[proofplan]
We construct the [Hilbert space](/page/Hilbert%20Space) by the GNS quotient associated to the positive functional $\mu$. The compact-support inequality makes left multiplication by each generator $X_i$ bounded on the quotient, with operator norm at most $R$. Since the generators are self-adjoint in the $*$-algebra, these bounded extensions are symmetric and therefore self-adjoint. Finally, the class of the unit polynomial is a unit cyclic vector, and the definition of the GNS [inner product](/page/Inner%20Product) gives the desired moment identity.
[/proofplan]
[step:Build the GNS pre-Hilbert space from the positive functional]
Let $\mathcal{A} := \mathbb{C}\langle X_1,\dots,X_d\rangle$. First note that positivity implies the $*$-compatibility identity
\begin{align*}
\mu(B^*) = \overline{\mu(B)}
\end{align*}
for every $B \in \mathcal{A}$. Indeed, for every $B \in \mathcal{A}$ and every $\lambda \in \mathbb{C}$, positivity gives
\begin{align*}
0 \leq \mu((1+\lambda B)^*(1+\lambda B)) = 1+\lambda\mu(B^*)+\overline{\lambda}\mu(B)+|\lambda|^2\mu(B^*B).
\end{align*}
The left-hand side is a real number for all $\lambda$. Taking the imaginary part and using first $\lambda=t \in \mathbb{R}$ and then $\lambda=it$ with $t \in \mathbb{R}$ gives respectively
\begin{align*}
\operatorname{Im}(\mu(B^*)+\mu(B))=0
\end{align*}
and
\begin{align*}
\operatorname{Re}(\mu(B^*)-\mu(B))=0.
\end{align*}
These two identities are exactly $\mu(B^*)=\overline{\mu(B)}$.
Define a sesquilinear form
\begin{align*}
(\cdot,\cdot)_\mu: \mathcal{A} \times \mathcal{A} \to \mathbb{C}, \qquad (P,Q)_\mu := \mu(Q^*P).
\end{align*}
This convention is linear in the first argument, matching the Hilbert-space convention in the statement. The $*$-compatibility identity gives conjugate symmetry:
\begin{align*}
(Q,P)_\mu = \mu(P^*Q)=\overline{\mu(Q^*P)}=\overline{(P,Q)_\mu}.
\end{align*}
Thus $(\cdot,\cdot)_\mu$ is a Hermitian sesquilinear form. Positivity of $\mu$ gives
\begin{align*}
(P,P)_\mu = \mu(P^*P) \geq 0
\end{align*}
for every $P \in \mathcal{A}$.
[claim:The null space is a linear subspace]
Define
\begin{align*}
\mathcal{N} := \{P \in \mathcal{A} : \mu(P^*P)=0\}.
\end{align*}
Then $\mathcal{N}$ is a complex linear subspace of $\mathcal{A}$.
[/claim]
[proof]
We first record the [Cauchy-Schwarz inequality](/theorems/432) for the positive form $(\cdot,\cdot)_\mu$. If $P,Q \in \mathcal{A}$ and $Q \notin \mathcal{N}$, then for every $\lambda \in \mathbb{C}$,
\begin{align*}
0 \leq (P+\lambda Q,P+\lambda Q)_\mu.
\end{align*}
Choosing
\begin{align*}
\lambda := -\frac{(P,Q)_\mu}{(Q,Q)_\mu}
\end{align*}
gives
\begin{align*}
0 \leq (P,P)_\mu - \frac{|(P,Q)_\mu|^2}{(Q,Q)_\mu}.
\end{align*}
Thus
\begin{align*}
|(P,Q)_\mu|^2 \leq (P,P)_\mu (Q,Q)_\mu.
\end{align*}
If $Q \in \mathcal{N}$, the same inequality follows by applying the preceding case to $Q+\varepsilon P$ for $\varepsilon > 0$ when necessary and then letting $\varepsilon \to 0$; equivalently, positivity of the quadratic polynomial in $\lambda$ forces $(P,Q)_\mu=0$.
