I have been looking around for an example of a general continuous (bounded) linear operator, who's spectrum is any compact set $K\subset\mathbb{C}$.

I have seen an example, where we take the set $\{\alpha_n\}$ to be a countable dense subset of $K$. Let $H$ be a hilbert space with orthonormal basis $\{e_n\}$, define the operator $T:H\to H$ such that:

$T(u)=\sum_n\alpha_n\langle u,e_n\rangle e_n.$ I can see that any eigenvalue of $T$ is some $\alpha_j\in\{\alpha_n\}$ since:

$Tu-\lambda u=0\Rightarrow \sum_n\alpha_n\langle u ,e_n \rangle e_n-\sum_n\lambda\langle u,e_n\rangle e_n=0$ Now each $i$-th term gives us:

$\alpha_iu_i-\lambda u_i=0\Rightarrow \lambda = \alpha_i$ so we know that every value in $\{\alpha_n\}$ is in the spectrum of $T$ so we see that the spectrum is at least a dense subset of $K$.

But this is where I am stuck and would appreciate help/ explaining what comes next. Thanks!

  • 1
    $\begingroup$ The spectrum is always closed, so $\sigma(T) \supset \overline{\{\alpha_n : n\in\mathbb{N}\}} = K$. On the other hand, it is not hard to see that if $\lambda\notin K$ then $\lambda I - T$ is invertible. $\endgroup$ – Daniel Fischer Apr 5 '14 at 15:00
  • $\begingroup$ Yes, but how do we go the other way i.e. $\sigma(T)\subset\overline{\{\alpha_n:n\in\mathbb{N}}\}$ $\endgroup$ – Ellya Apr 5 '14 at 15:02
  • $\begingroup$ As we then need to show that the resolvent exists, is bounded and is densely defined in $H$. I see existence and boundedness, but not necessarily densely defined? $\endgroup$ – Ellya Apr 5 '14 at 15:03
  • $\begingroup$ That's what the second sentence is about. If $\lambda \notin K$, then $\lambda \notin \sigma(T)$. $\endgroup$ – Daniel Fischer Apr 5 '14 at 15:03
  • $\begingroup$ Could you show me why? that is the bit I'm stuck on, thanks! $\endgroup$ – Ellya Apr 5 '14 at 15:04

By construction, every $\alpha_k$ is an eigenvalue of $T$, hence we have

$$K = \overline{\{\alpha_k : k\in\mathbb{N}\}} \subset \sigma(T).$$

On the other hand, if $\lambda\notin K$, then there is a $\delta > 0$ such that $\lvert\lambda - \alpha_k\rvert > \delta$ for all $k$, and then we find the inverse of $\lambda I - T$ explicitly:

$$\begin{align} (\lambda I - T)\left(\sum_{n\in\mathbb{N}} y_n\cdot e_n\right) &= \sum_{n\in\mathbb{N}} x_n e_n\\ \iff \sum_{n\in\mathbb{N}} (\lambda - \alpha_n)y_n e_n &= \sum_{n\in\mathbb{N}} x_n e_n\\ \iff \bigl(\forall n\in \mathbb{N}\bigr)\bigl( (\lambda-\alpha_n)y_n &= x_n\bigr)\\ \iff \bigl(\forall n\in\mathbb{N}\bigr)\biggl(y_n &= \frac{1}{\lambda-\alpha_n}x_n\biggr), \end{align}$$


$$(\lambda I - T)^{-1}\left(\sum_{n\in\mathbb{N}} x_n e_n\right) = \sum_{n\in\mathbb{N}} \frac{x_n}{\lambda-\alpha_n}e_n.$$

Since $\lvert\lambda - \alpha_n\rvert > \delta$ for all $n$, we have $\left\lvert\frac{1}{\lambda-\alpha_n}\right\rvert < \frac{1}{\delta}$, and so $(\lambda I - T)^{-1}$ is a globally defined continuous operator, hence $\lambda \notin \sigma(T)$, which yields the other inclusion $\sigma(T) \subset K$, altogether $\sigma(T) = K$.

  • $\begingroup$ One thing, the sums and sets should all be over $\mathbb{Q}$, since $K\subset \mathbb{C}$ and $\{\alpha_n\}$ is dense subset of $K$ i.e. all values in $K$ with rational coeffiencents. Anyway, would the last series converge $\forall x\in H$? $\endgroup$ – Ellya Apr 5 '14 at 15:22
  • 1
    $\begingroup$ Careful. $K$ need not contain any point with rational coefficients, so taking $\{\alpha_n\}$ to be the points in $K$ with rational coefficients doesn't work then. What one does is take any countable dense subset of $K$, and countable means we can index by $\mathbb{N}$. We could also index by other sets, and we can even take a nonseparable Hilbert space, e.g. $\ell^2(K)$ (if $K$ is countable, that is separable), with all $\alpha\in K$ as eigenvalues. $\endgroup$ – Daniel Fischer Apr 5 '14 at 15:38
  • 1
    $\begingroup$ Well, scrutinise the assignment once more, carefully, to see whether that says the operator should be compact. If it doesn't, go to your lecturer and tell him that compactness of the operator wasn't asked for. If it does say that, go to your lecturer and tell him that you overlooked that part. $\endgroup$ – Daniel Fischer Apr 18 '14 at 9:16
  • 1
    $\begingroup$ Okay thank you! this is just so frustrating, because I'm seeing alot of mistakes he's making, such as proofs in the notes, bu my exam is in less than a month, and I don't know if I should write the wrong things just to get the marks! $\endgroup$ – Ellya Apr 18 '14 at 18:48
  • 1
    $\begingroup$ Ouch! That's bad. Don't take my word for it, but I think it would be better to write the right things, and if the need arises, appeal to a higher authority. $\endgroup$ – Daniel Fischer Apr 18 '14 at 18:52

Suppose that a non-empty set $K$ in $\mathbb{C}$ is given and choose a dense sequence $(\lambda_n)$ in $K$. If $T$ is the corresponding diagonal operator, $T(e_n)=\lambda_n e_n$, then ${\rm sp}(T)=\overline{\{\lambda_n: n\in\mathbb{N}\}}=K$. Thus, the spectrum is an arbitrary non-empty compact set in general.

  • $\begingroup$ Hi @M.Kardel here $(\lambda_n)$ is a dense countable scalars series and it is bounded right? And we take H in general (or for example $H=l_2(Z)$) Can you explain how you concluded that $sp(T)$ equals K? $\endgroup$ – Maths1999_ Apr 16 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.