# Discrete or continuous spectrum?

1. Mar 15, 2013

### ShayanJ

Consider an unbounded self-adjoint operator defined in a hilbert space(its domain isn't the entire hilbert space,of course).Can its spectrum have discrete and continuous parts simultaneity?Does it have eigenvectors with finite norm?
Thanks

2. Mar 15, 2013

### micromass

Staff Emeritus
Take a "nice" (for example: piecewise continuous) function $f:\mathbb{R}\rightarrow \mathbb{R}$. We can consider the multiplication operator

$$M:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R}): g\rightarrow fg$$

An element $\lambda\in \mathbb{C}$ is not in the spectrum of $M$ if and only if $M-\lambda I$ is not invertible. But we have

$$(M-\lambda I)(g)(x) = f(x)g(x) - \lambda g(x) = ( f(x) - \lambda 1) g(x)$$

We can see from this (and we can prove), that $M-\lambda I$ is invertible only if $f(x)\neq \lambda$ for any $x\in \mathbb{R}$. So the spectrum of $M$ is $\overline(f(\mathbb{R}))$.

Now take for example, the function

$$f(x) = \left\{\begin{array}{l}x-1 ~\text{if}~ x\leq 0\\ 0~\text{if} ~x>0\end{array}\right.$$

Then the spectrum will be $(-\infty,-1]\cup \{0\}$, which consists of "continuous" and "discrete parts".

Now, $g$ is an eigenvalue of $M$ if $Mg = \lambda g$ for a certain $\lambda\in \mathbb{C}$. So $f(x)g(x) = \lambda g(x)$ for all $x\in \mathbb{R}$.

If we take the function $f$as above, then we see that

$$g(x) = \left\{\begin{array}{l} 0~\text{if} ~x\leq 0\\ 1~\text{if}~ 0<x<1\\ 0~\text{if}~ x\geq 1\end{array}\right.$$

is an eigenvector. Furthermore, this has finite norm.