Definition: (Resolution of Identity) A projection valued measure that is finitely additive and factors over intersections.

Theorem 1: Let $A$ be a normal operator in $H$ a Hilbert Space where everything is over the complex field. Then there exists an $(X, \Sigma, \mu)$ measure space and a unitary equivalence between $H$ and $L^2(X)$ so that $A$ is represented as a multiplication operator of an $L^\infty$ function.

I have already compared the versions of the spectral theorem for compact normal operators that involve no measure theory, as well as the interpretation of the bounded Borel functional calculus as a spectral theoretic result, and how this relates to the resolution of the identity. The latter two are basically identical principles, whereas the first is sort of a special case. The above theorem also seems to share the same basic idea as the bounded borel functional calculus and the existence and uniqueness resolution of identity corresponding to a normal operator. (This is usually called the spectral measure)

Question 1: I do not see how to even derive the above theorem from either of these two other versions of the spectral theorem, and would be especially interested in a proof of this sort because it would also show relationships between these supposedly intuitively similar ideas. I ask this because I suppose that the argument could probably be done quickly if the ideologies are basically the same, whereas were I to read separate arguments (which are hard to find on the web anyway) for this level of generality of theorem 1, I would then be left in the dark as to how it relates to anything resembling spectral theory. The subject of resolution of identity was quite technical, and so one would hope that there is a way to not repeat this technical stuff for what seems to be another type of analogy between spectral decomposition and measure theory.

Theorem 2: If $A$ is a bounded linear operator on $L^2(\mathbb{R}^N)$ commuting with translations, then there exists an $L^\infty$ function $M$ such that $Af=\mathcal{F}^{-1} M \mathcal{F}f$ for all $f \in L^2$

In both the theorems here, the multiplier has infinity norm equal to the norm of $A$.

Question 2: I observed that this actually implies, without assuming it, that in the context of theorem 2 $A$ is normal. Is there a direct and easy argument verifying this without the full blown effort to prove the theorem?

Question 3: since $A$ is normal, I see theorem 2 as saying that with additional hypotheses, you can pick a special unitary equivalence to a special $L^2$ space as the data realizing the conclusion of theorem 1. Is there a deeper relationship between theorem 1 and 2 here?

Question 4: We obviously have that the only reasonable uniqueness claim for theorem 2 is true. But for theorem 1, if we fix a particular measure space that is among those that work, and fix a unitary equivalence (both these seem necessary) can we at least claim uniqueness of the multiplier? It seems false without sigma finiteness, and so perhaps the following is not an interesting question but one could still ask if it's possible to select a measure space and unitary equivalence so that it is unique, even if the selection may not be sigma finite? Is it always possible to make a sigma finite selection? It would seem that the fact that $||A||$ being the same as the infinity norm of the multiplier in the context of theorem 1 is something you must be able to choose to have, but is not true of all otherwise valid choices. (For otherwise uniqueness of the multiplier would be guaranteed)

I am most interested in questions 1-3, but answers to any subset of the 4 would be greatly appreciated.


Let $A$ be a bounded normal operator on $X$ with specrum $\sigma$ and spectral resolution of the identity $E$. Then $A = \int_{\sigma}\lambda dE(\lambda)$. Choose any unit vector $x$. Then $d\mu_{x}(\lambda)=d\|E(\lambda)x\|^{2}$ is a Borel probability meausre on $\sigma$. For each bounded Borel function $f$ on $\sigma(A)$, define $$ \pi_{x}(f) = \int_{\sigma}f(\lambda)dE(\lambda)x. $$ Notice that $\pi_{x}(\lambda f(\lambda))=A\pi_{x}(f)$, so that the action of $A$ on the image under $\pi_{x}$ of all bounded Borel functions becomes multiplication by $\lambda$ on the bounded Borel functions. Then $\pi_{x}$ extends uniquely to an isometry $\pi_{x} : L^{2}_{\mu_{x}} \rightarrow X$ because $$ \|\pi_{x}(f)\|^{2} = \int_{\sigma} |f|^{2}\,d\mu_{x}. $$ The correspondence between $A$ and multiplication by $\lambda$ is preserved when completing the space to become Hilbert. If $x'$ is a unit vector which is orthogonal to $\pi_{x}(X)$, one obtains another $\pi_{x'}(X)$ which is orthogonal to $\pi_{x}(X)$.

