[proofplan] The lower bound follows by evaluating $\rho$ at the unit and using positivity to know that $\rho(1_A)$ is a non-negative real number. For the upper bound, we estimate $\rho(a)$ on the closed unit ball of $A$: the [Cauchy-Schwarz inequality](/theorems/432) for positive functionals bounds $|\rho(a)|^2$ by $\rho(a^*a)\rho(1_A)$, and the basic $C^*$-order estimate $0\le a^*a\le 1_A$ for $\|a\|_A\le 1$ bounds $\rho(a^*a)$ by $\rho(1_A)$. Taking the supremum over the unit ball gives $\|\rho\|_{A^*}\le \rho(1_A)$, and the two inequalities give the formula. The final assertion is then exactly the definition of a state as a positive linear functional of norm one. [/proofplan]

[step:Evaluate the positive functional at the unit to get the lower bound] Since $\rho$ is positive and $1_A=1_A^*1_A$ is positive, the scalar $\rho(1_A)$ is real and satisfies $\rho(1_A)\ge 0$. If $A\neq\{0\}$, then $\|1_A\|_A=1$, so the definition of the dual norm gives \begin{align*} \|\rho\|_{A^*}\ge |\rho(1_A)|=\rho(1_A). \end{align*} If $A=\{0\}$, then $1_A=0$, $\rho=0$, and the same inequality is immediate. Thus in all cases, \begin{align*} \|\rho\|_{A^*}\ge \rho(1_A). \end{align*} [/step]

[step:Control the value of $\rho$ on the unit ball]Let $a\in A$ satisfy $\|a\|_A\le 1$. By the Cauchy-Schwarz Inequality for Positive Functionals [citetheorem:8558], applied to the positive linear functional $\rho$ with the elements $a\in A$ and $b=1_A\in A$, we have \begin{align*} |\rho(1_A^*a)|^2\le \rho(a^*a)\rho(1_A^*1_A). \end{align*} Since $1_A^*=1_A$, this becomes \begin{align*} |\rho(a)|^2\le \rho(a^*a)\rho(1_A). \end{align*} The standard $C^*$-order estimate gives \begin{align*} 0\le a^*a\le \|a\|_A^2 1_A\le 1_A. \end{align*} Applying the positive linear functional $\rho$ to the positive element $1_A-a^*a$ gives \begin{align*} 0\le \rho(1_A-a^*a)=\rho(1_A)-\rho(a^*a), \end{align*} and therefore \begin{align*} \rho(a^*a)\le \rho(1_A). \end{align*} Combining the two inequalities yields \begin{align*} |\rho(a)|^2\le \rho(1_A)^2. \end{align*} Because $\rho(1_A)\ge 0$, it follows that \begin{align*} |\rho(a)|\le \rho(1_A). \end{align*}[/step]

[guided]We want an estimate for $|\rho(a)|$ that is uniform over all $a$ in the unit ball of $A$. The correct tool is the Cauchy-Schwarz Inequality for Positive Functionals [citetheorem:8558], because it converts the linear quantity $\rho(a)$ into the positive quantity $\rho(a^*a)$. Its hypotheses are satisfied: $A$ is a $C^*$-algebra, $\rho:A\to\mathbb C$ is positive by assumption, and $a,1_A\in A$. Applying it with $b=1_A$ gives \begin{align*} |\rho(1_A^*a)|^2\le \rho(a^*a)\rho(1_A^*1_A). \end{align*} Since $1_A^*=1_A$ and $1_A^*1_A=1_A$, this is \begin{align*} |\rho(a)|^2\le \rho(a^*a)\rho(1_A). \end{align*} It remains to bound $\rho(a^*a)$ by $\rho(1_A)$. This is where the assumption $\|a\|_A\le 1$ is used. The basic order estimate in a unital $C^*$-algebra says that \begin{align*} 0\le a^*a\le \|a\|_A^2 1_A. \end{align*} Since $\|a\|_A\le 1$, we get \begin{align*} 0\le a^*a\le 1_A. \end{align*} The inequality $a^*a\le 1_A$ means exactly that $1_A-a^*a$ is positive. Positivity of $\rho$ therefore gives \begin{align*} 0\le \rho(1_A-a^*a). \end{align*} Using linearity of $\rho$, this becomes \begin{align*} 0\le \rho(1_A)-\rho(a^*a), \end{align*} so \begin{align*} \rho(a^*a)\le \rho(1_A). \end{align*} Substituting this into the Cauchy-Schwarz estimate gives \begin{align*} |\rho(a)|^2\le \rho(1_A)^2. \end{align*} Because $\rho(1_A)\ge 0$, taking square roots gives \begin{align*} |\rho(a)|\le \rho(1_A). \end{align*} This proves the desired uniform estimate on the whole closed unit ball of $A$.[/guided]

[step:Take the supremum over the unit ball] By definition of the operator norm of the linear functional $\rho:A\to\mathbb C$, \begin{align*} \|\rho\|_{A^*}=\sup\{|\rho(a)|:a\in A,\ \|a\|_A\le 1\}. \end{align*} The preceding step proves that every term in this supremum is at most $\rho(1_A)$. Hence \begin{align*} \|\rho\|_{A^*}\le \rho(1_A). \end{align*} Together with the lower bound already proved, this gives \begin{align*} \|\rho\|_{A^*}=\rho(1_A). \end{align*} [/step]

[step:Translate the norm formula into the state condition] By definition, a state on a unital $C^*$-algebra is a positive linear functional of norm one. Since $\rho$ is already assumed positive, the norm formula gives \begin{align*} \|\rho\|_{A^*}=1\iff \rho(1_A)=1. \end{align*} Therefore $\rho$ is a state if and only if $\rho(1_A)=1$. [/step]