Finite Blaschke products with prescribed critical points, Stieltjes polynomials, and moment problems

TL;DR

This paper explores the equivalence between determining finite Blaschke products from critical points, charge interaction models, and moment problems, revealing new theoretical insights and algorithmic possibilities.

Contribution

It establishes the equivalence of three problems involving Blaschke products, charge configurations, and moment problems, and links these to polynomial solutions and energy minimization.

Findings

Proves the equivalence of three key problems in complex analysis and mathematical physics.

Connects Blaschke products with Stieltjes polynomials and moment problems.

Suggests new algorithmic approaches based on energy minimization.

Abstract

The determination of a finite Blaschke product from its critical points is a well-known problem with interrelations to other topics. Though existence and uniqueness of solutions are established for long, we present several new aspects which have not yet been explored to their full extent. In particular, we show that the following three problems are equivalent: (i) determining a finite Blaschke product from its critical points, (ii) finding the equilibrium position of moveable point charges interacting with a special configuration of fixed charges, (iii) solving a moment problem for the canonical representation of power moments on the real axis. These equivalences are not only of theoretical interest, but also open up new perspectives for the design of algorithms. For instance, the second problem is closely linked to the determination of certain Stieltjes and Van Vleck polynomials for a second order ODE and allows the description of solutions as global minimizers of an energy functional.

Equations

$$B(z)=c\prod _{k=1}^n\frac{z-a_k}{1-{\overline{a}}_{k}z}$$

$$T(z)=\mathrm{i}\frac{1+z}{1-z},$$

$$f(x)=-\frac{r_1}{x-x_1}-\frac{r_2}{x-x_2}-\cdots -\frac{r_n}{x-x_n}$$

$$g(x)=ax+b-\frac{s_1}{x-t_1}-\cdots -\frac{s_{n-1}}{x-t_{n-1}}$$

$$f^{\prime }(x)=\frac{r_1}{(x-x_1{)}^2}+\frac{r_2}{(x-x_2{)}^2}+\cdots +\frac{r_n}{(x-x_n{)}^2}=\frac{cP(x)}{Q(x{)}^2}$$

$$Q(x):=\prod _{k=1}^n(x-x_k).$$

$$P(x)=\prod_{k=1}^{n-1}(x-\zeta_{k})(x-\overline{\zeta}_{k})=\prod_{k=1}^{n-1}\big{(}(x-a_{k})^{2}+b_{k}^{2}\big{)}$$

$$cP(x)=\sum _{k=1}^n{r}_k\prod _{\begin{array}{c}j=1\\ j\ne k\end{array}}^n(x-x_j{)}^2.$$

$$r_k=\frac{cP(x_k)}{\prod _{\begin{array}{c}j=1\\ j\ne k\end{array}}^n(x_k-x_j{)}^2},k=1,2,\dots ,n.$$

$$Q_k(x):=\prod _{\begin{array}{c}j=1\\ j\ne k\end{array}}^n\frac{x-x_j}{x_k-x_j}=\frac{Q(x)}{Q^{\prime }(x_k)(x-x_k)},k=1,2,\dots ,n,$$

$$P(x)=\sum _{k=1}^{n}P(x_k)Q_k(x{)}^2.$$

$$P(x)=\sum _{k=1}^{n}P(x_k)h_k(x)+\sum _{k=1}^n{P}^{\prime }(x_k){\mathfrak{h}}_k(x),$$

$$h_k(x):=\left(1-\frac{Q^{\prime \prime }(x_k)}{Q^{\prime }(x_k)}(x-x_k)\right)Q_k(x{)}^2,{\mathfrak{h}}_k(x):=(x-x_k)Q_k(x{)}^2$$

$$P(x)=\sum _{k=1}^{n}P(x_k)Q_k(x{)}^2+\sum _{k=1}^n\left(P^{\prime }(x_k)-P(x_k)\frac{Q^{\prime \prime }(x_k)}{Q^{\prime }(x_k)}\right)(x-x_k)Q_k(x{)}^2.$$

$$Q^{\prime }(x_k)P^{\prime }(x_k)=Q^{\prime \prime }(x_k)P(x_k),k=1,2,\dots ,n.$$

$$Q^{\prime }{P}^{\prime }-Q^{\prime \prime }P=QR.$$

$$P{Q}^{\prime \prime }-P^{\prime }{Q}^{\prime }+RQ=0,$$

$$g^{\prime }(x)=a+\frac{s_1}{(x-t_1{)}^2}+\cdots +\frac{s_{n-1}}{(x-t_{n-1}{)}^2}=\frac{aP(x)}{S(x{)}^2}.$$

$$S(x):=\prod _{k=1}^{n-1}(x-t_k).$$

$$aP(x)=a\prod _{k=1}^{n-1}(x-t_k{)}^2+\sum _{k=1}^{n-1}{s}_k\prod _{\begin{array}{c}j=1\\ j\ne k\end{array}}^{n-1}(x-t_j{)}^2,$$

$$s_k=\frac{aP(t_k)}{\prod _{\begin{array}{c}j=1\\ j\ne k\end{array}}^{n-1}(t_k-t_j{)}^2},k=1,\dots ,n-1.$$

$$S_k(x):=\prod _{\begin{array}{c}j=1\\ j\ne k\end{array}}^{n-1}\frac{x-t_j}{t_k-t_j}=\frac{S(x)}{S^{\prime }(t_k)(x-t_k)},k=1,2,\dots ,n-1,$$

$$P(x)=S(x{)}^2+\sum _{k=1}^{n-1}P(t_k)S_k(x{)}^2.$$

$$P(x)-S(x{)}^2=\sum _{k=1}^{n-1}P(t_k)g_k(x)+\sum _{k=1}^{n-1}{P}^{\prime }(t_k){\mathfrak{g}}_k(x),$$

$$g_k(x):=\left(1-\frac{S^{\prime \prime }(t_k)}{S^{\prime }(t_k)}(x-t_k)\right)S_k(x{)}^2,{\mathfrak{g}}_k(x):=(x-t_k)S_k(x{)}^2.$$

$$P(x)-S(x{)}^2=\sum _{k=1}^{n-1}P(t_k)S_k(x{)}^2+\sum _{k=1}^{n-1}\left(P^{\prime }(t_k)-P(t_k)\frac{S^{\prime \prime }(t_k)}{S^{\prime }(t_k)}\right)(x-t_k)S_k(x{)}^2,$$

$$P^{\prime }(t_k)S^{\prime }(t_k)=P(t_k)S^{\prime \prime }(t_k),k=1,\dots ,n-1,$$

$$P{S}^{\prime \prime }-P^{\prime }{S}^{\prime }+\tilde{R}S=0,$$

$$A{Q}^{\prime \prime }+2B{Q}^{\prime }+CQ=0$$

$$A(x)=(x-a_0)(x-a_1)\cdots (x-a_p)$$

Full text

1 Introduction

A finite Blaschke product of degree $n$ is a rational function of the form

[TABLE]

whose (not necessarily distinct) zeros $a_{1},\dots,a_{n}$ are in the unit disc $\mathbb{D}$ and $c\in\mathbb{T}:=\partial\mathbb{D}$. A point $\xi$ where $B^{\prime}(\xi)=0$ is called critical point of the Blaschke product. Every Blaschke product of degree $n$ has exactly $n-1$ critical points $\xi_{k}$ in $\mathbb{D}$ (counted with multiplicities), and another $n-1$ critical points $1/\overline{\xi}_{k}$, symmetric to $\xi_{k}$ with respect to the unit circle $\mathbb{T}$ (see e.g. [24]). Note that if $a_{j}$ is a $k$-fold zero then $a_{j}$ and $1/\overline{a}_{j}$ are both criticals points of order $k-1$, and if [math] is a critical point then we also count $\infty$ as a critical point of the same multiplicity.

