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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.

Mariah
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  • I am not sure what you know about the properties of the measurable functional calculus. To prove the claim, you could use three things: (1) If $f\in\mathbb B$ (the space of all bounded measurable functions on $\sigma(A)$) and $f_n\to f$ pointwise a.e. and $|f_n|_\infty\le C$, then for all $x\in H$ we have $\Phi(f_n)x\to\Phi(f)x$ in $H$. (2) For $f\in\mathbb B$ there exists a uniformly bounded sequence of elementary (or simple) functions $(f_n)$ converging pointwise to $f$.... – amsmath Oct 10 '17 at 16:07
  • (3) If $E$ denotes the spectral measure of $A$ and $B$ commutes with $A$, then $E(\Delta)$ will commute with $A$ for every Borel set $\Delta\subset\sigma(A)$. – amsmath Oct 10 '17 at 16:08
  • @amsmath of the three you mentioned, I know of (1). but I've proved it with the assumption the point-wise convergence is everywhere (need the Dominated Convergence). Is there a version of the Dominated Convergence with the assumption of a.e pointwise limits? – Mariah Oct 10 '17 at 16:19
  • @amsmath oh, do you mean to use (2) and (1) both? – Mariah Oct 10 '17 at 16:28
  • Forget about the "a.e.". It is fine without. – amsmath Oct 10 '17 at 16:29
  • Sure I mean that. – amsmath Oct 10 '17 at 16:29
  • ok, I'm trying to show this for simple functions, but no success yet. I run into similar difficulties. Can you help?

    btw, doesn't (2) also hold for $f_n$ continuous?

     – Mariah Oct 10 '17 at 16:43
  • No, I don't think so. I cannot help if you don't tell me your problems. – amsmath Oct 10 '17 at 17:06
  • @amsmath well we need to prove that for a $f_n$ simple, given that $AB = BA$ we have $\Phi(f_n)B = B\Phi(f_n)$. Since I don't have $\Phi(f_n)$ explicitly I'm forced to used it through the inner product $(\psi, \Phi(f_n)\psi) = \int f du_\psi$. I'm still looking for a way to continue from here. – Mariah Oct 10 '17 at 17:23
  • Do you know how simple functions look like? I mean, how they can be represented... And also note that for a Borel set $\Delta$ you have that $\Phi(\chi_\Delta) = E(\Delta)$. – amsmath Oct 10 '17 at 17:25
  • @amsmath I wrote $f$ as $\sum \alpha_i\chi_{A_i}$. But I don't quite see how to relate something like this $(\psi, B\Phi(f)\psi)$ to $(\psi, \Phi(f)B\psi)$ – Mariah Oct 10 '17 at 17:29
  • Then $\Phi(f) = \sum_i\alpha_i E(A_i)$. And since the $E(A_i)$ commute with $A$, ...BTW, forget about the inner product. You don't need to use it. – amsmath Oct 10 '17 at 17:37
  • @amsmath you wrote $\Phi(\chi_\Delta) = E(\Delta)$. This is wrong no? It implies $\Phi(\chi_\Delta^2) = E(\Delta)^2 = \Phi(\chi_\Delta) = E(\Delta)$ Rather we have $\Phi(\chi)$ must be a projection.. – Mariah Oct 10 '17 at 18:43
  • The $E(\Delta)$ are projections! – amsmath Oct 10 '17 at 18:50
  • You have $\Phi(f) = \int f,dE$ and therefore $\Phi(\chi_\Delta) = \int\chi_\Delta,dE = E(\Delta)$. – amsmath Oct 10 '17 at 18:55
  • @amsmath can you define for me what $E$ is exactly? And how to get $\Phi(f) = \int f,dE$ from the construction of $\Phi$ as appears here on page 4: http://www.math.mcgill.ca/jakobson/courses/ma667/mendelsontomberg-spectral.pdf – Mariah Oct 10 '17 at 19:21
  • Ok, you can define $E(\Delta)$ by $\Phi(\chi_\Delta)$. Since $\Phi(fg) = \Phi(f)\Phi(g)$ you have indeed that $E(\Delta)$ is a projection: $\Phi(\chi_\Delta)^2 = \Phi(\chi_\Delta^2) = \Phi(\chi_\Delta)$. It is even an orthogonal projection since $E(\Delta)$ is selfadjoint: $\Phi(\chi_\Delta)^* = \Phi(\overline{\chi_\Delta}) = \Phi(\chi_\Delta)$. – amsmath Oct 10 '17 at 19:28
  • @amsmath alright, but then why does $AB = BA$ imply $E(\Delta)B = BE(\Delta)$? – Mariah Oct 10 '17 at 19:30
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    You can prove this in many ways. One is as follows: First, you have $B\Phi(f) = \Phi(f)B$ for continuous functions $f$ (you get that by approximating with polynomials). Now, take a closed interval $\Delta$. There exists a uniformly bounded sequence of continuous functions $(f_n)$ which converges pointwise to $\chi_\Delta$. This then gives you $BE(\Delta) = E(\Delta)B$.... – amsmath Oct 10 '17 at 19:47
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    Now, you can proceed with measure theoretic arguments. Define the set $\mathcal A := {\Delta\in\mathfrak B : E(\Delta)B = BE(\Delta)}$, where $\mathfrak B$ denotes the Borel sigma algebra on $\mathbb R$. $\mathcal A$ contains the closed intervals (which is a generator for $\mathfrak B$). You can now show, using the properties of $E(\cdot)$, that $\mathcal A$ is a Dynkin system which is closed under intersections. A theorem from measure theory then tells us that $\mathcal A = \mathfrak B$. – amsmath Oct 10 '17 at 19:50
  • @amsmath trying to find a reference for characteristic functions being approximated pointwise by uniformly bounded continuous functions. do you have one? – Mariah Oct 11 '17 at 18:49
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    Read my comment carefully. You only need that for intervals. And for those it should be clear how to do it (trapezium). – amsmath Oct 11 '17 at 18:51
  • @amsmath hey! I'll be happy for your input in my posted answer. – Mariah Oct 18 '17 at 12:51

