The spectral theorem states that for a self-adjoint linear operator $A$ with $\left\|A\right\|=1$ on a Hilbert space, there is a measure space $(X,\mathcal M,\mu)$ so that $A$ is unitarily equivalent to a multiplication operator on $L^2(I,d\mu)$ where $I:=[-1,1]$. If $H$ is separable, then $\mu$ can be taken to be a finite measure.

Now my question is: If $H$ is not separable, can we always take $\mu$ to be $\sigma$-finite?

  • $\begingroup$ you should probably say "on a Hilbert space $H$" $\endgroup$ – mathworker21 Oct 14 '18 at 23:32

If every real $x\in[0,1]$ is an eigenvalue, then the $\mu$ cannot be sigma finite.

  • $\begingroup$ For instance, let $H = \ell^2([0,1])$, i.e. $L^2$ of $[0,1]$ equipped with the discrete $\sigma$-algebra and counting measure. Then define $(Af)(x) = x f(x)$. $\endgroup$ – Nate Eldredge Oct 15 '18 at 4:30

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