Moments and orthogonal polynomials

Remarks

Orthogonality holds because, for $k < n$, applying $L$ to $x^k P_n$ replaces the last row by $(m_k, \ldots, m_{n+k})$, which equals row $k$, so the determinant vanishes. The leading coefficient is $1$ since the cofactor of $x^n$ is $\Delta_{n-1}$. Also $L(P_n^2) = \Delta_n / \Delta_{n-1} > 0$. Let $(m_n)$ be the moment sequence of a positive Borel measure $\mu$ on $\R$ with infinite support and all moments finite. Then $\Delta_n > 0$ for all $n$, and the $P_n$ are the monic orthogonal polynomials of $\mu$ in $L^2(\mu)$.

Example: for Lebesgue measure on $[-1, 1]$ one has $m_0 = 2$, $m_1 = 0$, $m_2 = \tfrac23$ and $\Delta_0 = 2$, $\Delta_1 = \tfrac43$. The formula gives $P_1(x) = \tfrac12 \det \begin{pmatrix} 2 & 0 \\ 1 & x \end{pmatrix} = x$ and $P_2(x) = \tfrac34 \left( \tfrac43 x^2 - \tfrac49 \right) = x^2 - \tfrac13$, the monic Legendre polynomials.

Non-example: the moments of the Dirac measure at $0$ are $m_0 = 1$ and $m_n = 0$ for $n \ge 1$. Then $\Delta_1 = \det \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = 0$, so the condition $\Delta_n > 0$ fails. Indeed $L(x^2) = 0$, and no monic polynomial of degree $1$ is orthogonal to $1$ with nonzero norm.