My question is essentially the same as this one, but I want to significantly expand the scope, because I am very confused about one of the central claims in that question. According to Lemma IX.8.7 of Conway's A Course in Functional Analysis, if $x\in B(H)$ is self-adjoint and has a separating vector, then the image of the Borel functional calculus is all of $W^*(x)$, its enveloping von Neumann algebra. My question is: how can this be true?
To explain my confusion, let's first step away from $B(H)$, and consider the C*-algebra $C(I)$ where $I$ is the unit interval. The von Neumann enveloping algebra of $C(I)$ is $C(I)^{**}$, its Banach space double dual. The inclusion $i:C(I)\rightarrow C(I)^{**}$ factors through the inclusion in $L^\infty(I)$: $$\require{AMScd} \begin{CD} C(I) @>i_1>> L^\infty(I) @>i_2>> C(I)^{**} \end{CD} $$ These inclusions are all injective. $i_2$ is $L^1(I)$-weak $\rightarrow$ $M(I)$-weak continuous. It seems well known that $i_2$ is not surjective, $C(I)^{**}=M(I)^*$ is much larger than $L^\infty(I)$. It seems a bit strange that $L^\infty(I)$, being a von Neumann algebra, is able to sit between $C(I)$ and its enveloping von Neumann algebra, but this is apparently possible.
By the universal property of the enveloping von Neumann algebra, there exists a unique weak$\rightarrow$weak continuous *-homomorphism $\rho:C(I)^{**}\rightarrow L^\infty(I)$ such that $i_1 = \rho i$. This will be a left inverse of $i_2$, but not an isomorphism.
Now let's map this picture into $B(H)$ using functional calculus. For definiteness, let $H=L^2(I)$, and let $x$ be the self-adjoint operator of "multiplication by $x$" so that the spectrum of $x$ is again the unit interval $I$. We have the following diagram: $$\require{AMScd} \begin{CD} C^*(x) @>j_1>> B(L^\infty) @>j_2>> W^*(x)\\ @A\Gamma AA @ABAA @AwAA\\ C(I) @>i_1>> L^\infty(I) @>i_2>> C(I)^{**} \end{CD} $$ The Gelfand representation $\Gamma$ is an isomorphism of $C(I)$ with $C^*(x)$. $C^*(x)$ is included in its von Neumann enveloping algebra $j:C^*(x)\rightarrow W^*(x)=\overline{C^*(x)}^{WOT}$. By the universal property of enveloping von Neumann algebras, there is a unique isomorphism $w:C(I)^{**}\rightarrow W^*(x)$ making this diagram commute. In the middle, the Borel functional calculus $B$ maps $L^\infty(I)$ to an algebra between $C^*(x)$ and $W^*(x)$, and also making this diagram commute. I'm aware that, in general, $B$ need not be injective, but my understanding is that this is due to using the wrong measure on the spectrum of $x$. In this case, the constant function $1\in L^2(I)$ is a separating vector, and the measure it generates on the spectrum of $x$ agrees with the usual Lebesgue measure, so with $L^\infty(I)$ already being taken modulo this measure, $B$ is injective. Given all of this, I am prepared to believe that the top row of this diagram is simply the isomorphic image of the bottom row. However, Conway's Lemma IX.8.7 asserts that $j_2$ is surjective, hence an isomorphism. This seems to contradict what was said above about $i_2$ not being an isomorphism.
