This is a problem from the book "A course in functional analysis" by John B. Conway (Chapter 8 Exercise 8.9).
If $\mathcal A$ is an unital $\textit C^*$-algebra. For $a\in \mathcal A$, $a\ge 0$ if and only if $f(a)\ge 0$ for every state $f$.
One side is trivial from definition. The problem is to prove the other direction. If I can prove $a$ is self-adjoint then I am done, because for every $\lambda$ in spectrum of $a$ there is a state such that $f(a)=\lambda$.
I couldn't show it is self adjoint.
Best Answer
This is basically Theorem 3.4.3. in Murphy's book.
Let $\pi : \mathcal{A} \to \mathbb{B}(H)$ be a faithful nondegenerate representation of $\mathcal{A}$ on a Hilbert space $H$. For a fixed unit vector $x \in H$ consider the functional $$f : A \to \Bbb{C}, \quad f(b)=\langle \pi(b)x,x\rangle.$$ It is easy to see that $f$ is a state so we have $0\le f(a) = \langle \pi(a)x,x\rangle$. In particular, we see that $$\langle \pi(a)x,x\rangle \ge 0, \quad \forall x \in H$$ which implies that $\pi(a)$ is self-adjoint and then $\pi(a) \ge 0$.
Now we have $\pi(a) = \pi(a)^* = \pi(a^*)$ so by faithfulness we get $a = a^*$. You can also get $a \ge 0$ automatically if you recall that $\pi(\mathcal{A}_+) = \pi(\mathcal{A})_+$ holds.