Let $A$ be a $C^*$-algebra. By Krein-Milman the set of pure states of $A$, $P(A)$, is always non-empty. For a step in a proof I need the existence of a faithful pure state $f\in P(A)$.
Does such a $f$ always exist?
Instead of asking for the existence of faithful pure states one can ask for existence of faithful irreducible $*$-representations. Sadly I have no idea if the answer is yes or no, so I appreciate your ideas.
"Pure" is kind of the opposite of "faithful".
As Shirly mentioned, if $A$ is abelian then all irreducible representations are 1-dimensional. So the only abelian C$^*$-algebra with a faithful pure state is $\mathbb C$.
And whenever $A$ is arbitrary, if $\phi$ is a faithful pure state, both faithfulness and pureness of the state survive to subalgebras. In particular, $\phi$ will also be pure and faithful on all abelian subalgebras (there are always many of this, take any normal $a\in A$ and consider $C^*(a)\subset A$). So all abelian subalgebras are trivial. This means that any selfadjoint $a\in A$ has single-point spectrum, i.e. it is a scalar multiple of the identity. As selfadjoints span $A$, this applies to all elements.
In summary, the only C$^*$-algebra with a faithful pure state is $\mathbb C$.