While it is straightforward to determine the critical points of a Blaschke product from its zeros (by computing the zeros of a polynomial), the reverse problem is much more challenging. The basic existence and uniqueness result is summarized in the following theorem.

Theorem 1**.**

Let $\xi_{1},\dots,\xi_{n-1}$ be $n-1$ points in $\mathbb{D}$. Then there is a Blaschke product $B$ of degree $n$ with critical points $\xi_{k}$. $B$ is unique up to post-composition with a conformal automorphism of $\mathbb{D}$.

This theorem has been proved (in chronological order) by Heins [15], Wang and Peng [39], Bousch [3], and Zakeri [40], using topological arguments. Stephenson [36, Theorem 21.1] obtains $B$ as the limit of discrete finite Blaschke products, i.e., he considers sequences of circle packings with prescribed branch set. Kraus and Roth [19], [20] describe an approach based on a solution of the Gaussian curvature equation (that works equally well for infinite Blaschke products), and ask for a procedure to actually compute $B$ from its critical points.

In this paper we show that the determination of Blaschke products with prescribed critical points is equivalent to two other classical problems of analysis. The first one is obtained after transforming the problem from the unit disk to the upper half plane in form of a second order ODE. We will have to look for its polynomial solutions known as Stieltjes and Van Vleck polynomials. Like in the case originally considered by Stieltjes this allows an electrostatic interpretation and the characterization of solutions as minimum points of an energy functional. This approach yields a new and (as we hope) very transparent proof of Theorem 1. Moreover, the polynomial encoding the given critical points is positive on the real axis and therefore an inner point of the convex cone of positive polynomials on $\mathbb{R}$. By mapping it to an inner point of the cone of power moments we demonstrate that the problem is also equivalent to the classical problem of finding canonical representations of given moments.

Algorithmic aspects of the different approaches and the results of numerical experiments will be discussed in a forthcoming paper.

2 Transformations

Let $B$ be Blaschke product of degree $n\geq 2$. As in [34, Lemma 3] it is convenient to transform the problem using the (inverse) Cayley transform

[TABLE]

which maps $\mathbb{D}$ and $\mathbb{T}$ onto the upper half-plane $\mathbb{H}:=\{x+\mathrm{i}y:y>0\}$ and the extended real line $\mathbb{R}\cup\{\infty\}$, respectively, and satisfies $T(1)=\infty$, $T(-1)=0$. Consequently, $f:=T\circ B\circ T^{-1}$ is a rational function that is real-valued on $\mathbb{R}$ (except at its poles). The transition between the unit disc and the upper half plane was also a key tool in the work of Gorkin and Rhoades [8] on boundary interpolatioin by finite Blaschke products.

Let us first assume that $B$ satisfies the normalization $B(1)=-1$. Then $f$ has a zero at infinity and therefore $f=p/q$ with real polynomials $p$ of degree $n-1$ and $q$ of degree $n$. Since, for an appropriate branch of the argument function, the mapping $\tau\mapsto\arg B(\mathrm{e}^{\mathrm{i}\tau})$, is continuous and strictly monotone from $[0,2\pi)$ onto some interval $[\varphi,\varphi+2n\pi)$, $f$ has $n$ simple poles at real numbers $x_{1}<x_{2}<\dots<x_{n}$ corresponding to the $n$ values of $\tau$ with $B(\mathrm{e}^{\mathrm{i}\tau})=1$. In each of the intervals $(x_{k},x_{k+1})$ as well as in $(-\infty,x_{1})$ and $(x_{n},\infty)$, the function $f$ is strictly increasing. Therefore, the partial fraction decomposition of $f$ has the form

[TABLE]

with positive numbers $r_{k}$. Conversely, if $f$ is a rational function of the form (1) with ordered poles $x_{1}<x_{2}<\dots<x_{n}$ and $r_{k}>0$, then $B:=T^{-1}\circ f\circ T$ is a rational function that maps $\mathbb{D}$ and $\mathbb{T}$ onto themselves and satisfies $B(1)=-1$. Hence $B$ is a Blaschke product. By the argument principle it has exactly $n$ zeros in $\mathbb{D}$, i.e., $B$ has degree $n$. Thus we have shown:

Lemma 1**.**

The mapping $B\mapsto f:=T\circ B\circ T^{-1}$ is a bijection between all Blaschke products $B$ of degree $n$ satisfying $B(1)=-1$ and all rational functions $f$ of the form (1) with ordered poles $x_{1}<x_{2}<\dots<x_{n}$ and numbers $r_{k}>0$ for $k=1,2,\dots,n$.

It will turn out useful to also consider Blaschke products with the side condition $B(1)=1$. Then $g:=T\circ B\circ T^{-1}$ has a pole at infinity and can be written in the form $g=p/q$ with real polynomials $p$ of degree $n$ and $q$ of degree $n-1$, respectively. As in the case considered above, one derives the existence of $n-1$ finite poles $t_{1}<\dots<t_{n-1}$ corresponding to the $n-1$ points $\tau\in(0,2\pi)$, where $B(\mathrm{e}^{\mathrm{i}\tau})=1$. In each of the intervals $(t_{k},t_{k+1})$ as well as in $(-\infty,t_{1})$ and $(t_{n-1},\infty)$ the function $g$ is strictly increasing, hence the partial fraction decomposition of $g$ has the form

[TABLE]

with $a>0$, $b\in\mathbb{R}$ and $s_{k}>0$. Proceeding as above, we get the following lemma.

Lemma 2**.**

The mapping $B\mapsto g:=T\circ B\circ T^{-1}$ is a bijection between all Blaschke products $B$ of degree $n$ satisfying $B(1)=1$ and all rational functions $g$ of the form (2) with $a>0$, $b\in\mathbb{R}$, $t_{1}<t_{2}<\dots<t_{n-1}$ and $s_{k}>0$ for $k=1,2,\dots,n-1$.

We first investigate the problem for functions of the form (1). It follows from the chain rule that $\xi_{k}\in\mathbb{D}$ is a critical point of $B$ if and only if $\zeta_{k}:=T(\xi_{k})\in\mathbb{H}$ is a critical point of $f$. Since Möbius maps preserve symmetries in circles and lines, also the points $\overline{\zeta}_{k}$ are critical points of $f$ corresponding to the critical points $1/\overline{\xi}_{k}$ of $B$. The derivative of $f$ has the form

[TABLE]

with a monic real polynomial $P$ of degree $2n-2$, $c:=r_{1}+r_{2}+\dots+r_{n}$ and

[TABLE]

We conclude that $P$ has the factorization

[TABLE]

where $a_{k}:=\mathrm{Re}\,\zeta_{k}$, $b_{k}:=\mathrm{Im}\,\zeta_{k}$. The polynomial $P$ is entirely determined by the location of the critical points and (3) shows that it satisfies the equation

[TABLE]

Evaluating (6) at $x=x_{k}$ we get

[TABLE]

Inserting (7) into (6), and introducing the Lagrange interpolation polynomials

[TABLE]

which satisfy $Q_{k}(x_{j})=\delta_{kj}$, we can rewrite equation (6) as

[TABLE]

Since $P$ has degree $2n-2$, the Lagrange-Hermite interpolation formula (cf. Chapter 14.1 of [38]) implies that for any pairwise distinct points $x_{1},x_{2},\dots,x_{n}$,

[TABLE]

where for $k=1,2,\dots,n$,

[TABLE]

are the fundamental polynomials of the first and second kind of Hermite interpolation. From formulas (10), (11) we get

[TABLE]

The polynomials $(x-x_{k})Q_{k}(x)^{2}$ are linearly independent, and comparing (9) and (12) we see that the representation (9) holds if and only if

[TABLE]

Since the $x_{1},x_{2},\dots,x_{n}$ are exactly the (simple) roots of $Q$, this condition means that the polynomial $Q^{\prime}P^{\prime}-Q^{\prime\prime}P$ is divisible by $Q$, i.e., there is a real polynomial $R$ such that

[TABLE]

Because $Q^{\prime}P^{\prime}-Q^{\prime\prime}P$ has exact degree $3n-4$ we see that $R$ has degree $2n-4$. One easily checks that these considerations can be reversed to obtain the following equivalence statement.

