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Suppose that $H$ is some (complex) Hilbert space and that $\{T_\alpha: \alpha \in I\}$ is some collection of bounded self-adjoint operators on $H$. A version of the spectral theorem states that for each $\alpha$, there exists a measure space $(X,M,\mu)$, a bounded real-valued function $\phi$ and a unitary operator $U: H \to L^2(\mu)$ such that

$$T_\alpha = U^*M_\phi U$$

where $M_\phi$ denotes the multiplication operator induced by $\phi$. I was wondering under which conditions on $\{T_\alpha\}$ we can find a measure space $(X,M,\mu)$ such that all of the $T_\alpha$ are unitarily equivalent to a multiplication on $L^2(\mu)$? Even better, under what conditions can we choose the same $U$ to work for all $T_\alpha$?

For example (although these are not bounded) all constant coefficient partial differential operators on $L^2(\mathbb{R}^n)$ are Fourier multipliers. What makes them special?

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up vote 3 down vote accepted

The situation that's usually considered is when you want to use the same $U$ for all $T_\alpha$. In this case, it's easy to see that it's necessary for the $T_\alpha$s to commute. As I recall, this is also sufficient: there's a spectral theorem for commuting families of self-adjoint operators. Unfortunately, I can't immediately find a reference so I could be missing a hypothesis. I've made this CW so somebody else could edit one in.

I don't know about the possibility of using a different $U_\alpha$ for each $T_\alpha$. It's not really clear to me what such a representation would be good for.

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Hm, that seems reasonable, since a commuting family would generate a commutative $C^*$ algebra and then we can proceed with the usual Gelfand stuff. Thanks! Both questions were inspired mainly by curiosity, not necessarily by usefulness. – user12014 Jun 9 '12 at 1:59

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