I'm trying to understand the proof of Kadison's function representation as it appears in Pedersen's $C^*$-algebra book (Theorem 3.10.3, see the linked image at the bottom). Let me give some definitions, most of which you can also find in the linked image. Given a $C^*$-algebra $A$, let $Q = Q(A)$ be the quasi-state space of $A$, that is, the set of positive linear functionals of norm less than or equal to one. Let $B_0(Q)$ be the Banach space of bounded, convex-linear functions $b:Q \rightarrow \mathbb{R}$ satisfying $b(0) = 0$ and let $A_0(Q)$ be the subspace of $B_0(Q)$ consisting of those functions that are continuous with respect to the weak* topology on $Q$.
The theorem establishes an order-preserving linear isometry $\hat{}$ of the self-adjoint elements of the enveloping algebra $A_\mathrm{sa}''$ onto $B_0(Q)$. Furthermore, the theorem states that the image of $A_\mathrm{sa}$ is precisely $A_0(Q)$. The isometry $\hat{}$ is defined by identifying an element of $A''$ with an element of the double dual $A^{**}$, then restricting it to $Q$.
I have understood all parts of the proof except the assertion that $A_0(Q) \subset (A_\mathrm{sa})^{\hat{}}$. Pedersen states that weak* continuity on $Q$ implies weak* continuity on the set of hermitian linear functionals of norm $\leq 1$ and weak* continuity on this set implies weak* continuity on the set $(A^*)_\mathrm{sa}$ of all hermitian linear functionals.
Why does weak* continuity on $Q$ imply weak* continuity on $(A^*)_\mathrm{sa}^1$? Why does this imply continuity on all hermitian linear functionals? Am I missing something simple?
I understand how to reduce the non-unital case to the unital case through unitization, so we can assume $A$ is unital.
It is tempting to try to argue this using Jordan decomposition of a hermitian linear functional. I am able to show that Jordan decomposition is weak* continuous on the set of hermitian linear functionals of norm one, but I think it is not weak* continuous on the set of all hermitian linear functionals, as this would imply weak* continuity of the norm. So I'm not sure this line of thinking will work.
