### The zeros of orthogonal polynomials, part II: OPRL vs OPUC

I finished up the previous post with some questions. We studied the zero distributions of orthogonal polynomials on the real line, and I was curious about the possible generalizations of Theorem 2. The ultimate generalization lies deeper then I can dig in one post, but this motivated me to write about orthogonal polynomials on the unit circle. Let’s recall the second question at the end of the previous post.

Question 2. What about measures supported on more general sets? For example, is there a similar theorem for measures on the unit circle?

In this post, I want to show the striking contrast between orthogonal polynomials on the real line (in short, OPRL) and on the unit circle (OPUC). We study the former ones first. For reference, the interested reader shall check [Freud] and [Ismail], two excellent books on orthogonal polynomials.

1. Zeros of OPRL. In this section, $\mu$ will be a Borel measure supported on the real line. As in the previous post, we shall impose two conditions on $\mu$.
(i) The support of $\mu$ contains infinitely many points, and
(ii) for all $k \in \{0,1,2,\dots\}$, we have $\int x^k d\mu(x) < \infty$.
Like before, $p_n(x,\mu) = p_n(x)$ will denote the $n$-th orthonormal polynomial. The dependence on the measure is not indicated, unless it is necessary.

Before we start studying the zeros, we need the following very useful and simple lemma.

Lemma 1. Let $\Pi(x)$ be a polynomial and suppose that $\Pi(x) > 0$ for every $x \in \textnormal{supp}(\mu)$. Then

$\int \Pi(x) d\mu(x) > 0$.

With this tool in our hands, we can start investigating the zeros of $p_n(x)$.

Theorem 1. The zeros of $p_n(x)$ are real and simple, moreover they lie in the convex hull of $\textnormal{supp}(\mu)$.

Proof. Suppose that $x_1, x_2, \dots, x_m$ are the zeros of $p_n$ lying in the convex hull of the support with odd multiplicity. If $m < n$, then by orthogonality we have

$\int p_n(x) \prod_{k=1}^{m}(x-x_k) d\mu(x) = 0$.

But the polynomial $p_n(x) \prod_{k=1}^{m}(x-x_k)$ is positive for every $x$ in the support! It follows that $m = n$, and this is what we needed to prove. $\Box$

Denote the zeros of $p_n$ with $x_{1,n} < x_{2,n} < \dots < x_{n,n}$. Another interesting observation is that the zeros of $p_n$ and $p_{n-1}$ interlace. We shall not prove it here, but we have the following.

Theorem 2. For every $k \in {1,2,\dots,n-1}$,

$x_{k,n} < x_{k,n-1} < x_{k+1,n}$

holds. $\Box$

This interlacing property is illustrated below with the zeros of the Chebyshev polynomials.

2. The moment problem. Before discussing some deeper results about the zeros of $p_n$, we have to mention the so-called moment problem. According to Barry Simon‘s excellent survey article on OPUC, the moment problem was a central question of mathematical analysis at the beginning of the 20th century. (We can also find out from the introduction of this article that the orthogonal polynomials are the Rodney Dangerfield of mathematics.)
It is clear that the measure $\mu$ determines the numbers

$m_k := \int x^k d\mu(x)$.

(Remember that according to our assumptions, this number is finite.) What about the other way? That is, if we are given a sequence of numbers $\{ m_k \}_{k=0}^{\infty}$, can we construct a measure $\mu$ such that $m_k = \int x^k d\mu(x)$ holds? It is easy to see that if $\mu$ is given, the determinants

$D_n(m_0, \dots, m_{2n}) = D_n := \begin{vmatrix} m_0 & m_1 & \dots & m_n\\ m_1 & m_2 & \dots & m_{n+1} \\ \vdots & \vdots & \ddots & \vdots \\ m_n & m_{n+1} & \dots & m_{2n} \end{vmatrix}$

must be positive. Indeed, taking the polynomial $\Pi_n(x) = \sum_{k=0}^{n} a_k x^k$ of degree $n$, we have

$\int \Pi_n(x)^2 d\mu(x) = \sum_{k,l=0}^{n} a_k a_l m_{k+l} > 0$,

which means that $D_n > 0$ for all $n \in \{0,1,2,\dots\}$. It turns out that this condition is not only necessary but also sufficient. That is, the following holds.

Theorem 3. If $\{ m_k\}_{k=0}^{\infty}$ is a sequence such that $D_n(m_0, \dots, m_n) > 0$, then there is a measure $\mu$ such that $m_k = \int x^k d\mu(x)$.

The problem is, although there is a measure $\mu$, unicity does not necessarily hold. We say that the moment problem is determinate for a given set of moments if and only if there is only one measure with the given moments.

For a (really) detailed discussion of the moment problem, check the classic book [Akhiezer]. I will definitely write a post dedicated to the moment problem later. One reason is that I want to understand in detail Barry Simon’s article from 1998 titled The classical moment problem as a self-adjoint finite difference operator (with an intimidating 122 pages), which described a connection between orthogonal polynomials and the spectral theory of Jacobi operators, and this connection proved to be very fruitful. I plan to study these tools, therefore it is necessary to understand the moment problem as deep as possible. For now, I conclude this topic and continue to study our much beloved zeros.

