# Borel operator calculus of compound function

I am studying operator calculus of complex-valued bounded Borel measurable function. In our textbook it is induced by Gelfand representation over $$A_N$$, which is the smallest C* algebra generated by normal operator $$N$$. It turns out that $$\phi (\psi(N))=(\phi \circ \psi) (N), \forall \phi, \psi \text{ continuous.}$$ which can be shown using Gelfand representation. Does it still correct in the case $$\phi, \psi$$ is bounded Borel measurable? I have seen the spectrum decomposition in this case, but don’t know if it can helps. Thanks in advance.

I know at least one variation that is true: Consider the following theorem from Murphy's excellent text "$$C^*$$-algebras and operator theory":

Theorem 2.5.7 (p73): Let $$u$$ be a normal operator on the Hilbert space $$H$$ and let $$g: \mathbb{C}\to \mathbb{C}$$ be a continuous function. Then $$(g\circ f)(u) = g(f(u))$$ for all $$f \in B_\infty(\sigma(u)).$$

• It helps! Thank you so much! – milky sausage Oct 30 at 10:10
• First I thought that one could get a more general result by approximating $g\in B_\infty(\sigma(f(u)))$ with continuous functions. When this is possible in the $\Vert\cdot \Vert_{L^\infty}$-norm, then the equality should persist. However, continuous functions are unfortunately not dense in Borel functions. – Jan Bohr Oct 30 at 10:13

The answer is affirmative for all Borel functions $$f$$ and $$g$$. Here is the reason.

First of all let me say that the most concrete form of the Spectral Theorem I know asserts that, given any normal operator $$T$$ on a separable Hilbert space $$H$$, there exists a $$\sigma$$-finite measure space $$(X, \mathscr A, \mu )$$, and a unitary operator $$U:H\to L^2(X)$$, such that $$T = U^{-1}M_\varphi U, \tag{1}$$ for a certain $$\varphi \in L^\infty (X)$$, where $$M_\varphi$$ refers to the pointwise multiplication operator $$\xi \in L^2(X)\mapsto \varphi \xi \in L^2(X).$$ In other words, every normal operator is unitarily equivalent to a multiplication operator.

This result is perhaps not so popular because its uniqueness part is a bit messy (see section (3.5) in Sunder, V. S., Functional analysis: spectral theory, Birkhäuser Advanced Texts. Basel: Birkhäuser. ix, 241 p. (1997). ZBL0919.46002.) but it is pretty useful, e.g. here.

Next let $$B(\sigma (T))$$ denote the algebra of all bounded Borel functions on $$\sigma (T)$$ and consider the *-homomorphism$$^{\dagger}$$ $$f\in B(\sigma (T)) \mapsto U^{-1}M_{f\circ \varphi }U \in \mathscr B(H). \tag{2}$$

It is easy to prove that this satisfies all of the properties of the Borel functional calculus, and hence this is the Borel calculus!

Given any $$f$$ in $$B(\sigma (T))$$, we thus have that $$f(T) = U^{-1}M_{f\circ \varphi }U. \tag{3}$$

Observe that (3) is precisely the expression of (1) for $$T'=f(T)$$ and $$\varphi '=f\circ \varphi$$, so the same reasoning above implies that the Borel functional calculus for $$f(T)$$ is $$g\in B(\sigma (f(T))) \mapsto U^{-1}M_{g\circ f\circ \varphi }U \in \mathscr B(H).$$

We then conclude that $$g(f(T)) = (g\circ f)(T),$$ for all $$g$$, as desired.

$$^{(\dagger)}$$ It should be noted that the spectrum of $$T$$ coincides with the essential range of the above function $$\varphi$$. Moreover, it is well known that $$\varphi (x)$$ lies in its essential range for almost all $$x$$, so the composition $$f\circ \varphi$$ in (1) is defined a.e. on $$X$$, and hence the multiplication operator $$M_{f\circ \varphi }$$ is well defined.

• Thanks for your inspiring answer! But H here is just a Hilbert space, not supposed to be separable, sorry for didn’t mentioned. Does it still hold true? – milky sausage Oct 31 at 15:02
• It is a bit awkward, and often also unnecessary from the point of view of applications, to mix measure theory with nonseparable spaces, but yes, I think the above also holds in general, except that the measure space $X$ will no longer be $\sigma$-finite. It will still be decomposable, so the main Theorems of analysis would still hold. – Ruy Oct 31 at 15:14
• thanks for your patience! You give me a new perspective to see the problem! – milky sausage Oct 31 at 15:17