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I was reading these short notes on the Borel functional calculus where the author discusses the uniqueness property of this calculus for both bounded and unbounded self-adjoint operators.

When it comes to unbounded self-adjoint operators, the proof goes like this. Suppose $\phi_{1}$ and $\phi_{2}$ are both $*$-homomorphisms satisfying the properties of the Borel functional calculus, and let: $$\mathcal{A} = \{f \in B_{b}(\mathbb{R}): \phi_{1}(f) = \phi_{2}(f)\}.$$ Using the terminology of the linked text, this is a subalgebra of $B_{b}(\mathbb{R})$, the space of complex-valued bounded measurable functions on $\mathbb{R}$, which is closed under pointwise limits of uniformly bounded sequences.

In the end of the text, the author states the following, concerning uniqueness for the Borel functional calculus for unbounded self-adjoint operators:

Uniqueness of the calculus is clear on the resolvent functions $r_{z}(x) = 1/(x-z)$ and finite linear combinations of those. Then Weierstrass theorem allows to extend the calculus uniquely to $C_{0}(\mathbb{R}) = \{f \in C(\mathbb{R}): \lim_{|x|\to \infty}f(x)\}$. After that, extend the calculus to $C_{b}(\mathbb{R})$ by approximating $f(x) = \lim_{n\to \infty}1_{|x|\le n}f(x)$ and using property (4). [...]

I really do not get the last statement. Suppose the Borel functional calculus is unique on $C_{0}(\mathbb{R})$, that is, $C_{0}(\mathbb{R}) \subset \mathcal{A}$. If $f \in C_{b}(\mathbb{R})$ (the space of bounded continuous complex-valued functions on $\mathbb{R}$), then $1_{|x|\le n}f$ is measurable and bounded, satisfies $\lim_{|x|\to \infty}1_{|x|\le n}f(x) = 0$ but it is not continuous, so it does not belong to $C_{0}(\mathbb{R})$. Hence, how can one conclude that $\phi_{1}(f) = \phi_{2}(f)$ if we only proved these agree on $C_{0}(\mathbb{R})$?

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  • $\begingroup$ One way to see this is to note that $C_0$ is dense in $C_b$ (on any locally compact space). We regard the latter not with the norm but with Buck's strict topology, i.e., the finest l.c. topology which agrees with compact convergence on the unit ball. It is a standard fact that the functional calculus is continuous for this topology. $\endgroup$
    – terceira
    Commented Mar 27, 2023 at 17:14
  • $\begingroup$ @terceira thanks for your comment. I have seen two different approaches to uniqueness of the Borel functional calculus: using that $C_{0}$ is dense in $C_{b}$ as you mentioned and using that every characteristic function $1_{E}$ of a Borel set $E$ is the limit of $C_{0}$ functions. However, I have found no source of proof of any of these statements. It seems that everybody knows it but nobody actually proves it explicitly. I really wanted to work out the details. Any reference is welcome! $\endgroup$
    – MathMath
    Commented Mar 27, 2023 at 17:27
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    $\begingroup$ Using the strict topology to prove the spectral theorem for unbounded s.a. operators is discussed in the monograph "Saks Spaces and Applications in Functional Analysis". $\endgroup$
    – terceira
    Commented Mar 27, 2023 at 18:28

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