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.
 A: 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 $f \in B_\infty(\sigma(u)).$
A: 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.
