Monic orthogonal polynomials exist uniquely through the quadrature order

By [F2], the pairing on the coefficient space ${\mathbb{R}}^m$ for degrees at most $m-1$ is an inner product. The list $e_0,\dots ,e_{m-1}$ is independent by [F4], so [F3] gives orthonormal coefficient vectors $u_0,\dots ,u_{m-1}$ whose first $j+1$ vectors span the same space as $e_0,\dots ,e_j$. The polynomial $P_{u_j}$ therefore has degree exactly $j$: it lies in the span of $1,x,\dots ,x^j$ but not in the span through degree $j-1$. Dividing by its nonzero leading coefficient, which is invertible by [F8], gives a monic $p_j$, orthogonal to every polynomial of degree less than $j$.

Put $G_{ik}=\int x^{i+k}d\mu$ for $0\le i,k<m$, and $b_i=\int x^{i+m}d\mu$. All entries are finite by [F1]. The matrix $G$ represents the inner product of [F2] in the standard basis. If $Gc=0$, then $c^{\mathsf{T}}Gc=⟨c,c{⟩}_{\mu }=0$, so [F2] gives $c=0$. Thus the linear map $G:{\mathbb{R}}^m\to {\mathbb{R}}^m$ has trivial kernel; [F5] makes it surjective. There is therefore a unique coefficient vector $c$ solving $Gc=-b$.

Define $p_m(X)=X^m+{\sum }_{k=0}^{m-1}{c}_k{X}^k$. Its leading coefficient is $1$, and the $i$th equation of $Gc=-b$ gives $\int p_m(x)x^{i}d\mu (x)=0$ for each $i<m$. Linearity [F6] extends this equality to every polynomial of degree less than $m$. If another monic polynomial had the same orthogonality, its difference from $p_m$ would have degree at most $m-1$ and pair to zero with itself; positive definiteness [F2] makes that difference zero.

Step 1.1 gives the unique monic polynomials below degree $m$, and steps 1.2–2.1 give the unique degree-$m$ polynomial using only moments through $2m-1$. This proves the assertion for every $j=0,\dots ,m$.