Back to the main track: zeros of OPRL. From now on, we assume that for our measure $\mu$, the moment problem is determinate. (That is, there is no other measure $\nu$ with $\int x^k d\nu(x) = \int x^k d\mu(x)$.) In this case, the zeros of the orthogonal polynomials behave very nicely in a sense.

Theorem 4. Let $\mu$ be a Borel measure supported on the real line with our usual assumptions. (See the beginning of this post.) Also suppose that the moment problem is determinate. Then for every $x_0 \in \textnormal{supp}(\mu)$ and $\epsilon > 0$, $p_n(x)$ has a zero in the interval $(x_0-\epsilon, x_0+\epsilon)$ for all large enough $n$. Moreover, if $(a,b) \cap \textnormal{supp}(\mu)$ is empty, then $p_n(x)$ has at most one zero in $(a,b)$ for all $n$. $\Box$

It is very interesting for me that a general (and to me, a bit esoteric) condition such as the unicity of the moment problem implies a deep result on the behavior of the zeros. Of course, Theorem 4 implies that the zeros of the orthogonal polynomials are dense in $\textnormal{supp}(\mu)$.

Now let’s take a brief look at the world of orthogonal polynomials on the unit circle.

3. Zeros of OPUC. Although I introduced the orthogonal polynomials for measures supported on the real line, it is not really necessary to restrict ourselves to $\mathbb{R}$. If we have a Borel measure $\mu$ which is supported on a subset of the complex plane (and of course satisfies our usual assumptions, i.e. contains infinitely many points in its support and all its moments are finite), we still have orthogonal polynomials. We shall study the case when the support is a subset of the unit circle. In this case I will denote the orthonormal polynomials with $\{ \varphi_n(z,\mu) \}_{n=0}^{\infty}$, to emphasize that we work with measures on the unit circle.
Probably the simplest example is if we take the normalized Lebesgue measure on the unit circle. That is, set

$d\mu(e^{it}) = \frac{1}{2\pi} dm(e^{it})$.

It is easy to see that $\{ z^k \}_{k=0}^{\infty}$ is orthonormal with respect to this measure. It is also easy to see that the only zero these polynomials have is $0$. The contrast with OPRL is striking. Not only a zero can have larger multiplicity, it can lie outside of the support for infinitely many orthogonal polynomials. (In this case, all of them.) Although they can behave strange, the zeros cannot be anywhere, as this next theorem illustrates.

Theorem 5. The zeros of $\varphi_n$ are contained in the closed unit disk $\overline{\mathbb{D}} = \{ z: |z| \leq 1 \}$.

Proof. Let $z_0$ be a zero of $\varphi_n(z)$. Then $\varphi_n(z) = (z-z_0) \Pi(z)$ for some polynomial $\Pi$ of degree one. It follows that $z \Pi(z) = \varphi_n(z) + z_0 \Pi(z)$. Then

$\| \Pi(z) \|^2 = \int \Pi(z) \overline{\Pi(z)} d\mu(z) = \int |z| \Pi(z) \overline{\Pi(z)} d\mu(z) = \int z \Pi(z) \overline{z \Pi(z)} d\mu(z) = \| z \Pi(z) \|^2 = \| \varphi_n(z) + z_0 \Pi(z)\|^2 = \| \varphi_n(z) \|^2 + |z_0|^2 \| \Pi(z) \|^2$,

which gives $\| \varphi_n \|^2 = (1-|z_0|^2) \| \Pi \|^2$. Since $\| \varphi_n\|^2 = 1$, it immediately follows that $|z_0| < 1$. $\Box$

Five more proofs (and this one) can be found in [Simon], Theorem 1.7.1. For the more general case, where the support of the measure is an arbitrary subset of the complex plane, see [Stahl-Totik], Theorem 2.1.1.

Although there can be some degenerate cases like the normed Lebesgue measure on the unit circle, there are also cases when the zeros are dense in the convex hull of the support. In general, we have the following theorem.

Theorem 6. If $S$ is a subset of the complex plane, then there is a measure $\mu$ such that $\textnormal{supp}(\mu) = S$ and the zeros of the orthogonal polynomials are dense in $S$. $\Box$

The detailed proof can be found in [Stahl-Totik], Example 2.1.2. and Theorem 2.1.3.

Since there are plenty of questions left from last time (I haven’t even answered the question what motivated me to write about the differences between OPRL and OPUC), I conclude this post with Theorem 6.

References.
[Akhiezer] Naum I. Akhiezer, The classical moment problem and some related questions in analysis, Hafner Publishing Co., New York 1965
[Freud] Géza Freud, Orthogonal polynomials, Pergamon Press, 1971
[Ismail] Mourad E. H. Ismail, Classical and quantum orthogonal polynomials in one variable, Cambridge University Press, Cambridge, 2005
[Simon] Barry Simon, Orthogonal polynomials on the unit circle, part I, AMS Colloquium Publications, 2004
[Stahl-Totik] Herbert Stahl and Vilmos Totik, General orthogonal polynomials, Cambridge University Press, 1992