This is the basic idea behind the multiplication version of the spectral theorem, with a lot of details to work out, including how to unite these mutually orthogonal cyclic subspaces $\pi_{x}(X)$, $\pi_{x'}(X)$, $\ldots$. If you have only a countable number of such subspaces, then I think you can imagine corresponding each $L^{2}_{\mu_{x}}$ space with an $L^{2}_{\mu}$ space on a square, and then you could tile the plane with such squares, gluing everything together using a composite measure; you lose the multiplication by $\lambda$, but you still get multiplication by a bounded function. The functions which vanish outside of one square correspond to one of the $\pi_{x}(X)$ spaces.

More general glueing: Using Zorn's lemma, it is possible to write $X$ as the orthogonal sum $X=\bigoplus_{x\in U}X_{x}$ where $U$ is a set of unit vectors and $X_{x}=\pi_{x}(L^{2}(S,\mu_{x}))$, where $S$ is a square containing the spectrum of $A$. Define a new measure space as a disjoint union $$ (\Omega,\mu) = \bigcup_{x \in U}(S_{x},\mu_{x}),\;\;\; S_{x}=S \mbox{ for all }x \in U. $$ Then $L^{2}_{\mu}(\Omega)$ is the orthogonal sum $$ L^{2}_{\mu}(\Omega)=\bigoplus_{x\in U}L^{2}_{\mu_{x}}(S). $$ For each $x\in U$, we have $\pi_{x}^{-1}A\pi_{x} f = xf(x)$. Define $a : \Omega\rightarrow \mathbb{C}$ by $a(\lambda)=\lambda$ on each $S_{x}$. Gluing together the maps $\pi_{x}$ gives a unitary $\pi : L^{2}_{\mu}(\Omega)\rightarrow X$ such that $\pi^{-1}A\pi f = af$ for all $f \in L^{2}_{\mu}(\Omega)$. There are many details to supply, but I believe this argument works for general uncountable $U$.

  • $\begingroup$ Thanks for the clear picture of what the connection is, but I am interested in the measure theoretic technicality you raised. So are we saying that theorem 1 as I stated it is only true for separable Hilbert spaces, and that for larger $H$ one must allow for unitary equivalence to an arbitrary cardinality direct sum of such $L^2$ spaces with the promise now that $A$ can be represented as a direct sum of multiplication operations? I can see that there is a way to take a disjoint union of countably many measure spaces in general, but not more. Is there some way here to do better? $\endgroup$ – Jeff May 19 '14 at 0:42
  • $\begingroup$ I'll explain more in the text. $\endgroup$ – DisintegratingByParts May 19 '14 at 10:07
  • $\begingroup$ First, what part of gluing together the pieces in the separable case is confusing? $\endgroup$ – DisintegratingByParts May 19 '14 at 10:31
  • $\begingroup$ I think you might be misunderstanding me. What I don't see is how to piece together uncountably many such $x$. It seems to me like if this happens one only gets a direct sum of $L^2$ spaces, as they cannot be united under a single measure space, not even up to unitary equivalence. Is this true? I.e. is theorem 1 false as stated? Or is there some way to construct the measure space without the countability assumption, so that one could indeed tile the plane? $\endgroup$ – Jeff May 19 '14 at 17:26
  • $\begingroup$ The tiling I mentioned is convenient for countable cases, but I do believe the result holds more generally. I've added more general details for you. $\endgroup$ – DisintegratingByParts May 20 '14 at 12:23

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