Now let $P_1,P_2 \in \mathcal{N}$ and let $\alpha_1,\alpha_2 \in \mathbb{C}$. By Cauchy-Schwarz,
\begin{align*}
(P_j,Q)_\mu = 0
\end{align*}
for every $Q \in \mathcal{A}$ and each $j \in \{1,2\}$. Taking $Q := \alpha_1P_1+\alpha_2P_2$, we obtain
\begin{align*}
(\alpha_1P_1+\alpha_2P_2,\alpha_1P_1+\alpha_2P_2)_\mu = 0.
\end{align*}
Hence $\alpha_1P_1+\alpha_2P_2 \in \mathcal{N}$, so $\mathcal{N}$ is a complex linear subspace.
[/proof]
Let $\mathcal{D} := \mathcal{A}/\mathcal{N}$, and write $[P]$ for the equivalence class of $P \in \mathcal{A}$. Define
\begin{align*}
([P],[Q])_{\mathcal{D}} := \mu(Q^*P).
\end{align*}
This is well-defined because Cauchy-Schwarz implies that every element of $\mathcal{N}$ is orthogonal to every element of $\mathcal{A}$. It is a positive definite inner product on $\mathcal{D}$. Let $H$ be the Hilbert-space completion of the pre-Hilbert space $\mathcal{D}$.
Define
\begin{align*}
\Omega := [1] \in \mathcal{D} \subset H.
\end{align*}
Since $\mu(1)=1$,
\begin{align*}
\|\Omega\|_H^2 = ([1],[1])_{\mathcal{D}} = \mu(1)=1.
\end{align*}
Thus $\Omega$ is a unit vector.
[guided]
The point of the GNS construction is to turn the positive functional $\mu$ into an inner product. Before using Cauchy-Schwarz, we must verify that positivity makes the form Hermitian. Let $B \in \mathcal{A}:=\mathbb{C}\langle X_1,\dots,X_d\rangle$. For every $\lambda \in \mathbb{C}$, positivity gives
\begin{align*}
0 \leq \mu((1+\lambda B)^*(1+\lambda B)) = 1+\lambda\mu(B^*)+\overline{\lambda}\mu(B)+|\lambda|^2\mu(B^*B).
\end{align*}
This number is real for every $\lambda$. Taking imaginary parts with $\lambda=t \in \mathbb{R}$ gives
\begin{align*}
\operatorname{Im}(\mu(B^*)+\mu(B))=0,
\end{align*}
and taking imaginary parts with $\lambda=it$ gives
\begin{align*}
\operatorname{Re}(\mu(B^*)-\mu(B))=0.
\end{align*}
Together these identities say
\begin{align*}
\mu(B^*)=\overline{\mu(B)}.
\end{align*}
Now define
\begin{align*}
(P,Q)_\mu := \mu(Q^*P)
\end{align*}
for $P,Q \in \mathcal{A}$. The order $Q^*P$ is chosen because Androma uses Hilbert inner products linear in the first variable, and with this convention the final identity will be
\begin{align*}
(P(A_1,\dots,A_d)\Omega,\Omega)_H = \mu(P).
\end{align*}
The identity $\mu(B^*)=\overline{\mu(B)}$, applied to $B=Q^*P$, gives
\begin{align*}
(Q,P)_\mu = \mu(P^*Q)=\overline{\mu(Q^*P)}=\overline{(P,Q)_\mu}.
\end{align*}
So $(\cdot,\cdot)_\mu$ is a positive Hermitian sesquilinear form. It may be degenerate, so we remove the zero-length vectors. Define
\begin{align*}
\mathcal{N} := \{P \in \mathcal{A} : \mu(P^*P)=0\}.
