# What is the use of Spectral Theorem?

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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– t.b. Jan 9 '12 at 15:23
Useful in what sense? Proving theorems? Gaining insight? Constructing aircraft? – user16299 Jan 9 '12 at 16:12
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 $B(H)$. – user16299 Jan 9 '12 at 16:23
@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? – Hui Yu Jan 10 '12 at 12:58
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 correctly – user16299 Jan 10 '12 at 18:19

As to why to Borel FC might be useful: well, it gives us a way of using what we know about bounded measurable functions to prove and construct things about (normal) operators on Hilbert space. For instance, the polar decomposition of a bounded operator $T$ on Hilbert space can only be proved, as far as I know, by using Borel functional calculus for the operator $T^*T$. In turn, the polar decomposition gives probably the quickest way to show that the unitary group of $B(H)$ is connected in the norm topology.
Note also that if $T$ is normal and belongs to a von Neumann algebra $M\subseteq B(H)$, then so does $f(T)$ for $f$ a bounded Borel function on $\sigma(T)$. So the functional calculus is - and needs to be - used when developing the structure theory of von Neumann algebras, because it allows us to construct elements with prescribed properties that still lie inside the specified von Neumann algebra.
Existence and uniqueness of the polar decomposition can be proved by hand by applying a suitable sequence of polynomials to $T^\ast T$ and using some facts on partial isometries that follow from Bessel and polarization. This is done in chapter 3.2 of Pedersen's Analysis Now, for example. This doesn't really detract from your point of course, but in that argument there's no spectral theory entering. – t.b. Jan 9 '12 at 18:00