Lemma 3**.**

Let $n\geq 2$, $x_{1}<x_{2}<\dots<x_{n}$, $r_{1},r_{2},\dots,r_{n}>0$, $\zeta_{1},\zeta_{2},\dots,\zeta_{n-1}\in\mathbb{H}$, and let $f$, $Q$, $P$ be defined by (1), (4), (5), respectively. Then the rational function $f$ has critical points $\zeta_{1},\zeta_{2},\dots,\zeta_{n-1}$ if and only if there is a real polynomial $R$ of (exact) degree $2n-4$ such that

[TABLE]

and the $r_{k}$ are given by (7) for some $c>0$.

The corresponding constructions for the function $g$ in (2) are similar: Its derivative is

[TABLE]

where $P$ is defined in (5) as before, and

[TABLE]

Now we obtain the equation

[TABLE]

in particular

[TABLE]

With the help of the Lagrange interpolation polynomials with nodes $t_{1},\dots,t_{n-1}$, given by

[TABLE]

equation (16) can be rewritten as

[TABLE]

Since $P$ and $S$ are monic, $P-S^{2}$ is at most of degree $2n-3$, and hence the Lagrange-Hermite interpolation formula applied to this polynomial tells us that

[TABLE]

where, for $k=1,2,\dots,n-1$,

[TABLE]

A straightforward computation yields

[TABLE]

and comparing this with (19) we get

[TABLE]

which is equivalent to the divisibility of $P^{\prime}S^{\prime}-PS^{\prime\prime}$ by $S$ (see (15)). The quotient $\tilde{R}$ is a polynomial of exact degree $2n-4$. We summarize these results in the next lemma.

Lemma 4**.**

Let $n\geq 2$, $a>0$, $b\in\mathbb{R}$, $t_{1}<t_{2}<\dots<t_{n-1}$, $s_{1},s_{2},\dots,s_{n-1}>0$, $\zeta_{1},\zeta_{2},\dots,\zeta_{n-1}\in\mathbb{H}$, and let $g$, $P$ and $S$ be given by (2), (5), and (15), respectively. Then the rational function $g$ has critical points $\zeta_{1},\zeta_{2},\dots,\zeta_{n-1}$ if and only if there is a real polynomial $\tilde{R}$ of degree $2n-4$ such that

[TABLE]

and the $s_{k}$ are given by (17).

In this way the determination of a Blaschke product with given critical points is reduced to the question for which real polynomials $R$ (resp. $\tilde{R}$) the second order ODE (13) (resp. (20)) has a polynomial solution $Q$ of degree $n$ (resp. $S$ of degree $n-1$) with simple real roots. This is a classical problem which will be considered in the next section.

3 Stieltjes and Van Vleck polynomials

Let $A$ and $B$ be given polynomials of degrees $p+1$ and $p$, respectively. A polynomial $C$ of degree $p-1$ is called Van Vleck polynomial, if the generalized Lamé equation

[TABLE]

has a polynomial solution $Q$ of preassigned degree $n$. The solutions $Q$ are called Stieltjes polynomials or Heine-Stieltjes polynomials. Under certain condition given below, Stieltjes [37] proved the existence of the polynomials that carry now his name (see also [38, Section 6.8]). Stieltjes assumed that

[TABLE]

has real roots $a_{0}<a_{1}<\dots<a_{p}$ and that the coefficients $\varrho_{k}$ in the partial fraction decomposition

[TABLE]

are all positive. Now consider partitions $n=n_{1}+n_{2}+\dots+n_{p}$ of $n$ into $p$ non-negative integer summands $n_{1},n_{2},\dots,n_{p}$. There are $\binom{n+p-1}{n}$ such partitions, and each partition corresponds to exactly one monic polynomial $Q$ with $n$ distinct real roots such that (21) holds for a suitable Van Vleck polynomial $C$. This polynomial $Q$ has exactly $n_{k}$ roots in $(a_{k-1},a_{k})$ for $k=1,2,\dots,p$ and can be characterized as follows: Put positive charges $\varrho_{k}$ at the fixed positions $a_{k}$ and movable unit charges on the real line such that each interval $(a_{k-1},a_{k})$ contains exactly $n_{k}$ of them. As usual in plane electrostatics, the force (repulsion or attraction) between two charges is assumed to be proportional to their magnitudes and to the inverse of their distance. Stieltjes proved that there is a unique equilibrium position of these $n$ positive movable unit charges; it is attained when the charges are located at the zeros of $Q$ and corresponds to the global minimum of the potential energy of each possible charge configuration.

In our equations (13) and (20) the polynomial $P$ does not have real zeros so that Stieltjes’ results are not directly applicable. The necessary adaptations to the situation at hand will be done in this section.

Assume first that $R$ is a Van Vleck polynomial for the equation (13) and that $Q$ has the form (4). The identity

[TABLE]

is well-known (see for instance formula (2.9) in [17]). Recall that the logarithmic derivative of $P$ has the representation

[TABLE]

Dividing (13) by $PQ^{\prime}$ and evaluating at $x_{k}$ using (22) and $Q(x_{k})=0$ we get

[TABLE]

Conversely, if the equations (23) are satisfied, the polynomial $P^{\prime}Q^{\prime}-PQ^{\prime\prime}$ has the zeros $x_{1},\dots,x_{n}$, so that $R:=(P^{\prime}Q^{\prime}-PQ^{\prime\prime})/Q$ is a polynomial which satisfies (13). Summarizing we get the following result:

Lemma 5**.**

Let $\zeta_{1},\dots,\zeta_{n-1}\in\mathbb{H}$ be given and define $P$ by (5). Then the polynomial $Q$ from (4) with $x_{1}<\dots<x_{n}$ is a Stieltjes polynomial for the equation (13) if and only if (23) holds. Similarly, the polynomial $S$ from (15) with $t_{1}<\dots<t_{n-1}$ is a Stieltjes polynomial for the equation (20) if and only if

[TABLE]

Equation (23) has the following electrostatic interpretation: Fix negative charges of size $-1/2$ at each of the $2n-2$ points $\zeta_{k}$ and $\overline{\zeta}_{k}$. Then $n$ (moveable) positive unit charges at the positions $x_{1},\dots,x_{n}$ on the real axis are in equilibrium in the field generated by all charges. Analogously, (24) describes the equilibrium positions of $n-1$ positive unit charges at $t_{1},\dots,t_{n-1}$ on the real axis in the presence of the same $2n-2$ negative charges at the points $\zeta_{k},\overline{\zeta}_{k}$.

The equilibrium of moveable positive unit charges in the presence of fixed negative charges in $\mathbb{C}\setminus\mathbb{R}$ was studied by Orive and García [29]. Their results are applicable to (24), and we will include a proof only to make our exposition self-contained (see Lemma 6 below). However, the equilibrium problem described by (23) is not covered by their results, since the number of movable unit charges surpasses the total of negative charges by $1$.

