Let $A$ be a normal operator on a Hilbert space $V$. The continuous functional calculus gives an isometry $\cdot(A): \mathcal{C}(\sigma(A))\to \mathcal{B}(V)$. There is a standard way to extend this to bounded Borel-measurable functions, but working through the argument I am confused why one has to stop there -- it seems that one trivially gets quite a bit more. Since I have not seen this anywhere I assume that this is wrong, but I am not sure where this is wrong and would appreciate any help in sorting this out.
Claim: the continuous functional calculus extends to a norm 1 bounded map $\cdot(A): (\mathcal{C}(\sigma(A)))'' \to \mathcal{B}(V)$, and furthermore we can repeat this argument any number of times so that the domain is the result of taking any even number of duals of $\mathcal{C}(\sigma(A))$.
I'll just do this for the double dual -- hopefully the inductive step is clear. Fix $u,v\in V$. Then, $f\mapsto (f(A)u, v)$ defines a $\tilde F_{u,v}\in (\mathcal{C}(\sigma(A)))'$ with norm at most $\|u\| \|v\|$, and hence an $F_{u,v} \in (\mathcal{C}(\sigma(A)))'''$ with norm at most $\|u\| \|v\|$. For any $\tilde f\in (\mathcal{C}(\sigma(A)))''$, one thus obtains a sesquilinear form $u\times v\mapsto F_{u,v}(\tilde f)$ which defines a bounded operator $\tilde f(A)$ with norm at most $\|\tilde f\|$. Also, $\tilde f(A)$ depends linearly on $\tilde f$.
I believe that this proves the claim. But, Double dual of the space $C[0,1]$ explains that the bounded Borel-measurable functions are contained strictly in $(\mathcal{C}(\sigma(A)))''$. So, the same argument that is usually used for bounded Borel-measurable functions seems to get much more.
Finally, Lax's Functional Analysis book on page 82 asserts that for any compact metric space $Q$, $(\mathcal{C}(Q))''$ is the space of bounded Borel-measurable functions on $Q$, which clearly disagrees with the linked post above. Is Lax simply wrong? He doesn't prove this result.
Thanks in advance!
