# Two versions of the spectral Theorem?

I'm studying the spectral Theorem (for bounded self-adjoint operators) by myself and I'm following Nik Weaver's nice book. Let me first introduce some notations first.

Notations: If $$\mathcal{H}$$ is a Hilbert space, $$\mathcal{B}(\mathcal{H})$$ is the (Banach space) of all bounded linear operators $$A: \mathcal{H} \to \mathcal{H}$$. If $$A \in \mathcal{B}(\mathcal{H})$$, $$\mbox{sp}(A)$$ is the spectrum of $$A$$.

Now, let $$(X, \mathcal{F},\mu)$$ be a $$\sigma$$-finite measure space. A measurable Hilbert bundle over $$X$$ is a disjoint union: $$\mathcal{X} = \bigcup_{n\in \mathbb{N}}(X_{n}\times \mathcal{H}_{n})$$ where $$\{X_{n}\}_{n\in \mathbb{N}}$$ is a measurable partition of $$X$$ and, for each $$0 \le n \le \infty$$, $$\mathcal{H}_{n}$$ is a Hilbert space with dimension $$n$$.

Finally, $$f: X \to \mathcal{H}$$ is weakly-measurable if the function $$x \mapsto \langle f(x),v\rangle$$ is measurable for every $$v \in \mathcal{H}$$. We denote $$L^{2}(X;\mathcal{H})$$ the set of all weakly measurable functions $$f: X \to \mathcal{H}$$ such that: $$||f|| := \int_{x}||f(x)||^{2}d\mu(x) < +\infty$$ modulo functions which are zero almost everywhere. This is a Hibert space with inner product: $$\langle f,g\rangle := \int_{x}\langle f(x),g(x)\rangle d\mu(x)$$ If $$f \in L^{2}(X;\mathcal{H})$$, $$M_{f}$$ is the operator multiplication by $$f$$. Also, $$L^{2}(X;\mathcal{X}) := \bigoplus_{n\in \mathbb{N}}L^{2}(\mathcal{X}_{n};\mathcal{H}_{n})$$.

Now, the statement of the spectral Theorem in this reference is as follows.

Theorem: Let $$\mathcal{B}(\mathcal{H})$$ be self-adjoint. Then there exits a probability measure $$\mu$$ on $$\mbox{sp}(A)$$, a measurable Hilbert bundle $$\mathcal{X}$$ over $$\mbox{sp}(A)$$ and an isometric isomorphism $$U: L^{2}(\mbox{sp}(A);\mathcal{X}) \to \mathcal{H}$$ such that $$A = UM_{x}U^{-1}$$.

However, I'm more interested in another version of this Theorem, which is stated in Dimock's book and goes like (with adapted notation)

Theorem: Let $$A \in \mathcal{B}(\mathcal{H})$$ be self-adjoint. Then, there exists a measure space $$(\mathcal{M},\mathcal{\Omega},\mu)$$, a bounded measurable function $$\tau: \mathcal{M}\to \mathbb{R}$$ and a unitary operator $$U: \mathcal{H}\to L^{2}(\mathcal{M},\mu)$$ such that $$A = UM_{\tau}U^{-1}$$.

Question: How can I obtain Dimock's version of the spectral Theorem from Weaver's version of it?

Let $$\mathcal{M}$$ be a disjoint union consisting of $$n$$ copies of $$X_n$$ for each $$n$$. The given measure on $$\mbox{sp}(A)$$ restricts to a measure on $$X_n$$ and thus induces a measure on $$\mathcal{M}$$. There is then an isomorphism $$L^2(\mathcal{M})\cong L^2(\mbox{sp}(A);\mathcal{X})$$: if you pick an orthonormal basis for each $$\mathcal{H}_n$$, then $$L^2(X_n;\mathcal{H}_n)$$ is just a direct sum of $$n$$ copies of $$L^2(X_n)$$, and when you take the direct sum of these over all $$n$$ you get $$L^2(\mathcal{M})$$. This isomorphism $$L^2(\mathcal{M})\cong L^2(\mbox{sp}(A);\mathcal{X})$$ turns multiplication by $$x$$ on $$\mbox{sp}(A)$$ to multiplication by the function $$\tau$$ on $$\mathcal{M}$$ which is given by the inclusion function $$X_n\to\mathbb{R}$$ on each copy of each $$X_n$$.
• Incidentally, Weaver's proof of Corollary 3.4.3 is wrong (specifically, Proposition 2.4.7 does not justify identifying $L^2(X;\mathcal{X})$ with $\bigoplus L^2(S_k,\mu|_{S_k})$). The result is correct but requires a more difficult proof, and is part of what is known as the Hahn-Hellinger theorem. Aug 23, 2020 at 2:59
• I don't know a reference for the full proof. You can find a more general theory (which applies even in the non-separable case) in the final chapter of Halmos's nice little book Introduction to Hilbert Space and the Theory of Spectral Multiplicity. Roughly speaking, this gives a "measurable Hilbert bundle" on the spectrum except that instead of having disjoint sets $X_n$ with respect to a single measure $\mu$, you have a family of pairwise orthogonal measures (and instead of indexing by $n\in\mathbb{N}\cup\{\infty\}$, it can be indexed by arbitrary cardinalities). Aug 23, 2020 at 3:49
• In the separable case, this family of pairwise orthogonal measures is countable and each measure is $\sigma$-finite, and so you can use the Lebesgue decomposition theorem to realize them as the restrictions of a single measure to disjoint subsets of the spectrum, thus obtaining a measurable Hilbert bundle in Weaver's sense. Aug 23, 2020 at 3:51