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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove or disprove the following assertions for a linear map $C$ from a Banach space $X$ into itself:

a) If C is compact then its spectral radius equals the maximum of the absolute value of $C$

I'm not sure what the definition of absolute value of an operator is. But the spectral value is defined as $|\sigma(M)| = max_{\lambda \in \sigma(M)} |\lambda|$

edit: I split it up to two questions

share|cite|improve this question
2  
I can't make sense of the question, and you also don't know what it means. Where did it come from? – Jonas Meyer Dec 12 '12 at 15:23
    
An old exam, maybe there was a miss print. – Johan Dec 12 '12 at 16:09
    
If the absolute value of $C$ is the norm of $C$, then the answer is no. It becomes YES though, assuming further that $C$ is self-adjoint. – Yiorgos S. Smyrlis Dec 30 '15 at 10:40

This is false, consider Volterra operator on $C[0,1]$. Radius $0$ but $ \mid Cf \mid \ge 1$, just take $ \chi_{[0,1]} $

share|cite|improve this answer

If your Banach space is $\mathbb R^2$ and $C=\left(\begin{matrix}0&1\\ 0&0 \end{matrix}\right)$, then $\lvert C\rvert=1$, while its spectral radius is $0$.

share|cite|improve this answer

Your Answer

 
discard

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

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