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

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  • $\begingroup$ It helps! Thank you so much! $\endgroup$ – milky sausage Oct 30 at 10:10
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    $\begingroup$ 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. $\endgroup$ – 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.

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  • $\begingroup$ 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? $\endgroup$ – milky sausage Oct 31 at 15:02
  • $\begingroup$ 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. $\endgroup$ – Ruy Oct 31 at 15:14
  • $\begingroup$ thanks for your patience! You give me a new perspective to see the problem! $\endgroup$ – milky sausage Oct 31 at 15:17

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