# Separable C* algebra has a cyclic representation that is isometric.

I'm studying Conway's Functional Analysis.

In chapter 8.5, we learn about GNS representation. Theorem 8.5.17 states that

$$\mathcal{A}$$ is separable then, we can select representation $$(\pi, \mathcal{H})$$ that $$\mathcal{H}$$ is separable.

And next line, every separable C*-algebra has a cyclic representation that is isometric.

I want to prove this, and Conway suggests that $$f_n$$ is countable weak* dense subset of $$S_\mathcal{A}$$(which is state space of $$\mathcal{A}$$), then if we set $$f = \sum 2^{-n}f_n$$ then $$\pi_f$$ is an isometry.($$\pi_f$$ is from GNS construction)

Since GNS Construction suggest that if we make $$\pi_f$$, then it is cyclic representation, so I need to prove that $$\pi_f$$ is isometry. But I cannot get any way to get $$\pi_f$$'s norm.

if $$e_f$$ is cyclic vector of $$\pi_f$$, $$\|p_f(a) \|^2 \geq \|\pi_f(a) e_f\|^2 = \langle \pi_f(a^* a) e_f, e_f\rangle = f(a^* a) = \sum 2^{-n}f_n(a^* a)$$

and we know that $$\|a\| = \sup\{f(a) : f \in S_\mathcal{A} \}$$ for positive element $$a$$ but I don't know that $$\sum 2^{-n}f_n(a^* a) \geq \|a^* a\| - \epsilon$$. How can i deal with this problem?

So $$f$$ is a state. Lets first show that it is faithful, meaning that $$f(a^*a)>0$$ whenever $$a\neq0$$:
Suppose $$a\in A$$ and $$a\neq0$$, then there is a state $$\varphi$$ with $$\varphi(a^*a)>0$$. Since $$f_n$$ is weak* dense you have a sub-sequence with $$f_{n_k}\to \varphi$$ in the weak* topology, but this topology is just so that you can recover $$f_{n_k}(a^*a)\to\varphi(a^*a)$$, hence there is some $$f_n$$ with $$f_n(a^*a)>0$$. In particular $$f(a^*a)>0$$.
Now lets show a general statement. Namely that if $$\pi_f$$ is the GNS representation onto a faithful state then $$\pi$$ is an isometry. For this statement we use the non-trivial statement that injective C* morphisms are isometries. So all we check is that $$\pi$$ doesn't have a kernel:
Suppose that $$\pi(a) =0$$, then $$\pi(a^*a)=\pi(a)^*\pi(a)=0$$. It follows that for the ground-state $$|f\rangle$$ you get $$\langle f,\pi(a^*a)f\rangle = f(a^*a) =0$$, implying that $$a=0$$ by faithfulness of $$f$$.