n}x_{n})y_{n} = 0, \end{align} for all $$x\in\overline{\text{im}(\lambda I-T_{\alpha})}$$. This implies $$\lambda=\alpha_{n}$$ for each $$n^{\text{th}}$$ element of $$x$$, which is impossible. Otherwise, $$x=0$$, however this contradicts our assumption that holds for all $$x\in\overline{\text{im}(\lambda I-T_{\alpha})}$$. Hence by contradiction, $$\overline{\text{im}(\lambda I-T_{\alpha})}=\ell^{2}$$ and $$\text{im}(\lambda I-T_{\alpha})^{\perp}=\{0\}$$. Therefore, $$\sigma_{r}(T_{\alpha})=\emptyset$$.

By above $$(\lambda I-T_{\alpha})$$ has trivial kernel and hence injective. Given $$\lambda\in\sigma_{c}(T_{\alpha})$$ we need $$(\lambda I-T_{\alpha})^{-1}$$ unbounded. From above this implies, \begin{align} |x|_{2}^{2}=\sum_{n=1}^{\infty}\frac{|y_{n}|^{2}}{|\lambda-\alpha_{n}|^{2}}\rightarrow\infty. \end{align} Hence, for all $$\epsilon>0$$, $$\exists N\in\mathbb{N}$$ such that for $$n>N$$, \begin{align} |\lambda-\alpha_{n}|<\epsilon. \end{align} Now $$(\alpha_{n})\in\ell^{\infty}$$ so they are uniformly bounded. Taking $$\lambda=\sup_{n\in\mathbb{N}}|\alpha_{n}|$$ we satisfy the criteria and hence $$(\lambda I-T_{\alpha})^{-1}$$ is unbounded.

Therefore, \begin{align} \sigma_{p}(T_{\alpha}) &= \{\lambda\in\mathbb{C}|\lambda=\alpha_{n}\text{ for some }n\in\mathbb{N}\},\\ \sigma_{c}(T_{\alpha}) &= \{\lambda\in\mathbb{C}|\lambda = \sup_{n\in\mathbb{N}}|\alpha_{n}|\},\\ \sigma_{r}(T_{\alpha}) &= \emptyset. \end{align}

• How did you go from\begin{align} \sum_{n=1}^{\infty}(\lambda x_{n}-\alpha_{n}x_{n})y_{n} = 0, \end{align} to concluding that $\lambda=\alpha_n$ for all $n$ in your argument? – DisintegratingByParts Apr 27 at 0:56
• I am working under the assumption $y\neq 0$. So in order for the sum on the left to be zero either $\lambda =\alpha_{n}$ for every $n$ or $x=0$. Both cases are not possible. – Zeta-Squared Apr 27 at 1:01
• Your conclusion that $\lambda=\alpha_x$ for all $n$ is not true. You could have $y_n=0$ for all but one $n$ and $\lambda=\alpha_n$ for the other $n$. And what would that look like? – DisintegratingByParts Apr 27 at 1:03
• Ah ok. I see. That however, would be in the point spectrum, and so not part of the residual. – Zeta-Squared Apr 27 at 1:10
• I was just pointing out that your conclusion was not correct. – DisintegratingByParts Apr 27 at 1:18

$$\lambda \in \sigma_r(T_{\alpha})$$ iff $$\lambda I-T_{\alpha}$$ is injective and there is a non zero element $$y$$ orthogonal to $$im(\lambda I-T_{\alpha})$$ which means $$\sum (\lambda x_i-\alpha_i x_i) y_i=0$$ for all $$x \in \ell^{2}$$. It should be easy to determine when this happens. For the continuous spectrum see what it means to have $$\|\lambda I-T_{\alpha}(x)\|^{2}\geq C \|x\|^{2}$$ for some $$C>0$$.

• Could you please clarify the orthogonality argument for the residual spectrum? – Zeta-Squared Apr 26 at 8:31
• @Jack A linear subspace is dense iff its orthogonal complement is $\{0\}$. This leads to the equation $\sum (\lambda x_i-\alpha_ix_i)y_i=0$ for all $x$. Take basis vectors $e_j$ for $x$ to see that $(\lambda -\alpha_i)y_i=0$ for all $i$. When does this happen for some $y \neq 0$? – Kabo Murphy Apr 26 at 8:35
• I am still confused. Would this mean $\lambda = (\alpha_{i})$? – Zeta-Squared Apr 26 at 8:55
• Some $y_i \neq 0$ so $\lambda =\alpha_i$ for this $i$. But remember to check if $\lambda I -T_{\alpha}$ is injective for this $\lambda$. – Kabo Murphy Apr 26 at 8:59
• So if I am understanding correctly, $\lambda I-T_{\alpha}$ will not be injective. In fact it will be the zero map? – Zeta-Squared Apr 26 at 9:02

A quick overview might be helpful, even though it does not really fit your requirement for an answer.

$$T_{\alpha}$$ is a bounded normal operator. That rules out all but continuous and point spectrum. Every $$\alpha_j$$ is in the point spectrum of $$T_{\alpha}$$, which is easily demonstrated by showing $$T_{\alpha}e_n = \alpha_n e_n$$ where $$e_n=\{0,0,\cdots,1,0,0,\cdots\},$$ and where $$1$$ is in the $$n$$-th position. Every cluster point of the set $$\{ \alpha_n \}$$ that is not in the set of eigenvalues is in the continuous spectrum, which is the approximate point spectrum.

• Why does $T_{\alpha}$ bounded rule out residual spectrum? Also, what do you mean by cluster point? Is this an accumulation point? EDIT: If we are talking about accumulation points, should we also be careful not include those points which are equal to an $\alpha_{n}$? – Zeta-Squared Apr 26 at 23:36
• "Bounded" doesn't rule out residual spectrum, but "bounded normal" does. – Andreas Blass Apr 27 at 0:23
• I have never heard of the term "bounded normal", what does this mean? – Zeta-Squared Apr 27 at 0:25
• @Jack : If $\alpha$ is a point of accumulation of eigenvalues, then it could be an eigenvalue if it is one of the $\alpha_n$ or it might not be. If the point of accumulation of eigenvalues is not an eigenvalue of your operator, then $(\alpha I-T)$ will not have a bounded inverse. – DisintegratingByParts Apr 27 at 0:47
• I understand now. Could you please make sure my argument for the residual being empty is good? – Zeta-Squared Apr 27 at 0:50