# Spectral Measures: Special Spectrum

Problem

Given a Hilbert space $\mathcal{H}$.

Denote eigenvalues by: $$\sigma_0(N):=\{\lambda\in\mathbb{C}:\mathcal{N}(\lambda-N)\neq(0)\}$$

Then arbitrary sets admit: $$\Lambda\subseteq\mathbb{C}:\quad\sigma_0(N)=\Lambda\quad\sigma(N)=\overline{\Lambda}$$

For some normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$

How can I construct such?

Application

For an open domain: $$\Lambda\subseteq\mathbb{C}:\quad\Lambda=\Lambda^\circ\implies E(\partial\Lambda)=0$$

So the implication fails: $$\Delta=\overline{\Delta}:\quad E(\sigma\cap\Delta)=0\nRightarrow\sigma\cap\Delta=\varnothing$$

It remains valid though: $$\Delta=\Delta^\circ:\quad E(\sigma\cap\Delta)=0\implies\sigma\cap\Delta=\varnothing$$

(As check for singular spectrum.)

Construction

Given the Hilbert space $\ell^2(\Lambda)$.

Construct spectral measure: $$E(A)\varphi:=\sum_{\lambda\in\Lambda}\delta_{\lambda\in A}\langle\delta_\lambda,\varphi\rangle\delta_\lambda$$

And its normal operator: $$N:=\int\lambda\mathrm{d}E(\lambda):\quad N^*N=NN^*$$

Then it has as desired: $$\sigma_0(N)=\Lambda\quad\sigma(N)=\overline{\Lambda}$$

Concluding the construction.

Reference

This builts up on: Constructions