I am trying to show that the statespace $S(A)$ of a $C^*$-algebra $A$ lies weak-* dense in the statespace $S(A'')$ of the enveloping von Neumann Algebra $A''$. I came across Goldstine's theorem and want to adapt it to the set $S^*=\{f\in A^* : f\text{ is positive and } ||f||\le 1\}$.
Let $i:A^*\rightarrow A^{***}$ be the canonical embedding and $S^{***}=\{f\in A^{***} : f \text{ is positive and } ||f||\le 1\}$. I now want to prove $\overline{i(S^*)}^{w^*}=S^{***}$.
Proof: $S^{***}\subset B_1(A^{***})$ is weak-* closed and thus weak-* compact (Banach-Alaoglu). $S^*$ is weak-* compact and convex. It follows that $\overline{i(S^*)}^{w^*}\subset S^{***}$. Now let $\rho\in S^{***}\setminus\overline{i(S^*)}^{w^*}$. Since $S^*$ is convex $\overline{i(S^*)}^{w^*}$ is weak-* closed and convex, thus by Hahn-Banach there is an $(i(\varphi)=)\widehat{\varphi}\in A^{**}=(A^{***},\sigma(A^{***},A^{**}))^*$, $\alpha\in \mathbb{R}$ and $\epsilon >0$ sucht that $$\Re(\widehat{\varphi}(\rho)) \le \alpha-\epsilon < \alpha \le \Re(\widehat{\varphi}(h)) \quad \forall h\in \overline{i(S^*)}^{w^*}.$$ We then get $$\Re(\rho(\varphi))\le \alpha-\epsilon < \alpha \le \Re(\widehat{\varphi}(\hat{x}) \quad \forall x\in S^{*} $$ and thus $$\Re(\rho(\varphi))\le \alpha-\epsilon < \alpha \le \Re(\varphi(x)) \quad \forall x\in S^*. $$
I now want to get a contradiction. If I could show, that $\overline{\varphi(S^*)}=B_{||\varphi||}(0)$ I would get a contradiction. For this I would need to know that for all $n\in \mathbb{N}$ there exists $y_n\in S^*$ such that $||\varphi||-\frac{1}{n}\le |\varphi(y_n)|\le ||\varphi||$. Do these $y_n$ exist? Is there another way to get a contradiction? Or is there even a better way to prove this once you have Goldstine's theorem?
Thank you in advance!