What is the use of Spectral Theorem?

Question

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!

Answer 1

What you describe is more or less the spectral theorem for normal operators on Hilbert space, which in turn yields the Borel functional calculus for such operators. See Chapter 12 of Rudin's Functional Analysis (2nd ed.), in particular 12.21-12.24.

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.

Answer 2

I like the application where one uses the spectral theorem to show that any irreducible unitary representation of an abelian group $G$, $\pi:G\to U(H)$, must be $1$ dimensional. I guess you would look at the spectrum of any fixed $\pi(x_0)$ and argue that if it had more than one point, you could use $P$ to find non-trivial projections which commute with every $\pi(x)$ (thus giving invariant subspaces). This last part uses the equivalence stated in the theorem. I don't know how to prove this without using the spectral theorem.

Also, the $L^2$-ergodic theorem has a neat proof using spectral theory. You can show $\frac{1}{n}(x+Ux+\dots U^{n-1}x)$ converges by looking at it as an integral on the spectrum of $U$ and using the dominated convergence theorem.