On page 95, Murphy states the following theorem (Theorem 3.4.3):
"Let $a$ be a self-adjoint element of a C*-algebra $A$. Then $a$ is positive if and only if $\tau(a)\geq0$ for all positive linear functionals $\tau$ on $A$."
Murphy gives the following proof:
"The forward implication is plain. Suppose conversely that $\tau(a)\geq0$ for all positive linear functionals $\tau$ on $A$. Let $(H,\varphi)$ be the universal representation of $A$, and let $x\in H$. Then the linear functional $$\tau_{x}\colon A\to\mathbb{C},\qquad\tau_{x}(b):=\langle\varphi(b)x,x\rangle,$$ is positive, so $\tau(a)\geq0$; that is, $\langle\varphi(a)x,x\rangle\geq0$. Since this is true for all $x\in H$, and since $\varphi(a)$ is self-adjoint, therefore $\varphi(a)$ is a positive operator on $H$. Hence, $\varphi(a)\in\varphi(A)^{+}$, so $a\in A^{+}$, because the map $\varphi\colon A\to\varphi(A)$ is a $*$-isomorphism."
Why do we need to assume that $a$ is self-adjoint? Because Theorem 2.3.5 in Murphy's book (see page 52-53) implies that ANY operator $u$ on $H$ is positive if and only if $\langle ux,x\rangle\geq0$ for all $x\in H$. So I think that the proof of Theorem 3.4.3 is still valid if we omit the assumption that $a$ (hence $\varphi(a)$) is self-adjoint. Am I right? If not, where did Murphy use the fact that $a$ is self-adjoint?
Any help will be greatly appreciated! Thanks in advance.
Yes, you are right. From $\langle \varphi(a)x,x\rangle\geq0$ for all $x$ it follows that $\varphi(a)$ is selfadjoint. Being a $*$-monomorphism, you get that $a$ is selfadjoint.