[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*}
[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]
[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]
