If $x \in \mathcal{A}$ is normal and $\lambda \in \sigma (x)$ is given. I want to show that there is a state $\varphi$ on $\mathcal{A}$ which is definite on x and satisfies $\varphi (x) = \lambda$
I've tried following the hint in the assignment saying that I should start by looking at $\mathcal{A}[x]$. So: I start by finding a state $\varphi_0$ on $\mathcal{A} [x]$ which satisfies that $\varphi_0 (x) = \lambda $ which is definite. Let $\mathcal{A} [x]$ denote the C*-algebra generated by x (Zhu p. 60). As x is normal $\mathcal{A} [x]$ is commutative (prop 10.1, Zhu p. 60). By thm 13.7 (Zhu p. 80) for each $\lambda \in \sigma (x)$ there exist a state $\varphi$ on $\mathcal{A}[x]$ which satisfies that $\varphi (x) = \lambda$. We shall denote this particular state as $\varphi_0 (x)$. Now I have a few questions:
1) Why is this definite? (i.e. $\varphi_0 (x^*x)= \vert \varphi_0 (x) \vert^2$ )
I can show that $\vert \varphi_0(x) \vert^2 =\vert \varphi_0(1^* x) \vert^2 \leq \varphi_0 (1^* 1) \varphi_0 (x^* x) = \varphi_0 (x^* x)$ (by Zhu prop 13.4, p. 78) but how do I achieve the other inequality?
2) How does it help me to look at $\mathcal{A}[x]$ first? Is it because $\mathcal{A}[x]$ and $\mathcal{A}$ has the same spectrum so I somehow can use the same state?
Instead of answering your questions (because I'm not sure if your line of reasoning will get you to a solution), I'll give you an alternative approach.
Rather than using theorem 13.7 to obtain $\varphi_0$, use the fact that $\mathcal A[x]\cong C(\sigma(x))$ is commutative, and that under this identification, $x$ maps to the identity function $z\mapsto z$ on $\sigma(x)$. Now consider the linear functional $\varphi_0:C(\sigma(x))\to\mathbb C$ given by $$\varphi_0(f)=f(\lambda)$$ As $\varphi_0$ is multiplicative, it is a definite state on $C(\sigma(x))$. Now use Hahn-Banach to extend to $\mathcal A$, and show that this extension is still positive. One of the results in section 13 of Zhu's book should make this a one-liner.