\end{align*}
To quotient by $\mathcal{N}$, we must know that $\mathcal{N}$ is a linear subspace and that changing representatives does not change inner products. Both facts come from Cauchy-Schwarz for the positive form $(\cdot,\cdot)_\mu$.
For $P,Q \in \mathcal{A}$, positivity gives
\begin{align*}
0 \leq (P+\lambda Q,P+\lambda Q)_\mu
\end{align*}
for every $\lambda \in \mathbb{C}$. If $(Q,Q)_\mu > 0$, choosing
\begin{align*}
\lambda := -\frac{(P,Q)_\mu}{(Q,Q)_\mu}
\end{align*}
yields
\begin{align*}
|(P,Q)_\mu|^2 \leq (P,P)_\mu (Q,Q)_\mu.
\end{align*}
If $(Q,Q)_\mu=0$, the same inequality forces $(P,Q)_\mu=0$ by applying positivity to $P+\lambda Q$ and varying $\lambda$. Therefore every null vector is orthogonal to every vector.
Now if $P_1,P_2 \in \mathcal{N}$ and $\alpha_1,\alpha_2 \in \mathbb{C}$, then each $P_j$ is orthogonal to every element of $\mathcal{A}$. In particular, it is orthogonal to $\alpha_1P_1+\alpha_2P_2$, so
\begin{align*}
(\alpha_1P_1+\alpha_2P_2,\alpha_1P_1+\alpha_2P_2)_\mu = 0.
\end{align*}
Hence $\mathcal{N}$ is a complex linear subspace.
We may therefore form the quotient [vector space](/page/Vector%20Space) $\mathcal{D}:=\mathcal{A}/\mathcal{N}$ and define
\begin{align*}
([P],[Q])_{\mathcal{D}} := \mu(Q^*P).
\end{align*}
This is independent of the chosen representatives because adding an element of $\mathcal{N}$ changes the expression only by inner products with null vectors, and those inner products vanish. The quotient is a pre-Hilbert space, and its completion is the Hilbert space $H$.
Finally, define the distinguished vector
\begin{align*}
\Omega := [1].
\end{align*}
The unital condition $\mu(1)=1$ gives
\begin{align*}
\|\Omega\|_H^2 = ([1],[1])_{\mathcal{D}} = \mu(1)=1,
\end{align*}
so $\Omega$ is a unit vector.
[/guided]
[/step]
[step:Define the generator operators by left multiplication]
For each $i \in \{1,\dots,d\}$, define a [linear map](/page/Linear%20Map)
\begin{align*}
T_i: \mathcal{D} \to \mathcal{D}, \qquad T_i[P] := [X_iP].
\end{align*}
This is well-defined. Indeed, if $P \in \mathcal{N}$, then the compact-support bound gives
\begin{align*}
\|[X_iP]\|_{\mathcal{D}}^2 = \mu((X_iP)^*X_iP)=\mu(P^*X_i^2P) \leq R^2\mu(P^*P)=0.
\end{align*}
Thus $X_iP \in \mathcal{N}$, so the definition of $T_i[P]$ is independent of the representative $P$.
[/step]
[step:Use compact support to extend the generators as bounded operators]
For every $P \in \mathcal{A}$ and every $i \in \{1,\dots,d\}$, the compact-support hypothesis gives
\begin{align*}
\|T_i[P]\|_{\mathcal{D}}^2 = \mu(P^*X_i^2P) \leq R^2\mu(P^*P)=R^2\|[P]\|_{\mathcal{D}}^2.
\end{align*}
Therefore
\begin{align*}
\|T_i v\|_{\mathcal{D}} \leq R\|v\|_{\mathcal{D}}
\end{align*}
for every $v \in \mathcal{D}$. Hence $T_i$ extends uniquely by continuity to a [bounded linear operator](/page/Bounded%20Linear%20Operator)
\begin{align*}
A_i: H \to H
\end{align*}
satisfying
\begin{align*}
\|A_i\|_{\mathcal{L}(H)} \leq R.
