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!

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

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.
• thank you very much. please what is the difference between $\sigma$-finite and finite? Dec 20, 2017 at 15:46