Other equilibrium problems involving fixed charges have recently been considered by Grünbaum [10], [11], Dimitrov and Van Assche [5], [6], and Grinshpan [9]; see Marcellán, Martínez-Finkelshtein, and Martínez-González [22] for an overview until 2007. More recent literature includes Martínez-Finkelshtein and Rakhmanov [23], McMillen, Bourget, and Agnew [25], Orive and Sánchez-Lara [30], [31], and Shapiro [35].

Example 1. Let $n=2$, i.e., there is only one critical point $\zeta:=\zeta_{1}$ given in $\mathbb{H}$. Then (24) with $t:=t_{1}$ simplifies to

[TABLE]

The only solution of this equation is $t=\operatorname{Re}\zeta$, the expected equilibrium position of one unit charge. The equations (23) read as follows

[TABLE]

From the first equation in (25) we get

[TABLE]

and after some elementary manipulations we arrive at

[TABLE]

Since this equation is invariant with respect to interchanging $x_{1}$ and $x_{2}$, the second equation of (25) yields the same conditon. Equation (26) has a simple geometric meaning; it says that, by Thales’ theorem, the points $\zeta,\overline{\zeta},x_{1}$ and $x_{2}$ lie on a circle.

\operatorname{Re}$$\operatorname{Im}$$t$$\zeta$$\overline{\zeta}$$x_{1}$$x_{2}$$t

The equilibrium positions $x_{1},x_{2}$ are therefore not uniquely determined: for each $x_{1}\in(-\infty,t)$ there is a corresponding $x_{2}\in(t,\infty)$, and vice versa. As we will see later (Theorem 2), the situation is similar for $n>2$.

As pointed out in [19], the uniqueness statement in Theorem 1 follows from Nehari’s generalization of Schwarz’ Lemma (see [26, corollary to Theorem 1]). While all proofs of the existence part of Theorem 1 in the literature are quite hard, the electrostatic interpretation allows us to provide a simple and transparent proof of existence and uniqueness.

The main argument for proving uniqueness is originally due to Sarason and Suarez [33] and has also been used in [9] and [29]. The formulation of [29, Theorem 2] can easily be extended to cover our situation here. In our exposition the proof is naturally based on the fundamental relation (19).

We start with the problem of $n-1$ movable unit charges at positions $t_{1},\dots,t_{n-1}$ on the real line and introduce their energy, which is (neglecting some physically motivated factor)

[TABLE]

Since $W$ does not change upon permutations of its variables, we can confine our considerations to the open subset $U$ of $\mathbb{R}^{n-1}$ where $t_{1}<\dots<t_{n-1}$.

Lemma 6**.**

The function $W$ attains a global minimum at a point $(t_{1},\dots,t_{n-1})$ in $U$. This point is the only critical point of $W$ in $U$ and corresponds to the unique solution of (24) with $t_{1}<t_{2}<\dots<t_{n-1}$.

Proof.

1. A point $T:=(t_{1},\dots,t_{n-1})$ is a critical point of $W$ if and only if $({\partial W}/{\partial t_{k}})(T)=0$ for $k=1,\dots,n-1$. A straightforward computation shows that this is equivalent to (24).

2. In order to prove that $W$ attains a global minimum in $U$, we observe that the first sum in (27) is bounded from below, while the second sum tends to $+\infty$ whenever $T$ approaches a finite boundary point of $U$ (which implies that $t_{j}-t_{j+1}\to 0$ for some $j$). So it only remains to study the behavior of $W(T)$ when $\|T\|\to\infty$ in some norm of $\mathbb{R}^{n-1}$. Let $C>0$ be a constant (depending on $\zeta_{1},\dots,\zeta_{n-1}$) such that

[TABLE]

Using the inequality

[TABLE]

we estimate

[TABLE]

and hence $W(T)\to\infty$ as $\|T\|_{1}\to\infty$. We conclude that $W$ attains a global minimum at a point $T\in{\cal O}$ which represents a solution of (24).

3. In order to show uniqueness (of the solution of (24) and hence the critical point of $W$ in $U$), we assume that $t_{1}<t_{2}<\dots<t_{n-1}$ and $\tilde{t}_{1}<\tilde{t}_{2}<\dots<\tilde{t}_{n-1}$ are two solutions of (24). For $k=1,\dots,n-1$ let

[TABLE]

By Lemma 5, both $S$ and $\tilde{S}$ are Stieltjes polynomials, i.e. they satify (20) for suitable Van Vleck polynomials. Going back from this equation in the calculations in Section 2 we infer they also satisfy equation (19), and hence we have

[TABLE]

The $2n-2$ polynomials $S_{1}^{2},\dots,S_{n-1}^{2},\tilde{S}_{1}^{2},\dots,\tilde{S}_{n-1}^{2}$ are of degree $2n-4$, hence they must be linearly dependent, i.e., there are real numbers $\alpha_{k},\beta_{k}$, not all vanishing, such that

[TABLE]

Multiplying this by $t\in\mathbb{R}$ and adding it to (28) we get

[TABLE]

for all $t\in\mathbb{R}$. Since $P(t_{k})>0,P(\tilde{t}_{k})>0$ we can choose $t$ such that (at least) one of the numbers $P(t_{k})+t\alpha_{k}$, $P(\tilde{t}_{k})+t\beta_{k}$ with $k\in\{1,\dots n-1\}$ vanishes while all the others remain non-negative. Assume e.g. $P(t_{k_{0}})+t\alpha_{k_{0}}=0$ for some $k_{0}\in\{1,\dots,n-1\}$. Since S\big{(}t_{k_{0}}\big{)}=0 and S_{k}\big{(}t_{k_{0}}\big{)}=0 for $k\not=k_{0}$, the left hand side of (29) vanishes at $x=t_{k_{0}}$. But the right hand side of this equation cannot have other zeros than $\tilde{S}^{2}$, hence $t_{k_{0}}=\tilde{t}_{j_{0}}$ for some $j_{0}\in\{1,\dots,n-1\}$.

If $n=2$ we are done. Otherwise the $2n-4$ polynomials of degree $2n-6$

[TABLE]

are also linearly dependent, so that we have a non-trivial relation of the form

[TABLE]

With (28) we have for $t\in\mathbb{R}$

[TABLE]

As above we find now another pair of equal zeros of $S$ and $\tilde{S}$. Proceeding inductively we finally get that both solutions of (24) are identical. ∎∎

The solution of (23) can now be reduced to (24) using an appropriate Möbius transformation. We summarize the results for both equations in the following theorem.

Theorem 2**.**

For arbitrary $\zeta_{1},\dots,\zeta_{n-1}\in\mathbb{H}$ we have:

1. (i)

The equation (24) has a unique solution with $t_{1}<t_{2}<\dots<t_{n-1}$. It realizes the global minimum of the potential energy $W$ defined in (27), and is (up to a rearrangement of the $t_{k}$) the only critical point of $W$. 2. (ii)

Let additionally $t_{0}:=-\infty$, $t_{n}:=+\infty$ and fix $k_{0}\in\{1,2,\dots,n\}$. For every $x_{k_{0}}\in(t_{k_{0}-1},t_{k_{0}})$ there are unique points $x_{k}\in(t_{k-1},t_{k})$ for $k=1,\dots,k_{0}-1,k_{0}+1,\dots,n$ such that equation (23) is satisfied. All $x_{k}$ depend continuously and monotonously on $x_{k_{0}}$, and we have $x_{k}\to t_{k-1}$ if $x_{k_{0}}\to t_{k_{0}-1}$ and $x_{k}\to t_{k}$ if $x_{k_{0}}\to t_{k_{0}}$.

