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?

  • 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

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.


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