$A^{**}$ is isometric to $A''$

Question

The picture bellow can be found in $C^*$-ALGEBRAS AND THEIR AUTOMORPHISM GROUPS by Pedersen.

In the picture, $S$ denotes all the states on the C* algebra, $(\pi_S ,H_S)$ denotes the universal representation of $A$ and $A''$ denotes $\pi_S(A)''$.

The proof is so short that I have several questions on it.

Each state of $A$ is a vector state in $(\pi_S ,H_S)$ and therefore a normal state on $A''$.

I know that for each state $\tau\in S$, $\tau(a)= \langle \pi_\tau(a)x_\tau, x_\tau\rangle$, where $(\pi_\tau, H_\tau)$ is the GNS representation associated with $\tau$ and $x_\tau$ is the cyclic vector for the representation. So “the normal state on $A''$” is $\tilde{\tau}:A''\to \mathbb{C}, u\mapsto \langle u x_\tau,x_\tau\rangle$? But I find nowhere to use the normalcy in the following.

Since by Jordan Decomposition each element of $A^*$ is a linear combination of elements from $S$ we can therefore define a map from $A^*$ into $A''_*$.

For each state $\tau\in S$, $\tilde{\tau}$ is weakly continuous, so it is $\sigma$-weakly continuous and the map $A^*\to A''_*, \sigma=\sum_{j=0}^3 k_j\tau_j\mapsto \sum_{j=0}^3 k_j\tilde{\tau_j}$ is well defined?

As $A$ is $\sigma$-weakly dense in $A''$ this map will be a linear isometry.

Is $A$ $\sigma$-weakly dense in $A''$? I only know that $A$ is weakly and strongly dense in $A''$, and here $A$ needs to be unital.

And I think it is sufficient to use SOT to imply isometry. Note that $\tilde{\tau}$ is weakly continuous on $A''$, and the weak closure of the closed unit ball of $A$ is equal to the closed unit ball of $A''$ by Kaplansky Density Theorem.

There's no need to answer all the questions, one of them is appreciated. Thx.

Answer 1

But I find nowhere to use the normalcy in the following.

If you don't have normalcy, you cannot think of $\tau$ as a normal state on $A''$, so you don't have the mapping $A^*\longmapsto (A'')_*$.

I only know that A is weakly and strongly dense in A′′, and here A needs to be unital.

And that's enough (you don't even need to consider the particular representation here). The $\sigma$-weak topology agrees with the wot on bounded sets. So you have, using Kaplanski $$ \overline{A_1}^{\sigma\text{-weak}}=\overline{A_1}^{wot}=\overline{A_1}^{sot}=(\overline{A}^{sot})_1=(A'')_1. $$ So, given $x\in A''$, you have $x/\|x\|\in (A'')_1$, and by the above this is a $\sigma$-weak limit of elements of $A$. That is, $A$ is $\sigma$-weak dense in $A''$.