Proof.

Assertion (i) has been proved in Lemma 6. In order to show the existence part of (ii) let $a>0$, $b\in\mathbb{R}$ be arbitrary and define $s_{1},\dots,s_{n-1}$ by (17). According to Lemma 4 and Lemma 5, the function $g$ from (2) has critical points $\zeta_{1},\dots,\zeta_{n-1}$. Since $g(x)\to\pm\infty$ as $x\to t_{k}\mp 0$ ($k=0,\dots,n)$ and $g$ is strictly increasing in each interval $(t_{k-1},t_{k})$ ($k=1,\dots,n)$, any such interval contains a unique solution $x_{k}$ of $g(x_{k})=g(x_{k_{0}})$. It is clear that these points have the claimed properties concerning continuity, monotonicity and limit behavior. The function $f$ defined by

[TABLE]

is rational and can be represented as a quotient $f=p/q$, where $p$ is a real polynomial of degree $n-1$, and $q$ is a real polynomial of degree $n$, respectively. Since $f$ has simple poles in $x_{1},\dots,x_{n}$, the partial fraction decomposition of $f$ has the form (1). From the monotonicity of $g$ we obtain that $f$ is increasing between two poles and thus $r_{k}>0$. Because $f$ and $g$ have the same critical points $\zeta_{1},\dots,\zeta_{n-1}$, we can combine Lemma 3 and Lemma 5 to obtain (23).

In order to show the uniqueness assertion of (ii) we let $\tilde{x}_{1}<\dots<\tilde{x}_{n}$ be any solution of (23) with $\tilde{x}_{k_{0}}=x_{k_{0}}$ and define $\tilde{r}_{k}>0$ by (7) for some $c>0$ and $x_{k}$ replaced by $\tilde{x}_{k}$. Let the function $\tilde{f}$ be of the form (1) with poles at $\tilde{x}_{k}$ and residues $\tilde{r}_{k}$. According to Lemma 3 and Lemma 5, $\tilde{f}$ has critical points $\zeta_{1},\dots,\zeta_{n-1}$. Since $\tilde{f}$ is real-valued and increasing between any two of its poles, it has zeros $\tilde{t}_{k}\in(\tilde{x}_{k},\tilde{x}_{k+1})$ for $k=1,\dots,n-1$. Hence $\tilde{g}:=-1/\tilde{f}$ has the form

[TABLE]

The critical points of $\tilde{g}$ are the same as those of $\tilde{f}$, hence we can again invoke Lemma 4 and Lemma 5 to conclude that $\tilde{t}_{1},\dots,\tilde{t}_{n-1}$ solve (24). From the uniqueness stated in (i) it follows that $\tilde{t}_{k}=t_{k}$ for $k=1,\dots,n-1$. Consequently, by (14) and (15),

[TABLE]

Hence $g^{\prime}=d\tilde{g}^{\prime}$ for some $d>0$, and therefore $g=d\tilde{g}+e$ with $e\in\mathbb{R}$. Using (30) we obtain

[TABLE]

and since $f$ and $\tilde{f}$ both have poles at $x_{k_{0}}$ it follows that $e=g(x_{k_{0}})$. Thus $\tilde{f}=df$, in particular $\tilde{x}_{k}=x_{k}$ for $k=1,\dots,n$. ∎∎

The proof of Theorem 2 also indicates how to construct rational functions $f$ and $g$ of the form (1) or (2) with given critical points $\zeta_{k}$ from the solutions of (23) and (24), respectively. Setting $B:=T^{-1}\circ f\circ T$ we obtain by Lemma 1 a Blaschke product with critical points $\xi_{k}=T^{-1}(\zeta_{k})$, and the same is true for $B:=T^{-1}\circ g\circ T$ by Lemma 2. This shows again the existence of a Blaschke product of degree $n$ with $n-1$ given critical points in $\mathbb{D}$. In the same way the uniqueness statement of Theorem 1 can be inferred from the uniqueness statements in Theorem 1 and we have thus provided an independent proof of this result.

Example 2. Consider the case where $\zeta_{1}=\dots=\zeta_{n-1}=\mathrm{i}$. Since $T^{-1}(\mathrm{i})=0$, we have $\xi_{1}=\dots=\xi_{n-1}=0\in\mathbb{D}$, i.e., this problem is equivalent to finding a Blaschke product $B$ of degree $n$ with critical point of order $n-1$ at $z=0$. An obvious solution with $B(1)=1$ is $B(z)=z^{n}$, and the solutions of $B(z)=1$ are the $n$th roots of unity $\eta_{k}=\mathrm{e}^{2\pi\mathrm{i}\,k/n}$ ($k=0,1,\dots,n-1$). The unique solution of (24) is then given by the images of $\eta_{1},\dots,\eta_{n-1}$ under the mapping $T$,

[TABLE]

The case where all $\zeta_{1},\dots,\zeta_{n-1}$ coincide at another point of $\mathbb{H}$ can be reduced to this cases by an appropriate automorphism $z\mapsto cz+d$ with $c>0,d\in\mathbb{R}$.

By now we know that the Stieltjes polynomials (4) (corresponding to the one-parameter family of solutions of (23)) and (15) (corresponding to the solution of (24)) are associated with certain Van Vleck polynomials satisfying (13) and (20), respectively. It turns out that all these polynomials are the same, so that we can speak about the Van Vleck polynomial for the critical points $\zeta_{k}$.

Lemma 7**.**

Let $\zeta_{1},\dots,\zeta_{n-1}\in\mathbb{H}$ and let $t_{1},\dots,t_{n-1}$ and $x_{1},\dots,x_{n}$ be the solutions described in Theorem 2. Then the Van Vleck polynomials $R$ and $\tilde{R}$, corresponding to solutions (4) and (15) of (13) and (20), respectively, coincide.

Proof.

Let $f$ be defined by (1) and (7) with $c>0$ and let $g$ be given by (2) and (17) with $a>0$ and $b\in\mathbb{R}$. As we have seen in the proof of Theorem 2, $f$ and $g$ are connected by the equation

[TABLE]

with constants $d>0$ and $e\in\mathbb{R}$ (i.e. they differ by an automorphism of $\mathbb{H}$ that maps [math] to $\infty$). Differentiating this equation we obtain

[TABLE]

and plugging in (3) and (14) we arrive at

[TABLE]

so that $a\,Q(x)^{2}f(x)^{2}=cd\,S(x)^{2}$. Since $Q(x)\sim x^{n}$, $S(x)\sim x^{n-1}$, $f(x)\sim-c\,x^{-1}$ as $x\to\infty$, we have shown the remarkable identity

[TABLE]

Differentiating (31) we get

[TABLE]

Recall from (3) that $f^{\prime}=c\,P/Q^{2}$, and thus

[TABLE]

Using (in this order) (20), (32), (13), (33), and (31), we get

[TABLE]

so that $R=\tilde{R}$. ∎∎

As a consequence of this lemma we obtain that for the Van Vleck polynomial $R$ the solution space of the Lamé equation

[TABLE]

consists only of polynomials, and that its general solution is given by

[TABLE]

Fixing $\lambda\neq 0$ and letting $\mu$ run through $\mathbb{R}$, the zeros of the polynomials (35) run through all equilibrium positions of $n$ points. Solutions with $\lambda=0$ correspond to the limit configuration where one charge escaped to infinity and only $n-1$ charges remain on the real line. The preceding result also confirms what was said in [29, Remark 3] about possible polynomial solutions of the Lamé equation.

