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:R→[0,+∞] such that ∫Rw(x)∣x∣kdx<+∞ for all k∈N and where w(x)>0 on some open set we construct a sequence of monic polynomials Pn(x)=xn+O(xn−1) such that
∫RPn(x)xkw(x)dx=0
for all k=0,…,n−1. We call Pn the nth monic orthogonal polynomial with respect to w.
We let hn=∫RPn(x)xnw(x)dx be the (squared) L2(w) norm of Pn. We may reformulate this as a Riemann-Hilbert problem.
is the Cauchy transform of the function f. Furthermore detXn(z)=1 identically. If n=0 the solution is X0(z)=(10CR(w)(z)1). △
That detXn(z)=1 can be seen directly from the RHP since detXn(z) has no jump across R and has continuous boundary values, and so by Morera’s theorem is entire. detXn(z)→1 as z→∞ 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 detXn(z)=1 we can introduce a “dual” Riemann-Hilbert problem Xn(z)=Xn(z)−T (inverse transpose). Xn solves the following RHP.
Dual Riemann-Hilbert problem for orthogonal polynomials:
Find a matrix valued function Xn:C∖R→C2×2 such that
Xn is analytic (entry-wise) on C∖R.
Xn has continuous non-tangential boundary values up to R from above (+) and below (−). We label these Xn±(x)=limϵ↓0Xn(x±iϵ) for x∈R.
These boundary values are related by the jump condition
Xn+(x)=Xn−(x)(1−w(x)01)
Finally, Xn is normalised at infinity by the scaling as z→∞
Xn(z)=(I+O(z−1))(z−n00zn).
Indeed, we know the unique solution of the dual RHP,
Let us now derive a pair of recursion relations. We note that Xn+1 satisfies properties 1-3 (of the Fokas-Its-Kitaev RHP), differing only on property 4. Thus if we let Δn(z)=Xn+1(z)Xn(z)−1 we see that Δn has no jump across the real axis, has continuous boundary values, and is analytic on C∖R. It is thus entire by Morera’s theorem. Let us expand this at infinity. Let Xn(z)=(I+Anz−1+O(z−2))(zn00z−n). Then
Δn(z)=zE1+An+1E1−AnE1+O(z−1)
where E1=(1000). However since Δn is entire the O(z−1) term is identically zero, so we have
Xn+1(z)=(zE1+An+1E1−E1An)Xn(z).
This is the famous “three term recurrence” for orthogonal polynomials, derived by complex analysis arguments. By a similar argument we find
Xn+1(z)=(zE2−An+1TE2+E2AnT)Xn(z)
where E2=(0001). Let us now consider quantity
Yn(z,w):=Xn(z)−1Xn(w)=Xn(z)TXn(w).
Using our recursion relations for Xn and Xn we may relate Yn and Yn+1 by Yn+1(z)=Xn−1(z)Δn(z)−1Δn(w)Xn(w). Δn(z)−1Δn(w) is a polynomial in two variables, moreover
Δn(z)−1Δn(w)=(z−w)(An+1)21E21+I
where E21=(0100). By our formula for solution Xn we find (An+1)21=−2πihn−1, and so
Remark: This way of writing the Christoffel-Darboux formula mirrors nicely with what happens for β=4. Here the relevant quantity that encodes eigenvalue correlation functions is the “pre-kernel,” written as a sum over skew-orthogonal polynomials. Namely,