3
$\begingroup$

Let $A\in\mathcal{L}(E)$ be a self-adjoint operator on a separable complex Hilbert space $E$. Then there exists a $\sigma$-finite measure space $(X,\mu)$, a bounded, measurable, real-valued function $\varphi$ on $X$, and a unitary map $U: E\rightarrow L^2(X, \mu)$ such that $$\left[UAU^*f\right](\lambda)=\varphi(\lambda)f(\lambda),\;\forall f\in L^2(X, \mu).$$

I want to know what happen if $E$ is not separable.

Thank you!

$\endgroup$
2
  • $\begingroup$ $\sigma$-finite $\ne$ finite $\endgroup$
    – daw
    Commented Dec 20, 2017 at 15:36
  • $\begingroup$ Thank you. What happen if $E$ is not separable $\endgroup$
    – Student
    Commented Dec 20, 2017 at 15:38

1 Answer 1

Reset to default
2
$\begingroup$

The spectral theorem tells you the existence of a measure space $(X,\mu)$. If $E$ would be separable then the measure space is $\sigma$-finite. If not: we do not know.

Take the null operator on your favorite non-separable space. Take the measurable space $X=\{0\}$, $\mu$ counting measure. Set $U:E\to L^2(\mu)$ to be zero, $\phi=0$. Now you have a spectral decomposition with a finite measure space despite the fact that $E$ is non-separable.

$\endgroup$
2
  • $\begingroup$ thank you very much. please what is the difference between $\sigma$-finite and finite? $\endgroup$
    – Student
    Commented Dec 20, 2017 at 15:46
  • $\begingroup$ en.wikipedia.org/wiki/%CE%A3-finite_measure $\endgroup$
    – daw
    Commented Dec 20, 2017 at 16:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .