[proofplan]
We first prove that $\rho$ takes real values on self-adjoint elements. The main input is the continuous functional calculus for self-adjoint elements, which gives a decomposition of a self-adjoint element into the difference of two positive elements. Once this real-valuedness is established, every element is decomposed into its self-adjoint real and imaginary parts, and complex linearity gives the Hermitian identity.
[/proofplan]
[step:Show that $\rho$ is real-valued on self-adjoint elements]
Let $h\in A$ satisfy $h=h^*$. Let $\sigma(h)\subset\mathbb R$ denote the spectrum of $h$ in the unitization of $A$. By the continuous functional calculus for self-adjoint elements, obtained from [citetheorem:8553] in the unitization and applied to the continuous functions $t\mapsto \max\{t,0\}$ and $t\mapsto \max\{-t,0\}$ on $\sigma(h)$, there exist positive elements $h_+,h_-\in A_+$ such that
\begin{align*}
h=h_+-h_-.
\end{align*}
The elements $h_+$ and $h_-$ lie in $A$ because both functions vanish at $0$, so the functional calculus in the unitization returns elements of the original nonunital algebra $A$.
Since $\rho$ is positive, $\rho(h_+)\in[0,\infty)$ and $\rho(h_-)\in[0,\infty)$. Therefore $\rho(h_+)$ and $\rho(h_-)$ are [real numbers](/page/Real%20Numbers). By complex linearity,
\begin{align*}
\rho(h)=\rho(h_+)-\rho(h_-).
\end{align*}
Thus $\rho(h)\in\mathbb R$.
[guided]
We want to prove that a positive functional behaves like an inner-product expression with respect to the involution. The first necessary fact is that self-adjoint elements should be sent to real numbers.
Let $h\in A$ be self-adjoint, so $h=h^*$. Let $\sigma(h)\subset\mathbb R$ denote the spectrum of $h$ in the unitization of $A$. The continuous functional calculus for self-adjoint elements, obtained from [citetheorem:8553] in the unitization, allows us to apply the real-valued continuous functions $t\mapsto \max\{t,0\}$ and $t\mapsto \max\{-t,0\}$ on $\sigma(h)$ to the element $h$. These functions vanish at $0$, so the resulting functional-calculus elements belong to the original algebra $A$, not merely to its unitization. Denote them by $h_+,h_-\in A$. They satisfy
\begin{align*}
h=h_+-h_-.
\end{align*}
Moreover $h_+$ and $h_-$ are positive elements of $A$, because those functions are nonnegative on $\sigma(h)$.
Now we use exactly the hypothesis that $\rho$ is positive. Since $h_+\in A_+$, the number $\rho(h_+)$ lies in $[0,\infty)$. Since $h_-\in A_+$, the number $\rho(h_-)$ also lies in $[0,\infty)$. In particular, both values are real. By complex linearity of $\rho$,
\begin{align*}
\rho(h)=\rho(h_+-h_-)=\rho(h_+)-\rho(h_-).
\end{align*}
The right-hand side is the difference of two real numbers, so $\rho(h)\in\mathbb R$.
[/guided]
[/step]
[step:Decompose an arbitrary element into self-adjoint parts]
Let $a\in A$, and let $i\in\mathbb C$ denote the imaginary unit. Define
\begin{align*}
h:=\frac{a+a^*}{2}
\end{align*}
and
\begin{align*}
k:=\frac{a-a^*}{2i}.
\end{align*}
Then
\begin{align*}
h^*=\frac{a^*+a}{2}=h
\end{align*}
and
\begin{align*}
k^*=\left(\frac{a-a^*}{2i}\right)^*=\frac{a^*-a}{-2i}=k.
\end{align*}
Thus $h$ and $k$ are self-adjoint. Also,
\begin{align*}
h+ik=\frac{a+a^*}{2}+i\frac{a-a^*}{2i}=a.
\end{align*}
Taking adjoints gives
\begin{align*}
a^*=h-ik.
\end{align*}
[/step]
[step:Use real-valuedness on the self-adjoint parts to prove the Hermitian identity]
By the first step applied to the self-adjoint elements $h$ and $k$, the numbers $\rho(h)$ and $\rho(k)$ are real. Since $a=h+ik$ and $a^*=h-ik$, complex linearity gives
\begin{align*}
\rho(a)=\rho(h)+i\rho(k)
\end{align*}
and
\begin{align*}
\rho(a^*)=\rho(h)-i\rho(k).
\end{align*}
Because $\rho(h),\rho(k)\in\mathbb R$,
\begin{align*}
\overline{\rho(a)}=\overline{\rho(h)+i\rho(k)}=\rho(h)-i\rho(k)=\rho(a^*).
\end{align*}
Since $a\in A$ was arbitrary, this proves
\begin{align*}
\rho(a^*)=\overline{\rho(a)}
\end{align*}
for every $a\in A$.
[/step]
