[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.[/step]

[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: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]