Potential theory and asymptotics of the zero distribution of orthogonal polynomials

This week I gave an introductory talk at the Tenth Summer School on Potential Theory about potential theoretic methods in the study of orthogonal polynomials. This post is based upon my notes for that lecture. The aim of that talk was to prove a theorem about the weak-* convergence of normalized counting measures for zeros of orthogonal polynomials.

A small note: This is a post which was migrated from my old blog. Actually, I gave this talk almost a year ago, but the mathematics still holds. (I sincerely hope so.)

1. Introduction. Our goal is to show that potential theoretic methods can be used effectively for studying orthogonal polynomials. The following is based upon [VanAssche] Chapter 1.2.

Let $\mu$ be a finite Borel measure on the real line. Suppose that its support contains infinitely many points and

$\int x^k d\mu(x) < \infty, \quad k \in \{ 0,1, \dots \}.$

As usual, we are going to study the orthogonal polynomials with respect to $\mu$. Let $P_n(x,\mu) = P_n(x)$ denote the $n$-th monic orthogonal polynomial with respect to $\mu$. It is easy to see that every polynomial can be written as a linear combination of (monic) orthogonal polynomials. They have a very important property: they are extremal in a sense, which is made precise in the following theorem.

Theorem 1.1. ($L^2$-extremality of monic orthogonal polynomials) Let $\mu$ be a Borel measure with finite moments containing infinitely many points in its support. If $P_n$ denotes the $n$-th monic orthogonal polynomial, then

$\| P_n \|_{L^2(\mu)} = \inf\{ \| Q_n \|_{L^2(\mu)}: q \textnormal{ is a monic polynomial of degree } n \}.$

Proof. Let $Q_n$ be a monic polynomial of degree $n$. It can be written as $Q_n = P_n + q_{n-1}$, where $q_{n-1}$ is some polynomial of degree $n-1$. It follows that

$\| Q_n \|_{L^2(\mu)}^2 = \int (P_n(x) + q_{n-1}(x))^2 d\mu(x)$

$= \int P_n(x)^2 + 2 P_n(x)q_{n-1}(x) + q_{n-1}(x)^2 d\mu(x)$

$= \| P_n \|_{L^2(\mu)}^2 + \| q_{n-1} \|_{L^2(\mu)}^2$

$\geq \| P_n \|_{L^2(\mu)}^2,$

and the minimum is clearly attained for $P_n$. ♦

A lot is known about the zeros of the orthogonal polynomials (for example, see my first and second post), some of them are collected in the following theorem.

Theorem 1.2. Let $P_n(x)$ be the $n$-th monic orthogonal polynomial with respect to $\mu$. Then the zeros of $P_n(x)$ are real and simple, moreover they are contained in the convex hull of $\textnormal{supp}(\mu)$. ♦

According to these facts, we can assume without the loss of generality that the zeros of $P_n(x)$ are denoted with $x_{1n}, \dots, x_{nn}$, moreover

$x_{1n} < x_{2n} < \dots < x_{nn}.$

For these points we can define the normalized counting measure $\mu_n$ as

$\mu_n := \frac{1}{n} \sum_{k=1}^{n} \delta_{x_{kn}},$

where $\delta_{x}$ is the Dirac probability measure concentrated at the point $\{ x \}$. Our aim is to prove the following theorem.

Theorem 1.3. (Erdős-Turán) Let $\mu$ be an absolutely continuous probability measure supported on $[-1,1]$ with $d\mu(x) = w(x) dx$ and let $P_n(x) = x^n + \dots$ denote the monic orthogonal polynomials. If $w(x) > 0$ almost everywhere on $[-1,1]$, then

$\| P_n \|_{L^2(\mu)}^{1/n} \to \frac{1}{2} \quad (n \to \infty),$

and

$\mu_n \xrightarrow{\phantom{w}w\phantom{w} } \nu \quad (n \to \infty),$

where $\nu$ is a measure on $[-1,1]$ defined as

$d\nu(x) = \frac{1}{\pi \sqrt{1-x^2}} dx,$

$\mu_n$ denotes the normalized counting measure of the zeros of $P_n$, and ”$\xrightarrow{w}$” denotes weak-* convergence of measures. ♦

This theorem was proven first by Pál Erdős and Pál Turán. The proof I am going to show was given by Walter Van Assche, and it can be found in his book. In the next part we shall collect some important tools and then we prove Theorem 3.

