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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}

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  • $\begingroup$ 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? $\endgroup$ – DisintegratingByParts Apr 27 at 0:56
  • $\begingroup$ 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. $\endgroup$ – Zeta-Squared Apr 27 at 1:01
  • $\begingroup$ 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? $\endgroup$ – DisintegratingByParts Apr 27 at 1:03
  • $\begingroup$ Ah ok. I see. That however, would be in the point spectrum, and so not part of the residual. $\endgroup$ – Zeta-Squared Apr 27 at 1:10
  • $\begingroup$ I was just pointing out that your conclusion was not correct. $\endgroup$ – DisintegratingByParts Apr 27 at 1:18
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$\lambda \in \sigma_r(T_{\alpha})$ 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 $C>0$.

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  • $\begingroup$ Could you please clarify the orthogonality argument for the residual spectrum? $\endgroup$ – Zeta-Squared Apr 26 at 8:31
  • $\begingroup$ @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 $y \neq 0$? $\endgroup$ – Kabo Murphy Apr 26 at 8:35
  • $\begingroup$ I am still confused. Would this mean $\lambda = (\alpha_{i})$? $\endgroup$ – Zeta-Squared Apr 26 at 8:55
  • $\begingroup$ 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 $\lambda$. $\endgroup$ – Kabo Murphy Apr 26 at 8:59
  • $\begingroup$ So if I am understanding correctly, $\lambda I-T_{\alpha}$ will not be injective. In fact it will be the zero map? $\endgroup$ – Zeta-Squared Apr 26 at 9:02
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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.

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  • $\begingroup$ 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 $\alpha_{n}$? $\endgroup$ – Zeta-Squared Apr 26 at 23:36
  • $\begingroup$ "Bounded" doesn't rule out residual spectrum, but "bounded normal" does. $\endgroup$ – Andreas Blass Apr 27 at 0:23
  • $\begingroup$ I have never heard of the term "bounded normal", what does this mean? $\endgroup$ – Zeta-Squared Apr 27 at 0:25
  • $\begingroup$ @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. $\endgroup$ – DisintegratingByParts Apr 27 at 0:47
  • $\begingroup$ I understand now. Could you please make sure my argument for the residual being empty is good? $\endgroup$ – Zeta-Squared Apr 27 at 0:50

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