Given a Borel function $f:\mathbb{R}\rightarrow\mathbb{R}\cup\{\infty\}$, functional calculus allows to calculate $F(x)$ for any unbounded selfadjoint operator $x$ on a Hilbert space $\mathcal{H}$, via $$F(x)=\int_{\mathbb{R}}f(\lambda)E(\lambda),$$ where $E(\lambda)$ is a spectral measure of $x$.

Question: given a function $F$ that maps unbounded selfadjoint operators to unbounded selfadjoint operators, what are the sufficient conditions to have an inverse of this calculus (i.e., to have a well-defined unique $f$, determined by $F$), and what is the specific formula that would generate $f$ from $F$?

More specifically:

  1. I am mostly concerned with a special case of the above question, with $F:E_1(\mathcal{N},\tau)^{\mathrm{sa}}\rightarrow E_2(\mathcal{N},\tau)^{\mathrm{sa}}$, where $E_1(\mathcal{N},\tau)^{\mathrm{sa}}$ and $E_2(\mathcal{N},\tau)^{\mathrm{sa}}$ are self-adjoint parts of two rearrangement invariant spaces over a semifinite von Neumann algebra $\mathcal{N}$ (acting on $\mathcal{H}$) with a faithful normal semifinite trace $\tau$. In this case $F$ maps within the selfadjoint part of a $*$-algebra of Nelson $\tau$-measurable operators affiliated with $\mathcal{N}$, so all unbounded operators in domain and codomain of $F$ are densely defined and closed;
  2. Is the heuristic formula $f(\lambda)=\sup_{\xi\in\mathcal{H}}\langle\xi,F(x)E(\lambda)\xi\rangle_{\mathcal{H}}$ correct? And, if (yes or no), then why?
  • 2
    $\begingroup$ $f$ is automatically unique if it exists (if by "unique" you mean determined $E$-a.e.). If you don't have degeneracies, then existence of $f$ is equivalent to $F(x)$ commuting with $x$, and obviously this condition is necessary no matter what. $\endgroup$ Nov 25, 2021 at 15:47

1 Answer 1


This is not a complete answer, but too long for a comment. Suppose that you have such a mapping $F$ between the unbounded s.a. operators. First consider its effect on multiples of the identity operator, say $\lambda I$. If it is implemented by such a real function $f$, then the image must be of the form $f(\lambda) I$. Hence, a first necessary condition is that $F$ map the span of the identity into itself and this shows that $f$ is uniquely determined by $F$ and shows how to recover it from the latter. This condition is, of course, far from being sufficient. The same argument sows that another necessary condition is that the same applies to each orthogonal projection $P$, i.e, $F(\lambda P)=f(\lambda) P$. Hence, we have the necessary condition: there is a function $f$ on the reals such that for each $P$ and each $\lambda$, $F(\lambda P)=f(\lambda) P$.

Before going further, let me mention two points. Firstly, you don´t mention any smoothness conditions on your $F$. A suitable condition would be continuity, but then this would require a specification of which topology you would use. There are several natural candidates but this would take me two far afield. Experience suggests that without some condition on $F$, pathology can raise its ugly head.

Finally, let me suggest that you consider the question in the more general context of a von Neumann algebra. This makes the problem easier rather more difficult since you can then start with toy examples by firstly considering special cases (the finite dimensional case, i.e., the algebra of $n \times n$ s.a. matrices, then the commutative cases, $n$-dimensional space, $\omega$ (the metrisable space of all real sequence) and $\cal S$, the space of equivalence classes of measurable functions on a suitable measure space). In such situations it is possible to give precise results, but I have not considered the situation of your query.


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