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Obviously the version for compact and self-adjoint linear operators on Hilbert Spaces is very useful since it decomposes the operators into orthogonal projections.

However, the following more general version for commutative $C^*$ subalgebras of $\mathcal{L}(\mathcal{H})$ do not seem so handy to me.

Let $\mathcal{A}\in\mathcal{L}(\mathcal{H})$ be a $C^*$ subalgebra containing $I$ and $\Sigma$ be its spectrum. Then for $u,v\in \mathcal{H}$, $f\mapsto \langle T_f u,v\rangle$ is a bounded linear functional on $C(\Sigma)$, where $T_f$ is the operator mapped to $f$ by Gelfand transformation, hence defines a measure $\mu_{u,v}$ on $\Sigma$.

Then there is a regular projection-valued measure $P$ on $\Sigma$ such that $T=\int \hat{T}dP$ for all $T\in\mathcal{A}$ and $T_{f}=\int fdP$ for all $f\in B(\Sigma)$, where $\hat{T}$ is the Gelfand transformation of $T$ and $T_{f}$ is defined by $\langle T_{f}u,v\rangle=\int fd\mu_{u,v}$. Moreover, if $S\in\mathcal{L}(\mathcal{H})$, the followings are equivalent:

i.$S$ commutes with every $T\in\mathcal{A}$.

ii. $S$ commmutes with $P(E)$ for every Borel $E$

iii. $S$ commutes with $\int f dP$ for every $f\in B(\Sigma)$.

Although it is also somewhat decomposition, but they involve integrals and regular measures that exist, but are not constructed in a step-by-step fashion like in the simpler version and about which we do not actually know much.

Thus I wonder how useful this theorem is and it would be great if some examples can be provided.

Thanks!

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    Related: http://math.stackexchange.com/questions/12547/what-is-the-spectral-theorem-for-compact-self-adjoint-operators-on-a-hilbert-spa2012-01-09
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    Useful in what sense? Proving theorems? Gaining insight? Constructing aircraft?2012-01-09
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    I'm going to go against a recent trend and remove the Banach algebras tag, because this question is very specific to the setting of subalgebras of $B(H)$.2012-01-09
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    @YemonChoi hi, Yemon! By usefulness I mean both proving theorems and gaining insights. I know some examples of how this spectral theorem might be used to prove things yet they do not seem deserve such a big theorem. Also, concerning the theorem you mentioned in your answer, I think mine is more general than the version in rudin and takes much more effort to prove. So my question is: why bother to prove this generalized version?2012-01-10
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    I am not convinced that what you have written is *really* more general than that in Rudin. I see that it applies to commutative C*-subalgebras whereas Rudin probably restricts attention to subalgebras generated by a single normal element, but the underlying principles are the same, if I recall correctly2012-01-10
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    As for "why bother to prove this generalized version" - it tells you that the same projection-valued measure works for every element in your algebra, so you now have *joint* functional calculus for a *set* of commuting normal operators.2012-01-10
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    An afterthought: representation theorems like this might not always have "uses" in the sense you seem to want. What they do is allow us to transport our understanding of one setting (in this case, functions and measures) to another setting (subalgebras of $B(H)$). The more complete our "dictionary", the better chance we have of making progress when doing research, because we have more ways of looking at the same problem.2012-01-11

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