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Let $T$ be an (unbounded) self-adjoint operator. Assume that there is a bounded operator $S$ such that $TS=ST.$ For which kind of $f$ do we have that $f(T)S=Sf(T)?$

My thought was that using a strategy exploiting Stone's formula for the resolvent and then the Stone-Weierstrass theorem one can show this for $f \in C_0(\mathbb{R}).$ (continuous functions tending to zero).

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  • $\begingroup$ Is $S$ self-adjoint as well, or at least normal? If yes, then your claim holds for any $f$ by the spectral theorem for commuting families of self-adjoint (or normal) operators. Otherwise I do not know any reference, but your strategy seems to work well. $\endgroup$ – Mateusz Kwaśnicki Aug 10 '17 at 9:22
  • $\begingroup$ Unfortunately no, $S$ should be some general operator. $\endgroup$ – Zinkin Aug 10 '17 at 9:32
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Any bounded Borel function $f: \mathbb{R} \to \mathbb{R}$. If $TS = ST$ then (taking adjoint of both sides) $S^*T = TS^*$. Therefore both ${\rm Re}(S) = \frac{1}{2}(S + S^*)$ and ${\rm Im}(S) = \frac{1}{2i}(S - S^*)$ commute with $T$, and since they are self-adjoint it follows from standard spectral theory that they commute with $f(T)$. Taking linear combinations, $S$ commutes with $f(T)$.

Edit: here is a possibly more direct proof. Suppose $TS = ST$, meaning that $S$ preserves the domain of $T$ and they commute on this domain. Then the same is true with $T + iI$ in place of $T$ (since the domain doesn't change), and that operator is invertible. Multiplying both sides of $(T + iI)S = S(T + iI)$ by $(T + iI)^{-1}$ (no problems here, $(T + iI)^{-1}$ is bounded and takes everything into the domain of $T + iI$) yields $S(T + iI)^{-1} = (T + iI)^{-1}S$.

Now $(T + iI)^{-1}$ is normal (its adjoint is $(T - iI)^{-1}$), so by standard spectral theory $S$ commutes with every bounded Borel function of $(T + iI)^{-1}$. But every bounded Borel function of $T$ is also a bounded Borel function of $(T + iI)^{-1}$, QED.

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  • $\begingroup$ May I ask what is the statement of the spectral theorem for commuting operators which is used here? My understanding (e.g. Proposition 3.2 of arxiv.org/pdf/1406.6966.pdf) is that the spectral theorem is pretty tricky to formulate when we're dealing with two unbounded operators. But maybe things are simpler when one of the two operators is bounded? $\endgroup$ – André Henriques Apr 15 '18 at 15:23
  • $\begingroup$ @AndréHenriques: fair enough. I think if $A$ and $B$ are self-adjoint, with $A$ bounded, and $AB = BA$ in the sense that $A$ preserves the domain of $B$ and they commute on this domain, then you can simultaneously realize them as multiplication operators. No doubt something like this is true even if both are unbounded. But I now see a more direct proof that doesn't use this, which I will add to my answer. $\endgroup$ – Nik Weaver Apr 15 '18 at 16:28
  • $\begingroup$ I would be very interested to have a reference for your claim: namely that if $A$ and $B$ are self-adjoint, with $A$ bounded, and $AB=BA$ in the sense that $A$ preserves the domain of $B$ and they commute on this domain, then one can simultaneously realize them as multiplication operators. I have tried (for my own independent reasons) to look online for a version of the spectral theorem that states exactly that, and I haven't been able to locate one. So your help would be very welcome. $\endgroup$ – André Henriques Apr 15 '18 at 16:51
  • $\begingroup$ It might be folklore. I don't know a reference. But the second proof I gave above is easily adapted to yield this conclusion --- if $S$ is self-adjoint and it commutes with the normal operator $(T + iI)^{-1}$, then they can be simultaneously realized as multiplication operators by usual bounded spectral theory. And $T$ would be a multiplication operator in this picture too. $\endgroup$ – Nik Weaver Apr 15 '18 at 18:00
  • $\begingroup$ I just asked a follow-up question: mathoverflow.net/questions/298129/… $\endgroup$ – André Henriques Apr 17 '18 at 22:27

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