Once two Stieltjes polynomials $Q$ and $S$ as solutions of (34) are known, the polynomial $P$ and the Van Vleck polynomial $R$ can be reconstructed as shows the following result.

Lemma 8**.**

Let $Q$ and $S$ be monic polynomial solutions of degree $n$ and $n-1$, respectively, of the Lamé equation (34). Then we have

[TABLE]

and

[TABLE]

Proof.

Since $Q$ and $S$ are linearly independent they form a fundamental system for the differential equation. Their Wronskian

[TABLE]

is therefore non-vanishing and Abel’s formula applied to (13) implies that for $x,x_{0}\in\mathbb{R}$

[TABLE]

Since $w$ and $P$ are easily seen to be monic, the two polynomials coincide and we have proved (36). Differentiating this equation we get

[TABLE]

and using (34), (36), we finally arrive at

[TABLE]

and (37) follows. ∎∎

4 Convex cones and moment problems

In this section we will show that a well-known problem in moment theory is also equivalent to the determination of a Blaschke product with given critical points. We start with some notions and facts from convex analysis, our main sources are [1], [2], [4], [16], [18], [21].

A convex cone $\mathcal{C}$ is a subset of a real vector space, such that $u,v\in\mathcal{C}$ implies $\lambda u+\mu v\in\mathcal{C}$ for all positive $\lambda$ and $\mu$. One standard example is the convex cone of real non-negative polynomials of degree at most $2n-2$,

[TABLE]

where we identified a polynomial with the vector of its coefficients. The interior of this convex cone consists of all positive polynomials; hence the polynomial $P$ from (5) belongs to the interior of ${\cal P}_{2n-1}$ provided that $\zeta_{k}\in\mathbb{H}$ for $k=1,\dots,n-1$. Recall that $P$ encodes the given data, i.e., the critical points of the Blaschke product.

Another important example is the convex cone of symmetric positive semi-definite matrices,

[TABLE]

The interior of $\mathcal{S}_{+}^{n}$ is the convex cone of symmetric positive definite matrices,

[TABLE]

Furthermore, we consider the moment cone

[TABLE]

where $M_{2n-1}$ is the set of nonnegative measures $\sigma$ on $\mathbb{R}$ such that

[TABLE]

The set ${\cal M}_{2n-1}$ is the conic hull of the moment curve

[TABLE]

i.e., ${\cal M}_{2n-1}$ is the smallest convex cone containing $C_{2n-1}$, cf. Theorem 2.1 in chapter V of [18]. Note that ${\cal M}_{2n-1}$ is not a closed subset of $\mathbb{R}^{2n-1}$ since points in the closure of this set can involve representations with “mass at infinity”. More precisely, $c$ belongs to the closure $\overline{\cal M}_{2n-1}$ of ${\cal M}_{2n-1}$ if and only if it has the representation

[TABLE]

with $\sigma\in M_{2n-1}$ and non-negative $\lambda$ (representing a mass at infinity). An alternative characterization of $\overline{\cal M}_{2n-1}$ is

[TABLE]

where

[TABLE]

denotes the Hankel matrix associated with $c$. Also, a point $c\in\mathbb{R}^{2n-1}$ is an inner point of ${\cal M}_{2n-1}$ if and only if $H(c)$ is positive definite. Using the linear mapping

[TABLE]

we can therefore write

[TABLE]

A representation (39) of a point $c$ in (the closure of) the moment cone is usually not unique. Therefore one searches for canonical representations of $c$ where the measure $\sigma\in M_{2n-1}$ is concentrated at a finite number of points. For example, if $\sigma$ is concentrated at points $t_{1},t_{2},\dots,t_{n-1}$ with masses $\sigma_{1},\dots,\sigma_{n-1}>0$ and mass $\lambda>0$ at infinity we have

[TABLE]

If $\sigma$ is concentrated at points $x_{1},x_{2},\dots,x_{n}$ with masses $\varrho_{1},\dots,\varrho_{n}>0$ and has no mass at infinity then we have

[TABLE]

For given moments $c_{k}$, the moment problem consists in finding the roots $t_{k},x_{k}$ and the corresponding weights $\sigma_{k}$, $\varrho_{k}$, featuring in the representations (42), (43), respectively. The following theorem is well known (see e.g. [1], [18]).

Theorem 3**.**

For any $c\in\operatorname{int}{\cal M}_{2n-1}$ the following assertions hold:

1. (i)

There is a representation (42) with uniquely determined roots $t_{1}<t_{2}<\dots<t_{n-1}$, weights $\sigma_{1},\dots,\sigma_{n-1}>0$ and $\lambda>0$. 2. (ii)

Let additionally $t_{0}:=-\infty$, $t_{n}:=+\infty$, and fix $k_{0}\in\{1,2,\dots,n\}$. Then for every $x_{k_{0}}\in(t_{k_{0}-1},t_{k_{0}})$ there are unique roots $x_{k}\in(t_{k-1},t_{k})$ for $k=1,\dots,k_{0}-1,k_{0}+1,\dots,n$ and weights $\varrho_{1},\dots,\varrho_{n}>0$ such that equation (43) is satisfied. All $x_{k}$ depend continuously and monotonically on $x_{k_{0}}$.

Before going on, we observe that the moment problem (43) can be rephrased as a matrix factorization problem. Let

[TABLE]

denote the Vandermonde matrix of the points $x_{1},\dots,x_{n}$. With the abbreviations $V:=V(x_{1},\dots,x_{n})$, $D:=\operatorname{diag}(\varrho_{1},\dots,\varrho_{n})$, and $H(c)$ from (40), the equations (43) can be rewritten as

[TABLE]

This representation is known as the Vandermonde factorization of the (positive definite) Hankel matrix $H(c)$ (see Heinig and Rost [13], [14]).

There is also a nice way to obtain the roots $t_{k},x_{k}$. The polynomial $D_{n-1}$ defined by

[TABLE]

is of (exact) degree $n-1$ and its roots are $t_{1},\dots,t_{n-1}$. To see this we observe that in view of $\det H(c_{0},\dots,c_{2n-4})>0$ the first $n-1$ columns of the matrix in (45) are linearly independent and in view of (42) each of them is a linear combination of the first $n-1$ columns of $V(t_{1},\dots,t_{n-1},x)$. Hence the spaces spanned by the first $n-1$ columns of these two matrices coincide, such that the two determinants $D_{n-1}(x)$ and $\mathrm{det}\,V(t_{1},\dots,t_{n-1},x)$ vanish for the same values of $x$.

Similarly, in view of $D_{n-1}(x_{k_{0}})\neq 0$ the first $n$ columns of the matrix in

[TABLE]

are linearly independent and by (43) we find that the zeros of $E_{n}(x)$ coincide with the zeros of $\det V(x_{1},\dots,x_{n},x)$. Hence the zeros of $E_{n}(x)$ are the roots described in Theorem 3 (ii).

There is an evident similarity between Theorem 2 and Theorem 3. As we will see in a moment, the solutions described in both theorems coincide if we map the polynomial $P\in\operatorname{int}{\cal P}_{2n-1}$ from (5) to an appropriate point $c\in\operatorname{int}{\cal M}_{2n-1}$. This mapping was investigated by Nesterov [27] and will be described below. Hachez and Nesterov [12] used it together with the Vandermonde factorization (44) to represent a positive polynomial as a weighted sum of squares of Lagrange interpolation polynomials as in formula (9). This yields another equivalent reformulation of the original Blaschke product problem.

