Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be a bounded linear operator in a Hilbert space $H$.

I had the misconception that the continuous spectrum of $A$ would necessarily have some "continuous" appearance: an interval, a union of intervals, or something like that. This is false as the following operator shows:

$$Vf(t)=\int_0^t f(s)\, ds,\qquad f \in L^2([0, 1])$$

(see Wikipedia) the spectrum of this operator is purely continuous and is reduced to $\{0\}$.

However this operator is not self-adjoint. This leads to the question:

Question Is it true that, if $A$ is self-adjoint and $0$ is an isolated spectral value, then $0$ is not in the continuous spectrum (and so it is an eigenvalue)?

This is true if $A$ is compact, I believe: in that case, if $0$ is an isolated spectral value then $A$ only has a finite number of eigenvalues and so, by Hilbert-Schmidt theorem, it is a finite rank operator. Then any spectral value is an eigenvalue. But what about the general case? I suspect that in the general setting things are not so easy.

share|improve this question
    
It is true more generally that an isolated point in the spectrum of a normal operator is an eigenvalue. See Pedersen, Analysis now, Prop. 4.4.5, p. 157. –  t.b. Sep 10 '11 at 17:54
    
@Theo: Indeed the operator $V$ above is not normal (I just checked :-) ). So, at least for this class of operators the continuous spectrum is really something "continuous", having no isolated points. I noticed that the converse is not true, that is, the set of eigenvalues needs not be "granular" in nature. For example, if I'm not mistaken, the operator $$Uf(t)=f(t-1), \qquad f \in L^2(\mathbb{R})$$ is unitary, hence normal, its spectrum consist of the whole unit circle and every spectral value is an eigenvalue. Strange! –  Giuseppe Negro Sep 10 '11 at 18:34
    
@Theo: Sorry to disturb you again, it is just to say that unfortunately the previous example I made is wrong! I erroneously assumed that $e^{i \theta t} \in L^2(\mathbb{R})$. And indeed, as I see in Rudin's Functional analysis, the point spectrum of a normal bounded operator in a separable Hilbert space is at most countable (Exercise 12-18.b pag.343). Anyway, your point in the last comment is very clear and it is exactly what I was looking for. It is important to know where terminology we use comes from. Thank you! –  Giuseppe Negro Sep 10 '11 at 19:00
    
Yes, sorry for the deletion of the comment and thanks for clarifying, but you had me confused for a moment (and I now really must run, but I wanted to clear it up): The point is of course that eigenvectors to distinct eigenvalues are orthogonal, so you can choose an orthonormal system of eigenvectors, and an orthonormal system can at most be countable. As I said before, the names point spectrum and continuous spectrum derive from how these parts of the spectrum typically look like. It does not describe the spectrum accurately, as you show with your examples. –  t.b. Sep 10 '11 at 19:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.