Spectral Theorem, $AB = BA \implies B\Phi(f) = \Phi(f)B$

Question

I'm studying the spectral theorem as appears in Reed and Simon's Functional Analysis.

Assume we have constructed the continuous functional calculus for a self adjoint bounded operator $A$ on a hilbert space $H$.

The construction of the Borel functional calculus is through a function $\Phi : \mathbb{B}(\mathbb{R}) \to L(H)$ from the bounded Borel functions to bounded linear operators on $H$, which extends the continuous functional calculus.

Please see this link for the explicit construction (page 4 of the pdf):

http://www.math.mcgill.ca/jakobson/courses/ma667/mendelsontomberg-spectral.pdf

I want to show $AB = BA \implies B\Phi(f) = \Phi(f)B$.

The claim is clearly true for polynomials (for $f(x) = x, \Phi(f) = A$), and hence for continuous functions, being the uniform limit of polynomials and as $\Phi$ is continuous under the supremum norm.

My attempt:

Step 1 is to prove $\forall x \in H (x, B\Phi(f) x) = (x, \Phi(f) B x)$.

Step 2 is to notice that the above implies $\forall x,y \in H$ $(x, B\Phi(f) y) = (x, \Phi(f) B y)$ by the polarization identity.

Step 3 the above of course implies that $\Phi(f) B = B\Phi(f)$.

So we only need a proof for 1:

Set $x \in H$.

Define $\phi_k = B^*x +i^kx$ for $k \in \{0,1,2,3\}$, $\psi_k = x +i^kBx$ for $k \in \{0,1,2,3\}$.

Define $\mu = \sum_{k = 0}^{3} |\mu_{\phi_k}| + |\mu_{\psi_k}|$. Where $\mu_{\phi_k}$, $\mu_{\psi_k}$ are the spectral measures associated to the corresponding vectors. As each is a regular Borel measure it follows that $\mu$ is a regular Borel measure.

Choose $\{f_n\} \subset C(\sigma(A))$ s.t $\int f_n d\mu \to \int f d\mu$. It follows that $\int f_n \mu_{\psi_k} \to \int f \mu_{\psi_k}$, and $\int f_n \mu_{\phi_k} \to \int f \mu_{\phi_k}$ for all $k \in \{0,1,2,3\}$.

So we have that (see the Mcgill link (page 4) for the second and sixth equalities):

$(x, B\Phi(f)x) = (B^*x, \Phi(f)x) \overset{definition}{=} \frac{1}{4} \sum_{k = 0}^{3} i^k \int f d\mu_{\phi_k} = lim_n \frac{1}{4} \sum_{k = 0}^{3} i^k \int f_n d\mu_{\phi_k} = lim_n(B^*x, \Phi(f_n)x) \overset{f_n \in C(\sigma(A))}{=} lim_n(x, \Phi(f_n)Bx) = lim_n \frac{1}{4} \sum_{k = 0}^{3} i^k \int f_n d\mu_{\psi_k} = \frac{1}{4} \sum_{k = 0}^{3} i^k \int f d\mu_{\psi_k} = (x, \Phi(f)Bx)$.

Btw, also posted this question on MSE, but the methods suggested draw from backgrounds I don't have yet, and I'd like to get feedback for this answer.

Answer 1

You know the result when $f$ is continuous. Now what is the Borel functional calculus? Given any bounded linear map $T: V \to W$ between Banach spaces we have a second dual map $T^{**}: V^{**} \to W^{**}$. So given the continuous functional calculus $T: C_0(\mathbb{R}) \to L(H)$, we automatically have a second dual map from $C_0(\mathbb{R})^{**}$ to $L(H)^{**}$.

However, $L(H)$ is a dual space, and whenever $W$ is a dual Banach space there is a natural (restriction) map from $W^{**}$ to $W$. In our situation, composing this restriction map with $T^{**}$ yields a map from $C_0(\mathbb{R})^{**}$ to $L(H)$. Every bounded Borel function appears as an element of $C_0(\mathbb{R})^{**}$ (remember the first dual of $C_0(\mathbb{R})$ is $M(\mathbb{R})$, and any bounded Borel function can be integrated against any $\mu \in M(\mathbb{R})$). So that is an abstract way to get the Borel functional calculus.

This immediately shows that the range of the Borel functional calculus is contained in the weak* closure of the range of the continuous functional calculus (in fact the two are equal). Indeed, since the unit ball of $V$ is weak* dense in the unit ball of $V^{**}$, it follows that for every bounded Borel $f$ there is a bounded net $(f_\lambda)$ in $C_0(\mathbb{R})$ such that $\Phi(f_\lambda) \to \Phi(f)$ weak*. And for bounded nets weak* convergence is just weak operator convergence. So you can conclude by checking that if $T_\lambda \to T$ boundedly weak operator and $ST_\lambda = T_\lambda S$ for all $\lambda$, then $ST = TS$.