We have two usual definitions for strictly positive elements in C*-algebras: Let $A$ be a C*-algebra
Definition (a) [MURPHY, C$^*$-algebras and Operator Theory] An element $a\in A_+$ is said to be strictly positive if $\overline{aAa}=A$, that is, the hereditary subalgebra generated by $a$ is dense in $A$.
One can easily show that the following equivalence holds: $$\overline{aA}=A\iff\overline{Aa}=A\iff\overline{aAa}$$
Definition (b) [TAKESAKI, Theory of Operator Algebras I] An element $a\in A$ is said to be strictly positive if $\tau(a)>0$ for every state $\tau\in A^*$. Equivalently, $a$ is strictly positive if $\tau(a)>0$ for every nonzero positive lineaer functional $\tau$ on $A$.
In his book Operator Algebras, Blackadar says that it can be shown that these two definitions are equivalent, using Hahn-Banach and the usual results about extension of positive functionals. I've already shown that $(a)\Rightarrow (b)$, so the other implication is the problem.
Any help is appreciated. Thank you.
Suppose that $\overline{aAa}\subsetneq A$. Choose $b\geq0$ in $A\setminus\overline{aAa}$ and use Hanh-Banach to construct a functional with $f(b)=1$ and $f(aca)=0$ for all $c\in A$. This functional is a linear combination of four states (see II.6.3.4 in Blackadar): at least one of them, say $f_1$, will satisfy that $f_1(b)>0$; by the Jordan Decomposition, we also have that $f_1(aca)=0$ for all $c\in A$ (because of the uniqueness of the Jordan decomposition, the Jordan Decomposition of $f|_{aAa}$ agrees with the Jordan decomposition of $f$).
But then $f(a^3)=0$. Using Kadison's Inequality twice, $$ 0\leq f(a)^2\leq f(a^2)\leq f(a^4)^{1/2}\leq\|a\|^{1/2}\,f(a^3)^{1/2}=0, $$ so $f_1(a)=0$.
Edit: this argument as written is wrong (concretely, the part about the Jordan decomposition of the restriction; for instance, it is perfectly possible for a state to not be a state when restricted to a certain subalgebra, and the uniqueness of the Jordan decomposition relies on a condition on the norm). The idea can still be made to work, but it requires more machinery. Eventually I will type it.