[proofplan] We encode the coefficients $c_0,\dots,c_N$ in the polynomial $p(t)=\sum_j c_jt^j$ and evaluate it at the self-adjoint element $a$. The Hankel quadratic form is exactly $\varphi(p(a)^*p(a))$. Positivity of the state then gives the desired non-negativity. [/proofplan] [step:Build the polynomial element from the coefficients] Fix an integer $N\geq 0$ and scalars $c_0,\dots,c_N\in\mathbb C$. Define \begin{align*} b:=\sum_{j=0}^{N}c_ja^j\in\mathcal A. \end{align*} Since $a^*=a$, induction gives $(a^i)^*=a^i$ for every integer $i\geq 0$. Therefore conjugate-linearity of the involution gives \begin{align*} b^*=\sum_{i=0}^{N}\overline{c_i}a^i. \end{align*} [/step] [step:Expand $b^*b$ and identify the Hankel form] Using distributivity and associativity in $\mathcal A$, we have \begin{align*} b^*b=\sum_{i=0}^{N}\sum_{j=0}^{N}\overline{c_i}c_ja^ia^j. \end{align*} Since powers of the same element multiply by adding exponents, $a^ia^j=a^{i+j}$. Hence \begin{align*} b^*b=\sum_{i=0}^{N}\sum_{j=0}^{N}\overline{c_i}c_ja^{i+j}. \end{align*} Applying the linear functional $\varphi$ and using $m_{i+j}=\varphi(a^{i+j})$, we obtain \begin{align*} \varphi(b^*b)=\sum_{i=0}^{N}\sum_{j=0}^{N}\overline{c_i}c_jm_{i+j}. \end{align*} [guided] The expression in the theorem is a quadratic form in the coefficients $c_0,\dots,c_N$. To connect it to positivity of the state, we must rewrite it as the value of $\varphi$ on a square. The element \begin{align*} b=\sum_{j=0}^{N}c_ja^j \end{align*} is the natural choice. Because $a$ is self-adjoint, each power $a^i$ is also self-adjoint, so \begin{align*} b^*=\sum_{i=0}^{N}\overline{c_i}a^i. \end{align*} Multiplying the two finite sums gives \begin{align*} b^*b=\sum_{i=0}^{N}\sum_{j=0}^{N}\overline{c_i}c_ja^ia^j. \end{align*} The factors are powers of the same algebra element, so $a^ia^j=a^{i+j}$. Therefore \begin{align*} b^*b=\sum_{i=0}^{N}\sum_{j=0}^{N}\overline{c_i}c_ja^{i+j}. \end{align*} Now apply the linear functional $\varphi:\mathcal A\to\mathbb C$ term by term. Since $m_{i+j}$ was defined to be $\varphi(a^{i+j})$, this gives \begin{align*} \varphi(b^*b)=\sum_{i=0}^{N}\sum_{j=0}^{N}\overline{c_i}c_jm_{i+j}. \end{align*} Thus the Hankel form is exactly the positive-square value $\varphi(b^*b)$. [/guided] [/step] [step:Apply positivity of the state] The state $\varphi$ is positive by hypothesis, so $\varphi(x^*x)\geq 0$ for every $x\in\mathcal A$. Applying this with $x=b$ gives \begin{align*} \varphi(b^*b)\geq 0. \end{align*} Combining this inequality with the identity from the previous step yields \begin{align*} \sum_{i=0}^{N}\sum_{j=0}^{N}\overline{c_i}c_jm_{i+j}\geq 0. \end{align*} This proves the claimed Hankel positivity. [/step]