2.1. Potential, energy, capacity, equilbrium measure. Let $K \subseteq \mathbb{C}$ be a compact subset of the complex plane and let $\mu$ be a finite Borel measure supported on $K$. The potential of $\mu$ is defined as

$U^{\mu}(z) := \int \log\frac{1}{|z-w|} d\mu(w).$

The potential is a lower semicontinuous function, i.e. $\liminf_{w \to z} U^{\mu}(w) \geq U^{\mu}(z)$.

The energy of the measure is defined as

$I(\mu) := \int \int \log \frac{1}{|z-w|} d\mu(z) d\mu(w).$

With the notation $\mathcal{M}_{1}(K) := \{ \mu: \textnormal{supp}(\mu) = K, \mu(K) = 1 \}$ (i.e. the Borel probability measures supported on $K$), the energy of the set $K$ is defined as

$I(K) := \inf_{\mu \in \mathcal{M}_1(K)} I(\mu).$

This quantity is also called Robin’s constant. The logarithmic capacity of $K$ is defined as

$\textnormal{cap}(K) := e^{-I(K)}.$

The capacity can be computed explicitly for some sets, for example, an interval $[a,b]$ on the real line has capacity $\textnormal{cap}([a,b]) = (b-a)/4$. Sets with zero logarithmic capacity are called polar sets. A property is said to hold quasi everywhere for a set $S$ if it holds for all points in $S \setminus E$, where $E$ is a polar set. (It is important that quasi everywhere implies almost everywhere in the Lebesgue sense, but not the other way around.)

Given a weakly convergent sequence of measures, the following well-known theorem holds.

Theorem 2.1. (Principle of descent and lower envelope theorem) Let $\mu$ and $\{ \mu_n \}_{n=1}^{\infty}$ be finite Borel measures on the complex plane with support lying in a compact set and assume that $\mu_n \xrightarrow{w} \mu$ as $n \to \infty$, where ”$\xrightarrow{w}$” denotes weak-* convergence. Then

$U^{\mu}(z) \leq \liminf_{n \to \infty} U^{\mu_n}(z), \quad z \in \mathbb{C}.$

(This is the ”principle of descent” part.) Moreover, we have

$U^{\mu}(z) = \liminf_{n \to \infty} U^{\mu_n}(z) \quad \textnormal{quasi everywhere on } \mathbb{C}.$

(This is the ”lower envelope” part.) ♦

For the proof, see [Saff-Totik] Chapter I.6 Theorems 6.8 and 6.9. Another important and frequently used theorem is the principle of domination.

Theorem 2.2. (Principle of domination) Let $\mu$ and $\nu$ be two positive and finite Borel measures with compact support on $\mathbb{C}$ with finite logarithmic energy and suppose that $\nu(\mathbb{C}) \leq \mu(\mathbb{C})$. If for some constant $c$ the inequality

$U^{\mu}(z) \leq U^{\nu}(z) + c$

holds $\mu$-almost everywhere, then it holds for all $z \in \mathbb{C}$. ♦

For the proof of the principle of domination, see [Saff-Totik] Chapter II, Theorem 3.2. If $K$ is not polar, then there is a unique measure denoted with $\nu_K$ such that

$I(\nu_K) = I(K).$

This measure is called the equilibrium measure of $K$. The equilibrium measure is known explicitly only for a few sets, for example the interval $[-1,1]$, on which the equilibrium measure $\nu_{[-1,1]}$ takes the form

$d\nu_{[-1,1]}(x) = \frac{1}{\pi \sqrt{1-x^2}} dx, \quad x \in [-1,1]$

which is the measure appearing in Theorem 2.4. According to the following well-known theorem, the potential of an equilibrium measure behaves nicely on the complex plane.

Theorem 2.3. (Frostman’s theorem) Let $K \subseteq \mathbb{C}$ be a compact set. Then
(i) $U^{\nu_K}(z) \leq I(K)$ for all $z \in \mathbb{C}$,
(ii) $U^{\nu_K}(z) = I(K)$ quasi everywhere on $K$. ♦

For the proof, see [Ransford] Theorem 3.3.4. These properties appearing in Frostman’s theorem characterize the equilibrium measure in a sense.

Theorem 2.4. (Characterization of the equilibrium measure) Let $K \subseteq \mathbb{C}$ be a compact set and let $\mu$ be a probability measure of finite energy on $K$. If $U^{\mu}(z) = I(K)$ quasi everywhere on $K$ then $\mu$ is the equilibrium measure $\nu_K$. ♦

For the proof, see [Saff-Totik] Section I.3, Theorem 3.3.

Exercise 2.5. Show that Theorem 2.4.is false if we do not require that $\mu$ is supported on $K$!

