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Let $A_1$ and $A_2$ be two commuting self-adjoint (or normal) operators on an infinite-dimensional complex Hilbert space $E$, then there exists a measure space $(X,\mathcal{E},\mu)$, two functions $\varphi_1,\varphi_2\in L^\infty(\mu)$ and a unitary operator $U:E\longrightarrow L^2(\mu)$, such that each $A_k$ is unitarily equivalent to multiplication by $\varphi_k$, $k=1,2$. i.e. $$UA_kU^*f=\varphi_kf,\;\forall f\in E,\,k=1,2.$$

It is well known that if $E$ is separable then $\mu$ can be taken $\sigma$-finite.

Now if $E$ is assumed to be not necessarily separable, according to this answer, $\mu$ can be taken localizable.

Does anyone have an exact reference?

Thanks.

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  • $\begingroup$ If, assitionally, $A_1,A_2$ are bounded, then this is a special case of Gelfand-Naimark theorem on commutative Banach algebras. $\endgroup$
    – Liviu Nicolaescu
    Commented Jun 1, 2018 at 13:13

2 Answers 2

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You can see for example section 1.4 - spectral theorem II(1.47) - in the book "A course in abstract harmonic analysis" by "Gerald B. Folland".

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  • $\begingroup$ Thank you for the reference but Theorem 1.47(Spectral Theorem II) only concerns a single operator T and I'm interesting with a commuting pair (or even tuple) of operators. $\endgroup$
    – Student
    Commented Jun 1, 2018 at 16:38
  • $\begingroup$ Are you sure! The theorem is about a commutative $C^*$algebra, and this is more general than your case, it is sufficient to consider the $C^*$-algebra generated by your two operators $A_1$ and $A_2$. $\endgroup$
    – MSMalekan
    Commented Jun 1, 2018 at 19:13
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According to @Meisam Soleimani Malekan answer, we have the follwoing theorem

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