A Riemann-Hilbert derivation of the Christoffel-Darboux formula

In my recent work on skew-orthogonal polynomials I was interested in deriving a Christoffel-Darboux type formula. To this end Thomas Bothner referred me to a paper of his in collaboration with Marco Bertola which contains a nice derivation of the Christoffel-Darboux formula using only complex analysis arguments (Theorem 2.8). I will present this argument in the simpler context of orthogonal polynomials rather than that of biorthogonal polynomials considered by the authors.

Let us recall the Riemann-Hilbert problem for orthogonal polynomials. Given a measurable function $w : \mathbb{R} \to [0,+\infty]$ such that $\int_\mathbb{R} w(x) \lvert x \rvert ^k \, \mathrm{d}x < +\infty$ for all $k \in \mathbb{N}$ and where $w(x) > 0$ on some open set we construct a sequence of monic polynomials $P_n(x) = x^n + \mathcal{O}(x^{n-1})$ such that

$$\int_\mathbb{R} P_n(x)x^k w(x)\, \mathrm{d}x = 0$$

for all $k=0, \dots, n-1$. We call $P_n$ the $n$th monic orthogonal polynomial with respect to $w$.

We let $h_n = \int_\mathbb{R} P_n(x)x^n w(x)\, \mathrm{d}x$ be the (squared) $L^2(w)$ norm of $P_n$. We may reformulate this as a Riemann-Hilbert problem.

Riemann-Hilbert problem for orthogonal polynomials (Fokas-Its-Kitaev, 1992):

Find a matrix valued function $X_n : \mathbb{C} \setminus \mathbb{R} \to \mathbb{C}^{2 \times 2}$ such that

1. $X_n$ is analytic (entry-wise) on $\mathbb{C} \setminus \mathbb{R}$.
2. $X_n$ has continuous non-tangential boundary values up to $\mathbb{R}$ from above ($+$) and below ($-$). We label these $X_n^\pm (x) = \lim_{\epsilon \downarrow 0} X_n(x\pm \mathrm{i} \epsilon)$ for $x \in \mathbb{R}$.
3. These boundary values are related by the jump condition $X_n^+(x) = X_n^-(x) \left( \begin{matrix} 1 & w(x) \\ 0 & 1\end{matrix} \right)$
4. Finally, $X_n$ is normalised at infinity by the scaling as $z \to \infty$

$$X_n(z) = \left( \mathbb{I}+\mathcal{O}(z^{-1}) \right) \left( \begin{matrix} z^n & 0 \\ 0 & z^{-n} \end{matrix} \right).$$

Proposition: The above RHP has a unique solution given by (for $n\geq 1$)

$$X_n(z) = \left( \begin{matrix} P_n(z) & C_\mathbb{R} \left( P_n w\right)(z) \\ - 2\pi \mathrm{i} h_{n-1}^{-1} P_{n-1}(z) & - 2\pi \mathrm{i} h_{n-1}^{-1} C_\mathbb{R} \left( P_{n-1} w\right)(z) \end{matrix} \right)$$

where

$$C_\mathbb{R}(f)(z) = \frac{1}{2\pi \mathrm{i}} \int_\mathbb{R} \frac{f(x)}{x-z}\, \mathrm{d}x$$

is the Cauchy transform of the function $f$. Furthermore $\det X_n(z) = 1$ identically. If $n=0$ the solution is $X_0(z) = \left( \begin{matrix} 1 & C_\mathbb{R} \left( w\right)(z) \\ 0 & 1 \end{matrix} \right)$. $\triangle$

That $\det X_n(z) = 1$ can be seen directly from the RHP since $\det X_n(z)$ has no jump across $\mathbb{R}$ and has continuous boundary values, and so by Morera’s theorem is entire. $\det X_n(z) \to 1$ as $z \to \infty$ and so by Liouville’s theorem is identically $1$. This implies that the RHP can have at most one solution. The reader may then verify that the above is a solution.

Because $\det X_n(z) = 1$ we can introduce a “dual” Riemann-Hilbert problem $\widehat{X_n}(z) = X_n(z)^{-\mathsf{T}}$ (inverse transpose). $\widehat{X_n}$ solves the following RHP.

Dual Riemann-Hilbert problem for orthogonal polynomials:

Find a matrix valued function $\widehat{X_n} : \mathbb{C} \setminus \mathbb{R} \to \mathbb{C}^{2 \times 2}$ such that

1. $\widehat{X_n}$ is analytic (entry-wise) on $\mathbb{C} \setminus \mathbb{R}$.
2. $\widehat{X_n}$ has continuous non-tangential boundary values up to $\mathbb{R}$ from above ($+$) and below ($-$). We label these $\widehat{X_n}^\pm (x) = \lim_{\epsilon \downarrow 0} \widehat{X_n}(x\pm \mathrm{i} \epsilon)$ for $x \in \mathbb{R}$.
3. These boundary values are related by the jump condition $\widehat{X_n}^+(x) = \widehat{X_n}^-(x) \left( \begin{matrix} 1 & 0 \\ -w(x) & 1\end{matrix} \right)$
4. Finally, $\widehat{X_n}$ is normalised at infinity by the scaling as $z \to \infty$

$$\widehat{X_n}(z) = \left( \mathbb{I}+\mathcal{O}(z^{-1}) \right) \left( \begin{matrix} z^{-n} & 0 \\ 0 & z^n \end{matrix} \right).$$

