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?
Best Answer
So $f$ is a state. Lets first show that it is faithful, meaning that $f(a^*a)>0$ whenever $a\neq0$:
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: