[proofplan] The forward implication is immediate from the definition of a positive linear functional: states take positive elements to nonnegative [real numbers](/page/Real%20Numbers). For the converse, we argue by contrapositive. If the self-adjoint element $a$ is not positive, the spectral characterization of positivity [citetheorem:8554] gives a negative point in $\sigma(a)$, where $\sigma(a)$ denotes the spectrum of $a$. Continuous functional calculus [citetheorem:8553] turns evaluation at this negative spectral point into a state on the unital subalgebra $C^*(1_A,a)$, and the state [extension theorem](/theorems/59) [citetheorem:8561] extends it to a state on $A$, contradicting nonnegativity on all states. [/proofplan] [step:Apply positivity of states to positive elements] Assume $a\ge 0$. Let $\rho\in S(A)$ be a state, so $\rho:A\to\mathbb C$ is a positive linear functional with $\rho(1_A)=1$. Since positive linear functionals are nonnegative on positive elements, the positivity of $a$ gives \begin{align*} \rho(a)\ge 0. \end{align*} Thus $\rho(a)\ge 0$ for every state $\rho\in S(A)$. [/step] [step:Choose a negative spectral point when $a$ is not positive] We prove the converse by contrapositive. Assume that $a=a^*$ and that $a$ is not positive. Let $\sigma(a)\subset\mathbb C$ denote the spectrum of $a$ in the unital $C^*$-algebra $A$. By the spectral characterization of positivity for self-adjoint elements in a unital $C^*$-algebra [citetheorem:8554], $a\ge 0$ is equivalent to $\sigma(a)\subset [0,\infty)$. Since $a$ is not positive, and since $\sigma(a)\subset\mathbb R$ by the spectral theorem for self-adjoint elements [citetheorem:8552], there exists \begin{align*} \lambda\in \sigma(a)\cap (-\infty,0). \end{align*} [guided] We now use the hypothesis $a=a^*$ in an essential way. Let $\sigma(a)\subset\mathbb C$ denote the spectrum of $a$ in $A$. For self-adjoint elements, positivity is detected by the spectrum: by the spectral characterization of positivity [citetheorem:8554], a self-adjoint element $x$ is positive exactly when its spectrum satisfies $\sigma(x)\subset [0,\infty)$. Applying this spectral characterization to the present element $a$, the assumption that $a$ is not positive means that the spectrum cannot be contained in $[0,\infty)$. Since $a=a^*$, the spectral theorem for self-adjoint elements [citetheorem:8552] gives $\sigma(a)\subset\mathbb R$, so a point of $\sigma(a)$ outside $[0,\infty)$ must be a negative real number. Therefore there exists a scalar \begin{align*} \lambda\in \sigma(a)\cap (-\infty,0). \end{align*} This negative spectral value will be converted into a state whose value on $a$ is negative. [/guided] [/step] [step:Construct an evaluation state on the algebra generated by $a$] Let $B:=C^*(1_A,a)\subset A$ denote the unital $C^*$-subalgebra generated by $1_A$ and $a$. Since $a=a^*$, the algebra $B$ is commutative. Let \begin{align*} \operatorname{id}_{\sigma(a)}:\sigma(a)&\to\mathbb C \end{align*} \begin{align*} t&\mapsto t \end{align*} denote the coordinate function on the compact spectrum $\sigma(a)$. By continuous functional calculus for self-adjoint elements [citetheorem:8553], there is a unital $*$-isomorphism \begin{align*} \Phi:C(\sigma(a))\to B \end{align*} such that $\Phi(\operatorname{id}_{\sigma(a)})=a$. Define the evaluation functional \begin{align*} \operatorname{ev}_{\lambda}:C(\sigma(a))&\to\mathbb C \end{align*} \begin{align*} f&\mapsto f(\lambda). \end{align*} This is a state on $C(\sigma(a))$: it is linear, positive because $f\ge 0$ implies $f(\lambda)\ge 0$, and unital because $\operatorname{ev}_{\lambda}(1)=1$. Define \begin{align*} \omega:B&\to\mathbb C \end{align*} \begin{align*} b&\mapsto \operatorname{ev}_{\lambda}(\Phi^{-1}(b)). \end{align*} Since $\Phi$ is a unital $*$-isomorphism and $\operatorname{ev}_{\lambda}$ is a state, $\omega$ is a state on $B$. Moreover, \begin{align*} \omega(a)=\operatorname{ev}_{\lambda}(\Phi^{-1}(a))=\operatorname{ev}_{\lambda}(\operatorname{id}_{\sigma(a)})=\lambda<0. \end{align*} [/step] [step:Extend the negative evaluation state to all of $A$] The subalgebra $B=C^*(1_A,a)$ is a unital $C^*$-subalgebra of $A$ with the same unit $1_A$. Since $\omega$ is a state on $B$, the [State Extension Theorem][citetheorem:8561] gives a state \begin{align*} \rho:A\to\mathbb C \end{align*} such that $\rho|_B=\omega$. Therefore \begin{align*} \rho(a)=\omega(a)=\lambda<0. \end{align*} Thus, if $a$ is not positive, there exists a state $\rho\in S(A)$ with $\rho(a)<0$. Taking the contrapositive, if $\rho(a)\ge 0$ for every state $\rho\in S(A)$, then $a\ge 0$. Together with the forward implication, this proves the equivalence. [/step]