In a unital C*-algebra $\mathcal{A}$, if $a\in\mathcal{A}$ is a positive element such that $\|a\|=1$, and, for every pure state $f$, $f(a)>0$. Show $a$ is invertible.
Can I conclude, for every state $g\in S(\mathcal{A})$, we always have $g(a)>0$? Since that's true for every pure state?
I'm trying to use that potentially problematic claim to prove invertibility of $a$.
Let $\mathcal{B}$ be the $C^*$-algebra generated by $a,1$. Then I claim that if $g$ is a pure state on $\mathcal{B}$, then also $g(a)> 0$.
To see this, choose a pure state $\overline{g}$ on $\mathcal{A}$ that restricts to $g$ on $\mathcal{B}$. Then $g(a) = \overline{g}(a)> 0$ by your assumption.
Thus, your condition is also satisfied for the commutative $C^*$-algebra $\mathcal{B}$. Thus, we may assume that $\mathcal{A}$ is commutative (by replacing it by $\mathcal{B}$).
But from there, the proof is easy: By the commutative Gelfand-Naimark theorem, we have $\mathcal{A}\cong C(X)$ where $X$ is some compact Hausdorff space. Then recall that the pure states on $C(X)$ are precisely the characters $\{\operatorname{ev}_x: x \in X\}$ so your assumption means that $$a(x) = \operatorname{ev}_x(a)>0$$ for any $x\in X$. Hence, $a$ is invertible.