A modified Christoffel function and its asymptotic properties

TL;DR

This paper introduces a regularized variant of the Christoffel function that asymptotically estimates the unknown density of a measure on a compact set, with explicit polynomial expressions and improved properties over the standard version.

Contribution

A new regularized Christoffel function variant is proposed, providing asymptotic density estimation with explicit polynomial formulas and similar growth properties as the standard function.

Findings

The reciprocal of the modified Christoffel function is a sum-of-squares polynomial.

Asymptotically, it estimates the measure's density at a point.

Computational complexity is comparable to standard Christoffel function calculations.

Abstract

We introduce a certain variant (or regularization) $\tilde{\Lambda}^\mu_n$ of the standard Christoffel function $\Lambda^\mu_n$ associated with a measure $\mu$ on a compact set $\Omega\subset \mathbb{R}^d$. Its reciprocal is now a sum-of-squares polynomial in the variables $(x,\varepsilon)$, $\varepsilon>0$. It shares the same dichotomy property of the standard Christoffel function, that is, the growth with $n$ of its inverse is at most polynomial inside and exponential outside the support of the measure. Its distinguishing and crucial feature states that for fixed $\varepsilon>0$, and under weak assumptions, $\lim_{n\to\infty} \varepsilon^{-d}\tilde{\Lambda}^\mu_n(\xi,\varepsilon)=f(\zeta_\varepsilon)$ where $f$ (assumed to be continuous) is the unknown density of $\mu$ w.r.t. Lebesgue measure on $\Omega$, and $\zeta_\varepsilon\in\mathbf{B}_\infty(\xi,\varepsilon)$ (and so $f(\zeta_\varepsilon)\approx f(\xi)$ when $\varepsilon>0$ is small). This is in contrast with the standard Christoffel function where if $\lim_{n\to\infty} n^d\Lambda^\mu_n(\xi)$ exists, it is of the form $f(\xi)/\omega_E(\xi)$ where $\omega_E$ is the density of the equilibrium measure of $\Omega$, usually unknown. At last but not least, the additional computational burden (when compared to computing $\Lambda^\mu_n$) is just integrating symbolically the monomial basis $(x^{\alpha})_{\alpha\in\mathbb{N}^d_n}$ on the box $\{x: \Vert x-\xi\Vert_\infty<\varepsilon/2\}$, so that $1/\tilde{\Lambda}^\mu_n$ is obtained as an explicit polynomial of $(\xi,\varepsilon)$.