Indeed, we know the unique solution of the dual RHP,

$$\widehat{X_n}(z) = \left( \begin{matrix} -2\pi \mathrm{i} h_{n-1}^{-1} C_\mathbb{R} \left( P_{n-1} w\right)(z) & 2\pi \mathrm{i} h_{n-1}^{-1} P_{n-1}(z) \\ -C_\mathbb{R} \left( P_n w\right)(z) & P_n(z) \end{matrix} \right).$$

Let us now derive a pair of recursion relations. We note that $X_{n+1}$ satisfies properties 1-3 (of the Fokas-Its-Kitaev RHP), differing only on property 4. Thus if we let $\Delta_n(z) = X_{n+1}(z) X_n(z)^{-1}$ we see that $\Delta_n$ has no jump across the real axis, has continuous boundary values, and is analytic on $\mathbb{C}\setminus \mathbb{R}$. It is thus entire by Morera’s theorem. Let us expand this at infinity. Let $X_n(z) = \left( \mathbb{I}+A_n z^{-1} + \mathcal{O}(z^{-2}) \right)\left( \begin{matrix} z^n & 0 \\ 0 & z^{-n}\end{matrix} \right)$. Then

$$\Delta_n(z) = z E_1 + A_{n+1}E_1 - E_1 A_n + \mathcal{O}(z^{-1})$$

where $E_1 = \left( \begin{matrix} 1 & 0 \\ 0 & 0\end{matrix} \right)$. However since $\Delta_n$ is entire the $\mathcal{O}(z^{-1})$ term is identically zero, so we have

$$X_{n+1}(z) = \left( z E_1 + A_{n+1}E_1 - E_1 A_n \right)X_{n}(z) .$$

This is the famous “three term recurrence” for orthogonal polynomials, derived by complex analysis arguments. By a similar argument we find

$$\widehat{X_{n+1}}(z) = \left( z E_2 - A_{n+1}^\mathsf{T} E_2 + E_2 A_n^\mathsf{T} \right)\widehat{X_{n}}(z)$$

where $E_2 = \left( \begin{matrix} 0 & 0 \\ 0 & 1\end{matrix} \right)$. Let us now consider quantity

$$Y_n(z,w) := X_n(z)^{-1}X_n(w) = \widehat{X_n}(z)^{\mathsf{T}}X_n(w) .$$

Using our recursion relations for $\widehat{X_n}$ and $X_n$ we may relate $Y_n$ and $Y_{n+1}$ by $Y_{n+1}(z,w) = X_n^{-1}(z) \Delta_n(z)^{-1} \Delta_n(w) X_n(w)$. $\Delta_n(z)^{-1} \Delta_n(w)$ is a polynomial in two variables, moreover

$$\Delta_n(z)^{-1} \Delta_n(w) = (z-w) (A_{n+1})_{21} E_{21} + \mathbb{I}$$

where $E_{21} = \left( \begin{matrix} 0 & 0 \\ 1 & 0\end{matrix} \right)$. By our formula for solution $X_n$ we find $(A_{n+1})_{21} = - 2\pi \mathrm{i} h_n^{-1}$, and so

$$Y_{n+1}(z,w) = Y_n(z,w) - \frac{2\pi \mathrm{i}}{h_n} (z-w) X_n^{-1}(z) E_{21} X_n(w)$$

If we now take the $(2,1)$ matrix element of both sides we find

$$Y_{n+1}(z,w)_{21} = Y_n(z,w)_{21} - 2\pi \mathrm{i} \frac{P_n(z)P_n(w)}{h_n} (z-w) .$$

If we now sum both sides from $n=0$ to $n=N-1$ we find

$$Y_{N}(z,w)_{21} = Y_0(z,w)_{21} - 2\pi \mathrm{i} (z-w) \sum_{n=0}^{N-1} \frac{P_n(z)P_n(w)}{h_n} .$$

Then from our solution $Y_0(z,w)_{21} = 0$ we find

$$\boxed{ \sum_{n=0}^{N-1} \frac{P_n(z)P_n(w)}{h_n} = -\frac{1}{2\pi i} \frac{(X_N(z)^{-1}X_N(w))_{21}}{z-w}}$$

which is the Christoffel-Darboux formula.

Remark: This way of writing the Christoffel-Darboux formula mirrors nicely with what happens for $\beta = 4$. Here the relevant quantity that encodes eigenvalue correlation functions is the “pre-kernel,” written as a sum over skew-orthogonal polynomials. Namely,

$\sum_{k=0}^{n-1} \frac{ P_{2k}(x) \mathrm{e}^{-V(x)} \frac{\mathrm{d}}{\mathrm{d}y}\left( P_{2k+1}(y) \mathrm{e}^{-V(y)} \right) - P_{2k+1}(x) \mathrm{e}^{-V(x)} \frac{\mathrm{d}}{\mathrm{d}y}\left( P_{2k}(y) \mathrm{e}^{-V(y)} \right)}{2 h_k}$ $= - \frac{\mathrm{e}^{-V(x)-V(y)}}{4\pi \mathrm{i}} \frac{(X_n(x)^{-1}X_n(y))_{21}}{x-y}$

where $P_k$ is the $k$th monic skew-orthogonal polynomial, $h_k$ is the skew-norm, and $X_n$ is a Riemann-Hilbert problem introduced in my recent paper.