Combining Frostman’s theorem with the principle of descent we can establish the main lemma on which we will base the proof of Theorem 1.3.

Lemma 2.6. (Brolin) Let $K$ be a compact subset of the complex plane, assume that $\textnormal{cap}(K) > 0$, or in other words, $K$ is not polar. Suppose that the equilibrium measure $\nu_K$ of $K$ is supported on $K$, i.e. $\textnormal{supp}(\nu_K) = K$. Let $\{ \mu_n \}_{n=1}^{\infty} \subset \mathcal{M}_1(K)$ be a sequence of probability measures on $K$ and suppose that they converge weakly to a probability measure $\mu \in \mathcal{M}_1(K)$. If

$\liminf_{n \to \infty} U^{\mu_n}(z) \geq I(K) \quad \nu_K \textnormal{-almost everywhere}$

holds, then the weak limit $\mu$ of $\mu_n$ is the equilibrium measure $\nu_K$.

Proof. According to Theorem 2.4, it is enough to prove that $U^{\mu}(z) = I(K)$ quasi everywhere on $K$, and our goal is to show this. The property

$I(K) \leq \liminf_{n \to \infty} U^{\mu_n}(z)$

implies that (since $\nu_K$ is a probability measure)

$I(K) \leq \int \liminf_{n \to \infty} u^{\mu_n}(z) d\nu_K(z).$

Using Fatou’s lemma, Frostman’s theorem (i.e. $U^{\nu_K}(z) \leq I(K)$ on $\mathbb{C}$) and Fubini’s theorem, we have

$I(K) \leq \int \liminf_{n \to \infty} U^{\mu_n}(z) d\nu_K(z)$

$\leq \liminf_{n \to \infty} \int U^{\mu_n}(z) d\nu_K(z)$

$= \liminf_{n \to \infty} \int U^{\nu_K}(z) d\mu_n(z)$

$\leq I(K),$

which gives

$\int \liminf_{n \to \infty} U^{\mu_n}(z) d\nu_K(z) = I(K).$

Since $\liminf_{n \to \infty} U^{\mu_n}(z) \geq I(K)$ and $\nu_K$ is a probability measure, it follows immediately that

$\liminf_{n \to \infty} V^{\mu_n}(z) = I(K) \quad \nu_K \textnormal{-almost everywhere.}$

The lower envelope theorem (see Theorem 2.1) gives $U^{\mu}(z) = \liminf_{n \to \infty} U^{\mu_n}(z)$ quasi everywhere, therefore

$U^{\mu}(z) = I(K) \quad \nu_K \textnormal{-almost everywhere}.$

Since $\nu_K$ is supported on $K$, for every fixed $z_0 \in K$ there is a $w_0$ arbitrary close to $z_0$ such that $U^{\mu}(w) = I(K)$. Using that $U^{\mu}$ is lower semicontinuous (i.e. $U^{\mu}(z) \leq \liminf_{w \to z} U^{\mu}(w)$), we have

$U^{\mu}(z) \leq I(K), \quad z \in K.$

On the other hand, the principle of domination (see Theorem 2.2) implies that $U^{\nu_K}(z) \leq U^{\mu}(z)$ for all $z \in K$, and, according to Frostman’s theorem, $U^{\nu_K}(z) = I(K)$ quasi everywhere on $K$, therefore

$U^{\mu}(z) \geq I(K) \quad \textnormal{quasi everywhere on } K.$

These imply that $\mu = \nu_K$, and this is what we wanted to show. ♦

2.2. Chebyshev polynomials. Let $K \subseteq \mathbb{C}$ be a compact subset of the complex plane containing infinitely many points and define the quantity

$m_n(K) := \inf \{ \| P \|_{K}: P \textnormal{ is a monic polynomial of degree } n\}.$

It is known that there exists a unique polynomial $Vyacheslav T_n(x,K) = T_n(x)$ of degree $Vyacheslav n$ such that $Vyacheslav \| T_n \|_K = m_n(K)$. (This property is the $Vyacheslav L^\infty$ analogue of Theorem 2.1. $Vyacheslav T_n$ is called the $Vyacheslav n$-th Chebyshev polynomial. It is also known, and shall prove useful for us, that

$\lim_{n \to \infty} m_n(K)^{1/n} = \lim_{n \to \infty} \| T_n \|_{K}^{1/n} = c(K).$

The Chebyshev polynomials on the interval $[-1,1]$ are very well known and have been intensively studied. They can be written in the form

$T_n(x) = \frac{\cos (n \arccos x)}{2^n}, \quad x \in [-1,1],$

which is indeed a monic polynomial. Incidentally, they are also the orthogonal on $[-1,1]$ for the measure $\frac{1}{\pi \sqrt{1-x^2}} dx$, which is the equilibrium measure for $[-1,1]$.

