Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Good evening,

I have a question concerning the relation between approximate point spectrum and the spectrum of an operator.

Let $T$ be a bounded linear operator of a complex Hilbert space $H.$ The approximate point spectrum of $T$ is the set of all values $\lambda \in \mathbb{C}$ such that there exists a sequence of unit vectors $u_n\in H$ so that $\|(T-\lambda)u_n\|\to 0$ as $n\to \infty.$ We denote this set by $\sigma_{ap}(T)$. We denote the wellknown spectrum of $T$ by $\sigma(T)$.

We know that $\sigma_{ap}(T)$ contains the topological boundary of $\sigma(T)$.

My question : Can we have $\sigma_{ap}(T)\subset\partial\sigma(T)$?

Any help is appreciated. Thanks in advance.

Duc Anh

share|improve this question
1  
What does "converse" mean. You could write (*) as $\sigma_{ap}(T) \supseteq \partial \sigma(T)$. Are you asking if the other inclusion, "$\subseteq$", is true? (It's not). –  Matthew Daws Apr 14 '12 at 8:14
    
you're right. I'm sorry for the unclear question. I've editted it as above. Do you have a counter-example for it? –  Đức Anh Apr 14 '12 at 9:27
2  
So you are asking if it's possible to have $\sigma_{ap}(T) = \partial \sigma(T)$? This isn't always true (left shift on $\ell^2$). But if $T$ is compact then $\sigma(T)$ is the point spectrum of $T$, a sequence converging to $0$ (or finite)-- so it is true if $T$ is compact. –  Matthew Daws Apr 14 '12 at 13:44
    
Thank you very much –  Đức Anh Apr 14 '12 at 16:00
add comment

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.