In order to describe a mapping between $\operatorname{int}{\cal P}_{2n-1}$ and $\operatorname{int}{\cal M}_{2n-1}$ we have to recall the notion of duality. If $K$ is a convex cone in a real Hilbert space $V$ with scalar product $\langle\cdot,\cdot\rangle$, its dual cone is $K^{*}:=\{x\in V:\langle x,y\rangle\geq 0\text{ for all }y\in K\}$. In the sequel we use scalar products for vectors $x=(x_{k}),y=(y_{k})\in\mathbb{R}^{p}$ and matrices $X=(x_{kj}),Y=(y_{kj})\in\mathbb{R}^{p\times q}$ defined by

[TABLE]

With respect to these scalar products we have the following duality relations:

[TABLE]

Let

[TABLE]

be the mapping which sends a matrix $X$ to the vector $y$ of its anti-diagonal sums. The mapping $H^{*}$ is dual to the mapping $H$ defined in (41) in the sense that

[TABLE]

Moreover, the cone of non-negative (positive) polynomials can be obtained as the image of the cone of non-negative (positive) definite matrices,

[TABLE]

The mapping $H^{*}$ has an interesting interpretation when we identify a vector $c\in\mathbb{R}^{2n-1}$ with the polynomial $c(x)=\langle c,u(x)\rangle=c_{0}+c_{1}x+\dots+c_{2n-2}x^{2n-2}$, and a matrix $X\in\mathbb{R}^{n\times n}$ with the polynomial in two variables

[TABLE]

where $u$ is the function defined in (38). Then $H^{*}$ is the operator of equating variables that maps $X=X(x,y)$ to the polynomial

[TABLE]

Nesterov [27] introduced the mapping

[TABLE]

which is defined for $c\in\mathbb{R}^{2n-1}$ whenever $H(c)$ is nonsingular. If $H(c)$ is even positive definite, the inverse matrix $H(c)^{-1}$ is also positive definite and we have for $x\in\mathbb{R}$

[TABLE]

i.e., $N(c)$ is (the coefficient vector of) a positive polynomial. Thus the Nesterov mapping $N$ maps $\operatorname{int}{\cal M}_{2n-1}$ to $\operatorname{int}{\cal P}_{2n-1}$. Nesterov [27] also remarked that $N$ is the (negative) gradient of the function

[TABLE]

which is a (strongly non-degenerate self-concordant) barrier functional for ${\cal M}_{2n-1}$, see [28] or [32] for definitions and basic properties. Hence the gradient of $h$ is a bijection of $\operatorname{int}{\cal M}_{2n-1}$ onto the interior $\operatorname{int}{\cal P}_{2n-1}$ of its dual cone ${\cal P}_{2n-1}$. We will give an alternative proof of this fact, demonstrating how the Nesterov mapping $N$ connects Theorem 2 and Theorem 3.

Theorem 4**.**

Let $p\in\operatorname{int}{\cal P}_{2n-1}$ be a positive polynomial with zeros $\zeta_{1},\dots,\zeta_{n-1}$ in the upper half plane. Then there is a unique $c\in\operatorname{int}{\cal M}_{2n-1}$ such that $p=N(c)$. If $x_{1},\dots,x_{n}$ and $t_{1},\dots,t_{n-1}$ satisfy (23) and (24), respectively, then the canonical representations (42) and (43) hold with the positive numbers

[TABLE]

where $p_{2n-2}$ is the leading coefficient of $p$. Conversely, if (42) and (43) are satisfied for $x_{1},\dots,x_{n}$, $t_{1},\dots,t_{n-1}$ and positive numbers $\sigma_{k},\varrho_{k},\lambda$, then also the equilibrium conditions (23) and (24) are true. Moreover, $\sigma_{k}$, $\varrho_{k}$, and $\lambda$ satisfy (47).

The proof of Theorem 4 is split into several lemmas. Without loss of generality we always assume that $x_{1}<\dots<x_{n}$ and $t_{1}<\dots<t_{n-1}$.

In the following we denote by $\operatorname{Bez}(v,w)$ the Bezoutian of $v,w\in\mathbb{R}^{n+1}$ (interpretable as polynomial of degree $\leq n$), which is the matrix $B=(b_{ij})\in\mathbb{R}^{n\times n}$ defined by

[TABLE]

see [7], [13], [14] for more information on this topic. Recall that the inverse of a non-singular Hankel matrix is a non-singular Bezoutian, and vice versa.

Lemma 9**.**

For each $p\in\operatorname{int}{\cal P}_{2n-1}$ there exists a vector $c\in\mathbb{R}^{2n-1}$ with $N(c)=p$.

Proof.

Let $p(x)=p_{2n-2}x^{2n-2}+\dots+p_{0}$ be a positive polynomial with zeros $\zeta_{k}\in\mathbb{H}$ and $\overline{\zeta}_{k}$ for $k=1,\dots,n-1$. Then $p(x)=p_{2n-2}P(x)$ with $p_{2n-2}>0$ and the monic polynomial $P$ from (5). Let $Q$ and $S$ be the polynomials (4) and (15) with zeros at the equilibrium points $x_{1}<\dots<x_{n}$ and $t_{1}<\dots<t_{n-1}$ according to Theorem 2. Let further $B=(b_{ij})=\operatorname{Bez}(Q,S)$ be the Bezoutian of $Q$ and $S$, i.e.,

[TABLE]

The application of $H^{*}$ yields in view of (46)

[TABLE]

We have shown in equation (36) of Lemma 8 that $P$ is the Wronskian of $S$ and $Q$, from which we conclude that $H^{*}B=P$. Since $B$ is a non-singular Bezoutian, its inverse is a Hankel matrix, hence there is $d\in\mathbb{R}^{2n-1}$ with $H(d)=B^{-1}$. This vector $d$ satisfies $N(d)=H^{*}\left(H(d)^{-1}\right)=H^{*}(B)=P$. Defining $c:=p_{2n-2}^{-1}d$ we get $N(c)=p_{2n-2}N(d)=p_{2n-2}P=p$. ∎∎

We will soon see that even $c\in\operatorname{int}{\cal M}_{2n-1}$. Before that we show:

Lemma 10**.**

If $x_{1},\dots,x_{n}$ satisfy (23) and $\varrho_{1},\dots\varrho_{n}$ are given by (47), then the canonical representation (43) holds.

Proof.

We show that for $k=0,\dots,2n-2$

[TABLE]

is equal to $c_{k}$. In view of (36) and $Q(x_{k})=0$ we have $Q^{\prime}(x_{k})S(x_{k})=P(x_{k})$ and hence

[TABLE]

Putting $y=x_{k}$ in (48) we find

[TABLE]

where $Q_{k}$ are the Lagrange interpolation polynomials defined in (8). Letting $x=x_{l}$ we obtain

[TABLE]

i.e., the matrix $Y$ with entries

[TABLE]

is a right inverse of the transposed Vandermonde matrix $V=V(x_{1},\dots,x_{n})^{\top}$. Since $V$ is a square matrix, $Y$ is also left inverse to $V$, so that for $k,l=0,\dots,n-1$

[TABLE]

So the Hankel matrix $H(e)$ is inverse to the Bezoutian $p_{2n-2}B=H(c)^{-1}$ and thus $e_{k}=c_{k}$ for $k=0,\dots,2n-2$. ∎∎

As a consequence of this lemma and (44) we get that $H(c)$ is similar to a diagonal matrix with positive diagonal elements and therefore positive definite. We have thus proved that $c\in\operatorname{int}{\cal M}_{2n-1}$ and hence the surjectivity of the Nesterov mapping $N:\operatorname{int}{\cal M}_{2n-1}\to\operatorname{int}{\cal P}_{2n-1}$.

