Let $\rho:C(X)\to B(H)$ is a representation where $X$ is compact, $T_2$. Prove that $\rho$ is injective if and only if $E(G)\ne0$ for all open $G$ where $E$ is corresponding spectral measure.
This is an exercise from Conway Functional Analysis book (Chapter IX, exercise 1.14)
We know that any such representation $\rho$ of $C(X)$ corresponds to unique spectral measure $E$ such that $\rho(u)=\int u\ dE\ \forall u\in C(X)$.
Suppose $\rho$ is injective. We can extend $\rho$ to $\tilde{\rho}:B(X)\to B(H)$ such that $\tilde{\rho}(\phi)=\int \phi\ dE\ \forall \phi\in B(X)$. Let $G$ be a non-empty open subset of $X$. Then by Uryshon's lemma, there is non-zero $u\in C(X)$ such that $0\le u\le\chi_G\implies E(G)=\tilde{\rho}(\chi_G)\ge \tilde{\rho}(u)=\rho(u)\ne 0$ (as $\rho$ is injective) $ \implies E(G)\ne 0$.
For the converse part, take $u\in C(X)$ such that $u\ne 0$. If $u\ge 0$, then there is open set $G$ such that $u(x)>0\ \forall x\in \overline{G}\implies u\ge c\chi_G$ for some $c>0$. Thus, $\rho(u)\ge c\tilde{\rho}(\chi_G)=cE(G)\ne 0$. But this shows only for non-negative $u$. I cannot prove for any $u\ne 0$.
Can anyone help me to finish the proof? Thanks for your help in advance.
You are actually done. For a representation, injectivity is equivalent to faithfulness. Because $$\|\phi(u)\|^2=\|\phi(u)^*\phi(u)\|=\|\phi(|u|^2)\|.$$