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Dimock's book states the spectral theorem for bounded operators as follows.

Theore: Let $T$ be a bounded self-adjoint operator on a Hilbert space $\mathcal{H}$. Then there exists a measure space $(\mathcal{M},\mu)$, a bounded measurable function $\tau: \mathcal{M}\to \mathbb{R}$ and a unitary operator $V:\mathcal{H}\to L^{2}(\mathcal{M},d\mu)$ such that $T = V^{-1}[\tau]V$ where $[\tau]$ is the operator multiplication by $\tau$.

The proof of the above result is omitted in Dimock's book but he refers to Reed and Simon's book for the proof. I've looked the latter, specifically in the section VII.2 - The spectral Theorem, but I did not find the exactly statement. There are some results such as Borel functional calculus and other spectral theorem results, but the one I think is the closest to Dimock's theorem is the following.

Theorem: Let $A$ be a bounded self-adjoint operator on a separable Hilbert space $\mathcal{H}$. Then there exists a finite measure space $(\mathcal{M},\mu)$, a bounded function $\tau$ on $\mathcal{M}$ and a unitary map $U: \mathcal{H}\to L^{2}(\mathcal{M},\mu)$ so that: \begin{eqnarray} (UAU^{-1}f)(m) = F(m)f(m) \tag{1}\label{1} \end{eqnarray}

As you can see, this is not Dimock's version: in Reed & Simon's version, $\mathcal{H}$ is separable and $(\mathcal{M},\mu)$ is finite. Furthermore, it does not state explicitly that $A = U^{-1}[\tau]U$.

Question: Does Dimock's version follows from the above result from Reed & Simon (or maybe another result from this book)? And in the affirmative case, how to prove it?

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  • $\begingroup$ Here's a link that may be helpful. mtaylor.web.unc.edu/files/2018/04/specthm.pdf $\endgroup$ Aug 31 '20 at 18:21
  • $\begingroup$ This was crossposted on MO: mathoverflow.net/questions/370558/… (this should not be re-deleted; posting links between cross-posted questions is best practice) $\endgroup$
    – Neal
    Aug 31 '20 at 22:32
  • $\begingroup$ @MathMath I came here from there. Just wanted to reassure you that my comment is not an indictment. :) $\endgroup$
    – Neal
    Aug 31 '20 at 22:34
  • $\begingroup$ @Neal oh, sure! I understand your point! No problem at all! :) $\endgroup$
    – MathMath
    Aug 31 '20 at 22:37
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There are lots of books that have this theorem, but to answer your question, yes, you can derive the nonseparable result from the separable one. Pick a nonzero vector $v \in H$ and let $H_v$ be the closed span of $\{v, Av, A^2v, \ldots\}$. This subspace is invariant for $A$, and since $A$ is self-adjoint its orthocomplement $H_v^{\perp}$ is also invariant for $A$. So we can now pick a nonzero vector $v' \in H_v^{\perp}$ and let $H_{v'}$ be the closed span of $\{v', Av', A^2v', \ldots\}$, and so on. The result is that you can decompose $H$ into a direct sum of separable subspaces each of which is invariant for $A$. (The argument can be made rigorous using Zorn's lemma.) Then apply the separable result to each summand and sum up.

As for $A = U^{-1}[\tau]U$, that really is an immediate consequence of $(UAU^{-1}f)(m) = \tau(m)f(m)$.

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  • $\begingroup$ Hey Nik! Thanks for the answer! I think I got the idea! I've checked some references besides Reed & Simon but all of them omitted the proof by saying that "you can use Zorn Lemma to prove it" and I didn't even know where to begin. Now it's pretty much clearer! Thanks a lot! $\endgroup$
    – MathMath
    Aug 31 '20 at 22:43
  • $\begingroup$ BTW: what should I do with the post on mathoverflow? Should I delet it? Don't know what is the best practice now hehe. $\endgroup$
    – MathMath
    Aug 31 '20 at 22:44
  • $\begingroup$ Yeah, I'd go ahead and delete the one on mathoverflow. (I don't object to your posting it there after not getting an answer here, but you probably should have waited a little longer ...) $\endgroup$
    – Nik Weaver
    Aug 31 '20 at 22:51
  • $\begingroup$ Yes! You are probably right. Thanks again! $\endgroup$
    – MathMath
    Aug 31 '20 at 22:52
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Suppose first that $\mathcal{H}$ is separable. Then R&S's version directly implies Dimock's. Note that the right side of (1) is precisely $([\tau] f)(m)$, so (1) can be written as $U A U^{-1} f = [\tau] f$ for all $f \in \mathcal{H}$, which is to say $U A U^{-1} = [\tau]$ as operators. Now just multiply by $U^{-1}$ on the left and $U$ on the right to obtain the desired $A = U^{-1} [\tau] U$. Thus we have exactly Dimock's conclusion, plus a little bit more: the conclusion is that $\mu$ is not only a measure but in particular a finite measure.

For the non-separable case, we have the following lemma: given a self-adjoint operator $A$ on $\mathcal{H}$, there is an orthogonal decomposition $\mathcal{H} = \bigoplus_{i \in I} \mathcal{H}_i$, for some index set $I$, where each $\mathcal{H}_i$ is separable and invariant under $A$. The proof is an exercise in transfinite induction or Zorn's lemma. The key idea is that if $\mathcal{H}_0$ is invariant under $A$, then by self-adjointness, $\mathcal{H}_0^\perp$ is also invariant under $A$. So if $\mathcal{H}_0^\perp$ is nonzero, then pick any $x \in \mathcal{H}_0^{\perp}$ and consider $\mathcal{H}_1$, the closed linear span of $\{x, Ax, A^2 x, \dots\}$, which by construction is separable, orthogonal to $\mathcal{H}_0$, and invariant under $A$. Replace $\mathcal{H}_0$ by $\mathcal{H}_0 \oplus \mathcal{H}_1$ and the induction continues.

Applying the separable case, there are measure spaces $(\mathcal{M}_i, \mu_i)$ and functions $\tau_i$ such that the restriction of $A$ to $\mathcal{H}_i$ is unitarily equivalent to $[\tau_i]$. Set $\mathcal{M}$ to be the disjoint union of all the $\mathcal{M}_i$, with $\mu, \tau$ given by pasting together the $\mu_i, \tau_i$ in the obvious way, and it follows that $A$ is unitarily equivalent to $[\tau]$.

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  • $\begingroup$ Amazing answer! Thanks so much!! $\endgroup$
    – MathMath
    Aug 31 '20 at 22:45

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