I have a question about the proof of Propostion 4.1 of the paper "On Nuclear C*-algebras" by Christopher Lance, it reads
Proposition 4.1. If G is an amenable discrete group then $C^*(G)$ has the CPAP.
The definition of CPAP is in the paper:
Definition 3.5. A $C^* $-algebra $A$ has the completely positive approximation property (CPAP) if the identity mapping on $A^* $ can be approximated in the topology of simple weak*-convergence by completely positive mappings of norm one and finite rank.
Lets assume that $G$ is countable and call the operators $T_n$. My first question is regarding this definition:
Question 1 By simple weak*-convergence is meant that for every $f \in A^* $, $T_nf \rightarrow f$ in the norm of $A^* $, or that for every $f \in A^* $ and $a \in A$, $(T_nf)(a) \rightarrow f(a)$?
Regarding the proof, we start by using the existence of a sequence $\sigma_n$ of functions of positive type on $G$ with finite support such that $\sigma_n \rightarrow$ $1$ pointwise (1 is the function that is identically 1). Then it is said we can view each $\sigma_n$ as a state on $C^*(G)$.
Question 2 Is the way that $\sigma_n$ is viewed as a state by defining $$ \sigma_n: C^*(G) \rightarrow \mathbb C \, \space u_g \rightarrow \sigma_n(g)? $$
After that, for every $\sigma \in E(C^* (G) ) $ (the set of states in $C^* (G)$) they consider a map $$ T_\sigma: E(C^* (G) ) \rightarrow E(C^* (G) ) $$ $$ \rho \rightarrow \rho\sigma $$ which is then extended to an operator in $B(C^* (G)^* )$. We can thus consider the family $T_{\sigma_n}$ and they prove that it is completely positive, has norm one and state that it is easy to show that they are of finite rank and tend to $1$ in the topology of weak*-convergence.
Question 3 I think there is a typo and when they say "tend to $1$", they mean "tend to the identity on $C^* (G)^* $". Is it true that they made a typo?
Question 4 This is the most important one, they claim that it is easy to show that $T_{\sigma_n}$ converge to the identity on $C^* (G)^* $ in the simple weak* topology but I am not being able to prove this.
Sorry for the long post and thank you in advance!
Q1: Judging from how the author uses the term in the proof of Lemma 3.2, it is safe to say that for linear maps $T,T_n:B\to A^*$ (where $A,B$ are Banach spaces) we have that $T_n\to T$ in the simple weak$^*$ topology when $T_n(b)\to T(b)$ in the weak-$^*$ topology of $A^*$ for all $b\in B$, or, equivalently when $|T_n(b)(a)-T(b)(a)|\to0$ as $n\to\infty$ for all $a\in A, b\in B$.
Q2: Yes, $\sigma_n$ is viewed as a state exactly as you say.
Q3: Yes, this is a typo, it should be the identity.
Q4: Since any functional over a $C^*$-algebra is written as a linear combination of at most four states, it suffices to show $\rho\in E(A)$ we have that $T_{\sigma_i}(\rho)\to\rho$ in the weak-$^*$ topology. By the remarks in the beginning of Section 4, we can see each element $\phi\in C^*(G)^*$ as an element of $\ell^\infty(G)$ via the map $\phi\mapsto(\phi(u_g))_{g\in G}$. This map carries over the weak-$^*$ topology of $C^*(G)^*$ to the topology of pointwise convergence in $\ell^\infty(G)$. Therefore, it suffices to show that the sequence (of sequences over $G$) $(T_{\sigma_i}(\rho)(u_g))_{g\in G}$ converges in the pointwise topology to (the sequence over $G$) $(\rho(u_g))_{g\in G}$. Fix $g\in G$. We have that $$|T_{\sigma_i}(\rho)(u_g)-\rho(u_g)|=|\sigma_i(u_g)\rho(u_g)-\rho(u_g)|=|\rho(u_g)|\cdot|\sigma_i(u_g)-1|\to0$$ since $(\sigma_i)_{i\in\mathbb{N}}$ converge to the constant sequence $1\in\ell^\infty(G)$ in the pointwise topology.
Comment: Please consult Brown-Ozawa, 2.5 and 2.6. It is much easier to read from there, it is a more modern approach.