3. The proof of the main theorem. Before we start with the proof of Theorem 2.4, we need one more lemma.

Lemma 3.1. (Ullman) Let $\mu$ be a positive measure on $K \subseteq \mathbb{R}$ and let $\{ f_n \}_{n=1}^{\infty}$ be a sequence of nonnegative $\mu$-measurable functions. Suppose that

$\limsup_{n \to \infty} \Big( \int f_n(x) d\mu(x) \Big)^{1/n} \leq 1.$

Then there exists a subsequence $\{ n_k \}_{k=1}^{\infty}$ such that

$\limsup_{k \to \infty} f_{n_k}(x)^{1/n_k} \leq 1 \quad \mu \textnormal{-almost everywhere.}$C

Proof. Let $\{ \varepsilon_n \}_{n=1}^{\infty}$ be a positive sequence of real numbers such that $\varepsilon_n \to 0$ and

$\Big( \int f_n(x) d\mu(x) \Big)^{1/n} \leq 1 + \varepsilon_n.$

Define

$g_n(x) := \frac{f_n(x)}{n(1 + \varepsilon_n)^n}.$

It is immediate that $\int g_n(x) d\mu(x) \to 0$, and therefore there exists a subsequence $n_k$ such that $\lim_{k \to \infty} g_{n_k}(x) = 0$ $\mu$-almost everywhere. (Recall that $f_n \xrightarrow{L^1(\mu)} f$ implies that every subsequence of $f_n$ has a subsequence which converges to $f$ $\mu$-almost everywhere.) But this means that for large $k$, we have

$f_{n_k}(x) \leq n_k (1 + \varepsilon_{n_k})^{n_k},$

which implies

$\limsup_{n \to \infty} f_{n_k}(x)^{1/n_k} \leq 1 \quad \mu \textnormal{-almost everywhere,}$

and this is what we needed to prove. ♦

Proof of Theorem 1.4. First we are going to show that $\lim_{n \to \infty} \| P_n \|_{L^2(\mu)}^{1/n} = 1/2 = \textnormal{cap([-1,1])}$. The $L^2$-extremal property of $P_n$ (see Theorem 1.1.) implies that

$\| P_n \|_{L^2(\mu)} \leq \Big( \int |T_n(x)|^2 d\mu(x) \Big)^{1/2} \leq \| T_n \|_{[-1,1]},$

where $T_n(x) = T_n(x,[-1,1])$ is the Chebyshev polynomial of $[-1,1]$. The convergence of the norm of the Chebyshev polynomials gives that

$\limsup_{n \to \infty} \| P_n \|_{L^2(\mu)}^{1/n} \leq \limsup_{n \to \infty} \| T_n \|_{[-1,1]}^{1/n} = \textnormal{cap}([-1,1]) = \frac{1}{2}.$

Now assume that

$\liminf_{n \to \infty} \| P_n \|_{L^2(\mu)}^{1/n} \leq \alpha < \textnormal{cap}([-1,1]).$

This means that there is a subsequence $n_k$ such that

$\lim_{k \to \infty} \Big( \int_{-1}^{1} |P_{n_k}(x)|^2 w(x) dx \Big)^{1/n_k} \leq \alpha^2.$

Using Lemma 3.1 with $f_n(x) := |P_n(x)|^2 w(x) \alpha^{-2n}$ we obtain a further subsequence $k_l$ such that

$\limsup_{l \to \infty} \big( |P_{k_l}(x)|^2 w(x) \big)^{1/k_l} \leq \alpha^2,$

which, since $w(x) > 0$, implies that $\limsup_{l \to \infty} |P_{k_l}(x)|^{1/k_l} \leq \alpha^2$ Lebesgue almost everywhere. Since the equilibrium measure $d\nu_{[-1,1]}(x) = \frac{1}{\pi \sqrt{1-x^2}} dx$ is mutually absolutely continuous with the Lebesgue measure, we have

$\limsup_{l \to \infty} |P_{k_l}(x)|^{1/k_l} \leq \alpha \quad \nu_{[-1,1]} \textnormal{-almost everywhere}.$

Now, it is easy to see that the potential of the normalized counting measure $\mu_n$ defined as

