Verify directly that $\{p_n\}$ are the orthogonal polynomials

Question

I have no idea about an exercise in the book by Percy Deift.

Let $\mu$ be a given positive Borel measure with bounded or unbounded support on $\mathbb{R}$. If the support is unbounded, it requires that $$A_n:=\int_{\mathbb{R}}x^n\mathrm{d}\mu(x)<+\infty,\ \forall n\geqslant 0.$$

For $n\geqslant 0$, set $D_n:=\mathrm{det}(A_{i+j})_{0\leqslant i,j\leqslant n}$. Write

$$p_n(x)=\frac{\left(D_n D_{n-1}\right)^{-1/2}}{n!}\int_{\mathbb{R}^n}\prod_{k=0}^{n-1}(x-x_k)\prod_{0\leqslant i<j\leqslant n-1}\left(x_i-x_j\right)^2\mathrm{d}\mu(x_0)\cdots\mathrm{d}\mu(x_{n-1}).$$ How to verify directly that $\{p_n\}$ are the orthogonal polynomials? Recall that $\{p_n\}$ are the orthogonal polynomials, iff

$$\int_{\mathbb{R}}p_n(x)p_m(x)\mathrm{d}\mu(x)=\delta_{nm},\ \forall n,m\geqslant 0;$$ $$p_n(x)=a_{n,n}x^n+a_{n,n-1}x^{n-1}+\cdots+a_{n,0},\ a_{n,n}>0.$$

Any help will be appreciated.

Answer 1

Don't know if this'll be direct enough for you, but one way to see this is by reverse-engineering the proof of Theorem 2.1.2 in

Ismail, M. (2005). Classical and quantum orthogonal polynomials in one variable. Cambridge University Press.

Namely, the hard part is seeing the following identity \begin{equation} \int_{\mathbb R^n}\dfrac{1}{n!}\prod_{k = 0}^{n-1} (x - x_k) \prod_{0 \leq i<j\leq n-1} (x_i - x_j)^2 d\mu(x_0) \cdots d\mu(x_{n-1}) = \int_{\mathbb R^n}\left| \begin{matrix} 1 & x_0 & \cdots & x_0^{n} \\ x_1 & x_1^2 & \cdots & x_1^{n + 1}\\ \vdots & \vdots & \ddots & \vdots \\ x_{n - 1}^{n - 1} & x_{n - 1}^n & \cdots & x_{n - 1}^{2n - 1} \\ 1 & x & \cdots & x^n \end{matrix}\right| d\mu(x_0) \cdots d\mu(x_{n-1}). \quad (*) \end{equation}

Using this identity we can see that the monic polynomials \begin{align*}P_n(x) &:= \dfrac{1}{n!D_{n - 1}} \int_{\mathbb R^n} \prod_{k = 0}^{n-1} (x - x_k) \prod_{0 \leq i<j\leq n-1} (x_i - x_j)^2 d\mu(x_0) \cdots d\mu(x_{n-1})\\ &= \dfrac{1}{D_{n - 1}} \int_{\mathbb R^n} \left| \begin{matrix} 1 & x_0 & \cdots & x_0^{n} \\ x_1 & x_1^2 & \cdots & x_1^{n + 1}\\ \vdots & \vdots & \ddots & \vdots \\ x_{n - 1}^{n - 1} & x_{n - 1}^n & \cdots & x_{n - 1}^{2n - 1} \\ 1 & x & \cdots & x^n \end{matrix}\right| d\mu(x_0) \cdots d\mu(x_{n-1}) \end{align*} satisfy $$\int x^k P_n(x) d \mu(x) = \delta_{nk} \frac{D_n}{D_{n - 1}}, \quad k \leq n. $$

So, since for a positive Borel measure we have $D_n \neq 0$, if we define $$p_n(x) := \sqrt{\frac{D_{n-1}}{D_n}}P_n(x)$$ we arrive at the required orthogonality.

To see why $(*)$ holds, recall the Vandermonde determinant formula $$ \prod_{0 \leq i<j\leq n-1} (x_i - x_j) = \left| \begin{matrix} 1 & x_0 & \cdots & x_0^{n - 1} \\ 1 & x_1 & \cdots & x_1^{n - 1}\\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n - 1} & \cdots & x_{n - 1}^{n - 1} \end{matrix}\right|, $$ which implies that $$ \prod_{k = 0}^{n-1} (x - x_k) \prod_{0 \leq i<j\leq n-1} (x_i - x_j)^2 = \left| \begin{matrix} 1 & x_0 & \cdots & x_0^{n} \\ 1 & x_1 & \cdots & x_1^{n}\\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n - 1} & \cdots & x_{n - 1}^{n} \\ 1 & x & \cdots & x^n \end{matrix}\right| \left| \begin{matrix} 1 & x_0 & \cdots & x_0^{n - 1} \\ 1 & x_1 & \cdots & x_1^{n - 1}\\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n - 1} & \cdots & x_{n - 1}^{n -1} \end{matrix}\right|. $$ Now we compute the second determinant using the definition of determinants, (here I think of elements of $S_n$ acting on $(0 \ 1 \ 2 \ \cdots \ n-1)$) $$ \left| \begin{matrix} 1 & x_0 & \cdots & x_0^{n - 1} \\ 1 & x_1 & \cdots & x_1^{n - 1}\\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n - 1} & \cdots & x_{n - 1}^{n -1} \end{matrix}\right| = \sum_{\sigma \in S_n} \mathrm{sign}(\sigma) x_0^{k_0} \cdots x_{n - 1}^{k_{n - 1}}, \quad k_j = \sigma(j).$$ By relabeling the variables $x_j \mapsto x_{k_j}$ we see that the integral of $$ \mathrm{sign}(\sigma) \int_{\mathbb R^n} \left| \begin{matrix} 1 & x_0 & \cdots & x_0^{n} \\ 1 & x_1 & \cdots & x_1^{n}\\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n - 1} & \cdots & x_{n - 1}^{n} \\ 1 & x & \cdots & x^n \end{matrix}\right| x_0^{k_0} \cdots x_{n - 1}^{k_{n - 1}} d\mu(x_0) \cdots d\mu(x_{n-1}) = \int_{\mathbb R^n} \left| \begin{matrix} 1 & x_0 & \cdots & x_0^{n} \\ 1 & x_1 & \cdots & x_1^{n}\\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n - 1} & \cdots & x_{n - 1}^{n} \\ 1 & x & \cdots & x^n \end{matrix}\right| x_1 x_2^2 \cdots x_{n - 1}^{n - 1} d\mu(x_0) \cdots d\mu(x_{n-1}) = \int_{\mathbb R^n}\left| \begin{matrix} 1 & x_0 & \cdots & x_0^{n} \\ x_1 & x_1^2 & \cdots & x_1^{n + 1}\\ \vdots & \vdots & \ddots & \vdots \\ x_{n - 1}^{n - 1} & x_{n - 1}^n & \cdots & x_{n - 1}^{2n - 1} \\ 1 & x & \cdots & x^n \end{matrix}\right| d\mu(x_0) \cdots d\mu(x_{n-1}). $$ Since there are $n!$ integrals on the right hand side (all of which are equal), dividing by $n!$ gives the identity we wanted.