1
$\begingroup$

As we know, the spectrum of an operator $T$ has a standard decomposition into three parts:

  1. a point spectrum, consisting of eigenvalues of $T$ ;
  2. a continuous spectrum, consisting of the scalars that are not eigenvalues but make the range of $T-\lambda$ a proper dense subset of the space;
  3. a residual spectrum, consisting of all other scalars in the spectrum.

My question is are there any spectrum has continuous spectrum but no residual spectrum? Or conversely, a spectrum has residual spectrum but no continuous spectrum?

$\endgroup$
4
  • 1
    $\begingroup$ Do you want examples of such operators $T$? $\endgroup$
    – mm-crj
    Dec 4 '18 at 11:52
  • 1
    $\begingroup$ You should search harder. Example of purely residual spectrum, and the other direction combine this with this. $\endgroup$ Dec 4 '18 at 11:57
  • $\begingroup$ By the way, it is called "continuous spectrum" and not "continue spectrum". $\endgroup$
    – MaoWao
    Dec 4 '18 at 12:22
  • $\begingroup$ Thanks! I will figure it out myself. $\endgroup$
    – Yuyi Zhang
    Dec 4 '18 at 13:09
2
$\begingroup$

We can look at this perhaps from a different point of view that will make things more clear to you.

We can define the spectrum $\sigma(T)$ of a bounded linear operator $T$ on a Hilbert space $H$ to be the set of all $\lambda \in \mathbb{C}$ such that the $(T- \lambda I)$ is not a bijection.

  • The point spectrum of $T$ consists of all $\lambda \in \sigma(T)$ such that $(T- \lambda I)$ is not one-to-one. In this case $\lambda$ is called an eigenvalues of $T$.

  • The continuous spectrum of $T$ consists of all $\lambda \in \sigma(T)$ such that $(T- \lambda I)$ is one-to-one but not onto, and $range(T − \lambda I )$ is dense in $H$.

  • The residual spectrum of A consists of all $\lambda \in \sigma(T)$ such that $(T- \lambda I)$ is one-to-one but not onto, and $range(T- \lambda I)$ is not dense in $H$.

Now I think you can construct the examples or understand the examples given by @user10354138 on your own.

$\endgroup$

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.