2
$\begingroup$

In general a spectrum of a linear operator has a decomposition into three parts: point spectrum, continuous spectrum and residual spectrum.

What I'm interested in is the topology of these parts of spectrum. Is the point spectrum always discrete? Is the continuous spectrum always continuous? (If yes can you refer to a proof?)

$\endgroup$
1
$\begingroup$

One source of nice examples is the operator of multiplication by $z$ on $L^2(\mu)$ where $\mu$ is a probability measure on a subset of the complex plane (a bounded subset if you want a bounded operator). These are normal operators, so there is no residual spectrum.

$\lambda$ is in the point spectrum iff $\mu(\{\lambda\}) > 0$. This is not necessarily discrete. For example, there are probability measures that assign positive probability to each rational in $[0,1]$, but to no irrationals. This is an example where the point spectrum is not discrete and the continuous spectrum is not open.

EDIT: For an example where the continuous spectrum is discrete, consider a case where $\mu$ assigns probability $1/n - 1/(n+1) > 0$ to $1/n$ for each natural number $n$. Then the continuous spectrum is $\{0\}$.

$\endgroup$
1
  • $\begingroup$ I'm sorry when I wrote open. I mean "not discrete", which means there are always another as closed as possible value in the spectrum. $\endgroup$ – quangtu123 Jan 6 '16 at 21:45

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.