positive elements in $C^*$ algebras and states

Question

I have problems to prove that an element $a $ is a $C^*$-algebra is positive if and only if $f(a) \geq 0$ for all states $f$.

The definitions I use:
-$f:A\to\mathbb{C}$ linear functional on a C*-algebra A is called positive, if $f(a^*a)\ge 0$ for every a in A.
-a state f is a positive functional with norm 1.
-$a\ge0$, this means a is selfadjoint and $\sigma(a)\subseteq[0,\|a\|]$.
The direction => is clear. But i stuck on <=. If i follow the hint, i use the representation theorem: I have a universal representation $(H,g)$ of A and is $x\in A$, a positive linear functional $f:A\to\mathbb{C}\; f(b)=\langle g(b)x,x\rangle $ such that $f(a)\ge 0$. This is similar as you can see in Murphys book, Theorem 3.4.3. But now, i can't say that $g(a)$ is selfadjoint, because a is not selfadjoint in general. How can i continue? Regards
An other hint is: First consider $A\cong C_0(X)$ commutative. But i dont know to prove this with this hint.
Maybe you could prove it without the representiation theorem.

Answer 1

Asyou remarked, one direction is trivial from definitions. For the other direction you can use the following well known facts:

1) An element is positive if and only if its spectrum is contained in $[0,\infty )$

2) If $\lambda$ belongs to spectrum of $a$ then there exists a state $f$ such that $f(a)=\lambda$.

Answer 2

Let $a\in A$ such that $\varphi(a)\geq0$ for all states $\varphi$. As $\varphi(\operatorname{Im} a)=\operatorname{Im} \varphi(a)=0$, we get $\operatorname{Im} a=0$ (by functional calculus, as the point evaluations are states) and so $a=a^*$. Knowing that $a$ is selfadjoint we can do the following. If it were not positive, it has $\lambda<0$ in its spectrum; using functional calculus, we can get a state (a character, actually) $\varphi$ with $\varphi(a)=\lambda<0$.