# Normal linear functionals on bicommutants of C*-algebras

I am going through the proof of the Sherman-Takeda theorem and Fillmore's book "A User's Guide on Operator Algebras" seems to have a nice approach, but something seems off to me:

We need to prove that if $$A$$ is a $$C^*$$-algebra then $$A^{**}$$ is isometrically isomorphic to $$\pi_u(A)''$$, where $$(H_u,\pi_u)$$ is the universal representation.

The idea is finding an isometric isomorphism $$e:A^*\to(\pi_u(A)'')_*$$ taking the adjoint map and employing the uniqueness of the predual in von Neumann algebras. To that end, Fillmore takes a state $$\rho\in S(A)$$ and says that $$\rho$$ extends uniquely to a vector state on $$\pi_u(A)''$$. I agree with this and I can see why this is true. The map $$e:A^*\to(\pi_u(A)'')_*$$ is then constructed as follows: an arbitrary functional of $$A^*$$ is written as a linear combination of (four) states, so the above observation allows us to extend any functional of $$A^*$$ to a linear combination of (four) vector states on $$\pi_u(A)''$$, so $$e$$ takes our functional to that extension. I understand why this map $$e:A^*\to(\pi_u(A)'')_*$$ is a linear isometry but I find it weird that it is onto. Fillmore's proof says that, if $$\tau\in(\pi_u(A)'')_*$$ is a normal linear functional, then the restriction $$\tau\vert_{\pi_u(A)}$$ is a functional of $$\pi_u(A)^*\cong A^*$$ so $$e(\tau\vert_{\pi_u(A)})=\tau$$. Even though this makes sense to me, there's this awkward part:

Doesn't this imply that all normal functionals on $$\pi_u(A)''$$ are linear combinations of vector states, hence (SOT) continuous? Is this true or is it a hint that something has gone wrong?

I feel awkward about this because I have the impression that it is very rare for SOT continuous functionals to be the same as the ultraweakly continuous ones.

Doesn't this imply that all normal functionals on $$\pi_u(A)''$$ are linear combinations of vector states, hence (SOT) continuous?
Ultraweak continuity on a von Neumann algebra $$M$$ is continuity for the unique dual space topology on $$M$$, and this is independent of any representation of $$M$$. In contrast, SOT continuity depends on the representation, and roughly speaking, if you pass to a bigger representation then there are more vectors, hence it is harder for a net of operators to SOT converge, hence it becomes easier to be a SOT-continuous linear functional.
If $$\rho$$ is any normal state on a von Neumann algebra $$M$$, then in the GNS representation it generates it becomes a vector state, right? So if you take the direct sum of these representations over all normal states, you get a representation in which every normal state is a vector state, and hence every normal state is SOT continuous.