# Classification of spectrum in functional analysis.

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?

• Do you want examples of such operators $T$? – mm-crj Dec 4 '18 at 11:52
• You should search harder. Example of purely residual spectrum, and the other direction combine this with this. – user10354138 Dec 4 '18 at 11:57
• By the way, it is called "continuous spectrum" and not "continue spectrum". – MaoWao Dec 4 '18 at 12:22
• Thanks! I will figure it out myself. – Yuyi Zhang Dec 4 '18 at 13:09

## 1 Answer

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.