\end{align*}
[/step]
[step:Prove that the bounded extensions are self-adjoint]
For $P,Q \in \mathcal{A}$, using $X_i^*=X_i$ gives
\begin{align*}
(T_i[P],[Q])_{\mathcal{D}} = \mu(Q^*X_iP).
\end{align*}
Also,
\begin{align*}
([P],T_i[Q])_{\mathcal{D}} = ([P],[X_iQ])_{\mathcal{D}} = \mu((X_iQ)^*P)=\mu(Q^*X_iP).
\end{align*}
Thus
\begin{align*}
(T_i[P],[Q])_{\mathcal{D}} = ([P],T_i[Q])_{\mathcal{D}}
\end{align*}
for all $P,Q \in \mathcal{A}$. Since $\mathcal{D}$ is dense in $H$ and $A_i$ is the bounded extension of $T_i$, continuity of the inner product gives
\begin{align*}
(A_i u,v)_H = (u,A_i v)_H
\end{align*}
for all $u,v \in H$. Hence $A_i$ is symmetric on all of $H$.
A bounded symmetric operator defined on the whole Hilbert space is self-adjoint: if $A_i^*$ denotes the Hilbert-space adjoint, then the identity above says $A_i \subset A_i^*$, while boundedness gives $\operatorname{Dom}(A_i)=H$ and $\operatorname{Dom}(A_i^*)=H$, so $A_i=A_i^*$. Therefore each $A_i$ is bounded and self-adjoint.
[/step]
[step:Identify polynomial evaluation with left multiplication]
We prove that for every polynomial $P \in \mathcal{A}$,
\begin{align*}
P(A_1,\dots,A_d)\Omega = [P].
\end{align*}
It is enough to check monomials and then use linearity. Let
\begin{align*}
W := X_{i_1}X_{i_2}\cdots X_{i_m}
\end{align*}
be a word in the generators, where $m \in \mathbb{N}$ and $i_1,\dots,i_m \in \{1,\dots,d\}$. Since operator products act on vectors from right to left, we prove the identity by induction on $m$. For $m=1$, the definition of $T_{i_1}$ gives
\begin{align*}
A_{i_1}\Omega = [X_{i_1}].
\end{align*}
Assume that, for some $m \geq 2$,
\begin{align*}
A_{i_2}\cdots A_{i_m}\Omega = [X_{i_2}\cdots X_{i_m}].
\end{align*}
Because $A_{i_1}$ extends $T_{i_1}$ and $[X_{i_2}\cdots X_{i_m}] \in \mathcal{D}$, applying $A_{i_1}$ gives
\begin{align*}
A_{i_1}A_{i_2}\cdots A_{i_m}\Omega = [X_{i_1}X_{i_2}\cdots X_{i_m}] = [W].
\end{align*}
The same identity holds for the empty word $1$, since $\Omega=[1]$. By linearity, it follows that
\begin{align*}
P(A_1,\dots,A_d)\Omega = [P]
\end{align*}
for every $P \in \mathcal{A}$.
[/step]
[step:Recover the functional as the vector state]
Let $P \in \mathcal{A}$. From the previous step,
\begin{align*}
P(A_1,\dots,A_d)\Omega = [P].
\end{align*}
Therefore, using the definition of the GNS inner product with $Q=1$,
\begin{align*}
(P(A_1,\dots,A_d)\Omega,\Omega)_H = ([P],[1])_{\mathcal{D}} = \mu(1^*P)=\mu(P).
\end{align*}
This proves the required moment identity for every $P \in \mathbb{C}\langle X_1,\dots,X_d\rangle$, with $\Omega$ a unit vector and $A_1,\dots,A_d$ bounded [self-adjoint operators](/page/Self-Adjoint%20Operators) satisfying $\|A_i\|_{\mathcal{L}(H)} \leq R$.
[/step]
