Let $ (\pi,\mathcal{H}) $ be a non-degenerate $ * $-representation of a $ C^{*} $-algebra $ A $, and let $ h \in \mathcal{H} $ with $ \| h \| = 1 $. Define $ f_{h}: A \to \Bbb{C} $ by $ {f_{h}}(a) \stackrel{\text{df}}{=} \langle \pi(a) h | h \rangle $ for all $ a \in A $.
Question. How can we show that $ f_{h} $ is a state?
We note that a linear functional $ \Phi: A \rightarrow \Bbb{C} $ is a state if:
- it is positive, i.e., $ \Phi(x^{*} x) \geq 0 $ for all $ x \in A $, and
- $ \| \Phi \| = 1 $.
Hence, we should prove that $ f_h(a^{*} a) \geq 0 $ for all $ a \in A $ and $ \| f_{h} \| = 1 $.
For $a\in A$, $$\phi(a^*a) = \langle \pi(a^*a)h,h\rangle = ||\pi(a)h||^2\geq 0$$ Which shows that $\phi$ is a positive linear functional. By theorem 3.3.3 of Murphy's C*-algebras and operator theory, if $\{u_i\}$ is approximate unit of C*-algebra $A$, then $\|\phi\|=\lim \phi(u_i)$. Using this we have $$\|\phi\|=\lim \phi(u_i)=\langle \pi(u_i)h,h\rangle = \langle h,h\rangle = 1 $$ $\pi$ is non-degenerate $(\overline{(\pi(A)H)} = H) $, so for every $\xi\in H$, $a\in A$ we have $$\pi(u_i)\pi(a)\xi \to \pi(a)\xi$$ $\pi(A)H$ is dense in $H$ which implies that $$\pi(u_i)\to id ~~(\text {sot})$$