$\mu_n = \frac{1}{n} \sum_{k=1}^{n} \delta_{x_{kn}},$

(where $x_{1n} < \dots < x_{nn}$ are the zeros of the monic orthogonal polynomial $P_n$) is

$U^{\mu_n}(z) = \log |P_n(z)|^{-1/n}.$

Let $l_m$ be a subsequence of $k_l$ such that $\mu_{l_m}$ weakly converges to some limit $\nu^*$. (Such a subsequence exists, since Helly’s selection theorem can be applied.) We get that

$\liminf_{m \to \infty} V^{\mu_{l_m}(x)} \geq \log \frac{1}{\alpha} > \log \frac{1}{\textnormal{cap}([-1,1])} = I([-1,1])$

$\nu_{[-1,1]}$-almost everywhere.The lower envelope theorem (see Theorem 2.2) gives

$U^{\nu^*}(x) = \liminf_{m \to \infty} U^{\mu_{l_m}(x)} > I(K) \quad \nu_{[-1,1]} \textnormal{-almost everywhere},$

which, according to Lemma 2.6, implies that $\nu^*$ is the equilibrium measure $\nu_{[-1,1]}$. But this is a contradiction, since then $U^{\nu^*}(x) = I([-1,1])$ quasi everywhere! Therefore

$\liminf_{n \to \infty} \| P_n \|_{L^2(\nu)}^{1/n} \geq \textnormal{cap}([-1,1]) = \frac{1}{2},$

which implies

$\lim_{n \to \infty} \| P_n \|_{L^2(\mu)}^{1/n} = \textnormal{cap}([-1,1]) = \frac{1}{2}.$

Now we prove that $\mu_n \xrightarrow{\phantom{w}w\phantom{w}} \nu_{[-1,1]}$. Let $\mu_{n_k}$ be an arbitrary subsequence of $\mu_n$. Since $\mu_n$ is a probability measure supported on a compact set, we can apply Helly’s selection theorem, which gives a subsequence $\mu_{k_l}$ of $\mu_{n_k}$ such that $\mu_{k_l} \xrightarrow{w} \mu^*$ for some $\mu^*$. Applying the same reasoning as before, we obtain that

$\liminf_{l \to \infty} V^{\mu_{k_l}}(x) \geq I([-1,1]) \quad \nu_{[-1,1]}\textnormal{-almost everywhere}$

and

$\liminf_{l \to \infty} V^{\mu_{k_l}}(x) = \liminf_{l \to \infty} U^{\mu^*}(x) \quad \textnormal{quasi everywhere},$

whose, with the application of Lemma 2.6 imply that $\mu^*$ is the equilibrium measure. Therefore $\mu_n \xrightarrow{w} \nu_{[-1,1]}$ holds. ♦

4. Generalizations of Theorem 1.4. The main theorem works not only for measures supported on $[-1,1]$, but for measures supported on compact subsets of the real line. To be more precise, the following theorem holds.

Theorem 4.1. Let $\mu$ be a Borel probability measure on the real line supported on a compact set $K$, which has positive capacity. Let $P_n(x) = x^n + \dots$ denote the $n$-th monic orthogonal polynomial with respect to $\mu$. Suppose that $\mu = \mu_1 + \mu_2$, where $\mu_1$ is absolutely continuous and $\mu_2$ is singular with respect to the equilibrium measure $\nu_K$. If $d\mu_1(x) = w(x) d\nu_K(x)$ and $w(x) > 0$ $\nu_K$-almost everywhere, then

$\| P_n \|_{L^2(\mu)}^{1/n} \to \textnormal{cap}(K) \quad (n \to \infty)$

and

$\mu_n \xrightarrow{\phantom{w}w\phantom{w}} \nu_K \quad (n \to \infty),$

where $\mu_n$ is the normalized counting measure of the zeros of $P_n$. ♦

It can be shown easily that if we do not require that $\mu$ is supported on the real line, then the statement of Theorem 4.1 regarding the weak-* convergence of the normalized counting measures is false.

References.

[Ransford] Thomas Ransford, Potential Theory in the Complex Plane, Cambridge University Press, 1995
[Saff-Totik] Edward B. Saff and Vilmos Totik, Logarithmic Potentials with External Fields, Springer-Verlag, 1997
[Stahl-Totik] Herbert Stahl and Vilmos Totik, General Orthogonal Polynomials, Cambridge University Press, 1992
[VanAssche] Walter Van Assche, Asymptotics for Orthogonal Polynomials, Lecture Notes in Mathematics, Springer-Verlag, 1987