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

Question

Consider the construction of the Borel Functional Calculus for a self adjoint operator $A$ as descibred here:

Continuity of the functional calculus form of the Spectral Theorem

or better yet, here:

http://www.math.mcgill.ca/jakobson/courses/ma667/mendelsontomberg-spectral.pdf on page 4.

Assume the continuous functional calculus was established.

I want to show: $AB = BA \implies B\Phi(f) = \Phi(f)B$ where $f \in \mathbb{B}(\mathbb{R})$ the bounded Borel functions on $\mathbb{R}$.

See my suggestion for a simpler proof below please.

Answer 1

The selfadjoint operators enjoy monotone convergence properties in the strong operator topology. Specifically, suppose $\{ A_n \}_{n=1}^{\infty}$ is a sequence of bounded selfadjoint operators on a Hilbert space $\mathcal{H}$ such that $$ A_1 \le A_2 \le A_3 \le \cdots \le A_{\infty} $$ for some selfadjoint operator $A_{\infty}$. Then $\lim_n A_n x=Ax$ exists for all $x$ and some selfadjoint operator $A$. Using this, you can extend the functional calculus for continuous functions $f$ to include characteristic functions $f=\chi_{E}$ of open, closed, and half-open intervals through strong limits. Hence, you can extend the commutativity $B\Phi(f)=\Phi(f)B$ to $f=\chi_{E}$, where $E$ is in the sigma algebra generated by such sets, which includes the Borel sigma algebra; indeed, you may extended to countable unions of sets because of the above monotone property, while intersections can be handled starting with $\Phi(f)\Phi(g)=\Phi(fg)$ for continuous functions. Thus $B\Phi(\chi_{E})=\Phi(\chi_{E})B$ must hold for all Borel sets $E$. From there, using monotone convergence properties, it follows that $B\Phi(f)=\Phi(f)B$ for all bounded Borel measurable functions $f$.

Answer 2

Here's my suggestion, will be happy for input:

Given $f \in \mathbb{B}(\mathbb{R})$ and a Borel measure $\mu$ there is a sequence of continuous functions $f_n$ s.t $\int f_nd\mu \underset{n \to \infty}{\to} \int f d\mu$.

$AB = BA$ implies $\Phi(f_n)B = B\Phi(f_n)$ since the claim is clearly true for polynomials, and hence for continuous functions, being the uniform limit of polynomials, and as $\Phi$ is continuous in the supremum norm.

Set $x \in H$. Let $\mu_x$ be the corresponding spectral measure and take $f_n$ as above.

Then $(x, \Phi(f_n)x) = \int f_n d\mu_x \underset{n \to \infty}{\to} \int f d\mu_x = (x,\Phi(f)x)$.

The polarization formula yields $(y, \Phi(f_n)x) \underset{n \to \infty}{\to} (y,\Phi(f)x)$. i.e. $\Phi(f_n) \underset{weakly}{\to} \Phi(f)$.

Hence we have $(y, \Phi(f_n)Bx) = (y,B\Phi(f_n)x) = (B^*y,\Phi(f_n)x) \to (B^*y, \Phi(f)x) = (y,B\Phi(f)x)$

While also $(y, \Phi(f_n)Bx) \to (y, \Phi(f)Bx)$

i.e $\Phi(f_n)B = B\Phi(f_n) \underset{weakly}{\to} \Phi(f)B$ and $\Phi(f_n)B = B\Phi(f_n) \underset{weakly}{\to} B\Phi(f)$

As the weak limit is unique $\Phi(f)B = B\Phi(f)$.

Answer 3

I think my last answer has a problem in it (see comments), I'll leave it there in case someone comes upon this. Here's my new suggested solution:

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 $k \in \{0,1,2,3\}$, $\psi_k = x +i^kB^*x$ 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 $k \in \{0,1,2,3\}$. See $\int_X f_n d\mu_1 \to \int_X fd\mu_1 $ and $\int_X f_n d\mu_2\to \int_X f d\mu_2$ for a bounded Borel function $f$

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

Comments are really appreciated!