3 Answers3

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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 $f$.

Disintegrating By Parts
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  • Considering the comments above - do yo think that Mariah understands what you have written? I guess that no. – amsmath Oct 10 '17 at 18:54
  • thank you for taking the time to answer. I don't, unfortunately, understand your solution. Do you think you have a way to explain this using just the basic construction of $\Phi$ as seen here on page 4: http://www.math.mcgill.ca/jakobson/courses/ma667/mendelsontomberg-spectral.pdf – Mariah Oct 10 '17 at 19:23
  • @Mariah : Using the additional lemma about selfadjoint operators helps, but you can work weakly, which is what I suggest. Start with continuous functions and use monotone functions to extend to all intervals (open, closed, half-open). Then consider the collection of all Borel subsets for which the desired identity $B\Phi(\chi_{E})=\Phi(\chi_{E})B$ holds, and show that it includes the sigma algebra containing the intervals. Conclude that this last identity holds for Borel subsets $E$. Then use monotone convergence of weak integrals to extend $B\Phi(f)=\Phi(f)B$ to all bounded Borel meas. $f$. – Disintegrating By Parts Oct 10 '17 at 19:42
  • @Mariah : Please note that you must restrict to Borel measurable sets and bounded Borel measurable functions. You cannot include Lebesgue measurable functions, for example. So be careful in your arguments. – Disintegrating By Parts Oct 10 '17 at 19:47
  • @DisintegratingByParts will be glad for your input in my answer! – Mariah Oct 18 '17 at 12:51
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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 $\Phi(f)B = B\Phi(f)$.

Mariah
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  • Hey Mariah. This looks really good! Congrats that you found a more elementary way to prove it! Actually, lines 2&3 are not quite correct. $f$ must be in $L^1(\mu)$ for this. But in your case you have that. – amsmath Oct 18 '17 at 18:07
  • @amsmath came back to this, when I say the polarization formula yields.. don't I need $\int f_n d\mu \underset{n \to \infty}{\to} \int f d\mu$ for all those spectral measure? I'm not sure how to justify that. – Mariah Oct 20 '17 at 19:21
  • @amsmath will appreciate if you have the time to look at my new solution. – Mariah Oct 21 '17 at 10:52
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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!

Mariah
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