Image of Borel functional calculus of a bounded normal operator

Question

Let $ H $ be a (not necessarily separable) Hilbert space and $ N $ be a bounded normal operator on $ H $. For a bounded Borel function $ \phi\colon \sigma(N) \to \mathbb{C} $ on the spectrum of $ N $, we consider Borel functional calculus $$ \phi(N) = \int_{\sigma(N)} \phi \,dE, $$ where $ E $ is the spectral measure of $ N $.

I want to find the image of Borel functional calculus $$ X = \{\phi(N) \mid \text{$ \phi\colon \sigma(N) \to \mathbb{C} $ is a bounded Borel function}\}. $$ To my understanding,

• $ X \subseteq W^*(N) $, where $ W^*(N) $ is the von Neumann algebra generated by $ N $ (Proposition IX.8.1 of Conway’s A Course in Functional Analysis (2nd edition)), and
• $ X = W^*(N) $ if $ H $ is separable (Lemma IX.8.7 of Conway).

I tried to generalize the proof of Conway’s Lemma IX.8.7 to non-separable case, but it seems that separability (or, more specifically, existence of a separating vector for $ W^*(N) $) is essential for Conway’s proof.

So, my question is:

Does $ X = W^*(N) $ hold if we don’t assume that $ H $ is separable? If not, how can we describe $ X $?

Answer 1

There is no equality in general. Consider $H=\ell^2[0,1]$, and $N=M_t$, the multiplication operator. This operator is diagonal with respect to the canonical basis $\{\delta_t\}_{t\in[0,1]}$. We have $\sigma(N)=[0,1]$ and $$ X=\{\phi(M_t):\ \phi\in B_b[0,1]\}=\{M_{\phi}:\ \phi\in B_b[0,1]\}=B_b[0,1], $$ the last equality under the identification $M_\phi\leftrightarrow\{\phi(t)\}_{t\in[0,1]}$.

As $N$ is diagonal with all entries distinct, we have that $\{N\}'$ consists of all diagonal operators, i.e. $\ell^\infty[0,1]$. And then $$ W^*(N)=\{N\}''=\ell^\infty[0,1]. $$ To see that the inclusion is proper, let $E\subset [0,1]$ be non-measurable. Then $1_E\in W^*(N)\setminus X$.

As for how to describe $X$, I think that $$ X=\{\phi(N):\ \phi\in B_b[0,1]\} $$ is a pretty good description. I cannot think of another.