Lemma 11**.**

If $t_{1},\dots,t_{n-1}$ satisfy (24) and $\lambda,\sigma_{1},\dots,\sigma_{n-1}$ are given by (47) (where $p_{2n-2}$ is the leading coefficient of $p$), then the canonical representation (42) holds.

Proof.

We set

[TABLE]

and show that $f_{k}=c_{k}$ for all $k$. Note that in view of (36) and $S(t_{k})=0$

[TABLE]

Setting $x=t_{k}$ in (48) we get

[TABLE]

where $S_{k}$ are the Lagrange interpolation polynomials defined in (18). Putting $y=t_{l}$ and using (50) we obtain

[TABLE]

which can be rewritten as

[TABLE]

where

[TABLE]

Since the $S_{k}$ are polynomials of degree $n-2$, comparing coefficients of $y^{n-1}$ in (51) yields that

[TABLE]

Defining $y_{in}:=p_{2n-2}b_{i,n-1}$ for $i=0,\dots,n-1$ this can be written as

[TABLE]

We also have (see (53) and (54))

[TABLE]

Since $b_{n-1,n-1}$ is the leading coefficient of the monic polynomial $H^{*}(B)=P$ we get

[TABLE]

This, together with the equations (52), (55), and (56), implies that the matrix $Y$ with entries $y_{ik}$ ($i=0,\dots,n-1$, $k=1,\dots,n$) is right inverse to the matrix

[TABLE]

Since both matrices are square, $Y$ is also a left inverse. Inserting the definitions of $y_{ki}$, taking into account that $b_{jk}=b_{kj}$, and recalling the definition (49) of $f_{k}$, we find for $k,l=0,\dots,n-1$

[TABLE]

Hence the Hankel matrix $H(f)$ is the inverse of the Bezoutian $p_{2n-2}B=H(c)^{-1}$ and thus $f_{k}=c_{k}$ for $k=0,\dots,2n-2$. ∎∎

Lemma 12**.**

Let $c\in\operatorname{int}{\cal M}_{2n-1}$ be any moment vector such that $N(c)=p$. If (42) and (43) are satisfied for $x_{1},\dots,x_{n},t_{1},\dots,t_{n-1}$ and positive numbers $\sigma_{k},\varrho_{k},\lambda$, then also (23), (24), and (47) hold.

Proof.

Since $H({c})^{-1}$ is a Bezoutian, there are monic polynomials ${Q}$ and ${S}$ with $H({c})^{-1}=\tilde{p}_{2n-2}\operatorname{Bez}({Q},{S})$ for some constant $\tilde{p}_{2n-2}\neq 0$ (that will turn out to be the leading coefficient of $p$). Since the Bezoutian ${B}:=\operatorname{Bez}({Q},{S})$ does not change if we add a multiple of ${Q}$ to ${S}$, we can assume that $\deg{S}<\deg{Q}\leq n$. Now we have

[TABLE]

Starting from

[TABLE]

plugging in

[TABLE]

and reversing the above computations we arrive at

[TABLE]

By (59), the polynomial $x\mapsto{Q}(x){S}({x}_{k})-{Q}({x}_{k}){S}(x)$ vanishes for $x=x_{l}$ and has leading coefficient ${S}({x}_{k})$, hence

[TABLE]

For fixed $x\in\mathbb{R}$, the polynomial

[TABLE]

vanishes for $y={x}_{k}$ and has leading coefficient $-{S}(x)$, hence we find

[TABLE]

By (61), the coefficient of $y^{n-1}$ in the polynomial $y\mapsto{B}({t}_{k},y)$ vanishes, hence

[TABLE]

for $k=1,\dots,n-1$, so that

[TABLE]

By (63) we therefore have

[TABLE]

and in particular (see (57))

[TABLE]

Hence $\tilde{p}_{2n-2}$ is the leading coefficient of $p$ and we can write $\tilde{p}_{2n-2}=p_{2n-2}$. If we put now

[TABLE]

and recall that $p(x)=p_{2n-2}P(x)$, we find

[TABLE]

i.e., ${g}$ is a real rational function of the form (2) having the critical points $\zeta_{k},\overline{\zeta}_{k}$. Also $s_{k}>0$ and $a>0$ in (2) since $g^{\prime}(x)>0$ and thus $g$ is strictly increasing in each interval without poles. Using Lemma 4 and Lemma 5 we conclude that (24) is satisfied. Consequently, the function $f(x):=-1/g(x)$ is of the form (1) with $r_{k}>0$, and has the same critical points $\zeta_{k},\overline{\zeta}_{k}$ as $g$. By Lemma 3 and Lemma 5, the equilibrium equation (23) is satisfied. From equation (60) with $k=l$ we get

[TABLE]

and from (59) with $k=l$ we obtain

[TABLE]

Finally, by (62) we have $\lambda=1/p_{2n-2}$, so that (47) has been verified. ∎∎

Lemma 13**.**

The Nesterov mapping $N:\operatorname{int}{\cal M}_{2n-1}\to\operatorname{int}{\cal P}_{2n-1}$ is bijective.

Proof.

The surjectivity of $N$ has already been shown. To prove its injectivity we assume that $N(c)=p$ and show that $c\in\operatorname{int}{\cal M}_{2n-1}$ is unique. By Theorem 3 there are numbers $x_{1}<t_{1}<x_{2}<\dots<t_{n-1}<x_{n}$ and $\sigma_{k},\varrho_{k},\lambda>0$ such that

[TABLE]

By Lemma 12, $(t_{1},\dots,t_{n-1})$ is a solution of the equilibrium equation (24). Since this equation has a unique solution by Lemma 6, we conclude that $t_{1},\dots,t_{n-1}$ are uniquely determined. Lemma 12 tells us that (47) holds, from which we get unique values of $\lambda$ and $\sigma_{1},\dots,\sigma_{n-1}$. Finally, (64) determines $c_{1},\dots,c_{2n-2}$ uniquely. ∎∎

With the preceding lemma the proof of Theorem 4 has been completed.

5 Concluding remarks

In this paper we have established a one-to-one relation between three problems:

• (i)

determining a finite Blaschke product from its critical points,

• (ii)

finding the equilibrium position of moveable unit charges on the real line in an electric field generated by a special configuration of negative point charges,

• (iii)

solving the moment problems (42), (43).

Algorithmically, the last problem requires the Vandermonde factorization of the associated Hankel matrix, but it is difficult to exploit this for solving problems (i) or (ii). While the transition between (i) and (ii) in both directions is based on simple transformations, the translation of (ii) into (iii) needs the construction of $c\in\operatorname{int}{\cal M}_{2n-1}$ with $N(c)=p$ for a positive polynomial $p$ with zeros at the given critical points $\zeta_{k},\overline{\zeta}_{k}$. Hence we have to invert the Nesterov mapping $N:\operatorname{int}{\cal M}_{2n-1}\rightarrow\operatorname{int}{\cal P}_{2n-1}$. This can be interpreted as finding a positive definite Bezoutian with prescribed anti-diagonal sums $p\in\operatorname{int}{\cal P}_{2n-1}$. Though this problem has a unique solution for every $p\in\operatorname{int}{\cal P}_{2n-1}$, we are not aware of an efficient procedure to find it. On the other hand, $N^{-1}$ can be computed indirectly, if problem (ii) can be solved: starting with (a coefficient vector of) a positive polynomial $p$, find the corresponding equilibrium positions $t_{k}$, and compute $\lambda$ and the weights $\sigma_{k}$ from (47). The solution $c\in\operatorname{int}{\cal M}_{2n-1}$ is then obtained by evaluating (42).

We shall discuss these algorithmic and numerical aspects in more detail in a forthcoming paper.