Let $H$ be Hilbert space and $A$, $B$ - self-adjoint (bounded or unbounded) operators on $H$. According to spectral theorem for every bounded Borel function $f: \mathbb{R}\to \mathbb{R}$ we have $$f(A) = \int d \mu_A(\lambda) ~f(\lambda) $$ $$f(B) = \int d \mu_B(\lambda) ~f(\lambda) $$ where $\mu_A$, $\mu_B$ are spectral measures of $A$ and $B$ respectively.

Let's assume that operators $A$ and $B$ commute (in case either of them is unbounded it means that all the projections in their associated spectral measure commute). I wonder whether it is possible to define $g(A,B)$ for any $g: \mathbb{R}^2\to \mathbb{R}$ by $$g(A,B) = \int d \mu_A(\lambda_1)d\mu_B(\lambda_2) ~g(\lambda_1,\lambda_2) .$$

According to Reed & Simon vol. 1, Thm VII.12 the above statement is true for $g(\lambda_1,\lambda_2)=\exp(i t_1\lambda _1 + i t_2 \lambda_2 )$, where $t_1$, $t_2$ are arbitrary real parameters.

If the Borel function $f:\mathbb{R}\to \mathbb{R}$ is unbounded and real-valued then $f(A)$ is self-adjoint on the domain consisting of $\psi\in H$ for which $$ \int (\psi,d\mu_A(\lambda)\psi) ~|f(\lambda)|^2<\infty$$.

Is it true that for $g:\mathbb{R}^2\to \mathbb{R}$ unbounded and real-valued, $g(A,B)$ is self-adjoint on the domain consisting of $\psi\in H$ for which $$ \int (\psi,d\mu_A(\lambda_1)d\mu_B(\lambda_2)\psi) ~|g(\lambda_1,\lambda_2)|^2<\infty.$$

For example: if $A$ and $B$ commute (in a sense of spectral projections) then we can define self-adjoint operator $A+B$ with domain consisting of $\psi\in H$ for which $$ \int (\psi,d\mu_A(\lambda_1)d\mu_B(\lambda_2)\psi) ~(\lambda_1+\lambda_2)^2<\infty.$$ Am I right?

  • $\begingroup$ The Borel functional calculus, applied to the von Neumann algebra generated by $A$ and $B$, implies that you can define $g(A, B)$ for any Borel $g$ (where $A, B$ are bounded self-adjoint). $\endgroup$ – Qiaochu Yuan Jun 17 '15 at 4:54
  • $\begingroup$ I haven't heard about the Borel functional calculus applied to the von Neumann algebra. Can you provide some references? Does it work if A and B do not commute? $\endgroup$ – user72829 Jun 19 '15 at 7:20
  • $\begingroup$ The Borel functional calculus can be applied to any commutative von Neumann algebra. This is a straightforward corollary of the classification of commutative von Neumann algebras (en.wikipedia.org/wiki/Abelian_von_Neumann_algebra). $\endgroup$ – Qiaochu Yuan Jun 19 '15 at 17:39
  • $\begingroup$ Do you mean 'wether' or 'weather'? :) $\endgroup$ – C-Star-W-Star Jun 23 '15 at 14:13

Yes, you are correct.

For reference: The Spectral Theorem For a Pair of Commuting Operators


  • $\begingroup$ The link seems to have died. $\endgroup$ – Jonas Dahlbæk May 24 '17 at 9:45

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