I recently learned about the definition of the continuous part of the spectrum of self adjoint operators on Hilbert spaces. For a self adjoint operator A on a Hilbert space $\mathcal{H}$ it goes like this:
\begin{align}\sigma_c(A) = \{\lambda \in \mathbb{R}: \text{ran}(A -\lambda \mathbb{1}) \neq \overline{\text{ran}(A -\lambda \mathbb{1})} \}
\end{align}
This is all fair and good. I can see that this is clearly defining some part of the spectrum because the resolvent does not exists. And i can also see that is not something that can be achieved on a finite dimensional vector space. But the question still remains for me if the set we just created is worthy of its name. A "continuous" spectrum should in my opinion really be made up out of open, half open or closed intervals.
But is this really always the case and can we prove this property?
Also: Since we only care about self adjoint operators restricting to the reals for the Eigenvalues is fine for now.
Edit: Terribly sorry. I got mixed up and instead wrote down the definition of the pure point spectrum. Also i actually originally referenced the "purely continuous part" which is also something a bit different as i understand. Thanks for pointing it out!
Best Answer
The continuous spectrum of an operator may not look like an interval at all. In fact, it can very well be the Cantor set! Here is an example:
Let $\mu $ be any positive Borel measure on the Cantor set, here denoted by $C$, such that $\mu $ has no atoms (that is, $\mu (\{x\})=0$, for all $x$ in $C$) and full support (that is, $\mu (U)>0$, for all nonempty open sets $U\subseteq C$).
See below for the construction of such a measure.
Consider the operator $T$ on $L^2(C,\mu )$ given by $$ T(\xi )|_x = h(x)\xi (x), \quad \forall \xi \in L^2(C,\mu ), \quad \forall x\in C, $$ where $h$ is the function defined on $C$ by $h(x)=x$.
Using this wikipedia page it follows that the spectrum of $T$ coincides with its continuous spectrum, which in turn coincides with the range of $h$, a.k.a $C$.
Here is one way to construct a measure $\mu $ on the Cantor set with the required properties. First of all recall that the Cantor set is homeomorphic to $\{0,1\}^{\mathbb N}$, also known as Bernouli's space. For a concrete homeomorphism take $ \varphi :\{0,1\}^{\mathbb N}\to C, $ given by $$ \varphi (x) = \sum_{n=1}^\infty x_n3^{-n}, $$ for every $x=(x_1,x_2,\ldots ) \in \{0,1\}^{\mathbb N}$.
Consider the uniform probability measure $\rho $ on $\{0,1\}$, given by $\rho (\{0\}) =\rho (\{1\}) =1/2$, and let $$ \nu =\prod_{n=1}^\infty \rho $$ be the corresponding product measure. Incidentally $\nu $ is known as the Bernouli measure.
It is well known (and easy to prove) that $\nu $ has no atoms and full support. Since $\varphi $ is a homeomorphism, it follows that the push forward measure $\mu := \varphi _*(\nu )$ has the same properties.